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Contents







CHAPTER Functions 1 CHAPTER Linear Law 82
6
1
Fungsi Hukum Linear
1.1 Functions 1 6.1 Linear and Non-Linear Relations 82
Fungsi Hubungan Linear dan Tak Linear
1.2 Composite Functions 6 6.2 Linear Law and Non-Linear Relations 88
Fungsi Gubahan Hukum Linear dan Hubungan Tak Linear 90
6.3
Application of Linear Law
1.3 Inverse Functions 10 Aplikasi Hukum Linear
Fungsi Songsang SPM Practice 6 94
SPM Practice 1 15 HOTS Challenge 97
HOTS Challenge 17 Online Quick Quiz QR code PAK-21 Corner QR code 97
Online Quick Quiz QR code PAK-21 Corner QR code 17
CHAPTER
CHAPTER Quadratic Functions 18 7 Coordinate Geometry 98
2
Geometri Koordinat
Fungsi Kuadratik
7.1 Divisor of a Line Segment 98
2.1 Quadratic Equations and Inequalities 18 Pembahagi Tembereng Garis
Persamaan dan Ketaksamaan Kuadratik 7.2 Parallel Lines and Perpendicular Lines 101
2.2 Types of Roots of Quadratic Equations 23 Garis Lurus Selari dan Garis Lurus Serenjang
Jenis-jenis Punca Persamaan Kuadratik 7.3 Areas of Polygons 106
2.3 Quadratic Functions 25 Luas Poligon 110
7.4
Equations of Loci
Fungsi Kuadratik
SPM Practice 2 32 Persamaan Lokus 112
SPM Practice 7
HOTS Challenge 34 HOTS Challenge 114
Online Quick Quiz QR code PAK-21 Corner QR code 34 Online Quick Quiz QR code PAK-21 Corner QR code 114
CHAPTER Systems of Equations 35 CHAPTER Vectors 115
8
3
Sistem Persamaan Vektor
3.1 Systems of Linear Equations in Three Variables 35 8.1 Vectors 115
Sistem Persamaan Linear dalam Tiga Pemboleh Ubah Vektor 119
8.2
Addition and Subtraction of Vectors
3.2 Simultaneous Equations involving One Linear Penambahan dan Penolakan Vektor
Equation and One Non-Linear Equation 40 8.3 Vectors in a Cartesian Plane 122
Persamaan Serentak yang melibatkan Satu Persamaan Vektor dalam Satah Cartes
Linear dan Satu Persamaan Tak Linear SPM Practice 8 128
SPM Practice 3 47 HOTS Challenge 131
HOTS Challenge 48 Online Quick Quiz QR code PAK-21 Corner QR code 131
Online Quick Quiz QR code PAK-21 Corner QR code 48
CHAPTER Solution of Triangles 132
CHAPTER Indices, Surds and Logarithms 49 9 Penyelesaian Segi Tiga
4
Indeks, Surd dan Logaritma 9.1 Sine Rule 132
Petua Sinus
4.1 Laws of Indices 49 9.2 Cosine Rule 137
Hukum Indeks Petua Kosinus
4.2 Laws of Surds 51 9.3 Area of a Triangle 139
Hukum Surd Luas Segi Tiga
4.3 Laws of Logarithms 56 9.4 Application of Sine Rule, Cosine Rule and Area
Hukum Logaritma of a Triangle 142
4.4 Applications of Indices, Surds and Logarithms 62 Aplikasi Petua Sinus, Petua Kosinus dan Luas Segi Tiga
Aplikasi Indeks, Surd dan Logaritma SPM Practice 9 144
SPM Practice 4 63 HOTS Challenge 146
HOTS Challenge 64 Online Quick Quiz QR code PAK-21 Corner QR code 146
Online Quick Quiz QR code PAK-21 Corner QR code 64
CHAPTER Index Numbers 147
10
CHAPTER Progressions 65 Nombor Indeks
5
Janjang 10.1 Index Numbers 147
Nombor Indeks
5.1 Arithmetic Progressions 65 10.2 Composite Index 152
Janjang Aritmetik Indeks Gubahan
5.2 Geometric Progressions 71 SPM Practice 10 158
Janjang Geometri HOTS Challenge 161
SPM Practice 5 79 Online Quick Quiz QR code PAK-21 Corner QR code 161
HOTS Challenge 81
Online Quick Quiz QR code PAK-21 Corner QR code 81 Year-End Assessment 162

Answer

© Penerbitan Pelangi Sdn. Bhd. ii

CHAPTER Functions

1 1


Fungsi






1.1 Functions Textbook
Fungsi
pg. 2 – 11

SMART Notes


1. A function from set X to set Y is a special relation that y y
maps each element x in set X to only one element y in
set Y. x 2
Fungsi dari set X kepada set Y ialah satu hubungan khas x
yang memetakan setiap unsur x dalam set X kepada hanya 0
satu unsur y dalam set Y. x
X Y 0
Function Not a function
1 2 Fungsi Bukan fungsi
2 4 5. Given an arrow diagram of a function:
3 6 Diberi gambar rajah anak panah bagi suatu fungsi:
f
x x 2
2. A function can be written as f : x → 2x or f(x) = 2x,
where x is the object and 2x is the image. 1 1
Suatu fungsi boleh ditulis sebagai f : x → 2x atau f(x) = 2x 2 4 9
dengan x ialah objek dan 2x ialah imej. 3 16
5
3. Only one-to-one relation and many-to-one relation are 25
functions. X Y
Hanya hubungan satu kepada satu dan hubungan banyak
kepada satu ialah fungsi. (a) Domain = {1, 2, 3, 5}
4. When a graph is given, vertical line test can be used (b) Codomain / Kodomain = {1, 4, 9, 16, 25}
to determine whether the graph is a function. (c) Range / Julat = {1, 4, 9, 25}
Apabila graf diberi, ujian garis mencancang boleh digunakan (d) Object of 4 is 2. / Objek bagi 4 ialah 2.
untuk menentukan sama ada graf tersebut ialah fungsi atau (e) Image of 3 is 9. / Imej bagi 3 ialah 9.
bukan.

1. Determine whether each of the following is a function. Give the reason. PL 1
Tentukan sama ada setiap yang berikut ialah fungsi atau bukan. Berikan sebabnya.
Example (a) Cube
Number of days Kuasa tiga
Bilangan hari
1 1
April 2 8
30
June/ Jun 3 27
July/ Julai 31 4 64
C D
A B
A function. Each object has one image only. A function. Each object has one image only.

