歷年高考集合與常用邏輯用語(yǔ)(含答案解析)(理科)

編制僅供參考審核批準(zhǔn)生效日期地址：電話：傳真:郵編:(2013山東,2,5分)已知集合A={0,1,2},則集合B={x-y|x∈A,y∈A}中元素的個(gè)數(shù)是()A.1B.3C.5D.9(2014課標(biāo)Ⅰ,1,5分)已知集合A={x|x2-2x-3≥0},B={x|-2≤x<2},則A∩B=()A.[-2,-1]B.[-1,2)C.[-1,1]D.[1,2)(2012江西,1,5分)若集合A={-1,1},B={0,2},則集合{z|z=x+y,x∈A,y∈B}中的元素的個(gè)數(shù)為()A.5B.4C.3D.2(2014課標(biāo)Ⅱ,1,5分)設(shè)集合M={0,1,2},N={x|x2-3x+2≤0},則M∩N=()A.{1}B.{2}C.{0,1}D.{1,2}(2011廣東,2,5分)已知集合A={(x,y)|x,y為實(shí)數(shù),且x2+y2=1},B={(x,y)|x,y為實(shí)數(shù),且y=x},則A∩B的元素個(gè)數(shù)為()A.0B.1C.2D.3(2014北京,1,5分)已知集合A={x|x2-2x=0},B={0,1,2},則A∩B=()A.{0}B.{0,1}C.{0,2}D.{0,1,2}(2011福建,1,5分)i是虛數(shù)單位,若集合S={-1,0,1},則()A.i∈SB.i2∈SC.i3∈SD.∈S(2014山東,2,5分)設(shè)集合A={x||x-1|<2},B={y|y=2x,x∈[0,2]},則A∩B=()A.[0,2]B.(1,3)C.[1,3)D.(1,4)(2011北京,1,5分)已知集合P={x|x2≤1},M={a}.若P∪M=P,則a的取值范圍是()A.(-∞,-1]B.[1,+∞)C.[-1,1]D.(-∞,-1]∪[1,+∞)(2014遼寧,1,5分)已知全集U=R,A={x|x≤0},B={x|x≥1},則集合?U(A∪B)=()A.{x|x≥0}B.{x|x≤1}C.{x|0≤x≤1}D.{x|0<x<1}(2011遼寧,2,5分)已知M,N為集合I的非空真子集,且M,N不相等,若N∩?IM=?,則M∪N=()A.MB.NC.ID.?(2014浙江,1,5分)設(shè)全集U={x∈N|x≥2},集合A={x∈N|x2≥5},則?UA=()A.?B.{2}C.{5}D.{2,5}(2013江蘇,4,5分)集合{-1,0,1}共有個(gè)子集.

(2014廣東,1,5分)已知集合M={-1,0,1},N={0,1,2},則M∪N=()A.{0,1}B.{-1,0,2}C.{-1,0,1,2}D.{-1,0,1}(2014四川,1,5分)已知集合A={x|x2-x-2≤0},集合B為整數(shù)集,則A∩B=()A.{-1,0,1,2}B.{-2,-1,0,1}C.{0,1}D.{-1,0}(2014陜西,1,5分)設(shè)集合M={x|x≥0,x∈R},N={x|x2<1,x∈R},則M∩N=()A.[0,1]B.[0,1)C.(0,1]D.(0,1)(2014大綱全國(guó),2,5分)設(shè)集合M={x|x2-3x-4<0},N={x|0≤x≤5},則M∩N=()A.(0,4]B.[0,4)C.[-1,0)D.(-1,0](2013廣東,1,5分)設(shè)集合M={x|x2+2x=0,x∈R},N={x|x2-2x=0,x∈R},則M∪N=()A.{0}B.{0,2}C.{-2,0}D.{-2,0,2}(2013浙江,2,5分)設(shè)集合S={x|x>-2},T={x|x2+3x-4≤0},則(?RS)∪T=()A.(-2,1]B.(-∞,-4]C.(-∞,1]D.[1,+∞)(2013遼寧,2,5分)已知集合A={x|0<log4x<1},B={x|x≤2},則A∩B=()A.(0,1)B.(0,2]C.(1,2)D.(1,2](2013北京,1,5分)已知集合A={-1,0,1},B={x|-1≤x<1},則A∩B=()A.{0}B.{-1,0}C.{0,1}D.{-1,0,1}(2013課標(biāo)全國(guó)Ⅱ,1,5分)已知集合M={x|(x-1)2<4,x∈R},N={-1,0,1,2,3},則M∩N=()A.{0,1,2}B.{-1,0,1,2}C.{-1,0,2,3}D.{0,1,2,3}(2013重慶,1,5分)已知全集U={1,2,3,4},集合A={1,2},B={2,3},則?U(A∪B)=()A.{1,3,4}B.{3,4}C.{3}D.{4}(2012山東,2,5分)已知全集U={0,1,2,3,4},集合A={1,2,3},B={2,4},則(?UA)∪B為()A.{1,2,4}B.{2,3,4}C.{0,2,4}D.{0,2,3,4}(2012浙江,1,5分)設(shè)集合A={x|1<x<4},集合B={x|x2-2x-3≤0},則A∩(?RB)=()A.(1,4)B.(3,4)C.(1,3)D.(1,2)∪(3,4)(2012北京,1,5分)已知集合A={x∈R|3x+2>0},B={x∈R|(x+1)(x-3)>0},則A∩B=()A.(-∞,-1)B.C.D.(3,+∞)(2011山東,1,5分)設(shè)集合M={x|x2+x-6<0},N={x|1≤x≤3},則M∩N=()A.[1,2)B.[1,2]C.(2,3]D.[2,3](2014江蘇,1,5分)已知集合A={-2,-1,3,4},B={-1,2,3},則A∩B=.

