莆田市2 0 1 3~2 0 1 4学年 高三3月教学质量检查

莆田第二十五中学2024-2025学年第一学期月考一高三物理试题一、单项选择题：本题共4小题，每小题4分，共16分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

1．在如图所示的照片中，关于物体或人在运动过程中的一些描述正确的是（）A．甲图是一列在12：35~20：07之间由厦门北开往武汉的动车，12：35是指动车运动的时间间隔B．乙图中的a点是电风扇扇叶边缘上一点，它在运动一周的过程中，其平均速度为零，故a点在任意时刻速度都为零C．丙图中400m跑步比赛的终点在同一直线上，但起点不在同一直线上，这样做的目的是使参加比赛的同学路程相同D．丁图是正在做精彩表演的芭蕾舞演员，台下的观众可以把她们视为质点2．新能源电动车已经走进我们的生活，逐渐为大家所青睐。

对某新能源电动车进行刹车测试时，该车以30m/s的速度开始做匀减速直线运动，运动最后1s内的位移大小为6m。

该车在匀减速直线运动过程中描述正确的（）A．加速度大小为6m/s2B．位移大小为37.5mC．减速的时间为5s D．平均速度大小为10m/s3．如图所示，是常见的杆秤的工作示意图。

三根长度均为5L的细绳上端连在一起，并固定在杆秤的左端，另一端与质量为m、直径为6L、质量分布均匀的秤盘相连，连接点将秤盘边缘三等分。

当在盘中放置质量为2m的物体，秤杆处于平衡状态时，秤盘静止在水平位置，这时每根细绳的拉力大小为（）A．53mgB．54mgC．56mgD．52mg4．春节贴“福”字是民间由来已久的风俗，新春佳节临近，某同学正写“福”字，他在水平桌面上平铺一张红纸，并在红纸左侧靠近边缘处用镇纸”压住以防止打滑，整个书写过程中红纸始终保持静止，则该同学在书写过程中（）A．提笔静止时，手对毛笔的摩擦力大小与握力成正比B．向下顿笔时，毛笔对红纸的压力大于红纸对毛笔的支持力C．向左行笔时，红纸对桌面的静摩擦力方向向左D．向右行笔时，红纸对“镇纸”的静摩擦力方向向右二、双项选择题：本题共4小题，每小题6分，共24分。

二0一三至二0一四学年度第第一学期期未考试（考查）质量分析城关二小二0一四年元月十八日城关二小2013—2014学年度第一学期教学质量分析期末考试已经结束了，本次期末考试由县教育局出题、中心学校组织全镇交叉监考、交叉评卷及分析。

此次考试既是对学生学习能力的考查，也是对教师教学能力的检测，同时也让我们从中寻找不足，以期新的提高。

在考试、评卷结束后，从学生的统考成绩看，我校的教学质量在不同程度上有所提高。

但也可看出在某些学科教学中，教师的教和学生的学还存在一些问题，特别是教师的教存在很多不足，有些年级学科考试（考查）成绩提高缓慢。

我们针对各年级、各学科的成绩做出了以下质量分析，为我校以后的教学工作指明新思路。

一、本学期统考成绩情况1、语文科（1）一年级平均分为8136分，及格率为87.83%，优分率为67.20%。

（2）二年级平均分为分65.93分，及格率为66.67%，优分率为30.36%。

（3）三年级平均分为47.63分，及格率为33.93%，优分率为3.57%。

（4）四年级平均分为62.96分，及格率为63.33%，优分率为18.10%。

（5）五年级平均分为61.40分，及格率为63.45%，优分率为9.14%。

(6)六年级平均分为69.14分，及格率76.30%，优分率为34.00%2、数学科(1)一年级平均分为77.78分，及格率为84.13%，优分率为55.56.%。

(2)二年级平均分为79.20分，及格率为88.10%，优分率为61.90%。

(3)三年级平均分69.497分，及格率为70.83%，优分率为47.02%。

(4)四年级平均分为66.15分，及格率为69.05%，优分率为36.67%。

(5)五年级平均分为67.44分，及格率68.55%，优分率为43.70%。

(6)六年级平均分为58.34分，及格率55%，优分率为25.00%3、品德与社会（1）三年级平均分为60.43分，及格率64.29%，优分率为5.95%。

福建省莆田市2021届高三第三次教学质量检测试卷化学试题本试卷满分100分。

考试用时75分钟。

注意事项：1.答题前，考生务必将自己的姓名。

考生号、考场号、座位号填写在答题卡上。

2.回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。

如需改动，用橡皮擦干净后，再选涂其他答案标号。

回答非选择题时，将答案写在答题卡上。

写在本试卷上无效。

3.考试结束后，将本试卷和答题卡一并交回。

4.本试卷主要考试内容：高考全部内容。

5.可能用到的相对原子质量：H1、C12、N14、O16、S32一、选择题：本题共10小题，每小题4分，共40分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

