Romanian Mathematical Olympiad

"$\Rightarrow$" If $f$ is useful, then $-f$ has properties (i) and (ii), hence it is useful. Consequently, we may look only at the useful functions $f : \mathbb{R} \to \mathbb{R}$, $f(x) = a x + b$, with $a > 0$, $b \in \mathbb{R}$.

Since $a > 0$, the inequality (i) is equivalent to $f(2) \le 2$ and $f(-2) \ge -2$, that is $2a + b \le 2$, (1) and $2a - b \le 2$, (2).

The inequality (ii) leads to the possibilities:

I) $f(-1) \le -1$ and $f(1) \le -1$ or

II) $f(-1) \ge 1$ and $f(1) \ge 1$

(if $f(-1) < 0 < f(1)$, then $-1 < -\frac{b}{a} < 1$ and $f(-\frac{b}{a}) = 0$, in contradiction with (ii)).

In case I we get $-a + b \le -1$, (3) and $a + b \le -1$, (4). Multiply (4) by 2 and add with (2) to get $4a + b \le 0$, that is $x_0 = -\frac{b}{a} \ge 4$.

In case II we get $a - b \le -1$, (5) and $-a - b \le -1$, (6). Multiply (5) by 2 and add with (1) to get $4a - b \le 0$, hence $x_0 = -\frac{b}{a} \le -4$.

From I) and II) follows $|x_0| \ge 4$.