The domain of the function $\mathrm{f}\left(\mathrm{x}\right)=\sqrt{{\mathrm{e}}^{{\sin }^{-1}\left({\log }_{16}{\mathrm{x}}^2\right)}} \mathrm{is}$

f (x) is defined if
$$\begin{gathered} \begin{array}{l}-1\le {\log }_{16}{\mathrm{x}}^2\le 1\Rightarrow {16}^{-1}\le {\mathrm{x}}^2\le {16}^1\\ \Rightarrow \frac{1}{16}\le {\mathrm{x}}^2\le 16\end{array} \\ \begin{array}{l}{\mathrm{x}}^2\ge \frac{1}{16}\Rightarrow \left(\mathrm{x}-\frac{1}{4}\right)\left(\mathrm{x}+\frac{1}{4}\right)\ge 0\\ \Rightarrow \mathrm{x}\le -\frac{1}{4}\text{ or }\mathrm{x}\ge \frac{1}{4} ........\left(1\right)\end{array} \\ {\mathrm{x}}^2\le 16\Rightarrow (\mathrm{x}-4)(\mathrm{x}+4)\le 0\Rightarrow -4\le \mathrm{x}\le 4 .......\left(2\right) \end{gathered}$$
From Eq. (1) and (2), we get domain of f
$=\left[-4,-\frac{1}{4}\right]\cup \left[\frac{1}{4},4\right]$