2021届福建省泉州市高三质量监测一模数学试题及答案

泉州市2021届高中毕业班质量检测（三）数学 试题 中考模拟试题doc幼小衔接 数学试题 下载云南高中历年会考数学试题下载N4真题下载党史题库下载 第1页（共2页）泉州市2021届高中毕业班质量监测（三）参考 答案 八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案 与评析高三数学填空题部分三、填空题：本题共4小题，每小题5分，共20分。13．6(2)x展开式中，二项式系数最大的项的系数为．（用数字填写答案）【命题意图】本小题主要考查二项式定理等基础知识，考查运算求解能力，体现基础性，导向对发展数学运算等核心素养的关注．【试题简析】依题意，二项式系数最大项为3333462160TCxx，其系数为160．故答案为160．14．甲问乙：“您有几个孩子”，乙说：“四个”．此时，一男孩过来．乙对甲说：“这是我小孩”，接着乙对该男孩说：“去把哥哥姐姐都叫过来，你们四人一起跟甲去趟学校”．根据上述信息，结合正确的推理，最多需要猜测次，才可以推断乙的四个小孩从长到幼的正确性别情况；第3次才猜对的概率为．（第一空2分；第二空3分）【命题意图】本小题主要考查古典概型等基础知识；考查阅读理解并提取信息进行推理论证的能力；体现基础性、创新性、应用性，导向对发展理性思维与数学应用等核心素养的关注．【试题简析】记iA为乙的第i个孩子是男性，依题意，四个孩子从长到幼的性别情况有1234(,,,)AAAA，1234(,,,)AAAA，1234(,,,)AAAA，1234(,,,)AAAA，1234(,,,)AAAA，1234(,,,)AAAA，共6种，最多需要猜测5次，便可以知道乙的四个小孩从长到幼的正确性别情况；第3次就猜对的概率为16．故答案为5；16．15．圆锥曲线光学性质（如图1所示）在建筑、通讯、精密仪器制造等领域有着广泛的应用．如图2，一个光学装置由有公共焦点12,FF的椭圆C与双曲线C构成，一光线从左焦点1F发出，依次经过C与C的反射，又回到点1F历时m秒；若将装置中的C去掉，则该光线从点1F发出，经过C两次反射后又回到点1F历时n秒．若C与C的离心率之比为13，则mn．保密★使用前泉州市2021届高中毕业班质量检测（三）数学试题第2页（共2页）图1图2【命题意图】本题考查椭圆定义、双曲线定义、离心率等基础知识；考查推理论证、运算求解等能力；考查数形结合、化归与转化等思想；体现综合性、创新性，导向对发展数学运算、逻辑推理、直观想象等核心素养的关注．【试题简析】设椭圆的长半轴长为1a，双曲线的实半轴长为2a，在图2左图中，由椭圆定义可得1212BFBFa……①由双曲线定义可得2122AFAFa……②①-②得111222AFABBFaa，所以1ABF的周长为1222aa．在图2右图中，光线从椭圆的一个焦点发出，被椭圆反射后经过椭圆的另一个焦点，即直线ED过点2F，所以1EDF的周长为14a，又因为椭圆与双曲线焦点相同，离心率之比为13，所以123aa，又两次所用时间分别为,mn，而光线速度相同，所以122212226214123aaaamnaa．16．若正数,xy满足216xyxy，则xy的最小值为_________．【命题意图】本小题主要考查不等式、函数与导数等基础知识；考查逻辑推理、运算求解等能力；考查函数与方程、化归与转化等思想；导向对发展逻辑推理、数学运算等核心素养的关注．【试题简析】令txy，则2216tyytyyty所以2216tyy，令216()fyyy，由216'()20fyyy，解得2y.0,2y时，'()0fy，()fy单调递减，2,y时，'()0fy，()fy单调递增；所以()fy的最小值为(2)12f，又对正数,xy有0txy，所以min23t．泉州市2021届高中毕业班质量检测（三）数学试题第1页（共5页）泉州市2021届高中毕业班质量监测（三）参考答案与评析高三数学解答题17-19题部分四、解答题：本题共6小题，共70分。解答应写出文字说明，证明过程或演算步骤。17．（10分）已知数列{}na，{}nb满足19a，1109nnaa，1nnba．（1）证明：{}nb是等比数列；（2）求数列{(1)lg}nnb的前n项和nS．【命题意图】本小题主要考查等比数列的定义与前n项和等基础知识；考查运算求解能力；考查函数与方程思想、化归与转化思想、分类与整合思想等．体现基础性和综合性，导向对发展数学运算等核心素养的关注．【试题解析】解法一：（1）依题意，1111nnnnbaba··············································································1分109110101011nnnnaaaa（非零常数）．·····························2分又11110ba．故{}nb为首项110b，公比10q的等比数列．··················3分（2）（1）由可知1110nnnbbq．···································································4分所以(1)lg(1)lg10(1)nnnnnncbn．··············································5分①当n为偶数时，(12)(34)[(1)]2nnSnn；··················7分②当n为奇数时，1(12)(34)[(2)(1)]2nnSnnn．9分故,21,.2nnnSnn为偶数，为奇数·························································（结论分）10分解法二：（1）同解法一；（2）由（1）可知1110nnnbbq．