九年(2010-2018年)高考真题文科数学精选(含解析)专题一 集合与常用逻辑用语第一讲 集合一、选择题1.(2018全国卷Ⅰ)已知集合{0,2}=A ,{21012}=--,,,,B ,则A B =A .{0,2}B .{1,2}C .{0}D .{21012}--,,,, 2.(2018浙江)已知全集{1,2,3,4,5}U =,{1,3}A =,则=U A A .∅ B .{1,3} C .{2,4,5} D .{1,2,3,4,5}3.(2018全国卷Ⅱ)已知集合{}1,3,5,7A =,{}2,3,4,5B =,则AB = A .{3} B .{5}C .{3,5}D .{}1,2,3,4,5,74.(2018北京)已知集合{|||2}A x x =<,{2,0,1,2}B =-,则AB = A .{0,1} B .{–1,0,1}C .{–2,0,1,2}D .{–1,0,1,2}5.(2018全国卷Ⅲ)已知集合{|10}A x x =-≥,{0,1,2}B =,则AB = A .{0} B .{1}C .{1,2}D .{0,1,2}6.(2018天津)设集合{1,2,3,4}A =,{1,0,2,3}B =-,{|12}C x x =∈-<R ≤,则()A B C =A .{1,1}-B .{0,1}C .{1,0,1}-D .{2,3,4}7.(2017新课标Ⅰ)已知集合{|2}A x x =<,{320}B x =->,则A .3{|}2AB x x =< B .A B =∅C .3{|}2A B x x =<D .A B =R 8.(2017新课标Ⅱ)设集合{1,2,3}A =,{2,3,4}B =错误!未找到引用源。
则A B = A .{1,2,3,4} B .{1,2,3} C .{2,3,4} D .{1,3,4}9.(2017新课标Ⅲ)已知集合{1,2,3,4}A =,{2,4,6,8}B =,则A B 中元素的个数为A .1B .2C .3D .410.(2017天津)设集合{1,2,6}A =,{2,4}B =,{1,2,3,4}C =,则()A B C =A .{2}B .{1,2,4}C .{1,2,4,6}D .{1,2,3,4,6}11.(2017山东)设集合{}11M x x =-<,{}2N x x =<,则M N = A .()1,1- B .()1,2- C .()0,2 D .()1,212.(2017北京)已知U =R ,集合{|22}A x x x =<->或,则U A = A .(2,2)- B .(,2)(2,)-∞-+∞ C .[2,2]- D .(,2][2,)-∞-+∞13.(2017浙江)已知集合{|11}P x x =-<<,{|02}Q x x =<<,那么P Q = A .(1,2)- B .(0,1) C .(1,0)- D .(1,2)14.(2016全国I 卷)设集合{1,3,5,7}A =,{|25}B x x =≤≤,则=A BA .{1,3}B .{3,5}C .{5,7}D .{1,7}15.(2016全国Ⅱ卷)已知集合{123}A =,,,2{|9}B x x =<,则A B =A .{210123}--,,,,,B .{21012}--,,,,C .{123},,D .{12},16.(2016全国Ⅲ)设集合{0,2,4,6,8,10},{4,8}A B ==,则A B =A .{48},B .{026},,C .{02610},,,D .{0246810},,,,,17.(2015新课标2)已知集合}21|{<<-=x x A ,}30|{<<=x x B ,则A B =A .)3,1(-B .)0,1(-C .)2,0(D .)3,2( 18.(2015新课标1)已知集合{32,},{6,8,10,12,14}A x x n n N B ==+∈=,则集合A B 中的元素个数为A .5B .4C .3D .219.(2015北京)若集合{|52}A x x =-<<,{|33}B x x =-<<,则A B =A .{|32}x x -<<B .{|52}x x -<<C .{|33}x x -<<D .{|53}x x -<< 20.(2015天津)已知全集{1,2,3,4,5,6}U =,集合{}2,3,5A =,集合{1,3,4,6}B =,则集合U A B =A .{3}B .{2,5}C .