Format 210mm X 297mm Extent= 224 pgs (11.58 mm) (70gsm paper) Status: CRC Date: 14/3
Cambridge IGCSE TM redue70% DA1302
ACE YOUR
ADDITIONAL
MATHEMATICS
Workbook
Cambridge IGCSE Ace Your Additional Mathematics is written to improve
TM
students’ approach to mathematics by providing them with high-quality
educational materials that uphold top academic standards, while allowing them
to acquire valuable techniques to excel in their examinations.
This workbook is written distinctively based on the Cambridge IGCSE 2022 Cambridge IGCSE TM ACE YOUR ADDITIONAL MATHEMATICS
and 2023–24 syllabuses for Additional Mathematics (0606) course. The scope,
sequence and level of the workbook has been constructed to match the
Cambridge IGCSE Year 10 and Year 11 syllabuses.
The workbook provides ample practice exercises for each topic. Answers
and fully-worked solutions are provided for all questions to enhance students’
understanding in how to master the approach to certain questions. It is highly
recommended as it serves as a good aid in evaluating students’ proficiency in Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
mathematical skills, concepts and processes, helping them achieve excellent
results in examinations.
About the Author TM
Kung Girly received her Bachelor of Laws (Honours) from the University of London and practiced law Cambridge IGCSE
for a few months. She then decided to venture into her true passion in education and teaching.
Throughout her teaching pathway, she has educated many students who sat for the IGCSE exams
with excellent results. She has also prepared and conducted seminars for exam-going students,
providing them the foundation and tips to excel in the examinations. She also provides coaching ACE YOUR
for students participating in international mathematics competitions.
Ms Kung has seven years of experience in teaching IGCSE Additional Mathematics, and currently
serves as a Director at Teras Murni Education Group. She is actively involved in educating students ADDITIONAL
to ensure they meet their academic potential.
MATHEMATICS
Workbook
www.dickenspublishing.co.uk
DA1302
ISBN: 978-1-78187-260-4
Suite G7-G8, Davina House, 137-149 Goswell Road,
London, EC1V 7ET, United Kingdom.
E-mail: [email protected] Kung Girly
Contents
Preface ------------------------------------------------------------------------------------------------------------------- iii
Assessment Overview --------------------------------------------------------------------------------------------------- iv
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List of Important Formulae ---------------------------------------------------------------------------------------------- v
Chapter 1 Functions ---------------------------------------------------------------------------------------------------- 1
Chapter 2 Quadratic Functions ------------------------------------------------------------------------------------ 13
Chapter 3 Equations, Inequalities and Graphs ------------------------------------------------------------------ 22
Chapter 4 Indices and Surds ---------------------------------------------------------------------------------------- 30
Chapter 5 Factors of Polynomials ---------------------------------------------------------------------------------- 37
Chapter 6 Logarithmic and Exponential Functions ------------------------------------------------------------ 46
Chapter 7 Straight Line Graphs ------------------------------------------------------------------------------------ 54
Chapter 8 Circular Measure ---------------------------------------------------------------------------------------- 68
Chapter 9 Trigonometry --------------------------------------------------------------------------------------------- 86
Chapter 10 Permutations and Combinations ---------------------------------------------------------------------- 98
Chapter 11 Series ----------------------------------------------------------------------------------------------------- 108
Chapter 12 Vectors ---------------------------------------------------------------------------------------------------- 120
Chapter 13 Differentiation and Integration ---------------------------------------------------------------------- 133
• Differentiation 133
• Integration 145
• Kinematics 160
Answers ------------------------------------------------------------------------------------------------------------------165
Note: The topic of ‘Simultaneous Equations’ is integrated with the other chapters.
xii Contents
04 content.indd 12 14/03/2022 12:28 PM
Preface
Cambridge IGCSE Ace Your Additional Mathematics is designed to help students achieve
TM
excellent grades in their examinations for the Cambridge IGCSE Additional Mathematics course.
This workbook provides a comprehensive and detailed coverage of all the topics in accordance
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with the latest 2022 and 2023–24 syllabuses for the Cambridge IGCSE Additional Mathematics
(0606) course.
= – 1
2 The contents of this workbook mainly focus on:
→ – Questions that follow the examination format of Cambridge IGCSE
= ABAB
– Papers 1 and 2: specially designed examination-style questions
– Step-by-step solutions provided for each question to allow students in identifying their
strengths and weaknesses in the course
– Detailed mathematical equations used in Cambridge IGCSE Additional Mathematics
= – 1
2 As such, this workbook presents an immense opportunity for students to develop the essential
1 knowledge, tools and skills to effectively prepare for their IGCSE examinations. The questions
= 2 2
= x are carefully designed and arranged in order of difficulty to enable students to gain confidence in
Additional Mathematics. We believe that this book both complements and supplements students’
learning in the classroom as they strive to build their foundation and achieve excellent grades in
mathematics. Students also develop essential mathematical skills for the advancement to Cambridge
International AS and A Levels, further education or any mathematics-related professions.
Preface iii
01 Preface AddMath.indd 3 14/03/2022 12:26 PM
1 Functions
Answer all questions.
If you need to show your work for any of the questions, write it clearly in the space provided below the
question.
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1. The function f is defined by f(x) = 3 − x + 8 for 0 < x < 3.
(a) Write down the range of f.
[2]
(b) Find f (x) and state its domain and range.
−1
[4]
2. A function f is such that f(x) = 5x − 4 for −10 < x < 5.
2
(a) Find the range of f.
[3]
(b) Write down a suitable domain for f for which f (x) exists.
−1
[1]
Chapter 1 Functions 1
Chapter 1 Add Math (1 to 12).indd 1 14/03/2022 12:16 PM
3. (a) Sketch the graph y = |2x + 6|, showing the coordinates of the points where the graph
meets the coordinate axes.
y
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x
[2]
(b) Solve 4 = |2x + 6|.
[2]
4. Solve the equation |x − 7x + 6| = 6 − x.
2
[3]
= – 1 5. Sketch the graph of y = |4x − 1| for −2 , x , 2.
2 y
10
8
6
4
2
x
–2 –1 0 1 2 3
–2
[2]
Cambridge IGCSE
TM
2 Ace Your Additional Mathematics
Chapter 1 Add Math (1 to 12).indd 2 14/03/2022 12:16 PM
(a) State the corresponding range of y.
[1]
(b) Find the range of values of c for which |4x − 1| = −4x + c has only one solution for
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−2 , x , 3.
[2]
6. On the same axes, sketch the graphs of y = |3x| and y = |x + 2|.
Show the coordinates of the points where the graphs meet the coordinate axes.
Find the coordinates of the intersection points of the two graphs.
y
x
[6]
Chapter 1 Functions 3
Chapter 1 Add Math (1 to 12).indd 3 14/03/2022 12:16 PM
8
7. It is given that f(x) = (x + 3) + 3 for which x . 0 and g(x) = for which x . 0.
2
Solve the equation fg(x) = 52. x
= – 1 [3]
2
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8. f(x) = ax + b, where a and b are constants, and a . 0.
It is given that f (x) = 49x − 40.
2
Find the values of a and b.
[3]
9. Sketch the graph of f(x) = 3 − x for which x . 0.
2
Sketch the inverse function of f(x) = 3 − x and find its equation.
2
y
x
[5]
10. Given the function f(x) = x + 3, and g(x) = x – 1 , x ≠ k.
2x – 1
(a) State the value of k.
[1]
Cambridge IGCSE
TM
4 Ace Your Additional Mathematics
Chapter 1 Add Math (1 to 12).indd 4 14/03/2022 12:16 PM
(b) Find
(i) gf(x).
[2]
(ii) fg(x).
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[2]
11. Given f(x) = 3x , x ≠ −1.
x + 1
(a) Find the function f .
−1
[3]
(b) State the value of x such that f does not exist.
−1
[1]
12. Given that f(x) = 5 − 2x and g(x) = 5x.
Find the value of
(a) f (−3).
−1
[2]
(b) (gf) (−5).
−1
[3]
Chapter 1 Functions 5
Chapter 1 Add Math (1 to 12).indd 5 14/03/2022 12:16 PM
13. Given the functions f(x) = 1 − 3x and g(x) = ax + b, where a and b are constants.
2
2
If the composite function gf(x) is given by gf(x) = x − x + 6, find the values of a and b.
2
3
[4]
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14. A function f is defined by f(x) = |3x − 4| − 2, for −2 < x < 4.
(a) Sketch the graph of y = f(x).
Show the coordinates of the points where the graph meets the coordinate axes.
y
= – 1
2
x
[3]
(b) State the range of f.
[1]
(c) Solve the equation f(x) = 3.
[2]
Cambridge IGCSE
TM
6 Ace Your Additional Mathematics
Chapter 1 Add Math (1 to 12).indd 6 14/03/2022 12:16 PM
(d) A function y is defined by g(x) = |3x − 4| − 2, for −2 < x < 4.
Write down a suitable domain for g for which g has an inverse.
−1
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[1]
15. The function f is defined by f(x) = |x − 10x + 9| for the domain 0 < x < 10.
2
(a) Sketch the graph of y = f(x).
Show the coordinates of the points where the graph meets the coordinate axes.
y
x
[3]
(b) Find the range of f.
[1]
(c) The function g is defined by g(x) = |x − 10x + 9| for the domain 3 < x < k.
2
Determine the value of k for which g (x) exists.
−1
[1]
Chapter 1 Functions 7
Chapter 1 Add Math (1 to 12).indd 7 14/03/2022 12:16 PM
1
16. The function f is defined by f(x) = 4 x − 1 2 2 − 5 for x R.
2
(a) Find the range of f.
[1]
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(b) State the coordinates of the turning point of the curves
(i) y = f(x).
= – 1
2
[1]
(ii) y = |f(x)|.
[1]
17. (a) Show that x + 4x + 9 can be written in the form of (x + a) + b and determine the values
2
2
of a and b.
[2]
(b) Sketch the graph of y = x + 4x + 9 and give the equation of its axis of symmetry.
2
y
x
[3]
Cambridge IGCSE
TM
8 Ace Your Additional Mathematics
Chapter 1 Add Math (1 to 12).indd 8 14/03/2022 12:16 PM
(c) The function f(x) = x + 4x + 9 has a domain set of all real numbers.
2
Find the range of f.
[1]
(d) Explain why f has no inverse with its given domain. Suggest a domain for f for which it
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has an inverse.
[2]
18. The function f is defined by f(x) = 2x − 1 for x R.
−1
(a) Sketch in a single diagram y = f(x) and y = f (x).
y
x
[3]
(b) The function g is defined by g(x) = −x − 2x for x R.
2
Express gf(x) in terms of x and find the maximum value of gf(x).
[3]
Chapter 1 Functions 9
Chapter 1 Add Math (1 to 12).indd 9 14/03/2022 12:16 PM
(c) The function h is defined by h(x) = −x − 2x for x > −1.
2
Express −x − 2x in the form of a − (x − b) and find the values of a and b.
2
2
[2]
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−1
(d) Express h (x) in terms of x.
= – 1
2 [2]
19. It is given that f(x) = 3x + 1 and g(x) = 2 − x.
(a) Find g(6).
