Page 14 - Modul A+1 Matematik Tambahan Tingkatan 4

(a) Lakarkan graf bagi f : x → x untuk domain 0  x  4. Pada satah yang sama, lakarkan graf bagi f (x). Seterusnya,
2
–1
nyatakan domain dan julat bagi f (x).
–1
Sketch the graph of f : x → x for the domain 0  x  4. On the same plane, sketch the graph of f (x). Hence, state the domain
2
–1
and range of f (x). BAB 1
–1
f(x)
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f(x) = x 2
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3 —
f (x) = √ x
–1
2
1
x
0 1 2 3 4
–1
Domain bagi f (x) ialah 0  x  4 dan julat bagi f (x) ialah 0  f (x)  2.
–1
–1
–1
6x 3 – 2x
(b) Fungsi songsang f (x) ditakrifkan oleh f : x → 5 + x , (c) Suatu fungsi f ditakrifkan sebagai f : x → 5x ,

–1
–1
x ≠ a. x ≠ 0, cari
The inverse function f (x) is defined by f : x → 6x , 3 – 2x
–1
–1
x ≠ a. 5 + x A function f is defined as f : x → 5x , x ≠ 0, find
(i) Nyatakan nilai a. (i) f (2),
–1
State the value of a. (ii) nilai-nilai x dengan keadaan f (x) = f (x).
–1
(ii) Cari f (3). the values of x such that f (x) = f (x).
–1
Find f (3).
(iii) Cari nilai ff (x). (i) f (2) ⇒ f(x) = 2
–1
–1
Find the value of ff (x). 3 – 2x
–1
5x = 2
(i) 5 + x ≠ 0 3 – 2x = 10x
x ≠ –5 3 = 12x
∴ a = –5 1
(ii) f (3) = x ⇒ f (x) = 3 x = 4
–1
–1
6x = 3 f (2) = 1
5 + x 4
6x = 15 + 3x (ii) Katakan 3 – 2x = y
3x = 15 5x
x = 5 3 – 2x = 5xy
∴ f (3) = 5 –2x – 5xy = –3
(iii) Katakan 6x = y 6x 2x + 5xy = 3
5 + x 5   x(2 + 5y) = 3
6x = 5y + xy ff (x) = 5 + x x = 3
–1
6x
6x – xy = 5y 6 – 5 + x 2 + 5y
x(6 – y) = 5y 30x Maka, f (x) = 3
–1
x = 5y =  5 + x  2 + 5x
–1
6 – y 6(5 + x) – 6x f (x) = f (x)
f (x) = 5x 5 + x 5 + x 3 – 2x = 3
6 – x 30x 30 5x 2 + 5x
= 5 + x ÷ 5 + x (3 – 2x)(2 + 5x) = 15x
= x 6 + 15x – 4x – 10x – 15x = 0
2
–10x – 4x + 6 = 0
2
x = –1 atau x = 0.6
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01_Modul A+ MateTam Tg4.indd 11 08/10/2021 11:18 AM