高考数学《集合》专项
练习(选择题含答案) -CAL-FENGHAI-(2020YEAR-YICAI)_JINGBIAN
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《集合》专项练习参考答案
1.(2016全国Ⅰ卷,文1,5分)设集合,,则A ∩B =( )
(A ){1,3} (B ){3,5} (C ){5,7} (D ){1,7}
【解析】集合A 与集合B 的公共元素有3,5,故}5,3{=B A ,故选B .
2.(2016全国Ⅱ卷,文1,5分)已知集合,则A ∩B =( )
(A ) (B ) (C ) (D ) 【解析】由29x <得33x -<<,所以{|33}B x x =-<<,因为{1,2,3}A =,所以{1,2}A B =,故选D .
3.(2016全国Ⅲ卷,文1,5分)设集合{0,2,4,6,8,10},{4,8}A B ==,则A B =( )
(A ){48}, (B ){026},, (C ){02610},,, (D )
{0246810},,,,,
【解析】由补集的概念,得{0,2,6,10}A B =,故选C .
4.(2016全国Ⅰ卷,理1,5分)设集合,
,
则A ∩B =( ) (A ) (B ) (C ) (D ) 【解析】对于集合A :解方程x 2-4x +3=0得,x 1=1,x 2=3,所以A ={x |1<x <3}(大于取两边,小于取中间).对于集合B :2x -3>0,解得x >
23.3{|3}2
A B x x ∴=<<.选D .
5.2016全国Ⅱ卷,理1,5分)已知(3)(1)i z m m =++-在复平面内对应的点在第四象限,则实数m 的取值范围是( )
(A )(31)
-, (B )(13)-,(C )(1,)∞+(D )(3)∞--, 【解析】要使复数z 对应的点在第四象限,应满足3010
m m +>??-,解得
31m -<<,故选A .
6.(2016全国Ⅲ卷,理1,5分)设集合{}{}(x 2)(x 3)0,T 0S x x x =--≥=>,则S ∩T =( )
(A) [2,3] (B)(-∞ ,2] [3,+∞) (C) [3,+∞) (D)(0,2] [3,+∞)
{1,3,5,7}A ={|25}B x x =≤≤{123}A =,
,,2{|9}B x x =<{210123}--,,,,,{21012}--,,,,{123},
,{12},2{|430}A x x x =-+<{|230}B x x =->3(3,)2--3(3,)2-3(1,)2
3(,3)2
3
7.(2016北京,文1,5分)已知集合{|24},{|3>5}A x x B x x x =<<=<或,则A B =( )
(A ){|2<<5}x x (B ){|<45}x x x >或 (C ){|2<<3}x x (D ){|<25}x x x >或
【解析】画数轴得,,所以,故选C .
8.(2016北京,理1,5分)已知集合,,则( )
(A )(B )(C )(D )
【解析一】对于集合A :(解绝对值不等的常用方法是两边同时平方)|x |<2,两边同时平方得x 2<4,解方程x 2=4得,x 1=-2,x 2=2,所以A ={x |-2<x <2}(大于取两边,小于取中间).所以A ∩B ={-1,0,1}.故选C .
【解析二】对于集合A :(绝对值不等式解法二:|x |<2?-2<x <2).A ={x |-2<x <2}.所以A ∩B ={-1,0,1}.故选C .
9.(2016上海,文理1,5分)设x ∈R ,则不等式31x -<的解集为_______.
【答案】(24),
【解析】试题分析:421311|3|<<-<-?<-x x x ,故不等式1|3|<-x 的解集为)4,2(.
【解析一】对不等式31x -<:(解绝对值不等的常用方法是两边同时平方)|x -3|<1,两边同时平方得(x -3)2<1,解方程(x -3)2=1得,x 1=2,x 2=4,所以A ={x |2<x <4}.
【解析二】对于集合A :(绝对值不等式解法二:|x -3|<1?-1<x -3<1,解得2<x <4).A ={x |2<x <4}.
10.(2016山东,文1,5分)设集合{1,2,3,4,5,6},{1,3,5},{3,4,5}U A B ===,则
()U A B = (A ){2,6} (B ){3,6} (C ){1,3,4,5} (D ){1,2,4,6}
【答案】A
11.(2016山东,理2,5分)设集合2{|2,},{|10},x A y y x B x x ==∈=- (A )(1,1)- (B )(0,1) (C )(1,)-+∞ (D )(0,)+∞ 【答案】C 【解析】对于集合A :∵y =2x >0,∴A ={y |y >0}.对于集合B :∵x 2-1=0,解得x =±1,∴B ={x |-1<x <1}(大于取两边,小于取中间).∴A ∪B =(1,)-+∞ (2,3)A B ={|||2}A x x =<{1,0,1,2,3}B =-A B ={0,1}{0,1,2}{1,0,1}-{1,0,1,2}- 4 12.(2016四川,文2,5分)设集合A ={x |1≤x ≤5},Z 为整数集,则集合A∩Z 中 元素的个数是 (A)6 (B)5 (C)4 (D)3 【答案】B 【解析】{1,2,3,4,5}A =Z ,由Z 为整数集得Z ={…-3,-2,-1,0,1,2,3…}.故A Z 中元素的个数为5,选B . 13.(2016四川,理1,5分)设集合{|22}A x x =-≤≤,Z 为整数集,则A Z 中元素的个数是( ) (A )3(B )4(C )5(D )6 【答案】C 【解析】由题意,知{2,1,0,1,2}A =--Z ,由Z 为整数集得Z ={…-3,-2,-1,0,1,2,3…}.故A Z 中元素的个数为5,选C . 14.(2016天津,文1,5分)已知集合}3,2,1{=A ,},12|{A x x y y B ∈-==,则A B = (A )}3,1{ (B )}2,1{ (C )}3,2{ (D )}3,2,1{ 【答案】A 【解析】∵},12|{A x x y y B ∈-==,∴当x =1时,y =2×1-1=1;当x =2时,y =2×2-1=3;当x =3时,y =2×3-1=5.