十年高考真題分類匯編(2010-2019) 數(shù)學(xué)專題01 集合 (含解析)

1、十年高考真題分類匯編（20102019）數(shù)學(xué)專題01集合1.(2019全國(guó)1理t1)已知集合m=x|-4x2,n=x|x2-x-60,則mn=()a.x|-4x3b.x|-4x-2c.x|-2x2d.x|2x3【答案】c【解析】由題意得n=x|-2x3,則mn=x|-2x0,b=x|x-10,則ab=()a.(-,1)b.(-2,1)c.(-3,-1)d.(3,+)【答案】a【解析】由題意,得a=x|x3,b=x|x1,所以ab=x|x-1,b=x|x2,則ab=()a.(-1,+)b.(-,2)c.(-1,2)d.【答案】c【解析】由題意,得ab=(-1,2),故選c.5.(2019全國(guó)3t

2、1)已知集合a=-1,0,1,2,b=x|x21,則ab=()a.-1,0,1b.0,1c.-1,1d.0,1,2【答案】a【解析】a=-1,0,1,2,b=x|-1x1,則ab=-1,0,1.故選a.6.(2019北京文t1)已知集合a=x|-1x1,則ab=()a.(-1,1)b.(1,2)c.(-1,+)d.(1,+)【答案】c【解析】a=x|-1x1,ab=(-1,+),故選c.7.(2019天津t1)設(shè)集合a=-1,1,2,3,5,b=2,3,4,c=xr|1x0,則ra=()a.x|-1x2b.x|-1x2c.x|x2d.x|x-1x|x2【答案】b【解析】a=x|x2,所以ra=

3、x|-1x2.10.(2018全國(guó)1文t1)已知集合a=0,2,b=-2,-1,0,1,2,則ab=()a.0,2b.1,2c.0d.-2,-1,0,1,2【答案】a【解析】由交集定義知ab=0,2.11.(2018全國(guó)2文t2,)已知集合a=1,3,5,7,b=2,3,4,5,則ab=()a.3b.5c.3,5d.1,2,3,4,5,7【答案】c【解析】集合a、b的公共元素為3,5,故ab=3,5.12.(2018全國(guó)3t1)已知集合a=x|x-10,b=0,1,2,則ab=()a.0b.1c.1,2d.0,1,2【答案】c【解析】由題意得a=x|x1,b=0,1,2,ab=1,2.13.(

4、2018北京t1)已知集合a=x|x|2,b=-2,0,1,2,則ab=()a.0,1b.-1,0,1c.-2,0,1,2d.-1,0,1,2【答案】a【解析】a=x|-2x2,b=-2,0,1,2,ab=0,1.14.(2018天津理t1)設(shè)全集為r,集合a=x|0x2,b=x|x1,則a(rb)=(a.x|0x1b.x|0x1c.x|1x2d.x|0x2【答案】b)【解析】rb=x|x1,a(rb)=x|0x1.故選b.15.(2018天津文t1)設(shè)集合a=1,2,3,4,b=-1,0,2,3,c=xr|-1x2,則(ab)c=()a.-1,1b.0,1c.-1,0,1d.2,3,4【答案

5、】c【解析】ab=-1,0,1,2,3,4.又c=xr|-1x2,(ab)c=-1,0,1.16.(2018浙江t1)已知全集u=1,2,3,4,5,a=1,3,則ua=()a.b.1,3c.2,4,5d.1,2,3,4,5【答案】c【解析】a=1,3,u=1,2,3,4,5,ua=2,4,5,故選c.17.(2018全國(guó)2理t2,)已知集合a=(x,y)|x2+y23,xz,yz,則a中元素的個(gè)數(shù)為()a.9b.8c.5d.4【答案】a【解析】滿足條件的元素有(-1,-1),(-1,0),(-1,1),(0,1),(0,0),(0,-1),(1,-1),(1,0),(1,1),共9個(gè)。18.

