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SPM
FORM FOCUS
ADDITIONAL 4∙5
MATHEMATICS KSSM SPM
FOCUS SPM KSSM Form 4 • 5 – a complete ADDITIONAL
REVISION
REVISION
REVISI
and precise series of reference books with special
üKeywords üSPM Tips features to enhance students’ learning as a whole.
üConcept Maps üRemember! This series covers the latest Kurikulum Standard MATHEMATICS SPM
Sekolah Menengah (KSSM) and integrates
Sijil Pelajaran Malaysia (SPM) requirements.
REINFORCEMENT
REINFORCEMENT A great resource for every student indeed! FORM
& ASSESSMENT ADDITIONAL MATHEMATICS 4∙5
& ASSESSMENT
üTry This! üSPM Model Paper REVISION
üSPM Practices üComplete Answers Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
REINFORCEMENT KSSM
EXTRA FEATURES ASSESSMENT
EXTRA FEATURES
üDaily Applications üSPM Highlights EXTRA
üHOTS Questions üQR Codes
üCalculator Corner üCloned SPM Questions
TITLES IN THIS SERIES
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CC038332
ISBN: 978-967-2720-66-9 Ooi Soo Huat
Moy Wah Goon NEW SPM ASSESSMENT
Yong Kuan Yeoh
FORMAT
PELANGI Moh Sin Yee 2021
CONTENTS
Mathematical Formulae v Chapter
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Form 4 6 Linear Law 108
Chapter 6.1 Linear and Non-Linear Relations 109
1 Functions 1 6.2 Linear Law and Non-Linear Relations 114
6.3 Applications of Linear Law 121
1.1 Functions 2 SPM Practice 6 124
1.2 Composite Functions 14
1.3 Inverse Functions 19
SPM Practice 1 27 Chapter 7 Coordinate Geometry 128
Chapter
2 Quadratic Functions 29 7.1 Divisor of a Line Segment 129
7.2 Parallel Lines and Perpendicular Lines 132
2.1 Quadratic Equations and Inequalities 30 7.3 Areas of Polygons 137
2.2 Types of Roots of Quadratic Equations 37 7.4 Equations of Loci 142
2.3 Quadratic Functions 39 SPM Practice 7 145
SPM Practice 2 52
Chapter Chapter
3 Systems of Equations 55 8 Vectors 148
3.1 Systems of Linear Equations in Three 8.1 Vectors 149
Variables 56 8.2 Addition and Subtraction of Vectors 155
3.2 Simultaneous Equations involving One 8.3 Vectors in a Cartesian Plane 166
Linear Equation and One Non-Linear SPM Practice 8 173
Equation 60
SPM Practice 3 65 Chapter
Chapter 9 Solution of Triangles 179
4 Indices, Surds and Logarithms 67
9.1 Sine Rule 180
4.1 Laws of Indices 68 9.2 Cosine Rule 184
4.2 Laws of Surds 70 9.3 Area of a Triangle 188
4.3 Laws of Logarithms 75 9.4 Application of Sine Rule, Cosine Rule and
4.4 Applications of Indices, Surds and Area of a Triangle 191
Logarithms 80 SPM Practice 9 194
SPM Practice 4 85
Chapter Chapter
5 Progressions 87 10 Index Numbers 196
5.1 Arithmetic Progressions 88 10.1 Index Numbers 197
5.2 Geometric Progressions 96 10.2 Composite Index 200
SPM Practice 5 104 SPM Practice 10 211
iii
00 Content SPM.indd 3 08/11/2021 4:10 PM
Form 5 Chapter
6 Trigonometric Functions 329
Chapter
1 Circular Measure 216 6.1 Positive Angles and Negative Angles 330
6.2 Trigonometric Ratios of any Angle 331
1.1 Radian 217 6.3 Graphs of Sine, Cosine and Tangent
5 Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
1.2 Arc Length of a Circle 219 Functions 336
1.3 Area of Sector of a Circle 225 6.4 Basic Identities 341
1.4 Application of Circular Measures 233 6.5 Addition Formulae and Double Angle
SPM Practice 1 236 Formulae 342
6.6 Application of Trigonometric Functions 344
Chapter SPM Practice 6 348
2 Differentiation 241
Chapter
2.1 Limit and its Relation to Differentiation 242 7 Linear Programming 351
2.2 The First Derivative 244
2.3 The Second Derivative 249 7.1 Linear Programming Model 352
2.4 Application of Differentiation 250 7.2 Application of Linear Programming 354
SPM Practice 2 261 SPM Practice 7 360
Chapter Chapter
3 Integration 264 8 Kinematics of Linear Motion 364
3.1 Integration as the Inverse of 8.1 Displacement, Velocity and Acceleration
Differentiation 265 as a Function of Time 365
3.2 Indefinite Integral 266 8.2 Differentiation in Kinematics of Linear
3.3 Definite Integral 271 Motion 373
3.4 Application of Integral 287 8.3 Integration in Kinematics of Linear
SPM Practice 3 290 Motion 379
8.4 Application of Kinematics of Linear
Motion
Chapter Permutation and SPM Practice 8 382
387
4 Combination 293
4.1 Permutation 294 SPM Model Paper 390
4.2 Combination 300
SPM Practice 4 304 Answers 399
Chapter
Probability Distribution 306
5.1 Random Variable 307
5.2 Binomial Distribution 311
5.3 Normal Distribution 316
SPM Practice 5 325
iv
00 Content SPM.indd 4 08/11/2021 4:10 PM
Chapter Learning Area : Algebra
Form 4
1 Functions
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KEYWORDS
• Absolute value function – Fungsi
nilai mutlak
• Arrow diagram – Gambar rajah
anak panah
• Composite function – Fungsi
gubahan
• Domain – Domain
• Function – Fungsi
• Function notation – Tatatanda
fungsi
• Horizontal line test – Ujian garis
mengufuk
• Image – Imej
• Inverse function – Fungsi songsang
Concept • Object – Objek
Map • Range – Julat
• Relation – Hubungan
• Vertical line test – Ujian garis
mencancang
Many natural and daily phenomena involve relationships between two variables. A function is a
special type of relationship such that the value of a variable is dependent on the value(s) of
• The height of a tree in a forest depends on variable such as the amount of sunlight received.
• The body mass of a person depends on many variables such as age, gender, height, diet,
one or more variable(s).
• The hourly salary of an employee depends on variable such as the number of working hours
activities and so on.
What are the other examples of functions in our daily lives?
of the employee.
1
Additional Mathematics SPM Chapter 1 Functions
1.1 Functions
1. There are four different types of relations.
y
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6
5
4
One-to-one 3 Each input value has
2 one output value.
1
x
–2 –1 0 1 2 3 4 5 6
–1
Form 4
–2
y
7
6
5
Many-to-one 4 Several input values
3 share one output
2 value.
1
x
–4 –3 –2 –1 0 1 2 3 4
Types of relations –1
y
4
One-to-many 2 An input value may
0 2 4 6 8 10 12 14 16 x have two or more
–2 output values.
–4
y
4
3
2
1 For two or more
Many-to-many –4 –3 –2 –1 0 1 2 3 4 x input values, there
–1 are two or more
–2
–3 output values.
–4
2. Out of the four relations above, only two can be considered as functions:
• One-to-one relation.
• Many-to-one relation.
2
Additional Mathematics SPM Chapter 1 Functions
A Explaining functions using graphical 1
representations and function The following arrow diagram shows the relationship
notations
between two sets X and Y.
(a) (b)
1. Assuming the price of a movie ticket is RM10.
Hence, the amount paid depends on the number Plus 1
of tickets purchased. 1 2 a 1
For example, 1 → 10, 2 → 20, 3 → 30 and Bhd. All Rights Reserved.
(a) The relationship between the number of 2 3 4 b c 4 5
tickets purchased and the total amount paid 3 5
can be represented in a graphical form as
follows. Set X Set Y Set X Set Y
(c) (d)
Squares Form 4
50 1 1 1 a
40 2 2 b
–2 4 3 c
30
Set Y 4
20
Set X Set Y Set X Set Y
10
Determine whether the relations are functions. Give
0 1 2 3 4 5 reasons for your answers.
Set X Solution
(a) Function. Each object has only one image even
(b) Note that each movie ticket purchased though element 5 does not have an object.
(input value) in set X is mapped to one and (b) Not a function. Object c has two images, which
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only one price (output value) in set Y. are 4 and 5.
(c) Function. Each object has only one image even
so on. though elements 2 and –2 have the same image.
This type of relation is known as function. (d) Not a function. Element 3 does not have an
image.
(c) If f denotes a function which maps set X to
set Y, then the function can be represented SPM Tips
by the following arrow diagram. There are two types of relations which can be considered
as functions.
f
x y Capital
1 10 Pahang Kuantan
2 20 Perlis Kota Bharu
3 30
4 40 Kelantan Kangar
Set X Set Y One-to-one relation
Material class
(d) Element 1 in set X is known as the object Steel
while element 10 in set Y is known as the Copper Metal
image of 1. The same applies for 20, 30, and Wood
40, which are the images of objects 2, 3, and Plastic Non-metal
4 respectively.
Many-to-one relation
Try Questions 1 – 2 in ‘Try This! 1.1’
3
Additional Mathematics SPM Chapter 1 Functions
Example of HOTS The vertical line intersects the
HOTS Question
graph at only one point.
Determine whether the relationship between the
number of tyres purchased and the total price in
the following advertisement represents a function.
Explain your answer. Graph of a The graph of relation f is a
function.
relation,
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Valid until end
of this month only f
The vertical line does not
intersect at any point, or
intersect at more than one
point in the graph.
Get your fourth tyre for
FREE!
The graph of relation f is not a
Form 4
Buy 1 RM250 function.
Buy 2 RM500
Buy 3 RM750
Buy 4 RM750
Solution 2
Function. Each object only has one image, i.e. buy 1
→ RM250, Buy 2 → RM500, Buy 3 → RM750, Buy Using the vertical line test, determine whether the
4 → RM750. following relations represented by each of these
graphs are y as a function of x. Justify your answers.
Try this HOTS question (a) y (b) y
The following table shows the formula in a
spreadsheet which is used to determine the areas
of circles with different radii. Determine whether O x
the relationship between the radius and the area x
of circle is a function. Explain your answer. O
(c) y
Radius Area
x
0
x = 1
Solution
Function. Each output only has one input. Solution
(a) This is not a graph of y as a function of x. The
vertical line intersects the graph at two different
points.
(a) Vertical line test
y
In this graph, for
1. The vertical line test can be used to determine any one input value
whether a graph of a relationship represents a x, there are more
function. than one output
value y.
x
O
4
Additional Mathematics SPM Chapter 1 Functions
(b) This is a graph of y as a function of x. The vertical 2. The function can be written as p : x → 4x
line intersects the graph at only one point. which is read as “function p maps x to 4x” or
p(x) = 4x which is read as “4x is the image of
y
In this graph, for object x under the function p”. Input value x is
any one input value known as the object while 4x is known as the
x, there is at most
x one output value y. image.
O
3. The letter f is frequently used to denote a
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function. f(x) represents the output of a function
(c) This is a graph of y as a function of x. The vertical f when the input is x.
line intersects the graph at only one point, except 4. Other letters can also be used to represent
for the line x = 1 which does not intersect at any certain quantities in a function, such as the
point in the graph. letter A for area and h for height. For example, Form 4
2
y h(t) = 9.8t – 3.6t can be used to represent the
In this graph, for height, h meters, reached by a ball t seconds after
any one input value
x, there is one being tossed.
x output value y.
0 5. When an equation involving two variables is
x = 1
used to represent a function, such as y = 2x + 3,
this equation can be written as f : x → 2x + 3 or
SPM Tips f(x) = 2x + 3.
Input Output Function
If there exists a vertical line which does not intersect
the graph at any value of x as in Example 2(c), then x f(x) f(x) = 2x + 3
the function is undefined for that value of x.
Generally, f(x) is the value of y for a function.
Hence, y = f(x).
Try Question 3 in ‘Try This! 1.1’
REMEMBER!
(b) Function notation
The notation f(4) represents the value of a function
1. Function notation is a method to represent when the input value is 4. 4 is not the output value.
functions. Likewise, f(4) does not mean f × 4.
For example,
• The perimeter, p, of a square 3
In with sides x units is four times
words x. (a) In the following diagram, set Y contains the
images of the objects in set X.
In • Perimeter, p = 4x.
formulae 4 2
9 3
16 4
In
function • p(x) = 4x. Set X Set Y
notation Write the relation between set X and set Y by
using function notation.
Hence, p(5) denotes the perimeter of a square
with sides 5 units. The notation p(5) represents (b) The volume, V, of a sphere with radius r units is
the output for the function when the input value 4 πr . Write the relation between the volume of the
3
is 5. 3
sphere and its radius by using function notation.
5
Additional Mathematics SPM Chapter 1 Functions
(c) Complete the following:
Solution
Equation Function In the first diagram, x = 0, f(x) = 2
f(0) = p(0) + q = 2 ......................a
y = 2x + 4
2
In the second diagram, x = 1, f(x) = 5
f(1) = p(1) + q = 5 ......................b
Solution From a, q = 2
(a) g : x → AB x or g(x) = AB x Substitute into b, p + 2 = 5
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4
4
(b) V : r → —πr 3 or V(r) = —πr p = 3
3
3 3
(c) f : x → 2x + 4 or f(x) = 2x + 4 Try this HOTS question
2
2
The following diagrams show the inputs of x into
Try Question 4 in ‘Try This! 1.1’ a function machine which produces output g(x).
