1202 Question Bank Additional Mathematics Form 5

©PAN ASIA PUBLICATIONS

Contents







iii – viii
Must Know


Chapter 1 Circular Measure 1 – 11
NOTES 1
Paper 1 1
Paper 2 8

Chapter 2 Differentiation 12 – 21
NOTES 12
Paper 1 13
Paper 2 18

Chapter 3 Integration 22 – 33
NOTES 22
Paper 1 24
Paper 2 30


Chapter 4 Permutation and Combination 34 – 42
NOTES 34
Paper 1 35
Paper 2 41

Chapter 5 Probability Distribution 43 – 54
43
NOTES
©PAN ASIA PUBLICATIONS
Paper 1 44
Paper 2 51

Chapter 6 Trigonometric Functions 55 – 65
NOTES 55
Paper 1 58
Paper 2 63

Chapter 7 Linear Programming 66 – 72
NOTES 66
Paper 1 67
Paper 2 69


Chapter 8 Kinematics of Linear Motion 73 – 81
NOTES 73
Paper 1 74
Paper 2 78


SPM Assessment 82 – 97



Answers 98 – 136








ii




00A_1202 QB AMath F5.indd 2 24/11/2021 4:28 PM

MUST


KNOW Important Facts







Relationships between Angle in Degrees, Angle in Limits and Its Relation to Differentiation
Radians, Arc Length and Area of Sector
If y = f (x), then

dy
δy
f (x + δx) – f (x)
A lim ——————— = lim —– = —– = f ʹ(x)
©PAN ASIA PUBLICATIONS
δx → 0 δx δx → 0 δx dx
r
s
θ
O where δx is a small change in x.
B

θ ° θ rad Arc length, s Area of sector
—— = –—— = –——————–––— = –————–—–—–
360° 2π 2πr (Circumference) πr (Area of circle)
2

Important Facts (Chapter 1) 1 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 7 @ Pan Asia Publications Sdn. Bhd.


Arc Length AB, Length of Chord AB, Area of Sector Stationary Point
AOB and Area of the Shaded Segment
dy
A point P(x, y) is a stationary point if —– = 0.
• Arc length AB, s = rθ A dx
• Length of chord AB: r The stationary point P(x, y) is
2
2
2
2
✤ AB = r + r – 2r cos θ ° (Cosine rule) O θ s d y
2
r
AB
✤ ——— = ———–———— • a maximum point if —–– , 0.
dx
2
180° – θ°
sin θ° sin ————– 2 B d y
1
2
2
dx
2
(Sine rule) • a minimum point if —–– . 0.
1 d y
2
2
• Area of sector AOB = —r θ • a point of inflection if —–– = 0.
2 dx 2
• Area of the shaded segment
= Area of sector AOB – Area of triangle AOB
1 1
2
2
= — r θ – — r sin θ
2 2
Important Facts (Chapter 1) 3 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 9 @ Pan Asia Publications Sdn. Bhd.
Techniques of Differentiation Small Changes and Approximation of
Certain Quantities
dy
• If y = ax , then —– = anx n – 1
n
dx dy • If y = f (x) and δx is a small change in x, then
• If y = a, where a is a constant, then —– = 0
dx
dy
dy δy = —– × δx
• If y = f (x) + g(x), then —– = f ʹ(x) + gʹ(x) dx
dx
dy
dy
du
• If y = g(u), where u = h(x), then —– = —– × —–
dx du dx • If y = f (x) and x = g(t), thus the rate of change of y is:
• If y = uv, where u = f (x) and v = g(x), then
dy
dy
dx
dy dv du —– = —– × —–
—– = u—– + v—– dt dx dt
dx dx dx
u
• If y = —, where u = f (x) and v = g(x), then
v
du dv • If x changes from x to x + δx, then:
v—– – u—– δx
dy dx dx ✤ The percentage change in x = —– × 100%
x
—– = —————–
dx v 2 δy
✤ The percentage change in y = —– × 100%
y
Important Facts (Chapter 2) 5 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 11 @ Pan Asia Publications Sdn. Bhd.
00B_1202 QB AMath F5.indd 3 02/12/2021 8:24 PM

MUST


KNOW Common Mistakes







Quotient Rule Converting Radians into Degrees and Vice Versa
x + 1 d x + 1 • Converting radians into degrees
2
1
Given y = —–—–, find —– —–—– .
2x – 1 dx 2x – 1 For example: 1.35 rad
Correct Wrong Correct Wrong
©PAN ASIA PUBLICATIONS
du dv dv du 180° π
v—– – u—– u—– – v—– 1.35 × ——– 1.35 × ——–
dy dx dx dy dx dx π 180°
Use —– = —————– • Use —– = —————–
dx v 2 dx v 2
where u = x + 1 and v = 2x – 1. where u = x + 1 and v = 2x – 1. • Converting degrees into radians
du dv For example: 46°
v—– + u—–
dy dx dx Correct Wrong
• Use —– = —————–
dx v 2
180°
π
where u = x + 1 and v = 2x – 1. 46° × ——– 46° × ——–
180°
π
Common Mistakes (Chapter 2) 8 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 1) 2 @ Pan Asia Publications Sdn. Bhd.

