MCQ
If the local maximum value of the function $f(x)=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^2 x}, \quad x \in\left(0, \frac{\pi}{2}\right)$, is $\frac{k}{e}$, then $\left(\frac{ k }{ e }\right)^8+\frac{ k ^8}{ e ^5}+ k ^8$ is equal to
  • A
    $e^5+e^6+e^{11}$
  • B
    $e^3+e^5+e^{11}$
  • $e ^3+ e ^6+ e ^{11}$
  • D
    $e^3+e^6+e^{10}$

Answer

Correct option: C.
$e ^3+ e ^6+ e ^{11}$
c
$\text { Let } y=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^2 x}$

$\ln y=\sin ^2 x \cdot \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right)$

$\frac{1}{y} y^{\prime}=\ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) 2 \sin x \cos x+\sin ^2 x \frac{2 \sin x}{\sqrt{3 e}} \frac{\sqrt{3 e}}{2}(-\operatorname{cosec} x \cot x)$

$\frac{ dy }{ dx }=0 \Rightarrow \ln \left(\frac{\sqrt{3 e }}{2 \sin x }\right) 2 \sin x \cos x -\sin x \cos x =0$

$\Rightarrow \sin x \cos x\left[2 \ln \left(\frac{\sqrt{3 e }}{2 \sin x }\right)-1\right]=0$

$\Rightarrow \ln \left(\frac{3 e }{4 \sin ^2 x }\right)=1 \Rightarrow \frac{3 e }{4 \sin ^2 x }= e \Rightarrow \sin ^2 x =\frac{3}{4}$

$\Rightarrow \sin x =\frac{\sqrt{3}}{2} \quad\left(\text { as } x \in\left(0, \frac{\pi}{2}\right)\right)$

$\Rightarrow \text { local max value }=\left(\frac{\sqrt{3 e }}{\sqrt{3}}\right)^{3 / 4}= e ^{3 / 8}=\frac{ k }{ e }$

$\Rightarrow k ^8= e ^{11}$

$\Rightarrow\left(\frac{ k }{ e }\right)^8+\frac{ k ^8}{ e ^5}+ k ^8= e ^3+ e ^6+ e ^{11}$

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