FOCUS KSSM Mathematics F2 (2023)

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CC032233
FORM
2 KSSM FOCUS





Mathematics








FOCUS KSSM Form 2 – a complete and precise series of reference books with Mathematics FORM
special features to enhance students’ learning as a whole. This series covers the 2 KSSM
latest Kurikulum Standard Sekolah Menengah (KSSM) and integrates Ujian Akhir
Dual Language Programme
Sesi Akademik (UASA) requirements. A great resource for every student indeed! Mathematics



REVISION REINFORCEMENT EXTRA • Ng Seng How
› Comprehensive Notes & ASSESSMENT FEATURES • Ooi Soo Huat
• Yong Kuan Yeoh
› Example and Solution › Formative Practices › Maths Info • Dr. Chiang K. W.
› Tips › Summative Practices › HOTS Challenge
› Common Mistakes
› UPSA Model Paper › Daily Application
› UASA Model Paper › TIMSS Challenge
› Answers › Digital Resources QR Code





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CC032233
ISBN: 978-629-7557-17-5
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Format: 190mm X 260mm TP Focus Tg123 2023 Maths F2_pgi CRC














Mathematics FORM







Dual Language Programme 2 KSSM



• Ng Seng How
• Ooi Soo Huat
• Yong Kuan Yeoh
• Dr. Chiang K. W.

Format: 190mm X 260mm TP Focus Tg123 2023 Maths F2_pgii CRC



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Mathematics PT3 Mathematical Formulae
Exclusive Features of this Book



Learning Area: Number and Operations
Infographic
visually highlights Chapter 1 Patterns and Sequences 5.1.2 circle (b) Draw a line passing through
Construct a circle and parts of a
Mathematics Form 2 Chapter 5 Circles
2
O and Q so that both ends
the key concepts KEYWORDS Construct a circle with on the circumference. Q O
of the line are points lying
(a) a radius of 1 cm,
of each chapter to • Pattern Solution: Try question 3 in Formative Practice 5.1
(b) a diameter of 2.4 cm.
• Sequence
• Term
• Even numbers
(a)
4
Access to
enhance students’ INFOGRAPHIC • Odd numbers 1 cm Construct a chord of length
• Pascal’s triangle
The diagram on the right
• Fibonacci numbers
• Algebraic expression
shows a circle with centre O.
learning. • Complete the sequence u Measure a distance of 1 cm on a pair of 1.6 cm passing through point Q. O Q Example
• Extend the sequence
Solution:
compasses.
u Measure a distance of
1.6 cm on a pair of provides
v Fix the point of the compasses at a point
compasses.
as the centre of the circle.
1.6 cm
v Fix the point of the
w Draw with the pencil on the compasses
through a complete turn. O Q
compasses at point Q. solution
1.6 cm CHAPTER
(b) u Measure a distance of
Chapter Introduction v Fix the point of the  Draw a line joining point Q and the 5 for example of
w Draw an arc that intersects
the circle.
1.2 cm on a pair of
compasses.
contains picture and w Draw the circle accordingly. 1.2 cm intersection point drawn in step w. questions in the
compasses at a point
TIPS
as the centre of the circle.
question to stimulate (i) Radius = Diameter (ii) Diameter = 2 × Radius circumference. subtopics.
TIPS
We can draw two chords of the same
length passing through a fixed point on the
2
interest and thinking Try question 2 in Formative Practice 5.1 Try question 4 in Formative Practice 5.1
3
about the chapter’s The diagram on the right shows centre is 110°. 5
Construct and shade a sector of a circle with a
radius of 1 cm and the angle subtended at the
content. Did you know, beehives have a matching pattern and often hexagonal? If the beehive is a circle with centre O. Copy and P Q O u Construct a circle of radius
Solution:
there will be waste of space. In your opinion, why triangle and square are not chosen?
circular, there will be empty spaces between the shapes when arrange next to each other and construct the diameter passing
through
(a) point P,
(b) point Q.
1 cm and mark the centre
(a) Solution: as O. N 110° M
1 Draw a line passing through 1 cm
O
v Draw the radius OM.
O so that one end of the
line is at P and the other w Draw another radius ON
O
end is at a point lying on so that ∠MON = 110°.
the circumference.
P  Shade the sector region OMN accordingly.
Mathematics Form 2 Chapter 6 Three-Dimensional Geometrical Shapes Summative Practice Mathematics Form 2 Chapter 8 Graphs of Functions
Try questions 5 and 6 in Formative Practice 5.1
8
(f) Section A 47 Full
solution
(e)
HOTS Challenge HOTS Challenge 11 cm 4 cm 3 cm 1. A 7 • • 3 B 2. Identify the functions from the list below and then
Praktis Formatif 1

complete the circle map given. i-THINK
provide questions 3.5 m 3 m 5 cm (g) (h) 4 cm 3 cm 10 • 9 • • 2 • 0 k = 3h r = 1 s m = – 8 [4 marks]
a = b + c
n 2
that stimulate A tent has a shape which is the combination of 7 cm 6 cm 4.5 cm 5 cm 11 cm C Many-to-one y = 5 p = q 2 + 1
type of the relation?
The diagram above shows a relation. What is the
a cylinder and a cone, as shown in the diagram
A One-to-one
above. In the structure of the tent, the height of
B One-to-many
student’s higher the cylinder is 3 m and the slanting height of the (i) 7 cm 2. Which of the following set of ordered pairs is not Functions Summative Practice
cone is 3.5 m. Encik Mutti has been provided with
D Many-to-many
a canvas of area 160 m 2 for the curved portions
of the cylindrical and conical shapes. What is the
a function?
order thinking maximum diameter of the tent that can be built? 5 cm 6 cm A P = {(5, 4), (4, 3), (3, 2), (2, 1)} provides ample
(Use π = 3.142)
B Q = {(2, 3), (2, 5), (7, 2), (8, 3)}
C R = {(2, 2), (3, 3), (4, 4), (5, 5)}
Solution:
D S = {(1, 4), (2, 4), (7, 5), (8, 1)}
Total surface area of curved face of cylinder and
following spheres based on the given radius.
skills. Let radius of tent = r 2. Calculate the surface area for each of the CHAPTER 3. 1 • 3 • P • 1 Q 3. Each of the following representations map x questions to test
(a) 12 cm (Use π = 3.142)
curved face of cone

onto y. Match the correct type of function.
= 2πrh + πrs
= 2πr(3)+ πr(3.5) (b) 17.5 cm Use π = 22  7 5 • 7 • • 9 • 25 x 3 6 9 [4 marks]
= 9.5πr 3. Calculate the surface area of each of the 6 9 • • 49 y 5 8 11 students’ mastery
Thus, 9.5πr = 160 160 following spheres. • 81
r = 9.5 × 3.142 (a) (b) 8 cm A 1 y = 7 x 2 Many-to-one
The diagram above shows the relation between
= 5.36 m 7 cm B 3 y function of the chapter.
set P and set Q. What is the object for 9?
Diameter = 2 × 5.36 C 9 CHAPTER
= 10.72 m (2 d.p.)
Therefore, the maximum diameter of the tent that (Use π = 3.142) D 81
can be built is 10.72 m. Use π = 22  7 Section B 0 x One-to-one 8
function
(c) 1. For each of the following representations, x is (10, –1), (20, –2),
Formative Practice 6.3 21 cm an independent variable and y is a dependent (30, –3), (40, –4)
1. Find the surface area of each of the following non-function. y [4 marks] Section C x
variable. Mark (✓) for a function and (✗) for
geometric shapes. (Use π = 3.142 if necessary)
(a)
1. (a)
–3
– 4
Formative Practice (a) 5 cm (b) 3 cm 4. Find the surface area of each of the following (b) 0 x The table above shows the value of x and 1
Use π = 22 
–2
–1
9
y
7
p
18
36
–36
x
(b)
provides questions 5 cm 5 cm 4 cm (d) 12 cm solids. Give your answers in mm 2 . 13 cm (c) y = 5x + 9 20 1 40 2 –40 2 60 3 (b) Given a function y = 8 . x 2 x [2 marks]
of P.
y
(a)
y for the function y = – 36 . State the value
15 cm
to test students’ (c) 9 cm 10 cm 7 cm 6 cm 8 cm (d) (–1, –2), (–3, –4), (–5, –6) (i) Complete the table of values below. 3
x
–2
–3
1
–1
2
y
understanding of 6 cm 3 cm 3 cm Use π = 22  7 69 [2 marks]
101
the subtopics and
reinforce learning. Mathematics Form 2 Chapter 10 Gradient of a Straight Line ANSWERS Answers
that have been derived, three cases of the
2. Thus, the formula of the gradient, m, of a 5. By using the formulae of the gradient Chapter 1 Patterns and Sequences 2. (a) The pattern of the sequence is add 5 help students
gradient of the straight lines are shown as
straight line passes through points (x 1 , y 1)


(a) The straight line that passes through
18, 33
to obtain each subsequent number.
and (x 2 , y 2 ) is follows. Formative Practice 1.1 (b) The pattern for this sequence is to check and
m = y 2 – y 1 origin. y 1. (a) The set of numbers begins with 4 number.
subtract 4 to obtain each subsequent
x 2 – x 1 and multiply by 5 to obtain each 64, 60
R(x, y) (b) The set of numbers begins with 1
subsequent number.
(c) The pattern for this sequence
and 3. Each consecutive number is
is divide by 2 to obtain each
subsequent number. (v) evaluate their
Maths Info MATHS INFO INFO The gradient, m = y – 0 = y x (c) The set of numbers begins with (d) The pattern for this sequence 7. The pattern of the geometrical shapes is
the sum of the two numbers before
x
it.
1 000, 125

O
Letter m is commonly used to represent gradient
2. Multiply by 4 and then multiply
the combination of shapes successively,
by 2 alternately to obtain each
is multiply by 3 to obtain each
subsequent number.
x – 0
which are square, triangle, rhombus and
subsequent number.
consists of of a straight line. x 2 – x 1 x 1 – x 2 x 1 – x 2 (b) The straight line that is parallel to x-axis. (d) The set of numbers begins with 3. 3. 9, 243 1 1 1 2 1 1 followed by triangle and then repeats. performance.
The value of y 2 – y 1 and y 1 – y 2 is the same.
Add 1, add 2, add 3, and so on to
obtain each subsequent number.
y
(i)
x 2 – x 1
extra info and Thus, the gradient, m = y 2 – y 1 = y 1 – y 2 . O S(x 1 , y) T(x 2 , y) x 2. (a) The set of shapes begins with two 1 1 1 5 4 3 10 10 6 3 4 5 1 1 1 (ii)
squares, followed by a triangle. This
pattern is then repeated.
y
(b) The set of letters A-B-C followed
3.
by a letter X. This pattern is then
Mathematics Form 2 Chapter 1 Patterns and Sequences

knowledge related b O B(0, b) A(a, 0) x (c) The straight line that is parallel to y-axis. 3. The number of leaves grow according 4. (a) The sequence begins with 5 and 8. (a) The pattern begins with two joined
repeated where the number of letter
The gradient, m = y – y = 0
X is added by 1 compared to the
previous one.
(v)
Alternative Method
Alternative Method
x 2 – x 1
multiply by –1 to obtain each
a
subsequent number. 5, –5, 5
to the set of odd numbers 1, 3, 5, 7, 9.
triangles which is formed by using
to the chapter. Based on the diagram above, This method is suitable to determine the 4. The pattern on the earthen jar is a 22 – 4(2) = 14 square is formed in between the
(b) The sequence begins with 9 and
The number of leaves begins with 1 and Sequence
Position of term
y
bigger term in a sequence.
5 matchsticks. In the next pattern,
add 11 to obtain each subsequent
add 2 for each time until the number of
V(x, y 2 )
leaves is 9.
3 matchsticks are added so that a
1
22 – 4(1) = 18
the gradient of line AB
(c) The sequence begins with 55 and
Position of term
Sequence
2
two triangles. The pattern then
subtract 6 to obtain each subsequent
Vertical distance
number. 31, 25, 19
3
combination of an equilateral triangle 22 – 4(3) = 10
consisting of 3 isosceles triangles.
of 3 matchsticks.
= Horizontal distance O 1 2 U(x, y 1 ) x 3(1) + 1 = 4 5. The scales on the thermometer is 2°C 22 – 4(4) = 6 (b) repeats with the number of square is
increasing by 1 unit for each addition
(d) The sequence begins with 488
number. 53, 64, 75 CHAPTER
and divide by –2 to obtain each
3(2) + 1 = 7
4
subsequent number. –61, 61 , – 61
1
= b – 0 The gradient, m = y 2 – y 1 = undefined for each mark from 0°C to 20°C. The 5. In this sequence, the numerator of the 4
2
…
…
3
x – x
3(3) + 1 = 10
pattern for the scale on thermometer is
0 – a
add 2 to the mark before it.
4
= – b a Try questions 2 and 3 in Formative Practice 10.1 3(4) + 1 = 13 6. The pattern for the arrangement of the 22 – 4n
n
fractions became the recurring digit in
the recurring decimal numbers.
…
…
The nth term = 22 – 4n
a is known as x-intercept, that is the 10.1.3 Make generalisation for the 3(n) + 1 CHAPTER 1. (a) A sequence because the list of 6. The number of squares follows the 27 TIMSS
tables and chairs is add 3 chairs for
9
4 = 0.444… ,
5 = 0.555…
x-coordinate of the point of intersection of
9
Substitute n = 12
each addition of one table. = 12 – 4(12)
The 12th term
n
a straight line with the x-axis. b is known as
gradient of a straight line
Formative Practice 1.2 = –26

The nth term = 3n + 1
y-intercept, that is the y-coordinate of the
pattern 7, 12, 17, …, that is adding 5
The 8th term = 3(8) + 1
Substitute
point of intersection of a straight line with
squares for each consecutive diagram.
= 25
One square is added to each horizontal
downwards to the right has a negative

Further explanation
numbers follows a pattern of multiply of Alternative
end of the letter I and the vertical column
value, whereas the gradient of a straight
or
by 2 to obtain each subsequent
number.
the y-axis.
Method Example 9
Formative Practice 1.3
× 3 n = 1, 2, 3, 4, 5, …, n
Position
line that inclines upwards to the right has
4. Thus, the formula of the gradient of 1. The gradient of a straight line that inclines n = 8 10 (b) Not a sequence because the list of across the letter I respectively. 1. Using numbers: 2, 3, 4, 5, … Digital Resources
INFO
3n = 3, 6, 9, 12, 15, …, 3n


pattern.
numbers does not follow a specific
a straight line, m, that passes through
a positive value.
+ 1 3n + 1 = 4, 7, 10, 13, 16, …, 3n + 1
Using words:
x-intercept and y-intercept is y Term (c) A sequence because the number of The straight line in Diagram 1 passes
unit cubes used to build the stairs
through two points and each subsequent
follows the following pattern:
m = – y-intercept m 0 m 0 +2 +3 TIPS +4 +5 +6 (iv) straight line passes through 1 point more help learners
than the previous straight line.
x-intercept
To determine the nth
x 1, 3, 6, 10, 15, 21 ... term of a sequence where Using algebraic expressions:
n + 1, where n represents the number
O
of nth straight line.
Further explanation of Alternative
the order of terms is decreasing, we subtract a
Tips x-intercept and y-intercept cannot be written in 2. The greater the absolute value of the multiple of n from a fixed number. For instance, better comprehend
Method Example 8
gradient of a straight line, the steeper the
the nth term of 15, 13, 11, 9, … is 17 – 2n.
INFO
TIPS
straight line.
Try question 3 in Formative Practice 1.3
points out the form of coordinates. TIPS × 1 + 1 = 115 4 175 concepts and
3
Term
important tips for difference × Position + or – Number = Term TIMSS deepen learning.
of term
students to take Determine the 12th term in the following In the diagram above, 9 matchsticks are used to *Video/Info
9
form 4 triangles in a row. How many triangles can
be formed in this way if 55 matchsticks are used?
sequence.
note of. Solution: 18, 14, 10, 6, 2, …
Extend the sequence by subtracting 4 until the
12th term. INFO Common mistake
–4 –4 –4 –4 –4 –4 –4 –4 –4 –4 –4
18, 14, 10, 6, 2, –2, –6, –10 –14, –18, –22, –26
The 12th term of the sequence is –26.
7
iii
0a F2 Exclusive Features.indd 3 06/03/2023 4:32 PM