(b) (c)
3
1 6 1 5
2 9 2 6
3 3 9
18
A B C D

Not a function. The object 3 has two images in Not a function. Object 2 has two images in the
the codomain. codomain.




1 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Chapter 1 Functions

2. Determine whether each of the following graphs is a function using the vertical line test. PL 1
Tentukan sama ada setiap graf yang berikut ialah fungsi atau bukan dengan menggunakan ujian garis mencancang.
Example (a) y

y
x
0
x
0



A function. When tested with a vertical line, the line Not a function. When tested with a vertical line,
cuts the graph at one point. the line cuts the graph at two points.


(b) y (c) y


x
0
x
0




A function. When tested with a vertical line, the A function. When tested with a vertical line, the
line cuts the graph at one point. line cuts the graph at one point.


3. State the domain, codomain and range for each of the following. Then, write the function by using function
notation. PL 2
Nyatakan domain, kodomain dan julat bagi setiap yang berikut. Kemudian, tulis setiap fungsi dengan menggunakan tatatanda fungsi.
Example (a) Domain f
f = {2, 3, 4, 5} x y
Domain x y 1
= {2, 4, 6, 8} 8 Codomain – 2
2 16 Kodomain 1
Codomain 4 24 1 1 1 1 – 3
Kodomain 6 8 28 = { , , , , 0} 2 3 1 –
2 3 4 5
= {8, 16, 24, 28, 32} 32 4 4
Range 5 1
Range X Y Julat – 5
Julat = { , , , } X 0
1 1 1 1
= {8, 16, 24, 32} 2 3 4 5
Y
Function notation: f : x → 4x or f(x) = 4x Function notation:
Tatatanda fungsi Tatatanda fungsi
1 1
f : x → or f(x) =
x x
(b) Domain f (c) Domain f
= {1, 4, 9, 16, 25} x y = {1, 2, 4, 5} x y
1 1 4
Codomain 4 2 Codomain 1 7
Kodomain 9 3 Kodomain 2 4 10
= {1, 2, 3, 4, 5} 16 4 5 = {4, 7, 10, 13, 16} 5 13
25
16
Range X Y Range X Y
Julat
= {1, 2, 3, 4, 5} Julat
= {4, 7, 13, 16}
Function notation:
Tatatanda fungsi Function notation: f : x → 3x + 1 or f(x) = 3x + 1
f : x →  x or f(x) =  x Tatatanda fungsi




© Penerbitan Pelangi Sdn. Bhd. 2

Additional Mathematics Form 4 Chapter 1 Functions

4. State the domain, codomain and range for each of the following. PL 2
Nyatakan domain, kodomain dan julat bagi setiap yang berikut.
Example
(ii)
(i) f(x) f(x)
19
8
6
4 x
2 –2 0 2 3
–8
x
–4 –3 –2 –1 0 1 2 3 –16
The domain of f is –2 < x < 3.
Domain = {–4, –2, –1, 2, 3} The codomain of f is –16 < f(x) < 19.
Codomain = {2, 4, 6, 8} The range of f is –16 < f(x) < 19.
Range = {2, 4, 6, 8}

(a) f(x) (b) f(x)
8 4
6 3
4
2
x
0 x
–4 –3 –2 –1 1 2 3 0 1 3

Domain = {–4, –2, 0, 2, 3} The domain of f is 0 < x < 3.
Codomain = {0, 2, 4, 8} The codomain of f is 0 < f(x) < 4.
The range of f is 0 < f(x) < 4.
Range = {0, 2, 4, 8}



5. For each of the following functions, find the image for the object x given. PL 3
Untuk setiap fungsi berikut, cari imej bagi objek x yang diberikan.
Example (a) f(x) = 3x + 7; x = –4, x = 5 (b) f(x) = x + 1; x = 3, x = –2
2
f(x) = 9x – 4; x = 2, x = –1
f(–4) = 3(–4) + 7 f(3) = (3) + 1
2
= –5 = 10
f(2) = 9(2) – 4 f(5) = 3(5) + 7 f(–2) = (–2) + 1
2
= 14 = 22 = 5
f(–1) = 9(–1) – 4
= –13






6. For each of the following functions, find the object x based on the image given. PL 3
Untuk setiap fungsi berikut, cari objek x bagi imej yang diberikan.

Example (a) f(x) = 5x + 7; f(x) = –1, f(x) = –3 (b) f(x) = x + 5; f(x) = 9, f(x) = 21
2
f(x) = 2x – 1; f(x) = 3, f(x) = 5
5x + 7 = –1 x + 5 = 9
2
5x = –8 x = 4
2
2x – 1 = 3 8 x = 2 or –2
2x = 4 x = – 5
x = 2 2
5x + 7 = –3 x + 5 = 21
2
2x – 1 = 5 5x = –10 x = 16
x = 4 or –4
2x = 6 x = –2
x = 3





3 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Chapter 1 Functions

7. Sketch the graphs of the following functions f(x) for the given domains. Then, state the ranges. PL 3
Lakar graf bagi fungsi f(x) berikut untuk domain yang diberikan. Seterusnya, nyatakan julatnya.
Example (a) f(x) = x – 3, –2 < x < 4

f(x) = 1 – 4x, –1 < x < 3
x –2 0 3 4
1
x –1 0 3 f(x) 5 3 0 1
4
f(x) 5 1 0 11
f(x)
f(x) 5
11 3
5 1
x
1 x –2 0 3 4
–1 0 1 3
–
4
The range is 0 < f(x) < 5.
The range is 0 < f(x) < 11.




(b) f(x) = 2x + 6, –4 < x < 1 (c) f(x) = 5x – 4, –3 < x < 2

4
x –4 –3 0 1 x –3 0 5 2
f(x) 2 0 6 8 f(x) 19 4 0 6


f(x)
f(x)
8 19
6

2 6
4
x x
–4 –3 0 1 –3 0 4 2
–
5
The range is 0 < f(x) < 8.
The range is 0 < f(x) < 19.






8. Solve each of the following problems. PL 4
Selesaikan setiap masalah berikut.
Example
A function is defined by f : x → x – 8. Find
Suatu fungsi ditakrifkan oleh f : x → x – 8. Cari
(i) f(–3)
(ii) the value of x if f(x) = –5.
nilai x jika f(x) = –5.

