Module & More Matematik Tambahan TG4

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Pembelajaran BERPANDU dan SISTEMATIK


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TAMBAHAN TINGKATAN




Additional Mathematics DWIBAHASA


Tee Hock Tian
Wong Kiat Hi KSSM



Latihan modular
Studi 1 Minit

Praktis SPM
Eksklusif
Klu Soalan Eksklusif
i-THINK/KBAT Format

Praktis Ekstra SPM Kod QR
Lembaran PBD Kod QR PELANGI SPM
ONLINE TEST
Jawapan Kod QR
https://qr.pelangibooks.com/?u=POTMT4 Terkini







Lembaran
PBD dengan
Jawapan
Bonus Digital

Kandungan






3.2 Persamaan Serentak yang melibatkan Satu Persamaan
Bab Fungsi 1 Linear dan Satu Persamaan Tak Linear.........................50
1 Functions Simultaneous Equations involving One Linear Equation and
One Non-Linear Equation
1.1 Fungsi ..................................................................................1
Functions Praktis SPM 3 ............................................................................57
1.2 Fungsi Gubahan .................................................................7 Praktis Ekstra SPM 3 Kod QR ............................................57
Composite Functions Sudut KBAT ..............................................................................59
Penerbitan Pelangi Sdn. Bhd.
1.3 Fungsi Songsang ..............................................................14 ........................................................................................ 59
Inverse Functions
Praktis SPM 1 ............................................................................19

Praktis Ekstra SPM 1 Kod QR ............................................19 Bab Indeks, Surd dan Logaritma 60
4
Indices, Surds and Logarithms
Sudut KBAT ..............................................................................21
........................................................................................ 21 4.1 Hukum Indeks ..................................................................60
Laws of Indices
4.2 Hukum Surd .....................................................................64
Laws of Surds
Bab Fungsi Kuadratik 22 4.3 Hukum Logaritma ...........................................................71
2 Quadratic Functions Laws of Logarithms
4.4 Aplikasi Indeks, Surd dan Logaritma ............................78
2.1 Persamaan dan Ketaksamaan Kuadratik ......................22 Applications of Indices, Surds and Logarithms
Quadratic Equations and Inequalities
2.2 Jenis-jenis Punca Persamaan Kuadratik .......................28 Praktis SPM 4 ............................................................................80
Types of Roots of Quadratic Equations Praktis Ekstra SPM 4 Kod QR ............................................80
2.3 Fungsi Kuadratik ..............................................................31 Sudut KBAT ..............................................................................82
Quadratic Functions
........................................................................................ 82
Praktis SPM 2 ............................................................................38
Praktis Ekstra SPM 2 Kod QR ............................................38
Sudut KBAT ..............................................................................43 Bab Janjang 83
........................................................................................ 43 5 Progressions

5.1 Janjang Aritmetik .............................................................83
Arithmetic Progressions
Bab Sistem Persamaan 44 5.2 Janjang Geometri .............................................................91
3 Systems of Equations Geometric Progressions
Praktis SPM 5 ..........................................................................101
3.1 Sistem Persamaan Linear dalam Tiga
Pemboleh Ubah ...............................................................44 Praktis Ekstra SPM 5 Kod QR ..........................................101
Systems of Linear Equations in Three Variables Sudut KBAT ............................................................................105
...................................................................................... 105



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Kan[M&M MateT Tg4].indd 3 20/12/2022 2:39 PM

Sudut KBAT ............................................................................164
Bab Hukum Linear 106 ...................................................................................... 164
6 Linear Law


6.1 Hubungan Linear dan Tak Linear ...............................106
Linear and Non-Linear Relations Bab Penyelesaian Segi Tiga 165
6.2 Hukum Linear dan Hubungan Tak Linear .................112 9 Solution of Triangles
Linear Law and Non-Linear Relations
6.3 Aplikasi Hukum Linear .................................................118 9.1 Petua Sinus ......................................................................165
Application of Linear Law Sine Rule

Praktis SPM 6 ..........................................................................122 9.2 Petua Kosinus .................................................................169
Cosine Rule
Praktis Ekstra SPM 6 Kod QR ..........................................122 9.3 Luas Segi Tiga .................................................................172
Sudut KBAT ............................................................................125 Area of a Triangle
Penerbitan Pelangi Sdn. Bhd.
...................................................................................... 125 9.4 Aplikasi Petua Sinus, Petua Kosinus dan Luas
Segi Tiga ..........................................................................174
Application of Sine Rule, Cosine Rule and Area of a Triangle
Praktis SPM 9 ..........................................................................175
Bab Geometri Koordinat 126
7 Coordinate Geometry Praktis Ekstra SPM 9 Kod QR ..........................................175
Sudut KBAT ............................................................................178
7.1 Pembahagi Tembereng Garis .......................................126
Divisor of a Line Segment ...................................................................................... 178
7.2 Garis Lurus Selari dan Garis Lurus Serenjang ...........130
Parallel Lines and Perpendicular Lines
7.3 Luas Poligon ...................................................................135
Areas of Polygons Bab Nombor Indeks 179
10
Index Numbers
7.4 Persamaan Lokus ...........................................................137
Equations of Loci
10.1 Nombor Indeks ..............................................................179
Praktis SPM 7 ..........................................................................140 Index Numbers
Praktis Ekstra SPM 7 Kod QR ..........................................140 10.2 Indeks Gubahan .............................................................185
Composite Index
Sudut KBAT ............................................................................143
Praktis SPM 10 .......................................................................190
...................................................................................... 143
Praktis Ekstra SPM 10 Kod QR ........................................190

Sudut KBAT ............................................................................193