{1,4,6}D .{2,3,5}21.(2015陕西)设集合2{|}M x x x ==,{|lg 0}N x x =≤,则M N = A .[0,1] B .(0,1] C .[0,1) D .(-∞,1]22.(2015山东)已知集合{}24A x x =<<,{}(1)(3)0B x x x =--<,则AB = A .()1,3 B .()1,4C .()2,3D .()2,423.(2015福建)若集合{}22M x x =-≤<,{}0,1,2N =,则M N 等于 A .{}0 B .{}1 C .{}0,1,2 D .{}0,124.(2015广东)若集合{}1,1M =-,{}2,1,0N =-,则M N =A .{}0,1-B .{}1C .{}0D .{}1,1-25.(2015湖北)已知集合22{(,)|1,,}A x y x y x y Z =+∈≤,{(,)|||2,B x y x =≤ ||2,,}y x y Z ∈≤,定义集合12121122{(,)|(,),(,)}A B x x y y x y A x y B ⊕=++∈∈,则A B ⊕中元素的个数为A .77B .49C .45D .3026.(2014新课标)已知集合A ={x |2230x x --≥},B ={x |-2≤x <2},则A B = A .[-2, -1] B .[-1,1] C .[-1,2) D .[1,2)27.(2014新课标)设集合M ={0,1,2},N ={}2|320x x x -+≤,则M N = A .{1} B .{2} C .{0,1} D .{1,2}28.(2014新课标)已知集合A ={-2,0,2},B ={x |2x -x -20=},则A B =A . ∅B .{}2C .{}0D .{}2-29.(2014山东)设集合},]2,0[,2{},21{∈==<-=x y y B x x A x 则=B AA . [0,2]B .(1,3)C . [1,3)D . (1,4)30.(2014山东)设集合2{|20},{|14}A x x x B x x =-<=≤≤,则AB =A .(0,2]B .(1,2)C .[1,2)D .(1,4)31.(2014广东)已知集合{1,0,1}M =-,{0,1,2}N =,则MN = A .{0,1} B .{1,0,2}- C .{1,0,1,2}- D .{1,0,1}-32.(2014福建)若集合{|24}P x x =<≤,{|3}Q x x =≥,则P Q 等于A .}{34x x ≤<B .}{34x x <<C .}{23x x ≤<D .}{23x x ≤≤33.(2014浙江)设全集{}2|≥∈=x N x U ,集合{}5|2≥∈=x N x A ,则U A =A .∅B . }2{C . }5{D . }5,2{34.(2014北京)已知集合2{|20},{0,1,2}A x x x B =-==,则AB = A .{0} B .{0,1}C .{0,2}D .{0,1,2}35.(2014湖南)已知集合{|2},{|13}A x x B x x =>=<<,则A B =A .{|2}x x >B .{|1}x x >C .{|23}x x <<D .{|13}x x <<36.(2014陕西)已知集合2{|0},{|1,}M x x N x x x R =≥=<∈,则MN = A .[0,1] B .[0,1) C .(0,1] D .(0,1)37.(2014江西)设全集为R ,集合2{|90},{|15}A x x B x x =-<=-<≤,则()R A B =A .(3,0)-B .(3,1)--C .(3,1]--D .(3,3)-38.(2014辽宁)已知全集,{|0},{|1}U R A x x B x x ==≤=≥,则集合()U A B =A .{|0}x x ≥B .{|1}x x ≤C .{|01}x x ≤≤D .{|01}x x <<39.(2014四川)已知集合2{|20}A x x x =--≤,集合B 为整数集,则AB = A .{1,0,1,2}- B .{2,1,0,1}--C .{0,1}D .{1,0}-40.(2014湖北)已知全集{1,2,3,4,5,6,7}U =,集合{1,3,5,6}A =,则U A = A .{1,3,5,6} B .{2,3,7} C .