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Mathematics
Learner’s Book 11
M. Bradley • J. Campbell • S. McPetrie
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PlatinumMathematicsGrade11
Grade 11 Mathematics textbook, 380 pages. Read online or get the free PDF download.
Publisher: Textbook
Mathematics · Grade 11. Textbook, 380 pages. Read online or download the PDF.
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TOPIC
1 Algebra: Exponents and surds
Unit 1: Laws of exponents (Revision)
From earlier grades you know the definition of an exponent and the laws
of exponents.
Definition: an = a × a × a … to n factors where n is a natural number
Laws where the exponents m and n are natural numbers:
KEY WORD 1 am × an = am + n 2 am ÷ an = am − n
When multiplying, add the When dividing, subtract the
exponent (index) – a
exponents of like bases. exponents of like bases.
number or variable that shows
Example: Example:
how many times a number
(the base) is multiplied by 2a3b2 × 5ab5 = 2 × 5× a3 + 1 × b2 + 5 26.36 ÷ 24.33 = 26 − 4.36 − 3
itself: = 10a4b7 = 22.33
base → 34→index or exponent = 4 × 27
=3×3×3×3 = 108
3 (am)n = amn 4 (ab)n = anbn
When raising a number with an When two or more numbers are
DID YOU KNOW? exponent to another exponent, multiplied and raised to an
1
2−2 = __ 1
3−2 = __ multiply the exponents. exponent, raise each one to the
4 9
1 1 exponent.
2−1 = __ 3−1 = __
2 3 Example: Example:
20 = 1 30 = 1 (a6)3 = a18 (2a4b2)3 = 21 × 3a4 × 3b2 × 3
21 = 2 31 = 3 Using Law 1: (a6)3 = a6.a6.a6 = 23a12b6
22 = 4 32 = 9
= a18 = 8a12b6
23 = 8 33 = 27
24 = 16 34 = 81
25 = 32 WORKED EXAMPLE 1 WORKED EXAMPLE 2
( )
26 = 64
14x7y5x6 4
Simplify: (–2a2b3)(4b4a)(–2a5b)2 Simplify: ________
12 2
21x y
REMEMBER SOLUTION SOLUTION
( )
7 5 6 4
14x y x
________
2.4 means 2 × 4 whereas 2,4 (–2a2b3)(4b4a)(–2a5b)2
21x12y2
4.
means 2___
10 Apply Law 3 to raise last bracket Apply Laws 1 and 2 before raising
to exponent 2. to exponent 4.
( 2x3y )
1 3 4
= [–2a2b3)(4b4a)(–2)2a10b2] = _____
Apply Law 1 24x4y12
= ______
4 3
= −2 × 4 × 4 × a2 + 1 + 10b3 + 4 + 2
16x4y12
= −32a13b9 = _______
81
4 Topic 1 Algebra: Exponents and surds
PLT MATHS LB 11 7th pgs (Real Book).indb 4 2012/07/02 2:20 PM
WORKED EXAMPLE 3 WORKED EXAMPLE 4
x+1 x−1
Simplify: 4 .3
_________
x−2
Simplify:
12
1 (3a3 – b)2 = 9a6 – 6a3b + b2
SOLUTION 2x+2
– 2x
________
2
x + 1.3x − 1 3.2x
4
_________
12x − 2
SOLUTION
Break down the numbers into prime
factor bases. Be careful how you apply the laws when
there are more than two terms.
(22)x + 1.3x − 1
= ___________
2 x−2 1 (3a3 – b)2 = 9a6 – 6a3b + b2
(2 .3)
x+2
Apply Laws 3 and 4. 2 2 – 2x
________
3.2x
2x + 2 x−1
=2 .3
__________
2x − 4 x − 2
2 .3 The ‘top’ has two terms, so factorise.
Apply Laws 1 and 2.
x 2
2 (2 – 1)
= ________
x | 2x + 2 = 2x.22
= 22x + 2 – (2x – 4).3x – 1 – (x – 2) = 26.31 3.2
= 192 2x(3)
= _____
3.2x
=1
EXERCISE 1
Simplify the expressions without using a calculator:
4 3
2(2a )
1 1 16 a3b2
__a5b12 × ___ 2 ______
4 5 16a12
3 6a + 4
9a
________ 4
4 3a 3
× 5a5
_________
5a2 (5a3)2
5 px − 1px + 1 6 −2(– 4a2 + b3)2
12x + 1.27x − 2
7 164x ÷ 42x 8 ___________
182x − 1
9 22 010 × 52 011
____________ 10 3x + 3x + 1
________
100 1 000 8
KEY WORD
negative exponent – a
Negative exponents and the meaning of a0 number raised to a negative
exponent and the reciprocal of
a0 = 1 because 5 ÷ 5 = 1 and 51 ÷ 51 = 50 = 1 | Using Law 2 that number with a positive
For example: (2x)0 = 1; 2x0 = 2 × 1 = 2; (a + b)0 = 1 exponent:
1 or ___
a−m = ___ 1 = am
1 and 50 ÷ 51 = 5−1 | Using Law 2
1 because 1 ÷ 5 = __
a−m = ___ m
a −m
a
m a 5
1 = __1 or ( a ) __a
__ –1 = b
Examples: 2 = __
−3 b
23 8
x10 = x10 − 12 = x−2 = __
___ 1
x12 x2
( )
−2
a −2 = a
__ ___ =b__2
b b−2 a2
When you simplify an expression with negative exponents, use the exponent laws
first and write the final answer with positive exponents. This is easier than trying to
get rid of the negative exponents first.
Unit 1 Laws of exponents (Revision) 5
PLT MATHS LB 11 7th pgs (Real Book).indb 5 2012/07/02 2:20 PM
WORKED EXAMPLE 1
2x −3
Simplify: _____
−3 (2x)
SOLUTION
2x−3 −3
_____ = 2x
_____ = 2 × 23 = 16
(2x)−3 2−3x−3
WORKED EXAMPLE 2
−3
Simplify: 2x
____
−2 2 x
SOLUTION
2x−3 8
____ = 21 + 2x−3 − 1 = 23x− 4 = __
−2
2 x x4
WORKED EXAMPLE 3
3x−1 + y
Simplify: _______
−1
| This problem has more than one term in the numerator.
x y
REMEMBER
SOLUTION
If the expression involves two
3x−1 + y ( __
3 + y)
or more terms, you cannot
invert from the top to the
_______
x−1y
= x
______
y
__
3 + xy
= ______
x ( )
3 + xy
× __xy = ______
y
x
bottom of the fraction.
or
−1 −1
3x + y 3x .x + y.x 3 + xy
_______ × x__ = __________ = ______
x y
x−1y x−1y.x
EXERCISE 2
Simplify the expressions without using a calculator. Give your answer with
positive exponents.
−2 −2
(3xy ) −1 −2 2
9 x y
1 12a0b3c−1
_________ 2 ________ 3 ________
−2 5 −3 −2
15a b c 3x y (3xy−1)−3
6(x2y)3 −2 −1
(x y)
____________ (2a−1 2
)
______ 92x.3−2
______
4 5 6
3x−3y5 2(a2)−1 81x − 1
( _____ )
2 −3 −1
x y −1 0
(2a )
______ a−2 + a−1
+ a0
___________
7 8 9 a
z−2 2(a0)−1
3−2 + 30
10 _______ 11 (2.3)−2 + 2.3−2 12 (2−1 − 5−1)2
3−1
xy
_______ 3.3x
________
13 14 (a3 + 3)−2 15
x−2 + y−2 3x + 3x − 1
(3x4)2
+ (2x2)4
____________ (8.2x+1
)(4.2x − 1)
______________ 12x + 1.27x − 2
___________
16 17 18
5.2 .2−5.x10
5 2 −3
x 182x − 1
6 Topic 1 Algebra: Exponents and surds
PLT MATHS LB 11 7th pgs (Real Book).indb 6 2012/07/02 2:20 PM
Exponential equations
Exponential equations have the unknown or variables in the exponent.
To solve equations with an unknown index, express each side with the same base and KEY WORD
use the property: if ax = ay then x = y | (a > 0, a ≠ 1) exponential equation – an
equation with unknowns or
variables in the exponent
WORKED EXAMPLE 1 WORKED EXAMPLE 2
Solve for x: 2x = 4x + 1 Solve for x: 8x + 1 = 32 ÷ 4x
SOLUTION SOLUTION
2x = 4x + 1 8x + 1 = 32 ÷ 4x
3 x+1
= (22)x + 1 (2 ) = 25 ÷ 22x
= 22x + 2 23x + 3 = 25 − 2x
x = 2x + 2 3x + 3 = 5 − 2x
x − 2x = 2 5x = 2
2
x = __
−x = 2 5
x = −2
EXERCISE 3
Solve for x:
1 73x + 1 = 49x – 1 2 32x + 1 = 27x – 2 3 4.82x + 1 = 32x + 2
4 5.9x + 1 = 15 5 2.32x + 1 = 54 6 ( __12 )x – 5 = 64
WORKED EXAMPLES
These equations have more than one term on each side.
• Rearrange the equations so the unknown exponents are on one side.
• Factorise by taking out a common factor.
• Solve the remaining exponential equation.
WORKED EXAMPLE 1 WORKED EXAMPLE 2
Solve for x: 2x + 2x + 1 = 24 Solve for x: 3x + 1 = 8 + 3x − 1
SOLUTION SOLUTION
2x + 2x + 1 = 24 3x + 1 = 8 + 3x − 1
2x + 2x.21 = 24 3x.31 − 3x.3−1 = 8
2x(1 + 2) = 24 3x(31 − 3−1) = 8
2x(3) = 24 ( )
1 =8
3x 3 − __
3
2x = 8
= 23 ( )
3 8 =8
x __
3
x=3 3x = 3
x=1
Unit 1 Laws of exponents (Revision) 7
PLT MATHS LB 11 7th pgs (Real Book).indb 7 2012/07/02 2:20 PM
EXERCISE 4
Solve for x:
1 3x − 3x − 2 = 24
2 2
3x + 1 = 3x + __
3
3 3x + 3x + 2 = 10
4 3 = 2.5x
5x + 1 − ___
25
5 2x + 2x + 2x = 96
6 12x + 12x = 288
7 2x + 2x – 2 = 5
8 22x + 4x + 1 = 80
EXERCISE 5
Solve for x:
1 1 = ___
8x. ____ 1
x−1
4 32
2 3x + 1 1
= 9 . ___
x
27
3
2 1
3x + 2 = 9.___x
27
4 1 = 25x
5x − 1.___
2
25
5 22x.4x + 1 = 1
6 12x.12x = 144
2 2
7 2x .2x − 2 + 1 = 65
1 −x
4x + 1 × __
4
___________
( )
1
8 = ___
22x + 1 32
9 1
81.9x − 1 = ___
27
2
10 102x .10x + 1 = 10 000
8 Topic 1 Algebra: Exponents and surds
PLT MATHS LB 11 7th pgs (Real Book).indb 8 2012/07/02 2:20 PM
Unit 2: Simplify expressions with
KEY WORD
rational exponents
rational exponents – an
m
__
Rational exponents and the meaning of a n exponent or index with
a rational number as the
1
__ 1
__ 1
__ __ __ __
32 × 32 = 31 32 = √3 because √3 .√3 = 3 exponent
1
__ 1
__ 1
__ 1
__ __ __ __ __ For example:
and a3 × a3 × a3 = a1 a3 = 3√a because 3√ a .3√ a .3√ a = a 1
__
___ • 83 is the exponential form
1
__ __ m
__
so an = √a and a n = √a where a > 0 and m and n are natural numbers
n n m of the cube root of 8
__
___
a n = ( am )n = n√am or a n = ( an ) = (n√a )m • √ 8 is its radical form
m
__ 1
__ m
__ 1 m
__ __ 3
3
__
• 54 is the exponential form
The laws for natural number exponents apply to rational exponents. of the 4th root of 125
___ ____
• √ 53 = √125 is its radical
4 4
form
WORKED EXAMPLE 1 AND SOLUTIONS
( 6√22 )6 = 22 | Because ( 226 ) = 22
___ 1 6
__
1
____ 1
__
2 √ −32 = −2
5
| Because (−25)5 = −2
__ 3
__ 1
__
3 √ 93 = 27 | Because 92 = [(32)3]2 = 33 = 27
Note: It is easier to write the number under the root as an exponent and to cancel
the exponents using Law 3 of the exponents.
WORKED EXAMPLE 2 AND SOLUTIONS
______
1 √64x16y8 = 4x8y4 | For square roots, divide the exponents by 2.
_______
2 √27x9y12 = 3x3y4 | For cube roots, divide the exponents by 3.
3
___________
3 √ 64x10 + 36x10 | Add the like terms first. DO NOT square root each term.
______
= √ 100x 10
= 10x 5
WORKED EXAMPLE 3
Simplify without using a calculator:
( ___
27 )
4
__
8 – 3
SOLUTION
( ___
27 ) ( __3 )– = 2___
4
__ 3 4
__ –4 4
8 – = 2 3 =3 81
__ = ___ 3
3 2 16 3 –4 4
WORKED EXAMPLE 4
Simplify:
2
__ 3
___
a3. √ a4
______
3
__ 4
___
a4. √a4
SOLUTION
2
__ 3
___ 2 4 6
√ a4
__ __ __
a3. 2
______ a .a
_____ __ a
a 3 __ 3 3
3
__ __ = 3__ 1__ = __4 = a = a
a4.4√ a a4.a4 a4
Unit 2 Simplify expressions with rational exponents 9
PLT MATHS LB 11 7th pgs (Real Book).indb 9 2012/07/02 2:20 PM
WORKED EXAMPLE 5
Prove that:
___
√ ax
3
___ = ____
x
______
( 3√ ax )2 a
SOLUTION
___
√ ax
3
___ = ____
x
______
( 3√ ax )2 a
___
√ ax 3
x ___ | Multiply by 1.
___ × ____
LHS = ______
3 ( √ ax )
2 3
√ ax
___ ___ ___
ax
3
√ x ax ax 3
√ 3
√
= x______
___ 3 = ______ = ____ = RHS
3( √ ax ) ax a
EXERCISE 6
1 Write in exponential form:
___ ____ ___
√ x2 √x5y6 5 √x15
3 4
1.1 1.2 4
1.3
2 Write in radical form:
3
__ 1
__ 2
__
2.1 x5 2.2 6x4 2.3 ( 3x2 )3
EXERCISE 7
1 Evaluate without using a calculator:
_______ ____
1.1 √ 10 000
4
1.2 √ 642
3
____ 2
– __
1.3 √ 163
4
1.4 ( 0,125 ) 3
( 2___
25 ) ( 111
25 )
1
__ 1
__
1.5 14 – 2 1.6 ___ – 2
__ __ 1
– __
1.7 (–2)0 + 3√8 + ( √ 3 )2 1.8 ( 0,0625 ) 4
1.10 81– 4( 273 – 92 )
2
–__ 3
__ 2
__ 3
__
1.9 ( 0,064 ) 3
1.11 49–2( 1253 – 90 )
1
__ 2
__ 1
__ 3
__ 1
__
1.12 16–2 + 164 – 64 –3
2 Simplify:
1
__ 2
__ 3
__
2.1 ( 27x27 )3 2.2 ( 64a6 )3 2.3 ( 81a16b8 )4
________ ____________ _____________
2.4 √ a8b16c20
4
2.5 √ 125a12b15c21
3
2.6 √ 15a15 + 17b15
5
( )
__
( ___
x )
3
__ ___ 1
__
16 √ x 2
___ √ x5 .x3
3
2.7 –8
4 2.8 3 2.9
– __
x 2
( )
___ ___ ___ ________
√ b2 2
3
____ √ x2 .3√ x4
3 √ 81a
4 –8 12
_________b
2.10 1 2.11 2.12 1 a2b –3
–1
__ __
b– 3 4
3 Simplify:
______ ________
√______ √_________
n n x x
3.1 n 8 .25 3.2 15 .3
2n x+1 x–2
4 9 .5
_________ ______ ____________
√ 8_________
x+1 x+2
3.3 5
16
.2
x 3.4 √(a + b)2 + √a2 + 2ab + b2
______ 1
1 – __
√(a + b)3 × (a + b)2
__
3.5 3.6 (x2 – 6x + 9) 2
10 Topic 1 Algebra: Exponents and surds
PLT MATHS LB 11 7th pgs (Real Book).indb 10 2012/07/02 2:20 PM
Unit 3: Solve equations with
rational exponents
Equations with rational exponents have the unknown in the base.
To solve equations with an unknown base, raise both sides of the equation to
the exponent which makes the exponent of x equal to 1.
To simplify the other side, write the number as a base with an exponent.
WORKED EXAMPLE 1 WORKED EXAMPLE 2
3
__ 5
__
Solve for x: x2 = 27 Solve for x: x3 = 32
SOLUTION SOLUTION
3
__ 5
__
x2 = 27 x3 = 32
3 2 2
__ 5 3 3
__
x2 × 3 = (33)3
__ __
x3 × 5 = (25)5
__ __
x = 32 x = 23
=9 =8
Equations with rational exponents that
have two solutions
Remember that x2 = 9 has two solutions.
Method 1 Method 2 Method 3
x2 − 9 = 0 x2 = 9 x2 = 32
__ 1
__ 1
__
(x − 3)(x + 3) = 0 x = ±√ 9 (x2)2 = ±(32)2
x = 3 or x = −3 = ±3 x = ±3
An equation in which the top of the rational exponent is even will have
two solutions.
WORKED EXAMPLE 1
2
__
Solve for x: x−3 = 9
SOLUTION
Method 1 Method 2
2
– __ 2
__
x 3–9=0 x− 3 = 9
( x−__13 − 3 )( x− __13 + 3 ) = 0 2 3
x−3 × −2 = ±(32) 2
__ __ 3
−__
1
__ 1
__
x− 3 = 3 or x− 3 = −3 x = ±3−3
1 or
x = 3−3 = ___ 1
x = −3−3 = −___ 1
= ±___
27 27 27
Unit 3 Solve equations with rational exponents 11
PLT MATHS LB 11 7th pgs (Real Book).indb 11 2012/07/02 2:20 PM
WORKED EXAMPLE 2
2
__
Solve for x: 5x− 5 = 20
SOLUTION
2
__
5x− 5 = 20 | The coefficient of x must be 1 before you can raise to an exponent.
2
−__
x 5=4 | Divide both sides by 5.
2 5 5
−__
x− 5 × − 2 = ±(22) 2
__ __
| Raise both sides to the exponent −__
5.
2
x = ±2−5
1
= ± ___ | There are two answers because the top of the rational
32
exponent is even.
EXERCISE 8
Solve for x:
1
__ 3
__
1 x4 = 3 2 x– 2 = 8
1
__ 1
__
3 7
x– 2 = __ 4 x– 3 + 4 = 0
2
2
__ 2
__
5 x– 3 – 25 = 0 6 x3 – 16 = 0
2
__
7 2x3 = 128 8 125x3 – 1 = 0
3
__
9 27x4 = 125 10 49x–2 – 1 = 0
3
__ 4
__
11 8x4 – 27 = 0 12 81x3 = 16
Always factorise equations that have more than one term on each side.
WORKED EXAMPLE 3
1
__ 1
__
Solve for x: x2 + 3x4 − 18 = 0
SOLUTION
1
__ 1
__
x2 + 3x4 − 18 = 0 | This is a trinomial with a 2:1 ratio of the exponents.
( x4 + 6 )( x4 − 3 ) = 0
1
__ 1
__
| Factorise the trinomial.
1
__ 1
__
x4 = −6 and x4 = 3 | Raise both sides to the exponent 4.
__ __
but 4√x = −6 has no solution because 4√ x > 0
1
__
x4 = 3
x = 34 = 81
12 Topic 1 Algebra: Exponents and surds
PLT MATHS LB 11 7th pgs (Real Book).indb 12 2012/07/02 2:20 PM
EXERCISE 9
Solve for x:
1
__ 1
__
1 x2 − 5x4 + 6 = 0
2
__ 1
__
2 x3 − 7x3 + 10 = 0
2
__ 1
__
3 x3 − x3 − 2 = 0
1
__ 1
__
4 2x2 − x4 − 3 = 0
1
__ 1
__ home work
5 3x2 − 5x4 − 2 = 0
__
6 5x + 7√x − 6 = 0
1
__
7 x − x2 −12 = 0
1
__
8 x−1 − x−2 = 20
__
9 2x + 7√x −15 = 0
5
__
___
x−3 = √95
6
10
Unit 3 Solve equations with rational exponents 13
PLT MATHS LB 11 7th pgs (Real Book).indb 13 2012/07/02 2:20 PM
Unit 4: Surds
KEY WORD
In the previous unit we worked mostly with numbers in the exponential form.
surds – irrational roots or
irrational numbers In this unit we work with numbers in their surd form.
The square root of a non- m
__
___
We can express all fractional exponents in surd form: a n = n√am
perfect square is an irrational
number and is called a surd.
For example:
__
• √___3 is the surd form of 32
1
__ Multiplying and dividing surds
• √12 written in __ simplified You can multiply surds with the same You can divide surds with the same roots.
√3 __ __
surd form is 2 roots. __
√a ÷ √b = n
√__ab
___ ___ ___ n n
• √20 ; √30 or √25 are
4 3 5
__ __ ___
√ a × √ b = √ ab
n n n
surds of different radical ___
___ ___ ____ ___ __ __
orders Example: √2a × √5a2 = √ 10a3 12 = √ 4
Example: √ 12 ÷ 3√ 3 = 3 ___
3
√3 3
Reducing surds
You should always write a surd as the product of a rational and an irrational number.
___ __ __ __
√ a2b = √ a2 √ b = a√ b
____ ____ ___ __ __
Examples: √128 = √ 64.2 = √ 64 .√ 2 = 8√2
___ _____ 3 ___ __ __
√ 54 = 3√ 27.3 = √ 33 .3√ 3 = 3.3√ 3
3
Adding surds
You can only add like surds, so you should reduce surds first.
__ __ __
√ a + √ a = 2√ a
__ __ __ __
√a + √b = √a + √b
__ ____ ___ ____ __ __ __
Example: √8 + √128 = √4.2 + √64.2 = 2√ 2 + 8√ 2 = 10√ 2
Rationalising denominators
When there is a surd in the bottom of a fraction, multiply the top and bottom by
the same surd to get rid of the surd at the bottom.
__ __
a__ × √
___ b = a√
___
__ b
____
√b √b b
__ __ __ __ __
1__ × √
Example: ___ 2 =√
___
__ 2 or 3 +
___ √ √5
__ 5 × ___
______ √
__ = 3 5 +5
_______
√2 √2 2 √5 √5 5
__ __
√ √
Example: 2 − b ×1−
______
__ b
______
__
1 + √b 1 − √b
__
√
=2−3 b+b
__________
1−b
When there are two terms in the bottom of a fraction, multiply the top and bottom
by two terms that make the bottom the difference of two squares.
__ __ __ __ __ __ __
√ √ 9 + 3√ 3 + 6√3 + 2(√3 )2 √ √
Example: 3 + 2 __3
_______ ×3+ 3 ____________________
______
__ = __ =9+9 3 + 6 15 +
__________ 9 3
= ________
3 − √3 2 3 + √3 9 − (√ 3 ) 9−3 6
14 Topic 1 Algebra: Exponents and surds
PLT MATHS LB 11 7th pgs (Real Book).indb 14 2012/07/02 2:20 PM
WORKED EXAMPLE 1
___ __ ___
Simplify: √ 18 – √8 + √50
SOLUTION
___ __ ___
√ 18 – √ 8 + √ 50
___ ___ ____
= √9.2 − √4.2 + √25.2
__ __ __
= 3√2 − 2√ 2 + 5√2
__
= 6√2
• Remember to reduce surds and add like terms or factorise if there are two
or more terms.
• If there are brackets, multiply them out. Look for perfect squares and the
difference of two squares.
WORKED EXAMPLE 2
___
20 √
Simplify: ____
___
√ 80
SOLUTION
___ ___ __
√ 20
____
___ =
√ 80 √___
80 √ 4 2
20 = __
1 = __
1
or
___ ___ __
√ 20 √ 4.5 2√__
5 __
____
___ = ______
____ = ____ =1
√ 80 √ 16.5 4√ 5 2
WORKED EXAMPLE 3
___
√
Simplify: 27 −3
_______
__
√ 3 −1
SOLUTION
___
√ 27
__ − 3
_______
√ 3 −1
__
√3 − 3
= 3 __
_______
√ 3 −1
__
3( √ 3 −1)
= ________
__
(√3 −1)
=3
WORKED EXAMPLE 4
__ ___
Simplify: (4 + √3 )2 − √48
SOLUTION
__ ___
(4 + √3 )2 − √ 48
__ __ ____
= 16 + 2(4)(√ 3 ) + (√3 )2 − √16.3
__ __
= 16 + 8√3 + 3 − 4√3
__
= 19 + 4√3
Unit 4 Surds 15
PLT MATHS LB 11 7th pgs (Real Book).indb 15 2012/07/02 2:20 PM
WORKED EXAMPLE 5
__ __ __ __ __ __
Simplify: (√ 2 + √ 8 + √ 7 )(√ 2 + √ 8 − √ 7 )
SOLUTION
__ __ __ __ __ __
(√2 + √8 + √7 )(√ 2 + √ 8 − √ 7 ) | Reduce surds.
__ __ __ __ __ __
= (√2 + 2√ 2 + √ 7 )(√ 2 + 2√ 2 − √ 7 ) | Add like surds.
__ __ __
= (3√2 + √ 7 )(3√ 2 − 7) | Multiply out difference of two squares.
= 9(2) − 7
= 11
WORKED EXAMPLE 6
__
___24 ___ = 2√ 6 .
Show that __________
√ 96 − √ 24
SOLUTION
__ __
√ √ __
6 _____
24
____________
LHS = ____ = __24
__________ 24__ × ___
__ = ____ __ = 12 6 = 2√ 6 = RHS
√ 16.6 − √ 4.6 4√6 − 2√6 2√6 √6 6
EXERCISE 10
1 Write in simplified surd form without using a calculator:
___ ___ ___
1.1 √ 32 1.2 √ 48 1.3 √ 72
____ ____ ___
1.4 √ 125 1.5 √ 128 1.6 √ 96
___ ___ __
1.7 √ 16
3
1.8 √ 96
3
1.9 √ x3
______ ________ ______
1.10 √ 20a4b3 1.11 √217a14b13 1.12 √50x5y8
_____ _____ _____
√x11y7 1.14 √22 011 1.15 √ 32 012
3 3
1.13
2 Simplify:
__ __ __ __
2.1 √7 × √5 2.2 2√5 × 3√ 5
___ __ __
√ 72
2.3 ____
__ 2.4 2√3 × 3√ 2
√8 ___
___ __ √ 98
2.5 2√12 × 3√ 8 2.6 ____
__
___ √8
√ 45 __ __ __
2.7 ____
___ 2.8 2√2 × 3√ 3 × 6√ 6
√ 10
____ ____ ____ ____
2.9 √ 12a × √ 9a5
3 3
2.10 √ 8a3 × 5√ 8a7
5
___ __ ___ ___
√ 12 × √ √ 3a3 × √ 6a5
2.11 __ 6
_________ 2.12 ___________
___
√8 2√ 2a
EXERCISE 11
1 Simplify without using a calculator:
___ __ ___ ___ ____ ____ ___
1.1 √ 32 – √ 8 + 2√ 18 1.2 5 √ 45 – √162 – √125 + √ 72
___ ___ ____ ___ ___ ___
1.3 √ 50 + √ 98 – √ 128 1.4 √ 48 + 2√ 12 – √ 75
___ ___ ___ __
√ 48 – √12 √ 72 + √
1.5 __________
___ 1.6 __ 8
_________
√ 75 √2
16 Topic 1 Algebra: Exponents and surds
PLT MATHS LB 11 7th pgs (Real Book).indb 16 2012/07/02 2:20 PM
___
2+ √ 28
1.7 ___ 5 ___
__________ 1.8 _______
__
√ 20 + √ 45 1 + √7
___ ___ ___
√ 18 √ 12 √ 20
1.9 ___ – ___
__________ 1.10 __ – 2
_______
√ 72 – √ 48 √5 – 1
_____ _____ _____ _____
√ 27m6
– √48m6
______________ √ 32x10
– √18x10
______________
1.11 _____ 1.12 ____
6 √ 12m 8 √ 50x
EXERCISE 12
Expand and simplify where possible:
__ __ __ __
1 2√3 (3 – 2√3 ) 2 (3 + √3 )(2 + √ 2 )
__ __ __
3 (2 + √2 )(2 – √2 ) 4 (√ 5 – 2)2
__ __
5 (1 – √ 3 )2 6 (1 + 3√ 3 )2
__ __ __ __
7 (7 + 2√ 3 )(7 – 2√3 ) 8 (√ 2 – √ 8 )2 + (0,5)–1
__ 1 __ 1
( 2 – ___
2 )
× ( 2 + ___
2 )
_______
___ _______
___ √7 __ √7 __
9 √4 – √12 × √4 + √12 10 2 2
EXERCISE 13
Rationalise the denominator:
__
1__
___ 3__
___ √2
___
1 2 3 __
√3 √3 √3
__ __
1+ √ √
4 6__
___ 5 __ 3
______ 6 4+ __ 2
______
√2 √3 √2
__ __ __
√3
______ 1 + √__
2
______ 1 + √__
3
______
7 __ 8 9
√3 – 1 1 – √2 3 – √2
EXERCISE 14
1 Show that:
__ __ __ __
1.1 8
______
1− 1
___
__
= 16 + 8√2 1.2 √__83 + 5√__23 − ___ 2 √3
13 2
1__ = ___
√6
__
√2
__________ _________ __ __ ___
___ __ ___ __ √8 − √ 3 − √ 12
1.3
5
√√37 + √5 .5√√37 – √5 = 2 1.4 _______________
___ __ 1
___ = __
√ 32 − 3√ 3 − √ 27 2
_____
__ √ 2 011
1.5 ___15 __ = √ 5
_________ 1.6 _____2
______________ 2
_____ = __
√ 80 − √ 5 √ 22 013 − √ 22 009 3
__ ___
√3 √ 27 ___ __ ____
+ 1__ 5 +
_______
1.7 = _______ 1.8 √ 12 + 2√ 3 + 31,5 = √ 147
4 − 2√3 2
__ __ _______
___
2 If m and n are rational numbers such that √ m + √n = √5 + √24 , calculate
a possible value for m2 + n2.
___ _____
3 Simplify: (1 + √ 3x2 )2 − √12x2
__ 3
__
4 If √ 2 = 1,414 show that 2,828 = 22.
Now prove without a calculator or long division that:
1
_______
______ = 0,707
√ 2,828
3
Unit 4 Surds 17
PLT MATHS LB 11 7th pgs (Real Book).indb 17 2012/07/02 2:20 PM
Solving equations with surds
REMEMBER
To solve equations with surds,
WORKED EXAMPLE 1
arrange the equation with the
____
surd on one side of the equals Solve for x: √x + 1 = 5
sign and the other terms on
the other side. Square both
SOLUTION
sides of the equation and
____
solve for the unknown. √x + 1 = 5
____
(√ x + 1 )2 = 52
(x + 1) = 25
REMEMBER x = 24
Check the solutions:
Always square sides and not
the terms:
for x = 24
______
(2 + 3) = 5 but
LHS = √ 24 + 1 = 5 = RHS
(2 + 3)2 = 25 22 + 32 ≠ 52
52 = 25 4 + 9 ≠ 25
Always check your solutions
when squaring both sides. WORKED EXAMPLE 2
Squaring can create extra or ____
Solve for x: 1 + √ x + 1 = x
extraneous solutions.
The first statement is false but
after squaring the statement, SOLUTION
it is true. ____
__ √x + 1 = x − 1
2 = −2 √ x = –2 ____
(2)2 = (−2)2
__
(√x )2 = (–2)2 (√ x + 1 )2 = (x − 1)2
22 = 4 x=4 x + 1 = x2 − 2x + 1
x2 − 3x = 0
x(x − 3) = 0
x = 0 or x = 3
Check the solutions:
_____
for x = 0: LHS = √0 + 1 = 1; RHS = 0 − 1 = −1 x ≠ 0
_____ __
for x = 3: LHS = √3 + 1 = √ 4 = 2; RHS = 3 − 1 = 2 x = 3
EXERCISE 15
REMEMBER
To solve surd equations: Solve for x:
____ ____
1 Isolate the surd. 1 √x + 2 = 4 2 √x − 3 + 3 = 0
2 Square both sides. __ __ __
__ √ x
___
3 Solve the equation. 3 3√x = 2√3 4 __ = 3√ 2
√2
4 CHECK the solution! _____ _____
5 √ x2 + 7 − 4 = 0 6 √ x2 − 9 = 4
____ ____
7 √x − 2 = 4 − x 8 √x + 5 − x = 3
______ ____
9 √ 3x + 13 − 1 = x 10 x + √5 − x + 1 = 0
_____ ____
11 √ 4 − 3x = 4 − 3x 12 x + 2√x − 1 = 4
18 Topic 1 Algebra: Exponents and surds
PLT MATHS LB 11 7th pgs (Real Book).indb 18 2012/07/02 2:20 PM
Surds in other topics
You have worked with surds in Grade 10 in coordinate geometry and trigonometry.
Remember the distance between two points A(x1;y1) and B(x2;y2) is
________________
AB = √ (x1 − x2)2 + (y1 − y2)2 y
In the diagram, A(–1;6) and B(5;0), then: A(–1;6)
________________ ______
AB = √ (−1− 5)2 + (6 − (0))2 = √36 + 36
The answer is in simplified surd form. Q(2;3)
To show that BC = 2PQ it is better to use simplified surd P(–3;1) B(5;0)
form than round off to two decimal places. x
_________________
PQ = √ (2 − (−3))2 + (3 − (1))2
______ ___
= √ 25 + 4 = √29 C(–5;–4)
__________________
BC = √ (5− (−5)) + (0 − (−4))
2 2
_______ ____
= √ 100 + 16 = √116
______ ___
= √ 4 × 29 = 2√29
y
You will work with the following types of surd equations when you make P
lengths equal to each other.
Points A(1;5) and B(7;3) are given. Another point P(5;a) lies on the line x = 5.
A(1;5)
Find the value of a if PA = PB.
______________ ______________ B(7;3)
√(5 − 1)2 + (a − 5)2 = √(5 − 7)2 + (a − 3)2
To solve this equation, square both sides.
O x
____
Remember that: (√a + b )2 = a + b x=5
(5 − 1)2 + (a − 5)2 = (5 − 7)2 + (a − 3)2
(4)2 + a2 −10a + 25 = (−2)2 + a2 − 6a + 9
16 −10a + 25 = 13 − 6a 1
−4a = −28 45°
a=7 30°
2 __ 1 __
√3 √2
Remember__the special angles in trigonometry.
3 √ 1__
sin 60° = ___ and tan 30° = ___ 60°
45°
2 √3
1
Now rationalise the denominator:
__ __
√
3 ___ √
1__ × ___
tan 30° = ___ __ = 3 which is the answer the calculator gives.
√3 √3 3
1__
sin 45° = cos 45° = ___
√2
Rationalise the denominator:
__ __
2 = ___
1__ × ___ 2 √ √
sin 45° = cos 45° = ___ __
√2 √2 2
Unit 4 Surds 19
PLT MATHS LB 11 7th pgs (Real Book).indb 19 2012/07/02 2:20 PM
Revision
1 Simplify using the laws of exponents:
Law 1: am × an = am + n
Law 2: am ÷ an = am – n
Law 3: (am)n = amn
Law 4: (ab)m = am × bm
Law 5: a0 = 1
3x3
1.1 15x4y6 × 2y2x0 1.2 ____ × (3x3)3
27x9
3 5
4x y
______ 213
× 214 − 20
________
1.3 1.4
12x9y0 225
2 4
2(x ) (x ) (−2x ) 4 3 3 2 x+1
1.5 _______________ 1.6 27x.9
________
16(x4)5 3 5x + 3
(3x4)3 + (2x3)4
____________
1.7 1.8 (2a4 − 3b3)2
81.3 x7
−4
1.9 18x.24x + 1
_________
123x.4−x
[3 × 9]
2 Simplify leaving your answers with positive exponents by using the law: a−n = 1
__
an
2.1 2−4 − 2−3 + 2−2 2.2 3x−4y
( 3a2_____ab )
2 −1
9(a b)
_______ −2 5 −2
2.3 2.4
6ab−1 −1
6(a−1b)2
_______
2.5 2.6 (3−1 − 2−1)−2
4(ab)−2
2.7 (2 + x−2)(2 − x−2) 2.8 (x + x−1)2 − (x − x−1)2
y
________
2.9
x−2y + y−1
[4 × 9]
3 Without using a calculator, simplify
___
or evaluate the expressions with rational
m
__
n
√ m
exponents using the property a = a n and the exponential laws.
____ __2 2
__
3.1 (√125 )3 3.2 32−5
5
__
____ 3
__
646 + √163 (0,16)−2
4
3.3 3.4
_______ ________
3.5 √27x6y12
3
3.6 √ 9x8 +16x8
___ ___
__ ___
√a
3 2 × 3√ a5
60 + √24 − √82 + (0,5)−2 __________
3
3.7 3.8 1
__
( )
a3
3.10 x2( x2 − x−2 )
1
−__ −2 1
__ 1
__ 1
__
3.9 b
____3
___
√ b2
3
__
(4 5−9 )
1 3
__ 1
__
−1 __ 4 √
3.11 ________
−1 −1
2 3.12 16 . 3 .42
_________
1 __
__ 2
27−2.83
[4 × 12]
20
PLT MATHS LB 11 7th pgs (Real Book).indb 20 2012/07/02 2:20 PM
4 Simplify by writing surds in their simplest surd form first.
Leave your answers with rational denominators. REMEMBER
__ ___ ___
4.1 √ 3 + √ 48 − √ 75
__ ___ ____ ___ We can reformat very large
4.2 √ 3 − √ 32 + √ 108 − √ 72
___ __ or very small numbers using
√ 50 + √ 2
_________
4.3 exponents. This form is called
6
___ ___ scientific notation.
3√18 − √
___ 50
___________
4.4 2350000000 in scientific
2√72
___ ___ ___ ___ notation is 2,35 × 109 and
60 √
4.5 (√15 × √20 ) − √ 75 − ____
__ 0,00000000000761 is
√5
___ __ __ 7,61 × 10–12
4.6 14
√ 50 + (√ 3 × √ 6 ) − ___
__ We write only one significant
√2
____ ____ number in front of the
√ 48x6 − √ 18x6
_____________
4.7 1 ___ decimal. The exponent of the
__ 4
2 × 3 2 . √ x8 base 10 indicates how many
__ ___ __ ___
4.8 (√3 x − √2x )(√3 x + √2x ) times you have multiplied or
divided by 10, that is, moved
__
√6 the decimal to the right or to
4.9 ___+ 1
_______
√ 24 + 2 the left.
___
√ 11
4.10 2 ___ –3
________
√ 11 + 2
[3 × 10]
5 Solve for x in equations with unknown exponents: __
5 √
5.1 2x − 3 × 16 = 4x 5.2 52x + 1 = ____
125
4x −x
5.3 2x − 3 × 64 = __ 5.4 3x2 − 1
= 27
____
8 3
5.5 32x + 1 − 32x = 18 5.6 3x + 2 = 42 − 5.3x
[3 × 6]
6 Solve for x in equations with rational exponents:
___ 2
__
6.1 √ x2 = 25
3
6.2 3x3 − 12 = 0
6.3 ( __3
( x + 2 ) x2 − 8 )=0 6.4
__
6 + 44√ x = 18
3
__ 1
__ 2
__ 1
__
6.5 3x4 = 24x4 6.6 3x3 − 5x3 −12 = 0
[3 × 6]
7 Solve for x in equations with surds:
_____ _____
7.1 √ 7x + 2 − 2 = 1 7.2 √ 3x − 2 − x = 0
__ ____
7.3 x − √x = 6 7.4 √2 − x = x + 4
_____ ____
7.5 √ 2 − 7x + 2 = x 7.6 2√ x − 3 + 3 = x
[5 × 6]
21
PLT MATHS LB 11 7th pgs (Real Book).indb 21 2012/07/02 2:20 PM
TOPIC
2 Equations and inequalities
Unit 1: Completing the square
Completing the square allows learners to:
KEY WORDS • solve quadratic equations, even if we cannot factorise them
square – the product of two
• develop a quadratic formula which we can use to solve quadratic equations
identical factors • determine the turning point of a parabola graph
For example: • determine the value of x for which an expression has a
• 1=1×1 minimum/maximum value
• 4=2×2 • determine the centre point of translated circles (Grade 12 Analytical geometry).
• 25 = 5 × 5
• x2 = x × x We can manipulate expressions so they take on the form of a perfect square. Although
• (x – 3)2 = (x – 3)(x – 3) the expressions will look different, they will not be different. Completing the square
quadratic equation – an enables us to predict the sign of an expression even if we do not know the specific
equation in which the highest
value(s) of x.
index is squared
For example:
Unless we know the value of x, we do not know whether x is positive or negative.
• x2 + 3x + 2 = 0
• x2 – x = 0
Neither do we know whether –x is positive or negative.
• 2x2 – 3x – 5 = 0
minimum – the smallest y
possible value
maximum – the biggest
possible value
x
A(0;–5)
P(2;–9)
y = x – 4x – 5
2
We know that:
• x2 is positive for all values of x ⇒ x2 ≥ 0
• –x2 is negative for all values of x ⇒ –x2 ≤ 0
If we write the expression x2 – 4x –5 as (x – 2)2 – 9 we can see that:
• (x – 2)2 ≥ 0 for all x ∊ ℝ and (x – 2)2 = 0 only when x = 2
• (x – 2)2 – 9 ≥ –9 for all x ∊ ℝ and (x – 2)2 – 9 = –9 only when x = 2
We say that this expression has a minimum value of –9 when x = 2.
In the same way, writing the expression –x2 + 4x + 5 as – (x – 2)2 + 9 we can see that:
• –(x – 2)2 ≤ 0 for all x ∊ ℝ and (x – 2)2 = 0 only when x = 2
• –(x – 2)2 + 9 ≤ 9 for all x ∊ ℝ and –(x – 2)2 + 9 = 9 only if x = 2
We say that this expression has a maximum value of 9 when x = 2.
22 Topic 2 Equations and inequalities
PLT MATHS LB 11 7th pgs (Real Book).indb 22 2012/07/02 2:20 PM
WORKED EXAMPLE 1
Consider the expression: x2+ 6x + 9
1 Solve for x if x2 + 6x + 9 = 0.
State whether x2 + 6x + 9 has a minimum or a maximum value.
3 State the value of x for which we can find this minimum or maximum value.
4 State the minimum or maximum value of the expression.
SOLUTIONS
1 x2 + 6x + 9 = 0 ⇒ (x + 3)(x + 3) = 0 and so x = –3 or x = –3
2 The expression has a minimum value. | a = 1 > 0
3 x2 + 6x + 9 = (x + 3)2 and it has a minimum value when x = – 3.
4 (x + 3)2 ≥ 0, so its minimum value is 0.
WORKED EXAMPLE 2
Given the expression: –x2 + 6x – 9
1 Solve for x if –x2 + 6x – 9 = 0.
2 State whether the expression has a minimum or a maximum value.
3 State the value of x for which we can find this minimum or maximum value.
4 State the minimum or maximum value of the expression.
y
A(3;0) REMEMBER
x
• 5x + 2 = 2x – 3
• x2 – 2x – 3 = 0
Inequalities have inequality
A(0;–9) C(6;–9)
signs.
• x > 3 means the value of x
has to be greater than 3
• x ≥ –2 means that
y = –x2 + 6x – 9 minimum value of x is –2
• x < 7 means that the value
of x must be less than 7
SOLUTIONS
• x ≤ 5 means that maximum
value of x is 5
1 –x2 + 6x – 9 = 0 ⇒ x2 – 6x + 9 = 0
Expressions have neither
(x – 3)(x – 3) = 0 and x = 3
equal signs nor inequality
2 The expression has a maximum value. | a = –1 < 0 signs.
3 –x2 + 6x – 9 = – (x2 – 6x + 9) = –(x – 3)2 has a maximum when x = 3. • x2 – 3x – 4
4 –(x – 3)2 ≤ 0, so its maximum value is 0. • 2x – 5
Unit 1 Completing the square 23
PLT MATHS LB 11 7th pgs (Real Book).indb 23 2012/07/02 2:20 PM
EXERCISE 1
1 Given the expression: x2 + 10x + 25
1.1 Solve for x if x2 + 10x + 25 = 0.
1.2 State whether the expression has a minimum or a maximum value.
1.3 State the value of x for which we can find this minimum or
maximum value.
1.4 State the minimum or maximum value of the expression.
2 Given the expression: x2 – 4x + 4
2.1 Solve for x if x2 – 4x + 4 = 0.
2.2 State whether the expression has a minimum or a maximum value.
2.3 State the value of x for which we can find this minimum or
maximum value.
2.4 State the minimum or maximum value of the expression.
KEY WORDS 3 Given the expression: –x2 – 12x – 36
3.1 Solve for x if –x2 – 12x – 36 = 0.
coefficient – a number that
3.2 State whether the expression has a minimum or a maximum value.
stands in front of a variable or
3.3 State the value of x for which we can find this minimum or
variables in a term
For example: maximum value.
the coefficient of 3x2 is 3 3.4 State the minimum or maximum value of the expression.
the coefficient of x is 1
4 Given the expression: –x2 + 8x – 16
roots of an equation – the
4.1 Solve for x if –x2 + 8x – 16 = 0.
solutions of the equation
real numbers – the set of 4.2 State whether the expression has a minimum or a maximum value.
all rational and irrational 4.3 State the value of x for which we can find this minimum or
numbers maximum value.
For example: 4.4 State the minimum or maximum value of the expression.
1 ; 2,5
Rational: –2; 0; __
__ 3
Irrational: √7 ; π
non-real numbers – There are many quadratic equations that we cannot solve by means of factorisation.
imaginary numbers or Completing the square enables us to solve equations which cannot be factorised.
complex numbers that are
not real Consider the equation: 2x2 – 12x + 3 = 0
__ ___
For example: √–2 , √–16
We can break up the process of completing the square to solve an equation into steps:
• Divide both sides by the coefficient of x2 ⇒ x2 – 6x + 1,5 = 0 | Divide by 2.
REMEMBER
• Add x and half the coefficient of x (with its sign) and square the result. Then
subtract the square of half the coefficient of x ⇒ (x – 3)2 – (–3)2 + 1,5 = 0
LHS means Left Hand Side • Isolate the perfect square on the LHS and simplify the RHS ⇒___ (x – 3)2 = 7,5
RHS means Right Hand Side indicating ± on the RHS ⇒ x – 3 = ±√7,5
• Square root both sides,___
• Solve for x ⇒ x = 3 ±√ 7,5 = 5,74 or x = 0,26 | Correct to two decimal places.
The roots of an equation are non-real (imaginary) when a perfect square equals
a negative value.
The roots of (x – 2)2 = –9 are non-real or imaginary and the same applies to x2 = –5.
The roots of an equation are real and rational when a perfect square equals
a perfect square.
The roots of (x – 3)2 = 16 are real and rational. The same applies to (x – 4)2 = 1
and x2 = 9.
24 Topic 2 Equations and inequalities
PLT MATHS LB 11 7th pgs (Real Book).indb 24 2012/07/02 2:20 PM
The roots are real and irrational when a perfect square equals a positive value
which is not a perfect square. If (x + 2)2 = 11 the roots are real and irrational. REMEMBER
The same applies if x2 = 7.
For decimal rounding correct
to one decimal place, look at
WORKED EXAMPLE 1 the second decimal digit.
• if it is less than 5, leave the
1 Solve for x by completing the square if (x – 3)2 – 5 = 0. first decimal unchanged.
2 State whether the roots are real and rational, real and irrational or non-real. For example: 2,74 ≈ 2,7
• if it is 5 or more, round up
SOLUTIONS the first decimal digit by 1.
For example: 2,76 ≈ 2,8
1 (x – 3)2 – 5 = 0 | Do not multiply out the brackets. For decimal rounding correct
(x – 3)2 = 5 | Isolate the perfect square on the LHS. to two decimal places, look at
__
x–3=± 5 √ | Square root both sides; remember ± on the RHS. the third decimal digit.
__ For example: 5,321 ≈ 5,32
x = 3 ±√ 5 | Solve for x, simplified surd form.
But: 5,326 ≈ 5,33
x = 5,24 or 0,76 | Correct to two decimal places.
2 These roots are real and irrational.
WORKED EXAMPLE 2
1 Solve for x by completing the square if x2 – 8x = 9.
2 State whether the roots are real and rational, real and irrational or non-real.
SOLUTIONS
1 x2 – 8x = 9
(x – 4)2 – 16 = 9 | (x + half the coefficient of x)2 – (half coefficient of x)2
= 25 | Isolate the perfect square on LHS and simplify RHS. REMEMBER
x–4=±5 | Square root both sides; remember ± on the RHS.
x = – 1 or x = 9 | Solve for x. complex numbers
imaginary numbers
2 The roots are real and rational.
ℝ real numbers
rational numbers
integers
0 whole numbers
WORKED EXAMPLE 3 natural numbers
′ irrational numbers
1 Solve for x by completing the square if 3x2 + 12x +15 = 0.
2 State whether the roots are real and rational, real and irrational or non-real.
ℝ
SOLUTIONS
0
′
1 3x2 + 12x + 15 = 0
x2 + 4x + 5 = 0 | Divide through by 3, the coefficient of x2.
(x + 2) – 2 + 5 = 0 | (x + half the coefficient of x)2 – (half coefficient of x)2
2 2
(x + 2)2 = –1 | Isolate the perfect square on LHS and simplify RHS.
__
x + 2 = ±√–1 | Square root both sides; remember ± on RHS.
__
x = – 2 ±i | Solve for x and note that √ –1 = i
2 The roots are non-real.
Unit 1 Completing the square 25
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The next worked example is a literal equation because it does not have numerical
coefficients. We follow the same steps in both numerical and literal equations.
WORKED EXAMPLE 4
Solve for x by completing the square if ax2 + bx + c = 0.
SOLUTION
ax2 + bx + c = 0
c =0
b x + __
x2 + __ | Divide through by a, the coefficient of x2.
a a
( x + ___
2a )
b 2– b 2c = 0 | (x + half the coefficient of x)2 – (half coefficient of x)2
___ + __
4a a 2
( x + ___
2a )
2
b 2=b – 4ac
_______ | Isolate perfect square on LHS; simplify the RHS.
2
4a
_______
b =± b√ 2
– 4ac
x + ___ ________ | Square root both sides; remember ± on RHS.
2a 2a
_______
√ 2
x = –b± b – 4ac
___________ | You will use this quadratic formula in Unit 2, page 30.
2a
EXERCISE 2
Solve for x by completing the square. Give your answers correct to two decimal places.
For Questions 1– 8 say whether the roots are real and rational, real and irrational or
non-real.
1 x2 – 6x –12 = 0 2 x2 + 8x + 5 = 0
3 2x2 + 4x – 8 = 0 4 5x2 – 20x + 20 = 0
5 2x2 + 5x – 12 = 0 6 3x2 – 24x = 27
7 2x2 – 12x + 12 = 0 8 x2 – 10x + 29 = 0
9 px2 + qx + r = 0 10 kx2 – mx – n = 0
11 2mx2 – mx + k = 0 12 3kx2 + 2nx – k = 0
26 Topic 2 Equations and inequalities
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Unit 2: Quadratic equations
REMEMBER
You can solve equations by factorisation or by using the quadratic formula.
Factors of 12 are numbers
There are three ways to solve quadratic equations. which divide into 12 without
1 Factorisation – usually the quickest method leaving a remainder:
x2 – 4x – 12 = 0 ⇒ (x – 6)(x + 2) = 0 and x = 6 or x = –2 • 1, 2, 3, 4, 6 and 12 are
factors of 12
2 Quadratic formula – if you cannot find the factors. • 1, 2, 3, 6, x and y are
(Proved in Worked example 2 on page 26 in Unit 1) factors of 6xy
x2 – 4x – 12 = 0 ⇒ a = 1, b = – 4 and c = –12 • (x – 2) and (x – 3) are
_______ factors of x2 – 5x + 6.
√ 2
x = –b ± b – 4ac
____________
Factorising reminders: always
2a _____________
–(–4) ± √ (–4)2 – 4(1)(–12) take out highest common
= ______________________ factor first, and then look for:
2(1)
___
√
4 ± 64 • a difference of squares
= _______
2 • a trinomial
• grouping in pairs.
=4 ± 8 = ___
_____ 12 or – 4
___
2 2 4
= 6 or –1
KEY WORDS
3 Completing the square – if you are told to or prefer to. (Covered in Unit 1)
degree (in algebra) – the
If x2 – 4x – 12 = 0, solve for x by completing the square.
highest index in an equation
x2 – 4x – 12 = 0 ⇒ (x – 2)2 – 16 = 0 and (x – 2)2 = 16
or term
x – 2 = 4 or x – 2 = – 4 and x = 6 or x = –2 For example:
Check the degree of the equation and look for the same number of solutions as • 3x – 7 = 5 + x
is a first degree equation
the degree:
• 5x2 – 2x – 3 = 0
• First degree equations are linear equations and usually have one solution. is a second degree
• Second degree equations are quadratic equations and usually have two solutions. equation
• x3 – 2x2 – 3x = 0
• Third degree equations are cubic equations and have up to three solutions. is a third degree equation
WORKED EXAMPLE 1
Solve for x if x2 – 5x = 0.
REMEMBER
• 0 × any real number = 0
SOLUTION • If A × B = 0 then A = 0 or
B=0
x2 – 5x = 0 | x is a common factor.
• If A + B = 0 and A = 0, then
x(x – 5) = 0 ⇒ x = 0 or x = 5 | If x is a common factor, x = 0 is a solution. B=0
• If x2 = 4, then x = ±2
WORKED EXAMPLE 2
Solve for x if 3x2 – 12x = 0.
SOLUTION
3x2 – 12x = 0 | The highest common factor is 3x.
3x(x – 4) = 0 ⇒ x = 0 or x = 4
Unit 2 Quadratic equations 27
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WORKED EXAMPLE 3
Solve for x if x2 – 9 = 0.
SOLUTION
x2 – 9 = 0 | This is the difference between two squares.
Method 1 Method 2
(x – 3)(x + 3) = 0 and x = 3 or x = –3 x2 – 9 and x = ±3
WORKED EXAMPLE 4
Solve for x if (x – 5)2 – 16 = 0.
SOLUTION
(x – 5)2 – 16 = 0 | This is a difference between squares.
Method 1 Method 2
[(x – 5) – 4][(x – 5) + 4] = 0 (x –5)2 = 16
(x – 9)(x – 1) = 0 x – 5 = ±4
x = 9 or x = 1 x = 5 + 4 or x = 5 – 4
x = 9 or x = 1
WORKED EXAMPLE 5
Solve for x if x2 – 5x − 6 = 0.
SOLUTION
x2 – 5x − 6 = 0 ⇒ (x – 6)(x + 1) = 0 and x = 6 or x = –1
WORKED EXAMPLE 6
Solve for x if x2 – x = 12.
SOLUTION
x2 – x – 12 = 0 ⇒ (x – 4)(x + 3) = 0 and x = 4 or x = –3
28 Topic 2 Equations and inequalities
PLT MATHS LB 11 7th pgs (Real Book).indb 28 2012/07/02 2:20 PM
WORKED EXAMPLE 7
Solve for x if 10x2 –35x + 15 = 0.
SOLUTION
10x2 –35x + 15 = 0 | 5 is a common factor, so divide through by 5.
2x2 –7x + 3 = 0 ⇒ (2x – 1)(x – 3) = 0
1 or x = 3
2x – 1 = 0 or x – 3 = 0 ⇒ x = __
2
WORKED EXAMPLE 8
Solve for x if 6x3 –19x2 + 10x = 0.
SOLUTION
6x3 –19x2 + 10x = 0 | It is important to notice that x is a common factor.
x(6x2 – 19x + 10) = 0 | Do not divide by x, or you will lose the solution x = 0.
2 or x = 2,5
x(3x – 2)(2x – 5) = 0 ⇒ x = 0 or x = __
3
WORKED EXAMPLE 9
____
Solve for x if √x + 5 = x + 3.
SOLUTION
____
√ x + 5 = x + 3 | Square both sides to eliminate the square root.
____
(√x + 5 )2 = (x + 3)2 | Remember to check your solutions at the end.
x + 5 = x2 + 6x + 9 ⇒ x2 + 5x + 4 = 0
(x + 1)(x + 4) = 0 ⇒ x = –1 or x ≠ –4 | Eliminate the solution only after the check.
__
If x = –1, LHS = √4 = 2 and RHS = –1 + 3 = 2, so LHS = RHS
__
If x = –4, LHS = √1 = 1 but RHS = – 4 + 3= –1, so LHS ≠ RHS
x = –1 is the only solution | Checking is essential, not optional.
Reminders about trinomial factorisation:
• Both brackets have the same sign as the middle term if the trinomial ends with
a + sign.
x2 + 6x + 8 = (x + 2)(x + 4) and x2 – 6x + 8 = (x – 2)(x – 4)
2
6x + 17x + 12 = (2x + 3)(3x + 4) and 6x2 – 17x + 12 = (2x – 3)(3x – 4)
2x2 + 11x + 15 = (x + 3)(2x + 5) and 2x2 – 11x + 15 = (x – 3)(2x – 5)
BUT
• The brackets have opposite signs if the trinomial ends with a – sign.
• The bigger value of the inner and outer products has the same sign as the middle
term of the trinomial, and the other value carries the opposite sign.
x2 + 5x – 6 = (x + 6)(x – 1) and x2 – 5x – 6 = (x – 6)(x + 1)
3x2 + x – 14 = (x – 2)(3x + 7) and 3x2 – x – 14 = (x + 2)(3x – 7)
2
6x + x – 12 = (3x – 4)(2x + 3) and 6x2 – x – 12 = (3x + 4)(2x – 3)
Unit 2 Quadratic equations 29
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EXERCISE 3
Solve for x by means of factorisation:
1 x2 – 10x + 21 = 0 2 2x2 – x – 6 = 0
3 4x2 + 8x + 3 = 0 4 3x2 + x – 2 = 0
5 3x2 – 3(x – 1) = 4(2x + 1) + 3 6 x2 + 5x = 0
7 x2 = x 8 2x2 – 5x = 3
9 x2 – 24 = 10x 10 (x + 3)(x – 2) = 6
11 (x + 2)(x – 5) + 6 = 0 12 x2 – 6x = 5(1 – 2x)
13 4x2 + 11x + 6 = 0 14 (2x – 3)(4x – 3) = 9
15 6(x2 + 1) = 13x 16 (2x + 3)(2x – 5) = 9
17 3
19 x = __
x2 + ___ 18 5 = 11
2x + __
12 2 x
2(2x – 1) 3x – 1 5
19 x(4x + 5) = ________ 20 ___ __ = ___
3 2 4 4x
21 5– 4x = ___
______ 1 22 x+ 1 = ___
______ 1
3 2x 14 – x 8x
5x(x – 3)
________ 9 = 13
23 = 3(x – 5) + 6 24 4x2 + __
2 2
x
25 6p2x2 – 19pqx + 10q2 = 0 26 2a2x2 + 5abx – 12b2 = 0
____ _____
27 √x – 1 + 3 = x 28 1 + √ 5x – 1 = 2x
The quadratic formula enables us to solve quadratic equations that we
cannot factorise.
WORKED EXAMPLE
Solve for x if 3x2 + 5x – 1 = 0 | We cannot solve this equation by factorisation.
SOLUTION
_______
√ 2
x = –b± b – 4ac
___________
2a
__________
–5±√ 52 – 4(3)(–1)
= _______________ | a = 3; b = 5 and c = –1
2(3)
___
–5±√ 37
= _______ | Simplified surd form.
6
= 0,18 or –1,85 | Correct to two decimal places.
• Show only your substitution step and final answers.
• Make sure you give your answer in surd form when you are asked to.
• Pay attention to the number of decimal places you must round off to.
30 Topic 2 Equations and inequalities
PLT MATHS LB 11 7th pgs (Real Book).indb 30 2012/07/02 2:20 PM
EXERCISE 4
Solve for x by using the quadratic formula, correct to two decimal places where
necessary.
1 2x2 – 7x + 3 = 0 2 7x2 + 4x – 12 = 0
3 5x2 + 9x + 3 = 0 4 3x(x – 3) = 4(1 – x)
5 2
5(2x – 1) – x = 7x(2 – x) 6 2(x – 1)(x + 1) = 5x
7 2 =2
x + ___ 8 x2 – 2px + 3p = 0
3x
If you are given no specific instructions for solving an equation or determining
the roots:
• factorise if possible
• use the quadratic formula if you cannot find the factors
• complete the square if you prefer to.
EXERCISE 5
Determine the roots of the equations:
1 x2 – 5x – 6 = 0 2 x2 – 10x – 200 = 0
3 (2x – 3)2 – 16 = 0 4 2x2 – 5x – 6 = 0
5 10x2 – 59x – 221 = 0 6 5x(2x – 3) = 7
7 4 3
____ – y = __ 8 y2 – 6y + 7 = 0
y–1 2
p ____
9 3 = __
_____ – __ 1 10 2x – √ x – 1 = 3
p–1 p 4
A quadratic equation may have an unknown constant as well as one unknown root.
In this case you will be given one of the roots.
REMEMBER
Substitute the root you know so that you can find the unknown constant. Then make
use of this result to determine the other root of the equation. The root of an equation is
a value which solves the
When you give your answer, state the other root. equation.
A constant is a value which
does not change.
WORKED EXAMPLE
Given that x = 5 is a root of the equation px2 − 13x + 15 = 0, determine the value of
p. Then determine the value of the other root.
SOLUTION
Method 1 Method 2
Substitute x = 5 into px2 − 13x + 15 = 0 x – 5 is a factor of px2 − 13x + 15 = 0
p(5)2 − 13(5) + 15 = 0
Factorise by inspection:
⇒ 25p – 65 + 15 = 0
(x – 5)(2x – 3) = 0 | The middle term
25p = 50 and p = 2
–13 = –3 – 10.
Now substitute p = 2 into the equation: 3
The other root is x = __
2
2x2 – 13x + 15 = 0 ⇒ (2x –3)(x – 5) = 0
3
The other root is x = __
2
Unit 2 Quadratic equations 31
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EXERCISE 6
1 x = –3 is one of the roots of the equation given by kx2 + 3x – 9 = 0.
1.1 Determine the value of k.
1.2 Then determine the value of the other root.
2 x = 2 is one of the roots of the equation 3x2 + px – 10 = 0.
2.1 Determine the value of p.
2.2 Then determine the value of the other root.
3 x = – 1 is one of the roots of the equation 5x2 + 3x + a = 0.
3.1 Determine the value of a.
3.2 Then determine the value of the other root.
4 x = 3 is one of the roots of the equation 4x2 + tx –15 = 0.
4.1 Determine the value of t.
4.2 Then determine the value of the other root.
We can solve real life problems with quadratic equations.
• Decide what you have been asked to find.
• Let what you are finding be x.
• Translate the words into mathematical statements.
• Create an equation relating the statements.
• Solve the equation.
WORKED EXAMPLE 1
A group of Grade 11 learners goes out for lunch to celebrate their exam results.
They agree to share the expenses equally. The total bill is R1 200. Ten learners do
not have any money, so the others learners pay an extra R20 each. Determine the
number of learners in the group.
SOLUTION
Let the number of learners who go to lunch be x.
If all the learners pay, then each learner pays 1 200 .
_____
x
Ten learners do not pay, so x – 10 learners each pay 1 200 .
______
x – 10
The learners who pay, pay R20 more, so the difference is R20.
1 200 – 1
______ 200 = 20 | Extra amount to pay is R20.
_____
x – 10 x
1 200 × __
x–1 (x – 10)
200 × _______ x(x – 10)
______
x
_____
x = 20 × ________
x – 10 (x – 10) x(x – 10)
1 200x – 1 200x + 12 000 = 20x2 – 200x ⇒ 20x2 – 200x –12 000 = 0
x2 – 10x – 600 = 0 ⇒ (x + 20)(x – 30) = 0
x ≠ 20, x = 30 only (x > 0)
There were 30 learners in the group.
Check the answer by comparing the amounts paid.
If 30 learners pay, each learner pays 1 200 = R40, but if only 20 learners pay, each
_____
30
learner pays 1 200 = R60.
_____
20
x = 30 is the correct solution because the payments differ by R20.
32 Topic 2 Equations and inequalities
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WORKED EXAMPLE 2
A man calculates that if his speed increases by 10 km/h, he will take 48 minutes
less to complete a journey of 240 km. Calculate his original speed.
SOLUTION
Let his original speed be x and the increased speed be x + 10.
240 .
Time taken to complete journey at original speed is ____
x
240 .
Time taken to complete journey at increased speed is ______
x + 10
Difference in times is 48 minutes which converts into 48 4 hours.
___ = __
60 5
240 – ______
____ 240 = __4
x x + 10 5
240(5)(x + 10) – 240(5)x 4x(x
_____________________ + 10)
= _________
5x(x + 10) 5x(x + 10)
1 200x + 12 000 – 1 200x = 4x2 + 40x
4x2 + 40x –12 000 = 0
x2 + 10x – 3 000 = 0 ⇒ (x + 60)(x – 50) = 0
x ≠ –60 or x = 50 (x > 0) | His original speed was 50km/h.
WORKED EXAMPLE 3
A bricklayer and his apprentice build a wall in 24 days. When each person works
separately, the apprentice takes 20 days longer than the bricklayer to complete
the job. Calculate the number of days each person takes to complete the job on
his own.
SOLUTION
Let the bricklayer take x days and the apprentice take (x + 20) days to build a wall.
1 and ______
The bricklayer and apprentice build __ 1 of the wall each day.
x x + 20
1+ 1
Working together, they build __ ______ of the wall each day.
x x + 20
1 of the wall.
Together they build the wall in 24 days, so in one day they build ___
24
1 + ______
⇒ __ 1 = ___
1
x x + 20 24
24(x + 20) + 24x
______________ x(x + 20)
= __________
24x(x + 20) 24x(x + 20)
24x + 480 + 24x = x2 + 20x
x2 – 28x – 480 = 0
(x – 40)(x + 12) = 0
x = 40 or x ≠ –12 (x > 0)
The bricklayer takes 40 days on his own and the apprentice 60 days on his own.
Unit 2 Quadratic equations 33
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EXERCISE 7
1 A piece of wire of length 300 millimetres is bent to form a rectangle with area
3 125 square millimetres. Calculate the dimensions of the rectangle.
2 The sum of the digits of a two-digit number is 13 and the product of the digits
is 36. Determine which two numbers fit this description.
3 Vula is a river guide on the Gariep River. Robert, a member of the group, is
injured. Vula paddles Robert to the nearest pick-up point, 12 kilometres away.
Vulu paddles back to his group. If the total paddling time for the return trip
(there and back) is five hours and the river flows at a constant speed of 1 km/h,
calculate the average speed that Vula paddles.
4 A motorcyclist travels from A to B at 40 km/h and from B to A at 60 km/h.
If the distance between A and B is x km, determine the average speed at which
the motorcyclist travels from A to B and back to A. (The average speed is not
50 km/h.)
5 Thabiso has a budget of R525 per month for petrol. The present price of petrol is
x cents per litre. If the price rises by 50 cents per litre, she can buy five litres less
petrol for R525. Calculate the present price of petrol.
6 6.1 Vusi sets out for a 40 kilometre run. After he runs for two hours he injures
his ankle. He takes an hour to walk the rest of the way. If he injured
himself after he had run 16 kilometres, he would have taken four hours to
complete the distance. Determine the average speed in kilometres per hour
at which Vusi ran.
6.2 Now determine the average speed in kilometres per hour at which
Vusi walked.
7 Two windmills work continuously and together can fill a reservoir in six days.
Working separately, one windmill takes nine days longer than the other to fill
the reservoir. Calculate how long it takes each windmill to fill the reservoir.
8 A reservoir is fed by two pipes of different diameter. The pipe with the larger
diameter takes three hours less than the smaller pipe to fill the reservoir. If both
pipes are opened simultaneously, the reservoir can fill in two hours. Calculate
how long it takes the pipe with the larger diameter to fill the reservoir on its
own.
9 A rectangular parking area has dimensions of 50 metres by 120 metres. If the
parking area is doubled by increasing both the length and the breadth by x
metres, determine the dimensions of the new parking area.
120 + x
120
new parking area 50 + x
original parking 50
34 Topic 2 Equations and inequalities
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Unit 3: Quadratic inequalities
REMEMBER
Inequalities are often easier to understand if we use graphs to determine their solutions.
To solve x2 – 9 > 0, look for
The equation x2 – 9 = 0 has one unknown and two solutions. These solutions all possible values of x which
represent the x-intercepts of the graph y = x2 – 9. make the expression x2 – 9
positive.
When you solve an inequality you give the boundaries within which the solutions lie.
but
To solve x2 – 9 < 0, look for
WORKED EXAMPLES AND SOLUTIONS all possible values of x which
make the expression x2 – 9
1 x2 – 9 > 0 negative.
(x – 3)(x + 3) > 0
| Number line solution
+ –3 – 3 +
x < –3 or x > 3 | Inequality notation
x ∊ (–∞;–3) ∪ (3;∞) | Interval notation
y y = x2 – 9
+ +
+ +
x
(–3;0) (3;0)
– –
– –
(0;–9)
x-axis divides the graph into + and – values
y-values on the x-axis are 0.
y-values above the x-axis are +.
y-values below the horizontal x-axis are –.
2 x2 – 9 < 0
(x – 3)(x + 3) < 0, x ∊ ℝ
| Number line solution
+ –3 – 3 +
–3 < x < 3 | Inequality notation
x ∊ (–3;3) | Interval notation
3 x2 – 9 ≤ 0
(x – 3)(x + 3) ≤ 0, x ∊ ℝ
| Number line solution
+ –3 – 3 +
–3 ≤ x ≤ 3 | Inequality notation
x ∊ [–3;3] | Interval notation
Unit 3 Quadratic inequalities 35
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EXERCISE 8
Solve for x and illustrate your answer on a number line.
Make a rough sketch of a parabola for each question to help you find the answer.
1 x2 – 16 < 0 2 x2 – 25 ≥ 0
3 4x2 – 9 < 0 4 x2 – 1 ≤ 0
EXERCISE 9
State the values of x for which the expressions are real:
______ ______
1 √ x2 – 16 2 √ 4x2 – 9
Graphic interpretation of the solution
of inequalities in one variable
WORKED EXAMPLE AND SOLUTIONS
y
4
y=x+2
2
y = –x2 + 4
x
–2 2
Use the graphs to help you complete the table.
(Solutions are in the columns for Inequality notation and Interval notation.)
Statement Inequality notation Interval notation
1.1 –x2 + 4 ≥ 0 –2 ≤ x ≤ 2 x ∊ [–2;2]
1.2 x + 2 > – x2 + 4 x < – 2 or x > 1 x ∊ (–∞;–2) ∪ (1;∞)
1.3 –x2 + 4 > x + 2 –2 < x < 1 x ∊ (–2;1)
1.4 –x2 + 4 ≤ 0 x ≤ – 2 or x ≥ 2 x ∊ (–∞;– 2] ∪ [2;∞)
1.5 –x2 + 4 ≤ x + 2 x ≤ – 2 or x ≥ 1 x ∊ (–∞;–2] ∪ [1;∞)
36 Topic 2 Equations and inequalities
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EXERCISE 10
1 Use the graphs to help you complete the table.
y
4
y = x2 + x – 6
2
x
–3 –2 –1 1 2 3
–2
–4
–6
y = –x2 + x + 2
Statement Inequality notation Interval notation
1.1 x2 + x – 6 > 0
1.2 –x2 + x + 2 > 0
1.3 –x2 + x + 2 ≥ x2 + x – 6
1.4 x2 + x – 6 ≤ 0
2 Use the graphs to help you complete the table.
y
5
y = –2x + 3
4
3
y = –x2 + 2x + 3
2
1
x
–2 –1 1 2 3 4
–1
–2
–3
–4
–5
Statement Inequality notation Interval notation
2.1 –x2 + 2x + 3 > 0
2.2 –x2 + 2x + 3 > –2x + 3
2.3 –2x + 3 ≥ – x2 + 2x + 3
2.4 –2x + 3 ≤ 0
Unit 3 Quadratic inequalities 37
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3 Use the graph to help you complete the table.
y
y = x 2 + 5x + 4 y = x + 4
4
x
–4 –1
Statement Inequality notation Interval notation
3.1 x2 – 5x + 4 ≥ 0
3.2 x+4>0
3.3 x2 – 5x2 + 4 > x + 4
3.4 x2 + 5x + 4 < 0
38 Topic 2 Equations and inequalities
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4 Use the graphs to help you answer the questions. Do not show calculations.
y
21
18
15
12
9
6
3 y=x+3
x
–4 –2 2 4 6
–3
–6
–9
y = –x2 – 2x –15
–12
–15
–18
4.1 Write down the value(s) of x for which x2 – 2x – 15 = 0.
4.2 For which values of x is x2 – 2x – 15 > 0?
4.3 For which values of x is x2 − 2x − 15 = x + 3?
4.4 For which values of x is x2 − 2x − 15 ≤ x + 3?
5 y
2
y = – 1__3x2 − 7__3x – 2 1
x
–7 –6 –5 –4 –3 –2 –1
–1
y = – 1__3x − 2
–2
5.1 1 x2 + __
Write down the value(s) of x for which __ 7 x + 2 = 0.
3 3
5.2 1 x2 – __
For which values of x is – __ 7 x – 2 < 0?
3 3
5.3 1 x − 2 = – __
For which values of x is – __ 1 x2 − __
7 x − 2?
3 3 3
5.4 (3
1 x + 2 − __
For which values of x is __ ) (3
1 x2 + __
7 x + 2 > 0?
3 )
Unit 3 Quadratic inequalities 39
PLT MATHS LB 11 7th pgs (Real Book).indb 39 2012/07/02 2:20 PM
Unit 4: Equations in two unknowns
(simultaneous equations)
WORKED EXAMPLE AND SOLUTION
If y = –2x2 + 9x – 9 and y = –x – 1,
y
solve simultaneously for x and y.
Show all necessary calculations.
x
–2x2 + 9x – 9 = – x – 1
2x2 – 10x + 8 = 0 y = –x – 1
B(1;–2)
x2 – 5x + 4 = 0
(x – 4)(x – 1) = 0
x = 4 or x = 1 y = –2x2 + 9x – 9 A(4;–5)
If x = 4, y = – 4 – 1 = –5 ⇒ (4;–5)
If x = 1, y = –1 – 1 = –2 ⇒ (1;–2)
Always substitute the value(s) you have found into the linear equation to find
the other value(s).
In Topic 5 you draw graphs of parabolas and straight lines. You then solve the
simultaneous equations graphically by determining the points of intersection of
the two graphs. A(4;–5) and B(1;–2) are the points of intersection of the graphs in
the Worked example.
EXERCISE 11
Solve simultaneously for x and y in each equation. Show all necessary calculations.
1 y = 3x2 – 2x – 8 and y = 5x – 2
2 y = –5x2 + 4x + 9 and y + 6x = 9
3 3 x2 + 12 and 3x + 2y = 12
y = −__
4
4 1 x2 – __
y = __ 1 x + 6 and x + 2y = 8
2 2
5 1 (x – 1)(x + 6) and 2x + 5y = 5
y = – __
2
6 y = x2 – 5x + 4 and y – 2x + 6 = 0
7 1 (x + 1)2 – 2
y = –x2 – 5x + 6 and y = __
2
8 1 (x – 5)2 + 6
y = (x – 2)2 – 3 and y = – __
2
9 y = –x2 – x + 4 and y = 2(x – 1)2 – 4
10 1 x2 + 4
y = (x – 3)2 – 5 and y = – __
2
In Topic 5 you draw the hyperbola and straight line graphs. You then determine the
point(s) of intersection of these graphs graphically. In the next exercise you determine
the point(s) of intersection between the hyperbola and the straight line graph
algebraically. You also determine the simultaneous solutions of other graphs.
40 Topic 2 Equations and inequalities
PLT MATHS LB 11 7th pgs (Real Book).indb 40 2012/07/02 2:20 PM
WORKED EXAMPLE
y
If y = 2x and y = 4
_____, solve simultaneously
x–1
for x and y. Show all necessary calculations. A
y = x –4
____
1
SOLUTION
x
2x = 4 _____ ⇒ 2x(x – 1) = 4 | Multiply both
x–1 B
2
2x – 2x – 4 = 0 sides by (x – 1).
x2 – x – 2 = 0 y = 2x
(x – 2)(x + 1) = 0
x = 2 or x = –1
If x = 2, y = 2(2) = 4 ⇒ A(2;4)
If x = –1, y = 2(–1) = –2 ⇒ B(–1;–2)
It is not always possible to make either x or y the subject of an equation.
If x2 − 5xy + y2 = 7and x − 4y = 5, you cannot make either x or y the subject of the first
equation. In this case, use a substitution method and focus on the linear equation,
making x the subject to avoid fractions. Notice that x = 5 + 4y is fraction free, whereas
y=x−
_____5 contains a fraction.
4
Start by substituting x = 5 + 4y into x2 − 5xy + y2 = 7 which gives
(5 + 4y)2 − 5y(5 + 4y + y)2 = 7.
Solve for y and then find the x value of each of the y values by substituting back into
the linear equation.
EXERCISE 12
Solve simultaneously for x and y in each equation. Show all necessary calculations.
1 3 −1
y = 3x + 5 and y = _____
x+2
2 1 +2
x + 4y = 9 and y = – _____
x−1
3 5 +1
5x + y + 14 = 0 and y = – _____
x+3
4 3 −2
x – 3y = 10 and y = _____
x−4
5 x2 – xy – 6y2 = 6 and x + 2y = 6
6 x2 – 3xy + 2y2 = 4 and 3x – 2y = 0
7 x2 – 5xy + y2 = 7 and x – 4y = 5
8 (x – 2y)(2x + y) = 12 and 7y – x + 9 = 0
9 4x2 – 7xy + 4y2 = 16 and y = 2x – 2
10 5x2 – 3xy + 3x – 2y2 + 4y = 5 and y + 2x = 1
11 4 | Factorise your equation by grouping.
y = (x + 2)2 – 5 and y = __
x
Unit 4 Equations in two unknowns (simultaneous equations) 41
PLT MATHS LB 11 7th pgs (Real Book).indb 41 2012/07/02 2:20 PM
Unit 5: Nature of roots
To solve a quadratic equation you can:
• factorise
• use the quadratic formula
• complete the square.
It is sometimes useful to know whether an equation can be solved or not.
If you can solve an equation, you may be interested in the nature of its roots.
• Are the roots real or non-real (imaginary)?
• If the roots are real, are they rational or irrational?
• If the roots are rational, are they equal roots or are there two different roots?
_______
√ 2
Consider the quadratic formula: x = −b ± b − 4ac
_____________
2a
If you know the value of b2 − 4ac, then you know that:
Roots are real if b2 − 4ac ≥ 0, but roots are non-real (imaginary) if b2 − 4ac < 0.
You can further classify real roots as:
rational and unequal if b2 − 4ac = perfect square (not including 0)
rational and equal if b2 − 4ac = 0 | 0 is the smallest perfect square.
irrational and unequal if b2 − 4ac = positive number which is not a perfect square
Only an equation with rational roots can be solved by factorisation.
If you know the value of b2 − 4ac, do not work it out again when you use the
quadratic formula.
WORKED EXAMPLES
1 Without solving for x, fully classify the roots of x2 − 2x + 2 = 0.
2 Solve for x in as many different ways as possible if x ∊ ℝ.
3 Without solving for x, fully classify the roots of x2 − 6x + 9 = 0.
4 Solve for x in as many different ways as possible, if x ∊ ℝ.
5 Without solving for x, fully classify the roots of 2x2 − 5x − 7 = 0.
6 Solve for x in as many different ways as possible, if x ∊ ℝ.
7 Without solving for x, fully classify the roots of 3x2 + 5x − 3 = 0.
8 Solve for x in as many different ways as possible if x ∊ ℝ.
SOLUTIONS
1 If 1x2 − 2x + 2 = 0, then a = 1, b = −2, c = 2
∴ b2 − 4ac = (−2)2 − 4(1)(2) = 4 – 8 = −4
∴ b2 − 4ac < 0
∴ Roots are non-real.
2 No real solutions are possible.
3 If 1x2 − 6x + 9 = 0, then a = 1, b = −6, c = 9
∴ b2 − 4ac = (−6)2 − 4(1)(9) = 36 − 36 = 0
∴ b2 − 4ac = 0
∴ Roots are real (b2 − 4ac is positive), rational (b2 − 4ac is a perfect square) and
equal (b2 − 4ac = 0).
⇒ Roots are real, rational and equal.
42 Topic 2 Equations and inequalities
PLT MATHS LB 11 7th pgs (Real Book).indb 42 2012/07/02 2:20 PM
4 Factors: (x − 3)(x − 3) = 0 ⇒ x = 3 or x = 3
_______
√ 2
Quadratic formula: x = −b ± b − 4ac
_____________
2a Note that these roots are
−(−6) ±0
________
= real, rational and equal.
2(1)
= 6
__
2
=3
Completing the square: x2 − 6x + 9 = 0 ⇒ (x − 3)2 = 0 ⇒ x = 3
5 If 2x2 − 5x − 7 = 0, then a = 2, b = 5, c = −7
∴ b2 − 4ac = (−5)2 − 4(2)(−7) = 25 + 56 = 81
∴ Roots are real (b2 − 4ac is positive), rational (b2 − 4ac is a perfect square) and
unequal (b2 − 4ac ≠] 0).
6 Factors: 2x2 − 5x − 7 = 0 ⇒ (2x − 7)(x + 1) = 0 and x = 3,5 or x = −1
_______
√ 2
−b ± b − 4ac ____
Quadratic formula: x = _____________ = 5±9 = ___
14 or ___
−4 = 3,5 or −1
2a 4 4 4
_______ ___
You already know √b2 − 4ac = √ 81 = 9
5 x = __
Completing the square: x2 − __ 5
7 ⇒ x − __
2 2 4
2
( )
25 = 56
7 + ___
= __
2 16
+ 25 = ___
_______
16
81
16
5 9
x− =± ⇒x=
__ __ 14
___ 7
__
= or x = −4
___ = −1
4 4 4 2 4
x = 3,5 or –1
7 If 3x2 + 5x − 3 = 0, then a = 3, b = 5, c = −3
∴ b2 − 4ac = (5)2 − 4(3)(−3) Note that these roots are
= 25 + 36 real, rational and unequal.
= 61
∴ Roots are real (b2 − 4ac is positive), irrational (b2 − 4ac is not a perfect
square) and unequal (b2 − 4ac ≠ 0).
___ ___
√ √
8 Quadratic formula: x = −b ± 61 −5±
________ 61
= _______ = 0,47 or −2,14
2a 6
5x = 1 ⇒
Completing the square: x2 + __
3 ___
( )
5 2 = 1 + ___
x + __
6
25 = ___
___ 36
61
36
Note that these roots
5 ±√ 61 −5±√ 61
__
x+ = _____ ⇒x= _______ = 0,47 or − 2,14 are real, irrational and
6 6 6
unequal.
EXERCISE 13
Fully classify the roots of each equation without solving the equation.
Solve each equation in as many different ways as possible if x ∊ ℝ.
1 5x2 – 2x + 1 = 0 2 7x2 – 3x – 4 = 0
3 7x2 – 3x + 4 = 0 4 2x2 – 3x + 5 = 0
5 4x2 – 12x + 9 = 0 6 3x2 + 5x + 1 = 0
7 25x2 + 20x + 4 = 0 8 x2 + 7x + 13 = 0
Unit 5 Nature of roots 43
PLT MATHS LB 11 7th pgs (Real Book).indb 43 2012/07/02 2:20 PM
Revision
1 Solve for x in each equation, correct to two decimal places where necessary.
1.1 (x – 3)(2x – 5) = 0 (2)
1.2 (x – 5)(x + 2) = 6 (5)
1.3 (x – 3)(x + 2) = 6 (5)
1.4 (x – 2)(x – 3) = (2x – 5)(x + 3) (7)
2
1.5 10x
______ =3 (6)
7x − 4
1.6 2x −3−
______ x− 1
__________ 1
= __________ (8)
x2 − 4 2x2 − 5x + 2 2x2 + 3x − 2
_____
1.7 √ 2x − 5 + 4 = x | Remember to check your answers. (6)
_____
1.8 x − √2x − 3 = 3 | Remember to check your answers. (6)
[45]
2 Solve for x by completing the square:
2.1 3x2 – 12x = 6 (5)
2.1 (2x – 5)(3x – 2) + 7x = 22 (6)
[11]
3 Consider the equations and match each one to the statement
which best describes its roots.
3.1 x2 – 7x – 18 = 0 (4)
3.2 25x2 + 20x + 4 = 0 (4)
3.3 2x2 + 7x + 4 = 0 (4)
3.4 3x2 – 5x + 3 = 0 (4)
A Real, rational and equal roots
B Real, rational and unequal roots
C Real, irrational and unequal roots
D Non-real or imaginary roots
[16]
4 Consider the equation: 2x2 – 5x = 9
4.1 Without solving for x, discuss the nature of the roots. (5)
4.2 Solve for x, correct to two decimal places, by:
4.2.1 completing the square (5)
4.2.2 using the quadratic formula. (4)
[14]
5 Solve for x and illustrate your answer on a number line.
5.1 x2 ≤ 64 (5)
5.2 81 – 4x2 ≥ 0 (5)
5.3 x2 – x – 12 > 0 (5)
5.4 x2 – 7x ≥ 0 (5)
[20]
44
PLT MATHS LB 11 7th pgs (Real Book).indb 44 2012/07/02 2:20 PM
6 Refer to the graphs and complete the table.
2 y
y = x –4x + 3 y = 2x + 3
15 • (6;15)
12
9
6
• 3
3
• •1 2 3• 4 5 6 x
–2 –1
– 1,5 1 3
Statement Inequality notation Interval notation
6.1 2x + 3 ≥ 0
6.2 (x – 1)(x – 3) ≤ 0
6.3 (x – 1)(x – 3) ≥ 2x + 3
6.4 (x – 1)(x – 3)(2x + 3) < 0 [20]
7 Solve for x and y if:
7.1 y = –x2 + 2x + 8 and 3x – y + 2 = 0 (6)
7.2 y – 4x + 14 = 0 and y = x2 – 3x – 4 (6)
7.3 y= 4
_____ and x + 2y + 4 = 0 (7)
x–5
7.4 y = – 2x2 + 10x – 8 and y = x2 – 2x – 8 (7)
[26]
8 The price of petrol is Rx per litre. After an increase in price of R1 per litre,
you can buy seven litres less petrol for R504. Calculate the original petrol
price. Show all your calculations. [6]
9 A group of learners raise money for a modelling and dance show in
a community hall. The total cost of producing the show is R6 000
and the learners plan to raise equal amounts of money. Eight learners
do not raise any money, so their friends raise an extra R25 each.
Determine the number of learners who participate in the show. [7]
10 The sum of two numbers is 40 and the sum of their squares is 818.
Determine the numbers. [7]
45
PLT MATHS LB 11 7th pgs (Real Book).indb 45 2012/07/02 2:20 PM
TOPIC
3 Number patterns
Unit 1: Linear patterns
Consider these number patterns:
1 3; 5; 7; 9; … If the pattern continues in the same way, the next three numbers
will be 11; 13; 15.
2 2; 4; 8; 16; … If the pattern continues in the same, the next three numbers will
be 32; 64; 128.
3 1; 4; 9; 16; … If the pattern continues in the same, the next three numbers will
be 25; 36; 49.
It is often easy to continue a pattern without using a formula. In the patterns above:
1 You add 2 to each term to get the next term. This is a linear pattern.
2 You double each term to get the next term. This is an exponential pattern.
3 The difference increases by 2 each time, so you add 3, then 5, then 7, then 9,
then 11 and finally 13. This is a quadratic pattern.
The problem with this method is that finding the 1 000th term requires you to work
out 999 terms to work out the 1 000th term! To avoid working out terms you do not
need, you develop a formula based on the position of the term.
Consider the same patterns again, but pay attention to each number’s position in
the pattern.
1 T1 = 3; T2 = 5; T3 = 7
KEY WORDS If you use the fact that each term increases by 2 and consider its position in the
pattern, then: T1 = 3 = 2 × 1 + 1; T2 = 2 × 2 + 1; T3 = 2 × 3 + 1; Tn = 2n + 1
n − the position of the term
nth term − the term in the 2 T1 = 2; T2 = 4; T3 = 8; T4 = 16
position n You know that each term is doubling, but you need to link the terms to their
positions.
T1 = 2 = 21; T2 = 4 = 2 × 2 = 22; T3 = 8 = 2 × 2 × 2 = 23;
T4 = 16 = 2 × 2 × 2 × 2 = 24
The formula for the nth term, Tn, is Tn = 2n.
3 Although you can see that the difference increases by 2 each time, you need to
think about these numbers carefully and to look for a connection between the
value of the number and its position.
T1 = 1 = 11; T2 = 4 = 2 × 2 = 22; T3 = 9 = 3 × 3 = 32; T4 = 16 = 4× 4 = 42
The formula for the nth term, Tn, is Tn = n2.
The linear pattern formula for 3; 5; 7; 9; … is Tn = 2n + 1 where 2 is the common
difference between the numbers. In the same way that you can continue the pattern
forwards with 11; 13 and 15, you can continue the pattern backwards and work out
that the term before 3 is 1. ⇒ T0 = 2 × 0 + 1 = 1
The general formula for the value of the nth term is Tn = an + b where a is the common
difference, b is the value of T0 and n is the position of the term in the pattern.
46 Topic 3 Number patterns
PLT MATHS LB 11 7th pgs (Real Book).indb 46 2012/07/02 2:20 PM
WORKED EXAMPLE 1 REMEMBER
Consider the pattern: 3; 11; 19; 27; …. If T2 = 11
1 Write the next three terms as if the pattern continues in the same way. Tn = 11 and n = 2
2 Write a formula for the nth term in the form Tn = … .
3 Determine the 200th term.
4 Which term has a value of 179?
SOLUTIONS
1 The terms increase by 8 each time, so the next three terms are 35; 43 and 51.
2 a = 8 and T0 = −5 ⇒ Tn = 8n − 5 | Check: T7 = 8(7) − 5 = 56 − 5 = 51
3 T200 = 8(200) − 5 = 1 600 −5 = 1 595
4 8n − 5 = 179 ⇒ 8n = 184 and n = 23, so T23 = 179
WORKED EXAMPLE 2 REMEMBER
Consider the pattern: 17; 13; 9; 5; … If the value is −255
you know Tn = −255
1 If the pattern continues in the same way, write down the next three terms. and you need to find n.
2 Write down a formula for the nth term in the form Tn = … . Find the 163rd term tells you
3 Determine the 163rd term. n = 163, so find T163.
4 Which term has a value of −255?
SOLUTIONS 2
1 The terms decrease by 4 each time, so the next three terms are 1; −3 and −7.
2 a = −4 and T0 = 21 ⇒ Tn = −4n + 21 | Check: T7 = −4(7) + 21 = −28 + 21 = −7
3 T163 = −4(163) + 21 = −631
4 −4n + 21 = −255 ⇒ −4n = −276 and n = 69, so T69 = −255
EXERCISE 1
1 Consider the pattern: 39; 34; 29; 24; 19; …
1.1 Write down the next three terms,
1.2 If the pattern continues in the same way, write a formula for the nth term
in the form Tn = … .
