MATHEMATICS

MATHEMATICS

N4
Mathematics
Lecturer Guide
Nigel Solomon and Jolandi Daniels

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 © Future Managers 2021

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 CONTENTS

Lecturer guidance vii

1. Subject aims vii
2. Admission requirements vii
3. Duration of course viii
4. Evaluation viii
5. Examination viii
6. General information ix
7. Subject matter ix
8. Workschedule x

Answers 1

Module 1: Determinants 1
Activity 1.1 2
Activity 1.2 2
Activity 1.3 5
Activity 1.4 7
Activity 1.5 8
Activity 1.6 10
Summative assessment: Module 1 17

Module 2: Complex numbers 21

Activity 2.1 24
Activity 2.2 25
Activity 2.3 27
Activity 2.4 30
Activity 2.5 31
Activity 2.6 37
Activity 2.7 37
Activity 2.8 40
Activity 2.9 41
Activity 2.10 44
Activity 2.11 47
Summative assessment: Module 2 51

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 Module 3: Sketch graphs 55
Activity 3.1 58
Activity 3.2 59
Activity 3.3 59
Activity 3.4 61
Activity 3.5 62
Activity 3.6 62
Activity 3.7 69
Activity 3.8 74
Activity 3.9 79
Activity 3.10 84
Activity 3.11 89
Activity 3.12 94
Activity 3.13 99
Activity 3.14 104
Summative assessment: Module 3 109

Module 4: Trigonometry 113

Activity 4.1 119
Activity 4.2 121
Activity 4.3 125
Activity 4.4 127
Activity 4.5 132
Activity 4.6 136
Activity 4.7 138
Activity 4.8 140
Activity 4.9 148
Summative assessment: Module 4 150

Module 5: Differential calculus 155

Activity 5.1 160
Activity 5.2 165
Activity 5.3 169
Activity 5.4 175
Activity 5.5 178
Activity 5.6 179
Activity 5.7 185
Activity 5.8 187
Activity 5.9 189
Activity 5.10 193
Activity 5.11 196
Summative assessment: Module 5 224

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 Module 6: Integral calculus 233
Activity 6.1 236
Activity 6.2 238
Activity 6.3 238
Activity 6.4 239
Activity 6.5 241
Activity 6.6 248
Activity 6.7 252
Activity 6.8 258
Activity 6.9 272
Summative assessment: Module 6 274

Exemplar examination paper 281

Exemplar examination paper Memorandum 284

Glossary 292

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N4 Mathematics - Lecturers Guide A4 Layout.indd 6 28/02/2022 11:37 am
 Lecturer guidance vii

1. Subject aims
1.1 General subject aims
Mathematics N4 aims to provide students with the skills to identify and calculate
mathematical problems in N4 and the content forms part of engineering
calculation problems from industry.
Furthermore, Mathematics N4 will equip students with relevant knowledge to
enable them to integrate meaningfully into their trade subjects and also form the
foundation for the N5–N6 syllabuses to finally achieve a National diploma.
Upon completion of this subject the student should be able to apply:
• the necessary knowledge of Mathematics to various engineering fields in
their respective working environments;
• higher cognitive skills pertaining to application, analysis, synthesis and
evaluation, logical and critical thought processes;
• their understanding in the interpretation of real world problems;
• promote Mathematics as a tool to be used to troubleshoot in different fields
of study; and
• certain theorems that are not examinable to be calculated.

1.2 Specific subject aims

The specific aims of Mathematics N4 is to conclude precalculus and introduce
differential and integral calculus thereby serving as a prerequisite for
Mathematics N5 and Mathematics N6.
Mathematics N4 strives to assist students to obtain trade-specific calculation
knowledge.
Other specific aims of Mathematics N4 also include:
• promote correct mathematical terminology;
• promote and focus on word problems and the problem-solving thereof, in
order to prepare the students for their relevant careers; and
• use technology in Mathematic and apply Mathematics to further technology.

2. Admission requirements
For admission to Mathematics N4, students must have passed:
• Grade 12 pure Mathematics
• NC(V) Level 4 Mathematics
• N3 Mathematics.

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 viii N4 Mathematics – Lecturer Guide

3. Duration of course
The duration of the subject is one trimester on full time, part time or distance
learning mode.

4. Evaluation
Students must be evaluated continually as follows:
4.1 ICASS trimester mark
• assessment marks are valid for a period of one year and are referred to
as ICASS trimester marks
• a minimum of 40% is required for a student to qualify for entry to the
final examination
• two formal class tests for full time and part time students (or two
assignments for distance learning students only).
4.2 Calculation of trimester mark will be as follows:
• weight of test or assignment 1 = 30% of the syllabus
• weight of test or assignment 2 = 70% of the syllabus.

5. Examination
A final examination will be conducted in April, August and November of each
year. The pass requirement is 40%.
The final examination will consist of 100% of the syllabus.
The duration of the final examination will be 3 hours.
The final examination will be a closed book examination.
Minimum pass percentage is 40%.
Assessments will be based on the cognitive domain of Bloom’s Taxonomy, that is
remember, understand, apply, analyse, evaluate and create.
The division of these aspects are as follows:

Remember Understand Apply Analyse Evaluate Create

20% 20% 20% 10% 20% 10%

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 Lecturer guidance ix

6. General information
Problems should be based on real-world scenarios allowing students to relate
theory to practice.
Emphasis of correct mathematical terminology should be encouraged and
promoted at all times.
A systematical approach to problem-solving should be adhered to.
Students should be encouraged to understand rather than memorise the basic
formulae applicable to Mathematics N4.
Calculators may be used to do mathematical calculations.
Answers to all calculations must be approximated correctly to three decimal
places, unless otherwise stated. Unless otherwise stated, approximations may
not be done during calculations. The final answer must be approximated to the
stipulated degree of accuracy.
The weight value of a module gives an indication of the time to be spent
on teaching the module as well as the relative percentage of the total marks
allocated to the module in the final examination (1 mark = 1,8 minutes).

7. Subject matter
Mathematics N4 strives to assist students to obtain trade-specific calculation
knowledge. Students should be able to acquire in-depth knowledge of the
following content:

Module Weighted value

1. Determinants 8
2. Complex numbers 12
3. Sketch graphs 10
4. Trigonometry 20
5. Differential calculus 25
6. Integral calculus 25
Total 100

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 x N4 Mathematics – Lecturer Guide

8. Workschedule
Week Module Topic Activities Hours

1 Module 1 1.1 Determinants 8 hours

Determinants 1.2 Determining the Activity 1.1
value of second order Activity 1.2
determinants
1.3 Determining the value of Activity 1.3
third order determinants Activity 1.4
Activity 1.5
Activity 1.6
Summative
assessment:
Module 1
2–3 Module 2 2.1 Defining complex 12 hours
Complex numbers (ℂ)
numbers 2.2 Working with complex Activity 2.1
numbers Activity 2.2
Activity 2.3
Activity 2.4
Activity 2.5
Activity 2.6
Activity 2.7
Activity 2.8
Activity 2.9
Activity 2.10
2.3 Solving complex Activity 2.11
equations with two Summative
variables assessment:
Module 2

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 Lecturer guidance xi

Week Module Topic Activities Hours

3–4 Module 3 3.1 Relations and functions 10 hours

Sketch graphs 3.2 Independent and
dependent variable
3.3 Domain and range Activity 3.1
3.4 Functions Activity 3.2
3.5 Symmetry Activity 3.3
3.6 Continuous and Activity 3.4
discontinuous functions
or non-functions
3.7 Inverse functions and Activity 3.5
relations
3.8 Sketch graphs of Activity 3.6
functions and relations Activity 3.7
Activity 3.8
Activity 3.9
Activity 3.10
Activity 3.11
Activity 3.12
Activity 3.13
Activity 3.14
Summative
assessment:
Module 3
4–6 Module 4 4.1 Reduction formulae 20 hours
Trigonometry 4.2 Negative and positive Activity 4.1
angles
4.3 Compound angles Activity 4.2
4.4 Co-ratios Activity 4.3
4.5 Double angles
4.6 Half angles Activity 4.4
4.7 Trigonometric identities Activity 4.5
Activity 4.6
Activity 4.7
4.8 Trigonometric graphs Activity 4.8
4.9 Draw reciprocal Activity 4.9
trigonometric sketch Summative
graphs of y = cosec x, assessment:
y = sec x and y = cot x for Module 4
− 2π ≤ x ≤ 2π

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 xii N4 Mathematics – Lecturer Guide

Week Module Topic Activities Hours

6–8 Module 5 5.1 Limits Activity 5.1 25 hours

Differential 5.2 The binomial theorem Activity 5.2
calculus 5.3 Differentiation as a rate Activity 5.3
of change
5.4 Standard derivatives Activity 5.4
5.5 Chain rule Activity 5.5
5.6 Product and quotient Activity 5.6
rules Activity 5.7
Activity 5.8
Activity 5.9
5.7 Second order derivatives Activity 5.10
Activity 5.11
Summative
assessment:
Module 5
9–10 Module 6 6.1 Understanding the 25 hours
Integral concept of integration
calculus 6.2 Applying standard forms
of integrals
6.3 Applying the rules for Activity 6.1
integration Activity 6.2
Activity 6.3
Activity 6.4
6.4 Integrate composite Activity 6.5
functions
6.5 Integrating polynomials Activity 6.6
6.6 Applying integration to Activity 6.7
determine the magnitude
of an area
6.7 Areas under a curve Activity 6.8
Summative
assessment:
Module 6
TOTAL 100 hours

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 MATHEMATICS

MODULE

1 Determinants
After they have completed this module, students should be able to:
• convert equations with either two or three variables into a determinant;
• calculate second and third order determinants using row elimination, followed
by the application of Cramer’s rule;
• state and calculate the minor of a third order determinant; and
• determine the co-factor of the minor.

Introduction
In previous years students learnt how to solve simultaneous equations with two or
three variables by using the elimination or substitution methods. Many problems in
engineering can be described with systems of linear equations.

The standard form of a linear equation with two variables is ax + by = c and a linear
equation with three variable is ax + by + cz = 0.

Students need the following pre-knowledge to successfully complete this module.

Pre-knowledge
Solve for x and y in the following linear simultaneous equations:
3y − 2x = 11 … Equation (1)
y + 2x = 9 … Equation (2)
Solution
From (2): y = 9 − 2x • Subtract 2x from both sides
Substitute this y-value into (1): • Substitution method
∴ 3(9 − 2x) − 2x = 11
27 − 6x − 2x = 11
27 − 8x = 11
− 8x = 11 − 27
− 8x = − 16
∴x=2

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 2 N4 Mathematics – Lecturer Guide

Calculate the value of y:

Substitute x = 2 into (2):
y = 9 − 2(2)
=9−4
∴y=5
Solutions: x = 2; y = 5 or (2; 5)

Activity 1.1 SB page 7

1. | 23 35 | = (3)(3) – (2)(5) = –1 2. | –2 5|
–5 12 = (–5)(5) – (–2)(12) = – 1

3. | –135 –52 | = (–13)(2) – (5)(–5) = –1 4. | 78 23 | = (8)(2) – (7)(3) = –5

5. | –12 12 | = (–1)(1) – (2)(2) = –5 6. | 23 51 | = (3)(5) – (2)(1) = 13
7. | –62 31 | = (–6)(3) – (2)(1) = –20 8. | 43 31 | = (3)(3) – (4)(1) = 5
9. | –11 23 | = (–1)(2) – (1)(3) = –5 | |
10. 6 2 = (6)(1) – (8)(2) = – 10
8 1

Activity 1.2 SB page 10

1. | |
D = 3 –5 = (3)(–3) – (4)(–5) = 11
4 –3

D = | 12 –5 | = (12)(–3) – (15)(–5) = 39
D
∴ x = __ 39 6
D = 11 = 3 11
x __ __
x 15 –3

D = | 3 12 | = (3)(15) – (4)(12) = – 3
D
∴ y = __ –3 3
D = 11 = – 11
y __ __
y 4 15

2.
3 4 | |
D = 5 2 = (5)(4) – (3)(2) = 14

= | –19 2 | = (–19)(4) – (–17)(2) = – 42

D
Da ∴ a = __ –42
D = 14 = –3
a ___
–17 4

= | 5 –19 | = (5)(–17) – (3)(–19) = – 28

D
Db ∴ b = __ –28
D = 14 = –2
b ___
3 –17

3. | |
D = 3 4 = (3)(–6) – (2)(4) = – 26
2 –6

D = | 5 4 | = (5)(–6) – (–1)(4) = – 26
D
∴ k = __ –26
D = –26 = 1
k ___
k –1 –6

D = | 3 5 | = (3)(–1) – (2)(5) = – 13
D
∴ l = __ – 13 _ 1
D = – 26 = 2
ℓ ___
l 2 –1

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 Module 1 • Determinants 3

4. | |
D = 2 3 = (2)(–4) – (1)(3) = – 11
1 –4

D = | 4 3 | = (4)(–4) – (5)(3) = – 31
D
∴ r = __ – 31 9
D = – 11 = 2 11
r ___ __
r 5 –4

D = | 2 4 | = (2)(5) – (1)(4) = 6
D
∴ s = __ 6 6
D = – 11 = – 11
s ___ __
s 1 5

5. | |
D = 9 –15 = (9)(–15) – (3)(–15) = – 90
3 –15

= | 8 –15 | = (8)(–15) – (–2)(–15) = – 150

D
Dx ∴ x1 = ___ – 150 2
D = – 90 = 1 3
x 1____ _
1 –2 –15

= | 9 8 | = (9)(–2) – (3)(8) = – 42
D
Dx ∴ x2 = ___ – 42 __ 7
D = – 90 = 15
x 2___
2 3 –2

6. | |
D = 2 –1 = (2)(–5) – (6)(–1) = – 4
6 –5

D = | 8 –1 | = (8)(–5) – (32)(–1) = – 8
D
∴ x = __ –8
D = –4 = 2
x __
x 32 –5

D = | 2 8 | = (2)(32) – (6)(8) = 16
D
∴ y = __ 16
D = – 4 = –4
y __
y 6 32

7.
5 2 | |
D = 2 –1 = (2)(2) – (5)(–1) = 9

= | 7 –1 | = (7)(2) – (4)(–1) = 18
D
Da ∴ a = __ 18
D = 9 =2
a __
4 2

= | 2 7 | = (2)(4) – (5)(7) = – 27
D
Db ∴ b = __ – 27
D = 9 = –3
b ___
5 4

8. | |
D = 3 –4 = (3)(1) – (2)(–4) = 11
2 1

D = | –5 –4 | = (–5)(1) – (4)(–4) = 11
D
∴ k = __ 11
D = 11 = 1
k __
k 4 1

D = | 3 –5 | = (3)(4) – (2)(–5) = 22
D
∴ l = __ 22
D = 11 = 2
ℓ __
l 2 4

9.
| |
D = –1_3 1 = (–1)(1) – (–_32 )(1) = _12
–2 1

–1 1 | |
Dr = –3 1 = (–3)(1) – (–1)(1) = – 2
D
∴ r = __ –2
D = 1 = –4
r __
_
2

– 2 –1 | |
Ds = –1_3 –3 = (–1)(–1) – (–_32 )(–3) = – _72
D
∴ s = __
– _7
D = 1 = –7
s __2
_
2

10.
–1 2 | |
D = –1 3 = (–1)(2) – (–1)(3) = 1

= | 2 3 | = (2)(2) – (0)(3) = 4
D
Dx ∴ x1 = ___ 4
D =1=4
x _
1

1 0 2

= | –1 2 | = (–1)(0) – (–1)(2) = 2
D
Dx ∴ x2 = ___ 2
D =1=2
x _
2

2 –1 0

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 4 N4 Mathematics – Lecturer Guide

11. 1 = 3 + 1_
_
x y
1_ – 6 = – _1
x y
1 – _1 = 3
_
x y
1 + _1 = 6
_
x y
Let a = 1_x and b = 1_y
∴a–b=3
a+b=6

|
|D| = 1 – 1
1 1 |
= 1(1) – (1)(– 1)
=1+1
=2

|D a| = 6 1 |3 –1
|
= 3(1) – 6(– 1)
=3+6
=9

|D b| = 1 6 |1 3|
= 1(6) – (1)(3)
=6–3
=3
∴_
a _
b _1
D = D = |D|
| a| | b|
a = 1_
_ b = 1_
_
9 2 3 2
∴ a = _92 ∴ b = _32
∴ x = 2_9 ∴ y = 2_3

12. _53 a + b = 3
a = 6 + 3_2 b
5a + b = 3
_
3
a – 3_2 b = 6

| |
5_ 1
|D| = 3
1 – _32

= 5_3(– 3_2) – 1(1)

= – 5_2 – 1
= – 1 5_2
= – 7_2

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 Module 1 • Determinants 5

|D a| = 6 – _3
2
|3 1 |
= 3(– 3_2) – 6(1)

= – 9_2 – 6
– 9 – 12
=_ 2
= – 21
_
2

|1 6|
5 3
_
|D b| = 3

= _35(6) – 1(3)
= 10 – 3
=7

∴_1 _
a ∴_1 _
b
|D| = D | a| |D| = D
| b|
1 =_
_ a 1 = b_
_
– 7_2 – 21
_
2 – 7_2 7

– 21
_
1
∴a=_2
7_ ∴b=7×_7_
–2 –2

= – 21
_ × –2
_ 7 = 7 × – 2_
=_
2 7 7_
–2 7

=3 = –2

Activity 1.3 SB page 14

1.
2 –4 1
1.1 1 –2 3
5 1 –1
| |
a) M11 = –2 3|
1 –1 | | |
b) M32 = 2 1
1 3
= (–2)(–1) – (1)(3) = (2)(3) – (1)(1)
M11 = – 1 M32 = 5

c) M22 = 2 1 |
5 –1 |
= (2)(–1) – (5)(1)
M22 = – 7

3 –2 1
1.2 4 –1 2
5 6 7
| |
a) M13 = 4 –1
5 6 | | |
b) M31 = –2 1
–1 2 |
= (4)(6) – (5)(–1) = (–2)(2) – (–1)(1)
M13 = 29 M31 = – 3

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 6 N4 Mathematics – Lecturer Guide

c) M23 = 3 –2
5 6 | |
= (3)(6) – (5)(–2)
M23 = 28

1.3 | 1 2 3
4 –1 –2
–3 –4 –5
|
a) M12 = | –34 –2
–5 | |
b) M33 = 1 2
4 –1 |
= (4)(–5) – (–3)(–2) = (1)(–1) – (4)(2)
M12 = – 26 M33 = – 9

c) M21 = | –42 –53 |

= (2)(–5) – (–4)(3)
M21 = 2

2. |
6 2 –3
2.1 2 3 –5
1 –1 1
|
a) |
M22 = 6 –3
1 1 | |
b) M23 = 6 2
1 –1 |
= (6)(1) – (1)(–3) = (6)(–1) – (1)(2)
M22 = 9 M23 = – 8

c) M11 = | –13 –51 |

= (3)(1) – (–1)(–5)
M11 = – 2

|
2 3 –1
2.2 3 5 2
1 –2 –3
|
a) |
M32 = 2 –1
3 2 | |
b) M31 = 3 –1
5 2 |
= (2)(2) – (3)(–1) = (3)(2) – (5)(–1)
M32 = 7 M31 = 11

c) M13 = 3 5|
1 –2 |
= (3)(–2) – (1)(5)
M13 = – 11

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 Module 1 • Determinants 7

2.3 | –1 –2 1
3 1 1
1 –1 2
|
a) M21 = –2 1
–1 2 | | | |
b) M12 = 3 1
1 2
= (–2)(2) – (–1)(1) = (3)(2) – (1)(1)
M21 = – 3 M12 = 5

c) M33 = –1 –2
3 1 | |
= (–1)(1) – (3)(–2)
M33 = 5

Activity 1.4 SB page 17

1. | 2 –1 3
1 –6 –1
3 6 2
|
1.1 C 21 = (– 1) 2 + 1 – 1 3
6 2 | | |
1.2 C23 = (–1)2 + 3 2 –1
3 6 |
3 5
= (– 1) [(– 1)(2) – (6)(3)] = (–1) [(2)(6) – (3)(–1)]
= –20 C23 = – 15

1.3 C32 = (–1)3 + 2 2 3

1 –1 | |
5
= (–1) [(2)(–1) – (1)(3)]
C32 = 5

2. | –2 0 1
–3 2 –5
4 –2 6
|
2.1 C21 = (–1)2 + 1 0 1
–2 6 | | |
2.2 C11 = (–1)1 + 1 2 –5
–2 6 |
3 2
= (–1) [(0)(6) – (–2)(1)] = (–1) [(2)(6) – (–2)(–5)]
C21 = – 2 C11 = 2

2.3 C31 = (–1)3 + 1 0 1

2 –5 | |
4
= (–1) [(0)(–5) – (2)(1)]
C31 = – 2

3. |
3 2 –1
3.1 3 –2 1
4 –5 –1
|
a) C32 = (–1)3 + 2 3 –1
3 1 | | |
b) C23 = (–1)2 + 3 3 2
4 –5 |
= (–1)5[(3)(1) – (3)(–1)] = (–1)5[(3)(–5) – (4)(2)]
C32 = – 6 C23 = 23

c) C12 = (–1)1 + 2 3 1
4 –1 | |
3
= (–1) [(3)(–1) – (4)(1)]
C12 = 7

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 8 N4 Mathematics – Lecturer Guide

3.2 | 1 1 4
2 –3 –1
–4 2 2
|
a) C22 = (–1)2 + 2 | –41 42 | b) C31 = (–1)3 + 1 | –31 –14 |
= (–1)4[(1)(2) – (–4)(4)] = (–1)4[(1)(–1) – (–3)(4)]
C22 = 18 C31 = 11

c) |
C13 = (–1)1 + 3 2 –3
–4 2 |
= (–1)4[(2)(2) – (–4)(–3)]
C13 = – 8

|
3 1 –2
3.3 4 –1 1
1 –3 –4
|
a) |
C11 = (–1)1 + 1 –1 1
–3 –4 | b) C21 = (–1)2 + 1 | –31 –2
–4 |
= (–1)2[(–1)(–4) – (–3)(1)] = (–1)3[(1)(–4) – (–3)(–2)]
C11 = 7 C21 = 10

c) |
C33 = (–1)3 + 3 3 1
4 –1 |
6
= (–1) [(3)(–1) – (4)(1)]
C33 = – 7

Activity 1.5 SB page 21

1. 1.1 | –13 –53 | = (3)(3) – (–1)(–5) = 4 1.2 | –12 70 | = (2)(0) – (–1)(7) = 7

1.3 2 | |
3 = (2)(–12) – (6)(3) = – 42
6 –12 | |
1.4 –2 4 = (–2)(–7) – (3)(4) = 2
3 –7

| |
1.5 –5 –3 = (–5)(–2) – (2)(–3) = 16
2 –2 | |
1.6 1 –2 = (1)(4) – (3)(–2) = 10
3 4

2. 2.1 |3 −1 2
1 −1 2
−2 3 1
| |
0 1 −3
2.2 2 4 − 1
4 −2 5
|
| | |
= 3 − 1 2 − (− 1) 1 2 + 2 1 − 1
3 1 −2 1 −2 3 | | | | |
= 0 − 1 2 − 1 + (−3) 2 4
4 5 4 −2 | |
= 3(− 1 − 6) + 1(1 + 4) + 2(3 − 2) = 0 − 1(10 + 4) − 3(− 4 − 16)
= 3(− 7) + 1(5) + 2(1) = –14 – 3(–20)
= –21 + 5 + 2 = –14 + 60
= –14 = 46

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 Module 1 • Determinants 9

2.3 |4 −4 5
3 1 4
−3 4 1
|
| |
= − (− 4) 3 4 + (1) 4 5 − 4 4 5
−3 1 −3 1 3 4 | | | |
= 4(3 + 12) + (4 + 15) − 4(16 − 15)
= 4(15) + 19 – 4
= 60 + 19 – 4
= 75

|
2 3 −1
2.4 2 − 1 2
3 −1 1
|
| |
= 3 3 − 1 − (−1) 2 − 1 + 1 2 3
−1 2 2 2 2 −1 | | | |
= 3(6 − 1) + 1(4 + 2) + 1(−2 − 6)
= 3(5) + 1(6) + 1(− 8)
= 15 + 6 – 8
= 13

3. |
3 –1 3
| | |
3.1 4 1 5 = (3) 1 5 + (4) –1 3 + (2) –1 3
2 1 3 13 13 15 | | | |
= 3[(1)(3) – (1)(5)] – 4[(–1)(3) – (1)(3)] + 2[(–1)(5) – (1)(3)]
= –6 + 24 – 16

|
3 –1 3
4 1 5 =2
2 1 3
|
3.2 | –2 1 0
|
0 3 –1 = (5)
5 0 6
| |
1 0 + (0) –2 0 + (6) –2 1
3 –1 0 –1 0 3 | | | |
= 5[(1)(–1) – (3)(0)] + 0 + 6 [(–2)(3) – (0)(1)]
= –5 + 0 – 36

| –2 1 0
|
0 3 –1 = –41
5 0 6

|
3 –2 1
| |
3.3 2 1 –5 = (–2) 2 –5 + (1) 3 1 + (1) 3 1
1 1 –6 1 –6 1 –6 2 –5 | | | | |
= 2[(2)(–6) – (1)(–5)] + 1[(3)(–6) – (1)(1)] – 1[(3)(–5) – (2)(1)]
= –14 – 19 + 17

|
3 –2 1
2 1 –5 = –16
1 1 –6
|
|
1 4 2
3 2 1
| 31 | |
3.4 –1 2 –1 = (–1) 4 2 + (2) 1 2 + (–1) 1 4
21 32 | | | |
= 1[(4)(1) – (2)(2)] + 2[(1)(1) – (3)(2)] + 1[(1)(2) – (3)(4)]
= 0 – 10 – 10

|
1 4 2
–1 2 –1 = –20
3 2 1
|

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 10 N4 Mathematics – Lecturer Guide

Activity 1.6 SB page 32

1. 1.1 |
1 2 2
D = 3 −1 4
3 2 −1
|
| | | | |
= 1 −1 4 − 2 3 4 + 2 3 −1
2 −1 3 −1 3 2 |
= 1(1 − 8) − 2(− 3 − 12) + 2(6 + 3)
= − 7 − 2(− 15) + 2(9)
= –7 + 30 + 18
= 41

|
4 2 2
D x = 25 − 1 4
−4 2 −1
|
| | | | |
= 4 − 1 4 − 2 25 4 + 2 25 − 1
2 −1 −4 −1 −4 2 |
= 4(1 − 8) − 2(− 25 + 16) + 2(50 − 4)
= 4(− 7) − 2(− 9) + 2(46)
= –28 + 18 + 92
= 82

|
1 4 2
D y = 3 25 4
3 −4 −1
|
| | |
= 1 25 4 − 4 3 4 + 2 3 25
−4 −1 3 −1 | |3 −4 |
= 1(− 25 + 16) − 4(− 3 − 12) + 2(− 12 − 75)
= − 9 − 4(− 15) + 2(− 87)
= − 9 + 60 − 174
= –123

|
1 2 4
D z = 3 − 1 25
3 2 −4
|
| | | | |
= 1 − 1 25 − 2 3 25 + 4 3 − 1
2 −4 3 −4 3 2 |
= 1(4 − 50) − 2(− 12 − 75) + 4(6 + 3)
= − 46 − 2(− 87) + 4(9)
= –46 + 174 + 36
= 164
D
∴ x = __ 82
D = 41 = 2
x __

D
∴ y = __ –123
D = 41 = –3
y ____

D
∴ z = __ 164
D = 41 = 4
z ___

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 Module 1 • Determinants 11

1.2 |
2 −3 5
D= 3 2 2
4 1 −4
|
| | |
= 2 2 2 − (− 3) 3 2 + 5 3 2
1 −4 4 −4 | | |
4 1
= 2(− 8 − 2) + 3(− 12 − 8) + 5(3 − 8)
= 2(− 10) + 3(− 20) + 5(− 5)
= − 20 − 60 − 25
= − 105

|4 −3 5
Da = 3 2 2
−6 1 −4
|
|
1 −4 | |
= 4 2 2 − (− 3) 3 2 + 5 3 2
−6 −4 | | −6 1 |
= 4(− 8 − 2) + 3(− 12 + 12) + 5(3 + 12)
= 4(− 10) + 3(0) + 5(15)
= − 40 + 75
= 35

|
2 4 5
Db = 3 3 2
4 −6 −4
|
|
−6 −4 | 4 −4|
= 2 3 2 − (4) 3 2 + 5 3 3 | |4 −6 |
= 2(− 12 + 12) − 4(− 12 − 8) + 5(− 18 − 12)
= − 4(− 20) + 5(− 30)
= 80 − 150
= − 70

|
2 −3 4
Dc = 3 2 3
4 1 −6
|
| | |
= 2 2 3 − (− 3) 3 3 + 4 3 2
1 −6 4 −6 | | |
4 1
= 2(− 12 − 3) + 3(− 18 − 12) + 4(3 − 8)
= 2(− 15) + 3(− 30) + 4(− 5)
= − 30 − 90 − 20
= − 140
D
∴ a = __ 35 1
D = –105 = – 3
a ____ _

D
∴ b = __ –70 2
D = –105 = 3
b ____ _

D
∴ c = __ –140 1
D = –105 = 1 3
c ____ _

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 12 N4 Mathematics – Lecturer Guide

2. |
1 1 1
|
2.1 |D| = 2 –1 3 = + (1) –1 3 – (1) 2 3 + (1) 2 –1
–1 2 2 2 2 –1 2 |–1 2 | | | | |
= 1[(–1)(2) – (2)(3)] – 1[(2)(2) – (–1)(3)] + 1[(2)(2) – (–1)(–1)]
= –8 – 7 + 3

| 1 1 1
∴ |D| = 2 –1 3 = –12
–1 2 2
|
|6 1 1
9 2 2
| 2 2 |
|Dx| = 9 –1 3 = + (6) –1 3 – (1) 9 3 + (1) 9 –1
9 2 9 2 | | | | |
= 6[(–1)(2) – (2)(3)] – 1[(9)(2) – (9)(3)] + 1[(9)(2) – (9)(–1)]
= –48 + 9 + 27

| 6 1 1
∴ |Dx| = 9 –1 3 = –12
9 2 2
|
| 1 6 1
| | |
|Dy| = 2 9 3 = + (1) 9 3 – (6) 2 3 + (1) 2 9
–1 9 2 9 2 –1 2 –1 9 | | | |
= 1[(9)(2) – (9)(3)] – 6[(2)(2) – (–1)(3)] + 1[(2)(9) – (–1)(9)]
= –9 – 42 + 27

| 1 6 1
∴ |Dy| = 2 9 3 = –24
–1 9 2
|
| 1 1 6
| |
|Dz| = 2 –1 9 = + (1) –1 9 – (1) 2 9 + (6) 2 –1
–1 2 9 2 9 –1 9 –1 2 | | | | |
= 1[(–1)(9) – (2)(9)] – 1[(2)(9) – (–1)(9)] + 6[(2)(2) – (–1)(–1)]
= –27 – 27 + 18

| 1 1 6
∴ |Dz| = 2 –1 9 = –36
–1 2 9
|
D
∴ x = __ –12
D = –12 = 1
x ___

D
∴ y = __ –24
D = –12 = 2
y ___

D
∴ z = __ –36
D = –12 = 3
z ___

2.2 |
4 –3 1
|
|D| = 1 2 –5 = + (3) –3 1 – (1) 4 1 + (2) 4 –3
3 1 2 2 –5 1 –5 |1 2 | | | | |
= 3[(–3)(–5) – (2)(1)] – 1[(4)(–5) – (1)(1)] + 2[(4)(2) – (1)(–3)]
= 39 + 21 + 22

|
4 –3 1
∴ |D| = 1 2 –5 = 82
3 1 2
|
|
–1 –3 1
|
|Dk| = 11 2 –5 = + (20) –3 1 – (1) –1 1 + (2) –1 –3
20 1 2 2 –5 11 –5 | 11 2 | | | | |
= 20[(–3)(–5) – (2)(1)] – 1[(–1)(–5) – (11)(1)] + 2[(–1)(2) – (11)(–3)]
= 260 + 6 + 62

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 Module 1 • Determinants 13

|
–1 –3 1
∴ |Dk| = 11 2 –5 = 328
20 1 2
|
|
4 –1 1
| | |
|Dl| = 1 11 –5 = + (3) –1 1 – (20) 4 1 + (2) 4 –1
3 20 2 11 –5 1 –5 |
1 11 | | |
= 3[(–1)(–5) – (11)(1)] – 20[(4)(–5) – (1)(1)] + 2[(4)(11) – (1)(–1)]
= –18 + 420 + 90

|
4 –1 1
∴ |Dl |= 1 11 –5 = 492
3 20 2
|
|
4 –3 –1
| | |
|Dm| = 1 2 11 = + (3) –3 –1 – (1) 4 –1 + (20) 4 –3
3 1 20 2 11 1 11 |
1 2 | | |
= 3[(–3)(11) – (2)(–1)] – 1[(4)(11) – (1)(–1)] + 20[(4)(2) – (1)(–3)]
= –93 – 45 + 220

|
4 –3 –1
∴ |Dm| = 1 2 11 = 82
3 1 20
|
D
∴ k = __ 328
D = 82 = 4
k ___

D
∴ l = __ 492
D = 82 = 6
ℓ ___

D
∴ m = __ 82
D = 82 = 1
m __

|
2 –1 4
| | |
2.3 |D| = 3 –2 1 = + (2) –2 1 – (3) –1 4 + (1) –1 4
1 –5 3 –5 3 –5 3 |
–2 1 | | |
= 2[(–2)(3) – (–5)(1)] – 3[(–1)(3) – (–5)(4)] + 1[(–1)(1) – (–2)(4)]
= –2 – 51 + 7

|
2 –1 4
∴ |D| = 3 –2 1 = –46
1 –5 3
|
|
6 –1 4
| | |
|Dr| = 2 –2 1 = + (6) –2 1 – (2) –1 4 + (9) –1 4
9 –5 3 –5 3 –5 3 |
–2 1 | | |
= 6[(–2)(3) – (–5)(1)] – 2[(–1)(3) – (–5)(4)] + 9[(–1)(1) – (–2)(4)]
= –6 – 34 + 63

|
6 –1 4
∴ |Dr| = 2 –2 1 = 23
9 –5 3
|
|
2 6 4
1 9 3
| | |
|Ds| = 3 2 1 = + (2) 2 1 – (3) 6 4 + (1) 6 4
9 3 9 3 | |
2 1 | |
= 2[(2)(3) – (9)(1)] – 3[(6)(3) – (9)(4)] + 1[(6)(1) – (2)(4)]
= –6 + 54 – 2

|
2 6 4
∴ |Ds| = 3 2 1 = 46
1 9 3
|

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 14 N4 Mathematics – Lecturer Guide

|
2 –1 6
1 –5 9
| –5 9 |
|Dt| = 3 –2 2 = + (2) –2 2 – (3) –1 6 + (1) –1 6
–5 9 |–2 2 | | | |
= 2[(–2)(9) – (–5)(2)] – 3[(–1)(9) – (–5)(6)] + 1[(–1)(2) – (–2)(6)]
= –16 – 63 + 10

|
2 –1 6
|
∴ |Dt| = 3 –2 2 = –69
1 –5 9
D
∴ r = __ 23 1
D = –46 = –2
r ___ _

D
∴ s = __ 46
D = –46 = –1
s ___

D
∴ t = __ –69 1
D = –46 = 12
t ___ _

3. |
2 4 –1
| |
3.1 |D| = 1 2 –3 = (2) 2 –3 + (4) 1 –3 + (–1) 1 2
3 –1 1 –1 1 3 1 |3 –1 | | | |
= 2[(2)(1) – (–1)(–3)] – 4[(1)(1) – (3)(–3)] – 1[(1)(–1) – (3)(2)]
= –2 – 40 + 7

|
2 4 –1
∴ |D| = 1 2 –3 = –35
3 –1 1
|
|2 4 –1
| | |
|Dx| = –4 2 –3 = (2) 2 –3 + (4) –4 –3 + (–1) –4 2
1 –1 1 –1 1 1 1 1 –1 | | | |
= 2[(2)(1) – (–1)(–3)] – 4[(–4)(1) – (1)(–3)] – 1[(–4)(–1) – (1)(2)]
= –2 + 4 – 2

| 2 4 –1
∴ |Dx| = –4 2 –3 = 0
1 –1 1
|
|
2 2 –1
| | |
|Dy| = 1 –4 –3 = (2) –4 –3 + (2) 1 –3 + (–1) 1 –4
3 1 1 1 1 3 1 3 1 | | | |
= 2[(–4)(1) – (1)(–3)] – 2[(1)(1) – (3)(–3)] – 1[(1)(1) – (3)(–4)]
= –2 – 20 – 13

|
2 2 –1
∴ |Dy| = 1 –4 –3 = –35
3 1 1
|
|
2 4 2
3 –1 1
| –1 1 | 3 1 |
|Dz| = 1 2 –4 = (2) 2 –4 + (4) 1 –4 + (2) 1 2
3 –1 | | | |
= 2[(2)(1) – (–1)(–4)] – 4[(1)(1) – (3)(–4)] + 2[(1)(–1) – (3)(2)]
= –4 – 52 – 14

|
2 4 2
∴ |Dz| = 1 2 –4 = –70
3 –1 1
|
D
∴ x = __ 0
D = –35 = 0
x ___

D
∴ y = __ –35
D = –35 = 1
y ___

D
∴ z = __ –70
D = –35 = 2
z ___

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 Module 1 • Determinants 15

3.2
5 –6 3
| | | |
|D| = 2 –3 2 = (–6) 2 2 + (–3) 5 3 + (–7) 5 3
3 –7 5 3 5 3 5 22 | | | |
= 6[(2)(5) – (3)(2)] – 3[(5)(5) – (3)(3)] + 7[(5)(2) – (2)(3)]
= 24 – 48 + 28
5 –6 3
|
∴ |D| = 2 –3 2 = 4
3 –7 5
|
– 9 –6 3
| –5 2
| |
–9 3
|
| x | – 5 –3 2 = (–6) –16 5 + (–3) –16 5 + (–7) –5 2
D =
–16 –7 5
1
–9 3
| | | |
= 6[(–5)(5) – (–16)(2)] – 3[(–9)(5) – (–16)(3)] + 7[(–9)(2) – (–5)(3)]
= 42 – 9 – 21
– 9 –6 3
|
∴ |Dx | = – 5 –3 2 = 12
–16 –7 5
1 |
| x|
D = 2 |
5 –9 3
–5 2 = (–9)
3 –16 5
2 35| 35 | |
2 2 + (–5) 5 3 + (–16) 5 3
22 | | | |
= 9[(2)(5) – (3)(2)] – 5[(5)(5) – (3)(3)] + 16[(5)(2) – (2)(3)]
= 36 – 80 + 64
5 –9 3
|
∴ |Dx | = 2 –5 2 = 20
3 –16 5
2 |
D =
5 –6 –9
2
3 –7 –16
3 | 2 –5
| |
5 –9
| x | –3 –5 = (–6) 3 –16 + (–3) 3 –16 + (–7) 2 –5 | 5 –9
| | | |
= 6[(2)(–16) – (3)(–5)] – 3[(5)(–16) – (3)(–9)] + 7[(5)(–5) – (2)(–9)]
= –102 + 159 – 49
5 –6 –9
|
∴ |Dx | = 2 –3 –5 = 8
3 –7 –16
3 |
D
∴ x1 = ___
x 12
D = 4 =3
__ 1

D
∴ x2 = ___
x 20
D = 4 =5
__ 2

D
∴ x3 = ___
x 8
D =4=2
_ 3

2 1 –1
3.3 D = 3 – 4 5
4 0 2
| |
| | | |
= 2 – 4 5 – 1 3 5 + (–1) 3 – 4
0 2 4 2 4 0 | |
= 2[–8] – [6 – 20] – [0 + 16]
= –16 – [–14] – 16
= –16 + 14 – 16
= –18

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 16 N4 Mathematics – Lecturer Guide

D =
1
2 1 –1
| I | 7 –4 5
9 0 2
| |
|
= 2 – 4 5 – 1 7 5 – (–1) 7 – 4
0 2 9 2 | | |
9 0 | |
= 2[–8] – [14 – 45] + [0 – (–36)]
= –16 – [–31] + 36
= 51
1 _ I
∴_
D= D
1
| I| 1

_1 I1
_
– 18 = 51
I1 = _51
– 18
= –2,833
3.4 _a1 + _b2 + 1_c = 0
1 = _1 – 2_ + 2
_
a b c

– _a1 + _b1 – 1_c + 3 = 0

Let _a1 = x, 1_6 = y and 1_c = z

∴ x + 2y + z = 0
x – y + 2z = 2
–x + y – z = –3
1 2 1
|D| = 1 – 1 2
–1 1 –1
| |
|
= – 1 – 1 2 – 2 1 + (– 1) 2 1
1 –1 1 –1 | |
–1 2 | | |
= –[1 – 2] – [–2 – 1] – 1[4 + 1]
= –1 – [–3] – 5
= –6 + 3
= –3
1 2 0
|D z| = 1 – 1 2
–1 1 –3
| |
|
= 1 – 1 2 – 1 2 0 + (– 1) 2 0
1 –3 1 –3 –1 2| | | | |
= [3 – 2] – [–6] – 1[4]
=1+6–4
=3
_1
∴ |D| = _ z
D | z|
1 = _z
_
–3 3
∴ z = –1
∴ 1_c = z
∴ 1_c = – 1
∴ c = –1

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 Module 1 • Determinants 17

Summative assessment: Module 1 SB page 33

1.
1 –1 | |
1.1 2 6 = (2)(–1) – (1)(6) = – 8 (2)

1.2 | −5 −1 1
9 1 −1 = −5
−1 1 −5
|
1 −5 | |
1 − 1 − (− 1) 9 − 1 + 1 9 1
−1 −5 |
−1 1 | | |
= − 5(− 5 + 1) + 1(− 45 + 1) + 1(9 + 1)
= − 5(− 4) − 44 + 10
= 20 − 44 + 10
= − 16 (3)

2.
–3 2 |
2.1 D = 2 – 1 0 – 2 1 0 + 1 1 – 1
3 2 3 –3| | | | |
= 2[–2] – 2[2] + [–3 + 3]
= –4 – 4
= –8 (2)

2.2 + 2 1
3 2 | |
= 2(2) – 3(1)
=4–3
=1 (2)

2.3 2 1
1 0 | |
= 2(0) – 1(1)
= –1 (2)

3. |D|
|
6 –7 1
= 1 –3 –8
–2 –1 3
|
–3 –8 | 1 –8|
= – 2 – 7 1 – (– 1) 6 1 + 3 6 – 7
1 –3 | | | |
= –2[56 + 3] + [–48 – 1] + 3[–18 + 7]
= –2[59] – 49 + 3[–11]
= –200

| I|
D =
3
1
6 – 7 0,5
–3 –9
–2 –1 4
| |
= –2
–9|––73
0,5 ( ) 6 0,5
– –1 |
1 –9 |
+ 4 6 –7
1 –3 | | |
= –2[63 + 3(0,5)] + [–56 – 0,5] + 4[–18 + 7]
= –2[64,5] – 56,5 + 4[–11]
= –36

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 18 N4 Mathematics – Lecturer Guide

1 I
∴_ _
|D| = D
3
| I| 3

_ 1 _I3
– 200 = – 36
∴ I 3 = 0,18 (8)

4. | |
D = –2 –3 = (–2)(–5) – (3)(–3) = 19
3 –5

|–2 –5 |
Dr = –5 –3 = (–5)(–5) – (–2)(–3) = 19
D
∴ r = __
r 19
D = 19 = 1
__

D = | –2 –5 | = (–2)(–2) – (3)(–5) = 19
D
∴ s = __
s 19
D = 19 = 1
__
(10)
s 3 –2

5. 5.1 | –x3 x2 | = 5x
(3)(2) – (–x)(x) = 5x
6 + x2 = 5x
x2 – 5x + 6 = 0
(x – 2)(x – 3) = 0
x – 2 = 0 or x–3=0
∴ x = 2 or x=3 (3)

5.2 | yy +– 34 –24 | = | –31 7y |

(y – 3)(4) – (y + 4)(–2) = (1)(y) – (–3)(7)
4y – 12 + 2y + 8 = y + 21
6y – 4 = y + 21
5y = 25
∴y=5 (3)

5.3
z 3 2
|
3 –2 5 = – 120
2 z 6
|
| | | |
+ (z) –2 5 – (3) 3 5 + (2) 3 –2 = – 120
z 6 2 6 2 z | |
z[(–2)(6) – (z)(5)] – 3[(3)(6) – (2)(5)] + 2[(3)(z) – (2)(–2)] = – 120
z[–12 – 5z] –3[18 – 10] + 2 [3z + 4] = – 120
– 12z – 5z2 – 54 + 30 + 6z + 8 = – 120
– 5z2 – 6z – 16 = – 120
– 5z2 – 6z + 104 = 0
5z2 + 6z – 104 = 0
(5z + 26)(z – 4) = 0
5z + 26 = 0 or z – 4 = 0
5z = – 26
∴ z = – 5_15 or ∴ z = 4
(4)

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 Module 1 • Determinants 19

6.
5 –6 3
6.1 |D| = 2 – 3 2
3 –7 5
| |
| |
= 5 – 3 2 – (–6) 2 2 + 3 2 – 3
–7 5 3 5 3 –7 | | | |
= 5[–15 + 14] + 6[10 – 6] + 3[–14 + 9]
= –5[–1] + 6[4] + 3[–5]
= 5 + 24 – 15
= 14 (3)
5 –6 –9
6.2 |D c| = 2 – 3 – 5 |
3 – 7 – 16
|
| |
= 5 – 3 – 5 – (– 6) 2 – 5 + (–9) 2 – 3
– 7 – 16 3 – 16 3 –7 | | | |
= 5[48 – 35] + 6[–32 + 15] – 9[–14 + 9]
= 5[13] + 6[–17] – 9[–5]
=8
_1
∴ |D| = _c
D | c|
1 = _c
_
14 8
∴ c = 0,571 (3)

7. 7.1 |D|
|
–2 –3 –3
| |
= 5 –1 –6 = + (–2) –1 –6 – (–3) 5 –6 + (–3) 5 –1
3 2 2 2 2 3 2 |
3 2 | | | |
= –2[(–1)(2) – (2)(–6)] + 3[(5)(2) – (3)(–6)] – 3[(5)(2) – (3)(–1)]
= –20 + 84 –39

|
–2 –3 –3
∴ |D| = 5 –1 –6 = 25
3 2 2
| (3)

7.2
–4 –3 –3
1 2 2
| –1 – 6
|
|Dk¡| = 3 –1 –6 = –4| 2 2 | – (3)| 2 2 | + 1| –1 – 6 |
–3 –3 –3 –3

= –4(–2 + 12) – 3(–6 + 6) + (18 – 3)

= –4(10) + 15
= –25 (3)

7.3
–2 –4 –3
| 3 1 2
|
|Dℓ| = 5 3 –6 = + (–3)| 3 1 | – (–6)| 3 1 | + (2)| 5 3 |
5 3 –2 –4 –2 –4

= –3[(5)(1) – (3)(3)] + 6[(–2)(1) – (3)(–4)] + 2[(–2)(3) – (5)(–4)]

= 12 + 60 + 28
–2 –4 –3
|
∴ |Dℓ| = 5 3 –6 = 100
3 1 2
| (3)

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 20 N4 Mathematics – Lecturer Guide

|
–2 –3 –4
7.4 |Dm| = 5 –1 3
3 2 1
|
[ | |] [ | |] [ |
= (5) (–1)2 + 1 –3 –4 + (–1) (–1)2 + 2 –2 –4 + (3) (–1)2 + 3 –2 –3
2 1 3 1 3 2 |]
= –5[(–3)(1) – (2)(–4)] − 1[(–2)(1) – (3)(–4)] – 3[(–2)(2) – (3)(–3)]
= – 25 – 10 – 15

|
–2 –3 –4
∴ |Dm| = 5 –1 3 = – 50
3 2 1
| (3)

|D |
–25 = –1
7.5 k = ___
k
|D|
= ___
25
|D | 100 = 4
l = ___
ℓ
|D|
= ___
25
|D |
–50 = –2
m = ___
m
|D|
= ___
25 (3)

TOTAL: [60]

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 MATHEMATICS

MODULE

2 Complex numbers
After they have completed this module, students should be able to:
• define an imaginary number;
• identify real and imaginary parts of a complex number in rectangular form;
• simplify complex powers;
• add, subtract and multiply complex numbers in rectangular form;
• define and determine the conjugate of a complex number;
• divide complex numbers in rectangular for using the conjugate;
• define the modulus and argument of the complex number and plot them on
an Argand diagram;
• convert a complex number from rectangular form to polar form and vice
versa, using a pocket calculator or any analytical method;
• multiply and divide complex numbers in polar form;
• state and apply De Moivre’s theorem to products, quotients and powers of
complex numbers; and
• solve complex equations in rectangular or polar form.

Introduction
Complex numbers were encountered previously when _______________
solving x-intercepts for
− b ± √b 2 − 4ac . When it yields a
a x + bx + c = 0 using the quadratic formula, x = _______________
2
2a
negative number inside the square root it implies that no solution exists in the
real number system, that is, the parabola does not intersect the x-axis.

To accommodate the square root of negative numbers the real number system needs to
be extended to include pure imaginary numbers. The expanded number system is called
the complex number system.

Before working with complex numbers, we need to revise the relationships between
several types of numbers.

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 22 N4 Mathematics – Lecturer Guide

Complex numbers (ℂ)

__
e.g. 3 + 2i ; 1___
+ √5 i ;
3 – 2√ −4

Real numbers (ℝ) Imaginary numbers (핀)

__ ___
e.g. _34 ; 0,05 ; √ 3 ; π e.g. √ – 2 ;
_____
√ −513 ;
3i ;
____
3√ −19 ;
___
Rational numbers (ℚ) Irrational numbers (ℚ´) 4
√−2 ;
___ __ ___
e.g. − _2 ; √−8 ; e.g. −√ 5 ;
3
3 __ i = √ −1
5√ 2 ;
0 ; 0,4 ; __
−3
___ √3 ;
√−8 ; 4 ; 5; __
3√ 5 ;
1,414…; _9 ; 317;
2 2,51476…;
Note: Fractions, π
terminating and Note: Non-terminating and
repeating decimal non-repeating (recurring)
numbers. decimal numbers.

Integers (ℤ)
e.g. {… ; – 3 ; – 2 ; – 1 ; 0 ; 1 ;
2 ; 3 ; …}

Whole numbers (ℕ0 )

e.g. {0 ; 1 ; 2 ; 3 ; 4 ; …}

Natural numbers (ℕ)

e.g. {1 ; 2 ; 3 ; 4 ; 5 ; …}

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 Module 2 • Complex numbers 23

Complex numbers can be represented in three forms, rectangular form, polar form and
exponential form.

Complex numbers

Rectangular form Polar form Exponential form

z = a + bi or z = r(cos θ + i sin θ) or z = re iθ or
z = a + bj z = r(cos θ + j sin θ) z = re jθ

In this module we only represent complex numbers in rectangular (standard) form and
polar (trigonometric) form with the imaginary unit represented as i instead of j as in
some texts.

Students need the following pre-knowledge to successfully complete this module.

