Via Afrika
Mathematics
Grade 12 Teacher’s Guide
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of the learners and inspiring flight – inspiring them to Grade 12 Study Guide
reach their full potential.
M. Malan, L.J. Schalekamp, E.C. Brown, L. Bruce,
— Brynn Taylor, Teacher G. du Toit, C.R. Smith, L.M. Botsane, J. Bouman, M. Pillay
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Via Afrika Mathematics
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1. The series was written to be aligned with CAPS. See page 5 to see how CAPS requirements are met.
2. A possible work schedule has been included. See page 6 to 9 to see how much time this could save you.
3. Each topic starts with an overview of what is taught, and the resources you need. See page 31 to find out how this will
help with your planning.
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is done.
5. Advice on how to introduce concepts and scaffold learning is given for every topic. See page 31 for an example.
6. All the answers have been given to save you time doing the exercises yourself. See page 32 for an example.
7. Also included is a CD filled with resources to assist you in your teaching and assessment. See the inside front cover.
The accompanying Learner’s Book is written in accessible language and contains all the content your learners need to master.
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Gr12 Mathematics Study Guide LR
Grade 12 Mathematics study guide, 174 pages. Read online or get the free PDF download.
Mathematics · Grade 12. Study guide, 174 pages. Read online or download the PDF.
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- Mathematics
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M. Malan
St udy G u id e
Via Afrika Mathematics
Grade 12
ISBN: 978-1-41546-335-2
Exponents and Surds
Contents
Introduction ......................................................................................................................... 3
Chapter 1 Number patterns, sequences and series ........................................ 4
OVERVIEW ............................................................................................................................... 4
Unit 1 Arithmetic sequences and series
Unit 2 Geometric sequences and series
Unit 3 The sum to n terms(Sn): Sigma notation
Unit 4 Convergence and sum to infinity
Mixed exercises .................................................................................................................... 8
Chapter 2 Functions ..................................................................................... 10
OVERVIEW ............................................................................................................................... 10
Unit 1 The definitions of a function
Unit 2 The inverse of a function
Unit 3 The inverse of y = ax + q
Unit 4 The inverse of the quadratic function y = ax2
Mixed exercises .................................................................................................................... 21
Chapter 3 Logarithms ................................................................................... 24
OVERVIEW ............................................................................................................................... 24
Unit 1 The definition of a logarithm
Unit 2 Solving exponential equations using logarithms
Unit 3 The graph of y = logbx where b > 1 and 0 < b < 1
Mixed exercises .................................................................................................................... 28
Chapter 4 Finance, growth and decay ............................................................ 29
OVERVIEW ............................................................................................................................... 29
Unit 1 Future value annuities
Unit 2 Present value annuities
Unit 3 Calculating the period
Unit 4 Analysing investments and loans
Mixed exercises .................................................................................................................... 35
Chapter 5 Compound angles ......................................................................... 37
OVERVIEW ............................................................................................................................... 37
Unit 1 Deriving a formula for cos(𝛼 − 𝛽)
Unit 2 Formula for cos(𝛼 + 𝛽) and 𝑠𝑖𝑖(𝛼 ± 𝛽)
Unit 3 Double angles
Unit 4 Identities
Unit 5 Equations
Unit 6 Trigonometric graphs and compound angles
Mixed exercises .................................................................................................................... 46
Chapter 6 Solving problems in three dimensions ........................................... 48
OVERVIEW ............................................................................................................................... 48
Unit 1 Problems in three dimensions
Unit 2 Compound angle formulae in three dimensions
Mixed exercises .................................................................................................................... 51
© Via Afrika ›› Mathematics Grade 12 1
Exponents and Surds
Chapter 7 Polynomials .................................................................................. 53
OVERVIEW ............................................................................................................................... 53
Unit 1 The Remainder Theorem
Unit 2 The Factor Theorem
Mixed exercises .................................................................................................................... 57
Chapter 8 Differential calculus ...................................................................... 58
OVERVIEW ............................................................................................................................... 58
Unit 1 Limits
Unit 2 The gradient of a graph at a point
Unit 3 The derivative of a function
Unit 4 The equation of a tangent to a graph
Unit 5 The graph of a cubic function
Unit 6 The second derivative (concavity)
Unit 7 Applications of differential calculus
Mixed exercises .................................................................................................................... 70
Chapter 9 Analytical geometry ...................................................................... 72
OVERVIEW ............................................................................................................................... 72
Unit 1 Equation of a circle with centre at the origin
Unit 2 Equation of a circle centred off the origin
Unit 3 The equation of the tangent to the circle
Mixed exercises .................................................................................................................... 77
Chapter 10 Euclidean geometry ..................................................................... 81
OVERVIEW ............................................................................................................................... 81
Unit 1 Proportionality in triangles
Unit 2 Similarity in triangles
Unit 2 Theorem of Pythagoras
Mixed exercises .................................................................................................................... 94
Chapter 11 Statistics: regression and correlation ........................................... 97
OVERVIEW ............................................................................................................................... 97
Unit 1 Symmetrical and skewed data
Unit 2 Scatter plots and correlation
Mixed exercises .................................................................................................................... 106
Chapter 12 Probability .................................................................................. 108
OVERVIEW ............................................................................................................................... 108
Unit 1 Solving probability problems
Unit 2 The counting principle
Unit 3 The counting principle and probability
Mixed exercises .................................................................................................................... 112
ANSWERS TO MIXED EXERCISES ...................................................................................................... 113
EXEMPLAR PAPER 1 ..................................................................................................................... 151
EXEMPLAR PAPER 2 .................................................................................................................... 168
© Via Afrika ›› Mathematics Grade 12 2
Exponents and Surds
Introduction to Via Afrika Mathematics Grade 12 Study Guide
Woohoo! You made it! If you’re reading this it means that you made it through Grade 11,
and are now in Grade 12. But I guess you are already well aware of that…
It also means that your teacher was brilliant enough to get the Via Afrika Mathematics
Grade 12 Learner’s Book. This study guide contains summaries of each chapter, and should
be used side-by-side with the Learner’s Book. It also contains lots of extra questions to
help you master the subject matter.
Mathematics – not for spectators
You won’t learn anything if you don’t involve yourself in the subject-matter actively. Do
the maths, feel the maths, and then understand and use the maths.
Understanding the principles
• Listen during class. This study guide is brilliant but it is not enough. Listen to your
teacher in class as you may learn a unique or easy way of doing something.
• Study the notation, properly. Incorrect use of notation will be penalised in tests
and exams. Pay attention to notation in our worked examples.
• Practise, Practise, Practise, and then Practise some more. You have to practise
as much as possible. The more you practise, the more prepared and confident you
will feel for exams. This guide contains lots of extra practice opportunities.
• Persevere. We can’t all be Einsteins, and even old Albert had difficulties learning
some of the very advanced Mathematics necessary to formulate his theories. If you
don’t understand immediately, work at it and practise with as many problems from
this study guide as possible. You will find that topics that seem baffling at first,
suddenly make sense.
• Have the proper attitude. You can do it!
The AMA of Mathematics
ABILITY is what you’re capable of doing.
MOTIVATION determines what you do.
ATTITUDE determines how well you do it.
“Pure Mathematics is, in its way, the poetry of logical ideas.” Albert Einstein
© Via Afrika ›› Mathematics Grade 12 3
1 Chapter
Number patterns, sequences and series
Overview
Unit 1 Page 10
Arithmetic sequences and • Formula for an arithmetic
series sequence
Unit 2 Page 14
Geometric sequences and • Formula for the nth term
Chapter 1 Page 8 series of a sequence
Number patterns,
sequences and Unit 3 Page 18
series The sum to 𝑖 terms (𝑆𝑛 ): Sigma • The sum to 𝑖 terms in an
notation arithmetic sequence
• The sum to 𝑖 terms in a
geometric sequence
Unit 4 Page 28
Convergence and sum to infinity • Convergence
REMEMBER YOUR STUDY APPROACH SHOULD BE:
1 Work through all examples in this chapter of your Learner’s Bok.
2 Work through the notes in this chapter of this study guide.
3 Do the exercises at the end of the chapter in the Learner’s Book.
4 Do the mixed exercises at the end of this chapter in this study guide.
© Via Afrika ›› Mathematics Grade 12 4
1 Chapter
Number patterns, sequences and series
TABLE 1: SUMMARY OF SEQUENCES AND SERIES
TYPE GENERAL TERM: 𝑇𝑛 SUM OF TERMS: 𝑆𝑛 EXAMPLES
A) 2 ; 5 ; 8 ; 11 ; ...
Arithmetic Sequence (AS) 𝑇𝑛 = 𝑎 + (𝑖 − 1)𝑑 𝑆𝑛
𝑖
(also named the linear = [2𝑎 + (𝑖 − 1)𝑑] 𝑑= +3 +3 +3
𝑎 = 𝑓𝑖𝑟𝑠𝑡 𝑡𝑐𝑟𝑚 𝑇1 2
sequence) or
𝑖 𝑇𝑛 = 2 + (𝑖 − 1)(3)
𝑑 = 𝑐𝑜𝑖𝑠𝑡𝑎𝑖𝑡 𝑑𝑖𝑓𝑓. 𝑆𝑛 = [𝑎 + 𝑙] = 2 + 3𝑖 − 3
Constant 2
st 𝑑 = 𝑇2 − 𝑇1 = 3𝑖 − 1
1 difference where
𝑜𝑟 𝑇3 − 𝑇2 etc.
𝑙 = the last term of
B) 1 ; -4 ; -9 ; ...
the sequence
𝑑 = -5 -5
𝑇𝑛 = 1 + (𝑖 − 1)(−5)
= 1 − 5𝑖 + 5
= −5𝑖 + 6
𝑎(𝑟 𝑛 − 1) A) 2 ; -4 ; 8 ; -16 ; ...
𝑆𝑛 =
Geometric Sequence (GS) 𝑇𝑛 = 𝑎𝑟 𝑛−1 𝑟−1
(also named exponential 𝑟 = x-2 x-2 x-2
𝑎 = 𝑓𝑖𝑟𝑠𝑡 𝑡𝑐𝑟𝑚 𝑇1 Or
sequence) 𝑛
𝑎(1 − 𝑟 )
𝑆𝑛 = 𝑇𝑛 = 2(−2)𝑛−1
𝑟 = 𝑐𝑜𝑖𝑠𝑡𝑎𝑖𝑡 1−𝑟
NOT CONVERGING as 𝑟 < −1
𝑟𝑎𝑡𝑖𝑜
𝑎
Or 𝑆∞ =
Constant 𝑇2 𝑇3
1−𝑟
B) 3 ;
3
;
3
;
3
; ...
ratio 𝑟= 𝑜𝑟 Where −1 < 𝑟 < 1 2 4 8
𝑇1 𝑇2
(Converging series) 1 1 1
𝑟= x x x
2 2 2
1 𝑛−1
𝑇𝑛 = 3 � �
2
CONVERGING as −1 < 𝑟 < 1
𝑇𝑛 = 𝑎𝑖2 + 𝑏𝑖 + 𝑐 3 ; 8 ; 16 ; 27 ; ...
Quadratic Sequence (QS)
𝑓= 1st difference 𝑓: 5 8 11
𝑠= 2nd difference
𝑠: 3 3
Constant Determine 𝑎, 𝑏 and 𝑐
nd
2 using simultaneous Setup three equations using
difference equations (see the first three terms:
example) 𝑇1 = 3:
3=𝑎+𝑏+𝑐 …(1)
Alternatively: 𝑇2 = 8:
𝑎 =𝑠÷2 8 = 4𝑎 + 2𝑏 + 𝑐 …(2)
𝑏 = 𝑓1 − 3𝑎 𝑇3 = 16:
𝑐 = 𝑇1 − 𝑎 − 𝑏 16 = 9𝑎 + 3𝑏 + 𝑐 …(3)
where Solving simultaneously leads
𝑓1 = first term of first to:
differences 𝑇𝑛 = 32𝑖2 + 12𝑖 + 1
© Via Afrika ›› Mathematics Grade 12 5
1
Chapter
Number patterns, sequences and series
TYPES OF QUESTIONS YOU STRATEGY TO ANSWER THIS TYPE EXAMPLE(S) OF THIS TYPE OF
CAN EXPECT OF QUESTION QUESTION
Identify any of the following Determine whether sequence has a See Table 1 above
three types of sequences: • constant 1st difference (AS)
Arithmetic (AS), Geometric • constant ratio (GS)
(GS) and Quadratic (QS) • constant 2nd difference (QS)
Determine the formula for the You need to find: See Table 1 above
general term, 𝑇𝑛 , of AS, GS • 𝑎 and 𝑑 for an AS
and QS (from Grade 11) • 𝑎 and 𝑟 for a GS
• 𝑎, 𝑏 and 𝑐 for a QS
Determine any specific term Substitute the value of 𝑖 into 𝑇𝑛 See Text Book :
for a sequence e.g. 𝑇30 Example 1, nr. 1 d and 2 d, p.8
(AS)
Example 1, nr. 1 b, 3 b, p.11
(AS)
Example 1, nr. 1, p. 15 (GS)
Determine the number of Substitute all known variables into See Text Book:
terms in a sequence, 𝑖, for an the general term to get an equation Example 1, nr.1 c, p.8
AS, GS and QS or with 𝒔 as the only unknown. Solve Example 1, nr.1 c, p.11
the position, 𝑖, of a specific for 𝑖. Example 1, nr. 3, p.15
given term or when the sum OR
of the series is given Substitute all known variables into
the 𝑆𝑛 -formula to get an equation
with 𝒔 as the only unknown. Solve
for 𝑖. Example 2, nr.3, p.20
Example 3, nr. 2, p.24
Remember:
𝑖 must be a natural number
(not negative, not a fraction)
When given two sets of For each set of information given, See Text Book:
information, make use of substitute the values of 𝑖 and 𝑇𝑛 or Example 1, nr. 3, p.11 (AS)
simultaneous equations to 𝑖 and 𝑆𝑛 . Example 1, nr.2, p.15 (AS)
solve: Example 3, nr.3, p.24 (GS)
𝒂 and 𝒅 (for an AS) You then have 2 equations which
𝒂 and 𝒓 (for a GS) you can solve simultaneously (by
substitution)
Determine the value of a For AS use constant difference: The first three terms of an AS
variable (𝑥) when given a 𝑇3 − 𝑇2 = 𝑇2 − 𝑇1 are given by
sequence in terms of 𝑥. 2𝑥 − 4; 𝑥 − 3; 8 − 2𝑥
For GS use constant ratio: Determine 𝑥:
𝑇2 𝑇3 8 − 2𝑥 − (𝑥 − 3) = 𝑥 − 3 −
= (2𝑥 − 4)
𝑇1 𝑇2
∴𝑥=5
© Via Afrika ›› Mathematics Grade 12 6
1
Chapter
Number patterns, sequences and series
For a series given in sigma Remember: ∑𝑛𝑘=1 𝑇𝑘 has 𝑖 terms
notation: The “counter” indicates the number (counter 𝑘 runs from 1 to 𝑖)
• Determine the number of of terms in the series ∑𝑛𝑘=0 𝑇𝑘 has (𝑖 + 1) terms
terms (counter runs from 0 to 𝑖; so
one term extra)
∑𝑛𝑘=5 𝑇𝑘 has (𝑖 − 4) terms
( four terms not counted )
• Determine the value of Remember the expression next to See Text Book:
the series, in other words, the ∑-sign is the general term, 𝑇𝑛 . Example 1, p.19
𝑆𝑛 . This will help you to determine 𝑎
and 𝑑 or 𝑟.
Determine the general term, 𝑇𝑘 and Example 1, p.19
Write a given series in sigma number of terms, 𝑖 and substitute
notation. into ∑𝑛
𝑘=1 𝑇𝑘
Determine the sum, 𝑺𝒔 , of an In some cases you have to first See Text Book:
AS and a GS (when the determine the number of terms, 𝑖
number of terms are given or using 𝑇𝑛 . Example 2, nr.1 & 2, p.20
not given) Substitute the values of 𝑎, 𝑖 and 𝑑/𝑟 Example 3, nr. 1, p.24
into the formula for 𝑆𝑛
Determine whether a GS is Converging if −1 < 𝑟 < 1
converging or not
Determine 𝑆∞ for a Substitute vales of 𝑎 and 𝑟 See Text Book:
converging GS Into formula for 𝑆∞ Example 1, nr. 1, p.29
Determine the value of a Determine 𝑟 in terms of 𝑥 See Text Book:
variable (𝑥) for which a series and use −1 < 𝑟 < 1 Example 1, nr. 3, p.29
will converge,
e.g. (2𝑥 + 1) + (2𝑥 + 1)2 +…
Apply your knowledge of Generate a sequence of terms from See Text Book:
sequences and series on an the information given. Identify the Exercise 5, nr. 6, p.30
applied example (often type of sequence.
involving diagram/s)
© Via Afrika ›› Mathematics Grade 12 7
1Chapter
Number patterns, sequences and series
Mixed Exercise on sequences and series
1 Consider the following sequence: 5; 9; 13; 17; 21; …
a Determine the general term.
b Which term is equal to 217?
2 a T5 of a geometric sequence is 9 and T9 is 729. Determine the constant ratio.
b Determine T10.
3 The following is an arithmetic sequence: 2 x − 4 ; 5 x ; 7 x − 4
a Determine the value of x .
b Determine the first 3 terms.
4 Consider the following sequence: 2 ; 7 ; 15 ; 26 ; 40 ; …
a Determine the general term.
b Which term is equal to 260?
5 How many terms are there in the following sequence?
17 ; 14 ; 11 ; 8 ; … ; -2785
6 Tom links balls with rods in arrangements as shown below:
Arrangement 1 Arrangement 2 Arrangement 3 Arrangement 4
1 ball, 4 rods 4 balls, 12 rods 9 balls, 24 rods 16 balls 40 rods
a Determine the number of balls in the nth arrangement.
b Determine the number of rods in the nth arrangement.
7 Determine the following:
30 10
1
a ∑ (8 − 5k)
k =1
b ∑ 4 (2)𝑘−1
k =2
8 Write the following in sigma notation: 1+5+9+…+21
9 The 5th term of an arithmetic sequence is zero and the 13th term is equal to 12.
Determine:
a the constant difference and the first term.
b the sum of the first 21 terms.
© Via Afrika ›› Mathematics Grade 12 8
1Chapter
Number patterns, sequences and series
10 The first two terms of a geometric sequence are: (𝑥 + 3) and (𝑥 2 − 9)
a For which value of 𝑥 is this a converging sequence?
b Calculate the value of 𝑥 if the sum of the series to infinity is 13.
99+97+95+⋯+1
11 Calculate the value of:
299+297+295+⋯+201
12 𝑆𝑛 = 3𝑖2 − 2𝑖. Determine 𝑇9 .
13 The first four terms of a geometric sequence are 7; 𝑥 ; 𝑦 ; 189.
a Determine the values of 𝑥 and 𝑦.
b If the constant ratio is 3, make use of a suitable formula to determine the number of
terms in the sequence that will give a sum of 206 668.
© Via Afrika ›› Mathematics Grade 12 9
Chapter 2
Functions
Overview
Unit 1 Page 40
The definition of a function • Relations and functions
• Type of relations
• Which relations are
functions?
• Definition of a function?
• Function notation
Unit 2 Page 44
Chapter 2 Page 36 The inverse of a function • The concept of inverses
Functions by studying sets of
ordered number pairs
Unit 3 Page 46
The inverse of 𝑦 = 𝑎𝑥 + 𝑞 • Graphs of 𝑓and 𝑓 −1 on
the same set of axes
Unit 4 Page 48
The inverse of the quadratic • Restricting the domain of
function 𝑦 = 𝑎𝑥 2 the parabola
REMEMBER YOUR STUDY APPROACH SHOULD BE:
1 Work through all examples in this chapter of your Learner’s Book.
2 Work through the notes in this chapter of the study guide.
3 Do the exercises at the end of the chapter in the Learner’s Book.
4 Do the mixed exercises at the end of this chapter in the study guide.
© Via Afrika ›› Mathematics Grade 12 10
Chapter 2
Functions
TYPES OF RELATIONS BETWEEN TWO VARIABLES
TYPE DESCRIPTION PROPERTIES TYPICAL EXAMPLES
NON-FUNCTIONS One-to-many • One 𝑥-value in domain has MORE • Inverse of a parabola
THAN ONE 𝑦-value (See Unit 4)
• Does NOT pass vertical line test
FUNCTIONS One-to-one • Each 𝑥-value has a unique 𝑦- • Straight line graph
value and its inverse
• No 𝑥- or 𝑦-value appear more • Hyperbola and its
than once in domain or range inverse
• Passes VERTICAL line test • Exponential graph
and its inverse, the
logarithmic function
Many-to- one • No 𝑥-value appears more than • Parabola
once in domain • Graph of the cubic
• More than one 𝑥-value maps onto function
the same 𝑦-value • Trigonometric graphs
• Passes VERTICAL line test
REVISION OF THE
STRAIGHT LINE GRAPH
Standard form: 𝒚 = 𝒎𝒙 + 𝒄
𝒎
• Gradient of line
• Indicates “steepness”
𝒄
and direction of line:
• 𝑦-intercept
𝑚 > 0 (+) • Where 𝑥 = 0
𝑚 < 0(−)
𝑚=0
𝑦 −𝑦
• 𝑚 = 𝑥2 −𝑥1
2 1
© Via Afrika ›› Mathematics Grade 12 11
Chapter 2
Functions
PARALLEL AND
PERPENDICULAR LINES
Let 𝑦 = 𝑚1 𝑥 + 𝑐1 and
𝑦 = 𝑚2 𝑥 + 𝑐2 be two lines.
If the lines are PARALLEL, then:
𝑚1 = 𝑚2
If the lines are PERPENDICULAR,
then: 𝑚1 × 𝑚2 = −1
TO DETERMINE THE EQUATION OF A STRAIGHT LINE
GIVEN: EXAMPLES
1
1. Gradient and a point A line has a gradient of and goes through the point (4;1):
2
1
𝑚=
2
1
Substitute point (4;1) into 𝑦 = 𝑥 + 𝑐
2
1
1 = (4) + 𝑐
2
𝑐 = −1
1
𝑦 = 𝑥−1
2
2. 𝒚-intercept and a point A line has a 𝑦-intercept 3 and goes through the point (-2;1):
𝑐=3
Substitute point (-2;1) into 𝑦 = 𝑚𝑥 + 3
1 = 𝑚(−2) + 3
𝑚=1
𝑦 =𝑥+3
3. Two points on the line A line goes through the points (4;-3) and (2;1).
𝑦 −𝑦 1−(−3)
𝑚 = 𝑥2 −𝑥1 = 4−(2) = 2
2 1
Substitute any one of the two points into 𝑦 = 2𝑥 + 𝑐
1=2(2)+c
𝑐 = −3
𝑦 = 2𝑥 − 3
4. A point or 𝒚-intercept plus a) A line is parallel to the line 𝑦 = −𝑥 + 3 and goes
information regarding through the point (5;-2).
relationship to another line Parallel lines have same gradients; so 𝑚 = −1
Sub (5;-2) into 𝑦 = −𝑥 + 𝑐
−2 = −(5) + 𝑐
𝑐=3
b) A line is perpendicular to the line 𝑦 = 2𝑥 − 1 and has a
𝑦-intercept of 4.
© Via Afrika ›› Mathematics Grade 12 12
Chapter 2
Functions
Perpendicular lines have gradients with a product of −𝟏.
1
𝑚 × 2 = −1 ∴𝑚=−
2
−1
𝑦= 𝑥+4
2
REVISION OF THE PARABOLA
EQUATION IN STANDARD FORM
𝒚 = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 (𝒂 ≠ 𝟎)
𝒄
𝒂
• 𝑦-intercept
Indicates shape of parabola
• Where 𝑥 = 0
𝒂 > 0 (+)
Concave up
𝒃
Remember:
Positive(+) people smile!
• Affects the axis of symmetry and
turning point (TP)
𝒂 < 0(−)
𝑏
• Equation of axis of symmetry: 𝑥 = −
Concave down 2𝑎
𝑏 4𝑎𝑎−𝑏2
• Coordinates of TP �− 2𝑎 ; 4𝑎
�
Remember: 𝒙-intercepts
Negative (−) people are sad!
• Also called roots/zeroes
• Substitute 𝑦 = 0
© Via Afrika ›› Mathematics Grade 12 13
Chapter 2
Functions
EQUATION IN
TURNING POINT FORM
𝒚 = 𝒂(𝒙 − 𝒑)𝟐 + 𝒒 (𝒂 ≠ 𝟎)
𝒂
Indicates shape of parabola
𝒑 and 𝒒
𝒂 > 0 (+)
Concave up
• Equation of axis of symmetry 𝒙 = 𝒑
Remember:
• Coordinates of turning point (𝒑; 𝒒)
Positive(+) people smile!
𝒂 < 0(−)
Concave down
Intercepts
• 𝑥-intercepts (make 𝑦 = 0)
• 𝑦-intercept (make 𝑥 = 0)
Remember:
Negative (−) people are sad!
DOMAIN: 𝑥 ∈ 𝑅
RANGE:
𝑦 ∈ (−∞; 𝑞) 𝑦 ∈ (𝑞; ∞)
© Via Afrika ›› Mathematics Grade 12 14
Chapter 2
Functions
DETERMINE THE EQUATION OF A PARABOLA
GIVEN: 2 ROOTS (𝒙-INTERCEPTS) PLUS 1 POINT GIVEN: TURNING POINT PLUS 1 POINT
FORM OF EQUATION: FORM OF EQUATION:
𝒚 = 𝒂(𝒙 − 𝒙𝟏 )(𝒙 − 𝒙𝟐 ) 𝑦 = 𝑎(𝑥 − 𝑝)2 + 𝑞
𝒙𝟏 and 𝒙𝟐 are the roots (𝑝; 𝑞) is die turning point of the parabola
EXAMPLE: EXAMPLE:
y y
5
(2;6)
x (−1;2)
−1 3
x
𝒙𝟏 = −𝟏 𝒙𝟐 = 𝟏 (𝑝; 𝑞) = (−1; 2)
𝒚 = 𝒂(𝒙 − 𝒙𝟏 )(𝒙 − 𝒙𝟐 ) 𝑦 = 𝑎(𝑥 − 𝑝)2 + 𝑞
𝒚 = 𝒂(𝒙 − (−𝟏))(𝒙 − 𝟏) 2
𝑦 = 𝑎�𝑥 − (−1)� + 2
𝒚 = 𝒂(𝒙 + 𝟏)(𝒙 − 𝟏) 𝑦 = 𝑎(𝑥 + 1)2 + 2
Now substitute the other point (𝟐; 𝟔): Now substitute the point (0;5):
𝟔 = 𝒂(𝟐 + 𝟏)(𝟐 − 𝟏) 5 = 𝑎(0 + 1)2 + 2
𝟔 = 𝒂(𝟏)(−𝟏) 5=𝑎+2
𝟔 = −𝟏𝒂 3=𝑎
−𝟐 = 𝒂
𝑦 = 3(𝑥 + 1)2 + 2
𝒚 = −𝟐(𝒙 + 𝟏)(𝒙 − 𝟏)
𝑦 = 3(𝑥 2 + 2𝑥 + 1) + 2
𝒚 = −𝟐�𝒙𝟐 − 𝟐𝒙 − 𝟏� 𝑦 = 3𝑥 2 + 6𝑥 + 3 + 2
𝒚 = −𝟐𝒙𝟐 + 𝟒𝒙 + 𝟔 (standard form) 𝑦 = 3𝑥 2 + 6𝑥 + 5 (standard form)
© Via Afrika ›› Mathematics Grade 12 15
Chapter 2
Functions
REVISION OF THE HYPERBOLA
𝒂
𝒚 = 𝒙−𝒑 + 𝒒
𝒒
𝒂 Horizontal asymptote
Indicates shape of hyperbola 𝒚=𝒒
(with respect to asymptotes)
𝒂 > 0 (+) 𝒑
Vertical asymptote Intercepts
𝒙=𝒑
• 𝑥-intercept (make 𝑦 = 0)
𝒂 < 0(−) • 𝑦-intercept (make 𝑥 = 0)
Domain: 𝑥 ∈ 𝑅; 𝑥 ≠ 𝑝
Range: 𝑦 ∈ 𝑅; 𝑦 ≠ 𝑝
2
EXAMPLE: 𝑦 = 𝑥−1 − 2
𝑦-intercept: Axes of symmetry:
2
𝑦= − 2 = −4
−1 Substitute(1; −2) into
Axes of symmetry (AS) 𝑥- intercept: 𝑦 = 𝑥 + 𝑘1 𝑎𝑖𝑑 𝑦 = −𝑥 + 𝑘2
2
0= − 2 ;𝑥 = 2
𝑥−1
• Two axes of symmetry −2 = 1 + 𝑘1 𝑎𝑖𝑑 − 2 = −1 + 𝑘2
• AS go through intersect of Asymptotes: 𝑘1 = −3 𝑎𝑖𝑑 𝑘2 = −1
𝑥 = 1 𝑎𝑖𝑑 𝑦 = −2 𝑦 = 𝑥 − 3 𝑎𝑖𝑑 𝑦 = −𝑥 − 1
asymptotes (𝑝; 𝑞)
y
• Equations: 𝑦 = 𝑥 + 𝑘1 and x
𝑦 = −𝑥 + 𝑘2 1 2
−2
• Substitute the point (𝑝; 𝑞) to
−4
calculate 𝑘1 and 𝑘2
© Via Afrika ›› Mathematics Grade 12 16
Chapter 2
Functions
REVISION OF THE
EXPONENTIAL GRAPH
𝒚 = 𝒂𝒙−𝒑 + 𝒒
𝒂
Indicates shape of hyperbola
𝒂>𝟏 𝒒
• Horizontal asymptote: 𝑦 = 𝑝
𝟎<𝒂<1 • Indicates that the graph 𝑦 = 𝑎 𝑥 was
translated (shifted) vertically up/down
𝒒 > 0: shifted upwards
𝒒 < 0: shifted downwards
𝒑
EXAMPLE: 𝑦 = 2𝑥+1 − 1
𝑥
Indicates that the graph 𝑦 = 𝑎 was Asymptote: 𝑦 = −1
translated (shifted) horizontally left/right 𝑥-intercept (𝑦 = 0): 2𝑥+1 − 1 = 0 ∴ 𝑥 = −1
0+1
𝒑 > 0: shifted left 𝑦-intercept: (𝑥 = 0): 𝑦 = 2 −1=1
𝒑 < 0: shifted right
y
Intercepts
1
• 𝑥-intercept (make 𝑦 = 0)
x
• 𝑦-intercept (make 𝑥 = 0) −1
−1
Domain: 𝑥 ∈ 𝑅
Range: 𝑦 ∈ (𝑞; ∞)
© Via Afrika ›› Mathematics Grade 12 17
Chapter 2
Functions
EXAMPLES OF SYMMETRICAL
EXPONENTIAL GRAPHS
SYMMETRICAL IN THE 𝑦 −axis
y
𝑥
1 𝑦 = 3𝑥
𝑦 = � � = 3−𝑥
3
x
SYMMETRICAL IN THE 𝑥 −axis
y 𝑦 = 3𝑥
x
𝑦 = −3𝑥
© Via Afrika ›› Mathematics Grade 12 18
Chapter 2
Functions
INTERSECTS OF TWO GRAPHS
To determine the coordinates of the
point where two graphs INTERSECT:
Use SIMULTANEOUS EQUATIONS
EXAMPLE
Determine the coordinates of the points of
intersection of 𝑓(𝑥) = 3𝑥 + 6 and
𝑔(𝑥) = −2𝑥 2 + 3𝑥 + 14
Equate the two equations and solve for 𝒙:
3𝑥 + 6 = −2𝑥 2 + 3𝑥 + 14
2𝑥 2 − 8 = 0
𝑥2 − 4 = 0
(𝑥 − 2)(𝑥 + 2) = 0
𝑥 = 2 or 𝑥 = −2
Substitute 𝒙-values back into one of equations
(choose the easier one):
If 𝑥 = 2 then 𝑦 = 3(2) + 6 = 12
So one point of intersection is (2; 12).
If 𝑥 = −2 then 𝑦 = 3(−2) + 6 = 0
The other point of intersection is (−2; 0) which
is also the 𝑥-intercept of both graphs.
© Via Afrika ›› Mathematics Grade 12 19
Chapter 2
Functions
THE INVERSE OF A FUNCTION
• The inverse of a function, 𝑓, is denoted by 𝑓 −1 .
• 𝑓 −1 is a reflection of 𝑓in the line 𝑦 = 𝑥
• To determine the equation of 𝑓 −1 , swop 𝑥 and 𝑦 in the
equation of 𝑓
• The 𝑥-intercept of 𝑓 is the 𝑦-intercept of 𝑓 −1
FUNCTION 𝒇 INVERSE OF EXAMPLES DIAGRAM
FUNCTION, 𝒇−𝟏
Straight line Straight line 𝑓: 𝑦 = 2𝑥 + 3 y
f
𝒇: 𝒚 = 𝒎𝒙 + 𝒄 Inverse: 2𝑦 + 3 = 𝑥
x
−1 1 3
𝑓 : 𝑦 = 2𝑥−2
Exponential Logarithmic 𝑓: 𝑦 = 3𝑥 y
f
graph function
Inverse:
𝒙 −1
𝒇: 𝒚 = 𝒂 𝑓 : 𝑦 = log 𝑎 𝑥
𝑓 −1 : 𝑦 = log 3 𝑥 x
Parabola The inverse of a 𝑓: 𝑦 = 2𝑥 2
y
parabola is NOT A
𝒇: 𝒚 = 𝒂𝒙𝟐 FUNCTION Inverse:
𝑥 = 2𝑦 2
NB: The DOMAIN of 1
the parabola has to 𝑦2 = 𝑥
2 x
be RESTRICTED to 1
𝑥 ≥ 0 or 𝑥 ≤ 0 so 𝑓 −1 : ±� 𝑥
2
that 𝑓 −1 is also a
function
© Via Afrika ›› Mathematics Grade 12 20
Chapter 2
Functions
Mixed Exercise on Functions
1 Determine the coordinates of the intercept of the following two lines:
2𝑥 − 3𝑦 = 17
3𝑥 − 𝑦 = 15
4 y
f
2 a Determine the equation of line 𝑓. 3
b Determine the equation of line 𝑔. P
c Determine the co-ordinates of point, P, 1
where the two lines intersect.
x
−4 −3 4
d Are these two lines perpendicular? (2;-1)
Give a reason for your answer.
e Write down the equation of the line g
which is parallel to line 𝑔 with a 𝑦-intercept −4
of -2.
