您好,欢迎访问三七文档
当前位置:首页 > 商业/管理/HR > 市场营销 > 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:13.242514.215.52716.6416.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1122331111nnnSabababab1111111146423341nn11111121244212481121nnnnn···································································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613A(,),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
本文标题:2019年3月厦门市高三质检数学(文)参考答案
链接地址:https://www.777doc.com/doc-3962551 .html