Pre-knowledge
Laws of exponents:
1. a m × a n = a m+n 2. a_mn = a m−n
a

3. (a m)n = a mn 4. (a.b)m = a m.b m

( a n ) = a nb
b
5. _
a m
_
a mb
6. a −n = _
1
an
_ _ _ _ _ _ _ _ _ n_
7. √ab = √a × √b or √ab = √a × √b
n n n
8. _a = _
√_
a or √b _a = _
√_a
√b _
n
n
√b √b
_ n _
9.
n
√a = a m _
m
n
10. ( √a ) = √a
m m n

Addition and subtraction of surds

_ _ _ _ _
√a + √a = 2 √a √a − √a = 0
_ _ _ _ _ _ _ _
√a + √b = √a + √b √a − √b = √a − √b

Multiplication and division of surds

_ _ _ _ _ _ _
a √b ÷ c √d = _ c √d
a √b × c √d = ac √bd b = a_ _b
a √_
c √d

• Perfect squares: 1 = 1 2; 4 = 2 2; 9 = 3 2; 16 = 4 2; 25 = 5 2; …
_ _ _ _ _ _ _ _
3 3
• √4 = √2 × 2 = 2; √9 = √3 × 3 = 3; √a 2 = √a × a = a; √27 = √3.3.3 = 3
_ 2 _ _ _
• (√6 ) = √6 .√6 = √36 = 6

opposite side
• sin θ = ___________
hypotenuse
adjacent side
• cos θ = _
hypotenuse r
y
opposite side
• tan θ = _
adjacent side
θ
• Theorem of Pythagoras: x 2 + y 2 = r 2 x

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 24 N4 Mathematics – Lecturer Guide

r1 θ 1
To divide ____
r θ
the moduli (r) will be divided and the arguments (θ) subtracted.
2 2

For example:
Given z1 = r1 θ 1 and z2 = r2 θ 2. Calculate z1 ÷ z2.
r1 θ 1
z1 ____
∴_
z2 = r θ
2 2

r
= _r21 θ 1 − θ 2 • Theorem

Activity 2.1 SB page 44

1. i 2 = –1 2. i 7 = (i 2)3i
= (– 1)3i
= –i

3. i 8 = (i 2)4 4. 3i 9 = 3(i 2)4i

= (– 1)4 = 3(– 1)4i
=1 = 3i

5. i 26 =(i 2)13 6. – i 11 = – (i 2)5i

= (– 1)13 = – (– 1)5i
= –1 =i

7. _5 = _5 × _
i
−i
i −i
8. _
1
=_
1
× _i
−i −i i

=_
− 5i
2 =_
i
2
−i −i

= _− 5i
− (−1)
=_
i
− (−1)
= – 5i =i

9. _
1
=_
1
×_
−i
10. – _
3
= –_
3
×_
−i
2i 2i −i 4i 4i − i

=_
−i
2 =_
3i
2
−2 i −4 i
= _ −i
=_
3i
− 2(−1) − 4(−1)

=_
−i _1
2 or – 2 i =_
3i _3
4 or 4 i

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 Module 2 • Complex numbers 25

Activity 2.2 SB page 47

____ ____
1. 1.1 √−49 1.2 √−45
___ ___ ___ ___
= √49 .√−1 = √9.5 .√−1
__
= 7i = 3√5 i
_____ ____
1.3 − √−100 1.4 √−32
____ ___ ___ ___
= − √100 .√−1 = √32 .√−1
____
= – 10i = √16.2 i
__
= 4√2 i
___ ___ ____ ___
1.5 √−9 × √−6 1.6 −√−64 .√−3
__ ___ __ ___ ___ ___ __ ___
= √9 .√−1 × √6 √−1 = − √64 .√−1 .√3 .√−1
__ __
= 3i × √6 i = – 8i√3 i
__ __
= 3√6 i2 = – 8√3 i2
__ __
= 3√6 (– 1) = – 8√3 (– 1)
__ __
= – 3√6 = 8√3
____ ___ _ _
– 12___
. √– 6
_______
√
1.7 1.8 ( √− 125 )(√5 )
√– 2 ____ ___ __
___ ___
12 √–__1 . √___
__ ___
6 √– 1
= (√25.5 .√−1 )(√5 )
= __________
√ __ __
√2 . √– 1 = 5√5 i√ 5
___ __ ___
√12__
i√ 6 i
= ______ = 5√25 i
√2 i
____ = 25i
= √____
12.6
2 i
___
= √36 i
= 6i
___ ___ ___ ___
1.9 (√−4 )5 1.10 √−5 (√−3 .√−4 )
__ ___ __ ___ __ ___ __ ___
= (√4 .√−1 )5 = √5 .√−1 (√3 √−1 )(√4 .√−1 )
__ __
= (2i)5 = √5 i(√ 3 i)(2i)
___
= 32i5 = 2√15 i 3
___
= 32(i2)2i = 2√15 (i 2)i
___
= 32(– 1)2i = 2√15 (−1)i
___
= 32i = – 2√15 i

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 26 N4 Mathematics – Lecturer Guide

____ ___ __ ____

√−12___
.√−4
1.11 ________ 1.12 7i2√3 (4i √ −27 )
√−6 __ ___ ___
___ ___ __ ___
√12 √−1 .√4 .√−1
= 7(– 1)√3 (4i √27 √−1 )
= __________
__ ___ __ ___
√6 .√−1 = – 7√3 (4i√ 9.3 i )
___ __ __
√12__.i.2i
= _____ = – 7√3 (4i.3√3 i )
√6 i
_
2
= – 7.4.3.3.i2
2 √12
=_ _
i
= – 252(– 1)
√6 i
___
= 2√__
12
i = 252
6
__
= 2√2 i
_____ _____ ____ ___
−6 + √−169
1.13 ________
3 1.14 (√−144 )(√−32 )(−√−2 )
____ ___
____ ___ ___ ___ __ ___
−6 + √169 √−1 = (√144 .√−1 )(√32 .√−1 )(−√2 .√−1 )
= _________ 3
____ _
−6 + 13i
= (12i)(√16.2 i)(– √2 i)
= ______ __ _
3 = (12i)(4√2 i)(– √2 i)
= – 2 + __
13
3i = 12.4.2.i3
or – 2 + 4,33i = 96(i2)i
= 96(– 1)i
= – 96i
___ ____ ___
2. 2.1 √−4 + √−16 – √81
__ ___ ___ ___
= √4 .√−1 + √16 .√−1 – 9
= 2i + 4i – 9
= 6i – 9 or – 9 + 6i
___ ____ ____ ___
2.2 − √−9 – √−25 + √−64 – √−1
__ ___ ___ ___ ___ ___
= − √9 √−1 – √25 √−1 + √64 .√−1 – i
= – 3i – 5i + 8i – i
= –i
___ ____
2.3 √−3 + √−48
__ ___ ____ ___
= √3 √−1 + √16.3 √−1
__ __
= √3 i + 4√3 i
__
= 5√3 i
____ ____ ____
2.4 √−12 + √−36 + √−28
___ ___ ___ ___ ___ ___
= √12 √−1 + √36 √−1 + √28 √−1
___ ___
= √4.3 i + 6i + √7.4 i
__ __
= 2√3 i + 6i + 2√7 i
__ __
= (2√3 + 6 + 2√7 )i

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 Module 2 • Complex numbers 27

____ _____
√−18
____
2.5 ___ + √−144
√−2
___ ___ ____ ___
√9.2 √−1
= ______
__ ___ + √144 .√−1
√2 .√−1
_
3 √_2 i
=_ + 12i
√2 i

= 3 + 12i

Activity 2.3 SB page 53

Number Rectangular Real Imaginary

form part (non-real) part

1. 1.1 – 5 + 3i – 5 + 3i –5 3i
1.2 π π + 0i π 0i
1.3 – 2i 0 – 2i 0 – 2i
1.4 3 – i9 3 – 9i 3 – 9i
3
____
1.5 4 + √−27 1 + 0i 1 0i
_____
1.6 5 – √−100 5 – 10i 5 – 10i
___
1.7 – 5√64 – 40 + 0i – 40 0i
1.8 – 8i – 1 – 1 – 8i –1 – 8i
1.9 51 51 + 0i 51 0i
____ ____
1.10 √– 81 − √−25 0 + 4i 0 4i

2. 2.1 i + 2 – 3i – 1 – 5i 2.2 (4 + i) – (4 – i)
= 1 – 7i =4+i–4+i
= 0 + 2i

2.3 (10 + 20i) – (– 11 – 12i) 2.4 (3 – 2i) – (4 – 3i) + (– 2 + 3i)

= 10 + 20i + 11 + 12i = 3 – 2i – 4 + 3i – 2 + 3i
= 21 + 32i = – 3 + 4i

2.5 (– 27 – 17i) – (– 19i) 2.6 (5 – 2i) – (3 – 4i) – (4 – i)i

= – 27 – 17i + 19i = 5 – 2i – 3 + 4i – 4i + i 2
= 2i – 27 = 5 – 2i – 3 + 4i – 4i + (– 1)
= – 27 + 2i = 1 – 2i

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 28 N4 Mathematics – Lecturer Guide

2.7 (4 – 5i) – (2i 4 – i 2)

= (4 – 5i ) – (2(i 2)2 – (– 1))
= (4 – 5i ) – (2(– 1)2 + 1)
= (4 – 5i ) – (2 + 1)
= 4 – 5i – 3
= 1 – 5i

2.8 30i 9 + 2i – 25i + 25i 2 + 50i 100

= 30(i 2)4i + 2i – 32i + 25(– 1) + 50(i 2)50
= 30(– 1)4i + 2i – 32i – 25 + 50(– 1)50
= 30i + 2i – 32i – 25 + 50
= 25 + 0i

3. 3.1 15 + 40i – (– 41 + 16i) 3.2 20i – 12 – (– 12 – 15i)

= 15 + 40i + 41 – 16i = 20i – 12 + 12 + 15i
= 56 + 24i = 0 + 35i

4. 4.1 – 5i(4 + 3i) 4.2 (1 + 3i)(5 + 5i)

= – 20i – 15i 2 = 5 + 5i + 15i + 15i 2
= – 20i – 15(– 1) = 5 + 20i + 15(– 1)
= – 20i + 15 = 5 + 20i – 15
= 15 – 20i = – 10 + 20i

4.3 (i – 1)(3 – 2i) 4.4 (3 + 2i)(2 – 4i)

= 3i – 2i 2 – 3 + 2i = 6 – 12i + 4i – 8i 2
= 3i – 2(– 1) – 3 + 2i = 6 – 8i – 8(– 1)
= 3i + 2 – 3 + 2i = 6 – 8i + 8
= – 1 + 5i = 14 – 8i

4.5 (3 + i)2 – (3 – i)2

= (3 + i)(3 + i) – (3 – i)(3 – i)
= 9 + 3i + 3i + i 2 – [9 – 3i – 3i + i 2]
= 9 + 6i + (– 1) – [9 – 6i + (– 1)]
= 9 + 6i – 1 – [9 – 6i – 1]
= 9 + 6i – 1 – [8 – 6i]
= 9 + 6i – 1 – 8 + 6i
= 0 + 12i
__ __
4.6 (2 + √5 i)(l – √5 i) 4.7 (_12 − _21 i)(1 – i)
__ __ ___
= 2 – 2√5 i + √5 i – √25 i 2 = _12 − _12 i – _12 i + _12 i 2
__
= 2 – √5 i – 5(– 1)
__ = _12 – i + _12 (– 1)
= 2 – √5 i + 5
__ =0–i
= 7 – √5 i

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 Module 2 • Complex numbers 29

4.8 2(3 – 4i)(i – 5) 4.9 (2 + i)2(3 – i)

= 2(3i – 15 – 4i 2 + 20i ) = (2 + i) (2 + i) (3 – i)
= 2[23i – 15 – 4(– 1)] = (4 + 2i + 2i + i 2)(3 – i)
= 2(23i – 15 + 4) = [4 + 4i + (– 1)](3 – i)
= 2(23i – 11) = (4 + 4i – 1)(3 – i)
= 46i – 22 = (3 + 4i)(3 – i)
= – 22 + 46i = 9 – 3i + 12i – 4i 2
= 9 + 9i – 4(– 1)
= 9 + 9i + 4
= 13 + 9i
4.10 (2 – i)(– 3 + 3i)(4 – 5i)
= (– 6 + 6i + 3i – 3i 2)(4 – 5i)
= (– 6 + 9i – 3(– 1))(4 – 5i)
= (– 3 + 9i)(4 – 5i)
= – 12 + 15i + 36i – 45i 2
= – 12 + 51i – 45(– 1)
= – 12 + 51i + 45
= 33 + 51i
5. I1 + I2 = I3
I2 = I3 – I1
= 25 + 11i – (18 – 13i)
= 25 + 11i – 18 + 13i
= 7 + 24i
6. 6.1 z1 + z2 – z3 6.2 (z1)(z3)
= 2 – 3i + (– 7i) – (– 4 + 2i) = (2 – 3i)(– 4 + 2i)
= 2 – 3i – 7i + 4 – 2i = – 8 + 4i + 12i – 6i 2
= 6 – 12i = – 8 + 16i – 6(– 1)
= – 8 + 16i + 6
= – 2 + 16i
6.3 2z2 – 2z1 6.4 z1.z2.z3
= 2(– 7i) – 2(2 – 3i) = (2 – 3i)(– 7i)(– 4 + 2i)
= – 14i – 4 + 6i = (– 14i + 21i 2)(– 4 + 2i)
= – 4 – 8i = [– 14i + 21(– 1)][– 4 + 2i]
= [– 14i – 21][– 4 + 2i]
= 56i – 28i 2 + 84 – 42i
= 14i – 28(– 1) + 84
= 14i + 28 + 84
= 14i + 112
= 112 + 14i

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 30 N4 Mathematics – Lecturer Guide

Activity 2.4 SB page 57

Complex number (z) Conjugate (z¯)

1. 1.1 – 7i 7i
1.2 – 3 + 3i – 3 – 3i
1.3 –i + 2 i+2
1.4 4i – 5 – 4i – 5
1.5 –5 – 5 – 0i
1.6 i –i
1.7 2i
– 6 − __
3 – 6 + __23 i
1.8 5 − 3i
____ _5 + __3 i
8 8 8
____
1.9 −√−64 8i
____
1.10 3i – √−25 2i

_
2. 2.1 z = – 2,3 – 2,5i 2.2 – z = – (– 2,3 + 2,5i)
= 2,3 – 2,5i
_ _
2.3 – z = – (– 2,3 – 2,5i) 2.4 z = – 2,3 + 2,5i
= 2,3 + 2,5i
_ _
3. 3.1 z1 + z2 – z3 + z 2 – z 3
= 2 + 5i + 3 – i – (– 2i – 4) + (3 + i) – (2i – 4)
= 2 + 5i + 3 – i + 2i + 4 + 3 + i – 2i + 4
= 16 + 5i
_ _ _
3.2 z 1.z3 3.3 z 2.z 3
= (2 – 5i)(– 2i – 4) = (3 + i)(2i – 4)
2
= – 4i – 8 + 10i + 20i = 6i – 12 + 2i 2 – 4i
= – 4i – 8 + 10(– 1) + 20i = 2i – 12 + 2(– 1)
= 16i – 8 – 10 = – 14 + 2i
= – 18 + 16i
_ _ _
3.4 z 1.z 3.z 4
= (2 – 5i)(2i – 4)(– 5i)
= (4i – 8 – 10i 2 + 20i)(– 5i)
= (24i – 8 – 10(– 1))(– 5i)
= (2 + 24i)(– 5i)
= – 10i – 120i 2
= – 10i – 120(– 1)
= 120 – 10i

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 Module 2 • Complex numbers 31

4. 4.1 (‾12 + 4i) + (14 + 3i) + (− 2 + 11i)

= 12
‾ + 4i + 14 + 3i + (− 2 + 11i)
‾
= 26 + 7i + (− 2 + 11i)
= 26 − 7i − 2 + 11i

4.2 (‾
= 24 + 4i
_ _ 2
− 1 + 2 √2 i) (1 + √2 i)
_ _ _
= (− 1 − 2 √2 i)(1 + √2 i)(1 + √2 i)
_ _ _
= (− 1 − 2 √2 i)(1 + √2 i + √2 i + 2 i 2)
_ _
= (− 1 − 2 √2 i)(1 + 2 √2 i − 2)
_ _
= (− 1 − 2 √2 i)(− 1 + 2 √2 i)
_ _
= 1 − 2 √2 i + 2 √2 i − 8 i 2
=1+8
= 9 + 0i
4.3 (−4 + 5i)(‾
−3 − i)(2 − 2i)
= (− 4 + 5i)(‾
− 6 + 6i − 2i + 2 i 2)

= (− 4 + 5i)(‾
= (− 4 + 5i)(‾
− 6 + 4i − 2)
− 8 + 4i)
= (− 4 + 5i)(− 8 − 4i)
= 32 + 16i − 40i − 20 i 2
= 32 − 24i + 20
= 52 − 24i

Activity 2.5 SB page 63

1. 1.1 _
−3 − 4i
− 12 1.2 _5i − _5i

=_
−3 _4i
− 12 – − 12 = _5i × _ii − _5i

= _14 + _13 i or 0,25 + 0,333i 5i _i

= __
2 − 5
i
5i
= ___ − _i
(−1) 5

= – 5i – _15 i

= – 5_15 i

= 0 – __
26
5 i or 0 – 5,2i

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 32 N4 Mathematics – Lecturer Guide

2 − 3i −3 − i
1.3 ____
i
1.4 ____
−3i
2 − 3i __
= ____
i
× −i
−i
−3 − i _
= ____
−3i
× 3i
3i
2 2
= ______
−2i + 3 i
2 = _____
−3i − i
2
−i −3i
−2i + 3(−1) −3i − (−1)
= ________
−(−1)
= _______
−3(−1)

= _____
−2i − 3
1 = _____
−3i + 1
3
−3i _
1
= – 3 – 2i = ___
3 +3

= – i + _13

= _13 – i
6
1.5 2i 8 − 3i 7 + __
4i
9
5i
2(i ) − 3(i ) i + 4(i 2)3
2 4 2 3
= ______________
2 4
5(i ) i
2(−1)4 − 3(−1)3i + 4(−1)3
= ________________
4
5(−1) i
2(1) − 3(−1)i + 4(−1)
= _____________
5(1)i

=_
2 + 3i − 4
5i

– 2 + 3i × _
=_ – 5i
5i – 5i
10i – 15 i
=_
2
– 25 i 2
10i – 15(– 1)
= _____________
– 25(– 1)
10i + 15
=_ 25

= _35 + __25 i or 0,6 + 0,4i

2. 2.1 z1 + z2 ÷ z3 2.2 z1.z¯2.z3

(−3 + 4i)
______ = (2 + 3i)(i – 3)(– 4i)
= 5 – 3i + i
= (2i – 6 + 3i 2 – 9i)(– 4i)
= 5 – 3i + (_____ −i )
−3 + 4i __
× −i
i = (– 7i – 6 + 3(– 1))(– 4i)
= 5 – 3i + (_____2 )
2
3i − 4i
−i
= (– 7i – 9)(– 4i)
= 5 – 3i + (_______) = 28i 2 + 36i
3i − 4(−1)
−(−1)
= – 28 + 36i
= 5 – 3i + (____
1 )
3i + 4

= 5 – 3i + 3i + 4
= 5 – 3i + 3i + 4
= 9 + 0i

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 Module 2 • Complex numbers 33

−i 3 − 5i
3. 3.1 ____
3 − 5i
3.2 ____
2+i

−i 3 + 5i 3 − 5i ___
= ____ × ____
3 − 5i 3 + 5i
= ____
2+i
× 22 −− ii
2 (3 − 5i)(2 − i)
= _____________
−3i − 5i
2 = _________
(2 + i)(2 − i)
9 + 15 i – 15 i − 25 i
− 3 i − 5(– 1) 2
=_
9 − 25 (−1)
= ___________
6 − 3i − 10i + 5 i
2
4−i
6 − 13i + 5(−1)
=_
− 3i + 5
9 + 25 = __________
4 − (−1)

=_
5 − 3i
34
1 − 13i
= _____
5

=_5 _3
34 – 34 i or = _15 – __
13
5 i or

= 0,147 – 0,088i = 0,2 – 2,6i

1 − 2i 1 − 7i
3.3 ____
i−5
3.4 _____
−2 − 4i
1 − 2i ____
= ____ × −5 − i
−5 + i −5 − i
1 − 7i
= _____ −2 + 4i
× _____
−2 − 4i −2 + 4i
(1 − 2i)(−5 − i) 2
= __________
(−5 + i)(−5 − i)
= ____________
−2 + 4i + 14i − 28i
2
4 − 16i
2
= ___________
− 5 − i + 10i + 2 i
2 = _________
−2 + 18i + 28
4 + 16
25 − i
− 5 + 9i + 2(−1)
___________ 26 + 18i
= 25 − (−1)
= ______
20

=_
− 5 + 9i − 2
26
= __
26 ___
18
20 + 20i

=_
− 7 + 9i
26
= __
13 __ 9
10 + 10 i

= −_
7
+_9
26 26
i = 1,3 + 0,9i

or – 0,269 + 0,346i
__
√2 + 5i__
______
3.5
−3i + √2
__
2 + 5i
√__
= _____
√2 − 3i
__ __
2 + 5i _____
√__
= _____ × √__2 + 3i
√2 − 3i √2 + 3i
__ __
2
2 + 3 √2 i + 5 √2 i + 15i
= _______________ 2
2 − 9i
__
2 + 8 √2 i − 15
= _________
2+9
__
= − __ + 8√2 i or
13 ___
11 11

= – 1,182 + 1,029i

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 34 N4 Mathematics – Lecturer Guide

4. 4.1 (3 – i)–2 4.2 _

2−i
2
(2 + i)
(2 − i)
= _____
1
2 =_
(2 + i.(2 + i)
(3 − i)

= ________
1
(3 − i)(3 − i)
=_
2−i
2
4 + 4i + i

= _______
1
2 =_ 2−i
4 + 4i − 1
9 − 6i + i

= ______
1
9 − 6i − 1
=_2−i 3 − 4i
× ____
3 + 4 i 3 − 4i
2
= ____
1 8 + 6i
× ____
8 − 6i 8 + 6i
=_
6 − 8i − 3i + 4i
2
9 − 16 i
8 + 6i
= ______ 2 =_
6 − 11 i − 4
9 + 16
64 − 36i
8 + 6i
= _____
64 + 36
=_
2 − 11 i
25

= ___
8 ___
6
100 + 100i =_
2 _
11
25 – 25 i or

= 0,08 + 0,06i = 0,08 – 0,44i

(3 − 2i)(−2 + i) (5 − 4i)(3 + i)
4.3 __________
3−i
4.4 _________
3
i
2 2
= ___________
−6 + 3i + 4i − 2i
3−i
= ___________
15 + 5i − 12i − 4i
2
(i )i
−6 + 7i − 2(−1)
= __________
3−i
= _______
15 − 7i + 4
(−1)i

−4 + 7i ___ 19 − 7i _i
= _____
3−i
× 33 ++ ii = _____
−i
×i
2 2
= ____________
−12 − 4i + 21i + 7i
2 = ______
19i − 7 i
2
9−i −i
19i − 7(−1)
= _________
−12 + 17i − 7
9+1 = _______
−(−1)

= – __
19 __
+ 17 i
10 10
= _____
19i + 7
1

= – 1,9 + 1,7i = 7 + 19i

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 Module 2 • Complex numbers 35

3
(2 + i)
2 − 3i ___
4.5 ____
4−i
– 21 +− ii 4.6 _________
(3 − 4i)(2 + i)

= (____ × 44 ++ ii ) – (___
2 + i 2 − i)
(2 + i)2
2 − 3i ___
4−i
1 − i ___
× 2−i = _____
3 − 4i

= (________ )–( )
2
_______
2 (2 + i)(2 + i)
8 − 10i − 3i
2
2 − 3i + i
2 = ________
3 − 4i
16 − i 4−i

= (_______ )–( 5 )
2
8 − 10i + 3 ______
2 − 3i − 1
= _______
4 + 4i + i
17 3 − 4i

= (______
17 ) – ( 5 )
11 − 10i 1 − 3i
____ = ______
4 + 4i − 1
3 − 4i

= __
11 __10 _1 _3
17 – 17 i – 5 + 5 i
3 + 4i ____
= ____ × 3 + 4i
3 − 4i 3 + 4i
2
= 0,447 + 0,012i = _________
9 + 24i + 16i
2
9 − 16i
9 + 24i + 16(−1)
= __________
9 − 16(−1)

−7 + 24i
= ______
25

= __
− 7 ___
24
25 + 25 i

= – 0,28 + 0,96i

2i 20 − i 19
4.7 ______
3i − 1
3i
4.8 _____2
(1 − i)
2 10 2 9
2(i ) − (i ) i 3i
= _________
3i − 1
= ________
(1 − i)(1 − i)
2(−1) 10 − (−1) 9i 3i
= __________
3i − 1
= _______
2
1 − 2i + i

= ____
2+i
3i − 1
3i
= ________
1 − 2i + (−1)

2+i − 1 − 3i 3i 2i
= _____
−1 + 3i
× _____
−1 − 3i
= ___ × __
−2i 2i
2 2
= __________
−2 − 6i − i − 3i
2 = ___
6i
2
1 − 9i −4i
6(−1)
= _______
−2 − 7i + 3
1+9 = _____
−4(−1)

1 − 7i
= ____
10 = __
−6
4

= __
1 ___
7
10 – 10i or = − _32 + 0i

= 0,1 – 0,7i

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 36 N4 Mathematics – Lecturer Guide

__ (2 − 2i)(3 + i) ____
√2 + 3i__ 4.10 _________ – 21++3ii
4.9 ______ −i + 2
−3i + √2
__ __ 2
2 + 3i _____ = __________
6 + 2i − 6i − 2i 2 + 3i
– ____
√__
= _____ × √2__ + 3i 2−i 1+i
√2 − 3i √2 + 3i
__ __ 6 − 4i − 2(−1)
2 + 3√2 i + 3 √2 i + 9 i 2
= _________ 2 + 3i
– ____
= ______________ 2
2−i 1+i
2 − 9i
= (___ × 82−−4ii ) – (____ × 11 −− ii )
__
2 + i ____ 2 + 3i ___
2 + 6√2 i + 9(−1)
= ___________
2 − 9(−1)
2+i 1+i

= (___________ )–( )
__ 2 2
2 + 6√2 i − 9
16 − 8i + 8i − 4i __________
2 − 2i + 3i − 3i
= ________ 11
2
4−i 2
1–i

= (____
5 )–( 2 )
_
− 7 + 6 √2 i
16 + 4 ______
2+i+3
=_ 11
__
6√2 = __
20 _ 5 _ 1
5 – 2 – 2i
= − __
7
+ ___
11 i 11

= __
15 _ 1
10 – 2 i

= 1_12 – _12 i
z ×z
5. zp = _
z + z + z3
1 2

1 2

(−3 + 2i)(−2 + 5i)

zp = _____________
(−3 + 2i) + (−2 + 5i)
+ (– 2 – 3i)
2
= ____________
6 – 15i − 4i + 10i
−3 + 2i − 2 + 5i
+ (– 2 – 3i)
6 − 19i + 10(−1)
= __________
−5 + 7i
+ (– 2 – 3i)
−4 − 19i _____
−5 − 7i
= ______
−5 + 7i
× −5 − 7i
+ (– 2 – 3i)
2
= ___________
20 + 123i + 133i
2 + (– 2 – 3i)
25 − 49i
20 + 123i + 133(−1)
= _____________
25 − 49(−1)
+ (– 2 – 3i)
–113 + 123i
= _______
74 + (– 2 – 3i)

=_
–113 ___
123i
74 + 74 – 2 – 3i

= –3,53 – 1,34i

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 Module 2 • Complex numbers 37

Activity 2.6 SB page 68

1. Im
z6 = 4i + 2
4

3 z5 = 3i
z2 = – 2 + 2i
2

Re
–4 –3 –2 –1 1 2 3 4

z3 = – 2 – i –1

–2 z1 = 4 – 2i
z4 = – 2i
–3

–4

2. Im

4
z2 = 3i _
3 z5 = 3i + 4
_
z3 = –3 + 2i _
2 z6 = 12 + 32i

1 _
z1 = 4 + i
z4 = i
Re
–4 –3 –2 –1 1 2 3 4
z4 = –i _
–1 z1 = 4 – i
–2 z6= 1 – 3i
2 2
z3 = –3 – 2i _
z2 = –3i
–3 z5 = 4 – 3i

–4

Activity 2.7 SB page 80

1. 1.1 z = – 2 + 3i 1.2 z = – 2
1.3 z = – i – 3 1.4 z = 2 – 4i
1.5 z = i + 6 1.6 z = – 5i

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 38 N4 Mathematics – Lecturer Guide

2. 2.1 z = 5 + 3i _________
r(mod) = √(5)2 + (3)2 Im
______
= √25 + 9
___
= √34 3
z = 5 + 3i
= 5,831
1
θ(arg) = tan–1(_ab ) 5,
83
)=
= tan–1 (_35 ) mo
d
r(
= 30,964° θ(arg) = 30,964°
Re
r |__
θ = 5,831 |_______
30,964° 5

2.2 z = 4i – 2
Im
∴ z = – 2 +__________
4i
r(mod) = √(−2)2 + (4)2 z = – 2 + 4i
______ 4
= √4 + 16
___ r (m
= √20 od
)=
= 4,472 4,
47
θ(arg) = 116,565°
θ(arg) = 180° – tan ( 2 )
–1 _4 2

= 180° – 63,435° ∞
Re
= 116,565° –2

r |__
θ = 4,472 |________
116,565°

2.3 z = – 5
Im
∴ z = – 5 +__________
0i
r(mod) = √(−5)2 + (0)2
___
= √25
=5
θ(arg) = 180° – tan–1 (_05 ) θ(arg) = 180°
z = – 5 + 0i
= 180° – 0° Re
= 180° – 5 r (mod) = 5
r |__
θ = 5 |____
180°

2.4 z = – 3i Im
∴ z = 0 – 3i__________
r(mod) = √(0)2 + (−3)2
__
= √9 θ(arg) = 270°
=3
Re
θ(arg) = 270° r (mod) = 3
r |__
θ = 3 |____
270° –3 z = 0 – 3i

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 Module 2 • Complex numbers 39

3
__
2.5 z = 3√8 – 5i
∴ z = 6 – 5i
__________ Im
r(mod) = √(6)2 + (−5)2 θ(arg) = 320,194°
_______
= √36 + 25 6
__ Re
___ ∞ 2
= √61 r (m √3
od
= 7,81 )=
7,
4 quadrant: θ(arg) = 360° – tan–1 (_56 )
th 81

= 360° – 39,806°
–5
= 320,194° z = 6 – 5i
∴ r |__
θ = 7,81 |________
320,194°
____________
__ __
√3
– 1 __
3. 3.1 r(mod) = √(−1) + (− √3 )
2 2
3.2 θ(arg) = 180° + tan 1
_____
= √1 + 3 = 180° + 60°
=2 = 240°
3.3 2 |____
240° 3.4 Im

θ(arg) = 240°
Re
–1
2
=
od)
m
r(
__
__ – √3
z = – 1 – √3 i

___________
4. 4.1 z = – 2 – 4i 4.2 r(mod) = √(−2)2 + (−4)2
___
= √20
= 4,472
4.3 θ(arg) = 180° + tan–1 (_42 ) 4.4 Negative argument = – 116,565°
= 243,435°

4.5 4,472 |________

243,435°

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 40 N4 Mathematics – Lecturer Guide

Activity 2.8 SB page 83

__
1. 1.1 √2 |___
45°
__
45°
= √2 (cos 45° + i sin 45°)
__
= √2 (__
1__
+ i.__
1__
) __
√2 √2 √2
=1+i 1

45°
1

1.2 2 cis 30°

= 2(cos 30° + i sin 30°) 30°
__
√3
= 2 (__ _1
2 + i. 2 )
__ 2 __
= √3 + i √3

60°
1
__
1.3 √3 (cos 60° + i sin 60°) 1.4 45°
|___
__ __
√3
= √3 (_12 + i.__ = 1(cos 45° + i sin 45°)
__ 2)
√3
= __ _3 = 1(__
1__ __
+ 1__ i)
2 + 2i √2 √2

= __
1__ __
+ 1__ i or
√2 √2
__ __
√2 √2
= __ + __
2 i 2
__
√3
1.5 __
2 cis 60°
__
√3
= __
2 (cos 60° + i sin 60°)
__ __
√3 _ √3
= __ 1 __
2 ( 2 + i. 2 )
__
√3
= __ _3
4 + 4i

2. 2.1 4 |____
135° 2.2 90°
|___
= 4(cos 135° + i sin 135°) = 1(cos 90° + i sin 90°)
= 4(– 0,707 + i.0,707) =0+i
= – 2,828 + 2,828i
__
2.3 2 |______
− 300° 2.4 √2 (cos 140° + i sin 140°)
__
= 2(cos(– 300°) + i sin(– 300°)) = √2 (– 0,766 + i.0,643)
= 2(0,5 + i.0,866) = – 1,083 + 0,909i
= 1 + 1,732i

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 Module 2 • Complex numbers 41

2.5 3,4 cis 330° 2.6 5 cis (– 220°)

= 3,4(cos 330° + i sin 330°) = 5(cos(– 220°) + i sin(– 220°))
= 3,4(0,866 + i (– 0,5)) = 5(0,766 + i (0,643))
= 2,944 – 1,7i = 3,83 + 3,215i

Activity 2.9 SB page 91

1. z1 = 6 cis 90° ; z 2 = 2 cis 30°

z1 = 6(cos 90° + i sin 90°)
= 0 + 6i
z2 = 2(cos 30° + i sin 30°)
_
= 2(____
2 + i. 2 )
√3 __1
_
= √3 + i
_
∴ z1 + z2 = 0 + 6i + √3 + i
_
= √3 + 7i
2. 30° × 2 |___
2.1 8 |___ 60° 2.2 (4 |___ 30° )( __12 |_____
50° )(6 |___ − 25° )
= 8 × 2 |________
30° + 60° = 4 × 6 × __12 |_____________
50° + 30° − 25°
= 16 |___
90°
55°
= 12 |___
_
√3 |______
86,34° 6,8 cis 40°
2.3 _______
_ 2.4 ________
4,2 cis – 130°
√2 |______
40,44°
_
6,8 |___
40°
√_3
= ____ | 86,34° − 40,44°
_____________ = _______
4,2 | − 130°
√2 ______
6,8
= 1,225 |_____
45,9° = ___
4,2 |____________
40° − (−130° )

= 1,619 |____
170°
2 cis 55° × 6,2 cis 67,3°
2.5 ________________
0,2 cis 22° × 4 cis (–220°)
55° × 6,2 |_____
2 |___ 67,3°
= _____________
0,2 |___
22° × 4 | − 220°
______
2 × 6,2 |_________
55° + 67,3°
= _____________
0,2 × 4 | 22° − 220°
_________
12,4
= _____
0,8 |______________
122,3° − (− 198°)
= 15,5 |______
320,3°
_
3. 3.1 √3 (cos 30° + i sin 30°) 3.2 45°
|___
_ _
= √3 (____
2 + i. 2 )
√3 __1 = 1(cos 45° + i sin 45°)
= 1(____
1_
+ i.____
1_
)
_
√3
= __32 + ____
2 i
√2 √2 _ _
√2 √2
= ____
1_ ____
+ 1_ i or ____2+ ____
2 i
√2 √2

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 42 N4 Mathematics – Lecturer Guide

_ _
3.3 √3 |___
60° + √2 |___
45°
_ _
= √3 (cos 60° + i sin 60°) + √2 (____ + 1_ i)
1_ ____
_
√2 √2
_ _
= √3 (__12 + ____
2 i) + √2 ( √2 + √2 i)
√3 ____
1_ ____
1_

_ _ _ _
√3 √9 2 2
= ____ ____
2 + 2 i+
____
√_
+ ____
√_
i
_
√2 √2
√3
= ____
2 + __3 i + 1 + i
2
_
= (____
2 + 1) + ( 2 + 1)i
√3 __3
_
√3 + 2 __
= ________
2 + 5i 2

3.4 3 |___
45° .2 cis −15° 3.5 ________
9 cis 90°
__1
3 cis 30°
= 3 × 2 |________
45° − 15°
= __9__1 |________
90° − 30°
= 6 |___
30° 3

= 6(cos 30° + i sin 30°) = 27 |___

60°
_
= 6(____
2 + i. 2 )
√3 __1 = 27(cos 60° + i sin 60°)
_
= 27(__12 + i.____
2)
_ √3
= 3 √3 + 3i
_
= ___ + 27 √3 i
27 _______
2 2

4. 4.1 2 cis 30° − 3 cis 70°

= 2(cos 30° + i sin 30°) − 3(cos 70° + i sin 70°)
= 1,732 + i − 1,026 − 2,819i
= 0,706 − 1,819i
4.2 0,59 cis 314,27° × 0,74 cis 16,3°
= 0,59 × 0,74 |_____________
314,27° + 16,3°
= 0,437 |_______
330,57°
= 0,437(cos 330,57° + i sin 330,57°)
= 0,381 − 0,215i
8 |___
45° .2 |___
60°
4.3 ________
3 − 75°
| _____
45° + 60°
8 × 2 |________
___________
= 3 _____
| − 75°
16 |____
105°
= ______
3 | − 75°
_____

= ___
16
3 |____________
105° − (− 75°)
180°
= 5,333 |____
= 5,333(cos 180° + i sin 180°)
= − 5,333 + 0i

4.4 ___________
2 cis 40° .3 cis 60°
3 | 20° .2 − 35°
___
2 |___
40° .3 |___
60°
= __________
3 | 20° .2 | − 35°
___ _____

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 Module 2 • Complex numbers 43

2 × 3 |________
40° + 60°
= ____________
3 × 2 |___________
20° + (− 35°)
6 |____
100°
= _____
6 | − 15°
_____

= |____________
100° − (− 15°)

= |____
115°
= (cos 115° + i sin 115°)
= − 0,423 + 0,906i
3 |____
130° .4 |___40°
4.5 __________
6 | 20° .8 | 80°
___ ___
3 × 4 |_________
130° + 40°
= ___________
6 × 8 | 20° + 80°
________
12 |____
170°
= ______
48 | 100°
____

= ___
12
| 170° − 100°
48 __________
70°
= 0,25 |___
= 0,25(cos 70° + i sin 70°)
= 0,086 + 0,235i

5. _________
2+i
10 cis 20°
_ Im
√5 |_______
26,565°
= ________
10 |___
20°
√5
_
1
z=2+i
= ____
10 26,565° − 20°
= 0,224 |______
6,565°
______________________ )
r(mod) = √(2) 2 + (− 1)2 od
_ r (m
= √5
θ(arg)
θ(arg) = tan −1(__12 ) Re
2
= 26,565°
_
∴ r |__θ = √5 |_______
26,565°

6. (−3 − i)(−2 + i)
(−3 − i) _________________________ (−2 + i) ______________________
r(mod) = √(− 3)2 + (− 1)2 r(mod) = √(− 2)2 + (1)2
_ _
= √9 + 1 = √4 + 1
_ _
= √10 = √5
= 3,162 = 2,236
θ(arg) = 180° + tan −1(__13 ) θ(arg) = 180° − tan −1(__21 )
= 198,435° = 153,435°

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 44 N4 Mathematics – Lecturer Guide

∴ r |__θ .r |__θ Im
= 3,162 . 2,236 |_________________
198,435° + 153,435°
z = –2 + i
= 7,071 |_______
351,78° 1

Re
–3 –2 –1

–1
z = –3 – i

Activity 2.10 SB page 97

1. 1.1 (3 |___
50° )4 1.2 (2,5 cis 60,3°)5
= 3 4 |______
50° × 4 = (2,5 |_____
60,3° )5
200°
= 81 |____ = 2,5 5 |________
60,3° × 5
= 97,656 |______
301,5°
_ (5 |___
85° ) 3

1.3 (√4,2 |_____

− 80° )
3
1.4 ______
_ (2 ___
| 20° )2
3
= (√4,2 ) |________
− 80° × 3 5 3 |______
85° × 3
= ________
2
2 ______
| 20° × 2
− 240°
= 8,607 |______
125 |____
255°
= _______
4 | 40°
___

= ____
125
| 255° − 40°
4 _________
= 31,25 215°
3
4 cis 45° × (3 cis 60°)
1.5 ____________________
2
(2 cis −50°)

= ____________________
4 cis 45° × 27 cis 180°
4 cis −100°
4 × 27 |_________
45° + 180°
= ____________
4 | − 100°
______

= ____
108
4 |_____________
225° − (− 100°)
325°
= 27 |____

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 Module 2 • Complex numbers 45

2. z1 = 21 cis 120°
z2 = 3 cis 80°
z3 = |______
− 108°
(z1) (z2)
3
________ (21 |____
120° ) (3 |___
80° )3
= ___________
(z3) ( |______
− 108° ) 2
2

(21 |____
120° )(27 |____
240° )
= ______________
| − 216° ______
21 × 27 |__________
120° + 240°
= _____________
| − 216°
______
567 |____
360°
= _______
| − 216°
______

= 567 |__________
360° + 216°
576°
= 567 |____
216°
= 567 |____

3.1 (_____
25° )
× (_____
190° )
4 −3
6 |___
50° 2 |___
60°
3. 3 |___ 4 |____
3.2 [2(cos 30° + i sin 30°)] 3
= (2 |___
30° )3
= (_____
25° )
× (_____
60° )
4 3
6 |___
50° 4 |____
190°
3 |___ 2 |___ = 2 3 |______
30° × 3
= (2 |___
25° )4 × (2 |____
130° )3 = 8 |___
90°
25° × 4 )(2 3 |_______
= (2 4 |______ 130° × 3 )
100° )(8 |____
= (16 |____ 390° )
100° + 390°
= 16 × 8 |__________
490°
= 128 |____
130°
= 128 |____

3.3 (________
2 cis 35° )
× (_________
4 cis 120° )
3 −2
3 cis 78° 3 cis 25°

= (________
2 cis 35° )
× (_________
3 cis 25° )
3 2
3 cis 78° 4 cis 120°

= (________) × ( 2| )
3 3 |______
78° × 3
________ 4 2 |_______
120° × 2
3
|2 ______
35° × 3 3 ______
25° × 2
27 |____
234° 16 |____
240°
= ______
8 | 105°
× ______
9 | 50°
____ ___

= (___
27
| 234° − 105° ) × (___
8 __________
16
| 240° − 50° )
9 _________

= (___
27
| 129° )(___
8 ____
16
| 190° )
9 ____

8 )( 9 ) __________
= (___
27 ___
16
| 129° + 190°

= 6 |____
319°

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 46 N4 Mathematics – Lecturer Guide

(¯ z2) 2
z1. ¯
4. 4.1 _____
z
× z3
4
2
(‾ − 105° )
__1
| 35° . 5 |______ _
3 ___
_____________
‾
= × √2 |___
45°
4 |____
210°
2
( 3 × 5 _________
35° − 105° )
__
1
‾ | _
= _____________
‾
× √2 |___
45°
4 |____
210°

( 3 |_____
− 70° )
2
__5 _
= _______
‾
× √2 |___
45°
4 |____
210°

( 3 |___
70° )
2
__5 _
‾
= _______ × √2 |___
45°
4 |______
− 210°

( 3 ) |______
2
__52 × 70° _
= _________ × √2 |___
4 |______
− 210°
45°
___
25
140°
|____ _
= ______
9
4 | − 210°
× √2 |___
45°
______
___
25 _
= ___
9
| 140° + 210° × √2 |___
4 __________ 45°
_
= ___
25
| 350° × √2 |___
36 ____
45°
_
= ___
25
× 2 | 350° + 45°
36 √ _________
_
25 √2
= _______ | 395°
36 ____
_
25 √2
= _______
36
35°
|___

(z1 + z3)
4

4.2 _________
_
z ÷ __
1
z 2 4
_
( 3 |___ 45° )
4
__1 35° + √2 |___

‾
= _____________ ÷ _____1
4 | 210°
| 5 ______
− 105° ____
_
[ 3(cos 35° + i sin 35°) + √2 (cos 45° + i sin 45°)]
4
__1

‾
= _____________________________ ÷ _____
1
4 |____
210°
5 |______
− 105°
_ _
[( 3 cos 35° + √2 cos 45°) + i( 3 sin 35° + √2 sin 45°)]
4
__1 __1

‾
= ________________________________ ÷ _____ 1
4 |____
210°
5 |______
− 105°

[1,273 + 1,191i] 4
= __________
5 | 105°
× 4 |____
210°
____

(1,743 |_______
43,097° ) 4

= ____________
5 | 105°
× 4 |____
210°
____
(1,743)4 |__________
4 × 43,097°
= ______________
5 | 105°
× 4 |____
210°
____
9,239 |________
172,390°
= __________
5 | 105°
× 4 |____
210°
____

9,239
= ______
5 |_____________
172,390° − 105° × 4 |____
210°
210°
= 1,848 67,390° × 4 |____
= 1,848 × 4 |____________
67,390° + 210°
= 7,391 |________
277,390°

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 Module 2 • Complex numbers 47

Activity 2.11 SB page 106

1. 1.1 x – yi = – 3 + i 1.2 – 3 + 2i = x + 4yi

∴ x = – 3; – y = 1 ∴ – 3 = x; 2 = 4y
y = –1 ∴ x = – 3 and y = _12
or (– 3; – 1) or (−3; _12 )

1.3 3x – 2i = – yi 1.4 – 2x – 8yi = – 10 + 16i

∴ 3x = 0; –2 = –y ∴ – 2x = – 10; – 8y = 16
∴ x = 0 and y = 2 ∴ x = 5 and y = – 2
∴ (0; 2) ∴ (5; – 2)
1.5 3x + 2yi – 5 = 4 + 6i
(3x – 5) + 2yi = 4 + 6i
∴ 3x – 5 = 4; 2y = 6
3x = 9; y=3
x=3
∴ (3; 3)
1.6 2x + 3 + i(y + 5) = x – y + ix + iy
(2x + 3) + i(y + 5) = (x – y) + i(x + y)
∴ 2x + 3 = x – y; y+5=x+y
x + y = – 3; x=5
∴ x + y = – 3 … substitute x = 5
5 + y = –3
y = –8 and x = 5
∴ (5; – 8)
1.7 4 + 5i = x + yi – (1 + i)
= x + yi – 1 – i
4 + 5i = (x – 1) + i(y – 1)
∴ 4 = x – 1; 5=y–1
∴ x = 5 and y = 6
∴ (5; 6)
1.8 (3 – 2i)2 = x – yi
(3 – 2i)(3 – 2i) = x – yi
9 – 12i + 4i 2 = x – yi
9 – 12i + 4(– 1) = x – yi
5 – 12i = x – yi
∴ 5 = x; – 12 = – y
∴ x = 5 and y = 12
∴ (5; 12)

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 48 N4 Mathematics – Lecturer Guide

1.9 (i + 1)2 + (3 + i)i = x + y + 4yi

(i + 1)(i + 1) + 3i + i 2 = (x + y) + 4yi
i 2 + 2i + 1 + 3i + i 2 = (x + y) + 4yi
(– 1) + 5i + 1 + (– 1) = (x + y) + 4yi
5i – 1 = (x + y) + 4yi
∴ 5 = 4y; –1 = x + y
y = _54 ; x = –1 – y
∴ x = – 1 – _54 … substitute y = _54
∴ x = – _94
∴ y = _54 or 1_14 and x = – _49 or – 2_14
∴ (− _94 ; _54 )
2. 2.1 i(x – iy) = i(y – i9) – 3x – i
xi – yi 2 = yi – 9i 2 – 3x – i
xi – y(– 1) = yi – 9(– 1) – 3x – i
xi + y = yi + 9 – 3x – i
xi + y = (9 – 3x) + i(y – 1)
∴x=y–1
y = 9 – 3x … (1)
y=x+1 … (2)
(2) = (1): x + 1 = 9 – 3x
4x = 8
x=2
∴y=x+1 … (2)
y=2+1
y=3
∴ (2; 3)
2.2 (1 + i)(x – iy) = (2 + 3i)2
x – iy + ix – i 2y = (2 + 3i)(2 + 3i)
x – iy + ix – (– 1)y = 4 + 12i + 9i 2
x – iy + ix + y = 4 + 12i + 9(– 1)
(x + y) + i(– y + x) = – 5 + 12i
∴ x + y = – 5 … (1); – y + x = 12 … (2)
∴ (1) + (2): 2x = 7
x = _72 or 3_12
∴ x + y = –5
_7 + y = – 5
2
y = – 5 – _72
∴ y = − __ 17 _1
2 or − 8 2
∴ (3_12 ; −8_12 )

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 Module 2 • Complex numbers 49

1+i
2.3 (5 – 2i)(x + yi) = ___
1−i
1 + i ___
5x + 5yi – 2xi – 2yi 2 = ___ × 1+i
1−i 1+i
2
5x + 5yi – 2xi – 2y(– 1) = _______
1 + 2i + i
2
1−i
1 + 2i + (−1)
________
(5x + 2y) + i(5y – 2x) =
1 − i2
2i
__
= 2
∴ (5x + 2y) + i(5y – 2x) = 0 + i
∴ 5x + 2y = 0; 5y – 2x = 1
5x = – 2y
x = − _25 y
∴ 5y – 2(− _25 ) y = 1
5y + _45 y = 1
__
29
y=1
5
y = __
5
29 or 0,172
∴ x = − _25(__
29 )
5

x = – __
2
29
= – 0,69
∴ (– 0,69; 0,172) or (– __
29 ; 29 )
2 __ 5

2y − i
2 + i ____
___
2.4 3−i
= x
2y
3+i
× 2 + i = __ – _1 i
3 + i ___
___
3−i x x

_______ 2y 2
6 + 5i + i
2 = __ – _1 i
x x
9−i
2y
6 + 5i + (−1) __
________
9 − (−1)
= – _1 i x x
2y
10 = – _1 i
5 + 5i __
____
x x
2y _
_1 + _1 i = __
2 2 – 1ix x
2y
∴ _12 = __
x;
_1 = _1
2 x

∴x=2
∴ 2y = _12 . 2

y = _12

∴ (2; _12 )

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 50 N4 Mathematics – Lecturer Guide

(3 + 5i)(2 − 5i)
3. 3.1 a + bi
(3 – 4i)2 = ____
2 3.2 (a + bi) = __________
1 − 3i
i
a + bi
____
(3 – 4i) (3 – 4i) = (−1) a + bi = ____________
6 − 15i + 10i − 25 i 2

1 − 3i
9 – 24i + 16i 2 = – a – bi 6 − 5i − 25(−1)
= __________
1 − 3i
9 – 24i + 16(– 1) = – a – bi
– 7 – 24i = – a – bi = _______
6 − 5i + 25
1 − 3i
∴ – 7 = – a; – 24 = – b 31 − 5i ____
= _____ × 11 ++ 3i
3i
1 − 3i
∴ a = 7 and b = 24 2
= ____________
31 + 93i − 5i − 15 i
2
1−9i
31 + 88i − 15(−1)
___________
= 1 − 9(−1)

a + bi = __
46 __88
10 + 10 i
∴ a = __
46
10 and b = __
88
10
a = 4,6 and b = 8,8
5
3.3 a – bi = ____
5−i
1+i
5 − (i 2)2i
= ______
1+i
5 − (−1)2i
= ______
1+i
5 − i ___
= ___ × 1−i
1+i 1−i
2
= _______
5 − 6i + i
2
1−i
5 − 6i + (−1)
________
= 1 − (−1)

= _42 – _62 i
∴ a – bi = 2 – 3i
a = 2; – b = – 3
b=3
2 − 3i ____
3.4 a + bi = ____
1+i
+ 11 −+ 2i
2i

a + bi = (___ × 21−+3ii ) + (____

1 − i ____
1−i 1 + 2i 1 − 2i )
1 − 2i ____
× 1 − 2i

a + bi = (________ )+( )
2 2
2 − 5i + 3 i ________
1 − 4i + 4i
2 2
1−i 1 − 4i

a + bi = (_________
1 − (−1) ) ( 1 − 4(−1) )
2 − 5i + 3(−1) 1 − 4i + 4(−1)
+ _________

2 − 2 i + (− 5 − 5 i)
a + bi = __
−1 _ 5 _3 _4

a + bi = − _12 − _52 i − _35 − _45 i

a + bi = − __
11 __33
10 – 10 i

∴ a = − __
11
10 and b = − __
33
10
= – 1,1 = – 3,3

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 Module 2 • Complex numbers 51

4. x + y + xi – yi = 8,944 63,436°
RHS: 8,944 63,436° = r(cos θ + i sin θ)
= 8,944(cos 63,436° + i sin 63,436°)
= 8,944(0,447 + i 0,894)
= 3,998 + 8,044i
≈ 4 + 8i
LHS: x + y + xi – yi = (x + y) + (x – y)i
∴ (x + y) + (x – y)i = 4 + 8i
x+y=4 … (1)
x–y=8 … (2)
2x = 12 … (3)
∴x=6
Substitute x = 6 into (1):
6+y=4
∴ y = –2
∴ (x; y) = (6 ; – 2)

5. |a +i b |
2a − b = − 5 − 7i
2
(a + b)(2) − (2a − b)(i) = − 5 − 7i
(2a + 2b) − (2a − b)i = − 5 − 7i
2a + 2b = − 5 and −(2a − b) = − 7
2a + 2(2a − 7) = − 5 b = 2a − 7
2a + 4a − 14 = − 5 = 2(_32 ) − 7
6a − 14 = − 5 =3−7
6a = 9 = –4
a = _32
a = _32 ; b = − 4

Summative assessment: Module 2 SB page 107

1. 1.1 12i 2 + 7 + i 11 + (2i )3 + 4i

= 12(– 1) + 7 + (i 2)5i + 8(i 2)i + 4i
= – 12 + 7 + (– 1)5i + 8(– 1)i + 4i
= – 12 + 7 – i – 8i + 4i
= – 5 – 5i (3)

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 52 N4 Mathematics – Lecturer Guide

____ ___
1.2 (5 – √−50 ) – (3 – √−8 )
____ ___ __ ___
= (5 – √25.2 . √−1 ) – (3 – √8 . √−1 )
__ ___
= 5 – 5√2 i – 3 + √4.2 i
__ __
= 5 – 5√2 i – 3 + 2√2 i
__
= 2 – 3√2 i or 2 – 4,243i (3)
___ _____ ___
1.3 √−9 – √−169 + √49
__ ___ ____ ___
= √9 . √−1 – √169 . √−1 + 7
= 3i – 13i + 7
= 7 – 10i (2)
−3 + 6i
2. z = _____
3
∴ z = – 1 + 2i
2.1 z¯ = – 1 – 2i
−3 − 6i
or z¯ = _____
3 (2)
2.2 and 2.5
Im

z = – 1 + 2i
2
r (m
od
).2
,2

θ(arg) = 116,565°
36

Re
–1

_ –2
z = – 1 – 2i
(3)
__________
2.3 r (mod) = √(−1)2 + (2)2
_____
= √1 + 4
__
= √5
= 2,236 (2)
2.4 θ(arg) = 180° – tan–1(_21 )
= 116,565° (2)
2.5 On Argand diagram (2)
2.6 z = 2,236 |________
116,565° (2)

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 Module 2 • Complex numbers 53

3. 3.1 (2 – 3i)(4 – 5i)

= 8 – 10i – 12i + 15i 2
= 8 – 22i + 15(– 1)
= 8 – 22i – 15
= – 7 – 22i (3)
−4 + 3i
3.2 _____
−i + 3
− 4 + 3i ___
= _____
3−i
× 33 ++ ii
2
= ____________
−12 − 4i + 9i + 3 i
2
9−i
−12 + 5i + 3(−1)
= ___________
9 − (−1)

= ________
−12 + 5i − 3
10
−15 + 5i
= ______
10

= − _32 + _12 i or – 1,5 + 0,5i (4)

__
4. √2 (cos 60° + i sin 60°)
__ __
30°
= √2 (_12 + i.__
2)
√3
__ __
√2 √6
= __ + __
2 i 2
2 __
√3

60°
1 (3)
5. 5.1 2 cis 120° + 3 cis 65°
= 2(cos 120° + i sin 120°) + 3(cos 65° + i sin 65°)
= 2(– 0,5 + i.0,866) + 3(0,423 + i.0,906)
= – 1 + 1,732i + 1,268 + 2,72i
= 0,268 + 4,452i (3)

5.2 ________
5__cis 145°
√2 cis −45°

=_
5_
145° − (−45°)
√2

= 3,536 190° or 3,536 − 170° (2)

_ V
6. VT = V 2
+V 3 1

3 + j2
_____________
= 2 + j3 + 3 + j
3 + (– 1)
= 5___________
( )
– j2 j + j

= ____________
2
5 – (– 1)j + j
5 – 2j
=_ 2 _
5 + 2j × 5 – 2j

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 54 N4 Mathematics – Lecturer Guide

10 – 4j
=_
25 – 4 j 2
10 – 4j
= ___________
25 – 4(– 1)
10 – 4j
=_
29

_–_
= 10 4
29 29 j
____________________
r = √(10
| |
29) (29)
_ 2+ _4 2
________________________
= √0,119 + 0,019
= 0,371

θ = tan –1(29
_)
_
4
_
10
29
–1
= tan 0,4
= 21,8°
∴ z = 0,371 cis 21,8° (5)
TOTAL: [40]

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 MATHEMATICS

MODULE

3 Sketch graphs
After they have completed this module, students should be able to:
• distinguish between a dependent and an independent variable;
• define a domain and range;
• state the difference and distinguish between functions and relations;
• identify the relevant functions that relate to their graphs;
• identify points of symmetry with reference to an axis or the lines y = ± x;
• calculate inverse functions and relations; and
• draw neat sketch graphs of the following functions/relations:
– ax + by + c = 0 _ _
– x + y = r , y = ± √r − x , x = ± √r 2 − y 2
2 2 2 2 2

– xy = c
2
x
_
2 y
– ±_=1
a2 b2
– y = k a nx, y = k e nx, y = k log a(nx) and y = k log e(nx):
– y = k a nx, with a > 0 and a and n positive integers
– y = k a nx, with 0 < a < 1 rational and n a positive integer
– y = k log e(nx), with n a positive integer
– y = k log a(nx), with a > 1 and a and n positive integers
– y = k log a(nx), with 0 < a < 1 rational and n a positive integer
– y = ax 2 + bx + c.

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 56 N4 Mathematics – Lecturer Guide

Introduction
A graph is a visual representation of the relations between certain variable quantities, or
the connections that exist between a set of points plotted with reference to a set of axes,
namely the x- and y-axes.

In this module we will focus on equations and graphs of the following relations:

1 Straight line ax + by + c = 0
_ _
2 Circle x 2 + y 2 = r 2; y = ± √r 2 − x 2 ; x = ± √r 2 − y 2
2
3 Ellipse _
x2 _
y
2 ± 2 = 1
a b

4 Rectangular hyperbola xy = c
5 Exponential y = k a nx; y = k e nx; y = k log a(nx) and y = k log e(nx)
6 Logarithmic y = a x 2 + bx + c
7 Parabola y = a x 3 + b x 2 + cx + d

Students need the following pre-knowledge to successfully complete this module.

Pre-knowledge
Cartesian plane
In previous years students learnt about the Cartesian plane as a two-dimensional
plane that is formed by the intersection of two perpendicular lines. The horizontal
line is known as the x-axis, and the vertical line is known as the y-axis.

The coordinates (x; y) on the Cartesian plane is an ordered number pair

representing a point. The horizontal distance of the point from the origin is x and
the vertical distance is y.
y
3 A(2; 3)

1 C(0; 1)

E(0; 0) B(3; 0)
0 x
–3 –2 –1 1 2 3

–1

–2
D(–3; –2)

–3

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 Module 3 • Sketch graphs 57

y = mx + c
m = gradient (slope) c = y-intercept

x- and y-intercept method:

Step 1: Find the x-intercept: let y = 0.
Step 2: Find the y-intercept: let x = 0.
Step 3: Plot the two points on a system of axes and draw the straight line graph.

Gradient (m):
y y

m = (+)
x x

m = (–1)

Gradient is positive: m > 0 Gradient is negative: m < 0

y y

m = (∞)
m = (0)
x x

There is no slope or gradient: m = 0 If m = ∞ the straight line is parallel to

the y-axis
y2 − y1
Gradient: m = _
x − x = tan θ
2 1

Parallel lines: m1 = m2
Perpendicular lines: m1.m2 = − 1

All the exponential laws as well as:

• a 0 = 1, for example 2 0 = 1

• _
1n
= a −n, for example _12 = 3 −2
a 3

• _
1−n
= a n, for example _ 2
−2 = 4 , and so on.
1
a 4

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 58 N4 Mathematics – Lecturer Guide

Common logarithm: log 100 • Base 10

Natural logarithm: log ex or ln x • Base e ≈ 2,718

Definition of a logarithm
If x = a y • Exponential form
then y = log ax • Logarithmic form; x > 0; a > 0 and a ≠ 1

For example: 4 3 = 64
∴ log 464 = 3
_

Quadratic formula: x = ___________

− b ± √b − 4ac 2

2a

Note
If a < 0, then there is a maximum turning point (∩) and if a > 0, then there is a
minimum turning point (∪).

