Ranger SPM 2022 - Additional Mathematics

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4.5 KSSM KC118332 PELANGI





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KC118332
ISBN: 978-967-2779-83-4 Tee Hock Tian
Moh Sin Yee

PELANGI Penerbitan Pelangi Sdn. Bhd. (89120-H) Dr. Pauline Wong Mee Kiong




02.indd 1 01/03/2022 3:53 PM

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Quick Revision SPM
Quick Revision




Form


4.5 KSSM



Additional




Mathematics





Chew Su Lian
Ong Yunn Tyug
Tee Hock Tian
Moh Sin Yee
Dr. Pauline Wong Mee Kiong


















PELANGI Penerbitan Pelangi Sdn. Bhd. (89120-H)







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First Published 2022

































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Formulae .................................................vi–viii

Form 4 4 Indices, Surds and
Logarithms 45
1 Functions 1
Concept Map ..............................................45
Concept Map .................................................1 4.1 Laws of Indices .............................46
1.1 Functions ........................................... 2 4.2 Laws of Surds ................................47
1.2 Composite Functions ....................... 5 4.3 Laws of Logarithms .......................51
1.3 Inverse Functions ........................... 7 4.4 Applications of Indices, Surds
Mastery SPM ............................................... 8 and Logarithms ..............................55
Example of HOTS Question ................. 11 Mastery SPM .............................................55
SPM Practice ..............................................12 Example of HOTS Question ................57
SPM Practice .............................................59
2 Quadratic Functions 14

Concept Map ...............................................14 5 Progressions 61
2.1 Quadratic Equations and
Inequalities ......................................15 Concept Map ...............................................61
2.2 Types of Roots of Quadratic 5.1 Arithmetic Progressions ..............62
Equations ..........................................18 5.2 Geometric Progressions ...............66
2.3 Quadratic Functions ......................19 Mastery SPM .............................................70
Mastery SPM .............................................27 Example of HOTS Question .................71
Example of HOTS Question ................28 SPM Practice .............................................73
SPM Practice .............................................29

3 Systems of Equations 33 6 Linear Law 77

i-THINK Gallery.......................................33 Concept Map ..............................................77
3.1 Systems of Linear Equations in 6.1 Linear and Non-Linear
Three Variables .............................34 Relations ..........................................78
3.2 Simultaneous Equations involving 6.2 Linear Law and Non-Linear
One Linear Equation and One Relations ...........................................81
Non-Linear Equation ....................37 6.3 Applications of Linear Law .........82
Mastery SPM .............................................39 Mastery SPM .............................................84
Example of HOTS Question .................41 Example of HOTS Question ................86
SPM Practice .............................................43 SPM Practice .............................................87



iii iii





0a Content Ranger Mate Tambahan SPM.indd 3 25/02/2022 9:04 AM

7 Coordinate Geometry 92 Mastery SPM ........................................... 147
Example of HOTS Question .............. 150
i-THINK Gallery.......................................92 SPM Practice ........................................... 152
7.1 Divisor of a Line Segment ..........93
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7.2 Parallel Lines and Perpendicular Form 5
Lines ..................................................94
7.3 Areas of Polygons .........................96 1 Circular Measure 162
7.4 Equations of Loci ...........................97
Mastery SPM .............................................98 Concept Map ............................................ 162
Example of HOTS Question ...............101 1.1 Radian ............................................. 163
SPM Practice ........................................... 102 1.2 Arc Length of a Circle ............... 163
1.3 Area of Sector of a Circle ..... 165
8 Vectors 105 1.4 Application of Circular
Measures ....................................... 166
i-THINK Gallery..................................... 105 Mastery SPM ........................................... 168
8.1 Vectors........................................... 106 Example of HOTS Question ...............171
8.2 Addition and Subtraction of SPM Practice ........................................... 173
Vectors........................................... 108
8.3 Vectors in a Cartesian Plane ..... 111 2 Differentiation 179
Mastery SPM ............................................112 Concept Map ............................................ 179
Example of HOTS Question ...............118 2.1 Limit and its Relation to
SPM Practice ............................................119 Differentiation ............................ 180
2.2 The First Derivative ...................181
9 Solution of Triangles 124 2.3 The Second Derivative .............. 184
2.4 Application of Differentiation .... 185
Concept Map ............................................ 124
9.1 Sine Rule ........................................ 125 Mastery SPM ........................................... 188
Example of HOTS Question ...............191
9.2 Cosine Rule .................................... 128 SPM Practice ........................................... 192
9.3 Area of a Triangle ...................... 130
9.4 Application of Sine Rule, Cosine 3 Integration 196
Rule and Area of a Triangle ..... 132
Mastery SPM ........................................... 133 Concept Map ............................................ 196
Example of HOTS Question .............. 135 3.1 Integration as the Inverse of
Differentiation ............................ 197
SPM Practice ........................................... 137 3.2 Indefinite Integral ..................... 198


10 Index Numbers 142 3.3 Definite Integral .........................200
3.4 Application of Integration ........207
Concept Map ............................................ 142 Mastery SPM ...........................................208
10.1 Index Numbers ............................ 143 Example of HOTS Question ...............211
10.2 Composite Index.......................... 144 SPM Practice ........................................... 212

iv






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4 Permutation and 7 Linear Programming 265
Combination 216
Concept Map ............................................265
Concept Map ............................................ 216 7.1 Linear Programming Model ........266
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4.1 Permutation ................................... 217 7.2 Application of Linear
4.2 Combination ...................................220 Programming .................................269
Mastery SPM ...........................................223 Mastery SPM ...........................................270
Example of HOTS Question ..............224 Example of HOTS Question ..............272
SPM Practice ...........................................225 SPM Practice ...........................................274


5 Probability Distribution 228 8 Kinematics of Linear
Motion 278
Concept Map ............................................228
5.1 Random Variable ..........................229 Concept Map ............................................278
5.2 Binomial Distribution .................. 231 8.1 Displacement, Velocity and
5.3 Normal Distribution ....................234 Acceleration as a Function of
Time ................................................279
Mastery SPM ...........................................238 8.2 Differentiation in Kinematics
Example of HOTS Question .............. 241 of Linear Motion ..........................279
SPM Practice ...........................................242 8.3 Integration in Kinematics of
Linear Motion ...............................282
6 Trigonometric Functions 247 8.4 Application of Kinematics of
Linear Motion ...............................282
Concept Map ............................................247 Mastery SPM ...........................................284
6.1 Positive Angles and Negative Example of HOTS Question ..............287
Angles .............................................248
6.2 Trigonometric Ratios of any SPM Practice ...........................................287
Angle ...............................................249
6.3 Graphs of Sine, Cosine and SPM Model Paper ...................................290
Tangent Functions ....................... 251
6.4 Basic Identities ...........................254
6.5 Addition Formulae and Double Answers ......................................................306
Angle Formulae ............................255
6.6 Application of Trigonometric
Functions .......................................256
Mastery SPM ...........................................258
Example of HOTS Question .............. 261
SPM Practice ...........................................262






v






0a Content Ranger Mate Tambahan SPM.indd 5 25/02/2022 9:04 AM

CHAPTER 2 QUADRATIC FUNCTIONS Form 4
Additional Mathematics SPM Chapter 2 Quadratic Functions






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Learning Area: Algebra


Form 4 CONCEPT MAP









Quadratic Functions






Quadratic Inequality Quadratic Equation Quadratic Functions
ax + bx + c = 0
2

Method of solving • General form
2
• Graph sketching f(x) = ax + bx + c
• Table • Vertex form
• Number line f(x) = a(x – h) + k
2
• Intercept form
f(x) = a(x – p)(x – q)
Forming equation
x – (α + β)x + αβ = 0
2
The effect of changes
in a, b and c towards
the shape and position
of graph.
f(x) = ax + bx + c
2
Type of roots Method of solving
• b – 4ac  0 • Completing the
2
Two different real square The effect of changes
roots • Formula in a, h and k towards
the shape and position
b
– 4
2
• b – 4ac = 0 x = –b ±ac of graph.
2
Two equal real roots 2a f(x) = a(x – h) + k
2
• b – 4ac  0
2
No real roots
1414





02 Ranger Add Mathematics Tg4.indd 14 25/02/2022 9:10 AM

Additional Mathematics SPM Chapter 2 Quadratic Functions

2.1 Quadratic Equations and Inequalities



1. The general form of the quadratic (b) Formula method
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equation is ax + bx + c = 0 where a, –b ±
2
b
4
ac

–
2
b and c are constants. x = 2a
2. Method for solving quadratic 3. If α and β are the roots of a quadratic
equations: equation then, x – (α + β)x + αβ = 0, Form 4
2
(a) Method of completing the where α + β is the sum of the roots
square and αβ is the product of the roots.
Make sure the coefficient of x is
2
1 before we start completing the
square.
4. Method to determine the range for a quadratic inequality:
(a) Graph sketching
For a quadratic equation in the form of (x – a)(x – b) = 0, where a  b:
If (x – a)(x – b)  0, then x  a or x > b.
If (x – a)(x – b)  0, then a  x  b.
(b) Number line
Select any integer x  a Select any integer x Select any integer x  b
to test whether the value that satisfies a < x < b to test whether the value
of (x – a)(x – b) is positive to test whether the value of of (x – a)(x – b) is positive
or negative. (x – a)(x – b) is positive or or negative.
negative.
a b x
x  a a  x  b x  b
(c) Table
Range of values of x
x  a a  x  b x  b
(x – a) – + +
(x – b) – – +
(x – a)(x – b) + – +



Example 1 Solution
Solve the following quadratic equations by (a) x – 8x + 4 = 0
2
using completing the square method. x – 8x = –4
2
(a) x – 8x + 4 = 0 2   2   2
2
–8
–8
(b) –2x – 7x + 5 = 0 x – 8x + 2 = –4 + 2
2
x – 8x + (–4) = –4 + (–4) 2
2
2
(x – 4) = 12
2
x – 4 = ±12
x = 0.5359 or x = 7.4641
15


02 Ranger Add Mathematics Tg4.indd 15 25/02/2022 9:10 AM

Additional Mathematics SPM Chapter 2 Quadratic Functions

(b) –2x – 7x + 5 = 0 Solution
2
7

–2 x + x – 5  = 0 (a) Sum of roots (SOR) = 4 + (–7)
2
2
2
7
x + x = 5 = –3
2
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2 2 Product of roots (POR) = 4 × (–7)
= –28
7
5
7
7
x + x +   2 = +   2 The quadratic equation is
2
2
4
4
2
Form 4  x + 7  2 = 89 x – (SOR)x + (POR) = 0
2
x – (–3)x + (–28) = 0
2
4
16
2
89
7
x + = ±  x + 3x – 28 = 0
4 16 (b) Sum of roots (SOR)
x = 0.6085 or x = –4.1085 1 3 13
 