Form 4
SPM Highlights Input x Input x Input x
2
4
0
f f f
7
49
6
–6 –2 2 14
36 Output g(x) Output g(x) Output g(x)
–7
If g(x) = ax + b, where a and b are constants,
2
Set A Set B determine the values of a and b.
The diagram above shows the relationship between Solution
set A and set B. a = 1, b = −2
(a) State the type of relation.
(b) Write the relation using function notation.
Solution (c) Reciprocal functions
(a) Many-to-one relation 1. A reciprocal function is expressed in the form of
(b) f : x → x 2 k
f(x) = —, x ≠ 0, where k is a constant.
x
k
2. The graph of function f(x) = — only approaches,
x
Example of HOTS but does not touch the line x = 0 (y-axis). This
HOTS Question
is because the solution of dividing any numbers
The following diagrams show the inputs of x into a with zero is undefined.
function machine which produces output f(x).
For example, the following diagram shows the
Input x Input x Input x 1
0 1 2 graph of function g(x) = .
x + 1
y
f f f
x = –1
2 5 8
Output f(x) Output f(x) Output f(x) x
0
If f(x) = px + q, where p and q are constants,
determine the values of p and q.
6
Additional Mathematics SPM Chapter 1 Functions
The expression x + 1 in the denominator of the y
equation cannot be zero.
Hence, x + 1 ≠ 0, which is x ≠ −1.
1 y = –x, x < 0 y = x, x 0
Therefore, the graph of function g(x) =
x + 1 x
approaches, but does not touch the line x = −1. 0
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SPM Tips
5
For x → –1, y → ∞, the line x = –1 is known as a vertical
asymptote. Likewise, for y → 0, x → ∞, the line y = 0 Which of the following statements is true regarding
(x-axis) is known as a horizontal asymptote. the graph of function f(x) = | x |? Mark ✓ for the true
statements and ✗ for the false statements. Form 4
(a) The graph has a W shape.
4 (b) The graph has a vertex at the origin, O.
For each of the following reciprocal functions, (c) The equation of the axis of symmetry is x = 0.
determine the value of p. Solution
3
(a) f(x) = , x ≠ p (a) ✗ (b) ✓ (c) ✓
x – 2
1
(b) g(x) = , x ≠ p
2x + 5 Try Question 6 in ‘Try This! 1.1’
Solution
(a) x – 2 ≠ 0
x ≠ 2
Therefore, p = 2. B Determining the domain and range of
a function
(b) 2x + 5 ≠ 0 1. In the arrow diagram below, set A is the set
2 x ≠ −5 of passengers who arrived at Kuala Lumpur
5
x ≠ – International Airport while set B is the set of
2 reasons for their visit to Malaysia.
5
Therefore, p = – .
2
Prasert Travelling
Try Question 5 in ‘Try This! 1.1’ Jusuf Habibie
Yashimoto Visiting relatives
Tan Chen Huat Work tour
Louise
(d) Absolute value functions Ahmed Wazir Further studies
Medial
1. An absolute value function, f(x) = | x | can be Pavitra tourisms
defined as Set A Set B
x if x > 0
| x | = –x if x , 0 (a) Set A, which is the set of all possible inputs,
is known as the domain of this mapping.
2. An absolute value or modulus of a real number (b) Set B, which comprises all possible reasons
is the numerical value of the number, without for the passengers’ visit to Malaysia, is
considering its sign. known as the codomain of this mapping.
For example, | 4 | = 4 (c) The seven interviewed passengers gave a set
and | –4 | = –(–4) = 4 of answers consisted of four reasons, which
3. The graph of function f(x) = | x | has a V shape are the output. The output formed the range
with the vertex at the origin, O, as shown in the for this mapping.
following diagram. The y-axis (x = 0) is the axis
of symmetry of the graph.
7
Additional Mathematics SPM Chapter 1 Functions
(d) Note that a range is the subset of the Discrete function
codomain which contains all mapped f(x)
images. y 2
Domain = set A
Codomain = set B
Range = {travelling, work tour, Range y 1
y 3
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further studies, medical
tourism} y 4
x
0 x 1 x 2 x 3 x 4
Domain = {x 1 , x 2 , x 3 , x 4 }
SPM Tips Range = {y 1 , y 2 , y 3 , y 4 }
Domain Codomain Continuous function
Form 4
f(x)
y 2
Range
Range
y 1
2. Note that 0 x
(a) the domain of a function is a set of possible x 1 Domain x 2
values of x which make the function defined. Domain : x 1 < x < x 2
(b) the codomain consists of the possible values Range : y 1 < y < y 2
of a functions. Absolute value function
(c) the range is a set of values of y obtained after
replacing all values of x into the functions. f(x)
For example, in the following arrow diagram,
Range
x f x + 1
0 x
–1 0
1 Domain
0
2
1
3
2 6
4
Determine the domain and range for each of the
Set X Set Y following functions.
(a)
Domain = {−1, 0, 1, 2} 0
–1 0
Codomain = {0, 1, 2, 3, 4} –2 1 1 4
2 9
Range = {0, 1, 2, 3} –3 16
3
3. The domain and range of a discrete function, (b) {(–4, 0), (–1, 3), (0, 2), (3, 3), (4, –2)}
continuous function and absolute value function
are shown in the following diagrams. (c) Number of children attending a reading camp
Age 4 5 6 7 8 9
Number of children 5 4 12 20 9 11
8
Additional Mathematics SPM Chapter 1 Functions
(d) y = 2x + 1
y = (x – 1) + 2 is a continuous function. There is no
2
(e) f(x) = (x – 1) + 2 value of x that causes the function to be undefined.
2
2
(f) f(x) = , x ≠ 3 (e) Domain is x R.
x – 3
Range is y > 2.
(g) f(x) = | x – 2 |
This graph of function is a parabola with a mininum
4
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(h) Volume of sphere: V = πr , r = radius of sphere. point at (1, 2). All values of y are equal to or larger
3
3 than 2.
Solution y
(a) Domain = {–3, –2, –1, 0, 1, 2, 3} 9
Range = {0, 1, 4, 9} 8
7 Form 4
Domain contains all of the first 6
coordinates in set of ordered pairs. 5
4
(b) Domain = {–4, –1, 0, 3, 4} 3
Range = {–2, 0, 2, 3} 2
1
Range contains all of the second
coordinates in set of ordered pairs. –2 –1 0 1 2 3 4 x
(c) Domain = {4, 5, 6, 7, 8, 9}
Range = {4, 5, 9, 11, 12, 20} The expression x – 3 in the denominator
cannot be zero. This is because 2 divided by
zero is undefined.
(d) Domain is x R.
Range is y R. (f) Domain is x R, x ≠ 3.
(Domain x R, x ≠ 3 is read as “domain includes
y = 2x + 1 is a straight line. There are no values of all real numbers except 3”.)
x or y that causes the function to be undefined at Range is y R, y ≠ 0.
any specific value.
y When the numerator 2 is divided with a non-zero
6 value, the quotient can never be zero. Hence, the
5 set of all real numbers, x (except x = 3), will give
rise to all real numbers in the range, y, except zero.
4
y
3
2 8
1
x 6
–3 –2 –1 0 1 2 3
–1
–2 4
–3
–4 2
x
–4 –2 0 2 4 6 8 10
–2
REMEMBER!
Real numbers are numbers in a set of all integers and –4
decimals, and can be represented by R.
–6
9
Additional Mathematics SPM Chapter 1 Functions
(g) Domain is x R. (c) y
Range is y > 0. y = f(x) (3, 7)
This absolute value function has a vertex V at (2, 0). All (2, 5)
values of y are greater or equal to 0.
y (1, 3)
6 (0, 1)
x
0
4
State a corresponding range for each domain
2
given.
x
–6 –4 –2 0 2 4 6 8 10 Solution
(a) When the domain is R, all values are possible
–2
for y.
Form 4
The range is y R.
This function represents the (b) When x is limited to x . 0 (positive values only),
(h) Domain is r . 0. volume of a sphere, so the
Range is V . 0. radius, r, must be positive. all values of y are greater than 1.
The range is y . 1.
Try Question 7 in ‘Try This! 1.1’ (c) The range is {1, 3, 5, 7}.
7 Try Question 9 in ‘Try This! 1.1’
For each of the following graphs of functions,
determine the domain and range.
Sketching graphs of absolute value
(b)
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y
y
13 18 functions
9
x
–3 0 2 5 Sketch the graph of y = 1 x – 1 for the domain
2 x 2
–9 0 1 3 7
–2 < x < 8. Hence, state the corresponding range.
Solution
(a) The domain is −3 < x < 2, the range is −9 < y < 13. Solution
(b) The domain is 1 < x < 7, the range is 2 < y < 18. Step I: Sketch the graph for y = 1 x – 1. The point
2
Try Question 8 in ‘Try This! 1.1’ of intersection at the x-axis and the y-axis
can be obtained by substituting x = 0 and
8 y = 0 into y = 1 2 x – 1.
The following diagrams show the representations of
y = f(x), where f(x) = 2x + 1 for different domains. y
(a) y (b) y 1
2
y = f(x) y = –x – 1
y = f(x) x
0 2
1 –1
1
x x
0 0
10
Additional Mathematics SPM Chapter 1 Functions
Step II: Reflect the negative part of the graph on the 2. Conversely, if the image is given, then the value
x-axis. of the object can be determined as well.
For example, when the value of the image is 4,
y
f(x) = 4
y = |–x – 1| 3x – 2 = 4
1
1 2 3 x = 6
x x = 2
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0 2
–1 Therefore, the object is 2.
10
Step III: When x = −2, y = 2. When x = 8, y = 3. The
2
graph is a V-shaped graph with the vertex at Given the function f(x) = x + 4x. Find the image Form 4
(2, 0). when the object is
(a) −2
y (b) 3t
y = |–x – 1| (c) x + 1
1
3 2
2 Solution
Range (a) f(x) = x + 4x
2
x f(−2) = (−2) + 4(−2)
2
–2 0 2 8 = −4
Domain
(b) f(x) = x + 4x
2
The range is 0 < y < 3. f(3t) = (3t) + 4(3t)
2
= 9t + 12t
2
Try Question 10 in ‘Try This! 1.1’
(c) f(x) = x + 4x
2
f(x + 1) = (x + 1) + 4(x + 1)
2
C Determining the image of a function = x + 2x + 1 + 4x + 4
2
when the object is given, and vice = x + 6x + 5
2
versa
Try Questions 11 – 13 in ‘Try This! 1.1’
1. To determine the image of a function when the
object is given, substitute the value of the object
in that function. 11
For example, for the function f(x) = 3x – 2, Given the function g(x) = 7x – 2. Find the object
then f(4) = 3(4) – 2 which has an image
= 10 (a) 19
4 is the object and 10 is the image. (b) 3x
2
(c) which is the same as the object under the
Input x = 4
(object) mapping of the function g.
Solution
(a) Let x be the object which has an image of 19.
Multiply the
input with 3 Therefore, g(x) = 19
and then 7x – 2 = 19
minus 2.
7x = 21
x = 3
Output = f(4)
(image)
11
Additional Mathematics SPM Chapter 1 Functions
(b) Let x be the object which has an image of 3x . 13
2
Therefore, g(x) = 3x 2
7x – 2 = 3x 2 The temperature of the surface of a lake is 22°C on a
3x – 7x + 2 = 0 specific day. A diver finds that the temperature of the
2
(3x – 1)(x – 2) = 0 water reduces by 1.2°C for every 6 m he dives to the
1 bottom of the lake.
x = or 2
3 (a) Using function notation, represent the temperature
of the water with respect to the depth of water.
(c) Since the image of the object x is x, (b) Using the function notation in (a), find the
g(x) = x temperature of the water at a depth of 45 m.
7x – 2 = x (c) The temperature of water at the bottom of the
6 x = 2 lake is 5.5°C. What is the depth, in m, of the lake?
1
x =
3 Solution
(a) Let d represents the depth, in m, from the surface
Form 4
Try Questions 14 – 16 in ‘Try This! 1.1’ of the lake, while T represents the temperature,
in °C, at a specific depth.
d
12 T(d) = 22 – 1.2 1 2 The temperature reduces
6
by 1.2°C for every 6 m.
Given f : x → | 2x – 6 |. Find 45
(a) f(−1) (b) T(45) = 22 – 1.2 1 2
6
(b) the values of x if f(x) = 4. = 13
The water temperature at a depth of 45 m is
Solution 13°C.
(a) f(x) = | 2x – 6 |
(c) T(d) = 5.5
f(−1) = |2(–1) – 6 |
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d
= | –8 | 22 – 1.2 1 2 = 5.5
6
= 8 d
1.2 1 2 = 22 – 5.5
6
(b) f(x) = 4 1.2d = (6)(16.5)
| 2x – 6 | = 4 99
2x – 6 = 4 or 2x – 6 = –4 d = 1.2
2 x = 4 + 6 2x = 2 = 82.5 m
2 x = 10 x = 1
x = 5 The depth of the lake is 82.5 m.
Alternative Method Try Question 19 in ‘Try This! 1.1’
y
8 Try This! 1.1
7
6 1. The arrow diagram below represents a function
5 which relates the time (p.m.) with temperature (°C).
4 f(x) = 4
Time (p.m.) Temperature (°C)
3
x = –1 12 35
2 1 5 30 27
1 2 33 31 28
x 3 32
–2 –1 0 1 2 3 4 5 6 4
6 31 34
29
Try Questions 17 – 18 in ‘Try This! 1.1’ Set A Set B
12
Additional Mathematics SPM Chapter 1 Functions
Mark (✓) for the characteristics of a function which (c) (d)
maps set A to set B. y y
(a) Each input in set A must be matched
with one output in set B.
x x
(b) There may be two or more inputs in O O
set A which can be mapped to the
same output in set B.
