Pls answer tuis ques fast..

Solution
Let $y={\left(\frac{1}{x}\right)}^{2x^2}$

$\Rightarrow \log y=-2x^2\log x$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=-4x\log x-2x$

$\Rightarrow \frac{dy}{dx}=-2x\left(2\log x+1\right)y$

$\Rightarrow \frac{dy}{dx}=-2x\left(\log x^2+1\right){\left(\frac{1}{x}\right)}^{2x^2}$

Now, the value of y is maximum or minimum when $\frac{dy}{dx}=0$
That is, when:-

$-2x\left(\log x^2+1\right){\left(\frac{1}{x}\right)}^{2x^2}=0$

$\Rightarrow \frac{4x\log x+2x}{x^{2x^2}}=0$

$$\begin{gathered} \Rightarrow 2x=-4x\log x \\ \Rightarrow \log x=-\frac{1}{2} \end{gathered}$$

$\Rightarrow x=\frac{1}{\sqrt{e}}$

Thus, the maximum value of y is:-

${\left({\displaystyle \frac{1}{ {\displaystyle \frac{1}{\sqrt{e}}} }}\right)}^{2{\left(\frac{1}{\sqrt{e}} \right)}^2}$

$={\left(\sqrt{e}\right)}^{\frac{2}{e}}$

$=e^{\frac{1}{e}}=\sqrt[e]{e}$
= (2) option