Bab Vektor 144 ...................................................................................... 193
8 Vectors
Kertas Pra-SPM ......................................................................195
8.1 Vektor ..............................................................................144
Vectors Lembaran PBD dan Jawapan
8.2 Penambahan dan Penolakan Vektor ...........................148 http://www.epelangi.com/Module&More2021/
Addition and Subtraction of Vectors MatematikTambahan/T4/LPBD.pdf
8.3 Vektor dalam Satah Cartes ...........................................152
Vectors in a Cartesian Plane
Jawapan
Praktis SPM 8 ..........................................................................159 http://www.epelangi.com/Module&More2021/
MatematikTambahan/T4/JawapanKeseluruhan.pdf
Praktis Ekstra SPM 8 Kod QR ..........................................159



iv





Kan[M&M MateT Tg4].indd 4 20/12/2022 2:39 PM

Bab Fungsi Matematik Tambahan Tingkatan 4 Bab 1 Fungsi
1 Functions







1.1 Fungsi Analisis Soalan SPM 2021
Kertas 1
Kertas 2
Functions
S9
–
STUDI 1 Minit Buku Teks: 2 – 11

Tiga Cara Perwakilan Hubungan / Three Ways to Represent Relations
Suatu hubungan memadankan unsur-unsur satu set dengan unsur-unsur set lain.
Penerbitan Pelangi Sdn. Bhd.
A relation matches the elements of one set with the elements of another set.
1 Gambar rajah anak panah 2 Graf / Graph 3 Pasangan bertertib
Arrow diagram Ordered pairs
Set A
separuh daripada {(1, 2), (2, 4), (3, 6), (4, 8)}
is half of 8
Set A Set B
6
1 2 4
2 4 2
3 6 0
4 8 1 2 3 4 Set B




Empat Jenis Hubungan / Four Types of Relations

1 Satu dengan satu 2 Banyak dengan satu 3 Satu dengan banyak 4 Banyak dengan banyak
One-to-one Many-to-one One-to-many Many-to-many
A B C D P Q M N


















x f y Tatatanda fungsi/Function notation:
f(x) = x atau / or f : x → x 2
2
3 9 Domain / Domain = {3, 4, 7}
16 Kodomain / Codomain = (9, 16, 49, 64}
4 Objek / Objects = 3, 4, 7
49
Imej / Images = 9, 16, 49
7 64 Julat / Range = {9, 16, 49}





1





01-[M&M Mate(T) Tg4] 2021.indd 1 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi

1. Tentukan sama ada setiap hubungan yang berikut ialah satu fungsi atau bukan. Beri sebab untuk jawapan
anda. TP 1
Determine whether each of the following relations is a function. Give the reason for your answer.
1
CONTOH (a)
BAB a p
q p 4
b q 8
r
c 12
s r

Suatu fungsi. Setiap objek hanya mempunyai satu Suatu fungsi. Setiap objek mempunyai hanya
imej. satu imej.
A function. Each object has only one image. A function. Each object has only one image.








(b) (c)
3 m 2 3
5 n 4 5
7 p 6 7
11 8

Penerbitan Pelangi Sdn. Bhd.
Bukan fungsi. 11 tidak mempunyai imej. Bukan fungsi. Objek 8 tidak mempunyai imej.
Not a function. 11 does not have any image. Not a function. The object 8 does not have any image.










(d) (e)
y y
Tip Penting
Gunakan ujian garis
mencancang.
Use the vertical line test. x
0
x
0 2
Bukan fungsi. Garis mencancang menyilang graf
Satu fungsi. Garis mencancang memotong graf di dua titik berlainan.
itu pada satu titik sahaja. Not a function. The vertical line cuts the graph at two
A function. The vertical line cuts the graph at only one different points.
point.








2





01-[M&M Mate(T) Tg4] 2021.indd 2 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi

(h) {(6 , p), (6, q) , (9, q) , (13, r)} (i) {(a , 2) , (b , 2) , (c, 4) , (d, 8), (f, 2)}

Bukan fungsi. Objek 6 mempunyai lebih daripada Fungsi. Setiap objek mempunyai hanya satu
satu imej, iaitu p dan q. imej. 1
Not a function. The object 6 has more than one image, A function. Every object has only one image.
BAB
that is, p and q.









Penerbitan Pelangi Sdn. Bhd.



2. Tulis setiap fungsi yang berikut menggunakan tatatanda fungsi. TP 2
Write each of the following by using the function notation.

(a)
CONTOH
g
f Tip Penting 2 4
Fungsi sering diwakili oleh
1 5 huruf kecil seperti f, g, h 3 6
dan k.
2 6 Functions are often 4 8
represented by small letters
such as f, g, h and k.
3 7 10

g : x → x + 4 g : x → 2x
g(x) = x + 4 g(x) = 2x






1 1 1 (c)
(b) {(3, ), (5, ), (6, )}.
9 25 36 h(x)
(4, 7)
(2, 5)
1 3
f : x → , x ≠ 0
x 2
1
f(x) = , x ≠ 0 x
x 2 0
h : x → x + 3
h(x) = 2x











3





01-[M&M Mate(T) Tg4] 2021.indd 3 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi

3. Lengkapkan jadual yang berikut. TP 1
Complete the following table.

1 Fungsi Domain Kodomain Objek Imej Julat
Function Domain Codomain Object Image Range
BAB CONTOH {0 < x < 6} {−3 < f(x) < 10} 0 < x < 6 –3 < f(x) < 13 {0 < f(x) < 13}
f(x)
13





1 x
0 2 6
–3

(a) {−2 < x < 4} {0 <f(x) < 10} −2 < x < 4 0 < f(x) < 10 {0 < f(x) < 10}
f(x)
10


6


2
x
–2 0 3 4 Penerbitan Pelangi Sdn. Bhd.
(b) {−1, 0, 1, 2, 3} {−2, 0, 2, 4, 6} −1, 0, 1, 2, 3 −2, 0, 2, 4, 6 {−2, 0, 2, 4, 6}
f(x)
6
5
4
3
2
1
x
–1 0 1 2 3
–1
–2
(c) {4, 10, 18} {2, 5, 9, 11} 4, 10, 18 2, 5, 9 {2, 5, 9}

4 2

10 5
18 9

11


(d) {(3, 10), (5, 26), {3, 5, 6, 9} {10, 26, 37, 3, 5, 6, 9 10, 26, 37, 82 {10, 26, 37,
(6, 37),(9, 82)} 82} 82}







4





01-[M&M Mate(T) Tg4] 2021.indd 4 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi

4. Lakar graf bagi setiap fungsi berikut. Seterusnya, nyatakan julat sepadan bagi domain yang diberikan. TP 4
Sketch the graph of the following functions. Hence, state the corresponding range for the given domain.

