If $f\left(x\right)=\left\{\begin{array}{l}\begin{array}{cc}[\cos \pi x],& 0\le x\le 1\end{array}\\ \begin{array}{cc}|2x-3|[x-2]& 1<x\le 2\end{array}\end{array}\right.$ then

the number of points at which it is discontinuous where  ($[.]$ denotes the greatest integer function)

Consider $x\in [0,1]$. From the graph, it is clear that $[\cos \pi x]$ is discontinuous at. $x=0,1/2$

Now, consider x∈(1,2].
f(x)=[x−2]∣2x−3∣
For x∈(1,2),[x−2]=−1, and for x=2,[x−2]=0.
Also, ∣2x−3∣=0 or x=3/2.
Therefore, $x= \frac{3}{2}$ and 2 may be the points at which $f\left(x\right)$ is discontiuous

$f\left(x\right)=\left\{\begin{array}{l}1 ,if x=0\\ 0 if 0<x\le 1/2\\ -1 ,if\frac{1}{2}<x\le 1\\ (2x-3) if 1<x\le 3/2\\ -(2x-3) if,3/2<x\le 2\\ 0 ,if x=2\end{array}\right.$

Thus $f\left(x\right)$ is continuous when $x\in [0,2]-\{0,\frac{1}{2},2\}$