IMC 2016 · Day 1 · P1

(a) Choose a convergent sequence $z_n$ of zeros and let $c = \lim z_n \in [a, b]$. By the continuity of $f$ we obtain $f(c) = 0$. We want to show that either $c = a$ or $c = b$, so $f(a) = 0$ or $f(b) = 0$; then the statement follows.

If $c$ was an interior point then we would have $f(c) = 0$ and $f'(c) = \lim\limits_{n \to \infty} \dfrac{f(z_n) - f(c)}{z_n - c} = \lim\limits_{n \to \infty} \dfrac{0 - 0}{z_n - c} = 0$ simultaneously, contradicting the conditions. Hence, $c = a$ or $c = b$.

(b) Let $$f(x) = \begin{cases} x \sin\dfrac{1}{x} & \text{if } 0 < x \le 1 \\ 0 & \text{if } x = 0. \end{cases}$$ This function has zeros at the points $\dfrac{1}{k\pi}$ for $k = 1, 2, \dots$, and it is continuous at 0 as well.

In $(0, 1)$ we have $$f'(x) = \sin\frac{1}{x} - \frac{1}{x} \cos\frac{1}{x}.$$ Since $\sin\frac{1}{x}$ and $\cos\frac{1}{x}$ cannot vanish at the same point, we have either $f(x) \ne 0$ or $f'(x) \ne 0$ everywhere in $(0, 1)$.