(必修一)集合与函数概念练习题
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无忧数学 ——集合与函数概念(必修一)第一章 集合第一节 集合的含义、表示及基本关系A 组1.已知A ={1,2},B ={x |x ∈A },则集合A 与B 的关系为________.解析:由集合B ={x |x ∈A }知,B ={1,2}.答案:A =B2.若∅{x |x 2≤a ,a ∈R },则实数a 的取值范围是________.解析:由题意知,x 2≤a 有解,故a ≥0.答案:a ≥03.已知集合A ={y |y =x 2-2x -1,x ∈R },集合B ={x |-2≤x <8},则集合A 与B 的关系是________.解析:y =x 2-2x -1=(x -1)2-2≥-2,∴A ={y |y ≥-2},∴B A .答案:B A4.已知全集U =R ,则正确表示集合M ={-1,0,1}和N ={x |x 2+x =0}关系的韦恩(Venn)图是________.解析:由N={x|x 2+x=0},得N ={-1,0},则N M .答案:②5.已知集合A ={x |x >5},集合B ={x |x >a },若命题“x ∈A ”是命题“x ∈B ”的充分不必要条件,则实数a 的取值范围是________.解析:命题“x ∈A ”是命题“x ∈B ” 的充分不必要条件,∴A B ,∴a <5. 答案:a <56.已知m ∈A ,n ∈B ,且集合A ={x |x =2a ,a ∈Z },B ={x |x =2a +1,a ∈Z },又C ={x |x =4a +1,a ∈Z },判断m +n 属于哪一个集合?解:∵m ∈A ,∴设m =2a 1,a 1∈Z ,又∵n ∈B ,∴设n =2a 2+1,a 2∈Z ,∴m +n =2(a 1+a 2)+1,而a 1+a 2∈Z ,∴m +n ∈B .B 组1.设a ,b 都是非零实数,y =a |a |+b |b |+ab |ab |可能取的值组成的集合是________. 解析:分四种情况:(1)a >0且b >0;(2)a >0且b <0;(3)a <0且b >0;(4)a <0且b <0,讨论得y =3或y =-1.答案:{3,-1}2.已知集合A ={-1,3,2m -1},集合B ={3,m 2}.若B ⊆A ,则实数m =________. 解析:∵B ⊆A ,显然m 2≠-1且m 2≠3,故m 2=2m -1,即(m -1)2=0,∴m =1.答案:13.设P ,Q 为两个非空实数集合,定义集合P +Q ={a +b |a ∈P ,b ∈Q },若P ={0,2,5},Q ={1,2,6},则P +Q 中元素的个数是________个.解析:依次分别取a =0,2,5;b =1,2,6,并分别求和,注意到集合元素的互异性,∴P +Q ={1,2,6,3,4,8,7,11}.答案:84.已知集合M ={x |x 2=1},集合N ={x |ax =1},若N M ,那么a 的值是________.解析:M ={x |x =1或x =-1},N M ,所以N =∅时,a =0;当a ≠0时,x =1a=1或-1,∴a =1或-1.答案:0,1,-15.满足{1}A ⊆{1,2,3}的集合A 的个数是________个.解析:A 中一定有元素1,所以A 有{1,2},{1,3},{1,2,3}.答案:36.已知集合A ={x |x =a +16,a ∈Z },B ={x |x =b 2-13,b ∈Z },C ={x |x =c 2+16,c ∈Z },则A 、B 、C 之间的关系是________.解析:用列举法寻找规律.答案:A B =C7.集合A ={x ||x |≤4,x ∈R },B ={x |x <a },则“A ⊆B ”是“a >5”的________.解析:结合数轴若A ⊆B ⇔a ≥4,故“A ⊆B ”是“a >5”的必要但不充分条件.答案:必要不充分条件8.设集合M ={m |m =2n ,n ∈N ,且m <500},则M 中所有元素的和为________.解析:∵2n <500,∴n =0,1,2,3,4,5,6,7,8.∴M 中所有元素的和S =1+2+22+…+28=511.答案:5119.设A 是整数集的一个非空子集,对于k ∈A ,如果k -1∉A ,且k +1∉A ,那么称k 是A 的一个“孤立元”.给定S ={1,2,3,4,5,6,7,8},由S 的3个元素构成的所有集合中,不含“孤立元”的集合共有________个.解析:依题可知,由S 的3个元素构成的所有集合中,不含“孤立元”,这三个元素一定是相连的三个数.