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Published by PENERBITAN PELANGI SDN BHD, 2021-03-04 00:40:21

Focus KSSM 2021 Tingkatan 5 - Add Maths DLP

Focus KSSM 2021 Tingkatan 5 - Add Maths DLP by Penerbitan Pelangi

PELANGI BESTSELLER

ADDITIONAL

MATHEMATICS SPM

5Form

KSSM

Moh Sin Yee • Yong Kuan Yeoh D ual L anguage
Moy Wah Goon • Ng Seng How • Ooi Soo Huat P rogramme

NEW SPM ASSESSMENT

FORMAT 2021

Format: 190mm X 260mm TP Focus F5 AddMaths BI pgi_CRC

ADDITIONAL SPM
MATHEMATICS
5Form
Moh Sin Yee
Yong Kuan Yeoh KSSM
Moy Wah Goon
Ng Seng How
Ooi Soo Huat

Penerbitan Pelangi Sdn. Bhd. (89120-H)
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ISBN: 978-967-2965-65-7
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First Published 2021

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CONTENTS

Mathematical Formulae v – vi

1Chapter Circular Measure 1

1.1 Radian 2

1.2 Arc Length of a Circle 4

1.3 Area of Sector of a Circle 10

1.4 Application of Circular Measures 18

SPM Practice 1 21

2Chapter Differentiation 26

2.1 Limit and its Relation to Differentiation 27
2.2 The First Derivative 29
2.3 The Second Derivative 34
2.4 Application of Differentiation 35
SPM Practice 2 46

3Chapter Integration 49

3.1 Integration as the Inverse of Differentiation 50

3.2 Indefinite Integral 51

3.3 Definite Integral 56

3.4 Application of Integral 72

SPM Practice 3 75

4Chapter Permutation and Combination 78

4.1 Permutation 79
85
4.2 Combination 89
SPM Practice 4

5Chapter Probability Distribution 91

5.1 Random Variable 92
5.2 Binomial Distribution 96
5.3 Normal Distribution 101
SPM Practice 5 110

iii

6Chapter Trigonometric Functions 114

7Chapter 6.1 Positive Angles and Negative Angles 115
8Chapter 6.2 Trigonometric Ratios of any Angle 116
6.3 Graphs of Sine, Cosine and Tangent Functions 121
6.4 Basic Identities 126
6.5 Addition Formulae and Double Angle Formulae 127
6.6 Application of Trigonometric Functions 129
SPM Practice 6 133

Linear Programming 136

7.1 Linear Programming Model 137

7.2 Application of Linear Programming 139

SPM Practice 7 145

Kinematics of Linear Motion 149

8.1 Displacement, Velocity and Acceleration as a Function of Time 150

8.2 Differentiation in Kinematics of Linear Motion 158

8.3 Integration in Kinematics of Linear Motion 164

8.4 Application of Kinematics of Linear Motion 167

SPM Practice 8 172

SPM Model Paper 175

Answers 184



iv

2Chapter Learning Area :  Calculus

Differentiation

Concept
Map

Keywords • Minimum point – Titik minimum
• Approximation – Penghampiran • Normal – Normal
• Chain rule – Petua rantai • Product – Hasil darab
• Chord – Perentas • Quotient – Hasil bahagi
• Curve – Lengkung • Rate of change – Kadar perubahan
• First derivative – Terbitan pertama • Second derivative – Terbitan kedua
• Fixed point – Titik tetap • Small change – Perubahan kecil
• Gradient – Kecerunan • Tangent – Tangen
• Gradient of chord – Kecerunan perentas • Turning point – Titik pusingan
• Limit – Had
• Maximum point – Titik maksimum

TsTtpDhqgooeeuirfeataawfdidnnmeoitrsweeriewntlandyftesto,rigwar1mrt0tieethaioa.haix3stnci7rmhqsehmupusieprmneilsesnps–cttast1i.taeonunDrndt,tsoo,UmetwtsashioanenritoionsmnhutteiuBhsugemeodhdmrloy.tetquaoarutensacdtnothienthertiddaetryerteump.dhntDineah.iefmsafhtaeshorineeprtntocatsdhiisaniasitebntidolagennetcsthcheiiasenrnusctpnahheleasoenodfrgub9teneh.sn5ru8oitnshugseagedthdcoitohsountaatdpfnpsihnc.eisdenAv1wtt0ehoir0teahgavmearecllyurseu,pernhtseai?csoitnf

26

Additional Mathematics  Form 5  Chapter 2 Differentiation    

2.1 Limit and its Relation to (b) Constructing Table
Differentiation
x –0.1 –0.01 –0.001 0 0.001 0.01 0.1
A Investigating and determining the f(x) –2.99 –2.9999 –2.999999 –3 –2.999999 –2.9999 –2.99
value of limit of a function when its
variable approaches zero The table above shows that when x → 0, f(x) → –3.
Therefore

lim x3 – 3x
x
1. The limit value a of a function f(x) when the 1 2 x → 0   = –3 Chapter 2
variable x approaches zero is written as
Investigation through graph
lim f(x) = x3 – 3x
a f(x) = x = x2 – 3 y
x → 0

2. The investigation and determination of the limit Observe that when x
value is done through graph and by constructing x → 0, f(x) → –3. Therefore 0
a table. –3
lim x3 – 3x
x
3. The determination of limit value can also be 1 2 x → 0   = –3

done through direct substitution such that the Direct substitution
pnrootdduecfitnoefdtchaenvnaolut ebea0a,c(caep=tecdonanstdantht)e,
which is lim x3 – 3x lim x(x2 – 3)
function x x
needs to be simplified to get the correct value. 1 2 1 2 x → 0  = x → 0  

= lim (x2 – 3)

x → 0
= (0)2 – 3
1 = –3
lim
Find the value of f(x) if
x → 0
(a) f(x) = x + 3,
x3 – 3x . REMEMBER!
(b) f(x) = x
Simplify the function first before doing direct substitution

Solution because
(a) Constructing Table
x3 – 3x 03 – 3(0)
x 0
1 2lim =

x → 0

x –0.1 –0.01 –0.001 0 0.001 0.01 0.1 produces result which is not defined, that is 0 and this
f(x) 2.9 2.99 2.999 3 3.001 3.01 3.1 is not the correct solution. 0

Consider the values of f(x) when x approaching Try Questions 1 – 3 in ‘Try This! 2.1’
zero either from the left (negative value) or from
the right (positive value). B Determining the first derivative
of a function f(x) by using the first
The table above shows that when x → 0, x + 3 → 3 principle
either from the left (→) or from the right ( ).
Therefore, 1. Differentiation is about the changes of a variable
with respect to the changes of another variable
lim (x + 3) = 3 (normally time). Generally, differentiation is
the process of determining the instantaneous
x → 0 changes happen to a given variable at a particular
moment.
Investigation through graph y
2. Graphically, assuming that the graph f(x) = x3
Observe that when representing the motion of an object, therefore
0, f(x) → 3. Therefore, the gradient on the curve at a particular moment
xli→m (x + 3) = 3 3 ot1fadnifderte2nctainatiboend. etermined through the process

x → 0 x

Direct substitution –3 0

xl i→m 0 (x + 3) = 0 + 3
= 3

27

  Additional Mathematics  Form 5  Chapter 2 Differentiation

y f(x) = x3 y = f(x)
t1 x (function of the curve)
t2
tangent O dy = f '(x)
dx
(first derivative of the function of the curve)
3. Tcahnhadentgg2reaissdinitehnteitmcohefawnthhgeeicshtaoinfsgft(ehxne)tssapsaettehdthetehreamstuoilsmtsohefnottwhtne1
Chapter 2 through the motion of the object at the instance 7. The use of the idea of limit in deriving the first
t1 or t2. derivative of a function f(x) is known as the
differentiation through first principle.

4. The gradient of the tangent which happens at the 2
iindestaaonfcelimt1 iatn, tdhta2t can be determined through the
is: Determine the first derivative of each of the following
function f(x) through first principle.
y (a) f(x) = 2x (b) f(x) = 5x2

chord Q (x2, y2) (c) f(x) = 3x2 – 8x (b) f(x) = 6
x

O P(xt11)(x1, y1)tangent x Solution
(a) Let y = f(x),
Observe that when the point Q is approaching that is y = 2x Assuming that x
point P, the chord PQ is gradually approaching y + dy = 2(x + dx) experiences a small
and overlapping with the tangent at point P. 2x + dy = 2x + 2dx change (dx) which
causes a small
change (dy) in y.

