QJEEAdv22Ma020c: JEE Advanced 2022 Mathematics - Class 12 JEEAdv22Ma020c Solution: Let be a positive real number. Let and be the functions defined …

$\mathop {\lim }\limits_{x \to {\alpha ^ + }} f(g(x)) = f\left( {\mathop {\lim }\limits_{x \to {\alpha ^ + }} g(x)} \right)$ [As $f(x)$ is continuous function so we can write this]

Now,

$\mathop {\lim }\limits_{x \to {\alpha ^ + }} g(x)$

$= \mathop {\lim }\limits_{x \to {\alpha ^ + }} {{2{{\log }_e}(\sqrt x - \sqrt \alpha )} \over {{{\log }_e}\left( {{e^{\sqrt x }} - {e^{\sqrt \alpha }}} \right)}}$

$= \mathop {\lim }\limits_{x \to {\alpha ^ + }} {{2{{\log }_e}(\sqrt x - \sqrt \alpha )} \over {{{\log }_e}\left( {{e^{\sqrt \alpha }}\left( {{{{e^{\sqrt x }}} \over {{e^{\sqrt \alpha }}}} - 1} \right)} \right)}}$

$= \mathop {\lim }\limits_{x \to {\alpha ^ + }} {{2{{\log }_e}(\sqrt x - \sqrt \alpha )} \over {{{\log }_e}\left[ {{e^{\sqrt \alpha }}({e^{\sqrt x - \sqrt \alpha }} - 1)} \right]}}$

$= \mathop {\lim }\limits_{x \to {\alpha ^ + }} {{2{{\log }_e}(\sqrt x - \sqrt \alpha )} \over {{{\log }_e}{e^{\sqrt \alpha }} + {{\log }_e}({e^{\sqrt x - \sqrt \alpha }} - 1)}}$

$= \mathop {\lim }\limits_{x \to {\alpha ^ + }} {{2{{\log }_e}(\sqrt x - \sqrt \alpha )} \over {\sqrt \alpha + {{\log }_e}\left[ {{{{e^{\sqrt x - \sqrt \alpha }} - 1} \over {(\sqrt x - \sqrt \alpha )}} \times (\sqrt x - \sqrt \alpha )} \right]}}$

$= \mathop {\lim }\limits_{x \to {\alpha ^ + }} {{2{{\log }_e}(\sqrt x - \sqrt \alpha )} \over {\sqrt \alpha + {{\log }_e}\left[ {\mathop {\lim }\limits_{x \to {\alpha ^ + }} \left( {{{{e^{\sqrt x - \sqrt \alpha }} - 1} \over {(\sqrt x - \sqrt \alpha )}} \times (\sqrt x - \sqrt \alpha )} \right)} \right]}}$

$= \mathop {\lim }\limits_{x \to {\alpha ^ + }} {{2{{\log }_e}(\sqrt x - \sqrt \alpha )} \over {\sqrt \alpha + {{\log }_e}(1 \times \sqrt x - \sqrt \alpha )}}$ [using $\mathop {\lim }\limits_{x \to 0} {{{e^2} - 1} \over x} = 1$ ]

$= \mathop {\lim }\limits_{x \to {\alpha ^ + }} {{2{{\log }_e}(\sqrt x - \sqrt \alpha )} \over {\sqrt \alpha + {{\log }_e}(\sqrt x - \sqrt \alpha )}}$

$= \mathop {\lim }\limits_{x \to {\alpha ^ + }} {2 \over {{{\sqrt \alpha } \over {{{\log }_e}(\sqrt x - \sqrt \alpha )}} + 1}}$

From graph you can see $\log _e^{{0^ + }} \to \, - \alpha$

$\therefore$ $\mathop {\lim }\limits_{x \to {\alpha ^ + }} {\log _e}(\sqrt x - \sqrt \alpha ) = \log _e^{{0^ + }} = - \alpha$

$= \mathop {\lim }\limits_{x \to {\alpha ^ + }} {2 \over {{{\sqrt \alpha } \over { - \alpha }} + 1}}$

$= {2 \over {0 + 1}}$

$= 2$

$\therefore$ $f\left( {\mathop {\lim }\limits_{x \to {\alpha ^ + }} g(x)} \right) = f(2) = \sin \left( {{{\pi \times 2} \over {12}}} \right) = \sin {\pi \over 6} = {1 \over 2}$