Answer:
Option A,B,C
Explanation:
(a) f(x)=sin[π6sinπ2(sinx)],xϵR
= sin(π6sinθ),θϵ[−π2,π2],
where θ=π2sinx
= sinα,αϵ[−π6,π6],
where α=π6sinθ
∴ 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.