= – + – = –
2 7 14
Example 2 Product of roots (POR)
3
1
3
 
Solve the equation –7x – 6x + 8 = 0 using = – × – = 14
2
7
2
the formula: The quadratic equation is
ac
–b ± 2
b
– 4
2
x = 2a x – (SOR)x + (POR) = 0

x – – 13  x + 3 = 0
2
Solution 14 14
14x + 13x + 3 = 0
2
a = –7, b = –6, c = 8
(–6)
–(–6) ± – 4(–7)(8)
2
x = 2(–7)
260
6 ± Alternative Method
x =
–14 (a) (x – α)(x – β) = 0
260
260
6 –  6 +  (x – 4)[x – (–7)] = 0
x = or x = (x – 4)(x + 7) = 0
–14 –14 x + 3x – 28 = 0
2
x = 0.7232 or x = –1.5803 (b) (x – α)(x – β) = 0
1
3
      = 0
x – –
x – –
2
7
 x + 1  x + 3   = 0
2
7
Checking answers using 13 3
2
calculator x + 14 x + 14   = 0
INFO 14x + 13x + 3 = 0
2
Example 3 Example 4
Form a quadratic equation using the given
roots. Given α and β are the roots of quadratic
2
(a) 4 and –7 equation x – 2x – 8 = 0. Form a quadratic
1
3
(b) – and – equation that has the roots 2α and 2β.
2 7
16
02 Ranger Add Mathematics Tg4.indd 16 25/02/2022 9:10 AM

Additional Mathematics SPM Chapter 2 Quadratic Functions

Solution
2
x – 2x – 8 = 0 New SOR: 2α + 2β = 2(α + β)
b
α + β = – = 2(2)
a
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= – (–2) = 4
1 New POR: (2α)(2β) = 4αβ
= 2 = 4(–8)
αβ = c TIPS = –32
a
= –8 Sum of roots, α + β = − b a The new quadratic equation is Form 4
2
1 c x – (2α + 2β)x + (2α)(2β) = 0
= –8 Product of roots, αβ = a x – 4x + (–32) = 0
2
x – 4x – 32 = 0
2

Example 5
Find the range of values of x for each of Table method
the following quadratic inequalities using Range of values of x
the method of graph sketching, number
line or table. x  –4 –4  x  3 x  3
(a) x + x – 12  0 (x – 3) – – +
2
(b) (2 – 3x)(x + 5)  x + 25 (x + 4) – + +
(x – 3)(x + 4) + – +
Solution
The range of values of x is < –4 and
(a) x + x – 12 > 0 x > 3.
2
Graph sketching method
Let x + x – 12 = 0 (b) (2 – 3x)(x + 5)  x + 25
2
2
(x – 3)(x + 4) = 0 –3x – 13x + 10  x + 25
2
x = 3 or x = –4 –3x – 14x – 15  0
y (–3x – 5)(x + 3)  0
+ + x Graph sketching method
–4 – 0 3 Let (–3x – 5)(x + 3) = 0
5
x = – or x = –3
3
The range of values of x is < –4 and
x > 3.
Number line method + x
– –3 5 –
x = –5, x = 0, x = 4, – 3
(–5 – 3)(–5 + 4) > 0 (0 – 3)(0 + 4) < 0 (4 – 3)(4 + 4) > 0
5
The range of values of x is –3  x  – .
x 3
+ –4 – 3 +
The range of values of x is < –4 and
x > 3.


17





02 Ranger Add Mathematics Tg4.indd 17 25/02/2022 9:10 AM

Additional Mathematics SPM Chapter 2 Quadratic Functions
Number line method Table method

x = –2, Range of values of x
x = –4, [–3(–2) – 5](–2 + 3) ≥ 0 x = 0,
[–3(–4) – 5](–4 + 3) ≤ 0 [–3(0) – 5](0 + 3) ≤ 0 5 5
x  –3 –3  x  – 3 x  – 3
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x
– + – (–3x – 5) + + –
–3 – 5
3 5 (x + 3) – + +
Form 4 The range of values of x is –3  x  – . (–3x – 5)(x + 3) – + – 5
3
The range of values of x is –3  x  – .
3

2.2 Types of Roots of Quadratic Equations



2
2
Discriminant b – 4ac  0 b – 4ac = 0 b – 4ac < 0
2
Two different real Two equal real roots No real roots
Type of roots
roots
Example 6 Example 7
Determine the type of roots for each of the The quadratic equation –5x – x = 2 – m
2
following quadratic equations. has two different real roots. Find the range
(a) x – 4x + 3 = 0 of values of m.
2
(b) –4x + 20x – 25 = 0
2
(c) 2x + 4x + 5 = 0 Solution
2
–5x – x – 2 + m = 0
2
Solution a = –5, b = –1, c = –2 + m
(a) x – 4x + 3 = 0 b – 4ac  0
2
2
a = 1, b = –4, c = 3 (–1) – 4(–5)(–2 + m)  0
2
b – 4ac = (–4) – 4(1)(3) 1 – 40 + 20m  0
2
2
= 4  0 20m  39
∴ The equation has two different real m  39
roots. 20
2
(b) –4x + 20x – 25 = 0
a = –4, b = 20, c = –25 Example 8
2
b – 4ac = 20 –4(–4)(–25) The quadratic equation hx + kx – 7 = 0
2
2
= 0 has two equal real roots. Express h in terms
∴ The equation has two equal real of k.
roots.
(c) 2x + 4x + 5 = 0 Solution
2
a = 2, b = 4, c = 5 b – 4ac = 0
2
2
b – 4ac = 4 – 4(2)(5) k – 4(h)(–7) = 0
2
2
2
= –24  0 k + 28h = 0 k 2
∴ The equation has no real roots. h = – 28
18
02 Ranger Add Mathematics Tg4.indd 18 25/02/2022 9:10 AM

Additional Mathematics SPM Chapter 2 Quadratic Functions

2.3 Quadratic Functions



1. Changes in
the values Effect towards the shape and position of graph
of a, b and f(x) = ax + bx + c
2
c
a  0 and The shape of the 0Reserved.
f(x)
a > 1
the value of a graph is and the Form 4
increases from 1. width of the graph a = 1
decreases. x
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a  0 and The shape of the f(x) a = 1
the value of a graph is and the
decreases from 1 width of the graph
and approaches 0. increases. 0 < a < 1
x
0
Only the
value of a a  0 and The shape of the f(x)
changes the value of a graph is and the
decreases from −1. width of the graph 0 x
decreases. a = –1



a < –1
a  0 and The shape of the f(x)
the value of a graph is and the
increases from −1 width of the graph 0 x
and approaches 0. increases.
–1< a < 0

a = –1
a  0 and b  0 The vertex of the f(x)
graph is on the left b > 0 b = 0
side of y-axis.



Only the 0 x
value of b
changes a  0 and b  0 The vertex of the f(x)
graph is on the right b = 0 b < 0
side of y-axis.

x
0


19





02 Ranger Add Mathematics Tg4.indd 19 25/02/2022 9:10 AM

Additional Mathematics SPM Chapter 2 Quadratic Functions

a  0 and b  0 The vertex of the f(x)
graph is on the right
side of y-axis. 0 x
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Only the b = 0 b > 0
value of b a  0 and b  0 The vertex of the
Form 4 changes graph is on the left f(x) x
side of y-axis.
0


b < 0 b = 0
c  0 The shape of the graph does not change but
the graph moves vertically upwards.
f(x)
f(x) c > 0
c > 0
x
c = 0 0
c = 0
Only the 0 x
value of c
changes c  0 The shape of the graph does not change but
the graph moves vertically downwards.
f(x) f(x)
x
c = 0 0
c = 0
x
0
c < 0 c < 0


2. Discriminant b – 4ac  0 b – 4ac = 0 b – 4ac  0
2
2
2
Two different real Two equal real roots No real roots
Type of roots
roots
a  0 a  0 a  0 a  0 a  0 a  0
a > 0 a < 0 a > 0 a < 0 a > 0 a < 0
x x
x x
Position of the x x
graph of f(x)
The graph intersects The graph touches The graph does not
x-axis at two x-axis at one point intersect or touch
different points. only. x-axis.




20





02 Ranger Add Mathematics Tg4.indd 20 25/02/2022 9:10 AM

Additional Mathematics SPM Chapter 2 Quadratic Functions

3. Expansion Factorisation or formula
Vertex form General form Intercept form
f(x) = a(x – h) + k f(x) = ax + bx + c f(x) = a(x – p)(x – q)
2
2
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Completing the Expansion
square
4. Changes in the values of Effect towards the shape and position of graph Form 4
a, h and k f(x) = a(x – h) + k
2
a  0 and The shape of the f(x) a = 1
the value of a graph is and the a > 1
increases from 1. width of the graph
decreases. 0 < a < 1

a  0 and The shape of the
the value of graph is and the 0 x
a decreases width of the graph
from 1 and increases.
Only the
value approaches 0.
of a The shape of the
changes a  0 and f(x)
the value of a graph is and the
decreases from width of the graph 0 x
−1. decreases.
–1 < a < 0
a  0 and The shape of the
the value of graph is and the a < –1
a increases width of the graph a = –1
from −1 and increases.
approaches 0.
The shape of the graph does not change but the graph moves horizontally
• to the right when the value of h increases
• to the left when the value of h decreases
f(x) When the value When the value of
x = h x = h of h decreases, h increases, the
Only the
value the graph moves f(x) graph moves to
to the left.
the right.
of h x
changes 0
0 x
When the value When the value of h
of h decreases, increases, the graph
the graph moves to the right.
moves to the lef x = h x = h





21





02 Ranger Add Mathematics Tg4.indd 21 25/02/2022 9:10 AM

Additional Mathematics SPM Chapter 2 Quadratic Functions

The shape of the graph does not change but the graph moves vertically
• upwards when the value of k increases
• downwards when the value of k decreases
f(x)
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Only the When the value k > 0 When the value f(x)
of k increases,
value of k increases, k = 0 the graph moves k > 0
the graph moves
of k
upwards.
Form 4 changes upwards. 0 x When the value of 0 x
When the value of
k decreases, the
k decreases, the
graph moves k < 0 graph moves k = 0
downwards.
downwards. k < 0
5. The quadratic function f(x) = a(x – h) + k has a vertex (h, k) and is symmetrical about
2
the line x = h.
• When a  0, the vertex (h, k) is the minimum point and TIPS
k is the minimum value.
• When a  0, the vertex (h, k) is the maximum point and h = − b and
k is the maximum value. 2a b 2
k = c −
6. Sketch the graph of a quadratic function 4a

Determine Determine Determine Determine Determine Draw a
the shape the position the the the symmetrical
of the of the graph. maximum intersection intersection smooth
graph. or minimum point with point with parabola.
Find the value point of the x-axis. y-axis.
Identify of b – 4ac. graph. Symmetrical
2
the value Solve Find the graph at
of a. Express f(x) = 0 value of f(0). x = h and
f(x) = a(x – h) + k passes
2
where (h, k) is the through
vertex. vertex.


