# Grade 12 Mathematics P1 Revision Guide 2026

> Grade 12 Mathematics Paper 1 study guide from the 2026, 26 pages. Read online or get the free PDF download.

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- Updated: 2026-10-22
- Subject: Mathematics
- Grade: Grade 12
- Document type: Study guide
- Year: 2026
- Paper: 1
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 26

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## About this document

Mathematics · Grade 12 · Paper 1 · Hlayiso study notes 2026. Complete revision notes for every type of P1 question: how to recognise it, the first line to write, 60 worked examples from standard to challenge level, and mark-saving tips. 26 pages. Read online or download the PDF.

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### Page 1

HLAYISO STUDY NOTES · 2026 · GRADE 12
Mathematics
Paper 1
Revision guide: every question type since 2018,
and the first line to write.
To the learner
Many of you have told us: “I should know what to do, but I don’t.” That feel-
ing is normal, and it can be fixed. In a hard question the mathematics is
usually something you already know; what is missing is the first line. For
every type of question you will learn how to recognise it, what to write
first, and how a full solution looks.
HOW TO USE IT
Cover the working, try the example
with a pencil, then read the reasons
in grey. Start with any section.
CONTENTS
IN THE EXAM
How to attack a hard paper
2
STRATEGY
1
Algebra: equations, inequalities and nature of roots
3
QUESTION 1 · 25 MARKS
2
Number patterns, sequences and series
7
QUESTIONS 2–3 · 25 MARKS
3
Functions and graphs
10
QUESTIONS 4–6 · 35 MARKS
4
Finance, growth and decay
14
QUESTION 7 · 15 MARKS
5
Differential calculus
18
QUESTIONS 8–10 · 35 MARKS
6
Probability and counting
22
LAST QUESTION · 15 MARKS
If you see … write first …
25
QUICK REFERENCE

### Page 2

BEFORE YOU START · THREE HOURS, 150 MARKS
How to attack a hard paper
Six habits that turn ``I don't know'' into marks
THE TIME RULE
1,2 minutes per mark. Twice the
time and still stuck? Write the
formula and the substitution, star
the question, move on.
Paper 1 counts 150 marks in three hours: 1,2 minutes per mark. A 4-
mark question deserves about five minutes. Keep this number in your
head; it stops one stubborn question from eating the time you need for
ten easy ones.
1 The first five minutes: read the whole paper
Mark each question with a tick (I can do it), a question mark (maybe) or a star
(difficult). You do not have to start “from 1.1 nje”: number your answers cor-
rectly and begin with the sure marks (derivatives by the rules, annuity formu-
las, asymptotes, sequence formulas, probability rules).
2 Collect method marks, even when you are stuck
SHOW THAT
You may use the given result in the
next part even if you could not
prove it.
Most memo marks are for method: the formula, the substitution, the standard
form. Write and substitute even if you cannot see the end; a blank is always
zero. A wrong answer in 6.1 used correctly in 6.2 still earns the marks for 6.2,
so always attempt the follow-up parts.
3 When you are stuck, do this in order
1 Underline the key words: tangent, converge, outstanding balance, real roots, in-
dependent, hence. Each points to a method.
2 Turn every fact into an equation: “passes through (2; 12)” is f (2) = 12;
“touches the x-axis at 2” is f (2) = 0 and f ′(2) = 0.
3 Draw something: a sketch, a number line, a time line, a tree or a Venn dia-
gram. Look at what the final answer must be.
4 “Show that”: work backwards from the given answer. A letter k: ask what
you would do if k were a number, and do exactly that.
4 Why hard papers feel hard
NAME THE DISGUISE
A parameter (k, p, t) instead of a
number; two topics joined;
working backwards. Name it and
the question becomes ordinary.
A challenge paper hides a method you already know behind a disguise. Name
the disguise (see the margin) and the question becomes ordinary.
5 The last ten minutes: check
Reject impossible answers (n /∈N, a probability above 1, a negative length, a
false root, 2x = −3). Round to two decimals, money to the cent, months in the
sensible direction. Give the form asked for: y = . . ., (x; y), an interval.
READING THE EXAMPLES
Working left of the green rule,
reasons in grey, answer
highlighted. Standard: a normal
exam question; Harder: the last
marks; Challenge: Hlayiso
challenge-paper level. Margin:
First line, Spot it, Watch out.
HLAYISO.COM
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### Page 3

1
QUESTION 1 · ABOUT 25 MARKS
Algebra
Equations, inequalities and nature of roots
IN THIS SECTION
1.1 Quadratics and hidden
quadratics
1.2 Surd equations
1.3 Exponential equations
1.4 Inequalities
1.5 Simultaneous equations
1.6 Nature of roots
1.7 Exponents and surds
1.8 Proofs, max and min, word
problems
Almost every equation in Question 1 ends as a quadratic: your job is to
find the substitution or the step that turns it into one.
1.1 Quadratic equations and hidden quadratics
Make one side zero and factorise; “two decimal places” means the formula.
The same block twice, once squared (x4 and x2, x and √x): let k = the block.
An x in a denominator: write the restriction, then multiply every term by the
LCD.
Example 1.1
STANDARD
FIRST LINE
ax2 + bx + c = 0, or ``x̸ = 0:
x4 + 18 = 11x2'', or ``Let k = . . .''.
WATCH OUT
(x + 3)(x −2) = 14 does not give
x + 3 = 14: ``each factor = 0''
needs a zero on the other side.
3x2 = 48 gives x = ±4; never divide
by x.
Solve for x:
(a) (x + 3)(x −2) = 14
(b) 2x2 = 5x + 4, correct to two
decimal places.
(a) x2 + x −20 = 0, so (x + 5)(x −4) = 0
multiply out first
x = −5 or x = 4
(b) x = 5 ±
√
(−5)2 −4(2)(−4)
2(2)
= 5 ±
√
57
4
2x2 −5x −4 = 0
x ≈3,14 or x ≈−0,64
round at the end only
Example 1.2
CHALLENGE
SPOT IT
(x2 −3x)2 −2(x2 −3x) −8 = 0:
k = x2 −3x, k = 4 or −2,
x ∈{−1; 1; 2; 4}.
x −5√x + 6 = 0: k = √x, so x = 4
or 9.
SPOT IT
``Hence'': a + 8
a = 6 gives a = 2 or
4; 2x + 23−x = 6 is the same with
a = 2x: x = 1 or 2.
Solve for x:
(a) x2 + 18
x2 = 11
(b)
x
x −1 −
2
x + 1 =
4
x2 −1
(a) x4 −11x2 + 18 = 0
x̸ = 0; multiply by x2
(x2 −2)(x2 −9) = 0:
x = ±
√
2 or x = ±3
a quadratic in x2
(b) x(x + 1) −2(x −1) = 4, x2 −x −2 = 0
x̸ = ±1; ×(x2 −1)
(x −2)(x + 1) = 0:
x = 2
x = −1 is not allowed
1.2 Surd equations
Get the root alone, square, solve, and check every answer in the original equa-
tion.
Example 1.3
STANDARD
SPOT IT
Two surds, √
x + 5 = 1 + √x:
square, the x terms cancel,
4 = 2√x, so x = 4 (check: 3 = 3).
WATCH OUT
Squaring creates false roots: the
check carries a mark. A root is
never negative, so √
2x + 9 = x −3
needs x ⩾3.
Solve for x: x −√
2x + 9 = 3.
√
2x + 9 = x −3, 2x + 9 = x2 −6x + 9
isolate, then square
x(x −8) = 0; x = 0: 0 −
√
9 = −3̸ = 3
reject x = 0
x = 8: 8 −
√
25 = 3, so x = 8
true
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### Page 4

