十年高考分类汇编（word版）.docx

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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|2x0,B=x|x-1-1,B=x|x2,则AB=()A.(-1,+)B.(-,2)C.(-1,2) D.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.0,1,26.(2019北京文T1)已知集合A=x|-1x1,则AB=()A.(-1,1) B.(1,2)C.(-1,+)D.(1,+)7.(2019天津T1

2、)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|x210.(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,-1,0,1,211.(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.1,2,3,4,5,712.(2018全国3T1)已知集合A=x|x-10,B=0,1,2,则AB=()A.0B.1C.1,2D.0,1,213.(2018北京T1)已知集合A=x|x|2,B

3、=-2,0,1,2,则AB=()A.0,1 B.-1,0,1C.-2,0,1,2D.-1,0,1,214.(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|0x215.(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.2,3,416.(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.1,2,3,4,517.(2018全国2理T2,)已知集合A=(x,

4、y)|x2+y23,xZ,yZ,则A中元素的个数为()A.9 B.8 C.5 D.418.(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.019.(2017全国1理T1)已知集合A=x|x1,B=x|3x1,则()A.AB=x|x1D.AB=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.1,521.(2017全国1文T1)已知集合A=x|x0,则()A.AB=xx32B.AB=C.AB=xx32D.AB=R22.(201

5、7全国2文T1)设集合A=1,2,3,B=2,3,4,则AB=()A.1,2,3,4B.1,2,3C.2,3,4D.1,3,423.(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.424.(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|-1x525.(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|1x326.(2017北京文T1)已知全集U=R,集合

6、A=x|x2,则UA=()A.(-2,2)B.(-,-2)(2,+)C.-2,2D.(-,-22,+)27.(2016全国1理T1)设集合A=x|x2-4x+30,则AB=()A.-3,-32B.-3,32C.1,32D.32,328.(2016全国2理T2)已知集合A=1,2,3,B=x|(x+1)(x-2)0,则ST=()A.2,3B.(-,23,+)C.3,+)D.(0,23,+)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.1,731.(2016全国2文T1)已知集合A=1,2,3,B=x|x29,则AB=()A.

7、-2,-1,0,1,2,3B.-2,-1,0,1,2C.1,2,3 D.1,232.(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.0,2,4,6,8,1033.(2016四川理T1)设集合A=x|-2x2,Z为整数集,则集合AZ中元素的个数是()A.3B.4C.5D.634.(2016天津理T1)已知集合A=1,2,3,4,B=y|y=3x-2,xA,则AB=()A.1B.4 C.1,3D.1,435.(2016山东理T2)设集合A=y|y=2x,xR,B=x|x2-10,则AB=()A.(-1,1)B.(0

8、,1)C.(-1,+)D.(0,+)36.(2016浙江理T1)已知集合P=xR|1x3,Q=xR|x24,则P(RQ)=()A.2,3B.(-2,3C.1,2)D.(-,-21,+)37.(2015全国2理T1)已知集合A=-2,-1,0,1,2,B=x|(x-1)(x+2)0,则AB=()A.-1,0 B.0,1C.-1,0,1D.0,1,238.(2015全国1文T1)已知集合A=x|x=3n+2,nN,B=6,8,10,12,14,则集合AB中元素的个数为()A.5B.4C.3D.239.(2015全国2文T1)已知集合A=x|-1x2,B=x|0x3,则AB=()A.(-1,3)B.

9、(-1,0) C.(0,2) D.(2,3)40.(2015陕西文T1)设集合M=x|x2=x,N=x|lg x0,则MN=()A.0,1B.(0,1 C.0,1)D.(-,141.(2015重庆理T1,)已知集合A=1,2,3,B=2,3,则()A.A=BB.AB=C.ABD.BA42.(2014全国1理T1)已知集合A=x|x2-2x-30,B=x|-2x2,则AB=()A.-2,-1B.-1,2)C.-1,1D.1,2)43.(2014全国2理T1)设集合M=0,1,2,N=x|x2-3x+20,则MN=()A.1B.2C.0,1D.1,244.(2014全国1文T1)已知集合M=x|-

10、1x3,N=x|-2x1,则MN=()A.(-2,1)B.(-1,1)C.(1,3)D.(-2,3)45.(2014全国2文T1)已知集合A=-2,0,2,B=x|x2-x-2=0,则AB=()A.B.2 C.0 D.-246.(2014辽宁理T1)已知全集U=R,A=x|x0,B=x|x1,则集合U(AB)=()A.x|x0B.x|x1C.x|0x1D.x|0x147.(2013全国2理T1)已知集合M=x|(x-1)24,xR,N=-1,0,1,2,3,则MN=()A.0,1,2B.-1,0,1,2C.-1,0,2,3D.0,1,2,348.(2013全国1文T1)已知集合A=1,2,3,

