IMC / 1994 / Problems / Day 1, P2

IMC 1994 · Day 1 · P2

Let $f \in C^1(a,b)$ , $\lim\limits_{x \to a+} f(x) = +\infty$ , $\lim\limits_{x \to b-} f(x) = -\infty$ and $f'(x) + f^2(x) \ge -1$ for $x \in (a,b)$ . Prove that $b - a \ge \pi$ and give an example where $b - a = \pi$ .

Solution (official)

From the inequality we get $\frac{d}{dx} \left( \operatorname{arctg} f(x) + x \right) = \frac{f'(x)}{1 + f^2(x)} + 1 \ge 0$ for $x \in (a,b)$ . Thus $\operatorname{arctg} f(x) + x$ is non-decreasing in the interval and using the limits we get $\frac{\pi}{2} + a \le -\frac{\pi}{2} + b$ . Hence $b - a \ge \pi$ . One has equality for $f(x) = \operatorname{cotg} x$ , $a = 0$ , $b = \pi$ .

IMC 1994 · Day 1 · P2

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