Q106 / 160 · TS EAMCET 2022 (20 Jul Shift 1) · Mathematics
$\lim _{x \rightarrow 0} \frac{\left(2^x-1\right)(1+\sin x)^{\frac{2}{\sin x}}}{\log (1+2 x)}=$
- A$e^2 \log 4$
- B$e \log \sqrt{2}$
- C$e^2 \log 2$
- Correct answerD$e^2 \log \sqrt{2}$
Solution
Given $\lim _{x \rightarrow 0} \frac{\left(2^x-1\right)(1+\sin x)^{\frac{2}{\sin x}}}{\log (1+2 x)}$
Differentiate w.r.t. ' $\mathrm{x}$ ' both sides
$$
f^{\prime}(x)=\frac{\left(e^{-x} \cos x-e^{-x} \sin x\right) \log x-e^{-x} \frac{\sin x}{x}}{\left(\log e^x\right)^2}
$$
Now take $f^{\prime}(x)=f(x) g(x)$
$$
\frac{\mathrm{e}^{-\mathrm{x}}(\cos \mathrm{x}-\sin \mathrm{x}) \log \mathrm{e}^{\mathrm{x}}}{\left(\log \mathrm{e}^{\mathrm{x}}\right)^2}-\frac{\mathrm{e}^{-\mathrm{x}} \sin \mathrm{x}}{\mathrm{x}\left(\log \mathrm{e}^{\mathrm{x}}\right)^2}=\frac{\mathrm{e}^{-\mathrm{x}} \sin \mathrm{x}}{\log \mathrm{e}^{\mathrm{x}}} \mathrm{g}(\mathrm{x})
$$
Divide bothside by $\frac{e^{-x} \sin x}{\log e^x}$
$$
\begin{aligned}
& g(x)=\frac{\frac{e^{-x}(\cos x-\sin x)}{\left(\log e^x\right)}}{\frac{e^{-x} \sin x}{\left(\log e^x\right)}}-\frac{e^{-x} \sin x}{\frac{e^{-x} \sin e^x}{2}} \\
& \log e^x \\
& g(x)=\frac{\cos x}{\sin x}-\frac{1}{x \log e^x} \\
& g(x)=\cot x-1-\frac{1}{x \log e^x}
\end{aligned}
$$
Differentiate w.r.t. ' $\mathrm{x}$ ' both sides
$$
\begin{aligned}
& \mathrm{g}^{\prime}(\mathrm{x})=-\operatorname{cosec} \mathrm{e}^2 \mathrm{x}+1-\frac{\log _{\mathrm{e}}^{\mathrm{x}}+1}{\mathrm{x}^2\left(\log _{\mathrm{e}}^{\mathrm{x}}\right)^2} \\
& \text { put } \mathrm{x}=\mathrm{e} \text { in } \mathrm{g}^{\prime}(\mathrm{x}) \\
& \mathrm{g}^{\prime}(\mathrm{e})=-\operatorname{cosec}^2(\mathrm{e})+\frac{\left(\log _{\mathrm{e}^{e+1}}\right)}{\left(\mathrm{e}^2\right)\left(\log _{\mathrm{e}^{\mathrm{e}}}\right)^2} \\
& \mathrm{~g}^{\prime}(\mathrm{e})=-\operatorname{cosec} \mathrm{ec}^2(\mathrm{e})+\frac{2}{\mathrm{e}^2 1}=2 \mathrm{e}^2-\operatorname{cosec} \mathrm{ec}^2(\mathrm{e})
\end{aligned}
$$
so, option (c) is correct
Hence, the correct answer is option: 'D'.