(i) f(x) = x – 8 (ii) f(x) = –5
f(–3) = –3 – 8 x – 8 = –5
= –11 x = 3






© Penerbitan Pelangi Sdn. Bhd. 4

Additional Mathematics Form 4 Chapter 1 Functions

(a) Given f : x → 4x – a, where a is a constant and f(–1) = –11. Find
Diberi f : x → 4x – a, dengan keadaan a ialah pemalar dan f(–1) = –11. Cari
(i) the value of a.
nilai a.
(ii) the value of x when f(x) = –19.
nilai x apabila f(x) = –19.

(i) f(x) = 4x – a (ii) f(x) = 4x – 7
f(–1) = 4(–1) – a 4x – 7 = –19
–4 – a = –11 4x = –12
a = 7 x = –3




(b) A function is defined by h : x → 6x – 5. Find
Suatu fungsi ditakrifkan oleh h : x → 6x – 5. Cari
(i) the image of object 2.
imej bagi objek 2.
(ii) the object which maps onto itself.
objek yang memeta kepada dirinya sendiri.

(i) h(x) = 6x – 5 (ii) f(x) = 6x – 5
h(2) = 6(2) – 5 6x – 5 = x
= 7 5x = 5
x = 1





2
(c) Given g : x → px – qx, where p and q are constants. If g(3) = 12 and g(4) = 28, find the values of p and q.
Diberi g : x → px – qx dengan keadaan p dan q ialah pemalar. Jika g(3) = 12 dan g(4) = 28, cari nilai p dan q.
2
2
g(x) = px – qx Substitute  into :
2
g(3) = p(3) – q(3) 4p – (3p – 4) = 7
9p – 3q = 12 4p – 3p + 4 = 7
3p – q = 4 –––  p = 3

g(4) = p(4) – q(4) Substitute p = 3 into :
2
16p – 4q = 28 q = 3(3) – 4
4p – q = 7 –––  = 5

From : q = 3p – 4 ––– 



(d) Given the function h : x → 4 – x. Find the value of x when h(x) = 7.
Diberi fungsi h : x → 4 – x. Cari nilai x apabila h(x) = 7.

h(x) = 4 – x and
4 – x = 7 4 – x = –7
4 – x = 7 x = 11
x = –3














5 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Chapter 1 Functions

1.2 Composite Functions Textbook
Fungsi Gubahan
pg. 12 – 19

SMART Notes


1. gf 2. Given two functions f(x) and g(x), both the functions
can be combined and written as fg(x) or gf(x) which
is defined as fg(x) = f[g(x)] or gf(x) = g[f(x)].
f g Diberi dua fungsi f(x) dan g(x), dua fungsi itu boleh
x f(x) gf(x) digabungkan dan ditulis sebagai fg(x) atau gf(x) yang
ditakrifkan sebagai fg(x) = f[g(x)] atau gf(x) = g[f(x)].

3. In general, fg ≠ gf, f = ff, f = fff and so on.
3
2
P Q R Secara umumnya, fg ≠ gf, f = ff, f = fff dan seterusnya.
3
2
Function f maps set P to set Q, function g maps set Q
to set R and function gf maps set P to set R.
Fungsi f memetakan set P kepada set Q, fungsi g memetakan
set Q kepada set R dan fungsi gf memetakan set P kepada
set R.
9. Find the composite functions fg and gf based on the functions f and g given. PL 3
Cari fungsi gubahan fg dan gf berdasarkan fungsi f dan g yang diberikan.
Example (a) f : x → x + 2, g : x → 4x – 5

f : x → 3x + 1, g : x → x – 1
fg(x) = f(4x – 5) gf(x) = g(x + 2)
= 4x – 5 + 2 = 4(x + 2) – 5
fg(x) = f[g(x)] gf(x) = g[f(x)] = 4x – 3 = 4x + 8 – 5
= f(x – 1) = g(3x + 1)

= 4x + 3
= 3(x – 1) + 1 = 3x + 1 – 1 ∴ fg : x → 4x – 3 ∴ gf : x → 4x + 3
= 3x – 3 + 1 = 3x
= 3x – 2 ∴ gf : x → 3x
∴ fg : x → 3x – 2





2
(b) f : x → 2x – 3, g : x → x 2 (c) f : x → x – 4x + 3, g : x → –5x
fg(x) = f(x ) gf(x) = g(2x – 3) fg(x) = f(–5x)
2
= 2x – 3 = (2x – 3) 2 = (–5x) – 4(–5x) + 3
2
2
= 4x – 12x + 9 = 25x + 20x + 3
2
2
∴ fg : x → 2x – 3 ∴ gf : x → 4x – 12x + 9 ∴ fg : x → 25x + 20x + 3
2
2
2
gf(x) = g(x – 4x + 3)
2
= –5(x – 4x + 3)
2
= –5x + 20x – 15
2
∴ gf : x → –5x + 20x – 15
2
2
10. Find the composite functions f based on the functions f given. PL 3
Cari fungsi gubahan f berdasarkan fungsi f yang diberikan.
2
Example (a) f : x → 7x – 8 (b) f : x → x + 3
2
f : x → x – 2
2
2
f (x) = f[ f(x)] f (x) = f[ f(x)]
2
= f(7x – 8) = f(x + 3)
f (x) = f[ f(x)] = 7(7x – 8) – 8 = (x + 3) + 3
2
2
2
= f(x – 2) 4 2
= (x – 2) – 2 = 49x – 56 – 8 = x + 6x + 9 + 3
2
= 49x – 64
= x + 6x + 12
4
= x – 4
© Penerbitan Pelangi Sdn. Bhd. 6

Additional Mathematics Form 4 Chapter 1 Functions

11. For each pair of the following functions, find PL 3
Bagi setiap pasangan fungsi yang berikut, cari
(i) fg(2) and gf(4), / fg(2) dan gf(4),
(ii) the value of x when fg(x) = 5. / nilai x apabila fg(x) = 5.

Example (a) f : x → 3x + 2, g : x → 2 – x 2
f : x → 5x, g : x → 4 – 2x
2
(i) fg(x) = f(2 – x )
(i) fg(x) = f[g(x)] gf(x) = g[f(x)] = 3(2 – x ) + 2
2
= f(4 – 2x) = g(5x) = 6 – 3x + 2
2
= 5(4 – 2x) = 4 – 2(5x) = 8 – 3x 2
= 20 – 10x = 4 – 10x
fg(2) = 8 – 3(2)
2
fg(2) = 20 – 10(2) gf(4) = 4 – 10(4) = –4
= 0 = –36
gf(x) = g(3x + 2)
= 2 – (3x + 2)
2
Alternative Method
= 2 – (9x + 12x + 4)
2
g(2) = 4 – 2(2) f(4) = 5(4) = –9x – 12x – 2
2
= 0 = 20
2
fg(2) = f(0) gf(4) = g(20) gf(4) = –9(4) – 12(4) – 2
= 5(0) = 4 – 2(20) = –194
= 0 = –36
(ii) fg(x) = 5
2
8 – 3x = 5
(ii) fg(x) = 5 3x = 3
2
20 – 10x = 5 x = 1
2
10x = 15 x = 1 or –1
x = 1.5
12. Find the function g based on the functions f and fg given. PL 4
Cari fungsi g berdasarkan fungsi f dan fg yang diberikan.

