Question

The maximum value of ${\left(\frac{1}{x}\right)}^{2x^2}$ is

• A. 1
• B. 9
• C. $e^{\frac{1}{e}}$
• D. None of these

Answer 1

Answer: (C)

$e^{\frac{1}{e}}$

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

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

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

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

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

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

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

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

$\Rightarrow$ $2x=-4x\log x$

$\Rightarrow$ $\log x=\frac{-1}{2}$

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

Its double derivative will be negative there hence value will be maximum

Thus, the maximum value of $y$ is,

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

$\Rightarrow$ $(\sqrt{e}{)}^{\frac{2}{e}}$

$\Rightarrow$ $e^{\frac{1}{2}\times \frac{2}{e}}$

$\Rightarrow$ $e^{\frac{1}{e}}$