A) \[(-\infty ,-1)\cup (-1,\infty )\]
B) \[(-\infty ,\infty )\]
C) \[(0,\infty )\]
D) \[(-\infty ,0\cup (0,\infty )\]
Correct Answer: B
Solution :
Since, \[f(x)=\frac{x}{1+|x|}\] Let \[f(x)=\frac{g(x)}{h(x)}=\frac{x}{1+|x|}\] It is clear that\[g(x)=x\]and\[h(x)=1+|x|\]are differentiate on\[(-\infty ,\infty )\]and \[(-\infty ,0)\cup (0,\infty ),\]respectively. Thus.\[f(x)\]is differentiate on \[(-\infty ,0)\cup (0,\infty )\] For \[x=0,\] \[LHD=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\frac{f(-h)-f(0)}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\frac{-h}{1+h}-0}{-h}=1\] \[RHD=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\frac{h}{1+h}-0}{h}=1\] Thus,\[f(x)\]is differentiable on \[(-\infty ,\infty )\].
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