Mathematical Formulae







MEASUREMENT AND GEOMETRY RELATIONSHIP AND ALGEBRA

Sum of interior angles of a polygon Distance = (x – x ) + (y – y ) 2
√
2
= (n – 2) × 180 o 2 1 2 1
Circumference of circle = pd = 2pr x + x y + y
Midpoint = 1 1 2 , 1 2 2
Area of circle = pr 2 2 2


Arc length Angle subtended at centre Average speed = Total distance travelled
Circumference = 360° Total time taken

Area of sector = Angle subtended at centre Vertical distance
Area of circle 360° Gradient = Horizontal distance

Surface area of prism y – y
2
= 2 × Cross-sectional area Gradient = x – x 1
+ (Cross-sectional perimeter × Height) 2 1

y-intercept
Surface area of pyramid Gradient = – x-intercept
= Base area + Total area of triangular faces

Surface area of cylinder = 2πr + 2πrh Mean = Sum of data
2
Number of data
Surface area of cone = πr + πrs
2
n(A)
Surface area of sphere = 4πr 2 P(A) = n(S)

Volume of prism = Cross-sectional area × Height
P(A9) = 1 – P(A)
1
Volume of pyramid = × Base area × Height
3

Volume of cylinder = πr h
2
1
2
Volume of cone = πr h
3
4
Volume of sphere = πr 3
3






iv





0b F2 Formulae Matematik.indd 4 06/03/2023 4:32 PM

CONTENTS









Chapter
LEARNING AREA Number and Operations
5 Circles 45
Chapter
1 Patterns and Sequences 1 5.1 Properties of Circles 46
5.2 Symmetrical Properties
of Chords 48
1.1 Patterns 2
1.2 Sequences 3 5.3 Circumference and
1.3 Patterns and Sequences 6 Area of a Circle 50
Summative Practice 1 9 Summative Practice 5 56



Chapter
LEARNING AREA Relationship and Algebra
6 Three-Dimensional
Chapter Geometrical Shapes 59
2 Factorisation and 12 6.1 Geometric Properties 60
of Three-Dimensional Shapes
Algebraic Fractions
6.2 Nets of Three-
2.1 Expansion 13 Dimensional Shapes 62
2.2 Factorisation 17 6.3 Surface Area of Three-
2.3 Algebraic Expressions and Laws Dimensional Shapes 64
of Basic Arithmetic Operations 21 6.4 Volume of Three-Dimensional
Summative Practice 2 23 Shapes 71
Chapter Summative Practice 6 77
3 Algebraic Formulae 25


3.1 Algebraic Formulae 26
Summative Practice 3 31 LEARNING AREA Relationship and Algebra

Chapter
Measurement and
LEARNING AREA 7 Coordinates 80
Geometry
Chapter 7.1 Distance in the Cartesian
4 Polygons 33 Coordinate System 81
7.2 Midpoint in the Cartesian
4.1 Regular Polygons 34 Coordinate System 84
4.2 Interior Angles and Exterior 7.3 The Cartesian
Angles of Polygons 36 Coordinate System 86
Summative Practice 4 42 Summative Practice 7 88







v





0c F2 Contents.indd 5 06/03/2023 4:33 PM

11.3 Reflection 132
Chapter
8 Graphs of Functions 90 11.4 Rotation 137
11.5 Translation, Reflection
and Rotation as an Isometry 143
8.1 Functions 91
8.2 Graphs of Functions 94 11.6 Rotational Symmetry 145
Summative Practice 8 101 Summative Practice 11 147

LEARNING AREA Statistics and Probability
Chapter
9 Speed and Acceleration 103 Chapter
12 Measures of Central
9.1 Speed 104 Tendencies 149
9.2 Acceleration 109 12.1 Measures of Central Tendencies 150
Summative Practice 9 111 Summative Practice 12 160



Chapter
Chapter
10 Gradient of a Straight Line 113 13 Simple Probability 163

10.1 Gradient 114 13.1 Experimental Probability 164
Summative Practice 10 120 13.2 Probability Theory involving
Equally Likely Outcomes 166
Measurement and 13.3 Probability of the Complement
LEARNING AREA
Geometry of an Event 169
13.4 Simple Probability 171
Chapter
11 Isometric Transformations 123 Summative Practice 13 173


11.1 Transformations 124 Answers 175
11.2 Translation 125





UPSA Model Paper UASA Model Paper
(Ujian Pertengahan Sesi Akademik) (Ujian Akhir Sesi Akademik)



https://qr.pelangibooks.com/?u=FocusMatF2UPSA https://qr.pelangibooks.com/?u=FocusMatF2UASA






Answers for UPSA and UASA Model Papers


https://qr.pelangibooks.com/?u=FocusMatF2AnsUjian






vi





0c F2 Contents.indd 6 06/03/2023 4:33 PM

Learning Area: Number and Operations

Chapter
1 Patterns and Sequences








KEYWORDS


• Pattern
• Sequence
Access to
INFOGRAPHIC • Term
• Even numbers
• Odd numbers
• Pascal’s triangle
• Fibonacci numbers
• Algebraic expression
• Complete the sequence
• Extend the sequence









































Did you know, beehives have a matching pattern and often hexagonal? If the beehive is
circular, there will be empty spaces between the shapes when arrange next to each other and
there will be waste of space. In your opinion, why triangle and square are not chosen?





1





F2 Chapter 1.indd 1 06/03/2023 4:34 PM

Mathematics Form 2 Chapter 1 Patterns and Sequences

(i) Each number in the Pascal’s
1.1 Patterns triangle is the direct sum of the
two numbers on the top where
1.1.1 Recognise and describe each row starts and ends with 1.
patterns (ii) There are many patterns of
numbers in rows and diagonals in
1. Pattern is an arrangement of shapes, this triangle. For instance,
colours, numbers, letters and many more • sum of numbers in each row will
CHAPTER
that enables us to identify and describe the form a list of numbers 1, 2, 4, 8,
1
behaviours of the next object or event. For 16, ….
instance, • list of numbers 1, 3, 6, 10, 15,
21, … is the triangular numbers.
(a) 2, 4, 6, 8, … is a set of even numbers.
+2 +2 +2 +2 (e) The pattern for the arrangement of
tables and chairs is adding two chairs
2, 4, 6, 8, … for one addition of table.
Set of even numbers starts with 2 and
the pattern is adding 2 to obtain each
subsequent number.

(b) 1, 3, 5, 7, … is a set of odd numbers.
+2 +2 +2 +2 2. When we are describing a pattern, we must
1, 3, 5, 7, … state the beginning of the patterns and
each change that happens in the pattern.
Set of odd numbers starts with 1 and
the pattern is adding 2 to obtain each
subsequent number. 1
(c) 1, 1, 2, 3, 5, 8, 13, 21, … is a set of Identify and describe the following set of
Fibonacci numbers. numbers.
(a) 22, 18, 14, 10, …
+ + +
(b) 3, 5, 8, 13, 21, …
1, 1, 2, 3, 5, 8, 13, 21, …
Solution:
+ + +
(a) –4 –4 –4 –4
In a set of Fibonacci numbers, the first
two numbers are 1. The pattern of each 22, 18, 14, 10, …
subsequent number is the sum of the
two numbers before it. The set of numbers starts with 22 and
subtract 4 to obtain each subsequent
(d) The set of numbers in the diagram number.
below is known as the Pascal’s triangle.
(b)
1 + +
1 1 2
1 + 1 3, 5, 8, 13, 21, …
1 1 2 2
1 + 2 + 1 +
1 2 1 4 2
1 + 3 + 3 + 1
1 3 3 1 8 2 The set of numbers starts with 3 and 5.
1 + 4 + 6 + 4 + 1
1 4 6 4 1 16 2 Each subsequent number is the sum of the
1 + 5 + ... + 5 + 1
1 5 10 10 5 1 32 2 two numbers before it.
1 + 6 + ... + 6 + 1
1 6 15 20 15 6 1 64 2
1 + 7 + ... + 7 + 1
1 7 21 35 35 21 7 1 128 Try question 1 in Formative Practice 1.1


2





F2 Chapter 1.indd 2 06/03/2023 4:35 PM

Mathematics Form 2 Chapter 1 Patterns and Sequences

2 4. The diagram below shows a pattern on an
earthen jar. Identify and describe the pattern
on the earthen jar. CHAPTER










5. Identify and describe the pattern for the scale 1
The diagram above shows a tiled wall in a of thermometer in the diagram below.
kitchen. Identify and describe the pattern of
the tiled wall.
Solution: 20ºC
The pattern of the tiled wall is the combination 10ºC
of triangles which consist of three trapeziums.
0ºC
Try questions 2 – 6 in Formative Practice 1.1



Formative Practice 1.1 6. The diagram below shows the arrangements of

1. Identify and describe the pattern for the trapezium tables surrounded by chairs. Identify
following sets of numbers. and describe the pattern for the arrangements
(a) 4, 20, 100, 500, … of tables and chairs.
(b) 1, 3, 4, 7, 11, …
(c) 2, 8, 16, 64, 128, …
(d) 3, 4, 6, 9, 13, … Table

Chair
2. Identify and describe the pattern for the
following sets of shapes or letters.
(a)
1.2 Sequences


1.2.1 Recognise sequences

(b) ABCXABCXXABCXXX 1. A list of succession of numbers created
from a set of geometric shapes, numbers
3. The diagram below shows the growth of shoot or objects is called a sequence.
of a tree. Identify and describe the pattern for For instance,
addition of leaves along the growth of the shoot.




The set of geometric shapes above is a
sequence because the geometric shape
starts with 3-sided polygon and the pattern
is adding one side for each subsequent
shape.





3





F2 Chapter 1.indd 3 06/03/2023 4:35 PM

Mathematics Form 2 Chapter 1 Patterns and Sequences

3 TIPS

Explain whether each of the following is a The reverse of division is multiplication. In
sequence. Example 4 (b), the first term can be obtained
(a) 15, 12, 10, 5, –3, … by multiplying 18 by 3.

(b) 54, 18, …
× 3
CHAPTER
First Second Third
1
Try questions 2 and 3 in Formative Practice 1.2


Fourth Fifth 5
Solution: Identify and describe the pattern for each of
(a) –3 –2 –5 –8 the following sequences. Hence, extend three
subsequent numbers in the sequence.
15, 12, 10, 5, –3, … (a) 64, 59, 54, 49, …
It is not a sequence because the list of (b) –1, 2, –4, 8, …
numbers does not follow a particular Solution:
pattern.
(b) It is a sequence because the set of (a) The sequence starts with 64 and subtract
geometrical shapes follows a particular 5 to obtain each subsequent number.
pattern. The pattern starts with a hexagon, –5 –5 –5 –5 –5 –5
followed by two triangles, and then repeats.
64, 59, 54, 49, 44, 39, 34
Try question 1 in Formative Practice 1.2
(b) The sequence starts with –1 and multiply
by –2 to obtain each subsequent number.
1.2.2 Identify and describe the pattern
of a sequence, complete and × (–2) × (–2) × (–2) × (–2) × (–2) × (–2)
extending the sequence
–1, 2, –4, 8, –16, 32, –64
4 TIPS

Identify and describe the pattern for each of A sequence of numbers can be obtained by
the following sequences. Then, complete the adding, subtracting, multiplying or dividing.
sequence.
(a) 9, 13, , 21, 25, , . Try questions 4 and 5 in Formative Practice 1.2
2
2
(b) , 18, 6 , , , , .
3 9
Solution: 6
(a) The pattern of the sequence is add 4 to Identify and describe the pattern for the
obtain each subsequent number. following sequence. Then, extend two diagrams
+4 +4 +4 +4 +4 +4 for the sequence.
9, 13, 17, 21, 25, 29 , 33
(b) The pattern of the sequence is divide by 3
to obtain each subsequent number. First Second Third

÷ 3 ÷ 3 ÷ 3 ÷ 3 ÷ 3 ÷ 3
54, 18, 6, 2, 2 , 2 , 2 Forth Fifth
3 9 27


4





F2 Chapter 1.indd 4 06/03/2023 4:35 PM

Mathematics Form 2 Chapter 1 Patterns and Sequences

Solution: Formative Practice 1.2
The pattern starts with triangle, rhombus and
followed by trapezium, and then repeats. 1. Explain whether each of the following is a
sequence.
Thus, the two subsequent diagrams are as (a) 5, 10, 40, 240, 1 920, …
follow. (b) 7, 11, 18, 23, 24, …

(c) CHAPTER

Sixth 1

2. Identify and describe the pattern for each of
Seventh
the following sequences. Hence, complete the
sequence.
Try questions 6 – 8 in Formative Practice 1.2
(a) 8, 13, , 23, 28,
(b) 76, 72, 68, , , 56
HOTS Challenge (c) 2 000, , 500, 250, , 125
Praktis Formatif 1
2
(d) , 27, 81, , 729, 2 187
What is the number needed to fill in the ‘empty’
triangle? 3. Complete the empty spaces in the following
Pascal’s triangle.
3 6 1
1 1
4 3 1 2 1
8 4 5 13 1 3 1

4 6 1
2 4 1 5 10 5 1

7 4. Identify and describe the pattern for each of
9 5 14 6 the following sequences. Hence, extend three
subsequent numbers in the sequence.
(a) 5, –5, 5, –5, , ,
Solution:
(b) 9, 20, 31, 42, , ,
3 6 (c) 55, 49, 43, 37, , ,

(d) 488, –244, 122, , ,
4 3
8 4 5 13 5. Identify and describe the pattern for the following
sequence. Hence, extend two subsequent
numbers of the sequence.
8 + 4 = 3 × 4 5 + 13 = 6 × 3 1 2 3
= 0.111 …, = 0.222 …, = 0.333 …
9 9 9
2 4 6. Identify and describe the pattern for the following
sequence. Hence, extend two subsequent
7 5 diagrams in the sequence.
9 5 14 6

9 + 5 = 2 × 7 14 + 6 = 4 × 5

(i) (ii) (iii)


5





F2 Chapter 1.indd 5 06/03/2023 4:35 PM

Mathematics Form 2 Chapter 1 Patterns and Sequences

Solution:
7. Identify and describe the pattern for the
following sequence. Then, complete diagrams Using numbers: 3, 4, 5, 6, …
(i), (ii) and (v) of the sequence.
Using words:
The first row has 3 cans and each row below it
has one can more than the row before it.
(i) (ii) (iii) Using algebraic expressions:
CHAPTER
Row Sequence
1
1 1 + 2 = 3
2 2 + 2 = 4
(iv) (v) (vi)
3 3 + 2 = 5
8. 4 4 + 2 = 6
5 5 + 2 = 7
… …
n n + 2
Mahad arranges matchsticks to form a few
patterns as shown in the diagram above. n + 2, where n represents the number of canned
(a) Identify and describe the pattern for the
sequence. drinks in the n row (from the top).
(b) Draw two subsequent patterns that will be
formed by Mahad. Try questions 1 and 2 in Formative Practice 1.3


1.3 Patterns and Sequences 1.3.2 Determine a term

1. Each element in a sequence is known as a
1.3.1 Make generalisation about the term.
pattern of a sequence
For instance, in the sequence
5, 10, 15, 20, 25, …
7
the first term is 5,
the second term is 10,
First row the third term is 15 and so on.
Second row

Third row 8
Determine the 8th term in the following

sequence.
4, 7, 10, 13, 16, …
Solution:
Extend the sequence by adding 3 until the 8th
The diagram above shows an arrangement of
canned drinks in a supermarket. The first row term.
has 3 cans. The second row has 4 cans, the third +3 +3 +3 +3 +3 +3 +3
row has 5 cans and so on. Make a generalisation 4, 7, 10, 13, 16, 19, 22, 25
for the pattern of the arrangement of canned
drinks using numbers, words and algebraic The 8th term in the sequence is 25.
expressions.