[1]
1
(b) Find the value of s if f(s − 2) = g(6).
3
[2]
(c) Find gf(x).
[2]
Cambridge IGCSE
TM
10 Ace Your Additional Mathematics
Chapter 1 Add Math (1 to 12).indd 10 14/03/2022 12:16 PM
(d) Sketch the graph of y = |fg(x)| for −2 < x < 4 and state the range of y.
y
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x
[5]
3
20. It is given that f(x) = 3x − 4, g(x) = ; x ≠ 0.
x
Find the expression for each of the following functions.
(a) ff(x)
[2]
(b) gf(x)
[2]
Chapter 1 Functions 11
Chapter 1 Add Math (1 to 12).indd 11 14/03/2022 12:16 PM
(c) f (x)
−1
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[2]
(d) Calculate the value of x such that gf(x) = ff(x). Write your answer in exact form.
[3]
21. Given that f(x) = kx − h and f (x) = 16x − 6.
2
(a) Find the values of h and k, where h and k are both integers.
[4]
2
(b) Find the value of x for which f(x ) = 16x.
Write your answer correct to 2 decimal places.
[4]
Cambridge IGCSE
TM
12 Ace Your Additional Mathematics
Chapter 1 Add Math (1 to 12).indd 12 14/03/2022 12:16 PM
2 Quadratic Functions
Answer all questions.
If you need to show your work for any of the questions, write it clearly in the space provided below the
question.
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1. Solve the simultaneous equations.
2x – y = 1
2
3x + y = xy + 15
[5]
2. Solve the simultaneous equations.
4y = 7(3 – x)
2
y
– 3 = 0
x
[5]
3. The quadratic equation 3ax + 4ax + a = 1 − x has two equal roots. Find the value of a.
2
2
[5]
Chapter 2 Quadratic Functions 13
Chapter 2 Add Math (13 to 21).indd 13 14/03/2022 12:16 PM
4. Find the values of z for which the line y = z − 6x and the curve y = x(2x + z) never meet.
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[5]
5. The curve x + y = 1 + xy cuts the line 2x − y = k at two distinct points. Find the set of values
2
2
of k.
[5]
6. The curve xy + y = x + 11 intersects the line y = 2x − 1 at points P and Q. Find the equation
2
2
of the perpendicular bisector of the line PQ.
= – 1
2
[8]
Cambridge IGCSE
TM
14 Ace Your Additional Mathematics
Chapter 2 Add Math (13 to 21).indd 14 14/03/2022 12:16 PM
7. Solve the simultaneous equations. Write your answer in exact form.
3x + 5y = 8
2y 3x
– = 6
x y
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[6]
8. Find the value of s for which the line y + 4 = sx is a tangent to the curve y = 1 − x and find
the x-coordinate of the point at which this tangent touches the curve. 3x
[7]
9. The quadratic equation 3ax + 2ax + a = 4 − x has two distinct roots. Find the range of values
2
2
of a.
[5]
10. Solve the quadratic equation x(2x − 5) = 2x + 1. Give your answer correct to 3 decimal places.
[3]
Chapter 2 Quadratic Functions 15
Chapter 2 Add Math (13 to 21).indd 15 14/03/2022 12:16 PM
11. Show that these two lines are perpendicular and find their intersection points.
3y + 4x = 3
3
y = x – 2
4
= – 1
2
[5]
12. The following information refers to the equations of two straight lines, PQ and RS.
PQ: y – 6 = 2ax
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x y
RS: = 1 –
6b 3
Given that the straight lines PQ and RS are perpendicular to each other, express a in terms of b.
[4]
Cambridge IGCSE
TM
16 Ace Your Additional Mathematics
Chapter 2 Add Math (13 to 21).indd 16 14/03/2022 12:16 PM
13. The diagram shows the graph of a quadratic function y = –(x + h) + k.
2
y
x
0
–2 (3,–2)
(b) the equation of the axis of symmetry.Bhd. All Rights Reserved.
Find
(a) h and k.
[4]
Penerbitan Pelangi Sdn
[1]
(c) the coordinates of the maximum point.
[1]
Chapter 2 Quadratic Functions 17
Chapter 2 Add Math (13 to 21).indd 17 14/03/2022 12:16 PM
14. The diagram shows the graph of a quadratic function f(x) = 2(x − a) + 3 where a is a constant.
2
y
(1, b)
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x
The curve y = f(x) has the minimum point (1, b). Find the values of a and b.
= – 1
2
[2]
15. Given that the graph of the quadratic function f(x) = 3x − 5x + p does not intersect the x-axis.
2
Find the range of values of p.
[3]
16. Solve the equation x (2√6 ) − 7x + √6 = 0 and express your answer in the form of a . Find
2
a and b, where they are both integers. √b
[3]
Cambridge IGCSE
TM
18 Ace Your Additional Mathematics
Chapter 2 Add Math (13 to 21).indd 18 14/03/2022 12:16 PM
17. Find the range of values of k for which the equation x + 2k = 3 + kx has real roots.
2
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[5]
18. A line y = mx − 8 intersects the curve y = x − 3x − 4 at two distinct points. Find the range
2
of values of m.
[5]
19. f(x) = 9 + 4x − 2x for 0 < x < 5
2
(a) Express 9 + 4x − 2x in the form of a − b(x + c) , where a, b and c are constants.
2
2
[2]
(b) Find the coordinates of the turning point of the function f(x). State whether it is a
maximum or minimum point.
[1]
Chapter 2 Quadratic Functions 19
Chapter 2 Add Math (13 to 21).indd 19 14/03/2022 12:16 PM
(c) Find the range of f(x).
[1]
(d) State, giving a reason, whether or not f(x) has an inverse.
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= – 1
2
[1]
20. f(x) = 2x + 3x – 7, where x > k.
2
(a) Find the smallest value of k for which f(x) has an inverse.
[3]
(b) Find the coordinates of the turning point and state if it is a maximum or minimum point.
[1]
(c) Sketch the graph f(x) = |2x + 3x − 7| and state the coordinates of the points where the
2
curve meets the coordinates axes.
[3]
Cambridge IGCSE
TM
20 Ace Your Additional Mathematics
Chapter 2 Add Math (13 to 21).indd 20 14/03/2022 12:16 PM
21. (a) Sketch the graph y = (x − 7)(x + 1) and state the coordinates of the points where the
curve meets the coordinates axes.
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[3]
(b) Find the set of values of m for which |(x − 7)(x + 1)| = m has two solutions.
[2]
22. (a) Sketch the graph y = −2 + 8x − 2x and state the coordinates of the points where the
2
curve meets the coordinates axes.
[5]
(b) Find the value of s for which |(–2 + 8x − 2x )| = s has three solutions.
2
[1]
Chapter 2 Quadratic Functions 21
Chapter 2 Add Math (13 to 21).indd 21 14/03/2022 12:16 PM
Answers
5. y
1 Functions
1. (a) f(x) = 3 – √x + 8 10
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f(0) = 3 – √8
f(3) = 3 – √11 8
3 – √11 < f(x) < 3 – 2√2 6
[2]
(b) Let f (x) = y 4
−1
f(y) = x
x = 3 − √y + 8 2
√y + 8 = 3 − x 1
y + 8 = (3 − x) x
2
f (x) = (3 − x) − 8 –2 –1 0 1 2 3
−1
2
= x − 6x + 1 –1
2
Domain: 3 − √11 < x < 3 − 2√2
Range: 0 < f (x) < 3 [2]
−1
[4] (a) y = |4(−2)−1|
2. (a) f(0) = −4 y = |−9|
f(−10) = 496 y = 9
−4 < f(x) < 496 0 , y , 9
[3] [1]
(b) x > 0 [1] (b) 1 , c , 12
[2]
3. (a) y
6. y
12 y = |3x|
10
y = |x + 2|
6
8
6
x
–3
4
[2] (1, 3)
(b) 2x + 6 = 4 or 2x + 6 = −4 (–0.5, 1.5) 2
x = −1 or x = −5
[2] x
–6 –4 –2 0 2 4 6
2
4. x − 7x + 6 = 6 − x
x − 6x = 0 3x = x + 2 or 3x = –x – 2
2
x = 0 or x = 6 2x = 2 or 4x = –2
x − 7x + 6 = x − 6 x = 1 or x = –0.5
2
x − 8x + 12 = 0 y = 3 or y = 1.5
2
x = 6 or x = 2 (1, 3) or (−0.5, 1.5)
x = 0, 2 and 6 [3] [6]
Answers 165
Answers Add Math.indd 165 14/03/2022 12:29 PM
–1
2
2
7. 1 8 x + 3 + 3 = 52 11. (a) Let f (x) = y
3y
= x
2
2
1 8 x + 3 = 49 y + 1
3y = xy + x
x
= 1 8 + 3 = 7 3y – xy = x
y(3 – x) = x
2 x = 2 x
= √3 [3] y = 3 – x
x
2
2
–1
= 1 8 x + 3 8. a(ax + b) + b = 49x – 40 f (x) = 3 – x [3]
a x + ab + b = 49x – 40
2
2
= a x = 49x (b) x = 3
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= a = 7 or a = −7 (rejected) [1]
ab + b = –40 12. (a) Let f (x) = y
–1
8b = –40 f(y) = x
b = –5 5 – 2y = x
[3] 2y = 5 – x
9. Let y = f(x) y = 5 – x
Let f (x) = z 2
–1
f(z) = x f (x) = 5 – x
–1
x = 3 – z 2
2
z = 3 – x f (–3) = 5 – (–3)
2
–1
z = √3 – x 2
f (x) = √3 – x f (–3) = 4
–1
–1
[2]
y (b) gf(x) = 5(5 – 2x)
3 = 25 – 10x
y = f(x) [gf(x)] = 25 – x
–1
10
(gf) (–5) = 25 – (–5)
–1
10
= 3
–1
y = f (x) [3]
13. a(1 – 3x) + b = a(1 – 6x + 9x ) + b
2
2
x = a – 6ax + 9ax + b
2
3
9a = 1
[5] a = 1
9
10. (a) 2k – 1 = 0 a + b = 6
2k = 1 b = 6 – 1
k = 1 9
2 [1] = 53
(b) (i) gf(x) = x + 3 – 1 9 [4]
2(x + 3) – 1 14. (a)
= x + 2 y
2x + 6 – 1
= x + 2
2x + 5 [2]
(ii) fg(x) = x – 1 + 3 2
2x – 1
= x – 1 + 3(2x – 1)
2x – 1
= x – 1 + 6x – 3 0 2 2 x
2x – 1 3
7x – 4
=
2x – 1
[2]
[3]
Cambridge IGCSE
TM
166 Ace Your Additional Mathematics
= 1
Answers Add Math.indd 166 2 14/03/2022 12:29 PM
= √3
2
2
= 1 8 + 3
x
=
=
(b) −2 < f(x) < 8 (c) f(x) > 5
[1] [1]
(c) |3x – 4| – 2 = 3 (d) It has no inverse because it is not a one-to-one
|3x – 4| = 5 function.
