资源简介 《不等式》专题15-1 集合涉二次、绝对值、分式不等式(14套,7页,含答案)设全集,集合,集合则下图中阴影部分表示的集合为( [endnoteRef:0] )A. B. C. D. [0: 答案:D;]已知集合A={x∈R|x2-3x≥0},B={-2,2},则(RA)∩B=([endnoteRef:1] )A. B. {-2} C. {2} D. {-2,2} [1: 答案:C;]已知集合,,则=( [endnoteRef:2] )A. B. C. D. [2: 答案:B;]若集合,,则([endnoteRef:3] )A. B. C.(-1,1) D.(-1,2) [3: 答案:C;]《不等式》专题15-2 集合涉二次、绝对值、分式不等式已知集合,则=([endnoteRef:4] )A. B. C. D. [4: 答案:C;]已知集合A={x∈R|x2-x-2<0},B={x∈Z|x=2t+1,t∈A},则A∩B=([endnoteRef:5] )A.{-1,0,1} B.{-1,0} C.{0,1} D.{0} [5: 答案:C;【解析】A={x∈R|x2-x-2<0}={x|-1<x<2},则x=2t+1∈(-1,5),所以B={0,1,2,3,4},所以A∩B={0,1},故选C.]设集合,若,则的取值范围是( [endnoteRef:6] )A. B. C. D. [6: 答案:D;]已知集合,集合,则( [endnoteRef:7] )A. B. C. D. [7: 【答案】A【解析】由题意得,,所以,故选A.]已知集合,,则=( [endnoteRef:8] )A. B. C. D. [8: 答案:A;]《不等式》专题15-3 集合涉二次、绝对值、分式不等式设全集则图中阴影部分表示的集合为 ([endnoteRef:9] )A. B. C. D.[9: 答案:B;]已知集合,则([endnoteRef:10] )A. B. C.或 D.或 [10: 答案:C;]已知全集,,,则等于( [endnoteRef:11] )A. B. C. D. [11: 答案:B;由,,又已知,]设全集为R,集合,,则=( [endnoteRef:12] )A. B. C. D. [12: 答案:B;【解析】由题知或,又集合,则,故选B.]《不等式》专题15-4 集合涉二次、绝对值、分式不等式设集合,,则([endnoteRef:13] )A. B. C. D. [13: 答案:A;]设全集,,,则如图中阴影部分表示的集合为( [endnoteRef:14] )A. B. C. D. [14: 答案:D;【解析】由题意可知,,阴影部分用集合表示为,而,故,,故选D.]已知集合,,则([endnoteRef:15] )A. B. C. D. [15: 答案:B;]已知,,,则( [endnoteRef:16])A. B. C. D. [16: 答案:D;]《不等式》专题15-5 集合涉二次、绝对值、分式不等式设集合A={x∈Z|-1≤x≤2},B={x|x2<1},则A∩B=([endnoteRef:17] )A. {-1,0,1} B. {0} C. {-1,0} D. {-1,0,1,2} [17: 答案:B;]设集合,集合, 则([endnoteRef:18] ).A. B. C. D. [18: 答案:A;【解析】由题意可得,,所以,故选A.]设集合A=,B=,则集合AB=( [endnoteRef:19] )A.(0,1] B.(0,1) C.(0,4) D.(0,4] [19: 答案:A;]已知集合A=,B=,则( [endnoteRef:20])A. B. C. D. [20: 答案:A解析:A=[﹣4,1],B=(﹣1,3),(A)B=(﹣1,1],故选A.]《不等式》专题15-6 集合涉二次、绝对值、分式不等式已知集合,,则([endnoteRef:21] )A. B. C. D. [21: 答案:C;]已知集合,集合,则([endnoteRef:22] )A. B. C. D. [22: 答案:C;]设,集合,,则=( [endnoteRef:23] )A. B. C.D. [23: 答案:B;]已知集合,,则( [endnoteRef:24] )A. B. C. D. [24: 答案:C;【解析】由,得,故选C.]《不等式》专题15-7 集合涉二次、绝对值、分式不等式设集合,,则( [endnoteRef:25] )A. B. C. D. [25: 答案:D;]已知集合,则等于([endnoteRef:26] )A. B. C. D. [26: 答案:A;]已知集合A={x|x2-5x+6≤0},B={x||2x-1|>3},则集合A∩B=( [endnoteRef:27] )(A){x|2≤x≤3} (B){x|2≤x<3} (C){x|2<x≤3} (D){x|-1<x<3} [27: 答案:C;]已知集合,,则( [endnoteRef:28] )A. B. C. D. [28: 答案:C;]《不等式》专题15-8 集合涉二次、绝对值、分式不等式若集合,,则( [endnoteRef:29] )A. B. C.[:] D. [29: 答案:C;【解析】∵,∴.]