JEE Advanced 2022 — Paper 1 — Question 2

$\lim _{x \rightarrow \alpha^{+}} f(g(x))=f\left(\lim _{x \rightarrow \alpha^{+}} g(x)\right)$

Now $\lim _{x \rightarrow \alpha^{+}} g(x)=\lim _{x \rightarrow \alpha^{+}} \frac{2 \ln (\sqrt{x}-\sqrt{\alpha})}{\ln \left(e^{\sqrt{x}}-e^{\sqrt{\alpha}}\right)} \quad\left(\frac{-\infty}{-\infty}\right)$ Apply D'L Hospital

$\lim _{x \rightarrow \alpha^{+}}+\frac{2 \cdot \frac{1}{\sqrt{x-\sqrt{\alpha}}} \cdot \frac{1}{2 \sqrt{x}}}{\frac{1}{e^{\sqrt{x}}-e^{\sqrt{\alpha}}} \cdot \mathrm{e}^{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}}$

$\lim _{x \rightarrow \alpha^{+}} \frac{2\left(\mathrm{e}^{\sqrt{x}}-\mathrm{e}^{\sqrt{\alpha}}\right)}{\mathrm{e}^{\sqrt{x}}(\sqrt{\mathrm{x}}-\sqrt{\alpha})}$

$\lim _{x \rightarrow \alpha^{+}} \frac{2 e^{\sqrt{\alpha}}\left(e^{\sqrt{x}-\sqrt{\alpha}}-1\right)}{e^{\sqrt{x}}(\sqrt{x}-\sqrt{\alpha})}=2$

Now $\mathrm{f}(\mathrm{x})=\sin \frac{\pi \mathrm{x}}{12}$

given $f(2)=\sin \frac{\pi(2)}{12}=\sin \frac{\pi}{6}=\frac{1}{2}=0.5$