3x – 4 = 5 or 3x – 4 = –5 x > −2
x = 3 or x = – 1 [2]
3 [2] 18. (a)
(d) x > 4 y y = f(x)
3 [1]
15. (a) y
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y = f (x)
–1
9 0.5
0.5 x
x [3]
1 9 (b) gf(x) = −(2x − 1) − 2(2x − 1)
2
[3] = –(4x – 4x + 1) – 4x + 2
2
(b) 0 < f(x) < 16 = –4x + 1
2
[1] Maximum value of gf(x) = 1
(c) k = 5 [3]
[1] (c) –x – 2x = (x + 2x + 1 – 1)
2
2
2
16. (a) f(x) > –5 = –(x + 1) + 1
[1] a = 1, b = –1 [2]
2
(b) (i) 1 1 2 , –5 (d) Let y = h (x)
–1
[1] h(y) = x 2
2
(ii) 1 1 2 , 5 –(y + 1) + 1 = x
(y + 1) = 1 – x
2
–1
[1] h (x) = –1 + √1 – x
[2]
17. (a) x + 4x + 2 – 2 + 9 19. (a) g(6) = 2 – 6
2
2
2
= (x + 2) + 5 = –4
2
a = 2, b = 5 [1]
[2] 1
(b) x = –2 (b) 3(s – 2) + 1 = (–4)
3
y 3s – 6 + 1 = – 4
14 3
3s – 5 = – 4
12 3
3s = 11
10 3
11
8 s =
9 [2]
6
(c) g(3x + 1) = 2 – (3x + 1)
4 = 2 – 3x – 1
= 1 – 3x
2 [2]
(d) fg(x) = 3(2 – x) + 1
x
–4 –2 0 2 fg(x) = –3x + 7
when x = –2, |fg(x)| = 13
[3]
when x = 4, |fg(x)| = 5
Answers 167
Answers Add Math.indd 167 14/03/2022 12:29 PM
(b) f(x ) = 16x
2
y
kx – h = 16x
2
(–4)x – (–2) – 16x = 0
2
12
–4 – 16x + 2 = 0
2
2x + 8x – 1 = 0
2
10 x = 0.12, –4.12
[4]
= 1 8
2 7 2 Quadratic Functions
= √3 6
1. y = 2x − 1
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= √(7) – 4 3x + (2x − 1) = x(2x − 1) + 15
2
2
2
3x + 4x − 4x + 1 − 2x + x − 15 = 0
2
2
2
5x − 3x − 14 = 0
2
2
2
= 1 8 2 + 3 2 (x − 2)(5x + 7) = 0
= x x = 2 or x = − 7
5
= –2 0 2 7 4 19
3 y = 3 or y = −
Range: 0 < fg(x) < 13 [5] 5
[5]
20. (a) ff(x) = 3(3x – 4) – 4
= 9x – 12 – 4 2. y = 3
= 9x – 16 x
[2] y = 3x
2
(b) gf(x) = 3 [2] y(3x) = 21 − 7x
2
3x – 4 36x + 7x − 21 = 0
2
(c) Let f (x) = y x = –(7) ± √(7) – 4(36)(–21)
–1
f(y) = x 2(36)
x = 3y – 4 x = 0.673 or x = −0.87
x + 4 = 3y y = 2.02 or y = −2.6
x + 4 [5]
y =
3 3. 3ax + 4ax + a + x − 1 = 0
2
2
2
f (x) = x + 4 (3a + 1)x + 4ax + a − 1 = 0
–1
3 [2] b − 4ac = 0
2
2
(d) 3 = 9x – 16 (4a) − 4(3a + 1)(a − 1) = 0
2
2
3x – 4 16a − 12a + 8a + 4 = 0
3 = 27x – 48x – 36x + 64 a + 2a + 1 = 0
2
2
27x – 84x + 61 = 0 (a + 1) = 0
2
2
14 + √13 14 – √13 a = −1
x = or x = [5]
9 9
[3]
4. z − 6x = x(2x + z)
2
21. (a) f (x) = k(kx – h) – h 2x + zx + 6x − z = 0
2
= k x – kh – h 2x + (z + 6)x − z = 0
2
2
k x = 16x b − 4ac , 0
2
2
k = ±4 (z + 6) − 4(2)(−z) , 0
2
–kh – h = 6 z + 12z + 8z + 36 , 0
2
4h + h = 6 z + 20z + 36 , 0
2
5h = 6 (z + 2)(z + 18) , 0
h = 6 −18 , z , −2
5 [5]
4h – h = –6
3h = –6
h = –2
k = –4, h = –2
[4]
Cambridge IGCSE
TM
168 Ace Your Additional Mathematics
Answers Add Math.indd 168 14/03/2022 12:29 PM
5. y = 2x − k [3(–13) + 3]x – 12x – 1 = 0
2
x + (2x − k) = 1 + x(2x − k) –36x – 12x – 1 = 0
2
2
2
x + 4x – 4xk + k – 1 – 2x + kx = 0 (6x + 1) = 0
2
2
2
2
2
3x – 3kx + k – 1 = 0 x = – 1
2
2
6
b – 4ac . 0 [7]
2
(–3k) – 4(3)(k – 1) . 0 9. (3a + 1)x + 2ax + a − 4 = 0
2
2
2
9k – 12k + 12 . 0 b – 4ac . 0
2
2
2
3k , 12 (2a) – 4(3a + 1)(a – 4) . 0
2
2
k , 4 2a – 11a – 4 , 0
2
2
–2 , k , 2 11 – 3√7 11 + 3√7
[5] 4 , a , 4
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6. x(2x − 1) + (2x − 1) = x + 11 [5]
2
2
2
x – x – 2 = 0 10. 2x – 5x – 2x – 1 = 0
2
2
(x – 2)(x + 1) = 0 2x – 7x – 1 = 0
x = 2 or x = –1 x = 3.637 or x = −0.137
y = 3 or y = –3 [3]
3
Midpoint of PQ 11. 3( x – 2) + 4x = 3
,
= 1 2 – 1 3 – 3 2 4 9x + 4x = 9
2
2
4
2
= 1 1 2 , 0 x = 36
25
m m = –1 Intersection point: x = 36 , y = – 23
25
25
2
1
m = –1 ÷ 1 –3 – 3 2 3y = 3 – 4x
2
–1 – 2
4
= – 1 y = – x + 1
3
2
3
m = , m = – 4
2
1 2
1
0 = – 1 1 + c 4 3
2 2 m × m = × – 4
3
c = 1 1 2 4 3
4 = –1
1
y = – x + 1 Both lines are perpendicular to each other.
2 4 [8] [5]
7. x = 8 – 5y 12. y = 2ax + 6
3 m = 2a
1
3 1 8 – 5y 2 y = 1 – x
2y – 3 = 6 3 6b
8 – 5y y x
3 y = – 2b + 3
6y – 8 – 5y = 6 1
8 – 5y y m = – 2b
2
–19y + 80y – 64 = 6
2
y(8 – 5y) m × m = –1
2
1
1
11y + 32y – 64 = 0 2a × – 1 2 = –1
2
–16 + 8√15 –16 – 8√15 2b
y = or y = a = b
11 11 [4]
x = 168 – 40√15 or y = 168 + 40√15 2
33 33 [6] 13. (a) −2 = −(0 + h) + k
−2 = −h + k
2
2
8. sx – 4 = 1 – x k = h – 2
3x 2
3sx – 12x = 1 – 3x –2 = –(3 + h) + k
2
2
2
2
(3s + 3)x – 12x – 1 = 0 −2 = −(3 + h) + h − 2
2
b – 4ac = 0 6h = −9 3
2
(–12) – 4(3s + 3)(–1) = 0 h = – 2
2
144 + 12s + 12 = 0
s = –13
Answers 169
Answers Add Math.indd 169 14/03/2022 12:29 PM
2
1
k = – 3 2 2 2 – 2 20. (a) 2x + 3x − 7
3
2
1
2
1 = 2 x + x – 7
= 2
4 2
31
[4] = 2 x + 3 2 – 9 4 – 7
(b) x = 0 + 3 = 1.5 [1] 4 2 16
1
2 = 2 x + 3 2 – 65
,
(c) Maximum point 1 3 1 2 [1] k = – 3 4 8
2 4
4
14. a = 1, b = 3 3 65 [3]
1
4
= 1 [2] (b) Minimum point – , – 8 2
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2 15. 3x − 5x + p = 0 [1]
2
= √3 b – 4ac , 0 (c)
2
(−5) − 4(3)(p) , 0 y
2
= √(7) – 25 − 12p , 0
2
12p . 25 – 3 , 65
4 8
2
2
= 1 8 2 + 3 p . 25 [3]
12
= –(–7) ± (−7) − 4(2√6 )(√6 )
2
= 16. x = 2(2√6) 7
x = 2 or x = 3
√6 2√6 x
3
a = 2 or a = (rejected) – 3 – 65 – 3 + 65
4
4
b = 6 2
[3] [3]
17. x – kx + 2k – 3 = 0 21. (a) y
2
b − 4ac . 0
2
(−k) − 4(1)(2k − 3) . 0 x
2
k − 8k + 12 . 0 –1 7
2
(k − 6)(k − 2) . 0
k , 2, k . 6
[5]
18. mx − 8 = x − 3x − 4 –7
2
x − (3 + m)x + 4 = 0
2
b − 4ac . 0
2
[−(3 + m)] − 4(1)(4) . 0 2 [3]
2
9 + 6m + m − 16 . 0 (b) x − 6x − 7
2
2
m + 6m − 7 . 0 = (x − 3) − 16
2
(m + 7)(m − 1) . 0 m > 16
m , −7, m . 1 [2]
[5] 22. (a) y = –2(x – 4x) – 2
2
2
19. (a) 9 + 4x − 2x = −2[(x − 2) − 4] − 2
2
2
= −2x + 4x + 9 = −2(x − 2) + 6
2
= −2[(x − 1) − 1] + 9 y
2
= −2(x − 1) + 11 (2, 6)
2
= 11 − 2(x − 1)
2
[2]
a = 11, b = 2, c = −1
(b) Maximum point (1, 11) [1] x
–2 2 – 3 2 + 3
(c) −21 > f(x) > 11 [1]
(d) No, it does not have an inverse because it is not
a one-to-one function. [1]
[5]
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TM
170 Ace Your Additional Mathematics
Answers Add Math.indd 170 14/03/2022 12:29 PM
(b) s = 6 [1] 5. –x < 8 – 9x 8 – 9x < x
8x < 8 8 < 10x
3 Equations, Inequalities and Graphs x < 1 x > 4
5
4 < x < 1
1. 5x – 6 = 10 – x 5 [2]
6x = 16 6. |3x + 2| . 7 – 2x
x = 8 3x + 2 , –(7 – 2x) 3x + 2 . 7 – 2x
3
x , –9 5x . 5
5x – 6 = –(10 – x) x . 1
4x = –4 [2]
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x = –1
2
2
[2] 7. (x – 5) . 1 1 x + 2
2
3
2. 8 = 28 1 4
2
2
x + 4 x – 1 x – 10x + 25 . x + x + 4
9
3
8x – 8 = 28x + 112 8 x – 34 x + 21 . 0
2
20x = –120 9 3
x = –6 8x – 102x + 189 . 0
2
8 = –28 (4x – 9)(2x – 21) . 0
9
x + 4 x – 1 x , and x . 21
8x – 8 = –28x – 112 4 2 [3]
36x = –104
x = – 26 8. |2 – x| < 3|2x + 4| – 5
9 [2] 2 – x < 3|2x + 4| – 5
7 – x < 3|2x + 4|
3. |x – 3| = 2 + |x + 1| 7 – x < |2x + 4|
x – 3 = 2 + |x + 1| 3 7 – x
|x + 1| = x – 5 2x + 4 > 3
x + 1 = x – 5 x + 1 = –x + 5 6x + 12 > 7 – x
0 = –6 (false) 2x = 4 7x > –5
x = 2 (rejected) 5
x – 3 = –2 – |x + 1| x > – 7
|x + 1| = 1 – x
x + 1 = 1 – x x + 1 = x – 1 2x + 4 > x – 7
2x = 0 0 = –2 (false) 3
x = 0 6x + 12 > x – 7
5x > –19
[4] 19
x > –
1
4. |1 – 2x| = |3x – 24| 5
6 2 – x < 5 – 3|2x + 4|
1
1 – 2x = (3x – 24) –3 – x < –3|2x + 4|
6 3 + x
6 – 12x = 3x – 24 |2x + 4| > 3
x = 2 3 + x
1 2x + 4 > 3
1 – 2x = – (3x – 24)
6 6x + 12 > 3 + x
–6 + 12x = 3x – 24 5x > –9
9
x = –2 x > – (rejected)
When x = 2, 5
y = |1 – 2(2)| 2x + 4 > 3 + x
y = 3 –3
–6x – 12 > 3 + x
When x = –2, –15 > 7x
y = |1 – 2(–2)| x < – 15 (rejected)
y = 5 7
5
[4] The solution is x > – or x < – 19 .
