2020-2021学年高考数学(理)考点:集合
1.集合与元素
(1)集合中元素的三个特征:确定性、互异性、无序性. (2)元素与集合的关系是属于或不属于,用符号∈或?表示. (3)集合的表示法:列举法、描述法、图示法. (4)常见数集的记法
集合 自然数集
正整数集 整数集 有理数集
实数集 符号
N
N *(或N +)
Z
Q
R
2.集合的基本关系
(1)子集:若对于任意的x ∈A 都有x ∈B ,则A ?B ; (2)真子集:若A ?B ,且A ≠B ,则A B ; (3)相等:若A ?B ,且B ?A ,则A =B ;
(4)?是任何集合的子集,是任何非空集合的真子集. 3.集合的基本运算
表示 运算 文字语言
集合语言 图形语言 记法
交集
属于A 且属于B 的所有元素组成的集合
{x |x ∈A ,且x ∈B }
A
∩B
并集 属于A 或属于B 的元素组成的集合
{x |x ∈A ,或x ∈B }
A ∪B
补集
全集U 中不属于A 的元素
组成的集合称为集合A 相对于集合U 的补集
{x |x ∈U ,x ?A }
?U A
概念方法微思考
1.若一个集合A 有n 个元素,则集合A 有几个子集,几个真子集. 提示 2n ,2n -1.
2.从A ∩B =A ,A ∪B =A 中可以分别得到集合A ,B 有什么关系? 提示 A ∩B =A ?A ?B ,A ∪B =A ?B ?A .
1.(2020?新课标Ⅲ)已知集合{(,)|A x y x =,*y N ∈,}y x ,{(,)|8}B x y x y =+=,则A
B 中
元素的个数为( ) A .2 B .3 C .4 D .6
【答案】C
【解析】集合{(,)|A x y x =,*y N ∈,}y x ,{(,)|8}B x y x y =+=, {(A
B x ∴=,*)|,}{(1,7)8,y x
y x y N x y ?∈=?+=?,(2,6),(3,5),(4,4)}.
A
B ∴中元素的个数为4.
故选C .
2.(2020?新课标Ⅲ)已知集合{1A =,2,3,5,7,11},{|315}B x x =<<,则A
B 中元素的个
数为( ) A .2 B .3 C .4 D .5
【答案】B
【解析】集合{1A =,2,3,5,7,11},{|315)B x x =<<, {5A B ∴=,7,11}, A
B ∴中元素的个数为3.
故选B .
3.(2020?新课标Ⅱ)已知集合{|||3A x x =<,}x Z ∈,{|||1B x x =>,}x Z ∈,则(A
B = )
A .?
B .{3-,2-,2,3}
C .{2-,0,2}
D .{2-,2}
【答案】D
【解析】集合{|||3A x x =<,}{|33x Z x x ∈=-<<,}{2x Z ∈=-,1-,0,1,2}, {|||1B x x =>,}{|1x Z x x ∈=<-或1x >,}x Z ∈, {2A
B ∴=-,2}.
故选D .
4.(2020?新课标Ⅰ)已知集合2{|340}A x x x =--<,{4B =-,1,3,5},则(A
B = )
A .{4-,1}
B .{1,5}
C .{3,5}
D .{1,3}
【答案】D
【解析】集合2{|340}(1,4)A x x x =--<=-,{4B =-,1,3,5},
则{1A B =,3},
故选D .
5.(2020?山东)设集合{|13}A x x =,{|24}B x x =<<,则(A
B = )
A .{|23}x x <
B .{|23}x x
C .{|14}x x <
D .{|14}x x <<
【答案】C
【解析】集合{|13}A x x =,{|24}B x x =<<, {|14}A
B x x ∴=<.
故选C .
6.(2020?浙江)已知集合{|14}P x x =<<,{|23}Q x x =<<,则(P
Q = )
A .{|12}x x <
B .{|23}x x <<
C .{|34}x x <
D .{|14}x x <<
【答案】B
【解析】集合{|14}P x x =<<,{|23}Q x x =<<, 则{|23}P
Q x x =<<.
