Question

The exhaustive domain of the function ${\sin }^{-1}\left[{\log }_2\right(x^2/2\left)\right]$ is

• A. $[-2,2]$
• B. $[1,2]$
• C. $[-2,-1]\cup [1,2]$
• D. $[0,2]$

Answer 1

The correct option is C $[-2,-1]\cup [1,2]$

Domain of ${\sin }^{-1}\left(x\right)$ is $[-1,1]$
$\therefore$ For ${\sin }^{-1}\left[{\log }_2\frac{x^2}{2}\right],$
${\log }_2\frac{x^2}{2}\in [-1,1]$
$-1\le {\log }_2\frac{x^2}{2}\le 1$ and
$\Rightarrow \frac{1}{2}\le \frac{x^2}{2}\le 2$
$\Rightarrow 1\le x^2\le 4$
$\therefore x\in [-2,-1]\cup [1,2]$