7 5 [3]
Answers 171
Answers Add Math.indd 171 14/03/2022 12:29 PM
9. y 12. (a) When x = 1
3(1) – 6(1) – 3(1) + 6 = 0
3
2
x – x – 2
2
3
2
x – 1 x – 2x – x + 2
3
2
x – x
2
15 –x – x
2
–x + x
–2x + 2
x
0.25 3 5 –2x + 2
y = |3x – 6x – 3x + 6|
2
3
= |3(x − 2x − x + 2)|
2
3
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= |3(x − 1)(x − x − 2)|
2
= |3(x − 1)(x + 1)(x − 2)|
[3]
y = x – 7x + 10
2
10. (a) y y = (x – 5)(x – 2)
y
10
x
1 2 4
–24 6
[3]
x
(b) y –1 1 2 5
[8]
(b) 3(x – 1)(x + 1)(x – 2) = (x – 5)(x – 2)
3x – 3 = x – 5
2
24 3x – x + 2 = 0 (rejected)
2
x 3(x – 1)(x + 1)(x – 2) = –(x – 5)(x – 2)
1 2 4
3x – 3 = –x + 5
2
3x + x – 8 = 0
2
x = 1.47
y = (1.475) – 7(1.475) + 10
2
[2] y = 1.85
or
11. y x = –1.81
y = (–1.808) – 7(–1.808) + 10
2
y = 25.92
Intersection points
(1.47, 1.85), (–1.81, 25.92) and (2, 0)
[4]
13. y = –(x + 1)(x – 1)(x – 2)
y = –(x – 1)(x – 2)
2
x
–5 0 5 a = –1, b = –1, c = –2
[3]
[3]
14. 2 = a(2)(–2)(–3)
a = 1
6
b = 2
c = –2
d = –3
Cambridge IGCSE
TM
172 Ace Your Additional Mathematics
Answers Add Math.indd 172 14/03/2022 12:29 PM
3
1
Gradient of straight line = – ÷ 3 Let x = y
2
2 5 5y + 32y – 21 = 0
2
5
3
= – 6 y = or y = –7
e = – 5 3 5 3
3
6 x = or x = –7 (rejected)
2
2
f = – 1 5
2 [4] x = 0.711 [7]
15. (a) x . 1 20. Let √x = y
[1] 20
(b) (x – 1)(x + 3) . –6 6y – y = 7
2
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–4.1 , x , –1.4, x . 0.5 6y – 7y – 20 = 0
2
[3] (3y + 4)(2y – 5) = 0
16. (a) –1.6 < x < 0.6, x > 3 y = – 4 or y = 5
[2] 3 2
4
(b) –1.5 till –1.6 , x , 1 √x = – (rejected) or √x = 5
x . 2.5 till 2.6 3 2
[4] = 6.25
[5]
17. (a) a = –1
b = 2
c = –1 4 Indices and Surds
d = 1
[2] 1. 3 3 + 3 ÷ 3 – 3 = k(3 )
n
n
n
n
(b) –1.1 < x < –0.9 till –0.8 ˙
1
2
1
0 < x < 0.8 till 0.9 3 3 + – 1 = k(3 )
n
n
[2] 3
1
k = 3 + – 1
18. Let x = y 7 3
2
–2y + 9y – y – 12 = 0 = 3
2
3
Let y = –1 [2]
–2(–1) + 9(–1) – (–1) – 12 = 0 2. 4 4 – 4 = 24
2
3
x
x
˙
x
2
–2y + 11y – 12 4 (4 – 1) = 24
x
2
3
y + 1 –2y + 9y – y – 12 4 = 8
2x
3
–2y – 2y 2 = 2
3
2
11y – y x = 1.5
2
11y + 11y [2]
2
– 12y – 12 3. 3(27 3 ) + 4(4 2) = 4
x
x
–2
x
˙
˙
x
x
–1
x
– 12y – 12 27 3 = 4 + (4 8)
˙
˙
x
–1
x
˙
(y + 1)(–2y + 11y – 12) = 0 27 3 = 4 (1 + 8)
2
(y + 1)(2y – 3)(4 – y) = 0 4 x = 3 –1
3 27 x 9
y = –1 or y = or y = 4
4
2 1 2 x = 1
x = –1 (rejected) or x = ± 3 or x = ±2 27 27
2
2 [7] p = 1, q = 27
[3]
1 3
2
2
19. x (5x + 32x – 21) = 0 4. (√5x + 1) = (5 – √x – 2)
2
3
1 5x + 1 = 25 – 10√x – 2 + x – 2
2
x = 0 (4x – 22) = (–10√x – 2)
2
2
x = 0 16x – 276x + 684 = 0
2
3 x = 14.25 (rejected) or x = 3
2
5x + 32x – 21 = 0
3
[4]
Answers 173
Answers Add Math.indd 173 14/03/2022 12:29 PM
5. 3 2 = 6 12. 8√3 – √3 + 8√3
1
6x
6x
2
˙
6 = 6 5
6x
2
6x = 2 79√3
1 = 5
x = [2]
3
[2] 4√2 – 3√3
– 1 13. h =
3
6. 4 2a – b = (8 b – 3 ) (√3 – √2) 2
2 4a – 2b = 2 3 – b = 4√2 – 3√3 × 5 + 2√6
4a – 2b = 3 – b 5 – 2√6 5 + 2√6
b = 4a – 3 20√2 + 8√12 – 15√3 – 6√18
= 1 5 – 3a =
2 2 = 4a – 3 25 – 24
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= √3 a = 1 = 20√2 + 16√3 – 15√3 – 18√2
b = 1 = √3 + 2√2
= √(7) – [4] [4]
2
x
x
x
7. (16x )(2 3 − a )(x ) = bx 14. 5(8 · 4) − 2(9 ÷ 9) = 8 x
−4
2a
−2 3 − a
x
x
2
2
4
2a
= 1 8 + 3 (2 )(x )(2 3 − a )(x −6 + 2a ) = bx −4 5(8 · 4)−(8 ) = 2(9 ÷ 9)
2
x 2
7− a 4a − 6
−4
x
1 2 x = bx 8 (20 – 1) = 9 1 2
9
= 2 2 4a – 6 = –4 9 x 171
= a = 0.5 1 2 = 2
8
b = 2 7 – 0.5 [3]
b = 90.51 15. (a) PR = PS + SR
2
2
2
[4] = (√20 + √5) + (√20 – √5)
2
2
1 x PR = √50
2
8. Let 5 = y = 5√2
3y + 5y – 2 = 0 [2]
2
(3y – 1)(y + 2) = 0 (b) tan ∠PRQ = √20 + √5
1
y = or y = –2 √20 – √5
3 20 + 2√100 + 5
1 x 1 1 x =
2
5 = or 5 = –2 20 – 5
2
3 = 3
x = –2 log 3 (rejected) or x = −1.365 [3]
5
[5] √5 + 2 √5 + 2
9. 27 x = 8 4 16. √5 – 2 × √5 + 2
x
˙
7 x 5 + 4√5 + 4
27 x = 4 = 5 – 4
7 8 x = 9 + 4√5
x
˙
27
1 2 x = 4 [3]
56
x = –1.9 17. 1 – √7 – 2(1 + √7 )
[3] (1 + √7 )(1 – √7 )
10. 5 + √6 – 5 + √6 = –3√7 – 1
1 – 7
(5 – √6)(5 + √6)
= 2√6 = 3√7 + 1
25 – 6 6
2√6 1 1
= = √7 +
19 2 6
[2] 1
11. 2√5 – 5√2 × 2√5 – 5√2 a = 6
2√5 + 5√2 2√5 – 5√2 b = 1
= 20 – 20√10 + 50 2 [3]
20 – 50
2
2
= 7 – 2√10 or 2√10 – 7 18. (a) PQ = (2√10) = 40
–3 3 QR = (1 + 3√3) = 28 + 6√3
2
2
2
7
2
2
= – + √10 PR = (√3 – 3) = 12 – 6√3
3 3 [3]
Cambridge IGCSE
TM
174 Ace Your Additional Mathematics
Answers Add Math.indd 174 14/03/2022 12:29 PM
QR + PR = PQ 24. 3x√15 = x√5 + √15
2
2
2
28 + 6√3 + 12 – 6√3 = 40 √5(3x√3) = √5(x + √3)
3x√3 – x = √3
It is proven that triangle PQR is a right-angled x(3√3 – 1) = √3
triangle using the Pythagorean theorem.
