AnsChap1.pdf

1 © Penerbitan Pelangi Sdn. Bhd.
Additional Mathematics Form 4 Chapter 1 Functions
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Notes
SMART
1. A function from set X to set Y is a special relation that
maps each element x in set X to only one element y in
set Y.

Fungsi dari set X kepada set Y ialah satu hubungan khas
yang memetakan setiap unsur x dalam set X kepada hanya
satu unsur y dalam set Y.

1
2
3
2
4
6
X Y
2. A function can be written as f : x → 2x or f(x) = 2x,
where x is the object and 2x is the image.
Suatu fungsi boleh ditulis sebagai f : x → 2x atau f(x) = 2x
dengan x ialah objek dan 2x ialah imej.
3. Only one-to-one relation and many-to-one relation are
functions.
Hanya hubungan satu kepada satu dan hubungan banyak
kepada satu ialah fungsi.
4. When a graph is given, vertical line test can be used
to determine whether the graph is a function.
Apabila graf diberi, ujian garis mencancang boleh digunakan
untuk menentukan sama ada graf tersebut ialah fungsi atau
bukan.
1 Functions
Fungsi
Functions
Fungsi
1.1 Textbook
pg. 2 – 11
0
y
x
0
y
x
Function Not a function
Fungsi Bukan fungsi
5. Given an arrow diagram of a function:
Diberi gambar rajah anak panah bagi suatu fungsi:

1
4
9
16
25
1
2
3
5
x
X Y
x2
f
(a) Domain = {1, 2, 3, 5}
(b) Codomain / Kodomain = {1, 4, 9, 16, 25}
(c) Range / Julat = {1, 4, 9, 25}
(d) Object of 4 is 2. / Objek bagi 4 ialah 2.
(e) Image of 3 is 9. / Imej bagi 3 ialah 9.
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
April
June/ Jun
July/ Julai
30
31
Number of days
Bilangan hari
A B
A function. Each object has one image only.
(a)
1
2
3
4
1
8
27
64
Cube
Kuasa tiga
C D
A function. Each object has one image only.
(b)
1
2
3
3
6
9
18
A B
Not a function. The object 3 has two images in
the codomain.
(c)
1
2
3
C
5
6
9
D
Not a function. Object 2 has two images in the
codomain.
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2
© Penerbitan Pelangi Sdn. Bhd.
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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
0
y
x
A function. When tested with a vertical line, the line
cuts the graph at one point.
(a)
0
y
x
Not a function. When tested with a vertical line,
the line cuts the graph at two points.
(b)
0
y
x
A function. When tested with a vertical line, the
line cuts the graph at one point.
(c)
0
y
x
A function. When tested with a vertical line, the
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
Domain
= {2, 4, 6, 8}
Codomain
Kodomain
= {8, 16, 24, 28, 32}
Range
Julat
= {8, 16, 24, 32}
Function notation: f : x → 4x or f(x) = 4x
Tatatanda fungsi
(a) Domain
= {2, 3, 4, 5}
Codomain
Kodomain
= { 1
2
, 1
3
, 1
4
, 1
5
, 0}
Range
Julat
= { 1
2
, 1
3
, 1
4
, 1
5
}
Function notation:
Tatatanda fungsi
f : x → 1
x
or f(x) = 1
x
(b) Domain
= {1, 4, 9, 16, 25}
Codomain
Kodomain
= {1, 2, 3, 4, 5}
Range
Julat
= {1, 2, 3, 4, 5}
Function notation:
Tatatanda fungsi
f : x → 
x or f(x) = 
x
(c) Domain
= {1, 2, 4, 5}
Codomain
Kodomain
= {4, 7, 10, 13, 16}
Range
Julat
= {4, 7, 13, 16}
Function notation: f : x → 3x + 1 or f(x) = 3x + 1
Tatatanda fungsi
8
16
24
28
32
2
4
6
8
x
X Y
y
f
1
–
2
1
–
3
1
–
4
1
–
5
0
2
3
4
5
x
X
Y
y
f
1
2
3
4
5
x
X Y
y
f
1
4
9
16
25
4
7
10
13
16
1
2
4
5
x
X Y
y
f

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4. State the domain, codomain and range for each of the following. PL 2
Nyatakan domain, kodomain dan julat bagi setiap yang berikut.
Example
(i)
2
0
–1 1 2 3
–2
–3
–4
4
6
8
x
f(x)
Domain = {–4, –2, –1, 2, 3}
Codomain = {2, 4, 6, 8}
Range = {2, 4, 6, 8}
(ii)
–16
–8
19
2
0
–2 3
x
f(x)
The domain of f is –2 < x < 3.
The codomain of f is –16 < f(x) < 19.
The range of f is –16 < f(x) < 19.
(a)
2
0
–1 1 2 3
–2
–3
–4
4
6
8
x
f(x)
Domain = {–4, –2, 0, 2, 3}
Codomain = {0, 2, 4, 8}
Range = {0, 2, 4, 8}
(b)
0 1
3
4
3
x
f(x)
The domain of f is 0 < x < 3.
The codomain of f is 0 < f(x) < 4.
The range of f is 0 < f(x) < 4.
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
f(x) = 9x – 4; x = 2, x = –1
f(2) = 9(2) – 4
= 14
f(–1) = 9(–1) – 4
= –13
(a) f(x) = 3x + 7; x = –4, x = 5
f(–4) = 3(–4) + 7
= –5
f(5) = 3(5) + 7
= 22
(b) f(x) = x2
+ 1; x = 3, x = –2
f(3) = (3)2
+ 1
= 10
f(–2) = (–2)2
+ 1
= 5
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
f(x) = 2x – 1; f(x) = 3, f(x) = 5
2x – 1 = 3
2x = 4
x = 2
2x – 1 = 5
2x = 6
x = 3
(a) f(x) = 5x + 7; f(x) = –1, f(x) = –3
5x + 7 = –1
5x = –8
x = –
8
5
5x + 7 = –3
5x = –10
x = –2
(b) f(x) = x2
+ 5; f(x) = 9, f(x) = 21
x2
+ 5 = 9
x2
= 4
x = 2 or –2
x2
+ 5 = 21
x2
= 16
x = 4 or –4

