2018届高三数学复习集合第一节集合夯基提能作业本文.docx

第一节集合A组基础题组1.(2014课标,1,5分)已知集合M=x|-1x3,N=x|-2x0,则MN=()A.0,1B.0,1,2 C.1,2D.-1,0,18.(2016辽宁沈阳模拟)设集合A=5,ba,a-b,B=b,a+b,-1,若AB=2,-1,则AB=()A.2,3B.-2,2,5C.2,3,5 D.-1,2,3,59.已知A=0,m,2,B=x|x3-4x=0,若A=B,则m=.10.已知集合A=x|-1x1,B=x|x2-2x0,则A(RB)=.11.已知集合A=x|1x5,C=x|-axa+3,若CA=C,则a的取值范围为.B组提升题组12.(2017山西大同模拟)已知全集为R,集合M=-1,0,1,5,N=x|x2-x-20,则M(RN)=()A.0,1 B.-1,0,1 C.0,1,5 D.-1,113.若集合A=xR|ax2+ax+1=0中只有一个元素,则a=()A.4 B.2 C.0 D.0或414.设集合M=x|-1x2,N=y|ya,若MN,则实数a的取值范围是()A.-1,2) B.(-,2 C.-1,+) D.(-1,+)15.(2016广西南宁模拟)已知全集U=xZ|0x0,若AB,则实数a的取值范围是()A.(-,-1) B.(-,-1 C.(-,-2) D.(-,-218.(2016辽宁沈阳二中月考)设x表示不大于x的最大整数,集合A=x|x2-2x=3,B=x|182x8,则AB=.答案全解全析A组基础题组1.BMN=x|-1x3x|-2x1=x|-1x1.2.A由题意可得B=1,3,5,AB=1,3,故选A.3.C化简A=y|y-1,因此AB=x|x2=B.4.A由题意知M=0,1,N=x|00=(0,+),MN=1,2.8.D由AB=2,-1,可得ba=2,a-b=-1或ba=-1,a-b=2,当ba=2,a-b=-1时,a=1,b=2,此时B=2,3,-1,所以AB=-1,2,3,5;当ba=-1,a-b=2时,a=1,b=-1,此时不符合题意,舍去.9.答案-2解析由题意知B=0,-2,2,若A=B,则m=-2.10.答案(-,12,+)解析由题意知B=x|x2-2x0=x|0x2,RB=(-,02,+),又A=-1,1,A(RB)=(-,12,+).11.答案a-1解析因为CA=C,所以CA.当C=时,满足CA,此时-aa+3,解得a -;当C时,要使CA,则有解得-32a-1.由,得a-1.B组提升题组12.A全集为R,N=x|x2-x-20=x|x-1或x2,RN=x|-1x-1,故选D.15.C由已知得U=1,2,3,4,5,6,7,N=2,6,又M=2,3,5,所以UN=1,3,4,5,7,UM=1,4,6,7,MN=2,3,5,6,MN=2,所以M(UN)=3,5,U(MN)=1,3,4,5,6,7,(UM)N=6,U(MN)=1,4,7,故选C.16.D由x2-2x-30,得(x+1)(x-3)0,则-1x3,又xN,故集合A=0,1,2,3.由题意知A*B中的元素有0+1=1,0+3=3,1+1=2,1+3=4,2+1=3(舍去),2+3=5,3+1=4(舍去),3+3=6,A*B=1,2,3,4,5,6,A*B中的所有元素之和为1+2+3+4+5+6=21.17.BA=x|y=lg(-x2+x+2)=x|-1xa.因为AB,所以a-1.18.答案-1,7解析x2-2x=3,x=x2-32,又xxx+1,-1x1-2或1+2x3