Focus SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 2 Differentiation
9. The sum of two positive numbers is 38. Determine
(b) dL = 3.4 – 3 = 0.4 the values of these two numbers such that the sum
When t = 1, x = 3 + 6(1) = 9 of the squares of these two numbers is minimum.
dL ≈ dL
dx dx 10. It is estimated that the cost RMK, to drive your
2 – 1 ≈ 0.4 own car from town A to town B follows the formula
3 dx 2 1
dx ≈ 1.2 K = 108V + V , such that V is the average speed
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in km h . Find the value of V if the cost is minimum.
–1
11. The diagram below shows a cuboid with length 3x
Try Questions 19 – 20 in ‘Try This! 2.4’ cm, width x cm and height h cm. Given that the sum
of the surface area of the cuboid is 96 cm .
2
Try This! 2.4
h cm
1. Find the gradient of the tangent to the curve
y = 2x – 5x at the point where the x-coordinate is 3x cm
2
3
–1. x cm
2. Find the equation of the tangent to the curve (a) Express the volume of the cuboid, V cm , in
3
1
y = x – 3 x + 2x at the point 1, 3 2 . Hence, find the terms of x.
2
3
2 2 (b) Hence, find the maximum volume of the cuboid.
coordinates of the other point such that the tangent to
the curve at the point is parallel to the tangent at the 12. Nizam plans to construct a closed cylinder with
1
point 1, 3 2 . radius r cm from a piece of metal with an area of
2
540 cm .
2
(a) Show that the height of the cylinder, h cm, is
3. Given that tangent to the curve y = 4x + px + q at 2
2
the point (–1, 10) is perpendicular to the straight line given by h = 270 – pr .
3y = x – 5. Find the values of p and q. pr
(b) Hence, find the values of r and h such that the
4. Find the equation of the normal to the curve 4y = x volume of the cylinder is maximum. Hence find
2
at the point (4, 4) and hence, find the coordinates of this maximum volume. [Use p = 3.142]
the point such that this normal intersects the curve
once again. 13. Given y = 8x – 5 x + 1 , find the rate of change
3
2
3 2
5
–1
5. M is a point on the curve y = such that the gradient of y when x increases with a rate of 3 unit s when
x
x = 1.
of the normal at point M is 5. The tangent and normal
at point M meet the y-axis at A and B respectively. If 14. Two positive variables h and k are related by
the x-coordinate of point M is positive, find k – 4
2
–1
(a) the coordinates of the midpoint of line AB, h = k . If k changes at a rate of 0.4 unit s , find
(b) the length of line AB. the rate of change of h when h = 3.
6. Find the turning points for y = x – 6x + 9x + 4 and
3
2
dy 15. A piece of ice cube with sides 3 cm melts at a rate of
determine the nature of each point by using and 0.03 cm s . Find the rate of change of its volume.
–1
d y . dx
2
Form 5
dx 2
16. A piece of metal in the shape of a sphere is heated
3 –1
7. Given that the graph of the function f(x) = x + h 2 has and expands at a rate of 0.6 cm s . Find
a turning point (3, q). x (a) the rate of change of its radius, and
(a) Find the values of h and q. (b) the rate of change of its surface area
3
(b) Determine the nature of this turning point. when its volume is 36p cm .
8. The curve y = x + px + qx + 1 has a turning point 17. Given y = 4 , find
2
3
(–1, 5). 2 – x
(a) Determine the values of p and q. (a) the small change in y when x increases from
(b) Find the other turning point and hence 1 to 1.05,
determine the nature of these two turning (b) the small change in x when y decreases from
2
points by using d y . 8 to 7.96.
dx 2
260

Additional Mathematics SPM Chapter 2 Differentiation

18. Given y = 1 . 20. The relation between the focal length, f m, of a
√x convex lens with its object length, u m and image
dy length, v m, is given by the formula:
(a) Express in terms of x.
dx
(b) Hence, calculate the approximate value of 1 = 1 + 1
1 . f v u
√80.98
If the focal length of the lens is 4 m, find the small
19. A filler funnel in a cone shape has a height of 8 cm. change in its object length when the image length
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If the radius of the base of cone changes from 6 cm changes from 28 to 27.8 m.
to 6.04 cm, find the approximate change in
(a) the volume of the cone,
(b) the curve surface area of the cone.






SPM Practice 2
SPM Practice



PAPER 1 (b) f"(–2) if f(x) = x 2x – 3x + 4 2 .
1
2
lim
1. (a) Find the value of x → 0 (8 – √x ). x
SPM 9. The gradient function of a curve is (3x – 5). Given
2018 (b) Differentiate the following function by using the the point Q(–2, 6) lies on the curve, find
first principle. (a) the gradient of the tangent at point Q,
y = 5 (b) the equation of the normal to the curve at
x 2
2 – 3x 2 point Q.
2. (a) Given f(x) = , find f '(x).
5x – 4 10. Given that the gradient of the normal to the curve
2
3
(b) Given f(x) = 18 , evaluate f'(2). y = 9x – mx at x = –3 is 4 .
x 4 3
(a) Find the value of m.
3. (a) Find d 1 1 2 . (b) Find the equation of the tangent to the curve at
dx 8x – 5 x = –3.
(b) Given r = 3x – 2 and y = – 5 2 . Find dy in terms 11. Given that hx – x is the gradient function of a
2
of x. r dx
curve such that h is a constant. If y – 7x + 5 = 0 is
4. (a) Differentiate x (1 + 5x) with respect to x. the equation of the tangent at point (–1, –12), find
6
3
(b) Given f(x) = (1 – 3x) , evaluate f'(–1). (a) the value of h.
5
5. Given the function h(x) = px – 8x + 6x, find (b) the equation of the normal to the curve at point
2
3
(a) h'(x), (–1, –12).
(b) the value of p if h'(–1) = 5. 12. Find the equations of two tangent lines from the point Form 5
d y dy 5 11
2
2
6. Given y = x(5 – x), express y + x + 24 in 1 , 2 to the curve y = 4 + 3x – x . HOTS
dx 2 dx 2 2 Applying
terms of x in the simplest form. Hence, solve the
equation 13. A car moves along a straight road. The
2
y d y + x dy + 24 = y. displacement, J metre, of the car from a fixed point
dx 2 dx X along the straight road is given by
7. (a) Given f(x) = (7x – 4) , find f"(x). J = 3t + 23 2
t – 3t – 3

6
3
(b) Given h(x) = kx – 5x + 6x, find the value k if 4
3
2
h"(–1) = 5. such that t is the time, in second, after passing
through point X. Find HOTS
8. Find the value of dJ Applying
SPM 4 (a) dt .
lim
x 2x – 3x +
2018 (a) x → 0 1 2 x 2 . (b) the maximum displacement of the car.
261

Additional Mathematics SPM Chapter 2 Differentiation

14. In an agricutural project, Ali is given the task of PAPER 2
fencing a piece of rectangular land with the size of 2
8x m × (6 – x) m. Determine the length, in m, the 1. (a) Given that y = 4x – 3x + 5, determine the first
fence that Ali needs to make, if the area of the land derivative by using the first principle.
is maximum. (b) Find
(i) d 1 1 2
15. The curve y = hx + 3x has a turning point at (2, k). dx 5x – 2
4
SPM Find (ii) d 1 x 2
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2019 (a) the value of h and k. dx 5x – 2
dy
(b) the value of when x = –1. (c) By using differentiation method, estimate the
dx 8
value of 3 .
16. The diagram below shows a piece of cardboard (2.01)
which is used to form a cylindrical-shape funnel 2. The gradient function of a curve is px – qx, such
2
with opening at both ends. that p and q are constants. Given the gradient of
the normal to the curve at the point x = 1 is –5 and
the point (5, –3) is the turning point of the curve.
(a) Find the value of p and q.
y cm
(b) Determine the nature of the turning point (5, –3)
by using the second derivative.
x cm
3. Give the equation of a curve is
3
2
Given that the perimeter of the cardboard is 50 cm. y = 4 – x(2 – x ).
Find the length x cm and width y cm such that the (a) Find the gradient function of the curve.
volume of the cylinder formed is maximum. (b) Find the coordinates of the turning points.
(c) Hence, determine whether each of the turning
17. The diagram below shows a closed right cylinder points is maximum or minimum.
with a fixed height of 8 cm and a changing radius
of base. 4. Given the equation of a curve is y = 3x(2x – 1) .
4
Find
(a) the gradient of the curve at the point where the
x-coordinate = 1.
8 cm (b) the equations of the tangent and normal to the
curve at the point where the x-coordinate = 1.

5. Given that the equation of a curve is y = – 4 2 .
Given that the radius of the base changes with a SPM dy x
rate of 0.3 cm s when the radius is 5 cm. 2017 (a) Find the value of dx when x = 5.
–1
Estimate (b) Hence, estimate the value of – 4 .
(a) the rate of change of the total surface area of (5.02) 2
the cylinder,
(b) the rate of change of the volume of the cylinder. 6. The diagram below shows a filter funnel in a cone-
shape with height 20 cm and a top circular surface
18. Two variables x and y are related by the equation radius of 8 cm. Ezza pours oil into the filter funnel
3
–1
yx = 25. Express in terms of p, the small change with a constant rate of 8 cm s .
2
in y when x changes from 5 to 5 – p, such that p is
Form 5
a small value.
x
19. Given that the equation of a curve is y = √x .
SPM dy Oil h
2017 (a) Find the value of dx when x = 16.
(b) Hence, estimate the value of √15.96 . Given that the surface radius of the oil in the filter
funnel is x cm and the height of the oil is h cm.
20. Given that A = 3q – 2q and x = 4q – 3. (a) Express the volume of the oil, V cm , in terms of
2
3
SPM dA h.
2018 (a) Express in terms of x.
dx (b) Hence, calculate the rate of change of the
(b) Find the small change in x when A changes height of the oil in the cone when the surface
from 4 to 3.9 when q = –1. radius is 2 cm.

