Question

The domain of the function $f\left(x\right)=\sqrt{{\sin }^{-1}\left(2x\right)+\frac{\pi }{6}}$ for real $x$, is

• A. $[-\frac{1}{4},\frac{1}{2}]$
• B. $[-\frac{1}{2},\frac{1}{2}]$
• C. $(-\frac{1}{2},\frac{1}{9})$
• D. $[-\frac{1}{4},\frac{1}{4}]$

Answer 1

The correct option is A $[-\frac{1}{4},\frac{1}{2}]$

For real values of $f\left(x\right)$

${\sin }^{-1}\left(2x\right)+\frac{\pi }{6}\ge 0$

${\sin }^{-1}\left(2x\right)+{\sin }^{-1}\left(\frac{1}{2}\right)\ge 0$

${\sin }^{-1}\left(2x\right)\ge -{\sin }^{-1}\left(\frac{1}{2}\right)$

$2x\ge \sin (-{\sin }^{-1}\left(\frac{1}{2}\right))$

$2x\ge -\frac{1}{2}$

$x\ge \frac{-1}{4}$

Also

${\sin }^{-1}\left(x\right)$ has a domain of $[-1,1]$

Hence for ${\sin }^{-1}2x$ it will be $[\frac{-1}{2},\frac{1}{2}]$

Hence the domain for $f\left(x\right)$ will be

$[\frac{-1}{4},\frac{1}{2}]$

Hence, option 'A' is correct.