# Grade 12 Mathematics P1 Cheat Sheet and Question Starter Notes 2026

> Grade 12 Mathematics Paper 1 study guide from the 2026, 60 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: 60

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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, 116 worked examples from standard to challenge level, and mark-saving tips. 60 pages. Read online or download the PDF.

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### Page 1

Grade 12 Mathematics
Paper 1 — Revision Notes
Hlayiso Study Notes 2026
To the learner. Many of you have told us: “I should know what to do, but I don’t.” That
feeling 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. These notes go through
every type of question that has appeared in Paper 1 since 2018. For each type you
will learn how to recognise it, what to write first, and how a full solution looks. Work
through the examples with a pencil, cover the solution, and try them yourself before
you read them. Start with any section — you do not have to begin at 1.1.
Contents
How to attack a hard paper
2
1
Algebra: equations, inequalities and nature of roots
4
2
Number patterns, sequences and series
20
3
Functions and graphs
28
4
Finance, growth and decay
37
5
Differential calculus
44
6
Probability and counting
53
If you see … write first …
59
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Hlayiso Study Notes 2026

### Page 2

Grade 12 Mathematics P1 — Revision Notes
How to attack a hard paper
How to attack a hard paper
Paper 1 counts 150 marks and you have three hours: that is 1,2 minutes per mark. A 4-mark
question deserves about five minutes. Keep this number in your head; it will stop one stubborn
question from eating the time you need for ten easy ones.
1. The first five minutes: read the whole paper
Page through every question before you write anything. Next to each question put a tick (I can do
it), a question mark (maybe) or a star (difficult). You are not obliged to start at 1.1. If you begin
“from 1.1 nje” and get stuck there, you lose both time and confidence. Number your answers
correctly and you may answer the questions in any order. Start with the marks you are sure of:
derivatives by the rules, annuity formulas, “write down the asymptotes”, arithmetic and geometric
formulas, probability rules.
2. Collect method marks, even when you are stuck
A memorandum gives most of its marks for method: the correct formula, the correct substitution,
the standard form, and finally the answer. So write the formula and substitute into it, even if you
cannot see the end. That alone is often one or two marks out of four. A blank page is always zero.
3. Consistent accuracy works for you
If your answer to 6.1 is wrong but you use it correctly in 6.2, the marker gives you the marks for
6.2. In a “show that” question you may use the given result in the next part even if you could not
prove it. So always attempt the follow-up questions.
4. When you are stuck, do this in order
1. Underline the key words: tangent, converge, outstanding balance, real roots, independent, hence.
Each one points to a method. The checklist at the back of these notes lists them.
2. Turn every piece of information into an equation. “Passes through (2; 12)” means f (2) = 12.
“Touches the x-axis at x = 2” means f (2) = 0 and f ′(2) = 0.
3. Draw something: a rough sketch, a number line, a time line for finance, a tree or Venn diagram.
4. Look at what the final answer must be: values of x, coordinates, a number of months?
5. In a “show that” question, work backwards from the given answer and ask what the line before
it must be.
6. Ask yourself: “If the letter k were a number, what would I do?” Then do exactly that, carrying k
along.
7. Still stuck after twice the time? Write the formula and substitution, star the question and move
on.
5. Why hard papers feel hard
A challenge paper hides a method you know behind three disguises: a parameter (k, p, t) instead
of a number, two topics joined together (a sequence inside a logarithm, a tangent inside a parabola
question), and working backwards (you are given the answer and must find the input). Name the
disguise and the question becomes ordinary.
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### Page 3

Grade 12 Mathematics P1 — Revision Notes
How to attack a hard paper
6. The last ten minutes: check
– Did you reject impossible answers? (n /∈N, a probability bigger than 1, a negative length, a
root that does not satisfy a surd equation, 2x = −3.)
– Did you round to two decimal places unless told otherwise, money to the cent, and a “number
of months” to a whole number in the sensible direction?
– Is the answer in the form asked for: y = . . ., coordinates (x; y), an inequality or interval?
How the examples are set out
Each worked example shows the working on the left and the reason for each line on the right,
in small italics. Read the reasons: they are what you say to yourself while you work. The final
answer is framed. Examples are graded: the first ones in each subsection are standard examination
questions, the later ones are at the level of the Hlayiso challenge paper.
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### Page 4

Grade 12 Mathematics P1 — Revision Notes
1 Algebra: equations, inequalities and nature of roots
1 Algebra: equations, inequalities and nature of roots
Question 1 is worth about 25 marks. It usually opens with ordinary quadratics and builds up to
equations in disguise. Almost every equation in this section ends as a quadratic: your job is to
find the substitution or the step that turns it into one.
1.1 Quadratic equations: factorising and the formula
Recognise it.
– The equation has an x2 term and no higher power: make one side zero and factorise.
– The question says “correct to two decimal places”: it will not factorise, so use the formula.
– A product equals a number that is not zero, such as (x + 3)(x −2) = 14: multiply out first.
Method.
1. Bring every term to one side so that the other side is 0 (standard form ax2 + bx + c = 0).
First line: ax2 + bx + c = 0
2. Factorise, or, if it does not factorise, write the formula from the information sheet and substi-
tute.
3. Set each factor equal to zero. Round only at the very end.
Example 1.1 (standard)
Solve for x:
(a) x(2x −5) = 0
(b) 3x2 = 48
(c) (x + 3)(x −2) = 14
WORKING
REASON
(a) x = 0 or 2x −5 = 0
already factorised: each factor = 0
x = 0 or x = 5
2
(b) x2 = 16
divide by 3
x = ±4
two roots: never forget −4
(c) x2 + x −6 = 14
multiply out; the right side is not 0
x2 + x −20 = 0
standard form
(x + 5)(x −4) = 0
factorise
x = −5 or x = 4
Example 1.2 (standard)
Solve for x, correct to two decimal places: 2x2 = 5x + 4.
WORKING
REASON
2x2 −5x −4 = 0
standard form: a = 2, b = −5, c = −4
x = −b ±
√
b2 −4ac
2a
= 5 ±
√
(−5)2 −4(2)(−4)
2(2)
write the formula, then substitute with
brackets
x = 5 ±
√
57
4
x ≈3,14 or x ≈−0,64
round at the end only
Watch out: (x + 3)(x −2) = 14 does not mean x + 3 = 14 or x −2 = 14. The “each factor equals zero”
rule only works when the other side is zero.
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### Page 5

Grade 12 Mathematics P1 — Revision Notes
1 Algebra: equations, inequalities and nature of roots
Watch out: Never divide both sides by x: you lose the root x = 0. Factorise instead.
1.2 Hidden quadratics and equations with fractions
Recognise it.
– The same block appears twice, once squared: x4 and x2, (x2 −3x)2 and (x2 −3x), x and √x,
1
x and
1
√x. Substitute k for the block.
– There is an x in a denominator: write down the restriction, then multiply every term by the
lowest common denominator.
– A question says “hence”: the previous answer is the tool. Look for the same shape in the new
equation.
Method.
1. Write the restriction(s) first, for example x̸ = 0 or x̸ = ±1.
2. Multiply every term by the denominator, or let k = the repeated block.
First line: “Let k = x2 −3x” or “x̸ = 0: x4 + 18 = 11x2”
3. Solve the quadratic in k, then go back to x. The value of k is not the answer.
4. Check every answer against the restrictions.
Example 1.3 (standard — same type as a challenge question)
Solve for x: x2 + 18
x2 = 11.
WORKING
REASON
x4 + 18 = 11x2
x̸ = 0; multiply every term by x2
x4 −11x2 + 18 = 0
standard form: a quadratic in x2
(x2 −2)(x2 −9) = 0
factorise as if x2 were k
x2 = 2 or x2 = 9
x = ±
√
2 or x = ±3
four answers
Example 1.4 (standard)
Solve for x: (x2 −3x)2 −2(x2 −3x) −8 = 0.
WORKING
REASON
Let k = x2 −3x: k2 −2k −8 = 0
the block appears squared and on its own
(k −4)(k + 2) = 0
x2 −3x = 4 or x2 −3x = −2
go back to x
(x −4)(x + 1) = 0 or (x −1)(x −2) = 0
two more quadratics
x ∈{−1 ; 1 ; 2 ; 4}
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### Page 6

