Functions: Self-inverse & \(f^n\) pattern — VJC 2025 H2 Math Prelim Paper 1
What this question tests
Question
The function \(\mathrm{f}\) is defined by \[ \mathrm{f}: x \mapsto \frac{x-1}{x},\quad x \in \mathbb{R},\ x \ne 0,\ x \ne 1. \]
(a) Show that \(\mathrm{f}^{2}(x) = \mathrm{f}^{-1}(x)\).
(b) Find \(\mathrm{f}^{3}(x)\) in simplified form.
(c) Find \(\mathrm{f}^{2030}(5)\).
Functions \(\mathrm{g}\) and \(\mathrm{h}\) are defined by \[ \mathrm{g}: x \mapsto \frac{x-1}{x},\quad x \in \mathbb{R},\ x \ge 1, \] \[ \mathrm{h}: x \mapsto -\sin ax,\quad x \in \mathbb{R}, \] where \(a\) is a positive constant.
(d) Find the value of \(a\) given that the range of \(\mathrm{hg}\) is \((-1,\,0]\).
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(a) Let \(y = \mathrm{f}(x) = \dfrac{x-1}{x} = 1 - \dfrac{1}{x}\). Then \[ \frac{1}{x} = 1 - y \quad\Longrightarrow\quad x = \frac{1}{1-y}, \] so \(\mathrm{f}^{-1}(x) = \dfrac{1}{1-x}\).
\[\begin{aligned} \mathrm{f}^2(x) &= \mathrm{f}\!\left(\frac{x-1}{x}\right) = \frac{\frac{x-1}{x} - 1}{\frac{x-1}{x}} = \frac{-1/x}{(x-1)/x} = \frac{-1}{x-1} = \frac{1}{1-x} = \mathrm{f}^{-1}(x).\quad \end{aligned}\](b) \[\begin{aligned} \mathrm{f}^3(x) &= \mathrm{f}(\mathrm{f}^2(x)) = \mathrm{f}(\mathrm{f}^{-1}(x)) = x. \end{aligned}\] (c) Since \(\mathrm{f}^3(x) = x\), the sequence cycles with period 3. Writing \(2030 = 3(676) + 2\): \[ \mathrm{f}^{2030}(5) = \mathrm{f}^2(5) = \frac{1}{1-5} = -\frac{1}{4}. \]
(d)
Range of \(\mathrm{g} = [0,1]\) = new domain of \(\mathrm{h}\)
Given: range of \(\mathrm{hg}\) is \((-1, 0]\) \[ \frac{\pi}{2a} = 1 \implies a = \frac{\pi}{2} \]