課時(shí)規(guī)范練1會(huì)合的觀點(diǎn)與運(yùn)算基礎(chǔ)穩(wěn)固組1.(2017北京,文1)已知全集U=R,會(huì)合A={x|x<-2或x>2},則?UA=( )A.(-2,2)B.(-∞,-2)∪(2,+∞)C.[-2,2]D.(-∞,-2]∪[2,+∞)?導(dǎo)學(xué)號(hào)24190701?2.已知會(huì)合A={x|(x-1)(x-2)(x-3)=0},會(huì)合B={x|y=-},則會(huì)合A∩B的真子集的個(gè)數(shù)是()A.1B.2C.3D.43.(2017山東青島模擬,文1)已知全集I=R,會(huì)合{log2,2},{)A=y|y=xx>B=x|y=-},則(A.A?BB.A∪B=AC.A∩B=?D.A∩(?IB)≠?4.(2017山東濰坊一模,文1)已知會(huì)合A={x|x=2n,n∈N*},B={x|≤},則A∩B=( )A.{2}B.{2,4}C.{2,3,4}D.{1,2,3,4}5.已知會(huì)合{0,2,4,6,8,10},{4,8},則?( )AA.{4,8}B.{0,2,6}C.{0,2,6,10}D.{0,2,4,6,8,10}6.(2017安徽安慶二模,文1)已知會(huì)合{4,-3,-2,-1,0,1},{∈R230},則∩( )M=-N=x|x+x<MN=A.{-3,-2,-1,0}B.{-2,-1,0}C.{-3,-2,-1}D.{-2,-1}?導(dǎo)學(xué)號(hào)24190702?7.(2017山東,文1)設(shè)會(huì)合{1|<1},{2},則∩()M=x||x-N=x|x<MN=A.(-1,1)B.(-1,2)C.(0,2)D.(1,2)8.(2017山西太原二模,文2)已知{1,2,4},{log2,x∈},則∪( )A=B=y|y=xAAB=A.{1,2}B.[1,2]C.{0,1,2,4}D.[0,4]9.已知會(huì)合A={1,2,3,4,5},B={(x,y)|x∈A,y∈A,x-y∈A},則B中所含元素的個(gè)數(shù)為.10.(2017江蘇,1)已知會(huì)合A={1,2},B={a,a2+3}.若A∩B={1},則實(shí)數(shù)a的值為.11.已知會(huì)合A={x|0<log4x<1},B={x|x≤},則A∩B=.12.已知A,B是全集I={1,2,3,4}的子集,A={1,2},則知足A?B的B的個(gè)數(shù)為.?導(dǎo)學(xué)號(hào)24190703?綜合提高組13.(2017全國(guó)Ⅲ,文1)已知會(huì)合A={1,2,3,4},B={2,4,6,8},則A∩B中元素的個(gè)數(shù)為( )14.(2017山東濰坊二模,文3)若會(huì)合M={x|x2-x<0},N={y|y=ax(a>0,a≠1)},R表示實(shí)數(shù)集,則以下選項(xiàng)錯(cuò)誤的選項(xiàng)是( )A.M∩(?RN)=?B.M∪N=RC.(?RM)∪N=RD.M∩N=M15.已知全集U=R,會(huì)合A={x|0≤x≤},B={y|≤y≤3},則(?UA)∪B=( )A.(2,3]B.(-∞,1]∪(2,+∞)C.[1,2)D.(-∞,0)∪[1,+∞)16.已知會(huì)合{4≤x≤6},[,],若?,則實(shí)數(shù)a-b的取值范圍是.A=x|B=abAB創(chuàng)新應(yīng)用組17.(2017浙江名校聯(lián)考)已知會(huì)合{},{12},且∪(?)R,則實(shí)數(shù)a的取值范圍是R( )a≤a<1a≥a>218.(2017河南平頂山模擬改編)已知會(huì)合{240},{22(1)210},若(?R)∩?,A=x|x+x=B=x|x+a+x+a-=BA=則a=.?導(dǎo)學(xué)號(hào)24190704?答案：1.C由于{2或2},因此?{2≤≤2}.應(yīng)選CU2.C化簡(jiǎn)會(huì)合得{1,2,3},會(huì)合{≥2},因此∩{2,3},A=B=x|xAB=則∩B的真子集有?,{2},{3}.應(yīng)選CA.3.A由于當(dāng)x>2時(shí),y=logx>1,因此A=(1,+∞).又由于B=[1,+∞),因此A?B,A∪2B=B,A∩B=A,A∩(?IB)=?,應(yīng)選A.4.B∵A={x|x=2n,n∈N*}={2,4,6,},B={x|≤2}={x|0≤x≤4},∴A∩B={2,4},應(yīng)選B.5.C依據(jù)補(bǔ)集的定義,知從會(huì)合A={0,2,4,6,8,10}中去掉會(huì)合B中的元素4,8后,剩下的4個(gè)元素0,2,6,10組成的會(huì)合即為?AB,即?AB={0,2,6,10},應(yīng)選C.6.D∵會(huì)合M={-4,-3,-2,-1,0,1},N={x∈R|x2+3x<0}={x|-3<x<0},∴M∩N={-2,-1},應(yīng)選D.7.C由|x-1|<1,得-1<x-1<1,即0<x<2.因此M={x|0<x<2},因此M∩N=(0,2).8.C∵A={1,2,4},B={y|y=log2x,x∈A}={0,1,2},∴A∪B={0,1,2,4}.應(yīng)選C.9.10由x∈A,y∈A,x-y∈A,得(x,y)可取值如下:(2,1),(3,1),(4,1),(5,1),(3,2),(4,2),(5,2),(4,3),(5,3),(5,4),故會(huì)合B中所含元素的個(gè)數(shù)為10.10.1由已知得1∈B,2?B,明顯a2+3≥3,因此a=1,此時(shí)a2+3=4,知足題意,故答案為1.11.(1,2]∵0<log4x<1,log41<log4x<log44,即1<x<4,∴A={x|1<x<4}.∵B={x|x≤2},∴A∩B={x|1<x≤2}.12.4由于A={1,2},且A?B,因此B={1,2}或B={1,2,3}或B={1,2,4}或B={1,2,3,4}.13.B由題意可得∩{2,4},則∩B中有2個(gè)元素.應(yīng)選BAB=A.14.B∵會(huì)合{20}{01},M=x|x-x<=x|<x<{x(0,a≠1)}{0},N=y|y=aa>=y|y>∩(?){01}∩{y|y≤0}?,故A正確;R∪(0,+∞),故B錯(cuò)誤;MN=(?)∪{≤0或x≥1}∪{y|y>0}R,故C正確;R∩{01}∩{y|y>0}{01},故D正確.應(yīng)選BMN=x|<x<=x|<x<=M.15.D由于?{2或0},{1≤≤3},因此(?)∪(,0)∪[1,).UU16.(,-2]會(huì)合{4≤2x≤16}{22≤2x≤24}{2≤≤4}[2,4]-∞A=x|=x|=x|x=.由于?,因此≤2,≥4.AB