Grade 12 Mathematics P1 — Revision Notes
1 Algebra: equations, inequalities and nature of roots
Example 1.5 (harder)
Solve for x:
(a) x −5√x + 6 = 0
(b) 6
x −5
√x + 1 = 0
WORKING
REASON
(a) Let k = √
x, so x = k2: k2 −5k + 6 = 0
x is the square of √x
(k −2)(k −3) = 0, so √
x = 2 or √
x = 3
both are positive, so both are valid
x = 4 or x = 9
square to get x
(b) Let k =
1
√x, so 1
x = k2: 6k2 −5k + 1 = 0
x > 0; 1
x is the square of
1
√x
(2k −1)(3k −1) = 0, so
1
√x = 1
2 or 1
3
√
x = 2 or √
x = 3
⇒
x = 4 or x = 9
take reciprocals, then square
Example 1.6 (standard — rejecting a root)
Solve for x:
x
x −1 −
2
x + 1 =
4
x2 −1.
WORKING
REASON
x̸ = 1 and x̸ = −1
restrictions: x2 −1 = (x −1)(x + 1)
x(x + 1) −2(x −1) = 4
multiply every term by (x −1)(x + 1)
x2 −x −2 = 0
(x −2)(x + 1) = 0
x = 2 or x = −1
x = −1 breaks the restriction: reject
x = 2
Example 1.7 (challenge level — “hence”)
(a) Solve for a: a + 8
a = 6.
(b) Hence solve for x: 2x + 23−x = 6.
WORKING
REASON
(a) a2 −6a + 8 = 0
a̸ = 0; multiply by a
(a −2)(a −4) = 0, a = 2 or a = 4
(b) 2x + 23
2x = 6
23−x = 23 · 2−x
This is (a) with a = 2x
the “hence”: same shape
2x = 2 or 2x = 4
x = 1 or x = 2
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### Page 7

Grade 12 Mathematics P1 — Revision Notes
1 Algebra: equations, inequalities and nature of roots
Example 1.8 (challenge level)
Solve for x: (x2 + x) +
2
x2 + x = 3. Give irrational answers correct to two decimal places.
WORKING
REASON
Let y = x2 + x: y + 2
y = 3
the block x2 + x appears twice
y2 −3y + 2 = 0, so (y −1)(y −2) = 0
y̸ = 0
x2 + x = 1 or x2 + x = 2
back to x
x2 + x −1 = 0: x = −1 ±
√
5
2
does not factorise: formula
x2 + x −2 = 0: (x + 2)(x −1) = 0
x ≈0,62; x ≈−1,62; x = 1; x = −2
Watch out: After a k-substitution, always go back to x. “k = 4” scores nothing as a final answer.
Watch out: Multiply every term by the denominator, including the terms that have no fraction.
1.3 Surd equations
Recognise it.
– The unknown is under a square root, for example x −√
2x + 9 = 3. Isolate the root, square,
and check.
Method.
1. Get the square root alone on one side.
First line: √
2x + 9 = x −3
2. Square both sides. Remember (x −3)2 = x2 −6x + 9, not x2 + 9.
3. Solve the quadratic.
4. Check every answer in the original equation and write “reject” next to the false one.
Example 1.9 (standard — same type as the prediction paper)
Solve for x: x −√
2x + 9 = 3.
WORKING
REASON
√
2x + 9 = x −3
isolate the surd
2x + 9 = x2 −6x + 9
square both sides
x2 −8x = 0, so x(x −8) = 0
x = 0 or x = 8
Check x = 0: 0 −
√
9 = −3̸ = 3
reject
Check x = 8: 8 −
√
25 = 3
true
x = 8
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### Page 8

Grade 12 Mathematics P1 — Revision Notes
1 Algebra: equations, inequalities and nature of roots
Example 1.10 (challenge level — two surds)
Solve for x: √
x + 5 = 1 + √x.
WORKING
REASON
x + 5 = 1 + 2√
x + x
square both sides; (1 + √x)2 has a middle
term
4 = 2√
x, so √
x = 2
the x terms cancel; one surd is left
x = 4
square again
Check:
√
9 = 3 and 1 +
√
4 = 3
true
x = 4
Watch out: Squaring creates false roots. The check is part of the answer and usually carries a mark.
Watch out: A square root is never negative: √
2x + 9 = x −3 needs x −3 ⩾0. That is a quick way to
reject roots.
1.4 Exponential equations
Recognise it.
– The unknown is in the exponent and both sides can be written with the same base: equate the
exponents.
– Several terms with the same base but different exponents, added or subtracted, such as 2x+1 +
2x−1: take out the common factor 2x.
– Both 32x (or 9x) and 3x appear, or both 3x and 3−x: let k = 3x.
– The unknown is the base, such as x
3
2 = 27: raise both sides to the reciprocal power.
Method.
1. Write every number as a power of a prime: 8 = 23, 27 = 33, 1
9 = 3−2,
√
3 = 3
1
2 .
2. Same base on both sides: equate the exponents.
First line: 23(x+1) = 22(2x)
3. Sum of terms: factorise.
First line: 2x (
2 + 1
2
) = 20
4. A square of a power appears: substitute.
First line: Let k = 3x: k2 −12k + 27 = 0
Example 1.11 (standard)
Solve for x: (a) 8x+1 = 42x
(b) 27x = 1
9
(c) x
3
2 = 27
(d) 5x = −25
WORKING
REASON
(a) 23x+3 = 24x
8 = 23 and 4 = 22
3x + 3 = 4x,
x = 3
same base: equate exponents
(b) 33x = 3−2, 3x = −2,
x = −2
3
1
9 = 3−2
(c) x = 27
2
3 =
(
3√
27
)2
,
x = 9
raise both sides to the power 2
3
(d)
no solution
5x > 0 for every real x
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### Page 9

Grade 12 Mathematics P1 — Revision Notes
1 Algebra: equations, inequalities and nature of roots
Example 1.12 (standard — common factor)
Solve for x: (a) 2x+1 + 2x−1 = 20
(b) 3x+2 −3x = 72
WORKING
REASON
(a) 2x · 2 + 2x · 2−1 = 20
split the exponents: 2x+1 = 2x · 2
2x
(
2 + 1
2
)
= 20
common factor 2x
2x = 8,
x = 3
20 ÷ 5
2 = 8
(b) 3x(9 −1) = 72, 3x = 9,
x = 2
common factor 3x
Example 1.13 (harder — k-substitution)
Solve for x: (a) 9x −4 · 3x+1 + 27 = 0
(b) 52x −5x −20 = 0
WORKING
REASON
(a) (3x)2 −12 · 3x + 27 = 0
9x = (3x)2 and 4 · 3x+1 = 12 · 3x
Let k = 3x: k2 −12k + 27 = 0
(k −3)(k −9) = 0
3x = 3 or 3x = 9,
x = 1 or x = 2
back to x
(b) Let k = 5x: k2 −k −20 = 0, (k −5)(k + 4) = 0
5x = 5 or 5x = −4
5x = −4 has no solution
x = 1
Example 1.14 (challenge level — bx and b−x together)
Solve for x: 2x+1 + 22−x = 9.
WORKING
REASON
2 · 2x + 4
2x = 9
22−x = 22
2x
Let k = 2x: 2k + 4
k = 9
2k2 −9k + 4 = 0
multiply by k (k > 0)
(2k −1)(k −4) = 0
2x = 1
2 or 2x = 4
x = −1 or x = 2
Watch out: 2x+1 + 2x is not 22x+1. You may only add exponents when powers are multiplied.
Watch out: 3 · 3x = 3x+1, not 9x.
1.5 Simultaneous equations, including unknowns in the exponents
Recognise it.
– One linear equation and one quadratic: make x or y the subject of the linear one and substitute.
– Both x and y are in exponents, such as 3x · 27y = 9 or 3x+y = 27: write each side with the same
base. Each exponential equation becomes a linear equation in x and y.
– A logarithm such as log3(x −y) = 2: use the definition to turn it into x −y = 32.
– A quadratic in which every term has degree 2, such as x2 −5xy + 4y2 = 0: try to factorise it
into two linear factors.
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### Page 10

Grade 12 Mathematics P1 — Revision Notes
1 Algebra: equations, inequalities and nature of roots
Method.
1. Turn every exponential or logarithmic equation into a linear one.
First line: 3x · 33y = 32 so x + 3y = 2
2. Make one letter the subject of the linear equation.
First line: x = 2 −3y
3. Substitute into the other equation, simplify to standard form and solve.
4. Find the other letter for each value, and write the answers as matched pairs.
Example 1.15 (standard)
Solve for x and y: y = 2x −3 and x2 + xy = 6.
WORKING
REASON
x2 + x(2x −3) = 6
y is already the subject: substitute
3x2 −3x −6 = 0, so x2 −x −2 = 0
divide by 3
(x −2)(x + 1) = 0
x = 2, y = 1 or x = −1, y = −5
use the linear equation for y
(2 ; 1) or (−1 ; −5)
matched pairs
Example 1.16 (standard — both unknowns in the exponents)
Solve for x and y: 3x+y = 27 and 2x−y = 1
2.
WORKING
REASON
3x+y = 33, so x + y = 3
same base: equate exponents
2x−y = 2−1, so x −y = −1
2x = 2
add the two linear equations
x = 1 and y = 2
Example 1.17 (challenge level — same type as the challenge paper)
Solve for x and y: 3x · 27y = 9 and x2 + 3y2 = 4.
WORKING
REASON
3x · 33y = 32
27 = 33, 9 = 32
x + 3y = 2, so x = 2 −3y
multiply powers: add exponents
(2 −3y)2 + 3y2 = 4
substitute into the quadratic
4 −12y + 9y2 + 3y2 = 4
12y2 −12y = 0, so 12y(y −1) = 0
y = 0, x = 2 or y = 1, x = −1
x = 2 −3y
(2 ; 0) or (−1 ; 1)
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### Page 11