Activity 3.1 SB page 116

1. Domain: {x: x ∈ ℝ} y 2. Domain: {x: –r ≤ x ≤ r, y

x ∈ ℝ} (0; r)
Domain: (–∞; ∞) Domain: [–r; r]

x 0 x
0 (–r; 0) (r; 0)
Range: {y, y ∈ ℝ} Range: {y: –r ≤ y ≤ r,
y ∈ ℝ} (0; –r)
Range: (–∞; ∞) Range: [–r; r]

3. Domain: {x: –r ≤ x ≤ r, 4. Domain: {x: –r ≤ x ≤ r, y

y
x ∈ ℝ} x ∈ ℝ}
(0; r)
Domain: [–r; r] Domain: [–r; r]

0 x
0 x (–r; 0) (r; 0)
Range: {y: 0 ≤ y ≤ r, (–r; 0) (r; 0) Range: {y: –r ≤ y ≤ 0,
y ∈ ℝ} y ∈ ℝ} (0; –r)
Range: [0; r] Range: [–r; 0]

5. Domain: {x: –r ≤ x ≤ 0, y 6. Domain: {x: x ∈ℝ, y

x ∈ℝ} x ≠ 0}
Domain: [–r; 0] (0; r)
Domain: (–∞; 0) ∪ (0; ∞)

(–r; 0) 0 x
0 x
Range: {y: –r ≤ y ≤ r, Range: {y: y ∈ℝ,
y ∈ℝ} y ≠ 0}
(0; –r)
Range: [–r; r] Range: (–∞; 0) ∪ (0; ∞)

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 Module 3 • Sketch graphs 59

7. Domain: {x: –a ≤ x ≤ a, y 8. Domain: {x: –a ≤ x ≤ a, y

x ∈ℝ} x ∈ℝ}
(0; b)
Domain: [–a; a] (0; b) Domain: [–a; a]

0 x
(–a; 0) (a; 0) 0 x
(–a; 0) (a; 0)
Range: {y: –b ≤ y ≤ b, (0; –b) Range: {y: –b ≤ y ≤ b,
y ∈ℝ} y ∈ℝ}
Range: [–b; b] Range: [–b; b] (0; –b)

9. Domain: {x: x > 0, y 10. Domain: {x: x ∈ℝ} y

x ∈ℝ}
Domain: (0; ∞) Domain: (–∞; ∞)

0 x
Range: {y: y ∈ℝ} Range: {y: y ∈ℝ} 0 x
Range: (–∞; ∞) Range: (–∞; ∞)

Activity 3.2 SB page 122

1. Function 2. Non-function
3. Function 4. Non-function
5. Function 6. Non-function
7. Non-function 8. Function
9. Function 10. Function

Activity 3.3 SB page 130

___
(2
1. 1.1 x-axis: ( –3; −6√2 ) 1.2 y = x: 13
_
; 2)
13
y = –x: (−_2
; −2)
y y

__
(–3; 6√2 ) y = –x y=x
5
13
(__
2
; 2)
y=0
0 x 0 x
10 –5 5
13
(–__
2
; –2)

__ –5
(–3; –6√2 ) ______
x = – √81 – y 2
xy = 13

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___ ___ ___

3√15
1.3 x-axis: (−_72 ; ____
8 ) 2. 2.1 x-axis: ( −3; −√10 ) ↔ ( −3; √10 )
___ ___
3√15
___
( 3; −√10 ) ↔ ( 3; √10 )
y-axis: (_72 ; − ____
8 ) ___ ___
y-axis: ( −3; −√10 ) ↔ ( 3; −√10 )
___ ___
( −3; √10 ) ↔ ( 3; √10 )
y y
2 2
5 x=0 –x – y + 19 = 0
y2
x 2 + __ x=0
__
=1 4 5
16 9 ___ ___
3 (–3; √10 ) (3; √10 )
___
3√15
(– _27 ; ____ ) 2
8
1
y=0 y=0
0 x 0 x
–5 –4 –3 –2 –1 1 2 3 4 5 –5 5
___
–1
___
3√15
(– _72 ; – ____
8
) –2 3√15
( _72 ; – ____ ) ___ ___
8
–3 (–3; –√10 ) (3; –√10 )
–4 –5
–5

__ __
2.2 x-axis: ( −8; −3√5 ) ↔ ( −8; 3√5 )
__ __
2.3 y = x: ( 52; −10 ) ↔ ( −10; 52 )
_ _

y-axis: ( −8; −3√5 ) ↔ ( 8; −3√5 ) y = –x: ( 52; −10 ) ↔ ( 10; –52 )

_ _

y y
2 2
16y - 9x = 144
y = –x y=x
3 x=0
y = – 4x _

10 y = _34x 10
__
(–8; 3√5 ) (–10; _52 )
y=0
0 x 0 x
–10 10 –10 10
__
(–8; –3√5 )
__ (10; – _52 )
(8; –3√5 )
–10 –10 (_25 ; –10)

–xy – 25 = 0

______
3. 3.1 y – √75 – x2 = 0
________
y – √75 – (–x2 ) = 0
______
y – √75 – x2 = 0
Since the equation is equivalent, the graph is symmetrical about the y-axis
The tests for symmetry about the x-axis, y = x and y = –x do not yield an
equivalent equation, therefore are not lines of symmetry.

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 Module 3 • Sketch graphs 61

3.2 3x2 + y2 = 2
3x2 + (–y2) = 2
3x2 + y2 = 2
Since the equation is equivalent, the graph is symmetrical about the x-axis.
3x2 + y2 = 2
3(–x2) + y2 = 2
3x2 + y2 = 2
Since the equation is equivalent, the graph is symmetrical about the y-axis.
The test for symmetry about the line y = x and the line y = –x do not yield an
equivalent equation, therefore are not lines of symmetry.
3.3 xy – 9 = 0
(y)(x) – 9 = 0
yx – 9 = 0
xy – 9 = 0
Since the equation is equivalent, the graph is symmetrical about the y = x.
xy – 9 = 0
(–y)(–x) – 9 = 0
yx – 9 = 0
xy – 9 = 0
Since the equation is equivalent, the graph is symmetrical about the y = –x.
The tests for symmetry about the x-axis and y-axis do not yield an equivalent
equation, therefore are not lines of symmetry.
2
x
3.4 _
5
– y2 = 1
2
x
_
5
– (–y)2 = 1
2
x
_
5
– y2 = 1

Since the equation is equivalent, the graph is symmetrical about the x-axis.
2
x
_
5
– y2 = 1
(–x)2
_
5
– y2 = 1
2
x
_
5
– y2 = 1

Since the equation is equivalent, the graph is symmetrical about the y-axis.
The tests for symmetry about the line x = y and line y = –x do not yield an
equivalent equation, therefore are not lines of symmetry.

Activity 3.4 SB page 134

1. Continuous 2. Continuous 3. Discontinuous

4. Continuous 5. Continuous 6. Continuous
7. Continuous 8. Discontinuous 9. Continuous
10. Continuous

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Activity 3.5 SB page 140

1. 1.1 No 1.2 Yes 1.3 No 1.4 Yes

2. 2.1 f(x) = 3x − 2 2.2 xy = 4
−1
f (x) = 3y − 2 ∴ y = 4_x
∴ x = 3y − 2
f −1(x) = 4_y
3y = x + 2
∴ x = 4_y
y = 1_3 x + 3_2
y = 4_x

2.3 x 2 + y 2 = 49 2.4 y = 3x
_____________
y = ± √49 − x 2 f −1(x) = 3 y
_____________
∴ x = 3y
∴ f −1(x) = ± √49 − y 2
_____________ y = log 3x
x = ± √49 − y 2
∴ x 2 = 49 − y 2
x 2 + y 2 = 49
2
y
_
2
2.5 4
+_
x
=125
2.6 y = − 3 x 2
___________
2.7 x = − y 2.8 x = √16 − y 2

Activity 3.6 SB page 147

1. a) i) y b) i) y = _2x
y = 2x

0 x

ii) (–∞; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

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 Module 3 • Sketch graphs 63

2. a) i) y b) i) x = –2

0 x
(0; –2)
y = –2

ii) (–∞; ∞) ii) (–2)

iii) (–2) iii) (–∞; ∞)
iv) Function iv) Non-function
v) Continuous v) Continuous

3. a) i) y b) i) y=4
x=4

(4; 0)
0 x

ii) (4) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (4)
iv) Non-function iv) Function
v) Continuous v) Continuous

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4. a) i) y b) i) y=x–3

10

(0; 3)

0 x
–10 (–3; 0) 10

y=x+3 –10

ii) (–∞; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

5. a) i) y b) i) y = –2x + 4

y = – _21x + 2 10

(0; 2)
0 x
–10 (4; 0) 10

–10

ii) (–∞; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

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 Module 3 • Sketch graphs 65

6. a) i) y b) i) y = _32 x + 6

10
–2x + 3y + 12 = 0

(6; 0) x
0
–10 10
(0; –4)

–10

ii) (–∞; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

7. a) i) y b) i) y = –_54 x – _
15
4

4x + 5y + 15 = 0 10

0 x
15
–10 (– _ ; 0) 10
4
(0; –3)

–10

ii) (–∞; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

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8. a) i) y b) i) y = _54 x – 5
y
x+ _ = 1
–_
5 4

10

(0; 4)

0 x
–10 (0; –5) 10

–10

ii) (–∞; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

9. a) i) y b) i) y = –_13 x + 2

10

(0; 6)

(2; 0)
0 x
–10 10

–10 x + _y = 1
_
2 6

ii) (–∞; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

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10. a) i) y b) i) y = _32 x + _72

10

(_72 ; 0)
0 x
–10 7 10
(0; – 3 )
_

–10
2 3
_
y+1
=_
x–2

ii) (–∞; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

11. a) i) y b) i) y = –_13 x – _
11
3

x
+4
_
= – _13
y–1 10

11
(– _
3
; 0)
0 x
–10 10

–10
(0; –11)

ii) (–∞; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

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12. a) i) y b) i) y = _12 x – 5

(0; 10)
10

(–5; 0)
0 x
–10 10

–10
2 1
_
y–4
–_
x+3
=0

ii) (–∞; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

13. a) i) y b) i) y = –x + 7
1 1
_
5–x
–_
2–y
=0

10

(0; 7)

0 x
–10 (7; 0) 10

–10

ii) (–∞; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

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 Module 3 • Sketch graphs 69

14. a) i) y b) i) y = –x

0 x

y = –x

ii) (–∞; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

Activity 3.7 SB page 151

1. a) i) y b) i) x2 + y2 = 16

5
(0; 4)

(–4; 0) (4; 0)
0 x
–5 5

x 2 + y 2 = 16
–5 (0; –4)

ii) [–4; 4] ii) [–4; 4]

iii) [–4; 4] iii) [–4; 4]
iv) Non-function iv) Non-function
v) Continuous v) Continuous

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2. a) i) y b) i) x2 + y2 = 23

_
(0; √23 ) 5
x 2 + y 2 = 23
_ _
(– √23 ; 0) ( √23 ; 0)
0 x
–5 5

_
–5 (0; – √23 )

_ _ _ _
ii) [−√23 ;√23 ] ii) [−√23 ;√23 ]
_ _ _ _
iii) [−√23 ;√23 ] iii) [−√23 ;√23 ]
iv) Non-function iv) Non-function
v) Continuous v) Continuous

3. a) i) y b) i) x2 + y2 = 49

(0; 7)

0 x
(–7; 0) –5 5 (7; 0)

x 2 – 49 = – y 2 –5

(0; –7)

ii) [–7; 7] ii) [–7; 7]

iii) [–7; 7] iii) [–7; 7]
iv) Non-function iv) Non-function
v) Continuous v) Continuous

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 Module 3 • Sketch graphs 71

4. a) i) y b) i) x2 + y2 = 19

_
(0; √19 ) 5
– x 2 – y 2 + 19 = 0

_ _
(– √19 ; 0) ( √19 ; 0)
0 x
–5 5

_
–5 (0; – √19 )

_ _ _ _
ii) [−√19 ;√19 ] ii) [–√19 ;√19 ]
_ _ _ _
iii) [−√19 ;√19 ] iii) [−√19 ;√19 ]
iv) Non-function iv) Non-function
v) Continuous v) Continuous

5. a) i) y b) i) x2 + y2 = 25

(0; 5)

0 x
(–5; 0) (5; 0)

4x 2 + 4y 2 = 100
(0; –5)

ii) [–5; 5] ii) [–5; 5]

iii) [–5; 5] iii) [–5; 5]
iv) Non-function iv) Non-function
v) Continuous v) Continuous

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6. a) i) y b) i) x2 + y2 = 45
_
(0; 3√5 )
–5x 2 + 225 –5y 2 = 0
5

_ 0 _x
(–3√5 ; 0) –5 5 (3√5 ; 0)

–5
_
(0; –3√5 )

_ _ _ _
ii) [−3√5 ; 3√5 ] ii) [−3√5 ;3√5 ]
_ _ _ _
iii) [−3√5 ; 3√5 ] iii) [−3√5 ; 3√5 ]
iv) Non-function iv) Non-function
v) Continuous v) Continuous

7. a) i) y b) i) x2 + y2 = 144

(0; 12)

10

0 x
(–12; 0) –10 10 (12; 0)

y2
x2 + _ –10
_
3 3
= 48
(0; –12)

ii) [–12; 12] ii) [–12; 12]

iii) [–12; 12] iii) [–12; 12]
iv) Non-function iv) Non-function
v) Continuous v) Continuous

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 Module 3 • Sketch graphs 73

8. a) i) y b) i) x2 + y2 = 7

5
_
y2
x2 + _
(0; √7 ) _
=1
7 7
_ _
(– √7 ; 0) ( √7 ; 0)
0 x
–5 5
_
(0; – √7 )

–5

_ _ _ _
ii) [−√7 ; √7 ] ii) [−√7 ; √7 ]
_ _ _ _
iii) [−√7 ; √7 ] iii) [−√7 ; √7 ]
iv) Non-function iv) Non-function
v) Continuous v) Continuous

9. a) i) y b) i) x2 + y2 = 36

(0; 6)

0 x
(–6; 0) –5 5 (6; 0)

_
x = ±√36 – y 2 –5

(0; –6)

ii) [–6; 6] ii) [–6; 6]

iii) [–6; 6] iii) [–6; 6]
iv) Non-function iv) Non-function
v) Continuous v) Continuous

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10. a) i) y b) i) x2 + y2 = 43
_
(0; √43 ) _
y = ±√43 – x 2
5

_ 0 _x
(–√43 ; 0) –5 5 (√43 ; 0)

–5
_
(0; –√43 )

_ _ _ _
ii) [−√43 ; √43 ] ii) [−√43 ; √43 ]
_ _ _ _
iii) [−√43 ; √43 ] iii) [−√43 ; √43 ]
iv) Non-function iv) Non-function
v) Continuous v) Continuous

Activity 3.8 SB page 155

_
1. a) i) y b) i) y = –√81 − x 2

(0; 9)

0 x
(–9; 0) 10

(0; –9) _
x = – √81 – y 2

ii) [–9; 0] ii) [–9; 9]

iii) [–9; 9] iii) [–9; 0]
iv) Non-function iv) Function
v) Continuous v) Continuous

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 Module 3 • Sketch graphs 75

_
2. a) i) y b) i) x = –√27 − y 2
_
y + √27 – x 2 = 0

_ 0 _ x
(–3√3 ; 0) (3√3 ; 0)

_
(0; –3√3 )

_ _ _
ii) [−3√3 ; 3√3 ] ii) [−3√3 ; 0]
_ _ _
iii) [−3√3 ; 0] iii) [–3√3 ; 3√3 ]
iv) Function iv) Non-function
v) Continuous v) Continuous
_
3. a) i) y b) i) x = +√100 − y 2

(0; 10)

0 x
(–10; 0) (10; 0)

_
y = +√100 – x 2
–10

ii) [–10; 10] ii) [0; 10]

iii) [0; 10] iii) [–10; 10]
iv) Function iv) Non-function
v) Continuous v) Continuous

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_
4. a) i) y b) i) y = +√32 − x 2
_
(0; 4√2 )
5
_
x – √32 – y 2 = 0

0 _x
–5 5 (4√2 ; 0)

–5
_
(0; –4√2 )

_ _ _
ii) [0; 4√2 ] ii) [–4√2 ; 4√2 ]
_ _ _
iii) [−4√2 ; 4√2 ] iii) [0; 4√2 ]
iv) Non-function iv) Function
v) Continuous v) Continuous
_
5. a) i) y b) i) y = +√169 − x 2

(0; 13)
10
_
x = +√169 – y 2

0 x
–10 10 (13; 0)

–10

(0; –13)

ii) [0; 13] ii) [–13; 13]

iii) [–13; 13] iii) [0; 13]
iv) Non-function iv) Function
v) Continuous v) Continuous

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 Module 3 • Sketch graphs 77

_
6. a) i) y b) i) x = +√75 − y 2

_
(0; 5√3 ) 10

_ _
(– 5√3 ; 0) (5√3 ; 0)
0 x
–10 10
_
y – √75 – x 2 = 0
–10

_ _ _
ii) [–5√3 ; 5√3 ] ii) [0; 5√3 ];
_ _ _
iii) [0; 5√3 ] iii) [–5√3 ; 5√3 ]
iv) Function iv) Non-function
v) Continuous v) Continuous
_

√ 25 − y
441
7. a) i) y b) i) x=– _ 2

√
441
y=– _
25
– x2 5

21 21
(– _
5
; 0) (_
5
; 0)
0 x
–5 5

21
–5 (0; – _
5
)

[ 5 5] [ 5
ii) 21 _
–_ ; 21 ii) 21
−_ ; 0]

iii) [−_
21
5
; 0] iii) [–_
5 5]
21 _
; 21

iv) Function iv) Non-function

v) Continuous v) Continuous

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_
8. a) i) y b) i) y = –√52 − x 2
_
(0; 2√13 )

5
_
x + √52 – y 2 = 0

_ 0 x
(–2√13 ; 0) –5 5

–5
_
(0; –2√13 )

_ _ _
ii) [−2√13 ; 0] ii) [–2√13 ; 2√13 ]
_ _ _
iii) [–2√13 ; 2√13 ] iii) [−2√13 ; 0]
iv) Non-function iv) Function
v) Continuous v) Continuous
_
9. a) i) y b) i) y = –√256 − x 2

(0; 16)

10 ______________
x + – √(16 + y)(16 – y)

0 x
(–16; 0) –10 10

–10

(0; –16)

ii) [–16; 0] ii) [–16; 16]

iii) [–16; 16] iii) [–16; 0]
iv) Non-function iv) Function
v) Continuous v) Continuous

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 Module 3 • Sketch graphs 79

_
10. a) i) y b) i) x = –√72 − y 2

________________
_ _
y + √(x + 6√2 )(6√2 – x) = 0
5

_ 0 _x
(– 6√2 ; 0) –5 5 (6√2 ; 0)

–5

_
(0; – 6√2 )

_ _ _
ii) [–6√2 ; 6√2 ] ii) [−6√2 ; 0]
_ _ _
iii) [−6√2 ; 0] iii) [–6√2 ; 6√2 ]
iv) Function iv) Non-function
v) Continuous v) Continuous

Activity 3.9 SB page 163

x 2 y2
1. a) i) y b) i) _
2 + _2 = 1
(3) (4)

5
y2
x2 + _
_
16
=1
9 4
(0; 3)
3

1
(–4; 0) (4; 0)
0 x
–5 –4 –3 –2 –1 1 2 3 4 5
–1

–2

–3
(0; –3)
–4

–5

ii) [–4; 4] ii) [–3; 3]

iii) [–3; 3] iii) [–4; 4]
iv) Non-function iv) Non-function
v) Continuous v) Continuous

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_____________
2. a) i) y b) i) x = 4_5 √25 – y 2

(0; 4) _
y = __45 √25 – x 2

0 x
(–5; 0) (5; 0)

ii) (–5; 5) ii) (0; 4)

iii) (4; 0) iii) (–5; 5)
iv) Function iv) Non-function
v) Continuous v) Continuous

x 2 y2
3. a) i) y b) i) _
2 +
_
2 = 1
(5) (2)
(0; 5)
25x 2 + 4y 2 = 100 5

1
(–2; 0) (2; 0)
0 x
–5 –4 –3 –2 –1 1 2 3 4 5
–1

–2

–3

–4

–5
(0; –5)

ii) [–2; 2] ii) [–5; 5]

iii) [–5; 5] iii) [–2; 2]
iv) Non-function iv) Non-function
v) Continuous v) Continuous

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 Module 3 • Sketch graphs 81

x 2 y2
4. a) i) y b) i) _
2 + _2 = 1
(2) (3)

4
4x 2 + 9y 2 = 36 (0; 2)
3

1
(–3; 0) (3; 0)
0 x
–4 –3 –2 –1 1 2 3 4
–1

–2

–3
(0; –2)
–4

ii) [–3; 3] ii) [–2; 2]

iii) [–2; 2] iii) [–3; 3]
iv) Non-function iv) Non-function
v) Continuous v) Continuous

x_ 2 y2
5. a) i) b) i) _
+_
_ 2 =1
y (√2 ) 2
(√3 )
2 _

_
(0; √2 )
3x + y = 2
2 2

1,2

0,8
_ _
0,4
√ √ ; 0)
2 2
(– _
3
; 0) ( _
3
0 x
–1,6 –1,2 –0,8 –0,4 0,4 0,8 1,2 1,6
–0,4

–0,8

–1,2
_
(0; –√2 )

_ _ _ _
[– √_ √_3 ]
2 2
ii) _
3
; _
ii) [−√2 ; √2 ]
_ _
iii) [–
√ √3 ]
2 2
iii) [–√2 ; √2 ] _
3
; _

iv) Non-function iv) Non-function

v) Continuous v) Continuous

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 82 N4 Mathematics – Lecturer Guide

x_ 2 y2
6. a) i) y b) i) _
+_ _ =1
(√6 ) 2 (√7 ) 2
_
(0; √6 )
2 6x 2 + 7y 2 = 42

_ 0 _ x
(–√7 ; 0) –2 –1 1 2 (√7 ; 0)

–1

–2
_
(0; –√6 )

_ _ _ _
ii) [–√7 ; √7 ] ii) [–√6 ; √6 ]
_ _ _ _
iii) [–√6 ; √6 ] iii) [–√7 ; √7 ]
iv) Non-function iv) Non-function
v) Continuous v) Continuous

x 2 y2
7. a) i) y b) i) _
+_ =1
(5) (3)
2
1
_ 1 2_

0,4

0,3
9x 2 + 25y 2 = 1
(0; _15 )
0,2

0,1
(– _13 ; 0) (_13 ; 0)
0 x
–0,4 –0,3 –0,2 –0,1 0,1 0,2 0,3 0,4
–0,1

–0,2
(0; – _15 )
–0,3

–0,4

[ 3 3] [ 5 5]
ii) − _1 ; _1 ii) − _1 ; _1

iii) [− _15 ; _15 ] iii) [− _13 ; _13 ]

iv) Non-function iv) Non-function

v) Continuous v) Continuous

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 Module 3 • Sketch graphs 83

x_2 y2
8. a) i) y b) i) _
(√3 ) 2
+_ =1
( √3 )
1_ 2
_
_
(0; √3 )
1 1 2
_
3
x2 + _
27
y = _19
1

1_
1_
(– _ ; 0) (_ ; 0)
√3 √3
0 x
–1 1

–1

_
(0; –√3 )

_ _
[
ii) −_ ; 1_ ];
1_ _
ii) [−√3 ; √3 ]
√3 √3
_ _
iii) [− _
√3 √3 ]
1_ _
iii) [−√3 ; √3 ] ; 1_

iv) Non-function iv) Non-function

v) Continuous v) Continuous
_____________
9. a) i) y b) i) x = 8_5 √25 – y 2

0 x
(–5; 0) (5; 0)

_
y = __85 √25 – x 2

(–8; 0)

ii) (–5; 5) ii) (–8; 0)

iii) (–8; 0) iii) (–5; 5)
iv) Function iv) Non-function
v) Continuous v) Continuous

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x 2 y2
10. a) i) y b) i) _
2 + _2 = 1
(4) (2)
(0; 4)
4
y 2 = 16 – 4x 2
3

1
(–2; 0) (2; 0)
0 x
–4 –3 –2 –1 1 2 3 4
–1

–2

–3

–4
(0; –4)

ii) [–2; 2] ii) [–4; 4]

iii) [–4; 4] iii) [–2; 2]
iv) Non-function iv) Non-function
v) Continuous v) Continuous

Activity 3.10 SB page 170

1. a) i) y b) i) xy = 13

xy = 13 5 _ _
(√13 ; √13 )

0 x
–5 5
_ _
(–√13 ; –√13 )
–5

ii) (–∞; 0) ∪ (0; ∞) ii) (–∞; 0) ∪ (0; ∞)

iii) (–∞; 0) ∪ (0; ∞) iii) (–∞; 0) ∪ (0; ∞)
iv) Function iv) Function
v) Discontinuous v) Discontinuous

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 Module 3 • Sketch graphs 85

2. a) i) y b) i) xy = –8

_ _ 5 xy = –8
(–2√2 ; 2√2 )

5
0 x
–5
_ _
(2√2 ; –2√2 )
–5

ii) (–∞; 0) ∪ (0; ∞) ii) (–∞; 0) ∪ (0; ∞)

iii) (–∞; 0) ∪ (0; ∞) iii) (–∞; 0) ∪ (0; ∞)
iv) Function iv) Function
v) Discontinuous v) Discontinuous

3. a) i) y b) i) xy = 10

10
y=_x 5 _ _
(√10 ; √10 )

0 x
–5 5

_ _
(–√10 ; –√10 )
–5

ii) (–∞; 0) ∪ (0; ∞) ii) (–∞; 0) ∪ (0; ∞)

iii) (–∞; 0) ∪ (0; ∞) iii) (–∞; 0) ∪ (0; ∞)
iv) Function iv) Function
v) Discontinuous v) Discontinuous

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 86 N4 Mathematics – Lecturer Guide

4. a) i) y b) i) xy = –6

x = – _6y
_ _ 5
(–√6 ; √6 )

0 x
–5 5
_ _
(√6 ; –√6 )
–5

ii) (–∞; 0) ∪ (0; ∞); ii) (–∞; 0) ∪ (0; ∞)

iii) (–∞; 0) ∪ (0; ∞) iii) (–∞; 0) ∪ (0; ∞)
iv) Function iv) Function
v) Discontinuous v) Discontinuous

5. a) i) y b) i) xy = 9

xy – 9 = 0

5
(3; 3)

0 x
–5 5

(–3; –3)
–5

ii) (–∞; 0) ∪ (0; ∞) ii) (–∞; 0) ∪ (0; ∞)

iii) (–∞; 0) ∪ (0; ∞) iii) (–∞; 0) ∪ (0; ∞)
iv) Function iv) Function
v) Discontinuous v) Discontinuous

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 Module 3 • Sketch graphs 87

6. a) i) y b) i) xy = –25

–xy – 25 = 0

10
(–5; 5)

0 x
–10 10

(5; –5)
–10

ii) (–∞; 0) ∪ (0; ∞) ii) (–∞; 0) ∪ (0; ∞)

iii) (–∞; 0) ∪ (0; ∞) iii) (–∞; 0) ∪ (0; ∞)
iv) Function iv) Function
v) Discontinuous v) Discontinuous

7. a) i) y b) i) x = 6_y

y – _6x = 0

0 x

ii) (–∞; 0) ∪ (0; ∞) ii) (–∞; 0) ∪ (0; ∞)

iii) (–∞; 0) ∪ (0; ∞) iii) (–∞; 0) ∪ (0; ∞)
iv) Function iv) Function
v) Discontinuous v) Discontinuous

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 88 N4 Mathematics – Lecturer Guide

8. a) i) y b) i) x + 5_y = 0

0 x

y + _5x = 0

ii) (–∞; 0) ∪ (0; ∞) ii) (–∞; 0) ∪ (0; ∞)

iii) (–∞; 0) ∪ (0; ∞) iii) (–∞; 0) ∪ (0; ∞)
iv) Function iv) Function
v) Discontinuous v) Discontinuous

9. a) i) y b) i) xy = 36

y
_
4
– _9x = 0

10
(6; 6)

0 x
–10 10

(–6; –6)
–10

ii) (–∞; 0) ∪ (0; ∞) ii) (–∞; 0) ∪ (0; ∞)

iii) (–∞; 0) ∪ (0; ∞) iii) (–∞; 0) ∪ (0; ∞)
iv) Function iv) Function
v) Discontinuous v) Discontinuous

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 Module 3 • Sketch graphs 89

10. a) i) y b) i) xy = –21

7
_
y
= – _3x

_ _ 10
(–√21 ; √21 )

0 x
–10 10
_ _
(√21 –√21 )
–10

ii) (–∞; 0) ∪ (0; ∞) ii) (–∞; 0) ∪ (0; ∞)

iii) (–∞; 0) ∪ (0; ∞) iii) (–∞;0) ∪ (0; ∞)
iv) Function iv) Function
v) Discontinuous v) Discontinuous

Activity 3.11 SB page 175

x 2 y2
1. a) i) y b) i) –_ 2 +
_
2 = 1
(2) (7)

y2
x2 – _
_
49
=14
10
y = – _27x y = _27x
(7; 0)
(0; 2)

0 x
–10 10
(0; –2)
(–7; 0)

–10

ii) (–∞; –7] ∪ [7; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (–∞; –7] ∪ [7; ∞)
iv) Non-function iv) Non-function
v) Discontinuous v) Discontinuous

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 90 N4 Mathematics – Lecturer Guide

x 2 y2
2. a) i) y b) i) _
2 – _2 = 1
(5) (3)

y = – _53x y = _53x

x +_=1
2 y2 10
–_
9 25

(0; 5)
(–3; 0) (3; 0)
0 x
–10 10
(0; –5)

–10

ii) (–∞; ∞) ii) (–∞; –5] ∪ [5; ∞)

iii) (–∞; –5] ∪ [5; ∞) iii) (–∞; ∞)
iv) Non-function iv) Non-function
v) Discontinuous v) Discontinuous

x 2 y2
3. a) i) y b) i) _
2 –
_
2 = 1
(1) (8)

x =1
y2 – _
2

64

10

y = – _18x (8; 0) y = _18x

(0; 1)
0 x
–20 20
(0; –1)
(–8; 0)

–10

ii) (–∞; ∞) ii) (–∞; –1] ∪ [1; ∞)

iii) (–∞; –1] ∪ [1; ∞) iii) (–∞; ∞)
iv) Non-function iv) Non-function
v) Discontinuous v) Discontinuous

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 Module 3 • Sketch graphs 91

x 2 y2
4. a) i) y b) i) –_ 2 +
__ =1
( )2
(1) √5
x2 –
_
y2 = 1
5

1_ 5 1_
y = –_ x y=_ x
√5 _ √5
(√5 ; 0)
(0; 1)

0 x
–5 5
_ (0; –1)
(–√5 ; 0)

–5

_ _
ii) (−∞; −√5 ] ∪ [√5 ; ∞) ii) (–∞; ∞)
_ _
iii) (–∞; ∞) iii) [−∞; −√5 ] ∪ [√5 ; ∞)
iv) Non-function iv) Non-function
v) Discontinuous v) Discontinuous

x 2 y2
5. a) i) y b) i) _
2 –
_
2 = 1
(3) (4)
16y – 9x = 144
2 2

y = – _34x 10 y = _34x
(0; 3)

(–4; 0) (4; 0)
0 x
–10 10

(0; –3)
–10

ii) (–∞; ∞) ii) (–∞; –3] ∪ [3; ∞)

iii) (–∞; –3] ∪ [3; ∞) iii) (–∞; ∞)
iv) Non-function iv) Non-function
v) Discontinuous v) Discontinuous

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x 2 y2
6. a) i) b) i) –_ 2 +
_
2 = 1
(6) (5)
y
36y 2 – 25x 2 = 900

y = – _65x y = _65x

10
(0; 6)

(–5; 0) (5; 0)
0 x
–10 10

(0; –6)
–10

ii) (–∞; –5] ∪ [5; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (–∞; –5] ∪ [5; ∞)
iv) Non-function iv) Non-function
v) Discontinuous v) Discontinuous

x 2 y2
7. a) i) y b) i) _
2 –
_
2 = 1
(3) (3)
–x 2 + y 2 – 9 = 0

y = –x 10 y=x
(0; 3)

(–3; 0) (3; 0)
0 x
–10 10

(0; –3)
–10

ii) (–∞;∞) ii) (–∞;–3] ∪ [3;∞)

iii) (–∞;–3] ∪ [3;∞) iii) (–∞;∞)
iv) Non-function iv) Non-function
v) Discontinuous v) Discontinuous

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 Module 3 • Sketch graphs 93

x 2 y2
8. a) i) y b) i) –__ 2 + _2 =1
_
( )
( √3 )
2
_ √2
–x + 3y = –2
2 2

1_ 1_
y = –_ x y=_ x
√3 √3
_

√)
2
(0; _
3

_ 0 _ x
(–√2 ; 0) (√2 ; 0)
_

√)
2
(0; – _
3

_ _
ii) (−∞; −√2 ] ∪ [√2 ; ∞) ii) (–∞; ∞)
_ _
iii) (–∞; ∞) iii) (−∞; −√2 ] ∪ [√2 ; ∞)
iv) Non-function iv) Non-function
v) Discontinuous v) Discontinuous

x 2 y2
9. a) i) y b) i) _
2 –
_
2 = 1
(3)
1
_ (1)

9y 2 – x 2 = 1

y = – _13x y = _13x
(0; _13 )

(–1; 0) (1; 0) x
0
–1 1
(0; – _31 )

–1

ii) (–∞; ∞) ii) (−∞; −_13 ] ∪ [_13 ; ∞)

iii) (−∞; −_13 ] ∪ [_31 ; ∞) iii) (–∞;∞)
iv) Non-function iv) Non-function
v) Discontinuous v) Discontinuous

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 94 N4 Mathematics – Lecturer Guide

x 2 y2
10. a) i) y b) i) –_ +_ =1
(5) (4)
1 2
_ 1 2 _

16x 2 – 25y 2 = 1

y = –_45x y = _45x
1

(0; _15 )

(– _14 ; 0) (_14 ; 0)
0 x
–1 1

(0; – _51 )

–1

ii) (−∞; −_14 ] ∪ [_14 ; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (−∞; −_14 ] ∪ [_14 ; ∞)
iv) Non-function iv) Non-function
v) Discontinuous v) Discontinuous

Activity 3.12 SB page 181

1. a) i) y b) i) y = log x
5

20

y = 5x

10

(0; 1)
0 x
–3 –2 –1 1 2 3

–50

ii) (–∞; ∞) ii) (0; ∞)

iii) (0; ∞) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

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 Module 3 • Sketch graphs 95

2. a) i) y b) i) y = log (– _3x )
2

–3 –2 –1 1 2 3
0 x
(0; –3)

–10 y = –3.2 x

–20

ii) (–∞; ∞) ii) (–∞; 0)

iii) (–∞; 0) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

3. a) i) y b) i) y = ln(6x)

y = _16 e x
1

(0; _16 )
0 x
–3 –2 –1 1 2 3

ii) (–∞; ∞); ii) (0; ∞);

iii) (0; ∞) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

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 96 N4 Mathematics – Lecturer Guide

4. a) i) y b) i) y = _12 log (− _4x )

3

0 x
–1 1
(0; –4)

–10
y = –4.3 2x

–20

ii) (–∞; ∞) ii) (–∞; 0)

iii) (–∞; 0) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

5. a) i) b) i) y = _13 ln(_2x )
y

–20

y = 2e 3x
–10

(0; 2)
0 x
–1 1

ii) (–∞; ∞) ii) (0; ∞)

iii) (0; ∞) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

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 Module 3 • Sketch graphs 97

6. a) i) y b) i) y = log 1(–x)
_
3

–3 –2 –1 1 2 3
0 x
(0; –1)

y = –(_13)
x

–10

–20

ii) (–∞; ∞) ii) (–∞; 0)

iii) (–∞; 0) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

7. a) i) y b) i) y = _12 log 3(_3x )

_
4

20

10
y = 3(_34)
2x

(0; 3)

0 x
–3 –2 –1 1 2 3

ii) (–∞; ∞) ii) (0; ∞)

iii) (0; ∞) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

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8. a) i) y b) i) y = _12 log 1(–2x)

_
5

0 x
–2 –1 1 1 2
(0; – 2 )
_

y = – _12(_15)
2x

–5

–10

ii) (–∞; ∞) ii) (–∞; 0)

iii) (–∞; 0) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

9. a) i) y b) i) y = – _12 ln(_5x )

–20

y = 5(_1 )
x

e2 –10

(0; 5)

0 x
–1 1

ii) (–∞; ∞) ii) (0; ∞)

iii) (0; ∞) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

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 Module 3 • Sketch graphs 99

10. a) i) b) i) y = _13 log 1(− _4x )

y _
2

0 x
–1 1
(0; –4)

y = – 4(_12)
3x
–10

–20

ii) (–∞; ∞) ii) (–∞; 0)

iii) (–∞; 0) iii) (–∞; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

Activity 3.13 SB page 188

1. a) i) y b) i) y = 10x

2
y = log x

(1; 0) x
0
10 20 30

ii) (0; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (0; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

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 100 N4 Mathematics – Lecturer Guide

2. a) i) y b) i) y = 2x

5 y = –log x
1
_
2

1
(1; 0)
0 x
10 20 30
–1

ii) (0; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (0; ∞)
iv) Function iv) Function
v) Continuous v) Continuous
x
3. a) i) b) i) y = e2
_

y = 2 log ex
5

(1; 0)
0 x
10 20 30

–5

ii) (0; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (0; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

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 Module 3 • Sketch graphs 101

4. a) i) y b) i) y = _13 e −x

1
(_13 ; 0)
0 x
10 20 30
–1

–2

–3

–4

–5 y = –log e(3x)

ii) (0; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (0; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

y = _14 .2 3
x
5. a) i) b) i)
_

20
y = 3 log 2(4x)

10

(_14 ; 0)
0 x
1 2 3 4 5 6 7 8

ii) (0; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (0; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

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 102 N4 Mathematics – Lecturer Guide

y = _12 .3 4
x
6. a) i) b) i)
_

y
y = –4 log (2x)1
_
3

10

(_12 ; 0)
0 x
5 10 15

ii) (0; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (0; ∞)
iv) Function iv) Function
v) Continuous v) Continuous
4x
7. a) i) b) i) y = _16 e 3
_

2 y = _34 ln(6x)

(_16 ; 0)
0 x
1 2 3

ii) (0; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (0; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

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 Module 3 • Sketch graphs 103

8. a) i) y b) i) y = _12 e −4x

(_12 ; 0)
0 x
5 10 15

y = –_14 ln(2x)

–1

ii) (0; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (0; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

9. a) i) y b) i) y = _13 .4 −5x

1
(_13 ; 0)
0 x
1 2 3 4 5 6 7 8 9 10
–0,1
–0,2
–0,3
–0,4
y = _15 log (3x)
1
–0,5
_
4

–0,6

ii) (0; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (0; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

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10. a) i) y b) i) y = _15 .3 −2x

(_15 ; 0)
0 x
1 2 3

–1
y = – _12 log 3(5x)

–2

ii) (0; ∞) ii) (–∞; ∞)

iii) (–∞; ∞) iii) (0; ∞)
iv) Function iv) Function
v) Continuous v) Continuous

Activity 3.14 SB page 202

1. a) i) y b) i) x = _12 y 2

10

y = _12x 2

0 x
–5 (0; 0) 5

ii) (–∞; ∞); ii) (0; ∞);

iii) (0; ∞) iii) (–∞; ∞)
iv) Function iv) Non-function
v) Continuous v) Continuous

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 Module 3 • Sketch graphs 105

2. a) i) y b) i) x = –3y2 – 15y
(– _52 ; _
75
4
) 20
y = –3x – 15x 2

10

(–5; 0)
0 x
–5 (0; 0)

x = – _52

75
ii) (–∞; ∞); ii) (−∞; _
4
]
75
iii) (−∞; _
4
] iii) (–∞; ∞)
iv) Function iv) Non-function
v) Continuous v) Continuous

3. a) i) y b) i) x = 9y2 – 16

10

(– _43 ; 0) ( _43 ; 0)
0 x
–1 1

–10

y = 9x 2 – 16
(0; –16)

ii) (–∞; ∞) ii) [–16; ∞)

iii) [–16; ∞) iii) (–∞; ∞)
iv) Function iv) Non-function
v) Continuous v) Continuous

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4. a) i) y b) i) x = –_14 y 2 + y – _32

1
–4 –3 –2 –1 1 2 3 4 5 6 7
0 x
–1
(0; – _32 )
–2 (2; – _12 ) y = –_14x 2 + x – _32
–3
–4
–5
x=2
–6
–7

ii) (–∞; ∞) ii) (–∞; –_12 ]

iii) (–∞; –_12 ] iii) (–∞; ∞)
iv) Function iv) Non-function
v) Continuous v) Continuous

5. a) i) y b) i) x = _16 y 2 – y + _56
1 5
y = 6x – x + 6
_ 2 _

10
x=3

(0; _56 ) (5; 0)

0 x
–5 10
(1; 0) (3; – _23 )

ii) (–∞; ∞) ii) [−_23 ; ∞)

iii) [−_23 ; ∞) iii) (–∞; ∞)
iv) Function iv) Non-function
v) Continuous v) Continuous

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 Module 3 • Sketch graphs 107

6. a) i) y b) i) x = –2y2 – 7y + 15

169
(– _74 ; _8
)
20
y = –2x – 7x + 15
2
(0; 15)

( _32 ; 0)
(–5; 0)
0 x
–7 –6 –5 –4 –3 –2 –1 1 2 3 4
x = – _74

169
ii) (–∞; ∞) ii) (−∞; _8
]
169
iii) (−∞; _8
] iii) (–∞; ∞)
iv) Function iv) Non-function
v) Continuous v) Continuous

7. a) i) y b) i) x = 3y2 – 2y – 7
y = 3x 2 – 2x – 7

x = _13
20

10

(–1,230; 0) (1,897; 0)
0 x
–3 –2 –1 1 2 3

(0; –7)
(_13 ; – _
22
3
)

[ 3
ii) (–∞; ∞) ii) 22
−_ ; ∞)

iii) [−_
22
3
; ∞) iii) (–∞; ∞)

iv) Function iv) Non-function

v) Continuous v) Continuous

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17 11
8. a) i) y b) i) x = –y2 + _5
y–_
5

(0,869; 0) (_ ; 69 )
17 _
10 100
0 x
–4 –3 –2 –1 1 2 3 4 5 6 7
11
(0; – _
5
)
(2,531; 0)
–10

17
x=_
10
–20

17 11
y = –x 2 + _
5
x–_5

69
ii) (–∞; ∞) ii) (–∞; _
100
]
69
iii) (−∞; _
100
] iii) (–∞; ∞)
iv) Function iv) Non-function
v) Continuous v) Continuous

x = 2(y − _4)
2
11 49
9. a) i) y b) i) –_
8

50 ) – 49
11 2 _
y = 2(x – _
4 8

40
11
x=_
4
30

20

10
(0; 9) (_29 ; 0)
0 x
–2 –1 1 2 3 4 5 6 7 8
–10 (1; 0) 11
(_ 49
; –_ )
4 8

[ 8
ii) (–∞; ∞); ii) 49
−_ ; ∞);

iii) [−_
49
8
; ∞) iii) (–∞; ∞)

iv) Function iv) Non-function

v) Cotinuous v) Continuous

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 Module 3 • Sketch graphs 109

10. a) i) b) i) x = –_13 (y + 4)2 + 3

y

(–4; 3)
(–1; 0)
(–7; 0)
0 x
–10 –5
(0; – _37 )

y = –_13(x + 4) 2 + 3
–5

x = –4
–10

ii) (–∞; ∞) ii) (–∞; 3]

iii) (–∞; 3] iii) (–∞; ∞)
iv) Function iv) Non-function
v) Continuous v) Continuous

Summative assessment: Module 3 SB page 203

1. 1.1 Domain: (–∞; ∞); Range: [_

95
24
; ∞) (4)

1.2 x = 6y2 – 11y + 9 (2)

1.3 Continuous (1)
1.4 Non-function (1)
2. 2.1 y

3 9x 2
_ 25y 2
225
+_=1
225

0 x
–5 5
Shape ✓
Labelling ✓
–3 x and y ✓
(3)
2.2 Both x-axis and y-axis. (1)

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3. 3.1 y = log 1(_3x ) _

(3)
2

3.2 y
f (x)
y=x

5
f –1(x)

(0; 3)

0 x
(3; 0) 5

(5)

f –1(f (x)) = log 1(_

3 )
3(_12 )x
3.3 _
2

= log 1 (_12 )x
_
2

= log 1 _12_
2

= x .1
f –1(f (x)) = x
f (f –1(x)) = 3(_12 ) (log (–3))
_x
_1
2

= 3. _3x
f (f –1(x)) = x (4)

Since f –1(f (x)) = f (f –1(x)), therefore f (x) = –3(_12 )x and f –1(x) = log 1(_3x ) are
_
2
inverses of each other in the line y = x.
4. 4.1 [–25; 25]; [–25; 0] (4)
4.2 y

_______________
y + √(25 + x)(–x + 25) = 0

(–25; 0) (25; 0)
0 x
–20 20

–20

(0; –25)
(5)

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 Module 3 • Sketch graphs 111

______________
4.3 y + √(25 + x)(–x + 25) = 0
___________________
y + √(25 + (–x))(–(–x) + 25) = 0
_____________
y + √(25 – x)(x + 25) = 0
______________
y + √(25 + x)(–x + 25) = 0
Since the equation is equivalent, the graph is symmetrical about the y-axis.
The tests for symmetry about the line x-axis, the line y = x and the line y = –x
does not yield an equivalent equation, therefore are not lines of symmetry. (3)
5. 5.1 y

y = – _76x x=0 y = _76x

(0; 7)

(–6; 0) (6; 0) y=0 x

0

y2
x2 – _
_
36 49
=1
(0; –7)

x-intercept: (–6; 0); (6; 0)

y-intercept: (0; –7); (0; 7) (6)
5.2 y = –_76 x; y = _76 x (3)
6. 6.1 y (3)

y = ex

0 x

Shape ✓
x and y ✓

6.2 Range = {0 < y < ∞} ✓ (1)

6.3 x = e y ✓ (1)
TOTAL: [50]

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 MODULE

4 Trigonometry
After they have completed this module, students should be able to:
• calculate special triangles pertaining to the four quadrants;
• apply the notion of compound angles such as sin(a ± b), cos(a ± b) and
tan(a ± b);
• apply the complementary angles specifically to trigonometric identities;
• apply factorisation of different types of trigonometric equations including
using identities;
• derive the following identities:
– compound angles
– double angles
– half angles;
• derive the co-ratios sin(90° ± θ), cos(90° ± θ) and tan(90° ± θ);
• use the square, invert and quotient identities and furthermore solve
trigonometric equations, simplify trigonometric expressions and prove
trigonometric identities; and
• draw neat sketch graphs of the following functions/relations:
– y = a sin(bx + c) + d for − π ≤ x ≤ π
– y = a cos(bx + c) + d for − π ≤ x ≤ π
– y = a tan(bx + c) + d for − π ≤ x ≤ π
– y = cosec x for − π ≤ x ≤ π
– y = sec x for − π ≤ x ≤ π
– y = cot x for − π ≤ x ≤ π.

Introduction
Trigonometry deals with the study of triangles and is used when precise distances,
lengths, heights and angles need to be calculated. The trigonometric skills in this
module will require students’ logical thinking skills.

Students need the following pre-knowledge to successfully complete this module.

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 114 N4 Mathematics – Lecturer Guide

Pre-knowledge
A
• Trigonometric ratios
opposite hypotenuse
sin θ = _ cosec θ = _
opposite
hypotenuse

se
nu
adjacent hypotenuse
cos θ = _ sec θ = _

te
opposite

po
hypotenuse adjacent

hy
opposite adjacent
tan θ = _ cot θ = _
opposite
adjacent
θ
B adjacent C

• Definition of the trigonometric ratios

y α
sin θ = _r sin α = x_r cosec θ = _yr cosec α = _xr e)
te nus
y y po opposite
cos θ = x_r cos α = _r sec θ = _xr sec α = _yr r (h y
adjacent
opposite
y y
tan θ = _x tan α = x_y cot θ = x_y cot α = _x θ
adjacent x

• Theorem of Pythagoras: x 2 + y 2 = r 2

• Signs of trigonometric functions: CAST diagram

Note: θ is the reference angle. Positive angle:
y 90°
II I (x; y)
S A r
y
θ
180° x 0°/360°
S in ratio + A ll ratios +
other ratios –
180° – θ θ
x
180° + θ 360° – θ 270°
T an ratio + C os ratio +
other ratios – other ratios –
Negative angle:
T C 90°
III IV

180°
x 0°/360°
–θ
r –y
(x; –y)
270°

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 Module 4 • Trigonometry 115

Quadrant Positive functions Negative functions

I all none
II sin, cosec cos, sec, tan, cot
III tan, cot sin, cosec, cos, sec
IV cos, sec sin, cosec, tan, cot

• Reference angle
y I II
y y y

θ θ θ
θ=α α
x x x x
α
α

III IV
θ = 180° – α θ = 180° + α θ = 360° – α

• Trigonometric identities

Reciprocal identities/ratios
1 1 1
cosec θ = _
sin θ
sec θ = _
cos θ
cot θ = _
tan θ

Quotient identities
sin θ cos θ
tan θ = _
cos θ
cot θ = _
sin θ

Square/Pythagorean identities

sin 2 θ + cos 2 θ = 1 tan 2 θ + 1 = sec 2 θ 1 + cot 2 θ = cosec2 θ

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• Special angles

45°
30° _
_
√2
2 1 90°
√3 1
0

60° 45° 0°
1 1 1

sin 30° = 1_2 1_

sin 45° = _ sin 0° = 0_1 = 0
_ √2
√3 1_
sin 60° = _2
cos 45° = _ sin 90° = 1_1 = 1
_ √2
√3
cos 30° = _2
tan 45° = 1_1 = 1 cos 0° = 1_1 = 1

cos 60° = 1_2 cos 90° = 0_1 = 0

1_
tan 30° = _ tan 0° = 0_1 = 0
3
√_
3
tan 60° = 1 √
_ tan 90° = 1_0 = ∞

• Standard trigonometric graphs

y y = sin x
1 y = sin x Amplitude 1
Range –1 ≤ y ≤ 1
0,5 Frequency 1
Period 360°
0 x
90° 180° 270° 360° Domain 0° ≤ x ≤ 360°
–0,5 Turning (90°; 1); (270°; –1)
points
–1 x-intercepts (0°; 0); (180°; 0);
(360°; 0)

y y = cos x
1 Amplitude 1
Range –1 ≤ y ≤ 1
0,5
Frequency 1
Period 360°
0 x
90° 180° 270° 360° Domain 0° ≤ x ≤ 360°
–0,5 Turning (0°; 1);
points (180°; –1);
y = cos x (360°; 1)
–1
x-intercepts (90°; 0); (270°; 0)

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 Module 4 • Trigonometry 117

y y = tan x
Amplitude undefined
2
y = tan x Range y ∈ℝ
1 Frequency 2
Period 180°
0 x
90° 180° 270° 360° Domain 0° ≤ x ≤ 360°,
–1 x ≠ 90°, x ≠ 270°
Asymptotes x = 90°; x = 270°
–2
x-intercepts (0°; 0); (180°; 0);
(360°; 0)

Reciprocal ratios
sin x = _ 1 ; cos x = _ 1 ; tan x = _ 1
cosec x sec x cot x

Quotient identities
sin x cos x
tan x = _
cos x
; cot x = _
sin x

Square identities
sin 2 x + cos 2 x = 1; 1 + tan 2 x = sec 2 x; 1 + cot 2 x = cosec 2 x

Double-angle identities (sum/difference of two angles)

sin(a ± b) = sin a.cos b ± cos a.sin b
cos(a ± b) = cos a.cos b +
¯ sin a.sin b
tan a ± tan b
tan(a ± b) = __________
¯ tan a.tan b
1+

Double-angle identities
sin 2x = 2 sin x.cos x
cos 2x = cos 2 x − sin 2 x; cos 2x = 1 – 2 sin2 x; cos 2x = 2 cos2 x – 1
2 tan x
tan 2x = _ 2
1 − tan x
and
cos 2 x = 1_2(1 + cos 2x); sin 2 x = _21(1 − cos 2x)

Half-angle identities
________ _

√_
(1 – cos x)
sin 1_2 x = ________
2
cos _21 x = 1_2(1 + cos x) ;
√
1 − cos x
tan _21 x = √_
1 + cos x
; tan 1_2 x = 1_
− cos x
sin x
; tan 21_ x = _ sin x
1 + cos x

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 118 N4 Mathematics – Lecturer Guide

Co-ratios (co-functions)
sin(90° − x) = cos x cos(90° − x) = sin x tan(90° − x) = cot x
sin(90° + x) = cos x cos(90° + x) = − sin x tan(90° + x) = − cot x
cosec(90° − x) = sec x sec(90° − x) = cosec x cot(90° − x) = tan x
cosec(90° + x) = sec x sec(90° + x) = − cosec x cot(90° + x) = − tan x

Factorisation
• Common factor, for example
sin θ.cos θ + sin θ = sin θ(cos θ + 1)
• Difference between squares, for example
cos 2 θ − sin 2 θ = (cos θ − sin θ)(cos θ + sin θ)
• Trinomials, for example
cos 2 θ + 2 cos θ + 1 = (cos θ + 1)(cos θ + 1) = (cos θ + 1)2

• Sum and difference of two cubes, for example

a 3 + b 3 = (a + b)( a 2 − ab + b 2)
a 3 − b 3 = (a − b)( a 2 + ab + b 2)
• Grouping, for example
a_
_b = a
_×d ad
_=_
_c b c bc
d
• Fractions: find common denominators and equivalent numerators,
for example
2 2
1 +_
_ cos x + sin x
1 =_
sin 2 x cos 2 x sin 2 x.cos 2 x
_______
− b ± √b − 4ac 2
• Quadratic formula: x = ___________ 2a

The basic graphs for the domain of 0° ≤ x ≤ 360° for:

• y = a sin x + d
• y = a cos x + d
• y = a tan x + d
where
• the a-value determines the shape and amplitude of the graph;
• the d-value shifts the graph vertically up or down.

