JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 25

Middle piece. On $(-1,1)$, $4x^2-1$ runs over $[-1,3)$. The floor jumps where $4x^2-1=0,1,2$: $x=\pm\frac12,\pm\frac1{\sqrt2},\pm\frac{\sqrt3}2$, six points. At $x=0$ it is $-1$ on both sides (no jump).

At $x=-1$. $f(-1)=|2+3-7|=2$ and the right limit is $[3^-]=2$: continuous.

At $x=1$. Left limit 2, but $f(1)=2+1=3$: discontinuous.

Total $6+1=7$.

Answer: 7