1. 化学与生产、生活密切相关，下列说法正确的是A. 75%的乙醇、含氯消毒剂、过氧乙酸均可有效灭活新型冠状病毒B. 金属汞一旦洒落在实验室地面或桌面时，必须尽可能收集，并深埋处理C. 光化学烟雾、臭氧空洞、温室效应的形成都与氮氧化物有关D. “凡火药，硫为纯阳，硝为纯阴”中的“硫”是指硫黄，“硝”是指硝酸2. 如何对肿瘤精准治疗一直是医疗领域需攻克的难题之一，我国科研人员开发出的一种医用亲水凝胶，能使药物的释放更为精确。

医用亲水凝胶生产过程中的一种中间体M的结构如图所示。

下列有关M的说法正确的是A. M分子间可通过缩聚反应合成高分子化合物B. 能发生消去反应C. 该分子亲水的原因是能形成分子内氢键D. 分子中所有的碳原子和氮原子可能共平面3. 下列实验操作描述正确且能达到实验目的的是A B C D苯萃取碘水后的分离操作配制1 mol/L的NaCl溶液时的定容操作探究接触面积对反应速率的影响验证Ksp(AgSO4)＞Ksp(Ag2S)A. AB. BC. CD. D4. 下列离子方程式书写正确的是A. 向硫酸铜溶液中加人少量过氧化钠：2Na2O2 +2Cu2+ +2H2O=4Na+ +2Cu(OH)2↓+O2↑B. 将氯气通入氢氧化钠溶液中：Cl2+OH-=Cl-+ClO- +H2OC. 等物质的量的CaCl2、Ba(OH)2和HCl溶液混合：Ca2++2OH-=Ca(OH)2 ↓D. 氧化亚铁溶于稀硝酸：FeO+2H+=Fe2+ +H2O5. 甲硫醇是一种重要的原料和化工试剂，硫化氢与甲醇合成甲硫醇的催化过程如图。

2023 届高考英语模拟试卷请考生注意：1.请用 2B 铅笔将选择题答案涂填在答题纸相应位置上，请用 0．5 毫米及以上黑色字迹的钢笔或签字笔将主观题的答案写在答题纸相应的答题区内。

写在试题卷、草稿纸上均无效。

2.答题前，认真阅读答题纸上的《注意事项》，按规定答题。

第一部分（共 20 小题，每小题 1.5 分，满分 30 分）1.He used to study in a village school, was rebuilt two years ago.A.whichB．thatC．where D．whose2.When the old man woke up, he found someone his TV in the living room.A．stole B．steal C．stealing D．stolen3.Not having worked out the program, leave the office.A．so he was forbidden to B．and he didn’t want toC．his little son couldn’t make him D．be couldn’t free himself to4.about t he m an w earing s unglasses d uring n ight t hat h e w as d etermined t o f ollow h im.A．So curious the detective wasB.So curious was the detectiveC.How curious was the detectiveD.How curious the detective was5.The witness an important detail when describing the accident.A.brought outB．kept offC．left out D．ran into6.—Sir, I’m late because my car broke down on the way.—. I’ve had enough of your excuses.A.Cut it out B．Suit yourself C．You can’t be seriousD．It makes sense7.—Oh, my God! What’s wrong with you?—I was going the normal speed limit when a truck appeared in front of me and I didn't have time to stop. A．between the linesB．around the clockC．out of the blue D．over the moon8.Mary liked to look back on her high—school days，she thought were the happiest in herlife．A．that B．what C．which D．when9.There was never any time for Kate to feel lonely, she was an only child.A.ever sinceB．now thatC．even though D．as though10．I quickly lowered myself, ducking my head to avoid looking directly into his eyes so that he wouldn't feel .A.challengedB.challengingC．to be challenged D．having challenged11．--It is really fun to hike and I often go hiking in the forest.--But hiking alone in the forest be very dangerous.A．can B．must C．shall D．will12.his overseas study, he was eventually employed by his dream company.pletingA.Having completedC．Completed D．To complete13.It is not surprising the new course on electronic games has attracted many s tudents.A．what B．whichC．that D．where14．--Tom, your foreign teacher speaks Chinese fluently!--Oh, she has lived in China for six years; otherwise she such good Chinese.A．didn’t speakB．would not have spokenC．would not speak D．hadn’t spoken15．I up my mind what I was going to say in the seminar, but it was cancelled.A．have made B．had madeC．was making D．would make16.—Did Max go to the concert with his family yesterday?—The report scheduled to be handed in tomorrow, he it.B．needn’t have attendedA．couldn’t have attendedC．wouldn’t attend D．shouldn’t attend17.The farmers are in bad need of rain,but a (an) large a mount o f r ainfall w ill c ause F loods. A．Eventually B．constantly C．gradually D．extremely18.To get a slim figure, Fanny has tried many ways, but in vain. So now she is reduced anything for supper.A．to not eatingB．not to eatC．not to eating D．to not e at19.Young couples will be happy to see their babies with good health and intelligence when they are born.B．blessingA．to blessC．blessed D．being blessed20.China’s BeiDou Navigation Satellite System, whose positioning will reach 2.5 meters by 2020, will soon provide services for more countries.A．accuracy B．categoryC．function D．reference第二部分阅读理解（满分40 分）阅读下列短文，从每题所给的A、B、C、D 四个选项中，选出最佳选项。