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 Module 4 • Trigonometry 119

Activity 4.1 SB page 216

1. 1.1 sin 140° 1.2 tan(–85°)

= sin(180° – 40°) = –tan 85°
= sin 40°

1.3 sec 265° 1.4 cos(_

36 )
67
π
= sec(180° + 85°)
= cos(2π − _
36 )
5
π
= –sec 85°
= cos(_
36 )
5
π

1.5 cot(− _
18 )
25
π 1.6 cosec(–465°)
= cosec(–105° – 360°)
= –cot(_
18 )
25
π
= cosec(–105° – (1)360°)
= –cot(π + _
18 )
7
π = cosec(–105°)
= –cosec 105°
= –cot(_
18 )
7
π = –cosec(180° – 75°)
= –cosec 75°

1.7 cot(− 315° ) 1.8 sec(− _4 )

25
π
= − cot 315°
= sec(− _14 π − 6π)
= − cot(360° − 45°)
= cot 45° = sec(− _14 π − 3.2π)

= sec(− _14 π)

= sec(_14 π)

2. 2.1 tan(–225°) 2.2 sin 240°

= –tan 225° = sin(180° + 60°)
= –tan(180° + 45°) = –sin 60°
_
= –tan 45° √3
= –_2
= –1
2.3 cos 120° 2.4 cosec 330°
= cos(180° – 60°) 1
=_
sin 330°
= –cos 60°
1
= – _12 = ___________
sin(360° − 30°)
1
=_
− sin 30°
1
=__1
−2

= –2

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 120 N4 Mathematics – Lecturer Guide

2.5 cot(− π_4) 2.6 sec(− _74 π)

= − cot _4π 1
=_
cos (− 4 π)
_7

= –1 1
=_
cos ( 4 π)
_7

1
=_
cos (2π − 4 π)
_1

1
=_
cos ( 4 π)
_1

1
=_
1_
_
√2
_
= √2
2.7 sin 150° 2.8 cot(− 30°)
= sin(180° − 30° ) 1
=_
tan(− 30°)
= sin 30°
1
= _21 =_
− tan 30°
1
=_
_1_
−
√3
_
= − √3
2.9 cosec(–120°)
= _______
1
sin(–120°)

= ______
1
– sin 120°

= ______ 1
– sin(180° – 60°)

= ______
1
– sin 60°

= ___
1__
√3
__ – 2

= – __
2__
√3

3. 3 cos 150° − sin(− 45°) + 2 tan 420° + sec 2(− 120°)

= 3 cos(180° − 30°) − [− sin 45°] + 2 tan(420° − 360°) + sec 2(180° − 60°)
= − 3 cos 30° + sin 45° + 2 tan 60° − sec 260°
_ _
= − 3( 2 ) + _
_
√3
1_
2
+ 2 √3 − (2)2
√
_ _
= − 3_
√3
2
1_
+_2
+ 2 √3 − 4
√

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 Module 4 • Trigonometry 121

Activity 4.2 SB page 221

1. 1.1 sin(α − 30° ) = sin αcos 30° − sin 30°cos α

1.2 cos(_π4 + 2λ) = cos(_π4 )cos 2λ − sin(_π4 )sin 2λ
tan ρ − tan 60°
1.3 tan(ρ − 60° ) = ____________
1 + tan ρ.tan 60°
1 1
1.4 cosec(105° + ω) = _ = ___________________
sin(105° + ω) sin 105° cos ω + sin ω cos 105°
1 1
1.5 sec(4θ – 15°) = _ = ___________________
cos(4θ − 15°) cos 4θ cos 15° + sin 4θ sin 15°
β

1.6 cot(75° + _3 ) = _
1 β 1 1 − tan 75° tan _
=_ = ___________ 3

tan (75° + 3 )
β
_
β
tan 75° + tan _ β
_
tan 75° + tan
___________
3
3
β
1 − tan 75° tan _3

2. 2.1 cos 40° cos 35° − sin 40° sin 35° = cos(40° + 35° )
2.2 sin 4x cos 55° + sin 55° cos 4x = sin(4x + 55°)
tan 2θ − tan 3ρ
___________
2.3 1 + tan 2θ tan 3ρ
= tan(2θ – 3ρ)
1
__________________ 1
2.4 =_ = cosec (3α – 2β)
sin 3α cos 2β − cos 3α sin 2β sin(3α − 2β)

( 1 − tan λ tan _π6 )

−1
1 − tan λ tan _π tan λ + tan _π
2.5 _6 = _6
tan λ + tan _π6

= (tan(λ + _π6 )) −1
1
=_
tan(λ + 6 )
π
_

= cot(λ + _π6 )
2.6 (cos 8x cos 2x + sin 8x sin 2x)–1 = (cos(8x – 2x))–1
= (cos 6x)–1
1
=_cos 6x
= sec 6x
3. 3.1 tan 75° = tan(30° + 45°)
tan 30° + tan 45°
= ____________
1 − tan 30° tan 45°
1__ _
__ +1
= ________
√3 1
1 − (__ ) (1)
1__ _ 1
√3
1__
__ +1
= ____
√3
1__
__
1−
√3
__
1+ √3
_____
= __
√3 − 1
__ __
1__+ √3 _____
= _____ × √__3 + 1
√3 − 1 √3 + 1
__
4 + 2√3
= _____
2
__
= 2 + √3

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3.2 sin 50° cos 55° + sin 55° cos 50°

= sin(50° + 55°)
= sin 105°
= sin(45° + 60°)
= sin 45° cos 60° + sin 60° cos 45°
_

√2 )(2) ( 2 )(√2 )
= (_1_ 1_ √3
+ _ _1_

_
=_ + √3_
1_ _
2 √2 2 √2
_
= 1_+√
_
2 2
3
√
_ _
1+√3 √_
2
=_ _ ×_
2 √2 √2
_ _
2 + √6
= √_ 4

3.3 sin 15° = sin(45° – 30°)

= sin 45° cos 30° – sin 30° cos 45°
__

√2 )( 2 )
= (__ − (_12 ).(__
√2 )
√3
1__ __ 1__

__

2√2 ( 2√2 )
√3__ 1__
= ___ − ___
__
√3 −__ 1
= _____
2√2
__ __
√3 −__ 1 2
= _____ × __
√__
2√2 √2
__ __
√6 − √2
= _____
4
__ __
√2 (√3 − 1)
= _______ 4
tan 63° − tan 48°
______________
3.4 1 + tan 63° tan 48°

= tan(63° − 48°)
= tan 15°
= tan(45° − 30°)
tan 45° − tan 30°
= 1______________
+ tan 45° tan 30°

(1) (√3 )
1_ 1_
_
−
=_
1 + (1_1)(_3)
1_
√
_
1_ − √_
3
= 1 + √3
_
3 −1
=_
√_
3 +1
√
_ _
3 −1 _
=_
√_
3 +1
× √_33 ++ 11
√ √

= ________
3 –__1
3 + 2√3 + 1

= ________
2 __
4 + 2√3

= ________
1 __
2 + √3

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 Module 4 • Trigonometry 123

1
____________________ 1
3.5 = ____________________
sin 33° sin 18° + cos 33° cos 18° cos 33° cos 18° + sin 33° sin 18°
1
=_
cos(33° − 18°)
1
=_
cos 15°
1
=_
cos(45° − 30°)
1
= ____________________
cos 45° cos 30° + sin 45° sin 30°
1
= _____________
__

( √2 )( 2 ) + ( √2 )( 2 )
1
__ √3
__
__ 1 _
__ 1 __

1
= ______
__
√3
___ 1
___
__ + __
2√2 2√2

1
= ____
__
√3 + 1
_____
__
2√2
__
2√2
= _____
__
√3 + 1
__ __
2√2 3 −1
= _____
__ × _____
√__
√3 + 1 √3 − 1
__ __
2√6 − 2√2
= _______
2
__ __
= √6 − √2
__ __
= √2 (√3 − 1)
3.6 sin(− 105°)
= − sin 105°
= − sin(60° + 45°)
= − (sin 60° cos 45° + sin 45° cos 60°)
_
= − [(_2 )( ) ( )(2)]
√3 1_
_ 1_ 1_
+ _
√2 √2
_
= − [_
2 2 2 2]
√3_ 1_
+_
√ √
_
√3_ 1_
= − 2_ −_
2 2 2
√ √
_ _
− √3 _
−1 _
=_ 2 2
× √_22
√ √
_ _ _ _
− √6 − √2 −(√6 + √2 )
=_ 4
or ___________
4

4. 4.1 To prove: sin(x − 180°) = − sin x

LHS = sin(x − 180°)
= sin x cos 180° − cos x sin 180°
= sin x (− 1) − cos x (0)
= − sin x
∴ LHS = RHS

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 124 N4 Mathematics – Lecturer Guide

__ __
√6 + √2
4.2 To prove: cos(− 15°) = _______ 4
LHS = cos(− 15° )
= cos(30° − 45° )
= cos 30° cos 45° + sin 30° sin 45°
_
= (_2 )( 2 ) (2)( 2 )
√3 _1_
+ 1_ _1_
√ √
_
√3_ 1_
=_ +_
2 2 2 2
√ √
_
√3 +_1
=_2 2 √
_ _
√3 +_1 2
=_2 2
×_
√_
2
√ √
_ _
√6 + √2
=_ 4
∴ LHS = RHS

4.3 To prove: cos 71° + cos 49° = cos 11°

LHS = cos 71° + cos 49°
= cos(60° + 11°) + cos(60° – 11°)
= (cos 60° cos 11° – sin 60° sin 11°) + (cos 60° cos 11° + sin 60° sin 11°)
= cos 60° cos 11° – sin 60° sin 11° + cos 60° cos 11° + sin 60° sin 11°
= 2 cos 60° cos 11°
= 2(_12 )cos 11°
= cos 11°
∴ LHS = RHS
__
4.4 To prove: sin 33° – sin 57° = – √2 sin 12°
LHS = sin 33° – sin 57°
= sin(45° – 12°) – sin(45° + 12°)
= (sin 45° cos 12° – sin 12° cos 45°) – (sin 45° cos 12° + sin 12° cos 45°)
= sin 45° cos 12° – sin 12° cos 45° – sin 45° cos 12° – sin 12° cos 45°
= –2 sin 12° cos 45°
= –2 sin 12°(__
1__
) √2
2__
= – __ sin 12°
√2
__
= – √2 sin 12°
∴ LHS = RHS

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 Module 4 • Trigonometry 125

Activity 4.3 SB page 225

1. 1.1 To prove: sec(90° – x) = cosec x 1.2 To prove: tan(_π2 − x) = cot x

LHS = sec(90° – x) LHS = tan(_π2 − x)
1
=_
cos(90° − x) sin( 2 − x)
π
_
=_
1
= _________________ cos(_π2 − x)
cos 90° cos x + sin 90° sin x
1 sin(_π ) cos x − sin x cos(_π )
= ___________
(0) cos x + (1) sin x = _________________
2 2
cos( 2 ) cos x + sin( 2 ) sin x
π
_ π
_
1
=_ (1) cos x − sin x (0)
sin x = ____________
(0) cos x + (1) sin x
= cosec x
cos x
=_
∴ LHS = RHS sin x
= cot x
∴ LHS = RHS
1.3 To prove: cot(_π2 + x) = –tan x 1.4 To prove: cosec(90° + x) = sec x
LHS = cosec (90° + x)
LHS = cot(_π2 + x)
1
=_
cos(_π2 + x) sin(90° + x)
=_
π
_ sin( 2 + x) 1
= _________________
sin 90° cos x + sin x cos 90°
cos(_π ) cos x − sin(_π ) sin x
= _________________
2 2
1
= ____________
sin( 2 ) cos x + sin x cos( 2 )
π
_ π
_
(1) cos x + sin x (0)
(0) cos x − (1) sin x
= ____________
(1) cos x + sin x (0)
=_ 1
cos x
–sin x
=_ = sec x
cos x
∴ LHS = RHS
= –tan x
∴ LHS = RHS
2. 2.1 sin 78° = sin(90° – 12°) 2.2 tan 51° = tan(90° – 39°)
= cos 12° = cot 39°
2.3 sec 27° = sec (90°– 63°)
= cosec 63°
sin(90° − x) sec(360° − x)
_______________________________
3. 3.1 cot(90° − x) sin(180° − x) cosec(180° + x)
cos x secx
= ________________
tan x sin x (− cosec x)

( 1 )(cos x)
cos
_x 1
_
= ____________
cos x)( 1 )(sin x)
−(_
sin x _sin x _1

_ 1
= − tan x
= − cot x

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 126 N4 Mathematics – Lecturer Guide

sin 2(180° + θ) + cos(180° − θ) cosec(90° − θ)

________________________________
3.2 tan(180° − θ) tan(90° − θ)
2
sin θ − cos θ sec θ
= ______________
− tan θ cot θ
2
sin θ − 1
=_ −1
−(1 − sin 2θ)
=_
−1

= 1 − sin 2θ
= cos 2θ

sin(90° − θ) cos(360° − θ)
______________________ = − 1
4. 4.1 To prove: cos(90° + θ) sin(180° + θ) − 1
sin(90° − θ) cos(360° − θ)
______________________
LHS = cos(90° + θ) sin(180° + θ) − 1

= ____________
cos θ cos θ
(–sin θ)(–sin θ) – 1
2
cos θ
=_
−(1 − sin θ)
2

2
cos θ
=_ 2
− cos θ

= −1
∴ LHS = RHS

tan 315° cos 390° cos(− θ) cos(90° − θ)

4.2 To prove: _____________________________
cos(360° − θ) sin 300° sin(180° − θ)
=1

tan 315° cos 390° cos(− θ) cos(90° − θ)

LHS = _____________________________
cos(360° − θ) sin 300° sin(180° − θ)

tan(360° – 45°) cos 30° cos(− θ) cos(90° − θ)

= _____________________________
cos(360° − θ) sin(360° – 60°) sin(180° − θ)

− tan 45° cos 30° cos θ sin θ

= ____________________
cos θ (− sin 60°) sin θ
_
√3
−_ cos θ sin θ
= _____________
2
_
cos θ (− 2 ) sin θ
√3
_

=1
∴ LHS = RHS

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 Module 4 • Trigonometry 127

Activity 4.4 SB page 235

1. sin 2x = 2 sin x cos x

1
∴ cosec 2x = _
sin 2x
1
=_
2 sin x cos x

cos 2x = cos2x – sin2x

= 2 cos2x – 1
= 1 – 2 sin2x
1
∴ sec 2x = _
cos 2x
1
=_
2 2
cos x − sin x
1
=_
2
2 cos x − 1
1
=_ 2
1 − 2 sin x
2 tan x
tan 2x = _ 2
1 − tan x
1
∴ cot 2x = _
tan 2x
1
=_
2 tan x
_
1 − tan 2x

2
1 − tan x
=_2 tan x

2. 2.1 sin 2θ = sin(θ + θ)

= sin θ cos θ + cos θ sin θ
= 2 sin θ cos θ
2.2 sin 2(60°) = 2 sin 60° cos 60°
_
= 2( 2 )(1_2) _
√3

_
= 2( 4 ) _
√3

_
_
√3
= 2

3. 3.1 cos 2α = cos(α + α)

= cos α cos α − sin α sin α
= cos 2α − sin 2α
= (1 − sin 2α) − sin 2α
= 1 − 2 sin 2α

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 128 N4 Mathematics – Lecturer Guide

3.2 cos 150° = cos 2(75°)

= 1 − 2 sin 275°
= 1 − 2 (sin 75°)2
= 1 − 2 (sin(30° + 45°))2
= 1 − 2 (sin 30° cos 45° + sin 45° cos 30°)2
_
= 1 − 2 [(1_2)(_2 ) ( 2 )( 2 )]
2
1_ 1_ √3
+ _ _
√ √
_
= 1 − 2 [_
2 2 2 2]
2
+ √3_
1_ _
√ √
_ 2
= 1 − 2[ 1+√
_ _
3
2 √2 ]
_
= 1 − 2[1_
+ 2 √3 + 3
8 ]
_
=1− 2[4_8 ]
+ 2 √3

_
− 8 − 4 √3
=1+_ 8
_
8 − 8 − 4 √3
=_ 8
_
− 4 √3
=_ 8
_
_
√3
=−2

3.3 cos(90° + α) = cos 90° cos α − sin 90° sin α

= (0) cos α − (1) sin α
= − sin α
3.4 cos 150° = cos(90° + 60°)
= − sin 60°
_
√3
= −_2

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 Module 4 • Trigonometry 129

4. 4.1 sin 37,5° cos 37,5° = _12 (2 sin 37,5° cos 37,5°)

= _12 sin[2(37,5°)]

= _12 sin 75°

= _12 sin (30° + 45°)

= _12 [sin 30° cos 45° + sin 45° cos 30°]

_
= _12 [(_12 )(_
1_
) + ( )( 2 )]
√3
1_ _
_
√2 √2
_
= _12 [_ + √3_ ]
1_ _
2√2 2√2
_
= _21 [_
1+_
√3
]
2√2
_
1+_
√3
=_
4√2
_ _
1+_
√3 2
=_ ×_
√_
4√2 √2
_ _
√2 + √6
=_ 8
_ _
(
√2 1 + √3 )
=_ 8

( 1 – tan 2(_ )
−1
1 − tan ( 12 )
2 π
_ 2 tan( 12 ) π
_
4.2 _ = _
2 tan(_
12 ) 12 )
π π

= (tan[2(_
12 )])
π −1

= (tan(_π6 )) −1
−1
= (_
1_
) √3
_
= √3

4.4 (_2 tan 22,5° )

2 −1
1 − tan 22,5° 2 tan 22,5°
4.3 cos 27,5° − sin 27,5° =_
1 − tan 222,5°
= cos 2(7,5°)
= tan[2(22,5°)]
= cos 15°
= tan 45°
= cos(45° − 30°)
=1
= cos 45° cos 30° + sin 45° sin 30°
_

2 )( 2 ) ( 2 )(2)
= (_
1_ √3
_ + _1_ 1_
√ √
_ _
√3_ √1_
=_ +_
2 2 2 2
√ √
_ _
_
√3 +_1 _
√2
= 2 √2
× _
√2
_ _
√
_2 (√3 + 1)
= 4
_ _
_
√6 + √2
= 4

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 130 N4 Mathematics – Lecturer Guide

2
sin x
5. tan2x = _ 2
cos x
1 − cos 2x
_
=_
2
1 + cos 2x
_
2

1 − cos 2x
=_
1 + cos 2x

6. 6.1 cos 215° 6.2 sin422,5°

1 + cos 2(15°)
= ___________ = (sin222,5°)2
2
= (___________ )
2
1 − cos [2(22,5°)]
1 + cos 30°
=_ 2
2
2
√3
1+_
_
= (_
1 − cos 45°
2 )
=_ 2

=( )
2 1_ 2
_ 1−_
_√2
= 2_
+ √3
4
2
_
= (_ )
2
√2 −_ 1
2√2
_
3 − 2√2
=_ 8

6.3 sin215° cos415°

= (sin215° cos215°) cos215°
= (sin 15° cos 15°)2 cos215°

= (_12 sin [2(15°)]) (_ )

1 + cos 2 15°2 [ ( )]
2

= (_12 sin 30°) (_ )

2
1 + cos 30°
2

= (_12 (_12 )) (_
2 )
_
√3
2 1+_2

16 ( 4 )
1 _2 + √3
=_
_
2 + √3
=_ 64

7. 7.1 sin 2x = 1_2(1 − cos 2x) 7.2 cos 2x = 1_2(1 + cos 2x)
1 − cos 2x 1 + cos 2x
=_ 2
=_ 2
___________ ___________
√_
1 − cos 2x
√_
_ 1 + cos 2x
_
∴ sin x = 2
∴ cos x = 2

sin _2x = √_
1 − cos x
2
cos _2x = √_
1 + cos x
2

7.3 tan 2x = tan(x + x)

tan x + tan x
= 1___________
− tan x tan x
2 tan x
=_ 2
1 − tan x

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 Module 4 • Trigonometry 131

_
1 − cos[2(_
12 )]
√
π

8. sin(_
12 )
π
= _
2
_
1 − cos(_π6 )
=√ _
2
__

=√
√3
1−_
_2
2
_ _
= √
2 − √3
_

_
4
_
√2 − √3
=_ 2

9. y

13
5

θ x
12

___________

√____2
1_ 1 + cos θ
_
9.1 sin 2θ = 2 sin θ cos θ 9.2 cos 2
θ =

13)(13)
= 2(_5 12
_
√
12
_
1 + 13
= ____
2
= 120
_ ____

=√
169 13 + 12
_
____
13
2
__

=√
25
_
__
13
2

5 __
=_____
_
13 2
√ √

= ___
5
___
√26

9.3 cos 4θ = 1 − 2 sin 22θ

2
= 1 − 2(120
169)
_

= 1 − 2(_ 2)
2
120
169

= –0,008368

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 132 N4 Mathematics – Lecturer Guide

Activity 4.5 SB page 241

1. sin(90° − x) = 2 cos x + 1_2

cos x − 2 cos x − _21 = 0

cos x − 2 cos x = 1_2

− cos x = 1_2

cos x = − 1_2

∴ x = 120° and x = 240°

2. 4 cos2x – 3 = –2 sin2x
4(1 – sin2x) – 3 = –2 sin2x
4 – 4 sin2x – 3 = –2 sin2x
– 4 sin2x + 1 = –2 sin2x
–2 sin2x = –1
sin2x = _12
1_
_
sin x = ±
√2

sin x = ⊕ _
1_
sin x = ⊖ _
1_
√2 √2
sin+; 1st sin+; 2nd sin–; 3rd sin–; 4th

x = xref x = π – xref x = π + xref x = 2π – xref

= sin (_
√2 )
= π – sin (_
√2 )
= π + sin (_
√2 )
1_–1 1_–1 1_–1
= 2π – sin–1(_
1_
)
√2
x = _π4 x=_3π
4
x=_5π
4
7π
x=_ 4

0 ≤ x ≤ 2π, therefore x = _π4 , x = _

3π
4
5π
,x=_4
7π
and x = _4
.

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 Module 4 • Trigonometry 133

3. _1 cos x = cos 2x
2
_1 cos x = 2 cos2x – 1
2

–2 cos2x + _12 cos x + 1 = 0

4 cos2x – cos x – 2 = 0
______________
–(–1) ± √(–1) 2 − 4(4)(–2)
_________________
cos x = 2(4)
_
1
_ ± √ 33
cos x = 8

_ _
1 + √33
_ 1 − √33
cos x = 8
cos x = _ 8

cos x = ⊕ 0,843 cos x = ⊖ 0,593

cos+; 1st cos+; 4th cos–; 2nd cos–; 3rd

x = xref x = 360° – xref x = 180° – xref x = 180° + xref

= cos (0,843) = 360° – cos (0,843) = 180° – cos (0,593) = 180° + cos–1(0,593)
–1 –1 –1

x = 32,534° x = 327,466° x = 126,375° x = 233,625°

0° ≤ x ≤ 360°, therefore x = 32,534°, x = 126,375°, x = 233,625° and x = 327,466°.

1
4. tan2x –_
sin(90° − x)
=1
1
(sec2x – 1) – _
cos x
=1
1
sec2x – 1 – _
cos x
=1
1
_ 1
–1–_ cos x
=1
cos 2x
1 1
–2 – _
cos x
+_ 2 =0
cos x
2
2 cos x + cos x – 1 = 0
(2 cos x – 1)(cos x + 1) = 0

2 cos x – 1 = 0 cos x + 1 = 0
2 cos x = + 1 cos x = ⊖1
cos x = ⊕_12
cos+; 1st cos+; 3rd cos–; 2nd cos–; 4th
x = xref x = 360° – xref x = 180° – xref x = 180° + xref
= cos–1(_12 ) = 360° – cos–1(_12 ) = 180° – cos–1(1) = 180° + cos–1(1)
x = 60° x = 300° x = 180° x = 180°
and x = –300° and x = –60° and x = –180° and x = –180°

–90° ≤ x ≤ 180°, therefore x = –60°, x = 60° and x = 180°

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 134 N4 Mathematics – Lecturer Guide

5. 3 cot2x – 4 cosec x = 1
3(cosec2x – 1) – 4 cosec x = 1
3 cosec2x – 3 – 4 cosec x = 1
–4 – 4 cosec x + 3 cosec2x = 0
4 3
–4 – _
sin x
+_2 =0
sin x
2
4 sin x + 4 sin x – 3 = 0
(2 sin x – 1)(2 sin x + 3) = 0

2 sin x – 1 = 0 2 sin x + 3 = 0
2 sin x = + 1 2 sin x = –3
sin x = ⊕_12 sin x ≠ − _32

sin+; 1st sin+; 2nd

x = xref x = 180° – xref

= sin–1(_21 ) = 180° – sin–1(_12 )
x = 30° x = 150°

0° ≤ x ≤ 180°, therefore x = 30° and x = 150°.

tan 2x + tan 30°
___________
6. 1 − tan 2x tan 30°
= – cot x

tan(2x + 30°) = – cot x

tan(2x + 30°) = tan(90° + x)
2x + 30° = 90° + x
x = 60°
0° ≤ x ≤ 180°, therefore x = 60°

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 Module 4 • Trigonometry 135

7. 1 − cos 2x
_
1 + cos 2x
– 2(_
cos x )
sin x
=4

tan2x – 2 tan x = 4
tan2x – 2 tan x – 4 = 0
______________
− (–2) ± √(–2) − 4(1)(–4)
2
__________________
tan x = 2(1)
_
2 ± 2√5
=_ 2
_
tan x =1 ± √5

_ _
tan x = 1 + √5 tan x = 1 – √5
tan x = ⊕3,236 tan x = ⊖1,236
tan+; 1st tan+; 3rd tan–; 2nd tan–; 4th

x = xref x = π + xref x = π – xref x = 2π – xref

= tan (3,236) = π + tan (3,236) = π – tan (1,236) = 2π – tan–1(1,236)
–1 –1 –1

x = 1,271 x = 4,413 x = 2,251 x = 5,393

and x = –5,012 and x = –1,871 and x = –4,032 and x = –0,891

–_π2 ≤ x ≤ π, therefore x = –0,891, x = 1,271 and x = 2,251

8. 7 cos2x – 1 = − _12 sin 2x

6 cos2x – 1 + cos2x = − _12 (2 sin x.cos x)

6 cos2x – (1 – cos2x) = –sin x.cos x
6 cos2x – sin2x = –sin x.cos x
2
sin x sin x
6–_ 2 = –
_
cos x
cos x
6 – tan2x = –tan x
–tan2x + tan x + 6 = 0
tan2x – tan x – 6 = 0
(tan x – 3)(tan x + 2) = 0

tan x – 3 = 0 tan x + 2 = 0
tan x = ⊕ 3 tan x = ⊖ 2
tan+; 1st tan+; 3rd tan–; 2nd tan–; 4th

x = xref x = 180° + xref x = 180° – xref x = 360° – xref

–1 –1 –1
= tan (3) = 180° + tan (3) = 180° – tan (2) = 360° – tan–1(2)
x = 71,656° x = 251,565° x = 116,656° x = 296,565°
and x = –288,435° and x = –108,435° and x = –243,435° and x = –63,435°

–180° ≤ x ≤ 180°, therefore x = –108,435°, x = –63,435°, x = 71,565° and

x = 116,565°

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 136 N4 Mathematics – Lecturer Guide

Activity 4.6 SB page 246

2
1 − tan x
_
1. (cos x – sin x)2 2. 2
1 + tan x
= cos2x – 2 sin x cos x + sin2x sin x 2
1−_
= (cos2x + sin2x) – 2 sin x cos x =_
cos 2x
sin x 2
1+_ 2
= 1 – sin 2x cos x
2 2
cos x − sin x
_
=_
cos 2x
2
cos x + sin x
_
2

cos 2x
2 2 2
= ________
cos x − sin x ________
2 × 2cos x 2
cos x cos x + sin x
2 2
= ________
cos x − sin x
2 2
cos x + sin x
cos 2x
=_ 1
= cos 2x

sin 2x
_ cos 2x
_
3. 1 − cos 2x
4. sin x − cos x
2 2
2 sin x cos x
=_ 2 = ________
cos x − sin x
sin x − cos x
1 − (1 − 2 sin x)
2 sin x cos x (cos x + sin x)(cos x − sin x)
=_ 2 = _________________
− (cos x − sin x)
1 − 1 + 2 sin x
2 sin x cos x
=_ 2
= –(cos x + sin x)
2 sin x
= –cos x − sin x
cos x
=_
sin x

= cot x

tan x
_ cot x
_ cot x
5. sin 2x − tan x
6. +_
cosec x − 1 cosec x + 1
sin x
_ cot x(cosec x + 1) + cot x(cosec x − 1)
= ____________
cos x = _______________________
(cosec x − 1)(cosec x + 1)
sin x
_
2 sin x cos x − cos x
cot x cosec x + cot x + cot x cosec x − cot x
sin x = __________________________
= _____________
2
2
cosec x − 1
2 sin x cos x − sin x
2 cosec x cot x
sin x
=_
= ___________
2
cot 2x
sin x(2 cos x − 1) 2 cosec x
_
= cot x
sin x
=_ sin x cos 2x
2 (_
1
)
1 = _____
sin x
=_
cos 2x
cos x
_
sin x

2 sin x
= sec 2x =_ ×_
sin x cos x

= 2____
1
cos x

= 2 sec x

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 Module 4 • Trigonometry 137

3 3
7. ________
sin x − cos x
8. sin 4x _
_ – cos 4x
cos x − sin x sin x cos x
2
(sin x − cos x)(sin 2x + sin x cos x + cos x) 2
= __________________________
− (sin x − cos x)
2 sin 2x cos 2x _
=_ sin x
– 2 cocos
s 2x − 1
x

= –(sin2x + sin x cos x + cos2x) 2(2 sin x cos x) cos 2x

= _____________ – ____________
2(2 cos 2x − 1) 2 − 1
sin x cos x
= –[(sin2x + cos2x) + sin x cos x]
2(4 cos 4x − 4 cos 2x + 1) − 1
4 sin x cos x cos 2x _________________
= ____________ –
= – [1 + _12 sin 2x] sin x cos x
4 2
8 cos x − 8 cos x + 2 − 1
= –1 – _12 sin 2x = 4 cos x cos 2x – _______________
cos x
4 2
8 cos x − 8 cos x + 1
= 4 cos x cos 2x – _____________
cos x
4 cos 2x cos 2x − (8 cos 4x − 8 cos 2x + 1)
= ________________________
cos x
4 cos 2x(2 cos 2x − 1) − (8 cos 4x − 8 cos 2x + 1)
= ____________________________
cos x
4 2 4 2
8 cos x − 4 cos x − 8 cos x + 8 cos x − 1
= _________________________
cos x
2
4 cos x − 1
=_ cos x
1
= 4 cos x – _
cos x
= 4 cos x – sec x

( 2 )
1 + cos 2x
_
cos 3x + cos x 1− + cos(90° − 2x)
9. _ 10. __________________
sin 3x + sin x 2 cos x + sin x
2
cos 2x cos x − sin 2x sin x + cos x 1 − cos x + sin 2x
= _____________________
sin 2x cos x + sin x cos 2x + sin x
= ____________
2 cos x + sin x
(2 cos 2x − 1)cos x − (2 sin x cos x)sin x + cos x (1 − cos 2x) + 2 sin x cos x
= ____________________________
2 = ________________
(2 sin x cos x) cos x + sin x(2 cos x − 1) + sin x 2 cos x + sin x
3 2 2
2 cos x − cos x − 2 sin x cos x + cos x sin x + 2 sin x cos x
= __________________________
2 2 = _____________
2 sin x cos x + 2 sin x cos x − sin x + sin x 2 cos x + sin x
3 2
2 cos x − 2 sin x cos x sin x(2 cos x + sin x)
= ______________ 2 = _____________
4 sin x cos x 2 cos x + sin x
2 cos 3x − 2(1 – cos2x) cos x = sin x
= ___________________
2
4 sin x cos x
3 3
2 cos x − 2 cos x + 2 cos x
= ________________ 2
4 sin x cos x
3
4 cos x − 2 cos x
=_ 2
4 sin x cos x
3
4 cos x 2 cos x
=_ 2 –
_
2
4 sin x cos x 4 sin x cos x
cos x _ 1
=_ –
sin x 2 sin x cos x
cos x _
=_ – 1
sin x sin 2x

= cot x – cosec 2x

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 138 N4 Mathematics – Lecturer Guide

Activity 4.7 SB page 251

2
1 + cos θ
_ tan θ
_
1. = tan 2θ
2 2. = sin θ cos θ
1 − sin θ 1 + tan 2θ
tan θ
1 − cos θ
LHS = _
2
LHS = _ 2
2
1 − sin θ 1 + tan θ
tan θ
sin θ
=_
2
=_ 2
2
cos θ sec θ
sin θ
= tan 2θ = RHS _
_
=_
cos θ
1
cos 2θ

2
sin θ _
=_
cos θ
× cos1 θ

= sin θ cos θ = RHS

_ _
3. (√2 cos x + 1)(√2 cos x − 1) = cos 2x 4. 2 cosec 2x = tan x + cot x
_ _
LHS = [√2 cos x + 1][√2 cos x − 1] RHS = tan x + cot x
_ _ sin x _
= 2 cos2x – √2 cos x +√2 cos x – 1 =_
cos x
+ cos x
sin x
= 2 cos2x – 1 sin x sin x + cos x cos x
= _______________
sin x cos x
= cos 2x
2
sin 2x + cos x
∴ LHS = RHS = _________
sin x cos x

1
=_
_1
2
sin 2x

= 2 cosec 2x
∴ RHS = LHS

cot x − tan x 2 − sec 2 _x

5. _ = cos 2x 6. _2 = cos x
cot x + tan x sec 2 _2x
cot x − tan x
LHS = _ 2 − sec 2 _x
cot x + tan x LHS = _2
sec 2 _2x
cos x _
_ − sin x
=_
sin x cos x
cos x _
_ sin x sec 2 _x
+ cos x 2
sin x =_ x –
2_
_2
2_
x
sec 2
sec 2
2 2
cos x − sin x
_
=_
sin x cos x
2 –1
=_
2 2
cos x + sin x
_
sin x cos x 1
_
2 2 cos 2 _2x
cos x − sin x
= ________
cos2x + sin2x
= 2 cos 2 _2x – 1
cos 2x
=_ 1 = cos x
= cos 2x ∴ LHS = RHS
∴ LHS = RHS

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 Module 4 • Trigonometry 139

sin(x − y) + sin(x + y)
7. 2 tan x = ______________
cos x cos y

sin(x − y) + sin(x + y)
RHS = ______________
cos x cos y

(sin x cos y − sin y cos x) + (sin x cos y + sin y cos x)

= ________________________________
cos x cos y

sin x cos y − sin y cos x + sin x cos y + sin y cos x

= ______________________________
cos x cos y

2 sin x cos y
=_
cos x cos y
2 sin x
=_cos x

= 2 tan x
∴ RHS = LHS

8. sin 4x = 4 sin x cos3x – 4 sin3x cos x

RHS = 4 sin x cos3x – 4 sin3x cos x
= (4 cos2x – 4 sin2x) sin x cos x
= 4(cos2x – sin2x) sin x cos x
= 2(cos2x – sin2x) 2 sin x cos x
= 2 cos 2x sin 2x
= 2 sin 2x cos 2x
= sin 4x
∴ RHS = LHS
2 cos 2x
9. 1 – tan2x = _
1 + cos 2x
10. 1 – cos 2x = tan x sin 2x
2 cos 2x
RHS = tan x sin 2x
RHS = _
cos x ( )
1 + cos 2x sin x _ 2 sin x cos x
=_ 1
2 2
2(cos x − sin x)
= __________
1 + (2 cos2x − 1) = 2 sin2x

= 2(_ )
2(cos2x − sin2x)
= __________
2 cos2x
1 − cos 2x
2
2
cos x − sin x2
= 1 – cos 2x
= ________
cos2x
∴ RHS = LHS
2 2
cos x _sin x
=_ 2 – 2
cos x cos x
= 1 – tan2x
∴ RHS = LHS

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 140 N4 Mathematics – Lecturer Guide

1 − tan x _
11. _ = cos 2x
1 + tan x 1 + sin 2x
1 − tan x
LHS = _
1 + tan x
sin x
1−_
=_cos x
sin x
1+_
cos x

cos x − sin x
_
=_ cos x
cos x + sin x
_
cos x

cos x − sin x
=_
cos x + sin x
cos x − sin x _
=_ × cos x + sin x
cos x + sin x cos x + sin x
2 2
cos x − sin x
= _________________
cos2x + 2 sin x cos x + sin2x
2 2
cos x − sin x
= __________________
(sin2x + cos2x) + 2 sin x cos x
cos 2x
=_
1 + sin 2x

∴ LHS = RHS

Activity 4.8 SB page 268

1.

y = sin x
y
1.1 Amplitude 1
1 y = sin x 1.2 Frequency 1

1.3 Period 360°

0,5
1.4 Domain 0° ≤ x ≤ 360°

1.5 Range −1 ≤ y ≤ 1
0 x
90° 180° 270° 360°
1.6 Turning (90°; 1);
–0,5 points (270°; –1)

1.7 x-intercepts (0°; 0);

–1 (180°; 0);
(360°; 0)

1.8 Asymptotes None

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 Module 4 • Trigonometry 141

y = cos x
y
1.1 Amplitude 1

1 1.2 Frequency 1
y = cos x
1.3 Period 360°
0,5
1.4 Domain 0° ≤ x ≤ 360°

1.5 Range −1 ≤ y ≤ 1
0 x
90° 180° 270° 360°
1.6 Turning (0°; 1);
–0,5 points (180°; –1);
(360°; 1)

–1 1.7 x-intercepts (90°; 0);

(270°; 0)

1.8 Asymptotes None

y = tan x

1.1 Amplitude Undefined

y 1.2 Frequency 2
y = tan x
1.3 Period 180°
2
1.4 Domain 0° ≤ x ≤ 360° ;
1 x ≠ 90° ;
x ≠ 270°

0 x 1.5 Range y∈R

90° 180° 270° 360°
1.6 Turning None
–1
points

–2 1.7 x-intercepts (0°; 0);

(180°; 0);
(360°; 0)

1.8 Asymptotes x = 90° ;

x = 270°

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 142 N4 Mathematics – Lecturer Guide

2.

Function Amplitude Frequency Period Horizontal Vertical

shift shift

2.1 y = 4 sin x − 2 4 1 360° No shift Shift

− 270° ≤ x ≤ 270° down 2

2.2 y = 3 cos 2x − 1 3 2 π No shift Shift

−π ≤ x ≤ π down 1

2.3 y = tan _2x Undefined 1 2π No shift No shift

−π ≤ x ≤ π

2.4 y = 2 cos(x − 45°) 2 1 360° Shift 45° No shift

0° ≤ x ≤ 360° right

2.5 y = 2 tan x − 4 Undefined 2 180° No shift Shift

− 180° ≤ x ≤ 180° down 4

2.6 y = 4 cos 2(x + 90°) 4 2 180° Shift 90° No shift

− 90° ≤ x ≤ 90° left

2.7 y = tan(x − _π4 ) + 1 Undefined 2 π Shift π_4 right Shift

− _π ≤ x ≤ π up 1
2

2.8 y = 3 sin(_23 x − 30°) + 1 3 _2

3 540° Shift 45° Shift
− 360° ≤ x ≤ 360° right up 1

2.9 y = − 3 tan(2x + 60°) Undefined 4 90° Shift 30° No shift

0° ≤ x ≤ 360° left

2.10 y = − 8 sin(x − _
12 )
π
8 1 2π π
Shift _
12
No shift
−_
3π _3π right
2 ≤x≤ 2

2.11 y = − cos(_2x − _
18 ) + 3
π
1 _1
2 4π Shift π_9 right Shift
− 2π ≤ x ≤ 2π up 3

3. 3.1 y = 4 sinx – 2, –270° ≤ x ≤ 270°

y
y = 4 sin x – 2
2

0 x
–270° –180° –90° 90° 180° 270°

–2

–4

–6

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 Module 4 • Trigonometry 143

3.2 y = 3 cos 2x – 1, –π ≤ x ≤ π
y

2 y = 3 cos 2x – 1

0 x
–π – _π2 π
_
2
π

–2

–4

3.3 y = tan 2_x; − π ≤ x ≤ π

y y = tan __2x

0 x
–π – _π2 π
_
2
π

–1

3.4 y = 2 cos(x − 45°); 0° ≤ x ≤ 360°

y
y = 2 cos(x − 45°)
2

0 x
–45° 45° 90° 135° 180° 225° 270° 315° 360°

–1

–2

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 144 N4 Mathematics – Lecturer Guide

3.5 y = 2 tan x – 4, –180° ≤ x ≤ 180°

y = 2 tan x – 4
y

0 x
–180° –90° 90° 180° 270°

–2

–4

–6

–8

3.6 y = 4 cos 2(x + 90°), –90°≤ x ≤ 90°

y
y = 4 cos 2(x + 90°)
4

0 x
–90° –45° 45° 90°

–2

–4

3.7 y = tan(x − _π4 ) + 1, − _π2 ≤ x ≤ π

y y = tan(x – _π4 ) + 1

0 x
3π
– _π2 – _π4 π
_
4
π
_
2
_
4
π

–2

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 Module 4 • Trigonometry 145

3.8 y = 3 sin(_23 x − 30°) + 1, –360° ≤ x ≤ 360°

y

y = 3 sin(_23x – 30°) + 1
4

–180° 360°
0 x
–360° –270° –90° 90° 180° 270°

–2

3.9 y = –3 tan(2x + 60°), 0° ≤ x ≤ 360°

y y = –3 tan(2x + 60°)

0 x
45° 90° 135° 180° 225° 270° 315° 360°

–2

–4

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 146 N4 Mathematics – Lecturer Guide

3.10 y = –8 sin(x − _
12 )
π 3π
,−_2
3π
≤x≤_2
y

y = –8 sin(x − _
12 )
π
8

0 x
3π 3π
–_2
–π – _π2 π
_
2
π _
2
–2

–4

–6

–8

3.11 y = –cos(_2x − _
18 )
π
+ 3, –2π ≤ x ≤ 2π

y = –cos(_2x − _
18 )
π
+3
4

3π
0 3π
–2π –_ –π – _π π
_ π _ 2π
2 2 2 2

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 Module 4 • Trigonometry 147

4.
Function Range Function/ Continuous/
Non-function Discontinuous

4.1 y = 4 sin x − 2 −6 ≤ y ≤ 2 Function Continuous

− 270° ≤ x ≤ 270°
4.2 y = 3 cos 2x − 1 −4 ≤ y ≤ 2 Function Continuous
−π ≤ x ≤ π
4.3 y = tan_2x y∈R Function Discontinuous
−π ≤ x ≤ π
4.4 y = 2 cos(x − 45°) −2 ≤ y ≤ 2 Function Continuous
0° ≤ x ≤ 360°
4.5 y = 2 tan x − 4 y∈R Function Discontinuous
− 180° ≤ x ≤ 180°
4.6 y = 4 cos 2(x + 90°) −4 ≤ y ≤ 4 Function Continuous
− 90° ≤ x ≤ 90°
4.7 y = tan(x − _π4 ) + 1 y∈R Function Discontinuous
− _π2 ≤ x ≤ π
4.8 y = 3 sin(_23 x − 30°) + 1 −2 ≤ y ≤ 4 Function Continuous
− 360° ≤ x ≤ 360°
4.9 y = − 3 tan(2x + 60°) y∈R Function Discontinuous
0° ≤ x ≤ 360°
4.10 y = − 8 sin(x − _
12 )
π
−8 ≤ y ≤ 8 Function Continuous
−_
3π _3π
2 ≤x≤ 2

4.11 y = − cos(_2x − _
18 ) + 3
π
2≤y≤4 Function Continuous
− 2π ≤ x ≤ 2π

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 148 N4 Mathematics – Lecturer Guide

Activity 4.9 SB page 273

1.
y = cosec x
2

1
y = sin x
0 x
–360° –300° –240° –180° –120° –60° 60° 120° 180° 240° 300° 360°

–1

–2

2.
y
y = sec x
6
5
4
3
2
1
–180° –90° 90° 180°
0 x
–π – __12 π __
1
2
π π
–1
–2
–3
–4
–5
–6

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 Module 4 • Trigonometry 149

3.
y

6
y = cot x
5
4
3
2
1
–180° –90° 90° 180°
0 x
–π – __21 π __
1
π π
–1 2

–2
–3
–4
–5
–6

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 150 N4 Mathematics – Lecturer Guide

Summative assessment: Module 4 SB page 274

_ _
√2 − √6
1. 1.1 To prove: cos 105° = _ 4
LHS = cos 105°
= cos(45° + 60°)
= cos 45° cos 60° – sin 45° sin 60°
_

√2 )( 2 ) √2 )( 2 )
= (_1_ _ 1
− (_1_ _√3

_
− √3_
1_ _
=_
2√2 2√2
_
1−_
√3
=_
2√2
_ _
1−_
√3 2
=_ ×_
√_
2√2 √2
_ _
√2 − √6
=_ 4

∴ LHS = RHS (4)

_
1.2 To prove: sin(x − _π4 ) = _
√2
2
(sin x − cos x)

LHS = sin(x − _π4 )

= sin x cos _π4 − sin _π4 cos x

= sin x (_
1_
) − ( ) cos x
1_
_
√2 √2

1_ (
=_ sin x − cos x)
√2
_
× √_2 (sin x − cos x)
1_ _
=_
√2 √2
_
√2
=_2
(sin x − cos x)

∴ LHS = RHS (4)

13 To prove: cos(x + y) cos(x – y) = (cos x cos y)2 – (sin x sin y)2
LHS = cos(x + y) cos(x – y)
= (cos x cos y – sin x sin y) (cos x cos y + sin x sin y)
= cos2x cos2y + sin x cos x sin y cos y – sin x cos x sin y cos y – sin2x sin2y
= cos2x cos2y – sin2x sin2y
= (cos x cos y)2 – (sin x sin y)2
∴ LHS = RHS (4)

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 Module 4 • Trigonometry 151

cot x – tan x
1.4 To prove: ________ = _1 sin 2x
cot2x + tan2x 2
cot x – tan x
LHS = ________
cot2x + tan2x
cot x – tan x
= __________________
(cot x + tan x)(cot x − tan x)
1
=_
cot x + tan x
1
=_
cos x _
_ sin x
sin x
+ cos x

= _______
1
cos 2x + sin x
_________
2

sin x cos x

1
=_
1
_
sin x cos x

= sin x cos x
= _12 sin 2x

∴ LHS = RHS (4)

_
2. sin 61° + sin 29° = √2 cos 16°
LHS = sin 61° + sin 29°
= sin(45° + 16°) + sin(45° – 16°)
= (sin 45° cos 16° + sin 16° cos 45°) + (sin 45° cos 16° – sin 16° cos 45°)
= sin 45° cos 16° + sin 16° cos 45° + sin 45° cos 16° – sin 16° cos 45°
= 2 sin 45° cos 16°
= 2 (_
1_
) cos 16°
√2
_2_
= cos 16°
√2
_
_2_ 2
= ×_
√_
cos 16°
√2 √2
_
2√2
=_2
cos 16°
_
= √2 cos 16°
∴ LHS = RHS (4)

3. cos(x + y) = cos x.cos y − sin x.sin y

5 )( 10 ) ( 5 )( 10 )
= (− _
2_ 3
−_
_ − _1_ _1
_
√ √ √ √
_ 6 1
= _ −_ _
√50 √50
_ 5
= _
√50
_ 5_
= 5 √2
__
_1_ √2
__
= or 2 (4)
√2

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 152 N4 Mathematics – Lecturer Guide

4. 4.1 cos 2x = 1 – 2 sin2x ∴ sin2x = _12 (1 − cos 2x)

cos 2x = 2 cos2x – 1 ∴ cos2x = _12 (1 + cos 2x)

2
sin x
But tan2x = _ 2
cos x
_1 (1 − cos 2x)
=_2
_1 (1 + cos 2x )
2

1 − cos 2x
tan2x = _
1 + cos 2x
(4)

1 − cos[2( 12 )]
π
_

( 12 ) 1 + cos[2(_
2 π
_ _________
4.2 tan =
12 )
π
]

1 − cos( _π )
=_6
1 + cos( 6 )
π
_

_
1 − (_2)
√3

=_ _
1+(2)
√3
_

_
√3
1−_
=_2
_
√3
1+_2
_
tan2(_
12 )
π 2 − √_
=_ 3
(4)
2 + √3
_
5. 5.1 cosec x + cot x = √3
_
cosec x = √3 – cot x
_
(cosec x)2 = (√3 – cot x)2
_
cosec2x = 3 – 2√3 cot x + cot2x
_
cosec2x = 3 – 2√3 cot x + (cosec2x –1)
_
cosec2x = 3 – 2√3 cot x + cosec2x –1
_
0 = 2 – 2√3 cot x
_
2√3 cot x = 2
1_
cot x = _
√3
_
∴ tan x = ⊕√3
tan +; 1st tan +; 3rd

x = xref x = 180° + xref

_ _
= tan–1(√3 ) = 180° + tan–1(√3 )
x = 60° x = 240°
_
2_ _
Verify the answers, cosec 60° + cot 60° = _ + 1_ = √3
√3 √3
_
2_ _
cosec 240° + cot 240° = –_ + 1_ ≠ √3
√3 √3
0° ≤ x ≤ 360°, therefore x = 60° (6)

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 Module 4 • Trigonometry 153

5.2 cos(x + _π3 ) = sin(x − _

18 )
π

cos(x + _π3 ) = cos[_π2 − (x − _

18 )]
π

x + _π3 = _π2 − (x − _
18 )
π

x + _π3 = _π2 − x + _
π
18

x + _π3 = − x + _
5π
9
2π
2x = _9

x = _π9 (4)

0 ≤ x ≤ _π2 , therefore x = _π9

6. 4 tan θ + 3 sec θ − 12 sin θ − 9 = 0

4(_
cos θ)
sin θ
+ 3(_
cos θ)
1
− 12 sin θ − 9 = 0
4 sin θ + 3 − 12 sin θ cos θ − 9 cos θ = 0
(4 sin θ + 3) − 3 cos θ(4 sin θ + 3) = 0
(4 sin θ + 3)(1 − 3 cos θ) = 0
sin θ = − 3_4 or cos θ = 1_3
∴ θ = 228,59° ; θ = 311,41° ; θ = 70,53° or θ = 289,47° (7)
7. cos 2A = 1 − 2 sin 2A
2 sin 2A = 1 − cos 2A
2 sin 2 A
_ = 1 − cos A
2

sin 2 A
_ = _1(1 − cos A)
2 2
_
√
1 − cos A
sin A
_=
2
_
2
(3)

8. 8.1 y = 5 cos(2x – 40°) + 2, –180° ≤ x ≤ 180°

y

8
y = 5 cos(2x – 40°) + 2
6

0 x
–180° –90° 90° 180°
–2

–4

(4)

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 154 N4 Mathematics – Lecturer Guide

8.2 y = cot x, –π ≤ x ≤ π
y

4 y = cot x

0 x
–π 2π
–_ –_π3 π
_ 2π
_ π
3 3 3
–1

–2

–3

–4
(4)

TOTAL: [60]

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 MODULE

5 Differential calculus
After they have completed this module, students should be able to:
• calculate limits that are indeterminate by making use of algebraic
expressions (the theorem of L’Hôspital may not be applied);
• use the binomial theorem in general terms;
• apply the binomial theorem with rational indices to expand a simple
binomial to four terms;
• define differentiation as a rate of change and derive the expression
f(x + ∆ x) − f(x) f(x + h) − f(x)
lim __________
∆x
or lim _________
h
from first principles with the aid
∆x→0 h→0
of a sketch as an introduction to differentiation; f(x) may only be in one of
the following forms: f(x) = ax n + b with ax n + bx n−1 + cx n−2 + … and n a
positive integer (exams will be limited to n ∈ {ℕ; n < 4});
dy
• determine __
dx
of the following standard forms:
– y=k
– y = kx n
– y = ka x
– y = ke x
– y = k ln x
– y = k log ax
– y = k sin x
– y = k cos x
– y = k tan x
– y = k cot x
– y = k sec x
– y = k cosec x;

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 156 N4 Mathematics – Lecturer Guide

• apply the chain rule to determine the first derivatives of ka nx, ke nx,
k log anx, k log enx, k sin(bx), k cos(bx), k tan(bx), k cot(bx), k sec(bx) and
k cosec(bx);
• apply the product and quotient rules for differentiation of differentiate
simple products and quotients (combinations of chain, product and
quotient rules may not be asked);
• determine the second derivatives of trigonometric functions, algebraic
terms and polynomials to determine maximum and minimum turning
points and points of inflection;
• draw neat sketch graphs of y = ax 3 + bx 2 + cx + d, where a, b, c and d are
integers; and
• sketch graphs indicating maximum and minimum values derived above.

Introduction
The word calculus describes a system of rules or reasoning used to do certain types of
calculations. It was developed independently by Sir Isaac Newton and Gottfried Leibniz
towards the end of the 17th century.

In calculus students will compare quantities that vary in a nonlinear way. Calculus
is generally used in science and engineering. It is the mathematics of rates of change.
Many concepts that they learnt about, such as velocity, acceleration and current in a
circuit, do not behave in a simple linear way. Quantities continuously change, so you
need calculus to interpret all the changes.

Differential calculus is a subfield of calculus and is the study of the rates at which
quantities change. The derivative of a function at a chosen input value describes the
rate of change of the function near that specific input value. Finding a derivative is
called differentiation.

The two main branches of calculus are differentiation and integration.

Students need the following pre-knowledge to successfully complete this module.

Pre-knowledge
• Sketching graphs (Module 3).
4 k 0
• Division by 0, for example __0 , __0 and __0 is undefined.
• __0 = 0, __0 = 0, and so on
k 3

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 Module 5 • Differential calculus 157

• Factorisation:

Common factor 3 x 2 y − 6x
= 3x(xy − 2)
Difference of two squares x2 − 9
= (x − 3)(x + 3)
Quadratic trinomial x 2 − 6x + 8
= (x − 2)(x − 4)
Sum of two cubes a 3 + b3
= (a + b)( a 2 − ab + b 2)
Difference of two cubes a 3 − b3
= (a − b)( a 2 + ab + b 2)

• x0 = 1

• FOIL: (x + h)(x + h) = x 2 + xh + xh + h 2 = x 2 + 2xh + h 2

• Factorising third-degree polynomials of the form

f(x) = ax 3 + bx 2 + cx + d:

Step 1: Find one factor by using the factor theorem.

• If f (a) = 0, then we know that (x – a) is a factor.
• We need the value for a so that f (a) = 0, because then (x – a) will be
a factor of f (x).
• Find a factor of f (x). Consider factors of the last (constant) term
first.

