\(y=[x]+|x-2|\) if number of discontinuous point are " \(p\) " and number of non-differentiable points are " \(q\) " the…

\(
f(x)= \begin{cases}-x ; & x \in(-2,-1) \\ 1-x ; & x \in[-1,0) \\ 2-x ; & x \in[0,1) \\ 3-x ; & x \in[1,2) \\ x ; & x \in[2,3)\end{cases}
\)

As we can see that function is discontinuous at \(-1,0,1\) and \(2\) . Therefore, function is non-differentiable at \(-1,0,1\) and 2.
Therefore, \(\mathrm{p}=4\) and \(\mathrm{q}=4\).
Then, \(\mathrm{p}+\mathrm{q}=8\).