1.6.2 Fungsi, SPM Praktis (Kertas 2 Soalan 6 - 10)

Soalan 6 (SPM 2018 – 8 markah):
Diberi bahawa g : x → 2x – 3 dan h : x → 1 – 3x.
(a) Cari
(i) h (5)
(ii) nilai k jika

g (k + 2) = \frac{1}{7} h (5),

(iii) hg(x).

(b) Seterusnya, lakarkan graf y = | hg(x) | untuk –1 ≤ x ≤ 3.
Nyatakan julat bagi y.

Penyelesaian:
(a)(i)

\begin{array}{l} h (x) = 1 - 3 x \\ h (5) = 1 - 3 (5) \\ = - 14 \end{array}

(a)(ii)

\begin{array}{l} g (x) = 2 x - 3 \\ g (k + 2) = \frac{1}{7} h (5) \\ 2 (k + 2) - 3 = \frac{1}{7} (- 14) \\ 2 k + 4 - 3 = - 2 \\ 2 k = - 3 \\ k = - \frac{3}{2} \end{array}

(a)(iii)

\begin{array}{l} g (x) = 2 x - 3, h (x) = 1 - 3 x \\ h g (x) = h (2 x - 3) \\ = 1 - 3 (2 x - 3) \\ = 1 - 6 x + 9 \\ = 10 - 6 x \end{array}

(b)
y = |hg(x)|,
y = |10 – 6x|
Julat bagi y : 0 ≤ y ≤ 16

Soalan 7:

Diberi f : x \to \frac{5 x}{m x - 3}, x \neq \frac{3}{2} dan g : x \to 3 - 4 x . Cari

(a) nilai bagi m, [2 markah]

(b) gf^-1(–2), [3 markah]

Penyelesaian:

(a)

\begin{array}{l} m x - 3 = 0 \\ m (\frac{3}{2}) - 3 = 0 \\ m = 3 \times \frac{2}{3} = 2 \end{array}

(b)

\begin{array}{l} Katakan y = \frac{5 x}{2 x - 3} \\ 2 x y - 3 y = 5 x \\ 2 x y - 5 x = 3 y \\ x (2 y - 5) = 3 y \\ x = \frac{3 y}{2 y - 5} \\ f^{- 1} (y) = \frac{3 y}{2 y - 5} \\ f^{- 1} (x) = \frac{3 x}{2 x - 5} \\ f^{- 1} (- 2) = \frac{3 (- 2)}{2 (- 2) - 5} = \frac{2}{3} \end{array}

\begin{array}{l} g f^{- 1} (- 2) = g (\frac{2}{3}) \\ = 3 - 4 (\frac{2}{3}) \\ = \frac{1}{3} \end{array}

(c)

hg (x) = 12x + 5

h [g(x)] = 12x + 5

h(3 – 4x) = 12x + 5

Katakan u= 3 – 4x

\begin{array}{l} x = \frac{3 - u}{4} \\ h (u) = 12 (\frac{3 - u}{4}) + 5 \\ h (u) = 9 - 3 u + 5 \\ h (u) = 14 - 3 u \\ h (x) = 14 - 3 x \end{array}

1.6.2 Fungsi, SPM Praktis (Kertas 2 Soalan 6 – 10)