集合
1.(2019全国Ⅰ文2)已知集合U则BIeUA?( ) A.
??1,2,3,4,5,6,7?,A??2,3,4,5?,B??2,3,6,7?,
?1,6?
B.
?1,7?
C.
?6,7?
D.
?1,6,7?
2.(2019全国Ⅱ文1)已知集合A={x|x??1},B?{x|x?2},则A∩B=( ) A.(–1,+∞) C.(–1,2)
B.(–∞,2) D.?
2
3.(2019全国Ⅲ文1)已知集合A?{?1,0,1,2},B?{xx?1},则AIB?( ) A.??1,0,1?
B.?0,1?
C.??1,1?
D.?0,1,2?
4.(2019北京文1)已知集合A={x|–1
(B)(1,2)
(C)(–1,+∞)
(D)(1,+∞)
5.(2019天津文1)设集合A???1,1,2,3,5?,B??2,3,4? ,C?{x?R|1?x?3} ,则
(AIC)UB?( )
(A){2}
(B){2,3}
(C){-1,2,3}
(D){1,2,3,4}
6.(2019江苏1)已知集合A?{?1,0,1,6},B?{x|x?0,x?R},则AIB? . 7.(2019浙江1) 已知全集U???1,0,1,2,3?,集合A??0,1,2?,B???1,0,1?,则eUAIB=( ) A.??1?
B.?0,1?
C.??1,2,3? D.??1,0,1,3?
?1,0,1,2},则AIB? 8.(2018全国卷Ⅰ)已知集合A?{0,2},B?{?2,A.{0,2} B.{1,2}
C.{0} ?1,0,1,2} D.{?2,9.(2018浙江)已知全集U?{1,2,3,4,5},A?{1,3},则eUA= A.?
B.{1,3}
C.{2,4,5}
D.{1,2,3,4,5}
10.(2018全国卷Ⅱ)已知集合A??1,3,5,7?,B??2,3,4,5?,则AIB? A.{3}
B.{5}
C.{3,5}
D.?1,2,3,4,5,7?
11.(2018北京)已知集合A?{x||x|?2},B?{?2,0,1,2},则AIB?
A.{0,1}
B.{–1,0,1} C.{–2,0,1,2} D.{–1,0,1,2}
12.(2018全国卷Ⅲ)已知集合A?{x|x?1≥0},B?{0,1,2},则AIB? A.{0}
B.{1}
C.{1,2}
D.{0,1,2}
13.(2018天津)设集合A?{1,2,3,4},B?{?1,0,2,3},C?{x?R|?1≤x?2},则
(AUB)IC?
A.{?1,1}
B.{0,1} C.{?1,0,1} D.{2,3,4}
14.(2017新课标Ⅰ)已知集合A?{x|x?2},B?{3?2x?0},则
323C.AUB?{x|x?} D.AUB?R
2A.AIB?{x|x?} B.AIB??
15.(2017新课标Ⅱ)设集合A?{1,2,3},B?{2,3,4},则AUB=
A.{1,2,3,4} B.{1,2,3} C.{2,3,4} D.{1,3,4}
16.(2017新课标Ⅲ)已知集合A?{1,2,3,4},B?{2,4,6,8},则AIB中元素的个数为
A.1 B.2 C.3 D.4
17.(2017天津)设集合A?{1,2,6},B?{2,4},C?{1,2,3,4},则(AUB)IC?
A.{2} B.{1,2,4} C.{1,2,4,6} D.{1,2,3,4,6} 18.(2017山东)设集合M?xx?1?1,则MIN? N?xx?2,A.??1,1? B.??1,2?
C.?0,2?
D.?1,2?
????19.(2017北京)已知U?R,集合A?{x|x??2或x?2},则eUA=
A.(?2,2) B.(??,?2)U(2,??) C.[?2,2] D.(??,?2]U[2,??) 20.(2017浙江)已知集合P?{x|?1?x?1},Q?{x|0?x?2},那么PUQ=
A.(?1,2) B.(0,1) C.(?1,0) D.(1,2)
答案
1.解析 因为U??1,2,3,4,5,6,7?,A?{2,3,4,,5}B?{2,3,6,7},
7}?6,7}, 则BIeUA?{6,所以CUA?{1,. 故选C.
2.解析 A?(?1,??),B?(??,2),AIB?(?1,2).故选C. 3.解析 因为A???1,0,1,2?,B?{x|x2剟1}?{x|?1所以AIB???1,0,1?.故选A.
4.解析 由数轴可知,AUB?xx?1.故选C.
5.解析 设集合A???1,1,2,3,5?,C?x?R1?x?3, 则AIC??1,2?. 又B??2,3,4?, 所以?AIC?UB??1,2?U?2,3,4???1,2,3,4?. 故选D.
6.解析 因为A???1,0,1,6?,B??x|x?0,x?R?, 所以AIB???1,0,1,6?Ix?1},
?????x|x?0,x?R???1,6?.
}.故选A. 7.解析 eUAIB?{?1UA?{?1,3},e8.A【解析】由题意AIB?{0,2},故选A.
9.C【解析】因为U?{1,2,3,4,5},A?{1,3},所以eUA={2,4,5}.故选C. 10.C【解析】因为A??1,3,5,7?,B??2,3,4,5?,所以AIB?{3,5},故选C.
11.A【解析】A?{x||x|?2}?(?2,2),B?{?2,0,1,2},∴AIB?{0,1},故选A. 12.C【解析】由题意知,A?{x|x?1≥0},则AIB?{1,2}.故选C.
13.C【解析】由题意AUB?{?1,0,1,2,3,4},∴(AUB)IC?{?1,0,1},故选C. 14.A【解析】∵B?{x|x?},∴AIB?{x|x?}, 选A. 15.A【解析】由并集的概念可知,AUB?{1,2,3,4},选A. 16.B【解析】由集合交集的定义AIB?{2,4},选B.
17.B【解析】∵AUB?{1,2,4,6},(AUB)IC?{1,2,4},选B.
323218.C【解析】M?{x|0?x?2},所以MIN?{x|0?x?2},选C. 19.C【解析】eUA?{x|?2≤x≤2},选C.
20.A【解析】由题意可知PUQ?{x|?1?x?2},选A.
三年高考(2017-2019)各地文科数学高考真题分类汇总:集合
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