Mathematics · 2021
JEE Main · 26 August 2021, Shift 2 · Q66
The local maximum value of the function f(x)=(2/x)^x^2, x>0, is:
The local maximum value of the function
$\displaystyle f(x)=\left(\frac{2}{x}\right)^{x^2}, x>0$, is :
Official answer
(2)
$\displaystyle (\mathrm{e})^{\frac{2}{\mathrm{e}}}$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.