Second Derivative Method Area of ∆AOB
One of the test to determine whether a stationary point is a maximum O r A
point or a minimum point. 1.45 rad
Correct Wrong B
Given a stationary point (a, b), Given a stationary point (a, b), Correct Wrong
then it is a maximum point if then it is a maximum point if Must convert an angle in Angle in radians is not
d y d y
2
2
—–– , 0. —–– . 0. radians into degrees first. converted into degrees.
dx 2 dx 2 180° Area of ΔAOB
Given a stationary point (a, b), Given a stationary point (a, b), 1.45 rad = 1.45 × ——– 1
π
2
then it is a minimum point if then it is a minimum point if = 83.08° = —r sin θ
2
d y d y Area of ΔAOB 1
2
2
—–– . 0. —–– , 0. 1 = —r sin (1.45)
2
dx 2 dx 2 = — r sin θ 2
2
2 1
2
1
= — r sin 83.08° = —r (0.0253)
2
2
2 = 0.0127r 2
= 0.4964r 2
Common Mistakes (Chapter 2) 10 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 1) 4 @ Pan Asia Publications Sdn. Bhd.
Indefinite Integral Chain Rule
2
∫ ———— dx = —–[3(2x – x) ] =
d
4
4
2
dx
(3x + 1)
Correct Wrong Correct Wrong
d 2 4 d 2 4
4
4
—–[3(2x – x) ]
2
∫ ———— dx ∫ ———— dx —–[3(2x – x) ] d 2 = 3(4)(2x – x) 4 – 1
2
dx
dx
(3x + 1)
(3x + 1)
2
4 – 1
2
dx
∫
∫
2
= 4(3x + 1) dx = 4(3x + 1) dx = 3(4)(2x – x) —–(2x – x) = 12(2x – x) 3
–2
–2
= 12(2x – x) (4x – 1)
d
2
3
—–(3x + 1) = 3
2
4(3x + 1) –2 + 1 dx 4(3x + 1) –2 + 1 = 12(4x – 1)(2x – x) 3 or
= —————– + c = —————– + c
(–2 + 1)(3) –2 + 1 d d
4
2
4(3x + 1) –1 4(3x + 1) –1 —–(3x + 1) = 3 —–[3(2x – x) ]
dx
dx
= ————– + c = ————– + c is missing
d
–3 –1 = —–[(6x – 3x) ]
2
4
4 4 dx
= – ——–—– + c = – —––— + c
3(3x + 1) 3x + 1
Common Mistakes (Chapter 3) 12 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 2) 6 @ Pan Asia Publications Sdn. Bhd.
00B_1202 QB AMath F5.indd 4 02/12/2021 8:24 PM

1 Circular Measure
Chapter





NOTES


1.1 Radian (a) —–— = —–— = —–————–——–
θ rad
Arc length, s
θ °
2πr (Circumference)
360°
©PAN ASIA PUBLICATIONS
2π
1. One radian is the measure of an angle subtended (b) Arc length AB, s = rθ
at the centre O of a circle where the arc length, s (c) Length of chord AB:
is the same as the radius of the circle, r, that is, (i) AB = r + r – 2r cos θ ° (Cosine rule)
2
2
2
2
s = r = 1 rad. r
AB
(ii) ——– = ————––— (Sine rule)
180° – θ °

A sin θ ° sin ———–– 2
r 2
s
1 rad
O 1.3 Area of Sector of a Circle
B
1. Given a circle with centre O and a radius of r units
where ∠AOB = θ rad (or θ °), then
2. 360° = 2π rad A
r
1.2 Arc Length of a Circle
O θ
1. Given a circle with centre O and a radius of r units
where ∠AOB = θ rad (or θ °) and the arc length AB is B
s units, then
A θ ° θ rad Area of the sector
(a) —–— = —–— = —–————–—–—––
2
r 360° 2π πr (Area of the circle)
s 1 2
θ (b) Area of sector AOB = —r θ
O 2
(c) Area of the shaded segment
B = Area of the sector AOB – Area of triangle AOB
1 1
= —r θ – —r sin θ °
2
2
2 2
PAPER 1


Section A

1. The perimeter of a sector AOB of a circle with centre Answer:
SPM O and an arc length of 5.4 cm is 23.4 cm. Find (a)
CLONE
(a) the radius of the circle, [3 marks]
(b) the angle AOB subtended at the centre of the
circle. [2 marks] (b)




PB Question 1: 1
SOS TIP PBPB (a) Perimeter of a sector = r + r + Arc length 1 SOS TIP

(b) Use s = rθ







01_1202 QB AMath F5.indd 1 02/12/2021 8:44 PM

2
2. The area of a sector KOL with centre O and a radius Given that OA : OP = 1 : 3, ∠POQ = —π rad and
3
of 11 cm is 160 cm . OA = 15 cm, find
2
(a) Find ∠KOL, in radians. [2 marks] (a) the perimeter of the area covered by the paper,
(b) If the sector KOL is folded to form a cone, find [3 marks]
the base radius of the cone. [3 marks] (b) the area of the paper used. [2 marks]
Answer: Answer:
(a) (a)




©PAN ASIA PUBLICATIONS


(b)
(b)






3. Given AOB is a sector of a circle with centre O and a
radius of r cm where ∠AOB = θ rad.
(a) Show that the length of the chord AB is 5. The diagram shows a sector AOB with centre O and a
θ
2r sin —. [2 marks] SPM radius of 15 cm.
2 CLONE
(b) Given r = 5 cm and ∠AOB = 1.2 rad, find the A
difference in length between the arc length AB
and the chord AB. [3 marks] B
15 cm
Answer: C
(a) θ

O
Given that C divides the line OB in the ratio 3 : 2 and
the length of chord AB is 10 cm. Find
(a) the angle θ, in radians, [2 marks]
(b) (b) the perimeter of the shaded region. [3 marks]
Answer:
(a)





4. The diagram shows a paper fan consisting of two
sectors, POQ and AOB. The shaded region is covered
by paper.
P (b)



A

O
B
Q

2 Question 2:
SOS TIP (b) Find the arc length of the sector. When the sector is folded, the arc length of the sector is the circumference of the base of the cone.
Question 5:
2
2
2
Use the cosine rule, a = b + c – 2bc cos θ
22


01_1202 QB AMath F5.indd 2 02/12/2021 8:44 PM

13. The diagram shows a rhombus inscribed in a sector 14. The diagram shows a circle with centre O. PT and
SPM AOC with centre O and a radius of r. SPM QT are tangents to the circle at points P and Q
CLONE CLONE
respectively.
O C
β rad
r P
r cm T
A
B θ
Given that ∠AOC = β rad and the area of the sector is O Q
20 cm , express each of the following in terms of r.
2
©PAN ASIA PUBLICATIONS
(a) The angle β. [3 marks]
(b) The perimeter, in cm, of the shaded region.
[2 marks] Given that the minor arc length PQ is 5 cm and
6
Answer: OT = — cm,
(a) θ
(a) express the radius, r of the circle in terms of θ,
[1 mark]
(b) find the area of the shaded region. [4 marks]
Answer:
(a)



(b)


(b)









Section B

15. The diagram shows a circular cross-section of a Answer:
container with centre O resting on two supports, each (a)
of height 25 cm. HOTS Applying




O
32 cm
A B

25 cm (b)
10 cm
The shortest distance of the container to the horizontal
surface is 10 cm. Given that the radius of the container
is 32 cm. Find
(a) the angle AOB, in radians, [4 marks]
(b) the area between the supports and the container.
[4 marks]