6





F2 Chapter 1.indd 6 06/03/2023 4:35 PM

Mathematics Form 2 Chapter 1 Patterns and Sequences


Alternative Method Alternative Method
This method is suitable to determine the Position of term Sequence
bigger term in a sequence.
1 22 – 4(1) = 18
Position of term Sequence 2 22 – 4(2) = 14 CHAPTER
1 3(1) + 1 = 4 3 22 – 4(3) = 10
2 3(2) + 1 = 7 4 22 – 4(4) = 6 1
3 3(3) + 1 = 10 … …
4 3(4) + 1 = 13 n 22 – 4n
… … The nth term = 22 – 4n
n 3(n) + 1 The 12th term = 12 – 4(12) Substitute n = 12
= –26
The nth term = 3n + 1
The 8th term = 3(8) + 1 Substitute n = 8
= 25
or Further explanation of Alternative
n = 1, 2, 3, 4, 5, …, n Position Method Example 9
× 3
3n = 3, 6, 9, 12, 15, …, 3n INFO
+ 1 3n + 1 = 4, 7, 10, 13, 16, …, 3n + 1
Term
TIPS

To determine the nth term of a sequence where
Further explanation of Alternative the order of terms is decreasing, we subtract a
Method Example 8 multiple of n from a fixed number. For instance,
the nth term of 15, 13, 11, 9, … is 17 – 2n.
INFO

TIPS Try question 3 in Formative Practice 1.3


3 × 1 + 1 = 4
TIMSS
Term Position + or – Number = Term
difference × of term



9
In the diagram above, 9 matchsticks are used to
Determine the 12th term in the following form 4 triangles in a row. How many triangles can
sequence. be formed in this way if 55 matchsticks are used?
18, 14, 10, 6, 2, …
Solution:
Extend the sequence by subtracting 4 until the Common mistake
12th term.
INFO
–4 –4 –4 –4 –4 –4 –4 –4 –4 –4 –4
18, 14, 10, 6, 2, –2, –6, –10 –14, –18, –22, –26
The 12th term of the sequence is –26.



7





F2 Chapter 1.indd 7 06/03/2023 4:35 PM

Mathematics Form 2 Chapter 1 Patterns and Sequences

1.3.3 Solve problems 2.


10 Daily Application
During an annual school sports, a school band
has performed various triangular formations at
the school field. The triangular formations of 2,
3, 4 and 5 bandsmen at the triangular bases are
CHAPTER
as shown in the diagram.
1
A grocery store exhibits cereal boxes in an
attractive arrangement to attract customers.
The diagram above shows the arrangement for
the first six top rows. Make a generalisation of
2 3 4 5 pattern of the number of cereal boxes by using
Determine the total number of bandsmen numbers, words and algebraic expressions.
needed to form a triangular formation of 8
bandsmen at the triangular base. 3. Determine the 12th term in each of the following
sequences.
Solution: (a) 1, 5, 9, 13, …
(b) 6, 11, 16, 21, …
Number of (c) 35, 32, 29, 26, …
bandsmen at (d) 40, 35, 30, 25, …
the triangular 2 3 4 5 6 7 8 (e) 190, 180, 160, 150, …
base
Total number 3 6 10 15 21 28 36 4. The diagram below shows part of a calendar
of bandsmen showing the dates for the month of January
year 2001. Determine the date of the last
+3 +4 +5 +6 +7 +8 Monday for the month of January year 2001.
The total number of bandsmen needed to form
a triangular formation of 8 bandsmen at the January 2001
triangular base is 36. Monday Tuesday Wednesday Thursday Friday Saturday Sunday
1 2 3 4 5 6 7
Try questions 4 – 8 in Formative Practice 1.3 8 9 10 11
Formative Practice 1.3 5. A diver can dive to a depth of –3 m after one

1. minute, –7 m after two minutes and –11 m after
three minutes. Assume that the diver keeps
diving following such pattern, determine the
position of the diver after 8 minutes.

6.




Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5
The diagram above shows the pattern of 1 layer 2 layers 3 layers
straight lines drawn passing through a number
of points. Make a generalisation of pattern of The diagram above shows the first three layers
the number of points passed through by each of a well built from unit cubes. How many layers
of the straight lines using numbers, words and of the well are built by using 256 unit cubes?
algebraic expressions.


8





F2 Chapter 1.indd 8 06/03/2023 4:35 PM

Mathematics Form 2 Chapter 1 Patterns and Sequences

7. The Form 2 students of SMK Taman Bahagia 8.
are planning to attend a Mathematics workshop. t n
The cost of transportation is RM15 per student 20
and the cost of participation is RM25 per
student. 15
(a) Complete the following table for the first 5
students. 10 CHAPTER
Number of Total cost 5
students (RM) n 1
0
1 40 1 2 3 4
2 80
The graph above shows the terms in a sequence
(b) Identify and describe the pattern that relates where t represents the nth term.
n
the number of students and the total cost. (a) Based on the graph, make a generalisation
Hence, write an algebraic expression for of pattern of the terms in the sequence
the total cost in terms of proper variables. by using numbers, words and algebraic
(c) If the total cost is RM1 760, how many expressions.
students will be attending the Mathematics (b) Hence, determine t of the sequence.
workshop? 10 Analysing
Analysing





Summative Practice 1 Full
solution


Section A 4.
1. Which of the following pattern is a sequence?
A 4, 19, 34, 48, …
B 145, 129, 113, 99, … 4 5 6 7 8 9 10 11 12 13 14 15 16
C 1 , 1 , 1 , 1 , … The diagram above shows a number line. What
4 8 16 32 are the next two numbers in the sequence of the
D –0.4, –0.8, –0.16, –0.32, … number line?
A 18, 23
2.
70, 64, 58, 52, … B 18, 24
C 19, 24
Describe the pattern for the set of numbers D 19, 25
above.
A Add 4 to the previous number. 5. Which of the following matches is incorrect?
B Subtract 6 from the previous number. A Starts with 100 1, 4, 7, 10,
C Divide the previous number by 3. and divide by 2. 13, 16, …
D Multiply the previous number by 4.

3. Which of the following is a set of Fibonacci B Starts with 100 1, 3, 9, 27,
numbers? and subtract 50. 81, …
A 1, 2, 4, 8, …
B 1, 2, 6, 24, … C Starts with 1 and 100, 50, 25,
C 1, 5, 6, 11, … add 7. 12.5, …
D 10, 20, 30, 40, …
D Starts with 1 and 100, 50, 0,
multiply by 3. –50, –100, …







9





F2 Chapter 1.indd 9 06/03/2023 4:35 PM

Mathematics Form 2 Chapter 1 Patterns and Sequences

Section B (iii) Hence, state the number of unused
cards. [1 mark]
1. (a) Mark (✓) for the correct statement and (7) for
the incorrect statement describing the pattern (b) Triangular numbers are a list of numbers
of the set of numbers 8, 5, 9, 6, 10, 7, … which can be represented by arrangements
[2 marks] of triangular dots. The diagram below shows
the first four terms in the triangular numbers.
(i) The set of numbers begins with Determine the 10th term. [2 marks]
8. Subtract 3 and then add 4
alternatively to obtain each of 1 3 6 10
CHAPTER
subsequent number.
1
(ii) The set of numbers begins with
8 and 5. Add 1 simultaneously
to both of the numbers to
obtain each of two subsequent Applying
numbers.
(b) (c) The first two numbers in a set of numbers
41 5 ? 13 are 3 and 5.
(i) Write two different sets of number
9 ? 21 17 patterns starting with the two numbers.
33 29 1 25 [2 marks]
(ii) Determine the 8th term of each of the
[2 marks]
set.
The diagram above shows a sequence of
unorganised numbers with two unknown 2. (a)
numbers. Find the two unknown numbers
and write the number sequence in ascending
order. [2 marks]
2. The diagram below represents patterned beads.

The diagram above shows an experiment
to measure the distance travelled in each
Mark (✓) for the pattern of letters that is suitable
to represent the arrangement of the beads and second of a trolley moving down a slope.
(7) for the pattern of letters that is not suitable The results of the experiment are shown in
the following table.
to represent the arrangement of the beads.
[4 marks] The distance travelled
Time (s)
A-B-A-B-B-C-C-B-D (cm)

A-B-C-B-B-C-C-A-D 1 3
2 5
D-C-D-C-C-A-A-C-B
3 7
D-C-C-B-B-A-A-C-D
4 9
Section C 5 11
1. (a)
First Second Third Forth Complete the table below based on the result
sequence sequence sequence sequence of the experiment. Applying [4 marks]
Write a generalisation of pattern for the
distance travelled by the trolley using
The diagram above shows a pattern formed
by arranging triangular cards. numbers words algebraic
(i) State the number of triangle used in the expression
fifth order. [1 mark]
(ii) Farid has 42 pieces of triangle cards.
Find the maximum sequence that can be
formed using the cards. [2 marks]

10





F2 Chapter 1.indd 10 06/03/2023 4:35 PM

Mathematics Form 2 Chapter 1 Patterns and Sequences

(b) 4. (a) A primary school in a rural area is facing a
decrease in registration of new students due
to the population migration to the city. It is
found that there is a uniform decrease of 110
students for every subsequent year. On the
A B C first year, the registration of new students is
(i) Identify and describe the pattern for the 2 100 students. CHAPTER
sequence in the diagram above. (i) Write an algebraic expression of the
[2 marks] number of students for a particular
(ii) Hence, extend two more diagrams in the year. [2 marks] 1
sequence. [2 marks] (ii) In which year the number of registration
(iii) Show that the sequence has more than of new students will be decreasing until
one pattern. [2 marks] the number is less than 800 students for
the first time? [2 marks]
Analysing
Analysing
3. Farid pulls apart blocks from a structure of brick (b)
game following a certain pattern.

January 2017
Sunday Monday Tuesday Wednesday Thursday Friday Saturday
1 2 3 4 5 6 7
8 9 10 11 12 13 14
15 16 17 18 19 20 21
22 23 24 25 26 27 28
29 30 31

The diagram above shows a calendar. Add
the three horizontal numbers and the three
vertical numbers in the shaded rectangles.
(i) What can you observe from the sum of
both sets of the three numbers?
[2 marks]
(ii) How does the sum of three numbers
relate to the number at the centre?
[2 marks]
(iii) Repeat (i) and (ii) with two other sets
of three numbers. Describe the pattern
of any two sets of three numbers that
exist in the same way on the calendar.
Analysing [2 marks]
(a) Write the pattern as a number sequence.
[2 marks]
(b) How many blocks are pulled apart by Farid
for each structure? [4 marks]
(c) How many blocks are left when Farid pulls
apart the blocks one more time?
[4 marks]
















11





F2 Chapter 1.indd 11 06/03/2023 4:35 PM

Learning Area: Relationship and Algebra

Chapter
8 Graphs of Functions




























KEYWORDS


• Origin
• Many-to-one
• Function
• Non-linear function
• Linear function
• Graph
• Relationship
• Table of values
• x-axis
• y-axis
• Ordered pair
• Dependent variable
• Independent variable
• Coordinate plane
• One-to-one
• Function notation
63.27




Access to
INFOGRAPHIC



The relationship between quantities is an important aspect in the field of Science. We can
study the relationship between two quantities or variables from the result of an experiment.
One of the methods to analyse the relationship between two variables in an experiment is by
using graph of functions. How can you analyse the relationship between the two variables by
using the graph of function?



90





F2 Chapter 8.indd 90 06/03/2023 4:44 PM

Mathematics Form 2 Chapter 8 Graphs of Functions

• Many-to-one relation – there are
8.1 Functions two or more inputs have the same
output
8.1.1 Explain the meaning of Hobby
functions Sabiah
Badminton
Li Heng
1. In a study, two quantities which are known Nora Swiming
as input and output respectively are related
by a specific relation.
(b) The relation that is not a function
2. Function is a relation which each input • One-to-many relation – there is
only has one output. an input that has more than one
outputs
3. In a function, all inputs are known as
domain while all outputs are known as Square root
range. 4 2
0
4. There are three ways to represent a relation, 0
namely arrow diagram, graph and ordered –2
pair. For instance, the relation between set
X = {2, 4, 5} and set Y = {4, 6, 7} is a function • Many-to-many relation – the
‘add 2’. This relation can be represented as combination of both many-to-one
follows: relation and one-to-many relation
(a) Arrow diagram Interest
X Y Matek History
Issac
Add 2 Mathematics
2 4 Samuel Science
Ayub
4 6

5 7
1
(b) Graph
Y A B CHAPTER
Square
7 3
9
6 2 8
5 4
4 –3
3 The arrow diagram shows a relation between
2
1 set A and set B.
0 X (a) State the type of relation shown in the arrow
1 2 3 4 5 diagram.
(c) Ordered pair (b) Represent the relation in ordered pair.
{(2, 4), (4, 6), (5, 7)} (c) State the range of the relation.
5. There are four types of relations: Solution:
(a) The relation that is a function (a) Many-to-one 9 has two
• One-to-one relation – each input inputs
has only one different output (b) {(–3, 9), (2, 4), (3, 9)}
(c) {4, 9} All the outputs
Multiply 3
1 3
2 6 Try questions 1 – 3 in Formative Practice 8.1
3 9



91





F2 Chapter 8.indd 91 06/03/2023 4:44 PM

Mathematics Form 2 Chapter 8 Graphs of Functions

8.1.2 Identify functions 3. The value of dependent variable is always
depends on the value of independent
1. The quantities involved in a function are variable.
variables. 4. A function maps the independent variable
2. There are two types of variables involved onto the corresponding dependent variable
in a function, dependent variable and under a specific relationship.
independent variable.

