JEE Main 2020 · 6 September · evening · Single correct · +4 / −1

For all twice differentiable functions f : R → R,
with f(0) = f(1) = f'(0) = 0

  1. (a) f''(x) ≠ 0, at every point x ∈ (0, 1)
  2. (b) f''(x) = 0, for some x ∈ (0, 1)
  3. (c) f''(0) = 0
  4. (d) f''(x) = 0, at every point x ∈ (0, 1)
Answer and solution

Answer (b) f''(x) = 0, for some x ∈ (0, 1)

Solution

Maths sir · 4 steps

Use Rolle's theorem twice: once on f to get a zero of f', then on f' to get a zero of f''.

  1. Rolle on f. f is differentiable and f(0) = f(1) = 0. So by Rolle's theorem there is some c ∈ (0, 1) with f'(c) = 0.
  2. Rolle on f'. Now f'(0) = 0 is given and f'(c) = 0, and f' is differentiable because f is twice differentiable. Rolle's theorem on [0, c] gives some d ∈ (0, c) with f''(d) = 0. Since (0, c) ⊂ (0, 1), this d lies in (0, 1).
  3. Test the others. Take f(x) = x³ - x². It satisfies f(0) = f(1) = 0 and f'(0) = 0. Here f''(x) = 6x - 2, so f''(0) = -2 ≠ 0, and f'' is zero only at x = 1/3. One function is enough to rule out (c) and (d).
  4. So. For every such function, f''(x) = 0 for some x ∈ (0, 1), which is option (b).

The idea to keep. When you know a zero of f and enough zeros of f', apply Rolle's theorem again on f'. Each step down the derivatives gives you one new zero.

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

Study Applications of Derivatives

Asked about 160 times in JEE Main since 2020. This question is from 2020.