MATHEMATICS P1 · REVISION GUIDE
1 · ALGEBRA
1.3 Exponential equations
Write every number as a prime power and equate exponents. Added terms
(2x+1 + 2x−1): take out 2x. Both 9x and 3x, or 3x and 3−x: let k = 3x.
Example 1.4
STANDARD
WATCH OUT
2x+1 + 2x̸ = 22x+1: add exponents
only when you multiply.
3 · 3x = 3x+1, not 9x. 5x = −25 has
no solution: 5x > 0.
Solve for x:
(a) 8x+1 = 42x
(b) x
3
2 = 27
(c) 2x+1 + 2x−1 = 20
(d)
3x+2 −3x = 72
(a) 23x+3 = 24x, 3x + 3 = 4x,
x = 3
8 = 23, 4 = 22
(b) x = 27
2
3 =
(
3√
27
)2
,
x = 9
both sides to the 2
3
(c) 2x
(
2 + 1
2
)
= 20, 2x = 8,
x = 3
common factor 2x
(d) 3x(9 −1) = 72, 3x = 9,
x = 2
common factor 3x
Example 1.5
CHALLENGE
SPOT IT
9x −4 · 3x+1 + 27 = 0 is
(3x)2 −12 · 3x + 27 = 0: with k = 3x,
x = 1 or x = 2.
SPOT IT
x, y natural,
(√
2
)x
+
(
3√
3
)y
= 25:
2
x
2 + 3
y
3 must be two whole
numbers; only 16 + 9 works: x = 8,
y = 6, xy = 48.
Solve for x:
(a) 52x −5x −20 = 0
(b) 2x+1 + 22−x = 9
(a) k = 5x: k2 −k −20 = 0, (k −5)(k + 4) = 0
52x = (5x)2
5x = 5,
x = 1
5x = −4 is impossible
(b) 2 · 2x + 4
2x = 9; k = 2x: 2k2 −9k + 4 = 0
22−x = 22
2x ; ×k
(2k −1)(k −4) = 0, 2x = 1
2 or 2x = 4
x = −1 or x = 2
1.4 Quadratic inequalities and ``when is it real?''
SPOT IT
x
−4
3
(x + 4)(x −3) < 0: between
> 0: outside
Example 1.6
STANDARD TO HARDER
FIRST LINE
One side zero, factorise, critical
values, sketch. ``Real'': under a
root ⩾0, denominator̸ = 0. A
fraction against 0signs of top and
bottom.
WATCH OUT
x2 ⩽9 gives −3 ⩽x ⩽3, not just
x ⩽3. A value that makes a
denominator zero is always
excluded.
(a) (x −2)(x + 3) > 6
(b) 3x −x2 ⩾−10
(c) For which x is
√
8 + 2x −x2
x + 1
real?
(d) x + 3
x −1 ⩽0
(a) x2 + x −12 > 0, (x + 4)(x −3) > 0
multiply out
x < −4 or x > 3
outside the roots
(b) x2 −3x −10 ⩽0, (x −5)(x + 2) ⩽0
×(−1): the sign turns
−2 ⩽x ⩽5
between the roots
(c) 8 + 2x −x2 ⩾0, so (x −4)(x + 2) ⩽0
under the root ⩾0
−2 ⩽x ⩽4, x̸ = −1
denominator̸ = 0
(d) −3 < x < 1 gives +
−< 0
critical values −3, 1
−3 ⩽x < 1
x = 1 never allowed
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### Page 5

MATHEMATICS P1 · REVISION GUIDE
1 · ALGEBRA
1.5 Simultaneous equations, including unknowns in
exponents
Linear and quadratic: make x or y the subject of the linear one and substitute.
Unknowns in exponents: same base, so each equation becomes linear. A log:
use the definition. Every term of degree 2: factorise.
Example 1.7
CHALLENGE
FIRST LINE
3x · 33y = 32, so x + 3y = 2.
SPOT IT
y = 2x −3, x2 + xy = 6:
3x2 −3x −6 = 0: (2; 1) or (−1; −5).
3x+y = 27, 2x−y = 1
2 : x + y = 3,
x −y = −1: x = 1, y = 2.
WATCH OUT
Give matched pairs. No calculator
logs: match the bases.
Solve for x and y:
(a) 3x · 27y = 9 and x2 + 3y2 = 4
(b) log3(x −y) = 2
and x2 −5xy + 4y2 = 0
(a) x + 3y = 2; (2 −3y)2 + 3y2 = 4
substitute x = 2 −3y
12y2 −12y = 0, so y = 0 or y = 1
the 4s cancel
(2 ; 0) or (−1 ; 1)
x = 2 −3y
(b) x −y = 32 = 9; (x −y)(x −4y) = 0
definition of log
x = 4y, 3y = 9:
y = 3 and x = 12
x −y = 9̸ = 0
1.6 Nature of roots and the discriminant
THE TABLE
∆< 0: non-real
∆= 0: real, equal, rational
∆> 0, a square: real, unequal,
rational
∆> 0, not a square: irrational
∆⩾0: real
“Nature of the roots” or “values of k”: write ∆= b2 −4ac in terms of k and ap-
ply the condition. “For all k”: ∆is a perfect square, or a square plus a positive
number. Roots given as 3±
√
k−2
4
: the part under the root is ∆.
Example 1.8
STANDARD TO HARDER
SPOT IT
2x2 −3x + 5 = 0: ∆= −31, non-real.
4x2 −12x + 9 = 0: ∆= 0, real,
equal, rational.
WATCH OUT
If b = −(k + 1), then b2 = (k + 1)2:
keep the brackets. Say ``non-real'',
not ``no roots''.
(a) For which k does x2 −(k + 1)x + 4 = 0 have equal roots? Find the root in
each case. (b) For which k does x2 + kx + k + 3 = 0 have non-real roots? (c)
Show that x2 −(p + 1)x + p −1 = 0 has real, unequal roots for all real p. (d)
One root of ax2 + bx + c = 0 is 1, the other is −4−√
16+20a
2a
. Find a, b, c.
(a) (k + 1)2 −4(1)(4) = 0:
k = 3 or k = −5
equal roots: ∆= 0
Root −b
2a = k + 1
2
: x = 2 or x = −2
(b) k2 −4(k + 3) < 0, (k −6)(k + 2) < 0
non-real: ∆< 0
−2 < k < 6
between
(c) ∆= (p + 1)2 −4(p −1) = p2 −2p + 5
brackets around b
= (p −1)2 + 4 > 0:
real and unequal for every p
a square +4
(d) −b = −4, b = 4; −4ac = 20a, c = −5
compare with the formula
a + 4 −5 = 0:
a = 1, b = 4, c = −5
x = 1 is a root
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### Page 6

MATHEMATICS P1 · REVISION GUIDE
1 · ALGEBRA
Example 1.9
CHALLENGE
SPOT IT
Integer roots from a fraction: split
it,
2k
k+1 = 2 −
2
k+1 , so k + 1 must
divide 2.
(k + 1)x2 −(3k + 1)x + 2k = 0, k̸ = −1. (a) Prove that the roots are rational
for every rational k. (b) Find all integers k for which both roots are integers.
(a) ∆= (3k + 1)2 −4(k + 1)(2k)
b = −(3k + 1), c = 2k
= k2 −2k + 1 = (k −1)2
a perfect square
∆is a perfect square and a,b,c are rational, so the roots are rational
(b) x = (3k + 1) ± (k −1)
2(k + 1)
: x =
2k
k + 1 or x = 1
√
∆= k −1
2k
k + 1 = 2 −
2
k + 1, so k + 1 ∈{−2; −1; 1; 2}
then
2
k + 1 is an integer
k ∈{−3 ; −2 ; 0 ; 1}
k = 0: roots 0; 1
1.7 Exponents and surds without a calculator
Example 1.10
STANDARD AND CHALLENGE
SPOT IT
√
48 +
√
75 −
√
27 =
4
√
3 + 5
√
3 −3
√
3 = 6
√
3.
(
√
12 +
√
3)2 = (3
√
3)2 = 27.
(a) Simplify 32025 −32023
32024 + 32022 .
(b) If x =
√
5 +
√
3, y =
√
5 −
√
3, find x
y + y
x.
(a) = 32022(33 −3)
32022(32 + 1) = 24
10 =
12
5
smallest power out
(b) xy = (
√
5)2 −(
√
3)2 = 2
difference of squares
x2 + y2 = (8 + 2
√
15) + (8 −2
√
15) = 16
the surds cancel
x
y + y
x = x2 + y2
xy
= 16
2 = 8
common denominator
1.8 Proofs, maximum and minimum, word problems
Example 1.11
HARDER AND CHALLENGE
SPOT IT
(2 + 1)(22 + 1)(24 + 1)(28 + 1):
multiply by (2 −1); it collapses to
216 −1 = 65 535.
Largest k with 2k | 20!:
10 + 5 + 2 + 1 = 18.
WATCH OUT
End a proof with a sentence.
(a) Prove that 3n+2 −3n+1 + 4 · 3n is divisible by 10 for every natural n. (b)
Simplify
(
1 + 1
3
) (
1 + 1
4
)
· · ·
(
1 + 1
n
)
and find all n ⩾3 for which it is an inte-
ger.
(a) = 3n (
32 −3 + 4
)
= 10 · 3n
take out 3n
a multiple of 10, so divisible by 10
(b) 4
3 · 5
4 · 6
5 · · · n + 1
n
= n + 1
3
it telescopes
n = 5; 8; 11; . . .
(n = 3m −1, m ⩾2)
n + 1 a multiple of 3
Example 1.12
STANDARD TO CHALLENGE
FIRST LINE
x2 −6x = (x −3)2 −9. The
maximum value is the y-value 14,
not x = 3.
WORD PROBLEMS
Perimeter 46 m, diagonal 17 m:
ℓ2 + (23 −ℓ)2 = 172: 15 m by 8 m.
30
v −
30
v+5 = 1: v = 10 km/h.
(a) Find the maximum value of 5 + 6x −x2.
(b) Find the maximum value of
12
x2 −4x + 7.
(c) Prove x2 −4xy + 5y2 ⩾0; when is it 0?
(a) = −
[
(x −3)2 −9
]
+ 5 = −(x −3)2 + 14
take out −1
maximum 14 at x = 3
−(x −3)2 ⩽0
(b) x2 −4x + 7 = (x −2)2 + 3 ⩾3:
max 4
smallest denominator
(c) = (x −2y)2 + y2 ⩾0:
0 only if x = y = 0
5y2 = 4y2 + y2
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### Page 7

