## Subject II

Determine $f: \mathbf{R} \rightarrow \mathbf{R}, f$ differentiable, knowing that $f$ admits a primitive $F: \mathbf{R} \rightarrow \mathbf{R}$, with the property that $F(x)=\frac{x f(x)}{2}$, for any $x \in \mathbf{R}$.
## Subject II

Since $F$ is the primitive of $f$, $f(x) = F'(x) = \frac{f(x) + x f'(x)}{2}, \forall x \in \mathbf{R}$.

It follows that $2 f(x) = f(x) + x f'(x)$, hence $f(x) - x f'(x) = 0, \forall x \in \mathbf{R}$.

If $x \in \mathbf{R} \backslash \{0\}, \frac{f(x) - x f'(x)}{x^2} = 0$, which means $\left(\frac{f(x)}{x}\right)' = 0 \forall x \in \mathbf{R} \backslash \{0\}$.

It follows that there exist $c_1, c_2 \in \mathbf{R}$, such that $\frac{f(x)}{x} = c_1, \forall x > 0$.

Then $f(x) = \begin{cases} c_1 x, & x > 0 \\ c_2 x, & x < 0 \end{cases}$

From the conditions that $f$ is continuous and differentiable on $\mathbf{R}$, it follows that $c_1 = c_2 = c \in \mathbf{R}$ and $a = 0$.

Then $f(x) = c x, \forall x \in \mathbf{R}$. $1 \text{p}$