4
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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
f(x) = 1 – 4x, –1 < x < 3
x –1 0
1
4
3
f(x) 5 1 0 11
0
–1
11
5
1
3
x
f(x)
1
–
4
The range is 0 < f(x) < 11.
(a) f(x) = x – 3, –2 < x < 4
x –2 0 3 4
f(x) 5 3 0 1
0 4
3
1
5
3
–2
x
f(x)
(b) f(x) = 2x + 6, –4 < x < 1
x –4 –3 0 1
f(x) 2 0 6 8
0
6
2
8
1
–4 –3
x
f(x)
(c) f(x) = 5x – 4, –3 < x < 2
x –3 0
4
5 2
f(x) 19 4 0 6
0
6
4
19
2
–3
x
f(x)
4
–
5
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
f(–3) = –3 – 8
= –11
(ii) f(x) = –5
x – 8 = –5
x = 3

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(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
f(–1) = 4(–1) – a
–4 – a = –11
a = 7
(ii) f(x) = 4x – 7
4x – 7 = –19
4x = –12
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
h(2) = 6(2) – 5
= 7
(ii) f(x) = 6x – 5
6x – 5 = x
5x = 5
x = 1
(c) Given g : x → px2
– 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 → px2
– qx dengan keadaan p dan q ialah pemalar. Jika g(3) = 12 dan g(4) = 28, cari nilai p dan q.
g(x) = px2
– qx Substitute  into :
g(3) = p(3)2
– q(3) 4p – (3p – 4) = 7
9p – 3q = 12 4p – 3p + 4 = 7
3p – q = 4 ––– 
p = 3
g(4) = p(4)2
– q(4) Substitute p = 3 into :
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

6
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Notes
SMART
1.
f(x)
x
f g
P Q R
gf(x)
gf
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.
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)].
Diberi dua fungsi f(x) dan g(x), dua fungsi itu boleh
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, f2
= ff, f3
= fff and so on.
Secara umumnya, fg ≠ gf, f2
= ff, f3
= fff dan seterusnya.
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
f : x → 3x + 1, g : x → x – 1
fg(x) = f[g(x)] gf(x) = g[f(x)]
= f(x – 1) = g(3x + 1)
= 3(x – 1) + 1 = 3x + 1 – 1
= 3x – 3 + 1 = 3x
= 3x – 2 ∴ gf : x → 3x
∴ fg : x → 3x – 2
(a) f : x → x + 2, g : x → 4x – 5
fg(x) = f(4x – 5) gf(x) = g(x + 2)
= 4x – 5 + 2 = 4(x + 2) – 5
= 4x – 3 = 4x + 8 – 5
= 4x + 3
∴ fg : x → 4x – 3 ∴ gf : x → 4x + 3
(b) f : x → 2x – 3, g : x → x2
fg(x) = f(x2
) gf(x) = g(2x – 3)
= 2x2
– 3 = (2x – 3)2
= 4x2
– 12x + 9
∴ fg : x → 2x2
– 3 ∴ gf : x → 4x2
– 12x + 9
(c) f : x → x2
– 4x + 3, g : x → –5x
fg(x) = f(–5x)
= (–5x)2
– 4(–5x) + 3
= 25x2
+ 20x + 3
∴ fg : x → 25x2
+ 20x + 3
gf(x) = g(x2
– 4x + 3)
= –5(x2
– 4x + 3)
= –5x2
+ 20x – 15
∴ gf : x → –5x2
+ 20x – 15
10. Find the composite functions f 2
based on the functions f given. PL 3
Cari fungsi gubahan f 2
berdasarkan fungsi f yang diberikan.
Example
f : x → x – 2
f 2
(x) = f[ f(x)]
= f(x – 2)
= (x – 2) – 2
= x – 4
(a) f : x → 7x – 8
f 2
(x) = f[ f(x)]
= f(7x – 8)
= 7(7x – 8) – 8
= 49x – 56 – 8
= 49x – 64
(b) f : x → x2
+ 3
f 2
(x) = f[ f(x)]
= f(x2
+ 3)
= (x2
+ 3)2
+ 3
= x4
+ 6x2
+ 9 + 3
= x4
+ 6x2
+ 12
Composite Functions
Fungsi Gubahan
1.2 Textbook
pg. 12 – 19

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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
f : x → 5x, g : x → 4 – 2x
(i) fg(x) = f[g(x)]
= f(4 – 2x)
= 5(4 – 2x)
= 20 – 10x
fg(2) = 20 – 10(2)
= 0
gf(x) = g[f(x)]
= g(5x)
= 4 – 2(5x)
= 4 – 10x
gf(4) = 4 – 10(4)
= –36
Alternative Method
g(2) = 4 – 2(2)
= 0
fg(2)= f(0)
= 5(0)
= 0
f(4) = 5(4)
= 20
gf(4) = g(20)
= 4 – 2(20)
= –36
(ii) fg(x) = 5
20 – 10x = 5
10x = 15
x = 1.5
(a) f : x → 3x + 2, g : x → 2 – x2
(i) fg(x) = f(2 – x2
)
= 3(2 – x2
) + 2
= 6 – 3x2
+ 2
= 8 – 3x2
fg(2) = 8 – 3(2)2
= –4
gf(x) = g(3x + 2)
= 2 – (3x + 2)2
= 2 – (9x2
+ 12x + 4)
= –9x2
– 12x – 2
gf(4) = –9(4)2
– 12(4) – 2
= –194
(ii) fg(x) = 5
8 – 3x2
= 5
3x2
= 3
x2
= 1
x = 1 or –1
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
f(x) = x + 3, fg(x) = 2x – 5
f [g(x)] = 2x – 5
g(x) + 3 = 2x – 5
g(x) = 2x – 8
∴ g : x → 2x – 8
(a) f(x) = x + 4, fg(x) = 3x + 9
f [g(x)] = 3x + 9
g(x) + 4 = 3x + 9
g(x) = 3x + 5
∴ g : x → 3x + 5
(b) f(x) = 3x – 6, fg(x) = 5x + 8
f [g(x)] = 5x + 8
3g(x) – 6 = 5x + 8
3g(x) = 5x + 14
g(x) = 5x + 14
3
∴ 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
g(x) = x + 2, fg(x) = 4x – 9
f[g(x)] = 4x – 9
f(x + 2) = 4x – 9
Let y = x + 2
x = y – 2
So, f(y) = 4(y – 2) – 9
= 4y – 8 – 9
= 4y – 17
Substitute y with x, f(x) = 4x – 17
∴ f : x → 4x – 17
(a) g(x) = 2x + 3, fg(x) = 7x – 4
f[g(x)] = 7x – 4
f(2x + 3) = 7x – 4
Let y = 2x + 3
x = y – 3
2
So, f(y) = 7y – 3
2  – 4
=
7y – 29
2
∴ f : x →
7x – 29
2
(b) g(x) = x2
– 9, fg(x) = 3x2
– 5
f[g(x)] = 3x2
– 5
f(x2
– 9) = 3x2
– 5
Let y = x2
– 9
x2
= y + 9
x = √y + 9
So, f(y) = 3(√y + 9)2
– 5
= 3(y + 9) – 5
= 3y + 27 – 5
= 3y + 22
∴ f : x → 3x + 22