2
QUESTIONS 2–3 · ABOUT 25 MARKS
Patterns
Number patterns, sequences and series
IN THIS SECTION
2.1 Quadratic patterns
2.2 Arithmetic series and sigma
2.3 Geometric series
2.4 Sum to infinity
2.5 Combined sequences and
proofs
First decide which kind of pattern you have. The test takes ten seconds
and tells you which formula to use.
TEST
KIND
GENERAL TERM
first differences constant
arithmetic
Tn = a + (n −1)d
second differences
constant
quadratic
Tn = an2 + bn + c, 2a = second
diff.
ratios T2
T1 = T3
T2 constant
geometric
Tn = arn−1
2.1 Quadratic patterns
Example 2.1
CHALLENGE
FIRST LINE
2a = second difference,
3a + b = T2 −T1, a + b + c = T1.
Unknown terms: Tn = an2 + bn + c
from the known terms. Here a, b, c
are not the first term and
differences.
SPOT IT
4; 9; 18; 31: 2a = 4, 3a + b = 5,
a + b + c = 4, so Tn = 2n2 −n + 3.
Given the quadratic pattern 2 ; p ; 16 ; q ; 54 ; . . . (a) Show that p = 6 and
q = 32 and find Tn. (b) Prove that 777 is not a term.
(a) a + b + c = 2, 9a + 3b + c = 16, 25a + 5b + c = 54
T1, T3, T5
8a + 2b = 14, 16a + 2b = 38: a = 3, b = −5, c = 4
subtract pairs
p = T2 = 6, q = T4 = 32;
Tn = 3n2 −5n + 4
(b) Tn = n(3n −5) + 4; n or 3n −5 is even
one factor is even
Tn is always even and 777 is odd:
777 is not a term
(or n ≈16,91 /∈N)
Example 2.2
STANDARD TO HARDER
FIGURES 1, 2, 3
FIRST LINE
Tn+1 −Tn = 4n −1; largest term at
n = −b
2a .
(a) 5 ; 8 ; 15 ; 26 ; . . .: between which terms is the first difference 199? (b)
Tn = −n2 + 12n −8: find the largest term and the first negative term. (c) How
many tiles in figure n (see the margin), and which figure uses 239?
(a) 3; 7; 11; . . .: 4n −1 = 199, n = 50
first differences
between T50 and T51
(b) n = −12
−2 = 6: T6 = 28
a < 0: a maximum
n2 −12n + 8 > 0: n > 6 +
√
28 ≈11,29
negative terms
T12 = −8
T11 = 3 > 0
(c) 5; 11; 19: a = 1, b = 3, c = 1
second difference 2
n2 + 3n + 1 = 239, (n + 17)(n −14) = 0
Tn = n2 + 3n + 1
figure 14
n = −17 /∈N
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### Page 8

