International Mathematics Competition
for University Students
2025

IMC2025: Day 2, Problem 6

Problem 6. Let \(\displaystyle f\colon (0,\infty) \to \mathbb{R}\) be a continuously differentiable function, and let \(\displaystyle b>a>0\) be real numbers such that \(\displaystyle f(a)=f(b) = k\). Prove that there exists a point \(\displaystyle \xi\in(a,b)\) such that

\(\displaystyle f(\xi)-\xi f'(\xi)=k . \)

Alberto Cagnetta, Università degli Studi di Udine

Solution. Observe that if we consider \(\displaystyle g(x) = f(x)/x\) and take its derivative, we get

\(\displaystyle g'(x) = \frac{xf'(x)-f(x)}{x^{2}}, \)

where the numerator is almost the expression we have in the problem.

Now we apply Cauchy's theorem to the functions \(\displaystyle g(x) = f(x) / x\) and \(\displaystyle h(x)=1/x\), well-defined over \(\displaystyle [a,b]\). This gives us the existence of a value \(\displaystyle \xi\in[a,b]\) such that

\(\displaystyle \frac{g(a)-g(b)}{h(a)-h(b)} = \frac{g'(\xi)}{h'(\xi)}. \)

Here,

\(\displaystyle \frac{g'(\xi)}{h'(\xi)} = \frac{\frac{\xi f'(\xi)-f(\xi)}{\xi^{2}}}{-\frac{1}{\xi^{2}}} = f(\xi)-\xi f'(\xi) \)

and

\(\displaystyle \frac{g(a)-g(b)}{h(a)-h(b)} = \frac{\frac{f(a)}{a}-\frac{f(b)}{b}}{\frac{1}{a}-\frac{1}{b}} = k, \)

which concludes the proof.