The function $\mathrm{f}$ is defined by \[\mathrm{f}:x\mapsto\sin 2x+\sqrt{3}\cos 2x,\quad 0\leq x

Question

The function $\mathrm{f}$ is defined by $$\mathrm{f}:x\mapsto\sin 2x+\sqrt{3}\cos 2x,\quad 0\leq x\leq k,$$ where $k>0$. Express $\mathrm{f}(x)$ in the form $R\sin(2x+\alpha)$, where $R>0$ and $0<\alpha<\frac{\pi}{2}$, giving the exact values of $R$ and $\alpha$.[2] Find the largest value of $k$ for which $\mathrm{f}^{-1}$ exists. For this value of $k$, find $\mathrm{f}^{-1}(x)$ and state its domain.[4] Using the value of $k$ found in part (b), sketch $y=\mathrm{f}(x)$ and $y=\mathrm{f}^{-1}(x)$ on the same diagram. Label the endpoints of both curves and indicate their relationship to the line $y=x$. [3] The functions $\mathrm{g}$ and $\mathrm{h}$ are defined by $$\mathrm{g}:x\mapsto\sin 2x+\sqrt{3}\cos 2x,\quad 0\sqrt{3},$$ where $k$ has the value found in part (b). Explain why $\mathrm{hg}$ exists. Find $\mathrm{hg}(x)$ and state its range in exact form.[3]

Accepted Answer

(a) $R=2,\ \alpha=\frac{\pi}{3}$. (b) $k=\frac{\pi}{12}$, $\mathrm{f}^{-1}(x)=\frac12[\sin^{-1}(x/2)-\frac{\pi}{3}]$, domain $[\sqrt{3},2]$. (d) $\mathrm{hg}(x)=\frac{1}{\sin 2x+\sqrt{3}\cos 2x-\sqrt{3}}$, range $[2+\sqrt{3},\infty)$.

Inverting a shifted sine and forming a reciprocal composite

Inverting a shifted sine and forming a reciprocal composite

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