1.6.3 Fungsi, SPM Praktis - Matematik Tambahan SPM

Soalan 5:

\begin{array}{l} Fungsi f ditakrifkan oleh f : x \to \frac{1 + x}{1 - x}, x \neq 1. \\ Cari f^{2}, f^{3}, f^{4} dan seterusnya tulis fungsi bagi f^{51} dan f^{52} . \end{array}

Penyelesaian:

\begin{array}{l} f (x) = \frac{1 + x}{1 - x}, x \neq 1 \\ f^{2} (x) = f [f (x)] = f (\frac{1 + x}{1 - x}) \\ = \frac{1 + (\frac{1 + x}{1 - x})}{1 - (\frac{1 + x}{1 - x})} = \frac{\frac{1 - x + 1 + x}{1 - x}}{\frac{1 - x - 1 - x}{1 - x}} \\ = \frac{2}{- 2 x} = - \frac{1}{x} \\ f^{3} (x) = f [f^{2} (x)] = f (- \frac{1}{x}) \\ = \frac{1 + (- \frac{1}{x})}{1 - (- \frac{1}{x})} = \frac{\frac{x - 1}{x}}{\frac{x + 1}{x}} \\ = \frac{x - 1}{x + 1} \\ f^{4} (x) = f [f^{3} (x)] = f (\frac{x - 1}{x + 1}) \\ = \frac{1 + (\frac{x - 1}{x + 1})}{1 - (\frac{x - 1}{x + 1})} = \frac{\frac{x + 1 + x - 1}{x + 1}}{\frac{x + 1 - x + 1}{x + 1}} \\ = \frac{2 x}{2} = x \\ f^{5} (x) = f [f^{4} (x)] = f (x) = \frac{1 + x}{1 - x} (berulang) \\ ∴ f^{51} (x) = f^{3} [f^{48} (x)] = f^{3} (x) \\ = \frac{x - 1}{x + 1} \\ f^{52} (x) = f^{4} [f^{48} (x)] = f^{4} (x) = x \end{array}

Soalan 6:
Dalam rajah di bawah, fungsi g memetakan set P kepada set Q dan fungsi h memetakan set Q kepada set R.

Cari
(a) dalam sebutan x, fungsi
(i) yang memetakan set Q kepada set P,
(ii) h(x).

(b) nilai x dengan keadaan gh(x) = 8x + 1.

Penyelesaian:
(a)(i)

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

(a)(ii)

\begin{array}{l} h g (x) = 12 x + 5 \\ h (3 x + 2) = 12 x + 5 \to g (x) = 3 x + 2 \\ Katakan u = 3 x + 2 \\ x = \frac{u - 2}{3} \\ h (u) = 12 (\frac{u - 2}{3}) + 5 \\ = 4 u - 8 + 5 \\ = 4 u - 3 \\ h (x) = 4 x - 3 \end{array}

(b)

\begin{array}{l} g h (x) = g (4 x - 3) \\ = 3 (4 x - 3) + 2 \\ = 12 x - 9 + 2 \\ = 12 x - 7 \\ 12 x - 7 = 8 x + 1 \\ 4 x = 8 \\ x = 2 \end{array}