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高三数学专题训练--集合的概念与运算

高三数学专题练习1 集合的概念与运算小题基础练①一、选择题1.[2018·全国卷Ⅱ]已知集合A={1,3,5,7},B={2,3,4,5},则A∩B=()A.{3} B.{5}C.{3,5} D.{1,2,3,4,5,7}答案:C解析:A∩B={1,3,5,7}∩{2,3,4,5}={3,5}.故选C.2.[2018·全国卷Ⅰ]已知集合A={x|x2-x-2>0},则∁R A=()A.{x|-1<x<2}B.{x|-1≤x≤2}C.{x|x<-1}∪{x|x>2}D.{x|x≤-1}∪{x|x≥2}答案:B解析:∵x2-x-2>0,∴ (x-2)(x+1)>0,∴x>2或x<-1,即A={x|x>2或x<-1}.在数轴上表示出集合A,如图所示.由图可得∁R A={x|-1≤x≤2}.故选B.3.[2019·河南中原名校质检]已知全集U={1,2,3,4,5,6},集合A={1,2,4},B={2,4,6},则A∩(∁U B)=()A.{1} B.{2}C.{4} D.{1,2}答案:A解析:因为∁U B={1,3,5},所以A∩(∁U B)={1}.故选A.4.[2019·河北衡水武邑中学调研]已知全集U=R,集合A ={x|0<x<9,x∈R}和B={x|-4<x<4,x∈Z}关系的Venn图如图所示,则阴影部分所表示集合中的元素共有()A .3个B .4个C .5个D .无穷多个答案:B解析:因为A ={x |0<x <9,x ∈R },所以∁U A ={x |x ≤0或x ≥9}.题图中阴影部分表示的集合为(∁U A )∩B ={x |-4<x ≤0,x ∈Z }={-3,-2,-1,0},故该集合中共有4个元素.故选B.5.[2019·惠州一调]已知集合U ={-1,0,1},A ={x |x =m 2,m ∈U },则∁U A =( )A .{0,1}B .{-1,0,1}C .∅D .{-1}答案:D解析:∵A ={x |x =m 2,m ∈U }={0,1},∴∁U A ={-1},故选D.6.[2019·河北省五校联考(二)]已知集合A ={x |x <1},B ={x |x 2-x -6<0},则( )A .A ∩B ={x |x <1} B .A ∪B =RC .A ∪B ={x |x <2}D .A ∩B ={x |-2<x <1}答案:D解析:∵x 2-x -6<0,∴-2<x <3,∴B ={x |-2<x <3},∴A ∪B ={x |x <3},A ∩B ={x |-2<x <1},故选D.7.[2019·江西赣州模拟]已知集合A ={x |-1≤lg x ≤1},B ={x |2x <4},则A ∩B =( )A.⎩⎪⎨⎪⎧⎭⎪⎬⎪⎫x ⎪⎪⎪ 110≤x <2 B .{x |0<x <2} C .{x |2<x ≤10} D .{x |0<x ≤10}答案:A解析:∵-1≤lg x ≤1,∴110≤x ≤10,∴A =⎩⎪⎨⎪⎧⎭⎪⎬⎪⎫x ⎪⎪⎪ 110≤x ≤10.由2x <4知x <2,∴B ={x |x <2},∴A ∩B =⎩⎪⎨⎪⎧⎭⎪⎬⎪⎫x ⎪⎪⎪110≤x <2.故选A. 8.[2019·广西桂林、百色、梧州、崇左、北海五市联合模拟]已知全集U=R,集合M={x|(x-1)(x+2)≥0},N={x|-1≤x≤2},则(∁U M)∩N=()A.[-2,-1] B.[-1,2]C.[-1,1) D.[1,2]答案:C解析:因为全集U=R,集合M={x|(x-1)(x+2)≥0}={x|x≤-2或x≥1},所以∁U M={x|-2<x<1}.又N={x|-1≤x≤2},所以(∁U M)∩N=[-1,1).故选C.二、非选择题9.[2018·江苏卷]已知集合A={0,1,2,8},B={-1,1,6,8},那么A∩B=________.答案:{1,8}解析:A∩B={0,1,2,8}∩{-1,1,6,8}={1,8}.10.