5. A function can be represented in the forms of ordered pair, table, graph and equation as shown
in the table below.
Ordered pair Table Graph Equation
(x, y)

(a) The dependent (b) The dependent (c) The dependent (d) The dependent
variable, y variable, y variable, y always variable, y is the
is twice the exceeds the equal to the square of the
independent independent independent independent
variable, x. variable, x by 3. variable, x. variable, x.
(1, 2), (2, 4), (3, 6), x 1 2 3 4 y y = x 2
(4, 8) 3
y 4 5 6 7
2

1
x
O 1 2 3
Function is a special relationship where each independent variable only has one corresponding
dependent variable.


6. There are two types of functions:

MATHS INFO
CHAPTER
INFO
One-to-one Many-to-one
8
function function
The variable in a linear equation is an unknown
Each value of More than that has fixed value while the variable involved in
independent one value of a function has varying values.
variable is independent
mapped onto a variables are 2
different value of mapped onto
dependent variable a same value Given x is an independent variable and y is a
respectively. of dependent dependent variable. Determine whether each
variables. of the following representations is a function.
Hence, give your justification based on the
7. A function that maps the value of x onto representation.
the value of y can be written in the function (a) (2, 6), (5, 15), (7, 21), (9, 27)
notation f(x) = y or f : x → y. For instance,
a function maps x onto 2x is written as (b) x 5 10 15 15
f(x) = 2x or f : x → 2x. y 2 2 4 6







92





F2 Chapter 8.indd 92 06/03/2023 4:44 PM

Mathematics Form 2 Chapter 8 Graphs of Functions

(c) y (d) y = x
2. x f(x)
2 16
x 3 24
O
h 48
Solution: 7 56
(a) Yes, because each value of x has only one The diagram shows a function represented in
corresponding different value of y. arrow diagram. Determine the value of h.
(b) No, because there is a value of x that has
two corresponding values of y, that is when 3. It is given that set K = {1, 3, 5} and set L =
x = 15, y = 4 and 6. {6, 8, 10}. The relation that maps from set K
(c) Yes, because each value of x has only one to set L is add 5. Represent the relation using
corresponding different value of y. (a) table
(d) No, because there is a positive value of x (b) ordered pair
that has two corresponding values of y. (c) graph
(d) equation
3
4. Given x is the independent variable and y is
Each of the following function representations the dependent variable. Determine whether
maps x onto y. Determine the type of function each of the following representations is a
for each of the following. function. Hence, give your justification based
on the representation.
(a) (–2, 1), (–1, 2), (0, 3), (1, 4) (a) (1, 2), (1, 3), (2, 4), (5, 6)
(b) y
(b) x 10 20 20 30

y 5 6 7 8
x
O (c)
Solution: y
(a) Each value of independent variables is
mapped onto a different value of dependent
variable respectively. Therefore, the type of O x CHAPTER
function is one-to-one. 6
(b) There are two independent variables (d) y = x 8
mapped onto the same value of dependent
variable. Therefore, the type of function is 5. Each of the following function representations
many-to-one. maps x onto y. Determine whether each of the
following is one to one function or many to one
Try questions 4 – 7 in Formative Practice 8.1 function.
(a) (–4, 5), (–3, 10), (–2, 15), (–1, 20)
Formative Practice 8.1 (b) x 2 4 6 8

1.
B
A
y 6 7 7 8
2
3 (c)
4 y
2
8
1 12
x
The arrow diagram shows a relation between O
set A and set B. State
(a) the type of relation. (d) f(x) = x 2
(b) the domain of the relation.
(c) the range of the relation.



93





F2 Chapter 8.indd 93 06/03/2023 4:44 PM

Mathematics Form 2 Chapter 8 Graphs of Functions

5. The flow map below shows the steps to
6. Determine the type of function of each of the
following. draw a graph of function.
(a) f(x) = 5x – 8 Identify the equation of the function.
(b) g(x) = 2x + 1
2
(c) m(x) = 9
x Construct the table of values based on
7. Temperature ( C) 20 40 60 80 the given range of values of x.
o
Volume (cm ) 50 100 150 200
3
The table above shows the results obtained
from an experiment. Given that the volume Form ordered pairs from the
of gas is always varying with its temperature. table of values.
If the relationship between the two variables
is written as a function, determine the
independent variable and dependent variable Plot each ordered pair on the
of the function. Hence, state the type of the Cartesian plane.
function. Applying
Connect all the ordered pairs to
draw the graph.
8.2 Graphs of Functions


8.2.1 Construct tables of values and 4
draw the graphs
Construct a table of values for the function
y = x – 3x + 5 for –2 < x < 3.
2
1. A function can be represented in the
form of graph by drawing a straight line Solution:
or curved line on the Cartesian plane by When x = –2, y = (–2) – 3(–2) + 5
2
connecting all ordered pairs. = 4 + 6 + 5
2. A linear function is represented by the = 15
2
graph of straight line while non-linear When x = –1, y = (–1) – 3(–1) + 5
function is represented by the graph of = 1 + 3 + 5
CHAPTER
curved line. = 9
8
When x = 0, y = (0) – 3(0) + 5
2

MATHS INFO = 0 + 0 + 5
INFO
= 5
2
Linear function y = ax When x = 1, y = (1) – 3(1) + 5
Non-linear y = ax –2 = 1 – 3 + 5
–1
function y = ax (Reciprocal function) = 3
2
y = ax (Quadratic function) When x = 2, y = (2) – 3(2) + 5
2
y = ax (Cubic function) = 4 – 6 + 5
3
where a is constant. = 3
When x = 3, y = (3) – 3(3) + 5
2
3. Before drawing the graph of function, a = 9 – 9 + 5
table of values needs to be constructed. = 5
2
4. When constructing the table of values, Table of values of the function y = x – 3x + 5 is
we need to substitute the suitable values x –2 –1 0 1 2 3
of independent variables (x) into the y 15 9 5 3 3 5
given function to obtain the values of
corresponding dependent variables (y).


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Mathematics Form 2 Chapter 8 Graphs of Functions

TIPS 6

Scientific calculator can be used to determine Draw the graph of the function
the values of y of the given function. 1
u Write the function y = x – 3x + 5. (a) y = , by using a scale of 2 cm to 1 unit on
2
x
Press ALPHA X x – 3 the x-axis and 2 cm to 0.5 unit on the y-axis
2
for –4 < x < 4.
ALPHA X + 5 CALC
Screen X? is shown. 14
(b) y = , by using a scale of 2 cm to 1 unit on
 Enter the value of x, for example x = –2, x 2
the x-axis and 2 cm to 2 units on the y-axis
by pressing (–) 2 =
for 1 < x < 6.
Screen which is shown
X – 3X + 5 Solution:
2
15 1
(a) y =
x
Try question 1 in Formative Practice 8.2 x –4 –3 –2 –1 0 1 2 3 4
y –0.25 –0.33 –0.5 –1 – 1 0.5 0.33 0.25
5

Draw the graph of function y = x – 4x + 3 by The value of y is undefined
2
using a scale of 2 cm to 1 unit on the x-axis and when x = 0.
2 cm to 2 units on the y-axis for –1 < x < 4.
Solution: y
Table of values of the function y = x – 4x + 3 is
2
1.0 1
x –1 0 1 2 3 4 y = x
y 8 3 0 –1 0 3 0.5

and the ordered pairs are (–1, 8), (0, 3), (1, 0), x CHAPTER
(2, –1), (3, 0) and (4, 3). –4 –3 –2 –1 0 1 2 3 4
–0.5
y
1 8
2 8 –1.0
6
3
4
2
y = x – 4x + 3 Common mistakes
2
1 INFO
x
–1 0 1 2 3 4
–2
14
(b) y =
x 2
x 1 2 3 4 5 6
u Draw the x-axis and y-axis.
The scale of x-axis is 2 cm to 1 unit.
The scale of y-axis is 2 cm to 2 units. y 14 3.5 1.56 0.88 0.56 0.39
 Plot each of ordered pair on the Cartesian plane.
 Without using a ruler, join all the ordered pairs to
draw a smooth curve for the non-linear function.


95





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Mathematics Form 2 Chapter 8 Graphs of Functions

Construct the table of values of the function
y
Q = 3m for 0 < m < 4.
14
m 0 1 2 3 4
12 Q 0 3 6 9 12

10 Q(kJ)
14
y = x 2 12
8
10
6 Q = 3m
Use a ruler
8 to draw
4 the straight
line passing
6
2 through all the
ordered pairs
4 for the linear
x
0 1 2 3 4 5 6 function.
2
Try question 2 in Formative Practice 8.2
0 1 2 3 4 m(kg)


MATHS INFO Try questions 3 and 4 in Formative Practice 8.2

INFO
Praktis Formatif 1
• The graph of a quadratic function is called a HOTS Challenge
parabola.
• The graph of a reciprocal function is called Daily Application
hyperbola.

1 x m
2
7 Daily Application
1 x m
Karmila carried out an experiment to investigate 2x m 2
CHAPTER
the relationship between the heat energy The diagram above shows a cuboid tank fully filled
8
content and the mass of a type of liquid at room with water. By using a scale of 2 cm to 1 m on
3
temperature. She found that the function that the x-axis and 2 cm to 5 m on the y-axis, draw a
relates both of the quantities is Q = 3m where graph representing the function of the volume of
3
Q is the heat energy content in kJ and m is the water, y m , in the tank for 0 < x < 4.
mass in kg. By using a scale of 2 cm to 1 kg on Solution:
the m-axis and 2 cm to 2 kJ on the Q-axis, draw Function of the volume of water,
the graph of function for the mass of the liquid y = 2x × 1 x × 1 x
which does not exceed 4 kg. 1 2 2 Form the function first.
Solution: y = 2 x 3

If mass does not exceed 4 kg,
then 0 < m < 4. x 0 1 2 3 4
y 0 0.5 4 13.5 32
Mass cannot be negative value,
thus m  0.







96





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Mathematics Form 2 Chapter 8 Graphs of Functions

9
3
y (m )
30 The graph of function below shows the area
of a film displayed on screen, y m , which is
2
25 projected from a projector, where x m is the
distance between the projector and the screen.
20 Based on the graph of function,
1
y = x 3
2 (a) what is the distance between the projector
15
and the screen if the area required of the
film displayed is 25 m ?
2
10
(b) explain the change of the area of the film
5 displayed on screen to the distance between
the projector and the screen.
x (m)
0 1 2 3 4 Solution:

Area, y(m ) 2
8.2.2 Interpret graphs of functions
30
1. We can interpret a graph of function to
make analysis of trends and prediction of 25
a situation.
20
8
15
The graph of function below shows the cost of
producing a type of smartphone by a company. 10
Based on the trend of the graph of function,
(a) determine the number of smartphones 5
produced with the cost of RM6 500,
(b) predict the cost to produce 60 units of 0 1 2 3 4 5 5.5 6 Distance, CHAPTER
x(m)
smartphones.
(a) 5.5 m
Solution: 8
(b) The greater the distance between the
Cost, y (RM) projector and the screen, the larger the area
12 000 of the film displayed on screen.

10 000 Try questions 5 – 7 in Formative Practice 8.2

8 000
8.2.3 Solve problem involving graphs
6 000 of functions
4 000
10
2 000
The graph shows the rate of the parking fee in
Quantity, two buildings, A and B, which are next to each
0 10 20 25 30 40 50 60 x (unit) other. Based on the graph,
(a) what is the parking fees for the first two
Extend the graph
(a) 25 units of smartphones. to determine the hours in both buildings?
(b) RM10 000 cost when x = 60.



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Mathematics Form 2 Chapter 8 Graphs of Functions

(b) state the time when the fee imposed by (c) Based on the graph, find
both buildings are the same. (i) the values of x when the area of the
(c) what is the difference of the parking fees land that is planted with vegetables is
between both buildings if Azmah’s car is 100 m .
2
parked for 10 hours? (ii) the maximum area of the land that is
Solution: planted with vegetables.
Solution:
Fee (RM) (a) Area of the shaded region,
A
6 A = Area of the rectangle – area of the
B
5.5 triangle
1
5 = 4x(16 – x) – × 10 × 4x
2
2
4 = 64x – 4x – 20x
3.5 = 44x – 4x 2
3
(b) x 0 1 2 3 4 5 6 7 8
2 A 0 40 72 96 112 120 120 112 96

2
1 A (m ) A = 44x – 4x 2
121
120
Time (hours)
0 2 4 6 8 10 100
80
(a) For building A, the fee of first two hours is
RM2 60
For building B, the fee of first two hours is
RM3.50 40
(b) 8 hours 20
(c) For the parking fee of 10 hours, the building
A is RM6 and the building B is RM5.50. 0 1 2 3 3.2 4 5 6 7 7.8 8 x(m)
Therefore, the difference of parking fees is
RM6 – RM5.50 = RM0.50
(c) (i) Based on the graph of function, when
CHAPTER
2
A = 100 m , x = 3.2 m or x = 7.8 m.
8
Try question 8 in Formative Practice 8.2
(ii) The graph of function reaches
11 Daily Application maximum when x = 5.5 m. Therefore,
the maximum area of the land that is
2
10 m planted with vegetables is 121 m .
4x m Alternative Method

Example 11 (a)
(16 – x) m
Area of shaded region
A rectangular land is divided into two parts as = area of trapezium
shown in the diagram above. Puan Jorana wants 1
to plant vegetables on the shaded region. = × (6 – x + 16 – x) × 4x
2
(a) Shows that the area of the shaded region, 1
2
A m , that is planted with vegetables is = × (22 – 2x) × 4x
2
A = 44x – 4x . = 44x – 4x 2
2
(b) By using a scale of 2 cm to 1 m on the x-axis
2
and 2 cm to 20 m on the L-axis, draw the
graph of function for 0 < x < 8. Try questions 9 – 11 in Formative Practice 8.2
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F2 Chapter 8.indd 98 06/03/2023 4:45 PM

Mathematics Form 2 Chapter 8 Graphs of Functions

Formative Practice 8.2 5. Time, y(second)

1. Construct a table of values for each of the 80
following functions based on the given range.
(a) y = 4x – 7, –1 < x < 4 60
(b) y = x + x – 3, –2 < x < 2
2
40
6
(c) y = , 1 < x < 5
x
15 20
(d) y = , 1 < x < 4
x 2
(e) y = 2x , –3 < x < 3 0 5 10 15 20 25 Average speed,
3
x(ms )
-1
2. Draw the graph of the function A car moves from the starting line to the
(a) y = x + 2, by using a scale of 2 cm to finishing line of 400 m track to investigate the
2 units on both axes for –4 < x < 6. relationship between the average speed and
(b) y = 2 – x, by using a scale of 2 cm to the time taken. The graph of function above
shows the results of the test where y seconds
1 unit on the x-axis and 2 cm to 2 units is the time taken and x m s is the average
–1
on the y-axis for –2 < x < 3. speed of the car.
9
(c) y = , by using a scale of 2 cm to 1 unit on (a) Find the time taken by the car when its
x
–1
the x-axis and 2 cm to 2 units on the y-axis for average speed is 16 m s .
–3 < x < 3. (b) Based on the trend of the graph of function,
explain the change of time respect to the
(d) y = 12 , by using a scale of 2 cm to 1 unit on average speed. Hence, predict the time
x 2
the x-axis and 2 cm to 2 units on the y-axis for taken by the car if its average speed is
–1
1 < x < 4. 30 m s .
(e) y = 5 – x , by using a scale of 2 cm to 6.
2
1 unit on the x-axis and 2 cm to 2 units y(m)
on the y-axis for –3 < x < 3. 500
(f) y = x , by using a scale of 2 cm to 1 unit
3
on the x-axis and 2 cm to 2 units on the 400
y-axis for –2 < x < 2. CHAPTER
300
3. Puan Aliya made a cake by following a recipe. 8
She found that the required mass of sugar, 200
G g, is represented by the function G = 5 T, 100
3
where T is the mass of flour needed. By using
a scale of 2 cm to 30 g on the T-axis and 2 0 x(m)
cm to 50 g on the G-axis, draw the graph of 100 200 300 400 500
function for 0 < T < 120.
Azlan rides a bicycle from town A to town B
along a straight road. The relationship between
4. An arrow shot by Luqman is not hitting the
target and falling to ground before reaching the distance of Azlan from town A, x m, and the
the target board. The vertical distance of the distance of Azlan from town B, y m, is shown
arrow from the horizontal ground, y m, is in the graph of function above.
given by y = 4x – x + 1, where x is the (a) What is the distance between Azlan and
2
horizontal distance, in m, of the arrow from the town B if Azlan has been riding 100 m
position of Luqman. Draw the graph of function from town A?
y = 4x – x + 1 for 0 < x < 4 by using a scale (b) What is the distance between town A and
2
of 2 cm to 1 m on both axes. town B?
(c) What is your observation about the trend
of the graph of function above?




