The function $\mathrm{f}$ is defined by \[\mathrm{f} : x \mapsto 3\sin \left( 2x - \frac{1}{6}\pi

Question

The function $\mathrm{f}$ is defined by $$\mathrm{f} : x \mapsto 3\sin \left( 2x - \frac{1}{6}\pi \right), \, 0 \le x \le k$$ Show that the largest exact value of $k$ such that $\mathrm{f}^{-1}$ exists is $\frac{1}{3}\pi$. Find $\mathrm{f}^{-1}(x)$.[3] It is given that $k = \frac{1}{3}\pi$. In a single diagram, sketch the graphs of $y = \mathrm{f}(x)$ and $y = \mathrm{f}^{-1}(x)$, labelling your graphs clearly.[3] The function $\mathrm{h}$ is defined by $\mathrm{h} : x \mapsto 3\sin \left( 2x - \frac{1}{6}\pi \right), \, 0 \le x \le \frac{1}{3}\pi, \, x \neq \frac{1}{12}\pi$. Another function $\mathrm{g}$ is defined by $\mathrm{g} : x \mapsto \left| 3 - \frac{1}{x} \right|, \, -3 \le x \le 3, \, x \neq 0$. Explain clearly why $\mathrm{gh}$ exists. Find $\mathrm{gh}(x)$ and its range.[3] Supposing $(\mathrm{gh})^{-1}$ exists for a restricted domain. Find the exact value of $x$ for which $(\mathrm{gh})^{-1}(x) = 0$. Show your working clearly.[2]

Accepted Answer

(a)${{\mathrm{f}}^{-1}}\left( x ight)=\frac{1}{2}\left( {{\sin }^{-1}}\frac{x}{3}+\frac{\pi }{6} ight)$ (c) ${{	ext{R}}_{	ext{gh}}}=\left[ 0,\infty ight)$ (d) $\frac{11}{3}$

2024 DHS Promo Q10

2024 DHS Promo Q10