Question

The minimum value of $x\log x$ is equal to

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

Answer 1

Answer: (C)

Let $y=x{\log }_{e}x$
On differentiating w.r.t. $x$, we get
$\frac{dy}{dx}=x\cdot \frac{1}{x}+\log x=(1+\log x)$
Again, differentiating, we get
$\frac{d^{2}y}{d{x}^2}=\frac{1}{x}$
Put, $\frac{dy}{dx}=0$ for maxima or minima.
$\Rightarrow 1+\log x=0$
$\Rightarrow x=\frac{1}{e}$
$\therefore {\left(\frac{d^{2}y}{d{x}^2}\right)}_{\left(x=\frac{1}{c}\right)}=e$
$\therefore y$ is minimum at $x=\frac{1}{e}$
$\therefore y_{\min }=\frac{1}{e}{\log }_e\left(\frac{1}{e}\right)=-\frac{1}{e}$