{2,4,7} D . {2,5,7}41.(2014湖北)设U 为全集,B A ,是集合,则“存在集合C 使得A C ⊆,U B C ⊆”是“∅=B A ”的A .充分而不必要条件B .必要而不充分条件C .充要条件D .既不充分也不必要条件42.(2013新课标1)已知集合A ={x |x 2-2x >0},B ={x |-5<x <5},则A .A ∩B =∅ B .A ∪B =RC .B ⊆AD .A ⊆B 43.(2013新课标1)已知集合{1,2,3,4}A =,2{|,}B x x n n A ==∈,则AB =A .{}14,B .{}23,C .{}916,D .{}12, 44.(2013新课标2)已知集合(){}2|14,M x x x R =-<∈,{}1,0,1,2,3N =-,则M N =A .{}0,1,2B .{}1,0,1,2- C .{}1,0,2,3- D .{}0,1,2,3 45.(2013新课标2)已知集合{|31}M x x =-<<,{3,2,1,0,1}N =---,则MN = A .{2,1,0,1}-- B .{3,2,1,0}--- C .{2,1,0}-- D .{3,2,1}---46.(2013山东)已知集合B A 、均为全集}4,3,2,1{=U 的子集,且(){4}U A B =,{1,2}B =,则U AB = A .{3} B .{4}C .{3,4}D .∅ 47.(2013山东)已知集合A ={0,1,2},则集合B ={}|,x y x A y A -∈∈中元素的个数是A .1B .3C .5D .948.(2013安徽)已知{}{}|10,2,1,0,1A x x B =+>=--,则()R C A B ⋂=A .{}2,1--B .{}2-C .{}1,0,1-D .{}0,149.(2013辽宁)已知集合{}{}4|0log 1,|2A x x B x x A B =<<=≤=,则A .()01,B .(]02,C .()1,2D .(]12, 50.(2013北京)已知集合{}1,0,1A =-,{}|11B x x =-≤<,则A B =A .{}0B .{}1,0-C .{}0,1D .{}1,0,1-51.(2013广东)设集合2{|20,}S x x x x R =+=∈,2{|20,}T x x x x R =-=∈,则S T =A .{0}B .{0,2}C .{2,0}-D .{2,0,2}-52.(2013广东)设整数4n ≥,集合{}1,2,3,,X n =,令集合{(,,)|,,S x y z x y z X =∈,且三条件,,x y z y z x z x y <<<<<<恰有一个成立},若(),,x y z 和(),,z w x 都在S 中,则下列选项正确的是A .(),,y z w S ∈,(),,x y w S ∉B .(),,y z w S ∈,(),,x y w S ∈C .(),,y z w S ∉,(),,x y w S ∈D .(),,y z w S ∉,(),,x y w S ∉53.(2013陕西)设全集为R , 函数()f x M , 则C M R 为A . [-1,1]B . (-1,1)C .,1][1,)(∞-⋃+∞-D .,1)(1,)(∞-⋃+∞-54.(2013江西)若集合{}2|10A x R ax ax =∈++=中只有一个元素,则a =A .4B .2C .0D .0或4 55.(2013湖北)已知全集为R ,集合112x A x ⎧⎫⎪⎪⎛⎫=≤⎨⎬ ⎪⎝⎭⎪⎪⎩⎭,{}2|680B x x x =-+≤,则R A C B =A .{}|0x x ≤B .{}|24x x ≤≤C .{}|024x x x ≤<>或D .{}|024x x x <≤≥或 56.(2012广东)设集合{1,2,3,4,5,6},{1,3,5}U M ==;则U C M =A .{,,}246B .{1,3,5}C .{,,}124D .U57.(2012浙江)设全集{}1,2,3,4,5,6U =,设集合{}1,2,3,4P =,{}3,4,5Q =,则U P Q ⋂=A .{}1,2,3,4,6B .{}1,2,3,4,5C .{}1,2,5D .{}1,258.(2012福建)已知集合{1,2,3,4}M =,{2,2}N =-,下列结论成立的是A .N M ⊆B .M N M =C .MN N = D .