Question

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

• A. ${\sin }^{-1}\left(\frac{x-2}{2}\right)-\frac{\pi }{6}$
• B. ${\sin }^{-1}\left(\frac{x-2}{2}\right)+\frac{\pi }{6}$
• C. ${\cos }^{-1}\left(\frac{x-2}{2}\right)+\frac{2\pi }{3}$
• D. $f^{-1}$ does not exist

Answer 1

The correct option is B ${\sin }^{-1}\left(\frac{x-2}{2}\right)+\frac{\pi }{6}$

$f\left(x\right)=\sqrt{3}\sin x-\cos x+2$
$=2\sin (x-\frac{\pi }{6})+2\cdots \left(1\right)$
Since, $\frac{\pi }{3}\le x\le \frac{2\pi }{3}$
$\Rightarrow \frac{\pi }{6}\le x-\frac{\pi }{6}\le \frac{\pi }{2}$
$\Rightarrow \frac{1}{2}\le \sin (x-\frac{\pi }{6})\le 1$
$\Rightarrow 3\le 2\sin (x-\frac{\pi }{6})+2\le 4$
$\Rightarrow$ Range of $f\left(x\right)$ is $[3,4]$

Also, $f\left(x\right)$ is strictly increasing in its domain.
$\therefore f\left(x\right)$ is invertible and inverse exists as it is one-one and onto.

From $y=f\left(x\right),x=f^{-1}\left(y\right)$
Substituting $x=f^{-1}\left(y\right)$ in $\left(1\right),$ we get
$y=2\sin \left(f^{-1}\right(y)-\frac{\pi }{6})+2$
$\Rightarrow \sin \left(f^{-1}\right(y)-\frac{\pi }{6})=\frac{y-2}{2}$
$\Rightarrow f^{-1}\left(y\right)-\frac{\pi }{6}={\sin }^{-1}\left(\frac{y-2}{2}\right)$
$\Rightarrow f^{-1}\left(y\right)={\sin }^{-1}\left(\frac{y-2}{2}\right)+\frac{\pi }{6}$
$\therefore f^{-1}\left(x\right)={\sin }^{-1}\left(\frac{x-2}{2}\right)+\frac{\pi }{6}$