Grade 12 Mathematics P1 — Revision Notes
1 Algebra: equations, inequalities and nature of roots
Example 1.18 (challenge level — a logarithm and a factorising quadratic)
Solve for x and y: log3(x −y) = 2 and x2 −5xy + 4y2 = 0.
WORKING
REASON
x −y = 32 = 9
definition: logb A = c ⇐⇒A = bc
(x −y)(x −4y) = 0
factorise the second equation
x −y = 0 is impossible, so x = 4y
x −y = 9̸ = 0
4y −y = 9, y = 3, x = 12
substitute into x −y = 9
x = 12 and y = 3
Example 1.19 (challenge level — natural-number exponents)
x and y are natural numbers such that
(√
2
)x
+
(
3√
3
)y
= 25. Determine xy.
WORKING
REASON
2
x
2 + 3
y
3 = 25
write roots as fractional exponents
Both terms must be whole numbers
an irrational root cannot cancel to give 25
2
x
2 ∈{2; 4; 8; 16}, 3
y
3 ∈{3; 9}
the values less than 25
16 + 9 = 25 is the only sum that works
try the pairs
x
2 = 4, x = 8 and y
3 = 2, y = 6
xy = 48
Watch out: Give answers as pairs. Writing x = 2 or x = −1 and then y = 0 or y = 1 separately loses the
matching.
Watch out: When both unknowns are in exponents, do not use logs on a calculator. Match the bases; you
get a linear system.
1.6 Quadratic inequalities and “for which values is it real?”
Recognise it.
– There is an inequality sign and an x2 term: make one side zero, find the critical values, use a
sketch or number line.
– “For which values of x is the expression real (defined)?”: the expression under a square root
must be ⩾0, and a denominator must not be 0. If the root is in the denominator, the expression
under it must be > 0.
– A fraction compared with 0, such as x+3
x−1 ⩽0: use the signs of numerator and denominator.
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### Page 12

Grade 12 Mathematics P1 — Revision Notes
1 Algebra: equations, inequalities and nature of roots
Method.
1. Bring everything to one side. If the x2 coefficient is negative, multiply by −1 and turn the sign
around.
First line: x2 + x −12 > 0
2. Factorise and write the critical values.
3. Sketch the parabola (roots on the x-axis) or draw a number line with signs. For a > 0: “< 0”
is between the roots; “> 0” is outside the roots.
x
−4
3
(x + 4)(x −3) < 0: between
> 0: outside
Example 1.20 (standard)
Solve for x:
(a) (x −2)(x + 3) > 6
(b) 3x −x2 ⩾−10
WORKING
REASON
(a) x2 + x −6 −6 > 0, x2 + x −12 > 0
multiply out; one side must be 0
(x + 4)(x −3) > 0
critical values −4 and 3
x < −4 or x > 3
“> 0”: outside the roots
(b) −x2 + 3x + 10 ⩾0
everything to the left
x2 −3x −10 ⩽0
multiply by −1: the sign turns
(x −5)(x + 2) ⩽0, −2 ⩽x ⩽5
“⩽0”: between the roots
Example 1.21 (harder — same type as a challenge question)
For which values of x is
√
8 + 2x −x2
x + 1
a real number?
WORKING
REASON
8 + 2x −x2 ⩾0
under the root: not negative
x2 −2x −8 ⩽0, (x −4)(x + 2) ⩽0
multiply by −1, turn the sign
−2 ⩽x ⩽4
between the roots
and x + 1̸ = 0
denominator not zero
−2 ⩽x ⩽4, x̸ = −1
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### Page 13

