1、第1课时并集、交集课后训练巩固提升A组1.已知集合M=1,2,3,4,5,6,集合N=xN|3x6,则NM等于()A.x|4x6B.x|1x6C.1,2,3,4,5,6D.4,5解析:N=4,5,MN=4,5.答案:D2.设集合A=x|-4x3,B=x|x2,则AB=()A.x|-4x3B.x|-4x2C.x|x2D.x|x3解析:由数轴得AB=x|-4x2.答案:B3.已知集合A=x|x0,则()A.AB=xx32B.AB=C.AB=xx32D.AB=R解析:A=x|x0=xx32,AB=x|x2,AB=xx0,3x+60,集合B=x|2x-10,3x+60,得-2x3,即A=x|-2x3.
2、解不等式2x-13,得x2,即B=x|x2,在数轴上分别表示集合A,B,如图所示.则AB=x|-2x2,AB=x|x3.10.已知集合A=x|x2+4x=0,B=x|x2+2(a+1)x+a2-1=0.(1)若AB=B,求实数a的值;(2)若AB=B,求实数a的值或取值范围.解:(1)A=-4,0.若AB=B,则B=A=-4,0,解得a=1.(2)若AB=B,则BA.若B为空集,则=4(a+1)2-4(a2-1)=8a+80,解得aa,若AB,则a的取值范围是()A.a8C.a-3D.-3a,要使AB,借助数轴可知a5,或x-1,T=x|axa+8,ST=R,则a的取值范围是()A.-3a-1
3、B.-3a-1C.a-3或a-1D.a-1解析:ST=R,a+85,a-1,-3a-1.答案:A5.已知集合A=0,2,a,B=1,a2.若AB=0,1,2,4,16,则a的值为.解析:A=0,2,a,B=1,a2,AB=0,1,2,4,16,a=4,a2=16或a2=4,a=16(舍去),解得a=4.答案:46.已知集合A=x|x5,B=x|axb,且AB=R,AB=x|5x6,则2a-b=.解析:如图所示,可知a=1,b=6,2a-b=-4.答案:-47.已知集合A=x|2x2-ax+b=0,B=x|bx2+(a+2)x+5+b=0,且AB=12,求AB.解:AB=12,12A,且12B.
4、2122-12a+b=0,b122+12(a+2)+5+b=0,解得a=-439,b=-269,A=x|18x2+43x-26=0=12,-269,B=x|26x2+25x-19=0=12,-1913.AB=12,-269,-1913.8.已知集合A=x|2a+1x3a-5,B=x|x16.(1)若AB=,求实数a的取值范围;(2)若A(AB),求实数a的取值范围.解:(1)若A=,则AB=成立.此时2a+13a-5,即a6.若A,如图,则2a+13a-5,2a+1-1,3a-516,解得6a7.经检验,a=6,a=7符合题意.综上,满足条件AB=的实数a的取值范围是a7.(2)因为A(AB),所以AB=A,即AB.显然A=满足条件,此时a6.若A,如图,则2a+13a-5,3a-516.由2a+13a-5,3a-516,解得a152.综上,满足条件A(AB)的实数a的取值范围是a152.3