8
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14. Solve the following problems. PL 4
Selesaikan masalah-masalah berikut.
Example
Given the functions f : x → 4x – 1 and g : x → (x – 3)2
. Find
Diberi fungsi f : x → 4x – 1 dan g : x → (x – 3)2
. Cari
(i) fg(x), (ii) fg(3).
(i) fg(x) = f[(x – 3)2
]
= 4(x – 3)2
– 1
= 4(x2
– 6x + 9) – 1
= 4x2
– 24x + 36 – 1
= 4x2
– 24x + 35
(ii) fg(3) = 4(3)2
– 24(3) + 35
= 36 – 72 + 35
= –1
(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)
= 3(x – 6) + 8
= 3x – 18 + 8
= 3x – 10
(ii) fg(–2) = 3(–2) – 10
= –16
(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 → x2
+ 4x + 4, (ii) gf : x → 2x2
– 3x + 4.
(i) fg(x) = x2
+ 4x + 4

f(x + 2) = x2
+ 4x + 4
Let y = x + 2
x = y – 2
So, f(y) = (y – 2)2
+ 4(y – 2) + 4
= y2
– 4y + 4 + 4y – 8 + 4
= y2
∴ f(x) = x2
(ii) gf(x) = 2x2
– 3x + 4

f(x) + 2 = 2x2
– 3x + 4
f(x) = 2x2
– 3x + 2
(c) Given f(x) = px + q and f 2
(x) = 4x + 6. Find the values of p and q.
Diberi f(x) = px + q dan f 2
(x) = 4x + 6. Cari nilai-nilai p dan q.
ff(x) = 4x + 6
f(px + q) = 4x + 6
p(px + q) + q = 4x + 6
p2
x + pq + q = 4x + 6
Comparing:
p2
= 4
p = ±√4
= 2 or –2
When p = 2,
pq + q = 6
2q + q = 6
3q = 6
q = 2
When p = –2,
pq + q = 6
–2q + q = 6
–q = 6
q = –6
(d) Functions g and h are defined as g : x → 5x – 6 and h : x →
3x
x – 2
. Find the value of x if gh(x) = h(x).
Fungsi g dan h ditakrifkan oleh g : x → 5x – 6 dan h : x →
3x
x – 2
. Cari nilai x jika gh(x) = h(x).
g(x) = 5x – 6
h(x) =
3x
x – 2
gh(x) = h(x)
g 3x
x – 2  =
3x
x – 2
5
3x
x – 2  – 6 =
3x
x – 2
15x
x – 2
– 6 =
3x
x – 2
15x – 6(x – 2) = 3x
15x – 6x + 12 = 3x
6x = –12
x = –2

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15. Solve each of the following. PL 5 Daily Application
Selesaikan setiap yang berikut.
Example
The temperature in Fahrenheit, F, can be converted to Celsius, C, using the function C(F) =
5
9
(F – 32).
The temperature in Kelvin, K, can be converted to Fahrenheit, F, using F(K) =
9
5
(K – 273) + 32. Suraya
obtained a result of temperature of 258 K in an experiment. Convert the temperature to Celsius, in °C.
Suhu dalam Fahrenheit, F, boleh ditukar kepada Celsius, C, dengan menggunakan fungsi C(F) =
5
9 (F – 32). Suhu dalam Kelvin, K,
boleh ditukar kepada Fahrenheit, F, dengan menggunakan fungsi F(K) =
9
5 (K – 273) + 32. Suraya memperoleh hasil suhu 258 K
dalam suatu eksperimen. Tukar suhu tersebut ke Celsius, dalam °C.
F(258) =
9
5
(258 – 273) + 32 C[F(258)] = C(5)
=
5
9
(5 – 32)
= –15
= 5
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 cm3
, in the container is given by the function
V(h) =
88
7
h.
Ketinggian air, h cm, dalam suatu bekas berbentuk silinder semakin meningkat mengikut masa, t s, dan fungsinya ialah
h(t) = 1.4t. Isi padu air, V cm3
, dalam bekas tersebut diberi oleh fungsi V(h) =
88
7 h.
(i) State the volume of water, V, as a function of time, t.
Nyatakan isi padu air, V, sebagai fungsi masa, t.
(ii) Find the volume of water, in cm3
, in the container after 5 seconds.
Cari isi padu air, dalam cm3
, dalam bekas tersebut selepas 5 saat.
(i) V[h(t)] = V(1.4t)
=
88
7
(1.4t)
= 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 cm3
.
(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

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Inverse Functions
Fungsi Songsang
1.3
Notes
SMART
1.
x
f
y
f –1
If f(x) = y, then the inverse function is f –1
(y) = x.
Jika f(x) = y, maka fungsi songsangnya ialah f –1
(y) = x.
2. Only one-to-one functions have inverse functions.
Hanya fungsi satu kepada satu mempunyai fungsi songsang.
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
hanya jika fg(x) = x, x dalam domain g dan gf(x) = x, x dalam
domain f.
4. If f and g are inverse functions of each other, then
Jika f dan g ialah fungsi songsang antara satu sama lain, maka
(a) domain of f = range of g and domain of g = range of f
domain f = julat g dan domain g = julat f
(b) graph g is the reflection of graph f on the line y = x
graf g adalah pantulan graf f pada garis y = x
5. Horizontal line test can be used to test the existence
of inverse functions.

