The local maximum value of the function f(x)= (2 x )^ x^ 2 , x>0,… - Vidyadip

MCQ

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

A
$(2 \sqrt{\mathrm{e}})^{\frac{1}{\mathrm{e}}}$
B
$\left(\frac{4}{\sqrt{\mathrm{e}}}\right)^{\frac{\mathrm{e}}{4}}$
✓
$(\mathrm{e})^{\frac{2}{\mathrm{e}}}$
D
$1$

✓

Answer

Correct option: C.

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

c
$f(x)=\left(\frac{2}{x}\right)^{x^{2}} ; x>0$

$\ell n f(x)=x^{2}(\ell \ln 2-\ell n x)$

$f^{\prime}(x)=f(x)\{-x+(\ell n 2-\ell n x) 2 x\}$

$f^{\prime}(x)=\underbrace{f(x)}_{+} \cdot \underbrace{x}_{+} \underbrace{(2 \ell n 2-2 \ell n x-1)}_{g(x)}$

$g(x)=2 \ell n^{2}-2 \ell n x-1$

$=\ell n \frac{4}{x^{2}}-1=0 \Rightarrow x=\frac{2}{\sqrt{e}}$

$\mathrm{LM}=\frac{2}{\sqrt{\mathrm{e}}}$

Local maximum value $=\left(\frac{2}{2 / \sqrt{\mathrm{e}}}\right)^{\frac{4}{\mathrm{e}}} \Rightarrow \mathrm{e}^{\frac{2}{\mathrm{e}}}$

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