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Mathematics Form 2 Chapter 8 Graphs of Functions


7. (c) when the difference in producing cost for
Distance, y(m)
both components with the same number
50 is RM250, determine the possible total
number of both components.
40
30 9.
y m
20
10 x m

Time, t (second) The diagram above shows the surface of a
0 1 2 3 4 5
rectangular table. Given that the surface area
of the table is 1 m .
2
Dilah launches a self-made water rocket at a (a) Based on the situation above, write a
field. The distance of water rocket from the function of y in the terms of x.
horizontal ground, y m, after t second is shown (b) Hence, draw the graph of function by using
by the graph of function above. a scale of 2 cm to 1 m on the x-axis and 2
(a) After how many seconds the water rocket cm to 0.2 m on the y-axis for 0 < x < 5.
will reach the maximum distance? State (c) From the graph, determine the value of y
the maximum distance. when x = 1.7 m.
(b) Based on the trend of the graph of function, Analysing
explain the change of the distance of the
water rocket from the horizontal ground in 10.
relation to the time. Hence, predict the time (12 – x) cm
when the water rocket hits the horizontal
ground.
2x cm
8.
Cost (RM) The diagram above shows a triangle.
2
(a) Show that the area, A cm , of the triangle
3 000 P
is A = 12x – x .
2
(b) By using a scale of 2 cm to 2 cm on the
2 500 2
Q x-axis and 2 cm to 10 cm on the A-axis,
draw the graph of function A = 12x – x
2
CHAPTER
2 000
for 0 < x < 10.
8
(c) Based on the graph of function, find
1 500
(i) the values of x when the area of the
triangle is 20 cm ,
2
1 000
(ii) the maximum area of the triangle.
500 Analysing
Number of 11. A cuboid container has a width of x cm. The
0 100 200 300 400 500 components length is 2 cm longer than its width and the
height is 1 cm lesser than its width.
A factory produces two types of electronic
3
components, P and Q. The graph shows the (a) Show that the volume, V cm , of the cuboid
3
2
producing cost for both components. Based on container is V = x + x – 2x.
the graph, (b) By using a scale of 2 cm to 1 cm on the
3
(a) stete the cost to produce 400 x-axis and 2 cm to 20 cm on the V-axis,
3
2
(i) component P, draw the graph of function V = x + x – 2x
(ii) component Q. for 0 < x < 4.
(b) determine the number of components when (c) Based on the graph of function, find the
the producing cost for both components volume of the container when x = 2.8 cm.
are the same. Analysing
100



F2 Chapter 8.indd 100 06/03/2023 4:45 PM

Mathematics Form 2 Chapter 8 Graphs of Functions

Summative Practice 8 Full
solution

Section A 2. Identify the functions from the list below and then
complete the circle map given.
1. A B i-THINK
[4 marks]
7 • • 3 1 8
9 • • 2 a = b + c r = s m = – n 2
10 • • 0 k = 3h y = 5 p = q + 1
2
The diagram above shows a relation. What is the
type of the relation?
A One-to-one
B One-to-many
C Many-to-one
D Many-to-many Functions
2. Which of the following set of ordered pairs is not
a function?
A P = {(5, 4), (4, 3), (3, 2), (2, 1)}
B Q = {(2, 3), (2, 5), (7, 2), (8, 3)}
C R = {(2, 2), (3, 3), (4, 4), (5, 5)}
D S = {(1, 4), (2, 4), (7, 5), (8, 1)} 3. Each of the following representations map x
onto y. Match the correct type of function.
3. P Q [4 marks]
1 • • 1 x 3 6 9
3 • • 9 y 5 8 11
5 • • 25
7 • • 49
9 • • 81 y = 7 Many-to-one
x 2 function
The diagram above shows the relation between CHAPTER
set P and set Q. What is the object for 9? y
A 1
B 3 One-to-one 8
C 9 0 x function
D 81
(10, –1), (20, –2),
Section B (30, –3), (40, –4)
1. For each of the following representations, x is
an independent variable and y is a dependent Section C
variable. Mark (✓) for a function and (✗) for
non-function. [4 marks] 1. (a) x – 4 –3 –2 –1 1
(a) y y 9 p 18 36 –36

The table above shows the value of x and
x y for the function y = – 36 . State the value
0 of P. x [2 marks]
(b) x 1 2 2 3 8
y 20 40 –40 60 (b) Given a function y = x 2 .
(i) Complete the table of values below.
(c) y = 5x + 9 x –3 –2 –1 1 2 3

(d) (–1, –2), (–3, –4), (–5, –6) y
[2 marks]


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F2 Chapter 8.indd 101 06/03/2023 4:45 PM

Mathematics Form 2 Chapter 8 Graphs of Functions

(ii) Based on the table in (i), draw the graph (c) y
8
of function y = for –3 < x < 3 by using
x
2
a scale of 2 cm to 1 unit on both axes. 5
[4 marks]
x
(iii) Based on the graph of function drawn, O
determine
• the value of y when x = 2.5, The diagram above shows a graph of
[1 mark] function. State the function of the graph in
• the value of x when y = 1.8. the diagram. [2 marks]
[1 mark]
3. (a) A stone is thrown up vertically. The distance
2. (a)
of the stone from the horizontal ground, h cm,
y is represented by the function h = 32t – 4t ,
2
where t is the time in second after the stone
80
is thrown. Draw a graph of function with
0 < t < 6 to determine the maximum distance
60
of the stone from the horizontal ground.
Hence, predict the time for the stone to hit
40
the horizontal ground. [4 marks]
Analysing
20
(b) Chong Kun carried out an experiment to
x
0 1 2 3 4 study the relationship between the pressure,
P kPa, and the volume of gas, V cm , in
3
an enclosed container. The relationship
The diagram above shows a graph of between the pressure and the volume of gas
function. Based on the given graph of 150
function, complete each of the following. is represented by the equation P = V .
[2 marks] (i) By using a scale of 2 cm to 10 cm on the
3
(i) When x = 1.5, y = . V-axis and 2 cm to 2 kPa on the P-axis,
(ii) The type of function shown by the graph draw the graph of function P = 150 for
is . V
(b) (i) Draw the graph of function for the volume 10 < V < 60. [4 marks]
of a cube with the side of x cm for (ii) Based on the graph of function drawn,
CHAPTER
0 < x < 4, by using a scale of 2 cm to determine the gas pressure when the
8
3
1 cm on the x-axis and 2 cm to 10 cm volume is 30 cm . Hence, explain how
3
on the y-axis. [4 marks] the pressure changes with the volume of
(ii) Based on the graph of function, gas. [2 marks]
determine Applying
• the volume of the cube when x = 3.7
cm, [1 mark]
• the side of the cube when its volume
is 20 cm . [1 mark]
3

















102





F2 Chapter 8.indd 102 06/03/2023 4:45 PM

ANSWERS






2. (a) The pattern of the sequence is add 5
Chapter
1 Patterns and Sequences to obtain each subsequent number.
18, 33
(b) The pattern for this sequence is
Formative Practice 1.1 subtract 4 to obtain each subsequent
number.
1. (a) The set of numbers begins with 4
and multiply by 5 to obtain each 64, 60
subsequent number. (c) The pattern for this sequence (v)
(b) The set of numbers begins with 1 is divide by 2 to obtain each
and 3. Each consecutive number is subsequent number. 7. The pattern of the geometrical shapes is
the sum of the two numbers before 1 000, 125 the combination of shapes successively,
it. (d) The pattern for this sequence which are square, triangle, rhombus and
(c) The set of numbers begins with is multiply by 3 to obtain each followed by triangle and then repeats.
2. Multiply by 4 and then multiply subsequent number.
by 2 alternately to obtain each 9, 243
subsequent number.
(d) The set of numbers begins with 3. 3. 1 (i)
Add 1, add 2, add 3, and so on to 1 1
obtain each subsequent number.
1 2 1

2. (a) The set of shapes begins with two 1 3 3 1 (ii)
squares, followed by a triangle. This 1 4 6 4 1
pattern is then repeated.
1 5 10 10 5 1
(b) The set of letters A-B-C followed
by a letter X. This pattern is then
repeated where the number of letter (v)
X is added by 1 compared to the 4. (a) The sequence begins with 5 and 8. (a) The pattern begins with two joined
previous one. multiply by –1 to obtain each triangles which is formed by using
subsequent number. 5, –5, 5 5 matchsticks. In the next pattern,
3. The number of leaves grow according (b) The sequence begins with 9 and 3 matchsticks are added so that a
to the set of odd numbers 1, 3, 5, 7, 9. add 11 to obtain each subsequent square is formed in between the
The number of leaves begins with 1 and number. 53, 64, 75 two triangles. The pattern then
add 2 for each time until the number of repeats with the number of square is
leaves is 9. (c) The sequence begins with 55 and increasing by 1 unit for each addition
subtract 6 to obtain each subsequent
number. 31, 25, 19 of 3 matchsticks.
4. The pattern on the earthen jar is a
combination of an equilateral triangle (d) The sequence begins with 488 (b)
consisting of 3 isosceles triangles. and divide by –2 to obtain each
subsequent number. –61, 61 , – 61
5. The scales on the thermometer is 2°C 2 4
for each mark from 0°C to 20°C. The 5. In this sequence, the numerator of the
pattern for the scale on thermometer is fractions became the recurring digit in
add 2 to the mark before it. the recurring decimal numbers.
6. The pattern for the arrangement of the 4 = 0.444… , 5 = 0.555…
tables and chairs is add 3 chairs for 9 9
each addition of one table.
6. The number of squares follows the TIMSS
pattern 7, 12, 17, …, that is adding 5
Formative Practice 1.2 squares for each consecutive diagram. 27
One square is added to each horizontal
1. (a) A sequence because the list of end of the letter I and the vertical column Formative Practice 1.3
numbers follows a pattern of multiply across the letter I respectively.
by 2 to obtain each subsequent 1. Using numbers: 2, 3, 4, 5, …
number.
(b) Not a sequence because the list of Using words:
numbers does not follow a specific The straight line in Diagram 1 passes
pattern. through two points and each subsequent
straight line passes through 1 point more
(c) A sequence because the number of than the previous straight line.
unit cubes used to build the stairs
follows the following pattern: (iv) Using algebraic expressions:
n + 1, where n represents the number
+2 +3 +4 +5 +6 of nth straight line.
1, 3, 6, 10, 15, 21 ...



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F2 Answers.indd 175 06/03/2023 4:53 PM

Mathematics Form 2 Answers

2. Using numbers: 2, 4, 6, 8, … 3. (a) 21, 18, 15, 12, 9
2. A-B-A-B-B-C-C-B-D ✓
Using words: ✗ (b) 3 bricks
The top row has two boxes and each A-B-C-B-B-C-C-A-D (c) 6 bricks
row below it has two boxes more than D-C-D-C-C-A-A-C-B ✓
th
the previous row. 4. (a) (i) 2 210 – 110n, where n is n year.
D-C-C-B-B-A-A-C-D ✗ (ii) 13 year.
th
Using algebraic expressions:
2n, where n represents the number of (b) (i) The sum of both sets of three
cereal boxes in n rows. Section C numbers are the same.
(ii) The sum of the three numbers is
3. (a) 45 1. (a) (i) 16 3 times the number at the centre.
(b) 61 (ii) 4 (ii) Let n be the number at the
(c) 2 (iii) 8 centre.
(d) –15 (b) 55 The pattern for the three
(e) 30 (c) (i) First set: 3, 5, 7, 9, 11, 13, … horizontal number is n – 1, n, n
+ 1 where the sum is 3n.
4. 29th Second set: 3, 5, 4, 6, 5, 7, … The pattern for the three vertical
(ii) First set: 17
5. –31 m numbers is n – 7, n, n + 7 where
Second set: 8 the sum is 3n.
6. 32
2. (a) Write a generalisation of pattern Chapter
7. (a) Number of Total cost for the distance travelled by the 2 Factorisation and
Algebraic Fractions
students (RM) trolley using
1 40 algrebraic
numbers words expression TIMSS
2 80 10pq + 25p + 14q + 35
3, 5, 7, Add 2 to 2n + 1,
3 120 9, 11 the previous n = 1, 2, Formative Practice 2.1
4 160 number 3, …
1. (a) 6ab + 18ac
5 200 (b) (i) The coloured squares and the (b) y + 2y – 8
2
white squares are joined end to
2
(b) The pattern relates the number of end alternately. The pattern of (c) a – b 2
students multiply by the sum of the number of coloured squares 2. (a) 2x + 16x
2
transportation cost and participation is 1, 3, … whereas the pattern
2
cost, which is RM40 to get the total of the white squares is 2, 4, … (b) a – 4a + 4
cost. (ii) (c) pt + 8t
Total cost = 40n, where n is the
number of students. 3. (a) kx + ky
2
(c) 40n = 1 760 (b) 8p – 40p
2
n = 1 760 D (c) –36y + 4ny
4
2
= 44 (d) 3m + 2mn – 8m
5
10
2
Therefore, 44 students will be (e) – 3 a b – ab + 15a
2
attending the Mathematics workshop.
4. (a) ab + 2a + b + 2
8. (a) Using number: 1, 4, 9, 16 (b) y – 8y + 15
2
Using words: (c) 2k + 9k – 56
2
The pattern for the sequence is (d) 15x – 34xy – 16y 2
2
the perfect square of the first four E 2
numbers. (b) The coloured squares and the (e) –18h + 38h – 4
Using algebraic expressions: white squares are joined end to 5. (a) x – 10x + 25
2
n , where n = 1, 2, 3, and 4. end alternately. The pattern for the (b) 49a + 14ab + b 2
2
2
number of coloured squares is 1, 3,
(b) 100 1, 3, … whereas the pattern for the (c) 36x – 6xy + 1 y 2
2
white squares is 2. (d) n – 64 4
2
Summative Practice 1 (e) 81x – 4y 2
2
2
6. (a) –a + 22a – 24
Section A 2
1. C 2. B 3. C 4. C 5. C (b) –5x + 11x – 8
54
2
2
Section B (iv) (c) 5 x + 48y – 52xy
(d) 28r – 29p + 4pr
2
2
1. (a) (i) ✓
2
7. (3x – 39x + 120) m 2
(ii) ✓
2
(b) 37, 45 8. (3x + 15x + 12) cm 3
2
2
Sequence: 1, 5, 9, 13, 17, 21, 25, (v) 9. (2y – 3y – 20) cm
29, 33, 37, 41, 45
176
F2 Answers.indd 176 06/03/2023 4:53 PM