Ujian garis mengufuk boleh digunakan untuk menguji
kewujudan fungsi songsang.
y = f(x)
0
y
x
f has inverse function
f mempunyai fungsi
songsang
y = h(x)
0
y
x
h does not have
inverse function
h tidak mempunyai
fungsi songsang
7. ff –1
(x) = f –1
f(x) = 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.
x y
f
2
3
9
11
5
6
(a) f(2) = 3 (b) f–1
(11) = 6
(c) f–1
(3) = 2 (d) f–1
(9) = 5
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
0 2
y
f
x
Has inverse function. When tested
with a horizontal line, the line cuts
the graph at one point.
(a)
0 3
–1
y
f
x
Does not have inverse
function. When tested with a
horizontal line, the line cuts
the graph at two points.
(b)
0
y
f
x
Has inverse function. When
tested with a horizontal line,
the line cuts the graph at one
point.
Textbook
pg. 20 – 29

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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
f(x) = x – 8, g(x) = x + 8
fg(x) = f(x + 8) gf(x) = g(x – 8)
= (x + 8) – 8 = (x – 8) + 8
= x = x
Since fg(x) = gf(x) = x, f and g are inverse functions
of each other.
(a) f(x) = 4 – 3x, g(x) =
4 – x
3
fg(x) = f 
4 – x
3 
= 4 – 3
4 – x
3 
= 4 – (4 – x)
= x
gf(x) = g(4 – 3x)
=
4 – (4 – 3x)
3
= 3x
3
= x
Since fg(x) = gf(x) = x, f and g are inverse
functions of each other.
19. Sketch graphs of f and f–1
on the same plane. Then, state the domain of f–1
. PL 3
Lakar graf bagi f dan f–1
pada satah yang sama. Seterusnya, nyatakan domain bagi f–1
.
Example
f : x → 4x, domain: 0 < x < 3
x 0 1 2 3
y 0 4 8 12
y = x
(3, 12)
(12, 3)
x
f –1
y
f
0
Domain of f –1
:
0 < x < 12
(a) f : x → –
5
x
, domain: 1 < x < 10
x 1 5 10
y –5 –1 –0.5
x
f
y
y = x
0
(–0.5, 10)
(–5, 1)
(1, –5)
(10, –0.5)
f –1
Domain of f –1
:
–5 < x < –0.5
20. Find the inverse function of each of the following functions. PL 3
Cari fungsi songsang bagi setiap fungsi berikut.
Example
f(x) = 4x + 5
Let y = f(x)
y = 4x + 5
4x = y – 5
x =
y – 5
4
Since f –1
(y) = x,
f –1
(y) =
y – 5
4
Substitute y with x,
f –1
(x) =
x – 5
4
∴ f –1
: x →
x – 5
4
(a) f(x) = 3x – 1
Let y = f(x)
y = 3x – 1
3x = y + 1
x = y + 1
3
Since f –1
(y) = x,
f –1
(y) = y + 1
3
f –1
(x) = x + 1
3
∴ f –1
: x → x + 1
3
(b) g(x) = 7 – 2x
Let y = g(x)
y = 7 – 2x
2x = 7 – y
x = 7 – y
2
Since g–1
(y) = x,
g–1
(y) = 7 – y
2
g–1
(x) = 7 – x
2
∴ g–1
: x → 7 – x
2
Range of f

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(c) g(x) =
x
4
+ 1
Let y = g(x)
y =
x
4
+ 1
x
4
= y – 1
x = 4(y – 1)
x = 4y – 4
Since g–1
(y) = x,
g–1
(y) = 4y – 4
g–1
(x) = 4x – 4
∴ g–1
: x → 4x – 4
(d) h(x) =
x – 1
2x + 1
Let y = h(x)
y =
x – 1
2x + 1
2xy + y = x – 1
y + 1 = x – 2xy
y + 1 = x(1 – 2y)
x =
y + 1
1 – 2y
Since h–1
(y) = x,
h–1
(y) =
y + 1
1 – 2y
h–1
(x) =
x + 1
1 – 2x
∴ h–1
: x →
x + 1
1 – 2x
(e) h(x) =
3
x – 1
Let y = h(x)
y =
3
x – 1
xy – y = 3
xy = 3 + y
x =
3 + y
y
x =
3
y
+ 1
Since h–1
(y) = x,
h–1
(y) =
3
y
+ 1
h–1
(x) =
3
x
+ 1
∴ h–1
: x →
3
x
+ 1
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
f –1
(x) =
4 + x
2
Let y = f –1
(x)
y =
4 + x
2
2y = 4 + x
x = 2y – 4
Since f (y) = x,
f (y) = 2y – 4
f (x) = 2x – 4
∴ f : x → 2x – 4
(a) f –1
(x) = 3x – 4
Let y = f –1
(x)
y = 3x – 4
3x = y + 4
x = y + 4
3
Since f (y) = x,
f (y) = y + 4
3
f (x) = x + 4
3
∴ f : x → x + 4
3
(b) f –1
(x) =
x + 6
5
Let y = f –1
(x)
y = x + 6
5
5y = x + 6
x = 5y – 6
Since f(y) = x,
f(y) = 5y – 6
f(x) = 5x – 6
∴ f : x → 5x – 6
(c) g–1
(x) =
x – 7
3
Let y = g–1
(x)
y = x – 7
3
3y = x – 7
x = 3y + 7
Since g(y) = x,
g(y) = 3y + 7
g(x) = 3x + 7
∴ g : x → 3x + 7
(d) g–1
(x) =
x + 4
x – 4
Let y = g–1
(x)
y = x + 4
x – 4
xy – 4y = x + 4
xy – x = 4y + 4
x = 4y + 4
y – 1
Since g(y) = x,
g(y) = 4y + 4
y – 1
g(x) = 4x + 4
x – 1
∴ g : x → 4x + 4
x – 1
(e) h–1
(x) =
5
x
– 2
Let y = h–1
(x)
y =
5
x
– 2
5
x
= y + 2
x = 5
y + 2
Since h(y) = x,
h(y) = 5
y + 2
h(x) = 5
x + 2
∴ h : x → 5
x + 2