Example
If f (x) = x3 – 7x2 – 10x + 16

Possible factors of 16 are:

± 1; ± 2; ± 4; ± 8; ± 16

Test for factors:

Let f (–1) Let f (1)

= (–1)3 – 7(–1)2 – 10(–1) + 16 = (1)3 – 7(1)2 – 10(1) + 16
= –1 – 7 + 10 + 16 = 1 – 7 – 10 + 16
= 18 =0
∴ 18 ≠ 0 ∴ (x – 1) is a factor because the
∴ (x + 1) is NOT a factor because the remainder = f (1) = 0
remainder = f (–1) ≠ 0

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 158 N4 Mathematics – Lecturer Guide

Step 2: Use inspection or long division to determine the values of a, b and c.

f (x) = ax3 + bx2 + cx + d
= (x ± k)(ax2 + bx + c)

Using the inspection method:

• Divide f (x) by (x – a) through inspection.
• If f (x) = x3 – 7x2 – 10x + 16 and (x – 1) is a factor as proven in Step 1, then x3 –
7x2 – 10x + 16 = (x – 1)(ax2 + bx + c).
• Use inspection to find ax and c.
first terms

(x – 1)(ax2 + …)
Multiply the first terms of each bracket and compare the answer to the term x3
in the given polynomial: f (x) = x3 – 7x2 – 10x + 16
You see that:
x.ax2 = ax3
∴ x.1x2 = 1x3
Therefore a = 1
last terms

(x – 1)(ax2 + bx + c)

Multiply the last terms of each bracket and compare the answer with the last
(constant) term of f (x), which is 16. You see that:
(–1)(c) = 16
∴ c = –16
• Substitute a = 1 and c = –16 in (x – 1)(ax2 + bx + c).
∴ (x – 1)(x2 + bx – 16)
• Find b using the coefficients of x.
∴ x3 – 7x2 – 10x + 16 = (x – 1)(x2 + bx – 16)

–x2
+
bx2

∴ –x2 + bx2 = –7x2 • (–1)(x2) + (x)(bx) = –7x2

bx2 = –7x2 + x2 • Solve for b
2 2
bx = –6x
∴ b = –6
• Substitute b = –6 in (x – 1)(x2 + bx – 16), f (x) = (x – 1)(x2 – 6x – 16)
Step 3: Factorise the quadratic expression completely.
f (x) = x3 – 7x2 – 10x + 16
= (x – 1)(x2 – 6x – 16) • b = –6
= (x – l)(x – 8)(x + 2) • Factorise (x2 – 6x – 16)

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 Module 5 • Differential calculus 159

Exponential laws:
a x × a y = a x+y
ax
a x ÷ a y = __
ay
= a x−y
(a x)y = a xy
a0 = 1
1
a −x = __
ax
(ab)x = a x b x
_x __1 y _
a y = (a x) y = √a x

Factorisation:
a+b _
____ a b
c
= +_ c c

Useful definitions:

Definition Example
n 2
x
__ 1 x 1
1 a
= __ x n a
__
3
= __ x 2 3

a
__ 3
__
2 xn
= ax −n = 3x − 2
x2
n 4
ax
___ a 2x 2
3 = _ xn
b b
___
3
= __ x 43

a
___ a 2 2
4 bx n
= _ x −n ___ = __ x −4
3
b 3x 4
_ _
n m
__ 5 __3
5 √a m = a n √2 3 = 2 5
_ _ _ _
__1 __3
6 √ax = √a x 2 √2x 3 = √2 x 2

a_ __
____ a m
__ 2_ __
____ 2 __2
7 n = m = ax − n __ 3 = __2 = 2x − 3
√x m x n
√x 2
x 3

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 160 N4 Mathematics – Lecturer Guide

Logarithmic
Logarithmic
laws
laws Logarithmic
Logarithmic
form
form
and
and
exponential
exponential
form
form

loglog
x +x log
a a
+ log
y =y log
a a
= log
xy xy
a a
loglog
N=
b b
N a=(logarithmic
a (logarithmic
form)
form)
x x
_ _
x −x log
loglog
a a
− log
y =y log
a a
= log
ay ay

x k x=k k=log
loglog
a a
k log
x x
a a N b=a b(exponential
N= a
(exponential
form)
form)
a =a 1= 1
loglog
a a

1 =1 0= 0
loglog
a a Note
Note
log___
log
b b
b =b ___
loglog =a a • • ln ln
e =e log
= log
e =e 1= 1
e e
a a log log
• • loglog
x =x ln
e e
= ln
x with
x with
e =e 2,71828
= 2,71828
• • loglog
x =x log
= logx x
10 10
(base
(base= 10)
= 10)

• • Factor
Factor
theorem
theorem

Activity
Activity
5.1
5.1 SBSB
page
page
286
286
2
_____
_____ x 2x
coscos
(x 3()x 3)
1. 1. limlim 2. 2. limlim
π 1 π−1sinx
− sinx
x→5x→5 x→_2x→_2
3 3
= (5)
= (5) 2π
_ 2 π_
coscos
= _____
= _____
2 2

= 125
= 125 − π_sinπ_
1 −1sin 2 2

=0_
=_ 0
1 −11− 1

= 0_0= 0_0

= no
= no
limit
limit

( x 2(+x 24x+ −4x6−) 6 )

2 2
_ x_ +x 2+ 2
(x 2(x+2 4x
3. 3. limlim + 4x
− 3)
− 3) 4. 4. limlim
4 4
x→–x→– x→3x→3
2 2
= (–4)
= (–4)+ 4(–4)
+ 4(–4)– 3– 3 (3) 2(3+) 22+ 2
=_
=_
= 16
= 16
– 16
– 16
– 3– 3 2 2
(3) (3+) 4(3)
+ 4(3)− 6− 6
_ 9_ +92+ 2
= –3
= –3 = = + −126− 6
9 +912
11 _
= = 11
_
15 15

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 Module 5 • Differential calculus 161

lim (_ )
x→1 √3x + 1 )
(
2
4x −9 2x − 9x + 4
5. lim __ 6. 2
x→_2
1 4x + 12x − 7

( 2 ) − 9( 2 ) + 4
2
_4(1) − 9 2 _1 _1
= _
___________
√3(1) + 1 =
4(_12 ) + 12(_12 ) − 7
2

_4_
−9
=
(4) 2
√3 + 1 2 _1 − _9 + 4
−5 =_
=__ 4(_14 ) + 6 − 7
√4
_1 − _9 + 4
−5
=_
2 =_
2 2
1+6−7

= − 2_12 = _00

lim (_2x 2 − 9x + 4
2
)
x→_ 4x + 12x − 7
1
2

= lim [_
(2x − 1)(2x + 7) ]
(2x − 1)(x − 4)
1
x→_
= lim [_
2x + 7 ]
2
x−4
1
x→_2

(2)
_1 − 4
=_
2(_12 ) + 7
_1 − 4
=_
2
1+7
− _72
=_
8
7
= −_
16
_ _

(x + 2x − 3 ) ( )
4
x − 81 √5 − √x
7. lim _
2 8. lim _ x−5
x→– 3 x→5
_ _
4
(−3) − 81 √5 − (5)
√
= ___________
2 =_
(5) − 5
(−3) + 2(−3) − 3
_ _
81 − 81 √5 − √5
=_ 9−6−3
=_ 5−5

= _00 = _00
_ _

x→– 3 x + 2x − 3 )
lim (_ 2
x 4 − 81
lim (_x−5 )
√5 − √x
x→5
_ _
= lim [_
x→– 3 (x − 1)(x + 3) ]
= lim [_ x−5 ]
2 2
(x + 9)(x − 9) − √x − √5 ( )
x→5
_ _
= lim [_____________ ] [ √x + √5 ]
−(√x − √5 )
2
(x + 9)(x + 3)(x − 3)
(x − 1)(x + 3)
= lim ____________
(
_ _ _
)(
_
)
x→– 3 x→5 √x − √5

= lim [_ ] [ √x + √5 ]
−1 _
= lim _
2
(x + 9)(x − 3) _
x→– 3 (x − 1) x→5
−1
((−3) 2 + 9)((−3) − 3)
= _____________ =_
_ _
((−3) − 1) √(5) + √5
−1 _
=_
(9 + 9)(−3 − 3) =__
(−3 − 1) √5 + √5
1_
=_
(18)(−6) = −_
(−4) 2√5

−108
=_−4
= 27

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 162 N4 Mathematics – Lecturer Guide

1 1 1 +1x+ x ) )
( ( x→2x→26 −6x−−xx−)x )
( (
2 2
_3x _
3+x 2x
+− 2x1− 1 2x_−
2x4− 4
9. 9. limlim 3 3 10.10.limlim_ 2 2
x→–x→–
3(−1 ) 2 +) 22(−1)
3(−1 + 2(−1) − 1− 1 2(2)2(2)
− 4− 4
= ____________
= ____________
3 3 =_
=_2 2
1 +1(−1 + (−1
) ) 6 −6(2) − (2)
− (2−) (2)
_ 3(1)_3(1)
− 2−−21− 1 _ 4_ −44− 4
== = 6=−62−−2(4) − (4)
1 +1(−1 )3 )3
+ (−1
3 −_
=_ = 32−+−2(−1)
1 +1(−1)
1− 1 _ _ 0 0
= 6=−62−−24− 4
3 −_
=_ =13−2−1−1−21−1 1 = _00= _00

( 6 (−6x−−xx−2 )x 2 )
= _00= _00 limlim_ 2x_−
2x4− 4
x→2x→2

( ( x 3) ) [ [ x+−x6)−]6) ]
3x 2_
_ 3+x 22x+− 2x1− 1 _ 2(x
_ 2(x
− 2)− 2)
limlim 3 = lim= lim 2 2
x→–x→–1 1 1 + 1x+ x→2x→2 −( x−( +
x

( ( 1) ) [ −(x
[ −(x − 2)−]2) ]
2 2
3x _
_ 3+x 2x +−2x1− 1 2(x2(x
− 2)− 2)
= lim= lim 3 3 = lim
= lim_ _
x→–x→– 1 1 x +x 1 + x→2x→2 + 3)(x
+ 3)(x

1 [ (x
1 [+(x1)(
+ x1)(−x x−+x1)+]1) ] [ −(x
[ −(x
+ 3)+]3) ]
(3x(3x
− 1)(x
− 1)(x
+ 1)+ 1) 2 2
= lim___________
= lim ___________
2 2 = lim
= lim_ _
x→–x→– x→2x→2

1 [ x1 [− ] 1]
3x −
3x 1− 1 2 2
= lim_
= lim 2
_
2 =_
=_ −((2)−((2)+ 3)+ 3)
x→–x→– x x−+x1+
_ 2 2
=_
=_
2 2
3(−1)3(−1)
− 1− 1 = =_
−(2−(2+ 3)+ 3)
(−1(−1
) −) (−1)
− (−1)
+ 1+ 1
_ 2_ 2
= _ −3_ −3
− 1− 1
= + 1++11+ 1 = = −(5)
−(5)
(1) (1)

= _=_ −4 −4 = _=2 _
−5 −5
2
1 +11++11+ 1

= −=_34− _43 = −=_25− _25

= −=1−_13 1_13

( 3( 3 ) )
2 2
_
x _−x 36
− 36 x −x2− 2
_________
_________
11.11.limlim 12.12.limlim_ __ _
x→6x→6x −x 216
− 216 √2x√−
x→2x→2 2x2−−2√−x √x
_
__ _
(6) 2(6−) 236− 36 x −x2− 2
=_
=_
3 3 = lim _________
= lim__________
__ × ×2x___________
√_ √−
2x2−+2√+x √x
_ ___________
__ _
(6) (6−) 216 − 216 x→2x→2 2x2−−2√−x √x √2x√−
√2x√− 2x2−+2√+x √x
_
__ _
_ 36_ −3636 − 36 )(√22x
)(√−2x2−+2√+x )√x )
(x −_____________
(x2−
= 216
= 216 − 216 − 216 = lim _____________
= lim
x→2x→2 (2x(−2x2−
) −2)x− x
0 _
_ 0 _
__ _
= 0= 0 )(√22x
)(√−2x2−+2√+x )√x )
(x −_____________
(x2−
_____________
= lim
= lim x −x2− 2

( x (−x 3216 ) )
x→2x→2
_ x2_ −x 236
− 36 _
_ _ _
limlim
x→6x→6
3
− 216 = lim
= lim
√2x
√2x
− 2− +2 √+x√x
x→2x→2

[ [ ] ]
(x +(x6)(x
_____________ + 6)(x
_____________ − 6)− 6) _ _ _ _
= lim
= lim 2 2 = √=2√+2 √+2√2
x→6x→6(x −(x6)(
− x6)(+x 6x
++6x36)
+ 36)
_ _

[ x 2[+x 26x+ +6x36+]36 ]

= lim_
= lim x +x6+ 6
_ = 2=√22√2
x→6x→6
(6) (6)
+ 6+ 6
=_
=_
2 2
(6) (6+) 6(6)
+ 6(6)
+ 36
+ 36
_ 6 + 66+ 6
= =_
36 +3636++ 3636
+ 36
_ 12 12
= =_
108108

= _1= _1
9 9

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 Module 5 • Differential calculus 163

lim (_ + x)
_1 + _1
14. lim (_
x→4 2 − √x )
2
x 2 x − 16
13. 2
_
x→– 2
1
_ + _1 (4) 2 − 16
_
(−2) 2 =__
= 2 + (−2) 2 − √(4)
_ 16 − 16
= _
− _12 + _12 2 − √4
=_ 0
2−2 =_ 2−2
= _00
= _00
lim (_
2 + x)
_1 + _1
lim (_
x→4 2 − √x )
x 2 x 2 − 16
_
x→– 2

= lim (_) = lim [_ ]

2+x
_
2x (x + 4)(x − 4)
_
x→– 2 2+x x→4 −(√x − 2)
_ _
= lim (_ = lim [_______________]
x→– 2 2x )
1 (x + 4)(√x + 2)(√x − 2)
_
x→4 −(√x − 2)
1 _
=_
2(−2) = lim [−(x + 4)(√x + 2)]
x→4 _
1
=_ −4
= –((4) + 4)(√(4) + 2)
_
= –(4 +4)(√4 + 2)
= − _14
= –(8)(2 + 2)
= –(8)(4)
= –32

16. lim (_
x 2 + 5x − 8 )
15. lim (_
x→∞ 2x + 3 )
2
3x + 2 6x − 4x + 2
x→∞ 3
3(∞) + 2 6(∞) 2 − 4(∞) + 2
=_
2(∞) + 3
= ___________
2
3(∞) + 5(∞) − 8
∞ ∞
=_
∞
=_∞

lim (_
x→∞ 3x + 5x − 8 )
lim (_
3x + 3 )
3x + 2 6x 2 − 4x + 2
2
x→∞

( x + x)
3x _
_ +2
2
6x 4x _
+ 2

(_
_ −_

− 8)
= lim _
x x
2x _
3 = lim _
x2 x2 x2
_
x→∞ 3x 5x _2
x→∞ +_ 2 2 2

(2 + x )
x x x
3 + _2x
6 − _4x + _
2
= lim __3
x→∞ ( 3 + _ − _2 )
x→∞ = lim _ x2
5 8
x x
2
3+_ 4 2
_ ∞ 6−_ +_
=
=_
(∞) (∞) 2
3
2+_ 5 8
∞ 3+_
(∞)
−_ 2
(∞)
_3+0
= _6−0+0
2+0 = 3+0−0
= _3
2 = _6
3
= 1_12 =2

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 164 N4 Mathematics – Lecturer Guide

( (3 3 ) ) ( (2 2 ) )
2 2 5 5
x −_
_ x 2x
−+2x2+ 2 3x _
_ 3−x 9x
−+9x2+ 2
17.17.limlim 18.18.limlim
x→∞
x→∞ x +x 2+ 2 x→∞ 2x
x→∞ 2+x 5x
++5x1+ 1
2
(∞)_
(∞ −) 22(∞)
− 2(∞)+ 2+ 2 ) 5 −) 59(∞)
3(∞___________
3(∞ − 9(∞)+ 2+ 2
=_ = 3 3 = ___________
= 2 2
(∞)(∞+) 2+ 2 2(∞2(∞
) +) 5(∞)+ 5(∞)+ 1+ 1
∞ _
_ ∞ ∞ _
_ ∞
= ∞= ∞ = ∞= ∞

( (x 3 +x 32+ 2) ) ( 2x( 2+x 25x+ +5x1+) 1 )

x 2 −_
_ x 22x
−+ 2x2+ 2 _3x 5_3−x 59x
−+ 9x2+ 2
limlim limlim 2
x→∞x→∞ x→∞x→∞
2 2 5 5
2x _
−+2x_
2 _
+ 2 3x _
3x _
− 9x−+_
9x_
2 _
+ 2

( (xx +_xx _x2+ _x2) )

_
x _
−x _ _

( 2xx( _ +)1 )
= lim
= lim _ _
x3 _
x 3x 3 x 3x 3 x 3
= lim
= lim_ _
x5 _
x 5 x 5 x 5x 5 x 5
3 3 2 2
2x _
x→∞x→∞ 3 3 3 3
x→∞x→∞ + 5x++_
5
5x_
1 _
5 5 5 5 5
x x xx x
_1 −_1_
2 _
−+2_ 2 _
+ 2 3 −3_ 9 _
−+9_ 2 _
+ 2

( (1 +1_+ 2) ) (( ))
= lim
= lim _x _ xx 2 x 2x 3 x 3
= lim
= lim_ _x 4 x 4x 5 x 5
2 _ 2 25 _
x→∞x→∞ x→∞_3 +_3_+
x→∞ +5_ 1 _
+ 1
x x 3 3
x x x 4 x 4x 5 x 5
1 _
_ −1 _
−2 _
+2 _
+2 _
2
−9 4_
3 −3_ +9 4_
+2 5_
2

=_
(∞) _
=_
=_
(∞)(∞) (∞) (∞) (∞)
2 2 3 3
(∞) (∞) (∞) (∞) 5

=
+2 3_
1 +1_ 2 2 _
_ +2 _
+5 _
+5 _
+1 _
1
3
(∞) (∞) (∞) 3(∞) 3(∞) 4(∞) 4(∞) 5(∞) 5
0 −_
=10+0−1+0+00+0 0
=_ 3 −_
= 30−+00+ 0
=_
0 +00++00+ 0

= _01= _01 = _30= _30

= 0= 0 = no
= no
limit
limit
_
_

( 2(−23x− 3x) ) ( ( x x ) )
√1 _
_ √+14+x 4x 2 2
_ _ ) 3 x−) 38− 8
(2 +(2x+
19.19.limlim 20.20.limlim
x→∞
x→∞ x→0x→0
_ _ 3
(2 +(2(0)
+ )(0)−) 38− 8
√= _
√ =_
=_
2 2
1 +14(∞
+ 4(∞
) )
=_2 −23(∞)
− 3(∞) (0) (0)
(2) 3(2−) 38− 8
=∞ _
=_ ∞ =_
=_
0 0
−∞−∞
8 −_
= ∞ _
−∞
−=_ =088−0 8
=_
∞ ∞
_
_

( 2(−23x− 3x) )
limlim√1 _
_ √+14+x 4x 2 2
= _00= _00

( ( x x ) )
x→∞
x→∞ _
_ ) 3 x−) 38− 8
(2 +(2x+

( ( − 3x) )
limlim_ _
(x 2 (x 2 ) )
√√
12 _
x 2 x_ +14+ 4 x→0x→0
= lim
= lim_ _

[ [ ] ]
x→∞ 2 −23x ((2 ((2
+ x)+−x)2)((2
− 2)((2
______________________
______________________ ) 2 x+) 22(2
+ x+ + 2(2
+ x)++x)4)+ 4)
x→∞
__ = lim
= lim

(( ))
x→0x→0 x x
√√
1 _
x2 +124+ 4
x _

[ [ ] ]
= lim_
x_
x
= lim (2 +(2x+−x2)((2
− 2)((2
_____________________ ) 2 x+) 22(2
+ x+
_____________________ + 2(2
+ x)++x)4)+ 4)
x→∞ 2 −23x
x→∞ − 3x = lim
= lim x x
__ x→0x→0
√_
√
[ [ ] ]
1 _ 1

(( ))
x _2x+ 42 + 4
_ x x + x+
x((2x((2 ) 2 x+) 22(2
_______________
_______________ + 2(2
+ x)++x)4)+ 4)
= lim
= lim
= lim_
x_
x
= lim 2 23x 3x
x→0x→0 x x
x→∞_ −__− _
x→∞
[(2[(2 + 4+] 4]
x xx x
_ _
= lim
= lim + x+) 2x+
) 2 2(2
+ 2(2
+ x)
+ x)
√2 _
√
1 _
+14+ 4

((_ −_23−)
3 )
_ x→0x→0
= lim_
2 2
x x
= lim = (2
= (2
+ (0))2 2
+ (0))
+ 2(2
+ 2(2
+ (0))
+ (0))
+ 4+ 4
x→∞x→∞
x x
_
_
= (2
= (2 + 20)+2 2(2
+ 0) + 2(2
+ 0)
+ 0)
+ 4+ 4
√=2_
√2
1 _
_ +1 4+ 4
=_
2 2
(∞) (∞)
2 2
__
(∞) (∞)
− 3− 3 = (2) + 2(2)
= (2) + 2(2)
+ 4+ 4
__
√0 _
=0 −√+0034−+34
=_ = 4=+4 4++4 4+ 4
_ _
√4 _ = 12
= 12
= √4
=_
−3 −3

=2 _
=_ 2
−3 −3

= −=_23− _23

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 Module 5 • Differential calculus 165

Activity 5.2 SB page 292

1. 1.1 The fourth term is T4,

∴ T4 = Tr+1
4=r+1
r = 3; n = 16
Thus,
T3+1 = _ n!
r!(n − r)!
x n–rh r
16!
T4 = _
3!(16 − 3)!
(x) 13(5) 3

T4 = 70 000x13

1.2 The eighth term is T8,

∴ T8 = Tr+1
8=r+1
r = 7; n = 14
14!
_
T7+1 = _
7!(14 − 7)!
(√2 ) 7(−x) 7
_
T8 = –27 456√2 x 7

1.3 The seventh term is T7,

∴ T7 = Tr+1
7=r+1
r = 6; n = 20
T6+1 = _ n!
r!(n − r)!
x n–rh r
20!
T7 = _
6!(20 − 6)!
(k 2) 14(5ℓ) 6

T7 = 605 625 000k28ℓ6

1.4 The nineth term is T9,

∴ T9 = Tr+1
9=r+1
r = 8; n = 11
T8+1 = _ n!
r!(n − r)!
x n–rh r

(x 3) 3(− _3x )
8
11!
T9 = _
8!(11 − 8)!

T9 = 1 082 565x

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 166 N4 Mathematics – Lecturer Guide

2. 2.1 (x + 2)7
(x) 7(2) 0 7(x) 6(2) 1 7.6(x) 5(2) 2 7.6.5(x) 4(2) 3
=_
0!
+_
1!
+_
2!
+_
3!
+…
7 6 5 4
x .1 _
=_1
+ 7.x1 .2 + _
7.6.x .4 _
2
+ 7.6.5.6x .8 + …

= x7 + 14x6 + 84x5 + 280x4 + …

2.2 (1 − _4x ) 15

[ ]
15
= (1) + (− _4x )

(1) ( 4 ) 15(1) ( 4 ) 15.14(1) ( 4 ) 15.14.13(1) ( 4 )

15 − _x 0
14 − _x 1
13 − _x 2
12 − _x 3

= _ + _ + ___________+ _____________ + …
0! 1! 2! 3!
2 3
15.1.− _4x 15.14.1._
x
15.14.13.1.− _
x
1.1 _ _ _
=_
16 64
1
+ 1 + 2
+ 6
+…
2 3
15x _
=1–_4
+ 105
16
x 455x
–_64
+…
1
2.3 _ 5
(1 + x)
= (1 + x)–5
(1) −5(x) 0 (−5)(1) −6(x) 1 (−5)(−6)(1) −7(x) 2 (−5)(−6)(−7)(1) −8(x) 3
=_
0!
+_
1!
+ ____________
2!
+ _______________
3!
+…
(−5)1x
11 _ _ ____________ (−5)(−6)1x 2 (−5)(−6)(−7)1x 3
=_1
+ 1 + 2
+ 6
+…
= 1 – 5x + 15x2 – 35x3 + …

2.4 (x − _15 )
− 12

= [(x) + (−_15 )]
− 12

(x) ( 5 ) (−12)(x) ( 5 ) (−12)(−13)(x) ( 5 ) (−12)(−13)(−14)(x) ( 5 )

0 1 2 3
1 1 1 1
−12 −_ −13 −_ −14 −_ −15 −_

= _ + __________ + _____________ + ________________ +…

0! 1! 2! 3!

(−12)(x )( 5 ) ____________
(−12)(−13)(x )( 25 ) _________________
(−12)(−13)(−14)(x )( 125 )
−12 −13 −_1 −14
1
_
−15
1
−_
x 1 __________
=_ + 1
+ 1
+ +… 2 6
1
_+_ 12 _78 _364
= + + +…
x 12 5x 13 25x 14 125x 15

2.5 (_3x + y) 10

(3) (3) (3) (3)

_x 10 0
10 _x 9(y) 1 10.9 _x 8(y) 2 10.9.8 _x 7(y) 3
(y)
=_+_+_+_+… 0! 1! 2! 3!
10 9 8 7
_
x
1 10_
x
y 10.9_
x
y2 10.9.8_
x
y3
=_ +_ +_ +_
59 049 19 683 6 561 2 187
1 1 2 6
+…
10 10x 9y 5x 8y 2 40x 7y 3
=_ x
+_+_
59 049 19 683 729
+_
729
+…

N4 Mathematics - Lecturers Guide A4 Layout.indd 166 28/02/2022 11:37 am

 Module 5 • Differential calculus 167

2.6 (4x + 3)6

(4x) 6.(3) 0 6(4x) 5(3) 1 6.5(4x) 4(3) 2 6.5.4(4x) 3(3) 3
=_
0!
+_
1!
+_
2!
+_
3!
+…
(4x) 61 6(4x) 5.3 6.5(4x) 4.9 6.5.4(4x) 3.27
=_
1
+_
1
+_
2
+_
6
+…

= 4 096x6 + 18 432x5 + 34 560x4 + 34 560x3 + …

Sum of the first four binomial coefficients
= 4 096 + 18 432 + 34 560 + 34 560
= 91 648
2.7 (x – 2y)9
= [(x) + (–2y)]9
(x) 9(−2y) 0 9(x) 8(−2y) 1 9.8(x) 7(−2y) 2 9.8.7(x) 6(−2y) 3
=_
0!
+_
1!
+_
2!
+ ___________
3!
+…
9
x .1 _ _ _ 9.x 8(−2y) 9.8x 74y 2 9.8.7x 6.−8y 3
=_1
+ 1 + 2 + 6
+…

= x9 – 18x8y + 144x7y2 – 672x6y3 + …

Sum of the first four binomial coefficients
= 1 – 18 + 144 – 672
= –545
2.8 (2a – 3b)–8
= [(2a) + (–3b)] –8
(2a) −8(−3b) 0 (−8)(2a) −9(−3b) 1 (−8)(−9)(2a) −10(−3b) 2 (−8)(−9)(−10)(2a) −11(−3b) 3
=_
0!
+ ___________
1!
+ _______________
2!
+ __________________
3!
+…
1
_ 1 1 1
.1 (−8)_ 9 (−3b) (−8)(−9)_ 10 9b
2
(−8)(−9)(−10)_ 3
11 (−27b )

= _ + _ + ____________ + ___________________ + …
8
256a 512a 1 024a 2 048a
1 1 2 6
2 3
1 3b 81b 405b
=_ 8 +
_
9 +
_
10 +
_
11 + …
256a 64a 256a 256a
_3
2.9 (x + 7) 4

4( 4) ( )(
_3 − _1 (x) −_54(7) 2 _3 − _1 − _54 )(x) −4(7) 3
_9
_3 _3 (x) −_14(7) 1
(x) 4(7) 0
_ _
4 ___________ _______________
4 4
= 0!
+ 1!
+ 2!
+ 3!
+…

4 ( 4) 4 ( 4) ( 4)
_3 _3 x −_14 7 _3 − _1 x −_54 49 _3 − _1 − _5 x −4 343
_9

x 4.1 _
=_ + _ + ______________ + …
4
+ 1 1 2 6
21_3 − _1 147 − _5 1 715 − _9
= x4 + _4
x 4 –_
32
x 4 +_128
x 4 +…
21_3147 1 715
= x4 + _ –_ +_ +… _1 _5 _9
4x 4 32x 4 128x 4
21_3147 1 715
= x4 + _ –_ 1
+_ 2
+… _1 _1 _1
4x 4 32x .x 4 128x .x 4
_
4 21 147 1 715
= √x 3 + _
4
_ –_ 4
_ +_
24
_ +…
4√x 32x √ x 128x √ x

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 168 N4 Mathematics – Lecturer Guide

_
(5 )
_x − 6 9
2.10
√ _1

[ ]
= (_5x − 6) 9
2

= ( 5 − 6) 2
_9
_
x
_9

[ ]
= (_5x ) + (−6)
2

( 5) 2 (5) 2 2 (5) 2 2 2 (5)

_x _92 (−6) 0 _9 _x _72 (−6) 1 _9 _7 _x _25 (−6) 2 _9 _7 _5 _x _32 (−6) 3
= _ + _ + _ + _ +…
0! 1! 2! 3!

( 5) 2 (5) 2 2 (5) 2 2 2 (5)

_x _92.1 _9 _x _72 (−6) _9 _7 _x _52 36 _9 _7 _5 _x _32 (−216)
= _ + _ + _ + ___________ +…
1 1 2 6

= (_5x ) 2 – 27(_5x ) 2 + _2 (5)

567 _
2 (5)
– 2835
_9 x 2 _ _7 _5 _3
_x 2 + …

= (_5x ) 4(_5x ) 2 – 27(_5x ) 3(_5x ) 2 + _2 (5) (5)

567 _
2 (5) (5)
– 2835
_1 x 2 _
x 2 _ _1 _1 _1
_x 1 _x 2 + …

_ _ _ _
4 3 2
=_ – 27x √_x + _
x √x_ _ 567x _√x _
– 567x_√x + …
625√5 125√5 50√5 2√5
_ _ _ _ _ _ _ _
4 3 2
√5 x √x 27√5 x √x _
=_ –_
3 125
+ 567√5 x √x – _
567√5 x√x
…
625 250 10
− _85
2.11 (4r + _2S )
(4r) −5(_2S ) (− 5 )(4r) ( 2 ) (− 5 )(− 5 )(4r) ( 2 ) (− 5 )(− 5 )(− 5 )(4r) ( 2 )
0 1 2 3
_8
_8 13
−_
5 _S _8 13
_ 18
−_5 _S _8 13
_ 18
_ 23
−_
5 _S
=_
0!
+ ___________
1!
+ _______________
2!
+ ____________________
3!
+…

(− 5 )(4r) 2 (− 5 )(− 5 )(4r) 4 (− 5 )(− 5 )(− 5 )(4r) 8

2 3
_8 _8 13
−_
5 _
S _8 13
_ 18
−_ S
5 _ _8 13
_ 18
_ 23
−_
5 _
S
(4r) −5.1
_ _ ______________ __________________
= 1
+ 1
+ 2
+ 6
+…
2 3
4S _8 13S 39S 13
_ 18
_ 23
_
= (4r) − 5 – _
5
(4r) − 5 + _25
(4r) − 5 – _
125
(4r) − 5 + …
2 3
1 4S _
=_ –_
5
. 1 +_
13S _
25
_8
. 1 –_
39S _
125
. 1 +… 13
_ 18
_ 23
_
(4r) 5 (4r) 5 (4r) 5 (4r) 5
2 3
1 4S _
=_
1
–_
5
. 21 + _
13S _
25
. 31 – _
39S _
125
. 41 + …
_3 _3 _3 _3
(4r) (4r) 5 (4r) (4r) 5 (4r) (4r) 5 (4r) (4r) 5
2 3
1_ _
=_ 5
– 2 5S_3 + _ 13S _ _
35
– 39S _
5
+…
4r √(4r) 3
20r √(4r) 1 600r √(4r) 3
32 000r 4 √(4r) 3

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 Module 5 • Differential calculus 169

1
2.12 _
_

√3x −
1 3 _
x2

1
=_ _1

( x 2)
1
3x − _
3

− _13

( )
1
= 3x − _2
x
− _13

[ ( )]
1
= (3x) + − _2
x

( x 2) (− 3 )(3x) (− x 2 ) (− 3 )(− 3 )(3x) (− x 2 ) (− 3 )(− 3 )(− 3 ) (3x) (− x 2 )

0 1 2 3
1_1
_1 −_43 1 _1 _4 −_73 1 _1 _4 _7 10
−_ 1
(3x) −3 −_ _ _ 3 _
= _ + ___________ + _______________ + __________________ +…
0! 1! 2! 3!

(− 3 )(3x) (− x 2 ) (− 3 )(− 3 )(3x) x4 (− 3 )(− 3 )(− 3 )(3x) (− x 6 )

_1
_1 −_43 1
_ _1 _4 1
−_73 _ _1 _4 _7 10
−_
3
1
_
(3x) −3.1
_ ____________ _____________ ____________________
= 1
+ 1
+ 2
+ 6
+…
1 2
+ 14 6 .(3x) − 3 + …
_1 _4 _7 10
_
= (3x) − 3 + _ 2 .(3x)
−3
+_ 4 .(3x)
−3 _
3x 9x 81x
1
_ _1 _ 1 _2 _ 1 _14 _ 1
= _1
+ 2. 4 + 4. 7 + 6. 10 +…
(3x) 3 3x (3x) _3 9x (3x) _3 81x (3x) _3
1 1 _ 1 2 _ 1 14 _ 1
=_ +_2. +_
_1 4. +_ 6. +… 1 1 1
(3x) 3 2 33x 3x.(3x) _3 9x (3x) (3x) _3 81x (3x) (3x) _3
1 1 2 14
=_
3
_ +_
33
_ +_
63
_ +_
93
_
√ 3x 9x √ 3x 81x √ 3x 2 187x √ 3x

Activity 5.3 SB page 300

1. f (x) = 5x – 3
f (x + h) = 5(x + h) – 3
f (x + h) = 5x + 5h – 3
f (x + h) – f (x) = (5x + 5h – 3) – (5x – 3)
= 5x + 5h – 3 – 5x + 3
f (x + h) – f (x) = 5h
f(x + h) − f(x) _
_
h
= 5h h
f(x + h) − f(x)
_
h
=5
f(x + h) − f(x)
f ʹ(x) = lim _
h h→0

= lim (5)
h→0

f ʹ(x) = 5

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 170 N4 Mathematics – Lecturer Guide

2. y = 2x2
∴ f (x) = 2x2
f (x + h) = 2(x + h)2
= 2(x + h)(x + h)
= 2(x2 + 2xh + h2)
f (x + h) = 2x2 + 4xh + 2h2
f (x + h) – f (x) = (2x2 + 4xh + 2h2) – (2x2)
= 2x2 + 4xh + 2h2 – 2x2
f (x + h) – f (x) = 4xh + 2h2
f(x + h) − f(x) _ 2
_
h
= 4xh + 2h h
h(4x + 2h)
=_
h
f(x + h) − f(x)
_
h
= 4x + 2h
f(x + h) − f(x)
f ʹ(x) = lim _
hh→0

= lim (4x + 2h)

h→0
= 4x + 2(0)
f ʹ(x) = 4x
dy
∴_
dx
= 4x

3. y=4
f(x) = 4x 0
f(x + h) = 4(x + h)0
=4
f(x + h) − f(x)
lim _________
h
h→0
4−4
= lim ____
h
h→0
0
= lim _h
h→0

= lim 0
h→0

∴ f ʹ(x) = 0
4. f (x) = 9x2 – 4
f (x + h) = 9(x + h)2 – 4
= 9(x + h)(x + h) – 4
= 9(x2 + 2xh + h2) – 4
f (x + h) = 9x2 + 18xh + 9h2 – 4
f (x + h) – f (x) = (9x2 + 18xh + 9h2 – 4) – (9x2 – 4)
= 9x2 + 18xh + 9h2 – 4 – 9x2+ 4
f (x + h) – f (x) = 18xh + 9h2

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 Module 5 • Differential calculus 171

f(x + h) − f(x) _ 2
_
h
= 18xh + 9h
h
h(18x + 9h)
=_
h
f(x + h) − f(x)
_
h
= 18x + 9h
f(x + h) − f(x)
f ʹ(x) = lim _
h
h→0

= lim (18x + 9h)

h→0
= 18x + 9(0)
f ʹ(x) = 18x

5. y = –3x2 + 7
∴ f (x) = –3x2 + 7
f (x + h) = –3(x + h)2 + 7
= –3(x + h)(x + h) + 7
= –3(x2 + 2xh + h2) + 7
f (x + h) = –3x2 – 6xh – 3h2 + 7
f (x + h) – f (x) = (–3x2 – 6xh – 3h2 + 7) – (–3x2 + 7)
= –3x2 – 6xh – 3h2 + 7 + 3x2 – 7
f (x + h) – f (x) = –6xh – 3h2
f(x + h) − f(x) _ 2
_
h
= −6xh – 3h
h
h(−6x − 3h)
=_
h
f(x + h) − f(x)
_
h
= –6x – 3h
f(x + h) − f(x)
f ʹ(x) = lim _
h
h→0

= lim (−6x − 3h)

h→0
= –6x – 3(0)
f ʹ(x) = –6x
dy
∴_
dx
= –6x

6. f (x) = x2 – 6x
f (x + h) = (x + h)2 – 6(x + h)
= (x + h)(x + h) – 6(x + h)
f (x + h) = x2 + 2xh + h2 – 6x – 6h
f (x + h) – f (x) = (x2 + 2xh + h2 – 6x – 6h) – (x2 – 6x)
= x2 + 2xh + h2 – 6x – 6h – x2 + 6x
f (x + h) – f (x) = 2xh – 6h + h2
f(x + h) − f(x) _ 2
_
h
= 2xh − 6h + hh
h(2x − 6 + h)
=_
h

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 172 N4 Mathematics – Lecturer Guide

f(x + h) − f(x)
_
h
= 2x – 6 + h
f(x + h) − f(x)
f ʹ(x) = lim _
h
h→0

= lim (2x − 6 + h)
h→0
= 2x – 6 + (0)
f ʹ(x) = 2x – 6

7. y = 2x2 + 3x – 8
∴ f (x) = 2x2 + 3x – 8
f (x + h) = 2(x + h)2 + 3(x + h) – 8
= 2(x + h)(x + h) + 3(x + h) – 8
= 2(x2 + 2xh + h2) + 3(x + h) – 8
f (x + h) = 2x2 + 4xh + 2h2 + 3x + 3h – 8
f (x + h) – f (x) = (2x2 + 4xh + 2h2 + 3x + 3h – 8) – (2x2 + 3x – 8)
= 2x2 + 4xh + 2h2 + 3x + 3h – 8 – 2x2 – 3x + 8
f (x + h) – f (x) = 4xh + 3h + 2h2
f(x + h) − f(x) _ 2
_
h
= 4xh + 3h + 2h h
h(4x + 3 + 2h)
=_
h
f(x + h) − f(x)
_
h
= 4x + 3 + 2h
f(x + h) − f(x)
f ʹ(x) = lim _
h
h→0

= lim (4x + 3 + 2h)

h→0
= 4x + 3 + 2(0)
f ʹ(x) = 4x + 3
dy
∴_
dx
= 4x + 3

8. f (x) = x3
f (x + h) = (x + h)3
(x) 3(h) 0 3(x) 2(h) 1 3.2(x) 1(h) 2 3.2.1(x) 0.(h) 3
=_
0!
+_
1!
+_
2!
+_
3!
3 2 1 2 3
x .1 _
=_1
+ 3.x1 .h + _
3.2.x .h
2
3.2.1.1.h
+_ 6

f (x + h) = x3 + 3x2h + 3xh2 + h3
f (x + h) – f (x) = (x3 + 3x2h + 3xh2 + h3) – (x3)
= x3 + 3x2h + 3xh2 + h3 – x3
f (x + h) – f (x) = 3x2h + 3xh2 + h3
f(x + h) − f(x) ___________
2 2 3
_
h
= 3x h + 3xh + h h
h(3x + 3xh + h 2)
2
= ___________
h

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 Module 5 • Differential calculus 173

f(x + h) − f(x)
_
h
= 3x2 + 3xh + h2
f(x + h) − f(x)
f ʹ(x) = lim _
h
h→0

= lim (3x 2 + 3xh + h 2)

h→0
= 3x2 + 3x(0) + (0)2
f ʹ(x) = 3x2

9. y = 5x3 – 4
∴ f (x) = 5x3 – 4
f (x + h) = 5(x + h)3 – 4

= 5[_ ]–4
(x) 3(h) 0 3(x) 2(h) 1 3.2(x) 1(h) 2 3.2.1(x) 0(h) 3
0!
+_
1!
+_
2!
+_
3!

= 5[_ ]–4
3 2 2 3
x .1 _
1
+ 3.x1 .h + _
3.2.x.h
2
3.2.1.1.h
+_ 6

= 5[x3 + 3x2h + 3xh2 + h3] – 4

f (x + h) = 5x3 + 15x2h + 15xh2 + 5h3 – 4
f (x + h) – f (x) = (5x3 + 15x2h + 15xh2 + 5h3 – 4) – (5x3 – 4)
= 5x3 + 15x2h + 15xh2 + 5h3 – 4 – 5x3 + 4
f (x + h) – f (x) = 15x2h + 15xh2 + 5h3
f(x + h) − f(x) _____________
2 2 3
_
h
= 15x h + 15xh + 5h h
h(15x + 15xh + 5h 2)
2
= _____________
h
f(x + h) − f(x)
_
h
= 15x2 + 15xh + 5h2
f(x + h) − f(x)
f ʹ(x) = lim _
h
h→0

= lim (15x 2 + 15xh + 5h 2)

h→0
= 15x2 + 15x(0) + 5(0)2
f ʹ(x) = 15x2
dy
∴_
dx
= 15x2

10. f (x) = –2x3 + 4x

f (x + h) = –2(x + h)3 + 4(x + h)
= –2[_ ] + 4(x + h)
(x) 3(h) 0 3(x) 2(h) 1 3.2(x) 1(h) 2 3.2.1(x) 0(h) 3
0!
+_
1!
+_
2!
+_
3!

= –2[_ ] + 4(x + h)
3 2 2 3
x .1 _
1
+ 3.x1 .h + _
3.2.x.h
2
3.2.1.1.h
+_ 6

= –2[x3 + 3x2h + 3xh2 + h3] + 4(x + h)

f (x + h) = –2x3 – 6x2h – 6xh2 – 2h3 + 4x + 4h
f (x + h) – f (x) = (–2x3 – 6x2h – 6xh2 – 2h3 + 4x + 4h) – (–2x3 + 4x)
= –2x3 – 6x2h – 6xh2 – 2h3 + 4x + 4h + 2x3 – 4x
f (x + h) – f (x) = –6x2h + 4h – 6xh2 – 2h3

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 174 N4 Mathematics – Lecturer Guide

f(x + h) − f(x) ________________

2 2 3
_
h
= − 6x h + 4h − 6xh − 2h h
h(−6x + 4 − 6xh − 2h 2)
2
= _______________
h
f(x + h) − f(x)
_
h
= –6x2 + 4 – 6xh – 2h2
f(x + h) − f(x)
f ʹ(x) = lim _
h
h→0

= lim (−6x 2 + 4 − 6xh − 2h 2)

h→0
= –6x2 + 4 – 6x(0) – 2(0)2
f ʹ(x) = –6x2 + 4

11. y = –6x4
∴ f (x) = –6x4
f (x + h) = –6(x + h)4
= –6[_ ]
(x) 4(h) 0 4(x) 3(h) 1 4.3(x) 2(h) 2 4.3.2(x) 1(h) 3 4.3.2.1(x) 0(h) 4
0!
+_
1!
+_
2!
+_
3!
+_
4!

= –6[_ ]
4 3 2 2 3 4
x .1 _
1
+ 4.x1 .h + _
4.3.x .h
2
4.3.2.x.h _
+_ 6
+ 4.3.2.1.1.
24
h

= –6[x4 + 4x3h + 6x2h2 + 4xh3 + h4]

f (x + h) = –6x4 – 24x3h – 36x2h2 – 24xh3 – 6h4
f (x + h) – f (x) = (–6x4 – 24x3h – 36x2h2 – 24xh3 – 6h4) – (–6x4)
= –6x4 – 24x3h – 36x2h2 – 24xh3 – 6h4 + 6x4
f (x + h) – f (x) = –24x3h – 36x2h2 – 24xh3 – 6h4
f(x + h) − f(x) ___________________
3 2 2 3 4
_
h
= −24x h − 36x h − 24xh − 6h h
h(−24x − 36x h − 24xh 2 − 6h 3)
3 2
= ____________________
h
f(x + h) − f(x)
_
h
= –24x3 – 36x2h – 24xh2 – 6h3
f(x + h) − f(x)
f ʹ(x) = lim _
h
h→0

= lim (–24x3 – 36x2h – 24xh2 – 6h3)

h→0
= –24x3 – 36x2(0) – 24x(0) – 6(0)3
f ʹ(x) = –24x3
dy
∴_
dx
= –24x3

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 Module 5 • Differential calculus 175

9_
12. y=_
4
4 √x
f(x) = _9 x − 1_4
4

f(x + h) = 9_4 (x + h)−4

1_

4[
x − 4 h 3 + …]
( 4)( 4)
− 1_ − 5_ (
− 4)(− 4)(− 4)
_1 5_ 9_

+ (− _14) x −4 h − 9_4
5_ 13
= _9 x − 1_4 + _ x h + 2 ___________ _

2! 3!

= 9_4[x −4 − 41_ x −4 h + _ x h + …]
5_ 9_ 13
1_
5 −4 2 _ 15 − 4 3 _

32
x h − 128
5_ 9_ 13
= 9_4 x −4 − _9 −4 45 − 4 2 135
1_ _

16
x h+_
128
x h −_512
x − 4 h3 + …
_9 1
−_4 9
__ 5
−_4
x − 16 x h + 128 x h − 512 x h + …. 4 x
45
___ 9
−_4 2 135
___ 13
−__
4
3 _9 1
−_4

lim ______________________________
4
h
h→0
9 − _54
= −_
16
x
9_ 0,563
= −_
4 or − _
4
_
5
16 √x 5
√x

Activity 5.4 SB page 306

1. 1.1 y=3 1.2 y = −x 8

dy _ dy _
∴ _
dx
= d (3) dx
∴ _
dx
= d (− x 8)
dx

=0 = −_ (x )
d 8
dx

= − 8(x 8−1)
= − 8x 7

1.3 f(x) = 4x 2 1.4 f(x) = _5x

f ʹ(x) = 8x = 51_ x

f ʹ(x) = 51_
5
1.5 y=_4 1.6 f(x) = a 2 x
x

= 5x −4 f ʹ(x) = a 2

f ʹ(x) = − 20x −5
− 20
= __
5
x
_
1.7 f(x) = 1_2 x −8 1.8 f(x) = − 3√x
_1
f ʹ(x) = (− 8) 1_2 x −9 = − 3x 2

f ʹ(x) = − 3_2 x −2
1_

= − 4x −9
4 3_
= −_9 = −_
x 2√x

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 176 N4 Mathematics – Lecturer Guide

2_
1.9 f(x) = _
4 1.10 y = mn2a3
√x
dy _
= 2
_
1_
_
dx
= d (m n 2 a 3)
dx
x4
− _14
= 2x =0
5_
f ʹ(x) = − 1_4(2) x −4
_5
= − 1_2 x −4
1_
= −_
4 5
2 √x
1.11 y = 2π 2
dy _
_
dx
= d (2π 2)
dx

=0
6 1
2. 2.1 y=_2 2.2 y = −_ 3
x 4x

= 6x −2
= − _14 x −3
dy
_ dy 3_ − 4
_
dx
= − 12x −3 dx
= x 4
3
= − 12
_
3 =_ 4
x 4x
3
_ _
2.3 y = − 2 √x 2.4 y = √5x
_1 _ 1_
= − 2x 3 = √5 x 2
_
(2)
dy _2 dy 1_
_
dx
= − 2_ x −3
3
_
dx
= √5 1_ x −2
_
2_ √5_
= −_
3 =_
3 √x
2 2√x
2
2.5 y= _
x_ 2.6 y = a 2 b2 x 2
√x
1_ dy
_
= x 2−2 dx
= 2a 2 b 2 x
_3
=x 2

dy 3_ 1_
_
dx
= x2 2
_
= 3_2 √x
_
3√x
=_ 2
3
πx 1
2.7 y=_6
2.8 y=_ −3
ax

= _π x 3 = _1 x 3
6 a

(a)
dy dy
_
dx
= 3 × π_ x 2 6
_
dx
= 3 _1 x2
2
= π_2 x 2 or _
πx
2
= _a3 x 2
2
3
= 3_
x
a
or _a x2

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 Module 5 • Differential calculus 177

1_
2.9 y = −_ 3
3√ x
3_
= − 3_1 x −2

2( 3)
dy 5_
_
dx
= − 3_ − 1_ x −2
1_
=_ 5
2√x

3.2 y = x 2(x − _1x ) + 5

2
3. 3.1 y = − x 3 − 3x 2 + 2x − 8
dy _
_
dx
= d (− x 3 − 3x 2 + 2x − 8)
dx = x 2(x − _1x )(x − _1x ) + 5
= −_
d 3
x − 3_
d 2
x + 2_
d
x−_
d
8
( x 2)
1
dx dx dx dx = x2 x2 − 1 − 1 + _ +5
= −(3x 2) − 3(2x) + 2(1) − (0)
( x2 )
1
= x2 x2 − 2 + _ +5
= − 3x 2 − 6x + 2
= x4 – 2x2 + 1 + 5
y = x4 – 2x2 + 6
dy _
_
dx
= d (x4) – _
dx
d
(2x2) + _
d
(6)
dx dx

= _
d 4
x – 2_
d 2
x +_
d
6
dx dx dx

= 4x3 – 2(2x) + 0
dy
_
dx
= 4x3 – 4x

2
_1 x 3 − 9
(x + 1)(x − 2)
y=_ y=_
3
3.3 3.4 2x − 6
x2
3 2 _1 (x 3 − 27)
x − 2x + x − 2
=_ _3
2
x = 2(x − 3)
3 2
2x
=_
x
–_ +_ –2
x _
_1 (x − 3)(x 2 + 3x + 9)
x2 x2 x2 x2
= _____________
3

= x – 2 + _1x – _
2
2
2(x − 3)
x
y = x – 2 + x–1 – 2x–2 = _16 (x 2 + 3x + 9)
dy _
_ = d (x) – _
d
(2) + _(x ) – _
d –1 d
(2x–2) y = _16 x 2 + _12 x + _32
dx dx dx dx dx

dx ( 6 ) dx ( 2 ) dx ( 2 )
= _
d
–_
d
2+_ x – 2_
d –1 d –2 dy _
dx
x dx dx dx
x _
dx
= d _1 x 2 + _ 1
d _
x +_ 3
d _

–2 –3
= 1 – 0 – x – 2(–2.x )
= _16 _
dx
x + 12 _
d 2 _ d
dx
x+_
d_ 3
dx 2
= 1 – x–2 + 4.x–3
dy
_ 1 4 = _16 (2x) + _12 (1) + 0
=1–_ +_
dx x2 x3 dy _
_
dx
= 1 x + _1
3 2

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 178 N4 Mathematics – Lecturer Guide

Activity 5.5 SB page 314

1. y = 1_2 e 4x 2. y = 3e − x
dy
_ dy
_
dx
= 2e 4x dx
= − 3e − x

3. y = 3 ln x 4. y = log 4x
dy 3_ dy _
_
dx
= x
_
dx
= 1 x ln 4
_
5. y = log 3x 2 6. y = − 5 log 2√x
_1
dy _ = − 5 log 2x 2
_
dx
= 2 x ln 3
dy
_
dx
= − 5_._
1
2 x ln 2
−5
=_
2x ln 2

7. y = 2(4 x) 8. y = − 4(2 −3x)

dy
_ dy
_
dx
= 2.4 x ln 4 dx
= − 3. − 4.2 −3x ln 2

= 12.2 −3x ln 2
3
_
9. f (x) = √ 125 x 10. f (x) = e ln4+x
_1
= [(5 3) x] = e ln4.e x
3

_1
= [5 3x] f (x) = 4.e x
3

f (x) = 5x f ʹ(x) = _
d
(4.e x)
dx
f ʹ(x) = _
d x
dx
5 = 4_
d x
e
dx
f ʹ(x) = 5x ln 5 f ʹ(x) = 4ex
_
11. f (x) = –5 ln x2 12. f (x) = 4 log √x 7
_1
= –5(2) ln x = 4 log(x 7) 2
_7
f (x) = –10 ln x = 4 log x 2
f ʹ(x) = _
d
dx
(–10 ln x) = 4(_72 ) log x
= –10_
d
dx
ln x f (x) = 14 log x
= –10._1x f ʹ(x) = _
d
(14 log x)
dx
10
f ʹ(x) = − _x = 14_
d
log x
dx
1
= 14._
x ln 10
14
f ʹ(x) = _
x ln 10

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 Module 5 • Differential calculus 179

( √x 4 )
13. y = _18 ln _ 1_
14. y = 6 log 1_x
3
dy _
8 ( (x 4) _12 ) dx( )
_1 ln _1 _ = d 6 log 1_x
= dx 3

(x 2)
= _18 ln _
1 = 6_
d
dx
log _1x
3

1
= 6. _
= _18 ln x −2 x ln 1_3
_ 6
=
= _18 (−2)ln x x ln 1_3

y = − _41 ln x

dx ( 4 )
dy _
_
dx
= d − _1 ln x

= − _41 _d
dx
ln x

= − _41 ._1x
dy
_ 1
dx
= −_ 4x

15. f(x) = − 3 log xy 16. f(x) = − 2 ln 3 x 2

= − 3 log x − 3 log y = − 2(ln 3 + ln x 2)
∴ f ʹ(x) = − 3_x = − 2 ln 3 − 2 ln x 2
= − 2 ln 3 − 4 ln x
∴ f ʹ(x) = − 4_x

17. f(x) = 2 log 54 x 3 18. y = − 3 log 4 1_2 x 5

= 2 log 54 + 2 log 5x 3 = − 3(log 4 1_2 + log 4x 5)
= 2 log 54 + 6 log 5x
= − 3 log 4 1_2 − 3 log 4x 5
6
∴ f ʹ(x) = ___ x ln 5
= − 3 log 4 1_2 − 15 log 4x
dy 15
∴ __ = ____
dx 4 ln x

Activity 5.6 SB page 319

1. 1.1 y = sin 5x 1.2 y = − 3 cos x

dy _
_ dy _
_
dx
= d (sin 5x)
dx dx
= d (− 3 cos x)
dx

= cos 5x._
d
dx
5x = − 3_
d
dx
(cos x)

= cos 5x.5 = − 3(− sin x)

= 5 cos 5x = 3 sin x

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 180 N4 Mathematics – Lecturer Guide

1.3 y = 4 tan 4x 1.4 y = − a 2 cosec x

dy _
_ dy _
_
dx
= d (4 tan 4x)
dx dx
= d (−a 2 cosec x)
dx

= 4_
d
dx
(tan 4x) = −a 2 _
d
dx
(cosec x)

= 4 sec 24x._
d
4x = − a 2(−cot x cosec x)
dx
= a 2 cot x cosec x
= 4 sec 24x.4
= 16 sec 24x

1.5 y = 1_2 sec πx 1.6 y = 31_ cot 3x

dx(2 ) dx(3 )
dy _ dy _
_
dx
= d _1 sec πx _
dx
= d 1_ cot 3x

= _1 _
d
(sec πx) = _1 _
d
(cot 3x)
2 dx 3 dx

= 1_2 sec πx tan πx._

d
dx
(πx) = 31_.− cosec 2 3x._
d
dx
(3x)

= π_2 sec πx tan πx = 31_.− cosec 2 3x.3

= − cosec 2 3x
2
1.7 y = sin 2θ 1.8 y=_
sin x
dy _
_ = d (sin 2θ)
dx(sin x)
dy _
_ 2
dθ dθ dx
= d _
= cos 2θ._
d
(2θ)
dθ =_
d
dx
(2 cosec x)
= 2 cos 2θ
= 2_
d
dx
(cosec x)

= 2(− cot x cosec x)

= − 2 cot x cosec x

2 cos x
1.9 y = 4 cos πx 1.10 f (x) = _
sin 2x
dy _
_
dx
= d (4 cos πx)
dx =_ 2 cos x
2 sin x cos x
= 4. _
d
(4 cos πx) 1
dx =_
sin x
= 4(− sin πx)._
d
(πx)
dx f (x) = cosec x
= 4(− sin πx) π
f ʹ(x) = _
d
dx
cosec x
= − 4π sin πx
f ʹ(x) = –cosec x cot x

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 Module 5 • Differential calculus 181

1
1.11 f (x) = ___________
4 4 1.12 y = 10 sin _2x .cos _2x
cos x − sin x
1
= _______________ = 5(2 sin _2x .cos _2x )
(cos x + sin x)(cos x − sin x)
2 2 2 2

1 y = 5 sin x
=_
1.cos x
dy _
_
1
= d (5 sin x)
= _ dx dx
cos x
= 5_d
dx
sin x
= sec x
dy
_ = 5 cos x
f ʹ(x) = _
d
dx
sec x dx

= sec x tan x
_
cos x
1.13 f (x) = _
_ 1.14 y = √4 + 4 tan 2x
2
√1 − cos x _
cos x = √4(1 + tan 2x)
=_ sin x _ _
= √4 .√1 + tan 2x
f (x) = cot x _
= 2√sec 2x
f ʹ(x) = _
d
dx
cot x
y = 2 sec x
f ʹ(x) = –cosec2x
dy _
_
dx
= d (2 sec x)
dx

= 2_
d
dx
sec x
dy
_
dx
= 2 sec x tan x
1
1.15 f (x) = _
___________
2
√3 cosec x − 3
1
=_____________
√3(cosec 2x − 1)
1
=_
_ _
√3 .√cosec x − 1
2

1
=_
_ _
√3 .√cot x
2

1
=_
_
√3 cot x

1_ _
=_ . cot1 x
√3

1_
f ʹ(x) = _ tan x
√3

dx ( √3 )
1_
f ʹ(x) = _
d _
tan x

1_ _
=_ d
tan x
dx
√3

1_
f ʹ(x) = _ sec 2x
√3

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 182 N4 Mathematics – Lecturer Guide

_
10
y = 3 cot x + _
3
2. 2.1 ex
− 4 √x − 8
1_
= 3 cot x + 10. e −x − 4. x 3 − 8

(4. x 3) − _
dy _
_ _1

dx
= d (3 cot x) + _
dx
d
(10. e −x) − _
dx
d d
(8) dx dx
1_
= 3_
d
dx
cot x + 10_
dx
e − 4_
d −x d 3 _
dx
d
x − dx 8

= 3(− cosec 2 x) + 10(e −x.− 1) − 4(1_3.x −3) − 0

_2

= − 3 cosec 2 x − 10
_−_
ex 3
4_
2
3 √x
_
2.2 y = 7 sin 2x – 3–x + _12 log√x − _52 x 4

= 7 sin 2x – 3–x + _12 log x 2 – _52 x 4

_1

= 7 sin 2x – 3–x + _12 (_12 )log x – _52 x 4

y = 7 sin 2x – 3–x + _14 log x – _52 x 4

dx ( 4 ) dx ( 2 )
dy _
_ 1 5 4
dx
= d (7 sin 2x) – _
dx
(3 ) + _
d –x d _
dx
log x – _
d _
x

= 7_
d
dx
sin 2x – _
dx
3 + 14 _
d –x _ d
dx
log x – _52 _
d 4
dx
x

= 7(cos 2x.2) – (3–x ln 3. –1) + _14 (_

x ln 10 ) 2
1
– _5 (4x 3)
dy ln 3 _
_
dx
= 14 cos 2x + _ + 1 – 10x3
3x 4x ln 10

_ _
2.3 y = _1 sec x – √2 8x –√3 ln 2x + 10log2
4
_1 _
= _1 sec x – (2 8x) –√3 ln 2x + 2
2

4
_
y = _14 sec x – 24x –√3 ln 2x + 2
_
dx ( 4 )
dy _
_ = d _1 sec x – _(2 ) – _
d 4x d
(√3 ln 2x) + _
d
(2)
dx dx dx dx
_d
= _14 _
d
dx
sec x – _
d
dx
2 4x – √3 _
dx
ln 2x + _
d
dx
2
_
= _14 (sec x tan x) – (24x ln 2.4) – √3 (_
2x )
1
.2 + 0
_
dy _
_
dx
= 1 sec x tan x – 4.24x ln 2 – _
4
√3
x

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 Module 5 • Differential calculus 183

2.4 y = − _12 cos 4x + 11 ln x + _

5
2 – e
x–ln 3
x

= − _12 cos 4x + 11 ln x + 5.x–2 – ex.e–ln 3

= − _12 cos 4x + 11 ln x + 5.x–2 –e x.e ln 3

−1

= − _12 cos 4x + 11 ln x + 5.x–2 – ex.e ln 3

_1

= − _12 cos 4x + 11 ln x + 5.x–2 – ex._13

y = − _12 cos 4x + 11 ln x + 5.x–2 – _13 .ex

dx ( 2 ) dx ( 3 )
dy _
_
dx
= d − _1 cos 4x + _
d
(11 ln x) + _
dx
d
(5x–2) – _ 1 x
d _
.edx

= − _12 _
d
dx
cos 4x + 11_
d
dx
ln x + 5_
dx
x – 13 _
d –2 _ d x
dx
e

= − _12 (–sin 4x.4) + 11(_1x ) + 5(–2.x–3) – _13 (ex)

dy 11 _
_ = 2 sin 4x + _ – 10 – _1 ex 3
dx x x3
_
12 1
y = −_ – 3e √9x + ln _
2
2.5 sin x 5 + 13
x