4 Question 14: 5
SOS TIP 44 Tangent PT = Tangent QT and ∠TPO = 90° 5 SOS TIP

Question 15:
(b) Area of the segment = Area of the sector AOB – Area of the triangle AOB






01_1202 QB AMath F5.indd 5 02/12/2021 8:44 PM

16. Two connecting gears are rotating simultaneoulsy. B
The smaller gear has a radius of 5 cm whereas the
larger gear’s radius is 9 cm.
(a) What is the angle subtended by the larger gear
if the smaller gear has made one complete A C
rotation? [4 marks]
(b) How many rotations will the smaller gear make if 2 θ rad r m
the larger gear makes one complete rotation?
[4 marks] P
Answer: Given that ∠APC = 2θ rad, express the width AC of the
©PAN ASIA PUBLICATIONS
(a) tunnel and the area of the wall filled up by the cement,
in terms of r and θ respectively. HOTS Analysing
[8 marks]
Answer:








(b)





19. The length of a chord PQ of a circle with centre O and
a radius of 3 cm is 4.4 cm. Calculate
(a) the angle subtended at the centre of the circle, in
radians, [4 marks]
(b) the area of the segment enclosed by the chord PQ
and the arc PQ. [4 marks]
17. A sector of a circle with angle θ and a radius of r cm
has an area of 5 cm and its perimeter is 9 cm. Find Answer:
2
the possible values of r and θ. (a)
[8 marks]
Answer:











(b)






18. The diagram shows the cross-section of a tunnel with
its width AC is the diameter of a semicircle ABC. The
shaded region is the wall of the tunnel and is fill up
with cement. The cross-section of the tunnel is in the
form of a sector with centre P and a radius of r m .
6 Question 17:
SOS TIP Relate the arc length to the area of the sector. 66










01_1202 QB AMath F5.indd 6 02/12/2021 8:44 PM

PAPER 2

Section A


1. The diagram shows a circle with centre O and a radius 4. The diagram shows a trapezium ABCD where AB is
SPM of 25 cm. parallel to DC. HOTS Applying
CLONE
A 5 cm B
A
α 6 cm
25 cm E
©PAN ASIA PUBLICATIONS
θ
O D π π C
B — rad 12 cm — rad
6 3
π
Given AB = 5 cm, DC = 12 cm, ∠ADC = — rad,
6
π
A sector with angle θ = 1.2 rad at the centre is ∠BCD = — rad and E is the midpoint of BC.
3
removed. Then, the end A is joined to B to form a Calculate
cone. Calculate (a) the perimeter of the shaded region, [4 marks]
(a) the base radius, in cm, of the cone, [3 marks] (b) the area of the shaded region. [4 marks]
(b) the angle α, in degrees. [4 marks]
5. The diagram shows a circle with centre O and a radius
of 6 cm and a rectangle ABCO with an area of 48 cm .
2
2. Point A is a fixed point on the circumference of a
circle with centre O and a radius of 10 cm. Point A
P moves along the circumference of the circle at a B
speed of 3 cm per second. Given the angle AOP is 6 cm E
θ rad, find
(a) the rate of change of θ, in radians per second, O D C
[3 marks]
(b) the rate of change of the area of sector AOP.
[3 marks]
Calculate
(a) ∠AOB, in radians, [3 marks]
3. The diagram shows one of the pattern of a mural (b) the area of the sector EOD. [4 marks]
SPM drawn by a pupil on the wall of a school canteen.
CLONE
C 6. A ball is floating on the surface of water such that the
highest point of the ball from the surface of the water
D is half of the radius, r cm, of the ball. HOTS Applying


O P Q
r cm
A B
10 cm O
Given AB = 10 cm is a chord of the major sector ACB
with centre O and a radius of 15 cm. AB is also the
diameter of a semicircle ADB. A small circle with
centre O inscribed in the semicircle. Calculate (a) Find the length of the chord PQ, in terms of r.
(a) the arc length ACB, [4 marks] [3 marks]
(b) the area of the shaded region. [4 marks] (b) Find the area, in cm , of the cross-section of the
2
ball below the water if r = 25 cm. [4 marks]


8 Question 3:
SOS TIP The diameter of the small circle is the same as the radius of the semicircle and AB is the tangent to the small circle.

Question 6:
(b) Cross-sectional area in the water = Cross-sectional area of a circle – Area of segment above the surface of water
88





01_1202 QB AMath F5.indd 8 02/12/2021 8:44 PM

Section B
13. Ani wants to make a cone-shaped cap for her 16. The diagram shows a rhombus ABCD with sides of
children’s party. The height of the cone is 25 cm and x cm and ∠A = θ rad. HOTS Analysing
the base radius of the cone is 8.5 cm. She takes a
piece of cardboard measuring 27 cm × 35 cm to cut B
out the net of the cone which is a sector of a circle. C
HOTS Analysing
(a) What is the angle, in radians, subtended at the A θ rad x cm
centre of the circle by the arc length of the sector
of the net? [4 marks] D
©PAN ASIA PUBLICATIONS
x
(b) Based on the calculation in (a), determine Four arcs each with radius — cm are drawn with
whether the cardboard is big enough to make a 3
cone-shaped cap. [6 marks] centres A, B, C and D respectively. Given the shaded
area is half of the area of the rhombus,
2
(a) show that sin θ = —π, [5 marks]
14. The diagram shows a semicircle PORQS with centre 9
O and a radius of 9 cm. HOTS Applying (b) find two possible values of θ. [5 marks]

9 cm O M R
P Q
θ 17. The diagram shows two sectors with centre O.
L
2 cm
M
S
RPS is an inscribed sector in the semicircle with 1.5 rad
centre P. A perpendicular line from S to PQ divides O
the radius of the semicircle into half at M. Find N P
(a) the angle θ, in radians, [3 marks]
(b) the arc length RS, [3 marks]
(c) the area of the shaded region. [4 marks] Given that ∠LOP = 1.5 rad, LM = NP = 2 cm and the
area of the sector LOP is 20.75 cm , find
2
(a) the radius OM, [4 marks]
15. The diagram shows the sector AOB with centre O, a (b) the arc length LP, [2 marks]
SPM π (c) the area of the shaded region. [4 marks]
CLONE radius OB = 8 cm and ∠AOB = — rad.
3
C
18. The diagram shows a sector of a circle with centre A.
A B HOTS Analysing
D E y
P
8 cm
C
π
— rad
3 B
3y + x = 9
O
OC is the bisector of ∠AOB and P is the midpoint of O A(4, 0) D x
OC. An arc DCE of a circle is drawn with centre P to
meet OA and OB at D and E respectively. Find The equation of BD is 3y + x = 9. Find
(a) the angle OPD, in radians, [3 marks] (a) the radius of the sector ABCD, [4 marks]
(b) the area of the shaded region. [7 marks] (b) the angle BAD, in radians, [2 marks]
(c) the area of the shaded region. [4 marks]