Through the idea of limit, the gradient of tangent dy = 2dx Simplify until dy is
dx
of the curve at point P, = lim y2 – y1 . dy = 2 obtained by substituting
m   x2 – x1 dx
x2 → x1 y = 2x.
lim dy
Awshseunmxi2n→g dxx1,=dxx2→– x01.and dy = y2 – y1, \ dx = 2 Use the idea of limit so
dx → 0

\ m = lim   y2 – y1 and dy = 2 that the first derivative of
x2 – x1 dx function f(x) is obtained.
x2 → x1
(b) Let y = f(x),
= lim   dy that is y = 5x2
dx y + dy = 5(x + dx)2
dx → 0 5x2 + dy = 5[x2 + 2xdx + (dx)2]
5x2 + dy = 5x2 + 10xdx + 5(dx)2
SPM Tips dy = 10xdx + 5(dx)2

dx and dy is read as “delta x” and “delta y”.

The symbol dy dy = 10x + 5dx
dx dx
5. is used to represent the gradient
\ lim dy = 10x + 0
m which is obtained through the idea of limit dx
dx → 0

and is known as gradient function, that is and dy = 10x
dx
dy lim dy
dx = dx REMEMBER!
dx → 0
Be careful with the expansion of (x + dx)2. Observe that
6. The gradient of function dy is the first derivative dx2 ≠ (dx)2.
dx
of function f(x), that is

28

Additional Mathematics  Form 5  Chapter 2 Differentiation    

(c) Let y = f(x), 2. Find the limit for each of the following when x → 0.

that is y = 3x2 – 8x (a) (b) y

y + dy = 3(x + dx)2 – 8(x + dx) y

3x2 – 8x + dy = 3[x2 + 2xdx + (dx)2] – 8x – 8dx Ox

3x2 – 8x + dy = 3x2 + 6xdx + 3(dx)2 – 8x – 8dx Ox
dy = 6xdx + 3(dx)2 – 8dx
y = –3x2 – 2 y = 3 – x3

dy = 6x + 3(dx) – 8 (c) 3 x2 + 5 (d) 2–x Chapter 2
dx 2 3+x

\ lim dy = 6x + 0 – 8 (e) 4x2 – 3x
dx 5x
dx → 0

and dy = 6x – 8 3. Find the values of each of the following.
dx
(a) lim (3 – 2x + x2)

x → 0

REMEMBER! (b) lim (2 – x)(5 – x)

Each of the x term must be added with dx because each x → 0
of them experience a small change.
(c) lim x4 + 5
x → 0 x2 – 10
x+8
(d) lim (x – 2)(x + 4)

x → 0

(d) Let y = f(x), (e) lim x3 – 5x4
6 4x3
x x → 0
dy =
that is y= 6 (f) lim x2 – 16
y+ + dx x → 0 x – 4

6 x 6 4. Differentiate each of the following function by using
x + dx
+ dy = x first principle. (b) f(x) = 1 – 2x2
(a) f(x) = 6x

dy = 6 – 6 (c) f(x) = (2x – 1)2 (d) f(x) = – 1
+ dx x x
x
6x – 6(x + dx) 5. Determine the first derivative of each of the following
dy = (x + dx)(x)
functions by using first principle.

dy = 6x – 6x – 6dx (a) y = 4(1 – 3x)2 (b) y = 6x2 – ­2x
(x + dx)(x)
–6 (c) y=– 2 (d) y= 5
+ dx)(x) 3 8x
dy =
dx (x

\ lim dy = –6 2.2 The First Derivative
dx (x + 0)(x)
dx → 0

and dy = 6
dx x2
A Deriving the formula of first
Try Questions 4 – 5 in ‘Try This! 2.1’ derivative inductively for the
function y = axn

Try This! 2.1 1. The first derivative differentiation involves the
dy
process of obtaining dx from y or f '(x) from f(x).

1. Determine the limit value of each of the following 2. If y = axn where a and n are constants, then
by using (i) table, (ii) graph, (iii) direct substitution
dy dy
method. dx dx

(a) lim (8x) = [axn] = f '(x) = naxn – 1

x → 0

(b) lim (x2 + 3) Observe that this formula is obtained inductively.

x → 0

29

  Additional Mathematics  Form 5  Chapter 2 Differentiation

3 Therefore, inductively,

Given y = 2xn, find dy for n = 1, 2 and 3. dy = 2 = 1(2x1 – 1)
dx dx

Hence, verify the first derivative formula dy = naxn –1 dy = 4x = 2(2x2 – 1)
inductively. dx dx

dy = 6x2 = 3(2x3 – 1)
dx
Solution   
Chapter 2 (a) When n = 1,
dy = n(axn – 1) = naxn – 1
dx
y = 2x

y + dy = 2(x + dx) B Determining the first derivative of
an algebraic function
2x + dy = 2x + 2dx

dy = 2dx 1. For any function f(x) or y written in the
form of axn where a and n are constants, f '(x) or
dy = 2
dx
dy = axn – 1.
\ lim dy = 2 dx
dx
dx → 0 2. When determining the first derivative of an
algebraic functions, make sure that each term
and dy = 2 is written in the form of axn before applying the
dx

When n = 2, formula.

y = 2x2 3. iddsxyusoerdf '(x) is known as a gradient function and it
to determine the gradient of the tangent
y + dy = 2(x + dx)2

2x2 + dy = 2[x2 + 2xdx + (dx)2] to a curve of y or f(x) at a point on the curve.

2x2 + dy = 2x2 + 4xdx + 2(dx)2 4. In summary:
•  function y = a (a = constant)
dy = 4xdx + 2x(dx)2

dy = dx[4x + 2xdx] dy = 0
dx
dy = 4x + 2xdx
dx •  function y = axn (a, n = constant)

\ lim dy = 4x dy = naxn – 1
dx dx
dx → 0

and dy = 4x •  function y = axn + bxm + …
dx (a, n, b, m = constant)

When n = 3, dy = naxn – 1 + mbxm – 1 + …
dx
y = 2x3

y + dy = 2(x + dx)3 4

2x3 + dy = 2[x3 + 3x2dx + 3x(dx)2 + (dx)3] Find the first derivative of each of the following

2x3 + dy = 2x3 + 6x2dx + 6x(dx)2 + 2(dx)3 function.

dy = 6x2dx + 6x(dx)2 + 2(dx)3 (a) y = 6x3
5
dy = 6x2 + 6xdx + 2x(dx)2 (b) y = 4x2
dx
(c) y = 3√x
\ lim dy = 6x2
dx Solution
dx → 0 (a) y = 6x3

and dy = 6x2 ddxy = 3[6x3 – 1]
dx

dy = 18x2
dx

30

Additional Mathematics  Form 5  Chapter 2 Differentiation    

(b) y = 5 f '(x) = –12x –3 – 8
4x2 12
5 = – x3 – 8
y = 4
x–2

3 4 dy = –2 5 x–2 – 1 REMEMBER!
dx 4
5 Terms are simplified to the form of axn before differentiation
= – 2 x –3 is done.

= – 5 Try Questions 1 – 4 in ‘Try This! 2.2’ Chapter 2
2x3

(c) y = 3√x 
1
y = 3x2
1 1 – 1] C Determining the first derivative of
ddxy = 2 [3x 2 composite function

= 3 – 1 1. A composite function is a function with brackets
2 2 that is impossible or not easy to expand.
x
2. The first derivative of composite function
= 3 y = un such that u = f(x) and n = integer is done
2√x  by using chain rule, that is

5

Differentiate each of the following function with dy = dy × du
respect to x. dx du dx

(a) f(x) = 4x3 – 8x + 3
(b) f(x) = (5x2 – 2)2 3. The limit idea is used to prove the truth of chain
6x – 8x4 rule such as:
(c) f(x) = x3
dy lim dy
Solution dx = dx ­
dx → 0
(a) f(x) = 4x3 – 8x + 3
lim dy du
du dx
dx → 0
f '(x) = 3(4x3 – 1) – 1(8x1 – 1) + 0(3x0 – 1) 1 2 = ×

= 12x2 – 8x0 + 0 = lim dy × lim du
= 12x2 – 8 du dx
dx → 0 dx → 0

(b) f(x) = (5x2 – 2)2 dy = lim dy × lim du
= (5x2 – 2)(5x2 – 2) dx du dx
= 25x4 – 10x2 – 10x2 + 4 du → 0 dx → 0
= 25x4 – 20x2 + 4
f '(x) = 100x3 – 40x dy = dy × du
dx du dx

1 2 dy dy du
3. Since du = nun – 1, then dx = nun – 1 dx .

REMEMBER! 1 2 The dy du is seen as the
equation dx = nun – 1 dx
• The derivative of a constant = 0.
• Terms with brackets which can be expanded should formula taken from the chain rule which is often
used for composite function.
be expanded first before differentiation is done.