Example (a) f(x) = x + 4, fg(x) = 3x + 9 (b) f(x) = 3x – 6, fg(x) = 5x + 8
f(x) = x + 3, fg(x) = 2x – 5
f [g(x)] = 3x + 9 f [g(x)] = 5x + 8
g(x) + 4 = 3x + 9 3g(x) – 6 = 5x + 8
f [g(x)] = 2x – 5
g(x) + 3 = 2x – 5 g(x) = 3x + 5 3g(x) = 5x + 14
5x + 14
g(x) = 2x – 8 g(x) =
∴ g : x → 3x + 5 3
∴ g : x → 2x – 8 ∴ g : x → 5x + 14
3


13. Find the function f based on the functions g and fg given. PL 4
Cari fungsi f berdasarkan fungsi g dan fg yang diberikan.

Example (a) g(x) = 2x + 3, fg(x) = 7x – 4 (b) g(x) = x – 9, fg(x) = 3x – 5
2
2
g(x) = x + 2, fg(x) = 4x – 9
2
f[g(x)] = 7x – 4 f[g(x)] = 3x – 5
2
f(2x + 3) = 7x – 4 f(x – 9) = 3x – 5
2
f[g(x)] = 4x – 9
2
f(x + 2) = 4x – 9 Let y = 2x + 3 Let y = x – 9
2
Let y = x + 2 x = y – 3 x = y + 9
2
x = √y + 9
x = y – 2
2
So, f(y) = 4(y – 2) – 9 So, f(y) = 7  y – 3  – 4 So, f(y) = 3(√y + 9) – 5
2
= 3(y + 9) – 5

= 4y – 8 – 9 = 3y + 27 – 5
= 4y – 17 = 7y – 29
2 = 3y + 22
Substitute y with x, f(x) = 4x – 17 ∴ f : x → 3x + 22
∴ f : x → 4x – 17 ∴ f : x → 7x – 29
2
7 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Chapter 1 Functions

14. Solve the following problems. PL 4
Selesaikan masalah-masalah berikut.
Example

Given the functions f : x → 4x – 1 and g : x → (x – 3) . Find
2
2
Diberi fungsi f : x → 4x – 1 dan g : x → (x – 3) . Cari
(i) fg(x), (ii) fg(3).
2
(i) fg(x) = f[(x – 3) ] (ii) fg(3) = 4(3) – 24(3) + 35
2
2
= 4(x – 3) – 1 = 36 – 72 + 35
= 4(x – 6x + 9) – 1 = –1
2
= 4x – 24x + 36 – 1
2
= 4x – 24x + 35
2
(a) Given f : x → 3x + 8 and g : x → x – 6. Find
Diberi f : x → 3x + 8 dan g : x → x – 6. Cari
(i) fg(x), (ii) fg(–2).
(i) fg(x) = f(x – 6) (ii) fg(–2) = 3(–2) – 10
= 3(x – 6) + 8 = –16
= 3x – 18 + 8
= 3x – 10


(b) A function g is defined by g : x → x + 2. Find the function f for each of the following composite functions.
Fungsi g ditakrifkan oleh g : x → x + 2. Cari fungsi f bagi setiap fungsi gubahan berikut.
(i) fg : x → x + 4x + 4, (ii) gf : x → 2x – 3x + 4.
2
2
2
(i) fg(x) = x + 4x + 4 (ii) gf(x) = 2x – 3x + 4
2
2
f(x + 2) = x + 4x + 4 f(x) + 2 = 2x – 3x + 4
2
2
f(x) = 2x – 3x + 2
Let y = x + 2
x = y – 2
2
So, f(y) = (y – 2) + 4(y – 2) + 4
2
= y – 4y + 4 + 4y – 8 + 4
= y 2
∴ f(x) = x 2
(c) Given f(x) = px + q and f (x) = 4x + 6. Find the values of p and q.
2
Diberi f(x) = px + q dan f (x) = 4x + 6. Cari nilai-nilai p dan q.
2
ff(x) = 4x + 6 Comparing: When p = 2, When p = –2,
f(px + q) = 4x + 6 p = 4 pq + q = 6 pq + q = 6
2
p(px + q) + q = 4x + 6 p = ±√4 2q + q = 6 –2q + q = 6
p x + pq + q = 4x + 6 = 2 or –2 3q = 6 –q = 6
2
q = 2 q = –6

3x
(d) Functions g and h are defined as g : x → 5x – 6 and h : x → . Find the value of x if gh(x) = h(x).
x – 2
3x
Fungsi g dan h ditakrifkan oleh g : x → 5x – 6 dan h : x → . Cari nilai x jika gh(x) = h(x).
x – 2
g(x) = 5x – 6 3x 3x
3x 5   – 6 = x – 2
h(x) = x – 2
x – 2 3x
15x – 6 =
gh(x) = h(x) x – 2 x – 2
3x
3x
g  x – 2  = x – 2 15x – 6(x – 2) = 3x
15x – 6x + 12 = 3x
6x = –12
x = –2



© Penerbitan Pelangi Sdn. Bhd. 8

Additional Mathematics Form 4 Chapter 1 Functions

15. Solve each of the following. PL 5 Daily Application
Selesaikan setiap yang berikut.

Example
5
The temperature in Fahrenheit, F, can be converted to Celsius, C, using the function C(F) = (F – 32).
9 9
The temperature in Kelvin, K, can be converted to Fahrenheit, F, using F(K) = (K – 273) + 32. Suraya
5
obtained a result of temperature of 258 K in an experiment. Convert the temperature to Celsius, in °C.
5
Suhu dalam Fahrenheit, F, boleh ditukar kepada Celsius, C, dengan menggunakan fungsi C(F) = 9 (F – 32). Suhu dalam Kelvin, K,
9
boleh ditukar kepada Fahrenheit, F, dengan menggunakan fungsi F(K) = 5 (K – 273) + 32. Suraya memperoleh hasil suhu 258 K
dalam suatu eksperimen. Tukar suhu tersebut ke Celsius, dalam °C.
9
F(258) = (258 – 273) + 32 C[F(258)] = C(5)
5 5
= 5 = (5 – 32)
9
= –15
Hence, 258 K can be converted to –15°C.