11、4,B=x|x=n2,nA,则AB=()A.1,4B.2,3C.9,16D.1,249.(2013全国2文T1)已知集合M=x|-3x1,N=-3,-2,-1,0,1,则MN=()A.-2,-1,0,1B.-3,-2,-1,0C.-2,-1,0D.-3,-2,-150.(2013上海理T15)设常数aR,集合A=x|(x-1)(x-a)0,B=x|xa-1.若AB=R,则a的取值范围为()A.(-,2) B.(-,2 C.(2,+) D.2,+)51.(2013广东理T8)设整数n4,集合X=1,2,3,n,令集合S=(x,y,z)|x,y,zX,且三条件xyz,yzx,zx0,B=x|5x5

12、,则()A.AB=B.AB=RC.BA D.AB55.(2012课标全国理T1)已知集合A=1,2,3,4,5,B=(x,y)|xA,yA,x-yA,则B中所含元素的个数为()A.3B.6C.8D.1056.(2012大纲理2)已知集合A=1,3,m,B=1,m,AB=A,则m=()A.0或3B.0或3C.1或3D.1或357.(2012全国文1)已知集合A=x|x2-x-20,B=x|-1x1,则()A.ABB.BAC.A=BD.AB=58.(2012大纲全国文T1,)已知集合A=x|x是平行四边形,B=x|x是矩形,C=x|x是正方形,D=x|x是菱形,则()A.ABB.CBC.DCD.A

13、D59.(2012湖北文T1)已知集合A=x|x2-3x+2=0,xR,B=x|0x5,xN,则满足条件ACB的集合C的个数为()A.1B.2C.3D.460.(2011全国文1)已知集合M=0,1,2,3,4,N=1,3,5,P=MN,则P的子集共有()A.2个B.4个C.6个D.8个61.(2011辽宁理T2)已知M,N为集合I的非空真子集,且M,N不相等,若N(IM)=,则MN=()A.MB.NC.ID.62.(2011广东理T8)设S是整数集Z的非空子集,如果a,bS,有abS,则称S关于数的乘法是封闭的.若T,V是Z的两个不相交的非空子集,TV=Z,且a,b,cT,有abcT;x,y

14、,zV,有xyzV,则下列结论恒成立的是()A.T,V中至少有一个关于乘法是封闭的B.T,V中至多有一个关于乘法是封闭的C.T,V中有且只有一个关于乘法是封闭的D.T,V中每一个关于乘法都是封闭的63.(2011福建文T12)在整数集Z中,被5除所得余数为k的所有整数组成一个“类”,记为k,即k=5n+k|nZ,k=0,1,2,3,4.给出如下四个结论:2 0111;-33;Z=01234;“整数a,b属于同一类”的充要条件是“a-b0”.其中,正确结论的个数是()A.1B.2C.3D.464.(2011福建理T1)i是虚数单位,若集合S=-1,0,1,则()A.iSB.i2SC.i3SD.2

15、iS65.(2010浙江理T1)设P=x|x4,Q=x|x24,则()A.PQ B.QPC.PRQD.QRP66.(2010天津理T9)设集合A=x|x-a|2,xR.若AB,则实数a,b必满足()A.|a+b|3B.|a+b|3C.|a-b|3D.|a-b|367.(2010全国T1)已知集合A=x|x|2,xR,B=x|x4,xZ,则AB等于()A.(0,2)B.0,2C.0,2D.0,1,268.(2018江苏T1)已知集合A=0,1,2,8,B=-1,1,6,8,那么AB=.69.(2017江苏T1)已知集合A=1,2,B=a,a2+3.若AB=1,则实数a的值为.70.(2013江苏

16、T4)集合-1,0,1共有个子集.71.(2013湖南,文T15)对于E=a1,a2,a100的子集X=ai1,ai2,aik,定义X的“特征数列”为x1,x2,x100,其中xi1=xi2=xik=1,其余项均为0.例如:子集a2,a3的“特征数列”为0,1,1,0,0,0.(1)子集a1,a3,a5的“特征数列”的前3项和等于;(2)若E的子集P的“特征数列”p1,p2,p100满足p1=1,pi+pi+1=1,1i99;E的子集Q的“特征数列”q1,q2,q100满足q1=1,qj+qj+1+qj+2=1,1j98,则PQ的元素个数为.72.(2012天津文T9,)集合A= xR|x-2

17、|5中的最小整数为.73.(2018北京理T20)设n为正整数,集合A=|=(t1,t2,tn),tk0,1,k=1,2,n.对于集合A中的任意元素=(x1,x2,xn)和=(y1,y2,yn),记M(,)=12(x1+y1-|x1-y1|)+(x2+y2-|x2-y2|)+(xn+yn-|xn-yn|).(1)当n=3时,若=(1,1,0),=(0,1,1),求M(,)和M(,)的值;(2)当n=4时,设B是A的子集,且满足:对于B中的任意元素,当,相同时,M(,)是奇数;当,不同时,M(,)是偶数.求集合B中元素个数的最大值;(3)给定不小于2的n,设B是A的子集,且满足:对于B中的任意两

18、个不同的元素,M(,)=0.写出一个集合B,使其元素个数最多,并说明理由.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.

