The domain of the function $\mathrm{f}\left(\mathrm{x}\right)={\cos }^{-1}\left(\frac{2-\left|\mathrm{x}\right|}{4}\right)+[\log (3-\mathrm{x}){]}^{-1} \mathrm{is}$

$$\begin{gathered} {\cos }^{-1}\left(\frac{2-\left|\mathrm{x}\right|}{4}\right)\text{ is defined for }-1\le \frac{2-\left|\mathrm{x}\right|}{4}\le 1 \\ \begin{array}{l}\Rightarrow -4\le 2-\left|\mathrm{x}\right|\le 4\Rightarrow -6\le -\left|\mathrm{x}\right|\le 2\\ \Rightarrow -2\le \left|\mathrm{x}\right|\le 6\Rightarrow \left|\mathrm{x}\right|\le 6\Rightarrow -6\le \mathrm{x}\le 6 ....\left(1\right)\end{array} \end{gathered}$$
$\frac{1}{\log (3-\mathrm{x})}$ is defined if log (3 – x) ≠ 0 and 3 – x > 0
$\Rightarrow 3-\mathrm{x}\ne {\mathrm{e}}^0=1\text{ and }\mathrm{x}<3\Rightarrow \mathrm{x}\ne 2\text{ and }\mathrm{x}<3$
From Eq. (1) and (2), we get domain of f
= [– 6, 6] ∩ ((– ∞, 3) \ {2}) = [– 6, 3)\{2}.