74th Romanian Mathematical Olympiad

Question

Let x_0 be a real number. Prove that there exists a useful function f so that f(x_0) = 0 if and only if |x_0| 4. We will call an affine function f : R R *useful* if it has the properties: (i) |f(x)| 2, for every real number x such that |x| 2; (ii) |f(x)| 1, for every real number x such that |x| 1.

Accepted Answer

"" 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 : R R, f(x) = a x + b, with a > 0, b R. Since a > 0, the inequality (i) is equivalent to f(2) 2 and f(-2) -2, that is 2a + b 2, (1) and 2a - b 2, (2). The inequality (ii) leads to the possibilities: I) f(-1) -1 and f(1) -1 or II) f(-1) 1 and f(1) 1 (if f(-1) < 0 < f(1), then -1 < -ba < 1 and f(-ba) = 0, in contradiction with (ii)). In case I we get -a + b -1, (3) and a + b -1, (4). Multiply (4) by 2 and add with (2) to get 4a + b 0, that is x_0 = -ba 4. In case II we get a - b -1, (5) and -a - b -1, (6). Multiply (5) by 2 and add with (1) to get 4a - b 0, hence x_0 = -ba -4. From I) and II) follows |x_0| 4.