©PAN ASIA PUBLICATIONS
Extra Features of This Book
CHAPTER
1 Quadratic Functions and
Equations in One Variable
SMART concept map
SMART SCOPE • Identify and describe the characteristics of quadratic expressions in Page 3
Important Learning Standards
one variable.
• Recognise quadratic function as many-to-one relation, hence, The entire content of the
describe the characteristics of quadratic functions. 3 chapter is summarised in the
Contains Learning Standards 1.1 Quadratic • Investigate and make generalisation about the effect of changing the 4
values of a, b and c on graphs of quadratic functions, f (x) = ax 2 + bx + c.
Functions and
Equations • Form quadratic functions based on situations, and hence relate to 5 form of a concept map.
the quadratic equations.
©PAN ASIA PUBLICATIONS
(LS) that need to be achieved • Explain the meaning of roots of a quadratic equation. 5
• Determine the roots of a quadratic equation by factorisation method. 6
in each chapter. • Sketch graphs of quadratic functions. 6
• Solve problems involving quadratic equations. 7 1 CHAP.
Concept 4 Form
2
Words
• Effect of change/ Kesan perubahan
• Horizontal line test/ Ujian garis mengufuk One variable, x Power of x is a whole number Highest power of x is 2
• Many-to-one relation/ Hubungan banyak kepada satu
• Maximum point/ Titik maksimum
• Method of factorisation/ Kaedah pemfaktoran
• Minimum point/ Titik minimum Many-to-one relations Characteristics of quadratic
Form • Quadratic equation/ Persamaan kuadratik expressions ax 2 + bx + c Life situations
• Quadratic function/ Fungsi kuadratik
4 Mathematics Chapter 1 Quadratic Functions and Equations in One Variable
• Rate of change/ Kadar perubahan are identify based on
• Root/ Punca Determine the roots of a quadratic equation
CHAP. Example 9 • Real root/ Punca nyata identify Quadratic Functions and Equations form Quadratic equation ax 2 + bx + c = 0
1 • Variable/ Pemboleh ubah Quadratic functions in One Variable Mathematics Chapter 1 Quadratic Functions and Equations in One Variable
by factorisation method
Determine whether each of the following values
of x is the root of the quadratic equation Solve quadratic equations by the method of describe sketch meaning determine
x 2 + x – 6 = 0. factorisation.
(a) x = 1 (b) x = 2 (c) x = –3 1 Write the quadratic equation in the form
Solution: ax 2 + bx + c = 0. Characteristics based on Quadratic graphs f (x) = ax 2 + bx + c Roots of quadratic equations
(a) When x = 1, x 2 + x – 6 = 1 2 + 1 – 6 2 Factorise ax 2 + bx + c = 0 in the form
(mx + p)(nx + q) = 0.
= –4 ≠ 0 3 State mx + p = 0 or nx + q = 0. 1
∴ x = 1 is not the root of the equation investigate the effect of
x 2 + x – 6 = 0. 4 Solve the two linear equations in 3 to obtain Curved- Axis of symmetry of graph change of a, b and c
(b) When x = 2, x 2 + x – 6 = 2 2 + 2 – 6 x = – p —– or x = – q —–. n shape is parallel to the y-axis
m
= 0
∴ x = 2 is the root of the equation Example 10 Maximum or c = 0 b = c = 0 b = 0
x 2 + x – 6 = 0. minimum point
(c) When x = –3, x 2 + x – 6 = (–3) 2 + (–3) – 6 Determine the roots of the following quadratic a . 0 a , 0 a , 0 a . 0 a , 0 a . 0
equations by the method of factorisation.
= 9 – 3 – 6 (a) x 2 + 5x = 14 y y y y y y
= 0 (b) (3x + 2)(x – 1) = 3x + 13
∴ x = –3 is the root of the equation O x O x x
x 2 + x – 6 = 0. Solution: x x O x
Try question 9 in Formative Zone 1.1 (a) x 2 + 5x = 14 x +7 +7x O O O
x 2 + 5x – 14 = 0 x –2 –2x
(x + 7)(x – 2) = 0 x 2 –14 +5x
BRILLIANT Tips x + 7 = 0 or or x – 2 = 0
x = –7
x = 2
y (b) (3x + 2)(x – 1) = 3x + 13
y = x 2 + x – 6 The graph y = x 2 + x – 6 3x 2 – 3x + 2x – 2 = 3x + 13
3x 2 – 4x – 15 = 0
cuts the x-axis at x = –3 (3x + 5)(x – 3) = 0 3x +5 +5x
–3 O 2 x and x = 2. Therefore, the x –3 –9x
roots of the quadratic 3x 2 –15 –4x
equation x 2 + x – 6 = 0 are 3x + 5 = 0 or x – 3 = 0
the x-intercepts of the x = – 5 — or x = 3 Form
graph y = x 2 + x – 6.
brilliant tips Sketch graphs of quadratic functions Try question 10 in Formative Zone 1.1 1.1 Quadratic Functions and (c) The expression 6t 2 + pt – 9 contains two CHAP.
4
3
Mathematics
Chapter 1 Quadratic Functions and Equations in One Variable
Graphs of quadratic functions of the form Equations variables p and t. Therefore, 6t 2 + pt – 9 is not 1
a quadratic expression in one variable.
1 —
Identify and describe the characteristics of (d) The expression y – 7y 2 contains one variable 1 —
y = a(x + m) 2
y = a(px + m)(qx + n)
y = ax 2 + bx
y = ax 2 + c
Useful tips to help (a) a . 0 (a) a . 0 (a) a . 0 quadratic expressions in one variable y. However, the power of y in the term 7y 2 is
(a) a . 0
1 —
not a whole number. Therefore, y – 7y 2 is not
y
y y y 1. Quadratic expression in one variable is an a quadratic expression in one variable.
algebraic expression of the form ax 2 + bx + c, a, b
and c are constants, a ≠ 0 and x is a variable.
students solve c x x 2. Characteristics of quadratic expressions in one Try question 2 in Formative Zone 1.1
x
m
O
n
O x O b – – a O –m variable: – –– p – –– q
(b) a , 0
problems in the (b) a , 0 y (b) a , 0 y (b) a , 0 y • Expressions contain only one variable. Spotlight portal
• The power of the variable is a whole number.
y
c O –m x • The highest power of the variable is 2. Quadratic equations in one variable
related subtopics. O x O b – – a x BRILLIANT – –– q n Tips https://bit.ly/3d1ZKeM
x
m O
– –– p
The variable x in quadratic expressions can also be
represented by other alphabet letters.
6 1.1.5 1.1.6 1.1.7 BRILLIANT Tips Scan the QR code to
If p represents a constant, then 6t 2 + pt – 9 is a
Example 1 quadratic expression in one variable, t.
4a 2 + b + 3 r 2 — – 2r –h 2 + 8h – 2 3t 2 + 5 — t browse website or video
2
Identify quadratic expressions in one variable Recognise quadratic function as many-
from the list of expressions above. Form to-one relation, hence, describe the
Chapter 2 Number Bases Mathematics 4 characteristics of quadratic functions related to the subtopics
Solution:
r 2
Example 12 2. Numbers in base two can be converted to 1. Quadratic function is a many-to-one relation. learned.
— – 2r, –h 2 + 8h – 2
2
numbers in base eight and vice versa based on
2. Characteristics of quadratic functions
calculator Convert the table above. Number in CHAP. • The graph has a curved shape.
Try question 1 in Formative Zone 1.1
(a) 1011101 2 to a number in base eight,
(b) 53 8 to a number in base two.
• It has a maximum point or a minimum point.
Example 2
• The axis of symmetry of the graph is parallel
base two
Solution:
(a) 1011101 2 = 1 × 2 6 + 0 × 2 5 + 1 × 2 4 + 1 × 2 3 Divide the digits of Replace each digit of the 2 to the y-axis.
Determine whether each of the following
expressions is a quadratic expression in one
+ 1 × 2 2 + 0 × 2 1 + 1 × 2 0 the number in base number in base eight
variable. Give your justification.
= 64 + 0 + 16 + 8 + 4 + 0 + 1 two into groups of with the corresponding BRILLIANT Tips
(a) 2m 2 – 9m + 5 three digits in base two.
= 93 10 8 93 Remainder three digits from
the right to the left.
(b) 5x 3 – 2x + 10 Ignore any zero in front
Explains how to use a = 135 8 8 8 11 … 5 3 Subsequently, replace two. Graphs of quadratic functions y = ax 2 + bx + c
(c) 6t 2 + pt – 9 of the number in base
the three digits with
1 …
0 … 1 the corresponding 1 —
digit in base eight.
(d) y – 7y 2
scientific calculator in (b) 53 8 = 5 × 8 1 + 3 × 8 0 2 43 Remainder Solution: Number in a . 0 a , 0 Form
= 40 + 3
base eight
Maximum point
= 43 10 2 21 … 1 (a) The expression 2m 2 – 9m + 5 contains one Chapter 1 Quadratic Functions and Equations in One Variable Mathematics 4
variable m and the highest power of m is
= 101011 2 2 2 10 … 1 0 2. Therefore, 2m 2 – 9m + 5 is a quadratic Minimum point CHAP.
Example 14
5 …
mathematics calculations. 2 2 … 1 Perform computations involving addition Sketch the graph for each of the following Sketch the graph for each of the following 1
Example 11
expression in one variable.
and subtraction of numbers in various bases
(b) The expression 5x 3 – 2x + 10 contains one
2 1 … 0 variable x. However, the highest power of x is Axis of symmetry Axis of symmetry quadratic functions. Mark the points where the
quadratic functions.
3. Therefore, 5x 3 – 2x + 10 is not a quadratic
0 … 1 1. Addition of number bases in various bases: (a) y = x 2 – 2 graph cuts the x-axis and the y-axis.
expression in one variable.
Alternative Method (a) Numbers in base two (b) y = –x 2 + 4 (a) y = (x – 1)(x – 3)
+ 0 2 1 2 (b) y = –2x 2 – 11x – 14
(a) 1011101 2 = 001 011 101 2 Solution:
0 2 0 2 1 2 (a) (b) y Solution:
= 1 3 5 8 y y = x 2 – 2
(b) 53 8 = 5 3 8 1.1.1 1 2 1.1.2 10 2 4 y = –x 2 3 + 4 (a) y y = (x – 1)(x – 3)
1 2
= 101 011 2 (b) Numbers in base three x 3
Calculator O O x
+ 0 3 1 3 2 3 –2
(a) Press: MODE MODE 3 BIN x
0 3 0 3 1 3 2 3 O 1 3
2 1 0 1 1 1 0 1 = OCT Try question 11 in Formative Zone 1.1
1 3 1 3 2 3 10 3 (b) y = –2x 2 – 11x – 14
(b) Press: MODE MODE 3 OCT
2 3 2 3 10 3 11 3 Example 12 = –(2x 2 + 11x + 14)
2 5 3 = BIN (c) Numbers in base four = –(2x + 7)(x + 2) y
Form Sketch the graph for each of the following
+
Try question 12 in Formative Zone 2.1 4 Mathematics Chapter 2 Number Bases quadratic functions. Mark the points where the 7 – – –2 O x
0 4
3 4
2 4
1 4
graph cuts the x-axis.
0 4 0 4 1 4 2 4 3 4 (a) y = x 2 + 3x (b) y = –2x 2 + 7x 2
–14
2 2
1 4• • • •
• • • •
• • • •
BRILLIANT Tips • • • • • • • • • • • • • • • • • • • • 1 4 2 4 2 4 3 4 10 4 3 4 10 4 11 4 9 × 8 + 3 (b) Place value 2 5 1 2 4 1 2 3 0 (a) Solution: 2 1 1 2 0 y 0 (b) y y = –2x 2 – 11x – 14
× 1
0
• • • •
Digit
• • • •
• • • •
2 4 • • • •
1. Table shows the digits for number in base eight • • • • • • • • • • • • • • • • 3 4 10 4 11 4 12 4 y = x 2 + 3x y = –2x 2 + 7x Try question 14 in Formative Zone 1.1
3 4
that correspond to the three digits for number • • • • • • • • ••• The place value of the underlined digit 0 = 2 3 x
The value of digit 0 = 0 × 2 3
in base two. CHAP. (d) Numbers in base five = 0 x O 7 – 2 Solve problems involving quadratic equations
9 eights and
Number 2 + 0 5 3 ones are 2 5 3 5 4 5 (c) Place value 7 2 7 1 7 0 –3 O
1 5
regrouped as
in base 0 1 2 3 4 5 6 7 0 5 0 5 1 sixty-fours, 2 5 3 5 4 5 Example 15
1 5
eight Digit 4 1 Try question 12 in Formative Zone 1.1
0
3 5
Number 1 5 1 5 1 eights and 3 4 5 10 5 In the diagram, PQRS is a rectangular plot of land.
2 5
ones.
The shaded region that is planted with brinjol has
Example 13
in base 000 001 010 011 100 101 110 111 2 5 2 5 3 5 4 5 10 5 11 5 The place value of the underlined digit 4 = 7 2 an area of 388 m 2 .
two The value of digit 4 = 4 × 7 2
3 5 4 5 10 5 11 5 12 5 = 196 Sketch the graph for each of the following S 30 m R
1 × 64
• • • •
• • • • • • • • • • • • • • • • • • • • 3 5 • • • • 4 5 10 5 11 5 12 5 + 1 × 8 13 5 (d) Place value 6 4 6 3 quadratic functions. Label its maximum point or x m
4 5 • • • •
6 2
6 1
• • • • • • • • • • • • • • • • • • • • • • • • + 3 × 1 Digit 2 0 minimum point. 6 0 N 20 m
• • • •
• • • •
5
1
(a) y = (x + 2) 2 1
(b) y = –2x 2 + 4x – 2
2.1.2 2.1.3 • • • • • • • • ••• 25 The place value of the underlined digit 2 = 6 4 P M
Solution:
The value of digit 2 = 2 × 6 4
75 = 1 × 64 + 1 × 8 + 3 × 1 = 2 592 (a) y = (x + 2) 2 y Determine the value of x. x m Q
Number in base eight to represent 75 is 113 8 . Try question 5 in Formative Zone 2.1 4 Solution:
Try question 4 in Formative Zone 2.1 (–2, 0) O x S 30 m R
tAgging ‘Try question ... 2. The place value for each digit of a number in Diagram shows number base blocks representing O y (1, 0) • x x m N 20 m
Example 6
(b) y = –2x 2 + 4x – 2
base a is a times greater than the place value of
= –2(x 2 – 2x + 1)
the digit on its right-hand side.
