Question

The maximum value of ${\left(\frac{1}{x}\right)}^x$ is
(a) e (b) e^e (c) e^1/e (d) ${\left(\frac{1}{e}\right)}^{1/e}$

Answer 1

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

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

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

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

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

Differentiating both sides with respect to x, we get

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

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

For maxima or minima,

$f\text{'}\left(x\right)=0$

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

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

$\Rightarrow \log x=-1$

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

Now,

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

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

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

$\Rightarrow f\text{'}\text{'}\left(\frac{1}{e}\right)<0$

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

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

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

Hence, the correct answer is option (c).