Example 9
Make generalisations on the shape and
The diagram below shows the graph of position of the graph when the value of a
f(x) = x – 5x + 6 where a = 1, b = –5 and changes to
2
c = 6. 1
(a)
f(x) 2
(b) 2

f(x) = x – 5x + 6 Hence, sketch the new graph of f(x).
2
6
x
0


22





02 Ranger Add Mathematics Tg4.indd 22 25/02/2022 9:10 AM

Additional Mathematics SPM Chapter 2 Quadratic Functions

Solution Solution
1
(a) When a changes from 1 to , the width (a) When b changes from 2 to –2, the
2 vertex is to the left of the y-axis. The
of the graph increases. The y-intercept shape of the graph and the y-intercept
f(x)Pelangi Sdn Bhd. All Rights Reserved.
does not change. do not change.
2
f(x) f(x) = x – 5x + 6 f(x)
2
15 f(x) = –x + 2x + 15 Form 4
f(x) = –x – 2x + 15
2
6 0 x
x
0
f(x) = 1 2 x – 5x + 6
2
(b) When c changes from 15 to 5, the graph
(b) When a changes from 1 to 2, the width moves 10 units vertically downwards.
of the graph decreases. The y-intercept The y-intercept is 5. The shape of the
does not change. graph does not change.
2
f(x) = 2x – 5x + 6 f(x)
f(x)
15 f(x) = –x + 2x + 15
2
f(x) = x – 5x + 6 5
2
6 0 x
x
0
f(x) = –x + 2x + 5
2


Example 10
The diagram below shows the graph of
Quiz
2
f(x) = –x + 2x + 15 where a = –1, b = 2 Quiz .
Quiz
Quiz
and c =15.
Penerbitan 15 0 f(x) = –x + 2x + 15 Example 11
In Example 10, if b = 0, the vertex is
at
2
x
Determine the type of roots for the
Make generalisations on the shape and
position of the graph when
f(x) = 0. Then, sketch the graph and make
(a) the value of b changes to –2.
a generalisation about the position of the
(b) the value of c changes to 5. following quadratic functions when
graph on the x-axis.
2
Hence, sketch the new graph of f(x). (a) f(x) = 3 – 6x + 2x
(b) f(x) = –9x + 12x – 4
2
23





02 Ranger Add Mathematics Tg4.indd 23 25/02/2022 9:10 AM

Additional Mathematics SPM Chapter 2 Quadratic Functions

Solution Example 13
(a) f(x) = 2x – 6x + 3 If the graph of f(x) = 2x + (1 – p)x + 8,
2
2
where p is a constant, does not intersect the
b – 4ac = (–6) – 4(2)(3) x-axis, find the range of values of p.
2
2
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= 12  0
This function has two different real Solution
Form 4 a  0, so the shape of the graph f(x) f(x) = 2x + (1 – p)x + 8 The graph does not
roots.
2
b – 4ac  0
2
intersect the x-axis.
and it intersects the x-axis at two
is
2
points. (1 – p) – 4(2)(8)  0
1 – 2p + p – 64  0
2
p – 2p – 63  0
2
x (p + 7)(p – 9)  0
+ +
x
2
(b) f(x) = –9x + 12x – 4 –7 – 9
b – 4ac = 12 – 4(–9)(–4)
2
2
= 0 –7  p  9
This function has two equal real roots.
a  0, so the shape of the graph f(x) Example 14
is and it intersects the x-axis at one
point. Express the quadratic function
x f(x) = 3 x – 3  2 – 147 in intercept form,
4
2
f(x) = a(x – p)(x – q) where a, p and q are
constants and q  p. State the values of
a, p and q.
Solution
Example 12 9

f(x) = 3 x – 3x +  – 147
2
If the graph of f(x) = 2x + 2x + 1 – k, 4 4
2
2
where k is a constant, intersects the x-axis = 3x – 9x – 30 General form
2
at two different points, find the range of = 3(x – 3x – 10)
values of k. = 3(x + 2)(x – 5) Intercept form
Compare with f(x) = a(x – p)(x – q).
Solution
Thus, a = 3, p = –2, q = 5
f(x) = 2x + 2x + 1 – k The graph intersects
2
b – 4ac  0 the x-axis at two
2
different points.
2 – 4(2)(1 – k)  0 Example 15
2
4 – 8 + 8k  0 Express f(x) = –2x – 10x – 11 in the form
2
8k  4 f(x) = a(x – h) + k where a, h and k are
2
k  1 constants. Hence, determine the values of
2 a, h and k.
24


02 Ranger Add Mathematics Tg4.indd 24 25/02/2022 9:10 AM

Additional Mathematics SPM Chapter 2 Quadratic Functions

Solution Solution

f(x) = –2 x + 5x + 11  (a) When a changes from –2 to –3, the
2
2
width of the graph decreases. The axis
5
5
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2

  = –2 x + 5x +     2 + 11  of symmetry and the maximum value
–
2
2
2
2
of the graph do not change.
  = –2 x + 5  2 – 25 + 11  f(x)

2
4
2

  = –2 x + 5  2 – 3  –3 0 2 x Form 4
4
2

  = –2 x + 5  2 + 3 f(x) = –2(x – 2) – 3
2
2
2
5 3
Thus, a = –2, h = – and k = .
2 2
2
f(x) = –3(x – 2) – 3
(b) When h changes from 2 to 4, the shape
Alternative Method of the graph does not change but the
graph moves 2 units horizontally to
a = –2, b = −10, c = –11 the right. The equation of the axis
h = – b ; k = c – b 2 of symmetry becomes x = 4 and the
2a 4a maximum value does not change.
(–10) (–10) 2
= – 2(–2) = –11 – 4(–2) f(x)
5 3
= – = x
2
2
–3 0 2 4
f(x) = –2(x – 4) – 3
2
Example 16
f(x) = –2(x – 2) – 3
2
Given the graph of f(x) = –2(x – 2) – 3
2
with conditions a = –2, h = 2 and k = –3. (c) When k changes from –3 to 6, the shape
f(x) of the graph does not change but the
graph moves 9 units vertically upwards.
x The maximum value becomes 6 and the
0 2
–3 axis of symmetry does not change.
f(x)
f(x) = –2(x – 2) – 3
2
6 f(x) = –2(x – 2) + 6
2
x
0 2
Make generalisations on the shape and –3
position of the graph when
(a) the value of a changes to –3.
(b) the value of h changes to 4.
(c) the value of k changes to 6. 2
Hence, sketch the new graph. f(x) = –2(x – 2) – 3
25
02 Ranger Add Mathematics Tg4.indd 25 25/02/2022 9:10 AM

Additional Mathematics SPM Chapter 2 Quadratic Functions

Example 18
Quiz The diagram below shows the graph of
Quiz
Quiz
Quiz
2
In Example 16, if the value of h the quadratic function f(x) = a(x – m) + n,
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changes from 2 to –2, the graph will where a, m and n are constants. The straight
be reflected in . line y = –48 is the tangent to the curve at
point K.
Form 4 Example 17 2 f(x)
Sketch the graph of f(x) = x – 3x + 2.
–6 0 2 x
Solution
a  0, the shape of the graph is . K
b – 4ac = (–3) – 4(1)(2) = 1  0
2
2
The graph of f(x) intersects the x-axis at (a) State the coordinates of the point K.
two different points. (b) Find the value of a.
f(x) = x – 3x + 2 (c) State the equation of the curve if the
2
graph is reflected in the y-axis.
   
= x – 3x + – 3 2 – – 3 2 + 2 Solution
2
2
2
9

= x – 3  2 – + 2 (a) Equation of axis of symmetry,
2
4
–6 + 2

= x – 3  2 – 1 x = 2
4
2
= –2
The minimum point is  3 , – 1  . Then, K(–2, –48)
4
2
2
When f(x) = 0, (b) f(x) = a(x + 2) – 48
x – 3x + 2 = 0 At (2, 0), 2
2
(x – 1)(x – 2) = 0 0 = a(2 + 2) – 48
48 = 16a
x = 1 or x = 2 a = 3
Thus, the graph intersects the x-axis at
x = 1 and x = 2. (c) When reflected in the y-axis, the value
of m changes from –2 to 2.
When x = 0, 2
f(0) = 0 – 3(0) + 2 = 2 Thus, f(x) = 3(x – 2) – 48
2
Thus, the y-intercept is 2.
f(x)
x = 3
2
2
x
0 1 2
3 2 , – 1 4


26





02 Ranger Add Mathematics Tg4.indd 26 25/02/2022 9:10 AM

Additional Mathematics SPM Chapter 2 Quadratic Functions



Mastery

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1. Solve 16 – x = 2x(2x + 3). Give the New SOR:
answer correct to three decimal places. (α + 2) + (β + 2) = α + β + 4
5
= + 4
Solution 3
16 – x = 2x(2x + 3) = 17 Form 4
16 – x = 4x + 6x New POR: 3
2
4x + 7x – 16 = 0 (α + 2)(β + 2) = αβ + 2α + 2β + 4
2
7
2
x = –7 ± – 4(4)(–16) 5
2(4) = –4 + 2   + 4
3
x = –7 ±305 = 10
8 3
x = –3.058 or x = 1.308 The new quadratic equation is
x – (SOR)x + (POR) = 0
2
2. Find the range of values of x where the 17 10
quadratic function f(x) = 12 – 4x – x is x – 3 x + 3 = 0
2
2
negative. 3x – 17x + 10 = 0
2
Solution 4. The diagram below shows a graph of
2
f(x) = 12 – 4x – x  0 quadratic function f(x) = –(x – h) + k
2
2
–x – 4x + 12  0 where h and k are constants.
x + 4x – 12  0 f(x)
2
(x – 2)(x + 6)  0
+ +
x
–6 – 2 –3 0 11 x

Find
x  –6 or x  2 (a) the equation of axis of symmetry,
3. Given that α and β are the roots of the (b) the maximum point.
quadratic equation 3x – 5x – 12 = 0. Solution
2
Form a new quadratic equation with the
roots α + 2 and β + 2. (a) Equation of axis of symmetry:
x = –3 + 11
Solution 2
= 4
3x – 5x – 12 = 0 (b) f(x) = –(x – 4) + k
2
2
b (–5) 5
α + β = – = – = At (11, 0),
a 3 3
2
c –12 0 = –(11 – 4) + k
αβ = = = –4 0 = –49 + k
a 3
k = 49
The maximum point is (4, 49).
27





02 Ranger Add Mathematics Tg4.indd 27 25/02/2022 9:10 AM

Additional Mathematics SPM Chapter 2 Quadratic Functions
5. Given a quadratic function Solution
f(x) = –3x + px – 8 has one maximum
2
point (4, q). Find the value of p and (a) When t = 2,
2
of q. h(2) = –4(2) + 24(2) + 28
= 60 m
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Solution
(b) h(t) = –4t + 24t + 28
2
p


2
f(x) = –3 x – x – 8 = –4(t – 6t) + 28 2 2
2
3
Form 4 = –3 x – x + – p 2 – – p 2 – 8 = –4[(t – 3) – 9] + 28
= –4[t – 6t +(–3) – (–3) ] + 28
2
p
    

