Page 20 - PBD Plus Matematik Tambahan T4 (EG)

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi
3. Rajah di sebelah y 4. Diberi bahwa f : x → 3x – 6 dan g : x → 1 – 3x.
menunjukkan graf It is given that f : x → 3x – 6 and g : x → 1 – 3x.
bagi fungsi y = f(x). 1 (a) Cari / Find
Diberi bahawa –1 0 3 x (i) g(–2),
f(x) = ax + b dengan (ii) nilai bagi p jika f(2p – 1) = 3[g(–2)],
keadaan a dan b ialah the value of p if f(2p – 1) = 3[g(–2)],
pemalar. Cari (iii) gf (x).
–1
The diagram shows the –5 [5 markah / 5 marks]
graph of function y = f(x). (b) Seterusnya, lakarkan graf y = |gf (x)| untuk
–1
It is given that f(x) = ax + b where a and b are constants. Find
(a) nilai a dan nilai b, SP 1.1.3 –6  x  0. –1
Hence, sketch the graph of y = |gf (x)| for –6  x  0.
the value of a and of b,
[4 markah / 4 marks] [2 markah / 2 marks]
(b) objek bagi 7, SP 1.1.3
the object of 7, Jawapan / Answer :
[2 markah / 2 marks] (a) (i) g(–2) = 1 – 3 (–2)
(c) nilai x jika f (x) = f (x). SP 1.2.2 SP 1.3.3 = 7
2
–1
the value of x if f (x) = f (x). (ii) f(2p – 1) = 3[g(−2)]
–1
2
[4 markah / 4 marks] 3(2p − 1) − 6 = 3(7)
Jawapan / Answer : 6p − 3 − 6 = 21
(a) f(3) = 1 6p =30
3a + b = 1 .................a p = 5
f(–1) = –5
–a + b = –5 .................b (iii) Katakan / Let y = 3x – 6
y + 6
x =
a – b: 4a = 6 3
3
–1
a = f (y) = x
2 = y + 6
3
3
– + b = –5 x + 6
f (x) =
2 2y + 7Penerbitan Pelangi Sdn. Bhd.
–1
b = – 7 3
2 gf (x) = g  x + 6 
–1
(b) f(x) = 7 3
3 x – = 7 = 1 – 3  x + 6 
7
2 2 3
3x – 7 = 14 = 1 – (x + 6)
x = 7 = –5 – x
3
(c) y = x – 7 (b) y
2 2
2y = 3x – 7 5
2y + 7 = 3x
1
x = x
3 –6 –5 O
f (y) = x
–1
2y + 7
=
3
2x + 7
∴ f (x) =
–1
3
3

ff(x) = f x – 7 
2
2
3 3 7 7

= x –  –
2 2 2 2
9 35
= x –
4 4
f (x) = f (x)
–1
2
7
2x + = 9x – 35
3 3 4 4
x = 7
13 © Penerbitan Pelangi Sdn. Bhd.