262

Additional Mathematics SPM Chapter 2 Differentiation

7. Given a formula H = 3 2 which is obtained from 10. (a) The diagram below shows a pattern in the
an experiment. 5 + g shape of a square PQRS with side of 8 cm.
(a) Express dH in terms of g. P Q
dg
(b) Ali read the measurement g from a faulty M
instrument with error and recorded the reading
as 2.98 instead of the real reading of 3. Find the
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estimated error in the measurement of H. S N R
x y
8. (a) The straight line – = 1 is the normal to
2 6 M and N are two points lie on the side PS
2
3
the curve y = x – 4x + 5x – 3 25 at point X. and SR respectively such that PM = x cm and
27
Find SN = 3x cm.
(i) the coordinates of point X. (i) Express the area of triangle QMN in terms
(ii) the equation of the tangent to the curve at of x.
point X. (ii) Hence, find the minimum area of triangle
(b) In an experiment concerning the focal length of QMN. State the value of x when this
a lens, an object is placed u cm from a lens minimum area occurs.
and the image distance v cm is measured. It (b) Given y = 5 and s = 8 – 3x. Find
is given that u and v are related by the formula 4s 2
1 + 1 = 1 . (i) the rate of change of the value s when x
u v 6 changes with a rate of 1 unit s ,
–1
(i) Express du in terms of v. 3
dv (ii) the approximate change in the value of y
(ii) Hence, find the rate of change of v when when x increase from 2.50 units to 2.56
u increases with a rate of 1.8 cm s when units.
–1
u = 5 cm.
11. The diagram below shows the front elevation
9. (a) The diagram below shows a hemispherical SPM of part of the rail of a roller coaster in an indoor
container with a radius of 12 cm. 2018 themepark inside a shoping mall.
12 cm
Ceiling

10 cm
h cm
Floor
x
Izzatul pours a type of liquid into the container
such that the height, h cm, of the liquid in the
container increases with a rate of 0.8 cm s . The curve part of the roller coaster is
–1
(i) Express the area, in cm , of the liquid represented by the equation y = 1 x – 1 x – 6x,
2
2
3
surface in the container in terms of h. 3 2
(ii) Hence, find the rate of change of the such that the point x is the origin. Find the shortest
surface area of the liquid when the height vertical distance, in m, from the rail to the ceiling of
of the liquid is 8 cm. the building. Form 5
(b) In the laboratory, it is discovered that the
4
duration of the oscillation, T seconds of an 12. The curve y = hx – 5x has a turning point (2, k).
SPM
oscillating object which is hung by using a 2019 Find the values of h and k.
string, l cm in length is given by
2
3
l 13. The curve y = 2x + 3x – 12x – 15 passes through
T = 2p point P(–3, –6) and has turning points A(1, –22)
10
(i) Find dT . and B. Find
dl (a) the gradient of the curve at point P.
(ii) Estimate the small change in the length (b) the equation of the normal to the curve at point
of the string when the duration of the P.
oscillation changes from 8p seconds to (c) the coordinates of point B and determine the
8.02p seconds. nature of this turning point B.
(iii) Find the small increase in the duration of
the oscillation when the length of the string
increases from 40 cm to 43 cm.
263

Chapter Learning Area : Geometry
Form 5
1 Circular Measure







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Concept
Map



Navigator
Watchmaker






Astronomer
Professions
involving circular
measure






Aeroplane designer


Architect
KEYWORDS


• Angle subtended – Sudut tercangkum
• Arc length – Panjang lengkok
• Area of sector – Luas sektor
• Chord – Perentas
• Degree – Darjah
• Radian – Radian
• Radius – Jejari
A variety of professions involve angle measurements as well as the calculation of the length of
• Segment – Tembereng

• A watchmaker determines the position of the gear component in the watch.
arc and the area of sector of a circle. For example,
• An astronomer uses space telescopes with the suitable angular resolution.
• An aeroplane designer measures the depth size of the circular shape of the plane.

What are some examples of other fields of profession that apply the concept of circular measure?
• An architect draws an arc with suitable angle when designing a house.
• A navigator uses sextant for navigation.






216
216

Additional Mathematics SPM Chapter 1 Circular Measure

1.1 Radian SPM Tips
× π
180°
A Relating angle measurement in
radian and degree
degree radian
1. Assume a circular wire is cut along the radius of
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circle. Each piece of the wire obtained will have 180°
an arc of the same length as the radius of the × π
circle.
3 2 1
4 1 Convert the angle 40° to the angle in radians expressed
Piece of wire (a) in terms of π,
5 6 (b) without involving π.

2. Observe that the angle subtended by each piece Solution
of wire at the centre of the circle is always 57.3° (a) 40° = 40° × π 2 rad
1
or one radian. 180°
s = 2 π rad
9

r θ radian Alternative Method
1 radian r
= 57.3º 180° = π rad
r
40° = q
3. One radian is defined as the measure of the angle 40° × π
subtended at the centre of a circle if the arc length q = 1 180° 2 rad
s is equal to the radius r, that is s = r. 2 2
= π rad = π c
4. The angle in radian is equal to the ratio of the 9 9
arc length to the radius of a circle which is SPM Tips
s
q = .
r Angles in radians is marked with abbreviation “rad”,
or superscript letter c or r.
5. The angle of one complete revolution in degrees
and radians are 360° and 2π radian respectively. 3.142
1
Therefore, the relationship between the angle (b) 40° = 40° × 180° 2 rad
measurements in degrees and radians can be = 0.6982 rad = 0.6982 r
expressed as 180° = π radian.
SPM Tips Form 5
360º = 2π radian
Generally, the value of π used is 3.142.

C
C C A L C U L A T O R Corner
6. Using the relationship of 180° = π radian, the
following method can be used to convert angles The following type of calculator can be used to convert the
in degrees to radians and vice versa. angle 40° to an angle in radian:
(a) Angles in degrees are converted to the angles Casio FX-570MS: Press MODE 4 times, choose 2 ,
in radians by multiplying π . press 4 0 , SHIFT , Ans , choose 1 , press = .
180° Casio FX-570ES: Press SHIFT , SETUP , choose 4 ,
(b) Angles in radians are converted to the angles press 4 0 , SHIFT , Ans , choose 1 , press = .
in degrees by multiplying 180° .
π Try Questions 1 – 2 in ‘Try This! 1.1’



217

Additional Mathematics SPM Chapter 1 Circular Measure

2
Solution
Convert 3.142
1
(a) 1.2 radian to degree by using π = 3.142, (a) 35° = 35° × 180° 2 rad
(b) 7 π radian to degree without using π = 3.142. = 0.6109 rad
4
Let change in speed = v
Solution v = (60 – 20) km h –1
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180 °
1
(a) 1.2 rad = 1.2 × 3.142 2 1 rad 0.6109 rad
–1
= 68.75° v = 65.48 km h for 1 radian
1
Minion Pro 10pt = — (b) Change in speed = 65.48 km h –1 × 1.8 rad
2 C C A L C U L A T O R Corner 1 rad
C
---------------------------------------- = 117.9 km h –1
Arial 8.5pt = 1 The following type of calculator can be used to convert the
angle 1.2 radian to an angle in degree:
2
---------------------------------------- Try this HOTS Question
Casio FX-570MS: Press MODE 4 times, choose 1 ,
Cth KBAT_Optima 9pt = 80 press 1 . 2 , SHIFT , Ans , choose 2 , press = . The diagram below shows the speedometer used to
measure the speed of a car in kilometre per hour.
1.25
---------------------------------------- Casio FX-570ES: Press SHIFT , SETUP , choose 3 ,
press 1 . 2 , SHIFT , Ans , choose 2 , press = .
---------------------------------------- 100 120 140 160 180 200
80 220 240 260
60
2
280
(b) 7 π rad = 1 7 π × 180 ° 40 20 300
4 4 π 0 km/h 320
= 315°
–1
Try Questions 3 - 4 in ‘Try This! 1.1’ It is given that the angle between 60 km h and
120 km h is 45°.
–1
(a) Calculate the change in speed, in km h , that
–1
Example of HOTS is represented by 1 radian.
HOTS Question
(b) If the pointer of the speedometer changes by
The diagram below shows the speedometer used to 2.1 radians, what is the change in speed, in
measure the speed of a motorcycle in kilometre per –1
hour. km h , experienced by the car?
[Use π = 3.142]
Answer:
120 140 160 –1
100 180 (a) 76.38 km h
80 200
60 220
40 240 (b) 160.4 km h –1
20 260
0 280
It is given that the angle between 20 km h and Try This! 1.1
–1
Form 5
60 km h is 35°.
–1
(a) Calculate the change in speed, in km h , that is 1. Convert each of the following angles to radians in
–1
represented by 1 radian. terms of π.
(a) 95°
(b) If the pointer of the speedometer changes by 1.8
radians, what is the change in speed, in km h , (b) 22.5°
–1
(c) 56°15’
experienced by the motorcycle?
2. Convert each of the following angles to radians by
[Use π = 3.142]
using π = 3.142.
(a) 121°
(b) 50.72°
(c) 82°2’
218