2
2
3
6
6
= –4(t – 3) + 36 + 28
2
p

2
= –3 x – p  2 – 36 2  – 8 = –4(t – 3) + 64
6
p

= –3 x – p  2 + 12 2 – 8 The height when the ball hits the
ground, h(t) = 0.
6
2
p = 4 –4(t – 3) + 64 = 0
2
6 4(t – 3) = 64
2
p = 24 (t – 3) = 16

t – 3 = +4
2
2
q = p – 8 = 24 – 8 = 40 t = –1 or t = 7
12 12
6. Find the values of m if the straight line Thus, the ball will hit the ground at
y = mx + 8 is the tangent to the curve the 7 second.
th
y = 2x + 3x + 10. (c) Maximum height = 64 m
2
Solution
HOTS Example 1
mx + 8 = 2x + 3x + 10
2
2x + (3 – m)x + 2 = 0 Muzafar has a rectangular piece of paper
2
b – 4ac = 0 Tangent to the with a length of (3x + 2) cm and a width of
2
3x cm. He cuts and removes a square with
curve means
(3 – m) – 4(2)(2) = 0 the straight sides of x cm from the paper. Find the range
2
9 – 6m + m – 16 = 0 line (tangent) of values of x if the area of the remaining
2
touches the
m – 6m – 7 = 0 curve at one paper is at least 90 cm .
2
2
(m + 1)(m – 7) = 0 point only.
m = –1 or m = 7 Solution
2
7. A ball is tossed upwards 3x(3x + 2) – x  90
2
2
from a position as in 9x + 6x – x – 90  0
2
the diagram beside. The 8x + 6x – 90  0
2
height h, in metres, of the 4x + 3x – 45  0
ball at t seconds is given Ground (4x + 15)(x – 3)  0
by a function + +
h(t) = –4t + 24t + 28. x
2
(a) What is the height of the ball when – 15 – 3
t = 2? 4
(b) In which second will the ball hit the 15
ground? x  – 4 (ignored) or x  3
(c) Calculate the maximum height of the Thus, x  3
ball from the horizontal ground.
28
02 Ranger Add Mathematics Tg4.indd 28 25/02/2022 9:10 AM

Additional Mathematics SPM Chapter 2 Quadratic Functions

HOTS Example 2 Solution
The diagram below shows a bucket in the Volume of cylinder
shape of a cylinder which is fully-filled with 22   2
21
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water and another container in the shape of = 7 × 2 × 18
a right prism with a rectangular base ABCD. = 6 237 cm 3
The trapezium ABFE is the uniform cross
section of the right prism. All the water Volume of water in the right prism = 6 237 cm 3
from the bucket is poured into the container 3 1 Form 4
3 × (14 + 10)x × (5x – 27) = 6 237
and filled of the container. 4 2
4 9x(5x – 27) = 6 237
H G x(5x – 27) = 693
2
x cm 5x – 27x – 693 = 0
14 cm D C
(–27)
2
E x = –(–27) ± – 4(5)(–693)
18 cm F (5x – 27) cm 2(5)
A B x = 27 ±14 589
21 cm 10 cm 10
x = –9.378 (ignored) or x = 14.78
22
Using π = , calculate the value of x.
7 Then, x = 14.78




SPM PRACTICE
SPM PRACTICE


Paper 1 5. Find the range of values of p if the
quadratic equation x – 6px + p = 0 has
2
1. Given the quadratic equation no roots.
x + px = 3x – 4 has two different roots, 2
2
find the range of values of k. 6. The quadratic equation px + qx + r = 0
6
3
2. Given 7 is one of the roots of the has roots – and 5 . Find the values of
7
4
quadratic equation 5(x + k) = 405, p, q and r.
2
where k is a constant. Find the values of 7. Given α and β are the roots of the
k. quadratic equation 2x – 9x + 9 = 0.
2
3. Given a quadratic equation Form a quadratic equation that has roots
β
kx – 3hx + 4k = 0 where k and h are α and .
2
constants, has only one x-intercept. Find 3 3
h : k.
8. (a) Express 4x(x – 5) = (3x – 1)(x + 1)
4. Given one of the roots of the equation in general form.
x + (3p – 4)x – p = 0, where p is a (b) Hence, solve the quadratic equation
2
2
constant, is negative to another root. and give the answers correct to 4
Find the value of the product of the significant figures.
roots.
29




02 Ranger Add Mathematics Tg4.indd 29 25/02/2022 9:10 AM

Additional Mathematics SPM Chapter 2 Quadratic Functions
9. The quadratic function f(x) = hx – 3x + k, 16. The diagram below shows a graph of a
2
where h and k are constants, has one quadratic function.
maximum point.
(a) Given h is an integer such that f(x)
–2  h  2, state the value of h. 8
(b) Using the answer in (a), find the 6
range of values of k when the curve x
of the quadratic function intersects
Form 4 the x-axis at two different points. 17. Given a quadratic function Reserved.
0
9
3
–
2
2


10. Given x – y = 3x, find the range of
2
vertex form, f(x) = a(x – h) + k.
2
values of x for y  4. Express the quadratic function in the
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11. Given 4 is one of the roots of the
2
quadratic equation (x – p) = 169, where f(x) = 3(x – 2) – 5 where a = 3, h = 2 and
2
p is a constant. Find the values of p. k = –5. If the graph of f(x) is reflected
in the x-axis, state the changes in the
12. Given a quadratic equation values of a, h and k.
ax + bx + c = 0, where a, b and c are 18. Given the equation 2x + 8x + 5 = 0 has
2
2
constants, has roots β and 3β. Express c roots p and q. Form a quadratic equation
in terms of a and b.
with the roots 2p + 3 and 2q + 3.
13. Given that straight line y = 6x + 1 does 19. Given a function f(x) = x – (1 – k)x + 1,
2
not intersect the curve f(x) = (2k – 3)x where k is a constant, is always negative
2
+ 4x – 6, where k is a constant. Find the when p  k  q. Find the value of p
range of values of k. and of q.
14. The diagram below shows the graph of Paper 2
5

f(x) = 8 x + 3  2 + . 1. The diagram below shows the graph of a
2
4
f(x) quadratic function f(x) which intersects
the x-axis at the origin and (4, 0).
c f(x)
x
0
x
0 4
Calculate the value of c.
15. Express f(x) = –2x – 6x + 9 in the (a) Find the value of a and of h when
2
form a(x – h) + k where a, h and k f(x) is expressed in the form of
2
are constants. Hence, determine the a(x – h) + 5.
2
maximum value or minimum value and (b) Hence, find the range of values of x
the corresponding value of x. 1
if f(x)  –56 .
4

30





02 Ranger Add Mathematics Tg4.indd 30 25/02/2022 9:10 AM

Additional Mathematics SPM Chapter 2 Quadratic Functions
2. The quadratic equation x – 2(6x – p) = 0, (a) State the coordinates of point M.
2
where p is a constant, has two different (b) Find the value of a.
real roots. One of the roots is five times (c) If the curve is reflected in the x-axis,
the root t, where t ≠ 0. state the new equation of the curve.
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(a) Find the value of t and of p.
(b) Hence, form a quadratic equation 7. The quadratic equation k – 3x = x – x + 3,
2
which has the roots of t – 17 and where k is a constant, has roots α and β.
t + 11. (a) Find the range of values of k if
α ≠ β.
3. Given a quadratic equation (b) Given that α + 3 and β + 3 are the Form 4
h(x + 36) = –4kx has two equal roots, roots of another quadratic equation
2
find the ratio h : k. Hence, solve the 4x – hx + 12 = 0, where h is a
2
equation.
constant. Find the value of h and of
4. The diagram below shows the graph of k.
f(x) = –2x – 5x + 12.
2
8. The quadratic equation x = 10(n – x),
2
f(x)
where n is a constant, has roots m and
12 –3m, where m ≠ 0.
(a) Find the value of m and of n.
x (b) Hence, form a quadratic equation
–4 0 3 n n
2 which has the roots and in the
4
5
2
(a) Make generalisations on the shape form of ax + bx + c = 0.
and position of the graph when the
value of b changes from –5 to 5. 9. Given a quadratic function
2
(b) Sketch the new graph. f(x) = –x + 7x – 10.
(c) State an equivalent transformation (a) Express f(x) in the form
2
on the change of the new graph f(x) = a(x + p) + q.
compared to the graph of f(x). (b) Find the minimum or maximum
value of the quadratic function f(x).
5. (a) Find the range of values of x for (c) Sketch the graph of
x – 8x + 12  0 and x – 8x  0. f(x) = –x + 7x – 10 for 0  x  7.
2
2
2
(b) Hence, solve the inequality (d) State the equation of the curve when
–12  x – 8x  0. the graph f(x) is reflected in the
2
6. The diagram below shows the graph of x-axis.
quadratic function f(x) = a(x + f ) + g
2
where a, f and g are constants. The 10. A rectangular piece of paper has a length
straight line y = 9 is a tangent to the and a width, in cm, of 3x and (12 – x)
curve at point M. respectively.
(a) Find the perimeter, in cm, of the
f(x) paper if the area of the paper is
M maximum.
(b) Hence, state the maximum area, in
cm , of the paper.
2
x
–1 0 5



31





02 Ranger Add Mathematics Tg4.indd 31 25/02/2022 9:10 AM

Additional Mathematics SPM Chapter 2 Quadratic Functions
11. A window in the shape of a parabola on (a) Find the range of values of p.
the curved side AED is represented by (b) If the curve is symmetrical about
the function f(x) = – 1 x + 16. HOTS x = 2, find the value of p.
2
HOTS
100
E 14. The diagram below shows a boy is
f(x) = (p – 3)x – 12x + 18Sdn Bhd. All Rights Reserved.
A D throwing a stone upwards. The path
of the stone produces a quadratic
function such that y is the height of the
Form 4 B Width C 50 cm stone and x is the horizontal distance of
HOTS
the stone. HOTS
y (m)
(a) Find the maximum height of the
window. 6
(b) Find the width of the window.
12. A piece of land in the shape of a right-
angled triangle has the longest side y m x (m)
while the other two sides with length 0 6 15
(x + 6) m and (3x – 2) m. Given that the
2
perimeter of the land is 100 m, find the Given f(x) = a(x – p) + q, where a, p
value of x. Hence, determine the area of and q are constants, has a maximum
the land. HOTS point (6, 6). Find
HOTS
(a) the values of a, p and q,
13. The diagram below shows the graph (b) (i) the height of the stone at the
of quadratic function time the stone threw off from
f(x) = ( p – 3)x – 12x + 18. the boy's hand,
2
f(x) (ii) the horizontal distance of the
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stone when the stone is at the
2
same height as (b)(i) again.
x
0























32





02 Ranger Add Mathematics Tg4.indd 32 25/02/2022 9:10 AM

CHAPTER 2 DIFFERENTIATION Form 5
Additional Mathematics SPM Chapter 2 Differentiation