3 The diagram shows the graphs of y = x 2 − 2 x − 3 y
and y = mx + c .
a Determine the lengths of OA, OB and OC.
b Determine the co-ordinates of the turning point D. A B x
c Determine m and c of the straight line.
d Use the graph to determine for which values C
of k for which the equation x 2 − 2 x + k = 0 would D
have only one real root.
C
4 The diagram shows the graph of f ( x) = −2( x + 1) + 8 .
2 E D
C is the turning point.
E is the mirror image of the y-intercept of 𝑓.
Determine:
a the length of AB. A B x
b the co-ordinates of C.
c the length of DE.
© Via Afrika ›› Mathematics Grade 12 21
Chapter 2
Functions
1 𝑥
5 Consider the function 𝑔(𝑥) = � � − 2.
2
a Make a neat drawing of 𝑔. Clearly show the asymptote and intercepts with the axes.
b Determine the domain of 𝑔.
c For which values of 𝑥 would 𝑔(𝑥) ≥ 0.
y
𝑎
6 The graph of 𝑓(𝑥) = ; 𝑥 ≠ 0 is shown.
𝑥
𝐴(−2; 2) is a point on the graph where it cuts the line 𝑦 = −𝑥.
a Determine the value of 𝑎 . A(−2;2)
x
b Write down the coordinates of B. f
B
c Graph 𝑓 is translated 2 units up and 1 unit right.
d Write down the equation of the new graph.
7 The graphs of the following are shown :
1
𝑓(𝑥) = −𝑥 2 − 2𝑥 + 8 and 𝑔(𝑥) = 𝑥 − 1
2
Determine:
a the coordinates for A E
y
b the coordinates for B and C F A
c the length of CD
R
d the length of DE which is parallel to the 𝑦-axis f
e the length of AF which is parallel to the 𝑥-axis
f the length of GH which is parallel to the 𝑦-axis H x
B D C
g the 𝑥-value for which RS would have a maximum length.
h the maximum length of RS. G
S
g
i the 𝑥-values for which 𝑓(𝑥) − 𝑔(𝑥) > 0.
y
8 The diagram alongside shows the graphs of the functions of (2 ;5 )
𝑎 f
𝑓(𝑥) = 𝑏 𝑥 + 𝑐 𝑎𝑖𝑑 𝑔(𝑥) = +𝑞
𝑥+𝑝
a Write down the equation of the asymptote of 𝑓.
B(−6 ;0 ) −2 x
b Determine the equation of 𝑓.
−1
c Write down the equations of the asymptotes of 𝑔.
g
d Determine the equation of 𝑔.
A(0 ;−3 )
e Determine the equations of the axes of symmetry of 𝑔 .
−4
f For which values of 𝑥 is 𝑓(𝑥) > 𝑔(𝑥)?
© Via Afrika ›› Mathematics Grade 12 22
Chapter 2
Functions
9 The graph of 𝑓(𝑥) = 2𝑥 2 is given.
y
a Determine the equation of 𝑓 −1 in
the form 𝑓 −1 : 𝑦 = ⋯
b How can one restrict the domain of 𝑓 so
that 𝑓 −1 will be a function?
f(x)=2x²
x
10 The graph of 𝑓(𝑥) = 𝑎 𝑥 is given. 3 y
The point A (-1; 3) lies on the graph. 2
1
a Determine the equation of 𝑓. x
−3 −2 −1 1 2 3
b Determine the equation of 𝑓 −1 in the form 𝑓 −1 : 𝑦 = ⋯ −1
c Make a neat drawing of the graph of 𝑓 −1 .
−2
d Determine the domain of 𝑓 −1 .
−3
11 A straight line graph has an 𝑥-intercept of -2 and a 𝑦-intercept of 3. Write down the
coordinates of the 𝑥- and 𝑦-intercepts of 𝑓 −1 .
© Via Afrika ›› Mathematics Grade 12 23
Chapter 3
Logarithms
Overview
Unit 1 Page 60
The definition of a logarithm • Changing exponents to
the logarithmic form
• Proofs of the logarithmic
laws
Unit 2 Page 64
Chapter 3 Page 58 Solve exponential equations • Using logarithms
Logarithms using logarithms
Unit 3 Page 66
The graph of 𝑦 = log 𝑏 𝑥 where • Inverse of
𝑏 > 1 and 0 < 𝑏 < 1 𝑦 = 𝑓(𝑥) = 2𝑥
• Inverse of the function
1 𝑥
𝑦 = 𝑓(𝑥) = � �
2
REMEMBER YOUR STUDY APPROACH SHOULD BE:
1 Work through all examples in this chapter of your Learner’s Bok.
2 Work through the notes in this chapter of this study guide.
3 Do the exercises at the end of the chapter in the Learner’s Book.
4 Do the mixed exercises at the end of this chapter in this study guide.
© Via Afrika ›› Mathematics Grade 12 24
Chapter 3
Logarithms
Definition of logarithm
If log 𝒃 𝒙 = 𝒚, then 𝒃𝒚 = 𝒙.
EXAMPLES: Converting from one form to another
Logarithmic form Exponential form
log 𝟏 𝟐𝟒𝟏 = 𝟓 𝟏𝟓 = 𝟐𝟒𝟏
log 𝟎,𝟓 𝟎, 𝟏𝟐𝟓 = 𝟏 𝟎, 𝟓𝟏 = 𝟎, 𝟏𝟐𝟓
log 𝟏𝟎 𝟏𝟎𝟎𝟎 = 𝟏 𝟏𝟎𝟏 = 𝟏𝟎𝟎𝟎
𝟏 𝟏
𝟏𝟐 = √𝟏
log 𝟏 √𝟏 =
𝟐
LOGARITHMIC LAW EXAMPLES
• log 𝑘 𝑎𝑏𝑐 = log 𝑘 𝑎 + log 𝑘 𝑏 + log 𝑘 𝑐
Law 1: 𝐥𝐥𝐥 𝒎 𝑨. 𝑩 = 𝐥𝐥𝐥 𝒎 𝑨 + 𝐥𝐥𝐥 𝒎 𝑩
• log 5 25.5 = log 5 25 + log 5 5 = 2 + 1 = 3
𝑦
• log 𝑚 𝑧 = log 𝑚 𝑦 − log 𝑚 𝑧
𝑨
Law 2: 𝐥𝐥𝐥 𝒎 𝑩 = 𝐥𝐥𝐥 𝒎 𝑨 − 𝐥𝐥𝐥 𝒎 𝑩
0,2
• log 5 = log 5 0,2 − log 5 25
25
= log 5 5−1 − log 5 25
= −1 − 2 = −3
• log 𝑦 𝑎3 = 3 log 𝑦 𝑎
Law 3: 𝐥𝐥𝐥 𝒙 𝑷𝒚 = 𝒚 𝐥𝐥𝐥 𝒙 𝑷
• log 5 0,04 = log 5 5−2 = −2 log 5 5 = −2
log 𝑎
• log 𝑏 𝑎 = log 𝑏
𝐥𝐥𝐥 𝒂
Law 4: 𝐥𝐥𝐥 𝒃 𝒂 = 𝐥𝐥𝐥 𝒃
log 5
• log 2 5 = log 2 = 2,32
Note that:
• log 𝑎 𝑎 = 1 (𝑎 ≠ 0)
• log 𝑎 1 = 0
• log 𝑎 = log 𝟏𝟎 𝑎
© Via Afrika ›› Mathematics Grade 12 25
Chapter 3
Logarithms
USING LOGARITHMS TO SOLVE EQUATIONS
We know that equations involving exponents can
be solved using exponential laws:
𝟐𝒙 = 𝟏𝟏𝟏
𝟐𝒙 = 𝟐𝟕 (prime factorise)
∴𝒙=𝟕
But, what if we cannot use prime factors?
𝟐𝒙 = 𝟏𝟏
𝐥𝐥𝐥 𝟐𝒙 = 𝐥𝐥𝐥 𝟏𝟏
𝒙𝐥𝐥𝐥 𝟐 = 𝐥𝐥𝐥 𝟏𝟏
𝐥𝐥𝐥 𝟏𝟏
𝒙= = 𝟑, 𝟕
𝐥𝐥𝐥 𝟐
© Via Afrika ›› Mathematics Grade 12 26
Chapter 3
Logarithms
THE INVERSE OF THE
𝑓: 𝑦 = 𝑎 𝑥
EXPONENTIAL GRAPH
is the logarithmic function
𝑓 −1 : 𝑦 = log 𝑎 𝑥 ; 𝑥 > 0
EXAMPLES
−1
𝑓(RED GRAPH) 𝑓 (BLUE GRAPH) DIAGRAM
y
1
𝒙
𝒚=𝟒 𝒚 = 𝐥𝐥𝐥 𝟒 𝒙 x
1
y
𝟏𝒙 𝒚 = 𝐥𝐥𝐥 𝟏 𝒙
𝒚= 1
𝟒 𝟒 x
1
y
𝒚 = −𝟒𝒙
−1
𝒚 = 𝐥𝐥𝐥 𝟒 (−𝒙)
x
−1
y
𝟏𝒙 𝒚 = 𝐥𝐥𝐥 𝟏 (−𝒙)
𝒚=− 𝟒
−1 x
𝟒
−1
© Via Afrika ›› Mathematics Grade 12 27
Chapter 3
Logarithms
Mixed Exercise on Logarithms
1 Make use of the definition of the logarithm to solve for 𝑥:
a log 3 𝑥 = 2
b log 1 𝑥 = 2
3
c − log 4 𝑥 = 2
d log 5 𝑥 = −2
e log 𝑥 3 = 6
f log 3 81 = 𝑥
1
g log 3 = 𝑥
9
9
2 The graph of 𝑓(𝑥) = 𝑎 𝑥 goes through the point �2; �.
4
a Determine the value of 𝑎.
b Determine the equation of 𝑓 −1.
c Determine the equation of 𝑔 if 𝑓 and 𝑔 are symmetrical in the 𝑦-axis.
d Determine the equation of ℎ, the reflection of 𝑓 −1 in 𝑥-axis.
3 The function 𝑓 is given by the graph 𝑓(𝑥) = log 2 𝑥.
a Determine the equations of the following graphs:
i 𝑔, the reflection of 𝑓 in the 𝑥-axis
ii 𝑝, the reflection of 𝑓 in the 𝑦-axis
iii 𝑞, the reflection of 𝑔 in the 𝑦-axis
iv 𝑓 −1 , the inverse of 𝑓
v 𝑔−1 , the inverse of 𝑔
vi ℎ, the translation of 𝑓 two units left
b Sketch the graphs of 𝑓, 𝑓 −1, 𝑔 and 𝑔−1 on the same system of axes.
y
c Determine the domain and range of 𝑓 −1 and 𝑔−1 .
B
4 The graph of 𝑦 = log 𝑏 𝑥 is shown in the diagram alongside.
a Determine the coordinates of point A. x
A
b How do we know that 𝑏 > 1.
3
c Determine 𝑏 if B is the point �8; �.
2
C
d Determine the equation of 𝑔, the inverse of this graph.
e Determine the value of 𝑎 if C is the point (𝑎; −2).
© Via Afrika ›› Mathematics Grade 12 28
4
Chapter
Finance, growth and decay
Overview
Unit 1 Page 78
Future value annuities • Deriving the future value
formula
Unit 2 Page 82
Present value annuities • Deriving the present
value formula
Chapter 4 Page 76
Finance, growth
Unit 3 Page 86
and decay
Calculating the period • Finding the value of 𝑖
Unit 4 Page 88
Analysing investments and • Outstanding balances on a
loans loan
• Sinking fund
• Pyramid schemes
REMEMBER YOUR STUDY APPROACH SHOULD BE:
1 Work through all examples in this chapter of your Learner’s Book.
2 Work through the notes in this chapter of this study guide.
3 Do the exercises at the end of the chapter in the Learner’s Book.
4 Do the mixed exercises at the end of this chapter.
© Via Afrika ›› Mathematics Grade 12 29
4
Chapter
Finance, growth and decay
HIRE PURCHASE AGREEMENTS
𝐴 = 𝑃(1 + 𝑖𝑖)
Example:
NB: Simple Kelvin buys computer equipment on hire purchase for R20 000.
Interest He has to put down 10% deposit and repays the amount monthly over 3 years.
The interest rate is 15% p.a.
Deposit = 10% of R20 000 = R2 000.
He has to repay 𝐴 = 18000(1 + 0,15 × 𝟏) = 𝑅26 100 in total.
36 monthly payments of R26 100÷ 36 =R725 each.
INFLATION / INCREASE IN PRICE OR VALUE
𝐴 = 𝑃(1 + 𝑖)𝑛
NB: Compound
Interest
𝑖= number of years
DEPRECIATION
Choose the correct formula!
Straight line method Reducing-balance method
𝐴 = 𝑃(1 − 𝑖𝑖) 𝐴 = 𝑃(1 − 𝑖)𝑛
𝑖= number of years
© Via Afrika ›› Mathematics Grade 12 30
4
Chapter
Finance, growth and decay
NOMINAL AND EFFECTIVE INTEREST RATES
NB: 𝒎 = the number of times per year
interest is added
Daily: 𝑚 = 365
𝑖𝑛𝑜𝑚 𝒎
�1 + 𝑖𝑒𝑓𝑓 � = �1 + � Monthly: 𝑚 = 12
𝒎
Quarterly: 𝑚=4
Half-yearly (semi-annually): 𝑚=2
EXAMPLE:
What is the effective rate if the nominal rate is 18% p.a. compounded quarterly?
In other words:
Which rate compounded annually will give me the same return as 18% compounded quarterly?
0,18 4
𝑖𝑒𝑓𝑓 = �1 + � −1
4
=0,1925186...
Effective rate = 19,25%
FUTURE VALUE ANNUITIES
Choosing the
𝑥[(1+𝑖)𝑛 −1] value of 𝒔 is very
𝐷= KEY WORDS:
𝑖 important!
• Regular investments
(monthly/quarterly etc.)
• Sinking funds
Example 1 • Annuity/pension
• Savings plan
First payment in one month’s time. Last payment in one year’s time.
Now
𝑖 = 12
First payment Last payment
© Via Afrika ›› Mathematics Grade 12 31
4
Chapter
Finance, growth and decay
Example 2
First payment immediately. Last payment in one year’s time.
Now
𝑖 = 13
First payment Last payment
Example 3
Assume investment pays out in one year’s time, but the first payment was made 2 months from now
and the last payment in one year’s time.
Now
𝑖 = 11
First payment Last payment
Example 4 (Watch out!)
First payment immediately, but last payment in 9 months’ time.
Now
𝑖 = 10
First payment Last payment
BUT, the investment still
𝒙[(𝟏+𝒔)𝟏𝟎 −𝟏] earns interest for another 3
𝑭= (𝟏 + 𝒔)𝟏
𝒔 months before paying out
© Via Afrika ›› Mathematics Grade 12 32
4
Chapter
Finance, growth and decay
FUTURE VALUE ANNUITIES
𝑥[1−(1+𝑖)−𝑛 ]
𝑃=
𝑖
KEY WORDS:
NB: There must always
be ONE GAP between
• Regular payments
the P-value and the first
payment! (monthly/quarterly etc.)
• Loan (NOT HIREPURCHASE)
• Bond/home loan
• Repayment of debt
• How long will money be enough to
Example 1 provide regular income?
Payment starts one month after the granting of the loan. Last payment in one year’s time.
GAP
Now
𝑖 = 12
First payment Last payment
𝑥[1−(1+𝑖)−12 ]
Granting of loan 𝑃=
𝑖
Example 2
Payment starts in 3 months’ time. Last payment in one year’s time.
GAP
Now
𝑖 = 12 − 2 = 10
First payment Last payment
Granting of loan
𝑥�1−(1+𝑖)−10 �
NB: Loan amount accumulates interest for 2 months: 𝑃(𝟏 + 𝒔)𝟐 = 𝑖
© Via Afrika ›› Mathematics Grade 12 33
4
Chapter
Finance, growth and decay
OUTSTANDING BALANCE OF LOAN
Option 2
Option 1
Use A- and F-formula
Use P-formula
𝒔 = number of payments
𝒔 = number of payments left
already made
Example
A loan of is being repaid over 20 years in monthly payments of R6 000. The interest rate is 15% p.a.
compounded monthly. What is the outstanding balance after 12½ years?
Option 1
Outstanding period = 7½ years = 90 months
0,15 −𝟗𝟎
6000�1−�1+ � �
12
𝐴𝑎𝑙𝑎𝑖𝑐𝑐 = 0,15
12
Option 2
Payments already made = 12½X12=150 payments already paid
Outstanding balance = 𝐴 − 𝐷
0,15 𝟏𝟓𝟎
0,15 𝟏𝟓𝟎 6000��1+ � −1�
12
Balance = 𝑃 �1 + � − 0,15 where P is the initial loan amount.
12
12
© Via Afrika ›› Mathematics Grade 12 34
4
Chapter
Finance, growth and decay
Mixed Exercise on Finance, growth and decay
1 Determine through calculation which of the following investments is the best, if R15 000 is
invested for 5 years at:
a 10,6% p.a. simple interest
b 9,6% p.a., interest compounded quarterly.
2 An amount of money is now invested at 8,5% p.a compounded monthly to grow to R95 000 in
5 years.
a Is 8,5% called the effective or nominal interest rate?
b Calculate the amount that must be invested now.
c Calculate the interest earned on this investment.
3 Shirley wants to buy a flat screen TV. The TV that she wants currently costs R8 000.
a The TV will increase in cost according to the rate of inflation, which is 6% per annum.
How much will the TV cost in two years’ time?
b For two years Shirley puts R2 000 into her savings account at the beginning of every
six month period (starting immediately). Interest on her savings is paid at 7% per
annum, compounded six-monthly. Will she have enough to pay for the TV in two
years’ time? Show all your calculations.
4 Calculate:
a the effective interest rate to 2 dec. places if the nominal interest rate is 7,85% p.a.,
compounded monthly.
b the nominal interest rate if interest on an investment is compounded quarterly, using
an effective interest rate of 9,25% p.a.
5 Equipment with a value(new) of R350 000 depreciated to R179 200 after 3 years, based on the
reducing balance method. Determine the annual rate of depreciation.
6 R20 000 is deposited into a new savings account at 9,75% p.a., compounded quarterly.
After18 months, R10 000 more is deposited. After a further 3 months, the interest rate
changes to 9,95% p.a., compounded monthly. Determine the balance in the account 3 years
after the account was opened.
7 A company recently bought new equipment to the value of R900 000 which has to be replaced
in 5 years’ time. The value of the equipment depreciates at 15% per year according to the
reduced-balance method. After 5 years the equipment can be sold second hand at the
reduced value. The inflation rate on the equipment is 18% per year.
© Via Afrika ›› Mathematics Grade 12 35
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Chapter
Finance, growth and decay
a The company wants to establish a sinking fund to replace the equipment in 5 years’
time. Calculate what the value of the sinking fund should be to replace the equipment.
b Calculate the quarterly amount that the company has to pay into the sinking fund to
be able to replace the equipment in 5 years’ time. The company makes the first
payment immediately and the last payment at the end of the 5 year period. The
interest rate for the sinking fund is 8% per year compounded quarterly.
8 Goods to the value of R1 500 is bought on hire purchase and repaid in 24 monthly payments
of R85. Calculate the annual interest rate that applied for the hire purchase agreement.
9 Peter makes a loan to buy a house. He pays back the loan over a period of 20 years in monthly
payments of R6 500. Peter qualifies for an interest rate of 12% per years compounded
monthly. He makes his first payment one month after the loan was granted.
a Calculate the amount Peter borrowed.
b Calculate the amount that Peter still owes on his house after he has been paying back
the loan for 8 years.
10 Megan’s father wants to make provision for her studies. He starts paying R1000 on a monthly
base into an investment on her 12th birthday. He makes the last payment on her 18th birthday.
She needs the money 5 months after her 18th birthday. The interest rate on the investment is
10% per annum compounded monthly. Calculate the amount Megan has available for her
studies.
11 Stephan starts investing R300 into an investment monthly, starting one month from now. He
earns interest of 9% per annum compounded monthly. For how long must he make these
monthly investments so that the total value of his investment is R48 000? Give your answer
as follows: …. years and …. Months
12 Carl purchases sound equipment to the value of R15 000 on hire purchase. The dealer expects
him to put down a 10% deposit. The interest rate is 12% per annum and he has to repay the
money monthly over 4 years. It is compulsory for him to insure the equipment through the
dealer at a premium of R30 per month. Calculate the total amount Carl has to pay the dealer
monthly.
13 Tony borrows money to the value of R400 000. He has to pay back the money in 16 quarterly
payments, but only has to make his first payment one year from now. The interest rate is 8%
per annum compounded quarterly. Calculate the quarterly payment Tony has to pay.
© Via Afrika ›› Mathematics Grade 12 36
5
Chapter
Compound angles
Overview
Unit 1 Page 108
Deriving a formula for • How to deriving a formula
𝑐𝑜𝑠(𝛼 − 𝛽) for 𝑐𝑜𝑠(𝛼 − 𝛽)
Unit 2 Page 112
Formulae for cos(𝛼 + 𝛽) and • Formula for cos(𝛼 + 𝛽)
sin(𝛼 ± 𝛽) • Formula for sin(𝛼 + 𝛽)
• Formula for sin(𝛼 − 𝛽)
Unit 3 Page 116
Double angles • Formula for sin2𝛼
• Formula for 𝑐𝑜𝑠2𝛼
Chapter 5 Page 102
Compound angles Unit 4 Page 120
Identities • Proving identities
• Finding the value(s) for
which the identity is not
defined
Unit 5 Page 124
Equations • Equations with compound
and double angles
Unit 6 Page 128
Trigonometric graphs and • Drawing and working with
compound angles graphs of compound
angles
REMEMBER YOUR STUDY APPROACH SHOULD BE:
1 Work through all examples in this chapter of your Learner’s Book.
2 Work through the notes in this chapter of this study guide.
3 Do the exercises at the end of the chapter in the Learner’s Book.
4 Do the mixed exercises at the end of this chapter in this study guide.
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5
Chapter
Compound angles
REVISION OF TRIGONOMETRY
BASIC TRIGONOMETRIC RATIOS
Ratio Inverse
𝑜 ℎ
𝑠𝑠𝑠𝑠 = 𝑐𝑐𝑐𝑐𝑐𝑐 = 𝒉
ℎ 𝑜 𝒄
𝑎 ℎ
𝑐𝑐𝑐𝑐 = 𝑠𝑠𝑠𝑠 = 𝜽 Γ̅
ℎ 𝑎
𝑜 𝑎 𝒂
𝑡𝑡𝑡𝑡 = 𝑐𝑐𝑐𝑐 =
𝑎 𝑜
Ratio Inverse
𝑜 ℎ
𝑠𝑠𝑠𝑠 = 𝑐𝑐𝑐𝑐𝑐𝑐 =
ℎ 𝑜
𝑎 ℎ
𝑐𝑐𝑐𝑐 = 𝑠𝑠𝑠𝑠 =
ℎ 𝑎 𝒓
𝑜 𝑎 𝒚
𝑡𝑡𝑡𝑡 = 𝑐𝑐𝑐𝑐 =
𝑎 𝑜 𝜽 Γ̅
𝒙
YOU HAVE TO KNOW IN WHICH QUADRANT AN ANGLES LIES AND WHICH
RATIO (AND ITS INVERSE) IS POSITIVE THERE:
2nd quadrant 𝒚 1st quadrant
(𝟏𝟏𝟎° − 𝜽) 𝜽
(−𝟏𝟏𝟎° − 𝜽) (𝟏𝟔𝟎° + 𝜽)
𝑥
(𝟏𝟏𝟎° + 𝜽) (𝟏𝟔𝟎° − 𝜽)
(𝜽 − 𝟏𝟏𝟎°) (−𝜽)
3rd quadrant 4th quadrant
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5
Chapter
Compound angles
REDUCTION FORMULAE
𝑦
𝑠𝑖𝑖(180° − 𝑠) 𝑠𝑖𝑖𝑠
𝑐𝑜𝑠(180° − 𝑠) −𝑐𝑜𝑠𝑠
𝑡𝑎𝑖(180° − 𝑠) −𝑡𝑎𝑖𝑠
𝑥
𝑠𝑖𝑖(180° + 𝑠) −𝑠𝑖𝑖𝑠 𝑠𝑖𝑖(360° − 𝑠) −𝑠𝑖𝑖𝑠
𝑐𝑜𝑠(180° + 𝑠) −𝑐𝑜𝑠𝑠 𝑐𝑜𝑠(360° − 𝑠) 𝑐𝑜𝑠𝑠
𝑡𝑎𝑖(180° + 𝑠) 𝑡𝑎𝑖𝑠 𝑡𝑎𝑖(360° − 𝑠) −𝑡𝑎𝑖𝑠
CO-RATIOS/CO-FUNCTIONS
Ratio Co-ratio
𝑠𝑠𝑠(90° − 𝜃) 𝑐𝑐𝑐𝑐 (90° − 𝑠)
Is in 1st quadrant
𝑐𝑐𝑐(90° − 𝜃) 𝑠𝑠𝑠𝑠
𝑡𝑡𝑡(90° − 𝜃) 𝑐𝑐𝑐𝑐
Ratio Co-ratio
𝑠𝑠𝑠(90° + 𝜃) 𝑐𝑐𝑐𝑐 (90° + 𝑠)
Is in 2nd quadrant
𝑐𝑐𝑐(90° + 𝜃) −𝑠𝑠𝑠𝑠
𝑡𝑡𝑡(90° + 𝜃) −𝑐𝑐𝑐𝑐
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5
Chapter
Compound angles
KNOW YOUR SPECIAL TRIANGLES!
√𝟐 𝟏
𝟒𝟓° Γ̅
𝟏𝟎°
𝟐
√𝟏
𝟔𝟎° Γ̅
𝟏
IDENTITIES
𝒔𝒔𝒔𝒔 𝟏 𝒄𝒄𝒄𝒄
𝒕𝒕𝒕𝒕 = and 𝒄𝒄𝒄𝒄 = =
𝒄𝒄𝒄𝒄 𝒕𝒕𝒕𝒕 𝒔𝒔𝒔𝒔
SQUARE IDENTITIES:
𝒔𝒔𝒔𝟐 𝜽 + 𝒄𝒄𝒄𝟐 𝜽 = 𝟏
From this follows that:
∴ 𝒄𝒄𝒄𝟐 𝜽 = 𝟏 − 𝒔𝒔𝒔𝟐 𝜽
∴ 𝒔𝒔𝒔𝟐 𝜽 = 𝟏 − 𝒄𝒄𝒄𝟐 𝜽
Note that the two identities above can both be
FACTORISED as differences of two squares:
𝒄𝒄𝒄𝟐 𝜽 = 𝟏 − 𝒔𝒔𝒔𝟐 𝜽 = (𝟏 − 𝒔𝒔𝒔𝒔)(𝟏 + 𝒔𝒔𝒔𝒔)
𝒔𝒔𝒔𝟐 𝜽 = 𝟏 − 𝒄𝒄𝒄𝟐 𝜽 = (𝟏 − 𝒄𝒄𝒄𝒄)(𝟏 + 𝒄𝒄𝒄𝒄)
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©Via Afrika >> Mathematics Grade 12 40
5
Chapter
Compound angles
COMPOUND ANGLE-IDENTITIES
𝒄𝒄𝒄(𝜶 − 𝜷) = 𝒄𝒄𝒄𝒄 𝒄𝒄𝒄𝒄 + 𝒔𝒔𝒔𝒔 𝒔𝒔𝒔𝒔
𝒄𝒄𝒄(𝜶 + 𝜷) = 𝒄𝒄𝒄𝒄 𝒄𝒄𝒄𝒄 − 𝒔𝒔𝒔𝒔 𝒔𝒔𝒔𝒔
𝒔𝒔𝒔(𝜶 + 𝜷) = 𝒔𝒔𝒔𝒔 𝒄𝒄𝒄𝒄 + 𝒄𝒄𝒄𝒄 𝒔𝒔𝒔𝒔
𝒔𝒔𝒔(𝜶 − 𝜷) = 𝒔𝒔𝒔𝒔 𝒄𝒄𝒄𝒄 − 𝒄𝒄𝒄𝒄 𝒔𝒔𝒔𝒔
DOUBLE ANGLE-IDENTITIES
𝒔𝒔𝒔 𝟐𝟐 = 𝟐 𝒔𝒔𝒔𝒔 𝒄𝒄𝒄 𝜶
𝒄𝒄𝒄 𝟐𝟐 = 𝒄𝒄𝒄𝟐 𝜶 − 𝒔𝒔𝒔𝟐 𝜶
= 𝟏 − 𝟐 𝒔𝒔𝒔𝟐 𝜶
= 𝟐𝒄𝒄𝒄𝟐 𝜶 − 𝟏
𝟐 𝒕𝒕𝒕 𝜶
𝒕𝒕𝒕 𝟐𝟐 =
𝟏 − 𝒕𝒕𝒕𝟐 𝜶
TIPS FOR PROVING IDENTITIES
• Work with LHS and RHS separately
• Write DOUBLE angles as SINGLE angles
• Watch out for SQUARE IDENTITIES
• Write everything in terms of 𝒔𝒔𝒔 and 𝒄𝒄𝒄
• When working with fractions, put EVERYTHING over the LCD
• Be on the look out for opportunities to FACTORISE, e.g.
𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 − 𝒔𝒔𝒔𝒔 = 𝒔𝒔𝒔𝒔(𝟐𝟐𝟐𝟐𝟐 − 𝟏)
𝒄𝒄𝒄𝟐 𝜶 − 𝒔𝒔𝒔𝟐 𝜶 = (𝒄𝒄𝒄𝒄 + 𝒔𝒔𝒔𝒔)(𝒄𝒄𝒄𝒄 − 𝒔𝒔𝒔𝒔)
𝟐𝒔𝒔𝒔𝟐 𝜶 + 𝒔𝒔𝒔𝒔 − 𝟏 = (𝟐𝟐𝟐𝟐𝟐 − 𝟏)(𝒔𝒔𝒔𝒔 + 𝟏)
• It is sometimes necessary to replace 𝟏 with 𝒔𝒔𝒔𝟐 𝜶 + 𝒄𝒄𝒄𝟐 𝜶
E.g. 𝒔𝒔𝒔𝟐𝜶 + 𝟏 = 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 + 𝒔𝒔𝒔𝟐 𝜶 + 𝒄𝒄𝒄𝟐 𝜶
= (𝒔𝒔𝒔𝒔 + 𝒄𝒄𝒄𝒄)𝟐
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©Via Afrika >> Mathematics Grade 12 41
5
Chapter
Compound angles
FINDING THE GENERAL SOLUTION
OF A TRIGONOMETRIC EQUATION
STEP EXAMPLES OF HOW TO APPLY STEP
Get trig ratio (sin/cos/tan) A 2 sin 3𝑥 = 0,4
alone on LHS 𝑠𝑖𝑖3𝑥 = 0,2
1
B 𝑐𝑜𝑠𝑥 = −0,2
One value alone on RHS 3
𝑐𝑜𝑠𝑥 = −0,6
C 2 tan(𝑥 − 10°) + 3 = 0
3
tan(𝑥 − 10°) = − 2
Now use RHS consisting of a: A 𝑠𝑖𝑖3𝑥 = +0,2
The + indicates the 1st and 2nd quadrant,
SIGN (+ or -) and a VALUE where sin is positive.
Reference ∠= 𝑠𝑖𝑖−1 (0,2) = 11,54°
B 𝑐𝑜𝑠𝑥 = −0,6
Indicates Get reference The – indicates the 2nd and 3rd quadrant,
Quadrant angle using: where cos is negative.
𝒔𝒔𝒔−𝟏 (+𝒗𝒂𝒍𝒖𝒆) Reference ∠= 𝑐𝑜𝑠 −1 (0,6) = 53,13°
𝒄𝒄𝒔−𝟏 (+𝒗𝒂𝒍𝒖𝒆) 3
C tan(𝑥 − 10°) = −
Or 𝒕𝒂𝒔−𝟏 (+𝒗𝒂𝒍𝒖𝒆) 2
The − indicates the 2nd and 4th quadrant,
where tan is negative.
