JEE Main 2026 · 6 Apr · Shift 2, Mathematics question 25

Where g jumps. $g(0^-)=-2$ and $g(0)=2$, so g is discontinuous only at 0. So $g\circ f$ can break where f crosses 0 or where f itself jumps.

Check each point.

• $x=0$: f jumps from 8 to −4, so $g\circ f$ jumps from $\sqrt{12}$ to $(-12)^{1/3}$.
• $x=-2$: $f=x^3+8$ crosses 0 from below, so $g\circ f$ jumps from −2 to 2.
• $x=2$: $f=x^2-4$ crosses 0 from below, likewise.

Count: 3.

Answer: 3