JEE Main 20246 Apr, Evening shiftMathematicsEasySets, Relations and Functions168× since 2002If the function f(x)=(1x)2x;x>0f(x)=\left(\frac{1}{x}\right)^{2 x} ; x>0f(x)=(x1)2x;x>0 attains the maximum value at x=1ex=\frac{1}{\mathrm{e}}x=e1 then :AOption A. eπ<πe\mathrm{e}^\pi<\pi^{\mathrm{e}}eπ<πeBOption B. e2π<(2π)e\mathrm{e}^{2 \pi}<(2 \pi)^{\mathrm{e}}e2π<(2π)eCOption C. (2e)π>π(2e)(2 e)^\pi>\pi^{(2 e)}(2e)π>π(2e)DOption D. eπ>πe\mathrm{e}^\pi>\pi^{\mathrm{e}}eπ>πeCheckSkip · show solution