Additional Mathematics SPM Chapter 1 Circular Measure
3. Convert each of the following angles to degrees by
using π = 3.142. REMEMBER!
(a) 1.8 radian
(b) 4 radian The value of θ must be in radians when applying the
3 formula s = rθ.
(c) 5.16 radian
4. Convert each of the following angles to degrees. 3
3
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(a) π radian
5 Calculate the arc length, s for each of the following
(b) 0.25π radian sectors.
(c) 12 π radian (a) (b)
7 s
s 230º
1.52 rad O
1.2 Arc Length of a Circle O 5.2 cm 2.75 cm

Solution
A Determining arc length, radius and (a) s = rq 1 3.142 2
angle subtended at the centre of a = 5.2 × 1.52 (b) 230° = 230° × 180° rad
circle = 7.904 cm = 4.015 rad
1. Assume a piece of plastic wrapper is cut to wrap s = rq
around a piece of cake that has the shape of sector = 2.75 × 4.015
of a circle. The length of the plastic wrapper cut = 11.04 cm
must be equal to the sum of the arc length, s and Try Questions 1 – 2 in ‘Try This! 1.2’
two times the radius, r.
Example of HOTS
HOTS Question
r
s
Nevis Swing in New Zealand is the biggest canyon
swing in the world with the cable length of 120 m.
r
The length of plastic wrapper = s + 2r A participant can swing up to a maximum angle of
2.5 rad each time the swing is made. Calculate the
maximum distance, in m, of the arc of the swing.
2. It is known that the ratio of the arc length to the
radius of a circle is equal to the angle subtended Solution
s
at the centre of the circle in radian, that is = q. s = rq
r = 120 × 2.5
= 300 m
s
r Try this HOTS Question
θ radian The rope length of a swing at a playground is
1.8 m. Form 5
r

3. By using the equation s r = θ, the following 1.8 m
formula can be derived to find the arc length of a
circle:

For a circle with radius r, the length of arc
subtending the angle θ at the centre of circle Wilson always plays the swing in the evening and
is s = rθ. rides up to 1.4 rad. Calculate the distance, in m,
the arc of the swing.
Answer: 2.52 m





219

Additional Mathematics SPM Chapter 1 Circular Measure

4 B Determining perimeter of segment
of a circle
Calculate the radius, r, for each of the following sectors.
(a) 1. Observe that the shaded region which is known
as the segment of the circle is surrounded by arc
4.8 rad s and chord p.
r O
15.15 cm
s
r p
θ
(b) O
7.1 cm
1
Minion Pro 10pt = —
2 89.6º
---------------------------------------- O r 2. For a circle with radius r, the chord length
Arial 8.5pt = 1 subtending the angle θ at the centre of the circle
2
---------------------------------------- Solution is p = 2r sin .
q
Cth KBAT_Optima 9pt = 80 (a) r = s 2 q
1.25
---------------------------------------- q 3. By using s = rq and p = 2r sin , the following
2
= 15.15 formula can be derived to determine the
---------------------------------------- 4.8 perimeter of segment of a circle.
= 3.156 cm
Perimeter of segment of a circle
1
(b) 89.6° = 89.6° × 3.142 2 rad = arc length, s + chord length, p
q
180°
= rθ + 2r sin
= 1.564 rad 2
1 q = Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
s
r =
XX q REMEMBER!
7.1
Penyelesaian = 1.564 Solution of triangle in circular measures can involve the
(a) XX use of sine rule and cosine rule, that is
= 4.540 cm a b
Sine rule : =
sin A sin B
Try Questions 3 - 4 in ‘Try This! 1.2’ Cosine rule : a = b + c – 2bc cos A
2
2
2
5
Calculate the angle θ, in radian, for sector AOB in the 6
diagram below. Calculate the perimeter, in cm, of the shaded segment
of each of the following circles.
7.1 cm
A [Use π = 3.142].
Form 5
θ B (a) (b)
O 4 cm 1.42 rad
O
Solution O
s 3.9 cm 44.7º 3.71 cm
r

= 7.1 Solution
4 (a) s = rq
= 1.775 rad = 3.9 × 1.42
= 5.538 cm
Try Question 5 in ‘Try This! 1.2’



220

Additional Mathematics SPM Chapter 1 Circular Measure

p = 2r sin q Perimeter of the shaded segment
2 = s + p
= 2 × 3.9 × sin 1 1.42 2 = 2.895 + 2.822
2
= 2 × 3.9 × sin 0.71 = 5.717 cm
= 5.084 cm
Try Questions 6 - 7 in ‘Try This! 1.2’
Perimeter of the shaded segment
= s + p
= 5.538 + 5.084 7
= 10.62 cm The diagram below shows a sector OFGH with radius
10 cm. It is given that the length of chord FH = 18 cm.
Alternative Method Calculate the perimeter, in cm, of the shaded region
Sine rule: FGH.
a = b
sin A sin B F G
p = 3.9 10 cm 18 cm
sin 1.42 3.142 – 1.42
sin 1 2 2 O H

p = 3.9 × sin 1.42 Solution
sin 0.861
= 5.083 cm For ∆OFH,
18 = 10 + 10 – 2(10)(10)(cos ∠FOH)
2
2
2
Cosine rule: 2 2 2
a = b + c – 2bc cos A cos ∠FOH = 10 + 10 – 18
2
2
2
p = (3.9) + (3.9) – (2)(3.9)(3.9)(cos 1.42) 2(10)(10)
2
2
2

= –0.62
p = 5.084 cm
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∠FOH = cos (–0.62)
–1
C
C C A L C U L A T O R Corner = 2.2395 rad
s = rq
The following type of calculator can be used to calculate = 10 × 2.2395
sin 0.71 radian.
= 22.395 cm
Casio FX-570MS: Press MODE 4 times, choose 2 ,
press sin 0 . 7 1 = Perimeter of the shaded segment
Casio FX-570ES: Press SHIFT , SETUP , choose 4 , = s + p
press sin 0 . 7 1 = = 22.395 + 18
= 40.395 cm
1
(b) 44.7° = 44.7° × 3.142 2 rad Try Question 8 in ‘Try This! 1.2’ Form 5
180°
= 0.7803 rad
s = rq C C A L C U L A T O R Corner
C
= 3.71 × 0.7803
= 2.895 cm The following type of calculator can be used to calculate
cos (–0.62):
−1
p = 2r sin q
2 Casio FX-570MS: Press MODE 4 times, choose 2 ,
= 2 × 3.71 × sin 1 44.7° 2 press SHIFT cos – 0 . 6 2 =
2
= 2 × 3.71 × sin 22.35° Casio FX-570ES: Press SHIFT , SETUP , choose 4 ,
= 2.822 cm press SHIFT cos – 0 . 6 2 =
221

Additional Mathematics SPM Chapter 1 Circular Measure

C Solving problems involving arc 9
length The diagram below shows a logo designed by Azlan for
a new electronic product. The logo consists of sector
8 OABC with centre O and sector BPQ with centre B.
The diagram below shows a sector ABC with centre B
B and a sector CDE with centre E. Sectors ABC and
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CDE have radii of 4 cm and 2.2 cm respectively. A P O
Q C
A 3
Given that OA = OP, OP = 3.2 cm,
C 2
D E 2.38 rad ∠BOP = 1.85 rad and ∠PBQ = 0.99 rad. Calculate the
1.43 rad perimeter, in cm, of the whole logo.
B
Solution
3
OA = OP
Calculate the perimeter, in cm, of the shaded region. 2
3
[Use π = 3.142] = (3.2)
2
Solution = 4.8 cm
s = rq s = rq ∠AOC major = 1.85 × 2
CD
AC
= 4 × 1.43 = 2.2 × 2.38 = 3.7 rad
= 5.72 cm = 5.236 cm s ABC = rq
= 4.8 × 3.7
∠BED = 3.142 – 2.38 = 17.76 cm
= 0.762 rad For ∆BOP,
BP = OP
For ∆BDE, sin ∠BOP sin ∠OBP

BD = DE BP = 3.2
sin ∠BED sin ∠DBE sin 1.85 sin 1 0.99 2
BD = 2.2 2
sin 0.762 sin 1.43 BP = 3.2 × sin 1.85
BD = 2.2 × sin 0.762 sin 1 0.99 2
sin 1.43 2
= 1.534 cm = 6.476 cm
AD = AB – BD s = rq
PQ
= 4 – 1.534 = 6.476 × 0.99
= 6.411 cm
Form 5
= 2.466 cm
AP = CQ
= 4.8 – 3.2
Perimeter of the shaded region = 1.6 cm
= s + s + AD
CD
AC
= 5.72 + 5.236 + 2.466 Perimeter of the whole logo
= 13.422 cm = s ABC + s + AP + CQ
PQ
= 17.76 + 6.411 + 1.6 + 1.6
Try Question 9 in ‘Try This! 1.2’ = 27.371 cm
Try Question 10 in ‘Try This! 1.2’



