2. Determine the cardinality of the set $A=\left\{x_{n} \in \mathbb{Q} \left\lvert\, x_{n}=\frac{n^{2}+2}{n^{2}-n+2}\right., n \in \mathbb{N}, 1 \leq n \leq 2015\right\}$.
Solution. We will determine how many terms of the sequence $\left(x_{n}\right)_{n}$ from set A coincide. Let $m, n \in \mathbb{N}, m \neq n$. We have $x_{n}=x_{m} \Leftrightarrow \frac{n^{2}+2}{n^{2}-n+2}=\frac{m^{2}+2}{m^{2}-m+2}$. After performing the calculations, we obtain $x_{n}=x_{m} \Leftrightarrow(n-m)(n m-2)=0$. Since $m \neq n$, it is necessary that $m n=2 \Leftrightarrow(m, n) \in\{(1,2),(2,1)\}$. In conclusion, the only terms that coincide are $x_{1}$ and $x_{2}$. Therefore, $\operatorname{card}(A)=2014$.

## Scoring.

| Determines the number of terms that coincide | $6 \mathrm{p}$ |
| :--- | :--- |
| Completion $\operatorname{card}(A)=2014$. | $1 \mathrm{p}$ |