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(c) One input in set A can be mapped to 4. (a) In each of the following diagrams, set B
two or more different outputs in set B. contains the images of some of the objects in
set A.
(i)
2. The following arrow diagrams show the relations x f(x)
between two sets, X and Y. 3 Form 4
(a) times 2 5 27
1 2 –5 125
–3
2 4
6 Set A Set B
4 8
(ii) x h(x)
Set X Set Y
0 1
1 2
3 4
(b) multiple of 5 6
2
Set A Set B
2 4
3 6 Write the relation between set A and set B
using function notation.
2
(b) The area, A of a circle with radius r units is πr .
Set X Set Y Write the relation between the area of the circle
and its radius using function notation.
(c) Complete the following.
(c) Equation Function
p a (i) y = 2x + 3x − 1
2
q
r b (ii) y = sin x
s
5. For each of the following reciprocal functions,
Set X Set Y
determine the value of k.
5 7
(a) f(x) = – , x ≠ k (b) g(x) = , x ≠ k
Determine whether the relations are functions. State x x + 3
your reasons. (c) h(x) = 3 , x ≠ k (d) p(x) = 1 – 1, x ≠ k
4x – 1 3x + 2
3. By using the vertical line test, determine whether the
relations in the following graphs represent y as a 6. Determine whether the following statements are true
function of x. Justify your answers. regarding the graph of function f(x) = |x|.
(a) The graph contains all positive values of x.
(a) (b) (b) The y-axis is the axis of symmetry of the graph.
y y (c) The entire graph lies above the x-axis.
x 7. Determine the domain and range for each of the
O following functions.
(a)
x x p
O
y q
z r
Set A Set B
13
Additional Mathematics SPM Chapter 1 Functions
(b) {(–2, 4), (–1, 1), (1, 1), (2, 4)} 13. Given a function h(t) = 5 − AB t . Find
(c) (a) h(9) (b) h(0.25) (c) h(4x )
2
Input value, x 1 2 3 4 5
Output value, y 12 24 36 48 60 14. Given a function g(x) = −2x + 15. Find the object
which has an image
(a) −5 (b) x 2
(d) y = −4x + 2 (e) y = (x + 2) + 3 (c) which is the same as the object under the
2
3
(f) y = (g) y = |x + 4| mapping of the function g.
(a) 3 Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
x + 5
2
2
(h) The total surface area of a cube: L = 6x , where 15. A function f is defined as f : x → x – 2x – 10 for
x is the length of each side of the cube. x . 0. Find the value of the object which is mapped
to itself.
8. For each of the following graphs of functions,
determine the domain and range. 16. Given a function g : x → |4x – 3|. Find
(a) (b) (a) the image for x = −1 and x = 2,
y y (b) the values of object which the image is 9,
15 9 (c) the values of x so that g(x) = x.
Form 4
7
17. Given f : x → |2x – 7|. Find
(a) f(2),
3 (b) the values of x so that f(x) = 5.
x x
0 8 –6 0 1 18. The following diagram shows the graph of function
y = f(x), where f(x) = |3x + 6|, for –3 < x < k.
9. The following diagrams show the representations of y
y = f(x), where f(x) = −x + 2, for different domains.
(a) (b) 9
y y
y = f(x) y = f(x) y = f(x)
h
x
2 0 k
x –3
0 0 x
(a) Find the values of h and k.
(c) y (b) Hence, state the corresponding range of f(x) for
y = f(x) the given domain.
(–4, 6)
(–3, 5) 19. The value V, in RM, of a car n years after purchased
(–2, 4)
(–1, 3) is given by
(0, 2) 76 000
V(n) = + 1 000
x n + 1
0 (a) What is the value of the car, in RM, when it was
State the corresponding range for the given domain. newly purchased?
(b) Determine the value of the car, in RM, 7 years
10. Sketch a graph for each of the following functions. after purchase.
Hence, state the corresponding range for each (c) In how many years will the value of the car
domain given. decrease to RM20 000?
(a) f(x) = |x| for the domain –2 < x < 2. (d) Is V(n) a function? Explain your answer.
(b) f(x) = |x + 2| for the domain –3 < x < 3.
(c) f(x) = |x – 5| for the domain 0 < x < 8.
(d) f(x) = |2x + 3| for the domain – 4 < x < 2.
11. Given a function f(x) = 5 − 2x. Find the image when 1.2 Composite Functions
the object is
1
(b) t (c) t + 1 A Describing the outcome of
2
composition of two functions
12. Given a function g(y) = 3y + 1. Find the image when
the object is 1. In the following arrow diagram, the output z is
(a) 0 (b) 3x (c) 2z − 1 ‘the value of input x multiplied by 2 and added
to 5’.
14
Additional Mathematics SPM Chapter 1 Functions
g f 14
x y y z
0 0 0 5 Fill in the empty boxes in each of the following arrow
1 2 2 7 diagrams.
2 4 4 9 (a)
3 6 6 11 f g
Set X Set Y Set Y Set Z x
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x z
0 5
1 7 Form 4
2 9 (b)
g f
3 11
Set X Set Z x
2. Note that the two functions involved in obtaining
output z are the combination of function g and f
as follows:
g : x → 2x ‘multiply by 2’ function
f : x → x + 5 ‘add 5’ function Solution
3. The outcome of the combined functions g and (a)
f can be represented by the composite function f g
fg(x), that is fg : x → 2x + 5. The function fg is x gf(x)
known as a composite function. f(x)
4. The composite function can directly give output
values as long as input values are given.
gf
SPM Tips
(b)
(a)
g f g f
x fg(x)
g(x)
x g(x) fg(x)
fg
fg
fg(x) means that the function g acts on x first, then Try Questions 1 – 2 in ‘Try This! 1.2’
only the function f acts on the outcome of function g.
(b) f g
B Determining the composite functions
x f(x) gf(x)
15
2
Given the functions f : x → x and g : x → 2x + 3.
Determine the composite functions for
gf
gf(x) means that the function f acts on x first, (a) fg, (b) gf,
then only the function g acts on the outcome of (c) f , (d) g .
2
2
function f.
15
01 Focus SPM Add Maths F4.indd 15 08/11/2021 1:58 PM
Additional Mathematics SPM Chapter 1 Functions
Solution Given g(y) = z,
(a) fg(x) = f [g(x)] g(y) = y 2
= f(2x + 3) g(m) = m = k
2
= (2x + 3) 2 g(5) = 5 = k
2
= 4x + 12x + 9 k = 5
2
2
= 25
(b) gf(x) = g[f(x)] (b) gf(x) = g(2x + 1)
= g(x ) 2
2
= 2x + 3 = (2x + 1)
2
= 4x + 4x + 1
2
(c) f (x) = ff(x) Try Question 5 in ‘Try This! 1.2’
2
= f [f(x)]
= f(x )
2
= (x ) C Determining the image of composite
2 2
Form 4
= x 4 functions given the object, and vice
SPM Tips versa
2
f (x) means ff(x), which means the function f acts 17
on the object x twice. f (x) ≠ [f(x)] 2
2
In the diagram on the x f y h z
right, the function f maps 9
2
(d) g (x) = gg(x) x to y and the function h
= g[g(x)] maps y to z.
= g(2x + 3) Find 5
= 2(2x + 3) + 3 (a) f(3), 3
= 4x + 6 + 3 (b) hf(3).
= 4x + 9
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Solution
Try Questions 3 – 4 in ‘Try This! 1.2’ (a) f(3) = 5
(b) hf(3) = h[f(3)]
= h(5)
16 = 9
Try Question 6 in ‘Try This! 1.2’
x f y g z
k
18
m
Given the functions f : x → (x – 1) – 2, x . –1 and
2
2
g : x → 2x + 8 , x . –1. Determine the composite
In the diagram above, the function f maps x to y such x + 1
that y = 2x + 1 and the function g maps y to z such function fg(5).
that z = y . Find
2
(a) the values of m and k, Solution
(b) the expression for gf(x). g(5) = 2(5) + 8 Function g acts on
5 + 1 5 first.
Solution = 3
(a) Given f(x) = y, fg(5) = f(3)
f(x) = 2x + 1 = (3 – 1) – 2
2
f(2) = 2(2) + 1 = 5 = 2
m = 5
Try Question 7 in ‘Try This! 1.2’
16
Additional Mathematics SPM Chapter 1 Functions
D Determining a related function given 20
composite function and another 1
function A function f is defined as f : x → , x ≠ 0.
x
2
3
4
1. When a composite function and one of its (a) Express f (x), f (x) and f (x) in the simplest
functions is given, the remaining function can forms.
41
be determined. (b) Hence, deduce the outcomes of f (x) and f (x).
42
Solution
19 (a) f (x) = f [f(x)]
2
Given g(x) = 2x – 1. Find the function h if 1
(a) gh(x) = 7 – 6x = f 3 4
x
3
(b) hg(x) = , x ≠ 0 1
2x = Form 4
Solution 1 x
(a) gh(x) = 7 – 6x = x
g[h(x)] = 7 – 6x
3
2
2h(x) – 1 = 7 – 6x f (x) = f [f (x)]
2 h(x) = 8 – 6x = f(x)
h(x) = 4 – 3x = 1
3 x
(b) hg(x) = 4 2 2
f (x) = f [f (x)]
2x 3Pelangi Sdn Bhd. All Rights Reserved.
2
h[g(x)] = 3 = f (x)
2x = x
h(2x – 1) = 3
2x Alternative Method
4
3
Let y = 2x – 1, f (x) = f [f (x)]
1
2 x = y + 1 = f 1 2
x = y + 1 x
2 = 1
3 1
h(y) = x
2 1 y + 1 2 = x
2
4
2
3 (b) Note that f (x) = x and f (x) = x
=
Therefore, h(x) = x + 1 , x ≠ –1 The same applies to f(x) = and f (x) =
1
1
y + 1
Penerbitan By deduction, f (x) = x when n is even and
3
x
x
n
1
Try Questions 8 – 13 in ‘Try This! 1.2’
f (x) = when n is odd.
n
x
1
Therefore, f (x) = and f (x) = x.
42
41
x
E Solving problems involving composite
functions
1. We can combine a function f with itself to form Try Questions 14 – 16 in ‘Try This! 1.2’
the composite function ff which is also written as
f . 21
2
2. The combination of f and f will form the The suggested retail price of a new hybrid car is RMp.
2
composite function f f or f f or f . A car company has advertised in the newspapers that
2
3
2
the car manufacturer offers a rebate of RM5 000 while
3. We can continue using this method to generate the car company itself gives a discount of 10%.
composite functions such as f , f and so on.
4
5
17
Additional Mathematics SPM Chapter 1 Functions
10% RM5 000 2. In the arrow diagrams below, the function f maps set
discount rebate A to set B, and the function g maps set B to set C.
RM86 000 RM77 400 RM72 400 f g
A B C
RM5 000 10%
rebate discount
RM86 000 RM81 000 RM72 900 x 3x + 2 12x + 5
(a) Using function notation, write down
(i) a function R, in terms of p, for the price of Determine, in terms of x,
the hybrid car after applying the rebate from (a) the function f,
the manufacturer. (b) the composite function gf.
(ii) a function S, in terms of p, for the price of 3. For each of the following pairs of functions,
the hybrid car after applying the discount determine the composite functions below.
from the car company. 2
(b) Interpret the composite functions RS(p) and (a) f : x → 3x – 1, g : x → 2x ,
Form 4
1
SR(p). (b) f : x → x , x ≠ 0, g : x → x + 5
(c) Find RS(86 000) and SR(86 000). Determine (i) fg, (ii) gf,
2
2
which function gives a lower retail price for the (iii) f , (iv) g .
hybrid car. Explain your answer. 4. For each of the following pairs of functions,
Solution determine the composite functions below.
2
(a) (i) R(p) = p – 5 000 (a) h : x → x + 1, g : x → AB x , x . 0
(ii) S(p) = 0.9p (b) h : x → |x|, g : x → x – 6
(i) hg, (ii) gh.
(b) RS(p) is the price of the hybrid car after a
10% discount is given, followed by a rebate of 5. In the diagram on the
RM5 000. right, the function f x f y g z
SR(p) is the price of the hybrid car after a rebate
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maps x to y such that
of RM5 000 is given, followed by a discount of y = x – 4, while the
10%. function g maps y to 6
z such that z = y – 1.
2
(c) RS(86 000) = R[0.9(86 000)] Find p
= R(77 400) (a) the values of p
and k,
= 77 400 – 5 000 (b) an expression for gf(x).
= RM72 400
SR(86 000) = S[R(86 000)] 6. In the diagram on the x h y g z 12
= S(86 000 − 5 000) right, the function h maps
x to y, while the function
= S(81 000) g maps y to z. Determine
= 0.9(81 000) (a) h(3), 8
= RM72 900 (b) gh(3). 3
Hence, RS(86 000) gives a lower price for the 2
hybrid car. This is because an initial discount of 7. Given the functions f : x → (x – 2) – 3, x . –2 and
3x + 5
10% will give rise to a smaller value compared g : x → x – 3 , x . 3. Determine the composite
to an initial rebate of RM5 000. function fg(5).
Try Question 17 in ‘Try This! 1.2’ 8. Given f(x) = x + 3 and fg(x) = 2x + 6, find the function g.
9. If f(x) = x + 1, find the function g such that
2
Try This! 1.2 fg(x) = 2x + 3x + 5.
10. Given a function f : x → 2x – 3. Another function, h,
1. Use arrow diagrams to represent the following such that fh : x → 4x – 9. Find the function h.
2
composite functions.
(a) gh(x) (b) hg(x) 11. Given g(x) = x + 2 and fg(x) = x + 9. Find the
(c) gg(x) (d) h (x) function f.
