JEE Main 2019 — Continuity and Differentiability Question with Solution

$f\left(x\right)=\left\{\begin{array}{c}\left|x\right|+\left[x\right];-1\le x<1\\ x+\left|x\right|;1\le x<2\\ \left|x\right|+\left[x\right];2\le x\le 3\end{array}\right.$

$=\left\{\begin{array}{c}-x-1;-1\le x<0\\ x+0;0\le x<1\\ 2x;1\le x<2\\ x+2;2\le x<3\\ 6;x=3 \end{array}\right.$

At $x=0, 1, 2, 3$ $f$ changes its definition.

$\therefore$ At $x=0 \mathrm{LHL}=-1, \mathrm{RHL}=0\Rightarrow \mathrm{f}$  is discontinuous at

$x=0$

$x=1 \mathrm{LHL}=1, \mathrm{RHL}=2\Rightarrow f$ is discontinuous at $x=1$

$x=2 \mathrm{LHL}=\mathrm{RHL}=f\left(2\right)=4\Rightarrow f$ is continuous at $x=2$

$x=3 \mathrm{LHL}=5, f\left(3\right)=6\Rightarrow f$ is discontinuous at $x=1$

$\therefore$ Points of discontinuity are $0, 1, 3.$