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)

$\sqrt[e]{e}$

Answer 2

$y =\left(\frac{1}{x}\right)^{2x^2} \Rightarrow \log y =2x^{2} \log \frac{1}{x}$
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, $\frac{d^{2}y}{dx^{2}}|_{x-e^{-1 /e}} <0$