故这样的集合共有6个.答案:610.已知A ={x ,xy ,lg(xy )},B ={0,|x |,y },且A =B ,试求x ,y 的值. 解:由lg(xy )知,xy >0,故x ≠0,xy ≠0,于是由A =B 得lg(xy )=0,xy =1.∴A ={x,1,0},B ={0,|x |,1x}. 于是必有|x |=1,1x=x ≠1,故x =-1,从而y =-1. 11.已知集合A ={x |x 2-3x -10≤0},(1)若B ⊆A ,B ={x |m +1≤x ≤2m -1},求实数m 的取值范围;(2)若A ⊆B ,B ={x |m -6≤x ≤2m -1},求实数m 的取值范围;(3)若A =B ,B ={x |m -6≤x ≤2m -1},求实数m 的取值范围.解:由A ={x |x 2-3x -10≤0},得A ={x |-2≤x ≤5},(1)∵B ⊆A ,∴①若B =∅,则m +1>2m -1,即m <2,此时满足B ⊆A .②若B ≠∅,则⎩⎪⎨⎪⎧ m +1≤2m -1,-2≤m +1,2m -1≤5.解得2≤m ≤3.由①②得,m 的取值范围是(-∞,3].(2)若A ⊆B ,则依题意应有⎩⎪⎨⎪⎧ 2m -1>m -6,m -6≤-2,2m -1≥5.解得⎩⎪⎨⎪⎧ m >-5,m ≤4,m ≥3.故3≤m ≤4,∴m 的取值范围是[3,4].(3)若A =B ,则必有⎩⎪⎨⎪⎧m -6=-2,2m -1=5,解得m ∈∅.,即不存在m 值使得A =B . 12.已知集合A ={x |x 2-3x +2≤0},B ={x |x 2-(a +1)x +a ≤0}.(1)若A 是B 的真子集,求a 的取值范围;(2)若B 是A 的子集,求a 的取值范围;(3)若A =B ,求a 的取值范围.解:由x 2-3x +2≤0,即(x -1)(x -2)≤0,得1≤x ≤2,故A ={x |1≤x ≤2}, 而集合B ={x |(x -1)(x -a )≤0},(1)若A 是B 的真子集,即A B ,则此时B ={x |1≤x ≤ a },故a >2.(2)若B 是A 的子集,即B ⊆A ,由数轴可知1≤a ≤2.(3)若A =B ,则必有a =2第二节 集合的基本运算A 组1.设U =R ,A ={x |x >0},B ={x |x >1},则A ∩∁U B =____.解析:∁U B ={x |x ≤1},∴A ∩∁U B ={x |0<x ≤1}.答案:{x |0<x ≤1}2.设集合A ={4,5,7,9},B ={3,4,7,8,9},全集U =A ∪B ,则集合∁U (A ∩B )中的元素共有________个.解析:A ∩B ={4,7,9},A ∪B ={3,4,5,7,8,9},∁U (A ∩B )={3,5,8}.答案:33.已知集合M ={0,1,2},N ={x |x =2a ,a ∈M },则集合M ∩N =________. 解析:由题意知,N ={0,2,4},故M ∩N ={0,2}.答案:{0,2}4.(原创题)设A ,B 是非空集合,定义A ⓐB ={x |x ∈A ∪B 且x ∉A ∩B },已知A ={x |0≤x ≤2},B ={y |y ≥0},则A ⓐB =________.解析:A ∪B =[0,+∞),A ∩B =[0,2],所以A ⓐB =(2,+∞).答案:(2,+∞)5.某班共30人,其中15人喜爱篮球运动,10人喜爱乒乓球运动,8人对这两项运动都不喜爱,则喜爱篮球运动但不喜爱乒乓球运动的人数为________.解析:设两项运动都喜欢的人数为x ,画出韦恩图得到方程15-x +x +10-x +8=30x =3,∴喜爱篮球运动但不喜爱乒乓球运动的人数为15-3=12(人).答案:126.已知集合A ={x |x >1},集合B ={x |m ≤x ≤m +3}.(1)当m =-1时,求A ∩B ,A ∪B ;(2)若B ⊆A ,求m 的取值范围.解:(1)当m =-1时,B ={x |-1≤x ≤2},∴A ∩B ={x |1<x ≤2},A ∪B ={x |x ≥-1}.(2)若B ⊆A ,则m >1,即m 的取值范围为(1,+∞)B 组1.若集合M ={x ∈R |-3<x <1},N ={x ∈Z |-1≤x ≤2},则M ∩N =________. 解析:因为集合N ={-1,0,1,2},所以M ∩N ={-1,0}.答案:{-1,0}2.已知全集U ={-1,0,1,2},集合A ={-1,2},B ={0,2},则(∁U A )∩B =________. 解析:∁U A ={0,1},故(∁U A )∩B ={0}.答案:{0}3.