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Learning Area: Calculus




CONCEPT MAP







Differentiation




First principles
dy lim δy Application of differentiation
dx = δx : 0 δx (a) If it is given a curve y = f(x),
lim
= δx : 0 f(x + δx) – f(x) then
(i) Equation of tangent at a point
δx
(x 1 , y 1 ) dy
First derivative y - y 1 = dx (x - x 1 )
dy (ii) Equation of normal at a point
(i) If y = ax , then dx = anx n – 1 (x 1 , y 1 )
n
d 1
(ii) dx [f(x) ± g(x)] y - y 1 = – dy (x - x 1 )
d
d
dx
= dx [f(x)] ± dx [g(x)] (b) Turning point is a maximum point
2
(iii) Chain rule dy = 0 and d y 2  0
dx
dx
dy = dy × du (c) Turning point is a minimum point
dx
du
dx
2
(iv) Multiplication rule dy d y
d dv du dx = 0 and dx 2  0
dx (uv) = u dx + v dx (d) Point of inflection
2
(v) Quotient rule dv dy = 0 and d y 2 = 0 Form 5
du
dx
dx
d u v dx – u dx (e) Small change and approximation
( ) =
dx v v 2 dy
δy  dx × δx
f(x + δx)  y + dy δx or
Second derivative dx dy
d y d dy d f(x + δx)  f(x) + dx δx
2
dx 2 = ( ) or f″(x) = dx [f′(x)]
dx dx
179
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Additional Mathematics SPM Chapter 2 Differentiation

2.1 Limit and Its Relation to Differentiation


1. In the diagram below, if P(x, y) and When δx approaches zero, (δx → 0),
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Q(x + δx, y + δy) are two points very then δy is the approximate value of
δx
close to one another on the curve the gradient of the tangent at P and
where δx and δy are very small δy
lim
increments, then the gradient of PQ is is written as δx : 0 δx . The process of
given by finding the value of the limit is known
(y + δy) – y δy as differentiation.
m PQ = (x + δx) – x = δx
2. Differentiation using the first
y Q(x + δx, y + δy) principles
y + δy
dy lim δy lim f(x + δx) – f(x)
Q 1 = δx : 0 δx =
Q 2 δy dx δx : 0 δx
Q
P (x, y) 3 3. The gradient function of a tangent
y
δx to a curve y = f(x) can be obtained
0 x x + δx x by differentiating the function, that is
dy
dx .


Example 1

lim
Find the value of x : –1 x + 3 TIPS
2x
Use factorisation or rationalising the numerator
Solution or denominator of the function so that the value
2 0
∞
lim x + 3 = –1 + 3 obtained is not undefined such as , or .
0 0
∞
x : –1 2x 2(–1)
= –1
Example 3
Example 2 The diagram below shows the points A and
2
B on the curve y = 2x .
2
lim
Find the value of x : 0 3x – x . y y = 2x 2
2x
Form 5 x : 0 3x – x = x : 0 x(3x – 1) A(x, y) δy
B(x + δx, y + δy)
Solution
2
lim
lim
2x
2x
lim
= x : 0 3x – 1 0 δx x
2
3(0) – 1 State
=
2 (a) the gradient of the chord AB,
1 (b) the gradient function of the curve by
= –
2 using the first principles.
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Additional Mathematics SPM Chapter 2 Differentiation

Solution Example 4
(a) The gradient of the chord AB Find the first derivative of y = 2 using
(y + δy) – y x + 1
= (x + δx) – x the first principles.
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δy
= Solution
δx
2
(b) y = f(x) = 2x 2 y = f(x) = x + 1
δy = f(x + δx) – f(x) δy = f(x + δx) – f(x)
= 2(x + δx) – 2x 2 2 2
2
= 2[x + 2xδx + (δx) ] – 2x 2 = x + δx + 1 – x + 1
2
2
= 4xδx + 2(δx) 2 2(x + 1) – 2(x + δx + 1)
δy = 4xδx + 2(δx) 2 = (x + δx + 1)(x + 1)
δx δx –2δx
= 4x + 2δx = (x + δx + 1)(x + 1)
dy = lim δy δy –2
dx δx : 0 δx δx = (x + δx + 1)(x + 1)
lim
= δx : 0 (4x + 2δx) dy = lim δy
= 4x dx δx : 0 δx
–2
lim
= δx : 0 (x + δx + 1)(x + 1)
–2
Quiz = (x + 1)(x + 1)
Quiz
Quiz
Quiz
Find the value of lim 3x – 1 . = –2 2
x : 0 x (x + 1)


2.2 The First Derivative



1. Given that a, n are constants, u = f(x) and v = g(x).

Function y First derivative Function y First derivative
a 0 uv dv du
u dx + v dx
ax n anx n – 1 u du dv Form 5
v v dx – u dx
v 2
u + v du dv h(u) where dy du
dx + dx u = f(x) du × dx







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02 Ranger Mate Tambahan Tg5.indd 181 25/02/2022 9:23 AM

Additional Mathematics SPM Chapter 2 Differentiation

Example 5 Solution
Find Given u = 3x + 1, then du = 3
2
(a) the first derivative for 3x – 2 with dy dx
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4
4
respect to x y = 5u , then du = 5(4)u 3
d 2 = 20u 3
x
(b)   + 
dx x By using the chain rule,
1
(c) yʹ(x) if y(x) = x – 2 dy dy du
4
2 3x 3 dx = du × dx
3
Solution = 20u × 3
= 60u 3 Substitute the expression of u.
d 3x – 2 d 3x 2 1
2
(a)   =  –  = 60(3x + 1) 3
dx 4 dx 4 2
= 3 (2)x 2 – 1 – 0
4 Example 7
= 3x
2 Differentiate each of the following with
d 2 d 1 respect to x.
2
x
(b)   +  = x + 2x  (a) y = (2x – 3x) 3
–1
2
dx x dx (b) y = (3 – x) (x + 2) 3
2
= 1 x 1 2 – 1 + 2(–1)x –1 – 1
2 Solution
1
–
3
2
= 1 x – 2x –2 (a) Let u = 2x – 3x and y = u
2
2 du dy
= 1 – 2 Then dx = 4x – 3 and du = 3u 2
2x x 2 dy dy du
= ×
d 1 2 dx du dx
2
(c) yʹ(x) =  x – x –3  = 3u (4x – 3)
4
dx 2 3 = 3(2x – 3x) (4x – 3)
2
2
1 2
= (4)x 4 – 1 – (–3)x –3 – 1
2 3 (b) Let u = (3 – x) and v = (x + 2) 3
2
= 2x + 2 Then du = 2(3 – x)(–1)
3
x 4 dx
= –2(3 – x)
dv = 3(x + 2) (1)
2
dx
= 3(x + 2)
2
TIPS By using the multiplication rule,
Form 5 Before differentiate, convert the expression in dy Factorize the
the form
dx
1
1
common terms
dv
to x
p
p
(ii) x to x
(i)
–p
x
= u
p
dx + v du (3 – x)(x +
dx
2) from this
2
= (3 – x) (3)(x + 2) expression.
2
2
Example 6 + (x + 2) (–2)(3 – x)
3
2
dy = (3 – x)(x + 2) [3(3 – x) – 2(x + 2)]
Given u = 3x + 1 and y = 5u , find dx = (3 – x)(x + 2) (5 – 5x)
4
2
in terms of x. = 5(3 – x)(x + 2) (1 – x)
2
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02 Ranger Mate Tambahan Tg5.indd 182 25/02/2022 9:23 AM

Additional Mathematics SPM Chapter 2 Differentiation

Example 8 (b) Let u = (x + 1) and v = (x + 3) 1 2
2
dy du
Find for each of the following. Then, = 2(x + 1) and
dx dx
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1
(a) y = x – 2 dv = (x + 3) – 1 2
3 – x 2 dx 2
(x + 1) 2
(b) y = = 1
x
 + 3 2x + 3
Solution v du – u dv
(a) Let u = x – 2 and v = 3 – x 2 dy = dx 2 dx
du dx v
Then, = 1 and 2
dx  + 3(2)(x + 1) – (x + 1)
x
dv 2 + 3
x
dx = –2x =  + 3 2
x
By using the quotient rule, 2(x + 3)(2)(x + 1) – (x + 1) 2
du dv =
x
dy v dx – u dx 2(x + 3) + 3
=
dx v 2 (x + 1)(4x + 12 – x – 1)
(3 – x )(1) – (x – 2)(–2x) =
2
= 2(x + 3)x + 3
(3 – x )
2 2
3 – x + 2x – 4x = (x + 1)(3x + 11)
2
2
= 2(x + 3)x + 3
(3 – x )
2 2
x – 4x + 3
2
= (3 – x )
2 2
Example 9
b
6
Alternative Method Given y = ax + and dy = 2x – x n – 1 ,
n
4
x
dx
x – 2 find the values of a, b and n.
y = = (x – 2)(3 – x )
2 –1
3 – x 2
Let u = x – 2 and v = (3 – x ) Solution
2 –1
dv
du = 1 and = –1(3 – x ) (–2x)
2 –2
4
dx dx y = ax + bx –1
2x
2 2
= (3 – x ) dy = 4ax – bx –2
3
By using the multiplication rule dx b
3
dy = (x – 2) 2x 2 2 + 1 = 4ax – x 2
dx  (3 – x ) 3 – x 2
2
= 2x(x – 2) + (3 – x ) By comparing, Form 5
(3 – x ) 4a = 2
2 2
2x – 4x + 3 – x 2 1
2
= a =
(3 – x ) 2
2 2
x – 4x + 3 b = 6
2
=
(3 – x ) n = 3
2 2
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Additional Mathematics SPM Chapter 2 Differentiation

2.3 The Second Derivative


dy d d dy d y
2
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1. Given y = f(x), then dx = dx [f(x)] is 2. dx dx dx 2
  can be written as
the first derivative. or fʺ(x) which also known as the
dy
Differentiating dx with respect to x second derivative.
d dy d d y d dy d
2
will give   or dx [fʹ(x)]. dx 2 =   or fʺ(x) = dx [fʹ(x)]
dx dx
dx dx
2
Example 10 d y 2 = –14(–3)(x – 2) –3 –1 (1)
d y dx = 42(x – 2) –4
2
Find for each of the following and
dx 2 2 = 42
hence, determine whether d y 2 is the same as (x – 2) 4
dx
dy
dy
–14
196
  2 .    (x – 2) 3 2 = (x – 2) 6
2
=
dx
dx
2x 3 1 d y dy 2
2
(a) y = – Then, 2  
≠
3 x dx dx
7
(b) y =
(x – 2) 2
Example 11
Solution d y dy
2
Given y = 3x(x – 1), express dx 2 + dx
2
2
(a) y = x – x –1 in terms of x. Hence, find the possible
3
3 d y dy
2
dy 2 values of x if + = 24.
= (3)x 3 – 1 – (–1)x –1 – 1 dx 2 dx
dx 3
= 2x + x –2 Solution
2
d y y = 3x – 3x
2
3
= 2(2)x 2 – 1 + (–2)x –2 – 1
dx 2 dy
2
= 4x – 2x –3 dx = 9x – 3
= 4x – 2 d y
2
x 3 dx 2 = 18x
dy
2
   2 x 1 2 2 d y 2 + dy = 18x + 9x – 3
= 2x +
2
dx
2
Form 5 Then, d y 2   2 dx d y + = 9x + 18x – 3
dx
2
dy
2
≠
dy
2
dx
dx
= 24
dx
dx
2
(b) y = 7(x – 2) –2 9x + 18x – 3 = 24
2
2
dy = 7(–2)(x – 2) –2 – 1 (1) 9x + 18x – 27 = 0
2
dx x + 2x – 3 = 0
= –14(x – 2) –3 (x + 3)(x – 1) = 0
= –14 x = –3 or 1
(x – 2) 3
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Additional Mathematics SPM Chapter 2 Differentiation

2.4 Application of Differentiation
Turning point


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1. f ʹ(x) is a gradient function or the 3. The equation of the normal to the
gradient function of a tangent to the curve y = f(x) at point (a, f(a)) is
curve f(x) at any point on the curve. 1
y – f(a) = – f ʹ(a) (x – a).
2. The equation of the tangent to
the curve y = f(x) at a point (a, f(a)) 4. Stationary point and turning point:
is given by y – f(a) = f ʹ(a) (x – a).

