The domain of the function
$\mathrm{f}\left(\mathrm{x}\right)={\sin }^{-1}\left\{{\log }_2\left(\frac{1}{2}{\mathrm{x}}^2\right)\right\} \mathrm{is}$

For f (x) to be defined, we must have
$-1\le {\log }_2\left(\frac{1}{2}{\mathrm{x}}^2\right)\le 1\Rightarrow 2^{-1}\le \frac{1}{2}{\mathrm{x}}^2\le 2^1 [\because \mathrm{the} \mathrm{base}=2>1]$
$$\begin{gathered} \Rightarrow 1\le {\mathrm{x}}^2\le 4 ........ \left(1\right) \\ \text{ Now, }1\le {\mathrm{x}}^2\Rightarrow {\mathrm{x}}^2-1\ge 0\text{ i.e. }(\mathrm{x}-1)(\mathrm{x}+1)\ge 0 \\ \Rightarrow \mathrm{x}\le -1\text{ or }\mathrm{x}\ge 1 ........ \left(2\right) \\ \text{ Also, }{\mathrm{x}}^2\le 4\Rightarrow {\mathrm{x}}^2-4\le 0\text{ i.e. }(\mathrm{x}-2)(\mathrm{x}+2)\le 0 \\ \Rightarrow -2\le \mathrm{x}\le 2 ........ \left(3\right) \end{gathered}$$
From Eq. (2) and (3), we get the domain of f
$$\begin{gathered} =\left(\right(-\mathrm{\infty },-1]\cup [1,\mathrm{\infty }\left)\right)\cap [-2,2] \\ =[-2,-1]\cup [1,2] \end{gathered}$$