∠R is 90°. x = √3
[2] 3√3 – 1
1 + 3√3 9 + √3
(b) tan ∠QPR = =
√3 – 3 26
= 1 + 3√3 × √3 + 3 a = 9, b = 3, c = 26
√3 – 3 √3 + 3 [4]
= √3 + 3 + 9 + 9√3 25. 1 + 1 + 1
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3 – 9 1 + x b + x a 1 + x c + x b 1 + x a + x c
–6 – 5√3 x c x c x a x a x b x b
=
3 1 1 1
[3] = + +
x + x + x a x + x + x b x + x + x c
b
a
c
a
b
c
19. Let y = 3 x x c x a x b
x 2
x
2
x
8(3 ) + (3 ) · (3 ) = 3 + 6 x + x + x b
c
a
2
8y + 9y − y + 6 = 0 = x + x + x c
b
a
y = 0.515 or y = −1.29 = 1
x
x
3 = 0.515 or 3 = −1.29 (rejected) [3]
x = −0.604
[5]
8
20. 2u + = 17 5 Factors of Polynomials
u
2u – 17u + 8 = 0 1. When x = −1
2
u = 1 or u = 8 x − 3x − x + 3
3
2
2 3 2
2 = 2 or 2 = 8 = (−1) − 3(−1) − (−1) + 3
–1
x
x
x = –1 or x = 3 = 0
2
[4] x – 4x + 3
x + 1 x – 3x – x + 3
3
2
21. 864 = 2 × 3 x + x
5
3
2
3
√864 = √2 × 3 –4x – x
3
5
2
= 12√2√3 –4x – 4x
2
[3] 3x + 3
22. 3 × 30√3 – 9 3x + 3
2 3 √3
3
2
= 15√3 – 9√3 x − 3x − x + 3 = 0
2
3 (x + 1)(x − 4x + 3) = 0
= 12√3 (x + 1)(x − 3)(x − 1) = 0
[2] x = 1 or x = −1 or x = 3
23. 3 – 2 2 3 [5]
4 2. When x = 2
2 3
2
2
2 4 −(2) − (m + 1)(2)+ m + m = 0
3 – (2 × 2 ) m − m − 6 = 0
3
3
2
= 4 (m − 3)(m + 2) = 0
2 3 m = 3 or m = −2
= – 1 [3]
4
2 3 3. 4x + 2x − 5x − 7
2
5
4
a = 2 = 16 = 4 1 2 5 + 2 1 2 4 – 5 1 2 2 – 7
1
1
4
1
[3] 2 2 2
= –8
[2]
Answers 175
Answers Add Math.indd 175 14/03/2022 12:29 PM
4. (a) 2x + 7x + 3 (b) 3x + 1
2
= (2x + 1)(x + 3) x – 5 3x + x – 15x – 5
3
2
2
2
3
1
1
When x = − , f – 1 2 = 0 3x + x – 15x
2
3
2 2 3x + x – 15x – 5
3
2
1
1
1
2 – 1 2 2 3 + A – 1 2 2 + B – 1 2 – 6 = 0 3x + x – 15x – 5
2
2
2
3
2
= 1 (3x + x − 15x − 5) ÷ (x − 5)
2 25 − A + 2B = 0 = 3x + 1
2
= √3 When x = −3, f(−3) = 0. f(x) = (3x + 1)(x − 5)
2(−3) + A(−3) + B(−3) − 6 = 0 [4]
3
2
= √(7) – 20 − 3A + B = 0 8. (a) f(−2) = (−2) − 3(−2) + 2
2
3
Solve the simultaneous equations to get: = 0 (shown)
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2
2
= 1 8 + 3 A = 3, B = −11 [6] (b) x + 0x – 2x + 1 [1]
2
2
3
1 (b) x + 2 x + 0x – 3x + 2
2
3
= 2 2 x – 2 x + 2x
3
2
2
3
2
= 2x + 7x + 3 2x + 3x – 11x – 6 3 2
2x + 7x + 3x x – 2x – 3x
2
3
3
2
– 4x – 14x – 6 x – 2x – 4x
2
3
2
– 4x – 14x – 6 x – 2x – 3x + 2
2
x – 2x – 3x + 2
3
2
f(x) = (x − 2)(2x + 1)(x + 3) f(x) = (x + 2)(x − 2x + 1)
2
[4] = (x + 2)(x − 1)
2
5. 3(1) + (1) − A(1) + B(1) − 6 = 0 [4]
2
3
4
3 + 1 − A + B − 6 = 0 9. (a) When x = 1, f(1) = 0
−A + B = 2 1 + a − b + 10 = 0
3(2) + (2) − A(2) + B(2) − 6 = 20 a − b = −11
3
4
2
48 + 8 − 4A + 2B − 6 = 20 When x = −1, f(−1) = 12
−2A + B = −15 −1 + a + b + 10 = 12
Solve the simultaneous equations to get: a + b = 3
A = 17, B = 19 Solve the simultaneous equations to get:
[5] a = −4, b = 7
[5]
6. When x − 3 = 0 (b) f(x) = 0
2
x = ±√3 (x − 1)(x − 3x − 10) = 0
2
When x = √3, (x − 1)(x − 5)(x + 2) = 0
Remainder x = 1 or x = 5 or x = −2
= 2(√3) + 5(√3) − 7(√3) + 1 [4]
4
6
2
= 79 10. f(2) = 25
When x = −√3, a(2) − 4(2) − b(2) + 3 = 25
2
3
Remainder 4a − b = 19
= 2(−√3) + 5(−√3) − 7(−√3) + 1 f(1) = 3
6
4
2
= 79 a(1) − 4(1) − b(1) + 3 = 3
3
2
[3] a − b = 4
7. (a) x − 5 = 0 Solve the simultaneous equations to get:
2
x = √5 or x = −√5 a = 5, b = 1
When x = √5 , f(√5 ) = 0. [5]
3(√5 ) + a(√5 ) + b(√5 ) − 5 = 0 11. When x = 2
2
3
2
3
When x = −√5 , f(−√5 ) = 0. (2) − (2) − 14(2) + 24 = 0
3
2
3(−√5 ) + a(−√5 ) + b(−√5 ) − 5 = 0 x + x + x – 12
3
2
3
x – 2 x – x – 14x + 24
2
Solve the simultaneous equations to get: x – 2x
3
2
a = 1, b = −15 x – x – 14x
2
3
[5]
x – 2x – 2x
2
3
x – 2x– 12x + 24
3
3
x – 2x– 12x + 24
Cambridge IGCSE
TM
176 Ace Your Additional Mathematics
Answers Add Math.indd 176 14/03/2022 12:29 PM
(x − x − 14x + 24)÷ (x − 2) 15. –x – 2x + 5x – 10x + 7
3
2
4
2
3
= x + x − 12 2x – 3 4x + 4x – 35x + 44x – 21
2
3
4
2
= (x − 3)(x + 4) 4x – 6x
4
3
4
4x 10x – 35x + 10
2
3
1 1 1
+ + 4x 10x – 15x + 10
3
2
4
a b c 4x 10x – 20x + 44x
3
2
4
= (x + 4)(x − 2) + (x − 3)(x − 2) + (x − 3)(x + 4) 4x 10x – 20x + 30x
4
3
2
(x − 3)(x + 4)(x − 2) 4x 10x – 35x + 14x – 21
2
3
4
2
= 3x − 2x − 14 4x 10x – 35x + 14x – 21
3
4
2
(x − 3)(x + 4)(x − 2) [6]
1 [3]
12. x = ± √2 16. –x – 3x – 3x + 3x + 5
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4
2
3
1
4
2
2
3
x – = 0 –x + 0x + 2 –x – 3x – 3x + 6x + 10
2
2 –x + 0x + 2x
4
3
2
2x – 1 = 0 –x – 3x – 5x + 6x + 10
2
3
2
4
8x + 2x +4x + x + 3 –x – 3x – 0x + 6x + 10
3
4
2
4
2
3
2x – 1 8x + 2x + 2x – x – 3 –x – 3x – 5x + 10
4
2
3
2
4
2
3
8x + 2x – 4x – x – 3 –x – 3x – 3x + 10
4
3
2
2
3
4
8x + 2x + 6x – x – 3 [3]
2
3
4
8x + 2x + 2x – x – 3
2
4
3
8x + 2x + 6x – x – 3 17. When x = –2,
3
2
4
3
2
8x + 2x + 6x – x – 3 f(−2) =(−2) − 4(−2) − 7(−2) + 10
4
3
2
= 0
4
(8x + 2x + 2x − x − 3) ÷ (2x − 1) = 4x + x + 3 x – 4x – 6x + 5
2
2
2
3
2
3
a = 2, b = −1, c = 4, d = 1, e = 3 x + 2 x – 4x – 7x + 10
2
3
[5] x + 2x
3
2
3
2
13. Using long division x – 6x – 7x
8x + 2x + x – 3x + 2 x – 6x – 12x
3
2
2
3
4
x + 0x – 2 x – 3x + 0x + 6x – 4 x – 4x – 5x + 10
2
2
3
4
3
2
x + 0x – 2x x – 4x – 5x + 10
2
3
4
2
3
x – 3x + 2x + 6x
2
3
4
2
x – 3x + 0x + 6x f(x) = (x + 2)(x − 6x + 5)
3
4
2
x – 3x + 2x + 6x – 4 = (x + 2)(x − 1)(x − 5)
2
3
4
x – 3x + 2x + 6x – 4 x = −2 or x = 1 or x = 5
3
4
2
(x − 3x + 6x − 4) ÷ (x − 2) a = 5, b = 1
4
2
3
= x − 3x + 2 [6]
2
=(x − 2)(x − 1) 18. (a) f(0) = –6
2
3
Factors of x − 3x + 6x − 4 p(0) + q(0) + r(0) + s = –6
4
3
= (x − 2)(x − 1)(x − 2) s = –6 (shown)
2
[4] [1]
(b) p + q + r − 6 = 0
14. −8p + 4q − 2r − 6 = 60
x – 5x + 12x – 7x + (26 – a)
3
2
4
x + 2x + a x – 5x + 12x + 8x – 9 −27p + 9q − 3r − 6 = 240
2
4
2
3
x + 2x + ax Solve the simultaneous equations to get:
4
2
3
x – 7x + (12 – a)x + 8x p = −9, q = 4, r = 11
3
2
4
x – 7x – 14x – 7ax [6]
2
4
3
2
x – 5x (26 – a)x + (8 + 7a)x – 9 (c) f(x) = −9x + 4x + 11x − 6
3
3
4
2
2
x – 5x (26 – a)x + (52 – 2a)x + a(26 – a) = (x − 1)(−9x − 5x + 6)
4
3
2
2
Remainder: (9a – 44)x – 9 – a(26 – a) = −(x − 1)(9x + 5x − 6)
(9a − 44)x = −71x a = 9, b = 5, c = −6
9a − 44 = −71 [3]
a = −3 19. (a) (−3) + a(−3) − 17(−3) + b = 0
3
2
b = −9 − a(26 − a) 9a + b = −24
3
2
= −9 − 26a + a (−1) + a(−1) − 17(−1) + b = 0
2
= −9 − 26(−3) + (−3) a + b = −16
2
= 78 [2]
[6]
Answers 177
Answers Add Math.indd 177 14/03/2022 12:29 PM
(b) Solve the simultaneous equations to get: 23. (2x − 1)(bx + cx + d)
2
a = −1, b = −15 = 2bx + (2c − b)x + (2d − c)x − d
3
2
f(x) = x − x − 17x − 15 Form the equations:
3
2
(x − x − 17x − 15)÷[(x + 3)(x + 1)] 2b = 4
2
3
= (x – x − 17x − 15) ÷ (x + 4x + 3) 2c − b = −8
3
2
2
= x – 5 2d − c = a
2
3
x – x – 17x – 5
2
x + 4x + 3 x – x – 17x – 15 −d = 18
2
3
3
2
x + 4x + 3x Solve the simultaneous equations to get:
x – 5x – 20x – 15 a = −33