(c) f(x) = 6x – 8x4 6
x3
6x 8x4 Differentiate each of the following function with
= x3 – ­ x3 respect to x by using the chain rule.
(a) y = (3x + 2)4
= 6 – 8x (b) y = 5(1 – 3x)6
x2
= 6x –2 – 8x

31

  Additional Mathematics  Form 5  Chapter 2 Differentiation

Solution y = u4 REMEMBER!
(a) Let u = 3x + 2   and
du = 3 dy = 4u3 Do not forget to differentiate the terms inside the brackets.
du
dx Try Questions 5 – 9 in ‘Try This! 2.2’

\  dy = dy × du
dx du dx
= 4u3 × 3
= 12u3
Chapter 2 = 12(3x + 2)3 D Determining the first derivative of
a function involving product and
(b) Let u = 1 – 3x   and y = 5u6 quotient of algebraic expressions
du = –3
dy = 30u5 1. The first derivative of the function y = uv such
dx du that u = f(x) and v = g(x) is obtained from the
formula
\  dy = dy × du
dx du dx
= 30u5 × (–3) d dv du
= –90u5 dx (uv) = u dx + v dx

= –90(1 – 3x)5 2. This formula can be verified by using the idea of
limit as follows:
7 Let y = uv …… 1
From the idea of limit,
Differentiate each of the following function with respect y + dy = (u + du)(v + dv)
y + dy = uv + udv + vdu + dudv …… 2
1 2to x by using the formula dy du .
dx = nun – 1 dx

(a) y= 4 1 (b) y = √1 – 5x Substitute 1 into 2,
2x – uv + dy = uv + udv + vdu + dudv
dy = udv + vdu + dudv
Solution 4 1
(a) y = 2x – Divide each term with dx,

y = 4(2x – 1)–1 dy = u dv + v du + dudv
dx dx dx dx
3 4 dd xy = – 1)–1 – 1] d (2x –
= –1[4(2x 1)–2(2) dx 1) Through limit,
–4(2x –
= –8(2x – 1)–2 dlxi →m 0 dy lim dv lim du lim dudv
8 dx ­ = u dx ­+v dx ­+ dx
= – – dx → 0 dx → 0 dx → 0

(2x 1)2 It is known that when dx → 0, dudv → 0.
Therefore,
SPM Tips
lim dy lim dv lim du
Assume 4(2x – 1)–1 as 4xn and differentiate for axn dx ­ = u dx ­ + v dx ­
as usual. dx → 0 dx → 0 dx → 0

and dy = u dv + v du
dx dx dx
(b) y = √1 – 5x
1 or d (uv) = dv + du
dx u dx v dx
y = (1 – 5x)2

ddxy = 1 (1 – 5x) 1 – 1(–5) byy=usuvin­ gsuthche
2 2
3. The first derivative of the function
that u = f(x) and v = g(x) is obtained
= – 5 (1 – 5x)– 1
2 2

= – 5 1 2formula d u = u du – v dv  .
2√1 – 5x dx v dx dx
v2

32

Additional Mathematics  Form 5  Chapter 2 Differentiation    

4. This formula can be verified by using the idea of From the product formula derived,
limit as follows:
u dy = u dv + v du
Let y = v …… 1 dx dx dx
= (3x)[6(2x + 1)2] + (2x + 1)3(3)
From the idea of limit, = 18x(2x + 1)2 + 3(2x + 1)3
+
y + dy = u + du ­ …… 2 = 3(2x + 1)2(6x + 2x + 1)
v dv
= 3(2x + 1)2(8x + 1)
Substitute 1 into 2,
+ Chapter 2
u + dy = u + du ­ (b) f(x) = ­3x–+23x
v v dv
+
dy = u + du ­ – u Let u = x + 3  and v = 3 – 2x
v dv v
v(u + du) – u(v + dv) ­ du = 1 dv = –2
dy = (v + dv)(v) dx dx

dy = uv + vdu – uv – udv ­ From quotient formula derived,
(v + dv)(v)
– v du – u dv
vdu + udv f '(x) = dx dx  
dy = v(v dv) ­ v2

Divide both sides with dx, = (3 – 2x)(1) – (x + 3)(–2) ­
(3 – 2x)2
dy = vdu – udv ­ ÷ dx 3 – 2x + 2x + 6­
dx v(v + dv) = (3 – 2x)2

= v du – u dv   = 9 ­
dx dx – 2x)2
v(v + dv) (3

Through limit, Try Questions 10 – 11 in ‘Try This! 2.2’

dlxi →m 0 dy = v lim du – u lim dv  
dx dx dx
dx → 0 dx → 0

v lim (v + dv) Try This! 2.2

dx → 0

It is known that when dx → 0, dv → 0 and 1. Differentiate each of the following with respect to x.
lim
(v + dv) = v. (a) y = 5x6 (b) y= 7
dx → 0 4x2

Therefore, lim du lim dv (c) y= √x 3
dx dx 4
v dx → 0 – u dx → 0 (d) y = 6x 2

dlxi →m 0 dy = v(v)   (e) y = – 9 (f) y= 3 √x 
dx 3√x  4

1 2and dy = d u = v du – u dv   2. Find f'(x) for each of the following functions.
dx dx v dx dx (a) f(x) = 4x3 + 5x2 – 8x
v2
(b) f(x) = 3x5 + ­83x2

8 (c) f(x) = 1 + 2 – 3 + 4
x x2 x3 x4
Differentiate each of the following function with
respect to x. (d) f(x) = x4(1 + 3x – x2)
(a) y = 3x(2x + 1)3
(b) f(x) = ­3x–+23x (e) f(x) = 3√x  (1 – 5√x )

(f) f(x) = (3 – √x )2

Solution 3. Given 12x3 = 3y, determine the value of ­ddyx if
(a) Let u = 3x   and 1
v = (2x + 1)3 (a) x=– 2

ddux = 3 ddxv == 3(2x + 1)2(2) (b) x= 4
6(2x + 1)2 3

33

  Additional Mathematics  Form 5  Chapter 2 Differentiation

4. Given the volume of a cone is V= 1 pr2h such that r is 2.3 The Second Derivative
3
the radius and h is the height of the cone. Determine
dV
(a) dr if h= 2 cm

(b) dh if r = 5 cm A Determining the second derivative
dV of an algebraic function

Chapter 2 5. Determine dy for each of the following functions. 1. Observe that
dx y = f(x)

(a) y = 12x – 1 22 dy = f '(x)
x dx
9x2 – 1 First derivative
(b) y= 1 + 3x ddx2y2 = f "(x) Second derivative

(c) y= (7 + 3x)2
x2
d2y
(d) y = x2(5 – x)2 2. The second derivative, dx2 is obtained through

6. Differentiate each of the following with respect to x. the similar processes of differentiation as in the

(a) f(x) = 316x – 3 23 first derivative.
x

(b) f(x) = 8x – 4 3. dy is the gradient function of the curve y = f(x)
point rate of change of the
(c) f(x) = (3x2 – 4)4 dx d2y
5 at a dx2
(x, y). is the
(d) y = 3
6x – 3
1 gradient of a curve with respect to x.
(e) y= 4(5 – 3x)2

(f) y = – 5x 9
x3 – 2x2
d2y
7. Given y = u6 and u = 5x + 8, find the value of ­ddyx when Find dx2 for each of the following functions.
x = –3.
(a) y = 3x5 (b) y = 8(2x – 3)4
8. If h(k2 – 2k) = 5, find the value of ­ddhk when k = 1.
and ­ddxy Solution
9. Given y = k(2x – 1)4 = 40(2x – 1)n, find the (a) y = 3x5 REMEMBER!
values of k and n.
dy = 15x4 ddde x2ryi2vecdafnroomnlyddbyxe.
dx
10. Determine the first derivative for each of the ddx2y2 = 60x3
following functions:
(a) y = x(2x – 1)4 (b) y = 8(2x – 3)4
(b) y = (x + 4) 2x – 3
(c) f(x) = (x – 5)3(2x + 7)2 dy = 32(2x – 3)3(2)
(d) f(x) = x (9x – 2)5 d x = 64(2x – 3)3
(e) f(x) = 3x(1 – x)(5 + x)4 dd x2y2
= 192(2x – 3)2(2)
1 2(f) f(x) = (2x3 – 3) ­1x – 2x 2 = 384(2x – 3)2

11. Differentiate each of the following with respect to x.
5x2
(a) f(x) = 2x – 1 10

(b) f(x) = √x + 1 Find the value of f "(0) for each of the following
√x  – 1 functions.
√2 – x
(c) f(x) = x2 (a) f(x) = x2(1 – 3x)
2x4 – 3x2
f(x) = (1 – 3x)4 (b) f(x) = x
(d) x3

(e) y= 3x – 8x2 Solution
2 + 5x – x2 (a) f(x) = x2(1 – 3x)
x3 – 2x + 6 = x2 – 3x3
(f) y= 8–x