2
18
Additional Mathematics SPM Chapter 1 Functions
12. Given g(x) = x + 2 and fg(x) = x + 4x + 5. Find the 2. Conversely, if Mimi wants to know how long
2
function f. she can talk for a specific total charge, she can
13. A function f is defined as f : x → 3x + 5. Find the find the inverse of the function by making T the
function g, such that the composite function gf is subject of the formula, such as
defined by gf : x → 9x + 33x + 31. C – 40
2
1 T(C) = 8
14. A function f is defined by f : x → – , x ≠ 0.
x
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(a) Express f (x), f (x) and f (x) in the simplest For example, for a total charge of 80 cent,
2
3
4
form. 80 – 40
(b) Hence, deduce the outcomes of f (x) and f (x). T(80) = 8
25
8
1 + x = 5 minutes
15. A function f is defined by f : x → , x ≠ 1.
1 – x
Generate an expression in a similar form for 3. The above example is the idea of the application Form 4
(a) f (b) f (c) f 37 of inverse functions in daily life.
2
4
1
16. A function f is defined by f : x → , x ≠ 0. 4. In function notation, an inverse function of f(x)
x 2
–1
(a) Express f (x), f (x) and f (x) in the simplest can be written as f (x).
4
3
2
form. 5. If f : x → y, then f : y → x
–1
(b) Hence, deduce the outcomes of f (x) and –1
30
f (x). or if f(x) = y, then f (y) = x
31
17. Encik Karim is a sales representative of a f
chemical company. He is paid a monthly salary of
RM4 500 and extra bonus of 6% on the total sales
that exceeds RM15 000. x y
f(x) = x – 15 000 f –1
g(x) = 0.06x
h(x) = 4 500 + x
SPM Tips
Based on the information above, and given that x
exceeded RM15 000 in a specific month, Note that the −1 in an inverse function f does not
–1
(a) which function represents Encik Karim’s bonus? mean the reciprocal of f, which is f ≠ 1 .
–1
fg(x) or gf(x)? f
(b) write a composite function which represents the
total income received by Encik Karim in that
month.
(c) determine Encik Karim’s total income for that 22
month if his total sales was RM28 000.
In the following arrow diagram, x h y
the function h maps x to y. 7
1.3 Inverse Functions Determine 6 a
(a) h (6)
–1
(b) the value of a if h (4) = 7. 2
–1
A Describing the inverse of a function
Solution
1. For Mimi’s overseas phone call to Malaysia, –1
the charge is 8 cent per minute and a service (a) h (6) = 2 If f(x) = y, then
–1
charge of 40 cent. The relationship between the (b) If h (4) = 7 f (y) = x
–1
total charge and talk time is given by a function then, h(7) = 4
C(T) = 8T + 40, where C(T) is the total charge,
in cent, for a phone call lasting T minutes. Based on the arrow diagram, h(7) = a.
Therefore, a = 4.
For example, if Mimi talks for 5 minutes,
C(T) = 8(5) + 40 Try Question 1 in ‘Try This! 1.3’
= 80 cent
19
Additional Mathematics SPM Chapter 1 Functions
23 24
In the following arrow diagram, the function f maps x Determine whether each of these functions have an
to y while the function h maps y to z. inverse function. Give reasons for your answers.
(a) (b)
x f y h z
9 x f y x g y
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1 p 1 p
5
2 q 2 q
3
3 r 3 r
Determine Set X Set Y Set X Set Y
(a) h(5),
(b) f (5), (c) h = {(1, 2), (1, 4), (2, 6), (2, 8)}
–1
(c) h (9),
–1
Form 4
(d) hf(3), Solution
(e) f h (9). Function Reason Conclusion
–1 –1
Solution (a) f Each element x in the f is a one-to-one
(a) h(5) = 9 domain is mapped to function, hence
only one element y in f has an inverse
(b) f (5) = 3 the codomain. function.
–1
(c) h (9) = 5
–1
(b) g The inverse of this g is not a one-
(d) hf(3) = h[f(3)] function maps one to-one function,
= h(5) = 9 element y in the hence g does not
(e) f h (9) = f [h (9)] codomain to three have an inverse
–1
–1
–1 –1
= f (5) = 3 elements x in the function.
–1
domain.
Try Question 2 in ‘Try This! 1.3’ (c) h Objects 1 and 2 each h is not a one-
have two images, to-one function,
1 → 2, 1 → 4, 2 → 6, hence h does not
B Making and verifying conjectures 2 → 8. have an inverse
related to the properties of inverse function.
functions
Try Question 3 in ‘Try This! 1.3’
(a) The properties of inverse functions
(I) Only one-to-one functions have inverse 25
functions
Determine whether the graphs of function of f and g
1. A function f which maps set X to set Y has below have inverse functions. Explain your answers.
an inverse function f if f is a one-to-one (a) y
–1
function. 4
2. For a one-to-one function, each element in the 3 f
domain is mapped to only one element in the 2
codomain. The inverse of this function also 1
maps each element in the codomain to only one –1 0 1 2 3 4 5 6 7 x
element in the domain.
20
Additional Mathematics SPM Chapter 1 Functions
4
(b) y g[f(x)] = g 1 x + 3 2 , x ≠ –3
6 4
5 = – 3
4 g 1 4 2
3 x + 3
2 = x + 3 – 3
1 = x, x ≠ –3
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x
0 1 2 3 4 5 6 Since fg(x) = x, x in the domain of g,
Solution and gf(x) = x, x in the domain of f,
(a) The graph of function f does not have an inverse therefore, g is the inverse of f.
function. f is not a one-to-one function.
(b) The graph of function g has an inverse function. Try Questions 6 – 7 in ‘Try This! 1.3’ Form 4
g is a one-to-one function.
(III) If f and g are inverse functions of each other,
Try Question 4 – 5 in ‘Try This! 1.3’ then
• The domain of f is equal to the range of g,
and
(II) f and g are inverse functions of each other if • The domain of g is equal to the range of f.
and only if
fg(x) = x, x in domain of g, 27
and gf(x) = x, x in domain of f.
Given f and g are inverse functions of each other such
1. Note that in the diagram below, fg(x) = x and that f(x) = AB 2
x + 1 – 5, x > –1 and g(x) = (x + 5) – 1,
gf(x) = x. x > –5. Write
g f (a) the range of f, (b) the range of g.
Solution
x g(x) x f(x)
(a) y > –5. The range of f is the domain of g.
(b) y > –1. The range of g is the domain of f.
f g
y
Since fg(x) and gf(x) are both similar to the
identity function x, hence f and g are mutually 3
inverse functions. g 2
Range 1
of g x
26 –6 –5 –4 –3 –2 –1 –1 0 1 2 3 4 5 6
By using the properties of inverse functions, show Domain of g –2 f
4 4 –3
that the functions f(x) = , x ≠ –3 and g(x) = – 3, –4 Range
x + 3 x of f
x ≠ 0 are mutually inverse functions. –5 Domain of f
Solution SPM Tips
2
f [g(x)] = f 1 4 – 3 , x ≠ 0
x
4 Domain f Range
of f
of f
=
2
1 4 – 3 + 3 x f(x)
x
4 Range Domain
= –1 f –1 –1
4 of f of f
x
= x, x ≠ 0 Try Question 8 in ‘Try This! 1.3’
21
Additional Mathematics SPM Chapter 1 Functions
(IV) If f and g are inverse functions of each other, Solution
then the graph of y = g(x) is the reflection of (a) y
the graph of y = f(x) along the line y = x. 5 y = f(x) y = x
y 4
y = x 3
7 2
6 y = g(x)
1
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5 x
4 –6 –5 –4 –3 –2 –1 –1 0 1 2 3 4 5 6
3
–2
2 –3
y = g(x) 1 –4
x
–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 –5
–1 y = f(x) –6
–2
Form 4
–3 (b) y
–4 y = x
5
–5 y = g(x)
4
3 y = f(x)
The line y = x is the axis of reflection. 2
1
x
–4 –3 –2 –1 0 1 2 3 4 5
28 –1
–2
Each of the following diagrams shows the graph of –3
y = f(x). Copy and draw the graph of y = g(x) where f –4
and g are inverse functions of each other.
(a)
y
Try Question 9 in ‘Try This! 1.3’
5 y = f(x)
4
3 (V) If a point (a, b) lies on the graph y = f(x), then
2 the point (b, a) lies on the graph y = g(x) where
1 f and g are inverse functions of each other.
x
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6
–1 29
–2
–3 The following table shows the points on the graph of
–4 y = f(x) and the corresponding points on the graph of
–5 y = g(x).
–6
Point on the graph of Point on the graph of
(b) y y = f(x) y = g(x)
5
4 (−3, −3) (−3, −3)
3 y = f(x)
2 (−1, 1) (a, −1)
1 (0, b) (3, c)
x
–4 –3 –2 –1 0 1 2 3 4 5
–1 Given f and g are inverse functions of each other.
–2 (a) Determine the values of a, b and c.
–3 (b) Why does the point (−3, −3) on the graph
–4
of y = f(x) remain unchanged on the graph of
y = g(x)?
22
Additional Mathematics SPM Chapter 1 Functions
Solution Solution (b) f has an inverse function.
(a) y (a) a = 1, b = 3, c = 0 There is no horizontal line that intersects with
y = f(x) the graph of f at more than one point.
5 y = x (b) The point (−3, −3) is located on the axis of
4 reflection y = x. y
3 4
2 y = g(x) Try Question 10 in ‘Try This! 1.3’ 3
1 2
x 1
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 (b) Horizontal line test
–1 0 x
–2 1. The reflection property of the inverse function –2 –1 –1 1 2 3
–3 graph allows a simple test to determine the –2
–4 existence of an inverse function. This test is –3
–5 known as the horizontal line test. Form 4
–6 SPM Tips
2. In the horizontal line test, a function f has
(b) y an inverse function if and only if there is no If there is no horizontal line that intersects with the
y = x
5 y = g(x) horizontal line that intersects with the graph of f graph of f at more than one point, then no value of y
4 at more than one point. will be mapped by more than one value of x. This is
an important property of a one-to-one function.
3 y = f(x)
2 30
1 Try Question 11 in ‘Try This! 1.3’
x By using the horizontal line test, determine whether
–4 –3 –2 –1 0 1 2 3 4 5 x Pelangi Sdn Bhd. All Rights Reserved.
–1 an inverse function for each of the following functions
–2 f exists. Justify your answers. C Determining the inverse functions
–3 (a) y
–4 1. The function f(x) = 2x + 3 can be shown in a
3 flowchart as follows.
2
Try Question 9 in ‘Try This! 1.3’ 1 f x × 2 2 x + 3 2x + 3
x
–3 –2 –1 0 1 2 3
–1
(V) If a point (a, b) lies on the graph y = f(x), then 2. If the operations which define the function
the point (b, a) lies on the graph y = g(x) where (b) y f(x) = 2x + 3 are reversed, the flowchart below is
f and g are inverse functions of each other. obtained.
4
f
3 x – 3 x – 3
29 2 x –3 ÷ 2 2
The following table shows the points on the graph of 1 x – 3
y = f(x) and the corresponding points on the graph of –2 –1 0 1 2 3 3. 2 is the inverse of the function f(x) = 2x + 3.
y = g(x). –1 Therefore, f (x) = x – 3 .
–1
–2
2
Point on the graph of Point on the graph of –3 4. The inverse of a function can be derived by
y = f(x) y = g(x) Penerbitan
Solution reversing the operations that defined the original
(−3, −3) (−3, −3) function.
(a) f does not have an inverse function.
(−1, 1) (a, −1) There are horizontal lines that intersect with the 5. For more complex functions, we can find their
inverse functions as shown in the following
(0, b) (3, c) graph of f at more than one point. examples.
y
Given f and g are inverse functions of each other. 3 31
(a) Determine the values of a, b and c. 2 f
(b) Why does the point (−3, −3) on the graph 1 Determine the inverse functions of
of y = f(x) remain unchanged on the graph of –3 –2 –1 0 1 2 3 x 2 – 3x
x +
3
y = g(x)? –1 (a) f(x) = 5 , (b) g(x) = AB 1
23
Additional Mathematics SPM Chapter 1 Functions
Solution 33
2 – 3x
(a) f(x) = The following diagram shows the graph of function
5 2
2 – 3x y = f(x), where f(x) = x for all real values of x.
y = Replace f(x) with y.
5 y
x = 2 – 3y Exchange the positions of x 6
5 and y. Then, solve for y. 5
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5x = 2 – 3y 4
3y = 2 – 5x 3 y = f(x)
y = 2 – 5x 2
3 1
2 – 5x 0 x
f (x) = Replace y with f (x) in –2 –1 1 2
–1
–1
3 the new equation.
(a) Explain why f is a function but does not have an
Form 4
inverse.
(b) g(x) = AB 1 (b) Given g(x) = x for x < k.
x +
3
2
–1
y = AB 1 (i) State the largest value of k so that g (x)
x +
3
exists.
x = AB 1 (ii) Based on the value of k in (b)(i), find g (x).
y +
3
–1
3
x = y + 1 Solution
x – 1 = y (a) Each value of x is mapped to one and only one
3
g (x) = x – 1 value of y, hence f is a function.
3
–1
f does not have an inverse because it is not a
one-to-one relation.
Try Question 12 in ‘Try This! 1.3’
(b) (i) If g (x) exists, the largest value of k = 0.