3
Reference ∠= 𝑡𝑎𝑖−1 �2� = 56,31°
A 2 sin 3𝑥 = 0,4
The angle in the trig equations will be equated to 𝑠𝑖𝑖3𝑥 = 0,2
the following in the respective quadrants: 1st : 3𝑥 = 11,54° + 𝑘360° ; 𝑘 ∈ 𝑍
𝑥 = 3,85° + 𝑘120° OR
1st = Ref ∠ 2nd : 3𝑥 = 180° − 11,54° + 𝑘360°
+𝑘360°; 𝑘 ∈ 𝑍 for sin/cos
2nd = 𝟏𝟏𝟎° −Ref ∠ 𝑥 = 56,15° + 𝑘120°
or
3rd = 𝟏𝟏𝟎° +Ref ∠
4th = 𝟏𝟔𝟎° −Ref ∠ +𝑘180°; 𝑘 ∈ 𝑍 for tan B 1
𝑐𝑜𝑠𝑥 = −0,2
3
𝑐𝑜𝑠𝑥 = −0,6
Then solve for 𝒙 2nd : 𝑥 = 180° − 53,13° + 𝑘360°; 𝑘 ∈ 𝑍
𝑥 = 126,87° + 𝑘360° OR
3rd : 𝑥 = 180° + 53,13° + 𝑘360°
𝑥 = 233,13° + 𝑘360°
C 2 tan(𝑥 − 10°) + 3 = 0
3
tan(𝑥 − 10°) = −
2
2nd : 𝑥 − 10° = 180° − 56,31° + 𝑘180°;
𝑘∈𝑍
𝑥 = 133,69° + 𝑘180°
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©Via Afrika >> Mathematics Grade 12 42
5
Chapter
Compound angles
EQUATIONS INVOLVING
TWO TRIGONOMETRIC FUNCTIONS
EXAMPLES COMMENTS
1 𝒔𝒔𝒔𝒙 = 𝒄𝒄𝒔𝒙
÷ by 𝑐𝑜𝑠𝑥 on both sides
𝒔𝒔𝒔𝒙 𝒄𝒄𝒔𝒙
=
𝒄𝒄𝒔𝒙 𝒄𝒄𝒔𝒙
𝒕𝒂𝒔𝒙 = 𝟏
𝒙 = 𝟒𝟓° + 𝒌. 𝟏𝟏𝟎° ; 𝒌𝝐𝒁
2 𝒔𝒔𝒔𝒙 = 𝒄𝒄𝒔𝟏𝒙 May NOT divide by 𝑐𝑜𝑠𝑥 both sides
𝒄𝒄𝒔(𝟗𝟎° − 𝒙) = 𝒄𝒄𝒔𝟏𝒙 Trig function on both sides should be the
same
𝟗𝟎° − 𝒙 = 𝟏𝒙 + 𝒌. 𝟏𝟔𝟎° ; 𝒌𝝐𝒁
−𝟒𝒙 = −𝟗𝟎° + 𝒌. 𝟏𝟔𝟎° Angles on LHS and RHS should either be the
𝒙 = 𝟐𝟐, 𝟓° − 𝒌. 𝟗𝟎° same or
or
𝟗𝟎° − 𝒙 = −𝟏𝒙 + 𝒌. 𝟏𝟔𝟎° ; 𝒌𝝐𝒁
𝟐𝒙 = −𝟗𝟎° + 𝒌. 𝟏𝟔𝟎° be in two different quadrants where 𝑐𝑜𝑠 have
𝒙 = −𝟒𝟓° + 𝒌. 𝟏𝟏𝟎° the same sign (1st and 4th quadrant)
3 𝐬𝐥𝐧(𝒙 + 𝟐𝟎°) = 𝐜𝐥𝐬(𝟐𝒙 − 𝟏𝟎°) Alternative: sin on both sides
𝒄𝒄𝒔[𝟗𝟎° − (𝒙 + 𝟐𝟎°)] = 𝒄𝒄𝒔(𝟐𝒙 − 𝟏𝟎°)
𝑠𝑖𝑖(𝑥 + 20°) = 𝑐𝑜𝑠(2𝑥 − 30°)
𝒄𝒄𝒔(𝟕𝟎° − 𝒙) = 𝒄𝒄𝒔(𝟐𝒙 − 𝟏𝟎°) 𝑠𝑖𝑖(𝑥 + 20°) = sin[90° − (2𝑥 − 30°)]
𝟕𝟎° − 𝒙 = 𝟐𝒙 − 𝟏𝟎° + 𝒌. 𝟏𝟔𝟎° 𝑠𝑖𝑖(𝑥 + 20°) = 𝑠𝑖𝑖(120° − 2𝑥)
−𝟏𝒙 = −𝟏𝟎𝟎° + 𝒌. 𝟏𝟔𝟎°
𝒙 = 𝟏𝟏, 𝟏𝟏° − 𝒌. 𝟏𝟐𝟎° ; 𝒌 ∈ 𝒁 𝑥 + 20° = 120° − 2𝑥 + 𝑘. 360°
or 3𝑥 = 100° + 𝑘. 360°
𝟕𝟎° − 𝒙 = −(𝟐𝒙 − 𝟏𝟎°) + 𝒌. 𝟏𝟔𝟎° 𝑥 = 33,33° − 𝑘. 120° ; 𝑘 ∈ 𝑍
𝒙 = −𝟒𝟎° + 𝒌. 𝟏𝟔𝟎° or
𝑥 + 20° = 180° − (120° − 2𝑥) + 𝑘. 360°
−𝑥 = 40° + 𝑘. 360°
𝑥 = −40° − 𝑘. 360°
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©Via Afrika >> Mathematics Grade 12 43
5
Chapter
Compound angles
EXAMPLES OF EQUATIONS INVOLVING
DOUBLE ANGLES
𝒄𝒄𝒔𝜽. 𝒄𝒄𝒔𝟏𝟒° + 𝒔𝒔𝒔𝜽. 𝒔𝒔𝒔𝟏𝟒° = 𝟎, 𝟕𝟏𝟓
𝒄𝒄𝒔(𝜽 − 𝟏𝟒°) = 𝟎, 𝟕𝟏𝟓
Ref ∠= 𝟒𝟒, 𝟏𝟔°
1st quadrant:
𝜽 − 𝟏𝟒° = 𝟒𝟒, 𝟏𝟔° + 𝒌. 𝟏𝟔𝟎°
𝜽 = 𝟓𝟏, 𝟏𝟔° + 𝒌. 𝟏𝟔𝟎° ; 𝒌 ∈ 𝒁
4th quadrant:
𝜽 − 𝟏𝟒° = −𝟒𝟒, 𝟏𝟔° + 𝒌. 𝟏𝟔𝟎°
𝜽 = −𝟏𝟎, 𝟏𝟔° + 𝒌. 𝟏𝟔𝟎° ; 𝒌 ∈ 𝒁
𝒔𝒔𝒔𝟐𝜽 + 𝟐𝒔𝒔𝒔𝜽 = 𝟎
𝟐𝒔𝒔𝒔𝜽𝒄𝒄𝒔𝜽 + 𝟐𝒔𝒔𝒔𝜽 = 𝟎
𝟐𝒔𝒔𝒔𝜽(𝒄𝒄𝒔𝜽 + 𝟏) = 𝟎
𝒔𝒔𝒔𝜽 = 𝟎 or 𝒄𝒄𝒔𝜽 = −𝟏
𝜽 = 𝒌. 𝟏𝟏𝟎° ; 𝒌 ∈ 𝒁 or 𝜽 = 𝟏𝟏𝟎° + 𝒌. 𝟏𝟔𝟎°
𝟐𝒔𝒔𝒔𝟐 𝜽 + 𝒔𝒔𝒔𝜽 = 𝟏
𝟐𝒔𝒔𝒔𝟐 𝜽 + 𝒔𝒔𝒔𝜽 − 𝟏 = 𝟎
∴ (𝟐𝒔𝒔𝒔𝜽 + 𝟏)(𝒔𝒔𝒔𝜽 − 𝟏) = 𝟎
𝟐𝒔𝒔𝒔𝜽 + 𝟏 = 𝟎 or 𝒔𝒔𝒔𝜽 + 𝟏 = 𝟎
𝟏
𝒔𝒔𝒔𝜽 = − or 𝒔𝒔𝒔𝜽 = −𝟏
𝟐
No solution 𝜽 = 𝟐𝟕𝟎° + 𝒌. 𝟏𝟔𝟎° ; 𝒌 ∈ 𝒁
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©Via Afrika >> Mathematics Grade 12 44
5
Chapter
Compound angles
PROBLEMS WITH COMPOUND-ANGLES TO
BE DONE WITHOUT A CALCULATOR
• Write given information in form where trig function is ALONE on LHS
• Select QUADRANT and draw TRIANGLE in correct quadrant
(2 sides of triangle will be known)
• Use the Theorem of PYTHAGORAS to determine 3rd side
• Now work with the expression of which you need to find the value:
write all compound or double angles in terms of SINGLE ANGLES
• Now SUBSTITUTE VALUES from diagram(s) and SIMPLIFY
Example:
𝟓
If 𝟏𝟏𝟏𝟏𝟏𝟏𝟏 + 𝟏𝟏 = 𝟎 and 𝜶 ∈ [𝟗𝟗°; 𝟐𝟐𝟐°] and 𝜷 = 𝟏𝟏 ; 𝜷 > 90°, determine
without the use of a calculator the value of:
a 𝒔𝒔𝒔(𝜶 − 𝜷) b 𝒄𝒄𝒄(𝜶 + 𝜷)
c 𝒔𝒔𝒔𝒔𝒔 d 𝒄𝒄𝒄𝒄𝒄
Solution:
𝟏𝟏 𝟓
𝒔𝒔𝒔𝒔 = − 𝟏𝟏 𝒄𝒄𝒄𝒄 = 𝟏𝟏
𝒔𝒔𝒔 negative in 3rd and 4th quad 𝒄𝒄𝒄 positive in 1st and 4th quad
But 𝜶 ∈ [𝟗𝟗°;𝛼 𝟐𝟐𝟐°], so 3rd quad But 𝜷 𝛽
> 90°,5so 4th quadrant
−12
13 13
𝒙 = −𝟓 𝒚 = −𝟏𝟏
−𝟏𝟏 𝟓 −𝟓 −𝟏𝟏 −120
a 𝒔𝒔𝒔(𝜶 − 𝜷) = 𝒔𝒔𝒔𝒔 𝒔𝒔𝒔𝒔 − 𝒄𝒄𝒄𝒄 𝒄𝒄𝒄𝒄 = � 𝟏𝟑 � �𝟏𝟏� − � 𝟏𝟏 � � 𝟏𝟏 � = 169
−𝟓 𝟓 −𝟏𝟏 −𝟏𝟏
b 𝒄𝒄𝒄(𝜶 + 𝜷) = 𝒄𝒄𝒄𝒄 𝒄𝒄𝒄𝒄 − 𝒔𝒔𝒔𝒔 𝒔𝒔𝒔𝒔 = � 𝟏𝟏 � �𝟏𝟏� − � 𝟏𝟏 � � 𝟏𝟏 � = −1
−𝟏𝟏 −𝟓 120
c 𝒔𝒔𝒔𝒔𝒔 = 𝟐𝟐𝟐𝟐𝟐 𝒄𝒄𝒄𝒄 = 𝟐 � 𝟏𝟏 � � 𝟏𝟏 � = 169
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©Via Afrika >> Mathematics Grade 12 45
5
Chapter
Compound angles
Mixed Exercise on Compound angles
1 Solve the following equations for . Give the general solution unless otherwise stated.
Answers should be given correct to 2 decimal places where exact answers are not possible.
a 2𝑐𝑜𝑠2𝑥 + 1 = 0
b 𝑠𝑖𝑖 x = 3 cos x for 𝑥 ∈ [90°; 360°]
c 𝑠𝑖𝑖𝑥 = cos 3𝑥
d 6 − 10𝑐𝑜𝑠𝑥 = 3𝑠𝑖𝑖2 𝑥 ; 𝑥 ∈ [−360°; 360°]
e 2 − 𝑠𝑖𝑖𝑥 𝑐𝑜𝑠𝑥 − 3𝑐𝑜𝑠 2 𝑥 = 0
f 3𝑠𝑖𝑖2 𝑥 − 8𝑠𝑖𝑖𝑥 + 16𝑠𝑖𝑖𝑥 𝑐𝑜𝑠𝑥 − 6𝑐𝑜𝑠𝑥 + 3𝑐𝑜𝑠 2 𝑥 = 0
2 Prove the following identities, stating any values of 𝑥 or 𝑠 for which the identity is not valid:
1
a cos x + tan x sin x =
cos x
sin θ cosθ 1
b − =
1 − cosθ sin θ sin θ
1 − cos 2 x
c = tan x sin x
cos x
sin 3 x + sin x cos 2 x
d = tan x
cos x
1 + tan x 1 + 2 sin x cos x
e =
1 − tan x cos 2 x − sin 2 x
1
f sin(45° + 𝑥) . sin(45° − 𝑥) = 2 𝑐𝑜𝑠2𝑥
sin 2𝜃−cos 𝜃 cos 𝜃
g =
sin 𝜃−cos 2𝜃 1+sin 𝜃
cos 𝑥−cos 2𝑥+2 1+cos 𝑥
h =
3 sin 𝑥−sin 2𝑥 sin 𝑥
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©Via Afrika >> Mathematics Grade 12 46
5
Chapter
Compound angles
3 Simplify:
a
( )
sin 180 0 − x tan (− x )
( ) (
tan 180 0 + x cos x − 90 0 )
b
( ) (
sin 180 0 + x tan x − 360 0 ) (without using a calculator)
( )
tan 360 0 − x cos 240 0 tan 225 0
4 Given that 𝑠𝑖𝑖17° = 𝑘, express in terms of 𝑘:
a cos 73°
b 𝑐𝑜𝑠(−163°)
c 𝑡𝑎𝑖197°
d 𝑐𝑜𝑠326°
5 Given that 5𝑐𝑜𝑠𝑥 + 4 = 0 , calculate, without the use of a calculator, the value(s) of :
a 5𝑠𝑖𝑖𝑥 + 3𝑡𝑎𝑖𝑥
b tan 2𝑥
3
6 If 3𝑠𝑖𝑖𝑥 = −1 ; 𝑥 ∈ [90°; 270°] and 𝑡𝑎𝑖𝑦 = ; 𝑦 ∈ [90°; 360°]. Determine without the use of a
4
calculator the value of:
a cos(𝑥 − 𝑦)
b 𝑐𝑜𝑠2𝑥 − 𝑐𝑜𝑠2𝑦
7 Simplify without the use of calculator:
a 𝑐𝑜𝑠 2 22,5° − 𝑠𝑖𝑖2 22,5°
b 𝑠𝑖𝑖22,5° 𝑐𝑜𝑠22,5°
c 2𝑠𝑖𝑖15°𝑐𝑜𝑠15°
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©Via Afrika >> Mathematics Grade 12 47
6
Chapter
Solving problems in three dimensions
Overview
Unit 1 Page 146
Problems in three dimensions • Trigonometry in real life
Chapter 6 Page 142
Solving problems in Unit 2 Page 150
three dimensions Compound angle formulae in • Using compound angle
three dimensions formulae in three
dimensions
REMEMBER YOUR STUDY APPROACH SHOULD BE:
1 Work through all examples in this chapter of your Learner’s Book.
2 Work through the notes in this chapter of this study guide.
3 Do the exercises at the end of the chapter in the Learner’s Book.
4 Do the mixed exercises at the end of this chapter in the study guide.
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©Via Afrika >> Mathematics Grade 12 48
Chapter 6
Solving problems in three dimensions
REVISION ON THE USE OF THE 𝒔𝒔𝒔𝒖𝒔 −, 𝒄𝒄𝒔𝒔𝒔𝒖𝒔 − and the
𝒂𝒓𝒆𝒂 −FORMULAE
INFORMATION GIVEN UNKNOWN FORMULA TO USE FORM OF FORMULA
𝒂 𝑏
2 angles and 1 side s 𝑠𝑖𝑖-rule = sin 𝐴
𝑠𝑖𝑛𝐴
(∠∠s) 𝒂 is unknown
𝑠𝑖𝑖-rule
2 sides and Watch out for 𝑠𝑖𝑖𝑨 𝑠𝑖𝑖𝐴
∠ =
𝑎 𝑏
a not- included ∠ ambiguous case!
(ss∠) ∠ can be acute or
𝑨 is unknown
obtuse
2 sides and
an included ∠ s 𝑐𝑜𝑠-rule 𝒂2 = 𝑏 2 + 𝑐 2 − 2𝑏𝑐𝑐𝑜𝑠𝐴
(s∠s) 𝒂 is unknown
3 sides 𝑐𝑜𝑠-rule 𝑏 2 + 𝑐 2 − 𝑎2
∠ 𝑐𝑜𝑠𝐴 =
(sss) 2𝑏𝑐
𝑨 is unknown
1
2 sides and Area Area-rule Area of ∆= 2 𝑎𝑏𝑠𝑖𝑖𝐶
an included ∠ Area is unknown
Area, side and ∠ s Area-rule 2 × 𝐴𝑟𝑐𝑎
𝒃=
𝑎𝑠𝑖𝑖𝐶
𝒃 is unknown
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©Via Afrika >> Mathematics Grade 12 49
6
Chapter
Solving problems in three dimensions
TIPS FOR SOLVING PROBLEMS IN THREE DIMENSIONS
• Where there are 3 triangles, start with the ∆ with the most information and
work via the 2nd ∆ to the 3rd ∆ which contains the unknown to be calculated.
• Indicate all RIGHT angles – remember they don’t always look like 𝟗𝟗° angles
• Shade the horizontal plane in the diagram (e.g. floor, ground)
• Be on the lookout for reductions like 𝒄𝒄𝒄(𝟗𝟗° − 𝜶) = 𝒔𝒔𝒔𝒔 and
𝒔𝒔𝒔(𝟏𝟏𝟏° − 𝜶) = 𝒔𝒔𝒔𝒔 to simplify expressions
• Use compound and double angle formulae to convert to single angles
• When writing out the solution – always indicate in which ∆ you are working
EXAMPLE
P, Q and R are in the same horizontal plane. TP is a vertical tower
5,9 m high. The angle of elevation of T from Q is 65°. 𝑃𝑄� 𝑅 = 𝑃𝑅� 𝑄.
a Calculate the length of PQ to the nearest meter.
b Hence show that 𝑅𝑄 = 5,5 𝑐𝑜𝑠𝑥.
c If it is further given that 𝑥 = 42°, calculate the area of ∆𝑃𝑄𝑅.
Solution:
5,9
a = 𝑡𝑎𝑖65°
𝑃𝑄
5,9
∴ 𝑃𝑄 = = 2,75 𝑚
𝑡𝑎𝑛65°
b 𝑄𝑃�𝑅 = 180° − 2𝑥
𝑅𝑄 𝑃𝑄
=
𝑠𝑖𝑛𝑃 𝑠𝑖𝑛𝑅
𝑅𝑄 2,75
=
sin(180°−2𝑥) 𝑠𝑖𝑛𝑥
𝑅𝑄 2,75
=
sin 2𝑥 𝑠𝑖𝑛𝑥
𝑅𝑄 2,75
=
2sin 𝑥𝑎𝑜𝑠𝑥 𝑠𝑖𝑛𝑥
∴ 𝑅𝑄 = 2 × 2,75𝑐𝑜𝑠𝑥
𝑅𝑄 = 5,5 𝑐𝑜𝑠𝑥
1
c Area of ∆𝑃𝑄𝑅 = × 𝑃𝑄 × 𝑄𝑅 × 𝑠𝑖𝑖𝑄
2
1
= × 2,75 × (5,5𝑐𝑜𝑠42°) × 𝑠𝑖𝑖42°
2
= 3,76 square units
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©Via Afrika >> Mathematics Grade 12 50
6
Chapter
Solving problems in three dimensions
Mixed Exercise on Problems in Three Dimensions
1 In the diagram alongside B, D and E are in the
same horizontal plane. 𝐴𝐷� 𝐷 = 120°
AB and CD are two vertical towers.
AB = 2CD = 2ℎ meter
The angle of elevation from E to A is 𝛼.
The angle of elevation from E to C is (90° − 𝛼).
a Determine the length BE in terms of ℎ and 𝛼.
b Show that the distance between the two towers can be given as:
ℎ√𝑡𝑎𝑛4 𝛼+2𝑡𝑎𝑛2 𝛼+4
𝐴𝐷 = 𝑡𝑎𝑛𝛼
c Hence determine the height of the tower CD, rounded to the nearest meter, if 𝛼 = 42° and
BD=400 m.
2 B, C and D are three points in the same horizontal plane and
AB is a vertical pole of length 𝑝 metres. The angle of elevation of
A from C is 𝑠 and 𝐴𝐶̂ 𝐷 = 𝑠. Also, 𝐶𝐴�𝐷 = 30° and 𝐴𝐷 = 8 𝑚.
a � 𝐴 in terms of 𝑠.
Express 𝐶𝐷
8sin(30°+𝜃)
b Hence show that 𝑝 = .
𝑎𝑜𝑠𝜃
3 In the diagram alongside, AB is a vertical flagpole 5 metres high.
AC an AD are two stays. B, C and D are in the same horizontal plane.
𝐴𝐷 = 12 𝑚, 𝐴𝐶̂ 𝐷 = 𝛼 and 𝐴𝐷 � 𝐶 = 𝛽.
13sin(𝛼+𝛽)
Show that 𝐶𝐷 =
𝑠𝑖𝑛𝛼
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©Via Afrika >> Mathematics Grade 12 51
6
Chapter
Solving problems in three dimensions
4 � 𝐶=𝑠.
In ∆𝐴𝐴𝐶 AD = ; DB = 𝑖 ; CD = 𝑝 and 𝐴𝐷
a Complete in terms of 𝑚 , 𝑝 and 𝑠: Area ∆𝐴𝐷𝐶 = ⋯
1
b Show that the area of ∆𝐴𝐴𝐶 = 2 𝑝(𝑚 + 𝑖) sin 𝑠 .
c If the area of ∆𝐴𝐴𝐶 = 12,6 𝑐𝑚2 ; 𝐴𝐴 = 5,9 𝑐𝑚
and 𝐷𝐶 = 8,1 𝑐𝑚, calculate the value(s) of 𝑠.
5 � = 90° , 𝐴𝐶̂ 𝐷 = 𝑠
In the diagram, 𝐷
𝐴𝐶̂ 𝐴 = 𝛼 ; AB = BC and 𝐴𝐷 = 𝑝 units.
a Express BC in terms of 𝑝 and 𝑠.
b Determine, without stating reasons,
the size of 𝐴�1 in terms of 𝛼.
𝑝.sin 2𝛼
c Hence , prove that AC = sin 𝜃.sin 𝛼
6 In the diagram PQ is a vertical building.
Q, R and S are points in the same horizontal plane.
The angle of elevation of P, the top of the building,
measured from R, is 𝛼.
𝑅𝑄�𝑆 = 30°
𝑄𝑆𝑅 = 150° − 𝛼
𝑄𝑆 = 12 𝑚
6(𝑎𝑜𝑠𝛼+√3𝑠𝑖𝑛𝛼)
a Show that 𝑄𝑅 =
𝑠𝑖𝑛𝛼
b Hence show that the height PQ of the building is given by
𝑃𝑄 = 6 + 6√3𝑡𝑎𝑖𝛼
c Hence calculate the value of 𝛼 if PQ = 23 m.
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©Via Afrika >> Mathematics Grade 12 52
7
Chapter
Polynomials
Overview
Unit 1 Page 158
The Remainder Theorem • The Remainder theorem
Chapter 7 Page 156
Polynomials
Unit 2 Page 160
The Factor Theorem • The Factor Theorem
REMEMBER YOUR STUDY APPROACH SHOULD BE:
1 Work through all examples in this chapter of your Learner’s Book.
2 Work through the notes in this chapter of the study guide.
3 Do the exercises at the end of the chapter in the Learner’s Book.
4 Do the mixed exercises at the end of this chapter in the study guide.
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©Via Afrika >> Mathematics Grade 12 53
7
Chapter
Polynomials
THE REMAINDER THEOREM
𝑓(𝑥) = (𝑎𝑎 + 𝑏). 𝑞(𝑥) + 𝑟(𝑥)
divisor quotient remainder
The remainder theorem can be used to calculate the remainder
when a polynomial 𝑓(𝑥)is divided by (𝑎𝑎 + 𝑏)
𝑏
∴ 𝑓 �− � = 𝑟(𝑥)
𝑎
Choosing the correct value to substitute is very important:
If you divide 𝒇 by 𝒈(𝒙) = Value to substitute into 𝒇(𝒙)
(𝒙 − 𝟐) 𝑓(2) =?
1
(𝟐𝟐 − 𝟏) 𝑓 � � =?
2
(𝒙 + 𝟑) 𝑓(−3) =?
2
(𝟑𝟑 + 𝟐) 𝑓 �− � =?
3
THE FACTOR THEOREM
𝑏
If 𝑓 �− 𝑎� = 0 then:
• (𝑎𝑎 + 𝑏) is a FACTOR of 𝑓(𝑥) and
• 𝑓(𝑥) is DIVISIBLE by (𝑎𝑎 + 𝑏)
When trying out 𝑥 −values that give 0, try them in the
following order: 1; −1; 2; −2; 3; −3 etc.
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©Via Afrika >> Mathematics Grade 12 54
Chapter7
Polynomials
DIFFERENT METHODS TO FACTORISE A CUBIC POLYNOMIAL (3RD DEGREE)
METHOD AND DESCRIPTION OF STEPS EXAMPLES
SUM AND DIFFERENCE OF CUBES 3
A) 𝑓(𝑥) = 𝑥 + 27
= (𝑥 + 3)(𝑥 2 − 3𝑥 + 9)
Cannot factorise further
B)𝑓(𝑥) = 8𝑥 3 − 1
= (2𝑥 − 1)(4𝑥 2 + 2𝑥 + 1)
Cannot factorise further
FACTORISE BY GROUPING 𝑓(𝑥) = 𝒙 + 𝟏𝒙𝟐 − 𝟒𝒙 − 𝟏𝟐
𝟏
= 𝑥 2 (𝒙 + 𝟏) − 4(𝒙 + 𝟏)
• Group terms in two pairs
= (𝒙 + 𝟏)(𝒙𝟐 − 𝟒)
• Take out common factor from each pair = (𝑥 + 3)(𝒙 + 𝟐)(𝒙 − 𝟐)
• Two sets of brackets now become common factor
• Factorise bracket further if possible
FACTORISE BY INSPECTION 𝑓(𝑥) = 2𝑥 3 − 𝟐𝒙𝟐 − 10𝑥 − 6
𝑓(−1) = 2(−1)3 − 2(−1)2 − 10(−1)
• Find one linear factor using factor theorem −6=0
• Find other factor (quadratic expression) by ∴ (𝑥 + 1) is a factor
inspection 𝑓(𝑥) = (𝑥 + 1)(𝒂𝑥 2 + 𝒃𝑥 + 𝒄)
Now find these coefficients
Start with 𝒂 and 𝒄:
1×𝑎 = 2 ∴𝑎 =2
1 × 𝑐 = −6 ∴ 𝑐 = 6
You now need to find 𝒃:
Multiply the two brackets; the two
𝑥 2 -terms need to give you −𝟐𝒙𝟐 :
𝑓(𝑥) = (𝑥 + 1)(𝟐𝑥 2 + 𝒃𝑥 + 𝟔)
𝑏𝑥 2 + 2𝑥 2 = −𝟐𝒙𝟐 ∴ 𝑏 = −4
2
∴ 𝑓(𝑥) = (𝑥 + 1)(2𝑥 − 4𝑥 + 6)
= (𝑥 + 1)(2𝑥 + 2)(𝑥 − 3)
SYNTHETIC OR LONG DIVISION 𝑓(𝑥) = 2𝑥 3 − 𝟐𝒙𝟐 − 10𝑥 − 6
𝑓(−1) = 2(−1)3 − 2(−1)2 − 10(−1)
• Find one linear factor using factor theorem −6=0
• Find other factor (quadratic expression) by long ∴ (𝑥 + 1) is a factor
division or synthetic division 𝑓(𝑥) = (𝑥 + 1)(𝒂𝑥 2 + 𝒃𝑥 + 𝒄)
(SEE NEXT PAGE) Find 𝑎, 𝑏, 𝑐 using synthetic
division
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©Via Afrika >> Mathematics Grade 12 55
7
Chapter
Polynomials
SYNTHETIC DIVISION
𝑓(−1) = 0 , so (𝑥 + 1)is a factor
𝑓(−1) = 0 𝑓(𝑥 ) = 2𝑥 3 − 2𝑥 2 − 10𝑥 − 6
−𝟏 𝟐 −𝟐 −𝟏𝟎 −𝟔
−𝟐 𝟒 𝟔
𝟐 −𝟒 −𝟔 𝟎
This method is called SYNTHETIC division, because we don’t really divide.
We actually multiply and add.
Note the following:
• The 𝑥-value of −1 that gave us the factor (𝑥 + 1) is written on the LHS
• The coefficients of the cubic polynomial are written in the top row
• The first coefficient, 2, is carried down to the last row
• Now starting from the left:
MULTIPLY along the dotted arrow repeat
and write the ANSWER in the block one row up and one column right steps
• Now ADD DOWN in the column (the two values underneath each other)
• You MUST get 0 in the last block
• The 3 values in the bottom row are the coefficients of the quadratic factor.
So, 𝑓(𝑥) = (𝑥 + 1)(2𝑥 2 − 4𝑥 − 6)
You can now complete the factorising:
𝑓(𝑥) = (𝑥 + 1)(2𝑥 + 2)(𝑥 − 3)
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©Via Afrika >> Mathematics Grade 12 56
Chapter7
Polynomials
Mixed Exercise on Polynomials
1 Factorise the following expressions completely:
a 27𝑥 3 − 8 b 5𝑥 3 + 40
c 𝑥 3 + 3𝑥 2 + 2𝑥 + 6 d 4𝑥 3 − 𝑥 2 − 16𝑥 + 4
e 4𝑥 3 − 2𝑥 2 + 10𝑥 − 5 f 𝑥 3 + 2𝑥 2 + 2𝑥 + 1
g 𝑥 3 − 𝑥 2 − 22𝑥 + 40 h 𝑥 3 + 2𝑥 2 − 5𝑥 − 6
i 3𝑥 3 − 7𝑥 2 + 4 j 𝑥 3 − 19𝑥 + 30
k 𝑥3 − 𝑥2 − 𝑥 − 2
2 Solve for 𝑥:
a 𝑥 3 + 2𝑥 2 − 4𝑥 = 0
b 𝑥 3 − 3𝑥 2 − 𝑥 + 6 = 0
c 2𝑥 3 − 12𝑥 2 − 𝑥 + 6 = 0
d 2𝑥 3 − 𝑥 2 − 8𝑥 + 4 = 0
e 𝑥3 + 𝑥2 − 2 = 0
f 𝑥 3 = 16 + 12𝑥
g 𝑥 3 + 3𝑥 2 = 20𝑥 + 60
3 Show that 𝑥 − 3 is a factor of 𝑓(𝑥) = 𝑥 3 − 𝑥 2 − 5𝑥 − 3 and hence solve 𝑓(𝑥) = 0.
4 Show that 2𝑥 − 1 is a factor of 𝑔(𝑥) = 4𝑥 3 − 8𝑥 2 − 𝑥 + 2 and hence solve 𝑔(𝑥) = 0.
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©Via Afrika >> Mathematics Grade 12 57
Chapter 8
Differential calculus
Overview
Unit 1 Page 170
Limits • Investigating limits
Unit 2 Page 172
The gradient of a graph at a • Function notation and the
point average gradient
• The gradient of a graph at a
point
Unit 3 Page 176
The derivative of a function • First principles
• Rules for differentiation
• The derivative at a point
Chapter 8 Page 166
Differential Unit 4 Page 182
calculus The equation of a tangent • Calculating the equation of a
to a graph tangent to a graph
Unit 5 Page 184
The graph of the cubic • Plotting a cubic function
function
Unit 6 Page 188
The second derivative • Change in concavity
(concavity)
Unit 7 Page 192
Application of differential • Modelling real-life problems
calculus
REMEMBER YOUR STUDY APPROACH SHOULD BE:
1 Work through all examples in this chapter of your Learner’s Book.
2 Work through the notes in this chapter of the study guide.
3 Do the exercises at the end of the chapter in the Learner’s Book.
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©Via Afrika >> Mathematics Grade 12 58
Chapter 8
Differential calculus
THE CONCEPT OF A LIMIT
Notation: 𝐥𝐥𝐥𝒙→𝟒 𝒇(𝒙)
We say: “The limit of 𝒇 as 𝒙 approaches 𝟒”
What does it mean?
The limit is the 𝒚-value (remember 𝒚 = 𝒇(𝒙)) which the function
approaches as the 𝒙-value approaches (gets closer to) a certain value
from the left or the right .
Examples: a Let 𝒇(𝒙) = 𝟐𝒙𝟐 + 𝟒
𝐥𝐥𝐥𝒙→𝟏 𝒇(𝒙) = 𝐥𝐥𝐥𝒙→𝟒 𝟐𝒙𝟐 + 𝟒
= 𝟐(𝟏)𝟐 + 𝟒 = 𝟔
Before calculating the limit, it is sometimes necessary to FACTORISE and
SIMPLIFY first:
An examples of this is:
𝑥 2 −9
lim𝑥→3 Substituting 𝑥 = 3 now will cause division by 0
𝑥−3
(𝑥−3)(𝑥+3)
= lim𝑥→3 First factorise the numerator and cancel out
𝑥−3
= lim (𝑥 + 3)
𝑥→3
= (3 + 3) Note that “lim” falls away in the step where you
=6 substitute
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©Via Afrika >> Mathematics Grade 12 59
Chapter 8
Differential calculus
AVERAGE GRADIENT BETWEEN 2 POINTS
From previous grades you know you can calculate the gradient between two
𝑦 −𝑦
points (𝑥1 ; 𝑦1 ) and (𝑥2 ; 𝑦2 ) using the formula: 𝑚 = 𝑥2−𝑥1
2 1
In the diagram below the points A(𝑥; 𝑓(𝑥)) and B(𝑥 + ℎ; 𝑓(𝑥 + ℎ)) are indicated.
𝑓(𝑥+ℎ)−𝑓(𝑥)
The AVERAGE GRADIENT between A and B is given by: 𝑚𝐴𝐴 =
ℎ
THE GRADIENT OF A GRAPH AT A POINT
By letting ℎ approach 0, the distance between point A and B will become smaller
and smaller. A and B will almost “become one point”.
