Verified Solution • 2025–26 Session moderate

QJEEM24Maa20e: JEE Main 2024 Mathematics - Class 11 JEEM24Maa20e Solution: If the function attains the maximum value at then :

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

Multiple Choice Options

  • A eπ<πe\mathrm{e}^\pi<\pi^{\mathrm{e}}
  • B e2π<(2π)e\mathrm{e}^{2 \pi}<(2 \pi)^{\mathrm{e}}
  • C (2e)π>π(2e)(2 e)^\pi>\pi^{(2 e)}
  • D eπ>πe\mathrm{e}^\pi>\pi^{\mathrm{e}}

Step-by-Step Verified Solution

f\left(\frac{1}{\pi}\right)&lt; f\left(\frac{1}{e}\right) \quad \text { as } \frac{1}{\pi}&lt;\frac{1}{e}

amp;⇒(11)2πlt;(11e)2eamp;⇒(π)2πlt;(e)2eamp;⇒πelt;eπ\begin{aligned}&amp; \Rightarrow\left(\frac{1}{1}\right)^{\frac{2}{\pi}}&lt;\left(\frac{1}{\frac{1}{e}}\right)^{\frac{2}{e}} \\&amp; \Rightarrow(\pi)^{\frac{2}{\pi}}&lt;(e)^{\frac{2}{e}} \\&amp; \Rightarrow \pi^e &lt; e^\pi\end{aligned}

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