−1
(ii) g(x) = x
2
32 y = x
2
x 2
Given the functions g : x → x – 8 and h : x → , x = y
3 2x – 3 ±AB x = y
x ≠ —, find
2
–1
Therefore, g (x) = –AB x, x > 0
(a) g (4), y
−1
y = g(x)
(b) hg (4). 4 y = x
−1
3
Solution 2
−1
(a) Let g (4) = a, 1
g(a) = 4 x
a − 8 = 4 –2 –1 0 1 2 3 4
–1
a = 12 –2
–1
–1
g (4) = 12 y = g (x)
–1
−1
(b) hg (4) = h[g (4)] SPM Tips
= h(12) Note that the graph of y = g(x) is not a one-to-one
12
= function for all values of x. However, if we take only half
2(12) – 3 of the graph into consideration and limit the domain to
x , 0, then an inverse function exists in the 4
th
4
= — quadrant where the values of y = g (x) are
−1
7
negative.
Try Questions 13 – 14 in ‘Try This! 1.3’
Try Question 15 in ‘Try This! 1.3’
24
Additional Mathematics SPM Chapter 1 Functions
6. When constructing a composite function f (a) (b)
with f or f with f, we will obtain an identity x f y a g b
–1
–1
function, x. The function f and its inverse f can p
–1
“cancel out” each other. a 4 x q
b 5 y r
34 c 6 z s t
A function f is defined as f(x) = 3x. Set X Set Y Set A Set B
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–1
(a) Find the inverse function, f .
(b) Hence, verify the truth that the composite (c) h = {(3, 12), (4, 16), (5, 20), (6, 24)}
–1
functions f [f (x)] and f [f(x)] are equal to the
–1
identity function, x. 4. Determine whether the graph of functions of f, g and
h in the following diagrams have inverse functions.
Solution Explain your answers. Form 4
(a) The function f multiplies each input with 3. To (a) y
reverse the operation that defined f(x), we divide 3
each input with 3. Hence, the inverse function of f 2
x
f(x) is f (x) = —. 1
–1
3 –3 –2 –1 0 1 2 3 x
x
(b) f [f (x)] = f 1 2 –1
–1
3
–2
x
= 3 1 2 = x –3
3
(b) y
Likewise, f [f(x)] = f (3x)
–1
–1
3x 5
= 4
3 g 3
= x 2
Therefore, f [f (x)] = f [f(x)] = x 1 x
–1
–1
–3 –2 –1 0 1 2 3
Try Question 16 in ‘Try This! 1.3’ –1
–2
Try This! 1.3 (c) y
3
1. In the arrow diagram on the x f y 2
right, the function f maps x to 10– 1 h
1
y. Determine 2 0 x
(a) f (8), 2 –1 –1 1 2 3 4 5 6 7 8 9
–1
(b) the value of p if –3 –2
f (p) = 2.
–1
8 –3
2. In the diagram on g 5. Determine whether the following graphs of functions
the right, the function f x f y z of f and g have inverse functions. Explain your
maps y to x while the 2 answers.
function g maps y to z. (a) y
Determine 0 2
(a) f (–2), f
–1
–1
(b) gf (–2), –2 1
(c) g (2), –2 –1 0 1 2 3 x
–1
(d) fg (2). –1
–1
–2
3. Determine whether each of the following functions –3
has an inverse function. Give reasons for your –4
answers. –5
25
Additional Mathematics SPM Chapter 1 Functions
(b) y 10. The following table shows the points on a graph of
function f and the corresponding points on the graph
3 g of function g.
2
1 Points on the graph of Points on the graph
x function f of function g
–2 –1 0 1 2 3 4 5 6
–1 (−5, 3) (a, −5)
–2
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–3 (b, 0) (c, −2)
(−1, −1) (−1, −1)
6. By using the properties of inverse functions, verify Given f and g are the inverse functions of each
5
the truth that the functions f(x) = x – 2 , x ≠ 2 and other.
g(x) = 5 x + 2, x ≠ 0 are the inverse functions of each (a) Determine the values of a, b and c.
other. (b) What can be said about the point (−1, −1) which
lies on the graphs of both f and g?
Form 4
7. By using the properties of inverse functions, 11. By using the horizontal line test, determine whether
determine which of the following is the inverse an inverse function exists for each of the following
1
function of f(x) = , x ≠ 2? functions f. Justify your answers.
x – 2 (a) y
1 1
g(x) = 2 – , x ≠ 0 h(x) = + 2, x ≠ 0 3
x x
2 f
1
8. Complete the table below. x
–3 –2 –1 0 1 2 3
Domain Range
(b) y
Function: x > 4 y > 5
f(x) = 2x – 3 6
Inverse function: 4
x + 3 2 f
f (x) =
–1
2 –4 –2 0 2 4 6 8 10 x
–2
9. Copy and draw the graphs of inverse functions of f
in each of the following diagrams. 12. Determine the inverse functions for each of the
(a) y following functions.
4 (a) f(x) = x (b) g(x) = 2x + 7
3 8 x – 1 3
2 f (c) h(x) = 5 (d) f(x) = 1 + x , x ≠ –1
1 x + 3
3
1 – x
x (e) g(x) = ABBBB (f) h(x) = x – 2 , x ≠ 2
–4 –3 –2 –1 0 1 2 3 4 5
–1
–2 5
13. A function w is defined as w(x) = , x ≠ 3 while
3 – x
the function v is defined as v(x) = 2x. Find
(b) y (a) w (2).
–1
–1
6 (b) vw (2).
5
4 f 14. The functions f and g are defined as f : x → 5x + 2
2x
3 and g : x → x + 2 , x ≠ –2 respectively. Find
2 (a) f (12)
–1
1 (b) g (−2)
–1
x
0 1 2 3 4 (c) gf (12)
–1
(d) fg (–2)
–1
26
Additional Mathematics SPM Chapter 1 Functions
15. The following diagram shows a graph of function 16. A function f is defined by f(x) = x + 5.
y = f(x), such that f(x) = (x – 1) – 3. (a) Find the inverse function, f .
2
–1
y (b) Hence, verify that the composite functions
f [f (x)] and f [f(x)] are equal to the identity
–1
–1
2 function.
1
x
–1 0 1 2 3 4
–1
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–2
–3
Find f (x) and state a suitable domain.
–1
SPM Practice 1 Form 4
SPM Practice
PAPER 1 (a)
1. Fahmi works at the marketing department of a y
perfume company. He intends to introduce a type of
perfume which is filled in a conical bottle where the
ratio of the height of the cone to the radius of the 1
cone is 5 : 1 as shown in the following diagram.
0.5
x
0 20° 40° 60° 80° 100° 120° 140° 160° 180°
h (b)
y
2
1
r x
–4 –3 –2 –1 0 1 2 3 4
By using function notation, write the volume of the
conical bottle
(a) as a function of the radius, r.
(b) as a function of the height, h. 4. The diagram below shows the graph of function
f(x) = |2x + 3| for the domain – 4 < x < 0.
2. A function t maps the temperature, in degrees
Celsius, to the temperature in degrees Farenheit. y
The function t is defined as t : C → 9C + 32, y = f(x)
5
C R. Find h
(a) t(–10),
(b) the value of C when t(C) = C.
–4 k 0 x
3. Complete the following table based on the given
function.
State
Function Domain Range (a) the values of h and k.
(b) the range of f(x) that corresponds to the given
(a) y = sin x 0° < x < 180° domain.
1
(b) y = x R
x + 1
2
27
Additional Mathematics SPM Chapter 1 Functions
5. The functions g and h are defined as PAPER 2
x
2
g(x) = – 1 and h(x) = x – x
2 1. The functions f and g are defined as
respectively. 2x
x + 1
(a) Find g (x). f(x) = x + 1 , x > 0 and g(x) = ABBBB, x . −1
–1
–1
(b) Solve for gh(x) = g (x).
respectively.
6. The following diagrams show four different graphs.
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In each case, the entire graph is shown and the (a) Find fg(8).
2
scale on both axes are the same. (b) Write an expression for f (x). Give your answer
ax
A y in the form of bx + c , where a, b and c are
constants.
(c) Write an expression for g (x). Hence, state its
–1
x
O domain and range.
(d) The following diagram shows the graph of
Form 4
y = g(x).
B y
y
7
x
O 6
5
4
3
C y 2 y = g(x)
1
x
–1 0 1 2 3 4 5 6 7 8 9
x
O
On the same axes, sketch the graph of
D y y = g (x).
−1
2. In the Beijing 2008 Olympics, Usain Bolt, or also
x known as ‘Lightning Bolt’, had set a world record
O
in a 100 m race at 9.69 seconds. However, in the
last 20 m of the race, Bolt was seen looking around,
and after knowing that he was leading the other
In the table below, mark ✓ for the correct relation for racers by quite a distance, Bolt slowed down and
each graph, if any. started celebrating his victory. His coach claimed
that if Bolt maintained his speed for the entire race,
A B C D he could have recorded a better run time of 9.52
Not a function seconds. A group of physicists, led by Hans Kristian
Eriksen, have investigated the claim. They found
One-to-one function that in most of Bolt’s races, his position, in m, can be
A function which is its own represented by a function s(t) = 11.8t – 12.5, where t
inverse is the time, in seconds.
A function which does not have (a) Suggest one reason why the domain of this
function need not necessarily begin at t = 0.
an inverse (b) Determine the inverse function of s(t).
(c) What is represented by the inverse function in
7. The functions f and g are defined as f : x → x – 3 (b)?
and g : x → x respectively. Find (d) Use the inverse function in (b) to determine
2
(a) gf(x), whether the claim by Bolt’s coach was true.
2
(b) h(x) such that hgf(x) = x – 6x + 3.
28
Learning Area : Calculus
Chapter Additional Mathematics SPM Chapter 2 Differentiation
2 Differentiation
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Concept
Map
KEYWORDS
• Approximation – Penghampiran • Minimum point – Titik minimum
• Chain rule – Petua rantai • Normal – Normal
• Chord – Perentas • Product – Hasil darab
• Curve – Lengkung • Quotient – Hasil bahagi
• First derivative – Terbitan pertama • Rate of change – Kadar perubahan
• Fixed point – Titik tetap • Second derivative – Terbitan kedua
• Gradient – Kecerunan • Small change – Perubahan kecil
• Gradient of chord – Kecerunan perentas • Tangent – Tangen
• Limit – Had • Turning point – Titik pusingan
• Maximum point – Titik maksimum
The world great sprinter, Usain Bolt recorded a short distance run of 9.58 seconds. Averagely, his
speed was 10.37 m s . Does this mean that he maintained this speed throughout his 100 m run?
To answer this question, we need to determine the changes in the running distance with respect
Differentiation helps us to study as indepth as possible the changes that happen to a certain
–1
quantity with respect to another quantity. Differentiation can also be used to find the values of
to time for each instant throughout the run.
gradient, maximum and minimum.
241
Additional Mathematics SPM 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
f(x) –2.99 –2.9999 –2.999999 –3 –2.999999 –2.9999 –2.99
A Investigating and determining the
value of limit of a function when its The table above shows that when x → 0, f(x) → –3.
variable approaches zero Therefore 3
lim
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1. The limit value a of a function f(x) when the x → 0 1 x – 3x 2 = –3
x
variable x approaches zero is written as
Investigation through graph
lim f(x) = a x – 3x
3
2
x → 0 f(x) = = x – 3
x y
2. The investigation and determination of the limit
value is done through graph and by constructing Observe that when 0 x
a table. x → 0, f(x) → –3. Therefore
3
lim
3. The determination of limit value can also be x → 0 1 x – 3x 2 = –3 –3
x
done through direct substitution such that the
a
product of the value , (a = constant), which is Direct substitution
0
2
3
lim
lim
not defined cannot be accepted and the function x → 0 1 x – 3x 2 = x → 0 1 x(x – 3) 2
x
x
needs to be simplified to get the correct value. lim
= x → 0 (x – 3)
2
2
1 = (0) – 3
= –3
lim
Find the value of x → 0 f(x) if
(a) f(x) = x + 3, REMEMBER!
x – 3x
3
(b) f(x) = .
x Simplify the function first before doing direct substitution
Solution because lim x – 3x 0 – 3(0)
3
3
(a) Constructing Table x → 0 1 x 2 = 0
x –0.1 –0.01 –0.001 0 0.001 0.01 0.1 produces result which is not defined, that is 0 0 and this
is not the correct solution.
f(x) 2.9 2.99 2.999 3 3.001 3.01 3.1
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
The table above shows that when x → 0, x + 3 → 3 of a function f(x) by using the first
either from the left (→) or from the right ( ). principle
Therefore,
1. Differentiation is about the changes of a variable
Form 5
lim with respect to the changes of another variable
x → 0 (x + 3) = 3
(normally time). Generally, differentiation is
Investigation through graph y the process of determining the instantaneous
Observe that when changes happen to a given variable at a particular
x → 0, f(x) → 3. Therefore, 3 moment.
lim
3
x → 0 (x + 3) = 3 x 2. Graphically, assuming that the graph f(x) = x
–3 0 representing the motion of an object, therefore
Direct substitution the gradient on the curve at a particular moment
lim
x → 0 (x + 3) = 0 + 3 t and t can be determined through the process
1
2
= 3 of diferentiation.
242
Additional Mathematics SPM Chapter 2 Differentiation
y y = f(x)
f(x) = x 3
(function of the curve)
t 2
tangent O t 1 x dy = f'(x)
dx
3. The gradient of the tangents at the moment t (first derivative of the function of the curve)
1
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and t is the changes of f(x) as the result of the
2
changes in time which is the speed that is shown 7. The use of the idea of limit in deriving the first
through the motion of the object at the instance derivative of a function f(x) is known as the
t or t . differentiation through first principle.
1 2
4. The gradient of the tangent which happens at the
instance t and t can be determined through the 2
2
1
idea of limit, that is: Determine the first derivative of each of the following
y function f(x) through first principle.