若全集U =R ,集合M ={x |-2≤x ≤2},N ={x |x 2-3x ≤0},则M ∩(∁U N )=________. 解析:根据已知得M ∩(∁U N )={x |-2≤x ≤2}∩{x |x <0或x >3}={x |-2≤x <0}.答案:{x |-2≤x <0}4.集合A ={3,log 2a },B ={a ,b },若A ∩B ={2},则A ∪B =________. 解析:由A ∩B ={2}得log 2a =2,∴a =4,从而b =2,∴A ∪B ={2,3,4}. 答案:{2,3,4}5.已知全集U =A ∪B 中有m 个元素,(∁U A )∪(∁U B )中有n 个元素.若A ∩B 非空,则A ∩B 的元素个数为________.解析:U =A ∪B 中有m 个元素,∵(∁U A )∪(∁U B )=∁U (A ∩B )中有n 个元素,∴A ∩B 中有m -n 个元素.答案:m -n6.设U ={n |n 是小于9的正整数},A ={n ∈U |n 是奇数},B ={n ∈U |n 是3的倍数},则∁U (A ∪B )=________.解析:U ={1,2,3,4,5,6,7,8},A ={1,3,5,7},B ={3,6},∴A ∪B ={1,3,5,6,7},得∁U (A ∪B )={2,4,8}.答案:{2,4,8}7.定义A ⊗B ={z |z =xy +x y,x ∈A ,y ∈B }.设集合A ={0,2},B ={1,2},C ={1},则集合(A ⊗B )⊗C 的所有元素之和为________.解析:由题意可求(A ⊗B )中所含的元素有0,4,5,则(A ⊗B )⊗C 中所含的元素有0,8,10,故所有元素之和为18.答案:188.若集合{(x ,y )|x +y -2=0且x -2y +4=0}{(x ,y )|y =3x +b },则b =________.解析:由⎩⎪⎨⎪⎧ x +y -2=0,x -2y +4=0.⇒⎩⎪⎨⎪⎧ x =0,y =2.点(0,2)在y =3x +b 上,∴b =2. 9.设全集I ={2,3,a 2+2a -3},A ={2,|a +1|},∁I A ={5},M ={x |x =log 2|a |},则集合M 的所有子集是________.解析:∵A ∪(∁I A )=I ,∴{2,3,a 2+2a -3}={2,5,|a +1|},∴|a +1|=3,且a 2+2a -3=5,解得a =-4或a =2,∴M ={log 22,log 2|-4|}={1,2}.答案:∅,{1},{2},{1,2}10.设集合A ={x |x 2-3x +2=0},B ={x |x 2+2(a +1)x +(a 2-5)=0}.(1)若A ∩B ={2},求实数a 的值;(2)若A ∪B =A ,求实数a 的取值范围.解:由x 2-3x +2=0得x =1或x =2,故集合A ={1,2}.(1)∵A ∩B ={2},∴2∈B ,代入B 中的方程,得a 2+4a +3=0⇒a =-1或a =-3;当a =-1时,B ={x |x 2-4=0}={-2,2},满足条件;当a =-3时,B ={x |x 2-4x +4=0}={2},满足条件;综上,a 的值为-1或-3.(2)对于集合B ,Δ=4(a +1)2-4(a 2-5)=8(a +3).∵A ∪B =A ,∴B ⊆A ,①当Δ<0,即a <-3时,B =∅满足条件;②当Δ=0,即a =-3时,B ={2}满足条件;③当Δ>0,即a >-3时,B =A ={1,2}才能满足条件,则由根与系数的关系得⎩⎪⎨⎪⎧ 1+2=-2(a +1)1×2=a 2-5⇒⎩⎪⎨⎪⎧a =-52,a 2=7,矛盾.综上,a 的取值范围是a ≤-3. 11.已知函数f (x )= 6x +1-1的定义域为集合A ,函数g (x )=lg(-x 2+2x +m )的定义域为集合B .(1)当m =3时,求A ∩(∁R B );(2)若A ∩B ={x |-1<x <4},求实数m 的值.解:A ={x |-1<x ≤5}.(1)当m =3时,B ={x |-1<x <3},则∁R B ={x |x ≤-1或x ≥3},∴A ∩(∁R B )={x |3≤x ≤5}.(2)∵A ={x |-1<x ≤5},A ∩B ={x |-1<x <4},∴有-42+2×4+m =0,解得m =8,此时B ={x |-2<x <4},符合题意.12.已知集合A ={x ∈R |ax 2-3x +2=0}.(1)若A =∅,求实数a 的取值范围;(2)若A 是单元素集,求a 的值及集合A ;(3)求集合M ={a ∈R |A ≠∅}.解:(1)A 是空集,即方程ax 2-3x +2=0无解.。