The average gradient then becomes the
𝑓(𝑥+ℎ)−𝑓(𝑥)
GRADIENT OF THE GRAPH AT A POINT = limℎ→0
ℎ
NOTATION: This is denoted by 𝒇′ (𝒙)
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©Via Afrika >> Mathematics Grade 12 60
Chapter 8
Differential calculus
𝑓(𝑥+ℎ)−𝑓(𝑥)
The formula 𝑓 ′ (𝑥) = limℎ→0 can be used to find any of the
ℎ
following from FIRST PRINCIPLES: Look out for the words:
FIRST PRINCIPLES
• The derivative of 𝑓 at any point
• The gradient of the tangent to graph 𝑓 at any point
• The gradient of the function 𝑓 at any point
• The rate of change of 𝑓 at any point
𝒇′ (𝒙) can also be determined using DIFFERENTIATION RULES
Function 𝒇 Derivative 𝑓 ′ (𝑥) Examples
• 𝑓(𝑥) = −5
𝒇(𝒙) = 𝒌 𝑓 ′ (𝑥) = 0
𝑓′(𝑥) = 0 • 𝑦=4
where 𝒌 is a constant 𝑑𝑦
=0
𝑑𝑥
• 𝐷𝑥 [𝑥 6 ] = 6𝑥 5
1
𝒇(𝒙) = 𝒙𝒔 ; 𝒙𝝐𝑹 𝑓 ′ (𝑥) = 𝑖𝑥 𝑛−1 • 𝑓(𝑥) = 3 = 𝑥 −3
𝑥
′ (𝑥) −3
𝑓 = −3𝑥 −4 =
𝑥4
4
• 𝑓(𝑥) = 2𝑥
𝑓 ′ (𝑥) = 2 × 4𝑥 4−1
𝒇(𝒙) = 𝒌𝒙𝒎 ; 𝒎 ∈ 𝑹 = 8𝑥 3
where 𝒌 is a constant 𝑓 ′ (𝑥) = 𝑘 × 𝑚𝑥 𝑚−1 1 5
1 5 5
−1
• 𝐷𝑥 � 𝑥 � = × 𝑥
2 2
2 2 2
5 3
= 𝑥2
4
When functions are added/subtracted, apply 2
• 𝐷𝑥 [5𝑥 − 4𝑥 + 6]
the rule to each function separately: = 5 × 2𝑥 − 4 + 0
= 10𝑥 − 4
𝑫𝒙 [𝒇(𝒙) ± 𝒈(𝒙)] = 𝑫𝒙 [𝒇(𝒙)] ± 𝑫𝒙 [𝒈(𝒙)]
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©Via Afrika >> Mathematics Grade 12 61
8
Chapter
Differential calculus
BEFORE YOU APPLY THE DIFFERENTIATION RULES,
MAKE SURE THERE ARE:
• No brackets :
a 𝑓(𝑥) = (𝑥 + 1)(2𝑥 − 1) = 2𝑥 2 + 𝑥 − 1
𝑓 ′ (𝑥) = 4𝑥 + 1
• No 𝑥 under a fraction line:
3𝑥 2 −2 3𝑥 2 2
b 𝑓(𝑥) = 𝑥
= 𝑥 − 𝑥 = 3𝑥 − 2𝑥 −1
2
𝑓 ′ (𝑥) = 3 − 2(−1)𝑥 −2 = 3 + 𝑥 2
𝑥 2 −𝑥−6 (𝑥−3)(𝑥+2)
c 𝑓(𝑥) = 𝑥+2
= =𝑥−3
𝑥+2
𝑓 ′ (𝑥) = 1
• No 𝑥 under a root sign:
1
d 𝑓(𝑥) = 3√𝑥 − 4𝑥 = 3𝑥 2 − 4𝑥
1 1
1 3
𝑓 ′ (𝑥) = 3 × 𝑥 2−1 − 4 = 𝑥 −2 − 4
NB: NOTE THE DIFFERENCE BETWEEN THE FOLLOWING:
𝑓(4) is the 𝒚-VALUE of the function at 𝑥 = 4
𝑓 ′ (4) is the GRADIENT of the function at 𝑥 = 4
As well as the gradient of the TANGENT at 𝑥 = 4
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©Via Afrika >> Mathematics Grade 12 62
Chapter 8
Differential calculus
THE CUBIC GRAPH 𝒚 = 𝒂𝒙𝟏 + 𝒃𝒙𝟐 + 𝒄𝒙 + 𝒅
𝒂 indicates THE SHAPE
STATIONARY POINTS
𝒂 > 0 (+)
TURNING POINTS POINTS OF
or
INFLECTION
𝑎
𝒂 < 0(−) Where 𝑓 ′ (𝑥) = 0 Where 𝑓 ′′ (𝑥) = 0
Local maximum
or
Local minimum
How do I determine whether it is:
A LOCAL MAXIMUM or A LOCAL MINIMUM?
𝑓 ′′ ( ) 0 𝑓 ′′ ( ) 0
What does 𝒇′ and 𝒇′′ tell me?
Negative (< 0) =0 Positive (> 0)
𝒇′ 𝑓 decreases 𝑓 turns 𝑓 increases
𝒇′′ Local maximum Point of inflection Local minimum
Concave down Concave up
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©Via Afrika >> Mathematics Grade 12 63
Chapter8
Differential calculus
𝒙 −INTERCEPTS/ROOTS AND SHAPE
• For 𝑥 −intercepts: solve 𝑓(𝑥) = 0
• A cubic graph can have either one(only) or two or three 𝑥 −intercepts
EXAMPLES:
a 𝑓(𝑥) = 𝑥 3 + 4𝑥 2 − 11𝑥 − 30
𝑓(3) = (3)3 + 4(3) − 11(3) − 30 = 0
-5 -2 3
∴ (𝑥 − 3) is a factor
𝑓(𝑥) = (𝑥 − 3)(𝑥 2 + 7𝑥 + 10)
= (𝑥 − 3)(𝑥 + 2)(𝑥 + 5)
The roots are −5; −2 and 3.
b 𝑓(𝑥) = 𝑥 3 − 3𝑥 + 2
𝑓(1) = (1)3 − 3(1) + 2 = 0
∴ (𝑥 − 1) is a factor
𝑓(𝑥) = (𝑥 − 1)(𝑥 2 + 𝑥 − 2)
𝑓(𝑥) = (𝑥 − 1)(𝑥 − 1)(𝑥 + 2) = (𝑥 − 1)2 (𝑥 + 2)
If TWO FACTORS are the SAME, then the 𝑥 −INTERCEPT is also a TURNING POINT.
The graph BOUNCES at 𝑥 = 1
-2 1
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©Via Afrika >> Mathematics Grade 12 64
Chapter 8
Differential calculus
EQUATION OF TANGENT TO GRAPH AT A SPECIFIC POINT
• Substitute 𝑥 −value into 𝒇(𝒙) to find coordinates of POINT of TANGENCY
• Determine 𝑓 ′ (𝑥) using differentiation rules
• Substitute 𝑥 −value into 𝒇′ (𝒙) to find GRADIENT of TANGENT, 𝒎
• Substitute gradient into 𝑦 = 𝒎𝑥 + 𝑐
• Substitute point of tangency in 𝑦 = 𝒎𝑥 + 𝑐 to find the value of 𝑐.
EXAMPLE
Determine equation of tangent to 𝑓(𝑥) = 2𝑥 3 − 5𝑥 2 − 4𝑥 + 3 at 𝑥 = 1
Substitute 𝑥 = 1: 𝑓(1) = 2(1)3 − 5(1)2 − 4(1) + 3 = −4
∴ Point of tangency is (1; −4)
𝑓 ′ (𝑥) = 6𝑥 2 − 10𝑥 − 4
𝑓 ′ (1) = 6(1)2 − 10(1) − 4 = −8
∴ Gradient of tangent at 𝑥 = 1 is −8 ; so 𝑦 = −8𝑥 + 𝑐
Substitute (1; −4) into 𝑦 = −8𝑥 + 𝑐: −4 = −8(1) + 𝑐
∴𝑐=4
Equation of tangent is 𝑦 = −8𝑥 + 4
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©Via Afrika >> Mathematics Grade 12 65
Chapter 8
Differential calculus
SKETCHING THE CUBIC GRAPH
EXAMPLE: 𝒇(𝒙) = 𝒙𝟏 − 𝒙𝟐 − 𝟏𝒙 + 𝟏𝟐
DESCRIPTION OF STEP STEP APPLIED TO THIS EXAMPLE
Determine shape (using 𝒂) 𝑎 = 1 (positive)
Determine 𝒚 −intercept 𝑦 = (0)3 − (0)2 − 8(0) + 12 = 12
Make 𝒙 = 𝟎
Determine 𝒙 −intercepts 𝑓(2) = 0
Solve 𝒇(𝒙) = 𝟎 ∴ (𝑥 − 2) is a factor
𝑓(𝑥) = (𝑥 − 2)(𝑥 2 + 𝑥 − 6)
𝑓(𝑥) = (𝑥 − 2)(𝑥 − 2)(𝑥 + 3)
Roots are −3 and 2.
𝑥 = 2 is also a turning point where graph bounces
Determine turning points and their 𝑓 ′ (𝑥) = 3𝑥 2 − 2𝑥 − 8 = 0
(3𝑥 + 4)(𝑥 − 2) = 0
𝒚 −values 4
𝑥 = − 3 or 𝑥 = 2
Solve 𝒇′(𝒙) = 𝟎
Substitute 𝒙 −values into 𝒇(𝒙) We already know from the previous step that (2; 0)
is one turning point (local minimum).
Let us now find the other TP’s 𝑦 −coordinates
−4 3 −4 2 −4
𝑓(𝑥) = � � − � � − 8 � � + 12 = 18,52
3 3 3
Local maximum at (−1,33 ; 18,52)
Make a neat drawing y
f
(−1,33 ; 18,52)
12
x
−3 2
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©Via Afrika >> Mathematics Grade 12 66
Chapter 8
Differential calculus
FINDING THE EQUATION OF A CUBIC GRAPH IN THE FORM
𝒚 = 𝒂𝒙𝟏 + 𝒃𝒙𝟐 + 𝒄𝒙 + 𝒅
INFORMATION GIVEN (CAN BE SHOW ON A STEPS
GRAPH OR NOT)
From the 𝑦 −intercept we already know that
𝑑 = −8.
But we are going to use the three roots:
𝑦 = 𝑎(𝑥 + 2)(𝑥 + 1)(𝑥 − 4)
Substitute the point (0; −8):
𝒚 −intercept: 𝒚 = −𝟏 −8 = 𝑎(0 + 2)(0 + 1)(0 − 4)
and 𝒙 −intercepts: 𝒙 = −𝟐; −𝟏 and 𝟒 −8 = −8𝑎
𝑎=1
∴ 𝑦 = 1(𝑥 + 2)(𝑥 + 1)(𝑥 − 4)
Removing the brackets gives:
𝑦 = 𝑥 3 − 𝑥 2 − 10𝑥 − 8
We were given two roots (of which one is also
a turning point) and the other turning point.
y
NB: The graph BOUNCES at 𝑥 = 1. This factor
will therefore have to be squared.
1 x
4
𝑦 = 𝑎(𝑥 − 1)2 (𝑥 − 4)
Substitute the other turning point (3; −4):
( 3;−4)
−4 = 𝑎(3 − 1)2 (3 − 4)
−4 = −4𝑎
𝑎=1
∴ 𝑦 = 𝑎(𝑥 − 1)2 (𝑥 − 4)
Removing the brackets gives:
𝑦 = 𝑥 3 − 6𝑥 2 + 9𝑥 − 4
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©Via Afrika >> Mathematics Grade 12 67
8
Chapter
Differential calculus
SPECIAL APPLICATIONS OF DERIVATIVES
RATES OF CHANGE
• Distance/Displacement 𝑠(𝑡)
• Speed/Velocity 𝑠 ′ (𝑡)
• Acceleration 𝑠 ′′ (𝑡)
EXAMPLE
The displacement of a moving object is described by the equation
𝑠(𝑡) = 10𝑡 − 𝑡 2 where 𝑠, represents displacement in metres and 𝑡, time in
seconds.
a Determine the displacement after 2 seconds.
b What time will it take for the object to reach a maximum displacement?
c Determine the velocity of the object after 3 seconds.
d Determine the acceleration of the object. Is it going faster or slower?
SOLUTIONS
a 𝑠(2) = 10(2) − (2)2 = 16 𝑚
b 𝑠 ′ (𝑡) = 10 − 2𝑡 = 0
∴ 10 − 2𝑡 = 0
∴𝑡 =5𝑠
c 𝑠 ′ (3) = 10 − 2(3) = 4 𝑚. 𝑠 −1
d 𝑠 ′′ (𝑡) = −2 𝑚. 𝑠 −2
The object is going slower because the acceleration is negative.
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©Via Afrika >> Mathematics Grade 12 68
Chapter 8
Differential calculus
USING FIRST DERIVATIVE TO DETERMINE
MINIMUM OR MAXIMUM
• For area, 𝐴(𝑥), to be a min/max, solve 𝐴′ (𝑥) = 0
• For volume, 𝑉(𝑥), to be a min/max, solve 𝑉 ′ 4(𝑥) = 0
• For cost, 𝐶(𝑥), to be a minimum, solve 𝐶 ′ (𝑥) = 0
• For profit, 𝑃(𝑥), to be a maximum, solve 𝑃′(𝑥) = 0
EXAMPLE
The volume of water in a water reservoir is given by: 𝑉(𝑡) = 60 + 8𝑡 − 3𝑡 2
where 𝑉(𝑡) is the volume in thousands of litres and 𝑡 is the number of days
water is pumped into the reservoir.
a Determine the rate of change of the volume after 3 days.
b When will the volume of water in the reservoir be a maximum?
c What will the maximum level of water in the reservoir be?
SOLUTIONS
a 𝑉 ′ (𝑡) = 8 − 6𝑡
∴ 𝑉 ′ (3) = 8 − 6(3) = −10 thousand liters/day
b 𝑉 ′ (𝑡) = 8 − 6𝑡 = 0
∴ 8 − 6𝑡 = 0
4
𝑡 = 3 = 1,3 days
4 4 4 2
c 𝑉 �3� = 60 + 8 �3� − 3 �3� = 58,67 thousand litres
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©Via Afrika >> Mathematics Grade 12 69
Chapter 8
Differential calculus
Mixed Exercise on Differential Calculus
1 Determine 𝑓′ from first principles if
a 𝑓(𝑥) = 1 − 𝑥 2
b 𝑓(𝑥) = −3𝑥 2
2 Determine:
𝑑𝑦 1
a 𝑑𝑥
if 𝑦 = √𝑥 − 2𝑥 2
2𝑥 2 −𝑥−15
b 𝐷𝑥 � 𝑥−3
�
3 Determine the equation of the tangent to the curve 𝑓(𝑥) = −2𝑥 3 + 3𝑥 2 + 32𝑥 + 15 at
the point 𝑥 = −2.
4 Sketch the graph with the following properties showing all the key points on the
graph:
𝑓 ′ (𝑥) < 0 when 1 < 𝑥 < 5
𝑓 ′ (𝑥) > 0 when 𝑥 < 1 and 𝑥 > 5
𝑓 ′ (5) = 0 and 𝑓 ′ (1) = 0
𝑓(0) = −6 and 𝑓(3) = 0
𝑓 ′′ (3) = 0
5 (2; 9) is a turning point of the graph 𝑓(𝑥) = 𝑎𝑥 3 + 5𝑥 2 + 4𝑥 + 𝑏. Determine the
values of 𝑎 and 𝑏 in the equation of 𝑓.
6 The diagram below represents the graph of 𝑦 = 𝑓 ′ (𝑥), the derivative of 𝑓.
4 y
x
a Write down the 𝑥-values of the turning point of 𝑓. −2 2 4 6 8
b Write down the 𝑥-value of the point of inflection of 𝑓. −4
c For which values of 𝑥 will 𝑓(𝑥) decrease? −8
−12
y=f '(x)
−16
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©Via Afrika >> Mathematics Grade 12 70
Chapter8
Differential calculus
7 The graph of 𝑓(𝑥) = 𝑥 3 + 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 is drawn. The curve has turning points at B
and (1; 0). The points (−1; 0) and (1; 0) are 𝑥-intercepts.
y
a Show that 𝑎 = −1; 𝑏 = −1 and 𝑐 = 1. B
b Determine the coordinates of B. x
(−1;0) (1;0)
8 The distance covered in metres by an object is given as 𝑠(𝑡) = 𝑡 3 − 2𝑡 2 + 3𝑡 + 5.
Determine:
a an expression for the speed of the object at any time 𝑡.
b the time at which the speed of the object is at a minimum.
c the time at which the acceleration of the object will be 8 𝑚. 𝑠 −2.
9 The sketch below shows a rectangular box with base ABCD.
AB = 2𝑥 metres and BC = 𝑥 metres. The volume of the box is 24 cubic metres.
Material to cover the top (PQRS) of the box costs R25 per square metre.
Material to cover the base ABCD and the four sides costs R20 per square metre.
a Show that the height(ℎ) of the box is given by ℎ = 12𝑥 −2 .
b Show that the total cost (C) in rand is given by: 𝐶(𝑥) = 90𝑥 2 + 1440𝑥 −1 .
c Determine the value of 𝑥 for which the cost will be a minimum.
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©Via Afrika >> Mathematics Grade 12 71
Chapter9
Analytical geometry
Overview
Unit 1 Page 204
Equation of a circle with centre • Finding the equation of a
at the origin circle
• Symmetrical points on a
circle
Chapter 9 Page 202 Unit 2 Page 208
Analytical Equation of a circle centred off • Finding the equation of
geometry the origin a circle with any given
centre
• General form
Unit 3 Page 214
The equation of the tangent to • Lines on circles
the circle • Equation of a tangent
REMEMBER YOUR STUDY APPROACH SHOULD BE:
1 Work through all examples in this chapter of your Learner’s Book.
2 Work through the notes in this chapter of the study guide.
3 Do the exercises at the end of the chapter in the text book.
4 Do the mixed exercises at the end of this chapter in the study guide.
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©Via Afrika >> Mathematics Grade 12 72
Chapter9
Analytical geometry
REVISION OF CONCEPTS FROM PREVIOUS GRADES
CONCEPT FORMULA / METHOD
Distance between two points 𝑨(𝒙𝟏 ; 𝒚𝟏 ) and 𝑑 = �(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2
𝑩(𝒙𝟐 ; 𝒚𝟐 )
𝑥1 + 𝑥2 𝑦1 + 𝑦2
Coordinates of midpoint � ; �
2 2
𝑦2 − 𝑦1
Average gradient between two points 𝑚=
𝑥2 − 𝑥1
𝑨(𝒙𝟏 ; 𝒚𝟏 ) and 𝑩(𝒙𝟐 ; 𝒚𝟐 )
Or when given the angle of inclination, 𝑠, use
Gradient of straight line through 𝑨 and 𝑩 𝑚 = 𝑡𝑎𝑖𝑠
Equation of a straight line 𝑦 = 𝑚𝑥 + 𝑐
Or 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
Angle of inclination, 𝜽 𝑚 = 𝑡𝑎𝑖𝑠
NB: Angle between line and POSITIVE If 𝑚 > 0(+) , then 𝑠 is
𝒙 −axis an acute angle
(smaller than 90°)
If 𝑚 < 0(−) , then 𝑠 is
an obtuse angle
(bigger than 90° but )
To prove that points 𝑨, 𝑩 and 𝑪 are collinear Prove that 𝑚𝐴𝐴 = 𝑚𝐴𝐸
(i.e. arranged in a straight line) Or 𝑚𝐴𝐴 = 𝑚𝐴𝐸
Or 𝑚𝐴𝐸 = 𝑚𝐴𝐸
Parallel lines Two lines 𝑦 = 𝑚1 𝑥 + 𝑐1 and 𝑦 = 𝑚2 𝑥 + 𝑐2
are parallel if 𝑚1 = 𝑚2
Perpendicular lines Two lines 𝑦 = 𝑚1 𝑥 + 𝑐1 and 𝑦 = 𝑚2 𝑥 + 𝑐2
are perpendicular if 𝑚1 × 𝑚2 = −1
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©Via Afrika >> Mathematics Grade 12 73
9
Chapter
Analytical geometry
OTHER DEFINITIONS/CONCEPTS YOU HAVE TO KNOW
DEFINITION EXAMPLES
A (1;4)
C(3;-2)
K
B(-3;4)
Altitude of a triangle = To determine the equation of altitude AK:
line from one vertex perpendicular to • Determine gradient of BC
opposite side • Determine gradient of AK
• Determine equation of AK (substitute A)
−2 − 4
𝑚𝐴𝐸 = = −1
3 − (−3)
But BC⏊AK, so 𝑚𝐴𝐾 = 1
Substitute point A(1;4):
𝑦 − 4 = −1(𝑥 − 1)
Equation of AK:
𝑦 = −𝑥 + 5
K(1;6)
M(4;2)
A
L(-4;-4)
Median = line joining vertex of triangle to To determine the equation of median KA:
midpoint of opposite side • Determine coordinates of midpoint A
• Determine gradient of KA
• Determine equation of KA
−4 + 4 −4 + 2
𝐴� ; �
2 2
𝐴(0; −1)
6 − (−1)
𝑚𝐾𝐴 = =7
1−0
𝑦 − 6 = 7(𝑥 − 1)
𝑦 = 7𝑥 − 1
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©Via Afrika >> Mathematics Grade 12 74
Chapter9
Analytical geometry
C(4;6)
B(-2;4)
Perpendicular bisector = the line through To determine equation of perpendicular
the midpoint of a line and bisector of BC:
perpendicular to that line • Determine gradient of BC
• Determine gradient of bisector
• Determine equation of bisector
6−4 1
𝑚𝐴𝐸 = =
4 − (−2) 3
Product of gradients must be −1:
𝑚𝑝𝑒𝑟𝑝 𝑏𝑖𝑠𝑒𝑎𝑡𝑜𝑟 = −3
𝑦 − 6 = −3(𝑥 − 4)
𝑦 = −3𝑥 + 18
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©Via Afrika >> Mathematics Grade 12 75
Chapter9
Analytical geometry
THE CIRCLE : (𝒙 − 𝒂)𝟐 + (𝒚 − 𝒃)𝟐 = 𝒓𝟐
(centre-radius form)
SUMMARY ON CIRCLES
Equation of circle with radius 𝒓
𝑥2 + 𝑦2 = 𝑟2
and centre at the origin
Equation of circle with radius 𝒓
(𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2
and centre (𝒂; 𝒃)
Example A: Determine the radius and centre of the circle
with equation (𝑥 + 1)2 + (𝑦 − 3)2 = 16
(𝑥 + 1)2 + (𝑦 − 3)2 = 16 can be written as
To determine radius and (𝑥 − (−1))2 + (𝑦 − (3))2 = 16
centre of circle when given
equation Centre: (−1; 3) Radius = √16 = 4
Example B: Determine the radius and centre of the circle
with equation 𝑥 2 + 𝑦 2 + 4𝑥 + 6𝑦 − 10 = 0
We are going to use COMPLETION OF THE SQUARE
• Constant term to RHS; group 𝑥- and 𝑦-terms
𝑥 2 + 4𝑥 + ⋯ + 𝑦 2 + 6𝑦 + ⋯ = 10
1 2
• Complete square for 𝑥 and 𝑦 - add �2 × 𝑐𝑜𝑐𝑓𝑓𝑖𝑐𝑖𝑐𝑖𝑡�
𝑥 2 + 4𝑥 + 𝟒 + 𝑦 2 + 6𝑦 + 𝟗 = 10 + 𝟒 + 𝟗
• Write in centre-radius form
(𝑥 + 2)2 + (𝑦 + 3)2 = 23
Centre: (−2; −3) Radius = √23
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©Via Afrika >> Mathematics Grade 12 76
Chapter 9
Analytical geometry
NB: Radius is PERPENDICULAR to tangent
Equation of tangent to circle at
given point
Determine the equation of the tangent to the circle
(𝒙 + 𝟏)𝟐 + (𝒚 − 𝟏)𝟐 = 𝟏𝟔 through the point (𝟐; 𝟓).
The steps are:
• Determine centre of circle
• Determine gradient of RADIUS
• Determine gradient of TANGENT:
Remember: 𝒎𝒓𝒂𝒅 × 𝒎𝒕𝒂𝒔 = −𝟏
• Determine equation of tangent by substituting point
of tangency
The centre of circle (𝒙 + 𝟏)𝟐 + (𝒚 − 𝟏)𝟐 = 𝟏𝟔 is (−𝟐; 𝟏).
Radius joins centre (−𝟏; 𝟏) with point of tangency
(𝟐; 𝟓).
𝟓−𝟏 𝟐 𝟏
𝒎𝒓𝒂𝒅 = = =
𝟐 − (−𝟐) 𝟒 𝟐
∴ 𝒎𝒕𝒂𝒔𝒈𝒆𝒔𝒕 = −𝟐
Equation of tangent: 𝒚 − 𝟓 = −𝟐(𝒙 − 𝟐)
𝒚 = −𝟐𝒙 + 𝟗
Mixed Exercise on Analytical Geometry
1 A (−2; 1), B(𝑝; −4), C(5; 0) and D (3; 2) are the vertices of trapezium ABCD in a
Cartesian plane with 𝐴𝐴ǁ𝐶𝐷.
a Show that 𝑝 = 3.
b Calculate AB:CD in simplest form.
c If N (𝑥; 𝑦) is on AB and NBCD is a parallelogram, determine the coordinates of
N.
d Determine the equation of the line passing through B and D.
e What is the angle of inclination of line BD?
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©Via Afrika >> Mathematics Grade 12 77
9
Chapter
Analytical geometry
f Calculate the area of parallelogram NBCD.
g R (−1; 𝑞), A and C are collinear. Calculate the value of 𝑞.
2 𝑥 2 + 4𝑥 + 𝑦 2 + 2𝑦 – 8 = 0 is the equation of a circle with centre M in a Cartesian
plane.
a Prove that the circle passes through the point N(1; −3)
b Determine the equation of PN, the tangent to the circle at N.
c Calculate 𝑠, the angle of inclination of the tangent, rounded off to one decimal
place.
d Determine the coordinates of the point where the tangent in 2 b intersects the
𝑥-axis.
e Calculate the coordinates of the point(s) where the circle with centre M cuts
the 𝑦-axis.
3 In the diagram, P, R and S are vertices of ∆PRS . P is a point on the y-axis. The
coordinates of R is (-6; -12). The equation of PR is 3𝑥 – 𝑦 + 6 = 0.
The median SM and the altitude RN intersect at the origin O.
a Calculate the gradient of RO.
b Calculate the gradient of PS.
c Determine the equation of PS.
d Calculate the inclination of PS rounded of to one
decimal digit.
e ( )
If the coordinates of N are 2n; 3 5 + n , determine
3
the value of 𝑖.
f Calculate the coordinates of S.
4 The equation of a circle is 𝑥 2 + 𝑦 2 + 4𝑥 – 2𝑦 – 4 = 0.
a Determine the coordinates of M, the centre of the circle, as well as the length
of the radius.
b Calculate the value of p if N(𝑝; 1) with 𝑝 > 0, is a point on the circle.
c Write down the equation of the tangent to the circle at N.
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©Via Afrika >> Mathematics Grade 12 78
9
Chapter
Analytical geometry
5 A (-3; 3), B(2; 3), C(6; - 1) and D(𝑥; 𝑦) are the vertices of quadrilateral ABCD in a
Cartesian plane.
a Determine the equation AD.
b Prove that the coordinates of D are
( ) if D is equidistant from B and C.
3 3
;−
2 2
c Determine the equation of BD.
d Determine the size of θ , the angle between BD
and BC, rounded off to one decimal digit.
e Calculate the area of ∆BDC rounded off to the nearest square unit.
6 In the diagram, points A(2; 3), B(𝑝; 0) and C(5; - 3) are the vertices of ∆ABC in a
Cartesian plane. AC cuts the 𝑥-axis at D.
a Calculate the coordinates of D.
b Calculate the value of p if BC = AC and 𝑝 < 0.
c Determine the angle of inclination of
straight line AC, rounded off to one
decimal place.
d If 𝑝 = −1, calculate the size of  , rounded
off to one decimal digit.
7 In the Cartesian plane the equation of a circle with centre M is given by:
𝑥 2 + 𝑦 2 + 6𝑦 – 7 = 0
Determine, by calculation, whether the straight line 𝑦 = 𝑥 + 1 is a tangent to
the circle, or not.
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©Via Afrika >> Mathematics Grade 12 79
9
Chapter
Analytical geometry
8 In the diagram, centre C of the circle lies on the straight line 3𝑥 + 4𝑦 + 7 = 0.
The straight line cuts the circle at D and E(−1; −1). The circle touches the y-axis at
P(0; 2).
a Determine the equation of the circle
in the form: (𝑥 – 𝑖)2 + (𝑦 – 𝑞)2 = 𝑟 2
b Determine the length of diameter DE.
c Determine the equation of the perpendicular
bisector of PE.
d Show that the perpendicular bisector of PE and
straight line DE intersect at C.
9 In the diagram, P, R(4; −4), S and T (0; 4) are the vertices of a rectangle.
P and S lie on the 𝑥 – axis. The diagonals intersect at W.
a Show that the coordinates of S are (2 + 2√5; 0).
b Determine the gradient of TS rounded off to
two decimals.
c Calculate 𝑅𝑇�𝑆 rounded off to two decimals.
10
a Show that the equation of the tangent to the circle 𝑥 2 + 𝑦 2 – 4𝑥 + 6𝑦 + 3 = 0
at the point (5; -2) is 𝑦 = −3𝑥 + 13
b If T(𝑥; 𝑦) is a point on the tangent in 10.a, such that its distance from the centre of the
circle is 20 units, determine the values of 𝑥 and 𝑦.
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©Via Afrika >> Mathematics Grade 12 80
10
Chapter
Euclidean geometry
Overview
Unit 1 Page 244
Proportionality in triangles • Ratio
• Theorem 1
Chapter10 Page 236 Unit 2 Page 250
Euclidean Similarity in triangles • Theorem 2
geometry • Theorem 3
Unit 3 Page 256
Theorem of Pythagoras • Prove of Theorem of
Pythagoras
REMEMBER YOUR STUDY APPROACH SHOULD BE:
1 Work through all examples in this chapter of your Learner’s Book.
2 Work through the notes in this chapter of the study guide.
3 Do the exercises at the end of the chapter in the Learner’s Book.
4 Do the mixed exercises at the end of this chapter in the study guide.
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©Via Afrika >> Mathematics Grade 12 81
Chapter10
Euclidean geometry
REVISION of geometry
From previous years
CONGRUENCY
SSS
AAS
SAS
(included angle)
RHS
SIMILARITY
AAA
SSS
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©Via Afrika >> Mathematics Grade 12 82
10
Chapter
Euclidean geometry
PROPERTIES OF SPECIAL QUADRILATERALS
PARALLELOGRAM
• Both pairs of opposite sides are parallel
• Both pairs of opposite side are equal
• Both pairs of opposite angles are equal
• Diagonals bisect each other
RECTANGLE
All properties of parallelogram
Plus:
• Both diagonals are equal in length
• All interior angles are equal to 90°
RHOMBUS
All properties of parallelogram
Plus:
• All sides are equal
• Diagonals bisect each other perpendicularly
• Diagonals bisect interior angles
SQUARE
All properties of a rhombus
Plus:
• All interior angles are 90°
• Diagonals are equal in length
KITE
• Two pairs of adjacent sides are equal
• Diagonal between equal sides bisects other
diagonal
• One pair of opposite angles are equal
(unequal sides)
• Diagonal between equal sides bisects
interior angles (is axis of symmetry)
• Diagonals intersect perpendicularly
TRAPEZIUM
• One pair of opposite sides are parallel
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©Via Afrika >> Mathematics Grade 12 83
10
Chapter
Euclidean geometry
HOW TO PROVE THAT A QUADRILATERAL
IS A PARALLELOGRAM
Prove any ONE of the following (most often by congruency):
• Prove that both pairs of opposite sides are parallel
• Prove that both pairs of opposite sides are equal
• Prove that both pairs of opposite angles are equal
• Prove that the diagonals bisect each other
HOW TO PROVE THAT A
PARALLLELOGRAM IS A RHOMBUS
Prove ONE of the following:
• Prove that the diagonals bisect each other
perpendicularly
• Prove that any two adjacent sides are equal in length
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©Via Afrika >> Mathematics Grade 12 84
Chapter 10
Euclidean geometry
MIDPOINT THEOREM
The line segment joining the midpoints of two sides of a triangle, is
parallel to the 3rd side of the triangle and half the length of that side.
1
If AD = DB and AE = EC, then DE ǁ BC and DE = 2BC
CONVERSE OF MIDPOINT THEOREM
If a line is drawn from the midpoint of one side of a triangle parallel to
another side, that line will bisect the 3rd side and will be half the length of
the side it is parallel to.
1
If AD = DB and DE ǁ BC, then AE = EC and DE = 2BC.
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©Via Afrika >> Mathematics Grade 12 85
Chapter 10
Euclidean geometry
REVISION OF CIRCLE GEOMETRY (FROM GRADE 11)
Theorem 1 Converse of Theorem 1
If AC = CB in circle O, then OC ⏊ AB. If OC ⏊ chord AB, then AC = BC.
Theorem 2
The angle at the centre of a circle subtended by an arc/a chord is double the angle at the
circumference subtended by the same arc/chord. AO� B = 2 × AC� B
Theorem 3 Converse of Theorem 3
The angle on the circumference subtended by If 𝐶̂ = 90°, then AB is the diameter of the circle.
the diameter, is a right angle. (The angle in a
semi-circle is 90°).
Theorem 4 Converse of Theorem 4
The angles on the circumference of a circle, If a line segment subtends equal angles at two
subtended by the same arc or chord, are equal. other points, then these four points lie on the
circumference of a circle.
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©Via Afrika >> Mathematics Grade 12 86
10
Chapter
Euclidean geometry
Corollaries of Theorem 4 Equal chords of equal circles
Equal chords subtend equal Equal chords subtend equal subtend equal angles at the
angles at the circumference angles at the centre of the circle. circumference.
of the circle.
Theorem 5 Converse of Theorem 5
The opposite angles of a cyclic quadrilateral are If the opposite angles of a quadrilateral
supplementary. are supplementary, then it is a cyclic
𝐴̂ + 𝐶̂ = 180° quadrilateral.
𝐴� + 𝐷� = 180°
Theorem 6 Converse of Theorem 6
The exterior angle of a cyclic quadrilateral is equal If the exterior angle of a quadrilateral is equal to the
to the opposite interior angle. opposite interior angle, then it is a cyclic
quadrilateral.
Theorem 7 Converse of Theorem 7
The tangent to a circle is perpendicular to the radius If a line is drawn perpendicularly to the radius
at the point of tangency. through the point where the radius meets the
circle, then this line is a tangent to the circle.
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©Via Afrika >> Mathematics Grade 12 87
10
Chapter
Euclidean geometry
Theorem 8
If two tangents are drawn from the same point outside a circle, then they are equal in length.
Theorem 9 (Tan chord theorem) Converse of Theorem 9
The angle between the tangent to a circle and a If a line is drawn through the endpoint of a chord to
chord drawn from the point of tangency, is equal to form an angle which is equal to the angle in the
the angle in the opposite circle segment. opposite segment, then this line is a tangent.
Acute angle Obtuse angle
THREE WAYS TO PROVE THAT A QUADRILATERAL
IS A CYCLIC QUADRILATERAL
Prove that:
• one pair of opposite angles are supplementary
• the exterior angle is equal to the opposite interior angle
• two angles subtended by a line segment at two other vertices
of the quadrilateral, are equal.
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©Via Afrika >> Mathematics Grade 12 88
Chapter 10
Euclidean geometry
Example 1
In the diagram alongside< O is the centre of circle DABMC.
BC and DM are diameters.
AC and DM intersect at T.
OT =3DT
ABǁDM
a Prove that T is the midpoint of AC.
b Determine the length of MC in terms of DT.
c � in terms of 𝑂�2.