2
CONTOH (a) y = |x − 1|, −3 < x < 3 1
f(x) = |1 − 2x|, −1 < x < 3
f(x)
1 BAB
x –1 0 1 2 3 (–3, 8) (3, 8)
2 8
7
f(x) 3 1 0 1 3 5
6
5
f(x)
4
5 3
Penerbitan Pelangi Sdn. Bhd.
2
4
1
3 x
–3 –2 –1 0 1 2 3
2

1 Julat/ Range: 0 < f(x) < 8

x
–1 0 1 2 3


Julat/ Range: 0 < f(x) < 5


5. Selesaikan setiap yang berikut. TP 5
Solve each of the following.
CONTOH

3
Fungsi f ditakrifkan sebagai f : x → 4x + , x ≠ 0. Cari
x
Function f is defined as f : x → 4x + 3 , x ≠ 0. Find
(i) f(−2) x
1
(ii) imej bagi di bawah f.
4
the image of 1 under f.
4
(iii) nilai-nilai f apabila imej ialah 13.
the values of f when the image is 13.
Penyelesaian:
3
(i) f(x) = 4x + (iii) f(x) = 13
x 3
3 1 4x + = 13
f(–2) = –8 – = –9 x
2 2 4x + 3 = 13x
2
2
3 4x – 13x + 3 = 0
(ii) f(x) = 4x +
x (4x – 1)(x – 3) = 0
1
1
1
f 1 2 = 4 1 2 + 3 x = , x = 3
1
4
4
4
4
= 1 + 12 = 13
5


01-[M&M Mate(T) Tg4] 2021.indd 5 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi


CONTOH
Fungsi g ditakrifkan sebagai g : x → |2 – 8x|, cari
1 Function g is defined as g : x → |2 – 8x|, find
1
BAB (i) g 1 2
2
(ii) nilai-nilai x yang memetakan hubungan dirinya.
the values of x which maps to itself.
Penyelesaian:
(i) g(x) = |2 – 8x| (ii) g(x) = x
1
g 1 2 = |2 − 4| |2 – 8x| = x
2 = |−2| 2 – 8x = x 2 – 8x = –x
x – 2 = 16 Penerbitan Pelangi Sdn. Bhd.
= 2
2 = x + 8x 2 = 8x – x
9x = 2 7x = 2
2 2
x = x =
9 7


12
(a) Fungsi f ditakrifkan sebagai f : x → , x ≠ 2. (b) Diberi g(x) = u3x – 7u, cari
Cari x – 2 Given g(x) = u3x – 7u,
The function f is defined as f : x → 12 , x ≠ 2. Find (i) nilai bagi g(−2) dan g(5).
– 2
(i) nilai bagi f(−4) dan f(10). the value of g(−2) and g(5).
the value of f(−4) and f(10). (ii) objek dengan keadaan imej ialah 14.
3 the objects for which the image is 14.
(ii) nilai-nilai x dengan keadaan f(x) = .
the values of x for which f(x)= . 4
(i) g(x) = u3x – 7u
(i) f(x) = 12 g(–2) = u–6 – 7u = 13
x – 2 g(5) = u15 – 7u = 8
f(–4) = 12 = –2
–6
f(10) = = 3 (ii) g(x) = 14
2 u3x – 7u = 14
3x – 7 = 14 3x – 7 = –14
3x = 21 3x = –7
(ii) f(x) = 3 7
4 x = 7 x = –
12 = 3 3
x – 2 4
3(x – 2) = 48
x = 18




















6





01-[M&M Mate(T) Tg4] 2021.indd 6 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi

(c) Diberi h(x) = 2x + mx − 30 dengan keadaan m (d) Fungsi f ditakrifkan sebagai f(x) = 7 + |3x|. Cari
2
ialah satu pemalar dan h(2) = −34. Cari The function f is defined as f(x) = 7 + |3x|. Find
Given that h(x) = 2x + mx − 30 for which m is a constant (i) nilai bagi f(−1) dan f(6).
2
and h(2) = −34. Find the values of f(−1) and f(6). 1
(i) nilai m. (ii) objek-objek dengan keadaan imejnya ialah
the value of m. 19.
(ii) nilai-nilai x yang memetakan kepada dirinya. the objects for which the image is 19. BAB
the values of x which map to itself.
2
(i) h(x) = 2x + mx − 30 (i) f(x) = 7 + |3x|
h(2) = 8 + 2m − 30 = −34 f(−1) = 7 + |−3| = 10
2m = −34 + 22 f(6) = 7 + |18| = 25
m = −6
(ii) f(x) = 19
Penerbitan Pelangi Sdn. Bhd.
(ii) h(x) = x 7 + |3x| = 19
2
2x − 6x − 30 = x |3x| = 12
2
2x − 7x − 30 = 0 3x = 12 3x = –12
(x + 5)(x − 6) = 0 x = 4 x = –4
5
x = − , x = 6
2


1.2 Fungsi Gubahan
Composite Functions
STUDI 1 Minit Buku Teks: 12 – 19






Fungsi Gubahan Composite Functions A x f gf B y g C z • Fungsi f memetakan x kepada y dan fungsi g memetakan y
kepada z.
Function f maps x onto y, and function g maps y onto z.

• Fungsi gubahan gf memetakan x terus kepada z.

Composite function gf maps x directly onto z.
gf(x) = g[f(x)]

6. Huraikan fungsi gubahan untuk setiap kes yang berikut. TP 2
Describe the composite function for each of the following case.

CONTOH (a)
p q r s
f g h

a b c d e
p q r s

h k
u v

h = qp k = rs
u = gf k = hg






7





01-[M&M Mate(T) Tg4] 2021.indd 7 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi

(b) (c)
f g g f f g

1
w x y z p q r s
BAB

h k u v


h = gf k = gg / g u = ff / f 2 v = fg
2


7. Bagi setiap pasangan fungsi yang berikut, cari (i) fg(x) dan (ii) gf(x). TP 3
For each of the following pairs of functions, find (i) fg(x) and (ii) gf(x).

CONTOH (a) f(x) = 5 – 3x
g(x) = 2x + 1
f(x) = x 2
g(x) = 3x + 1
(i) fg(x) = f(2x + 1)
Penyelesaian: = 5 – 3(2x + 1)
(i) fg(x) = f(3x + 1) = 2 – 6x
2
= (3x + 1)
= 9x + 6x + 1 (ii) gf(x) = g(5 – 3x)
2
= 5 – 2x
= 2(5 – 3x) + 1

= 11 – 6x
(ii) gf(x) = g(x ) Penerbitan Pelangi Sdn. Bhd.

