Page 29 - Hybrid PBD 2022 Form 4 Additional Mathematics

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi
SPM

PRAKTIS PdPR Jawapan
Bab 1 PRAKTIS PdPR Bab 1

Kertas 1 4. Dalam rajah di bawah, fungsi f memetakan y
kepada x dan fungsi g memetakan y kepada z.
In the diagram below, the function f maps y to x and the
1. Diberi fungsi m(x) = 3x dan n(x) = h – kx dengan function g maps y to z.
keadaan h dan k ialah pemalar. Ungkapkan h x f y g z
dalam sebutan k dengan keadaan nm(1) = 4. r
It is given the functions m(x) = 3x and n(x) = h – kx where h and
k are constants. Express h in terms of k such that nm(1) = 4.
(b) g(x) Penerbitan Pelangi Sdn Bhd. All Rights Reserved
[3 markah / 3 marks] q
Jawapan / Answer :
nm(x) = h – k(3x) p
= h – 3kx
nm(1) = 4 (a) Cari gf (p).
–1
h – 3k(1) = 4 Find gf (p).
–1
h – 3k = 4 (b) Ungkapkan q dalam sebutan p jika diberi
h = 4 + 3k f(y) = 3y – 5 , y ≠ 0.
y 3y – 5
Express q in terms of p if it is given f(y) = y , y ≠ 0.
[3 markah / 3 marks]
2. Diberi fungsi f(x) = px + 1, g(x) = 3x – 5 dan Jawapan / Answer :
fg(x) = 3px + q. Ungkapkan p dalam sebutan q. (a) f(q) = p dan / and g(q) = r
–1
It is given the functions f(x) = px + 1, g(x) = 3x – 5 and f (p) = q
fg(x) = 3px + q. Express p in terms of q. gf (p) = g(q)
–1
[3 markah / 3 marks] = r
Jawapan / Answer :
fg(x) = p(3x – 5) + 1 (b) f(q) = p
= 3px – 5p + 1 3q – 5 = p
q
3px – 5p + 1 = 3px + q 3q – 5 = pq
–5p + 1 = q 3q – pq = 5
1 – q = 5p q(3 – p) = 5
1 – q q = 5
p =
5 3 – p

Kertas 2

3. Diberi fungsi f(x) = 3x + 4 dan fg(x) = 6x + 7, cari 1. (a) Rajah di bawah menunjukkan fungsi
Given functions f(x) = 3x + 4 and fg(x) = 6x + 7, find x
(a) fg(4) f(x) = – b dengan keadaan a dan b ialah
a
pemalar.
[3 markah / 3 marks] The diagram below shows the function f(x = x a – b,
Jawapan / Answer : where a and b are constants.
(a) fg(4) = 6(4) + 7 x f x – b
= 31 a
6
(b) fg(x) = 6x + 7
3g(x) + 4 = 6x + 7 10 –4
3g(x) = 6x + 3 –12
g(x) = 2x + 1
Hitung nilai a dan nilai b.
Calculate the value of a and of b.
[5 markah / 5 marks]

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