[2019·南昌模拟]已知集合A={1,2,3},B={(x,y)|x∈A,y∈A,x+y∈A},则集合B的子集的个数为________.答案:8解析:∵集合A={1,2,3},集合B={(x,y)|x∈A,y∈A,x +y∈A},∴B={(1,1),(1,2),(2,1)},∴集合B有3个元素,∴集合B的子集个数为23=8.11.[2019·石家庄质检]已知集合A={x|-2<x<4},B={x|y =lg(x-2)},则A∩(∁R B)=________.答案:(-2,2]解析:由题意得B={x|y=lg(x-2)}=(2,+∞),∴∁R B=(-∞,2],∴A∩(∁R B)=(-2,2].12.[2019·辽宁省五校联考]已知集合P={x|x2-2x-8>0},Q={x|x≥a},P∪Q=R,则a的取值范围是________.答案:(-∞,-2]解析:集合P={x|x2-2x-8>0}={x|x<-2或x>4},Q={x|x≥a},若P∪Q=R,则a≤-2,即a的取值范围是(-∞,-2].课时增分练①一、选择题1.[2018·全国卷Ⅱ]已知集合A={(x,y)|x2+y2≤3,x∈Z,y∈Z},则A中元素的个数为()A.9 B.8C.5 D.4答案:A解析:将满足x2+y2≤3的整数x,y全部列举出来,即(-1,-1),(-1,0),(-1,1),(0,-1),(0,0),(0,1),(1,-1),(1,0),(1,1),共有9个.故选A.2.[2019·湖南省名校联考]已知全集U=R,集合A={x|x2-3x≥0},B={x|1<x≤3},则如图所示的阴影部分表示的集合为()A.[0,1) B.(0,3]C.(0,1] D.[1,3]答案:C解析:因为A={x|x2-3x≥0}={x|x≤0或x≥3},B={x|1<x≤3},所以A∪B={x|x>1或x≤0},所以图中阴影部分表示的集合为∁U(A∪B)=(0,1],故选C.3.设集合A={x|-3≤x≤3,x∈Z},B={y|y=x2+1,x∈A},则集合B中元素的个数是()A.3 B.4C.5 D.无数个答案:B解析:∵A={x|-3≤x≤3,x∈Z},∴A={-3,-2,-1,0,1,2,3},∵B={y|y=x2+1,x∈A},∴B={1,2,5,10},故集合B 中元素的个数是4,选B.4.[2019·四川广元第三次高考适应性统考(三诊)]已知集合A ={x |x 2-4x <0},B ={x |x <a },若A ⊆B ,则实数a 的取值范围是( )A .(0,4]B .(-∞,4)C .[4,+∞)D .(4,+∞)答案:C解析:由已知可得A ={x |0<x <4}.若A ⊆B ,则a ≥4.故选C.5.[2019·贵州遵义南白中学联考]已知集合A ={x |x 2+x -2<0},B ={x |log 12x >1},则A ∩B =( )A.⎝ ⎛⎭⎪⎫0,12 B .(0,1) C.⎝ ⎛⎭⎪⎫-2,12 D.⎝ ⎛⎭⎪⎫12,1 答案:A解析:由题意,得A ={x |-2<x <1},B =⎩⎪⎨⎪⎧⎭⎪⎬⎪⎫x ⎪⎪⎪ 0<x <12,所以A ∩B =⎩⎨⎧x ⎪⎪⎪⎭⎬⎫0<x <12=⎝ ⎛⎭⎪⎫0,12.故选A. 6.[2019·河北唐山模拟]已知集合A ={x ∈N |x <3},B ={x |x =a -b ,a ∈A ,b ∈A },则A ∩B =( )A .{1,2}B .{-2,-1,1,2}C .{1}D .{0,1,2}答案:D解析:A ={x ∈N |x <3}={0,1,2},B ={x |x =a -b ,a ∈A ,b ∈A }.由题意知,当a =0,b =0时,x =a -b =0;当a =0,b =1时,x =a -b =-1;当a =0,b =2时,x =a -b =-2;当a =1,b =0时,x =a -b =1;当a =1,b =1时,x =a -b =0;当a =1,b =2时,x =a -b =-1;当a =2,b =0时,x =a -b =2;当a =2,b =1时,x =a -b =1;当a =2,b =2时,x =a -b =0,根据集合中元素的互异性,B ={-2,-1,0,1,2},∴A ∩B ={0,1,2}.