34

Additional Mathematics  Form 5  Chapter 2 Differentiation    

f '(x) = 2x – 9x2 2.4 Application of Differentiation
f "(x) = 2 – 18x
f "(0) = 2 – 18(0) A Interpreting the gradient of tangent
= 2 to a curve at different points

(b) f(x) = 2x4 – 3x2 1. The first derivation of a function y = f(x), that is
= 2x3 –x3x uddsxyth=efv'(axlu) eisotfhtehegrgardadieinent tfuonf cthtieontanwgheincthtogitvhees
f '(x) = 6x2 – 3 Chapter 2
f "(x) = 12x curve y = f(x) at a point on the curve.
f "(0) = 0
y
Try Questions 1 – 7 in ‘Try This! 2.3’

Try This! 2.3 A

y = f(x)

1. Dfuentcetriomnins.e ­ddxy and ­ddx2y2 for each of the following 0 tangent x m = dy
dx

(a) y = 8x4 (b) y = 7x2 – 8x3 2. The gradient of the tangent line drawn at a point
(c) y = ­2x – √x 
(e) y = 5(4x – 3)6 (d) y = x3(1 – 2x) A(x1, y1) on the curve is obtained by substituting

(f) y = ­­6x – 5x3 the value of x1 into dy = f(x).
x2 dx

2. Find f '(x) and f "(x) for each of the following functions. 3. The accuracy of the value of gradient m, can
be tested/verified by drawing a graph manually
(a) f(x) = 7x2 – 5x + ­1x on a graph paper or by using computer apps/
(b) f(x) = 4x2(x + 2)2 programs.

(c) f(x) = ­x 6 3
+
(d) f(x) = ­13–xx
11
3. Find the value of f"(0) for each of the following.
(a) f(x) = (9 – 3x)4 Find the gradient of the tangent to the curve y = 4x3 – 2
(b) f(x) = 4x(x – 3)2 at the given points.
x+3
(c) f(x) = x–3 (a) Point A(1, 2)

4. Given the function f(x) = 3x5 + mx3 – 4. Find (b) Point B1 1 , – 3 2
(a) the second derivative of the function f. 2 2
(b) the value of m if f"(–1) = –5.
Solution
5. Given the function y = (x – 3)2 .
dy d 2y (a)   y = 4x3 – 2 (b)   y = 4x3 – 2
(b) Find dx and dx2 .
ddxy = 12x2 dy = 12x2
2y x dy (3x d 2y = 12(1)­2 1 2 dx =
(c) Show that – dx + – 9) dx2 = 0. 12 1 2­
= 12 2
6. Given the function y = x(4 – x).
dy d 2y = 121 1 2
(a) Show that y – 1 x dx + x dx2 = 0. = 4
2 3
x
(b) Find tdhye +po2sxsddib x2ly2e.values of when
y=x dx
12
7. Given the function y = 3(x – 1)2.
d 2y dy Calculate the gradient of the tangent to the curve
(a) Express y dx2 + dx in terms of x. 6 + 2x
y = x at the point where the y-coordinate is –2.
(b) Solve the equation y d 2y + dy = 0.
dx2 dx

35

  Additional Mathematics  Form 5  Chapter 2 Differentiation

Solution Solution
When y = –2, From y = 3x2 – 9x + 4

y = 6 + 2x y = 6 + 2x dy = 6x – 9
x x dx

–2 = 6 + 2x = 6x–1 + 2 At the point (2, –2),
x
–2x = 6 + 2x 1 2At the point – 3 , –2 , Gradient of tangent, m1 = dy = 6(2) – 9
2 dx = 3
4x = –6 dy
Chapter 2 3 1 2\ dx = –6x–2 Equation of tangent line,
x = – 2 = 6
– 3 y –y(––2y)1 == 3m(1x(x––2)x1)
– 2 2

24 y + 2 = 3x – 6
= – 9 y = 3x – 8

= – 8 Since m31 × mm22 = –1
3 × = –1

Try Question 1 in ‘Try This! 2.4’ Therefore, the gradient of normal, m2 = – 1
3
Equation of normal line,
B Determining the equation of y – y1 = m2(x – x1)
tangent and normal to a curve at a 1
point y – (–2) = – 3 (x – 2)

1. The equation of the tangent to a curve y = f(x) y + 2 = – 1 x + 2
at a given point –(xy1,1 y=1) is determined by using 3 3
the formula y m1(x – x1), such that
y = – 1 x + 2 –2
m1 = dy = f '(x1). 3 3
dx
=– 1 x – 4
3 3
y y = f(x)

(x1, y1) C Solving problems involving tangent
and normal
O tangent x

2. The equation of the normal to a curve y = f(x) at 1. There are a few things involved in the problems
involving the tangent and normal, which are

a given point (x1, y1) on the curve is determined (a) problems about the curve y = f(x) such that
by using the formula – = m2(x – x1), such that the information about the point and the
m1 × m2 = –1. y y1 equation of tangent or normal are given,

y (b) problems tahbeouint fofirnmdiantigonthaebopuotintthe(xc1u, ryv1e)
m such that
y = f(x) and the equation of tangent or
2 m1 normal are given,

(x1, y1) (c) problems which involve the equations of
tangent or normal or both such that the
Normal information about the curve y = f(x) and the
Ox point (x1, y1) are given.

13 2. Normally, the problems involving tangent and
Find the equations of the tangent and normal to the normal involve layers of systematic solving
curve y = 3x2 – 9x + 4 at the point (2, –2). steps.

36

Additional Mathematics  Form 5  Chapter 2 Differentiation    

14   Example of HOTS Question
Given a curve y = 5x2 – 3x – 2. Find the coordinates
of the point on the curve such that the gradient of the A research done on the movement of an object
tangent at the point is 7. Hence, find the equation of found out that it follows the function f(x) = ax3 + bx2
the tangent. + c such that a, b, and c are constants. Given that
the curve of the function f(x) passes through points
Solution y = 5x2– 3x –2 (–1, 0) and (0, 5) when the graph of the function f(x) Chapter 2
y = 5x2 – 3x – 2 (x1, y1) is drawn on the Cartesian plane. The tangents to the
curve f(x) at the points where x = 0 and x = 1 are
dy = 10x – 3 m=7 parallel to the x-axis. Determine the function f(x) by
dx finding the values of a, b and c.
7 = 10x – 3
10x = 10 Solution
x = 1 Given the point (–1, 0),
f(x) = ax3 + bx2 + c
Therefore, y = 5(1)2 – 3(1) – 2 0 = a(–1)3 + b(–1)2 + c
= 0 0 = –a + b + c …… 1
The coordinate of the point is (1, 0). For the point (0, 5),
5 = a(0)3 + b(0)2 + c
Equation of tangent, c = 5
yy––y01 == 7m(1x(x––1)x1) Therefore, 0 = a + b + 5
a – b = 5 …… 2
y = 7x – 7
f(x) = ax3 + bx2 + c
SPM Tips f'(x) = 3ax2 + 2bx

Make a rough sketch for easier understanding. For x = 1,
0 = 3a(1)2 + 2b(1)
15 3a + 2b = 0 …… 3
2 × 3, 3a – 3b = 15 …… 4
Given that the gradient of the curve y = hx2 + kx at the 3 – 4, 5b = –15
point (2, 8) is 6. Find the values of h and k. b = –3
From 2, a – (–3) = 5
Solution a + 3 = 5
Given the gradient = 6, a = 2
y = hx2 + kx \  f(x) = 2x3 – 3x2 + 5

dy = 2hx + k REMEMBER!
dx
6 = 2h(2) + k The gradient of the tangent line parallel
4h + k = 6 …… 1 to the x-axis is 0.

Given the point (2, 8),
y = hx2 + kx
8 = h(2)2 + k(2) Try this HOTS Question
8 = 4h + 2k …… 2
The motion of an object follows the equation
2 – 1: k = 2 y = x2 – 6x + 5. The tangent to the curve at a point
P on the curve is parallel to the straight line which
From 1, 4h + 2 = 6 joins the point A(1, 0) and B(7, 12). Given that the
4h = 4 normal line on the point P intersects the curve at
h = 1 two points. Find the equation of the normal at point
P and the coordinates of the points of intersection
SPM Tips between the normal and the curve.

1 2Answer:1 3 7
The gradient of the curve refers to the gradient y=– 2 x – 1; 2 , – 4
of the tangent.

