山东省青岛市2019届高三3月教学质量检测（一模）数学（理）试题

青岛市高三年级教学质量检测数学（理科）答案第1页（共6页）2019年青岛市高三年级教学质量检测数学（理科）参考答案及评分标准一、选择题：本大题共12小题．每小题5分，共60分．CAADCCBABBBD二、填空题：本大题共4小题，每小题5分，共20分．13．3514．415．(1465)16．3三、解答题：共70分．解答应写出文字说明，证明过程或演算步骤．第17题~21题为必考题，每个试题考生都必须作答．第22、23题为选考题，考生根据要求解答．（一）必考题：共60分．17.（本小题满分12分）解：（1）因为在数列{}na中，11nnnSSa，所以11nnnSSa，所以11nnaa，数列{}na是公差为1的等差数列又11a，1(1)1nann··································································3分设等比数列{}nb的公比为q，因为134bb，所以23111,42bqqb若12q，则21212bbbq，符合题意若12q，则21212bbbq，不符合题意所以1111()()222nnnb·············································································6分（2）因为11111(1)1nnaannnn所以1111111......1122311nWnnn········································8分又11(1())1221()(0,1)1212nnnT，所以11nT···········································11分所以1nnWT·····························································································12分青岛市高三年级教学质量检测数学（理科）答案第2页（共6页）18．（本小题满分12分）证明：（1）取BC的中点O，连结AO，DO因为5BDCD，所以DOBC，222DOCDOC又因为DO平面BCD，平面DBC平面ABCBC，平面BCD平面ABC所以DO平面ABC················2分因为AE平面ABC，所以//AEDO又因为2DOAE所以四边形AODE为平行四边形所以//EDAO因为三角形ABC为等边三角形，所以AOBC又因为AO平面ABC，平面BCD平面ABCBC，平面BCD平面ABC所以AO平面BCD，················································································5分所以ED平面BCD，又因为ED平面EBD所以平面EBD平面BCD；········································································6分（2）由（1）得，AO平面BCD，AODO，又DOBC，AOBC，分别以,,OBAOOD所在的直线为x轴，y轴，z轴，建立如图所示的空间直角坐标系Oxyz，则(0,3,0),(1,0,0),(0,0,2),(0,3,2)ABDE···············································7分设平面ABE的一个法向量为111,,)mxyz（，因为(1,3,0),(1,3,2)ABBE则1111130320mABxymBExyz，取13x，得(3,1,0)m设平面BED的一个法向量为222(,,)nxyz，因为(1,0,2),(1,3,2)BDBE则2222220320nBDxznBExyz，取22x，得(2,0,1)n所以15cos,5||||mnmnmn······························································11分设二面角AEBD的平面角为，则15cos5····································12分ABCDEOxyz青岛市高三年级教学质量检测数学（理科）答案第3页（共6页）19．（本小题满分12分）解：（1）由22221xyab，令xc得，2bya所以22||bPQa，22112||||22622APBQbSABPQaba四边形，23b······························3分又离心率12cea，222abc所以24a所以椭圆C的方程为：22143xy·································································5分（2）由题知，(1,0)F，直线l的方程为(1)ykx由22143(1)xyykx得，2222(43)84120kxkxk设11(,)Pxy，22(,)Qxy则2122843kxxk，212241243kxxk，121226()243kyykxxkk,设PQ的中点为N，则22243(,)4343kkNkk·····················································7分则MN的方程为222314()4343kkyxkkk令0y得，22(,0)43kMk所以223(1)||43kMFk···················································································9分又221212||1()4PQkxxxx2222222284(412)12(1)1()434343kkkkkkk··································11分所以||1||4MFPQ为定值·················································································12分青岛市高三年级教学质量检测数学（理科）答案第4页（共6页）20．