The local maximum value of the function $f(x)=\left(\frac{2}{x}\right)^{x^{2}}, x>0$, is:

(B) $(\mathrm{e})^{\frac{2}{\mathrm{e}}}$

(A) $(2\sqrt{\mathrm{e}})^{\frac{1}{\mathrm{e}}}$

(C) $\left(\frac{4}{\sqrt{\mathrm{e}}}\right)^{\frac{\mathrm{e}}{4}}$

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Application of Derivatives

The local maximum value of the function

f (x) = (\frac{2}{x})^{x^{2}}, x > 0

, is:

Answer: (B)

Taking logarithms,

ln f (x) = x^{2} (ln 2 - ln x)

for

x > 0

. Differentiating,

\frac{f ^{'} ( x )}{f ( x )} = 2 x (ln 2 - ln x) - x = x [2 ln \frac{2}{x} - 1]

f (x) > 0

throughout,

f^{'} (x) = 0

requires

ln \frac{2}{x} = \frac{1}{2}

, that is

x = \frac{2}{e}

. The bracket

2 ln \frac{2}{x} - 1

is strictly decreasing in

x

, so

f^{'}

is positive to the left of this point and negative to its right, which makes it a maximum. At that point

\frac{2}{x} = e

and

x^{2} = \frac{4}{e}

, so

f (\frac{2}{e}) = (e)^{4/ e} = e^{2/ e}

The local maximum value is

e^{2/ e}

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option