已知集合,B={0,1,2,4},则=([endnoteRef:30] )A. {0,1} B.{0,1,2} C.{1,2} D.{0,1,2,4}[30: 答案:A;【解析】解:;.故选:A.可求出集合A,然后进行交集的运算即可.考查描述法、列举法的定义,以及交集的运算.]已知集合,.若,则的取值范围为( [endnoteRef:31] )A. B. C. D. [31: 答案:A;]集合,,则( [endnoteRef:32] )A. B. C. D. [32: 答案:C]已知集合,,则等于([endnoteRef:33] )A. B. C. D. [33: 答案:B;]《不等式》专题15-9 集合涉二次、绝对值、分式不等式设集合,,则集合([endnoteRef:34] )A. B. C. D. [34: 答案:A;]已知集合,,则( [endnoteRef:35])A. B. C. D. [35: 答案:C;]若全集U={x∈R|x2≤4} A={x∈R||x+1|≤1}的补集CuA为( [endnoteRef:36] )A |x∈R |0<x<2| B |x∈R |0≤x<2| C |x∈R |0<x≤2| D |x∈R |0≤x≤2| [36: 【答案】C]已知全集( [endnoteRef:37] )A.{3} B.{0,3,5} C.{3,5} D.{0,3} [37: [答案]D[解析]全集U={0,1,2,3,4},则CuA={0,3}[考点]分式不等式及集合运算.]《不等式》专题15-10 集合涉二次、绝对值、分式不等式已知集合,则 RA=([endnoteRef:38] )A. B. C. D. [38: 答案:B;解答:或,则.]已知集合,,,则=( [endnoteRef:39])A. B. C. D. [39: 答案:A;]设集合,若,则的取值范围是( [endnoteRef:40] )A. B. C. D. [40: 答案:D;]已知集合为整数集,则集合中所有元素的和等于 [endnoteRef:41] . [41: 答案:3;]设集合,( [endnoteRef:42] )A. B. C. D. [42: 答案:A;]《不等式》专题15-11 集合涉二次、绝对值、分式不等式已知集合 A ={} ,B = {},则( [endnoteRef:43] )A.(0,+∞) B.(0,1) C.[0,1) D. [1, +∞) [43: 答案:C;]设集合,,则( [endnoteRef:44])(A) (B) (C) (D) [44: 答案:B;]集合中最小整数为[endnoteRef:45] . [45: 【答案】 【解析】不等式,即,,所以集合,所以最小的整数为。]已知集合M={x|},N={x|},则M∩N=( [endnoteRef:46] )A.{x|-1≤x<0} B.{x |x>1} C.{x|-1《不等式》专题15-12 集合涉二次、绝对值、分式不等式设集合, 则集合等于([endnoteRef:47] )A. B. C. D. [47: 答案:A;]已知集合,,则([endnoteRef:48] )A. B. C. D. [48: 答案:B;]已知集合P={x︱x2≤1},M={a}.若P∪M=P,则a的取值范围是( [endnoteRef:49] )A.(-∞, -1] B.[1, +∞) C.[-1,1] D.(-∞,-1] ∪[1,+∞)[49: 答案:C;]若集合,,则 [endnoteRef:50] 。 [50: ]已知集合,则 [endnoteRef:51] 。 [51: 答案:解析:由题知,,故.]《不等式》专题15-13 集合涉二次、绝对值、分式不等式已知集合,,则([endnoteRef:52] )A. B. C. D. [52: 答案:B;]已知集合,则满足条件的集合的个数为([endnoteRef:53] )A. B. C. D. [53: 答案:C;解析:∵,又,∴集合的个数为个,故选C.]已知全集U={非零整数},集合A={x||x+2|>4, xU}, 则CA=( [endnoteRef:54] )A、{-6,-5,-4,-3,-2,-1,0,1,2} B、{-6,-5,-4,-3,-2,-1,1,2}C、{-5,-4,-3,-2,0,-1,1} D、{-5,-4,-3,-2,-1,1} [54: 答案:B;]若集合则A∩B是([endnoteRef:55] )A. B. C. D. [55: 答案 D解析 集合,∴ 选D]《不等式》专题15-14 集合涉二次、绝对值、分式不等式已知集合,集合,则([endnoteRef:56] )A. B. C. D. [56: 答案:A;]已知集合,,则如图中阴影部分所表示的集合为( [endnoteRef:57] )A. B. C. D. [57: 答案:D;]已知全集U=R,集合M={x||x-1|2},则[endnoteRef:58]( )(A){x|-13} (D){x|x-1或x3} [58: 【答案】C【解析】因为集合,全集,所以,故选C.【命题意图】本题考查集合的补集运算,属容易题.]设集合,则=[endnoteRef:59]( )A. B. C. D. [59: 答案 B解:..故选B.] 展开更多...... 收起↑ 资源预览