JEE Main 202126 Aug, Evening shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002The local maximum value of the function f(x)=(2x)x2f(x) = {\left( {{2 \over x}} \right)^{{x^2}}}f(x)=(x2)x2, x > 0, isAOption A. (2e)1e{\left( {2\sqrt e } \right)^{{1 \over e}}}(2e)e1BOption B. (4e)e4{\left( {{4 \over {\sqrt e }}} \right)^{{e \over 4}}}(e4)4eCOption C. (e)2e{(e)^{{2 \over e}}}(e)e2DOption D. 1CheckSkip · show solution
