A) \[x=e\]
B) \[x=\frac{1}{e}\]
C) \[x=1\]
D) \[x=\sqrt{e}\]
Correct Answer: B
Solution :
Let \[y={{x}^{x}}\]Þ \[\log y=x.\log x,\ \ \ \ (x>0)\] Differentiating \[\frac{dy}{dx}={{x}^{x}}(1+\log x)\]; \[\therefore \frac{dy}{dx}=0\] Þ \[\log x=-1\]Þ\[x={{e}^{-1}}=\frac{1}{e}\] \[\therefore \]Stationary point is \[x=\frac{1}{e}\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}={{x}^{x}}{{(1+\log x)}^{2}}+{{x}^{x}}.\frac{1}{x}\] When \[x=\frac{1}{e},\ \ \frac{{{d}^{2}}y}{d{{x}^{2}}}={{\left( \frac{1}{e} \right)}^{(1/e)-1}}>0\] Therefore y is minimum at \[x=\frac{1}{e}\]and minimum value \[={{\left( \frac{1}{e} \right)}^{1/e}}={{e}^{-1/e}}\].
You need to login to perform this action.
You will be redirected in
3 sec
