Question

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

• A. $\frac{1}{e}$
• B. ${\left(\frac{1}{2}\right)}^2$
• C. $e^{\frac{1}{e}}$
• D. ${\left(\frac{1}{e}\right)}^e$

Answer 1

The correct option is C $e^{\frac{1}{e}}$

$\mathrm{Let}y={\left(\frac{1}{x}\right)}^x\Rightarrow \log y=x\log \left(\frac{1}{x}\right)=-x\log x\mathrm{Differentiating}\mathrm{both}\mathrm{sides}w.r.t.x,\mathrm{we}\mathrm{get}\Rightarrow \frac{1}{y}\frac{\mathrm{dy}}{\mathrm{dx}}=-\log x-x\times \frac{1}{x}=-\left(\log x+1\right)\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=-{\left(\frac{1}{x}\right)}^x\left(\log x+1\right)\mathrm{Now},\frac{\mathrm{dy}}{\mathrm{dx}}=0\Rightarrow -{\left(\frac{1}{x}\right)}^x\left(\log x+1\right)=0\Rightarrow \log x+1=0\Rightarrow x=\frac{1}{e}x=\frac{1}{e}\mathrm{is}\mathrm{the}\mathrm{critical}\mathrm{point}\mathrm{and}\mathrm{can}\mathrm{be}\mathrm{represented}\mathrm{on}\mathrm{the}\mathrm{number}\mathrm{line}\mathrm{as}$
$\therefore x=\frac{1}{e}\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{maxima}.\mathrm{Maximum}\mathrm{value}\mathrm{is}{\left(\frac{1}{\frac{1}{e}}\right)}^{\frac{1}{e}}={\left(e\right)}^{\frac{1}{e}}.$