10 Question 13:
SOS TIP The radius of the sector of the net is the inclined side of the cone. Sketch the net of the cone as well as the cone before calculating the length and width of the

cardboard.
Question 16:
Area of ABCD = Length of AD × Height of B to AD
1010




01_1202 QB AMath F5.indd 10 02/12/2021 8:44 PM

SPM Assessment






PAPER 1

Time: 2 hours
©PAN ASIA PUBLICATIONS
Section A
[64 marks]
Answer all questions.

1. (a) Diagram 1 on the answer space shows a part of 2. (a) Given that the range of y = f (x) + 1 is
the graph of a function y = f (x) for the domain –2 < y < 3, find the range of f (x). [1 mark]
0 < x < 3. On the same axes, sketch the (b) Given that f : x → x + 2 and gf : x → x + 4x + 2,
2
corresponding part of the graph y = f (x) and find
–1
1
state its domain. [2 marks] (i) g(2),
(b) The function F maps (x, y) onto (x – y, x + 2y) and (ii) the values of x if fg(x) = 9. [4 marks]
A is the point (2, 3). F maps A onto B and B onto Answer:
C. Find the coordinates of points B and C. (a)
[2 marks]
(c) If f : x → 3 – 4x, find f (–3). [1 mark]
–1
Answer:
(a) y


3

2
y = f(x)
1

x (b) (i)
0 1 2 3 4
Diagram 1


(b)










(ii)


(c)












82





09_1202 QB AMath F5.indd 82 04/12/2021 9:25 AM

Section B
[16 marks]
Answer any two questions from this section.

13. (a) A rectangular container without a lid is made up of thin aluminium sheet. The sides of the base are 2x cm and
3x cm and the height is h cm. If the total surface area is 200 cm ,
2
20
3x
(i) show that h = —– – —–, [2 marks]
x
5
(ii) find the dimensions of the container such that the volume is maximum,
[2 marks]
©PAN ASIA PUBLICATIONS
(iii) hence, find the maximum volume of the container.
[1 mark]
(b) If water drips into the container in (a) at a constant rate of 21 cm s , find the rate of change of the height of water
–1
3
in the container when h = 1 cm.
[3 marks]
Answer:
(a) (i)













(ii)













(iii)














(b)














87





09_1202 QB AMath F5.indd 87 04/12/2021 9:25 AM

PAPER 2

Time: 2 hours 30 minutes

Section A
[50 marks]
Answer all questions.

1. The sum of the digits of three-digit number is 16. The unit digit is 2 more than the sum of the other two digits. The
tens digit is 5 more than the hundreds digit. What is the number?
©PAN ASIA PUBLICATIONS
[7 marks]



2. (a) Diagram 1 shows the net of an open box.


x cm x cm



(15 – 2x) cm





5 cm
Diagram 1
(i) Show that the volume of the box, in cm , is given by V = 75x – 10x .
3
2
[2 marks]
(ii) Hence, find the value of x, in cm, such that the volume of the box is maximum. State the maximum volume
of the box.
[3 marks]
(b) Find the possible range of values of x if the volume of the box is between 90 cm and 125 cm .
3
3
[2 marks]


3. In Diagram 2, A, B and C are the vertices of a right-angled triangle.

C
D α
F
H
x cm




A
G E B
Diagram 2
Given that ∠ACB = α and BC = x cm.
(a) Express BD and DE in terms of x and α.
[2 marks]
(b) Show that BD, DE and EF form the first three terms of a geometric progression and state the common ratio.
[3 marks]
(c) Find the length of HG in terms of x and α.
[1 mark]
(d) Given that x = 8 cm and α = 60°, find the sum to infinity of the geometric progression.
[2 marks]
90





09_1202 QB AMath F5.indd 90 04/12/2021 9:25 AM

Section B
[30 marks]
Answer any three questions from this section.

——–——.
8. (a) (i) Find lim 3 + 2x – x 2 [1 mark]
x → 3 x – 3
dy
(ii) Given that x = t + t and y = 2t + 1, where t . 0, find —– in terms of y and hence, find the approximate
2
dx
change in y if x decreases from 2 to 1.98 when t = 1. [3 marks]
2
(b) Diagram 4 shows a curve y = x(x – 3) intersects with the straight line y = 4x at O, A and B.
©PAN ASIA PUBLICATIONS
y
B


y = 4x

y = x(x – 3) 2


A


x
O
Diagram 4
(i) Find the coordinates of A and B. [2 marks]
(ii) Calculate the area of the shaded region. [4 marks]



→ → →
9. In Diagram 5, OP = 2x, OQ = 3y and QR = x – y. The lines PQ and OR intersect at X.
~
~
~
~
Q

X R


O

P
Diagram 5
→ → → →
Given PX = hPQ and OX = kOR,
→
(a) Express PX in terms of h, x and y. [2 marks]
~
→ ~
(b) Show that OX = 2(1 – h)x + 3hy. [2 marks]
~
~
(c) Find the values of h and k. [3 marks]
2
(d) If the area of the triangle QOX is 24 units , find the area of the triangle XOP. [3 marks]










93





09_1202 QB AMath F5.indd 93 04/12/2021 9:25 AM

Section C
[20 marks]
Answer any two questions from this section.

12. (a) In Diagram 7, ABC is a triangle where AB = 13 cm, AC = 8.5 cm and ∠ABC = 38°.
B


13 cm

©PAN ASIA PUBLICATIONS
A C
8.5 cm
Diagram 7
Calculate
(i) the angle BAC where ACB is an obtuse angle,
(ii) the length of BC.
[4 marks]
(b) A new triangle is formed such that AB, AC and the size of the angle ABC remain unchanged.
(i) Sketch the new triangle. [1 mark]
(ii) Calculate the area of the new triangle. [3 marks]
(iii) Hence, find the shortest distance from A to BC. [2 marks]


13. Table 2 shows the price indices, the changes in price index and the quantities required for the four ingredients, A, B,
C and D used to make a product.