故选B .
7.(2020?海南)某中学的学生积极参加体育锻炼,其中有96%的学生喜欢足球或游泳,60%的学生喜欢足球,82%的学生喜欢游泳,则该中学既喜欢足球又喜欢游泳的学生数占该校学生总数的比例是( ) A .62% B .56%
C .46%
D .42%
【答案】C
【解析】设只喜欢足球的百分比为x ,只喜欢游泳的百分比为y ,两个项目都喜欢的百分比为z ,
由题意,可得60x z +=,96x y z ++=,82y z +=,解得46z =.
∴该中学既喜欢足球又喜欢游泳的学生数占该校学生总数的比例是46%.
故选C .
8.(2020?海南)设集合{2A =,3,5,7},{1B =,2,3,5,8},则(A
B = )
A .{1,3,5,7}
B .{2,3}
C .{2,3,5}
D .{1,2,3,5,7,8}
【答案】C
【解析】因为集合A ,B 的公共元素为:2,3,5 故{2A
B =,3,5}.
故选C .
9.(2020?天津)设全集{3U =-,2-,1-,0,1,2,3},集合{1A =-,0,1,2},{3B =-,0,2,3},则()(U A B =? ) A .{3-,3} B .{0,2} C .{1-,1} D .{3-,2-,1-,1,3 }
【答案】C
【解析】全集{3U =-,2-,1-,0,1,2,3},集合{1A =-,0,1,2},{3B =-,0,2,3}, 则
{2U
B =-,1-,1},
(){1U A B ∴=-?,1},
故选C .
10.(2020?北京)已知集合{1A =-,0,1,2},{|03}B x x =<<,则(A
B = )
A .{1-,0,1}
B .{0,1}
C .{1-,1,2}
D .{1,2}
【答案】D
【解析】集合{1A =-,0,1,2},{|03}B x x =<<,则{1A B =,2},
故选D .
11.(2020?新课标Ⅰ)设集合2{|40}A x x =-,{|20}B x x a =+,且{|21}A
B x x =-,则(a =
) A .4- B .2- C .2 D .4
【答案】B
【解析】集合2{|40}{|22}A x x x x =-=-,1
{|20}{|}2
B x x a x x a =+=-,
由{|21}A
B x x =-,可得1
12
a -=,
则2a =-. 故选B .
12.(2020?新课标Ⅱ)已知集合{2U =-,1-,
0,1,2,3},{1A =-,0,1},{1B =,2},则()(U
A B =
)
A .{2-,3}
B .{2-,2,3}
C .{2-,1-,0,3}
D .{2-,1-,0,2,3}
【答案】A
【解析】集合{2U =-,1-,0,1,2,3},{1A =-,0,1},{1B =,2}, 则{1A B =-,0,1,2}, 则
(){2U
A
B =-,3},
故选A .
13.(2019?全国)设集合2{|20}P x x =->,{1Q =,2,3,4},则P
Q 的非空子集的个数为(
) A .8 B .7 C .4 D .3
【答案】B
【解析】{|P x x =或;
{2P Q ∴=,3,4};
P
Q ∴的非空子集的个数为:1233337C C C ++=个.
故选B .
14.(2019?天津)设集合{1A =-,1,2,3,5},{2B =,3,4},{|13}C x R x =∈<,则()
(A
C B =
) A .{2} B .{2,3} C .{1-,2,3} D .{1,2,3,4}
【答案】D
【解析】设集合{1A =-,1,2,3,5},{|13}C x R x =∈<, 则{1A
C =,2},
{2B =,3,4}, ()
{1A
C B ∴=,2}{2?,3,4}{1=,2,3,4};
故选D .
15.(2019?浙江)已知全集{1U =-,0,1,2,3},集合{0A =,1,2},{1B =-,0,1},
则()(U A B =
) A .{1}- B .{0,1} C .{1-,2,3} D .{1-,0,1,3}
【答案】A 【解析】{1U
A =-,3},
()
U A B ∴
{1=-,3}{1-?,0,}l {1}=-
故选A .