Stationary point
dy
dx = 0

Turning point


Maximum point Minimum point Point of inflection
dy
dx = 0
dy dy
P dx < 0 dy dy = 0 dx > 0
dy > 0 dy < 0 dy < 0 dy > 0 dx = 0 dx
dx dx dx dx P P
P dy < 0 dy > 0
dy dx dx
dx = 0
dy dy
(i) dx : + → – dy (i) dx : – → –
d y (i) dx : – → + dy
2
(ii) 2  0 dx : + → +
dx d y
2
(ii) dx 2  0 (ii) d y 2 = 0
2
dx
5. If y = f(x), then the change of y and x with respect to time t can be related by
dy dy dx
dt = dx × dt

6. If δx and δy are small changes of x and y respectively, then the approximate change
of y is given by Form 5
dy
δy  dx × δx

7. The approximate value of y is given by
dy dy
f(x + δx)  y + dx δx or f(x + δx)  f(x) + dx δx






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Additional Mathematics SPM Chapter 2 Differentiation

Example 12 For turning points, dy = 0
dx
2
The diagram below shows a graph of 3x – 4x + 1 = 0
function y = (x – 1)(x – 3). (3x – 1)(x – 1) = 0
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1
y x = or x = 1
3
1 1 3 1 2 1
When x = , y =   – 2   + + 3
3
3
3
3
A B
0 x = 85
27
2
3
Find When x = 1, y = 1 – 2(1) + 1 + 3
(a) the gradient of the tangent at B where = 3
the curve intersects the x-axis,  1 85 
,
(b) the equation of the tangent at B, Thus, the turning points are 3 27 and
(c) the equation of the normal at B. (1, 3).
To determine the type of the turning points,
Solution
dy dy 1 dy 1
(a) y = (x – 1)(x – 3) dx at x = 0 dx at x = 3 dx at x = 2
= x – 4x + 3
2
dy + 0
= 2x – 4 –
dx
Using B(3, 0), the gradient of the 1 85
,
tangent at B Hence,  3 27  is a maximum point.
= 2(3) – 4
= 2 dy di x = 1 dy di x = 1 dy di x = 3
(b) Equation of the tangent at B dx 2 dx dx 2
y – 0 = 2(x – 3) – 0 +
y = 2x – 6
1
(c) Gradient of the normal at B = –
2 Hence, (1, 3) is a minimum point.
Equation of the normal at B
1
y – 0 = – (x – 3)
2
1 3 Alternative Method
y = – x +
2 2 Using the second derivative to determine the
type of turning points:
Form 5 Example 13 When x = , 2 2 = 6  – 4
2
d y
2 = 6x − 4
dx
Find the coordinates of the turning points
1 d y
1
3
3 dx
on the curve y = x – 2x + x + 3. Hence,
2
3
determine the type of the turning points. 1 85 = –2  0
Hence,  ,  is a maximum point.
3 27
Solution d y
2
When x = 1, 2 = 6(1) – 4
y = x – 2x + x + 3 dx = 2  0
3
2
dy = 3x – 4x + 1 Hence, (1, 3) is a minimum point.
2
dx
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Additional Mathematics SPM Chapter 2 Differentiation

1
Example 14 Area = 6(6) – (6) 2
The diagram below shows a garden besides = 36 – 18 2
Joshua’s house. = 18 m 2
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B


Example 15
x m
Garden A boy blows a spherical balloon and its
volume increases with a rate of 10 cm /s.
3
Find the rate of change of its surface area
A y m C
when the radius, r is 4 cm.
BC is a wall from Joshua’s house and the
garden is fenced by 12 m of fencing. Solution
4
(a) Show that the area of the garden is Volume of a sphere, V = πr 3
1
A = 6x – x m . dV 3
2
2
2 = 4πr 2
(b) Find the length of AB and AC so that dr
the area of the garden is maximum. Given dV = 10 cm /s
3
Hence, find the maximum area of the dt
garden. dV = dV × dr

dt dr dt
Solution 2 dr
10 = 4πr × dt
(a) x + y = 12 dr = 10
y = 12 – x dt 4πr 2 ................a
1
Area of the garden, A = xy Surface area of a sphere, A = 4πr 2
2 dA
1
= x(12 – x) dr = 8πr
2
1
= 6x – x 2 dA = dA × dr
2 dt dr dt
dr
(b) For maximum area, = 8πr × dt ................
dA = 0
dx Substitute a into ,
6 – x = 0 dA 10
x = 6 dt = 8πr × 4πr 2
20
y = 12 – 6 = r Form 5
= 6 When r = 4 cm,
Hence, AB = AC = 6 m dA = 20
d A dt 4
2
= –1  0, the area is maximum. = 5 cm /s
2
dx 2



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Additional Mathematics SPM Chapter 2 Differentiation

Example 16
1
Given that y = 3x + , and x changes from 2 to 1.98, find
x
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(a) the approximate change in y,
(b) the approximate value of y.
Solution
1 1
(a) y = 3x + x (b) When x = 2, y = 3(2) + 2
= 3x + x –1 = 13
dy 2
= 3 – x –2
dx dy 1
When x = 2 and δx = 1.98 – 2 When x = 2, dx = 3 – 2 2
= –0.02 = 2.75
δy  dy × δx f(x + δx)  y + dy δx
dx
dx

δy = 3 – x 1 2 × (–0.02) = 13 + 2.75(–0.02)
2

= 3 – 2 1 2 × (–0.02) = 6.445
= 2.75(–0.02)
= –0.055





Mastery


1. The diagram below shows a plate which Solution
is made up of a rectangle ABCD and
semicircles at both ends. The area of the (a) Let y = diameter of the semicircle
rectangle is 200 cm . Perimeter of the plate,
2
P = 2x + 2πr
A B
= 2x + πy
Area of ABCD = 200
xy = 200
Form 5 D x cm C y = 200
x

(a) Show that the perimeter, P of the Hence, P = 2x + πy 200
plate is P = 2x + 200π . = 2x + π  x 
x
(b) Find the minimum perimeter P and = 2x + 200π
the corresponding value of x. x
(b) P = 2x + 200πx –1
dP 200π
= 2 –
dx x 2
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Additional Mathematics SPM Chapter 2 Differentiation
For the perimeter to be minimum, (b) Equation of the tangent
dP = 0 4 4

dx y – = 4 x – 3 
3
2 – 200π = 0 y = 4x – 16 + 4
(a) the coordinates of point A,Sdn Bhd. All Rights Reserved.
x
2
3
3
2
2x = 200π y = 4x – 4
x = 10π
3. The equation of a curve is y = x + 2x – 3.
2
200π
∴ P = 210π  + dy
10π (a) Find the value of dx when x = –2.
= 20 + 20 (b) Calculate the approximate change in
π
π
= 40 cm y in terms of h when x changes from
π
–2 to (–2 + h) where h is a small
CAUTION! value.
Solution
For the perimeter to be minimum,
2
d P 400π (a) y = x + 2x – 3
2
dx 2 = x 3 dy
When x = 10, dx = 2x + 2
π
d P 400π  0 dy
2
dx 2 = 10π  3 When x = –2, dx = 2(–2) + 2
Hence, the perimeter is minimum = –2
(b) When x = –2
and δx = –2 + h – (–2)
2. The point A lies on the curve = h
1
y = (3x – 2) . Given that the gradient δy  dy × δx
2
3
dx
1
of the normal at A is – . Find δy = –2h
4
Penerbitan Pelangi
1
(b) the equation of the tangent at point 4. Given that y = x 3 .
A. dy
(a) Find the value of when x = 2.
Solution dx 3
1 (b) Hence, estimate the value of 1.98 3 .
(a) y = (3x – 2) 2
3
dy 2 Solution
dx = (3x – 2)(3) 1
3
–3
= 2(3x – 2) (a) y = x 3 = x
Gradient of tangent = 4 dy = –3x –4 Form 5
2(3x – 2) = 4 dx
4
x = When x = 2,
3 dy 3
1 4 2 = – 4
– 2
Therefore, y =     dx 2
3
3 3 = – 3
= 4 16
3
,

Then, A  4 4 
3 3
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Additional Mathematics SPM Chapter 2 Differentiation
1 (b) (i) The equation of the tangent:
(b) When x = 2, y =
2 3 y = –5x + 2
= 1 Then, the gradient = –5
δx = 1.98 – 2 8 kx + 2 = –5
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= –0.02 At x = –2, (–2)k + 2 = –5
1 dy –2k = –7
1.98 3  y + dx δx k = 7
1

= + – 3  (–0.02) 2
8 16 (ii) Equation of the normal
1
= 0.12875 y – 10 = (x + 2)
3 = 3(0.12875) 5 52
1
1.98 3 = 0.38625 y = x + 5
5
5. (a) Given y = 2x – x + 3, solve the 6. A cylindrical container is open at
2
d y dy 2
2
+
equation 2   – 2y = 5. the top has a base radius of j cm. The
dx dx total surface area of the container is
(b) Given (kx + 2) is the gradient 300π cm .
2
function of a curve where k is (a) Show that the volume, I cm , of the
3
a constant. y + 5x – 2 = 0 is the πj 2
equation of a tangent to the curve at container is I = 2 (300 – j ).
point (–2, 10). Find (b) Find the value of j when dI = 0.
(i) the value of k, dj
(ii) the equation of the normal at the (c) Show that the value of j obtained
point (–2, 10). in (b) will make the volume of
the container maximised. Hence,
Solution calculate the maximum volume of
(a) y = 2x – x + 3 the container.
2
dy
= 4x – 1 Solution
dx
d y (a) Let h = height of the container
2
= 4
dx 2 Surface area of the container
2
d y dy 2 L = πj + 2πjh
2
dx 2   – 2y = 5 300π = π( j + 2jh)
+
2
dx
4 + (4x – 1) – 2(2x – x + 3) = 5 2jh = 300 – j 2
2
2
4 + 16x – 8x + 1 – 4x + 2x – 6 = 5 h = 300 – j 2
2
2
2j
12x – 6x – 6 = 0
2
Form 5 (2x + 1)(x – 1) = 0 I = πj h 300 – j 2
2x – x – 1 = 0
2
2
1
x = – , 1
2 = πj 2  2j 
πj
= 2 (300 – j )
2
190