= –12 cosec x – 3e3x + ln x–5 + 13

y = –12 cosec x – 3e3x – 5 ln x + 13

dy _
_
dx
= d (–12 cosec x) – _
dx
d
(3e3x) – _
d
(5 ln x) + _
dx
d
(13)
dx dx

= –12_
d
dx
cosec x – 3_
dx
e – 5_
d 3x d
dx
ln x + _
d
dx
13

= –12(–cosec x cot x) – 3(e3x.3) – 5(_1x ) + 0

dy
_
dx
= 12 cosec x cot x – 9e3x – _5 x

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 184 N4 Mathematics – Lecturer Guide

log x
1_ _ x
2.6 y = –2 tan x + _
4
5
–_
2 + ln6 6
x √x

= –2 tan x + _14 log 5x – _

2
1
+_1
ln 6
.6 x _1
x .x 2

= –2 tan x + _14 log 5x – _

1
+_1
ln 6
.6 x
_5
x2

y = –2 tan x + _14 log 5x –x − 2 + _1 _5

ln 6
.6 x

(x − 2) + _
dx ( 4 ) dx ( ln 6 )
dy _
_ 1 _5 1
dx
= d (–2 tan x) + _
dx
d _
log 5x – _
d d _
.6 x dx

tan x + _14 _ x + ln16 _

_5
= –2_
d
dx
d
dx
log 5x – _
d −2 _
dx
d x
dx
6

x ln 5 ) ( 2
= –2(sec2x) + _14 (_ 1
– − _5 .x − 2) + _1 _7

ln 6
(6 x ln 6)

1
= –2 sec2x + _ + _5 ._
4x ln 5 2
1
+ 6x _7
x2
1
= –2 sec2x + _ + _5 ._
4x ln 5 2 3
1
+ 6x _1
x .x 2
dy
_ 1 5 _
= –2 sec2x + _ +_ + 6x
4x ln 5
dx 2x 3√x

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 1.
Rewrite as du
_ dy
_ _ dy
dy _ du
Function u = g(x) dx
= g′(x) y = f (u) du
= f ′(u) dx
= ×_du dx
y = f [g(x)]

N4 Mathematics - Lecturers Guide A4 Layout.indd 185

x x
−_ x
__ _
10 x dy 10
_
du _ _ _
dy _
Activity 5.7

y = (_13 ) y = 3 10 u=_
10 dx
= 110
y = 3u du
= 3u ln 3 dx
= 3 ln 310

7 du
_ dy
_ dy
_
y=_
3x y = 7.e–3x u = –3x dx
= –3 y = 7.eu du
= 7.eu dx
= –21.e–3x
e
_ 5x 5x
_ 5x dy _
y = √e 5x y=e2 u=_ _
du _
=5 y = eu _ = eu _
dy _
= 5 .e 2
2 dx 2 du dx 2

du
_ dy
_ dy
_
y = eln 2+9x y = 2.e9x u = 9x dx
=9 y = 2.eu du
= 2.eu dx
= 18.e9x

u = _8x _
du _
=1 y = 6 log u _
dy _
= 6 _
dy _
= 48
dx 8 du u ln 10 dx x ln 10
y = 6 log(_8x ) y = 6 log(_8x )

_ du
y = ln (√4x ) y = _12 ln 4x u = 4x _ =4 _
dy _
= 1 _
dy _
= 1
dx
y = _12 ln u du 2u dx 2x

1 _
du _ dy
_ dy
_
y = cot _ y = cot(_2x ) u = _2x dx
=1 2
y = cot u du
= –cosec2 u dx
= –_12 cosec2(_2x )
( 2x −1 )

3x 3x 3x _
du _ dy
_ dy
_ 3x
y = 4 cos(_π2 + _
4)
y = –4 sin _
4
u=_
4 dx
=3 4
y = –4 sin u du
= –4 cos u dx
= –3 cos _ 4

_ −1 _ _ _ _ _ _
1_ 1_ du
_ 1_ dy
_ 1_ dy
_ √6 sec (√3 x) tan (√3 x)
y=_ (cos (π − √3x 2 )) y = −_ sec(√3 x) u = √3 x dx
= √3 y = −_ sec u du
= −_ sec u tan u dx
= − _______________ 2
√2 √2 √2 √2

du
_ dy
_ dy
_
y = 3 cos(π + 6x) y = –3 cos 6x u = 6x dx
=6 y = –3 cos u du
= 3 sin u dx
= 18 sin 6x
Module 5 • Differential calculus

SB page 328
185

28/02/2022 11:38 am
 186 N4 Mathematics – Lecturer Guide

2. 2.1 y = 3 ln (cos x) 2.2 y = 8.elog x

Let u = cos x then y = 3 ln u Let u = log x then y = 8.eu
Therefore, Therefore,
dy dy
_
du
dx
= –sin x _
du
= 3._1u
_
dx
= 1
du _
x ln 10
_
du
= 8.eu
dy _
_ dy dy _
dy
dx
= ×_du
du dx
_ = ×_
dx
du
du dx

= 3._u1 × –sin x = 8.eu × _ 1

x ln 10
u
3 sin x 8e
= −_ u
=_
x ln 10
3 sin x dy _ logx
= −_cos x
_ = 8e
dx x ln 10
dy
_
dx
= –3 tan x

2.3 y = 4 cot(7x + 2) 2.4 y = sec(4x – 2)

Let u = 7x + 2 then y = 4 cot u Let u = 4x – 2 then y = sec u
Therefore, Therefore,
_
du dy
_ _
du dy
_
dx
=7 du
= 4. –cosec2u dx
=4 du
= sec u.tan u
dy _
_ dy dy _
dy
dx
= ×_du
du dx
_ = ×_
dx
du
du dx

= 4. –cosec2u × 7 = sec u.tan u × 4

= –28 cosec2u = 4 sec u.tan u

dy
_ dy
_
dx
= –28 cosec2(7x + 2) dx
= 4 sec (4x – 2) tan (4x – 2)
_
2.5 y = (2x – 1)4 2.6 y = √x 2 + x − 1
_1
Let u = 2x – 1 then y = u4 y = (x 2 + x − 1) 2
_1
Therefore, Let u = x2 + x – 1 then y = u 2
_
du dy
_ Therefore,
dx
=2 du
= 4u3
dy _
= 1u − 2
_1
dy _
dy _
du
= 2x + 1 _
_ = ×_du
dx dx 2
dx du dx
dy _
_ dy
3 = ×_du
= 4u × 2 dx du dx

= _12 u − 2 × (2x + 1)
_1
3
= 8u
2x +_1
dy
_ =_
dx
= 8(2x – 1)3 2√u
dy
_ 2x + 1
dx
=_
_
2√ x + x − 1
2

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 Module 5 • Differential calculus 187

_ _
3 4
2.7 y = √x 2 − 1 2.8 y = √ tan x
_1 _1
y = (x 2 − 1) 3 y = (tan x) 4
_1 _1
Let u = x2 – 1 then y = u 3 Let u = tan x then y = u 4
Therefore, Therefore,
dy _ dy _
= 1u − _23
= 1u − 4
_3
_
du
= 2x _ _
du
= sec2x _
dx du 3 dx du 4
dy _
_ dy dy _
dy
dx
= ×_du
du dx
_
dx
= ×_du
du dx

= _13 u − _23
× 2x = _14 u − _34
× sec2x
2
2x_ sec_
=_
3 =_4
x
3 √u
2 3
4 √u
dy
_ 2x
_ dy
_ se_
cx 2

dx
= _
3
dx
=_
4
3 √(x 2 − 1) 2 3
4 √ tan x

Activity 5.8 SB page 331

1. y = ( x 2 + 3x)(2x − 2)
dy
_
dx
= (x 2 + 3x)(2) + (2x − 2)(2x + 3)
= 2x 2 + 6x + (4x 2 − 4x + 6x − 6)
= 2x 2 + 6x + 4x 2 + 2x − 6
= 6x 2 + 8x − 6

2. y = x5.3x
Let f (x) = x5 and g(x) = 3x
Thus f ʹ(x) = 5x4 and gʹ(x) = 3x ln 3
dy
_
dx
= f (x)_
d
g(x) + g(x)_
d
dx
f (x) dx

= (x )(3 ln 3) + (3 )(5x4)
5 x x

dy
_
dx
= x4.3x(x ln 3 + 5)

3. y = ex.log x
Let f (x) = ex and g(x) = log x
1
Thus f ʹ(x) = ex and gʹ(x) = _
x ln 10
dy
_
dx
= f (x)_
d
g(x) + g(x) _
dx
d
f (x) dx

= (e )(_
x
x ln 10 )
1
+ (log x)(ex)

ex (_ + log x)
dy
_ 1
dx
= x ln 10

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 188 N4 Mathematics – Lecturer Guide

4. y = ln x sin x
Let f (x) = ln x and g(x) = sin x
Thus f ʹ(x) = _1x and gʹ(x) = cos x
dy
_
dx
= f (x)_
d
g(x) + g(x) _
dx
d
f (x) dx
= (ln x)(cos x) + (sin x)(_1x )
dy
_ sin x
dx
= cos x ln x + _ x

5. s = t 3 cos t
Let f (t) = t 3 and g(t) = cos t
Thus f ʹ(t) = 3t 2 and g ʹ(x) = − sin t

dt ( )
_=
ds
f (t) _
d
g(t) + g(t) _
d
f t
dt dt
= (t 3)(− sin t) + (cos t)(3t 2)
= t 2(−t sin t + 3 cos t)

6. y = tan x.ex
Let f (x) = tan x and g(x) = ex
Thus f ʹ(x) = sec2x and gʹ(x) = ex
dy
_
dx
= f (x)_
d
g(x) + g(x) _
dx
d
f (x) dx
= (tan x)(e ) + (e )(sec2x)
x x

dy
_
dx
= ex(tan x + sec2x)
_
7. y = (ln x − 1).√x
_1
y = (ln x − 1).x 2
_1
Let f (x) = ln x – 1 and g(x) = x 2
Thus f ʹ(x) = _1x and gʹ(x) = _12 .x − 2
_1

dy
_
dx
= f (x)_
d
g(x) + g(x) _
dx
d
f (x) dx

= (ln x – 1)(_12 x − 2) + (x 2)(_1x )

_1 _1

_1
ln x _
−1 _
=_ + xx
2

2√x
_ _
√x (ln x − 1)
=_+_ 2x
√x
x
_ _
√x (ln x − 1) + 2√x
= ____________
2x
_ _ _
√x ln x − √x + 2√x
= ____________
2x
_ _
√x ln x + √x
=_ 2x
_
dy _
_ √x (ln x + 1)
dx
= 2x

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 Module 5 • Differential calculus 189

( )
1
8. x= 2+_4 log 6t
x

= (2 + x −4) log 6t
Let f (t) = 2 + x −4 and g(t) = log 6t
1
Thus f ʹ(t) = − 4 x −5 and g ʹ(t) = _
t ln 6
_
dx
= f (t) _
d
g(t) + g(t) _
d
f (t)
dt dt dt

= (2 + x −4)(_
t ln 6) (
1
+ log 6t)(− 4.x −5)
2 1 4 log t
=_
t ln 6
+_
4 −_5
6

tx ln 6 x

9. y = 4x(cot x – x2)
Let f (x) = 4x and g(x) = cot x – x2
Thus f ʹ(x) = 4x ln 4 and gʹ(x) = –cosec2x – 2x
dy
_
dx
= f (x)_
d
g(x) + g(x)_
d
f (x) dx dx
= (4 )(–cosec x – 2x) + (cot x – x2)(4x ln 4)
x 2

dy
_
dx
= 4x(–cosec2x – 2x + cot x ln 4 – x2 ln 4)
_
10. y = (√x 3 − sec x) cosec x
y = (x 2 − sec x) cosec x
_3

_3
Let f (x) = x 2 – sec x and g(x) = cosec x
Thus f ʹ(x) = _32 .x 2 − sec x tan x and gʹ(x) = – cosec x cot x
_1

dy
_
dx
= f (x)_
d
g(x) + g(x)_
d
f (x) dx dx
= (x − sec x)(–cosec x.cot x) + (cosec x)(_32 x 2 − sec x tan x)
_3 _1
2

= − x 2 cosec x cot x + cosec2x + _32 x 2 cosec x − sec 2x

_3 _1

_ _
dy
_ 3√x cosec x
dx
= −x √x cosec x cot x + cosec2x + _ – sec2x 2

Activity 5.9 SB page 333

4
1 −_x
1. y=_
√x

x 2(− 4x 3) − (1 − x 4) 21_ x −2
_1 1_

dy
_ = ____________
(x 2)
2
dx 1_

7_ 1_ 7_
− 4x 2 − 1_2 x −2 + 21_ x 2
= ______________ x
7_
− 7_2 x − 1_2 x −2
1_
2

= ___________
x
5_ 3_
= − 7_2 x 2 − 1_2 x −2
_
= − 7_2 √x 5 − _
1_
3
2√x

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 190 N4 Mathematics – Lecturer Guide

7
2. y=_x
ln x

Let f (x) = x7 and g(x) = ln x

Thus f ʹ(x) = 7x6 and gʹ(x) = _1x
g(x)_
d
f(x) − f(x)_
dy ______________
d
g(x)
_ =
dx dx
dx (g(x)) 2

(x)
(ln x)(7x 6) − (x 7) _1
= _____________
(ln x) 2
6 6
7x ln x − x
=_ 2
ln x
6
dy _
_ x (7 ln x − 1)
=
dx ln 2x

4 cos x
3. y=_ 10 x

Let f (x) = 4 cos x and g(x) = 10x

Thus f ʹ(x) = –4 sin x and gʹ(x) = 10x ln 10
g(x)_
d
f(x) − f(x)_
dy ______________
d
g(x)
_ =
dx dx
dx (g(x)) 2

(10 x)(−4 sin x) − (4 cos x)(10 x ln 10)

= ______________________
x 2
(10 )

4.10 x(− sin x − cos x.ln 10)

= _________________
x 2
(10 )
dy ______________
_ 4(− sin x − cos x ln 10)
dx
= 10 x

log r
4. x=_5
3

Let f (r) = log 3r and g(r) = r 5

1
Thus f ʹ(r) = _
r ln 3
and g ʹ(r) = 5r 4
g(r) _
d
f (r) − f (r) _
d
g(r)
_ = _______________
dx dr dr
dr (g(r)) 2

(rln 3)
r5 3
1
_ − (log r)(5r 4)
= ___________ 2
(r 5)
4
_
r
− 5r 4 log r
= ____________
ln 3 3
10
r

(ln 3 3 )
1
_
r4 − 5 log r
= ___________
r 10
1
_ − 5 log r
=_
ln 3 3
6
r

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 Module 5 • Differential calculus 191

5. y = (2 ln x)–1
1
y=_
2 ln x

Let f (x) = 1 and g(x) = 2 ln x

Thus f ʹ(x) = 0 and gʹ(x) = _2x
g(x)_
d
f(x) − f(x)_
dy ______________
d
g(x)
_ =
dx dx
dx (g(x)) 2

(x)
(2 ln x)(0) − (1) _2
= ____________
(2 ln x) 2

− _2x
= _2
(2 ln x)
dy
_ 1
= −_
dx 2x ln 2x

6. r = cosec θ
1
=_
sin θ

Let f(θ) = 1 and g(x) = sin θ

Thus f ʹ(θ) = 0 and g ʹ(θ) = cos θ
g(θ) _
d
f(θ) − f(θ) _
d
g(θ)
_
dr ____________
=
dθ dθ
dθ (g(θ)) 2

(sin θ)(0) − (1)(cos θ)

= ________________
2
(sin θ)
− cos θ
=_ 2
sin θ

= −_ 1 _
.cos θ
sin θ sin θ

= − cosec θ cot θ
x
7. y=_
3
e
x +1
Let f (x) = ex and g(x) = x3 + 1
Thus f ʹ(x) = ex and gʹ(x) = 3x2
g(x)_
d
f(x) − f(x)_
dy ______________
d
g(x)
_ =
dx dx
dx (g(x)) 2

(x 3 + 1)(e x) − (e x)(3x 2)
= ______________
3 2
(x + 1)
x 3 2
dy _
_ e (x − 3x + 1)
=
dx (x + 1) 2
3

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 192 N4 Mathematics – Lecturer Guide

2
4−x
8. y=_2x

Let f (x) = 4 – x2 and g(x) = 2x

Thus f ʹ(x) = –2x and gʹ(x) = 2x ln 2
g(x)_
d
f(x) − f(x)_
dy ______________
d
g(x)
_ =
dx dx
dx (g(x)) 2

(2 x)(−2x) − (4 − x 2)(2 x ln 2)
= _________________
x 2
(2 )

2 x(x 2 ln 2 − 2x − 4 ln 2)
= _______________
x 2
(2 )

dy ____________
2
_
dx
= x ln 2 − 2x − 4 ln 2 2x

3
_
√t
9. s=_ sec t − t
_1

=_ t 3

sec t − t
1_
Let f (t) = t 3 and g(t) = sec t − t
2_
Thus f ʹ(t) = 1_3.t −3 and g ʹ(t) = sec t.tan t − 1

g(t) _
d
f(t) − f (t) _
d
g(t)
_ = _______________
ds dt dt
dt (g(t)) 2

(sec t − t)(1_3 t − 3) − (t 3)(sec t tan t − 1)

2_ 1_

= ___________________
(sec t − t)2
1_ − 2_3 1_ _1 _1
t sec t − 1_3 t 3 − t 3 sec t tan t + t 3
3________________
=
(sec t − t)2
1_ − 2_3 1_ 1_
t sec t − t 3 sec t tan t + 2_ t 3
= ______________
3 3
2
(sec t − t)
_1 t 1_3(t − 1 sec t − 3 sec t tan t + 2)
= _______________
3
(sec t − t)2

1_
t 3(sec t − 3t sec t tan t + 2t)
= ______________
2
3t (sec t − t)

3
_
√t (sec t − 3t sec t tan t + 2t)
= ______________
3t (sec t − t)2

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 Module 5 • Differential calculus 193

x
e + sin x
10. y=_ log x 4
x
Let f (x) = e + sin x and g(x) = log 4x
1
Thus f ʹ(x) = ex + cos x and gʹ(x) = _
x ln 4
g(x)_
d
f(x) − f(x)_
dy ______________
d
g(x)
_ =
dx dx
dx (g(x)) 2

( x ln 4 )
1
(log x)(e x + cos x) – (e x + sin x) _
= _______________________
4

(log 4x) 2
x
sin x
e xlog 4x + cos x log 4x − _ e
−_
= _____________________
x ln 4 x ln 4
2
(log 4x)

x x
x e log 4x ln 4 + x cos x log 4x ln 4 − e − sin x
dy ____________________________
_ =
dx x ln 4(log 4x) 2

Activity 5.10 SB page 338

y = x2(x – _x ) + 5
1 2
1. y = –2x3 + 5x2 + 3x – 12 2.
dy
_ dy
_
dx
= –6x2 + 10x + 3 = 4x3 – 4x
dx
d 2y
_ d 2y
=_
d
(–6x2 + 10x + 3) _ =_
d
(4x3 – 4x)
dx 2 dx dx 2 dx

=_
d
dx
(–6x2) + _
d
dx
(10x) + _
d
dx
(3) =_
d
(4x3) – _
d
(4x)
dx dx

= –6_
d 2
dx
x + 10_
d
dx
x+_
d
dx
3 = 4_
d 3
x – 4_
d
x
dx dx

= –6(2x) + 10(1) + 0 = 4(3x2) – 4(1)

2 2
dy
_ dy
_
= –12x + 10 = 12x2 – 4
dx 2 dx 2

2
_1 t 3 − 9
(x + 1)(x – 2)
y = _________
3_
3. 4. s= 2t − 6
x2
dy
_ 1 4
=1–_ +_ _ = 1_ t
ds
dt 3
+ 21_
dx x2 x3

dt(3 2)
dy
_ d2 s _
= 1 – x–2 + 4x–3 ∴ _ = d _1 t + 1_
dx dt 2
2

dt(3 ) dt(2)
dy _
_ = d (1 – x–2 + 4x–3) d 1_
=_ t + _d 1_
dx 2 dx

=_
d
dx
(1) – _
dx
(x ) + _
d –2 d
dx
(4x–3) = 1_3 dt
_
d d 1_
t+_
dt 2

=_
d
dx
1–_
dx
x + 4_
d –2 d –3
dx
x = 1_3(1) + 0
2
d s 1_
_
= 0 – (–2x–3) + 4(–3x–4) = 3
dt 2
2
dy _
_ = 2 –_
12
dx 2 x3 x4

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 194 N4 Mathematics – Lecturer Guide

_
10
y = 3 cot x + _
3
5. – 4 √ x– 8
_x
e2
dy
_ 5 _ 4_
dx
= –3 cosec2x – _ _x – 3
e 2 3 √x 2
dy
= –3(cosec x)2 – 5e − 2 – _4 x − 3
_x _2
_
dx 3

dx ( )
2
dy _
− 3(cosec x) 2 − 5e– 2 − _4 x − 3
_x _2
_ d
2 =
dx 3

dx ( 3
(5e 2) – _ x )
4 −3 − _x _2
=_
d
dx
(–3(cosec x)2) – _
d
dx
d _

e – _43 _
_x _2
= –3_
d
dx
(cosec x)2 – 5_
d −2
dx
d −3
dx
x

= –3(2 cosec x.–cosec x cot x) – 5(e − 2.− _12 ) – _43 (− _32 x − 3)

_x _5

2
dy
_ 5 8_
= 6 cosec2x cot x + _ _x +
_
dx 2 3
2e 2 9x √x 5

_
6. y = 7 sin 2x – 3–x + _12 log√x – _52 x4
dy
_ ln 3 _1
dx
= 14 cos 2x + _ x + – 10x3
3 4x ln 10
dy
_ 1
dx
= 14 cos 2x + ln 3.3–x + _ .x–1 – 10x3 4 ln 10
2
dy _
_ 1
= d (14 cos 2x + ln 3.3–x + _ .x–1 – 10x3) 4 ln 10
dx 2 dx

dx ( 4 ln 10
=_
d
dx
(14 cos 2x) + _
d
dx
(ln 3.3–x) + _
d _ 1
.x −1) – _
d
dx
(10x3)
1 _
= 14_
d
dx
cos 2x + ln 3_
d
dx
3 –x + _ x – 10_
d −1
4 ln 10 dx
d 3
dx
x
1
= 14(–sin 2x.2) + ln 3(3–x ln 3. –1) + _
4 ln 10
(− 1x −2) – 10(3x2)
2 2
dy
_ (ln 3) 1
= –28 sin 2x – _ – _
2 – 30x2 3 x
dx 4x 2 ln 10

_ _
7. y = _14 sec x – √2 8x – √3 ln 2x + 10log 2
_
dy _
_
dx
= 1 sec x tan x – 4.24x ln 2 – _
4
√3
x
dy _ _
_
dx
= 1 sec x tan x – 4.24x ln 2 – √3 x −1
4
_
dx ( 4
2
dy _
_ 1
d _
2 = sec x tan x – 4.24x ln 2 – √3 x −1)
dx
_
dx ( 4
=_
d _ 1
sec x tan x) – _
d
dx
(4.24x ln 2) – _
d
dx
(√3 x −1)
_
= _14 _
d
dx
sec x tan x – 4 ln 2_
dx
2 – √3 _
d 4x d −1
dx
x
_
= _14 [sec x_
d
dx
tan x + tan x _
d
dx
sec x] – 4 ln 2_
d
dx
2 4x – √3 _
d −1
dx
x
_
= _14 [sec x sec2x + tan x sec x tan x] – 4 ln 2(24x ln 2.4) –√3 (–1.x–2)
_
= _14 [sec3x + sec x tan2x] – 16.24x(ln 2)2 + _
√3
2
x

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 Module 5 • Differential calculus 195

_
= _1 [sec3x + sec x (sec2x – 1)] – 16.24x(ln 2)2 + _
4
√3
x2
_
= _14 [sec3x + sec3x – sec x] – 16.24x(ln 2)2 + _
√3
2
x
_
= _14 [2 sec3x – sec x] – 16.24x(ln 2)2 + _
√3
2
x
_
d 2y
_ = _12 sec3x – _14 sec x – 16.24x(ln 2)2 + _
√3
dx 2 2
x

8. y = – _12 cos 4x + 11 ln x + _
5
2 –e
x – ln 3
x
dy
_ 11
_ 10
_ _1 ex
= 2 sin 4x + – – 3
dx x x3
dy
_
dx
= 2 sin 4x + 11x–1 – 10x–3 – _1 ex 3
2
dy _
_ = d (2 sin 4x + 11x–1 – 10x–3 – _1 ex) 3
dx 2 dx

=_
d
dx
(2 sin 4x) + _
d
dx
(11x–1) – _
d
dx
(10x–3) – _ ( 1 ex)
d _
dx 3

= 2_
d
dx
sin 4x + 11_ x – 10_
d –1
dx dx
x – 13 _
d –3 _ d x
dx
e

= 2(cos 4x4) + 11(–1x–2) – 10(–3x–4) – _13 (ex)

2
dy 11 _
_ = 8 cos 4x – _ + 30 – _1 ex 3
dx 2 x2 x4
_
12 1
y = −_ – 3e √9x + ln _
2
9. sin x 5 + 13
x
dy
_
dx
= 12 cosec x cot x – 9e3x – _5 x
dy
_
dx
= 12 cosec x cot x – 9e3x – 5.x–1
2
dy _
_ = d (12 cosec x cot x – 9e3x – 5.x–1)
dx 2 dx

=_
d
dx
(12 cosec x cot x) – _
d
dx
(9e3x) – _
d
dx
(5.x–1)

= 12_
d
dx
cosec x cot x – 9_
dx
e – 5_
d 3x d –1
dx
x

= 12[cosec x_
d
dx
cot x + cot x_
d
dx
cosec x] – 9_ e – 5_
d 3x
dx
d –1
dx
x

= 12[cosec x. –cosec2x + cot x. –cosec x cot x] – 9(e3x.3) – 5(–1x–2)

5
= 12[–cosec3x – cosec x cot2x] – 27e3x + _2
x
5
= 12[–cosec3x – cosec x(cosec2x – 1] – 27e3x + _2
x
5
= 12[–cosec3x – cosec3x + cosec x] – 27e3x + _2
x
5
= 12[–2 cosec3x + cosec x] – 27e3x + _2
x
d 2y
_ 5
= –24 cosec3x + 12 cosec x – 27e3x + _
dx 2 2
x

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 196 N4 Mathematics – Lecturer Guide

1_ _log x x
10. y = –2 tan x + _
4
5
–_
2 + ln6 6
x √x
dy
_ _ 1 5 _
dx
= –2 sec2x + 4x ln 5
+_
3 + 6x
2x √x
dy 1
+ _25 .x − 2 + 6x
_7
_ = –2(sec x)2 + _ .x–1
dx 4 ln 5
2
dy _ 1
.x–1 + _5 .x − 2 + 6x)
_7
_ = d (–2(sec x)2 + _
dx 2 dx 4 ln 5 2

dx ( 2
dx ( 4 ln 5
1
.x −1) + _ .x ) + _
5 −2 _7
=_
d
dx
(–2(sec x)2) + _
d _ d _ d x
dx
(6 )
1 _
x + 52 _
_7
= –2_
d
dx
(sec x)2 + _ d –1 _
4 ln 5 dx
d −2 _
dx
d x
x + dx 6
1
(–1x–2) + _52 (− _72 x − 2) + 6x ln 6
_9
= –2(2 sec x sec x tan x) + _
4 ln 5
2
dy
_ 1 35 _
= –4 sec2x tan x – _ –_ + 6x ln 6
dx 2 4x 2 ln 5 4x 4√x

Activity 5.11 SB page 360

1. 1.1 a) y = x 3 + 3x 2
dy
_
dx
= 3x 2 + 6x

0 = 3x 2 + 6x
0 = 3x(x + 2)
∴ x = 0 or x = − 2
∴ y = 0 or y = (− 2)3 + 3(− 2)2
∴ y=4
Thus, the turning points of y = x 3 + 3x 2 are (0; 0) and (–2; 4)
b) Nature of turning points:
2
d y
_ = 6x + 6
dx2

For x = 0:
2
d y
_ = 6(0) + 6 = 6 > 0 ∴ Minimum (0; 0)
dx2

For x = − 2:
2
d y
_ = 6(− 2) + 6 = − 6 < 0 ∴ Maximum (–2; 4)
dx2
2
d y
_
c) =0
dx2

6x + 6 = 0
6x = − 6
x = −1
∴ y = (− 1) 3 + 3(− 1)2
= −1 + 3
=2
Point of inflection: (–1; 2)

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 Module 5 • Differential calculus 197

d) x-intercept: y = 0 y
0 = x 2(x + 3) y = x3 – 3x2
6
∴ x = 0 or x = − 3
(–2; 4)
y-intercept: x = 0 4

∴ y=0 2
(–1; 2)
(–3; 0)
x
–3 –2 –1 (0; 0) 1 2 3
–2

–4

–6

1.2 a) y = x 3 − 4x 2 + 4x
dy
_
dx
= 3x 2 − 8x + 4

0 = 3x 2 − 8x + 4
_
− b ± √b 2 − 4ac
x = ____________
2a

∴ x = 2_3 or x = 2
∴ y = 1,19 or y = 0
Thus, the turning points of y = x 3 44x 2 + 4x are (2_3; 1,19) and (2; 0)
2
d y
_
b) = 6x − 8
dx2

For x = 2_3:
2

(3)
− 8 = − 4 < 0 ∴ Maximum (2_3; 1,19)
d y
_ = 6 2_
dx2

For x = 2:
2
d y
_ = 6(2) − 8 = 4 > 0 ∴ Minimum (2; 0)
dx2
2
d y
_
c) = 6x − 8 = 0
dx2
6x = 8
x = 34_
∴ y = (4_3) − 4(4_3) + 4(4_3)
3 2

64
=_
27
− 4(16
9)
_ + 16
_
3

= 0,593
Point of inflection: (4_3; 0,593)

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 198 N4 Mathematics – Lecturer Guide

d) x-intercepts: y = 0 y
0 = x 3 − 4x 2 + 4x y = x3 – 4x2 + 4x
0 = x(x 2 − 4x + 4) 3

0 = x(x − 2)(x − 2) 2
∴ x = 0 or x = 2 2
(__3 ; 1,19)
1 4
(__3 ; 0,593)
y-intercept: x = 0 (2; 0)
x
∴ y=0 –3 –2 –1 (0; 0) 1 2 3
–1

–2

–3
1.3 y = x3 – x2 – 3x + 3
a) y = x3 – x2 – 3x + 3
dy _
_
dx
= d (x3 – x2 – 3x + 3)
dx

= _
d 3
x –_
d 2 _ d
x – dx 3x + _
d
3
dx dx dx

= _
d 3
x –_
d 2
x – 3_
d
x+_
d
3
dx dx dx dx

= 3x2 – 2x – 3.1 + 0
dy
_
dx
= 3x2 – 2x – 3
dy
But _
dx
= 0 at the turning points,
3x2 – 2x – 3 = 0
_
− b ± √b 2 − 4ac
x=_ 2a
______________
− (− 2) ± √(− 2) 2 − 4(3)(− 3)
_________________
= 2(3)
_
2 ± √40
x=_ 6
_ _
2 + √40 2 − √40
x=_ 6
and x=_ 6

x = 1,387 x = –0,721

Substitute x = –0,721 and x = 1,387 into y = x3 – x2 – 3x + 3,

y = x3 – x2 – 3x + 3
= (–0,721)3 – (–0,721)2 – 3(–0,721) +3
y = 4,268
y = x3 – x2 – 3x + 3
= (1,387)3 – (1,387)2 – 3(1,387) +3
y = –0,417
Thus, the turning points of y = x3 – x2 – 3x + 3 are (–0,721; 4,268) and
(1,387; –0,417)

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 Module 5 • Differential calculus 199

2
dy _
_ = d (3x2 – 2x – 3)
dx 2 dx

=_
d
dx
3x2 – _
d
dx
2x – _
d
dx
3

= 3_
d 2
dx
x – 2_
d
dx
x–_
d
dx
3

= 3.2x – 2.1 – 0
2
dy
_ = 6x – 2
dx 2

b)

( 3 ; 1 27 )
Stationary point (–0,721; 4,268) _1 25
_ (1,387; –0.417)
2 2 2
d 2y dy dy dy
Sign of _2 _ = –6,325 < 0
2
_ =02
_ = +6,325 > 0
dx dx dx dx 2

Conclusion Maximum Test fails Minimum

x = _13

Interval –0,721 < x < _13 _1 < x < 1,387

3

Test value x=0 x = _23

dy dy dy
Sign of _____
dx
_
dx
= –3 < 0 _ = –3 < 0
dx

Conclusion Decreasing Decreasing

Inflection point

d 2y
c) But _2 = 0 at the point of inflection,
dx
6x – 2 = 0
6x = 2
x = _13
Substitute x = _13 into y = x3 – x2 – 3x + 3,
y = x3 – x2 – 3x + 3

= (_13 ) – (_13 ) – 3(_13 ) + 3

3 2

25
y = 1_
27
Thus, the point of inflection of y = x3 – x2 – 3x + 3 is (_13 ; 1_
27 )
25
.

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 200 N4 Mathematics – Lecturer Guide

d) y
(–0,721; 4,268)
4

0; 3

2
( 3 ; 1 27 )
1
_ 25
_
y = x – x – 3x + 3
3 2

x
–4 –3 –2 –1 1 2 3 4
–1 (1,387; –0,417)

–2

1.4 y = –x3 – 6x2 – 3x + 5

a) y = –x3 – 6x2 – 3x + 5
dy _
_
dx
= d (–x3 – 6x2 – 3x + 5)
dx

= _
d
(–x3) – _d
6x2 – _d
3x + _d
5
dx dx dx dx

= –_
dx
x – 6_
d 3
dx
x – 3_
d 2 d
dx
x+_ d
dx
5

= –(3x2) – 6.2x – 3.1 + 0

dy
_
dx
= –3x2 – 12x – 3
dy
But _
dx
= 0 at the turning points,

–3x2 – 12x – 3 = 0
x2 + 4x + 1 = 0
_
− b ± √b 2 − 4ac
x=_ 2a
___________
− (4) ± √(4) − 4(1)(1)
2
= ______________
2(1)
_
− 4 ± √12
x=_ 2
_ _
− 4 + √12 − 4 − √12
x=_ 2
and x=_ 2

x = –0,268 x = –3,732

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 Module 5 • Differential calculus 201

Substitute x = –3,732 and x = –0,268 into y = –x3 – 6x2 – 3x + 5,

y = –x3 – 6x2 – 3x + 5
= –(–3,732)3 – 6(–3,732)2 – 3(–3,732) + 5
y = –15,392
y = –x3 – 6x2 – 3x + 5
= –(–0,268)3 – 6(–0,268)2 – 3(–0,268) + 5
y = 5,392
Thus, the turning points of y = –x3 – 6x2 – 3x + 5 are (–3,732; –15,392) and
(–0,268; 5,392).
2
dy _
_ = d (–3x2 – 12x – 3)
dx 2 dx

= _d
(–3x2) – _d
12x – _d
3
dx dx dx

= –3_ d 2
dx
x – 12_d
dx
x–_ d
dx
3

= –3.2x – 12.1 – 0
d 2y
_ = –6x – 12
dx 2

b)

Stationary point (–3,732; –15,392) (–2; –5) (–0,268; 5,392)

2 2 2
d 2y dy dy dy
Sign of _2 _ = +10,392 > 0
2
_ =02
_ = –10,392 < 0
dx dx dx dx 2

Conclusion Minimum Test fails Maximum

x = –2

Interval –3,732 < x < –2 –2 < x < –0,268

Test value x = –3 x = –1
dy dy dy
Sign of _____
dx
_
dx
= +6 > 0 _
dx
= +6 > 0

Conclusion Increasing Increasing

Inflection point

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 202 N4 Mathematics – Lecturer Guide

d 2y
c) But _2 = 0 at the point of inflection,
dx
–6x – 12 = 0
–6x = 12
x = –2
Substitute x = –2 into y = –x3 – 6x2 – 3x + 5,
y = –x3 – 6x2 – 3x + 5
= –(–2)3 – 6(–2)2 – 3(–2) + 5
y = –5
Thus, the point of inflection of y = –x3 – 6x2 – 3x + 5 is (–2; –5).
d) y

(–0,268; 5,392) (0; 5)

4
y = –x – 6x –3x + 5
3 2

x
–6 –5 –4 –3 –2 –1 1 2
–2

–4
(–2; –5)
–6

–8

–10

–12

–14

(–3,732; –15,392) –16

1.5 y = 2x3 + x2 – 4x – 3
a) y = 2x3 + x2 – 4x – 3
dy _
_
dx
= d (2x3 + x2 – 4x – 3)
dx

=_
d
dx
2x3 + _
d 2 _
dx
d
x – dx 4x + _
d
dx
3

= 2_
d 3 _
dx
d 2
x + dx x – 4_
d
dx
x+_
d
dx
3

= 2.3x2 + 2x – 4.1 – 0
dy
_
dx
= 6x2 + 2x – 4

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 Module 5 • Differential calculus 203

dy
But _
dx
= 0 at the turning points,

6x2 + 2x – 4 = 0
3x2 + x – 2 = 0
(3x – 2)(x + 1) = 0
3x – 2 = 0 and x + 1 = 0
3x = 2 x = –1
x = _23
Substitute x = –1 and x = _23 into y = 2x3 + x2 – 4x – 3,
y = 2x3 + x2 – 4x – 3
= 2(–1)3 + (–1)2– 4(–1) – 3
y=0
y = 2x3 + x2 – 4x – 3
= 2(_23 ) + (_23 ) – 4(_23 ) – 3
3 2

17
y = − 4_
27

Thus, the turning points of y = 2x3 + x2 – 4x – 3 are (–1; 0) and (_23 ; − 4_

27 )
17
.
2
dy _
_ = d (6x2 + 2x – 4)
dx 2 dx

=_
d
dx
6x2 + _
d
dx
2x – _
d
dx
4

= 6_
d 2
dx
x + 2_
d
dx
x–_
d
dx
4

= 6.2x + 2.1 – 0
2
dy
_ = 12x + 2
dx 2

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 204 N4 Mathematics – Lecturer Guide

b)

(− 6 ; − 2 54 ) ( 3 ; − 4 27 )
Stationary point (–1; 0) _1 17
_ _2 17
_

2 2 2
d 2y dy dy dy
Sign of _2 _ = –10 < 0
2
_ =0 2
_ = +10 > 0
dx dx dx dx 2

Conclusion Maximum Test fails Minimum

x = –_16

Interval –1 < x < –_16 –_16 < x < _23

Test value x = –_12 x = _16

dy dy dy
Sign of _____
dx
_
dx
= –3_1 < 0
2
_
dx
= –3_1 < 0
2

Conclusion Decreasing Decreasing

Inflection point

2
dy
c) But _2 = 0 at the point of inflection,
dx
12x + 2 = 0
12x = –2
x = − _61

Substitute x = − _16 into y = 2x3 + x2 – 4x – 3,

y = 2x3 + x2 – 4x – 3

= 2(− _16 ) + (− _16 ) – 4(− _61 ) – 3

3 2

17
y = − 2_
54

Thus, the point of inflection of y = 2x3 + x2 – 4x – 3 is (− _16 ; − 2_

54 )
17
.

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 Module 5 • Differential calculus 205

d) y

(–1; 0)
x
–4 –3 –2 –1 1 2 3 4

–1

–2
( 54 )
–_1 ; –2_
17
6

(0; –3)

–4
y = 2x3 + x2 – 4x – 3

( 3 ; –4 27 )
2
_ 17
_
–5

1.6 y = –3x3 + 11x2 – 8x – 4

a) y = –3x3 + 11x2 – 8x – 4
dy _
_
dx
= d (–3x3 + 11x2 – 8x – 4)
dx

=_
d
(–3x3) + _
dx
d
11x2 – _
d
8x – _
dx
d
4 dx dx

= –3_x + 11_
d 3
dx
x – 8_
d 2 d
x–_
d
4
dx dx dx
2
= –3.3x + 11.2x – 8.1 – 0
dy
_
dx
= –9x2 + 22x – 8
dy
But _
dx
= 0 at the turning points,

–9x2 + 22x – 8 = 0
9x2 – 22x + 8 = 0
(9x – 4)(x – 2) = 0
9x – 4 = 0 and x–2=0
9x = 4 x=2
x = _49

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Substitute x = _49 and x = 2 into y = –3x3 + 11x2 – 8x – 4,

y = –3x3 + 11x2 – 8x – 4

= –3(_49 ) + 11(_49 ) – 8(_49 ) – 4

3 2

157
y = − 5_
243

y = –3x3 + 11x2 – 8x – 4
= –3(2)3 + 11(2)2 – 8(2) – 4
y=0

Thus, the turning points of y = –3x3 + 11x2 – 8x – 4 are (_49 ; − 5_

243 )
157
and (2; 0).
2
dy _
_ = d (–9x2 + 22x – 8)
dx 2 dx

=_
d
(–9x2) + _
dx
d
22x – _
dx
d
8 dx

= –9_x + 22_
d 2 d
x–_
dx
d
8
dx dx

= –9.2x + 22.1 – 0
2
dy
_ = –18x + 22
dx 2

b)

( 9 ; − 5 243 ) (1 9 ; − 2 243 )
Stationary point _4 157
_ _2 200
_ (2; 0)
2 2 2
d 2y dy dy dy
Sign of _2 _ = +14 > 0 _ =0 _ = –14 < 0
dx dx 2 dx 2 dx 2

Conclusion Minimum Test fails Maximum

x = 1_29

Interval _4 < x < 1_2 1_29 < x < 2

9 9

Test value x=1 x = 1_49

dy dy dy
Sign of _____
dx
_
dx
= +5 > 0 _
dx
= +5 > 0

Conclusion Increasing Increasing

Inflection point

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 Module 5 • Differential calculus 207

d 2y
c) But _2 = 0 at the point of inflection,
dx
–18x + 22 = 0
–18x = –22
11
x=_9

x = 1_29
11
Substitute x = _
9
into y = –3x3 + 11x2 – 8x – 4,
y = –3x3 + 11x2 – 8x – 4

= –3(_9)
+ 11(_9)
– 8(_
9)
3 2
11 11 11
–4
200
y = − 2_
243

Thus, the point of inflection of y = –3x3 + 11x2 – 8x – 4 is (1_29 ; − 2_

243 )
200
.

d) y

y = –3x3 + 11x2 – 8x – 4
1

(2; 0)
x
–4 –3 –2 –1 1 2 3 4

–1

–2

(1_29 ; –2_
200
)
–3 243

(0; –4)

–5

(_49 ; –5_
157
)
–6 243

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1.7 y = x3 – x2 – 8x + 12
a) y = x3 – x2 – 8x + 12
dy _
_
dx
= d (x3 – x2 – 8x + 12)
dx

= _
d 3
x –_
d 2 _ d
x – dx 8x + _
d
12
dx dx dx

= _
d 3
x –_
d 2
x – 8_
d
x+_
d
12
dx dx dx dx

= 3x2 – 2x – 8.1 + 0
dy
_
dx
= 3x2 – 2x – 8
dy
But _
dx
= 0 at the turning points,

3x2 – 2x – 8 = 0
(3x + 4)(x – 2) = 0
3x + 4 = 0 and x–2=0
3x = –4 x=2
x = − _43

x = − 1_13
Substitute x = − _43 and x = 2 into y = x3 – x2 – 8x + 12,
y = x3 – x2 – 8x + 12
= (− _43 ) – (− _43 ) – 8(− _43 ) + 12
3 2

14
y = 18_
27
y = x3 – x2 – 8x + 12
= (2)3 – (2)2 – 8(2) + 12
y=0
Thus, the turning points of y = x3 – x2 – 8x + 12 are (− 1_13 ; 18_
27 )
14
and (2; 0).
2
dy _
_ = d (3x2 – 2x – 8)
dx 2 dx

= _d
3x2 –_
d
2x – _
d
8
dx dx dx

= 3_ d 2
dx
x – 2_
d
dx
x–_
d
dx
8

= 3.2x – 2.1 – 0
2
dy
_ = 6x – 2
dx 2

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 Module 5 • Differential calculus 209

b)

(− 1 3 ; 18 27 ) ( 3 ; 9 27 )
Stationary point _1 14
_ _1 7
_ (2; 0)
2 2 2
d 2y dy dy dy
Sign of _2 _ = –10 < 0
2
_ =02
_ = +10 > 0
dx dx dx dx 2

Conclusion Maximum Test fails Minimum

x = _13

Interval –1_13 < x < _13 _1 < x < 2

3

Test value x = –_13 x=1

dy dy dy
Sign of _____
dx
_
dx
= –7 < 0 _
dx
= –7 < 0

Conclusion Decreasing Decreasing

Inflection point

d 2y
c) But _2 = 0 at the point of inflection,
dx
6x – 2 = 0
6x = 2
x = _13

Substitute x = _13 into y = x3 – x2 – 8x + 12,

y = x3 – x2 – 8x + 12

= (_13 ) – (_13 ) – 8(_13 ) + 12

3 2

7
y = 9_
27

Thus, the point of inflection of y = x3 – x2 – 8x + 12 is (_13 ; 9_

27 )
7
.

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d) y
(–1_13 ; 18_
14
27
)
18

16
14

y = x3 – x2 – 8x + 12 (0; 12)

10
(_31 ; 9_7
27
)
8
6
4
2
(2; 0)
x
–4 –3 –2 –1 1 2 3 4
–2

–4

1.8 y = –x3 – 2x2 + 15x + 36

a) y = –x3 – 2x2 + 15x + 36
dy _
_
dx
= d (–x3 – 2x2 + 15x + 36)
dx

=_
d
(–x3) – _
dx
d
2x2 + _
d
dx
15x + _
d
36dx dx

= –_d 3
dx
x – 2_
d
dx
x2 + 15_d
dx
x + _36
d
dx

= –(3x2) – 2.2x + 15.1 + 0

dy
_
dx
= –3x2 – 4x + 15
dy
But _
dx
= 0 at the turning points,

–3x2 – 4x + 15 = 0
3x2 + 4x – 15 = 0
(3x – 5)(x + 3) = 0
3x – 5 = 0 and x + 3 = 0
3x = 5 x = –3
x = _53

x = 1_23

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 Module 5 • Differential calculus 211

Substitute x = –3 and x = _53 into y = –x3 – 2x2 + 15x + 36,

y = –x3 – 2x2 + 15x + 36
= –(–3)3 – 2(–3)2 + 15(–3) + 36
y=0
y = –x3 – 2x2 + 15x + 36

= –(_53 ) – 2(_53 ) + 15(_53 ) + 36

3 2

22
y = 50_
27

Thus, the turning points of y = –x3 – 2x2 + 15x + 36 are (–3; 0) and (1_23 ; 50_
27 )
22
.
2
dy _
_ = d (–3x2 – 4x + 15)
dx 2 dx

= _d
(–3x2) – _d
4x + _d
15
dx dx dx

= –3_ d 2
dx
x – 4_
d
dx
x+_ d
dx
15

= –3.2x – 4.1 + 0
2
dy
_ = –6x – 4
dx 2

b)

(− 3 ; 25 27 ) (1 3 ; 50 27 )
Stationary point (–3; 0) _2 11
_ _2 22
_

2 2 2
d 2y dy dy dy
Sign of _2 _ = +14 > 0
2
_ =0 2
_ = –14 < 0
dx dx dx dx 2

Conclusion Minimum Test fails Maximum

x = –_23

Interval –3 < x < − _23 − _23 < x < 1_23

Test value x = –1 x = –_13

dy dy dy
Sign of _____
dx
_
dx
= +16 > 0 _
dx
= +16 > 0

Conclusion Increasing Increasing

Inflection point

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d 2y
c) But _2 = 0 at the point of inflection,
dx
–6x – 4 = 0
–6x = 4

x = − _23

Substitute x = − _23 into y = –x3 – 2x2 + 15x + 36,

y = –x3 – 2x2 + 15x + 36

= –(− _23 ) – 2(− _23 ) + 15(− _23 ) + 36

3 2

11
y = 25_
27

Thus, the point of inflection of y = –x3 – 2x2 + 15x + 36 is (− _23 ; 25_

27 )
11
.

d) y
(1_23 ; 50_
22
27
)
50
y = –x3 – 2x2 + 15x + 36
40
(0; 36)

30
(–_25 ; 11
25_
27
)

20

10

(–3; 0)
x
–5 –4 –3 –2 –1 1 2 3 4

1.9 y = 4x3 – 8x2 – 7x + 17

a) y = 4x3 – 8x2 – 7x + 17
dy _
_
dx
= d (4x3 – 8x2 – 7x + 17)
dx

= _
d
4x3 –_
d
8x2 – _
d
7x + _
d
17
dx dx dx dx

= 4_d 3
dx
x – 8_
d 2
dx
x – 7_
d
dx
x+_
d
dx
17

= 4.3x2 – 8.2x – 7.1 + 0

dy
_
dx
= 12x2 – 16x – 7

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 Module 5 • Differential calculus 213

dy
But _
dx
= 0 at the turning points,
12x2 – 16x – 7 = 0
_
− b ± √b 2 − 4ac
x=_ 2a
________________
− (− 16) ± √(− 16) − 4(12)(− 7)
2
= ___________________
2(12)
_
16 ± √592
=_ 24
_ _
16 + √592 16 − √592
x=_ 24
and x = _ 24

x = 1,681 x = –0,347
Substitute x = –0,347 and x = 1,681 into y = 4x3 – 8x2 – 7x + 17,
y = 4x3 – 8x2 – 7x + 17
= 4(–0,347)3 – 8(–0,347)2 – 7(–0,347) + 17
y = 18,299
y = 4x3 – 8x2 – 7x + 17
= 4(1,681)3 – 8(1,681)2 – 7(1,681) + 17
y = 1,627
Thus, the turning points of y = 4x3 – 8x2 – 7x + 17 are (–0,347; 18,299) and
(1,681; 1,627).
2
dy _
_ = d (12x2 – 16x – 7)
dx 2 dx

= _d
12x2 –_
d
16x – _
d
7
dx dx dx

= 12_ d 2
dx
x – 16_
d
dx
x–_
d
dx
7

= 12.2x – 16.1 – 0
2
dy
_ = 24x – 16
dx 2

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b)

( 3 ; 9 27 )
Stationary point (–0,347; 18,299) _2 26
_ (1,681; 1,627)
2 2 2
d 2y dy dy dy
Sign of _2 _ = –24,331 < 0
2
_ =02
_ = +24,331 > 0
dx dx dx dx 2

Conclusion Maximum Test fails Minimum

x = _23

Interval –0,347 < x < _23 _2 < x < 1,681

3

Test value x = _13 x=1

dy dy dy
Sign of _____
dx
_
dx
= –11 < 0 _
dx
= –11 < 0

Conclusion Decreasing Decreasing

Inflection point

d 2y
c) But _2 = 0 at the point of inflection,
dx
24x – 16 = 0
24x = 16
x = _23
Substitute x = _23 into y = 4x3 – 8x2 – 7x + 17,
y = 4x3 – 8x2 – 7x + 17

= 4(_32 ) – 8(_23 ) – 7(_23 ) + 17

3 2

26
y = 9_
27

Thus, the point of inflection of y = 4x3 – 8x2 – 7x + 17 is (_23 ; 9_

27 )
26
.

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 Module 5 • Differential calculus 215

d) y
(–0,347; 18,299)
(0; 17)
16

14

12

10 (_32 ; 9_
26
)
27
y = 4x3 – 8x2 – 7x + 17
8

2
(1,681; 1,627)
x
–4 –3 –2 –1 1 2 3 4
–1
–2

1.10 y = –5x3 + 9x2 – 2x + 8

a) y = –5x3 + 9x2 – 2x + 8
dy _
_
dx
= d (–5x3 + 9x2 – 2x + 8)
dx

=_
d
(–5x3) + _
dx
d
9x2 – _
dx
d
2x + _
d
8 dx dx

= –5_x + 9_
d 3
dx
x – 2_
d 2 d
dx
x+_
d
8 dx dx
2
= –5.3x + 9.2x – 2.1 + 0
dy
_
dx
= –15x2 + 18x – 2
dy
But _
dx
= 0 at the turning points,

–15x2 + 18x – 2 = 0
15x2 – 18x + 2 = 0 _
− b ± √b 2 − 4ac
x=_ 2a
_______________
− (− 18) ± √(− 18) 2 − 4(15)(2)
__________________
= 2(15)
_
18 ± √204
x=_ 30
_ _
18 + √204 18 − √204
x=_ 30
and x = _ 30

x = 1,076 x = 0,124

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Substitute x = 0,124 and x = 1,076 into y = –5x3 + 9x2 – 2x + 8,

y = –5x3 + 9x2 – 2x + 8
= –5(0,124)3 – 9(0,124)2 – 2(0,124) + 8
y = 7,881
y = –5x3 + 9x2 – 2x + 8
= –5(1,076)3 – 9(1,076)2 – 2(1,076) + 8
y = 10,039
Thus, the turning points of y = –5x3 + 9x2 – 2x + 8 are (0,124; 7,881) and
(1,076; 10,039).
2
dy _
_ = d (–15x2 + 18x – 2)
dx 2 dx

= _d
(–15x2) + _d
18x – _d
2
dx dx dx

= –15_ d 2
dx
x + 18_d
dx
x–_ d
dx
2

= –15.2x + 18.1 – 0
2
dy
_ = –30x + 18
dx 2

b)

( 5 ; 8 25 )
Stationary point (0,124; 7,881) _3 24
_ (1,076; 10,039)
2 2 2
d 2y dy dy dy
Sign of _2 _ = +14,283 > 0
2
_ =02
_ = –14,283 < 0
dx dx dx dx 2

Conclusion Minimum Test fails Maximum

x = _35

Interval 0,124 < x < _35 _3 < x < 1,076

5

Test value x = _25 x = _45

dy dy dy
Sign of _____
dx
_
dx
= +2,800 > 0 _
dx
= +2,800 > 0

Conclusion Increasing Increasing

Inflection point

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 Module 5 • Differential calculus 217

d 2y
c) But _2 = 0 at the point of inflection,
dx
–30x + 18 = 0
–30x = –18

x = _35

Substitute x = _53 into y = –5x3 + 9x2 – 2x + 8,

y = –5x3 + 9x2 – 2x + 8

= –5(_35 ) + 9(_35 ) – 2(_35 ) + 8

3 2

24
y = 8_
25

Thus, the point of inflection of y = –5x3 + 9x2 – 2x + 8 is (_35 ; 8_

25 )
24
.
d) y

14

12

(1,076; 10,039)
10

(_35 ; 8_
24
)
(0; 8) 25

(0,124; 7,881)

4
y = –5x3 + 9x2 – 2x + 8

2
x
–4 –3 –2 –1 1 2 3 4
–2

1.11 f(x) = x3 + 2x2 – 10x – 30

a) y = x3 + 2x2 – 10x – 30
dy _
_
dx
= d (x3 + 2x2 – 10x – 30)
dx

= _
d 3
x +_
d
2x2 – _
d
10x – _
d
30
dx dx dx dx

= _
d 3
x + 2_
d 2
x – 10_
d
x–_
d
30
dx dx dx dx

= 3x2 + 2.2x – 10.1 – 0

dy
_
dx
= 3x2 + 4x – 10

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dy
But _
dx
= 0 at the turning points,

3x2 + 4x – 10 = 0
_
− b ± √b 2 − 4ac
x=_ 2a
______________
− (4) ± √(4) 2 − 4(3)(− 10)
________________
= 2(3)
_
− 4 ± √136
x=_ 6
_ _
− 4 + √136 − 4 − √136
x=_ 6
and x=_ 6

x = 1,277 x = –2,610
Substitute x = –2,610 and x = 1,277 into y = x3 + 2x2 – 10x – 30,
y = x3 + 2x2 – 10x – 30
= (–2,610)3 + 2(–2,610)2 – 10(–2,610) – 30
y = –8,055
y = x3 + 2x2 – 10x – 30
= (1,277)3 + 2(1,277)2 – 10(1,277) – 30
y = –37,426
Thus, the turning points of y = x3 + 2x2 – 10x – 30 are (–2,610; –8,055) and
(1,277; –37,426).
2
dy _
_ = d (3x2 + 4x – 10)
dx 2 dx

= _d
3x2 +_
d
4x – _
d
10
dx dx dx

= 3_ d 2
dx
x + 4_
d
dx
x–_
d
dx
10

= 3.2x + 4.1 – 0
2
dy
_ = 6x + 4
dx 2

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 Module 5 • Differential calculus 219

b)

(− 3 ; − 22 27 )
Stationary point (–2,610; –8,055) _2 20
_ (1,277; –37,426)
2 2 2
d 2y dy dy dy
Sign of _2 _ = –11,622 < 0
2
_ =0 2
_ = +11,622 > 0
dx dx dx dx 2

Conclusion Maximum Test fails Minimum

x = –_23

Interval –2,610 < x < − _23 − _23 < x < 1,277

Test value x = –1 x = − _13

dy dy dy
Sign of _____
dx
_
dx
= –11 < 0 _
dx
= –11 < 0

Conclusion Decreasing Decreasing

Inflection point

d 2y
c) But _2 = 0 at the point of inflection,
dx
6x + 4 = 0
6x = –4
x = − _23
Substitute x = − _23 into y = x3 + 2x2 – 10x – 30,
y = x3 + 2x2 – 10x – 30

= (− _32 ) + 2(− _23 ) – 10(− _32 ) – 30

3 2

20
y = − 22_
27

Thus, the point of inflection of y = x3 + 2x2 – 10x – 30 is (− _23 ; − 22_

27 )
20
.

