IMC 2005 · Day 2 · P7

Write $f(x) = \left( x + \frac{b}{2} \right)^2 + d$ where $d = c - \frac{b^2}{4}$. The absolute minimum of $f$ is $d$.

If $d \ge 1$ then $f(x) \ge 1$ for all $x$, $M = \emptyset$ and $|M| = 0$.

If $-1 < d < 1$ then $f(x) > -1$ for all $x$, $$-1 < \left( x + \frac{b}{2} \right)^2 + d < 1 \iff \left| x + \frac{b}{2} \right| < \sqrt{1 - d}$$ so $$M = \left( -\frac{b}{2} - \sqrt{1-d},\ -\frac{b}{2} + \sqrt{1-d} \right)$$ and $$|M| = 2\sqrt{1-d} < 2\sqrt{2}.$$ If $d \le -1$ then $$-1 < \left( x + \frac{b}{2} \right)^2 + d < 1 \iff \sqrt{|d| - 1} < \left| x + \frac{b}{2} \right| < \sqrt{|d| + 1}$$ so $$M = \left( -\sqrt{|d|+1}, -\sqrt{|d|-1} \right) \cup \left( \sqrt{|d|-1}, \sqrt{|d|+1} \right)$$ and $$|M| = 2 \left( \sqrt{|d|+1} - \sqrt{|d|-1} \right) = 2\, \frac{(|d|+1) - (|d|-1)}{\sqrt{|d|+1} + \sqrt{|d|-1}} \le 2\, \frac{2}{\sqrt{1+1} + \sqrt{1-0}} = 2\sqrt{2}.$$