Answer:
Option A,B,C
Explanation:
(a) $f(x)=\sin \left[ \frac{\pi}{6}sin\frac{\pi}{2}(\sin x)\right],x\epsilon R$
= $\sin (\frac{\pi}{6}\sin \theta),\theta\epsilon\left[-\frac{\pi}{2},\frac{\pi}{2}\right],$
where $\theta =\frac{\pi}{2}\sin x$
= $\sin \alpha,\alpha \epsilon\left[-\frac{\pi}{6},\frac{\pi}{6}\right],$
where $\alpha= \frac{\pi}{6}\sin\theta$
$\therefore$ $f(x)\epsilon \left[-\frac{1}{2},\frac{1}{2}\right]$
Hence, range of $f(x)\epsilon \left[-\frac{1}{2},\frac{1}{2}\right]$
So, option (a) is correct.
(b) $f\left\{g(x)\right\}=f(t),t\epsilon \left[- \frac{\pi}{2},\frac{\pi}{2}\right]$
$\Rightarrow$ $f(t)\epsilon \left[-\frac{1}{2},\frac{1}{2}\right]$
$\therefore$ Option (b) is correct.
(c) $\lim_{x \rightarrow 0}\frac{f(x)}{g(x)}$
$=\lim_{x \rightarrow 0}\frac{\sin\left[\frac{\pi}{6}\sin(\frac{\pi}{2}\sin x)\right]}{\frac{\pi}{2}(\sin x)}$
$=\lim_{x \rightarrow 0}\frac{\sin\left[\frac{\pi}{6}\sin(\frac{\pi}{2}\sin x)\right]}{\frac{\pi}{6}\sin(\frac{\pi}{2}\sin x)}$
$\frac{\frac{\pi}{6}\sin(\frac{\pi}{2}\sin x)}{(\frac{\pi}{2}\sin x)}$
= $1\times \frac{\pi}{6} \times 1=\frac{\pi}{6}$
$\therefore$ Option (c) is correct.
(d) g{ f(x)}=1
$\Rightarrow$ $\frac{\pi}{2}\sin\left\{f(x)\right\}=1$
$\Rightarrow$ $\sin\left\{f(x)\right\}=\frac{2}{\pi}$ .........(i)
But $ f(x)\epsilon \left[-\frac{1}{2},\frac{1}{2}\right]\subset\left[ -\frac{\pi}{6},\frac{\pi}{6}\right]$
$\therefore$ $\sin \left\{ f(x)\right\}\epsilon \left[-\frac{1}{2},\frac{1}{2}\right]$ ..............(ii)
$\Rightarrow$ $\sin \left\{ f(x)\right\}\neq\frac{2}{\pi}$
[from Eqs(i) and (ii) ]
i.e, no solution
$\therefore$ Option (d) is not correct.