222

Additional Mathematics SPM Chapter 1 Circular Measure

Example of HOTS SPM Highlights
HOTS Question
Measat-3b is a Malaysian communications satellite The diagram below shows two sectors KON and LOM
launched into the geostationary orbit, 35 786 km of two circles with a common centre O.
above the Earth. The satellite takes 24 hours to
complete one orbit.
O
Measat-3b L M
satellite
35 786 km K N
The angle subtended at the centre O by the major arc
6 400 km KN is 8θ and the perimeter of the whole diagram is
30 cm. Given ON = r cm, ON = 3OM and ∠LOM = 3q.
Express r in terms of q.
It is given that the approximate radius of the Earth
is 6 400 km. By assuming that the orbit is circular,
calculate the distance, in km, travelled by Measat-3b Solution s = rq

satellite within an hour. KN = r × 8q
[Use π = 3.142] = 8rq
s = rq
Solution LM = r × 3q

q = 1 hour 3
2π 24 hours = rq r 2
2 × 3.142 KL = MN = r – = r
q = 3 3
24
= 0.2618 rad Perimeter of the whole diagram = 30 cm
s + s + KL + MN = 30

s = rq KN LM 2 2
r +
r = 30
rq + rq +

8
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= (35 786 + 6 400) × 0.2618
3
3
= 42 186 × 0.2618 9 rq + 4 r = 30
= 11 044.29 km 3
The distance travelled = 11 044.29 km 27rq + 4r = 30
3
r(27q + 4) = 90
90
Try this HOTS Question r =
27q + 4
Measat-5 is a Malaysian communication
satellite that rotates around the Earth through a
geostationary orbit which is located 35 786 km
from the Earth’s surface. The satellite takes 24 Try This! 1.2
hours to complete one orbit.
1. Calculate the arc length s for each of the following Form 5
By assuming that one orbit is circular, calculate sectors.
the distance, in km, travelled by Measat-5 satellite (a) (b)
within three hours if the approximate radius of the s 2 cm
Earth is 6 400 km. O
[Use π = 3.142]. 0.5 rad 5.28 cm 4.28 rad
Answer: 33 137.10 km O s
(c)
2.89 cm
O
2.3 rad
s




223

Additional Mathematics SPM Chapter 1 Circular Measure
2. Using π = 3.142, calculate the arc length s for each (c)
of the following sectors.
(a) (b) s 3.13 cm O θ
s
2 π rad 174.8º 12.7 cm
O 5
4.4 cm O
3.88 cm 6. Calculate the perimeter, in cm, of the shaded
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segment for each of the following circles.
(c) (a) (b)
1
Minion Pro 10pt = — 302.5º 6.5 cm
2 s 1.12 rad
---------------------------------------- O O 2.52 rad O
Arial 8.5pt = 1 2.24 cm 3.88 cm
2
1
Arial 8pt = (c)
2
---------------------------------------- 3. Calculate the radius r for each of the following O 4.06 cm
sectors.
Cth KBAT_Optima 9pt = 80 (a) 11.38 cm 1.9 rad
1.25
---------------------------------------- 2.24 rad r
O 7. Calculate the perimeter, in cm, of the shaded
----------------------------------------
segment for each of the following circles.
(b) [Use π = 3.142]
0.28 rad (a) (b)
2.32 cm
O 4.28 cm
r O 4.78 cm O 138.2º
(c) 105.7º
r
O
4.2 rad (c)
15.93 cm
33.3º 5.1 cm
4. Calculate the radius r for each of the following
sectors by using π = 3.142. O
(a) 4.1 cm (b) 8.8 cm

r 105.3º 8. Given the radius and the length of chord AB.
41º r Calculate the perimeter, in cm, of the shaded
O segment for each of the following circles.
O [Use π = 3.142]
(c) (a) A (b) 9.65 cm
r A B
Form 5
O 9 π rad
5 8.3 cm
4.95 cm
O B 12.2 cm
17.1 cm O
5. Calculate the angle q, in radian, for each of the
following sectors. (c) B
(a) O (b) 16.18 cm

θ 9 cm
5.2 cm O θ O
10.11 cm A
5 cm
2.7 cm


224

Additional Mathematics SPM Chapter 1 Circular Measure
9. (a) The diagram below shows sector MON with 2. It is known that the ratio of the area of a sector
centre O and sector PQN with centre Q. Sectors A to the area of a circle πr is equal to the ratio of
2
MON and PQN have radii of 5.8 cm and 3.9 cm the angle subtended at the centre of the circle, θ,
respectively. HOTS
Applying to the angle 2π which is A = q .
M πr 2 2π
P 2.7 rad 3. By using the equation A = q , the following
2.36 rad N πr 2 2π
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O Q formula can be derived to find the area of sector
Calculate the perimeter, in cm, of the shaded of a circle.
region. [Use π = 3.142]
(b) The diagram below shows sector OTU with For a circle with radius r, the area of sector of
centre O and a quadrant STV with centre S. a circle subtending the angle θ at the centre of
O HOTS Analysing the circle is A = r q.
1
2
2
V S
U
T SPM Tips
Given that the radius of sector OTU is 9 cm The value of q must be in radians when applying the
and S is the midpoint of the straight line OST. 1
2
Calculate the perimeter, in cm, of the shaded formula A = 2 r q.
region. [Use π = 3.142]
10. The diagram below shows a logo designed by
a logo designer for a type of online game. The
logo consists of a sector SROT with centre S 10
and a sector OMN with centre O. Given that
RM = 2MS, RM = 4 mm, ∠MOS = 0.18 rad and Calculate the area, A for each of the following sectors.
∠MSN = 1.46 rad. (a) (b)
R T HOTS Analysing O
M N 4.5 rad 4.52 cm 55.8º
S O 4.96 cm



O
Calculate the perimeter, in mm, for the whole logo. Solution
1
[Use π = 3.142] (a) A = r q
2
2
1
2
1.3 Area of Sector of a Circle = × (4.52) × 4.5
2
= 45.765 cm 2 Form 5
A Determining area of sector, radius
1
and angle subtended at the centre of (b) 55.8° = 55.8° × 3.142 2 rad
a circle 180°
= 0.9740 rad
1. A slice of pizza with the shape of a sector of a 1
2
circle is taken from one piece of pizza. The area A = r q
2
of a slice of pizza is equal to the area of a sector of
a circle. = 1 × (4.96) × 0.9740
2
2
= 11.98 cm
2
r
θ radian
Try Questions 1 - 2 in ‘Try This! 1.3’


225

Additional Mathematics SPM Chapter 1 Circular Measure

Example of HOTS 11
HOTS Question
Calculate the radius, r for each of the following sectors.
At 6 p.m. every day, the lawn sprinkler at Pak Samad’s
farm rotates back and forth automatically in the (a) (b)
angle of 2 radians to water the plants up to 9 m away. 4.52 rad
O 113.8º
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O
1
Minion Pro 10pt = — 9 m Area of sector Area of sector
2
2
---------------------------------------- 2 rad = 133.95 cm = 24.09 cm 2
Arial 8.5pt = 1 Solution
2
1 2A
Arial 8pt = 2 (a) r =
2 Calculate the maximum area, in m , of Pak Samad’s q
farm that the sprinkler could water.
---------------------------------------- 2 × 133.95
=
Cth KBAT_Optima 9pt = 80 Solution 4.52
1.25
---------------------------------------- A = r q = 7.699 cm
1 2
2
1
---------------------------------------- = (9) (2) (b) 113.8° = 113.8° × 3.142 2 rad
1
2
2 180°
= 81 m 2 = 1.986 rad
1 Try this HOTS Question r = 2A
q
XX A water tank truck will do road cleaning work at 2 × 24.09
Penyelesaian 7 p.m. every day. At a certain time, the truck waters = 1.986
the road at a distance of 3.5 m covering a sector
(a) XX
that subtends the angle 1.8 radians at the nozzle = 4.925 cm
of sprinkler.
Try Questions 3 - 4 in ‘Try This! 1.3’

12
Calculate the angle, θ, in radian, for the sectors below.
1.8 rad
3.5 m 17.4 cm O
θ


Calculate the maximum area, in m , of the road Area of sector = 60.8 cm 2
2
Form 5
watered by the water tank truck at that time.
Solution
Answer: 11.025 m 2
q = 2A
r 2
= 2 × 60.8
(17.4) 2
= 0.4016 rad

Try Question 5 in ‘Try This! 1.3’





226

Additional Mathematics SPM Chapter 1 Circular Measure

B Determining the area of segment of 13
a circle
Calculate the area, in cm , of the shaded segment for
2
1. Observe that the area of segment of a circle each of the following circles.
represented by the shaded region can be
determined by subtracting the area of triangle [Use π = 3.142]
from the area of sector of the circle. (a)
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2 rad
r p O 3.5 cm
θ
O
r

(b)
2. For a circle with radius r, the area of triangle can 60º
1
be determined by using the formula A = r sin 23 cm
2
2 O
1
q or A = s(s – p)(s – r) , such that s = (2r + p).
2
2
3. By using A = 1 r q and A = 1 r sin q or
2
2
2 2 Solution
A = s(s – p)(s – r) , the following formulae can 1 2
2
be derived to determine the area of segment of (a) A = r (q – sin q)
2
a circle. 1 2
(a) Area of segment of a circle = (3.5) (2 – sin 2)
2
= Area of sector – Area of triangle = 6.681 cm
2
1
= 1 r q – r sin q
2
2
2 2
1
= 1 r (q – sin q) (b) 60° = 60° × 3.142 2 rad
2
2 180°
= 1.047 rad
(b) Area of segment of a circle
1
= Area of sector – Area of triangle A = r (q – sin q)
2
2
= 1 r q – s(s – p)(s – r)
2
2
2 = 1 (23) (1.047 – sin 60°)
2
1
such that s = (2r + p) 2
2 = 47.87 cm
2
Form 5
REMEMBER! SPM Tips
Solution of triangles in circular measure can involve the If the angle θ is given in the unit of degrees, use it directly
use of two different types of formulae, which is to find the sine value. Converting degrees to radians will
1 only reduce the accuracy of the sine value.
A = ab sin C, or
2
A = s(s – a)(s – b)(s – c)
such that s = 1 (a + b + c)
2
Try Questions 6 - 7 in ‘Try This! 1.3’