1.3 Determine the 22nd term.
1.4 Which term has a value of −251?
2 Consider the pattern: −11; −8; −5; −2; …
2.1 Write down the next three terms if the pattern continues in the same way.
2.2 If the pattern continues in the same way, write a formula for the nth term
in the form Tn = … .
2.3 Determine the 100th term.
2.4 Which term has a value of 178?
Unit 1 Linear patterns 47
PLT MATHS LB 11 7th pgs (Real Book).indb 47 2012/07/02 2:20 PM
Unit 2: Quadratic patterns
You often refer to quadratic patterns as second difference patterns that have a
general formula: Tn = an2 + bn + c
A linear pattern has a constant first difference, but a quadratic pattern has a
constant second difference. Other patterns, such as exponential patterns, do not
have a constant difference.
WORKED EXAMPLE
The T0 method
Consider the pattern 1; 4; 9; 16; … and expand it as follows:
Step 1: Write T1 = 1; T2 = 4; T3 = 9; T4 = 16; T5 = 25 as indicated below.
Step 2: Work out the first differences for 3; 5; 7 and 9 and record the ansyours in the
gaps as shown.
Step 3: Repeat the process, working out the second differences, which all equal 2.
Step 4: A common second difference confirms that this is a quadratic or second
difference pattern.
Work backwards one row at a time, filling in the number to the left of
each row.
These numbers have been enclosed in a square to show that they youre filled
in afterwards.
When you fill in the number under T0, you have the constant term.
T0 T1 T2 T3 T4 T5
0 1 4 9 16 25
d1 1 3 5 7 9
d2 2 2 2 2
d
Tn = an2 + bn + c with a = __2 ; T0 = c; T1 = a + b + c
2
You know that a = 1 and c = 0, but you still need to find b.
You know that T1 = 1 and by substitution into T1 = a + b + c you determine b.
T1 = a + b + c ⇒ 1 = 1 + b + 0 and so b = 0 and Tn = n2.
EXERCISE 2
For each number pattern, write down the values of the 6th and 7th terms. Use the
T0 method to determine the formula for the nth term, Tn. Work out the position
of the last value in each sequence.
1 0; 3; 10; 21; 36; … 17 020 2 1; 12; 29; 52; 81; … 9 857
3 −3; 1; 7; 15; 25; … 865 4 2; 6; 12; 20; 30; … 992
5 5; 3; −3; −13; −27; … −8 707 6 1; 3; 6; 10; 15; … 1 540
7 −1; −1; 0; 2; 5; … 1 769 8 1; 10; 24; 43; 67; … 439
9 0; −6; −17; −33; −54; … −873 10 0; −2; −2; 0; 4; … 4 690
48 Topic 3 Number patterns
PLT MATHS LB 11 7th pgs (Real Book).indb 48 2012/07/02 2:20 PM
WORKED EXAMPLE
The second differences method
If we consider the quadratic formula Tn = an2 + bn + c and work out
the first six terms we get:
T1 = a(1)2 + b(1) + c = a + b + c
T2 = a(2)2 + b(2) + c = 4a + 2b + c
T3 = a(3)2 + b(3) + c = 9a + 3b + c
T4 = a(4)2 + b(4) + c = 16a + 4b + c
T5 = a(5)2 + b(5) + c = 25a + 5b + c
T6 = a(6)2 + b(6) + c = 36a + 6b + c
Now arrange these terms as shown:
T1 T2 T3 T4 T5 T6
a+b+c 4a + 2b + c 9a + 3b + c 16a + 4b + c 25a + 5b + c 36a + 6b + c
3a + b 5a + b 7a + 2b 9a + b 11a + b
2a 2a 2a 2a
Compare the result with the numbers for the pattern 1; 4; 9; 16; 25; …
T1 T2 T3 T4 T5 T6
1 4 9 16 25 36
3 5 7 9 11
2 2 2 2
Starting from the bottom row, write equations and solve each one as you work
your way to the top.
Step 1: 2a = 2 ⇒ a = 1
Step 2: 3a + b = 3, we know a = 1, so 3(1) + b = 3 ⇒ b = 0
Step 3: a + b + c = 3, we know that a = 1 and b = 0, so 1 + 0 + c = 1 and c = 0
The formula: Tn = n2
EXERCISE 3
Consider each number pattern and write down the next two terms. Using the second
differences method, determine the formula for each of the sequences below.
Now determine how many terms there are in each sequence if the value of the last
term is given.
1 1; 4; 11; 22; 37; … 667 2 8; 7; 4; −1; −8; … −617
3 −4; 0; 10; 26; 48; … 1 896 4 7; 1; −8; −20; −35; … −2 099
5 −9; −3; 10; 30; 57; … 3 432 6 19; 10; −7; −32; −65; … −1 085
Unit 2 Quadratic patterns 49
PLT MATHS LB 11 7th pgs (Real Book).indb 49 2012/07/02 2:20 PM
REMEMBER WORKED EXAMPLE
1 2 3 Pictures often contain patterns. You must be able to extract information from the
pictures in the same way as you do from numbers.
4 5 6
7 8 9
4, 5 and 6 are in the same
horizontal row.
A hexagon has six sides. Figure 1 Figure 2 Figure 3 Figure 4
Consider the four figures and use them to help you complete the table. It would
take a long time to draw Figure 5, and this is not necessary.
Figure 1 Figure 2 Figure 3 Figure 4 Figure 5 Figure n
No. of
horizontal 1 3 5 Tn = 2n − 1
rows
No. of
hexagons in 1 3 5 Tn = 2n − 1
longest row
No. of green
1 1
hexagons
No. of red
0 6 n≥2
hexagons
No. of yellow
0 n≥3
hexagons
No. of blue
0 n≥4
hexagons
No. of
hexagons in 0
outer ring
Total no. of
1 7
hexagons
No. of
hexagons excl. 0 6
green
Think carefully about your answers to Figures 1− 4 and then find the pattern. There
are often different patterns within one figure. Make use of each pattern as soon as
you discover it.
50 Topic 3 Number patterns
PLT MATHS LB 11 7th pgs (Real Book).indb 50 2012/07/02 2:20 PM
SOLUTION
Fig. 1 Fig. 2 Fig. 3 Fig. 4 Fig. 5 Fig. n
Number of
1 3 5 Tn = 2n − 1
horizontal rows
No. of hexagons
1 3 5 Tn = 2n − 1
in longest row
No. of green
1 1 1 1 1 1
hexagons
No. of red
0 6 6 6 6 6 (n ≥ 2)
hexagons
No. of yellow
0 0 12 12 12 12 (n ≥ 3)
hexagons
No. of blue
0 0 0 18 18 18 (n ≥ 4)
hexagons
No. of hexagons
0 0 12 18 24 Tn = 6(n − 1)
in outer ring
Total no. of
1 7 19 37 61 Tn = 3n2 − 3n + 1
hexagons
No. of hexagons
0 6 Tn = 3n2 − 3n
excl. green
Outer rings: 0; 6; 12; 18 …
Linear pattern with common difference = 6 and T0 = − 6 ⇒ Tn = 6n − 6 = 6(n − 1)
Total number of hexagons:
T0 T1 T2 T3 T4 T5
1 1 7 19 37 61
d1 0 6 12 18 24
d2 6 6 6 6
Using the T0 method:
2a = 6 ⇒ a = 3 and c = T0 = 1. T1 = a + b + c, so 1 = 3 + b + 1 and b = −3
Tn = 3n2 − 3n + 1
Using the second differences method:
Step 1: 2a = 6 ⇒ a = 3
Step 2: 3a + b = 6, so 3(3) + b = 6 and b = − 3
Step 3: a + b + c = 1, so 3 + (− 3) + c = 1 and c = −1
Tn = 3n2 − 3n + 1
If the green hexagon is excluded, then the total number of hexagons is one less
than before.
Use the last result and subtract 1 from the formula.
Unit 2 Quadratic patterns 51
PLT MATHS LB 11 7th pgs (Real Book).indb 51 2012/07/02 2:20 PM
EXERCISE 4
1
REMEMBER
A cube has 6 faces.
Face
You can see three faces in
Figure 1 Figure 2 Figure 3
this cube:
• one blue face Consider the figures. The cubes are stacked against two walls so we can see
• two white faces no faces on the other side. Copy and complete the table.
The stack of cubes has
three layers. Fig. 1 Fig.2 Fig. 3 Fig. 4 Fig. 5 Fig. n
Total number of blue faces 1
top layer Total number of white faces 2
middle layer Total number of faces visible 3
bottom layer Number of layers 1
Number of cubes in bottom layer 1
2
REMEMBER
The figure is a pentagon.
Figure 1 Figure 2 Figure 3 Figure 4 Figure 5
Figures 1−5 show the first five pentagonal numbers. A pentagon has five sides
and each polygon is pentagonal in shape. In each figure the red beads are joined
by black rods.
Each red dot is a bead.
Each line is a rod. Complete the table.
Fig. 1 Fig. 2 Fig. 3 Fig. 4 Fig. 5 Fig. n
Number of red beads 1 5 12
Number of black rods 0 5 13
Number of beads and rods 1 10 25
Not all patterns are linear or quadratic. You can solve exponential and cubic patterns
by inspection, so you do not use special formulae or rules. In each number pattern
question, decide what type of pattern you have. Use the checklist to help you to
identify patterns:
• Numbers with a common difference form a linear pattern.
• Numbers with a common second difference form a quadratic pattern.
• Numbers with a common third difference form a cubic pattern.
52 Topic 3 Number patterns
PLT MATHS LB 11 7th pgs (Real Book).indb 52 2012/07/02 2:20 PM
• Numbers with no common differences can be broken up into prime factors.
• If you cannot find a pattern, check whether adding or subtracting a constant
amount to each term results in a pattern of exponents or cubes.
WORKED EXAMPLES
Consider each number pattern and determine the nth term, Tn:
1 2; 4; 8; 16; 32; …
2 1; 8; 27; 64; 125; …
3 2; 8; 26; 80; 242; …
4 5; 10; 20; 40; 80; …
SOLUTIONS
1 2; 2 × 2; 2 × 2 × 2; …
21; 22; 23; 24; 25
Tn = 2n
2 1; 2 × 2 × 2; 3 × 3 × 3; 4 × 4 × 4; …
13; 23; 33; 43; …
Tn = n3
3 3 − 1; 9 − 1; 27 − 1; 81 − 1; 243 − 1, …
31 − 1; 32 − 1; 33 − 1; 34 − 1; 35 − 1; …
Tn = 3n − 1
Adding 1 to each given term gives you an exponent of 3, so express each term
as an exponent of 3 and then −1.
4 5; 5 × 2; 5 × 4; 5 × 8; 5 × 16
5 × 20; 5 × 21; 5 × 22; 5 × 23; 5 × 24
Tn = 5 × 2n − 1
Each index is 1 less than the position of the term. Instead of an index of n,
use n − 1.
The next exercise is a mixed exercise with linear and quadratic patterns as well as
unexpected patterns for which you find a rule by trial and error!
EXERCISE 5
If each pattern continues in the same way, write down the next two terms in
the pattern.
Determine a rule for the patterns. Then determine how many terms there are
in the sequence.
1 1; 5; 25; 125; 625; … 390 625 2 14; 12; 4; −10; −30; … −2 146
3 1; 8; 27; 64; 125; … 1 000 4 6; 12; 24; 48; 96; … 3 072
5 −2; −1; 4; 13; 26; … 18 334 6 13; 3; −15; −41; −75; … −635
7 2; 6; 18; 54; 162; … 13 122 8 0; 7; 26; 63; 124; … 9 999
Unit 2 Quadratic patterns 53
PLT MATHS LB 11 7th pgs (Real Book).indb 53 2012/07/02 2:20 PM
Revision
1 Consider the pattern: 7; 3; −1; −5; −9; … −121
If the pattern continues in the same way:
1.1 Write down the next two numbers in the pattern. (2)
1.2 Determine the formula for the nth term, Tn . (5)
1.3 Determine the value of the 19th term. (3)
1.4 Which term has a value of −121? (3)
[13]
2 Consider the pattern: −14; −15; −12; −5; 6; … 4 080
If the pattern continues in the same way:
2.1 Write down the next two numbers in the pattern. (2)
2.2 Determine the formula for the nth term, Tn . (5)
2.3 Determine the value of the 25th term. (3)
2.4 Which term has a value of 4 080? (5)
[15]
3 Consider the pattern: 14; 14;12; 8; 2; … −96
If the pattern continues in the same way:
3.1 Write down the next two terms in the pattern. (2)
3.2 Determine the formula for the nth term, Tn. (5)
3.3 If the pattern is continued to include more terms, find the value
of the 51st term. (3)
3.4 How many terms are there in the sequence? (4)
[14]
4 Consider the quadratic pattern: −9; −6; 1; x; 27; …
4.1 Give the value of the second difference. (2)
4.2 Determine an expression for the second difference in terms of x. (2)
4.3 Determine the value of x. (1)
4.4 Determine the 9th term in the sequence. (6)
4.5 Which term in the sequence has a value of 397? (4)
[15]
5 Consider the number patterns that follow. In each case state the value of
the unknown term and write the formula for the nth term, Tn .
5.1 7; 14; 21; 28; a; 42; … (2)
5.2 −7; −2; 3; b; 13; … (3)
5.3 14; 28; 56; c; 224; … (5)
5.4 3; 21; 147; d; 7 203; … (4)
5.5 6; 18; 36; e; 90; 126, … (6)
[20]
54
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6 A children’s construction kit is designed so that flat, square tiles connect to
each other to create cubic storage boxes, which are open at the top. In the figures REMEMBER
each structure has been turned upside down so it is easier to count the tiles in
the base. flat tile
vertical
flat tile
The yellow surface is a flat tile.
The structure is upside down,
Figure 1 Figure 2 Figure 3 so the yellow tile is the base
tile. You can see two pink tiles.
Refer to the figures and complete the table.
This cube has:
Figure 1 2 3 4 5 Structure n • four vertical pink tiles
Number of tiles in the base of • one yellow base tile
1 4 9 • there is no tile on the top.
the structure
Number of vertical tiles in the open top
4 12
structure
Total number of tiles in
5 16 33 56
the structure [11]
7
Figure 1 Figure 2 Figure 3 Figure 4 Figure 5
Figures 1− 5 show the first five heptagonal numbers. A heptagon has seven sides
and each polygon is heptagonal in shape. In each figure the red beads are joined
by blue rods. Complete the table.
Fig. 1 Fig. 2 Fig. 3 Fig. 4 Fig. 5 Fig. n
Number of red beads 1 7 18
Number of blue rods 0 7 19
Number of beads and rods 1 14 37
[15]
55
PLT MATHS LB 11 7th pgs (Real Book).indb 55 2012/07/02 2:20 PM
TOPIC
4 Analytical geometry
Unit 1: Equations of a line
KEY WORD
From Grade 10 you know these formulae for working with points in the
Cartesian plane – a two- Cartesian plane: ________________
dimensional plane formed
by a horizontal number
• The distance between points A(x1;y1) and B(x2;y2) = √(x2 − x1)2 + (y2 − y1)2
• The midpoint between points A(x1;y1) and B(x2;y2) = ( 1 2 )
line representing x-values x +x y +y
______2 ; 1
______2
and a vertical number line 2
(y – y )
representing y-values, which _______
• The gradient of a line between points A(x1;y1) and B(x2;y2) = (x2 1
– x1)
intersect at the point where 2
both values are 0; ordered rise
(This is the same as considering ____
run from point A to point B, taking the direction of
pairs or coordinates refer to
the line into account to determine whether the gradient is positive or negative.)
the position of points in the
plane
= Positive gradient = Negative gradient
REMEMBER It is important to remember that:
Rise represents the change in • parallel lines have equal gradients
y values (the number of units • perpendicular lines have gradients which are negative reciprocals.
upwards), and run represents (This implies that their product = –1).
the change in x values (the 3 and –
For example: gradients of __ 3×–
4 are perpendicular because __
___ 4 = –1
___
4 3 4 3
number of units across) to get
from one point to another. You know that the standard form equation of a straight line is y = mx + c.
Another formula that we can also use for straight lines, particularly when we do not
know the y-intercept (the value of c) is:
KEY WORD
reciprocal – the multiplication y – y1 = m(x – x1) where m represents the gradient of the line.
inverse of a number is
obtained by interchanging
the numerator and the (x1;y1) represents the coordinates of any point on the graph.
denominator of that number
WORKED EXAMPLE
A line passes through the points A(–1;3) and B(5;0).
1 Determine the equation of line AB.
2 Determine the equation of a line parallel to AB, and passing through
the point (2;–3).
3 Determine the equation of the line perpendicular to AB and passing
through the origin.
56 Topic 4 Analytical geometry
PLT MATHS LB 11 7th pgs (Real Book).indb 56 2012/07/02 2:20 PM
SOLUTIONS
1 m= 0 – 3 = ___
_______ –3 = – __
1
5 – (–1) 6 2
y – y1 = m(x – x1) using (x1;y1) = (5;0)
1 (x – 5)
y – 0 = – __
2
5 or 2y + x = 5
1 x + __
y = – __
2 2
2 1 | Parallel lines have equal gradients.
m = – __
2
y – y1 = m(x – x1) using (x1;y1) = (2;–3)
1 (x – 2)
y + 3 = – __
2
1x + 1 – 3
y = – __
2 The Worked example
1 x – 2 or 2y + x = –4
y = – __ would have been
2
simpler if you realised
3 m = 2 | Perpendicular gradients; – __
1 × 2 = –1 that if the line passes
2
through the origin
y – y1 = m(x – x1) using (x1;y1) = (0;0)
c = 0, so y = mx + c
y – 0 = 2(x – 0) becomes y = 2x.
y = 2x
Important terminology A
The median of a triangle is a line from the
vertex to the midpoint of the opposite side.
So, AD is a median of △ABC.
Note: D is perpendicular to BC only when B C
AB = AC. D
KEY WORDS
The altitude (height) of a triangle is a line A
vertex – a point on a triangle
from any vertex, perpendicular to the
where the sides meet, so a
opposite side.
triangle has three vertices
So, AE is an altitude of △ABC. perpendicular – lines are
perpendicular when the angle
Note: E will be the midpoint of BC only if between the lines is 90°
AB = AC. B C
E equilateral triangle – a
triangle with all three sides
The perpendicular bisector of a line passes A equal and all three angles
through the midpoint of the line, and is equal
perpendicular to the line.
F
So, FG is the perpendicular bisector of BC.
Note: FG will pass through point A only if AB = AC.
B C
If AB = AC, then the altitude and
perpendicular bisectors are the same line.
G
In an equilateral triangle, this will be true for
all three possible medians, altitudes and perpendicular bisectors of the triangle.
Unit 1 Equations of a line 57
PLT MATHS LB 11 7th pgs (Real Book).indb 57 2012/07/02 2:20 PM
WORKED EXAMPLE
Given points P(–2;6), Q(5;9) and R(3;–3).
1 Determine the equation of the y
Q(5;9)
median of △PQR from point P.
2 Determine the equation of the P(–2;6)
altitude of △PQR from point R.
3 Determine the equation of the
perpendicular bisector of QR.
4 Determine the point of intersection
x
of the median and the altitudes
found in Questions 1 and 2.
5 Determine the coordinates of point
R(3;–3)
S so that PQSR is a parallelogram.
SOLUTIONS
1 Midpoint of QR = 5 + 3;9
_____
( 2
– 3 = (4;3)
_____
2 )
the gradient of the median is: m = 6 –3 = 3
______ 1
___ = – __
–2 – 4 –6 2
1 (x + 2)
y – 6 = – __ | Using point P(–2;6); you could also use midpoint (4;3).
2
1 x – 1 + 6 y = – __
y = – __ 1 x + 5 or 2y + x = 10
2 2
2 3
Gradient of PQ = __
7
perpendicular gradient = – 7
__
3
7 (x – 3)
y + 3 = – __ | Using point P(3;–3)
3
y=–7 7 x + 4 or 3y + 7x = 12
__x + 7 – 3 y = – __
3 3
3 12 = 6
Gradient of QR = ___
2
1
perpendicular gradient = – __
6
1 (x – 4)
y – 3 = – __ | Using midpoint of QR, (4;3)
6
y = –1
__x + 2 1 x + ___
__ + 3 y = – __ 11 or 6y + x = 22
6 3 6 3
4 The point of intersection will be the simultaneous solution
of the two equations.
1 x + 5 = – __
– __ 7x + 7
2 3
–3x + 30 = –14x + 42
12
11x = 12 x = ___
11
Substitute this into either equation.
( ) 3 + 5 = 58
–6 + 5 = ___
1 ___
y = – __
2 11 11
___
11
6 ;___
So the point of intersection is – ___ 58
( 11 11 )
58 Topic 4 Analytical geometry
PLT MATHS LB 11 7th pgs (Real Book).indb 58 2012/07/02 2:20 PM
5 We can use various methods to determine point S, but remember that
we must read points P, Q, S and R in this order for the vertices of the
parallelogram. If the parallelogram were formed with point S in the third
quadrant, it would be named PQRS.
Method 1
Find the equations of QS and RS. Then find the simultaneous solution to
these equations, as point S would be where those two lines intersect.
Gradient of QS = gradient of PR | Opposite sides of a parallelogram are parallel.
And gradient RS = gradient of PQ
9
gradient of QS = – __ 3
and gradient of RS = __
5 7
9 (x – 5)
y – 9 = – __ 3 (x – 3)
y + 3 = __
5 7
9 x + ___
= – __ 45 + 9 3 x – __
= __ 9–3
5 5 7 7
9 x + ___
= – __ 90 3 x – ___
= __ 30
5 5 7 7
9 x + ___
– __ 3 x – ___
90 = __ 30 3 (10) – ___
y = __ 30 = 0
5 5 7 7 7 7
–63x + 630 = 15x – 150
–78x = –780
x = 10
So S = (10;0)
But we know the properties of a parallelogram, so we can determine point S
using quicker methods:
Method 2
The diagonals of a parallelogram bisect each other, so we can find the
coordinates of the midpoint of diagonal QR. This is also the midpoint of PS,
so we can find point S using the midpoint formula.
Midpoint of QR = 5 + 3;9
_____
( 2
– 3 = (4;3)
_____
2 )
So, midpoint of PS is also (4;3).
–2 6+y
+ x = 4 and _____
______ =3
2 2
–2 + x = 8 and 6 + y = 6
x = 10 and y = 0
S = (10;0)
Unit 1 Equations of a line 59
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Method 3
Because the opposite sides of y
a parallelogram are parallel
Q(5;9)
and equal in length, the
quickest method to use is the P(–2;6)
rise
____
run , which is the same for rise of 9
opposite sides.
rise of 9
By inspection we can find the run of 5 S(10;0)
x
coordinates of S:
From point P to point R there run of 5
is a rise of 9 and a run of 5, so R(3;–3)
we determine 9 units down
and 5 units across from point
Q to find point S.
Important pointers learnt from Worked examples
To find the equation of a median:
• Find the coordinates of the midpoint of the line through which the median
will pass.
• Find the gradient of the median, using the coordinates of the required vertex
and the midpoint.
• Use the formula y – y1 = m(x – x1) where (x1; y1) can be either the midpoint or
the relevant vertex.
To find the equation of an altitude:
• Find the gradient of the line on which the altitude stands.
• Then find the perpendicular gradient by inverting and changing to the
opposite sign, so that the gradients have a product of –1.
• Use the formula y – y1 = m(x – x1) and the coordinates of the relevant vertex
as (x1;y1).
To find the equation of a perpendicular bisector:
• Find the gradient of the line to be bisected perpendicularly, and then find
the perpendicular gradient.
• Find the coordinates of the midpoint of the bisected line.
• Use the formula y – y1 = m(x – x1) and the coordinates of the midpoint, (x1; y1).
To find the coordinates of the point of intersection of two lines:
• Find a simultaneous solution by equating the two equations.
• Solve for x.
• Substitute the value of x into either equation to solve for y.
To find the fourth coordinate of a parallelogram, given the other three points:
rise
• Use the method of inspection: determine the ____
run of one of the given sides of
the parallelogram.
rise
• Use the same ____
run on the opposite side.
60 Topic 4 Analytical geometry
PLT MATHS LB 11 7th pgs (Real Book).indb 60 2012/07/02 2:20 PM
EXERCISE 1
1 In the figure K(x;y), L(–2;–1) and M (4;3) are the vertices of △KLM.
The equation of line KL is y = 5x + 9 and the equation of line KM is
5y + x – 19 = 0.
y
K(x;y)
M(4;3)
N
x
O
L(–2;–1)
1.1 Show that the coordinates of K are (–1;4).
1.2 Determine the equation of the median, KN, of triangle KLM, and express it
in the form y = mx + c.
1.3 Find the gradient of LM.
1.4 Then prove that KN is the perpendicular bisector of LM.
1.5 What can you deduce about triangle KLM without further calculations?
1.6 If L, M and the point J(7; p) are collinear, calculate the value of p.
1.7 Write down the coordinates of a point Q which will make KLQM a KEY WORD
parallelogram.
collinear – points that lie on
1.8 Determine the equation of a line parallel to KM and passing through
the same line; the gradients
point N.
between any two of these
1.9 Determine the coordinates of P, the point of intersection of the line found points will be the same
in Question 1.8, and line KL.
1.10 Prove that KM = 2 PN.
2 A(–3;3), B(3;–3) and C(6;12) are the vertices of triangle ABC and AD is a median
with D on BC. Determine:
2.1 the coordinates of D
2.2 the equation of AD
2.3 the equation of the altitude of triangle ABC, from point B
2.4 the equation of the perpendicular bisector of AC
2.5 the equation of line BC.
Unit 1 Equations of a line 61
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Unit 2: Inclination of a line
The angle of inclination of a line is the angle between that line and the positive
KEY WORDS x-axis.
angle of inclination – the y y y
angle that a line makes with A
the positive x-axis B B
obtuse angle – any angle
between 90° and 180°
θ θ θ
x x x
B
A
A
Figure 1 Figure 2 Figure 3
y
The angle of inclination will always be between 0°
and 180°. We determine this angle using the formula: B
tan θ = m
opposite
We can understand why when we consider the figure
θ
on the right, which shows that: x
adjacent
oppositerise
tan θ = ________ = ____
run = m
adjacent
A
Note: When the gradient is negative, the angle of inclination is obtuse (between 90°
and 180°) as shown in Figure 2.
To understand why, consider the graph of y = tan θ shown below.
Notice that the graph is below the x-axis (that is, has negative y values) when the
angle is between 90° and 180°.
An angle of inclination is between 0° and 180° so when we know that tan θ is
negative, θ would have to be an obtuse angle (any angle between 90° and 180°).
y
x
90° 180° 270° 360°
62 Topic 4 Analytical geometry
PLT MATHS LB 11 7th pgs (Real Book).indb 62 2012/07/02 2:20 PM
To find the angle of inclination in the figure:
3=–1
tan θ = –__ The reference angle is
3 y
always the acute angle that
reference angle (RA) = 45°
the line makes with the B
θ = 135° x-axis. We can find it by
(–1;3)
ignoring the negative sign,
and finding tan–1. RA θ
x
(2;0)
A
WORKED EXAMPLES
Determine the angle of inclination of these lines, giving your answers correct
to two decimal places.
WORKED EXAMPLE 1 WORKED EXAMPLE 2
y y
(1;5)
θ x α
3 x
(–4;–2)
–5
SOLUTION SOLUTION
tan θ = 5
__ θ = 59,04° 7 α = 54,46°
tan α = __
3 5
WORKED EXAMPLE 3 WORKED EXAMPLE 4
y y
(–7;5)
2 θ
β x
x (0;–3)
3
SOLUTION SOLUTION
2 reference angle = 33,69°
tan β = – __ 8 reference angle = 48,81°
tan ϕ = – __
3 7
β = 146,31° ϕ = 131,19°
Unit 2 Inclination of a line 63
PLT MATHS LB 11 7th pgs (Real Book).indb 63 2012/07/02 2:20 PM
The angle between two lines
Because we can find the angle of inclination of lines, we can also find the angle
between two lines.
• Find the angle of inclination of both lines that form the required angle.
• Draw a line parallel to the x-axis, passing through the vertex of the required angle.
• Using Euclidean geometry to find equal angles, determine the required angle
through simple calculations.
WORKED EXAMPLES y
A(–2;8)
Determine these angles, giving your answers
correct to one decimal place:
1 ^C
AB C(3;2)
2 ^B
AC x
B(–9;–10)
SOLUTIONS
y
1 First find the angles of inclination of lines
A(–2;8)
AB and BC:
18
tan θ = gradient AB = ___
7
∴ θ = 68,7° C(3;2)
12 = 1 θ α
tan α = gradient BC = ___ x
12
∴ α = 45°
Now draw a line through point B,
parallel to the x-axis, and label the equal
corresponding angles θ and α. B(–9;–10)
y
It is now clear to see that the value of
A(–2;8)
AB^C = θ – α
= 68,7° – 45°
= 23,7°
C(3;2)
θ α
x
θ
α
B(–9;–10)
64 Topic 4 Analytical geometry
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2 First find the angles of inclination
y
of lines AC and CB. A(–2;8)
To visualise the angle of
inclination of line AC, we extend
line AC so that it it intersects the C(3;2)
x-axis. α β
6 x
tan β = gradient AC = – __
5
reference angle = 50,2°
β = 129,8°
We already know that α = 45°
B(–9;–10)
If we draw a line through C,
y
parallel to the x-axis, we can A(–2;8)
find the equal alternate and
corresponding angles α and β.
So, AC^ B = 180° – β + α β
C
= 50,2° + 45° α
α β
x
= 95,2°
B(–9;–10)
In this example it may have been y
A(–2;8)
easier to consider this method:
Since AC ^ B is the exterior angle of
△CMN,
AC^ B = α + CN ^M C(3;2)
θ β
= α + 180° – β x
M N
= 45° + 180° – 129,8°
= 95,2°
B(–9;–10)
Unit 2 Inclination of a line 65
PLT MATHS LB 11 7th pgs (Real Book).indb 65 2012/07/02 2:20 PM
EXERCISE 2
1 Determine the angle of inclination of line AB where the gradient of AB is given:
1.1 m=2
1.2 m = –5
1.3 5
m = __
3
1.4 m = – 0,72
2 Determine the angle of inclination of these lines, correct to one decimal place:
2.1 1x – 8
y = __
2
2.2 5y + 2x = 10
2.3 y+x=3
3 Given points P(–1;–1), Q(4;2), R(7;–5) and S(–3;–7).
y
Q(4;2)
x
P(–1;–1)
R(7;–5)
S(–3;–7)
Determine (correct to one decimal place):
3.1 the acute angle between QR and the x-axis.
3.2 the angle of inclination of PQ
3.3 PQ^R
3.4 the angle of inclination of SP
3.5 ^R
SP
3.6 P^
SR
66 Topic 4 Analytical geometry
PLT MATHS LB 11 7th pgs (Real Book).indb 66 2012/07/02 2:20 PM
Revision
1 A triangle with vertices A(–1;7), B(7;5) and C(5;–3) is given.
1.1 Show that AB^ C = 90° (4)
1.2 Prove that △ABC is isosceles. (4)
1.3 Determine the area of △ABC. (2)
[10]
2 P(–3;1), Q(2;5) and R(7;a) are three points in the Cartesian plane.
Find the value(s) of a in each case if:
2.1 P, Q and R are collinear (4)
2.2 PQ ^ R = 90° (4)
2.3 PR = QR. (4)
[12]
3 P(3;5), Q(–1;–5), and R(4;1) are the vertices of △PQR.
3.1 Calculate the length of PQ (leave answer in simplest surd form). (3)
3.2 Find the coordinates of M, the midpoint of PQ. (2)
3.3 Find the equation of the perpendicular bisector of PQ. (3)
3.4 If ME ∥ QR with E on PR, use analytical methods to calculate
the coordinates of E. (7)
3.5 Find the equation of the median of △PQR drawn from Q. (3)
3.6 ^ R.
Calculate the size of PQ (4)
3.7 Determine the coordinates of S, so that PQRS is a parallelogram. (2)
[24]
4 Prove that the points A(1;6), B(1+ 3k;6 – k) and C(1 + 3n;6 – n) are
collinear for all real values of k and n. [4]
5 Points A(–4;2), B(–2;–1) and C(1;1) are given.
5.1 Show that AB ⊥ BC. y (3)
5.2 Determine the area of △ABC. (4)
5.3 Determine the coordinates of D if ABCD is a square. (2)
5.4 Determine the equation of the median of △ABC from point A to BC. A(–4;2) (4)
5.5 Determine the equation of the perpendicular bisector of BC. C(1;1) (4)
5.6 Determine the coordinates of the point of intersection of the x
perpendicular bisector found in Question 5.5, and line AC. (7)
5.7 Determine BA ^ C using analytical methods. B(–2;–1)
(6)
[30]
6 The figure shows a straight line y
with equation 5y + 2x = 10, which
meets the x-axis at A and the y-axis
at B. The line BC is perpendicular B
to AB, and meets the x-axis at C.
Showing all your working, find the
x
area of △ABC. C A
[8]
67
PLT MATHS LB 11 7th pgs (Real Book).indb 67 2012/07/02 2:21 PM
TOPIC 1: REVISION CONTINUED
7 In the figure A(4;2), B(8;8), C and D are the vertices of a parallelogram
whose diagonals intersect at M(3;7).
y
C
B(8;8)
D M(3;7)
A(4;2)
x
7.1 Determine the coordinates of C and D. (6)
7.2 ^ B = 90°.
Prove that AM (3)
7.3 What kind of quadrilateral is ABCD? Give a reason for your answer. (4)
[13]
8 Points P(–2;5), Q(5;9) and S(–1;–3) are given. R(m; n) is a fourth point in
the plane.
y
Q(5;9)
P(–2;5)
R
x
S(–1;–3)
8.1 Write down the values of m and n if PQRS is a parallelogram. (2)
8.2 Show that in this case PQRS is a rhombus. (3) REMEMBER
8.3 Write down the coordinates of the midpoint of QS. (1)
A rhombus is a parallelogram
8.4 Calculate the area of the rhombus PQRS. (4) which has all sides equal, and
8.5 ^ R.
Calculate the size of PQ (2) diagonals bisecting at right
8.6 Determine the equation of the line through S parallel to PR. (4) angles.
8.7 Determine the equation of the perpendicular bisector of PR. (4)
8.8 Show that the perpendicular bisector in Question 8.7 passes
through S. (2)
[22]
9 The coordinates of △ABC are A(1;7), B(3;7) and C(–5;–1). Calculate:
9.1 the perimeter of △ABC (4)
9.2 the area of △ABC (4)
9.3 the magnitude of BC^A (6)
9.4 the coordinates of D so that ABCD is a parallelogram (2)
[16]
68
PLT MATHS LB 11 7th pgs (Real Book).indb 68 2012/07/02 2:21 PM
TOPIC 1: REVISION CONTINUED
10 Points A(–1;–3), B(6;5) and C are the vertices of a triangle. y
The equation of AC is y = –3x – 6. B(6;5)
Determine, giving answers to two decimal places:
10.1 the angle of inclination of line AB C (2)
10.2 the angle of inclination of line AC (3)
10.3 CA ^ B. (2)
x
[7]
11 Points P(3;7), R(1;1), S and T(9;10) are the vertices of A(–1;–3)
a parallelogram PRST.
11.1 Write down the coordinates of S. (2)
11.2 Find the equation of the line perpendicular to RS, passing through P. (4)
11.3 Calculate the coordinates of the point of intersection of RS and the
line you found in Question 11.2. (6)
[12]
12 A(–3;2), B(4;8), C and D(6;4) are the vertices of a parallelogram.
12.1 Find the coordinates of C. (2)
12.2 Show that ABCD is a rhombus. (3)
12.3 Write down the coordinates of M, the midpoint of AC. (2)
12.4 Find the equation of BD. (3)
12.5 Show that M lies on BD. (3)
12.6 Describe in words the property of a parallelogram that you have
verified in your answers to Questions 12.3 to 12.5. (2)
12.7 Calculate the size of BA^ C. (3)
[18]
13 In the figure, the equation of line AB is x – 3y + 6 = 0 and M is the
point (3;1).
y
B
A
M(3;1)
x
13.1 Determine the gradient of line AB. (1)
13.2 Determine the equation of the straight line MN which is
perpendicular to AB and passes through point M. (4)
13.3 Determine the equation of the line EF which is parallel to
AB and passes through point M. (3)
13.4 Calculate the perpendicular distance between lines AB and EF.
(Leave your answer in surd form.) (7)
[15]
69
PLT MATHS LB 11 7th pgs (Real Book).indb 69 2012/07/02 2:21 PM
Formal assessment: Investigation: Ratios and the Golden Ratio
Task 1
If x2 – 3xy + 2y2 = 0
(x – 2y)(x – y) = 0
x = 2y or x = y
x = __
__ 2 or __xy = __
1
y 1 1
A: For each equation:
1 x2 – 2xy + y2 = 0 (4)
2 2 2 (4)
5x – 3xy – 8y = 0
3 2 2 (4)
8x – 5xy – 13y = 0
• Solve for x in terms of y.
• Find the value of the ratio __xy .
• Write the ratio __xy in decimal form, correct to three decimal places
where possible.
[12]
B: For each pair of simultaneous equations:
1 x + y = 13 x + y = 21 (4)
and and
2 5x – 8y = 0 8x – 13y = 0 (4)
• Solve for x and y.
• Find the value of the ratio __xy .
• Write the ratio __xy in decimal form correct to three decimal places where possible.
[8]
Task 2
You may have noticed that the numbers 1, 2, 5, 8 and 13 have appeared as solutions
in Task 1. These numbers form a sequence called the Fibonacci sequence.
In the Fibonacci sequence each number after the second term is the sum of the two
before it.
The table shows the first eight terms of the sequence (T1 = term 1):
T1 T2 T3 T4 T5 T6 T7 T8
1 1 2 3 5 8 13 21
70 Term 1 Assessment
PLT MATHS LB 11 7th pgs (Real Book).indb 70 2012/07/02 2:21 PM
1 Write down the next three terms of the sequence (3)
2 Find a new sequence by finding the ratios
T2 ___
___ T T T T11
; 3 ; ___4 ; ___5 up to ___ . (2)
T1 T2 T3 T4 T10
3 Find the ratios in Question 2 correct to three decimal places. (2)
4 Describe what you notice. (1)
[8]
Task 3
Fibonacci or ‘Leonard
We can generate a sequence of numbers by continued square roots: of Pisa’ lived from
______ __________
______ _______________
____________
______
__ __ __ __
T1 = √x ; T2 = √1 + √x ; T3 = √ 1 + √1 + √ x ; T4 = √ 1 + √1 + √ 1 + √ x ;
about 1170 to 1240
___________________
_______________
____________
______
T5 = √1 + √1 + √1 + √1 + √x
__
1 Use a calculator (or Excel) to find the first ten terms of this sequence
for x = 1. Give your answers correct to four decimal places.
Calculator tip:
__ _______
For T1 press √1 = ; For T2 press √1 + Ans =
_______
For T3 press √ 1 + Ans = and so on, for each successive term.
Write down each term’s value in a table. (2)
2 Find the first 11 terms of the same sequence for x = 2 and x = 3.
Record these values in the same table. (4)
3 Describe what you notice about the values of the terms. (2)
[8]
Task 4
We can generate a sequence of numbers by continued fractions:
1 ; T = 1 + _____
T1 = 1 + __ 1 ; T = 1 + ________
1 ; T4 = 1 + 1 1
__________; T = 1 + ____________
x 2 1 3 1 1 5 1
1 + __
x 1 + _____
1
1 + ________
1
1 + __________
1
1 + __
x 1 + _____
1
1 + ________
1
1 + __
x 1 + _____
1
1 + __
x
1 Use a calculator (or Excel) to find the first ten terms of this sequence for x = 1.
Give your answers correct to four decimal places.
Calculator tip:
1 = ; For T press 1 + 1
For T1 press 1 + __ ____ =
1 2 Ans
1 = … and so on, for each successive term
For T3 pres 1 + ____
Ans
Write down each term’s value in a table. (2)
2 Find the first ten terms of the same sequence for x = 2 and x = 3.
Record these values in the same table. (4)
3 Describe what you notice about the values of the terms. (2)
[8]
Investigation: Ratios and the Golden Ratio 71
PLT MATHS LB 11 7th pgs (Real Book).indb 71 2012/07/02 2:21 PM
Investigation continued
Task 5
We have seen the number 1,618 (1,61803398 to be more accurate), in three
different situations.
• the ratio of the terms of the Fibonacci sequence
• continued square roots
• continued fractions.
This number is called ‘phi’ or φ. It is an important number and is also called
the Golden ratio. φ has many special properties. Let φ = 1,61803398
1 1
Write down the value of 1 + __
φ. (1)
2 What do you notice? (2)
__
1 √5
φ to solve for φ and show that φ = 1 +
Use the fact that φ = 1 + __
3 ___ . (3)
__ 2
√5
4 What kind of number is 1 + ___ ? (1)
2
__
√5
___
5 Determine the value of 1 + correct to nine decimal places. (1)
2
[8]
Task 6
φ
The Golden rectangle is a rectangle in which the ratio of the sides is __
1
or approximately 1,618.
Throughout the ages this ratio has been recognised as aesthetically
pleasing. Many buildings and works of art have the Golden rectangle
in them. Two examples are the Parthenon in Greece and the United
Nations building in New York, where the ratio of the width of the
building compared with the height of every ten floors is the Golden ratio.
The Parthenon
Rectangle 1 has sides x and y with the ratio of length to breadth
of x:y.
y x–y
Rectangle 1 Rectangle 2
Rectangle 2 is formed by creating a square of length y on the one side of
Rectangle 1. The remaining Rectangle 2 has sides of length x – y and y
with the ratio of length to breadth of y:x –y.
If the two rectangles have the same ratio of length to breadth, then they The United Nations
y
are Golden rectangles. In other words: __xy = ____
x–y Head Quarters
72 Term 1 Assessment
PLT MATHS LB 11 7th pgs (Real Book).indb 72 2012/07/02 2:21 PM
1 Show that the equation on page 72 is the same as x2 − xy − y2 = 0. (2)
2 If y = 1, solve for x in the equation. Leave your answer in surd form.
(Why can we discard one of the values of x?) (3)
3 What do you notice about this number? (1)
4 Use the equation x x = φ, the Golden ratio.
2 − xy − y2 = 0 to show that __ (2)
y
[8]
Task 7
Draw a rectangle on A4 size paper with dimensions
16,2 cm by 10 cm. Notice the ratio of length:breadth is
1,62 which is nearly 1,618, the Golden ratio. You have
created a Golden rectangle.
B F C
A E D
Study the pictures and use your rectangle to create your own
version. Each new rectangle is formed by drawing a square on
one side on the larger rectangle. If you join the dotted lines,
the diagonals of the squares, by a smooth curve, they form
a spiral.
Make your own version of this spiral and colour your diagram
in creatively. Hand this in with your answers. [5]
y y
Task 8 A4
Most paper is cut to internationally agreed sizes: A0; A1; A2; A3; A4 and A5. x A3
This paper has the property that A1 is half the size of A0, A4 is half the size A5 A5
of A3 and A5 is half the size of A4, and so on.
Find the ratio of the sides of A series paper. Use the diagram to show that the
__
ratio of the sides of the paper is √ 2 :1 [5]
Total: 70 marks
Investigation: Ratios and the Golden Ratio 73
PLT MATHS LB 11 7th pgs (Real Book).indb 73 2012/07/02 2:21 PM
Formal assessment: Term 1 Test
1 Simplify each expression as far as possible, showing necessary workings.
Do not use a calculator.
1.1 2x+3
× 4x
_________ (4)
2 × 8x + 1
(4 5−9 )
−1 1
__
1.2 ________ 2 (4)
−1 −1
___ ___ ___
1.3 ( √18 – √72 + √50 )2 (4)
___ ___
___ ___
1.4 ( √3x − √2x )( √3x + √2x ) (3)
____ ____
√ 75k8 − √ 12k8
______________
1.5 ____ (4)
√ 27k6
[19]
2 Solve for x in each equation:
2.1 11x + 8 = (x + 2)(3x + 4) (4)
2.2 3(2x2 − 5) = x (4)
2.3 2x − 5 3
__ = __ (5)
x 2
2 − 7 = x (correct to two decimal places)
2.4 3x (4)
2 3
2.5 x
_________ 4 = _____
+ _____ (6)
x2 − 2x − 8 x − 4 x + 2
2.6 2x2 − x − 3 > 0 (4)
2.7 1
2 = ___
x
(2)
32
2.8 2.25x + 1 = 250 (4)
x = 5x
2.9 12 (1)
3
__
2.10 x2 = 27 (2)
2
− __
2.11 3x 3 = 75 (3)
______
2.12 √ 10x − 1 = 3 (2)
______
2.13 3 − 2x + √ 10x − 1 = 0 (6)
[47]
3 Solve for x and y:
3.1 x − 8 = 2y and (x − 3)(y − 2) = y2 − 10 (7)
3.2 10 and y = __
y = _____ 5x + 5 (6)
x−2 3
3.3 9x + 2.27y + 1 = 31−x and x2 + y2 + xy = 7 (7)
[20]
4 4.1 Solve for x by completing the square if 3x2 + 12x − 21 = 0 (4)
4.2 State whether the roots in Question 4.1 are real and rational,
real and irrational or non-real. (2)
4.3 Find the nature of the roots of the equation 2x2 + 10x + 8 = 0
without solving for x. (3)
74
PLT MATHS LB 11 7th pgs (Real Book).indb 74 2012/07/02 2:21 PM
____
4.4 Given: P = √_____ 3
5 − __
x+1 x
4.4.1 For which value(s) of x will P be undefined? (2)
4.4.2 For which value(s) of x will P be non-real? (2)
[13]
5 Given the sequence: 2; 11; 20; 29; …
5.1 Determine the sixth term, T6. (1)
5.2 Write down a formula for the general term, Tn. (2)
[3]
6 Given the sequence: 6; 12; 20; 30; …
6.1 Find the next two terms. (2)
6.2 Find a formula in terms of n for the nth term. (4)
6.3 Which term is equal to 506? (3)
[9]
7 7.1 Copy and complete the fourth and fifth lines the pattern:
12 = 1
12 + 22 = 1 + 4 = 5
12 + 22 + 32 = 1 + 4 + 9 = 14
12 + 22 + 32 + … = 1 + 4 + 9 + … = …
12 + 22 + 32 + … + … = 1 + 4 + 9 + … + … = … (2)
7.2 Given that the last number in the nth row is 1
__n3 + bn2 + cn.
6
Find the values of b and c. (6)
[8]
8
Shape 1 Shape 2 Shape 3 Shape 4
8.1 Find the number of green dots in the fifth and sixth shapes
of the pattern. (2)
8.2 Find the number of orange dots in the fifth and sixth shapes of the
pattern. (2)
8.3 Find a formula for the nth term of green dots. (2)
8.4 Find a formula for the nth term of the orange dots. (4)
8.5 How many green dots will there be in the ninth pattern? (1)
8.6 How many orange dots will there be in the eleventh pattern? (2)
8.7 Which pattern number will have 132 orange dots? (3)
[16]
75
PLT MATHS LB 11 7th pgs (Real Book).indb 75 2012/07/02 2:21 PM
TOPIC 1: TEST CONTINUED
9 The figure represents △ABC with A(3;8), B(17;6)and C(5;–6).
The altitude AD cuts the median BM at P.
y
A(3;8)
B(17;6)
P
D
M
x
C(5;–6)
Determine:
9.1 the length of AC (2)
9.2 the coordinates of point M, the midpoint of AC (2)
9.3 the equation of median BM (3)
9.4 the gradient of line BC (2)
9.5 the equation of altitude AD (4)
9.6 the angle of inclination of line BC (2)
9.7 ^ B.
the size of AC (3)
[18]
10 A(a;b), B(–12;1) and C(8;1) are the vertices of △ABC. M is
the midpoint of BC and N(3;6) is the midpoint of AC.
10.1 Draw a rough diagram to illustrate the information you are given. (1)
10.2 Calculate the coordinates of M. (2)
10.3 Calculate the values of a and b. (2)
10.4 Calculate the size of NM ^ C. (3)
10.5 Calculate the length of MN in simplified surd form. (2)
10.6 Prove that NM ∥ AB. (2)
10.7 Without further calculation, write down the length of AB. (2)
10.8 Write down the equation of the perpendicular bisector of BC. (2)
[16]
11 Given P(3;5), Q(k;6), R(–3;7) and S(–2;2). Determine the value of k if:
11.1 PR ∥ QS (3)
11.2 PQ ⊥ RS (3)
11.3 QR = QS. (3)
[9]
76
PLT MATHS LB 11 7th pgs (Real Book).indb 76 2012/07/02 2:21 PM
12 In the figure, OABC is a parallelogram with points A(–3;3), B(4;4)
and O, the origin. The diagonals intersect at D.
y
B(4;4)
A(–3;3)
D
C
x
O
12.1 Determine the coordinates of D. (2)
12.2 Show that the equation of AC is 5y + x – 12 = 0 (4)
12.3 Find the gradient of AB. (1)
12.4 Find the equation of OC. (2)
12.5 Find the coordinates of C. (2)
12.6 Find the size of the angle which AB (produced) makes with the
positive x-axis. (3)
[14]
Total: 192 marks
77
PLT MATHS LB 11 7th pgs (Real Book).indb 77 2012/07/02 2:21 PM
Term 1 summary
Topic 1
Exponents and surds
Laws for exponents: Rules for surds:
__ ___
Solving equations with:
__
1 am × an = am + n 1 n√ a × √ b = √ ab | Multiplying
n n
Unknown exponents:
__
__ n __ If ax = ay then x = y | a > 0, a ≠ 1
2 am ÷ an = am – n √
a | Dividing
2 n√ a ÷ √ b = n __
___ __
b Rational exponents:
__ __
3 (am)n = amn 3 √ a2b = √ a2 √ b = a√ b | Reducing
a
__ b
__
If xb = y c then x = ( y c )a | x > 0, x ≠ 1
4 (ab)m = ambm __ __ __ Surd equations:
4 √ a + √ a = 2√ a | Adding like surds ____
1
5 a–n = __ __ __ __ __ If √ x + a = x + b then x + a = ( x + b )2
n √b
a__ × ___ a√ b or ______
2 − √__ 2 − √__
b × ______
b
___ a __ 5 ___ __ = ____
√
n m
6 a = an
m
√b √b b 2 + √b 2 − √b and solve the quadratic equation.
__
√
4 − 4 b + b | Rationalising the denominator
7 a0 = 1 = __________
4−b
Topic 2
Equations and inequalities
Solve quadratic equations in standard form ax2 + bx + c = 0 by:
• factorising (always try this first)
• completing the square (only if asked to)
_______
−b±√b2 − 4ac
• using the formula x = ____________ (answers usually given correct to two decimal places).
2a
Write all quadratic equations with fractions in standard form by:
• factorising the denominator and numerator (if possible)
• multiplying both sides of the equation by the LCD to get rid of fractions.
Solve simultaneous equations by:
• substitution of one equation (usually the linear one) into the other (usually the quadratic equation)
• equating the two y values (usually done with graphs in y form).
Solve quadratic inequalities by:
• factorising y
• using a number line or using a graph.
Statement Inequality notation Interval notation
– – + + + + – –
–x2 + 4 ≥ 0 −2 ≤ x ≤ 2 x ∊ [−2;2] x
–2 2
Determine the nature of roots by:
• looking at both roots (solutions to equation) to decide if they are real or non-real.
If real decide if they are rational or irrational and equal or unequal.
• looking at b2 − 4ac (part of the quadratic formula under the square root)
b2 − 4ac < 0 ⇒ roots will be non-real
b2 − 4ac > 0 ⇒roots will be real
b2 − 4ac = 0 ⇒roots will be equal (and real)
b2 − 4ac = perfect square ⇒ roots will be rational (and real)
78 Term 1 summary
PLT MATHS LB 11 7th pgs (Real Book).indb 78 2012/07/02 2:21 PM
Topic 3
Number patterns
Patterns can be given as a sequence of numbers or a set of
figures which change according to a rule.
Linear patterns
Linear patterns have a constant first difference. Tn = an + b
where a is the common difference, b the value of T0 and n the
position of the term in the pattern.
Quadratic patterns
Quadratic patterns have a constant second difference.
Tn = an2 + bn + c, 2a = common second difference,
3a + b = first difference = T2 – T1
a + b + c = T1 or c = T0
Topic 4
Analytical geometry
Given points A(x1;y1) and B(x2;y2) in the Cartesian plane:
________________
• The distance between A and B = √ ( x2 – x1 )2 + ( y1 – y2 )2
• The midpoint, M, between points A and B is M = 1 (
x + x2 y1
______
2
+y
;______2
2 )
y − y1
______
• The gradient of a line between points A and B is m = x2
2 − x1
• The equation of the straight line through A and B is y = mx + c or y − y1 = m( x − x1 )
• The angle of inclination that line AB makes with the positive x-axis is found by letting tan θ = m
Parallel lines have equal gradients: m1 = m2
The product of the gradients of perpendicular lines is equal to –1: m1 × m2 = –1
We can find equations of perpendicular bisectors, altitudes and medians using all the above formulae.
Term 1 summary 79
PLT MATHS LB 11 7th pgs (Real Book).indb 79 2012/07/02 2:21 PM
Term
2
80
PLT MATHS LB 11 7th pgs (Real Book).indb 80 2012/07/02 2:21 PM
TOPIC 5 Functions: Effects of parameters
Unit 1 The effects of the parameters a, p
and q on parabolas 82
Unit 2 The effects of the parameters a, p
and q on hyperbolas 92
Unit 3 The effects of the parameters a, p
and q on exponential graphs 101
Unit 4 Real life applications 108
Unit 5 The average gradient between two
points on a curve 111
Revision 114
TOPIC 6 Functions: Trigonometric graphs
Unit 1 Revision of basic trigonometric graphs
using point by point plotting 116
Unit 2 The effects of the parameter k on
some trigonometric functions (graphs
of the functions defined by y = sin kx,
y = cos kx and y = tan kx) 123
Unit 3 Horizontal shifts 125
Unit 4 Determine the equations of
trigonometric graphs 130
Unit 5 Sketch graphs which have a change
in period and a horizontal shift 132
Revision 136
TOPIC 7 Trigonometry
Unit 1 Revision of Grade 10 trigonometry 138
Unit 2 Identities 142
Unit 3 Reduction formulae 146
Unit 4 Trigonometric equations – specific
and general solutions 155
Revision 163
Formal assessment: Assignment: Term 2 test 166
Term summary 168
Formal assessment: Examination Practice Paper 1 170
Examination Practice Paper 2 172
81
PLT MATHS LB 11 7th pgs (Real Book).indb 81 2012/07/02 2:21 PM
TOPIC
2 5 Functions: Effects of parameters
Unit 1: The effects of the parameters a, p
KEY WORDS
and q on parabolas
parameter – a variable that
restricts or gives a particular We define trigonometric graphs by parametric equations which group the graphs
form or shape to the equation into families. When we assign values to the parameters and substitute them into
it characterises the equation, it becomes a specific equation with a specific shape that is restricted
function – a relationship by these values. Parameters do not change the type of graph, but define the
between two variables,
characteristics of a particular graph within a family of graphs.
usually x and y, where for
every value of x there is one We will investigate the effect of the parameters a, p and q on the graph of the
corresponding value for y function defined by y = f (x) = a(x + p)2 + q. But firstly, we will revise the effects of a
increase – to grow bigger and q on the parabola given by y = f (x) = ax2 + q by comparing the following graphs:
decrease – to grow smaller
1 x2 as well as y = −x2; y = −4x2 and y = − 1
f(x) = x2; g(x) = 4x2; h(x) = __ __x2.
reflect – create a mirror 4 4
image or reflection
f (1) = 12 = 1⇒(1;1) 2
g(2) = 2 = 4⇒(2;4) 2
g(3) = 3 = 9⇒(3;9)
y-intercept – the point
where the graph cuts the g(x) = 4x2
x-axis (where x = 0)
g(1) = 4(1)2 ⇒(1;4) g(2) = 4(2)2 = 16 ⇒ (2;16) g(3) = 4(3)2 = 36 ⇒ (3;36)
1 x2
h(x) = __
4
1 (1)2 = 1
h(1) = __ __ (1;0,25) 1 (2)2 = 1 (2;1)
h(2) = __ 9 (3;2,25)
1 (3)2 = __
h(3) = __
4 4 4 4 4
We can conclude that:
• the greater the value of a, the steeper the curve
• when a is negative the graph reflects in the x-axis
• changing the value of a has no impact on the y-intercept.
Parabolas y
6 y = 1__4x2
y = 4x2 y = x2
5
4
3
2
1
x
–4 –3 –2 –1 0 2 3 4
–1
–2
–3
–4
–5
y = –4x2 y = –x2 y = – 1__4x2
–6
Axis of symmetry
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WORKED EXAMPLE KEY WORD
Consider the function: f (x) = 2x2 reflection – a mirror image
1 Sketch f, showing the coordinates of three points. about a line of symmetry
2 If f is shifted up 1 unit, the translated graph is g. Draw g on the same system
of axes.
3 State the equation of g.
4 Sketch h, the reflection of f in the x-axis, on the same system of axes.
5 State the equation of h.
6 Sketch k, the reflection of g in the x-axis, on the same system of axes.
7 State the equation of k.
SOLUTIONS
1 f (0) = 0 ⇒ (0;0) y
g f
f(1) = 2(1)2 = 2 ⇒ (1;2)
f (–1) = 2(–1)2 = 2 ⇒ (–1;2) (–1;3) (1;3)
REMEMBER
2 Add 1 to each y-value. (–1;2) (1;2) The vertical axis is the y-axis.
(0;0) → (0;1) The equation of the y-axis is
(1;2) → (1;3) given by x = 0.
(–1;2) → (–1;3) 1 The horizontal axis is the
3 g(x) = f (x) + 1 = 2x2 + 1 x-axis.
x The equation of the x-axis is
4 Leave the x-coordinates 0
given by y = 0.
unchanged, but change
The axes cross at the origin,
the sign of each y-value: –1 (0;0)
(0;0) → (0;0) Coordinates
(–1;–2) (1;–2)
(1;2) → (1;–2) Give both the x- and
(–1;2) → (–1;–2) y-coordinates in the form
5 h(x) = –f (x) = –2x2 (–1;–3) (1;–3) (2;3).
k h
6 Leave the x-coordinates If the first coordinate is 0,
unchanged, but change then you have a y-intercept:
(0;3) cuts the y-axis at 3
the sign of each y-value:
If the second coordinate is 0,
(0;1) → (0;–1) (1;3) → (1;–3) (–1;3) → (–1;–3)
then you have an x-intercept:
7 k(x) = –g(x) = –2x2 – 1 (2;0) cuts the x-axis at 2
Vertical shifts require movement up or down only and do not change the
x-coordinates.
Reflections in the x-axis do not change x-coordinates, but change the signs
of the y-coordinates.
• f (x) + 2 will increase each y-coordinate by 2 and the graph will shift up 2 units.
• f (x) – 3 will decrease each y-coordinate by 3 and the graph will shift down 3 units.
• –f (x) will reflect f in the x-axis.
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EXERCISE 1
REMEMBER
1 1 x2
Consider the function: f (x) = – __
On the ‘same set of axes’ 2
means you will draw more 1.1 Sketch the graph of f, showing the coordinates of three points.
than one graph on the same 1.2 If g is obtained by shifting f up 2 units, sketch g on the same system
sketch. of axes as f.
1.3 State the equation of g.
1.4 Sketch h, the reflection of f in the x-axis, on the same system of axes
REMEMBER as f and g.
1.5 State the equation of h.
To find an x-intercept, 1.6 Sketch k, the reflection of g in the x-axis, on the same system of axes
substitute y = 0.
as f, g and h.
To find a y-intercept,
1.7 State the equation of k.
substitute
2 Consider the function: f (x) = –8x2
2.1 Sketch the graph of f, showing the coordinates of three points.
2.2 If g is obtained by shifting f down 1 unit, sketch g on the same system
of axes as f.
2.3 State the equation of g.
2.4 Sketch h, the reflection of f in the x-axis, on the same system of axes
as f and g.
2.5 State the equation of h.
2.6 Sketch k, the reflection of g in the x-axis, on the same system of axes
as f, g and h.
2.7 State the equation of k.
3 1 x2
Consider the function: f (x) = __
3
3.1 Sketch the graph of f, showing the coordinates of three points.
3.2 If g is obtained by shifting f down three units, sketch g on the same
system of axes as f.
3.3 State the equation of g.
3.4 Sketch h, the reflection of f in the x-axis, on the same system of axes
as f and g.
3.5 State the equation of h.
3.6 Sketch k, the reflection of g in the x-axis, on the same system of axes
as f, g and h.
3.7 State the equation of k.
4 Consider the function: f (x) = –3x2
4.1 Sketch the graph of f, showing the coordinates of three points.
4.2 If g is obtained by shifting f down 2 units, sketch g on the same system
of axes as f.
4.3 State the equation of g.
4.4 Sketch h, the reflection of f in the x-axis, on the same system of axes
as f and g.
4.5 State the equation of h.
4.6 Sketch k, the reflection of g in the x-axis, on the same system of axes.
4.7 State the equation of k.
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Horizontal shifts of the parabola y = x2
x –6 –5 –4 –3 –2 –1 0 1 2 3 Shift
2
f y=x 36 25 16 9 4 1 0 1 4 9 None
g y = (x + 2)2 16 9 4 1 0 1 4 9 16 25 2 left
h y = (x – 1)2 49 36 25 16 9 4 1 0 1 4 1 right
2
k y = (x + 4) 4 1 0 1 4 9 16 25 36 49 4 left
• Adding to x increases its value so the pattern of answers shifts to the left by the
amount added.
• Subtracting from x reduces its value so the pattern shifts to the right by the amount
subtracted.
• This pattern of horizontal shifting applies to all graphs, not only to the parabola.
The graphs in the table are sketched in the figure below. Notice that they are all
identical in shape.
The x-intercept in each graph clearly indicates how the graph has shifted.
KEY WORD
y = (x + 4)2 y = (x + 2)2 y y = x2 y = (x – 1)2
x-intercept – the point where
the graph cuts the x-axis
(where y = 0)
x
–5 –4 –3 –2 –1 1 2
Each graph has equal roots and touches the x-axis, without crossing it.
There are three forms of each graph:
• y = (x + 2)2 is the turning point form
• y = (x + 2)(x + 2) is the x-intercept form
• y = x2 + 4x + 4 is standard form.
EXERCISE 2
Write each equation in the turning point form. State the turning point and then
describe the shift of each parabola from the origin (0;0).
1 y = x2 + 6x + 9
2 y = x2 – 8x + 16
3 y = x2 – 4x + 4
4 y = –x2 – 2x – 1
Unit 1 The effects of the parameters a, p and q on parabolas 85
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Horizontal and vertical shifts
Consider the parabola equation y = (x + p)2 + q:
• changing p results in a horizontal shift
• changing q results in a vertical shift
• p and q do not influence each other.
x –6 –5 –4 –3 –2 –1 0 1 2 3 Shift
f y = x2 36 25 16 9 4 1 0 1 4 9 None
2
g y = (x + 2) – 4 12 5 0 –3 –4 –3 0 5 12 21 2 left, 4 down
2 + 2 51 38 27 18 11 6 3 2 3 6 1 right, 2 up
h y = (x – 1)
k y = (x + 4)2 – 1 3 0 –1 0 3 8 15 24 35 48 4 left, 1 down
The graphs in the figure are drawn from the table. Notice that all the graphs are
identical in shape.
y
k g f h 6
5
4
3
2
(1;2)
1
KEY WORD x
–6 –5 –4 –3 –2 –1 1 2 3
vertical shift – with straight –1
(–4;–1)
up or down, no sideways –2
movement
–3
horizontal shift – with
sideways, with no up or down –4
(–2;–4)
movement
axis of symmetry – a line
that divides a graph into • y = x2 has a minimum value of 0 at (0;0) and an axis of symmetry given by x = 0.
two identical halves and
• y = (x + 2)2 – 4 has a minimum value of – 4 at (–2;– 4) and an axis of symmetry
about which a function is x = –2.
symmetrical
• = (x – 1)2 + 2 has a minimum value of 2 at (1;2) and an axis of symmetry x = 1.
y
• y = (x + 4)2 – 1 has a minimum value of –1 at (– 4;–1) and an axis of symmetry
x = – 4.
EXERCISE 3
Write each equation in standard form and in the x-intercept form. State the turning
point of each graph and then describe the shift of each parabola from the origin (0;0).
1 y = (x – 4)2 – 9
2 y = 4(x + 2)2 – 1
3 y = –(x + 5)2 + 16
4 y = –(x – 3)2 + 4
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Sketching a parabola
Check the sign of a:
• If a > 0, then the parabola ‘smiles’ and the arms go up.
• If a < 0, then the parabola ‘frowns’ and the arms go down.
Find the y-intercept by making x = 0 and solving for y:
• If y = 2x2 – 3x – 5, then the constant will be the y-intercept, so y = –5.
• If y = (x – 5)2 – 4, then y = (0 – 5)2 – 4 = 25 – 4 = 21.
Find the x-intercepts, make y = 0 and then solve the equation.
• Most graphs will have rational roots and so we can find the roots by factorisation:
y = 2x2 – 3x – 5 ⇒ 0 = (2x – 5)(x + 1) and x = 2,5 or x = –1
x-intercepts (–1;0) and (2,5;0)
• Use the quadratic formula for graphs with irrational roots:
y = x2 – 2x – 4 ⇒ x2 – 2x – 4 = 0 cannot be solved by factors.
quadratic formula:
___
a = 1, b = –2 and c = – 4 ⇒ b2 – 4ac = (–2) – 4(1)(– 4) = 20
( ) √
– –2 ± 20
x = __________
( )
= 3,24 or –1,24
21
x-intercepts (–1,24;0) and (3,24;0)
• Parabolas do not necessarily have x-intercepts and in this case we cannot find
roots:
y = x2 – x + 2 ⇒ x2 – x + 2 = 0 cannot be solved by factors.
Test b2 – 4ac from the quadratic formula: a = 1, b = –1 and
c = 2 ⇒ b2 – 4ac = (–1)2 – 4(1)(2) = –7
No x-intercepts because b2 – 4ac = 0
Find the turning point: REMEMBER
• Every parabola has a turning point.
b. a > 0 means that a is positive.
The x-coordinate is the axis of symmetry, given by x = – ___
2a a < 0 means that a is
The y-coordinate is the minimum value if a > 0, then the parabola ‘smiles’ . negative.
The y-coordinate is the maximum value if a < 0, then the parabola ‘frowns’ . minimum – smallest possible
Find the y-coordinate by substituting the x-coordinate. value
x +x maximum – largest possible
______2
• If you know the x-intercepts, then the axis of symmetry is 1 2 value
–1 + 2,5 1,5 axis of symmetry – divides
x-intercepts (–1;0) and (2,5;0) ⇒ axis of symmetry is x = _______ = ___ = 0,75 graph into two identical
2 2
b halves
• If the roots are irrational or if there are no roots, then use the formula x = – ___
2a
– –2 = 1
–b = _____ ( )
y = x2 – 2x – 4 has a = 1 and b = –2, so the axis of symmetry is x = ___ ( )
2a 21
( )
y = x2 – x + 2 has a = 1 and b = –1, so the axis is symmetry is x = –_____
–1 = __
( )
1
21 2
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WORKED EXAMPLE
1 1 (x – 4)2 + 2 and h(x) = x2 – 4x + 6 on the same system of axes, clearly
Sketch f (x) = x2 – 5x – 6, g(x) = – __
2
indicating the intercepts with the axes and the coordinates of the turning points.
2 Use your graphs to determine two values of x for which f (x) = g(x).
3 Solve algebraically for x if f (x) = g(x).
4 Determine h(4) – g(4).
5 Use your graph to determine two values of x for which g(x) – f(x) = 0.
SOLUTIONS
1 f(x) = x2 – 5x – 6 1 (x − 4)2 + 2
g(x) = −__ h(x) = x2 – 4x + 6
2
y-intercept: (0; – 6) y-intercept: (0;6)
y-intercept: (0;– 6)
x-intercepts: (6;0) and (–1;0) x-intercepts: None
1 (0 − 4)2 + 2 = −6
− __
x2 – 5x – 6 = 0 2 No factors ⇒ check b2 – 4ac
(x – 6)(x + 1) = 0 x-intercepts: (0;0) and (4;0) a = 1, b = – 4 and c = 6
x = 6 or x = –1 b – 4ac = (– 4)2 – 4(1)(6) = –8
1 (x − 4)2 + 2 = 0
− __
2 No roots
Turning point: (2,5;–12,25)
(x − 4)2 = 4
x = –1 + 6 = __
______ 5 = 2,5 Turning point: (2;2)
2 2 x − 2 = 2 or x − 2 = −2 (– 4)
b = – ____ = 2
y = (2,5)2 – 5(2,5) – 6 = –12,25 x = 4 or x = 0 x = – ___
2a 2(1)
Turning point: (4;2) y = (2)2 – 4(2) + 6 = 2
y
h
f
6 (4;6)
(2;2) (4;2)
x
–1 2 6
–6
g
(2,5;–12,25)
2 x = 0 or x = 6
3 1 (x – 4)2 + 2 ⇒ – 2x2 + 10x + 12 = x2 – 8x + 16 – 4
x2 – 5x – 6 = – __
2
3x – 18x = 0 ⇒ 3x(x – 6) = 0 and x = 0 or x = 6
2
4 h(4) – g(4) = 6 – 2 = 4 units
5 x = 0 or x = 6
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WORKED EXAMPLE REMEMBER
1 f(x) = 2x2 – 4x – 6 and g(x) = 2x – 6 If A × B > 0, then A and B
1.1 Draw neat sketch graphs of f and g on the same system of axes. Indicate are both positive or both
negative.
the intercepts with the axes, as well as the coordinates of any turning
If C × D < 0, then C and D
points.
must be opposite in sign.
1.2 Use your graphs to answer the questions: f (x) = g(x) means that the
1.2.1 For which value(s) of x is f(x) = g(x)? y-value of f is equal to the
1.2.2 For which value(s) of x is f(x) < g(x)? y-value of g.
1.2.3 For which value(s) of x is f(x).g(x) ≥ 0? f (x) < g(x) means that the
y-value of f is less than the
SOLUTIONS y-value of g.
f (x).g(x) ≥ 0 means that the
1 f(x) = 2x2 – 4x – 6 product of the y-values of f
y-intercept: (0;– 6) and g is positive or 0.
x-intercepts: (–1;0) and (3;0) If a > 0, the arms of the
parabola go up and the
1.1 2x2 – 4x – 6 = 0 f (x) = 2x2 – 4x – 6 y parabola has a minimum
⇒ x2 – 2x – 3 = 0 y-value.
(x – 3)(x + 1) = 0, If a < 0, the arms of the
so x = 3 or x = –1 parabola go down and
the parabola has a maximum
Turning point: (1;–8) y-value.
3 + (–1) 2=1 x
x = _______ = __ –1 3
2 2
y = 2(1)2 – 4(1) – 6 = –8
g(x) = 2x – 6
y-intercept: (0;– 6)
x-intercept: (3;0)
2x – 6 = 0 REMEMBER
2x = 6 ⇒ x = 3 –6
x-intercepts and roots
1.2 1.2.1 x = 0 and x = 3
represent the point where the
1.2.2 0<x<3 g(x) = 2x – 6 (1;–8) graph cuts the x-axis (where
1.2.3 x ∊ [–1;∞) y = 0)
EXERCISE 4
Label each graph clearly and indicate the intercepts with the axes as well as any
turning point(s).
1 1.1 Sketch f (x) = –x2 + 4x + 12 and g(x) = 4x + 8 on the same system of axes.
1.2 For which value(s) of x is f (x) = g(x)?
1.3 For which value(s) of x is f (x) ≥ g(x)?
2 2.1 1 x2 – x – 4 and g(x) = x – 4 on the same system of axes.
Sketch f (x) = __
2
2.2 For which value(s) of x is g(x) – f (x) = 2?
2.3 For which value(s) of x is f (x) ≤ g(x) ?
3 3.1 Sketch f (x) = –(x + 1)2 + 9 and g(x) = –3x + 6 on the same system of axes.
3.2 For which value(s) of x is f (x). g(x) ≤ 0?
Unit 1 The effects of the parameters a, p and q on parabolas 89
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Finding the equation of a parabola
We can represent a parabola in three forms:
• y = ax2 + bx + c | Standard form
• y = a(x – x1)(x – x2) | x-intercept formula, with x1 and x2 being the x-intercepts
• y = a(x – p)2 + q | TP formula, with TP (p;q)
If the x-intercepts and a random point are given, use the x-intercept formula.
WORKED EXAMPLE 1
Determine the equation of a parabola which has points (2;0), (– 4;0) and (4;8).
SOLUTION
y = a(x – x1)(x – x2) where x1 = 2, x2 = – 4, and the substitution point is ( 4;8 )
Substitute x-intercepts: y = a(x – 2)[x – (– 4)] ⇒ y = a(x – 2)(x + 4)
1
Now substitute the point: 8 = a(4 – 2)(4 + 4) ⇒ 8 = 16a and a = __
2
1 (x – 2)(x + 4) in x-intercept form or y = 1
The equation is y = __ __x2 + x – 4 in standard
2 2
form.
If the turning point and a random point are given, use the turning point formula.
WORKED EXAMPLE 2
Determine the equation of the parabola if it has a turning point (1;2) and passes
through (2;–1).
SOLUTION
y = a(x – p)2 + q where the turning point is (1;2) and the substitution point
is (2;–1).
Substitute the turning point: (1;2): y = a(x – 1)2 + 2
Now substitute the point: (2;–1) ⇒ –1 = a(2 – 1)2 + 2 and a = –3
The equation is y = –3(x – 1)2 + 2 in turning point form or y = –3x2 + 6x – 1 in
standard form.
EXERCISE 5
1 Determine the equation of the parabola which passes through the points:
1.1 (– 4;0), (1;0) and (0;12) 1.2 ( __12;0 ), (–2;0) and ( – __12;6 )
1.3 ( __13;0 ), (3;0) and (0;1) 1.4 ( __32;0 ), (– 4;0) and (2;–3)
90 Topic 5 Functions: Effects of parameters
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2 2.1 Determine the equation of the parabola which has a turning point (3;4)
and passes through the point (4;2).
2.2 Determine the equation of the parabola which has a turning point (1;–2)
and passes through the point (2;1).
2.3 Determine the equation of the parabola with a turning point (–2;4) and a
y-intercept of 2.
2.4 Determine the equation of the parabola which has a turning point __5 ;– __
2 8
9
( )
and passes through the point (5;2).
EXERCISE 6
Complete the table and then draw each graph on its own system of axes.
• State each function in the three ways: y = ax2 + bx + c, y = a(x – x1)(x – x2) and
y = a(x – p)2 + q
• p is associated with the horizontal shift and q indicates the vertical shift.
• The first two rows have been completed for you as Worked examples with the
given information high-lighted in blue.
• Roots (x-intercepts) tell us where a graph passes through the x-axis and y = 0.
• Not all parabolas have x-intercepts.
Function Shift Turning Min/Max Axis of Roots
point symmetry y-intercept
1
y = x2 – 6x + 5
3 right Minimum (1;0)
y = (x – 1)(x – 5) (3;–4) x=3 (0;5)
4 down y = –4 (5;0)
y = (x – 3)2 – 4
2
y = 2x2 + 4x + 6
1 left Minimum
No factorised form (1;4) x = –1 (0;6) None
4 up y=4
2 + 4
y = 2(x + 1)
3
(– 4;9) (0;–7)
4
y = –(x + 5)2 + 16
5
y = 3(x – 4)2 – 12
6
3 right
(0;10)
8 down
Unit 1 The effects of the parameters a, p and q on parabolas 91
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Unit 2: The effects of the parameters a, p
and q on hyperbolas
We will investigate the effect of the parameters a, p and q on the graph of the function
a
defined by f (x) = _____
KEY WORDS x + p + q.
asymptote – a line that a The effect of a on the hyperbola is the same as the effect of a on the parabola.
graph gets close to, but never
touches
• The greater the value of a, the steeper the curve..
xy = 4 has two asymptotes, • If a is negative the graph reflects in the x-axis.
x = 0 and y = 0
If x = 0 or y = 0, then xy = 4 Consider the hyperbola: y = f (x) = a
__ + q
x
is undefined. 1 . Changing q causes a
If q = 0 and a = 1, we have the basic hyperbola graph y = __
x
1
__
vertical shift upwards if q > 0, for example, y = x + 2 shifts up 2 units, but a vertical
1 – 1, shifts down 1 unit.
shift downwards if q < 0, for example, y = __
x
Reminders about the hyperbola y = 1
__
x
• asymptotes y
x = 0 (vertical asymptote) hyperbola
y=x
y = 0 (horizontal asymptote) y = 1__x
4
• lines of symmetry
y=x 3
symmetry line
y = –x 2
1
• Domain: x ∊ ℝ, x ≠ 0 horizontal asymptote y = 0
Range: y ∊ ℝ; y ≠ 0 x
–8 –7 –6 –5 –4 –3 –2 –1 1 2 3 4 5 6 7 8
–1
–2
–3 symmetry line
vertical
–4
asymptote
x=0 y = –x
Reminders about the hyperbola y = 1
__ + 2
x y
• asymptotes
x = 0 (vertical asymptote) y = 1__x + 2
5
y = 2 (horizontal asymptote)
4
• lines of symmetry
y=x+2 horizontal 3 hyperbola
y = –x + 2 asymptote
2
y=2
• Domain: x ∊ ℝ, x ≠ 0
Range: y ∊ ℝ, y ≠ 2 1
x
–4 –3 –2
–1 1 2 3 4
–1
symmetry symmetry
line –2 line
y=x+2 vertical
–3
asymptote
x=0
92 Topic 5 Functions: Effects of parameters
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Reminders about the hyperbola y = 1
__ – 1
x
y
• asymptotes
x = 0 (vertical asymptote) 4
y = –x –1 y = x –1
y = –1 (horizontal asymptote) 3
• lines of symmetry 2
y=x–1 symmetry symmetry
1
line line
y = –x – 1 x
• Domain: x ∊ ℝ, x ≠ 0 –4 –3 –2 –1
–1
1 2 3 4
Range: y ∊ ℝ; y ≠ –1 y = –1
y = 1__x – 1 –2
horizontal
–3 asymptote
–4
–5 vertical
asymptote
hyperbola –6
x=0
1
Horizontal shifts of the hyperbola y = __
x
x –5 –4 –3 –2 –1 0 1 2 3 4 Shift
1
y = __ 1
– __ 1
– __ 1
– __ 1
– __ –1 Undefined 1 1
__ 1
__ 1
__ None
f x 5 4 3 2 2 3 4
g y= 1
_____ 1
– __ 1
– __ –1 Undefined 1 1
__ 1
__ 1
__ 1
__ 1
__ 2 left
x+2 3 2 2 3 4 5 6
h y= 1
_____ 1
– __ 1
– __ 1
– __ 1
– __ 1
– __ –1 Undefined 1 1
__ 1
__ 1 right
x–1 6 5 4 3 2 2 3
k y= 1
_____ 1
– __ –1 Undefined 1 1
__ 1
__ 1
__ 1
__ 1
__ 1
__ 3 left
x+3 2 2 3 4 5 6 7
• Adding to x increases its value so the pattern of answers shifts to Hyperbolae and horizontal shifts
the left by the amount added. y
• Subtracting from x reduces its value so the pattern shifts to the
right by the amount subtracted. 3
2 h(x) = x –1
____
• This pattern of horizontal shifting applies to all graphs. 1
1
• Note that there is no vertical shift and the horizontal
asymptote is y = 0. x
–5 –4 –3 –2 –1 1 2 3 4
–1
–2
Vertical asymptotes
–3
• vertical asymptote of f (x) = __1x is x = 0 k(x) = x +1
_____
3 –4
1 is x = –2
• vertical asymptote of g(x) = _____
x+2 1
g(x) = _____ 1
__
x + 2 f (x) = x
1 is x = 1
• vertical asymptote of h(x) = _____
x–1
1 is x = –3
• vertical asymptote of k(x) = _____
x+3
Notice that:
f (1) = g(–1) = h(2) = k(–2) = 1
shift 2 left
shift 1 right
shift 3 left
Unit 2 The effects of the parameters a, p and q on hyperbolas 93
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Horizontal and vertical shifts of the hyperbola
1
y = _____
x–p+q
Changing p results in a horizontal shift, but changing q results in a vertical shift.
Symmetry lines using Intercepts with
k
y = _____
x–p+q Shift Asymptotes
shifts substitution the axes
1 y= 4
_____ + 3 2 right x=2 y=x Substitute (2;3) into If x = 0
x–2
3 up y=3 becomes y=x+c y = –2 + 3 = 1
Both symmetry lines will y = (x – 2) + 3 3=2+c
pass through (2;3) =x+1 c=1
y=x+1
y = –x Substitute (2;3) into If y = 0
becomes y = –x + c 4 +3
0 = _____
x–2
y = –(x – 2) + 3 3 = –2 + c = 4 + 3(x – 2)
= –x + 5 c=5 = 4 + 3x – 6
y = –x + 5
5 –2
y = – _____
2 x+1
1 left x = –1 y=x Substitute (–1;–2) into If x = 0
2 down y = –2 becomes y=x+c y = –2 + 3 = 1
Both symmetry lines will y = (x + 1) – 2 –2 = –1 + c
pass through (–1;–2) =x–1 c = –1
y=x–1
y = –x Substitute (–1;–2) into If y = 0
becomes y = –x + c 5 –2
= – _____
x+1
y = –(x + 1) – 2 –2 = 1 + c = –5 – 2(x + 1)
= –x – 3 c = –3 = –5 – 2x – 2
y = –x – 3 2x = 7
x = –3,5
WORKED EXAMPLES
1 6 +3
If f (x) = _____
x+1
1.1 State the equations of the asymptotes of f.
1.2 Determine the equations of the axes of symmetry of f.
1.3 Determine the coordinates of the x-and y-intercepts of f.
1.4 Draw f(x) and indicate the intercepts, the asymptotes and the
symmetry lines.
2 If g(x) is the reflection of f(x) in the x-axis, state the equation of g(x).
Answer all the questions in 1. for g(x), replacing f with g.
3 If h(x) is the reflection of f(x) in the y-axis, state the equation of h(x).
Answer all the questions in 1. for h(x), replacing f with h.
SOLUTIONS
1 1.1 6 +3
f (x) = _____
x+1
Vertical asymptote when x + 1 = 0 ⇒ x = –1 and horizontal asymptote
when y = 3
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1.2 Axes of symmetry: y = x + c and y = –x + c
Using shifts:
y = (x + 1) + 3 = x + 4 ⇒ y = x + 4
y = – (x + 1) + 3 = –x + 2 ⇒ y = –x + 2
Using substitution of the point (–1;3):
y = x + c ⇒ 3 = –1 + c and c = 4 ⇒ y = x + 4
y = –x + c ⇒ 3 = –( – 1 ) + c and c = 2 ⇒ y = –x + 2
1.3 To find the x-intercept, substitute y = 0.
6 + 3 = 0 ⇒ 6 + 3(x + 1) = 0 | × (x + 1)
_____
x+1
6 + 3x + 3 = 0 ⇒ 3x = –9 and x = –3
⇒ x-intercept (–3;0)
To find the y-intercept, substitute x = 0.
6 +3⇒6+3=9
y = _____
0+1
⇒ y-intercept (0;9)
1.4 y
12
x = –1
11
10
9
8
7
6 6 +3
f (x) = _____
x+1
5
4
y=3
3
2
1
x
–8 –7 –6 –5 –4 –3 –2 –1 1 2 3 4 5 6
–1
–2
–3
y=x+4 –4 y = –x + 2
2 A reflection in the x-axis leaves x as it is, but changes the sign of y.
(x + 1 )
6 + 3 = – _____
g(x) = –f (x) = – _____ 6 –3
x+1
2.1 Vertical asymptote when x + 1 = 0 ⇒ x = –1 and horizontal asymptote
when y = –3.
2.2 Axes of symmetry: y = x + c and y = –x + c
Using shifts:
y = (x + 1) – 3 = x – 2 ⇒ y = x – 2
y = –(x + 1) – 3 = –x – 4 ⇒ y = –x – 4
Using substitution of the point (–1;–3):
y = x + c ⇒ – 3 = –1 + c and c = –2 ⇒ y = x – 2
y = –x + c ⇒ – 3 = –(–1 + c) and c = – 4 ⇒ y = –x – 4
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2.3 To find the x-intercept, substitute y = 0:
6 – 3 = 0 ⇒ – 6 – 3(x + 1) = 0 | × (x + 1)
– _____
x+1
– 6 – 3x – 3 = 0 ⇒ 3x = –9 and x = –3
⇒ x-intercept (–3;0)
To find the y-intercept, substitute x = 0:
6 – 3 ⇒ – 6 – 3 = –9
y = – _____
0+1
⇒ y-intercept (0;–9)
2.4 y
y = –x – 4 3
x = –1
2
1
x
–8 –7 –6 –5 –4 –3 –2 –1 1 2 3 4
–1
–2
(–1;–3) y = –3
–3
−4
−5 6 – 3
y = _____
x+1
−6
–7
–8
–9
–10
y=x–2
–11
–12
3 A reflection in the y-axis leaves y alone, but changes the sign of x.
6 + 3 = _____
h(x) = f (–x) = ______ –6 + 3
–x + 1 x–1
3.1 vertical asymptote: ⇒ x = 1
horizontal asymptote: y = 3
3.2 Axes of symmetry: y = x + c and y = –x + c
Using shifts:
y = (x – 1) + 3 = x + 2
y = –(–x – 1) + 3 = –x + 4
Using substitution of the point (1;3):
y = x + c ⇒ 3 = 1 + c and c = 2 ⇒ y = x + 2
y = –x + c ⇒ 3 = –1 + c and c + 4 ⇒ y = –x + 4
3.3 To find the x-intercept, substitute y = 0:
– 6 + 3 = 0 ⇒ – 6 + 3(x – 1) = 0
_____ | × (x – 1)
x–1
– 6 + 3x – 3 = 0 ⇒ 3x = 9 and x = 3
⇒ x-intercept (3;0)
To find the y-intercept, substitute x = 0:
y= – 6 +3⇒6+3=9
_____
0+1
⇒ y-intercept (0;9)
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3.4 y
12
x=1
11
y = –x + 4 y=x+2
10
9
8
7
6
5
4
(1;3) y=3
3
2
1
x
–6 –5 –4 –3 –2 –1 1 2 3 4 5 6 7 8
–1
–2
–3
y = x– 6 +3
____
–1
–4
EXERCISE 7
Sketch each graph on a separate system of axes. In each case:
• State the equations of the asymptotes.
• Determine the equations of the axes of symmetry.
• Determine the coordinates of the x-intercept.
• Determine the coordinates of the y-intercept.
• Clearly indicate the intercepts with the axes, the asymptotes and the
symmetry lines.
1 1 –3
f (x) = _____
x+2
2 g(x) if g(x) is the reflection of f(x) in the x-axis
3 h(x) if h(x) is the reflection of f(x) in the y-axis
4 4 –1
r(x) = – _____
x–3
5 t(x) if t(x) is the reflection of r(x) in the x-axis
6 v(x) if v(x) is the reflection of r(x) in the y-axis
7 3 +2
w(x) = – _____
x–1
8 z(x) if z(x) is the reflection of w(x) in the x-axis
9 d(x) if d(x) is the reflection of w(x) in the y-axis
Unit 2 The effects of the parameters a, p and q on hyperbolas 97
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Finding the equation of a hyperbola
y= xa
____ + q is the general equation of a hyperbola.
–p
The asymptotes are x = p and y = q.
The symmetry lines are y = x + c and y = –x + c.
Assume that the domain is x ∊ ℝ, x ≠ p, unless you are told otherwise.
KEY WORDS If the domain is restricted, then state the restriction when you state the equation.
domain – the set of all the
x-values of the graph You may be given the equations of the asymptotes and a random point and asked to
• exclude any values of x find the equation of the hyperbola and the equations of the symmetry lines.
which make y undefined
• check to see whether the
graph is restricted or not
range – the set of all the
WORKED EXAMPLE 1
y-values of the graph
A hyperbola has two asymptotes, x = 2 and y = –1, and it passes through
• exclude any values which
the point (4;1).
make y undefined
• check to see whether the Determine the equation of the hyperbola and state the equations of its
graph is restricted or not symmetry lines.
SOLUTION
a –1
Substitute the p and q values: y = _____
x–2
a – 1 ⇒ 2 = __
Now substitute: (4;1): 1 = _____ a and a = 4
4–2 2
4 –1
The equation is: y = _____
x–2
Both symmetry lines pass through (2;–1), the point of intersection of the
asymptotes:
Symmetry line 1: y = x + c ⇒ –1 = 2 + c and c = –3, so y = x – 3
Symmetry line 2: y = –x + c ⇒ –1 = –2 + c and c = 1, so y = –x + 1
WORKED EXAMPLE 2
y
The equations of the graphs alongside are:
a
_____ + q and g(x) = mx + c
f (x) = x + p
x
1 Determine the values of a, p, q, m and c. 1
Briefly explain how you found each solution.
–2
2 State the domain and the range of f.
3 State the equation of each graph.
98 Topic 5 Functions: Effects of parameters
PLT MATHS LB 11 7th pgs (Real Book).indb 98 2012/07/02 2:21 PM
SOLUTION
1 p = –1 because the vertical asymptote is x = 1.
q = 0 because the horizontal asymptote is the y-axis.
a ⇒ –2 = _____
To find a substitute: (0;–2) into y = _____ a and so a = 2.
x–1 0–1
2=2
m is the gradient of the straight line, so m = __
1
c = –2 because this is the y-intercept of the graph.
2 Domain of f: x < 1, x ∊ ℝ and range of f: y < 0, y ∊ ℝ
3 2 , x < 1 and g(x) = 2x – 2
f (x) = _____
x–1
EXERCISE 8
1 A hyperbola has two asymptotes, x = –1 and y = 4, and it passes through the
point (1;3). Determine the equation of the hyperbola and state the equations of
its symmetry lines.