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22. Solve each of the following problems. PL 5
Selesaikan masalah-masalah berikut.
Example
Given the function f : x → 4x – 7 and g : x → x2
– 5, find
Diberi fungsi f : x → 4x – 7 dan g : x → x2
– 5, cari
(i) f –1
(x), (ii) gf –1
(x), (iii) gf –1
(5)
(i) Let y = f(x)
y = 4x – 7
4x = y + 7
x =
y + 7
4
Since f –1
(y) = x,
f –1
(y) =
y + 7
4
f –1
(x) =
x + 7
4
(ii) gf –1
(x) = g
x + 7
4 
= 
x + 7
4 
2
– 5
= 
x2
+ 14x + 49
16  – 5
= x2
+ 14x + 49 – 80
16
= x2
+ 14x – 31
16
(iii) gf –1
(5) = 52
+ 14(5) – 31
16
= 64
16
= 4
(a) Given g : x → 2x + p
p – 4
, p ≠ 4 and g(4) = 5, find
Diberi g : x →
2x + p
p – 4
, p ≠ 4 dan g(4) = 5, cari
(i) the value of p / nilai p,
(ii) g–1
(x),
(iii) the value of x if g–1
(x) = 4 / nilai x jika g–1
(x) = 4.
(i) g(4) =
2(4) + p
p – 4
5 =
8 + p
p – 4
5(p – 4) = 8 + p
5p – 20 = 8 + p
4p = 28
p = 7
(ii) g(x) =
2x + 7
3
Let y = g(x)
y =
2x + 7
3
3y = 2x + 7
x =
3y – 7
2
g–1
(x) =
3x – 7
2
(iii)
g–1
(x) =
3x – 7
2
4 =
3x – 7
2
8 = 3x – 7
3x = 15
x = 5
(b) It is given g : x → 8x – 3 and h : x →
5
x
, x ≠ 0. Find
Diberi bahawa g : x → 8x – 3 dan h : x →
5
x
, x ≠ 0. Cari
(i) g–1
(x), (ii) hg–1
(x), (iii) g–1
h(10).
(i) Let y = g(x)

y = 8x – 3
8x = y + 3

x =
y + 3
8
Since g–1
(y) = x,
g–1
(y) =
y + 3
8
g–1
(x) =
x + 3
8
(ii) hg–1
(x) = h
x + 3
8 
=
5
x + 3
8
= 40
x + 3
(iii) h(10) =
5
10
=
1
2
g–1
h(10) = g–1
1
2 
=
1
2
+ 3
8
=
7
16
g–1
(y) = x

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(c) Given the function h(x) = ax + b, where a . 0, a and b are constants and h2
(x) = 49x – 40, find
Diberi fungsi h(x) = ax + b, dengan keadaan a . 0, a dan b ialah pemalar dan h2
(x) = 49x – 40, cari
(i) the value of a and of b, / nilai a dan b,
(ii) h–1
(–3),
(iii) the value of x if (h–1
)2
(x) = 1 / nilai x jika (h–1
)2
(x) = 1.
(i) Given h(x) = ax + b
h2
(x) = hh(x)
= h(ax + b)
= a(ax + b) + b
= a2
x + ab + b
Compare with 49x – 40,
a2
= 49
a = 
49
= 7
ab + b = –40
7b + b = –40
8b = –40
b = –5
(ii) h(x) = 7x – 5
Let y = h(x)
y = 7x – 5
7x = y + 5
x =
y + 5
7
Since h–1
(y) = x,
h–1
(y) =
y + 5
7
h–1
(x) =
x + 5
7
∴ h–1
(–3) =
–3 + 5
7
=
2
7
(iii) (h–1
)2
(x) = 1
x + 5
7
+ 5
7
= 1

x + 5
7
+ 5 = 7

x + 5
7
= 2
x + 5 = 14
x = 9
(d) It is given that f –1
(x) =
3x – 6
7 + 4x
and f (x) =
mx + 6
3 + nx
. Find HOTS Analysing
Diberi bahawa f –1
(x) =
3x – 6
7 + 4x
dan f (x) =
mx + 6
3 + nx
. Cari
(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.
(i) f –1
(x) =
3x – 6
7 + 4x
Let y = f –1
(x)
y =
3x – 6
7 + 4x
7y + 4xy = 3x – 6
7y + 6 = 3x – 4xy
7y + 6 = x(3 – 4y)
x =
7y + 6
3 – 4y
Since f(y) = x,
f(y) =
7y + 6
3 – 4y
f(x) =
7x + 6
3 – 4x
Compare with
f (x) =
mx + 6
3 + nx
7x + 6
3 – 4x
=
mx + 6
3 + nx
∴ m = 7, n = –4
(ii) f(4x + 3) = 7, f(x) =
7x + 6
3 – 4x
7 =
7(4x + 3) + 6
3 – 4(4x + 3)
7 =
28x + 21 + 6
3 – 16x – 12
7 =
28x + 27
–9 – 16x

