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《2014届高考数学一轮复习第1章第2讲集合的基本运算课件理.ppt》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库。
1、第一章集合、常用逻辑用语Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.集合的基本运算Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.1.如图所示,U是全集,A和B是U的子集,则阴影部分所表示的集合是__________
2、.(∁UA)∩B{0,1,2,4}Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.3.(2012·广东肇庆期末卷)已知集合M={x
3、x2-2x-3=0},N={x
4、-25、x2-2x-3=0}={-1,3},所以M∩N={-1,3}.{-1,3}Evaluationonly.CreatedwithAspose.Sli6、desfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.4.设集合A={5,log2(a+3)},B={a,b},若A∩B={2},则A∪B=_______.解析:由题意知log2(a+3)=2,解得a=1,所以b=2,所以A∪B={1,2,5}.5.满足{1,3}∪A={1,3,5,7}的集合A有______个.{1,2,5}4Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfil7、e5.2.0.0.Copyright2004-2011AsposePtyLtd.集合的运算Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.集合的运算与不等式联系时,可借助数8、轴将问题形象化.此题应注意点评Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.【变式练习1】已知集合A={a2,a+1,-3},B={a-3,a-2,a2+1},若A∩B={-3},求A∪B.【解析】由A∩B={-3}得-3∈B,又a2+1≥1,故只有a-3,a-2可能等于-3.(1)当a-3=-3时,a=0,此时A={0,1,-3},B={-3,-2,1},A∩B≠{-9、3},故a=0舍去;(2)当a-2=-3时,a=-1,此时A={1,0,-3},B={-4,-3,2},满足A∩B={-3}.从而A∪B={-4,-3,0,1,2}.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.集合与简单不等式的综合应用Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.010、.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.点评本题所给的集合都是确定的数集,重点是考查集合基本运算掌握的熟练程度,主要方法是:首先化简集合,即将集合化简到可以用数轴能直观感知的数集,然后在数轴上描绘出集合元素的取值范围(或用Venn图),再根据集合交、并、补的意义求出所要求的集合,最后的结果用区间表11、示即可.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3
5、x2-2x-3=0}={-1,3},所以M∩N={-1,3}.{-1,3}Evaluationonly.CreatedwithAspose.Sli
6、desfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.4.设集合A={5,log2(a+3)},B={a,b},若A∩B={2},则A∪B=_______.解析:由题意知log2(a+3)=2,解得a=1,所以b=2,所以A∪B={1,2,5}.5.满足{1,3}∪A={1,3,5,7}的集合A有______个.{1,2,5}4Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfil
7、e5.2.0.0.Copyright2004-2011AsposePtyLtd.集合的运算Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.集合的运算与不等式联系时,可借助数
8、轴将问题形象化.此题应注意点评Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.【变式练习1】已知集合A={a2,a+1,-3},B={a-3,a-2,a2+1},若A∩B={-3},求A∪B.【解析】由A∩B={-3}得-3∈B,又a2+1≥1,故只有a-3,a-2可能等于-3.(1)当a-3=-3时,a=0,此时A={0,1,-3},B={-3,-2,1},A∩B≠{-
9、3},故a=0舍去;(2)当a-2=-3时,a=-1,此时A={1,0,-3},B={-4,-3,2},满足A∩B={-3}.从而A∪B={-4,-3,0,1,2}.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.集合与简单不等式的综合应用Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0
10、.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.点评本题所给的集合都是确定的数集,重点是考查集合基本运算掌握的熟练程度,主要方法是:首先化简集合,即将集合化简到可以用数轴能直观感知的数集,然后在数轴上描绘出集合元素的取值范围(或用Venn图),再根据集合交、并、补的意义求出所要求的集合,最后的结果用区间表
11、示即可.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3
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