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1.1集合的概念與運(yùn)算A組基礎(chǔ)題組1.(2018浙江,1,4分)已知全集U=1,2,3,4,5,A=1,3,則UA=() A.B.1,3C.2,4,5D.1,2,3,4,5答案C本小題考查集合的運(yùn)算.U=1,2,3,4,5,A=1,3,UA=2,4,5.2.已知全集U=R,集合M=x|x2-2x-30,N=y|y=x2+1,則M(UN)=()A.x|-1x1B.x|-1x1C.x|1x3D.x|1x3答案A由題意得,M=x|-1x3,N=y|y1,則M(UN)=x|-1x1.3.已知集合M=x|x2+x-120,N=y|y=3x,x1,則集合x|xM且xN=()A.(0,3B.-4,3C.-4,0)D.-4,0答案D易得M=-4,3,N=(0,3,則x|xM且xN=-4,0,故選D.4.已知集合A=x|x2-3x0,B=1,a,且AB有4個(gè)子集,則實(shí)數(shù)a的取值范圍是()A.(0,3)B.(0,1)(1,3)C.(0,1)D.(-,1)(3,+)答案BAB有4個(gè)子集,AB中有2個(gè)元素,aA,a2-3a0,0aa,且(UA)B=R,則實(shí)數(shù)a的取值范圍是()A.(-,1)B.(-,1C.1,+)D.(1,+)答案A因?yàn)锳=x|x1,所以UA=x|x1,又(UA)B=R,所以a2C.a2D.a2答案D因?yàn)?-,a(-,2),所以a2,故選D.7.若集合A=x|y=lg(3x-x2),B=yy=1+4x+1,xA,則ARB=()A.(0,2B.(2,3)C.(3,5)D.(-2,-1)答案AA=(0,3),B=(2,5),A(RB)=(0,2.故選A.8.設(shè)U為全集,對(duì)集合A,B定義運(yùn)算“*”,A*B=U(AB),若X,Y,Z為三個(gè)集合,則(X*Y)*Z=()A.(XY)UZB.(XY)UZC.(UXUY)ZD.(UXUY)Z答案BX*Y=U(XY),對(duì)于任意集合X,Y,Z,(X*Y)*Z=U(XY)*Z=UU(XY)Z=(XY)UZ,故選B.9.已知集合A=xR|x-1x=0,則滿足AB=-1,0,1的集合B的個(gè)數(shù)是()A.2B.3C.4D.9答案C解方程x-1x=0,得x=1或x=-1,所以A=1,-1,又AB=-1,0,1,所以B=0或0,1或0,-1或0,1,-1,共有4個(gè).10.設(shè)A=x|x2-4x+30,B=x|ln(3-2x)0,則圖中陰影部分表示的集合為()A.-,32B.1,32C.1,32D.32,3答案BA=x|x2-4x+30=x|1x3,B=x|ln(3-2x)0=x|03-2x1=x|1x32,圖中陰影部分表示的集合為AB=x|1x32,故選B.11.已知集合A=1,2,B=a,a2+3.若AB=1,則實(shí)數(shù)a的值為.答案1解析B=a,a2+3,AB=1,a=1或a2+3=1,aR,a2+31,故a=1.經(jīng)檢驗(yàn),滿足題意.12.設(shè)集合A=x|(x-a)21,且2A,3A,則實(shí)數(shù)a的取值范圍為.答案1a2解析由題意得(2-a)21,(3-a)21,即1a3,a2或a4,所以1a2.13.(2019嘉興一中月考)已知集合A=2x,y-1x,1,B=x2,x+y,0,若A=B,則x+y=.答案2解析由題意,得A中必有零,又x0,所以y-1x=0,所以y=1.此時(shí)A=2x,0,1,B=x2,x+1,0.因?yàn)锳=B,所以2x=x2,x+1=1或2x=x+1,x2=1,即x=0(不合題意,舍去)或x=1.所以x+y=2.B組提升題組1.(2018福建廈門二模)已知集合A=-1,0,1,2,B=x|x=2n,nZ,則AB=() A.2B.0,2C.-1,0,2D.答案B因?yàn)榧螦=-1,0,1,2,B=x|x=2n,nZ,所以AB=0,2,故選B.2.已知集合A=1,2,B=x|x2-(a+1)x+a=0,aR,若A=B,則a=()A.1B.2C.-1D.-2答案BA=1,2,B=x|x2-(a+1)x+a=0,aR,由A=B,可得1、2是方程x2-(a+1)x+a=0的兩個(gè)根, 由根與系數(shù)的關(guān)系可得1+2=a+1,12=a,解得a=2,故選B.3.已知全集U=R,集合A=x|2x0,則A(UB)=()A.x|x0C.x|0x1答案A由2x1得x0,所以A=x|x0得x1,所以B=x|x1,所以UB=x|x1,所以A(UB)=x|x0,故選A.4.集合A=x|2x2-3x0,xZ,B=x|12x32,xZ,若ACB,則滿足條件的C的個(gè)數(shù)為()A.3B.4C.7D.8答案DA=x|2x2-3x0,xZ=x0x32,xZ=0,1,B=x|12x32,xZ=x|0x5,xZ=0,1,2,3,4,易知C=AM,其中M為集合2,3,4的子集,由子集個(gè)數(shù)公式可得C的個(gè)數(shù)為23=8.5.已知集合M=(x,y)|y=f(x),若對(duì)于任意(x1,y1)M,存在(x2,y2)M,使得x1x2+y1y2=0成立,則稱集合M是“垂直對(duì)點(diǎn)集”,給出下列兩個(gè)集合:M=(x,y)|y=1x;M=(x,y)|y=sinx+1.則以下選項(xiàng)正確的是()A.是“垂直對(duì)點(diǎn)集”,不是“垂直對(duì)點(diǎn)集”B.不是“垂直對(duì)點(diǎn)集”,是“垂直對(duì)點(diǎn)集”C.都是“垂直對(duì)點(diǎn)集”D.都不是“垂直對(duì)點(diǎn)集”答案B由題意知M為函數(shù)y=f(x)圖象上的點(diǎn)組成的集合,對(duì)于任意(x1,y1)M,存在(x2,y2)M,使得x1x2+y1y2=0成立,即函數(shù)y=f(x)的圖象上存在點(diǎn)P(x1,y1),Q(x2,y2)滿足OPOQ(O為坐標(biāo)原點(diǎn)).易知函數(shù)y=1x的圖象上,不存在點(diǎn)P(x1,y1),Q(x2,y2)滿足OPOQ(O為坐標(biāo)原點(diǎn)),且對(duì)于函數(shù)y=sinx+1圖象上任意一點(diǎn)P(x1,y1),都存在點(diǎn)Q(x2,y2)滿足OPOQ(O為坐標(biāo)原點(diǎn)),即不是“垂直對(duì)點(diǎn)集”,是“垂直對(duì)點(diǎn)集”,故選B.