2
2
= 3x + 1




(b) f(x) = x + 3 (c) f(x) = 1 – x
2
g(x) = 5x – 2 4
g(x) =
x
(i) fg(x) = f(5x – 2)
2
4
2
= 5x – 2 + 3 (i) fg(x) = f 1 2
= 5x + 1 x
2
4
= 1 – , x ≠ 0
(ii) gf(x) = g(x + 3) x
= 5(x + 3) – 2 (ii) gf(x) = g(1 – x)
2
2
= 5(x + 6x + 9) – 2 4
2
= 5x + 30x + 43 = 1 – x , x ≠ 1








8





01-[M&M Mate(T) Tg4] 2021.indd 8 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi

8. Bagi setiap pasangan fungsi yang berikut, cari TP 3
For each pair of the following functions, find
(i) fg(3) (ii) gf(4)
1
CONTOH (a) f : x → 3x + 2, g : x → 2 – x
2
f : x → 5x, g : x → 4 – 2x BAB
(i) g(3) = 2 – (3) = –7
2
Penyelesaian: fg(3) = f(–7)
(i) g(3) = 4 – 2(3) = –2 = 3(–7) + 2
fg(3) = f(–2) = –19
= 5(–2)
(ii) f(4) = 3(4) + 2 = 14
= –10
gf(4) = g(14)
3 Penerbitan Pelangi Sdn. Bhd.
2
1 Gantikan 4 ke dalam f, anda = 2 – (14)
akan dapat nilai 20. = –194
(ii) f(4) = 5(4) = 20 Insert 4 into f, you will get 20.
gf(4) = g(20) 2 Gantikan 20 ke dalam g.
= 4 – 2(20) Insert 20 into g.
= –36












3 x
(b) f : x → x – 3, g : x → , x ≠ 0 (c) f : x → 2 – 8x, g : x → , x ≠ –5
x x + 5
3 3
(i) g(3) = =
(i) g(3) = = 1 3 + 5 8
3
fg(3) = f(1) fg(3) = f 1 2
8
= 1 – 3 3
= –2 = 2 – 8 1 2
8
= –1
(ii) f(4) = 4 – 3 = 1
gf(4) = g(1)
(ii) f(4) = 2 – 8(4) = –30
= gf(4) = g(–30)
1
= 3 = –30
–30 +5
6
=
5














9





01-[M&M Mate(T) Tg4] 2021.indd 9 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi

9. Bagi setiap pasangan fungsi yang berikut, cari TP 3
For each pair of the following functions, find
(i) f (x) (ii) g (x) (iii) (2) (iv) (3)
2
f
2
g
2
2
1
CONTOH (a) f(x) = 3x – 1, g(x) = 4x + 2
BAB f(x) = 2x, g(x) = x – 5
2
(i) f (x) = f(3x – 1)
Penyelesaian: = 3(3x – 1) – 1
(i) f (x) = f(2x) = 9x – 4
2
= 2(2x)
2
= 4x (ii) g (x) = g(4x + 2)
= 4(4x + 2) + 2
2Penerbitan Pelangi Sdn. Bhd.
(ii) g (x) = g(x – 5) = 16x + 10
2
= (x – 5) – 5
2
= x – 10 (iii) f (2) = 9(2) – 4
= 18 – 4
2
(iii) f (2) = 4(2) = 14
= 8
(iv) g (3) = 16(3) + 10
2
2
(iv) g (3) = 3 – 10 = 48 + 10
= –7 = 58
(b) f(x) = x – 6, g(x) = x + 1 (c) f : x → 2 + 5x, g : x → x , x ≠ 1
2
x – 1
2
(i) f (x) = f(x – 6) (i) f (x) = f(2 + 5x)
2
= (x – 6) – 6 = 2 + 5(2 + 5x)
= x – 12 = 12 + 25x
x
(ii) g (x) = g(x + 1) x – 1
2
2
2
= (x + 1) + 1 (ii) g (x) = x
2
2
= x + 2x + 1 ( x – 1 ) – 1
2
2
x
(iii) f (2) = 2 – 12 x – 1
2
= –10 =
x – (x – 1)
x – 1
(iv) g (3) = 3 + 2(3) + 3 = x
2
2
= 81 + 18 + 1
= 100
(iii) f (2) = 12 + 25(2)
2
= 62
(iv) g (3) = 3
2






10





01-[M&M Mate(T) Tg4] 2021.indd 10 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi

10. Selesaikan setiap yang berikut. TP 4
Solve each of the following.

CONTOH (a) Diberi f : x → x + 3, g : x → 9 – 2x dan fg(x) = 6. 1
Given f : x → x + 3, g : x → 9 – 2x and fg(x) = 6.
Diberi f : x → 2x + 4, g : x → x – 2 dan fg(x) = 2.
Given f : x → 2x + 4, g : x → x – 2 and fg(x) = 2. BAB
fg(x) = f(9 – 2x)
Penyelesaian: = 9 – 2x + 3
fg(x) = f(x – 2) 1 Cari fungsi gubahan fg(x). = 12 – 2x

Find the composite function fg(x).
= 2(x – 2) + 4 fg(x) = 12 – 2x = 6
= 2x 2x = 6
fg(x) = 2x = 2 2 Samakan fg(x) dengan 2. x = 3
Equate fg(x) with 2.

x = 1
Penerbitan Pelangi Sdn. Bhd.
3
(b) Diberi f : x → x – 3, g : x → 3 – 5x dan gf(x) = –2. (c) Diberi f : x → , g : x → 2x + 1 dan gf(x) = –1.
Given f : x → x – 3, g : x → 3 – 5x and gf(x) = –2. x
Given f : x → 3 , g : x → 2x + 1 and gf(x) = –1.
x
gf(x) = g(x – 3)
3
= 3 – 5(x – 3) gf(x) = g 1 2
= 18 – 5x x
3
gf(x) = 18 – 5x = –2 = 2 1 2 + 1
x
5x = 20 6 + x
x = 4 = x , x ≠ 0
6 + x
gf(x) = = –1
x
6 + x = –x
2x = –6
x = –3


