资源描述
2019版数学精品资料(人教版)学业分层测评(五)补集及综合应用(建议用时:45分钟)学业达标一、选择题1若全集U0,1,2,3且UA2,则集合A的真子集共有()A3个B5个C7个D8个【解析】A0,1,3,真子集有2317.【答案】C2已知全集UR,Ax|x0,Bx|x1,则集合U(AB)()Ax|x0Bx|x1Cx|0x1Dx|0x2,Tx|x23x40,则(RS)T_.【解析】集合Sx|x2,RSx|x2,由x23x40,得Tx|4x1,故(RS)Tx|x1【答案】(,17已知集合A、B均为全集U1,2,3,4的子集,且U(AB)4,B1,2,则AUB_.【解析】U1,2,3,4,U(AB)4,AB1,2,3,又B1,2,3A1,2,3又UB3,4,AUB3【答案】38设全集UR,集合Ax|x0,By|y1,则UA与UB的包含关系是_【解析】UAx|x0,UBy|y1x|x1UAUB.【答案】UAUB三、解答题9(2016宁波高一检测)设AxZ|x|6,B1,2,3,C3,4,5,求:(1)A(BC);(2)AA(BC)【解】A5,4,3,2,1,0,1,2,3,4,5,(1)由BC3,A(BC)A5,4,3,2,1,0,1,2,3,4,5(2)由BC1,2,3,4,5,A(BC)5,4,3,2,1,0,AA(BC)5,4,3,2,1,010设全集为R,Ax|3x7,Bx|2x10,求:(1)AB;(2)RA;(3)R(AB)【解】(1)Ax|3x7,Bx|2x10,ABx|3x7(2)又全集为R,Ax|3x7,RAx|x3或x7(3)ABx|2x10,R(AB)x|x2或x10能力提升1(2016石家庄高一检测)若全集U1,2,3,4,5,6,M2,3,N1,4,则集合5,6等于()AMNBMNC(UM)(UN)D(UM)(UN)【解析】全集U1,2,3,4,5,6,M2,3,N1,4,MN1,2,3,4,则(UM)(UN)U(MN)5,6故选D.【答案】D2已知全集U1,2,3,4,5,集合Ax|x23x20,Bx|x2a,aA,则集合U(AB)中元素个数为()A1B2C3D4【解析】A1,2,B2,4,AB1,2,4,U(AB)3,5【答案】B3已知全集U2,3,a2a1,A2,3,若UA1,则实数a的值是_【解析】U2,3,a2a1,A2,3,UA1,a2a11,即a2a20,解得a1或a2.【答案】1或24(2016哈尔滨师大附中高一检测)设全集UR,集合Ax|x2或x5,Bx|x2求(1)U(AB);(2)记U(AB)D,Cx|2a3xa,且CDC,求a的取值范围. 【导学号:97030024】【解】(1)由题意知,Ax|x2或x5,Bx|x2,则ABx|x2或x5,又全集UR,U(AB)x|2x5(2)由(1)得Dx|2x5,由CDC得CD,当C时,有a2a3,解得a1;当C时,有解得a.综上,a的取值范围为(1,).
展开阅读全文