IMC 2006 · Day 1 · P1

(a) False. Consider function $f(x) = x^3 - x$. It is continuous, $\operatorname{range}(f) = \mathbb{R}$ but, for example, $f(0) = 0$, $f\left( \frac{1}{2} \right) = -\frac{3}{8}$ and $f(1) = 0$, therefore $f(0) > f\left( \frac{1}{2} \right)$, $f\left( \frac{1}{2} \right) < f(1)$ and $f$ is not monotonic.

(b) True. Assume first that $f$ is non-decreasing. For an arbitrary number $a$, the limits $\lim\limits_{a-} f$ and $\lim\limits_{a+} f$ exist and $\lim\limits_{a-} f \le \lim\limits_{a+} f$. If the two limits are equal, the function is continuous at $a$. Otherwise, if $\lim\limits_{a-} f = b < \lim\limits_{a+} f = c$, we have $f(x) \le b$ for all $x < a$ and $f(x) \ge c$ for all $x > a$; therefore $\operatorname{range}(f) \subset (-\infty, b) \cup (c, \infty) \cup \{f(a)\}$ cannot be the complete $\mathbb{R}$.

For non-increasing $f$ the same can be applied writing reverse relations or $g(x) = -f(x)$.

(c) False. The function $g(x) = \arctan x$ is monotonic and continuous, but $\operatorname{range}(g) = (-\pi/2, \pi/2) \ne \mathbb{R}$.