JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Continuity and Differentiability

On the first piece, for $x\le -1$ the rule $|2x^2-3x-7|$ is a composition of continuous functions and so is continuous throughout, taking the value $|2+3-7|=2$ at the junction $x=-1$. Approaching $x=-1$ from the right the middle rule $[4x^2-1]$ has $4x^2-1$ just below $3$, so its greatest-integer value is also $2$; the two sides agree, so the function is in fact continuous at $x=-1$. Inside $(-1,1)$ the quantity $u=4x^2-1$ falls from near $3$ to its minimum $-1$ at $x=0$ and climbs back to near $3$, and $[u]$ jumps exactly where $u$ passes through an integer. Solving $4x^2-1=2,1,0$ gives the crossings $x=\pm \frac{\sqrt{3}}{2},\pm \frac{1}{\sqrt{2}},\pm \frac{1}{2}$, which are six points of discontinuity; at $x=0$, where $u=-1$ sits at the bottom, $[u]=-1$ on both sides and the function stays continuous. At $x=1$ the left-hand value from $[4x^2-1]$ is $2$ while the third rule gives $|1+1|+|1-2|=3$, so there is a jump there. For $x\ge 1$ the rule $|x+1|+|x-2|$ equals $2x-1$ on $[1,2)$ and the constant $3$ beyond, and these match at $x=2$, contributing nothing more. Collecting the six interior jumps together with the single jump at $x=1$ leaves $7$ points of discontinuity.