Question

Let $f:[-\frac{\pi }{3},\frac{2\pi }{3}]\to [0,4]$ be a function defined by $f\left(x\right)=\sqrt{3}\sin x-\cos x+2$ then $f^{-1}\left(x\right)$ equals

• A. $\frac{2\pi }{3}-{\cos }^{-1}\left(\frac{x-2}{2}\right)$
• B. ${\sin }^{-1}\left(\frac{x-2}{2}\right)+\frac{\pi }{3}$
• C. ${\sin }^{-1}\left(\frac{x-2}{2}\right)$
• D. ${\sin }^{-1}\left(\frac{x+2}{2}\right)+\frac{\pi }{6}$

Answer 1

The correct option is A $\frac{2\pi }{3}-{\cos }^{-1}\left(\frac{x-2}{2}\right)$

$y=2(\frac{\sqrt{3}}{2}\sin \left(x\right)-\frac{1}{2}\cos \left(x\right))+2$

$\Rightarrow \frac{y}{2}-1=\frac{\sqrt{3}}{2}\sin \left(x\right)-\frac{1}{2}\cos \left(x\right)$

$\Rightarrow \frac{y-2}{2}=-\cos (\frac{\pi }{3}+x)$

$\Rightarrow \frac{y-2}{2}=\cos (\pi -(\frac{\pi }{3}+x))$

$\Rightarrow \frac{y-2}{2}=\cos (\frac{2\pi }{3}-x)$

$\Rightarrow {\cos }^{-1}\left(\frac{y-2}{2}\right)=\frac{2\pi }{3}-x$

$x=\frac{2\pi }{3}-{\cos }^{-1}\left(\frac{y-2}{2}\right)$
Therefore,
$f^{-1}\left(x\right)=\frac{2\pi }{3}-{\cos }^{-1}\left(\frac{x-2}{2}\right)$