············································（通项公式分）4分所以(1)lg(1)lg10(1)nnnnnncbn．·······························（运算分）5分保密★使用前泉州市2021届高中毕业班质量检测（三）数学试题第2页（共5页）211(1)2(1)(1)(1)(1)nnnSnn①231(1)1(1)2(1)(1)(1)(1)nnnSnn②·········（方法分）6分①-②得212(1)(1)(1)(1)nnnSn······························（方法分）7分1(1)[1(1)](1)1(1)nnn····················（等比数列求和公式分）9分111(1)(1)11(1)()(1)222nnnnn故111()(1)442nnnS．···················································（结论分）10分18．（12分）脱贫攻坚取得的全面胜利是中国共产党领导全国人民创造的又一个彪炳史册的人间奇迹．某地区有一贫困村坐落于半山平台，村民通过悬崖峭壁间的藤条结成的“藤梯”往返村子，因而被称为“悬崖村”．当地政府把“藤梯”改成钢梯，使之成为村民的“脱贫天梯”，实现了“村民搬下来，旅游搬上去”，做到了长效脱贫．如图，为得到峭壁上的,AB两点的距离，钢梯的 设计 领导形象设计圆作业设计ao工艺污水处理厂设计附属工程施工组织设计清扫机器人结构设计 团队在崖底的,PQ两点处分别测得1APQ，1BPQ，APB，2AQP，2BQP，且PQs．（1）用12,,s 表 关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf 示AP；（2）已知117，2150，90.0s米，51.3，又经计算得250.0AP米，求AB．参考数据：sin130.225，cos130.974，sin51.30.780，cos51.30.625．【命题意图】本小题以“悬崖村”的脱贫事件为背景，以修建钢梯的测量为问题情境，考查正弦定理、余弦定理，解三角形等基础知识；考查抽象概括能力，空间想象能力，运算求解能力与应用意识和创新意识；考查转化与化归思想，函数与方程思想；考查基本活动经验；导向对数学抽象，数学建模，数学运算核心素养的关注．泉州市2021届高中毕业班质量检测（三）数学试题第3页（共5页）【试题解析】（1）如图，在APQ△中，根据正弦定理得212sinsinπAPPQ，····················3分化简得212sinsinsAP；·······································································5分（2）在BPQ△中，根据正弦定理得212sinsinπBPPQ，····························6分可得212sin900.5200sin1800.225sBP，·······································7分又在ABP△中，根据余弦定理得2222cosABAPBPAPBP············10分代入得2400006250022002500.62540000AB，所以200AB米．·············································································12分19．（12分）永春老醋以其色泽鲜艳、浓香醇厚的独特风味，与山西陈醋、镇江香醋、保宁药醋并称中国四大名醋．为提高效率、改进品质，某永春老醋生产公司于2018年组织技术团队进行发酵工艺改良的项目研究．2020年底，技术团队进行阶段试验成果检验，为下阶段的试验提供数据参考．现从改良前、后两种发酵工艺生产的成品醋中，各随机抽取100件进行指标值M的检测，检测分两个步骤，先检测是否合格，若合格，再进一步检测是否为一等品．因检测设备问题，改良后的成品醋有20件只进行第一步检测且均为合格，已完成检测的180件成品醋的最终结果如下表所示．指标区间[2,1)[1,0)[0,1)[1,2)[2,3)[3,4)来源改良前改良后改良前改良后改良前改良后改良前改良后改良前改良后改良前改良后个数315230263134241572附：成品醋的品质采用指标值M进行评价，评价标准如下表所示．泉州市2021届高中毕业班质量检测（三）数学试题第4页（共5页）[0,1)M[1,3)M[0,3)M一等品二等品三等品合格不合格（1）现从样本的不合格品中随机抽取2件，记来自改良后的不合格品件数为X，求X的分布列；（2）根据以往的数据，每销售一件成品醋的利润y（单位：元）与指标值M的关系为5,[0,1),3,[1,3),2,[0,3).MyMM若欲实现“改良后成品醋利润比改良前至少增长20%”，则20件还未进一步检测的样本中，至少需要几件一等品？【命题意图】本小题主要考查条件概率、独立性检验、数学期望等基础知识；考查数据处理能力、应用意识和创新意识等；考查统计与概率思想；导向对发展逻辑推理、数学运算、数学建模、数据 分析 定性数据统计分析pdf销售业绩分析模板建筑结构震害分析销售进度分析表京东商城竞争战略分析 等核心素养的关注．【试题解析】（1）依题意，已检测的不合格品样本共有20个，其中改良前的有15个，改良后的有5个．······················································2分0,1,2X．