MATHEMATICS P1 · REVISION GUIDE
2 · PATTERNS
2.2 Arithmetic series and sigma notation
Example 2.3
STANDARD
FIRST LINE
Tn = a + (n −1)d;
Sn = n
2 [2a + (n −1)d] or n
2 (a + ℓ).
Sigma: first term, last term, top −
bottom +1 terms.
SPOT IT
Three terms 2x + 1; 3x + 2; 5x −1:
T2 −T1 = T3 −T2 gives x = 4.
5 + 9 + 13 + . . . = 945:
(2n + 45)(n −21) = 0, n = 21.
3 + 8 + 13 + . . . + 248: (a) How many terms? (b) Find the sum. (c) Find the
sum of the even terms.
(a) 3 + (n −1)5 = 248,
n = 50
Tn = 248
(b) S50 = 50
2 (3 + 248) = 6 275
Sn = n
2 (a + ℓ)
(c) 8; 18; . . . ; 248: 8 + (m −1)10 = 248, m = 25
every second term
S25 = 25
2 (8 + 248) = 3 200
Example 2.4
HARDER AND CHALLENGE
SPOT IT
Saving R200, then R50 more each
month: n2 + 7n −400 > 0,
n > 16,80: the 17th month.
WATCH OUT
From n = 11 to n = 40 there are 30
terms, not 29.
(a)
40
∑
n=11
(4n −3)
(b) Write 2 + 5 + 8 + . . . + 59 in sigma notation.
(c) Solve
m
∑
k=1
(2k + 3) = 320.
(d) Find k if
k
∑
r=1
log3
(
9r
3
)
= 144.
(a) 41 + . . . + 157: 30
2 (41 + 157) = 2 970
40 −11 + 1 = 30 terms
(b) 3n −1 = 59, n = 20:
∑20
n=1(3n −1)
(c) m
2 [2(5) + (m −1)2] = 320
a = 5, d = 2
m2 + 4m −320 = 0, (m + 20)(m −16) = 0,
m = 16
(d) log3
32r
3 = 2r −1: 1 + 3 + 5 + . . .
simplify Tr first
Sk = k2 = 144,
k = 12
k = −12 not allowed
2.3 Geometric sequences and series
Example 2.5
STANDARD
FIRST LINE
Two terms given: divide to remove
a. n in an exponent: isolate the
power, then logs.
WATCH OUT
When 0 < r < 1, log r < 0: dividing
by it turns ``<'' into ``>''.
(a) Which term of 3; 6; 12; . . . is 1 536? (b) T3 = 18 and T6 = 486: find a and r.
(c) How many terms of 2 + 6 + 18 + . . . exceed 10 000?
(a) 3 · 2n−1 = 1 536, 2n−1 = 29,
n = 10
(b) ar5
ar2 = 27, r = 3, 9a = 18:
a = 2, r = 3
divide to remove a
(c) 2(3n −1)
3 −1
> 10 000, 3n > 10 001
Sn, a = 2, r = 3
n > log 10 001
log 3
≈8,38:
9 terms
round up
Example 2.6
HARDER
A ball is dropped from 8 m and each bounce rises to 75% of the previous
height. (a) How high is the 4th bounce? (b) After which bounce is the rise first
below 1 m? (c) Sipho says the ball travels more than 50 m. Is he right?
(a) 8(0,75)4 ≈2,53 m
bounce n: 8(0,75)n
(b) 8(0,75)n < 1, n > log 0,125
log 0,75 ≈7,23
log 0,75 < 0: turn
after the 8th bounce
(c) 8 + 2 ·
6
1 −0,75 = 56 m:
yes, 56 > 50
every rise is fallen again
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MATHEMATICS P1 · REVISION GUIDE
2 · PATTERNS
2.4 Sum to infinity and convergence
Example 2.7
CHALLENGE
FIRST LINE
``Converges'': −1 < r < 1,
S∞=
a
1−r . ``Differs from S∞by less
than'': S∞−Sn = arn
1−r .
SPOT IT
54 + 18 + 6 + . . . =
54
1−1
3 = 81.
0,˙2˙7 =
0,27
1−0,01 =
3
11 .
15 + 15(3x −2) + 15(3x −2)2 + . . . (a) For which x does it converge? (b)
S∞= 24. Find the smallest n for which Sn differs from S∞by less than 0,001.
(a) −1 < 3x −2 < 1:
1
3 < x < 1
r in the middle
(b)
15
1 −r = 24, r = 3
8; S∞−Sn = 24rn
Sn = 24(1 −rn)
24(0,375)n < 0,001, n > 10,28
logs; the sign turns
11 terms
2.5 Combined sequences, Sn formulas and proofs
Example 2.8
CHALLENGE
SPOT IT
Arithmetic terms that are
geometric: equal ratios. Sn given:
Tn = Sn −Sn−1. Alternating terms:
two sequences.
The 1st, 2nd and 5th terms of an arithmetic sequence (d̸ = 0) are the first three
terms of a geometric sequence. (a) Show that d = 2a. (b) The sum of the first 8
arithmetic terms is 192. Find the 6th geometric term.
(a) a + d
a
= a + 4d
a + d , (a + d)2 = a(a + 4d)
equal ratios
d2 = 2ad, d̸ = 0, so d = 2a
(b) 8
2 [2a + 7(2a)] = 64a = 192, a = 3
d = 6
3 ; 9 ; 27: T6 = 3(3)5 = 729
r = 3
Example 2.9
HARDER
(a) Sn = 3n2 + 2n. Find Tn.
(b) 1; 3; 4; 9; 7; 27; 10; 81; . . . continues in the same
way. Find the sum of the first 30 terms.
(a) 3n2 + 2n −
[
3(n −1)2 + 2(n −1)
]
= 6n −1
Tn = Sn −Sn−1
(b) 15
2 [2 + 14(3)] = 330
odd places: d = 3
3(315 −1)
3 −1
= 21 523 359; total 21 523 689
even places: r = 3
Example 2.10
STANDARD
Prove (a) Sn = n
2 [2a + (n −1)d] for an arithmetic series and (b) Sn = a(rn−1)
r−1
for a geometric series, r̸ = 1.
(a) Sn = a + (a + d) + . . . + [a + (n −1)d]
forwards
Sn = [a + (n −1)d] + . . . + (a + d) + a
backwards
2Sn = n [2a + (n −1)d]:
Sn = n
2 [2a + (n −1)d]
add: equal pairs
(b) rSn = ar + ar2 + . . . + arn
multiply by r
rSn −Sn = arn −a:
Sn = a(rn−1)
r−1
subtract; r̸ = 1
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3
QUESTIONS 4–6 · ABOUT 35 MARKS
Functions
Functions and graphs
IN THIS SECTION
3.1 The hyperbola
3.2 Parabola, tangents and lines
3.3 Exponential and log graphs
3.4 Inverses and transformations
3.5 Two graphs together
Most of the marks come from a few moves that work for every graph.
Learn the left column as a reflex.
YOU ARE ASKED FOR
WRITE FIRST
y-intercept / x-intercepts
put x = 0 / put y = 0 and solve
a point lies on the graph /
intersections
f (2) = 12 / f (x) = g(x)
average gradient from a to b /
inverse
f (b)−f (a)
b−a
/ swap x and y
3.1 The hyperbola
Split x+3
x−1 into
a
x+p + q. Asymptotes x = −p, y = q; axes of symmetry y =
±(x + p) + q. “The line does not cut the graph”: equate, standard form, ∆< 0.
Example 3.1
STANDARD
SKETCH
−3
−3
f
y = x
x
y
FIRST LINE
x + 3
x −1 = (x −1) + 4
x −1
f (x) = x + 3
x −1. (a) Show f (x) =
4
x −1 + 1. (b) Give the asymptotes, domain
and range. (c) Find the intercepts and axes of symmetry; sketch f.
(a) (x −1) + 4
x −1
= x −1
x −1 +
4
x −1 = 1 +
4
x −1
split the numerator
(b)
x = 1 and y = 1 ; domain x̸ = 1; range y̸ = 1
(c) (0; −3) and (−3; 0)
original form
y = x and y = −x + 2
through (1; 1), m = ±1
Example 3.2
CHALLENGE
For which values of m will the line y = mx NOT intersect the graph of f (x) =
x + 3
x −1?
mx2 −mx = x + 3, mx2 −(m + 1)x −3 = 0
equate; x̸ = 1
∆= (m + 1)2 + 12m < 0, m2 + 14m + 1 < 0
no real roots
−7 −4
√
3 < m < −7 + 4
√
3
roots −7 ± 4
√
3
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MATHEMATICS P1 · REVISION GUIDE
3 · FUNCTIONS
Example 3.3
HARDER
SPOT IT
x · h(x) ⩽0: zeros of each factor (0,
4), the asymptote (2), then signs
interval by interval.
The asymptotes of h(x) =
a
x+p + q meet at (2; −1) and h passes through (3; 1).
(a) Find a, p, q. (b) Give the axis of symmetry with negative gradient. (c) For
which x is x · h(x) ⩽0?
(a) h(x) =
a
x −2 −1; 1 = a
1 −1,
a = 2
p = −2, q = −1
(b) y = −(x −2) −1:
y = −x + 1
gradient −1
(c) h > 0 only for 2 < x < 4
x-intercept 4
0 ⩽x < 2 or x ⩾4
x = 2 excluded
3.2 The parabola, tangents and lines
Roots given: y = a(x −x1)(x −x2), axis halfway. “y = 8x + k is a tangent”:
f ′(x) = 8. “Maximum length MN” (vertical): top −bottom, then maximise.
g(x) = f (x + t): every x moves t left.
Example 3.4
CHALLENGE
SPOT IT
Both roots positive: the line lies
between the tangent and the line
through f's y-intercept.
f (x) = a(x + 3)(x −1) has its turning point on y = 2x −6. (a) Show a = 2. (b)
y = 8x + k is a tangent to f: find k. (c) For which k does y = 8x + k cut f twice,
both with positive x? (d) g(x) = f (x + t): for which t are both x-intercepts of g
negative? (e) Find the average gradient of f from x = 0 to x = 2.
(a) x = −3 + 1
2
= −1, y = 2(−1) −6 = −8
on the line
a(2)(−2) = −8, so a = 2
substitute (−1; −8)
(b) f ′(x) = 4x + 4 = 8, x = 1, f (1) = 0
gradient 8
0 = 8(1) + k,
k = −8
(c) 2x2 −4x −(6 + k) = 0: two points if k > −8
∆= 64 + 8k > 0
