Question

Domain of definition of the function $f\left(x\right)=\sqrt{{\sin }^{-1}\left(2x\right)+\frac{\pi }{6}}$ for real valued $x$, 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 $[-\frac{1}{4},\frac{1}{2}]$

Here $f\left(x\right)=\sqrt{{\sin }^{-1}\left(2x\right)+\frac{\pi }{6}}$ to find domain we must have

${\sin }^{-1}\left(2x\right)+\frac{\pi }{6}\ge 0$ ( but $-\frac{\pi }{2}\le {\sin }^{-1}\theta \le \frac{\pi }{2}$)
As $-\frac{\pi }{6}\le {\sin }^{-1}\left(2x\right)\le \frac{\pi }{2}\Rightarrow \sin (-\frac{\pi }{6})\le 2x\le \sin \left(\frac{\pi }{2}\right)$
$\Rightarrow -\frac{1}{2}\le 2x\le 1\Rightarrow -\frac{1}{4}\le x\le \frac{1}{2}\Rightarrow x\in [-\frac{1}{4},\frac{1}{2}]$