## Problem 2:

$$
\begin{gathered}
f\left(x_{1}\right)-f\left(x_{2}\right)=\frac{a x_{1}^{3}+1}{x_{1}}-\frac{a x_{2}^{3}+1}{x_{2}}=a x_{1}^{2}-a x_{2}^{2}+\frac{1}{x_{1}}-\frac{1}{x_{2}} \\
=a\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}\right)-\frac{x_{1}-x_{2}}{x_{1} x_{2}}= \\
\frac{\left[a\left(x_{1}+x_{2}\right) x_{1} x_{2}-1\right]\left(x_{1} x_{2}\right)}{x_{1} x_{2}}
\end{gathered}
$$

## (1p)

If \( a = 0 \) and \( f\left(x_{1}\right) = f\left(x_{2}\right) \), it follows that \( x_{1} = x_{2} \), so the function \( f \) is injective (1p).

If \( a \neq 0 \), there exist \( a, x_{1}, x_{2}, x_{1} \neq x_{2} \) such that \( a\left(x_{1}+x_{2}\right) x_{1} x_{2} - 1 = 0 \), i.e., \( f\left(x_{1}\right) = f\left(x_{2}\right) \) and \( f \) would not be injective. We justify this statement:

Let \( s = x_{1} + x_{2} \) and \( p = x_{1} x_{2} \), we get \( asp - 1 = 0 \) or \( sp = \frac{1}{a} \).

The quadratic equation with roots \( x_{1} \) and \( x_{2} \) is \( x^{2} - s x + p = 0 \).

Choose \( s = \frac{2}{\sqrt[3]{a}} \) and \( p = \frac{2}{2 \sqrt[3]{a^{2}}} \). Then the discriminant of the equation

\[
\Delta = s^{2} - 4p = \frac{4}{\sqrt[3]{a^{2}}} - \frac{2}{\sqrt[3]{a^{2}}} = \frac{2}{\sqrt[3]{a^{2}}} > 0,
\]

so the roots \( x_{1} \) and \( x_{2} \) are real and \( x_{1} \neq x_{2} \) (3p).

We conclude that \( f(x) = \frac{1}{x}, f: \mathbb{R}^{*} \rightarrow M \).

The equation \( f(x) = y \) has the solution \( x = \frac{1}{y}, y \in \mathbb{R}^{*} \), so the function is surjective, and thus bijective.

(1p)

Since \( y \in \mathbb{R}^{*} \), it follows that \( M = \mathbb{R}^{*} \).

(1p)
## Problem 2:

Let $\boldsymbol{f}: \mathbb{R}^{*} \rightarrow \boldsymbol{M}, \boldsymbol{M}$ be a subset of $\mathbb{R}, \boldsymbol{f}(\boldsymbol{x})=\frac{a x^{3}+1}{x}, a \in \mathbb{R}$. Determine $a$ and $M$ such that the function $f$ is bijective.