Question

The domain of the function ${\sin }^{-1}\left[{\log }_2\left(\frac{x^2}{2}\right)\right]$ 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 $[-1,1]$
${\log }_2\frac{x^2}{2}\in [-1,1]\Rightarrow -1\le {\log }_2\frac{x^2}{2}\le 1\Rightarrow \frac{1}{2}\le \frac{x^2}{2}\le 2\Rightarrow 1\le x^2\le 4\therefore x\in [-2,-1]\cup [1,2]$