Mathematics Form 2 Answers

Formative Practice 2.2 (d) 7s Chapter
s – 5 3 Algebraic Formulae
1. (a) 1, 2, a, 2a (e) 4x – 25y 2
2
(b) 1, 7, y, y , 7y, 7y 2 (x + 1) 2 PISA
2
(c) 1, 2, 5, m, n, 10, 2m, 2n, 5m, 5n, 3. (a) 2x + 16x + 9
2
mn, 10m, 10n, 2mn, 5mn, 10mn (a) 40 cm
(b) –3xy – 17x + y (b) 1.95 km/h
2
2
2. (a) 5p(p + 2); factors = 1, 5, p, 5p, (c) –2c + 14c
(p + 2), 5(p + 2), p(p + 2), (d) 8 – 3y Formative Practice 3.1
2
2
2
5p(p + 2) 5x + 2
2
(b) (a – 2)(b – 7); factors = 1, (a – 2), (e) 5a + b 1. (a) V = pqr
(b – 7), (a – 2)(b – 7) a – b 1
(c) 2(n + 4)(n – 1); factors = 1, 2, 20n + 19n (b) A = 2 st
2
(n + 4), (n – 1), 2(n + 4), 2(n – 1), 4. (n + 2)(3n – 1) (c) P = 4a
(n + 4)(n – 1), 2(n + 4)(n – 1)
(d) (h + 3k) ; factors = 1, (h + 3k), 2 2. (a) y = 9 + x 2
2
(h + 3k) Summative Practice (b) P = W
2
t
3. (a) 14a (a + 4) Section A (c) E = mc 2
3
2
(b) n(3 – 5m + 8n) 1. B 2. B 3. A 4. D 3. z = 1.5x + y
(c) 3p (5pq – 2p + 11) Section B
2
2
4. N = 2m + 6
(d) (x + 1)(y + 4) 1.
(e) (2a – 5)(6b + 7) 5a 5. H = 50 – 0.5d
(f) (2k – h)(3k – 1) 5a(a + b) 5 6. (a) y = x 3
(b) y = 3x
4. (a) (a + 3)(a + 5)
(b) (h – 2)(h – 3) 1 5a(a + b) a 7. (a) 11
(c) (n + 2)(n – 6) (b) p = 2n + 1
(d) (3x + 7)(x – 2) 5a(a + b) (a + b) 8. (a) h 1 2 3 4 5
(e) (2y – 1)(5y – 9) a(a + b) d 5 8 11 14 17
(f) (7p – 2q)(4p + q)
(b) d = 3h + 2
5. (a) (f + 9)(f – 9) 2. 9. (a) e = h – 4
(b) (2m + 7)(2m – 7) (x + y) 2 x + 2xy + y 2 (b) c = e – 2d
2
(c) 7(2p + 3q)(2p – 3q) (x + y)(x – y) x – 2xy + y 2 (c) b = l + d
2
(d) (y + 5) 2 (x – y) 2 x – y 2 (d) p = r – h
2
2
(e) (p – 3q) 2 x(x + y) x + xy (e) m = 4nr
2
(f) 5a(a + 2b)(a – 2b) (f) I = V
Section C R
6. (5n + 2) minutes
2
1. (a) (i) –2p + 2 (g) r = s
7. (n + 5) pens 2t
(ii) 25y + 4x – 20xy
2
2
8. (8l + 22) cm (h) n = 3v
(b) (3xy – 3x – y + 1) cm q
2
9. y  6 (c) 12p + 13p + 7 (i) w = 2 – y
2
7
2. (a) (i) 23kp + 5k – 12p 2 

Formative Practice 2.3 (ii) 4m + 16n – 30m (j) u = v – 2as
2
2
3
(b) (k + 8) 3 4j k
1. (a) 9xy – 3x – 4y – 20 (k) I = 3
(b) –65b 2 (c) 11n + 8 gT 2
5
p + p + 2 (l) L = k 2
2
(c) 2 3. (a) (i) (x – 2)(x + 6)
p – 2p v – u
(ii) (m + 1)(n + 4) 10. t =
4x + 13 g
(d) t – 2
y + 3 (b) 7t – 1
3m + m + 3 11. (a) 47
2
(e) (c) No, because x + 5 is not the factor (b) –1
mn
for xy + 3x + 8y + 24.
2
2. (a) –2f + 16f + 2f – 16f 4. (a) (i) –7x – 28x – 105 12. (a) 11
3
4
2
6p 6
(b) (ii) 12ab – 23a + 7b – 28
q – 2 2 (b) –5
3ab – 15b (b) 8k h + 8kh – 20k 13. (a) 1
2
(c) 3h – 5h
b + 2 (b) 2
(c) (9a + 2b) m
177
F2 Answers.indd 177 06/03/2023 4:53 PM

Mathematics Form 2 Answers

14. (a) 48 (c) r = 3p – 4q – 8 (b)
(b) 1 2(q + 2)
15. (a) 4 2. (a) (i) b = 2 + c 60° 60° 60°
3.2 cm
(b) 2 (ii) –9 60° 60°
5 (b) (i) J = 3p + 34 60°
16. (a) –
3 (ii) 24 years old
1 (c) 2
(b) –
6
17. (a) –8 6. (a)
20 Chapter
(b) – 4 Polygons
17 45°
45° 45°
18. (a) 242 45° 45°
(b) 25 Formative Practice 4.1 45° 45°
45°
19. 157.1 cm 3 67.5°
1. • Has 8 sides of equal length
20. (a) A = 2πr + 2πrh • All interior angles are congruent
2
(b) 40 cm • Has 8 axes of symmetry 3 cm
21. 4 cm 2. (a) (b)
A – 2y 2 x x
22. (a) h =
4y x x 40° 40°
(b) 5 40° 40°
x x 40° 40°
23. p = 8 + 4h + 44 + h 40° 40°
 2
x x 40°
q + 10
24. k = 70°
4
g + 5 Regular octagon 2.8 cm
25. h =
5
(b) Formative Practice 4.2
Summative Practice 3 x x x
1. (a) 540°
Section A x x (b) 1 620°
1. A 2. D 3. C 4. C 5. B x x (c) 3 240°
6. B 7. B 8. B 9. A 10. D
2. (a) 1 080°
Section B Regular heptagon
(b) 720°
1. (a) P = 8q (c) 540°
4 (c)
(b) = 1
5m – c x x 3. (a) 168°
4 = 5m – c
(b) 30°
5m = 4 + c x x
4 + c 4. (a) 108°
m = x
5 (b) 128.57°
(c) 140°
Regular pentagon
2. (a) (d) 135°
1
Formulae v = u + at —(a + b)t = s √p + q = r 3. (a) 9 5. (a) 162°
2
Subject (i) v (ii) s (iii) r (b) 6 (b) 150°
(c) 7
(b) (c) 156°
Variable Possible values (d) 168.75°
4. (a) True
r = Number of –5, 0 , 5 , 10.5 (b) True
students who 6. (a) 115°
attend extra (c) False (b) 103°
class (d) True (c) 124°
5. (a) 7. (a) 72°
Section C (b) 36°
3 72°
1. (a) (i) z = x – 6y 2 72° 2.5 cm (c) 22.5°
4
(ii) –30 72° 72° 8. (a) 76°
 72° (b) 22°
p
2
(b) h =
 2
(c) 79.5°
178
F2 Answers.indd 178 06/03/2023 4:53 PM

Mathematics Form 2 Answers

9. (a) 12 2. (a) (i) 36° (b)
(b) 30 (ii) 120°
(c) 60 (b) 56° O
10. (a) 12 (c) 20 Q P
(b) 24 3. (a) (i) 36°
(c) 30 (ii) 65° (c)
11. (a) 34 (b) 9 P
(b) 18 (c) x = 36°, y = 126° O Q
(c) 23 4. (a) 110°
(d) 36 (b) 8
12. 72° (c) x = 27°, y = 126° 4. (a), (b)
13. 72° M
Chapter 2.6 cm
14. 140° 5 Circles 3.3 cm
15. 12 3.7 cm
16. 70° Formative Practice 5.1
17. 127° 1. (a) (i) Centre
(ii) Chord 5.
Summative Practice 4 (iii) Radius 85° O
(iv) Diameter 4 cm
Section A (v) Minor sector
1. A 2. A 3. D 4. A
(vi) Minor arc
Section B 6. (a)
(vii) Minor segment
1. (a) (i) False (b) (i) Points lying on circumference 145°
O
(ii) True are equidistant from O. 3.4 cm
(b) (i) 15 (ii) Length of line PS is twice that
(ii) 30 of line OQ.
2. (a) (b)
2. (a) (i) 360°
(ii) 3m 2.2 cm 76° O
(b) (i) 900° 2.5 cm
(ii) 1 260°
3. (a) 54°
(b) (b) Formative Practice 5.2
Sum of interior
Polygon 3.1 cm 1. (a) True
angles
(b) False
Nonagon 1 260°
(c) True
20 sided 3 240° (d) True
polygon
(c) 3. 5 cm
Section C 3.3 cm 4. 48.99 cm
1. (a) (i) 24°
5. It is known that OP = OQ = OS (radii)
(ii) 40 and OQ = OS = QR (sides of the
(b) rhombus). Therefore, OSR and OQR
(d) are equilateral triangles. Thus, ∠POS =
U T ∠QOR = ∠SOR = 60°. Hence, arc PS
2.3 cm = arc SR = arc RQ.
V 45° S
45° 45° TIMSS
45° 45° 1
45° 45° 113 cm 2
W 45° R 7
67.5° 3. (a) Q
P Q Formative Practice 5.3
P O
(c) (i) 36° 1. (a) C = 2πm
(ii) 2 angle P and 1 angle Q (b) C = πm



179





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Mathematics Form 2 Answers

2. (a) A = πq 2 16. (a) 16.87 cm Chapter
πp 2 6 Three-Dimensional
(b) A = (b) 16.46 cm Geometrical Shapes
4
(c) 13.04 cm
3. (a) 44 cm (d) 12.62 cm Formative Practice 6.1
(b) 66 cm
17. 188 cm
(c) 66 cm 1. (a) Right prism. A triangular cross-
(d) 14.67 cm 18. 31.5 cm 2 section and three other faces are
rectangles.
19. 141.43 m
4. (a) 50.27 cm (b) Right cylinder. Two parallel circular
(b) 47.13 cm 20. 32 cm 2 faces and one curved face.
(c) 49.02 cm 21. (a) 132 cm (c) Right square pyramid. A square base
and four other faces are isosceles
(d) 39.59 cm (b) 462 cm 2 triangles.
5. (a) 17.5 cm 22. (a) 110 cm (d) Sphere. A curved face.
(b) 31.5 cm (b) 275 cm 2 (e) Right pentagonal pyramid. A
pentagonal base and five other
(c) 22.75 cm faces are isosceles triangles.
Summative Practice 5
6. (a) 26.1 cm 2. (a) 6, 6, 10
(b) 25.14 cm Section A (b) 6, 8, 12
(c) 0.9548 m or 95.48 cm 1. C 2. A 3. B 4. C (c) sphere
(d) cube, cuboid, prism, pyramid
7. (a) 616 cm 2 Section B (e) prism, cylinder
(b) 13.86 cm 2 1. (a) (i) ✗
(c) 98.56 cm 2 (ii) ✓ Formative Practice 6.2
(d) 46.59 cm 2 (b) (i) longest chord
(ii) centre 1. (a) 1, 1
8. (a) 201.09 cm 2
(b) 283.57 cm 2 2. (a) OP, OR, OT, OV
(c) 415.53 cm 2 (b) QU, QS, SV
(d) 514.79 cm 2 3. (a) (i) chord PQ = chord QR
(ii) PS = SQ
9. (a) 3.5 cm (b) 2, 1
(b) (i) arc STUP
(b) 21 cm
(c) 21 cm (ii) arc UPQR
4. (a) Minor segment
10. (a) 18 cm (b)
(b) 11 cm Radius Diameter Circumference Area
(c) 7.2 cm
(i) 4 cm 8 cm 25.14 cm 50.29 cm 2
11. (a) 5.13 cm (ii) 9 cm 18 cm 56.57 cm 254.57 cm 2 (c) 1, 5
(b) 19.27 cm
(c) 35.19 cm Section C
1. (a) 18 cm
12. (a) 51.5°
(b) 164.47 cm 2
(b) 341.5° (c) 24 cm
(c) 150.5°
(d) 45.8° 2. (a) (d) 2, 4
S
13. (a) 155.63 cm
(b) 57.29 cm O Q
(c) 57.77 cm
(d) 33.42 cm
P
14. (a) 21.39 cm 2 2. (a)
(b) 218.3 cm 2 (b) 102 cm 5 cm
(c) 122.5 cm 2 (c) 63 cm
2
15. (a) 20° 3. (a) 127.29 m 4 cm
(b) 38° (b) 479 1 cm 2 6 cm
9
(c) 125° (c) 45π cm 2 5 cm
(d) 78°



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Mathematics Form 2 Answers

(b) 3. (a) 616 cm 2 11. 4 356.91 m 3
(b) 804.35 cm 2 12. (a) 10 039.14 mm 3
(c) 1 386 cm 2 (b) 79.05 g
4. (a) 51 816 mm
2
6 cm (b) 22 284 mm 2 Summative Practice 6
6 cm (c) 47 506 mm 2
Section A
5. (a) 15
8 cm 1. A 2. C 3. B 4. B 5. C
(b) 13
(c) 7 Section B
(c) (d) 8 1. (a) mp
(e) 4 (b) 16πp
7 cm 1
(f) 11 (c) V
(g) 5 3
44 cm (d) 3V
10 cm 6. 277 2. (a) False
7. 2 067 cm 2 (b) True
7 cm (c) True
8. 704 cm 2
(d) True
9. 240.31 cm 2
3. (a) (i) Triangular prism
10. (a) 1 771.3 cm 2
3. (a) (ii) Cylinder
4 cm (b) 17.71 m 2
5 cm (b) (i) Sphere
(ii) 4
4 cm Formative Practice 6.4
6 cm Section C
7 cm 1. (a) 1 281.28 cm 3 3
(b) 252 cm 3 1. (a) 1 488 cm
Quadrilateral prism (b) (i) 10.5 cm
(c) 90 cm 3
(d) 513.33 cm 3 (ii) 11 434.5 cm 3
(b) (c) p = 13; 686.85 cm 3
(e) 1 437.33 cm 3
12 cm (f) 90 cm 3 2. (a) 2 262.24 cm
3
(g) 1 732.5 cm 3 (b) 36 cm
5 cm 3 2
(h) 26.67 cm (c) (i) 339.34 mm
Cone (i) 154 cm 3 (ii) 452.45 mm 3
2. (a) 246 cm 3. (a) If the height of the pyramid is
3
(c) preserved, the volume is proportional
(b) 1 005.71 cm 3
7 cm 3 to the base area. Therefore, the
(c) 720 cm base area will also be doubled.
3. 2 310 cm 3 (b) 10.18 cm
5 cm 4. (a) 141.17 cm 3 (c) (i) 9.5 cm
(b) 0.656 (ii) 269 cm 2
Pentagonal pyramid 3
5. (a) 56 cm 2 4. (a) 460 cm
Formative Practice 6.3 (b) 270 cm 2 (b) (i) 9.77 cm 3
(c) 50 mm 2 (ii) 1 691.51 cm
1. (a) 150 cm 2 (c) 160π cm
2
(b) 192 cm 2 6. (a) 4
(c) 198 cm 2 (b) 12
(d) 245.08 cm 2 (c) 3 Chapter Coordinates
7
(e) 251.36 cm (d) 3
2
(f) 60 cm 2 (e) 6
(f) 13 Formative Practice 7.1
2
(g) 78.59 cm
(h) 225.41 cm 2 7. 6.8 cm 1. K(–2, 4), L(4, 3), M(6, –2), N(–3, –1),
(i) 107.97 cm 2 8. 905.14 mm 3 P(3, 0), Q(0, –2)
2. (a) 1 809.79 cm 2 9. 36 cm 3 2. (c) ✓
(b) 3 850 cm 2 10. 39.82 cm 3 3. (a) 5 units
(b) 7 units



