Question

The maximum value of ${\left(\frac{1}{x}\right)}^x,(x>0)$ is

• A. $e$
• B. $(e{)}^{\frac{1}{e}}$
• C. ${\left(\frac{1}{e}\right)}^e$
• D. $\frac{1}{e}$

Answer 1

The correct option is B $(e{)}^{\frac{1}{e}}$

$y={\left(\frac{1}{x}\right)}^x$
$\Rightarrow \log y=-x\log x$
Differentiate w.r.t. $x$
$\Rightarrow \frac{1}{y}\cdot y^{\prime }=-[x\cdot \frac{1}{x}+\log x]$
$\Rightarrow y^{\prime }=-y(1+\log x)$
From $y^{\prime }=0$
$\Rightarrow \log x=-1\Rightarrow x=e^{-1}(\therefore y\ne 0)$
Since, sign of $f^{\prime }\left(x\right)$ changes from positive to negative as $x$ crosses $e^{-1}$ from left to right, therefore $x=e^{-1}$ is a point of local maxima.
There is only one critical point of local maxima.
Hence, function is maximum at $x=e^{-1}$
Maximum value is $f\left(e^{-1}\right)=(e{)}^{\frac{1}{e}}$