Question

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

• A. True
• B. False

Answer 1

The correct option is A True

$y={\left(\right(\frac{1}{x}\left)\right)}^{x}logy=xlog\left(\frac{1}{x}\right)\left(\frac{1}{y}\right)\left(\frac{dy}{dx}\right)=\left(\frac{d(-xlogx)}{dx}\right)\left(\frac{1}{y}\right)\left(\frac{dy}{dx}\right)=-[logx+x\times (\frac{1}{x}\left)\right]\left(\frac{dy}{dx}\right)=-y[logx+1]=0forpointofinclinationlogx+1=0logx=-1\therefore x=\left(\frac{1}{e}\right)\therefore maximumvalue={\left(\right(\frac{1}{\left(\frac{1}{e}\right)}\left)\right)}^{\left(\frac{1}{e}\right)}=e^{\left(\frac{1}{e}\right)}$