Let f be any continuous function on [0, 2] and twice differentiable on (0, 2). If f(0) = 0, f(1) = 1 and f(2) = 2, then

Solution

Maths sir · 4 steps

Use the mean value theorem twice to get two equal slopes, then Rolle's theorem on f' gives f'' = 0.

1. MVT on each half. On [0, 1] there is some c₁ ∈ (0, 1) with f'(c₁) = (f(1) - f(0))/(1 - 0) = 1. On [1, 2] there is some c₂ ∈ (1, 2) with f'(c₂) = (f(2) - f(1))/(2 - 1) = 1.

   f'(c₁) = f'(c₂) = 1, 0 < c₁ < 1 < c₂ < 2

2. Rolle on f'. Since f is twice differentiable on (0, 2), f' is continuous on [c₁, c₂] and differentiable on (c₁, c₂). It takes equal values at both ends, so Rolle's theorem gives some c ∈ (c₁, c₂) with f''(c) = 0.
3. Test the others. f(x) = x fits all the data with f'(x) = 1 and f''(x) = 0 everywhere, which rules out (c) and (d). f(x) = x + sin π x also fits, with f''(x) = -π² sin π x, which is not zero everywhere, so (a) fails.
4. So. The only statement true for every such f is that f''(x) = 0 for some x ∈ (0, 2), option (b).

The idea to keep. Three points on a function give two MVT slopes. If those slopes are equal, apply Rolle's theorem to f' and you get a point where f'' = 0.

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