If the local maximum value of the function f( | vedclass

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(C) Let $y=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^2 x}$.Taking natural logarithm on both sides: $\ln y = \sin^2 x \cdot \ln \left(\frac{\sqrt

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$e^3+e^6+e^{11}$

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(C) Let $y=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^2 x}$.Taking natural logarithm on both sides: $\ln y = \sin^2 x \cdot \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right)$.Differentiating with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = 2 \sin x \cos x \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) + \sin^2 x \cdot \frac{2 \sin x}{\sqrt{3 e}} \cdot \frac{\sqrt{3 e}}{2} \cdot (-\csc x \cot x)$.Simplifying the derivative: $\frac{1}{y} \frac{dy}{dx} = \sin x \cos x \left[ 2 \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) - 1 \right]$.Setting $\frac{dy}{dx} = 0$,we get $2 \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) = 1$,which implies $\ln \left(\frac{3 e}{4 \sin^2 x}\right) = 1$.Thus,$\frac{3 e}{4 \sin^2 x} = e$,so $\sin^2 x = \frac{3}{4}$.Since $x \in (0, \frac{\pi}{2})$,$\sin x = \frac{\sqrt{3}}{2}$.The local maximum value is $f(x) = \left(\frac{\sqrt{3 e}}{2 \cdot \frac{\sqrt{3}}{2}}\right)^{3/4} = (\sqrt{e})^{3/4} = e^{3/8}$.Given $e^{3/8} = \frac{k}{e}$,we have $k = e^{1 + 3/8} = e^{11/8}$.Then $k^8 = (e^{11/8})^8 = e^{11}$.Substituting these values: $\left(\frac{k}{e}\right)^8 + \frac{k^8}{e^5} + k^8 = (e^{3/8})^8 + \frac{e^{11}}{e^5} + e^{11} = e^3 + e^6 + e^{11}$.

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

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