Price index for the year 2019 based Changes in price index from the year
Ingredient Quantity (g)
on the year 2015 2019 to 2020
A 125 Increased by 20% 500

B 120 Decreased by 15% 200

C 90 No change 200

D 150 Increased by 10% 100

Table 2
(a) Calculate
(i) the price of 1 kg of ingredient A in the year 2015 if the price in the year 2019 is RM15,
(ii) the price index of ingredient B in the year 2020 based on the year 2015.
[4 marks]
(b) Calculate the composite index for the cost of the product in the year 2020 based on the year 2019.
[2 marks]
(c) If the cost of the product in the year 2019 is RM50, find the corresponding cost in the year 2020.
[2 marks]
(d) The composite index for the cost of making the product from the year 2020 to the year 2021 is expected to
increase at the same rate as the rate of increase from the year 2015 to the year 2019. Find the expected composite
index for the cost of making the product for the year 2021 based on the year 2015.
[2 marks]










95





09_1202 QB AMath F5.indd 95 04/12/2021 9:25 AM

©PAN ASIA PUBLICATIONS













Answers
























































98





10_1202 QB AMath F5.indd 98 10/01/2022 4:49 PM

CHAPTER 1 5. A (b) OL = 18 cos 41.44°
Height above PQ
Paper 1 B = 18 – 18 cos 41.44°
15 cm
Section A C θ = 4.51 cm 1
1. A 9 cm 9. 60 seconds = 33— revolutions
3
O 100 1 5
3
OC
9
60
r rθ (a) —— = — 1 second = —— × —– = — revolution
3
10
5
θ OB 5 3 (a) — × 2π = —–π radians per second
O B OC = — × 15 9 9
r 5
P = 2r + rθ = 23.4 = 9 cm (b) r = 20 cm
10
(a) 2r = 23.4 – 5.4 AB = 15 + 15 – 2(15) cos θ Speed = —–π × 20
2
2
2
2
15 + 15 – 10
2
2
2
r = 9 cm cos θ = ——————– 9 –1
(b) 9θ = 5.4 2(15) 2 = 69.81 cm s –1
θ = 0.6 rad θ = 38° 57ʹ = 0.698 m s
∴ ∠AOB = 0.6 rad θ = 0.68 rad 10.
2
2
2
(b) AC = 15 + 9 – 2(15)(9) cos 38° 57ʹ
2.
AC = 9.8 cm D C
Perimeter of the shaded region 5 50 cm
K = 6 + 15(0.68) + 9.8 A —π rad B
9
= 26 cm 24 cm
θ
11 cm O
L 6.
O (a) Area traversed by the wiper
1
(a) —(11) θ = 160 1 2 5 1 2 5
2
 2
 2
2 = —(74) —π – —(24) —π
θ = 2.64 rad O 2 9 2 9
∴ ∠KOL = 2.64 rad = 4 276.06 cm 2
(b) 12 cm (b) Perimeter of the area traversed by
C the wiper
5
5
 2
 2
A B = 24 —π + 2(50) + 74 —π
9
9
(a) Area of ΔAOB = 72
1 = 271.04 cm
—(12)AB = 72
2
r AB = 12 cm 11. (a) Perimeter of the shaded region
2
 2
π
5
∠AOB = 45° = — rad = 10 —π + 10 sin 72°
Circumference, KL = 11(2.64) 4 + (10 – 10 cos 72°)
= 29.04 cm (b) Area of minor sector AOC = 28.99 cm
1
O ©PAN ASIA PUBLICATIONS
π
2πr = 29.04
(b) Area of the shaded region
 4 2

= —(12) —
2
r = 4.62 cm 2 = —(10) —π
1
2 2
= 18π cm 2 2  2
5
3.
1
B – —(10 cos 72°)(10 sin 72°)
7. P 2
A 12 = 48.14 cm 2
r cm R 11 1
12.
10 2 T
θ rad
9 3
O O π
AB
θ
6
(a) sin — = —— 8 4 Q U –– rad
2 2r 7 6 5 r
θ
2r sin — = AB O
2 1 2
(b) ∠AOB = 1.2 rad = 68.75° ∠POQ = — × 360° = 120° (a) Area = 20 cm 1
3
1
2 π
2  2
2
Difference ∠ROP = — × 30° + 30° = 50° 20 = —r — – —r sin 30°
2
2
6
= Arc length AB – Length of the 3 2 π sin 30°
2 2

chord AB Total angle = 170° = 2.97 rad 20 = r —– – ———–
12
= 5(1.2) – 2(5) sin 34.38° 8. r = ——————–
20
2
= 0.353 cm π sin 30°
2
4. P —– – ———–
12
O r = 41.17 cm
41.44° (b) Length of the chord UT
18 cm
A = 2(41.17) sin 15°
15 cm L
R = 21.31 cm
2
— π rad B Perimeter of the segment
3 30 cm (a) Arc length = 100 cm, r = 18 cm
π
 2
Q = 21.31 + 41.17 —
(a) Perimeter rθ = 100 = 42.87 cm 6
100
2
2
 3 2
 3 2
= 15 —π + 2(30) + 45 —π θ = ——
18
= 185.66 cm θ = 5.56 rad
θ = 318.56°
(b) Area of the paper
1 2 2 1 2 2
 3 2
 3 2
= —(45) —π – —(15) —π
2 2
= 600π cm 2
99
10_1202 QB AMath F5.indd 99 10/01/2022 4:49 PM

