Question

If $f(x)={\cos }^{-1}\left(\frac{2-|x|}{4}\right)+\log (3-x){]}^{-1}$, then its domain is given by

• A. [-6, 2) $\cup$ (2, 3)
• B. [- 2, 6]
• C. [- 6, 2)
• D. {- 2, 2) $\cup$ (2, 3)

Answer 1

Answer: (A)

The domain of $co{s}^{-1}\left(\frac{2-|x|}{4}\right)$ is given by
$-1\le \frac{2-|x|}{4}\le \Rightarrow -4\le 2-|x|\le 4$
$\Rightarrow -6\le -|x|\le 2\Rightarrow -2\le |x|\le 6\Rightarrow |x|\le 6$
$\therefore$ domain of this function is -6 $\le x\le 6$.
Also the domain of $\frac{1}{\log (3-x)}$ is given by 3-x > 0
and x $\ne$ 2 ($\because 3-x\ne 1)$
i,e., x= $\ne$ and x < 3
$\therefore$ required domain is $-6\le x<3$
and $x\ne 2$ i.e., [-6, 2) U (2, 3)