2020年莆田市高中毕业班教学质量检测试卷理科综合能力测试第I卷(选择题)一、选择题：本题共7小题，每小题6分，在每小题给出的四个选项中，只有一项是符合题目要求的。

1.《唐本草》和《本草图经》中记载：“绛矾，本来绿色，……正如瑁璃烧之赤色”“取此物(绛矾)置于铁板上，聚炭，……吹令火炽，其矾即沸，流出，色赤如融金汁者是真也”。

其中不涉及的物质是()A. FeSO4·7H2OB. SC. Fe2O3D. H2SO4【答案】B【解析】【详解】由信息可知，绿矾为硫酸亚铁的结晶水合物，即FeSO4·7H2O，加热发生氧化还原反应，Fe元素的化合价升高，S元素的化合价降低，Fe的氧化物只有氧化铁为红色，则“色赤”物质可能是Fe2O3，故生成氧化铁、二氧化硫、三氧化硫和水蒸气，三氧化硫与水结合可生成硫酸，不涉及的物质是S，答案选B。

【点睛】硫为淡黄色固体，在题目中并未涉及，做本题时要关注各物质的物理性质。

2.“84消毒液”的主要成分是NaClO。

下列说法错误的是()A. 长期与空气接触会失效B. 不能与“洁厕剂”(通常含盐酸)同时使用C. 1L0.2mol/LNaClO溶液含有0.2molClO-D. 0.1molNaClO起消毒作用时转移0.2mole-【答案】C【解析】【详解】A．空气中含有二氧化碳和水蒸气，长期与空气接触发生反应生成次氯酸，次氯酸不稳定分解，导致消毒液失效，故A正确；B．“84消毒液”的主要成分是NaClO ，“洁厕剂”通常含盐酸，二者同时使用发生氧化还原反应生成氯气而失效，不能同时使用，故B正确；C．1L0.2mol/LNaClO溶液中溶质的物质的量为0.2mol，NaClO属于强碱弱酸盐，ClO-水解生成HClO，导致溶液中ClO-的物质的量小于0.2mol，故C错误；D．NaClO具有氧化性，起消毒作用时Cl元素化合价降低，由+1价变为-1价，则0.1molNaClO转移0.2mole-，故D正确；答案选C。