02 Ranger Mate Tambahan Tg5.indd 190 25/02/2022 9:23 AM

Additional Mathematics SPM Chapter 2 Differentiation
πj 3
(b) I = 150πj – 2 Solution
dI 3πj 2 (a)
dj = 150π – 2
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dI h cm
When = 0,
dj x x cm 6 cm
cm
3πj 2
150π – 2 = 0 t cm
3πj 2
2 = 150π 2 2 cm
j = 150π × 3π h x
2
= 100 6 = 2
j = 10 cm h = 3x
d I Thus, the height of the cylinder
2
(c) = –3πj t = 6 – h
dj 2
When j = 10 cm, = 6 – 3x
d y The volume of the cylinder
2
= –3π(10)  0
dx 2 V = πx t
2
Hence, the volume of the container = πx (6 – 3x)
2
is maximum when j = 10 cm. = 6πx – 3πx 3
2
π(10)
I = 2 [300 – (10) ] (b) dV = 12πx – 9πx 2
2
dx
= 1 000π cm 3 For the volume to be maximum
dV
HOTS Example 1 dx = 0
The diagram below shows a cone with a 12πx – 9πx = 0
2
radius of 2 cm and a height of 6 cm on a 3πx(4 – 3x) = 0
horizontal table. A cylinder with a radius of 4 – 3x = 0
x cm lies inside the cone and touches the x = 4
surface of the cone and its axis of symmetry 3
2
coincides with the axis of symmetry of the d V = 12π – 18πx
cone. Given that the volume of the cylinder dx 2 4
is V cm . When x = ,
3
3
d V 4
2
dx 2 = 12π – 18π  
3
= –12π  0
6 cm Form 5
4
Hence, V is maximum when x = .
3
x cm 4 2 4 3
3
3
2 cm V = 6π   – 3π  
(a) Show that V = 6πx – 3πx . = 32 π cm 3
2
3
(b) When the value of x changes, find the 9
maximum volume of the cylinder in
terms of π.
191




02 Ranger Mate Tambahan Tg5.indd 191 25/02/2022 9:23 AM

Additional Mathematics SPM Chapter 2 Differentiation


SPM PRACTICE
SPM PRACTICE

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2
Paper 1 13. Given that y = px + qx and
dy
d y
2
 
2
= 4
– 32y where p and q are
3x – 1
dx
dx
2
lim
1. Find the value of x : 0 x + 2 . constants, find the possible values of
2. Find the value of lim 3 + 11x – 4x 2 . p and q.
x : 3 x – 3 14. Find the gradient function for the curve
dy y = (x + x )(2 – x) and hence, find the
2
3. Find dx for y = 2 – x using the first
2
principles. equation of the tangent at x = 1.
15. Find the gradient function for the curve
dy 2 3
4. Find dx for y = – + 1 using the first y = (x + 2) 2 and hence, find the equation
x
principles. of the normal at x = –1.
d 1 16. The gradient of tangent to the curve

5. Find  (8x – x) . a
4
2
dx 4 y = 2 – bx – 1 at point (–1, –2) is 4.
x
6. Find the first derivative for y = (1 – x) 3 Find the values of a and b.
with respect to x. x + 2 17. The equation of a curve is
3x y = x – 3x + 5. Find the coordinates
3
2
7. Find f ʹ(x) for f(x) = .
x – 2 of the points on the curve such that the
dy 2(x – 7) tangent at the points are parallel to the
8. Given the value of dx for y = x x-axis.
is 7 when x = k. Find the possible 11
2 18. Given y = 2x + is the equation of
2
values of k. the tangent to a curve y = –2x + 8x + 1
2
9. Find f ʹ(–3) for f (x) = (4 – 2x) (x + 3). at point (h, k). Find the value of h and
2
2
of k.
d y
2
10. Find for each of the following. 19. Given that the curve y = ax + bx has a
3
dx 2
(a) y = 1 – 2 stationary point at (2, 4), find the values
x
x 2 4 of a and b.
5 – x
Form 5 11. Given y = 2x 1 , show that 20. The sum of two numbers x and y is 10.
–
(b) y =
3x
2
The product of the square of x and y is P.
Find the value of x so that P is maximum.
d y x 21. A ball is thrown upwards from a
dy
2
2x + 3 = 0. building. After t seconds, the height,
dx 2 dx
s meter from the initial point is given by
12. Given y = x , show that s = 15t – 5t .
4
2
4y d y – dy 2 = 0. (a) When will the ball reach its
2
   
3 dx 2 dx maximum height?
(b) State the maximum height.
192
02 Ranger Mate Tambahan Tg5.indd 192 25/02/2022 9:23 AM

Additional Mathematics SPM Chapter 2 Differentiation
22. The volume of a cube increases at a rate dy
of 3 cm /s. If the volume of the cube is (a) dx  0
3
125 cm , find the rate of change of dy
3
(a) its side, (b) dx = 0
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(b) its surface area. dy
23. Given that a circle has a radius of r cm, (c) dx  0
a circumference of C cm and an area of
A cm . 2
2
dC 1 Paper
(a) Show that = .
dA r x 2
(b) The area of the circle increases at 1. Given that y = px + r where p and r are
a rate of 3 cm /s, find the rate of non-zeros. 2 2
2
change of the circumference when (a) Show that d y 2 = 2r 3 .
the radius is 4 cm. 2 dx (px + r)
(b) If y = x has a stationary point
8 dy px + r
24. Given y = x 2 , find the value of dx when at (3, 3), find the values of p and
x = –2. Hence, find the approximate r. Hence, show that the point is a
value for 8 . minimum point.
(–2.01) 2
2
25. Given y = 5t + t and x = 2 – 2t. 2. The tangent to the curve y = x – 4 at
2
dy x = a intersects the x-axis and y-axis
(a) Find in terms of x. at P and Q respectively. Given a is a
dx
HOTS
(b) If t changes from 3.01 to 3, find the positive integer. HOTS
approximate change in y. (a) Show that the area of the triangle
26. A coil of wire of length 25 cm is shaped OPQ where O is the origin is
2
2
into a sector of a circle with radius r cm (a + 4) .
4a
and the angle at the centre is θ radians. (b) Hence, find the minimum area of the
If the radius of the sector increases by triangle.
4% when r = 6, find the corresponding
percentage change in the area, A cm , of 3. A rectangle ABCD is inscribed in a
2
the sector. semicircle with a radius of 5 cm and
27. The diagram below shows a graph centre O as shown in the diagram below.
y = f(x) and the line y = g(x).
B C
y
D
x cm Form 5
B C
A D
O
x
0 A Given that the width of the rectangle is
x cm.
Point A lies on the straight line while (a) Show that the area of the rectangle is
points B, C and D lie on the curve. State 2x cm .
x
2
2
25 –
the point/points which satisfy/satisfies (b) Hence, determine the maximum area
the following conditions. of the rectangle.
193




02 Ranger Mate Tambahan Tg5.indd 193 25/02/2022 9:23 AM

Additional Mathematics SPM Chapter 2 Differentiation
4. The normal to the curve y = x + cx at 9. The cost of making x units of handicrafts
3
1

point (2, d) has a gradient of . Find is RM  1 x + 50x + 50 and will be sold
2
2 2
(a) the value of c and of d, for RM 80 – x each. Find
1


(b) the equation of the tangent to the 4
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curve at x = –1. (a) the profit function, P from the sale
5. An open cuboid with a square base has of x units of handicrafts,
a volume of 750 cm . The prices per (b) the value of x so that the profit
3
cm of a piece of iron sheet to make function is maximum and hence,
2
the sides and the base are RM2 and find the maximum profit.
RM3 respectively. Find the dimensions 10. The diagram shows the water in a
of the cuboid so that the price to make hemispheric bowl of radius 8 cm.
the cuboid is minimum. Hence, find the 8 cm
minimum price to make 50 of similar
cuboids.
6. Given that the equation of a curve is
y = 6 . h cm
x 2
(a) Find dy when x = 3.
dx
(b) Hence, find the approximate value for Given that the rate of change of height
HOTS
6 of water in the bowl is 0.2 cm/s. HOTS
3.01 2 correct to 4 significant figures. (a) Express the surface area of the water,
A cm , in the bowl in terms of h.
2
7. An empty cone with base radius of (b) Hence, find the rate of change of the
10 cm and height of 10 cm is being surface area of the water in terms of
filled with water at a rate of 4π cm /s. π when h = 5 cm.
3
HOTS
Find the rate of change of HOTS 11. The diagram below shows a piece of
(a) the height at 18 seconds, brick which is rectangular in shape.
(b) the surface area of the water at The total surface area of the brick is
18 seconds.
2
HOTS
300 cm . HOTS
8. The diagram below shows a mold with
length 5 m and its cross section is an x cm h cm
HOTS
equilateral triangle. HOTS
Form 5 80 cm (a) Show that h = 50 – 2x .
2x cm
x
3
3
Water is poured into the mold at a rate (b) If the volume of the brick is V cm ,
express V in terms of x.
of 1003 cm /s. Find (c) If the value of x of the above brick
3
(a) the rate of change of height of the can be changed, find the maximum
water in the mold at 25 minutes, volume of the brick. Prove that the
(b) the volume of the water at that time. volume of the brick is maximum.




194





02 Ranger Mate Tambahan Tg5.indd 194 25/02/2022 9:23 AM

Additional Mathematics SPM Chapter 2 Differentiation
12. A curve has a function 14. A drop of water drops on a carpet and it
y = 2x – 3x – 36x + 40. spreads into a circle. HOTS
3
2
HOTS
dy (a) Find the change of the area of the water
(a) Find .
dx circle if the radius increases from
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(b) State the coordinates of the 3 cm to 3.02 cm.
stationary points on the curve. (b) Calculate the rate of change of the
Hence, determine the type of the radius of the water circle if the rate
stationary points. of changes of its area increases
at 0.35 cm s when the radius is
–1
2
13. A container in the shape of a cone has 2.5 cm.
a base radius of 6 cm and a height of
12 cm is held with its tip at the bottom.
Water drips into the cone at a rate of
2 cm s . Find the rate of change of
3
–1
height of the water when the height is
3 cm.



























