Let f and g be twice differentiable even functions on (-2, 2) such that f( 1/4 ) = 0, f( 1/2 ) = 0, f(1) = 1 and g( 3/4 ) = 0, g(1) = 2. Then, the minimum number of solutions of f(x)g''(x) + f'(x)g'(x) = 0 in ( - 2,2) is equal to ________.

Solution

Maths sir · 5 steps

Spot that the left side is the derivative of f(x)g'(x), then count zeros of that product and use Rolle.

1. Find the product. Let h(x) = f(x) g'(x). By the product rule, h'(x) = f'(x)g'(x) + f(x)g''(x), which is exactly the left side of the equation. So we need the zeros of h'.
2. Zeros of f. f is even, so f(1/4) = f(1/2) = 0 also gives f(-1/4) = f(-1/2) = 0. That is four zeros of h.
3. Zero of g'. g is even, so g' is odd, and an odd function satisfies g'(0) = -g'(0), giving g'(0) = 0. So h(0) = 0 as well. The values g(3/4) and g(1) are not needed.

   h(x) = 0 at x = -1/2, -1/4, 0, 1/4, 1/2

4. Apply Rolle. h is differentiable and has five distinct zeros, so Rolle's theorem gives a zero of h' in each of (-1/2, -1/4), (-1/4, 0), (0, 1/4) and (1/4, 1/2).
5. So. The equation f(x)g''(x) + f'(x)g'(x) = 0 has at least 4 solutions in (-2, 2).

The common slip: 3. Forgetting that g'(0) = 0 for an even g leaves only four zeros of h, which gives 3 instead of 4.

The idea to keep. When an equation looks like u'v + uv', rewrite it as (uv)' = 0. Then count the zeros of uv: n distinct zeros give at least n - 1 zeros of the derivative by Rolle.

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