Question

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

• A. $e$
• B. $\sqrt[e]{ee}$
• C. $1$
• D. none of these

Answer 1

The correct option is B $\sqrt[e]{ee}$

Let
$y=(\frac{1}{x}{)}^{2x^2}$
Or
$\ln y=-2x^2\ln x$
Now
$y^{\prime }=-y[4x\ln x+2x]$

$=-\left(\frac{1}{x}{)}^{2x^2}\right(4x\ln x+2x)$

Now
$(\frac{1}{x}{)}^{2x^2}$ cannot be equal to zero.

Hence $4x\ln \left(x\right)+2x=0$
Or
$2x\left(2\ln \right(x)+1)=0$
Or
$x=0$ or $\ln x=\frac{-1}{2}$ $\Rightarrow x=e^{-\frac{1}{2}}$
Now $f\left(0\right)$ is not defined.
Hence
$y=e^{\frac{1}{e}}$.