Given function

$f(x)=\left(\frac{\sqrt{3e}}{2\sin x}\right)^{\sin^2 x}, \qquad x\in\left(0,\frac{\pi}{2}\right).$

We need its local maximum value.

Take logarithm

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

Put $t=\sin^2 x.$ Since $x\in\left(0,\frac{\pi}{2}\right)$, we have $\sin x\in(0,1)$, hence $t\in(0,1).$

Also, $\sin x=\sqrt t.$ Therefore

Simplify the logarithm expression

Since

Differentiate

$\phi'(t)=\ln\left(\frac{\sqrt{3e}}{2}\right)-\frac12(\ln t+1).$

Set $\phi'(t)=0$: $\ln\left(\frac{\sqrt{3e}}{2}\right)-\frac12(\ln t+1)=0.$

Multiply by $2$: $2\ln\left(\frac{\sqrt{3e}}{2}\right)-\ln t-1=0.$

Hence $\ln t=2\ln\left(\frac{\sqrt{3e}}{2}\right)-1.$

Now, $2\ln\left(\frac{\sqrt{3e}}{2}\right)=\ln\left(\frac{3e}{4}\right).$ Therefore $\ln t=\ln\left(\frac{3e}{4}\right)-1=\ln\left(\frac{3e}{4}\right)-\ln e=\ln\left(\frac34\right).$

So $t=\frac34.$

Check maximum

$\phi''(t)=-\frac{1}{2t}<0 \qquad (t>0),$ so this critical point gives a maximum.

Find the maximum value

Since $t=\sin^2 x=\frac34$, we have $\sin x=\frac{\sqrt3}{2}.$

Then

Compute the required expression

We need $\left(\frac{k}{e}\right)^8+\frac{k^8}{e^5}+k^8.$

Since $\frac{k}{e}=e^{3/8},$ we get $\left(\frac{k}{e}\right)^8=e^3.$

Also, $k=e^{11/8}\implies k^8=e^{11}.$

Therefore, $\frac{k^8}{e^5}=e^{11-5}=e^6.$

So the expression becomes $e^3+e^6+e^{11}.$

Match with options

This is Option C.

Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

Hence, they agree.