= –2(x – 1) 2
For example, the place values for each digit in a number in a certain base. Determine the number –2 (20 – x) m P (30 – x) m M x m Q
and represent it in terms of number value.
the number 32014 5 are shown in the following
in Formative Zone ...’ table. Place value 5 4 5 3 5 2 5 1 5 0 (a) (b) Try question 13 in Formative Zone 1.1 y = –2x 2 + 4x – 2 Area of the shaded region = 388 m 2 example
Digit 3 2 0 1 4
3. The value of a digit in the number x a is
determined by multiplying the digit with the Solution: 1.1.7 1.1.8 7
place value of its corresponding digit.
The tagging is located at the end Example 5 (a) Place value 6 1 6 0 Examples with complete
Digit 1 4
Based on the table of place values, determine
of the example guides the student the value of the underlined digit in each of the 14 6 = 1 × 6 1 + 4 × 6 0 solutions to enhance students’
= 6 + 4
following numbers.
(a) 1502 8 (b) 110010 2 = 10 10
(c) 410 7 (d) 20511 6 (b) Place value 5 2 5 1 5 0
to answer the corresponding Solution: Digit 3 2 2 understanding of the chapters
(a) Place value 8 3 8 2 8 1 8 0 322 5 = 3 × 5 2 + 2 × 5 1 + 2 × 5 0
questions in Formative Zone. Digit 1 5 0 2 = 75 + 10 + 2 learned.
= 87 10
The place value of the underlined digit 5 = 8 2 Try question 6 in Formative Zone 2.1
The value of digit 5 = 5 × 8 2
= 320
22 2.1.1
iv
Extra Features Spotlight A+ Mate F4.indd 4 15/03/2021 3:14 PM
Form
Chapter 1 Quadratic Functions and Equations in One Variable Mathematics 4
1.1 CHAP. 1 alternative method
1. 9y 2 + 16 x 2 – 2y 2 + 4 6. Diagram shows the graph y = x 2 . On the
diagram, draw the graphs y = x 2 + 4 and
7
—– – 20 1 — k 2 5 y = x 2 – 3.
v 2
Identify quadratic expressions in one variable y
from the list of expressions above. C1 8
2. Determine whether each of the following 6 Provides alternative solutions to
expressions is a quadratic expression in one
formative zone variable. Give your justification. C2 y = x 2 4 2 x certain questions.
(b) 1
—– – 6
(a) 7c 2 + 3
y 2
(c) a 2 + 4b 2 + 9
(d) 3p 2 – 10p + 5
3. Given the quadratic function y = –x 2 + 4. C2 –2 –1 –2 0 1 2
(a) Explain why the quadratic function is a Make generalisation about the effect of
many-to-one relation.
(b) Sketch the graph y = –x 2 + 4 by describing change for the values of c on the graph
y = x 2 + c.
C4
Questions to test 4. On one diagram, draw the graph y = ax 2 for 7. Diagram shows the graphs y = x 2 , y = x 2 – 6x
the characteristics of the quadratic function.
a = 1 — and a = – 1 — where –3 < x < 3. Hence, and y = x 2 + 6x. Form 4 Mathematics Chapter 1 Quadratic Functions and Equations in One Variable
2
2
students’ understanding at determine the relation between the graphs y = x 2 + 6x y y = x 2 y = x 2 – 6x CHAP.
y = 1 —x 2 and y = – 1 —x 2 .
Alternative Method
2 2 C4 0 x 1 30 × 20 – 1 —(30 – x)(20 – x) – 1 —x(20) = 388 y = ax 2 + bx + c
2
2
6
the end of each subtopic. 5. Draw the graphs y = 3 —x 2 and y = 4x 2 on the Make a generalisation about the effect of 600 – 1 —(600 – 50x + x 2 ) – 10x = 388 x = 1, y = 0: 25a – 5b + c = 0 ...............
3
–6
–3
a + b + c = 0 ...............
2
2
x = –5, y = 0:
following diagram.
2
y change for the values of b on the graph 600 – 300 + 25x – 1 —x 2 – 10x = 388 x = –2, y = –18: 4a – 2b + c = –18 ...........
– ,
24a – 6b = 0
16 y = x 2 + bx. C4 300 + 15x – 1 —x 2 = 388 4a – b = 0 ................................
2
– , 3a – 3b = –18
14 8. In the diagram, PQRS is a trapezium. 1 —x 2 – 15x + 88 = 0 – , a – b = –6 ..............................
2
3a = 6
12 S R x –8 –8x x 2 – 30x + 176 = 0 a = 2
(x – 8)(x – 22) = 0
10 (x + 5) cm x –22 –22x x = 8 or x = 22 From , b = 8
©PAN ASIA PUBLICATIONS
From , 2 + 8 + c = 0
8 x 2 +176 –30x c = –10
P Q y = 2x 2 + 8x – 10
6 When x = 22, 20 – x = 20 – 22 = –2. Length of NP
The length of PQ is 4 cm more than RS. Try question 16 in Formative Zone 1.1
4 C3 is a positive quantity. Therefore, x = 8.
2 y = x 2 (a) Form a quadratic function to represent the Try question 15 in Formative Zone 1.1
area of the trapezium PQRS.
–2 –1 0 1 2 x (b) Hence, find a quadratic equation in the Example 17
Hence, complete the following generalisation form ax 2 + bx + c = 0, given the area of BRILLIANT Tips A quadratic function y = ax 2 + bx + c has the
following information.
Form about the effect of change for the values of a the trapezium PQRS is 323 cm 2 . 1. Represent the desired quantity to be found by a • The axis of symmetry is x = 2.
C4
4 Mathematics Chapter 1 Quadratic Functions and Equations in One Variable values of x is the root of the quadratic equation suitable symbol such as x. • The graph cuts the x-axis at x = –1.
9. Determine whether each of the following
on the graph y = ax 2 .
When the value of a increases, the graph
• The graph has a maximum point.
x 2 – 4x – 5 = 0. C2
given information.
y = ax 2
CHAP. 1 10. Determine the roots for each of the following . 15. In the diagram, KLMN is a plot of land in the 2. Form a quadratic equation in terms of x from the Sketch the graph y = ax 2 + bx + c.
(a) x = 2
3. Solve the quadratic equation by the method of
spm simulation hots quadratic equations by the method of shape of a trapezium. (b) x = –1 4. Check the roots of the quadratic equation to Solution: y
factorisation.
(c) x = 5
C3
factorisation.
M
determine the value of x that represents the
(a) x 2 – 3x – 4 = 0
(b) 2x 2 + x = 10
quantity in the problem.
(c) (4x + 1)(x – 2) = –5
(d) (3x + 4)(x – 2) = 16 – 3x
questions 11. Sketch the graph for each of the following 15 m N Q x m 9 Diagram shows the graph of a quadratic function. –1 O 2 5 x
Example 16
quadratic functions. C2
(a) y = x 2 + 5
(b) y = 3x 2 + 4
y
P L
(d) y = –2x 2 – 3
(c) y = –x 2 + 1
K x m
12. Sketch the graph for each of the following Given KL = 10 m and LM = 25 m. The shaded –5 O 1 x
quadratic functions. Mark the points where region is a fish pond with an area of 188 m 2 . Try question 17 in Formative Zone 1.1
the graph cuts the x-axis. C2 Calculate the possible values of x. C5 (–2, –18)
(a) y = x 2 – 2x
Provide a complete solutions with 16. Diagram shows the graph of the quadratic Determine the quadratic function in the form BRILLIANT Tips
(b) y = –x 2 – 4x
function y = ax 2 + bx + c.
y = ax 2 + bx + c.
(c) y = 2x 2 – x
(d) y = –4x 2 + 8x y (2, 48) Solution:
the examiner’s comments for the y = a(x – 1)(x + 5) 1. The maximum point lies on the axis of
13. Sketch the graph for each of the following
symmetry.
When x = –2, y = –18,
quadratic functions. Label its maximum point
2. The distances of the points of intersection of the
or minimum point. C2
–18 = a(–2 – 1)(–2 + 5)
(a) y = 3(x – 2) 2 –2 O 6 x –18 = a(–3)(3) graph and the x-axis from the axis of symmetry
are equal.
–18 = –9a
SPM simulation HOTS questions. 17. A quadratic function y = ax 2 + bx + c has the a = 2 x = 1 ⇒ x – 1 is a factor
BRILLIANT Tips
(b) y = –(x + 4) 2
Determine the values of a, b and c.
(c) y = –x 2 + 6x – 9
C5
y = 2(x – 1)(x + 5)
(d) y = 4x 2 + 24x + 36
= 2x 2 + 8x – 10
14. Sketch the graph for each of the following following information. C5 = 2(x 2 + 4x – 5) x = –5 ⇒ x + 5 is a factor
quadratic functions. Mark the points where the
graph cuts the x-axis and the y-axis. C4 • The minimum point is (–3, –2).
• One x-intercept of the graph
(a) y = –(x – 2)(x + 4) y = ax 2 + bx + c is –2.
(b) y = (2x – 5)(x – 1) 8 1.1.8
(c) y = 3x 2 – 17x + 24 (a) Sketch the graph.
(d) y = –4x 2 – x + 18 (b) Determine the values of a, b and c.
Form
4 Mathematics Chapter 3 Logical Reasoning
SPM Simulation HOTS Questions EXAMINER’S
COMMENT
1. An aquarium in the shape of a cuboid has length (x + 9) cm, width x cm and height 50 cm. The volume Paper 1
of the aquarium is 31 500 cm 3 . Calculate the value of x. C3
1. Which of the following sentences is a statement? 6. Which of the following shows a false statement
Examiner’s Comments: C1 A What is the value of 100? C2 is changed to a true statement by using the
Volume = 31 500 CHAP. B The square of 8 is 64. word “no” or “not”?
(x + 9)(x)(50) = 31 500 50 cm 3 C Multiply both sides of the inequality –y –2 A The symbol π is a rational number.
x 2 + 9x = 630 by –1. The symbol π is not a rational number.
x 2 + 9x – 630 = 0 D Factorise the quadratic expression x 2 + 4x – 5. B x 2 = 8 is a quadratic equation.
(x – 21)(x + 30) = 0 x 2 = 8 is not a quadratic equation.
x = 21 or x = –30 x cm 2. I The lowest common multiple of 3 and 12 C 20 000 has one significant figure.
Since x . 0, x = 21. (x + 9) cm C2 is 12. 20 000 does not have one significant figure. summative zone
II The factors of 14 are 2 and 7. D Right-angled triangles have an angle of 90°.
III Write the formula for the area of Right-angled triangles do not have an angle
parallelogram ABCD. of 90°.
IV The sum of two odd numbers is an even
10 number. 7. Which of the following compound statements is
true?
Which of the following are statements? C4 A 5 – 6 = 1 or –8 + 3 = –11
A I, II and III B 0 ÷ 1 = 0 or 1 ÷ 0 = 0 Questions of various
B I, II and IV C 4 × (–2) = –6 or (–3) × (–1) = –3
C I, III and IV D 3 2 = 6 or 2 3 = 9
D II, III and IV levels of thinking skill
SPM MODEL PAPER 3. Which of the following statements is true? 8. C4 I 0.0028 = 2.8 × 10 –2 or
7.03 × 10 4 = 70 300
A 17 × 10 –3 = 0.017
C4
B 38 000 = 3.8 × 10 3 II 1 : 2 = 2 : 3 or 40 : 56 = 5 : 8
3
Paper 1 / Kertas 1 C 2 2 + 3 2 = 5 2 III (x – 3)(2x + 1) = 2x 2 – 5x – 3 are provided to evaluate
Instruction: Answer all questions. Time: 1 hour 30 minutes D 7 3 or x 2 – 4x + 4 = (x – 2) 2
Arahan: Jawab semua soalan. Masa: 1 jam 30 minit 10 5 IV 30 ÷ 0.01 = 300 or 0.006 × 1 000 = 60
4. Determine the statement that is true. Determine the false compound statements. the understanding of
1. Round off 3.04856 correct to three significant
A All quadrilaterals have four sides of the same
5. Express 243 7 as a number in base three.
C4
spm model paper figures. C 3.048 Ungkapkan 243 7 sebagai satu nombor dalam asas tiga. B All quadratic equations have two positive A I and III each chapter.
length.
B II and III
Bundarkan 3.04856 betul kepada tiga angka bererti.
A 10210 3
C 12010 3
B 11210 3
C II and IV
A 3.04
roots.
D 12110 3
B 3.05
D III and IV
D 3.049
D Some cuboids have a square base.
A 101111 2
560 × 10 5
2. ————— = 6. 1011101 2 – 100110 2 = C 110111 2 C Some proper fractions are greater than 1. 9. Determine the true compound statement.
3 = 0.6 and 0.7 = 7
B 110011 2
C4
0.007
D 111011 2
A 8 × 10 6 7. In Diagram 3, QT is a tangent to the circle PQR 5. Which of the following is true? A 5 10
C4
Statement
SPM format questions B 8 × 10 7 with centre O at Q. PORT is a straight line. A All obtuse angles lie Truth value B 6 : 12 = 1 : 2 and 3 : 2 = 9 : 4
C 8 × 10 8
C 3 –1 = 1 and 49 = 7 –2
False
Dalam Rajah 3, QT ialah tangen kepada bulatan PQR
3
D 8 × 10 9
dengan pusat O pada Q. PORT ialah satu garis lurus.
between 90° and 180°.