16.(2019?新课标Ⅲ)已知集合{1A =-,0,1,2},2{|1}B x x =,则(A
B = )
A .{1-,0,1}
B .{0,1}
C .{1-,1}
D .{0,1,2}
【答案】A
【解析】因为{1A =-,0,1,2},2{|1}{|11}B x x x x ==-, 所以{1A
B =-,0,1},
故选A .
17.(2019?新课标Ⅱ)已知集合{|1}A x x =>-,{|2}B x x =<,则(A
B = )
A .(1,)-+∞
B .(,2)-∞
C .(1,2)-
D .?
【答案】C
【解析】由{|1}A x x =>-,{|2}B x x =<,
得{|1}{|2}(1,2)A
B x x x x =>-<=-.
故选C .
18.(2019?新课标Ⅱ)设集合2{|560}A x x x =-+>,{|10}B x x =-<,则(A
B = )
A .(,1)-∞
B .(2,1)-
C .(3,1)--
D .(3,)+∞
【答案】A
【解析】根据题意,2{|560}{|3A x x x x x =-+>=>或2}x <,
{|10}{|1}B x x x x =-<=<,
则{|1}(,1)A B x x =<=-∞;
故选A .
19.(2019?新课标Ⅰ)已知集合{1U =,2,3,4,5,6,7},{2A =,3,4,5},{2B =,3,6,
7},则(U
B
A = )
A .{1,6}
B .{1,7}
C .{6,7}
D .{1,6,7}
【答案】C
【解析】{1U =,2,3,4,5,6,7},{2A =,3,4,5},{2B =,3,6,7}, {1U C A ∴=,6,7},
则{6U
B
A =,7}
故选C .
20.(2019?北京)已知集合{|12}A x x =-<<,{|1}B x x =>,则(A
B = )
A .(1,1)-
B .(1,2)
C .(1,)-+∞
D .(1,)+∞
【答案】C 【解析】
{|12}A x x =-<<,{|1}B x x =>,
{|12}{|1}(1,)A
B x x x x ∴=-<<>=-+∞.
故选C .
21.(2019?新课标Ⅰ)已知集合{|42}M x x =-<<,2{|60}N x x x =--<,则(M
N = )
A .{|43}x x -<<
B .{|42}x x -<<-
C .{|22}x x -<<
D .{|23}x x <<
【答案】C
【解析】{|42}M x x =-<<,2{|60}{|23}N x x x x x =--<=-<<, {|22}M
N x x ∴=-<<.
故选C .
22.(2018?全国)已知全集{1U =,2,3,4,5,6},{1A =,2,6},{2B =,4,5},则()
(U A B =
)
A .{4,5}
B .{1,2,3,4,5,6}
C .{2,4,5}
D .{3,4,
5}
【答案】A
【解析】由全集{1U =,2,3,4,5,6},{1A =,2,6}, 得
{3U
A =,4,5},{2
B =,4,5},
则(){3U A B =,4,5}{2?,4,5}{4=,5}.
故选A .
23.(2018?新课标Ⅱ)已知集合22{(,)|3A x y x y =+,x Z ∈,}y Z ∈,则A 中元素的个数为(
) A .9 B .8 C .5 D .4
【答案】A
【解析】当1x =-时,22y ,得1y =-,0,1, 当0x =时,23y ,得1y =-,0,1, 当1x =时,22y ,得1y =-,0,1, 即集合A 中元素有9个, 故选A .
24.(2018?天津)设集合{1A =,2,3,4},{1B =-,0,2,3},
{|12}C x R x =∈-<,则()
(A B C =
)
A .{1-,1}
B .{0,1}
C .{1-,0,1}
D .{2,3,4}
【答案】C 【解析】{1A =,2,3,4},{1B =-,0,2,3},
(){1A
B ∴=,2,3,4}{1-?,0,2,3}{1=-,0,1,2,3,4},
又{|12}C x R x =∈-<, ()
{1A
B C ∴=-,0,1}.
故选C .
