2019年3月厦门市高三质检数学(文)参考答案

1ODEFACBS厦门市2019届高中毕业班第一次质量检查参考答案文科数学一、选择题：1-5:ABDBD6-10:DBAAC11-12:AC12.设11(,)Bxy，22(,)Cxy，直线:(1)1(0)ABykxk，代入2xy得：210xkxk，所以11xk，则21(1)yk；同理：21xk，22(+1)yk，所以21201.2yyyk由题设知：220,44(2)0,kkkk得：0k且2k，所以01y且05y，选C.二、填空题：13．242514．215．52716．6416．解：因为//平面SBC，平面ABCl,平面SBC平面ABCBC,所以l//BC．取AB中点D，连接DE，DF，则DE//BC，所以l//DE．所以异面直线l和EF所成角即为DEF或其补角．取BC中点O，则SOBC，AOBC，又SOAOO，所以BC平面SOA，又SA平面SOA，所以BCSA，所以DEDF．在RtDEF中，3DE，5DF，所以22EF，36cos422DEF．所以异面直线l和EF所成角的余弦值为64．三、解答题：17．本题主要考查等差数列的基本量运算，考查裂项求和法；考查运算求解能力；考查分类与整合思想等。满分12分。解：（1）依题意12ba，因为16b，所以26a，······························································1分又2d，所以2262222naandnn.···················································2分因为321123nnbbbban，所以3121231nnbbbban，2n两式相减得：12nnnbaan，2n，················································································4分2BCADFEMHGBCADFE即2nbn，2n，·····················································································································5分所以6,12,2nnbnn······················································································································6分（2）1n时，1111114624Sab；··················································································7分2n时，1111112224141nnabnnnnnn···············································8分1122331111nnnSabababab1111111146423341nn11111121244212481121nnnnn···································································11分当1n时，1124S满足上式.综上，21121nnSn.···························································12分18．本题考查直线与平面、平面与平面的位置关系，几何体的体积以及平面几何的性质与计算等基础知识；考查空间想象能力、推理论证能力、运算求解能力；考查数形结合思想、化归与转化思想。满分12分。解：（1）取AB，DE中点G，H，连接CG，FH，HG，则平行四边形CFHG即为所求截面.··························································································································································2分理由如下：因为AD，BE，CF均垂直于平面ABC，所以AD//BE//CF.因为2AD，4BE，所以ABED为梯形.又G，H分别为AB，DE中点，所以HG//BE，=3HG，所以HG//CF，HGCF，所以CFHG为平行四边形，········································································································4分因为ACBC，G为AB中点，所以CGAB.又AD平面ABC，CG平面ABC，所以ADCG.又ABADA，所以CG平面ABED.又CG平面CFHG，所以平面CFHG平面ABED，所以平行四边形CFHG即为所作的截面.···················································································6分（2）法一：过点A作AMBC于点M，因为BE平面ABC，AM平面ABC，所以BEAM，又BCBEB，BC，BE平面BCFE，所以AM平面BCFE.····························7分在ABC中，ACBC，120ACB，43AB，得=4ACBC，所以144sin120432ABCS.3A'B'ACBC'DFEPQBCADFE因为3=sin604232AMAC，·················································································9分所以111283[(34)4]233323DBCFEBCFEVSAM，1183432333DABCABCVSAD.所以8328312333ABCDEFDABCDBCFEVVV.······················································12分法二：将多面体ABCDEF补成直三棱柱ABCABC，其中4AD，2BE，3CF，6AA，则12ABCDEFABCABCVV.······················8分在ABC中，ACBC，120ACB，43AB，得=4ACBC，所以144sin120432ABCS.·····················································10分所以436243ABCABCABCVSAA，所以123ABCDEFV.·····························12分法三：在多面体ABCDEF中作直三棱柱ABCDPQ，则ABCDEFABCDPQDEFQPVVV.·································································································7分在ABC中，ACBC，120ACB，43AB，得=4ACBC，所以144sin120432ABCS.设BC边上的高为AM，则3=sin604232AMAC.因为BE平面ABC，AM平面ABC，所以BEAM又BCBEB，BC，BE平面BCFE，所以AM平面BCFE.····························9分所以43283ABCDPQABCVSAD.111[(12)4]2343332DEFQPEFQPVSAM.所以8343123ABCDEFABCDPQDEFQPVVV.······················································12分19．本小题主要考查散点图、回归直线、函数最值等基础知识；考查数据处理能力，运算求解能力；考查统计概率思想。满分12分。解：（1）由散点图知，选择回归类型dycx更适合.·····················································2分（2）对dycx两边取对数，得lnlnlnycdx，即lnvcdu.·························4分由表中数据得，^122130.5101.51.51==46.5-101.51.53niiiniiuvnuvdunu.4所以^1ln-=1.5-1.51,3cvdu所以^ce.所以年研发费用x与年销售量y的回归方程为13yex.····················································8分（3）由（2）知，1327zxxx（)，求导得2'391zxx（），令2'391=0zx，得27x，·····························································································10分函数1327zxx在027（，）上单调递增，在27+（，）上单调递减，所以当27x时，年利润z取最大值5.4亿元.答：要使得年利润取最大值.预计下一年度投入2.7亿元.·························································12分20．本题考查直线方程、直线平行的判定与直线与圆锥曲线的位置关系等知识；考查运算求解能力等；考查数形结合思想、化归与转化思想等。满分12分。解：（1）设点11(,)Axy，当AB垂直于x轴时，可得：11x，则163y，····················1分所以613A（，），所以63ADk，·························································································2分所以直线AD的方程为：6(1)3yx···················································································4分（2）法一：设11(,)Axy，22(,)Bxy，①当直线AB的斜率不存在时，即1x，得：63BEyy或63BEyy，所以//CDBE·································································································································5分②当直线AB的斜率存在时，设:(1)(0)ABykxk代入椭圆，得：2222(13)6330,kxkxk所以