2 A hyperbola has two asymptotes, x = 2 and y = 1, and it passes through the point
(5;2). Determine the equation of the hyperbola and state the equations of its
symmetry lines.
3 A hyperbola has two asymptotes, x = –3 and y = –2, and it passes through the
point (–2;–3). Determine the equation of the hyperbola and state the equations
of its symmetry lines.
4 A hyperbola has two asymptotes, x = 4 and y = –3, and it passes through the
point (–1;– 4). Determine the equation of the hyperbola and state the equations
of its symmetry lines.
5 f (x) = x +k
_____ – q and g(x) = ax2 + bx + c. A is the turning point
p y
of the parabola.
5.1 State the domain and range of f.
5.2 Determine the values of k, p and q. x= 1
6
5.3 State the equations of the symmetry lines of f.
5.4 State the coordinates of A, B and C.
5.5 Does g have a maximum or a minimum value?
f
Briefly explain your answer.
5.6 Determine the equation of g by making use of: A y=2
5.6.1 the turning point formula
5.6.2 x-intercept method.
5.7 For which value(s) of x is f (x).g(x) < 0? x
B C
5.8 State the value of f (0) – g(0).
5.9 g is translated in such a way that the turning point of
the translated graph is (0;6).
5.9.1 Describe the translation in words.
g
5.9.2 State the equation of the translated graph.
5.10 State the equation of the graph obtained by translating
f 1 unit to the left and 2 units down.
Unit 2 The effects of the parameters a, p and q on hyperbolas 99
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EXERCISE 9
Complete the table and sketch the graphs, clearly indicating the asymptotes,
intercepts with the axes and symmetry lines. Plot all the points in the table on the
graphs as well.
Intersection of Symmetry Intercepts Point on graph
k +
y = _____
x–p q Shifts Asymptotes Asymptotes Lines with axes x y
2 –3
y = _____ 3
1 x–1
x = –1
2 1 3
y=2
x=3 (5;0)
3 –1
y = –2
4 (2;1) 3
(0;2)
4 –1
y = – _____ –1
5 x+5
1 +4
y = _____ 5
6 x–2
EXERCISE 10
1 x2 + __
Given f (x) = – __ 1 x + 3 and g(x) = _____
2 +1
4 4 x+1
1 State the equations of the asymptotes and the symmetry lines of g.
2 Draw f and g on the same system of axes, indicating the x- and y-intercepts with
the axes, asymptotes and symmetry lines.
3 Determine the coordinates of the points of intersection of f and the symmetry
line of g which has a positive gradient.
4 State the domain of g and the range of f.
5 Solve algebraically for x and y if f (x) = g(x).
6 For which value(s) of x is f(x) ≥ g(x)?
7 State the equation of the graph obtained by translating g 2 units right and 1 unit
down. Write your answer in the form y = …
8 Write down the equation of g(2x) in the form y = …
9 Write g(–x) in the form y = … and describe this transformation in words.
10 If f is translated so that its turning point lies on the intersection of the symmetry
lines of g, state the equation of the translated graph in the form y = …
11 How must g be translated so that the x and y axes become the asymptotes of g?
12 Determine an expression for g( __ 1 ) and state the value(s) of x for which it is
x
undefined.
13 Determine an expression for g(x – 4) and state the value(s) of x for which it is
undefined.
14 Show algebraically that: g(x) + g( __1 ) = 4g(–x)g(x – 4)
x
100 Topic 5 Functions: Effects of parameters
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Unit 3: The effects of the parameters a, p
and q on exponential graphs
You will investigate the effect of the parameters a, p Exponential graphs y
and q on the graph of the function defined by y = 4 × 2x y = 2x
7
y = f (x) = abx + p + q where b > 0, b ≠ 1.
6
Although the exponential graph y = abx is different from y = 1__4 × 2x
5
a,
the parabola y = ax2 and the hyperbola y = __
x 4
the effects of a are exactly the same.
3
• The greater the value of a, the steeper the curve. 2
• If a is negative the graph reflects in the x-axis. 1 1
__ y=0
4 x
–3 –2 –1 – _1_ 1 2 3 4
–1 4 horizontal
–2 asymptote
–3
–4
–5
y = – __14 × 2x
–6
–7
Reminders about the exponential
( )
y = –4 × 2x y = –2x
graphs y = 2x and y = 1 __ x
2
• These graphs share the same horizontal asymptote,
y = 0.
• Exponential graphs do not have any lines of symmetry. y
• Domain: x ∊ ℝ and Range: y > 0, y ∊ ℝ
• y = 2x is an increasing function because as x 6
increases, y increases y = 2–x y = 2x
• y = ( __12 ) is a decreasing function because as
x 5
x increases, y decreases 4
• y = 2x is the reflection of y = ( __2 ) in the y-axis
1 x
3
(2)
1 x = 2–x
because __
(–1;2) 2 (1;2)
• For a reflection in y-axis, leave y alone and change sign of x.
1
• For a reflection in x-axis, leave x alone and change sign of y. (–1;0,5) (1;0,5)
x
Notice in the figure on the right that: –3 –2 –1 1 2 3
(–1;–0,5) –1 y=0
If y = 2x is reflected in the x-axis, the reflected graph is y = –2x.
If y = 2x is reflected in the y-axis, the reflected graph is y = 2–x. –2 (1;–2)
–3
–4
–5
–6
y = –2x
Unit 3 The effects of the parameters a, p and q on exponential graphs 101
PLT MATHS LB 11 7th pgs (Real Book).indb 101 2012/07/02 2:21 PM
WORKED EXAMPLE
1 State the asymptote and y-intercept of y = 3x.
2 Consider the exponential graph:
f (x) = 3x + 1
2.1 State the asymptote of f.
2.2 State the y-intercept of f and the coordinates of two other points.
3 If g(x) is the reflection of f in the x-axis:
3.1 state the equation of g
3.2 state the equation of the asymptote of g
3.3 state the y-intercept g and the coordinates of two other points.
4 On the same set of axes, sketch f and g and show the asymptotes, the
y-intercept and two other points on each graph.
SOLUTION
y
1 The asymptote is y = 0 and the y-intercept
is (0;1). 6
y = 3x + 1
2 2.1 The asymptote of is y = 1. 5
2.2 The y-intercept is (0;2) and two other
( )
1 and (1;4).
points are −1;1__
3
4 (1;4)
3
3 3.1 g(x) = −f (x) = −(3x + 1) = −3x − 1
3.2 The asymptote is y = −1. ( –1;11__3 ) 2
3.3 The y-intercept is (0;−2) and
( )
1
two other points are −1;−11 __
3
and (1;−4) x
–2 –1 1 2 3 4
4 See the graph alongside. –1
y = –1
( −1;−11__3 ) –2
–3
–4 (1;–4)
–5
y = –(3x) – 1
–6
There is a table function on most calculators which will generate points for you.
Most scientific calculators will work according to the detailed instructions below:
Consider f (x) = –3x – 1 and follow the instructions below to generate a table of values:
• Select MODE.
• Press the number next to TABLE and your calculator will display f(X) =
• Press ‘–3’.
• Now Press ‘x□’, then ‘ALPHA’ followed by ‘X’.
• Use the replay button to take the cursor down to its normal position.
• Press ‘–1’.
• Your display will now show f (X) = –3x – 1
102 Topic 5 Functions: Effects of parameters
PLTMATHSLB11LB_05.indd 102 2012/07/14 2:47 PM
• Press ‘=’ and your display will change to ‘Start?’.
• Press ‘–3’ = (or any other start value of your choice).
• Your display will now change to End?.
• Press ‘3=‘ (or any other end value of your choice).
• Your display will now show ‘Step?’.
• Press ‘1=‘ and a table of values will appear.
• Choose the points you want to plot and ignore the rest.
• If you want to edit your equation, or change the start, end or step values press ‘AC’.
• Return to normal mode by selecting ‘MODE’, then press the number next to
‘COMP’.
• Do not depend on the table for the asymptotes or the x-intercept.
EXERCISE 11
1 If f (x) = 5×2x, sketch f (x), –f (x) and f (–x) on the same system of axes.
2 If g(x) = 5x, sketch g(x), –g(x) and g(–x) on the same system of axes.
( 3 ) + 2, sketch h(x), −h(x) and h(–x) on the same system of axes.
1 x
3 If h(x) = __
4 If t(x) = 3 × 2x – 1, sketch t(x), –t(x) and t(–x) on the same system of axes.
Horizontal shifts of the exponential graph y = 2x
y
k g f h
(–4;1) (–2;1) (0;1)
(1;1)
x
x –5 –4 –3 –2 –1 0 1 2 3 4 Shift
y = 2x 1
___ 1
___ 1
__ 1
__ 1
__ 1 2 4 8 16 None
f 32 16 8 4 2
y = 2x + 2 1
__ 1
__ 1
__ 1 2 4 8 16 32 64 2 left
g 8 4 2
y = 2x – 1 1
___ 1
___ 1
___ 1
__ 1
__ 1
__ 1 2 4 8 1 right
h 64 32 16 8 4 2
y = 2x + 4 1
__ 1 2 4 8 16 32 64 128 256 4 left
k 2
Unit 3 The effects of the parameters a, p and q on exponential graphs 103
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Horizontal and vertical shifts of the exponential
REMEMBER
graph y = abx + p + q
If an unknown value appears
more than once in the same
equation, it has the same WORKED EXAMPLE
value each time it appears.
x0 = 1 or all x ∊ ℝ , x ≠ 0 Calculations required for sketching the shifted exponential graph:
If y = abx + p + q, q is not the
y-intercept. y = abx + p + q Shift Asymptote y-intercept (x = 0) x-intercept (y = 0)
Find the y-intercept by letting
2 right y = 5.2–2 + 3 = 4,25 No x-intercept
x = 0. 1 y = 5.2x – 2 + 3 y=3
3 up (0;4,25) y – 3, y ≠ 0
The asymptote is given by
y=q No x- intercept
2 left y = – 4.32 – 1 = –37
You may need to use 2 y = – 4.3x + 2 – 1 y = –1 – 4.3 x + 2 ≠ 1
1 down (0; – 37)
simultaneous equations to y < –1, y ≠ 0
determine the values of a, 1 left y = 2.5 – 2 = 8 5x + 1 = 1 = 50
b, p and q. Refer back to 3 y = 2.5x + 1 – 2 y = –2
2 down (0;8) x = –1 ⇒ (–1;0)
Topic 2 if you need to revise
(2) + 1
1 x y = 3(1) + 1 = 4 No x-intercept
4 y = 3 __ 1 up y=1
(0;4) y > 1, y ≠ 0
( 3 )x – 1 + 6 6 up
1 1 right y = –2(3) + 6 x=0
5 y = –2 __ y=6
= 0 ⇒ (0;0) (0;0)
Finding the equation of an exponential graph
in the form y = abx + p + q
All exponential graphs have a horizontal asymptote and if y = abx + p + q, then y = q is
the equation of the horizontal asymptote.
WORKED EXAMPLE y
h
h(x) = abx + p + q and passes through C(3;4) and D(2;8).
D(2;8)
The asymptote of h is given by y = 2.
If a = q, determine the values of b, p and q.
C(3;4)
y=2
x
SOLUTION
Substitute the points (3;4), (2;8) and q = 2.
q = 2 ⇒ y = 2bx + p + 2
(3;4) ⇒ 4 = 2b3 + p + 2 and 2 = 2b3 + p, so 3 + p = 0 and p = –3
1
(2;8) ⇒ 8 = 2b2 – 3 + 2 and 6 = 2b–1, so 3 = b- 1 and b = __
3
⇒ h(x) = 2 1( )
__ x – 3 + 2
3
104 Topic 5 Functions: Effects of parameters
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EXERCISE 12
y y
f l
(1;5)
y=2 y=2
2
g 1
x x
–1
y = –1 (–1;–3)
x=1
h
–8
1 1.1 If f (x) = abx + c, determine the values of a, b and c
1.2 If g(x) = –2dx + p + e, determine the values of d, e and p
1.3 If h(x) = jrx + t, determine the values of j, r and t
1.4 v
If l(x) = _____
x – w + z, determine the values of v, w, and z.
EXERCISE 13
Complete the table by filling in the missing values. Then use the information in the
table to draw each graph on a separate set of axes. Clearly indicate the intercepts with
the axes, the asymptotes and the coordinates of at least two points.
Point 1 Point 2
y = abx + p + q Shift Asymptote Intercepts x y x y
1 y = 3.2x + 1 – 6 –1 6
2 y = –5x – 2 + 1 –1 –4
3 ( 3 )x + 1 – 2
1
y = 2. __ 1 4
4 y = 2–x – 1 –2 –0,5
y = –5.3x – 1 + 5 2 4
4__
5 9
6 ( 5 )x – 1 – 2
1
y = 2 __ 2 48
Unit 3 The effects of the parameters a, p and q on exponential graphs 105
PLT MATHS LB 11 7th pgs (Real Book).indb 105 2012/07/02 2:21 PM
7 f (x) = ax2 + bx + c, g(x) = x d
____ + j and h(x) = vkx + q
+e
y
h A(2;8)
f
x = –3
x
y = –2 –1
–1
g
B(–4;–10)
7.1 Determine the values of a, b and c if the turning point of f is A(2;8) and f
passes through the point B(–4;–10).
7.2 Determine the values of d, e and j if g passes through B(–4;–10) and has
asymptotes x = –3 and y = –2.
7.3 Determine v, k and q if h passes through (–1;0) and (0;–1) and has any
asymptote given by y = –2.
8 Sketched in the figure are the graphs f (x) = ax2 + bx + c;
g(x) = x k
____ + q and h(x) = mx + n.
–p
y
f
g h
x
A –2 4
–1 Q
x = –2 B R
P(–1;–5)
T
A and B are the x- and y-intercepts of g respectively.
Graphs f, g and h intersect at P(–1;–5) and graphs g and h intersect at Q.
106 Topic 5 Functions: Effects of parameters
PLT MATHS LB 11 7th pgs (Real Book).indb 106 2012/07/02 2:21 PM
8.1 State the equation of the axis of symmetry of f.
8.2 For which value(s) of x is f (x) an increasing function?
8.3 Determine the equations of the symmetry lines of g.
8.4 For which value(s) of x is f (x) ≤ h(x)?
8.5 Determine the equations of graphs f, g and h.
8.6 Determine the coordinates of A and B.
8.7 Determine the coordinates of Q.
8.8 For which value(s) of x is h(x) ≤ g(x)?
8.9 State the values of x for which f(x). g(x) ≤ 0.
8.10 RT is parallel to the y-axis, with R on h and T on f.
8.10.1 Determine an expression for the length of RT.
8.10.2 Determine the maximum length of RT.
8.11 For which values of x is g an increasing function?
4 −1
8.12 Consider the function: g( x ) = − _____
x+2
8.12.1 Determine the equation of v if v is the reflection of g in the y-axis.
8.12.2 Determine the symmetry lines of v.
8.12.3 Sketch v on a new set of axes, clearly indicating the intercepts with
the axes and at least one other point.
8.12.4 Sketch and label the symmetry lines of v.
8.13 Given that w is obtained by translating g 2 units to the right and 1 unit up.
8.13.1 State the equation of w.
8.13.2 State the asymptotes of w.
8.13.3 State the equations of the symmetry lines of w.
8.13.4 Determine the shortest distance between w and the origin.
Unit 3 The effects of the parameters a, p and q on exponential graphs 107
PLT MATHS LB 11 7th pgs (Real Book).indb 107 2012/07/02 2:21 PM
Unit 4: Real life applications
WORKED EXAMPLE
Vincent has money to invest and considers two options:
Option 1: Simple interest at a rate of 13,5% using the formula A = P(1 + in)
Option 2: Compound interest at a rate of 7,5% p.a. compounded annually using
the formula A = P(1 + i)n
1 Vincent wants know how long it will take for the two options to
become equal.
1.1 Set up an equation which would help him to work out how long it
would take.
1.2 Explain why is makes no difference how much money he invests at
the start.
2 Consider the formula: A = 1 + in | Ignore P as it does not affect the calculation.
2.1 Rewrite the formula to express it in terms of x and y.
2.2 Is the graph a straight line graph or an exponential graph?
3 Consider the formula: A = (1 + i)n | Ignore P as it does not affect the
calculation.
3.1 Determine 1 + i and give your answer in its simplest form
3.2 Rewrite the formula to express it in terms of x and y.
3.3 Is the graph a straight line graph or an exponential graph?
4 Sketch both graphs on the same system of axes, clearly labelling the graphs
and indicating the points used to plot the graph.
5 Use your graph to estimate the number of years it will take for the
investments to be equal. Give your answer to the nearest year.
SOLUTIONS
1 1.1 P( 1 + in ) = P( 1 + i )n ⇒ 1 + in = ( 1 + i )n
1.2 The initial amount cancels out when the equation is set up to compare
both options.
2 2.1 y = 1 + 0,135x
2.2 Straight line
3 3.1 1,075
3.2 y = 1,075x
3.3 Exponential graph
108 Topic 5 Functions: Effects of parameters
PLT MATHS LB 11 7th pgs (Real Book).indb 108 2012/07/02 2:21 PM
4 Simple interest and compound interest
y
y = 1,075x y = 1 + 0,135x
4
(18;3,68)
(20;3,7)
3
(15;2,96)
2 (12;2,38)
(9;1,92)
(6;1,54)
1 (3;1,24)
x
5 10 15 20
5 It will take 16 years to the nearest year.
EXERCISE 14
1 South African death rates since 1997
1 200 000
1 000 000
y = 39 378x + 281 235
Number of deaths
800 000
600 000
y = –1 709x2 + 56 468x + 249 902
400 000
200 000
0
0 5 10 15 20 25 30 35
Number of years since 1996
In the graph, death rates have been plotted and two possible models suggested.
1.1 If the linear model proves to be correct, briefly discuss the long-term
implications.
1.2 After how many years would the maximum number of deaths be reached
if the parabolic model is correct? Give your answer correct to two decimal
places.
1.3 According to the parabolic model, what will the maximum number of
deaths be? (Remember to give your answer as a whole number.)
1.4 State the year in which the maximum number of deaths occurs.
1.5 After which year does the number of deaths begin to decrease?
2 Jack is in Grade 11 and takes a gap year after school. He plans to start a savings
account with R10 at the end of the first month. For the next 11 months, he will
save double the amount of each previous month.
2.1 Calculate how much money he will add to his account at the end of the
fourth month.
2.2 Draw a graph which shows the first five monthly deposits.
Unit 4 Real life applications 109
PLT MATHS LB 11 7th pgs (Real Book).indb 109 2012/07/02 2:21 PM
2.3 Is the graph continuous or discrete? Justify your answer.
2.4 Does the graph start at 0 months or 1 month? Explain your answer.
2.5 Do you think Jack will be able to maintain his monthly payments for
12 months? Justify your answer.
2.6 Determine the equation for your graph.
2.7 Check your equation by making use of your calculator.
3 This question is designed to show you that you can set up your own model from
raw data.
The data in the table was used to generate the graphs you used in Question 2.
Number of Number of deaths
years since per year in South
Year 1996 Africa
1997 1 316 559
1998 2 365 109
1999 3 381 037
2000 4 414 768
2001 5 500 082
2002 6 554 199
2003 7 572 620
2004 8 593 337
2005 9 605 408
3.1 Using your calculator or Excel, generate each of these models:
f (X) = A + BX; g(X) = A + BX + CX2; h(X) = A.BX
3.2 Determine the number of years since 1996 if the year is 2020.
3.3 Now determine the anticipated number of deaths in 2020 for each
different model.
3.4 Briefly discuss your findings and comment on the different outcomes.
110 Topic 5 Functions: Effects of parameters
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Unit 5: The average gradient between
two points on a curve
EXERCISE 15
REMEMBER
Consider the sketch of the parabola f (x) = –x2 + 2x + 8 and eight different Gradient of straight line:
straight line graphs. Each straight line passes through point A and one other y − y1
_______
m = x2
point on the parabola. 2 − x1
y Function notation:
y = f( x )
If x = 3, then
y = f( 3 ) = − ( 3 )2 + 2( 3 ) + 8
F = −9 + 6 + 8
G E
=5
H D (3;5) lies on f.
I C
x
–5 –4 –3 –2 –1 1 2 3 4 5 6 7
A B
1 Determine the coordinates of the points A, B, C, D, E, F, G, H and I.
2 Consider AB and take note that xB – xA = 5 – (–3) = 8 units:
y
x
A(–3;–7) B(5;–7)
The shaded area enclosed by the parabola and the line passing through A and B
is a large area, as shown in the figure.
Now consider AE and notice that xE – xA = 2 – (–3) = 5 units:
y
E(2;8)
x
A(–3;–7)
The shaded area enclosed by the parabola and the line passing through A and
E is smaller than the shaded area enclosed by the parabola and the line passing
through A and B.
Unit 5 The average gradient between two points on a curve 111
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Complete the statements by filling in the missing words:
The … the points are …, the greater is the enclosed area.
The … the points are …, the … is the enclosed area.
We use the gradient of the straight line between two points on a curve to
determine the average gradient of the curve between those points.
y –y
______
3 Use the formula m = x2 1
– x and determine the average gradient of the parabola
2 1
f (x) = –x2 + 2x + 8 between the points:
3.1 A and B 3.2 A and C
3.3 A and D 3.4 A and E
3.5 A and F 3.6 A and G
3.7 A and H 3.8 A and I
4 Which average gradient that you determined in Question 3 best approximates
the gradient of the parabola at point A? Briefly explain your answer.
EXERCISE 16
1 4 –1
Consider the hyperbola: g(x) = _____
x–3
1.1 Determine:
1.1.1 g(–1)
1.1.2 g(0)
1.1.3 g(1)
1.1.4 g(2)
1.2 Determine the average gradient of g between:
1.2.1 x = –1 and x = 0
1.2.2 x = –1 and x = 1
1.2.3 x = –1 and x = 2
1.3 Sketch g(x) = 4_____ – 1 and indicate your solutions to Questions 1.1 and
x–3
1.2 on your sketch.
1.4 Which average gradient in 1.2 best represents the gradient of g at x = −1?
2 Consider the exponential graph: h(x) = 2 – x
2.1 Determine:
2.1.1 h(–2)
2.1.2 h(–1)
2.1.3 h(0)
2.1.4 h(1)
2.2 Determine the average gradient of h between:
2.2.1 x = –2 and x = –1
2.2.2 x = –2 and x = 0
2.2.3 x = –2 and x = 1
2.3 Sketch h(x) = 2– x and indicate your solutions to Questions 2.1, and 2.2
on the sketch.
2.4 Which average gradient in 2.2 best represents the gradient of h at x = −2?
112 Topic 5 Functions: Effects of parameters
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3 Consider the parabola: k(x) = x2 – 4x – 12
3.1 Determine:
3.1.1 k(–1)
3.1.2 k(0)
3.1.3 k(2)
3.1.4 k(4)
3.2 Determine the average gradient of h between:
3.2.1 x = –1 and x = 0
3.2.2 x = –1 and x = 2
3.2.3 x = –1 and x = 4
3.3 Sketch k(x) = x2 – 4x – 12 and indicate your solutions to Questions 3.1
and 3.2 on the sketch.
3.4 Which average gradient in 3.2 best represents the gradient of k at x = −1?
4 Determine the average gradient of the curve:
4.1 p(x) = – 2
_____ + 5 between x = 1 and x = 2
x–3
4.2 q(x) = 2 × 3x – 1 between x = –1 and x = 1
4.3 r(x) = –2x2 – 5x + 7 between x = –2 and x = 1
Unit 5 The average gradient between two points on a curve 113
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Revision
1 y
f g h
F(–2;3)
x
E C
O
D
B
A
1 – 1; h(x) = __
Sketched in the figure are g( x ) = – _____ 1 (x – 2)2 – 2; and f (x) = ax + q.
x–1 2
A and B are the points of intersection of g and h. D and E are the points of
intersection between the vertical asymptote of g and graphs h and f respectively.
O and C are the x-intercepts of h. E and F are points on f.
1.1 Determine the values of a and q. (4)
1.2 State the coordinates of the turning point of h. (2)
1.3 State the equation of the axis of symmetry of h. (1)
1.4 Determine the length of OC. (2)
1.5 If P(–4;2) lies on h, state the coordinates of Q
if Q is the reflection of P in the axis of symmetry of h. (2)
1.6 State the equations of the asymptotes of g. (2)
1.7 Determine the length of ED. (3)
1.8 Determine the coordinates of B. (7)
1.9 An axis of symmetry of the graph of g is a straight line defined as
y = mx + c where m > 0. Write down the equation of this line in the
form y = … (2)
1 (x – 2)2 – 2
1.10 Consider the graph: h( x ) = __
2
1.10.1 Describe in words the translation which must take place
so that the turning point of graph h is shifted to the origin. (2)
1.10.2 State the equation of the translated graph in the form y = … (2)
[29]
2 2 – 3; g(x) = – __
Consider the graphs defined by: f (x) = – _____ 1 (x + 6)(x – 4)
x+1 4
and h(x) = –2x – 5
2.1 State the equations of the asymptotes of f. (2)
2.2 Determine the equations of the symmetry lines of f and g. (6)
2.3 Draw f, g and h on the same system of axes. Clearly label all
intercepts with the axes as well as the coordinates of any turning
points. Draw and clearly label all symmetry lines and asymptotes. (10)
2.4 Solve for x if f (x) = h(x). (6)
2.5 For which value(s) of x is – 1
__(x + 6)(x – 4) > 0? (2)
4
2.6 For which values of x is (x).h(x) < 0? (4)
2.7 If f (x) is translated 2 units to the right and 1 unit up, state
the equation of the translated graph in the form y = … (2)
114
PLT MATHS LB 11 7th pgs (Real Book).indb 114 2012/07/02 2:21 PM
2.8 If h(x) is reflected in the y-axis, state the equation of the reflected
graph in the form y = … (2)
2.9 If g(x) is translated 3 units left and 2 units down, determine the
equations of the translated graph. State your answer in the
form y = ax2 + bx + c. (4)
2.10 Determine g(–2) – f (–2). (3)
2.11 Determine the average gradient of f(x) between:
2.11.1 x = –7 and x = –2 (3)
2.11.2 x = –5 and x = –2 (3)
2.11.3 x = –3 and x = –2 (3)
2.12 Consider the gradients you found in Question 2.11 and briefly
explain which gradient best represents the gradient of f at x = –2. (2)
2.13 Solve for x, correct to two decimal places, if g(x) = h(x). (4)
2.14 Determine the average gradient of g(x) between:
2.14.1 x = – 4 and x = 0 (3)
2.14.2 x = – 4 and x = –3 (3)
2.15 Which gradient from your solutions to Question 2.14 best
represents the gradient of g(x) at x = – 4? Briefly explain your answer. (2)
[64]
3. Nyati makes up his own graph application question for a maths project.
He decides to compare the height of a stone that is thrown up into the air,
against the time taken by the stone to reach that height.
After several trials he records these results, with t being the time in seconds
and h the height of the stone in metres.
t 0 1 2 3 4 5
h 15 20 21 18 11 0
3.1 How long did it take for the stone to hit the ground? (1)
3.2 Draw a graph to represent the information in the table. (4)
3.3 Determine an equation which models the relationship and
state your answer in the form h = … (5)
3.4 After how many seconds did the stone reach its maximum height? (2)
3.5 What was the maximum height reached? (1)
[13]
115
PLT MATHS LB 11 7th pgs (Real Book).indb 115 2012/07/02 2:21 PM
TOPIC
2 6 Functions: Trigonometric graphs
Unit 1: Revision of trigonometric graphs
KEY WORDS
using point-by-point plotting
period – the number of
degrees needed to complete The three basic trigonometric graphs are:
a wave pattern or cycle • y = sin x, y = cos x and y = tan x
amplitude – half the total
distance between the Trigonometric graphs covered in Grade 10 include:
minimum and maximum
• y = asin x + p, y = acos x + p and y = atan x + p
values
• always positive
You achieve point by point plotting by substituting x-values into an equation to find
• maximum – minimum
__________________
the y-values.
2
Use the table function on your calculator to generate the table of points which you
need to plot. There are detailed calculator instructions in Unit 5 of this topic.
WORKED EXAMPLE
1 Use a calculator to complete the table for y = sin x.
x y x y x y x y x y x y
–360° –330° –300° –270° –240° –210°
When you cover Topic 7,
you will learn how to work –180° –150° –120° –90° –60° –30°
out the values in tables
0° 30° 60° 90° 120° 150°
using special triangles.
180° 210° 240° 270° 300° 330°
360°
The range is 2 Give the range of y.
y ∊ (minimum;maximum)
3 Sketch the graph y = sin x by plotting the points from the table.
4 Give the period of the graph.
5 Give the amplitude of the graph.
6 Describe how the graph y = sin x must be shifted to become y = cos x.
__
1 = 0,5 √ 3
__ ___ ≈ 0,87
2 2
SOLUTIONS
1 x y x y x y x y x y x y
__ __
1
__ √3
___ √3
___ 1
__
–360° 0 –330° –300° –270° 1 –240° –210°
2 2 2 2
__ __
1 √3 √3 1
–180° 0 –150° – __ –120° – ___ –90° –1 –60° – ___ –30° – __
2 2 2 2
__ __
1
__ √ 3
___ √ 3
___ 1
__
0° 0 30° 60° 90° 1 120° 150°
2 2 2 2
__ __
1 √3 √3 1
180° 0 210° – __ 240° – ___ 270° –1 300° – ___ 330° – __
2 2 2 2
360° 0
116 Topic 6 Functions: Trigonometric graphs
PLTMATHSLB11LB_06.indd 116 2012/07/07 11:23 AM
2 Range: y ∊ [–1;1]
3 y
(–270°;1) y = sin x (90°;1)
1
x
–360° –270° –180° –90° 90° 180° 270° 360°
–1
(–90°;–1) (270°;–1)
4 Period = 360°
5 Amplitude = 1
6 y = sin x must shift 90° to the left to become y = cos x.
EXERCISE 1
REMEMBER
1 1.1 Copy the table provided for the Worked example and complete it for
An asymptote is a line which
y = cos x. cannot be touched or crossed
1.2 Give the range of y. by the graph to which it is an
1.3 Sketch the graph y = cos x by plotting the points from the table. asymptote.
1.4 Give the period of the graph.
1.5 Give the amplitude of the graph.
1.6 Describe the shift required for the graph y = cos x to become y = sin x
2 2.1 Copy the table below and complete it for y = tan x.
x y x y x y x y
–360° –315° –270° –225°
–180° –135° –90° –45°
0° 45° 90° 135°
180° 225° 270° 315°
360°
2.2 For which value(s) of x is tan x undefined?
2.3 y = tan x is undefined at x = 90°, which is one of the vertical asymptotes
of the tan graph. State the equations of three other vertical asymptotes of
y = tan x for x ∊ [–360°;360°].
2.4 Sketch y = tan x for x ∊ [–360°;360°], indicating the intercepts with the axes
and the asymptotes.
2.5 State the range and the amplitude of y = tan x. REMEMBER
2.6 Give the period of y = tan x. Division by zero is undefined.
Unit 1 Revision of trigonometric graphs using point-by-point plotting 117
PLTMATHSLB11LB_06.indd 117 2012/07/07 11:23 AM
Important facts about the three basic trigonometric graphs
x ∊ [360°;360°] y = sin x y = cos x y = tan x
Amplitude 1 1 undefined
Period (P) 360° 360° 180°
Asymptotes none none x = ±90°
x = ±270°
Range y ∊ [–1;1] y ∊ [–1;1] y ∊ (–∞;∞)
Interval spacing 90° 90° 45°
y
y = tan x
(–270°;1) (45°;1) (90°;1) (360°;1)
1
y = sin x
y = cos x
x
–360° –315° –270° –225° –180° –135° –90° –45° 45° 90° 135° 180° 225° 270° 315° 360°
–1
(–180°;–1) (–90°;–1) (180°;–1) (270°;–1)
–2
x = –270° x = –90° x = 90° x = 270°
y = sin x and y = cos x each have a period of 360° because one wave pattern
takes 360°.
• Both graphs have turning points and intercepts which are spaced at 90°
intervals.
• Intervals help us to set up the scale on the x-axis.
• If you generate a table of values using your calculator, select a step of 90°.
1.
Consider the graph given by y = –cos x + __
2
y
(–180°;1,5) (180°;1,5)
1,5 maximum
is 1,5 units
value
1
amplitude = 1
(–270°;0,5) (–90°;0,5)
0,5 (90°;0,5) (270°;0,5)
amplitude = 1
x
–360° –300° –240° –180° –120° –60° 60° 120° 180° 240° 300° 360°
The period is 360° because it takes 360°
to complete one wave pattern minimum value
–0,5 is –0,5 units
(–360°;–0,5) (360°;–0,5)
• the amplitude is 1 and the vertical shift is __12 unit up
118 Topic 6 Functions: Trigonometric graphs
PLTMATHSLB11LB_06.indd 118 2012/07/07 11:23 AM
• the period is 360° and the interval spacing is 90°
• the vertical shift creates a need for smaller steps of 60° or 30°
• the range is y ∊ [– 0,5;1,5]
• the maximum value is y = 1,5 and the minimum value is y = – 0,5.
It may not always be possible to determine the x-intercepts without complicated
calculations. It is acceptable to give the coordinates of the shifted x-intercepts in
these cases, rather than the x-intercepts.
y = tan x has a period of 180° because one wave pattern takes 180°.
• The equations of the asymptotes are x = –270°; x = –90°; x = 90° and x = 270°.
• Each asymptote is positioned in the middle of the period.
• (45°;1) is halfway between the (0°;0) and the asymptote x = 90°.
• Important values occur every 45°, so the interval spacing for y = tan x is 45°. KEY WORD
• If you generate a table of values using your calculator, select a step of 45°.
interval spacing – the
number of degrees between
Consider the influence of a and q on the graphs:
critical points
y = asin x + q y = acos x + q y = atan x + q
a influences the vertical stretch of these graphs whereas q influences the vertical shift.
Graph Amplitude Vertical shift REMEMBER
y = sinx 1 None Amplitude = half the total
y = −3sinx + 1 3 Up 1 distance between the
1 sinx − 2 1 minimum and maximum
y = __ __ Down 2
2 2 values.
y = − 5cosx + 2 5 Up 2 Amplitude is always positive,
y = 2cosx − 1 2 Down 1 even if a is negative.
tan graphs have neither a
y = 3tanx − 1 Undefined Down 1
minimum nor maximum
value, and therefore
y = tanx has a period of 180° because one wave pattern takes 180°.
amplitude is undefined.
• The equations of the asymptotes are x = −270°; x = − 90°; x = 90° and x = 270° tan 45° = 1, so when x = 45°,
• Each asymptote is positioned in the middle of the period. y = atan x + q will result in the
• (45°;1) is exactly halfway between the (0°;0) and the asymptote x = 90°. coordinates (45°;a + q).
• Important values occur every 45°, so the interval spacing for y = tanx is 45°
• If you generate a table of values using your calculator, select a step of 45°.
WORKED EXAMPLE
Given f (x) = −tanx and g(x) = tanx + 1 or x ∊ [−90°;270°].
1 State the equation(s) of the asymptotes of f and g.
2 Draw f and g on the same system of axes.
3 State the range of g.
4 State the period of f.
5 State the amplitude of g.
6 Show that f (−26,57°) = g(−26,57).
7 State one other value of x for which f (x) = g(x).
8 State two values of x for which g(x) − f (x) = 1.
9 State two values of x for which g(x) − f (x) = 3.
10 For which values of x is f (x).g(x) > 0?
Unit 1 Revision of trigonometric graphs using point-by-point plotting 119
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SOLUTIONS
1 Asymptotes of f and g: x = −90°, x = 90° and x = 270°
2 f(x) = – tan x y g(x) = tan x + 1
3
2 (45° ;2) (225° ;2)
1 (180° ;1)
(– 45° ;1) (135° ,1)
x
– 90° – 45° 45° 90° 135° 180° 225° 270°
–1
(45° ;–1) (225° ;–1
–2
–3
–4
x = – 90° x = 90° x = 270°
3 Range of g: y ∊ (−∞;∞)
4 Period of f is 180°.
5 Amplitude of g is undefined.
6 f(−26,57°) = −tan(−26,57°) = 0,50 and g(−26,57°) = tan(−26,57°) + 1 = 0,50
7 −26,57° + 180° = 153,43°
8 x = 0° and x = 180°
9 x = 45° and x = 225°
10 x ∊ (−45°;0°) ∪ (135°;180°)
WORKED EXAMPLE
Given f (x) = sin x and g(x) = 2sin x for x ∊ [–360°; 360°]
1 Draw f and g on the same system of axes.
2 State the amplitudes of f and g.
3 State the ranges of f and g.
4 State the periods of f and g.
5 What is the maximum value of g?
6 For which values of x is f (x) = g(x)?
7 For which values of x is f (x) ≥ g(x)?
8 For which value(s) of x is f (x) – g(x) = 1?
9 For which values of x is f (x) an increasing function?
120 Topic 6 Functions: Trigonometric graphs
PLTMATHSLB11LB_06.indd 120 2012/07/07 11:23 AM
SOLUTIONS
y
1 (90°;2)
(–270°;2)
2
(–270°;1) (90°;1)
g(x) = 2sin x
x
–360° –180° 180° 360°
(–90°;–1) (270°;–1)
1 unit f (x) = sin x 1 unit
–2
(–90°;–2) (270°;–2)
2 The amplitude of f is 1 unit and the amplitude of g is 2 units.
3 The range of f is y ∊ [–1;1] and the range of g is y ∊ [–2;2].
4 Both f and g have a period of 360°.
5 Maximum value of g is 2 units.
6 x = 0° or ± 180° or ±360°
Note: f (x) = g(x) where the graphs intersect, as indicated by green dots on
the x-axis.
7 x ∊ [–180°;0] ∪ [180°;360°] | The thick orange lines indicate where sin x ≥ 2sin x
and the thick blue lines on the x-axis indicate where the solution is read.
8 x = –90° or x = 270° | There are four x-values for which the graphs differ by
1 unit, but only two of the values occur when sin x ≥ 2sin x and are indicated by
thick green vertical lines.
9 x ∊ [–360°; – 270°] ∪ [ – 90°;90°] ∪ [270°;360°] | The black dashes on the
orange graph indicate where sin x is an increasing function.
EXERCISE 2
For each question, draw the graphs on the same system of axes. State the period, REMEMBER
amplitude and range for each graph. If there are asymptotes, sketch and label them. In an increasing function,
the y-values increase as x
1 h(x) = 2cos x and g(x) = cos x for x ∊ [–180°;360°]
increases.
1.1 For which value(s) of x is g(x) = 0?
In a decreasing function,
1.2 For which values of x is h(x) > 0? the y-value decreases as x
1.3 For which values of x is h(x) an increasing function? increases.
To reflect a graph in the
2 f (x) = tan x and g(x) = 2tan x for x ∊ [–180°;180°]
x-axis, leave x and change
2.1 For which value(s) of x is f (x) = g(x)?
the sign of y.
2.2 For which value(s) of x is f (x) – g(x) = 1? To reflect a graph in the
3 f (x) = –2sin x + 1 and g(x) = __3 cos x for x ∊ [–90°;270°] y-axis, leave y and change
2 the sign of x.
3.1 For which values of x is f (x) – g(x) = 3?
3.2 State one value of x for which g(x) – f (x) = 1.
3.3 For which values of x is g(x) ≥ 0?
3.4 If f is reflected in the x-axis, state the new equation in the form y = …
Unit 1 Revision of trigonometric graphs using point-by-point plotting 121
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4 Given the graphs f (x) = –tan x and g(x) = __ 1 tan x for x ∊ [– 45°;135°]
2
4.1 For which value(s) of x is g undefined?
4.2 Determine g(45°) – f(45°).
4.3 If f is reflected in the x-axis, state the new equation in the form y = …
4.4 If g is reflected in the y-axis, state the new equation in the form y = …
EXERCISE 3
Complete the table for the graphs y = asin x + q, y = acos x + q and y = atan x + q.
Draw each graph on a separate set of axes for the stated domain. Indicate all x-and
y-intercepts, the coordinates of all turning points and endpoints.
End
Equation a Amplitude Range q Shift Domain points
1 y = 2sin x + 1 x ∊ [−180°;180°]
3 cos x – 1
y = __ x ∊ [−180°;180°]
2 2
3 y = tan x + 2 x ∊ [−90°;90°]
4 1 sin x – 1
y = __ x ∊ [−90°;270°]
2
5 1 cos x + __
y = – __ 1 x ∊ [−90°;270°]
3 2
6 y = –2tan x x ∊ [−135°;135°]
7 y = –sin x + 2 x ∊ [–270°; 90°]
8 y = –3cos x – 1 x ∊ [−180°;180°]
122 Topic 6 Functions: Trigonometric graphs
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Unit 2: The effect of the parameter k on
some trigonometric functions
We will consider the effect of the parameter k on the functions defined by y = sin kx,
y = cos kx and y = tan kx.
WORKED EXAMPLE REMEMBER
1 Complete the table. (Use the table function on your calculator if you The amplitude is half the
need help.) total distance between the
minimum and maximum
x –180° –135° –90° –45° 0° 45° 90° 135° 180° values. It is:
f (x) = sin 2x
• always positive
• the result of
g(x) = cos 2x maximum – minimum .
__________________
2
2 Sketch f and g on the same system of axes for x ∊ [180°;180°]. In trigonometric graphs,
a period is the number of
3 State the amplitudes of f and g. degrees needed to complete
4 State the periods of f and g. one cycle of a pattern.
The range is represented by
5 State the ranges of f and g.
y ∊ [minimum;maximum].
6 For which value(s) of x is f (x) – g(x) = 1? An interval spacing is the
7 If f (x) = sin 2x is moved up 1 unit, state the equation of the shifted graph number of degrees between
in the form y = … critical points.
SOLUTIONS
1
x –180° –135° –90° –45° 0° 45° 90° 135° 180°
f (x) = sin 2x 0 1 0 –1 0 1 0 –1 0
g(x) = cos 2x 1 0 –1 0 1 0 –1 0 1
2 y
(–180°;1) (–135°;1) (45°;1) (180°;1)
1
g(x) = cos (2x) f(x) = sin (2x)
x
–180° –135° –90° –45° 45° 90° 135° 180°
–1
(–90°;–1) (–45°;–1) (90°;–1) (135°;–1)
3 The amplitude is 1 for both f and g.
4 Both f and g have a period of 180°.
5 Range for both f and g is given by y ∊ [–1;1].
6 f (x) – g(x) = 1 when x = –135°, –90°, 45° or 90°.
7 y = f (x) + 1 = sin 2x + 1
Unit 2 The effect of the parameter k on some trigonometric functions 123
PLTMATHSLB11LB_06.indd 123 2012/07/07 11:23 AM
EXERCISE 4
1 1.1 Complete the table correct to one decimal place where necessary.
x –360° –270° –180° –90° 0° 90° 180° 270° 360°
1x
f (x) = sin __
2
1x
g(x) = cos __
2
1.2 Sketch f and g on the same system of axes for x ∊ [–360°;360°].
1.3 State the amplitude, the period and the range for each graph.
1.4 State two values of x for which g(x) − f (x) = 1.
1.5 State three values of x for which f (x) − g(x) = 1.
1.6 State two values of x for which f (x) = g(x).
2 1 x for x ∊ [–180°;180°].
Consider the graphs of f (x) = tan2x and g(x) = tan __
2
2.1 Write down the period of f (x) = tan 2x and state the step size.
2.2 1 x and state the step size.
Write down the period of g(x) = tan __
2
2.3 State the equations of the asymptotes of f and g for x∊ [–180°;180°].
2.4 Sketch f and g on the same system of axes for x ∊ [–180°;180°].
2.5 State one value of x for which g(x) – f (x) = 1.
2.6 1 x = 1.
State one value of x for which tan 2x – tan __
2
2.7 State the range of f.
2.8 State the amplitude of g.
EXERCISE 5
Complete the table and then sketch each graph on a new set of axes.
REMEMBER • In the table, k is the coefficient of x in each graph equation.
Period
• Show all x- and y-intercepts and the coordinates of all turning points and
360° endpoints.
y = asin kx has P = ____
k
360°
• a graph has asymptotes, sketch and label them.
If
y = acos kx has P = ____
k
180°
k Period Step Domain End Points
y = atan kx has P = ____
k
1 y = cos 3x x ∊ [–30°;180°]
Steps (interval spacing)
P 2 1x
y = tan __ x ∊ [–360°;360°]
S = __ 4
4
Asymptotes (tan graph only)
3 1x
y = sin __ x ∊ [–360°;360°]
are in the middle of the 3
period.
If the period is 180°, 4 y = cos2x x ∊ [–270°;90°]
x = 90° there will be a vertical
5 y = tan3x x ∊ [–60°;90°]
asymptote.
6 y = sin 2x x ∊ [–90°;210°]
7 1x
y = cos __ x ∊ [–360°;360°]
2
8 1x
y = tan __ x ∊ [–360°;360°]
3
124 Topic 6 Functions: Trigonometric graphs
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Unit 3: Horizontal shifts
Horizontal shifts of y = sin x
x –90° –60° –30° 0° 30° 60° 90° 120° 150° Shift
__ __ __
√ 3 1 1 √ 3 √ 3 1
f (x) = sin x –1 – ___ – __ 0 __ ___ 1 ___ __ None
2 2 2 2 2 2
__ __ __
1
__ √ 3 √ 3 1
__ 1 √ 3
g(x) = sin (x + 90°) 0 ___ 1 ___ 0 – __ – ___ 90° left
2 2 2 2 2 2
__ __ __ __
√3 √ 3 1 1 √3 √3
h(x) = sin (x – 30°) – ___ –1 – ___ – __ 0 __ ___ 1 ___ 30° right
2 2 2 2 2 2
__ __
1 1 √ 3 √ 3 1 1
k(x) = sin (x + 60°) – __ 0 __ ___ 1 ___ __ 0 – __ 60° left
2 2 2 2 2 2
• Adding to x increases its value so the pattern of answers shifts to the left by the
amount added.
• Subtracting from x reduces its value so the pattern shifts to the right by the amount
subtracted.
The graphs in the table are sketched in the figure. They all have the same period and
amplitude, but some have shifted horizontally.
y
(0°;1) (30°;1) f (90°;1) (120°;1) h
0,87 (150°;0,87)
k (150°;0,5)
G x
– 90 ° – 60° – 30° 0° 30° 60° 90° 120° 150°
(– 90°;– 0,5) – 0,5 (150°;– 0,5)
(– 90°;– 0,87) g
(150°;– 0,87)
(– 90°;–1) (– 60°;–1)
It is easy to generate points for plotting shifted graphs point by point. Make sure
that you know the period of the graph and the step, or the interval spacing between
critical points.
For all trigonometric graphs remember to show:
• x- and y-intercepts
• coordinates of turning points and coordinates of endpoints.
The domain determines the x-coordinates of the endpoints. Substitute these
coordinates into the graph equation to determine the y-coordinates. If you are
__
3
sketching y = cos x for x ∊ [−30°;240°], the endpoint on the left will be −30°;___
__
( √
2 )
because y = cos (−30°) =
2
√ 3
___ 1 ,
≈ 0,87, and the endpoint on the right will be 240°;−__ ( 2 )
1
__
because y = cos 240° = − .
2
Unit 3 Horizontal shifts 125
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EXERCISE 6
1 1.1 Draw f (x) = cos x, g(x) = cos (x + 60°) and h(x) = cos (x – 90°) on the same
REMEMBER set of axes for x ∊ [–90°;150°].
The amplitude is half the Use your graphs to answer these questions:
total distance between the 1.2 State one value of x for which h(x) – f (x) = 1.
minimum and maximum 1.3 For which values of x do f, g and h all increase as x increases?
values. It is: 1.4 For which values of x is g(x).h(x) ≥ 0?
• always positive f (x)
• the result of 1.5 For which values of x is ____ ≤ 0?
g(x)
maximum – minimum .
__________________
2 1.6 If t( x ) = –sin x is the graph obtained by shifting g(x), how must g(x) be
A period is the number of shifted?
degrees needed for a wave
2 2.1 Draw f (x) = tan x, g(x) = tan (x + 45°) and h(x) = tan (x – 15°) on the same
pattern.
set of axes for x ∊ [–135°;90°].
The range is represented by
y ∊ [minimum;maximum]. Use your graphs to answer the questions below:
An interval spacing is the 2.2 State one value of x for which g(x) – f (x) = 1.
number of degrees between
2.3 For which values of x ∊ [–135°;90°] do f, g and h all increase as x increases.
critical points.
2.4 For which values of x is g(x).h(x) ≥ 0?
f (x)
2.5 For which values of x is ____ ≤ 0?
g(x)
3 3.1 Sketch f (x) = tan (x – 90°), g(x) = –2cos (x + 45°) and h(x) = –sin (x – 45°) on
the same system of axes for x ∊ [–135°;225°].
3.2 State the amplitude of g.
3.3 State the period of f.
3.4 State the range of h.
3.5 For which values of x is f (x) = 0?
3.6 State one value of x for which f (x) = h(x).
3.7 State the maximum value of g(x) – h(x) and give the corresponding x-value.
3.8 For which values of x is f (x).g(x) ≤ 0?
3.9 Evaluate: h(15°) – g(15°).
3.10 State three values of x for which h(x) – f (x) = 1.
3.11 If h is reflected in the x-axis, state the equation of the reflected graph in
the form y = …
3.12 If g is shifted 45° to the left, state the equation of the shifted graph in the
form y = …
3.13 For which values of x is h(x) a decreasing function?
3.14 For which values of x is f (x) ≥ 0?
3.15 If h is shifted 45° to the right, state the equation of the shifted graph in the
form y = …
4 4.1 Sketch f (x) = sin(x – 30°) – 1
__ and g(x) = 2cos x on the same set of axes for
2
x ∊ [–120°;120°].
4.2 What is the period of f ?
4.3 What is the amplitude of g?
4.4 What is the range of f?
4.5 State one value of x for which f (x) = g(x).
4.6 For which value of x is g(x) – f (x) = 3?
4.7 Determine g(–60°) – f (–60°).
126 Topic 6 Functions: Trigonometric graphs
PLTMATHSLB11LB_06.indd 126 2012/07/07 11:23 AM
4.8 For which values of x do both f (x) and g(x) increase as x increases?
4.9 For which values of x is f (x). g(x) > 0? REMEMBER
4.10 If g is reflected in the x-axis, give the new equation in the form y = …
In an increasing function,
4.11 If f is shifted 1 unit up, give the new equation in the form y = …
the y-values increase as x
4.12 If f is shifted 30° to the left, state the new equation in the form y = … increases.
4.13 1 is obtained by shifting f (x), describe the shift.
If h( x ) = –cos x – __
2 In a decreasing function,
4.14 If t( x ) = –2sin x is obtained by shifting g(x), describe the shift. the y-value decreases as x
increases.
5 Consider the functions f (x) = cos (x – 60°) and g(x) = tan (x – 15°) for
To reflect a graph in the
x ∊ [–210°;150°] x-axis, leave x and change the
5.1 Sketch the graphs of f and g for x ∊ [–210°;150°] on the same system sign of y.
of axes. To reflect a graph in the
5.2 State the amplitude of f. y-axis, leave y and change the
5.3 What is the period of g? sign of x.
5.4 For which values of x is f (x) an increasing function?
5.5 For which value(s) of x is g( x ) undefined?
5.6 For which value(s) of x is f (x).g(x) ≥ 0?
5.7 State one value of x for which f( x ) = g( x ).
5.8 For which values of x ∊ [–120°;150°], is f( x ) ≥ g(x)?
5.9 For which value of x is g(x) – f (x) = 2?
5.10 State two values of x for which f (x) – g(x) = 1.
5.11 If h( x ) = –tan x is obtained by shifting and reflecting g( x ),
5.11.1 fully describe the shift and reflection
5.11.2 state the equations of the asymptotes of h(x) or x ∊ [–210°; 150°].
5.12 Describe the shift if t(x) = –sin x is obtained by shifting f (x):
5.12.1 to the right
5.12.2 to the left.
A summary of trigonometric graphs
y = asin kx y = asin (x + p) y = asin x + q
y = acos kx y = acos (x + p) y = acos x + q
y = atan kx y = atan (x + p) y = atan x + q
All sine and cosine graphs have exactly the same properties as each other:
• Amplitude = a if a > 0, but amplitude = –a if a < 0
360°
• Period = ____
k
The tan graph has unique properties:
180°
• Amplitude is undefined for all tan graphs and period = ____ k
• Asymptotes are always positioned midway through each wave pattern.
Properties common to the trigonometric graphs:
• Step or interval spacing = __P4, where P = period
• If there is a vertical shift, you may need to adjust the interval spacing, so consider
steps of 15°, 30°, 45° or 60°
• Vertical shift will be up if q > 0, but down if q < 0.
• Horizontal shift will be left if you add to x, but right if you subtract from x.
Unit 3 Horizontal shifts 127
PLTMATHSLB11LB_06.indd 127 2012/07/07 11:23 AM
Asymptote with smaller
Graph Amplitude Period Steps Vertical shift Horizontal shift
position (x) value
y = 2sin 3x 2 120° 30° None None None
y = –3cos x + 2 3 360° 90° None 2 up None
y = tan 2x – 1 Undefined 90° 22,5° x = 45° 1 down None
y = sin (x – 30°) 1 360° 90° None None 30° right
y = 2cos (x + 60°) 2 360° 90° None None 60° left
y = 2tan 1
__x Undefined 360° 90° 180° None None
2
y = –2sin (x + 45°) 2 360° 90° None None 45° left
y = cos (x – 90°) – 2 1 360° 90° None 2 down 90° right
y = tan (x – 45°) Undefined 180° 45° x = 135° None 45° right
y=3
__cos (x + 15°) 3
__ 360° 90° None None 15° left
2 2
y = sin (x – 75°) + 1 1 360° 90° None 1 up 75° right
EXERCISE 7
There are eleven graphs in the table above and five of them are sketched here.
y
f
g
2
h
1
x
–105 ° –90° –75° –60° –45° –30° –15° 0 15° 30° 45° 60° 75° 90° 105° 120° 135° 150° 165°
–1
k
–2
–3
t
128 Topic 6 Functions: Trigonometric graphs
PLTMATHSLB11LB_06.indd 128 2012/07/07 11:23 AM
Complete the table.
Endpoints (correct to
Equation two decimal places) Turning points Range
f
g
h
k
t
Unit 3 Horizontal shifts 129
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Unit 4: Determine the equations
of trigonometric graphs
EXERCISE 8
1 y
(300°;1) (360°;1)
REMEMBER 1
f g
h
Period 0,5
360°
y = asin kx has P = ____
k x
360° –360° –180°–150° 30° 180° 360°
y = acos kx has P = ____
k
–0,5
180°
y = atan kx has P = ____
k
–1
(–360°;–1) (120°;–1)
Steps (interval spacing)
P
S = __
4 1.1 If f (x) = asin (x + b), write down the values of a and b.
Asymptotes (tan graph only)
1.2 If f (x) = dcos (x + c), write down the values of c and d.
are in the middle of the
period. 1.3 What is the period of g?
If the period is 180°, 1.4 If g(x) = psin qx, write down the values of p and q.
x = 90° there will be a vertical 1.5 If h(x) = vcos wx, write down the values of v and w.
asymptote. 1.6 1 (x + t), write down the values of r and t.
If h(x) = rsin __
2
1.7 State two value(s) of x for which f (x) = h(x)
1.8 State one value of x for which f (x) − h(x) = 1.
REMEMBER
2 y
The amplitude is half the h
2
total distance between the
minimum and maximum g
values. It is: 1
• always positive f
• the result of
maximum – minimum .
__________________ x
2 –360° 360°
In trigonometric graphs,
a period is the number of –1
degrees needed to complete
one cycle of a pattern.
The range is represented by 2.1 State the period of f.
y ∊ [minimum;maximum]. 2.2 State the period of g.
An interval spacing is the 2.3 State the amplitude of f.
number of degrees between 2.4 State the amplitude of g.
critical points. 2.5 If g(x) = sin ax + b, determine the values of a and b.
2.6 If f (x) = dcos (x + p), determine the values of d and p.
2.7 If f (x) = msin nx, determine the values of m and n.
2.8 If h(x) = vtan wx, write down the values of v and w.
2.9 1 (x + c) + t, determine the values of c and t.
If g(x) = cos __
2
2.10 For which value of x is g(x) – f (x) = 2?
130 Topic 6 Functions: Trigonometric graphs
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3 y
g
3
2
f
h
1
x
–270° –195° 0 75° 135° 255° 270°
–1
–2
–3
3.1 State the period of f.
3.2 State the amplitude of f.
3.3 State the period of g.
3.4 State the amplitude of g.
3.5 What is the period of h?
3.6 If f (x) = asin (x + b), determine the values of a and b.
3.7 If f (x) = dcos (x + c), determine the values of c and d.
3.8 If g(x) = vtan wx, determine the values of v and w.
3.9 If h(x) = pcos qx, determine the values of p and q.
3.10 For which values of x will f (x).g(x) ≤ 0?
3.11 For which value of x is f(x) = g(x)?
3.12 State two values of x for which g(x) − h(x) = 1. REMEMBER
EXERCISE 9 Period
360°
y = asin kx has P = ____
k
1 If f (x) = cos x for x ∊ [–360°;360°], is moved up 1 unit and 20° to the right: 360°
y = acos kx has P = ____
k
1.1 state the equation of the shifted graph in the form y = …
180°
y = atan kx has P = ____
1.2 give the range of the shifted graph. k
2 If g(x) = sin x for x ∊ [–360°;360°], is moved 1 unit down and 90° to the left: Steps (interval spacing)
P
S = __
2.1 state the equation of the shifted graph in the form y = … 4
2.2 state the minimum value of the shifted graph. Asymptotes (tan graph only)
are in the middle of the
3 h(x) = atan bx has a period of 360° and passes through the point (90°;3) period.
determine the values of a and b. If the period is 180°,
x = 90° there will be a vertical
4 If f (x) = tan x is shifted 45° to the right and 2 units down, state the equation of asymptote.
the shifted graph in the form y = …
5 f (x) = –sin ax + b and g(x) = ctan dx. Both f and g have a period of 720° and both
pass through the point (–180°;2). The range of f is y ∊ [0;2].
5.1 Determine the values of a, b, c and d.
5.2 Sketch f and g on the same set of axes for x ∊ [–360°;360°].
Unit 4 Determine the equations of trigonometric graphs 131
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Unit 5: Sketch graphs which have a change
in period and a horizontal shift
In the equations y = asin k(x + p), y = acos k(x + p) and y = atan k (x + p):
• p indicates the horizontal shift
• pk indicates a change in period.
If the equations are simplified as shown below, then kp does not indicate the
horizontal shift y = asin (kx + kp), y = acos (kx + kp) and y = atan (kx + kp)
Factorisation solves the problem:
y = cos (2x – 60°) ⇒ y = cos 2(x – 30°) and the horizontal shift is 30°.
(2 )
1 x – 25° ⇒ y = sin __
y = sin __ 1 (x – 50°) and the horizontal shift is 50°.
2
REMEMBER WORKED EXAMPLE
Horizontal shifts (See calculator instructions on the next page.)
• to shift right, subtract
from x 1 Consider the graphs f (x) = cos 3(x – 15°) and g(x) = sin (2x – 90°)
• to shift left, add to x 1.1 Complete the table.
Graph Amplitude Period Steps Horizontal shift
Period calculations
• both sine and cos graphs f (x) = cos 3(x – 15°)
360°
P = ____
k
180°
• tan graphs P = ____
k g(x) = sin 2(x + 45°)
1.2 Draw the graphs on the same system of axes for x ∊ [– 45°;135°].
SOLUTIONS
(There are detailed calculator instructions on the next page.)
1.1 Graph Amplitude Period Steps Horizontal shift
f (x) = cos 3(x – 15°) 1 360° = 120°
____ 120° = 30°
____ 15° right
3 4
g(x) = sin 2(x + 45°) 2 360° = 180°
____ 180° = 45°
____ 45° left
2 4
1.2
y
(15°;1)
g(x) = sin 2(x + 45°) 1 (135°;1)
f(x) y = cos 3(x – 15°)
0,7
x
–45° –30° –15° 15° 30° 45° 60° 75° 90° 105° 120° 135°
–1
(–45°;–1) (75°;–1)(90°;–1)
132 Topic 6 Functions: Trigonometric graphs
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General calculator instructions for most scientific calculators used in schools
• Select ‘MODE’ and then press the number which appears next to the word ‘TABLE’. REMEMBER
The display shows ‘F(X) = ‘
Period
– Remember that F(X) = y 360°
y = asin kx has P = ____
– Your equation must be in the form f (x) = … or g(x) = … or h(x) = … or y = … k
360°
y = acos kx has P = ____
• the variable is entered using ‘ALPHA’, followed by ‘X’ k
– The variable may be θ, α, β, x, or any other letter, except y 180°
y = atan kx has P = ____
k
• Enter the equation by selecting the appropriate keys.
• Remember that sin, cos and tan automatically appear with an open bracket; close Steps (interval spacing)
the brackets at the end by pressing the key ‘)’. P
S = __
4
• Once you have entered your equation, press ‘=’, and once only. If this does not
work for your calculator, press ‘=‘ a second time, ignoring ‘g(X) =‘. Then follow the
prompts to enter the start and end values, followed by the step size.
• To edit your equation or change the start, end or step values press ‘AC’, then ‘=’.
When you have finished with the table, return to normal mode by selecting ‘MODE’,
followed by the number which appears next to ‘COMP’.
Generate a table of values for f (x) = cos [3(x – 15°)] for x ∊ [– 45°;135°] 360° = 120°
P = ____
3
• Without adjusting the step size, the table below will be generated. 120 = 30°
S = ____
4
Calculator display Points to plot
X F (X) (x;y)
1 –45 –1 (–45°;–1)
2 –15 0 (–15°;0)
3 15 1 (15°;1)
4 45 0 (45°;0)
5 75 –1 (75°;–1)
6 105 0 (105°;0)
7 135 1 (135°;1)
• It is important to note we have not found the y-intercept.
• Press ‘AC’, followed by ‘=’ and then repeat ‘=’ until you get back to Step.
• Change the step size to 15 by pressing ‘15’ followed by ‘=’
• An expanded version of the table will come up which now includes the row:
4 0 0,7071 (0°;0,71)
Unit 5 Sketch graphs which have a change in period and a horizontal shift 133
PLTMATHSLB11LB_06.indd 133 2012/07/07 11:23 AM
360° = 180°
Now generate a table for g(x) = sin (2x – 90°) for x ∊ [– 45°;135°] P = ____
2
180 = 45°
S = ____
Calculator display Points to plot 4
X F (X) (x;y)
1 –45 0 (–45°;0)
2 0 –1 (0°;–1)
3 45 0 (45°;0)
4 90 1 (90°;1)
5 135 0 (135°;0)
This table contains all the critical points which you need.
EXERCISE 10
Complete the table and sketch each pair of graphs on the same set of axes.
Graph Amplitude Period Steps Asymptotes Shift Domain and endpoints
1.1 f (x) = cos 2(x + 30°) x ∊ [210°;105°]
(–210°; )(105°; )
g(x) = sin 2(x – 60°) (–210°; )(105°; )
2.1 1 (x + 45°)
f (x) = tan __ x ∊ [–45°;270°]
2
(–45°; )(270°; )
g(x) = sin 3(x + 15°) (–45°; )(270°; )
3.1 f (x) = tan (2x + 60°) x ∊ [–120°;60°]
(–120°; )(60°; )
g(x) = 2cos (x + 30°) (–120°; )(60°; )
4.1 1 sin (x + 30°)
f (x) = __ x ∊ [–210°;150°]
2
(–210°; )(150°; )
g(x) = cos (2x – 120°) (–210°; )(150°; )
1.2 Refer to your solution to 1.1 and answer the questions that follow.
1.2.1 State four value(s) of x for which f (x) – g(x) = 1.
1.2.2 For which values of x is f (x).g(x) < 0?
2.2 Refer to your solution to 2.1 and answer the questions:
2.2.1 State the value(s) of x for which f (x) – g(x) = 1.
2.2.2 For which values of x is f (x).g(x) > 0?
2.2.3 For which value(s) of x is f (x) undefined?
3.2 Refer to your solution to 3.1 and answer the questions:
3.2.1 State the value(s) of x for which f (x) = g(x)
3.2.2 State the value(s) of x for which g(x) – f (x) = 2?
134 Topic 6 Functions: Trigonometric graphs
PLTMATHSLB11LB_06.indd 134 2012/07/07 11:23 AM
4.2 Refer to your solution to 4.1 and answer the questions:
4.2.1 State the value(s) of x for which g(x) – f (x) = 1,5.
4.2.2 State the value(s) of x for which g(x) – f (x) = 0,5.
4.2.3 State the value(s) of x for which f (x) – g(x) = 1.
4.2.4 Determine f (0°) – g(0°).
4.2.5 For which value(s) of x is f (x) > 0?
EXERCISE 11
1 1 cos (x + 30°) and g(x) = sin (2x – 120°).
Consider the graphs defined by f (x) = –__
2
1.1 Sketch f (x) and g( x ) on the same system of axes for x ∊ [–210°;150°].
1.2 Use your graphs to answer the questions:
1.2.1 State two values of x for which f (x) = g(x).
1.2.2 State two values of x for which f (x) – g(x) = 0,5.
1.2.3 For which values of x is f (x) ≤ 0?
( )
1 x + 45° and g(x) = 2sin (x – 30°). REMEMBER
2 Consider the graphs defined by f (x) = tan __
2
2.1 Sketch f (x) and g(x) on the same system of axes for x ∊ [–180°;180°]. Period
360°
y = asin kx has P = ____
2.2 Use your graphs to answer the questions below: k
2.2.1 State two values of x for which g(x) – f (x) = 2. 360°
y = acos kx has P = ____
k
2.2.2 For which values of x ∊ [–180°;0°] is f (x).g(x) < 0? 180°
y = atan kx has P = ____
3 Consider the functions f (x) = asin (x + p), g(x) = tan k(x + d), h(x) = c and j(x) = e k
x ∊ [–240°;270°].
Steps (interval spacing)
y P
S = __
4
j h Asymptotes (tan graph only)
g are in the middle of the
T (210°;2) period.
f If the period is 180°,
Z
x = 90° there will be a vertical
asymptote.
x
R V
U W
(30°;–2)
3.1 Write down the coordinates of R, T, U, V, W and Z.
3.2 Write down the values of a, p, k, d, c and e.
3.3 State the length of TU.
3.4 If f (x) is shifted 30° to the left, state the equation of the shifted graph.
3.5 How must f (x) be shifted if the equation of the shifted graph is y = –asin x?
3.6 If g(x) is shifted so that the new graph equation is y = tan kx, describe
the shift.
3.7 For which value(s) of x is g(x) – f (x) = 3?
Unit 5 Sketch graphs which have a change in period and a horizontal shift 135
PLTMATHSLB11LB_06.indd 135 2012/07/07 11:23 AM
Revision
1 Write down the period, amplitude and range and the coordinates of the
y-intercept of each graph. REMEMBER
1.1 f (x) = 2sin (x + 30°)
Period
1.2 g(x) = −cos 2x 360°
y = asin kx has P = ____
1.3 h(x) = tan 2x k
360°
y = acos kx has P = ____
1.4 k(x) = sin (2x – 20°) – 1 k
1x 180°
y = atan kx has P = ____
1.5 p(x) = 3tan __ k
2
1.6 1 cos (x – 20°)
q(x) = __ [24]
2 Steps (interval spacing)
P
S = __
2 2.1 If h(x) = cos x is translated 90° to the right, state the equation of the 4
translated graph. Simplify your answer. (3) Asymptotes (tan graph only)
are in the middle of the
2.2 If h(x) = cos x is translated 90° to the left, will you get the same result
period.
as 2.1? Explain your answer. (3)
If the period is 180°,
[6} x = 90° there will be a vertical
3 y asymptote.
E
G
x
−180° F D 180°
H
A B
C
The graphs f (x) = 2cos (x + 30°) and g(x) = tan __ 1 (x + 30°) – 1, x ∊ [–180°;150°]
2
are sketched.
3.1 Write down the coordinates of A, B, C, D, E, F, G and H. (16)
3.2 State the equation of the asymptote of g. (1)
3.3 For which values of x is
3.3.1 f (x) ≥ g(x) ? (2)
3.3.2 f (x).g(x) ≥ 0? (2)
3.4 State the period and describe the shift of g. (3)
3.5 If f (x) is translated 60° to the left, state the equation of the
translated graph. Simplify your answer. (3)
3.6 If g(x) is reflected in the y-axis, state the equation of the
reflected graph. (2)
3.7 For which values of x is f (x) ≤ g(x) ? (3)
3.8 For which values of x is f (x).g(x) ≥ 0? (3)
3.9 If f (x) is reflected in the x-axis, state the equation of the
reflected graph. (2)
3.10 If g(x) is reflected in the x-axis, state the equation of the
reflected graph and simplify your answer. (3)
[40]
136
PLTMATHSLB11LB_06.indd 136 2012/07/07 11:23 AM
4 Consider the functions defined by:
(2 )
3 x – 90° for x ∊ [–180°;180°]
f (x) = –2cos (x + 30°) and g(x) = sin __
4.1 State the period, amplitude, range and endpoints of f. (6)
4.2 State the period, amplitude, range and endpoints of g. (6)
4.3 Sketch f and g on the same system of axes. Indicate the intercepts
with the axes and the coordinates of the turning points and endpoints. (6)
4.4 Determine g(–120°) – f (–120°). (1)
4.5 State one value of x for which f (x) = g(x). (1)
__
4.6 For which value(s) of x is f (x) – g(x) = √ 3 ? (2)
4.7 For which values of x is g(x) ≤ 0? (2)
4.8 Write g(x) as a cos graph in the form y = ... (3)
[27]
5 y
(–315°;3) (45°;3)
C f
(–360°;1) (360°;1)
(270°;0)
x
–270° A (285°;0)
B
g (225°;–1)
–2 D
Consider the graphs in the figure and use them to answer the questions.
5.1 State the amplitudes and periods of both f and g. (4)
5.2 If f (x) = asin (x + b) + c and g(x) = dcos ex, write down the values of
a, b, c, d and e. (5)
5.3 Write down the coordinates of A and B. (4)
5.4 Determine the length of CD. (3)
5.5 If f (x) = pcos (x + q) + v, write down the values of p, q and v. (6)
5.6 If g(x) = wsin k(x + z), write down the values of w, k and z. (6)
5.7 For which values of x is f (x) an increasing function? (6)
5.8 For which values of x is f (x).g(x) ≤ 0? (6)
5.9 State the range of −f (x). (2)
5.10 State the range of g(–x). (1)
5.11 For which values of x is f (x) < 0? (4)
[47]
137
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TOPIC
2 7 Trigonometry
Unit 1: Revision of Grade 10 Trigonometry
REMEMBER
We work in the Cartesian plane for angles bigger than 90°.
The positive angle θ always rotates anti-clockwise from the positive x-axis about the
hypotenuse
opposite origin.
θ If angle θ rotates clockwise from the positive x-axis about the origin, it is negative.
adjacent ______
OP = r = √x2 + y2 (radius of the circle, r > 0)
We define the three We now define the trigonometric functions as:
trigonometric ratios for the
y y 90°
angle θ in a right-angled sin θ = _r ; cos θ = __xr ; tan θ = __x y
triangle as: Quadrant 2 Quadrant 1
o , cos θ = a
If 90° < θ < 180°, point P is in
sin θ = __ __, and P(x;y)
h h quadrant 2 and x < 0, y > 0.
o
tan θ = __ r
a y –x = −,
So, sin θ = _r = + , cos θ = __
r y
y θ
and tan θ = __ 180° x O x 0° or 360°
–x = –.
If 180° < θ < 270°, point P is in
quadrant 3 and x < 0, y < 0.
–y –x
So, sin θ = __
r–y = –, cos θ = r = –
__ Quadrant 3 Quadrant 4
and tan θ = __–x = +
270°
If 270° < θ < 360° point P is in quadrant 4 and x > 0, y < 0.
–y x –y
So, sin θ = __
r = –, cos θ = r = + and tan θ = x = –
__ __
You can see the ratios change signs
90°
as θ moves from 0° to 360°: y
sin θ is positive in Quadrants 1 and Quadrant 2 Quadrant 1
2.
cos θ is positive in Quadrants 1 and
4.
θ x
tan θ is positive in Quadrants 1 and 180° x 0° or 360°
3. O
y
r
P(x;y)
Quadrant 3 Quadrant 4
270°
Use these diagrams to help you
remember the signs of the trigonometric ratios:
y y y
2 1 2 1 2 1
CAST shows
r r
y y Sin All S A you in which
quadrant the
–x x
x x x ratios are
–y –y Tan Cos T C POSITIVE
r r
3 4 3 4 3 4
138 Topic 7 Trigonometry
PLTMATHSLB11LB_07.indd 138 2012/07/14 2:50 PM
EXERCISE 1
1 In each case say which quadrant the given angle is in.
1.1 223° 1.2 153°
1.3 292° 1.4 –130°
1.5 – 400° 1.6 –51°
1.7 –182° 1.8 225°
1.9 520° 1.10 730°
2 Work out the sign of each ratio. (First use the CAST diagram, then check
your answer with a calculator.)
2.1 cos 115° 2.2 tan 150°
2.3 cos (– 40°) 2.4 sin (–30°)
2.5 tan (–170°) 2.6 sin 200°
2.7 cos 330° 2.8 tan 190°
2.9 sin 300° 2.10 cos (–60°)
3 In each case say which quadrant θ lies in if:
3.1 sin θ > 0 and cos θ < 0
3.2 sin θ < 0 and cos θ > 0
3.3 sin θ > 0 and tan θ < 0
3.4 cos θ > 0 and tan θ > 0
3.5 cos θ > 0 and tan θ < 0
3.6 sin θ < 0 and tan θ > 0
y
WORKED EXAMPLE
Find the length OP and then write down
the values of sin θ, cos θ, tan θ.
θ
x
O
P(–5;–12)
SOLUTION
___________
OP = √(–5)2 + (–12)2 = 13 | Pythagoras’ Theorem y
–5 , tan θ = ____
–12 , cos θ = ___
sin θ = ____ –12 = ___
12
13 13 –5 5
–5 θ
x
O
–12
13
P
Unit 1 Revision of Grade 10 Trigonometry 139
PLT MATHS LB 11 7th pgs (Real Book).indb 139 2012/07/02 2:21 PM
EXERCISE 2
1 In each figure, find the length OP (in surd form if necessary). Then write down
the values of sin θ, cos θ, tan θ.
1.1 y 1.2 y
P(7;24)
P(–3;4)
θ θ
x x
O
O
1.3 y 1.4 y
θ
x θ
O x
O
P(–5;–10) P(5;–3)
2 In each figure, find the length of r, x or y (in surd form if necessary). Then write
down the three trigonometric ratios for each given angle (for example: sin 600°,
tan 135°, cos 210° or sin 300°). Give your answers with rational denominators.
2.1 y 2.2 y
__
P(1;√ 3 )
P(–2;2)
135°
60°
x x
O O
2.3 y 2.4 y
210° 300°
x x
O O
–1 __
2 –√ 3
2
140 Topic 7 Trigonometry
PLT MATHS LB 11 7th pgs (Real Book).indb 140 2012/07/02 2:21 PM
WORKED EXAMPLE
1 8 and sin θ < 0, find the value of:
If cos θ = ___
17
1.1 tan θ
1.2 sin θ
_____
cos θ
1.3 sin2 θ + cos2 θ
2 If 7sin θ + 3 = 0 and 90° < θ < 270°, find the values of:
2.1 cos θ
2.2 1 + tan2 θ
SOLUTIONS
You are required to draw a diagram. Make sure that it is in the correct quadrant,
and that the lengths you mark have a – sign where appropriate.
1 8 and sin θ < 0
cos θ = ___ y
17
θ lies in Quadrant 4
_______
y = –√172 – 82 = –15 θ 8
x
–15 O
1.1 tan θ = ____
8
1.2 sin θ = ____
_____ –15 ÷ ___
8 = ____
–15 –15
cos θ 17 17 8 17
1.3 ( ) ( )
–15 2 + ___
sin2 θ + cos2 θ = ____
17
8 2
17
= 225 + 64 = 1
________ P
289
y
2 θ lies in Quadrant 3
–3
sin θ = ___
7 ___
________ _____ ___ –√ 40 θ
x = –√72 – (–3)2 = –√49 – 9 = –√40 O
x
___
– 40√
2.1 cos θ = _____
7 –3
2.2 1 + tan2 θ = 1 + –3
_____
___
( –√40 )2 9 = ___
= 1 + ___
40
49
40
7
EXERCISE 3 REMEMBER
1 If 5 sin θ + 4 = 0 and cos θ > 0 , find the value of tan θ.cos θ. Steps to solve ‘from one ratio
10 – _____
12 and 90° < x < 270°, find the value of _____ 5 . to another’ type questions:
2 If tan θ = ___
5 cos θ cos θ 1 Isolate the trigonometric
3 If 7tan θ = 3 and cos θ > 0, find the value of sin θ + cos θ .
___________ ratio.
2sin θ
2 Decide in which quadrant
4 If 5cos x + 3 = 0 and 180° < x < 360°, find the value of 3tan x + 25 sin2 x. the angle lies.
7 and 180° < x < 360°, find the value of 25 cos x – _____
7 . 3 Draw the triangle in the
5 If sin x = ___ tan x
25 correct quadrant.
6 3 , tan y = __
If cos x = __ 5 , x < 90° and y < 90°, prove that 74sin2 y + ______
7 = 34. 4 Use Pythagoras’ Theorem
4 7 2 tan x
to find the third side.
7 If sin θ = k and 90° < θ < 180°, determine the value of tan θ in terms of k. 5 Answer the question.
Unit 1 Revision of Grade 10 Trigonometry 141
PLT MATHS LB 11 7th pgs (Real Book).indb 141 2012/07/02 2:21 PM
Unit 2: Identities
KEY WORD
sin θ and sin2 θ + cos2 θ = 1 are two standard trigonometric identities that you
tan θ = _____
cos θ
trigonometric identity – an
equality which is true for need to prove and learn to use.
all values of an unknown
An identity consists of two sides: Left Hand Side (LHS) and Right Hand Side (RHS).
variable, for which both sides
To prove an identity you must prove LHS = RHS.
of the identity are defined (so
no zero denominators) To prove the standard identities, use the basic trigonometric definitions in the
It differs from a trigonometric Cartesian plane. You can prove other identities using the two standard identities
equation which is true only
to simplify each side.
for certain values of the
unknown variable.
sin x
_____
Proof of tan x = cos Proof of sin2 θ + cos2 θ = 1
x
y
P(x;y)
r
y
θ x
O x
y
LHS = tan θ = __ LHS = sin2 θ + cos2 θ
x
θ
RHS = sin = ( _r ) + ( __xr )2
_____ y 2
cos θ
y
_ y2 2
r
= __ = __2 + x__2
x
__ r r
r
y y2 + x2
= __ × __r = ______ | x2 + y2 = r2 Pythagoras
r x 2r
y 2
= __ = r__2 = 1 = RHS
x r
LHS = RHS
In the Worked examples and exercises that follow, you do not use the trigonometric
definitions to prove identities or simplify expressions.
To prove identities follow these steps:
sin θ or _____
1 = cos θ so that all the ratios are in terms of sin and cos.
• Make tan θ = cos
_____
θ
_____
tan x sin θ
• Consider the LHS or RHS as an algebraic expression and use algebraic
manipulations to simpifly further.
• If there are fractions, find the LCD and add.
• If there are fractions over fractions, simplify as you would in algebra.
• Factorise where possible.
• Use the square identity where possible in any of the forms:
sin2 θ + cos2 θ = 1 or sin2 θ = 1 – cos2 θ or cos2 θ = 1 – sin2 θ
• Simplify both sides of the identity as far as possible.
142 Topic 7 Trigonometry
PLT MATHS LB 11 7th pgs (Real Book).indb 142 2012/07/02 2:21 PM
WORKED EXAMPLE
Simplify the expression using the identities proved before:
(1 + tan2 x)(1 – sin2 x)
SOLUTION
( 1 + _____
cos x )
2
sin x (1 – sin x) | Identity: tan x = _____
2
2 sin x
cos x
= ( cos x )
2 2
+ sin x (cos x) | Identity: 1 – sin x = cos x
____________ 2 2 2
2
cos x
1 × cos2 x
= _____
2
| Identity: sin2 x + cos2 x = 1
cos x
=1
EXERCISE 4
Simplify the expressions as far as possible.
1 sin x
_________ 2 tan2 x(1 – sin2 x)
cos x.tan x
2
(1 – cos x)
3 __________ 4 8sin2 θ + 8cos2 θ
sin x
5 1 – tan x.cos x.sin x 6 (3 – 3sin θ)(3 + 3sin θ)
7 1 – cos
_____
sin x
x
_____
tan x
8 ( _____ )
1 + tan x ( sin x.cos x )
tan x
9 sin x.cos x
_______________ 10 1
tan2 x – _____
1 + cos2x – sin2 x 2
cos x
WORKED EXAMPLE 1
Prove the identity.
sin x4x – cos4 x = 1 – 2cos2 x
SOLUTION
LHS = (sin2 x – cos2 x)(sin2 x + cos2 x)
= (sin2 x – cos2 x)(1)
= (1 – cos2 x) – cos2 x
= 1 – 2cos2 x
= RHS
Unit 2 Identities 143
PLT MATHS LB 11 7th pgs (Real Book).indb 143 2012/07/02 2:21 PM
WORKED EXAMPLE 2
Prove the identity.
tan x + sin x = sin x
___________
1
1 + _____
cos x
SOLUTION
sin x + sin x
_____
__________ × cos
LHS = cos x x
_____
1 cos x
1 + _____
cos x
= sin x + sin x.cos x
_______________
cos x + 1
sin x(1 + cos x)
_____________
=
(1 + cos x)
= sin x
= RHS
EXERCISE 5
Prove the identities.
1 sin x.cos x.tan x = 1 – cos2 x
2 cos3 x + cos x.sin2 x = cos x
3 (sin x + cos x)2 + (sin x – cos x)2 = 2
4 (1 – sin2 x)(1 + tan2 x) = 1
5 1 – sin x = cos
_____ x
_____
sin x tan x
6 cos x – cos3 x
________________ = tan2 x
cos x – cos x.sin2 x
________
7 sin θ – √ 1 – sin2 θ = 0
_____
tan θ
8 1
tan2 θ + 1 = _____
cos θ
2
9 sin x + cos x = cos x
___________
1 + tan x
10 1 = _________
tan θ + _____ 1
tan θ sin θ.cos θ
11 tan2 x – sin2 x = sin2 x.tan2 x
12 cos x + tan x = _____
________ 1
1 + sin x cos x
13 tan θ – sin θ.cos θ = tan θ
_______________
sin2 θ
2
14 tan2 x.cos2 x + sin
______
2
x =1
tan x
15 1
________ 1
+ ________ 2
= _____
1 – sin x 1 + sin x cos2 x
144 Topic 7 Trigonometry
PLT MATHS LB 11 7th pgs (Real Book).indb 144 2012/07/02 2:21 PM
More complex identities
WORKED EXAMPLE
Prove the identity: 1 + cos x = sin
________ x
________
sin x 1 – cos x
You are not able to simplify the LHS or RHS. Therefore multiply one side by
1+ cos x (=1) or 1 –
________ cos x (=1) to create a workable expression, without changing
________
1 + cos x 1 – cos x
its value.
SOLUTION
LHS = 1 + cos x × 1 –
________ cos x | Notice you are multiplying by 1.
________
sin x 1 – cos x
1 – cos2 x
= _____________ | Notice the difference of 2 squares.
sin x(1 – cos x)
2
= sin x
_____________ | Notice the identity 1 – cos2 x = sin2 x.
sin x(1 – cos x)
= sin x = RHS
________
1 – cos x
EXERCISE 6
Prove the identities.
1 cos x =1+
________ sin x
________
1 – sin x cos x
2 cos2 x – cos x – sin2 x = _____
__________________ 1 – _____
1
2sin x.cos x + sin x tan x sin x
3 sin x – ___________ + cos x = 2sin
1 x.cos x
___________
sin x + cos x sin x + cos x
4 sin2 x + sin x – cos2 x
__________________ = sin x +1
________
2sin x.cos x – cos x cos x
5 1 – cos x – ________
________ sin x = 0
sin x 1 + cos x
6 _____ = sin x
________ – 2
1 +1
________
sin x + 1 cos2 x sin x – 1
7 1 = tan
tan θ + _____ _____θ
tan θ 2
sin θ
_______
8 sin x
_________
(1 + cos x)
= √ 1 – cos x
________
1 + cos x
9 sin2 θ + tan2θ co ( sin θ )
s θ + cos2 θ = 2
_____
2
2
10 (2sin x – cos x)(2sin x + cos x) = 5sin2 x – 1
Unit 2 Identities 145
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Unit 3: Reduction formulae
In Unit 1 you learnt that as the positive angle θ rotates anti-clockwise from the
positive x-axis about the origin, the signs of x and y change and therefore the
trigonometric ratios change sign. This is also true for negative angles rotating
clockwise from the positive x-axis about the origin.
If θ is an acute angle that lies between the rotating arm and the positive or negative
x-axis, we name the angles in the different quadrants as follows:
Quadrant 1 Quadrant 2 Quadrant 3 Quadrant 4
θ 180° – θ 180° + θ 360° – θ
y y y
180° – θ 180° + θ 360° – θ
KEY WORD r r r r
y y y y
θ θ –x θ x θ x
reduction formulae – x x θ x θ x
–x
–y –y
trigonometric identities that r r
express the trigonometric
ratios of an angle of any size
in terms of the trigonometric
ratios of an acute angle
Example: These are the reduction formulae:
sin (180° + θ) = –sin θ and y –y –y
sin (180° – θ) = _r = +sin θ sin (180° + θ) = __
r = –sin θ sin (360° – θ) = __
r = –sin θ
cos (360° – θ) = cos θ
–x = –cos θ
cos (180° – θ) = __ –x = –cos θ
cos (180° + θ) = __ cos (360° – θ) = x__r = +cos θ
r r
y –y –y
tan (180° – θ) = __
–x = –tan θ tan (180° + θ) = __
–x = +tan θ tan (360° – θ) = __
x = –tan θ
y
2 1 Notice that the ratios stay the same but the signs change according to the CAST
diagram (see Unit 1). We always write ratios as a function of θ (an acute angle) only.
Sin All
x WORKED EXAMPLE
Tan Cos
Simplify as far as possible.
3 4
3cos (180° + x)sin (360° – x)
________________________ sin2(180° + θ) + cos2(180° – θ)
_________________________
1 2
sin (360° + x)cos (180° – x) tan (180° – θ).cos (360° – θ)
SOLUTIONS
3cos (180° + x)sin (360° – x)
________________________ (–sin θ) + (–cos θ)
2
________________
2
1 2
sin (360° + x)cos (180° – x) (–tan θ).(cos θ)
= si–nsinθθ+ cos θ
3(–cos x)(–sin x) 2 2
= ______________ ____________
sin x (–cos x) ______.(cos θ)
cos θ
= –3
=– 1
_____
sin θ
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EXERCISE 7
REMEMBER
Simplify the expressions as far as possible.
Follow these steps to simplify
cos (180° + x)tan (180° – x)sin (180° + x)
__________________________________ algebraic expressions with
1
sin (180° – x)sin x reduction formulae:
1 Decide in which quadrant
2sin (180° – x)cos (360° – x)
________________________ the angle lies.
2
sin (180° + x)cos (180° – x) 3 Use the CAST diagram to
determine the sign of the
sin (360° – x).tan (180° – x).cos (360° – x)
___________________________________
ratio.
3 4 Remember that a squared
sin (180° – x).sin (180° + x)
ratio is always positive.
cos2(180° − x).ta n2(360° + x) 5 Rewrite the ratio as a ratio
4 ________________________
2
sin (180° − x) of the acute angle (θ).
6 Simplify the remaining
expression.
sin2(180° + x)
5 1 – _______________ 7 Use identities where
1 – cos (180° + x)
appropriate.
Negative angles
θ – 360° –180° – θ θ – 180° –θ
y y
x x
–θ
θ
θ – 180°
θ – 360°
θ θ x
x y y
–180° – θ
sin (θ – 360°) = sin θ sin (–180° – θ) = sin θ sin (θ – 180°) = –sin θ sin (–θ) = –sin θ
cos (θ – 360°) = cos θ cos (–180° – θ) = –cos θ cos (θ – 180°) = –cos θ cos (–θ) = +cos θ
tan (θ – 360°) = tan θ tan (–180° – θ) = –tan θ tan (θ – 180°) = tan θ tan (–θ) = –tan θ
EXERCISE 8
Simplify the expressions as far as possible.
2sin (180° – x).cos ( – x)
_________________________
1
sin ( – 180° – x).cos (x – 180°)
sin ( 180° + x ) . cos ( – x )
_____________________
2 + sin (x – 360°).sin (x – 180°)
tan ( – x)
2( )
sin 180° + x
______________________
3 1– ( ) ( )
cos x – 180° .cos x – 360°
1 + sin ( – x)
____________
4
cos2(x – 180°)
tan (x – 180°) + sin (180° + x).cos ( – x)
_________________________________
5
sin2(360° – x)
Unit 3 Reduction formulae 147
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The co-ratios are 90° – θ and 90° + θ
REMEMBER
Co-ratios are also called complementary ratios.
Complementary angles add
up to 90°. (40° + 50° = 90° y
and 20° + 70° = 90° are two
pairs of complementary
90° + θ
angles.) θ θ
a θ x
b
90° – θ
REMEMBER
c
Any function of (180° ± θ) or
(360° ± θ) is numerically equal Look at the figure and you will see that:
to the SAME function of θ.
b
sin (90° – θ) = __ c
a = cos θ and cos (90° – θ) = a = sin θ
__
Any function of (90° ± θ) is
numerically equal to the
The angle (90° + θ) lies in Quadrant 2.
co-function of θ.
sin (90° + θ) = cos θ and cos (90° + θ) = –sin θ | because cos θ is negative in Quadrant 2
WORKED EXAMPLES
Examples: Simplify the expressions as far as possible.
