JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Continuity and Differentiability

Write the function out piece by piece. On $[-1,0)$ the greatest integer $[x]=-1$ and $|x|=-x$, so $f(x)=-x-1$; on $[0,1)$ it is $f(x)=x$; on $[1,2)$, $x+|x|=2x$; and on $[2,3)$, $[x]=2$, so $x+[x]=x+2$, while at the right end $f(3)=3+[3]=6$. Now test every junction and integer in the domain. At $x=0$ the left limit is $-1$ but the right limit is $0$, so $f$ breaks there. At $x=1$ the left limit is $1$ while the right limit is $2$, another break. At $x=2$ the left limit $2x\to 4$ equals the right limit $x+2\to 4$ and $f(2)=4$, so it is continuous. At the right end $x=3$ the values just below give $x+2\to 5$, but $f(3)=6$, so there is a jump there as well. The function is therefore discontinuous at three points, $x=0$, $x=1$ and $x=3$.