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

Question

The function $\mathrm{f}$ is defined by $$\mathrm{f}:x\mapsto\sqrt{3}\cos x-\sin x,\quad -\pi\leq x\leq\pi.$$ Express $\mathrm{f}(x)$ in the form $R\cos(x+\alpha)$, where $R>0$ and $0<\alpha<\frac{\pi}{2}$, giving the exact values of $R$ and $\alpha$.[2] Hence state the range of $\mathrm{f}$ and sketch the graph of $y=\mathrm{f}(x)$, indicating the coordinates of the turning points and the intersections with the coordinate axes.[3] The function $\mathrm{g}$ is defined by $$\mathrm{g}:x\mapsto\sqrt{3}\cos x-\sin x,\quad -\alpha\leq x\leq b,$$ where $b>-\alpha$. Find the largest value of $b$ for which $\mathrm{g}^{-1}$ exists.[2] Using this value of $b$, find $\mathrm{g}^{-1}(x)$, stating its domain and range.[3]

Accepted Answer

(a) $R=2,\ \alpha=\frac{\pi}{6}$. (b) $[-2,2]$. (c) $b=\frac{5\pi}{6}$. (d) $\mathrm{g}^{-1}(x)=\cos^{-1}(x/2)-\frac{\pi}{6}$, domain $[-2,2]$, range $[-\frac{\pi}{6},\frac{5\pi}{6}]$.

Restricting a shifted cosine for an inverse

Restricting a shifted cosine for an inverse

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