2023届福建省莆田市高中毕业班第四次教学质量检测英语试卷学校:___________姓名：___________班级：___________考号：___________一、阅读理解Activities for National Manufacturing DayThe STEM (Science, Technology, Engineering and Mathematics) Careers Coalition (职业联盟) offers students and families new STEM resources to explore the world of modern manufacturing. With over 4 million manufacturing roles predicted to be available in the next ten years, the new resources from the coalition connect today’s students to the manufacturing careers of tomorrow.National Manufacturing Day occurs on annually the first Friday of October to motivate today’s students to become the next generation of innovators leading the manufacturing industry. In order to support this celebration, the coalition offers three different collections to all students, educators, and families on modern manufacturing that support learning wherever it takes place. The following resources ate available to all at no cost：Manufacturing-focused Career ProfilesExplore the profiles of manufacturing professionals using the skills they learned at school to become the kind of problem solvers that make a difference. Make the connection from the classroom to careers and find out how all students can make it in manufacturing. Hands-on Student ActivitiesFrom innovating food packaging to discovering the principles of accurate design to exploring water sustainability, at-home, hands-on learning activities help students take STEM skills from theory to practice.Classroom ActivitiesAccess the no-cost, standard classroom activities to support future career success for all students. Each, activity features important STEM skills and a step-by-step guide for promoting activities into classroom.Virtual Interactive Field TripsEach virtual experience features accompanying educator resources and student activities that make it easy for teachers to put these manufacturing events into the learning experiences they design and deliver daily.All resources are available at no cost at stemcareerscoalition. org.1．What is the purpose of the STEM Careers Coalition?A．To count the vacant jobs in manufacturing.B．To guide students to work on manufacturing.C．To report the current state of manufacturing.D．To offer students pre-job training in manufacturing.2．Which activity will introduce manufacturing masters to students?A．Classroom Activities.B．Hands-on Student Activities C．Virtual Interactive Field Trips.D．Manufacturing-focused Career Profiles. 3．What do the activities have in common?A．The activities are free of charge.B．All of the activities are virtual. C．Only students practice in the activities.D．The activities are only experienced online.Wildlife photographer Sam Turley and his wife Vera found a two-week-old blesbok (南非白面大羚羊)—whom they named Meme—on her own. She was spotted wandering around on her own after her mother passed away.The pair took Meme in and raised the animal themselves for seven months before releasing her back out into the wild. Neither of them had ever raised a blesbok before. So it required a huge amount of patience and compassion to keep her fit and healthy but it deserved the effort. Meme would follow Vera wherever she went and sometimes the couple would have a blesbok, a dog and a cat all walking together. In the right areas, the blesbok is a really common antelope to see and is therefore often overlooked and under-appreciated. It was a privilege to be let into the blesbok’s secret world and they learned more about the blesbok than they thought was ever possible.After releasing Meme back out into the wild, the couple were convinced they would never see her again, until several months later when they were driving around a reserve and happened to come across a herd of blesboks.Vera called out Meme’s name in the hope of finding Meme. Blesboks all look very similar to one another, which makes it very difficult to distinguish individuals. When Vera started to talk, the herd raised their heads and looked over. Vera continued to call Meme and as all the other blesboks lost interest, one individual held her gaze and ran straight up to her.In fact, most research today suggests procrastination is a behavioral pattern, which means it’s something you can change, regardless of whether you’re impulsive. What works best is still under research. Some scientists have reported success with time management. But the evidence for that is all over the map. That’s because poor time management is a symptom not a cause of procrastination.For some procrastinators, seemingly obvious tips can work. Researchers advise students to simply put down their smartphones. Silencing notifications or studying in the library rather than at home can regulate distractions and keep people on task. But that won’t be enough for many people. Stubborn procrastinators may benefit from cognitive behavioral therapy (认知行为疗法). This type of therapy, which involves managing thoughts and emotions and trying to change behavior, seemed to be the most helpful. Still, not many studies have examined treatments, and there’s room for improvement.8．What does procrastination mean to longtime procrastinators?A．A personal advantage.B．A habitual behavior.C．A working motivation.D．An excuse for laziness.9．What do experts think of procrastinators?A．They are inactive.B．They are efficient.C．They are anxious.D．They are careful.10．What causes people to procrastinate?A．Their temporary relief from the stress.B．Their imperfection in personalities.C．Their poor time management at work.D．Their high ability to deal with hard work.11．What may be the best procrastination treatment according to the text?A．Avoiding smartphones.B．Proper studying places.C．Good time management.D．Cognitive behavioral therapy.From the moment we are born, we interact with the world through movement. We move our lips to smile or to talk. We extend our hand to touch. How does our brain remember this wide range of motions? How does it learn new ones? How does it make the calculations necessary for us to grab a glass of water, without dropping it, squashing it, or missing it?Technion Professor Jackie Schiller from the Ruth and Bruce Rappaport Faculty ofMedicine and her team examined the brain at a single-neuron (单神经元) level to shed light on this mystery. They found that calculation happens not just in the interaction between neurons, but within each individual neuron. It turns out that each of these cells is not a simple switch, but a complicated calculating machine.This discovery, published recently in the Science, promises changes not only to our understanding of how the brain works, but better understanding of conditions ranging from Parkinson’s disease to autism. And if that isn’t enough, these same findings are expected to advance machine learning, providing inspiration for new architectures.“We used to think of each neuron as a sort of whistle, which either rings, or doesn’t,” Prof. Schiller explains. “Instead, we are looking at a piano. Its keys can be struck at the same time, or in order, producing an infinity (无限) of different tunes.” This complex music playing in our brains is what enables us to learn and perform an infinity of different, complex and precise movements.These same findings can also serve as an inspiration for the machine learning community. Deep neural networks, as their name suggests, attempt to create software that learns and functions somewhat similarly to a human brain. Although their advances constantly make the news, these networks are primitive compared to a living brain. A better understanding of how our brain actually works can help in designing more complex neural networks, enabling them to perform more complex tasks.12．Why does the author raise the questions in paragraph 1?A．To pursue answers.B．To raise his puzzles.C．To introduce the topic.D．To share his interest13．What does paragraph 3 mainly talk about?A．The result of the research.B．The function of a single neuron.C．The way the research was conducted.D．The aspects the discovery can be applied to.14．What does Schiller stress in her explanation in paragraph 4?A．The way that neurons work.B．The music that neurons play.C．The difficulties that researchers encounter.D．The process where neurons produce memory.15．What might be the best title for the text?A．Individual Neuron Works like a PianoB．Calculation Happens Between Neurons like a MachineC．Complex Neural Networks in the Brain Enable People to LearnD．The Mystery of Learning and Remembering Movement Has Been Uncovered 二、七选五F．The task that is easily done at the first attempt is undoubtedly very easy.G．He was able to do it again through years of hard work and perseverance.三、未知For his entire life, Sergio Peralta dreamed about playing catch. When he was born, his right hand didn’t fully ________. Ever since he was a child, his classmates have asked about his hand, and some have even ________ him. Computer science teacher Jeff Wilkins noticed Peralta was the only student who ________ his mouse to the left side of his keyboard. He then saw Peralta didn’t have a right hand.When Wilkins ________ Peralta and his mother about a prosthetic hand (假手), they expressed interest but knew building one could be ________ for a high school class. In early November, Wilkins ________ assigned three of his students to the project. “I didn’t want to get his ________ up. I’d rather under-promise and over-deliver than over-promise and under-deliver on something like this,” Wilkins said. The group did so while keeping their________ a secret. They measured classmates’ hands to calculate Peralta’s ________ fit.After working for about a week, the students used the school’s 3D printer to ________ a model. Students said they ________ Peralta wouldn’t like or use the hand, but as soon as he ________ the model, he could bend his fingers. Then Wilkins threw him a yellow rubber ball again and again. Students yelled in ________ when he finally caught the ball. Peralta said he only ________ the prosthetic hand when be sleeps. “This just showed me a different way to ________ the community,” said Jaramillo, a senior who involved in the class project. 21．A．raise B．develop C．change D．connect 22．A．convinced B．defeated C．teased D．admired 23．A．cycled B．reserved C．moved D．kicked 24．A．approached B．promised C．observed D．advertised 25．A．rewarding B．inspiring C．approving D．challenging 26．A．randomly B．deliberately C．temporarily D．secretly 27．A．fears B．hopes C．demands D．doubts 28．A．attention B．memories C．progress D．conflicts 29．A．typical B．ideal C．normal D．familiar 30．A．create B．select C．deliver D．save31．A．remembered B．worried C．argued D．regretted 32．A．turned up B．searched for C．put on D．threw away 33．A．satisfaction B．excitement C．hesitation D．anxiety 34．A．removes B．returns C．researches D．replaces 35．A．fit B．move C．view D．help四、用单词的适当形式完成短文五、其他应用文46．假定你是李华，新西兰某高中校长Mr Smith将来你校参观并洽谈交换生项目，期间会出席你校举办的外语合唱节（Chorus Festival of Foreign Languages）。

