高考数学十年真题分类汇编(2010-2019)专题01集合

1、十年高考真题分类汇编（20102019）数学专题01 集合1.(2019全国1理T1)已知集合M=x|-4x2,N=x|x2-x-60,则MN=()A.x|-4x3B.x|-4x-2C.x|-2x2D.x|2x3【答案】C【解析】由题意得N=x|-2x3,则MN=x|-2x0,B=x|x-10,则AB=()A.(-,1)B.(-2,1)C.(-3,-1)D.(3,+)【答案】A【解析】由题意,得A=x|x3,B=x|x1,所以AB=x|x-1,B=x|x2,则AB=()A.(-1,+)B.(-,2)C.(-1,2) D.【答案】C【解析】由题意,得AB=(-1,2),故选C.5.(2019全国3T1)已知集合A=-1,0,1,2,B=x|x21,则AB=()A.-1,0,1B.0,1C.-1,1 D.0,1,2【答案】A【解析】A=-1,0,1,2,B=x|-1x1,则AB=-1,0,1.故选A.6.(2019北京文T1)已知集合A=x|-1x1,则AB=()A.(-1,1) B.(1,2)C.(-1,+)D.(1,+)【答案】C【解析】A=x|-1x1,AB=(-1,+),故选C.7

2、.(2019天津T1)设集合A=-1,1,2,3,5,B=2,3,4,C=xR|1x0,则RA=()A.x|-1x2 B.x|-1x2C.x|x2D.x|x-1x|x2【答案】B【解析】A=x|x2,所以RA=x|-1x2.10.(2018全国1文T1)已知集合A=0,2,B=-2,-1,0,1,2,则AB=()A.0,2B.1,2C.0D.-2,-1,0,1,2【答案】A【解析】由交集定义知AB=0,2.11.(2018全国2文T2,)已知集合A=1,3,5,7,B=2,3,4,5,则AB=()A.3B.5C.3,5D.1,2,3,4,5,7【答案】C【解析】集合A、B的公共元素为3,5,故AB=3,5.12.(2018全国3T1)已知集合A=x|x-10,B=0,1,2,则AB=()A.0B.1C.1,2D.0,1,2【答案】C【解析】由题意得A=x|x1,B=0,1,2,AB=1,2.13.(2018北京T1)已知集合A=x|x|2,B=-2,0,1,2,则AB=()A.0,1 B.-1,0,1C.-2,0,1,2D.-1,0,1,2【答案】A【解析】A=x|-2x2,B=-2,

3、0,1,2,AB=0,1.14.(2018天津理T1)设全集为R,集合A=x|0x2,B=x|x1,则A(RB)=()A.x|0x1B.x|0x1C.x|1x2D.x|0x2【答案】B【解析】RB=x|x1,A(RB)=x|0x1.故选B.15.(2018天津文T1)设集合A=1,2,3,4,B=-1,0,2,3,C=xR|-1x2,则(AB)C=()A.-1,1 B.0,1C.-1,0,1D.2,3,4【答案】C【解析】AB=-1,0,1,2,3,4.又C=xR|-1x2,(AB)C=-1,0,1.16.(2018浙江T1)已知全集U=1,2,3,4,5,A=1,3,则UA=()A. B.1,3C.2,4,5D.1,2,3,4,5【答案】C【解析】A=1,3,U=1,2,3,4,5,UA=2,4,5,故选C.17.(2018全国2理T2,)已知集合A=(x,y)|x2+y23,xZ,yZ,则A中元素的个数为()A.9 B.8 C.5 D.4【答案】A【解析】满足条件的元素有(-1,-1),(-1,0),(-1,1),(0,1),(0,0),(0,-1),(1,-1),(1,0),(1

4、,1),共9个。18.(2017全国3理T1,)已知集合A=(x,y)|x2+y2=1,B=(x,y)|y=x,则AB中元素的个数为A.3B.2C.1D.0【答案】B【解析】A表示圆x2+y2=1上所有点的集合,B表示直线y=x上所有点的集合,易知圆x2+y2=1与直线y=x相交,故AB中有2个元素.19.(2017全国1理T1)已知集合A=x|x1,B=x|3x1,则()A.AB=x|x1D.AB=【答案】A【解析】3x1=30,x0,B=x|x0,AB=x|x0,AB=x|x1.故选A.20.(2017全国2理T2)设集合A=1,2,4,B=x|x2-4x+m=0.若AB=1,则B=()A.1,-3B.1,0C.1,3D.1,5【答案】C【解析】由AB=1,可知1B,所以m=3,即B=1,3.21.(2017全国1文T1)已知集合A=x|x0,则()A.AB=xx32B.AB=C.AB=xx32D.AB=R【答案】A【解析】A=x|x2,B=xx32, AB=x|x2,AB=xx32,故选A.22.(2017全国2文T1)设集合A=1,2,3,B=2,3,4,则AB=()A.1,2

5、,3,4B.1,2,3C.2,3,4D.1,3,4【答案】A【解析】因为A=1,2,3,B=2,3,4,所以AB=1,2,3,4,故选A.23.(2017全国3文T1)已知集合A=1,2,3,4,B=2,4,6,8,则AB中元素的个数为()A.1B.2C.3D.4【答案】B【解析】由题意可得AB=2,4,则AB中有2个元素.故选B.24.(2017天津理T1)设集合A=1,2,6,B=2,4,C=xR|-1x5,则(AB)C=()A.2B.1,2,4 C.1,2,4,6D.xR|-1x5【答案】B【解析】A=1,2,6,B=2,4,AB=1,2,4,6.C=xR|-1x5,(AB)C=1,2,4.25.(2017北京理T1)若集合A=x|-2x1,B=x|x3,则AB=()A.x|-2x-1B.x|-2x3C.x|-1x1D.x|1x3【答案】A【解析】AB=x|-2x-1,故选A.26.(2017北京文T1)已知全集U=R,集合A=x|x2,则UA=()A.(-2,2)B.(-,-2)(2,+)C.-2,2D.(-,-22,+)【答案】C【解析】因为A=x|x2,所以UA=x|-2x

6、2.27.(2016全国1理T1)设集合A=x|x2-4x+30,则AB=()A.-3,-32B.-3,32C.1,32D.32,3【答案】D【解析】A=(1,3),B=32,+,所以AB=32,3,故选D.28.(2016全国2理T2)已知集合A=1,2,3,B=x|(x+1)(x-2)0,xZ,则AB=()A.1 B.1,2C.0,1,2,3D.-1,0,1,2,3【答案】C【解析】由题意可知,B=x|-1x0,则ST=()A.2,3B.(-,23,+)C.3,+)D.(0,23,+)【答案】D【解析】S=x|x2或x3.因为T=x|x0,所以ST=x|0x2或x3,故选D.30.(2016全国1文T1)设集合A=1,3,5,7,B=x|2x5,则AB=()A.1,3B.3,5C.5,7D.1,7【答案】B【解析】AB=3,5,故选B.31.(2016全国2文T1)已知集合A=1,2,3,B=x|x29,则AB=()A.-2,-1,0,1,2,3B.-2,-1,0,1,2C.1,2,3 D.1,2【答案】D【解析】B=x|-3x3,AB=1,2.故选D.32.(2016全国3文T1)设集合A=0,2,4,6,8,10,B=4,8,则AB=()A.4,8B.0,2,6C.0,2,6,10D.0,2,4,6,8,10【答案】C【解析】根据补集的定义,知从集合A=0,2,4,6,8,10中去掉集合B中的元素4,8后,剩下的4个元素0,

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