【题文】已知集合A={x∈R|≤0},B={x∈R|(x-2a)(x-a2-1)<0}.若A∩B=?,则实数a的取值范围是( )A.(2,+∞)B.[2
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【题文】已知集合A={x∈R|
![](http://img.shitiku.com.cn/uploads/allimg/20200322/20200322171237-23964.png)
≤0},B={x∈R|(x-2a)(x-a
2-1)<0}.若A∩B=?,则实数a的取值范围是( )
A.(2,+∞) | B.[2,+∞) |
C.{1}∪[2,+∞) | D.(1,+∞) |
答案
【答案】C
解析
【解析】由
![](http://img.shitiku.com.cn/uploads/allimg/20200322/20200322171237-23964.png)
≤0,得A={x∈R|-1<x≤4},B={x∈R|(x-2a)(x-a
2-1)<0}={x∈R|2a<x<a
2+1}.若B≠?,则在数轴上可以看出2a≥4,所以a≥2;若B=?,只能a=1.综上选C.
举一反三
【题文】已知集合A={x|0<ln
![](http://img.shitiku.com.cn/uploads/allimg/20200322/20200322171205-13485.png)
<1},B={x|
![](http://img.shitiku.com.cn/uploads/allimg/20200322/20200322171205-44227.png)
<x<
![](http://img.shitiku.com.cn/uploads/allimg/20200322/20200322171206-66195.png)
},则A∩B=( )
【题文】若全集
![](http://img.shitiku.com.cn/uploads/allimg/20200322/20200322171015-43839.png)
,集合
![](http://img.shitiku.com.cn/uploads/allimg/20200322/20200322171015-38743.png)
,
![](http://img.shitiku.com.cn/uploads/allimg/20200322/20200322171016-94575.png)
,则
![](http://img.shitiku.com.cn/uploads/allimg/20200322/20200322171016-25141.png)
( )
A.{2} | B.{1,2} | C.{1,2,4} | D.{1,3,4,5} |
【题文】已知集合
![](http://img.shitiku.com.cn/uploads/allimg/20200322/20200322170946-36758.png)
,则
![](http://img.shitiku.com.cn/uploads/allimg/20200322/20200322170946-25006.png)
Z=
.
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