A function $\mathrm{f}$ is said to self-inverse if \(\mathrm{f}\left( x \right)={{\mathrm{f}}

Question

A function $\mathrm{f}$ is said to self-inverse if $\mathrm{f}\left( x \right)={{\mathrm{f}}^{-1}}\left( x \right)$ for all $x$ in the domain of $\mathrm{f}$. The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by $\mathrm{f}:x\mapsto \frac{ax+7{{a}^{2}}}{x-a},x\in \mathbb{R},x\ne a,0 < a < 1$ $\mathrm{g}:x\mapsto \left( x-2 \right)\left( x+1 \right),x\in \mathbb{R},x>3.$ Explain why ${{\mathrm{f}}^{-1}}$ exists and show that $\mathrm{f}$ is self-inverse.[4] Hence, evaluate ${{\mathrm{f}}^{241}}\left( \frac{1}{a} \right)$ in terms of $a$, showing your working clearly.[2] Show that $\mathrm{fg}$ exists. Find the exact range of $\mathrm{fg}$ in terms of $a$.[3]

Accepted Answer

(i) $\mathrm f^{-1}(x)=\dfrac{ax+7a^2}{x-a}$, $x\ne a$, so $\mathrm f^{-1}=\mathrm f$ (ii) $\dfrac{a+7a^3}{1-a^2}$ (iii) $\mathrm{fg}$ exists and $R_{\mathrm{fg}}=\left(a,\dfrac{4a+7a^2}{4-a}\right)$

2021 SAJC P2 Q1

2021 SAJC P2 Q1

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