{2}M N = 59.(2012新课标)已知集合2{|20}A x x x =--<,{|11}B x x =-<<,则A .AB B .B AC .A B =D .A B =∅60.(2012安徽)设集合A ={|3213x x --},集合B 为函数)1lg(-=x y 的定义域,则A ⋂B=A .(1,2)B .[1,2]C .[ 1,2)D .(1,2 ]61.(2012江西)若集合{1,1}A =-,{0,2}B =,则集合{|,,}z z x y x A y B =+∈∈中的元素的个数为A .5B .4C .3D .262.(2011浙江)若{|1},{|1}P x x Q x x =<=>-,则A .P Q ⊆B .Q P ⊆C .R C P Q ⊆D .R Q C P ⊆63.(2011新课标)已知集合M ={0,1,2,3,4},N ={1,3,5},P M N =⋂,则P 的子集共有A .2个B .4个C .6个D .8个64.(2011北京)已知集合P =2{|1}x x ≤,{}M a =.若PM P =,则a 的取值范围是A .(-∞, -1]B .[1, +∞)C .[-1,1]D .(-∞,-1][1,+∞) 65.(2011江西)若全集{1,2,3,4,5,6},{2,3},{1,4}U M N ===,则集合{5,6}等于A .M N ⋃B .M N ⋂C .()()n n C M C N ⋃D .()()n n C M C N ⋂66.(2011湖南)设全集{1,2,3,4,5}U M N =⋃=,{2,4}U M C N ⋂=,则N =A .{1,2,3}B .{1,3,5}C .{1,4,5}D .{2,3,4}67.(2011广东)已知集合A ={(,)|,x y x y 为实数,且221}x y +=,B ={(,)|,x y x y 为实数且1}x y +=,则A ⋂B 的元素个数为A .4B .3C .2D .168.(2011福建)若集合M ={-1,0,1},N ={0,1,2},则M ∩N 等于A .{0,1}B .{-1,0,1}C .{0,1,2}D .{-1,0,1,2}69.(2011陕西)设集合{}22||cos sin |,M y y x x x R ==-∈,1{|||N x x i =-<}i x R ∈为虚数单位,,则M N ⋂为A .(0,1)B .(0,1]C .[0,1)D .[0,1]70.(2011辽宁)已知M ,N 为集合I 的非空真子集,且M ,N 不相等,若N I M =∅,则=N MA .MB .NC .ID .∅ 71.(2010湖南)已知集合{}1,2,3M =,{}2,3,4N =,则A .M N ⊆B .N M ⊆C .{}2,3M N =D .{}1,4M N =72.(2010陕西)集合A ={}|12x x -≤≤,B ={}|1x x <,则()R A B ⋂=A .{}|1x x >B .{}|1x x ≥C .{}|12x x <≤D .{}|12x x ≤≤73.(2010浙江)设P ={x ︱x <4},Q ={x ︱2x <4},则A .P Q ⊆B .Q P ⊆C .R P Q ⊆D .R Q P ⊆ 74.(2010安徽)若集合121log 2A x x ⎧⎫⎪⎪=≥⎨⎬⎪⎪⎩⎭,则A =RA .2(,0],2⎛⎫-∞+∞ ⎪ ⎪⎝⎭B .22⎛⎫+∞ ⎪ ⎪⎝⎭C .2(,0][,)2-∞+∞D .2)2+∞ 75.(2010辽宁)已知,A B 均为集合U ={1,3,5,7,9}的子集,且{3}AB =,{9}U B A =,则A =A .{1,3}B .{3,7,9}C .{3,5,9}D .{3,9}二、填空题 76.(2018江苏)已知集合{0,1,2,8}A =,{1,1,6,8}B =-,那么A B = .77.(2017江苏)已知集合{1,2}A =,2{,3B a a =+},若{1}AB =,则实数a 的 值为____.78.(2015江苏)已知集合{}123A =,,,{}245B =,,,则集合AB 中元素的个数为 .79.(2015湖南)已知集合U ={}1,2,3,4,A ={}1,3,B ={}1,3,4,则A (U B )= .80.(2014江苏)已知集合A ={4,3,1,2--},}3,2,1{-=B ,则=B A .81.