Školjka

%V0 Let $\mathbb{R}^+$ be the set of all non-negative real numbers. Given two positive real numbers $a$ and $b,$ suppose that a mapping $f: \mathbb{R}^+ \mapsto \mathbb{R}^+$ satisfies the functional equation: $$f(f(x)) + af(x) = b(a + b)x.$$ Prove that there exists a unique solution of this equation.

%V0 Does there exist a function $f : \mathbb N \to \mathbb N$, such that $f(f(n)) =n + 1987$ for every natural number $n$? (IMO Problem 4) Proposed by Vietnam.

%V0 Let $a,b,A,B$ be given reals. We consider the function defined by $$f(x) = 1 - a \cdot \cos(x) - b \cdot \sin(x) - A \cdot \cos(2x) - B \cdot \sin(2x).$$ Prove that if for any real number $x$ we have $f(x) \geq 0$ then $a^2 + b^2 \leq 2$ and $A^2 + B^2 \leq 1.$

%V0 Find all functions $f : \mathbb{N}\rightarrow \mathbb{N}$ satisfying following condition: $$f(n+1)>f(f(n)) \quad \forall n \in \mathbb{N}$$

%V0 $G$ is a set of non-constant functions $f$. Each $f$ is defined on the real line and has the form $f(x)=ax+b$ for some real $a,b$. If $f$ and $g$ are in $G$, then so is $fg$, where $fg$ is defined by $fg(x)=f(g(x))$. If $f$ is in $G$, then so is the inverse $f^{-1}$. If $f(x)=ax+b$, then $f^{-1}(x)= \frac{x-b}{a}$. Every $f$ in $G$ has a fixed point (in other words we can find $x_f$ such that $f(x_f)=x_f$. Prove that all the functions in $G$ have a common fixed point.

%V0 For what real values of $x$ is $$\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=A$$ given a) $A=\sqrt{2}$; b) $A=1$; c) $A=2$, where only non-negative real numbers are admitted for square roots?