·························································································3分2015522021(0)38CCPXC；····································································4分1115522015(1)38CCPXC；·····································································5分021552201(2)19CCPXC．·····································································6分故X的分布列为：X012P21381538119（2）由样本估计总体的思想，···········································································7分改良前成品醋利润的数学期望30551553(2)2.85100100100；··············8分若要使“改良后成品醋利润比改良前至少增长20%”，则改良后的利润至少应为2.85(120%)3.42.·····································9分假设改良后20个还未进行进一步检测的样本中，一等品有x个，泉州市2021届高中毕业班质量检测（三）数学试题第5页（共5页）则，改良后的一等品有26x个，二等品有69x个.改良后成品醋利润的数学期望2669553(2)100100100xx．···············10分依题意，2669553(2)3.42100100100xx≥．·································11分求得7.5x≥，又xN，故20个还未进行进一步检测的样本中，一等品至少需要8个．·······12分泉州市2021届高中毕业班质量检测（三）数学试题第1页（共11页）泉州市2021届高中毕业班质量监测（三）参考答案与评析高三数学解答第20至22题部分20．（12分）如图，在四棱锥PABCD中，二面角PADC是直二面角，AD为等腰直角三角形PAD的斜边，2ADCD，1ABBC，5BD，M为线段PC上的动点．（1）当PMMC时，证明：PA//平面MBD；（2）若平面MBD平面ABCD，求二面角BMDC的余弦值．【命题意图】本题考查空间几何点线面位置关系、线面垂直的性质和判定、面面垂直的判定、点面距离的求法等基础知识；考查空间想象能力、推理论证能力、运算求解能力；考查化归与转化的思想；考查直观想象、逻辑推理和数学运算等核心素养．【试题解析】（1）连结AC交BD于N，连结MN，········································································1分因为2ADCD，1ABBC，所以BD为AC的垂直平分线，则ANCN，························································2分又因为PMMC，所以MN为PAC的中位线，则PA//MN，······························3分又因为PA平面MBD，MN平面MBD，......4分所以PA//平面MBD．·······················································································4分保密★使用前泉州市2021届高中毕业班质量检测（三）数学试题第2页（共11页）（2）解法一：取AD的中点O，因为平面PAD平面ABCD，所以OP平面ABCD，··············································································5分过O作AD的垂线作为x轴，分别以,ODOP所在的直线为y轴，z轴建立如图空间直角坐标系，则(0,0,1)P，(0,1,0)A，(0,1,0)D，···························································6分由已知22214ABADBD，得ABAD，故(1,1,0)B，························7分假设(,,0)Cxy，因为(,1,0)ACxy，(1,2,0)BD，(1,1,0)BCxy，(,1,0)DCxy，由00ACBDBCDC，得2222010xyxxy，解得81(,,0)55C，···························8分8186,,1,(,,0)5555PCDC，···························································9分假设平面MCD的一个法向量是1111,,,xyzn则1100PCDCnn，即11111850430xyzxy，令1113,4,4,xyz取13,4,4,n··························10分因为平面MBD平面ABCD，且ACBD，所以平面MBD的一个法向量是84,,0,55AC取22,1,0,n························11分假设二面角BMDC的平面角为，因为二面角BMDC的平面角为锐角，泉州市2021届高中毕业班质量检测（三）数学试题第3页（共11页）则二面角BMDC的余弦值1212102205cos41415nnnn．........12分