(a) f(x) = 2x (b) f(x) = 5x
2
Q (x y )
chord 2, 2 (c) f(x) = 3x – 8x (b) f(x) = 6
2
x
P tangent Solution
(x y )
1, 1
O x 1 x (a) Let y = f(x), Assuming that x
(t )
1 that is y = 2x experiences a small
Observe that when the point Q is approaching y + dy = 2(x + dx) change (dx) which
point P, the chord PQ is gradually approaching causes a small
and overlapping with the tangent at point P. 2x + dy = 2x + 2dx change (dy) in y.
dy = 2dx
Through the idea of limit, the gradient of tangent dy = 2 Simplify until dy is
dx
of the curve at point P, m = lim y – y 1 . dx obtained by substituting
2
x → x
2 1 x – x 1 lim dy y = 2x.
2
Assuming dx = x – x and dy = y – y , \ dx → 0 dx = 2
1
1
2
2
when x → x , dx → 0. and dy = 2 Use the idea of limit so
that the first derivative of
2
1
y – y
lim
\ m = x → x 1 x – x 1 1 dx function f(x) is obtained.
2
2
2
= lim dy (b) Let y = f(x), 2
that is y = 5x
dx → 0 dx
y + dy = 5(x + dx)
2
SPM Tips 5x + dy = 5[x + 2xdx + (dx) ]
2
2
2
5x + dy = 5x + 10xdx + 5(dx)
2
2
2
dx and dy is read as “delta x” and “delta y”.
dy = 10xdx + 5(dx) Form 5
2
dy = 10x + 5dx
dy dx
5. The symbol is used to represent the gradient
dx lim dy
m which is obtained through the idea of limit \ dx → 0 dx = 10x + 0
and is known as gradient function, that is dy
and = 10x
dx
dy = lim dy
dx dx → 0 dx REMEMBER!
2
6. The gradient of function dy is the first derivative Be careful with the expansion of (x + dx) . Observe that
dx ≠ (dx) .
2
2
dx
of function f(x), that is
243
Additional Mathematics SPM Chapter 2 Differentiation
(c) Let y = f(x), 2. Find the limit for each of the following when x → 0.
that is y = 3x – 8x (a) (b) y
2
y + dy = 3(x + dx) – 8(x + dx) y
2
3x – 8x + dy = 3[x + 2xdx + (dx) ] – 8x – 8dx O x
2
2
2
3x – 8x + dy = 3x + 6xdx + 3(dx) – 8x – 8dx O x
2
2
2
dy = 6xdx + 3(dx) – 8dx y = –3x – 2 y = 3 – x 3
2
2
dy = 6x + 3(dx) – 8
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dx (c) 3 x + 5 (d) 2 – x
2
2 3 + x
lim dy
2
\ dx → 0 dx = 6x + 0 – 8 (e) 4x – 3x
5x
and dy = 6x – 8 3. Find the values of each of the following.
dx
(a) lim (3 – 2x + x )
2
x → 0
REMEMBER! (b) x → 0 (2 – x)(5 – x)
lim
4
Each of the x term must be added with dx because each (c) lim x + 5
of them experience a small change. x → 0 x – 10
2
(d) lim x + 8
x → 0 (x – 2)(x + 4)
lim x – 5x
(d) Let y = f(x), (e) x → 0 3 4x 3 4
that is y = 6 lim x – 16
2
x (f) x → 0 x – 4
y + dy = 6
x + dx 4. Differentiate each of the following function by using
6 + dy = 6 first principle.
x x + dx (a) f(x) = 6x (b) f(x) = 1 – 2x 2
1
6
2
dy = 6 – (c) f(x) = (2x – 1) (d) f(x) = – x
x + dx x 5. Determine the first derivative of each of the following
dy = 6x – 6(x + dx) functions by using first principle.
(x + dx)(x) (a) y = 4(1 – 3x) (b) y = 6x – 2
2
2
dy = 6x – 6x – 6dx x
(x + dx)(x) (c) y = – 2 (d) y = 5
dy = –6 3 8x
dx (x + dx)(x)
lim dy
–6
\ dx → 0 dx = (x + 0)(x) 2.2 The First Derivative
and dy = 6
dx x 2
A Deriving the formula of first
Form 5
derivative inductively for the
Try Questions 4 – 5 in ‘Try This! 2.1’
function y = ax n
Try This! 2.1 1. The first derivative differentiation involves the
dy
process of obtaining from y or f'(x) from f(x).
dx
1. Determine the limit value of each of the following 2. If y = ax where a and n are constants, then
n
by using (i) table, (ii) graph, (iii) direct substitution
method.
lim
n
(a) x → 0 (8x) dy = dy [ax ] = f'(x) = nax n – 1
dx
dx
(b) lim (x + 3)
2
x → 0 Observe that this formula is obtained inductively.
244
Additional Mathematics SPM Chapter 2 Differentiation
3 Therefore, inductively,
dy
n
Given y = 2x , find dy for n = 1, 2 and 3. dx = 2 = 1(2x 1 – 1 )
dx dy
Hence, verify the first derivative formula dy = nax n – 1 dx = 4x = 2(2x 2 – 1 )
inductively. dx dy 2 3 – 1
dx = 6x = 3(2x )
formula.Rights Reserved.
Solution
(a) When n = 1, dy = n(ax n – 1 ) = nax n – 1
y = 2x dx
y + dy = 2(x + dx)
2x + dy = 2x + 2dx B Determining the first derivative of
an algebraic function
dy = 2dx
dy = 2 1. For any function f(x) or y written in the
form of ax where a and n are constants, f'(x) or
n
dx dy n – 1
= ax
.
lim dy
dx
\ dx → 0 dx = 2 2. When determining the first derivative of an
and dy = 2 algebraic functions, make sure that each term
dx is written in the form of ax before applying the
n
When n = 2,
y = 2x 2 3. dy or f'(x) is known as a gradient function and it
dx
y + dy = 2(x + dx) 2 is used to determine the gradient of the tangent
2
2x + dy = 2[x + 2xdx + (dx) ] to a curve of y or f(x) at a point on the curve.
2
2
2x + dy = 2x + 4xdx + 2(dx)
2
2
2
dy = 4xdx + 2x(dx) 4. In summary:
2
• function y = a (a = constant)
dy = dx[4x + 2xdx] dy
dy = 4x + 2xdx dx = 0
dx • function y = ax (a, n = constant)
n
lim dy
dy
Penerbitan Pelangi Sdn Bhd. All dx = nax n – 1
= 4x
\
dx → 0
dx
n
m
and dy = 4x • function y = ax + bx + …
dx (a, n, b, m = constant)
dy
When n = 3, dx = nax n – 1 + mbx m – 1 + …
y = 2x 3
y + dy = 2(x + dx) 3 4 Form 5
2x + dy = 2[x + 3x dx + 3x(dx) + (dx) ] Find the first derivative of each of the following
3
3
2
3
2
2x + dy = 2x + 6x dx + 6x(dx) + 2(dx) function.
3
3
2
3
2
dy = 6x dx + 6x(dx) + 2(dx) (a) y = 6x
2
3
2
3
dy = 6x + 6xdx + 2x(dx) (b) y = 5
2
2
4x
2
dx (c) y = 3√x
lim dy
\ dx → 0 dx = 6x 2 Solution
3
and dy = 6x (a) y = 6x
2
dx dy = 3[6x 3 – 1 ]
dx
2
dy = 18x
dx
245
Additional Mathematics SPM Chapter 2 Differentiation
–3
(b) y = 5 f'(x) = –12x – 8
4x 2 = – 12 – 8
5
y = x x 3
–2
4
dy = –2 3 5 x –2 – 1 4 REMEMBER!
dx 4 Terms are simplified to the form of ax before differentiation
n
5
= – x is done.
–3
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2
= – 5
2x 3 Try Questions 1 – 4 in ‘Try This! 2.2’
(c) y = 3√x
1
y = 3x 2 C Determining the first derivative of
dy = [3x 1 2 – 1 ] composite function
1
dx 2 1 1. A composite function is a function with brackets
3
–
2
= x that is impossible or not easy to expand.
2
= 3 2. The first derivative of composite function
2√x y = u such that u = f(x) and n = integer is done
n
by using chain rule, that is
5 dy dy du
Differentiate each of the following function with dx = du × dx
respect to x.
(a) f(x) = 4x – 8x + 3 3. The limit idea is used to prove the truth of chain
3
(b) f(x) = (5x – 2) rule such as:
2
2
(c) f(x) = 6x – 8x 4 dy lim dy
x 3 dx = dx → 0 dx
Solution lim dy du
(a) f(x) = 4x – 8x + 3 = dx → 01 du × dx 2
3
f'(x) = 3(4x 3 – 1 ) – 1(8x 1 – 1 ) + 0(3x 0 – 1 ) lim dy lim du
= 12x – 8x + 0 = dx → 0 du × dx → 0 dx
2
0
= 12x – 8
2
dy = lim dy × lim du
dx du → 0 du dx → 0 dx
(b) f(x) = (5x – 2) dy dy du
2
2
= (5x – 2)(5x – 2) dx = du × dx
2
2
= 25x – 10x – 10x + 4
4
2
2
= 25x – 20x + 4 3. Since dy = nu n – 1 , then dy = nu n – 1 du .
2
4
1 2
f'(x) = 100x – 40x du dx dx
3
1 2
Form 5
The equation dy = nu n – 1 du is seen as the
REMEMBER! dx dx
formula taken from the chain rule which is often
• The derivative of a constant = 0.
• Terms with brackets which can be expanded should used for composite function.
be expanded first before differentiation is done.
6
(c) f(x) = 6x – 8x 4 Differentiate each of the following function with
x 3 respect to x by using the chain rule.
= 6x – 8x 4 4
x 3 x 3 (a) y = (3x + 2)
6
= 6 – 8x (b) y = 5(1 – 3x)
x 2
= 6x – 8x
–2
246
Additional Mathematics SPM Chapter 2 Differentiation
Solution
(a) Let u = 3x + 2 and y = u REMEMBER!
4
du = 3 dy = 4u
3
dx du Do not forget to differentiate the terms inside the brackets.
\ dy = dy × du
dx du dx Try Questions 5 – 9 in ‘Try This! 2.2’
= 4u × 3
3
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= 12u
3
= 12(3x + 2) D Determining the first derivative of
3
a function involving product and
quotient of algebraic expressions
(b) Let u = 1 – 3x and y = 5u
6
du = –3 dy = 30u 1. The first derivative of the function y = uv such
5
dx du that u = f(x) and v = g(x) is obtained from the
\ dy = dy × du formula
dx du dx
d
= 30u × (–3) dx (uv) = u dv + v du
5
dx
dx
= –90u
5
= –90(1 – 3x)
5
2. This formula can be verified by using the idea of
7 limit as follows:
Let y = uv …… 1
Differentiate each of the following function with respect From the idea of limit,
1 2
to x by using the formula dy = nu n – 1 du . y + dy = (u + du)(v + dv)
dx dx y + dy = uv + udv + vdu + dudv …… 2
(a) y = 4 (b) y = √1 – 5x
2x – 1 Substitute 1 into 2,
uv + dy = uv + udv + vdu + dudv
Solution dy = udv + vdu + dudv
(a) y = 4
2x – 1 Divide each term with dx,
–1
y = 4(2x – 1) dy = u dv + v du + dudv
dy = –1[4(2x – 1) –1 – 1 ] 3 d (2x – 1) dx dx dx dx
4
dx dx Through limit,
= –4(2x – 1) (2)
–2
lim du
lim dudv
lim dy
lim dv
= –8(2x – 1) dx → 0 dx = u dx → 0 dx + v dx → 0 dx + dx → 0 dx
–2
= – 8
(2x – 1) 2 It is known that when dx → 0, dudv → 0.
SPM Tips Therefore,
lim dy = u lim dv + v lim du Form 5
n
–1
Assume 4(2x – 1) as 4x and differentiate for ax dx → 0 dx dx → 0 dx dx → 0 dx
n
as usual. dy dv du
and = u + v
dx dx dx
(b) y = √1 – 5x d dv du
1 or dx (uv) = u dx + v dx
2
y = (1 – 5x)
dy = (1 – 5x) 1 2 – 1 (–5) 3. The first derivative of the function y = u such
1
v
dx 2 that u = f(x) and v = g(x) is obtained by using the
5 – 1
= – (1 – 5x) 2 du dv
2 d u v dx – u dx
1 2
= – 5 formula dx v = v 2 .
2√1 – 5x
247
02 [Focus AddMath F5]_ENG 2021.indd 247 08/11/2021 3:14 PM
Additional Mathematics SPM Chapter 2 Differentiation
4. This formula can be verified by using the idea of From the product formula derived,
limit as follows: dy = u dv + v du
u
Let y = …… 1 dx dx dx
v = (3x)[6(2x + 1) ] + (2x + 1) (3)
3
2
From the idea of limit, = 18x(2x + 1) + 3(2x + 1)
2
3
y + dy = u + du …… 2 = 3(2x + 1) (6x + 2x + 1)
2
v + dv = 3(2x + 1) (8x + 1)
2
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Substitute 1 into 2,
u + dy = u + du (b) f(x) = x + 3
v v + dv 3 – 2x
dy = u + du – u Let u = x + 3 and v = 3 – 2x
v + dv v du dv
dy = v(u + du) – u(v + dv) dx = 1 dx = –2
(v + dv)(v)
dy = uv + vdu – uv – udv From quotient formula derived,
(v + dv)(v) v du – u dv
dy = vdu – udv f'(x) = dx 2 dx
v(v + dv) v
Divide both sides with dx, = (3 – 2x)(1) – (x + 3)(–2)
2
(3 – 2x)
dy = vdu – udv ÷ dx
dx v(v + dv) = 3 – 2x + 2x + 6
2
v du – u dv (3 – 2x)
9
= dx dx =
v(v + dv) (3 – 2x) 2
Through limit, Try Questions 10 – 11 in ‘Try This! 2.2’
v lim du – u lim dv
lim dy = dx → 0 dx dx → 0 dx
dx → 0 dx 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. 7
6
dx → 0 (a) y = 5x (b) y = 4x 2
Therefore, (c) y = √x (d) y = 6x 3 2
v lim du – u lim dv 4
lim dy = dx → 0 dx dx → 0 dx (e) y = – 9 (f) y = 3 √x
dx → 0 dx v(v) 3 √x 4
v du – u dv 2. Find f'(x) for each of the following functions.
