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If the function $f(x)=\left(\frac{1}{x}\right)^{2 x} ; x>0$ attains the maximum value at $x=\frac{1}{\mathrm{e}}$ then :
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Verified Answer
The correct answer is:
$\mathrm{e}^\pi>\pi^{\mathrm{e}}$
Let $y=\left(\frac{1}{x}\right)^{2 x}$
$\begin{aligned}
& \ell \text { ny }=2 x \ell n\left(\frac{1}{x}\right) \\
& \ell \text { ny }=-2 x \ell \ln x \\
& \frac{1}{y} \frac{d y}{d x}=-2(1+\ell n x)
\end{aligned}$
for $\mathrm{x}>\frac{1}{\mathrm{e}} \mathrm{f}^{\mathrm{n}}$ is decreasing
$\text {so, } \mathrm{e} < \pi$
$\begin{aligned}
& \left(\frac{1}{\mathrm{e}}\right)^{2 \mathrm{e}}>\left(\frac{1}{\pi}\right)^{2 \pi} \\
& \mathrm{e}^\pi>\pi^{\mathrm{e}}
\end{aligned}$
$\begin{aligned}
& \ell \text { ny }=2 x \ell n\left(\frac{1}{x}\right) \\
& \ell \text { ny }=-2 x \ell \ln x \\
& \frac{1}{y} \frac{d y}{d x}=-2(1+\ell n x)
\end{aligned}$
for $\mathrm{x}>\frac{1}{\mathrm{e}} \mathrm{f}^{\mathrm{n}}$ is decreasing
$\text {so, } \mathrm{e} < \pi$
$\begin{aligned}
& \left(\frac{1}{\mathrm{e}}\right)^{2 \mathrm{e}}>\left(\frac{1}{\pi}\right)^{2 \pi} \\
& \mathrm{e}^\pi>\pi^{\mathrm{e}}
\end{aligned}$
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