2cos (90° – x) – sin (360° – x)
_________________________
1
2sin (90° – x) – cos (180° + x)
cos2 (90° – x) + sin2 (90° + x)
________________________
2
tan (180° + x).sin (90° + x)
SOLUTIONS
2cos (90° – x) – sin (360° – x)
_________________________
1
2sin (90° – x) – cos (180° + x)
2sin x – (–sin x)
= ______________
2cos x – (–cos x)
= 3sin x
______
3cos x
= tan x
2 2
cos (90° – x) + sin (90° + x)
2 ________________________
tan (180° + x).sin (90° + x)
2 2
= si–tan
n x+ cos x
____________
x.cos x
1
= ____________ –1
= _____
–sin x
_____ × cos x sin x
cos x
148 Topic 7 Trigonometry
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EXERCISE 9
Simplify the expressions as far as possible.
tan (180° – x).sin (360° – x).cos (90° – x)
__________________________________
1
sin (180° + x).cos (90° + x).tan (180° + x)
sin (180° + θ).cos (360° – θ).tan (–θ)
______________________________
2
2cos (90° + θ).sin (90° + θ)
2sin (90° – x) + cos (180° – x)
_________________________
3
sin (90° – x) – cos (x – 180°)
4 cos (90° – β) + cos (β + 90°) + cos (180° – β) – cos (β + 180°) + cos (–β)
5 [ cos( 90º + x ) cos (–x) tan (360° – x) ] – [cos(360° – x) sin(x – 90º)]
6 [sin (180º – x) + sin (90º – x)][cos(90° – x) + cos(–x)]
tan (180° + x).sin (90° – x) cos (180°
______________________ – x)
7 – ____________
cos (90° – x) sin (90° + x)
8 cos (90° – θ).tan (180° + θ).cos (360° – θ) – cos (–θ).sin (θ – 90°)
Numerical reductions
You have applied the reduction formulae to algebraic expressions where the angle is
a variable, for example x or θ. Now in numerical examples you apply the reduction
formulae to reduce a trigonometric ratio with an angle greater than 90° to a ratio with
an acute angle.
The acute angle, called the central angle, is the angle (θ) that lies between the rotating
arm and the x-axis.
WORKED EXAMPLES REMEMBER
Write each function as a ratio of acute angles: Follow these steps when
1 sin 215° 2 tan 120° 3 cos 320° reducing trigonometric ratios
with numerical angles.
y y y
1 Decide on the quadrant
and hence the +/– sign
(use the CAST diagram).
2 Keep the function
215° central
x 320° (trigonometric ratio)
35° O angle 60° 120°
central x the same.
angle x O 40° central
O 3 Replace the angle with the
angle ‘central’ angle (acute angle
between arm and x-axis).
SOLUTIONS
1 sin 215° = –sin 35°
Explanation: sin 215° = sin (180° + 35°) = –sin 35° sin (180° + θ) = –sin θ
2 tan 120° = –tan 60°
Explanation: tan 120° = tan (180° – 60°) = –tan 60° tan (180° – θ) = –tan θ
3 cos 320° = cos 40°
Explanation: cos 320° = cos (360° – 40°) = cos 40° cos (360° – θ) = cos θ
Unit 3 Reduction formulae 149
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EXERCISE 10
REMEMBER
1 Rewrite each ratio as a ratio of an acute angle:
To find the central or acute
angle θ quickly:
1.1 sin 122º 1.2 cos 156º
Quadrant 2: 1.3 tan 133º 1.4 sin 166°
125° = 180° – θ 1.5 tan 145º 1.6 cos 99º
θ = 180° – 125° = 55° 1.7 sin 245° 1.8 cos 225°
Quadrant 3: 1.9 tan 191° 1.10 tan 209°
235° = 180° + θ 1.11 cos 230º 1.12 sin 216°
θ = 235° – 180° = 55°
1.13 cos 350° 1.14 cos 300°
Quadrant 4:
1.15 tan 288º 1.16 sin 302º
305° = 360° – θ
θ = 360° – 305° = 55°
1.17 sin 295° 1.18 tan 322°
Numerical reductions with negative angles
You know that sin (–θ) = –sin θ , cos (–θ) = cos θ and tan (–θ) = –tan θ.
For this to be true, θ does not have to be acute.
WORKED EXAMPLES
Write each ratio as a ratio of positive acute angles
1 sin (–200°) 2 cos (–120°)
SOLUTIONS
1 y 2 y
central x
angle O
central 60°
20° x –120°
angle
O
–200°
sin (–200°) = sin 20° cos( –120°) = – cos60°
Explanation: Explanation:
sin (–200°) = –sin 200° cos (–120°) = cos 120°
= –sin (180° + 20°) = cos (180° – 60°) = –(cos 60°)
= –(–sin 20°) = sin 20° or –120° lies in Quadrant 3
or –200° lies in Quadrant 2 where where cos θ < 0 and the central
sin θ > 0 and the central angle is 20° angle is 60°
sin (–200°) = sin 20° cos (–120°) = –cos 60°
150 Topic 7 Trigonometry
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EXERCISE 11
1 Rewrite each ratio as a ratio of a positive acute angle.
1.1 sin (–112°) 1.2 cos (–125º)
1.3 tan (–93º) 1.4 sin (–106°)
1.5 tan (–135º) 1.6 sin (–204°)
1.7 cos (–265°) 1.8 tan (–189°)
1.9 tan (–237°) 1.10 cos (–251º)
1.11 cos (–340°) 1.12 cos (–292°)
1.13 tan (–286º) 1.14 sin (– 327º)
1.15 sin (–355°)
Numerical reductions with algebraic ratios
WORKED EXAMPLES REMEMBER
If cos 20° = p, find these ratios in terms of p: 70° 1 Draw a diagram. The given
angle is acute, so you
1 cos 160° do not use the Cartesian
1 _____
2 sin (–70°) √1 – p2 plane.
3 cos 200° 2 Find the third side using
4 sin 340° Pythagoras’ Theorem.
20° 3 Reduce the ratios to ratios
SOLUTIONS p with acute angles (check
your signs).
1 cos 160° = –cos 20° = − p | 160° = 180° – 20° 4 Use the diagram to help
p you write down the
2 sin (–70°) = –sin70° = − __ required ratio.
1
3 cos 200° = –cos 20° = −p | 200° = 180° + 20°
_____
√1 – p2 | 340° = 360° – 20°
4 sin 340° = –sin 20° = – _______
1
EXERCISE 12
1 If cos 52° = k, find each ratio in terms of k.
1.1 cos 128° 1.2 sin 38°
1.3 tan 218° 1.4 sin 308° Hint:
1.5 cos 232° 1.6 tan (–142°) Draw a
diagram.
2 If 2tan 25º = p, find each ratio in terms of p.
2.1 tan 155º 2.2 sin 205º
2.3 cos 335º 2.4 tan 65º
2.5 cos (–115°) 2.6 sin 295º
3 If sin 43° = a, express each ratio in terms of a.
3.1 sin 223° 3.2 cos 317°
3.3 sin (–137°) 3.4 cos 133°
Unit 3 Reduction formulae 151
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Numerical reductions with special angles
You can apply the reduction formulae to the special angles you learnt in Grade 10.
Quadrant Special angles in all four quadrants
1: θ 30° 45° 60° 0° 90°
2: 180° – θ 150° 135° 120° 180° 90°
3: 180° + θ 210° 225° 240° 180° 270°
4: 360° – θ 330° 315° 300° 360° 270°
REMEMBER
The figures show the triangles with special angles. The table of values is useful, but remember,
and use, the triangles.
30° 45° 60° 0° 90°
__
1
__ 1__
___ √3
___
sin 0 1 60° __ 45°
2 √2 2 √2
__ __ 2 1 1
√3
___ √3
___ 1
__
cos 1 0
2 2 2
__ 30° 45°
1__
___ 1 √3 0 Undefined __
tan √3 √3 1
REMEMBER WORKED EXAMPLES
1 Reduce the ratios to ratios Evaluate the expressions without using a calculator.
with acute angles (check
1 cos 240°.sin 315°.tan 120°
______________________
your signs). sin 150°.tan 210°.cos 225°
2 Use the special angle
2 cos2 330° – sin2 210°
__________________
triangles to write down the tan2 135°.sin 90°
required ratio.
3 Simplify and evaluate the
SOLUTIONS
remaining expression.
(–cos 60°).(–sin45°).(–tan60°)
_________________________
1
(sin 30°).(tan 30°).(–cos45°)
( –__12 )( –√1 2)
__
__ ( –√ 3 )
___
__ √__
______________ = 3 =3
√ 3 ×___
( __12 )( √1
___
) (
__ –___
3
1__
√2 ) 1
cos2 30° – (–sin 30°)2
__________________
2
(–tan 45°)2.sin 90° You can ignore the negative
__
( ) –( )
√3 2 1
signs for the squared ratios:
___ __ 2
= 2 2
__________ 3 – __
= __ 1 = __
1 tan2 135° = tan2 45° = 12 = 1
2
(–1) .(1) 4 4 2
152 Topic 7 Trigonometry
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EXERCISE 13
Evaluate the expressions without using a calculator.
1 4tan 330° cos 135° sin 300°
2 3tan 315° sin 225° sin 120°
_______________________
cos 315° cos 210°
3 sin 240° tan 135°.tan 120°
_______________________
sin 150° sin 210°.cos 330°
4 sin 210° cos180°.cos2 45°
_____________________
1 + tan2 330°
5 sin 135°.cos 315° + cos 360° tan 225°.sin 90°
( )
__
6 sin 240° √3 – tan 120° sin 210°
_______________
cos 180°
Numerical reductions with co-ratios
You know that sin (90° – θ) = cos θ and cos (90° – θ) = sin θ.
To apply these formulae to numerical examples, work with acute angles only:
sin 40° = cos 50° because sin (90° – 50°) = cos 50°
cos 10° = sin 80° because cos (90° – 80°) = sin 80°
If you are given angles bigger than 90°, reduce them to acute angles where the ratio
stays the same, and then use the co-ratio formulae:
sin 145° = sin 35° = cos 55°
cos 250° = –cos 70° = –sin 20°
sin 302° = –sin 58° = –cos 32°
WORKED EXAMPLES REMEMBER
Simplify the expressions as far as possible. 1 Reduce angles to acute
angles.
1 tan 210°.sin 480°.sin 170°
______________________
cos 225°.sin 315°.cos 260° 2 Look out for pairs of ratios
that have complementary
2 cos 70°.sin 150° + cos 240° cos 250°
______________________________
sin 20° angles.
3 Use co-ratios to make the
SOLUTIONS angles the same.
1 tan 30°.sin 60°.sin 10°
________________________
(–cos 45°).(–sin 45°).cos 80°
__
1__ . √
___ ___3
.sin 10°
√3 2
= _________________
1 1
(– ___
__ ).(– ___
__ )sin 10°
√2 √2
1 ÷ __
= __ 1=1
2 2
cos 70°.sin 30° + (–cos 60°)(–cos 70°)
_______________________________
2
sin 20°
= cos 70°.sin 30° + sin 30° cos 70°
___________________________
cos 70°
= 2cos 70°.sin 30° = 2sin 30° = 2 __
______________
cos 70°
1 =1
(2)
Unit 3 Reduction formulae 153
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EXERCISE 14
1 Write each ratio as a ratio of 15° or 70°.
1.1 sin 20° 1.2 sin 290°
1.3 sin 160° 1.4 cos 75°
1.5 cos 105° 1.6 cos 285°
1.7 cos 340° 1.8 cos 430°
2 Simplify and evaluate the expressions. Where possible, do not use a calculator.
2.1 sin 340°
________
cos 430°
2.2 sin2 (100°) ÷ cos2 370°
2.3 4sin 248°.tan 158°
________________
sin 202°
2.4 2cos 143°.cos 217°
________________
sin 127°.sin (–53°)
2.5 sin 147°.tan 114°
_______________
3cos 57°.tan 246°
2.6 cos 149°.sin 239° – cos 301° sin 211°
sin ( –75° ).tan 420°
____________________
2.7
sin 60°.sin 195°.tan 75°
2.8 sin 145° – sin (–235°).cos 325°
_________
cos (–55°)
154 Topic 7 Trigonometry
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Unit 4: Trigonometric equations – specific
and general solutions
In Grade 10 you learnt to solve trigonometric equations for angles less than 90°.
In Grade 11 you learn to solve for angles greater than 90° and for negative angles.
Solving equations with one ratio for angles in
the interval [0º;360º]
WORKED EXAMPLES REMEMBER
Solve for x: Follow these steps for solving
1 2sin x = 1 2 3tan x + 4 = 0 trigonometric equations with
y y one ratio.
1 Solve for the given ratio.
2 1 2 1 2 Determine in which two
quadrants the solution lies.
Use the CAST diagram.
30° 30° 53,13° This will be where the
x x
O O 53,13° given ratio is either positive
or negative.
3 4 3 Draw a diagram in
3 4
the Cartesian plane to
represent this.
4 Find the central or
SOLUTIONS ‘calculator’ angle by
pressing shift sin–1 or
1 2sin x = 1 2 3tan x + 4 = 0 cos–1 or tan–1 of the
1
∴ sin x = __ 4
∴ tan x = – __ POSITIVE ratio.
2 3
5 Solve for the unknown
Calculator angle = 30° Calculator angle = 53,13° angle:
Quadrant 1:
Quadrant 1: Quadrant 2:
x = calculator angle
x = 30° x = 180° – 53,13° = 126,87°
Quadrant 2:
Quadrant 2: Quadrant 4: x = 180° – calculator angle
x = 180° – 30° = 150° x = 360° – 53,13° = 306,87° Quadrant 3:
x = 180° + calculator angle
Quadrant 4:
x = 360° – calculator angle
EXERCISE 15
Solve for x where 0° ≤ x ≤ 360°:
1 sin x = 0,68 2 tan x = –2,41
3 cos x = 0,454 4 sin x = – 0,237
5 tan x = 3,526 6 cos x = – 0 ,813
7 3sin x + 1 = –1 8 5tan x – 7 = 2
tan2 x
9 4cos2 x – 3 = 0 10 ______ –1=0
3 __
11 4sin2 x = 2 12 2sin x + √ 3 ≤ 0
Unit 4 Trigonometrtic equations – specific and general solutions 155
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Solving equations with one ratio for angles in
the interval [–360º;360º]
WORKED EXAMPLES
Solve for x:
__
1 2sin (x + 20°) = √ 3 2 3cos (x – 25°) + 2 = 0
SOLUTIONS
Notice that the rotating angle is x + 20° for Worked example 1 and x – 25° for
Worked example 2.
y y
2 1 x – 25°
x + 20° 2 1
x + 20° x – 25°
60° 60° 48,19°
x x
O 48,19° O
x – 25°
3 4 3 4
x + 20°
To solve the equation follow the steps given in the Remember box on the
previous page. __
2sin (x + 20°) = √ 3 3cos (x – 25°) + 2 = 0
__
√
3 2
sin (x + 20°) = ___ cos (x – 25°) = – __
2 3
Calculator angle = 60° Calculator angle = 48,19°
Quadrant 1 (positive angle): Quadrant 2 (positive angle):
x + 20° = 60° (x – 25°) = 180° – 48,19°
x = 40° x = 131, 81° + 25° = 156,81°
Quadrant 1 (negative angle): Quadrant 2 (negative angle):
x + 20° = –360° + 60° (x – 25°) = –180° – 48,19°
x = –320° x = –228,19° + 25° = –203,19°
Quadrant 2 (positive angle): Quadrant 3 (positive angle):
x + 20° = 180° – 60° (x – 25°) = 180° + 48,19°
x = 100° x = 228,19° + 25° = 253,19°
Quadrant 2 (negative angle): Quadrant 3 (negative angle):
x + 20° = –180° – 60° (x – 25°) = –180° + 48,19°
x = –260° x = –131,81° + 25° = –106,81°
156 Topic 7 Trigonometry
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EXERCISE 16
Solve for x where – 360° ≤ x ≤ 360°.
__
1 3 √
1 sin (x – 22º) = – __ 2 cos (x + 15°) = ___
2 __ 2
3 tan (x + 40º) = –1 4 √ 5 cos (x – 15º) = 2
5 cos x + cos 71º = sin 17º 6 3 tan x + sin 27º = 3
__
7 √ 2 sin (x – 27º) – 1 = 0 8 2sin (x + 50º) + 4tan 20º = 0
9 5cos (x + 15°) – 2 = 0 10 3 tan (x – 10°) + 10 = 3
tan (x + 50°)
___________
11 3sin (x + 32º) = 2sin 32° 12 +1=3
3
The concept of a general solution
You need not restrict the solutions to trigonometric equations to an interval of
[0º;360º] or [–360º; 360º]’
1:
Look at the solutions to sin x = __
2
1 , but so does sin 390° = sin (30° + 360°) = sin 30° = __
sin 30° = __ 1
2 2
1
sin 750° = sin (30° + 2 × 360°) = __
2
1
and sin 1 110° = sin (30° + 3 × 360°) = sin 30° = __
2
1 , but so does sin 510° = sin (150° + 360°) = sin 30° = __
sin 150° = __ 1
2 2
1 and sin 1 230° = sin (150° + 3 × 360°) = sin 30° = __
sin 870° = sin (150° + 2 × 360°) = __ 1
2 2
You can summarise the solutions to trigonometric equations as the general solution
1:
of the equation sin x = __
2
x = 30° + n.360°, n ∊ ℤ or x = (180° – 30°) + n.360°, n ∊ ℤ
1.
You can see the solution graphically by drawing the graphs of y = sin x and y = __
2
1 in the interval [– 450º;450º].
The graphs show that there are five solutions to sin x = __
2
y
1
0,5
E D A B C
x
–450° –360° –270° –180° –90° 90° 180° 270° 360° 450°
–0,5
–1
Unit 4 Trigonometrtic equations – specific and general solutions 157
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Finding the general solution to equations with
one ratio
WORKED EXAMPLE
KEY WORDS
Find the general solution to the equation sin 2x = – 0 ,757.
general solution – the
formula which lists all possible
solutions to a trigonometric
SOLUTIONS
equation; takes into account Method 1 Method 2
the period of the ratios so Using the positive ratio and ‘calculator Using the given ratio (+ or –) and the
the angle can be positive or
angle’ is a positive acute angle. ‘calculator angle’ is acute, obtuse or
negative
negative.
solution – a value of the Calculator angle = 49,2°
angle which satisfies a given (key in positive ratio) Calculator angle = – 49,2°
trigonometric equation (sin < 0 in Quadrant 3 and Quadrant 4) (key in given, negative ratio)
Quadrant 3: Use 2x = calculator angle
2x = 180° + 49,2° + n.360° or 2x = 180° – calculator ang
REMEMBER = 229,2° + n.360°(÷ by 2) 2x = – 49, 2° + n.360°(÷ by 2)
x = 114,16° + n.180° x = –24,6° + n.180°
For sin x = a the general
solution is: Quadrant 4: 2x = 180° – (– 49,2°) + n.360°
x = sin–1(a) + n.360°, n ∊ ℤ or 2x = 360° – 49,2° + n.360° = 229,2° + n.360°
x = 180° – sin – 1(a) + n.360°, = 310,8° + n.360° x = 114,6° + n.180°, n ∊ ℤ
n∊ℤ x = 155,4° + n.180°, n ∊ ℤ
For cos x = b the general
Notice that x = 155,4° + (–1).180° = –24,6°, which is the same solution
solution is:
x = cos–1(b) + n.360°, n ∊ ℤ or for Method 2.
x = –cos – 1(b) + n.360°,
n∊ℤ
For tan x = c the general
solution is:
EXERCISE 17
x = tan–1(c) + n.180°, n ∊ ℤ
Find the general solutions to these equations.
1 sin (x – 16º) = 0,616
2 cos 2x = 0,789
3 cos 3x = – 0,123
4 tan (x + 56°) = 7,56
5 x = –1,421
tan __
2
6 sin (2x + 44º) = – 0,708
7 3cos (x – 15°) + 1 = – 0,456
8 2tan ( 2x – 10°) = 10,67
158 Topic 7 Trigonometry
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Solving equations with more than one ratio
WORKED EXAMPLES KEY WORDS
1 Solve for x if cos x – 4cos x sin x = 0 and –180° ≤ x ≤ 360°. specific solutions – solutions
2 Solve for x if 3cos2 x – 5sin x = 1 and write down the general solution. which satisfy a given
trigonometric equation in a
3 Find the general solution to the equation 3cos x – sin x = 0.
restricted interval, such as
– 360° ≤ x ≤ 360°
SOLUTIONS
1 cos x + 4cos xsin x = 0 | Factorise by taking out a common factor.
cos x(1 + 4sin x) = 0 REMEMBER
1
cos x = 0 or sin x = –__
4 1 Convert all ratios to sine
x = 90° + n.360° or calculator angle = 14,5°
and cosine to only one
x = –90° + n.360° or x = 180° + 14,5° + n.360° ratio if possible.
or x = 360° – 14,5° + n.360° n ∊ ℤ Use the identities
Specific solution: Start with n = 0, then n = ±1, n = ±2 … sin x,
_____
tan x = cos x
Continue to work out specific values of x until your answers fall outside cos2 x = 1 – sin2 x,
– 180° ≤ x ≤ 360°. sin2 x = 1 – cos2 x.
Then x = 90°; 270°; –90°; 194,5°; –65,5°; 345,5°; –14,5° 2 Simplify and factorise
to change into simpler
2 3cos2 x – 5sin x = 1
equations.
3(1 – sin2 x) – 5sin x = 1 | cos2 x = 1 – sin2 x
3 Find the general solution
3 – 3sin2 x – 5sin x – 1 = 0 and solve for the given
3sin2 x + 5sin x – 2 = 0 | trinomial interval.
(3sin x – 1)(sin x + 2) = 0
1 or sin x = –2
sin x = __
3
calculator angle = 19,5° or No solution
REMEMBER
Quadrant 1: x = 19,5° + n.360°
Qudrant 2: x = 160,5° + n.360, n ∊ ℤ If you use Method 2:
Key in the given ratio either
3 3cos x – sin x = 0 positive or negative (what is
3cos x = sin x | ÷ both sides by cos x given). The calculator angle
sin x tan x = 3
_____
3 = cos x could be acute, negative or
calculator angle = 71,57° obtuse. Learn these patterns
x = 71,57° + n.180°, n ∊ ℤ and DO NOT change them
for any situation:
sin x = …
x = calc + n.360°
EXERCISE 18 or 180° – calc + n.360°, n ∊ ℤ
cos x = …
Solve the equations: (GS = general solution) x = calc + n.360°
or – calc + n.360°, n ∊ ℤ
1 3sin x – 2cos x = 0, find GS
tan x = …
2 3sin2 x – 2sin x = 0 and x ∊ [–360°;360°] x = calc + n.180°, n ∊ ℤ
3 3tan2 x + tan x = 2 – 180° ≤ x ≤ 360°
4 5sin2 x – 3sin x – 2 = 0, find GS
5 4cos2 x – sin x = 1 and x ∊ [–180°;360°]
6 3cos2 x = 2(sin x + 1), find GS
7 3sin2 x – 5sin xcos x = 0 and x ∊ [–180°;180°]
8 10sin2 x + cos x = 8, find GS
9 4tan2 x – 8tan x + 3 = 0, find GS
Unit 4 Trigonometrtic equations – specific and general solutions 159
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10 1 – 5sin2 x = 3sin x + 1 and x ∊ [–180°;180°]
REMEMBER 11 6sin2 x + 7sin x + 2 = 0 and x ∊ [–360°;360°]
__
12 3cos2 x + √ 3 sin x cos x = 0, find GS
1 Convert all ratios to sine
13 8sin x cos x + 4cos x + 6 sin x + 3 = 0, find GS
and cosine or only one
ratio if possible.
14 4cos2 x – sin2 x = 3cos x + 1 and x ∊ [–180°;180°]
Use the identities 15 3sin2 x + 2sin x.cos x = 4cos x + 6 sin x and x ∊ [0°;360°]
sin x,
_____
tan x = cos x
cos2 x = 1 – sin2 x,
sin2 x = 1 – cos2 x. Solving equations without a calculator
2 Simplify and factorise Look at these examples for x ∊ [0°;360°]:
to change into simpler
If sin x = sin 41°: calculator ang = 41° so x = 41° or x = 180° – 41° = 139°
equations.
If cos x = –cos 58°: calculator ang = 58° so x = 180° – 58° = 122° or x = 180° + 58° = 238°
3 Find the general solution
and solve for the given
If cos x = sin 24°, then cos x = sin 66° (co-ratio) so x = 66° or x = 360° – 66° = 294
interval. We can extend the concept of solving without a calculator to equations with algebraic
expressions for the angles.
WORKED EXAMPLE 1
Solve for x if sin (3x – 20°) = sin (x + 10°) and –360° ≤ x ≤ 360°.
SOLUTIONS
REMEMBER
Find the general solution first:
If you use Method 2: Quadrant 1 or Quadrant 2
Key in the given ratio – 3x – 20° = x + 10° + n.360° 3x – 20° = 180° – (x + 10°) + n.360°
positive or negative (what is 2x = 30° + n.360° 4x = 190° + n.360°
given). The calculator angle x = 15° + n.180°, n ∊ ℤ x = 47,5° + n.90°, n ∊ ℤ
could be acute, negative or
obtuse. Learn these patterns For – 360° ≤ x ≤ 360° use integral values for n
and DO NOT change them x = 15°; 15° + (1)180° = 195°; 15° + (–1)180° = –165°; 15° – (2)180° = –375°
for any situation: x = 47,5°; 47,5° + (1)90° = 137,5°; 47,5° + (2)90° = 227,5°; 47,5° + (3)90° = 317,5°
sin x = … x = calc + n.360° x = 47,5°; 47,5° – (1)90° = –42,5°; 47,5° – (2)90° = –132,5°; 47,5° – (3)90° = 222,5°;
or 180° – calc + n.360°, n ∊ ℤ
47,5° – (4)90° = 312,5°
cos x = … x = calc + n.360°
or – calc + n.360°, n ∊ ℤ
WORKED EXAMPLE 2
tan x = … x = calc + n.180°,
n∊ℤ
Find the general solution of cos (2x + 30°) = sin x.
SOLUTIONS
Use co-ratio: sin x = cos (90° – x) and find the general solution:
cos (2x + 30°) = cos (90° – x)
Quadrant 1 or Quadrant 4
2x + 30° = 90° – x + n.360° 2x + 30° = –(90° – x) + n.360°
3x = 60° + n.360° 2x + 30° = –90° + x + n.360°
x = 20° + n.120°, n ∊ ℤ x = –120° + n.360°, n ∊ ℤ
For Quadrant 4 you may use:
2x + 30° = 360° – (90° – x) + n.360°
x = 240° + n.360°
(which is the same as –120°)
160 Topic 7 Trigonometry
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EXERCISE 19
Solve the equations without a calculator: (GS = general solution)
1 cos x = sin 32° and x ∊ [0°;360°]
2 tan2x = tan 128° and x ∊ [–360°;360°]
3 sin (x – 40°) = cos 50° and x ∊ [–360°;360°]
4 tan x = tan (2x – 25°), find GS
5 cos (2x – 25°) = cos (38° – x) and x ∊ [–90°;180°]
6 sin (x + 10º) = sin ( 50º + 2x ), find GS
7 sin (2x – 5º) = cos ( x – 35º ), find GS
8 cos x = – cos 42° and x ∊ [–180°;360°]
9 cos (2x – 10°) = sin (x – 40°) and x ∊ [–180°;360°]
10 sin (x + 10º) = –sin (30º – 2x), find GS
EXERCISE 20
Solve for x in each equation.
1 3cos x + 2 = 1 and x ∊ [0°;360°]
2 sin x = 0,112 and x ∊ [–360°;360°]
_____
2 REMEMBER
3 tan (x – 10°) + 1 = 3 and x ∊ [–180°;180°]
1 If sin A = sin B, then A = B
4 15sin (x + 14º) = 14 and x ∊ [0°;360°]
or A = 180° – B
5 2cos (2x + 20°) = 1, find GS
__ 2 If cos A = cos B, then A = B
6 x = √ 2 and x ∊ [–360°;360°]
2sin __
2 or A = 360° – B (or – B)
7 4tan (2x – 40°) + 13 = 3, find GS 3 If tan A = tan B then A = B
or A = 180° + B
8 cos (x – 12°) = sin 56° and x ∊ [0°;360°]
4 If sin A = cos B then
9 13cos x + 12sin x = 0, find GS
sin A = sin (90° – B) and
10 4cos2 xsin x – 3sin x = 0, find GS A = (90° – B) or
11 3cos2 x + 11sin x + 1 = 0 and x ∊ [–360°;360°] A = 180° – (90° – B)
12 5tan2 x + 7tan x = 6 and x ∊ [–180°;360°]
13 cos (2x + 20°) = cos (30° – x), find GS
14 2sin xcos x – cos2 x – 2sin x + cos x = 0 and x ∊ [–180°;180°]
Finding the values for which identities are invalid
or undefined
We have worked through trigonometric equations, so now we consider the values
for which identities or invalid or undefined. KEY WORD
Remember from Unit 2 that an identity is an equation which is true for all values of trigonometric equation
– an equation involving
the unknown variable, for which both sides of the identity are defined.
trigonometric ratios which is
• An identity is undefined if the denominator
___
is zero. true only for certain values of
• If an identity is in the form A = √f(x) , it is not valid for the values of x which the unknown variable
make f (x) < 0
• If an identity involves tan x, it is not valid for x = 90° + n.180°
sin x and cos x = 0 when x = 90° + n.180°
_____
tan x = cos x
When proving identities, you may be asked to find the values of the variable (angle)
for which the identity is undefined or invalid.
Unit 4 Trigonometrtic equations – specific and general solutions 161
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WORKED EXAMPLE
1 For which values of x is the identity tan x + sin x = sin x undefined?
___________
1 1 + _____
cos x
________
2 For which values of x does the identity sin θ – √ 1 – sin2 θ = 0 hold true?
_____
tan θ
SOLUTIONS
1 1
_____ = 0
Undefined when tan x is undefined or when 1 + cos x
1
_____ = 0 cos x = –1 x = 180° + n.360°
x = 90° + n.180° or 1 + cos x
2 Undefined or invalid when:
2.1 tan θ = 0 | θ = 0° + n.180
2.2 tan θ is undefined when θ = 90° + n.180°
2.3 sin2 θ ≤ 0 sin2 θ ≥ 1 which is never true
1 – ________
√ 1 – sin2 θ is defined for x ∊ ℝ
EXERCISE 21
In Exercises 4 and 5, you proved identities but did not state restrictions. Now find
the values of the variable x or θ for which the following identities are not valid.
1 1 – sin x = cos
_____ x
_____
sin x tan x
2 cos x – cos3 x
________________ = tan2 x
cos x – cos x.sin2 x
3 tan2 θ + 1 = 1
_____
2
cos θ
4 cos x + tan x = _____
________ 1
1 + sin x cos x
5 1
________ + 1 2
________ = _____
1 – sin x 1 + sin x cos2 x
6 cos x =1+
________ sin x
________
1 – sin x cos x
7 cos2 x – cos x – sin2 x = _____
__________________ 1 – _____
1
2sin x.cos x + sin x tan x sin x
8 sin x – ___________ + cos x = 2sin
1 x.cos x
___________
sin x + cos x sin x + cos x
9 _____ = sin x
________ – 2
1 +1
________
sin x + 1 cos2 x sin x – 1
_______
10 sin x
_________ =
(1 + cos x) √________
1 – cos x
1 + cos x
162 Topic 7 Trigonometry
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Revision
1 y
P(–5;4)
θ
x
O
1.1 Use the diagram to find the length OP. (1)
1.2 Now write down the value of:
1.2.1 sin θ (1)
1.2.2 41cos2 θ + 1 (2)
1.2.3 tan (180° – θ) (2)
[6]
2 y
θ
x
O
__
P(√ 3 ;–1)
2.1 Use the diagram to find the length OP. (1)
2.2 Now write down the value of:
2.2.1 cos θ.sin θ (3)
2.2.2 tan2 θ + 1 (2)
2.2.3 θ (2)
[8]
3 If 6sin θ = 5 and cos θ < 0, find, without a calculator, the value of:
3.1 1 – 2cos2 θ (4)
3.2 cos θ.tan θ (3)
[7]
4 If 4cos θ = 3 and 7tan β = 5; 90° < θ < 360° and β > 90°,
7 + 74sin2 β = 34
prove without using a calculator that _____ [6]
tan θ
2
5 tanA.sinA
4 and A + B = 90°, use a sketch to find the value of _________
If cos A = __ [5]
7 cosB
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TOPIC 7: REVISION CONTINUED
6 Prove each identity.
6.1 sin2 x + sin x.cos x = 1
_________
tan x
(4)
6.2 1
cos θ + cos θ.tan2 θ = _____ (5)
cos θ
6.3 ( 1
cos θ tan θ + _____
tan θ ) 1
= _____
sin θ
(5)
6.4 cos2 x + (sin x – tan x)(sin x + tan x) = 1 – tan2 x (6)
6.5 1 [(sin x + cos x)2 – 1]
sin xcos x = __ (5)
2
[25]
7 Simplify each expression.
2cos (90° + x) – sin (360° – x)
_________________________
7.1 (6)
3sin (90° – x) – cos (180° – x)
sin (180° + x)
______________________________
7.2 (5)
tan (180° – x) tan 45°.cos (360° – x)
2cos (–180° – x).sin (–x)
_______________________
7.3 (5)
sin (x – 180°).cos (x – 360°)
cos 2( 180° + x )
7.4 _____________________ +1 (5)
sin( 180° – x ).cos( x – 90° )
7.5 [ sin ( 90º + x ) sin ( – x) tan (180° – x) ] – [ sin ( 90° – x ) cos ( x – 180º ) ] (5)
[26]
8 Draw a diagram for each question.
8.1 If tan 28° = p, write down the following in terms of p.
8.1.1 tan 152° (2)
8.1.2 cos (–28°) (2)
8.1.3 sin 62° (2)
8.2 If 2sin 41° = a, write down the following in terms of a.
8.2.1 sin 319° (2)
8.2.2 cos 49° (2)
8.2.3 cos 139° (2)
[12]
164
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9 Simplify without the use of a calculator:
9.1 sin 330° + cos 250°
________ (5)
sin 160°
9.2 sin 120°.cos 135°.tan 240°
______________________ (6)
cos 150°.tan 150°.sin 315°
9.3 1__ sin 120° + __
___ sin 55°
1 tan 45° cos2 225° – ________ (6)
√3 2 cos 145°
9.4 sin2 300° + sin 240° cos 150°
________________________ (6)
tan 225°.sin 270°
9.5 sin 225°.cos 135° + 2sin 210°.sin 90° (4)
[27]
10 Solve these equations.
10.1 sin (θ + 10°) = 0,757 and 0° ≤ x ≤ 360° (3)
__
10.2 √2 sin 3 θ + 1 = 0 and – 180° ≤ x ≤ 180° (4)
10.3 3cos θ = 2sin2 θ – 3. Find the general solution. (7)
10.4 tan (2 θ – 30°) = tan 42°; –360° ≤ θ ≤ 360° (6)
10.5 sin (x + 20°) = cos 2x. Find the general solution. (6)
[26]
11 Find the values of x and θ for which the identities in Question 6
are invalid or undefined.
11.1 sin2 x + sin x.cos x = 1
_________
tan x
(3)
1
11.2 cos θ + cos θ.tan2 θ = _____ (4)
cos θ
11.3 (
cos θ tan θ + 1
_____
tan θ
= 1
_____
sin θ) (4)
[11]
165
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Formal assessment: Term 2 assignment
1 ( 2 )x – 1 – 3
1
f( x ) = 3 __ 1 ( x + 2 )2 – 3
g( x ) = __
3
4 +1
h( x ) = – _____
x+3
1.1 State the equations of the asymptotes of f and h. (3)
1.2 Draw f, g and h on the same set of axes. Indicate the x- and
y-intercepts and the asymptotes of both f and h. (12)
1.3 Use your graphs to help you answer the questions.
1.3.1 State one value of x for which f(x) = g(x) = h(x) (1)
1.3.2 State one value of x for which f(x) ≠ g(x) = h(x) (1)
1.3.3 For which values of x is f(x) ≥ 0? (2)
1.3.4 For which values of x > –3 is f(x) ≤ g(x)? (2)
h(x)
1.3.5 For which values of x is ____ ≥ 0? (4)
g(x)
1.4 If y = –x + c is one of the symmetry lines of h, determine the value
of c. (2)
1.5 If g is reflected in the y-axis, the reflected graph is given by y = … (2)
1.6 State h(–x) in the form y = … and describe the translation in words. (3)
2 1–1
Given f(x) = x2 – 1, g(x) = –2x + 1 – 1 and h(x) = __
x
2.1 State the equations of the asymptotes of g and h. (2)
2.2 State the equations of the symmetry lines of f and h. (5)
2.3 Draw f, g and h on the same system of axes. Indicate all intercepts
with the axes, the asymptotes of both g and h and the symmetry
lines of f and h. (12)
2.4 Determine f (1) – g(1). (2)
2.5 Determine g(–1) – h(–1). (2)
2.6 State the ranges of f, g and h and the domain of h. (4)
2.7 State the values of x and y for which f(x) = g(x). (2)
2.8 For which value(s) of x is f(x) ≥ g(x)? (3)
2.9 Solve algebraically for x and y if f(x) = h(x). (5)
2.10 For which values of x is f(x) ≤ h(x)? (2)
2.11 Write down the equation of f (2x) in the form y = … (3)
2.12 If h is translated 3 units left and 2 units up, the translated
graph is given by y = … (3)
2.13 If g is translated 2 units right and 1 unit down, the translated
graph is given by y = … (3)
2.14 If g is reflected in the x-axis, write the reflected equation in
the form y = … (2)
3 Draw a sketch for y = ax2 + bx + c, where: REMEMBER
3.1 a > 0, b < 0 and c < 0 (4)
The quadratic
3.2 a < 0, b < 0 and c > 0 (4) formula
_______
3.3 a < 0, b < 0, c < 0 and b2 − 4ac (6) √ 2
x = −b± b − 4ac
____________
3.4 a > 0, b > 0, c > 0 and b2 − 4ac = 0 (6) 2a
3.5 a < 0, b < 0, c < 0 and b2 − 4ac > 0 (6) b2 − 4ac < 0
3.6 a > 0, b = 0 and c > 0 (4) ⇒ non-real roots
b2 − 4ac > 0
⇒ real, unequal roots
b2 − 4ac = 0
⇒ real, equal roots
166
PLT MATHS LB 11 7th pgs (Real Book).indb 166 2012/07/02 2:22 PM
4 Consider the functions g(x) = psin qx and h(x) = vcos wx for x ∊ [–120°;90°].
y
1
A E
h 0,92 F
B 0,71
g C
D
0,38
x
–120° –97,5° –75° 67,5° 90°
4.1 State the period of both g and h. (2)
4.2 Determine the values of p and q in g(x) = psin qx. (2)
4.3 Determine the values of v and w in h(x) = vcos wx. (2)
4.4 State the coordinates of D, E and F. (6)
4.5 For which values of x is h(x) ≥ g(x)? (5)
4.6 For which values of x does h(x) increase as g(x) decreases? (3)
5 y
(60°;2)
h
E
g (30°;1)
x
C –30° 120° D
A
B
5.1 State the period, the amplitude and the range of h and g. (8)
5.2 If h(x) = asin (x + b) and g(x) = pcos (x + q), determine the values
of a, b, p and q. (6)
5.3 If h(x) = dcos (x + c) and g(x) = vsin (x + w), determine the values
of c, d, v and w. (6)
5.4 State the coordinates of A, B, C, D and E. (10)
5.5 For which values of x is g(x) ≥ 0? (2)
5.6 For which values of x is h(x).g(x) > 0? (5)
5.7 For which values of x is h(x) a decreasing function? (3)
5.8 For which values of x is g(x) an increasing function? (3)
Total: 175 marks
167
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Term 2 summary
Topic 5
The parabola
Standard form: y = f(x) = ax2 + bx + c
x-intercept form: y = f(x) = a(x – xint1)(x = xint2)
Turning point form : y = f( x ) = a( x – p )2 + q
The hyperbola: y = f( x ) = x a
____ + q
–p
The exponential function: y = f (x) = abx – p + q where b > 0, b ≠ 1
The effect of a, p and q is the same for all three graphs.
The effect of a
• Increasing a (so that a > 1) results in a vertical stretch.
• Decreasing a (so that 0 < a < 1) results in a vertical shrink.
• Changing the sign of a (a + to –) results in a reflection in the x-axis.
The effect of p and q
• Changing p results in a horizontal shift.
If p > 0, (x – p) shifts the graph to the right and (x + p) shifts the graph to the left.
• Changing q results in a vertical shift.
• If q > 0, graph moves up.
• If q < 0, graph moves down.
• p and q do not influence each other.
Graph Turning point Asymptotes Domain (x); Range (y) Axis of symmetry
f( x ) = ax2 + bx + c or
x∊ℝ x=p
f( x ) = a( x – p )2 + q or (p;q) None –b
y ∊ ℝ, y ≥ p if a > 0 x = ___
f(x) = a(x – x1)(x – x2) 2a
y=q x ∊ ℝ, x ≠ p y = (x – p) + q
y = f( x ) = x a
____ + q
–p None
x=p y ∊ ℝ, y ≠ q y = –(x – p) + q
x∊ℝ
f( x ) = abx – p + q None y=q y ∊ ℝ, y > q (if a > 0) None
y ∊ ℝ, y < q (if a < 0)
• The average gradient of the curve between the two points on a curve is the same as the gradient of the
______ y –y
straight line between two points on a curve: m = x2 –x
1
2 1
168 Term 2 summary
PLT MATHS LB 11 7th pgs (Real Book).indb 168 2012/07/02 2:22 PM
Topic 6
Trigonometric graphs
Graph x ∊ [–360°;360°] Amplitude Period (P) Asymptotes Range Interval spacing
y = sin x 1 360° none y ∊ [–1;1] 90°
y = cos x 1 360° none y ∊ [–1;1] 90°
y = tan x Not applicable 180° x = ±90° y ∊ (–∞;∞) 45°
x = ±270°
a if a > 0 360° P
y = asin kx + q ____ none y ∊ [–a + q;a + q] S = __
– a if a < 0 k 4
a if a > 0 360°
y = acos kx + q ____ none y ∊ [–a + q;a + q] S = P
__
– a if a < 0 k 4
180°
____ P + n.P
__ y ∊ (–∞;∞) S=P
__
y = atan kx + q Not applicable k 2 4
y = asin (x ± p) p results in a horizontal shift
y = acos (x ± p) to the left if p > 0
y = atan (x ± p) to the right if p < 0
Topic 7
Trigonometry
Trigonometry
Signs of ratios in all four quadrants θ is the acute angle between
the arm and the x-axis
y y y
2 1 2 1
CAST shows r r 180° – θ θ
S A you in which y y
quadrant the –x x
x x x
ratios are
T C POSITIVE
–y –y
180° + θ 360° – θ
r r
3 4 3 4
Identities
tan θ = sin θ and sin2 θ + cos2 θ = 1
_____
cos θ
Prove other identities using the formula above and algebra to show LHS = RHS.
Reduction formulae
sin (180° – θ) = + sin θ sin (180° + θ) = –sin θ
cos (180° – θ) = –cos θ cos (180° + θ) = –cos θ
tan (180° – θ) = –tan θ tan (180° + θ) = + tan θ
sin (360° – θ) = –sin θ sin (–θ) = –sin θ
cos (360° – θ) = + cos θ cos (–θ) = + cos θ
tan (360° – θ) = –tan θ tan (–θ) = –tan θ
Co-ratios
sin (90° – θ) = cos θ; cos (90° – θ) = sin θ; sin (90° + θ) = cos θ; cos (90° + θ) = –sin θ
Trigonometric equations
Standard pattern for general solutions: 'calc' is obtained by pressing
sin x = … x = ‘calc’ + n.360° or 180° – ‘calc’ + n.360°, n ∊ ℤ shift sin–1 (or cos–1 or tan–1)
cos x = … x = ‘calc’ + n.360° or – ‘calc’ + n.360°, n ∊ ℤ of the given ratio.
tan x = … x = ‘calc’ + n.180°, n ∊ ℤ
Term 2 summary 169
PLT MATHS LB 11 7th pgs (Real Book).indb 169 2012/07/02 2:22 PM
Formal assessment
Examination practice Paper 1
Time 2 hours
Question 1
Solve for x:
1.1 3(x2−4) = 4(x−3) (4)
1.2 2 − 4x − x2 = 0 (correct to two decimal places) (4)
____
1.3 √ 1 + x + 5=x (4)
2
1.4 3x −x≤0
______ (4)
3
x2−6 = 32x
1.5 4 (5)
3
__
1.6 27
2x4 = ___ (3)
4
[24]
Question 2
2.1 Consider the equation: x2 + 4xy − 21y2 = 0
2.1.1 Calculate the value of the ratio __xy , y ≠ 0. (4)
2.1.2 Now calculate the values of x and y if x + 3y = 3. (4)
____
2.2 If f(x) = √ x7_____
− x determine:
+3
2.2.1 f(−2) (2)
2.2.2 for which values of x, f(x) = 0 (2)
2.2.3 for which values of x, f(x) is real. (2)
2.3 Given f(x) = 2x, find a simplified expression for: f(x) + f( 1 1 + f(−x)
__) + ___
x (4)
f(x)
[18]
Question 3
Simplify each of the following expressions. Do not use a calculator in this
question. Leave your answers with positive exponents and rational denominators
where necessary.
( )
__
√ b3 −1
____
3.1 1
__
(3)
b−2
x +y −1
3.2 ______
−1
(4)
y +x
3.3 ___9
_________
__ (4)
√ 48 − √ 3
____ ____
√ 45x5
− √20x3
_____________
_____
3.4 (4)
3 √ 125x
[15]
Question 4
4.1 Tsepo is doing a Life Sciences project in which he observes the number of
cockroaches entering a drain pipe per day.
Day Number of cockroaches
1 1
2 10
3 23
4 40
170 Exam practice paper 1
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4.1.1 Determine how many cockroaches he can expect to find on
day 5 and day 6. (2)
4.1.2 Determine an algebraic formula to calculate how many cockroaches
there are on the nth day. (5)
4.2 Examine the tiling pattern shown in the diagrams.
Stage 1 Stage 2 Stage 3
Stage 1 Stage 2 Stage 3 Stage 4 Stage n
A: Number of striped tiles 3 5 7 9
B: Number of black tiles 1 4 9 16
C: Number of white tiles 2 6 12 20
D: Total number of tiles 6 15 28 45
4.2.1 Find the rule for stage n of patterns A, B and C. (4)
4.2.2 Will any stage require 21 striped tiles? Justify your answer with
calculations. (2)
4.2.3 At what stage will a total of 120 tiles be needed to complete
the pattern? (3)
4.2.4 If the dimensions of a tile are 0,5 m × 0,5 m, write a formula
for the total area in m2 of the tiling pattern at any given stage n. (2)
[18]
Question 5
The sketch shows the parabola g(x) = x2 − 2x − 3 and the straight line f(x) = mx + c. y
f(x) and g(x) intersect B and C. g
P
5.1 Write g(x) = x2 − 2x − 3 in the form g(x) = a(x − p) + q. 2
(3)
5.2 Determine the lengths of OC and AB. (4)
f
5.3 Give the coordinates of D, the turning point of g(x). (2)
5.4 Find the values of m and c. (2)
Q A B
5.5 State the values of x for which g(x) ≤ f(x). (2) x
O
5.6 Give the new equations of the graph of g(x) if:
5.6.1 g(x) is moved two units to the right and 1 unit down (3)
5.6.2 g(x) is reflected in the x-axis. (2) C
[18]
D
R
Question 6
−6 − 1
Consider the function: f(x) = _____
x−3
6.1 Sketch the graph f(x) showing clearly the asymptotes and intercepts
with the axes. (5)
6.2 For which values of x is f(x) > 0? (2)
[7]
Total: 100 marks
Exam practice paper 1 171
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Formal assessment
Examination practice Paper 2
Time: 2 hours
Question 1
In the figure, A(1;−1), B(0;2) and C(6;4) are the vertices of a triangle.
y
C(6;4)
B(0;2)
x
A(1;–1)
1.1 Determine the gradient of BC (2)
1.2 Determine the equation of line BC. (2)
1.3 Prove that B ^ =90°. (3)
1.4 Write down the coordinates of D if ABCD is a rectangle. (2)
1.5 Write down the coordinates of the midpoint of AC. (2)
1.6 Determine the equation of the perpendicular bisector of line AC. (4)
1.7 Find the size of angle BA ^ C. (4)
1.8 Determine the area of the △ABC. (4)
1.9 If P (a;−1), B and C are collinear, find the value of a. (3)
___
1.10 If Q (−1;b) is a point such that QB = √26 , find the value of b, b > 0 (4)
[30]
Question 2
2 and cos θ > 0, determine without finding the value of θ
2.1 If sin θ = − __
3
and with the aid of a sketch, the value of:
2.1.1 cos θ (3)
2.1.2 tan2 θ × 5sin θ (3)
1 , use a sketch to find the following in terms of p:
2.2 If cos 20° = __
p
2.2.1 sin 70° (2)
2.2.2 cos 200° (2)
2.2.3 tan 160° (2)
[12]
172 Exam practice paper 2
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Question 3
3.1 Evaluate without the use of a calculator: sin 210°.sin 225°
_______________________ (7)
2
cos 120° tan 150°cos 315°
sin 130°.cos (360° − θ).cos (−θ)
3.2 Simplify: sin(180° + θ).cos(90° + θ) − __________________________ (7)
cos 220°
[14]
Question 4
4.1 Prove that: cos A + cos (180°− A).sin2 A = cos3 A (4)
2
(sin x − cos x)
1 − 2tan x
4.2 Given: _____________
2
= _____
2
1−sin x cos x
4.2.1 Prove the above identity. (5)
4.2.2 For which values of x in the interval [0°;360°] is the above identity
undefined? (3)
[12]
Question 5
You may use calculators in this question; give answers correct to TWO decimal places.
5.1 Solve for x if tan (x – 10°)
___
= sin 72° and x ∊ [0°; 360°]. (3)
15 √
5.2 Solve for x if sin 2x = ____ , –270° ≤ x ≤ 180°. (5)
8
2
5.3 Determine the general solution of 8sin x + 6cos x = 9. (7)
[15]
Question 6
6.1 Draw sketch graphs of f(x) = cos (x−30°) and g(x) = sin 3x for −180° ≤ x ≤ 180°.
Label all the turning points and intercepts with the axes. (7)
6.2 Read from your graphs the solution to the equation cos (x−30°) = sin 3x. (2)
6.3 Confirm these solutions by solving the equation in 6.2 (6)
6.4 Write down the values of x for which g(x) ≥ f(x). (2)
[17]
Coordinate Geometry: 30;
Trigonometry: 70
Total: 100 marks
Exam practice paper 2 173
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Term
3
174
PLT MATHS LB 11 7th pgs (Real Book).indb 174 2012/07/02 2:22 PM
TOPIC 8 Measurement
Unit 1 Revision of Grade 10 measurement 176
Revision 185
TOPIC 9 Euclidean geometry and
measurement
Unit 1 Geometry revision 188
Unit 2 Circles, perpendicular lines through
the centre, chords, midpoints 190
Unit 3 Angle at centre theorem and cyclic
quadrilaterals 194
Unit 4 Tangents 202
Revision 211
TOPIC 10 Trigonometry: sine, cosine and
area rules
Unit 1 Proof and application of sine, cosine
and area rule 214
Unit 2 Two-dimensional problems using
sine, cosine and area rules 223
Revision 232
TOPIC 11 Finance, growth and decay
Unit 1 Simple and compound decay 236
Unit 2 The effect of different periods of
compound growth and decay 242
Revision 250
TOPIC 12 Probability
Unit 1 Addition and complementary rules;
Dependent and independent events 252
Unit 2 Venn diagrams 256
Unit 3 Using tree diagrams to solve problems
for events not necessarily independent 262
Unit 4 Contingency tables 267
Revision 273
Formal assessment: Term 3 Test 1 276
Term 3 Test 2 278
Term Summary 282
175
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TOPIC
2 8 Measurement
Unit 1: Revision of Grade 10 measurement
KEY WORDS
This work was covered in Grade 10, but is also examined in Grades 11 and 12.
congruent – exactly the same
perpendicular height – We have explained the important concepts and summarised the formulae.
perpendicular distance
Right prisms and right cylinders:
between bases
• always have a pair of congruent bases which are parallel to each other
• the bases are perpendicular to the lateral surfaces
• the base of a right prism is a polygon and the base of a right cylinder is a circle.
REMEMBER
Area rectangle
= length × breadth
Area triangle
1 base × perpendicular
= __
2 rectangular prism triangular prism right cylinder
height
Area circle = πr2
In the figures:
Circumference circle = 2πr
Lateral surface area cylinder • the shaded surfaces are the bases
= 2πrh • the unshaded surfaces are rectangles which make up the lateral surface area
• the height of a right prism is the perpendicular height or perpendicular distance
between the bases
• the lateral surface area of a right cylinder is a rolled-up rectangle with dimensions
2πr and h.
Total surface area = 2 × area of base + sum of areas of all lateral surfaces
Surface area is two-dimensional and if:
• one dimension is multiplied by a constant factor of k, then surface area increases
by a factor of k. 6x²
• if both dimensions are multiplied by a constant factor of k, then the surface area
x
2
3x²
x²
2
increases by a factor of k . 2x 6x²
x 3x
2x²
Volume = area of base × perpendicular height 3x²
Volume = 6x³
2x
Volume is three-dimensional and if: 3x
• one dimension is multiplied by a constant Volume = 6x³
factor of k, then the volume increases by a
REMEMBER
factor of k 51x²
The dimensions of a rectangle • two dimensions are multiplied by a constant 3 x
18x²
are its length and breadth. factor of k, then the volume increases by a 51x² 27x²
The dimensions of a right
factor of k2 3x 6x
prism are its length, breadth
and height. • three dimensions are multiplied by a 18x² 9x
27x²
The dimensions of a right
constant factor of k, then the volume
6x Volume = 162x³
cylinder are its radius and increases by a factor of k3. 9x
height.
Each dimension has been multiplied by a constant factor of three with
Volume this
= 162 x³ effect:
• each surface area has increased by a factor of 9, which is 3 2
• the volume has increased by a factor of 27, which is 33.
176 Topic 8 Measurement
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Spheres and hemispheres sphere
4 πr 3. KEY WORDS
• Volume of sphere = __
3
d hemisphere – half a sphere
• Total surface area sphere = 4πr 2. r
• Volume of hemisphere = __23πr 3.
• Total surface area of hemisphere = 3πr 2.
• The area of the flat section = πr 2. r radius
• The area of the curved section = 2πr 2.
Right pyramids and right cones
• Only one base
• The perpendicular height measures the distance between the vertex of the pyramid
or cone and the centre of its base.
• The slant height runs from the side of the base to the top of the pyramid or cone
and is used to determine surface area.
right cone
• The base of a right pyramid is a polygon, the base of a
right cone is a circle.
s
• Volume = __13area of base × perpendicular height h
• Total surface area = sum of areas of all surfaces
• Area of pyramid = area of base + area of lateral surfaces
• Area of cone = πr2 + πr s
Pythagoras’ Theorem is used in right cones and
right pyramids
• A right-angled triangle is always formed
between h, s and r or h, s and x.
s h
• h is the perpendicular height to the base. h s
• s is the slant height.
r
• r is the radius of the circular base of a x
right cone.
• x is the shortest distance between the
perpendicular height and the edge of the base of a pyramid.
Unit 1 Revision of Grade 10 measurement 177
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Summary of formulae:
(h represents the perpendicular height and s the slant height)
Solid Shape Volume Total surface area
Right prism Area of base × h area of base + areas of lateral surfaces
s h
Right pyramid 1 (Area of base) × h
__ area of base + areas of all triangles
3
x
Right cylinder πr2h 2πr2 + 2πrh
Right cone h s 1 πr2h
__ πr2 + πrs
3
r
Sphere 4 πr3
__ 4πr2
3
Hemisphere r radius 2 πr3
__ 3πr2 (includes flat circular surface)
3
178 Topic 8 Measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 178 2012/07/02 2:22 PM
Pyramids, right prisms and hemispheres are often combined to form complex shapes.
WORKED EXAMPLE 1 K REMEMBER
90
1 AB = 29 units, D is the midpoint Area rectangle = l × b
A
H b⊥h
of CE, DL = 21 units and 29 Area △ = _____
2
AK = 90 units. Area circle = πr2
B J
1.1 Determine the area of the G Area semicircle = π r2
___
2
base of the solid in terms of π. Circumference circle = 2πr
C E
1.2 Determine the total surface D Circumference semicircle:
21
area of the figure. • πr + 2r, including diameter
L • πr, not including diameter
1.3 Determine the volume of the figure.
LSA is lateral surface area.
LSA = perimeter of base ×
2 If the solid is enlarged so that AB = 58 units, DL = 42 units and ⊥h
AK = 180 units: TSA is total surface area.
2.1 state the constant factor by which each dimension is multiplied Effects of multiplying each
2.2 state the factor by which each surface area increases dimension by a constant
2.3 determine the total surface area of the enlarged solid factor of k:
2.4 state the factor by which the volume increases • TSA increases by a factor
of k2.
2.5 determine the volume of the enlarged solid.
• Volume increases by a
factor of k3.
SOLUTION
BG = CE = 42 units | CD = DL, radii
_______ ____
√
AF = 29 −21 = √400 = 20 | Pythagoras' theorem
2 2
1 1.1 Calculations for the area of the base:
1 ( 42 )( 20 ) = 420
Area of triangle = __
2
Area rectangle = (20)(42) = 840
1 πr2 = __
Area semi-circle = __ 1 π(21)2 = 220,5π
2 2
Area of base = 420 + 840 + 220,5π = 1 260 + 220,5π
1.2 Area of base = 1 260 + 220,5π
Lateral surface area = 2(29)(90) + 2(20)(90) + π(21)(90)
= 8 820 + 1 890π
Total surface area = 2(1 260 + 220,5π) + 8 820 + 1 890π
= 11 340 + 2 331π
= 18 663,05 units2
1.3 Volume = (1 260 + 220,5π) × 90 = 175 744,91 units3
2 2.1 Each dimension is multiplied by a constant factor of 2.
2.2 Each surface area increases by a factor of 22 = 4.
2.3 TSA = 4(18 663,05) = 74 652,20 units2
2.4 The volume increases by a factor of 23 = 8.
2.5 The volume = 8(175 744,91) = 11 247 675,24 units3
Unit 1 Revision of Grade 10 measurement 179
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WORKED EXAMPLE 2
Consider the cylindrical flask in the figure with its
hemispherical top. The radius of the cylinder is 6 cm 6
and the height of the flask is 25 cm. Give all answers
correct to two decimal places. 25
1 Determine the volume of the flask in cm3 and in litres.
2 Determine the total surface area of the flask.
3 If the dimensions are trebled:
3.1 state the factor of enlargement of the total surface
area and then determine the surface area
3.2 state the factor of enlargement of the volume and
then determine the volume.
SOLUTION
1 The height of the cylinder without the hemispherical top is 19 cm.
2 πr3 = π( 6 )2(19) + __
V = πr2h + __ 2 π( 6 )3
3 3
= 684π + 144π = 828π = 2 601,24 cm3 = 2,60 ℓ
2 TSA = πr2 + 2πrh + 2πr2 = 3πr2 + 2πrh
= 3π( 36 ) + 2π( 6 )( 19 ) = 336π = 1 055,58 cm2
3 3.1 The factor of enlargement is 32 = 9 and
A = 9( 336π ) = 3 024π = 9 500,18 cm2.
3.2 The factor of enlargement is 33 = 27 and
V = 27( 828π ) = 22 356π = 70 233,45 cm3 = 70,23 ℓ
WORKED EXAMPLE 3
H = 49 cm, h = 12 cm and r = 9 cm
Give all answers correct to two decimal places. 40 cm s
1 Calculate the volume and the surface area h 90 cm
r
of the solid. Give the volume in cm3 and
in litres.
2 If the dimensions are increased by a factor H
of 1,5 determine the volume and surface
area of the enlarged solid.
180 Topic 8 Measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 180 2012/07/02 2:22 PM
SOLUTION REMEMBER
1 Add the volumes of each of the three shapes. Volume conversions:
1 πr 2h = __
1 π( 9 )2( 12 ) = 324π 1 litre = 1 dm3 = 1 000 cm3
Vcone = __
3 3 If each of the dimensions is
Vcylinder = πr 2h = π( 9 )2( 40 ) = 3 240π increased by a factor of k:
2 πr 3 = __
Vhemisphere = __ 2 π( 9 )3 = 486π • the total surface area
3 3 increases by a factor of k2
V = 4 050π = 12 723,45cm3 = 12,72 ℓ
_______ ____ • the volume increases by a
2 s = √122 + 92 = √225 = 15 cm | Pythagoras’ Theorem factor of k3.
TSA = 2πr2 + 2πrH + πrs If each of the dimensions is
= 2π(9)2 + 2π(9)(40) + π(9)(15) increased by a factor of 1,5:
= 162π + 720π + 135π • the total surface area
increases by a factor of 1,52
= 1 017π cm2 = 3 195 cm2
• the volume increases by a
Volume = ( 1,5 )3( 4 050π ) = 13 668,75π = 42 941,64 cm3 = 42,94 ℓ factor of 1,53.
TSA = ( 1,5 )2( 1 017π ) = 2 288,25π = 7 188,75 cm2
WORKED EXAMPLE 4 A
A rectangular pyramid is
combined with a rectangular
right prism. The pyramid has 15
I
a height of 15 m, which is
indicated by AD. The prism
B D
has a height of 4 m and the H
G
length and breadth of both 40 J C
the prism and the pyramid 4
are 40 m and 16 m respectively. E F
16
1 1.1 Determine the volume of the solid.
1.2 State the length of BD and then determine the slant height AB.
1.3 Determine the area of the triangular face AIJ.
1.4 State the length of DC and then determine the slant height AC.
1.5 Determine the area of the triangular face AJG.
1.6 Determine the total surface area of the solid.
2 If each dimension is multiplied by a constant factor of five:
2.1 determine the factor by which the volume is increased
2.2 determine the volume of the enlarged solid
2.3 determine the factor by which the total surface area is increased
2.4 determine the total surface area of the enlarged solid.
Unit 1 Revision of Grade 10 measurement 181
PLT MATHS LB 11 7th pgs (Real Book).indb 181 2012/07/02 2:22 PM
REMEMBER SOLUTION
Complex solids are formed by 1 1.1 Volume of solid = Volume of prism + volume of pyramid
joining two or more separate
V = ( 40 )( 16 )( 4 ) + __
1 ( 40 )( 16 )( 15 ) A
solids. The surface area of a 3
complex solid is always less = 2 560 + 3 200
than the total surface areas of = 5 760 m3
the separate solids. 17
1.2 BD = __
1 JG = 8 m
2
AB2 = 152 + 82 = 289 | Pythagoras’ Theorem I 40
J
____
AB = √ 289 = 17 m
A
1.3 Area △AIJ = b × h = 40
_____ × 17 = 340 m2
_______
2 2
1.4 DC = __
1 IJ = 20 m
2
25
AC2 = 152 + 202 = 62 | Pythagoras’ Theorem
____
AB = √ 625 = 25 m J G
16
1.5 Area △AJG = b × × 25 = 200 m2
_______
h = 16
_____
2 2
1.6 TSA = ( 40 )( 16 ) + 2( 40 )( 4 ) + 2( 16 )( 4 ) + 2( 340 ) + 2( 200 )
= 640 + 320 + 128 + 680 + 400
= 2 168 m2
2 2.1 The volume is increased by a factor of 53 = 125.
2.2 V = 125( 5 760 ) = 720 000 m3
2.3 The TSA is increased by a factor of 52 = 25.
2.4 TSA = 25( 2 168 ) = 54 200 m2
EXERCISE 1
Determine the volume and the surface area for each figure.
1 2 5
29 cm
21 cm
12
8 cm
3 4
25 cm
7m
48 m
65 cm
48 m
182 Topic 8 Measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 182 2012/07/02 2:22 PM
5 6
REMEMBER
13 12
Vhemisphere = __23πr3
15 cm Vcylinder = πr2h
39 cm
Vcone = __13πr2h
36 TSAhemisphere = 3πr 2
100 cm
39 TSAcylinder = 2πr 2 + 2πrh
TSAcone = πr 2 + πrs
T
7 8
25 m
J N
5m
6m
K OM = 24 units
O M
24 m OL = 32 units
24 m
OJ = 7 units
6m
LR = 100 units
R
24 m
L
9 10
77
r = 24
The solid above is a regular
octahedron. Each edge is 10 units.
11
15 cm
EXERCISE 2
Where necessary, round off answers to two decimal places.
1. A right cone has a volume of 3 392,92 units3 and a radius of 9 units.
1.1 Write down the formula for the volume of a cone.
1.2 Determine the height of the cone.
1.3 Determine the slant height of the cone.
1.4 Determine the surface area of the cone.
Unit 1 Revision of Grade 10 measurement 183
PLT MATHS LB 11 7th pgs (Real Book).indb 183 2012/07/02 2:22 PM
1.5 If the radius doubles, but the height remains the same:
REMEMBER 1.5.1 describe the effect this will have on the volume of the cone
1.5.2 determine the volume of the cone.
Remember that the surface
1.6 If the height trebles, but the radius remains the same:
area of any complex solid
is less than the sum of the
1.6.1 describe the effect this will have on the volume of the cone
surface areas of the shapes 1.6.2 determine the volume of the cone.
which combined to form it. 1.7 If all the dimensions are halved, determine:
Do not add the areas of any 1.7.1 the volume of the cone
surfaces at which the two 1.7.2 the total surface area of the cone.
solids are joined.
• A hemisphere and a cone 2. A cylinder has a volume of 33 238,05 units3 and a radius of 23 units.
that are joined by their flat 2.1 Write down the formula for the volume of a cylinder.
circular surfaces will each 2.2 Determine the height of the cylinder.
lose that area from the 2.3 Determine the total surface area of the cylinder.
total surface area of the 2.4 If the height of the cylinder is multiplied by __1 , but the
4
complex solid.
radius remains the same:
• A cylinder and a cone 2.4.1 say how this will affect the volume
that are joined by their
flat circular surfaces will
2.4.2 determine the volume.
each lose that circular 2.5 If the radius halves, but the height remains constant:
surface area. 2.5.1 say how this will affect the volume
• A square-based pyramid 2.5.2 determine the volume.
and a right prism will be 2.6 If all the dimensions are multiplied by __ 1:
3
joined by the base of the 2.6.1 say how this will effect the volume
pyramid so they will each
2.6.2 determine the volume
lose the area of the square
2.6.3 say how this will effect the total surface area
surface which joins them.
2.6.4 determine the total surface area.
3 A cylinder has a total surface area of 2 777,17 units2 and a radius of 17 units.
3.1 Write down the formula for the total surface area of a cylinder.
3.2 Determine the height of the cylinder.
3.3 Determine the volume of the cylinder.
4. A sphere has a volume of 333 038,14 units3.
4.1 Write down a formula for the volume of a sphere.
4.2 Determine the radius of the sphere.
4.3 Determine the total surface area of the sphere.
4.4 If the radius of the sphere is doubled:
4.4.1 state the factor by which the volume
is increased
4.4.2 determine the volume
4.4.3 state the factor by which the total surface area is increased
4.4.4 determine the total surface area.
184 Topic 8 Measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 184 2012/07/02 2:22 PM
Revision
1 The base of a triangular right prism is an equilateral triangle.
Each side of the equilateral triangle is 15 m and the height of
the prism is 25 m.
E
A
F
D
25 m
B 15 m C
1.1 Determine the volume of the prism. (6)
1.2 Determine the surface area of the prism. (4)
1.3 If the dimensions are multiplied by a constant factor of 4:
1.3.1 determine the volume (3)
1.3.2 determine the surface area. (3)
[16]
2 Consider the figure which indicates a 10 m long section of a concrete pipe.
The outer radius is 100 cm and the inner radius is 90 cm.
2.1 Calculate how thick the walls of the pipe are. (2)
2.2 Express both radii in metres. (2)
2.3 Determine the volume of concrete in the pipe, correct to
three decimal places. (6)
2.4 What is the total surface area of the pipe, correct to three
decimal places? (6)
2.5 If the dimensions are doubled:
2.5.1 state the factor of enlargement of the volume (2)
2.5.2 determine the volume (2)
2.5.3 state the factor of enlargement of the total surface area (2)
2.5.4 determine the total surface area. (2)
[24]
3 Consider the figure of a right cone which has a height of 35 m and
a slant height of 37 m.
3.1 Determine the radius of the circle. (3)
3.2 Determine the volume of the cone. (3)
3.3 Determine the surface area of the cone. (5)
3.4 If the height is multiplied by five but the radius remains the same,
state the factor by which the volume will be enlarged. (2)
3.5 If r is multiplied by three but the height remains the same, state
the factor of enlargement of the volume. (2)
[15]
185
PLT MATHS LB 11 7th pgs (Real Book).indb 185 2012/07/02 2:22 PM
TOPIC 8: REVISION CONTINUED
4 The rectangular right pyramid has a base with dimensions 96 units by 42 units.
The height of the pyramid is 20 units. REMEMBER
D Vhemisphere = __23πr3
Vcylinder = πr2h
Vcone = __13πr2h
20
H C TSAhemisphere = 3πr 2
F G 42 TSAcylinder = 2πr 2 + 2πrh
TSAcone = πr 2 + πrs
A E B
96
4.1 Determine the volume of the pyramid. (4)
4.2 Determine the slant heights DE and DG. (6)
4.3 Determine the lateral surface area of the pyramid. (5)
4.4 Determine the total surface area. (4)
4.5 Determine the factor of enlargement of the volume if the shorter side of
42 units is doubled, but the other dimensions remain constant. (2)
[21]
5 Determine the volume and surface area of a sphere with a radius of 24 cm. (6)
6 Determine the volume and surface area of a hemisphere with a diameter
of 40 cm. (6)
7 Consider a sphere with a volume of 212 174,79 units3
7.1 Write down the formula for the volume of a sphere. (1)
7.2 Use the formula to determine the radius. (3)
7.3 Determine the surface area of the sphere. (3)
7.4 Determine the volume of the sphere if the radius is divided by five. (3)
[10]
8
120 m
28 m
The diameters of the outer and inner shells of a semi-circular right
prism are 30 m and 28 m respectively. The length of the prism is 120 m.
8.1 Determine the area of the base in terms of π. (4)
8.2 Determine the volume of the prism. (3)
8.3 Determine the total surface area of the prism. (4)
8.4 If the dimensions are halved, determine the total surface area of
the prism. (4)
[15]
186
PLT MATHS LB 11 7th pgs (Real Book).indb 186 2012/07/02 2:22 PM
9 Consider the right cone which has a volume of 1 005,31 cm3 and
a height of 15 cm.
9.1 Write down the formula for the volume of the cone. (1)
9.2 Determine the radius of the base. (3)
9.3 Determine the slant height of the cone. (3)
9.4 Determine the lateral surface area in terms of π. (3)
9.5 Determine the total surface area. (4)
9.6 Determine the volume of the cone if the dimensions are multiplied
by a constant factor of 2,5. (3)
[17]
10 Consider the square-based right pyramid. The volume or the pyramid
is 4 320 units3 and the height is 40 units.
10.1 Write down the formula for the volume of a pyramid. (1)
10.2 Determine the dimensions of the base. (3)
10.3 Determine the slant height of the pyramid. (3)
10.4 Determine the lateral surface area of the pyramid. (4)
10.5 Determine the total surface area of the pyramid. (3)
10.6 Determine the volume of the pyramid if its height is doubled. (3)
[17]
11 A cylinder has a volume of 9 424,78 units3 and a height of 30 units.
11.1 Write down the formula for the volume of a cylinder. (1)
11.2 Determine the radius of the cylinder. (3)
11.3 Determine the total surface area of the cylinder. (4)
2 , determine the volume of the cylinder.
11.4 If the radius is multiplied by __ (3)
3
11.5 If all the dimensions are halved, determine the total surface area. (3)
[14]
187
PLT MATHS LB 11 7th pgs (Real Book).indb 187 2012/07/02 2:22 PM
2 9 Euclidean geometry and measurement
Unit 1: Geometry revision
Summary
Q
KEY WORDS y
P a b T
congruent – identical in all
respects d c x z
included angle – angle e f x + y + z = 180° | ∠ sum △
between two known sides
V h g W
hypotenuse – the side
opposite the right angle in a R
right-angled triangle
a=c | vertically opposite ∠s y
a + b = 180° | straight ∠
a=e | corresponding ∠s, PT ∥ VW
REMEMBER z x
d=f | alternate ∠s, PT ∥ VW
• Supplementary angles all d + e = 180° | co-interior ∠s, PT ∥ VW x = y + z | exterior ∠ of △
add up to 180°. a + b + c + d = 360° | revolution
• Complementary angles all
A R TOPIC
add up to 90°
• In a revolution, angles add c p
up to 360° b q
B a C Q r P
REMEMBER
c2 = a2 + b2 | Pythagoras’ Theorem If p2 = q2 + r2 ⇒ P = 90°
To prove two lines parallel, | converse Pythagoras’ Theorem
show one of the following:
• alternate angles are equal A D
• corresponding angles are
equal
• co-interior angles are
supplementary. B C F E
△ABC ≡ △DEF (SSS)
Congruent: SSS – three pairs equal sides
G K P R S T A D
y y
x x x x
H J L M Q U B C F E
△GHJ ≡ △KML (SAS) △PQR ≡ △TUS (RHS) △ABC ≡ △DEF (AAS)
Congruent: SAS – two pairs equal sides Congruent: RHS – right angle, Congruent: AAS – two sets equal angles
and included angle hypotenuse, side and a pair of corresponding sides equal
188 Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 188 2012/07/02 2:22 PM
Quadrilateral properties
KEY WORDS
A parallelogram is a quadrilateral with both pairs of opposite sides parallel.
diagonal – a straight line
• both pairs of opposite sides are equal
joining opposite vertices
• one pair of opposite sides is both equal and parallel bisect – to cut or divide into
• two pairs of opposite angles are equal equal parts
vertex – point where two
• both diagonals bisect each other straight lines meet to form
• area = base × perpendicular height an angle
perpendicular – at right
A rectangle is a parallelogram with a right angle. angles or at 90°
symmetry line – a fold line
• a parallelogram with a right angle
that makes the shapes the
• a parallelogram with equal diagonals same on both sides
• area = length × breadth
A rhombus is a parallelogram with a pair of adjacent sides equal.
• all four sides are equal d1
• diagonals bisect each other at 90°
d2
• both diagonals bisect the corner angles
d1 × d2
• area = base × perpendicular height or area = _______
2
A square is a rhombus with a right angle.
• a rectangle with a pair of adjacent sides equal
• equal diagonals bisect each other at 90°
• both diagonals bisect corner angles into 45° and 45°
d × d2
_______
• area = side × side or area = 1 2 b1 REMEMBER
A trapezium is a quadrilateral with one pair of opposite • A rectangle has all
sides parallel. the properties of a
h
parallelogram as well as its
• an isosceles trapezium has equal sides which are not parallel own special features.
• area = __12 h(b1 + b2) where h is the distance between the b2 • A square has all
parallel sides, b1 and b2 the properties of a
parallelogram, a rectangle
and a rhombus.
A kite is a quadrilateral with two pairs of adjacent sides equal.
• A rhombus has all
• one diagonal bisects the other diagonal at 90° the properties of a
parallelogram.
• one of the diagonals is a symmetry line
d × d2
_______
• area = 1 2
where d1 and d2 are diagonals
Unit 1 Geometry revision 189
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Unit 2: Circles, perpendicular lines through
the centre, chords and midpoints
A theorem is a formal proof of a geometric fact. You do not need to prove a
KEY WORDS theorem every time you use it, but you can be asked to prove a theorem as a theory
question. Rather than learning each theorem by heart, make sure that you know the
theorem – formal proof of a
construction and the method.
geometric statement
construction – a line which
needs to be added to a sketch
chord – line with both Theorem 1
endpoints on the circle The line drawn from the centre of a circle,
converse – reversing the logic perpendicular to a chord, bisects the chord.
and proving a theorem in
reverse Given: Circle with centre O and chord PQ.
⊥ – symbol for perpendicular OR ⊥ PQ, with R on PQ. O
radius – line from centre of Required to prove: PR = RQ
circle to circumference Construction: OP and OQ 1 2
P Q
R
Proof:
In △OPR and △OQR:
^ =R
1) R ^ = 90° | OR ⊥ PQ
1 2
2) OP = OQ | radii
3) OR is common
△OPR ≡ △OQR | RHS
PR = RQ | △OPR ≡ △OQR
⇒ OR bisects PQ.
Most theorems have a converse. In a converse theorem, the logic is reversed.
Converse (Theorem 1)
The line drawn from the centre of a circle to the midpoint of a chord will be
perpendicular to the chord.
Although you apply the converse theorems in application questions, you will not
prove any of the converse theorems.
WORKED EXAMPLE
OE ⊥ AB, CF = FD, OE = 63 cm, FE = 3 cm, AB = (2x − 18) cm and CD = 2x cm.
Determine the length of the radius.
SOLUTION
OF ⊥ CD | line from centre to midpoint chord
EB = x − 9 | OE ⊥ AB O
FD = x | CF = FD
2 2
OD = 60 + x 2 | Pythagoras in △OFD F
C D
OB2 = 632 + (x − 9)2 | Pythagoras in △OEB
3 600 + x2 = 3 969 + x2 − 18x + 81 | Radii A E B
______
18x = 450 ⇒ x = 25 and radius = √4 225 = 65 cm
190 Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 190 2012/07/02 2:22 PM
Theorem 2
The perpendicular bisector of a chord passes through the centre of the circle.
Given: A circle with chord AB. Q
T is the midpoint of AB and QT ⊥ AB.
Required to prove: QT passes through the centre of the
circle. R
Construction: RA and RB where R is any point on QT.
Proof:
In △RAT and △RBT
^ =T
1) T ^ = 90° | QT ⊥ AB 1 2
1 2 A T B
2) AT = BT | given
3) RT is common
⇒ △RAT ≡ △RBT | SAS
RA = RB | △RAT ≡ △RBT
This proves that any point on QT is equidistant from A and B. The centre of the circle KEY WORDS
is also equidistant from A and B, so QT passes through the centre of the circle.
equidistant – the same
distance
A R intersect – to meet or cross
WORKED EXAMPLE at a point
A circle with radius 50 units has chords AB = 60 units
and CD = 28 units. TU and RS are the perpendicular T
P
bisectors of AB and CD respectively. Determine the O
shortest distance of each chord from the centre of
the circle. U
B
Q
C D
SOLUTION S
TU and RS intersect at O, so O is the centre of
A R
the circle. Join OB and OD.
AP = PB = 30 | OP ⊥ AB 30
OP2 = 502 − 302 = 1 600 | Pythagoras in △OPB T
P
OP = 40 units O
CQ = QD = 14 | OQ ⊥ CD 30
OQ2 = 502 − 142 = 2 304 | Pythagoras in △OQD 50 50 U
OQ = 48 units
B
14 Q 14
AB is 40 units from the centre and CD is C D
48 units from the centre of the circle. S
Reminders about Pythagoras’ Theorem
In a right-angled triangle, the square on the hypotenuse is equal to the sum of the
squares on the other two sides. If you know all three sides of a triangle but do not
know whether the triangle is right-angled or not, use the converse of Pythagoras’
Theorem. If the square on the longer side is equal to the sum of the squares on the
two shorter sides, then the triangle is right-angled and the right angle will lie between
the two shorter sides.
Unit 2 Circles, perpendicular lines through the centre, chords and midpoints 191
PLT MATHS LB 11 7th pgs (Real Book).indb 191 2012/07/02 2:22 PM
A triangle with sides 20, 21 and 29 is right-angled because 202 + 212 = 841 and
292 = 841, but a triangle with sides 6, 7 and 8 is not right-angled because
62 + 72 = 85 but 82 = 64 and 85 ≠ 64.
REMEMBER WORKED EXAMPLES
D
To prove two lines parallel, A, B, C and D are points on the circumference F
show one of the following: of the circle with centre O.
• alternate angles are equal AB = 40 units, OE = 15 units and OF = 7 units. O C
• corresponding angles are 1 Determine the radius of the circle.
equal
2 Determine the length of the chord DC.
• co-interior angles are
A E B
supplementary.
SOLUTIONS
D
1 AE = EB = 20 units | OE ⊥ AB
OA2 = 202 + 152 = 625 | Pythagoras in △AOE F
OA = 25 units
The radius is 25 units O C
2 DF2 = 252−72 = 576 | Pythagoras in ∠DOF
DF = 24 units = FC | OF ⊥ DC
A E B
DC = 48 units
EXERCISE 1
O is the centre of the circle. Sketches are not drawn to scale. Give reasons unless you
are told otherwise.
D
1 DC is perpendicular to AB and cuts AB at E.
DE = 18 units and AB = 24 units.
1.1 Determine, with reasons, the length of
AE.
O
1.2 If the radius of the circle is x units,
express OE
in terms of x. A B
E
1.3 Calculate, with reasons, the value of x.
C
2 2.1 If RS = 48 units, OP = 7 units and
TU = 40 units. R
2.1.1 Determine the radius of the circle. P
2.1.2 Determine the length of OV.
2.2 If OP = 16 units, OV = 25 units and the circle has a O
S
radius of 65 units, show that the sum of the chords
is 246 units. V
T U
192 Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 192 2012/07/02 2:23 PM
3 OE ⊥ AB, EB = 7 units, OE = 24 units and G
D
DF = FC = 15 units.
Determine, with reasons: F
3.1 AO 15 C
3.2 FO O
3.3 FG 24
A E 7 B
4 A circle with radius 65 units has two chords, PQ T
KEY WORDS
and RT. PQ is 104 units in length and RT is
120 units in length. P concentric – circles that share
4.1 Copy the sketch and show how you can the same centre
locate the centre of the circle by adding two
construction lines, one passing through chord
PQ and the other through chord TR. Q R
Name the centre of the circle O.
4.2 Determine the distance between the centre of the circle and:
4.2.1 chord PQ
4.2.2 chord TR.
A REMEMBER
5 AE = x units, CF = x − 4 units, EO = 7 units and C
EF = 22 units. EOF is a straight line. b1
5.1 Prove that AB ∥ CD.
5.2 State the length of OF. E F
5.3 Determine AO in terms of x. O h
5.4 Determine CO in terms of x.
5.5 Why is AO = CO? D b2
5.6 Solve for x. Area of trapezium
B
5.7 State the lengths of AB and CD. = __
1 h(b + b )
2 1 2
5.8 Determine the area of ABDC.
6 O is the centre of two concentric circles. The radii of these
circles are 65 units and 34 units. OG ⊥ AB and CD = 60 units.
6.1 Determine the length of:
6.1.1 CG
6.1.2 OG
6.1.3 AB
6.1.4 AC
6.2 If OB^ F = θ:
6.2.1 determine ^E2 in terms of θ. O
6.2.2 prove that EH ∥ FB. 3 H
6.3 Determine the length of FB, in simplified surd form. 1
6.4 Join AO and AF. A C G 2 D 0 B
6.4.1 Prove that AF = BF
E
6.4.2 What type of quadrilateral is AFBO?
6.4.3 Justify your answer to 6.4.1 in two different ways.
6.4.4 Calculate the area of AFBO.
F
Unit 2 Circles, perpendicular lines through the centre, chords and midpoints 193
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Unit 3: Angle at centre theorem
and cyclic quadrilaterals
Theorem 3
The angle subtended by an arc at the centre of the circle is double the size of the
KEY WORDS angle subtended by the same arc at any point on the circumference of the circle.
subtended by an arc – C C
formed by arc
1 2
arc – a part of the 1 2 A
circumference of a circle B
reflex angle – angle greater
O
1 2
1 2 1 2 C
than 180° 1
O O 2
A
A B
B
Figure 1 Figure 2 Figure 3
Given: A, B and C are three points on the circle with centre O.
Construction: CO
^ B = 2AC
Required to prove: In Figures 1 and 3 prove AO ^ B, but in Figure 2 prove reflex
AO^ B = 2AC^ B.
Proof:
Note that in Figure
^ B refers to the Figure 1 and Figure 2:
2 AO ^ = x and C ^ =y
Let C 1 2
reflex angle. ^ = x and B ^=y
⇒A | radii
^
AOC = 180° − 2x | ∠ sum △AOC
BO^ C = 180° − 2y | ∠ sum △BOC
^ ^ ^ |
AOB + O1 + O2 = 360° sum of ∠s around a point
∴ AO ^ B + 180° −2x + 180° – 2y = 360°
∴ A^ OB = 2(x + y) = 2A ^CB | sum of ∠s around a point
^ ^
Figure 3: Let C1 = x and C2 = y
^ = x and B
A ^=x+y | radii
^
AOC = 180° − 2x | ∠ sum of △AOC
^
BOC = 180° − (2x + 2y) | ∠ sum of △BOC
AO^ B = AO^ C − BO^ C = 2y = 2AC
^B
V
The angle subtended by an arc at the centre of the circle is double the size of the angle
subtended by the same arc at any point on the circumference of the circle.
This theorem does not have a converse.
O Theorem 3 proves that the angle subtended by an arc at the centre of the circle
P
is twice the size of the angle subtended by the same arc at any point on the
circumference of the circle.
If the angle at the circumference of a circle is half the size of an angle subtended by
R T S the same arc at a point inside the circle, it is not necessarily true that this point will be
the centre of the circle.
In the figure alongside, both O and P lie on the circumference of the smaller circle
194 Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 194 2012/07/02 2:23 PM
which passes through the centre of the larger circle and intersects the larger circle at R
and S.
^ =P
O ^ | ∠s on chord RS of the smaller circle
^ = __
V 1O ^ | ∠ at centre of larger circle
2
V = __
1P ^ ^ =P
|O ^)
2
But, P is not the centre of the circle.
Although it is always true that the angle subtended by an arc at the centre of a circle is
twice the angle subtended by the same arc on the circumference, it is not necessarily
true that the converse applies.
REMEMBER
≡ is the symbol for
Theorem 4 congruent.
In Theorem 3 we proved that the angle subtended by an arc at the centre of the Congruent means identical in
all respects.
circle is double the size of the angle subtended by the same arc at any point on the
circumference of the circle.
Using this result we can prove that the angles subtended by a chord of the circle, on
the same side of the chord, are equal. G
x
F F
O O
x x
G
2x 2x
x
A B A B
F
Given: A, F, G and B are points on the circle with centre O.
Construction: Join AO and OB.
^=G
Required to prove: F ^
Proof:
AO^ B = 2F
^ | ∠ at centre
AO^ B = 2G
^ | ∠ at centre
^
F=G ^
The corollary of Theorem 4 is that equal chords U
subtend equal angles. KEY WORDS
In △OPQ and △OSR: x
corollary – deduction based
1) OP = OS | radii
T on the result of a theorem
2) OQ = OR | radii
O
3) PQ = RS | given x
⇒ △OPQ ≡ △OSR | SSS 2x 2x
P
PO^ Q = RO
^ S | △OPQ ≡ △OSR
PO^ Q = 2U ^ S = 2T
^ and RO ^ | ∠ at centre S
^
U=T ^
Q R
Unit 3 Angle at centre theorem and cyclic quadrilaterals 195
PLT MATHS LB 11 7th pgs (Real Book).indb 195 2012/07/02 2:23 PM
WORKED EXAMPLE
O is the centre of the circle and A, B, C, D
and E lie on the circle.
^ D = 21° and AB
BA ^ E = 60°.
E
Determine, with reasons, v, w, x, y and z.
x v
A O
SOLUTION
w
21°
v = 21° | ∠s on chord BD
w = 42° | ∠ at centre
O
60 z
x = 318° | revolution
y
y = 159° | ∠ at centre B D
C
z = 60° | ∠s on chord AE
EXERCISE 2
O is the centre of the circle in each figure. In each case determine the value of the
lower case letters a, b, c, … . Give reasons for each statement.
1 N 2 3
P
R Q
c
32° f e a b
Ob g O
O
a j i V 70° 22°
cd e
h g
f d
32° a b R
J c
M M T N
K T
4 5
M E
21° u
A
27° D
y
t x v 40°
O
d
z
f e b B w
N a R c Q C
g
P
196 Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 196 2012/07/02 2:23 PM
Theorem 5
KEY WORDS
The opposite angles of a cyclic quadrilateral are supplementary.
cyclic quadrilateral – a
F E four-sided figure with all
F four vertices lying on the
circumference of a circle
D
E supplementary – add up to
180°
G 2
G 2 revolution – the sum of the
O 1 1
angles around a point
O
produce – lengthen a
straight line; for example, AB
produced to C means that
ABC is a straight line
D exterior angle – outside
Figure 1 Figure 2 angle of a polygon formed
by a line which has been
Given: D, E, F and G are 4 points on the circle with centre O. extended
Construction: EO and GO. interior angle – inside angle
^ +F
Required to prove: D ^ = 180° and DE ^ F = 180°
^ F + DG
Proof: (The proof is the same for both figures.)
Let D^ =x
^
O2 = 2x | ∠ at centre
^ = 360° − 2x
O | revolution
1
^
F = 180° − x | ∠ at centre
^ ^
D + F = 180°
FE ^ F = 180° | ∠ sum quad
^ D + DG
Converse (Theorem 5)
A quadrilateral is cyclic if its opposite angles are supplementary.
Corollary: The exterior angle of a cyclic quadrilateral is equal to the interior
opposite angle.
REMEMBER
WORKED EXAMPLE
A D
• Angles in any triangle add
A, B, C and D lie on the circle and BC is z x
up to 180°.
produced to E. B^ = 85° and DC ^ E = 100°.
• Angles in any quadrilateral
Determine, with reasons, x, y and z. add up to 360°.
100° • Angles around any point
E add up to 360°.
y
85° C
B • Exterior angle of triangle
SOLUTIONS = sum of two interior
opposite angles.
x + 85° = 180° | opposite ∠s cyclic quadrilateral
• Exterior angle of cyclic
⇒ x = 95° quadrilateral = one interior
y = 80° | straight ∠ opposite angle.
z + 80° = 180° | opposite ∠s cyclic quadrilateral
⇒ z = 100°
or z = 100° | exterior ∠ cyclic quadrilateral
Unit 3 Angle at centre theorem and cyclic quadrilaterals 197
PLT MATHS LB 11 7th pgs (Real Book).indb 197 2012/07/02 2:23 PM
There are three ways to prove that a quadrilateral is cyclic. D
1 If a line subtends equal angles at two different points on the A y
x
same side of the line, then the four points are cyclic.
If x = y, then ABCD is a cyclic quadrilateral.
B C
T
2 If one pair of opposite angles of a quadrilateral is x
supplementary, then the quadrilateral is cyclic.
If x + y = 180°, then TUVW is a cyclic quadrilateral. W
y
U V
3 If the exterior angle of a quadrilateral equals the interior T
opposite angle, then the quadrilateral is cyclic.
y
If x = y, then PQRT is a cyclic quadrilateral.
