A function $\mathrm{g}$ is said to be self-inverse if $\mathrm{g}(x) = \mathrm{g}^{-1}(x)$

Question

A function $\mathrm{g}$ is said to be self-inverse if $\mathrm{g}(x) = \mathrm{g}^{-1}(x)$ for all $x$ in the domain of $\mathrm{g}$. The function $\mathrm{g}$ is defined by$$\mathrm{g} : x \mapsto \frac{\alpha x + \beta}{x - 2}, \quad x \in \mathbb{R}, x \neq 2,$$where $\alpha$ and $\beta$ are constants. Find the values of $\alpha$ and $\beta$ for $\mathrm{g}$ to be a self-inverse function.[3] Given that $\mathrm{g}$ is self-inverse, find the value of $x$ where $\mathrm{g}^{2023}(x) = -\frac{\beta}{2}$.[2] The functions $\mathrm{f}$ and $\mathrm{h}$ are defined by $$\mathrm{f} : x \mapsto ex + 1, \quad x > 0.$$$$\mathrm{h} : x \mapsto 3x^3 - 9x^2 - 12x + 36, \quad x \geq -2.$$ Are $\mathrm{f}^2(x)$ and $[\mathrm{f}(x)]^2$ equivalent? Explain your answer.[2] By considering $\mathrm{f}^2(x)$ and $\mathrm{f}^3(x)$, show that $$\mathrm{f}^n(x) = \frac{e^{n+1}x + e^n(1 - x) - 1}{e - 1}.$$[3] Explain why the composite function $\mathrm{hf}$ exists.[2] Find the range of $\mathrm{hf}$.[2]

Accepted Answer

(a)(i) $\alpha=2$, $\beta\in\mathbb R\setminus\{-4\}$ (a)(ii) $x=0$ (b)(i) not equivalent (b)(ii) $f^n(x)=\dfrac{\mathrm e^{n+1}x+\mathrm e^n(1-x)-1}{\mathrm e-1}$ (b)(iii) $hf$ exists (b)(iv) $\left[18-\dfrac{14\sqrt{21}}3,\infty\right)\approx[-3.39,\infty)$

2023 HCI C1 BT Q12

2023 HCI C1 BT Q12

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