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

$e^{1/e}$

The correct answer is Option (3) → $e^{1/e}$ ##

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

Taking log on both sides

$\Rightarrow \log y = x \cdot \log \frac{1}{x}$

$∴\frac{1}{y} \frac{dy}{dx} = x \cdot \frac{1}{1/x} \left( -\frac{1}{x^2} \right) + \log \frac{1}{x} \cdot 1$

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

$∴\frac{dy}{dx} = \left( \log \frac{1}{x} - 1 \right) \cdot \left( \frac{1}{x} \right)^x \quad \left[ ∵y = \left( \frac{1}{x} \right)^x \right]$

Now, $\frac{dy}{dx} = 0$

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

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

$∴x = \frac{1}{e}$

Hence, the maximum value of $f\left( \frac{1}{e} \right) = (e)^{1/e}$.