Let  $f\left(x\right)=\{\begin{array}{l}\frac{5e^{1/x}+2}{3-e^{1/x}},x\ne 0\\ 0,x=0\end{array}$. Then match the following lists. 

	List-I		List-I
a)	$y=f\left(x\right)$  is	p)	continuous at x = 0
b)	$y=xf\left(x\right)$ is	q)	discontinuous at x = 0
c)	$y=x^{2}f\left(x\right) is$	r)	differentiable at x = 0
d)	$y=x^{-1}f\left(x\right) is$	s)	non-differentiable at x = 0

$$\begin{gathered} a) f\left(x\right)=\{\begin{array}{cc}\frac{5e^{1/x}+2}{3-e^{1/x}},& x\ne 0\\ 0,& x=0\end{array} \\ f\left(0^{+}\right)=\underset{h\to 0}{\lim }\frac{5e^{1/h}+2}{3-e^{1/h}} \\ =\underset{h\to 0}{\lim }\frac{5+2e^{-1/h}}{3e^{-1/h}-1}=-5 \end{gathered}$$

Hence,  $f\left(x\right)$ is discontinuous and non-differentiable at $x=0$

$$\begin{gathered} b) g\left(x\right)=xf\left(x\right)=\{\begin{array}{cc}x\frac{5e^{1/x}+2}{3-e^{1/x}},& x\ne 0\\ 0,& x=0\end{array} \\ f\left(0^{+}\right)=\underset{h\to 0}{\lim }h\frac{5e^{1/h}+2}{3-e^{1/h}} \\ =\underset{h\to 0}{\lim }h\frac{5+2e^{1/h}}{3e^{-1/h}-1}=0\times (-5)=0 \\ f\left(0^{-}\right)=\underset{h\to 0}{\lim }h\frac{5e^{-1/h}+2}{3-e^{-1/h}}=0\times (2/3)=0 \end{gathered}$$.

Hence, $f\left(x\right)$ is continuous at $x=0$

$$\begin{gathered} Rg\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{g(0+h)-g\left(0\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{g\left(h\right)-0}{h} \\ =\underset{h\to 0}{\lim }f\left(h\right)=\underset{h\to 0}{\lim }\frac{5e^{1/h}+2}{3-e^{1/h}} \end{gathered}$$

$$\begin{gathered} =\underset{h\to 0}{\lim }\frac{5+2e^{-1/h}}{3e^{-1/h}-1} \\ =\frac{5+0}{0-1}=-5 \\ \therefore LG\text{'}\left(0\right)\ne RF\text{'}\left(0\right) \end{gathered}$$

Hence,  $f\left(x\right)$ is not differentiable, but continuous at $x=0$.
c) For  $x^{2}f\left(x\right),$ Let  $F\left(x\right)=x^{2}f\left(x\right)$
clearly, $F\left(x\right)=x\left(xf\left(x\right)\right)$  is continuous at $x=0$
 $$\begin{gathered} \therefore LF\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{F(0-h)-F\left(0\right)}{-h} \\ =\underset{h\to 0}{\lim }\frac{h^{2}f(-h)-0}{-h}=0 \\ RF\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{F(0+h)-F\left(0\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{h^{2}f\left(h\right)-0}{h}=0 \\ \therefore LF\text{'}\left(0\right)=RF\text{'}\left(0\right) \end{gathered}$$

Hence, $F\left(x\right)$ is differentiable at $x=0$
d) Clearly, from the above discussion, $y=x^{-1}f\left(x\right)$ is discontinuous and, hence, non-differentiable at $x=0$