Question

The minimum value of x log_e x is equal to

(a) e

(b) 1/e

(c) $-$1/e

(d) 2/e

(e) $-$e

Answer 1

$$\begin{gathered} \left(\mathrm{c}\right)\frac{-1}{e} \\ \mathrm{Here}, \\ f\left(x\right)=x{\log }_{e}x \\ \Rightarrow f\text{'}\left(x\right)={\log }_{e}x+1 \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow {\log }_{e}x+1=0 \\ \Rightarrow {\log }_{e}x=-1 \\ \Rightarrow x=e^{-1} \\ \mathrm{Now}, \\ f\text{'}\text{'}\left(x\right)=\frac{1}{x} \\ \Rightarrow f\text{'}\text{'}\left(e^{-1}\right)=e>0 \\ \mathrm{So},x=e^{-1}\mathrm{is}\mathrm{a}\mathrm{local}\mathrm{minima}. \\ \mathrm{Hence},\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}f\left(x\right)=f\left(e^{-1}\right). \\ \Rightarrow e^{-1}{\log }_e\left(e^{-1}\right)=-e^{-1}=\frac{-1}{e} \end{gathered}$$