Grade 12 Mathematics P1 — Revision Notes
1 Algebra: equations, inequalities and nature of roots
Example 1.22 (harder — a fraction and zero)
Solve for x: x + 3
x −1 ⩽0.
WORKING
REASON
Critical values: x = −3 (top = 0), x = 1 (bottom = 0)
Signs: x < −3: −
−> 0; −3 < x < 1: +
−< 0; x > 1: +
+ > 0
test each interval
−3 ⩽x < 1
x = −3 allowed; x = 1 never allowed
Watch out: x2 ⩽9 gives −3 ⩽x ⩽3, not just x ⩽3. Always sketch.
Watch out: An endpoint that makes a denominator zero is always excluded, even with ⩽or ⩾.
1.7 Nature of roots and the discriminant
Recognise it.
– “Determine the nature of the roots” or “without solving”: calculate ∆= b2 −4ac.
– “For which values of k are the roots equal / real / non-real?”: write ∆in terms of k and apply
the condition.
– “Show that the roots are real (or rational) for all values of k”: show ∆is a perfect square, or a
square plus a positive number.
– “The roots are x = 3±
√
k−2
4
”: the part under the root is the discriminant.
Discriminant
Nature of the roots
∆< 0
non-real
∆= 0
real, equal and rational
∆> 0 and a perfect square
real, unequal and rational
∆> 0 and not a perfect square
real, unequal and irrational
∆⩾0
real
Method.
1. Write the equation in standard form and read off a, b and c with their signs. Keep brackets
around expressions.
First line: ∆= b2 −4ac =
[−(k + 1)
]2 −4(1)(4)
2. Simplify ∆and apply the condition (= 0, ⩾0, < 0, …). This usually gives an equation or a
quadratic inequality in k.
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Grade 12 Mathematics P1 — Revision Notes
1 Algebra: equations, inequalities and nature of roots
Example 1.23 (standard)
Without solving, determine the nature of the roots of (a) 2x2 −3x + 5 = 0
(b) 4x2 −12x + 9 = 0.
WORKING
REASON
(a) ∆= (−3)2 −4(2)(5) = −31
∆< 0
non-real roots
(b) ∆= (−12)2 −4(4)(9) = 0
∆= 0
real, equal, rational roots
Example 1.24 (standard — equal roots)
For which values of k does x2 −(k + 1)x + 4 = 0 have equal roots? Determine the root in each
case.
WORKING
REASON
∆= (k + 1)2 −4(1)(4) = 0
equal roots: ∆= 0
(k + 1)2 = 16, k + 1 = ±4
k = 3 or k = −5
Equal root: x = −b
2a = k + 1
2
when ∆= 0 the formula gives −b
2a
k = 3: x = 2;
k = −5: x = −2
Example 1.25 (harder — a quadratic inequality in k)
For which values of k does x2 + kx + k + 3 = 0 have non-real roots?
WORKING
REASON
∆= k2 −4(1)(k + 3) < 0
non-real: ∆< 0
k2 −4k −12 < 0
(k −6)(k + 2) < 0
critical values −2 and 6
−2 < k < 6
“< 0”: between
Example 1.26 (harder — for all values)
Show that x2 −(p + 1)x + p −1 = 0 has real, unequal roots for all real values of p.
WORKING
REASON
∆=
[−(p + 1)
]2 −4(1)(p −1)
brackets around b
= p2 + 2p + 1 −4p + 4 = p2 −2p + 5
= (p −1)2 + 4
complete the square
(p −1)2 ⩾0, so ∆⩾4 > 0 for all p
a square is never negative
∆> 0 for every p, so the roots are real and unequal
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Grade 12 Mathematics P1 — Revision Notes
1 Algebra: equations, inequalities and nature of roots
Example 1.27 (challenge level — same type as the challenge paper)
Given (k + 1)x2 −(3k + 1)x + 2k = 0, k̸ = −1.
(a) Prove that the roots are rational for every rational value of k.
(b) Determine all integer values of k for which both roots are integers.
WORKING
REASON
(a) ∆= (3k + 1)2 −4(k + 1)(2k)
a = k + 1, b = −(3k + 1), c = 2k
= 9k2 + 6k + 1 −8k2 −8k
= k2 −2k + 1 = (k −1)2
a perfect square for every k
∆is a perfect square and a,b,c are rational, so the roots are rational
(b) x = (3k + 1) ± (k −1)
2(k + 1)
use
√
∆= k −1 in the formula
x =
4k
2(k + 1) =
2k
k + 1 or x =
2k + 2
2(k + 1) = 1
one root is always 1
2k
k + 1 = 2(k + 1) −2
k + 1
= 2 −
2
k + 1
split the fraction
k + 1 must divide 2: k + 1 ∈{−2; −1; 1; 2}
only then is
2
k+1 an integer
k ∈{−3 ; −2 ; 0 ; 1}
check: k = 0 gives x2 −x = 0, roots 0; 1
Example 1.28 (harder — reading the formula backwards)
One root of ax2 + bx + c = 0 is x = 1 and the other is x = −4 −√
16 + 20a
2a
. Determine a, b, c and
the other root.
WORKING
REASON
x = −b −
√
b2 −4ac
2a
compare with the formula
−b = −4, so b = 4
b2 −4ac = 16 + 20a, so −4ac = 20a, c = −5
a̸ = 0
a(1)2 + 4(1) −5 = 0, so a = 1
x = 1 is a root: substitute
a = 1, b = 4, c = −5; other root x = −5
−4−
√
36
2
= −5
Watch out: If b = −(k + 1), then b2 = (k + 1)2. Losing the brackets is the most common error in this
topic.
Watch out: “Non-real” is the correct word. Do not write “no roots” or “imaginary”.
1.8 Exponents and surds without a calculator
Recognise it.
– Huge powers such as 32025: factorise out the smallest power. Everything cancels to a small
fraction.
– x =
√
5 +
√
3 and y =
√
5 −
√
3 (conjugates): find xy and x2 + y2 first; xy is always rational.
– “Prove that … is divisible by” or “is even for all n”: factorise out the smallest power of the base.
Method.
1. Take out the smallest power as a common factor in numerator and denominator.
First line: 32022(33 −3)
32022(32 + 1)
2. With conjugates use (√a +
√
b)(√a −
√
b) = a −b and (√a ±
√
b)2 = a + b ± 2
√
ab.
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Grade 12 Mathematics P1 — Revision Notes
1 Algebra: equations, inequalities and nature of roots
3. Simplify surds with square factors:
√
48 =
√
16 · 3 = 4
√
3.
Example 1.29 (standard — same type as the prediction paper)
Without using a calculator, determine the value of 32025 −32023
32024 + 32022 .
WORKING
REASON
= 32022(33 −31)
32022(32 + 1)
smallest power 32022 out of every term
= 27 −3
9 + 1 = 24
10
12
5
Example 1.30 (standard)
Simplify without a calculator: (a)
√
48 +
√
75 −
√
27
(b)
(√
12 +
√
3
)2
WORKING
REASON
(a) 4
√
3 + 5
√
3 −3
√
3 = 6
√
3
48 = 16 · 3, 75 = 25 · 3, 27 = 9 · 3
(b)
(
2
√
3 +
√
3
)2
=
(
3
√
3
)2
= 27
simplify inside the bracket first
Example 1.31 (challenge level — same type as the challenge paper)
Given x =
√
5 +
√
3 and y =
√
5 −
√
3. Without a calculator, determine x
y + y
x.
WORKING
REASON
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
common denominator xy
8
Example 1.32 (challenge level)
A rectangular box has faces with areas 2
√
3 cm2, 6
√
2 cm2 and 2
√
6 cm2. Without a calculator,
determine its volume.
WORKING
REASON
Let the edges be a, b, c: ab = 2
√
3, bc = 6
√
2, ac = 2
√
6
name what you do not know
(abc)2 = 2
√
3 · 6
√
2 · 2
√
6 = 24
√
36 = 144
multiply the three areas: each edge appears
twice
V = abc = 12 cm3
Watch out:
√
a + b̸ = √a +
√
b and (√a +
√
b)2̸ = a + b: the middle term 2
√
ab is always there.
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Grade 12 Mathematics P1 — Revision Notes
1 Algebra: equations, inequalities and nature of roots
1.9 Proofs, products and other algebra
These appear as the last question of Question 1 and look unfamiliar. Each one still uses a single
idea.
Example 1.33 (harder — divisibility)
Prove that 3n+2 −3n+1 + 4 · 3n is divisible by 10 for every natural number n.
WORKING
REASON
= 3n (
32 −3 + 4
)
take out the smallest power 3n
= 3n(10) = 10 · 3n
10 · 3n is a multiple of 10, so the expression is divisible by 10
Example 1.34 (challenge level — telescoping product)
Simplify
(
1 + 1
3
) (
1 + 1
4
) (
1 + 1
5
) · · ·
(
1 + 1
n
)
and determine all values of n (n ⩾3) for which the
product is an integer.
WORKING
REASON
= 4
3 · 5
4 · 6
5 · · · n + 1
n
write each bracket as one fraction
= n + 1
3
each numerator cancels the next
denominator
n + 1 must be a multiple of 3
n = 5; 8; 11; . . . (n = 3m −1, m ⩾2)
Example 1.35 (challenge level — repeated difference of squares)
Calculate (2 + 1)(22 + 1)(24 + 1)(28 + 1) without a calculator.
WORKING
REASON
Multiply by (2 −1) = 1: (2 −1)(2 + 1) = 22 −1
multiplying by 1 changes nothing
(22 −1)(22 + 1) = 24 −1, (24 −1)(24 + 1) = 28 −1
each step is a difference of squares
(28 −1)(28 + 1) = 216 −1
65 535
Example 1.36 (challenge level — counting prime factors)
Determine the largest value of k such that 2k is a factor of 20! = 1 × 2 × 3 × · · · × 20.
WORKING
REASON
Multiples of 2 up to 20: 10
each gives at least one factor 2
Multiples of 4: 5; of 8: 2; of 16: 1
each gives one extra factor 2
10 + 5 + 2 + 1
k = 18
Watch out: For a proof, end with a sentence that answers the question. “= 10 · 3n” alone does not say why
it is divisible by 10.
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Grade 12 Mathematics P1 — Revision Notes
1 Algebra: equations, inequalities and nature of roots
1.10 Maximum and minimum values: completing the square
Recognise it.
– “Maximum or minimum value of the expression”, “range of y”, “show it can never be nega-
tive”: complete the square.
– The expression is a fraction with a quadratic in the denominator, such as
12
x2−4x+7: the fraction
is largest when the denominator is smallest.
Method.
1. Make the x2 coefficient 1 (take out a common factor if needed).
2. Add and subtract
(
b
2
)2
.
First line: x2 −6x + . . . = (x −3)2 −9 + . . .
3. A square is never negative: (x −3)2 ⩾0, with equality when x = 3.
Example 1.37 (standard)
Determine the maximum value of 5 + 6x −x2 and the value of x at which it occurs.
WORKING
REASON
= −(x2 −6x) + 5
take out −1
= −
[(x −3)2 −9
] + 5 = −(x −3)2 + 14
complete the square
−(x −3)2 ⩽0
a negative square is at most 0
maximum 14 when x = 3
Example 1.38 (harder)
Determine the maximum value of
12
x2 −4x + 7.
WORKING
REASON
x2 −4x + 7 = (x −2)2 + 3 ⩾3
complete the square in the denominator
The fraction is largest when the denominator is smallest
12 is positive
maximum 12
3 = 4 (at x = 2)
Example 1.39 (challenge level)
Prove that x2 −4xy + 5y2 ⩾0 for all real x and y. When is it equal to zero?
WORKING
REASON
= x2 −4xy + 4y2 + y2
split 5y2 = 4y2 + y2
= (x −2y)2 + y2
a perfect square plus a square
Both squares are ⩾0, so the sum is ⩾0
= 0 only when y = 0 and x = 0
both squares must be 0
Watch out: The maximum value is the y-value (14), not the x-value (3). Read which one is asked.
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Grade 12 Mathematics P1 — Revision Notes