1 2
and dy = d u = dx dx (a) f(x) = 4x + 5x – 8x
3
2
dx dx v v 2
3
Form 5
(b) f(x) = 3x + 8x 2
5
8 (c) f(x) = + 2 – 3 + 4
1
x
x
x
2
3
Differentiate each of the following function with (d) f(x) = x (1 + 3x – x ) x 4
4
2
respect to x. (e) f(x) = 3√x (1 – 5√x )
(a) y = 3x(2x + 1) (f) f(x) = (3 – √x )
3
2
(b) f(x) = x + 3
3 – 2x 3 dy
Solution 3. Given 12x = 3y, determine the value of dx if
(a) Let u = 3x and v = (2x + 1) (a) x = – 1
3
du = 3 dv = 3(2x + 1) (2) 4 2
2
dx dx = 6(2x + 1) (b) x = 3
2
248
Additional Mathematics SPM Chapter 2 Differentiation
1
4. Given the volume of a cone is V = pr h such that r is 2.3 The Second Derivative
2
3
the radius and h is the height of the cone. Determine
(a) dV if h = 2 cm
dr
dh A Determining the second derivative
(b) if r = 5 cm
dV of an algebraic function
5. Determine dy for each of the following functions. 1. Observe that
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dx y = f(x)
1
(a) y = 2x – 1 2 2 dy = f'(x) First derivative
x
2
(b) y = 9x – 1 dx
1 + 3x d y = f"(x)
2
(7 + 3x) 2 dx 2 Second derivative
(c) y =
x 2
2
(d) y = x (5 – x) 2 2. The second derivative, d y 2 is obtained through
2
dx
6. Differentiate each of the following with respect to x. the similar processes of differentiation as in the
1
(a) f(x) = 3 6x – 3 x 2 3 first derivative.
(b) f(x) = 8x – 4 3. dy is the gradient function of the curve y = f(x)
2
(c) f(x) = (3x – 4) 4 dx
5 d y
2
3 at a point (x, y). is the rate of change of the
(d) y = dx 2
6x – 3 gradient of a curve with respect to x.
(e) y = 1
4(5 – 3x) 2
5x
(f) y = – 9
x – 2x 2 2
3
dy Find d y for each of the following functions.
7. Given y = u and u = 5x + 8, find the value of when dx 2
6
x = –3. dx (a) y = 3x (b) y = 8(2x – 3)
4
5
dh
8. If h(k – 2k) = 5, find the value of when k = 1.
2
dk Solution
dy (a) y = 3x
5
9. Given y = k(2x – 1) and = 40(2x – 1) , find the
4
n
values of k and n. dx dy = 15x REMEMBER!
4
dx
10. Determine the first derivative for each of the 2 d y
2
following functions: d y = 60x dx 2 can only be
3
(a) y = x(2x – 1) dx 2 derived from dy .
4
(b) y = (x + 4) 2x – 3 (b) y = 8(2x – 3) dx
4
(c) f(x) = (x – 5) (2x + 7) dy 3
2
3
(d) f(x) = x (9x – 2) 5 dx = 32(2x – 3) (2)
3
(e) f(x) = 3x(1 – x)(5 + x) 4 d y = 64(2x – 3) Form 5
2
1
2
1
(f) f(x) = (2x – 3) – 2x 2 2 dx 2 = 192(2x – 3) (2)
3
x
= 384(2x – 3)
2
11. Differentiate each of the following with respect to x.
5x 2
(a) f(x) = 10
2x – 1
(b) f(x) = √x + 1 Find the value of f"(0) for each of the following
√x – 1 functions.
√2 – x (a) f(x) = x (1 – 3x)
2
(c) f(x) =
x 2 2x – 3x 2
4
(1 – 3x) 4 (b) f(x) =
(d) f(x) = x
x 3
(e) y = 3x – 8x 2 Solution
2
2 + 5x – x 2 (a) f(x) = x (1 – 3x)
x – 2x + 6 = x – 3x
3
2
3
(f) y =
8 – x
249
Additional Mathematics SPM Chapter 2 Differentiation
f'(x) = 2x – 9x
2
f"(x) = 2 – 18x 2.4 Application of Differentiation
f"(0) = 2 – 18(0)
= 2
A Interpreting the gradient of tangent
4
(b) f(x) = 2x – 3x 2 to a curve at different points
x
= 2x – 3x 1. The first derivation of a function y = f(x), that is
3
f'(x) = 6x – 3 dy = f'(x) is the gradient function which gives
2
dx
f"(x) = 12x us the value of the gradient of the tangent to the
f"(0) = 0 curve y = f(x) at a point on the curve.
Try Questions 1 – 7 in ‘Try This! 2.3’ xReserved.
y
A
Try This! 2.3
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y = f(x)
2
1. Determine dy and d y 2 for each of the following 0 tangent m = dy
functions. dx dx dx
(a) y = 8x (b) y = 7x – 8x 2. The gradient of the tangent line drawn at a point
4
3
2
x
(c) y = – √x (d) y = x (1 – 2x) A(x , y ) on the curve is obtained by substituting
3
2 1 1 dy
(e) y = 5(4x – 3) (f) y = 6x – 5x 3 the value of x into = f(x).
6
x 2 1 dx
2. Find f '(x) and f"(x) for each of the following functions. 3. The accuracy of the value of gradient m, can
1
2
(a) f(x) = 7x – 5x + be tested/verified by drawing a graph manually
x
(b) f(x) = 4x (x + 2) on a graph paper or by using computer apps/
2
2
(c) f(x) = 6 programs.
x + 3
3x
(d) f(x) =
1 – x 11
3. Find the value of f"(0) for each of the following.
3
4
(a) f(x) = (9 – 3x) Find the gradient of the tangent to the curve y = 4x – 2
(b) f(x) = 4x(x – 3) at the given points.
2
(c) f(x) = x + 3 (a) Point A(1, 2)
x – 3
4. Given the function f(x) = 3x + mx – 4. Find (b) Point B 1 1 , – 3 2
5
3
(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) . 3 3
2
dy d y (a) y = 4x – 2 (b) y = 4x – 2
2
(b) Find and . dy dy
2
2
dx dx 2 = 12x = 12x
Form 5
2
(c) Show that 2y – x dy + (3x – 9) d y = 0. dx = 12(1) 2 dx 1 2
dx dx 2 = 12 1 2
= 12 2
6. Given the function y = x(4 – x).
1
1 dy d y = 12 1 2
2
(a) Show that y – x + x = 0. 4
2 dx dx 2
(b) Find the possible values of x when = 3
2
y = x dy + 2x d y .
dx dx 2 12
7. Given the function y = 3(x – 1) . Calculate the gradient of the tangent to the curve
2
2
(a) Express y d y + dy in terms of x.
dx 2 dx y = 6 + 2x at the point where the y-coordinate is –2.
2
(b) Solve the equation y d y + dy = 0. x
dx 2 dx
250
Additional Mathematics SPM Chapter 2 Differentiation
Solution Solution
When y = –2, From y = 3x – 9x + 4
2
y = 6 + 2x y = 6 + 2x dy = 6x – 9
x x dx
–1
–2 = 6 + 2x = 6x + 2 At the point (2, –2),
x 1 3 2 dy
–2x = 6 + 2x At the point – , –2 , Gradient of tangent, m = dx = 6(2) – 9
2
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1
4x = –6 \ dy = –6x –2 = 3
x = – 3 dx 6 Equation of tangent line,
2 = – 2 y – y = m (x – x )
1
1
1
1 – 3 2 2 y – (–2) = 3(x – 2)
y + 2 = 3x – 6
= – 24 y = 3x – 8
9
= – 8 Since m × m = –1
2
1
3 3 × m = –1
2
Try Question 1 in ‘Try This! 2.4’ Therefore, the gradient of normal, m = – 1
2 3
Equation of normal line,
B Determining the equation of y – y = m (x – x )
1
1
2
tangent and normal to a curve at a 1
point y – (–2) = – (x – 2)
3
1. The equation of the tangent to a curve y = f(x) y + 2 = – x + 2
1
at a given point (x , y ) is determined by using 3 3
1
1
the formula y – y = m (x – x ), such that y = – x + – 2
1
2
1
1
1
dy 3 3
m = = f'(x ).
1 dx 1 1 4
= – x –
3 3
y
y = f(x)
(x y ) C Solving problems involving tangent
1, 1
x and normal
O tangent
1. There are a few things involved in the problems
involving the tangent and normal, which are
2. The equation of the normal to a curve y = f(x) at (a) problems about the curve y = f(x) such that
a given point (x , y ) on the curve is determined the information about the point and the
1
1
by using the formula y – y = m (x – x ), such that equation of tangent or normal are given, Form 5
1
2
1
m × m = –1. (b) problems about finding the point (x , y )
1
2
1
1
such that the information about the curve
y y = f(x) and the equation of tangent or
m 2 m 1
normal are given,
(x y ) (c) problems which involve the equations of
1, 1
Normal tangent or normal or both such that the
x
O information about the curve y = f(x) and the
point (x , y ) are given.
1
1
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 = 3x – 9x + 4 at the point (2, –2). steps.
2
251
Additional Mathematics SPM Chapter 2 Differentiation
14 Example of HOTS
HOTS Question
Given a curve y = 5x – 3x – 2. Find the coordinates
2
of the point on the curve such that the gradient of the A research done on the movement of an object
found out that it follows the function f(x) = ax + bx
3
2
tangent at the point is 7. Hence, find the equation of + c such that a, b, and c are constants. Given that
the tangent. the curve of the function f(x) passes through points
Solution (–1, 0) and (0, 5) when the graph of the function f(x)
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2
y = 5x – 3x – 2 y = 5x – 3x –2 is drawn on the Cartesian plane. The tangents to the
2
curve f(x) at the points where x = 0 and x = 1 are
dy = 10x – 3 (x y ) parallel to the x-axis. Determine the function f(x) by
dx 1, 1 finding the values of a, b and c.
7 = 10x – 3 m = 7
10x = 10 Solution
x = 1 Given the point (–1, 0), 2
3
f(x) = ax + bx + c
Therefore, y = 5(1) – 3(1) – 2 0 = a(–1) + b(–1) + c
2
3
2
= 0 0 = –a + b + c …… 1
The coordinate of the point is (1, 0). For the point (0, 5),
3
2
Equation of tangent, 5 = a(0) + b(0) + c
c = 5
y – y = m (x – x ) Therefore, 0 = a + b + 5
1
1
1
y – 0 = 7(x – 1) a – b = 5 …… 2
y = 7x – 7
f(x) = ax + bx + c
3
2
SPM Tips f'(x) = 3ax + 2bx
2
For x = 1,
Make a rough sketch for easier understanding. 0 = 3a(1) + 2b(1)
2
3a + 2b = 0 …… 3
2 × 3, 3a – 3b = 15 …… 4
15 3 – 4, 5b = –15
b = –3
Given that the gradient of the curve y = hx + kx at the From 2, a – (–3) = 5
2
point (2, 8) is 6. Find the values of h and k. a + 3 = 5
a = 2
Solution \ f(x) = 2x – 3x + 5
2
3
Given the gradient = 6,
y = hx + kx
2
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 = hx + kx Try this HOTS Question
2
Form 5
8 = h(2) + k(2)
2
8 = 4h + 2k …… 2 The motion of an object follows the equation
y = x – 6x + 5. The tangent to the curve at a point
2
2 – 1: k = 2 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
The gradient of the curve refers to the gradient Answer: y = – x – 1; 1 3 , – 7 2
of the tangent. 2 2 4
Try Questions 2 – 5 in ‘Try This! 2.4’
252
Additional Mathematics SPM Chapter 2 Differentiation
(a) Tangent sketching method
D Determining the turning points and
their nature
(i) Maximum
1. Turning points or stationary points, are the
points on a curve where the gradient of the
tangent at the points are 0. The points have either (ii) Minimum
maximum or minimum value.
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(a) Maximum point (iii) Inflection or
point
y
0
positive negative (b) Second derivative method
2
x d y
O (i) Maximum , 0
dx 2
The value of the gradient of the tangent (ii) Minimum d y . 0
2
changes from positive to zero to negative, dx 2
from left to right.
(iii) Inflection d y = 0
2
(b) Minimum point point dx 2
y
16
negative positive Find the turning points of the function y = x – 6x + 9x + 2
3
2
0 and determine the nature of each turning point.
x
O
Solution
The value of the gradient of the tangent y = x – 6x + 9x + 2
3
2
changes from negative to zero to positive,
from left to right. dy = 3x – 12x + 9
2
dx
2. If the value of the stationary point is neither At the turning point,
maximum nor minimum, it is called an 0 = 3x – 12x + 9
2
inflection point. x – 4x + 3 = 0
2
y y (x – 3)(x – 1) = 0
negative
positive x = 3, x = 1
x x
O O When x = 3,
positive negative Form 5
y = (3) – 6(3) + 9(3) + 2
3
2
The value of the gradient of the tangent changes = 27 – 54 + 27 + 2
from positive or negative to zero to positive or = 2
negative back in both directions.
