Let f: ℝ → ℝ be a thrice differentiable function such that f(0)=0, f(1)=1, f(2)=-1, f(3)=2 and f(4)=-2. Then, the minimum number of zeros of (3 f′ f^(′ ′)+f f^(′ ′ ′))(x) is __________.

Solution

Maths sir · 5 steps

Spot that this expression is half the third derivative of f², then count zeros downwards using Rolle.

1. Find the hidden derivative. d/dx(ff') = f'² + ff''. Differentiating again gives 2f'f'' + f'f'' + ff''' = 3f'f'' + ff'''. Also ff' = 1/2(f²)', so with g = f²:

   3f'f'' + ff''' = 1/2 g'''(x), g = f²

2. Zeros of f. f(0) = 0 is one zero. The sign changes 1 → -1, -1 → 2 and 2 → -2 give one zero each in (1,2), (2,3) and (3,4). So f has at least 4 zeros: 0 < α < β < γ.
3. Zeros of g'. g' = 2ff' vanishes wherever f does, which gives 4 zeros. Rolle's theorem on f gives a zero of f' strictly inside each of (0,α), (α,β) and (β,γ). These 7 points are all different, so g' has at least 7 zeros.
4. Apply Rolle twice. Between consecutive zeros of g' there is a zero of g'', so g'' has at least 6 zeros. In the same way, g''' has at least 5 zeros.
5. So. 3f'f'' + ff''' = 1/2g''' has at least 5 zeros.

The common slip: 3. Missing that f(0) = 0 is itself a zero leaves only 3 zeros of f. The count then runs 5 → 4 → 3 and gives 3, not 5.

The idea to keep. When an expression mixes f, f', f'', f''', try to write it as an exact derivative. Then count the zeros of the base function, and remember that each derivative loses at most one zero by Rolle.

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