Page 18 - PBD Plus Matematik Tambahan T4 (EG)

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi
(b) Rajah di bawah menunjukkan graf bagi 7 . Diberi bahawa f : x → 2x + a dan g : x → 3b − x
fungsi f(x) = 2x − 1. SP 1.3.2 dengan keadaan a dan b ialah pemalar dan
The diagram below shows the graph of function fg(x) = gf(x). Ungkapkan a dalam sebutan b.
f(x) = 2x − 1. Given that f : x → 2x + a and g : x → 3b − x such that a and b
y are constants and fg(x) = gf(x). Express a in terms of b. SP 1.2.5
6 [3 markah / 3 marks]
(3, 5) y = x
4 Jawapan / Answer :
f fg(x) = f(3b – x)
2
f –1 = 2(3b – x) + a
x = 6b + a – 2x
–4 –2 0 2 4 6
–2 gf(x) = g(2x + a)
= 3b – (2x + a)
(–1, –3)
–4 = 3b – a – 2x
Pada graf itu, lakarkan graf bagi y = f (x). fg(x) = gf(x)
−1
On the graph, sketch the graph of y = f (x).
−1
[3 markah / 3 marks] 6b + a – 2x = 3b – a – 2x
2a = –3b
3b
Jawapan / Answer : a = – 2
1
(a) (i) x = 1, = 1
1
1
x = 4, = 0.25 8. (a) Diberi f(x) = |2x – 3|, cari nilai-nilai x yang
4 memberikan imej 1. SP 1.1.3
x = 10, 1 = 0.1 Given that f(x) = |2x – 3|, find the values of x which gives
10 image of 1.
1
f(x) = , x ≠ 0 [2 markah / 2 marks]
x
(b) Diberi h(x) = 12x – 5 dan k(x) = 3x – 2, cari
hk (x). SP 1.2.2 SP 1.3.3
(ii) f(p) = 0.04 Penerbitan Pelangi Sdn. Bhd.
–1
1 = 0.04 Given that h(x) = 12x – 5 and k(x) = 3x – 2, find hk (x).
–1
p [3 markah / 3 marks]
p = 25 Jawapan / Answer :
(a) |2x − 3| = 1
6. Diberi h : x → x − 2 dan k : x → 2x + 5. Cari nilai bagi 2x − 3 = −1 , 2x − 3 = 1
2x = 4
Given h : x → x − 2 and k : x → 2x + 5. Find the value of 2x = 2 x = 2
x = 1
−1
(a) h (3), SP 1.3.3
(b) hk(–1). SP 1.2.3 (b) Katakan / Let y = 3x − 2
[4 markah / 4 marks] x = y + 2
3
–1
Jawapan / Answer : k (y) = x
(a) Katakan / Let y = x − 2 = y + 2
x = y + 2 3
−1
−1
h (y) = x k (x) = x + 2
3
= y + 2 x + 2
−1
∴h (x) = x + 2 hk (x) = h  3 
−1
h (3) = 3 + 2 = 12  x + 2  − 5
−1
= 5 3
= 4x + 3
(b) hk(x) = h(2x + 5)
= (2x + 5) − 2
= 2x + 3
hk(−1) = 2(−1) + 3
= 1

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