227

Additional Mathematics SPM Chapter 1 Circular Measure

Example of HOTS 14
HOTS Question
French drain using perforated circular pipes is a The diagram below shows sector MON of a circle. The
very effective rainwater management in solving length of chord MN = 11 cm.
landscaping water issues. The diagram below shows
the cross section of the circular pipe used in the
French drain. The pipe is partially filled with water N 14 cm
that accumulates as rainwater flows into it. 11 cm
O
1
Minion Pro 10pt = — M
2
----------------------------------------
Arial 8.5pt = 1 O 5 cm Find the area, in cm , of the shaded segment.
2
2 2 rad
1
Arial 8pt =
2 Solution
---------------------------------------- If the length of the circular pipe is 20 cm, calculate For ΔMNO,
the volume, in cm , of the water collected.
3
2
2
2
Cth KBAT_Optima 9pt = 80 11 = 14 + 14 – 2(14)(14)(cos ∠MON)
1.25
---------------------------------------- Solution 14 + 14 – 11 2
2
2
Cross sectional area of the accumulated water cos ∠MON =
---------------------------------------- = r (q – sin q) 2(14)(14)
1 2
2 = 0.6913
1
= (5) (2 – sin 2)
2
–1
2 ∠MON = cos (0.6913)
2
= 13.63 cm
1 = 0.8075 rad
Volume of the accumulated water
1
XX Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
A
= r q
2
= Cross sectional area × Length of circular pipe sector 2
Penyelesaian = 13.63 × 20
1
(a) XX = 272.6 cm 3 = × 14 × 0.8075
2
2
= 79.135 cm
2
Try this HOTS Question
1
The circular pipes are widely used in sewage s = (2r + p)
2
treatment system. The diagram below shows the
sewage collected in the circular pipe. 1
= [2(14) + 11]
2
= 19.5 cm
O 4 m
1.8 rad A triangle = s(s – p)(s – r) 2
Sewage
Form 5
= 19.5(19.5 – 11)(19.5 – 14)
2
Calculate the volume, in m , of the sewage = 70.81 cm
2
3
collected in the circular pipe with the length of
1.5 km. Area of shaded segment = A sector – A triangle
Answer: 9 913.5 m 3 = 79.135 – 70.81
= 8.325 cm 2
Try Question 8 in ‘Try This! 1.3’
228

Additional Mathematics SPM Chapter 1 Circular Measure
SPM Highlights C Solving problems involving areas of

sectors
The diagram below shows a circle and a sector of
a circle with a common centre O. The radius of the 15
circle is r cm.
The diagram below shows sector AOB with centre O
M and sector PQR with centre P. Sector AOB has radius
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of 8 cm, OQ = OP = 3 cm and ∠AOB is a right angle.
P
θ A
O Q B
N
Q R
P
O
Given that PM = 6 and arc length of MN and PQ are
2
21 cm and 7 cm respectively. Calculate Calculate the area, in cm , of the shaded region.
(a) the values of r and q, [Use π = 3.142]
(b) the area, in cm , of the shaded region.
2
[Use π = 3.142] Solution
1
A sector AOB = r q
2
Solution 2
1
(a) Arcs MN and PQ subtend the same angle θ at = × 8 × 1 3.142 2
2
the centre, therefore 2 2
1
s s = × 8 × 1.571
2
MN = PQ 2
r + 6 r = 50.27 cm
2
21 = 7
r + 6 r A = × 3 × 3
1
21r = 7(r + 6) triangle OPQ 2
21r = 7r + 42 = 4.5 cm
2
14r = 42 For ∆OPQ
r = 3 cm ∠QPR = ∠POQ + ∠OQP
q = s 1.571
r = 1.571 + 1 2 2
= 7 rad = 2.3565 rad
3
2
1 PQ = OP + OQ 2
2
(b) A = r sin q 2 2
triangle MON 2 = 3 + 3
7
= 1 × (6 + 3) × sin 1 2 = 4.243 cm
2
2 3
1
= 29.28 cm A sector PQR = r q Form 5
2
2
2
A = 1 r q 1
2
2
sector POQ 2 = × (4.243) × 2.3565
1 7 2
2
= × 3 × = 21.21 cm
2
2 3
= 10.5 cm Area of the shaded region
2
Area of the shaded region = A – A – A
= A – A sector AOB triangle OPQ sector PQR
triangle MON sector POQ = 50.27 – 4.5 – 21.21
= 29.28 – 10.5 2
= 18.78 cm = 24.56 cm
2
Try Question 9 in ‘Try This! 1.3’
229

Additional Mathematics SPM Chapter 1 Circular Measure

16
The diagram below shows the area on the rear-view Solution
mirror with the shape of a sector of a circle with centre (a) s = 12
BD
O that can be cleaned by the windshield wiper. r × x = 12
12
r = cm
x

1 1
Minion Pro 10pt = —
Minion Pro 10pt = — 2 rad 52 cm (b) Since BC and DC are tangents, thus ∠ABC
2 2 O 18 cm and ∠ADC are right-angled.
---------------------------------------- By using π = 3.142, calculate the area, in cm , of the BC = DC
----------------------------------------
2

Arial 8.5pt = 1 1 xReserved.
=
Arial
8.5pt

2 2 rear-view mirror that can be cleaned by the windshield 13 2 12 2
–
x
x
1 1 wiper. = 1 2 1 2
Arial 8pt =
Arial 8pt =
2 2
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Solution 169 144
---------------------------------------- 1 = 2 – x 2
----------------------------------------
A large sector = r q
2
2
80
Cth KB A T_Optima 9pt = 80 1 2 = 25
Cth KBAT_Optima 9pt =
1.25
1.25
----------------------------------------
---------------------------------------- = 2 × (18 + 52) × 2 x 2
= 1 × 70 × 2 = 5 cm
2
----------------------------------------
---------------------------------------- 2 x
= 4 900 cm Area of triangle ABC = Area of triangle ADC
2
1
A = r q 1 12 5
2
small sector 2 = × 1 2 1 2
×
1 1 = 1 × 18 × 2 2 x x
2
2
XX
XX 2 = 30 2 cm
x
= 324 cm
2
Penyelesaian
Penyelesaian Area of quadrilateral ABCD
(a) XX Area of the rear-view mirror that can be cleaned = Area of triangle ABC + Area of triangle ADC
(a)

X
X
= A large sector – A small sector
= 4 900 – 324 = 30 2 + 30 2
x
x
= 4 576 cm
2
= 60 cm
2
Try Question 10 in ‘Try This! 1.3’ x 2
1
Area of sector ABD = r q
2
SPM Highlights 2
2
12
= 1 × 1 2 × x
The diagram below shows a circle with centre A. 2 x
= 72 cm
2
x
Form 5
A
Area of the shaded region
x rad D
= Area of quadrilateral ABCD
B C – Area of sector ABD
60 72
BC and DC are tangents to the circle at points B and = x 2 – x
D respectively. Given that the minor arc length BD is
13 60 – 72x 2
12 cm and AC = cm. Express in terms of x, = x 2 cm
x
(a) the radius, r, of the circle,
(b) the area, A, of the shaded region.
230

Additional Mathematics SPM Chapter 1 Circular Measure
SPM Highlights
Solution
The diagram below shows a rhombus KLMN inscribed s
in sector KLM with centre L and radius h cm. (a) ∠KOL = r
L = 10
K λ rad M 15
N 2
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= rad
3
Given the area of sector KLM is 12.5 cm , express in 1
2
2 2
1 2
terms of h (b) Area of sector KOL = 2 (15 ) 3
(a) l,
2
(b) the perimeter, in cm, of the shaded region. = 75 cm
Solution 1 2 3.142
(a) A = 12.5 Area of sector LOM = (15 ) 1 2
KLM 2 4
1 2
2
h l = 12.5 = 88.37 cm
2
l = 25 rad tan ∠NOM = MN
h 2 OM
(b) s KLM = rq tan 45° = MN
= hl 15
= h × 25 MN = 15 tan 45°
h 2 = 15 cm
= 25 cm
h 1
Area of triangle NOM = (15)(15)
Perimeter of the shaded region 2
25 = 112.5 cm 2
= h + h +
h Area of the shaded region
2
= 2h + 25 cm
h = Area of sector KOL + (Area of triangle NOM –
Area of sector LOM)
= 75 + (112.5 – 88.37)
= 99.13 cm
2
SPM Highlights

The diagram below shows sector KOM with Try This! 1.3
centre O.
2
O 1. Calculate the area, in cm , for each of the following
15 cm
K sectors.
45 Form 5
(a) (b)
1.9 cm
L O 3.76 rad
2.44 rad O
N M 1.7 cm
Given that the arc length of KL is 10 cm.
Calculate
(a) ∠KOL, in radian, (c)
(b) the area, in cm , of the shaded region.
2
[Use π = 3.142] O 2 rad
2.22 cm




231

Additional Mathematics SPM Chapter 1 Circular Measure
2. Calculate the area, A, for each of the following (c)
sectors by using π = 3.142.
(a) (b) θ
O
2.2 cm 2.16 cm 1.42 cm
O O 2
191 Area of sector = 4.12 cm
4 π rad
2
3 6. Calculate the area, in cm , of the shaded segments
for each of the following circles.
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(c) (a) (b)
3.6 cm O 6.1 cm 20.1 cm
47.6
1.2 rad O O
2.5 rad

3. Calculate the radius, r, for each of the following (c)
sectors.
(a) (b)
0.64 rad O
4.3 rad 2.86 rad 10.4 cm
O
O
7. By using π = 3.142, calculate the area, in cm , of the
2
shaded segment for each of the following circles.
Area of sector Area of sector (a) (b)
= 28.56 cm 2 = 144 cm 2 40.25
(c) O 13.2 cm O O 33.5 cm
0.9 rad 93.4