2024年莆田市初中毕业班质量检查试卷语文（满分： 150分；考试时间： 120分钟）注意：本试卷分为“试题”和“答题卡”两部分，答题时请按答题卡中的“注意事项”要求认真作答，答案写在答题卡上的相应位置。

一、积累与运用（23分）1. 根据语境，补写出古代诗文名句。

（1）孟子在《鱼我所欲也》中提出了一个著名的观点，即在生与义不可兼得的情况下，应当做到“________”。

（2）李商隐《夜雨寄北》中，诗人想象回家团聚后剪烛西窗、共话今宵情景的两句是“________，________” 。

（3）古文中有一些短语已经变成了成语，在后世流传。

如范仲淹在《岳阳楼记》中用“______”形容事物多样、壮观；欧阳修在《醉翁亭记》中以“______”比喻经过挫折后出现转机。

（4）辛弃疾《破阵子·为陈同甫赋壮词以寄之》中“______，_______”两句，描绘战马奔驰、万箭齐发的战斗情景，壮烈豪迈。

（5）秋瑾《满江红（小住京华）》结尾长叹“_______”，借用典故表达词人难觅知音的苦闷。

【答案】①. 舍生而取义者也②. 何当共剪西窗烛③. 却话巴山夜雨时④. 气象万千⑤. 峰回路转⑥. 马作的卢飞快⑦. 弓如霹雳弦惊⑧. 青衫湿【解析】【详解】本题考查名句名篇默写。

默写题作答时，一是要透彻理解诗文的内容；二是要认真审题，找出符合题意的诗文句子，三是答题内容要准确，做到不添字、不漏字、不写错字。

注意本题中的易错字：剪、窗、烛、霹、雳、弦、湿。

2. 阅读下面的文字，按要求作答。

红色资源是红色文化的①（zǎi）体，红色基因是红色文化世代传承的密码。

在我国广袤无垠的大地上，红色资源（），见证了我们党艰辛而辉煌的奋斗②（lì）程。

每一位革命英雄、每一种革命精神、每一件革命文物，都让红色基因库不断丰富扩③（róng）。

一个个革命博物馆、纪念馆、党史馆、烈士陵园，都在讲述着党的故事、革命的故事、英雄的故事；一条条红色旅游线路，一个个红色遗迹遗址，都在生动（）着红色文化。

2024届福建省莆田市高三下学期高考英语质量检测试题（三模）注意事项：1.答题前，考生务必将自己的姓名、考生号、考场号、座位号填写在答题卡上。

2.回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号的黑。

如需改动，用橡皮擦干净后，再选涂其他答案标号。

回答非选择题时，将答案写在答题卡上。

写在本试卷上无效。

3.考试结束后，将本试卷和答题卡一并交回。

第一部分听力（共两节，满分30分）做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的答案涂到答题卡上。

第一节（共5小题；每小题1.5分，满分7.5分）听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅再一遍。