6、(2017全國(guó)3理t1,)已知集合a=(x,y)|x2+y2=1,b=(x,y)|y=x,則ab中元素的個(gè)數(shù)為a.3b.2c.1d.0【答案】b【解析】a表示圓x2+y2=1上所有點(diǎn)的集合,b表示直線y=x上所有點(diǎn)的集合,易知圓x2+y2=1與直線y=x相交,故ab中有2個(gè)元素.19.(2017全國(guó)1理t1)已知集合a=x|x1,b=x|3x1,則()a.ab=x|x1d.ab=【答案】a【解析】3x1=30,x0,b=x|x0,ab=x|x0,ab=x|x1.故選a.20.(2017全國(guó)2理t2)設(shè)集合a=1,2,4,b=x|x2-4x+m=0.若ab=1,則b=()a.1,-3b.1,0c

7、.1,3d.1,5【答案】c【解析】由ab=1,可知1b,所以m=3,即b=1,3.21.(2017全國(guó)1文t1)已知集合a=x|x0,則()a.ab=x|x32b.ab=c.ab=x|x32d.ab=r【答案】a【解析】a=x|x2,b=x|x3,ab=x|x2,ab=x|x3,故選a.2222.(2017全國(guó)2文t1)設(shè)集合a=1,2,3,b=2,3,4,則ab=()a.1,2,3,4b.1,2,3c.2,3,4d.1,3,4【答案】a【解析】因?yàn)閍=1,2,3,b=2,3,4,所以ab=1,2,3,4,故選a.23.(2017全國(guó)3文t1)已知集合a=1,2,3,4,b=2,4,6,8,

8、則ab中元素的個(gè)數(shù)為()a.1b.2c.3d.4【答案】b【解析】由題意可得ab=2,4,則ab中有2個(gè)元素.故選b.24.(2017天津理t1)設(shè)集合a=1,2,6,b=2,4,c=xr|-1x5,則(ab)c=()a.2b.1,2,4c.1,2,4,6d.xr|-1x5【答案】b【解析】a=1,2,6,b=2,4,ab=1,2,4,6.c=xr|-1x5,(ab)c=1,2,4.25.(2017北京理t1)若集合a=x|-2x1,b=x|x3,則ab=()a.x|-2x-1b.x|-2x3c.x|-1x1d.x|1x3【答案】a【解析】ab=x|-2x-1,故選a.26.(2017北京文t

9、1)已知全集u=r,集合a=x|x2,則ua=()a.(-2,2)c.-2,2b.(-,-2)(2,+)d.(-,-22,+)【答案】c【解析】因?yàn)閍=x|x2,所以u(píng)a=x|-2x2.27.(2016全國(guó)1理t1)設(shè)集合a=x|x2-4x+30,則ab=()a.(-3,-3)2c.(1,3)2b.(-3,3)2d.(3,3)2【答案】d【解析】a=(1,3),b=(3,+),所以ab=(3,3),故選d.2228.(2016全國(guó)2理t2)已知集合a=1,2,3,b=x|(x+1)(x-2)0,xz,則ab=()a.1b.1,2c.0,1,2,3d.-1,0,1,2,3【答案】c【解析】由題意

10、可知,b=x|-1x0,則st=()a.2,3b.(-,23,+)c.3,+)d.(0,23,+)【答案】d【解析】s=x|x2或x3.因?yàn)閠=x|x0,所以st=x|0x2或x3,故選d.30.(2016全國(guó)1文t1)設(shè)集合a=1,3,5,7,b=x|2x5,則ab=()a.1,3b.3,5c.5,7d.1,7【答案】b【解析】ab=3,5,故選b.31.(2016全國(guó)2文t1)已知集合a=1,2,3,b=x|x29,則ab=()a.-2,-1,0,1,2,3b.-2,-1,0,1,2c.1,2,3d.1,2【答案】d【解析】b=x|-3x3,ab=1,2.故選d.32.(2016全國(guó)3文t

11、1)設(shè)集合a=0,2,4,6,8,10,b=4,8,則ab=()a.4,8b.0,2,6c.0,2,6,10d.0,2,4,6,8,10【答案】c【解析】根據(jù)補(bǔ)集的定義,知從集合a=0,2,4,6,8,10中去掉集合b中的元素4,8后,剩下的4個(gè)元素0,2,6,10構(gòu)成的集合即為ab,即ab=0,2,6,10,故選c.33.(2016四川理t1)設(shè)集合a=x|-2x2,z為整數(shù)集,則集合az中元素的個(gè)數(shù)是()a.3b.4c.5d.6【答案】c【解析】由題意,az=-2,-1,0,1,2,故其中的元素個(gè)數(shù)為5,選c.34.(2016天津理t1)已知集合a=1,2,3,4,b=y|y=3x-2,x

