A) Continuous and differentiable at \[x=0\]
B) Neither continuous nor differentiable at \[x=0\]
C) Continuous but not differentiable at \[x=0\]
D) Not continuous but differentiable at \[x=0\]
Correct Answer: C
Solution :
We have, \[f(x)=\left\{ \begin{matrix} {{e}^{-x}}, & x\ge 0 \\ {{e}^{x}}, & x<0 \\ \end{matrix} \right.\] Clearly, \[f(x)\] is continuous and differentiable for all non zero x. Now \[\underset{x\to 0-}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,{{e}^{x}}=1\],\[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x){{e}^{-x}}=1\] Also, \[f(0)={{e}^{0}}=1\]. So, \[f(x)\] is continuous for all x. (LHD at \[x=0)\] \[={{\left( \frac{d}{dx}({{e}^{x}}) \right)}_{x=0}}=1\] (RHD at \[x=0)\] \[f(x)\] So, \[\underset{x\to 7}{\mathop{\lim }}\,\frac{2-\sqrt{x-3}}{{{x}^{2}}-49}\] is not differentiable at \[L\,{f}'\,(1)=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{f(1-h)-f(1)}{-h}\]. Hence \[f(x)={{e}^{-\,|\,x\,|}}\] is everywhere continuous but not differentiable at \[x=0\].
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