【答案】
分析:根据不等式组
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193744542469584/SYS201311011937445424695022_DA/0.png)
恰有三个整数解,可得出a的取值范围;联立一次函数及反比例函数解析式,利用二次函数的性质判断其判别式的值的情况,从而确定交点的个数.
解答:解:不等式组的解为:a≤t≤
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193744542469584/SYS201311011937445424695022_DA/1.png)
,
∵不等式组恰有3个整数解,
∴-2<a≤-1.
联立方程组
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193744542469584/SYS201311011937445424695022_DA/2.png)
,
得:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193744542469584/SYS201311011937445424695022_DA/3.png)
x
2-ax-3a-2=0,
△=a
2+3a+2=(a+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193744542469584/SYS201311011937445424695022_DA/4.png)
)
2-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193744542469584/SYS201311011937445424695022_DA/5.png)
=(a+1)(a+2)
这是一个二次函数,开口向上,与x轴交点为(-2,0)和(-1,0),对称轴为直线a=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193744542469584/SYS201311011937445424695022_DA/6.png)
,
其图象如下图所示:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193744542469584/SYS201311011937445424695022_DA/images7.png)
由图象可见:
当a=-1时,△=0,此时一元二次方程有两个相等的根,即一次函数与反比例函数有一个交点;
当-2<a<-1时,△<0,此时一元二次方程无实数根,即一次函数与反比例函数没有交点.
∴交点的个数为:1或0.
故答案为:1或0.
点评:本题考查了二次函数、反比例函数、一次函数、解不等式、一元二次方程等知识点,有一定的难度.多个知识点的综合运用,是解决本题的关键.