−(6 + k)
2
> 0, k < −6:
−8 < k < −6
product of the roots
(d) 1 −t < 0:
t > 1
roots −3 −t, 1 −t
(e) f (2) −f (0)
2 −0
= 10 −(−6)
2
= 8
Example 3.5
HARDER
GRAPH
A
C
D
M
N
g
f
x
y
WATCH OUT
A maximum at y = 9 gives range
y ⩽9, not y ⩾9.
f (x) = −x2 + 4x + 5 and the line g meet at A(−1; 0) and C(4; 5); MN is verti-
cal, M on f, N on g. (a) Find the turning point D and the range of f. (b) Find
g. (c) For which x is f ⩾g and f increasing? (d) Find the maximum length of
MN. (e) Find the tangent to f parallel to g.
(a) x = −4
−2 = 2, f (2) = 9:
D(2; 9); y ⩽9
x = −b
2a
(b) m = 5 −0
4 + 1 = 1:
g(x) = x + 1
through A
(c) −1 ⩽x ⩽4 and x < 2:
−1 ⩽x ⩽2
left of D
(d) MN = −x2 + 3x + 4; −2x + 3 = 0, x = 3
2
top minus bottom
MNmax = −9
4 + 9
2 + 4 =
25
4 units
(e) −2x + 4 = 1, x = 3
2, f
(3
2
)
= 35
4
gradient 1
35
4 = 3
2 + c:
y = x + 29
4
y = x + c
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MATHEMATICS P1 · REVISION GUIDE
3 · FUNCTIONS
3.3 Exponential and logarithmic graphs
y = bx+p + q has asymptote y = q. Inverse (reflection in y = x): swap x and y,
and y = logb x ⇐⇒
x = by. Domain of f −1 = range of f. A log inequality:
use the definition, then add the domain.
Example 3.6
CHALLENGE
FIRST LINE
Swap: x = 3y+1 −3, so
3y+1 = x + 3.
WATCH OUT
logb x exists only for x > 0: a log
inequality answer needs this lower
limit.
h(x) = 3x+p + q passes through the origin and (1; 6). (a) Show p = 1, q = −3.
(b) Find h−1. (c) Give the domain of h−1. (d) Solve h−1(x) ⩽1.
(a) 3p + q = 0 and 3p+1 + q = 6
both points
3 · 3p −3p = 6, 3p = 3: p = 1, q = −3
subtract
(b) y + 1 = log3(x + 3):
y = log3(x + 3) −1
definition of log
(c)
x > −3
range of h
(d) log3(x + 3) ⩽2, x + 3 ⩽9:
−3 < x ⩽6
include the domain
Example 3.7
HARDER
SKETCH
1
1
f
g
y = x
x
y
f (x) = bx passes through
(
−2; 1
16
)
. (a) Show b = 4. (b) g is the reflection of f
in y = x: give g. (c) h is f moved 1 right and 2 down: give h and its x-intercept.
(d) For which x is g(x) · h(x) ⩾0?
(a) b−2 = 1
16, b2 = 16, b = 4; (b)
g(x) = log4 x
b > 0
(c)
h(x) = 4x−1 −2 ; 22x−2 = 21,
x = 3
2
right 1, down 2
(d) g < 0 on 0 < x < 1; h < 0 for x < 3
2
from the sketch
0 < x ⩽1 or x ⩾3
2
same sign; x > 0
Example 3.8
STANDARD
Determine the point of intersection of f (x) = 2x and g(x) =
(
1
2
)x−1
.
2x = 2−(x−1) = 21−x
both with base 2
x = 1 −x, x = 1
2; y = 2
1
2 =
√
2:
(
1
2;
√
2
)
equate exponents
3.4 Inverses, transformations, domain and range
NEW FUNCTION
WHAT HAPPENS TO THE GRAPH
(x; y) BECOMES
f (x −a) /
f (x) + b
shift a right / b up
(x + a; y) /
(x; y + b)
−f (x) /
f (−x)
reflect in the x-axis / in the y-axis
(x; −y) / (−x; y)
−f (−x) /
f −1(x)
both axes (180◦about O) / in y = x
(−x; −y) / (y; x)
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MATHEMATICS P1 · REVISION GUIDE
3 · FUNCTIONS
Example 3.9
STANDARD AND HARDER
SPOT IT
Inverse of y = 2x2, x ∈R: one x
gives two y-values (x = 2: y = ±1),
so it is not a function. Restrict to
x ⩽0 or x ⩾0.
(a) g(x) = 3x −6: find g−1, where g meets g−1, and solve g−1(x) · g(x) = 0.
(b) Find f −1 for f (x) = 2x2, x ⩽0, and p−1 for p(x) = 2√x, x ⩾0.
(a) x = 3y −6:
g−1(x) = x
3 + 2
swap; y the subject
3x −6 = x:
(3; 3) ;
x = −6 or x = 2
on y = x; a factor is 0
(b) x = 2y2, y ⩽0:
f −1(x) = −
√
x
2, x ⩾0
domain of f: x ⩽0
x = 2√y:
p−1(x) = x2
4 , x ⩾0
the restriction stays
Example 3.10
HARDER
WATCH OUT
f (x −2) moves the graph right. A
hyperbola reflected in y = q equals
its reflection in x = −p.
(a) P(1; −3) lies on f (x) = x2 −4x. Give its image on f (x −2) + 1, −f (x),
f (−x), f −1, and the range of g(x) = f (x −2) + 1. (b) Describe two ways to
transform f (x) =
4
x−1 + 1 to g(x) =
4
x+1 −1.
(a)
(3; −2) ;
(1; 3) ;
(−1; −3) ;
(−3; 1)
use the table
range of g:
y ⩾−3
turning point (4; −3)
(b) f (x + 2) −2 =
4
x + 1 + 1 −2 = g(x)
a translation
2 units left and 2 down
or −f (−x) = g(x):
reflect in both axes
180◦about O
3.5 Two graphs together: inequalities from graphs
Example 3.11
CHALLENGE
GRAPH
f
g
x
y
WATCH OUT
Endpoints only for ⩽or ⩾, never at
an asymptote.
f (x) = x2 −7 and g(x) = −6
x meet at x = −3, 1 and 2 (margin). For which x is
f (x) ⩾g(x)?
x3 −7x + 6 = (x −1)(x −2)(x + 3) = 0
check the intersections
f above g: x < −3, 0 < x < 1, x > 2
asymptote at x = 0
x ⩽−3 or 0 < x ⩽1 or x ⩾2
x = 0 never included
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4
QUESTION 7 · ABOUT 15 MARKS
Finance
Finance, growth and decay
IN THIS SECTION
4.1 Growth and effective rates
4.2 Depreciation
4.3 Future value, sinking funds
4.4 Loans and balances
ALWAYS
i and n per period: 9% p.a. monthly
gives i = 0,09
12 , n in months. Keep
the whole calculator value.
Every finance question uses the formulas on the information sheet. The
skill is choosing the right one: draw a time line first.
SITUATION
FORMULA
CLUE WORDS
one amount grows or
shrinks
A = P(1 ± ni),
A = P(1 ± i)n
invest, depreciate
equal payments, value
at the end
F = x[(1+i)n−1]
i
save, sinking fund
equal payments, value
now
P =
x[1−(1+i)−n]
i
loan, bond, balance
4.1 Simple and compound growth, changing rates,
effective rates
“How long?”: n is in an exponent, so use logs. A rate change: work in stages.
Effective rate: 1 + ieff =
(
1 + inom
m
)m
.
Example 4.1
STANDARD
R12 000 for 4 years at 7,5% p.a.: (a) simple; (b) compounded monthly. (c) At
9% p.a. monthly, when does money double? (d) R5 000 grows to R8 000 in 6
years (annually): find the rate. (e) 10% p.a. monthly as an effective rate. (f)
The quarterly nominal rate equal to 9% p.a. effective.
(a) 12 000(1 + 0,075 × 4) = R15 600
A = P(1 + ni)
(b) 12 000
(
1 + 0,075
12
)48
= R16 183,19
n = 4 × 12
(c) n =
log 2
log 1,0075 = 92,77:
93 months
2P = P(1,0075)n
(d) 1 + i =
6√
1,6:
8,15% p.a.
isolate the bracket
(e)
(
1 + 0,1
12
)12
−1 = 10,47%
(f) 1,09 =
(
1 + i
4
)4
:
8,71% p.a.
i = 4
(
4√
1,09 −1
)
Example 4.2
STANDARD
Lerato has R20 000 to invest for 5 years. Option A: 9% p.a. simple interest. Op-
tion B: 8,4% p.a. compounded monthly. Which is better?
A: 20 000(1 + 0,09 × 5) = R29 000
simple
B: 20 000
(
1 + 0,084
12
)60
= R30 394,73
n = 60 months
Option B, by R1 394,73
answer in words
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MATHEMATICS P1 · REVISION GUIDE
4 · FINANCE
Example 4.3
CHALLENGE
R30 000 earns 7% p.a. compounded monthly for 2 years, then 10% p.a. com-
pounded quarterly. After how many months is it first worth more than
R50 000?
30 000
(
1 + 0,07
12
)24
= R34 494,18
stage 124 months
50 000 = 34 494,18(1,025)q, q = 15,03
stage 2quarters
16 quarters: 24 + 16 × 3 = 72 months
15 quarters: R49 957,86
4.2 Depreciation
Example 4.4
STANDARD
WATCH OUT
Depreciation uses (1 −i). Writing
(1 + i) loses all the marks.
(a) R250 000, 12% p.a. straight-line: the value after 5 years. (b) R520 000 falls
to R230 000 in 6 years, reducing balance: the rate. (c) At 15% p.a. reducing bal-
ance, after how many full years is an asset first worth less than half its cost?
(a) 250 000(1 −0,12 × 5) = R100 000
A = P(1 −ni)
(b) 1 −i =
6
√
230 000
520 000 = 0,8729,
i = 12,71%
A = P(1 −i)n
(c) 0,5 = (0,85)n, n = log 0,5
log 0,85 = 4,27
after 5 years
4.3 Future value annuities and sinking funds
Deposits, value at the end: F. Sinking fund target = future cost −trade-in. De-