(2014重庆)设全集{|110}U n N n =∈≤≤,{1,2,3,5,8}A =,{1,3,5,7,9}B =,则()U A B ⋂= .82.(2014福建)若集合},4,3,2,1{},,,{=d c b a 且下列四个关系:①1=a ;②1≠b ;③2=c ;④4≠d 有且只有一个是正确的,则符合条件的有序数组),,,(d c b a 的个数是_________.83.(2013湖南)已知集合{2,3,6,8},{2,3},{2,6,8}U A B ===,则()U A B = .84.(2010湖南)若规定{}1210,,...,E a a a =的子集{}12,,...,n i i i a a a 为E 的第k 个子集,其中k =12111222n i i i ---++⋅⋅⋅+,则(1){}1,3,a a 是E 的第____个子集;(2)E 的第211个子集是_______.85.(2010江苏)设集合{1,1,3}A =-,2{2,4}B a a =++,{3}AB =,则实数a =__.专题一 集合与常用逻辑用语第一讲 集合答案部分1.A 【解析】由题意{0,2}A B =,故选A .2.C 【解析】因为{1,2,3,4,5}U =,{1,3}A =,所以=U A {2,4,5}.故选C .3.C 【解析】因为{}1,3,5,7A =,{}2,3,4,5B =,所以{3,5}A B =,故选C . 4.A 【解析】{|||2}(2,2)A x x =<=-,{2,0,1,2}B =-,∴{0,1}AB =,故选A . 5.C 【解析】由题意知,{|10}A x x =-≥,则{1,2}AB =.故选C . 6.C 【解析】由题意{1,0,1,2,3,4}A B =-,∴(){1,0,1}A B C =-,故选C .7.A 【解析】∵3{|}2B x x =<,∴3{|}2AB x x =<, 选A . 8.A 【解析】由并集的概念可知,{1,2,3,4}AB =,选A . 9.B 【解析】由集合交集的定义{2,4}AB =,选B . 10.B 【解析】∵{1,2,4,6}A B =,(){1,2,4}A BC =,选B .11.C 【解析】{|02}M x x =<<,所以{|02}MN x x =<<,选C . 12.C 【解析】{|22}U A x x =-≤≤,选C .13.A 【解析】由题意可知{|12}P Q x x =-<<,选A .14.B 【解析】由题意得,{1,3,5,7}A =,{|25}B x x =,则{3,5}A B =.选B . 15.D 【解析】易知{|33}B x x =-<<,又{1,2,3}A =,所以{1,2}AB =故选D . 16.C 【解析】由补集的概念,得{0,2,6,10}A B =,故选C .17.A 【解析】∵(1,2)A =-,(0,3)B =,∴(1,3)A B =-.18.D 【解析】集合{|32,}A x x n n N ==+∈,当0n =时,322n +=,当1n =时, 325n +=,当2n =时,328n +=,当3n =时,3211n +=,当4n =时, 3214n +=,∵{6,8,10,12,14}B =,∴A B 中元素的个数为2,选D .19.A 【解析】{|32}AB x x =-<<. 20.B 【解析】{2,5}U B =,∴U A B {2,5}.21.A 【解析】∵{0,1}M =,{|01}N x x ≤=<,∴M N =[0,1]. 22.C 【解析】因为{|13}B x x,所以(2,3)A B =,故选C . 23.D 【解析】∵{0,1}M N . 24.B 【解析】{1}M N =.25.C 【解析】由题意知,22{(,)1,,}{(1,0),(1,0),(0,1),(0,1)}A x y x y x y =+≤∈=--Z ,{(,)||2,||2,,}B x y x y x y =≤≤∈Z ,所以由新定义集合A B ⊕可知,111,0x y =±=或110,1x y ==±.当111,0x y =±=时,123,2,1,0,1,2,3x x +=---,122,1,0,1,2y y +=--,所以此时A B ⊕中元素的个数有:7535⨯=个;当110,1x y ==±时,122,1,0,1,2x x +=--,123,2,1,0,1,2,3y y +=---,这种情形下和第一种情况下除12y y +的值取3-或3外均相同,即此时有5210⨯=,由分类计数原理知,A B ⊕中元素的个数为351045+=个,故应选C .26.A 【解析】{}|13A x x x =-≤或≥,故A B =[-2, -1]. 