Question

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

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

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
$3|\sin x|-\sin x-2\ge 0$
If $\sin x>0$ then $2\sin x-2\ge 0$
$\Rightarrow \sin x\ge 1$
$\Rightarrow \sin x=1\Rightarrow x=2n\pi +\frac{\pi }{2}$
If $\sin x<0$ then $-4\sin x-2\ge 0$
$\Rightarrow -1\le \sin x\le -\frac{1}{2}$
$\Rightarrow x\in \left[2n\pi +\frac{7\pi }{6},2n\pi +\frac{11\pi }{6}\right]$
$x\in \left[2n\pi +\frac{7\pi }{6},2n\pi +\frac{11\pi }{6}\right]\cup \left\{2m\pi +\frac{\pi }{2}\right\},n,m\in I$