A) \[\left( -\infty ,\infty \right)\]
B) \[\left( 0,\infty \right)\]
C) \[\left( -\infty ,0 \right)\cup \left( 0,\infty \right)\]
D) none of these
Correct Answer: A
Solution :
\[\left| x \right|=x,x>0,\left| x \right|=-x,x<0,\left| x \right|=0,x=0\]\[\therefore f\left( x \right)=\frac{x}{1-x},x<0;\] \[f\left( x \right)=0,x=0\,\,and\,\,\,f\left( x \right)=\frac{x}{1+x},x>0\]\[\operatorname{LHD}\,\,at\,\,\left( x=0 \right)=\underset{h\to 0}{\mathop{lim}}\,\frac{f\left( 0-h \right)-f\left( 0 \right)}{-h}\] \[=\underset{h\to 0}{\mathop{lim}}\,\frac{\frac{-h}{1+h}-0}{-h}=1\]\[\operatorname{RHD}\,at\,\left( x=0 \right)=\underset{h\to 0}{\mathop{lim}}\,\frac{f\left( 0+h \right)-f\left( 0 \right)}{h}\]\[=\underset{h\to 0}{\mathop{lim}}\,\frac{\frac{h}{1+h}-0}{h}=1\] \[\therefore \] fis differentiable at x=0. Also, for all other values of\[\operatorname{x}\ne 0\], the function is differentiable. Hence, the function is differentiable in\[\left( -\infty ,\infty \right)\]
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