IMC 2023 · Day 2 · P7

Hint: Find a function $h \in V$ such that $h' - h$ is constant, then apply Rolle's theorem to $f - h$. Alternatively, you can apply Cauchys's mean value theorem with some auxiliary functions.

First consider the function $$h(x) = \frac{e^x - 1}{e - 1}, \quad \text{which has the property that} \quad h'(x) = \frac{e^x}{e - 1}.$$ Note that $h \in V$ and that $h'(x) - h(x) = 1/(e - 1)$ is constant. As such, $\alpha = 1/(e - 1)$ is the only possible value that could possibly satisfy the condition from the problem. For $f \in V$ arbitrary, let $$g(x) = f(x) e^{-x} + h(-x), \quad \text{with } g(0) = 0 \text{ and also } g(1) = e^{-1} + \frac{e^{-1} - 1}{e - 1} = 0.$$ We compute that $$g'(x) = f'(x) e^{-x} - f(x) e^{-x} - h'(-x).$$ Now apply Rolle's Theorem to $g$ on the interval $[0, 1]$; it yields some $\xi \in (0, 1)$ with the property that $$g'(\xi) = 0 \ \Longrightarrow \ f'(\xi) e^{-\xi} - f(\xi) e^{-\xi} - \frac{e^{-\xi}}{e - 1} = 0 \ \Longrightarrow \ f'(\xi) = f(\xi) + \frac{1}{e - 1},$$ showing that $\alpha = 1/(e - 1)$ indeed satisfies the condition from the problem.

Notice that the expression $f'(x) - f(x)$ appears in the derivative of the function $F(x) = f(x) \cdot e^{-x}$: $F'(x) = \bigl( f'(x) - f(x) \bigr) e^{-x}$.

Apply Cauchy's mean value theorem to $F(x)$ and the function $G(x) = -e^{-x}$. By the theorem, there is some $\xi \in (0, 1)$ such that $$\begin{gather*} \frac{F'(\xi)}{G'(\xi)} = \frac{F(1) - F(0)}{G(1) - G(0)} \\ f'(\xi) - f(\xi) = \frac{e^{-1} - 0}{-e^{-1} + 1} = \frac{1}{e - 1}. \end{gather*}$$ This proves the required property for $a = \dfrac{1}{e - 1}$.

Now we show that no other $\alpha$ is possible. Choose $f$ and $F$ in such a way that $\frac{F'(x)}{G'(x)} = f'(x) - f(x) = \frac{1}{e-1}$ is constant. That means $$\begin{gather*} F'(x) = \frac{G'(x)}{e - 1} = \frac{e^{-x}}{e - 1}, \\ F(x) = \frac{1 - e^{-x}}{e - 1}, \\ f(x) = F(x) \cdot e^x = \frac{e^x - 1}{e - 1}. \end{gather*}$$ With this choice we have $f(0) = 0$ and $f(1) = 1$, so $f' \in V$, and $f'(x) - f(x) \equiv \dfrac{1}{e-1}$ for all $x$, so for this function the only possible value for $\alpha$ is $\dfrac{1}{e - 1}$.