Let `f(x)= {{:((x)/(1+|x|)",", |x| ge1), ((x)/(1-|x|)",", |x| lt 1):},` then domain of `f'(x)` is:

Options

A. discontinuous and non-differentiable at `x=-1,1,0`
B. discontinuous and non-differentiable at x=-1, whereas continuous and differentiable at x=0,1
C. discontinuous and non-differentiable at x=-1,1 wheras continuous and differentiable at x=0.
D. none of these

Answer

C

Explanation

We have
`f(x)={{:(,(x)/(1+|x|),|x|ge1),(,(x)/(1-|x|),|x| lt 1):}`
`Rightarrow f(x)={{:(,(x)/(1-x),x le -1),(,(x)/(1+x'),-1 lt x lt 0),(,(x)/(1-x),0 lt x lt 1),(,(x)/(1+x),x ge 1):}`
Clearly, f(x) is discontinuous and hence and non-differerntiable at `x=pm1`.
It can be easily seen that f(s) is continuous and differential at x=0