(2014重慶,11,5分)設(shè)全集U={n∈N|1≤n≤10},A={1,2,3,5,8},B={1,3,5,7,9},則(?UA)∩B=.

(2011天津,13,5分)已知集合A={x∈R||x+3|+|x-4|≤9},B=,則集合A∩B=.

答案和解析[答案]C[解析]①當(dāng)x=0時(shí),y=0,1,2,此時(shí)x-y的值分別為0,-1,-2;②當(dāng)x=1時(shí),y=0,1,2,此時(shí)x-y的值分別為1,0,-1;③當(dāng)x=2時(shí),y=0,1,2,此時(shí)x-y的值分別為2,1,0.綜上可知,x-y的可能取值為-2,-1,0,1,2,共5個(gè),故選C.[答案]A[解析]由不等式x2-2x-3≥0解得x≥3或x≤-1,因此集合A={x|x≤-1或x≥3},又集合B={x|-2≤x<2},所以A∩B={x|-2≤x≤-1},故選A.[答案]C[解析]集合{z|z=x+y,x∈A,y∈B}={-1,1,3},故選C.[答案]D[解析]由已知得N={x|1≤x≤2},∵M(jìn)={0,1,2},∴M∩N={1,2},故選D.[答案]C[解析]解法一:A為圓心在原點(diǎn)的單位圓,B為過(guò)原點(diǎn)的直線,故有2個(gè)交點(diǎn),故選C.解法二:由可得或故選C.[答案]C[解析]A={0,2},B={0,1,2},∴A∩B={0,2}.故選C.[答案]B[解析]i2=-1,-1∈S,故選B.[答案]C[解析]A={x||x-1|<2}={x|-1<x<3},B={y|y=2x,x∈[0,2]}={y|1≤y≤4},∴A∩B={x|-1<x<3}∩{y|1≤y≤4}={x|1≤x<3}.[答案]C[解析]由P∪M=P,有M?P,∴a2≤1,∴-1≤a≤1,故選C.[答案]D[解析]A∪B={x|x≥1或x≤0},因此?U(A∪B)={x|0<x<1}.故選D.[答案]A[解析]∵N∩?IM=?,∴N?M.又M≠N,∴N?M,∴M∪N=M.故選A.[答案]B[解析]∵A={x∈N|x≥}={x∈N|x≥3},∴?UA={x∈N|2≤x<3}={2},故選B.[答案]8[解析]集合{-1,0,1}的子集有?,{-1},{0},{1},{-1,0},{-1,1},{0,1},{-1,0,1},共8個(gè).[答案]C[解析]由集合的并集運(yùn)算可得,M∪N={-1,0,1,2},故選C.[答案]A[解析]由x2-x-2≤0得-1≤x≤2,故集合A中的整數(shù)為-1,0,1,2.所以A∩B={-1,0,1,2}.[答案]B[解析]∵N=(-1,1),∴M∩N=[0,1),故選B.[答案]B[解析]M={x|x2-3x-4<0}={x|-1<x<4},則M∩N={x|0≤x<4}.故選B.[答案]D[解析]化簡(jiǎn)兩個(gè)集合,得M={-2,0},N={0,2},則M∪N={-2,0,2},故選D.[答案]C[解析]?RS={x|x≤-2},又T={x|-4≤x≤1},故(?RS)∪T={x|x≤1},選C.[答案]D[解析]A={x|0<log4x<1}={x|log41<log4x<log44}={x|1<x<4},A∩B=(1,2],故選D.[答案]B[解析]∵A={-1,0,1},B={x|-1≤x<1},∴A∩B={-1,0},故選B.[答案]A[解析]化簡(jiǎn)得M={x|-1<x<3},所以M∩N={0,1,2},故選A.[答案]D[解析]A∪B={1,2,3},?U(A∪B)={4}.故選D.[答案]C[解析]由題意知?UA={0,4},又B={2,4},∴(?UA)∪B={0,2,4},故選C.[答案]B[解析]B={x|-1≤x≤3},A∩(?RB)={x|3<x<4},故選B.[答案]D[解析]∵A=xx>-,B={x|x<-1或x>3},∴A∩B={x|x>3},故選D.[答案]A[解析]∵M(jìn)={x|-3<x<2},N={x|1≤x≤3},∴M∩N={x|1≤x<2}.[答案]{-1,3}