《集合》双空题专题训练(一)

1、集合双空题专题训练（一）1. 已知全集U =1,2,3,4,5,集合A = 1,2,3, OB = 2,5,则集合B=,B =.1. 1,3,41,3【解析】由补集立义可知：B = l,3,4, .AB = 1,3.故答案为：1,3,4 ： 1,3.2. 已知全集U =1,2,3,4,5,6,7,集合 A = 1,2,3 , B=(2, 3 4),则 ACB =，C/A =2. 2,3 4,5,6,7解析】全集 =1,2,34,5,6,7,集合力= 1,2,3, B = 2,3,4,所以 ACB = 2,3； Q7A = 4,5,6,7.3. 设集合 A = , Q = xxat若PUQ =

2、R,则实数。的取值范围 是，若PcQ = Q,则实数的取值范围是.4. (P,-2(4.+)【解析】P = IX .v2 -2x-8 = xx -2,Q = xxa,若 PuQ = R 则。4.5. 已知全集f=-l,0,1,2,3, A = 0,l,2, B = -1,O,1,则CtzA =,ArlB =.5. -l,3)0,l)【解析.t = -l,0,l,2,3,A = 0,l,2 ,. CuA = -1,3, V B = -l,O,l , .AB = O,16. 设集合A = (x,y)lyx-l, B = (x,y)ly-x+.(1) 若ArB0,则实数“的取值范围是(2) 当 =

3、3时，若(x,y)AB,则2x + y的最大值是.6. 1,-Ho) 5【解析】(1)考虑不等式yx-l, y-x+a表示的平面区域，它们如图 所示，英中I对应不等式y-i表示的平面区域，I【对应不等式y-表示的平 而区域.若ArB0,则al.试卷第5页，总6页（2）当a = 3时，AB对应的平而区域如下图所示:把初始直线2x + y = O平移至M时,y = x-l% = -X + 3x = 2Z = 2x+y可取最大值.令即M(2,l),故Z7. 已知集合 A = x-2x,B = xx = t-a,t.若 = 0,则Ar13 =:若AB,则实数的取值范围是.7. xlxl 2,3【解析】

4、当幺=0时，B = xlx = r2,r = xlxB = xOx.当 (-2,l)时，x = t2-a -aA-a),由 B,得a 2,解得2S*3,4-a 1,8. 已知全集U =1,2,3,4,5, A = l,3, B = l,2,5,则Ar3 = , (CUA)Q.A = 2J7),则 4= , B=3,5,11,137,11,13,19【解析】由题意,全集 =2,3,5,7,11,13,17,19, AnQrB = 3,5, A 集合 A 中含有3, 5 ,集合3 中没有3, 5, BCC/ = 7,19,集合A中没有7, 19,集合3中含有7, 19，Y CCoA = 2,17,

5、 AUB = 3,5,7,11,13,19,. A = 3,5,ll,13, B = 7,11,13,19,设全集U = R, = xl-5x5), B = (XlOX7,则QyAUoB=-41,则 AB=, AUB=,jvxv2 x卜一2【解析】T A = x F4 v = x-2 VX V2,3 = xxl , . 4c3 = xl Vx V 2 , AkJB = xx-2.设M=(x,y)lm + ny = 4,且(2),(一2,5) 贝m =, n=10.11.11.12.12.13.13.14.14.15.【解析】(2,1),(-2,5)M,则彳2n + n = 4-2/?+ 5/?

6、 = 4已知集合 M =(x9y)x + y = a9 N = (x,y)x-y = b 9 若 MCN = (3,-1),则C(AUB) =15. -xO或5xlx-5或7【解析】由题意知OA = xlx-5或x25,0B = xlxv O 或 X n 7.则。AUcB= xlx0 或 5;由题意知 AjB = x -5 X , B = x F+r + bS,且 AJB = R t4c3 = x3 vxS4,贝IJd=t b =16. -3-4【解析】A = (XIX2 -2x-3 = (-,-1)k(3,o),.4U8 = R,AcB = (3,4,：.B = xx2 +tu+b = -1

7、,4, .-l,4为方程2+ax+b = 0 的两个解，a = -3V 方=一417. (1)若3M-l,3m,m2-l,则实数加=； (2)若2xx-a0f 则实数“的取值范围是.17. 4 或 2aa2解析】(1)由加 _1 = 3,得加=4,此时 3n = 12. -l = 15.符合题意.由 3m = 3,得 TM = L 此时 Hl-1 = m -1=0 故舍去由 w2 -1 = 3 得加=2 ,当 HI = 2 时，W-I=I 2m = 6 符合题意:.当WI =2时，?一 1=3, 3n = 6 符合题意, 综上所述，W= 4或2.(2)因为2exx-a0,所以2不满足不等式x-

8、a0,即2满足不等式x-a0, 所以2-0,即眇2.所以实数“的取值范围是Ia I a2.1&设M=(, y)ly = 2-, d0,N=(, y)I(x-1)2= t/2, 0, 则MnNH0时，实数“的最大值是,最小值是.18. 22+222-2【解析】. - = yj2a2-x2f ：. x2+y2 = 2ay 0) 它表示以原点 O为圆心，以施/为半径的上半i.(x-l)2+(y-3)2=Z它表示以点A (1,J)为圆心, 以为半径的圆.MNh0时，半圆与圆相交或相切，所以2a-aOAy2a + a,（当半圆与圆内切时- = IOAI,当半圆与圆外切时,QAI=d + d),所以26-

9、d7l2+ry2a + af所以y2a-a2y2a + a实数的最大值是22+2 的最小值是22-2 19. 若全集U = xx-4, A = x-3x2, B = x-x2,【解析J -A = (Xl-3x2 f3 = xl-lx3,. AB = -1,2, U = xx-4 , .QrA = xl-4x2,20. 设全集 U = R,集合 A = xx,则ArIB =:MUB=.20. (-oo,-l) (P,l)U2,w)【解析】因为集合B = xl2-2x 3 = xlxv-1 或 x3,集合A = xx2,所以AB= (-oo,-l),又CUA = xx2,所以 (CtZA)UB = (o-I)2,oo).21. 设全集U = 2,3,2d-3, 4 = 2,b, C = 5,则=,b =21. 4 3【解析】由题意，全集U = 2,3,2-3,集合A = 2,b,因为CUA = 5,可得 23 = 5if C ，解得a = b = 3b =