（本小题满分12分）解：（1）由乙流水线样本的频率分布直方图可知，合格品的个数为100(10.04)96所以，22列联表是：甲流水线乙流水线总计合格品9296188不合格品8412总计100100200································································2分所以222()200(924968)1.4182.072()()()()10010018812nadbcKabacbdcd····4分所以，在犯错误的概率不超过0.15的前提下，不能认为产品的包装合格与两条自动包装流水线的选择有关····························································································5分（2）乙流水线的产品生产质量指标z服从正态分布2(200,12.2)N所以生产质量指标的数学期望为200，数值波动的标准差为12.2因为()0.6826Pz,(22)0.9544Pz所以(2)(0)(02)PzPzPz11()(2)22PzPz1(0.68260.9544)0.81852即:(20012.220012.22)(187.8224.4)0.8185PzPz所以质量指标落在(187.8,224.4)的概率为0.8185··············································9分（3）若以频率作概率，则从甲流水线任取一件产品是不合格品的概率0.08p设“任取两件产品，至少有一件合格品”为事件A则A为“任取两件产品，两件均为不合格品”，且22()0.080.0064PAp所以()1()10.00640.9936PAPA所以任取两件产品至少有一件为合格品的概率为0.9936·····································12分21．（本小题满分12分）解：（1）由题知:xaxxfln1)(······························································1分令()1ln,()(0)axgxxaxgxxx······················································2分所以，当0a时，0)(xxaxg，即)(xg在),0(上单调递减····················3分又因为0)1()1(gf，所以，当01x时，()0fx；当1x时，()0fx所以，()fx在(0,1)上单调递增，在(1,)上单调递减，所以0)1()(fxf所以)(xf只有一个零点·················································································5分青岛市高三年级教学质量检测数学（理科）答案第5页（共6页）（2）由（1）知：当0a时，()fx的极大值等于0，符合题意···························6分①当10a时，因为当(0,)xa时，()0gx；当(,)xa时，()0gx；且(1)0g，111()110aaageee故存在11(,)axea，满足111()0,(0,),()0;(,),()0;fxxxfxxxafx又(,1),()0;(1,),()0xafxxfx；所以，此时1x是)(xf的唯一极大值点，且0)1(f，符合题意.······················8分②当1a时，因为;0)(),,1(;0)(),1,0(xgxxgx且0)1(g，所以0)(xg，即)(xf在),0(上单调递减无极值点，不合题意························9分③当1a时，因为当(0,)xa时，()0gx；当(,)xa时，()0gx；且0)1(g，21)(aeegaa；令aeaaW1)(2，则2(1)()0aaWae；所以1)1()(WaW，所以aea21，即0)(aeg······································································10分又因为21aaae，故存在),(0aeax，满足0)(),,(;0)(),,(,0)(000xfexxxfxaxxfa，···································11分此时1x是)(xf的唯一极小值点，0xx是)(xf的唯一极大值点，0)1()(0fxf因此不合题意综上可得：1a·························································································12分（二）选考题：共10分．请考生在第22、23两题中任选一题作答．如果多做，则按所做的第一题记分．22．（本小题满分10分）选修44：坐标系与参数方程解：（1）因为曲线1C的参数方程为25cos15sinxy（其中为参数）所以1C的普通方程为22(2)(1)5xy①····················································2分在极坐标系中，将cossinxy代入①得24cos2sin0,化简得，1C的极坐标方程为4cos2sin②············································5分（2）因为直线l的极坐标方程为3(R)4，且直线l与曲线1C和曲线2C分别交于,MN，则可设1233(,),(,)44MN青岛市高三年级教学质量检测数学（理科）答案第6页（共6页）将13(,)4M代入②得133224cos2sin4()224422·······7分将23(,)4N代入曲线2:4sinC得2324sin42242················9分所以12|||||||2||22|32MN··············································10分23．（本小题满分10分）选修45：不等式选讲解:（1）当1a时，()|2||1|fxxxxx①当1x时，()(2)(1)30fxxxxxx，解得3x所以31x··························································································2分②当12x时，()(2)(1)10fxxxxxx，解得1x所以11x······························&