Express 𝐷
Solution:
a 𝐴̂1 = 90° ∠ in semi ⊙
𝑇�1 = 90° int. ∠s suppl
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©Via Afrika >> Mathematics Grade 12 89
10
Chapter
Euclidean geometry
REVISING THE CONCEPT OF PROPORTIONALITY
A 6 cm B 4 cm C
D 9 cm E 6 cm F
AB : BC = 6 : 4 = 3 : 2
DE : EF = 9 : 6 = 3 : 2
Although, AB : BC = DE : EF it does NOT mean that AB = DE, AC = DF or BC = EF.
GRADE 12 GEOMETRY
Theorem 1 Converse of Theorem 1
A line drawn parallel to one side of a triangle If a line divides two sides of a triangle
that intersects the other two sides, will divide proportionally, then the line is parallel
the other two sides proportionally. to the third side of the triangle.
𝐴𝐴 𝐴𝐴 𝐴𝐴 𝐴𝐴
If DE ǁ BC then = If = then DE ǁ BC.
𝐷𝐷 𝐸𝐸 𝐷𝐷 𝐸𝐸
or AD : DB = AE : EC
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©Via Afrika >> Mathematics Grade 12 90
10
Chapter
Euclidean geometry
Theorem 2 (Midpoint Theorem) Converse of Theorem 2
(Special case of Theorem 1)
The line segment joining the midpoints of two If a line is drawn from the midpoint of
sides of a triangle, is parallel to the 3rd side one side of a triangle parallel to another
of the triangle and half the length of that side. side, that line will bisect the 3rd side and
will be half the length of the side it is
parallel to.
1
If AD = DB and AE = EC, then DE ǁ BC and DE = BC. If AD = DB and DE ǁ BC, then AE = EC
2
1
and DE = BC.
2
Theorem 3 Converse of Theorem 3
The corresponding sides of two equiangular If the sides of two triangles are
proportional, triangles are proportional. then the triangles are equiangular.
𝐴𝐴 𝐵𝐵 𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐴𝐴
If ∆𝐴𝐴𝐴 ||| ∆𝐷𝐷𝐷 then = = If = = then ∆𝐴𝐴𝐴 |||∆𝐷𝐷𝐷
𝐷𝐷 𝐸𝐸 𝐷𝐷 𝐷𝐷 𝐸𝐸 𝐷𝐷
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©Via Afrika >> Mathematics Grade 12 91
10
Chapter
Euclidean geometry
Theorem 4
The perpendicular drawn from the vertex of the right angle of a right-angled triangle,
divides the triangle in two triangles which are similar to each other and similar to the
original triangle.
Corollaries of Theorem 4
∆𝐴𝐴𝐴|||∆𝐷𝐷𝐷 ∆𝐴𝐴𝐴|||∆𝐷𝐷𝐷 ∆𝐷𝐷𝐷|||∆𝐷𝐷𝐷
𝐴𝐴 𝐵𝐵 𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐴𝐴 𝐷𝐷 𝐵𝐵 𝐷𝐷
∴ = = ∴ = = ∴ = =
𝐷𝐷 𝐵𝐵 𝐷𝐷 𝐷𝐷 𝐴𝐴 𝐷𝐷 𝐷𝐷 𝐴𝐴 𝐷𝐷
∴ 𝐴𝐴2 = 𝐵𝐵. 𝐵𝐵 ∴ 𝐴𝐴 2 = 𝐶𝐶. 𝐶𝐶 ∴ 𝐴𝐴 2 = 𝐵𝐵. 𝐷𝐷
Theorem 5 (The Theorem of Pythagoras)
Using the corollaries of Theorem 4, it can be proven that:
𝐵𝐵 2 = 𝐴𝐴2 + 𝐴𝐴 2
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©Via Afrika >> Mathematics Grade 12 92
10
Chapter
Euclidean geometry
Example
4
Given: 𝐴𝐷: 𝐷𝐴 = 2: 3 and 𝐴𝐷 = 3 𝐷𝐶.
Instruction: Determine the ratio of 𝐶𝑃: 𝑃𝐷.
Solution:
𝐴𝐴 5 5
In ∆𝐴𝐴𝐷 𝐾𝐴
=2 ∴ 𝐴𝐷 = 2 𝐾𝐷
4
But it was given that 𝐴𝐷 = 3 𝐷𝐶
4 5
∴ 3 𝐷𝐶 = 2 𝐾𝐷
𝐴𝐸 5 4 15
𝐾𝐴
=2÷3 = 8
𝐸𝑃 𝐸𝐴 15
In ∆𝐶𝐷𝐾 𝑃𝐴
= 𝐴𝐾 = 8
∴ 𝐶𝑃: 𝑃𝐷 = 15: 8
TIPS TO SOLVE A GEOMETRY PROBLEM
• READ-READ-READ the information next to the diagram thoroughly
• TRANSFER all given information on the DIAGRAM
• Look for KEYWORDS, e.g.
TANGENT: What do the theorems say about tangents?
CYCLIC QUADRILATERAL: What are the properties of a cyclic quad?
• Set yourself “SECONDARY” GOALS, e.g.
- To prove that two sides of triangle are equal (primary goal), first prove
that there are two equal angles (secondary goal)
- To prove that a line is a tangent, the secondary goal can be to prove that
the line is perpendicular to a radius
• For questions like: Prove that 𝐴̂1 = 𝐶̂2 . Start with ONE PART. Move to the
OTHER PART step-by-step stating reasons.
� 𝟏 = 𝐴̂2 ; 𝐴̂2 = 𝐶̂1 ; 𝐶̂1 = 𝑪
E.g. 𝑨 �𝟐 ; ∴ 𝑨 �𝟐
�𝟏 = 𝑪
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©Via Afrika >> Mathematics Grade 12 93
10
Chapter
Euclidean geometry
Mixed Exercise on Euclidian Geometry
1 In the diagram, TBD is a tangent to circles BAPC and BNKM at B.
AKC is a chord of the larger circle and is also a tangent to the smaller circle at K.
Chords MN and BK intersect at F. PA is produced to D.
BMC, BNA and BFKP are straight lines.
Prove that:
a MN ǁ CA
b ∆𝐾𝑀𝑁 is isosceles
𝐴𝐾 𝐴𝑀
c 𝐾𝑃
= 𝑀𝐸
d DA is a tangent to the circle passing through
points A, B and K.
2 In the diagram below, chord BA and tangent TC of circle ABC are produced to meet at
R. BC is produced to P with RC=RP. AP is not a tangent.
Prove that:
a ACPR is a cyclic quadrilateral.
b ∆𝐶𝐴𝐴|||∆𝑅𝑃𝐴
𝐸𝐴.𝑅𝐴
c 𝑅𝐶 = 𝐴𝐸
d 𝑅𝐴. 𝐴𝐶 = 𝑅𝐶. 𝐶𝐴
e Hence prove that 𝑅𝐶 2 = 𝑅𝐴. 𝑅𝐴
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©Via Afrika >> Mathematics Grade 12 94
Chapter10
Euclidean geometry
3 In the diagram alongside, circles ACBN and AMBD
Intersect at A and B.
CB is a tangent to the larger circle at B.
M is the centre of the smaller circle.
CAD and BND are straight lines.
Let 𝐴̂3 = 𝑥
a � in terms of 𝑥.
Determine the size of 𝐷
b Prove that:
i CB ǁ AN
ii AB is a tangent to circle ADN.
4 In the diagram below, O is the centre of circle ABCD.
DC is extended to meet circle BODE at point E.
OE cuts BC at F. Let𝐷�1 = 𝑥.
a Determine 𝐴̂ in terms of 𝑥.
b Prove that:
i BE=EC
ii BE is NOT a tangent
to circle ABCD.
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©Via Afrika >> Mathematics Grade 12 95
10
Chapter
Euclidean geometry
5 In the diagram alongside, medians AM and CN
of ∆𝐴𝐴𝐶 intersect at O.
BO is produced to meet AC at P.
MP and CN intersect in D.
ORǁMP with R on AC.
𝑁𝐴
a Calculate, giving reasons, the numerical value of 𝑁𝐸 .
𝑅𝑃
b Use 𝐴𝑂: 𝐴𝑀 = 2: 3, to calculate the numerical value of 𝑃𝐸 .
6 In the diagram, AD is the diameter of circle ABCD.
AD is extended to meet tangent NCP in P.
Straight line NB is extended to Q and intersect AC
in M with Q on straight line ADP.
AC⏊NQ at M.
a Prove that NQǁCD.
b Prove that ANCQ is a cyclic quadrilateral.
c i Prove that ∆𝑃𝐶𝐷|||∆𝑃𝐴𝐶.
ii Hence, complete: 𝑃𝐶 2 = ⋯
d Prove that 𝐴𝐶 2 = 𝐶𝐷. 𝑁𝐴
e If it is further given that PC=MC, prove that
𝐴𝑀2 𝐴𝑃.𝐴𝑃
1 − 𝐴𝐸 2 = 𝐸𝐴.𝑁𝐴
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©Via Afrika >> Mathematics Grade 12 96
11
Chapter
Statistics: regression and correlation
Overview
Unit 1 Page 266
Symmetrical and skewed data • Symmetrical data
• Skewed data
Unit 2 Page 270
Scatter plots and correlation • Bivariate data
• Correlation
Chapter 11 Page 262
• Examples of scatter
Statistics: regression and
plots and correlation
correlation
• Drawing scatter plots
• The least squares
method
• The correlation
coefficient
• Using a calculator to
find the regression line
REMEMBER YOUR STUDY APPROACH SHOULD BE:
1 Work through all examples in this chapter of your text book.
2 Work through the notes in this chapter of the study guide.
3 Do the exercises at the end of the chapter in the text book.
4 Do the mixed exercises at the end of this chapter in the study guide.
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 97
11
Chapter
Statistics: regression and correlation
UNGROUPED DATA GROUPED DATA
Mode = most frequent number Modal class = interval with highest frequency
∑ 𝑥𝑖 ×𝑓𝑖
Estimated mean =
Measures of central tendency
𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 ∑ 𝑓𝑖
Mean = where 𝑥𝑖 = midpoint of class 𝑖
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑎𝑙𝑢𝑒𝑠
and 𝑓𝑖 = frequency of class 𝑖
NB: Data has to be arranged in ascending order
𝑄2 , Median = Middle value Median class interval = class/interval in which
(for an odd number of values) middle value lies
(𝑛+1)
Position of 𝑄2 = 2
𝑆𝑢𝑚 𝑜𝑓 𝑡𝑤𝑜 𝑚𝑖𝑑𝑑𝑙𝑒 𝑣𝑎𝑙𝑢𝑒𝑠
Or 2
(for an even number of values)
Percentiles (divide data into 100 equal parts)
30
E.g. the position of 𝑃30 = 100 (𝑖 + 1)
Measures of dispersion (indicates
𝑄1 , Lower quartile = Middle value of all the values Position of 𝑄1 =
(𝑛+1)
4
below the median (excluding median)
𝑄3 , Upper quartile = Middle value of all the values Position of 𝑄3 =
3(𝑛+1)
4
above the median (excluding median)
spread of data)
Range = Maximum – Minimum
Inter quartile range (IQR) = 𝑄3 − 𝑄1
𝑄3 −𝑄1
Semi Inter quartile range =
2
Five point summary (used to draw box-and-whisker
diagram): Min, 𝑄1 , Median, 𝑄3 , Max
DISTRIBUTION OF DATA
SYMMETRICAL DISTRIBUTION ASYMMETRICAL DISTRIBUTIONS
NORMAL DISTRIBUTION NEGATIVELY SKEWED POSITIVELY SKEWED
𝒎𝒆𝒂𝒔 = 𝒎𝒆𝒅𝒔𝒂𝒔 = 𝒎𝒄𝒅𝒆 𝑚𝑐𝑎𝑖 − 𝑚𝑐𝑑𝑖𝑎𝑖 < 0 𝑚𝑐𝑎𝑖 − 𝑚𝑐𝑑𝑖𝑎𝑖 > 0
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©Via Afrika >> Mathematics Grade 12 98
11
Chapter
Statistics: regression and correlation
OGIVE
Ogive = cumulative frequency graph
80 Ages
70
60
NB: When drawing the ogive:
50
• plot the (upper class boundary ; cumulative frequency) 40
• the graph has to be grounded 30
• the shape of the graph has to be smooth rather
20
10
than consist of “connected dots” Q1 Q3 Freq
5 10 15 20 25 30
THE OGIVE CAN BE USED TO DETERMINE THE MEDIAN AND QUARTILES.
MEASURES OF DISPERSION AROUND THE MEAN
VARIANCE 𝜎 2
Variance, 𝜎 2 , is an indication of how far each value in the data set is from the mean, 𝑥̅ .
∑(𝑥𝑖 −𝑥̅ )2
𝜎2 = (for population)
𝑛
STANDARD DEVIATION 𝜎
Standard deviation (SD), 𝜎: SD = √𝑣𝑎𝑟𝑖𝑎𝑖𝑐𝑐
The larger the standard deviation, the larger the deviation from the mean would be.
A normal distribution is shown below:
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©Via Afrika >> Mathematics Grade 12 99
Chapter11
Statistics: regression and correlation
USING A TABLE TO CALCULATE VARIANCE
AND STANDARD DEVIATION
UNGROUPED DATA
First calculate the mean,𝑥̅ then the following columns.
DATA VALUES, 𝑥 (𝑥 − 𝑥̅ ) (𝑥 − 𝑥̅ )2
Calculate the total of this column,∑(𝑥 − 𝑥̅ )2
∑(𝑥−𝑥̅ )2
Variance = 𝑛
Standard variance = √𝑣𝑎𝑟𝑖𝑎𝑖𝑐𝑐
GROUPED DATA
∑ 𝑓×𝑚
First calculate the estimated mean, 𝑥̅ = ∑𝑓
Class Frequency Midpoint
𝑓×𝑚 𝑚 − 𝑥̅ (𝑚 − 𝑥̅ )2 𝑓 × (𝑚 − 𝑥̅ )2
Interval 𝑓 𝑚
Calculate the total of this column
∑ 𝑓(𝑥−𝑥̅ )2
Variance = 𝑛
Standard variance = √𝑣𝑎𝑟𝑖𝑎𝑖𝑐𝑐
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©Via Afrika >> Mathematics Grade 12 100
Chapter 11
Statistics: regression and correlation
USING CASIO 𝒇𝒇-82ZA PLUS CALCULATOR
TO CALCULATE STANDARD DEVIATION
MODE
2 : STAT To switch on the frequency column
1 : 1 – VAR when calculating the SD for a frequency
table, first do the following:
Enter the data points: Push = after each data point
AC Shift Setup; Down arrow (on big REPLAY
button); 3: STAT; 2: ON
SHIFT STAT (above the 1 button)
4 : VAR
3: 𝜎𝑥𝑖
To clear screen: MODE 1: COMP
DETERMINING OUTLIERS
Inter quartile range, IQR = 𝑄3 − 𝑄1
An outlier is identified if it is:
• Less than 𝑄1 − 𝐼𝑄𝑅 × 1,5 or
• Larger than 𝑄3 + 𝐼𝑄𝑅 × 1,5
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©Via Afrika >> Mathematics Grade 12 101
Chapter 11
Statistics: regression and correlation
SCATTER DIAGRAMS (SCATTER PLOTS)
FOR BIVARIATE DATA
Scatter diagrams are used to graphically determine whether there is an association between
two variables.
By investigation one can determine which of the following curves (regression functions)
would best fit the diagram:
Linear (straight line) Quadratic (parabola) Exponential function
USING A CALCULATOR TO DETERMINE THE EQUATION OF THE
REGRESSION LINE (LEAST SQUARES REGRESSION LINE)
The standard form of a straight line equation is: 𝑦 = 𝑚𝑥 + 𝑐
where 𝑚 is the gradient and 𝑐 is the 𝑦-intercept.
NB: On the calculator the regression line is determined in the form:
𝒚 = 𝑨 + 𝑩𝒙
(In this form 𝐴 = the gradient of the line and 𝐴 = the 𝑦-intercept)
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©Via Afrika >> Mathematics Grade 12 102
11Chapter
Statistics: regression and correlation
On the calculator press:
MODE 2
2: A+Bx
Enter the data points (column X and Y): Push = after each data point
Press AC.
SHIFT STAT
5: REG
1: A = (to determine the 𝑦- intercept of the line)
SHIFT STAT
5: REG
2: B = (to determine the gradient of the line)
SHIFT STAT
5: REG
3: 𝑟 = (to determine correlation coefficient)
EXAMPLE
𝑥 5 10 15 20 25 30 35 40
𝑦 20 223 25 35 30 40 50 55
Using the calculator, the equation for the line of best fit (or regression line) can be
determined giving:
𝑦 = 1𝑥 + 12,25
100 y
95
90
85
80
75
70
65
60
NB: The line of best fit ALWAYS
55 goes through the point (𝑥̅ ; 𝑦�).
50
45
40 In this case it goes through the
point ( 23 ; 35 )
35
30
25
20
15
10
5 x
5 10 15 20 25 30 35 40 45
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©Via Afrika >> Mathematics Grade 12 103
11
Chapter
Statistics: regression and correlation
CORRELATION
The strength of the relationship between the two variables represented in a scatter diagram,
depends on how close the points lie to the line of best fit. The closer the points lie to this
line, the stronger the relationship or correlation.
Correlation (tendency of the graph) can be described in terms of the general distribution of
data points, as follows:
Strong positive Fairly strong positive Perfect positive No correlation
Correlation
Strong negative Fairly strong negative Perfect negative
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©Via Afrika >> Mathematics Grade 12 104
11
Chapter
Statistics: regression and correlation
CORRELATION COEFFICIENT
The correlation between two variables can also be described in terms of a number, called the
correlation coefficient. The correlation coefficient, 𝑟, indicates the strength and the
direction of the correlation between two variables. This number can be anything between
−1 and 1.
𝒓 Interpretation
𝟏 Perfect positive relationship
𝟎, 𝟗 Strong positive relationship
Fairly strong positive
𝟎, 𝟓
relationship
𝟎, 𝟐 Weak positive relationship
𝟎 No relationship
−𝟎, 𝟐 Weak negative relationship
Fairly weak negative
−𝟎, 𝟓
relationship
−𝟎, 𝟗 Strong negative relationship
−𝟏 Perfect negative relationship
Example
Refer to the previous example again.
For the given data set 𝑟 = 0,958 which means that there is a strong positive relationship
between the two variables.
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©Via Afrika >> Mathematics Grade 12 105
11
Chapter
Statistics: regression and correlation
Mixed Exercise on Statistics
1 A national soccer team has participated against teams of other countries in a
competition for the past 14 years. Their results were as follows:
YEAR MATCHES WINS DRAWS LOSSES GOALS GOALS
PLAYED FOR AGAINST
1999 5 3 2 0 11 3
2000 3 1 1 1 2 22
2001 5 3 1 1 10 4
2002 4 2 0 2 8 6
2003 7 2 3 2 5 4
2004 7 6 1 0 14 5
2005 5 2 0 3 8 7
2006 7 5 1 1 15 4
2007 6 1 2 3 9 11
2008 4 2 1 1 4 2
2009 3 1 1 1 2 3
2010 3 1 0 2 5 10
2011 1 0 0 1 2 3
2012 5 4 0 1 18 9
a Determine the quartiles for:
i the matches played
ii the wins
iii the goals scored against the soccer team.
b Draw a box and whisker plot for the goals against the soccer team and comment on
the distribution of the data.
c Calculate the mean of the number of matches played.
d Calculated the standard deviation of the number of matches played.
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©Via Afrika >> Mathematics Grade 12 106
11
Chapter
Statistics: regression and correlation
2 Fifty people were asked what percentage of their December holiday expenses were
related to transport costs. The responses were as follows:
PERCENTAGE FREQUENCY (f)
10 < 𝑥 ≤ 20 6
20 < 𝑥 ≤ 30 14
30 < 𝑥 ≤ 40 16
40 < 𝑥 ≤ 50 11
50 < 𝑥 ≤ 60 3
a Draw an ogive to represent the data above.
b Use your ogive to determine the median percentage of the holiday expenses spent on
travel expenses.
c Calculate the estimated mean.
d Calculate the standard deviation of the data.
3 An athlete’s ability to take and use oxygen is called his VO2 max. The following table
shows the VO2 max and the distance eleven atheletes can run in an hour.
VO2 max 20 55 30 25 40 30 50 40 35 30 50
Distance(km) 8 18 13 10 11 12 16 14 13 9 15
a
Represent the data on a scatter graph.
b Determine the equation of the line of best fit.
c Draw the line of best fit on the scatter graph.
d Use your line of best fit to predict the VO2 max of an athlete that runs 19 km.
e Determine the correlation coefficient of the data and comment on the correlation.
4 Five number 4; 8; 10; 𝑥 and 𝑦 have a mean of 10 and a standard deviation of 4. Find 𝑥
and 𝑦.
5 The standard deviation of five numbers is 7,5. Each number is increased by 2. What
will the standard deviation of the new set of numbers be? Explain your answer.
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 107
Chapter 12
Probability
Overview
Unit 1 Page 282
Solving probability problems • Venn diagrams
• Tree diagrams
• Two-way
contingency tables
Chapter 12 Page 280 Unit 2 Page 288
Probability The counting principle • The fundamental
counting principle
Unit 3 Page 292
The counting principle and • Using the counting
probability principle to calculate
probability
REMEMBER YOUR STUDY APPROACH SHOULD BE:
1 Work through all examples in this chapter of your Learner’s Book.
2 Work through the notes in this chapter of the study guide.
3 Do the exercises at the end of the chapter in the Learner’s Book.
4 Do the mixed exercises at the end of this chapter in the study guide.
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 108
Chapter12
Probability
SUMMARY OF THEORY ON PROBABILITY
CONCEPT/DEFINITION MATHEMATICAL NOTATION/RULE EXAMPLE
Probability = the chance that an 𝑃 Values of probability can range from 0
event will occur to1
• For an event, K, that is certain NOT to
happen 𝑃(𝐾) = 0
• For an event, K that is CERTAIN to
happen 𝑃(𝐾) = 1
Sample Space = the set of all 𝑆
possible outcomes
The number of elements in the 𝑖(𝑆) If 𝑆 = {2; 4; 6} then 𝑖(𝑆) = 3
sample space
General rule for A and B inside the 𝑃(𝐴𝑜𝑟𝐴) = 𝑃(𝐴) + 𝑃(𝐴) − 𝑃(𝐴𝑎𝑖𝑑𝐴)
sample space S
Intersection 𝐴 𝑎𝑖𝑑 𝐴 or 𝐴 ∩ 𝐴
Union 𝐴 𝑜𝑟 𝐴 or 𝐴 ∪ 𝐴
Inclusive events have elements in 𝑃(𝐴 ∩ 𝐴) ≠ 0
A B
common
Mutually exclusive/disjoint events 𝑃(𝐴 ∩ 𝐴) = 0
DON’T INTERSECT,
∴ 𝑃(𝐴𝑜𝑟𝐴) = 𝑃(𝐴) + 𝑃(𝐴)
i.e. have NO elements in common
Exhaustive events = together they ∴ 𝑃(𝐴 ∩ 𝐴) = 1
contain ALL elements of S
Complement of A = all elements Complement of A = A’
which are NOT in A
Complementary events = mutually 𝑃(𝑖𝑜𝑡 𝐴) = 1 − 𝑃(𝐴)
𝑃(𝐴′ ) = 1 − 𝑃(𝐴)
exclusive and exhaustive
Or
(everything NOT in A, is in B)
𝑃(𝐴′ ) + 𝑃(𝐴) = 1
Independent events = outcome of 𝑃(𝐴 ∩ 𝐴) = 𝑃(𝐴) × 𝑃(𝐴) Tossing a coin and throwing a die
1st event DOES NOT influence the
outcome of 2nd event
Dependent events = outcome of 1st 𝑃(𝐴 ∩ 𝐴) ≠ 𝑃(𝐴) × 𝑃(𝐴) Choosing a ball from a bag, not
event DOES influence the outcome replacing it, then choosing a 2nd ball
of 2nd event
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©Via Afrika >> Mathematics Grade 12 109
Chapter 12
Probability
FACTORIAL NOTATION
The product 5 × 4 × 3 × 2 × 1 can be written as
5!
∴ 𝑖! = 𝑖 × (𝑖 − 1) × (𝑖 − 2) × … × 3 × 2 × 1
The Fundamental Counting Principle
RULE EXAMPLE
RULE 1 a) You have 3 pants and 4 shirts.
Where there are 𝒎 ways to do one thing and 𝒔 That means you have 3 × 4 = 12
ways to do another, then there are 𝒎 × 𝒔 ways to different outfits.
do both
RULE 2 b) 5 children have to be seated on 5
Where 𝒔 different things have to be placed in 𝒔 chairs in the front row of a class. The
positions, the number of arrangements is 𝒔! number of ways they can be seated is
5!=120
RULE 3 c) 8 students participated in a 100 m
Where 𝒔 different things have to be placed in 𝒓 race. The first three positions can be
𝒔!
positions, the number of arrangements is (𝒔−𝒓)! occupied in
8! 8!
(8−3)!
= 5! = 8 × 7 × 6 = 336 ways.
RULE 4 d) 3 girls and 4 boys have to be seated
When seating 𝒃 boys and 𝒈 girls in a row, the on 7 chairs, with girls together and boys
number of arrangements are: together.
• Boys and girls in any order: (𝒃 + 𝒈)! ways Number of ways = 2 × 3! × 4! = 288
• Boys together and girls together: 𝟐 × 𝒃! × 𝒈! e)5 Maths books and 2 Science books
ways have to be place on a shelf, but the
• Only girls together: (𝒃 + 𝟏)! × 𝒈! arrangements Maths books have to be placed together.
• Only boys together: (𝒈 + 𝟏)! × 𝒃! Number of ways = (2 + 1)! × 5! = 360
arrangements
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©Via Afrika >> Mathematics Grade 12 110
Chapter 12
Probability
LETTER ARRANGEMENTS
When making new words from the letters in a given word , one has to distinguish between:
Treating repeated letters as Treating repeated letters as IDENTICAL.
DIFFERENT letters.
The following rule applies:
The normal counting
principle (Rule 2) For 𝑛 letters of which 𝑚1 are identical, 𝑚2 are
identical, … and 𝑚𝑛 are identical, the number of
applies here.
arrangements is given by:
𝑛!
𝑚1 ! × 𝑚2 ! × … × 𝑚𝑛 !
Examples:
1 How many different arrangements can be made with the letters of the word
MATHEMATICS, if repeated letters are treated as different letters.
The letters are regarded as 11 different letters.
Number of arrangements 11!
2 How many different arrangements can be made with the letters of the word
MATHEMATICS, if repeated letters are treated as identical.
The letters are regarded as 11 different letters.
11!
Number of arrangements = 2!×2!×2! = 6 652 800 (The M, A and T repeat)
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©Via Afrika >> Mathematics Grade 12 111
Chapter 12
Probability
Mixed Exercise on Probability
1 How many different 074- cell phone numbers are possible if the digits may not
repeat?
2 How many different 082- cell phone numbers are possible if the digits may only be
integers?
3 What is the probability that you will draw a queen of diamonds from a pack cards?
4 How many different arrangements can be made with the letters of the word
TSITSIKAMMA, if:
a repeating letters are regarded as different letters
b repeating letters are regarded as identical.
5 Four different English books, three different German books and two different
Afrikaans books are randomly arranged on a shelf.
Calculate the number of arrangements if:
a the English books have to be kept together
b all books of the same language have to be kept together
c the order of the books does not matter.
6 In how many different ways can a chairman and a vice-chairman be chosen from a
committee of 12 people?
7 The letters of the word MATHEMATICS have to be rearranged. Calculate the probability
that the “word” formed will not start and end with the same letter.
8 In how many different ways can the letters of the word MATHEMATICS rearranged so
that
a the H and the E stay together.
b the E keep its position.
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©Via Afrika >> Mathematics Grade 12 112
Answers to Mixed Exercises
Chapter 1: Number patterns, sequences and series
1 a 𝑇𝑛 = 𝑎 + (𝑖 − 1)𝑑
𝑎=5; 𝑑 =4
𝑇𝑛 = 5 + (𝑖 − 1)4 = 4𝑖 + 1
b 217 = 4𝑖 + 1
4𝑖 = 216
∴ 𝑖 = 54
2 a 9 = 𝑎𝑟 4
729 = 𝑎𝑟 8
729 𝑎𝑟 8
9
= 𝑎𝑟 4
𝑟 4 = 81
𝑟 = ±3
b 𝑇10 = 𝑟 × 𝑇9
𝑇10 = ±2187
3 a 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
�5𝑥 − (2𝑥 − 4)� = �(7𝑥 − 4) − 5𝑥�
5𝑥 − 2𝑥 − 7𝑥 + 5𝑥 = −4 − 4
𝑥 = −8
b −20 ; −40 ; −60
4 a 𝑇𝑛 = 𝑎𝑖2 + 𝑏𝑖 + 𝑐
3
𝑎=2
3 1
𝑏 = 5 − 3 �2� = 2
3 1
𝑐 =2−2−2=0
3 1
𝑇𝑛 = �2� 𝑖2 + �2� 𝑖 Note: alternative methods can be used
3 1
b 260 = �2� 𝑖2 + �2� 𝑖
3𝑖2 + 𝑖 − 520 = 0
(3𝑖 + 40)(𝑖 − 13) = 0
𝑖 = 13
13𝑡ℎ 𝑡𝑐𝑟𝑚 𝑖𝑠 𝑐𝑞𝑢𝑎𝑙 𝑡𝑜 260.
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©Via Afrika >> Mathematics Grade 12 113
Answers to Mixed Exercises
5 𝑇𝑛 = 𝑎 + (𝑖 − 1)𝑑
𝑎 = 17 ; 𝑑 = −3
−2785 = 17 + (𝑖 − 1)(−3)
−2802 = (𝑖 − 1)(−3)
934 = (𝑖 − 1)
𝑖 = 935
The sequence has 935 terms.
6 a 𝑇𝑛 = 𝑖2
b 𝑇𝑛 = 𝑎𝑖2 + 𝑏𝑖 + 𝑐
𝑎 =4÷2=2
𝑏 = 8 − 3(2) = 2
𝑐 =4−2−2 =0
∴ 𝑇𝑛 = 2𝑖2 + 2𝑖
7 a 𝑇1 = 3 ; 𝑇2 = −2 ; 𝑇3 = −7
𝑛
𝑆𝑛 = [2𝑎 + (𝑖 − 1)𝑑]
2
𝑎 = 3 ; 𝑑 = −5
30
𝑆30 = 2 [2(3) + (30 − 1)(−5)]
𝑆30 = −2085
1
b 𝑇1 = 2 ; 𝑇2 = 1 ; 𝑇3 = 2
1 9
�2 −1�
𝑆9 = 2
2−1
𝑆9 = 255,5
8 𝑖=6
𝑇𝑛 = 1 + (𝑖 − 1)4 = 4𝑖 − 3
1 + 5 + 9 + ⋯ + 21 = ∑6𝑘=1 4𝑘 − 3
9 a 𝑇5 = 0 ; 𝑇13 = 12
0 = 𝑎 + 4𝑑 …(1)
12 = 𝑎 + 12𝑑 …(2)
(2)-(1): 12 = 8𝑑
3
𝑑=2
3
𝑎 = −4 �2� = −6
21 3
b 𝑆21 = 2 �2(−6) + (21 − 1) �2��
𝑆21 = 189
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©Via Afrika >> Mathematics Grade 12 114
Answers to Mixed Exercises
10 a For it to be a converging sequence −1 < 𝑟 < 1.
𝑇 �𝑥 2 −9�
𝑟 = 𝑇2 = 𝑥+3
1
(𝑥+3)(𝑥−3)
𝑟=
𝑥+3
𝑟 =𝑥−3
∴ −1 < 𝑥 − 3 < 1
2<𝑥<4
𝑎
b 𝑆∞ = 1−𝑟
(𝑥+3)
13 =
1−(𝑥−3)
(𝑥+3)
13 = (−𝑥+4)
13(−𝑥 + 4) = (𝑥 + 3)
−13𝑥 + 52 = 𝑥 + 3
−13𝑥 − 𝑥 = 3 − 52
−14𝑥 = −49
7
𝑥=2
11 For series in numerator:
99 = 1 + (𝑖 − 1)2
𝑖 = 50 terms
50
𝑆50 = 2 [2(1) + (50 − 1)2] = 2 500
For series in denominator:
299 = 201 + (𝑖 − 1)2
𝑖 = 50 terms
50
𝑆50 = 2 [2(201) + (50 − 1)2] = 12 500
2 500 1
Value = 12 500 = 5
12 𝑇9 = 𝑆9 − 𝑆8
𝑆9 = 3(9)2 − 2(9) = 225
𝑆8 = 3(8)2 − 2(8) = 176
∴ 𝑇9 = 225 − 176 = 49
13 a Let 𝑟 =constant ratio
7𝑟 3 = 189
𝑟 3 = 27
𝑟=3
𝑥 = 7 × 3 = 21
𝑦 = 21 × 3 = 63
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©Via Afrika >> Mathematics Grade 12 115
Answers to Mixed Exercises
7(3𝑛 −1)
b 206 668 =
3−1
7(3𝑛 −1)
206 668 =
2
𝑛
413 336 = 7(3 − 1)
59 048 = 3𝑛 − 1
3𝑛 = 59 049
3𝑛 = 310
∴ 𝑖 = 10
Chapter 2: Functions
1 2𝑥 − 3𝑦 = 17 … (1)
3𝑥 − 𝑦 = 15 … (2)
(2) × 3: 9𝑥 − 3𝑦 = 45 … (3)
(1) − (3) : − 7𝑥 = −28
𝑥=4
Substitute into (1):
2(4) − 3𝑦 = 17
𝑦 = −3
Intercept is (4; −3)
2 a 𝑦 = 𝑚𝑥 + 3
Substitute (−3; 0)
0 = 𝑚(−3) + 3
𝑚=1
∴ 𝑦 =𝑥+3
b 𝑦 = 𝑚𝑥 + 1
Substitute (2; −1):
−1 = 𝑚(2) + 1
𝑚 = −1
𝑔: 𝑦 = −𝑥 + 1
c 𝑥 + 3 = −𝑥 + 1
2𝑥 = −2
𝑥 = −1
Substitute 𝑥 = −1:
𝑦 = −1 + 3 = 2
∴ 𝑃(−1; 2)
d Yes, because the products of their gradients is −1.