2
4
x – 5x – 20x – 15 b = 2
4
2
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f(x) = (x + 3)(x + 1)(x – 5) c = −3
[5] d = −18
(c) y [5]
24. f(1) = 12
1 + a + b − 4a + 12 = 12
−3a + b = −1
f(4) = 60
3
2
4
x (4) + a(4) + b(4) − 4a(4) + 12 = 60
–3 –1 5
3a + b = −13
–15 Solve the simultaneous equations to get:
a = −2, b = −7
[5]
[3]
3
(d) x + x − 4x − 4 = x − x − 17x − 15
3
2
2
2
2x + 13x + 11 = 0 6 Logarithmic and Exponential Functions
,
A(–1, 0); B 1 –11 –945 2 1. y = 4
2
8
b
[3]
b = log y
4
20. (−1) + a(−1) − 7(−1) + 6 = (−2) + a(−2) − 7(−2) + 6 x 2
3
2
2
3
3a = 0 log 4y
4
a = 0 = 2 log x − (log 4 + log y)
[3] 4 4 4
= 2a − (1 + b)
21. f(k) = 2 = 2a − 1− b
k + k(k) − 5k = 2 [5]
3
k + k − 5k − 2 = 0 2. 2 + 8(log 4) = log x
3
2
When k = 2, Let u = log x 4
x
(2) + (2) − 5(2) − 2 = 0 4 1
3
2
(k + k − 5k − 2) ÷ (k − 2) 2 + 8 1 2 = u
3
2
u
= k + 3k + 1 u − 2u − 8 = 0
2
2
k = 2 or k = –3 + √5 or k = –3 – √5 (u − 4)(u + 2) = 0
2 2 [6] u = 4 or u = −2
1
22. 3x − 11x − 6x + 8 = 0 x = 256 or x = 16
2
3
3x + 3x + x – 2 [5]
2
3
x – 4 3x – 11x – 6x + 8
2
3
3x – 12x 3. log x = log (2x + 3)
2
3
9
3
3x – 11x – 6x log (2x + 3)
3
2
3
3x – 11x – 4x log x = log 3 2
3
2
3
3x – 11x – 2x + 8 3
3
2
3x – 11x – 2x + 8 1
3
2
x = (2x + 3) 2
(x − 4)(3x + x − 2) = 0 x − 2x − 3 = 0
2
2
(x − 4)(3x − 2)(x + 1) = 0 x = 3 or x = −1 (rejected)
x = 4 or x = –1 or x = 2 [4]
3
[5]
Cambridge IGCSE
TM
178 Ace Your Additional Mathematics
Answers Add Math.indd 178 14/03/2022 12:29 PM
4. 3 + log a = log (b + 4) 10. log(6x – 19x + 5) – 2 log x = 1
2
2 8
1 log(6x − 19x + 5) − log x = log 10
2
2
log 2 + log a = log (b + 4) 3
3
2 2 2 6x – 19x + 5
2
1 = 10
8a = (b + 4) 3 x 2
1 (4x − 1)(x + 5) = 0
a = (b + 4) 3 x = −5 (rejected) or x = 1
8 [4] 4
[5]
log √x
5. log y – 3 = log 3
4
3
3 log 3 2 3 11. log (xy ) = 10
a
3
y log x + 3 log y = 10
a
a
1 = 3 3 2
4
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log (x y ) = 16
(√x) 2 1 3 log x + 2 log y = 16
a
y = 81x 4 a a
[5] Solve the simultaneous equations to get: = 1
log y = 2 2
a
1 log x = 4
6. log (x + y) − log 2 = (log xy) a = √3
2
a a a log √xy
2
x + y = (xy) 1 2 1 a = √(7) –
2 = [log x + log y]
2
a
a
1 x + y 2 2 = xy = [4 + 2] = 1 8 + 3
1
2
2
2
x + 2xy + y = 4xy = 3 2 2
2
2
1
x + y = 2xy (proven) [7] = 2 2
2
2
[4] =
2
3
log y 12. log x – log y = log xy + log 2
2
2
2
2
7. log x + 2 log 3 2 = log 3 x 3 = (xy)(2 )
3
–2
2
3
3
3
log x + log y = log 3 −2 y y = ± x
3
3
3
xy = 3 −2 2 [5]
= 1
9 [3] 13. (x − 4)e – 4(x − 4) = 0
2
3 −
x
2
2
2
3 −
4
3
x
2
8. (a) log c (x − 4) e – 4 = 0
3
ab 2
log c
3 −
x
= 3 3 log a + log b 4 x − 4 = 0 or e – 4 = 0
2
c
2
c
c
x
1
= 3 1 x + y 2 x = ±2 or 3 − = ln 4
x = 1.24
= 3 [3]
x + y [3] 14. (a) log p 3
ab
3 log p
a1 2
(b) log b – log 1 c = log a + log b
p
a
log b p p
= c – log c = 3
–1
log a a x + y [3]
c
= y + 1
x x ab 3
y + 1 (b) log a 1 p 2
=
x
[3] log p 1 ab 3 2
= p
9. log x + 2 log y = 4 log a
3 9 p
log y 2 log a + log b – log p
3
log x + 3 = log 3 = p p p
4
3 log 3 2 3 log a
3
log (xy) = log 3 x + 3y – 1 p
4
3 xy = 81 3 = x
[3] [3]
Answers 179
Answers Add Math.indd 179 14/03/2022 12:29 PM
3
2
15. 6x + 8x – 4 = 10 (b) √x y 2
x 2 1
3
2
6x + 8x − 4 = 10x 2 = (x )(y )
2
1
x − 2x + 1 = 0 = (2p) (8 )
2
3
q 2
(x − 1) = 0 p
2
x = 1 = 2 · 2
3
6q
= 1 [4] = 2 p 3 + 6q
2 16. 8 · 5 x − 1 + 2 = 5 x + 1 [3]
x
−1
x
1
= √3 8 · 5 · 5 + 2 = 5 · 5 1
x
Let 5 = y 21. Let e = y
2
x
= √(7) – 8y 13y = 14 + 19y
2
2
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2
5 + 2 = 5y 13y − 19y − 14 = 0
7
2
2
= 1 8 + 3 17y = 10 y = – 13 or y = 2
10
2
1 y = 17 1 2 x 7 1 2 x
= 2 2 10 e = – 13 (rejected) or e = 2
x
= 5 = 17 1 x = ln2
x = −0.3297 2 x = 2 ln 2
[4] [4]
17. y = 2 , log y = b 22. 2x log e = log √2 − 3x log(2e + 1)
b
2
3
x
2
4
3
log 2 3 4 2x log e + 3x log(2e + 1) = log √2
3
8y
3
4
3
= log x − log 8y x[2 log e + 3 log(2e + 1)] = log √2
2
3
4
2
2
3
= 2 log x − [log 2 + log y] x = log √2
3
2
2
2
= 2a − 3 − b 2 log e + 3 log(2e + 1)
3
4
[3] = 0.0502
2
18. 9e − e x = 27 − 3x [3]
3x
2
3x
e (9 − x ) = 3(9 − x )
2
2
3x
3x
2
e = 3 or 9 − x = 0 23. (a)
x = ln 3 or x = ±3 f(x)
3 [3] 8
19. Let y = log x 7
5
3 log 5
2 log x = 1 + log x 6 f(x) = 2In (4x – 3)
5
5
2y = 1 + 3 5 5
y
2y – y – 3 = 0 4 e + 3
2
x
2
3
–1
y = or y = –1 3 f (x) = 4
2
x = √5 or x = 1 2
3
5 [4] 1
20. (a) log 16y x
x
= log 16 + log y 0 1 2 3 4 5 6 7 8
x
x
4 log 2 log y [2]
= 2 + 2
−1
2
2 log x (b) Let f (x) = y
= 4 + 3q 2 ln (4y − 3) = x
p p x
= 4 + 3q ln (4y − 3) = ln e 2
p x 2
[4] f (x) = e + 3
–1
4
[3]
Cambridge IGCSE
TM
180 Ace Your Additional Mathematics
Answers Add Math.indd 180 14/03/2022 12:29 PM
24. (a) 3. (a) m = 9 – 5 = 2
f(x) 2 – 0
y = mx + c
4 5 = 2(0) + c
3.5 c = 5
f(x) = 5 log (x + 2) y = 2x + 5
3
–3 = 2d + 5
2.5 d = –4
[3]
2
(b) m m = –1
1.5 x 1 2 1
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f (x) =10 – 2 m = –
5
–1
1 2 2
y = mx + c
0.5 1
2
1
–3 = – (–4) + c
2
0 0.5 1 1.5 2 2.5 3 3.5 4 x –3 = 2 + c
c = –5
1
[2] y = – x – 5
(b) Let f (x) = y 2 [2]
−1
5 log(y + 2) = x
x 2 2
log 10 = log(y + 2) (c) AB = √(2 – 0) + (9 – 5)
5
x = √20
5
f (x) = 10 − 2 = 2√5
−1
[3]
BC = √[0 – (–4)] + [5 – (–3)]
2
2
7 Straight Line Graphs = √80
= 4√5
2
1. (a) 2(0) − p(−4) = 8 AB : BC
−p(−4) = 8 2√5 : 4√5
p = 2 1 : 1
e = 1, f = 2
2
2x − p(0) = 8 [3]
2
2x = 8
2
x = 4 4. Point C,
x = ±2
3
Q(2, 0), S(−2, 0) x = (4 – 4) × 5 4 + 4
4
[2] = 9
(b) Area of ΔQRS 5
3
= 4 × 4 = 8 unit y = (4 – 0) × 4 4 + 8
2
2 [1] = 13
2. 3x + 2y = 12 C(9, 13)
2y = 12 − 3x m = 8 – 4
y = 12 – 3x 1 4 – 0
2 = 1
3
y = – x + 6
2 m m = –1
2
1
Gradient = − 3 m = –1
2
2 1
y = mx + c = –1
3
2 = − (−1) + c 13 = –1(9) + c
2 c = 22
1
c = y = –x + 22
2 [3]
3
y = – x + 1
2 2 [3]
Answers 181
Answers Add Math.indd 181 14/03/2022 12:29 PM
2
2
1
3
2
1
5. Area = x x x x 1 AC = √(x – 3) + (y – 0)
2
2
2 y y y y 1 1 = x – 6x + 9 + y
1
2
3
x – 6x + y + 8 = 0 …… (1)
2
2
1
5.5 = 1 –2 d 1
2 3 –1 4 3 2x – 2 = y
11 = |−1 − 8 + 3d − (−6)−(−d) − 4)| 3 2
11 = |–7 + 4d| 4x – 12 = 3y
4x – 12
= 1 −7 + 4d = 11 y = 3 …… (2)
2 d = 4.5 Substitute equation (2) into (1):
= √3 –7 + 4d = –11 4x – 12 2
2
d = –1 x – 6x + 3 3 4 + 8 = 0
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= √(7) – [2] 9x – 54x + 16x – 96x + 144 + 72 = 0
2
2
2
2
6. y = x – 9x – 10 25x – 150x + 216 = 0
2
2
2
= 1 8 + 3 dy = 2x – 9 x = –b ± √b – 4ac
2
2
1 dx 2a
= 2 2 Gradient of the tangent line at point A: –(–150) ± √(–150) – 4(25)(216)
2
= m m = −1 = 2(25)
2
1
m = –1 ÷ 1 18 12
1 2 x = or x =
= –2 5 5
4
2x – 9 = –2 y = or y = – 4
5
5
7
x =
2 C 1 18 4 2 (invalid, as it is not on AB) or
,
5
5
7
7
y = 1 2 2 – 9 1 2 – 10 C 1 12 –4 2 .
