JEE MAIN - Mathematics (2023 - 12th April Morning Shift - No. 10)
If the local maximum value of the function $$f(x)=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^{2} x}, x \in\left(0, \frac{\pi}{2}\right)$$ , is $$\frac{k}{e}$$, then $$\left(\frac{k}{e}\right)^{8}+\frac{k^{8}}{e^{5}}+k^{8}$$ is equal to
$$e^{3}+e^{6}+e^{10}$$
$$e^{3}+e^{5}+e^{11}$$
$$e^{3}+e^{6}+e^{11}$$
$$e^{5}+e^{6}+e^{11}$$
Explanation
$$
\text { Let } y=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^2 x}
$$
$$ \ln \mathrm{y}=\sin ^2 \mathrm{x} \cdot \ln \left(\frac{\sqrt{3 \mathrm{e}}}{2 \sin \mathrm{x}}\right) $$
$$ \frac{1}{y} y^{\prime}=\ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) 2 \sin x \cos x+\sin ^2 x \frac{2 \sin x}{\sqrt{3 e}} \frac{\sqrt{3 e}}{2}(-\operatorname{cosec} x \cot x) $$
For maxima or minima, $$ \frac{d y}{d x}=0 $$
$$ \Rightarrow \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) 2 \sin x \cos x-\sin x \cos x=0 $$
$$ \Rightarrow \sin x \cos x\left[2 \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right)-1\right]=0 $$
$$ \Rightarrow \ln \left(\frac{3 \mathrm{e}}{4 \sin ^2 \mathrm{x}}\right)=1 $$
$$ \Rightarrow \frac{3 e}{4 \sin ^2 x}=e $$
$$ \Rightarrow \sin ^2 x=\frac{3}{4} $$
$$ \Rightarrow \sin \mathrm{x}=\frac{\sqrt{3}}{2} \quad\left(\text { as } \mathrm{x} \in\left(0, \frac{\pi}{2}\right)\right) $$
$$ \Rightarrow \text { Local max value }=\left(\frac{\sqrt{3 \mathrm{e}}}{\sqrt{3}}\right)^{3 / 4}=\mathrm{e}^{3 / 8}=\frac{\mathrm{k}}{\mathrm{e}} $$
$$ \Rightarrow \mathrm{k}^8=\mathrm{e}^{11} $$
$$ \therefore $$ $$ \left(\frac{k}{e}\right)^8+\frac{k^8}{e^5}+k^8=e^3+e^6+e^{11} $$
$$ \ln \mathrm{y}=\sin ^2 \mathrm{x} \cdot \ln \left(\frac{\sqrt{3 \mathrm{e}}}{2 \sin \mathrm{x}}\right) $$
$$ \frac{1}{y} y^{\prime}=\ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) 2 \sin x \cos x+\sin ^2 x \frac{2 \sin x}{\sqrt{3 e}} \frac{\sqrt{3 e}}{2}(-\operatorname{cosec} x \cot x) $$
For maxima or minima, $$ \frac{d y}{d x}=0 $$
$$ \Rightarrow \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) 2 \sin x \cos x-\sin x \cos x=0 $$
$$ \Rightarrow \sin x \cos x\left[2 \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right)-1\right]=0 $$
$$ \Rightarrow \ln \left(\frac{3 \mathrm{e}}{4 \sin ^2 \mathrm{x}}\right)=1 $$
$$ \Rightarrow \frac{3 e}{4 \sin ^2 x}=e $$
$$ \Rightarrow \sin ^2 x=\frac{3}{4} $$
$$ \Rightarrow \sin \mathrm{x}=\frac{\sqrt{3}}{2} \quad\left(\text { as } \mathrm{x} \in\left(0, \frac{\pi}{2}\right)\right) $$
$$ \Rightarrow \text { Local max value }=\left(\frac{\sqrt{3 \mathrm{e}}}{\sqrt{3}}\right)^{3 / 4}=\mathrm{e}^{3 / 8}=\frac{\mathrm{k}}{\mathrm{e}} $$
$$ \Rightarrow \mathrm{k}^8=\mathrm{e}^{11} $$
$$ \therefore $$ $$ \left(\frac{k}{e}\right)^8+\frac{k^8}{e^5}+k^8=e^3+e^6+e^{11} $$
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