已知命題p:|x-2|<a(a>0),命題q:|x2-4|<1,若p是q的充分不必要條件,則實數(shù)a的取值范圍是 .
【答案】
分析:解絕對值不等式,化簡命題p和命題q,根據(jù)p是q的充分不必要條件得到 2-a≥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057915908/SYS201310241806370579159012_DA/0.png)
,且2+a≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057915908/SYS201310241806370579159012_DA/1.png)
,a>0.
由此求出實數(shù)a的取值范圍.
解答:解:命題p:|x-2|<a(a>0),即2-a<x<2+a.
命題q:|x
2-4|<1,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057915908/SYS201310241806370579159012_DA/2.png)
<x<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057915908/SYS201310241806370579159012_DA/3.png)
,或-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057915908/SYS201310241806370579159012_DA/4.png)
<x<-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057915908/SYS201310241806370579159012_DA/5.png)
.
由題意得,命題p成立時,命題q一定成立,但當命題q成立時,命題p不一定成立.
∴2-a≥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057915908/SYS201310241806370579159012_DA/6.png)
,且2+a≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057915908/SYS201310241806370579159012_DA/7.png)
,a>0.解得 0<a≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057915908/SYS201310241806370579159012_DA/8.png)
,
故答案為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057915908/SYS201310241806370579159012_DA/9.png)
.
點評:本題考查絕對值不等式的解法,充分條件、必要條件的定義,判斷 2-a>
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057915908/SYS201310241806370579159012_DA/10.png)
,且2+a<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057915908/SYS201310241806370579159012_DA/11.png)
,是解題的難點.