13. O r C (b) Area of the segment (b) Area of the segment
1
1
1
1
2
2
2
2
β = —(32) (2.02) – —(32) sin 115.82° = —(3) (1.65) – —(3) sin 94.33°
2
2
2
2
r 2
= 573.35 cm 2 = 2.94 cm
β Area of rectangle AOB 20.
A = 25(32 sin 57.91°) B
B 5.6 cm Q
1 = 677.77 cm 2 4 cm
2
(a) 20 = —r β Area required = 677.77 – 573.35 A 4 cm
2
40
β = —– = 104.42 cm 2 P r θ
r 2 16. O
(b) Perimeter of the shaded region (a) rθ = 4
9 cm (r + 4)θ = 5.6
= 2r + rβ 5 cm 4 + 4θ = 5.6
©PAN ASIA PUBLICATIONS
40
 r 2
= 2r + r —– O A 4θ = 1.6
2

r 2
40
= 2r + —– cm θ = 0.4 rad
4
0.4
14. r = OP = —— = 10 cm
P (a) Circumference of the small gear (b) ∠POQ = 0.4 rad
= 2π(5) (c) Area of ABQP
r T = 10π cm 1 1
2
2
10π = 9θ = —(14) (0.4) – —(10) (0.4)
2
2
O 10
θ Q θ = —–π rad = 19.2 cm 2
— 9
2
(b) Circumference of the big gear 21. O 6 cm
= 2π(9) P Q
(a) rθ = 5 = 18π cm 6 cm 6 cm
5 Number of rotations 9 cm
r = — cm 18π S R
θ = ——
10π
6
(b) OT = — = 1.8 rotations T
θ
∠SOR
4.5
6
5
θ
cos — = — ÷ — 17. (a) sin ——— = —–
2 θ θ 2 6
5 θ θ r cm ∠SOR
= — × — ——— = 48.59°
θ 6 2
5
= — ∠SOR = 97.18°
6 ∠SOR = 1.7 rad
θ = 67.11° (b) Area of the segment STR
1
1
1
θ = 1.17 rad —r θ = 5 ................. 1 = —(6) (1.7) – —(6) sin 97.18°
2
2
2
2
Area of ∆POT 2r + rθ = 9 ................. 2 2 2 2
1
6
2  θ 2
= —r — sin 33.56° rθ = 9 – 2r .......... 3 = 12.74 cm
1 5 6 From 1, 22. M 6 cm O N
2  θ 2 θ 2
= — — — sin 33.56° 1 —– rad 60°
π
2
1 5 6 —r(9 – 2r) = 5 3 120°
2  1.172 1.172
= — —— —— sin 33.56° 9r – 2r = 10
2
2
= 6.06 cm 2 2r – 9r + 10 = 0 60°
(2r – 5)(r – 2) = 0
Area of the shaded region 5 P Q
1
2
= 2(6.06) – —r θ 1 5 2 r = — or r = 2 (a) Arc length PQ
2
2
1
2  2 2
5
1
2
2
2  1.172
2
= 2(6.06) – — —— — — θ = 5 or —(2) θ = 5 = π(6) – Arc length MP
– Arc length QN
8
5
π
π
= 1.44 cm 2 θ = — rad θ = — rad = 6π – 6 — – 6 —
 2  2
2
5
Section B 18. AC = 2r sin θ = 2π cm 3 3
15. Area (b) Perimeter of the shaded region
π 1 1 π π
 2  2
2
2
2
= —(r sin θ) – —r (2θ) – —r sin 2θ 4 = 12 + 6 — + 6 — + 2(6 cos 60°)
2 3 2 2 3 3
O 1 2 2 1
 2 2
2
32 cm = —r (π sin θ – 2θ + sin 2θ) = 12 + 4π + 12 —
19. = 18 + 4π
A B
D = 30.57 cm
25 cm
10 cm
O Paper 2
(a) OD = 32 + 10 – 25 θ 3 cm Section A
= 17 cm 1. (a) Circumference = 2πr
∠AOB P 4.4 cm Q
32 cos ——— = 17 (2π – 1.2)(25) = 2πr
2
(2π – 1.2)(25)
2.2
∠AOB (a) sin — = —— r = ——————
θ
——— = 57.91° 2π
2 2 3
θ
∠AOB = 115.82° — = 47° 10’ r = 2.02 cm
∠AOB = 2.02 rad 2
θ = 94.33°
θ = 1.65 rad
100
10_1202 QB AMath F5.indd 100 10/01/2022 4:49 PM

(b) Perimeter of the shaded region (b) Perimeter of the shaded region
π
 2
 2
π
= 6 — + 5 + 1.73 + 1.73 — = (2h cos θ – h)(2θ) + (2h cos θ)θ
α 6 3 + (h – (2h cosθ – h))
+ (6 – 1.73) = 4θh cos θ – 2θh + 2θh cos θ + 2h
25 cm = 15.95 cm – 2h cos θ
(b) Area of the shaded region = 6θh cos θ – 2θh + 2h – 2h cos θ
1 1
2 π
2  2
r = —(5 + 12)(3) – —(6) — = h(6θ cos θ – 2θ + 2 – 2 cos θ)
2
6
π
1
2 π
 2
2.02
6
α
sin — = ——– – —(1.73) — If θ = —,
2
2 25 3 Perimeter of the shaded region
3  2
 2
α = 14.51 cm 2 π π
— = 4° 38ʹ = h 6 — cos 30° – 2 — + 2
2 5. A 6 6
α = 9° 16ʹ B – 2 cos 30° 4
2. A 6 cm E = h(1.94)
= 1.94h cm
C
O D 8. M
θ rad
O N
P r 8 cm
10 cm 3 cm
(a) Area = 48 cm 2 O π
6(AB) = 48 —– rad P
6
(a) 10θ = 3 AB = 8 cm Q
3
8
θ = —– tan ∠AOB = —
10 6 π
 2
θ = 0.3 rad s –1 ∠AOB = 53.13° (a) r — = 3
6
18
(b) Area of sector AOP ∠AOB = 0.93 rad r = —– cm
1
= —r θ (b) ∠EOD = 90° – 53.13° π
2
π
 2
2 = 36.87° OM — = 8
= 50θ Area of the sector EOD 6 48
3
 10 2
36.87°
= 50 —– = ——— × π(6) 2 OM = —–
π
360°
= 15 cm s –1 = 11.58 cm 2 OM = 15.28 cm
2
18
48
5
π
3. sin θ = —– 6. NM = —– – —– = 9.55 cm
π
15
1
θ = 19° 28ʹ P —r cm Q (b) MQ = 2(OM sin 15°)
2


= 2(15.38) sin 15°
∠AOB = 360° – 2(19° 28ʹ) = 29.52 cm
= 321.07° r cm Perimeter of the shaded region
O
= 5.6 rad
6 ©PAN ASIA PUBLICATIONS
= 29.52 + 9.55 × 2 + 3