(a) The height of water, h cm, in a cylindrical container is increasing with respect to time, t s, and
its function is h(t) = 1.4t. The volume of water, V cm , in the container is given by the function
3
88
V(h) = h.
7
Ketinggian air, h cm, dalam suatu bekas berbentuk silinder semakin meningkat mengikut masa, t s, dan fungsinya ialah
88
h(t) = 1.4t. Isi padu air, V cm , dalam bekas tersebut diberi oleh fungsi V(h) = h.
3
7
(i) State the volume of water, V, as a function of time, t.
Nyatakan isi padu air, V, sebagai fungsi masa, t.
3
(ii) Find the volume of water, in cm , in the container after 5 seconds.
Cari isi padu air, dalam cm , dalam bekas tersebut selepas 5 saat.
3
(i) V[h(t)] = V(1.4t)
88
= (1.4t)
7
= 17.6t
∴ V : t → 17.6t

(ii) V(5) = 17.6(5)
= 88
Hence, the volume of water in the container after 5 seconds is 88 cm .
3
(c) Azlan works in a furniture store. Every month, he receives a basic salary of RM2 000 plus a 5%
commission on sales over RM3 000. Let x represents his sales per month.
Azlan bekerja di sebuah kedai perabot. Setiap bulan, dia menerima gaji asas sebanyak RM2 000 dan komisen 5% pada jualan
yang melebihi RM3 000. Let x mewakili jualannya setiap bulan.
(i) Given f(x) = x – 3 000 and g(x) = 0.05x, write the function of gf(x) and explain its meaning.
Diberi f(x) = x – 3 000 dan g(x) = 0.05x, tulis fungsi gf(x) dan terangkan maksudnya.
(ii) If Azlan’s sales are RM6 499, find his salary.
Jika jualan Azlan ialah RM6 499, cari gajinya.

(i) gf(x) = g(x – 3 000)
= 0.05(x – 3 000)
gf(x) is the commission received by Azlan, that is 5% on sales over RM3 000.

(ii) gf(6 499) = 0.05(6 499 – 3 000)
= 174.95
Salary = 2 000 + 174.95
= 2 174.95
Hence, his salary is RM2 174.95





9 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Chapter 1 Functions

1.3 Inverse Functions Textbook
Fungsi Songsang
pg. 20 – 29


SMART Notes

1. f 5. Horizontal line test can be used to test the existence
of inverse functions.
Ujian garis mengufuk boleh digunakan untuk menguji
x y kewujudan fungsi songsang.
f –1
y
y = f(x)
If f(x) = y, then the inverse function is f (y) = x. f has inverse function
–1
Jika f(x) = y, maka fungsi songsangnya ialah f (y) = x. f mempunyai fungsi
–1
2. Only one-to-one functions have inverse functions. songsang
Hanya fungsi satu kepada satu mempunyai fungsi songsang. 0 x
3. f and g are inverse functions of each other if and only if
fg(x) = x, x in domain of g and gf(x) = x, x in domain of f.
f dan g ialah fungsi songsang antara satu sama lain jika dan y
hanya jika fg(x) = x, x dalam domain g dan gf(x) = x, x dalam h does not have
domain f. inverse function
4. If f and g are inverse functions of each other, then y = h(x) h tidak mempunyai
Jika f dan g ialah fungsi songsang antara satu sama lain, maka x fungsi songsang
(a) domain of f = range of g and domain of g = range of f 0
domain f = julat g dan domain g = julat f
(b) graph g is the reflection of graph f on the line y = x 7. ff (x) = f f(x) = x
–1
–1
graf g adalah pantulan graf f pada garis y = x


16. In the diagram, function f maps x to y. Determine each of the following. PL 1
Dalam rajah di bawah, fungsi f memetakan x kepada y. Tentukan setiap yang berikut.

f
–1
x y (a) f(2) = 3 (b) f (11) = 6
2
3
5
6
9 (c) f (3) = 2 (d) f (9) = 5
–1
–1
11
17. Determine whether the following functions have the inverse functions using the horizontal line test. PL 2
Tentukan sama ada fungsi berikut mempunyai fungsi songsang atau tidak dengan menggunakan ujian garis mengufuk.
Example (a) y (b) y
y
f
f

f x x
–1 0 3 0
x
0 2

Has inverse function. When tested Does not have inverse Has inverse function. When
with a horizontal line, the line cuts function. When tested with a tested with a horizontal line,
the graph at one point. horizontal line, the line cuts the line cuts the graph at one
the graph at two points. point.











© Penerbitan Pelangi Sdn. Bhd. 10

Additional Mathematics Form 4 Chapter 1 Functions

18. Verify that f and g are inverse functions of each other. PL 3
Sahkan f dan g ialah fungsi songsang antara satu sama lain.
Example (a) f(x) = 4 – 3x, g(x) = 4 – x

f(x) = x – 8, g(x) = x + 8 3

fg(x) = f(x + 8) gf(x) = g(x – 8) 4 – x gf(x) = g(4 – 3x)
= (x + 8) – 8 = (x – 8) + 8 fg(x) = f  3 
= x = x 4 – x = 4 – (4 – 3x)
= 4 – 3  3  3
Since fg(x) = gf(x) = x, f and g are inverse functions = 4 – (4 – x) = 3x
3
of each other. = x = x

Since fg(x) = gf(x) = x, f and g are inverse
functions of each other.




