Question

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

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

Answer 1

Answer: (D)

$f(g(x))=\sqrt{3|\sin x|-\sin x-2}$
$\therefore 3|\sin x|-\sin x-2\ge 0$
$\sin x>0\sin x<0$
$\sin x\ge 1$
$\sin x\le \left(-\frac{1}{2}\right)$
$\Rightarrow \sin x=1$
$x=2m\pi +\frac{\pi }{2},m\in I$ ......(i)
$x\in \left[\frac{7\pi }{6},\frac{11\pi }{6}\right]\Rightarrow x\in \left[2n\pi +\frac{7\pi }{6},2n\pi +\frac{11\pi }{6}\right],n\in I$....(ii)
$\therefore \text{ Union of (i) and (ii) is the domain of fog(x). }$