Let f(x) = begin{cases} frac{5 e^{1/x} + 2}{3 | vedclass

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(C) Given $f(x) = \begin{cases} \frac{5 e^{1/x} + 2}{3 - e^{1/x}}, & x 
eq 0 \ 0, & x = 0 \end{cases}$.Then $x f(x) = \begin{cases} \frac{x(5 e^{1/x

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Continuous and not differentiable

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(C) Given $f(x) = \begin{cases} \frac{5 e^{1/x} + 2}{3 - e^{1/x}}, & x 
eq 0 \ 0, & x = 0 \end{cases}$.Then $x f(x) = \begin{cases} \frac{x(5 e^{1/x} + 2)}{3 - e^{1/x}}, & x 
eq 0 \ 0, & x = 0 \end{cases}$.Check continuity of $x f(x)$ at $x = 0$:$	ext{L.H.L} = \lim_{h 	o 0^+} \frac{(-h)(5 e^{-1/h} + 2)}{3 - e^{-1/h}} = \frac{0(0 + 2)}{3 - 0} = 0$.$	ext{R.H.L} = \lim_{h 	o 0^+} \frac{h(5 e^{1/h} + 2)}{3 - e^{1/h}} = \lim_{h 	o 0^+} \frac{h e^{1/h}(5 + 2e^{-1/h})}{e^{1/h}(3e^{-1/h} - 1)} = \lim_{h 	o 0^+} \frac{h(5 + 0)}{0 - 1} = 0$.Since $	ext{L.H.L} = 	ext{R.H.L} = f(0) = 0$,$x f(x)$ is continuous at $x = 0$.Check differentiability of $f(x)$ at $x = 0$:$	ext{L.H.D} = \lim_{h 	o 0^+} \frac{f(-h) - f(0)}{-h} = \lim_{h 	o 0^+} \frac{\frac{5 e^{-1/h} + 2}{3 - e^{-1/h}} - 0}{-h} = \lim_{h 	o 0^+} \frac{2}{3(-h)} = -\infty$.Since the limit is not finite,$f(x)$ is not differentiable at $x = 0$.Thus,$x f(x)$ is continuous and $f(x)$ is not differentiable.

List-$I$	List-$II$
$A$. If $y = \|x\| + \|x - 2\|$, then at $x = 2$, $\frac{dy}{dx} =$	$I$. $2$
$B$. If $f(x) = \|\cos 2x\|$, then $f'(\frac{\pi}{4} +) =$	$II$. $0$
$C$. If $f(x) = \sin(\pi[x])$, where $[x]$ is the greatest integer function, then $f'(1-) =$	$III$. $-2$
$D$. If $f(x) = \log\|x - 1\|$, $x \neq 1$, then $f'(\frac{1}{2}) =$	$IV$. does not exist

List-$I$	List-$II$
$(A) \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$	$(I) \cos x-\sin x$
$(B) \tan ^{-1}\left(\frac{1-x}{1+x}\right)$	$(II) \frac{-1}{1+x^2}$
$(C) e^{\log (\sin x+\cos x)}$	$(III) \frac{2}{1+x^2}$
$(D) \sqrt{1-\sin 2 x} \text{ for } (0 < x < \frac{\pi}{4})$	$(IV) \cos x+\sin x$
	$(V) -\sin x-\cos x$

Let $f(x) = \begin{cases} \frac{5 e^{1/x} + 2}{3 - e^{1/x}}, & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then at $x = 0$,$x f(x)$ and $f(x)$ are respectively:

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