(−1 × 1 = −1)
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©Via Afrika >> Mathematics Grade 12 116
Answers to Mixed Exercises
e 𝑦 = −𝑥 − 2
3 a Let 𝑦 = 0:
0 = 𝑥 2 − 2𝑥 − 3
0 = (𝑥 − 3)(𝑥 + 1)
∴ 𝑥 = 3 𝑜𝑟 𝑥 = −1
𝐴(−1; 0) and 𝐴(3; 0)
Let 𝑥 = 0:
𝑦 = (0)2 − 2(0) − 3
𝑦 = −3
∴ 𝐶(0; −3)
𝑂𝐴 = 1 𝑢𝑖𝑖𝑡
𝑂𝐴 = 3 𝑢𝑖𝑖𝑡𝑠
𝑂𝐶 = 3 𝑢𝑖𝑖𝑡𝑠
−𝑏 2
b 𝑥 = 2𝑎 = 2(1) = 1
Substitute 𝑥 = 1:
𝑦 = (1)2 − 2(1) − 3 = −4
𝐷(1; −4)
c 𝑐 = −3
−3−0
𝑚 = 0−3
𝑚=1
d For the graph to have only one real root it has to move 4 units up.
𝑦 = 𝑥 2 − 2𝑥 − 3 + 4 = 𝑥 2 − 2𝑥 + 1
∴𝑘=1
4 a Let 𝑦 = 0:
0 = −2(𝑥 + 1)2 + 8
0 = −2𝑥 2 − 4𝑥 + 6
0 = (−2𝑥 + 2)(𝑥 + 3)
𝑥 = 1 𝑜𝑟 𝑥 = −3
𝐴(−3; 0) and 𝐴(1; 0)
𝐴𝐴 = 4 units
b 𝐶(−1; 8)
c 𝑥 = 0, 𝑦 = 6
𝐷(0 ; 6) 𝐷(−2; 6)
∴ 𝐷𝐷 = 2 units
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©Via Afrika >> Mathematics Grade 12 117
Answers to Mixed Exercises
5 a y
x
−1
−1
−2
b 𝑥∈𝑅
c 𝑥 ≤ −1
𝑎
6 a Substitute the point A into the equation 𝑦 = 𝑥
𝑎
2 = −2
𝑎 = −4
b 𝐴(2; −2)
−4
c 𝑦 = 𝑥−1 + 2
7 a 𝑦 = −(0)2 − 2(0) + 8 = 8
𝐴(0; 8)
b 0 = −𝑥 2 − 2𝑥 + 8
0 = (−𝑥 + 2)(𝑥 + 4)
𝑥 = 2 𝑜𝑟 𝑥 = −4
𝐴(−4; 0) and 𝐶(2; 0)
c 𝐷(−1; 0)
𝐶𝐷 = 3 𝑢𝑖𝑖𝑡𝑠
d 𝑥 = −1
𝑦 = −(−1)2 − 2(−1) + 8
𝑦 = −1 + 2 + 8 = 9
𝐷(−1; 9)
𝐷𝐷 = 9 𝑢𝑖𝑖𝑡𝑠
e 𝐴(0; 8)
𝐷(−2; 8)
𝐴𝐷 = 2 𝑢𝑖𝑖𝑡𝑠
1
f −𝑥 2 − 2𝑥 + 8 = 2 𝑥 − 1
−2𝑥 2 − 4𝑥 + 16 = 𝑥 − 2
−2𝑥 2 − 5𝑥 + 18 = 0
(2𝑥 + 9)(−𝑥 + 2) = 0
−9
𝑥 = 2 𝑜𝑟 𝑥 = 2
−9
𝑥 = 2 at H
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©Via Afrika >> Mathematics Grade 12 118
Answers to Mixed Exercises
−9 1
Substitute 𝑥 = 2 into the equation 𝑦 = 2 𝑥 − 1
1 −9
𝑦 = 2� 2 �− 1
13
𝑦=−4
−9 −13
∴ 𝐺�2 ; 4 �
𝐺𝐻 = 3,25 𝑢𝑖𝑖𝑡𝑠
1
g 𝑓(𝑥) − 𝑔(𝑥) = −𝑥 2 − 2𝑥 + 8 − �2 𝑥 − 1�
5
= −𝑥 2 − 2 𝑥 + 9
Minimum at turning point:
5
5
𝑥 = −2 = − 4
2
5 2 5 −5 169
h 𝑅𝑆𝑚𝑎𝑥 = − �− 4� − 2 � 4 � + 9 = 16
i 𝑓(𝑥) − 𝑔(𝑥) > 0 ∴ 𝑓(𝑥) > 𝑔(𝑥)
9
−2 < 𝑥 < 2
8 a 𝑦 = −4
b 𝑦 = 𝑏𝑥 + 𝑐
𝑐 = −4
𝑦 = 𝑏𝑥 − 4
Substitute the point (2; 5) into the equation:
5 = 𝑏2 − 4
𝑏2 = 9
𝑏=3
𝑦 = 3𝑥 − 4
c 𝑦 = −1 ; 𝑥 = −2
𝑎
d 𝑦 = 𝑥+2 − 1
Substitute the point 𝐴(0; −3):
𝑎
−3 = 0+2 − 1
𝑎
−3 = 2 − 1
𝑎
2
= −2
𝑎 = −4
−4
𝑦 = 𝑥+2 − 1
e Substitute (−2; −1) into 𝑦 = 𝑥 + 𝑘1 and 𝑦 = −𝑥 + 𝑘2
−1 = −2 + 𝑘1 −1 = 2 + 𝑘2
𝑘1 = 1 𝑘2 = −3
𝑦 =𝑥+1 𝑦 = −𝑥 − 3
f 𝑥 > −2; 𝑥 ≠ 0
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©Via Afrika >> Mathematics Grade 12 119
Answers to Mixed Exercises
9 a 𝑦 = 2𝑥 2
𝑥 = 2𝑦 2
𝑥
𝑦2 = 2
𝑥
𝑦 = ±�2
b 𝑥 ≤ 0 𝑜𝑟 𝑥 ≥ 0
10 a 𝑦 = 𝑎𝑥 c f
y
y=x
Substitute point 𝐴:
3 = 𝑎−1
1
𝑎=3
1 𝑥 x
𝑦 = �3�
b 𝑓 −1 : 𝑦 = log �1� 𝑥 d 𝑥>0
3
11 𝑥 −intercept: (3; 0)
𝑦 −intercep𝑡: (0; −2)
Chapter 3: Logarithms
1 a 𝑥 = 32 = 9
1 2 1
b 𝑥 = �3� = 9
c log 4 𝑥 = −2
1 1
𝑥 = (4)−2 = (4)2 = 16
1 1
d 𝑥 = (5)−2 = (5)2 = 25
e 𝑥 3 = 106
𝑥 = 102
𝑥 = 100
f 81 = 3𝑥
3𝑥 = 34
𝑥=4
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©Via Afrika >> Mathematics Grade 12 120
Answers to Mixed Exercises
1
g 9
= 3𝑥
3𝑥 = 3−2
𝑥 = −2
9 9
2 a Substitute �2; 4�: 4
= 𝑎2
3
𝑎=2
b 𝑓 −1 : 𝑦 = log �3� 𝑥
2
3 −𝑥
c 𝑔(𝑥) = �2�
d ℎ(𝑥) = − log �3� 𝑥
2
3 a i) 𝑔(𝑥) = − log 2 𝑥 b
ii) 𝑝(𝑥) = log 2 (−𝑥)
iii) 𝑞(𝑥) = − log 2 (−𝑥)
iv) 𝑓 −1 : 𝑦 = 2𝑥
v) 𝑔−1 : 𝑦 = 2−𝑥
vi) ℎ(𝑥) = log 2 (𝑥 + 2)
c For 𝑓 −1 𝑎𝑖𝑑 𝑔−1 :
Domain 𝑥 ∈ 𝑅 ;
Range 𝑦 > 0
4 a 𝑦 −coordinate = 0
0 = log 𝑏 𝑥
𝑥 = 𝑏0 = 1
𝐴(1; 0)
b Because graph is increasing as 𝑥 increases.
3
c Substitute 𝐴: 2 = log 𝑏 8
3
8 = 𝑏2
2
2 3 3
(8) = �𝑏 �
3 2
2 2
𝑏 = (8)3 = (23 )3 = 22 = 4
d 𝑔(𝑥) = 4𝑥
e Substitute 𝑦 = −2: −2 = log 4 𝑥
1
𝑥 = 4−2 = 16
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©Via Afrika >> Mathematics Grade 12 121
Answers to Mixed Exercises
Chapter 4: Finance, growth and decay
1 a 𝐴 = 𝑃(1 + 𝑖. 𝑖)
𝐴 = 15 000�1 + (0,106)(5)�
𝐴 = 𝑅22 950
b 𝐴 = 𝑃(1 + 𝑖)𝑛
20
𝐴 = 15 000�1 + (0,024)�
𝐴 = 𝑅24 104,07
It is better to invest it at 9.6% p.a , interest compounded quarterly.
2 a Nominal interest rate
b 𝐴 = 𝑃(1 + 𝑖)𝑛
0,085 60
95 000 = 𝑃 �1 + 12
�
95 000
𝑃= 0,085 60
�1+ �
12
𝑃 = 𝑅62 202,48
c 𝑅95 000 − 𝑅62 202,48 = 𝑅32 797,52
3 a 𝐴 = 𝑃(1 + 𝑖)𝑛
𝐴 = 8 000(1 + 0.06)2
𝐴 = 𝑅8 988,80
𝑥[(1+𝑖)𝑛 −1]
b 𝐷= [1 + 𝑖]
𝑖
0,07 4
2 000�� � −1� 0,07
2
𝐷= �1 + 2
�
0.035
𝐷 = 𝑅8 724,93
She will NOT have enough money to buy the TV in two years.
𝑖 𝑚
4 a 1 + 𝑖𝑒𝑓𝑓 = �1 + 𝑛𝑜𝑚 �
𝑚
0.0785 12
1 + 𝑖𝑒𝑓𝑓 = �1 + 12
�
𝑖𝑒𝑓𝑓 = 0.08138 …
𝐷𝑓𝑓. 𝑟𝑎𝑡𝑐 = 8.14%
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©Via Afrika >> Mathematics Grade 12 122
Answers to Mixed Exercises
𝑖 𝑚
b 1 + 𝑖𝑒𝑓 = �1 + 𝑛𝑜𝑚 �
𝑚
𝑖 4
1 + 0,0925 = �1 + 𝑛𝑜𝑚 �
4
4 𝑖
�1,0925 = �1 + 𝑛𝑜𝑚 �
4
𝑖𝑛𝑜𝑚
1.022 − 1 = 4
𝑖𝑛𝑜𝑚 = 0.0894 …
Nom. rate= 8,95% p.a. compounded quarterly
5 𝐴 = 𝑃(1 + 𝑖)𝑛
179 200 = 350 000(1 − 𝑖)3
0,512 = (1 − 𝑖)3
3
1 − 𝑖 = �0,512
𝑖 = 0.2
Dep. rate= 20%
0.0975 7 0.0975 0.0995 15
6 𝐴 = �20 000 �1 + 4
� + 10 000 �1 + 4
�� �1 + 12
�
𝐴 = �23 672.43 + 10 243.75�(1.13185 … )
𝐴 = 𝑅38 388,36
OR
0.0975 6 0.0975 0.0995 15
𝐴 = �20 000 �1 + 4
� + 10 000� �1 + 4
� �1 + 12
�
𝐴 = �23 109.142 + 10 000�(1.024375)(1.13 … )
𝐴 = 𝑅38 388,36
7 a 𝐴 = 900 000(1 − 0.15)5
𝐴 = 𝑅399 334,78
𝐴 = 900 000(1 + 0.18)5
𝐴 = 𝑅2 058 981,98
𝑅2 058 981,98 − 𝑅399 334,78 = 𝑅1 659 647,20
𝑥�(1+0.02)61 −1�
b 1 659 647,20 = 0,02
0,02×1659647,20
𝑥 = [(1+0,02)61 −1]
𝑥 = 𝑅14 144,81
8 𝐴 = 𝑃(1 + 𝑖. 𝑖) 2 yrs = 24 months
(24 × 85) = 1 500(1 + 𝑖. 2)
𝑖 = 0.18
rate= 18%
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©Via Afrika >> Mathematics Grade 12 123
Answers to Mixed Exercises
𝑥[1−(1+𝑖)−𝑛 ]
9 a 𝑃=
𝑖
�6 500��1−(1+0,01)−240 �
𝑃= 0,01
𝑃 = 𝑅590 326,21
�6 500��1−(1+0,01)−96 �
b 𝑃= 0,01
𝑃 = 𝑅399 930,07
0,01 73
1 000��1+ � −1�
12
10 𝐷= 0,01
12
𝐷 = 𝑅99 915,81
𝐴 = 𝑃(1 + 𝑖)𝑛
0,01 5
𝐴 = 99 915,81 �1 + 12 �
𝐴 = 𝑅104 147,21
𝑥[(1+𝑖)𝑛 −1]
11 𝐷=
𝑖
0,09 𝑛
300��1+ � −1�
12
48 000 = 0,09
12
2,2 = (1,0075)𝑛
𝑖 = log1,0075 2,2
𝑖 = 106
8 years and 10 months
12 𝐴 = 𝑃(1 + 𝑖. 𝑖)
𝐴 = 13 500�1 + (0.12)(4)�
𝐴 = 𝑅19 980
Repayment = 𝑅19 980 ÷ 48 = 𝑅416,25
Including insurance= 𝑅416,25 + 𝑅30 = 𝑅446,25
13 𝐴 = 400 000(1 + 0,02)4
𝐴 = 𝑅432 972,86 (amount owing after 1 year)
𝑥[1−(1+𝑖)−𝑛 ]
𝑃=
𝑖
𝑥�1−(1+0,02)−16 �
432 972,86 = 0,02
𝑥 = 𝑅31 888,51
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©Via Afrika >> Mathematics Grade 12 124
Answers to Mixed Exercises
Chapter 5: Compound angles
1 a 2 cos 2 x = −1
1
∴ 𝑐𝑜𝑠2𝑥 = −
2
2𝑥 = ±120° + 𝑘. 360°; 𝑘 ∈ 𝑍
∴ 𝑥 = ±60° + 𝑘. 180°; 𝑘 ∈ 𝑍
b 𝑠𝑖𝑖𝑥 = 3𝑐𝑜𝑠𝑥
𝑠𝑖𝑛𝑥
=3
𝑎𝑜𝑠𝑥
𝑡𝑎𝑖𝑥 = 3
∴ 𝑥 = 71,47° + 𝑘. 180°; 𝑘 ∈ 𝑍
c 𝑠𝑖𝑖𝑥 = 𝑐𝑜𝑠3𝑥
∴ cos(90° − 𝑥) = 𝑐𝑜𝑠3𝑥
90° − 𝑥 = ±3𝑥 + 𝑘. 360°
4𝑥 = 90° + 𝑘. 360° or 2𝑥 = −90° + 𝑘. 360°
𝑥 = 22,5° + 𝑘. 90° 𝑥 = −45° + 𝑘. 180°; 𝑘 ∈ 𝑍
d 3
6 − 10𝑐𝑜𝑠𝑥 − 3(1 − 𝑐𝑜𝑠 𝑥) = 0
∴ 3𝑐𝑜𝑠 3 𝑥 − 10𝑐𝑜𝑠𝑥 + 3 = 0
∴ (3𝑐𝑜𝑠𝑥 − 1)(𝑐𝑜𝑠𝑥 − 3) = 0
1
∴ 𝑐𝑜𝑠𝑥 = or 𝑐𝑜𝑠𝑥 = 3 (no solution)
3
∴ 𝑥 = ±70,53° + 𝑘. 360°; 𝑘𝜖𝑍
For 𝑥 ∈ [−360°; 360°] 𝑥 ∈ {−289,47°; −70,53°; 289,47°}
e 2(𝑠𝑖𝑖2 𝑥 + 𝑐𝑜𝑠 2 𝑥) − 𝑠𝑖𝑖𝑥𝑐𝑜𝑠𝑥 − 3𝑐𝑜𝑠 2 𝑥 = 0
2𝑠𝑖𝑖2 𝑥 − 𝑠𝑖𝑖𝑥𝑐𝑜𝑠𝑥 − 𝑐𝑜𝑠 2 𝑥 = 0
(2𝑠𝑖𝑖𝑥 + 𝑐𝑜𝑠𝑥)(𝑠𝑖𝑖𝑥 − 𝑐𝑜𝑠𝑥) = 0
1
𝑡𝑎𝑖𝑥 = − or 𝑡𝑎𝑖𝑥 = 1
2
𝑥 = −26,57° + 𝑘. 180° or 𝑥 = 45° + 𝑘. 180°; 𝑘 ∈ 𝑍
f 3(𝑠𝑖𝑖2 𝑥 + 𝑐𝑜𝑠2 𝑥) − 8𝑠𝑖𝑖𝑥 + 16𝑠𝑖𝑖𝑥𝑐𝑜𝑠𝑥 − 6𝑐𝑜𝑠𝑥 = 0
∴ 3 − 6𝑐𝑜𝑠𝑥 − 8𝑠𝑖𝑖𝑥 + 16𝑠𝑖𝑖𝑥𝑐𝑜𝑠𝑥 = 0
3(1 − 2𝑐𝑜𝑠𝑥) − 8𝑠𝑖𝑖𝑥(1 − 2𝑥𝑜𝑠𝑥) = 0
(1 − 2𝑐𝑜𝑠𝑥)(3 − 8𝑠𝑖𝑖𝑥) = 0
1 3
𝑐𝑜𝑠𝑥 = or 𝑠𝑖𝑖𝑥 =
2 8
∴ 𝑥 = ±60° + 𝑘. 360° or 𝑥 = 22,02° + 𝑘. 360° or 𝑥 = 157,98° + 𝑘. 360°; 𝑘 ∈ 𝑍
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©Via Afrika >> Mathematics Grade 12 125
Answers to Mixed Exercises
sin x
2 a LHS= cos x + × sin x
cos x
cos 2 x + sin 2 x
=
cos x
1
=
cos x
=RHS
Not valid for x = 90 0 + k .180 0 ; k ∈ Z
sin 2 θ − cos θ (1 − cos θ ) sin 2 θ + cos 2 θ − cos θ 1 − cos θ 1
b LHS = = = =
(1 − cos θ )sin θ (1 − cos θ )sin θ (1 − cos θ )sin θ sin θ
=RHS
Not valid for θ = k .180 0 ; k ∈ Z
1 − cos 2 x sin 2 x sin x
c LHS = = = . sin x = tan x. sin x = RHS
cos x cos x cos x
Not valid for x = 90 0 + k .180 0 ; k ∈ Z
d LHS =
(
sin x sin 2 x + cos 2 x
=
)sin x
= tan x =RHS
cos x cos x
Not valid for x = 90 0 + k .180 0 ; k ∈ Z
sin x
1+
LHS = cos x = cos x + sin x × cos x
×
cos x + sin x
sin x cos x cos x − sin x cos x + sin x
1−
e cos x
cos x + 2 sin x cos x + sin 2 x 1 + 2 sin x cos x
2
= = = RHS
cos 2 x − sin 2 x cos 2 x − sin 2 x
Not valid for x = ±45 0 + k .180 0 ; k ∈ Z
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©Via Afrika >> Mathematics Grade 12 126
Answers to Mixed Exercises
f 𝐿𝐻𝑆 = sin(45° + 𝑥). sin(45° − 𝑥)
= (sin 45°. cos 𝑥 + sin 𝑥. 𝑐𝑜𝑠45°) × (sin 45°. cos 𝑥 − sin 𝑥 . cos 45°)
√2 √2 √2 √2
= � 2 cos 𝑥 + 2 sin 𝑥� � 2 cos 𝑥 − 2 sin 𝑥�
2 2
√2 √2
= � 2 cos 𝑥� − � 2 sin 𝑥�
1 1
= 𝑐𝑜𝑠 2 𝑥 − 𝑠𝑖𝑖2 𝑥
2 2
1
= (𝑐𝑜𝑠 2 𝑥 − 𝑠𝑖𝑖2 𝑥)
2
1
= cos 2𝑥
2
= 𝑅𝐻𝑆
sin 2𝜃−cos 𝜃
g 𝐿𝐻𝑆 =
sin 𝜃−cos 2𝜃
2 sin 𝜃.cos 𝜃−cos 𝜃
=
sin 𝜃−(1−2𝑠𝑖𝑛2 𝜃)
cos 𝜃(2 sin 𝜃−1)
= 2𝑠𝑖𝑛2 𝜃+sin 𝜃−1
cos 𝜃(2 sin 𝜃−1)
= (2
sin 𝜃−1)(sin 𝜃+1)
cos 𝜃
=
sin 𝜃+1
= 𝑅𝐻𝑆
cos 𝑥−cos 2𝑥+2
h 𝐿𝐻𝑆 =
3 sin 𝑥−sin 2𝑥
cos 𝑥−�2𝑎𝑜𝑠2 𝑥−1�+2
= 3 sin 𝑥−2 sin 𝑥.cos 𝑥
−2𝑎𝑜𝑠2 𝑥+cos 𝑥+3
=
3 sin 𝑥−2 sin 𝑥.cos 𝑥
(−2 cos 𝑥+3)(cos 𝑥+1)
= sin 𝑥(3−2 cos 𝑥)
cos 𝑥+1
=
sin 𝑥
= 𝑅𝐻𝑆
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©Via Afrika >> Mathematics Grade 12 127
Answers to Mixed Exercises
3 a
( )
sin 180 0 − x tan (− x )
=
sin x(− tan x )
= −1
( ) (
tan 180 + x cos x − 90
0 0
)
tan x(sin x )
b
( ) (
sin 180 0 + x tan x − 360 0
=
)
sin x. tan x
= 2 sin x
( )(
tan 360 − x − cos 60 tan 45
0 0 0
)( )
− tan x(− 0,5)(1)
4 a ( )
cos 730 = cos 90 0 − 17 0 = sin 17 0 = k
b cos(−163°) = 𝑐𝑜𝑠163° = −𝑐𝑜𝑠17° = −√1 − 𝑘 2
𝑘
c 𝑡𝑎𝑖197° = 𝑡𝑎𝑖17° =
√1−𝑘 2
d 𝑐𝑜𝑠326° = 𝑐𝑜𝑠34° = 𝑐𝑜𝑠2(17°) = 1 − 2𝑠𝑖𝑖2 17° = 1 − 2𝑘 2
4
5 a 𝑐𝑜𝑠𝑥 = − 5
3 3 − 3 − 3
5 sin x + 3 tan x = 5 + 3 or = 5 + 3
5 − 4 5 −4
9 3 9 3
= 3− = or = −5 + =−
4 4 4 4
2 tan x
b tan 2 x =
1 − tan 2 x
3
2
−4 3 16 24 3 16 24
∴ tan 2 x = 2
=− × =− or tan 2 x = × =
3 2 6 7 2 7 7
1−
−4
6 a
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©Via Afrika >> Mathematics Grade 12 128
Answers to Mixed Exercises
cos(𝑥 − 𝑦) = cos 𝑥. cos 𝑦 + sin 𝑥. sin 𝑦
−2√2 −4 −1 −3
= × + ×
3 5 3 5
8√2 1
= +
15 5
8√2+3
=
15
b cos 2𝑥 − cos 2𝑦
= 1 − 2𝑠𝑖𝑖2 𝑥 − (1 − 2𝑠𝑖𝑖2 𝑦)
= 2𝑠𝑖𝑖2 𝑦 − 2𝑠𝑖𝑖2 𝑥
−3 2 −1 2
= 2� � − 2� �
5 3
18 2 112
= − =
25 9 225
1
7 a cos 2(22,5°) = cos 45° =
√2
1 1
b × 2 sin 22,5°. cos 22,5° = × sin 2(22,5°)
2 2
1 √2
= sin 45° = 4
2
1
c sin 2(15°) = sin 30° =
2
Chapter 6: Solving problems in three dimensions
2ℎ 2ℎ
1 a 𝐼𝑖 ∆𝐴𝐴𝐷 ∶ tan 𝛼 = 𝐴𝐴 ∴ 𝐴𝐷 = tan 𝛼
ℎ
b 𝐼𝑖 ∆𝐶𝐷𝐷: tan(90° − 𝛼) = 𝐴𝐴 ∴ 𝐷𝐷 = ℎ tan 𝛼
𝐼𝑖 ∆𝐴𝐷𝐷:
𝐴𝐷2 = 𝐴𝐷 2 + 𝐷𝐷2 − 2(𝐴𝐷)(𝐷𝐷). cos 𝐷
= (2ℎ. cot 𝛼)2 + (ℎ. tan 𝛼)2 − 2(2ℎ. cot 𝛼)(ℎ. tan 𝛼) cos 120°
1
= 4ℎ2 . 𝑐𝑜𝑡 2 𝛼 + ℎ2 𝑡𝑎𝑖2 𝛼 − 4ℎ2 (cot 𝛼 . tan 𝛼) �− 2�
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©Via Afrika >> Mathematics Grade 12 129
Answers to Mixed Exercises
= ℎ2 (4𝑐𝑜𝑡 2 𝛼 + 𝑡𝑎𝑖2 𝛼 + 2)
4
= ℎ2 �𝑡𝑎𝑛2 𝛼 + 𝑡𝑎𝑖2 𝛼 + 2�
ℎ2 �𝑡𝑎𝑛4 𝛼+2𝑡𝑎𝑛2 𝛼+4�
= 𝑡𝑎𝑛2 𝛼
ℎ√𝑡𝑎𝑛4 𝛼+2𝑡𝑎𝑛2 𝛼+4
𝐴𝐷 = tan 𝛼
𝐴𝐴 tan 𝛼
c ℎ = √𝑡𝑎𝑛4
𝛼+2𝑡𝑎𝑛2 𝛼+4
509.tan 42°
= √𝑡𝑎𝑛4
42°+2𝑡𝑎𝑛2 42°+4
𝐶𝐷 = 182,90 𝑚
2 a � 𝐴 = 180° − 𝑠 − 30° = 150° − 𝑠
𝐶𝐷
𝑝
b tan 𝑠 = 𝐸𝐴 ∴ 𝑝 = 𝐶𝐴. tan 𝑠
𝐸𝐴 8
sin(150°−𝜃)
= sin 𝜃
8.sin(150°−𝜃)
𝐶𝐴 =
sin 𝜃
8.sin�180°−(150°−𝜃)�
= sin 𝜃
8.sin(30°+𝜃)
=
sin 𝜃
8.sin(30°+𝜃) 8.sin(30°+𝜃)
𝑝=� � tan 𝑠 =
sin 𝜃 cos 𝜃
3 𝐴𝐷 = 13(𝑃𝑦𝑡ℎ𝑎𝑔𝑜𝑟𝑎𝑠)
𝐴̂ = 180° − (𝛼 + 𝛽)
𝐸𝐴 13
∴ sin[180°−(𝛼+𝛽)] = sin 𝛼
𝐸𝐴 13
∴ sin(𝛼+𝛽) = sin 𝛼
13 sin(𝛼+𝛽)
∴ 𝐶𝐷 =
sin 𝛼
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©Via Afrika >> Mathematics Grade 12 130
Answers to Mixed Exercises
1
4 a 𝐴𝑟𝑐𝑎 ∆𝐴𝐷𝐶 = 2 𝑚. 𝑝 sin(180° − 𝑠)
1
b 𝐴𝑟𝑐𝑎 ∆𝐴𝐷𝐶 = 2 𝑖. 𝑝 sin 𝑠
𝐴𝑟𝑐𝑎 ∆𝐴𝐴𝐶 = 𝐴𝑟𝑐𝑎 ∆𝐴𝐷𝐶 + 𝐴𝑟𝑐𝑎 ∆𝐴𝐷𝐶
1 1
= 2 𝑚𝑝 sin(180° − 𝑠) + 2 𝑖𝑝 sin 𝑠
1 1
= 2 𝑚𝑝. sin 𝑠 + 2 𝑖𝑝 . sin 𝑠
1
= 2 𝑝(𝑚 + 𝑖) sin 𝑠
1
c 12,6 = 2 (8,1)(5,9) sin 𝑠
sin 𝑠 = 0,527306968 …
𝑠 = 31,82° 𝑂𝑅 𝑠 = 180° − 31,82° = 148,18°
𝑝
5 a sin 𝑠 = 𝐴𝐸
𝑝
∴ 𝐴𝐶 = sin 𝜃
b 𝐴�1 = 180° − 2𝛼
𝐴𝐸 𝐴𝐸
c sin 𝐴�1
= sin 𝐴
𝑝
𝐴𝐸
sin(180°−2𝛼)
= sin
sin 𝜃
𝛼
𝑝 sin(180°−2𝛼) 𝑝.sin 2𝛼
𝐴𝐶 = = sin 𝜃.sin 𝛼
sin 𝜃.sin 𝛼
6 a 𝑅� = 180° − 30° − (150° − 𝛼) = 𝛼
12 𝑄𝑅
sin 𝑅�
= sin(150°−𝛼)
12 𝑄𝑅
sin 𝛼
= sin(30°+𝛼)
12 sin(30°+𝛼)
𝑄𝑅 =
sin 𝛼
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©Via Afrika >> Mathematics Grade 12 131
Answers to Mixed Exercises
12(sin 30°.cos 𝛼+cos 30°.sin 𝛼)
=
sin 𝛼
1 √3
12� .cos 𝛼+ sin 𝛼�
2 2
=
sin 𝛼
6�cos 𝛼+√3 sin 𝛼�
= sin 𝛼
b 𝑃� = 180° − 90° − 𝛼 = 90° − 𝛼
𝑄𝑅 𝑃𝑄
sin 𝑃�
= sin 𝛼
6�cos 𝛼+√3 sin 𝛼�
𝑃𝑄
sin 𝛼
= sin 𝛼
sin(90°−𝛼)
sin 𝛼.6�cos 𝛼+√3 sin 𝛼�
𝑃𝑄 sin(90° − 𝛼) = sin 𝛼
6 cos 𝛼 6√3 sin 𝛼
𝑃𝑄 = cos 𝛼 + cos 𝛼
𝑃𝑄 = 6 + 6√3 tan 𝛼
c 23 = 6 + 6√3 tan 𝛼
17 = 6√3 tan 𝛼
𝑡𝑎𝑖𝛼 = 1,64
𝛼 = 58,56°
Chapter 7: Polynomials
1 a 27𝑥 3 − 8 = (3𝑥 − 2)(9𝑥 2 + 6𝑥 + 4)
b 5𝑥 3 + 40 = 5(𝑥 3 + 8) = 5(𝑥 + 2)(𝑥 2 − 2𝑥 + 4)
c 𝑥 3 + 3𝑥 2 + 2𝑥 + 6
= 𝑥 2 (𝑥 + 3) + 2(𝑥 + 3)
= (𝑥 + 3)(𝑥 2 + 2)
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©Via Afrika >> Mathematics Grade 12 132
Answers to Mixed Exercises
d 4𝑥 3 − 𝑥 2 − 16𝑥 + 4
= 𝑥 2 (4𝑥 − 1) − 4(4𝑥 − 1)
= (4𝑥 − 1)(𝑥 2 − 4)
= (4𝑥 − 1)(𝑥 − 2)(𝑥 + 2)
e 4𝑥 3 − 2𝑥 2 + 10𝑥 − 5
= 2𝑥 2 (2𝑥 − 1) + 5(2𝑥 − 1)
= (2𝑥 − 1)(2𝑥 2 + 5)
f 𝑥 3 + 2𝑥 2 + 2𝑥 + 1
= (𝑥 3 + 1) + (2𝑥 2 + 2𝑥)
= (𝑥 + 1)(𝑥 2 − 𝑥 + 1) + 2𝑥(𝑥 + 1)
= (𝑥 + 1)(𝑥 2 + 𝑥 + 1)
g 𝑥 3 − 𝑥 2 − 22𝑥 + 40
= (𝑥 − 2)(𝑥 2 + 𝑥 − 20)
= (𝑥 − 2)(𝑥 + 5)(𝑥 − 4)
h 𝑥 3 + 2𝑥 2 − 5𝑥 − 6
= (𝑥 − 2)(𝑥 2 + 4𝑥 + 3)
= (𝑥 − 2)(𝑥 + 3)(𝑥 + 1)
i 3𝑥 3 − 7𝑥 2 + 4
= (𝑥 − 1)(3𝑥 2 − 4𝑥 − 4)
= (𝑥 − 1)(3𝑥 + 2)(𝑥 − 2)
j 𝑥 3 − 19𝑥 + 30
= (𝑥 − 2)(𝑥 2 + 2𝑥 − 15)
= (𝑥 − 2)(𝑥 + 5)(𝑥 − 3)
k 𝑥3 − 𝑥2 − 𝑥 − 2
= (𝑥 − 2)(𝑥 2 + 𝑥 + 1)
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©Via Afrika >> Mathematics Grade 12 133
Answers to Mixed Exercises
2 a 𝑥(𝑥 2 + 2𝑥 − 4) = 0
𝑥 = 0 𝑜𝑟 𝑥 = −1 ± √5
b (𝑥 − 2)(𝑥 2 − 𝑥 − 3) = 0
1±√3
𝑥 = 2 𝑜𝑟 𝑥 = 2
c (2𝑥 3 − 12𝑥 2 ) − (𝑥 − 6) = 0
2𝑥 2 (𝑥 − 6) − (𝑥 − 6) = 0
(𝑥 − 6)(2𝑥 2 − 1) = 0
1
𝑥 = 6 𝑜𝑟 𝑥 = ±�2
d (2𝑥 3 − 𝑥 2 ) − (8𝑥 − 4) = 0
𝑥 2 (2𝑥 − 1) − 4(2𝑥 − 1) = 0
(𝑥 2 − 4)(2𝑥 − 1) = 0
1
𝑥 = 2 𝑜𝑟 𝑥 = −2 𝑜𝑟 𝑥 = 2
e (𝑥 − 1)(𝑥 2 + 2𝑥 + 2) = 0
𝑥=1
f (𝑥 + 2)(𝑥 2 − 2𝑥 − 8) = 0
(𝑥 + 2)(𝑥 − 4)(𝑥 + 2) = 0
𝑥 = −2 𝑜𝑟 𝑥 = 4
g (𝑥 3 − 20𝑥) + (3𝑥 2 − 60) = 0
𝑥(𝑥 2 − 20) + 3(𝑥 2 − 20) = 0
(𝑥 2 − 20)(𝑥 + 3) = 0
𝑥 = ±2√5 𝑜𝑟 𝑥 = −3
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©Via Afrika >> Mathematics Grade 12 134
Answers to Mixed Exercises
3 𝑓(3) = 33 − 32 − 5(3) − 3
= 27 − 9 − 15 − 3 = 0
(𝑥 − 3) is a factor
(𝑥 − 3)(𝑥 2 + 2𝑥 + 1) = 0
(𝑥 − 3)(𝑥 + 1)2 = 0
𝑥 = 3 𝑜𝑟 𝑥 = −1
1 1 3 1 2 1
4 𝑔 �2� = 4 �2� − 8 �2� − 2 + 2
1 1
= −2− +2=0
2 2
(2𝑥 − 1)(2𝑥 2 − 3𝑥 − 2) = 0
(2𝑥 − 1)(2𝑥 + 1)(𝑥 − 2) = 0
1 1
𝑥 = 2 𝑜𝑟 𝑥 = − 2 𝑜𝑟 𝑥 = 2
Chapter 8: Differential calculus
𝑓(𝑥+ℎ)−𝑓(𝑥)
1 a 𝑓 ′ (𝑥) = limℎ→0
ℎ
1−(𝑥+ℎ)2 −�1−𝑥2 �
= limℎ→0 ℎ
1−�𝑥 2 +2𝑥ℎ+ℎ2 �−�1−𝑥2 �
= limℎ→0 ℎ
−2𝑥ℎ−ℎ2
= limℎ→0 ℎ
ℎ(−2𝑥−ℎ)
= limℎ→0
ℎ
= lim (−2𝑥 − ℎ)
ℎ→0
= −2𝑥
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©Via Afrika >> Mathematics Grade 12 135
Answers to Mixed Exercises
𝑓(𝑥+ℎ)−𝑓(𝑥)
b 𝑓 ′ (𝑥) = limℎ→0
ℎ
−3(𝑥+ℎ)2 −�−3𝑥2 �
= limℎ→0 ℎ
−3�𝑥 2 +2𝑥ℎ+ℎ2 �−�−3𝑥2 �
= limℎ→0 ℎ
−6𝑥ℎ−3ℎ2
= limℎ→0 ℎ
ℎ(−6𝑥−3ℎ)
= limℎ→0
ℎ
= lim (−6𝑥 − 3ℎ)
ℎ→0
= −6𝑥
1
1 1
2 a 𝑦 = √𝑥 − 2𝑥 2 = 𝑥 2 − 2 𝑥 −2
1
𝑑𝑦 1 1
𝑑𝑥
= 2 𝑥 −2 − 2 × −2𝑥 −3
1
1
= 2 𝑥 −2 + 𝑥 −3
1 1
= 2 𝑥 + 𝑥3
√
2𝑥 2 −𝑥−15
b 𝐷𝑥 � 𝑥−3
�
(2𝑥+5)(𝑥−3)
= 𝐷𝑥 � 𝑥−3
�
= 𝐷𝑥 [2𝑥 + 5] = 2
3 𝑓(𝑥) = −2𝑥 3 + 3𝑥 2 + 32𝑥 + 15
𝑓(−2) = −2(−2)3 + 3(−2)2 + 32(−2) + 15 = −21
𝑓 ′ (𝑥) = −6𝑥 2 + 6𝑥 + 32
𝑓 ′ (−2) = −6(−2)2 + 6(−2) + 32 = −4
Sub (−2; −21) into 𝑦 = −4𝑥 + 𝑐
−21 = −4(−2) + 𝑐 𝑐 = −29 𝑦 = −4𝑥 − 29
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©Via Afrika >> Mathematics Grade 12 136
Answers to Mixed Exercises
4
Point of
inflection
5 (2; 9) is a point on the graph and a turning point
∴ 𝑓(2) = 9 and 𝑓 ′ (2) = 0
𝑓 ′ (𝑥) = 3𝑎𝑥 2 + 10𝑥 + 4
0 = 3𝑎(2)2 + 10(2) + 4
∴ 𝑎 = −2
9 = (−2)(2)3 + 5(2)2 + 4(2) + 𝑏
∴ 𝑏 = −3
6 a Turning point where 𝑓 ′ (𝑥) = 0 ∴ 𝑥 = −2 and 𝑥 = 5
′′ (𝑥)
b Point of inflections is where 𝑓 = 0 , therefor where graph of 𝑓′ turns
c 𝑓 will decrease where its gradient 𝑓 ′ is negative (𝑓 ′ < 0)
−2 < 𝑥 < 5
7 a The graph bounces at 𝑥 = 1 and has an 𝑥-intercept at 𝑥 = −1
∴ 𝑓(𝑥) = (𝑥 + 1)(𝑥 − 1)2
𝑓(𝑥) = (𝑥 + 1)(𝑥 2 − 2𝑥 + 1) = 𝑥 3 − 𝑥 2 − 𝑥 + 1
∴ 𝑎 = −1; 𝑏 = −1; 𝑐 = 1
b B is a turning point where 𝑓 ′ (𝑥) = 0
3𝑥 2 − 2𝑥 − 1 = 0
(3𝑥 + 1)(𝑥 − 1) = 0)
1
At B 𝑥 = − 3
1 1 3 1 2 1 32
𝑓 �− 3� = �− 3� − �− 3� − �− 3� + 1 = 27
1 32
𝐴 �− 3 ; 27�
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©Via Afrika >> Mathematics Grade 12 137
Answers to Mixed Exercises
8 a If 𝑠(𝑡) is distance, then 𝑠 ′ (𝑡) is speed.
