第一章集合与函数的概念一、选择题1.设全集U={1,2,3,4,5,6},设集合P={1,2,3,4},Q{3,4,5},则P∩(CUQ)=()A.{1,2,3,4,6}B.{1,2,3,4,5}C.{1,2,5}D.{1,2}2.设集合A={x|1x4},B={x|x2-2x-3≤0},则A∩(CRB)=()A.(1,4)B.(3,4)C.(1,3)D.(1,2)3.设集合{,}Aab,{,,}Bbcd,则AB()A.{}bB.{,,}bcdC.{,,}acdD.{,,,}abcd4.已知全集{0,1,2,3,4}U,集合{1,2,3}A,{2,4}B,则()UABð为()A.{1,2,4}B.{2,3,4}C.{0,2,4}D.{0,2,3,4}5.已知全集U={0,1,2,3,4,5,6,7,8,9},集合A={0,1,3,5,8},集合B={2,4,5,6,8},则()()UUCACB()A.{5,8}B.{7,9}C.{0,1,3}D.{2,4,6}6.已知集合A={x|x2-x-20},B={x|-1x1},则()A.ABB.BAC.A=BD.A∩B=7.若全集U={x∈R|x2≤4}A={x∈R||x+1|≤1}的补集CuA为()A.|x∈R|0x2|B.|x∈R|0≤x2|C.|x∈R|0x≤2|D.|x∈R|0≤x≤2|8.设集合21,0,1,|MNxxx,则MN()A.1,0,1B.0,1C.1D.09.已知集合2|320,,|05,AxxxxRBxxxN,则满足条件ACB的集合C的个数为()A.1B.2C.3D.410.(集合)设集合1,2,3,4,5,6U,1,3,5M,则UCM()A.2,4,6B.1,3,5C.1,2,4D.U11.已知集合1,2,3,4,2,2MN,下列结论成立的是()A.NMB.MNMC.MNND.2MN12.已知集合|Axx是平行四边形,|Bxx是矩形,|Cxx是正方形,|Dxx是菱形,则()A.ABB.CBC.DCD.AD13.已知集合320AxRx,(1)(3)0BxRxx,则AB=()A.(,1)B.2(1,)3C.2(,3)3D.(3,)14.已知集合{1,2,3,4,5}A,{(,),,}BxyxAyAxyA;,则B中所含元素的个数为()A.3B.6C.D.15.集合{|lg0}Mxx,2{|4}Nxx,则MN()A.(1,2)B.[1,2)C.(1,2]D.[1,2]16.已知全集0,1,2,3,4U,集合1,2,3,2,4AB,则UCAB为()A.1,2,4B.2,3,4C.0,2,4D.0,2,3,417.已知全集U={0,1,2,3,4,5,6,7,8,9},集合A={0,1,3,5,8},集合B={2,4,5,6,8},则)()(BCACUU为()A.{5,8}B.{7,9}C.{0,1,3}D.{2,4,6}18.设集合M={-1,0,1},N={x|x2≤x},则M∩N=()A.{0}B.{0,1}C.{-1,1}D.{-1,0,0}19.(集合)设集合1,2,3,4,5,6U,1,2,4M,则UCM()A.UB.1,3,5C.3,5,6D.2,4,620.已知集合1,3,,1,,AmBmABA,则m()A.0或3B.0或3C.1或3D.1或321.已知集合320AxRx,(1)(3)0BxRxx,则AB=()A.(,1)B.2(1,)3C.2(,3)3D.(3,)22.若集合A={-1,1},B={0,2},则集合{z︱z=x+y,x∈A,y∈B}中的元素的个数为()A.5B.4C.3D.2二、填空题23.集合|25AxRx中最小整数位_________.24.若集合}012|{xxA,}1|{xxB,则BA=_________.25.已知集合={||+2|3}AxRx,集合={|()(2)0}BxRxmx,且=(1,)ABn,则=m__________,=n___________.26.设全集{,,,}Uabcd,集合{,}Aab,{,,}Bbcd,则)()(BCACUU_______.27.若集合}012|{xxA,}21|{xxB,则BA=_________.28.已知集合[1,2,},{2,5}.AkB若{1,2,3,5},AB则k______.29.已知集合{124}A,,,{246}B,,,则AB____.1.【答案】D2.【解析】A=(1,4),B=(-1,3),则A∩(CRB)=(3,4).【答案】B3.[答案]D4.解析:}4,2,0{)(},4,0{BACACUU.答案选C.5.【答案】B.6.【解析】A=(-1,2),故BA,故选B.7.C【解析】{|22}Uxx,{|20}Axx,则{|02}UCAxx.8.【答案】B9.D【解析】求解一元二次方程,得10.解析:A.2,4,6UCM.11.【答案】D12.答案B13.【答案】D14.【解析】选D5,1,2,3,4xy,4,1,2,3xy,3,1,2xy,2,1xy共10个15.故选C.16.C.17.【答案】B18.【答案】B19.解析:C.3,5,6UCM.20.答案B21.【答案】D22.C23.解析:运用排除法,奇函数有1yx和||yxx,又是增函数的只有选项D正确.24.【答案】D25.B26.【答案】B27.【解析】3不等式52x,即525x,73x,所以集合}73{xxA,所以最小的整数为3.28.[解析]),(21A,)1,1(B,A∩B=)1,(21.29.【答案】1,130.[答案]{a,c,d}31.[解析]),(21A,)3,1(B,A∩B=)3,(21.32.333.【答案】1,2,4,6.