Question

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

• A. $e^{-1/2}$
• B. $\sqrt[e]{e}$
• C. $1$
• D. $e^2$

Answer 1

Answer: (B)

$y={\left(\frac{1}{x}\right)}^{2x^2}\Rightarrow \log y=2x^2\log \frac{1}{x}$
Differentiate w.r.t. x, we get
$\Rightarrow \frac{1}{y}.\frac{dy}{dx}=4x\log \frac{1}{x}+2x^2\left(-\frac{1}{x}\right)$
For maxima and minima
$\frac{dy}{dx}=0\Rightarrow -4x\log x-2x=0$
$\Rightarrow 2\log x+1\Rightarrow x=e^{-\frac{1}{e}}$
Now, $\frac{d^{2}y}{d{x}^2}{|}_{x-e^{-1/e}}<0$