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d) y
x
–5 –4 –3 –2 –1 1 2 3 4

–5
(–2,610; –8,055)

–10

–15

–20

(–_23 ; –22_
20
27
)
–25

(0; –30)

y = x3 + 2x2 – 10x – 30
–35

(1,277; –37,426)

1.12 f(x) = –x3 – 4x2 – 3x – 2

a) y = –x3 – 4x2 – 3x – 2
dy _
_
dx
= d (–x3 – 4x2 – 3x – 2)
dx

= _
d
(–x3) – _d
4x2 – _d
3x – _d
2
dx dx dx dx

= –_
dx
x – 4_
d 3
dx
x – 3_
d 2 d
dx
x–_ d
dx
2

= –3x2 – 4.2x – 3.1 – 0

dy
_
dx
= –3x2 – 8x – 3
dy
But _
dx
= 0 at the turning points,

–3x2 – 8x – 3 = 0
3x2 + 8x + 3 = 0
_
− b ± √b 2 − 4ac
x=_ 2a
___________
− (8) ± √(8) − 4(3)(3)
2
= ______________ 2(3)
_
− 8 ± √28
x=_ 6
_ _
− 8 + √28 − 8 − √28
x=_ 6
and x=_ 6

x = –0,451 x = –2,215

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 Module 5 • Differential calculus 221

Substitute x = –2,215 and x = –0,451 into y = –x3 – 4x2 – 3x – 2

y = –x3 – 4x2 – 3x – 2
= –(–2,215)3 – 4(–2,215)2 – 3(–2,215) – 2
y = –4,113
y = –x3 – 4x2 – 3x – 2
= –(–0,451)3 – 4(–0,451)2 – 3(–0,451) – 2
y = –1,369
Thus, the turning points of y = –x3 – 4x2 – 3x – 2 are (–2,215; –4,113) and
(–0,451; –1,369).
2
dy _
_ = d (–3x2 – 8x – 3)
dx 2 dx

= _d
(–3x2) – _d
8x – _d
3
dx dx dx

= –3_ d 2
dx
x – 8_
d
dx
x–_ d
dx
3

= –3.2x – 8.1 – 0
2
dy
_ = –6x – 8
dx 2

b)

(− 1 3 ; − 2 27 )
Stationary point (–2,215; –4,113) _1 20
_ (–0,451; –1,369)
2 2 2
d 2y dy dy dy
Sign of _2 _ = +5,292 > 0
2
_ =0 2
_ = –5,292 < 0
dx dx dx dx 2

Conclusion Minimum Test fails Maximum

x = –1_13

Interval –2,215 < x < − 1_13 − 1_13 < x < –0,451

Test value x = − 1_23 x = –1

dy dy dy
Sign of _____
dx
_
dx
= +2 > 0 _
dx
= +2 > 0

Conclusion Increasing Increasing

Inflection point

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d 2y
c) But _2 = 0 at the point of inflection,
dx
–6x – 8 = 0
–6x = 8

x = − _43

x = − 1_13

Substitute x = − _43 into y = –x3 – 4x2 – 3x – 2,

y = –x3 – 4x2 – 3x – 2

= –(− _43 ) – 4(− _43 ) – 3(− _43 ) – 2

3 2

20
y = − 2_
27

Thus, the point of inflection of y = –x3 – 4x2 – 3x – 2 is (− 1_13 ; − 2_

27 )
20
.

d) y

x
–5 –4 –3 –2 –1 1 2 3
(–0,451; –1,369)
(0; –2)
(–1_13 ; –2_
20
27
)

–4
(–2,215; –4,113)

–6

y = –x3 – 4x2 – 3x – 2
–8

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 Module 5 • Differential calculus 223

2. 2.1
y
y = x3

x
–2 –1 1 2

–1

–2

2.2
y
y = x3 + 8

x
–2 –1 1 2
–1

x-intercept: y = 0
0 = x3 + 8
x 3 = –8
x = –2
y-intercept: x = 0
y=8

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 224 N4 Mathematics – Lecturer Guide

Summative assessment: Module 5 SB page 361

_
1.1 lim (_ lim (_ )
x→2 16 − 5x )
3
√2 + 7x 27 + x
1. 1.2 2
x→– 3 x −9
_
27 + (− 3) 3
√
=_
2 + 7(2) =_2
16 − 5(2) (− 3) − 9
_
27 − 27
√2 + 14
=_ =_9−9
16 − 10
_
√16
=_ = _00
6

lim (_ )
3
= _46 27 + x
2
x→– 3 x −9
= _2
= lim (_
x→– 3 x − 9 )
3
(3) 3
x + 27
2

= lim [____________
(x + 3)(x − 3) ]
(x + 3)(x 2 − 3x + 9)
x→– 3

= lim [_ x−3 ]
2
x − 3x + 9
x→– 3

(− 3) 2 − 3(− 3) + 9
= ___________
(− 3) − 3
9+9+9
=_−3 − 3
27
=_
−6

= − _29

= − 4_12 (3)

1.3 lim (_
x→∞ x (5x + 2) )
4 2
3x + x
3

3(∞) 4 + (∞) 2
=_
3
(∞) (5(∞) + 2)
∞
=_
∞

lim (_ )
4 2
3x + x
3
x→∞ x (5x + 2)

= lim (_ )
4 2
3x + x
4 3
x→∞ 5x + 2x
4 2
3x

(_
_ +_

)
x

= lim _
x4 x4
4 3
5x 2x
x→∞ +_4 4
x x

(5+ x)
1
3+_
= lim __2
2
x

x→∞

1
3+_
=_
2
(∞)
2
5+_
(∞)
3+0
=_
5+0

= _35 (3)

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 Module 5 • Differential calculus 225

_1
2. 2.1 (2 − 3x) − 2
_1
= [(2) + (− 3x)] − 2
(− 2 )(2) (− 3x) (− 2 )(− 2 )(2) (− 3x) (− 2 )(− 2 )(− 2 )(2) (− 3x)
_1 _1 − _32 1 _1 _3 − _52 2 _1 _3 _5 − _72 3
(2) − 2(− 3x) 0
_ ___________ _______________ ___________________
= 0!
+ 1!
+ 2!
+ 3!
+…

(− 2 )(2) (− 3x) (− 2 )(− 2 )(2) (9x ) (− 2 )(− 2 )(− 2 )(2) (− 27x )

_1 _1 − _32 _1 _3 − _52 2 _1 _3 _5 − _72 3
(2) − 21
_ ___________ _______________ ____________________
= 1
+ 1
+ 2
+ 6
+…
2 3
1_ _
=_ + 3x_ + _
27x_ _
+ 135x_ + …
√2 4√2 32√2 128√2
_ _ _ _
2 3
√2 3√ 2 x _
=_ +_
2
+ 27√2 x + _
8
135√2 x
+… 64 256
(4)

2.2 There are 13 terms in the expansion.

Thus, the middle term is T7.
The seventh term is T7,
∴ T7 = Tr+1
7=r+1
r=6
T6+1 = _ n!
r!(n − r)!
x n–rh r
12!
T7 = _
6!(12 − 6)!
(z2)6(4y2)6

T7 = 3 784 704z12y12 (4)

3. 3.1 f (x) = –x3 – 2x2

f (x + h) = –(x + h)3 – 2(x + h)2

= –[_ ] – 2(x + h)(x + h)

(x) 3(h) 0 3(x) 2(h) 1 3.2(x) 1(h) 2 3.2.1(x) 0(h) 3
0!
+_
1!
+_
2!
+_
3!

= –[_ ] – 2(x + 2xh + h )

3 2 1 2 3
x .1 _
1
+ 3.x1 .h + _
3.2.x .h
2
3.2.1.1.h
+_ 6
2 2

= –[x3 + 3x2h + 3xh2 + h3] – 2(x2 + 2xh + h2)

= –x3 – 3x2h – 3xh2 – h3 – 2x2 – 4xh – 2h2
f (x + h) = –x3 – 2x2 – 3x2h – 4xh – 3xh2 – 2h2 – h3
f (x + h) – f (x) = (–x3 – 2x2 – 3x2h – 4xh – 3xh2 – 2h2 – h3) – (–x3 – 2x2)
= –x3 – 2x2 – 3x2h – 4xh – 3xh2 – 2h2 – h3 + x3 + 2x2
f (x + h) – f (x) = –3x2h – 4xh – 3xh2 –2h2 – h3
f(x + h) − f(x) ___________________ 2 2 2 3
_
h
= − 3x h − 4xh − 3xh − 2h − h h
h(− 3x − 4x − 3xh − 2h − h 2)
2
= __________________
h
f(x + h) − f(x)
_
h
= –3x2 – 4x – 3xh –2h – h2
f(x + h) − f(x)
f ʹ(x) = lim _
h h→0

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 226 N4 Mathematics – Lecturer Guide

= lim (− 3x 2 − 4x − 3xh − 2h − h 2)
h→0

= –3x2 – 4x – 3x(0) –2(0) – (0)2

f ʹ(x) = –3x2 – 4x (5)

3.2 f (x) = –x3 – 2x2

f ʹ(x) = _
d
dx
(–x3) – _
d
dx
(2x2)

= –_
d 3
dx
x – 2_
d 2
dx
x

= –(3x2) – 2(2x)

f ʹ(x) = –3x2 – 4x (2)

4. 4.1 y = –5 sin(_π2 − _5x )

y = –5 cos_5x

Let u = _5x then y = –5 cos u

Therefore,
dy
_
dx
=1
du _
5
_
du
–5. –sin u
dy _
_ dy
dx
= × _
du
du
dx

= –5. –sin u × _15

5 sin u
=_ 5
5 sin _x
=_
5
5
dy
_
dx
= sin _5x (4)

4.2 y = ex.sin x

Let f (x) = ex and g(x) = sin x

Thus f ʹ(x) = ex and gʹ(x) = cos x

dy
_
dx
= f (x) _
d
g(x) + g(x) _
dx
d
f (x) dx

= (e )(cos x) + (sin x)(ex)

x

dy
_
dx
= ex (cos x + sin x) (4)

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 Module 5 • Differential calculus 227

ln_x
4.3 y = _
√x
_1
Let f (x) = ln x and g(x) = x 2

Thus f ʹ(x) = _1x and gʹ(x) = _12 x − 2

_1

g(x)_
dy ______________
d
f(x) − f(x)_
d
g(x)
_ =
dx dx
dx (g(x)) 2

(2 )
(x 2) _1 − (ln x) _1 x − 2
(x)
_1 _1

= _____________
(x 2)
_1 2

x − 2 − _1 x − 2 ln x
_1 _1

=_
2
x

( ) 1 − _1 ln x
− _1
x 2

=_
2
x
1 − _12 ln x
=__
x√x
_
( )
1
√x 1 − _ ln x
=_
2

x2
_
dy _ ( )
_ = √x 2 − ln x (4)
dx 2x 2
1
5. 5.1 y = tan x + log3x – 4e–x – _5
x
y = tan x + log3x – 4e – x–5 –x

dy _
_
dx
= d (tan x) + _
d
dx
(log3x) – _
d
(4e–x ) – _
d –5
(x )dx dx dx

=_
d
dx
tan x + _
d
dx
log3x – 4_ e –_
d –x
dx
d –5
dx
x
1
= sec2x + _
x ln 3
– 4(e–x. –1) – (–5x–6)
dy
_ 1 4 5
= sec2x + _ +_ +_ x ln 3 ex
dx x6

dx ( x ln 3 ) dx ( e ) dx ( x )
2
dy _
_ 1 4 5
2 =
d
(sec2x) + _
d _
dx
+_
d _
x +_
d _
6
dx

dx ( ln 3
=_
d
dx
(sec2x) + _
d _ 1
.x −1) + _
d
dx
(4e–x) + _
d
dx
(5x–6)

1 _
=_
d
dx
(sec x) 2 + _
ln 3 dx
x + 4_
d −1
e + 5_
d –x
dx
d –6
dx
x

1 (
= (2 sec x sec x tan x) + _
ln 3
−1x −2) + 4(e–x. –1) + 5(–6x–7)
2
dy 1 4 _
_ = 2 sec2x tan x – _ –_ – 30 ex
(6)
dx 2 x 2 ln 3 x7

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 228 N4 Mathematics – Lecturer Guide

_
y = ln(_6x ) – √e x + cosec x – 2 log x
2
5.2 2

= ln(_6x ) – (ex) 2 + cosec x – x2

_1

y = ln(_6x ) – e 2 + cosec x – x2
_x

(e ) + _
dx dx ( ( 6 )) dx
dy _
_ d _2x
_x – _
= d
ln d
dx
(cosec x) – _
d 2
dx
(x )

ln (_6x ) – _
_x
=_
d
dx
d 2 _
dx
d
e + dx cosec x – _
d 2
dx
x

(6 )
= _1_x ._16 – (e 2._12 ) + (–cosec x cot x) – 2x
_x

dy _
= 1 – _1 e 2 – cosec x cot x – 2x
_x
_
dx x 2

dx ( x ) dx ( 2 )
2
dy _
_ 1
d _ 1 _2x
2 = –_
d _
e –_ d
(cosec x cot x) – _
d
(2x) dx dx
dx

dx ( 2 ) dx
1 2 _x
=_ (x ) – _
d –1
dx
d _
e –_
d
(cosec x cot x) – _
d
dx
(2x)

x – 12 _
_x
=_
d –1 _
dx
d 2 _
dx
d
e – dx cosec x cot x – 2_
d
dx
x

x – 12 _
_x
=_
d –1 _
dx
d 2
dx
e –[cosec x_
d
dx
cot x + cot x_
d
dx
cosec x] – 2_
d
dx
x

= –1.x–2 – _12 (e 2._12 ) –[cosec x. –cosec2x + cot x. – cosec x cot x] – 2.1

_x

1 _1 2 _x
= −_ 3 2
2 – e – [–cosec x – cosec x cot x] – 2
4
x
1 _1 2 _x
= −_ 3 2
2 – e – [–cosec x – cosec x(cosec x –1)] – 2
4
x
1 _1 2 _x
= −_ 3 3
2 – e – [–cosec x – cosec x + cosec x] – 2
4
x
1 _1 2 _x
= −_ 3
2 – e – [–2 cosec x + cosec x] – 2
4
x
d 2y
_ 1 _1 2 _x
2 = −_ 4
3
2 – e + 2 cosec x – cosec x – 2 (6)
dx x

6. 6.1 y = x3 – 4x2 + x + 6
dy _
_
dx
= d (x3 – 4x2 + x + 6)
dx

=_
d 3 _
x – d 4x2 + _
dx
d
x+_
d
dx
6 dx dx

=_x – 4_
d 3
dx
d 2 _
x + dx+_
d
6 dx dx dx
2
= 3x – 4.2x + 1 + 0
dy
_
dx
= 3x2 – 8x + 1

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 Module 5 • Differential calculus 229

dy
But _
dx
= 0 at the turning points,

3x2 – 8x + 1 = 0
_
− b ± √b 2 − 4ac
x=_ 2a
_____________
− (− 8) ± √(− 8) 2 − 4(3)(1)
________________
= 2(3)
_
8 ± √52
x=_ 6
_ _
8 + √52 8 − √52
x=_ 6
and x = _ 6

x = 2,535 x = 0,132
Substitute x = 0,132 and x = 2,535 into y = x3 – 4x2 + x + 6,
y = x3 – 4x2 + x + 6
= (0,132)3 – 4(0,132)2 + (0,132) + 6
y = 6,065
y = x3 – 4x2 + x + 6
= (2,535)3 – 4(2,535)2 + (2,535) + 6
y = –0,879
Thus, the turning points of y = x3 – 4x2 + x + 6 are (0,132; 6,065)
and (2,535; –0,879). (6)
dy
_
6.2 dx
= 3x2 – 8x + 1
d 2y
_ =_
d
(3x2 – 8x + 1)
dx 2 dx

=_
d
dx
3x2 – _
d
dx
8x + _
d
dx
1

= 3_
d 2
dx
x – 8_
d
dx
x+_
d
dx
1

= 3.2x – 8.1 + 0
2
dy
_ = 6x – 8
dx 2

Stationary point (0,132; 6,065) (2,535; –0,879)

2 2
d 2y dy dy
Sign of _2 _ = –7,211 < 0
2
_ = +7,211 > 0
dx dx dx 2

Conclusion Maximum Minimum

(4)

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 230 N4 Mathematics – Lecturer Guide

2
dy
6.3 _2 = 6x – 8
dx
d 2y
But _2 = 0 at the point of inflection,
dx
6x – 8 = 0
6x = 8

x = _43

x = 1_13

Substitute x = _43 into y = x3 – 4x2 + x + 6,

y = x3 – 4x2 + x + 6

= (_43 ) – 4(_43 ) + (_43 ) + 6

3 2

16
y = 2_
27

Thus, the point of inflection of y = x3 – 4x2 + x + 6 is (1_13 ; 2_

27 )
16
. (3)

6.4 x-intercept(s), y = 0:
y = x3 – 4x2 + x + 6
0 = x3 – 4x2 + x + 6
From the factor theorem y = 0 when x = –1. Therefore, x + 1 is a factor of
y = x3 – 4x2 + x + 6.
Thus, the quadratic is
x2 – 5x + 6 = 0
Therefore,
0 = x3 – 4x2 + x + 6
0 = (x + 1)(x2 – 5x + 6)
0 = (x + 1)(x – 2)(x – 3)
x + 1 = 0 or x – 2 = 0 or x–3=0
x = –1 x=2 x=3
(–1; 0) (2; 0) (3; 0)
y-intercept, x = 0:
y = x3 – 4x2 + x + 6
= (0)3 – 4(0)2 + (0) + 6
y=6
(0; 6) (4)

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 Module 5 • Differential calculus 231

6.5 y

(0; 6) (0,132; 6,065)

3
y = x3 –4x2 + x + 6 (1_13 ; 2_
16
)
27
2

1
(–1; 0) (2; 0) (3; 0)
x
–3 –2 –1 1 2 3 4 5
–1
(2,535; –0,879)
–2 (5)

TOTAL: [70]

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 232 N4 Mathematics – Lecturer Guide

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 MODULE

6 Integral calculus
After they have completed this module, students should be able to:
• understand the concept of integration as a summation function (definite
integral) and as a process of anti-differentiation (indefinite integral);
• apply standard forms of integrals as a process of anti-differentiation;
• integrate functions given on the formula sheet:
– k x n, n real with n ≠ − 1
– _k, k a nx, k e nx with a ≥ 0, k, n ∈ ℝ
x
– k sin(bx) and k cos(bx) with b and k ∈ ℝ;
• integrate polynomials consisting of terms of the above forms;
• apply integration to determine the magnitude of an area included by a
curve and the x-axis, or by a curve, the x-axis and the ordinates x = a and
x = b, where a and b are integers;
• using the definite integral with two limits to calculate the area bounded
by the graph, the x-axis and values given to define the area; areas include
areas above the x-axis, areas below the x-axis and joined areas above and
below the x-axis:
A = ∫ a y dx or A total = ∫ a y dx + ∫ c y dx; and
b b d

• calculate the intersection points of two curves, and sketch the two graphs
on the same system of axes indicating the area bounded by the two
intersection points calculated and show the representative strip used to
calculate the area.

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 234 N4 Mathematics – Lecturer Guide

Introduction
In this module students will analyse and represent mathematical and contextual
situations using integrals and find areas under curves by using integration rules.
Integration can also be used to find areas, volumes and central points.

Integration and differentiation are the two main branches of calculus. Integration is an
important concept in mathematics and it is the inverse process of differentiation.

The term ‘integral’ may also be referred to as the anti-derivative. Integrals and
derivatives have numerous applications in science and engineering, for example to
calculate surface areas and to formulate physical laws of electrodynamics.

Integral calculus is the division of calculus that is concerned with:

• the inverse operation of differentiation – this will be dealt with when we solve
indefinite integrals; and
• the summing of the values of a function over a particular range. One of the main
uses is to calculate the areas of irregular shapes. This will be dealt with when we
solve definite integrals.

To summarise: Integration is the process of finding the definite or indefinite integral of

a function and it is the inverse of differentiation (anti-differentiation).

Students need the following pre-knowledge to successfully complete this module.

Pre-knowledge
• Laws of exponents

Law Deductions/Definition

x m × x n = x m+n x −m = _1
xm
x m ÷ x n = x m−n
1
_ = xm
(x m)n = x mn x −m

(x)
m
(xy) = x × y _x − m = _y
(y)
m m m

(y)
m
_x m = _
x
ym x0 = 1
m
_ n
_
x n = √x m ; x > 0; n > 0

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 Module 6 • Integral calculus 235

• Logarithmic laws
• Trigonometric identities
• Standard derivatives

f(x) f′(x)

k 0
kx n n.k x n−1

ka x k a x ln a
k a nx n.k a nx ln a

ke x ke x
k e nx n.k e nx

k ln x _k
x

k ln(nx) _k
x

k log ax _
k
x ln a

k log a(nx) _
k
x ln a

k sin(nx) nk cos(nx)
k cos(nx) − nk sin(nx)
k tan(nx) nk sec 2(nx)
k cot(nx) − nk cosec 2(nx)
k sec(nx) nk sec(nx).tan(nx)
k cosec(nx) − nk cosec(nx).cot(nx)

• Rules for standard integrals

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 236 N4 Mathematics – Lecturer Guide

It is essential to be able to sketch and recognise the following graphs dealt with in
Module 3: Sketch graphs.

Graph Equation
1. Straight line y = mx + c or ax + by + c = 0
_ _
2. Circle or semi-circles x 2 + y 2 = r 2; y = ± √r 2 − x 2 ; x = ± √r 2 − y 2

3. Rectangular hyperbola xy = c or y = _xc or x = _yc

2
x
_
2 y
4. Ellipse +_=1
a2 b2
2
x
_
2 y
5. Central hyperbola −_=1
a2 b2

6. Exponential graph y = k a nx; y = k e nx

7. Logarithmic graph y = k log a(nx); y = k log e(nx)
8. Parabola y = ax 2 + bx + c
9. Cubic graph y = ax 3 + b x 2 + cx + d
10. Trigonometric curves y = a sin(bx + c) + d; y = a cos(bx + c) + d;
y = a tan(bx + c) + d
y = cosec x; y = sec x; y = cot x

Activity 6.1 SB page 373

1. ∫ x5 dx 2. ∫ 4x2 dx
6 3
4x
=_
x
6
+c =_3
+c

or _16 x6 + c or _43 x3 + c

3. ∫ 2 dx 4. ∫ da
= 2x + c =a+c
2
5. ∫_2 dx 6. ∫ 3 dy
x
= 3y + c
= ∫ 2x–2 dx
−1
2x
=_−1
+c

= − _2x + c

7. ∫ 0,23 dr 8. ∫ – 3x4 dx
5
= 0,23r + c − 3x
=_ +c
5

or = − _53 x5 + c

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 Module 6 • Integral calculus 237

9. ∫_
dx
3 10. 3∫ x3 dx
x
4
1
= ∫_3 dx = 3._
x
4
+c
x
= ∫ x–3 dx = _34 x4 + c
−2
=_
x
−2
+c
1
= −_ 2 + c
2x
_
11. – 2 ∫ 3x dx 12. ∫ √x 2 dx
2
3x
= – 2._2
+c = ∫ x dx
2
= – 3x2 + c =_
x
2
+c

or _12 x2 + c
_
1
13. ∫ 3√x 3 dx 14. ∫ _ 2 dx
3x
_3
= ∫ 3.x 2 dx = ∫ _13 x–2 dx
_3 + 1

= 3._x −2+1
+c
2

3_ +1
= _13 ._x
−2 + 1
+c
2
−1
= _13 ._
x
_5

= 3._
x
+c
2

−1
+c
5_
2
1
= −_ +c
= 3._2 x 2 + c
_5 3x
5
_
= _65 √x 5 + c
4
15. ∫ _5 dx 16. ∫ xπ dx
x
π+1
= ∫ 4x–5 dx =_
x
π+1
+c
−4
4x
=_−4
+c
1
= −_4 + c
x

17. ∫ π dx 18. ∫ − _14 t dt

= πx + c 2
= − _14 ._
t
2
+c

= − _18 t2 + c

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 238 N4 Mathematics – Lecturer Guide

_
19. ∫ a dx 20. ∫ √5x dx
= ∫ ax0 dx _ _1
= ∫ √5 x 2 dx
= ax + c _ _3

= √ 5 ._
x
+c
2

3
_
2
_2 _
_ 3
= √5 . 3 √x + c
_ _
2√5
=_3
√x 3 + c

Activity 6.2 SB page 375

1. ∫ _7x dx 2. ∫ 2x–1 dx
= 7 ln x + c = 2 ln x + c

3. ∫ _1r dr 4. 1
∫_
3x
dx
= ln r + c = _13 ln x + c

5. 4∫ _
dx
4x
6. ∫ _a2 da

= 4._14 ln x + c = 2 ln a + c

= ln x + c

7. ∫ − _3x dx 8. ∫_
dx
x
= – 3 ln x + c = ln x + c

9. ∫ _xk dx 10. ∫ – 3x–1 dx

= k ln x + c = – 3 ln x + c

Activity 6.3 SB page 378

1. ∫ 5x dx 2. ∫ 43x dx
x 3x
5 4
=_
ln 5
+c =_
3 ln 4
+c

3. ∫ 7–x dx 4. ∫ 4(2)3x dx
−x 3x
7 4.2
=_
− ln 7
+c =_
3 ln 2
+c
1
= –_
7 x ln 7
+c

∫(_12 ) da
a 1
5. 6. ∫_
10 −x
dx
a

(2)
1_
= ∫ 10x dx
= ___ +c
ln _1 10 x
2
=_
ln 10
+c

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 Module 6 • Integral calculus 239

7. ∫ 5.45x dx 8. ∫ πx dx
5x x
5.4 π
=_
5 ln 4
+c =_
ln π
+c
x
4
=_
ln 4
+c

9. ∫ 8.3–4x dx 10. 3∫ 2–5x dx

−4x
8.3
=_
−5x

− 4 ln 3
+c =_3.2
+c
− 5 ln 2
−4x
− 2.3
=_ ln 3
+c 3
=_
5x +c
− 5.2 ln 2
−2
=_ +c
3 4x ln 3

11. ∫ (1_5) dx
2x
12. ∫ 24t dt
4t
2
(5)
1 2x
_ =_ +c
= _ +c 4 ln 2
2 ln(_1)
5

13. ∫ e −x dx 14. ∫ 2e 7x dx
−x
e
=_
−1
+c = 2∫ e 7x dx
x
1
= −_
ex
+c = 2∫(e 7) dx
7x
e
= 2._
7 lne
+c
7x
e
= 2._
7.1
+c

= 2_7 e 7x + c

15. ∫ e x+ln 3 dx

= ∫ e x.e ln 3 dx

= e ln 3 ∫ e x dx

= 3∫ e x dx

= 3e x + c

Activity 6.4 SB page 381

1. 1.1 ∫ sin 5x dx 1.2 ∫–2 sin 2_x dx

cos 5x 2 cos _x
= −_ 5
+c =_ 2
+c
1_
2

= 4 cos _2x + c

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 240 N4 Mathematics – Lecturer Guide

1.3 ∫ 3 cos 3x dx 1.4 ∫ 4 cos 2θ dθ

3 sin 3x
=_ 3
+c = 4_
sin 2θ
2
+c

= sin 3x + c = 2 sin 2θ + c

1.5 ∫ 3 sin ax
_ dx
4
1.6 ∫–2 cos _21 x dx
ax
3 cos _ 2 sin 1_ x
= −_
a_
4
+c = −_
1_
2
+c
4 2
12
= −_
a
ax
cos _
4
+c = − 4 sin 1_2 x + c
1
1.7 ∫ _
sec 4x
dx 1.8 ∫ 2 sin x cos x dx

= ∫ cos 4x dx = ∫ sin 2x dx
sin 4x cos 2x
=_ 4
+c = −_ 2
+c
4
1.9 −∫ _
cosec x
dx

= −∫ 4 sin x dx

= 4 cos x + c
_________________
2. 2.1 ∫ √(1 + cos x)(1 – cos x) dx Differentiate to check:
_
= ∫ √1 – cos 2x dx _
d
(– cos x + c)
dx

= ∫ sin x dx = –_
d
dx
cos x + _
d
dx
c

= – cos x + c = – (– sin x) + 0
= sin x
2.2 ∫ (cos 2 _2x – sin 2 _2x ) dx _
d
dx
(sin x + c)
= ∫ cos x dx =_
d
dx
sin x + _
d
dx
c
= sin x + c = cos x + 0
= cos x
sec x sin 2x _
2.3 ∫ ______________ 2 dx d
dx
(– cos x + c)
(1 – cos 2x)(1 + cot x)
1
_ .2 sin x cos x
= ∫ ___________
cos x
2 2 dx = –_
d
dx
cos x + _
d
dx
c
2sin x cosec x

2 sin x
= ∫_ 2 _ 1 dx = – (– sin x) + 0
2sin x.
sin 2x

= ∫ sin x dx = sin x
= – cos x + c

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 Module 6 • Integral calculus 241

2.4 ∫ (_ 1 + tan x )
cos x + sin x
dx =_
d
dx
(sin x + c)

( )
cos x + sin x
=∫ _ sin x
_
dx =_
d
dx
sin x + _
d
dx
c
1 + cos x

( )
cos x + sin x
=∫ _cos x + sin x
_
dx = cos x + 0
cos x

= ∫ (_
cos x + sin x _
1 x + sin x )
× cos cos x
dx = cos x

= ∫ cos x dx
= sin x + c

Activity 6.5 SB page 389

dx ( 28
.e + c)
4x
1 4x
1. 1.1 ∫ _
e
7
dx Verify: _
d _

1_
= _71 ∫ e 4x dx =_ d 4x _ d
28 dx
.e + dx c
1 4x
u = 4x =_
28
.e .4 + 0

du = 4 dx = _17 .e 4x
_1 du = dx
4

∴ _71 ∫ e 4x dx

= _17 ∫ e u._14 du
1
=_
28
∫ e u du
1 u
=_
28
.e + c
1 4x
=_
28
.e + c

(2.e 2 + c)
1
_ _x
1.2 ∫ e 2x dx –1
Verify: _
d
dx
_x
= 2_
d 2 _ d
_x
= ∫ e 2 dx dx
e + dx c

= 2.e 2._12 + 0
_x
u = _2x

du = _12 dx
_x
= e2
2du = dx
_x
∴ ∫e2

= ∫ e u.2 du
= 2∫ e u du
= 2eu + c
_x
= 2e 2 + c

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 242 N4 Mathematics – Lecturer Guide

dx ( 36
1.3 ∫ _ 1
– 9x + ln 4
dx Verify: _
d _
.e + c)
1 9x
e
1_
= ∫ e 9x –ln 4 dx =_ d 9x _
36 dx
d
.e + dx c
1 9x
= ∫ e 9x.e –ln 4 dx =_
36
.e .9 + 0
= ∫ e 9x.e ln 4 dx = _14 .e 9x
–1

_1
= ∫ e 9x.e ln 4 dx
= ∫ e 9x._14 dx
= _14 ∫ e 9x dx
u = 9x
du = 9 dx
_1 du = dx
9

∴ _14 ∫ e 9x dx

= _14 ∫ e u._19 du
1
=_
36
∫ e u du
1 u
=_
36
.e + c
1 9x
=_
36
.e + c

dx ( 3 ln 5
. 5 + c)
3x
2_
1.4 ∫ 2.5 3x dx Verify: _
d _

2 _
= 2∫ 5 3x dx =_ d 3x _
3 ln 5 dx
d
5 + dx c
2
u = 3x =_
3 ln 5
.5 3x ln 5.3 + 0

du = 3 dx = 2.53x
_1 du = dx
3

∴ 2∫ 5 3x dx

= 2∫ 5 u._13 du

= _23 ∫ 5 u du
u
= _23 ._5
ln 5
+c
3x
= _23 ._5
ln 5
+c

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 Module 6 • Integral calculus 243

dx ( 80
1.5 _18 ∫ sin 10x dx Verify: _
d _
– 1 .cos 10x + c)
1_
u = 10x = –_ d
80 dx
cos 10x + _
d
dx
c
1
du = 10 dx = –_
80
.– sin 10x.10 + 0
1
_
10
du = dx = _18 sin 10x

∴ _18 ∫ sin 10x dx

= _81 ∫ sin u._

1
10
du
1
=_
80
∫ sin u du
1
=_
80
.– cos u + c
1
= –_
80
.cos 10x + c

dx ( 7
7x
1.6 2∫ cos _
9
dx Verify: _
d _ 18 7x
.sin _
9
+ c)
7x 18 _ 7x _
u=_
9
=_ d
7 dx
sin _
9
d
+ dx c

du = _79 dx 18
=_7
7x _
.cos _ .7 + 0
9 9
_9 du = dx 7x
= 2 cos _
7 9
7x
∴ 2∫ cos _
9
dx

= 2∫ cos u._97 du
18
=_
7
∫ cos u du
18
=_
7
.sin u + c
18 7x
=_
7
.sin _
9
+c

dx ( 3
1.7 ∫ cos (_π2 + 3x) dx Verify: _
d _ 1
cos 3x + c)
1_
= – ∫ sin 3x dx =_ d
3 dx
cos 3x + _
d
dx
c

u = 3x = _13 .– sin 3x.3 + 0

du = 3 dx = – sin 3x
_1 du = dx
3

∴ – ∫ sin 3x dx

= – ∫ sin u._13 du

= – _13 ∫ sin u du

= – _13 .– cos u + c

= _13 cos 3x + c

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 244 N4 Mathematics – Lecturer Guide

dx ( 2
1.8 ∫ (cos 4x – sin 4x) dx Verify: _
d _ 1
sin 2x + c)

= ∫ [(cos 2x + sin 2x)(cos 2x – sin 2x)] dx

1_
=_ d
2 dx
sin 2x + _
d
dx
c

= ∫ [1.cos 2x] dx = _12 .cos 2x.2 + 0

= ∫ cos 2x dx = cos 2x
u = 2x
du = 2 dx
_1 du = dx
2

∴ ∫ cos 2x dx

= ∫ cos u._12 du

= _21 ∫ cos u du

= _21 .sin u + c

= _12 sin 2x + c

dx ( 3
2. 8
2.1 ∫ _
3x dx Verify: _
d _
– 8 .e –3x + c)
e

= 8∫ e –3x dx = – _83 _
d – 3x _
dx
d
e + dx c

u = – 3x = – _83 .e –3x.– 3 + 0

du = – 3 dx = 8.e –3x
– _13 du = dx

∴ 8∫ e –3x dx

= 8∫ e u.– _13 du

= – _83 ∫ e u du

= – _83 .e u + c

= – _83 .e –3x + c
_
dx ( 6
– _5 .e – 5 + c)
5 6x
_
2.2 ∫ √ e –6x dx Verify: _
d

_1
= ∫ (e –6x) 5 dx = – _56 _
6x
_

dx
e +_
d –5 d
dx
c

= – _56 .e – 5 .– _65 + 0
6x
_ 6x
_
= ∫ e – 5 dx
6x 6x
_
u = –_
5
= e–5

du = – _65 dx

– _56 du = dx

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 Module 6 • Integral calculus 245

6x
_
∴ ∫ e – 5 dx

= ∫ e u.– _56 du

= – _56 ∫ e u.du

= – _56 .e u + c

= – _56 .e – 5 + c
6x
_

dx ( 4 ln 3
+ c)
– 4x
2.3 ∫ (3 x) –4 dx Verify: _
d
– _1 ._
3

1 _
= ∫ 3 –4x dx = –_ d – 4x _
4 ln 3 dx
d
3 + dx c
1
u = – 4x = –_
4 ln 3
.3 –4x ln 3.– 4 + 0

du = – 4 dx = 3–4x
– _14 du = dx
= ∫ 3 u.– _14 du

∴ ∫ 3 –4x dx

= – _14 ∫ 3 u du
u
= – _14 ._3
ln 3
+c
– 4x
= – _14 ._
3
ln 3
+c
_
dx ( 12 ln 6
. 6 + c)
2x
1 _
2.4 ∫ √6 4x –2 dx Verify: _
d _

_1
= ∫ (6 4x –2) 2 dx
1 _
=_ d 2x _
12 ln 6 dx
d
6 + dx c
1
= ∫ 6 2x –1 dx =_
12 ln 6
.6 2x(ln 6)(2) + 0

= ∫ 6 2x.6 –1 dx = _16 .6 2x

= ∫ 6 2x._16 dx

= _61 ∫ 6 2x dx
u = 2x
du = 2 dx
_1 du = dx
2

∴ _16 ∫ 6 2x dx

= _16 ∫ 6 u._12 du
1
=_
12
∫ 6 u du
u
1 _
=_ .6 +c
12 ln 6
2x
1 _
=_ .6 +c
12 ln 6

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 246 N4 Mathematics – Lecturer Guide

dx ( 5
2.5 ∫ sin(_π2 – _
2)
5x
dx Verify: _
d _ 2 5x
sin _
2
+ c)
5x
= ∫ cos _
2
dx = _25 _
d
dx
5x _
sin _
2
d
+ dx c
5x
u=_
2
= _25 .cos _
5x _
.5 + 0
2 2

du = _52 dx 5x
= cos _
2

_2 du = dx
5
5x
∴ ∫ cos _
2
dx

= ∫ cos u._25 du

= _52 ∫ cos u du

= _52 .sin u + c

= _25 sin _
5x
2
+c

dx ( 8
sin 2x
2.6 ∫ _
sec 2x
dx Verify: _
d _
– 1 cos 4x + c)
sin 2x
= ∫__ 1 dx = – _18 _
d
dx
cos 4x + _
d
dx
c
cos 2x

= ∫ sin 2x.cos 2x dx = – _18 .– sin 4x.4 + 0

= ∫ _12 sin 4x dx = _12 sin 4x

= _21 ∫ sin 4x dx
u = 4x
du = 4 dx
_1 du = dx
4

∴ _21 ∫ sin 4x dx

= _12 ∫ sin u._14 du

= _18 ∫ sin u du

= _18 .– cos u + c

= – _18 cos 4x + c

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 Module 6 • Integral calculus 247

dx ( 27
2.7 _13 ∫ (cos 7x cos 2x – sin 7x sin 2x) dx Verify: _
d _ 1
sin 9x + c)
1_
= _13 ∫ cos(7x + 2x) dx =_ sin 9x + _
d d
27 dx dx
c
1
= _13 ∫ cos 9x dx =_
27
.cos 9x.9 + 0

u = 9x = _13 cos 9x
du = 9 dx
_1 du = dx
9

∴ _13 ∫ cos 9x dx

= _13 ∫ cos u._19 du

1
=_
27
∫ cos u du
1
=_
27
.sin u + c
1
=_
27
sin 9x + c

dx ( 8
2.8 ∫ 4 sin 2x cos 2x cos 4x dx Verify: _
d _
– 1 cos 8x + c)

= ∫ 2(2 sin 2x cos 2x) cos 4x dx = – _18 _

d
dx
cos 8x + _
d
dx
c

= ∫ 2 sin 4x cos 4x dx = – _18 .– sin 8x.8 + 0

= ∫ sin 8x dx = sin 8x
u = 8x
du = 8 dx
_1 du = dx
8

∴ ∫ sin 8x dx

= ∫ sin u._18 du

= _18 ∫ sin u du

= _18 .– cos u + c

= – _18 cos 8x + c

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 248 N4 Mathematics – Lecturer Guide

Activity 6.6 SB page 394

1
1. 1.1 ∫ (x4 + 3x) dx 1.2 ∫ (x5 – _3 + 5) dx
x
5 2
3x
=_
x
+_
5
+c 2 = ∫ (x – x–3 + 5) dx
5

6 −2
=_
x
6
–_
x
−2
+ 5x + c

= _16 x6 + _
1
2 + 5x + c
2x

1.3 ∫ (x2 + y2) dx 1.4 ∫ (x2 + y2) dy

3
=_
x
3
+ y2x + c = x2y + _ +c
y3
3
or _13 x3 + y2x + c or x2y + _13 y3 + c
_
1.5 ∫ (x2 + y2) da 1.6 ∫(_2x – 2√x + πx) dx
= x2a + y2a + c
= ∫(_2x – 2x 2 + πx) dx
_1

_3 2
2x πx
= 2 ln x – _ +_ +c
2

3_ 2
2
_
4
= 2 ln x – _3 √x 3 + _12 πx2 + c

1.7 ∫ (8 – _4x – _
1
3x
– 23x) dx
12 − 1
= ∫ (8 – _ 3x
– 23x) dx
11
= ∫ (8 – _
3x
– 23x) dx
3x 3x
11
= 8x – _
3
ln x – _2
3 ln 2
+c or = 8x – 4 ln x – _13 ln x – _2
3 ln 2
+c
3x
11 2
= 8x – _3
ln 2 – _
3 ln 2
+c
_
1.8 ∫ (3.4–2x – x–1 – _
x 4)
3 1
dx 1.9 ∫ (√x 3 + 4x–1 – _ 3 – a) dx
2x
– _12 x–3
_3
∫ (3.4 – 2x –1 –4
– x – 3x ) dx = ∫ (x + 4x 2
–1
– a) dx
−2x −3
3. 4 3x
=_ – ln x – _
_5 −2

− 2 ln 4 −3
+c =_
x
+ 4 ln x – _12 ._
2 x
– ax + c
5_ −2
2
3 1
=_ – ln x + _3 + c
_
= _25 √x 5 + 4 ln x + _
1
2x
− 2. 4 ln 4 x
2 – ax + c
4x

1.10 ∫(2x–2 + _
1
– 3 + _3x ) dx
x2

= ∫(3x–2 – 3 + _3x ) dx or ∫(2x–2 + x–2 – 3 + _3x ) dx

−1 −1 −1
2x
3x
=_−1
– 3x + 3 ln x + c =_−1
+_
x
−1
– 3x + 3 ln x + c

= − _3x – 3x + 3 ln x + c = – _2x – _1x – 3x + 3 ln x + c

= – _3x – 3x + 3 ln x + c

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 Module 6 • Integral calculus 249

1.12 ∫ [_ + 6x – 3.24x + (_13 ) ] dx

2_ –x
1.11 ∫ (2x – π2x + xπ – c) dx
√x
x 2x π+1
2 π
=_ –_ +_
x
= ∫(2x −2 + 6x – 3.24x + (_13 ) ) dx
ln 2 2 ln π π + 1
– cx + c1 _1 –x

−x

(3)
_1 2 4x
1
_
2x 2 _
=_ + 6x – _
3. 2
+_ +c
1_
2
2 4 ln 2 − ln (_1)
3
_ 3. 2 1 4x
= 4√x + 3x2 – _ –_ +c
(3 ) ln (3)
x
4 ln 2 1_ 1_

2.2 ∫(bx + _
bx )
1
2
5x + 2x − 1
2. 2.1 ∫ _ x
dx dx

= ∫(5x + 2 – _1x ) dx =_
bx 2
+ _b1 ln x + c
2
2
5x
=_2
+ 2x – ln x + c

5
t + 3t − 1
2.3 ∫ (x2 + 3)(x – 4) dx 2.4 ∫ _ 5 dt
t
= ∫ (x – 4x +3x – 12) dx
3 2
= ∫ (1 + 3t–4 – t –5) dt
4 3 2
4x 3x
=_
x
–_ +_
−3 −4

4 3 2
– 12x + c 3t
=t+_ –_
t
+c
−3 −4
or _14 x4 – _43 x3 + _32 x2 – 12x + c = t – _13 + _
1
4 + c
t 4t
_
2.5 ∫ (3 – √x )2 dx 2.6 ∫ x(x2 – _3x ) dx
_ _
= ∫ (3 – √x )(3 – √x ) dx = ∫ (x3 – 3) dx
_ _
= ∫ (9 – 3√x – 3√x + x) dx 4

_1
=_
x
4
– 3x + c
= ∫ (9 – 6x 2 + x) dx
_3 2
or _14 x4 – 3x + c
6x
= 9x – _ +_
x
+c
2

3_ 2
2
_
= 9x – 4√x 3 + _12 x2 + c

3.1 ∫(x_ ) dx
2
− 9x +8
3. _ 3.2 ∫(e −x − e x)2 dx
√x

= ∫(_ ) dx
= ∫(e −x − e x)(e −x − e x) dx
2
− 9x_ + _
x_ _ 8_
√x √x √x
= ∫(e −2x − 2e 0 + e 2x) dx
= ∫(x − 9x 2 + 8x −2) dx
_3 _1 1_
2

_3 1_
−1_2
= ∫(e −2x − 2 + e 2x) dx
= ∫ x dx − 9∫ x dx + 8∫ x dx
2 2
−2x 2x
e e
_3+1 _1+1 −1_2+1
=_
−2
− 2x + _2
+c
x x x
= __ − 9._ + 8._
2 2
+c 1
3
+11_ 1_
+1 −2 + 1 = −_ _1 2x + c
2 2 2x − 2x + 2 e
2e
2_ 2
_ _ _
= 5
x √x − 6x √x + 16√x + c

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 250 N4 Mathematics – Lecturer Guide

3.3 ∫(2e −4x + 2a 2x + πx) dx 3.4 ∫(3 sin 2x + 4.3 2x − 6_x) dx

−4x 2x 2 2x
= 2_
e
−4
2a
+ 2_ln a
+ π_
x
2
+c = − 3_
cos 2x _
2
+ 24.3
ln 3
− 6 ln x + c
2x 2x
1
= −_4x +
_a
+ 1_ π x 2 + c
ln a 2
= – 23_ cos 2x + _
2.3
ln 3
− 6 ln x + c
2e

3.5 ∫(x π − π x − _nx + n_x) dx 3.6 ∫(_ ) dx

2
1 − x sin 2x
x2

= ∫(_ ) dx
2
π+1 n1+1 1 x sin 2x
_
2 −
x
π
x
=_ −_ − _1._
x
π + 1 ln π n n + 1
− n ln x + c x 2
x
1

= ∫(x − sin 2x) dx

−2

−1
x cos 2x
=_
−1
+_ 2
+c

= − 1_x + 1_2 cos 2x + c

3.7 ∫(3 cos 1_2 x + b sin ax) dx 3.8 ∫(sin − 3.10 −3x) dx
2 2
_ x + cos x
sec x

=_
3 sin 1_ x
2 cos ax
− b_ +c = ∫(sec
_ 1
x
− 3.10 −3x) dx
1_ a
2
= ∫(cos x − 3.10 −3x) dx
= 6 sin 1_2 x − _ab cos ax + c
−3x
3.10
= sin x − _
− 3 ln 10
+c
−3x
10
= sin x + _
ln 10
+c
1
= sin x + (_) 3x
+c
ln 10 10

4.1 ∫ (4 sin x cos x – _6x – 2e x) dx

= 2∫ 2 sin x cos x dx – ∫ _6x_ _ dx – 2∫ e x dx

= 2∫ sin 2x dx – ∫__x6 dx – 2∫ e x dx
u = 2x
du = 2 dx
_1 .du = dx
2

= 2∫ sin u._12 du – ∫ _6x dx – 2∫ e x dx

= ∫ sin u du – ∫ _6x dx – 2∫ e x dx
= – cos u – 6 ln x – 2.ex + c
= – cos 2x – 6 ln x – 2ex + c

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 Module 6 • Integral calculus 251

4.2 ∫ (e –4x – 9 sin 2x) dx

= ∫ [e –4x – 9(_12 – _12 cos 2x)] dx

= ∫ [e –4x – _92 + _92 cos 2x] dx

= ∫ e –4x dx – ∫ _92 dx + _92 ∫ cos 2x dx

↓ ↓
u = – 4x v = 2x
du = – 4 dx dv = 2 dx
– _14 du = dx _1 dv = dx
2

= ∫ e u – _14 du – ∫ _92 dx + _92 ∫ cos v _21 dv

= – _14 ∫ e u du – ∫ _92 dx + _49 ∫ cos v dv

= – _14 e u – _29 x + _94 sin v + c

= – _14 e –4x – _92 x + _49 sin 2x + c

dx ( 4
Verify: _
d _
– 1 e –4x – _92 x + _94 sin 2x + c)

= – _14 _
dx
e – 92 _
d – 4x _ d
dx
x + _94 _
d
dx
sin 2x + _
d
dx
c

= – _14 .e –4x.– 4 – _92 .1 + _94 cos 2x.2 + 0

= e –4x – _92 + _92 cos 2x

4.3 ∫ (_2 sin x – 6 )

3
4 sin x – 108
dx

= ∫ [_
2(sin x – 3) ]
4(sin 3x – 27)
dx

= ∫ [___________________] dx
4(sin x – 3)(sin 2x + 3 sin x + 9)
2(sin x – 3)

= ∫ [2(sin 2x + 3 sin x + 9] dx

= ∫ [2(_21 – _12 cos 2x + 3 sin x + 9)] dx

= ∫ [2(– _12 cos 2x + 3 sin x + _

2 )]
19
dx

= ∫ [– cos 2x + 6 sin x + 19] dx

= – ∫ cos 2x dx + 6∫ sin x dx + ∫ 19 dx
↓
u = 2x
du = 2 dx
_1 du = dx
2

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 252 N4 Mathematics – Lecturer Guide

= – ∫ cos u._12 du + 6∫ sin x dx + ∫ 19 dx

= – _12 ∫ cos u du + 6∫ sin x dx + ∫ 19 dx

= – _12 .sin u + 6.– cos x + 19x + c

= – _12 sin 2x – 6 cos x + 19x + c

dx ( 2
Verify: _
d _
– 1 sin 2x – 6 cos x + 19x + c)

= – _12 _
d
dx
sin 2x – 6_
d
dx
cos x + 19_
d
dx
x+_
d
dx
c

= – _12 .cos 2x.2 – 6.– sin x + 19.1 + 0

= – cos 2x + 6 sin x + 19

Activity 6.7 SB page 400

1. 1.1 ∫ 20 2x dx 1.2 ∫ 31 (5x2 – 1) dx

= [_ = [_ – x] 31
2 ]
2 2 5x 3
2x
3
0

= (_ – (3)) – (_ – (1))
5 (3) 3 5 (1) 3
= [x2] 20 3 3
= (2)2 – (0)2
= (42) – (_23 )
=4
= 41_13
1.3 ∫ 21 (x3 + 2x2 – 5) dx

= [_ – 5x]
4 3 2
2x
x
4
+_3 1

= (_ – 5(2)) – (_ – 5(1))
2 4
2 (2)
1
3 4
2 (1)3
4
+_
3 4
+_
3

= (4 + _
16
3
– 10) – (_14 + _23 – 5)

= (– 6 + _3 ) ( 12
16 3+8
– _ – 5)

= – _23 + 4_1
12
– 8 + 49 _
=_ 12
= 41
12
5
= 3_
12
_
1.4 ∫ 42 (√x – _
1
+ 4x3) dx
x3
_1
= ∫ 42 (x 2 – x–3 + 4x3) dx

= [_
3_ – − 2 + 4 ]
_3 −2 4
4
x2 _ x 4x
_
2
2

= (_23 (4) 2 1
+ (4)4) – (_23 (2) 2 + _
1
)
_3 _3
+ _ 4
2 + (2)
2 (4) 2 2 (2)
= (261,365) – (18,011)
= 243,354

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 Module 6 • Integral calculus 253

_5
1.5 ∫ 53 (x −3) dx

= [_
− 2_ ]
−_32 5
x
3 3

= (− _32 (5) −3) – (− _32 (3) −3)

_2 _2

= (– 0,513) – (– 0,721)
= 0,208

1.6 ∫0,5 _1x dx

1

1
= [ln x]0,5
= (ln 1) – (ln 0,5)
= 0,693

1.7 ∫ 0 5–2x dx
_1
2

= [_ ]
__
−2x
5 2

− 2 ln 5
0

= (_
− 2 ln 5 ) ( − 2 ln 5 )
−2( 21_ ) −2(0)
5 5
– _

= (_
− 2 ln 5 ) ( − 2 ln 5 )
−1
5 1
– _

= (_
− 10 ln 5 ) ( 2 ln 5 )
1
+ _ 1
= – 0,062 + 0,311

= 0,249
1.8 ∫ 21 (3x –1 + 2x) dx
2
= [3 ln x + _
ln 2 ]
x
2
1

= (3 ln 2 + _
ln 2 ) ( ln 2 )
2
2 2
– 3 ln 1 + _

ln 2 ) ( ln 2 )
= (3 ln 2 + _4
– 0+_2

= 4,965
1.9 ∫−12(4x – 1)2 dx
= ∫−12(4x – 1)(4x – 1) dx
= ∫−12(16x2 – 8x + 1) dx

= [_ + x]
3 2 1
16 x 8x
3
–_2
−2

= (_ (1 ) – 4(12) + (1)) – (_
16 3
3
16
3
(– 23) – 4(– 22) + (– 2))

= (_
16
3
– 4 + 1) – (− _
128
3
– 16 – 2)

= 2,333 – (– 60,667)
= 62,9997
= 63

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 254 N4 Mathematics – Lecturer Guide

t 2 + 2t − 1
1.10 ∫ 32 _ 2 dt
t

= ∫ (1 + _2t – t –2) dt
3
2

= [t + 2 ln t – _
−1]
−1 3
t
2

= [t + 2 ln t + _1t ]
3

= (3 + 2 ln 3 + _13 ) – (2 + 2 ln 2 + _12 )

= 5,531 – 3,886
= 1,645

1.11 ∫ 30 (_3 – _13 ) dy

y

= [_13 ._ – _13 y]
3
y2
2
0

= [_ – _13 y]
3
y2
6
0

= (_ – _13 (3)) – (_ – _13 (0))

2 2
3 0
6 6

= (_96 – 1)

= 0,5
1.12 ∫ 32 (_x )
x+2 2
dx
(x + 2)(x + 2)
= ∫ 32 _2 dx
x

= ∫ 32 (_ ) dx
2
x + 4x + 4
2
x

= ∫ 32 1 + 4x–1 + 4x–2 dx

= [x + 4 ln x + 4._
−1 ]
−1 3
x
2

= [x + 4 ln x – _4x ]
3

= (3 + 4 ln 3 – _43 ) – (2 + 4 ln 2 – _42 )

= 6,061 – 2,773
= 3,288

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 Module 6 • Integral calculus 255

20
w = ∫ 0,2 _
0,5
2. V
dV
0,5
= [20 ln V]0,2

= (20 ln 0,5) – (20 ln 0,2)

= – 13,863 – (32,189)
= 18,326
_
23 2 5 x –1
3. 3.1 ∫
–1 √ _x dx 3.2 ∫ _ _ dx
3 √17

3
_
2 √2 1 51
=∫ _3 _ dx =_
_ ∫ _ dx
x
–1 √ x √17 3

_ 2 _1
= √2 ∫ x – 3 dx
3
_ [ln | x|]
1 5
–1 =_ 3
√17

[ – _13 + 1 ]
_ x –_13 + 1 2
_ [(ln 5) – (ln 3)]
1
3
= √2 _ =_
√17
–1
= 0,124
_ 3 _2 2
√2 [_2 x 3]
3
=
–1
_
= √2 [(_32 (2) 3) – (_32 (– 1) 3)]
3 _2 _2

= 1,110

–1
π
_ _____________________
3.3 ∫ e x + ln 3 dx 3.4 ∫ _π √(1 + cos x)(1 – cos x) dx
4

–3 12

π
_ _
–1 x
= ∫ e .e ln 3
= ∫ _π √1 – cos 2x dx
4
dx
–3 12

π
_
–1
= e ln 3∫ e x dx = ∫ _π sin x dx
4

–3 12

–1
= 3∫ e x dx = [– cos x] _4π
π
_
–3
12

= [(– cos ( 4 )) –
_π
(– cos (12))]
π
_
= 3[e x] –3
–1

= 3[(e (–1)) – (e (–3))] = 0,259

= 0,954

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 256 N4 Mathematics – Lecturer Guide