When x = 1,
3. To determine the turning point, let dy = 0. y = (1) – 6(1) + 9(1) + 2
3
2
dx
= 1 – 6 + 9 + 2
4. To determine the nature of the turning point,
either maximum, minimum or inflection point, = 6
the following two methods can be applied: \ The turning points are (3, 2) and (1, 6).
253
Additional Mathematics SPM Chapter 2 Differentiation
Determining the nature of the turning points: Solution
Tangent sketching method y = x + ax + b
3
At (3, 2): At (0, 4), 4 = (0) + a(0) + b
3
Choose one value Choose one value b = 4
where x , 3 where x . 3
2
x 2 3 4 Also, dy = 3x + a
dx
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dy 0 = 3(0) + a
2
dx –3 0 9 a = 0
Sign – zero + \ y = x + 4
3
Tangent dy 2
sketching dx = 3x
2
Graph and d y = 6x
sketching dx 2 = 6(0)
\ (3, 2) is a minimum point. = 0
\ The point (0, 4) is an inflection point.
At (1, 6):
x 0 1 2 Try Questions 6 – 8 in ‘Try This! 2.4’
dy
dx 9 0 –3 E Solving problems involving
maximum and minimum values and
Sign + zero –
interpreting the solutions
Tangent
sketching 1. Problems involving the maximum value and
minimum value can be solved by using the
Graph following steps:
sketching (a) Based on the information given in the
problem, recognise the main function
\ (1, 6) is a maximum point. (area, volume etc) that needs to be formed
or derived so that the differentiation can be
carried out. This main function can be easily
Alternative Method
recognised through the word “maximum”
Second derivative method or “minimum” stated in the problem.
d y = 6x – 12 (b) Obtain this main function and express it in
2
dx 2 terms of one variable. Normally, there would
x = 3, be one condition stated in the problem so that
d y = 6(3) – 12 the variables can be mutually substituted.
2
dy
Form 5
dx 2 = 6 . 0 (c) Let f'(x) or dx = 0 to obtain the value of
x = 1, the variable which causes the function to be
maximum or minimum.
d y = 6(1) – 12 (d) If there is more than one variables obtained,
2
dx 2 = –6 , 0 investigate the nature of the values by using
d y (second derivative method) or the
2
17 dx 2 dy
Given that the graph of the function y = x + ax + b, gradient of tangent dx .
3
where a and b are constants, has a turning point (0, 4). (e) Find the maximum value or the minimum
Determine the values of a and b. Hence, determine the value of the function according to the
nature of the turning point. requirement of the problem.
254
Additional Mathematics SPM Chapter 2 Differentiation
18 Solution
A wire with length 84 cm is bent to form a rectangle.
Find the maximum area of the rectangle that can be height = h
formed. width = x
Solution length = 2x
Area of rectangle = A From the analysis of the problem, the production
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Let A = length × width Width = y cost depends on the total surface area of the box
= x × y produced.
= xy Length = x
Therefore, A = total surface area
Length of wire = perimeter of the rectangle = 2[(2x)(x)] + 2(xh) + 2[(2x)(h)]
x + x + y + y = 84
2x + 2y = 84 h
2x = 84 – 2y x x h
x = 42 – y 2x 2x
2
Substitute into A, \ A = 4x + 2xh + 4xh
A = (42 – y)(y) Volume of the box given = 72 cm
3
= 42y – y Length × width × height = 72
2
dA = 42 – 2y 2x × x × h = 72
dy 72
At the turning point, h = 2x 2
0 = 42 – 2y = 36
2y = 42 x 2
y = 21 Substitute h = 36 into A,
x 2
2
36
36
Also, d A = –2 , 0 A = 4x + 2x 1 2 + 4x 1 2
2
dy 2 x 2 x 2
2
y = 21 will caused the value of A to be maximum. = 4x + 72 + 144
x
x
2
SPM Tips = 4x + 216
x
dA –2
A can also be expressed in terms of x to determine \ dx = 8x – 216x
the value of x.
At the turning point, 8x – 216 = 0
2
A max = 42(21) – (21) x 8x = 216
2
= 441 cm x 2
2
3
8x = 216
\ The maximum area of the rectangle that can be x = 27
3
formed is 441 cm . x = 3 Form 5
2
2
Try Questions 9 – 12 in ‘Try This! 2.4’ Also, d L = 8 + 432x
–3
dx 2
d L . 0
2
When x = 3,
Example of HOTS dx 2
HOTS Question
x = 3 will cause the value of A to be minimum.
A factory plans to produce a closed cuboid-shape 216
box such that the volume of the box produced is A = 4(3) + = 108 cm 2
2
min
72 cm . Given that the length of the box is twice its 3
3
width. The production cost of the box is estimated The minimum production cost of a box produced is
to be RM0.15 per cm . Find the minimum production = 108 × 0.15
2
cost that can be obtained in the production of a = RM16.20
box.
255
Additional Mathematics SPM 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 cm . The factory estimates that the production between the variables y and x. The changes of
3
2
cost would be RM0.18 per cm . 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,
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Answer:
229 dy = dy × dx
dt dx dt
2. The above relationship exists between dy ,
dt
dy and dx follows the chain rule which helps
SPM Highlights dx dt
in solving 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 dt
= × 1 = 2
5 m dx dt dx dx dx
dt dt
Horizontal ground level dx 1 dy 1 dx
or = × 1 = 2
dt dy dt dy dt
The curve part of the roller coaster track is represented dt dx
by an equation y = 1 x – 3 x , with point p as the
2
3
64 16 dy dx
origin.Find the shortest vertical distance, in m, from the 3. dt . 0 or dt . 0 denotes the increase in the
track to ground level. dy
rate of change of the variable, while dt , 0
Solution or dx , 0 denotes the decrease in the rate of
y = 1 x – 3 x dt
3
2
64 16
dy 3 3 change of the variable.
2
= x – x
dx 64 8
At the turning point, 19
3
3 x – x = 0 Two variables x and y are related by the equation
2
64 8 14
4
3 x 3 1 x – 1 = 0 y = 2x + x . If y increases with a rate of 8.5 units per
8 8
x = 8, x = 0 second, find the rate of change of x when x = 2. State the
d y 3 3 type of rate of change of x at that particular moment.
2
Form 5
Also, = x –
dx 2 32 8 Solution
2
For x = 8, d y . 0 (y is minimum.) Given dy = 8.5 (Positive value because y is increasing.)
dx 2 dt
–1
2
For x = 0, d y , 0 (y is maximum (ignored)) and y = 2x + 14x
dx 2 dy
–2
1 3 = 2 – 14x
Substitute x = 8, y = (8) – (8) 2 dt
3
64 16 = 2 – 14
= –4 x 2
\ The shortest vertical distance is 5 – 4 = 1 m. When x = 2, dy = 2 – 14
dt (2) 2
= – 3
2
256
Additional Mathematics SPM Chapter 2 Differentiation
From dy = dy × dx G Solving problems involving rates of
dt dx dt change for related quantities and
3
8.5 = – × dx interpreting the solutions
2 dt
\ The rate of change of x, dx = – 17 units per second. 1. The rate of change of variable (or quantity y)
dt 3 can be determined by using the formula
x experiences a rate of decreasing at the moment when dy dy dx
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x = 2 because dx , 0. dt = dx × dt if y = f(x) and x = f(t).
dt
2. Observe that dy is the rate of change of y and
20 dx dt
The relationship between two variables p and q is dt is the rate of change of x.
p q = 4. Given that p decrease with a rate of 3 unit s ,
2
–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
Given dp = –3 (Negative because the rate decreased) between these quantities are the main focus.
dt
From p q = 4,
2
q = 4 21
p 2 The area of the wave of water on the surface of a water
–2
q = 4p tank expands with a rate of 5 cm s . Find the rate of
2 –1
dq = –8p change of its cicular radius when the area of the circle
–3
dp is 6.25π cm .
2
= – 8
p 3 Solution
Given q = 64, therefore p q = 4 dA 2 –1
2
p (64) = 4 Given dt = 5 cm s
2
2
p = 1 Also, A = πr
2
16 dA
p = ± 1 dr = 2πr
4 When A = 6.25π
1
When p = ± , πr = 6.25π
2
4
dq = – 8 or dq = – 8 r = 2.5 cm
1
dp 1 2 3 dp 1 – 1 2 3 From dA = dA × dr
4
4
= –512 = 512 When r = 2.5, dt dr dt Form 5
Therefore, dq = dq × dp dr
dt dp dt 5 = 2π(2.5) × dt
dq dq dr 5
When = –512, = –512 × (–3) dt = 5π
dp dt = 1 536 unit s 1 –1
–1
π
When dq = 512, dq = 512 × (–3) = cm s
dp dt = –1 536 unit s \ The radius of the circular wave increases with a rate
–1
(q may increase or decrease at the rate of 1 536 units 1
–1
2
per second.) of cm s when the area of the circle is 6.25π cm .
π
Try Questions 13 – 14 in ‘Try This! 2.4’
257
Additional Mathematics SPM 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 5. dx . 0 or dy . 0 denotes the small increase in x
of 15 cm s . Find the rate of change of side length, in or y, dx , 0 or dy , 0 denotes the small decrease
2 –1
3
–1
cm s , when the volume of the cube is 125 cm .
in x or y.
Solution
x 22
2
x Given y = 3x + 6x, find the small change in y when x
x changes from 4 units to 4.02 units.
Let the length of the side of the cube = x cm
2
Therefore the area of the cube, A = 6x Solution
2
3
When the volume of the cube = 125 cm , From y = 3x + 6x
x = 125 dy = 6x + 6
3
x = 5 cm dx
dA When x = 4,
\ dx = 12x dy
= 12(5) dx = 6(4) + 6
= 60 = 30
and dA = 15 cm s dx = 4.02 – 4
2 –1
dt = 0.02
dA dA dx dy
From = × Therefore, dy = × dx
dt dx dt dx
dx
= 30 × 0.02
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15 = 60 ×
dt = 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 √63.98 by using
3
4
when the volume of the cube is 125 cm . differentiation.
3
Solution
3
Try Questions 15 – 16 in ‘Try This! 2.4’ Let y = √x
1
= x
3
H Interpreting and determining small \ dy = 3x
1
Penerbitan ≈ dy when dx → 0. Choose a value of x close to 63.98, such that its cube
changes and approximations of
2
dx
3
lim dy
3
1. From dx → 0 dx = dy , where dx is the small Let 3 √63.98 = √64 + dy
dx
SPM Tips
changes in x and dy is the small changes in y, it is
Form 5
dy
found out that
dx
dx
dy
dy
dy
≈
2. formula is used to estimate the approximate root value can be obtained easily. dy is the change or
can be written as dy ≈
× dx. This
the approximate difference between √63.98 and
3
dx
dx
dx
3
3
√64 , that is dy = √63.98 – √64 .
3
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
dx
dy ≈ dy × dx is not the exact change in y. The ≈ 1 × dx
dx
2
value is just an approximation to the real value 3x
3
of change.
258
Additional Mathematics SPM Chapter 2 Differentiation
1
≈ 2 × (–0.02) dx = 63.98 – 64 I Solving problems involving small
3
3(64) = –0.02 changes and approximations of
3
≈ 1 × (–0.02) When y = √64 , certain quantities
48 x = 64 1. The formula dy ≈ dy or dy ≈ dy × dx is used in
≈ – 1 or –0.0004167 dx dx dx
2 400 solving problem which involves a small changes,
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either in x or y.
1
1
Therefore, √63.98 ≈ √64 + – 2 400 2 2. Recognising two quantities involved in the
3
3
≈ 4 – 1 problem and forming the function or relation
2 400 between the two quantities are the main things
≈ 3.9996 in solving the problems.
Try Questions 17 – 18 in ‘Try This! 2.4’
24
SPM Highlights Find the approximate change in the surface area of a
sphere when the radius of the sphere decreases from
5 3.0 cm to 2.97 cm.
It is given that the equation of a curve is y = x 2 . Solution
(a) Find the value of dy when x = 3. Area = 4π (radius)
2
dx
2
(b) Hence, estimate the value of 5 . y = 4πx
(2.98) 2
and dy = 8πx
Solution dx = 8π(3)
(a) y = 5x = 24π
–2
When x = 3, dy = –10x dy
–3
dx Therefore, dy ≈ × dx
= – 10 dx
x 3 ≈ 24π × (2.97 – 3.0)
= – 10 ≈ 24π × (–0.03)
(3) 3 ≈ –0.72π cm
2
2
= – 10 \ The surface area of the sphere decreases 0.72π cm .
27 (a negative value is obtained)
5 5
(b) Let = + dy
(2.98) 2 (3) 2 SPM Highlights
such that x = 3 and dx = 2.98 – 3
= –0.02 It is given that L = 4t – t and x = 3 + 6t.
2
Small changes in y, (a) Express dL in terms of t.
dy dx
dy ≈ × dx (b) Find the small change in x when L changes from Form 5
dx
10 3 to 3.4 when t = 1.
≈ – × (–0.02)
27 Solution
1 2
≈ or 0.007407 (a) L = 4t – t x = 3 + 6t
135 dL dx
5 5 1 dt = 4 – 2t dt = 6
Therefore, ≈ +
(2.98) 2 (3) 2 135 dL dL 1
5 1 dx = dt × dx
≈ +
9 135 dt
≈ 76 or 0.5630 = (4 – 2t) × 1
135 2 – t 6
=
3
259
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