Problem 10.1. Consider the inequality $\left|x^{2}-5 x+6\right| \leq x+a$, where $a$ is a real parameter.

a) Solve the inequality for $a=0$.

b) Find the values of $a$ for which the inequality has exactly three integer solutions.

Stoyan Atanassov
10.1. a) We consider two cases. If $x \in(-\infty, 2] \cup[3, \infty)$, then the inequality becomes $x^{2}-6 x+6 \leq 0$, 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 $x^{2}-4 x+6 \geq 0$, which is satisfied for every $x \in(2,3)$. Thus $x \in[3-\sqrt{3}, 3+\sqrt{3}]$.

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

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

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. $a \in[-2,1)$.