例：How much is the shirt?A. £ 19.15.B. £ 9.18.C. £ 9.15.答案是C。

1. What is the man going to do?A. Hold a party.B. Go to the bookstore.C. Celebrate his birthday.2. When will the man clean up his room?A. In just half an hour.B. After seeing the movie.C. Before going to the cinema.3. How does the man usually get to his company?A. By subway.B. By car.C. By bike.4. What was the matter with Alice today?A. She did badly in the math test.B. She forgot to check her homework.C. She made many mistakes in all tests.5. How much should the woman pay in total?A. $ 30.B. $ 18.C. $ 6.第二节（共15 小题；每小题1.5分，满分22.5分）听下面5段对话或独白。

绝密★启用前试卷类型：A2023-2024学年福州市高三年级第三质量检测评分参考数学2024.4一、单项选择题：本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的．1．已知复数z 满足（i 是虚数单位），则z =A ．1-B ．1C ．i-D ．i解析：∵i i 1i z +=+，∴i 1z =，即i z =-，故选C.2．已知角α的顶点在坐标原点，始边与x 轴非负半轴重合，cos α=，(,2)P m 为其终边上一点，则m =A ．4-B ．4C ．1-D ．1解析：∵cos α=，∴2tan 2m α==，∴1m =，故选D .解析：结合该函数为偶函数，及()03f =可判断应选A.4．在菱形ABCD 中，若||||AB AD AB -= ，且AD 在AB 上的投影向量为AB λ，则λ=A ．12-B ．12C ．22-D ．22解析：由已知AB AD AB -=知该菱形中AB AD BD ==，∴由D 向AB 作垂线，垂足即为AB 中点，∴12λ=，故选B .5．已知5log 2a =，2log b a =，1(2bc =，则A.c b a >>B.c a b>> C.a b c >> D.b c a>>解析：∵55log 2log 51a =<=，∴2log 0b a =<，1(12b c =>，∴c a b >>，故选B.6．棱长为1的正方体1111ABCD A B C D -中，点P 为1BD 上的动点，O 为底面ABCD 的中心，则OP 的最小值为 A.33B.63C.66D.32解析：在正方体中，易知AC BD ⊥，1AC DD ⊥，且1BD DD D = ，∴AC ⊥平面1BDD ，易知当OP ⊂平面1BDD ，且1OP BD ⊥时，OP 的长度最小，在1RT BDD △中,不难求得66OP =，故选C.7．若直线y ax b =+与曲线e xy =相切，则a b +的取值范围为A ．(,e]-∞B ．[2,e]C ．[e,)+∞D ．[2,)+∞解析：设切点为00(,e )x x ，则0e ,x a =∴切线方程为000e ()e x x y x x =-+，则00(1)e x b x =-，∴00(2)e x a b x +=-，设00()(2)e x f x x =-，则00()(1)e x f x x '=-，易知函数()(1)e f x f ≤=，又(2)02f =<，故可判断选A.（由图象知当且仅当切线与曲线相切于()1,e 时，11e e a b a b +=⨯+==最大，亦可知选A.）8．已知函数()2sin cos )f x x x x ωωω=+(0)ω>在π(0,)3上单调递增，且对任意的实数a ，()f x 在(,π)a a +上不单调，则ω的取值范围为A ．5(1,]2B ．5(1,]4C ．15(,22D ．15(,]24解析：∵π()2sin cos )2sin(2)3f x x x x x ωωωω=+=-+∵()f x 在π(0,3上单调递增，∴πππ2332ω⋅-≤，∴54ω≤，∵对任意的实数a ，()f x 在区间(,π)a a +上不单调，∴()f x 的周期2πT <，∴2π2π2T ω=<，∴12ω>，∴1524ω<≤，故选D ．二、多项选择题：本题共3小题，每小题6分，共18分．在每小题给出的四个选项中,有多项符合题目要求．全部选对的得6分，部分选对的得部分分，有选错的得0分．题号91011答案ABDACDBC9．双曲线2222:13x y C a a-=(0)a >的左、右焦点分别为1F ，2F ，且C 的两条渐近线的夹角为θ，若12||2F F e =（e 为C 的离心率），则解析：易知该双曲线实半轴为a ，半焦距为2a ，∴离心率22ae a==，∴焦距44a =，即1a =，∴选项A 正确，选项C 错误；易知C 的两条渐近线的斜率为3k a=±=，∴这两条渐近线的倾斜角分别为π3和2π3，∴C 的两条渐近线的夹角为π3，∴选项B ，D 正确；综上所述，应选ABD .10.定义在R 上的函数()f x 的值域为(,0)-∞，且(2)()()0f x f x y f x y ++-=，则A ．(0)1f =-B ．2(4)[(1)]0f f +=C ．