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JEE Main 20246 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2024 · 6 April · Evening Shift · Mathematics · Application of Derivatives

If the function f(x)=((1)/(x))^ 2 x ; x>0 attains the maximum value...

If the function f(x)=(1x)2x;x>0f(x)=\left(\frac{1}{x}\right)^{2 x} ; x>0 attains the maximum value at x=1ex=\frac{1}{\mathrm{e}} then :

Options

  1. Aeπ<πe\mathrm{e}^\pi < \pi^{\mathrm{e}}
  2. Beπ>πe\mathrm{e}^\pi>\pi^{\mathrm{e}}
  3. C(2e)π>π(2e)(2 e)^\pi>\pi^{(2 e)}
  4. De2π<(2π)e\mathrm{e}^{2 \pi} < (2 \pi)^{\mathrm{e}}

Correct answer

B. eπ>πe\mathrm{e}^\pi>\pi^{\mathrm{e}}

Step-by-step solution

Let y=(1x)2xy=\left(\frac{1}{x}\right)^{2 x}ℓ ny =2xℓn(1x)ℓ ny =−2xℓln⁡x1ydydx=−2(1+ℓnx)\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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