D (–1) 2 = –1 and –4 = –2
3. Diagram 1 shows a cylindrical water pipe with
according to the latest SPM Q B All triangles have the True
radius 2 m.
same area.
Rajah 1 menunjukkan sebatang paip air yang
berbentuk silinder dengan jejari 2 m. 27° C Some even numbers are True
P O R x° T prime numbers.
2021 assessment format D Some multiples of 3 are False
7 m Diagram 3/ Rajah 3 SPM MODEL PAPER divisible by 5.
Diagram 1/ Rajah 1
cover all the chapters in Calculate the volume, in cm 3 , of the water pipe. Find the value of x. 58
Hitung isi padu, dalam cm 3 , bagi paip air itu.
Cari nilai x.
3 Use/ Guna π = 22 4 —– 7 A 27 C 46
D 63
B 36
Forms 4 and 5. A 1.76 × 10 7 C 1.76 × 10 8 8. In Diagram 4, T is the midpoint of PQ. It is
D 8.8 × 10 8
B 8.8 × 10 7
4. In Diagram 2, PQRSTU is a regular hexagon. given that cos x° = 3 — 5 and sin y° = 7 — 8 . answers
PQH and PRL are straight lines such that Dalam Rajah 4, T ialah titik tengah bagi PQ. Diberi
PH = PL. bahawa kos x° = 3 — dan sin y° = 7 —.
Dalam Rajah 2, PQRSTU ialah sebuah heksagon 5 8
sekata. PQH dan PRL ialah garis lurus dengan R
keadaan PH = PL.
T S
L y° Complete answers
P Q
R T
U 64° K 6 cm x°
x° are provided.
P Q H S Diagram 4/ Rajah 4
Diagram 2/ Rajah 2 Find the length, in cm, of PR.
Calculate the value of x. Cari panjang, dalam cm, bagi PR. Scan the QR Code
Hitung nilai x. A 12 C 16
A 121 C 133 B 14 D 20
B 129 D 138 ANSWERS Complete answers
http://bit.ly/3vmOen4
429 429 provided to get
FORM 4
Chapter 1 Quadratic Functions and Equations in One 6. y = x 2 + 4 8 y
Variable the steps to the
KMODEL Spotlight A+ Mathematics F5.indd 429 05/03/2021 2:50 PM 6
Question Bank 1. 9y 2 + 16, 1 k 2 1.1 y = x 2 4 2 solution.
5
2. (a) Yes; One variable, c and the highest power of c x
is 2.
Chapter 1 Quadratic Functions and Equations in One Variable (b) No; The power of y is not a whole number. –2 –1 –2 O 1 2
1. Which of the following characteristics of quadratic function is not correct for f(x) = x 2 + 9? (c) No; Two variables, a and b. y = x 2 – 3
(d) Yes; One variable, p and the highest power of p
A The graph f(x) = x 2 + 9 has a minimum point at (0, 9). is 2. Graph y = x 2 + 4 is the translation of the graph y = x 2
B f is a many to one function. four units upwards, graph y = x 2 – 3 is the translation
C The axis of symmetry for the graph f(x) = x 2 + 9 is x = 0. 3. (a) Two values x = –1 and x = 1 are mapped to one of the graph y = x 2 three units downwards.
D The graph f(x) = x 2 + 9 cuts the x-axis at x = ±3. (b) value y = 3. y 7. The axis of symmetry x = – b of the graph y = x 2 + bx
2. Which of the following graphs is correct about the value of a on the graph of the quadratic function y = ax 2 ? 4 changes with the value of b. 2
A y C y 2 y = –x 2 + 4 8. (a) (x 2 + 12x + 35) cm 2
(b) x 2 + 12x – 288 = 0
y = x 2 y = 4x 2 x 9. (a) No (b) Yes (c) Yes
–3 –2 –1 O 1 2 3
y = x 2
y = 3x 2 –2 –4 10. (a) x = 4 or x = –1 (b) x = – 5 or x = 2
2
question bank O x O x 4. Maximum point (0, 4), axis of symmetry x = 0 11. (a) 4 y (b) 3 y FORM 4 ANSWERS
(c) x = 3 or x = 1
(d) x = 8 or x = –3
y
B y D y 4 y = x 2 1 – 2 5 O x 4 O x
y = x 2 y = x 2 2
Drill questions with –3 –2 –1 –2 O 1 2 y =– x 2 3 1 – 2 x (c) y (d) y
O x O x –4 1 x –3 O x
complete answers can be y = – x 2 1 – 2 y = –5x 2 Graphs y = 1 x 2 and y = – 1 x 2 are congruent and O
2
2
reflected on the x-axis. y
5.
y
obtained by scanning the 3. The diagram below shows a right-angled triangle. 16 12. (a) y (b)
(x + 10) cm (x + 5) cm 14 12 O 2 x –4 O x
QR Code on the cover of 25 cm 10 y = 4x 2
Based on the given information, find the quadratic equation that can be formed. 8 6 y = x 2 3 – 2 (c) y (d) y
the book. A x 2 + 15x – 250 = 0 C 2x 2 + 15x – 500 = 0 4 O x O 2 x
D 2x 2 + 15x – 750 = 0
B x 2 + 30x – 500 = 0
4. Which of the following is not a quadratic equation? 2 y = x 2 1 – 2
O
A p 2 + 8 = 3p C x 2 + 3 = 4 increases steeper 1 2 x
–2
–1
x
B k = 10 – 3k 2 D 13 + 9w – 3w 2 = 0 451
7
5. Given 4 is a root of the quadratic equation ax 2 + 14x – 8 = 0, determine the value of a.
A –4 C 3
B –3 D 4
1
v
Extra Features Spotlight A+ Mate F4.indd 5 15/03/2021 3:14 PM
CONTENTS
Formulae viii Relationship and Algebra
FORM 4 Revision
date
Relationship and Algebra Chapter 6 Linear Inequalities in Two 114
Revision Variables
©PAN ASIA PUBLICATIONS
date
6.1 Linear Inequalities in Two 116
Chapter 1 Quadratic Functions and 1 Variables
Equations in One Variable 6.2 Systems of Linear 120
1.1 Quadratic Functions and 3 Inequalities in Two
Equations Variables
Summative Zone 11 Summative Zone 126
Chapter 7 Graphs of Motion 137
Number and Operations 7.1 Distance-Time Graphs 139
7.2 Speed-Time Graphs 142
Chapter 2 Number Bases 18
Summative Zone 146
2.1 Number Bases 20
Summative Zone 32 Statistics and Probability
Chapter 8 Measures of Dispersion for 157
Discrete Mathematics Ungrouped Data
8.1 Dispersion 159
Chapter 3 Logical Reasoning 38
8.2 Measures of Dispersion 161
3.1 Statements 40 Summative Zone 172
3.2 Argument 47 Chapter 9 Probability of Combined Events 181
Summative Zone 58
9.1 Combined Events 183
Chapter 4 Operations on Sets 70 9.2 Dependent Events and 185
Independent Events
4.1 Intersection of Sets 72 9.3 Mutually Exclusive 189
4.2 Union of Sets 76 Events and Non-Mutually
4.3 Combined Operations on 79 Exclusive Events
Sets 9.4 Application of Probability 194
Summative Zone 83 of Combined Events
Summative Zone 198
Chapter 5 Network in Graph Theory 92
Number and Operations
5.1 Network 94
Summative Zone 105 Chapter 10 Consumer Mathematics: 207
Financial Management
10.1 Financial Planning and 209
Management
Summative Zone 215
vi
Contents.indd 6 15/03/2021 3:15 PM
FORM 5
Relationship and Algebra Revision
date
Revision
date Chapter 6 Ratios and Graphs of 346
Chapter 1 Variation 220 Trigonometric Functions
1.1 Direct Variation 222 6.1 The Value of Sine, Cosine 348
1.2 Inverse Variation 227 and Tangent for Angle q,
©PAN ASIA PUBLICATIONS
0° q 360°
1.3 Combined Variation 230
6.2 The Graphs of Sine, Cosine 357
Summative Zone 235 and Tangent Functions
Summative Zone 369
Chapter 2 Matrices 241
2.1 Matrices 243
Statistics and Probability
2.2 Basic Operation on Matrices 246
Summative Zone 263 Chapter 7 Measures of Dispersion for 379
Grouped Data
Number and Operations 7.1 Dispersion 381
7.2 Measures of Dispersion 392
Chapter 3 Consumer Mathematics: 269
Insurance Summative Zone 400
3.1 Risk and Insurance Coverage 271
Summative Zone 279 Relationship and Algebra
Chapter 8 Mathematical Modeling 414
Chapter 4 Consumer Mathematics: 283 8.1 Mathematical Modeling 416
Taxation Summative Zone 425
4.1 Taxation 285
Summative Zone 297
SPM Model Paper 429
Measurement and Geometry
Answers 451 – 495
Chapter 5 Congruency, Enlargement and 301
Combined Transformations
5.1 Congruency 303
5.2 Enlargement 307
5.3 Combined Transformation 315
5.4 Tessellation 323
Summative Zone 329
vii
Contents.indd 7 15/03/2021 3:15 PM
Formulae
• Speed = change of distance • An ungrouped data set is changed uniformly by
change of time multiplying a constant, c.
= the gradient of the distance-time graph (a) The range for the new data set
= c × the range for the original data set
• Distance travelled = area under speed-time graph (b) The interquartile range for the new data set
= c × the interquartile range for the original
• Area of trapezium data set
1
= × sum of the length of parallel sides × height (c) The standard deviation for the new data set
2
• Acceleration = change of speed = c × the standard deviation for the original
change of time data set
= the gradient of the speed-time graph (d) The variance for the new data set
= c × the variance for the original data set
2
• Average speed = total distance travelled
total time taken • P(A) = n(A)
n(S)
sum of data
• Mean, x = number of data • P(Aʹ) = 1 – P(A)
• Mean, x = sum of (midpoint × frequency) • A and B are two independent events
sum of frequencies
P(A ∩ B) = P(A) × P(B)
• Range = largest value of data – smallest value of data
• A and B are two non-mutually exclusive events
n
• If is an integer, s, then the first quartile, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
4
x + x
• A and B are two mutually exclusive events
.
s
Q = ©PAN ASIA PUBLICATIONS
s + 1
1 2 P(A ∪ B) = P(A) + P(B)
n
• If is not an integer but lies between s and s + 1, • Monthly net income
4
then the first quartile, Q = x . = total monthly incomes – total monthly
1 s + 1
expenditures – monthly fixed saving
• Second quartile, Q = median
2
• If 3n is an integer, s, then the third quartile, • Premium
4 x + x = face value of policy × (premium rate per RMx)
RMx
Q = s 2 s + 1 . • Amount of required insurance
3
• If 3n is not an integer but lies between s and s + 1, = (percentage of co-insurance) × (insurable value of
4
then the third quartile, Q = x . property)
3 s + 1
d –b
1
• Interquartile range = Q – Q 1 • A = ad – bc –c a
–1
3
x
x
• Variance, σ = ∑(x – ) 2 or σ = ∑x 2 – 2 PAʹ
2
2
N N • Scale factor, k = PA
x
• Standard deviation, σ = ∑(x – ) 2 or σ = ∑x 2 – • Area of image = k × area of object
x
2
2
N
N
• For ungrouped data in frequency table, variance,
x
σ = ∑f(x – x) 2 2 ∑fx 2 – , standard
or σ =
2
2
N N
x
deviation, σ = ∑f(x – x) 2 or σ = ∑fx 2 – .
2
N
N
viii
Formula Spotlight A+ Mate Tg4.indd 8 15/03/2021 3:52 PM
CHAPTER
1 Quadratic Functions and
Equations in One Variable
SMART
©PAN ASIA PUBLICATIONS
Important Learning Standards Page
• Identify and describe the characteristics of quadratic expressions in
one variable. 3
• Recognise quadratic function as many-to-one relation, hence, 3
describe the characteristics of quadratic functions.
• Investigate and make generalisation about the effect of changing the 4
values of a, b and c on graphs of quadratic functions, f (x) = ax + bx + c.
2
1.1 Quadratic
Functions and • Form quadratic functions based on situations, and hence relate to
Equations the quadratic equations. 5
• Explain the meaning of roots of a quadratic equation. 5
• Determine the roots of a quadratic equation by factorisation method. 6
• Sketch graphs of quadratic functions. 6
• Solve problems involving quadratic equations. 7
Words
• Effect of change/ Kesan perubahan
• Horizontal line test/ Ujian garis mengufuk
• Many-to-one relation/ Hubungan banyak kepada satu
• Maximum point/ Titik maksimum
• Method of factorisation/ Kaedah pemfaktoran
• Minimum point/ Titik minimum
• Quadratic equation/ Persamaan kuadratik
• Quadratic function/ Fungsi kuadratik
• Rate of change/ Kadar perubahan
• Real root/ Punca nyata
• Root/ Punca
• Variable/ Pemboleh ubah
1
01 Maths F4.indd 1 19/01/2021 9:22 AM
Form
4 Mathematics Chapter 1 Quadratic Functions and Equations in One Variable
CHAP.