_π

3.5 ∫ _π cos 3x dx
3

12
π
_

= – ∫ _π sin 3x dx
3

12

u = 3x
du = 3 dx
_1 du = dx
3
π
_

= – ∫ _π sin u._13 du
3

12
π
_

= – _31 ∫ _π sin u du
3

12
π
_
= – _31 [– cos u] _3π
12

π
_
= – _13 [ – cos 3x] _3π
12

= – _13 [(– cos (3. 3 )) –

_π
(– cos (3.12))]
π
_

= – 0,569

3π
_

3.6 _13 ∫ _π (cos 7x.cos 2x – sin 7x.sin 2x) dx

4

2
3π
_

= _13 ∫ _π cos (7x + 2x) dx

4

3π
_

= _13 ∫ _π cos 9x dx
4

u = 9x
du = 9 dx
_1 du = dx
9
3π
_

= _13 ∫ _π cos u._19 du

4

3π
_
1
=_ ∫ cos u du
4

27 _π
2
3π
_
1
=_
27
[sin u] _π4
2

3π
_
1
=_
27
[ sin 9x] _π4
2

27 [(
1
=_ sin (9._4 )) (
3π
– sin (9._π2 ))]

= – 0,011

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 Module 6 • Integral calculus 257

3.7 ∫ (_
x)
1 42 2_
3x
–_ 2 +
_ dx
1 x √

= ∫ (_13 ._1x – 4.x –2 + 2.x – 2) dx

2 _1

[ – 2 + 1]
–2 + 1 – _1 + 1
2
= _13 .ln | x | – 4._
x
–2 + 1
+ 2._
x 2
_1
1

_
= [_13 ln | x | + _4x + 4√x ]
2

1
_ _
= [(_13 ln (2) + _4
(2)
+ 4√(2) ) – (_13 ln (1) + _4
(1)
+ 4√(1) )]

= – 0,112

4π

3.8 ∫ _π (_2 sin x – 6 )

_ 3
4 sin x – 108
5
dx
5

= ∫ _π (___________________) dx
4π
_
5 4[(sin x – 3)(sin 2x + 3 sin x + 9)]
5
2(sin x – 3)
4π
_

= ∫ _π [2(sin 2x + 3 sin x + 9)] dx

5

4π

= ∫ _π [2(_12 – _12 cos 2x + 3 sin x + 9)] dx

_
5

4π

= ∫ _π [2(– _12 cos 2x + 3 sin x + _2 )]

_
5 19
dx
5

4π
_

= ∫ _π [– cos 2x + 6 sin x + 19] dx

5

= [– _ + 6.– cos x + 19x] _π

4π
_
sin 2x 5

2
5

= [(– _12 sin (2._5)

4π
– 6 cos (_5)
4π
+ 19(_5 )) ( 2
4π
– – _1 sin (2._π5 ) – 6 cos (_π5 ) + 19(_π5 ))]

= 46,473

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 258 N4 Mathematics – Lecturer Guide

Activity 6.8 SB page 415

1. ∆A = y∆x 2. ∆A = y∆x

A = ∫ 20 y dx A = ∫ 21 x3 dx

= [_
4]
2
= ∫ 20 (– x2 + 2x) dx
4
x
1
= [− _ 2 ]
3 2 2
2x
x
+_
= (_
4) (4)
4 4
2 1
3
0 – _

= [− _13 x3 + x2]
2
= 3,75 units2
0

= (− _13 (2)3 + (2)2) – (− _13 (0)3 + (0))

= − _83 + 4
= 1,333 units2

3. 3.1 y = 2x + 2 y

x-intercept: y = 0 6 y = 2x + 2
0 = 2x + 2
(x; y)
2x = – 2
∴ x = –1 2

y-intercept: x = 0 –1 2
x
∆x
∴y=2

3.2 ∆A = y∆x

A = ∫ 20 y dx

= ∫ 20 (2x + 2) dx

= [_ + 2x]
2 2
2x
2
0
2 2
= [x + 2x] 0

= ((2)2 + 2(2)) – (0 + 2(0))

= 8 units2

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 Module 6 • Integral calculus 259

4. y = – x2 + 2x + 3 y
x-intercept: y = 0
(1; 4)
0 = – x2 + 2x + 3 4
y = – x2 + 2x + 3
2 3
0 = x – 2x – 3
0 = (x – 3)(x + 1)
∴ x = 3; x = – 1
x
–1 ∆x 3
y-intercept: x = 0
∴y=3
−b
TP: x = _
2a
−2
=_
2(−1)

=1
y = – (1)2 + 2(1) + 3
=4
∴ TP: (1; 4)
∆A = y∆x
A = ∫−31 y dx

= ∫−31(– x2 + 2x + 3) dx

= [− _ + 3x]
3 2 3
2x
x
3
+_2
−1

= (− _13 (3)3 + (3)2 + 3(3)) – (− _13 (– 1)3 + (– 1)2 + 3(– 1))

= (9) – (– 1_23 )

= 10_23 units2 or 10,667 units2

5. ∆A = y∆x

A = ∫ 10 (x2 – x) dx

= [_ 2]
3 2 1
x
3
–_
x
0

= (_ 2 )
3
(1) (1) 2
3
–_ – (0)

= |_13 – _12 |

= 0,167 units2

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 260 N4 Mathematics – Lecturer Guide

6. 6.1
y

xy = 3 3

1
∆x x
–3 –2 –1 1 2 3
–1

–2
(x; y)
–3

6.2 ∆A = y∆x
A = – ∫ 31 y dx

= – ∫ 31 (− _3x ) dx

= – [– 3 ln x] 31
= – [(– 3 ln 3) – (– 3 ln 1)]
= – [(– 3,296) – (0)]
= 3,296 units2

7. 7.1 f (x) = x3 – 6x2 + 8x y

y-intercept: x = 0 (0,845; 3,079)
3
∴y=0 (x; y)
y = x3 – 6x2 + 8x
2
x-intercept: y = 0
x3 – 6x2 + 8x = 0 1
x(x2 – 6x + 8) = 0 x
–1 ∆x 1 2 3 4 5
x(x – 4)(x – 2) = 0
–1
x = 0 or x = 4 or x = 2
–2
Turning points
f ʹ(x) = 3x2 – 12x + 8 = 0 –3
_______________
√
− (−12) ± (−12) 2 − 4(3)(8)
x = ___________________ –4
2(3)
_
12 ± √48
=_ 6

= 3,155 or 0,845
f (3,155) = – 3,079
f (0,8450) = 3,079

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 Module 6 • Integral calculus 261

7.2 ∆A = y∆x
A1 = ∫ 20 y dx

= ∫ 20 (x3 – 6x2 + 8x) dx

= [_ – 2x3 + 4x2]
4 2
x
4
0

= (_ – 2(2)3 + 4(2)2) – (0)

4
(2)
4

= 4 units2
A2 = – ∫ 42 y dx

= – ∫ 42 (x3 – 6x2 + 8x) dx

= – [_ – 2x3 + 4x2] 42
4
x
4

= – [(_41 (4)4 – 2(4)3 + 4(4)2) – (_14 (2)4 – 2(2)3 + 4(2)2)]

= – [(0) – (4)]
= 4 units2
∴ Total area = A1 + A2
= (4 + 4) units2
= 8 units2
Alternative:
A1 = 4 units2 × 2
Total A = 8 units2

8. 8.1 f (x) = (x + 2)2 – 1 y = (x + 2)2 – 1 y

x-intercept: y = 0
0 = (x + 2)2 – 1
3
0 = x2 + 4x + 4 – 1
0 = x2 + 4x + 3
0 = (x + 3)(x + 1)
∆x x
∴ x = – 3; x = – 1 –3 –1
y-intercept: x = 0 (x; y) –1
y=3
−b
TP: x = _
2a
−4
=_
2

= –2
y = (– 2)2 + 4(– 2) + 3
=4–8+3
= –1
∴ TP: (– 2; – 1)

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 262 N4 Mathematics – Lecturer Guide

8.2 ∆A = y∆x
A = – ∫−−13(x2 + 4x + 3) dx

= – [_ + 2x2 + 3x]
3 −1
x
3
−3

= – [(_13 (– 1)3 + 2(– 1)2 + 3(– 1)) – (_13 (– 3)3 + 2(– 3)2 + 3(– 3))]

= – [(− _13 + 2 – 3) – (– 9 + 18 – 9)]

= – [(– 1,333) – (0)]

= – [– 1,333]
= 1,333 units2
_π

9.1 Area = ∫ 2 cos x dx

4
9.
0
_π

= 2∫ cos x dx
4

0
π
_
= 2[sin x] 04

= 2[sin(π_4) − sin 0]

= 2[_
1_
− 0]
√2

= 1,414 units2
dy
9.2 Find __
dx
to get the TP.
d
__
dx
= − 2 sin x
y = 2 cos x
dy
For turning points, let __
dx
= 0:
− 2 sin x = 0
x = sin −10
x = 0° or x = 180°
y-value when x = 0°:
y = 2 cos 0°
=2
y-value when x = 180°:
y = 2 cos 180°
= −2
∴ Turning points: (0° ; 2) and (180° ; − 2)

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 Module 6 • Integral calculus 263

y
d
_
9.3 = − 2 cos x
dx 2
= − 2 cos 0°
= −2
2
d
_y
or: = − 2 cos x
dx 2

= − 2 cos 180°
=2
Maximum: (0° ; 2)
Minimum: (180° ; − 2)

10. 10.1 y = 3 sin x for 0° ≤ x ≤ 360°

y

3 y = 3 sin x

x
90° 180° 270° 360°

–3

∆ A = y∆ x
180°
A = 2∫ 0° 3 sin x dx
180°
= 2[− 3 cos x] 0°

= − 6[cos 180° − cos 0°]

= − 6[ − 1 − 1]
= 12 units2

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 264 N4 Mathematics – Lecturer Guide

10.2 y = cos 2x, the x-axis and the ordinates x = π_2 and 2π
y

1 y = cos 2x

x
π
__ 3π
__ 3π
__
2 4 π 2
2π

–1

∆ A = y∆ x
3π
_

A = − ∫ π_ cos 2x dx
4

− [sin2 ] π_
3π
_
_ 2x 4
=
2

= − 1_2[sin 2(3π
4) (2)]
_ − sin 2 π_

= − 1_2[sin_
3π
2
− sin π] or = − 1_2[sin 270° − sin 180°]

= − 1_2[− 1 − 0]

= 1_2

∴ A = 1_2 × 6

A T = 3 units2

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 Module 6 • Integral calculus 265

11. 11.1 y = _13 x − 3

x = – 10 y
x-intercepts, y = 0:
y = _13 x − 3 10
0 = _1 x − 3
3
− _1 x = −3 _x
1 –3
3 ∆x y= 3
x
x=9 (9; 0)
∴ (9; 0) (0; – 3)
(x; y)
y-intercept, x = 0: – 10
y = _13 x − 3
= _13 (0) − 3
y = −3
∴ (0; −3)
∆A = |y∆x|

|
A = ∫ y dx
0
−10
| • Area below the x-axis

= ∫ (_31 x − 3) dx
| |
0
−10

|
= [_16 x 2 − 3x] |
0

−10

= |[_16 (0) 2 − 3(0)] − [_16 (−10) 2 − 3(−10)]|

A = 46_23 units2

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 266 N4 Mathematics – Lecturer Guide

11.2 xy = −12
x = –6 x = –2 y
12
∴ y = −_ x

x y
(x; y) 5
– 7,000 1,714
– 6,000 2,000 x
∆x 5
– 5,000 2,400
– 4,000 3,000 xy = – 12
–5
– 3,000 4,000
– 2,000 6,000
– 1,000 12,000
0,000 ERROR
1,000 – 12,000
2,000 – 6,000
3,000 – 4,000
4,000 – 3,000
5,000 – 2,400
6,000 – 2,000
7,000 – 1,714

∆A = y∆x

A = ∫ −6 y dx
−2
• Area above the x-axis

= ∫ −6 (− _x)
−2 12
dx

= [−12 ln | x |] −6
−2

= [−12 ln | (−2) |] − [−12 ln | (−6) | ]

A = 13,183 units2
11.3. y = x2 + 3x − 4
x-intercepts, y = 0:
y = x2 + 3x − 4
0 = x2 + 3x − 4
0 = (x − 1)(x + 4)
x−1=0 or x+4=0
x=1 x = −4
∴ (1; 0) ∴ (−4; 0)

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 Module 6 • Integral calculus 267

y-intercept, x = 0: y
y = x2 + 3x − 4
5
= (0)2 + 3(0) − 4
y = −4
∴ (0; −4)
∆x
Axis of symmetry, x = − _b
2a
: x
(– 4; 0) (1; 0)
x = −_b
2a
(3) (0; – 4)
= −_
2(1)
(x; y)
(– _3 ; – _
25
4
)
2
x = −1_12 – 10

Turning point(s):
y = x2 + 3x − 4

= (− _32 ) + 3(− _23 )−4

2

y = − 6_14

∴ (− _32 ; − 6_14 )

∆A = |y∆x|

|
A = ∫ y dx
1
−4
| • Area below the x-axis or −∫ y dx
1
−4

= |∫ (x 2 + 3x − 4) dx|
1
−4

|
= [_13 x 3 + _32 x 2 − 4x] |
1

−4

= |[_13 (1) 3 + _32 (1) 2 − 4(1)] − [_31 (− 4) 3 + _32 (− 4) 2 − 4(− 4)]|

A = 20_56 units2
11.4 y = (x − 4)2 + 3 y x=8
x-intercepts, y = 0:
y = (x − 4)2 + 3
0 = (x − 4)2 + 3
25
−(x − 4)2 = 3 y = (x – 4)2 + 3

_(x − 4)2 = −3 (0; 19)

_ (8; 19)
√ (x − 4 ) 2
= √ −3 (x; y)
∴ There are no real x-intercepts.
y-intercept, x = 0: (4; 3)
y = (x − 4)2 + 3 x
∆x
10
= ((0) − 4)2 + 3
y = 19
∴ (0; 19)

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 268 N4 Mathematics – Lecturer Guide

Axis of symmetry:
y = (x − 4)2 + 3
∴x−4=0
x=4
Turning point(s):
y = (x − 4)2 + 3
Since the coefficient of the bracket is positive, therefore y = 3 is a minimum
value.
∴ (4; 3)
∆A = y∆x
8
A = ∫ y dx • Area above the x-axis
0

= ∫ [(x − 4) 2 + 3] dx
8

= ∫ [(x 2 − 8x + 19] dx
8

= [_13 x 3 − 4x 2 + 19x]
8

= [_13 (8) 3 − 4(8) 2 + 19(8)] − [_13 (0) 3 − 4(0) 2 + 19(0)]

A = 66_23 units2
11.5 y = x3 + 3x2 − 2
x-intercepts, y = 0:
y = x3 + 3x2 − 2
0 = x3 + 3x2 − 2
Consider the factors of the constant term.
x = ±1 and x = ±2
f (x) = x3 + 3x2 − 2
f (−1) = (−1)3 + 3(−1)2 − 2
f (−1) = 0
∴ x + 1 is a factor
2
x + 2x − 2
______________
x + 1| x 3 + 3x 2 + 0x − 2
x 3 + 1x 2 ↓
_
2x 2 + 0x
2x 2 + 2x
_
−2x − 2
_
−2x −2
_ ∙ ∙

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 Module 6 • Integral calculus 269

∴ (x + 1)(x2 + 2x − 2) = 0
x+1=0 or x2 + 2x − 2 = 0
_
−b ± √b 2 − 4ac
x = −1 x=_ 2a
____________
−(2) ± √(2) − 4(1)(−2)
2
∴ (−1; 0) = ______________
2(1)
_
−2 ± 2√3
x=_ 2
_ _
x = −1 +√3 or x = −1 − √3
x = 0,732 x = −2,732
∴ (0,732; 0) ∴ (−2,732; 0)
y-intercept, x = 0:
x = –2 y
y = x3 + 3x2 − 2
4
0 = (0)3 + 3(0)2 − 2 y = x3 + 3x2 – 2
3
y = −2 (– 2; 2)
2
∴ (0; −2)
1
(– 2,732; 0) ∆x (0,732; 0)
Turning point(s): x
–4 –3 1 2 3 4
y = x3 + 3x2 − 2 (x; y)
dy
∴ _ = 3x2 + 6x = 0
dx
(0; – 2)
–3
3x(x + 2) = 0 (– 1; 0)
–4
3x = 0 or x+2=0
x=0 x = −2
Substitute x = 0 and x = −2 into y = x3 + 3x2 − 2.
y = (0)3 + 3(0)2 − 2 = −2 ∴ (0; −2)
y = (−2)3 + 3(−2)2 − 2 = 2 ∴ (−2; 2)

∆A = |y∆x|

|
A = ∫ y dx
0
−1
| • Area below the x-axis or A = −∫ y dx
−1
0

= |∫ (x 3 + 3x 2 − 2) dx|
0
−1

|
= [_14 x 4 + x 3 − 2x] |
0

−1

= |[_14 (0) 4 + (0) 3 − 2(0)] − [_14 (−1) 4 + (−1) 3 − 2(−1)]|

A = 1_14 units2
Therefore,
Atotal = 2 × A
= 2 × _54
Atotal = 2_12 units2

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 270 N4 Mathematics – Lecturer Guide

11.6 y = − (x + 3)(x − 1)2

x-intercepts, y = 0:
y = − (x + 3)(x − 1)2
0 = − (x + 3)(x − 1)2
0 = − (x + 3) or 0 = (x − 1)2
0 = −x − 3 0=x−1
x = −3 1=x
∴ (−3; 0) ∴ (1; 0)
y-intercept, x = 0: x = – 3,5 y
y = − (x + 3)(x − 1)2
= − ((0) + 3)((0) − 1)2 10
y = −3 y = –(x + 3)(x – 1) 2

∴ (0; −3)
(– 3;0) ∆x (1; 0)
Turning point(s): x
–4 2
y = − (x + 3)(x − 1)2 (0; – 3)
y = − x3 − x2 + 5x − 3
dy (x; y) – 10
∴_
dx
= −3x2 − 2x + 5 = 0 (– _53 ; – _
256
27
)

3x2 + 2x − 5 = 0
(3x + 5)(x − 1) = 0
3x + 5 = 0 or x−1=0
3x = −5 x=1
x = −1_32
Substitute x = −1_23 and x = 1 into y = − (x + 3)(x − 1)2.

y = − ((− _53 ) + 3)((− _53 ) − 1) = − 9_ ∴ (− 1_23 ; − 9_

27 )
2
13 13
27

y = − ((1) + 3)((1) − 1) 2 = 0 ∴ (1; 0)

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 Module 6 • Integral calculus 271

Area below the x-axis,

∆A = |y∆x|

|
A = ∫ y dx
1
−3
|
= |∫ [−(x + 3)(x − 1) ] dx |
1 2
−3

= |∫ [−x − x + 5x − 3] dx |
1 3 2
−3

|
= [− _14 x 4 − _13 x 3 + _52 x 2 − 3x] |
1

−3

= |[− _14 (1) 4 − _13 (1) 3 + _52 (1) 2 − 3(1)] − [− _14 (−3) 4 − _13 (−3) 3 + _52 (−3) 2 − 3(−3)]|

A = 21_13 units2
Area above the x-axis,
∆A = y∆x

A = ∫ −3,5 y dx
−3

= ∫ −3,5 [− (x + 3)(x − 1) 2] dx
−3

= ∫ −3,5 [− x 3 − x 2 + 5x − 3] dx
−3

= [− _14 x 4 − _13 x 3 + _52 x 2 − 3x]

−3

−3,5

[−4(−3) − 3(−3)]
= _1 4
− _1 (−3) 3 + _5 (−3) 2
3 2

− [−_14 (−3,5) 4 − _13 (−3,5) 3 + _52 (−3,5) 2 − 3(−3,5)]

67
A = 2_
192
units2
Therefore,
Atotal = 21_13 + 2_67
192
131
Atotal = 23_
192
units2

11.7 y = x3 − 9x2 + 24x + 4

x-intercepts, y = 0:
y = x3 − 9x2 + 24x + 4
0 = x3 − 9x2 + 24x + 4
The x-intercept will have to be computed by approximation.

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 272 N4 Mathematics – Lecturer Guide

y-intercept, x = 0: y (x = 4)

y = x3 − 9x2 + 24x + 4 (2; 24) (x; y)

y = x3 – 9x2 + 24x + 4
= (0)3 − 9(0)2 + 24(0) + 4 20
(4; 20)
y=4
∴ (0; 4)
Turning point(s):
(0; 4)
y = x3 − 9x2 + 24x + 4
dy ∆x
∴_
dx
= 3x2 − 18x + 24 = 0 x
2 –1 1 2 3 5
x − 6x + 8 = 0
(x − 2)(x − 4) = 0
x − 2 = 0 or x − 4 = 0
x=2 x=4
Substitute x = 2 and x = 4 into y = x3 − 9x2 + 24x + 4.
y = (2)3 − 9(2)2 + 24(2) + 4 = 24 ∴ (2; 24)
y = (4)3 − 9(4)2 + 24(4) + 4 = 20 ∴ (4; 20)
∆A = y∆x
A = ∫ 0 y dx
4
• Area above the x-axis

= ∫ 0 [x 3 − 9x 2 + 24x + 4] dx
4

= [_14 x 4 − 3x 3 + 12x 2 + 4x]

4

= [_14 (4) 4 − 3(4) 3 + 12(4) 2 + 4(4)] − [_14 (0) 4 − 3(0) 3 + 12(0) 2 + 4(0)]

A = 80 units2

Activity 6.9 SB page 424

1. A = ∫ −1 [(− x 2 + 6) − (x 2 − 2x + 2)] dx
2

= ∫ −1 [− x 2 + 6 − x 2 + 2x − 2] dx
2

= ∫ −1 [− 2 x 2 + 2x + 4] dx
2

2
= [− _ ]
3 2
2x + _
2 x + 4x
3 2 −1

= [− _23 x 3 + x 2 + 4x]
2

−1

= (− _32 (2)3 + (2)2 + 4(2)) − (− 2_3 (− 1)3 + (− 1)2 + 4(− 1))

= (− 16 ) (3 + 1 − 4)
2
_+4+8 − _
3
= (− 16 ) (3 − 3)
_ + 12 − 2
3
_

= 9 units 2

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 Module 6 • Integral calculus 273

2. x 2 = 3x − 2 y y1 = x2
0 = x 2 − 3x + 2 y2 = 3x – 2
4
0 = (x − 2)(x − 1) (2; 4)
x = 2; x = 1
(x; y2)
y = 4; y = 1
(x; y1)
∴ (2; 4) and (1; 1)
(1; 1)

A = ∫ 1 (3x − 2 − x 2) dx
2
x
3 2
1 ∆x 2
= [_ 3]
2
3 x − 2x − _
x
2 1

= (3_
(2)2
2 − 2(2 ) − (_
2)3
3 ) − (
(1)2
3_
2 − 2(1) − 3 )
(_
1)3 –2

= (6 − 4 − 38_) − (3_2 − 2 − 1_3)

= 0,167 units 2

3. − x 2 − 2x = x y

∴ x 2 + 3x = 0 2
x(x + 3) = 0 g(x) = –x2 – 2x f (x) = x
1
x = 0; x = − 3 (x; y2)
x
y = 0; y = − 3 –3 –2 –1 1 2
∴ (0; 0) and (( − 3; − 3) –1
(x; y1)
A = ∫ −3 [(− x 2 − 2x) − x] dx
0
–2
∆x

= ∫ −3(− x 2 − 3x) dx
0
–3
(–3; –3)
2 0
= [− _ 2 ]
3
3x
x −_
3 −3

= 0 − (−(_3 − 2 )
− 3) 3 (− 3)3 2
_

= 0 − (27 2)
_ − 27
3
_

= 4 1_2 or 4,5 units 2

y
4. Points of intersection: y1 = 6 – x
− 1_4 x 2 + 9 = 6 − x 8
(x; y2)

− 1_4 x 2 + 9 − 6 + x = 0 y2 = – __14 x2 + 9
− 1_4 x 2 + x + 3 = 0 (x; y1)
2
x − 4x − 12 = 0
∆x
(x − 6)(x + 2) = 0
∴ x = 6; x = − 2
x
∴ y = 0; y = 8 –6 6
∴ (6; 0); (−2; 8)

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 274 N4 Mathematics – Lecturer Guide

A = ∫ −2 [− 1_4 x 2 + 9 − (6 − x)] dx
6

= ∫ −2 [− 41_ x 2 + 3 + x] dx
6

2 6
= [− 1_4 . _ 2]
3
x + 3x + _
x
3 −2

= (− _1 (6) 3 + 3(6) + (_
12
6)2
2 ) − (− 1 (− 2) 3 + 3(− 2) + (_
_
12 2 )
− 2)2

= (− 18 + 18 + 18) − (0,667 − 6 + 2)
= 21,333 units 2

Summative assessment: Module 6 SB page 424

1. 1.1 ∫ [(x 4 – 2 + x)3x –1] dx

= ∫ [3x 3 – 6.x –1 + 3] dx

= 3∫ x 3 dx – 6∫ _1x dx + ∫ 3 dx
3+1
= 3._
x
3+1
– 6 ln x + 3x + c

= _34 x 4 – 6 ln x + 3x + c (5)
_
1.2 ∫(_
1
+ 3_ sin 3x + 3√x − 10.10 x + _ 1
− 2x 0) dx
x3 2 sec 2x
x
−1 1_ 10.10
_ + 1_ sin 2x − 2x + c
3_
=_ 2 − 2 cos 3x + 2x −
2
ln 10 2
(7)
2x

1.3 ∫(_ ) dx
2 2
cos x + sin x
cos 2x
1
= ∫_ 2 dx
cos x

= ∫ sec 2x dx
= tan x + c (3)

1.4 ∫(3 cos 4x − 3e 2x + 4.3 −x) dx

= 3∫ cos 4x dx − 3∫ e 2x dx + 4∫ 3 −x dx
2x −x
= 3_4 sin 4x − _
3e
2
− 4._3
ln 3
+c

= 3_4 sin 4x − 3_2 e 2x − _ 4

(ln 3)3 x
+c (5)

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 Module 6 • Integral calculus 275

∫ 1 [(_1x + 3x – 4)] dx
4
2. 2.1

= ∫ 1 [(1 – _x + 3)(x – 12)] dx

4 4

= ∫ 1 [3x – _x – 11] dx
4 4

= [3._ – 4 ln x – 11x]
1+1 4
x
1+1 1

[ 2 – 4 ln x – 11x]
4
3x 2
_
=
1

= (_ – 4 ln(4) – 11(4)) – (_ – 4 ln |(1)| – 11(1))

3(4) 2 3(1) 2
2 2

= – 16,045 (5)

2.2 ∫ 1 (5_x + 2) dx
3

3
= [5 ln x + 2x] 1
= (5 ln(3) + 2(3)) − (5 ln(1) + 2(1))
= 9,493 (3)
_π
2.3 ∫ 0 2 sin θ cos θ dθ
2

π
_
= ∫ 02 sin 2θ dθ
_π

= [− _ 2 ]
cos 2θ 2

= − _12[cos 2(π_2) − cos 2(0)]

= − _21[cos π − cos 0]

= − _21[− 1 − 1]

= − _21[ − 2]
=1 (4)
π
_

2.4 ∫ _π(2 sin 3x – 2.8 5x + _

x 3)
9 2
dx
– 3
π
_

= ∫ _π(2 sin 3x – 2.8 5x + 2x –3) dx

9

–3
π
_

= [2._ – 3 + 1 ] – _π
5x –3 + 1 9
– cos 3x 8
3
– 2._
5 ln 8
+ 2._
x
3
π
_

[– 3 x 2 ] – _π
= _2 cos 3x – _ 2
.8 5x – 1 9
_
5 ln 8
3

[ ] [ ]
2
cos(3._π9 ) 2 5(_π9 ) 1
– – _23 cos (3.– _π3 ) – _ 2
.8 5(– 3 ) – _
1
π
_
= –_ – _ . 8 – _
(9) (– 3 )
3 5 ln 8 π
_ 2
5 ln 8 π 2
_

= – 15,545 (5)

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 276 N4 Mathematics – Lecturer Guide

_
S = ∫ 1 (√t − 5)2 dt
4
3.
_ _
= ∫ 1 (√t − 5)(√t − 5) dt
4

_ _
= ∫ 1 (t − 5√t − 5√t + 25) dt
4

_
= ∫ 1 (t − 10√t + 25) dt
4

= [_ + 25t]
2 _3 4
t 20t
−_
2

2 3 1

= (_ + 25(4)) − (_ + 25(1))
_3 3_
4 2
20(4) (1)2 20(1)2
−_ −_
2

2 3 2 3

= 8 − 53,333 + 100 − _21 + 6,667 − 25

= 35,834 (3)

4. 4.1 A = ∫ 21 y dx

= ∫ 21 (– x + 2) dx

= [– _ + 2x]
2 2
x
2
1

= (– _ + 2(2)) – (– _ + 2(1))
(2) 2 (1) 2
2 2

= 0,5 units2 (4)

4.2 A1 = ∫ 0 (x3 + x2 – 12x) dx

−4

= [_ – 6x2]
4 3 0
x
4
+_
x
3
−4
= 0 – (– 53,33)
= 53,33 units2

A2 = ∫ 30 (x3 + x2 – 12x) dx

= [_ – 6x2]
4 3 3
x
4
+_
x
3
0
= |– 24,75 – 0|
= |– 24,75|
= 24,75 units2
AT = 78,08 units2 (6)

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 Module 6 • Integral calculus 277

5. 5.1 y = 3.e2x y

x y 25
– 1,500 0,149
20
– 1,000 0,406
– 0,500 1,104 15

0,000 3,000 10
x = –1
0,500 8,155
(x; y) 5 x = 0,5
1,000 22,167
x
– 1,5 ∆x 0 1

∆A = y∆x
0,5
A = ∫ –1 y dx • Area above the x-axis
0,5
= ∫ –1 3.e 2 x dx
0,5
= [_2 ]
2x
3e
–1

= [_ 2 ] [ 2 ]
2(0,5) 2(– 1)
3e 3e
– _

A = 3,874 units2 (6)

5.2 y = – (x – 2)2 – 1 y
x-intercepts, y = 0: x=2
∆x
y = – (x – 2)2 – 1 x
0
0 = – (x – 2)2 – 1 –1
(x – 2)2 = – 1
_ y = –(x – 2)2 – 1
_ –2 (x; y)
√(x – 2) 2 = √– 1
∴ There are no real x-intercepts. –3
_
x – 2 = ±√– 1 –4
x – 2 = ±i
–5
x = 2±i
y-intercept, x = 0:
y = – (x – 2)2 – 1
= – ((0) – 2)2 – 1
y = –5
∴ (0; – 5)
Axis of symmetry:
y = – (x – 2)2 – 1
∴x–2=0
x=2

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 278 N4 Mathematics – Lecturer Guide

Turning point(s):
y = – (x – 2)2 – 1
Since the coefficient of the bracket is negative, therefore y = – 1 is a maximum
value.
∴ (2; – 1)
∆A = |y∆x|

A = | ∫ 0 y dx|
2
• Area below the x-axis

= | ∫ 0 [– (x – 2) 2 – 1] dx|
2

= | ∫ 0 [– x 2 + 4x – 5] dx|
2

|
= [– _ + 2x 2 – 5x] |
3 2
x
3 0

|
= [– _
3
+ 2(2) 2 – 5(2)] – [– _
(2) 3
3
+ 2(0) 2 – 5(0)]
(0) 3
|
A = 4_23 units2 (6)

5.3 y = x(x2 – 3)
y

_ _ x
–2 – √3 –1 0 1 √3 2 2,5

–1

–2

x-intercepts, y = 0:
y = x(x2 – 3)
0 = x(x2 – 3)
_ _
0 = x(x + √3 )(x – √3 )
_ _
0=x or 0 = x + √3 or 0 = x – √3
_ _
– √3 = x √3 = x
_ _
∴ (0; 0) ∴ (– √3 ; 0) ∴ (√3 ; 0)
y-intercept, x = 0:
y = x(x2 – 3)
y = (0)((0)2 – 3)
y=0
∴ (0; 0)

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 Module 6 • Integral calculus 279

Turning point(s):
y = x(x2 – 3)
y = x3 – 3x
dy
_
dx
= 3x2 – 3 = 0
x2 – 1 = 0
(x + 1)(x – 1) = 0
x+1=0 or x – 1 = 0
x = –1 x=1
Substitute x = – 1 and x = 1 into y = x(x2 – 3).
y = (– 1)((– 1)2 – 3) = 2 ∴ (– 1; 2)
2
y = (1)((1) – 3) = – 2 ∴ (1; – 2)
Area below the x-axis,
∆A = |y∆x|
_

A = | ∫ 0 y dx|
√3

= | ∫ 0 [x(x 2 – 3)] dx|

√3

= | ∫ 0 [x 3 – 3x] dx|
√3

| |
_

[ 4 – 2 ]0
√3
x4 _
_ 3x 2
=
_ _
= [_|
(√3 ) 4 _
2 ] [ 4
3(√3 ) 2
2 ] |
4 2
(0) 3(0)
4
– – _–_

A = 2_14 units2

Area above the x-axis,

∆A = y∆x
2,5
A = ∫ √_3 y dx

= ∫ √_3 [x(x 2 – 3)] dx

2,5

= ∫ √_3 [x 3 – 3x] dx
2,5

2,5
= [_ 2 ]
4 2
3x
x
4
–_ _
√3
_ _
= [_ 2 ] [ 4 2 ]
(2,5) 4
√3 3(2,5) 2
3 √3 ( )4 ( )2
4
–_ – _ –_
41
A = 2_
64
units2

Therefore,
ANET = 2_14 + 2_
41
64
57
ANET = 4_
64
units2 (8)

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 280 N4 Mathematics – Lecturer Guide

6. x2 = x + 2 y
y1 = x
2
x2 − x − 2 = 0 y2 = x + 2
4
(x − 2)(x + 1) = 0
∴ x = 2; x = − 1 3 (x; y2)
∴ y = 4; y = 1
2
∴ (2; 4); (− 1; 1)
1
A = ∫ −1 [(x + 2) − x 2] dx
2

(x; y1)
= ∫ −1(x + 2 − x 2) dx x
2
–2 –1 ∆x 1 2

= [__ 3]
2
x2 x3
2
+ 2x − __
–1

= (___ 3 ) ( 2 3 )
(2) 2
(2)3 (− 1)2 (− 1)3
2
+ 2(2) − ___ − ____ + 2 (− 1) − ____

= (2 + 4 − _3 ) − (_2 − 2 + _3 )
8 1 1

= 4,5 units 2 (10)

TOTAL: [80]

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 Exemplar examination paper 281

Exemplar examination paper

Time: 3 hours Marks: 100

INSTRUCTIONS AND INFORMATION

1. Answer ALL the questions in full. Show ALL the calculations and
intermediary steps and simplify where possible.
2. For graph work, the values of the intercept(s) with the system of axes and
turning point(s) MUST be shown on the graph.
3. ALL final answers must be rounded to THREE decimal places.
4. Questions may be answered in any order but subsections must NOT be
separated.
5. A formula sheet is attached to this question paper. You are NOT compelled to
use the formulae and the list is NOT necessarily complete.

QUESTION 1
1.1 Given: x + 3y – 2z = – 13
2x – 6y + 3z = 32
3x – 4y – z = 12
Solve for y by using determinants. (8)
1.2 Simplify
_____ the following
____ ___ and leave your answer in a + bj form:
√–144 + √169 – √–1 . (2)
1.3 Given: z = – 1 – 4j
1.3.1 Calculate the modulus and the argument of z. The argument must be
positive. Show ALL steps. (3)
1.3.2 Express z in polar form. (1)
1.3.3 Represent z on an Argand diagram. (1)
1.4 Solve for x and y:
3 – 2j 1 – 3j
x + yj = _ _
1 – j – 1 + 3j (5)
[20]

QUESTION 2
2.1 Prove that: _ 2 sin x + 1
sin 2x + cos x = sec x (3)
2.2 If tan x = _5 3_
12 and tan y = 4 , with both x and y acute angles, determine without
using a calculator, sin(x – y). (3)
2.3 Given: 2 cos 2θ + 5 sin θ = – 1; 0° ≤ θ ≤ 360°
Solve for θ. (5)
2.4 Simplify (sec θ + tan θ)2 (3)
2.5 Derive a formula for tan 2θ. (3)
2.6 Determine cot 105° WITHOUT using a calculator. (3)
[20]

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QUESTION 3
3.1 3.1.1 Sketch the graph of y = – x 2 – 2x + 3 (3)
3.1.2 Is the graph of y = – x 2 – 2x + 3 continuous? (1)
3.1.3 What is the domain of the graph y = – x 2 – 2x + 3? (1)
3.2 Sketch the graph of x = ln y (3)
3.3 Sketch the graph of 3x 2 = 9y 2 + 27 (2)
[10]

QUESTION 4
4.1 Differentiate from first principles: y = – x 2 – x + 1 (5)
4.2 Given: y = 4 cos(7x + 2)
Differentiate by the use of the chain rule or function of a function. (4)
Given: (2y – 3)
2 4
4.3
Use the binomial theorem to expand to FOUR terms. (5)
4.4 Differentiate with respect to x:
y = 2 sec 3x – 4 cosec 2x + 6 ln x + p (4)
3
4.5 Given: y = 2x – 8x
Calculate, using differentiation, the coordinates of the maximum and minimum
turning points. Distinguish between the maximum and minimum turning points
by the use of the second derivative. (7)
[25]

QUESTION 5
5.1 Determine:
_
π

∫ 02 (sin θ + cos θ) dθ (4)

5.2 Integrate:
∫ (2 cos 2x + 6k – _
e –x x 2)
1 + 10.10 x – _
1 dx (6)
5.3 5.3.1 Sketch, and clearly indicate the area bounded by the graph of y = 2.3 2x;
the x-axis, the y-axis and the line x = 2. Also indicate the representative
strip to be used to calculate the area. (3)
5.3.2 Calculate, using integration, the value of the area in QUESTION 5.3.1. (4)
5.4 5.4.1 Sketch the graphs of y = – x + 3 and xy = 2 on the same system of
axes. Clearly indicate the area bounded by the two graphs. Show the
representative strip used to calculate the area. (4)
5.4.2 Calculate, using integration, the value of the area shown in
QUESTION 5.4.1. (4)
[25]
TOTAL: 100

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 Exemplar examination paper 283

Possible formula sheet

a x = b ⇔ log a x = log b tan(a ± b) = _tan a ± tan b
1 ∓ tan a tan b

ln x = log ex sin 2x + cos 2x = 1

(r θ)n = r n nθ 1 + cot 2x = cosec 2x

a + bj = c + dj ⇔ a = c and b = d 1 + tan 2x = sec 2x

sin(a ± b) = sin a cos b ± sin b cos a sin 2A = 1_2 – _21 cos 2A

cos(a ± b) = cos a cos b ∓ sin a sin b cos 2A = _21 + 1_2 cos 2A

dy
_
y dx

ax n nax n–1
ka x ka x ln a
k ln x _k
x

sin x cos x
cos x – sin x
tan x sec 2x
cot x – cosec 2x
sec x sec x tan x
cosec x – cosec x cot x

y = u(x). v(x) ∫ a_x dx = a ln x + c

dy
∴_ ∫ ka x dx = _
x

dx
= u(x)v′(x) + u′(x)v(x) ka
ln a
+c
( )
y = u_ x
v(x)
∫ sin x dx = – cos x + c
dy ( ) ( )
x u′ x – u x v′ x ( ) ( )
∴_ = v___________ ∫ cos x dx = sin x + c
dx v(x) 2 [ ]

A OX = ∫ a y dx
dy _
_ dy du b
dx
= ×_ du dx

A = ∫ a (y2 – y1) dx
b
∫ ax n dx = _
n+1
ax
n+1 + c

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Exemplar examination paper Memorandum

QUESTION 1

| |
1 3 –2
1.1 |D| = 2 –6 3 ✓ • Expand using any row or column
3 –4 –1
= +(1) – 6
–4 | |
3 – (3) 2 3 + (–2) 2 – 6 ✓
–1 3 –1 |
3 –4 | | |
• Positional sign (element) | minor
of element for the entire row or
column
= 1(6 + 12) – 3(– 2 – 9) – 2(– 8 + 18)
= 18 + 33 – 20
= 31 ✓

| |
1 – 13 – 2
|D y| = 2 32 3 ✓ • | D y|: replace the ‘y’ column by
3 12 – 1 inserting the constant column in its
place

|
12 – 1 3 –1 |
= (+ 1) 32 3 – (– 13) 2 3 + (– 2) 2 32 ✓
3 12 | | | |
= 1(– 32 – 36) + 13(– 2 – 9) – 2(24 – 96)
= – 68 – 143 + 144
= –67 ✓
D
∴y=_y
D ✓

=_
– 67
31
= – 2,161 ✓ (8)

_____ ____ ___ ___

1.2 √– 144 + √169 – √– 1 • √– 1 = j
___ ___
= 12√– 1 + 13 – √– 1 ✓
= 12j + 13 – j
= 13 + 11j ✓ (2)

1.3 1.3.1 z = – 1 – 4j • Modulus = radius; argument = angle

______
r = √x 2 + y 2
__________
= √(– 1)2 + (– 4)2
___
= √17
= 4,123 ✓
θ = tan –1(_x)
y

= tan –1(4_1)
= 75,964° ✓

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 Exemplar examination paper 285

Note Im

To know in which quadrant, make a sketch of z

on an Argand diagram. Re
–1
Now you see that it is in the 3rd quadrant.
Therefore, 180 + 75,964 = 255,964°.

–4

z = 4,123 255,964°
Modulus = 4,123
Argument = 255,964° ✓ (3)
1.3.2 z = 4,123 255,964° (1)
1.3.3 Im

1
Re
–1

–4

(1)
(
3 – 2j ) 1 – 3j
1.4 x + yj = _ _
1 – j – 1 + 3j ( )

= (_ )( ) _
– (1 + 3j)(1 – 3j) ✓
3 – 2j 1 + j 1 – 3j 1 – 3j
• Rationalise each fraction by
(1–j 1+j )( ) ( )( )
multiplying by the conjugate of the
denominator
3 + j – 2j 2 1 – 6j + 9j 2
=_
1 – j2
–_
1 – 9j 2
✓ • Multiply out and add like terms
3 + j – 2(– 1) 1 – 6j + 9(– 1)
=_
1 – (– 1)
–_
1 – 9(– 1)
• j 2 = –1
5+j – 8 – 6j
=_ _
2 – 10
( ) ( )
5 5 + j – 1 – 8 – 6j
= ____________
10
25 + 5j + 8 + 6j
=_
10
33 + 11j
=_
10
_+_
= 33 11
10 10 j ✓

∴ x = 33
_ and
10
_
y = 11
10
= 3,3 ✓ = 1,1 ✓ • Equate real terms and imaginary
terms. (5)
[20]

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 286 N4 Mathematics – Lecturer Guide

QUESTION 2
2.1 LHS = _ 2 sin x + 1
sin 2x + cos x

= 2____________
2 sin x + 1
sin xcos x + cos x ✓ • Double angle identity sin 2x = 2 sin xcos x
=_ 2 sin x + 1
cos x(2 sin x + 1)
✓ • Factorise

=_ 1
cos x ✓

= sec x = RHS (3)

2.2 sin(x – y) y
= sin x cos y – cos x sin y ✓
=_
5 4_ 12 _ 3_
13 . 5 – 13 . 5 ✓ 13
5
_ – 36
= 20 _
65 65 x x
12
= –_
16
65 ✓ (3)
y

5
3
y x
4

2.3 2 cos 2θ + 5 sin θ = – 1

2(1 – sin 2θ) + 5 sin θ + 1 = 0 ✓ • Change to same ratio
cos 2θ = 1 – sin 2θ
2 – 2 sin 2θ + 5 sin θ + 1 = 0
– 2 sin 2θ + 5 sin θ + 3 = 0 • (× –1): LHS and RHS
2 sin 2θ – 5 sin θ – 3 = 0 ✓½ • Trinomial
(2 sin θ + 1)(sin θ – 3) = 0 ✓½
∴ 2 sin θ = – 1 ✓½ • sin θ = 3 ✓½
sin θ = – 1_2 • Not possible: maximum of
sin θ is 1
∴ in 3rd and 4th quadrants
θ = 180° + sin –10,5 and θ = 360° – sin –10,5
= 210° ✓ = 330° ✓ (5)

2.4 (sec θ + tan θ)2

= (sec θ + tan θ)(sec θ + tan θ)
= sec 2θ + 2 sec θ tan θ + tan 2θ ✓

cos 2θ (cos θ) (cos θ) cos 2θ

2
=_ 1 + _ 2 _
sin θ
+_
sin θ
• Change to sin θ and cos θ
2
= ___________
1 + 2 sin θ + sin θ
cos 2θ
✓ • Common denominator
2
= ___________
1 + 2 sin θ + sin θ
1 – sin 2θ
• cos 2θ = 1 – sin 2θ

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 Exemplar examination paper 287

(
1 + sin θ 1 + sin θ )( )
= ____________
(1 – sin θ)(1 + sin θ)
• Factorise top and bottom

= 1_ + sin θ
1 – sin θ
✓ (3)

2.5 tan 2θ = tan(θ + θ) ✓

=_
tan θ + tan θ
1 – tan 2θ
✓ • Compound angles

=_2 tan θ
1 – tan 2θ
✓ (3)

2.6 cot 105°

= cot(60° + 45°) ✓ 45°
30° __
__
2 √3 √2
1

60° r 45° r
1 1

( )
= cos
_ 60° + 45°
sin(60° + 45°)
Alternatively

= _______________
cos 60° cos 45° – sin 60° sin 45°
sin 60° cos 45° + cos 60° sin 45° ✓½
cot 105° = _ 1
tan 105°

_1 × __ √3
1__ – __ × __
1__
__
=_ 1
tan(45° + 60°)
= _________
2 √2
__
2 √2
✓½
√3
__ __
1 1_ × __
1__
× __ + _ 1
2 √2 2 √2 = ___________
tan 45° + tan 60°
__ 1 – tan 45°tan 60°
___ √3__
1__ – ___
= ______
2√__2 2√2
√3__
___ = ___________
1 – tan 45° tan 60°
2√2
+ ___
1__
2√2
tan 45° + tan 60°
__ __
1– 1 √( )
3
= ______
1–√
__3
____ __
= ___
2__√2
• Everything over 1 + √3
√3 +__ 1
____
2√2 LCM 1 – √__
__
3 ____
__

__ __
= ____ × 1 – √__3
1 + √3 1 – √3
• Rationalise
1–√
__3 × ____
= ____ 2__√2
• Invert denominator
2√2 √3 + 1 __
1_______
– 2√3 + 3
1__– √3
__ __
3 –1
= 1–3
= ____
√3 + 1
× √3 – 1 • Rationalise ____
√__
__
4 – 2√3
__
√3 – 3 – 1 + √3
__ = _____
–2
= __________
3–1
__
2√3 – 4
__ = – 2 + √3
= _____
2
__
__
(√3 – 2)
2______ = √3 – 2
= 2 • Factorise
__
= √3 – 2 ✓

(3)
[20]

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 288 N4 Mathematics – Lecturer Guide

QUESTION 3
3.1 3.1.1 y = – x 2 – 2x + 3 • Exponent of y = 1; exponent of x = 2
∴ parabola
x-intercepts (y = 0)
– x 2 – 2x + 3 = 0
x 2 + 2x – 3 = 0
(x + 3)(x – 1) = 0
∴ x = – 3 or x = 1
y-intercept (x = 0)
y=3
– –2 ( )
Turning point: x = _
–b _
2a = – 2 = – 1
y = f (– 1) = –(– 1)2 – 2(– 1) + 3 = 4
y
✓
4
(–1, 4)
3 ✓

1
✓½ ✓½
x
(–3, 0) (1, 0)
(3)

3.1.2 Continuous ✓ (1)

3.1.3 – ∞ ≤ x ≤ ∞ or x ∈ ℝ ✓ (1)

3.2 x = ln y y
x = log e y
ex = y ✓ ✓
y = e x or x = ln y
✓
1
x
(3)

3.3 3x 2 = 9y 2 + 27 y
2
_
3x 9 2 27
27 =_ y +_ 27 27 √3
__ ✓
2 3x 2 = 9y 2 + 27
y
x_2 _
9 – =1 3 ✓
x
–3 3

__
–√ 3 (2)
[10]

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 Exemplar examination paper 289

QUESTION 4
4.1 y = –x 2 – x + 1
f(x) = – x 2 – x + 1
f(x + h) = –(x + h)2 – (x + h) + 1 ✓
= –(x 2 + 2xh + h 2) – x – h + 1
= – x 2 – 2xh – h 2 – x – h + 1 ✓
f(x + h) – f(x) = – x 2 – 2xh – h 2 – x – h + 1 – (– x 2 – x + 1) ✓
= – x 2 – 2xh – h 2 – x – h + 1 + x 2 + x – 1
= – 2xh – h 2 – h ✓
(x + h) – f(x)
f_
h
= – 2x – h – 1
f(x + h) – f(x)
_ = – 2x – 1 ✓
lim
h→0 h
• Substitute h = 0 (5)

4.2 y = 4 cos(7x + 2) Let u = 7x + 2

y = 4 cos u then _
du
dx
=7 ✓
dy
_
du
= – 4 sin u ✓
dy _
_ dy du
dx
= ×_ du dx
= – 4 sin u. 7 ✓
= – 4 sin(7x + 2). 7
= – 28 sin(7x + 2) ✓ (4)

(2y – 3)
2 4
4.3
✓4 ✓ 4(3) (2y 2)2 (– 3)2 ✓ 4(3)(2)(2y 2) (– 3)3 ✓
= (2y ) + 4(2y ) (– 3) + _
2 2 3
2! +_ 3! +…
( ) ( ) 12 4y 4 (9) 24 2y 2 (– 27)
= 16y 8 + (– 12)(8y 6) + _2 +_3×2 +…
= 16y 8 – 96y 6 + 216y 4 – 216y 2 + … ✓ (5)

4.4 y = 2 sec 3x – 4 cosec 2x + 6 ln x + p

dy
_ ✓ ✓ ✓
dx
= 2.3 sec 3x tan 3x – 4.2(– cosec 2x cot 2x) + 61_x + 0
= 6 sec 3x tan 3x + 8 cosec 2x cot 2x + 6_x ✓ (4)

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 290 N4 Mathematics – Lecturer Guide

4.5 y = 2x 3 – 8x
dy
For turning points _
dx
= 6x 2 – 8 = 0 ✓
∴ Critical values of x are 6x 2 = 8
x 2 = 8_6 = 4_3
_
x = ± √34_

∴ x = ± 1,155 ✓
d2 y
For the nature of the turning points, examine _
dx 2
2
d y
_ dy
_
dx 2
= 12x ✓ • dx
= 6x 2 – 8 ✓
2
d y
_
At x = 1,155 dx 2
= 12(1,155) ✓½
positive ∴ it will have a minimum turning point
Min T.P (1,155; f(1,155))
= (1,155; –6,158) ✓ • substitute y = 2(1,155)3 – 8(1,155)
= –6,158
2
d y
_
At x = – 1,155 dx 2
= 12(–1,155) ✓½
negative ∴ it will have a maximum turning point
Max T.P (–1,155; f(–1,155))
= (–1,155; 6,158) ✓ • substitute y = 2(– 1,155)3 – 8(– 1,155)
= 6,158 (7)
[25]

QUESTION 5
_
π

5.1 ∫ 02 (sin θ + cos θ) dθ

= [– cos θ + sin θ] 90°
0
✓
= [– cos 90° + sin 90°] – [– cos 0° + sin 0°] ✓
= [0 + 1] – [– 1 + 0] ✓
=1+1
=2 ✓ (4)

5.2 ∫(2 cos 2x + 6k – _

e –x x 2)
1 + 10.10 x – _
1 dx

= ∫(2 cos 2x + 6k – e x + 10.10 x – x –2) dx

✓ ✓ x✓ _ ✓ –1 ✓
10.10 x _
=_
2 sin 2x
2 + 6kx – e + ln 10
– x– 1 + c
x
= sin 2x + 6kx – e x + _
10.10
ln 10
+ 1_x + c ✓ (6)

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 Exemplar examination paper 291

5.3 5.3.1
y
y = 2.3 2x
✓

2 ✓
✓
x
∆x 2
(3)

5.3.2 Area of strip: ∆ A = y∆ x

A = ∫ 0 y dx
2

✓
= ∫ 0 2.3 2x dx ✓
2

2
= [_
2 ln 3]
2x
2.3
✓
0

= (_
ln 3) (ln 3)
2(2) 2(0)
3
– _
3

= 72,819 units 2 ✓ (4)

5.4 5.4.1 y = –x + 3 y = 2_x

∴ – x + 3 = 2_x y
✓ y = 2_x ✓
– x 2 + 3x – 2 = 0 y = –x + 3
3
x 2 – 3x + 2 = 0
(x – 2)(x – 1) = 0
(x; y2) ✓ Strip
∴ x = 1 or x = 2
(x; y1) ✓ Area

x
1 ∆x 2 3

(4)
5.4.2 Area = ∫ 1 (y2 – y1) dx
2

= ∫ 1 (– x + 3 – 2_x) dx ✓
2

= [– x_2 + 3x – 2 ln x] ✓
2 2

= (– 2 + 6 – 2 ln 2) – (– 1_2 + 3 – 0)
= 0,114 units 2 ✓ (4)
[25]
TOTAL: 100

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 292 N4 Mathematics – Lecturer Guide

Glossary

A
Acute angle – an angle that lies between 0 and π_4 radians, or between 0° and 90°
Angle of inclination – the angle between the line and the positive x-axis
Area – the amount of space inside the boundary of a flat (two-dimensional)
object
Argand diagram – a graphical representation of complex numbers
Axis – a horizontal line and a vertical line that intersect at the origin are used to
describe the position of points in a coordinate system
Axis of symmetry – a line that divides the graph into two symmetrical halves

C
Cartesian coordinate system – a system in which the location of a point is given
by coordinates that represent its distance from the axes
Centre point – the point inside the circle that is the same distance from all
points on the circle
Coefficient – a number that is multiplied by a variable
Complex numbers (ℂ) – numbers that can be expressed in the form a + bi,
where a and b are real numbers; the imaginary part is i; in the expression a + bi,
a is the real part and b is the imaginary part of the complex number
Congruent – has exactly the same shape and size
Coordinates – a set of values that shows the exact position of a point in relation
to the axes
Co-terminal angle – an angle in standard position (angle with the initial side on
the positive x-axis) that have a common terminal side

D
Delta – the symbol used to indicate a change in value or the differences in values
Dependent variable – the output value of a function that is dependent on the
input value
Derivative – the gradient of a curve at any point, or the gradient of the tangent
to the curve at that point; the instantaneous rate of change of a function
Determinant – a function of which the input is a square matrix and the output is
a number
Diameter – the distance from one side of the circumference of a circle, through
the centre of the circle, to the other side
Differentiate – the process of finding the derivative of a function

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 Glossary 293

Differentiating by first principles – finding the derivative of a function using

the definition

E
Equation – a mathematical statement that two equations have the same value
Exponent – shows how many times something is multiplied by itself

F
Frequency – the number of cycles completed by a graph over a given interval
Function – a rule or relationship for which any input value results in one unique
output value

G
Gradient – the slope or steepness and is the ratio of vertical change to horizontal
change

I
Imaginary number – a complex number that can be written as a real number
multiplied by an imaginary unit i, defined by i 2 = − 1
Imaginary numbers (핀) – the square root of negative numbers
Inclination – the slope or gradient of a line
Independent variable – the input value of a function that does not depend on
anything
Inflection point – a point where the concavity of a curve changes; can be a
stationary point, but not always
Integers (ℤ) – positive and negative numbers
Irrational numbers (ℚʹ) – any number that cannot be written as a fraction in
the form a_b, b ≠ 0 with a, b ∈ ℤ

L
Limit – the value that a function or sequence approaches as the input
approaches some value
Logarithm – a number that is the exponent by which another fixed value, the
base, must be raised to produce that number, so a logarithm is an exponent

M
Midpoint – the point exactly halfway along a line segment, which divides the
line segment into two equal pieces

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N
Natural numbers (ℕ) – numbers used for counting; starts at 1

P
Phase shift – indicates how far the function is shifted horizontally from the
usual position

Q
Quadratic equation – any equation in the form ax 2 + bx + c = 0, where x
represents an unknown (variable) and a, b and c are constants, with a ≠ 0

R
Radian – the angle at the centre of an angle subtended by an arc of the same
length as the radius
Radius – the distance from the centre to any point on a circle
Rational numbers (ℚ) – any number that can be written as a fraction in the
form a_b, b ≠ 0 with a, b ∈ ℤ
Real numbers (ℝ) – rational and irrational numbers

S
Secant – a line that intersects a circle at two points
Second derivative – the derivative of the derivative; it tells us about the
concavity of the graph
Simultaneous equations – two or more equations with the same variables; when
there are at least as many equations as variables they may be solvable
Stationary point – a point where the derivative of the function is zero; i.e. the
gradient at a stationary point is zero

T
Tangent – a line that intersects a circle at only one point
Turning point – the point at which the graph has a maximum or minimum
value

V
Variable – a letter that represents or stands for a number

W
Whole numbers (ℕ0) – natural numbers plus the number 0

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