1. Determine the functions $f:(0, \infty) \rightarrow(0, \infty)$ that satisfy the relation: $f\left(\frac{x}{f(y)}\right)=\frac{x}{f(x \sqrt{y})}$, for all $x, y \in(0, \infty)$.

Supliment G.M. 11/2015
Solution: $f(1) \stackrel{\text { not }}{=} a, y=1 \Rightarrow f\left(\frac{x}{a}\right)=\frac{x}{f(x)}$

$x=a \Rightarrow f(1)=\frac{a}{f(a)} \Rightarrow f(a)=1$.

Take $y=a$ in the relation from the hypothesis and we have $f(x)=\frac{x}{f(x \sqrt{a})}, \forall x \in(0, \infty)$

In relation (2), take, in turn, $x=\sqrt{a} \Rightarrow f(\sqrt{a})=\frac{\sqrt{a}}{f(a)} \Rightarrow f(\sqrt{a})=\sqrt{a}$ and $x=1 \Rightarrow a=\frac{1}{f(\sqrt{a})} \Rightarrow a=\frac{1}{\sqrt{a}} \Rightarrow a=1$.

We return to relation (1) and obtain $f(x)=\frac{x}{f(x)} \Leftrightarrow f^{2}(x)=x \Leftrightarrow f(x)=\sqrt{x}, \forall x \in(0, \infty)$, a function that satisfies the relation from the statement.

Scoring:

| $f(1)=a, y=1 \Rightarrow f\left(\frac{x}{a}\right)=\frac{x}{f(x)}$ | $1 \text{p}$ |
| :--- | :--- |
| $x=a \Rightarrow f(1)=\frac{a}{f(a)} \Rightarrow f(a)=1$ | $1 \text{p}$ |
| $f(x)=\frac{x}{f(x \sqrt{a})}, \forall x \in(0, \infty)$ | $1 \text{p}$ |
| Determine $f(1)$ | $2 \text{p}$ |
| Finalization: determine the function | $2 \text{p}$ |