181





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Mathematics Form 2 Answers

4. (a) 4 units 4. (a) (–8, 2) (c) L
(b) 4 units (b) 17.89 units
(c) 4 units 5. (a) 7 10 8
(d) 9 units (b) 36 units 6
(e) 5 units 6. (a) (5, 2) 4
(f) 6 units (b) 9.49 units 2
(g) 4 units 0 K
(h) 12 units 7. A(2, 5), D(–2, 1) 1 2 3 4 5
5. (a) 5 units 8. (a) (1, 3) (d) L = K + 5
(b) 10 units (b) P 4. (a) No, because when x = 1, there are
(c) 13 units two corresponding values of y.
(d) 25 units Summative Practice 7 (b) No, because when x = 20, there are
(e) 12.08 units Section A two corresponding values of y.
(f) 12.21 units 1. C 2. D 3. B 4. C 5. D (c) Yes, because each value of x has
(g) 16.16 units 6. C 7. A 8. B only one corresponding value of y.
(h) 21.26 units (d) Yes, because each value of x has
Section B only one corresponding value of y.
6. 2 units
1. (i) ✗ 5. (a) One-to-one function
7. 4, –12 (ii) ✓ (b) Many-to-one function
8. 21.32 units (iii) ✗ (c) One-to-one function
9. (a) 100.32 km (iv) ✓ (d) Many-to-one function
(b) Town Gaya and Town Molek 2. (i) Yes 6. (a) One-to-one function
10. (a) (–7, 6) (ii) No (b) Many-to-one function
(b) 10.77 units (iii) Yes (c) One-to-one function
(c) 28 units (iv) No
7. Independent variable: temperature
Formative Practice 7.2 Section C Dependent variable: volume
1. (a) (i) p = –5, q = 16 Type: one-to-one function
1. (a) No
(b) Yes (ii) 52 cm
(b) (i) 6 Formative Practice 8.2
2. (a) (5, 3)
(ii) 13 units
(b)  5 2 , –4  (c) (i) 3 1. (a) x –1 0 1 2 3 4
(c) (–7, 4) (ii) 20 units y –11 –7 –3 1 5 9
(d) (1, 7) 2. (a) ±6
(e) (3, 9) (b) 42 units 2
(f) (4, –3) (c) (i) 8 km (b) x –2 –1 0 1 2
(g) (3, 2) (ii) 24 km
(h)  7 2 , – 11  y –1 –3 –3 –1 3
2
Chapter
3. (13, 4) 8 Graphs of Functions
4. (–2, 2) (c) x 1 2 3 4 5
5. (4, 5) Formative Practice 8.1 y 6 3 2 1.5 1.2
6. 2, –10 1. (a) one-to-one
7. m = 10, n = 5 (b) {1, 2, 3} (d) x 1 2 3 4
(c) {4, 8, 12}
8. (1, 4)
Formative Practice 7.3 2. 6 y 15 3.75 1.67 0.94
1. (a) (4, 4) 3. (a) K 1 3 5
(b) (5, 7), (–1, 7) (e) x –3 –2 –1 0 1 2 3
L 6 8 10
2. (a) 7.81 units
(b) 30 units 2 y –54 –16 –2 0 2 16 54
(b) {(1, 6), (3, 8), (5, 10)}
3. (a) 14
(b) 30 units



182





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Mathematics Form 2 Answers

2. (a) y (f) y 9. (a) y = 1
8 8 x
6 6 (b) y(m)
y = x + 2 1.0
4 4
y = x 3 0.8
2 2 y = 1
0.6 x
x x 0.58
– 4 – 2 0 2 4 6 –2 –1 0 1 2 0.4
–2 – 2
0.2
– 4 x(m)
0 1 2 3 4 5
(b) y – 6
4 – 8 (c) y = 0.59
2 y = 2 – x
1
x 3. 10. (a) A = (2x)(12 – x)
– 2 – 1 0 1 2 3 G 2
–2 = x(12 – x)
200
= 12x – x 2
150 (b)
2
(c) y A (cm )
5
10 100 G = T 40 A = 12x – x 2
3
36
8 9 50 30
y = 20
6 x T
4 0 30 60 90 120 10
2 0 2 4 6 8 10 x (cm)
x 4. y(m)
–3 –2 –1 0 1 2 3 y = 4x – x + 1
2
– 2 5 (c) (i) 2 cm and 10 cm
– 4 4 (ii) 36 cm 2
– 6 3 11. (a) V = x(x + 2)(x – 1)
= (x + 2x)(x – 1)
2
– 8 2 = x – x + 2x – 2x
2
2
3
= x + x – 2x
3
2
–10 1
(b) 3
x(m) V (cm )
0 1 2 3 4 80
(d) y
12 60 V = x + x – 2x
2
3
5. (a) 25 seconds 40
10 (b) The faster the average speed, the 24
lesser the time taken. 13 m s . 20
–1
8
x (cm)
6 12 6. (a) 400 m 0 1 2 3 4
x 2 (b) 500 m
4
(c) If the distance between Azlan and (c) 24 cm 3
2 town A is increasing, then the
distance between Azlan and town B
x
0 1 2 3 4 will be decreasing. Summative Practice 8
7. (a) 3 seconds. 45 m.
(e) y (b) The distance of the rocket from the Section A
6 horizontal ground is increasing and 1. A 2. B 3. B
reaching maximum after 3 seconds
4 y = 5 – x 2 and then it is decreasing. 6 seconds. Section B
2 1. (a) ✗
8. (a) (i) RM2 500
x (ii) RM2 000 (b) ✗
–3 –2 –1 0 1 2 3 (c) ✓
–2 (b) 200
(c) 100 or 300 (d) ✓
– 4
183
F2 Answers.indd 183 06/03/2023 4:53 PM

Mathematics Form 2 Answers

2. (a) (b) (i) Chapter
9 Speed and Acceleration
Volume, y(cm ) 3
p = q + 1 Formative Practice 9.1
2
60 y = x 3
51
50 1. (a) The car travels 55 km in 1 hour.
1
r = Functions (b) The ball moves 8 m in 1 second.
s 40
k = 3h 30 2. (a) Uniform speed
8 (b) Non-uniform speed
m = – 2
n 20 (c) Non-uniform speed
10 (d) Uniform speed

Length of sides,
0 1 2 2.7 3 4 x(cm) 3. (a) Uniform speed
(b)
(b) Uniform speed
(c) Non-uniform speed
x 3 6 9 (ii) Volume of cube = 51 cm , (d) Non-uniform speed
3
y 5 8 11
Side of cube = 2.7 cm
4. (a) 2.88 km/h
7 Many-to-one (c) y = x + 5
3
y = (b) 150 cm/s
x 2 function
y 3. (a) (c) 20 m/s
5. (a) 1.6 m/s
One-to-one h (cm) (b) 40 km/h
function 70
0 x h = 32t – 4t 2 6. 15 km
60
(10, –1), (20, –2),
(30, –3), (40, –4) 50 7. 50 minutes
40 8. (a) 54 km/h
Section C 30 (b) 6.67 minutes
1. (a) 12 20 9. 8.31 km/h
(b) (i) 10 10. 1.44 km
t
x –3 –2 –1 0 1 2 3 0 1 2 3 4 5 6 (second) 11. (a) 1542 hours
(b) 82.84 km/h
y 0.89 2 8 undefined 8 2 0.89 The maximum distance of the stone
from the horizontal ground is 64 cm. 12. 87.5 km/h
The time for the stone to hit the
(ii) ground is 8 seconds. 13. 28 km
y
(b) (i) 14. No
8
7 P (kPa) 15. (a) 1510 hours
16 (b) 70 km/h
6
y = 8 14
5 x 2 Formative Practice 9.2
12
4 150
10 P = 1. (a) The speed of the runner is increasing
3 V at 1 m/s every second.
8
2 (b) The speed of the runner is
6 decreasing at 3 m/s every second.
1 1.2 5
–2.1 2.1 x 4 2. (a) 2 160 km/h per minute
–3 –2 –1 0 1 2 3 2 (b) 0.08 m/s 2
3
V (cm ) 3. 1.25 m/s 2
(iii) When x = 2.5, y = 1.2, 0 10 20 30 40 50 60
When y = 1.8, x = –2.1 and 2.1 4. 45 km/h per minute
(ii) 5 kPa. When the volume of gas
2. (a) (i) 10 is increasing, the gas pressure 5. 18.72 s
(ii) one-to-one is decreasing.
6. 24 km/h



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Mathematics Form 2 Answers

7. (a) 2 m/s 4. (a) CD Section B
(b) 104.17 s (b) AB 1. (a)
8. 2.33 s 5. (a) AB
(b) EF
9. Engine A; because the engine gives a
higher acceleration. 6. (a) l and l
3
2
(b) l and l 6 Inclines
1
10. (a) 6 s 7. (a) 2 upwards to
(b) 2.5 m s –2 7 the right
(b) – 9
11. 13.82 km/h
8. (a) 2
7
Summative Practice 9 (b) 7
8 Inclined
(c) 5 downwards
Section A 4 to the right
1. C 2. B
9. (a) –3
Section B
(b) 3
1. (a) ✓ (c) –4 (b) (i) Straight line AB is steeper than
(b) ✗ (d) –3 the straight line AC.
(c) ✗ (ii) Absolute value of gradient AB is
(d) ✓ 10. (a) 2 greater than AC.
(b) 1 2. (a) l
2. (a) Acceleration 15 (b) l , l
7
(c) – 1 5
(b) Acceleration 7 (c) l , l 4
3
3
(c) Deceleration (d) – (d) l , l 6
2
2
(d) Deceleration 1 3.
(e) 3
Section C
1. (a) (i) 5 km (f) 1
(ii) 24 seconds 11. (a) 2
(b) 1824 hours (b) 2
(c) 7
(c) 1.2 minute
2 Section C
(d)
2. (a) (i) 48 km/h 3 1. (a) (i) EF, GH
(ii) 4 140 km (e) 1 (ii) (0, –1)
(b) 15 minutes (f) 3 2 (b) 12
(c) 3.2 m/s 2 1 5
12. (0, – ) (c) (i) 1 200 cm
2
3. (a) 64.8 km/h 3
13. –3 (ii)
(b) Wahida, 1.33 s 4
14. 5
(c) The movement of the car is in the 2. (a) (i) Negative gradient
safe state while passing the corner 15. 5 (ii) The value of y will decrease
because the speed is 64 km/h. because the gradient of the
16. –2
( 65 km/h). straight line is negative.
17. 4 (b) 15
2 (c) (i) –5
18. (a) –
Chapter Gradient of a Straight 5 (ii) (–13, –12)
10
Line (b) 16
5
Formative Practice 10.1 19. a = –4, b = –1 Chapter Isometric Transformations
11
1. PQ and QR incline upwards to the right. 20. 9.22 units
RS inclines downwards to the right. Formative Practice 11.1
y
2. (a) m = – Summative Practice 10 1. (a) The object is flipped. Position and
x
q orientation change but shape and
(b) m =
p Section A size are preserved.
1. D 2. C 3. C 4. B 5. A (b) The object is shrunk. Position
3. (a) l and l 7 and size change but shape and
4
(b) l and l 6 6. D 7. A 8. C 9. B orientation are preserved.
2
185
F2 Answers.indd 185 06/03/2023 4:53 PM

Mathematics Form 2 Answers

(c) The object is moved. Position (iii) (d) (i)
changes but shape, size and
orientation are preserved. N
(d) The object is rotated. Position Q N
changes but shape, size and
orientation are preserved.
2. (a) Position changes but shape, size k
and orientation are preserved. N N
(b) Position and orientation change but P
shape and size are preserved.
(c) Position changes but size, shape ⎯→  
–1
and orientation are preserved. PQ or 5 or (ii) 5 units to the right followed by 4
(d) Position and size change but shape (x, y) → (x – 1, y + 5) units upwards
and orientation are preserved.
(b) (i) (iii) Q
3. (a) Position and size change but shape
and orientation are preserved.
(b) Position changes but shape, size N
and orientation are preserved. N
4. (a) Congruent. Object and image have P
the same shape and size. k N
(b) Similar. Object and image have the N
same shape but different size.
(c) Congruent. Object and image have
the same shape and size. (ii) 5 units to the right followed by 3
5
⎯→
4
(d) Congruent. Object and image have units downwards PQ or   or
the same shape and size. (x, y) → (x + 5, y + 4)
5. (a) Congruent (iii)
5. (i)
(b) Similar
N
Formative Practice 11.2 (e) (a)
P
1. (a) Yes. Each point on the object is N
moved towards the same direction P
through the same distance. (b)
(b) No. Object and image are not Q (d)
congruent. (c)
5
⎯→
–3
2. P is not an image. Orientation has PQ or   or
changed.
(x, y) → (x + 5, y – 3) (ii) (a) 7 units to the right followed by 4
Q is not an image. Shape has changed. (c) (i) units upwards
R is an image. Every point on M is (b) 6 units to the right followed by 1
moved towards the same direction N unit downwards
through the same distance.
(c) 3 units to the right followed by 4
S is not an image. Points on M are units downwards
not moved towards the same direction N k (d) 5 units to the left followed by 2
through the same distance. units downwards
3. (a) True (e) 4 units to the left followed by 4
(b) False units upwards
(c) True (ii) 2 units to the left followed by 4
units downwards 7
4
4. (a) (i) (iii) (a)  
(iii)
P 6
N  
N (b) –1
3
k (c)  
–4
Q N
N
–5
–2
(d)  
–4
4
–2
(ii) 1 unit to the left followed by 5 PQ or   or (e)  
⎯→
–4
units upwards
(x, y) → (x – 2, y – 4)
186

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Mathematics Form 2 Answers

6. (a) (i) (d) (i) y (b)
y
4 P
4 P
P
2 2 k
k x P
x –2 O 2 4
–4 –2 O 2 4 –2 P
–2
P
Translation by the movement of
Translation by the movement of arrow k.
arrow k. (c)
(ii) Translation of moving 1 unit to
(ii) Translation of moving 3 units the right followed by 4 units P
to the right followed by 5 units upwards.
downwards.
1
4
(iii) Translation   or translation P
3
–5
(iii) Translation   or translation (x, y) → (x + 1, y + 4).
(x, y) → (x + 3, y – 5).
7. (a)
(b) (i) y r
4 (d)
P
2 P
k
x P
–4 –2 O 2 4 P
–2 P
–4
P
(b)
Translation by the movement of
arrow k. P
9. (a) (5, 4)
(ii) Translation of moving 2 units P (b) (–4, 7)
to the left followed by 7 units (c) (1, –6)
upwards.
(d) (4, –6)
–2
7
(iii) Translation   or translation
(x, y) → (x – 2, y + 7). 10. (a)
(c) P
(c) (i)
y R
P P Q
4
2 R
k P
x
–4 –2 O 2 4
P –2

(b)
Translation by the movement of 8. (a)
arrow k.
R
Q
(ii) Translation of moving 2 units
to the left followed by 5 units P
downwards. R
–2 P P
–5
(iii) Translation   or translation
(x, y) → (x – 2, y – 5).