25.(2018?天津)设全集为R ,集合{|02}A x x =<<,{|1}B x x =,则()(R A B =? ) A .{|01}x x < B .{|01}x x << C .{|12}x x < D .{|02}x x <<
【答案】B 【解析】
{|02}A x x =<<,{|1}B x x =,
{|1}R B x x ∴=<, (){|01}R A
B x x ∴=<<.
故选B .
26.(2018?新课标Ⅰ)已知集合{0A =,2},{2B =-,1-,0,1,2},则(A
B = )
A .{0,2}
B .{1,2}
C .{0}
D .{2-,1-,0,1,2}
【答案】A
【解析】集合{0A =,2},{2B =-,1-,0,1,2}, 则{0A
B =,2}.
故选A .
27.(2018?新课标Ⅱ)已知集合{1A =,3,5,7},{2B =,3,4,5},则(A
B = )
A .{3}
B .{5}
C .{3,5}
D .{1,2,3,4,5,7}
【答案】C
【解析】集合{1A =,3,5,7},{2B =,3,4,5}, {3A
B ∴=,5}.
故选C .
28.(2018?新课标Ⅰ)已知集合2{|20}A x x x =-->,则(R
A = )
A .{|12}x x -<<
B .{|12}x x -
C .{|1}{|2}x x x x <->
D .{|1}{|2}x x x x -
【答案】B
【解析】集合2{|20}A x x x =-->, 可得{|1A x x =<-或2}x >, 则:
{|12}R
A x x =-.
故选B .
29.(2018?新课标Ⅲ)已知集合{|10}A x x =-,{0B =,1,2},则(A
B = )
A .{0}
B .{1}
C .{1,2}
D .{0,1,2}
【答案】C 【解析】{|10}{|1}A x x x x =-=,{0B =,1,2},
{|1}{0A
B x x ∴=?,1,2}{1=,2}.
故选C .
30.(2018?北京)已知集合{|||2}A x x =<,{2B =-,0,1,2},则(A
B = )
A .{0,1}
B .{1-,0,1}
C .{2-,0,1,2}
D .{1-,0,1,2}
【答案】A
【解析】{|||2}{|22}A x x x x =<=-<<,{2B =-,0,1,2}, 则{0A
B =,1},
故选A .
31.(2018?浙江)已知全集{1U =,2,3,4,5},{1A =,3},则(U
A = )
A .?
B .{1,3}
C .{2,4,5}
D .{1,2,3,4,5}
【答案】C
【解析】根据补集的定义,
U
A 是由所有属于集合U 但不属于A 的元素构成的集合,由已知,有
且仅有2,4,5符合元素的条件.
{2U
A =,4,5}
故选C .
32.(2020?上海)已知集合{1A =,2,4},集合{2B =,4,5},则A
B =_________.
【答案】{2,4}
【解析】因为{1A =,2,3},{2B =,4,5}, 则{2A
B =,4}.
故答案为:{2,4}.
33.(2020?江苏)已知集合{1A =-,0,1,2},{0B =,2,3},则A
B =_________.
【答案】{0,2}
【解析】集合{0B =,2,3},{1A =-,0,1,2}, 则{0A
B =,2},
故答案为:{0,2}.
34.(2020?上海)集合{1A =,3},{1B =,2,}a ,若A B ?,则a =_________. 【答案】3
【解析】3A ∈,且A B ?,3B ∴∈,3a ∴=, 故答案为:3.
35.(2019?上海)已知集合(,3)A =-∞,(2,)B =+∞,则A
B =_________.
【答案】(2,3)
【解析】根据交集的概念可得(2,3)A B =.
故答案为:(2,3).
36.(2019?江苏)已知集合{1A =-,0,1,6},{|0B x x =>,}x R ∈,则A
B =_________.
【答案】{1,6} 【解析】{1A =-,0,1,6},{|0B x x =>,}x R ∈,
{1A
B ∴=-,0,1,6}{|0x x >,}{1x R ∈=,6}.
故答案为:{1,6}.
37.(2019?上海)已知集合{1A =,2,3,4,5},{3B =,5,6},则A
B =_________.
【答案】{3,5}
【解析】集合{1A =,2,3,4,5}, {3B =,5,6}, {3A
B ∴=,5}.