11. Cari fungsi g. TP 4
Find the function g.

CONTOH (a) f(x) → x + 2, fg(x) = 7 – 3x

f(x) = x + 1, fg(x) = x + 7 f(x) = x + 2
Penyelesaian: fg(x) = g(x) + 2
f(x) = x + 1 7 – 3x = g(x) + 2
1 Gantikan g(x) ke dalam f(x). g(x) = 7 – 3x – 2
fg(x) = g(x) + 1 Insert g(x) into f(x). = 5 – 3x
x + 7 = g(x) + 1 2 Samakan kedua-dua fg(x). g : x → 5 – 3x
Equate both fg(x).
g(x) = x + 7 – 1
= x + 6 3 Selesaikan untuk g(x).
Solve for g(x).
g : x → x + 6










11





01-[M&M Mate(T) Tg4] 2021.indd 11 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi

(b) f(x) → 2x – 1, fg(x) = 9 – 4x 16 – x
(c) f(x) = 4 – x, fg(x) =
8
f(x) = 2x – 1 f(x) = 4 – x
1
fg(x) = 2g(x) – 1 fg(x) = 4 – g(x)
9 – 4x = 2g(x) – 1 16 – x
BAB 2g(x) = 9 – 4x + 1 8 = 4 – g(x)
2g(x) = 10 – 4x g(x) = 4 – 1 16 – x 2
10 – 4x 8
g(x) = 32 – (16 – x)
2 =
= 5 – 2x 8
g : x → 5 – 2x = 16 + x
8
g : x → 16 + x
y = Penerbitan Pelangi Sdn. Bhd.
8



CONTOH (d) f(x) = 5 – x, gf(x) = 18 – x
f(x) = x + 2, gf(x) = 7 – 3x gf(x) = 18 – x
g(5 – x) = 18 – x
Penyelesaian: Katakan y = 5 – x
gf(x) = 7 – 3x 1 Gantikan f(x) = x + 2 ke x = 5 – y
g(x + 2) = 7 – 3x dalam gf(x). g(y) = 18 – (5 – y)
Katakan/ Let y = x + 2 Insert f(x) = x + 2 into gf(x). = 13 + y
x = y – 2 2 Gantikan y = x + 2 dan g(x) = 13 + x
g(y) = 7 – 3(y – 2) x = y – 2 ke dalam g(x + 2). g : x → 13 + x
Insert y = x + 2 and
= 7 – 3y + 6 x = y – 2 into g(x + 2).
= 13 – 3y
g(x) = 13 – 3x 3 Selesaikan untuk g(y) ≈ g(x).
Solve for g(y) ≈ g(x).

g : x → 13 – 3x


6 (f) f(x) = x + 2, gf(x) = 8 – 3x 2
(e) f : x → , x ≠ 0, gf : x → 2x + 3
x
gf(x) = 8 – 3x 2
gf(x) = 2x + 3 g(x + 2) = 8 – 3x 2
6
g 1 2 = 2x + 3 Katakan y = x + 2
x
x = y – 2
6 g(y) = 8 – 3(y – 2)
2
Katakan
x = 8 – 3(y – 4y + 4)
2
6 2
x = = 8 – 3y + 12y – 12
= –3y + 12y – 4
2
6
2
g(y) = 2 1 2 + 3 g(x) = –3x + 12x – 4
y
g : x → –3x + 12x – 4
2
12
= + 3
y
12
g(x) = + 3
x
12
g : x → + 3, x ≠ 0
x
12



01-[M&M Mate(T) Tg4] 2021.indd 12 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi

12. Selesaikan setiap yang berikut. TP 5
Solve each of the following.

CONTOH 1
Diberi fungsi f : x → 9 – 2x, g : x → ax + b dan fg : x → 1 – 6x. Cari nilai bagi a dan b.
Given the function f : x → 9 – 2x, g : x → ax + b and fg : x → 1 – 6x. Find the value of a and of b. BAB
Penyelesaian:
fg(x) = f(ax + b) 1 Cari fungsi gubahan fg(x).
= 9 – 2(ax + b) Find composite function fg(x).
= 9 – 2b – 2ax
9 – 2b – 2ax = 1 – 6x 2 Samakan kedua-dua fg(x).
Equate both fg(x).
3 Bandingkan pemalar.
Compare the constant.

Bandingkan pemalar: 9 – 2b = 1, –2a = –6
Compare the constants: 2b = 8 a = 3
b = 4


(a) Diberi fungsi f : x → 4x + k, g : x → x – 3 dan fg : x → mx – 8. Cari nilai bagi k dan m.
Given the function f : x → 4x + k, g : x → x – 3 and fg : x → mx – 8. Find the value of k and of m.

fg(x) = f(x – 3)
= 4(x – 3) + k

= 4x – 12 + k
gf(x) = g(5x – 1) Penerbitan Pelangi Sdn. Bhd.
4x – 12 + k = mx – 8
Bandingkan pemalar:
Compare the constants:
4x = mx, –12 + k = –8
m = 4 k = 4





(b) Diberi fungsi f : x → 5x – 1, g : x → 2x dan gf : x → ax + b. Cari nilai bagi a dan b.
Given the function f : x → 5x – 1, g : x → 2x and gf : x → ax + b. Find the value of a and of b.


