JEE MAIN - Mathematics (2024 - 6th April Evening Shift - No. 1)
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 :
$$\mathrm{e}^\pi<\pi^{\mathrm{e}}$$
$$\mathrm{e}^{2 \pi}<(2 \pi)^{\mathrm{e}}$$
$$(2 e)^\pi>\pi^{(2 e)}$$
$$\mathrm{e}^\pi>\pi^{\mathrm{e}}$$
Explanation
$$f\left(\frac{1}{\pi}\right)< f\left(\frac{1}{e}\right) \quad \text { as } \frac{1}{\pi}<\frac{1}{e}$$
$$\begin{aligned} & \Rightarrow\left(\frac{1}{1}\right)^{\frac{2}{\pi}}<\left(\frac{1}{\frac{1}{e}}\right)^{\frac{2}{e}} \\ & \Rightarrow(\pi)^{\frac{2}{\pi}}<(e)^{\frac{2}{e}} \\ & \Rightarrow \pi^e < e^\pi \end{aligned}$$
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