Statement: If f is continuous on [a, b] and k is any value between f(a) and f(b), then there exists at least one c in (a, b) such that f(c) = k.

Special case for roots: If f is continuous on [a, b] and f(a) * f(b) < 0 (signs differ), then there exists c in (a, b) such that f(c) = 0.

How JEE uses IVT:

1. "Prove that the equation $x^5$ - 3x + 1 = 0 has a root in (0, 1)."

   • f(0) = 1 > 0, f(1) = -1 < 0. Since f is continuous and changes sign, root exists.

2. "Find the number of roots of f(x) = 0 in an interval."

   • Use IVT to show existence, and f'(x) analysis (monotonicity) to show at most one root in each interval.

Limitations: IVT only guarantees existence, not uniqueness. It also requires continuity — does not work for discontinuous functions. IVT does not tell you the value of c, only that it exists.

Common mistake: Students try to use IVT when the function is discontinuous in the interval. Always verify continuity first.