19、【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.(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|

20、-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,

21、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=-2,0,1,2,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|

22、-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,

23、1),(0,1),(0,0),(0,-1),(1,-1),(1,0),(1,1),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)

24、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.1,2,3,4B.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,

25、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|-

26、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|-2x2.27.(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

27、,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【

28、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0,2,6,10构成的集合即为AB,即AB=0,2,6,10,故选C.33.(2016四川理T1)设集合A=x|-2x2,Z为整数集,则集合AZ中元素的个数是()A.3B.4C.5D.6【答案】C【解析】由题意,AZ=-2,-1,0,1,2,故其中的元素个数为5,选C.34.(

29、2016天津理T1)已知集合A=1,2,3,4,B=y|y=3x-2,xA,则AB=()A.1B.4 C.1,3D.1,4【答案】D【解析】由题意知集合B=1,4,7,10,则AB=1,4.故选D.35.(2016山东理T2)设集合A=y|y=2x,xR,B=x|x2-10,B=x|-1x-1,选C.36.(2016浙江理T1)已知集合P=xR|1x3,Q=xR|x24,则P(RQ)=()A.2,3B.(-2,3C.1,2)D.(-,-21,+)【答案】B【解析】Q=xR|x-2,或x2,RQ=xR|-2x2.P(RQ)=xR|-2x3=(-2,3.故选B.37.(2015全国2理T1)已知集

30、合A=-2,-1,0,1,2,B=x|(x-1)(x+2)0,则AB=()A.-1,0 B.0,1C.-1,0,1D.0,1,2【答案】A【解析】B=x|-2x1,AB=-1,0.38.(2015全国1文T1)已知集合A=x|x=3n+2,nN,B=6,8,10,12,14,则集合AB中元素的个数为()A.5B.4C.3D.2【答案】D【解析】由条件知,当n=2时,3n+2=8,当n=4时,3n+2=14.所以AB=8,14.故选D.39.(2015全国2文T1)已知集合A=x|-1x2,B=x|0x3,则AB=()A.(-1,3)B.(-1,0) C.(0,2) D.(2,3)【答案】A【解

31、析】由题意,得AB=x|-1x3,即AB=(-1,3).40.(2015陕西文T1)设集合M=x|x2=x,N=x|lg x0,则MN=()A.0,1B.(0,1 C.0,1)D.(-,1【答案】A【解析】M=0,1,N=x|0x1,MN=x|0x1,即为0,1.41.(2015重庆理T1,)已知集合A=1,2,3,B=2,3,则()A.A=BB.AB=C.ABD.BA【答案】D【解析】因为A=1,2,3,B=2,3,所以BA.42.(2014全国1理T1)已知集合A=x|x2-2x-30,B=x|-2x2,则AB=()A.-2,-1B.-1,2)C.-1,1D.1,2)【答案】A【解析】由已

32、知,可得A=x|x3或x-1,则AB=x|-2x-1=-2,-1.故选A.43.(2014全国2理T1)设集合M=0,1,2,N=x|x2-3x+20,则MN=()A.1B.2C.0,1D.1,2【答案】D【解析】N=x|x2-3x+20=x|1x2,MN=0,1,2x|1x2=1,2.故选D.44.(2014全国1文T1)已知集合M=x|-1x3,N=x|-2x1,则MN=()A.(-2,1)B.(-1,1)C.(1,3)D.(-2,3)【答案】B【解析】由已知得MN=x|-1x1=(-1,1),故选B.45.(2014全国2文T1)已知集合A=-2,0,2,B=x|x2-x-2=0,则AB=()A.B.2 C.0 D.-2【答案】B【解析】易得B=-1,2,则AB=2,故选B.46.(2014辽宁理T1)已知全集U=R,A=x|x0,B=x|x1,则集合U(AB)=()A.x|x0B.x|x1C.x|0x1D.x|0x1【答案】D【解析】AB=x|x0或x1,U(AB)=x|0x1.故选D.47.(2013全国2理T1)已知集合M=x|(x-1)24,x