The function $\mathrm{f}$ is defined by \(\mathrm{f}:x\mapsto \left\{ \begin{matrix} \cos \frac{\p

Question

The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \left\{ \begin{matrix} \cos \frac{\pi x}{2},\,\,0\le x<1 \ 1-x,\,\,\,\,\,\,\,\,1\le x\le 2 \ \end{matrix} \right.$ Sketch the graph of $\mathrm{f}$, indicating clearly the coordinates of the axial intercepts and endpoints.[3] Show that $\mathrm{f}$ has an inverse, and define it in similar form.[5] Find the range of values of $x$ for which $\mathrm{f}{{\mathrm{f}}^{-1}}(x)={{\mathrm{f}}^{-1}}\mathrm{f}(x)$.[2] The function $\mathrm{g}$ is defined by $\mathrm{g}:x\mapsto {{\mathrm{e}}^{x}}$, $0\le x\le \ln 2$. Find the range of $\mathrm{fg}$.[2]

Accepted Answer

(i) ${(0,1)}$, ${(1,0)}$, ${(2,-1)}$ (ii) ${\mathrm{f}^{-1} : x \mapsto \begin{cases} \frac{2}{\pi}\cos^{-1}x, & 0 < x \leq 1 \ 1-x, & -1 \leq x \leq 0 \end{cases}}$ (iii) ${0 \leq x \leq 1}$ for ${\mathrm{f}\mathrm{f}^{-1}(x) = \mathrm{f}^{-1}\mathrm{f}(x)}$ (iv) ${\mathrm{R}_{\mathrm{fg}} = [-1,0]}$

2020 HCI Promo Q11

2020 HCI Promo Q11

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