P
x
V Q R
EXERCISE 3
1 Complete the statements by filling in the missing words:
REMEMBER 1.1 The line drawn from the centre of a circle, perpendicular to a chord, …
1.2 The line drawn from the centre of a circle to the midpoint of a chord …
To prove two lines parallel,
1.3 If PQ is the perpendicular bisector of chord AB, then PQ passes through …
show one of the following:
1.4 If PQ and JK are the perpendicular bisectors of any two non-parallel chords
• alternate angles are equal
on the same circle, then PQ and JK will intersect each other and the centre
• corresponding angles are
equal of that circle will lie on their …
• co-interior angles are 1.5 The angle subtended by a chord at the centre of the circle is …
supplementary 1.6 The angles subtended by a chord in the same segment of the circle …
Four points are concyclic if 1.7 The angle subtended by a diameter on the circumference of the circle is
they lie on the same line. always equal to …
1.8 If a chord subtends a right angle on the circumference of a circle, then the
chord is …
1.9 The opposite angles of a cyclic quadrilateral …
1.10 If the opposite angles of a quadrilateral are supplementary, then …
1.11 If a line subtends equal angles at two points on the same side of itself,
then …
1.12 If a line subtends unequal angles at two points on the same side of itself,
then …
1.13 If the exterior angle of a quadrilateral is equal to the interior opposite
angle, then …
1.14 If the exterior angle of a quadrilateral is not equal to the interior opposite
angle, then …
1.15 Equal chords subtend …
198 Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 198 2012/07/02 2:23 PM
2 A, B, C, D, E and F all lie on the circle. F
DC^ E = 17°, BC
^ E = 86° and AE ^ C = 57° E
u 1
2
2.1 Is AE ∥ BC? Justify your answer. 57°
2.2 Complete the statements: A x D
2.2.1 t = …° (∠s on chord ED) w
2.2.2 FD^ C = FE ^ C (…)
^ y
2.2.3 BAE = … (∠s on chord BE)
2.2.4 AE^ F = ... (∠s on chord AF) 17°
v
2.3 Determine u, v, w, x, y and z in that order. B
86°
C
REMEMBER
WORKED EXAMPLE E Equal chords subtend equal
A 2 angles.
CD is a diameter of the circle with centre O. 1 Conversely, equal angles are
^ E and DG ⊥ AB.
FD bisects CD F 3 subtended by equal chords.
12
1 Prove that DEFG is a cyclic quadrilateral. 2
^ F = x, determine, with reasons, x
2 If ED 1 3 1
C D
three other angles equal to x. G 4 O
3 Prove that EFGD is a kite.
4 If AB = 30 cm and CG = 9 cm,
determine the length of the radius.
5 Determine the length of EC. B
(Leave your answer in simplified
surd form).
SOLUTIONS
1 CE^ D = 90° | ∠ on diameter
^
OGB = 90° | given
DEFG is a cyclic quadrilateral | exterior ∠ = int opp ∠
(Do not use the word ‘cyclic’ in your reason when proving that a quadrilateral
is cyclic.)
2 ^ = x | FD bisects CD
D ^E
1
^
E1 = x | ∠s on chord FG
^ = x | ∠s on chord EF
G 2
3 EF = FG | ED^ F = FD^ G and EF and FG are chords on the circle EFGD
^ = 90° − x
E | adjacent complementary ∠s
2
^
G3 = 90° − x | straight ∠
ED = EG ^ opposite ∠s △DEG
| ^E2 = G 3
DEFG is a kite | two pairs of adjacent sides equal
Unit 3 Angle at centre theorem and cyclic quadrilaterals 199
PLT MATHS LB 11 7th pgs (Real Book).indb 199 2012/07/02 2:23 PM
4 AG = GB = 15 cm | OG ⊥ AB
Let the radius be k, then OG = x − 9
k2 = 152 + (k − 9)2 | Pythagoras in △OGB
k2 = 225 + k2 − 18k + 81
18k = 306° ⇒ k = 17
The radius is 17 cm
5 CD = 34cm | diameter
GD = 25cm = ED | proved in 3
EC2 = 342−252 | Pythagoras in △ECD
____ ___
EC = √531 = 3√ 59 cm
EXERCISE 4
1 A, B, C, D, E, F, G and H all lie on the circumference of the circle. Determine,
with reasons, w, x, y and z.
H
x
E D
w
z
A
88°
79°
B y G
84°
C F
2 Two circles intersect at W and Z. U and V lie on the smaller circle, Y and X lie on
^ = 101° and Y
the larger circle. V ^ = 82°.
Y
Z
82°
U a
f
d
101°
c e
V b
W
X
2.1 Determine, with reasons, a, b, c, d, e and f.
2.2 Is UY ∥ VX? Justify your answer.
2.3 Is UV ∥ YX? Justify your answer.
2.4 Is UVXY a cyclic quadrilateral? Justify your answer.
200 Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 200 2012/07/02 2:23 PM
3 ^ F. BD and AE
A, B, C, D and E lie on the circle. AB = AF and DE bisects CE
produced meet at F and F ^ = x.
A
E
3 1 F
2 x
2
1
D
K
1
2
B
C
3.1 Name two cyclic quadrilaterals.
3.2 Prove that DE = DF.
3.3 ^ C.
Prove that BF bisects AB
3.4 ^
Prove that ABC = C.^
3.5 Is ABCF a cyclic quadrilateral? Justify your answer.
3.6 Is ABKE a cyclic quadrilateral? Justify your answer.
3.7 Is AF ∥ BC? Justify your answer.
4 A, B, C and D are points on the circumference of the circle. COE is a straight line
such that E lies on AB. DC = BC and C^ = x.
1
D
1
A 2 3
1 F
E 2 1
1 2
2
O x
1
2
C
B
4.1 ^ in terms of x.
^ and D
Determine the size of B2 2
4.2 Show that EC bisects BC^ D.
4.3 Prove that DF = FB.
4.4 Show that AEOD is a cyclic quadrilateral.
4.5 Draw ED and prove that EBOD is a kite.
Unit 3 Angle at centre theorem and cyclic quadrilaterals 201
PLT MATHS LB 11 7th pgs (Real Book).indb 201 2012/07/02 2:23 PM
Unit 4: Tangents
Theorem 6
Two tangents drawn to a circle from the same point outside the circle are equal
KEY WORD in length.
Given: Point R lies outside the circle. P
tangent – a straight line
which touches a circle, but RP and RQ are tangents to the circle
does not pass through it with centre O.
Required to prove: RP = RQ
Construction: OP, OQ and OR O
R
Proof:
In △RPO and △RQO
1) OP = OQ | radii
^ ^
2) P = Q = 90° | radius ⊥ tangent
3) OR is common Q
△RPO ≡ △RQO | RHS
⇒ RP = RQ
Theorem 7
The angle between the tangent to a circle and the chord drawn from the point of
REMEMBER contact is equal to the angle subtended by the chord in the alternate segment.
Given: A, B and C are points on the circle with
C
centre O. DA is a tangent to the circle at A.
^D = C
Required to prove: In Figure 1, BA ^ and in Figure B
0 1
A 2, reflex B ^
AD = ^C
B Construction: OA and OB
2
1
Proof:
^D = C
^ D
Figure 1: BA A
• The shortest distance
between a point and a Let ^
C=x Figure 1
line is the perpendicular O = 2x | ∠ at centre
distance.
^ =B
A ^ = 90° − x | ∠ sum isosceles △AOB, radii B
2 1
• A radius is always ^ =x=C
A ^ | Radius ⊥ tangent
1
perpendicular to a tangent 1
O
at the point of contact. C
^D = C
Figure 2: BA ^ 1
2
• A chord divides a circle into
two segments.
^
Let C = x 2
• In the figure above, the red Reflex O^ = 2x | ∠ at centre 1
1
segment is the alternate ^
O2 = 360° − 2x | Sum of ∠s around a point
segment. ^ =B ^ = x − 90° | ∠ sum isosceles △AOB, radii D A
A 2 1 Figure 2
• Complementary angles add ^ = 90°
A | Radius ⊥ tangent
1
up to 90°.
BA^D = x = C
^
Converse (Theorem 7)
If the angle between a line and a chord equals
the angle subtended by the chord in the alternate segment, then the line is a tangent
to the circle.
202 Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 202 2012/07/02 2:23 PM
Circle geometry summary with reasons
O O
A K B A R B
AK = KB | OK ⊥ AB OR ⊥ AB
| line from centre to midpoint chord
E
F A R
D P Q
A 0
0 D
B C
B
^ ^ | Equal chords AB & CD
E=F C
BO^ C = 2A^ | ∠ at centre
^= ^
A D | ∠s on chord BC
^ ^
Reflex O = 2R | ∠ at centre
D D
E G
O E G
^ = 90° | ∠ on diameter
D F
^+G
E ^ = 180°
| opposite ∠s cyclic quadrilateral
D
G
O
E
F
H
^H = D
^ | exterior ∠ cyclic quadrilateral A
GF B
C
^ C = 90°
OB | Radius ⊥ tangent
F D
G
E
1 2
D E A 3
DEFG is a cyclic quadrilateral B
| DE subtends equal ∠s at F & G
C
^ =E
B ^ | tan AB, chord BD
1
B3 = ^
^ D | tan BC, chord BE
Unit 4 Tangents 203
PLT MATHS LB 11 7th pgs (Real Book).indb 203 2012/07/02 2:23 PM
T
M A
x
P
180°– x
B
N V
MNVT is a cylic quadrilateral PA = PB | tangents from common point P
| opposite ∠s supplementary
A
C
D
A B B
D C E
^ B = 90°
AB is the diameter | AC
ABCD is a cyclic quadrilateral
^ E = interior opposite A
| exterior DC ^
WORKED EXAMPLE
PQ and RQ are tangents to the circle at V and
^ = 69° R
W respectively. T
T
1 Prove that OVQW is a kite.
2 Prove that OVQW is a cyclic quadrilateral. 69°
3 Determine, with reasons, the sizes of a, b, W
O g d
c, d, e, f, g and h. P e a
U
SOLUTION f
c
1 OV = OW | radii V
VQ = WQ | tangents from common
point Q b
OVQW is a kite | two pairs of adjacent sides equal h
2 OW^ Q = 90° | radius ⊥ tangent
^ Q
OVQ = 90° | radius ⊥ tangent
⇒ OW ^ Q + OV
^ Q = 180°
OVQW is a cyclic quadrialteral | opposite ∠s supplementary
3 a = 138° | ∠ at centre
b = 42° | opposite ∠s cyclic quadrilateral OVQW
c = 69° | tangent VQ, chord VW
d = 69° | tangent WQ, chord VW or equal tangents from common point Q
e = 69° | ∠s on chord VQ or OQ symmetry line of kite OVQW
f = 90° | diagonals kite OVQW
g = 21° | radius ⊥ tangent
h = 21° | ∠s on chord VO
204 Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 204 2012/07/02 2:23 PM
EXERCISE 5
KEY WORDS
O is the centre of the circle in each figure. In each case determine the value of
opposite sides – AD and
a, b, c, ... . Clearly state your reasons. BC are opposite sides (no
1 2 3 common vertex)
D P adjacent sides – AD and DC
Q S
52ºf P are adjacent sides (common
d z x vertex D)
x C ^ and D
^
w y g V opposite angles – B
T
0 a Q
are opposite angles
ee
F u O 70 º symmetry line – fold line
z B b
31°
27° c A D
t R 50º y w
v R U
A T
E S
C
B
WORKED EXAMPLE
E
ABED is a cyclic quadrilateral. FDCG is
REMEMBER
a tangent to the circle at D. BEC is a
straight line. AD ∥ BC, AD = AE and AD ^ E = x. A To prove a cyclic quadrilateral,
2 show one of the following:
1 Write down, with reasons, five 1 1
• opposite angles are
other angles equal to x. x D supplementary
2 Prove that ABCD is a parallelogram. 2
• a line subtends equal
3 Prove that ABED is an isosceles trapezium. angles on the same side
• an exterior angle equals an
SOLUTIONS B 1 2 interior opposite angle.
3 1
^ | isosceles △AED, AE = AD E
1 E2 = x 2 C
D^ =x | tangent FD, chord AD
1
^
C1 = x | corresponding ∠s, AD ∥ BC
^E = x | alternate ∠s, AD ∥ BC
3
BA ^D = x | exterior ∠ cyclic quadrilateral ABED G
2 ^D = D
BA ^ =x | proved in 1
1
AB∥FG | alternate ∠s equal
ABCD is a parallelgram | AB ∥ DC and AD ∥ BC
3 DC = DE |^E3 = ^
C1 = x
DC = AB | opposite sides parallelogram
⇒ AB = DE
ABED is an isosceles trapezium | AD ∥ BE and AB = DE
Remember to think about what you have already done as you start each new
question. In the Worked example above, Questions 2 and 3 both used the solution
from the previous question.
Unit 4 Tangents 205
PLT MATHS LB 11 7th pgs (Real Book).indb 205 2012/07/02 2:23 PM
EXERCISE 6
1 Complete the statements by filling in the missing words:
1.1 Tangents drawn to a circle from a common point outside the circle are …
1.2 The radius of any circle always meets a tangent to the circle at … at the
point of contact.
1.3 The angle between a tangent and a chord is … to the angle subtended
by the same chord in the alternate segment.
1.4 If the angle between a line and a chord is equal to the angle subtended
by the chord in the alternate segment, then the line is …
2
REMEMBER
A
To prove two lines parallel,
show one of the following: 54°
• alternate angles are equal G
v E
• corresponding angles are
u
equal
• co-interior angles are B 80° y x
supplementary.
F z w
t
C
D
Two circles touch at C. DE is a tangent to both circles at C. A and B lie on
^ = 54°; B
the larger circle, F and G lie on the smaller circle. A ^ = 80°.
2.1 Determine, with reasons, t, u, v, w, x, y and z.
2.2 Is DE ∥ BA? Justify your answer.
2.3 Is BA ∥ FG? Justify your answer.
3
N
60° M
x
3x 2x
J K L
^ N = 3x, NK
JK ^ M = x, MK ^ L = 2x and N
^ = 60°. M, K and N are points on the circle.
JKL is a straight line.
3.1 Determine the value of x.
3.2 Prove that:
3.2.1 NK is a diameter of the circle
3.2.2 JKL is a tangent to the circle at K.
206 Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 206 2012/07/02 2:23 PM
4 ^ F = 35°, OH
DF is a tangent to the circle at E. KE ^ G = 31° and
F
EK = HK. O is the centre of the circle. Determine: E
4.1 ^
E 35°
4 D
4.2 ^
K 1 4
2 3
^H K
4.3 DE
4.4 ^
G
4.5 ^
O 1
4.6 OE ^D 1
O 3
4.7 ^
E1 2
31°
4.8 What type of quadrilateral is EOHK? Justify your answer. H
4.9 Prove that OK perpendicularly bisects EH.
4.10 If KO is produced, will it pass through G? Justify your answer. G
5 PN is the diameter of the circle with NRM a tangent to the circle M
^ = x. O is the centre of
at N. P, N, T and Q lie on the circle and M x
the circle.
5.1 Name four angles equal to 90°.
5.2 Determine two other angles equal to x. Q
3
5.3 Prove that MQTR is a cyclic quadrilateral. 1 R
1
5.4 Is PNRQ a cyclic quadrilateral? Justify your answer. 2 1
2 2
T
1
1
2 2
P 3 N
O
6 EF is a tangent to the circle at C. CDG is a straight line. A, B, C and D are points G
on the circle. BC^ E = 53°, CB
^ D = 60° and AD
^ G = 105°.
6.1 Is BD ∥ EF? Justify your answer.
6.2 Complete the following statements by filling in the A 105° D
missing words: w y
6.2.1 The angle between a tangent and a chord is … p
6.2.2 The opposite angles of a cyclic quadrilateral are … x
B 60°
6.2.3 The exterior angle of a cyclic quadrilateral is …
6.3 Determine angles p, q, v, w, x and y in that order.
Give reasons for your answer.
v q F
53°
C
E
Unit 4 Tangents 207
PLT MATHS LB 11 7th pgs (Real Book).indb 207 2012/07/02 2:23 PM
7 A, B, C and D lie on the circle. PCQ is a tangent, A
E is the midpoint of AC, A ^ = 18° and C^ = 43°.
2 1
O is the centre of the circle. 1 23
7.1 Determine the size of:
7.1.1 ^
A O
^
3 E
7.1.2 C 1 2
3
7.1.3 AO^C B 2
^ D
7.1.4 D 3
7.1.5 PC^A 4 2
5 1
7.1.6 AB^C
P C Q
7.2 Prove that: O^ =O ^
1 2
7.3 Determine the size of C ^ .
5
8 AC and BC are tangents to the circle. A, B, A
P and Q lie on the circle and A ^
PC = 30°. C
2 3 1
O is the centre of the circle. 1
2 2
8.1 Prove that AOBC is a kite. Q1
4 3
8.2 Prove that AOBC is a cyclic 1 2
30° 3
quadrilateral. O 4
8.3 Determine the size of PA^ Q, O
^ , 1 2 3
2 P 1
AC^ B and AQ^ B.
B
8.4 Prove that APBC is a rhombus.
9
O V P
S
x 1 2 3 1 2
2
T 1 3
4 2
3 4
1
Q
R
SV is a diameter of the circle, with centre O. SV is extended to P. PR is a tangent
to the circle at Q. RO intersects QS at T. VQ ∥ OR and ^S = x.
9.1 Write down, with reasons, three other angles equal to x.
9.2 Prove that QRSO is a cyclic quadrilateral.
9.3 Write down, with reasons, the size of PQ ^ S in terms of x.
9.4 Prove that T is the midpoint of QS.
208 Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 208 2012/07/02 2:23 PM
EXERCISE 7
This is a mixed exercise. O is the centre of each circle.
1
P
10°
O T
d e a
f g
Q c V b S
R
P, Q, R, S and T lie on the circumference of the circle and OR ⊥ QS.
1.1 Prove that OQRS is a kite.
1.2 Determine the size of a, b, c, d, e, f and g.
1.3 If QS = 48 units and VR = 2 units, determine the length of OQ, the radius
of the circle.
2
P U
i g
d
31°
c
Q
α mO V x k T
20°
l n y z f
e
j h b
w
R
S
QR produced, meets tangent US at S. SU touches the circle at T. PT ^ U = 31° and
RQ^ T = 20°.
2.1 Determine the size of:
2.1.1 ^T
QP
2.1.2 ^T
QR
2.1.3 ^
QTU
2.2 Determine the angles a, b, c, d, e, f, g, h, i, j, k, l, m, n, w, x, y and z
in that order.
Unit 4 Tangents 209
PLT MATHS LB 11 7th pgs (Real Book).indb 209 2012/07/02 2:23 PM
S
3 PR is a diameter of the circle with centre O.
PR and chord SQ intersect at T. SR, QR and 1
2
PS are chords on the circle. OQ is joined
and RT ⊥ SQ.
3.1 Prove that PQRS is a kite. 1 O T1 2
^ = 29°, calculate, with reasons, 1 R
3.2 If P P 2 4 3
1 2 1 2
1
the size of:
3.2.1 ^
R 1
3.2.2 Q2 1 23
3.2.3 ^
O 1
Q
C
4 ABC is a tangent to the circle at B. E is the A B
midpoint of DF.
4.1 Prove that BOEC is a cyclic quadrilateral. D
4.2 Prove that OC2 = OD2 + EC2 − EF2
4.3 Calculate BC if DF = 80 units, OB = 50 units
E
and DC = 10 units. O
F
5 O is the centre of two concentric circles.
K
The radii of these circles are 37 units and 13 units.
OH ⊥ DE and OH = 12 units. KE is a tangent to
the smaller circle and touches it at M. M
5.1 How far apart are the circumferences of
O
the circles?
5.2 Determine the length of:
5.2.1 FH D F H G E
5.2.2 DH
5.2.3 DF
5.3 State the lengths of OM and OE.
5.4 Determine the length of KE, in simplified surd form.
5.5 What type of quadrilateral is OHEM?
210 Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 210 2012/07/02 2:23 PM
Revision
1
O
A D B
C
O is the centre of the circle.
1.1 If AB = 54 units and DC = 3 units, determine with reasons,
the length of the radius. (5)
1.2 What type of quadrilateral is OBCA? Fully justify your answer. (3)
1.3 ^ B in terms of x.
^ C = x, determine with reasons, the size of AO
If AB (4)
[12]
2
O
A F B
9
C D
E
AB and CD are parallel chords of a circle with centre O. AB is 40 units,
CD is 14 units and the distance between AB and CD is 9 units. If OF is
x units, determine, with reasons:
2.1 OF^A (2)
2.2 an expression, in terms of x,
2.2.1 for the radius OA. (2)
2.2.2 for the radius OC. (2)
2.3 Use the results from 2.2 to determine the length of the radius. (5)
[11]
3 O is the centre of the circle, Q, R, T and P lie
on the circle and TOP is a straight line.
^ Q = 2x + 8° and R
PO ^ = 4x − 14°.
P
O
Determine, with reasons, the value of x.
2x = 8°
T
4x – 14°
Q
R
[4]
211
PLT MATHS LB 11 7th pgs (Real Book).indb 211 2012/07/02 2:23 PM
TOPIC 9: REVISION CONTINUED
4 P
REMEMBER
2 ______ _______ ___
1 √ 32 + 42 = √ ( 9 + 16 ) = √ 25
1 Q =5
__ __
√ 32 + √ 42 = 3 + 4 = 7
2
1 2
______ __ __
T √ 32 + 42 ≠ √ 32 + √ 42
O
S
1 2
R
P, Q, R and S are points on the circumference of the circle with
centre O. SOQ and PTR are straight lines. PO ∥ SR and OP ^ R = 24°.
Determine, with reasons:
4.1 R^ (2)
1
4.2 ^
R (2)
2
4.3 ^
O (2)
2
4.4 P^ (2)
2
4.5 ^S (2)
4.6 ^
Q (2)
2
[12]
5 A
F
1
2
2 C E
1 3 1
x 2
1
2
2 D
1
G
A and B lie on the smaller circle, D and E lie on the bigger circle and
C and F are the points of intersection of the two circles. ABG, FCG,
EDG and AFE are straight lines. ^B1 = x.
5.1 Determine, with reasons, two other angles equal to x. (4)
5.2 Prove that BCDG is a cyclic quadrilateral. (2)
5.3 Join BD and then prove that
5.3.1 GB^D = E
^. (4)
5.3.2 ^
BDG = A^ (4)
5.3.3 What type of quadrilateral is ABDE? Justify your answer. (2)
[16]
212
PLT MATHS LB 11 7th pgs (Real Book).indb 212 2012/07/02 2:23 PM
6
A F B
2 3 1 2
1 2
1
1
x 2
O
2 1
3 5
4 G
2
3 2 1
1
E D C
The diameter AG is produced to C. EC is a tangent to the circle at D.
A, D, G and F lie on the circle and DC = BC. AFB is a straight line
^ = x.
and A 1
6.1 Prove that ADCB is a cyclic quadrilateral. (4)
6.2 Prove that AC bisects BA^ D. (2)
6.3 Prove that DG = FG. (2)
6.4 ^
Determine, with reasons, the size of C (3)
2
6.5 ^ ^
Prove that G1 = O2. (6)
6.6 Prove that FBCG is a cyclic quadrilateral. (3)
6.7 For what value of x will DC be a tangent to the circle through
F, B, C and G? (3)
[23]
D
7
O
1 2
x C
A
B
AB and AD are tangents to the circle from A. O is the centre of the
^ = x. Prove that O
circle which passes through B, C and D. C ^ − ^A = 180°. [8]
2
213
PLT MATHS LB 11 7th pgs (Real Book).indb 213 2012/07/02 2:23 PM
TOPIC
210 Trigonometry: Sine, cosine
and area rules
Unit 1: Proof and application of the sine,
cosine and area rules
All triangles have three angles and three sides. If you are given three of any of
KEY WORDS them, as long as one is a length, you can solve the triangle (find all the other
sides and angles).
solve a triangle – to calculate
the lengths and sizes of the
unknown sides and angles Solving right-angled triangles – Revision
of Grade 10
For a right-angled triangle use the three trigonometric definitions:
hypotenuse o , cos θ = __
sin θ = __ o
a and tan θ = __
opposite h h a
θ
adjacent WORKED EXAMPLE
Find the size of the unknown sides and angles of both triangles.
A E
1 2
50 cm d
c 32 cm
REMEMBER
To find a side: 35° θ
C B F D
unknown side = sin, cos or tan
____________ a 58 cm
known side
of the known angle
or SOLUTION
known side = sin, cos or tan
____________
unknown side
of the known angle
1 In △ABC: 2 In △DEF
c = sin 35°
___ 32 = tan θ
___
To find an angle: 50 58
known side 1 = sin, cos or tan
___________
known side 2
c = 50sin 35° θ = shift tan−1 32
___
( 58 )
of the unknown angle = 28,68 cm = 28,89°
c = tan 35°
__ 58 = cos 28, 89°
___
a d
c
a = _______ | substitute c from above 58
d = __________
tan 35° cos 28,89°
= 40,96 cm = 66, 24 cm
| or use Pythagoras
214 Topic 10 Trigonometry: Sine, cosine and area rules
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EXERCISE 1
Solve the triangles.
1 A 2 A 3 A 4 A
33 cm
13 cm 12 cm 60 cm
24 cm C 33° B
69° C B B C
50 cm
C B
Solving triangles with no right angles
For triangles that have no right angle you cannot use the trigonometric
definitions. You therefore need to know and prove two new rules called REMEMBER
the sine and cosine rules. We label triangles with no
right angles in this way:
The sine rule and its proof side AB = c; side AC = b and
a = _____ sin A = _____
c or _____
b = _____ sin B = sin __
C side BC = a
In any △ABC: _____ a c
sin A sin B sin C b
Proof: A
A A
c b
c b c
h
h b B a C
D
B C B D
a C 180° – c
Acute-angled triangle Obtuse-angled triangle
Let AD = h Let AD = h
= height of △ABC with base BC = height of △ABC with base BC
h and sin C = __
sin B = __ h h and sin (180° − C) = __
h
c b sin B = __
c b
h = csin B and h = bsin C but sin C = sin (180° − C)
Equate h on both sides: h = csin B and h = bsin C
csin B = bsin C Equate h on both sides:
Divide both sides by bc csin B = bsin C
C sin B = b sin
_______ ______C Divide both sides by bc
bc bc
c sin
______B = b sin
______C
sin
_____B = sin C
_____
c bc bc
b
Let CE = h sin B = sin
_____ C
_____
c
b
= height of △ABC with base AB Let AD = h
Repeat the steps above to get: = height of △ABC with base AB
sin B = sin
_____ A
_____
b a Repeat the steps above to get:
sin A sin
_____ B sin
_____
a = b = c
C
_____ sin B = sin
_____ A
_____
a
b
sin A sin
_____ B sin
_____ C
_____
a = b = c which is the same as
a b = c
_____ = _____ _____
sin A sin B sin C
Unit 1 Proof and application of the sine, cosine and area rules 215
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REMEMBER WORKED EXAMPLE
Use the sine rule if you know A
Solve for x in each triangle.
a side–angle pair (that is, an 1 x is the unknown side use:
angle and the side opposite x 15 x 15 cm
______ = ______
it) and one other given side sin 66° sin 55°
or angle. x = 15sin 66°
________
sin 55° 55° 66°
To find a side: B C
= 16, 73 cm
unknown side
_______________________
sine of the angle opposite it A
2 x is the unknown angle use:
= known side
_______________________
sine of the angle opposite it
sin 36°
sin x = ______
_____ x
25 19
25sin 36° 19 cm
To find an angle: sin x = ________
19
sine of the unknown angle
_________________________ x = 50,66° or x = 129,34°
known side opposite the angle
Notice there are two possibilities for x 36°
sine of the known angle B C
= _________________________ 25 cm
known side opposite the angle x = 180° − 50,66° = 129,34°
This is called an ambiguous case. Always check for ambiguity when solving for
angles. In many cases the diagram will show that no ambiguity occurs as the sum
of two of the angles will be greater than 180°.
The sine rule may be ambiguous if you know two sides and a non-included angle.
(SSA)
EXERCISE 2
Find the value of x in each triangle. The diagrams are not drawn to scale. Give all
answers correct to two decimal places.
1 A 2 A
18 cm
33 cm x
28° 54°
B C
x
35° 54°
B C
3 A 4 A
x
68 cm 108°
x
52° 42°
B C B C
71 cm 163 cm
216 Topic 10 Trigonometry: Sine, cosine and area rules
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5 A 6 A
100°
15 cm
80 cm
x
B C
18 cm x 67°
B C
65 cm
The cosine rule and its proof
For triangles that do not have a known side-angle pair you cannot use the sine rule.
You therefore need to know and prove the cosine rule.
In any △ABC: a2 = b2 + c2−2bccos A or b2 = a2 + c2−2accos B or c2 = a2 + b2 − 2abcos C
A A
c b
C h
h
b
a–x x B
180° – c
B C a c x D
D (a + x)
a
Acute-angled triangle Obtuse-angled triangle
Let AD = h Let AD = h
= height of △ABC with base BC = height of △ABC with base BC
Let DC = x BD = a − x Let DC = x BD = a + x
Apply Pythagoras’ Theorem to △ABD to Apply Pythagoras’ Theorem to △ABD to
find c2 find c2
c2 = h2 + (a − x)2 c2 = h2 + (a + x)2
= h2 + a2−2ax + x2 = h2 + a2 + 2ax + x2
= a2 + (h2 + x2) − ax = a2 + (h2 + x2) + 2ax
= a2 + b2 − 2ax = a2 + b2 + 2ax
b2 = h2 + x2 | Pythagoras in △ADC b2 = h2 + x2 | Pythagorus in △ADC
x = cos C
__ x = cos (180°− C)
__
b b
x = bcos C x = −bcos C | cos (180°− C) = −cos C
c2 = a2 + b2−2abcos C c2 = a2 + b2 − 2abcos C
Unit 1 Proof and application of the sine, cosine and area rules 217
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REMEMBER WORKED EXAMPLE
A
To find a side: Solve for x in each triangle.
(unknown side)2 = (known A
1 2
side 1)2 + (known side 2)2
− 2(known side 1)(known side 47 cm 36 cm 29 cm
x
2) cos (∠ between 2 known
sides)
48° x
To find an angle: B C B C
55 cm 25 cm
Change the subject of the
formula of the cosine rule to
get: SOLUTIONS
b2 + c2 − a2
cos A = __________
2bc x is the unknown side:
or cos B = a + c
2 2
− b2
__________ x2 = 472 + 552 − 2(47)(55)cos 48°
2ac _____________ ___
x = √ 1 774, 594765... | Press √ Ans do not round off
a2 + b2 − c2
or cos C = __________
2ab x = 42, 13 cm
Use the cosine rule if you 2 x is the unknown angle:
know two sides and the 362 = 252 + 292−2(25)(29)cos x
included angle (SAS) or three 2 2 2
angles (AAA). cos x = 25 + 2 9 − 36
_____________
2(25)(29)
= 0,11724 | Do not round off; use shift cos−1(Ans)
x = 83,27°
EXERCISE 3
Find the value of x in each triangle. The diagrams are not drawn to scale. Give all
answers correct to two decimal places.
1 A 2 A
x 18 cm
33 cm x
54°
44° B C
B C 28 cm
54 cm
3 A 4 A
x
58 cm 14,2 cm
10,8 cm
42° x
B C B C
63 cm 16,3 cm
218 Topic 10 Trigonometry: Sine, cosine and area rules
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5 A 6 A
x
10 cm 15 cm
60 cm 60 cm
B C
20 cm
42°
B x C
The area rule and its proof
1 base × height. If you are not given
You have learnt that the area of a triangle is A = __
2
or cannot calculate the height, you need to know and use the area rule.
1 bcsin A or Area △ABC = 1
Area △ABC = __ 1 absin C
__acsin B or Area △ABC = __
2 2 2
Proof:
Let AD = h = height of △ABC with base BC
1a × h
Area △ ABC = __
2
h = sin B or __
__ h = sin C
c b
h = csin B or h = bsin C
Substitute h:
1 acsin B or A = __
Area = __ 1 absin C
2 2
Let CE = h = height of △ABC with base AB
Repeat the steps above to get:
1 acsin B or A = __
Area = __ 1 acsin A
2 2
A
c b
h
B C
D a
Acute-angled triangle
Use the area rule if you know two sides and the included angle (SAS).
Unit 1 Proof and application of the sine, cosine and area rules 219
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WORKED EXAMPLE
Find the area of △ABC. A
SOLUTION
Area △ABC = 1 1 (65)(33)sin B
__acsin B = __
2 2
B, find ^
To find ^
65 cm
A first by using the Sine Rule
sin A = ______
_____ sin 54°
33 65 54°
B C
(33)sin 54°
__________ 33 cm
sin A =
65
^ = 24,25°
A
^ = (180° − 24,25° − 54°)
B
= 101,75°
1 (65)(33)sin 101,75°
Area △ABC = __
2
= 1 050,03 cm2
EXERCISE 4
Find the areas of each triangle. The diagrams are not drawn to scale.
Give answers correct to two decimal places.
1 A 2 A
18 cm 18 cm
40 cm
52°
36° B C
B C
44 cm
3 A 4 A
85°
37°
B C 47° 47°
C
28 cm B 60 cm
5 A 6 A
25 cm
18 cm
105 cm
B C 30°
28 cm B C
123 cm
220 Topic 10 Trigonometry: Sine, cosine and area rules
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Application of the sine, cosine and area rules
WORKED EXAMPLE
C
Find the value of x in the figure alongside.
SOLUTION
In right-angled △PBC: 15
15 = sin 75°
___
PC
15 = 15,53 cm
PC = ______
sin 75°
55° 75°
A x B
In △APC: A ^
P
CP = 20° | ext angle △APC
x
______ PC
= ______
sin 20° sin 55°
15,53sin 20°
x = ___________ = 6,48 cm
sin 55°
EXERCISE 5
1 The diagrams below are not drawn to scale. All units are in cm. Give the answers
correct to two decimal places.
1.1 Find the area and perimeter of 1.2 ABCD is a trapezium. Find the
quadrilateral ABCD. length of DC and area of ABCD.
A 15
A B
4 110°
70°
B
12
7
67° 35°
D C
D C
1.3 ABCD is a rectangle. Find the 1.4 ABCD is a rhombus. Find:
^ E.
size of AC 1.4.1 D ^
1.4.2 Area of ABCD
1.4.3 Length of BD
A 3 E B A B
7 12 10
x
D 14
C D 10 C
Unit 1 Proof and application of the sine, cosine and area rules 221
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1.5 Find: 1.6 Find the values of x and y.
^C
1.5.1 the size of DB
B
1.5.2 the length of AC
1.5.3 the area of ABDC
27°
D
y
24 mm 48° 62°
A x D 11
C
40 mm
C B
40°
45 mm
A
2 In each case find the length of BC in terms of the given letters and symbols.
2.1 B 2.2 A
x x
0 β
A 0 C B C
y
2.3 A 2.4 C
β
x
0 0
B C α
A B
a D
2.5 B 2.6 A x F
0 0
α E
α β
A C
β B
a
E
C
222 Topic 10 Trigonometry: Sine, cosine and area rules
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Unit 2: Two-dimensional problems using
REMEMBER
sine, cosine and area rules
The angles of elevation and
In Grade 10 you learnt about the angles of elevation and depression in depression are the angles
two-dimensional problems. You will now apply the sine and cosine rules between the line of sight and
to two-dimensional problems. the horizontal.
Application of triangle rules using Angle of
elevation
numerical values Horizontal
Angle of
depression
WORKED EXAMPLE
In the diagram AB represents a vertical Horizontal
A x
cliff which is 120 m high. An observer 65°
Angle of
at A sees two boats on the sea below. 72° depression
The angle of depression from A to boat
120 m
C is 72° and from A to boat D is 65°.
The boats are in the same horizontal Angle of
plane as B, the foot of the cliff. AX elevation Horizontal
is parallel to BCD. B C D
Calculate the horizontal distance between the two boats.
REMEMBER
SOLUTION
1 Use geometry to fill in as
^ D = 72° − 65° = 7°
CA many angles as you can.
2 Start with the triangle
^ B = 72° | alt angles AX ∥ BD and AD
AC ^ B = 65° | alt angles AX ∥ BD
which has a given length.
120 = sin 72° AC = ______
120 = 126,18 m 3 Find a side which links the
In right-angled △ABC: ____
AC sin 72° two triangles (keep your
CD = ______
AC CD = ACsin 126,18.sin 7°
7° = ____________
In △ACD: _____ ________ = 16,97 m answer in the calculator).
sin 7° sin 65° sin 65° sin 65°
4 Use this side to solve sides
or angles in the other
triangle.
5 Repeat the process if
EXERCISE 6
necessary.
1 A man standing at a point C looks up at
A
an angle of 35° to the top (A) of the cliff
AB
150 m away from him. He turns around
to walk 300 m in the opposite direction
D
(away from AB) at an angle of inclination 300 m
of 15° 35° 15°
to point C. A, B, C and D are in the same B C
150 m
vertical plane. Calculate, to the nearest
metre, the distance between A and D.
Unit 2 Two-dimensional problems using sine, cosine and area rules 223
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2 The oil rig in the photograph has a width BC of 50 m. If the angle of elevation
from B to A, the top of the rig, is 32° and from C to A is 55°, calculate the height
AD of the rig.
A
32° 55°
B D C
50 m B
3 A straight road with a slope of 12° leads directly
to the foot C of a tree BC. A and T, two points
on the road, are 8 m apart. The angle of elevation C
65°
from A to the top of the tree is 40°. If B ^
TC = 65°, 8m
calculate the height, BC, of the tree. T
12°
40°
A D
A
4 AD is a telephone pole and AB and AC are two wires 33°
59°
connected to the top and bottom of stay CB, which
is inclined at 64° to the horizontal line DC. The angle
B
of depression of B from A is 33° and of C from A is 59°.
If the length of stay BC is 3 m, calculate the length 3m
of the pole AD.
64°
D
C
5 The drivers of two trucks which are
T
parked 5 m apart both measure the
angle of elevation from where they
are parked, to the top of the same
building. The angle of elevation of A,
the top of the building, from the truck
at point C is 65° and from the truck at
point D is 52°. Calculate the height, AB,
of the building. 52° 65°
D 5m C B
224 Topic 10 Trigonometry: Sine, cosine and area rules
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Applications of triangle rules using symbols
to do proofs
WORKED EXAMPLE 1
A
^ C = θ and AC = k.
In △ADC, AD = DC, AD
^
In △ABD, ABD = β. k
kcos __θ
2
Prove that AB = ______
sin β β θ
B C
D
SOLUTION
Find other angles:
^ D = DA
In isosceles △ADC: AC ^ C = 180° − θ = 90° − θ
_______ __
2 2
^ B = 180° − θ
In △ADB: AD
Now use the sine rule to find the linking side AD:
AD
___________ k (
ksin 90°− __θ
2
____ AD = ___________
) kcos __θ
______2
= =
(
sin 90° − __θ
2 )
sin θ sin θ sin θ
In △ADB use sine rule to find AB:
AB AD AB = ADsin
____________ = _____ _______θ [sin (180° − θ) = sin θ]
sin (180° − θ) sin β sin β
AB =
( kcos θ__
)
2 sin θ
______
sin θ
___________ | substitute AD from above
sin β
kcos __θ
2
AB = ______
sin β
WORKED EXAMPLE 2
In the diagram AD^ C = θ, AC = DC = r, A
BD = 2r, AC = k and AB = 2k
2k k
1 Use △ADC to express cos θ in r
terms of r and k.
θ
B C
2 Use △ABD to express cos θ in 2r D
terms of r and k.
3 1.
Use your answers to show cos θ = __
4
Unit 2 Two-dimensional problems using sine, cosine and area rules 225
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SOLUTIONS
1 In △ADC use the cosine rule:
r + r − k = _______
2r − k 2 2 2 2 2
k2 = r2 + r2 − 2r2cos θ cos θ = __________
2 2
2r 2r
2 ^ B = 180°− θ
In △ADB AD
Now use the cosine rule:
(2k)2 = r 2 + (2r)2 − 2(2r) r cos (180°− θ)
4k2 = r 2 + 4r2 + 4r2 cos θ
2 2 2 2 2
cos θ = 4k − r 4k − 5r [cos (180° – θ) = – cos θ]
− 4r = ________
____________
2 4r 2 4r
3 Make the two cos θs equal to each other:
2
2r 2 − k2 = 4k
_______ − 5r 2 | × both sides by 4r 2
________
2
2r 4r 2
4r2 − 2k2 = 4k2 − 5r2
9r2 = 6k2 | solve for k2
2
k2 = 3r
___
2
2 2
Now substitute k2 into cos θ = 2r −k
_______
2 2r
___2
2x 2 − 3r 1r 2
__
cos θ = ________2 = 2 1
___ = __
2r 2 2r 2 4
EXERCISE 7
1 Use the given information in the diagrams to show that:
ksin β 2θsin α θ
1.1 d = _____________ 1.2 d = ksin
___________ 1.3. d = 8sin
_________
sin (α + β)tan θ sin θsin β sin (α + θ)
A A A
d
k k k
0 α
0
C 0 B C
B α β B α C 60°
d 4
d β
D D
D
E
2 A field has the shape of a trapezium with
a
a right-angled triangle added onto the
A D
one side.
x
AB ^ C = DC
^ C = 90°, BA ^ E = x, AC
^ D = y and DE = a.
2.1 ^ C in terms of x and y.
Find AD y
acos (y − x) x
2.2 Prove that: AC = __________
sin x B C
226 Topic 10 Trigonometry: Sine, cosine and area rules
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3 A cableway AB connects the top of two A
mountains across an intervening valley. α
The length of AB is d. From A the angle of β
d
depression of B is α and the angle of
depression of the bottom of the valley at C B
is β. From B the angle of depression of the 0
bottom of the valley at C is θ .
3.1 ^ B = 180°− (θ + β)
Show that: AC
(θ + α)
3.2 Prove that: AC = dsin _________ C D
sin (θ + β)
P A
4 ^ P = AC
In the diagram, QC = CB = x CQ ^ B = θ,
^
ACP = β and PQ ∥ AC
Q 0
4.1 Express AC in terms of x and θ.
4.2 Express PC in terms of x, θ and β.
x
4.3 Hence show that the area of
β
.tan θ
2 0
△APC = x_______
2 C x B
5 △XYZ is an isosceles triangle with z = y. X
2(Area of △XYZ)
5.1 Show that: y2 = _______________
sin X
5.2 Now use the Cosine Rule and your answer
z y
to Question 5.1 to show that:
2
Area of △XYZ = x sin X
___________
4(1 − cos X)
Y x Z
6 ^ C. AB
In △ABC, BD bisects AB ^ D = DB ^ C = y. CB = a and AB = b.
^ C = x and BA
C
D a
x
y x
A B
b
6.1 Prove that: DC = asin x
_________
sin (x + y)
6.2 Prove that:AD = bsin x
_________
sin (x + y)
6.3 Find the ratio DC:AD.
Unit 2 Two-dimensional problems using sine, cosine and area rules 227
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Applications of triangle rules in navigation
You can use trigonometry to solve surveying and navigation problems.
REMEMBER
• In navigation, a bearing WORKED EXAMPLE
is the direction from one
object to another. An aeroplane flies from Q for 300 km at a bearing S
• Bearings are given as of 310°. It changes it course to a bearing of 200°. 200°
angles rotated clockwise
from the north. 1 How far is the plane from its starting point 450 km
300 km
• Angles may also be given after travelling 450 km on this new course?
in terms of North, South, 2 What is the bearing of the plane from its Q
East and West. starting point (that is, what is the bearing 310°
North of Q from P?) P
West East
SOLUTION
North
The diagram alongside shows the acute triangles in △PSQ. N2
South
1 PQ2 = 3002 + 4502 − 2(300)(450)cos 70° S 130°
___________
N PQ = √ 200 154,5613 = 447,39 km
A sin ^Q
SP
2 _______ = sin 70°
______ 450 km 20° 50° N1
300 447,39
N3 300 km 50°
50° ^ Q = 300sin
sin SP 70° = 0,63
_________
447,39
P
^ Q = 39,06° | Not ambiguous SP
SP ^Q ≠ 140,94° Q
310°
A has a bearing of ^
SPN3 = 20°
50° from P bearing of Q from P is 59,06° P
A has a bearing
N 50’E
N
A EXERCISE 8
50° Note: The diagrams accompanying these questions are not drawn to scale.
310° 1 A ship travels 80 km on a bearing of 28°, and
P
then travels for 190 km on a bearing of 155°.
How far is ship from its starting point?
A has a bearing of
310° from P North
A has a bearing
N 50’W B 155°
190 km
80 km
North
C
18°
A
228 Topic 10 Trigonometry: Sine, cosine and area rules
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2 Lighthouse A is situated directly north North
of lighthouse B. From lighthouse A the
bearing of a ship 1 745 m away is 125° 62".
A 125° 62"
From lighthouse B the bearing of the ship
is 32° 43". Find the distance between the 1 745 m
lighthouses in km.
S
32° 43"
B
3 T is a tree on one side of a field. Its bearing T
from A, which is on the opposite side of the North
field, is 42°. B is a point 85 m from A on the
same side of the field as A. The bearing of the w
tree from B is 330°. Calculate the width of
the field.
A 85 B
Applications of triangle rules in circle geometry
You can use trigonometry to find the lengths of chords and angles in geometric
problems. Apply your knowledge of circle geometry and trigonometry in the exercises
that follow.
WORKED EXAMPLE 1
A
O is the centre of the circle. BC = 15 units and
B^
AC = 52°. Calculate the radius of the circle.
52°
SOLUTION O
^ C = 104° | angle at centre
BO
Let BO = CO = r B C
15
Use cosine rule in △BOC:
152 = r 2 + r 2 − 2(r)(r)cos 104°
225 = 2r 2(1 − cos 104°)
___________
r= √_____________
225
2(1 − cos 104°)
= 9,52 units
Unit 2 Two-dimensional problems using sine, cosine and area rules 229
PLT MATHS LB 11 7th pgs (Real Book).indb 229 2012/07/02 2:23 PM
WORKED EXAMPLE 2
AB is the diameter of the circle and C is a point on the circumference. DC is a
tangent to the circle at C and meets AB at D.
If AB = d and B ^ dsin θ 2
AC = θ, prove that: BD = ______ A
cos 2θ
d 0
SOLUTION
^ D = θ tan
BC | chord theorem B
^ A = 90°
BC | angle in semicircle
D
^ A = 180° − (90° + θ + θ)
CD C
= 90°− 2θ | sum angles in △ACD
BC = sin θ BC = dsin θ
___
d
BD =
____ BC
____________
sin θ sin (90° − 2θ)
BC sin θ
BD = ____________ dsin θ.sin θ = ______
= __________ dsin θ 2
sin (90° − 2θ) cos 2θ cos 2θ
EXERCISE 9
1 If O is the centre of the circle in each case, find x.
1.1 A 1.2 B 1.3 D
x
x
C
O O A
8
10
B x A 44° 16° 3
D 15 C
118° 9
D B
C
2 ABC is a secant of the circle, and CT is the A
tangent at T. AT = 16 cm, AB = 12 cm and
^ = 41°.
A 41°
2.1 Calculate the length TB correct to 12
two decimal places. 16
2.2 Calculate the size of the angle AT ^ B.
2.3 Calculate the length of secant ABC. B
2.4 Calculate the area of △TBC.
T
C
230 Topic 10 Trigonometry: Sine, cosine and area rules
PLT MATHS LB 11 7th pgs (Real Book).indb 230 2012/07/02 2:23 PM
3 O is the centre of the circle, AOD is a diameter. B
AB = 14 cm, BC = 17 cm and AC = 9 cm.
3.1 ^ correct to one
Calculate the size of B
decimal place. O
A D
3.2 Calculate the radius of the circle, correct
to one decimal place.
C
4 AB is the diameter of the circle and C
is a point on the circumference. DC is A
a tangent to the circle at C and meets
^ C = θ: 0
AB at D. If DB = BC = a and BA
4.1 Prove that: DC2 = 2a2(1 + cos 2θ)
___ B
4.2 If DC = √ 12 and a = 2, show a
that θ = 30°.
D
C
5 AB is a tangent to circle O with radius r.
AO^C = θ
5.1 Find the area of △AOB in terms
of r and tan θ. O C B
0
5.2 Find the area of △AOC in terms
of r and sin θ. x
5.3 Use your results to show that
tan θ > sin θ where θ is acute. A
6 In the sketch ABC is a cyclic quadrilateral.
^D = x A
AB = AD = a and BC = CD = b and BA
6.1 Prove that: BC2 = 2a2(1 − cos x) x
6.2 Find BD2 in terms of b and
a a
a trigonometric ratio of x.
6.3 a=1+
Use your results to show that: __ cos x
________
b sin x
B D
b b
C
Unit 2 Two-dimensional problems using sine, cosine and area rules 231
PLT MATHS LB 11 7th pgs (Real Book).indb 231 2012/07/02 2:23 PM
Revision
None of the diagrams are drawn to scale.
1 Solve for x in each triangle.
1.1 A 1.2 A
x
x
21 cm
38° 64°
B C 47°
35 cm B C
28 cm
1.3 A 1.4 A
x
x 18 cm 35 cm
24 cm
62°
B C B C
28 cm 38 cm
1.5 A 1.6 A
21 cm x = Area Δ
x = Area Δ 28° 42°
54° B C
B C 88 cm
28 cm
[3 × 6 = 18]
2 ^ S = 54°.
In the diagram, PQ = PS and QP ∥ RS. PS = 18 cm and RS = 15 cm. PQ
P
18 cm
54°
Q S
15 cm
R
Calculate (to two decimal places if appropriate):
2.1 the magnitude of Q ^PS (3)
2.2 the length of QS (3)
2.3 the area of quadrilateral PQRS. (3)
[9]
232
PLT MATHS LB 11 7th pgs (Real Book).indb 232 2012/07/02 2:23 PM
3 A, B, C, D lie on a circle. AB = 7 cm, BC = 6 cm, AC = 10 cm, AD = 4 cm.
A
7
B 4
D
10
6
C
Calculate, correct to two decimal places:
3.1 ^
the size of B (3)
3.2 the size of AC^ D. (3)
[6]
4 A farmer wants to calculate the height, PQ, of P
his windmill from the ground to the centre, P,
of its wheel. He measures the angle of elevation
from point R to point P to be 65,22° and from
another point S, 2 metres from R, to point P to
be 52,43°. Calculate the height PQ of the windmill.
65,22° 52,43°
Q R 2m S
[4]
A
5 A sail boat has sails consisting of two triangles
joined together by mast AB, 44 °
2,8 m 30 ° 3,75 °
as shown in the diagram.
5.1 Calculate the height of the mast AB. (3)
5.2 Calculate the area of △AEB. E (3)
C D
5.3 Find the length of side BE. (3)
1,70 m
[9]
B
P
6 Use the diagrams below to prove that:
6.1 PS = xsin θ tan α
__________
sin 2θ
α
S Q
0
x
R (4)
233
PLT MATHS LB 11 7th pgs (Real Book).indb 233 2012/07/02 2:23 PM
TOPIC 10: REVISION CONTINUED
6.2 BD = 2a sin θ
_______ A
cos β
0
a
150°
B C
ß
D (4)
[8]
7 In the figure, ECD is the wall of a building
with a flagpole BC at an angle to it. The B
E
angles of elevation of B from C and A are θ
and α respectively and the angle of elevation
of C from A is β. AD = m.
msin (α – β)
Prove that: BC = ______________
sin (α + θ) cos β θ
C
α
β
A D
m [5]
8 O is the centre of the circle with diameter AC and B another point on the
circumference.
B
A θ C
radius = a O
If the radius of the circle is a, show that:
8.1 BC = 2a cos θ (2)
8.2 Area △ABC = 2a2 sin θ cos θ (3)
[5]
234
PLT MATHS LB 11 7th pgs (Real Book).indb 234 2012/07/02 2:23 PM
9 A ship, Q, sails from P, on a bearing of 43° for 205 km. Another ship S also sails
from P on a bearing of 142° for a distance of 148 km.
Q
N
205
P
148
S
9.1 Calculate the distance between the two ships S and Q (4)
9.2 What is the bearing of Q from S? (4)
[8]
pqsin x cos y
10 Use the figure to prove that: Area = ___________
2sin (x + y)
P
x
S
p
y
Q q R
[5]
235
PLT MATHS LB 11 7th pgs (Real Book).indb 235 2012/07/02 2:23 PM
TOPIC
11
2 Finance, growth and decay
Unit 1: Simple and compound decay
What is meant by decay
Certain items depreciate in value over time. Each item and also different brands of
KEY WORD the same type of item, depreciate in different ways and at different rates. There are
two types of depreciation, or decay:
depreciate – to lose value
• simple decay, or the straight line method of depreciation
• compound decay, or the reducing balance method of depreciation.
Simple decay
In simple decay, the loss of value on an item each year is always a percentage of the
original value.
We can explain this by considering this situation:
Peter bought a new car for R250 000. If the car depreciates in value at 15% p.a. on
REMEMBER the straight line method, the table below shows the value of the car over a five-year
period. In this situation, we calculate 15% of R250 000, and that amount remains the
p.a. stands for ‘per annum’, constant loss of value each year.
which means each year.
Age of car in years Value of car in Rand 0,15 × 250 000
= 37 500
0 250 000
1 212 500 250 000 – 37 500
REMEMBER 2 175 000 212 500 – 37 500
3 137 500 175 000 – 37 500
To calculate 15% of a value
we multiply that value by 4 100 000 137 000 – 37 500
( 15 = 0,15 .
0,15: 15% = ____
100 ) 5 62 500 100 000 – 37 500
It is clear that, under this method, a time will come when the value of Peter’s car will
KEY WORD be R0. In reality this is not likely to happen as a car will always have some value, even
if it is scrap value. It is important to understand that this is only a model, and that in
scrap value – the value of
reality there may well be other issues that affect these values.
an item after it has passed its
useful life in its original form,
and certain parts can be sold
as scrap
236 Topic 11 Finance, growth and decay
PLTMATHSLB11LB_12.indd 236 2012/07/07 11:27 AM
If we plot a graph of the value of Peter’s car over time, we see that it results in a
straight line with a gradient of –37 500: REMEMBER
Gradient = ___
rise
run and this
gradient is negative because
the function values are
250 000
decreasing.
200 000
Value of car
150 000
100 000
50 000
x
1 2 3 4 5 6 7 KEY WORD
Age of car in years
appreciation – growth
in value; the opposite of
We can see that according to this model, the car will have reached R0 after
depreciation
approximately 6,6 years.
In Grade 10 you studied simple appreciation, which showed a similar linear
situation. In that case, where the function values were increasing, the line had a
positive gradient. REMEMBER
Instead of completing a table or drawing a graph, we can determine the value over Function values are the y
time after simple decay by applying the formula: A = P(1 – in) values obtained from each
value of x.
A = the final amount; P = the present or principal value, n = the number of years
and i = the rate of interest
So, we could determine the value of Peter’s car after 5 years using this formula: REMEMBER
A = 250 000(1 – 0,15 × 5) = R62 500 The formula for simple
appreciation is: A = P(1 + in)
WORKED EXAMPLES
1 A piece of equipment is expected to depreciate at 9% p.a. on the straight line REMEMBER
method. If its value is R11 352, 50 after 3 years, calculate its original value. If interest is 12%, then
2 Calculate how long it would take an item valued at R30 000 to depreciate to a 12 = 0,12
i = ____
100
value of R10 000 at 8, 5% p.a. on the straight line method. Give your answer
to the nearest year.
3 The total number of Aids deaths in South Africa reached 314 196 in 2005.
REMEMBER
Over the next 4 years there was a decline in the number of deaths, so that
in 2010 there were 270 107 reported Aids deaths in South Africa. Using the Round off money to two
straight line method of depreciation, determine the rate per annum at which decimal places ONLY at the
the Aids deaths had decreased over these 4 years, correct to one decimal place. end of the calculation. This
will make your answer more
accurate.
Unit 1 Simple and compound decay 237
PLTMATHSLB11LB_12.indd 237 2012/07/07 11:27 AM
SOLUTIONS
1 9 = 0,09 and n = 3
A = P(1 – in) where the final amount (A) = 11 325,50; i = ____
100
11 325,50 = P(1 – 0,09 × 3)
11 325,50 = 0,73P
P = R15 551,37
2 A = P(1 – in) where the final amount (A) = 10 000, the original amount
8,5
P = 30 000 and i = ____ = 0,085
100
10 000 = 30 000(1 – 0,085 × n)
1 = 1 – 0,85n)
= __
3
2 = –0,85n
– __
3
n ≈ 8 years
3 A = P(1 – in) where the final amount (A) = 270 107, the original amount
P = 314 196 and n = 4
270 107 = 314 196(1 – i × 4)
( 314 196 )
_______
270 107 – 1 ÷ (–4) = i
i = 0, 03508 = 3,5%
Compound decay
In compound decay, the loss of value of an item each year is calculated as a
percentage of the value that it was the previous year. Under this method, the loss is
greatest in the first year, becoming less as time passes. So it will be impossible for the
value of the item to become zero.
Consider this situation:
Thabile bought a car for R250 000 at the same time as Peter. His car depreciated at
15% p.a. on the reducing balance method. The value of his car over the first 5 years is
shown in the table.
Age of car in years Value of car in Rands 0,15 × 250 000
= 37 500
0 250 000
1 212 500 250 000 – 37 500
2 180 625 212 500 – 31 875 0,15 × 212 500 = 31 875
3 153 531,25 180 625 – 27 093,75 0,15 × 180 625 = 27 093,75
4 130 501,5625 153 531,25 – 23 029,6875 0,15 × 153 531,25 = 23 029,6875
5 110 926,33 130 501,5625 – 19 575,23438 0,15 × 130 501,5625
= 19 575,23438
238 Topic 11 Finance, growth and decay
PLTMATHSLB11LB_12.indd 238 2012/07/07 11:27 AM
The calculations in the right-hand boxes show that each year the decay, or loss of
value, is 0,15 × the value of the previous year. We can clearly see that these decay
values get smaller each year, which means that the rate at which the car is decreasing
in value each year reduces over time. This is evident when we consider the graph of
this situation:
y
REMEMBER
250 000 The gradient is shown by
the steepness of the graph,
so when the line becomes
200 000 less steep, the gradient is
decreasing.
compound decay
150 000
Value of car
100 000
simple decay
50 000
x
1 2 3 4 5 6 7
Age of car (years)
If we compare the compound decay in Thabile’s case to the simple decay in Peter’s
case, we can see how the loss in value of Thabile’s car is the same as that of Peter’s
in the first year. Thereafter, as the gradient decreases, the loss in value of Thabile’s
car becomes increasingly less and less as time passes. Although decreasing instead REMEMBER
of increasing, this situation is similar to the increasing gradient that you studied in
For compound appreciation:
compound appreciation in Grade 10, which resulted in exponential growth. We used
A = P(1 + i)n
an exponential formula for calculations involving compound appreciation.
The formula for finding the value of an item after compound decay is similar:
A = P( 1 − i )n
A = the final amount; P = the Present or Principal value, n = the number of years and
i = the rate of interest
So, we could determine the value of Thabile’s car after 5 years using this formula:
A = 250 000(1 − 0,15)5 = R110 926,33
Unit 1 Simple and compound decay 239
PLTMATHSLB11LB_12.indd 239 2012/07/07 11:27 AM
WORKED EXAMPLES
1 Cheryl leaves a bowl of soup to cool. The temperature of the soup decreases
on a reducing balance method at a rate of 4% per minute, and it is now 65 °C.
Calculate the temperature of the soup when she served it 6 minutes ago.
2 An investor bought shares in a certain company, but found that his money
had become half of it’s original value after a period of 4 years. What was the
annual rate of decay, on the reducing balance method, for the shares that he
had bought? Give your answer to two decimal places.
SOLUTIONS
1 A = P( 1 − i )n
65 = P(1 – 0,04)6
65
_________ =P
6
(1 – 0,04)
P = 83 °C (to the nearest degree)
Note: i) Although 65 is the present temperature, we are looking for the
temperature before it had decreased to 65, so 65 is the value A and we
need P.
ii) As the interest rate and the time period are both in minutes, we can
substitute them into the formula with no adaptations.
2 As we do not know the original amount of money we call it x, so the final
amount after 4 years will be half of x, that is, 0,5x.
A = P( 1 − i )n
0,5x = x(1 – i)4
Divide
____
by x, 0, 5 = (1 – i)4
√0,5 –1 = –i
4
i = 0,1591 = 15,91% p.a.
EXERCISE 1
1 Determine the value of R13 000 after it has depreciated for 3 years at 6% p.a:
1.1 on the straight line method
1.2 on a reducing balance.
2 A farmer bought a tractor. Five years later it had a book value of R168 345,22.
KEY WORD Determine the original value of the tractor if the annual rate of depreciation was
14% p.a.
book value – the depreciated
2.1 on the straight line method
value of a vehicle at any
specified time 2.2 on the reducing balance method.
3 A machine costs R48 000 and has a scrap value of R8 000 after 10 years. Find the
annual rate of depreciation, correct to two decimal places, if it is calculated on
the:
3.1 straight line method
3.2 reducing balance method
240 Topic 11 Finance, growth and decay
PLTMATHSLB11LB_12.indd 240 2012/07/07 11:27 AM
4 How long will it take an item bought for R12 800 to depreciate to R8 600 if
depreciation is calculated at 11,5% p.a. on the straight line method? Give your
answer to the nearest year.
5 An item is currently valued at R21 500. What was its value 3 years ago if it has
depreciated at a rate of 16,5% p.a. simple decay over the past 3 years?
6 Calculate the original value of a motor car that has a book value of R151 471,31
at the end of 8 years if depreciation is calculated at 13,5% p.a. on the reducing
balance method.
7 Four friends, Tim, Thabo, Tracy and Thandi each received a lump sum of
R12 000. Tim invested his money in shares for a new company called Newco,
Tracy invested her money in shares for a company called Womco, and Thandi
bought shares in a company called Steepco. Thabo did not invest with the stock
market and invested his money in an investment that guaranteed a growth in his
money of 5,5% p.a. simple interest. After 3 years Newco shares had depreciated
at a rate of 2% p.a. on the straight line method, Womco shares had appreciated
at a rate of 4,8% p.a. compound interest, and Steepco shares had depreciated at
a rate of 2,8% p.a. on the reducing balance method.
Calculate the value of each of the four friends’ investments at the end of 3 years.
Then state who had received the best return on their investment, and who had
received the worst return.
8 Rhino poaching is a serious problem, and has resulted in rhinos becoming an
endangered species, particularly the black rhino. Statistics suggest that there were
60 000 black rhino in 1970, but that only 4 200 remained in 2011.
8.1 Determine the annual rate of depreciation that these statistics represent if
the rate is:
8.1.1 simple decay
8.1.2 compound decay.
8.2 Using your answer to 8.1.2, determine the number of black rhino there
would be in 2050 if this rate of compound decay continues.
A black rhino
Unit 1 Simple and compound decay 241
PLTMATHSLB11LB_12.indd 241 2012/07/07 11:27 AM
Unit 2: The effect of different periods of
compound growth and decay
In reality, compound interest is often compounded more often than once a year. Let
KEY WORD us consider the implications of this. Suppose that R10 000 is invested at an interest
rate of 12% p.a. for one year, but the interest is to be compounded quarterly. This
compounded quarterly 1 of 12% (= 3%) of the total at that
– interest is applied to an
means that each quarter year (every 3 months), __
4
investment every 3 months, stage will be added to the investment. We can follow the growth of the investment in
1 of the quoted annual
and __
4 the table below:
rate is applied to the total at
each stage Quarters passed Value of investment
0 10 000
1 10 300 10 000 + 0,03(10 000) [or 10 000 × 1,03]
2 10 609 10 300 + 0,03(10 300) [or 10 300 × 1,03]
REMEMBER
3 10 927,27 10 609 + 0,03(10 609) [or 10 609 × 1,03]
To find a percentage increase: 10 927,27 + 0,03(10 927,27)
4 11 255,0881
increase
_____________ × 100 [or 10 927,27 × 1,03]
original amount
Therefore the percentage
So, at the end of the year, the investment is worth R11 255,09. This is an increase of
1 255,09
increase = ________ × 100 R1 255,09.
10 000
= 12,5509%
In effect, therefore, the investment has had a growth of 12,5509% p.a. instead of the
quoted 12%. We call the quoted rate of 12% the nominal interest rate per annum,
KEY WORDS and the actual growth of 12,5509% the effective interest rate per annum. The more
often the interest is compounded, the more the investment will benefit from the fact
nominal interest rate – the
annual interest rate that is that compound interest is calculated by multiplying the growing total by the rate of
quoted growth each time interest is added to the investment.
effective interest rate – the
When interest is compounded more often than once a year, the effective interest rate
actual rate achieved per
will be higher than the nominal interest rate due to the effect of compound interest
annum
which adds interest to a growing total.
Notation
• We use i(m) to represent the nominal interest rate where the value of m represents
how often interest is compounded each year.
• We use ieff to represent the effective interest rate per year.
So, in the situation we have discussed, we can say i(4) = 12% p.a. and ieff = 12,5509%
We saw that R10 000 became R11 255 (to the nearest Rand) in one year based on the
interest rate of 12% p.a. compounded quarterly. We could have calculated this final
value using the formula for compound interest:
A = P(1 + i)n
A = 10 000(1 + 0,125509)1 = R11 255,09 | Using the effective annual interest rate for
one year
242 Topic 11 Finance, growth and decay
PLTMATHSLB11LB_12.indd 242 2012/07/07 11:27 AM
OR
REMEMBER
A = P(1 + i)n
(0,12 4
)
A = 10 000 1 + ____ = R11 255,09 | Using the quarterly interest rate for four quarters
4 Whenever we use this
( )
0,12 4 formula, the value that the
So, 10 000(1 + 0,125509) = 10 000 1 + ____ interest is divided by will
4
divide by 10 000 1 + 0,125509 = 1 +
0,12 4
____
4 ( )
where 0,125509 = ieff and 0,12 = i(4)
always be the same as the
value of the exponent. This
formula always looks at the
We can generalise these calculations into the following formula: 1 + ieff = ( 1 +
m )
i m m
_____ comparison of interest rates
per year, regardless of how
It is important to remember that this formula shows the comparison between the long the investment may last.
nominal and effective interest rates per annum, so the length of time that the money
is invested is irrelevant.
We could have used this formula to find the effective interest rate in the above
situation, where interest was quoted as 12% compounded quarterly, that is,
12 = 0,12
i(4) = 12% = ____
100
( 4 ) = ( 1 + ____
4 )
4 4 0,12 4
1 + ieff = 1 + i
_____
( 0,12 4
ieff = 1 + ____
4 )
– 1 = 0,12550881 ≈ 12,5509%
WORKED EXAMPLES
1 Determine the effective annual interest rate of 9,5% compounded monthly,
correct to two decimal places.
REMEMBER
SOLUTION The value in the bracket refers
9,5 to how many times per year
1 i(12) = 9,5% = ____ = 0,095
100 interest will be compounded.
1 + ieff = ( 1 + i__
m) = 1 +
m m
i___
( (12) 12
12 )
( 0,09512 )
= 1 + _____
12
KEY WORDS
= ( 1 + _____ )
0,095 12
– 1 = 0,0992475 ≈ 9,92% per annum – for the year
12
compounded semi-annually
2 Determine the nominal interest rate, compounded semi-annually, which – semi-annually means every
results in an effective annual interest rate of 11,5% p.a. (correct to one half year (every six months),
decimal place). which means that interest is
twice a year, so this is notated
SOLUTION as i(2)
11,5
2 ieff = 11,5% = ____ = 0,115 Find i(2)
100
1 + ieff = ( 1 + i__
m)
m m
1 + 0,115 = 1 + i__ ( 2 2
2 )
_____
( √1,115 – 1 ) × 2 = i 2
i 2 = 0,11187 ≈ 11,2% p.a.
Unit 2 The effect of different periods of compound growth and decay 243
PLTMATHSLB11LB_12.indd 243 2012/07/07 11:27 AM
3 Khabela deposits R7 000 into an account offering an interest rate of 8,5% p.a.
compounded monthly. Calculate how much he will have in his account at
the end of 4 years.
SOLUTION
8,5 0,085
3 8,5% = ____ = 0,085, so 0,085 per year = _____ per month
100 12
A = P(1 + i)n; n = 4 years = 4 × 12 months = 48 months
REMEMBER (
A = 7 000 1 + _____
0,085 48
12 )= R9 822,85
When interest is compounded
It is important to remember that the value of n must be the same unit of frequency
m times per year:
as that of the compounding interest. So the annual interest rate is divided by the
i is divided by m and n is
frequency per year, and the number of years is multiplied by the frequency per
multiplied by m.
year.
4 Keenan invested R15 000, and after 8 years his investment was worth
R32743,12. Determine the interest rate that he had received if interest was
compounded semi-annually.
SOLUTION
4 As interest is compounded every six months we are looking for i(2) p.a.,
(2)
i . As interest is applied
REMEMBER so every half-year the interest applied will be ___
2
i(2) is a notation that shows half-yearly, the value of n must be in terms of half-years.
that the nominal interest rate So n = 8 years = 16 half-years.
is compounded twice a year.
This does NOT mean that the So A = P(1 + i)n where A = 32 743,12 and P = 15 000
value of i is squared. i
32 743,12 = 15 000 1 + ___
_________
( )
(2) 16
2
√ 32
16 743,12
_________
15 000
– 1 = ___
2
i(2)
REMEMBER i 2 = 0,05 × 2 ≈ 0,1 = 10% p.a.
Every three months means
four times per year, so this is 5 Kirsten wanted to buy herself a car, but she first had to save more money. She
i(4) not i(3) deposited the money into an account offering 9% p.a. interest, compounded
every three months. After 2 years she had R102 582,31. Calculate how much
interest she earned on her investment.
REMEMBER
SOLUTION
Interest earned refers to the
amount of money that was 5 Firstly we calculate how much money Kirsten deposited into the account:
gained, that is, how much 9 = 0,09
A = P(1 + i)n where A = 102 582,31; i (4) = ____
100
more you had at the end
compared with what you
and n = 2 years = 8 quarter years
started with. This is not the
same as the RATE of interest,
( 0,09 8
) 0,09
102 582,31 = P 1 + ____ ; interest is 0,09 per year so ____ per quarter year
4 4
102 582,31
which is a percentage. __________ = P P = R85 855,07
Interest earned = A – P ( 1 + ____
4 )
0,09 8
Therefore the interest that Kirsten earned
= R102 582,31 – R85 855,07 = R16 727,24
244 Topic 11 Finance, growth and decay
PLTMATHSLB11LB_12.indd 244 2012/07/07 11:27 AM
Time lines
The formulae we have studied so far have involved a single transaction at a time.
When more than one transaction occurs which are linked to each other, it is useful to
picture the process over time with the help of a time line. If we list each transaction
on a time line, we will ensure that no mistakes are made with regard to interest gained
or lost in the transactions.
WORKED EXAMPLE 1
Henry deposited R4 000 into an account. The interest rate for the first 2 years was
6,5% p.a. compounded quarterly, 7% p.a. compounded semi-annually for the next REMEMBER
3 years and 8,5% p.a. effective thereafter. Calculate how much he will have saved
8,5% p.a. effective means
after 9 years.
that interest is compounded
once a year.
SOLUTION
After the first two years, the amount accumulated will be:
(
A = P( 1 + i )n = 4 000 1 + _____ )
0,065 8
4
0,065
If the rate is 0,065 per year, then it is _____
4
per quarter, and n = 2 years, which is
8 quarter years.
The answer to this becomes the present value for the next 3 years when the interest
rate has changed, so that the value at the end of these 3 years will be:
(
A = P(1 + i)n = 4 000 1 + _____ ) (
0,065 8
4
0,07 6
1 + ____
2 ) 0,07
A rate of 0,07 per year = ____
2
per half-year, and n = 3 years
= 6 half-years
P
The answer to this becomes the present value for the next 4 years, when the
interest rate changes again, this time to 8,5% effective, which means that it is
compounded annually. So the value at the end of these 4 years will be:
(
A = P(1 + i)n = 4 000 1 + _____ )(
0,065 8
4 )
0,07 6
1 + ____ (1 + 0,085)4
2
P
It is important that we do not do separate calculations along the way and round off
answers. This will lead to the final result being inaccurate. Always consider the full
situation and do one calculation, rounding off to two decimal places at the end.
As we can see from the calculations above, each time there is a change in interest
rate we multiply by another bracket. Each bracket shows the rate of interest and
time period for that stage of the investment.
Unit 2 The effect of different periods of compound growth and decay 245
PLTMATHSLB11LB_12.indd 245 2012/07/07 11:27 AM
We can visualise and summarise the situation using data on a timeline:
i 2 = 0,7 i = 0,085
T0 i = 0,065
4
T2 ( 0,065 8
4 000 1 + _____
4
) T5 ( )(
0,065 8
4 000 1 + _____
4 )
0,07 6
1 + ____
2
T9
4 000 Final amount
Notice how we can use a dotted line to signify a change in interest and write the
interest rate for each time period above the time line. When you draw your own
time line, you do not have to write the accumulated amounts along the way each
time. The final amount at the end of 9 years will be:
(
A = P(1 + i)n = 4 000 1 + _____
0,065 8
4 )(0,07 6
)
1 + ____ (1 + 0,085)4 = R7 752,21
2
WORKED EXAMPLE 2
Sunette took out a loan and paid it back in full with three payments: She paid
R10 000 after 2 years, R8 000 one year later, and R49 353,56 five years after the
loan was granted. If interest was charged at 7% p.a. compounded monthly, what
was the original value of her loan?
SOLUTION
Represent this information on a time line:
T0 i(12) = 0,07 T2 T3 T5
REMEMBER
Loan = x
1 = a–m
___
am
( )
0,07 –24
10 000 1 + ____
12
10 000
REMEMBER ( )
0,07 –36
8 000 1 + ____
12
8 000
( )
Interest earned refers to the 0,07 –60
49 353,56 1 + ____ 49 353,56
amount of money that was 12
gained, that is, how much
more you ended up with
It is important to understand that we can do calculations relating to money
than what you started with.
This is not the same as the
transactions over time at any stage on the time line, as long as all of the
RATE of interest, which is a transactions are compared at the same moment in time. As the value that we are
percentage. looking for in this question is at T0, it makes sense to find the value of each future
deposit at T0.
Part of each deposit made covers a portion of the loan debt, as well as some of the
interest that has to be paid on the loan. By finding the value at T0 for each deposit,
we find the portion of each deposit that covers the loan only. Therefore the sum of
these values at T0 will be the value of the loan.
246 Topic 11 Finance, growth and decay
PLTMATHSLB11LB_12.indd 246 2012/07/07 11:27 AM
When working back in time, we find the P value when the A value is known. So, REMEMBER
make P the subject of the formula:
Whenever you are taking
A = P(1 + i)n values back in time, divide by
the interest bracket, so the
∴ A
______ = P
n
(1 + i) exponent will be negative.
∴ A(1 + i)–n = P
Note: When working back in time on a time line we are not depreciating, so the
formula is NOT A(1 – i)n. We are, however, finding the value BEFORE any interest
growth, so we DIVIDE the future value by the interest bracket, resulting in the
bracket being raised to a negative exponent.
Therefore, in the example, we calculate Sunette’s original loan:
REMEMBER
(
x = 10 000 1 + ____ )
0,07 –24
12 ( )
0,07 –36
+ 8 000 1 + ____
12 ( 0,07 –60
+ 49 353,56 1 + ____
12 )= R50 000
When money is withdrawn
0,07 from an account, the loss to
Note: To simplify this calculation, first calculate 1 + ____, then enter:
12 the account is not only the
10 000 × ANS–24 + 8 000 × ANS–36 + 49 353,56 × ANS–60 = R50 000 value of the withdrawal, but
also the potential interest
that the money would have
earned if it had not been
WORKED EXAMPLE 3 withdrawn. So the value of
the withdrawal together with
Mrs Naidoo opened a savings account and the following transactions occurred: its potential interest growth,
will have to be subtracted
She deposited R15 000 immediately.
from the total in the account.
Three years later she withdrew Rx.
After a further four years she deposited R21 500.
Seven years later (fourteen years after the initial investment), she had R25 735,50
in her account.
Interest is calculated at 6% p.a. compounded annually for the first five years,
and 10% p.a. compounded quarterly for the next nine years. Draw a time line to
represent the above and calculate the value of x.
SOLUTION
T0 i = 0,06 T3 T5 T7 i 4 = 0,1 T14
15 000 (
15 000(1 + 0,06)5 1 + ___ )
0,1 36
4
–x (
–x(1 + 0,06)2 1 + ___ )
0,1 36
4
21 500 ( )
0,1 28
21 500 1 + ___
4
25 735,50
Unit 2 The effect of different periods of compound growth and decay 247
PLTMATHSLB11LB_12.indd 247 2012/07/14 2:54 PM
Although the more instinctive way of calculating this is to work out how much
the original deposit had accumulated after 3 years, and to then subtract the
withdrawal. Find how much this total had accumulated to be at year 7, add
the next deposit, and then allow this total to grow with interest to year 14, this
[stop-start’] approach can be very tedious, and the calculations can become rather
unwieldy. It is far better to consider each transaction as separate values, and to
create an equation with these separate amounts once you have taken them all to
the same moment in time.
The most obvious moment in time would be to take each transaction to the end,
that is, to term 14, and the equation would then be:
(
0,1 36
15 000(1 + 0,06)5 1 + ___
4 )
0,1 36 0,1 28
– x(1 + 0,06)2 1 + ___ + 21 500 1 + ___ = 25 735,50
4 4 ( ) ( )
5
15 000(1 + 0,06) 1 +
0,1 36
___
4 (
+ 21 500 1 +
0,1 28
___
4 )
– 25 735,50 ( )
( 0,1
= x(1 + 0,06)2 1 + ___
4
36
)
15 000(1 + 0,06)5 1 + ___
4 ( 4 ) + 21 500( 1 + ___
0,1 36
__________________________________________________ =x
0,1
) – 25 735,50 28
(1 + 0,06) ( 1 + ___ )
0,1 36 2
4
It would have been easier, however, to take all of the values to term 5 where the
interest changes, as this would require no changes of interest, and therefore no
double interest brackets.
T0 i = 0,06 T3 T5 T7 i(4) = 0,1 T14
15 000 15 000(1 + 0,06)5
–x –x(1 + 0,06)2
21 500 1 + ___ ( 0,1 –8
4 ) 21 500
(
25 735,50 1 + ___ )
0,1 –36
4
25 735,50
So, at year 5:
REMEMBER 0,1 –8
15 000(1 + 0,06)5 – x(1 + 0,06)2 + 21 500 1 + ___
4
0,1 –36
= 25 735,50 1 + ___
4 ( ) ( )
When money is withdrawn
from an account, the loss to
5
15 000(1 + 0,06) + 21 500 1 +
0,1 –8
___
4
– 25 735,50 1 +
0,1 –36
___
4 (
= x(1 + 0,06)2 ) ( )
the account is not only the
value of the withdrawal, but
15 000(1 + 0,06)5 + 21 500 1 + ___
4 4
__________________________________________________ =x
( ) – 25 735,50( 1 + ___
0,1 –8 0,1
) –36
2
(1 + 0,06)
also the potential interest
that the money would have
earned if it had not been x = R24 154,27
withdrawn. So we subtract Always remember that as long as all of the values are taken to the same moment
the value of the withdrawal,
in time, we can the calculate at the most convenient moment on the time line. If
together with its potential
there is a change in the rate of interest, then the moment when the interest rate
interest growth, from the
total in the account. changes is usually the most effective moment to choose.
248 Topic 11 Finance, growth and decay
PLTMATHSLB11LB_12.indd 248 2012/07/07 11:27 AM
EXERCISE 2
1 Determine the effective annual interest rate, to one decimal place, for these
interest rates:
1.1 10,5% p.a. compounded quarterly
1.2 6,5% p.a. compounded daily
2 If an effective annual interest rate is 8%, determine the nominal interest rate p.a.
(correct to two decimal places) if interest was compounded:
2.1 every six months
2.2 every three months.
3 Khadija invested R25 000 into an account offering interest at 10,3% p.a.
compounded quarterly.
3.1 Determine how much she has in her account after 6 years.
3.2 How much interest has she received after 6 years?
3.3 What is the effective annual interest rate that Khadija obtained (to one
decimal place)?
4 Wouter has R8 000 to invest over a period of 10 years. If he requires R15 000 at
the end of the ten-year period, what annual interest rate, compounded monthly,
will he need? Give your answer correct to one decimal place.
5 Vicky invests R10 000 for a period of 10 years. During the first 3 years, the
interest rate is 9% p.a. compounded monthly. Thereafter, interest changes
to 12% p.a. compounded semi-annually. Calculate the future value of the
investment after 10 years.
6 Mr Jacobs invested R60 000. Four years later he withdrew R5 000 from his
account. After a further two years he deposited R8 000. Interest was 10% p.a.
compounded half-yearly. Use a time line to determine how much he had in his
account after a total of 10 years.
7 Mrs Mohamed opens a savings account and these transactions take place:
• She deposits R8 000 immediately, and a further R6 000 five years later.
• Two years after the deposit of R6 000 she withdraws R10 000.
• Interest is calculated at 10% p.a. compounded annually for the first two years,
and 9,5% p.a. compounded quarterly thereafter.
Draw a time line to represent the above and calculate the amount of money that
she will have saved after 10 years.
8 Uthmaan purchased a car five years ago. After paying a deposit, he took out a
loan for the balance that he owed. He paid off the loan with two payments: R30
000 after 2 years and a final payment of R113 582,40 which he made 5 years after
taking out the loan. Interest on the loan was 10% p.a. compounded monthly
during the first 3 years, and 11,5% p.a. effective for the remaining 2 years. Draw
a time line and determine the original price of the car.
9 Dumisani borrows R150 000 and plans to repay the loan over a period of 5 years
as follows:
• R5 000 after one year, R10 000 two years later, and a final payment of Rx at
the end of the five-year period. Interest is charged as follows:
• 14% p.a. for the first 3 years and then 10,5% p.a. compounded quarterly for
the last 2 years.
Calculate the value of Dumisani’s final payment by drawing a time line.
Unit 2 The effect of different periods of compound growth and decay 249
PLTMATHSLB11LB_12.indd 249 2012/07/07 11:27 AM
Revision
1 Determine the rate of depreciation for an item to become half of its
original value after 5 years on the straight line method of depreciation. [4]
2 Determine the length of time (to two decimal places), that it will take
for an investment of R450 000 to be depreciate to R220 000, if the rate
of simple decay is 12% p.a. [4]
3. What nominal annual interest rate, compounded monthly, would give
the same return on your investment as 9% p.a. effective? Give the answer
to one decimal place. [4]
4 Sarah will need R800 000 to buy a flat in 5 years time.
4.1 Calculate how much she must deposit now into an account
offering 10% p.a. compounded monthly, to have the necessary
funds in 5 years time. (4)
4.2 What was the effective interest rate, to 2 decimals, that Sarah
received each year? (3)
[7]
5 Fatima invested R10 000 into an investment offering 8% p.a. compounded
quarterly. At the same time Bongi invested R10 000 into shares which
depreciated at a rate of 4, 5% p.a. on the reducing balance method. Calculate
how much better off Fatima is than Bongi after 5 years. [6]
6 Thomas deposits R12 500 into an account at an interest rate of 9, 5%
compounded quarterly. Two years later, he deposits another R10 000.
Three years after that, he withdraws R8 000. Calculate how much will
he have in the account at the end of 8 years. [8]
7 Marc buys a fridge on hire purchase at an interest rate of 12 % p.a.
After paying a 10% deposit, he continues to pay monthly payments REMEMBER
of R300 for 3 years.
Inflation uses compound
7.1 Calculate the cash price of the fridge if he had bought it immediately. (5)
interest.
7.2 Calculate how much interest he paid. (2)
[7]
8 Calculate how long it would take for an investment to become one third
of its original value at a rate of simple decay of 12% p.a. Give your answer
correct to one decimal place. [4]
9 Determine the compound interest rate, correct to two decimal places,
Emile would require to triple his investment over a period of 10 years? [4]
10 I plan to purchase two airline tickets in two years time. At the moment
the tickets cost R15 200.
10.1 Calculate the cost of each ticket in two years’ time if an inflation
rate of 6,5% p.a. is taken into consideration. (3)
10.2 Calculate how much I need to invest now at 7, 5% p.a. interest
compounded quarterly, to provide for the purchase of these two
airline tickets in two years’ time. (4)
[7]
250
PLTMATHSLB11LB_12.indd 250 2012/07/07 11:27 AM
11 Nicola makes two deposits into an account: a first amount of R5 000
and R2 000 four years later. The interest was 5% p.a. compounded
annually for the first 5 years, and then increased to 6% p.a. compounded
half-yearly. Calculate the total value of these deposits after 7 years. [8]
12 Zafar receives a loan, which he plans to pay back in three payments:
R4 000 in 2 years’ time, R8 000 in 4 years’ time and R6 000 in 5 years’
time. The interest rate is 10% p.a. compounded semi-annually.
Calculate the amount of money that he receives as a loan now. [6]
13 Mr Sandler deposits a lump sum into an account now to allow for
withdrawals when each of his three children starts tertiary education.
He calculates that he will need R30 000 for the first withdrawal in
5 years’ time, R32 500 for the second withdrawal 2 years later, and
R36 000 for the final withdrawal 10 years from now. If the interest rate
is 12% p.a. compounded monthly, calculate how much he deposits now? [8]
14 Mpho deposits R15 000 into a savings account to save for a holiday in
3 years time. The account offers her 10% p.a. compounded monthly.
14.1 How much will she have accumulated at the end of 3 years? (4)
14.2 What was the effective interest rate that Mpho received each year
(to two decimal places)? (3)
14.3 How much interest had Mpho received at the end of 3 years? (2)
[9]
15 Marisha invested R2 500 in a savings account. Calculate how much she
will have in the account at the end of 5 years if the interest rate is 7,5% REMEMBER
compounded quarterly for the first year, 6% compounded monthly for
Hire purchase uses simple
the second year and 8,3% compounded semi-annually thereafter. [7]
interest.
16 Kevin has a scooter that cost him R8 000. He wishes to replace it with
the equivalent model in 5 years’ time. If depreciation on his scooter is
at 15% per annum on the reducing balance method, and inflation is at
5,3% p.a., calculate:
16.1 the expected cost of the equivalent model in 5 years’ time (3)
16.2 the expected book value of his scooter in 5 years’ time (3)
16.3 the amount of money Kevin is expecting to have to pay in if he
sells his old scooter to offset the cost of the new one. (2)
[8]
17 At the beginning of 2005 Joseph invested R10 000 into an account with
an interest rate of 6,5% compounded every 3 months. At the beginning
of 2007 an extra R5 000 was added to the account. At the beginning of
2010 he had to withdraw a certain amount from the account.
17.1 How much did he withdraw if the final amount yielded at the
end of year 2011 was R20 169,13? (10)
17.2 What was the effective annual interest rate, correct to two
decimal places? (3)
[13]
251
PLTMATHSLB11LB_12.indd 251 2012/07/07 11:27 AM
TOPIC
212 Probability
Unit 1: Addition and complementary rules;
Dependent and independent events
Revision of Grade 10 probability
The probability of an event taking place is always:
total number of ways an event can occur
_________________________________________
total number of possible outcomes for the event.
For example, the probability of getting a 4 when you throw a die is __ 1 . There is
6
only one 4 on a die, so the number of ways of getting a 4 is 1. The total number of
KEY WORDS possibilities is 6, as there are 6 numbers altogether on a die. This means that one in
every six throws is likely to be a 4. This is the theoretical outcome, and although it
theoretical outcome – the may not always be the case in an experiment, the more throws in an experimental
result, in theory, that is
situation, the closer the experimental outcome will be to the theoretical outcome.
expected for a certain event
to happen
The probability of an event taking place will range between being impossible,
experiment – a trial situation
involving chance (probability = 0) to certain (probability = 1, or 100%).
These are some notations that we use in probability questions:
• P(A) means the probability of event A happening.
• P(A′) means the probability of event A not happening.
• P(A ∩ B) means the probability of event A and event B happening.
• P(A ∪ B) means the probability of event A or event B happening.
• P(A | B) means the probability of event A happening, assuming that event B is
taken as given.
Addition rule
In general, the addition rule is true in all cases where we consider two events:
• P(A or B) = P(A) + P(B) – P(A and B)
If events A and B are mutually exclusive, it follows that P(A and B) = 0
REMEMBER
Therefore, the addition rule for mutually exclusive events is:
Mutually exclusive events
are events that have no • P(A or B) = P(A) + P(B) if A and B are mutually exclusive
intersection or overlap, so it is
impossible for them to both
WORKED EXAMPLES
occur.
1 If a die is rolled, determine the probability of it being a six or a one.
2 If a card is drawn from a pack of playing cards, determine the probability of it
being a ten or a heart.
252 Topic 12 Probability
PLT MATHS LB 11 7th pgs (Real Book).indb 252 2012/07/02 2:24 PM
SOLUTIONS
1 Call event A ‘rolling a 6’ and event B ‘rolling a 1’. These are mutually
exclusive so:
1 + __
P(rolling a 6 or rolling a 1) = P(A or B) = P(A) + P(B) = __ 1=2__ = 1
__
6 6 6 3
2 Call event A ‘drawing a ten’ and event B ‘drawing a heart’. We cannot use
the addition rule in this example, because events A and B are not mutually
exclusive as there is an overlap. It is possible for the card to be a ten as well as
a heart (there is a ten of hearts). Therefore we use the general rule:
1 + __
P(A or B) = P(A) + P(B) – P(A and B) = ___ 1 – ___
1 = ___
4
13 4 52 13
Complementary rule
We say A and B are complementary events if both these statements are true: KEY WORD
• They are mutually exclusive. exhaustive events – events
• P(A) + P(B) = 1 (This implies that the two events are exhaustive, as a probability which use up the full sample
of 1 is 100%.) space, that is, all possible
outcomes
This would then mean that P(B) = P(A′) = 1 – P(A)
WORKED EXAMPLES
Given the set of numbers: {3; 4; 5; 6; 7; 8}
Are these events complementary:
1 ‘choosing an even number’ and ‘choosing a prime number’
2 ‘choosing an odd number’ and ‘choosing a factor of 8’
3 ‘choosing an odd number’ and ‘choosing a factor of 24’?
SOLUTIONS
1 The even numbers in the given set are A = {4; 6; 8} and the prime numbers
are B = {3; 5; 7). So, these events are mutually exclusive.
3 = __
and P(A) = __ 3 = __
1 and P(B) = __ 1 P(A) + P(B) = __
1 + __
1=1
6 2 6 2 2 2
these events are complementary.
2 The odd numbers in the given set are A = {3; 5; 7} and the factors of 8 are
B = {4; 8}. So, these events are mutually exclusive.
3 = __
and P(A) = __ 1 and P(B) = __
2 = __
1 P(A) + P(B) = __
1 + __ 5≠1
1 = __
6 2 6 3 2 3 6
these events are not complementary.
3 The odd numbers in the given set are A = {3; 5; 7} and the factors of 24 are
B = {3; 4; 6; 8}. These events are not mutually exclusive as 3 is an element
of both sets, so there is an overlap.
these events are not complementary.