解法二：（1）同解法一；（2）因为22214ABADBD，得ABAD，又因为平面PAD平面ABCD，············································································5分故以点A为原点，AB，AD所在的直线为x轴，y轴，在平面PAD内过点A作AD的垂线为z轴建立如图空间直角坐标系，则0,0,0A，1,0,0B，0,2,0D，0,1,1P，····················································6分84,,055C，······································································································7分于是84,,055AC，0,1,1PD，86,,055DC，·······································8分假设平面MCD的一个法向量是1111,,,xyzn则1100PDDCnn，即11110430yzxy，令1113,4,4,xyz取13,4,4,n·········································10分因为平面MBD平面ABCD，且ACBD，所以平面MBD的一个法向量是84,,0,55AC取22,1,0,n·································11分假设二面角BMDC的平面角为，则1212102205cos41415nnnn．因为二面角BMDC的平面角为锐角，所以其大小的余弦值220541．·····················12分泉州市2021届高中毕业班质量检测（三）数学试题第4页（共11页）解法三：（1）同解法一；（2）以PA所在直线为x轴，PD所在的直线为y轴，过P作AB平行线为z轴，································································································································5分则2,0,0A，0,2,0D，0,0,0P，2,0,1B，··········································6分32228,,555C，·····························································································7分于是22228,,555AC，0,2,0PD，32328,,555PC，···················8分假设平面MCD的一个法向量是1111,,,xyzn则1100PDPCnn，即1111033420yxyz，令11142,0,3,xyz取142,0,3,n·················10分因为平面MBD平面ABCD，且ACBD，所以平面MBD的一个法向量是22228,,555AC，取21,1,22,n·············11分假设二面角BMDC的平面角为，则121242622205cos414110nnnn．因为二面角BMDC的平面角为锐角，所以其大小的余弦值220541．·····················12分解法四：（1）同解法一；（2）取AD的中点O，因为平面PAD平面ABCD，所以OP平面ABCD，过O作AD的垂线作为x轴，分别以,ODOP所在的直线为y轴，z轴建立如图空间直角坐标系，····5分由已知22214ABADBD，得ABAD，故(1,1,0)B，泉州市2021届高中毕业班质量检测（三）数学试题第5页（共11页）又(0,0,1)P，(0,1,0)A，(0,1,0)D，(1,2,0)BD，······································6分假设00(,,0)Cxy，因为00(,1,0)ACxy，00(1,1,0)BCxy，00(,1,0)DCxy，由00ACBDBCDC，得002200022010xyxxy，解得81(,,0)55C，8186,,1,(,,0)5555PCDC，······························································7分假设,,Mxyz，,,1PMxyz，设81,,55PMPC，解得81,,155M，81,1,155DM，因为平面MBD平面ABCD，且ACBD，则ACDM，又因为84,,055AC，所以8481604,,0,1,105555255ACDM，解得13，于是812,,15153M，则8162,,15153DM，······························8分假设平面MCD的一个法向量是1111,,,xyzn则1100DCDMnn，即111114304850xyxyz，令1113,4,4,xyz取13,4,4,n···························10分依题意可取平面MBD的一个法向量为84,,0,55AC取22,1,0,n··················11分假设二面角BMDC的平面角为，则1212102205cos41415nnnn．因为二面角BMDC的平面角为锐角，所以其大小的余弦值220541．．··············12分21．（12分）已知椭圆22:143xyC的左、右顶点分别为,AB，右焦点为F，折线1(0)xmym与C交于,MN两点．