IVT for Root Existence: If f is continuous and f(a)*f(b) < 0, then f(c) = 0 for some c in (a,b).

Strategy for Root Problems:

1. Define g(x) = equation rearranged to g(x) = 0
2. Find a and b where g changes sign
3. Conclude root exists by IVT

Fixed Point Theorem: If f:[a,b]->[a,b] is continuous, then f(c) = c for some c. Proof: g(x) = f(x) - x. g(a) = f(a) - a >= 0. g(b) = f(b) - b <= 0. IVT gives g(c) = 0.

Number of Solutions: Combine IVT (at least n roots via sign changes) with derivative analysis (at most k roots via monotonicity or Rolle's).

Common JEE Equations:

• $e^x$ = kx (number of solutions depends on k)
• x = a*sin(x) (fixed point problem)
• Polynomial = transcendental (use IVT + derivative bounds)