Unit 1 Addition and complementary rules; Dependent and independent events 253
PLT MATHS LB 11 7th pgs (Real Book).indb 253 2012/07/02 2:24 PM
Dependent and independent events and the
Note:
product rule
• Every time that you toss a
Two events are independent when the result of the first event does not affect the
coin, the result ‘heads’ or
result of the second event. Examples of such events are:
‘tails’ is not dependent on
the result of the previous • When drawing a card from one pack and a card from another pack, the two events
throw. are independent of each other. The possibilities of drawing whichever card is
• Every time a child is
drawn from the first pack will not be affected by the card that is drawn from the
born the possibility of it
second pack.
being a boy or girl is not
dependent on the sex of • If a card is drawn from one pack and replaced, and then a second card is drawn,
the previous child born. the possibilities of whichever card is drawn second will not be affected by the card
that was drawn first, so these two events are independent.
• A marble is chosen from a bag that contains 2 green and 5 yellow marbles, and
then replaced. If a second marble is then chosen at random, the possibilities of its
colour does not depend on the colour of the first marble removed.
The following situations show events that are NOT independent:
• If a bag contains 3 red and 2 blue marbles, the probability of drawing a red marble
3 . If the marble is not replaced, the probability of drawing a red marble on a
is __
5
second draw is __ 2 = __ 3 if the first marble had been
1 if the first marble was red, but __
4 2 4
blue.
So the probability of drawing a red marble a second time when the first marble is
not replaced, is dependent on the outcome of the first draw.
• you draw two cards from the same pack without replacement, the probability of
If
what you draw the second time will depend on what you drew the first time, so the
two draws are not independent of each other.
The product rule for independent events is: P(A and B) = P(A) × P(B)
REMEMBER The product rule for dependent events is: P(A and B) = P(A) × P(B) | A
P(B | A) means the
probability of B happening,
WORKED EXAMPLES
given that event A has already
happened.
1 If you roll two dice, what is the probability of getting a six on both dice?
2 A and B are two events such that P(A) = __2 and P(A ∩ B) = ___4.
5 15
If it is given that A and B are independent events, find P(B).
3 A bag contains four R2 coins, six R5 coins and three R1 coins. If two coins are
chosen randomly and not replaced, what is the probability that both coins
are R5 coins?
254 Topic 12 Probability
PLT MATHS LB 11 7th pgs (Real Book).indb 254 2012/07/02 2:24 PM
SOLUTIONS
1 As the result on each die is not affected by the result on the other die, these
are independent events. We want to ‘get a 6 on the first die’ (call that event
A) and ‘get a 6 on the second die’ (call that event B).
1 × __
So, P(A and B) = P(A) × P(B) = __ 1 = ___
1
6 6 36
2 As these are independent events, P(A ∩ B) = P(A) × P(B)
4 = __
___ 2 × P(B)
15 5
4 ÷ __
P(B) = ___ 2 = ___ 5 = __
4 × __ 2
15 5 15 2 3
3 As the first coin is not replaced, the outcome of the second event depends
on what coin was drawn first, so these are dependent events. The probability
6 but if this is not replaced, there will
of choosing R5 in the first draw is ___
13
be only five R5 coins left, out of a total of 12 coins left in the bag, so the
probability of choosing a second R5 coin will be ___ 5.
12
6 × 5
Therefore P(A and B) = P(A) × P(B) | (A) = ___ 5
__ = ___
13 12 26
We will discuss further applications of independent events in the context of Venn
diagrams, tree diagrams and contingency tables.
EXERCISE 1
1 A coin is tossed and a die is rolled. Determine the probability that the outcome
will be:
1.1 a head on the coin and a 5 on the die
1.2 a tail on the coin or a prime number on the die
1.3 a head on the coin and not a 6 on the die.
2 Are the following events complementary? Give a reason for each answer.
2.1 Scoring less than 4 when rolling a die, and scoring more than 4 when
rolling a die.
2.2 Choosing all consonants from the word MATHS, and choosing an A
from MATHS.
2.3 Drawing a red card and drawing a king from a pack of cards.
3 Are the following events independent or dependent?
3.1 Selecting a card and then choosing a second card without replacing the
first card
3.2 Rolling a die and tossing a coin
4 A bag consists of five green marbles and eight blue marbles. If one marble is
drawn, then replaced, and a second marble is drawn, determine the probability
that:
4.1 both marbles are blue
4.2 the first marble is blue and the second marble is green
4.3 the first marble is green or the second marble is green.
5 Repeat questions 4.1 and 4.2, assuming that the first marble is not replaced
before drawing the second marble.
Unit 1 Addition and complementary rules; Dependent and independent events 255
PLT MATHS LB 11 7th pgs (Real Book).indb 255 2012/07/02 2:24 PM
Unit 2: Venn diagrams
When more than one event takes
Outcomes that apply to place, we can represent the outcomes
What is written
both A and B of these events in various ways.
here shows
the total of the You studied the first representation
A B sample space in Grade 10, and this assists you
with solving problems. In this
representation, we use circles to
Anything
written here represent each event. If it is possible
shows outcomes for both events to take place, the
that are part circles will intersect. The area
of the sample enclosed in the overlap will show
space, but are A∩B when both events happened. We
not relevant to place the circles inside a rectangle,
either A or B Outcomes that apply Outcomes that apply where the rectangle represents the
to A only to B only
sample space.
The sketches that follow show possible Venn diagrams. In each case the notation
below the diagram describes what the shaded region represents.
A B A B A B
A∪B B′ A ∩ B′
A B A B
KEY WORD
sample space – the full set of C C
data values
A∩B∩C (A ∪ B ∪ C)′
A B A B
C C
C ∩ (B ∪ A)′ A ∩ B ∩ C′
256 Topic 12 Probability
PLT MATHS LB 11 7th pgs (Real Book).indb 256 2012/07/02 2:24 PM
WORKED EXAMPLE 1 A B
piano guitar
The Venn diagram shows the results of asking learners whether they play the
guitar or the piano.
1 Determine the probability that a learner chosen at random from the group: 10 7 15
1.1 plays the piano, but not the guitar
1.2 does not play either instrument
1.3 plays the guitar
18
1.4 plays the piano, given that he/she plays the guitar?
2 Give reasons to explain why the events ‘playing piano’ and ‘playing
guitar’ are not mutually exclusive, exhaustive or independent events.
SOLUTION
1 1.1 From the Venn diagram, there are10 learners who play the piano but not the guitar, so:
10 = 1
P(piano ∩ not guitar) = P(A ∩ B′ ) = ___ __
50 5
1.2 There are 18 learners who do not play either instrument:
18 = ___
P(no instrument) = ___ 9
50 25
1.3 There are 7 + 15 = 22 learners who play the guitar:
22 = ___
P(plays guitar) = P(B) = ___ 11
50 25
1.4 From the 22 learners who play the guitar, we can see that there are seven who also play the piano:
7
P(plays piano given plays guitar) = P(A | B) = ___
22
2 P( A ∪ B ) = 32
___
50
17 + 22
P(A) + P(B) = ___ 39
___ = ___
50 50 50
∴ P( A ∪ B ) ≠ P(A) + P(B)
i.e. P( A ∩ B ) ≠ 0
Therefore these events are not mutually exclusive.
17 + ___
P( A ) + P( B ) = ___ 39 ≠ 1
22 = ___
50 50 50
i.e. P( A ∪ B )´ ≠ 0
Therefore these events are not exhaustive.
7 = 0,14
P( A ∩ B ) = ___
50
17 × ___
P( A ) × P( B ) = ___ 187 = 0,1496
22 = _____
50 50 1 250
P( A ∩ B ) ≠ P( A ) × P( B )
Therefore these events are not independent.
WORKED EXAMPLE 2
Fifty learners were asked if they had ever broken an arm, a leg or their nose.
21 had broken a leg 28 had broken an arm
8 had broken their noses 9 had broken an arm and a leg
5 had broken an arm and their nose 6 had broken their nose and a leg
10 had not broken any of these
1 Display this information in a Venn diagram. Then determine how many
learners had broken an arm, a leg and their nose.
Unit 2 Venn diagrams 257
PLT MATHS LB 11 7th pgs (Real Book).indb 257 2012/07/02 2:24 PM
2 Determine how many learners had only broken a leg.
3 A learner is randomly chosen from those surveyed. Find the probability that:
3.1 the learner had broken an arm only
3.2 the learner had broken an arm given that he or she had broken a leg
3.3 the learner had not broken their nose
3.4 the learner had broken a nose and a leg
3.5 the learner had broken an arm or a leg.
SOLUTION
1 Always start by filling in the central area, which 50
Broken Broken
represents the number of learners who had broken all leg arm
three body parts. We do not know this value, so we call
it x. Then look for the given information that refers to
the number in each overlapped area. Nine learners had
6+x 9–x 14 + x
broken an arm and leg, so there are 9 – x in the rest of
the overlap for a broken arm and leg. Similarly, fill in all
x
the other overlap areas.
6–x 5–x
We know that 21 learners had broken a leg. So the
broken leg circle must contain 21 in total. So 21 – (9 –
x) – x – (6 – x) = 6 + x in the non-overlapping part of the x–3
circle. Similarly, for the other circles we get 14 + x and
Broken
x – 3 for the non-overlapping part. nose
10
We place the ten learners who have not broken any of
the three body parts in the square, but outside the circles.
Now solve for x, knowing that the total number of people is 50.
6 + x + 9 – x + 14 + x + 6 – x + x + 5 – x + x – 3 + 10 = 50
x=3
We can now complete the Venn diagram without any variables.
2
Broken Broken 50
leg arm
9 6
17
3
3 2
0
Broken
nose 10
Nine learners had broken a leg only.
3.1 17
P(broken arm only) = ___
50
258 Topic 12 Probability
PLT MATHS LB 11 7th pgs (Real Book).indb 258 2012/07/02 2:24 PM
3.2 3
9 = __
P(broken arm | broken leg) = ___
21 7 REMEMBER
Using the broken leg circle as the total sample space, we see that nine
The sample space is all
learners (6 + 3) have also broken an arm.
possible outcomes.
3.3 P(Not broken nose) = ___ 21 | There are 8 learners in the broken nose
42 = ___
50 25
circle: 3 + 3 + 2, so there are 42 who have not broken a nose, out of the
total 50 learners
3.4 3 | There are 6 learners in the intersection between
6 = ___
P(Nose ∩ Leg) = ___
50 25
Nose and Leg
3.5 4 | There are 40 learners in total in the Arm and Leg
40 = __
P(Arm ∪ Leg) = ___
50 5
circles: 9 + 6 + 3 + 3 + 17 + 2
WORKED EXAMPLE 3
Eighty percent of the learners at a local high school play sport and 62,5 percent belong to a society. Fifty percent of
the learners play a sport and belong to a society.
1 Draw a Venn diagram to represent this information.
2 What is the probability that a learner who does not belong to a society, plays sport?
3 What is the probability that a learner belongs to a society or plays a sport?
4 What is the probability that a learner belongs to no society nor plays a sport?
5 Determine whether playing sport and belonging to a society are independent events in this high school.
i.e. = 100%
SOLUTION
Society Sport 1
Notice that in this example, percentages instead of actual numbers of
learners are given. We can change these percentages to decimal values,
0,5
and enter the probability of an event into each area of the Venn diagram, 0,3
0,125 (or
instead of the actual values. (or 12,5%) (or 30%)
50%)
1 Fill in the intersection, which is 50%, or 0,5. Then, knowing that
0,075
62,5% (or 0,625) belong to a society, we can determine that 10% (or
0,1) belong to a society but do not play a sport. (62,5% – 50% = 12,5%).
Similarly, 80% – 50% = 30% (or 0,3).
So 0,3 play a sport but do not belong to a society.
the number of learners who play sport
2 We need to find ______________________________________________.
the number of learners who do not belong to a society
This is P(sport | does not belong to a society) = 0,3
3 P(society ∪ sport) = = 0,125 + 0,5 + 0,3 = 0,925
4 P(society or sport)′ = 0,075
5 P(Society ∩ Sport) = 0,5
P(Society × P(Sport) = 0,625 × 0,8 = 0,5 = P(Society ∩ Sport)
∴ these are independent events
Unit 2 Venn diagrams 259
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WORKED EXAMPLE 4
We know the following facts about a group of 32 learners:
• 12 of them like hamburgers
• 16 of them like hotdogs
• 8 of them like chips, but not hamburgers and hotdogs
• 7 of them like chips and hamburgers
• 13 of them like chips and hotdogs
• 2 of the learners who like hotdogs and hamburgers also like chips
• All the learners like either hotdogs, hamburgers or chips
1 Draw a Venn diagram to represent this information, using A for hamburgers, B for hotdogs and C for chips.
2 Explain with reasons why:
2.1 A, B and C are exhaustive events
2.2 A, B and C are not complementary events
2.3 B and C are independent events
A 32
2.4 A and B are not independent events.
B
SOLUTION 2
1 The second last piece of information tells us that 2 learners like all three, so we fill in
8
that first at the intersection of all 3 circles. The third piece of information tells us that C
there are 8 who like chips only, so we fill that into the chips only part of the diagram.
Next, because 7 like chips and hamburgers, 7 – 2 = 5 must like chips and A 32
hamburgers but not hotdogs. B
Similarly, as 13 like chips and hotdogs, 13 – 2 = 11 must like chips and hotdogs but
2
not hamburgers 5 11
Since we do not know how many learners like hamburgers and hotdogs but not
8
chips, we let that value be x, and complete the rest of the diagram: C
Hamburgers only: 12 – 5 – 2 – x = 5 – x
Hotdogs only: 16 – 2 – 11 – x = 3 – x A 32
We know from the last piece of information given that none of the x B
32 learners who fall outside of any of the 3 circles, we can say:
2
5 – x + x + 2 + 5 + 11 + 8 + 3 – x = 32 5 11
Therefore x = 2
8
The Venn diagram is now complete: C
2.1 P( A ∪ B ∪ C )´ = 0 so A, B and C are exhaustive.
32 = 1 A 32
2.2 P( A ∪ B ∪ C ) = ___
32
16 + ___
12 + ___ 26 = ___
54 ≠ P( A ∪ B ∪ C ) 3 2 1 B
P( A ) + P( B ) + P( C ) = ___
32 32 32 32
2 ≠0 2
Also P( A ∩ B ∩ C ) = ___ 5 11
32
So A, B and C are not mutually exclusive. 8
12 + 16 26 = ___
54 ≠ 1 C
Also P( A ) + P( B ) + P( C ) = ___ ___ + ___
32 32 32 32
So A, B and C are not complementary.
260 Topic 12 Probability
PLTMATHSLB11LB_13.indd 260 2012/07/07 11:31 AM
13
2.3 P( B ∩ C ) = ___
32
16 × ___
P( B ) × P( C ) = ___ 13
26 = ___
32 32 32
∴ P( B ∩ C ) = P( B ) × P( C )
So B and C are independent events.
2 = ___
2.4 P( A ∩ B ) = ___ 1
32 16
3
16 = ___
12 × ___
P( A ) × P( B ) = ___
32 32 16
∴ P( A ∩ B ) ≠ P( A ) × P( B )
So A and B are not independent events.
EXERCISE 2
1 A group of 120 people visiting a music shop were interviewed and asked if they
had listened to Freshly Ground or Steve Hofmeyer. The results show that 73
had listened to Freshly Ground, 57 had listened to Steve Hofmeyr and 28 had
listened to both.
Using F for Freshly Ground and S for Steve Hofmeyr, draw a Venn diagram to
represent this data. Use your Venn diagram when you answer the questions.
1.1 How many people interviewed had listened to neither Freshly Ground nor
Steve Hofmeyr?
1.2 Determine the probability that a person chosen at random from the group
interviewed had listened to:
1.2.1 Freshly Ground and not Steve Hofmeyr
1.2.2 either Freshly Ground or Steve Hofmeyr or both
1.2.3 both Freshly Ground and Steve Hofmeyr.
1.3 Determine:
1.3.1 P(F ∪ S)’ 1.3.2 P(S ∩ F’) 1.3.3 P(F ∪ S’)
1.4 Explain why listening to Freshly Ground and listening to Steve Hofmeyer
are not mutually exclusive, exhaustive or independent events, based on
this survey.
2 Let U (the universal set) be all integer values from 1 to 30. Let A, B and C be
subsets of U, so that: A = {x: x is a factor of 30}; B = {x: x is an odd number} and KEY WORD
C = {x: x is a perfect square}
universal set – the sample
2.1 List the elements of sets A, B and C. space, which is the full set of
This means ‘the number
2.2 List the elements of A ∩ B. data values
of elements in the set that
2.3 List the elements of B ∩ C. follows’.
2.4 Determine n(A ∩ B ∩ C).
2.5 Copy and complete the Venn diagram and indicate the number of
elements that would be included in each section of the Venn diagram. Use A B U
the Venn diagram to answer the following questions.
2.6 What is the probability that a number chosen randomly from 1 to 30,
will be:
2.6.1 a factor of 30
2.6.2 an odd number that is not a perfect square
2.6.3 not a perfect square C
2.6.4 not a factor of 30, an odd number or a perfect square?
2.7 Show that, while being a factor of 30 and being an odd number from 1 to
30 are not mutually exclusive events, they are independent events.
Unit 2 Venn diagrams 261
PLT MATHS LB 11 7th pgs (Real Book).indb 261 2012/07/02 2:24 PM
3 Fifty people were surveyed at a shopping mall and asked whether they like to
A B
shop at ‘Perfect Purchases’ or at ‘Shopaholics’ supermarkets. The Venn diagram
0,6 0,3 shows the results of the survey. A represents ‘Perfect Purchases’ and B represents
‘Shopaholics’, showing the probability of each choice.
3.1 Suggest what is implied by the two circles that do not overlap.
x
3.2 How many people said that they shop at ‘Perfect Purchases’?
3.3 Determine the value of x.
3.4 Determine:
3.4.1 P(A ∩ B) 3.4.2 n(A ∪ B)′ 3.4.3 P(A ∪ B ′)
3.5 Determine whether A and B are complementary events showing your
reasoning.
4 A group of 70 learners were asked about their subject choice. Their responses
showed that: 32 take Physical Sciences, 43 take Mathematics, 25 take Life
Sciences and 6 take none of these three subjects. Also, 18 take Physical Sciences
and Mathematics but not Life Sciences; 12 take Life Sciences only and 5 take
Physical Science and Life Sciences.
4.1 Draw a Venn diagram to represent this information. Use the letters S, M
and L to represent Physical Sciences, Mathematics and Life Sciences.
4.2 Determine:
4.2.1 the number of learners who take Physical Sciences, Mathematics
and Life Sciences
4.2.2 P(M ∩ L)′ 4.2.3 P(S | M) 4.2.4 nP(S ∩ M ∩ L′)
4.2.5 the probability that a learner who takes Life Sciences does not take
Mathematics
4.2.6 the probability that a learner taking Life Sciences and Mathematics
does not take Physical Sciences.
5 A veterinarian surveyed 50 of his clients, and found that 26 have dogs, 24 have
cats and 12 have birds. Ten clients have dogs and cats, four have dogs and birds,
one has a cat and a bird, and two have dogs, cats and birds.
5.1 Draw a Venn diagram to represent this information using the letters D, C
and B to represent dogs, cats and birds.
5.2 Determine:
5.2.1 n(D ∪ C ∪ B) ′ 5.2.2 n(D ∩ C ′) 5.2.3 P(B | C)
5.2.4 P'(D ∩ C ∩ B) ′.
6 A group of 50 people were asked whether they like dark chocolate, milk chocolate or
D M white chocolate. The probabilities of each are shown in the Venn diagram.
0,14 0,06 0,44
6.1 Determine the probability that a person did not like any of the three types
0,1
of chocolates.
0,02 0,04
0,08
6.2 Determine the number of people from those surveyed who:
6.2.1 like all three types of chocolate
W 6.2.2 do not like any of these types of chocolate
6.2.3 like dark and white chocolate but not milk chocolate
6.2.4 like milk chocolate only
6.2.5 like dark chocolate and milk chocolate.
262 Topic 12 Probability
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Unit 3: Using tree diagrams to solve problems
REMEMBER
for events not necessarily independent
Simultaneously means at the
When more than one event takes place consecutively or simultaneously, it is useful to same time.
represent them as a tree diagram. We represent each event by a column of branches, Consecutively means one
and the number of branches is determined by the number of possible outcomes for after the other.
that event. For example, if a die is thrown, there are six possible outcomes, numbers
1 to 6, which we represent by six different lines (or ‘branches’) drawn from a starting
point. If a coin is then tossed (with two possible outcomes, heads or tails), we draw
the possible ‘branches’ of heads or tails after the previous lines. We draw the tree
diagram as shown below.
1
__
2 H
1
1
__ T
2
1
__
6 1
__
2 H
2
1
__ 1
__ T
6 2
1
__
1
__ 2 H
6 3
1
__ T
2
1
__
6 1
__
2 H
4
1
__ 1
__ T
6 2
1
__
1
__ 2 H
6 5
1
__ T
2
1
__
2 H
6
1
__ T
2
Notice that we write the probability of each ‘branch’ alongside that branch.
This representation helps us decide which values we multiply and which values we
add. We always multiply probabilities along the branches. If there is more than one
set of branches that we can follow to achieve the result, add the final answers for each
pathway. So, in the example, if we want the probability of getting a prime number
and tossing a head, we follow the pathways: 2 then H; or 3 then H; or 5 then H.
So the probability of getting a prime number and tossing a head is:
1 × __
__ 1 + __
1 × __
1 + __
1 × __
1 = ___
1 + ___
1 + ___ 3 = __
1 = ___ 1
6 2 6 2 6 2 12 12 12 12 4
Note: When each branch of all the events have equal probabilities, as they do in this
case, we can see that 3 of the 12 possible final outcomes satisfy the conditions set, so
3 = __
the probability of achieving the required outcome is ___ 1.
12 4
It is also useful to remember that the events at the ends of the branches are mutually
exclusive and exhaustive. Therefore the sum of the probabilities of each branch will
equal 1.
Unit 3 Using tree diagrams to solve problems for events not necessarily independent 263
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WORKED EXAMPLE 1
Ken and Claire plan to have three children.
1 Draw a tree diagram to represent the possible combinations of the sexes
of their children.
2 Determine the probability that they will have 3 girls.
3 Determine the probability that they will have 2 boys and 1 girl.
4 Determine the probability that they will have 2 girls, given that their
first child is a boy.
5 Determine the probability that they will have at least one boy.
SOLUTIONS
1
__
2 B Each column of branches represents the
1 possible results of a new child that is born, so
1
__ B
2 1
__ G the first child could be a boy or a girl (B or G).
2
B If the first child is a boy, the second child could
1
__ be a boy or a girl, so the first B is followed by
1 1
__ 2 B
__ 2
2 G the branches B or G. If the first child had been
1
__ G a girl, there is still the option of the second
2 child being a B or G. The same options of B or
1
__ G apply for the third child.
2 B
1
__ B
2 1
__
2 1
__ G
2
G
1
__
1
__ 2 B
2 G
1
__ G
2
2 1 × __
P(3 girls) = __ 1 × __
1 = __
1 (There is only one pathway that has three girls: GGG.)
2 2 2 8
3 There are three different pathways that result in two boys and one girl:
BBG or BGB or GBB (Shown on the tree diagram with dotted lines.)
1
__
2 B So, P(2 boys and 1girl)
1
__ B = BBG + BGB + GBB
2 1
__ G 1 × __
1 × __
1 + __
1 × __
1 × __
1 + __
1 × __
1 × __
1
2 = __
B 2 2 2 2 2 2 2 2 2
1
__ 1 + __
= __ 1 + __ 3
1 = __
1 1
__ 2 B 8 8 8 8
__ 2
2 G
1
__ G
2
1
__
2 B
1
__ B
2 1
__
2 1
__ G
2
G
1
__
1
__ 2 B
2 G
1
__ G
2
Or, because each branch has equal probabilities, three of the eight possible
outcomes satisfy the requirement of having 2 boys and 1 girl.
264 Topic 12 Probability
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4 1
__
2 B If we are given that the first child is a boy,
1
__ B we consider only the branches that follow
2 1
__ G that outcome. So there are only four possible
2
further outcomes, with only one resulting in
1
__ 2 girls.
1
__ 2 B
2
G
1
__ G
2
1.
So the probability of 2 girls is __
4
Or, we can see that the only possible pathway to follow the first boy is GG,
which is 1 1 = __
__ × __ 1.
2 2 4
5 There are many possible outcomes that will result in at least one boy. In
such a situation, it is quicker to determine the answer by realising that the
probability of anything happening will always be 1 minus the probability of
the required outcome not happening. (P(A) = 1 – P(not A)) REMEMBER
So P(at least one boy) = 1 – P(no boy) where the only way of having no boy is
A probability of 1 represents
GGG, so P(at least one boy) = 1 – 1 7
__ = __ certainty, or 100%.
8 8
WORKED EXAMPLE 2
Simon either walks or cycles to school. When he cycles to school, the probability
that he is on time is 0,9 but if he walks to school the probability that he is on time
is 0,8. He cycles to school 60% of the time.
1 Draw a tree diagram to represent this information.
2 Determine the probability that he is late for school on any day chosen at
random.
3 Are ‘walking to school’ and ‘being late’ independent events for Simon?
SOLUTIONS
1 The first event that will affect the answer is whether Simon walks or cycles
on the day chosen at random. It he cycles to school 60% of the time, he must
walk to school for the other 40% of days. This means that the probability of 0,8 On time
60 = 0,6 and the probability of him walking is ____
him cycling is ____ 40 = 0,4. Each Walk
100 100 0,4 Late
option lead to the possibility of him arriving at school on time or late. Use 0,2
the given probabilities for him arriving on time, and subtract them from 1 to
get the probabilities of him being late. Now draw the tree diagram as shown 0,6 0,9 On time
Cycle
on the right. Late
0,1
2 The probability of Simon arriving late for school is shown by the dotted
pathways. We calculate the probability as follows:
P(Late) = 0,4 × 0,2 + 0,6 × 0,1 = 0,08 + 0,06 = 0,14
3 P (walking and being late) = 0,08
P( walking ) × P( being late ) = 0,4 × 0,14 = 0,56
∴ P (walking and being late) ≠ P( walking ) × P( being late )
So walking to school and being late are not independent events for Simon.
Unit 3 Using tree diagrams to solve problems for events not necessarily independent 265
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WORKED EXAMPLE 3
Based on their past achievements, a hockey coach knows that his team has a 65%
chance of winning their matches if they play a home game, and a 25% chance of
a draw. If they play an away game, they have a 55% chance of winning and a 40%
chance of losing. They play 50% of their matches at home. Draw a tree diagram,
and then determine the probability that they will lose their next match if it is not
known whether it is a home or away match.
SOLUTION
The first event that will affect the answer is whether they play at home or away.
The options are that they could win, lose or draw. So the tree diagram will have
two columns of branches, one for each possible outcome of these events:
65% win So, the pathways that result in a loss
Home 25% will be Home Lose and Away Lose as
draw
50% shown by the dotted lines.
10% lose
55% win
50% 5%
Away draw
Notice that we can write probabilities
40% lose as fractions or percentages.
1
P(Lose) = 50% × 10% + 50% × 40% = 0,5 × 0,1 + 0,5 × 0,4 = __
4
WORKED EXAMPLE 4
A bag contains 6 white balls, 4 black balls and 5 green balls. A ball is drawn at
random and not replaced. A second ball is drawn.
1 Show that drawing a white ball in the first draw, and drawing a white ball in
the second draw are not independent events.
2 Draw a tree diagram and use it to determine that:
2.1 both balls are green
2.2 the first ball is black and the second ball is green
2.3 the two balls are white and green in any order.
SOLUTIONS
1 6 , but once that white ball has been removed and not
P(white first) = ___
15
replaced, there will be only 5 white balls left, and only 14 balls in total.
Therefore P(white second) = ___ 5 if a white ball had been drawn first,or ___
6 if a
14 14
different colour had been drawn first. Therefore the probability of the second
event is affected by the first, and therefore the events are not independent.
It is important to realise that these two events are not independent of each
other. As the first ball is not replaced after it has been drawn, there will be one
less ball of a certain colour for the second choice. So the result of the second
draw will be affected by the colour of the ball that was drawn first.
266 Topic 12 Probability
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5
___ W The probabilities of these branches are based
14
4
___ on the assumption that the first ball drawn was
14
W B white. So there are only 5 white balls, 4 black
balls and 5 green balls left to draw. The total
5
___
6
___ 14 G number of balls left in the bag is 14.
15
6
___ W The probabilities of these branches are based
4
___
14
3
___ on the assumption that the first ball drawn was
15 14
B B black. So there are only 3 black balls, 6 white
balls and 5 green balls left to draw. The total
5
___
14 G number of balls left in the bag is 14.
5
___
15 6
___ W
14 The probabilities of these branches are based
4
___ on the assumption that the first ball drawn was
G 14
B green. So there are only 4 green balls, 6 white
4
___ balls and 4 black balls left to draw. The total
14 G number of balls left in the bag is 14.
2.1 5 × ___
P(both green) = ___ 4 = ___
2
15 14 21
5 = ___
4 × ___ 2 In any order, so WG
2.2 P(black then green) = ___
15 14 21 or GW.
2.3 6 × ___
P(white and green) = ___ 5 + ___
5 × ___
6 = __
2 P(WG) + P(GW)
15 14 15 14 7
WORKED EXAMPLE 5
A company buys stationery from two suppliers, A and B, and places twice as many
orders with A as with B. The probability that either supplier A or B will deliver an
order on time is 0,8.
1 Draw a tree diagram to represent this information.
2 Determine whether ordering from supplier A and the order not being
delivered on time are independent events.
SOLUTIONS
1
0,8 on time
–23 A
0,2 not on time
0,8 on time
–13
B
0,2 not on time
2 2 × 0,2 = ___
P( A and not on time ) = __ 2
3 15
2 × __
P( A ) × P( not on time ) = __ 2 × 0,2 + 1
3 (3 6 = ___
__ × 0,2 = ___ 2
3 ) 45 15
∴ P( A and not on time ) = P( A ) × P( not on time )
So these are independent events.
Unit 3 Using tree diagrams to solve problems for events not necessarily independent 267
PLT MATHS LB 11 7th pgs (Real Book).indb 267 2012/07/02 2:24 PM
EXERCISE 3
1 An ice cream shop sells ‘combos’ which consist of a cone with one scoop of
ice cream and a topping. The cones are plain or chocolate, the ice cream flavours
are vanilla, chocolate or banana, and the toppings are nuts, chocolate sprinkles
or smarties. Draw a tree diagram to represent these choices, and then answer the
questions.
1.1 Determine how many different ‘combos’ have smarties as a topping.
1.2 Thandi does not like chocolate. How many ‘combos’ can she choose from?
1.3 If Nina is allergic to nuts, determine the probability that she will have an
ice cream with chocolate sprinkles, given that she will not choose nuts as
a topping.
2 A bag contains 6 blue, 5 red and 9 white marbles. A marble is drawn and not
replaced, and another marble is then drawn. Draw a tree diagram to represent
this information, and use it to answer the questions.
2.1 Explain why drawing a second blue marble is not independent on drawing
a blue marble first.
2.2 What are the probabilities that:
2.2.1 both marbles are white
2.2.2 both marbles are blue
2.2.3 one blue and one red marble is chosen
2.2.4 neither of the marbles chosen is red?
3 There are eight houses in a school, five for day scholars and three for boarders.
Each house has chosen a Head of House. From these Heads of House, the learners
must select a Head prefect and a Deputy Head prefect. Assume that each Head
of House has an equal chance of being elected. Draw a tree diagram showing the
probabilities of each position being filled by a boarder or a day scholar. Then
answer the questions.
3.1 Determine the probability that the Head prefect of the school will be
a day scholar.
3.2 Is it true to say that the election of the Deputy is independent of the
election of the Head prefect? Give a reason.
3.3 What is the probability that both positions are filled by boarders?
3.4 What is the probability that at least one of the positions is filled by
a boarder?
268 Topic 12 Probability
PLT MATHS LB 11 7th pgs (Real Book).indb 268 2012/07/02 2:24 PM
Unit 4: Contingency tables
Contingency tables are statistical tables that represent the relationships between two
or more variables. We show the frequencies of each variable in rows and columns. So,
if a survey of 100 people was done to analyse the relationship between gender and
right- or left-handedness, the results could be:
This represents the total of the first row,
that is, the total number of males surveyed.
This represents the total
Right-handed Left-handed Totals of the second row, that
Male 41 12 53 is, the total number of
females surveyed.
Female 38 9 47
Totals 79 21 100
This represents the
total number of people
This represents the total of the This represents the total of the surveyed, and is the total
first column, that is, the total second column, that is, the total of the right-hand column
number of right-handed people. number of left-handed people. and the bottom row.
According to this survey, the probability of a male chosen at random being left-
12 ≈ 0,23, as 12 of the 53 males were left-handed. The probability of a
handed is ___
53
9 ≈ 0,19.
female being left-handed is ___
47
This survey suggests that males have a higher probability of being left-handed
than females, but we cannot draw such conclusions from a single survey of only
100 people.
Also notice that the events listed in the rows and those listed in the columns must be
complementary events, otherwise the total in the bottom right-hand corner will not REMEMBER
be the total sample space.
Complementary events are
mutually exclusive (have no
overlap), and exhaustive (use
WORKED EXAMPLE 1
up all possibilities).
A group of Mathematics and Physical Sciences learners at a college were
interviewed about the subjects they studied. The college offered two different
Mathematics courses and two different Physical Sciences courses. No learner was
enrolled in more than one Mathematics or Physical Sciences course at any one
time. The results of the survey are recorded in the table.
Mathematics A Mathematics B Total
Physical Sciences A 6 20
Physical Sciences B 13
Total 23
1 Copy and complete the table.
2 How many learners were interviewed?
Unit 4 Contingency tables 269
PLT MATHS LB 11 7th pgs (Real Book).indb 269 2012/07/02 2:24 PM
3 Find the probability that a learner chosen at random from those interviewed:
3.1 was enrolled in Mathematics A
3.2 was enrolled in Physical Sciences B and Mathematics B
3.3 was enrolled in Physical Sciences B, given that they were enrolled in
Mathematics A.
3.4 Five years later, a similar survey was conducted on 100 learners. It was
found that the probability of a learner:
• enrolled in Mathematics A and Physical Sciences A was 0,3
• enrolled in Mathematics A was 0,8
• not enrolled in either Mathematics A or in Physical Sciences A
was 0,1.
3.4.1 Copy and complete the contingency table.
Mathematics A Not Mathematics A Total
Physical Sciences A
Not Physical Sciences A
Total
3.4.2 Determine the probability that a learner surveyed takes
Mathematics A but not Physical Sciences A.
3.4.3 Determine the probability that a learner surveyed takes
Physical Sciences A but not Mathematics A.
3.4.4 How many learners surveyed were enrolled in Physical
Sciences A at that time?
3.4.5 Determine the probability that a learner surveyed does either
Mathematics A or Physical Sciences A.
3.4.6 Compare the probability for Mathematics A and for Physical
Sciences A in each table. Discuss whether the popularity for
each of these subjects has changed or not over the five years.
SOLUTIONS
1 By adding across rows and down columns, we can fill in the missing values
as follows:
20 – 6 = 14
Mathematics A Mathematics B Total
Physical Sciences A 14 6 20
Physical Sciences B 13 17 30
Total 27 23 50
13 + 17 = 30
23 – 6 = 17
14 + 13 = 27 20 + 30 = 50
and
27 + 23 = 50
2 50 learners were interviewed.
270 Topic 12 Probability
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3 3.1 P(Maths A) = total in Maths A
____________________ = 27
___
total number surveyed 50
3.2 17
P(Science B and Maths B) = P(Science B ∩ Maths B) = ___
50
number doing Science B and Maths A ___
3.3 P(Science B | Maths A) = ________________________________ = 13
number doing Maths A 27
3.4 Notice that this time the values in the table represent probabilities and
not an exact number of learners. So the value in the bottom right-hand
block represents the total probability, which is 1. First fill in the given
values and the 1 in the bottom right-hand block (shown in bold). Then
add the rows and columns, and fill in the values.
0,2 – 0,1 = 0,1 0,2 – 0,1 = 0,1
3.4.1
Maths A Not Maths A Total
Science A 0,3 0,1 0,4
Not Science A 0,5 0,1 0,6
Total 0,8 0,2 1
0,8 – 0,3 = 0,5 1 – 0,4 = 0,6
1 – 0,8 = 0,2
0,3 + 0,1 = 0,4
3.4.2 P(Maths A, not Science A) = 0,5. The values in the table are
probabilities, so we only give the probability in the block that
is relevant to the question.
3.4.3 P(Science A, not Maths A) = 0,1
3.4.4 P(Science A) = 0,4 0,4 × 100 = 40 learners REMEMBER
3.4.5 P(Maths A or Science A)
The definition: P(A ∪ B) =
= P(Maths A ∪ Science A)
P(A) + P(B) – P(A ∩ B)
= P(Maths A) + P(Science A) – P(Maths A ∩ Science A)
= 0,8 + 0,4 – 0,3 = 0,9
Or realise that:
P(Maths A or Science A) = 1 – P(not Maths A, not Science A)
= 1 – 0,1 = 0,9
3.4.6 Five years ago, P(Maths A) was 27
___ ≈ 0,54 and P(Science A)
50
20 = 0,4
was ___
50
Now, P(Maths A) = 0,8 and P(Science A) = 0,4
So over the five years, Maths A has become more popular,
whereas Science A has retained the same popularity.
Unit 4 Contingency tables 271
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WORKED EXAMPLE 2
A group of people were asked whether they had watched Gcina Mhlope’s plays or
whether they had listened to her story-telling. Based on the results of the survey,
the contingency table was drawn up showing the probabilities of these events.
Watched plays Not watched plays Total
Heard stories 0,6
Not heard stories 0,26
Total 0,35
1 Copy and complete the table.
REMEMBER 2 Show that watching her plays and hearing her story-telling are independent
events.
For independent events:
P(A and B) = P(A) × P(B) SOLUTIONS
1 Watched plays Not watched plays Total
Heard stories 0,21 0,39 0,6
Not heard stories 0,14 0,26 0,4
Total 0,35 0,65 1
2 P(Plays and Stories) = 0,21
P(Plays) × P(Stories) = 0,35 × 0,6 = 0,21 = P(Plays and Stories)
Therefore watching the plays and hearing the stories are independent events.
WORKED EXAMPLE 3
A study was conducted to investigate the relationship between eye colour and
hair colour. A group of 250 people were observed. The results are:
• 12 people had red hair, 64 had blonde hair and 82 had black hair
• 35 of the blondes had blue eyes, and 21 had green eyes
• 68 of the brunettes had brown eyes
• 38 of those with black hair had brown eyes, and 41 had green eyes
• 5 of the redheads had blue eyes
• 116 people had brown eyes and 86 had green eyes
1 Copy and complete the contingency table.
Blonde hair Brunette Black hair Red hair Total
Blue eyes
Brown eyes
Green eyes
Total
272 Topic 12 Probability
PLT MATHS LB 11 7th pgs (Real Book).indb 272 2012/07/02 2:24 PM
2 Based on these results, determine the probability of a person having:
2.1 blonde hair and green eyes
2.2 black hair and brown eyes
2.3 blue eyes, given that they are brunette
2.4 red hair, given that they have green eyes.
SOLUTIONS
1 First fill in the given values (shown in bold). Then calculate the missing
values by adding each column and row to equal the total at the end of each
row and column.
Blonde hair Brunette Black hair Red hair Total
Blue eyes 35 5 3 5 48
Brown eyes 8 68 38 2 116
Green eyes 21 19 41 5 86
Total 64 92 82 12 250
2 2.1 21
P(Blonde and green eyes) = ____
250
2.2 38 = ____
P(Black hair and brown eyes) = ____ 19
250 125
2.3 5
P(Blue eyes | Brunette) = ___
92
number of redheads with green eyes 5
2.4 P(Redhead | Green eyes) = _______________________________ = ___
total with green eyes 86
EXERCISE 4
1 A stationery distributor investigated the number of defective pens that
they obtain from two different suppliers. They recorded their findings for a
consignment of pens from both suppliers in a contingency table.
Supplier A Supplier B Total
Defective 18
Not defective 144
Total 350 500
1.1 Copy and complete the table.
1.2 Determine the probability that a pen chosen at random from the whole
consignment will be defective.
1.3 Determine the probability that a pen chosen from supplier A will be
defective.
1.4 Determine the probability that a pen chosen from supplier B will not be
defective.
1.5 Based on these findings, which supplier should the distributor order from
in the future?
Unit 4 Contingency tables 273
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2 A group of 50 learners discuss the chances of their first rugby and hockey teams
winning their respective matches the following day. The opinions of the learners
were as follows:
• 30 thought that their rugby team would win
• 35 thought that their hockey team would win
• 6 thought that both their hockey and rugby teams would lose
• nobody thought that there would be a draw in either game.
2.1 Copy and complete the contingency table.
Hockey will win Hockey will lose Total
Rugby will win
Rugby will lose
Total
2.2 Based on these opinions, determine the probability that:
2.2.1 both the hockey and rugby teams will win their matches.
2.2.2 the hockey team will win and the rugby team will lose.
2.3 Show that according to these opinions, winning hockey and winning
rugby are independent events.
274 Topic 12 Probability
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Revision
1 Consider this list of events:
a Scoring an even number when rolling a die
b Scoring more than 3 when rolling a die
c Scoring less than 5 when rolling a die
d Scoring a 1 when rolling a die
e Scoring an odd number when rolling a die
Which pair of the events are:
1.1 mutually exclusive but not complementary (2)
1.2 exhaustive but not complementary (2)
1.3 complementary? (2)
[6]
2 A group of 100 people visiting a bookshop were asked whether they had
read Coconut by Kopano Matlwa or Rainmaker by Don Pinnock.
• 67 had read Coconut
• 59 had read Rainmaker
• 22 had read neither.
2.1 Represent this data in a Venn diagram. (5)
2.2 Determine how many of the people visiting the store had read
both books. (3)
2.3 Determine the probability that a person had not read either of
the books. (2)
2.4 Determine the probability that a person chosen at random had
read Rainmaker only. (2)
2.5 Determine the probability that a person chosen at random had
read Coconut, given that they had read Rainmaker. (2)
[14]
3 A card is drawn from a pack of playing cards, replaced, and a second card
is drawn.
3.1 Use a tree diagram to show the probabilities of drawing red or black
cards for both drawings. (4)
3.2 Determine the probability that both of the cards drawn were red. (2)
3.3 Determine the probability that the cards were different colours. (4)
3.4 Determine the probability that the second card was black. (4)
3.5 Determine P(both black). (2)
3.6 Determine the probability that at least one card was red. (3)
[19]
275
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TOPIC 12: REVISION CONTINUED
4 Grade 9 learners are considering their subjects for Grade 10. They have to
choose one of Physical Sciences or Geography, and one of Life Sciences or
Accountancy. Complete the contingency table which shows the results of
their choices.
Physical Sciences Geography Total
Life Sciences 15 30
Accountancy 10
Total 50
4.1 Copy and complete the table. (5)
4.2 Determine the probability that a learner chosen at random will do
Accountancy and Physical Sciences. (2)
4.3 Determine the probability that a learner will take Geography, given
that they take Life Sciences. (2)
4.4 Show that taking Geography and taking Accountancy are
independent events. (4)
[13]
5 The probability of a team winning a match is 2
__. If the team play two
3
matches, determine the probability that they lose both matches, assuming
that there is no draw.
Use a tree diagram to answer the question. [4]
6 A shelf in a bookshop had books written by South African authors.
Sarah found four books written by Mark Mathabane, nine by
Dalene Matthee and ten by James McClure. She took a book off the
shelf at random, did not replace it, and then took another.
6.1 Represent this situation in a tree diagram, showing the probabilities
of each branch. (9)
6.2 Determine the probability that:
6.2.1 the books that she chose were by Mark Mathabane and
Dalene Matthee in any order (4)
6.2.2 the books that she chose were by the same author (4)
6.2.3 neither of the books were by Dalene Matthee (6)
6.2.4 at least one of the books was by James McClure. (4)
[27]
276
PLT MATHS LB 11 7th pgs (Real Book).indb 276 2012/07/02 2:24 PM
7 At a garden nursery, 60 people were asked whether their gardens contained
shrubs, climbers or bulbs. Forty-one people said that their garden contained
shrubs, 25 said that their gardens contained climbers, and 21 said that their
gardens contained bulbs. The Venn diagram shows the rest of the results of
the survey where S stands for shrubs, C for climbers and B for bulbs.
S C 60
2x
8
x
5
B
5
7.1 Determine the value of x. (6)
7.2 Determine:
7.2.1 nP(S ∩ C) (3)
7.2.2 P(S ∩ C ∩ B)′ (3)
7.2.3 P(C ∪ B) (3)
7.2.4 P(S ∩ C ′ ) (3)
7.2.5 P(B | S) (2)
7.2.6 P(S ∪ C ∪ B)′ (2)
[22]
8 A group of learners were asked whether they travel to school by bus or
by train. Some learners use both train and bus to get to school, some use
either a bus or a train, and others use neither a bus nor a train. The
contingency table shows the probabilities for this group of learners.
Travel by bus Not travel by bus Total
Travel by train 0,1 0,6
Not travel by train
Total 0,7
8.1 Copy and complete the table. (6)
8.2 Determine the probability that a learner chosen at random does
not travel by train. (1)
8.3 Determine the probability that a learner chosen at random travels
by train and not by bus. (1)
8.4 Determine whether travelling by train and travelling by bus are
independent events. (4)
8.5 Determine the probability that a learner travels by bus or by train. (4)
[16}
277
PLT MATHS LB 11 7th pgs (Real Book).indb 277 2012/07/02 2:24 PM
Formal assessment: Term 3 Test 1
1
A H
B J
F G 100
15
49
C
D E
L
The solid drawn above has a height of 100 units. The length of AL is
49 units, BC is 15 units and AF = DL.
1.1 Determine the area of the base, in terms of π. (4)
1.2 Determine the perimeter of the base, in terms of π. (3)
1.3 Determine the volume of the solid. (3)
1.4 Determine the total surface area. (5)
[15]
2
G
E
H K
A
D
B C
A paper weight is made up of a square-based right pyramid and a squared-based
right prism. It has a slant height of 34 units. AF = 32 units and ED = 7 units.
2.1 Determine the volume of the paper weight. (5)
2.2 Determine the total surface area of the paper weight. (5)
[10]
3 CD = 60 units, AB = 28 units and OH is 8 units longer than OG. OG = x units.
B
A H
O
C G D
3.1 Express, giving reasons, the lengths of these sides in terms of x:
3.1.1 OH (1)
3.1.2 AO (3)
3.1.3 CO. (3)
3.2 Solve for x. (3)
3.3 State the length of the radius. (1)
[11]
278
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4
O P
3
2 1
x
5
4 1
Q 3
1 U
2 2
2 1
3 1 2
R T S
SQ and SP are tangents to the circle at Q and P. PQ produced meets the line
^ =x
through S at R. SR is parallel to UQ. PU produced meets SR at T and P1
4.1 Find five angles equal to x. (10)
4.2 Prove that PQST is a cyclic quadrilateral. (4)
4.3 Prove that PQ^ T = 3TP^ S. (4)
4.4 Prove that RQ = PS. (3)
4.5 Determine QO ^ P in terms of x. (4)
4.6 Join OU and then prove that OU is the perpendicular bisector of QP. (4)
4.7 Is OQUP a cyclic quadrilateral? Justify your answer. (3)
4.8 Determine the value of x if OQUP is a rhombus. (2)
[34]
Total: 70 marks
279
PLT MATHS LB 11 7th pgs (Real Book).indb 279 2012/07/02 2:24 PM
Formal assessment: Term 3 Test 2
1 Determine the total surface area of the shape in the diagram
(correct to one decimal place).
35 mm
10 mm
[5]
2 The box shown in the diagram contains two wooden shapes with
dimensions as shown. Determine:
8 cm
30 cm 20 cm
12 cm
15 cm 10 cm 40 cm
50 cm
2.1 the volume of each wooden shape (4)
2.2 the percentage of the volume of the box that is taken up by the
two wooden shapes (to two decimal places). (4)
[8]
3 In the figure A, P, Q and R are points on the circumference of the circle,
centre O. AB is a tangent to the circle at A. OR and AQ intersect at T and
QT = TA. QA ^ B = 56°.
Q
R
B
T
O
56°
A
P
3.1 Give a reason for each statement:
3.1.1 OA^ B = 90° (1)
3.1.2 ^
OTA = 90° (1)
3.2 Determine (with reasons) the size of each angle:
3.2.1 AO^R (2)
3.2.2 ^
AQR (2)
3.2.3 AP^Q (2)
3.3 Is PQ ∥ AB? Justify your answer. (4)
3.4 If AQ is 8 units and the OA = 5 units, determine the length of TR,
giving reasons. (3)
[15]
280
PLT MATHS LB 11 7th pgs (Real Book).indb 280 2012/07/02 2:24 PM
4 The figure shows circle ABCF with DC a tangent to the circle at C.
ABCD is parallelogram with side DA meeting the circle at F. E is a
^ = x.
point on AC so that EF ∥ CD. D
B A
2
1
1 1
E F
2
2
3 2
1 x
C D
4.1 Prove CF = CD. (5)
4.2 Determine A^ in terms of x, giving reasons. (4)
1
4.3 Prove that ECDF is a cyclic quadrilateral. (5)
[14]
5 A farmer wants to subdivide a paddock ABCD by erecting a fence from
B to E and from E to C. The dimensions of ABCD are:
AB = 28 m; ED = 14 m; and B ^
AE = 35°
The farmer plans to use 41 m of fencing at an angle of 25° to BC.
Let the point where the fencing reaches line AD be E.
C
B
25°
28 41
35°
A x
E 14 D
5.1 If x represents the angle between the fence BE and the
boundary AE (that is, BE ^ A), determine the value of x, correct to the
nearest degree. (3)
5.2 Explain why there is no ambiguity with regard to the answer for 5.1. (1)
5.3 If the farmer erects a fence from E to C, so that CE ⊥ BE, determine
the length of fencing that he will require (that is, determine the
length of CE), correct to the nearest metre. (3)
5.4 Determine the area of the paddock ABCD, to two decimal places. (6)
[13]
281
PLT MATHS LB 11 7th pgs (Real Book).indb 281 2012/07/02 2:24 PM
Test 2 continued
6 In the diagram △AB ^ C = y.
^ C has ABC = x; AC = b and BA
^ C = θ.
Line CD is drawn perpendicular to BC, so that CD = a and BD
sin x = __
_____
Prove that: cos b
y a
A
y b
x
B C
a
0
D
[7]
7. 7.1 Tom hears that the value of his car depreciates at 13% p.a. on a
straight line basis. If after 4 years the value of his car is R35 640,
calculate the original price of his car. (3)
7.2 If Tom’s car had depreciated on a reducing balance method,
calculate the rate of depreciation if his car had depreciated to
R35 640 after 4 years. (Use your answer to 7.1 as the original
price of the car.) (3)
[6]
8. 8.1 Uthmaan deposits R8 000 into a savings account. Four years later
he withdraws R5 500 from the account. Interest is calculated at
7,8% p.a. compounded quarterly for the first three years,
and 9% p.a. compounded annually thereafter. Calculate how much
money will be in the account at the end of 6 years. (8)
8.2 What was the effective annual interest rate that Uthmaan received
over the first 3 years? (3)
[11]
9 Forty-two percent of the students at a local college live in the college
residence and fifty-five percent work part time. Ten percent of the students
work part time and live in residence.
9.1 Draw a Venn diagram showing the relationships. (4)
9.2 Determine the probability that a student who does not live in
residence works part-time. (1)
9.3 Determine the probability that a student chosen at random does
not live in residence or work part-time. (1)
[6]
282
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10 Conrad investigates the effect of using a car phone on the speed that a person
drives. He interviewed 750 people in his area, and the results of his survey are
shown in the contingency table.
Speeding violation No speeding violation Total
Car phone user 25 250
Not a car phone user 450
Total 750
10.1 Copy and complete the contingency table. (5)
10.2 Calculate these probabilities using the table:
10.2.1 P(person had no speeding violation) (1)
10.2.2 P(person is a car phone user or person had no
speeding violation) (3)
10.2.3 P(person is a car phone user given person had a
speeding violation) (2)
10.3 Prove that, according to this survey, being a car phone user and
getting a speeding violation are independent events. (4)
[15]
Total: 100 marks
283
PLT MATHS LB 11 7th pgs (Real Book).indb 283 2012/07/02 2:24 PM
Term 3 summary
Topic 8
Measurement
Summary of formulae h represents the perpendicular height and s the slant height.
Solid Volume Total surface area
Right prism
area of base × h area of base + areas of lateral surfaces
Right pyramid
s h 1 (area of base ) × h
__ area of base + areas of all triangles
h s 3
r x
Right cylinder
π r2 h 2πr2 +2πrh
Right cone
s h 1 πr2 h
__ π r2 +π r s
h s
3
r x
Sphere
sphere
4 πr3
__ 4πr2
d 3
r
Hemisphere
2 πr3
__ 3πr2 (includes flat circular surface)
r radius 3
284 Term 3 summary
PLT MATHS LB 11 7th pgs (Real Book).indb 284 2012/07/02 2:24 PM
Topic 10
Euclidean geometry and measurement
E
F
O
O D
A R B A
A K B
OR ⊥ AB (line from centre to midpoint B C
AK = KB (OK ⊥ AB) chord) ^ = F^ (equal chords AB & CD)
E
A R
P Q D
O D O
D
B E
C E G G
BO^ C = 2A^ (∠ at centre) O
^
A= ^ D (∠s on chord BC) F
^ ^
Reflex O = 2R (∠ at centre) ^ = 90° (∠ on diameter)
D ^ = 180° (opp ∠s cyclic quad)
^+G
E
F
G
D O
G
E D E
F A B
H C DEFG is a cyclic quad
^H = D
GF ^ (ext ∠ cyclic quad) ^ C = 90° (rad ⊥ tan)
OB (DE subtends equal ∠s at F & G)
D
E A
T
M
A 1 2 P
3 x
B
180° = x B
C
^ =E
B ^ (tan AB, chord BD) N V PA = PB
1
^ ^ (tan BC, chord BD)
B3 = D MNVT is a cylic quad (opp ∠’s suppl) (tangents from common point P)
A
D
C
B C E
A D B ABCD is a cyclic quad
^ B = 90°)
AB is the diameter (AC ^ E = int opp A
(ext DC ^)
Term 3 summary 285
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Term 3 summary continued
Topic 10
Trigonometry: Sine, cosine and area rules
When solving for lengths or angles of triangles, remember:
• If it is a right-angled triangle, use the trigonometric ratios: sin θ = o a ; tan θ = o
__; cos θ = __ __
h h a
There are two cases of non-right-angled triangles that require the cosine rule. These are:
• SAS (that is, two sides and the angle between them). In this case we use:
a2 = b2 + c2 – 2bc cos A or b2 = a2 + c2 – 2ac cos B or c2 = a2 + b2 – 2ab cos C
• SSS (that is, three sides and no angles). In this case we use:
2 2 2 2 2 2 2 2 2
cos A = b + c – a or cos B = a + c
_________ – b or cos C = a + b
_________ –c
_________
2bc 2ac 2ab
All other cases of non-right-angled triangles, use the sine rule:
a = b
• When looking for a side, use: ____ c
____ = _____
sin A sin B sin C
• When looking for an angle, use: sin A
____ sin
____
a = b = c
B sin C
_____
• When you are given two sides and a non-included angle, look out for the ambiguous case. This occurs
when the side opposite the given angle is the shorter of the two given sides.
To determine the area of a triangle we can use:
1 base × height for triangles where the perpendicular height is known or
• area = __
2
1 ab sin c or __
area △ABC = __ 1 ac sin b or __
1 bc sin a
2 2 2
Topic 11
Finance, growth and decay
• Growth takes place with:
simple interest: A = P(1 + in) or
compound interest: A = P(1 + i)n
Decay takes place with:
simple decay (called the straight line method): A = P(1 – in)
compound decay (called the reducing balance method): A = P(1 – i)n
• If the annual interest that is quoted (the nominal interest rate) is compounded more frequently than
once a year, the effective interest rate will be higher than the nominal interest rate, and is determined
using the formula:
(
1 + 1eff = 1 + i___
m )
(m) m
• To determine the amount accumulated after an investment has been growing with compound interest
that is compounded k times per year: divide the quoted interest rate by k and multiply the number of
years by k.
• When more than one transaction occurs, draw a time-line to visualise what has happened over time.
Remember to take all values to any ONE moment in time, before adding or subtracting values. Use the
logic that ‘total of money in = total of money out’.
• When taking values back in time, you are finding the P value for a known A value, so the formula
becomes A = P(1 + i)–n
286 Term 3 summary
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Topic 12
Probability
• The probability of an event taking place is always:
total number of ways an event can occur
_______________________________________
total number of possible outcomes for the event
Notation:
• P(A) means ‘the probability of event A happening’.
• P(A′) means ‘the probability of event A not happening’.
• P(A ∩ B) means ‘the probability of event A and event B happening’.
• P(A ∪ B) means ‘the probability of event A or event B happening’.
• P(A | B) means ‘the probability of event A happening, assuming that event B
is taken as given’.
• For all events A and B: P(A or B) = P(A) + P(B) – P(A and B)
• P(A or B) = P(A) + P(B) if A and B are mutually exclusive.
• For complementary events (mutually exclusive and exhaustive):
P(B) = P(A′) = 1 – P(A)
• For independent events: P(A and B) = P(A) × P(B)
• Use Venn diagram, tree diagrams and contingency tables to represent data
and to assist in determining the probabilities of various outcomes.
Term 3 summary 287
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Term
4
288
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TOPIC 13 Statistics
Unit 1 Histograms and frequency polygons 288
Unit 2 Ogive curves 294
Unit 3 Variance and standard deviation
of ungrouped data 299
Unit 4 Symmetric and skewed data 304
Unit 5 Identification of outliers 307
Revision 312
Formal assessment: Term 4 Test 316
Term Summary 318
Formal assessment: Examination practice Paper 1 320
Examination practice Paper 2 323
289
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TOPIC
13 Statistics
Unit 1: Histograms and frequency polygons
Histograms
Histograms are similar to bar graphs, but have important differences:
KEY WORDS • In histograms the bars are adjacent to each other with no gaps between the
rectangles whereas in bar graphs the bars are sometimes separate rectangles.
histogram – a graph with
rectangles that show the
• Bar graphs always have bars of equal widths whereas histograms sometimes have
bars of different widths.
frequency distribution of
grouped data with no space • The height of the bar in bar graphs represents the frequency of each category
between the groups, so the whereas in histograms the height of the bar represents the relative frequency or
rectangles touch each other; frequency density. This is relevant when the bars are different widths and the area
we try to keep the groups of of each bar becomes important.
equal width so that the height
of the rectangle (frequency
distribution) is the same as WORKED EXAMPLE
the frequency
bar graph – a graph with
A teacher analyses the ages of a group of children in a playground and tabulates
rectangles that show the the data. Draw a bar graph to represent the data in the table.
frequency with which Ages (years) Frequency
different types of data values
occur; these data values need 0<x≤2 1
not be in consecutive groups, 2<x≤4 3
so the rectangles do not
4<x≤5 6
touch each other
5<x≤7 8
7 < x ≤ 11 2
The problem with drawing a bar graph from the data in the table is that the age
groupings are not equally spaced, so the widths of the bars will not be the same.
Most of the groupings have a width of 2, so we adapt the frequency of the groups
which do not have a width of 2 to compare the area of each bar correctly with one
another.
Age group 4–5 has a width of 1, so we double the frequency from 6 to 12. Age
group 7–11 has a width of 4, so we halve the frequency from 2 to 1.
We now have this table:
Ages (in years) Frequency Relative frequency (Area)
0<x≤2 1 1
2<x≤4 3 3
4<x≤5 6 12
5<x≤7 8 8
7 < x ≤ 11 2 1
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When drawing a histogram, we place the groups on the x-axis and the relative
frequency on the y-axis:
Histogram showing ages of children
14
12
Relative frequency
10
8
6
4
2
0
0–2 2–4 4–5 5–7 6–11
Number of children
We work on this principle:
area of bar (that is, group width × relative frequency) = frequency × standard width
We chose 2 as the standard width (the most common grouping width), so the area
of each bar is its frequency × 2.
REMEMBER
Be aware that histograms that have different group widths occur, but we deal only
The area of each bar of a
with histograms where the group widths are the same, and therefore the relative histogram is important, so it
frequency (value represented on the y-axis) will always be the same as the frequency. is easier if the width of each
group is constant. We use this
principle when looking at the
Frequency polygons area of a frequency polygon.
A frequency polygon enables us to represent the information in a frequency table
by means of line graphs. If we draw a line from the mid-class value of each bar in
a histogram and a line to the mid-class of the groupings before and after the given KEY WORD
groupings, we create a frequency polygon.
frequency polygon – a
Note: polygon that demonstrates
• It is important to complete the polygon on each end. the frequency of each group
• The area of the polygon will equal the total area of the histogram. in a set of grouped data, and
that displays the spread of
the data
WORKED EXAMPLE
The frequency table shows the Mathematics marks obtained by Grade 11
Mathematics learners in a test out of 50 marks. REMEMBER
Mark obtained (out of 50) Number of learners A polygon is a closed shape
10 ≤ x < 20 25 that is made up of three
or more lines. Examples of
20 ≤ x < 30 74
polygons are triangles (three
30 ≤ x < 40 66 lines), quadrilaterals (four
40 ≤ x < 50 35 lines) and pentagons (five
lines).
Unit 1 Histograms and frequency polygons 291
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1 Draw a histogram to represent the results shown in the table.
2 Draw a frequency polygon on the same set of axes as the histogram.
SOLUTIONS
1 80 Histogram showing learners’ marks
60
Frequency
40
20
0
10 20 30 40 50 60
Marks
2 Mark obtained Number of learners
(out of 50) Mid-class value (frequency)
0 ≤ x < 10 4,5 0
10 ≤ x < 20 14,5 25
20 ≤ x < 30 24,5 74
30 ≤ x < 40 34,5 66
40 ≤ x < 50 44,5 35
50 ≤ x < 60 54,5 0
Notice that the mid-class value for the group 0 ≤ x < 10 is not 5. This is discrete
data so the highest possible score less than 10 for a test out of 50, is 9. So we must
KEY WORD find the mid-point from 0 to 9.
discrete data – data with a We can now plot a point for each mid-class value at the height of the relevant
restricted number of values; frequency (which should be in the middle of the possible scores), and then join
includes units of quantity (for these points with lines.
example, number of people,
number of cars, with only To complete the polygon, consider groups of the same width before and after the
whole number values) given grouping. We recognise that the frequency of those groups is zero.
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80 Histogram showing learners’ marks
60
Frequency
40
20
0
4,5 10 14,5 20 24,5 30 34,5 40 44,5 50 54,5 60
Marks
To understand why the area of the polygon is the same as the total area of the bars,
consider the triangles that are not common to the polygon and the bars (see triangles
A–K in the graph).
80 Histogram showing learners’ marks
E
D
F
60 G
Frequency
40 H
C
J
20 B
A K
0
4,5 10 14,5 20 24,5 30 34,5 40 44,5 50 54,5 60
Marks
△A is part of the polygon but not part of the histogram.
△B is part of the histogram but not part of the polygon.
But area △A = area △B
Similarly, △C = △D; △E = △F; △G = △H and △J = △K
So △B + △D + △E + △ G + △ J = △A + △C + △F + △H + △K
Therefore the total area of the histogram equals the area of the polygon.
It is not necessary to draw a histogram before we draw a frequency polygon. All we
require is the mid-class value of each group and the frequency of each group.
Unit 1 Histograms and frequency polygons 293
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WORKED EXAMPLE
A traffic official records the speed at which vehicles travel on a road on a certain
day in a frequency table:
Speed of vehicle (km/h) Number of vehicles
60 ≤ x < 70 12
70 ≤ x < 80 18
80 ≤ x < 90 32
90 ≤ x < 100 47
100 ≤ x < 110 52
110 ≤ x < 120 68
120 ≤ x < 130 45
130 ≤ x < 140 16
Draw a frequency polygon to represent this data.
SOLUTION
Firstly we need to complete the given table to include the mid-class values
as shown in the next table.
Speed of vehicle (km/h) Mid-class value Number of vehicles
50 ≤ x < 60 55 0
60 ≤ x < 70 65 12
70 ≤ x < 80 75 18
80 ≤ x < 90 85 32
90 ≤ x < 100 95 47
100 ≤ x < 110 105 52
110 ≤ x < 120 115 68
120 ≤ x < 130 125 45
130 ≤ x < 140 135 16
140 ≤ x < 150 145 0
KEY WORD
Note that the mid-class value for the group 50 ≤ x < 60 is 55 even though 60 is not
continuous data – data
included in the group. This is continuous data, which means that the highest
without breaks and that can
possible score less than 60 can be so close to 60, that we choose the mid-point
have any value; includes all
units of measurement (for
between 50 and 60. We can now plot a point for each mid-class value at the height
example, height, mass) of the relevant frequency, and join these points with lines.
294 Topic 13 Statistics
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y
Frequency polygon showing vehicle speeds
70
60
50
40
Frequency
30
20
10
x
50 60 70 80 90 100 110 120 130 140 150
Speed (km/h)
Unit 1 Histograms and frequency polygons 295
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Unit 2: Ogive curves
An ogive curve is also called a cumulative frequency curve. Remember that:
KEY WORD • Ogive curves involve cumulative frequency and not individual frequencies.
• The graph is a smooth curve, and not straight lines that are joined together.
cumulative frequency • Unlike frequency polygons, which work with mid-class values, this curve uses the
– a running total of the end points of each group.
frequencies
• While the curve is connected to the x-axis at the end point of the previous group
(the lower limit of the first group), the curve does not return to a point on the
x-axis like the frequency polygon does.
WORKED EXAMPLE
A teacher is concerned about the results obtained by her class in the examinations.
She feels that the learners are spending too much time watching television instead
of doing homework or studying. She conducts a survey to discover how many
hours of television the learners watch each week. The results of this survey are
shown in the table.
Number of hours of television
watched each week Number of learners (f )
0≤x<5 30
5 ≤ x < 10 6
10 ≤ x < 15 29
15 ≤ x < 20 27
20 ≤ x < 25 26
25 ≤ x < 30 20
30 ≤ x < 35 18
35 ≤ x < 40 7
40 ≤ x < 45 2
1 Draw an ogive curve to represent the data in the table.
2 Use the ogive curve to determine approximate values for the median, as well
as the upper and lower quartiles of the data.
3 Determine the modal group and the approximate mean of the data by using
the frequency table.
4 Could the mode have been determined by looking at the ogive curve?
5 Determine an approximate value for the interquartile range.
SOLUTIONS
1 We first need to determine the cumulative frequency so we can draw an ogive
curve. Extend the given table to represent this data.
296 Topic 13 Statistics
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Number of hours of television Number of Cumulative
watched each week learners ( f ) frequency
0≤x<5 30 30
5 ≤ x < 10 6 36 30 + 6
10 ≤ x < 15 29 65 36 + 29
15 ≤ x < 20 27 92 65 + 27
20 ≤ x < 25 26 118 92 + 26
25 ≤ x < 30 20 138 118 + 20
30 ≤ x < 35 18 156 138 + 18
35 ≤ x < 40 7 163 156 + 7
40 ≤ x < 45 2 165 163 + 2
Total f = 165
Notice that the total frequency should always be the same as the final
cumulative frequency value.
Now plot the end points of each group with the cumulative frequency. Join
the points with a smooth curve to create an ogive curve. The curve starts at the
end of the preceding group, which will be the start of the first group given.
y Ogive curve showing learners’
hours of television viewing
150
Cumulative frequency
100
50
x
10 20 30 40 50
Number of hours watching television
Unit 2 Ogive curves 297
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2 In Grade 10 you learnt that the median is the middle data value. This means
1 (n + 1)th value where n represents the
that the median, (Q2), will be the __
2
1 (165
total number of data values. So, in this case, the median will be the __
2
th rd
+ 1) value, that is, the 83 value. We can use the ogive curve to find an
approximation of this value by:
• finding 83 on the y-axis (as close as possible)
• reading across to the curve, and then down to the x-axis to find the
approximate corresponding x value.
1 (n + 1)th data value, that is,
The lower quartile, (Q1), will be the __
4
1 (165 + 1)th = the 41,5th value.
the __
4
3 (n + 1)th data value, that is,
The upper quartile, (Q3), will be the __
4
3 (165 + 1)th = the 124,5th value.
the __
4
These values are difficult to read off accurately. Statistics often works with
estimates and is usually based on a sample of data values, which is part of a
greater number of data values. The emphasis is on determining trends rather
than a high level of accuracy.
To determine the approximate upper and lower quartile values, we read the
values from the ogive curve in a similar manner to how we read the median.
y Ogive curve showing learners’
hours of television viewing
150
Cumulative frequency
124,5
100
83
50
41,5
x
10 20 30 40 50
Q1 Q2 Q3
Number of hours watching television
The approximate values read off from the graph are:
Q1(lower quartile) ≈ 11; Q2 (median) ≈ 18,5; Q3(upper quartile) ≈ 26,5
298 Topic 13 Statistics
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3 The modal group is easy to see as it is the group with the highest frequency
(group 0 ≤ x < 5). So the modal time is 0 to 5 hours.
To determine the approximate mean, we find the mid-class values, multiply
each value by the frequency, and divide the total by the total frequency
(covered in Grade 10).
Number of hours of television Number of Mid-class Mean
watched each week learners (f ) value (m) m×f
0≤x<5 30 2,5 75
5 ≤ x < 10 6 7,5 45
10 ≤ x < 15 29 12,5 362,5
15 ≤ x < 20 27 17,5 472,5
20 ≤ x < 25 26 22,5 585
25 ≤ x < 30 20 27,5 550
30 ≤ x < 35 18 32,5 585
35 ≤ x < 40 7 37,5 262,5
40 ≤ x < 45 2 42,5 85
Totals 165 3 022,5
3 022,5
Therefore the approximate mean = _______ = 18,32 (to two decimal places)
165
4 We cannot read the mode from the ogive curve, but we can observe that the
modal group is the first group. The gradient of the ogive is steepest at this
stage, which shows that the change in frequency is highest over that portion
of the graph.
5 The interquartile range = upper quartile – lower quartile
= 26,5 – 11
= 15,5
EXERCISE 1
Exercise 1 has
1 The table shows the heights of a group of learners. questions that
cover Units 1
Height (in cm) Frequency
and 2.
140 ≤ x < 150 15
150 ≤ x < 160 27
160 ≤ x < 170 18
170 ≤ x < 180 10
1.1 Copy and complete the frequency table on the next page. Use the
information in the table to answer the questions that follow.
Unit 2 Ogive curves 299
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Height (in cm) Frequency (f ) Mid-class value (m) Mean m × f
140 ≤ x < 150 15
150 ≤ x < 160 27
160 ≤ x < 170 18
170 ≤ x < 180 10
Totals
1.1.1 Determine the approximate mean height (to two decimal places).
1.1.2 What is the modal group?
1.1.3 Determine the range.
1.2 Draw a histogram to represent the data in the frequency table.
1.3 On the same set of axes, draw a frequency polygon to represent this data.
1.4 Draw an ogive curve on a new set of axes.
1.5 Use your ogive curve to determine an approximate value for:
1.5.1 the median
1.5.2 the lower quartile
1.5.3 the upper quartile
1.5.4 the interquartile range
1.5.5 the 90th percentile
1.6 Draw a box-and-whisker diagram to represent the spread of this data.
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Unit 3: Variance and standard deviation
of ungrouped data
Statisticians have derived formulae to find the variance and standard deviation of
a set of data values. It is important to understand these formulae and to know how KEY WORDS
to use them. However, we can also calculate these values using a calculator, which
variance – the average of the
is a quicker method. Unless a question requires the use of the formulae, choose the
squared differences from the
calculator method.
mean
standard deviation – the
square root of the variance
Using formulae
Standard deviation is a way of measuring how spread out a set of data values are.
It is often useful to know whether most of the data values lie close to the mean, or
whether they are widely spread out with many values much higher or much lower
than the mean.
The graph shows two sets of data both with a mean of 100. But the blue data values
have a far greater standard deviation than the red data values. It is clear that the
blue data values are far more spread out than the red data values. So, a low standard
deviation indicates that the data values tend to be very close to the mean, whereas a
high standard deviation indicates that the data values are spread out and have a large
range.
The matric results of two schools are represented on the frequency polygons shown
below. Although the two schools have the same mean (58%), the results of the
learners in the school shown in blue (standard deviation of 16) are far more widely
spread than those of the learners from the school shown in red (standard deviation
of 9).
y Frequency polygon showing standard deviations of matric results
standard deviation of red = 9
standard deviation of blue = 16
Mean = 58%
Frequency
x
58
Marks obtained (%)
The crucial aspect that we are interested in when determining the standard deviation,
is to find out how far each data value falls from the mean. This process leads us to
determine the variance (the average of the squared differences from the mean). The
standard deviation is therefore the square root of the variance.
Unit 3 Variance and standard deviation of ungrouped data 301
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The method we use to find the variance is:
( )
__ ___________________
• Determine the mean value for the set of data: x = sum of data values
number of data values
__
• Subtract the mean from each data value: (xi − x) where xi represents each value of x
__
• Square each of these differences: (xi − x)2. This will ensure that all values will be
positive, as we are not interested in whether the data values are higher or lower
than the mean, but only how far they are from the mean. Also, by squaring we
allow the bigger differences to stand out, as these have an impact on the standard
deviation. n
• Add these squared differences: ∑( xi − x )2
__
i=1 n
• Divide the total of the squared differences by the number of data values:∑( xi − x2 )
i=1
This result will represent the average of the squared differences from
√
__________
the mean, that is, the variance. n
∑( xi − __x )2
• The formula to find the standard deviation is therefore: σ = i = 1
__________
n
WORKED EXAMPLE
A game ranger measured the heights of different buck and recorded the results.
The heights at the shoulders are: kudu 150 cm; waterbuck 130 cm; duiker 100 cm;
steenbok 50 cm and dik-dik 35 cm
Heights of different buck
150
125
Heights (cm)
100
75
50
25
Determine the variance and standard deviation for this set of data.
SOLUTION
First we need to find the mean:
1 500 + 1 300 + 1 000 + 500 + 350 = _____
Mean = _____________________________ 4 650 = 930
5 5
Now we can determine the variance using a table:
__ __
Heights in mm ( xi ) ( xi − x ) ( xi − x )2
1 500 570 324 900
1 300 370 136 900
1 000 70 4 900
500 –430 184 900
350 –580 336 400
Total 988 000
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n
Therefore the variance = σ2 = ∑(xi − __x)2 = 988 000 = 197 600
_______
5
i=1
Notice: the symbol for standard deviation is σ. As the variance is the square of the
standard deviation, we can use the symbol σ2 for variance.
_______
Therefore the standard deviation = σ = √197 600 = 444,5 mm.
We can now determine which heights are within one standard deviation of the mean:
• Add the standard deviation value of 444,5 to the mean value 930 to get the upper
limit (1 374,5).
• Subtract the standard deviation value of 444,5 from the mean 930 to get the lower
limit (485,5)
• So any height that falls between 485,5 mm and 1 374,5 mm lies within one
standard deviation from the mean.
Any height that does not fall within one standard deviation of the mean would be
regarded as being particularly tall (the 1 500 mm buck) or short (the 350 mm buck).
Using a calculator
When there is a large set of data values, use a calculator to get the variance and/or
standard deviation. It would be tedious to go through the process necessary to apply
the formulae. Using the example about the heights of five different buck, we can use
our calculators to determine these answers.
• Go to MODE: choose the STAT option.
• Choose the VAR option: You will now have a table on your screen.
• Type in each of the data values, pressing = after each entry.
• You should now have this information on your screen:
X
1 600
2 480
3 170
4 410
5 300
• Press AC.
• Go to SHIFT 1 (the other STAT option).
• Choose the VAR option.
• Choose the xσn option, followed by =
• You should have the answer 147,7 on the screen, which is the standard deviation.
• If you want the variance, then square the standard deviation.
If you have data that occurs more often than once, first set up your calculator to
include a frequency column.
Unit 3 Variance and standard deviation of ungrouped data 303
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WORKED EXAMPLE
The table shows the number of questions answered correctly by a class of learners
in a general knowledge test consisting of ten questions.
Number of questions
answered correctly Number of learners
0 3
1 2
2 1
3 5
4 7
5 6
6 2
7 4
8 5
9 2
10 1
1 Determine the mean number of questions answered correctly.
2 Determine the standard deviation (correct to one decimal place).
SOLUTIONS
1 Using a calculator:
• Go to SET UP.
• Scroll down until you see the option STAT.
• Now choose frequency ON. (You will now have created a frequency
column when you get to your table.)
• Clear (AC).
• Go to MODE and choose STAT.
• Choose VAR.
• Insert the two columns of values into the table, using the top central key
to scroll up, down , left and right. Remember to press = after each entry.
• Clear (AC).
• Go to the other STAT option (Shift 1).
• Choose the __VAR option.
• Choose the x option followed by =
This will show that the mean is 4,9 questions answered correctly.
2 You have already set up the table of values, so:
• Clear (AC).
• Go to Shift STAT again and choose VAR again.
• Choose the xσn option, followed by =
This will show that the standard deviation is 2,6.
304 Topic 13 Statistics
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EXERCISE 2
1 A teacher asked a group of learners how long in minutes it took them to
complete their mathematics homework. They gave these answers:
12; 19; 33; 40; 24; 25; 15; 38
1.1 Determine the mean number of minutes taken by the learners to complete
their homework.
1.2 Determine the variance and standard deviation to two decimal places by
_________
√
completing the table n
∑(xi − __x)2
below using the formula: σ = __________
i=1
n
__› __›
Time taken in minutes ( xi − x ) ( xi − x )2
12
15
19
24
25
33
38
40
1.3 How many data values fall within one standard deviation of the mean?
2 A learner does a survey on 23 cars travelling past him. He counts the number
of passengers in each car. His results are recorded in the table.
Number of passengers Frequency
1 2
2 9
3 5
4 4
5 3
Using your calculator, determine (correct to two decimal places):
2.1 the standard deviation
2.2 the variance
2.3 the mean.
Unit 3 Variance and standard deviation of ungrouped data 305
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Unit 4: Symmetric and skewed data
We describe how the data values are distributed throughout the range, relative to
the median.
• Symmetrical data values are balanced on either side of the median (the middle
value). If we draw a histogram (or frequency polygon) to represent symmetrical
data, we obtain a symmetrical shape.
• y
10
8
6
4
2
x
2 4 Mode 6 8 10
Median
Mean
A symmetrical frequency polygon A symmetrical histogram
This histogram, while not perfectly symmetrical, is close to being an exact image
on either side. When we have perfect symmetry as in the frequency polygon
with the middle value having the highest frequency, the median (middle value)
and modal values will be the same. When we work out the mean, it is also the
same value.
If we represent this spread in a box-and-whisker diagram, we can clearly see
the symmetry:
• Skewed data values are more y
spread out on one side than 20
on the other. Due to the high
frequency of many of the
smaller values, the median 15
(middle value) is closer to the
Frequency
left. The values are more spread
out on the right, so we say that 10
this set of data is skewed to the
right (or the data is positively
skewed). Notice that the high 5
data values on the right will
cause the mean to be higher
than the median. x
100 200 300 400 500 600
Ratio
Median
Mean
306 Topic 13 Statistics
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A box-and-whisker diagram of data values that are skewed to the right would be:
A frequency polygon can also show data that is skewed to the right:
y
10
8
6
4
2
x
2 4 6 8 10
In the histogram on the right, the mode (most frequent value) is the same y
as the median, but the data values are more spread out on the left. We say 24
that this set of data values is skewed to the left (or negatively skewed). 22
20
The fact that there are some very low values will cause the mean to be 18
lower than the median. 16
Frequency
14
A box-and-whisker diagram of this situation would be:
12
10
8
6
We can conclude: 4
2
• If the mean = the median, then the data is symmetrical. x
• If the mean > the median, then the data is skewed to the right. 45 46 47 48 49 50 51 52 53 54 55
• If the mean < the median, then the data is skewed to the left. Age
It is possible for three sets of data values to have the same range and the Mean Median
same interquartile range when one set is symmetrical, one is skewed to
the right and one is skewed to the left. See the box-and-whisker plots:
Skewed to the right
Symmetrical
Skewed to the left
Notice that the widths of the boxes show the interquartile range and the widths of
the whiskers show the range, which are the same. But the central line which shows
the median is different, showing that some data values are skewed.
Unit 4 Symmetric and skewed data 307
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Unit 5: Identification of outliers
KEY WORD
An outlier is a data value that does not follow the pattern or trend of the rest of the
outlier – a data value that data. We can sometimes see this clearly when the data values have been plotted in a
does not follow the trend of
scatter plot.
the rest of the data
y
20
x
15
10
5
x xx x
x xx x xx x x xxx xx
x x xx
0 x
100 200 300 400 500
In this scatter plot, the data points follow a similar pattern and create a linear shape,
except for the point at approximately (350;16). This point clearly lies outside the
pattern created by the rest of the data values and we call it an outlier. If we omit
this outlier, the rest of the data values will have a strong positive correlation, and all
strongly follow a similar pattern that creates a straight line with a positive gradient.
In Grade 12 we will sketch scatter plots and discuss further the various types of
correlation that may be present. For now we must be able to visualise outliers by
recognising that they are points on a scatter plot which do not follow the pattern
created by the rest of the data values.
It is possible to determine whether a data value is an outlier without drawing
a scatter plot.
Any value that is either:
less than Q1– 1,5 × IQR, or greater than Q3 + 1,5 × IQR is an outlier.
WORKED EXAMPLE
The table indicates the CPI (consumer price index) inflation rate in South Africa
between 2005 and 2010.
Year CPI rate
2005 3,4
2006 4,7
2007 7,1
2008 11,5
2009 7,1
2010 4,3
1 Determine the five-number summary for this set of data.
2 Draw a box-and-whisker plot to represent the dispersion of this data.
3 Calculate whether there are any outliers.
308 Topic 13 Statistics
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SOLUTION
1 Arrange these values in ascending order: 3,4; 4,3; 4,7; 7,1; 7,1; 11,5
The median is therefore between 4,7 and 7,1 which is 5,9.
( 4 )th data value, which is 4,075.
7
Q1 = __
Q3 = ( ___
4 )
21 th data value, which is 8,2.
So: Minimum = 3,4; Q1 = 4,075; Q2 (median) = 5,9; Q3 = 8,2;
Maximum = 11,5
2
3,4 4,075 8,2 11,5
3 4 5 6 7 8 9 10 11 12
3 IQR = 8,2 – 4,075 = 4,125
1,5 × 4,125 = 6,1875
Q1 – 6,1875 = – 2,11 and Q3 + 6,1875 = 14,3875
Therefore, although the value of 11,5 in 2008 was higher than the others
(causing the data to be skewed to the right), it was not high enough to be
classified as an outlier.
If the CPI in 2008 had been 14,5, then that data value would have been an
outlier, and the box-and-whisker diagram would look like this:
3 4 5 6 7 8 9 10 11 12 13 14
Notice that a very long whisker is often an indication that there is an outlier.
This outlier causes the data to be skewed to the right, even though the
median falls almost in the middle of the interquartile range.
Unit 5 Identification of outliers 309
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EXERCISE 3
Exercise 3 has
1 Raeez studied the in-migration figures for some municipalities of South Africa as
questions that
cover Units 4 shown in the table.
and 5. Municipality Province In-migration figures in 2006
Nkangala DM Mpumalanga 1 452
Siyanda DM Northern Cape 1 504
Ehlanzeni DM Mpumalanga 2 465
Nelson Mandela MM Eastern Cape 6 715
DID YOU KNOW? Waterberg DM Limpopo 11 694
In-migration refers to people Umgungundlovu DM KwaZulu-Natal 13 149
who are not born in a Overberg DM Western Cape 14 965
particular area, but move and Bojanala Platinum DM North West 20 168
settle there.
Ekurhuleni MM Gauteng 140 252
Zimbabwe
Botswana Mozam-
bique
Limpopo
Namibia
Mpumalanga
North West Gauteng
Swazi -
land
Free State KwaZulu-
Natal
Northern Cape Lesotho
Eastern Cape
Western Cape
A provincial map of South Africa
1.1 Determine the five-number summary for the set of data in the table.
1.2 Draw a box-and-whisker plot to demonstrate this dispersion.
1.3 Describe the dispersion of this set of data.
1.4 Which aspect of this box-and-whisker diagram suggests that there is
an outlier?
1.5 Prove that the value of 140 252 is an outlier.
1.6 Give a possible reason for why the Ekurhuleni MM municipality had an
in-migration rate much higher than any of the other municipalities.
Raeez found a list which included more municipalities. This list is shown in the
table on the next page, with the figures arranged in ascending order.
310 Topic 13 Statistics
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Municipality Province In-migration in 2006
Nkangala DM Mpumalanga 1 452
Siyanda DM Northern Cape 1 504
Ehlanzeni DM Mpumalanga 2 465
Southern DM North West 4 914
Nelson Mandela MM Eastern Cape 6 715
Waterberg DM Limpopo 11 694
Umgungundlovu DM KwaZulu-Natal 13 149
Overberg DM Western Cape 14 965
West Coast DM Western Cape 17 211
Metsweding DM Gauteng 18 560
Boland DM Western Cape 18 770
Bojanala Platinum DM North West 20 168
Eden DM Western Cape 22 983
eThekwini MM KwaZulu-Natal 27 277
West Rand DM Gauteng 42 674
City of Johannesburg MM Gauteng 120 330
City of Cape Town MM Western Cape 129 400
City of Tshwane MM Gauteng 137 685
Ekurhuleni MM Gauteng 140 252
1.7 Determine whether Ekurhuleni’s data value is still an outlier. Explain why
your answer reflects the analysis of in-migration figures better than the
result in Question 1.5.
1.8 Complete the frequency table which divides the values from the table
into groups.
In-migration groups Frequency
0–30 000
30 000–60 000
60 000–90 000
90 000–120 000
120 000–150 000
1.9 Draw a histogram to represent the grouped data in the table.
1.10 Give a possible reason for the four municipalities at the upper end of the
range all having in-migration figures that are far higher than the rest.
2 Many concerned South Africans who are aware of the problems of water shortage
hold this opinion: ‘Another South African town runs out of water as two of its
dams dry up. Urgent attention is needed to prevent this happening to more
towns and cities across the country. Our nation’s use of water will have to change
to prevent total water scarcity.’
Refer to the table on dam levels and answer these questions:
2.1 Determine the mean percentage of water in the dams listed in the table.
2.2 Determine the standard deviation for water in the dams.
2.3 How many data values are within one standard deviation of the mean?
2.4 Determine the interquartile range.
Unit 5 Identification of outliers 311
PLT MATHS LB 11 7th pgs (Real Book).indb 311 2012/07/02 2:25 PM
2.5 Are there any outliers in this set of data? Justify your answer.
2.6 Compare the median and mean values. State whether the data is
symmetrical, skewed to the left or skewed to the right.
The table shows the percentage filled of some dams around the country in
March 2010.
Dam River % full
Middel-Metabo Dam Middel-Metabo River 6,9
Berg River Dam Berg River 83,8
Eikenhof Dam Palmiet River 46,8
Misverstand Berg River 100
Steenbras Dam lower Steenbras River 46
Steenbras Dam Upper Steenbras River 69,2
Voelvlei Dam Voëlvlei River 65
Wemmershoek Dam Wemmers River 52,2
Groendal Dam Swartkops River 40,1
Ernest Robertson Dam Groot Brak River 87,2
Garden Route Dam Swart River 32,2
Hartebeestkuil Dam Hartenbos River 46,8
Impofu Dam Krom River 49,7
Kromrivier Dam Krom River 12,6
Wolwedans Dam Groot Brak River 30,6
Korentepoort Dam Korinte River 26,9
Duiwelshok Dam Duiwenhoks River 26,9
Kommandodrift Dam Tarka River 39,0
Bridle Drift Dam Buffalo River 32,8
Gariep Dam Orange River 95,0
Elandskloof Dam Elands River 36,4
Roode Els Berg Dam Sanddriftskloof River 36,9
Theewaterskloof Dam Riviersonderend 70,6
Gamka Dam Gamka River 0,0
Kammanassie Dam Kammanassie River 9,4
Leeugamka dam Leeu River 12,1
Miertjieskraal Dam Brand River 0,0
Stompdrift Dam Olifants River 33,9
Beervlei Dam Groot River 0,0
Haarlem Dam Groot River 23,2
Kouga Dam Kouga River 33,9
Loerie Dam Loerie Spruit 40,9
Darlington Dam Sondags River 42,8
Grassridge Dam Groot Brak River 41,2
Katrivier Dam Kat River 48,6
Katse Dam Malibamatso River 98,2
Vanderkloof Dam Orange River 98,6
Vaal Dam Vaal River 99,8
Source: Water Rhapsody
312 Topic 13 Statistics
PLT MATHS LB 11 7th pgs (Real Book).indb 312 2012/07/02 2:25 PM
Revision
1 Fifteen households were surveyed in suburb A to find out how much each one
spent on electricity for a ten-day period. The results in rand are:
90 102 50 125 141 220 196 78 137 142 123 157 118 165 121
1.1 Determine:
1.1.1 the median (2)
1.1.2 the lower quartile (1)
1.1.3 the upper quartile. (1)
1.2 Draw a box-and-whisker diagram to illustrate this data. (4)
1.3 Calculate the mean expenditure on electricity for the 15 households. (3)
1.4 Determine the standard deviation for the data to two decimal places. (2)
1.5 Calculate the percentage of households whose expenditure on
electricity falls within one standard deviation of the mean. (3)
1.6 The results of the same survey conducted in suburb B are shown
in the box-and-whisker diagram.
20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360 380 400 420
Compare and comment on the dispersion of household expenditures
on electricity in suburbs A and B. (3)
1.7 Which suburb is more likely to have an outlier in their data values?
Justify your answer. (2)
[21]
2 The bar chart shows the distribution of the marks scored for a test by
a group of learners.
Distribution of marks
16
14
12
Frequency
10
8
6
4
2
0
1 2 3 4 5
Marks out of 5
2.1 Show clearly that the mean mark is 3. (2)
2.2 Copy and complete the table using the data on the bar chart.
__ __ __
xi f xi – x (xi – x)2 fi × (xi – x)2
1
2
3
4
5
Total – – (4)
313
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TOPIC 13: REVISION CONTINUED
2.3 How calculate the standard deviation of the mark distribution,
correct to two decimal places. (3)
2.4 The same group of learners wrote a second test. The standard
deviation for the second test was 2,12. Compare and discuss the
results of the two tests. (2)
[11]
3 A squad of 20 rugby players had their masses recorded as a stem-and-leaf plot.
Stem Leaf
6 5 7
7 8 8 9
8 0 0 1 8 9
9 0 5 9
10 5 8 8
11 4 8 8
12 9
3.1 Write down the five-number summary for the data. (5)
3.2 Use your five-number summary to draw a box-and-whisker
diagram to represent the data. (3)
3.3 Use your box-and-whisker diagram to help you decide whether
the statements are true or false. If false, write down the correct statement.