= 2(5x – 1)
= 10x – 2
10x – 2 = ax + b

Bandingkan pemalar:
Compare the constants:
10x = ax, b = –2
a = 10










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01-[M&M Mate(T) Tg4] 2021.indd 13 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi


1.3 Fungsi Songsang
Inverse Functions
1
STUDI 1 Minit Buku Teks: 20 – 30
BAB

Suatu fungsi f yang berhubungan Langkah-langkah menentukan f : CONTOH/EXAMPLE:
–1
–1
satu dengan satu mempunyai Steps to determine f : f(x) = 2x + 6
–1
fungsi songsangan f . 1 Ungkapkan fungsi f(x) = y
A function f with a one-to-one relation sebagai x dalam sebutan y. Katakan/Let f(x) = y
has an inverse function f . 2x + 6 = y
–1
Rewrite the function f(x) = y as y – 6
x in terms of y. x =
f 2
–1
2 Tulis x sebagai f (y). Oleh kerana / Since f (y) = x,
–1
–1
Write x as f (y). –1 y
x y f (y) =
3 Gantikan pemboleh ubah y
–1
f –1 dengan x. Jadi / So, f (x) = x – 6
Replace variable y with x. 2


13. Cari nilai bagi setiap yang berikut. TP 3
Find the values for each of the following.
(a) Dalam gambar rajah anak panah yang berikut,
CONTOH Penerbitan Pelangi Sdn. Bhd.
fungsi f memetakan x kepada y. Cari
Dalam gambar rajah anak panah yang berikut, In the following arrow diagram, the function f maps x
fungsi f memetakan x kepada y. Cari to y. Find
In the following arrow diagram, the function f maps x to y. f
Find x y
f 4
x y 2
4
5 1 3
–
4
2 –2
–6
–1
–1
–8
(i) f(1)
(i) f(2) (ii) f (−1)
−1
−1
(ii) f (−8) (iii) f (2)
−1
(iii) f (−2) −1 3
−1
(iv) f(4) (iv) f 1 2
4
Penyelesaian: (i) f(1) = −1
(i) f(2) = 4 (ii) f(1) = −1, f (−1) = 1
−1
−1
(ii) f(−1) = −8, f (−8) = −1
−1
(iii) f(4) = 2, f (2) = 4
(iii) f(5) = −2, f (−2) = 5 3
−1
1 2
(iv) f(−6) = , f −1 3 = −6
(iv) f(2) = 4, f (4) = 2 4 4
−1
14



01-[M&M Mate(T) Tg4] 2021.indd 14 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi

3x
CONTOH (b) Diberi g(x) = x – 2 , x ≠ 2. Cari
8
Fungsi f ditakrifkan oleh f(x) = , x ≠ 1. Cari 3x
x – 1 Given that g(x) = x – 2 , x ≠ 2. Find 1
8
Function f is defined by f(x) = , x ≠ 1. Find
x – 1 (i) f(4) (ii) f (3)
(iii) f (9) (iv) f (10) BAB
−1
−1
(i) f(3) (ii) f(5) (iii) f (2) (iv) f (4)
–1
–1
Penyelesaian: 3(4)
(i) f(4) = (ii) f(3) =
8 8 4 – 2
(i) f(3) = (ii) f(5) =
3 – 1 5 – 1 = 6 = 9
= 4 = 2
f
–1
–1
(iii) f (9) = k (iv) (10) = k
f(k) = 9 f(k) = 10
(iii) f (2) = k (iv) f (4) = x 3k 3k
–1
–1
f(k) = 2 f(x) = 4 k – 2 = 9 k – 2 = 10
8 = 2 8 = 4 3k = 9k – 18 3k = 10k – 20
k – 1 x – 1 6k = 18 7k = 20
k = 5 x = 3 k = 3 20
k =
7

14. Tentukan sama ada setiap fungsi f berikut mempunyai fungsi songsang atau tidak. Beri satu sebab untuk
jawapan anda. TP Penerbitan Pelangi Sdn. Bhd.
3
Determine whether each of the following function f has an inverse function or not. Give a reason for your answer.
CONTOH (a)
f
f(x) x y


p –3
q 5

r 7
x
0

f bukan fungsi satu dengan satu kerana garis f ialah fungsi satu dengan satu. f mempunyai
mengufuk memotong graf itu pada dua titik. f tidak fungsi songsang.
ada fungsi songsang. f is a one-to-one function. f has an inverse function.
f is not a one-to-one function because the horizontal line
cuts the graph at two points. f has no inverse function.














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01-[M&M Mate(T) Tg4] 2021.indd 15 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi

(b) (c)
f(x) f(x)
1

BAB


x
x 0
0


f ialah fungsi satu dengan satu. f mempunyai f bukan fungsi satu dengan satu kerana garis
Penerbitan Pelangi Sdn. Bhd.
fungsi songsang. mengufuk memotong graf itu pada dua titik. f
f is a one-to-one function. f has an inverse function. tidak ada fungsi songsang.
f is not a one-to-one function because the horizontal line
cuts the graph at two points.. f has no inverse function.





15. Rajah berikut menunjukkan graf bagi fungsi satu dengan satu, f. Dalam setiap kes, lakar graf bagi f . TP 4
−1
The following diagrams show the graph of one-to-one function, f. In each case, sketch the graph of f .
−1
CONTOH (a)
f(x)
f(x) y = x
5
(3, 5) y = x f


(5, 3) Tip Penting
(1, 0) → (0, 1) –3 5 x
f (5, 3) → (3, 5) 0
(0, 1)
f –1
x
0 (1, 0) –3



(b) (c)
f(x) f(x)
y = x
y = x 8

f –1
4 f –1
f
3
x f
0 4
x
0 3 8









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01-[M&M Mate(T) Tg4] 2021.indd 16 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi

16. Cari fungsi songsang bagi setiap fungsi yang berikut. TP 3
Find the inverse function for each of the following functions.

CONTOH (a) f : x → 3 – 4x 1
–1
f : x → x – 8 Katakan f (x) = y
1 Katakan f (x) = y Let f(y) = x BAB
–1
Penyelesaian: Let f –1 (x) = y 3 – 4y = x
Katakan/ Let f (x) = y 3 – x
–1
f(y) = x 2 Jadikan y sebagai perkara y = 4
rumus.
y – 8 = x Make y into subject of the 3 – x
–1
y = x + 8 equation. f (x) = 4
\ f (x) = x + 8 3 Gantikan y ke dalam
–1
–1
f : x → x + 8 f (x) = y –1
–1
Insert y into f (x) = y
Penerbitan Pelangi Sdn. Bhd.
(b) g : x → 5x 2
(c) h : x → – , x ≠ 1
x – 1
Katakan g (x) = y Katakan h (x) = y
–1
–1
Let g(y) = x Let h(y) = x
5y = x 2
x = x
y = y – 1
5 2
x y – 1 =
x
–1
g (x) =
5 2
y = + 1
x
2
h (x) = + 1, x ≠ 0
–1
x
17. Tentukan fungsi gubahan berikut yang melibatkan fungsi sonsang. TP 4
Determine the following composite function involving inverse function.
CONTOH
2
Diberi dua fungsi f(x) → x – 3 dan g(x) = , x ≠ –1. Cari
x + 1
Given two functions f(x) → x – 3 and g(x) = 2 , x ≠ 1. Find
x + 1
(i) f g (ii) gf –1
–1
Penyelesaian:
–1
Katakan/ Let f (x) = y (i) f g(x) = f –1 1 2 2 (ii) gf = g(x + 3)
–1
–1
f(y) = x x + 1 = 2
y – 3 = x = 2 + 3 x + 3 + 1
y = x + 3 x + 1 = 2 , x ≠ –4
\ f (x) = x + 3 2 + 3(x + 1) x + 4
–1
=
x + 1
5 + 3x
= , x ≠ –1
x + 1







