1.6.1 Fungsi, SPM Praktis (Kertas 2 Soalan 1 - 5)

Soalan 1:

Fungsi f dan g ditakrifkan sebagai f : x → x– 1 dan

g : x \to \frac{3 - x}{x + 4}

. Cari
(a) nilai gf(3),
(b) nilai fg(-1 ),
(c) fungsi gubahan fg,
(d) fungsi gubahan gf,
(e) fungsi gubahan g² ,
(f) fungsi gubahan f².

Penyelesaian:

Soalan 2:

Diberi f: x → hx + k dan f² : x → 4x + 15.
(a) Cari nilai h dan k.
(b) Ambil nilai h> 0, cari nilai-nilai x di mana f (x²) = 7x

Penyelesaian:

(a)

Langkah 1: cari f² (x)

Diberi f (x) = hx + k

f² (x) = ff (x) = f (hx + k)

= h (hx + k) + k

= h²x + hk + k

Langkah 2: bandingkan dengan f²(x) yang diberi

f² (x) = 4x + 15

h²x + hk + k = 4x + 15

h²= 4

h = ± 2

Apabila, h = 2

hk + k = 15

2k + k = 15

k = 5

Apabila, h = –2

hk + k = 15

–2k + k = 15

k = –15

(b)

h > 0, h = 2, k = 5

Diberi f (x) = hx + k

f (x) = 2x + 5

f (x²) = 7x

2 (x²) + 5 = 7x

2x² – 7x+ 5 = 0

(2x – 5)(x – 1) = 0

2x – 5 = 0 atau x – 1= 0

x = 5/2 atau x = 1

Soalan 3:
Fungsi f dan g ditakrifkan oleh

\begin{array}{l} f : x \mapsto 2 x - 3 \\ g : x \mapsto \frac{2}{x}; x \neq 0 \end{array}

Ungkapkan dalam bentuk yang serupa
(a) ff,
(b) gf,
(c) f² , Hitungkan nilai x supaya ff(x) = gf(x).

Penyelesaian:
(a)

\begin{array}{l} f f (x) = f [f (x)] \\ = f (2 x - 3) \\ = 2 (2 x - 3) - 3 \\ = 4 x - 9 \\ Jadi, f f : x \mapsto 4 x - 9 \end{array}

(b)

\begin{array}{l} g f (x) = g [f (x)] \\ = g (2 x - 3) \\ = \frac{2}{2 x - 3} \\ Jadi, g f : x \mapsto \frac{2}{2 x - 3} \end{array}

(c)

\begin{array}{l} Katakan f^{- 1} (x) = y, \\ maka f (y) = x \\ 2 y - 3 = x \\ y = \frac{x + 3}{2} \\ maka f^{- 1} (x) = \frac{x + 3}{2} \\ f^{- 1} : x \mapsto \frac{x + 3}{2} \\ Apabila f f (x) = g f (x), \\ 4 x - 9 = \frac{2}{2 x - 3} \\ (4 x - 9) (2 x - 3) = 2 \\ 8 x^{2} - 30 x + 27 = 2 \\ 8 x^{2} - 30 x + 25 = 0 \\ (4 x - 5) (2 x - 5) = 0 \\ 4 x - 5 = 0 atau 2 x - 5 = 0 \\ x = \frac{5}{4} atau x = \frac{5}{2} \end{array}

Soalan 4:
Fungsi f dan g ditakrifkan oleh

\begin{array}{l} f (x) = 3 x - 2 \\ g (x) = \frac{3}{x}, x \neq 0 \\ Cari \\ (a) f^{- 1} (2), \\ (b) g f (- 3), \\ (c) fungsi h jika diberi h f (x) = 3 x + 2, \\ (d) fungsi k jika diberi f k (x) = 4 x - 7. \end{array}

Penyelesaian:
(a)

\begin{array}{l} Katakan f^{- 1} (2) = x, \\ maka f (x) = 2 \\ 3 x - 2 = 2 \\ 3 x = 4 \\ x = \frac{4}{3} \\ f^{- 1} (2) = \frac{4}{3} \end{array}

(b)

\begin{array}{l} g f (- 3) = g [3 (- 3) - 2] \\ = g (- 11) \\ = - \frac{3}{11} \end{array}

(c)

\begin{array}{l} h [f (x)] = 3 x + 2 \\ h (3 x - 2) = 3 x + 2 \\ Katakan y = 3 x - 2 \\ Maka x = \frac{y + 2}{3} \\ h (y) = 3 (\frac{y + 2}{3}) + 2 \\ = y + 2 + 2 \\ = y + 4 \\ Maka h (x) = x + 4 \end{array}

(d)

\begin{array}{l} f [k (x)] = 4 x - 7 \\ 3 k (x) - 2 = 4 x - 7 \\ 3 k (x) = 4 x - 5 \\ k (x) = \frac{4 x - 5}{3} \end{array}

1.6.1 Fungsi, SPM Praktis (Kertas 2 Soalan 1 – 5)