Page 24 - Hybrid PBD 2022 Form 4 Additional Mathematics

Matematik Tambahan Tingkatan 4 Bab 1 Fungsi
9. Nyatakan fungsi f bagi setiap yang berikut berdasarkan fungsi g dan fungsi gubahan yang diberikan.
State the function f of each of the following based on the given function g and the composite function. SP 1.2.4 TP4

(i) g(x) = x − 6 dan / and gf(x) = x − 3. (ii) g(x) = x + 5 dan / and fg(x) = 2x + 15.

gf(x) = x – 3 Katakan / Let y = g(x)
f(x) – 6 = x – 3 y = x + 5
f(x) = x + 3 x = y – 5
fg(x) = 2x + 15
f(y) = 2(y – 5) + 15
= 2y – 10 + 15
= 2y + 5
f(x) = 2x + 5
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(a) g(x) = x + 9 dan / and gf(x) = x − 7. (b) g(x) = 2x – 5 dan / and fg(x) = 2x + 1.

gf(x) = x – 7 Katakan / Let fg(x) = 2x + 1
f(x) + 9 = x – 7 y = g(x) f(y) = 2  y + 5  + 1
f(x) = x – 16 y = 2x – 5 2
x = y + 5 = y + 6
2 f(x) = x + 6

10. Selesaikan setiap yang berikut. SP 1.2.5
Solve each of the following.

(a) Diberi fungsi f(x) = 7 – 2x, g(x) = ax + b dan fg(x) = 1 – 4x, cari nilai a dan nilai b. TP3
Given the function f(x) = 7 – 2x, g(x) = ax + b and fg(x) = 1 – 4x, find the value of a and of b.
fg(x) = f(ax + b) Secara perbandingan / By comparison
= 7 – 2(ax + b) 7 – 2b – 2ax = 1 – 4x
= 7 – 2ax – 2b 7 – 2b = 1 , –2a = –4

b = 3 a = 2

(b) Fungsi f ditakrifkan oleh / Function f is defined by
f : x → x , x ≠ –1. TP4
x + 1
(i) Ungkapkan f (x), f (x) dan f (x) dalam bentuk termudah.
2
3
4
Express f (x), f (x) and f (x) in the simplest form.
2
3
4
(ii) Kemudian, deduksikan f (x). / Hence, deduce f (x).
5
5
(i) f (x) = f [f(x)] f (x) = f [f (x)] f (x) = f [f (x)]
2
2
4
3
3
x x x
= x + 1 = 2x + 1 = 3x + 1
x + 1 x + 1 x + 1
x + 1 2x + 1 3x + 1
= x = x = x
2x + 1 3x + 1 4x + 1
(ii) Pengangka sentiasa x dan penyebut ialah 2x + 1, 3x + 1 dan 4x + 1, maka penyebut seterusnya ialah
5x + 1 .
The numerator is always x and the denominators are 2x + 1, 3x + 1 and 4x + 1, thus the next denominator is 5x + 1.
∴ f (x) = x
5
5x + 1

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