QJEEAdv15Mad042: JEE Advanced 2015 Mathematics - Class 11 JEEAdv15Mad042 Solution: Let for all and g(x) = for all x R. Let denote f(g(x)) and…

(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.