故答案为:{3,5}.
38.(2019?上海)已知集合[A t =,1][4t t ++,9]t +,0A ?,存在正数λ,使得对任意a A ∈,都有
A a
λ
∈,则t 的值是_________.
【答案】1或3-
【解析】当0t >时,当[a t ∈,1]t +时,则
[4t a
λ
∈+,9]t +,
当[4a t ∈+,9]t +时,则[t a
λ
∈,1]t +,
即当a t =时,
9t a
λ
+;当9a t =+时,
t a
λ
,即(9)t t λ=+; 当1a t =+时,
4t a
λ
+,当4a t =+时,
1t a
λ
+,即(1)(4)t t λ=++,
(9)(1)(4)t t t t ∴+=++,解得1t =.
当104t t +<<+时,当[a t ∈,1]t +时,则[t a
λ
∈,1]t +.
当[4a t ∈+,9]t +,则[4t a
λ
∈+,9]t +,
即当a t =时,
1t a
λ
+,当1a t =+时,
t a
λ
,即(1)t t λ=+,
即当4a t =+时,
9t a
λ
+,当9a t =+时,
4t a
λ
+,即(4)(9)t t λ=++,
(1)(4)(9)t t t t ∴+=++,解得3t =-.
当90t +<时,同理可得无解. 综上,t 的值为1或3-. 故答案为:1或3-.
39.(2018?江苏)已知集合{0A =,1,2,8},{1B =-,1,6,8},那么A
B =_________.
【答案】{1,8} 【解析】{0A =,1,2,8},{1B =-,1,6,8},
{0A
B ∴=,1,2,8}{1-?,1,6,8}{1=,8},
故答案为:{1,8}.
40.(2018?上海)已知集合{|02}A x x =<<,{|11}B x x =-<<,则A
B =_________.
【答案】{|01}x x << 【解析】{|02}A x x =<<,{|11}B x x =-<<,
{|01}A
B x x ∴=<<.
故答案为:{|01}x x <<.
1.(2020?汉阳区校级模拟)设全集{|25U x x =-<,}x Z ∈,{0A =,2,3,4},{2B =-,1-,0,1,2},则图中阴影部分所表示的集合为( )
A .{0,2}
B .{3,4}
C .{0,3,4}
D .{2-,1-,0,1,2}
【答案】B
【解析】全集{|25U x x =-<,}{2x Z ∈=-,1-,0,1,2,3,4},{0A =,2,3,4},{2B =-,
1-,0,1,2},
{3U C B ∴=,4},
∴图中阴影部分所表示的集合为:
(){3U A
C B =,4}.
故选B .
2.(2020?金凤区校级四模)已知集合{|21}A x x =-<<,2{|30}B x x x =-,则(A
B = )
A .(0,1)
B .(2-,3]
C .[0,1)
D .(1,3]
【答案】C 【解析】{|21}A x x =-<<,{|03}B x x =,
[0A
B ∴=,1).
故选C .
3.(2020?泸州四模)已知集合{(,)|}A x y y x ==,2{(,)|}B x y y x ==,则A
B 的元素个数为(
) A .0 B .1 C .2 D .4
【答案】C
【解析】集合{(,)|}A x y y x ==,2{(,)|}B x y y x ==,
{(A
B x ∴=,2
)|}{(0,0)y x
y y x =?=?=?,(1,1)}, A
B ∴的元素个数为2.
故选C .
4.(2020?龙凤区校级模拟)集合{|13}A x x =-<<,2{|60B x x x =+-<,}x Z ∈,则(A
B =
) A .(1,2)- B .(3,3)-
C .{0,1}
D .{0,1,2}
【答案】C
【解析】集合{|13}A x x =-<<,
2{|60B x x x =+-<,}{|32x Z x x ∈=-<<,}{2x Z ∈=-,1-,0,1}, {0A
B ∴=,1}.
故选C .
5.(2020?运城模拟)已知集合2{|3}A x x =<,2{|3}B x x x =<,则(A
B = )
A .(
B .