12、a,則ab=()a.1b.4c.1,3d.1,4【答案】d【解析】由題意知集合b=1,4,7,10,則ab=1,4.故選d.35.(2016山東理t2)設(shè)集合a=y|y=2x,xr,b=x|x2-10,b=x|-1x-1,選c.36.(2016浙江理t1)已知集合p=xr|1x3,q=xr|x24,則p(rq)=()a.2,3b.(-2,3c.1,2)d.(-,-21,+)【答案】b【解析】q=xr|x-2,或x2,rq=xr|-2x2.p(rq)=xr|-2x3=(-2,3.故選b.37.(2015全國(guó)2理t1)已知集合a=-2,-1,0,1,2,b=x|(x-1)(x+2)0,則ab=()

13、a.-1,0b.0,1c.-1,0,1d.0,1,2【答案】a【解析】b=x|-2x1,ab=-1,0.38.(2015全國(guó)1文t1)已知集合a=x|x=3n+2,nn,b=6,8,10,12,14,則集合ab中元素的個(gè)數(shù)為()a.5b.4c.3d.2【答案】d【解析】由條件知,當(dāng)n=2時(shí),3n+2=8,當(dāng)n=4時(shí),3n+2=14.所以ab=8,14.故選d.39.(2015全國(guó)2文t1)已知集合a=x|-1x2,b=x|0x3,則ab=()a.(-1,3)b.(-1,0)c.(0,2)d.(2,3)【答案】a【解析】由題意,得ab=x|-1x3,即ab=(-1,3).40.(2015陜西文t

14、1)設(shè)集合m=x|x2=x,n=x|lgx0,則mn=()a.0,1b.(0,1c.0,1)d.(-,1【答案】a【解析】m=0,1,n=x|0x1,mn=x|0x1,即為0,1.41.(2015重慶理t1,)已知集合a=1,2,3,b=2,3,則()a.a=bb.ab=c.abd.ba【答案】d【解析】因?yàn)閍=1,2,3,b=2,3,所以ba.42.(2014全國(guó)1理t1)已知集合a=x|x2-2x-30,b=x|-2x2,則ab=()a.-2,-1b.-1,2)c.-1,1d.1,2)【答案】a【解析】由已知,可得a=x|x3或x-1,則ab=x|-2x-1=-2,-1.故選a.43.(2

15、014全國(guó)2理t1)設(shè)集合m=0,1,2,n=x|x2-3x+20,則mn=()a.1b.2c.0,1d.1,2【答案】d【解析】n=x|x2-3x+20=x|1x2,mn=0,1,2x|1x2=1,2.故選d.44.(2014全國(guó)1文t1)已知集合m=x|-1x3,n=x|-2x1,則mn=()a.(-2,1)b.(-1,1)c.(1,3)d.(-2,3)【答案】b【解析】由已知得mn=x|-1x1=(-1,1),故選b.45.(2014全國(guó)2文t1)已知集合a=-2,0,2,b=x|x2-x-2=0,則ab=()a.b.2【答案】bc.0d.-2【解析】易得b=-1,2,則ab=2,故選b

16、.46.(2014遼寧理t1)已知全集u=r,a=x|x0,b=x|x1,則集合u(ab)=()a.x|x0b.x|x1c.x|0x1d.x|0x1【答案】d【解析】ab=x|x0或x1,u(ab)=x|0x1.故選d.47.(2013全國(guó)2理t1)已知集合m=x|(x-1)24,xr,n=-1,0,1,2,3,則mn=()a.0,1,2b.-1,0,1,2c.-1,0,2,3d.0,1,2,3【答案】a【解析】m=x|-1x3,n=-1,0,1,2,3,所以mn=0,1,2,故選a.48.(2013全國(guó)1文t1)已知集合a=1,2,3,4,b=x|x=n2,na,則ab=()a.1,4b.2

17、,3c.9,16d.1,2【答案】a【解析】b=1,4,9,16,ab=1,4.49.(2013全國(guó)2文t1)已知集合m=x|-3x1時(shí),集合a=x|x1或xa,由ab=r,可知a-11,即a2.故1a2.當(dāng)a=1時(shí),集合a=r,顯然ab=r.故a=1,滿足題意.當(dāng)a1時(shí),集合a=x|x1或xa,由ab=r,可知a-1a顯然成立,故a1.綜上可知,a的取值范圍是a2.故選b.51.(2013廣東理t8)設(shè)整數(shù)n4,集合x=1,2,3,n,令集合s=(x,y,z)|x,y,zx,且三條件xyz,yzx,zxy恰有一個(gè)成立.若(x,y,z)和(z,w,x)都在s中,則下列選項(xiàng)正確的是()a.(y,

