If the total maximum value of the function f(x)=( √3e 2 sinx)^ sin^2x,x∈ (0, π 2), is k e, then ( k e)^8+ k^8 e^5+k^8 is equal to

Given function is $f(x)={(\frac{\sqrt{3}e}{2\mathrm{sin}x})}^{{\mathrm{sin}}^{2}x}$f(x)=(2sinx3​e​)sin2x

For maxima or minima ${f}^{'}(x)=0$f′(x)=0

$${f}^{'}(x)=f(x)[2\mathrm{sin}x\mathrm{cos}x\times \mathrm{ln}(\frac{\sqrt{3}e}{2\mathrm{sin}x})+{\mathrm{sin}}^{2}x\frac{1\times 2\mathrm{sin}x}{\sqrt{3}e}\times \frac{\sqrt{3}e}{2}(-\frac{1}{{\mathrm{sin}}^{2}x}\times \mathrm{cos}x)]$$f′(x)=f(x)[2sinxcosx×ln(2sinx3​e​)+sin2x3​e1×2sinx​×23​e​(−sin2x1​×cosx)]

$$\Rightarrow f(x)[\mathrm{sin}2x\mathrm{ln}(\frac{\sqrt{3}e}{2\mathrm{sin}x})-\mathrm{sin}x\mathrm{cos}x]=0$$⇒f(x)[sin2xln(2sinx3​e​)−sinxcosx]=0

Now on equation we get, $\mathrm{sin}2x=0$sin2x=0 (not possible)

So, $\mathrm{ln}(\frac{\sqrt{3}e}{2\mathrm{sin}x})=+\frac{1}{2}$ln(2sinx3​e​)=+21​

$$\Rightarrow \frac{\sqrt{3}\times \sqrt{e}}{2\mathrm{sin}x}={e}^{\frac{1}{2}}$$⇒2sinx3​×e​​=e21​

$$\Rightarrow \mathrm{sin}x=\frac{\sqrt{3}}{2}$$⇒sinx=23​​

$$\Rightarrow {f}_{\mathrm{max}}={(e)}^{\frac{3}{8}}=\frac{{e}^{\frac{11}{8}}}{e}\Rightarrow k={e}^{\frac{11}{8}}$$⇒fmax​=(e)83​=ee811​​⇒k=e811​

$$\Rightarrow {(\frac{k}{e})}^{8}+\frac{{k}^{8}}{{e}^{5}}+{k}^{8}={e}^{3}+{e}^{6}+{e}^{11}$$⇒(ek​)8+e5k8​+k8=e3+e6+e11