IMC 1998 · Day 1 · P4

Define the function $$g(x) = \frac{1}{2} f^2(x) + f'(x).$$ Because $g(0) = 0$ and $$f(x) \cdot f'(x) + f''(x) = g'(x),$$ it is enough to prove that there exists a real number $0 < \eta \le 1$ for which $g(\eta) = 0$.

a) If $f$ is never zero, let $$h(x) = \frac{x}{2} - \frac{1}{f(x)}.$$ Because $h(0) = h(1) = -\frac{1}{2}$, there exists a real number $0 < \eta < 1$ for which $h'(\eta) = 0$. But $g = f^2 \cdot h'$, and we are done.

b) If $f$ has at least one zero, let $z_1$ be the first one and $z_2$ be the last one. (The set of the zeros is closed.) By the conditions, $0 < z_1 \le z_2 < 1$.

The function $f$ is positive on the intervals $[0, z_1)$ and $(z_2, 1]$; this implies that $f'(z_1) \le 0$ and $f'(z_2) \ge 0$. Then $g(z_1) = f'(z_1) \le 0$ and $g(z_2) = f'(z_2) \ge 0$, and there exists a real number $\eta \in [z_1, z_2]$ for which $g(\eta) = 0$.

Remark. For the function $f(x) = \dfrac{2}{x+1}$ the conditions hold and $f \cdot f' + f''$ is constantly 0.