MCQ
If the function $f(x)=\left(\frac{1}{x}\right)^{2 x} ; x>0$ attains the maximum value at $\mathrm{x}=\frac{1}{\mathrm{e}}$ then :
  • A
     $\mathrm{e}^\pi<\pi^{\mathrm{e}}$
  • B
     $\mathrm{e}^{2 \pi}<(2 \pi)^{\mathrm{e}}$
  • $\mathrm{e}^\pi>\pi^{\mathrm{e}}$
  • D
    $(2 \mathrm{e})^\pi>\pi^{(2 \mathrm{e})}$

Answer

Correct option: C.
$\mathrm{e}^\pi>\pi^{\mathrm{e}}$
c
Let $y=\left(\frac{1}{x}\right)^{2 x}$

$ \ell \text { ny }=2 x \ell n\left(\frac{1}{x}\right) $

$ \ell n y=-2 x \ell n x $

$ \frac{1}{y} \frac{d y}{d x}=-2(1+\ell n x)$

for $\mathrm{x}>\frac{1}{\mathrm{e}} \mathrm{f}^{\mathrm{n}}$ is decreasing

so, $ \mathrm{e}<\pi $

$ \left(\frac{1}{\mathrm{e}}\right)^{2 \mathrm{e}}>\left(\frac{1}{\pi}\right)^{2 \pi} $

$ \mathrm{e}^\pi>\pi^{\mathrm{e}}$

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