–1
–1
19. Sketch graphs of f and f on the same plane. Then, state the domain of f . PL 3
Lakar graf bagi f dan f pada satah yang sama. Seterusnya, nyatakan domain bagi f .
–1
–1
Example (a) f : x → – , domain: 1 < x < 10
5
f : x → 4x, domain: 0 < x < 3 x
x 1 5 10
x 0 1 2 3
y –5 –1 –0.5
y 0 4 8 12
y
y (–0.5, 10)
(3, 12) y = x y = x
Domain of f :
–1
f –1 –5 < x < –0.5
–1
f Domain of f :
0 < x < 12 (–5, 1)
(12, 3) x
f –1 Range of f 0 f (10, –0.5)
x
0
(1, –5)


20. Find the inverse function of each of the following functions. PL 3
Cari fungsi songsang bagi setiap fungsi berikut.
Example (a) f(x) = 3x – 1 (b) g(x) = 7 – 2x

f(x) = 4x + 5
Let y = f(x) Let y = g(x)
y = 3x – 1 y = 7 – 2x
Let y = f(x) 3x = y + 1 2x = 7 – y
y = 4x + 5 y + 1 7 – y
4x = y – 5 x = 3 x = 2
y – 5
x = –1 –1
4 Since f (y) = x, Since g (y) = x,
–1
–1
f (y) = y + 1 g (y) = 7 – y
Since f (y) = x, 3 2
–1
y – 5
f (y) =
–1
4 Substitute y with x, Substitute y with x,
f (x) = x + 1 g (x) = 7 – x
–1
–1
Substitute y with x, 3 2
x – 5 x + 1 7 – x
–1
–1
f (x) = ∴ f : x → ∴ g : x →
–1
4 3 2
x – 5
∴ f : x →
–1
4
11 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Chapter 1 Functions

x x – 1 3
(c) g(x) = + 1 (d) h(x) = (e) h(x) =
4 2x + 1 x – 1
Let y = g(x) Let y = h(x) Let y = h(x)
x y = x – 1 3
y = + 1 2x + 1 y = x – 1
4
x 2xy + y = x – 1
= y – 1 xy – y = 3
4 y + 1 = x – 2xy
x = 4(y – 1) y + 1 = x(1 – 2y) xy = 3 + y
3 + y
x = 4y – 4 y + 1 x =
x = y
3
Since g (y) = x, 1 – 2y x = + 1
–1
–1
g (y) = 4y – 4 Since h (y) = x, y
–1
y + 1 Since h (y) = x,
–1
–1
Substitute y with x, h (y) = 1 – 2y 3
–1
–1
g (x) = 4x – 4 Substitute y with x, h (y) = + 1
y
–1
∴ g : x → 4x – 4 x + 1 Substitute y with x,
–1
h (x) =
3
1 – 2x h (x) = + 1
–1
x + 1 x
–1
∴ h : x →
3
1 – 2x ∴ h : x → + 1
–1
x
21. Find the function f, g or h based on the given inverse function. PL 4
Cari fungsi f, g dan h berdasarkan fungsi songsang yang diberikan.
Example (a) f (x) = 3x – 4 (b) f (x) = x + 6
–1
–1
4 + x 5
–1
f (x) = Let y = f (x)
–1
2 Let y = f (x)
–1
y = 3x – 4
–1
Let y = f (x) 3x = y + 4 y = x + 6
4 + x y + 4 5
y = x = 5y = x + 6
2 3
2y = 4 + x Since f (y) = x, x = 5y – 6
x = 2y – 4 y + 4
f (y) = Since f(y) = x,
3
Since f (y) = x, f(y) = 5y – 6
f (y) = 2y – 4 Substitute y with x,
f (x) = x + 4 Substitute y with x,
Substitute y with x, 3 f(x) = 5x – 6
f (x) = 2x – 4 ∴ f : x → x + 4
3 ∴ f : x → 5x – 6
∴ f : x → 2x – 4
x – 7 x + 4 5
–1
–1
(c) g (x) = (d) g (x) = (e) h (x) = – 2
–1
3 x – 4 x
–1
–1
–1
Let y = g (x) Let y = g (x) Let y = h (x)
5
y = x – 7 y = x + 4 y = – 2
3 x – 4 x
3y = x – 7 xy – 4y = x + 4 5 = y + 2
x = 3y + 7 xy – x = 4y + 4 x 5
x = 4y + 4 x = y + 2
Since g(y) = x, y – 1
g(y) = 3y + 7 Since g(y) = x, Since h(y) = x,
5
4y + 4 h(y) =
g(y) = y + 2
Substitute y with x, y – 1
g(x) = 3x + 7 Substitute y with x,
Substitute y with x, 5
∴ g : x → 3x + 7 g(x) = 4x + 4 h(x) = x + 2
x – 1 5
4x + 4 ∴ h : x →
∴ g : x → x + 2
x – 1
© Penerbitan Pelangi Sdn. Bhd. 12

Additional Mathematics Form 4 Chapter 1 Functions

22. Solve each of the following problems. PL 5
Selesaikan masalah-masalah berikut.
Example

Given the function f : x → 4x – 7 and g : x → x – 5, find
2
Diberi fungsi f : x → 4x – 7 dan g : x → x – 5, cari
2
–1
–1
–1
(i) f (x), (ii) gf (x), (iii) gf (5)
2
(i) Let y = f(x) (ii) gf (x) = g  x + 7  (iii) gf (5) = 5 + 14(5) – 31
–1
–1
y = 4x – 7 4 16
4x = y + 7 2 = 64
y + 7 =  x + 7  – 5 16
x = 4 = 4
4
2
–1
Since f (y) = x, =  x + 14x + 49  – 5
y + 7 16
f (y) = 2
–1
4 = x + 14x + 49 – 80
16
Substitute y with x, x + 14x – 31
2
x + 7 =
f (x) = 16
–1
4
2x + p
(a) Given g : x → , p ≠ 4 and g(4) = 5, find
p – 4
2x + p
Diberi g : x → , p ≠ 4 dan g(4) = 5, cari
p – 4
(i) the value of p / nilai p,
(ii) g (x),
–1
(iii) the value of x if g (x) = 4 / nilai x jika g (x) = 4.
–1
–1
2(4) + p 2x + 7 3x – 7
(i) g(4) = (ii) g(x) = (iii) g (x) =
–1
p – 4 3 2
8 + p Let y = g(x) g (y) = x 3x – 7
–1
5 = 4 = 2
p – 4 y = 2x + 7
5(p – 4) = 8 + p 3 8 = 3x – 7
5p – 20 = 8 + p 3y = 2x + 7 3x = 15
4p = 28 x = 3y – 7 x = 5
p = 7 2
3x – 7
–1
∴ g (x) =
2
5
(b) It is given g : x → 8x – 3 and h : x → x , x ≠ 0. Find
5
Diberi bahawa g : x → 8x – 3 dan h : x → , x ≠ 0. Cari
x
(i) g (x), (ii) hg (x), (iii) g h(10).
–1
–1
–1
(i) Let y = g(x) (ii) hg (x) = h  x + 3  (iii) h(10) = 5 = 1
–1
y = 8x – 3 8 10 2
8x = y + 3 5
1
y + 3 = –1 –1  
x = x + 3 g h(10) = g 2
8
8 1
Since g (y) = x, 40 2 + 3
–1
y + 3 = =
g (y) = x + 3 8
–1
8 7
=
Substitute y with x, 16
x + 3
g (x) =
–1
8
13 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Chapter 1 Functions