1 x
Life situations Quadratic equation ax 2 + bx + c = 0 meaning determine Roots of quadratic equations a . 0 y O
Highest power of x is 2 based on b = 0 a , 0 y x O
form investigate the effect of change of a, b and c b = c = 0 a . 0 y x O
Power of x is a whole number Characteristics of quadratic expressions ax 2 + bx + c identify Quadratic Functions and Equations in One Variable a , 0 y x O x
Quadratic graphs f (x) = ax 2 + bx + c
Concept ©PAN ASIA PUBLICATIONS
a , 0
y
O
One variable, x identify sketch c = 0 a . 0 y x O
Many-to-one relations are Quadratic functions based on Axis of symmetry of graph is parallel to the y-axis
describe
Characteristics Maximum or minimum point
Curved- shape
2
01 Maths F4.indd 2 19/01/2021 9:22 AM
Form
4
Chapter 1 Quadratic Functions and Equations in One Variable Mathematics
1.1 Quadratic Functions and (c) The expression 6t + pt – 9 contains two CHAP.
2
2
Equations variables p and t. Therefore, 6t + pt – 9 is not 1
a quadratic expression in one variable.
1
—
2
(d) The expression y – 7y contains one variable
Identify and describe the characteristics of 1
—
quadratic expressions in one variable y. However, the power of y in the term 7y is
2
1
—
2
1. Quadratic expression in one variable is an not a whole number. Therefore, y – 7y is not
algebraic expression of the form ax + bx + c, a, b a quadratic expression in one variable.
2
and c are constants, a ≠ 0 and x is a variable. Try question 2 in Formative Zone 1.1
©PAN ASIA PUBLICATIONS
2. Characteristics of quadratic expressions in one
variable:
• Expressions contain only one variable.
• The power of the variable is a whole number.
• The highest power of the variable is 2. Quadratic equations in one variable
https://bit.ly/3d1ZKeM
BRILLIANT Tips
The variable x in quadratic expressions can also be
represented by other alphabet letters. BRILLIANT Tips
If p represents a constant, then 6t + pt – 9 is a
2
Example 1 quadratic expression in one variable, t.
r
2
5
2
2
2
4a + b + 3 — – 2r –h + 8h – 2 3t + —
2 t
Recognise quadratic function as many-
Identify quadratic expressions in one variable
from the list of expressions above. to-one relation, hence, describe the
Solution: characteristics of quadratic functions
r 2 1. Quadratic function is a many-to-one relation.
— – 2r, –h + 8h – 2
2
2 2. Characteristics of quadratic functions
Try question 1 in Formative Zone 1.1 • The graph has a curved shape.
• It has a maximum point or a minimum point.
Example 2
• The axis of symmetry of the graph is parallel
Determine whether each of the following to the y-axis.
expressions is a quadratic expression in one
variable. Give your justification.
2
(a) 2m – 9m + 5 BRILLIANT Tips
(b) 5x – 2x + 10
3
2
(c) 6t + pt – 9 Graphs of quadratic functions y = ax + bx + c
2
1 —
(d) y – 7y 2
Solution: a . 0 a , 0
2
(a) The expression 2m – 9m + 5 contains one Maximum point
variable m and the highest power of m is
2. Therefore, 2m – 9m + 5 is a quadratic Minimum point
2
expression in one variable.
3
(b) The expression 5x – 2x + 10 contains one
Axis of symmetry Axis of symmetry
variable x. However, the highest power of x is
3
3. Therefore, 5x – 2x + 10 is not a quadratic
expression in one variable.
1.1.1 1.1.2 3
01 Maths F4.indd 3 19/01/2021 9:22 AM
Form
4
Chapter 2 Number Bases Mathematics
Convert numbers from one base to another Solution:
using various methods (a) 57 = 32 + 16 + 8 + 1
10 = 2 + 2 + 2 + 2 0
5
4
3
1. The conversion of numbers from one base = 1 × 2 + 1 × 2 + 1 × 2 + 0 × 2
3
4
2
5
to base ten can be carried out by using place + 0 × 2 + 1 x 2 0
1
values. = 111001 CHAP.
2 2
Example 7 Place value 2 5 2 4 2 3 2 2 2 1 2 0
Digit 1 1 1 0 0 1
Convert each of the following numbers to a
number in base ten. (b) 2 57 Remainder
(a) 11010 (b) 123 4 2 28 … 1
2
(c) 4031 (d) 652
5 8 2 14 … 0
Solution:
2 7 … 0
(a) Place value 2 4 2 3 2 2 2 1 2 0 2 3 … 1
2 1 … 1
Digit 1 1 0 1 0
0 … 1
11010
2 57 = 111001
3
4
= 1 × 2 + 1 × 2 + 0 × 2 + 1 × 2 + 0 × 2 0 10 2
2
1
= 16 + 8 + 0 + 2 + 0 Calculator
= 26
10
Press: MODE MODE 3 DEC
(b) Place value 4 2 4 1 4 0
2 5 7 = BIN
Digit 1 2 3
1
2
123 = 1 × 4 + 2 × 4 + 3 × 4 0 Try question 8 in Formative Zone 2.1
4
= 16 + 8 + 3
= 27 Example 9
10
By using place values, convert 65 to a number in
(c) ©PAN ASIA PUBLICATIONS
5
5
5
5
0
2
1
3
Place value
10
(a) base five, (b) base eight, (c) base three.
Digit 4 0 3 1
Solution:
4031 = 4 × 5 + 0 × 5 + 3 × 5 + 1 × 5 0 (a) 65 = 50 + 15
2
3
1
5
= 500 + 0 + 15 + 1 10 = 2 × 25 + 3 × 5
= 516 2 1 0
10 = 2 × 5 + 3 × 5 + 0 × 5
= 230
(d) Place value 8 2 8 1 8 0 5
Place value 5 2 5 1 5 0
Digit 6 5 2
Digit 2 3 0
1
2
652 = 6 × 8 + 5 × 8 + 2 × 8 0 (b) 65 = 64 + 1
8
= 384 + 40 + 2 10 = 8 + 1
2
= 426
2
1
10 = 1 × 8 + 0 × 8 + 1 × 8 0
= 101
Try question 7 in Formative Zone 2.1 8
Place value 8 2 8 1 8 0
2. The conversion of numbers in base ten to a Digit 1 0 1
number in another base can be carried out by (c) 65 = 54 + 9 + 2
10
using place values and division. = 2 × 27 + 9 + 2
1
2
= 2 × 3 + 1 × 3 + 0 × 3 + 2 × 3 0
3
Example 8 = 2102 3
Place value 3 3 3 2 3 1 3 0
Convert 57 to a number in base two by using
10 2 1 0 2
(a) place values, (b) division. Digit
Try question 9 in Formative Zone 2.1
2.1.2 23
02 Spotlight A+ MM F4.indd 23 19/01/2021 9:29 AM
Form
4 Mathematics Chapter 4 Operations on Sets
(b) Solve problems involving the union of sets
ξ
●6 ●19 ●18
A Example 14
●7 B ●17
●10
●5 A residential area has 105 families. A total of 40
●8 ●20
●15 ●16 families own motorcycles. The number of families
that own cars only is twice the number of families
●9
●11 ●12 ●13 ●14 that own motorcycles only. The number of families
that own motorcycles and cars are the same as the
©PAN ASIA PUBLICATIONS
CHAP. Try question 4 in Formative Zone 4.2 number of families that do not own motorcycles
or cars.
4 (a) Draw a Venn diagram to show the information
above.
(b) Hence, determine the number of families that
own motorcycles or cars.
Operations on sets
https://bit.ly/2IRwsBs Solution:
(a) = {families in the residential area}
A = {families that own motorcycles}
B = {families that own cars}
Example 13
ξ
Given the universal set = {x : 4 < x , 16, A B
x is an integer}, P = {x : x is a perfect square}, x y 2x
Q = {x : x is a multiple of 3} and R = {x : x is a prime y
number}.
(a) Determine n[(P < Q < R)].
(b) Shade the region that represents (P < Q < R) n(A) = 40
on a Venn diagram. x + y = 40 ....... a
n() = 105
Solution: x + y + 2x + y = 105
(a) = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} 3x + 2y = 105 ..... b
P = {4, 9} a × 2, 2x + 2y = 80 ....... c
Q = {6, 9, 12, 15} b – c, x = 25
R = {5, 7, 11, 13}
P < Q < R = {4, 5, 6, 7, 9, 11, 12, 13, 15} From a, 25 + y = 40
(P < Q < R) = {8, 10, 14} y = 15
n[(P < Q < R)] = 3 ξ
A B
Alternative Method
25 15 50
n() = 12
n(P < Q < R) = 9 15
n[(P < Q < R)] = 12 – 9
= 3 (b) n(A < B) = 25 + 15 + 50
(b) = 90
ξ The number of families that own motorcycles
P Q or cars is 90.
●6
●4 ●9 ●12 Try question 6 in Formative Zone 4.2
●15
●8 ●14
●5 ●7
●10 ●11
●13 R
Try question 5 in Formative Zone 4.2
78 4.2.2 4.2.3
ENG04 Spotlight Matematik F4.indd 78 14/01/2021 9:53 PM
Form
Chapter 4 Operations on Sets Mathematics 4
4.3
1. The Venn diagram below shows the elements of 3. The Venn diagram below shows three sets A, B,
sets A, B, C and the universal set . C and the universal set .
ξ ξ A
A B
●b C ●c B C
●f ●a
●d ●g
●e
©PAN ASIA PUBLICATIONS
●h On a separate Venn diagram, shade the region CHAP.
List down the elements of C4 that represents C4 4
(a) (A < C > B), (b) [A < (B > C)].
(a) A > B < C, (b) (A < C) > B, 4. In the Venn diagram below, is the universal
(c) A < B > C, (d) (A > C) < B. set. P, Q and R are three sets. The number of
2. In the Venn diagram below, the number of elements in each set are shown.
elements of sets A, B, C and the universal set
are shown. ξ Q
P
ξ R
B x 2 y 26
A
x
25 6 7 10 13
C
5 Given n() = 109 and n(Q) = 71. C5
4
(a) Find the values of x and y.
Determine C4 (b) Hence, determine
(a) n[(A > C < B)], (b) n{[A < (B > C)]}. (i) n(P > Q < R),
(ii) n[(P > R) < Q].
SPM Simulation HOTS Questions COMMENT
EXAMINER’S
1. The diagram below shows a Venn diagram with Answer: C
the universal set , sets P, Q and R.
Examiner’s Tips:
ξ Candidates should not guess the answer for this
P Q question. Actually, candidates are required to
R
shade the region that represents each set in the
options.
ξ ξ
P Q P Q
Which of the following represents the shaded R R
region? C4
A (P < Q) > R C (Q > R) > P
B (Q < R) > P D (P > Q) < R
(P < Q) > R (Q < R) > P
Examiner’s Comments:
ξ
ξ ξ
P Q P Q P R Q
R R
Q > R (Q > R) > P (P > Q) < R
81
ENG04 Spotlight Matematik F4.indd 81 14/01/2021 9:53 PM
Form
Chapter 4 Operations on Sets Mathematics 4
Paper 1
1. Given set A = {1, 3, 4, 5, 7, 9} and set 6. The diagram below shows a Venn diagram with
C2 B = {2, 3, 4, 5, 6, 8, 10}. Which of the following C4 the universal set = P < Q < R.
©PAN ASIA PUBLICATIONS
describes the intersection of sets A and B? R
A Set of integers from 3 to 5. P Q CHAP.
B Set of integers from 3 to 6. ●3 ●1 ●5 ●7 ●2 4
C Set of integers between 3 and 5. ●6 ●4
D Set of integers between 3 and 6.
List down all the elements of the set (P > Q > R).
2. Given set A = {5, 7, 11, 13} and set A {1, 3, 6}
C2 B = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. Determine B {1, 2, 4, 7}
A > B in the set builder notation. C {2, 3, 4, 6, 7}
A {x : 4 < x , 12, x is an odd number} D {1, 2, 3, 4, 6, 7}
B {x : 4 < x < 13, x is a whole number}
C {x : 5 < x < 12, x is an even number} 7. A group of 49 customers bought at least one
D {x : 5 < x < 13, x is a prime number} C3 type of fruits from mangosteen, cempedak and
rambutan. The Venn diagram below shows the
3. Given set F = {letters of the word KOMPAS} and number of customers who bought mangosteen
C2 set G = {letters of the word PEMBARIS}. Which and cempedak only.
of the following Venn diagrams describes F > G? P Q
A C R
F G F G 10 18
●K ●A ●B ●K ●A ●B
●P ●E ●I ●M ●E
●I
●P
●M
●S ●O ●R ●O ●S Given P = {customers who bought mangosteen},
●R Q = {customers who bought cempedak} and
B D R = {customers who bought rambutan}. The
F ●B G F G number of customers who bought mangosteen
●K ●A ●E ●K ●A ●B
●M ●O ●I ●S and cempedak is three times the number of
●P ●O ●I ●P customers who bought rambutan. Determine the
●S ●R ●M ●R ●E
number of customers who bought all the three
fruits.
4. Given set A = {2, 4, 6, 8, 10, 12}, set A 5 C 19
C4 B = {1, 3, 4, 6, 8, 9} and set C = {3, 4, 5, 7, 8, 10, 11}. B 7 D 21
Find A > B > C.
A {4, 8} 8. A school has 33 teachers. An interview was
B {6, 8} C3 carried out to find out the number of teachers
C {4, 6, 8} who have visited Hokkaido in Japan and Jeju
D {4, 8, 10} Island in Korea. In the interview, it was found
that each teacher has visited at least one of the
5. Given the universal set = {x : 4 < x < 20, x is tourist places. A total of 18 teachers have visited
C4 an even number}, set M = {multiples of 4} and Hokkaido and 26 teachers have visited Jeju
set N = {even numbers between 4 and 16}. Island. Find the number of teachers who have
Determine (M > N). visited Hokkaido and Jeju Island.
A {4, 6, 10, 14, 16, 18} A 8
B {4, 6, 10, 14, 16, 20} B 9
C {4, 6, 10, 14, 16, 18, 20} C 11
D {4, 6, 8, 10, 12, 14, 16, 20} D 13
83
ENG04 Spotlight Matematik F4.indd 83 14/01/2021 9:53 PM
Form
4 Mathematics Chapter 4 Operations on Sets
25. The universal set = P < Q < R. Which of the 26. The Venn diagram below shows the relation
C4 following shaded regions in the Venn diagram C5 among three sets A, B and C such that the
represents [(P < Q) > R]? universal set = A < B < C.