posits starting now: one more payment, or ×(1 + i).
Example 4.5
STANDARD
FIRST LINE
Time line: first payment, last
payment, and when the value is
wanted. Months 1 to 120n = 120.
WATCH OUT
F is the value on the day of the last
payment. Wanted later? Grow it
further.
R1 500 at the end of every month for 10 years at 8% p.a. monthly. (a) The value
after 10 years. (b) Left 2 more years without deposits. (c) Instead the first de-
posit is now and the last one month before the end: the value after 10 years.
(a) F = 1 500
[(1 + i)120 −1
]
i
= R274 419,05
i = 0,08
12
(b) 274 419,05(1 + i)24 = R321 862,80
one amount grows
(c) 274 419,05(1 + i) = R276 248,51
one month more each
Example 4.6
CHALLENGE
A tractor costs R400 000 and is replaced in 6 years. Prices rise 7% p.a.; the old
one depreciates 18% p.a. (reducing balance) and is traded in. Monthly pay-
ments into a sinking fund at 10,2% p.a. monthly start in one month and end
after 6 years. Find the payment.
400 000(1,07)6 = R600 292,14
new price
400 000(0,82)6 = R121 602,67
trade-in
478 689,47 = x
[(1,0085)72 −1
]
0,0085
target; i = 0,102
12
x = R4 847,62
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MATHEMATICS P1 · REVISION GUIDE
4 · FINANCE
4.4 Loans, outstanding balance, extra and final
payments
Loan: P. First payment after 4 months: grow the loan 3 months first. Balance
= PV of the payments left. Lump sum, new instalment, missed payments, new
rate: find the balance then, adjust it, start again. n not whole: its integer part is
the number of full payments.
Example 4.7
STANDARD AND CHALLENGE
TIME LINE FOR (C)
0
1
2
3
4
51
· · ·
R250 000
P
x
x
WATCH OUT
First payment in month 4grow for
3 months, not 4. The P formula
puts the first payment one period
after P.
(a) R1 200 000 over 20 years, monthly from one month on, at 10,5% p.a.
monthly: find the instalment and (b) the total interest. (c) R250 000 at 15%
p.a. monthly, 48 instalments, the first 4 months after the loan: show the instal-
ment is R7 221,88.
(a) 1 200 000 = x
[
1 −(1 + i)−240]
i
, i = 0,105
12
n = 240
x = R11 980,56
(b) 240 × 11 980,56 −1 200 000 = R1 675 334,40
paid minus borrowed
(c) 250 000(1,0125)3 = R259 492,68
3 months, i = 0,0125
259 492,68 = x
[
1 −(1,0125)−48]
0,0125
:
x = R7 221,88
an ordinary loan
Example 4.8
CHALLENGE
FIRST LINE
Balance =
x[1−(1+i)−left]
i
; then
(1 + i)−n = 1 −Pi
x .
SPOT IT
Loan after a 15% deposit on
R300 000300 000 × 0,85 =
R255 000.
For the loan in Example 4.7(c): after the 20th instalment an extra R40 000 is
paid in and the instalment stays the same. (a) How many more full instal-
ments? (b) How much interest was paid up to the 20th instalment?
(a) 7 221,88
[
1 −(1,0125)−28]
0,0125
= R169 732,36
28 payments left
169 732,36 −40 000 = R129 732,36
minus the lump sum
(1,0125)−n = 1 −129 732,36 × 0,0125
7 221,88
same x, new n
n = 20,47:
20 more full instalments
then a smaller one
(b) 20 × 7 221,88 = R144 437,60 paid
debt fell R80 267,64
144 437,60 −80 267,64 = R64 169,96
paid minus reduction
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MATHEMATICS P1 · REVISION GUIDE
4 · FINANCE
Example 4.9
CHALLENGE
WATCH OUT
Full payments: round down;
months to reach a target: round up.
For the loan in Example 4.7(a): after the 60th payment the instalment rises to
R14 000. (a) How many months sooner is it paid off? (b) Find the final pay-
ment. (c) If instead the rate rose to 12% p.a. after the 60th payment, find the
new instalment.
(a) 11 980,56
[
1 −(1 + i)−180]
i
= R1 083 822,30
180 left, i = 0,105
12
1 083 822,30 = 14 000 [1 −(1 + i)−n]
i
, n = 129,86
new x
180 −130 = 50 months sooner
129 full + 1 final
(b) 1 083 822,30(1 + i)129 −14 000
[(1 + i)129 −1
]
i
balance after 129
that ×(1 + i) = R12 007,62
one more month
(c) 1 083 822,30 = x
[
1 −(1,01)−180]
0,01
i = 0,12
12
x = R13 007,69
Example 4.10
HARDER
For the loan in Example 4.7(a): after the 60th payment the owner misses the
next three payments. From month 64 a new instalment is paid so that the loan
still ends after 20 years. Find the new instalment.
Balance after 60 payments: R1 083 822,30
as in Example 4.9
1 083 822,30(1 + i)3 = R1 112 522,30
3 months unpaid
1 112 522,30 = x
[
1 −(1 + i)−177]
i
240 −63 = 177 left
x = R12 384,16
i = 0,105
12
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5
QUESTIONS 8–10 · ABOUT 35 MARKS
Calculus
Differential calculus
IN THIS SECTION
5.1 First principles and rules
5.2 Tangents
5.3 Cubic graphs
5.4 Increasing and inequalities
5.5 Optimisation
5.6 Rates and working backwards
First principles and the rules are the safest marks in the paper: learn
them until they are automatic.
5.1 First principles and the rules
First principles: f (x + h) first, simplify f (x + h) −f (x) until every term has h,
then the limit. The rules: rewrite brackets, roots and divisions as powers of x
first.
Example 5.1
STANDARD AND CHALLENGE
FIRST LINE
f ′(x) = lim
h→0
f (x + h) −f (x)
h
WATCH OUT
Write ``lim'' on every line until h
has gone. Using the rules here
scores nothing.
From first principles:
(a) f (x) = −2x2 + 3x
(b) f (x) = 3x2 + 2
x
(a) f (x + h) −f (x) = −4xh −2h2 + 3h
expand f (x + h) first
f ′(x) = lim
h→0(−4x −2h + 3) = −4x + 3
divide by h
(b)
2
x + h −2
x = 2x −2(x + h)
x(x + h)
=
−2h
x(x + h)
common denominator
f (x + h) −f (x) = 6xh + 3h2 −
2h
x(x + h)
every term has h
f ′(x) = lim
h→0
(
6x + 3h −
2
x(x + h)
)
divide by h
f ′(x) = 6x −2
x2
let h →0
Example 5.2
STANDARD AND CHALLENGE
SPOT IT
f (x) = (x + 2)(3x −1): multiply out
first, f ′(x) = 6x + 5.
WATCH OUT
1
3x2 = 1
3 x−2, not 3x−2.
(a) dy
dx: y = 4x3 −2
x2 + 5√x (b) dy
dx: y =
x3+27
x+3
(c) dy
dx: y =
(x+2)2−4
√x
(d)
Dx
[(
2√x + 1
x
)2]
(e) dy
dx: xy = 3x2 −5 (f) dy
da: y = ax3 + a
(a) y = 4x3 −2x−2 + 5x
1
2
powers first
dy
dx = 12x2 + 4x−3 + 5
2x−1
2
(b) y = (x + 3)(x2 −3x + 9)
x + 3
= x2 −3x + 9
sum of cubes
dy
dx = 2x −3
x̸ = −3
(c) y = x2 + 4x
x
1
2
= x
3
2 + 4x
1
2
divide each term
dy
dx = 3
2x
1
2 + 2x−1
2
(d) 4x + 4x−1
2 + x−2:
4 −2x−3
2 −2x−3
middle: 2 · 2x
1
2 · x−1
(e) y = 3x −5x−1:
3 + 5
x2
(f)
x3 + 1
y the subject; a varies
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MATHEMATICS P1 · REVISION GUIDE
5 · CALCULUS
5.2 Tangents
Point from f, gradient from f ′, then y −y1 = m(x −x1). Parallel to y = 9x + 2:
f ′(x) = 9. Smallest gradient: f ′′(x) = 0. Through a point not on the curve:
P(p; f (p)). Tangent to two graphs: ∆= 0 for each.
Example 5.3
STANDARD AND HARDER
FIRST LINE
m = f ′(2) and f (2); angle of
inclination: tan θ = m.
(a) Tangent to y = x3 −2x2 + 1 at x = 2, and its inclination. (b) Points on
y = x3 −3x where the tangent is parallel to y = 9x + 2. (c) Why has no tangent
to p(x) = 2x3 + x a negative gradient? (d) Tangent to f (x) = x3 −3x2 + 5x
with the smallest gradient.
(a) y(2) = 1; y′ = 3x2 −4x, m = y′(2) = 4
point (2; 1)
y = 4x −7 ; tan θ = 4,
θ = 75,96◦
(b) 3x2 −3 = 9, x = ±2:
(2; 2) and (−2; −2)
gradient 9
(c) p′(x) = 6x2 + 1 ⩾1 > 0 for all x
x2 ⩾0
(d) f ′′(x) = 6x −6 = 0, x = 1; m = 2, f (1) = 3
minimum of f ′
y = 2x + 1
Example 5.4
CHALLENGE
(a) The tangent to y = x3 −3x2 at P(p; p3 −3p2) passes through (0; 1), not
on the curve. Find P. (b) y = 2x −1 is a tangent to f (x) = x2 and to g(x) =
−x2 + bx −5, touching g at a positive x. Find b.