7(–9 – 16x) = 28x + 27
–63 – 112x = 28x + 27
–140x = 90
x = –
9
14

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Paper 1
1. The diagram shows the composite function fg that maps a
to c.
Rajah di bawah menunjukkan fungsi gubahan fg yang
memetakan a kepada c.
b
a c
fg
State / Nyatakan
(a) the function that maps a to b,
fungsi yang memetakan a kepada b,
(b) f –1
(c).
[2]
Ans: (a) g
(b) f –1
(c) = b
2. Given the functions f(x) = 5x + 1 and g(x) = mx – n, where
m and n are constants, express m in terms of n such that
fg(2) = 1.
Diberi fungsi f(x) = 5x + 1 dan g(x) = mx – n, dengan keadaan
m dan n ialah pemalar, ungkapkan m dalam sebutan n dengan
keadaan fg(2) = 1.
[3]
Ans: m =
1
2 n
3. It is given the function f : x → 8x – 4. Find
Diberi fungsi f : x → 8x – 4. Cari
(a) the value of x if f(x) maps onto itself,
nilai x jika f(x) memeta kepada dirinya sendiri,
(b) the value of h if f(3 – h) = 4h.
nilai h jika f(3 – h) = 4h.
[4]
Ans: (a) x =
4
7
(b) h =
5
3
4. It is given the function g : x → 3x – 9. Find
Diberi fungsi g : x → 3x – 9. Cari
(a) g–1
(x),
(b) the value of p if g2

4p
3  = 24.
nilai p jika g2

4p
3  = 24.
[4]
Ans: (a) g–1
(x) =
x + 9
3
(b) p = 5
SPM
2015
SPM
2015
SPM
2016
SPM
2017
5. Given the functions m : x → px + 3, h : x → 4x – 1 and
mh(x) = 4px + q, express p in terms of q.
Diberi fungsi m : x → px + 3, h : x → 4x – 1 dan
mh(x) = 4px + q, ungkapkan p dalam sebutan q.
[3]
Ans: p = 3 – q
6. It is given the functions
Diberi fungsi
g : x → 4x + 1
fg : x → 16x2
+ 8x – 5
Find / Cari
(a) g–1
(x),
(b) f(x).
[3]
Ans: (a) g–1
(x) =
x – 1
4
(b) f(x) = x2
– 6
7. The diagram shows the graph of the function
f : x → |1 – 3x| for the domain –1 < x < 3.
Rajah di bawah menunjukkan graf bagi fungsi
f : x → |1 – 3x| untuk domain –1 < x < 3.
0
8
(–1, 4)
f(x)
3
x
State / Nyatakan
(a) the object of 8,
objek bagi 8,
(b) the image of 2,
imej bagi 2,
(c) the domain of 0 < f(x) < 2.
domain bagi 0 < f(x) < 2.
[3]
Ans: (a) 3
(b) 5
(c) – 1
3
< x < 1
8. Given the functions g : x → 3x – 1 and h : x → 6x, find
Diberi fungsi g : x → 3x – 1 dan h : x → 6x, cari
(a) hg(x),
(b) the value of x if hg(x) = 1
3
g(x).
nilai x jika hg(x) =
1
3 g(x).
[4]
Ans: (a) hg(x) = 18x – 6
(b) x = 1
3
SPM
2016
SPM
2016
SPM
2017
SPM Practice 1

16
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9. The diagram shows the relation between set P, set Q and
set R.
Rajah di bawah menunjukkan hubungan antara set P, set Q
dan set R.
P
Q
R
fg : x → 9x + 12
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.
Diberi bahawa set P dipetakan kepada set Q oleh fungsi 5x + 6
dan dipetakan kepada set C oleh fg : x → 9x + 12.
(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
menggunakan tatatanda fungsi.
(b) Find the function which maps set Q to set R.
Cari fungsi yang memetakan set Q kepada set R.
[4]
Ans: (a) g : x → 5x + 6
(b) f : x →
9x + 6
5
10. The diagram shows the function g : x → x – 2p, where p is
a constant.
Rajah di bawah menunjukkan fungsi g : x → x – 2p, dengan
keadaan p ialah pemalar.
6
x
g
10
x – 2p
Find / Cari
(a) the value of p,
nilai p,
(b) g–1
(x),
(c) the value of x if g–1
(x) = 5.
nilai x jika g–1
(x) = 5.
[5]
Ans: (a) p = –2
(b) g–1
(x) = x – 4
(c) x = 9
11. Given the functions f(x) = 9 – 5x and g(x) =3x, find
Diberi fungsi f(x) = 9 – 5x dan g(x) = 3x, cari
(a) the value of fg(–1),
nilai bagi fg(–1),
(b) gf–1
(x).
[4]
Ans: (a) 24
(b) gf–1
(x) =
27 – 3x
5
SPM
2018
12. Given the function g : x → 2x – 7, find
Diberi fungsi g : x → 2x – 7, cari
(a) g(6),
(b) the value of p if 2g–1
(p) = g(6).
nilai p jika 2g–1
(p) = g(6).
[4]
Ans: (a) 5
(b) p = –2
2
1. In the diagram, function f maps set P to set Q and function
g maps set Q to set R.
Dalam rajah di bawah, fungsi f memetakan set P kepada set Q
dan fungsi g memetakan set Q kepada set R.
x
P
f g
2x + 5
Q
6x + 11
R
Find/ Cari
(a) in terms of x, the function
dalam sebutan x, fungsi
(i) which maps set Q to set P,
yang memetakan set Q kepada set P,
(ii) g(x),
[5]
(b) the value of x such that fg(x) = 3x + 4.
nilai x dengan keadaan fg(x) = 3x + 4.
[3]
Ans: (a) (i) f –1
(x) =
x – 5
2
; (ii) g(x) = 3x – 4
(b) x =
7
3
= 2
1
3
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
defined as h(y) =
60
y + q
, y ≠ –q.
Rajah di bawah menunjukkan pemetaan dari set X ke set Y yang
ditakrifkan sebagai g(x) = px + 3 dan pemetaan dari set Y ke
set Z yang ditakrifkan sebagai h(y) = 60
y + q
, y ≠ –q.
4
X
g h
11
Y
15
Z
Find/ Cari
(a) the values of p and q,
nilai-nilai p dan q,
[4]
(b) the function that maps set X to set Z.
fungsi yang memetakan set X kepada set Z.
[3]
Ans: (a) p = 2, q = –7
(b) hg(x) =
60
2x – 4
, x ≠ 2