1 Algebra: equations, inequalities and nature of roots
1.11 Word problems
Method.
1. Name the unknowns with letters and write down what each letter means.
2. Turn each sentence into an equation. Perimeter, area, Pythagoras and “time = distance
speed ” are the
usual ones.
3. Solve, and reject answers that make no sense in the context (negative lengths or speeds).
Example 1.40 (standard — same type as the prediction paper)
A rectangle has a perimeter of 46 m and a diagonal of 17 m. Calculate its length and breadth.
WORKING
REASON
2ℓ+ 2b = 46, so b = 23 −ℓ
perimeter
ℓ2 + b2 = 172
Pythagoras on the diagonal
ℓ2 + (23 −ℓ)2 = 289
substitute
2ℓ2 −46ℓ+ 240 = 0, ℓ2 −23ℓ+ 120 = 0
(ℓ−15)(ℓ−8) = 0
15 m by 8 m
both roots give the same rectangle
Example 1.41 (harder)
A learner cycles 30 km. If she cycled 5 km/h faster, the trip would take one hour less. Determine
her speed.
WORKING
REASON
Let her speed be v km/h: 30
v −
30
v + 5 = 1
time = distance
speed
30(v + 5) −30v = v(v + 5)
multiply by v(v + 5)
v2 + 5v −150 = 0, (v + 15)(v −10) = 0
v = 10 km/h
v = −15: a speed cannot be negative
Example 1.42 (standard — “show that”)
A closed box has a square base of side x cm and height y cm. Its total surface area is 200 cm2. Show
that x2 + 2xy −100 = 0.
WORKING
REASON
Area = 2x2 + 4xy
top and bottom: 2x2; four sides: 4xy
2x2 + 4xy = 200
x2 + 2xy −100 = 0
divide by 2: the required result
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Grade 12 Mathematics P1 — Revision Notes
2 Number patterns, sequences and series
2 Number patterns, sequences and series
Questions 2 and 3 are worth about 25 marks together. Before you do anything, decide which kind
of pattern you have:
Test
Kind
General term
first differences constant
arithmetic
Tn = a + (n −1)d
second differences constant
quadratic
Tn = an2 + bn + c, with 2a = second difference
ratios T2
T1 = T3
T2 constant
geometric
Tn = arn−1
2.1 Quadratic patterns: the general term
Recognise it.
– The second differences are constant: the pattern is quadratic.
– Some terms are unknown (p, q): use the condition “second differences are equal”, or set up
Tn = an2 + bn + c from the known terms.
– “Prove that 777 is not a term”: solve Tn = 777 and show n is not a natural number, or use a
parity (odd/even) argument.
Method.
1. Write the first and second differences under the pattern.
2. Use the three equations: 2a = second difference, 3a + b = T2 −T1, a + b + c = T1.
First line: 2a = 4, so a = 2
Example 2.1 (standard)
Given the quadratic pattern 4 ; 9 ; 18 ; 31 ; . . . Determine the general term and T20.
WORKING
REASON
First differences: 5 ; 9 ; 13; second difference: 4
write them under the terms
2a = 4, a = 2
3a + b = 5, b = −1
3a + b is the first first-difference
a + b + c = 4, c = 3
a + b + c = T1
Tn = 2n2 −n + 3; T20 = 783
check: T3 = 18 −3 + 3 = 18
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Grade 12 Mathematics P1 — Revision Notes
2 Number patterns, sequences and series
Example 2.2 (challenge level — same type as the challenge paper)
Given the quadratic pattern 2 ; p ; 16 ; q ; 54 ; . . .
(a) Show that p = 6 and q = 32.
(b) Determine Tn.
(c) Prove that 777 is not a term of the
pattern.
WORKING
REASON
(a) Tn = an2 + bn + c: a + b + c = 2; 9a + 3b + c = 16;
25a + 5b + c = 54
use the three known terms T1, T3, T5
8a + 2b = 14 and 16a + 2b = 38
subtract pairs to remove c
8a = 24: a = 3, b = −5, c = 4
p = T2 = 12 −10 + 4 = 6; q = T4 = 48 −20 + 4 = 32
the required values
(b)
Tn = 3n2 −5n + 4
from (a)
(c) Tn = n(3n −5) + 4
If n is even, n(3n −5) is even; if n is odd, 3n −5 is even
one factor is always even
So Tn is always even; 777 is odd
777 is not a term
(or: 3n2 −5n −773 = 0 gives n ≈16,91, not a natural number)
Watch out: a, b, c in Tn = an2 + bn + c are not the first term and differences. Use the three equations.
2.2 First differences, largest terms, and patterns in context
Recognise it.
– “Between which two terms is the first difference 199?”: the first difference is itself an arithmetic
sequence; find its general term.
– “Which term is the largest?” or “from which term are the terms negative?”: a quadratic Tn
with a < 0. Use the turning point or an inequality.
– A diagram of tiles, dots or matches: count the first three figures, then treat the counts as a
pattern.
Method.
1. First difference between Tn and Tn+1: Tn+1 −Tn, or write the first differences as an arithmetic
sequence.
First line: Tn+1 −Tn = 4n −1
2. Largest term: n = −b
2a, then round to the nearest natural number and check both neighbours.
Example 2.3 (standard — same type as the prediction paper)
Given the quadratic pattern 5 ; 8 ; 15 ; 26 ; . . . with Tn = 2n2 −3n + 6. Between which two
consecutive terms is the first difference equal to 199?
WORKING
REASON
First differences: 3 ; 7 ; 11 ; . . .
arithmetic: a = 3, d = 4
nth first difference: 3 + (n −1)4 = 4n −1
this is Tn+1 −Tn
4n −1 = 199, n = 50
between T50 and T51
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Grade 12 Mathematics P1 — Revision Notes
2 Number patterns, sequences and series
Example 2.4 (harder)
The pattern 3 ; 12 ; 19 ; 24 ; . . . has Tn = −n2 + 12n −8. Determine the largest term, and the first
term that is negative.
WORKING
REASON
Turning point: n = −b
2a = −12
−2 = 6
a < 0: a maximum
T6 = −36 + 72 −8 = 28, so the largest term is 28
−n2 + 12n −8 < 0, so n2 −12n + 8 > 0
negative terms
n > 6 +
√
28 ≈11,29
critical values 6 ±
√
28
T12 = −8 is the first negative term
check: T11 = 3 > 0
Example 2.5 (harder — pattern from a diagram)
Tiles are arranged as shown. Determine a formula for the number of tiles in figure n, and which
figure uses 239 tiles.
Figure 1
Figure 2
Figure 3
WORKING
REASON
Tiles: 5 ; 11 ; 19; first differences 6 ; 8; second 2
count the figures
a = 1, 3a + b = 6 so b = 3, c = 5 −4 = 1
Tn = n2 + 3n + 1
(each figure is an (n + 1) × (n + 2)
rectangle minus one tile)
n2 + 3n + 1 = 239, n2 + 3n −238 = 0
(n + 17)(n −14) = 0
n = −17 is not natural
figure 14
2.3 Arithmetic sequences and series
Recognise it.
– “How many terms?”: use Tn = a + (n −1)d with the last term.
– “Sum of the series”, or “how many terms give a sum of 945?”: use Sn = n
2 [2a + (n −1)d], or
Sn = n
2(a + ℓ) when you know the last term ℓ.
– “Sum of the even terms” (or every second term): those terms form a new arithmetic sequence.
– Three terms in terms of x that are arithmetic: T2 −T1 = T3 −T2.
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Grade 12 Mathematics P1 — Revision Notes
2 Number patterns, sequences and series
Example 2.6 (standard — same type as the prediction paper)
Given the arithmetic series 3 + 8 + 13 + . . . + 248. (a) How many terms are there? (b) Calculate
the sum. (c) Calculate the sum of the terms that are even numbers.
WORKING
REASON
(a) 3 + (n −1)5 = 248,
n = 50
Tn = a + (n −1)d with Tn = 248
(b) S50 = 50
2 (3 + 248) = 6 275
Sn = n
2 (a + ℓ)
(c) Even terms: 8 ; 18 ; 28 ; . . . ; 248
every second term: a = 8, d = 10
8 + (m −1)10 = 248, m = 25
S25 = 25
2 (8 + 248) = 3 200
Example 2.7 (harder)
(a) How many terms of 5 + 9 + 13 + . . . must be added to give 945?
(b) 2x + 1 ; 3x + 2 ; 5x −1 are the first three terms of an arithmetic sequence. Determine x.
WORKING
REASON
(a) n
2 [2(5) + (n −1)4] = 945
Sn formula with a = 5, d = 4
n(2n + 3) = 945, 2n2 + 3n −945 = 0
(2n + 45)(n −21) = 0,
n = 21
n must be natural
(b) (3x + 2) −(2x + 1) = (5x −1) −(3x + 2)
common difference
x + 1 = 2x −3,
x = 4
terms 9 ; 14 ; 19
Example 2.8 (harder — context)
Thandi saves R200 in January. Each month she saves R50 more than the month before. In which
month will her total savings first exceed R10 000?
WORKING
REASON
n
2 [2(200) + (n −1)50] > 10 000
total savings is Sn; a = 200, d = 50
n(350 + 50n) > 20 000, n2 + 7n −400 > 0
n > −7 +
√
1 649
2
≈16,80
positive critical value
the 17th month
S16 = 9 200; S17 = 10 200
2.4 Sigma notation
Recognise it.
– ∑40
n=11(4n −3): write out the first term, the last term and the number of terms (top −bottom
+1).
– “Write in sigma notation”: find Tn and the value of n for the last term.
– The unknown is the top number: write Sn in terms of it and solve.
– A log or exponent inside the sigma: simplify the general term first. It usually becomes arith-
metic.
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Grade 12 Mathematics P1 — Revision Notes
2 Number patterns, sequences and series
Example 2.9 (standard)
(a) Calculate
40
∑
n=11
(4n −3).
(b) Write 2 + 5 + 8 + . . . + 59 in sigma notation.
(c) Solve for m:
m
∑
k=1
(2k + 3) = 320.
WORKING
REASON
(a) 41 + 45 + . . . + 157; 40 −11 + 1 = 30 terms
substitute n = 11 and n = 40
S30 = 30
2 (41 + 157) = 2 970
(b) Tn = 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
Example 2.10 (challenge level — same type as the challenge paper)
Determine k if
k
∑
r=1
log3
(9r
3
)
= 144.
WORKING
REASON
log3
(32r
3
)
= log3 32r−1 = 2r −1
simplify the general term first
Series: 1 + 3 + 5 + . . .
a = 1, d = 2
k
2 [2 + (k −1)2] = k2 = 144
k = 12
k = −12 is not allowed
Watch out: The number of terms from n = 11 to n = 40 is 30, not 29.
2.5 Geometric sequences and series
Recognise it.
– Each term is a fixed multiple of the one before (“90% of the previous swing”, “doubles”):
geometric.
– “Which term is …” or “how many terms must be added to exceed …”: the unknown n is in an
exponent, so use logs.
– Two terms are given, such as T3 and T6: divide the equations to remove a.
Method.
1. Write Tn = arn−1 or Sn = a(rn−1)
r−1
with the values substituted.
2. Isolate the power, then n = log(...)
log r . If you divide by a negative log (because 0 < r < 1), the
inequality turns.
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Grade 12 Mathematics P1 — Revision Notes
2 Number patterns, sequences and series
Example 2.11 (standard)
(a) Which term of 3 ; 6 ; 12 ; . . . is 1 536?
(b) In a geometric sequence T3 = 18 and T6 = 486.