()()1f x f x -=D ．()()2f x f x +-≤-解析：令0x y ==，则()()2000f f+=，∵函数()f x 的值域为(,0)-∞，∴(0)1f =-，选项A 正确；令1x =，0y =，则2(2)[(1)]f f =-，令2x =，0y =，则24(4)[(2)][(1)]f f f =-=-，∴选项B 错误；令0x =，则(0)()()0f f y f y +-=，∴()()(0)1f y f y f -=-=，即()()1f x f x -=，∴选项C 正确；∵()0f x ->，()0f x -->，∴[()()]2f x f x -+-≥∴()()2f x f x +-≤-，故选项D 正确；综上所述，应选ACD .11.投掷一枚质地均匀的硬币三次，设随机变量1,1,(1,2,3)n n n X n ⎧==⎨-⎩第次投出正面,第次投出反面,．记A 表示事件“120X X +=”，B 表示事件“21X =”，C 表示事件“1231X X X ++=-”，则A ．B 和C 互为对立事件B ．事件A 和C 不互斥C ．事件A 和B 相互独立D ．事件B 和C 相互独立解析：考查选项A ，事件B 和C 均会出现“反，正，反”的情况，故选项A 错误；考查选项B ，事件A 和C 均会出现“反，正，反”的情况，故选项B 正确；考查选项C ，易知12211()(22P A C ==，1()2P B =，事件AB 为前两次投出的硬币结果为“反，正”，则1()4P AB =，∴1()()()4P AB P A P B ==，故选项C 正确；考查选项D ，由选项AC 可知311()(28P BC ==，1()2P B =，在事件C 中三次投出的硬币有一次正面，两次反面，则23313()(28P C C ==，∴()()()P BC P B P C ≠，故选项D 错误；综上所述，应选BC ．三、填空题：本题共3小题，每小题5分，共15分．12.160；13.2；14.22mm +；1或2．12.62()x x+的展开式中常数项为．解析：易知该二项展开式通项为662()r r r C x x-，∴当3r =时，得到常数项为160，故应填160．13．某圆锥的体积为π3，其侧面展开图为半圆，则该圆锥的母线长为．解析：设该圆锥的母线长为l ，底面圆半径为r ，根据侧面展开图为半圆得2ππr l =，即2l r =，又根据圆锥体积得1ππ33r =，解得1r =，2l =，故应填2．14.设n T 为数列{}n a 的前n 项积，若n n T a m +=，其中常数0m >．则2a =（结果用m 表示）；若数列1{}nT 为等差数列，则m =．解析：易知112m T a ==，∴12221)(2m a a a a m =+=+，解得222a m m =+，故应填22m m +；（方法一）211111111111111n n n n n n n n T T m a m a m a m ma a m m m a ---------=-=-=-----+(2)n ≥，若数列1{}n T 为等差数列，则2111n n m ma a ----为常数d ，①若0d =，则11n a -=(2)n ≥恒成立，即1n a =(1)n ≥恒成立，∴2m =；②若0d ≠，则1211n n dm dm a a --=--，∴2,,11dm dm ==⎧⎨⎩解得1,1,d m ==⎧⎨⎩综上所述，若数列1{}nT 为等差数列，则1m =，或2m =，故应填1或2．（方法二）∵1{}n T 为等差数列，∴111n n d T T -=+(2)n ≥，易知112T m =，且12(1)n n d T m=+-，当2n ≥时，∵n n T a m +=，∴1n n n T T m T -+=，∴111n n m T T -=+，∴由12(1)n n d T m =+-，可得22(1)1(2)m n d n d m+-=++-，∴2(1)1(2)m dn m d m-=-++-对于任意n 恒成立，∴1,21(2)0,m m d m =⎧⎪⎨-++-=⎪⎩或0,21(2)0,d m d m =⎧⎪⎨-++-=⎪⎩解得1,1,m d =⎧⎨=⎩或0,2,d m =⎧⎨=⎩综上所述，若数列1{}nT 为等差数列，则1m =，或2m =，故应填1或2．四、解答题：本题共5小题，共77分．解答应写出文字说明、证明过程或演算步骤．15．（13分）ABC △中，角A ，B ，C 的对边分别是a ，b ，c ，且sin sin a C c B =，2π3C =.（1）求B 的大小；（2）若ABC △的面积为4，求BC 边上中线的长.解：（1）∵sin sin a C c B =，∴由正弦定理，得sin sin sin sin A C C B =，…………2分∵0πC <<，∴sin 0C >，∴sin sin A B =，………………………………………3分∵0πA <<，0πB <<，∴A B =，……………………………………………………5分∵πA B C ++=，且2π3C =，∴π6B =.……………………………………………6分（2）依题意1sin 42ab C =，………………………………………………………………7分∵A B =，∴a b =，………………………………………………………………8分212πsin 23a ==，解得a =，…………………………………………10分设边BC 的中点为D ，∴32CD AC ==∴在ACD △中，由余弦定理知2222cos AD AC CD AC CD C=+-⋅⋅332π2132cos4234=+-⨯=，………………………………………………………12分∴BC 边上中线的长为212.