A
R A B
P Q C
4 x 5 11
B 15
R
©PAN ASIA PUBLICATIONS
P Q Given that n(A) = n(B > C), determine
CHAP. n[(A < C) > B].
4 A 12
C B 15
R C 23
P Q D 27
D
R
P Q
Paper 2
1. Given set A = {6, 8, 10, 12, 16, 24} and set B = {7, 8, 9, 11, 13, 16, 24}. Complete the following diagram.
C2
Representation of
A > B
(a) Description (b) Listing (c) Set builder notation (d) Venn diagram
2. The Venn diagram below shows the relation among sets A, B and C.
C2
A C
●2 B ●1 ●3
●5 ●4 ●6
●7 ●9 ●12
●10 ●8 ●16
●11
Mark 3 or 7 for each of the following intersection sets.
(a) A > B
{ x : 5 < x < 7, x is an odd number}
(b) A > C
Set of perfect squares
(c) B > C
86
ENG04 Spotlight Matematik F4.indd 86 14/01/2021 9:53 PM
CHAPTER
5 Network in Graph Theory
SMART
Important Learning Standards Page
©PAN ASIA PUBLICATIONS
• Identify and explain a network as a graph. 94
• Compare and contrast
(i) directed graphs and undirected graphs. 96
5.1 Network (ii) weighted graphs and unweighted graphs.
• Identify and draw subgraphs and trees. 97
• Represent information in the form of networks. 100
• Solve problems involving networks. 100
Words
• Network/ Rangkaian
• Graph/ Graf
• Point/ Bintik
• Vertex/ Bucu
• Edge/ Tepi
• Degree/ Darjah
• Simple graph/ Graf mudah
• Directed graph/ Graf terarah
• Undirected graph/ Graf tak terarah
• Loop/ Gelung
• Multiple edges/ Berbilang tepi
• Weighted graph/ Graf berpemberat
• Unweighted graph/ Graf tak berpemberat
• Subgraph/ Subgraf
• Graph theory/ Teori graf
92
ENG05 Spotlight Matematik F4.indd 92 14/01/2021 9:56 PM
Form
Chapter 5 Network in Graph Theory Mathematics 4
Undirected graphs
compare Unweighted graphs
Directed graphs compare Weighted graphs
Multigraphs identify Graphs CHAP.
5
draw
Simple graphs identify identify Subgraphs draw identify Trees
Network in Graph Theory
Concept ©PAN ASIA PUBLICATIONS
is
identify
represents Social networks
represents
Road networks Network
represents Transportation networks
93
ENG05 Spotlight Matematik F4.indd 93 14/01/2021 9:56 PM
Form
4 Mathematics Chapter 5 Network in Graph Theory
1. Difference between directed graph and
BRILLIANT Tips undirected graph.
Graph
In a graph that has loops and multigraph, the degree
of a vertex is determined in the same way as simple
graphs. Directed graph Undirected graph
C C
Example 7
A B A B
©PAN ASIA PUBLICATIONS
A, B, C and D are four vertices of a graph. Without
using graph, determine the degree for each of the
vertices with the following edges. D D
(a) {(A, B), (A, D), (B, D), (C, D), (C, D), (C, D)}
(b) {(A, D), (A, D), (B, B), (B, C), (B, C), (B, D), (C, C)} Each edge is marked Each edge is not
CHAP. Solution: with an arrow. marked with an arrow.
5 (a) Vertex A : 2 edges {(A, B), (A, D)} The edge A • • B The edge A • • B
º Degree (A) = 2 is written in the is written in the form
ordered pair (B, A). (A, B).
Vertex B : 2 edges {(A, B), (B, D)} (A, B) ≠ (B, A) (A, B) = (B, A)
º Degree (B) = 2
The edges of the graph The edges of the graph
Vertex C : 3 edges {(C, D), (C, D), (C, D)} are {(A, C), (A, D), are {(A, B), (A, C),
º Degree (C) = 3
(B, A), (B, C), (D, A)} (A, D), (A, D), (B, C)}
Vertex D : 5 edges {(A, D), (B, D), (C, D), (C, D),
(C, D)} BRILLIANT Tips
º Degree (D) = 5
(b) Vertex A : 2 edges {(A, D), (A, D)}
º Degree (A) = 2 1. For the directed graph, (A, A) represents a directed
loop.
Vertex B : 4 edges {(B, B), (B, C), (B, C), (B, D)}
º Degree (B) = 4 or
Vertex C : 3 edges {(B, C), (B, C), (C, C)} A A
º Degree (C) = 3 2. For the undirected graph, (A, A) represents an
undirected loop.
Vertex D : 3 edges {(A, D), (A, D), (B, D)}
º Degree (D) = 3
A
Try question 7 in Formative Zone 5.1
Example 8
BRILLIANT Tips Match the following.
(a)
The degree of a vertex is found by determining the
number of edges that involve the vertex.
Directed graph
Compare and contrasting (b)
(i) directed graphs and undirected graphs
(ii) weighted graphs and unweighted graphs
Undirected
(c)
graph
96 5.1.1 5.1.2
ENG05 Spotlight Matematik F4.indd 96 14/01/2021 9:56 PM
Form
Chapter 5 Network in Graph Theory Mathematics 4
Solution:
(a) BRILLIANT Tips
The weight on the edge of a weighted graph may
represents distance, time, cost, price and so on.
Directed graph Example 10
(b) State whether each of the following graphs
represents a weighted graph or an unweighted
graph.
©PAN ASIA PUBLICATIONS
(a) (b)
3
5
(c) Undirected 10
graph 2
7
(c) 11 CHAP.
2 5
6 5 4
Try question 8 in Formative Zone 5.1 2
3
Example 9 Solution:
A graph has vertices {A, B, C, D} and edges {(A, A), (a) A weighted graph
(A, B), (A, D), (B, C), (C, A), (C, B)}. (b) An unweighted graph
(a) Mark 3 to show the correct graph. (c) A weighted graph
Try question 10 in Formative Zone 5.1
Directed graph Undirected graph
Give your justification. Identify and draw subgraphs and trees
(b) Hence, draw a diagram to represent the
graph. 1. Subgraph of a graph is the graph consisting of
Solution: the particular vertex and edge of the original
graph.
(a) Directed graph 3
{(B, C)} and {(C, B)} are directed edges that BRILLIANT Tips
connect vertex B and vertex C.
(b) 1. If H is the subgraph of graph G, then
C (a) the vertices of H are also the vertices of G,
A (b) the edges with its two vertices of H are also
the edges with its two vertices of G.
2. A subgraph may have shapes that are different
D B from the original graph but its vertices and edges
must be the same.
Try question 9 in Formative Zone 5.1 3. Each graph is a subgraph of itself.
2. Difference between weighted graph and B C B
unweighted graph. B
Graph A C
B A C
C Subgraphs A
Weighted graph Unweighted graph A of graph B
5 B B C
2 C C B A
7 9 A B A C
C B
B A
Each edge is associated Each edge is not A C C
with a number called associated with a A A
weight. number.
5.1.2 5.1.3 97
ENG05 Spotlight Matematik F4.indd 97 14/01/2021 9:56 PM
Form
4 Mathematics Chapter 5 Network in Graph Theory
Example 11 Example 13
The diagram on the right shows D C The diagram below shows a graph G.
graph G. D
Determine whether each of the
following is a subgraph of G. A B C
(a) (b) C
D
B B A E
©PAN ASIA PUBLICATIONS
Graph H has the following vertices and edges.
Determine whether H is a subgraph of G. Give
A B D your justification.
(c) (d) A (a) Vertices of H = {A, B, C, D}
D C B
Edges of H = {(A, B), (A, C), (B, C), (C, D)}
CHAP. (b) Vertices of H = {A, B, C, E}
Edges of H = {(A, B), (A, C), (C, E)}
5 A B D Solution:
Solution: (a) H is a subgraph of G because the vertices and
(a) A subgraph of G. edges of H are the same as G.
(b) A subgraph of G although its shape is different (b) H is not a subgraph of G because {(C, E)} is
from G but it has the same vertices {B, C, D} not an edge of G.
and edges {(B, C), (B, D), (C, D)}. Try question 13 in Formative Zone 5.1
(c) Not a subgraph of G because G does not
contain the edge {(A, D)}.
(d) A subgraph of G although its shape is different
from G but it has the same vertices {A, B, D}
and edge {(B, D)}. Example 14
Try question 11 in Formative Zone 5.1
Draw four possible subgraphs for each of the
following graphs G.
Example 12 (a) (b)
The diagram below shows a directed graph G. D E
D D
E C
C F A B
A B E
B
Solution:
A C
(a)
State whether each of the following directed
graphs is a subgraph of G. Give your justification. D D E D E D
(a) A (b)
A C C
D B E F C F F
A B B
E
(b)
C D D D D
Solution: C B C C
A B
(a) Not a subgraph of G because the edge
{(C, D)} is not in its graph. E E E E
(b) A subgraph of G because the vertices {A, B,
D, E} and edges {(A, B), (A, E), (D, E), (E, D)} Try question 14 in Formative Zone 5.1
are the same as graph G.
Try question 12 in Formative Zone 5.1
98 5.1.3
ENG05 Spotlight Matematik F4.indd 98 14/01/2021 9:56 PM
Form
4 Mathematics Chapter 5 Network in Graph Theory
Represent information in the form of Solution:
networks Langkawi Kota Kinabalu
Kota Bharu
Example 18 Pulau Sibu Tawau
Pinang Kuching
The diagram below shows some roads in an area.
Kuala
Lumpur Johor Bahru
Try question 19 in Formative Zone 5.1
©PAN ASIA PUBLICATIONS
Example 20
Five friends Asiah, Khairul, Natifah, Roslan and
Zeti made conversation by video calls in the
By using the road junction as vertex and the road following situation.
CHAP. as edge, draw an undirected graph to represent
5 the network of the roads. Asiah called Khairul and Roslan but not Natifah
Solution: or Zeti. Zeti only called Roslan and Natifah. Zeti
had never called Khairul although Khairul always
called Zeti. Often Natifah and Roslan will call
one another as Khairul and Natifah.
Draw a directed graph to represent the
conversation by video calls described in the
situation.
Solution:
Natifah
Try question 18 in Formative Zone 5.1
Khairul
Example 19 Zeti
The diagram below shows some locations of Roslan Asiah
airports in Malaysia.
Try question 20 in Formative Zone 5.1
Langkawi Kota Kinabalu
Kota Bharu
Sibu Solve problems involving networks
Pulau Tawau
Pinang Kuching
Example 21
Kuala
Lumpur Johor Bahru The diagram below shows a network of roads
connecting five places A, B, C, D and E.
A company Air X scheduled 10 flights on a day
as follows. 200 D
• Pulau Pinang to Johor Bahru B 340 C 100 270
• Kuching to Tawau 400 120
• Tawau to Kuching
• Kuching to Kuala Lumpur A 350 E
• Kota Kinabalu to Kuala Lumpur
• Kuala Lumpur to Langkawi The weights stated on each edge represent the
• Langkawi to Pulau Pinang distances, in m, between two places. A path is
• Johor Bahru to Sibu a route along the edges of the graph from one
• Pulau Pinang to Kota Bharu vertex to the other vertices without repeating any
• Kuala Lumpur to Kuching one vertex.
By listing down all the possible paths, determine
Draw a directed graph to represent the information the shortest path from A to C and state its length.
of the flights.
100 5.1.4 5.1.5
ENG05 Spotlight Matematik F4.indd 100 14/01/2021 9:56 PM
Form
4 Mathematics Chapter 7 Graphs of Motion
3. Solve problems involving speed-time graphs
Speed Positive gradient represents
acceleration (Speed keeps
on increasing) Example 12
O Time The diagram below shows the speed-time graph
Zero for the movement of two particles P and Q in a
gradient
Speed period of 30 seconds.
Interpretation represents Speed (m s )
–1
of speed-time zero
graph acceleration 20 B
©PAN ASIA PUBLICATIONS
O Time (Uniform
speed) 16 C
Speed u D
Negative gradient
represents deceleration A
O Time (Speed keeps on decreasing) O 18 F 30 E Time (s)
Example 11 Graph AB represents the movement of particle P
whereas graph CDE represents the movement of
The speed-time graph below shows Maria’s run in particle Q.
a sports practice. (a) Determine the value of u.
Speed (m s ) (b) By using the value of u obtained in (a), find
–1
10 A B the acceleration of particle Q in the first
CHAP. 18 seconds.
7 (c) Calculate the difference between the distance
travelled by particles P and Q in the period of
30 seconds.
C
O Time (s)
5 15 18 Solution:
(a) Find the acceleration, in m s , for the (a) Gradient AD = gradient AB
–2
(i) first 5 s, u = 20
(ii) duration of time from 5 s until 15 s, 18 30
(iii) last 3 s. u = 20 × 18
(b) Hence, describe Maria’s run in the period of 30
18 s. = 12
(b) Acceleration = gradient CD
Solution: = – 16 – 12
(a) (i) Acceleration = gradient OA 4 18
= 10 = – 18
5
2
= 2 m s –2 = – m s –2
(ii) Acceleration = gradient AB 9
= 0 m s –2 (c) Distance travelled by particle P
(iii) Acceleration = gradient BC = area of triangle ABE
1
= – 10 = 2 × 30 × 20
18 – 15 = 300 m
10
= – m s –2 Distance travelled by particle Q
3
BRILLIANT Tips = area of trapezium ACDF + area of triangle
DEF
1
Negative acceleration is known as deceleration. = 1 × (16 + 12) × 18 + × 12 × 12
2 2
= 252 + 72
(b) For the first 5 s, Maria ran from 0 m s until
–1
10 m s with acceleration 2 m s . After = 324 m
–1
–2
that, she ran with speed 10 m s for 10 s Difference between the distance travelled by
–1
particles P and Q = 324 – 300
and decreased her speed with deceleration = 24 m
1
3 m s until finally stopped.