18、z,w)s,(x,y,w)sb.(y,z,w)s,(x,y,w)sc.(y,z,w)s,(x,y,w)sd.(y,z,w)s,(x,y,w)s【答案】b【解析】由(x,y,z)s,不妨取xyz,要使(z,w,x)s,則wxz或xzw.當(dāng)wxz時(shí),wxyz,故(y,z,w)s,(x,y,w)s.當(dāng)xzw時(shí),xyz0,b=x|5x5,則()a.ab=b.ab=rc.bad.ab【答案】b【解析】集合a=x|x2,由圖象可以看出ab=r,故選b.55.(2012課標(biāo)全國(guó)理t1)已知集合a=1,2,3,4,5,b=(x,y)|xa,ya,x-ya,則b中所含元素的個(gè)數(shù)為()a.3b.6c.8d.10【

19、答案】d【解析】由xa,ya,x-ya,得(x,y)可取如下:(2,1),(3,1),(4,1),(5,1),(3,2),(4,2),(5,2),(4,3),(5,3),(5,4),故集合b中所含元素的個(gè)數(shù)為10.56.(2012大綱理2)已知集合a=1,3,m,b=1,m,ab=a,則m=()a.0或3b.0或3c.1或3d.1或3【答案】b【解析】ab=a,ba,m=3或m=m.m=3或m=0或m=1.當(dāng)m=1時(shí),與集合中元素的互異性不符,故選b.57.(2012全國(guó)文1)已知集合a=x|x2-x-20,b=x|-1x1,則()a.abb.bac.a=bd.ab=【答案】b【解析】由題意可

20、得a=x|-1x2,而b=x|-1x1,故ba.58.(2012大綱全國(guó)文t1,)已知集合a=x|x是平行四邊形,b=x|x是矩形,c=x|x是正方形,d=x|x是菱形,則()a.abb.cbc.dcd.ad【答案】b【解析】正方形組成的集合是矩形組成集合的子集,cb.59.(2012湖北文t1)已知集合a=x|x2-3x+2=0,xr,b=x|0x5,xn,則滿足條件acb的集合c的個(gè)數(shù)為()a.1b.2c.3d.4【答案】d【解析】a=1,2,b=1,2,3,4.又acb,c=1,2或1,2,3或1,2,4或1,2,3,4,故選d.60.(2011全國(guó)文1)已知集合m=0,1,2,3,4,

21、n=1,3,5,p=mn,則p的子集共有()a.2個(gè)b.4個(gè)c.6個(gè)d.8個(gè)【答案】b【解析】p=mn=1,3,p的子集有22=4個(gè).61.(2011遼寧理t2)已知m,n為集合i的非空真子集,且m,n不相等,若n(im)=,則mn=()a.mb.nc.id.【答案】a【解析】作出滿足條件的韋恩(venn)圖,易知mn=m.62.(2011廣東理t8)設(shè)s是整數(shù)集z的非空子集,如果a,bs,有abs,則稱s關(guān)于數(shù)的乘法是封閉的.若t,v是z的兩個(gè)不相交的非空子集,tv=z,且a,b,ct,有abct;x,y,zv,有xyzv,則下列結(jié)論恒成立的是()a.t,v中至少有一個(gè)關(guān)于乘法是封閉的b.t

22、,v中至多有一個(gè)關(guān)于乘法是封閉的c.t,v中有且只有一個(gè)關(guān)于乘法是封閉的d.t,v中每一個(gè)關(guān)于乘法都是封閉的【答案】a【解析】令t=n,v=zn,則t對(duì)乘法封閉,而v對(duì)乘法不封閉排除d.令t=-1,0,1,v=zt,則t,v都對(duì)乘法封閉,排除b,c.故選a.63.(2011福建文t12)在整數(shù)集z中,被5除所得余數(shù)為k的所有整數(shù)組成一個(gè)“類”,記為k,即k=5n+k|nz,k=0,1,2,3,4.給出如下四個(gè)結(jié)論:20111;-33;z=01234;“整數(shù)a,b屬于同一類”的充要條件是“a-b0”.其中,正確結(jié)論的個(gè)數(shù)是()a.1b.2c.3d.4【答案】c【解析】對(duì)于:2011=5402+1