27.D 【解析】{}|12N x x =≤≤,∴M N ={1,2}. 28.B 【解析】∵{}1,2B =-,∴A B ={}2.29.C 【解析】|1|213x x -<⇒-<<,∴(1,3)A =-,[1,4]B =.∴[1,3)AB =. 30.C 【解析】∵(0,2)A =,[1,4]B =,所以A B =[1,2).31.C 【解析】{}{}{}1,0,10,1,21,0,1,2M N ⋃=-⋃=-,选C .32.A 【解析】P Q =}{34x x ≤<.33.B 【解析】由题意知{|2}U x N x =∈≥,{|A x N x =∈,所以U A ={|2x N x ∈<≤,选B .34.C 【解析】∵{}{}2|200,2A x x x =-==.∴AB =={}0,2. 35.C 【解析】A B ={|23}x x <<.36.B 【解析】∵21x <,∴11x -<<,∴MN ={}|01x x <≤,故选B . 37.C 【解析】{}|3,3A x x =-<,{}|15R B x x x =->≤或, ∴()R A B ={}|31x x --≤≤.38.D 【解析】由已知得,{=0A B x x ≤或}1x ≥,故()U A B ={|01}x x <<.39.A 【解析】{|12}A x x =-≤≤,Z B =,故AB ={1,0,1,2}-. 40.C 【解析】{}2,4,7U A =.41.C 【解析】“存在集合C 使得,U A C B C ⊆⊆”⇔“∅=B A ”,选C . 42.B 【解析】A =(-∞,0)∪(2,+∞),∴AB =R ,故选B . 43.A 【解析】{}1,4,9,16B =,∴{}1,4AB =. 44.A 【解析】∵(1,3)M =-,∴{}0,1,2M N =.45.C 【解析】因为{31}M x x =-<<,{3,2,1,0,1}N =---, 所以M N {2,1,0}=--,选C .46.A 【解析】由题意{}1,2,3AB =,且{1,2}B =,所以A 中必有3,没有4, {}3,4U B =,故U A B ={}3.47.C 【解析】0,0,1,2,0,1,2x y x y ==-=--;1,0,1,2,1,0,1x y x y ==-=-;2,0,1,2,2,1,0x y x y ==-=.∴B 中的元素为2,1,0,1,2--共5个.48.A 【解析】A :1->x ,{|1}R A x x =-≤,(){1,2}R A B =--,所以答案选A49.D 【解析】由集合A ,14x <<;所以(1,2]AB =. 50.B 【解析】集合B 中含-1,0,故{}1,0A B =-.51.A 【解析】∵{}2,0S =-,{}0,2T =,∴ST ={}0. 52.B 【解析】特殊值法,不妨令2,3,4x y z ===,1w =,则()(),,3,4,1y z w S =∈,()(),,2,3,1x y w S =∈,故选B .如果利用直接法:因为(),,x y z S ∈,(),,z w x S ∈,所以x y z <<…①,y z x <<…②,z x y <<…③三个式子中恰有一个成立;z w x <<…④,w x z <<…⑤,x z w <<…⑥三个式子中恰有一个成立.配对后只有四种情况:第一种:①⑤成立, 此时w x y z <<<,于是(),,y z w S ∈,(),,x y w S ∈;第二种:①⑥成立, 此时x y z w <<<,于是(),,y z w S ∈,(),,x y w S ∈;第三种:②④成立, 此时y z w x <<<,于是(),,y z w S ∈,(),,x y w S ∈;第四种:③④成立, 此时z w x y <<<,于是(),,y z w S ∈,(),,x y w S ∈.综合上述四种情况,可得(),,y z w S ∈,(),,x y w S ∈.53.D 【解析】()f x 的定义域为M =[-1,1],故R M =(,1)(1,)-∞-⋃+∞,选D54.A 【解析】当0a =时,10=不合,当0a ≠时,0∆=,则4a =.55.C 【解析】[)0,A =+∞,[]2,4B =,∴[0,2)(4,)R AB =+∞. 