2
2
,
= –117 5 5 [10]
4
Equation of the normal line: (b) m m = –1
1
2
y = mx + c m = –1 ÷ 4
2
–117 = 1 7 + c 3 3
1 2
4 2 2 = – 4
c = –31
4
1 2
1 – = – 3 12 + c
y = x – 31 5 4 5
2
[5] c = 1
3
7. (a) Point A: y = – x + 1
4
2x – 2 = 0 [2]
3 2
2x = 2 8. (a) m = –10 – 3
1
3 0 – 7
x = 3 = 1
A(3, 0) m m = –1
1
2
m = –1
2
Point B: y = mx + c
2(0) – 2 = y –3 = (–1)(7) + c
3 a c = 4
y = –2a \ y = –x + 4
B(0, –2a) [2]
1 0 0 7 0
Distance AB: (b) Area =
5 = √(3 – 0) + (0 – (–2a)) 2 0 –10 –3 0
2
2
1
5 = √9 + 4a = |–70|
2
2
25 – 9 = 4a = 35 unit
2
2
a = 2 or a = –2 (invalid) [2]
Cambridge IGCSE
TM
182 Ace Your Additional Mathematics
Answers Add Math.indd 182 14/03/2022 12:29 PM
9. (a) x = 5 – 2y …… (1) Gradient of PQ = Gradient of SR
y + xy = 6 …… (2) 1 = c – 1
2
12 b – –4
Substitute equation (1) into (2): b + 4 = 12c – 12
y + (5 − 2y)(y) = 6
2
y + 5y – 2y – 6 = 0 b = 12c – 16 …… (2)
2
2
y – 5y + 6 = 0 Substitute equation (2) into (1):
2
(y – 3)(y – 2) = 0 9(12c − 16) = 56 + 8c
y = 3 or y = 2 108c − 144 = 56 + 8c
x = –1 or x = 1 100c = 200
P(−1, 3) and Q(1, 2) c = 2
[3] b = 12(2) – 16
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(b) Length of PQ = 8
= √(–1 – 1) + (3 – 2) [8]
2
2
= √5 unit 8 + 1 9 – 5
[1] 11. (a) Midpoint of AB = 1 2 , 2 2
3 – 2 1
(c) m = = – 9
2
1 –1 – 1 2 = 1 2 , 2
m m = –1
1 2 1 9 – (–5)
m = –1 ÷ – m = 8 – 1
1
2
= 2 2 = 2
,
Midpoint = 1 –1 + 1 3 + 2 2 m m = –1 1
2
1
2
2
2
1
= 0, 5 2 2 m = – 2
y = mx + c y = mx + c 9
1
1
2
5 = 0 + c 2 = – 2 21 2 + c
2 17
c = 5 c =
2 4
1
y = 2x + 5 y = – x + 17
2 [3] 2 4 [5]
1
10. (a) Midpoint = 1 16 + (–4) , 11 + 1 2 (b) 0 = – x + 17
2
2
2
4
= (6, 6) x = 17
[1] 2
2
(b) PS = PQ D 1 17 , 0
2
2
√(–4 – a) + (1 – 10) = √(a – 16) + (10 – 11) 2
2
2
2
2
2
16 + 8a + a + 81 = a – 32a + 256 + 1 CD = 1 17 – (–3) + (0 – (–5))
2
2
40a = 160 2
a = 4 = 12.54 units
[2]
Gradient of PS = 10 – 1 (c) Area
4 – (–4) 8 1 –3 17 8
= 9 = 1 2
8 2 9 –5 –5 0 9
2
Gradient of QP = 11 – 10 (8 × −5) + (1 × −5) + (–3 × 0) + 1 17 × 9 –
16 – 4 = 1 17 2
1
= 1 2 (9 × 1) − (−5 × –3) − −5 × 2 2 − (0 × 8)
12 1 153 85
Gradient of PS = Gradient of QR = 2 –40 – 5 + 2 – 9 – 15 + 2
9 = 11 – c = 1 |50|
8 16 – b 2
144 – 9b = 88 – 8c = 25 unit
2
9b = 56 + 8c …… (1) [3]
Answers 183
Answers Add Math.indd 183 14/03/2022 12:29 PM
1
1 4 0 k 4
12. (a) Area = 14. log y = – log x + 1
2 8 0 3 8 4 1
1 (4 × 0) + (0 × 3) + (k × 8) –
–
4
38 = log y = log x + log 10
2 (8 × 0) – (0 × k) – (3 × 4) – 1
1 y = x (10)
4
38 = |8k – 12|
2 – 1
2
8k – 12 = 76 or 8k – 12 = –76 y = 100x 2
k = 11 (invalid) or k = –8 a = 100, b = – 1
[3] 2 [5]
(b) Equation of line AB:
m = 8 – 3 15. m = 18 – 6 = 3
7 – 3
= 1 4 – (–8)
2 = 5 6 = 3(3) + c
= √3 12 c = –3
y
8 = 5 (4) + c x = 3√x – 3
= √(7) – 12 3 2
2
c = 19 y = 3x – 3x
3
3
2
2
= 1 8 + 3 y = 5 x + 19 c = 3, d = , e = –3 [4]
2
2
1 12 3
= 2 2 Point C: 8 – 3
= 5 19 16. m = 6 – 1 = 1
y = (1) +
12 3 y = mx + c
= 27 ln y = m ln x + c
4 3 = (1)(1) + c
1
C 1, 27 2 c = 2
4
ln y = ln x + 2
2
ln y = ln x + ln e
27
2
1
2
BC = (–8 – 1) + 3 –
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2
ln y = ln (xe )
4
2
2
= 1521 ln y = x(e ) [5]
16
= 39 units 5.1 – 3.4 1
4 17. (a) m = 4 – 0.6 = 2
1
CA = (4 – 1) + 8 – 27 2 2 3.4 = 0.5(0.6) + c
2
c = 3.1
4
1
2
= 169 lg y = x + 3.1
2
16 1 2 + 3.1
x
2
= 13 units lg y = lg 10
4 1 2
x
y = 10 × 10
2
3.1
BC : CA
3.1
:
39 13 A = 10 = 1258.93
4 4 1 2
39 : 13 B = 10 = 3.16
3 : 1 1 [5]
n = 3 (b) y = [10 2 (2) 2 ] × 10
3.1
[6] = 125892.5
13. m = 6 – (–3) = –3 = 126000
–3 – 0 [2]
2
x
2
= –3 1 2 – 3 1.2 – 0.6 3
y 3x 18. (a) m = = –
x
2
= –6 – 9x 0.1 – 0.5 2
y 3x 1.2 = −1.5(0.1) + c
3x = y(–9 – 6x) c = 1.35 2
3
y = – x 3 ln y = −1.5x + 1.35
3 + 2x [3] [3]
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TM
184 Ace Your Additional Mathematics
Answers Add Math.indd 184 14/03/2022 12:29 PM
(b) ln y = ln e –1.5x + 1.35
2
y = e −1.5x 2 · e 8 Circular Measure
1.35
y = 3.86r
–1.5
[3] 1. (a) ∠QOR = π – 2 1 2
π
15 – 5 6
19. (a) m = = 5 2π
3 – 1 =
5 = 5(1) + c 3 [1]
c = 0
2
2
lg y = 5x (b) RSQ = OR
lg y = lg 10 sin ∠QOR sin ∠RQO
2
5x
2
2
y = 10 RSQ = 5
2
5x
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5x 2 2π π
y = 10 sin 3 sin 6
2
5
a = , b = 0 RSQ = 5√3 cm
2
[5] Perimeter
(b) When x = 1 = 5 × 2π + 5√3 × π
2 3 2
5( 2 ) = 24.08 cm
1 2
y = 10 2
5 Area
= 10 1 5√3 2 1 2 2π 1 2π
8
= 4.22 = 2 1 2 2 (π) – 3 2 (5) 1 2 – (5)(5) sin 3 4
3
2
[2] 75 25 25√3
(c) lg 3 = 5x = 8 π – 1 3 π – 4 2
2
2
x = 2 lg 3 = 25 π + 25√3
2
5 24 4
2
x = ± 2 lg 3 = 14.1 cm
5 [6]
= ±0.44
[2] 2. (a) OQ = 6 cm ÷ 2
= 3 cm
20. (a) m = 7 – (–3) = –5 QR = OR – OQ
2
2
2
0 – 2
2
2
7 = –5(0) + c = 6 – 3
c = 7 QR = 3√3 cm
1
e = –5 1 2 + 7 PT = PO + OT
y
x
= PT – PO
OT
e
ln = ln (–5x + 7) = 8 cm − 6 cm
–1
y
1
y = ln 7 – 5 x 2 = 2 cm
QR
[5] tan q = QO + OT
5
(b) 7 – = 0
x = 3√3
7 = 5 3 + 2
x q = 0.805 rad
x . 5
7 [2] [3]
(b) RT = RQ + QT 2
2
2
1 15 2
2
2
(c) y = ln 7 – 5 = (3√3) + 5
6 RT = 2√13 cm
= ln 5 SR = ST – RT
[2] = 8 – 2√13
5
4
(d) e = 7 – x cos ∠ROQ = 3
5 = 7 – e 6
4
x ∠ROQ = π
1
x = 5 3
7 – e 4 [2]
Answers 185
Answers Add Math.indd 185 14/03/2022 12:29 PM
Perimeter 5. Q
= SP + PR + RS
1
π + (8 – 2√13)
= (8)(0.805) + 6 1 2 T
3
= 13.51 cm
[5]
(c) Area of sector TSP O P
1
2
= (PT) (q)
2
= 1 (8) (0.805) S
2
2
= 25.76 cm 2 R
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Area of sector ORP Let radius of circle = r
= 1 (OP) (∠ROQ) sin ∠TOP = TP
2
2 OP
1 2 1 π r
= (6) 1 2 sin =
π
2 3 6 30 – r
= 18.85 cm 30 – r = 2r
2
Area of triangle 3r = 30
r = 10 cm
= 1 (RT)(OT) sin q
2π
2 Arc TS = 10 1 2
= 1 (2√13)(2) sin 0.805 20π 3
2 = cm
= 5.198 cm 3
2
OT = OP – PT
2
2
2
Area of shaded region = 20 – 10
2
2
= (25.76 − 18.85 − 5.198) cm OT = 10√3 cm
2
= 1.712 cm 2
[4] Perimeter of shaded region
1 = 10√3 + 10√3 + 20π
2
3. (a) 60 = (8) (q) 3
2 = 55.58 cm
q = 1.875 rad Area of sector TPS
[1]
1
1 2
(b) Perimeter = AO + OB + BA = (10) 2 2π
3
2
1
= 8 + 8 + (8) (π − 1.875) 100
2
2 = π cm
2
= 56.531 cm 3
[3] Area of OTPS
4. ∠QOR = 2 × 30° = Area of triangle OTP × 2
1
= 60° = × 10√3 × 10 × 2
2
(Angle at the centre is twice the angle at = 100√3 cm
2
circumference)
Area of shaded region
Area of sector QOR 100
1
1 2
= (5) 2 1 π = 100√3 – 3 π
2 3 = 68.49 cm
2
= 25 π cm [7]
2
6
Area of triangle QOR 6. (a) Major arc PQ
3
2
1
1 1 = 8 2π – π
3 1 24
= (5)(5) sin π 7
2 3 = 39.49 cm
= 25√3 cm [1]
2
4
Area of shaded region (b) Area of sector POQ
2 3
1
π
