Q. The domain of the function $f\left(x\right)=4\sqrt{\left(cos\right)^{- 1} \left(\frac{1 - \left|x\right|}{2}\right)}$ is

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A

$\left(- \in fty , - 3\right)\cup\left(3 , \in fty\right)$

B

$\left[- 3,3\right]$

C

$- \in fty , - 3\right]\cup\left[3 , \in fty$

D

$\phi$

Solution:

$\cos ^{-1}\left(\frac{1-|x|}{2}\right)$ is defined if
$-1\leq \frac{1 - \left|x\right|}{2}\leq 1$
$\Rightarrow -2\leq 1-\left|x\right|\leq 2\Rightarrow -3\leq -\left|x\right|\leq 1$
$\Rightarrow -1\leq \left|x\right|\leq 3$
$\left|x\right|\geq -1$ is true for all real values of $x$ .
$\left|x\right|\leq 3\Rightarrow -3\leq x\leq 3$
Also, $4\sqrt{\left(cos\right)^{- 1} \left(\frac{1 - \left|x\right|}{2}\right)}$ is defined if
$\left(cos\right)^{- 1} \left(\frac{1 - \left|x\right|}{2}\right)\geq 0$
$\Rightarrow \frac{1 - \left|x\right|}{2}\geq cos 0=1$
$\Rightarrow 1-\left|x\right|\geq 2$
$\Rightarrow \left|x\right|\leq -1$ (not possible)
$\therefore $ Domain of $f\left(x\right)=\left[- 3,3\right]$

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