1. Determine the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy the relation $f(f(x) y)+x y=2 f(x) f(y)$, for all $x, y \in \mathbb{R}$.

G.M. 9/2013, modified statement
Solution. $y=0 \Rightarrow f(0)=2 f(x) f(0), \forall x \in \mathbb{R} \Rightarrow f(0)=0$ or $f(x)=\frac{1}{2}, \forall x \in \mathbb{R}$. Since the constant function does not satisfy the relation given in the statement, we must have $f(0)=0$. Consider $y=1$ in the given relation and we obtain

$$
f(f(x))+x=2 f(x) f(1), \forall x \in \mathbb{R}
$$

We show that $f$ is an injective function. Let $a, b \in \mathbb{R}$ such that $f(a)=f(b) \Rightarrow f(f(a))=f(f(b))$. Using relation (1) for $x=a$ and $x=b$ we immediately obtain that $a=b$.

By swapping $x$ with $y$ in the hypothesis relation we get:

$$
\begin{gathered}
f(f(y) x)+x y=2 f(x) f(y), \forall x, y \in \mathbb{R}, \text { so } \\
f(f(x) y)=f(f(y) x), \forall x, y \in \mathbb{R}
\end{gathered}
$$

from which, using the injectivity of the function, we deduce: $f(x) y=x f(y), \forall x, y \in \mathbb{R}$.

For $\mathrm{y}=1$ we get $f(x)=x f(1)$ and substituting into the initial relation results in

$$
x y \cdot f^{2}(1)+x y=2 x y \cdot f^{2}(1), \forall x, y \in \mathbb{R} \Leftrightarrow x y=x y \cdot f^{2}(1), \forall x, y \in \mathbb{R} \Rightarrow f^{2}(1)=1 \Rightarrow f(1)= \pm 1
$$

| ![](https://cdn.mathpix.com/cropped/2024_06_07_aa15727edf4b5156a708g-2.jpg?height=67&width=1453&top_left_y=1288&top_left_x=175) | $1 \mathrm{p}$ |
| :---: | :---: |
| $f$ is an injective function | $2 \mathrm{p}$ |
| Deduce $f(x) y=x f(y), \forall x, y \in \mathbb{R} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots$ | $2 \mathrm{p}$ |
| Finalization-find the functions. | $2 \mathrm{p}$ |

In conclusion, the required functions are $f_{1}, f_{2}: \mathbb{R} \rightarrow \mathbb{R}, f_{1}(x)=x, f_{2}(x)=-x, \forall x \in \mathbb{R}$.

Grading.