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Mathematics Form 2 Answers

11. (a) 16. (2, 14) (b)
Q P
17. p = 5, q = 8
R
Formative Practice 11.3
P
M M
S 1. (a) They do not overlap. It is not a
reflection.
(b) They overlap. It is a reflection.
(c) They overlap. It is a reflection. Q
(b)
Q (d) They do not overlap. It is not a (c)
reflection. P
P
2. (a) True
R (b) True M
(c) False
S
(d) False M
3. Reflection on the line parallel to the
x-axis that passes through point (0,1). Q
12. (a)
4. 7. (a) (5, 3)
P P (b) (1, –3)
K
(c) (5, 5)
T (d) (–3, 5)
8. (a)
r Q
J
R
Q
(b)
Reflection on line PQ.
5. (a) Reflection on the line passing P R
P through (0, 1) and parallel to the
T x-axis.
(b)
P (b)
P
N R

(c)
M R Q
Q

T Reflection on line PQ.
(c) Reflection on the line passing 9. (a) (4, –3)
through (1, 0) and parallel to the (b) (0, –3)
P
y-axis.
(c) (–4, 7)
6. (a) (d) (3, 4)
13. (a) (–3, 4) M 10. (3, 1)
(b) (3, 2) 11. y
(c) (0, 0) P Q B
(d) (–6, 1) 4
14. (a) (5, –3) M 2 P
(b) (–4, 1) A x
(c) (–4, 0) –2 O 2
C
(d) (7, 0) –2
–3
–4
15.   or (x, y) → (x – 3, y – 4) or other
equivalent methods.




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Mathematics Form 2 Answers

12. (11, 2) 4. (a) y (d)
13. A 4
A B D C O T
2
P B
C D x T
–4 –2 O 2 A 4
P
A –2
8.
Rotation of 180° about (0, 2). y
14. 53° (b) y 4 A
D 4 C
TIMSS C 2 B C
Rotation 12.5 clockwise at O. A 2
o
B A –4 –2 O 2 4 x
D
Formative Practice 11.4 –2 O 2 4 x A B –2
1. (a) A rotation –2 B C –4
(b) A rotation
(c) Not a rotation Rotation of 90° anticlockwise about 9. (a) (6, –5)
(–1, –1).
2. (a) True 5. (a) Rotation of 90° clockwise about (–8, (b) (1, –3)
(c) (4, 6)
(b) True 4). (d) (–5, 7)
(c) False (b) Rotation of 180° about (2, 4).
(d) True (c) Rotation of 90° anticlockwise about 10.
(–4, –5).
3. (a)
(d) Rotation of 90° anticlockwise about M
(3, –3).
6. H
P
M I P
M
O I J
11. (–1, –1)
Rotation of 90° clockwise about O.
(b) 12. (a)
H
M M J O
O
7. (a)
W
T
Rotation of 180° about O.
(c)
O
120° M T (b)
M
(b)
O
T
Rotation of 120° anticlockwise about O W
O.
(d)
P 60° T (c)

(c) W
M M O
T T
O

Rotation of 60° clockwise about P.



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Mathematics Form 2 Answers

(d) Formative Practice 11.6 3. (a) Q
W 1. (a) It has rotational symmetry
M
(b) It has rotational symmetry O
O
(c) It does not have rotational symmetry
N
(d) It does not have rotational symmetry
P
2. (a) Rotational symmetry of order six
3
–3
13. (a) (9, –1) Translation   .
(b) (10, 2) (b) Rotational symmetry of order two Reflection on line PQ.
(c) (1, –1) (c) Rotational symmetry of order four Rotation of 180° about O.
(d) (4, –7) (d) Rotational symmetry of order two (b) (i) (6, –1)
14. (5, 2) 3. (a) It does not have rotational symmetry (ii) (–2, –1)
(b) It has rotational symmetry of order (c)
15. (a) R eight A
(b) 56° (c) It has rotational symmetry of order D
two M
16. (4, –2)
(d) It has rotational symmetry of order C B
four B T C
17. Rotation of 135° anticlockwise about O.
Rotation of 225° clockwise about O.
Summative Practice 11 D M
18. 70 A
Section A
1. A 2. C 3. C 4. D
Formative Practice 11.5 Chapter Measures of Central
12
Section B Tendencies
1. 30 cm 2 1. (a) (i) Congruent
Formative Practice 12.1
(ii) Two
2. Same length
(b) 1. (a) mode = 3.46
3. (a) ✓ Reflection (b) mean = 3.57
(b) (c) median = 3.55
Rotation
(c) ✓ 2. (a) mode = RM7 000
mean = RM9 250
2. (a) (i) ✓
4. Yes. An isometry preserves size and median= RM8 250
shape. (ii) (b) No. The monthly salaries for 8 out
of 9 officers are less than calculated
(b) (i) Rotation
5. Yes. Both reflection and rotation are mean. The extreme value of
isometries. (ii) Reflection RM22 000 has given a significant
effect on the mean.
Section C
6. (a) False 3. (a) RM10 000
(b) True 1. (a) (i) P: Rotation (b) RM9 800
(c) True Q: Reflection (c) RM10 000
(ii) • All corresponding sides of M 4. (a) new mode = 21, new mean = 25.8,
7. (a) 30 and N are of the same length new median = 18
(b) 105 cm 2 • All corresponding interior. (b) new mode = 13.5, new mean = 15.9,
angles of M and N are equal. new median = 12
8. 69 (b) (i) Move 300 m to the left followed (c) new mode = 21, new mean = 25.8,
by 200 m downwards. new median = 18
9. (ii) 360.56 m (d) new mode = 4.5, new mean = 6.9,
S (c) Rotation of 110° anticlockwise new median = 3
D R about Q.
A 5. 2.25
2. (a) 144° 6. (a) Mean will be shifted to a value lesser
Q (b) Under a rotaion, the object and than 7.
B P image has similar orientation.
C (b) Mean will be shifted to a value lesser
Shape E and shape F had different than 7.
orientation. (c) Mean will be shifted to a value more
10. x = 108, y = 89 (c) 13.74 seconds than 7.



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Mathematics Form 2 Answers

7. (a) 15. (a) 2 kg Section C
Lifespan
(hour) Tally Frequency (b) 2.875 kg 1. (a)
(c) 2.5 kg
163 – 166 /// 3 Time Frequency, Midpoint, fx
(minutes) f x
16. (a) 3.6 cm, 3.9 cm, 5.1 cm
167 – 170 //// // 7
(b) 4.15 cm 6 – 10 2 8 16
171 – 174 //// //// 10
(c) 4.15 cm 11 – 15 4 13 52
175 – 178 //// / 6
17. (a) Mean leap of Chun Wei = 2.65 m 16 – 20 5 18 90
179 – 182 //// 4 Mean leap of Raj = 2.71 m
Raj shows a better performance. 21 – 25 4 23 92
1 (b) Median leap of Chun Wei = 2.78 m
(b) 33 % 26 – 30 3 28 84
3 Median leap of Raj = 2.75 m
8. (a) 31 – 35 6 33 198
If median is considered, Chun Wei
shows a better performance. Median
Score Tally Frequency leap of Chun Wei is more larger than 36 – 40 2 38 76
Raj.
90 – 94 /// 3 41 – 45 4 43 172
(c) Range leap of Chun Wei = 0.58 m
95 – 99 //// 5
Range leap of Raj = 0.49 m (b) 31 – 35 minutes
100 – 104 //// / 6 Range leap for Raj is smaller, (c) 26 minutes
therefore Raj is more consistent
105 – 109 //// // 7 in each of his jumps compared to
Chun Wei. Raj will be chosen to 2. (a) Number
110 – 114 //// //// 10 represent the school. of 0 1 2 3 4 5 6
serving
115 – 119 //// //// 9
Summative Practice 12 Number
120 – 124 //// 4 of 7 6 4 5 4 3 1
student
125 − 129 //// / 6 Section A
1. A 2. C 3. C 4. D
(b) 29 students (b) mode = 0, mean = 2.2, median = 2
(c) 100 Section B (c) The number of servings of fruits and
9. (a) 4 students 1. Score Frequency vegetable taken by students did not
(b) RM16 – 20 meet the requirement suggested.
Most of the students took 2 or lesser
(c) RM19.38 1 3 servings of fruits and vegetables
2 5
10. (a) 50 – 59 km h –1 each day.
(b) 52.2 km h –1 3 4
3. (a) (i) 51 – 60 kg
(c) 2 : 3 4 2 (ii) 55.5 kg
11. Median is more appropriate because the 5 1 2
set has no mode and the extreme value, (iii) 36 %
3
RM155, which is larger than the other 5
values will influence the mean more than (b) (i) 4 cm
the median. 2. (ii) 11 cm
Mode 2.7 3.6
12. The most appropriate measure is mode. (iii) 48 cm
There are more tourists from Singapore Mean 3.0
compared to other countries. Mode of 4. (a) (i) Mode represent the scores that
composition of tourists is Singapore. Median 2.75
Median and mean do not give any occur most frequently, therefore
mode is not an appropriate
significant information about the measure to describe the
composition of tourists. 3. (a) (i) All whole number except 5 and
6 distribution of marks for the two
13. The most appropriate measure is mode. (ii) n  5, n is whole number groups of students. Since the
The most popular colour will win the scores are distributed uniformly
votes. Median and mean do not give any (b) and there is no extreme value
significant information on the choice of in the set, therefore mean and
colour. 9 median are appropriate to be
Mode used.
14. (a) RM100 10
Mean
(b) RM87.23 7
(c) RM50


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Mathematics Form 2 Answers

(ii) Arif Bijaksana 3. (a) (ii) Gʹ = {February, April, June,
September, November}
Group Group
Outcomes A B C D E
Mean 12 12 (d) (i) Event of the blood donor is not
from blood group O.
Probability 0.2 0.2 0.2 0.2 0.2
Median 12.5 12 (ii) Hʹ = {A, B, AB}
(b)
2
Range 7 4 2. (a)
Outcomes A B C D E 9
4
The values of mean and median (b)
9
for the two groups are almost the Experimental
same. Since the range for Bijaksana probability 0.25 0.19 0.14 0.22 0.19 (c) 5
Group is smaller than Arif Group, the 9
performance of Bijaksana Group is (c) The number of trials in this 3
more consistent than Arif Group. experiment is 36 times, which is not 3. (a) 10
large enough.
1
(b) 4, 4, 5, 6, 11 or 4, 4, 5, 7, 10 or (b)
4, 4, 5, 8, 9 2
4. (a) 4
(c) A value which is bigger than mean or (c) 5
median is removed from the original Outcomes 1 2 3 4 5 6
data. 4. 5
Probability 0.167 0.167 0.167 0.167 0.167 0.167 8
Chapter
13 Simple Probability 5. 0.55
(b) 0.175; 0.165 1
Formative Practice 13.1 (c) When the number of trials is 6. 2
getting bigger, the experimental
probability converging to the
1. (a) 0.03 theoretical probability. 7. The weather forecast of not raining
tomorrow is 75%; probability = 0.75
(b) 0.44 2
5. 1
9 7 8. (a)
2. Probability of Ann wins = 5
20 7
4
11 6. (a) (b)
Probability of Badrul wins = 15 5
20 8
3. 0.25 (b) 15
1 Formative Practice 13.4
4. (a) 0.75, 0.62. (c) 30
(b) Assuming that the shape of the can 1. (a) {TTT, TTH, THT, THH, HTT, HTH,
is unchanged, the difference in the 7. (a) 1 HHT, HHH}
results is due to the chances of the 20 1
can landing on the side or vertically. 1 (b)
(b) 2
4
5. 0.47 (c) 1 1
2. 12
1
6. (a) 0.32, 0.3316, 0.3336, 0.331 (d)
(b) The experimental probability 2 3. 1
approaching 0.33 (to the nearest 2 11 12
decimal places) when the number (e) 20 4. (a) 40
of simulations is large enough.
8. 3 8 (b) 11 or 0.275
TIMSS 40
A 5. (a) 48
Formative Practice 13.2 Formative Practice 13.3 (b) 0.56
1. (a) {Sunday, Monday, Tuesday, 1. (a) (i) Event of the traffic light not (c) 1 5
Wednesday, Thursday, Friday, turning green when a car
Saturday} passing through a junction. 1
(ii) Eʹ = {red, yellow} 6. (a)
3
(b) E = {Sunday, Saturday, Tuesday,
Thursday} (b) (i) Event of not getting a prime (b) 0
number when a dice is rolled.
2. (a) {CC, CA, CT, AC, AA, AT, TC, TA, (ii) Fʹ = {1, 4, 6} 7. 9
TT} 25
(c) (i) Event of not getting 31 days in
(b) (i) X = {CC, AA, TT} a month when a month is picked 8. 4
(ii) Y = {CT, AT, TC, TA} at random from a year. 25
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Mathematics Form 2 Answers

Section C (b) (i) 0  m  1
Summative Practice 13
1. (a) (i) S = {(Black, Moon), (Black, (ii) 1 – m
Section A Sun), (Black, Cloud), (c) 2
(Black, Star), (Black, 375
1. B 2. D 3. D 4. D 5. C
Rainbow), (White, Moon),
Section B (White, Sun), (White, 3. (a) (i) 1
Cloud), (White, Star), 6
1. (a) Getting a prime (White, Rainbow)} 1
number from a Certain (ii) 10
perfect squared (ii) X = {Black, Sun), (Black, Star),
1
list. (White, Sun), (White, Star)} (b)
4
(b) Getting a (b) (i) 30
picture from a (ii) 1 (c) P(Kumar) = 0.28
50 cents coin 10 P(Safwan) = 0.30
toss. Possible
(c) 3 Safwan choosen;
(c) Getting multiple 8 because P(Safwan)  P(Kumar)
of 2 from a list
of multiple of 4. 2. (a) (i)
(d) Getting number First rotation
30 from a list
of numbers Impossible 1 3 5 7
from 1 to 20.
1 (1, 1) (3, 1) (5, 1) (7, 1)
Second rotation 5 (1, 5) (3, 5) (5, 5) (7, 5)
2. 3 (1, 3) (3, 3) (5, 3) (7, 3)
0 0.5 1

C D B E A 7 (1, 7) (3, 7) (5, 7) (7, 7)

(ii) 1









































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