If the local maximum value of the function f(x)=(√3e2 x)^ ^2x,x (0,( )/(2)) , is (k)/(e), …

Question

If the local maximum value of the function $f(x)={(\frac{\sqrt{3e}}{2\sin x})}^{{\sin }^{2}x},x\in (0,\frac{\pi }{2})$ , is $\frac{k}{e}$, then ${(\frac{k}{e})}^{8}+\frac{{k}^{8}}{{e}^{5}}+{k}^{8}$ is equal to

Accepted Answer

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

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

$\frac{1}{y}{y}^{'}=\ln (\frac{\sqrt{3e}}{2\sin x})2\sin x\cos x+{\sin }^{2}x\frac{2\sin x}{\sqrt{3e}}\frac{\sqrt{3e}}{2}(-\csc x\cot x)$

For maxima or minima, $\frac{dy}{dx}=0$

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

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

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

$\Rightarrow \frac{3e}{4{\sin }^{2}x}=e$

$\Rightarrow {\sin }^{2}x=\frac{3}{4}$

$\Rightarrow \sin x=\frac{\sqrt{3}}{2}\, (x\in (0,\frac{\pi }{2}))$

$\Rightarrow ={(\frac{\sqrt{3e}}{\sqrt{3}})}^{3/4}={e}^{3/8}=\frac{k}{e}$

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

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

Solutions · 1