= 25 π – 25√3 = 2 (8) 1 2
7
6 4 2
= 2.26 cm 2 = 43.08 cm
[4]
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TM
186 Ace Your Additional Mathematics
Answers Add Math.indd 186 14/03/2022 12:29 PM
Area of triangle POQ (b) Area of segment
3
1
= 1 (8) sin π 2 = Area of triangle POQ + area of major sector POQ
2
1
2 7 = 12(8)(8) sin 1.35 + (8) (2π − 1.35)
2
= 31.198 cm 2
2
= 189.09 m
2
Area of shaded region [4]
= (43.08 − 31.198) cm
2
= 11.88 cm 9. Perimeter
2
[5] = QO + OP + PQ
1
r
3
7. tan ∠BOA = AB = r + r + r 1 2
OA = 2r + r 2
AB = 5 tan 0.8 3
= 5.148 cm Area
1
1 2
OB = OA + AB = r 2 1 r
2
2
2
OB = 25 + 26.504 2 3
2
1
OB = 7.177 cm = r
3
6
CB = OB − OC
1
2
r
= 7.177 − 5 2r + = 2 1 2
r
3
= 2.177 cm 3 2 3 6
r
CA = (5)(0.8) 2r + = r 3
3
= 4 cm 6r + r = r
3
2
2
Perimeter of ABC r – r – 6 = 0
= AB + BC + CA (r – 3)(r + 2) = 0
= 5.148 + 2.177 + 4 r = 3 or r = –2 (rejected)
= 11.325 cm [4]
OD 10. Area of AOD
cos ∠BOA =
OA = (6) 2 2 π
1
1 2
OD = 5 cos 0.8 2 3
= 12π cm
2
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= 3.484 cm
Area of triangle DOA Area of ABCD
1
= (OD)(OA) sin ∠BOA = 12π × 4
2 3 2
1
= (3.484)(5) sin 0.8 = 16π cm
2 Area of BOC
= 6.247 cm = Area of AOD + Area of ABCD
2
2
Area of sector COA = 28π cm
1
1
= (5) (0.8) Area of BOC = (OB) 2 2 π
2
1 2
2 2 3
2
= 10 cm 2 π
28π = (OB) 1 2
3
Area of ACD OB = 2√21 cm or OB = –2√21 (invalid)
= 10 − 6.247
= 3.753 cm 2 Perimeter of the shaded region
[9] = AB + DC + AD + BC
2
2
π
π + 2√21
2
2
2
8. (a) 10 = 8 + 8 − 2(8)(8) cos ∠POQ = (2√21 – 6) + (2√21 – 6) + 6 1 2 1 2
3
3
cos ∠POQ = 7 = 38.1 cm
32 [5]
∠POQ = 1.35 rad
11. (a) Area of secor QOR
Perimeter of segment = 1 (6) (1.3)
2
= Arc PQ + Line PQ 2
= 8(2π − 1.35) + 10 = 23.4 cm
2
= 49.47 m [2]
[4]
Answers 187
Answers Add Math.indd 187 14/03/2022 12:29 PM
(b) QP = OP – OQ – 2(OP)(OQ) cos ∠QOP 13. (a) 28 = 20 + 20 − 2(20)(20) cos ∠POQ
2
2
2
2
2
2
QP = 6 + 6 − 2(6)(6) cos (π − 1.3) ∠POQ = 1.55 rad
2
2
2
QP = 9.55 cm Reflex angle ∠POQ = 2π − 1.55
∠QOS = 4.733 rad
∠QPS =
2 [3]
= 1.3 rad (b) Area of triangle POQ
2 = 1 (20)(20) sin 1.55
= 0.65 rad 2
= 200 cm 2
Area of sector QPS [2]
= 1 (9.55) (0.65) (c) Area of the unshaded region
2
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2 = Area of circle − Area of major sector POQ
= 29.64 cm − Area of triangle POQ
2
[5] 2 1 2
(c) Area of segment PQ = π(20) − (20) (4.733) – 200
2
= Area of sector POQ − Area of triangle POQ = 110 cm
2
1 1 [3]
= (6) (π – 1.3) – (6)(6) sin(π – 1.3)
2
2 2 14. (a) 13 = 20 + 20 − 2(20)(20) cos ∠BOC
2
2
2
2
= 15.805 cm ∠BOC = 0.662 rad
Area of shaded region ∠OBC = π – ∠BOC
= Area of semicircle PQR – Area of sector QPS 2
− Area of segment PQ = 1.2398 rad
= 1 (6) π – 29.64 – 15.805 Perimeter of sector BAC
2
2 = BA + BC + AC
= 11.1 cm = 13 + 13 + (13)(1.2398)
2
[4] = 42.12 cm
[4]
12.
P Q (b) Area of segment BC
= Area of sector BOC − Area of triangle BOC
1 1
= 2 (20) (0.662) − (20)(20) sin (0.662)
2
2
U = 9.46 cm
2
R Area of sector BAC
2
T = 1 (13) (1.2398)
2
= 104.76 cm
2
S Area of shaded region
= Area of segment BC + Area of sector BAC
PQ = TS = √180 – 20 = 9.46 + 104.76
2
2
= 80√5 cm = 114.22 cm
2
cos ∠QRU = 20 [5]
180 1
2
∠QRU = 1.459 rad 15. 20 = r (1)
2
Arc length QS r = 2√10 cm (length of rectangle)
= 100(2π − 1.459 × 2) Height of rectangle
= 336.519 cm
= (2√10) + (2√10) – 2(2√10)(2√10) cos (1)
2
2
Arc length PT = 6.064 cm
= 80[2π − 2(π − 1.459)]
= 233.44 cm Area of rectangle
= 38.35 cm
2
Length of rope [4]
= 80√5 × 2 + 336.519 + 233.44 16. (a) q = 1.3 rad × 2
= 927.73 cm = 2.6 rad
[6] [1]
Cambridge IGCSE
TM
188 Ace Your Additional Mathematics
Answers Add Math.indd 188 14/03/2022 12:29 PM
(b) PQ = 5 + 5 − 2(5)(5) cos (2.6) 19. (a) tan ∠ROS = 5
2
2
2
PQ = 9.636 cm 1.5
[2] tan ∠ROS =1.2793 rad
(c) PRQ = (12)(1.3) ∠POR = π − 2(1.2793)
= 15.6 = 0.583 rad
PSQ = (5)(2.6) [2]
= 13
(b) OR = 5 + 1.5
2
2
2
Difference in length OR = 5.22 cm
= 15.6 − 13 Perimeter of shaded region
= 2.6 cm = PQR + PR
[2] = (5.22)(0.583) + 3
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π
17. (a) BC = 10 × = 6.04 cm
3 [2]
= 10π cm (c) Area of triangle POR
3 1
Perimeter = 2 × 3 × 5
= 10π × 3 = 7.5 cm
2
3
= 10π cm (or 31.42 cm) Area of sector POR
[2] = 1 (5.22) (0.583)
2
(b) Area of sector ABC 2
1 2
2
= 1 (10) 2 π = 7.943 cm
2 3
50π Area of shaded region
= cm = 7.943 − 7.5
2
3 = 0.44 cm
2
sin π = Height of triangle [4]
3 10
20. (a) OQ = OP – PQ
2
2
2
Height of triangle = 5√3 cm
= 2 – 1
2
2
Area of triangle OQ = √3
= 1 × 10 × 5√3 √3
2 SQ = 2 m
= 25√3 cm SP = SQ + PQ
2
2
2
2
√3
2
Area = Area of segment × 3 + Area of triangle = 1 2 2 + 1
2
2
= 3 1 50π – 25√3 + 25√3 SP = √7 m
3
= 70.48 cm 2 [2]
2
[5] –1 PQ
(b) ∠PSQ = tan 1 2
18. (a) ∠PQR = 2 × ∠PQO SQ
1 2
1
= 2 × tan –1 9 1 2
12
= 1.287 rad = tan –1 √3
[1] 2
(b) Area of sector PQO = 0.857 rad
= 1 (12) (0.6435) [1]
2
2 (c) ∠POQ = tan –1 PQ
1 2
= 46.33 cm OQ
2
1 2
Area of triangle PQO = tan –1 1
2
= 1 (12)(12) sin 0.6435 = 0.464 rad
2
= 43.20 cm Area of shaded region
2
= Area of sector POR − Area of triangle POR
Area of shaded region 1 1
= 2 × (46.33 − 43.20) = 2 (0.464 × 2) − (2)(√3)
2
=6.26 cm = 3.59 m
2
2
[4]
[3]
Answers 189
Answers Add Math.indd 189 14/03/2022 12:29 PM
21. (a) sin ∠OAE = OE
OA y
OA = 10
7
sin 1 36 π 2
= 17.434 cm 3
AD = 17.434 + 10
= 27.434 cm
DC = 27.434 × 1.222
= 33.52 cm
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[3]
(b) tan ∠OAE = OE
AE x
AE = 10 0 π 2π
7
tan 1 36 π 2 [3]
= 14.281 cm 2. 1 – sec A
sec A
Area of EOBA 1 1
= 2 × Area of triangle OEA = 1 – cos A
= 2 × 3 10 × 14.281 4 cos A 1
2
= 142.81 cm = cos A – cos A
2
2
Area of major sector EOB = cos A – 1
cos A
2
1
= 1 (10) 2 25 π –sin A
2
2 18 =
= 218.166 cm sin A
2
tan A
Area of sector DCA = −sin A tan A
1
2
= 1 (27.434) 2 7 π [4]
2 18 3. 4 sin q + 2 cos q = 0
= 459.75 cm 4 sin q = −2 cos q
2
Area of shaded region sin q = – 1
= 459.75 − 142.81 − 218.166 cos q 2
= 98.78 cm tan q = – 1
2
[4] 2
q = 153.43° or 333.435°
[4]
9 Trigonometry cos 2B 1
2
4. + = 3
sin 2B sin 2B
2
2
1. (a) and (b) cos 2B + 1 = 3 sin 2B
2
2
1 1 −sin 2B + 1 = 3 sin 2B
2
2
xy =
2 sin 2B = 1
2
y = 1 2
2x 1
8 solutions sin 2B = ± √2
[3] −360° < 2B < 360°
2B = ±45°, ±135°, ±225°, ±315°
B = ±22.5°, ±67.5°, ±112.5°, ±157.5°
[4]
Cambridge IGCSE
TM
190 Ace Your Additional Mathematics
Answers Add Math.indd 190 14/03/2022 12:29 PM
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