Let g: R → R be a non constant twice differentiable function such that g′(1/2)=g′(3/2). If a real valued function f is defined as f(x)=1/2[g(x)+g(2-x)], then

Solution

Maths sir · 4 steps

Find three zeros of f' first, then Rolle's theorem on f' gives two zeros of f''.

1. Differentiate f. By the chain rule, the derivative of g(2-x) is -g'(2-x). So f'(x)=1/2[g'(x)-g'(2-x)].
2. Three zeros of f′. At x=1/2 you get 1/2[g'(1/2)-g'(3/2)]=0, by the given condition. At x=3/2 the same two terms appear in the other order, so it is also 0. At x=1 you get 1/2[g'(1)-g'(1)]=0.

   f'(1/2)=f'(1)=f'(3/2)=0

3. Rolle on f′. g is twice differentiable, so f' is continuous and differentiable. Apply Rolle's theorem to f' on [1/2,1] and on [1,3/2]. You get a zero of f'' in (1/2,1) and another in (1,3/2).
4. So. f''(x)=0 for at least two values of x in (0,2), which is option (a).

The idea to keep. When f' vanishes at several points, apply Rolle's theorem to f' itself. If f' has n zeros, f'' has at least n-1 zeros between them.

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