Determine a and r.
(c) How many terms of 2 + 6 + 18 + . . . must be added for the sum to exceed 10 000?
WORKING
REASON
(a) 3 · 2n−1 = 1 536, 2n−1 = 512 = 29,
n = 10
(b) ar5
ar2 = 486
18 , r3 = 27, r = 3
divide to remove a
a(3)2 = 18,
a = 2, r = 3
(c) 2(3n −1)
3 −1
> 10 000, 3n > 10 001
Sn with a = 2, r = 3
n > log 10 001
log 3
≈8,38,
9 terms
round up
Example 2.12 (harder — same type as the prediction paper)
A ball is dropped from a height of 8 m. After each bounce it rises to 75% of the previous height.
(a) How high does it rise after the 4th bounce? (b) After which bounce does it first rise less than
1 m? (c) Sipho says the ball will travel more than 50 m before it stops. Is he correct?
WORKING
REASON
(a) 8(0,75)4 ≈2,53 m
after bounce n: 8(0,75)n
(b) 8(0,75)n < 1, (0,75)n < 0,125
n > log 0,125
log 0,75 ≈7,23
log 0,75 < 0: the sign turns
after the 8th bounce
(c) Down 8 m, then up and down: 2(6 + 4,5 + . . .)
each rise is also fallen again
Total = 8 + 2 ·
6
1 −0,75 = 8 + 48 = 56 m
S∞with a = 6, r = 0,75
Yes: 56 m > 50 m
Watch out: When 0 < r < 1, log r is negative. Dividing by it turns “<” into “>”.
2.6 Sum to infinity and convergence
Recognise it.
– “Converges”, “sum to infinity”, “if the process continues forever”: geometric with −1 < r < 1
and S∞=
a
1−r.
– “For which values of x does the series converge?”: r is an expression in x; solve −1 < r < 1.
– “Smallest n so that the sum differs from S∞by less than …”: S∞−Sn = arn
1−r, then logs.
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Grade 12 Mathematics P1 — Revision Notes
2 Number patterns, sequences and series
Example 2.13 (standard)
(a) Calculate 54 + 18 + 6 + . . . (to infinity).
(b) Write 0,˙2˙7 as a fraction using a series.
WORKING
REASON
(a) r = 18
54 = 1
3, S∞=
54
1 −1
3
= 81
−1 < r < 1, so it converges
(b) 0,27 + 0,0027 + . . .: a = 0,27, r = 0,01
S∞= 0,27
0,99 =
3
11
Example 2.14 (challenge level — same type as the challenge paper)
Given 15 + 15(3x −2) + 15(3x −2)2 + . . .
(a) For which values of x does the series converge?
(b) The sum to infinity is 24. Determine the smallest number of terms that must be added so that
the sum differs from the sum to infinity by less than 0,001.
WORKING
REASON
(a) r = 3x −2: −1 < 3x −2 < 1
convergence condition
1 < 3x < 3,
1
3 < x < 1
(b)
15
1 −r = 24, 1 −r = 5
8, r = 3
8
S∞=
a
1−r
S∞−Sn = 24 −15 (1 −rn)
1 −r
= 24rn
15
1−r = 24, so Sn = 24(1 −rn)
24(0,375)n < 0,001
n >
log
(
0,001
24
)
log 0,375 ≈10,28
sign turns: log 0,375 < 0
11 terms
x itself (x = 19
24) is not needed
Watch out: Write the convergence condition with r in the middle: −1 < r < 1. Then solve both sides at
once.
2.7 Combined sequences, Sn formulas and proofs
Recognise it.
– Terms of an arithmetic sequence form a geometric sequence: write them as a, a + d, … and set
the ratios equal.
– Sn is given as a formula in n: Tn = Sn −Sn−1 (and T1 = S1).
– The terms alternate between two patterns (odd and even positions): split into two sequences.
– “Prove that Sn = . . .”: learn the two standard proofs below; they are often 4 free marks.
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Grade 12 Mathematics P1 — Revision Notes
2 Number patterns, sequences and series
Example 2.15 (challenge level — same type as the challenge paper)
The first, second and fifth 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 terms of the arithmetic sequence is 192. Deter-
mine the 6th term of the geometric sequence.
WORKING
REASON
(a) T1 = a, T2 = a + d, T5 = a + 4d
a + d
a
= a + 4d
a + d , (a + d)2 = a(a + 4d)
geometric: equal ratios
a2 + 2ad + d2 = a2 + 4ad, d2 = 2ad
d(d −2a) = 0, d̸ = 0, so d = 2a
(b) 8
2 [2a + 7(2a)] = 64a = 192, a = 3, d = 6
Geometric: 3 ; 9 ; 27 ; . . ., r = 3
T1 = 3, T2 = 9, T5 = 27
T6 = 3(3)5 = 729
Example 2.16 (harder)
(a) For a series, Sn = 3n2 + 2n. Determine Tn.
(b) The sequence 1 ; 3 ; 4 ; 9 ; 7 ; 27 ; 10 ; 81 ; . . . continues in the same way. Calculate the sum
of the first 30 terms.
WORKING
REASON
(a) Tn = Sn −Sn−1 = 3n2 + 2n −
[
3(n −1)2 + 2(n −1)
]
= 6n −1
check: T1 = 5 = S1
(b) Odd positions: 1 ; 4 ; 7 ; . . . (d = 3); even positions:
3 ; 9 ; 27 ; . . . (r = 3)
two sequences, 15 terms each
S15 = 15
2 [2 + 14(3)] = 330
arithmetic part
S15 = 3(315 −1)
3 −1
= 21 523 359
geometric part
21 523 689
add
Example 2.17 (standard — the two proofs)
Prove that (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).
WORKING
REASON
(a) Sn = a + (a + d) + . . . + [a + (n −1)d]
write the series forwards
Sn = [a + (n −1)d] + [a + (n −2)d] + . . . + a
and backwards
2Sn = [2a + (n −1)d] + . . . (n times)
add: every pair has the same sum
Sn = n
2 [2a + (n −1)d]
(b) Sn = a + ar + ar2 + . . . + arn−1
rSn = ar + ar2 + . . . + arn−1 + arn
multiply by r
rSn −Sn = arn −a
subtract: the middle terms cancel
Sn = a(rn −1)
r −1
divide by r −1̸ = 0
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Grade 12 Mathematics P1 — Revision Notes
3 Functions and graphs
3 Functions and graphs
Questions 4 to 6 carry about 35 marks. Most of the marks come from a small set of moves that
work for every graph.
You are asked for
Write first
y-intercept
put x = 0
x-intercept(s)
put y = 0 and solve
a point lies on the graph
substitute the point: f (2) = 12
points of intersection of f and g
f (x) = g(x)
average gradient from x = a to x = b
f (b) −f (a)
b −a
inverse of f
swap x and y, then make y the subject
3.1 The hyperbola y =
a
x+p + q
Recognise it.
– f (x) = 3x+1
x−1 (a fraction with x at the top and bottom): rewrite it as
a
x+p + q by splitting the
numerator.
– Asymptotes: x = −p and y = q. Domain x̸ = −p; range y̸ = q.
– Axes of symmetry pass through the point where the asymptotes meet, with gradients 1 and
−1: y = (x + p) + q and y = −(x + p) + q.
– “For which values of m (or k) does the line not cut the graph?”: equate, get a quadratic, use
∆< 0.
Method.
1. To split the fraction, write the numerator in terms of the denominator.
First line: x + 3
x −1 = (x −1) + 4
x −1
2. To find a, p, q from a graph: read p and q from the asymptotes, then substitute a point to find
a.
3. To sketch: asymptotes (dotted), intercepts, and one more point on each branch.
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Grade 12 Mathematics P1 — Revision Notes
3 Functions and graphs
Example 3.1 (standard — same type as the challenge paper)
Given f (x) = x + 3
x −1. (a) Show that f (x) =
4
x −1 + 1. (b) Write down the asymptotes, domain
and range. (c) Determine the intercepts and the axes of symmetry, and sketch f.
WORKING
REASON
(a) x + 3
x −1 = (x −1) + 4
x −1
= x −1
x −1 +
4
x −1 = 1 +
4
x −1
write x + 3 as (x −1) + 4
(b)
x = 1 and y = 1
denominator 0; the constant q
Domain x ∈R, x̸ = 1; range y ∈R, y̸ = 1
(c) f (0) = −3: (0; −3); x + 3 = 0: (−3; 0)
use the original form for intercepts
y = (x −1) + 1 = x and y = −(x −1) + 1 = −x + 2
through (1; 1), gradients ±1
−3
−3
f
x
y
Example 3.2 (challenge level — same type as the challenge paper)
For which values of m will the line y = mx NOT intersect the graph of f (x) = x + 3
x −1?
WORKING
REASON
mx = x + 3
x −1, mx2 −mx = x + 3
equate; x̸ = 1
mx2 −(m + 1)x −3 = 0
standard form
No intersection: ∆= (m + 1)2 + 12m < 0
no real roots
m2 + 14m + 1 < 0; critical values m = −7 ± 4
√
3
formula: −14±
√
192
2
−7 −4
√
3 < m < −7 + 4
√
3 (−13,93 < m < −0,07)
between the roots
Check m = 0: y = 0 cuts f at x = −3
so m = 0 is correctly excluded
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Grade 12 Mathematics P1 — Revision Notes
3 Functions and graphs
Example 3.3 (harder — same type as the prediction paper)
The asymptotes of h(x) =
a
x+p + q meet at (2; −1) and the graph passes through (3; 1). (a) De-
termine a, p and q. (b) Write down the axis of symmetry with a negative gradient. (c) For which
values of x is x · h(x) ⩽0?
WORKING
REASON
(a) p = −2, q = −1: h(x) =
a
x −2 −1
vertical asymptote x = 2
1 =
a
3 −2 −1,
a = 2
substitute (3; 1)
(b) y = −(x −2) −1,
y = −x + 1
through (2; −1), gradient −1
(c) x-intercept:
2
x −2 = 1, x = 4
critical values: 0, 2, 4
h > 0 for 2 < x < 4; h < 0 for x < 2 or x > 4
read from the sketch
x · h(x) ⩽0: one factor positive, one negative (or zero)
0 < x < 2: (+)(−); x > 4: (+)(−)
0 ⩽x < 2 or x ⩾4
x = 2 excluded; x = 0, x = 4 give 0
Example 3.4 (harder — describing a transformation)
f (x) =
4
x−1 + 1 and g(x) =
4
x+1 −1. Describe two different ways in which f can be transformed
to g.
WORKING
REASON
f (x + 2) −2 =
4
x + 1 + 1 −2 = g(x)
try a translation first
translate 2 units left and 2 units down
−f (−x) = −
(
4
−x −1 + 1
)
=
4
x + 1 −1 = g(x)
−f (−x): reflect in both axes
reflect in the y-axis, then in the x-axis
(a rotation of 180◦about O)
Watch out: For intercepts use the original form x+3
x−1 if it is easier; for asymptotes use
a
x+p + q.
Watch out: Reflecting a hyperbola in its horizontal asymptote y = q gives the same graph as reflecting it in
its vertical asymptote x = −p.
3.2 The parabola, tangents and lines
Recognise it.
– x-intercepts are given: use y = a(x −x1)(x −x2). The turning point is given: use y = a(x −
p)2 + q. Then substitute one more point for a.
– The axis of symmetry lies halfway between the x-intercepts: x = x1+x2
2
.
– “The line y = 8x + k is a tangent”: f ′(x) = 8 gives the point of contact (or equate and use
∆= 0).
– “Maximum length of MN” with MN vertical: length = top graph −bottom graph, then max-
imise.
– g(x) = f (x + t): the graph moves t units left; every x-value decreases by t.
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Grade 12 Mathematics P1 — Revision Notes
3 Functions and graphs
Example 3.5 (standard — same type as the challenge paper)
The graph of f (x) = a(x + 3)(x −1) has its turning point on the line y = 2x −6. Show that a = 2.
WORKING
REASON
Axis of symmetry: x = −3 + 1
2
= −1
halfway between the roots
y = 2(−1) −6 = −8: turning point (−1; −8)
it lies on the line
a(2)(−2) = −8, −4a = −8, a = 2
