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.

Column $I$	Column $II$
$(A)$ $f(x) = x\|x\|$	$(p)$ continuous in $(-1, 1)$
$(B)$ $f(x) = \sqrt{\|x\|}$	$(q)$ differentiable in $(-1, 1)$
$(C)$ $f(x) = x + [x]$	$(r)$ strictly increasing in $(-1, 1)$
$(D)$ $f(x) = \|x - 1\| + \|x + 1\|$	$(s)$ not differentiable at least at one point in $(-1, 1)$

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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