JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Continuity and Differentiability

$f(x)=2x^2+x+[x^2]-[x]$ on $[-1,2]$. Since $2x^2+x$ is continuous everywhere, discontinuities of $f$ can only occur where $[x]$ or $[x^2]$ jumps.
At $x=-1$ (right end only, domain starts here): $f(-1)=2-1+1-(-1)=3$; but just to the right, $x^2<1$ so $[x^2]=0$ and $[x]=-1$, giving the limit $2-1+0+1=2\ne f(-1)$ — discontinuous.
At $x=0$: just to the left, $[x^2]=0,[x]=-1$, limit $=0+0+0+1=1$; but $f(0)=0+0+0-0=0\ne 1$ — discontinuous.
At $x=1$: both one-sided limits and $f(1)$ equal $3$ — continuous.
At $x=\sqrt{2}$: $[x^2]$ jumps from $1$ to $2$ while $[x]=1$ on both sides, so the left/right limits differ from $f(\sqrt{2})$ — discontinuous.
At $x=\sqrt{3}$: similarly $[x^2]$ jumps from $2$ to $3$ with $[x]=1$ unchanged — discontinuous.
At $x=2$ (left end only): the left-hand limit equals $f(2)=12$ — continuous.
Discontinuities occur at $x=-1,0,\sqrt{2},\sqrt{3}$ — 4 points.