substitute into f
Example 3.6 (challenge level — same type as the challenge paper)
f (x) = 2(x + 3)(x −1) = 2x2 + 4x −6.
(a) The line y = 8x + k is a tangent to f. Determine k.
(b) For which values of k does y = 8x + k cut f at two distinct points, both with positive x-
coordinates?
(c) g(x) = f (x + t). For which values of t are both x-intercepts of g negative?
(d) Calculate the average gradient of f between x = 0 and x = 2.
WORKING
REASON
(a) f ′(x) = 4x + 4 = 8, x = 1; f (1) = 0
gradient of tangent = 8
0 = 8(1) + k,
k = −8
point (1; 0) on the line
(b) 2x2 + 4x −6 = 8x + k, 2x2 −4x −(6 + k) = 0
equate
Two points: k > −8
line above the tangent (∆= 64 + 8k > 0)
When k = −6 the roots are x = 0 and x = 2
the line passes through f’s y-intercept
(0; −6)
Both positive: product −(6 + k)
2
> 0, so k < −6
below that position
−8 < k < −6
(c) Intercepts of g: x = −3 −t and x = 1 −t
every x moves t to the left
1 −t < 0,
t > 1
the larger one must be negative
(d) f (2) −f (0)
2 −0
= 10 −(−6)
2
= 8
Example 3.7 (harder — same type as the prediction paper)
f (x) = −x2 + 4x + 5 and the line g meet at A(−1; 0) and C(4; 5). MN is a vertical line segment
with M on f and N on g, between A and C. (a) Find the turning point D and the range of f.
(b) Find the equation of g. (c) For which x is f (x) ⩾g(x) and f increasing? (d) Calculate the
maximum length of MN. (e) Determine the equation of the tangent to f parallel to g.
A
C
D
M
N
g
f
x
y
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Grade 12 Mathematics P1 — Revision Notes
3 Functions and graphs
WORKING
REASON
(a) x = −b
2a = −4
−2 = 2, f (2) = 9:
D(2; 9); range y ⩽9
(b) m =
5 −0
4 −(−1) = 1,
g(x) = x + 1
through A(−1; 0)
(c) f ⩾g for −1 ⩽x ⩽4; f increasing for x < 2
between the intersections; left of D
−1 ⩽x ⩽2
both conditions
(d) MN = (−x2 + 4x + 5) −(x + 1) = −x2 + 3x + 4
top minus bottom
d
dx(MN) = −2x + 3 = 0, x = 3
2
maximise
MNmax = −9
4 + 9
2 + 4 = 25
4 units
(e) f ′(x) = −2x + 4 = 1, x = 3
2; f
(3
2
)
= 35
4
parallel to g: gradient 1
y = x + c: 35
4 = 3
2 + c,
y = x + 29
4
Watch out: The range of a parabola with a maximum at y = 9 is y ⩽9, not y ⩾9. Look at which way it
opens.
3.3 Exponential and logarithmic graphs, and inverses
Recognise it.
– y = bx+p + q: horizontal asymptote y = q; it is increasing when b > 1 and decreasing when
0 < b < 1.
– “Reflection in the line y = x” or “inverse”: swap x and y. The inverse of y = bx is y = logb x.
– Domain of the inverse = range of f; and a point (a; b) on f gives (b; a) on f −1.
– Inequalities such as h−1(x) ⩽1: rewrite with the definition of the log, then remember the
domain.
Method.
1. Definition of a logarithm: y = logb x ⇐⇒x = by (b > 0, b̸ = 1, x > 0).
First line: x = 3y+1 −3 (swap), then 3y+1 = x + 3
2. Laws you may need: logb(AB) = logb A + logb B, logb
A
B = logb A −logb B, logb An =
n logb A, logb b = 1, logb 1 = 0.
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Grade 12 Mathematics P1 — Revision Notes
3 Functions and graphs
Example 3.8 (harder — same type as the challenge paper)
The graph of h(x) = 3x+p + q passes through the origin and through (1; 6).
(a) Show that p = 1 and q = −3. (b) Determine h−1 in the form y = . . . (c) Write down the
domain of h−1. (d) For which x is h−1(x) ⩽1?
WORKING
REASON
(a) 3p + q = 0 …(1); 3p+1 + q = 6 …(2)
substitute both points
(2)−(1): 3 · 3p −3p = 6, 3p = 3, p = 1
3p+1 = 3 · 3p
q = −31 = −3
from (1)
(b) x = 3y+1 −3, 3y+1 = x + 3
swap x and y
y + 1 = log3(x + 3),
y = log3(x + 3) −1
definition of log
(c)
x > −3
range of h is y > −3
(d) log3(x + 3) ⩽2, x + 3 ⩽9, x ⩽6
h−1 is increasing
−3 < x ⩽6
include the domain
Example 3.9 (harder — same type as the prediction paper)
f (x) = bx passes through
(−2; 1
16
)
. (a) Show that b = 4. (b) g is the reflection of f in y = x; write
down g. (c) h is f translated 1 unit right and 2 units down. Find h and its x-intercept. (d) For
which x is g(x) · h(x) ⩾0?
WORKING
REASON
(a) b−2 = 1
16, b2 = 16, b = 4
b > 0
(b)
g(x) = log4 x
inverse of 4x
(c)
h(x) = 4x−1 −2
right: x →x −1; down: −2
4x−1 = 2, 22x−2 = 21,
x = 3
2
(d) g < 0 for 0 < x < 1, g > 0 for x > 1; h < 0 for x < 3
2,
h > 0 for x > 3
2
signs from the sketch below
Same sign (or zero):
0 < x ⩽1 or x ⩾3
2
g exists only for x > 0
1
1
f
g
y = x
x
y
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Grade 12 Mathematics P1 — Revision Notes
3 Functions and graphs
Example 3.10 (standard)
Determine the point of intersection of f (x) = 2x and g(x) =
( 1
2
)x−1.
WORKING
REASON
2x = 2−(x−1) = 21−x
write both with base 2
x = 1 −x, x = 1
2; y = 2
1
2 =
√
2
equate exponents
(1
2;
√
2
)
Watch out: logb x only exists for x > 0. Every inequality answer with a log needs this lower limit.
3.4 Inverses of lines and parabolas
Recognise it.
– Inverse of a straight line: swap and make y the subject. A line and its inverse meet on y = x.
– Inverse of y = ax2: y = ±
√x
a. It is only a function if the domain of f is restricted to x ⩾0 or
x ⩽0.
– “Is f −1 a function?”: one x-value may give only one y-value (vertical line test on f −1, horizontal
line test on f).
Example 3.11 (standard)
g(x) = 3x −6. (a) Determine g−1. (b) Find the point where g and g−1 intersect. (c) Solve
g−1(x) · g(x) = 0.
WORKING
REASON
(a) x = 3y −6,
g−1(x) = x
3 + 2
swap, then make y the subject
(b) On y = x: 3x −6 = x, x = 3, (3; 3)
they meet on the mirror line
(c) g−1(x) = 0: x = −6; g(x) = 0: x = 2
a product is 0 if a factor is 0
x = −6 or x = 2
Example 3.12 (harder)
f (x) = 2x2, x ⩽0. (a) Determine f −1. (b) Explain why the inverse of y = 2x2, x ∈R, is not a
function. (c) p(x) = 2√x, x ⩾0: determine p−1.
WORKING
REASON
(a) x = 2y2, y2 = x
2, y = ±
√
x
2
swap and solve
f −1(x) = −
√
x
2, x ⩾0
y ⩽0 because the domain of f was x ⩽0
(b)
for one x (e.g. x = 2) there are two y-values, ±1: one-to-many
(c) x = 2√y,
p−1(x) = x2
4 , x ⩾0
the restriction stays
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Grade 12 Mathematics P1 — Revision Notes
3 Functions and graphs
3.5 Transformations, reflections, domain and range
New function
What happens to the graph
Point (x; y) becomes
f (x −a)
shift a units right
(x + a; y)
f (x) + b
shift b units up
(x; y + b)
−f (x)
reflect in the x-axis
(x; −y)
f (−x)
reflect in the y-axis
(−x; y)
−f (−x)
reflect in both axes (180◦about O)
(−x; −y)
f −1(x)
reflect in the line y = x
(y; x)
Example 3.13 (standard)
P(1; −3) lies on f (x) = x2 −4x. Give the image of P on each of: (a) g(x) = f (x −2) + 1 (b)
h(x) = −f (x) (c) k(x) = f (−x) (d) f −1. Also give the range of g.
WORKING
REASON
(a) (3; −2)
2 right, 1 up
(b) (1; 3)
(c) (−1; −3)
(d) (−3; 1)
use the table
Turning point of f: (2; −4), so of g: (4; −3)
f has range y ⩾−4
Range of g: y ⩾−3
Watch out: f (x −2) moves the graph to the right. The sign inside the bracket works backwards.
3.6 Two graphs together: inequalities from graphs
Recognise it.
– “For which x is f (x) ⩾g(x)?”: find the intersections; read where f is above g.
– “f (x) · g(x) < 0” or “x · f (x) ⩾0”: find where each factor is zero, then compare signs interval
by interval.
– Watch for asymptotes: a hyperbola changes sign at its vertical asymptote without touching the
axis.
Example 3.14 (challenge level — parabola and hyperbola)
The graphs of f (x) = x2 −7 and g(x) = −6
x are shown. They intersect at x = −3, x = 1 and
x = 2. For which x is f (x) ⩾g(x)?
f
g
x
y
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Grade 12 Mathematics P1 — Revision Notes
3 Functions and graphs
WORKING
REASON
x2 −7 = −6
x ⇐⇒x3 −7x + 6 = 0 ⇐⇒
(x −1)(x −2)(x + 3) = 0
check the given intersections
Left of −3: f above g; between −3 and 0: g above
g has an asymptote at x = 0
0 < x < 1: f above (g →−∞); 1 < x < 2: g above; x > 2: f
above
read interval by interval
x ⩽−3 or 0 < x ⩽1 or x ⩾2
x = 0 is never included
Watch out: When you read an inequality from a graph, include the endpoints only for ⩽or ⩾, and never at
an asymptote.
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Grade 12 Mathematics P1 — Revision Notes
4 Finance, growth and decay
4 Finance, growth and decay
Question 7 is worth about 15 marks. Every finance question is solved with the formulas on the
information sheet. The skill is choosing the right one, and the habit that makes this easy is drawing
a time line first.
Situation
Formula
Clue words
one amount grows / 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, annuity
equal payments, value now
P = x [1 −(1 + i)−n]
i
loan, bond, instalment, balance
Always write i and n for the compounding period: 9% p.a. compounded monthly gives i = 0,09
12
and n in months. Keep at least four decimals (or the whole calculator value) until the end.
4.1 Simple and compound growth, changing rates, effective rates
Recognise it.
– One sum of money, no regular payments: A = P(1 + ni) (simple) or A = P(1 + i)n (com-
pound).
– “How long?” or “after how many months?”: n is the unknown in an exponent, so use logs.
– The rate changes part of the way through: work in stages; the end value of one stage is P of the
next.
– “Effective” rate: the rate that, compounded once a year, gives the same growth:
1 + ieff =
(
1 + inom
m
)m
.
Example 4.1 (standard)
R12 000 is invested for 4 years at 7,5% p.a. (a) What is it worth with simple interest? (b) With
interest compounded monthly? (c) At 9% p.a. compounded monthly, how many months does it
take for an investment to double? (d) R5 000 grows to R8 000 in 6 years, compounded annually.
Find the rate.
WORKING
REASON
(a) A = 12 000(1 + 0,075 × 4) = R15 600
simple: A = P(1 + ni)
(b) A = 12 000
(
1 + 0,075
12
)48
= R16 183,19
n = 4 × 12
(c) 2P = P(1,0075)n, n =
log 2
log 1,0075 = 92,77
P cancels: any amount
93 months
the first whole month after it doubles
(d) 8 000 = 5 000(1 + i)6, 1 + i =
6√
1,6
isolate the bracket
i = 0,0815,
8,15% p.a.
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### Page 38