56.A 【解析】U M ={,,}246. 57.D 【解析】{}3,4,5Q =,∴U Q ={}1,2,6,∴U P Q ={}1,2.58.D 【解析】由M ={1,2,3,4},N ={-2,2},可知-2∈N ,但是-2∉M ,则N ⊄M ,故A 错误.∵M N ={1,2,3,4,-2}≠M ,故B 错误.M∩N ={2}≠N ,故C 错误,D 正确.故选D .59.B 【解析】A =(-1,2),故B ⊂≠A ,故选B .60.D 【解析】{3213}[1,2]A x x =-≤-≤=-,(1,)(1,2]B A B =+∞⇒=.61.C 【解析】根据题意容易看出x y +只能取-1,1,3等3个数值.故共有3个元素.62.D 【解析】{|1}P x x =< ∴{|1}R P x x =≥,又∵{|1}Q x x =>, ∴R Q P ⊆,故选D .63.B 【解析】{1,3}P MN ==,故P 的子集有4个. 64.C 【解析】因为P M P =,所以M P ⊆,即a P ∈,得21a ≤,解得11a -≤≤,所以a 的取值范围是[1,1]-.65.D 【解析】因为{1,2,3,4}MN =,所以()()U U M N =()U M N ={5,6}. 66.B 【解析】因为U M N ⊂,所以()()()U U U U N N M N M == =[()]U U N M ={1,3,5}.67.C 【解析】由2211x y x y ⎧+=⎨+=⎩消去y ,得20x x -=,解得0x =或1x =,这时1y =或0y =,即{(0,1),(1,0)}A B =,有2个元素.68.A 【解析】集合{1,0,1}{0,1,2}={0,1}MN =-. 69.C 【解析】对于集合M ,函数|cos 2|y x =,其值域为[0,1],所以[0,1]M =,根据复<21x <,所以(1,1)N =-,则[0,1]M N =.70.A 【解析】根据题意可知,N 是M 的真子集,所以MN M =. 71.C 【解析】{}{}{}1,2,32,3,42,3MN ==故选C. 72.D 【解析】{}{}|1,|12R R B x x A B x x ==≥≤≤73.B 【解析】{}22<<x x Q -=,可知B 正确, 74.A 【解析】不等式121log 2x ,得12112201log log ()2x >⎧⎪⎨⎪⎩,得22x , 所以R A =2(,0],2⎛⎫-∞+∞ ⎪ ⎪⎝⎭. 75.D 【解析】因为{3}A B =,所以3∈A ,又因为{9}U B A =,所以9∈A ,所以选D .本题也可以用Venn 图的方法帮助理解.76.{1,8}【解析】由集合的交运算可得A B ={1,8}.77.1【解析】由题意1B ∈,显然1a =,此时234a +=,满足题意,故1a =.78.5【解析】{1,2,3}{2,4,5}{1,2,3,4,5}A B ==,5个元素.79.{1,2,3}【解析】{2}U B ,A (U B )={1,2,3}.80.{}1,3-【解析】=B A {}1,3-.81.{}7,9【解析】{}1,2,3,4,5,6,7,8,9,10U =,{}4,6,7,9,10U A =,{}()7,9U A B =.82.6【解析】因为①正确,②也正确,所以只有①正确是不可能的;若只有②正确,①③④都不正确,则符合条件的有序数组为(2,3,1,4),(3,2,1,4);若只有③正确,①②④都不正确,则符合条件的有序数组为(3,1,2,4);若只有④正确,①②③都不正确,则符合条件的有序数组为(2,1,4,3),(3,1,4,2),(4,1,3,2).综上符合条件的有序数组的个数是6.83.{}6,8【解析】()U A B ={6,8}{2,6,8}{6,8}=.84.【解析】(1)5 根据k 的定义,可知1131225k --=+=;(2)12578{,,,,}a a a a a 此时211k =,是个奇数,所以可以判断所求集中必含元素1a ,又892,2均大于211,故所求子集不含910,a a ,然后根据2j (j =1,2,⋅⋅⋅7)的值易推导出所求子集为12578{,,,,}a a a a a .85.1【解析】考查集合的运算推理.3∈B ,23a +=,1a =.。