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Matematik Tambahan Tingkatan 4 Bab 1 Fungsi

1 1
(a) Diberi dua fungsi f(x) = 6 – 3x dan g(x) = , x ≠ . Cari
2 – 1 2
Given two function f(x) = 6 – 3x and g(x) = 1 , x ≠ 1 . Find
1 2x – 1 2
(i) f g (ii) gf –1
–1
BAB Katakan f (x) = y (i) f g(x) = f 1 1 2 (ii) gf (x) = g 1 6 – x 2
–1
–1
–1
–1
f(y) = x 2 – 1 3
6 – 3y = x 1 = 1
6 – x 6 – 1 2x – 1 2 2 1 6 – x 2 – 1
y = = 3
3 3 1
f (x) = 6 – x = 6(2x – 1) – 1 = 12 – 2x – 3
–1
3 3(2x – 1) 1 2
Penerbitan Pelangi Sdn. Bhd.
12 – 7 3
= , x ≠ 3 9
6 – 3 = , x ≠
9 – 2x 2
(b) Diberi fungsi f : x → 7 – 4x. Cari
Given the function f : x → 7 – 4x. Find
–1
(i) f f –1 (ii) f f
Katakan f (x) = y (i) ff (x) = f 1 7 – x 2 (ii) f f(x) = f (7 – 4x)
–1
–1
–1
–1
f(y) = x 4 7 – (7 – 4x)
7 – 4y = x = 7 – 4 1 7 – x 2 = 4
(7 – x) 4 7 – 7 + 4x
y = =
4 = 7 – 7 + x 4
f (x) = 7 – x = x = 4x
–1
4 4
= x





3
(c) Diberi f(x) = , x ≠ 0 dan g(x) = 4 + x. Cari
2x
It is given that f(x) = 3 , x ≠ 0 and g(x) = 4 + x. Find
2x
(i) f g(x) (ii) f g(g )
–1
3 (ii) Katakan / Let y = 4 + x
(i) f g(x) =
2g(x) x = y – 4
–1
3 g (y) = y – 4
= –1
2(4 + x) g (x) = x – 4
(7 – x) fg(g ) = 3
–1
= , x ≠ –4 8 + 2g (x)
–1
4
3
=
8 + 2(x – 4)
3
= , x ≠ 0
2x



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01-[M&M Mate(T) Tg4] 2021.indd 18 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi

Praktis SPM 1 Praktis Ekstra
SPM
SPM 1
1

Kertas 1 3x – 9 = 4
y BAB
1. (a) Dalam rajah di bawah, fungsi f memetakan x = 4 + 3
Buku set X kepada set Y dan fungsi g memetakan 3y
Teks 4
–1
ms.12-30 set Y kepada set Z. Maka / Then, k (x) = + 3, x ≠ 0.
In the diagram below, function f maps set X to set Y 3y
and function g maps set Y to set Z.
(b)
f g
X Y Z y
Penerbitan Pelangi Sdn. Bhd.
y = h(x)
x 3x – 7 g(y)
(0, 3)
x
0 (1, 0)
4
Diberi g(y) = , y ≠ 2, cari fungsi yang
y – 2
memetakan set Z kepada set X dalam sebutan
x.
x
x
Given that g(y) = 4 , y ≠ 2, find the function 2. Diberi fungsi g(x) = − 3 dan fg(x)= − 7, cari
y – 2 4 2
which maps set Z to set X in terms of x. Buku x x
Teks Given function g(x) = − 3 and fg(x) = − 7, find
(b) Rajah menunjukkan suatu graf fungsi ms.20-30 4 2
–1
y = h(x). (a) g (x),
The diagram shows the graph of the function (b) f(x),
y = h(x). (c) fg (4).
–1
y
y = h(x) (a) Katakan / Let g(x) = y
x
− 3 = y
4
x = 4y + 12
–1
(0, 3) g (x) = 4x + 12
x
0 g –1
(b) fg(g ) = – 7
–1
Pada rajah yang sama, lakarkan graf h (x). 2
–1
–1
On the same diagram, sketch the graph of h (x). f(x) = 4x + 12 – 7
4
(a) Katakan / Let gf(x) = k(x) f(x) = 2x – 1
4
Maka / Then, k(x) =
f(x) – 2 (c) g (4) = 4(4) + 12
–1
–1
4 g (4) = 28
k(x) =
3x – 7 – 2 fg (4) = f(28)
–1
4
k(x) = , x ≠ 3 = 2(28) – 1
3x – 9
= 55
Katakan / Let k(x) = y
y = 4
3x – 9
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01-[M&M Mate(T) Tg4] 2021.indd 19 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi

