Match the items in Column-I with those in Column-II and select the correct option.

Column-I	Column-Ii
a) $f\left(x\right)=x^2.\log x$	p) f(x) has one point of minima
b) $f\left(x\right)=x.{\log }_{e}x$	q) f(x) has one point of maxima
c) $f\left(x\right)=\frac{\log x}{x}$	r) f(x) increases in $(0,e)$
d) $f\left(x\right)=x^{-x}$	s) f(x) decreases in $(0,\frac{1}{e})$

a) $f\left(x\right)=x^2\log x\to f\left(x\right)=x(2\log x+1)=0\Rightarrow x=\frac{1}{\sqrt{e}}$ which is the point of minima as derivatives changes in from negative to positive

The function increases in $(0,\frac{1}{\sqrt{e}})$

b) $y=x.\log x\Rightarrow \frac{dy}{dx}=x\frac{1}{x}+1.\log x=1+\log x\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{1}{x}$

$\frac{dy}{dx}=0\Rightarrow \log x=-1\Rightarrow x=\frac{1}{e}\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{1}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$e$}\right.}=e>0$ at $x=\frac{1}{e}$

y is min for $x=\frac{1}{e}$

c) $f\left(x\right)=\frac{\log x}{x},f^{\text{'}}\left(x\right)=\frac{1-\log x}{x^2}=0\Rightarrow x=e,$Derivative changes sign from positive to negative at x=e, hence it is the point of maxima.

d) $f\left(x\right)=x^{-x},f^{\text{'}}\left(x\right)=-x^{-x}(1+\log x)=0\Rightarrow x=\frac{1}{e}$, which is clearly point of maxima.