JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Continuity and Differentiability

The function $f(x)=|[x]|+\sqrt{x-[x]}$ is potentially discontinuous at integer points. In the interval $(-2,1)$, the integers are $-1$ and $0$. At $x=-1$: $f(-1^{+})=|-1|+0=1$ and $f(-1^{-})=|-2|+1=3$, so discontinuous. At $x=0$: $f(0^{+})=0+0=0$ and $f(0^{-})=|-1|+1=2$, so discontinuous. At $x=1$ (boundary, not in open interval). Thus there are 2 points of discontinuity.