Question

Let $f(x)=x^2 \log x, x>0$. Then the minimum value of $f$ is

• A. $\frac{1}{\sqrt{e}}$
• B. $2 e$
• C. $-2 e$
• D. $\sqrt{e}$
• E. $\frac{-1}{2 e}$

Answer 1

Answer: (E)

$ f(x)=x^2 \log x, x>0 $
$ f^{\prime}(x)=x^2 \times \frac{1}{x}+\log x \times 2 x $
$ =x(1+2 \log x) $
$ =x\left(1+\log x^2\right) $
$ f^{\prime}(x)=0$
As $x \neq 0,1+\log x^2=0$
$ \log x^2=-1 $
$ x^2=e^{-1} $
$ x=\frac{1}{\sqrt{e}} $
$ f^{\prime \prime}(x)=1+\frac{2}{x} $
$ f^{\prime \prime}\left(\frac{1}{\sqrt{6}}\right)=1+2 \sqrt{e}>0 $
$ \text { Hence, } f(x) \text { is minimum at } x=\frac{1}{\sqrt{6}}$
$ f\left(\frac{1}{\sqrt{e}}\right)=\frac{1}{e} \log e^{-1 / 2}=\frac{-1}{2 e}$