Try Questions 2 – 5 in ‘Try This! 2.4’

37

  Additional Mathematics  Form 5  Chapter 2 Differentiation

D Determining the turning points and (a) Tangent sketching method
their nature (i) Maximum

1. Turning points or stationary points, are the (ii) Minimum
points on a curve where the gradient of the
tangent at the points are 0. The points have either (iii) Inflection
maximum or minimum value. point

Chapter 2 (a) Maximum point or

y

positive 0 (b) Second derivative method
negative

Ox (i) Maximum d 2y ,0
(ii) Minimum dx2
(iii) Inflection
The value of the gradient of the tangent d 2y .0
changes from positive to zero to negative, point dx2
from left to right.
d 2y =0
dx2
(b) Minimum point
16
y Find the turning points of the function y = x3 – 6x2 + 9x + 2
and determine the nature of each turning point.
negative 0 positive
O x

The value of the gradient of the tangent Solution
changes from negative to zero to positive,    y = x3 – 6x2 + 9x + 2
from left to right.
dy = 3x2 – 12x + 9
dx

At the turning point,
0 = 3x2 – 12x + 9
2. If the value of the stationary point is neither x2 – 4x + 3 = 0
maximum nor minimum, it is called an (x – 3)(x – 1) = 0
inflection point. x = 3,   x = 1

yy When x = 3,
negative y = (3)3 – 6(3)2 + 9(3) + 2
= 27 – 54 + 27 + 2
O positive Ox =2
positive x negative

The value of the gradient of the tangent changes When x = 1,
from positive or negative to zero to positive or y = (1)3 – 6(1)2 + 9(1) + 2
negative back in both directions. =1–6+9+2
=6
3. To determine the turning point, let dy = 0.
dx \  The turning points are (3, 2) and (1, 6).

4. To determine the nature of the turning point, Determining the nature of the turning points:
either maximum, minimum or inflection point, Tangent sketching method
the following two methods can be applied: At (3, 2):

38

Additional Mathematics  Form 5  Chapter 2 Differentiation    

Choose one value Choose one value Solution
where x , 3 where x . 3 y = x3 + ax + b

x 2 34 At (0, 4), 4 = (0)3 + a(0) + b
–3 09
dy – zero + b = 4
dx
Also, dy = 3x2 + a
Sign dx
Tangent
sketching 0 = 3(0)2 + a Chapter 2

a = 0

Graph \  y = x3 + 4
sketching
dy = 3x2
dx

\ (3, 2) is a minimum point. and dd x2y2 = 6x
= 6(0)
At (1, 6):
0 1 2 = 0
x 9 0 –3
+ zero – \  The point (0, 4) is an inflection point.
dy
dx Try Questions 6 – 8 in ‘Try This! 2.4’
Sign
Tangent E Solving problems involving
sketching maximum and minimum values and
interpreting the solutions
Graph
sketching 1. Problems involving the maximum value and
minimum value can be solved by using the
\ (1, 6) is a maximum point. following steps:
(a) Based on the information given in the
Alternative Method problem, recognise the main function
(area, volume etc) that needs to be formed
Second derivative method or derived so that the differentiation can be
carried out. This main function can be easily
d 2y = 6x – 12 recognised through the word “maximum”
dx2 or “minimum” stated in the problem.
(b) Obtain this main function and express it in
x = 3, terms of one variable. Normally, there would
be one condition stated in the problem so that
d 2y = 6(3) – 12 the variables can be mutually substituted.
dx2 = 6. 0

x = 1, (c) Let f '(x) or wddhxyic=h 0 to obtain the value of
the variable causes the function to be
d 2y = 6(1) – 12 maximum or minimum.
dx2 = –6 , 0 (d) If there is more than one variables obtained,

17 investigate the nature of the values by using

Given that the graph of the function y = x3 + ax + b, gddr x2ay2die(nsteocof ntadngdeenrtivdayti.ve method) or the
where a and b are constants, has a turning point (0, 4). (e) Find the maximumdvxalue or the minimum
Determine the values of a and b. Hence, determine the
nature of the turning point. value of the function according to the
requirement of the problem.

39

  Additional Mathematics  Form 5  Chapter 2 Differentiation

18 Solution

A wire with length 84 cm is bent to form a rectangle. height = h
Find the maximum area of the rectangle that can be width = x
formed.

Solution length = 2x
Area of rectangle = A
Chapter 2 Let A = length × width Width = y From the analysis of the problem, the production
= x × y Length = x cost depends on the total surface area of the box
= xy produced.

Length of wire = perimeter of the rectangle Therefore,  A = total surface area
x + x + y + y = 84 = 2[(2x)(x)] + 2(xh) + 2[(2x)(h)]
2x + 2y = 84
2x = 84 – 2y x h
x = 42 – y 2x xh
Substitute into A,
A = (42 – y)(y)   \ A = 4x2 + 2xh + 4xh 2x
= 42y – y2
ddAy = 42 – 2y Volume of the box given = 72 cm3

At the turning point, Length × width × height = 72
0 = 42 – 2y 2x × x × h = 72
2y = 42 72
y = 21 h = 2x2

= 36
x2
36
Substitute h = x2 into A,

Also, d 2A = –2 0 1 2 1 2A = 4x2 + 36 36
dy2 , 2x x2 + 4x x2

y = 21 will caused the value of A to be maximum. = 4x2 + 72 + 144
x x
216
SPM Tips = 4x2 + x

A can also be expressed in terms of x to determine \  dA = 8x – 216x–2
the value of x. dx

At the turning point, 8x – 216 = 0
x2
Amax = 42(21) – (21)2 216
= 441 cm2 8x = x2

\ The maximum area of the rectangle that can be 8x3 = 216
formed is 441 cm2.
x3 = 27

x = 3

Try Questions 9 – 12 in ‘Try This! 2.4’ Also, d 2L = 8 + 432x–3
dx2
d 2L
  Example of HOTS Question When x = 3, dx2 .0

A factory plans to produce a closed cuboid-shape x = 3 will cause the value of A to be minimum.
box such that the volume of the box produced is
72 cm3. Given that the length of the box is twice its Amin = 4(3)2 + 216 = 108 cm2
width. The production cost of the box is estimated 3
to be RM0.15 per cm2. Find the minimum production
cost that can be obtained in the production of a The minimum production cost of a box produced is
box. = 108 × 0.15
= RM16.20

40

Additional Mathematics  Form 5  Chapter 2 Differentiation    

Try this HOTS Question F Interpreting and determining the
rates of change of related quantities
A factory intends to produce a cylindrical container
with an opening at one end with a volume of 1. A function y = f(x) expresses the relationship
27π cm3. The factory estimates that the production between the variables y and x. The changes of
cost would be RM0.18 per cm2. Determine the the variables y and x with respect to time t are
maximum number of containers that can be dy dx
produced if the production cost is RM3 500. written as dt and dt . Hence,

Answer: dy = dy × dx Chapter 2
229 dt dx dt

2. The above relationship exists between heddlpyt s,
dy saonlvdinddgxt follows the chain rule which
  SPM Highlights idnx
many rate of change of variables
The diagram below shows the side view of a part of a problems. This relationship can also be written
roller coaster track in a park. as:

p dy = dy × 1 1 21 = dt
or dx dt dx dx dx
5m
dt dt
Horizontal ground level 1 1 21
dx = dy × dy dy = dx
dt dt dt

The curve part of the roller coaster track is represented dt dx

by an equation y = 1 x3 – 3 x2, with point p as the dy dx
64 16 dt dt
3. .0 or . 0 denotes the increase in the
origin.Find the shortest vertical distance, in m, from the
track to ground level. rate of change of the variable, while dy ,0
dt
Solution dx
y = 1 x3 – 3 x2 or dt , 0 denotes the decrease in the rate of
64 16
dy change of the variable.
dx = 3 x2 – 3 x
64 8

At the turning point, 19

3 x2 – 3 x = 0 Two variables x and y are related by the equation
64 8 14
= 2x + x . If increases with a rate of 8.5 units per
3 x 3 1 x – 14 = 0 y y
8 8
x = 8, x = 0 second, find the rate of change of x when x = 2. State the
type of rate of change of x at that particular moment.
Also, d 2y = 3 x – 3
dx2 32 8 Solution

For x = 8,  d 2y . 0  (y is minimum.) Given ddyty = 8.5 (Positive value because y is increasing.)
dx2 and = 2x + 14x–1

For x = 0,  d 2y , 0 (y is maximum (ignored)) dy
dx2 dt
= 2 – 14x–2
Substitute x = 8, y = 1 (8)3 – 3 (8)2 = 2 – 14
64 16 x2

= –4

\ The shortest vertical distance is 5 – 4 = 1 m. When x = 2, dy = 2 – 14
dt (2)2
3
= – 2

41

  Additional Mathematics  Form 5  Chapter 2 Differentiation

From dy = dy × dx G Solving problems involving rates of
dt dx dt change for related quantities and
3 interpreting the solutions
8.5 = – 2 × dx
dt
17
\ The rate of change of x, dx =– 3 units per second. 1. The rate of change of variable (or quantity y)
dt can be determined by using the formula

x experiences a rate of decreasing at the moment when ddyt dy dx
dx = dx dt if y = f(x) and = f(t).
Chapter 2 x = 2 because dt , 0. × x