3. Diberi fungsi p(x) = mx + 3, q(x) = 2x − 7 dan pq(x) (a) (i) Cari / Find f(4),
Buku = 2mx + n, ungkapkan m dalam sebutan n. (ii) Seterusnya, cari nilai p jika g(p + 3)
Teks
ms.12-19 Given the functions p(x) = mx + 3, q(x) = 2x − 7 and = f(4).
1 pq(x) = 2mx + n, express m in terms of n. 1
Hence, find the value of p if g(p + 3) = f(4).
8
BAB pq(x) = m[q(x)] + 3 (b) (i) Tentukan gf(x)
Determine gf(x).
= m(2x – 7) + 3
= 2mx – 7m + 3 (ii) Seterusnya lakarkan graf y = ugf(x)u untuk
–1 < x < 2. Nyatakan julat bagi y.
Membandingkan dengan pq(x) = 2mx + n, Hence, sketch the graph of y = ugf(x)u for
Comparing with pq(x) = 2mx + n, –1 < x < 2. State the range of y.
n = –7m + 3
3 – n
m = (a) (i) f(4) = 4 – 5(4) = –16
7
1
(ii) g(p + 3) = f(4)
8
1
3(p + 3) – 5 = (–16)
4. Diberi fungsi f(x) = 3x − 7, cari 8
Buku Given the function f(x) = 3x − 7, find 3p + 9 – 5 = –2
Teks –1 3p = –6
ms.12-30 (a) f (x), p = –2
4s
(b) nilai k jika f 2 1 2 = 20.
3
4s (iii) gf(x) = g(4 – 5x)
the value of k if f 1 2 = 20.
2
3 = 3(4 – 5x) – 5
= 12 – 15x – 5
(a) Katakan / Let 3x – 7 = y = 7 – 15x
y + 7 Penerbitan Pelangi Sdn. Bhd.
x =
3 (b)
x + 7 7
f (x) = x –1 0 2
−1
3 15
(b) f (x) = 3(3x – 7) – 7 y 22 7 0 23
2
f (x) = 9x – 28
2
4s
4s
f 2 1 2 = 9 1 2 – 28 y
3 3
20 = 12s – 28
s = 4 23
22
7
x
–1 0 7 2
Kertas 2 ––
15
Klu Soalan Julat/Range y : 0 < y < 23
Pintasan-x iaitu nilai x apabila y = 0 perlu dicari sebagai titik di mana
suatu graf linear dipantul menjadi bentuk-V.
The x-intercept, which is value of x when y = 0, must be determined as it is the point
at which the linear graph is reflected to form a V-shape.
1. Fungsi f dan g masing-masing ditakrifkan oleh f : x
Buku → 4 – 5x dan g : x → 3x – 5.
Teks The functions f and g are defined as f : x → 4 – 5x and
ms. 2-19
g : x → 3x – 5.



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Matematik Tambahan Tingkatan 4 Bab 1 Fungsi

Klu Soalan x = 4 – u
2
Titik pusingan (x, y) bagi graf kuadratik ditentukan dengan mengambil g(u) = (4 – u) – 6(4 – u) + 8
2
–1
x = purata pintasan-x iaitu 2 + 4 = 3 dan y = |gf (3)|. = 16 – 4u – 4u + u – 24 + 6u + 8
2 2 1
The turning point (x, y) of a quadratic graph is determined by taking x = average of = u – 2u
2
the x-intercepts which is 2 + 4 = 3 and y = |gf (3)|. g(x) = x – 2x
–1
2
Membanding dengan g(x) = ax + bx BAB
2
2. Fungsi f dan g diberi oleh f(x) = 4 – x dan Compairing with g(x)= ax + bx,
2
2
Buku g(x) = ax + bx, dengan keadaan a dan b ialah a = 1, b = –2
Teks pemalar.
ms. 2-30
The functions f and g are given by f(x) = 4 – x and (b) y = |gf (x)|
–1
g(x) = ax + bx where a and b are constants. = |x – 6x + 8|
2
2
–1
(a) Diberi fungsi gubahan gf → x – 6x + 8, = |(x – 2)(x – 4)|
2
tentukan nilai a dan nilai b. Apabila/ When y = 0: x = 2, x = 4
Given the composite function gf (x) = x – 6x + 8, Apabila/ When x = 0: y = 8
2
–1
determine the value of a and of b. Apabila/ When x = 7: y = |(7 – 2)(7 – 4)| = 15
(b) Lakar graf y = |gf (x)| untuk domain 0 < x < 7
−1
dan nyatakan julat yang sepadan dengannya. f(x)
−1
Sketch the graph of y = |gf (x)| for the domain
0 < x < 7 and state the corresponding range. 15
(a) f(x) = 4 – x
Mengambil / Taking y = f(x), 8
4 – x = y
4 – y = x
f (x) = 4 – x (3, 1)
–1
x
0
Diberi / Given: Penerbitan Pelangi Sdn. Bhd. 2 4 7
–1
2
gf (x) = x – 6x + 8
g(4 – x) = x – 6x + 8 Julat/Range : 0 < f(x) < 15
2
Mengambil / Taking,
4 – x = u
























21





01-[M&M Mate(T) Tg4] 2021.indd 21 20/12/2022 1:43 PM

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi
Sudut KBATKBAT KBAT

EKSTRA
1

BAB Rajah di atas menunjukkan pemetaan y kepada x di bawah fungsi f(y) x y z
1
= 5y – 3 dan pemetaan y kepada z di bawah fungsi g(y) = my , y ≠ .
4
Diagram above shows the mapping of y onto x by the function f(y) = 5y – 3 and 1
m 1 2 –2
the mapping of y onto z by the function of g(y) = 4y – 1 , y ≠ 4 .
(a) Cari nilai m.
Find the value of m.
(b) Cari fungsi yang memetakan x kepada y.
Find the function which maps x onto y.
(c) Cari fungsi yang memetakan x kepada z.
Find the function which maps x onto z.
(d) Lakar rajah untuk menunjukkan pemetaan fg(1).
Sketch a diagram to show the mapping of fg(1).


m (c) Fungsi yang memetakan x kepada z ialah gf (x).
–1
(a) g(y) =
4y – 1 The function that mapes x onto z is gf (x).
–1
Daripada rajah / From the diagram, x + 3
–1
g(1) = 2 gf (x) = g 1 5 2
m –6
= –2 Penerbitan Pelangi Sdn. Bhd.
=
m = –6 4 1 x + 3 2 – 1
5
–6
(b) f(y) = 5y – 3 =
Katakan / Let f (y) = x 1 4x + 12 – 5 2
–1
f(x) = y 5
5x – 3 = y 30 7
–1
+ 3 gf (x) = – , x ≠ –
x = 4x + 7 4
5
y + 3
–1
f (y) = (d) fg(1) = f (–2)
5
x + 3 = 5(–2) – 3
Jadi / So, f (x) = = –13
–1
5
1020c










Jawapan
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m*mAtT4*M



22





01-[M&M Mate(T) Tg4] 2021.indd 22 20/12/2022 1:43 PM

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