Question

At $x=\frac{1}{\sqrt{e}}$, the function $x^2\log \frac{1}{x}$ has maximum value.

• A. True
• B. False

Answer 1

The correct option is A True

We have,

$f\left(x\right)=x^2\log \frac{1}{x}$ $............\left(1\right)$

On differentiating w.r.t $x$, we get

$f^{\prime }\left(x\right)=2x\log \frac{1}{x}+x^2\times x\times \frac{-1}{x^2}$

$f^{\prime }\left(x\right)=2x\log \frac{1}{x}-x$ $............\left(2\right)$

Put $f^{\prime }\left(x\right)=0$ for maximum and minimum value

$2x\log \frac{1}{x}-x=0$

$x(2\log \frac{1}{x}-1)=0$

$2\log \frac{1}{x}=1$

$\log \frac{1}{x}=\frac{1}{2}$

$\frac{1}{x}=e^{\frac{1}{2}}$

$x=\frac{1}{\sqrt{e}}$ $.............\left(3\right)$

Again on differentiating w.r.t $x$, we get

$f"\left(x\right)=2\log \frac{1}{x}+2x\times x\times \frac{-1}{x^2}-1$

$f"\left(x\right)=2\log \frac{1}{x}-2-1$

$f"\left(x\right)=2\log \frac{1}{x}-3$

On putting $x=\frac{1}{\sqrt{e}}$, we get

$f"\left(\frac{1}{\sqrt{e}}\right)=2\log \sqrt{e}-3$

$f"\left(\frac{1}{\sqrt{e}}\right)=\log e-3$

$f"\left(\frac{1}{\sqrt{e}}\right)=1-3=-2<0$

$f"\left(x\right)<0$ at $x=\frac{1}{\sqrt{e}}$

Therefore,

$f\left(x\right)$ is maximum at $x=\frac{1}{\sqrt{e}}$

Hence, this is the answer.