高考数学《集合》专项练习(选择题含答案)
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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30
10
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,+∞)
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
{1,3,5,7}A ={|25}B x x =≤≤{1
23}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,)23(,3)2
(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}-
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 ==∈=-<R 则A ∪B =( )
(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,)-+∞
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 )=( )
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 . 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 =∴
=。