Let f : (a, b) → R be twice differentiable function such that f(x) = ∫ₐˣ g(t)dt for a differentiable function g(x). If f(x) = 0 has exactly five distinct roots in (a, b), then g(x)g'(x) = 0 has at least :

Solution

Maths sir · 4 steps

Since f' = g, Rolle gives roots of g; then apply Rolle to g², whose derivative is 2gg'.

1. Roots of g. By the fundamental theorem of calculus, f'(x) = g(x). Let the five roots of f be x₁ < x₂ < x₃ < x₄ < x₅. Rolle's theorem on each of the 4 gaps gives a root of g strictly inside each gap: c₁ < c₂ < c₃ < c₄.
2. Look at g squared. Let h(x) = g(x)², which is differentiable because g is. Then h is zero at c₁, c₂, c₃, c₄, and its derivative is exactly the product we want.

   h'(x) = 2 g(x) g'(x)

3. Count the roots. h' is zero at the 4 points c₁, c₂, c₃, c₄, because g is zero there. Rolle's theorem on h gives 3 more roots, one strictly inside each of (c₁, c₂), (c₂, c₃), (c₃, c₄). These 3 lie strictly between the cᵢ, so all 4 + 3 roots are distinct.
4. So. g(x)g'(x) = 0 has at least seven roots in (a, b), which is option (c).

The idea to keep. When a question asks for roots of gg', think of it as the derivative of 1/2g². The zeros of g count once, and Rolle's theorem adds one more root in each gap between them.

Answer from the NTA final answer key. Solution by PrepWiser.