泉州市2021届高中毕业班质量检测（三）数学试题第6页（共11页）（1）当2m时，求MFNF的值；（2）直线AM与BN交于点P，证明：点P在定直线上．【命题意图】本题主要考查直线与椭圆的位置关系、弦长计算、两直线的位置关系等基础知识；考查运算求解能力、应用意识和创新意识等；考查数形结合、化归与转化等思想；体现综合性、创新性，导向对发展逻辑推理、数学运算、直观想象、数学抽象等核心素养的关注．【试题解析】解法一：（1）由已知可得(1,0)F，设点M关于x轴的对称点为1M，则1MFMF，···························································································1分如图，不妨设直线21xy与椭圆相交于1,MN两点，设11122(,),(,)MxyNxy，联立2221143xyxy，可得223(21)4120yy，即2161290yy，················3分所以121239,416yyyy，··········································································4分故22111212()()MFNFMFNFMNxxyy22121291512()455164yyyy．································5分（2）由已知可得(2,0),(2,0)AB，111(,)Mxy，1122(,),(,)MxyNxy，不妨设直线1myx与椭圆相交于点NM,1，联立221143xmyxy，可得223(1)4120myy，即22(34)690mymy，······6分泉州市2021届高中毕业班质量检测（三）数学试题第7页（共11页）所以12122269,3434myyyymm，·······························································7分且)(232121yyymy.···················································································8分直线11:(2)2yAMyxx，直线22:(2)2yBNyxx，···············································································9分联立两直线方程，消去y可得)1()3()2()2(2221122112myymyyxyxyxx，·························10分即33)(233)(233322121221112221yyyyyyyymyyymyxx，·············································11分所以23(2)xx，1x，即点P在定直线1x上．··················································································12分解法二：（1）同上.（2）由已知可得(2,0),(2,0)AB，111(,)Mxy，1122(,),(,)MxyNxy，不妨设直线1myx与椭圆相交于点NM,1，联立221143xmyxy，可得223(1)4120myy，即22(34)690mymy，···6分所以12122269,3434myyyymm，····························································7分直线11:(2)2yAMyxx，泉州市2021届高中毕业班质量检测（三）数学试题第8页（共11页）直线22:(2)2yBNyxx，···········································································8分由点M在椭圆上，可知22113(4)4yx，所以11112324yxxy，所以直线1123:(2)4xAMyxy，··································································9分联立两直线方程，消去y可得121223(2)(2)42xyxxyx，即12123242(2)(2)yyxxxx，········································································10分即1212122121212123242(2)(2)(1)(1)()1yyyyyyxxxxmymymyymyy，所以22232994296344xxmmm，···················································11分所以23(2)xx，1x，即点P在定直线1x上．···············································································12分解法三：（1）同上．