𝑠 ′ (𝑡) = 3𝑡 2 − 4𝑡 + 3
b Speed is a minimum where 𝑠 ′′ (𝑡) = 6𝑡 − 4 = 0
2
𝑡=3
c 6𝑡 − 4 = 8
𝑡 = 2𝑠
9 a Volume = 2𝑥 2 ℎ = 24
12
ℎ = 𝑥 2 = 12𝑥 −2
b 𝐶(𝑥) = 2𝑥 2 × 25 + 2𝑥 2 × 20 + 2 × 𝑥ℎ × 20 + 2 × 2𝑥ℎ × 20
= 90𝑥 2 + 120𝑥ℎ
= 90𝑥 2 + 120𝑥(12𝑥 −2 )
= 90𝑥 2 + 1440𝑥 −1
c 𝐶 ′ (𝑥) = 180𝑥 − 1440𝑥 −2 = 0
1440
180𝑥 − 𝑥 2 = 0
180𝑥 3 − 1440 = 0
𝑥3 = 8
𝑥=2
Chapter 9: AnalyticalgGeometry
1 a 𝑚𝐴𝐴 = 𝑚𝐸𝐴
1−(−4) 0−2
= 5−3
−2−𝑝
5 −2
−2−𝑝
= 2 = −1
2+𝑝=5
𝑝=3
b 𝐴𝐴 = �(3 − (−2))2 + (−4 − 1)2 = 5√2
𝐶𝐷 = �(5 − 3)2 + (0 − 2)2 = 2√2
𝐴𝐴: 𝐶𝐷 = 5√2: 2√2 = 5: 2
c 𝑚𝑁𝐴 = 𝑚𝐸𝐴
𝑦+4
𝑥−3
= −1 ∴ 𝑦 = −𝑥 − 1 … (1)
𝑚𝑁𝐴 = 𝑚𝐴𝐸
𝑦−2
𝑥−3
=2 ∴ 𝑦 = 2𝑥 − 4 … (2)
(1)-(2): 0 = 3𝑥 − 3
𝑥 = 1 𝑦 = −2
𝑁(1; −2)
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©Via Afrika >> Mathematics Grade 12 138
Answers to Mixed Exercises
d B and D have the same 𝑥-coordinate, so it is a vertical line with equation 𝑥 = 3.
e Angle of inclination of a vertical line is 90°.
f Area of parallelogram = base x perpendicular height
𝐴𝑟𝑐𝑎 𝑜𝑓 𝑁𝐴𝐶𝐷 = 𝐶𝐷 × 𝑝𝑐𝑟𝑝 ℎ𝑐𝑖𝑔ℎ𝑡
= 2√2 × 6 = 12√2
g 𝑚𝐴𝑅 = 𝑚𝐴𝐸
𝑞−1 1−0
−2+1
= −2−5
𝑞−1 1
−1
= −7
8
∴𝑞=7
2 a Substitute 𝑥 = 1 and 𝑦 = −3 in LHS. If LHS=0, then the point 𝑁(1; −3) lies on the
circle.
𝐿𝐻𝑆 = 𝑥 2 + 4𝑥 + 𝑦 2 + 2𝑦 – 8
= (1)2 + 4(1) + (−3)2 + 2(−3)– 8 = 0
∴ 𝑁 lies on the circle
b First determine the centre of the circle:
𝑥 2 + 4𝑥 + 4 + 𝑦 2 + 2𝑦 + 1 = 8 + 4 + 1
(𝑥 + 2)2 + (𝑦 + 1)2 = 13
Centre of circle is 𝑀(−2; −1)
−1+3 2
𝑚𝑀𝑁 = −2−1 = − 3
MN⏊PN (radius⏊tangent)
3
∴ 𝑚𝑃𝑁 = 2
3
Substitute 𝑁(1; −3): 𝑦 = 2 𝑥 + 𝑐
3 9
−3 = 2 (1) + 𝑐 ∴ 𝑐 = −2
3 9
𝑦 = 2𝑥 − 2
3
c 𝑠 = 𝑡𝑎𝑖−1 �2� = 56,3°
d 𝑥 −intercept where 𝑦 = 0:
3 9
0 = 2𝑥 − 2 ∴𝑥=3
e 𝑦 −intercepts are where 𝑥 = 0:
(0)2 + 4(0) + 𝑦 2 + 2𝑦 – 8 = 0
∴ 𝑦 2 + 2𝑦 – 8 = 0
(𝑦 + 4)(𝑦 − 2) = 0
The points are (0; −4) and (0; 2).
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©Via Afrika >> Mathematics Grade 12 139
Answers to Mixed Exercises
−12
3 a 𝑚𝑅𝑂 = −6 = 2
b PS⏊RN (RN is altitude of ∆)
𝑚𝑃𝑆 × 𝑚𝑅𝑁 = −1
1
∴ 𝑚𝑃𝑆 = − 2
c 𝑃(0; 6) (𝑦 −intercept of PR)
1
∴ 𝑦 = −2𝑥 + 6
1
d 𝑡𝑎𝑖−1 �2� = 26,57°
Inclination of PS = 180° − 26,57° = 153,43°
3
e Substitute 𝑁(2𝑖; 3 5 + 𝑖) into equation of PS
3 1
3 5 + 𝑖 = − 2 (2𝑖) + 6
3
3 5 + 𝑖 = −𝑖 + 6
2 12
2𝑖 = 2 5 = 5
6
𝑖=5
f Find equation of SM. SM is the median, so M is the midpoint of PR.
−6+0 −12+6
𝑀� 2
; 2
� = (−3; −3)
𝑚𝑀𝑆 = 1 so equation of SM: 𝑦 = 𝑥
Solve equations of SM and PS simultaneously to calculate coordinates of S
1
𝑥 = −2𝑥 + 6 ∴ 𝑥 = 4; 𝑦 = 4
𝑆(4; 4)
4 a 𝑥 2 + 4𝑥 + 𝑦 2 – 2𝑦 = 4
𝑥 2 + 4𝑥 + 4 + 𝑦 2 – 2𝑦 + 1 = 4 + 4 + 1
(𝑥 + 2)2 + (𝑦 − 1)2 = 9
Centre 𝑀(−2; 1) radius= 3
b Substitute 𝑁(𝑝; 1) into equation of circle.
𝑝 2 + 12 + 4(𝑝) – 2(1) – 4 = 0
𝑝2 + 4𝑝 − 5 = 0
(𝑝 + 5)(𝑝 − 1) = 0
∴ 𝑝 = 1 as 𝑝 > 0
c Radius through N is horizontal.
Therefore the tangent will be vertical.
Equation of tangent: 𝑥 = 1
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©Via Afrika >> Mathematics Grade 12 140
Answers to Mixed Exercises
3−0
5 a 𝑚𝐴𝐴 = −3−0 = −1
AD goes through origin: So, equation is 𝑦 = −𝑥
b 𝐴𝐷2 = 𝐷𝐶 2
(𝑥 − 2)2 + (𝑦 − 3)2 = (𝑥 − 6)2 + (𝑦 + 1)2
Substitute 𝑦 = −𝑥
(𝑥 − 2)2 + (−𝑥 − 3)2 = (𝑥 − 6)2 + (−𝑥 + 1)2
𝑥 2 − 4𝑥 + 4 + 𝑥 2 + 6𝑥 + 9 = 𝑥 2 − 12𝑥 + 36 + 𝑥 2 − 2𝑥 + 1
16𝑥 = 24
3 3
𝑥=2 ∴ 𝑦 = −2
3
3−(− )
c 𝑚𝐴𝐴 = 3
2
=9
2−
2
Substitute 𝐴(2; 3) into 𝑦 = 9𝑥 + 𝑐
3 = 9(2) + 𝑐 ∴ 𝑐 = −15
𝑦 = 9𝑥 − 15
d Inclination of BD= 𝑡𝑎𝑖−1 (9) = 83,7°
3−(−1)
𝑚𝐴𝐸 = = −1
2−6
Inclination of BC= 135°
∴ 𝑠 = 135° − 83,7° = 51,3°
3 2 3 2 √82
e 𝐴𝐷 = ��2 − 2� + �3 + 2� = 2
𝐴𝐶 = �(3 + 1)2 + (2 − 6)2 = 4√2
1
𝐴𝑟𝑐𝑎 𝑜𝑓 ∆𝐴𝐷𝐶 = 2 𝐴𝐷 × 𝐴𝐶 × 𝑠𝑖𝑖𝑠
1 √82
= 2 × 2 × 4√2 × 𝑠𝑖𝑖51,3°
= 10 𝑠𝑞 𝑢𝑖𝑖𝑡𝑠
6 a First determine equation of AC
3−(−3)
𝑚𝐴𝐸 = = −2
2−5
Substitute (2; 3): 3 = −2(2) + 𝑐 ∴ 𝑦 = −2𝑥 + 7
7 7
𝑥 − 𝑖𝑖𝑡𝑐𝑟𝑐𝑐𝑝𝑡(𝑦 = 0): 𝑥 = 2 𝐷 �2 ; 0�
b 𝐴𝐶 2 = 𝐴𝐶 2
(𝑝 − 5)2 + (0 + 3)2 = (5 − 2)2 + (−3 − 3)2
𝑝2 − 10𝑝 + 25 = 9 + 36
𝑝2 − 10𝑝 − 20 = 0
10±√180
𝑝= = 5 ± 3√5
2
𝑝 = 5 − 3√5
c 𝑚𝐴𝐸 = −2
𝐼𝑖𝑐𝑙𝑖𝑖𝑎𝑡𝑖𝑜𝑖 𝑜𝑓 𝐴𝐶 = 180° − 𝑡𝑎𝑖−1 (2) = 116,6°
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©Via Afrika >> Mathematics Grade 12 141
Answers to Mixed Exercises
d 𝐴(−1; 0)
3−0
𝑚𝐴𝐴 = 2+1 = 1
Inclination of 𝐴𝐴 = 45°
𝐴̂ = 𝑖𝑖𝑐𝑙𝑖𝑖𝑎𝑡𝑖𝑜𝑖 𝑜𝑓 𝐴𝐶 − 𝑖𝑖𝑐𝑙𝑖𝑖𝑎𝑡𝑖𝑜𝑖 𝑜𝑓 𝐴𝐴
= 116,6° − 45°
= 71,6°
7 The line will be a tangent if it intersects the circle in only one point.
Substitute 𝑦 = 𝑥 + 1 into equation of circle and solve for 𝑥.
There should be only one solution.
𝑥 2 + (𝑥 + 1)2 + 6(𝑥 + 1) − 7 = 0
𝑥 2 + 𝑥 2 + 2𝑥 + 1 + 6𝑥 + 6 − 7 = 0
2𝑥 2 + 8𝑥 = 0
𝑥 = 0 or 𝑥 = −4
The line is NOT a tangent.
8 a 𝑦 = 2 at C. Substitute into 3𝑥 + 4𝑦 + 7 = 0
3𝑥 + 4(2) + 7 = 0
3𝑥 = −15
𝑥 = −15
∴ 𝐶(−5; 2) and the radius is 5.
(𝑥 + 5)2 + (𝑦 − 2)2 = 25
b length of 𝐷𝐷 = 10
2+1
c 𝑚𝑃𝐴 = 0+1 = 3
1
𝑚𝑝𝑒𝑟𝑝 𝑏𝑖𝑠𝑒𝑎𝑡𝑜𝑟 = − 3
0−1 2−1 1 1
Midpoint of PE= � 2 ; 2 � = �− 2 ; 2�
1
Substitute midpoint into 𝑦 = − 3 𝑥 + 𝑐
1 1 1
2
= − 3 �− 2� + 𝑐
1
𝑐=3
1 1
𝑦 = −3𝑥 + 3
1 1
d 3𝑥 + 4 �− 3 𝑥 + 3� + 7 = 0
4 4
3𝑥 − 3 𝑥 + 3 + 7 = 0
5 25
3
𝑥=− 3
𝑥 = −5
1 1
𝑦 = − 3 (−5) + 3 = 2 The lines intersect at (−5; 2)
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©Via Afrika >> Mathematics Grade 12 142
Answers to Mixed Exercises
9 a Let the coordinates of S be (𝑥; 0)
ST⏊SR
4 4
𝑚𝑆𝑇 × 𝑚𝑆𝑅 = −𝑥 × 𝑥−4 = −1
𝑥(𝑥 − 4) = 16
𝑥 2 − 4𝑥 − 16 = 0
4±�(−4)2 −4(−16)
𝑥= = 2 ± 2√5
2
But S is on positive 𝑥 −axis, so 𝑆�2 + 2√5; 0�
4−0
b 𝑚𝑆𝑇 = 0−�2+2√5� = −0,62
c Inclination of TS= 180° − 𝑡𝑎𝑖−1 (0,62) = 148,20°
4+4
𝑚 𝑇𝑅 = 0−4 = −2
Inclination of TR= 180° − 𝑡𝑎𝑖−1 (2) = 116,57°
𝑅𝑇�𝑆 = 148,20° − 116,57° = 31,63°
10 a 𝑥 2 + 𝑦 2 – 4𝑥 + 6𝑦 + 3 = 0
𝑥 2 − 4𝑥 + 𝑦 2 + 6𝑦 = −3
𝑥 2 − 4𝑥 + 4 + 𝑦 2 + 6𝑦 + 9 = −3 + 4 + 9
(𝑥 − 2)2 + (𝑦 + 3)2 = 10
Centre is (2; −3)
−2+3 1
𝑚𝑟𝑎𝑑𝑖𝑢𝑠 = 5−2 = 3
𝑚𝑡𝑎𝑛𝑔𝑒𝑛𝑡 = −3
Substitute (5; −2) into 𝑦 = −3𝑥 + 𝑐
−2 = −3(5) + 𝑐
𝑐 = 13
∴ 𝑦 = −3𝑥 + 13
b �(𝑥 − 2)2 + (𝑦 + 3)2 = √20
(𝑥 − 2)2 + (𝑦 + 3)2 = 20
Substitute 𝑦 = −3𝑥 + 13 into equation above:
(𝑥 − 2)2 + (−3𝑥 + 13 + 3)2 = 20
(𝑥 − 2)2 + (−3𝑥 + 16)2 = 20
𝑥 2 − 4𝑥 + 4 + 9𝑥 2 − 96𝑥 + 256 = 20
10𝑥 2 − 100𝑥 + 240 = 0
𝑥 2 − 10𝑥 + 24 = 0
(𝑥 − 6)(𝑥 − 4) = 0
𝑥 = 6 or 𝑥 = 4
𝑦 = −3(6) + 13 = −5 or 𝑦 = −3(4) + 13 = 1
𝑇(6; −5) or 𝑇(4; 1)
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©Via Afrika >> Mathematics Grade 12 143
Answers to Mixed Exercises
Chapter 10: Euclidian geometry
1 a 𝐴�1 = 𝑀�1 tan chord
𝐴�1 = 𝐶̂
∴𝑀 �1 = 𝐶̂
∴ 𝑀𝑁||𝐶𝐴 corr ∠s
b �1 = 𝑀
𝐾 �2 alt ∠s
�1 = 𝑁
𝐾 �2 tan chord
∴ ∆𝐾𝑀𝑁 is isosceles
c �4 = 𝑁
𝐾 �2 alt ∠s
𝑁�2 = 𝐴�3 ∠s in same segment
𝐴̂3 = 𝐴�3 ∠s in same segment
�
∴ 𝐾4 = 𝐴3 ̂
∴ 𝑁𝐾||𝐴𝑃 alt ∠s=
𝐴𝑁 𝐴𝐾
∴ 𝑁𝐴 = 𝐾𝑃 line || to one side of ∆
𝐴𝑁 𝐴𝑀
But 𝑁𝐴 = 𝑀𝐸 line || to one side of ∆
𝐴𝐾 𝐴𝑀
∴ 𝐾𝑃 = 𝑀𝐸
d 𝐴̂3 = 𝐴�3 ∠s in same segment
𝐴�3 = 𝐴�2 equal chords subt equal ∠s
̂
∴ 𝐴3 = 𝐴2 �
∴ 𝐷𝐴 is a tangent to the circle through A, B and K
2 a 𝐶̂3 = 𝐶𝑃� 𝑅 ∠s opp equal sides
̂ ̂ ̂
𝐶3 + 𝐶2 = 𝐴1 + 𝐴� ext ∠ of ∆
𝐶̂2 = 𝐴� tan chord
̂
∴ 𝐶3 = 𝐴1̂
∴ 𝐴̂1 = 𝐶𝑃�𝑅 both = 𝐶̂3
ACPR is a cyclic quadrilateral (ext ∠ of quad)
b In ∆𝐶𝐴𝐴 and ∆𝑅𝑃𝐴:
𝑃�2 = 𝐶̂2 ∠s in same segment
= 𝐴� proven in 2 a
∴ 𝐴� = 𝑃�2
𝐶̂1 = 𝐴𝑅� 𝑃 ext ∠ of cyclic quad
𝐴̂1 = 𝐴̂3 3rd ∠ of ∆
∴ ∆𝐶𝐴𝐴|||∆𝑅𝑃𝐴 ∠∠∠
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©Via Afrika >> Mathematics Grade 12 144
Answers to Mixed Exercises
𝑅𝑃 𝑅𝐴
c 𝐸𝐴
= 𝐸𝐴 from 2 b
𝐸𝐴.𝑅𝐴
𝑅𝑃 = 𝐸𝐴
but 𝑅𝑃 = 𝑅𝐶
𝐸𝐴.𝑅𝐴
∴ 𝑅𝐶 = 𝐸𝐴
d In ∆𝑅𝐴𝐶 and ∆𝑅𝐶𝐴:
𝐶̂2 = 𝐴� tan chord
𝑅�1 is common
𝑅𝐶̂ 𝐴 = 𝑅𝐴̂𝐶 3rd angle
∴ ∆𝑅𝐴𝐶|||∆𝑅𝐶𝐴 ∠∠∠
𝐴𝐸 𝑅𝐸
𝐸𝐴
= 𝑅𝐴 ∆𝑠 |||
𝑅𝐴. 𝐴𝐶 = 𝑅𝐶. 𝐶𝐴
𝐸𝐴 𝐸𝐴
e 𝑅𝑃
= 𝑅𝐴 from 2 b
𝐸𝐴 𝐸𝐴
𝑅𝐸
= 𝑅𝐴 RC=RP
𝐸𝐴.𝑅𝐴
𝐴𝐶 = 𝑅𝐸
…(i)
𝑅𝐸.𝐸𝐴
From 2 d 𝐴𝐶 = 𝑅𝐴
𝐸𝐴.𝑅𝐴 𝑅𝐸.𝐸𝐴
∴ 𝑅𝐸
= 𝑅𝐴
2
∴ 𝑅𝐶 = 𝑅𝐴. 𝑅𝐴
3 a 𝐴�2 = 𝐴̂3 = 𝑥 ∠s opp equal sides
𝑀�1 = 180° − 2𝑥 sum ∠s of ∆
∴𝐷 � = 2𝑥
𝑀 �
bi 𝐶̂ = 21 ∠ at centre =2x∠circ
= 90° − 𝑥
𝐶𝐴� 𝐷 = 180° − (90° − 𝑥 + 2𝑥) sum ∠s of ∆
= 90° − 𝑥
𝑁�1 = 𝐶̂ = 90° − 𝑥 ext ∠ of cyclic quad
∴ 𝐶𝐴� 𝐷 = 𝑁 �1
∴ 𝐶𝐴||𝐴𝑁 corr ∠s
b ii � �
𝐶𝐴 𝐴 = 𝐷 = 2𝑥 tan chord
𝐶𝐴� 𝐴 = 𝐴̂2 alt ∠s
𝐴̂2 = 𝐷
�
∴AB is a tangent (∠betw line&chord= ∠sub chord)
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©Via Afrika >> Mathematics Grade 12 145
Answers to Mixed Exercises
4 a 𝐴�3 = 𝐷�1 = 𝑥 ∠s in same segment
𝐴�3 = 𝐷�2 = 𝑥 ∠s opp = sides
�
𝐴𝑂𝐷 = 180° − 2𝑥 sum ∠s of ∆
𝐴̂ = 90° − 𝑥 ∠ at centre =2x∠circ
bi 𝐶̂1 = 90° − 𝑥 ext ∠ of cyclic quad
𝐷�2 = 180° − (𝑥 + 90° − 𝑥) sum ∠s of ∆
= 90°
In ∆𝐴𝐷𝐷 and ∆𝐶𝐷𝐷:
𝐷�1 = 𝐷�2 = 90° adj ∠s str line
BF = FC
FE is common
∆𝐴𝐷𝐷 ≡ ∆𝐶𝐷𝐷 s∠s
BE = EC (≡)
ii 𝐴�1 = 90° − 𝑥 sum ∠s of ∆
∴ 𝐴�1 = 𝐴̂
∴ BE is not a tangent �𝐴�1 + 𝐴�2 ≠ 𝐴̂�
5 a P is midpoint of AC medians concur
AB||PM midpt theorem
In ∆𝐴𝑁𝐶:
𝑁𝐴 𝐴𝑀 𝐴𝑃
𝑁𝐸
= 𝐴𝐸 = 𝐴𝐸 line || 1 side of ∆
𝐴𝑀 1
= 2𝐴𝑀 = 2
b In ∆𝐴𝑀𝑃:
𝐴𝑂 2𝑂𝑀
𝑂𝑀
= 𝑂𝑀
𝑅𝑃 𝑅𝑃
𝑃𝐸
= 𝐴𝑃 BP is a median
𝑂𝑀
= 𝐴𝑀 line || 1 side of ∆
𝑂𝑀
= 3𝑂𝑀
1
=3
6 a 𝐶̂2 = 90° ∠ in semi ⊙
𝑀�2 = 90° AM⏊NM
∴ 𝑁𝑄||𝐶𝐷 corr ∠s=
b 𝐶̂1 = 𝑁
� || lines, corr ∠s
𝐴̂2 = 𝐶̂1 tan chord
=𝑁�
∴ ANCQ is a cyclic quad ∠s subt by same line segm
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 146
Answers to Mixed Exercises
ci In ∆𝑃𝐶𝐷 and ∆𝑃𝐴𝐶:
𝐶̂1 = 𝐴̂2 tan chord
𝑃� is common
𝐷�1 = 𝐴𝐶̂ 𝑃 3rd ∠
∴ ∆𝑃𝐶𝐷 ||| ∆𝑃𝐴𝐶 ∠∠∠
ii 𝑃𝐶 2 = 𝐴𝑃. 𝐷𝑃
d In ∆𝑁𝐴𝐶 and ∆𝐴𝐶𝐷:
𝑁� = 𝐴̂2 ∠s in same segm
= 𝐴�2 ∠s in same segm
𝐶̂4 = 𝐴̂1 tan chord
=𝐷 �2 ∠s in same segm
𝐴�1 = 𝐴𝐶̂ 𝐷 3rd ∠
∴ ∆𝑁𝐴𝐶 ≡ ∆𝐴𝐶𝐷 ∠∠∠
𝐴𝐸 𝐸𝐴
∴ 𝑁𝐴 = 𝑁𝐴
𝐴𝐶 2 = 𝐶𝐷. 𝑁𝐴
𝐴𝑀2 𝐴𝐸 2 −𝐴𝑀2
e 1 − 𝐴𝐸 2 = 𝐴𝐸 2
𝑀𝐸 2
= 𝐴𝐸 2 Pyth.
𝑃𝐸 2
= 𝐴𝐸 2
𝐴𝑃.𝐴𝑃
= 𝐸𝐴.𝑁𝐴
Chapter 11: Statistics: regression and correlation
1 a
Lower Q Median Upper Q
Matches played 3 5 6
Wins 1 7 3
Goals scored against 3 4,5 9
b
Positively skewed (skewed to the right)
65
c 14
= 4,64
d Standard deviation = 1,72
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 147
Answers to Mixed Exercises
2 a
% transport costs
60
50
40
30
20
10
0
10 20 30 40 50 60
% transport costs
b Median = ±32%
c
Class midpoint Frequency FreqxMidpoint
15 6 90
25 14 350
35 16 560
45 11 495
55 3 165
TOTAL 50 1660
1660
Estimated mean 𝑥̅ = 50 = 33,2%
d
Class midpt 𝑥𝑖 Freq 𝑓 𝑥̅ − 𝑥𝑖 (𝑥̅ − 𝑥𝑖 )2 𝑓(𝑥̅ − 𝑥𝑖 )2
15 6 -18,2 331,24 1987,44
25 14 -8,2 67,24 941,36
35 16 1,8 3,24 51,84
45 11 11,8 139,24 1531,64
55 3 21,8 475,24 1425,72
TOTAL 50 5938
5938
Standard deviation= � 50 = 10,90
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 148
Answers to Mixed Exercises
3 a&c 20 Di sta nce
15
10
5
VO2
10 20 30 40 50 60
b 𝑦 = 0,2432𝑥 + 3,6834
d Substitute 𝑦 = 19 then 𝑥 = 62,98 (VO2)
e 𝑟 = 0,8985 …
Strong positive correlation
4
𝑁𝑢𝑚𝑏𝑐𝑟 (𝑁𝑢𝑚𝑏𝑐𝑟 − 𝑚𝑐𝑎𝑖)2
4 36
8 4
10 0
𝑥 (𝑥 − 10)2
𝑦 (𝑦 − 10)2
Mean= 10
4+8+10+𝑥+𝑦
∴ 5
= 10
Which simplifies to: 𝑥 + 𝑦 = 28 … . (1)
Standard deviation = 4
36+4+0+(𝑥−10)2 +(𝑦−10)2
∴� =4
5
Which simplifies to: (𝑥 − 10)2 + (𝑦 − 10)2 = 40 … (2)
Substitute 𝑦 = 28 − 𝑥 from (1) into (2):
𝑥 2 − 28𝑥 + 192 = 0
(𝑥 − 12)(𝑥 − 16) = 0
𝑥 = 12 or 𝑥 = 16
𝑦 = 16 or 𝑦 = 12
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 149
Answers to Mixed Exercises
5 The standard deviation will remain 7,5.
If all the numbers are 2 bigger, then the mean will also be 2 bigger.
The difference between each number and the mean will therefore remain the same
leaving the standard deviation unchanged.
Chapter 12: Probability
1 7 spaces that have to be filled using 7 digits without repetition(as 0,7 and 4 may not be
used again)
∴ 7! = 5040
2 7 spaces have to be filled – 10 digits are available for each space
∴ 107 = 10 000 000
1
3 P(Queen of diamonds)= 52
4 a 11!
11!
b 2!2!2!2!2!
= 1 247 400 (5 letters repeat)
5 a Regard the 4 English books as a unit. The number of arrangements for the English
books is 4!=24
Total number of arrangements= 4! × 6! = 17 280
b 4! × 3! × 2! × 3! = 1728
c 9! = 362 880
6 12 × 11 = 132
11!
7 First calculate the total number of words: 2!2!2! = 4 989 600
Now calculate how many of these WILL start and end on the same letter.
It can start and end with M, A or T
9!
∴ 2!2! = 90720
90 720 54
P(not start and end on same letter)= 1 − 4 989 600 = 55
10!×2 10!
8 a 2!2!2!
= 907 200 b 2!2!2!
= 453 600
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 150
Exemplar Paper 1
Exemplar Paper 1 (3 hours; 150 marks)
1 a Solve for 𝑥:
2
i 𝑥 + 2 = 𝑥+1 (4)
ii 𝑥 − √𝑥 = 6 (4)
(𝑥 2 +4)(2−𝑥)
iii 𝑥+2
≥0 (6)
iv 5𝑥−2 + 5𝑥+1 = 126 (5)
b Consider the equation: 𝑓(𝑥) = 2𝑥 3 + 𝑝𝑥 2 + 𝑏𝑥 − 9𝑝
If (2𝑥 + 𝑝) is a factor of 𝑓(𝑥) and 𝑝 ≠ 0, determine the value(s) of 𝑏. (5)
c 2 is a root of 2𝑥 2 − 3𝑥 − 𝑝 = 0. Determine the value of 𝑝 and hence the
other root. (4)
[28]
2 a The sum of the first 20 terms of an arithmetic progression is 410, while the
sum of the next 30 terms is 2865. Determine the first three terms of the
progression. (7)
b 3; 𝑥; 15; 𝑦; 35 is a quadratic sequence.
i Determine the values if 𝑥 and 𝑦. (4)
ii Determine formula for 𝑇𝑛 . (4)
c Find 𝑖 such that ∑𝑛𝑘=7(2𝑘 − 3) is equal to the sum of the first 6 terms of the
sequence −24; 48; −96; … (7)
d For which value(s) of 𝑥 will the following series be convergent?
(𝑥 + 2) + (𝑥 + 2)2 + (𝑥 + 2)3 + ⋯ (2)
[24]
3 a Melissa decides to save R1 200 per month for a certain period. The bank offers
her an interest rate of 12% p.a. compounded monthly for this period.
