$\text{ Let }f:[-1,3]\to \mathrm{R}\text{ be defined as }$$f\left(x\right)=\left\{\begin{array}{ll}\left|x\right|+\left[x\right],& -1\le x<1\\ x+\left|x\right|,& 1\le x<2\\ x+\left[x\right],& 2\le x\le 3\end{array}\right.$where [t] denotes the greatest integer less than or equal to t. Then, the number of points at which f is discontinuous are

$\begin{array}{c}\text{Given function is }\\ f\left(x\right)=\left\{\begin{array}{ll}\left|x\right|+\left[x\right],& -1\le x<1\\ x+\left|x\right|,& 1\le x<2\\ x+\left[x\right],& 2\le x\le 3\end{array}\right.\\ =\left\{\begin{array}{ll}-x-1,& -1\le x<0\\ x,& 0\le x<1\\ 2x,& 1\le x<2\\ x+2,& 2\le x<3\\ 6& x=3\end{array}\right.\\ \Rightarrow f(-1)=0,f\left(-1^{+}\right)=0\\ f\left(0^{-}\right)=-1,f\left(0\right)=0,f\left(0^{+}\right)=0\\ f\left(1^{-}\right)=1,f\left(1\right)=2,f\left(1^{+}\right)=2\\ f\left(2^{-}\right)=4,f\left(2\right)=4,f\left(2^{+}\right)=4\\ f\left(3^{-}\right)=5,f\left(3\right)=6\end{array}$

Therefore, f(x) is discontinuous at 0,1,3

Hence the number of discontinuous points of the function $f\left(x\right)$ is 3