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69th Belarusian Mathematical Olympiad

Belarus algebra

Problem

Find all functions

f (x) : R \to R

satisfying for all

x \in R

the inequality

x = \frac{3}{4} f (∣ x ∣) + ∣ f (x) ∣.

(I. Voronovich)

Solution

Answer:

f (x) = - 4 x

for all

x \geq 0

and

f (x) = \pm 2 x

for all

x < 0

(signs are chosen independently).

Final answer

All functions f: R -> R such that f(x) = -4x for x >= 0, and for each x < 0 one may choose independently f(x) = 2x or f(x) = -2x.

Techniques

Existential quantifiers