The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by \({\mathrm{f}:x\rightarrow\frac{x-2}{

Question

The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by ${\mathrm{f}:x\rightarrow\frac{x-2}{x+2},\quad x\in\mathbb{R},\; x\neq -2,}$ ${\mathrm{g}:x\rightarrow -x^2,\quad x\in\mathbb{R},\; x < -\sqrt{2},}$ Find ${{\mathrm{f}}^{-1}}\left( x \right)$ and state its domain.[3] Find an expression for $\mathrm{fg}\left( x \right)$ and hence, or otherwise, find ${{\left( \mathrm{fg} \right)}^{-1}}\left( 3 \right)$.[4] It is given that ${\mathrm{h}(x)=\begin{cases} -4x-12 & \text{for}\hspace{0.5em} -4 \leq x \leq -2, \ x\left|x\right| & \text{for}\hspace{0.5em} -2 < x \leq -2 \end{cases}}$ and that $\mathrm{h}\left( x \right)=\mathrm{h}\left( x+6 \right)$ for all real values of $x$. Evaluate $\mathrm{h}\left( -4 \right)$ and $\mathrm{h}\left( 12 \right)$.[2] Sketch the graph of $y=\mathrm{h}\left( x \right)$ for $-4\le x\le 6$ and explain why $\mathrm{h}$ has no inverse.[4]

Accepted Answer

(a)(i) $\mathrm{f}^{-1}(x)=\frac{2+2x}{1-x}\,,\; x \neq 1$ (ii) $(\mathrm{fg})(x)=\frac{-x^2-2}{-x^2+2}$; $(\mathrm{fg})^{-1}(3)=-2$ (b)(i) ${\mathrm{h}(-4)=4\;; \mathrm{h}(12)=0}$ (ii) The line $y=0$ cuts $y=\mathrm{h}(x)$ at two or more points. Hence $\mathrm{h}$ does not have an inverse.

2022 JPJC P1 Q9

2022 JPJC P1 Q9

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