北京市2020版高考数学第一章1第一节集合夯基提能作业本.docx

第一节集合A组基础题组1.(2018北京顺义二模,1)设集合A=x|x2+3x+2=0,B=-2,-1,0,1,2,则AB=()A.-2,-1B.-2,1C.1,2D.-2,-1,0,1,2答案A集合A=x|x2+3x+2=0=-2,-1,B=-2,-1,0,1,2,则AB=-2,-1,故选A.2.(2018北京房山一模,1)若集合M=-1,0,1,2,N=y|y=2x+1,xM,则集合MN等于()A.-1,1B.1,2C.-1,1,3,5D.-1,0,1,2答案A由已知得N=-1,1,3,5,所以MN=-1,1,故选A.3.(2019北京朝阳高三期末,1)已知集合A=xN|1x3,B=2,3,4,5,则AB=()A.2,3B.2,3,4,5C.2D.1,2,3,4,5答案D4.(2017北京,1,5分)若集合A=x|-2x1,B=x|x3,则AB=()A.x|-2x-1 B.x|-2x3C.x|-1x1 D.x|1x3答案A由集合的交集运算可得AB=x|-2x1,B=x|log2x1,则AB=()A.x|x1B.x|1x2D.x|x0答案CB=x|log2x1=x|x2,又A=x|x1,AB=x|x2,故选C.6.(2017北京朝阳一模,1)已知集合A=x|-1x3,B=xZ|x22,那么AB=()A.(2,4)B.(2,4C.1,+)D.(2,+)答案CA=x|1x4,B=x|x2,AB=x|x1.8.(2018北京西城一模,1)若集合A=x|3x+20,B=x|x2-2x-30,则AB=()A.x|x3答案D由题意得A=,B=x|x3,所以AB=x|x3,故选D.9.设全集U=R,集合A=x|0x2,B=x|x1,则集合(UA)B=()A.(-,0)B.(-,0C.(2,+)D.2,+)答案B由题意得UA=(-,02,+),又B=(-,1),(UA)B=(-,0,故选B.10.集合A=x|xa,B=x|x2-5x0,若AB=B,则a的取值范围是()A.a5 B.a4 C.a5D.a4答案AB=x|x2-5x0=x|0x0,则下列结论正确的是()A.MNB.M(RN)C.(RM)ND.(RM)(RN)答案B由题意得N=x|x3,所以RN=x|-1x3,又M=x|0x2,所以M是RN的子集,故选B.13.(2017北京丰台二模,7)S(A)表示集合A中所有元素的和,且A1,2,3,4,5,若S(A)能被3整除,则符合条件的非空集合A的个数是()A.10B.11C.12D.13答案BA1,2,3,4,5,且S(A)能被3整除,按A中元素个数分类如下:1个:3;2个:1,2,1,5,2,4,4,5;3个:1,2,3,1,3,5,2,3,4,3,4,5;4个:1,2,4,5;5个:1,2,3,4,5,共11个,故选B.14.已知集合A=1,B=-1,2m-1.若AB,则实数m的值为.答案1解析若AB,则必有1=2m-1,解得m=1,验证可知符合集合元素的互异性.15.已知集合A=x|x2或x-1,B=x|axb,若AB=R,AB=x|22或x-1,B=x|axb,AB=R,AB=x|22,B=x|(x-1)(x-3)1B.x|2x3C.x|1x2或x1答案B由(x-1)(x-3)0得1x3,即B=x|1x2,AB=x|2x3.故选B.17.(2018北京石景山一模,1)设集合A=x|(x+1)(x-2)0,集合B=x|1x3,则AB=()A.x|-1x3B.x|-1x1C.x|1x2D.x|2x3答案AA=x|-1x2,B=x|1x3,AB=x|-1x1,B=x|x1,则AB=()A.(1,2B.(1,+)C.(1,2)D.1,+)答案D由log2x1,得x2,所以AB=1,+),故选D.19.(2018北京朝阳一模,1)已知全集为实数集R,集合A=x|x2-3x1,则(RA)B=()A.(-,03,+)B.(0,1C.3,+)D.1,+)答案C集合A=x|x2-3x0=x|x(x-3)0=x|0x1=x|2x20=x|x0,所以RA=x|x0或x3,所以(RA)B=x|x3,故选C.易错警示求补集时注意全集的范围.20.已知集合U=R,A=x|x2-5x-60,B=x|2x1,则图中阴影部分表示的集合是()A.x|2x3B.x|-1x0C.x|0x6D.x|x-1答案C由x2-5x-60,解得-1x6,所以A=x|-1x6.由2x1,解得x0,所以B=x|x0.又图中阴影部分表示的集合为(UB)A,因为UB=x|x0,所以(UB)A=x|0x6,故选C.21.(2017北京朝阳二模,14)已知两个集合A,B,满足BA.若对任意的xA,存在ai,ajB(ij),使得x=1ai+2aj(1,2-1,0,1),则称B为A的一个基集,若A=1,2,3,4,5,6,7,8,9,10,则基集B中元素个数的最小值是.答案4解析设B中元素a1a2an,aiaj,则1ai+0aj有n种,1ai+1aj有n种,1ai-1aj有种,-1ai+1aj有种,n+n+10,n2+n10,n3.当n=3时,共12种,最多有两种不符合题意,设B=a1,a2,a3,a1a2a3,则2a310,2a210,a35,a25.a3=5时,a3+a2=9,a2=4,a3+a1=7或a2+a1=7,a1=2或3,B=5,4,3(舍)或B=5,4,2(舍);a3=6时,若a2=5,则a3+a1=7或a2+a1=7,a1=1或2,B=6,5,2(舍)或B=6,5,1(舍),若a2=4,则a1+a3=9,a1=3,B=6,4,3(舍);a3=7时,a1+a310,a13,若a1=3,则3a25,无法构成9,若a1=2,则a2+a3=10或2a2=10,a2=3或5,B=7,5,2(舍)或B=7,3,2(舍),若a1=1,则a2+a3=10或2a2=10,a2=3或5,B=7,5,1(舍)或B=7,3,1(舍);a3=8时,a1+a810,a1=1或2,若a1=1,则a2+a3=10或2a2=10,a2=2或5,B=8,5,1(舍)或B=8,2,1(舍),若a1=2,则2a25,无法构成9;a3=9时,a1=1,110,a3+a210