Grade 12 Mathematics P1 — Revision Notes
4 Finance, growth and decay
Example 4.2 (standard)
(a) Convert 10% p.a. compounded monthly to an effective annual rate. (b) Find the nominal rate,
compounded quarterly, that is equivalent to an effective rate of 9% p.a.
WORKING
REASON
(a) 1 + ieff =
(
1 + 0,1
12
)12
, ieff = 10,47%
(b) 1,09 =
(
1 + i
4
)4
, i
4 =
4√
1,09 −1
i = 8,71% p.a. compounded quarterly
Example 4.3 (challenge level — same type as the challenge paper)
R30 000 is invested. For the first 2 years it earns 7% p.a. compounded monthly; after that 10%
p.a. compounded quarterly. After how many months, from the start, is the investment first worth
more than R50 000?
WORKING
REASON
After 2 years: 30 000
(
1 + 0,07
12
)24
= R34 494,18
stage 1: n = 24 months
50 000 = 34 494,18(1,025)q
stage 2: i = 0,10
4 , q quarters
q =
log
(
50 000
34 494,18
)
log 1,025
= 15,03
16 quarters are needed
interest is only added at the end of a quarter
24 + 16 × 3 = 72 months
after 15 quarters it is only R49 957,86
Example 4.4 (standard — comparing options)
Lerato has R20 000 to invest for 5 years. Option A: 9% p.a. simple interest. Option B: 8,4% p.a.
compounded monthly. Which is better?
WORKING
REASON
A: 20 000(1 + 0,09 × 5) = R29 000
B: 20 000
(
1 + 0,084
12
)60
= R30 394,73
Option B, by R1 394,73
always answer the question in words
4.2 Depreciation
Recognise it.
– “Straight-line” (or “simple decay”): the same rand amount is lost every year, A = P(1 −ni).
– “Reducing balance”: a percentage of the current value is lost every year, A = P(1 −i)n.
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