The maximum value

Solution:

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

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

$\Rightarrow \log f(x)=\log \left(\frac{1}{x}\right)^{x}$

$\Rightarrow \log f(x)=x \log \left(\frac{1}{x}\right)$

$\Rightarrow \log f(x)=-x \log x$

Differentiating both sides with respect to x, we get

$\frac{1}{f(x)} \times f^{\prime}(x)=-\left(x \times \frac{1}{x}+\log x \times 1\right)$

$\Rightarrow f^{\prime}(x)=-\left(\frac{1}{x}\right)^{x}(1+\log x)$                .....(1)

For maxima or minima,

$f^{\prime}(x)=0$

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

$\Rightarrow 1+\log x=0 \quad\left[\left(\frac{1}{x}\right)^{x}>0\right]$

$\Rightarrow \log x=-1$

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

Now,

$f^{\prime \prime}(x)=-\left[\left(\frac{1}{x}\right)^{x} \times \frac{1}{x}+(1+\log x) \times\left\{-\left(\frac{1}{x}\right)^{x}(1+\log x)\right\}\right]$            [Using (1)]

$\Rightarrow f^{\prime \prime}\left(\frac{1}{e}\right)=-\left[(e)^{\frac{1}{e}} \times e+\left(1+\log \frac{1}{e}\right) \times\left\{-(e)^{\frac{1}{e}}\left(1+\log \frac{1}{e}\right)\right\}\right]$                $\left(\log \frac{1}{e}=\log e^{-1}=-1\right)$

$\Rightarrow f^{\prime \prime}\left(\frac{1}{e}\right)=-e^{\frac{1}{e}+1}-0=-e^{\frac{1}{e}+1}$

$\Rightarrow f^{\prime \prime}\left(\frac{1}{e}\right)<0$

So, $x=\frac{1}{e}$ is the point of local maximum of $f(x)$.

$\therefore$ Maximum value of $f(x)=\left(\frac{1}{\frac{1}{e}}\right)^{\frac{1}{e}}=e^{\frac{1}{e}}$

Thus, the maximum value of $\left(\frac{1}{x}\right)^{x}$ is $e^{\frac{1}{e}}$.

Hence, the correct answer is option (c).