(a) y −(p3 −3p2) = (3p2 −6p)(x −p)
tangent in terms of p
(0; 1): 2p3 −3p2 + 1 = 0, (p −1)2(2p + 1) = 0
factor theorem
P(1; −2) or P
(
−1
2; −7
8
)
(b) x2 −(2x −1) = (x −1)2: touches f
a perfect square
x2 −(b −2)x + 4 = 0; (b −2)2 −16 = 0
equate with g; ∆= 0
contact x = b −2
2
> 0:
b = 6
5.3 Cubic graphs
Root given: factor theorem. “Touches the x-axis at −1”: f (−1) = 0 and
f ′(−1) = 0. Turning points f ′ = 0; inflection f ′′ = 0 (halfway). Graph of f ′: its
roots are the turning points of f. “f (x) = k has n roots”: slide y = k across the
sketch.
Example 5.5
STANDARD
(a) Show that x = 2 is a root of x3 −4x2 + x + 6 = 0 and factorise. (b) g(x) =
ax3 + bx2 + cx + d cuts the x-axis at −2, 1, 3 and the y-axis at 12. Find g.
(a) f (2) = 8 −16 + 2 + 6 = 0: (x −2) is a factor
factor theorem
(x −2)(x2 −2x −3) = (x −2)(x −3)(x + 1)
divide
(b) g(x) = a(x + 2)(x −1)(x −3), 6a = 12
a = 2
g(x) = 2x3 −4x2 −10x + 12
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MATHEMATICS P1 · REVISION GUIDE
5 · CALCULUS
Example 5.6
CHALLENGE
SKETCH
−1
5
−5
(3; −32)
(1; −16)
y = k
f
x
y
SPOT IT
Only one real root: y = k above the
maximum or below the minimum,
k > 0 or k < −32.
f (x) = x3 + px2 + qx −5 touches the x-axis at x = −1. (a) Show p = −3,
q = −9. (b) Find the turning points and the inflection, and sketch f. (c) For
which k has f (x) = k two distinct positive roots? (d) Where is g(x) = f (x + 2)
decreasing?
(a) f ′(−1) = 0: 3 −2p + q = 0
touching
f (−1) = 0: p −q = 6; so p = −3, q = −9
solve together
(b) f ′(x) = 3x2 −6x −9 = 3(x −3)(x + 1) = 0
(−1; 0) and (3; −32)
substitute into f
f ′′(x) = 6x −6 = 0:
(1; −16)
halfway
f (x) = (x + 1)2(x −5): intercepts −1, 5, (0; −5)
for the sketch
(c) three roots for −32 < k < 0
slide y = k
middle root positive below −5:
−32 < k < −5
k = −5: root x = 0
(d) f decreases on −1 < x < 3:
−3 < x < 1
g is f moved 2 left
Example 5.7
CHALLENGE
GRAPH OF f ′
−1
3
9
f ′
x
y
WATCH OUT
The turning point of f ′ is the
inflection of f, not a turning point
of f.
The graph of f ′ for f (x) = ax3 + bx2 + cx + d cuts the x-axis at −1 and 3 and
the y-axis at 9; f (0) = 5. (a) Find f. (b) Find the turning points. (c) Find the
inflection, the tangent there and its inclination. (d) For which k has f (x) = k
three real roots? (e) Where is f concave up?
(a) f ′(x) = m(x + 1)(x −3), −3m = 9, m = −3
f ′(0) = 9
−3x2 + 6x + 9 = 3ax2 + 2bx + c
compare
f (x) = −x3 + 3x2 + 9x + 5
d = f (0)
(b)
minimum (−1; 0), maximum (3; 32)
f ′: −to + at −1
(c) x = 1, f (1) = 16, m = f ′(1) = 12
halfway
y = 12x + 4 ; tan θ = 12,
θ = 85,24◦
(d)
0 < k < 32
between turning points
(e) f ′′(x) = −6x + 6 > 0:
x < 1
concave up: f ′′ > 0
5.4 Increasing, decreasing and inequalities with f
and f ′
Example 5.8
HARDER AND CHALLENGE
SPOT IT
Increasing: f ′ > 0. Concave down:
f ′′ < 0. A product: zeros of each
factor, then signs.
(a) f (x) = x3 −6x2 + 9x: where is f increasing, concave down, and where is
x · f ′(x) < 0? (b) f (x) = 2(x + 3)(x −1): where is f (x) · f ′(x) > 0?
(a) f ′(x) = 3(x −1)(x −3) > 0:
x < 1 or x > 3
outside the roots
f ′′(x) = 6x −12 < 0:
x < 2
concave down
x · f ′(x) < 0:
x < 0 or 1 < x < 3
zeros 0, 1, 3
(b) f < 0 on −3 < x < 1; f ′ < 0 for x < −1
f ′(x) = 4x + 4
−3 < x < −1 or x > 1
same signs
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MATHEMATICS P1 · REVISION GUIDE
5 · CALCULUS
5.5 Optimisation
Write the quantity in ONE variable (use the constraint to remove the other),
differentiate, set equal to zero.
Example 5.9
CHALLENGE
FIRST LINE
πr2h = 250π, so h = 250
r2 .
SPOT IT
Box from a 30 cm sheet, corners x:
V = x(30 −2x)2,
V′ = (30 −2x)(30 −6x) = 0, x = 5,
V = 2 000 cm3.
An open cylinder holds 250π cm3. The base costs R0,06 per cm2, the side R0,03
per cm2. (a) Show C(r) = 0,06πr2 + 15π
r . (b) Find the minimum cost.
(a) C = 0,06(πr2) + 0,03(2πrh)
one base; side 2πrh
= 0,06πr2 + 0,06πr · 250
r2 = 0,06πr2 + 15π
r
h = 250
r2
(b) C′(r) = 0,12πr −15π
r2
= 0, r3 = 125
r = 5: C(5) = 4,5π = R14,14
Example 5.10
CHALLENGE
CROSS-SECTION
r
h
4
10
A cylinder stands inside a cone (base radius 4 cm, height 10 cm), its top edge
touching the cone. (a) Show V = 10πr2 −5
2πr3. (b) Find the maximum vol-
ume.
(a) 10 −h
r
= 10
4 , h = 10 −5
2r
similar triangles
V = πr2
(
10 −5
2r
)
= 10πr2 −5
2πr3
(b) V′(r) = 20πr −15
2 πr2 = 0, r = 8
3
r̸ = 0
V = 640π
27
≈74,47 cm3
h = 10
3
5.6 Rates of change and working backwards
Example 5.11
STANDARD AND HARDER
SPOT IT
Velocity: h′(t). Initially: t = 0.
Derivative given: which function
has it? Then use the extra fact.
(a) h(t) = 30t −5t2: velocity at 2 s, maximum height, velocity on landing. (b)
Line g: g(1) = 5, g′(3) = 2. (c) f (x) = 3x2 + bx + c, f ′(1) = 9, f (1) = 0. (d)
p′(x) = 4x −8 and the minimum of p is −3.
(a) h′(t) = 30 −10t, h′(2) = 10 m/s
velocity
t = 3: h(3) = 45 m ; t = 6: h′(6) = −30 m/s
h′ = 0; h = 0
(b) g(x) = 2x + c, 5 = 2 + c:
g(x) = 2x + 3
gradient 2
(c) 6 + b = 9, 3 + 3 + c = 0:
b = 3, c = −6
f ′(x) = 6x + b
(d) p(x) = 2x2 −8x + c; p(2) = −3
minimum: p′(x) = 0
p(x) = 2x2 −8x + 5
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6
LAST QUESTION · ABOUT 15 MARKS
Probability
Probability and counting
IN THIS SECTION
6.1 The rules and tables
6.2 Venn and tree diagrams
6.3 Counting
6.4 Restrictions and other
questions
Write every rule you use: the rule itself often earns the first mark. In
counting, draw boxes for the positions before you multiply.
WORDS IN THE QUESTION
RULE
“or”
P(A or B) = P(A) + P(B) −P(A and B)
mutually exclusive /
independent
P(A and B) = 0 / P(A) × P(B)
“not”, “at least one”
P(not A) = 1 −P(A)
n different objects / p and q
alike
n! /
n!
p! q!
6.1 The rules, unknown probabilities and two-way
tables
Example 6.1
STANDARD AND CHALLENGE
TABLE FOR (C)
Pass Fail Total
Male
48
32
80
Female
72
48
120
Total
120
80
200
WATCH OUT
A probability above 1 is rejected:
say so.
(a) P(A) = 0,3, P(B) = 0,5, P(A or B) = 0,65: mutually exclusive? indepen-
dent? (b) A, B independent, P(A) = x, P(B) = x + 0,1, P(A or B) = 0,58:
find x and P(exactly one). (c) Is passing independent of gender (table in the
margin)?
(a) 0,65 = 0,3 + 0,5 −P(A and B)
addition rule
P(A and B) = 0,15̸ = 0:
not mutually exclusive
0,3 × 0,5 = 0,15:
independent
P(A) × P(B)
(b) 0,58 = x + (x + 0,1) −x(x + 0,1)
independent
x2 −1,9x + 0,48 = 0, x = 1,9 ± 1,3
2
:
x = 0,3
reject x = 1,6 > 1
0,58 −0,3 × 0,4 = 0,46
or minus P(A and B)
(c)
80
200 × 120
200 = 0,24 = 48
200:
independent
totals vs the cell
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MATHEMATICS P1 · REVISION GUIDE
6 · PROBABILITY
6.2 Venn diagrams and tree diagrams
Example 6.2
CHALLENGE
FIRST LINE
Venn: fill from the middle
outwards; a letter for the unknown
region; an equation from the total.
VENN DIAGRAM
M
S
D
31
25
4
5
10
6
9
10
100 learners: 50 in Maths (M), 40 in Science (S), 30 in Debating (D); 5 in all
three; 6 in S and D only; 10 in none. M and D are independent. (a) Show that
15 are in M and D. (b) Find M and S only. (c) Find P(exactly one club).
(a)
50
100 × 30
100 = 0,15: n = 15
independent
(b) M and D only: 10; D only: 30 −10 −5 −6 = 9
middle outwards
(35 −x) + (29 −x) + 9 + x + 10 + 6 + 5 + 10 = 100
total; x = M and S only
x = 4
(c) 31 + 25 + 9 = 65:
P = 0,65
Example 6.3
