A) \[(-\infty ,\infty )\] only
B) \[(0,\infty )\] only
C) \[(-\infty ,0)\cup (0,\infty )\]only
D) \[(-\infty ,0)\] only
Correct Answer: A
Solution :
| [a] Given \[f(x)=\frac{x}{1+\left| x \right|}\] |
| \[=\left\{ \begin{matrix} \frac{x}{1-x},x<0 \\ \frac{x}{1+x},x\ge 0 \\ \end{matrix} \right.\] |
| \[\left( \because \left| x \right|=\left\{ \begin{matrix} x & if & x\ge 0 \\ -x & if & x<0 \\ \end{matrix} \right. \right)\] |
| \[\therefore LHD=f'({{0}^{-}})=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0-h)-f(0)}{-h}\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(-h)-f(0)}{-h}\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\frac{-h}{1+\left| -h \right|}-0}{-h}\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\frac{-h}{1+h}-0}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{1+h}=1\] |
| and RHD \[=f'({{0}^{+}})\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h}\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\frac{h}{1+h}-0}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{1+h}=1\] |
| Since, LHD = RHD |
| \[\therefore f(x)\] is differentiable at \[x=0\] |
| Hence, \[f(x)\] is differentiable in \[(-\infty ,\infty )\]. |
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