Bulgarian Mathematical Competitions

Solution:

a) We consider two cases. If $x \in (-\infty, 2] \cup [3, \infty)$, then the inequality becomes $x^{2}-6x+6 \leq 0$, whence $x \in [3-\sqrt{3}, 3+\sqrt{3}]$. Therefore the solutions of the inequality are $x \in [3-\sqrt{3}, 2] \cup [3, 3+\sqrt{3}]$.

If $x \in (2,3)$, then the inequality becomes $x^{2}-4x+6 \geq 0$, which is satisfied for every $x \in (2,3)$. Thus $x \in [3-\sqrt{3}, 3+\sqrt{3}]$.

b) If the inequality $|x^{2}-5x+6| \leq x+a$ has an integral solution $x$ then $x \in (-\infty, 2] \cup [3, \infty)$ and therefore

$$x^{2}-6x+6-a \leq 0$$

This inequality has a solution if and only if $a \geq -3$ and in this case we have that $x \in [3-\sqrt{a+3}, 3+\sqrt{a+3}]$. This interval contains the number $3$ and is symmetric with respect to $3$. Therefore it contains exactly three integers if and only if $1 \leq \sqrt{a+3} < 2$, i.e. $a \in [-2,1)$.