Question

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

• A. $\{2n\pi +\frac{\pi }{2}\}$
• B. $(2n\pi +\frac{7\pi }{6},2n\pi +\frac{11\pi }{6})$
• C. $\{2n\pi +\frac{7\pi }{6}\}$
• D. $(2n\pi +\frac{7\pi }{6},2n\pi +\frac{11\pi }{6})\bigcup _{n,m\in I}^{}(2n\pi +\frac{\pi }{2})$

Answer 1

Answer: (D)

$(2n\pi +\frac{7\pi }{6},2n\pi +\frac{11\pi }{6})\bigcup _{n,m\in I}^{}(2n\pi +\frac{\pi }{2})$

We have $f\left(x\right)=\sqrt{3\left|x\right|-x-2}$ and $g\left(x\right)=\sin x$

$\therefore f\circ g\left(x\right)=\sqrt{3\left|\sin x\right|-\sin x-2}$ which is defined,

if $3\left|\sin x\right|-\sin x-2\ge 0$

If $\sin x>0,$ then $2\sin x-2\ge 0$

$\Rightarrow \sin x\ge 1$

$\Rightarrow \sin x=1\Rightarrow x=\frac{\pi }{2}$

If $\sin x<0,$ then $-4\sin x-2\ge 0\Rightarrow -1\le -\frac{1}{2}$

$\therefore x\in (2n\pi +\frac{7\pi }{6},2n\pi +\frac{11\pi }{6})\bigcup _{n,m\in I}^{}(2n\pi +\frac{\pi }{2})n,m\in I$