International Mathematics Competition
for University Students
2023

IMC2023: Day 2, Problem 7

Problem 7. Let \(\displaystyle V\) be the set of all continuous functions \(\displaystyle f\colon[0,1] \to \mathbb{R}\), differentiable on \(\displaystyle (0,1)\), with the property that \(\displaystyle f(0)=0\) and \(\displaystyle f(1)=1\). Determine all \(\displaystyle \alpha\in\mathbb{R}\) such that for every \(\displaystyle f\in V\), there exists some \(\displaystyle \xi\in(0,1)\) such that

\(\displaystyle f(\xi) + \alpha = f'(\xi). \)

Mike Daas, Leiden University

Solution 1. First consider the function

\(\displaystyle h(x) = \dfrac{e^x-1}{e-1}, \quad \text{which has the property that} \quad h'(x) = \dfrac{e^x}{e-1}. \)

Note that \(\displaystyle h \in V\) and that \(\displaystyle h'(x) - h(x) = 1/(e-1)\) is constant. As such, \(\displaystyle \alpha = 1/(e-1)\) is the only possible value that could possibly satisfy the condition from the problem. For \(\displaystyle f \in V\) arbitrary, let

\(\displaystyle g(x) = f(x)e^{-x} + h(-x), \quad \text{with} \quad g(0) = 0 \quad \text{and also} \quad g(1) = e^{-1} + \dfrac{e^{-1}-1}{e-1} = 0. \)

We compute that

\(\displaystyle g'(x) = f'(x)e^{-x} - f(x)e^{-x} - h'(-x). \)

Now apply Rolle's Theorem to \(\displaystyle g\) on the interval \(\displaystyle [0,1]\); it yields some \(\displaystyle \xi \in (0,1)\) with the property that

\(\displaystyle g'(\xi) = 0 \implies f'( \xi ) e^{-\xi} - f( \xi )e^{-\xi} - \dfrac{e^{-\xi}}{e-1} = 0 \implies f'( \xi ) = f( \xi ) + \dfrac{1}{e-1}, \)

showing that \(\displaystyle \alpha = 1/(e-1)\) indeed satisfies the condition from the problem.

Solution 2. Notice that the expression \(\displaystyle f'(x)-f(x)\) appears in the derivative of the function \(\displaystyle F(x)=f(x)\cdot e^{-x}\): \(\displaystyle F'(x)=\big(f'(x)-f(x)\big)e^{-x}\).

Apply Cauchy's mean value theorem to \(\displaystyle F(x)\) and the function \(\displaystyle G(x)=-e^{-x}\). By the theorem, there is some \(\displaystyle \xi\in(0,1)\) such that

\(\displaystyle \dfrac{F'(\xi)}{G'(\xi)} = \dfrac{F(1)-F(0)}{G(1)-G(0)} \)

\(\displaystyle f'(\xi)-f(\xi) = \dfrac{e^{-1}-0}{-e^{-1}+1} = \dfrac{1}{e-1}. \)

This proves the required property for \(\displaystyle a=\dfrac{1}{e-1}\).

Now we show that no other \(\displaystyle \alpha\) is possible. Choose \(\displaystyle f\) and \(\displaystyle F\) in such a way that \(\displaystyle \dfrac{F'(x)}{G'(x)}=f'(x)-f(x)=\dfrac{1}{e-1}\) is constant. That means

\(\displaystyle F'(x) = \dfrac{G'(x)}{e-1} = \dfrac{e^{-x}}{e-1}, \)

\(\displaystyle F(x) = \dfrac{1-e^{-x}}{e-1}, \)

\(\displaystyle f(x) = F(x)\cdot e^x = \dfrac{e^x-1}{e-1}. \)

With this choice we have \(\displaystyle f(0)=0\) and \(\displaystyle f(1)=1\), so \(\displaystyle f'\in V\), and \(\displaystyle f'(x)-f(x)\equiv\dfrac{1}{e-1}\) for all \(\displaystyle x\), so for this function the only possible value for \(\displaystyle \alpha\) is \(\displaystyle \dfrac{1}{e-1}\).