20 2. Observe that dy is the rate of change of y and
dt
dx is the rate of change of x.
The relationship between two variables p and q is dt
p2q = 4. Given that p decrease with a rate of 3 unit s–1,
find the rate of change of q at the moment when q = 64. 3. In the process of solving problems involving
rate of change of related quantities, recognising
Solution the quantities of x and y and the relationship
dp between these quantities are the main focus.
Given dt = –3 (Negative because the rate decreased)

From p2q = 4,
4
q = p2 21

q = 4p–2 The area of the wave of water on the surface of a water
tank expands with a rate of 5 cm2 s–1. Find the rate of
dq = –8p–3 change of its cicular radius when the area of the circle
dp = is 6.25π cm2.
– 8
p3

Given q = 64, therefore p2q = 4 Solution

p2(64) = 4 dA = 5 cm2 s–1
1 dt
p2 = 16 Given A = πr2
Also,
p = ± 1
4 ddAr = 2πr

When p = ±  1 ,
4
When A = 6.25π
dq = – 8 3  or dq = – 8
1 dp 1 3 πr2 = 6.25π
1 2 1 2dp – 4
4 r = 2.5 cm

= –512 = 512 From dA = dA dr
dt dr dt
Therefore, dq = dq × dp ×
dt dp dt
When r = 2.5,
When dq = –512, dq = –512 × (–3) dr
dp dt = 1 536 unit s–1 5 = 2π(2.5) × dt

When ddpq = 512, ddqt = 512 × (–3) dr = 5
= –1 536 unit dt 5π
s–1
= 1 cm s–1
(q may increase or decrease at the rate of 1  536 units π
per second.)
\ The radius of the circular wave increases with a rate
1
Try Questions 13 – 14 in ‘Try This! 2.4’ of π cm s–1 when the area of the circle is 6.25π cm2.

42

Additional Mathematics  Form 5  Chapter 2 Differentiation    

  SPM Highlights 4. The smaller the value of dx, the more accurate
the approximate value of dy.
The surface area of a cube increase at a constant rate
of 15 cm2 s–1. Find the rate of change of side length, in 5. dx . 0 or dy . 0 denotes the small increase in x
cm s–1, when the volume of the cube is 125 cm3. or y, dx , 0 or dy , 0 denotes the small decrease
in x or y.

Solution

x 22 Chapter 2
Given y = 3x2 + 6x, find the small change in y when x
xx changes from 4 units to 4.02 units.

Let the length of the side of the cube = x cm Solution
Therefore the area of the cube, A = 6x2 From y = 3x2 + 6x

When the volume of the cube = 125 cm3, dy = 6x + 6
x3 = 125 When x = 4, dx
x = 5 cm
\ ddAx = 12x
= 12(5) ddxy = 6(4) + 6
= 30
and ddAt = 60 dx = 4.02 – 4
= 15 cm2 s–1 = 0.02

From  dA = dA × dx Therefore, dy = dy × dx
dt dx dt dx
dx
15 = 60 × dt = 30 × 0.02
= 0.6 unit
dx 15
dt = 60 \ y experiences small changes for 0.6 unit.

= 1 cm s–1
4
23
\ The length of side increase with a rate of 1 cm s–1 Estimate the approximate value of 3√63.98 by using
4 differentiation.
when the volume of the cube is 125 cm3.

Solution
Let  y = 3√x 
Try Questions 15 – 16 in ‘Try This! 2.4’

H Interpreting and determining small 1
changes and approximations of
certain quantities = x3
1
\  dy = 2
dx
3x 3

1. From lim dy = iddsxyth, ewshmearlel dx is the small Let 3√63.98 = 3√64 + dy
adnxd dy changes in y, it is
dx → 0 SPM Tips
changes in x
dy dy
found out that dx ≈ dx when dx → 0. Choose a value of x close to 63.98, such that its cube
root value can be obtained easily. dy is the change or
2. dy ≈ dy can be written as dy ≈ dy × dx. This the approximate difference between 3√63.98 and
dx dx dx 3√64 , that is dy = 3√63.98 – 3√64 .
formula is used to estimate the approximate
change or the small change in y when there is a
small change dx in x. Small changes in y,

3. Observe that the value of dy obtained from dy ≈ dy × dx
ddisxy ju×stdaxnisapnportotxhiemeaxtiaocnt dx
dy ≈ change in y. The 1
value to the real value ≈ 2 × dx
3x 3
of change.

43

  Additional Mathematics  Form 5  Chapter 2 Differentiation

≈ 1 2 × (–0.02) dx = 63.98 – 64 I Solving problems involving small
3(64)3 = –0.02 changes and approximations of
certain quantities
≈ 1 × (–0.02) When y = 3√64 ,
48 x = 64 1. The formula dy ≈ dy or dy ≈ dy × dx is used in
dx dx dx
≈ – 1 or –0.0004167 solving problem which involves a small changes,
2 400
either in x or y.
Chapter 2 1 2Therefore, 3√63.98 ≈ 3√64 + – 1
2 400 2. Recognising two quantities involved in the
1 problem and forming the function or relation
≈4– 2 400 between the two quantities are the main things
in solving the problems.
≈ 3.9996

Try Questions 17 – 18 in ‘Try This! 2.4’ 24
Find the approximate change in the surface area of a
  SPM Highlights sphere when the radius of the sphere decreases from
3.0 cm to 2.97 cm.
y 5
It is given that the equation of a curve is = x2 . Solution
(a) Find the value of ­ddyx when x = 3. Area = 4π (radius)2
y = 4πx2
(b) Hence, estimate the value of 5 .
(2.98)2
and dy = 8πx
Solution dx = 8π(3)
= 24π
(a) y = 5x –2 dy
dx
When x = 3, = –10x –3 Therefore, dy ≈ dy × dx
dx
= – 10 ≈ 24π × (2.97 – 3.0)
x3

=– 10 ≈ 24π × (–0.03)
(3)3 ≈ –0.72π cm2

= – 10 \ The surface area of the sphere decreases 0.72π cm2.
27 (a negative value is obtained)

(b) Let 5 = 5 + dy   SPM Highlights
(2.98)2 (3)2

such that x = 3 and dx = 2.98 – 3 It is given that L = 4t – t2 and x = 3 + 6t.

= –0.02 dL
dx
Small changes in y, (a) Express in terms of t.

dy ≈ dy × dx (b) Find the small change in x when L changes from
dx
3 to 3.4 when t = 1.
≈ – 10 × (–0.02)
27
Solution
≈ 1   or  0.007407 (a)   L = 4t – t2 x = 3 + 6t
135 dL dx
dt = 4 – 2t dt = 6
Therefore, 5 ≈ 5 + 1
(2.98)2 (3)2 135 dL dL 1
dx = dt × dx
5 1
≈ 9 + 135 dt

≈ 76   or  0.5630 = (4 – 2t) × 1
135 6
2–t
= 3

44

Additional Mathematics  Form 5  Chapter 2 Differentiation    

(b) dL = 3.4 – 3 = 0.4 9. The sum of two positive numbers is 38. Determine
When t = 1, x = 3 + 6(1) = 9 the values of these two numbers such that the sum
dL dL
dx dx of the squares of these two numbers is minimum.

≈ 10. It is estimated that the cost RMK, to drive your
own car from town A to town B follows the formula
2 – 1 ≈ 0.4
3 dx
K= 108V 2 + 1 , such that V is the average speed
dx ≈ 1.2 V
in km h–1. Find the value of V if the cost is minimum.
Chapter 2
Try Questions 19 – 20 in ‘Try This! 2.4’ 11. The diagram below shows a cuboid with length 3x
cm, width x cm and height h cm. Given that the sum
of the surface area of the cuboid is 96 cm2.

Try This! 2.4 h cm

1. Find the gradient of the tangent to the curve 3x cm
y = 2x3 – 5x2 at the point where the x-coordinate is
–1. x cm

2. Find the equation of the tangent to the curve (a) Express the volume of the cuboid, V cm3, in
terms of x.
1 2y = x3 –3 x2 + 2x the 3 . Hence, find the
2 at point 1, 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
radius r cm from a piece of metal with an area of
1 2point 1, 3 .
2 540 cm2.

3. Given that tangent to the curve y = 4x2 + px + q at (a) Show that the height of the cylinder, h cm, is
the point (–1, 10) is perpendicular to the straight line πr2
3y = x – 5. Find the values of p and q. given by h = 270 – .
πr
(b) Hence, find the values of r and h such that the
4. Find the equation of the normal to the curve 4y = x2
at the point (4, 4) and hence, find the coordinates of volume of the cylinder is maximum. Hence find
the point such that this normal intersects the curve
once again. this maximum volume. [Use π = 3.142]

13. Given y = 8x3 – 5 x2 + 1 , find the rate of change
3 2
of y when x increases with a rate of 3 unit s–1 when
5. M is a point on the curve y = ­5x such that the gradient 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 k2
(a) the coordinates of the midpoint of line AB, h= – 4 . If k changes at a rate of 0.4 unit s–1, find
(b) the length of line AB. k
the rate of change of h when h  = 3.

