Let $$f(x) = \sin \left( {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \right)$$ for all $$x \in R$$ and g(x) = $${{\pi \over 2}\sin x}$$ for all x$$\in$$R. Let $$(f \circ g)(x)$$ denote f(g(x)) and $$(g \circ f)(x)$$ denote g(f(x)). Then which of the following is/are true?
(a) $$f(x) = \sin \left[ {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \right],\,x \in R$$
$$ = \sin \left( {{\pi \over 6}\sin \theta } \right),\,\theta \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]$$,
where, $$\theta = {\pi \over 2}\sin x$$
$$ = \sin \alpha ,\alpha \in \left[ { - {\pi \over 6},{\pi \over 6}} \right]$$,
where, $$\alpha = {\pi \over 6}\sin \theta $$
$$\therefore$$ $$f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$$
Hence, range of $$f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$$
So, option (a) is correct.
(b) $$f\{ g(x)\} = f(t),t \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]$$
$$ \Rightarrow f(t) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$$
$$\therefore$$ Option (b) is correct.
(c) $$\mathop {\lim }\limits_{x \to 0} {{f(x)} \over {g(x)}}$$
$$ = \mathop {\lim }\limits_{x \to 0} {{\sin \left[ {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \right]} \over {{\pi \over 2}(\sin x)}}$$
$$ = \mathop {\lim }\limits_{x \to 0} {{\sin \left[ {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \right]} \over {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)}}\,.\,{{{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \over {\left( {{\pi \over 2}\sin x} \right)}}$$
$$ = 1 \times {\pi \over 6} \times 1 = {\pi \over 6}$$
$$\therefore$$ Option (c) is correct.
(d) $$g\{ f(x)\} = 1$$
$$ \Rightarrow {\pi \over 2}\sin \{ f(x)\} = 1$$
$$ \Rightarrow \sin \{ f(x)\} = {2 \over \pi }$$ ..... (i)
But, $$f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right] \subset \left[ { - {\pi \over 6},{\pi \over 6}} \right]$$
$$\therefore$$ $$\sin \{ f(x)\} \in \left[ { - {1 \over 2},{1 \over 2}} \right]$$ ..... (ii)
$$ \Rightarrow \sin \{ f(x)\} \ne {2 \over \pi }$$, [from Eqs. (i) and (ii)]
i.e. No solution.
$$\therefore$$ Option (d) is not correct.
Let $$f:( - 1,1) \to R$$ be such that $$f(\cos 4\theta ) = {2 \over {2 - {{\sec }^2}\theta }}$$ for $$\theta \in \left( {0,{\pi \over 4}} \right) \cup \left( {{\pi \over 4},{\pi \over 2}} \right)$$. Then the value(s) of $$f\left( {{1 \over 3}} \right)$$ is(are)
$$f(\cos 4\theta ) = {2 \over {2 - {{\sec }^2}\theta }}$$
Let $$\cos 4\theta = t$$
$$ \Rightarrow 2{\cos ^2}2\theta - 1 = t \Rightarrow {\cos ^2}2\theta = {2 \over 3}$$
For $$t = {1 \over 3}$$ we have $${\cos ^2}2\theta = {2 \over 3}$$
$$\cos 2\theta = \sqrt {{2 \over 3}} $$ or $$\cos 2\theta = - \sqrt {{2 \over 3}} $$
$$f(\cos 4\theta ) = {2 \over {2 - {1 \over {{{\cos }^2}\theta }}}} = {{2{{\cos }^2}\theta } \over {2{{\cos }^2}\theta - 1}} = {{1 + \cos 2\theta } \over {\cos 2\theta }} = 1 + {1 \over {\cos 2\theta }}$$
Hence, $$f\left( {{1 \over 3}} \right) = 1 + \sqrt {{3 \over 2}} $$ or $$1 - \sqrt {{3 \over 2}} $$