23、,20111.對(duì)于:-3=5(-1)+2,-32,故不正確;對(duì)于:任意一個(gè)整數(shù)z被5除,所得余數(shù)共分為五類,z=01234,故正確;對(duì)于:若整數(shù)a,b屬于同一類,則a=5n1+k,b=5n2+k,a-b=5n1+k-5n2-k=5(n1-n2)=5n,a-b0,若a-b0,則a-b=5n,即a=b+5n,故a與b被5除的余數(shù)為同一個(gè)數(shù),a與b屬于同一類,所以“整數(shù)a,b屬于同一類”的充要條件是“a-b0”,故正確.正確結(jié)論的個(gè)數(shù)是3.64.(2011福建理t1)i是虛數(shù)單位,若集合s=-1,0,1,則()a.isb.i2sc.i3sd.2si【答案】b【解析】i2=-1,而集合s=-1,0,1

24、,i2s.65.(2010浙江理t1)設(shè)p=x|x4,q=x|x24,則()a.pqb.qpc.prqd.qrp【答案】b【解析】p=x|x4,q=x|-2x2,qp.66.(2010天津理t9)設(shè)集合a=x|x-a|2,xr.若ab,則實(shí)數(shù)a,b必滿足()a.|a+b|3b.|a+b|3c.|a-b|3d.|a-b|3【答案】d【解析】a=x|a-1xb+2或xb-2,xr.若ab,則需滿足a+1b-2或a-1b+2,即a-b-3或a-b3,|a-b|3.67.(2010全國(guó)t1)已知集合a=x|x|2,xr,b=x|4,xz,則ab等于()a.(0,2)b.0,2c.0,2d.0,1,2【

25、答案】d【解析】a=x|-2x2,b=0,1,2,3,16,ab=0,1,2.68.(2018江蘇t1)已知集合a=0,1,2,8,b=-1,1,6,8,那么ab=.【答案】1,8【解析】由題設(shè)和交集的定義可知,ab=1,8.69.(2017江蘇t1)已知集合a=1,2,b=a,a2+3.若ab=1,則實(shí)數(shù)a的值為.【答案】1【解析】由已知得1b,2b,顯然a2+33,所以a=1,此時(shí)a2+3=4,滿足題意,故答案為1.70.(2013湖南,文t15)對(duì)于e=a1,a2,a100的子集x=ai1,ai2,aik,定義x的“特征數(shù)列”為x1,x2,x100,其中xi1=xi2=xik=1,其余項(xiàng)

26、均為0.例如:子集a2,a3的“特征數(shù)列”為0,1,1,0,0,0.(1)子集a1,a3,a5的“特征數(shù)列”的前3項(xiàng)和等于;(2)若e的子集p的“特征數(shù)列”p1,p2,p100滿足p1=1,pi+pi+1=1,1i99;e的子集q的“特征數(shù)列”q1,q2,q100滿足q1=1,qj+qj+1+qj+2=1,1j98,則pq的元素個(gè)數(shù)為.【答案】（1）2（2）17【解析】(1)a1,a3,a5的特征數(shù)列為1,0,1,0,1,0,0,前3項(xiàng)和為2.(2)根據(jù)題意知,p的特征數(shù)列為1,0,1,0,1,0,則p=a1,a3,a5,a99有50個(gè)元素,q的特征數(shù)列為1,0,0,1,0,0,1,則q=a1

27、,a4,a7,a10,a100有34個(gè)元素,6pq=a1,a7,a13,a97,共有1+97-1=17個(gè).71.(2013江蘇t4)集合-1,0,1共有個(gè)子集.【答案】8【解析】由于集合-1,0,1有3個(gè)元素,故其子集個(gè)數(shù)為23=8.72.(2012天津文t9,)集合a=xr|x-2|5中的最小整數(shù)為.【答案】-3【解析】|x-2|5,-3x7,最小整數(shù)為-3.73.(2018北京理t20)設(shè)n為正整數(shù),集合a=|=(t1,t2,tn),tk0,1,k=1,2,n.對(duì)于集合a中的任意元素=(x1,x2,xn)和=(y1,y2,yn),記m(,)=1(x1+y1-|x1-y1|)+(x2+y2-|x2-y2|)+(xn+yn-|xn-yn|).2(1)當(dāng)n=3時(shí),若=(1,1,0),=(0,1,1),求m(,)和m(,)的值;(2)當(dāng)n=4時(shí),設(shè)b是a的子集,且滿足:對(duì)于b中的任意元素,當(dāng),相同時(shí),m(,)是奇數(shù);當(dāng),不同時(shí),m(,)是偶數(shù).求集合b中元素個(gè)數(shù)的最大值;