If $f\left(x\right)=|x-1|.(\left[x\right]-[-x]),$   then where [.] represents  greatest integer function). 

$f\left(x\right)=|x-1|.(\left[x\right]-[-x])$

$\left[x\right]-[-x]$ is discontinuous at x = 1 

But  $g\left(x\right)=|x-1|$ is continuous and 

$g\left(1\right)=0Hence\left(f\right)iscontinuousatx=1Also,$

$f\left(x\right)=\left\{\begin{array}{ccc}(1-x)(0-(-1)),& x\in & (0,1)\\ 0(1-(-1)),& x=1& \\ (x-1)(1-(-2)),& x\in & (1,2)\end{array}\right.\begin{array}{} \\ \\ \end{array}$

$=\left\{\begin{array}{cc}1-x,& x\in (0,1]\\ 3(x-1),& x\in (1,2)\end{array}\right.$

  Clearly f(x) is continuous at x = 1 and non-differentiable at x = 1 as 

 $f\text{'}\left(1^{-}\right)=-1andf\text{'}\left(1^{+}\right)=3$