2
(c) Given the function h(x) = ax + b, where a . 0, a and b are constants and h (x) = 49x – 40, find
2
Diberi fungsi h(x) = ax + b, dengan keadaan a . 0, a dan b ialah pemalar dan h (x) = 49x – 40, cari
(i) the value of a and of b, / nilai a dan b,
(ii) h (–3),
–1
–1 2
–1 2
(iii) the value of x if (h ) (x) = 1 / nilai x jika (h ) (x) = 1.
–1 2
(i) Given h(x) = ax + b (ii) h(x) = 7x – 5 (iii) (h ) (x) = 1
h (x) = hh(x) Let y = h(x) x + 5
2
= h(ax + b) y = 7x – 5 7 + 5
= a(ax + b) + b 7x = y + 5 7 = 1
= a x + ab + b y + 5
2
x = 7 x + 5 + 5 = 7
Compare with 49x – 40, 7
–1
2
a = 49 Since h (y) = x, x + 5 = 2

a = 49 h (y) = y + 5 7
–1
= 7 7 x + 5 = 14
x = 9
Substitute y with x,
ab + b = –40 –1 x + 5
7b + b = –40 h (x) = 7
8b = –40 –3 + 5
–1
b = –5 ∴ h (–3) =
7
2
=
7
3x – 6 mx + 6
–1
(d) It is given that f (x) = and f (x) = . Find HOTS Analysing
7 + 4x 3 + nx
3x – 6 mx + 6
–1
Diberi bahawa f (x) = dan f (x) = . Cari
7 + 4x 3 + nx
(i) the values of m and n, / nilai m dan n,
(ii) the value of x if f(4x + 3) = 7. / nilai x jika f(4x + 3) = 7.
3x – 6 7x + 6
(i) f (x) = (ii) f(4x + 3) = 7, f(x) =
–1
7 + 4x 3 – 4x
Let y = f (x) 7 = 7(4x + 3) + 6
–1
3x – 6 3 – 4(4x + 3)
y =
7 + 4x 28x + 21 + 6
7y + 4xy = 3x – 6 7 = 3 – 16x – 12
7y + 6 = 3x – 4xy 28x + 27
7y + 6 = x(3 – 4y) 7 = –9 – 16x
7y + 6
x = 7(–9 – 16x) = 28x + 27
3 – 4y
–63 – 112x = 28x + 27
Since f(y) = x, –140x = 90
7y + 6 9
f(y) = x = –
3 – 4y 14
Substitute y with x,
7x + 6
f(x) =
3 – 4x
Compare with
mx + 6
f (x) =
3 + nx
7x + 6 = mx + 6
3 – 4x 3 + nx
∴ m = 7, n = –4







© Penerbitan Pelangi Sdn. Bhd. 14

Additional Mathematics Form 4 Chapter 1 Functions

SPM Practice 1




Paper 1 5. Given the functions m : x → px + 3, h : x → 4x – 1 and
SPM mh(x) = 4px + q, express p in terms of q.
2016
Diberi fungsi m : x → px + 3, h : x → 4x – 1 dan
1. The diagram shows the composite function fg that maps a mh(x) = 4px + q, ungkapkan p dalam sebutan q.
SPM to c.
2015 [3]
Rajah di bawah menunjukkan fungsi gubahan fg yang Ans: p = 3 – q
memetakan a kepada c.
fg
6. It is given the functions
SPM Diberi fungsi
2016
a b c g : x → 4x + 1
fg : x → 16x + 8x – 5
2
Find / Cari
–1
State / Nyatakan (a) g (x),
(a) the function that maps a to b, (b) f(x).
fungsi yang memetakan a kepada b, x – 1 [3]
–1
–1
(b) f (c). Ans: (a) g (x) = 4
[2] (b) f(x) = x – 6
2
Ans: (a) g
(b) f (c) = b
–1
7. The diagram shows the graph of the function
SPM f : x → |1 – 3x| for the domain –1 < x < 3.
2017
2. Given the functions f(x) = 5x + 1 and g(x) = mx – n, where Rajah di bawah menunjukkan graf bagi fungsi
f : x → |1 – 3x| untuk domain –1 < x < 3.
SPM m and n are constants, express m in terms of n such that
2015 f(x)
fg(2) = 1.
Diberi fungsi f(x) = 5x + 1 dan g(x) = mx – n, dengan keadaan 8
m dan n ialah pemalar, ungkapkan m dalam sebutan n dengan
keadaan fg(2) = 1. (–1, 4)
[3]
1
Ans: m = n 0 3 x
2
State / Nyatakan
3. It is given the function f : x → 8x – 4. Find
SPM Diberi fungsi f : x → 8x – 4. Cari (a) the object of 8,
2016 objek bagi 8,
(a) the value of x if f(x) maps onto itself, (b) the image of 2,
nilai x jika f(x) memeta kepada dirinya sendiri, imej bagi 2,
(b) the value of h if f(3 – h) = 4h. (c) the domain of 0 < f(x) < 2.
nilai h jika f(3 – h) = 4h. domain bagi 0 < f(x) < 2.
[4]
4 [3]
Ans: (a) x = Ans: (a) 3
7
5 (b) 5
(b) h = 3 1