A: 25% of the players weigh more than 108 kg.
B: There are more players whose weight lies in the third quartile
than in the second quartile.
C: The data is skewed to the right. (3)
[11]
4 A group of learners wrote a standardised Mathematics test that was scored
out of 60. The results were represented in a cumulative frequency graph or
ogive curve.
Cumulative frequency curve of Mathematics results
250
200
Cumulative frequency
150
100
50
0
0 5 10 15 20 25 30 35 40 45 50 55 60 65
Mark
314
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4.1 How many learners wrote the test? (1)
4.2 How many learners scored at most 20 out of 60? (1)
4.3 What was the median test score? (1)
4.4 Determine the interquartile range. (Show your working.) (2)
4.5 Which was the modal group score? (1)
4.6 Determine whether there were any outliers in this set of data. (3)
4.7 Use your calculator to determine an approximate value for:
4.7.1 the mean (show some working) (4)
4.7.2 the standard deviation. (2)
[15]
5 The table shows the results from a survey of cell phone expenditure for
100 learners.
Expenditure (in rand) Frequency Cumulative frequency
x < 50 2
50 ≤ x < 100 22
100 ≤ x < 150 58
150 ≤ x < 200 14
200 ≤ x < 250 4
5.1 Copy and complete the table. (1)
5.2 Draw an ogive curve for this data. (4)
5.3 Use your graph to estimate the median cell phone expenditure for
this group of learners, indicating where you read off your answer. (2)
5.4 Draw a frequency polygon for this data. (4)
5.5 Determine an approximate mean for this data. (4)
5.6 Discuss the dispersion of this data. (2)
[17]
6 Grade 11 learners conducted a survey on weather y
patterns in their area. They measured the Scatter plot showing rainfall and temperature
100
average rainfall per month against the average
temperature per month over seven months. 90
They represented their results 80
on a scatter plot. 70
Rainfall in mm
6.1 What can you see from the graph about the 60
general trend of rainfall 50
patterns as the temperature increases? (1)
40
6.2 Which reading would you consider an
30
outlier? (1)
[2] 20
10
0 x
10 20 30 40 50 60 70 80
Temperature in °C
315
PLT MATHS LB 11 7th pgs (Real Book).indb 315 2012/07/02 2:25 PM
Formal assessment: Term 4 Test
1 The percentage marks for a class of 19 learners for a Mathematics test were:
81; 80; 74; 75; 57; 55; 91; 88; a; 76; 61; 60; 83; 89; b; 69; 80; 90; 80
The box-and-whisker diagram shows the five point summary:
55 b c 88 a
1.1 If the range of the data is 40 and the inter-quartile range is 23,
determine the values of a, b and c. (4)
1.2 Describe the skewness of the distribution. (2)
[6]
2 Snowskiers pass a point on a ski slope at a speed in km/h.
Their speeds are in the table.
Speed (km/h) Frequency ( f i) Cumulative frequency
0 ≤ x < 10 10
10 ≤ x < 20 20
20 ≤ x < 30 45
30 ≤ x < 40 71
40 ≤ x < 50 21
2.1 Copy and complete the table. (2)
2.2 Draw a cumulative frequency curve (ogive). (4)
2.3 Indicate clearly on your cumulative frequency curve the lower
quartile (Q1), median (Q 2) and upper quartile (Q 3). (3)
2.4 Use your graph to estimate the number of skiers who passed the
point with speeds greater than 35 km/h. (2)
[11]
316
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Formal assessment: Term 4 Test
3 A cricket coach recorded the number of minutes the batsmen in his
team needed to reach 50 runs during the most recent cricket season.
His results are tabled below:
Minutes Midpoint of Number of __ __
needed interval ( fi ) batsmen (xi ) fi.xi (x – x)2 f.(x – x)2
50 < x ≤ 60 12
60 < x ≤ 70 18
70 < x ≤ 80 28
80 < x ≤ 90 37
90 < x ≤ 100 5
3.1 Complete the columns for the midpoints and the f.x and then
calculate an estimate of the mean. Give your answer for the mean
correct to one decimal place. (4)
3.2 Complete the remainder of the table to calculate the standard
deviation, correct to one decimal place. (5)
[9]
4 Consider the given data which represents the individual distances travelled
on a particular day by a group of cyclists:
23; 32; 32; 23; 29; 32; 27; 27; 36; 28; 11; 27; 27; 40; 36
4.1 Determine the five-number summary, and hence draw
a box-and-whisker diagram to represent this data. (7)
4.2 Use your answer in 4.1 to discuss the dispersion of this data. (3)
4.3 Use your calculator to determine, correct to one decimal place:
4.3.1 the mean number of kilometres travelled (3)
4.3.2 the standard deviation. (3)
4.4 Determine the percentage of trips that is within one standard
deviation of the mean. (3)
4.5 4.5.1 By grouping the given data, complete the table:
Kilometres travelled Frequency Midpoint
10 < x ≤ 20 1
20 < x ≤ 30 8
30 < x ≤ 40 6 (2)
4.5.2 Use the grouped data to sketch a frequency polygon. (4)
[25]
Total: 51 marks
317
PLT MATHS LB 11 7th pgs (Real Book).indb 317 2012/07/02 2:25 PM
Term 4 summary
Topic 13
Statistics
We can represent grouped data graphically in various ways including:
• Histograms
Masses of fish
15
Frequency
10
5
10–12 12–14 14–16 16–18
Mass (kg)
• A bar graph
– The area of each bar is relevant; your groups should be equal in width if possible.
– The bars must be joined.
– Frequency polygons (drawn with or without a histogram)
Vehicles in a particular route
2,5
2,0
Frequency
1,5
1
0,5
0
0–6 7–12 13–18 19–24
Hours
• A frequency polygon on a histogram
20
D
Frequency
C
B
10
E
F
A
M N
0 32 37 42 47 52 57 62 67 72
Marks
– Plot the midpoint of each group against the frequency of the group.
– Join each point as a jagged line graph.
– Plot the midpoint of the group before and after each given group on the x-axis,
to secure the polygon to the x-axis.
– The area enclosed by the polygon will equal the total area of the histogram.
318 Term 4 summary
PLTMATHSLB11LB_14.indd 318 2012/07/07 11:40 AM
• Cumulative frequency curve (ogive curve)
40
35
30
25
Frequency
20
15
10
5
0
10,5 15,5 20,5 25,5 30,5 35,5 40,5 45,5
Length (mm)
– Plot the end of each group against the cumulative frequency at the end of each
group.
– Join each point as a curved graph.
– Plot the end point of the group before the given groups on the x-axis to secure
the beginning of the curve to the x-axis.
You can analyse and understand the spread or dispersion of data better by
considering:
Range:
• When there is a large range, the data is widely spread between the maximum and
minimum values
• When there is a small range, the data values are all closely bunched.
Symmetrical or skewed data:
• When the mean is greater than the median, the data is skewed to the right.
• When the mean is less than the median, the data is skewed to the left
• When the mean equals the median, the data is symmetrical.
Standard deviation:
• This value reflects how close most of the data is to the mean
_________
√
n
∑(x − _x)
i
2
__________
i=1
• We can find the standard deviation by using the formula: n
or using a calculator.
Outliers: Any data value that is less than Q1 – 1,5 × IQR or greater than
Q1 + 1,5 × IQR is an outlier.
Term 4 summary 319
PLTMATHSLB11LB_14.indd 319 2012/07/07 11:40 AM
Formal assessment
Examination practice Paper 1
Time: 3 hours
The following formulae are given for you to use:
_______
√ 2
x = −b± b − 4ac
____________
2a
y2 – y
m = ______1
x2 – x1 ; y = mx + c; y – y1 = m(x – x1)
A = P(1 + in); A = P(1 – in); A = P(1 + i)n; A = P(1 – i)n
Question 1
1.1 Solve for x:
1.1.1 2x2 + 3 = 7x (4)
1.1.2 1
_____ – x– 2
_________ + 7x + 10
________ (6)
x – 2 x2 + 2x + 4 2(8 – x3)
_____
1.1.3 8 – 2√2x + 5 = x (6)
2
__
1.1.4 3x3 – 12 = 0 (3)
1.1.5 4x + 1 = 322x – 1 (4)
1.2 Solve simultaneously for x and y if x2 – 2xy – 3y2 = 0 and 3x + y – 2 = 0. (8)
1.3 Consider the expression x2 – 4x – 12 and determine the value(s) of x for which:
1.3.1 x2 – 4x – 12 < 0 (4)
1.3.2 2 1
__________ is undefined. (2)
x – 4x – 12
1.4 Determine the nature of the roots of 2x2 – 3x – 8 = 0. (3)
1.5 Fully simplify:
________
√ 2_________
399 396
1.5.1 +2 (3)
396
2
___ ___ ___
___
1.5.2 ( √5x – √3x )( √5x + √3x ) (2)
[45]
Question 2
2.1 Consider the sequence 3; 8; 13; 18; 23; …
2.1.1 Write down the next two terms in the sequence if the pattern
continues in the same way. (2)
2.1.2 Write down a formula for the nth term in the form Tn = … (2)
2.2 Consider the sequence –2; 3; 12; 25; 42; …
2.2.1 Write down the next two terms in the sequence if the pattern
continues in the same way. (2)
2.2.2 Determine a formula for the nth term in the form Tn = … (5)
2.2.3 Which term in the sequence has a value of 558? (5)
2.3 Determine the value of x if 3; 12; x; 54; ... is a quadratic sequence. (5)
2.4 Consider the sequence 6; 18; 54; 162; 486 …
2.4.1 Write down the next term in the sequence if the pattern
continues in the same way. (1)
2.4.2 Write down a formula for the nth term in the form Tn = … (3)
2.4.3 Hence write down a formula for the nth term of the sequence
11; 23; 59; 167; 491; … if the pattern continues in the same way. (1)
[26]
320 Exam practice paper 1
PLT MATHS LB 11 7th pgs (Real Book).indb 320 2012/07/02 2:25 PM
Question 3
4 + 2, g(x) = − __
Given f (x) = _____ 1 (x + 1)2 + 8 and h(x) = __ 5.
1 x + __
x–1 2 2 2
3.1 State the range of g. (2)
3.2 Determine g(−1) – h(−1). (2)
3.3 Solve for x if f (x) = h(x). (6)
3.4 State the equations of the asymptotes of f. (2)
3.5 Determine the x- and y-intercepts of f. (3)
3.6 Sketch f, g and h on the same system of axes. Show all the x-intercepts and
y-intercepts and the coordinates of any turning point(s). (13)
3.7 State the equation of v(x) if v is the reflection of f in the x-axis. (2)
3.8 w(x) = g(x – 5) – 4 , write down the equation of w(x). (2)
[32]
Question 4
y
g
(–1;16)
f
(6;8)
4
y=1
x
–2 3
Consider the sketch which is not drawn to scale.
f (x) = ax2 + bx + c passes through (−2;0), (3;0) and (6;8).
y = 1 is the asymptote of g(x) = kpx + q which passes through (–1;16) and (0;4).
4.1 Determine the equation of f and write your answer in the form
f (x) = ax2 + bx + c. (5)
4.2 Determine the equation of g. (5)
4.3 For which value(s) of x is f (x) an increasing function? (2)
4.4 For which value(s) of x is f (x).g(x) ≥ 0? (2)
[14]
Question 5
5.1 If an investment doubles in value over 8 years, calculate the rate of interest
per annum (to two decimal places), if the interest was:
5.1.1 simple interest paid annually (3)
5.1.2 compounded annually. (3)
Exam practice paper 1 321
PLT MATHS LB 11 7th pgs (Real Book).indb 321 2012/07/02 2:25 PM
5.2 A company buys a bus at a cost of R2 750 000. Calculate the book value
of the truck after 8 years if the rate of depreciation on a reducing
balance method is expected to be 9% p.a. (3)
5.3 Mr Smith has two sons who are 5 years and 13 years respectively.
He wishes to give each son R21 000 on his 21st birthday. Calculate
how much he should invest now if interest is 11,5% p.a. compounded
semi-annually. (7)
[16}
Question 6
Thirty people were asked whether they had travelled by train, car or bus that day.
Some people had used one form of transport, some two and one person all three
forms. Seven people had used no transport that day. The details of this survey are
recorded in the Venn diagram.
T C
6 4 5
1
2 x
3
B 7
6.1 Find the value of x. (2)
6.2 How many people used both bus and train? (2)
6.3 How many people have not used a train as transport? (2)
6.4 Determine:
6.4.1 P(T ∩ C)′ (2)
6.4.2 P(B | T) (2)
6.5 Two taxi companies, Call-a-cab and Terrific Taxis operate in a certain
suburb. Tracy calls Terrific Taxis 70% of the time when she needs a taxi,
and Call-a-cab on every other occasion. Terrific Taxis arrives late 10% of
the time and Call-a-cab arrives late 15% of the time.
6.5.1 Draw a tree diagram to represent this information, showing the
probability of each branch. (4)
6.5.2 Determine the probability that she calls Call-a-cab and it arrives
on time. (3)
[17]
Total: 150 marks
322 Exam practice paper 1
PLT MATHS LB 11 7th pgs (Real Book).indb 322 2012/07/02 2:25 PM
Formal assessment
Examination practice Paper 2
Time: 3 hours
The following formulae are given for you to use:
y –y
______
m = x2 1
– x ; y = mx + c; y – y1 = m( x – x1 )
2 1
( )
________________ x + x2 y1 +y
d = √ ( x2 – x1 )2 + ( y2 – y1 )2 ; M 1
_______ ;______2 ; m = tan θ
2 2
n
∑( xi − x )
_________
√
__ 2
∑x __ ∑fx
n ___
__________; SD = ∑ ( xi − x )2
›
__
x = _____; x = _____; var = i = 1
n n
_______
n n
i=1
n(A)
P(A) = ____
n(s)
Question 1
The table shows the number of registered voters in each province for the 2011
Municipal elections in South Africa:
Province Registered voters Rounded to the nearest hundred thousand
Northern Cape 572 140 600 000
Free State 1 386 521 1 400 000
North West 1 576 898 1 600 000
Mpumalanga 1 718 309 1 700 000
Limpopo 2 340 799 2 300 000
Western Cape 2 706 736 2 700 000
Eastern Cape 3 111 535 3 100 000
KwaZulu-Natal 4 648 733 4 600 000
Gauteng 5 592 676 5 600 000
Source: Electoral commission
1.1 Use the values rounded to the nearest hundred thousand in the last column
to answer the questions.
1.1.1 Determine the median of the given data. (1)
1.1.2 Determine the interquartile range for the data. (3)
1.1.3 Draw a box-and-whisker diagram on a diagram sheet like the one
below to represent the data.
y
Cumulative frequency
x
Percentage of voter turnout (3)
Exam practice paper 2 323
PLT MATHS LB 11 7th pgs (Real Book).indb 323 2012/07/02 2:25 PM
1.1.4 Use your box-and-whisker diagram a diagram sheet comment on
the symmetry of the data. (1)
1.1.5 Calculate the mean number of registered voters per province
(to two decimal places). (2)
1.1.6 Calculate the standard deviation for this data. (2)
1.1.7 Determine whether there are any outliers. (2)
1.2 The percentage of voter turnout at these elections, as recorded in each
province, is shown in the frequency table:
Percentage of voter turnout Number of provinces
50 < x ≤ 55 2
55< x ≤ 60 4
60 < x ≤ 65 3
1.2.1 Draw an ogive curve on the diagram sheet like the one below,
to represent the voter turnout. (4)
600 000 1 600 000 2 600 000 3 600 000 4 600 000 5 600 000
1.2.2 Given that the mean voter turnout for these elections was 57,6%,
use your ogive to determine the number of provinces that had
a voter turnout higher than the mean. Show on your ogive how
you read off your answer. (2)
[20]
Question 2
In the diagram below, △ABC is given with vertices A(6;1), B(3;–5) and C(–11;2).
y
C(–11;2)
A(6;1)
x
B(3;–5)
^ C = 90°
2.1 Prove: AB (3)
2.2 Determine the equation of the median AE of △ABC where E lies on BC. (5)
2.3 Determine the coordinates of D, so that ABCD is a parallelogram. (2)
2.4 Prove that AC = BD. (4)
2.5 What kind of parallelogram is ABCD? Give a reason for your answer. (2)
2.6 Determine the coordinates of F, the point of intersection of AC and BD. (2)
2.7 Determine the equation of the perpendicular bisector of BC. (4)
324 Exam practice paper 2
PLT MATHS LB 11 7th pgs (Real Book).indb 324 2012/07/02 2:25 PM
2.8 Show that the perpendicular bisector found in 2.6 passes through point F. (2)
^ D (to one decimal place).
2.9 Determine the size of AB (5)
[29]
Question 3
3__ where θ ∊ [90°; 270°]. Determine, using a diagram,
3.1 Given tan θ = ___
√7
the value of:
3.1.1 sin θ (4)
3.1.2 4cos2 θ – sin(180° + θ) (4)
tan 120°.cos180°
3.2 Simplify without using a calculator: _______________________________ (8)
sin 160°.cos 70° − cos 20°.sin (−110°)
3.3 3.3.1 State the values of θ ∊ [−180°;180°] for which the expression
cos θ + tan θ will be undefined.
________ (3)
1 + sin θ
3.3.2 Prove that: cos θ + tan θ = 1
________ _____ (5)
1 + sin θ cos θ
3.4 Solve the equations, correct to two decimal places:
3.4.1 3tan(2x + 10°) + 5 = 0 for 0° ≤ 2x + 10° ≤ 360° (5)
3.4.2 Find the general solution for 4sin x cos x − 3sin2 x = 0. (5)
[34]
Question 4
Given the functions f(x) = sin 2x and g(x) = tan x − 1.
4.1 Sketch the graphs of f and g on the same system of axes for
x ∊ [−180°; 180°], clearly labelling endpoints, turning points,
intercepts with the axes and asymptotes. (6)
4.2 Use your graphs to determine the value(s) of x for which:
4.2.1 sin 2x < 0 (2)
f(x)
4.2.2 ____ ≥ 0 (2)
g(x)
4.2.3 If h(x) = f(x) − 3, write down the range of h. (2)
[12]
Question 5
ABC is a triangle and D is a point on BC. AD is joined. DC = m, BD = 2m,
^ D = 30°, AB
BA ^ C = α and AD
^ C = θ.
A
30°
α θ
B 2m m C
D
5.1 Prove: Area △ADC = 2 m2 sin α sin θ (4)
5.2 If it is now given that m = 12 cm and θ = 70°, determine the length of
AC to the nearest centimetre. (3)
[7]
Exam practice paper 2 325
PLT MATHS LB 11 7th pgs (Real Book).indb 325 2012/07/02 2:25 PM
Question 6
The diagram represents a pencil with a hemispherical eraser on one end and a conical
sharpened point on the other end.
The total length of the pencil is 13 cm, and the cylindrical portion of the pencil, with
a diameter of 1 cm, is 10 cm long.
10 cm
1 cm
13 cm
Determine, to two decimal places:
6.1 the slant height of the conical end (4)
6.2 the total volume of the pencil (3)
6.3 the surface area of the cylindrical section. (5)
[12]
Question 7
Q
R
A B
P
7.1 In the diagram, AB is a tangent to the circle at P. Q and R are points on
the circumference of the circle. PQ, QR and RP are joined.
Prove the theorem that states R ^PB = R ^
QP. (5)
7.2 In the diagram AE is a diameter of the circle with centre O and ABC and
^ = 50°. FG is a tangent to the circle at A.
EDC are straight lines with C
DA and BE intersect at H. BO is joined.
C
50°
F
B
1 3
1
2 A
2 3
2
D 1 1 H 2
1
O
2 1 G
E
326 Exam practice paper 2
PLT MATHS LB 11 7th pgs (Real Book).indb 326 2012/07/02 2:25 PM
7.2.1 Give a reason why AB ^ E = 90°. (1)
7.2.2 Hence prove that BCDH is a cyclic quadrilateral. (3)
7.2.3 Determine the size of angles H ^ ,A^ and E giving reasons. (6)
1 2 2
7.2.4 ^ ^ ^ ^
If A1 = x, determine E1, O2 and O2 in terms of x, giving reasons. (7)
7.2.5 Prove that: A ^ =B^ (3)
1 2
[25]
Question 8
AF is a diameter of circle O with chords AB and DF. M is a point on the circumference
so that OM bisects AB at C. OB is joined, with OB ∥ DF.
D
F
O 1
B
1 2
C
A
M
^ = 2B
8.1 Prove that F ^. (6)
8.2 If it is given that CM = 8 mm and AC = 12 mm, determine the length
of OC, giving reasons. (5)
[11]
Total: 150 marks
Exam practice paper 2 327
PLT MATHS LB 11 7th pgs (Real Book).indb 327 2012/07/02 2:25 PM
Optional assessment: Pattern investigation
Task 1 Triangular numbers Note:
1.1 The diagram on the right represents the 3rd pattern of arranging squares in a This is an individual
triangular shape. Copy and complete the table below: investigation. Complete
Number of layers 1 2 3 4 5 6 all questions on a
separate answer sheet.
Number of squares in bottom layer 1 2 3
Total number squares 1 3 6 (2)
1.2 You know the bottom row of numbers in the table as the triangular numbers.
Use the method of differences to show that the formula for the nth triangular
1 n(n + 1).
number is Tn = __ (5)
2
1.3 Use this formula to find the number of squares needed for 10 layers. (1)
1.4 How many layers would use a total of 91 squares? (3)
[11]
Task 2 Square pyramid numbers
2.1 If the diagram on the right represents the 3rd pattern of stacking boxes (each box
is an identical cube) in a square-base pyramid form, copy and complete the table:
Number of layers 1 2 3 4 5 6
Number of boxes in base 1 4
Total number of boxes 1 5 14
(2)
2.2 How many boxes are there in a base that is n boxes wide (find Tn for row 2)? (1)
2.3 The formula for the nth square-based pyramid number (that is, find Tn for row 3)
n3 + __
is __ n2 + __
n.
3 2 6
Show how you could have used the numbers in the bottom row of the table to
establish that the formula is a 3rd degree expression. (2)
[5]
Task 3 Another look at the square bases
Each small square in the diagram has side of length 1 unit. Within the larger squares
there are smaller squares of side 1 unit, 2 units or 3 units, and so on. So within a
3-unit square there are 91-unit squares as well as some 2-unit squares, and so on.
328 Term 4 Assessment
PLT MATHS LB 11 7th pgs (Real Book).indb 328 2012/07/02 2:25 PM
3.1 By counting in the diagram, complete the table:
Total width Total width Total width Total width 4 Total width
Size of big square →
1 unit 2 units 3 units units 5 units
No. 1-unit squares 1 16
No. of 2-unit squares
No. of 3-unit squares 1
No. of 4-unit squares
No. of 5-unit squares
Total no. of squares 55
(5)
3.2 Do you notice anything familiar about the numbers in the bottom row? (1)
3.3 On the basis of what the table seems to show, suggest how we might calculate Hint:
the total number of squares (of all sizes) that can be found in a 9-by-9 grid. (2)
Hint: look at the table in
Question 2.1.
At the moment the only way we can do this calculation is with a calculator: do it
now, and see what you get for the 9-by-9 situation. (1)
If you look more closely at the calculation, you may find a formula.
3.4 Copy and complete the table:
From Total row in table above Multiply by 6 (call this Mk) Find value of Wk (in rational form)
1-by-1 1 6 M1 = 1 × 2 × W1; W1 =
2-by-2 M2 = 2 × 3 × W2; W2 =
3-by-3 M3 = 3 × 4 × W3; W3 =
4-by-4 M4 = 4 × 5 × W4; W4 =
5-by-5 55 330 M5 = 5 × 6 × W5; W5 = 11
…
9-by-9 M9 =
(8)
3.5 Decide how many squares (of all sizes) there would be in a 9-by-9 situation (1)
3.6 Does this answer match the answer you obtained earlier? (1) REMEMBER
3.7 Now try to write down a formula in terms of n for the total number of squares
Remember that the last
that can you can find in the n-by-n situation. (3)
column of the table
3.8 Now show that this formula is the same as the formula given in task 2.2. (2)
above gives numbers
[24] that are six times too big.
Pattern Investigation 329
PLT MATHS LB 11 7th pgs (Real Book).indb 329 2012/07/02 2:26 PM
Pattern Investigation continued
Task 4 Compare the triangular, square
and pyramid numbers
4.1 Summarise your findings of the previous tasks in the table. Copy and complete
more columns, and use them to answer the questions that follow.
Number of layers 1 2 3 4 5 6 7 8 9 10 11 12 13
Number of squares in a triangular shape 1 3 6 28
Number of squares in a square base 1 4 9 49
Total number of cubes in a square pyramid form 1 5 14 140
(6)
We can see that one square or box (of 1 unit) can be arranged as a triangular number,
a square number or a square pyramid number.
4.2 What is the next number (of squares or boxes) that can be arranged in a square
pyramid and also in a triangular shape? (2)
4.3 What number (of boxes or squares) can be arranged as either a square base or a
triangular shape? (2)
4.4 A large number of boxes is stacked in a square pyramid. The pyramid is
disassembled; some of the boxes are arranged in a square and the rest in a
triangle. How many boxes must there have been?
In other words, we want to find a square pyramid number that equals the sum of a
square number plus a triangular number. Find two such square pyramid numbers. (6)
4.5 If the boxes were built as towers as shown below, copy and complete the table for
the number of boxes used to make each tower.
Figure 1 Figure 2 Figure 3
Copy and complete the table. (6)
Figure number 1 2 3 4 5 6 n
Number of boxes in base
Total number of boxes used
How many boxes will a tower with 1 029 boxes in the base use? (3)
[25]
330 Term 4 Assessment
PLT MATHS LB 11 7th pgs (Real Book).indb 330 2012/07/02 2:26 PM
Answers
Note: Some answers are too complicated to provide
in this section. Please ask your teacher to check your
solutions after consulting the Teacher’s Guide.
TOPIC 1: EXERCISE 1 TOPIC 1: EXERCISE 6
4 a8b14 _
2 _
_ 3
5 __
15
1 __ 2 1 3 3a2 1.1 x3 1.2 x4y2 1.3 5x 4
5
__ __ ____
4 3 a2
___ 5 p2x 2.1 √x3
5
2.2 6√
4
x 2.3 √( 9x4 )
3
5
6 –32a4 + 16a2b3 – 2b6 7 212x
8 __x
3
TOPIC 1: EXERCISE 7
8 ___ 9 21 010 × 51 011 10
27 2 1 1.1 10 1.2 16
TOPIC 1: EXERCISE 2 1.3 8 1.4 4
4a c2 2
_____ y3
__ __
3x
1 2 3 y 1.5 _
5 1.6 _
5
5b2 27 8 6
11
4 ___
2x 5 2 6 9 1.7 6 1.8 2
3
y
y3
___ _
1 + a2
_______
1+a __
25 _
7 8 9 1.9 1.10 – 2
x2z2 2 a3 4 3
10 __
10 11 _
1 12 ___
49 __
1.11 24 1.12 8
3 4 100 7
x 3y3
13 _____ 14 _________
1 15 _
9 2 2.1 3x 9 2.2 16a4
y + x2
2 a6 + 6a3 + 9 4
16 5
__ 17 28 + x 18 8
___ 2.3 27a12b6 2.4 a2b4c5
x2 27
2.5 5a4b5c7 2.6 2a3
TOPIC 1: EXERCISE 3
2.7 8x6 2.8 x4
1 x = –3 2 x=7 3 x=5
2.9 x2 2.10 b2
4 _
x = –1 5 x=1 6 x = –1
2
2.11 x2 _
2.12 3
4
TOPIC 1: EXERCISE 4 3 3.1 50 3.2 5
__
3
1 x=3 2 x = –1 3 x=0
3.3 2 3.4 2(a + b)
4 x = –2 5 x=5 6 x=2
3.5 (a + b)2 3.6 _____
1
7 x=2 8 x=2 (x – 3)
TOPIC 1: EXERCISE 5 TOPIC 1: EXERCISE 8
1 x = 81 2 _
x=1 3 x = __
4
1 x = –7 2 x=4 4 49
4 x = – 64 5 x = ±___
1 6 x = ±64
3 x = 0 or x = –3 4 x = 3 or x = –1 125
7 x = 16 8 x= 1__ 9 ___
x = 625
_ 5 81
5 x = –1 6 x = –2 or x = 1
2
10 x=7 11 __
x = = 81 12 __
x= 8
16 27
7 x = 2 or x = –2 8 x = –3
9 x = –5
__ 10 x = –3
__ or x = 1
TOPIC 1: EXERCISE 9
2 2
1 x = 16 or x = 81 2 x = 8 or x = 125
3 x = 8 or x = –1 4 x = __
81 or x = 1
16
Answers 331
PLT MATHS LB 11 7th pgs (Real Book).indb 331 2012/07/02 2:26 PM
__ __
5 x = 16 6 x = __
9 5 4 – 2√3 6 10 + 6√3
25
7 x = 16 8 x = 25 7 37 8 4
9 _
x=9 10 _
x=1 9 2 10 _
3
4 3 2
TOPIC 1: EXERCISE 10 TOPIC__ 1: EXERCISE 13 __
__ __ √ __ √
1 1.1 4√ 2 1.2 4√3 1 3
___ 2 √3 3 6
___
3 3
__ __ __
__ __ √3 +3
______
1.3 6√2 1.4 5√5 4 3√ 2 5 6 2√ 2 + 1
__ 3
__
__ __ 7 3 + √3
______ 8 –3 – 2√ 2
1.5 8√2 1.6 4√6 2
__ __ __
__ ___ 9 3 + √2 + 3√ 3 + √6
_______________
2.√2 2√12
3 3
1.7 1.8 7
__ ___
1.9 x√x 1.10 2a2b√5b
TOPIC 1: EXERCISE 14
___ ___ __
1.11 28 7 6√
a b 2b 1.12 5x y 2 4√
2x 1 1.1 _____
8 = 16 + 8√ 2
___ 1 – ___
1__
__ √2
__
1.13 x3y2.√
3
x2y 1.14 21 005√2 2 √
LHS = _____
8 × ___
__
__ 1 1 – ___
__ √2
1.15 3670.√
3
9 √2
__ __
√2 √2 + 1
___ = 8__
_____ ______
× __
2 2.1 √ 35 2.2 30 √2 – 1 √2 + 1
__ __
6√6 8(2) + 8√ 2
2.3 3 2.4 = ________
2–1
__ __
2.5 24√6 2.6 _
7
2 = 16 + 8√2 = RHS
___
3 __ __ __
√3 √3 √_23
2.7 __ 2.8 216 2 – ___
1__ = __
√2 1.2 _
8 +5 _ 13
___
√6 2
__
2.9 3a2.√
3
4 2.10 a2 √
5
64 ___ __
LHS = √3 ___
4.2 + 5
√_23 – ____
1
___
√ 2.3
2.11 3 2.12 3a3 __ __ __
√3
_ +5 2
=2 2
√2
__ × ___
______
_ – __1
√3
__
√ 2. √ 3 √2
__ __
TOPIC 1: EXERCISE 11
__ __ √3√2
_ – ____
=7 2 __
2√ 3
1 1.1 8√ 2 1.2 7√5 __ __
__ __ √3
2 –_
=7 _ 1 _
2
2 √3
1.3 4√2 1.4 3√3 __
1.5 _
2 1.6 8
= √ 2_3 ( 7 – 1_2 )
5 __
3( 2 )
= √_ 2 __13
1.7 ___
1__ 1.8 2
√5 __
1.9 _
1 1.10 2
__ 2
= 13
2 3√ _ = RHS
2 _________
___ __ _________
___ __
_ _ 1.3 √
5
√ 37 + √ 5 .√ √ 37 – √ 5 = 2
5
1.11 –1 1.12 x
___________________
___ __ ___ __
2 5
LHS = √
5
(√ 37 + √ 5 )(√ 37 – √5 )
_____________
___ __
=√
5
(√ 37 )2 – (√ 5 )2)
TOPIC 1: EXERCISE 12 _____
__ __ __ __ =√
5
37 – 5
1 6√3 – 12 2 6 + 3√2 + 2√3 + √6 ___
= 5√ 32
__
3 2 4 9 – 4√5 = 2 = RHS
332 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 332 2012/07/02 2:26 PM
__ __ ___ _______
___
√8 – √ 3 – √ 12 __ __
1.4 ______________
___ __ ___ = 1 _ 2 √ m + √ n = √ 5 + √ 24
√ 32 – 3√ 3 – √ 27 2 _______
___
__ __
__ __
√ 2 – √ 3 – 2√ 3
__ (√ m + √ n )2 = (√5 + √ 24 )2
LHS = 2 __
______________
__ __ ___ ___
4√2 – 3√3 – 3√ 3 m + 2√ mn + n = 5 + √ 24
__ __ __
√ 2 – 3√ 3 ___
= 2 __
_________ __ m + n + 2√mn = 5 + 2√6
4√2 – 6√ 3
__ __ m + n = 5 and mn = 6
√
= 2 2__–
___________3√3__ (m + n)2 = 25
2(2√2 – 3√3 )
m2 + 2mn + n2 = 25
=_
1 = RHS
m2 + n2 = 25 – 2mn
2
__ m2 + n2 = 25 – 2(6)
1.5 ________
___15 __ = √ 5
√ 80 – √ 5 m2 + n2 = 13
LHS = ________
15 __
__ ___ ____
4√5 – √5 3 (1 + √3x___
) – √12x2 ___
2 2
__
____ 5
__ × ___
√ = 1 + 2√ 3x2 + 3x2 – 2√ 3x2
= 15 __
3√5 √5
__
__ _
3 __
√5 __ 4 RHS = 22 = √ 23 = 2√ 2
= 15
_____ = √ 5 = RHS
3×5 = 2(1,414) = 2,828 = LHS
_____
√ 2 011
2 ______
1
_____ = ____
1__
1.6 ____________
_____ _____ = _
2
√2,828 √2_2
3 3 3
√ 22 013 – √ 22 009 3
______
2 .2 √ 2 010 ____ = __
= 1 1
_______________
LHS = ______
√2
2 012
______
2 008
.2 – √2 .2 ( 22 )3
_ _
3 1
22
_
1
__ __
__
1 005√ 2 ___
1 √ 2 =√
___
= __ × __
1,414
2 = _____
___ = 0,707
= 1 006 2
______________
__ _ √2 √2 2 2
2 √ 2 – 21 004√ 2
__
2 __ 2
____________
= 1 004
1 005√ TOPIC 1: EXERCISE 15
2 2 √ 2 (2 – 1)
1 x = 14 2 No solution 3 x=4
__
3
=_
2 = RHS
3
__ ___ 4 x = 36 5 x = ±3 6 x = ±5
√3
______+1_=5+ √ 27
______
1.7
4 – 2√ 3 2 7 x=3 8 x = –1 9 x=4
__ __
√ 3 + 1 4 + 2√ 3 _
LHS = _______ × ______
4 – 2√3 4 + 2√ 3 10 x = –4 11 _ or x = 1 12
x=4 x=2
__ __ __ 3
4√3 + 2(√3 )2 + 4 + 2√3
= __________________
__
2 2
4 – (2√3 )
__ __ TOPIC 1: REVISION
√ √
=6 3+ 6+4=6 3
_________ + 10
_______
16 – 12 4 1 1.1 30x4y8 1.2 3x3
__ __
2(3√3 + 5) (3√3 + 5) 5
= ________ = ________ 1.3
y
___ 1.4 3
4 2 3x6
__
(3√3 + 5) x
__
6
_
1
RHS = ________ 1.5
2
1.6
3
2
LHS = RHS 1.7 43x5
___ __ ____
1.8 √ 12 + 2√ 3 + 31,5 = √ 147 1.8 4a8 – 12a4b3 + 9b6 1.9 24
__ __ _3
LHS = 2√3 + 2√3 + 32 __
7 3y
__
__
2 2.1 2.2
__ 16 x4
= 4√3 + √33 b2
__ __ 2.3 ___
3 2.4 _____
144
= 4√3 + 3√3 2a3 a12
__ 3b4
2.5 ___ 2.6 __
36
= 7√ 3 2 25
____ ______ __
RHS = √147 = √49 × 3 = 7√3 2.7 __
4– 1 2.8 4
4 x
2 2
LHS = RHS x y
_____
2.9
y2 + x2
Answers 333
PLT MATHS LB 11 7th pgs (Real Book).indb 333 2012/07/02 2:26 PM
3 3.1 5 3.2 1
__ TOPIC 2: EXERCISE 2
4
1 x = 7,58 or x = –1,58; Real, irrational
3.3 40 3.4 125
____
8 2 x = – 0,68 or x = –7,32; Real, irrational
3.5 3x2y4 3.6 5x4 3 x = 1,24 or x = –3,24; Real, irrational
4 x = 2; Real, rational
3.7 5 3.8 a2
5 _ or x = – 4 Real, rational
x=3
2
3.9 b2 3.10 x – 1 6 x = 9 or x = –1; Real, rational
3.11 6
__ 3.12 36 7 x = 4,73 or x = 1,27; Real, irrational
5 ___
__ __ 8 x = 5 ±√– 4 ; Non-real or imaginary roots
4 4.1 0 4.2 7√3 – 10√2 ______
–q±√ q – 4pr
2
__ 9 x = __________
4.3 √2 4.4 1
__ 2p
________
3 √ 2 + 4kn
10 m± m
x = __________
__ __
4.5 3√3 4.6 √2 2k
________
√ 2
– 8mk
11 x = m± m
__________
4m
4.7 1 4.8 x _______
√ 2 + 3k2
___ 12 x = –n± n
__________
_
1 3k
4.9 4.10 4 – √11
2
5 5.1 x=1 5.2 _
x = –7 TOPIC 2: EXERCISE 3
4
1 x = 3 or x = 7 2 x = –1,5 or x = 2
5.3 x = –6 5.4 x = 0 or x = –3
3 x = –1,5 or x = – 0,5 4 2
x = –1 or x = __
5.5 x=1 5.6 x=1 3
5 1 or x = 4
x = – __ 6 x = 0 or x = –5
6 6.1 x = ±125 6.2 x = ±8 3
7 x = 0 or x = 1 8 x = – 0,5 or x = 3
6.3 x = –2 or x = 4 6.4 x = 81
9 x = 12 or x = –2 10 x = – 4 or x = 3
6.5 x = 64 or x = 0 6.6 x = –___
64 or x = 27
27
11 x = 4 or x = –1 12 x = –5 or x = 1
7 7.1 x=1 7.2 x = 1 or x = 2
13 3 or x = –2
x = – __ 14 x = 0 or x = 2,25
7.3 x=9 7.4 x = –2 4
15 3 or x = __
x = __ 2 16 x = 3 or x = –2
7.5 Both solutions are invalid. 2 3
17 9 2
x = –__ or x = __ 18 x = 0,5 or x = 5
7.6 x = 3 or x = 7 4 3
19 1 or x = – __
x = – __ 2 20 5 or x = 1
x = – __
4 3 6
TOPIC 2: EXERCISE 1
21 3 or x = 1
x = __ __ 22 x=7
__ or x = –2
1 1.1 x = –5 1.2 Minimum 4 2 8
23 6 or x = 3
x = __
1.3 x = –5 1.4 0 5
2 2.1 x=2 2.2 Minimum 24 x = 1,5 or x = –1,5 or x = 1 or x = –1
2q 5q 3b or x = – __
4b
2.3 x=2 2.4 0 25 x = __ or x = __ 26 x = __ a
3p 2p 2a
3 3.1 x = –6 3.2 Maximum 27 x = 5 only 28 x = 2 only
3.3 x = –6 3.4 0
4 4.1 x=4 4.2 Maximum
4.3 x=4 4.4 0
334 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 334 2012/07/02 2:26 PM
TOPIC 2: EXERCISE 4 TOPIC 2: EXERCISE 9
1 3 or _
1 2 1,05 or –1,63 1 x ≤ – 4 or x ≥ 4 or x ∊ (– ∞;– 4] ∪ [4;∞)
2
2 x ≤ –1,5 or x ≥ 1,5 or x ∊ (– ∞;–1,5] ∪ [1,5;∞)
3 – 0,44 or –1,36 4 2,26 or – 0,59
5 1,31 or – 0,64 6 2,85 or – 0,35 TOPIC 2: EXERCISE 10
______ 1
7 1,58 or 0,42 8 p±√ p2 – 3p Inequality notation Interval notation
1.1 x < −3 or x > 2, x ∊ ℝ x ∊ (−∞;−3) ∪ (2;∞)
1.2 −1 < x < 2, x ∊ ℝ x ∊ (−1;2)
TOPIC 2: EXERCISE 5
1.3 −2 ≤ x ≤ 2, x ∊ ℝ x ∊ [−2;2]
1 x = 6 or x = –1 2 x = –10 or x = 20 1.4 −3 ≤ x ≤ 2, x ∊ ℝ x ∊ [−3;2]
3 x = 3,5 or x = – 0,5 4 3,39 or – 0,89
2. Inequality notation Interval notation
5 __
17 or __
13 6 1,87 or – 0,37
2 5 2.1 −1 < x < 3, x ∊ ℝ x ∊ (−1;3)
2.2 0 < x < 4, x ∊ ℝ x ∊ (0;4)
7 2,11 or –2,61 8 4,41 or 1,59
2.3 x ≤ 0 or x ≥ 4, x ∊ ℝ x ∊ (−∞;0] ∪ [4;∞)
9 There is no real solution. 2.4 x ≥ 1,5, x ∊ ℝ x ∊ [1,5;∞)
10 x = 2 only
3 Inequality notation Interval notation
3.1 x ≤ –2 or –1 ≤ x ≤ 2, x ∊ ℝ x ∊ (−∞;−2] ∪ [−1;2]
TOPIC 2: EXERCISE 6 3.2 x > −2, x ∊ ℝ x ∊ (−2;∞)
1 1.1 k=2 3.3 x < −2 or 0 < x < 1, x ∊ ℝ x ∊ (−∞;−2) ∪ (0;1)
1.2 The other root is x = 1,5
3.4 –2 < x < −1 or x > 2, x ∊ ℝ x ∊ (−2;−1) ∪ (2;∞)
2 2.1 p = –1
2.2 5
The other root is x = – __
3 4 4.1 x = –3 or x = 5
3 3.1 a = –2
2 4.2 x < –3 or x > 5, x ∊ ℝ; x ∊ (– ∞;–3) ∪ (5;∞)
3.2 The other root is x = __
5 4.3 x = –3 or x = 6
4 4.1 t = –7
4.4 –3 ≤ x ≤ 6, x ∊ [–3;6]
4.2 The other root is x = –1,25
5. 5.1 x = – 6 or x = –1
TOPIC 2: EXERCISE 7 5.2 x < – 6 or x > –1, x ∊ ℝ; x ∊ (– ∞;– 6) ∪ (–1;∞)
5.3 x = –6 or x = 0
1 length 125 mm; breadth 25 mm
2 The numbers are 49 and 94. 5.4 –6 < x > 0, x ∊ ℝ; x ∊ (–6;0)
3 5 km/h
4 48 km/h TOPIC 2: EXERCISE 11
5
6.
x ≠ –750 or x = 700
6.1 16 km/h
1 ( – 2_3;– 16
3 )
__ and (3;13) 2 (0;9) and (2;–3)
6.2 8 km/h 3 (4;0) and (–2;9) 4 (2;3) and (–2;4)
7 9 days and 18 days respectively
8 larger pipe takes 3 hours 5 ( _45;__
25 )
17 and (–5;3) 6 (5;4) and (2;–2)
9 150 m by 80 m
7 (–5;6) and (1;0) 8 (5;6) and (1;–2)
TOPIC 2: EXERCISE 8 9 (2;–2) and (–1;4) 10 (0;4) and (4;– 4)
1 – 4 < x < 4 or x ∊ (– 4;4)
2 x ≤ –5 or x ≥ 5 or x ∊ (– ∞;–5] ∪ [5;∞]
3 –1,5 < x < 1,5 or x ∊ (–1,5;1,5)
4 –1 ≤ x ≤ 1 or ∊ [–1;1]
Answers 335
PLT MATHS LB 11 7th pgs (Real Book).indb 335 2012/07/02 2:26 PM
TOPIC 2: EXERCISE 12 6
Inequality notation Interval notation
1 (–1;2) and (–3;– 4) 2 (3;1,5) and (–1;2,5) 6.1 x ≥ – 2, x ∊ ℝ x ∊ [–2;∞)
3 (–2;– 4) and (– 4;6) 4 (7;–1) and (1;–3) 6.2 x ≤ – 2 or 1≤ x ≤ 3, x ∊ ℝ x ∊ (– ∞;–2] ∪ [1;3]
6.3 –2 ≤ x ≤ 0 or x ≥ 4, x ∊ ℝ x ∊ [–2;0] ∪ [4;∞)
5 (4;1) 6 (2;3) and (–2;–3)
6.4 1 < x < 3, x ∊ ℝ x ∊ (1;3)
7 (29;6) and (1;–1) 8 (2;–1) and (–5;–2)
7. 7.1 (–3;–7) and (2;8)
9 (0;–2) and (3;4) 10 (1;–1) and (–1;3) 7.2 (5;6) or (2;– 6)
7.3 (4;– 4) and (–3;– 0,5)
11 (– 4;–1), (1;4) and (–1;– 4) 7.4 (0;– 8) and (4;0)
8 The original price was R8 per litre.
9 48 learners participated
TOPIC 2: EXERCISE 13
10 17 and 23
1 The roots are non-real.
2 The roots are real, rational and unequal.
4 or x = 1
x = – __ TOPIC 3: EXERCISE 1
7
3 The roots are non-real. 1 1.1 14; 9; 4 1.2 Tn = –5n + 44
4 The roots are non-real.
5 The roots are real, rational and equal. 1.3 T22 = – 66 1.4 T59 = –251
x = 1,5 only
2 2.1 4; 7; 10 2.2 Tn = 3n – 14
6 The roots are real, irrational and unequal.
– 0,23 or –1,43 2.3 T100 = 286 2.4 T64 = 178
7 The roots are real, rational and equal.
– 0,4 only
8 The roots are non-real. TOPIC 3: EXERCISE 2
1 T6 = 55 and T7 = 78; Tn = 2n2 – 3n + 1; n = 93
TOPIC 2: REVISION 2 T6 = 116 and T7 = 157; Tn = 3n2 + 2n – 4; n = 57
1 1.1 x = 3 or x = 2,5 3 T6 = 37 and T7 = 51; Tn = n2 + n – 5; n = 29
1.2 x = 5,77 or –2,77 4 T6 = 42 and T7 = 56; Tn = n2 + n; n = 31
1.3 x = 4 or x = –3 5 T6 = – 45 and T7 = – 67; Tn = –2n2 + 4n + 3; n = 67
1.4 x = 2,48 or – 8,48
6 T6 = 21 and T7 = 28; Tn = _
1 n2 + _
1 n; n = 55
1.5 The roots are non-real. 2 2
1.6 x=_ 7 or x = 1 7 T6 = 9 and T7 = 14; Tn = _
1 n2 – _
3 n; n = 61
3 2 2
1.7 x = 7 only 8 T6 = 96 and T7 = 130; Tn = _
5 n2 + _
3 n – 3; n = 13
2 2
1.8 x = 6 only 9 _n2 + 3
T6 = – 80 and T7 = –111; Tn = – 5 _n + 1; n = 19
2 2
2 2.1 x = 4,45 or x = – 0,45
10 T6 = – 80 and T7 = –111; Tn = n2 – 5n + 4; n = 71
2.2 x = 2,73 or x = – 0,73
3 3.1 B Real, rational and unequal roots.
3.2 A Real, rational and equal roots. TOPIC 3: EXERCISE 3
3.3 C Real, irrational and unequal roots. 1 56; 79 Tn = 2n2 – 3n + 2; n = 19
3.4 D Non-real or imaginary roots. 2 –17; –28 Tn = –n2 + 2n + 7; n = 26
4 4.1 The roots are real, irrational and unequal. 3 76; 110 Tn = –n2 + 2n + 7; n = 26
4.2 4.2.1 3,71 or –1,21
4 –53; –74 Tn = –1,5n2 – 1,5n + 10; n = 37
4.2.2 3,71 or −1,21
5 5.1 – 8 ≤ x ≤ 8; x ∊ ℝ, x ∊ [– 8;8] 5 91; 132 Tn = 3,5n2 – 4,5n – 8; n = 32
5.2 – 4,5 ≥ x ≥ 4,5; x ∊ ℝ, x ∊ [– 4,5;4,5] 6 –106; –155 Tn = – 4n2 + 3n + 20; n = 17
5.3 x < –3 or x > 4; x ∊ ℝ, x ∊ (– ∞;–3) ∪ (4;∞)
5.4 x < 0 or x > 7; x ∊ ℝ, x ∊ (– ∞;0) ∪ (7;∞)
336 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 336 2012/07/02 2:26 PM
TOPIC 3: EXERCISE 4 TOPIC 3: REVISION
1 Fig 1 Fig 2 Fig 3 Fig 4 Fig 5 Fig n 1 1.1 T6 = –13 and T7 = –17
1.2 Tn = – 4n + 11
Total 1.3 T19 = – 4( 19 ) + 11 = –65
number n( n + 1)
______ 1.4 n = 33
1 3 6 10 15 2
of blue
faces 2 2.1 T6 = 21 and T7 = 40
Total 2.2 Tn = 2n2 – 7n – 9
number 2.3 T25 = 2( 25 )2 – 7( 25 ) – 9 = 1 066
2 6 12 20 30 n(n + 1) 2.4 n = 47
of white
faces 3 3.1 T6 = –6 and T7 = –16
Total 3.2 Tn = – n2 + 3n + 12
3n( n + 1 )
_______
number 3 9 18 30 45 2 3.3 T51 = –( 51 )2 + 3( 51 ) + 12 = –2 436
of faces 3.4 n = 12
Number
1 2 3 4 5 n 4 4.1 4
of layers
4.2 x – 8 or 28 – 2x
Number
4.3 x = 12
of cubes
n( n + 1)
______ 4.4 Tn = 2n2 – 3n – 8
in 1 3 6 10 15 2 4.5 n = 15; T15 = 397
bottom
layer 5 5.1 a = 35 and Tn = 7n
5.2 b = 8 and Tn = 5n – 12
Fig Fig Fig Fig Fig 5.3 c = 112 and Tn = 7 × 2n or Tn = 14 × 2n – 1
2 Fig n
1 2 3 4 5 5.4 d = 1 029 and Tn = 3 × 7n – 1
Number 5.5 e = 60; Tn = 3n2 + 3n
of red 1 5 12 22 35 1,5n2 – 0,5n
beads 6
Number Structure 1 2 3 4 5 Structure n
of black 0 5 13 24 38 2 + 0,5n – 2 Number of
1,5n
rods tiles in the
1 4 9 16 25 n2
Number base of the
of beads 1 10 25 46 73 3n2 – 2 structure
and rods Number
of vertical
4 12 24 40 60 2n(n + 1)
tiles in the
TOPIC 3: EXERCISE 5 structure
1 Tn = 5n – 1 There are nine terms in the sequence. Total number
2 Tn = –3n2 + 7n + 10; n = 28 of tiles in the 5 16 33 56 85 3n2 + 2n
structure
3 T6 = 216 and T7 = 343 There are ten terms in
the sequence.
4 T6 = 192 and T7 = 384; Tn = 3 × 2n; n = 10
TOPIC 4: EXERCISE 1
5 T6 = 43 and T7 = 64; Tn = 2n2 – 5n + 1; n = 97
1 1.1 Substitute y = 5x + 9 into
6 T6 = –117 and T7 = –167; Tn = – 4n2 + 2n + 15; 5y + x – 19 = 0
n = 13 5(5x + 9) + x – 19 = 0
7 T6 = 486 and T7 = 1 458; Tn = 2 × 3n – 1; n = 9 25x + 45 + x – 19 = 0
8 T6 = 215 and T7 = 342; Tn = n3 – 1; n = 10 26x = –26
x = –1
y = 5(–1) + 9 = 4
K = (–1;4)
Answers 337
PLT MATHS LB 11 7th pgs (Real Book).indb 337 2012/07/02 2:26 PM
______________
1.2 y = –3 5
__x + __ 1.2 AB = √ (7 + 1)2 + (5 – 7)2
2 2 ______ ___
1.3 2
__ = √ 64 + 4 = √ 68
3 ______________
3×2 BC = √ (7 – 5)2 + (5 + 3)2
1.4 – __ __ = –1
______ ___
2 3
= √ 4 + 64 = √ 68
KN ⊥ LM and N is the midpoint of LM, so
AB = BC
KN is the perpendicular bisector of LM.
△ABC isosceles
1.5 KL = KM
1.3 34 units2
1.6 p=5
1.7 Q = (3;–2) 2 2.1 a=9 2.2 a = –5
__
4
1.8 y = –1
__x + __6
5 5 2.3 51
a = – ___
1.9 (
3 ;__
P = – __
22
3
) ___
8
_______________ ___ 3 3.1 2√29 3.2 (1;0)
1.10 KM = √ (–1 – 4)2 + (4 – 3)2 = √ 26
(2 )
_______________
y=5 5 7 ;3
E = __
√( – __32 – 1 )2 + ( __32 – 1 )2 3.3 __x – __ 3.4
PN = 2 2
___ ___
___ y = 16
__x – 29
__
= √ ___ 26 1
26 = ____ √ 3.5 3.6 18°
= __ × √26 9 9
4 2 2
KM = 2PN 3.7 (8;11)
2 2.1 _; 9
D= 9 _ (2 2) 4 _______
mAB = 6 – k __ = –1
– 6 = –k
1 + 3k – 1 3k
__
3
2.2 y=1 18
__x + ___
5 5 mAC = _______
6 – n – 6 = __
–n = __
–1
1 + 3n – 1 3n 3
2.3 y = –x mAB = mAC A, B and C are collinear.
2.4 y = –x + 9 5 5.1 mAB = – _
3 and m = _
2
BC 2 3
2.5 y = 5x – 18 –_
3×_
2 = –1
2 3
AB ⊥ BC
TOPIC 4: EXERCISE 2 __ units
13
5.2
2
5.3 D = (–1;4)
1 1.1 θ = 63,4° 1.2 θ = 101,3°
5.4 y = –4
__ x – 2
__ 5.5 y = –3
__ x – 3
__
1.3 θ = 59,0° 1.4 θ = 144,2° 7 7 2 4
5.6 ( – 3__2;3__2 ) 5.7 45°
2 2.1 θ = 26,6° 2.2 θ = 158,2°
6 __
29 units2 7.1 D = (–2;6)
5
2.3 θ = 135°
7.2 mAM = – _
5 and m =_
1
3 3.1 66,8° 3.2 31,0° MB 1 5
–5 × _ ^ B = 90°
1 = –1 AM
5
3.3 82,2° 3.4 71,6°
7.3 mAD = __
–2 and m = _ ^ B = 90°
3 DA
AB 3 2
3.5 81,8° 3.6 60,3° ABCD is a square because it is a
parallelogram with diagonals bisecting
at right angles, and has right angles at its
TOPIC 4: REVISION
vertices.
1 1.1 2 = – __
mAB = ___ 8=4
1 and m = __
–8 BC 4 2
8 8.1 R = (6;1)
1 × 4 = –1 AB ⊥ BC
– __
4 8.2 Prove that the diagonals are perpendicular.
^ C = 90°
AB _ = –1
mPR = – 4 _ and m = 12
__ = 2
8 2 QS 6
_
1
– × 2 = –1 PR ⊥ QS
2
338 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 338 2012/07/02 2:26 PM
8.3 (2;3) 8.4 60 units2 TERM 1 INVESTIGATION
TASK 1A
8.5 67,39° 8.6 1 x – __
y = –__ 7
1 _ = 1 or x
x _=1 2 1,6 or x_y = 1
2 2 y y
8.7 y = 2x – 1
3 1,625 or x_y = 1
8.8 Subs x = –1 into y = 2x – 1
y = 2(–1) – 1 = – 3 TASK 1B
1 8 = 1,6
x = __
__ 2 y = 21 – 8 = 13
the line passes through (–1;–3) y 5
__
9 9.1 12 + 8√2 9.2 8 units2 3 13 = 1,625
x = ___
__
y 8
9.3 8,13° 9.4 D = (–7;–1)
TASK 2
10 10.1 59,62° 10.2 108,43° 1 34; 55; 89
2 _
1; _
2; _
3; _
5; _
8 ; __
13 ; __
21 ; __
34 ; __
55 ; __
89
10.3 59,62° 1 1 2 3 5 8 13 21 34 55
3 1; 2; 1,5; 1,667; 1,6; 1,625; 1,615; 1,619; 1,618;
11 11.1 S = (7;4) 11.2 y = –2x + 13
1,618
11.3 (5;3) 4 The numbers in the sequence get closer and closer
to 1,618.
12 12.1 (13;10)
TASK 3
12.2 While we could show that the sides
1 and 2
are equal, it is quicker to show that the
diagonals are perpendicular. x T1 T2 T3 T4 T5
mBD = – _
4 = –2 1 1 1,41421 1,55377 1,59805 1,61185
2
mAC = __
8 =_ 1 2 1,41421 1,55377 1,59805 1,61185 1,61612
16 2
3 1,73205 1,65289 1,62877 1,62135 1,61906
_ = –1
–2 × 1
2 x T6 T7 T8 T9 T10
BD ⊥ AC 1 1,61612 1,61744 1,617851 1,617978 1,618017
So ABCD is a rhombus (parallelogram 2 1,61744 1,61785 1,617978 1,618017 1,618029
with diagonals perpendicular must be a
3 1,61835 1,61813 1,618064 1,618043 1,618037
rhombus or a square, which is a special case
rhombus).
TASK 4
12.3 (5;6) 12.4 y = – 2x + 16 1 and 2
x T1 T2 T3 T4 T5
12.5 Substitute x = 5 into the equation of BD
y = –2(5) + 16 = 6 1 2 1,5 1,66667 1,6 1,625
So M(5;6) lies on DB 2 1,5 1,66667 1.6 1,625 1,61538
3 1,333333 1,75 1,57143 1,63636 1,61111
12.6 As we have shown that the midpoint of
diagonal AC lies on diagonal BD, we have x T6 T7 T8 T9 T10
shown that diagonal AC is bisected by 1 1,61538 1,61905 1,6176471 1,61818 1,61798
diagonal BD. 2 1,61905 1,61765 1,6181818 1,61798 1,61806
^ C = 14°
12.7 BA 3 1,62069 1,61702 1,6184211 1,61789 1,61809
13 13.1 1
mAB = __ 13.2 y = – 3x + 10
3
___ 3 The terms of the sequence converge or get closer
√
_x
13.3 y = 1
3
13.4 3 10
_____
and closer to 1,618
5
Answers 339
PLT MATHS LB 11 7th pgs (Real Book).indb 339 2012/07/02 2:26 PM
TASK 5 5 5.1 T6 = 47
1 1,618033992 5.2 9n – 7
2 They are equal to one another 6 6.1 42; 56
_____________ __
–(–1)±√ (–1)2 – 4 (1)( –1) _____
√5 6.2 Tn = n2 + 3n + 2 = (n + 1)(n + 2)
3 x = __________________ = 1±
2(1) 2 6.3 n = 21
__
1± 5 √
But the ratio is positive φ = _____ 7 7.1 12 + 22 + 32 + 42 = 1 + 4 + 9 + 16 = 30
2
__
1+ √5 12 + 22 + 32 + 42 + 52 = 1 + 4 + 9 + 16 + 25
4 ______ is an irrational number
2 = 55
5 1,618033989
7.2 b = 1; c = – 1
__
6
TASK 6
8 8.1 T5 = 18 and T6 = 22
1 0
__ 8.2 T5 = 20 and T6 = 30
1± √5
______
2 8.3 Tn = 4n – 2
2
3 The number is the Golden Ratio φ. 8.4 Tn = n2 – n = n(n – 1)
_______ __ 8.5 T9 = 34
4 _= –b±√
x b2 – 4ac x
____________ ; _y = 1 ±
√5
______
y 2a 2 8.6 T11 = 110
TASK 8 8.7 n = 12 or n = – 11
__
√2
9 9.1 14,14 9.2 M = (4;1)
9.3 y= 5
__x – 7
__ 9.4 mBC = 1
13 13
TERM 1: ASSESSMENT TEST
9.5 y = –x + 11 9.6 θ = 45°
1 1.1 1
__ 1.2 6
__
2 5 9.7 ^ B = 53,13°
AC
1.3 8 1.4 x y
10 10.1 A(a;b)
1.5 k
2 2.1 1
x = 0 or x = __ 2.2 3
5 or x = – __
x = __
3 3 2
2.3 x = 2 or x = – 5
__
4 N(3;6)
2.4 x = 1,70 or x = –1,37
2.5 x = –5 2.6 x>_
3 or x < –1
2
2.7 x = –5 2.8 _
x=1 B(–12;1) M C(8;1)
2 x
2.9 x=0 2.10 1
2.11 x = 1
___ 2.12 x = 1
125
10.2 (–2;1) 10.3 (3;6)
2.13 x = 5
__
^ C = 45°
10.4 NM 10.5 5√2
3 3.1 (x;y) = (0;8) or (–1;6)
3.2 (x;y) = ( – 4;– __ )
5 or (3;10) 10.6 mMN = 1; mAB = 1
3
__
3.3 (x;y) = (–1;3) or (3;–1) 10.7 10√ 2 10.8 x = –2
___
4 4.1 x = – 2 ± √11 11 11.1 k = –14 11.2 k = 8
4.2 Roots are real and irrational.
4.3 Roots are real and rational. 11.3 k = 5
4.4 4.4.1 x = – 1 or x = 0
12 12.1 D = (2;2) 12.2 5y + x – 12 = 0
4.4.2 x > –1
340 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 340 2012/07/02 2:26 PM
3
y
1
12.3 mAB = __ 12.4 y = _
1x
7 7 f g
12.5 C = (7;1) 12.6 θ = 8,13° 3
(–3;3) (3;3)
TOPIC 5: EXERCISE 1
1
y
–3 0 3
h x
k
(–2;2) 2 (2;2)
(–3;–3) (3;–3)
–3
k
–2 0 2
x h
3.1 Refer to graph.
3.2 Refer to graph.
(–2;–2) –2 (2;–2)
3.3 g(x) = __
1 x2 – 3
3
g 3.4 Refer to graph.
3.5 h(x) = –f(x) = – __
1 x2
3
f
1.1 Refer to graph. 3.6 Refer to graph.
1.2 Refer to graph. 3.7 k(x) = –g(x) = – __
1 x2 + 3
3
1.3 g(x) = – __
1 x2 + 2
4
y
2 k
1.4 Refer to graph. h
1.5 h(x) = –f(x) = __
1 x2h(x) = –f(x) = 12x2 (–1;5) (1;5)
2
1.6 Refer to graph. (–1;3) (1;3)
k(x) = –g(x) = __
1 x2 – 2 2
1.7
2 x
2 y
(–1;–3) –2 (1;–3)
k
h
(–1;9) (1;9)
(–1;8) (1;8) (–1;–5) (1;–5)
f
g
4.1 Refer to graph.
x
4.2 Refer to graph.
4.3 g( x ) = –3x2 – 2
(–1;–8) (1;–8) 4.4 Refer to graph.
(–1;–9) (1;–9)
g
f 4.5 h( x ) = –f( x ) = 3x2
4.6 Refer to graph.
2.1 Refer to graph.
4.7 k( x ) = –g( x ) = 3x2 + 2
2.2 Refer to graph.
2.3 g( x ) = – 8x2 – 1
2.4 Refer to graph. TOPIC 5: EXERCISE 2
2.5 h( x ) = –f( x ) = 8x2 1 y = ( x + 3 )2; TP ( –3;0 ); Shift 3 left
2.6 Refer to graph. 2 y = ( x – 4 )2; TP ( 4;0 ); Shift 4 right
2.7 k( x ) = –g( x ) = 8x2 + 1 3 y = ( x – 2 )2; TP ( 2;0 ); Shift 2 right
4 y = –( x + 1 )2; TP(–1;0); Shift 1 left
Answers 341
PLT MATHS LB 11 7th pgs (Real Book).indb 341 2012/07/02 2:26 PM
TOPIC 5: EXERCISE 3 3.2 x < – 4 and x = 2
1 y = x2 – 8x + 7 = ( x – 7 )( x – 1 )
TP(4;–9); Shift 4 right, 9 down TOPIC 5: EXERCISE 5
2 y = 4x2 + 16x + 15 = ( 2x + 3 )( 2x + 5 ) 1 1.1 y = –3x2 – 9x + 12
TP (–2;–1); Shift 2 left, 1 down
1.2 y = – 4x2 – 6x + 4
3 y = –x2 – 10x – 9 = –( x + 9 )( x + 1 ) 10 x + 1
TP ( –5;16 ); Shift 5 left, 16 up
1.3 y = x2 – ___
3
1.4 5x + 6
y = –x2 – __
4 y = –x2 + 6x – 5 = –( x – 5 )( x – 1 ) 2
TP(3;4); Shift 3 right, 4 up
2 2.1 y = – 2x2 + 12x – 14
2.2 y = 3x2 – 6x + 1
TOPIC 5: EXERCISE 4 1 x2 – 2x + 2
2.3 y = – __
1 1.1 2
g(x) = 4x + 8 5x + 2
1 x2 – __
y
2.4 y = – __
2 2
16 - (2;16)
f(x) = –x² + 4x + 12
12 - TOPIC 5: EXERCISE 6
1 y y = x² – 6x + 5
8-
4- 5
- - -
- - - - x
–4 –2 2 4 6
x
1 5
1.2 x = –2 and x = 6
1.3 x ∊ [ –2;2 ] (3;–4)
2 2.1 y
4- 1 2 y = 2x² + 4x + 6 y
f(x) = x² – x – 4
2
- - -
- - - - x
–4 –2 2 4 6
–4 - (–2;6) 6
g(x) = x – 4
(1;–4,5)
(–1;4)
2.2 x = 2 only
2.3 x ∊ [ 0;4 ]
x
3 3.1
g(x) = –3x + 6 y
(–1;9)
- 8
6-
- - -
- - - x
–4 –2 2 4
f(x) = –(x + 1)² + 9
342 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 342 2012/07/02 2:26 PM
3 y 6 y y = –2x² – 12x + 10
(–4;9)
10
x
–7 –1
x
1 5
–7
y = –x² – 8x – 7
(3;–8)
4 y
(–5;16)
TOPIC 5: EXERCISE 7
1 y
f
x = –2
x
–9 –1
– 53
1
x
–9 –5
-1
y = –x² – 10x – 9
y = –3 (–2;–3) 2,5
5 y
y = 3x² – 24x + 36
(–3;–4)
36 –5
y=x–1 y = –x – 5
2 6
x 2 y
g
y=x+5
(4;–12)
y = –x + 1
5
(–3;4)
y=3
(–2;3)
2,5
1
–5 – 53
x
1
x = –2
Answers 343
PLT MATHS LB 11 7th pgs (Real Book).indb 343 2012/07/02 2:26 PM
3 y 6 y
v
h x = –3
y = –x – 1
x=2 (–1;1) 1
3
x
5
x –4 1
1 3 5 (–3;–1)
–1 y = –1
–2,5 (–5;–3)
(2;–3)
–4
y = –3 y = –x – 4
(3;–4)
y=x+2
–5
y=x–5 7 y
x=1 y=x+1
4 y
5
r
x=3 w y=2
2 3 x
1 (1;1) –1 2,5
3
x
–1 2 4
(3;–1)
y=–1
y = –x + 3
(5;–3)
–4
8 y
y=x–4
y = –x + 2
5 y 2,5
3
t x
x=3 –1
y = –2 –1
4 (1;–2)
z –3
(5;3)
–5
y=1 (3;1)
x x=1 y = –x – 1
–1 1 2 4
3 y=x–3
(1;–1)
–2
y=x–2 y = –x + 4
344 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 344 2012/07/02 2:26 PM
9 y
y=x+3
TOPIC 5: EXERCISE 9
x = –1 1 y
5- 2
y= –3
5 4- x–1
3-
3 2-
1 y=2 1-
–3 - - 0- - - - - x
–4 –2 5 2 4 6
x
–2,5 1 –1 - 3
–2 - (3;–2)
y = –3
–3 -
–4 -
y = –x + 1 –5 - y = –x – 2
y=x–4 –6 -
d
–7 -
x=1
–8 -
TOPIC 5: EXERCISE 8
2 y
1 y = – _____
2 +4
x = –1
x+1 6-
Symmetry lines: y = x + 5 and y = –x + 3 5-
2 y = _____
3 +1 4-
y=
2
+2
x–2 3- x+1
y=2
Symmetry lines: y = x – 1 and y = –x + 3 2-
(1;3)
3 y = – _____
1 –2 1-
x+3 - - 0-
- - - - - - - - - - -
Symmetry lines: y = x + 1 and y = –x – 5
x
–6 –5 –4 –3 –2 –1 - 1 2 3 4 5 6
–1
4 y = _____
5 –3
–2 -
x–4
Symmetry lines: y = x – 7 and y = –x + 1 –3 -
y = –x + 3 –4 -
5 5.1 x ∊ ℝ , x ≠ 1; y ∊ ℝ , y ≠ 2 –5 - y = –x + 1
5.2 k = – 4, p = 1 and q = 2 –6 -
5.3 y = x + 1; y = –x + 3
5.4 A(1;2); B(0;0); C(2;0) 3 y
4
y= –2
5.5 maximum; y ≤ 2 for all y ∊ ℝ 5- x–3
5.6 5.6.1 y = –2( x – 1 )2 + 2 = –2x2 + 4x 4-
3-
5.6.2 y = –2x( x – 2 ) = –2x2 + 4x
2-
5.7 x ∊ (–∞ ; 0) ∪ (1; 2) ∪ (3; ∞) 1-
5.8 6 units - 0-- - - - - - - - - - - - x
–3 –2 –1 - 1 2 3 4 5 6 7 8 9
5.9 5.9.1 1 left, 4 up –1
3;–2) (7;1)
5.9.2 y = –2x2 + 6 –2 -
5.10 y = – 4
___
x
–3 - y = –2
–4 - y = –x + 1
–5 -
y = –x – 5 -
–6
–7 -
–8 - x=3
Answers 345
PLT MATHS LB 11 7th pgs (Real Book).indb 345 2012/07/02 2:26 PM
4 y TOPIC 5: EXERCISE 10
5-
x=2 1 Asymptotes: x = –1 and y = 1
4 - (1;3)
Symmetry lines: y = x + 2 and y = –x
3-
2- 2
y=1
y
(2;1)
1-
- 0-- - - - - - - - - - - - x x = –1 4- ( 1 ;3 1 )
–3 –2 –1 - 1 2 3 4 5 6 7 8 9 2 16 y=x+2
–1
y = –x 3-
–2 - f
–3 - 2-
–4 - y=1
g
- 1-
y = x – 1 –5
–6 - y = –x + 3
- - - - - 0-
- - - - - - - - x
–7 - –5 –4 –3 –2 –1 1 2 3 4 5 6 7
2
–8 - y= +1 –1-
x–2
y –2-
5 4
y= –1 –3-
x+5
4 –4-
–5-
y = –x – 6 x= 1
2
–6-
–7-
- - - - - - - - - - - - - - x
–11 –10 –9 –8 –7 –6 –4 –3 –2 –1 1 2 3 ( – 4;–2 ); ( 1;3 )
( –5;1)
y=–1
(–1;2) 4 g: x ∊ ℝ , x ≠ – 1; f: y ∊ (–∞;3,0625] or
–1,3 (
y ≤ 3___
1 ,y∊ℝ
16 ]
5 ( 0;3 ); ( –3;0 ) or ( 3;1,5 )
–4
y=x+4 6 x ∊ [–3;–1) ∪ [0;3] or –3 ≤ x < –1 or 0 ≤ x ≤ 3;
x∊ℝ
7 y = _____
2
–6 x–1
x=–5 8 y = ______
2 +1
2x + 1
–8
9 Reflection in the y-axis.
6 1 10 y = – __ __x + __
1 x2 – 1 3
y= +4 4 2 4
y x–2 11 1 unit to the right; 1 unit down.
8- 12 g( __ ) 3x + 1; undefined if x = –1 or 0
1 = ______
x 1+x
7- 13 g(x – 4) = _____
2 + 1; undefined if x = 3
x–3
6- 14 g(x) + g( __
x )
1 = 4g(–x)g( x – 4 )
5- (3;5)
y=4 LHS = _____
2 + 1 + ______
3x + 1 = _______________
2 + x + 1 + 3x + 1
x+1 1+x x+1
4-
(2;4) + 4 = 4( x
______
= 4x + 1)
_______ =4
3- x+1 x+1
2- y=x+6 RHS = 4g(–x)g( x – 4 )
1- [ x–1
_____ + 1 _____
=4– 2 2 +1
][ x – 3 ]
0-
( x – 1 )( x – 3 )
_________
= 4 –2 + ________
x–1 2+ x–3
- - - - - - - - - - - - x
–4 –3 –2 –1 - 1 2 3 4 5 6 7
–1
= 4( x –
x – 1 )( x – 3 )
_____
3 x– _____
1 =4
–2 -
y=x+2 –3 - ⇒ LHS = RHS
346 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 346 2012/07/02 2:26 PM
TOPIC 5: EXERCISE 11 4 y
t(x) t(–x)
1 y
f(–x) 20 -
f(x) (–1;5) (1;5)
2
15 - y=1
10 -
x
(–1;10) (1;10) y = –1
–2
5-
(–1;2,5) (1;2,5) (1;–5)
- - -
- - - x
–2 –1 1 2
–5 - –t(x)
(–1;–2,5)
–10 - (1;–10)
TOPIC 5: EXERCISE 12
–15 -
12 12.1 a = 3, b = 2 and c = –1
–20 - –f(x)
12.2 d = 5, e = 2 and p = 1
1 and t = 2
12.3 j = –1, r = __
2 y
5
g(–x) g(x)
20 - 12.4 v = 2, w = 1 and z = 2
15 -
TOPIC 5: EXERCISE 13
10 -
1 y
(–1;5) 5- (1;5)
y = 3 2x + 1 – 6
(–1;0,2) (1;0,2)
-
- - - - - - - x (1;6)
–3 –2 –1 1 2 3
–5 - (1;–5)
–10 - x
0
–15 - (–1;–3)
–20 - –g(x)
y = –6
3 h(x)
h(–x)
y
2 y
124
(–1,5) (1,5) (1; 125 ) y=1
3
0
x
y=2
x
y = –2
(3;–4)
(1;– 37 )
(–1;–5) –3
y = –5x – 2 + 1
–h(x)
Answers 347
PLT MATHS LB 11 7th pgs (Real Book).indb 347 2012/07/02 2:26 PM
3 7 7.1 1 , b = 2 and c = 6
a = −__
y = 2 13 x + 1 – 2 y 2
7.2 d = 8, e = 3 and j = −2
7.3 1 , q = −2 and v = 1
k = __
(–2;–4) 2
8 8.1 x=1
8.2 x > 1 or x ∊ (1;∞)
8.3 y=x+c⇒y=x+1
–1
x 8.4 x ∊ [ –1;4 ] or –1 ≤ x ≤ 4, x ∊ ℝ
– 43 (1;– 16
9
) 8.5 4 –1
f( x ) = x2 – 2x – 8; g( x ) = – _____
x+2
y = –2 8.6 A( –6;0 ); B( 0;–3 )
8.7 Q( 2;–2 )
8.8 x ∊ ( –∞;–2 ) ∪ [–1;2]
4 y = 2 –x –1 y 8.9 x ∊ [ –∞;–6 ] ∪ [ 4;∞ )
8.10 8.10.1 RT = – x2 + 3x + 4
(–2;3) 8.10.2 RT = 6,25
8.11 x ∊ ( –∞;–2 ) ∪ ( –2;∞ ) or x ∊ ℝ , x ≠ –2
8.12 8.12.1 [__ 4 – 2] – 1
x
(1;–0,5) 8.12.2 y = x – 3; y = –x + 1
x
y = –1 8.12.3 Teacher to provide
8.12.4 Teacher to provide
8.13 8.13.1 w(x = –__ 4
x
5 y
8.13.2 x = 0 and y = 0
8.13.3 y = x and y = –x
__
8.13.4 √ 2
2
y=5
TOPIC 5: EXERCISE 14
1
33
0 1
x 1 1.1 The death rate will continue to increase at
a constant rate. In the long term this would
have a catastrophic impact on life in South
Africa and would ultimately mean that the
battle against HIV/AIDS has been lost.
(2;–10)
1.2 16,52 years
y = –5 3
x–1
+5 1.3 Maximum = 716 349
1.4 during the year 2012
6 y 1.5 The number of deaths begins to decrease
after the end of 2012.
8
y = 2 ( 15 ) x – 1 – 2
x
0
(2;–1,6)
y = –2
348 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 348 2012/07/02 2:26 PM
2 2.1 R80 TOPIC 5: EXERCISE 15
2.2 1 A( –3;–7 ); B( 5;–7 ); C( 4;0 ); D( 3;5 ); E( 2;8 ); F( 1;9 );
200 -
G( 0;8 ); H( –1;5 ) and I( –2;0 )
180 -
2 The further apart the points are from each
other, the greater is the enclosed area. The closer
160 -
(5;160)
the points are to each other, the smaller is the
enclosed area.
140 -
3 3.1 mAB = 0 3.2 mAC = 1
120 -
3.3 mAD = 2 3.4 mAE = 3
100 -
3.5 mAF = 4 3.6 mAG = 5
3.7 mAH = 6 3.8 mAI = 7
80 -
(4;80)
4 mAI = 7 because the line through A and I, although
60 -
not a tangent, is much closer to the parabola than
the line between A and any of the other points.
40 -
(3;40)
20 -
(2;20)
TOPIC 5: EXERCISE 16
(1;10) 1 1.1 1.1.1 g(–1) = –2
7
- - - - - -
1 2 3 4 5 6
1.1.2 g(0) = – __
3
2.3 It is discrete. Each point represents the
1.1.3 g(1) = –3
month in which the deposit was made and
1.1.4 g(2) = –5
the amount of the deposit.
1.2 1.2.1 1
m = – __
2.4 The graph starts when the first deposit 3
is made, which is at the end of the first 1.2.2 m = –1
__
2
month. 1.2.3 m = –1
2.5 No. His 12th payment will be R20 480. 1.3
2.6 y = 5.2x or y = 10.2x – 1
y
4
y= –1
2.7 A = 5 and B = 2 x–3
3 3.1 f (x) = 281 234, 5833 + 39 377,95x
g(x) = 249 902,5 + 56 468,17727x
7
– 1 709,022727x2 x
h(x) = 303 433,3851 × 1,089710572x
3.2 24 years 1 y = –1
–2 3
(–1;–2)
3.3 f (24); 1 226 305 deaths
3
g(24); 620 742 deaths
h(24); 235,724 deaths (2;–5)
3
3.4 The linear model, f, shows that a steady
increase is anticipated in the long term.
The parabolic model, g, produces the
most encouraging result and indicates 1.4 m = – __
1
3
that a steady decline in the death rate is 2 2.1 2.1.1 h( –2 ) = 4
anticipated as the maximum would have 2.1.2 h( –1 ) = 2
been reached in 2012. 2.1.3 h( 0 ) = 1
The exponential model, h, shows that 1
2.1.4 h(1) = __
an alarmingly rapid rate of increase is 2
anticipated in the death rate. 2.2 2.2.1 m = –2
2.2.2 m = – __
3
2
2.2.3 m = – __
7
6
Answers 349
PLT MATHS LB 11 7th pgs (Real Book).indb 349 2012/07/02 2:26 PM
2.3 y 2 2.1 x = –1 and y = –3
y = 2 –x
2.2 f: y = x – 2 and y = –x – 4; g: x = –1
2.3 y
(–2;4)
h(x) = –2x – 5
3
(–1;6,25) 6
1
2
x
2.4 m = –2 y = –x – 4
3 3.1 3.1.1 k(–1) = –7 –1,6 g(x) = – 14 (x + 6)(x+4)
–2,5
3.1.2 k(0) = –12 –6 –4 2 4
x
(–2;–1)
3.1.3 k(2) = –16
(–7;–2,3) (–5;–2,5)
3.1.4 k(4) = –12 (–3;–2)
y = –3
3.2 3.2.1 m = –1 f(x) = – 2 – 3
x+1
3.2.2 m = –1 –5
3.2.3 m = –1 x = –1
3.3 y y = x² – 4x – 12 y=x–2
- - - - - -
2.4 x = 0 or 2
x
–4 –2 2 4 6
5 2.5 x ∊ (–6;4) or –6 < x < 4, x ∊ ℝ
(–1;–7)
5 2.6 x ∊ (–∞;–6) ∪ (–2,5;4)
–12 (4;–12) or x < –6 or –2,5 < x < 4
2.7 y=– 2 _____ – 2
(2;–16) x–1
2.8 y = 2x – 5
3.4 m = –5
2.9 y = – __ __
1 x2 – 2x – 1
4 4.1 m=1 4 4
2.10 g(–2) – f (–2) = 7
4.2 m = __
8
3 1
2.11 2.11.1 m = __
4.3 m = –3 3
2.11.2 m = __
1
2
TOPIC 5: REVISION 2.11.3 m = 1
2.12 closest point to (–2;–1) is the point (–3;–2);
1 1.1 q = –1 1.2 A(2;2) m = –1
2.13 x = 10,28 or – 4,28
1.3 x=2 1.4 4 units
2.14 2.14.1 m = __1
2
1.5 Q(8;2) 1.6 x = 1 and y = –1 2.14.2 5
m = __
4
1.7 ED = 1 1.8 B(3;–1,5) 2.15 (3;5,25) is much closer to (– 4;4) than ( 0;6 ) is
1.9 y=x–2 3 3.1 5 seconds
1.10 1.10.1 2 units left and 2 units up
1.10.2 y = __
1 x2
2
350 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 350 2012/07/02 2:26 PM
h
3.2 (1;20)
(2;21) 1.1 x = –90°, 90°or 270°
1.2 x ∊ (–90°;90°) ∪ (270°;360°)
(3;18) –90° < x < 90° or 270° < x ≤ 360°
1.3 x ∊ (–180°;0°) ∪ (180°;360°)
(0;15)
or (–180°;0) ∪ (180°;360°)
2 period 180°; amplitude is undefined; y ∊ (–∞;∞) or
(4;11)
y ∊ ℝ; Asymptotes for both graphs: x = –90° and
x = 90°
y
2-
g f
(45°;2)
1-
(45°;1)
t
(5;0)
3.3 y = –2x2 + 7x + 15 - - -- - - x
3.4 t = 1,75 seconds –180° –90° 90° 180°
3.5 Maximum height = 21,125 m
–1-
TOPIC 6: EXERCISE 1 x = –90 –2-
1 1.1 Teacher to provide x = 90
1.2 Range: y ∊ [ –1; 1 ] 2.1 x = –180°, 0° or 180°
1.3 Teacher to provide 2.2 x = – 45° or x = 135°
1.4 Period is 360°
3 period 360°; amplitude of f is 2; amplitude of
1.5 Amplitude = 1
g is __
3 or 1,5.
1.6 90° to the right 2
Range of f: y ∊ [–1; 3]
2 2.1 Teacher to provide
2.2 tan x is undefined at ±90° and ±270°
3 ;__
Range of g: y ∊ –__ 5
[ 2 2]
2.3 x = –270°, x = –90° and x = 270° Range of g: y ∊ [ – 1,5; 1,5 ]
2.4 Teacher to provide
y
2.5 Range y ∊ (–∞;∞), but the amplitude is
(–90°;3) (270°;3)
undefined. 3-
f
2.6 Period 180°
2-
(0;1,5)
TOPIC 6: EXERCISE 2 1-
g
1 360° ; Amplitude of h is 2; Amplitude of g is 1;
-
Range of h: y ∊ [–2;2]; Range of g: y ∊ [–1;1]
- - - - - x
–90° 90° 180° 270°
–1 -
y
(360°;2) (90°;–1)
2-
h (180°;–1,5)
(360°;1)
1- g 3.1 x = –90° or x = 270°
3.2 x = 90°
- - -- - - - - x 3.3 x ∊ [–90°;90°] or x = 270°
–180° –90° 90° 180° 270° 360°
3.4 y = –f (x) = 2sin x – 1
–1-
(–180°;1) (180°;–1)
–2-
(–180°;–2) (180°;–2)
Answers 351
PLT MATHS LB 11 7th pgs (Real Book).indb 351 2012/07/02 2:26 PM
4 period 180°; undefined amplitude. 3
y
x = –90
Range y ∊ (–∞;∞) or y ∊ ℝ 2-
Asymptote x = 90° (–45;1)
1-
y
-
–90° 90°
x
g
2- –1-
–2-
1- x = 90
(–45°;1) (135°;1)
f
4 y
- - x
90° (135°;–0,5)
- - - - - x
–90° 90° 180° 270°
(–45°;0,5)
–1 - (90°;–0,5)
0,5
(180°;–1)
–2 - –1
x = 90 –1,5
(–90°;–1,5) (270°;–1,5)
4.1 x = 90°
4.2 1,5
4.3 y = –f (x) = tan x 5 y
4.4 y = g(–x) = __1 tan (–x) = – __
1 tan x 1-
2 2 (180°; 56 )
(–90°;1)
TOPIC 6: EXERCISE 3 0,5 -
(90°;12 )
1 y 1
(90°;3) - -6
- - - - x
3- –90° 90° 180° 270°
2- –0,5 -
(–180°;1)
1-
(180°;1) 6
g y
- - -- - - x
–180° –90° 90° 180° (–45°;2)
2-
–1- (–135°;2)
(–90°;–1) 1-
2 y - -
–135°–90°–45°
- - - - -
45° 90° 135°
- x
0,5 –1-
- - - - - x (–135°;2) (45°;–2)
–180° –90° 90° 180° –2-
(–90°;–1) –1- (90°;–1) x = –90 x = 90
–2-
(–180°;–2,5) (180°;–2,5)
352 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 352 2012/07/02 2:27 PM
7 TOPIC 6: EXERCISE 5
y
(–90°;3)
3
1 y
(–180°;2) 2 (120°;1)
1-
y = cos 3x
1
(–270°;1) (90°;1)
- - - - - - - x
- - - - - x –30° 30° 60° 90° 120° 150°
–270° –180° –90° 90°
–1 -
(60°;–1) (180°;–1)
8 y
(–180°;2) (180°;2) 2
2 y
1 y = tan 14 x
x
–180° –90° 90° 180°
–1
(–90°;–1) (90°;–1) (180°;1)
–2
–3 –180°
- -
180°
x
–4 (180°;–1)
TOPIC 6: EXERCISE 4 x = –360° x = 360°
1 1.1 Teacher to provide 3 Teacher to provide
1.2 Teacher to provide
1.3 Amplitude of f and g: 1 4 y
1.4 x = –180° or 0° (–180°;1)
1- y = cos 2x
Period of f and g: 720°
1.5 x = –360°, 180° or 360°
Range of f and g: y ∊ [–1;1]
- - - - - - - - - x
–270°–225°–180°–135° –90° –45° 45° 90°
1.6 x = –270°, 90°
–1-
P = ____
180° = 90°; S = ___
90° = 225° (–270°;–1) (–90°;–1) (90°;–1)
2 2.1
2 4
2.2 P = ____
180° = 360°; S = ____
360° = 90°
5
( __12 ) 4 y = sin 2x
y
(45°;1) (210°;0,87)
2.3 f: x = –135°; x = – 45°; x = 45°; x = 135° 1-
g: x = –180°; x = 180°
- - - - - - -
2.4 Teacher to provide
x
–90°–45° 45° 90° 135°180°
2.5 x = 90°
–1-
g(90°) – f (90°) = tan __
1 (90°) – tan 2(90°) (–45°;–1) (135°;1)
2
= tan 45° – tan 180° = 1 – 0 = 1
2.6 x = ±90°
If x = –90°, then tan 2(–90°) – tan __
1 (–90°)
2
= tan (–180°) – tan (– 45°) = 0 – (–1) = 1
2.7 y ∊ (–∞;∞)
2.8 The amplitude is undefined.
Answers 353
PLT MATHS LB 11 7th pgs (Real Book).indb 353 2012/07/02 2:27 PM
6 y y = sin 2x 3.15 y = –sin [(x – 45°) – 45°]
(45°;1) (210°;0,87) = –sin (x – 90°) = cos x
1-
4 4.1 Teacher to provide
- -
–90°–45°
- - -
45° 90° 135°180°
- - x 4.2 The period of f is 360°.
4.3 g has an amplitude of 2.
(–45°;–1)
–1-
(135°;1) 4.4 y ∊ [–1,5;0,5]
4.5 –12°
7 y 4.6 x = 0°
4.7 2,5 units
1- y = cos 12 x
4.8 x ∊ (–60°;0°)
- - x 4.9 x ∊ [–120°;–90°) ∪ (60°;90°)
–180° 180° 4.10 y = –g( x ) = –2cos x
–1- 4.11 y = sin (x – 30°) + __
1
2
(–360°;–1) (360°;–1) 4.12 y = sin x – __
1
2
4.13 60° to the right
8 Teacher to provide
4.14 90° to the left
TOPIC 6: EXERCISE 6 5 5.1 Teacher to provide
1 1.1 Teacher to provide 5.2 f has an amplitude of 1.
1.2 x = 90° 5.3 g has a period of 180°.
1.3 x ∊ (–90°;–60°) 5.4 x ∊ (–120°;60°)
1.4 x ∊ [0°;30°] or 0° ≤ x ≤ 30° 5.5 x = –75° or x = 105°
1.5 x = –90° or x ∊ [30°;90°] 5.6 x ∊ [–210°;–165°] ∪ (–75°;–30°] ∪ [15°;10°)
1.6 30° to the left. 5.7 x = 60°
5.8 x ∊ (–75°;60°] ∪ (10°;150°]
2 2.1 Teacher to provide
5.9 x = 120° Check: g(120°) – f(120°) = tan
2.2 x = 0°
(–135°) – cos (–180°) = 2
2.3 x ∊ (–135°;–90°) ∪ (–90°;–75°) ∪ (–75°;45°) ∪
5.10 x = –210° or – 30° or 150° (any two)
(45°;90°)
5.11 5.11.1 15° to the left; reflected in the x-axis
2.4 x ∊ (–75°;– 4°] ∪ [15°;45°)
(order does not matter)
2.5 x ∊ (–135°;–90°) ∪ [– 45°;0°] ∪ (45°; 90°)
5.11.2 x = ± 90°
3 3.1 Teacher to provide 5.12 5.12.1 210° to the right
3.2 g has an amplitude of 2. 5.12.2 150° to the left
3.3 f has a period of 180°.