6. Find the turning points for y = x3 – 6x2 + 9x + 4 and
dy 15. A piece of ice cube with sides 3 cm melts at a rate of
determine the nature of each point by using dx and 0.03 cm s–1. Find the rate of change of its volume.
d 2y
dx2 .

7. Given that the graph of the function f(x) = x + h has 16. A piece of metal in the shape of a sphere is heated
a turning point (3, q). x2 and expands at a rate of 0.6 cm3 s–1. Find

(a) Find the values of h and q. (a) the rate of change of its radius, and

(b) Determine the nature of this turning point. (b) the rate of change of its surface area
when its volume is 36π cm3.

8. The curve y = x3 + px2 + qx + 1 has a turning point 17. Given y = 4 x , find

(–1, 5). 2
(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 ndda x2yt2u.re of these two turning (b) the small change in x when y decreases from
points by using
8 to 7.96.

45

  Additional Mathematics  Form 5  Chapter 2 Differentiation

18. Given y = √1x . 20. The relation between the focal length, f m, of a
dy convex lens with its object length, u m and image
(a) Express dx in terms of x. length, v m, is given by the formula:

(b) Hence, calculate the approximate value of 1 = 1 + 1
1 f v u
√80.98 .

Chapter 2 19. A filler funnel in a cone shape has a height of 8 cm. If the focal length of the lens is 4 m, find the small
If the radius of the base of cone changes from 6 cm change in its object length when the image length
to 6.04 cm, find the approximate change in changes from 28 to 27.8 m.
(a) the volume of the cone,
(b) the curve surface area of the cone.

SPM Practice 2

PAPER 1 1 2(b) 4
f"(–2) if f(x) = x 2x2 – 3x + x .

1. (a) Find the value of lim (8 – √x ). 9. The gradient function of a curve is (3x – 5). Given
the point Q(–2, 6) lies on the curve, find
SPM x → 0 (a) the gradient of the tangent at point Q,
(b) the equation of the normal to the curve at
2018 (b) Differentiate the following function by using the point Q.

first principle.

y= 5
x2

2. (a) Given f(x) = 2 – 3x2 , find f '(x). 10. Given that the gradient of the normal to the curve
5x – 4
4
(b) Given f(x) = 18 , evaluate f '(2). y = 9x2 – mx3 at x = –3 is 3 .
x4
(a) Find the value of m.

1 2 d (b) Find the equation of the tangent to the curve at
3. (a) Find dx 1 . x = –3.
(b) 8x – 5
5 dy
Given r = 3x – 2 and y = – r2 . Find dx in terms 11. Given that hx2 – x is the gradient function of a
of x. curve such that h is a constant. If y – 7x + 5 = 0 is
the equation of the tangent at point (–1, –12), find
4. (a) Differentiate x3(1 + 5x)6 with respect to x. (a) the value of h.
(b) Given f(x) = (1 – 3x)5, evaluate f '(–1). (b) the equation of the normal to the curve at point
(–1, –12).
5. Given the function h(x) = px3 – 8x2 + 6x, find
(a) h'(x), 12. Find the equations of two tangent lines from the point
(b) the value of p if h'(–1) = 5.

6. Given y = x(5 – x), express y­ddx2y2 + x dy + 24 in 1 25, 11 to the curve y = 4 + 3x – x2. HOTS
dx 2 2
terms of x in the simplest form. Hence, solve the Applying

equation y d 2y + x dy + 24 = y. 13. A car moves along a straight road. The
dx2 dx displacement, J metre, of the car from a fixed point
X along the straight road is given by

7. (a) Given f(x) = (7x – 4)6, find f "(x). J = 3t3 + 23 t 2 – 3t – 3
(b) Given h(x) = kx3 – 5x2 + 6x, find the value k if 4
h"(–1) = 5.
such that t is the time, in second, after passing
through point X. Find
8. Find the value of dJ HOTS Applying
dt
SPM lim 4 (a) .
x
1 22018 x → 0
(a) x 2x2 – 3x + . (b) the maximum displacement of the car.

46

Additional Mathematics  Form 5  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
8x m × (6 – x) m. Determine the length, in m, the 1. (a) Given that y = 4x2 – 3x + 5, determine the first
fence that Ali needs to make, if the area of the land
is maximum. derivative by using the first principle.

(b) Find d
dx
1 2(i)
15. The curve y = hx4 + 3x has a turning point at (2, k).   1
5x – 2

d x
dx 5x – 2
SPM Find 1 2(ii)  
2019 (a) the value of h and k.
dy Chapter 2
(b) the value of dx when x = –1. (c) By using differentiation method, estimate the
8
16. The diagram below shows a piece of cardboard value of (2.01)3 .
which is used to form a cylindrical-shape funnel
2. The gradient function of a curve is px2 – qx, such
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
y cm the point (5, –3) is the turning point of the curve.
(a) Find the value of p and q.
x cm (b) Determine the nature of the turning point (5, –3)
by using the second derivative.
Given that the perimeter of the cardboard is 50 cm.
Find the length x cm and width y cm such that the 3. Give the equation of a curve is
volume of the cylinder formed is maximum.
y= 3 – x(2 – x2).
17. The diagram below shows a closed right cylinder 4
with a fixed height of 8 cm and a changing radius
of base. (a) Find the gradient function of the curve.

(b) Find the coordinates of the turning points.

(c) Hence, determine whether each of the turning

points is maximum or minimum.

8 cm 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.

(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 .
x2
Given that the radius of the base changes with a SPM dy
rate of 0.3 cm s–1 when the radius is 5 cm. 2017 (a) Find the value of dx when x = 5.
Estimate
(a) the rate of change of the total surface area of (b) Hence, estimate the value of – 4 .
the cylinder, (5.02)2
(b) the rate of change of the volume of the cylinder.
6. The diagram below shows a filter funnel in a cone-
18. Two variables x and y are related by the equation shape with height 20 cm and a top circular surface
yx2 = 25. Express in terms of p, the small change radius of 8 cm. Ezza pours oil into the filter funnel
in y when x changes from 5 to 5 – p, such that p is with a constant rate of 8 cm3 s–1.
a small value.

19. Given that the equation of a curve is y = √x . x
dy
SPM (a) Find the value of dx when x = 16. Oil h
2017
Given that the surface radius of the oil in the filter
(b) Hence, estimate the value of √15.96 . funnel is x cm and the height of the oil is h cm.
(a) Express the volume of the oil, V cm3, in terms of
20. Given that A = 3q – 2q2 and x = 4q – 3. h.
dA (b) Hence, calculate the rate of change of the
SPM (a) Express dx in terms of x. height of the oil in the cone when the surface
2018 radius is 2 cm.

(b) Find the small change in x when A changes
from 4 to 3.9 when q = –1.

47

  Additional Mathematics  Form 5  Chapter 2 Differentiation

7. Given a formula H = 3 which is obtained from 10. (a) The diagram below shows a pattern in the
an experiment. 5 + g2 shape of a square PQRS with side of 8 cm.

(a) Express dH in terms of g. PQ
dg M
(b) Ali read the measurement g from a faulty

instrument with error and recorded the reading

as 2.98 instead of the real reading of 3. Find the S NR
estimated error in the measurement of H.

Chapter 2 8. (a) The straight line x – y = 1 is the normal to M and N are two points lie on the side PS
the curve y = x3 2 6 and SR respectively such that PM = x cm and
– 4x2 + 5x 25 X. SN = 3x cm.
– 3 27 at point (i) Express the area of triangle QMN in terms
of x.
Find (ii) Hence, find the minimum area of triangle
(i) the coordinates of point X. QMN. State the value of x when this
minimum area occurs.
(ii) the equation of the tangent to the curve at
point X.

(b) In an experiment concerning the focal length of (b) Given y = 5 and s = 8 – 3x. Find
a lens, an object is placed u cm from a lens 4s2
and the image distance v cm is measured. It
is given that u and v are related by the formula

1 + 1 = 1 . (i) the rate of change of the value s when x
u v 6
du 1
(i) Express dv in terms of v. changes with a rate of 3 unit s­–1,

(ii) Hence, find the rate of change of v when (ii) the approximate change in the value of y
u increases with a rate of 1.8 cm s–1 when when x increase from 2.50 units to 2.56
u = 5 cm.
units.

9. (a) The diagram below shows a hemispherical 11. The diagram below shows the front elevation
container with a radius of 12 cm.
SPM of part of the rail of a roller coaster in an indoor
2018
12 cm themepark inside a shoping mall.