(c)
2
Area of sector = 6.2 cm
4. Given π = 3.142, calculate the radius, r, for each of 6.12 cm 142.8
the following sectors. O
(a) (b)
O O
45.6
2
54.4 8. Calculate the area, in cm , of the shaded segment
for each of the following circles.
[Use π = 3.142]
Area of sector Area of sector (a)
= 4.97 cm 2 = 24.06 cm 2 N
4.14 cm 4 cm
(c) 1.2 rad
M O
O

277.8 (b)
Form 5
Area of sector = 123.56 cm O
2
12.46 cm
5. Calculate the angle q, in radian, for each of the
following sectors. 17.6 cm N
(b) M
(c)
5.5 cm θ M 19.98 cm
θ O O 8.75 cm O
39.3 cm
Area of sector Area of sector N
= 20 cm 2 = 124.6 cm 2


232

Additional Mathematics SPM Chapter 1 Circular Measure
9. (a) The diagram below shows a sector POQ
with centre O and sector PMN with centre M.
Given that the radius of sector POQ is 24.2 cm
and ∠POQ is a right angle. M and R are the
midpoints of OP and OQ respectively. 1.8 cm
O 0.8 cm
R
N Q
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M
P Calculate INFO
Calculate the area, in cm , of the shaded region. (a) the rotating distance, in cm, of the small gear,
2
[Use π = 3.142] (b) the rotating angle, in radian, of the big gear,
(b) The diagram below shows sector ABC with (c) the area, in cm , of the sector covered by the big
2
centre B and sector AOD with centre O. Given gear.
that the radius of sector ABC is 9 cm. ∠ABC [Use π = 3.142]
and ∠AOD are 0.87 radian and 1.15 radian
respectively. Solution
A 9 cm 3.142
1
B (a) 300° = 300° × 2 rad
O 180°
= 5.237 rad
Rotating distance of small gear = rq
C
D = 0.8 × 5.237
Calculate the area, in cm , of the shaded region. = 4.19 cm
2
[Use π = 3.142]
10. The diagram below shows a pair of windshield (b) Rotating distance = Rotating distance
wipers of a car that can rotate back and forth at the of large gear of small gear
angle of 1.5 radian. HOTS 1.8 × q = 4.19
Applying 4.19
q =
Wiper blade 1.8
62 cm = 2.328 rad
(c) Area of sector covered = 1 r q
2
2
= 1 (1.8) (2.328)
2
If the windshield wiper blade has a length of 60.9 2
cm and radius of its sector is 62 cm, calculate the = 3.771 cm 2
area of windscreen, in cm , that can be cleaned by
2
the pair of wipers. Try Question 1 in ‘Try This! 1.4’
[Use π = 3.142]
18 Form 5
1.4 Application of Circular Measures The diagram below shows the transition curve on one
side of the road where ∠AOB subtended by the chord
AB at centre O. Given that the length of chord AB is 30
A Solving problems involving circular m and the length of both radii OA and OB are 42 m.
measure

17 A B
The following diagram shows part of the gear network
system in a printer consisting of one small gear and one
large gear with radii 0.8 cm and 1.8 cm respectively. At O
a certain time, the small gear rotates at 300°.




233

Additional Mathematics SPM Chapter 1 Circular Measure
Calculate Solution
(a) ∠AOB, in radian, (a) 22.5° = 22.5° × 3.142 2 rad
1
(b) the arc length, in m, bounded by the chord AB, 180°
= 0.3928
(c) the area of segment, in m , bounded by the arc and s = rq
2
chord AB. = 12 × 0.3928
[Use π = 3.142] = 4.714 cm

Solution Length of the plastic wrapper = s + 2r
1 1
Minion Pro 10pt = — (a) For ∆AOB, = 4.714 + 2(12)
Minion Pro 10pt = —
2 2 2 2 2 = 28.714 cm
----------------------------------------
---------------------------------------- AB = OA + OB – 2(OA)(OB) cos ∠AOB
2
2
2
Arial
=
8.5pt


Arial 8.5pt = 1 1 30 = 42 + 42 – 2(42)(42) cos ∠AOB (b) Upper surface area of the slice of cake
2 2 42 + 42 – 30 2 1
2
2
2
1 1 cos ∠AOB = = r q
Arial 8pt = 2(42)(42) 2
Arial 8pt =
2 2 3Rights Reserved.
∠AOB = 0.7304 rad = 1 (12) (0.3928)
2
----------------------------------------
---------------------------------------- 2
(b) The arc length = rq = 28.282 cm
2
80
T_Optima 9pt =
A
Cth KB
Cth KBAT_Optima 9pt = 80 = 42 × 0.7304
1.25
1.25
---------------------------------------- (c) Volume of the slice of cake
----------------------------------------
= 30.68 m = Surface area of the slice of cake × Thickness of cake
Penerbitan Pelangi Sdn Bhd. All
----------------------------------------
1
---------------------------------------- (c) Area of segment = r (q – sin q) = 28.282 × 10
2
2 = 282.82 cm
= 1 (42) (0.7304 – sin 0.7304) Try Question 3 in ‘Try This! 1.4’
2
2
= 55.77 m
2
20
Try Question 2 in ‘Try This! 1.4’ The diagram below shows a round picture frame
with centre O and radius of 12.5 cm placed on top
19 of a picture holder. P and R are the midpoints of the
straight lines OA and OB respectively. The height of the
The diagram below shows a slice of cake in the form of picture holder is 8 cm.
sector of a circle VOW with centre O and radius 12 cm. Q
Given that the thickness of the cake is 10 cm.
O
P 100 R
V
8 cm
W A
22.5 B
O 38.2 cm
Calculate
(a) the obtuse angle POR, in radian,
Form 5
(b) the perimeter, in cm, of the whole diagram,
2
A worker needs to wrap the slice of cake with plastic (c) the area, in cm , of the picture holder.
wrapper. Calculate [Use π = 3.142]
(a) the length, in cm, of the plastic wrapper needed Solution
to wrap the entire vertical surface of the slice of (a) 100° = 100° × 3.142 2 rad
1
cake, 180°
(b) the upper surface area, in cm , of the slice of cake, = 1.746 rad
2
(c) the volume, in cm , of the slice of cake. Obtuse angle POR = 2(3.142) – 1.746
3
[Use π = 3.142] = 4.538 rad
234

Additional Mathematics SPM Chapter 1 Circular Measure

Alternative Method Solution
Obtuse angle POR = 360° – 100° (a)
= 260° A
1
= 260° × 3.142 2 rad
180°
= 4.538 rad O
C B
(b) Arc length PQR = rq
= 12.5 × 4.538 Arc length of 3 semicircles = 15π cm
= 56.725 cm Arc length of 1 semicircle = 5π cm
Perimeter = 56.725 + 12.5 + 12.5 + 38.2 r × π = 5π
= 119.925 cm semicircle
r semicircle = 5 cm O
(c) For ∆POR, d semicircle = 5 × 2 120 r
PR = OP + OR – 2(OP)(OR) cos ∠POR = 10 cm C B
2
2
2
= (12.5) + (12.5) – 2(12.5)(12.5)(cos 100°) 10 cm
2
2
PR = 19.15 cm In ∆OCB, CB = 10 cm
1 r logo 10
Area of trapezium APRB = (19.15 + 38.2)(8) =
1
2 sin 180° – 120° 2 sin 120°
2
= 229.4 cm 2
1
Area of segment = r (q – sin q) r logo = 10 × sin 30°
sin 120°
2
2 = 5.774 cm
= 1 (12.5) (1.746 – sin 100°)
2
2 (b)
= 59.47 cm 2
Area of picture holder
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X
= Area of trapezium APRB – Area of segment
= 229.4 – 59.47 θ
= 169.93 cm Y Z
2

Try Question 4 in ‘Try This! 1.4’
Area of equilateral triangle XYZ,
1
SPM Highlights A = 2 (5)(5) sin 60°
1
= 10.83 cm
2
STEM Club of SMK Ceria organised a competition to 3.142
1
design a logo for the club. The diagram below shows q = 60° × 180° 2 rad
the circular logo designed by Fauziah.
= 1.047 rad Form 5
Area of segment XY,
A = 1 (5 )(1.047 – sin 60°)
2
2 2
= 2.262 cm
2
Total area of the shaded region
= 3A + (6 – 3)A
2
1
The three shaded regions are congruent. Given that = 3(10.83) + 3(2.262)
the perimeter of the shaded region is 15π cm. Calculate
2
(a) the radius, in cm, of the logo, = 39.276 cm
(b) the total area, in cm , of the shaded region.
2
[Use π = 3.142]
235

Additional Mathematics SPM Chapter 1 Circular Measure

Try This! 1.4 3. The diagram below shows the cross section of a
slice of cake in the form of a sector of circle AOB
1. The diagram below shows two gears rotating with centre O and radius of 10 cm.
simultaneously in a model. The diameters of small B
gear and large gear are 5 cm and 12 cm respectively. A
At a certain time, the large gear rotates at 140°. 36 O


12 cm

Minion Pro 10pt = —
Minion Pro 10pt = — 1 1 If the thickness of the slice of cake is 8 cm, calculate
2 2
----------------------------------------
---------------------------------------- 5 cm (a) the length, in cm, of the plastic wrapper
wrapping the entire vertical surface of the slice
Arial 8.5pt =
Arial 8.5pt = 1 1 of cake,
2 2 2
1 1 Calculate (b) the upper surface area, in cm , of the slice of
8pt