2
—π rad
C 3 = 51.62 cm
1
2
(c) Area of MOQ = —(15.28) sin 30°
2
D (a) Length of the chord PQ = 58.37 cm 2
1
r – —r
2
O = 2ABBBBBB 2 Area of the shaded region
15 cm 4 1 18 2 π
 π 2  2
ABBB
3
θ 2.5 cm = 2 —r 2 = 58.37 – — —– —
6
2
A B 4 = 49.78 cm 2
5 cm 5 cm
2AB 3
(a) Arc length ACB = 15(5.6) = ——r 9. A
2
= 84 cm = AB 3r cm
(b) Area of the segment (b) Area of the circle = πr 2 60° a
38° 56ʹ
= ———– × π(15) Area under water a
2
360° = πr – Area of the segment O B
2
1
– —(15) sin 38° 56’ 1 1 60° a
2
3
2 = πr – —r θ – —r sin 120° 4
2
2
2
= 5.75 cm 2 2 2 P
AB 3
1
1
2 2
3  2
 24
Area of the shaded region = πr – —r —π – —r —– 1
2
2
—a
π
= π(15) – 5.75 – —(5) + π(2.5) 2 2 3 2 2 (a) cos 30° = ——
2
2
2
2
2πr
AB 3
2
= 681.47 cm 2 = —–— + —–r 2 OP
3 4 AB 3 1
 2
4. A 5 cm B If r = 25 cm, OP —– = —a
2
2
a
Area under water = 1 579.63 cm 2 OP = —–
6 cm 3 cm E AB 3
π h 7. P
—– rad 1.73 cm OP = OA
D C Arc length AB with centre O
π
12 cm —– rad A
a
3 = —– —–
2π
3
(a) h = 6 sin 30° = 3 cm  2 2
AB 3
3
2πa
AB 3
sin 60° = —– O θ rad 2 θ rad = –—–– × –––
BC h cm T Q
BC = 3.46 cm (a) (i) OP = 2h cos θ 3AB 3 AB 3
2AB 3πa
BE = 1.73 cm (ii) TQ = 2h cos θ – h = —–—–
9
101
10_1202 QB AMath F5.indd 101 10/01/2022 4:49 PM

125
75
2 4
= —– + 75 – —–– – 0 (b) f : (x, y) → (x – y, x + 2y) (α + 1)(β + 1) = –1
3 2
αβ + (α + β) + 1 = –1

f : (2, 3) → (–1, 8)
1
= 50 m f : (–1, 8) → (–9, 15) αβ = –1 – 1 – —
∴ B(–1, 8), C(–9, 15) 2
5
25. (a) v = 2t(10 – 3t) (c) f (x) = 3 – 4x = – —
v = 20t – 6t 2 Let y = 3 – 4x = –3 1 1 2
dv
a = —– 6 = 4x If — and — are roots,
α
β
dt
3
1
1
α + β
a = 20 – 12t x = — — + — = ——––
2
For maximum velocity, 3 α β αβ
1
–1
a = 0 ∴ f (–3) = — —
2
2
20 – 12t = 0 = ——
5
12t = 20 2. (a) y = f (x) + 1 – —
2
–2 < y < 3
©PAN ASIA PUBLICATIONS
5
t = — –2 < f (x) + 1 < 3 = – —
1
3
5
5
–3 < f (x) < 2
When t = —, (b) f (x) = x + 2 —– = ——– = – —
1
2
1
3
5
5
5
 3 22
 3 2
2
v = 2 — 10 – 3 — gf (x) = x + 4x + 2 αβ – — 5
2
50
v = —– gf (x) = g(x + 2) Therefore, the equation is
Let
y = x + 2
3 1 2
2
Therefore, the maximum velocity of x = y – 2 2 x + —x – — = 0
5
5
50
2
the particle is —– m s . g(y) = (y – 2) + 4(y – 2) + 2 5x + x – 2 = 0
–1
2
3 = y – 4y + 4 + 4y – 8 + 2
∫
2
(b) s = v dt (i) = y – 2 2 5. (a) Midpoint AC = (3, 1)
g(x) = x – 2
y
∫
s = (20t – 6t ) dt g(2) = 2 – 2 = 2
2
2
s = 10t – 2t + c (ii) 2 fg(x) = 9 A(1, 4)
2
3
When t = 0, s = 0 f (x – 2) = 9 D(h, k)
2
10(0) – 2(0) + c = 0 x – 2 + 2 = 9
2
3
2
c = 0 x = 9 B(–3, 0) O x
Hence, s = 10t – 2t . x = ±3 C(5, –2)
2
3
When t = 2, s = 10(2) – 2(2) 3 3. (a) x + 3 = t(x + 1)
2
2
2
h – 3
s = 24 x – tx + 3 – t = 0 ——– = 3
2
2
When t = 3, s = 10(3) – 2(3) 3 b – 4ac , 0 2
2
2
3
s = 36 (–t) – 4(1)(3 – t) , 0 h = 9
2
3
k + 0
Total distance travelled t + 4t – 12 , 0 ——– = 1
2
= s – s 2 (t – 2)(t + 6) , 0 2
k = 2
3
= 36 – 24 –6 , t , 2 ∴ h = 9, k = 2
= 12 m (b) y = 2x
2
2 – (–2)
4
(c) When the particle passes through y = mx + c (b) m = ———– = — = 1
PC
point P, (mx + c) = 2x 9 – 5 4
2
s = 0 m x + 2mcx + c – 2x = 0 Gradient  to DC = –1
2
2 2
10t – 2t = 0 m x + (2mc – 2)x + c = 0 Therefore, the equation is
3
2
2 2
2
2t (5 – t) = 0 b – 4ac = 0 y – 4 = –(x – 1)
2
2
t = 0 or t = 5 (2mc – 2) – 4m c = 0 y = –x + 5
2
2 2
Therefore, t = 5 s. 4m c – 8mc + 4 – 4m c = 0 6. (a) 2 2x + 2 + 1 = 5(2 )
x
2 2
2 2
(d) When the particle reverses its –8mc = –4 2 ⋅2 + 1 – 5(2 ) = 0
2x 2
x
–4
direction of motion, m = —–– 2 ⋅ 2 – 5(2 ) + 1 = 0
2
2x
x
v = 0 –8c 4(2 ) – 5(2 ) + 1 = 0
x
2x
1
2t(10 – 3t) = 0 m = —– (4(2 ) – 1)(2 – 1) = 0
x
x
10
x
x
t = 0 or t = —– 2c 4(2 ) = 1 or 2 = 2 0
3
–2
x
10 4. (a) h(t) = 20t – 5t 2 2 = 2 x = 0
Therefore, t = —– s. = 5t(4 – t) x = –2
3
xy
h(t) (b) (i) log ABB
4
1 log xy
2 
SPM ASSESSMENT = — ——–– 2
2
log 4
20 2
1
Paper 1 h(t) = 20t – 5t 2 = —(log x + log y)
4 2 2
1
Section A = —(p + q)
1. (a) y 4
16y
0 4 t (ii) log ——
8 x 2
3
y = x 16y
–1
y = f (x) log ——
x
When t = 2, = ————
2
2
2 h(2) = 20(2) – 5(2) 2 1 3
= 20 = —(log 16 + log y – 2 log x)
3
2
2
2
0 < h(t) < 20
1
1 = —(4 + q – 2p)
5
y = f(x) (b) (α + 1) + (β + 1) = — 3
2
x 5
0 1 2 3 4 α + β = — – 2
2
1
0 < x < 2 = —
2
132
10_1202 QB AMath F5.indd 132 10/01/2022 4:49 PM