–2
3 Try question 6 in Formative Zone 7.2
Try question 5 in Formative Zone 7.2
144 7.2.3 7.2.4
ENG07 Spotlight Matematik F4.indd 144 16/03/2021 8:58 AM
Form
Chapter 9 Probability of Combined Events Mathematics 4
9.1 Combined Events Example 1
In an experiment, Fahmi chooses a number at
Describe combined events and list out the random from {1, 3, 4, 6} and Ganan chooses a
possible combined events number at random from {3, 6, 9}. Given A is the
event Fahmi chooses an even number and B is the
1. Combined event is the event that is produced event Ganan chooses an odd number.
from the union or intersection of two or more Describe, in words and listing, the outcomes of
events. each of the combined events.
(a) A and B (b) A or B
©PAN ASIA PUBLICATIONS
Combined Event A or Event A and Solution:
event event B event B A = Event Fahmi chooses an even number
= {(4, 3), (4, 6), (4, 9), (6, 3), (6, 6), (6, 9)}
Notation A < B A > B B = Event Ganan chooses an odd number
A occurs or B = {(1, 3), (1, 9), (3, 3), (3, 9), (4, 3), (4, 9), (6, 3), (6, 9)}
Meaning occurs or both A Both A and (a) The combined event A and B is the event
B occur
Fahmi chooses an even number and Ganan
and B occur chooses an odd number.
A > B = {(4, 3), (4, 9), (6, 3), (6, 9)}
ξ ξ (b) The combined event A or B is the event Fahmi
A B A B
Venn chooses an even number or Ganan chooses
diagram an odd number.
A < B = {(1, 3), (1, 9), (3, 3), (3, 9), (4, 3), (4, 6),
(4, 9), (6, 3), (6, 6), (6, 9)}
Try question 1 in Formative Zone 9.1
BRILLIANT Tips Example 2
Combined event for three events A, B and C. A number is chosen at random from {10, 12,
(a) A < B < C 13, 15, 18, 20, 23, 24}. Given A is the event of
choosing a prime number, B is the event of
ξ choosing a multiple of 3 and C is the event of CHAP.
A B 9
choosing a number with the sum of digits greater
than or equal to 5.
List down all the outcomes for each of the
following combined events.
(a) A and C (b) B and C
(c) A or C (d) B or C
C Solution:
A = Event of choosing a prime number
(b) A > B > C = {13, 23}
B = Event of choosing a multiple of 3
ξ = {12, 15, 18, 24}
A B C = Event of choosing a number with the sum of
the digits greater than or equal to 5
= {15, 18, 23, 24}
(a) A and C = A > C
= {23}
(b) B and C = B > C
= {15, 18, 24}
C (c) A or C = A < C
= {13, 15, 18, 23, 24}
(d) B or C = B < C
= {12, 15, 18, 23, 24}
Try question 2 in Formative Zone 9.1
9.1.1 183
ENG09 Spotlight Matematik F4.indd 183 10/02/2021 11:17 AM
CHAPTER
10 Consumer Mathematics:
Financial Management
SMART
Important Learning Standards Page
©PAN ASIA PUBLICATIONS
10.1 Financial • Describe effective financial management process. 209
Planning and
Management • Construct and present personal financial plans to achieve short-term
and long-term financial goals, and hence evaluate the feasibility of 211
the financial plans.
Words
• Financial planning/ Perancangan kewangan
• Financial management/ Pengurusan kewangan
• Financial goal/ Matlamat kewangan
• Financial plan/ Pelan kewangan
• Short-term/ Jangka pendek
• Long-term/ Jangka panjang
• Needs/ Keperluan
• Wants/ Kehendak
• Cash flow/ Aliran tunai
• Active income/ Pendapatan aktif
• Passive income/ Pendapatan pasif
• Fixed expenditure/ Perbelanjaan tetap
• Variable expenditure/ Perbelanjaan tidak tetap
• Net salary/ Gaji bersih
• Balance income/ Baki pendapatan
• Monthly net income/ Pendapatan bersih bulanan
207
ENG10 Spotlight Matematik F4.indd 207 14/01/2021 10:17 PM
Form
4 Mathematics Chapter 10 Consumer Mathematics: Financial Management
R Realistic T Time-bound
guided by
A Attainable SMART concept Effective process of financial management Setting financial goals Evaluating financial status Creating financial plan Carrying out financial plan Review and revising the progress
M Measurable S Specific describe
Variable Expenditures Financial Management
Concept ©PAN ASIA PUBLICATIONS
Fixed
CHAP. Credit card
10 construct
Debts
Loan based on , 1 year Short-term
Emergency fund involve Personal financial plan Financial goals
Savings . 5 years evaluate influenced by
Fixed saving Long-term Feasibility of financial plans Inflation, health, sources of incomes and expenditures
Passive Incomes
Active
208
ENG10 Spotlight Matematik F4.indd 208 14/01/2021 10:17 PM
Form
Chapter 10 Consumer Mathematics: Financial Management Mathematics 4
10.1 Financial Planning and Example 1
Management Match each of the following.
Describe effective financial management Pay car installment for every
process month.
1. Financial planning is a process of evaluating Give six months salary as Needs
the current and future financial status of an yearly bonus.
individual or organisation by using the sources
©PAN ASIA PUBLICATIONS
of incomes and expenditures to determine the Provide lunch to all workers. Wants
cash flow, asset value and financial plan of the
future.
2. Financial management is a process involving Buy personal Takaful
insurance.
the use of sources of incomes and assets on
the expenditures, savings, investments and Solution:
protection to fulfil the financial goals of an
individual or organisation. Pay car installment for every
month.
3. The process of financial management involves
five main steps. Give six months salary as
yearly bonus. Needs
Setting financial
1 goals Provide lunch to all workers. Wants
5
Evaluating financial status Financial progress Buy personal Takaful
insurance.
Review and
revising the
process
2 management Try question 1 in Formative Zone 10.1
(b) Evaluating financial status.
4 • The current financial status can be
Carrying out determined based on cash flow statement CHAP.
and net asset value statement.
Creating
3 financial plan • The cash flow statement gives information 10
financial plan
on how the sources of incomes are
acquired and spent.
(a) Setting financial goals.
• The financial goal is usually set at the BRILLIANT Tips
beginning of a year.
• When setting a financial goal, all Positive cash flow occurs when the total incomes is
expenditures have to be given priority to exceeding the total expenditures whereas negative
needs than wants. cash flow occurs when the total incomes is less than
• Financial goals are categorised as short- the total expenditures.
term (less than one year), medium-term Total Positive
(one year up to five years) and long-term incomes cash flow
(more than five years). . total
expenditures
Cash flow
Total
incomes
, total Negative
expenditures cash flow
10.1.1 209
ENG10 Spotlight Matematik F4.indd 209 14/01/2021 10:17 PM
Form
4 Mathematics Chapter 10 Consumer Mathematics: Financial Management
• The asset value statement gives • Financial plan is usually prepared
information associated with current monthly for a term of one year.
assets (properties) and liabilities (debts).
• Assets are cash and investments such as Example 3
savings, shares, unit trusts and properties. The information below shows the incomes and
• Liabilities are loans, debts of credit cards, expenditures of Nasrul in May.
rental and utility bills. Active incomes: RM3 470
• The financial status needs to be examined Passive incomes: RM650
at least six months once to check the Fixed expenditures: RM2 860
©PAN ASIA PUBLICATIONS
achievement of financial goals. Variable expenditures: RM1 280
(a) Find
Example 2 (i) the total incomes of Nasrul in May,
(ii) the total expenditures of Nasrul in May.
The information below shows the total incomes (b) Hence, state whether Nasrul is a person who
and total expenditures of Alwi and Basri in a is wise in the financial management in May.
month.
Give your justification.
Alwi : Total incomes = RM2 340 Solution:
Total expenditures = RM2 070
Basri : Total incomes = RM2 580 (a) (i) Total incomes = 3 470 + 650
Total expenditures = RM2 750 = RM4 120
(ii) Total expenditures = 2 860 + 1 280
Determine the monthly cash flow of Alwi and Basri. = RM4 140
Hence, state the individual who has the better (b) The total expenditures exceeded the
monthly financial status. Give your justification. total incomes in May. Therefore, Nasrul is
Solution: not a person who is wise in the financial
Monthly cash flow of Alwi = 2 340 – 2 070 management in May.
= RM270 Try question 3 in Formative Zone 10.1
Monthly cash flow of Basri = 2 580 – 2 750
= –RM170 (d) Carrying out financial plan.
The monthly cash flow of Alwi is positive. • When carrying out a financial plan,
Therefore, the monthly financial status of Alwi is payment for fixed expenditures has
better. to be given priority to avoid charge or
Try question 2 in Formative Zone 10.1 additional interest that needs to pay on
CHAP. late payment.
10 (c) Creating financial plan. • Expenditures that are planned are
• Financial plan is an important component required to make comparison to the
in driving the achievement of financial actual expenditures for a particular
goal based on the expected financial month to control wastage.
value and the actual financial value. • If the monthly net income is surplus,
• The financial value is determined the money can be used for saving and if
based on incomes, savings, debts and deficit, the expenditures of the month are
expenditures. required to examine for the subsequent
month.
Example 4
BRILLIANT Tips Miss Wong is a secondary school teacher who
1. Incomes are comprised of active incomes (salary, receives a net salary of RM3 500 a month. She
allowance, commission) and passive incomes conducts tuition class and obtains an income
(rental, interest, dividend). of RM980 a month. Miss Wong also gets a
2. Expenditures are comprised of fixed expenditures monthly commission of RM560 from direct sale of
(house rental, car installment, insurance payment) products. Each month, Miss Wong spends RM750
and variable expenditures (utility bills, medical on food, RM1 000 on house rental, RM300 on car
cost, car fuel). fuel and RM200 on handphone. Find the monthly
net income of Miss Wong.
210 10.1.1
ENG10 Spotlight Matematik F4.indd 210 14/01/2021 10:17 PM
Form
5
Chapter 3 Consumer Mathematics: Insurance Mathematics
3.1 Risk and Insurance Coverage 10. Compensation that is paid by the insurance
company is based on the principle of indemnity.
Explain the meaning of risk and the According to the principle of indemnity, the
importance of insurance coverage, and insurance company will only recover the
hence determine the types of life insurance policyholder as in the original state, that is,
and general insurance for protecting a the condition before the loss incurred.
variety of risks
11. Compensation can be paid in one of the
1. When an individual experiences an unexpected following ways:
©PAN ASIA PUBLICATIONS
event or disaster, such a situation is known as • Payment
risk. • Recovery or replacement
12.
2. Risk is the possibility of the occurrence of an Types of insurance
unavoidable disaster.
CHAP.
3. Examples of risks are road accidents, home theft, 3
home fires and so on. Life insurance General insurance
4. Taking out insurance is a measure to protect risk. A Life insurance
5. Insurance is an economic mechanism that 1. Life insurance is an insurance policy that
transfers risk from one party, that is, the provides protection to the policyholder in
insurance policyholder to another party, that is, the event of the occurrence of something
the insurance company. undesirable such as death, critical illness, loss of
ability and hospitalisation.
transfer risk
2. Types of life
insurance Explanation
Pays a premium
Insurance Insurance Whole life Premium has to be paid as
policyholder company insurance long as the policyholder is
Pays compensation still alive.
Endowment Premium must be paid
6. Advantages of taking out insurance to life insurance by the policyholder for
policyholders: a certain period of time
• Receive compensation in the event of the and will be refunded to
occurrence of an insurable risk. the policyholder upon
• No need to withdraw money from own funds reaching maturity together
to cover the risk. with the bonuses collected.
7. The amount of compensation paid by the Term life Premium must be paid
insurance company depends on the total amount insurance by the policyholder for
of insurance taken by the policyholder. a specified duration of
time but does not have a
8. The risk that is insured must occur during the maturity value.
period when the insurance is in force.
B General insurance
9. The compensation that is paid by the insurance
company to the policyholder depends on the 1. General insurance is an insurance policy that
amount of the actual loss and does not exceed protects an individual from any losses and
the amount of the loss. damages incurred such as the loss of property
PB 3.1.1 271
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Form
5 Mathematics Chapter 3 Consumer Mathematics: Insurance
due to theft or fire, death due to accident or (f) Motor insurance
injury and liability arising from third parties – Provides coverage against any loss or
caused by such individual. damage related to the use of motor
vehicle
2. Characteristics of general insurance:
• Needs to be renewed annually 4. Four types of motor insurance policies:
• Using the principle of indemnity (a) Act policy
– Protects liability of the third party
3. Types of general insurance: for death or bodily injury (excluding
passengers)
©PAN ASIA PUBLICATIONS
(b) Third party policy
Motor – Additional protection on the act policy
insurance which provides additional protection to
Medical the third party
and health Fire
CHAP. insurance – This policy covers losses on property that
3 insurance is suffered by the third party
Types of (c) Third party, fire and theft policy
general – Additional protection on the third
insurance party policy which provides additional
Travel Personal protection to policyholder
insurance accident – This policy covers losses to own vehicle
insurance caused by accidental fire or theft
Robbery (d) Comprehensive policy
insurance – Additional protection on the third party,
fire and theft policy which provides
additional protection to policyholder
(a) Fire insurance – This policy covers losses and damages to
– Provides protection against fire risk own vehicle due to accident
(b) Personal accident insurance 5. Group insurance:
– Provides coverage to policyholder who • Provides coverage to a group of individuals,
suffers bodily injury, disability or death usually employees of a company such as bank
resulting directly from accident. This employees or pupils in schools and students
insurance is different from life insurance, in institutes of higher education
and medical and health insurance. • Employees who are covered under this policy
(c) Robbery insurance receive financial protection in the events of
– Provides protection due to robbery death, disability, hospitalisation and surgery
(d) Travel insurance in accordance to the claim limits set out in the
– Protects policyholder against losses policy
incurred during travel such as death and
permanent disability, loss of luggage,
passport and money, medical expenses Investigate, interpret and perform
and others. calculations involving insurance rates and
(e) Medical and health insurance premiums
– Pays compensation to policyholder who
requires health care due to health and 1. Premium is the amount of money payable by
illness treatments the policyholder to the insurance company to
cover the risk.