Ceiling

h cm

10 cm

Izzatul pours a type of liquid into the container x Floor
such that the height, h cm, of the liquid in the

container increases with a rate of 0.8 cm s–1.

(i) Express the area, in cm2, of the liquid The curve part of the roller coaster is
surface in the container in terms of h.
represented by the equation y = 1 x3 – 1 x2 – 6x,
(ii) Hence, find the rate of change of the 3 2
such that the point x is the origin. Find the shortest
surface area of the liquid when the height

of the liquid is 8 cm. vertical distance, in m, from the rail to the ceiling of

(b) In the laboratory, it is discovered that the the building.
duration of the oscillation, T seconds of an

oscillating object which is hung by using a 12. The curve y = hx4 – 5x has a turning point (2, k).
string, l cm in length is given by Find the values of h and k.
SPM
T = 2π l 2019
10
dT 13. The curve y = 2x3 + 3x2 – 12x – 15 passes through
(i) Find dl . point P(–3, –6) and has turning points A(1, –22)
and B. Find
(ii) Estimate the small change in the length (a) the gradient of the curve at point P.
(b) the equation of the normal to the curve at point
of the string when the duration of the P.
oscillation changes from 8π seconds to (c) the coordinates of point B and determine the
8.02π 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.

48

PELANGI   Additional Mathematics  Form 5  Answers

ANSWERS

1Chapter Circular Measure 8. (a) 1.619 cm2 (b) 44.07 cm2
(c) 483.44 cm2 (b) 6.81 cm2

Try This! 1.1 9. (a) 214.3 cm2

10. 5 764.185 cm2

1. (a) 19 π rad (b) 1 π rad Try This! 1.4
36 8
1. (a) 14.664 cm (b) 5.8656 rad
(c) 5 π rad (c) 18.33 cm2 (b) 46.68 m
16 (b) 31.42 cm2
2. (a) 0.9335 rad (b) 205.5 cm
2. (a) 2.112 rad (b) 0.8853 rad (c) 162.2 m2

(c) 1.432 rad 3. (a) 26.28 cm
(c) 251.36 cm3
3. (a) 103.1° (b) 76.38°
(c) 295.6° 4. (a) 4.102 rad
(c) 332 cm2
4. (a) 108° (b) 45°
(c) 308.6°

Try This! 1.2 SPM Practice 1

1. (a) 2.64 cm (b) 8.56 cm Paper 1
(c) 6.647 cm

2. (a) 5.530 cm (b) 11.84 cm 1. j= 24
(c) 11.83 cm 6t + 1

3. (a) 5.080 cm (b) 8.286 cm 2. (a) 3k rad (b) 24k2 – 27k cm2
(c) 3.793 cm 2

4. (a) 5.729 cm (b) 4.788 cm 3. (a) r = 4 cm (b) 8 + 4a2 cm
(c) 3.024 cm a a

4. 0.785 rad

5. (a) 0.5192 rad (b) 2.022 rad 5. (a) 20.95 cm (b) 4.667 rad
(c) 4.058 rad (b) (πr 2 − 5r) cm2
6. (a) a = 5 rad
r
6. (a) 8.4676 cm (b) 28.76 cm
(c) 14.319 cm 7. 2.265 m2

7. (a) 16.439 cm (b) 18.317 cm 8. (a) 1.231 rad (b) 7.201r cm
(c) 5.89 cm
9. (a) q = 8–r rad (b) 56r – 7r 2 cm2
2r 2
8. (a) 18.146 cm (b) 19.571 cm
(c) 36.29 cm 10. 32.12 m2

9. (a) 27.65 cm (b) 16.77 cm 11. (a) q = 34° (b) 314 cm3

10. 39.625 mm Paper 2

Try This! 1.3 1. (a) r = 3.857 cm, q = 0.5185 rad
(b) 70.783 cm2

1. (a) 4.4042 cm2 (b) 5.4332 cm2 2. (a) 4 cm (b) 35.196 cm
(c) 4.9284 cm2 (b) 7.778 cm2
(b) 10.03 cm 3. (a) 1.974 rad (b) 58.52 cm2
2. (a) 10.14 cm2 (b) 7.775 cm
(c) 5.384 cm2 (b) 3.255 rad 4. (a) 18.63 cm3 (b) 4.025 cm2
(b) 384.1 cm2
3. (a) 3.645 cm (b) 31.69 cm2 5. (a) 83π2 cm
(c) 3.712 cm
(b) Area of major sector PRQM = Area of major sector PSQN
4. (a) 3.235 cm
(c) 7.139 cm = 1 × π2 × 4π
2 3
5. (a) 1.322 rad
(c) 4.086 rad = 2π3
3
6. (a) 4.986 cm2
(c) 2.315 cm2  1 2 1 π π
Area of segment = 2 × π2 × 3 – sin 3
7. (a) 55.04 cm2
(c) 35.36 cm2 = π2 1 π – 1 23 2
2 3

= π3 –  π23
6 4

184

  Additional Mathematics  Form 5  Answers   PELANGI

Area of shaded region 3. (a) 3 1 (b) 10
= (2 × Area of major sector PRQM) – (4 × Area of segment) 2 (d) –1
(c) –
 π23
2π3 π3 4
1 2 1 2 =2 3 –4 6 – (e) 1 (f) 4
4
= 4π3 2π3
3 – 3 + π23 4. (a) 6 (b) –4x
(c) 8x – 4
= 2π3 + π23 (d) 1
3 x2
 2π3 + 3π23
3
1 2 = cm2 5. (a) –24 + 72x (b) 12x + 2
x2

6. (a) 1.26 rad (b) 77.67 cm (c) 0 (d) – 5
8x2
(c) 48.54 cm2

7. (a) 167.51 cm (b) 775.66 cm2 Try This! 2.2

8. (a) 425.5 m2 (b) 48 510 m3 1. (a) 30x5 (b) – 7
(b) 1.896 rad 2x3
9. (a) 12 cm
(c) 384.17 cm2 (c) 1 1
8√x 
(d) 9x   2

10. (a) 280.84 cm (b) 3 153.14 cm2 (e) 3 ­ 3
4 8√x 
11. The volume of a slice of cake B per RM (48 cm3 per RM) x3 (f)
is greater than the volume of a slice of cake A per
RM(25.14 cm3 per RM). Therefore, the price of a slice of 2. (a) 12x2 + 10x – 8 (b) 15x4 – 3
4x3
cake B is more reasonable.
1 4 9 16 (d) 4x3 + 15x4 – 6x5
(c) – x2 – x3 + x4 – x5

2Chapter Differentiation (e) 3 – 15 (f) – 3 +1
2√x  √x 

Try This! 2.1 3. (a) 3 (b) 21 1
3
4. (a) 34 πr (b) 235p
1. (a) (i) x –0.1 –0.01 –0.001 0 0.001 0.01 0.1 5. (a) 8x – ­x23
f(x) –0.8 –0.08 –0.008 0 0.008 0.08 0.8 (b) 3

lim 8x (c) –   98 – 42 (d) 5 – 2x
x3 x2
\  x → 0 = 0
   6. 3 3 2 4
(ii) y (iii) xl i→m 0 8x = 8(0) (a) 9 6+ x2 6x – x (b) √8x – 4
= 0

(c) 24x (3x2 – 4)3 (d) = – 9
5
3
(6x – 3) 2
Ox
(e) 3 (f) 5
2(5 – 3x)3 3
2(x – 2) 2

∴ lim 8x = 0 7. –504 210

x → 0

(b) (i) x –0.1 –0.01 –0.001 0 0.001 0.01 0.1 8. 0
f(x) 3.01 3.0001 3.000001 3 3.000001 3.0001 3.01
9. k = 5, n = 3

\  lim (x2 + 3) = 3 10. (a) (10x – 1)(2x – 1)3
3x + 1
x → 0 (b) √2x – 3

(ii) y (c) (2x + 7)(x – 5)2(10x + 1)

(d) (99x – 2)(9x – 2)4
2√x 
(e) 3(5 + x)3[5 – 5x – 6x2]
3
Ox (f) 40x4 – 24x2 – 24x + 6 + 2
x3

\ lim (x2 + 3) = 3 11. (a) 1(02xx(x– –1)12) (b) –   1 – 1)2
√x (√x 
x → 0

( iii) xl i→m 0 (x2 + 3) = 02 + 3 (c) 3x – 8 (d) –3(1 + xx)(41 – 3x)3
= 3 2x3√2 – x 

2. (a) –2 (b) 3 (e) 6 – 32x – 37x2 (f) –2(x3(8– 12x2 + 5)
(2 + 5x – x2)2 – x)2
(c) 5 (d) 2
3
3
(e) –   5

185

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