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3. (a)
Given the functions f : x → 3x – 4 and g : x → 2 – 5x,
find
Diberi fungsi f : x → 3x – 4 dan g : x → 2 – 5x, cari
(i) g(4),
(ii) the value of n if f(n + 1) =
1
6
g(4),
nilai n jika f(n + 1) = 1
6
g(4),
(iii) gf(x).
[5]
(b) Hence, sketch the graph of y = ugf(x)u for –1 < x < 3.
State the range of y.
Seterusnya, lakar graf y = ugf(x)u untuk –1 < x < 3.
Nyatakan julat bagi y.
[3]
Ans: (a) (i) –18; (ii) n = –
2
3
; (iii) gf(x) = 22 – 15x
(b) Range: 0 < y < 37
SPM
2018
4. Given f : x →
x + 3
7
and g : x →
4
x – 2
, x ≠ 2, find
Diberi f : x → x + 3
7
dan g : x → 4
x – 2
, x ≠ 2, cari
(a) g–1
(x),
[2]
(b) g–1
f(x),
[2]
(c) the value of x if (f –1
)2
(x) = 25.
nilai x jika (f–1
)2
(x) = 25.
[3]
Ans: (a) g–1
(x) =
4
x
+ 2, x ≠ 0
(b) g–1
f(x) =
34 + 2x
x + 3
, x ≠ –3
(c) x = 1
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?
Apply concepts in functions and inverse functions.
Gunakan konsep dalam fungsi dan fungsi songsang.
Answer Guide
Ans: (a) 2 710 800 Rp
(b) f–1
(x) =
x
3 388.50
; The inverse function obtained converts money in Indonesian Rupiah (Rp) to Malaysian
Ringgit (RM).
2. Given f : x →
mx
n – x
, x ≠ n, g : x → 2x – m and the composite function gf : x →
9x – 15
5 – x
, find HOTS Applying
Diberi f : x → mx
n – x
, x ≠ n, g : x → 2x – m dan fungsi gubahan gf : x → 9x – 15
5 – x
, cari
(a) the values of m and n,
nilai m dan n,
(b) the inverse functions of f and g,
fungsi songsang bagi f dan g,
(c) the values of x if g = fg.
nilai x jika g = fg.
Apply concepts in composite functions and inverse functions.
Gunakan konsep dalam fungsi gubahan dan fungsi songsang.
Answer Guide
Ans: (a) m = 3, n = 5
(b) f–1
(x) =
5x
3 + x
, g–1
(x) =
x + 3
2
(c) x =
5
2
or
3
2
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Additional Mathematics Form 4 Answers
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CHAPTER 1 Functions
SPM Practice
Paper 1
1. (a) The function that maps a to b is g.
(b) f -1
(c) = b
2. g(x) = mx – n
g(2) = m(2) – n
= 2m – n
fg(2) = 1
5(2m – n) + 1 = 1
10m – 5n + 1 = 1
10m = 5n
m =
1
2
n
3. (a) f(x) = x
8x – 4 = x
7x = 4
x =
4
7
(b) f(x) = 8x – 4

f(3 – h) = 8(3 – h) – 4
4h = 24 – 8h – 4
4h = 20 – 8h
12h = 20
h =
5
3
4. (a) Let y = g(x)
y = 3x – 9
3x = y + 9
x =
y + 9
3
Since g-1
(y) = x,
g-1
(y) =
y + 9
3
g-1
(x) =
x + 9
3
(b) g2
(x) = gg(x)
= g(3x – 9)
= 3(3x – 9) – 9
= 9x – 27 – 9
= 9x – 36

g2
1
4p
3 2 = 91
4p
3 2 – 36
24 = 12p – 36
12p = 60
p = 5
5. m(x) = px + 3

h(x) = 4x – 1

mh(x) = m(4x – 1)
= p(4x – 1) + 3
Comparing:
4px + q = 4px – p + 3
q = –p + 3
p = 3 – q
6. (a) Let y = g(x)
y = 4x + 1
4x = y – 1
x =
y – 1
4
Since g-1
(y) = x,
g-1
(y) =
y – 1
4
g-1
(x) =
x – 1
4
(b) fg : x → 16x2
+ 8x – 5
f(x) = fgg-1
(x)
= fg[g-1
(x)]
= fg1
x – 1
4 2
= 161
x – 1
4 2
2
+ 81
x – 1
4 2 – 5
= 161
x2
– 2x + 1
16 2 + 2(x – 1) – 5
= x2
– 2x + 1 + 2x – 2 – 5
= x2
– 6
7. (a) 3
(b) f(2) = 1 – 3(2)
= –5
= 5
(c) When f(x) = 2,

1 – 3x = 2
1 – 3x = 2 and 1 – 3x = –2
3x = -1 3x = 3
x = –
1
3
x = 1
Domain is –
1
3
< x < 1.
8. (a) g(x) = 3x – 1
h(x) = 6x
hg(x) = h(3x – 1)
= 6(3x – 1)
= 18x – 6

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(b) hg(x) =
1
3
g(x)
18x – 6 =
1
3
(3x – 1)
3(18x – 6) = 3x – 1
54x – 18 = 3x – 1
51x = 17
x =
1
3
9. (a) g : x → 5x + 6
(b)
fg(x) = 9x + 12

g(x) = 5x + 6
Let y = g(x)

y = 5x + 6
5x = y – 6

x =
y – 6
5
f(y) = 91
y – 6
5 2 + 12
=
9y – 54 + 60
5
=
9y + 6
5
f(x) =
9x + 6
5
f : x →
9x + 6
5
10. (a) g(x) = x – 2p
g(6) = 6 – 2p

