JEE Main 2025 Apr 7 Shift 1, Mathematics Q21: Continuity of Function

Both greatest-integer terms jump only where their argument crosses an integer. So $f$ can only be discontinuous where at least one of $\frac{x^2}{2}$ or $\sqrt{x}$ becomes an integer.

Where $\frac{x^2}{2}$ is an integer in $[0,4]$: $x^2=2k$ for $k=0,1,2,\dots ,8$, giving $x=0,\sqrt{2},2,\sqrt{6},\sqrt{8},\sqrt{10},\sqrt{12},\sqrt{14},4$.

Where $\sqrt{x}$ is an integer in $[0,4]$: $x=0,1,4$.

Union of candidate jump points in $[0,4]$:

$\{0,1,\sqrt{2},2,\sqrt{6},\sqrt{8},\sqrt{10},\sqrt{12},\sqrt{14},4\}$ — ten points.

Check the boundary:

At $x=0$ (right endpoint of the domain from the right) and $x=4$ (left endpoint from the left), $f$ can only be one-sided; if the one-sided limit matches $f$ at that endpoint, we do not count it as a discontinuity within the domain. Direct checking shows $f$ is continuous at $x=0^{+}$ and at $x=4^{-}$.

Counting the interior discontinuities: removing $x=0$ and $x=4$ from the list leaves eight points, each of which is a genuine discontinuity for $f$.

$\therefore$ Answer = $\boxed{8}$.