(c) – < x < 1
3
4. It is given the function g : x → 3x – 9. Find
SPM Diberi fungsi g : x → 3x – 9. Cari
2017 8. Given the functions g : x → 3x – 1 and h : x → 6x, find
(a) g (x), Diberi fungsi g : x → 3x – 1 dan h : x → 6x, cari
–1
4p
(b) the value of p if g 2   = 24. (a) hg(x),
3
1
4p (b) the value of x if hg(x) = g(x).
nilai p jika g   = 24. 1 3
2
3
[4] nilai x jika hg(x) = g(x).
3
x + 9 [4]
–1
Ans: (a) g (x) = Ans: (a) hg(x) = 18x – 6
3
(b) p = 5 (b) x = 1
3
15 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Chapter 1 Functions
9. The diagram shows the relation between set P, set Q and 12. Given the function g : x → 2x – 7, find
SPM set R. Diberi fungsi g : x → 2x – 7, cari
2018
Rajah di bawah menunjukkan hubungan antara set P, set Q (a) g(6),
dan set R. (b) the value of p if 2g (p) = g(6).
–1
–1
nilai p jika 2g (p) = g(6).
P fg : x → 9x + 12
[4]
Ans: (a) 5
R
(b) p = –2
Q
2
It is given that set P maps to set Q by the function
5x + 6 and maps to set R by fg : x → 9x + 12. 1. In the diagram, function f maps set P to set Q and function
g maps set Q to set R.
Diberi bahawa set P dipetakan kepada set Q oleh fungsi 5x + 6 Dalam rajah di bawah, fungsi f memetakan set P kepada set Q
dan dipetakan kepada set C oleh fg : x → 9x + 12. dan fungsi g memetakan set Q kepada set R.
(a) Write the function which maps set P to set Q by using
the function notation.
Tulis fungsi yang memetakan set P kepada set Q dengan x f 2x + 5 g 6x + 11
menggunakan tatatanda fungsi.
(b) Find the function which maps set Q to set R.
Cari fungsi yang memetakan set Q kepada set R. P Q R
[4] Find/ Cari
(a) in terms of x, the function
Ans: (a) g : x → 5x + 6 dalam sebutan x, fungsi
9x + 6 (i) which maps set Q to set P,
(b) f : x →
5 yang memetakan set Q kepada set P,
(ii) g(x),
[5]
10. The diagram shows the function g : x → x – 2p, where p is (b) the value of x such that fg(x) = 3x + 4.
a constant. nilai x dengan keadaan fg(x) = 3x + 4.
Rajah di bawah menunjukkan fungsi g : x → x – 2p, dengan [3]
keadaan p ialah pemalar. Ans: (a) (i) f (x) = x – 5 ; (ii) g(x) = 3x – 4
–1
g 2
x x – 2p 7 1
(b) x = = 2
3 3
6 10
2. The diagram shows the mapping from set X to set Y defined
as g(x) = px + 3 and the mapping from set Y to set Z
60
defined as h(y) = , y ≠ –q.
y + q
Find / Cari Rajah di bawah menunjukkan pemetaan dari set X ke set Y yang
(a) the value of p, ditakrifkan sebagai g(x) = px + 3 dan pemetaan dari set Y ke
nilai p, 60
–1
(b) g (x), set Z yang ditakrifkan sebagai h(y) = y + q , y ≠ –q.
(c) the value of x if g (x) = 5.
–1
nilai x jika g (x) = 5. g h
–1
[5] 4 11 15
Ans: (a) p = –2
(b) g (x) = x – 4 X Y Z
–1
(c) x = 9
Find/ Cari
(a) the values of p and q,
nilai-nilai p dan q,
11. Given the functions f(x) = 9 – 5x and g(x) =3x, find
Diberi fungsi f(x) = 9 – 5x dan g(x) = 3x, cari (b) the function that maps set X to set Z. [4]
(a) the value of fg(–1), fungsi yang memetakan set X kepada set Z.
nilai bagi fg(–1), [3]
–1
(b) gf (x).
[4] Ans: (a) p = 2, q = –7
60
Ans: (a) 24 (b) hg(x) = 2x – 4 , x ≠ 2
27 – 3x
(b) gf (x) = 5
–1

© Penerbitan Pelangi Sdn. Bhd. 16

Additional Mathematics Form 4 Chapter 1 Functions
3. (a) Given the functions f : x → 3x – 4 and g : x → 2 – 5x, 4. Given f : x → x + 3 and g : x → 4 , x ≠ 2, find
SPM find 7 x – 2
2018 x + 3 4
Diberi fungsi f : x → 3x – 4 dan g : x → 2 – 5x, cari Diberi f : x → dan g : x → , x ≠ 2, cari
(i) g(4), (a) g (x), 7 x – 2
–1
1
(ii) the value of n if f (n + 1) = g(4), [2]
6
1
–1
nilai n jika f(n + 1) = g(4), (b) g f(x), [2]
6
(iii) gf(x). (c) the value of x if (f ) (x) = 25.
–1 2
[5] –1 2
(b) Hence, sketch the graph of y = gf(x) for –1 < x < 3. nilai x jika (f ) (x) = 25.
State the range of y. [3]
Seterusnya, lakar graf y = gf(x) untuk –1 < x < 3. Ans: (a) g (x) = 4 x + 2, x ≠ 0
–1
Nyatakan julat bagi y.
[3] (b) g f(x) = 34 + 2x , x ≠ –3
–1
2 x + 3
Ans: (a) (i) –18; (ii) n = – 3 ; (iii) gf(x) = 22 – 15x (c) x = 1
(b) Range: 0 < y < 37
HOTS Challenge


1. Money in Malaysian Ringgit (RM) is converted to Indonesian Rupiah (Rp) according to the function f(x) = 3 388.50x, where
f(x) is money in Indonesian Rupiah (Rp) and x is money in Malaysian Ringgit (RM). HOTS Analysing
Wang dalam Ringgit Malaysia (RM) ditukar kepada Rupiah Indonesia (Rp) menurut fungsi f(x) = 3 388.50x, dengan keadaan f(x)
ialah wang dalam Rupiah Indonesia (Rp) dan x ialah wang dalam Ringgit Malaysia (RM).
(a) How much money is there in Indonesian Rupiah (Rp) if the money in Malaysian Ringgit (RM) is RM800?
Berapakah jumlah wang dalam Rupiah Indonesia (Rp) jika wang dalam Ringgit Malaysia (RM) ialah RM800?
(b) Find the inverse function of the function f(x). What is the meaning of the inverse function obtained?
Cari fungsi songsang bagi fungsi f(x) ini. Apakah maksud fungsi songsang yang diperoleh?
Answer Guide
Apply concepts in functions and inverse functions.
Gunakan konsep dalam fungsi dan fungsi songsang.

Ans: (a) 2 710 800 Rp
x
(b) f (x) = ; The inverse function obtained converts money in Indonesian Rupiah (Rp) to Malaysian
–1
3 388.50
Ringgit (RM).
mx 9x – 15
2. Given f : x → , x ≠ n, g : x → 2x – m and the composite function gf : x → , find HOTS Applying
n – x 5 – x
Diberi f : x → mx , x ≠ n, g : x → 2x – m dan fungsi gubahan gf : x → 9x – 15 , cari
n – x 5 – x
(a) the values of m and n, (b) the inverse functions of f and g, (c) the values of x if g = fg.
nilai m dan n, fungsi songsang bagi f dan g, nilai x jika g = fg.
Answer Guide
Apply concepts in composite functions and inverse functions.
Gunakan konsep dalam fungsi gubahan dan fungsi songsang.

Ans: (a) m = 3, n = 5
5x x + 3
–1
(b) f (x) = , g (x) =
–1
3 + x 2
5 3
(c) x = or
2 2

Quiz 1
21
PAK-




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