Arial 8pt = (a) the rotating distance, in cm, of the large gear, cake,
Arial

=
2 2
(b) the rotating angle, in radian, of the small gear, (c) the volume, dalam cm , of the slice of cake.
3
----------------------------------------
---------------------------------------- (c) the area, in cm , of the sector covered by the [Use π = 3.142]
2
small gear.
80
Cth KB A T_Optima 9pt = 80 [Use π = 3.142] 4. The diagram below shows a round shaped souvenir
Cth KBAT_Optima 9pt =
MKN with centre O and radius of 20 cm placed
1.25
1.25
----------------------------------------
---------------------------------------- 2. The diagram below shows a transition curve on one above a souvenir holder. The straight lines OJ and
section of rail track where ∠POQ subtended by the OL passing through points M and N respectively.
---------------------------------------- chord PQ at centre O. Given that the length of chord The straight line JL is the tangent to the circle MKN
----------------------------------------
PQ is 45 m and the length of both radii OP and OQ at point K.
are 50 m respectively.
1 1 P Q O 20 cm O
XX Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
XX
125
N
M
Penyelesaian J K L
Penyelesaian

(a) XX
(a)
X
X
Calculate
Calculate (a) the obtuse angle MON, in radian,
(a) ∠POQ, in radian, (b) the perimeter, in cm, of the whole souvenir
(b) the arc length, in m, bounded by the chord PQ, including its holder,
(c) the area of segment, in m , bounded by the arc (c) the area, in cm , of the souvenir holder.
2
2
and chord PQ. [Use π = 3.142]
[Use π = 3.142]
SPM Practice 1
SPM Practice
Form 5
PAPER 1
1. The diagram below shows two sectors AOD and The angle subtended at centre O by the major arc
SPM BOC with a common centre O. AD is 15t radian and the perimeter of the whole
2017
C diagram is 96 cm. Given OA = j cm, OB = 3OA and
B ∠BOC = 3t radian, express j in terms of t.
D
A
O
236

Additional Mathematics SPM Chapter 1 Circular Measure
2. The diagram below shows a circle with centre O. 5. The diagram below shows a box being elevated
SPM to the upper level of a building using a pulley with
2018
P diameter of 15 cm. HOTS Applying
3k cm
Q θ rad O

R
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PQ and RQ are the tangents to the circle at points
P and R respectively. Given that the minor arc
length PR is 9 cm and OQ = 5k cm, express in
terms of k Calculate
(a) the angle q, (a) the height, in cm, of the box when the pulley is
(b) the area, A, of the shaded region. rotated by 160°,
(b) the rotating angle, in radian, of the pulley if the
box is elevated 0.35 m.
3. The diagram below shows a sector BCD with [Use π = 3.142]
SPM centre C and radius r cm inscribed in the rhombus
2019
ABCD. 6. The diagram below shows a circle with centre O
B and radius r cm.

A rad C
2
J
D O
2 rad
K
Given the area of sector BCD is 8 cm , express in
2
terms of α,
(a) the radius, r, in cm, If the major arc length JK is 10 cm, express
(b) the perimeter, in cm, of the shaded region. (a) α in terms of r,
(b) the area, in cm , of minor sector JOK in terms of
2
π and r.
4. The diagram below shows a way to treat a broken
femur to reduce pain and bleeding. The angle 7. The diagram below shows the front view of a tent
between the leg and torso is an inclined angle θ at a comping area. DE and FG are the arcs of
whereas the acute angle between the leg and the circles with centres G and D respectively whereas
bed is the reference angle, α. EOF is a sector of a circle with centre O.

HOTS
Analysing

E F Form 5
O


θ
30 30
D 2 m G
If the inclined angle q is 135°, calculate the value of Calculate the area, in m , for the whole diagram.
2
α, in radian. [Use π = 3.142]
[Use π = 3.142]









237

Additional Mathematics SPM Chapter 1 Circular Measure
8. The diagram below shows a small circle and a Calculate the area, in m , of the field that cannot be
2
large circle with centres B and C respectively. The grazed by the cow.
radius of the large circle is twice the radius of the [Use π = 3.142]
small circle. Both circles touching each other at
point E. Straight line AD is the common tangent to 11. A cheese cake baked by Reyna has a radius of
both circles at point A and point D. HOTS 11.5 cm and thickness of 8 cm. This cake is divided
Analysing into slices that will be sold at Kedai Kek Istimewa.
B The plastic wrapper used to wrap the vertical
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A
surface of a piece of cake has the measurement of
E 30 cm × 10 cm. Reyna needs to cut the cake with
rad maximum angle, q at the centre of the cake, so that
D C the plastic wrapper is sufficient to wrap the entire
vertical surface of the cake. HOTS
Calculate Creating
(a) q, in degrees, to the nearest integer,
Given the radius of the small circle is r cm, calculate (b) the volume, in cm , of each slice of cake based
3
Arial 8.5pt = 1 (a) α, in radian, on the answer for (a) .
2 (b) the perimeter, in cm, of the shaded region. [Use π = 3.142]
[Use π = 3.142]

9. The diagram below shows sector POQ and PAPER 2
sector OABCD of concentric circle with centre 1. The diagram below shows a circle and a sector
O. Points B and C are the midpoints of the SPM of circle with a common centre O. The radius of
straight lines OP and OQ respectively. Given that 2017 sector AOB is r cm and ∠AOB is q rad.
∠AOB : ∠BOC : ∠COD = 1 : 3 : 1 and the perimeter
of the whole diagram is 32 cm.
D
P A A
B C O
B
O
C
Q D
Given that the arcs length AB and CD are 2 cm
If ∠AOB = θ radian and the radius of sector and 9 cm respectively and BC = 13.5 cm.
OABCD = r cm, express in terms of r,
(a) q, Calculate
(b) the area, in cm , for the whole diagram. (a) the values of r and q,
2
(b) the area, in cm , of the shaded region.
2
10. The diagram below shows a square shaped grass [Use π = 3.142]
field with the length of each side is 8 m. A cow is
tied to a post with a 4 m rope. The post is located 2. The Nature Lovers Club of SMK Setia organised
2 m from the edge of the field as shown in the SPM a competition to design a logo for the club. The
diagram. 2018 diagram below shows a circular logo designed by
HOTS Dahlan. A square is inscribed in the logo.
Form 5
Analysing
8 m
D C A
A
A
8 m A
E 4 m Given that the area of the region marked A is
2 m (32 – 8π) cm . Calculate
2
A B (a) the radius, in cm, of the logo,
2 m
(b) the perimeter, in cm , of the region marked A.
2
[Use π = 3.142]

238

Additional Mathematics SPM Chapter 1 Circular Measure
3. The diagram below shows a sector AOC with 6. The diagram below shows a circle with centre O
SPM centre O. and radius of 5 cm. The perimeter of circle bounded
2019
A by semicircles APB, BQC, CRD, DSE and ETA with
diameters AB, BC, CD, DE and EA respectively.
O
7.5 cm S
D
E R
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C B
P T O
C
A
Given that ∠AOC is 150° and the arc length BC is B Q
4.83 cm. Calculate P
(a) ∠AOB, in radian, Calculate
2
(b) the area, in cm , of the shaded region. (a) ∠AOB, in radian, correct to three significant
[Use π = 3.142]
figures,
(b) the perimeter, in cm, of the shaded region,
4. The diagram below shows a piece of paper that (c) the area, in cm , of the shaded region.
2
has a shape of equilateral triangle with the side [Use π = 3.142]
length of 10 cm. Teacher Bala gives the papers to
the pupils to make cones. HOTS 7. The diagram below shows the chain drive of
Creating 1
a bicycle. The bicycle chain drive consists of Arial 8.5pt = 2
sprocket with radius 12 cm and hub with radius
3.5 cm located on the rear wheels of the bicycle.

pedal
crank A wheel
arm B
Devyn wants to use the paper as maximum as F C hub
possible to make a cone. Calculate sprocket D
3
(a) the maximum volume, in cm , of the cone made E
by Devyn, chain
(b) the area, in cm , of the paper unused by Devyn.
2
[Use π = 3.142] (a) Given that the radius of the rear wheel of the
bicycle is 35 cm, how far, in cm, can the bicycle
move if the pedals of the sprocket is rotated at
5. The diagram below shows two circles with centres 80°?
M and N overlap with each other. Both circles have (b) If the lengths of both chains AB and DE are
the same radii, which is π cm.
32 cm respectively, calculate the area of the
region ABCDEF, in cm , of the system of the Form 5
2
P
bicycle chain drive.
[Use π = 3.142]
R M N S
8. The following diagram shows the cross section of
Q a tunnel with length of 1.8 km and radius of 12 m
built to reduce the rate of traffic congestion and
(a) Calculate the perimeter of the whole diagram in
terms of π. also solve the problems of flash flood in major
(b) Show that the area of the shaded region is cities. The segment PQ acts as a trench with the
concrete road over it.
3
2
2
1 2π + 3π √3 2 cm .
3


239

Additional Mathematics SPM Chapter 1 Circular Measure
10. The diagram below shows the frame of a wau
helang with the symmetrical axis on the straight
line QV. PXR and PYR are sectors of circle with
radius of 50 cm with centres Q and T respectively.
P Q
TUVW is a sector of circle of radius 20 cm with
15 m centre T. PTW and RTU are straight lines.
Q
Calculate
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2
(a) the cross-sectional area, in m , of the tunnel X
above the highway, P R
(b) the maximum volume, in m , of the water that
3
can be stored in the trench during flash flood. Y
[Use π = 3.142] T
U W
9. The diagram below shows the cross-section of a V
ball with centre O floating on the surface of water.
Points A and B are on the water surface and the Given ∠PQR and ∠PTR are 115°, calculate
lowest point, C is 5 cm below the water surface. (a) the perimeter, in cm, of the shaded region,
(b) the area, in cm , of the shaded region.
2
[Use π = 3.142]
O 11. The diagram below shows two slices of chocolate
A B cake with different size and price.
5 cm HOTS
C Evaluating
30
Given the volume of the spherical ball is 2 304π
cm , calculate 8 cm 5 cm
3
(a) the radius, in cm, of the ball, 12 cm 12 cm 8 cm
(b) ∠AOB, in radian, Slice of cake A Slice of cake B
(c) the cross-sectional area, in cm , of the ball RM12 per slice RM10 per slice
2
above the water surface.
[Use π = 3.142] Ethyn could not choose the slice of cake that
is more reasonably priced. You as a smart
consumer can help Ethyn buy a slice of cake at
a more reasonable price. Justify your reason
mathematically.