20
2
3

5 2
dh = – —– – — dx (ii) tan 2A 2x – 15x + 25 . 0
2 tan A
(2x – 5)(x – 5) . 0
2
x
= ————
dx
dV
dV
2
5
3
—– = —– × —– 1 – tan A — , x , — and 5 , x , 6
dh dx dh 2 — 2 2
3
 2 2
 – —– – —2
54x
1
2
2

= 120 – —— ————–– = ———— 2 3. (a) BD = x sin α
5
20
3
3
 2 2
x 2 5 1 – — DE = DB sin α
= (x sin α)(sin α)

3
When h = 1, = ——— = x sin α
2
9
3x
20
—– – —– = 1 1 – — (b) EF = DE sin α
4
x
5
3
12
100 – 3x = 5x = – —– = x sin α 2 3
2
3x + 5x – 100 = 0 5 ∴ x sin α, x sin α, x sin α
2
x sin α
2
p
(3x + 20)(x – 5) = 0 15. P(spoilt) = —–, n = 5 r = ———– = sin α
— ©PAN ASIA PUBLICATIONS
x sin α
x = 5 10 x sin α
3
1
dV
54(5)
2
2
2

x sin α
—– = 120 – —–––– ————–– (a) p P(X = 0) = 0.1681 r = ———– = sin α
0 q
5
0  10 2  10 2
dh 5  – —– – —2 5 C —– —– = 0.1681 (c) T = HG 5
20
3
5
5
5
2
= x sin α
q
5
a
7 2

= –150 – — —– = 0.7 (d) S = ——–
10
1 – r
∞
x sin α
= 107.14 Therefore, p = 3 q = 7 = ————
dh
dV
 2
—– = 107.14 —– (b) 7 good, 3 spoilt 1 – sin α
dt dt If 3 spoilt bulbs are arranged If x = 8, α = 60°
8 sin 60°
dV
Given —– = 21 cm s together, then number of ways S = ————–
3 –1
dt ∞ 1 – sin 60°
dh
8!
 2
21 = 107.14 —– = —– = 8 ——
8AB 3
7!
dt
2
dh
—– = 0.196 cm s –1 (c) p + q + 0.2 + 2p + q = 1 S = ———–
∞
AB 3
dt 3p + 2q = 0.8 ...........1 1 – —–
14. (a) BC = 10 sin θ P(X < 2) = p + q = 0.55 – 0.2 .....2 2
8AB 3(2 + AB 3)
AB = 10 cos θ 2p + 2q = 0.7 ................3 S = ———————
(i) Area, A 1 – 3: p = 0.1 ∞ (2 – AB 3)(2 + AB 3)
1
= —(10) sin θ (10 cos θ) q = 0.35 – 0.1 = 0.25 S = 24 + 16AB 3
2 ∞
= 50 sin θ cos θ 4. (a) y = ab x – 2
= (25 sin 2θ) cm 2 Paper 2 log y = log a + (x – 2) log b
10
10
10
25AB 3 Section A Y = log y, X = x – 2, m = log b,
10
10
(ii) 25 sin 2θ = ——– c = log a
2 1. x y z 10
AB 3 x + y + z = 16 .................................1 (b)
sin 2θ = —–
2 z – 2 = x + y x – 2 –2 –1 1 3 4 5
x + y – z = –2 .................................2 log y –0.9 –0.6 0 0.6 0.9 1.2
2 y – 5 = x 10
3 log y
y – x = 5 ...................................3 10
2θ 1 + 2: 2x + 2y = 14
1 1.0
2θ = 60°, 120° x + y = 7 ......................4 0.8
θ = 30°, 60° –x + y = 5 .......................................5
4 + 5: 2y = 12
π
π
θ = — rad, — rad y = 6 0.6
6
3
(b) 2x – x – 3 = 0 x = 1 0.4
2
(2x – 3)(x + 1) = 0 1 + 6 + z = 16 0.2
3 z = 9
x = — or x = –1 Therefore, the number is 169. (x – 2)
2
3 –2 –1 0 1 2 3 4 5
tan A = —, tan B = –1 2. (a) (i) V = 5(15 – 2x)(x) –0.2
2
Sum of roots: V = 75x – 10x 2 –0.4
1
dV
tan A + tan B = — (ii) —– = 75 – 20x = 0 –0.6
2
Product of roots: dx x = —– –0.8
75
3
tan A tan B = – — 20
2 15
(i) tan (A + B) x = —– (c) From the graph,
4
tan A + tan B 15 0.9
= —————— When x = —–, m = —– = 0.3
1 – tan A tan B 4 2 3
15
15
 4 2
 4 2
1 V = 75 —– – 10 —– log b = 0.3
10
2 b = 2
5
= ————– V = 140— cm 3
3
2 2

8
1 – – — (b) 90 , 75x – 10x , 140 log a = c = –0.3
10
a = 0.5
2
1
— 10x – 75x + 90 , 0 ∴ a = 0.5, b = 2
2
2
= —— 2x – 15x + 18 , 0
2
5
1
— 5. (a) —(3 – cos 2x)
2 (2x – 3)(x – 6) , 0 2
3
1
1
2
= — — < x , 6 = —(3 – (1 – 2 sin x))
5 2 2
75x – 10x , 125 1
2
2
10x – 75x + 125 . 0 = —(2 + 2 sin x)
2
2
134
10_1202 QB AMath F5.indd 134 10/01/2022 4:49 PM

©PAN ASIA PUBLICATIONS