∴{1,3,5},{1,3}B A B ==.选A . 15.(2016天津,理1,5分)已知集合}{ 4,3,2,1=A ,}{A x x y y B ∈-==,23,则=B A (A )}{1 (B )}{4 (C )}{ 3,1 (D )}{4,1 【答案】D 【解析】∵}{A x x y y B ∈-==,23,∴当x =1时,y =3×1-2=1;当x =2 时,y =3×2-2=4;当x =3时,y =3×3-2=7;当x =4时,y =4×3-2=10. ∴{14710}{14}B =A B =,,,,,.选D . 16.(2016浙江,文1,5分)已知全集U ={1,2,3,4,5,6},集合P ={1, 3,5},Q ={1,2,4},则U P Q ()=( ) A .{1} B .{3,5} C .{1,2,4,6} D .{1,2,3,4,5} 【答案】C 17.(2016浙江,理1,5分)已知集合P ={x ∈R |1≤x ≤3},Q ={x ∈R |x 2≥4},则P ∪(C R Q )=( ) 5 A .[2,3] B .(-2,3] C .[1,2) D .(?∞,?2]∪[1,+∞) 【答案】B 【解析】对于集合Q :∵x 2=4,解得x =±2,∴B ={x |x ≤-2或x ≥2}(大于取两边,小于取中 间). 18.(2016江苏,文理1,5分)已知集合{1,2,3,6},{|23},A B x x =-=-<<则=A B _______. 【答案】{}1,2- 【解析】{}{}{}1,2,3,6231,2A B x x =--<<=-.故答案应填:{}1,2- 19.(2015全国Ⅰ卷,文1,5分)已知集合A ={x |x =3n +2,n ∈N},B ={6, 8,10,12,14},则集合A∩B 中元素的个数为( ) A .5 B .4 C .3 D .2 【答案】D 【解析】由已知得A ={2,5,8,11,14,17,…},又B ={6,8,10,12,14},所以A∩B ={8,14}. 20.(2015全国Ⅱ卷,文1,5分)已知集合A ={x |-1<x <2},B ={x |0<x <3},则A ∪B =( ) A .(-1,3) B .(-1,0) C .(0,2) D .(2,3) 【答案】A 【解析】因为A =(-1,2),B =(0,3),所以A ∪B =(-1,3),故选A . 21.(2014全国Ⅰ卷,文1,5分)已知集合M ={x |-1<x <3},N ={x |-2<x < 1},则M∩N =( ) A .(-2,1) B .(-1,1) C .(1,3) D .(-2,3) 【答案】B 【解析】M∩N ={x |-1<x <3}∩{x |-2<x <1}={x |-1<x <1}. 22.(2014全国Ⅱ卷,文1,5分)已知集合A ={-2,0,2},B ={x |x 2-x -2=0},则A∩B =( ) A .? B .{2} C .{0} D .{-2} 【答案】B 【解析】∵集合A ={-2,0,2},B ={x |x 2-x -2=0}={2,-1},∴A∩B ={2},故选B . 23.(2013全国Ⅰ卷,文1,5分)已知集合A ={1,2,3,4},B ={x |x =n 2, n ∈A},则A∩B =( ) A .{1,4} B .{2,3} C .{9,16} D .{1,2} 【答案】A 【解析】∵B ={x |x =n 2,n ∈A}={1,4,9,16},∴A∩B ={1,4},故选A . 6 24.(2013全国Ⅱ卷,文1,5分)已知集合M ={x |-3<x <1},N ={-3,-2, -1,0,1},则M∩N =( ) A .{-2,-1,0,1} B .{-3,-2,-1,0} C .{-2,-1,0} D .{-3,-2,-1} 【答案】C 【解析】由题意得M∩N ={-2,-1,0}.选C . 25.(2012全国卷,文1,5分)已知集合A ={x |x 2-x -2<0},B ={x |-1<x <1},则( ) (A )A ?≠B (B )B ?≠A (C )A =B (D )A∩B =? 【答案】B 【解析】A ={x |-1<x <2},B ={x |-1<x <1},则B ?≠A ,故选B . 26.(2011全国卷,文1,5分)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 【答案】B 【解析】由题意得P =M∩N ={1,3},∴P 的子集为?,{1},{3},{1,3},共4个. 27.(2010全国卷,文1,5分)已知集合,则 (A )(0,2)(B )[0,2](C )|0,2|(D )|0,1,2| 【解析】,,选D 28.(2009全国卷,文2,5分)设集合A ={4,5,7,9},B ={3,4,7,8, 9},全集,则集合中的元素共有( ) (A)3个 (B )4个 (C )5个 (D )6个 【解析】,.故选A . 29.(2008全国卷,文1,5分)已知集合M ={x |(x +2)(x -1)<0},N ={x |x +1<0},则M∩N =( ) A.(-1,1) B.(-2,1) C.(-2,-1) D.(1,2) 【答案】C 【解析】易求得{}{}|21,|1=-<<=<-M x x N x x ∴{}|21=-<<-M N x x 30.(2007全国卷,文1,5分)设{|210}S x x =+>,{|350}T x x =-<,则S T ?= A .? B .1{|}2x x < C .5{|}3x x > D .15{|}23 x x -<< 【答案】D . 2,,4,|A x x x R B x x Z =≤∈=∈A B ={}|22,{0,1,2}A x x B =-≤≤={}0,1,2A B =U A B =()U A B {3,4,5,7,8,9}A B ={4,7,9}(){3,5,8}U A B A B =∴=