A) \[{{e}^{2}}\]
B) \[\frac{1}{e}\log \frac{1}{\sqrt{e}}\]
C) \[{{e}^{2}}\log \sqrt{e}\]
D) None of these
Correct Answer: A
Solution :
\[f(x)={{x}^{2}}\log x\] Þ \[{f}'(x)=(2\log x+1)x\] Now \[{f}'(x)=0\] Þ \[x={{e}^{-1/2}},\,0\] \[\because \] \[0<{{e}^{-1/2}}<1\] \[\because \] None of these critical points lies in the interval [1, e] \ So we only complete the value of \[f(x)\] at the end points 1 and e. We have \[f(1)=0,\,f(e)={{e}^{2}}\] \ Greatest value = \[{{e}^{2}}.\]
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