（2）由已知可得(2,0),(2,0)AB，111(,)Mxy，1122(,),(,)MxyNxy，不妨设直线1myx与椭圆相交于点NM,1，联立221143xmyxy，可得223(1)4120myy，即22(34)690mymy，······6分所以12122269,3434myyyymm，·······························································7分设直线1,AMAN的斜率分别为12,kk，则泉州市2021届高中毕业班质量检测（三）数学试题第9页（共11页）12121212122121212121222(2)(2)(3)(3)3()9yyyyyyyykkxxxxmymymyymyy2229191827364mmm，······························································8分所以直线,AMAN的斜率满足1214kk，又设直线BN的斜率为3k，则222223222232244yyykkxxx···································································9分所以直线,AMBN的斜率满足1313kk····························································10分故直线331:(2),:(2)3AMykxBNykx·················································11分联立解得1x，即点P在定直线1x上．···············································································12分22．（12分）已知函数()esinxfxaxx．（1）若()fx在(0,2π)单调递减，求实数a的取值范围；（2）证明：对任意整数a，()fx至多1个零点．【命题意图】本小题主要考查函数与方程，不等式，导数的应用等基础知识，考查逻辑推理，运算求解能力，体现综合性，导向对发展数学抽象、数学运算、逻辑推理、直观想象等核心素养的关注．【试题解析】解法一：（1）()ecos1xfxax.····························································1分【当0a≥时，()0fx≤显然成立.泉州市2021届高中毕业班质量检测（三）数学试题第10页（共11页）········································（无持续求解，只写这一结论，可补这1分.)2分】()fx在(0,2π)单调递减对(0,2π)x，恒有()0fx≤(0,2π)x，恒有ecos1xax≥，·····················2分令()ecos1xgxx（[0,2π]x），························································3分则π()ecos1sine2sin14xxgxxxx，令()0gx，解得3π0,22xxx（或或．············································4分则当3π0,2x时，()0gx，()gx单调递减；当3π,2π2x时，0gx，()gx单调递增.·························5分又(0)(2π)0gg，所以当[0,2π]x时，max()0gx.所以0a≥.·························································································6分（2）令()sinxxx，则()cos10xx≤，所以()x单调递减，又因为(0)0，所以0x≥时，sin0xx≤；0x时，sin0xx.·······························7分令()e(sin)xFxaxx，则()Fx与()fx零点一致.······························8分当0x≥时，()e(sin)cos10xFxxxx≤，所以()Fx递减，()(0)FxFa≤.·······························9分当0x时，有e(sin)e(1)xxaaxxax≤，令()e(1)0xGxaxx，因为()e0xGxx，()(,0)Gx在递增，所以0()(0)e(10)1GxGaa.······························10分故()1aFxa.综上，当0a≥时，()Fx在0x≥有唯一零点，在0x恒正不存在零点；··········11分当1a≤时，()10Fxa≤，不存在零点.·························12分泉州市2021届高中毕业班质量检测（三）数学试题第11页（共11页）即对任意整数a，()Fx至多1个零点，所以()fx至多1个零点．解法二：（1）同解法一（2）当0a≥时，()ecos10xfxax≤恒成立，()fx为递减函数，所以()fx至多1个零点.···························7分令()sinxxx，则()cos10xx，所以()x单调递减，又因为(0)0则0x≥时，sin0xx≤；0x时，sin0xx.·························8分当1a≤时，()esinesinxxfxaxxxx≤.························9分令()esinxxxx．当0x≥时，()esinsin0xxxxxx≤.·······················10分当0x时，()ecos1xxx；()esin1sin0xxxx≤≤,所以()x在0x时单调递减，此时()(0)1x，所以()x在0x时单调递增，所以()(0)=1x≤.综上所述，当1a≤时，()0x.·····································11分所以当1a≤时，()esin0xfxxx≤.····································12分所以，对任意整数a，函数()fx至多1个零点．