C .(
D .(0,3)
【答案】B
【解析】
{}{}22|3(|3(0,3)A x x B x x x =<==<=,
∴(0,3)A
B =.
故选B .
6.(2020?南岗区校级模拟)若全集U R =,集合{|(6)}A x y lg x ==-,{|21}x B x =>,则图中阴影部分表示的集合是( )
A .(2,3)
B .(1-,0]
C .[0,6)
D .(-∞,0]
【答案】D
【解析】全集U R =,集合{|(6)}{|6}A x y lg x x x ==-=<,
{|21}{|0}x B x x x =>=>, {|0}U B x x ∴=.
∴图中阴影部分表示的集合为:
(){|0}U A
B x x =.
故选D .
7.(2020?香坊区校级一模)已知集合2{|2}A x Z x x =∈-,{1B =,}a ,若B A ?,则实数a 的取值集合为( )
A .{1-,1,0,2}
B .{1-,0,2}
C .{1-,1,2}
D .{0,2}
【答案】B
【解析】2{|2}{|12}{1A x Z x x x Z x =∈-=∈-=-,0,1,2},因为B A ?, 若B A ?,则1a =-或0或2. 则实数a 的取值的集合为{1-,0,2} 故选B .
8.(2020?东湖区校级模拟)已知集合{||21|3}A x x =-,2{|(6)}B x y lg x x ==--,则
(R
A B =
) A .(1,3)- B .? C .(2,3) D .(2,1)--
【答案】B
【解析】因为{||21|3}{|2A x x x x =-=或1}x -, 所以(1,2)R
A =-,2{|(6)}{|3
B x y lg x x x x ==--=>或2}x <-, 则
R
A
B =?.
故选B .
9.(2020?天津二模)已知全集{1U =-,0,1,2,3},集合{0A =,1,2},{1B =-,0,1},
则()(U
A B =
) A .{1}- B .{0,1} C .{1-,2,3} D .{1-,0,1,3}
【答案】C
【解析】{1U =-,0,1,2,3},{0A =,1,2},{1B =-,0,1}, {0A B ∴==,1}, (){1U
A
B ∴
=-,2,3}.
故选C .
10.(2020?兴庆区校级四模)若集合2{|430}A x x x =-+=,{2B =,3,4},则(A
B = )
A .{1}
B .{2}
C .{3}
D .{1,2,3,4}
【答案】C 【解析】{1A =,3},{2B =,3,4},
{3}A
B ∴=.
故选C .
11.(2020?镜湖区校级模拟)已知集合{||2A x x =,}x Z ∈,2{|60}B x x x =--<,则(A
B =
)
A .{2-,1-,0,1,2,3}
B .{2-,1-,0,1,2}
C .{1-,0,1,2}
D .{2-,1-,0,1}
【答案】C
【解析】{2A =-,1-,0,1,2},{|23}B x x =-<<,
{1A
B ∴=-,0,1,2}.
故选C .
12.(2020?河南模拟)集合{|2A x x =>,}x R ∈,2{|230}B x x x =-->,则(A
B = )
A .(-∞,1)(3-?,)+∞
B .(3,)+∞
C .(2,)+∞
D .(2,3)
【答案】B 【解析】{|2}A x x =>,{|1B x x =<-或3}x >,
(3,)A
B ∴=+∞.
故选B .
13.(2020?安徽模拟)已知集合{|(1)}A x y ln x ==-,{|21}x B x =>,则(A
B = )
A .[1,)+∞
B .(1,)+∞
C .(0,)+∞
D .(0,1)
【答案】B
【解析】集合{|(1)}{|1}A x y ln x x x ==-=>,
{|21}{|0}x B x x x =>=>, {|1}(1,)A
B x x ∴=>=+∞.
故选B .
14.(2020?庐阳区校级模拟)设集合{|23}A x lnx ln =,{|6}B x x =,则(A
B = )
A .{|03}x x <
B .{|6}x x
C .{|06}x x <
D .{|36}x x
【答案】B
【解析】集合{|23}{|03}A x lnx ln x x ==<, {|6}B x x =, {|6}A
B x x ∴=.
故选B .