Question
If $$f\left( x \right) = \sqrt {3\left| x \right| - x - 2} $$ and $$g\left( x \right) = \sin \,x,$$ then domain of definition of $$fog\left( x \right)$$ is :
A.
$$\left\{ {2n\pi + \frac{\pi }{2}} \right\},\,n\, \in \,I$$
B.
$$\mathop \cup \limits_{n\, \in \,I} \left\{ {2n\pi + \frac{{7\pi }}{6},\,2n\pi + \frac{{11\pi }}{6}} \right\}$$
C.
$$\left\{ {2n\pi + \frac{{7\pi }}{6}} \right\},\,n\, \in \,I$$
D.
$$\left\{ {\left( {4m + 1} \right)\frac{\pi }{2}:m\, \in \,I} \right\}\mathop \cup \limits_{n\, \in \,I} \left[ {2n\pi + \frac{{7\pi }}{6},\,2n\pi + \frac{{11\pi }}{6}} \right]$$
Answer :
$$\left\{ {\left( {4m + 1} \right)\frac{\pi }{2}:m\, \in \,I} \right\}\mathop \cup \limits_{n\, \in \,I} \left[ {2n\pi + \frac{{7\pi }}{6},\,2n\pi + \frac{{11\pi }}{6}} \right]$$
Solution :
For $$\left( {fog} \right)\left( x \right)$$ to exists range of $$g \subseteq $$ domain of $$f.$$
$$\eqalign{
& \therefore \,{\text{Domain of }}f \Rightarrow 3\left| x \right| - x - 2 \geqslant 0 \cr
& \Rightarrow 3\left| x \right| - x \geqslant 2 \cr
& {\text{When }}x \geqslant 0 \Rightarrow x \geqslant 1 \cr
& {\text{When}}\,x < 0 \Rightarrow x < - \frac{1}{2} \cr
& \therefore \,\sin \,x \geqslant 1{\text{ and }}\sin \,x < - \frac{1}{2}{\text{ for }}f\left\{ {g\left( x \right)} \right\}{\text{ to exists}}{\text{.}} \cr
& {\text{i}}{\text{.e}}{\text{., }}\sin \,x = 1{\text{ and }} - 1 \leqslant \sin \,x < - \frac{1}{2} \cr
& \therefore \,x = \left( {4m + 1} \right)\frac{\pi }{2}{\text{ and }}2n\pi + \frac{{7\pi }}{6} \leqslant x \leqslant 2n\pi + \frac{{11\pi }}{6} \cr
& {\text{i}}{\text{.e}}{\text{., }}\left\{ {\left( {4m + 1} \right)\frac{\pi }{2}:m\, \in \,I} \right\}\mathop \cup \limits_{n\, \in \,I} \left[ {2n\pi + \frac{{7\pi }}{6} \leqslant x \leqslant \,2n\pi + \frac{{11\pi }}{6}} \right] \cr} $$