272 3.1.1 3.1.2 273
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Form
5
Chapter 3 Consumer Mathematics: Insurance Mathematics
(a) Comprehensive policy for Peninsular
Malaysia: BRILLIANT Tips
Basic premium The premium rate for third party, fire and theft
= Rate for the first RM1 000 + RM26 for policy is 75% of the basic premium of the
each RM1 000 or part thereof on value comprehensive policy.
exceeding RM1 000
Gross premium for third party policy:
(b) Comprehensive policy for Sabah and RM
©PAN ASIA PUBLICATIONS
Sarawak: Basic premium 151.20
Minus NCD 25% 37.80
Basic premium Gross premium 113.40
= Rate for the first RM1 000 + RM20.30
for each RM1 000 or part thereof on Try question 3 in Formative Zone 3.1
value exceeding RM1 000 CHAP.
3
Example 3 BRILLIANT Tips
Mr Hafiz has a car of model X to use in Peninsular 1. No Claim Discount (NCD) clause will be issued
Malaysia. The information of the car is as follows. if no claim is made against the motor insurance
during the coverage period before the policy
Sum insured : RM94 000 renewal is made.
Age of vehicle : 5 years 2. NCD will reduce the total premium payment for
Engine capacity : 1 799 cc the insurance policyholder.
NCD : 25% 3. The benefit of NCD will be lost if the policyholder
has made a claim for a personal or a third party
Calculate the gross premium of Mr Hafiz's car damage.
under the comprehensive policy, the third party,
fire and theft policy, and the third party policy.
Solution:
Gross premium for comprehensive policy: Solve problems involving insurance
including deductible and co-insurance
RM
The first RM1 000 339.10
RM26 × RM93 (each RM1 000 2 418.00 A Deductible
balance) 1. Deductible is a provision in an insurance
Basic premium 2 757.10 contract that requires the policyholder to borne
Minus NCD 25% 689.28 a small portion of the losses incurred.
Gross premium 2 067.82
2. Under this provision, a certain amount that has
BRILLIANT Tips been set will be deducted from the amount of
compensation that should have been paid by the
RM94 000 – RM1 000 insurance company.
——————————– = RM93
RM1 000
3. Deductible is a provision that exists in the
Gross premium for third party, fire and theft policy: medical and health insurance as well as the
property insurance.
RM
Basic premium 2 067.83 4. Deductible is not used in the life insurance.
Minus NCD 25% 516.96
Gross premium 1 550.87
274 3.1.2 3.1.3 275
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Form
5
Chapter 4 Consumer Mathematics: Taxation Mathematics
4.1 Taxation purchase of certain goods or use of services.
For example, cigarette tax.
Explain the purpose of taxation • Government policy implementation tool.
• Taxes are used as a tool to control inflation
1. Tax is a method used by the government to and economic recession.
collect revenue from individuals, companies 5. As responsible citizens, we must pay taxes that
and other entities, to channel in the country’s are levied on time.
development projects besides providing various
public facilities for the well-being of citizens. 6. Paying of taxes is not something that is a burden,
in fact it helps to improve the development of the
©PAN ASIA PUBLICATIONS
2. In Malaysia, taxes are usually collected by nation.
the Inland Revenue Board (IRB), the Royal
Malaysian Customs Department (RMCD) and
several other government departments. Describe various taxes, hence the
consequences of tax evasion from legal and
3. Taxes are usually collected in the form of money financial aspects
based on certain conditions.
4. The purpose and the importance of taxes that are 1. In Malaysia, tax collection is managed by two
levied by the government: main agencies under the Ministry of Finance,
• To smoothen the administration of the namely the Inland Revenue Board (IRB), the
country and to help citizens through the Royal Malaysian Customs Department (RMCD)
development of projects. and the Royal Malaysian Excise.
CHAP.
2. Types of taxes in Malaysia:
Tax 4
Income
Government Citizens tax
– Salary to public – Construction of
sector employees hospitals
– Other expenses – Construction of Service tax Road tax
for managing schools
the government – Free textbooks
administration Types of
taxes
• Property
Entrepreneurs Sales tax assessment
tax
pay taxes to
Government Quit rent
Tax is channeled to underprivileged group A Income tax
through various types of assistance
1. Income tax is a tax paid by individuals,
companies and other entities that earn
• Taxes paid are used as economic stimulants or more than a certain amount after deducting
movers. tax exemptions and tax relief allowed by the
• Taxes can be levied on certain goods or government for the period of the concern year
services so that the citizens can reduce the of assessment.
4.1.1 4.1.2 285
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Form
5 Mathematics Chapter 4 Consumer Mathematics: Taxation
2. D Quit rent
1. Besides property assessment tax, property
Fine of RM1 000 owners also need to pay quit rent.
up to RM10 000
Make omission or imprisonment 2. Quit rent is a tax that is levied on private land.
on any income or both and
in Income Tax penalty of 200% 3. Quit rent is paid by the landowner to the state
Return Form of the underpaid authority through the State Land Office.
(ITRF) taxes (Income Tax
3. ©PAN ASIA PUBLICATIONS
Act 1967 (Act 53) 4. Consequences faced by the landowner for failing
Section 113(1)(a)) to pay quit rent:
• Arrears will be charged in addition to quit
Income rent.
tax • Notice of claim will be issued. If quit rent is
not settled within 3 months from the notice of
Fine of RM200 instruction, the land may be seized (National
up to RM20 000 Land Code 1965 Section 100).
Failure to submit or imprisonment
annual report if of not exceeding E Sales tax
eligible for tax 6 months or both 1. Sales tax is levied on final goods that are sold
payment (Income Tax Act
CHAP. 1967 (Act 53) by domestic firms and imported goods from
4 Section 112(1)) abroad.
2. Sales tax is a level of indirect tax that is
administered by the Royal Malaysian Customs
B Road tax Department through the provision of the Sales
Tax Act 2018.
1. Motor vehicle owner must pay road tax before
the vehicle is legally allowed to drive on the road. 3. Sales Tax Registration Threshold when the firm’s
sales value exceeds RM500 000 per year.
2.
Vehicle owners
Motor vehicle can be fined of not 4. Sales tax is set at various rates, i.e. 5% or 10%.
owners who fail to exceeding RM2 000
renew the expired in accordance with 5. However, not all goods that are manufactured in
road tax (no the Road Transport Malaysia are subject to sales tax.
insurance) Act 1987 (Act 333)
Section 23(1) 6. Consequences faced by the business firms/
companies if evading from paying sales tax:
C Property assessment tax First offence of tax evasion
1. Property assessment tax is levied on properties
such as industrials and vacant lands. Minimum fine of 10 times up to 20 times
2. Property assessment tax is also known as the amount of sales tax or imprisonment of
assessment rate. not exceeding 5 years or both (Sales Tax Act
2018 (Act 806) Section 86(1) and 86(2)).
Property owners who Property will
do not pay property be seized or
assessment tax auctioned
286 4.1.2
C04 SPotlightA+ Mathematics F5.indd 286 03/03/2021 4:50 PM
Form
5
Chapter 4 Consumer Mathematics: Taxation Mathematics
Tax payable for the assessment year of 2019 9 231
Minus: Monthly tax deduction (PCB) 1 050
Tax balance for the assessment year of 2019 8 181
Try question 3 in Formative Zone 4.1
BRILLIANT Tips Rebate
©PAN ASIA PUBLICATIONS
• Tax rebate deducted from the tax levied
A separate tax assessment is more economical than to determine the tax payable for the
a joint tax. A joint assessment will only result in the assessment year.
reduction of the total tax payable if either husband • Divided into two, namely zakat or
or wife has a low amount of income. fitrah tax rebate and individual tax
rebate of RM400 for the taxpayer who
has a chargeable income not exceeding
RM35 000.
8. Differences between tax relief and rebate:
Tax relief 9. Calculation of road tax:
• Road tax is levied based on the engine capacity
• Expenditures that are allowed to be of the vehicle used by the owner.
deducted from the total annual income to • The higher is the engine capacity of vehicle, CHAP.
reduce the amount of income to be taxed. the higher is the road tax rate being levied. 4
Example 4
The table below shows the road tax rates for private cars in Peninsular Malaysia.
Road tax rate
Engine capacity
Base rate Progressive rate
1 000 cc and below RM20.00 -
1 001 cc – 1 200 cc RM55.00 -
1 201 cc – 1 400 cc RM70.00 -
1 401 cc – 1 600 cc RM90.00 -
1 601 cc – 1 800 cc RM200.00 + RM0.40 for each cc exceeding 1 600 cc
1 801 cc – 2 000 cc RM280.00 + RM0.50 for each cc exceeding 1 800 cc
Mr Yusuf and his wife bought two cars with engine capacity of 850 cc and 1 850 cc respectively for private use
in Pahang. Calculate the road tax that is levied on both the cars.
Solution:
Car road tax (850 cc) = RM20.00
Car road tax (1 850 cc) = Base rate + Progressive rate
= RM280 + (1 850 – 1 800) × RM0.50
= RM280.00 + RM25
= RM305.00
Try question 4 in Formative Zone 4.1
4.1.3 293
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Form
5 Mathematics Chapter 6 Ratios and Graphs of Trigonometric Functions
6.1 The Value of Sine, Cosine and (b) y
Tangent for Angle θ,
0° < θ < 360° 140° 140° lies in
x quadrant II.
O
Make and verify conjecture about the
value of sine, cosine and tangent for
angles in quadrants II, III and IV with the (c) y
corresponding reference angle
©PAN ASIA PUBLICATIONS
1. A Cartesian plane is divided into four parts 230° 230° lies in
(called quadrant) by the x-axis and the y-axis. O x quadrant III.
The quadrants are named as quadrant I,
quadrant II, quadrant III and quadrant IV in the
anticlockwise direction. y
y (d)
90°
310° lies in
310° quadrant IV.
Quadrant Quadrant O x
II I
0°
x
O
180° 360°
Quadrant Quadrant
III IV Try question 1 in Formative Zone 6.1
270°
3. y
2. y
P(x, y)
P 1 y
θ x
θ O x Q
x
O
PQ y
sin θ = —–— = —– = y
1
OP
CHAP. An angle θ is measured by rotating the line OP ∴ sin θ = y-coordinate
6 in the anticlockwise direction from the positive OQ x
x-axis at the origin. cos θ = —–— = —– = x
OP 1
∴ cos θ = x-coordinate
Example 1 y
PQ
tan θ = —–— = —–
Sketch on separate diagrams to represent the OQ x
angles (a) 50°, (b) 140°, (c) 230° and (d) 310°. y-coordinate
Hence, determine the quadrant for each of the ∴ tan θ = —–—————
x-coordinate
angles lies.
Solution: BRILLIANT Tips
(a) y BC
C • sin θ = —––
Hypotenuse AC
AB
50° lies in Opposite side • cos θ = —––
50° quadrant I. θ AC
BC
O x • tan θ = —––
A B AB
Adjacent side
348 6.1.1
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Form
5
Chapter 8 Mathematical Modeling Mathematics
Compound interest model r A(t) = P 1 1 + —2 nt , n where A(t) is the amount of savings after t years, P is the principal of the savings, r is the yearly interest rate, n is the number of interests being compound per year and t is the number of years
if a . 0 and 0 , b , 1
Exponential model y = a(b) x Exponential function shows decay such that Exponential decay model is y = a(1 − r) t , where a is the initial value, r is the decay rate and t is the time
if a . 0 and b . 1 such that
Exponential function shows growth Exponential growth model is y = a(1 + r) t , where a is the initial value, r is the growth rate and t is the time
Mathematical Modeling
Solve the three equations by using
involving
Three coordinates (x 1 , y 1 ), (x 2 , y 2 )
parabola and the vertex (h, k):
parabola and the x-intercept:
A coordinate (x 1 , y 1 ) on the
A coordinate (x 1 , y 1 ) on the
Concept ©PAN ASIA PUBLICATIONS
y = a(x − p)(x − q)
y = a(x − h) 2 + k
y = ax 2 + bx + c
Quadratic model
and (x 3 , y 3 ):
y = ax 2 + bx + c
given
Linear model y = mx + c The gradient m and the y-intercept, c: y = mx + c The gradient m and a coordinate (x 1 , y 1 ): y − y 1 = m(x − x 1 ) Two coordinates (x 1 , y 1 ) and (x 2 , y 2 ): y = mx + c and y − y 1 = m(x − x 1 )
given CHAP.
8
PB 415
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Form
5 Mathematics Chapter 8 Mathematical Modeling
This situation involves the linear function because B Quadratic modeling
the first differences are a constant, that is, RM39. 1. Quadratic modeling is related to a quadratic
The first differences is the gradient, m, that is, the function. The basic equation for quadratic
rate of change of y. function is y = ax + bx + c, where a is the leading
2
Hence, y = 39x + c. coefficient, b is the center coefficient and c is the
When (0, 200), y-intercept.
200 = 39(0) + c 2. The value of the leading coefficient a can
c = 200 determine the shape of the curve of a parabola.
y = 39x + 200
©PAN ASIA PUBLICATIONS
(a) When a , 0,
When x = 60, • the parabola opens downward.
y = 39(60) + 200 • the function has a maximum value known
y = 2 540 as vertex (h, k).
