Let $f\left(x\right)=\sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)\right)$ for all and $g\left(x\right)=\frac{\pi }{2}\sin x$ for all . Let $\left(fog\right)\left(x\right)$ denote $f\left(g\right(x)$ and $\left(gof\right)\left(x\right)$ denote $g\left(f\right(x\left)\right)$. Then which of the following is (are) true?

The correct option is C Range of fog is $[-\frac{1}{2},\frac{1}{2}]$
$f\left(x\right)=\sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)\right)$

$\frac{-\pi }{2}\le \frac{\pi }{2}\sin x\le \frac{\pi }{2}$

$-1\le \sin \left(\frac{\pi }{2}\sin x\right)\le 1$

$\sin \left(\frac{-\pi }{6}\right)\le \sin \left(\frac{\pi }{6}\left(\sin \left(\frac{\pi }{2}\sin x\right)\right)\right)\le \sin \frac{\pi }{6}$

$\frac{-1}{2}\le \sin \left(\frac{\pi }{6}\left(\sin \left(\frac{\lambda }{2}\sin x\right)\right)\right)\le \frac{1}{2}$

$f\left(g\left(x\right)\right)=\sin \left(\frac{\pi }{6}\left(\sin \left(\frac{\pi }{2}\sin \left(\frac{\pi }{2}\sin x\right)\right)\right)\right)$

$-1\le \sin \left(\frac{\pi }{2}\sin x\right)\le 1$

$\frac{-\pi }{2}\le \frac{\pi }{2}\sin \left(\frac{\pi }{2}\sin x\right)\le \frac{\pi }{2}$

$-1\le \sin \left(\frac{\pi }{2}\left(\sin \left(\frac{\pi }{2}\sin x\right)\right)\right)\le 1$

$-\frac{\pi }{6}\le \frac{\pi }{6}\sin \left(\frac{\pi }{2}\left(\sin \left(\frac{\pi }{2}\sin x\right)\right)\right)\le \frac{\pi }{6}$

$\frac{-1}{2}\le \sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin \left(\frac{\pi }{2}\sin x\right)\right)\right)\le \frac{1}{2}$

${lim}_{x\to 0}\frac{\sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)\right)}{\frac{\pi }{2}\sin x}\times \frac{\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)}{\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)}$

$\Rightarrow {lim}_{x\to 0}\frac{\pi }{6}\frac{\sin \left(\frac{\pi }{2}\sin x\right)}{\frac{\pi }{2}\sin x}=\frac{\pi }{6}$


Range of $g\left(f\right(x\left)\right)\text{is}[\frac{\pi }{2}\sin \left(\frac{-1}{2}\right),\frac{\pi }{2}\sin \frac{1}{2}]$

$[-\frac{\pi }{2}\sin \frac{1}{2},\frac{\pi }{2}\sin \frac{1}{2}]$

Hence 1 does not belong to this range.