Question

If $f(x)=\sqrt{3|x|-x-2}$ and $g(x)=\sin \,x$, then domain of definition of $fog(x)$ is

• A. $\left\{2 n \pi+\frac{\pi}{2}\right\}_{n \in I}$
• B. $\bigcup_{n \in I}\left(2 n \pi+\frac{7 \pi}{6}, 2 n \pi+\frac{11 \pi}{6}\right)$
• C. $\left\{2 n \pi+\frac{7 \pi}{6}\right\}_{n \in I}$
• D. $\left\{(4 m+1) \frac{\pi}{2}: m \in I\right\} \bigcup_{n \in I}\left[2 n \pi+\frac{7 \pi}{6}, 2 n \pi+\frac{11 \pi}{6}\right]$

Answer 1

Answer: (D)

$f(x)=\sqrt{3|x|-x-2}$ and $g(x)=\sin\, x$
for $fog(x)=\sqrt{3|\sin\, x|-\sin \,x-2}$ which is defined if
$3|\sin \,x|-\sin \,x-2 \geq 0$
If $\sin\, x > 0$ then $2\, \sin \,x-2 \geq 0$
$ \Rightarrow \sin \,x \geq 1$
$\Rightarrow \sin\, x=1 \Rightarrow x=2 n \pi+\frac{\pi}{2}$
If $\sin x < 0$ then $-4\, \sin \,x-2 \geq 0$
$\Rightarrow -1 \leq \sin \,x \leq-\frac{1}{2} $
$\Rightarrow x \in\left[2 n \pi+\frac{7 \pi}{6}, 2 n \pi+\frac{11 \pi}{6}\right]$
$x \in\left[2 n \pi+\frac{7 \pi}{6}, 2 n \pi+\frac{11 \pi}{6}\right] \cup\left\{2 m \pi+\frac{\pi}{2}\right\}, n, m \in I$