3.4 Range of h: y ∊ [ –1;1 ] TOPIC 6: EXERCISE 7
3.5 x = –90° or 90°
Teacher to provide
3.6 x = – 45°
3.7 3 units at x = 135°
3.8 x ∊ [–90°;0°) ∪ [45°;90°] ∪ [180°;225°] TOPIC 6: EXERCISE 8
3.9 h(15°) – g(15°) = –sin (–30°) + 2cos 60° 1 1.1 f (x) = –sin (x – 35°) ⇒ a = –1 and b = –35°
= __
1 + 2 __
2 (2)
1 = 1__
1
2
1.2 f (x) = cos (x + 60°) ⇒ c = 60° and d = 1
1.3 720°
3.10 x = –135°, 45° or 225° 1
1.4 p = 1; q = __
3.11 y = sin(x – 45°) 2
1.5 1 ; w = __
v = – __ 1
3.12 y = –2cos (x + 90°) = 2sin x 2 2
3.13 x ∊ (– 45°;135°) 1.6 1 ; t = –180°
r = __
2
3.14 x ∊ [–90°;0°) ∪ [90°;180°) 1.7 x = ±360°
1.8 x = 0°
354 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 354 2012/07/02 2:27 PM
2 2.1 Period of f is 360° 3 3.1 Teacher to provide
2.2 Period of g is 720° 3.2 3.2.1 x = –120°, 0° or 60°
2.3 Amplitude of f is 1 3.2.2 x = –30°
2.4 Amplitude of g is 1
4 4.1 Teacher to provide
2.5 g(x) = sin__ __ and b = 1
1x + 1 ⇒ a = 1
4.2 4.2.1 x = –120°
2 2
2.6 f (x) = cos (x + 90°) ⇒ d = 1 and p = 90° 4.2.2 x = 60°
2.7 f (x) = –sin x ⇒ m = –1 and n = 1 4.2.3 x = –210°, –30° or 150°
1 4.2.4 f (0°) – g(0°) = 0,75
2.8 v = –1 and w = __
2 4.2.5 x ∊ (30°;150°)
2.9 e = –180° and t = 1
2.10 x = 180° TOPIC 6: EXERCISE 11
3 3.1 Period of f is 360° 1 1.1 Teacher to provide
3.2 Amplitude of f is 2 units 1.2 1.2.1 x = 120° or 60°
3.3 Period of g is 540° 1.2.2 x = –210° or 150°
3.4 The amplitude of g is undefined. 1.2.3 x ∊ [–120°;60°]
3.5 540° 2 2.1 Teacher to provide
3.6 f (x) = 2sin (x + 1°) ⇒ a = 2 and b = 1° 2.2 2.2.1 x = ±180°
3.7 f (x) = 2cos (x – 7°)3.8 2.2.2 x ∊ [–180°;150°) ∪ (–90°;0°]
g(x) = tan __
x ⇒ v = 1 and w = __
1
3 3 3 3.1 R(–60°;0); T(150°;2); U(–150°;–1);
3.9 2 ⇒ p = –1 and = 2
h(x) = –cos__ __
3 3 V(120°;0); W(210°;–1) and Z(30°;1)
3.10 x ∊ [– 270°; –195] ∪ [–1°; 0°] ∪ [165°; 270°) 3.2 a = –2, p = 60°, k = __
1 , d = 60°, c = 120°
2
3.11 x = 135° and e = –240°
3.12 x = 0° or x = 135° 3.3 TU is 3 units.
3.4 y = –2cos x or y = –2sin (x + 90°)
TOPIC 6: EXERCISE 9 3.5 y = –2sin [(x – 240°) + 60°] = –2sin (x – 180°)
1 1.1 y = cos(x – 20°) + 1 = 2sin x
1.2 Range y ∊ [0;2] 3.6 If must be shifted 60° to the right.
3.7 x = 30°
2 2.1 y = sin(x + 90°) – 1 = cos x – 1
2.2 Minimum is –2 TOPIC 6: REVISION
3 1x
a = 3 ⇒ y = 3tan __
2 1 1.1 The period of f is 360°, the amplitude 2, the
range is given by y ∊ [–2;2] and y-intercept
4 y = tan (x – 45°) – 2
(0;1)
5 5.1 a=1__, b = 1, c = –2 and d = __
1
2 4 1.2 The period of g is 180°, the amplitude 1, the
5.2 Teacher to provide range is given by y ∊ [–1;1] and y-intercept
(0;–1)
TOPIC 6: EXERCISE 10 1.3 The period of h is 90°, the amplitude
1 1.1 Teacher to provide undefined, the range is given by y ∊ (–∞;∞)
1.2 1.2.1 x = –210°, –165°, –30° or 15° and y-intercept (0°;0)
1.2.2 x ∊ (–210°;–165°) ∪ (–120;–75°) 1.4 The period of k is 180°, the amplitude 1, the
∪ (–30°;15°) ∪ (60°;105°) range is given by y ∊ [–2;0] and y-intercept
2. 2.1 Teacher to provide (0°;–1,34)
2.2 2.2.1 x = ±45°
2.2.2 x ∊ (–15°;45°) ∪ (105°;15°)
∪ (165°;225°)
2.2.3 x = 13°
Answers 355
PLT MATHS LB 11 7th pgs (Real Book).indb 355 2012/07/02 2:27 PM
1.5 The period of p is 360°, the amplitude 5.10 y ∊ (−2;1)
undefined, the range is given by y ∊ (–∞;∞) 5.11 x ∊ (−195°;−75°) ∪ (165°;285°)
and y-intercept (0°;0)
1.6 The period of q is 360°, the amplitude __
1 , the
TOPIC 7: EXERCISE 1
2
__
1
[
__
1
range is given by y ∊ – ; and
2 2 ]
y-intercept (0°;0,47) 1 1.1 Q3 1.2 Q2
2 2.1 y = cos (x – 9°) = sin x
1.3 Q4 1.4 Q3
2.2 No, because y = cos (x + 9°) = –sin x
1.5 Q4 1.6 Q4
3 3.1 A(–180°;–1,73); B(150°;–2); C(–180°;–3,73);
D(60°;0); E(–30°;2); F(–120°;0); G(0°;1,73); 1.7 Q2 1.8 Q3
H(0°; 0,27)
3.2 x = 150° 1.9 Q2 1.10 Q1
3.3 x ∊ [–180°;60°]
3.4 The period 360°, the shift is 30° to the left. 2 2.1 negative 2.2 negative
3.5 f (x) = 2cos(x + 90°) = −2sin x 2.3 positive 2.4 negative
3.6 1 (−x + 30) − 1 = −tan__
g(−x) = tan__ 1 (x − 30) − 1
2 2
2.5 positive 2.6 negative
3.7 x ∊ [30°;150°)
3.8 x ∊ [−180°;−120°] and x = 60° 2.7 positive 2.8 positive
3.9 −f (x) = −2cos(x + 30°)
3.10 −g(x) = − tan__1 (x + 30°) − 1 2.9 negative 2.10 positive
2
= −tan__1 (x + 30°) + 1
2 3 3.1 Q2 3.2 Q4
4 4.1 Period of f is 360°
3.3 Q2 3.4 Q1
Amplitude of f is 2
Range of f: y ∊ [– 2;2] 3.5 Q4 3.6 Q3
__ __
Endpoints: (–180°;√3 ) and (180°;√ 3 )
4.2 Period of g is 240°
Amplitude of g is 1 TOPIC 7: EXERCISE 2
Range of g: y ∊ [–1;1]
(–180°;0) and (180°;0) 1 1.1 sin θ = ___
24 1.2 __
sin θ = 4
25 5
4.3 Teacher to provide cos θ = ___
7 cos θ = ___
–3
25 5
4.4 g(–120°) – f (–120°) = 1
tan θ = ___
24 ___
tan θ = 4
4.5 x = 6° 7 –3
4.6 x = ± 180° __ __
√ √
1.3 sin θ = ____
–10__ = ___ __ = –2
5
–2 × ___ 5
_____
4.7 x ∊ [–60°;60°] or – 60° ≤ x ≤ 60° 5√ 5 15 √5 5
(2 ) ( )
__ __
4.8 g( x ) = sin __
3 x – 9° = –sin 9° – __
3x
cos θ = ____
–5 –1 √
__ × ___
__ = ___ 5
__ = – √
____5
2
5√ 5 √5 √5 5
= –cos __
3x
2 tan θ = ____
–10 = 2
–5
5 5.1 f has an amplitude of 2 and a period of 360°
g has an amplitude of 2 and a period of 1.4 sin θ = ____
–3
___
√ 34
1 080°
cos θ = ____
5___
5.2 a = 2, b = 45°, c = 1, d = –2 and e = __
1
√ 34
3
5.3 A(–75°;0) and B(–135°;–1) tan θ = ___
–3
__
5
5.4 CD = 3 + √2 = 4,14 __
3 √
5.5 p = 2, q = −45° and q = 1 2 2.1 sin 60° = ___
1 and z = 270° 2
5.6 w = −2, k = __
3 cos 60° = __
1
5.7 x ∊ [−360°;−315°) ∪ (−135°;45°) ∪ (225°;360°] 2 __
√3 __
5.8 x ∊ [−270°;−195°] ∪ [−75°;165°] ∪ [270°;285°] tan 60° = ___ = √3
1
5.9 y ∊ (−3;1)
356 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 356 2012/07/02 2:27 PM
__ __
√ √ 2
2.2 sin 135° = ____
2 __ = ___ __ = ___
2
1__ × ___ 2 cos3 x + cos x.sin2 x = cos x
2√2 √ 2 √2
__
2
__
√ – √2
LHS = cos x (cos2 x + sin2 x) = cos x (1) = RHS
cos 135° = ____
–2 1 __ = ____
2
__ × ___
__ = ___
3 Teacher to provide
2√2 √ 2 √2 2
tan 135° = ___
2 = –1
4 (1 – sin2 x)(1 + tan2 x) = 1
–2
2
______
LHS = (cos2 x)(1 + sin x)
2.3 sin 210° = ___
–1 2 cos x
2
__
–√
____3 cos2 x + sin2 x = 1 = RHS
cos 210° = _____
1 – sin x = _____
cos x
2
__ __
5 tan x
sin x
√3 √3
tan 210° = ____
–1__ × ___
__ = ___
LHS = _____
1 – sin x = ________
1 – sin x = ______
cos x 2 2
–√ 3 √3 3 sin x sin x sin x
__ 2
–√
____3 _____
RHS = cos _____
x × cos ______
x = cos x
2.4 sin 300° = 1 sin x sin x
2
LHS = RHS
cos 300° = __
1
2
__
√3 cos x – cos3x
________________ = tan2 x
tan 300° = –____ 6
cos x – cos x.sin2 x
1
cos x (1 – cos2 x) 2
LHS = _______________
2
______
= sin 2
x = tan2 x
TOPIC 7: EXERCISE 3 cos x (1 – sin x) cos x
4 LHS = RHS
1 – __ 2 –13
5
________
3 5
__ 4 20 7 θ – √ 1 – sin2 θ = 0
_____
sin
3 tan θ
______
5 0 _____θ × cos
LHS = sin _____
x – √ cos2 θ
1 sin x
6 If cos x = __
3 , tan y = __
5 , x < 90° and y < 90° , = cos x – cos x = 0 = RHS
4 7
2
LHS = 74sin y + ______
7
8 tan2 θ + 1 = ______
1
tan2 x 2 cos θ
(
____
= 74 5 ) ( )
___ + _____
√ 74
2
√
7__
7 2
___ LHS = ______
cos θ
2
+ 1 = sin θ +
sin2 θ
2
cos θ = 1
2
______________ ______ = RHS
cos θ
2
2
cos θ
3
( 74 )2
=74 ___
25 + 7 × __
9
7
9 ___________
sin x + cos x = cos x
1 + tan x
= 25 + 9 = 34 = RHS LHS = ___________
sin x + cos x = ___________
sin x + cos x
1 + _____
sin x
cos x
___________
cos x + sin x
cos x
k
________ _____________
cos x
7 = (sin x + cos x) ×
√
– 1 – k2 (sin x + cos x)
= cos x = RHS
TOPIC 7: EXERCISE 4
1 1 sin2 x 3 sin x 10 tan θ + _____
1 = _________
1
2 tan θ sin θ.cos θ
4 8 5 cos2 x 6 9cos2 θ _____θ = sin2 θ +
θ + cos
_____
sin cos2 θ
____________
LHS =
7 sin x 8 1 9 _____
tan x cos θ sin θ sin θ.cos θ
2
10 –1 = _________
1
sin θ.cos θ
= RHS
TOPIC 7: EXERCISE 5
_____
sin x = sin2 x 11 tan2 x – sin2 x = sin2 x.tan2 x
1 LHS = sin x.cos x. cos x 2 2 2 2
RHS = 1 – cos2 x LHS = ______
sin x – sin2 x = _________________
2
sin x – cos x.sin x
2
cos x cos x
= sin2 x si n2 x (1 – cos2 x) 2 x.sin2 x
= _______________ __________
= sin
LHS = RHS cos2 x cos2 x
= sin2 x.tan2 x = RHS
Answers 357
PLT MATHS LB 11 7th pgs (Real Book).indb 357 2012/07/02 2:27 PM
12 ________
cos x + tan x = _____
1 3 sin x – ___________
1 ___________
+ cos x = 2sin x.cos x
1 + sin x cos x sin x + cos x sin x + cos x
cos2 x + sin x.(1 + sin x)
LHS = ________
cos x + _____
sin x = _____________________
cos x
sin x (sin x + cos x) – 1 + cos x (sin x + cos x)
LHS = ______________________________________
1 + sin x (1 + sin x).cos x sin x + cos x
2 x + sin x + sin2 x
__________________ (1 + sin x)
= co s = ______________ 2 2
(1 + sin x).cos x (1 + sin x).cos x = ____________________________________
sin x + sin x.cos x – 1 + cos x.sin x + cos x
sin x + cos x
= _____
1
cos x = RHS 2sin x.cos x + (sin2 x + cos2 x) – 1
= ____________________________
sin x + cos x
13 tan θ – sin θ.cos θ = tan θ
_______________
sin2 θ ___________
= 2sin ________________
x.cos x = 2sin x.cos x + 1 – 1 = RHS
sin x + cos x sin x + cos x
θ – sin θ.cos θ
_____
sin sin θ – sin θ.cos2 θ
________________
cos θ
_______________ θ
______________
cos 2 2
LHS = = 4 __________________
sin x + sin ________
x – cos x = sin x +1
sin2 θ sin2 θ 2sin x.cos x – cos x cos x
sin θ (1 – cos2 θ)
_______________________θ
sin θsi n 2
sin2 x + sin x – (1 – sin2 x)
= θ
_____________
cos ________
= cos θ
= tan θ.sin2 θ
_________ LHS = _______________________
2sin x.cos x – cos x
sin2 θ sin2 θ sin2 θ
2
= tan θ = RHS = ________________
2sin x + sin x – 1 (factorise)
2sin x.cos x – cos x
(2sin x – 1)(sin x + 1)
14 tan2 x.cos2 x + ______
sin x = 1 2
= ___________________ = sin
________
x +1
cos x
2
tan x cos x (2sin x – 1)
2 2
LHS = ______
sin x × cos2 x + sin2 x × ______
2
cos x
2 5 ________
1– ________
cos x – sin x =0
cos x sin x sin x 1 + cos x
= sin2 x + cos2 x = 1 = RHS (1 – cos x)(1 + cos x) – sin2 x
LHS = _________________________
sin x (1 + cos x)
15 ________
1 + ________
1 = ______
2 (1 – cos2 x) – sin2 x
1 – sin x 1 + sin x cos2 x = _________________
sin x (1 + cos x)
__________________
LHS = 1 + sin x + 1 – sin x = ________
2 = ______
2
2 2
(1 – sin x)(1 + sin x) 2 21 – sin x cos x ______________
= sin x – sin x = 0 = RHS
sin x (1 + cos x)
TOPIC 7: EXERCISE 6 6 ________
1 – ______ ________
2 = sin x +1
sin x + 1 cos2 x sin x – 1
1 ________
cos ________
x =1+ sin x
1 – sin x cos x cos2 x – 2 (sin x + 1)
cos x (1 + sin x) RHS = __________________
LHS = ________
cos x × ________
1 + sin x = ______________ 2
(sin x + 1) cos x
1 – sin x 1 + sin x 2
1 – sin x (1 – sin2 x) – 2 (sin x + 1)
cos x (1 + sin x) = ______________________
= ______________
2
________
= 1 cos
+ sin x = RHS
x
(sin x + 1) cos2 x
cos x 2
__________________
= 1 – sin x – 2sin x – 2
2 cos2 x – cos x – sin2 x = _____
__________________ 1 – _____
1 2
(sin x + 1) cos x
2sin x.cos x + sin x tan x sin x
– (sin2 x + 2sin x + 1)
cos2 x – cos x – (1 – cos2 x) = ___________________
LHS = _______________________
2
(sin x + 1)(1 – sin x)
2sin x.cos x + sin x
–(sin x + 1)(sin x + 1)
2 = _________________________
________________ (sin x + 1)(1 – sin x)(1 + sin x)
= 2cos x – cos x – 1 (Factorise)
2sin x.cos x + sin x
– (sin x + 1)
(2cos x + 1)(cos x – 1) (cos x – 1) = ___________ = sin x
________
+ 1 = RHS
(1 – sin x) sin x – 1
= ____________________ = __________
sin x (2cos x + 1) sin x
7 _____ = tan
tan θ + 1 _____θ
_____
x – _____ (cos x – 1)
1 = __________ tan θ sin θ
RHS = cos
2
= LHS
sin x sin x sin x
_____θ = sin θ +
θ + cos
_____
LHS = sin cos θ
____________ 2 2
cos θ sin θ cos θ.sin θ
= _________
1
cos θ.sin θ
sin θ × _____
RHS = _____ 1 = _________
1 = LHS
cos θ 2 sin θ cos θ.sin θ
358 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 358 2012/07/02 2:27 PM
_______
8 _________
sin x
(1 + cos x)
= √ 11 +–
________
cos x
cos x TOPIC 7: EXERCISE 11
_______________ 1 1.1 –sin 68° 1.2 –cos 55°
RHS = √ 11 +–
________ ________
cos x × 1 +
cos x
cos x
1 + cos x 1.3 tan 87° 1.4 –sin 74°
_________ _________
√ √ (1 +sin
cos2 x =
__________
1– 2
___________
x 1.5 tan 45° 1.6 sin 24°
=
(1 + cos x)2 cos x) 2 1.7 –cos 85° 1.8 –tan 9°
__________
sin x 1.9 –tan 57° 1.10 –cos 71°
= = LHS
(1 + cos x) 1.11 cos 20° 1.12 cos 68°
( sin θ )
______θ + cos2 θ = 2 1.13 tan 74° 1.14 sin 33°
2
9 sin2 θ + tan2 θ cos2
1.15 sin 5°
( cos θ )( sin θ )
2 θ
______
LHS = (sin2 θ + cos2 θ) + sin
2
θ
______
cos
2 2
= 1 + 1 = 2 = RHS TOPIC 7: EXERCISE 12
1 1.1 –k 1.2 k
10 (2sin x – cos x)(2sin x + cos x) = 5sin2 x – 1
_____
LHS = 4sin2 x – cos2 x = 4sin2 x – (1 – sin2 x) 1.3 _______
k
_____ 1.4 –√1 – k2
= 4sin2 x – 1 + sin2 x = 5sin2 x – 1 = RHS √ 1 – k2
1.5 –k 1.6 _______
k
_____
√ 1 – k2
TOPIC 7: EXERCISE 7 p p
_______
2 2.1 –__ 2.2 – _____
1 –1 2 2 3 –1 2 √4 + p2
4 1 5 cos x 2.3 – _______
4
_____ 2.4 tan 65º = __
2
p
√4 + p
2
p
_______
2.5 – _____ 2.6 – _______
2
_____
TOPIC 7: EXERCISE 8 √4 + p2 √4 + p
2
______
1 –2 2 1 3 ______
1 √1 – a2
________
cos x2 3 3.1 –a 3.2
1
4 1
________ 5 tan x
1 + sin x 3.3 –a 3.4 –a
TOPIC 7: EXERCISE 9 TOPIC 7: EXERCISE 13
1 1 2 _____θ
– tan 3 1
__ __ __
2 2 1 – √2 2 –3 3 4√ 3
4 cos β 5 1 3
___ 3
__ 5
4 5 6 – __
6 1 + 2sin x.cos x 16 2 4
7 2 8 1
TOPIC 7: EXERCISE 14
1 1.1 cos 70° 1.2 –sin 70°
TOPIC 7: EXERCISE 10
1 1.1 sin 58° 1.2 –cos 24° 1.3 cos 70° 1.4 sin 15°
1.3 –tan 47° 1.4 sin 14°
1.5 –tan 35° 1.6 –cos 81° 1.5 –sin15° 1.6 sin 15°
1.7 –sin 65° 1.8 –cos 45°
1.7 sin 70° 1.8 cos 70°
1.9 tan 11° 1.10 tan 29°
1.11 –cos 50° 1.12 –sin 36° 2 2.1 –1 2.2 1
1.13 cos 10° 1.14 cos 60°
2.3 –4 2.4 –2
1.15 –tan 72° 1.16 –sin 58°
2.5 1
– __ 2.6 1
1.17 –sin 65° 1.18 –tan 38° 3
2.7 2 2.8 sin2 35°
Answers 359
PLT MATHS LB 11 7th pgs (Real Book).indb 359 2012/07/02 2:27 PM
TOPIC 7: EXERCISE 15 TOPIC 7: EXERCISE 18
1 137,16° 2 92,54° 1 x = 33,69° + n.180°
3 297° 4 346,29° 2 x = 0° + n.360°; x = 180° + n.360°;
5 254,17° 6 215,61° x = 0°; 180°; 360°; 41,81°; –318,19°;
138,19°; –221,81°
7 318,19° 8 240,95°
3 x ≤ 360° x = 33,69°; 213,69°; –146,31°; – 45°; 135°
9 x = 30°; 150°; 210°; 330° 4 x = –23,58° + n.360° or x = 90° + n.360°;
10 x = 60°, 120°, 240°, 300° x = 203,58° + n.360°, n ∊ ℤ
5 x = 48,59°; 131,41°; 270°; –90°
11 x = 45°,135°,225°,315°
6 x = 19,47° + n.360° or n = 270° + n.360°
12 Special angle = 60°; Q3: x = 240°; Q4: x = 300° x = 169,53° + n.360°, n ∊ ℤ
7 x = 0°; 180°; –180°; 59,03°; –120,97°
8 x = –113,58° + n.360° or
TOPIC 7: EXERCISE 16 x = –60° + n.360°, n ∊ ℤ
1 x = 232°; x = –128° x = 352°; x = –8° 9 x = 56,31° + n.180° or 26,57° + n.180°, n ∊ ℤ
2 x = 15°; x = –345°; x = 315°; x = – 45° 10 x = 0°; 180°; –180°; –36,87°; –143,13°
3 x = 95°; x = –265°; x = 275° x = –85° 11 x = – 41,81°; 318,19°; 221,81°; –138,19°; –30°; 330°;
4 x = 41,57°; x = –318,43°; x = 348,43°; x = –11,57° 210°; –150°
5 x = 91,90°; x = –268,10°; x = 268,10°; x = –91,90° 12 x = 90° + n.180° or x = –60° + n.360°
6 x = 40,32°; x = –319,68°; x = 220,32°; x = –139,68° 13 x = –30° + n.360° or x = 138,59° + n.360°, n ∊ ℤ
7 x = 72°; x = 342°; x = 162°; x = –198° x = 210° + n.360° or x = –138,59° + n.360°, n ∊ ℤ
8 x = 176,71°; x = –183,29° x = 263,29°; x = –90,71° 14 x = ± 113,58°; 0°
9 x = 51,42°; x = –308,58°; x = 278,58°; x = –81,42° 15 x = 146,31°; 326,31°
10 x = 123,20°; x = –236,80°; x = 303,2°; x = –76,80°
11 x = –11,31°; x = –371,31°; x = 127,31°; x = –232,69°
TOPIC 7: EXERCISE 19
12 x = 30,54°; x = –329,46°; x = 210,54°; x = –149,46°
1 x = 58°; 302°
2 x = 64°; 154°; 244°; 334°; –26°; –116°; –206°;
TOPIC 7: EXERCISE 17 –296°; –386°
1 x–16º = 38,02° + n.360°; x = 54,02° + n.360° n ∊ ℤ; 3 x = 80°; –280°; 180°; –180°
x = 157,98° + n.360° n ∊ ℤ 4 x = 25° + n.180°, n ∊ ℤ
2 x = 18,95° + n.180°; x = –18,95° + n.180° 5 x = 21°; 141°; –13°
3 x = 32,36° + n.120°, n ∊ ℤ; 6 x = 40° + n.120°, n ∊ ℤ
x = 87,64° + n.120°, n ∊ ℤ; x = 32,36° + n.120°, n ∊ ℤ 7 x = 60° + n.360°
x = –32,36° + n.120°, n ∊ ℤ 8 x = 138°; 222°; –138°
4 x = 26,46° + n.180°, n ∊ ℤ 9 x = 44,67°; 166,67°; 286,67°; –73,33°;
5 x = 250,28° + n.360°, n ∊ ℤ; –120°; 240°
x = –109,73° + n.360°, n ∊ ℤ 10 x = 40° + n.360°, n ∊ ℤ
6 x = 90,54° + n.180°, n ∊ ℤ
x = 135,46° + n.180°, n ∊ ℤ
TOPIC 7: EXERCISE 20
x = 90,54° + n.180°, n ∊ ℤ
7 x = 134,03° + n.360°; x – 15° = 240,97° + n.360°
1 x = 109,47° or 250,53°
2 x = 12,94°; –347,06°; 167,06°; –192,94°
x = 255,97° + n.360°; x = 134,03° + n.360°;
3 x = 73,43°; –106,57°
x = –104,03° + n.360°
4 x ∊ [0°;360°] x = 54,96°; 97,07°
8 x = 44,69° + n.90°, n ∊ ℤ
5 x = – 40° + n.180°, n ∊ ℤ
6 x ∊ [–360°;360°] x = 90°; 270°
7 x = –14,10° + n.90°, n ∊ ℤ
8 x = 46° or x = 338°
9 x = – 47,29° + n.180° n ∊ ℤ
360 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 360 2012/07/02 2:27 PM
10 x = ± 150° + n.360°, n ∊ ℤ 6.3 ( _____ = 1
cos θ tan θ + 1 _____
tan θ ) sin θ
11 x = –19,47°; 340,53°; 199,47°; –160,53°
12 x = 30,96°; 210,96°; –149,04°; –63,43°; 116,57° ( cos θ
_____θ
θ + cos
_____
LHS = cos θ sin
sin θ )
13 x = 0° + n.360°, n ∊ ℤ
14 x = 26,57°; –153,43°; 0° ( cos θ.sin θ )
sin θ + cos θ
= cos θ ____________
2 2
TOPIC 7: EXERCISE 21
( cos θ.sin θ )
= cos θ _________
1 = _____
1 = RHS
sin θ
7 7.1 _____
– tan x 7.2 1
1 x = 0° + n.180° or x = 90° + n.180° 4 1
______
7.3 –2 7.4
2 x = 90° + n.180° sin2 x
3 x = 90° + n.180° 7.5 1
4 x = 90° + n.180° or x = 270° + n.360° 8 8.1 8.1.1 –p 8.1.2 _______
1
_____
√1 + p2
5 x = 90° + n.180°
8.1.3 _______
1
_____
6 x = 90° + n.180°
√1 + p2
7 x = 0° + n.180° or x = ±120° + n.360°
8.2 8.2.1 – __
a 8.2.2 __
a
or x = 90° + n.360° 2 2
_____
8 x = 135° + n.180° –√ 4 – a2
________
8.2.3 2
9 x = 90° + n.180
10 x = ± 180 + n.360° 9 9.1 –3
__ 9.2 3
2
7
__ 9.4 – __
3
TOPIC 7: REVISION 9.3
4 2
___
1 OP = √41
____ 9.5 – __
1
1.2 1.2.1 sin θ = 4___ 2
√ 41
1.2.2 = 26 10 10.1 θ = 39,20° or θ = 120,80°
1.2.3 – __
4
(5) 10.2 x = 75°; – 45°; –165°; –15°; 105°
10.3 θ = ± 120° + n.360°
2 2.1 OP = 2 __
√3 or θ = ± 180° + n.360°; n ∊ ℤ
2.2 2.2.1 – ___
4 10.4 θ = 36°; 126°; 216°; 306°; –54°;
2.2.2 __
4 –144°; –234°; –324°
3
10.5 x = 23,33° + n.120°; n ∊ ℤ
2.2.3 θ = 330°
x = –70° + n.360°; n ∊ ℤ
3 5 and is in Quad 2
sin θ = __
6 11 11.1 x = 0° + n.180° or x = 90° + n.180°, n ∊ ℤ
3.1 1 – 2cos2 θ = ___
7
11.2 x = 90° + n.180°, n ∊ ℤ
18
11.3 x = 0° + n.180° or x = 90° + n.180°, n ∊ ℤ
3.2 cos θ.tan θ = __
5
6
4 34 TERM 2: EXAMINATION PRACTICE
5 tan A.sin A = ___
__________ 33 PAPER 1
cos B 16
1 1.1 x = 0 or x = __
4
_________ 3
6 6.1 sin2 x + sin x.cos x = 1
tan x 1.2 x = 0,45 or x = –4,45
LHS = sin2 x + _________
sin x.cos x × _____
cos x
1.3 x=8
1 sin x
= sin2 x + cos2 x = 1 = RHS 1.4 0 ≤ x ≤ __
1
3
_____
cos θ + cos θ.tan2 θ = 1
6.2
cos θ 1.5 x = – __
3 or x = 4
2
LHS = cos θ + cos θ. 2 θ = cos θ + sin θ
2 2
______
si n _____ ___
81
cos θ 1.6
cos θ 16
__
x = 7 or __
x=3
= cos θ + sin2 θ = 1
2
____________ _____ = RHS 2 2.1 2.1.1
cos θ cos θ y y
2.1.2 y = ___
3 or y = __
1
10 2
Answers 361
PLT MATHS LB 11 7th pgs (Real Book).indb 361 2012/07/02 2:27 PM
2.2 2.2.1 3 2.3 5
___ 3 3.1 –3
2x
2.2.1 x=7 3.2 1
2.2.2 x≤7
1
__ y
__ 4. 4.1 cos A + cos (180° – A).sin2A = cos3 A
3 3.1 b6 3.2 x
__
LHS = cos A – cos A sin2A
(3x – 2)
_______
3.3 √3 3.4 = cos A(1 – sin2A)
5
= cos A.cos2A
4 4.1 4.1.1 day 5 = 61, day 6 = 86 = cos3A = RHS
4.1.2 Tn = 2n2 + 3n – 4 or Tn = 2n2 + 3n – 4
(sin x – cos x)2
4.2 4.2.1 A : 2n + 1 4.2 4.2.1 _____________ = _____
1 – 2tanx
2
1 – sin x cos2x
B: n2
C: n(n + 1) = n2 + n RHS = _____ ______
1 – 2sin
2
x
cos x
cos x
4.2.2 n = 10
_____________
= 1 – 2sin x.cos x
4.2.3 n=7 2
cos x
4.2.4 n2 + ___
___
2 3n + __
1 sin2 x – 2sin x cos x + cos2 x
4 4 4 LHS = ________________________
cos2x
5 5.1 (x – 1)2 – 4 5.2 AB = 4; OC = 3
_____________
= 1 – 2sin x cos x
2
cos x
5.3 D = (1;–4)
5.4 m =1; c = −3; y = x − 3 LHS = RHS
5.5 0≤x≤3 4.2.2 x = 90° or 270°
5.6 5.6.1 y = x2 – 6x + 4 5 5.1 233,56°
5.2 104,48°
5.6.2 y = –x2 + 2x + 3
5.3 x = 75,52° + n.360° or
6.1 Teacher to provide x = 60° + n.360° = –75,52° + n.360° or
x = –60° + n.360°; n ∈ ℤ
6.2 –1 < x < 3
6 6.1 Teacher to provide
PAPER 2 6.2 x = 30°; 120°; –60°; –150°
1 1.1 __
1
6.3 x = 30° + n.180°
3
6.4 –60° ≤ x ≤ 120°
1.2 y = __
1x + 2
3
1.3 ^ = 90°
1 × –3 = –1; BC ⊥ AB; B
mBC × mAB__
3 TOPIC 8: EXERCISE 1
1.4 D = (7;1) 1 Volume = 18 849,56 cm3
1.5 ( __72;__32 ) TSA = 4 084,07cm2
2 Volume cylinder = 942,48 units3
1.6 y = –x + 5
TSA = 534,07 units2
1.7 63,43° 3 Total volume = 71 994,83 cm3
TSA = 9 032,08 cm2
1.8 10 units2
4 Volume pyramid = 5,376 m2
1.9 a = –9 TSA = 4 704 m2
5 Volume truncated cone = 11 623,89 units3
1.10 b = 7
__ TSA = 3 392,92 units2
5 √
2 2.1 2.1.1 cos θ = ___ 6 Volume = 50 186,94 units3
3
2.1.2 ___
–8 TSA = 7 869,69 units2
3
7 Volume = 4 416 m3
2.2 2.2.1 1
sin 70° = __
p TSA = 1 776 m2
2.2.2 ___
–1 8 Volume = 93 600 units3
p
_____ TSA = 14 892 units2
√p 2 – 1
_______
2.2.3 p
362 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 362 2012/07/02 2:27 PM
9 Volume = 471,40 units3 2.2 0,9 cm
TSA = 346,41 units2 2.3 5,969 m3
10 Volume = 124 859,46 units3 2.4 119,18 m2
TSA = 13 420,88 units2 3 3.1 12
11 Volume of sphere = 1 767,15 cm3 3.2 5 277,88 m2
TSA = 706,86 units2 3.3 1 847,26 units2
4 4.1 26 880 units2
TOPIC 8: EXERCISE 2 4.2 DE = 29; DG = 52
1 1.1 Volume of cone = __1 πr2h 4.3 4 968 units2
3 4.4 9 000 units2
1.2 40 units
1.3 s = 41 5 Volumesphere = 57 905,84 units3
1.4 Surface area of cone = 1 413,72 units2 TSA = 7 238,23 units2
1.5 1.5.1 It increases by a factor of 4. 6 Volumehemisphere = 16 755,16 cm3
1.5.2 V = 13 571,68 units3 TSA = 3 769,91 cm2
1.6 1.6.1 It will increase by a factor of 3. 4 πr3
__
7 7.1
3
3
1.6.2 V = 10 178,76 units 7.2 37 units
1.7 1.7.1 V = 424,115 ≈ 424,12 units3 7.3 17 203,36 units2
1.7.2 TSA = 335,43 units2
8 8.1 14,5π
8.2 5 466,37 units3
2 2.1 Volume of cylinder = πr2h
8.3 11 263,85 m2
2.2 20 units
9 9.1 1 πr2h
V = __
2.3 TSA = 6 214,07 units2 3
2.4 1.
2.4.1 The volume will be multiplied by __ 9.2 8 cm
4
9.3 17
2.4.2 V = 8 309,51 units3 9.4 136π
2.5 2.5.1 1.
The volume will be multiplied by __
4 9.5 628,32 cm2
2.5.2 V = 8 309,51 units3 1 are of base × ⊥
1 10 10.1 Volume of pyramid = __
2.6 2.6.1 The volume will be multiplied by ___ 3
27 height
2.6.2 V = 1 231,04 units 3
10.2 18 × 18
2.6.3 It will be multiplied by 1
__.
9 10.3 41
2.6.4 TSA ≈ 698,45 units 2
10.4 1 476 units2
10.5 1 800 units2
3 3.1 Surface area of cylinder = 2πr2 + 2πrh
3.2 9 units 11 11.1 Volume of cylinder = πr2h
3.3 Volume = 8 171,28 units3 11.2 10 units
11.3 2 513,27 units2
4 4.1 Volume of sphere = __
4 πr3
3
4.2 43 units TOPIC 9: EXERCISE 1
4.3 TSA = 23 235,22 units2
1 1.1 12 units
4.4 4.4.1 8
1.2 OE = 18 − x units
4.4.2 V = 2 664 305,12 units3
1.3 13 units
4.4.3 4
4.4.4 92 940,88 units2
2 2.1 2.1.1 OS = 25
2.1.2 OV = 15 units
TOPIC 8: REVISION 2.2 TU = 120 units
1 1.1 2 435,70 m2
1.2 1 319,86 units2 3 3.1 AO = 25 units
3.2 OF = 20 units
2 2.1 10 cm
3.3 FG = 5 units
Answers 363
PLT MATHS LB 11 7th pgs (Real Book).indb 363 2012/07/02 2:27 PM
4 4.1 perpendicular bisectors of PQ and RT 1.9 Supplementary
4.2 4.2.1 39 units 1.10 Concylic quadrilateral
4.2.2 25 units 1.11 Four points are cyclic
1.12 The four points do not form a cyclic
5 5.1 EO ⊥ AB; OF ⊥ CD; AB∥CD quadrilateral
5.2 OF = 15______
units 1.13 The quadrilateral is a cyclic quadrilateral
5.3 √
AO = x2 + 49 1.14 The quadrilateral is not a cyclic quadrilateral
___________
5.4 CO = √( x − 4 )2 + 225 5.5 1.15 Equal angles on the circumference
5.5 radii
5.6 x = 24 2 2.1 No, co-interior angles not supplementary.
5.7 AB = 48 units and CD = 40 units 2.2 2.2.1 t = 17°
5.8 968 units2 2.2.2 FD^ C = FE
^C
2.2.3 ^
BAE = BFE^
6 6.1 6.1.1 CG = GD = 30 units ^ F = AB
^F
2.2.4 AE
6.1.2 OG = 16 units
2.3 u = 77°; v = 103°; w = 77° ; x = 104°; y = 20°;
6.1.3 AB = 126 units
z = 20°
6.1.4 AC = 33 units
6.2 6.2.1 E^ =θ
2
6.2.2 corresponding angles equal TOPIC 9: EXERCISE 4
____
6.3 FB = 7√130 units 1 w = 101°; x = 96°; y = 92°; z = 60°
6.4 6.4.1 △AGF ≡ △BGF
6.4.2 kite 2 2.1 a = 101°; b = 79°; c = 82°; d = 98°; e = 98°;
6.4.3 Two pairs adjacent sides equal f = 79°
OF is the perpendicular bisector 2.2 No, because Y ^+X ^ = 161° and so the
of AB co-interior angles are not supplementary.
6.4.4 4 095 units2 2.3 Yes, because ^V+ ^X = 180° and so the
co-interior angles are supplementary.
TOPIC 9: EXERCISE 2 2.4 No, it is NOT a cyclic quadrilateral because
1 a = 64°; b = 296°; c = 148° opposite angles are not supplementary.
2 a = 90° ; b = 32°; c = 58°; d = 58°; e = 29°;
3 3.1 ABCE; ABCD
f = 29°; g = 29° ; h = 29°; i = 122°; j = 122° ^ = x; E
^ = x; ED = FD
3.2 B
3 a = 35°; b = 35°; c = 70°; d = 35°; e = 35°; f = 110° 1
^ = x; B
1
^ = x; BF bisects AB
^C
3.3 E
g = 22° 2
^ ^
2
3.4 ABC = C
4 a = 27°; b = 63°; c = 27°; d = 54°; e = 90°; f = 36°
3.5 Opposite ∠s are not supplementary
g = 63°
3.6 Opposite ∠s not supplementary
5 t = 40°; u = 40°; v = 80°; w = 21°; x = 42°; y = 29°
3.7 AF ∥ BC
z = 29°
4 4.1 ^ = 90° − 2x
D2
TOPIC 9: EXERCISE 3 4.2 Teacher to provide
1. 1.1 Bisects the chord 4.3 Teacher to provide
1.2 Perpendicular to the chord 4.4 Teacher to provide
1.3 Centre of the circle 4.5 Teacher to provide
1.4 Point of intersection
1.5 Twice the size of the angle subtended by the TOPIC 9: EXERCISE 5
same chord on the circumference of
1 t = 59°; u = 59°; v = 63°; w = 63°; x = 58°; y = 59°;
the circle
z = 62°
1.6 Are equal
2 a = 104°; b = 52°; c = 90°; d = 38°; e = 19°; f = 71°;
1.7 90°
g = 123°
1.8 Diameter
3 w = 70° ; x = 50°; y = 60°; z = 120°
364 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 364 2012/07/02 2:27 PM
TOPIC 9: EXERCISE 6 TOPIC 9: EXERCISE 7
1 1.1 Equal in length. 1 1.1 QV = VS (OV ⊥ QS)
1.2 90° OQRS is a kite (OR bisects QS
1.3 Equal perpendicularly)
1.4 Tangent to the circle. 1.2 QR = SR (adjacent sides kite)
a = 10° (equal chords QR and SR)
2 2.1 t = 54°; v = 126°; w = 80°; x = 80°; y = 100°; b = 10° (∠s on chord QR)
z = 46° c = 10° (∠s on chord RS)
2.2 No, alternate angles unequal d = 20° (∠ at centre)
2.3 Yes, corresponding angles equal e = 20° (∠ at centre)
3 3.1 x = 30°; straight ∠ f = 70° (exterior ∠ of △OQV)
3.2 3.2.1 NM ^ K = 90°; ∠sum △NMK g = 70° (isosceles △OQS, radii)
3.2.2 Teacher to provide 1.3 OQ is 145 units.
4 and 5 Teacher to provide 2 2.1 2.1.1 Q^
PT = 90° (∠ on diameter)
2.1.2 ^ T = 90° (∠on diameter)
QR
6 6.1 No, alternate angles unequal ^
2.1.3 QTU = 90° (radius ⊥ tangent)
6.2.1 equal to the angle subtended by
2.2 a = 31° (tangent UT, chord PT)
the chord in the alternate segment.
b = 31° (tangent UT, chord PT or ∠s on
6.2.2 supplementary.
chord PT)
6.2.3 equal to the opposite interior angle
c = 62° (∠ at centre)
6.3 p = 53°; q = 60°; v = 67°; w = 113°; x = 45°;
d = 20° (∠s on chord RT)
y = 22°
e = 20° (tangent ST, chord RT)
7 7.1 7.1.1 ^ + 18° = 43°; tangent CQ,
A f = 70° (∠ △sum QTR)
3
chord CD; A ^ = 25° g = 39° (∠ sum isosceles △OPT, radii)
3
7.1.2 ^ = 18°; isosceles △OAC, radii
C h = 39° (radii)
3
7.1.3 AO^ C = 144° ; ∠ sum △AOC i = 31° (radii)
7.1.4 ^
D = 72°; ∠ at centre j = 20° (radii)
7.1.5 PC^ A = 72°; tangent PC, chord CA k = 59° (adjacent complementary)
7.1.6 AB^ C = 108°; (opposite ∠s cyclic l = 140° (∠ sum △OQR)
quadrilateral) m = 118° ( ∠ sum △OQP)
7.2 Teacher to provide n = 40° (straight ∠)
7.3 Teacher to provide w = 70° (∠ sum △QTS)
x = 101° (∠ sum △PTV)
8 Teacher to provide y = 101° (vertically opposite)
9 9.1 Q^ = x (tan PQ, chord VQ) z = 79° (straight ∠)
4
^ = x (corresponding ∠s, OR ∥ VQ)
R 3 3.1 PR bisects SQ perpendicularly
Q2 = x ( isosceles △OSQ, radii) 3.2 3.2.1 P^SR = 90° (∠ on diameter)
9.2 ^ = ^S = x (from 10.1)
R ^ = 61° (∠ sum △PSR)
R 1
OQ subtends equal angles at S and R; QRSO 3.2.2 ^ = 61° PR symmetry line of kite)
R 2
is a cyclic quadrilateral. Q^ +Q ^ = 61° (∠ sum isosceles
^ V = 90° (∠ on diameter) 2 3
9.3 SQ △OQR, radii)
Q4 = x (proved in 10.1) Q^ = 29° (∠s on chord SR)
SQ^ P = 90° + x 3
^ = 32°
Q 2
9.4 ^ = 90° (corresponding ∠s, RO ∥ QV)
T ^ = 58° (∠ sum △QOR)
2 3.2.3 O 1
ST = TQ (OT ⊥ SQ)
4 4.1 Teacher to provide
4.2 Teacher to provide
4.3 BC = 30 units
Answers 365
PLT MATHS LB 11 7th pgs (Real Book).indb 365 2012/07/02 2:27 PM
5 5.1 They are 24 units apart. TOPIC 10: EXERCISE 1
5.2 5.2.1 FH = 5 units
1 CB = 9,21 cm
5.2.2 DH = 35 units
AC = 25,71 cm
5.2.3 DF = 30 units
A^ = 21°
5.3 OM = 13 units (radius of small circle)
OE = 37 units 2 ^ = 67,38°
C
(radius of larger circle) ^ = 22,62°
A
__
5.4 KE = 40√ 3 (OM ⊥ KE) CB = 5 cm
3 ^ = 57°
C
5.5 OHEM is a cyclic quadrilateral
AC = 21,43 cm
TOPIC 9: REVISION CB = 39,35 cm
4 ^ = 50,19°
C
1 1.1 r = 123
^ = 39,81°
A
1.2 OC perpendicularly bisects AB;
OACB is a kite AC = 78,11 cm
1.3 AO^ B = 4x
^A = 90°
TOPIC 10: EXERCISE 2
2 2.1 OF _______ 1 23,40 cm
2.2.1 AO = √ x__________
2 + 400
2 37,97 cm
2.2.2 OC = √( x + 9 )2 + 49 (4) 3 x = 55,36 cm or 124,64°
____
2.3 OA = √625 = 25 units 4 85,69 cm
3 x = 36° 5 x = 55,15°
^ = 24° (alternate ∠s, PO ∥ SR) 6 x = 64,59°
4 4.1 R 1
4.2 ^
R2 = 66° (on diameter)
^ = 132° (at centre)
TOPIC 10: EXERCISE 3
4.3 O 2 1 x = 37,96 cm
4.4 P^ = 24° (sum isosceles △POQ, PO = OQ)
2 2 x = 22,70 cm
4.5 ^S = 48° (∠s on chord RQ)
3 x = 75,15° or x = 104,85° (ambiguous case)
4.6 ^ = 42° (∠ sum △RSQ)
Q 2 4 x = 40,75°
5 5.1 ^ = x (exterior ∠cyclic quadrilateral)
F 5 x = 104,47°
1
D1 = x (exterior ∠cyclic quadrilateral) 6 x = 89,18 cm
5.2 BCDG is a cyclic quadrilateral
(exterior ∠ = interior opposite ∠) TOPIC 10: EXERCISE 4
5.3 5.3.1 Teacher to provide 1 517,25 cm2
5.3.2 Teacher to provide 2 157,19 cm2
5.3.3 ABDE is a cyclic quadrilateral. 3 200,86 cm2 (if AB rounded off)
(Exterior ∠ = interior opposite ∠) 200,83 cm2 (if AB answer from calculator)
4 965,16 cm2 (if AB rounded off)
6 6.1 DC subtends equal angles at A and B 965,13 cm2 (if AB answer from calculator)
6.2 ^ = x (equal chords DC and BC)
A 2 5 Area = 221,19 cm2
AC bisects BA^ D (A^ =A ^ )
1 2 6 Area = 658,18 cm2
6.3 ^ ^
DG = FG (A1 = A2)
6.4 D2 = 90° TOPIC 10: EXERCISE 5
(∠ on diameter) 1 1.1 Perimeter ABCD = 31,84 cm
^ = 90°(∠s on chord AB)
C 2 Area ABCD = 53,49 cm2
6.5 Teacher to provide 1.2 DC = 26,11 cm
6.6 exterior ∠ = interior opposite ∠ Area ABCD = 289,73 cm2
6.7 x = 30° 1.3 AC^ E = 5,91°
7 Teacher to provide 1.4 BD = 16 cm
1.5 986,21 mm2
1.6 x = 13,10 cm
366 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 366 2012/07/02 2:27 PM
______________
2 2.1 BC = √ x2 + y2 − 2xycos θ 1.6 Area △ABC = 1 294,40cm2
xsin ( θ + β ) (or 1 294,26 if AC is rounded)
2.2 BC = __________
sin θ ^ S = 72°
2 2.1 QP
2.3 BC = 2xcos θ
2.2 QS2 = 21,16 cm
asin ( α + β )sin α
2.4 BC = _______________ 2.3 282,46 cm2
sin β
α
_________
asin 3 3.1 ^
B = 100,29°
2.5 BC =
sin βsin θ ^ D = 23,18°
3.2 AC
xsin β
2.6 BC = _________
cos θsin α 4 PQ = 6,5m
TOPIC 10: EXERCISE 6 5 5.1 AB = 4,12 m
1 AD = 440,63 m 5.2 Area △AEB = 5,37 m2
2 AD = 21,73 m 5.3 BE = 2,97 m
3 BC = 7,64 m 6 Teacher to provide
4 AD = 5,82 m
5 AB = 15,87m 7 Teacher to provide
8 Teacher to provide
TOPIC 10: EXERCISE 7
9 9.1 SQ = 270,96 km
Teacher to provide
9.2 The bearing of Q from S is 10,35°.
TOPIC 10: EXERCISE 8 10 Teacher to provide
1 155,58 km
2 2,74 km TOPIC 11: EXERCISE 1
3 57,51 m 1 1.1 R10 660
1.2 R10 797,59
TOPIC 10: EXERCISE 9
2 2.1 R561 150,73
1 1.1 x = 14,13 units
2.2 R357 856,19
1.2 20,04 units
1.3 ^ = 79,33°
D 3 3.1 8,33%
3.2 16,40%
2 2.1 TB = 10,5 units
2.2 ^ B = 48,59°
AT 4 3 years
2.3 21,07 units 5 R42 574,26
2.4 47,60 units2
6 R483 283,00
3 3.1 ^ = 31,9°
B
3.2 r = 8,51 units 7 For Tim: R11 294,30
For Thabo: R13 980
4 4.1 Teacher to provide For Tracy: R11 019,96
4.2 Teacher to provide For Thandi: R10 683,33
5 5.1 Area △OAB = 1 __r2tan θ Thabo got the best return and Thandi got
2
5.2 Area △OAC = __ 1 r2sin θ the worst.
2
5.3 tan θ > sin θ 8 8.1 8.1.1 2,27%
6 6.1 2a2 (1 − cos x) 8.1.2 6,28%
6.2 2b2 (1 + cos x) 8.2 335 rhinos left in 2050.
6.3 Teacher to provide
TOPIC 11: EXERCISE 2
TOPIC 10: REVISION 1 1.1 10,9%
1 1.1 32,16 cm 1.2 6,7%
1.2 x = 77,20° or x = 102,8° 2 2.1 7,85%
1.3 x = 25,19 2.2 7,77%
1.4 x = 77,73°
3 3.1 R46 018,94.
1.5 Area △ABC = 237,85 cm2
Answers 367
PLT MATHS LB 11 7th pgs (Real Book).indb 367 2012/07/02 2:27 PM
3.2 R21 018,94 12 R12 389
3.3 10,7%
13 T0 T5 i(12) = 0,12 T7 T10
4 6,3% p.a. compounded monthly
x
5 R29 587,21 P1 –30 000
P2 –32 500
6 R162 038,22 P3 –36 000
7 T0 T2 T5 i(4) = 0,0951T7 T10 R41 510,55
ieff = 0,1
8 000 14 14.1 R20 222,73
6000 14.2 10,47% p.a.
–10 000
14.3 R5 222,73
x
15 R3 648,87
R16 856,70
16 16.1 R10 356,95
8 T0
i(12) = 0,1 T2 T3 ieff = 0,115
T5
16.2 R3 549,64
x
30 000
16.3 R6 807,31
113 582,40
17 17.1 R961,52
17.2 6,67% p.a.
R92 348,44
TOPIC 12: EXERCISE 1
T0 T1 T3 T5
9 ieff = 0,14 i(4) = 0,105
___
1
1 1.1
Lening 12
= 150 000 5 000
10 000 1.2 __
3
x 4
1.3 ___
5
12
R253 122,40 2 2.1 No, because the events are not exhaustive
(4 is not included in either event).
TOPIC 11: REVISION 2.2 Yes, because the events are mutually
1 10% p.a. exclusive and exhaustive.
2.3 No, because the king of hearts belongs to
2 4,26 years
both events, which means that the events
3 8,6% are not mutually exclusive.
4 4.1 R486 230,87 3 3.1 These are dependent events; without
4.2 10,47% replacement means that when selecting
the second card, there will be one less card
5 R 6 915,88
to choose from, and the probabilities will
6 R33 454,56 depend on the type of card that was selected
in the first draw.
7 7.1 R8 823,53 = cash price
3.2 These are independent events; the outcome
7.2 R2 858,82
of tossing the coin (heads or tails) will not
8 5,6 years be affected by the number that was thrown
9 11,61% p.a. on the die.
4 4.1 ____
64
10 10.1 R17 240,22 per ticket 169
10.2 R29 718,84 4.2 ____
40
169
11 T0 T2 T5 T7 4.3 105
____
ieff = 0,05 i(2) = 0,06 169
5 000 5 5.1 P(blue and blue) = P(blue) × P(blue | blue
2 000 8 × ___
first) = ___ 7 = ___
14
x 13 12 39
5.2 P(blue and green) = P(blue) × P(green | blue
R9 545,90 5 = 10
8 × ___
first) = ___ ___
13 12 39
368 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 368 2012/07/02 2:27 PM
TOPIC 12: EXERCISE 2 3.2 0,6 × 50 = 30 3.3 x = 0,1
3.4 3.4.1 P(A ∩ B) = 0
1 120
3.4.2 P(A ∪ B)′ = P(not A or B) = 0,1
F S
∴ n(A ∪ B)′ = 0,1 × 50 = 5
3.4.3 P(A ∪ B′) = P(A or not B) = 0,6 + 0,1
29
= 0,7
45 28
3.5 Teacher to provide
4 4.1 70
S M
x
1.1 45 + 28 + 29 + x = 120 9 18 17 – x
∴ x = 18
x
1.2 1.2.1 ____
45 = __
3
120 8
5–x 8
1.2.2 ___________
45 + 28 + 29 = ___
17
120 20
1.2.3 ____
28 = ___
7
120 30 12
1.3 1.3.1 P( F ∪ S )′ = P(not F or S) L
6
= ____
18 = ___
3
120 20 70
4.2 S M
1.3.2 P( S ∩ F’ ) = P(S and not F)
= ____
29
18
120 9 12
1.3.3 P(F ∪ S’) = P(F or not S)
5
___________
= 45 + 28 + 18 = ____
91
120 120 0 8
1.4 Teacher to provide
2. 2.1 A = {1; 2; 3; 5; 6; 10; 15; 30} 12
B = {1; 3; 5; 7; 9; 11; 13; 15; 17; 19; 21; 23; L
25; 27; 29} 6
C = {1; 4; 9; 16; 25} 4.2.1 5 learners take Maths, Physical
2.2 A ∩ B = {1; 3; 5; 15} Sciences and Life Sciences
2.3 B ∩ C = {1; 9; 25} 4.2.2 ___
57 4.2.3 ___
23
70 43
2.4 n(A ∩ B ∩ C) = 1 ___
12
4.2.4 18 4.2.5
2.5 25
A B 30 ___
8
4.2.6
13
5.1
D C 50
4 3 9
1 8
14 13
0 2
2
2 1
2
C
9
7
2.6 2.6.1 ___
8 = ___
4 2.6.2 ___ = ___
12 6
B
30 15 30 15
3
2.6.3 1 – ___
5 = __
5 2.6.4 ___
9 = ___
3
30 6 30 10
5.2.1 n(D ∪ C ∪ B)´ = 3
2.7 Teacher to provide
5.2.2 n(D ∩ C´) = 14 + 2 = 16
3 3.1 None of the shoppers interviewed like to shop 3
5.2.3 P(B | C) = ___
at both ‘Perfect Purchases’ and ‘Shopaholics’. 50
48 = ___
5.2.4 P(D ∩ C ∩ B)´ = ___ 24
50 25
Answers 369
PLTMATHSLB11LB_16.indd 369 2012/07/14 3:02 PM
6 6.1 As the total probability = 1, we can calculate: 1.5 Supplier A has marginally less probability of
1 − (0,14 + 0,06 + 0,1 + 0,02 + 0,04 + 0,44 supplying defective pens.
+ 0,08) = 0,12
2 2.1 Hockey
6.2 6.2.1 0,1 × 50 = 5 Hockey
6.2.2 0,12 × 50 = 6
will Total
will win
lose
6.2.3 0,02 × 50 = 1
6.2.4 0,44 × 50 = 22 Rugby will win 21 9 30
6.2.5 0,16 × 50 = 8 Rugby will lose 14 6 20
Total 35 15 50
TOPIC 12: EXERCISE 3
1 Teacher to provide 2.2.1 ___
21
50
1.1 6 ___
14 = ___
7
2.2.2
1.2 2 50 25
___
6 = __
1 2.3 P(win hockey and win rugby) = ___ 21 ; P(win
1.3
12 2 hockey) = ___ 30 50
35 ; P(win rugby) = ___
50 50
2 2.1 When drawing the first blue marble there ___
35 × ___
30 = ___
21 P(win hockey) × P(win
50 50 50
are 6 blue marbles out of a total of 20 rugby) = P(win hockey and win rugby)
6 . Because
marbles, so the probability is ___
20 Therefore winning hockey and winning
the marble is not replaced, the probability
rugby are independent events.
of drawing a blue marble with the second
drawing will depend on the colour of the
TOPIC 12: REVISION
first marble drawn. If it was blue, then
there will be 5 out of 19 possibilities of 1 1.1 a and d, or b and d
drawing another blue marble, but if the first 1.2 b and c
marble had not been blue, the probability 1.3 a and e
of drawing a blue marble with the second 2 2.1 100
drawing would be ___6.
20 Coconut Rainmaker
2.2.1 ___
18 2.2.2 ___
3
95 38
2.2.3 ___
3 2.2.4 21
___ 67 – x x 59 – x
19 38
3 3.1 __
5
8
3.2 No. Once the head prefect has been chosen 22
there will only be 7 learners left to choose 2.2 x = 48
from for the deputy, and the number of
2.3 ____
22 = ___
11
boarders and day scholars left to choose 100 50
from will depend on whether the head 2.4 ______
59 − x = ____
11
100 100
prefect was a boarder or day scholar.
2.5 ___
48
___
3 9
___ 59
3.3 3.4
28 14 3 3.1
1
__ Red
TOPIC 12: EXERCISE 4 2
1 1.1 1
__ Red Black
Supplier Supplier 2 1
Total __
A B 2
Defective 12 6 18 1
__ Red
1
__ 2
2
Not defective 338 144 482 Black Black
1
__
Total 350 150 500 2
3.2 __
1 3.3 __
2 = __
1
1.2 ____
18 = ____
9 1.3 ____
12 = ____
6 4 4 2
500 250 350 175
1.4 ____
144 = ___
24
150 25
370 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 370 2012/07/02 2:27 PM
3.4 __
2 = __
1 8 8.1 Teacher to provide
4 2
8.2 0,4
3.5 __
1
4 8.3 0,5
3.6 __
3 8.4 P(P) = 0,6 P(P) = 0,3 and P(B and T) = 0,1
4 0,6 × 0,3 = 0,18 so P(T) × P(B) ≠ P(T and B)
4 4.2 ___ __
10 = 1
50 5 so they are not independent events.
4.3 ___ __
15 = 1 8.5 P(B or T) = P(B) + P(T) – P(B and T)
30 2
0,3 + 0,6 – 0,1 = 0,8
4.4 __
1
5
5 1
__ TOPIC 13: EXERCISE 1
3 Win
1
__ Win 1 1.1 1.1.1 158,29 cm
3 1
__ Lose 1.1.2 150 ≤ x < 160
3
1.1.3 40 cm
1
__
1
__ 3 Win 1.4 Teacher to provide
3
Lose 1.5 1.5.1 157 cm
1
__ Lose
3 1.5.2 151 cm
1.5.3 166 cm
P(LL) = __
1 × __
1=1__
3 3 9 1.5.4 15 cm
1.5.5 174 cm
6 6.1
3
___ Mathabane 1.6 Teacher to provide
22 ___
9
22
Mathabane Matthee
TOPIC 13: EXERCISE 2
10
___
4
___ 22 McClure 1 1.1 25,75
23
8
___ Mathabane _____
√_____
9
___ 22 4
___ 759,5
23 22 Matthee
1.2
Matthee 8
10
___
22 McClure Time taken __ __
(xi – x) (xi – x)2
10
___ in minutes
23 4
___ Mathabane
22 9
___ 12 –13,75 189,0625
McClure 22
Matthee
15 –10,75 115,5625
9
___
22 McClure 19 –6,75 45,5625
6.2 6.2.1 ____
36 24 –1,75 3,0625
253
25 –0,75 0,5625
6.2.2 ____
87
253 33 7,25 52,5625
6.2.3 ____
91
253 38 12,25 150,0625
6.2.4 ____
175 40 14,25 203,0625
253
759,5
7 7.1 x=4
Standard deviation = 9,74; Variance = 94,94
7.2 7.2.1 16
1.3 four data values
7.2.2 ___
13
15 2 2.1 1,19
7.2.3 ___
17 2.2 1,42
30
2.3 2,87
7.2.4 ___
5
12
7.2.5 ___
12
41
7.2.6 ___
1
12
Answers 371
PLT MATHS LB 11 7th pgs (Real Book).indb 371 2012/07/02 2:27 PM
TOPIC 13: EXERCISE 3 1.6 In suburb B, the data values are mostly
1 1.1 Minimum value = 1 452 bunched on the left (mostly low values)
Maximum value = 140 252 with a long whisker on the right, so the data
Lower quartile = 1 984,5 is skewed to the right, with a range of 370
Median = 11 694 and an IQR of 80. In suburb A, the range is
Upper quartile = 17 566,5 narrower (170), and so is the IQR (55).
1.2 Teacher to provide The spread is closer to being symmetrical,
1.3 very wide range (138 800), and is skewed to but the mean is slightly less than the
the right. median, with the median closer to the lower
1.4 very long whisker on the right quartile than the upper quartile, so the data
1.5 IQR = 17 566,5 – 1 984,5 = 15 582 is skewed to the right.
17 566,5 + 1,5(15 582) = 40 939,5 1.7 Suburb B is more likely to have an outlier
So Ekurhuleni (140 252) is an outlier. because of the long whisker to the right.
1.6 many job opportunities 2 2.1 1 × 5 + 2 × 10 + 3 × 15 + 4 × 10 + 5 × 5 = 3
___________________________________
45
1.7 So the City of Johannesburg, the City
2.2 __ __ __
of Cape Town, the City of Tshwane and xi f (xi – x) (xi – x)2 xi × (xi – x)2
Ekurhuleni are all outliers. This is a bigger
1 5 –2 4 20
sample, including a wider variety of
municipalities, so it will reflect a better result. 2 10 –1 1 10
1.8 3 15 0 0 0
In-migration
Frequency 4 10 1 1 10
groups
0–30 000 14 5 5 2 4 20
30 000–60 000 0 Total 45 … … 60
60 000–90 000 1
2.3 1,15
90 000–120 000 0 2.4 The results in the second test were very
120 000–150 000 4 widely spread, with a greater discrepancy
1.9 Teacher to provide between results than in the first test.
1.10 all cities, many people migrate towards 3 3.1 Minimum value = 65;
them in search of employment. Maximum value = 129
2 2.1 45,16 So LQ = 79,25; Median = 89,5; UQ = 108
2.2 29,16 3.2 Teacher to provide
2.3 24 values 3.3 A: True
2.4 IQR = 39,15 B: False: Each quartile represents 25% of the
2.5 No outliers. data. The values in the third quartile
2.6 median < mean, so the data is skewed to are spread out more than in the second
the right quartile.
4 4.1 200
TOPIC 13: REVISION 4.2 60
1 1.1 1.1.1 125 4.3 27
1.1.2 102 4.4 36 – 18 = 18
1.1.3 157 4.5 20 to 30
1.2 Teacher to provide 4.6 It is impossible to get less than –9 or greater
1.3 131 than 63 for a test out of 60, so there were no
1.4 42,03 outliers.)
1.5 ˙%
73,3
372 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 372 2012/07/02 2:27 PM
4.7 1.1.3 x = 2 only
Mid-class Cumulative
Frequency 3
__
value Frequency 1.1.4 x = ± ( 22 )2 = ± 8
5 20 20
1.1.5 x=7
__ = 0,875
8
15 60 40
25 120 60
1.2 3 ;__
(1;−1) and __ 1
(5 5)
35 170 50 1.3 1.3.1 x ∊ (−2;6)
45 190 20 1.3.2 x = −2 or x = 6
1.4 The roots are real, irrational and unequal.
55 200 10
1.5 1.5.1 3
4.7.1 27 1.5.2 2x
4.7.2 12,88 2 2.1 2.1.1 28; 33
5 5.1 2.1.2 Tn = 5n − 2
Cumulative
Expenditure Frequency 2.2 2.2.2 Tn = 2n2 − n − 3
frequency
2.2.3 n = 17 or −16,5
x < 50 2 2
2.3 54 − x ; 66 − 2x
50 ≤ x < 100 22 24 3x = 87 and x = 29
100 ≤ x < 150 58 82 2.4 2.4.1 6; 18; 54; 162; 486; 1 458
150 ≤ x < 200 14 96 2.4.2 Tn = 2 × 3n
200 ≤ x < 250 4 100 2.4.3 Tn = 2 × 3n + 5
3 3.1 y ≤ 8, y ∊ ℝ or y ∊ (−∞;8)
5.2 y 3.2 g(−1) − h(−1) = 6
100 - 3.3 x = 1 and y = 2
-
- 3.4 x = 3 or –3
-
80 - 3.5 (−1;0)
Cumulative frequency
-
- 3.6
- y
60 - f
-
50,5 - (–1;8)
-
40 -
-
- h
-
20 -
-
-
- y=2 2,5
0-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- x
50 100 150 200 250
median
Cell phone expenditure (R) x
–5 –1 3
5.3 Median read off from the ogive is
–2
approximately R122.
5.4 Teacher to provide
5.5 123
5.6 almost symmetrical
6 6.1 The rainfall decreases with increase in x=1
temperature. g
6.2 The reading 30 °C and 80 mm rainfall
3.7 4 −2
v( x ) = − _____
x−1
TERM 4: EXAMINATION PAPER 1 3.8 1 ( x − 4 )2 + 4
w( x ) = − __
2
1 1.1 1.1.1 x = 1,5 or x = 2 4 4.1 1 x2 − __
y = __ 1x − 2
3 3
5
(5)
1.1.2 ____________ 1 x+1
2(x2 + 2x + 4) 4.2 g( x ) = 3 __
Answers 373
PLTMATHSLB11LB_16.indd 373 2012/07/14 3:04 PM
4.3 [2 )
1 ;∞ or x ≥ __
x ∊ __ 1, x ∊ ℝ
2
2 y
4.4 x ∊ [ −∞;−2 ] ∪ [ 3;∞ ) or x ≤ − 2 3 D
or x ≥ 3, x ∊ ℝ
6
5. 5.1 5.1.1 i = 12,5% F A(6;1)
C(–11;2)
5.1.2 i = 9,05% θ α x
5.2 A = R1 293 194,45 6
E
5.3 R12 094,38
B(3;–5) 3
6 6.1 x=2
6.2 3 people
2.1 ^ C = 90°
AB
6.3 17
5 1 x − __
y = __ 1
6.4 6.4.1 P(T ∩ C)′ __ 2.2
4 2
6
6.4.2 3
P(B | T) = ___ 2.3 D = (–8; 8)
13 ____ ____
6.5 6.5.1 2.4 AC = √ 290 ; BD = √290 = AC
0,15 arrives late 2.5 ABCD is a rectangle; parallelogram with
equal diagonals or parallelogram with
Calls Call a cab
90° vertex
0,3
0,85 arrives on time 2.6 3
5 ;__
F = − __ ( 2 2)
arrives late 13
0,1 2.7 y = 2x + ___
0,7 2
Calls Terrific Taxis 2.8 3
y = __
2
2.9 AB^ D = 66,8°
0,9 arrives on time
6.5.2 P(Call a Cab and On Time) = 51
____ 3 3.3 3.1.1 sin θ = − 3
__
4
200
y
TERM 4: EXAMINATION PAPER 2
1 1.1 1.1.1 Median = 2 300 000 3__
tan θ = ___
1.1.2 LQ = 1 500 000; UQ = 3 850 000 ; √7
IQR = 2 350 000
1.1.3 Teacher to provide – 7 0
1.1.4 The data is skewed to the right.
x
1.1.5 Mean = 2 622 222,22
–3
1.1.6 Standard deviation = 1 512 744,22 4
1.1.7 No outliers.
1.2 1.2.1 Teacher to provide
y
10 3.1.2 1
__
√3
___
3.2
8 2
Cumulative frequency
3.3 3.3.1 1
_____
6 cos θ
3.3.2 RHS
4 3.4 3.4.1 x=0
3.4.2 x = n.180° or
2 x = 53,13° + n.360°, n ∊ ℕ
x
50 55 57,6 60 65
Percentage of voter turnout
1.2.2 There were four provinces below the
mean, so there were five provinces
above the mean.
374 Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 374 2012/07/02 2:28 PM
4 4.1 Teacher to provide
4.2 4.2.1 −90° < x < 0° or 90° < x < 180°
−180° ≤ x < −135° or 0° ≤ x < 45°
or x = 180°
4.2.2 including –180°, 0° and 180°;
excluding –135° and 45°
4.2.3 –4y ≤ –2
5 5.1 area △ADC = 2(m)2 sin α) sin θ
5.2 AC = 29 cm
6 6.1 s = 2,55 cm
6.2 2 π( 0,5 )2 + π( 0,5 )2( 10 ) + __
V = __ 1 ( π )( 0,5 )2( 2,5 )
3 3
6.3 SA = 31,42 cm2
23 π = 9,03 cm3
___
8
7 7.1 ^P
^ B = RQ
RP
7.2 7.2.1 Angle in semi-circle
7.2.2 BCDH is a cyclic quadrilateral
C
B F
1 3
2
1
2 A
3 4
2 3
2
D1 1 H 12
0
21 G
E
7.2.3 ^ = 40°
A
7.2.4 ^ = 50° − x
A 3
7.2.5 ^ =B
A 1 2
8 8.1 ^ = 2B
F ^
8.2 OC = x – 8 = 5 mm
Answers 375
PLT MATHS LB 11 7th pgs (Real Book).indb 375 2012/07/02 2:28 PM
Glossary
corollary a deduction based on the result of a theorem
A
amplitude half the total distance between the minimum cumulative frequency a running total of the frequencies
and maximum values; it is always positive cyclic quadrilateral a four-sided figure with all four points
angle of inclination the angle that a line makes with the lying on the circumference of a circle
positive direction of the x-axis
appreciation growth in value; the opposite of depreciation D
arc a part of the circumference of a circle decrease to grow smaller
asymptote a line that cannot be crossed by a graph; it tends degree (in algebra) the highest index in an equation or term
towards, but never touches, a graph
depreciate to lose value
axis of symmetry a line that divides a graph into
two identical halves and about which a function is diagonal a straight line joining opposite vertices
symmetrical discrete data data with a restricted number of values;
includes units of quantity (for example, number of
B people), with only whole number values
bar graph a graph with rectangles that show the frequency domain the set of all x-values of a graph or function;
with which different types of data values occur; data excludes any values of x which make y undefined
values need not be in consecutive groups, so the bars do
not touch each other
bisect to cut or divide into two equal parts
E
effective interest rate the actual rate achieved per annum
book value the depreciated value of a vehicle at any
specified time equidistant the same distance
equilateral triangle a triangle with all three sides equal and
C all three angles equal
Cartesian plane a two-dimensional plane formed by exhaustive events events which use up the full sample
a horizontal number line representing x-values and
space, that is, all possible outcomes
a vertical number line representing y-values, which
intersect at the point where both values are 0; ordered experiment a trial situation involving chance
pairs or coordinates refer to the position of points in exponent (index) a number or variable that shows how
the plane many times a number (the base) is multiplied by itself
chord a line with both endpoints on the circle
exponential equation an equation with unknowns or
coefficient a number that stands in front of a variable or variables in the exponent
variables in a term
exterior angle outside angle of a polygon formed by one of
collinear points that lie on the same line; the gradients
the sides which has been extended
between any two of these points will be the same
compounded quarterly interest is applied to an investment
every 3 months, and __ 1
of the quoted annual rate is
F
4
applied to the total at each stage frequency polygon a polygon that demonstrates the
compounded semi-annually semi-annually means every frequency of each group in a set of grouped data, and
half year (every six months), which means that interest is that displays the spread of the data
twice a year function a relationship between two variables, usually x and
concentric circles that share the same centre y, where for every value of x there is one corresponding
congruent identical in all respects value for y
construction a line which needs to be added to a sketch
continuous data data without breaks and that can have any G
value; includes all units of measurement (for example, general solution the formula which lists all possible
height, mass) solutions to a trigonometric equation; takes into account
converse reversing the logic and proving a theorem in the period of the trigonometric function so an angle can
reverse be positive or negative
376 Glossary
PLT MATHS LB 11 7th pgs (Real Book).indb 376 2012/07/02 2:28 PM
perpendicular height the perpendicular distance between
H
bases or between the vertex of a polygon and the base
hemisphere half a sphere
opposite it
hence sometimes used to show that you should make use of
produce to lengthen a straight line; for example, AB
the answer that you have found
produced to C means that ABC is a straight line
histogram a graph with bars that show the frequency
distribution of grouped data with no space between Q
the groups, so the bars touch each other; groups are
quadratic equation an equation in which the highest index
usually equal in width so that the height of the rectangle
is squared
(frequency distribution) is the same as the frequency
horizontal sideways
R
horizontal shift when sideways, with no up or down radius a line from the centre of circle to its circumference
movement
range the set of all y-values of a graph; the y-values of the
hypotenuse the side opposite the right angle in a right- graph which exclude any values which make y undefined
angled triangle
rational exponent an exponent or index with a rational
number as the exponent
I real numbers the set of all rational and irrational numbers
included angle the angle between two known sides
reciprocal the multiplicative inverse of a number; obtained
increase to grow bigger by interchanging the numerator and the denominator of
interior angle inside angle that number
intersect to meet or cross at a point; to overlap (sets) reduction formulae trigonometric identities that express
interval spacing the number of degrees between critical the trigonometric ratios of an angle of any size in terms of
points on a trigonometric graph the trigonometric ratios of an acute angle
reflect to create a mirror image or reflection
M reflection a mirror image about a line of symmetry
maximum the largest possible value reflex angle an angle greater than 180°
minimum the smallest possible value revolution the sum of the angles around a point
roots (of an equation) the solutions of the equation
N
n the position of the term in a pattern S
nth term the term in the position n sample space the full set of data values
negative exponent a number raised to a negative exponent scientific notation the form in which you use exponents to
and the reciprocal of that number with a positive write very large or very small numbers
exponent scrap value the value of an item after it has passed its useful
nominal interest rate the annual interest rate that is quoted life in its original form, and certain parts can be sold as
non-real numbers imaginary numbers or complex numbers scrap
that are not real numbers solution the value of x or any other variable which solves
an equation; a value of the angle which satisfies a given
O trigonometric equation
obtuse angle any angle between 90° and 180° solve a triangle to calculate the lengths and sizes of the
unknown sides and angles
outlier a data value that does not follow the trend of the
rest of the data specific solutions solutions which satisfy a given
trigonometric equation in a restricted interval
P square the product of two identical factors
parameter a variable that restricts or gives a particular form standard deviation the square root of the variance
or shape to the equation it characterises subtended by an arc formed by an arc
per annum for the year supplementary angles that add up to 180°
period the number of degrees needed to complete a wave surds irrational roots or irrational numbers; the square root
pattern or cycle of a non-perfect square is an irrational number
perpendicular at right angles or at 90°; lines are symmetry line a fold line that makes the shapes the same
perpendicular when the angle between the lines is 90° on both sides
Glossary 377
PLT MATHS LB 11 7th pgs (Real Book).indb 377 2012/07/02 2:28 PM
T
tangent a straight line which touches a circle, but does not
pass through it
theorem a formal proof of a geometric statement
theoretical outcome the result, in theory, which is expected
for a certain event to happen
trigonometric equation an equation involving
trigonometric ratios which is true only for certain values
of the unknown variable
trigonometric identity an equality which is true for all
values of an unknown variable, for which both sides of
the identity are defined (so no zero denominators)
U
universal set the sample space, which is the full set of data
values
V
variance the average of the squared differences from the
mean
vertex a point on a triangle where the sides meet, so a
triangle has three vertices; a point where two straight
lines meet to form an angle
vertical straight up or down
vertical shift straight up or down, with no sideways
movement
vertical shrink grows smaller without changing the
y-intercept
vertical stretch grows taller without changing the
y-intercept
X
x-intercept, root the point where the graph cuts the x-axis
(where y = 0)
Y
y-intercept the point where the graph cuts the y-axis
(where x = 0)
378 Glossary
PLT MATHS LB 11 7th pgs (Real Book).indb 378 2012/07/02 2:28 PM
Index
A D
altitudes 57−8, 79 data values 256, 259−61, 288, 296, 299−301, 305, 307,
amplitude 116−7, 119, 121−4, 130−1, 135−7, 169, 376 308, 311, 313, 319
angles decay 175, 236−40, 242, 250, 286
corresponding angles 64−5, 188, 192, 198, 206 decimal 19, 21, 25, 30, 44, 250
acute angles decimal places 19, 24, 30, 44
central 149−50, 155 degrees 116, 119, 123−4, 126, 130, 135−6, 376
co-interior 188, 192, 198, 206 depreciation 236−7, 240–1, 250, 282, 322, 376
complementary 148, 153, 188 annual rate of 240−2
equal 64, 188, 195, 198−9, 205 straight line method of 236−7, 250
exterior 65, 188, 195, 198−199, 203, 205 depression 223−4, 227
negative 146−150, 155−6 angle of 223−4, 227
opposite 189, 197−8, 205 difference 14−6, 27, 46, 48, 52, 79, 108, 288, 290, 301
positive 138, 146, 156 common 7, 27, 29, 46, 48, 51−2, 79, 127, 159, 191,
reference 63, 65 205−6
applications of triangle rules 225, 228−9 first 48, 79
area rectangle 176, 179 squared 301−302
area rules 175, 214, 220, 223, 286 dispersion 308, 310, 313
asymptotes 92−4, 97−107, 124, 130−1, 134−6, 376 distance 19, 34, 56, 69, 79, 107, 116, 119, 123−4, 126,
average gradient 81, 111−3, 115, 168 130, 136, 176−7, 189, 191, 193, 202−3, 129–135, 317,
axes 83−4, 188−9, 91, 94−7, 100, 102−3, 105 376
axes of symmetry 94−6, 376 domain 92−3, 98−101, 122, 124−5, 134, 166
B E
base 4, 7, 11, 21, 55, 176−9, 184−5 effects of parameters 81−2, 84, 86, 88, 90, 94, 96, 98,
bearing 228−9, 235 elevation, angle of 223−4, 233−234
box-and-whisker diagram 300, 306–310; 313–314, 316–317 events 175, 252−6, 259, 262–7, 269−5, 277, 283, 287
298, 304−5, 308−9, 312, 316 complementary 253, 267, 285
exclusive 252
C independent 175, 252−4, 257, 259–61, 270, 272, 274−5,
calculator instructions 116, 132−3 281, 285
CAST diagram 139, 146−7, 149, 155 exponential equations 7, 376
circle 22, 138, 176, 177, 179, 185, 190−9 exponential graphs 81, 101, 103−7
circumference 176, 179, 190, 192, 194−5, 197−9 shifted 104
concentric 193, 210 exponents 3−14, 16, 18, 20−1, 53, 78, 85, 87, 89, 91, 170
diameter 34, 179, 186, 198−200 negative 5−6
geometry 3, 22, 56, 64, 79, 175, 188, 223, 229 positive 5−6, 20, 170
reflex angle 194
circumference 176, 179, 190, 192, 194−5, 197−8, 200, F
209−10, 212 factorisation 24, 27, 29−30, 42, 87
coefficient 12, 24−6 factors 4, 22, 27, 31, 42−3, 53, 87−8, 253
common factor 7, 27, 29, 159 Fibonacci sequence 70, 72
complementary rule 175, 252−3 finance 175, 236, 238, 240−8, 286
compound decay 175, 236−9, 241, 286 first degree equations 27
compound growth, periods of 175, 242−5, 249 form
compound interest 108−9, 241−2, 250, 284 exponential 9−10, 14, 239
congruent 176, 188, 195 radical 9−10
constant factor 176, 179, 181, 185, 187 standard 56, 78, 85−6, 90, 168
contingency table 269−72, 273– 5, 283, 287 formula 22, 26−7, 30−1, 42−4, 46−9, 51−2, 54, 56, 59−60,
converse 188, 190−1, 195, 197, 202, 376 62, 75, 78−9, 81, 87, 90, 99, 108, 112, 146−9, 151−3,
cos graphs 132 158, 166, 169, 171, 176−8, 183−7, 218, 237, 239−43,
cos, cosine 19, 18, 116−119, 127, 159−60, 175, 214−5, 245, 284, 299−303, 319−23
217−8, 221, 223, 226, 286 general 46, 48
cosine and area rules 175, 214, 223, 284 x-intercept 90
cumulative frequency 296−8, 314, 316, 319 formulae 52, 56, 79, 81, 146−53, 69, 176, 178, 245, 182,
cyclic quadrilateral 175, 194, 197−9, 376 286, 301, 303, 305, 319
Index 379
PLTMATHSLB11LB_17.indd 379 2012/07/07 11:54 AM
frequency 244, 287−98, 303−4, 307, 311, 314−9, 323−4 midpoint 56−60, 67−9, 79, 172, 175, 179, 190−1, 193,
relative 290−1 198, 203, 208, 210, 285, 317−8
total 297, 299 chord 175, 190−3, 195−6, 198−9, 202−7, 210−11,
frequency distribution 290, 377 229−30, 285
frequency polygons 289−96, 300–1, 306−7, 315, 317−318, formula 59
376 mode 102−3, 109, 115, 133, 136−7, 251, 296, 299, 303−4,
frequency table 291, 294, 299−300, 311, 324 306−7
functions 81−2, 84, 86−112, 116−8, 120−8, 130−8, 167,
325, 376 N
trigonometric 81, 123, 138 number 3−5, 9, 11, 14, 17, 21, 24−5, 30, 32−3, 34−6, 42,
44−56, 70−75, 78−9, 102−3, 108, 123, 133, 140, 171,
G 237, 239, 241, 244, 252−7
gradient 56−69, 76−7, 79, 81, 99−100, 111−15, 168, 172, imaginary 24−5
237, 239, 299, 308 irrational 14, 24−5
graph, amplitude, period, steps, asymptotes, shift, domain natural 4, 9, 25
134 rational 17, 25
graph, amplitude, period, steps, horizontal shift 132 real 24−5
graph, amplitude, vertical shift 119 number patterns 3, 46, 48, 50, 54, 79
graph equation 124−5, 135
graph, turning point, asymptotes, domain 168 O
graphs 35−8, 40, 45, 78, 81−2, 85−9, 93, 98, 100−1, obtuse 62, 158−60, 215, 217
103−8, 111, 114, 116−22, 124−28, 130−7, 157, 166, obtuse-angled triangle 215, 217
168−9, 173, 290−1, 325 ogive curve 287, 294−98, 314–315, 319, 324
bar 290 opposite sides 59−60, 189, 205
reflected 101, 115, 121−2, 126−7, 136, 166, 171 parallel 56−61, 64−5, 67−9, 77, 79, 107, 176, 188, 189,
sketching 87, 104, 125, 133, 135 198, 205−6, 211, 223, 279
H P
histograms 290−4, 300, 306, 311, 318, 377 parabola 22, 36, 40, 81−3, 85−7, 89−92, 99, 101, 111−13,
horizontal 35, 50−1, 56, 81, 83−6, 91−6, 99, 101, 103−4, 168, 171
125, 127, 129, 132−3, 168−9, 223−4, 377 parallel 57−61, 64−5, 67−9, 79, 176, 188−9, 192, 198,
horizontal asymptote 92−6, 99, 101, 104 205−6, 211, 223, 279
horizontal shift 81, 85−6, 91, 93−4, 103, 125, 127, 129, parallelogram 58—61, 67−9, 77, 189, 205, 281, 324
132−3, 135, 168−9, 377 parameters 81−4, 86−8, 92
hyperbola 40, 81, 92−5, 97−9, 101, 112, 168 pattern 3, 46−54, 75, 79, 85, 93, 116, 118−9, 123−7,
135−6, 159−60, 169, 171, 308, 315, 320
of answers shifts 85, 93, 125
I cubic 27, 52, 55
inequalities 3, 22−4, 26, 28−42, 78
exponential 48
interest 108−9, 237, 239−51, 282, 286, 301−302, 321−2
second difference 7, 48−9, 51−2, 54, 79
changes 245, 248−9
perpendicular 56−8, 60−1, 67−9, 76, 79, 172, 175, 176−8,
interest rate 240, 242−6, 248−51, 282, 286
189, 190−3, 198, 202, 207, 277, 279, 324−5, 377
effective 242−3, 250−1, 286
bisector 57−8, 60−1, 67−8, 76, 79, 172, 191, 198, 279,
effective annual 242−3, 249, 252, 282
324
nominal 242−3, 249, 286, 377
gradient 57, 58, 60
interior opposite angle 197−8, 205
height 176−8, 189, 286, 288
interquartile range 294, 297−8, 307, 311, 315, 323 lines 56, 79, 175, 190−1, 193
intersection 40, 58, 60−1, 67, 69, 98, 100, 114, 212, 252, polygon 52, 55, 176−7, 197, 290−301, 306−7, 315, 317−8
259−60, 324 positive ratio 155, 158
prime number 253−4, 263
L probability 252−5, 257−75, 280, 287, 322
laws 3−5, 7, 9, 20, 78, 85, 87−9, 91 of event 252, 273, 287
of exponents 3−5, 7, 20, 85, 87, 89, 91 Pythagoras’ Theorem 139, 141−2, 151, 177, 179, 181−2,
linear equations 27 188, 190−2, 200, 214, 217
linear patterns 3, 46−7, 79
literal equations 26
Q
quadrant 59, 138−9, 141, 146−50, 155−6, 158−61, 169,
M 377
median 57−8, 60−1, 67, 76, 79, 294, 296, 298, 306−9, 312, quadratic equation 22, 24, 27, 30–2, 42, 77, 377
315, 316, 319, 323−4 quadratic patterns 3, 48−53, 79
380 Index
PLTMATHSLB11LB_17.indd 380 2012/07/07 11:54 AM
quadrilateral 68, 175, 189, 193−4, 197−201, 203−5, 207, axis of 82, 86−7, 91, 107, 114, 168
110−13, 221, 231−2, 279, 281, 291 symmetry lines 94, 97−100, 107, 114, 166
opposite angles 189, 197−8, 205, 207
R
radius 138, 176−8, 180, 183−7, 190−3, 199−200, 202−4, T
206, 209, 211, 229, 231, 234, 278, 284, 377 tan graphs 119, 127, 132
range 92−3, 98−101, 116−24, 126, 129−31, 135−7, 166−9, tangents 175, 202, 204−9, 213, 279, 285, 378
252, 294, 197−8, 301, 306, 307, 309, 311, 315–6, 319, theorem 139, 141, 151, 175, 177, 179, 181−2, 188, 190−1,
321, 323, 325, 377 194−5, 197, 201−2, 217, 230, 326, 378
rational exponents 3, 9, 11, 21, 78, 377 time line 245−49, 286
ratios 3, 70−1, 73, 138, 140, 142, 146, 148−9, 151−3, tree diagram 175, 263−8, 275−6, 287, 322
158−61, 169, 286, 377 triangle rules 223, 225, 228−9
given 19, 23−4, 31−2, 53, 58, 66, 75−9, 82, 87, 100, triangles 116, 152, 178, 214, 217, 223, 228, 233, 284, 286,
104, 106, 107, 119−23, 139−40, 151, 153−60, 166, 291, 293
170−1, 173, 190−1, 194−5, 197, 199, 202, 214 acute-angled 215, 217
required 151−2 equilateral 57, 185
reduction formulae 81, 146−7, 149−52, 169, 377 non-right-angled 286
reflection 82−4, 94−7, 101−2, 107, 114, 127, 168, 321, 377 right-angled 138, 177, 188, 191−2, 214, 221, 223, 226,
revolution 188, 196−7, 377 286
roots 3, 9, 14, 24−6, 31−2, 42−4, 71−2, 74, 78, 85, 87−9, trigonometric definitions 142, 214−5
91, 166, 230, 377 trigonometric equations 81, 155, 157, 159, 161, 169, 378
cube 9, 52−3, 55 trigonometric graphs 81−2, 116−25, 127, 130, 169
equal 42, 44, 85, 166 equations of 81
irrational 14, 87, trigonometric ratios 138, , 140, 146, 149, 161, 286
rational 42, 87 trinomial 12, 27, 29, 159
unequal 44, 166 turning points 88−9, 114, 118, 121−2, 124−5, 137, 173,
325
S
sample space 253, 256, 258, 260–1, 269, 376 U
sequence 48−9, 53−4, 70−2, 72, 79, 320 undefined 75, 92, 98, 100, 114−8, 120, 127, 134, 152,
of numbers 71, 79 161−2, 173, 320, 325
simultaneous equations 3, 40−1, 70, 78, 104
sine 132, 159, 160, 175, 214−23, 225, 286 V
using 175, 223 variance 289–302, 305, 378
sine rule 215−7, 220, 225, 286 Venn diagram 175, 256−61, 275, 277, 282, 287, 322
standard deviation 287, 301−305, 311, 313−6, 317, 319, vertex, vertices 57–61, 64, 67−9, 76, 82, 172, 177, 189,
324, 377 197, 205, 324
of ungrouped data 289, 300 vertical asymptote 82, 92−6, 99, 114, 117, 224, 130−1,
statistics 237, 241, 289−90, 298, 318 135−6
surd equations 18−9, 78, 236 vertical shift 83, 86, 91−4, 104, 118−9, 227, 168, 378
surd form 14, 16, 19, 21, 25, 30, 67, 69, 73, 76, 140, 193,
199, 210 X
surds 3−4, 14−9, 21, 78, 377 x-intercepts 35, 87−91, 114, 119, 321, 378
surface area 176−87, 278, 280, 284, 326
lateral 176, 179, 186−7 Y
symmetry 82−3, 86−7, 91−101, 107, 114, 166, 189, 204−5, y-intercepts 82−4, 87−9, 94, 99, 100, 102, 104, 106, 122,
306, 324 124−5, 166, 321, 378
Index 381
PLTMATHSLB11LB_17.indd 381 2012/07/07 11:54 AM
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