9. Let $A$ be the set of real numbers $x$ satisfying the inequality $x^{2}+x-110<0$ and $B$ be the set of real numbers $x$ satisfying the inequality $x^{2}+10 x-96<0$. Suppose that the set of integer solutions of the inequality $x^{2}+a x+b<0$ is exactly the set of integers contained in $A \cap B$. Find the maximum value of $\lfloor|a-b|\rfloor$.
9. Answer. 71
Note that $x^{2}+x-110=(x-10)(x+11)$. Thus the set of real numbers $x$ satisfying the inequality $x^{2}+x-110<0$ is $-11<x<10$.
Also note that $x^{2}+10 x-96=(x-6)(x+16)$. Thus the set of real numbers $x$ satisfying the inequality $x^{2}+10 x-96<0$ is $-16<x<6$.
Thus $A=\{x:-11<x<10\}$ and $B=\{x:-16<x<6\}$, implying that
$$
A \cap B=\{x:-11<x<6\} \text {. }
$$

Now let $x^{2}+a x+b=\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1} \leq x_{2}$. Then the set of integer solutions of $x^{2}+a x+b<0$ is
$$
\left\{k: k \text { is an integer, } x_{1}<k<x_{2}\right\} \text {. }
$$

By the given condition,
$$
\left\{\begin{array}{l}
\left\{k: k \text { is an integer, } x_{1}<k<x_{2}\right\}=\{k: k \text { is an integer, }-11<k<6\} \\
=\{-10,-9, \cdots, 5\} .
\end{array}\right.
$$

Thus $-11 \leq x_{1}<-10$ and $5<x_{2} \leq 6$. It implies that $-6<x_{1}+x_{2}<-4$ and $-66 \leq x_{1} x_{2}<-50$.
From $x^{2}+a x+b=\left(x-x_{1}\right)\left(x-x_{2}\right)$, we have $a=-\left(x_{1}+x_{2}\right)$ and $b=x_{1} x_{2}$. Thus $4<a<6$ and $-66 \leq b<-50$. It follows that $54<a-b<72$.
Thus max $\lfloor|a-b|\rfloor \leq 71$.
It remains to show that it is possible that $\lfloor|a-b|\rfloor=71$ for some $a$ and $b$.
Let $a=5$ and $b=-66$. Then $x^{2}+a b+b=(x+11)(x-6)$ and the inequality $x^{2}+a b+b<0$ has solutions $\{x:-11<x<6\}$. So the set of integer solutions of $x^{2}+a b+b<0$ is really the set of integers in $A \cap B$.
Hence $\max \lfloor|a-b|\rfloor=71$.