Question

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

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

Answer 1

Answer: (C)

Let $y={\left(\frac{1}{x}\right)}^x$
$\Rightarrow logy=xlog\left(\frac{1}{x}\right)$
$\Rightarrow logy=-xlogx$
Differentiating w.r.t. $x$., we get
$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\left(1+logx\right)$
$\Rightarrow \frac{dy}{dx}=-y\left(1+logx\right)$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=-\frac{dy}{dx}\left(1+logx\right)-\frac{y}{x}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=y{\left(1+logx\right)}^2-\frac{y}{x}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}={\left(\frac{1}{x}\right)}^x{\left(1+logx\right)}^2-\frac{1}{x^{\left(x+1\right)}}$
For maximum value, $\frac{dy}{dx}=0$
$-y\left(1+logx\right)=0$
$\Rightarrow 1+logx=0\left(\because y\ne 0\right)$
$\Rightarrow logx=-1$
$\Rightarrow x=e^{-1}$
$\Rightarrow x=1/e$
Also ${\left[\frac{d^{2}y}{d{x}^2}\right]}_{x=\frac{1}{e}}=e^{1/e}{\left(1+log\frac{1}{e}\right)}^2-e^{\left(1/e+1\right)}$
$=e^{1/e}{\left(1-loge\right)}^2-e^{1/e+1}$
$=-e^{1/e+1}<0$
So, $x=1/e$ is a point of local maxima.
Hence, local maximum value $y={\left(e\right)}^{1/e}$.