STANDARD AND HARDER
TREE DIAGRAM
R
B
R
20
56
B
R
B
6
56
5
8
3
8
4
7
3
7
5
7
2
7
SPOT IT
Tree: multiply along the branches,
add between them. Without
replacement the second-stage
numbers change.
(a) 5 red and 3 blue balls; two are drawn without replacement. Find the prob-
ability of the same colour, and of at least one blue. (b) 4 blue and n red balls;
the probability of two blue (without replacement) is 2
15. Find n.
(a) 5
8 · 4
7 + 3
8 · 2
7 = 26
56 =
13
28
RR or BB
1 −P(RR) = 1 −20
56 =
9
14
``at least one''
(b)
4
n + 4 ·
3
n + 3 = 2
15, n2 + 7n −78 = 0
one fewer each time
(n + 13)(n −6) = 0:
n = 6
6.3 Counting: letters, codes and digits
Example 6.4
HARDER
FIRST LINE
Draw a box per position. Restricted
positions first; overlapping
conditions: split into cases.
SPOT IT
LIMPOPO
7!
2! 2! = 1 260. 4-digit
PINs: 104 = 10 000; without
repetition 10 · 9 · 8 · 7 = 5 040.
4-digit numbers from 1; 2; 3; 4; 5; 6, no repetition. (a) How many? (b) How
many are odd and greater than 4 000? (c) P(1 and 2 next to each other).
(a) 6 × 5 × 4 × 3 = 360
(b) first digit 5: 1 × 4 × 3 × 2 = 24
last digit 1 or 3
first digit 4 or 6: 2 × 4 × 3 × 3 = 72;
96
last digit 1, 3 or 5
(c) 3 × 2 × (4 × 3) = 72, P = 72
360 =
1
5
``12'' as one unit
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MATHEMATICS P1 · REVISION GUIDE
6 · PROBABILITY
6.4 Arrangements with restrictions, and other
questions
Example 6.5
CHALLENGE
FIRST LINE
Together: glue into one unit, then
arrange inside. No two together:
place the others, use the gaps.
Apart: 1 −P(together).
WATCH OUT
n(S) is the count with no restriction
(here 8!).
3 teachers (one is Mr Dube) and 5 learners (one is Sipho) sit in a row of 8. (a)
How many ways with the teachers together? (b) The probability that no two
teachers are adjacent? (c) The probability that Mr Dube and Sipho are not ad-
jacent? (d) How many ways with a learner at each end?
(a) 6! × 3! = 4 320
6 units; inside 3!
(b) P = 5! × 6 × 5 × 4
8!
= 14 400
40 320 =
5
14
learners, then 6 gaps
(c) 1 −2 × 7!
8!
= 1 −1
4 =
3
4
complement
(d) 5 × 4 × 6! = 14 400
ends first
Example 6.6
HARDER AND CHALLENGE
SPOT IT
4 boys and 4 girls alternating:
2 × 4! × 4! = 1 152.
(a) PRETORIA: all arrangements, those starting with P, those with the four
vowels together. (b) Thabo (hits 0,4) and Lindiwe (0,5) shoot in turn, Thabo
first; the first hit wins. Find the probability that Thabo wins. (c) A game costs
R5; two dice are thrown and a total of 10 or more wins. 300 people play per
hour and the owner keeps at least 60% of the income. What is the largest prize
per winner?
(a) 8!
2! = 20 160 ; 7!
2! = 2 520
R twice
5!
2! × 4! = 1 440
vowels as one unit
(b) 0,4 + 0,3(0,4) + 0,32(0,4) + . . . =
0,4
1 −0,3 =
4
7
r = 0,6 × 0,5
(c) (4;6), (5;5), (6;4), (5;6), (6;5), (6;6): P = 6
36 = 1
6
list the outcomes
prizes at most R600; 1
6 × 300 = 50 winners
40% of R1 500
R12 per winner
600 ÷ 50
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QUICK REFERENCE
If you see … write first …
The night before: cover the right-hand column and say the first line aloud
IF YOU SEE …
WRITE FIRST …
1 ALGEBRA
a quadratic that is not = 0 / “two decimal
places”
standard form ax2 + bx + c = 0 / the formula, with
brackets
x4 and x2; x and √x; a repeated block
“Let k = . . .”; then go back to x
x in a denominator / a surd equation
restriction, multiply every term by the LCD / isolate,
square, check
2x+1 + 2x−1 / 9x and 3x; 2x and 2−x
2x(2 + 1
2) / “Let k = 3x” (reject k ⩽0)
x and y in exponents / log3(x −y) = 2
same base ⇒linear equations / x −y = 32
linear + quadratic simultaneous
make x or y the subject of the linear one
inequality with x2 / “when is
√
A
B real”
one side 0, factorise, critical values, sketch / A ⩾0,
B̸ = 0
nature of roots, values of k
∆= b2 −4ac = . . .
“rational for all k” / “real for all p”
∆= (. . .)2 / ∆= (. . .)2+ positive
large powers; “divisible by” / √a ±
√
b
take out the smallest power / xy = a −b and x2 + y2
maximum / minimum of an expression
complete the square
2 SEQUENCES AND SERIES
second differences constant
2a = . . ., 3a + b = . . ., a + b + c = . . .
unknown terms p, q / “is not a term”
Tn = an2 + bn + c from the known terms / n /∈N, or
odd/even
“first difference is 199” / “largest term”
Tn+1 −Tn = . . . / n = −b
2a
“how many terms” / a sum is given
Tn = a + (n −1)d or arn−1 / Sn formula = the sum
∑notation / a log inside the ∑
first term, last term, number of terms / simplify Tr first
n in an exponent
n = log(...)
log r
(watch the sign when r < 1)
“converges” / “sum to infinity” / S∞−Sn
−1 < r < 1 / S∞=
a
1−r /
arn
1−r
Sn as a formula / arithmetic terms that are
geometric
Tn = Sn −Sn−1 /
Tb
Ta = Tc
Tb
3 FUNCTIONS
3x+1
x−1
/ axes of symmetry of a hyperbola
3(x−1)+4
x−1
= 3 +
4
x−1 / y = ±(x + p) + q
line does not cut / touches a graph
equate, standard form, ∆< 0 / ∆= 0
parabola with given roots / its turning point
y = a(x −x1)(x −x2) / x = −b
2a or f ′(x) = 0
“tangent with gradient m” / “maximum length
MN”
f ′(x) = m / top −bottom, then maximise
g(x) = f (x + t) / “passes through (2; 12)”
every x moves t left / f (2) = 12
inverse / domain of f −1
swap x and y; y = logb x ⇐⇒x = by / range of f
f (x) ⩾g(x) / x · f (x) ⩽0, f (x) · g(x) > 0
intersections, read the graph / zeros of each factor, signs
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MATHEMATICS P1 · REVISION GUIDE
QUICK REFERENCE
IF YOU SEE …
WRITE FIRST …
average gradient
f (b)−f (a)
b−a
4 FINANCE
any finance question / depreciation
a time line; i and n per period / (1 −i), never (1 + i)
“effective rate” / the rate changes
1 + ieff =
(
1 + i
m
)m
/ stage by stage
saving towards a target, sinking fund
F = x[(1+i)n−1]
i
; target = cost −trade-in
loan, instalment / first payment after k months
P =
x[1−(1+i)−n]
i
/ grow the loan k −1 months first
outstanding balance
PV of the payments that are left
lump sum, bigger instalment, missed payments
new balance, then solve for n (logs) or the new x
5 CALCULUS
“first principles”
f (x + h) first, then limh→0
f (x+h)−f (x)
h
brackets, roots or division
rewrite as powers of x, then differentiate
tangent at a point / through a point not on the
curve
m = f ′(x1), y −y1 = m(x −x1) / P(p; f (p)), tangent in p
“touches the x-axis at a”
f (a) = 0 and f ′(a) = 0
turning points / inflection, concavity
f ′(x) = 0 / f ′′(x) = 0; sign of f ′′
graph of f ′ given / f (x) = k has n roots
roots of f ′ = turning points of f / slide y = k across the
sketch
optimisation / rate, velocity
constraint ⇒one variable, derivative = 0 / the
derivative; “initially” t = 0
6 PROBABILITY AND COUNTING
“or” / “independent”
P(A) + P(B) −P(A and B) / P(A and B) = P(A) × P(B)
“at least one” / three overlapping groups
1 −P(none) / Venn diagram from the middle outwards
one event after another
tree diagram; without replacement changes the numbers
repeated letters / digit conditions
n!
p! q! / restricted positions first; split into cases
“together” / “no two next to each other”
glue into one unit, × arrangements inside / place the
others, use the gaps
You have done the work. Read each question slowly, write the first line, and trust your
method. Good luck.
HLAYISO.COM
26

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