1. Given $A=\left\{x \mid x^{2}-2 x-3>0\right\}, B=\left\{x \mid x^{2}+a x+b \leqslant 0\right\}$. If $A \cup$ $B=\mathbf{R}, A \cap B=(3,4]$, then the correct conclusion is $(\quad)$.
A. $a=3, b=4$
B. $a=3, b=-4$
C. $a=-3, b=4$
D. $a=-3, b=-4$
1. D.
$A=(-\infty,-1) \cup(3,+\infty)$, From the problem, we get $B=[-1,4]$, then $-a=-1+4, b=-1 \cdot 4$.