Determine how long Melissa has to make this monthly payment if she wants to
have a lump sum of R200 000. (5)
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 151
Exemplar Paper 1
b Richard plans to buy a house on a 20 year mortgage and can only afford to pay
R5 000 per month. If the interest rate is currently 12% per annum compounded
monthly, determine the size of the mortgage he can take, if he starts paying
one month after the mortgage was approved. (3)
c An amount of R300 000 is to be used to provide quarterly withdrawals for the
next 10 years. The withdrawal amount will remain fixed and the first
withdrawal will be in 3 months’ time. An interest rate of 15% p.a. compounded
quarterly applies. Determine the value of each quarterly withdrawal. (4)
[12]
1 1
4 In the diagram 𝑓 is the graph of 𝑦 = − 2 𝑥 2 + 2 𝑥 + 𝑘 cuts the 𝑥 −axis at B and C
3
and the 𝑦 −axis at D. 𝑔 is the graph of 𝑦 = 𝑎𝑥 − 2 and cuts the 𝑥 −axis at B. ℎ is
the graph of 𝑦 = 𝑚 𝑥 and cuts the 𝑦 −axis at D. QR and ST are parallel to the 𝑦 −axis.
1
𝐴 �𝑥; 4� is a point on ℎ and vertically above C.
y
h
g
Q S
D
A
x
P B F C
T
f
R
a Determine the values of 𝑘 and 𝑚. (6)
b Determine the value of 𝑎. (2)
c Calculate the length of QR if OP = 2 units. (4)
d Determine the length op OF, if ST = 4 units. (4)
e Determine the equation of ℎ−1. (2)
f Write down the domain of ℎ−1 . (2)
[20]
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 152
Exemplar Paper 1
𝑎
5 The functions 𝑓(𝑥) = 𝑥+𝑏 + 𝑐 and 𝑔(𝑥) = 2𝑥 − 13 intersect each other. The
asymptotes of 𝑓(𝑥) intersect in(6; −8). 𝑓(𝑥) goes through (7; −4).
y
y = 2x - 13
x
·
(7;-4)
(6;-8)
a Determine the values of a, 𝑏 and 𝑐. (4)
b Determine the co-ordinates of the intersects of 𝑓 and 𝑔. (5)
c For which values of 𝑥 would 𝑔(𝑥) ≥ 𝑓(𝑥)? (3)
d Determine the equation of the dotted line which is the axis of symmetry of the
hyperbola. (3)
[15]
6 Determine:
𝑥 2 −1
a lim𝑥→1 1−𝑥 (3)
b 𝑓 ′ (𝑥) from first principles if 𝑓(𝑥) = −2𝑥 2 . (4)
1
c 𝑔′ (𝑡) if 𝑔(𝑡) = 2√𝑡 + 2𝑡 2 ; 𝑡 ≠ 0 (4)
[11]
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 153
Exemplar Paper 1
7 The figure shows the graph of 𝑓(𝑥) = 2𝑥 3 + 𝑎𝑥 2 + 𝑏𝑥 + 3. The curve has a local
minimum turning point F at (2; −9).
y
D
E
A B C x
F (2;−9)
a Show that 𝑎 = −5 and 𝑏 = −4. (6)
b If it is given that 𝐴(−1; 0), calculate the coordinates of B and C. (5)
c Determine the equation of the tangent to the graph at 𝑥 = 3. (4)
[15]
8 A container firm is designing an open-top rectangular box that will hold 108 𝑐𝑚3 .
The box has a square base with sides 𝑥 and height ℎ.
ℎ
𝑥
a Show that the total outside surface area of the 𝑥
432
box will be 𝑆 = 𝑥 2 + 𝑥 . (4)
b For which value of 𝑥 and ℎ will the outer surface area
be a minimum. (5)
[9]
9 a A six-member working group is to be selected from five teachers and nine
students. If the working group is randomly selected, what is the probability
that it will include at least two teachers? (4)
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 154
Exemplar Paper 1
b 𝑃(𝐴 𝑜𝑟 𝐴) = 0,6 and 𝑃(𝐴) = 0,2
i Find 𝑃(𝐴) given that events A and B are mutually exclusive. (2)
ii Find 𝑃(𝐴) given that events A and B are independent. (4)
c i In how many ways can the letters of the word PROBABILITY
be arranged to form different “words” – the word “probability” itself
is included? (3)
ii In how many ways can the letters of the word PROBABILITY
be arranged to form different “words” if the R and O have to be kept
together? (3)
[16]
MEMORANDUM: Exemplar Paper 1
2
1 a i 𝑥 + 2 = 𝑥+1
(𝑥 + 2)(𝑥 + 1) = 2
𝑥 2 + 3𝑥 = 0
𝑥(𝑥 + 3) = 0
∴ 𝑥 = 0 𝑜𝑟 𝑥 = −3
ii 𝑥 − √𝑥 = 6
Let = √𝑥 , then 𝑘 2 = 𝑥
𝑘2 − 𝑘 − 6 = 0
(𝑘 − 3)(𝑘 + 2) = 0
𝑘 = √𝑥 = 3 or 𝑘 = √𝑥 = −2
𝑥=9 Not valid
(𝑥 2 +4)(2−𝑥)
iii 𝑥+2
≥0
(𝑥 2 + 4) > 0 for all values of 𝑥 ∈ 𝑅
(2−𝑥)
(𝑥+2)
≥0
(𝑥−2)
(𝑥+2)
≤0
∴ −2 < 𝑥 ≤ 2
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 155
Exemplar Paper 1
iv 5𝑥−2 + 5𝑥+1 = 126
5𝑥 . 5−2 + 5𝑥 . 51 = 126
1
5𝑥 �25 + 5� = 126
126
5𝑥 � 25 � = 126
5𝑥 = 52
𝑥=2
𝑝 −𝑝 3 𝑝 2 𝑝
b If (2𝑥 + 𝑝) is a factor, then 𝑓 �− 2� = 2 � 2 � + 𝑝 �− 2� + 𝑏 �− 2� − 9𝑝 = 0
𝑝3 𝑝3 𝑏𝑝
− 4 + 4 − 2 − 9𝑝 = 0
× 2) 𝑏𝑝 = −18𝑝
÷ 𝑝) 𝑏 = −18
c Substitute 𝑥 = 2: 2(2)2 − 3(2) − 𝑝 = 0
∴𝑝=2
2𝑥 2 − 3𝑥 − 2 = 0
(2𝑥 + 1)(𝑥 − 2) = 0
1
∴ 𝑥 = − 2 is the other root
2 a 𝑆20 = 410
𝑆50 = 𝑆20 + 𝑠𝑢𝑚 𝑜𝑓 𝑖𝑐𝑥𝑡 30 𝑡𝑐𝑟𝑚𝑠 = 410 + 2865 = 3275
20
410 = 2 [2𝑎 + 19𝑑]
41 = 2𝑎 + 19𝑑 … . (1)
50
3275 = 2 [2𝑎 + 49𝑑]
131 = 2𝑎 + 49𝑑 … . (2)
(2)-(1): 30𝑑 = 90
∴ 𝑑 = 3 en 𝑎 = −8
b i 𝑥 = 8 ; 𝑦 = 24
ii T: 3 ; 8 ; 15 ; 24 ; 35
𝑓: 5 ; 7 ; 9 ; 11
𝑠: 2 ; 2 ; 2
𝑎 =2÷2=1 𝑏 = 5 − 3(1) = 2 𝑐 =3−1−2 =0
2
𝑇𝑛 = 𝑖 + 2𝑖
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 156
Exemplar Paper 1
c −24; 48; −96; … is a geometric series with 𝑎 = −24 and 𝑟 = −2
−24�(−2)6 −1�
𝑆6 = −2−1
= 504
∑𝑛𝑘=7(2𝑘 − 3) = 11 + 13 + 15 + ⋯ + (2𝑖 − 3)
𝑛
504 = [2(11) + (𝑖 − 1)(2)]
2
2
𝑖 + 10𝑖 − 504 = 0
(𝑖 + 28)(𝑖 − 18) = 0
∴ 𝑖 = 18
d 𝑟 =𝑥+2
For convergent series −1 < 𝑟 < 1
−1 < 𝑥 + 2 < 1
−3 < 𝑥 < −1
0,12 𝑛
1200[�1+ � −1]
3 a 200 000 = 12
0,12
12
5
(1,01)𝑛 =
3
5
𝑙𝑜𝑔
𝑖 = 𝑙𝑜𝑔1,01 = 51,33755
3
Melissa must make at least 52 payments
0,12 −240
5000�1−�1+ � �
12
b 𝑃= 0,12 = 𝑅454 097,08
12
0,15 −40
𝑥�1−�1+ � �
4
c 300000 = 0,15
4
0,15
300000×
𝑥= 4
0,15 −40
= 𝑅14957,84
�1−�1+ � �
4
4 a D is the 𝑦 −intercept of 𝑓 and ℎ .
Substitute 𝑥 = 0 into 𝑦 = 𝑚 𝑥
∴𝑦=1
∴𝑘=1
To find 𝑚 we need the coordinates of A.
First find the roots of 𝑓 os we can get the 𝑥-value of A.
1 1
− 2 𝑥2 + 2 𝑥 + 1 = 0
𝑥2 − 𝑥 − 2 = 0
(𝑥 − 2)(𝑥 + 1) = 0
𝑥 = 2 at C and A
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 157
Exemplar Paper 1
1
Sub 𝐴 �2; 4� into 𝑦 = 𝑚 𝑥
1
4
= 𝑚2
1
∴𝑚=2
3 3
b Sub 𝐴(−1; 0): 0 = 𝑎(−1) − 2 ∴ 𝑎 = −2
c 𝑥 = −2 at Q and R
𝑄𝑅 = 𝑦𝑄 − 𝑦𝑅
3 3 1 1
= − 2 (−2) − 2 − [ − 2 (−2)2 + 2 (−2) + 1]
7
=2
1 1 3 3
d − 2 𝑥2 + 2 𝑥 + 1 + 2 𝑥 + 2 = 4
1 3
− 2 𝑥 2 + 2𝑥 − 2 = 0
𝑥 2 − 4𝑥 + 3 = 0
(𝑥 − 3)(𝑥 − 1) = 0
∴ 𝑂𝐷 = 1
e ℎ−1 = log 1 𝑥
2
f 𝑥 > 0; 𝑥 ∈ 𝑅
5 a Asymptotes go through (6; −8)
∴ 𝑏 = −6 and 𝑐 = −8
𝑎
Substitute (7; −4) into 𝑓(𝑥) = 𝑥−6 − 8
𝑎
−4 = 7−6 − 8
∴𝑎=4
4
b 𝑥−6
− 8 = 2𝑥 − 13
4
𝑥−6
= 2𝑥 − 5
4 = (2𝑥 − 5)(𝑥 − 6)
2𝑥 2 − 17𝑥 + 26 = 0
(2𝑥 − 13)(𝑥 − 2) = 0
13
𝑥 = 2 or 𝑥 = 2
𝑦 = 0 or 𝑦 = −9
13
Intersects are � 2 ; 0� and (2; −9)
13
c 𝑥 ∈ [2; 6) or 𝑥 ∈ [ 2 ; ∞)
d Substitute (6; −8) into 𝑦 = 𝑥 + 𝑐
−8 = 6 + 𝑐 ∴ 𝑐 = −14 𝑦 = 𝑥 − 14
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 158
Exemplar Paper 1
𝑥 2 −1 (𝑥−1)(𝑥+1)
6 a lim𝑥→1 1−𝑥 = lim𝑥→1
−(𝑥−1)
= lim𝑥→1 −(𝑥 + 1) = −2
𝑓(𝑥+ℎ)−𝑓(𝑥)
b 𝑓 ′ (𝑥) = limℎ→0
ℎ
−2(𝑥+ℎ)2 −�−2𝑥 2 �
= lim ℎ
ℎ→0
−2𝑥ℎ−2ℎ2
= lim
ℎ→0 ℎ
= lim − 2𝑥 − ℎ
ℎ→0
= −2𝑥
1
1 1
c 𝑔(𝑡) = 2√𝑡 + 2𝑡 2 = 2𝑡 2 + 2 𝑡 −2
1
1 1
𝑔′ (𝑡) = 2 × 2 𝑡 −2 + 2 × −2𝑡 −3
1 1
= − 𝑡3
√𝑡
7 a 𝑓(2) = −9 and 𝑓 ′ (2) = 0
𝑓 ′ (𝑥) = 6𝑥 2 + 2𝑎𝑥 + 𝑏
0 = 6(2)2 + 2𝑎(2) + 𝑏
4𝑎 + 𝑏 = −24 … . (1)
−9 = 2(2)3 + 𝑎(2)2 + 𝑏(2) + 3
2𝑎 + 𝑏 = −14…(2)
(1)-(2): 2𝑎 = −10
∴ 𝑎 = −5
𝑏 = −2𝑎 − 14
= −2(−5) − 14 = −4
b From (a) it follows that 𝑓(𝑥) = 2𝑥 3 − 5𝑥 2 − 4𝑥 + 3
If 𝑥 = −1 is a root, then (𝑥 + 1) is a factor of 𝑓
𝑓(𝑥) = 2𝑥 3 − 5𝑥 2 − 4𝑥 + 3
= (𝑥 + 1)(2𝑥 2 − 7𝑥 + 3
= (𝑥 + 1)(2𝑥 − 1)(𝑥 − 3)
1
𝑥 = −1, or 𝑥 = 2 or 𝑥 = 3
1
𝐴 �2 ; 0� and 𝐶(3; 0)
c 𝑓 ′ (𝑥) = 6𝑥 2 − 10𝑥 − 4
𝑓 ′ (3) = 6(3)2 − 10(3) − 4 = 20
Sub (3; 0) into 𝑦 = 20𝑥 + 𝑐
Eq of tangent: 𝑦 = 20𝑥 − 60
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 159
Exemplar Paper 1
8 a Volume = 108
𝑥 2 ℎ = 108
108
∴ ℎ = 𝑥2
𝑆 = 𝑥 2 + 4𝑥ℎ
108
= 𝑥 2 + 4𝑥 � 𝑥 2 �
432
= 𝑥2 + 𝑥
b 𝑆 will be a minimum where 𝑆 ′ (𝑥) = 0
𝑆 = 𝑥 2 + 432𝑥 −1
432
𝑆 ′ (𝑥) = 2𝑥 − 𝑥 2 = 0
𝑥 3 = 216
108
𝑥 = 6 𝑚 and ℎ = (6)2 = 3 𝑚
14!
9 a Total number of different six-member groups= 7! = 17 297 280
9!
Number of groups with no teacher= 3! = 60 480
Number of groups with one teacher only= 5 × 9 × 8 × 7 × 6 × 5 = 75 600
Total number of groups with less than two teachers= 136 080
Total number of groups with two or more teachers
= 17 297 280 − 136 080 = 17 161 200
17 161 200
P(two or more teachers)= 17297280 = 0,99
bi 𝑃(𝐴𝑜𝑟𝐴) = 𝑃(𝐴) + 𝑃(𝐴) for mutually exclusive
0,6 = 0,2 + 𝑃(𝐴)
𝑃(𝐴) = 0,4
ii 𝑃(𝐴 𝑎𝑖𝑑 𝐴) = 𝑃(𝐴) × 𝑃(𝐴) for independentevents
𝑃(𝐴 𝑜𝑟 𝐴) = 𝑃(𝐴) + 𝑃(𝐴) − 𝑃(𝐴 𝑎𝑖𝑑 𝐴)
𝑃(𝐴 𝑜𝑟 𝐴) = 𝑃(𝐴) + 𝑃(𝐴) − 𝑃(𝐴) × 𝑃(𝐴)
0,6 = 0,2 + 𝑃(𝐴) − 0,2𝑃(𝐴)
0,4 = 0,8𝑃(𝐴)
𝑃(𝐴) = 0,5
11!
ci 2!2!
= 9 979 200
10!
ii 2!2!
= 907 200
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©Via Afrika >> Mathematics Grade 12 160
Exemplar Paper 2
Exemplar Paper 2 (3 hours; 150 marks)
1 Given the following box-and-whisker plot:
a Which quarter has the smallest spread of data?
What is the spread? (2)
b Determine the inter quartile range. (2)
c Are there more data in the interval 5-10 or in the interval 10-13?
How do you know this? (2)
d Which interval has the fewest data in it? Is it 0-2, 2-4, 10-12 or
12-13? How do you know it? (2)
[8]
2 A factory produces and stockpiles metal sheets to be shipped to a motor
vehicle manufacturing plant. The factory only ships when there is a minimum
of 3254 sheets in stock at the beginning of that day. The table shows the day
and the number of sheets in stock at the beginning of that day.
Day 1 2 3 4 5 6
Sheets 854 985 1054 1195 1204 1384
a Determine the equation of the least squares regression line for this set of
data rounding coefficients to three decimal places. (3)
b Use this equation to determine the day the sheets will be shipped. (3)
[6]
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 161
Exemplar Paper 2
3 The ogive below represents the results of a survey amongst first year students on the
average time per day they spend exercising. Answer the questions that follow.
a How many students participated in the survey? (1)
b Approximately how many students spend more between 10 and 20 minutes
per day exercising? (1)
c Use the ogive to determine the median time spent on daily exercise. (2)
[4]
4 In the diagram, KC is a diameter of the circle
and 𝐾(1; 4); 𝐶(7; 2) and 𝐴(𝑥; 𝑦) are points
on the circle.
Determine:
a the equation of the circle (5)
1
b point B if the gradient of KB= 2 (9)
[14]
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 162
Exemplar Paper 2
5 In the diagram, ABCD is a quadrilateral
with 𝐴(4; 12), 𝐴(1; 3), 𝐶(4; 2) and D(8;4).
a Determine the gradients of BC and CD. (4)
b Show that AB ⏊ BC. (3)
c Prove that ABCD is a cyclic
quadrilateral. (4)
d Determine the equation of the circle ABCD. (7)
[18]
6 Given the vertices 𝐴(2; 3), 𝐴(5; 4), 𝐶(4; 2) and 𝐷(1; 1) of parallelogram ABCD.
Determine:
a the coordinates of M, the point of intersection of diagonals AC
and BD (2)
b the equation of the median PM of ∆DMC (5)
[7]
5
7 a If 𝑠𝑐𝑐𝐴 = 4 and 180° < 𝐴 < 360°, determine the following without the use of a
calculator:
i 𝑐𝑜𝑠𝐴 (1)
ii 𝑠𝑖𝑖2𝐴 (4)
𝑎𝑜𝑠𝑒𝑎73°
b If 𝑠𝑖𝑖17° = 𝑘, express 𝑎𝑜𝑠343° in terms of 𝑘. (3)
[8]
8 a Determine the value of the following without using a calculator:
𝑐𝑜𝑠69°. 𝑐𝑜𝑠9° + 𝑐𝑜𝑠81°. 𝑐𝑜𝑠21° (4)
b Consider the following identity:
1+𝑎𝑜𝑠𝑥+𝑎𝑜𝑠2𝑥 1
𝑠𝑖𝑛𝑥+𝑠𝑖𝑛2𝑥
= 𝑡𝑎𝑛𝑥
i For which values of 𝑥 will the identity be undefined? (4)
i Prove the identity. (4)
[12]
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 163
Exemplar Paper 2
9 a Solve the following equations for the interval [−90°; 90°]:
i 2𝑡𝑎𝑖𝑥 = −0,6842 (2)
ii 𝑠𝑖𝑖2𝑥. 𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑖𝑥. 𝑐𝑜𝑠2𝑥 = 0,5 (2)
b Determine the general solution of:
1
cos �2 𝑥 + 15°� = sin(2𝑥 − 15°) (5)
[9]
10 The graphs of 𝑦 = 𝑎𝑠𝑖𝑖𝑥 and 𝑦 = 𝑐𝑜𝑠𝑏𝑥 are drawn over the interval
y
2
1
x
−180 −150 −120 −90 −60 −30 30 60 90 120 150 180
−1
−2
a Write down the values of 𝑎 and 𝑏. (2)
b Use your graph to determine approximate values of 𝑥; 𝑥 ∈ [−180°; 180] for
1
which 𝑐𝑜𝑠 2 𝑥 − 𝑠𝑖𝑖𝑥 = 2 (5)
[7]
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 164
Exemplar Paper 2
11 In the diagram below, Q is the base of a vertical tower PQ, while R and S are points
in the same horizontal plane as Q. The angle of elevation of P, the top
of the tower, as measured from R, is 𝑥. Furthermore, 𝑅𝑄� 𝑆 = 𝑦, 𝑄𝑆 = 𝑎 𝑚𝑐𝑡𝑟𝑐𝑠
and the area of ∆𝑄𝑅𝑆 = 𝐴 𝑚2.
2𝐴𝑡𝑎𝑛𝑥
a Show that 𝑃𝑄 = 𝑎𝑠𝑖𝑛𝑦 (5)
b Calculate the value of 𝑦 if 𝑃𝑄 = 76,8𝑚; 𝑎 = 87,36; 𝐴 = 480,9𝑚2 and
𝑥 = 46,5°. (3)
[8]
12 a Write down the converse of the following theorem:
The angle between a tangent to a circle and a chord drawn through
the point of contact, is equal to an angle in the alternate segment. (2)
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 165
Exemplar Paper 2
b The diagonal AC of quadrilateral ABCD bisects 𝐴𝐶̂ 𝐷 while AD is
a tangent to the circle ABC at point A. Prove that AB is a tangent to
circle ACD.
(5)
c Two circles intersect at A and B. AB is produced to P. PQ is a tangent to the
smaller circle at Q. QB produced meets the larger circle at R. PR cuts the
larger circle at X. AX and AQ are drawn.
Prove that:
i Points A, X, P and Q are on the circumference of the same circle.
(5)
ii PQ is a tangent to the circumscribed circle of ∆𝑄𝑅𝑋. (3)
[15]
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 166
Exemplar Paper 2
13 a In ∆𝐴𝐴𝐶 and ∆𝐷𝐷𝐷, 𝐴̂ = 𝐷
� and 𝐴� = 𝐷� .
𝐴𝐴 𝐴𝐸 𝐴𝐸
Prove the theorem that 𝐴𝐴 = 𝐴𝐸 = 𝐴𝐸 . (8)
b In the diagram A, B, C and E are points on a circle.
AE bisects 𝐴𝐴̂𝐶 and BC.
AE intersect in D.
Prove that:
i ∆𝐴𝐴𝐷///∆𝐴𝐷𝐶 (4)
ii 𝐴𝐴. 𝐴𝐶 = 𝐴𝐷2 + 𝐴𝐷. 𝐷𝐶 (7)
[19]
𝐴𝑅 3
14 In ∆𝐴𝐴𝐶 , P is the midpoint of AC, RS//BP and 𝐴𝐴 = 5.
CR and BP intersect at T.
Determine, giving reasons, the following ratios:
𝐴𝑆
a 𝑆𝑃
(4)
𝐴𝑆
b 𝑆𝐸
(3)
𝑅𝑇
c 𝑇𝐸
(3)
𝐴𝑟𝑒𝑎 ∆𝑇𝑃𝐸
d 𝐴𝑟𝑒𝑎 ∆𝑅𝑆𝐸
(6)
[15]
TOTAL: 150
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 167
Exemplar Paper 2
MEMORANDUM: Exemplar Paper 2
1 a Fourth quarter. Spread = 13 − 12 = 1
b IQR= 12 − 2 = 10
c More data in 10-13
Median = 10 and Max=13. Therefore 50% of the data lies in interval 10-13
25% of data lies between 2-10. Therefore less than 50% in 5-10
d 2-4 has fewest data
0-2, 2-10 ,10-12 and 12-13 all represent 25% of the data
2-4 will only be a part of 25% (less than 25%)
2 a 𝑦 = 767,867 + 98,514𝑥
b 3254 = 767,867 + 98,514𝑥
98,514𝑥 = 2486,133
𝑥 = 25,236
Shipping will be done on the 26th day.
3 a 100
b 80-20=60
c 14 minutes
1+7 4+2
4 a Midpoint= � 2 ; 2 � = (4; 3)
Radius = �(4 − 1)2 + (3 − 4)2 = √10
(𝑥 − 4)2 + (𝑦 − 3)2 = 10
b KB ⏊ BC (𝐴� = 90°; angle in semi circle)
𝑚𝐴𝐸 = −2
𝑦−4 1
∴ 𝑚𝐾𝐴 = 𝑥−1 = 2 2(𝑦 − 4) = 𝑥 − 1 …..(1)
𝑦−2
∴ 𝑚𝐴𝐸 = 𝑥−7 = −2 (𝑦 − 2) = −2(𝑥 − 7) … … (2)
Solving equations (1) and (2) simultaneously yields:
𝑥 = 5 ;𝑦 = 6
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 168
Exemplar Paper 2
3−2 1 4−2 1
5 a 𝑚𝐴𝐸 = 1−4 = − 3 𝑚𝐸𝐴 = 8−4 = 2
12−3 1
b 𝑚𝐴𝐴 = 4−1 = 3 ∴ 𝑚𝐴𝐴 × 𝑚𝐴𝐸 = − 3 × 3 = −1
∴ AB ⏊ BC
12−4 1
c 𝑚𝐴𝐴 = 4−8 = −2 ∴ 𝑚𝐸𝐴 × 𝑚𝐴𝐴 = 2 × −2 = −1
∴ AD ⏊ CD
𝐴� = 90° from 5 b � = 90°
𝐷
𝐴� + 𝐷
� = 180°
ABCD is a cyclic quad (opp angles supp)
d AC is diameter of circle( angles in semi circle =90°)
4+4 12+2
Midpoint of AC= � 2 ; 2
� = (4; 7)
Radius = 12 − 7 = 5
(𝑥 − 4)2 + (𝑦 − 7)2 = 25
2+4 3+2 5
6 a M� 2 ; 2 � = �3; 2� (diagonals bisect each other)
b Median PM join M with point P on DC, where P is the midpoint of DC
1+4 1+2 5 3
P� 2 ; 2 � = �2 ; 2�
1 4
7 a i 𝑐𝑜𝑠𝐴 = 𝑠𝑒𝑎𝐴 = 5
ii 𝑠𝑖𝑖2𝐴 = 2𝑠𝑖𝑖𝐴𝑐𝑜𝑠𝐴
−3 4
=2× 5 ×5
−24
= 25
b
1
𝑎𝑜𝑠𝑒𝑎73° 𝑎𝑜𝑠𝑒𝑎17° 1
17° = cos 17° = �𝑘𝑘2−1 =
𝑎𝑜𝑠343° 𝑘√𝑘 2 −1
1
8 a 𝑐𝑜𝑠69°. 𝑐𝑜𝑠9° + 𝑐𝑜𝑠81°. 𝑐𝑜𝑠21° = 𝑠𝑖𝑖21°. 𝑐𝑜𝑠9° + 𝑠𝑖𝑖9°. 𝑐𝑜𝑠21°
= sin(21° + 9°)
1
= sin 30 ° = 2
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 169
Exemplar Paper 2
b i Undefined at asymptotes of tan 𝑥: 𝑥 = 90° + 𝑘. 180°; 𝑘 ∈ 𝑍
Also undefined where 𝑠𝑖𝑖𝑥 + 𝑠𝑖𝑖2𝑥 = 0
𝑠𝑖𝑖𝑥 + 2𝑠𝑖𝑖𝑥𝑐𝑜𝑠𝑥 = 0
𝑠𝑖𝑖𝑥(1 + 2𝑐𝑜𝑠𝑥) = 0
1
∴ 𝑠𝑖𝑖𝑥 = 0 or 𝑐𝑜𝑠𝑥 = − 2
𝑥 = 𝑘. 360° or 𝑥 = ±120° + 𝑘. 360°; 𝑘 ∈ 𝑍
1+𝑎𝑜𝑠𝑥+𝑎𝑜𝑠2𝑥 1+𝑎𝑜𝑠𝑥+2𝑎𝑜𝑠2 𝑥−1
ii 𝑠𝑖𝑛𝑥+𝑠𝑖𝑛2𝑥
= 𝑠𝑖𝑛𝑥+2𝑠𝑖𝑛𝑥𝑎𝑜𝑠
𝑎𝑜𝑠𝑥(1+2𝑎𝑜𝑠𝑥)
=
𝑠𝑖𝑛𝑥(1+2𝑎𝑜𝑠𝑥)
𝑎𝑜𝑠𝑥
= 𝑠𝑖𝑛𝑥
1
= 𝑡𝑎𝑛𝑥
9 a i 2𝑡𝑎𝑖𝑥 = −0,6842
𝑡𝑎𝑖𝑥 = −0,3421
𝑥 = −18,89°
ii 𝑠𝑖𝑖2𝑥. 𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑖𝑥. 𝑐𝑜𝑠2𝑥 = 0,5
sin(2𝑥 − 𝑥) = 0,5
𝑠𝑖𝑖𝑥 = 0,5
𝑥 = 60°
1
b cos �2 𝑥 + 15°� = sin(2𝑥 − 15°)
1
cos �2 𝑥 + 15°� = cos[90° −(2𝑥 − 15°)]
1
cos �2 𝑥 + 15°� = cos[90° −(2𝑥 − 15°)]
1
cos �2 𝑥 + 15°� = cos(105° − 2𝑥)
1
�2 𝑥 + 15°� = (105° − 2𝑥) + 𝑘. 360°
1
or �2 𝑥 + 15°� = −(105° − 2𝑥) + 𝑘. 360°
𝑥 = 36° + 𝑘. 144°; 𝑘 ∈ 𝑍 𝑥 = 80° − 𝑘. 240°; 𝑘 ∈ 𝑍
10 a 𝑎 = 2; 𝑏 = 2
1
b 𝑐𝑜𝑠 2 𝑥 − 𝑠𝑖𝑖𝑥 = 2
2𝑐𝑜𝑠 2 𝑥 − 2𝑠𝑖𝑖𝑥 = 1
2𝑐𝑜𝑠 2 𝑥 − 1 = 2𝑠𝑖𝑖𝑥
∴ It is where the two graphs meet.
𝑥 = 20° 𝑜𝑟 160°
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 170
Exemplar Paper 2
𝑃𝑄
11 a 𝑡𝑎𝑖𝑥 = 𝑄𝑅 ∴ 𝑃𝑄 = 𝑄𝑅𝑡𝑎𝑖𝑥
1
Area of ∆𝑄𝑅𝑆 = 2 𝑄𝑆. 𝑄𝑅𝑠𝑖𝑖𝑄𝑅� 𝑆
1
∴ 𝐴 = 2 𝑎 × 𝑄𝑅𝑠𝑖𝑖𝑦
2𝐴
𝑄𝑅 = 𝑎𝑠𝑖𝑛𝑦
2𝐴 2𝐴𝑡𝑎𝑛𝑥
𝑃𝑄 = 𝑎𝑠𝑖𝑛𝑦 𝑡𝑎𝑖𝑥 = 𝑎𝑠𝑖𝑛𝑦
2(480,9)𝑡𝑎𝑛46,5°
b 76,8 = 87,36𝑠𝑖𝑛𝑦
2(480,9)𝑡𝑎𝑛46,5°
𝑠𝑖𝑖𝑦 = 87,36(76,8)
= 0,151064
𝑦 = 8,69° 𝑜𝑟 171,31°
12 a If a line is drawn through the endpoint of a chord to form an angle which is
equal to the angle in the opposite segment, then this line is a tangent.
b 𝐴̂2 = 𝐴� = 𝑥 tan chord
𝐶̂1 = 𝐶̂2 = 𝑦 given
𝐴̂1 = 180° − (𝑥 + 𝑦) sum of angles of ∆𝐴𝐴𝐶
� = 180° − (𝑥 + 𝑦) sum of angles of ∆𝐴𝐷𝐶
𝐷
∴ 𝐴̂1 = 𝐷
�
∴ AB is a tangent to the circle
c i 𝑋�1 = 𝐴�2 angles in same segm
= 𝐴̂2 + 𝑄�3 ext angle of triangle
But
𝐴̂2 = 𝑄�1 + 𝑄�2 tan chord
∴ 𝑋�1 = 𝑄�1 + 𝑄�2 + 𝑄�3
∴ A,X,P,Q concyclic (ext angle = opp int angle)
ii 𝑄�1 = 𝐴̂1 AXPQ cyclic quad
= 𝑅� angles in same segm
∴ PQ is a tangent
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 171
Exemplar Paper 2
13 a Book work
b i In ∆ABD and ∆AEC:
𝐴̂1 = 𝐴̂2 given
𝐴� = 𝐷� angles in same segm
∴∆ABD⦀∆AEC (AAA)
ii In ∆ABD and ∆CED:
𝐴� = 𝐷� proven
�1 = 𝐷
𝐷 �2 vert opp ⊾s
∴∆ABD⦀∆CED (AAA)
𝐴𝐴 𝐴𝐴
∴𝐴𝐴 = 𝐴𝐸
∴𝐴𝐴. 𝐴𝐶 = 𝐴𝐷. 𝐴𝐷
= (𝐴𝐷 + 𝐷𝐷)𝐴𝐷
= 𝐴𝐷2 + 𝐴𝐷. 𝐷𝐷
𝐴𝐴 𝐴𝐴
But AD.DE=BD.DC (𝐴𝐸 = 𝐴𝐴 )
∴𝐴𝐴. 𝐴𝐶 = 𝐴𝐷2 + 𝐴𝐷. 𝐷𝐶
𝐴𝑅 3
14 a 𝐴𝐴
= 5 given
Let 𝐴𝑅 = 3𝑘 and 𝐴𝐴 = 5𝑘
𝐴𝑆 3
∴ 𝑆𝑃 = 2 RS//BP
b Let 𝐴𝑆 = 3𝑚 and 𝐴𝑃 = 5𝑚
but AP=PC (given)
∴ 𝐴𝑃 = 𝑃𝐶 = 5𝑚
𝐴𝑆 3𝑚 3
∴ 𝑆𝐸 = 7𝑚 = 7
𝑅𝑇 2𝑚
c 𝑇𝐸
= 5𝑚 RS//TP
2
=5
1
𝐴𝑟𝑒𝑎∆𝑇𝑃𝐸 𝑇𝐸.𝑃𝐸.𝑠𝑖𝑛𝐴𝐸̂ 𝑅
d = 21
𝐴𝑟𝑒𝑎∆𝑅𝑆𝐸 𝑅𝐸.𝑆𝐸.𝑠𝑖𝑛𝐴𝐸̂ 𝑅
2
𝑇𝐸 𝑃𝐸 5 5 25
= 𝑅𝐸 . 𝑆𝐸 = 7 . 7 = 49
_____________________________________________________________________________________
©Via Afrika >> Mathematics Grade 12 172
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