Form 5
















195





02 Ranger Mate Tambahan Tg5.indd 195 25/02/2022 9:23 AM

Jawapan
Answers


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Form 4 12. – 1
5
13. 2 4
5
CHAPTER 1 Functions 14. 7
Paper 2
Quiz on page 3 1. (a) The relation is a function because every
(a) {1, 2, 3, 4} object has only one image.
2
(b) {3, 5, 6, 7, 9} (b) f : x → x or f(x) = x 2
(c) {3, 5, 7, 9} (c) (i) {-1, 1, 2, 3}
(d) 1 (ii) {1, 4, 9}
(e) 7 (iii) -1, 1
5
2. (a) (i) –
Quiz on page 8 2
(a) -3  x  7 (ii) -1
(b) -2  f (x)  3 3. (a) 18
–1
(b) (ii) 361
SPM PRACTICE 5
(iii) –
2
Paper 1 4. (a) p = 1, q = –6, r = 5
1. –11  f(x)  7 (b) (i) 11 – 6x
2. p = –6, q = 13 (ii) 4
3. function, one-to-one relation 3
4. 5. (a) 1 1
,
y 4 6
(b) (i) 6 – 3
y = f (x)  x + 1
–1
2 (ii) -1
x
0 2 6. (a) 14 – 3x
5
(b) (i) 4
(ii) 0  x  2 8
5. (a) – 7 y 9
5
(b) 6 m(x) = |9x –13|
13
6. (a) 2
(b) 1
7. (a) x + 4 0 4 8 x
(b) x – 13 1 9 2 9
3
8. 6x 2 7. (a) – 5
9. –5  f(x)  1 2 2
10. 7 (b) (i)
3
10 x + 2
2
11. 2 (ii) 3
3

306





4 Jawapan Ranger Mate Tambahan.indd 306 25/02/2022 9:32 AM

Additional Mathematics SPM Answers
8. (a) (i) -1  x  1 18. x + 2x – 5 = 0
2
(ii) 0  f(x)  7 19. p = –1, q = 3
(iii) When it is tested with horizontal line
test, the line cuts the graph at two Paper 2
points. The type of the function is not 5
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one-to-one. Thus, f (x) does not exist. 1. (a) a = – , h = 2 (b) –5  x  9
-1
4
2
(b) (i) 4 2. (a) t = 2, p = 10 (b) x + 2x – 195 = 0
x 3. 1 : 3; –6
9. (a) x + 8 4. (a) When b changes from –5 to 5, the vertex
2 is to the right of the y-axis. The shape of
(b) 3x + 2 the graph and y-intercept do not change.
6 (b)
1
(c) 7 f(x)
3
CHAPTER 2 Quadratic Functions 12
–4 0 3 x
Quiz on page 23 2
y-axis (c) The new graph is a reflection of the
Quiz on page 26 graph f(x) in the y-axis.
y-axis 5. (a) x  2 or x  6; 0  x  8
(b) 0  x  2 or 6  x  8
SPM PRACTICE 6. (a) (2, 9) (b) –1
2
Paper 1 (c) f(x) = (x – 2) – 9
1. p  –1 or p  7 7. (a) k  2 (b) h = 16, k = 3
2. 2, –16 8. (a) m = 5, n = 15 (b) 16x – 54x + 45 = 0
2
3. 4 : 3 2

4. – 16 9. (a) f(x) = – x – 7  2 + 9
9 2 4
1
5. 0  p  (b) Maximum value = 9
9 4
6. p = 28, q = –143, r = –123 (c)
7. 2x – 3x + 1 = 0 f(x) 7 , 9
2
8. (a) x – 22x + 1 = 0 (b) 0.04555, 21.95 2 4
2
9. (a) –1 (b) k  – 9 0 2 5 7 x
10. x  –1 or x  4 4
11. –9, 17
12. c = 3b 2 –10
16a

13. k  10 (d) f(x) = x – 7  2 – 9
7 2 4
14. 7 10. (a) 48 cm (b) 108 cm 2

15. f(x) = –2 x + 3 2  2 + 27 11. (a) 66 cm (b) 80 cm
2
2
27 3 12. 13.34; 367.65 m
Maximum value = ; x = –
2 2 13. (a) p  5 (b) 6
16. f(x) = – 8  x + 3  2 + 8 14. (a) a = – 6 , p = 6, q = 6
9 2 81
1
17. The value of a changes to –3, the value h (b) (i) 3 m (ii) 12 m
does not change and the value of k changes 3
to 5.
307





4 Jawapan Ranger Mate Tambahan.indd 307 25/02/2022 9:32 AM

Additional Mathematics SPM Answers
3
CHAPTER 3 Systems of Equations 9. 3
9
10.
5 5 – 3
11. log 10 25
Quiz on page 42 4
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2x + y + z = 24 12. –2
14. x = 0.93
SPM PRACTICE 15. x = 2, x = 64
Paper 2 Paper 2
1. x = 1, y = 2, z = -1 1. x = 3, y = – 1
2. m = -3, n = 2, p = –5 2
2
1
3. x = 3.40, y = -0.267; x = 0.881, y = 1.41 2. (b) x = – , y =
6
3
4.  1 4 , 6 1  , (4, 14) 3. (b) x = 9, x = 81
2
5. Price of a piece of curry puff = RM0.70 (c) x = –2, x = 0
Price of a packet of nasi lemak = RM1.70 4. (a) 0.3679 (b) 0.1353
Price of a glass of orange juice = RM0.60 5. (a) 100 g (b) 155.2 g (c) 19 hours
6. Length = 54.4 m, width = 12.5 m
7. 18 and 51 6. x = 9 + 13 , x = –5 + 13
8. x = 8, y = 20 7. 3 + 2 34 6
2
1
9. A(-3, 5), B  7 3 , -3  8. 1 + 2
2
10. a = 30.5°, b = 5.5°, c = 133° 9. –6 + 5
2
11. x = 11.5 10. 4 –   cm
3
 17 , 4  1 3 + y
12. - 29 29 11. (a) x + 2y (b) x – 2 (c)
13. x = 24, y = 36 9 2 3 2
14. Mass of the block X = 8 kg 12. x = , y =
2
4
Mass of the block Y = 2 kg 1
13. (a) ln a = 3p, ln b = p (b)
15. (b) a = 48.2 and b = 51.2. Both average 3
speed, a and b did not exceed 60 km h , 14. (a) c  0, c ≠ 1 (b) q = 1
-1
thus, the owner of the car will not receive p 2
summons. 15. (a) I = 170. 170°C is the initial temperature.
(b) 4.159
CHAPTER 4 Indices, Surds and Logarithms CHAPTER Progressions
Quiz on page 54 5
x = 3.597 Quiz on page 66
5 050
SPM PRACTICE
SPM PRACTICE
Paper 1
1. x = 2 Paper 1
2. q = p 2 1. (a) 8, 20, 32, … (b) 12
16
5 6 2
3. 3 x y 2. (a) 168 (b) 39
4. y = 4 3. (a) k = h + 1 (b) 13h + 33
3
6. 11 4. (a) 3 (b) 103, 107, 111
5 5
7. – +  3 5. (a) 60 (b) –17
2 2
308





4 Jawapan Ranger Mate Tambahan.indd 308 25/02/2022 9:32 AM

y
x





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Form 4


Geometric progressions,
CHAPTER 2 T = ar n – 1
n
Quadratic Functions
n
S = a(r – 1) for | r | . 1 or
r – 1
n
2
n
x = –b ± √b – 4ac S = a(1 – r ) for | r | , 1
2a n 1 – r
a
CHAPTER 4 Indices, Surds and S = 1 – r for | r | , 1
∞
Logarithms
a × a = a m + n CHAPTER
n
m
a ÷ a = a m – n 7 Coordinate Geometry
m
n
(a ) = a
mn
m n
√a × √b = √ab Divisor of a line segment,
a 1 nx + mx 2 ny + my 2 2
1
1
√a ÷ √b = (x, y) = ,
b m + n m + n
log mn = log m + log n
a a a Area of a triangle
1
log m = log m – log n = u(x y + x y + x y ) – (x y + x y
n
a
a
a
2
2 1
3 2
3 1
1 2
2 3
log m = n log m + x y )u
n
a a 1 3
log b
log b = log a Area of a quadrilateral
c
a
c
1
= u(x y + x y + x y + x y )
1 2
3 4
2 3
4 1
2
CHAPTER 5 Progressions – (x y + x y + x y + x y )u
4 3
2 1
1 4
3 2
Area of a polygon
Arithmetic progressions, 1 x x … x x
2
T = a + (n – 1)d = 2 y 1 y … y n y 1
n 1 2 n 1
n
S = [2a + (n – 1)d]
n 2
n
S = [a + l ]
2
n
vi
0b Rumus Ranger Mate Tambahan SPM.indd 6 25/02/2022 9:06 AM

Additional Mathematics SPM Formulae

CHAPTER 8 Vectors Form 5


CHAPTER 1
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u r u = √x + y 2 Circular Measure
2
~
r
^ r ~ Arc length,
~ =
u r u s = rq
~
Area of sector,
1
L = r q
2
2
CHAPTER 9 Area of triangle
Solution of Triangles
1
2
= r sin q
2
a = b = c
sin A sin B sin C CHAPTER
a = b + c – 2bc cos A 2 Differentiation
2
2
2
b = a + c – 2ac cos B
2
2
2
2
2
c = a + b – 2ab cos C y = uv, dy = u dv + v du
2
Area of a triangle dx dx du dx dv
1 1 1 v – u
= ab sin C = bc sin A = ac sin B u dy dx dx
2 2 2 y = , = v 2
v dx
Heron’s formula dy dy du
= √s(s – a)(s – b)(s – c), dx = du × dx
a + b + c
s =
2
CHAPTER 3 Integration
CHAPTER 10 Area under a curve
Index Numbers
b
= y dx or
Q a
I = 1 × 100 b
Q
0 = x dy
a
– ∑I w
I = i i Generated volume
∑w i
= b πy dx or
2
a
b
= πx dy
2
a
vii





0b Rumus Ranger Mate Tambahan SPM.indd 7 25/02/2022 9:06 AM

Additional Mathematics SPM Formulae

CHAPTER 4 Permutation and Combination CHAPTER 6 Trigonometric Functions



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n! 2 2
n P = sin A + cos A = 1
r (n – r)! sec A = 1 + tan A
2
2
n! cosec A = 1 + cot A
2
2
n C = (n – r)! r! sin 2A = 2 sin A cos A
r
cos 2A = cos A – sin A
2
2
Formula involving identical objects = 2 cos A – 1
2
= n! 2
p! q! r!… = 1 – 2 sin A
tan 2A = 2 tan A
2
1 – tan A
CHAPTER 5 sin (A ± B) = sin A cos B ± cos A sin B
cos (A ± B) = cos A cos B sin A sin B
Probability Distribution
tan (A ± B) = tan A ± tan B
P(X = r) = C p q , p + q = 1 1 tan A tan B
n
r n – r
r
Mean, µ = np
s = √npq
X – µ
Z =
s






























viii





0b Rumus Ranger Mate Tambahan SPM.indd 8 25/02/2022 9:06 AM

Form
4.5 KSSM KC118332 PELANGI




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• Kedai Buku Online • Perpustakaan Online • Chew Su Lian
W.M: RM19.95 / E.M: RM19.95 Ong Yunn Tyug
KC118332
ISBN: 978-967-2779-83-4 Tee Hock Tian
Moh Sin Yee

PELANGI Dr. Pauline Wong Mee Kiong
Penerbitan Pelangi Sdn. Bhd. (89120-H)



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