If f(x)=xe−1|x|+1x,x≠00,x=0 then f is - Infinity Learn by Sri Chaitanya

A
discontinuous everywhere
B
continuous and differentiable at $x = 0$
C
differentiable everywhere
D
continuous but not differentiable at x= 0

Solution:

$f (x) = \{\begin{array}{cc} x & x < 0 \\ x e^{- \frac{2}{x}} & x > 0 \\ 0 & x = 0 \end{array}$

$f$ is clearly a continuous function

$\begin{array}{l} f^{'} (x) = \{\begin{array}{cc} 1 & , x < 0 \\ e^{- \frac{2}{x}} + \frac{2}{x} e^{- \frac{2}{x}} & , x > 0 \end{array} \\ lim_{h \to 0^{-}} \frac{f (0 + h) - f (0)}{h} = lim_{h \to 0^{-}} \frac{h}{h} = 1 \\ lim_{h \to 0^{+}} \frac{f (0 + h) - f (0)}{h} = lim_{h \to 0^{+}} \frac{h e^{- \frac{2}{h}}}{h} = lim_{h \to 0^{+}} e^{- \frac{2}{h}} = 0 \end{array}$

$f$ is not differentiable at $x = 0 .$

If $f (x) = \{\begin{array}{cc} x e^{- \{\frac{1}{| x |} + \frac{1}{x}\}} & , x \neq 0 \\ 0 & , x = 0 \end{array}$ then $f$ is

Solution:

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