高二理科数学下册期末考试4

学而思教育·学习改变命运思考成就未来！高考网学而思教育·学习改变命运思考成就未来！高考网高二理科数学下册期末考试高二数学试题（理科）（总分：150分考试时间：120分钟）一、选择题：本大题共10小题，每小题5分，共50分．在每小题给出的四个备选项中，只有一项是符合题目要求的．1．在(23)nx的展开式中，各项系数和为（）A．1B．2C．–1D．–1或12．复数数列{}na满足1ai，21nnaai（2n，i为虚数单位），则5a()A．0B．iC．iD．1i3．某班从6名学生中选出4人分别参加数、理、化、生四科竞赛且每科只有1人，其中甲、乙两人不能参加生物竞赛．则不同的选派方法共有（）A．96种B．180种C．240种D．280种4．某单位有职工160人，其中有业务人员120人，管理人员24人，后勤服务人员16人．为了解职工的某种情况，要从中抽取一个容量为20的样本，将这次调查记做①；从某中学高二年级的18名体育特长生中选出3人调查学习负担情况，将这次调查记做②．那么完成上述二项调查应采用的抽样方法是（）A．①用随机抽样法，②用系统抽样法B．①用分层抽样法，②用随机抽样法C．①用系统抽样法，②用分层抽样法D．①用分层抽样法，②用系统抽样法5．正方体1111DCBAABCD中，A1B与D1B1所成的角是（）A．60B．30C．45D．以上都不对6．某校决定面向社会招聘3名微机专业技术人才．两个好朋友一起去应聘．该校主管人事的领导通知他们面试时间的时候透露：“你们二人同时被招聘的概率为170”．据此推断面试总人数为（）A．70个B．21个C．42个D．35个7．已知A，B是球O上两点，若∠AOB=4，且A、B的球面距离为24，则球的体积为（）A．83B．823C．82D．88．袋中编号为1，2，3，4，5的五只小球，从中任取3只球，以表示取出的球的最大号A1B1C1D1DCBA学而思教育·学习改变命运思考成就未来！高考网学而思教育·学习改变命运思考成就未来！高考网码，则E的值是（）A．5B．4.75C．4.5D．49．设连续掷两次骰子得到的点数分别为m、n，则直线myxn与圆22(3)1xy相离的概率是（）A．1136B．2136C．3136D．413610．正方体ABCD—A1B1C1D1的棱长为1，点M在棱AB上，且点A与点M不重合，点P是平面ABCD内的动点，且点P到A1D1的距离与点P到点M的距离的平方差为1，则点P的轨迹是（）A．圆B．双曲线C．直线D．抛物线二、填空题：本大题共5小题，每小题5分，共25分．11．设～N(0，1)，则(0)P_________________．12．6人排成一排照像，其中甲、乙两人中间恰有一人的排法总数是．13．24(1)(12)xx的展开式中含2x的系数为_________________．14．如右图，它满足①第n行首尾两数均为n②表中的递推关系如杨辉三角，则第n行(n≥2)的第二个数是．15．如图，在多面体ABCDEF中，EF=2，且EF∥面ABCD，其余棱长均为1，则BF与平面CDEF所成的角的正切值为_______________．三、解答题：本大题共6小题，共75分，解答应写出文字说明、证明过程或演算步骤．16．(13分)某学生骑自行车上学途中要经过4个交叉路口，在各交叉路口遇到红灯的概率是14（各交叉口遇到红灯的事件相互独立）．(1)求这名学生在上学途中3次遇到红灯的概率；(2)求这名学生在途中最多遇到1次红灯的概率．图-16FEDCBA学而思教育·学习改变命运思考成就未来！高考网学而思教育·学习改变命运思考成就未来！高考网．(13分)数学研究性学习小组共13个人，其中男同学8人，女同学5人．(1)从这13人中选出正、副组长各1人，有多少种选法？(2)从这13人中选出3人准备作报告，在选出的3个人里至少要有一名女同学，一共有多少种不同的选法？18．(13分)在二项式322(3)nxx的展开式中，各项的系数和比各项的二项式系数和大992，试求该二项式展开式中系数最大的项．19．(12分)如图，在直三棱柱ABC—A1B1C1中，90ACB，AC=BC=CC1=2．(1)求证：AB1BC1；(2)求点B到平面AB1C1的距离；(3)求二面角C1—AB1—A1的大小．20．(12分)某社区文化站举行一次象棋比赛，经优胜劣汰，最后由甲、乙二人决赛．根据他们过去比赛的情况统计知，单局比赛甲胜乙的概率为0.6．本次比赛采用五局三胜制，即先胜三局者获胜．设各局比赛相互间没有影响．求：(1)前三局甲领先的概率；(2)本场比赛乙以3∶2取胜的概率；(3)令为本场比赛的局数，求的分布列和数学期望．21．(12分)等比数列{an}的前n项和为Sn，已知对任意的*nN，点（n，Sn）均在函数BC1A1B1CA学而思教育·学习改变命运思考成就未来！高考网学而思教育·学习改变命运思考成就未来！高考网(01xybrbb，且，b、r均为常数)的图象上．(1)求r的值；(2)当b=2时，记22(log1)nnba．证明：对任意*nN，不等式12121111nnbbbnbbb成立．（命题人：郑莹莹审题人：周先凤）学而思教育·学习改变命运思考成就未来！高考网学而思教育·学习改变命运思考成就未来！高考网—2009学年度下期期末考试高二数学试题参考答案(理科)一、选择题：本大题共10小题，每题5分，共50分．1．D2．C3．C4．B5．A6．B7．B8．C9．C10．D二、填空题：本大题共5小题，每题5分，共25分．11．1212．19213．2514．222nn15．2三、解答题：本题共6小题，共75分．16．解：(1)334133()4464C····································································································6分(2)1344133189()()444256C······················································································13分17．解：(1)213156A···················································································································6分(2)33138230CC········································································································13分18．解：令x=1，得系数和为(13)4nn·················································································2分二项式系数和为2n·········································································································4分由题42992nn∴n=5····························································································································5分设第r+1项即104325231555()(3)33rrrrrrrrrTCxxCxC系数最大∴1155115531336133351rrrrrrrrCCrrCCrr······································································10分∴141844r∴r=4····························································································································12分∴最大项2635405Tx·······························································································13分19．(1)证明：∵11BCCB，1ACCB∴11CBACB面∴11CBAB·······································································································3分(2)解：由体积法1111ABBCBACBVV·······················································································4分1122BBCSAC，，111123222ABACCB，，，1122ACBS···········6分学而思教育·学习改变命运思考成就未来！高考网学而思教育·学习改变命运思考成就未来！高考网∴2h即B到11ABC距离为2········································································7分(3)过C1作C1D⊥A1B1于D，则C1D⊥面A1B，过D作DE⊥AB1于E，连结C1E，则∠C1ED即为所求二面角的平面角·································9分易知1623CDDE，················································10分在Rt△C1DE中11tan3CDCEDDE∴160CED故所求二面角平面角为60·····························································································12分20．(1)设“前三局甲胜三局”为事件A，“甲胜两局”为事件B，“前三局甲领先”为事件C，则C=A+B3()(0.6)0.216PA·········································································································1分223()(0.6)0.40.432PBC·························································································2分()()()0.648PCPAPB·························