故选D.7.[2019·浙江教育绿色评价联盟模拟]已知集合P ={x ∈R |-2<x ≤3},Q =⎩⎪⎨⎪⎧⎭⎪⎬⎪⎫x ∈R ⎪⎪⎪⎪ 1+x x -3≤0,则( )A .P ∩Q ={x ∈R |-1<x <3}B .P ∪Q ={x ∈R |-2<x <3}C .P ∩Q ={x ∈R |-1≤x ≤3}D .P ∪Q ={x ∈R |-2<x ≤3}答案:D解析:由1+x x -3≤0,得(1+x )(x -3)≤0且x ≠3,解得-1≤x <3,故P ∩Q ={x ∈R |-1≤x <3},P ∪Q ={x ∈R |-2<x ≤3}.故选D.8.已知全集U =R ,集合A ={x |x 2-2x ≤0},B ={y |y =sin x ,x ∈R },则图中阴影部分表示的集合为( )A .[-1,2]B .[-1,0)∪(1,2]C .[0,1]D .(-∞,-1)∪(2,+∞)答案:B解析:由题意可知阴影部分对应的集合为[∁U (A ∩B )]∩(A ∪B ),A ={x |x 2-2x ≤0}={x |0≤x ≤2}=[0,2],B ={y |y =sin x ,x ∈R }={y |-1≤y ≤1}=[-1,1],∴A ∩B =[0,1],A ∪B =[-1,2],∴[∁U (A ∩B )]∩(A ∪B )=[-1,0)∪(1,2],故选B.二、非选择题9.[2019·无锡五校联考(一)]已知集合A ={x |(x -1)(x -a )≥0},B ={x |x ≥a -1},若A ∪B =R ,则实数a 的最大值为________.答案:2解析:当a >1时,A =(-∞,1]∪[a ,+∞),B =[a -1,+∞),若A ∪B =R ,则a -1≤1,∴1<a ≤2;当a =1时,易得A =R ,此时A ∪B =R ;当a <1时,A =(-∞,a ]∪[1,+∞),B =[a -1,+∞),若A ∪B =R ,则a -1≤a ,显然成立,∴a <1.综上,实数a 的取值范围是(-∞,2],则实数a 的最大值为2.10.[2019·内蒙古呼和浩特质量普查调研]已知集合A ={x |0<x <2},集合B ={x |-1<x <1},集合C ={x |mx +1>0},若A ∪B ⊆C ,则实数m 的取值范围是________.答案:⎣⎢⎡⎦⎥⎤-12,1 解析:由A ={x |0<x <2},B ={x |-1<x <1},得A ∪B ={x |-1<x <2},∵集合C ={x |mx +1>0},A ∪B ⊆C ,①当m <0时,x <-1m ,∴-1m ≥2,∴m ≥-12,∴-12≤m <0;②当m =0时,成立;③当m >0时,x >-1m ,∴-1m ≤-1,∴m ≤1,∴0<m ≤1,综上所述,-12≤m ≤1.11.[2019·江西玉山一中月考(二)]已知集合A ={x |3≤3x ≤27},B ={x |log 2x >1}.(1)分别求A ∩B ,(∁R B )∪A ;(2)已知集合C ={x |1<x <a },若C ⊆A ,求实数a 的取值范围. 解析:(1)∵3≤3x ≤27,即31≤3x ≤33,∴1≤x ≤3,∴A ={x |1≤x ≤3}.∵log 2x >1,即log 2x >log 22,∴x >2,∴B ={x |x >2}. ∴A ∩B ={x |2<x ≤3}.∵∁R B ={x |x ≤2},∴(∁R B )∪A ={x |x ≤3}.(2)由(1)知A ={x |1≤x ≤3},C ⊆A .当C 为空集时,满足C ⊆A ,a ≤1;当C 为非空集合时,可得1<a ≤3.综上所述,a 的取值范围为(-∞,3].实数a 的取值范围为(-∞,3].。

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