Question

Determine domain of the function ${\sin }^{-1}\left[{\log }_2\left(\frac{1}{2}{x}^2\right)\right]$

• A. $[-2,-1]\cup [1,2].$
• B. $[-2,1]\cup [1,2].$
• C. $[-2,-1]\cup [1,3].$
• D. $[-2,0]\cup [1,2].$

Answer 1

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

We have $-1\le {\log }_2\left(\frac{1}{2}{x}^2\right)\le 1$
$\Rightarrow 2^{-1}\le \frac{1}{2}{x}^2\le 2\Rightarrow 1\le x^2\le 4$
$\Rightarrow 1\le x\le 2$ or $-2\le x\le -1.$Thus the required domain is $[-2,-1]\cup [1,2].$