A) \[(-\infty ,\infty )\]
B) \[[0,\infty ]\]
C) \[(-\infty ,\,0)\cup (0,\infty )\]
D) \[(0,\infty )\]
Correct Answer: A
Solution :
Let \[h(x)=x,x\in (-\infty ,\infty )\]; \[g(x)=1+|x|,\,\,x\in (-\infty ,\infty )\] Here h is differentiable in \[(-\infty ,\infty )\] but \[|x|\] is not differentiable at \[x=0\]. Therefore g is differentiable in \[(-\infty ,0)\cup (0,\infty )\] and \[g(x)\ne 0,x\in \] \[(-\infty ,\infty )\], therefore \[f(x)=\frac{h(x)}{g(x)}=\frac{x}{1+|x|}\] It is differentiable in \[(-\infty ,0)\cup (0,\infty )\] for \[x=0\] \[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)-f(0)}{h-0}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\frac{h}{1+|h|}-0}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{1+|h|}=1\] Therefore f is differentiable at \[x=0\], so f is differentiable in \[(-\infty ,\infty )\].
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