6 – 2p = 10
–2p = 4
p = –2
(b) g(x) = x + 4
Let y = g(x)
y = x + 4
x = y – 4
Since g-1
(y) = x,
g-1
(y) = y – 4
g-1
(x) = x – 4
(c)
g–1
(x) = x – 4
5 = x – 4
x = 9
11. (a) fg(x) = f(3x)
= 9 – 5(3x)
= 9 – 15x
fg(–1) = 9 – 15(–1)
= 24
(b) Let y = f(x)

y = 9 – 5x
5x = 9 – y

x =
9 – y
5
Since f -1
(y) = x,
f -1
(y) =
9 – y
5
f -1
(x) =
9 – x
5
gf -1
(x) = g1
9 – x
5 2
= 31
9 – x
5 2
=
27 – 3x
5
12. (a) g(x) = 2x – 7
g(6) = 2(6) – 7
= 5
(b) Let y = g(x)

y = 2x – 7
2x = y + 7

x = y + 7
2
Since g-1
(y) = x,
g-1
(y) = y + 7
2
g-1
(x) = x + 7
2
Given 2g-1
(p) = g(6)
21
p + 7
2 2 = 5
p + 7 = 5
p = –2
Paper 2
1. (a) (i) f(x) = 2x + 5
Let y = f(x)
y = 2x + 5
2x = y – 5
x =
y – 5
2
Since f -1
(y) = x,
f -1
(y) =
y – 5
2
f -1
(x) =
x – 5
2
(ii) f(x) = 2x + 5

gf(x) = 6x + 11

g(x) = gff -1
(x)

= gf1
x – 5
2 2

= 61
x – 5
2 2 + 11

= 3x – 15 + 11

= 3x – 4

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(b) fg(x) = 3x + 4
f(3x – 4) = 3x + 4
2(3x – 4) + 5 = 3x + 4
6x – 8 + 5 = 3x + 4
3x = 7
x =
7
3
= 2
1
3
2. (a) g(x) = px + 3
g(4) = 4p + 3
4p + 3 = 11
4p = 8
p = 2
h(y) =
60
y + q
h(11) =
60
11 + q
15 =
60
11 + q
11 + q = 4
q = –7
(b)
g(x) = 2x + 3
h(y) =
60
y – 7
hg(x) = h(2x + 3)
=
60
2x + 3 – 7
=
60
2x – 4
, x ≠ 2
3. (a) (i) g(x) = 2 – 5x
g(4) = 2 – 5(4)
= 2 – 20
= –18
(ii) f(n + 1) =
1
6
g(4)
3(n + 1) – 4 =
1
6
(–18)
3n + 3 – 4 = –3
3n = –2
n = –
2
3
(iii) gf(x) = g(3x – 4)
= 2 – 5(3x – 4)
= 2 – 15x + 20
= 22 – 15x
(b) y = gf(x)
y = 22 – 15x
x −1 0
22
15
3
y 37 22 0 23
23
3
22
15
0
–1
37
y
x
22
Range of y: 0 < y < 37
4. (a) Let y = g(x)
y =
4
x – 2
xy – 2y = 4
xy = 4 + 2y
x =
4 + 2y
y
x =
4
y
+ 2
Since g -1
(y) = x,
g-1
(y) =
4
y
+ 2
g-1
(x) =
4
x
+ 2, x ≠ 0
(b) g-1
f(x) = g-1
1
x + 3
7 2
=
4
x + 3
7
+ 2
=
28 + 2(x + 3)
x + 3
=
28 + 2x + 6
x + 3
=
34 + 2x
x + 3
, x ≠ –3
(c) Let y = f(x)

y =
x + 3
7
7y = x + 3

x = 7y – 3
Since f -1
(y) = x,
f -1
(y) = 7y – 3
f -1
(x) = 7x – 3
f –1
f –1
(x) = f –1
(7x – 3)
= 7(7x – 3) – 3
= 49x – 21 – 3
= 49x – 24
If (f –1
)2
(x) = 25
49x – 24 = 25
49x = 49
x = 1

4
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HOTS Challenge
1. (a) f(x) = 3388.50x
f(800) = (3388.50)(800)
= 2 710 800 Rp
(b) Let y = f(x)
y = 3 388.50x
x =
y
3 388.50
Since f -1
(y) = x,
f -1
(y) =
y
3 388.50
f -1
(x) =
x
3 388.50
The inverse function obtained converts money in
Indonesian Rupiah (Rp) to Malaysian Ringgit (RM).
2. (a) gf(x) = g1
mx
n – x 2
9x – 15
5 – x
= 21
mx
n – x 2 – m
=
2mx
n – x
– m
=
2mx – m(n – x)
n – x
=
2mx – mn + mx
n – x
=
3mx – mn
n – x
Comparing:

5 – x = n – x

n = 5
and
3mx – mn = 9x – 15
3m = 9
m = 3
(b) Let y = f(x)
y =
3x
5 – x
5y – xy = 3x
5y = 3x + xy
5y = x(3 + y)
x =
5y
3 + y
Since f -1
(y) = x,
f -1
(y) =
5y
3 + y
f -1
(x) =
5x
3 + x
Let y = g(x)
y = 2x – 3
2x = y + 3
x =
y + 3
2
Since g-1
(y) = x,
g-1
(y) =
y + 3
2
g-1
(x) =
x + 3
2
(c) fg(x) = f(2x – 3)
=
3(2x – 3)
5 – (2x – 3)
=
6x – 9
5 – (2x – 3)
=
6x – 9
8 – 2x
g(x) = fg(x)
2x – 3 =
6x – 9
8 – 2x
(2x – 3)(8 – 2x) = 6x – 9
16x – 4x2
– 24 + 6x = 6x – 9
4x2
– 16x + 15 = 0
(2x – 5)(2x – 3) = 0
x =
5
2
or
3
2

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