Form 5


















240

Additional Mathematics SPM Answers
ANSWERS






Form 4

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Chapter (b) y
1 Functions
5
Try This! 1.1 Range: 0 ≤ f(x) ≤ 5
2
1. (a) 3 (b) 3 (c) 7 1
2. (a) Function. Each object has only one image, even 0 x
though the element 6 does not have an object. –3 –2 3
(b) Not a function. Object 2 has three images, which are
2, 4, and 6. (c) y
(c) Not a function. Element s does not have an image.
3. (a) This graph is not a graph of y as a function of x. There
are vertical lines which intersect the graph at two 5 Range: 0 ≤ f(x) ≤ 5
different points.
(b) This graph is a graph of y as a function of x. Each 3
vertical line intersects the graph at one point only.
(c) This graph is a graph of y as a function of x. Each
vertical line intersects the graph at one point only.
(d) This graph is not a graph of y as a function of x. All x
vertical lines intersect the graph at two different points, 0 5 8
except at the y-axis where the vertical line intersects
at one point only.
(d)
3
4. (a) (i) f : x → x or f(x) = x 3 y
(ii) h : x → x + 1 or h(x) = x + 1
(b) A : r → πr or A(r) = πr 7
2
2
2
(c) (i) f : x → 2x + 3x – 1 or f(x) = 2x + 3x – 1
2
(ii) g : x → sin x or g(x) = sin x 5
1
2
5. (a) 0 (b) –3 (c) — (d) – —
4 3
3
6. (a) False (b) True (c) True
7. (a) Domain = {x, y, z}, range = {p, q}
(b) Domain = {–2, –1, 1, 2}, range = {1, 4}
(c) Domain = {1, 2, 3, 4, 5}, range = {12, 24, 36, 48, 60} –4 3 0 2 x
(d) Domain is x ∈ , range is y ∈ . – – 2
(e) Domain is x ∈ , range is y > 3.
(f) Domain is x ∈ , x ≠ –5, range is y ∈ , y ≠ 0.
(g) Domain is x ∈ , range is y > 0. 11. (a) –1 (b) 5 – t (c) 3 – 2t
(h) Domain is x . 0, range is L . 0. 12. (a) 1 (b) 9x + 1 (c) 6z – 2
8. (a) Domain 0 < x < 8, range 3 < y < 15. 13. (a) 2 (b) 4.5 (c) 5 – 2x
(b) Domain –6 < x < 1, range 0 < y < 9.
14. (a) 10 (b) –5, 3 (c) 5
9. (a) When the domain is , all values of y are possible. 15. 5
The range is y ∈ . 3 3
(b) When x is limited to negative values, all values of y will 16. (a) 7, 5 (b) – —, 3 (c) 1, —
2
5
be at least 2. The range is y > 2. 17. (a) 3
(c) The range is {2, 3, 4, 5, 6}. (b) 1, 6
10. (a) y 18. (a) h = 3, k = 1
(b) 0 < f(x) < 9
19. (a) RM77 000
Range: 0 ≤ f(x) ≤ 2 (b) RM10 500
(c) 3
x
–2 0 2 (d) V(n) is a function. Each input n will give rise to one
and only one output value of V(n).
399

Additional Mathematics SPM Answers

2
Try This! 1.2 16. (a) f (x) = x 2 2 (b) f (x) = x 2 30
30
f (x) = x (–2) 3 f (x) = x (–2) 31
31
3
1. (a) 4 2 4
h g f (x) = x
17. (a) gf(x)
x gh(x) (b) h[gf(x)] = 3 600 + 0.06x, x . 15 000
h(x) (c) RM5 280
Try This! 1.3
gh 1
2 Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
1. (a) –3 (b) p = 10
2
(b) 2. (a) 0 (b) 2 (c) 0 (d) –2
g h
3. (a) f has an inverse function. f is a one-to-one function.
(b) g does not have an inverse function. The inverse of g
x hg(x) maps two elements in the codomain to one element a
g(x)
in the domain.
(c) h has an inverse function. h is a one-to-one function.
4. (a) The graph of function f has an inverse function. f is a
hg one-to-one function.
(b) The graph of function g does not have an inverse
(c) g g function. g is not a one-to-one function.
(c) The graph of function h has an inverse function. h is a
one-to-one function.
x gg(x)
g(x) 5. (a) The graph of function f has an inverse function. f is a
one-to-one function.
(b) The graph of function g does not have an inverse
function. g is not a one-to-one function.
gg
2
6. f [g(x)] = f 1 5 x + 2 , x ≠ 0
(d) 5
h h =
5
x + 2 – 2
x h (x)
2
h(x) = 5
5
x
= x, x ≠ 0
h 2
and g[f(x)] = g 1 5 2 , x ≠ 2
2. (a) 3x + 2 (b) 12x + 5 x – 2
3. (a) (i) 6x – 1 (ii) 18x – 12x + 2 = 5 + 2
2
2
(iii) 9x – 4 (iv) 8x 4 5
1 1 x – 2
(b) (i) , x ≠ –5 (ii) + 5, x ≠ 0
x + 5 x = x – 2 + 2
(iii) x, x ≠ 0 (iv) x + 10 = x, x ≠ 2
2
4. (a) (i) x + 1 (ii)  Since fg(x) = x, where x is in the domain of g, and
x + 1
(b) (i) |x – 6| (ii) |x| – 6 gf(x) = x, where x is in the domain of f, therefore g is an
inverse function of f.
2
5. (a) p = 10, k = 35 (b) gf(x) = x – 8x + 15
2
6. (a) 8 (b) 12 7. f [h(x)] = f 1 1 x + 2 , x ≠ 0
7. 61 1
=
8. 2x + 3 1 + 2 – 2
x
9. 2x + 3x + 4
2
1
10. 2x – 3 = = x, x ≠ 0
2
1
11. x + 7 x
2
12. x + 1 1
h[f(x)] = h 1 2 , x ≠ 2
13. x + x + 1 x – 2
1 1
4
2
3
14. (a) f (x) = x, f (x) = – , x ≠ 0, f (x) = x = + 2
x 1
1 x – 2
(b) f (x) = x, f (x) = – , x ≠ 0
8
25
x = x – 2 + 2 = x, x ≠ 2
1
15. (a) – , x ≠ 0 (b) x Since fh(x) = x, where x is in the domain of h, and
x hf(x) = x, where x is in the domain of f, therefore h(x) is
1 + x
(c) , x ≠ 1 the inverse of f(x).
1 – x
400

Extent : 304pg (13.74 mm) = 0.72mm (16pg70gsm) + 11.52 mm (288pg 60gsm) + 1.5mm Format 190mm X 260mm Extent : 456pg (19.70 mm) (60gsm-2C) Status CRC Date 15/11
Fomula:
16 pg (70gsm) = 16 pgs X 0.09 ÷2 = 0.72 mm PELANGI BESTSELLER
288 pg (60gsm) = 288 pgs X 0.08 ÷2 = 11.52 mm CC038332



SPM




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FORM FOCUS
ADDITIONAL 4∙5

MATHEMATICS KSSM SPM





FOCUS SPM KSSM Form 4 • 5 – a complete ADDITIONAL
REVISI and precise series of reference books with special
REVISION
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üKeywords üSPM Tips features to enhance students’ learning as a whole.
üConcept Maps üRemember! This series covers the latest Kurikulum Standard MATHEMATICS SPM
Sekolah Menengah (KSSM) and integrates
Sijil Pelajaran Malaysia (SPM) requirements.
REINFORCEMENT A great resource for every student indeed! FORM
REINFORCEMENT
& ASSESSMENT ADDITIONAL MATHEMATICS 4∙5
& ASSESSMENT
üTry This! üSPM Model Paper REVISION
üSPM Practices üComplete Answers
KSSM
REINFORCEMENT

EXTRA FEATURES
EXTRA FEATURES ASSESSMENT
üDaily Applications üSPM Highlights EXTRA
üHOTS Questions üQR Codes
üCalculator Corner üCloned SPM Questions

TITLES IN THIS SERIES
• Bahasa Melayu • Matematik • Mathematics FORM 4•
• Karangan • Matematik Tambahan • Additional Mathematics
• English • Sains • Science
• Bahasa Cina • Biologi • Biology
Purchase • Sejarah • Fizik • Physics 5
eBook here! • Pendidikan Islam • Kimia • Chemistry
• Pendidikan Seni Visual • Prinsip Perakaunan
KSSM






Dual L anguage
P rogramme
W.M: RM36.95 / E.M: RM37.95
W.M: RM??.?? / E.M: RM??.??
Ng Seng How
CC038332
ISBN: 978-967-2720-66-9 Ooi Soo Huat
Moy Wah Goon NEW SPM ASSESSMENT
Yong Kuan Yeoh FORMAT
PELANGI Moh Sin Yee 2021