Thus, the membership fee to be paid for 5 years y
is RM2 540.
10
Try questions 3 & 4 in Formative Zone 8.1
(0, 6)
5
Example 6 –10 –5 0 5 10 x
The graph below shows the journey distance, –5
y km, of a helicopter for x seconds.
y (km)
(b) When a . 0,
• the parabola opens upward.
8
(2, 7) • the function has a minimum value known
6 as vertex (h, k).
4 y
2
15
x (s)
–2 0 2 4 10
–2
5
(a) Based on the graph, write the equation (0, 1)
involved. –10 –5 0 5 10 x
(b) The total journey for the helicopter to reach
the designated destination is 18 200 km. How 3. Strategies that can be proposed in the
much time is required by the helicopter to construction of a quadratic model to solve a
reach the destination?
problem:
Solution: (a) Identify the variables involved along with
y – y their units.
2
1
(a) m = ——–— (b) Identify the important information of
x – x
2 1 the situation such as the maximum point,
7 – 0
m = ———
2 – 0 minimum point and related points.
m = 3.5 (c) Identify the solution. Typically, the solution
Thus, y = 3.5x will involve the use of a table of values to
obtain the formula for the function of model
(b) y = 3.5x
18 200 = 3.5x to be solved.
x = 5 200 (d) Construct a formula for the function
CHAP. Thus, the time required by the helicopter to involved.
8 reach the destination is 5 200 seconds. (e) Solve the function using the constructed
formula.
420 8.1.2 421
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Form
5
Chapter 8 Mathematical Modeling Mathematics
(a) Exponential growth model Use the constructed model to solve the problem.
The exponential function shows growth if y = 19 000(1 + 0.15) 9
a . 0 and b . 1.
The exponential growth model: = 66 839.6495
y ≈ 66 840
y = a(1 + r) t Thus, the population of the city in the 9th year is
66 840 people.
where a = the initial value
r = the growth rate Try questions 7 & 8 in Formative Zone 8.1
t = the time
y
Example 12
3
The value of a car worth RM350 000 will be
2
depreciated every year by 11%.
1 (a) If y represents the value of the car, in RM, for
t years, model the situation.
t (b) What is the value of the car after 9 years?
–2 –1 0 1
Solution:
(b) Exponential decay model (a) The initial value is 350 000, that is,
The exponential function shows decay if a = 350 000.
a . 0 and 0 , b , 1. The decay rate is 11%, that is, r = 0.11.
The exponential decay model: Thus, the exponential decay model is
y = 350 000(1 – 0.11) t
y = a(1 – r) t y = 350 000(0.89) t
9
where a = the initial value (b) y = 350 000(0.89)
= 122 624.7413
r = the decay rate y ≈ 122 624.74
t = the time
©PAN ASIA PUBLICATIONS
Thus, the value of the car after 9 years is
y RM122 624.74.
Try question 9 in Formative Zone 8.1
3
2
8. The compound interest model is the interest
1 calculated with the reference to the original
principal as well as the accumulated interest
t
–1 0 1 2 from the previous retention period.
9. The compound interest model:
r
1
Example 11 A(t) = P 1 + — 2 nt
n
The population of a city is expected to increase by
15% annually. The population of today is 19 000 where A(t) = the amount of savings after t years
people. State the model involved and hence find P = the principal of the savings
the population of the city in the 9th year. r = the yearly interest rate
n = the number of interests being
Solution: compound per year
The initial value is 19 000, that is, a = 19 000. t = the number of years
The growth rate is 15%, that is, r = 0.15.
Hence, the exponential growth model is
y = 19 000(1 + 0.15) t CHAP.
8
422 8.1.2 423
C08 SpotlightA+ Mathematics F5.indd 423 03/03/2021 4:59 PM
SPM MODEL PAPER
Paper 1 / Kertas 1
Instruction: Answer all questions. Time: 1 hour 30 minutes
Arahan: Jawab semua soalan. Masa: 1 jam 30 minit
1. Round off 3.04856 correct to three significant 5. Express 243 as a number in base three.
7
figures. Ungkapkan 243 sebagai satu nombor dalam asas tiga.
©PAN ASIA PUBLICATIONS
7
Bundarkan 3.04856 betul kepada tiga angka bererti. A 10210 C 12010 3
3
A 3.04 C 3.048 B 11210 D 12110 3
3
B 3.05 D 3.049
6. 1011101 – 100110 =
2
2
560 × 10 5 A 101111 C 110111
2. ————— = 2 2
0.007 B 110011 D 111011 2
2
A 8 × 10 6
B 8 × 10 7. In Diagram 3, QT is a tangent to the circle PQR
7
C 8 × 10 8 with centre O at Q. PORT is a straight line.
D 8 × 10 9 Dalam Rajah 3, QT ialah tangen kepada bulatan PQR
dengan pusat O pada Q. PORT ialah satu garis lurus.
3. Diagram 1 shows a cylindrical water pipe with Q
radius 2 m.
Rajah 1 menunjukkan sebatang paip air yang
berbentuk silinder dengan jejari 2 m. 27° x°
P T
O R
7 m SPM MODEL PAPER
Diagram 1/ Rajah 1 Diagram 3/ Rajah 3
Calculate the volume, in cm , of the water pipe. Find the value of x.
3
Hitung isi padu, dalam cm , bagi paip air itu. Cari nilai x.
3
22
3 Use/ Guna π = —– 4 A 27 C 46
B 36
7
D 63
A 1.76 × 10 7 C 1.76 × 10
8
B 8.8 × 10 7 D 8.8 × 10 8 8. In Diagram 4, T is the midpoint of PQ. It is
3 7
given that cos x° = — and sin y° = —.
4. In Diagram 2, PQRSTU is a regular hexagon. 5 8
PQH and PRL are straight lines such that Dalam Rajah 4, T ialah titik tengah bagi PQ. Diberi
7
3
PH = PL. bahawa kos x° = — dan sin y° = —.
Dalam Rajah 2, PQRSTU ialah sebuah heksagon 5 8
sekata. PQH dan PRL ialah garis lurus dengan R
keadaan PH = PL.
T S
L y°
P Q
R K T
U 64° 6 cm
x°
x°
S
P Q H Diagram 4/ Rajah 4
Diagram 2/ Rajah 2 Find the length, in cm, of PR.
Calculate the value of x. Cari panjang, dalam cm, bagi PR.
Hitung nilai x. A 12 C 16
A 121 C 133 B 14 D 20
B 129 D 138
429
429
KMODEL Spotlight A+ Mathematics F5.indd 429 05/03/2021 2:50 PM
Mathematics SPM Model Paper
Paper 2 / Kertas 2
Time: 2 hours 30 minutes / Masa: 2 jam 30 minit
Section A/ Bahagian A
[40 marks/ 40 markah]
Instruction: Answer all questions in this section.
Arahan: Jawab semua soalan dalam bahagian ini.
1. (a) The Venn diagram in Diagram 1.1 in the answer space shows sets P and Q such that the universal set
ξ = P Q. Shade the set Pʹ.
Gambar rajah Venn dalam Rajah 1.1 pada ruang jawapan menunjukkan set P dan set Q dengan keadaan set semesta
ξ = P Q. Lorek set Pʹ.
(b) The Venn diagram in Diagram 1.2 in the answer space shows sets H, J and K such that the universal set
ξ = H J K. Shade the set (H J) K.
Gambar rajah Venn dalam Rajah 1.2 pada ruang jawapan menunjukkan set H, set J dan set K dengan keadaan set
semesta ξ = H J K. Lorek set (H J) K.
[4 marks/ 4 markah]
Answer/ Jawapan:
(a) (b)
P Q H K J
SPM MODEL PAPER ©PAN ASIA PUBLICATIONS
Diagram 1.2/ Rajah 1.2
Diagram 1.1/ Rajah 1.1
2. Diagram 2 shows a composite solid consisting of a right prism and a cuboid.
Rajah 2 menunjukkan sebuah gabungan pepejal yang terdiri daripada sebuah prisma tegak dan sebuah kuboid.
6 cm
3 cm
8 cm
4 cm
5 cm
Diagram 2/ Rajah 2
3
Calculate the volume, in cm , of the composite solid.
Hitung isi padu, dalam cm , bagi gabungan pepejal itu.
3
[4 marks/ 4 markah]
Answer/ Jawapan:
436 437
KMODEL Spotlight A+ Mathematics F5.indd 436 05/03/2021 2:50 PM
ANSWERS Complete answers
http://bit.ly/3vmOen4
FORM 4
6. y
Chapter 1 Quadratic Functions and Equations in One y = x + 4
2
Variable 8
6
1.1 4
1 y = x 2
1. 9y + 16, k 2 2
2
5
2. (a) Yes; One variable, c and the highest power of c
is 2. –2 –1 O 1 2 x
(b) No; The power of y is not a whole number. –2
(c) No; Two variables, a and b. y = x – 3
2
(d) Yes; One variable, p and the highest power of p Graph y = x + 4 is the translation of the graph y = x 2
2
is 2. four units upwards, graph y = x – 3 is the translation
2
3. (a) Two values x = –1 and x = 1 are mapped to one of the graph y = x three units downwards.
2
value y = 3. b
2
(b) y 7. The axis of symmetry x = – of the graph y = x + bx
2
4 changes with the value of b.
2
y = –x + 4 8. (a) (x + 12x + 35) cm 2
2
2 (b) x + 12x – 288 = 0
2
x 9. (a) No (b) Yes (c) Yes
–3 –2 –1 O 1 2 3 5
–2 10. (a) x = 4 or x = –1 (b) x = – or x = 2
2
3 8
–4 (c) x = or x = 1 (d) x = or x = –3
4 3 FORM 4 ANSWERS
Maximum point (0, 4), axis of symmetry x = 0 11. (a) y (b) y
y
4. ©PAN ASIA PUBLICATIONS
4 5 4
1
–
y = x 2 O x O x
2 2
x
–3 –2 –1 O 1 2 3 (c) y (d) y
1
–2 y =– x 2
–
2 1 x
–4 O
–3
x
O
1 1
Graphs y = x and y = – x are congruent and
2
2
2 2
reflected on the x-axis.
5. y 12. (a) y (b) y
16
14 x
x –4 O
O 2
12
y = 4x 2
10
8
(c) y (d) y
3
–
6 y = x 2
2
4 x
O 1 x O 2
2
2 y = x –
2
x
–2 –1 O 1 2
increases steeper
451
Answer Spotlight A+ Math F4.indd 451 15/03/2021 2:12 PM
Question Bank
Chapter 1 Quadratic Functions and Equations in One Variable
1. Which of the following characteristics of quadratic function is not correct for f(x) = x + 9?
2
A The graph f(x) = x + 9 has a minimum point at (0, 9).
2
B f is a many to one function.
C The axis of symmetry for the graph f(x) = x + 9 is x = 0.
2
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2
D The graph f(x) = x + 9 cuts the x-axis at x = ±3.
2. Which of the following graphs is correct about the value of a on the graph of the quadratic function y = ax ?
2
A y C y
y = 4x 2
y = x 2
y = x 2
y = 3x 2
x
O
x
O
B y D y
y = x 2 y = x 2
x x
O O
y = –5x 2
1
–
y = – x 2
2
3. The diagram below shows a right-angled triangle.
(x + 10) cm
(x + 5) cm
25 cm
Based on the given information, find the quadratic equation that can be formed.
2
A x + 15x – 250 = 0 C 2x + 15x – 500 = 0
2
B x + 30x – 500 = 0 D 2x + 15x – 750 = 0
2
2
4. Which of the following is not a quadratic equation?
2
2
A p + 8 = 3p C x + 3 = 4
x
B k = 10 – 3k 2 D 13 + 9w – 3w = 0
2
7
5. Given 4 is a root of the quadratic equation ax + 14x – 8 = 0, determine the value of a.
2
A –4 C 3
B –3 D 4
1
Answers
1. y 9. a = –1 0
Shape
2
f(x) = 0, 3x – x = 0
x(3 – x) = 0
x = 0 or x = 3
(0, 9) y
x
O
Answer: D
2. Answer: C O 3 x
2
3. (x + 10) + (x + 5) = 25 2
2
2
2
x + 20x + 100 + x + 10x + 25 = 625
2x + 30x – 500 = 0
2
x + 15x – 250 = 0 Answer: A
2
Answer: A 10. y = a(x + 4)(x – 1)
2
2
4. x + 3 = 4 = ax + 3ax – 4a
2
a = 1, y = x + 3x – 4
x
2
The power of x is –1. a = 2, y = 2x + 6x – 8
2
Answer: C a = 3, y = 3x + 9x – 12
Answer: D
2
5. x = 4, a(4) + 14(4) – 8 = 0
16a + 56 – 8 = 0 11. y
16a = –48
y = (x + 3)
2
2
y = (x – 1)
a = –3
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Answer: B
6. 2x – 13x + 15 = 0 1
2
(2x – 3)(x – 5) = 0 x
2x – 3 = 0 or x – 5 = 0 –3 –1 O 1
3
x = or x = 5 –2 2
2
Answer: A y = (x – 1) – 2
7. (y + 3) = 4
2
2
y + 6y + 9 = 4 12. (a) y
2
y + 6y + 5 = 0
(y + 1)(y + 5) = 0 16
y = –1 or y = –5 y = –x + 16
2
Answer: B
8. x + 8x + 16 = 0
2
(x + 4) = 0 O x
2
x = –4 or x = –4 (same sign) –4 4
x + 7x + 6 = 0
2
(x + 1)(x + 6) = 0 (b) y
x = –1 or x = –6 (same sign) 2
x – 2x – 8 = 0 y = x + 5x
2
(x + 2)(x – 4) = 0
x = –2 or x = 4 (different sign)
x – 11x + 24 = 0 x
2
(x – 3)(x – 8) = 0 –5 O
x = 3 or x = 8 (same sign)
Answer: C
4
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