……………………………………………………………13分16．（15分）如图，在三棱柱111ABC A B C -中，平面11ACC A ⊥平面ABC ，12AB AC BC AA ====，1A B =.（1）设D 为AC 中点，证明：AC ⊥平面1A DB ；（2）求平面11A AB 与平面11ACC A 夹角的余弦值．（第16题图）解：（1）∵D 为AC 中点，且2AB AC BC ===，∴在ABC △中，有BD AC ⊥，且BD =……………………………………………1分∵平面11ACC A ⊥平面ABC ，且平面11ACC A 平面ABC AC =，∴BD ⊥平面11ACC A ，………………………………………………………………………2分∵1A D ⊂平面11ACC A ，∴1BD A D ⊥，……………………………………………………3分∵1A B =，BD =1A D ，……………………………………………………4分∵1AD =，12AA =，1A D =，∴由勾股定理，有1AC A D ⊥，……………………………………………………………6分∵AC BD ⊥，1A D BD D = ，∴AC ⊥平面1A DB ，…………………………………………………………………………7分（2）如图所示，以D 为原点，DA ，DB ，1DA 所在直线分别为x 轴，y 轴，z 轴建立空间直角坐标系D xyz -，可得(1,0,0)A，1A，B ，………………………………………………9分∴1(AA =-，(AB =-，…………………………………………………10分设平面11A AB 的法向量为(,,)x y z =n ，则由10,0,A A B A ⎧⋅=⎪⎨⋅=⎪⎩n n得0,0,x x ⎧-+=⎪⎨-+=⎪⎩令x =1y =，1z =，∴=n ，…………………………………………12分由（1）可知，BD ⊥平面11ACC A ，∴平面11ACC A的一个法向量为(0,BD =，…………………………………………13分记平面11A AB 与平面11ACC A 的夹角为α，∴5cos ||5||BD BD α⋅==n |n |，∴平面11A AB 与平面11ACC A 夹角的余弦值为5．………………………………………15分17．（15分）从一副扑克牌中挑出4张Q 和4张K ，将其中2张Q 和2张K 装在一个不透明的袋中，剩余的2张Q 和2张K 放在外面．现从袋中随机抽出一张扑克牌，若抽出Q ，则把它放回袋中；若抽出K ，则该扑克牌不再放回，并将袋外的一张Q 放入袋中.如此操作若干次，直到将袋中的K 全部置换为Q.（1）在操作2次后，袋中K 的张数记为随机变量X ，求X 的分布列及数学期望；（2）记事件“在操作1n +()n *∈N 次后，恰好将袋中的K 全部置换为Q ．”为n A ，记()n n P P A =.（i ）在第1次取到Q 的条件下，求总共4次操作恰好完成置换的概率；（ii ）试探究1n P +与n P 的递推关系，并说明理由.解：（1）由题意X 的取值可能为0，1，2，……………………………………………1分当0X =时，即第一次取出K ，第二次也取出K ，∴211(0)22318P X ==⨯=++，…………………………………………………………2分当1X =时，即第一次取出Q ，第二次取出K ，或第一次取出K ，第二次取出Q ，∴2223135(1)22222231488P X ==⨯+⨯=+=++++，……………………………3分当2X =时，即第一次取出Q ，第二次也取出Q ，∴221(2)22224P X ==⨯=++，…………………………………………………………4分∴X 的概率分布列为…………………………………………………………………5分∴X 的数学期望1519()0128848E X =⨯+⨯+⨯=．……………………………………6分（2）（i ）记事件“第1次取到Q ”为B ，事件“总共4次操作恰好完成置换”为C ，则1()2P B =，………………………………………………………………………………7分依题意，若第1次取出Q ，则剩余的3次操作，须将袋中K 全部置换为Q ，①若第2次亦取出Q ，则第3次和第4次均须取出K ，X 012P185814其概率为1221122+22+23+132⨯⨯⨯=；………………………………………………………8分①若第2次取出K ，则第3次须取出Q ，第4次须取出K ，其概率为1231322+23+13+164⨯⨯⨯=；………………………………………………………9分∴13()53264(|)1()322P CB P C B P B +===，即在第1次取到Q 的条件下，总共4次操作恰好完成置换的概率为532．…………………………………………………………………………10分（ii ）（方法一）由题可知若事件1n A +发生，即操作2n +次后，恰好将袋中的K 全部置换为Q ，①当第1次取出Q ，则剩余的1n +次操作，须将袋中K 全部置换为Q ，概率为212+22n n P P ⨯=；……………………………………………………………………12分②当第1次取出K ，则从第2次起，直到第1n +次均须取出Q ，且第2n +次取出K ，概率为23113(()2+23+13+184n n⨯⨯=⨯；………………………………………………………14分∴1+113(284n n n P P +⨯=．…………………………………………………………………15分（方法二）由题可知若事件1n A +发生，即操作2n +次后，恰好将袋中的K 全部置换为Q ，则一定有第2n +次（最后一次）取出K ，①当第1n +次（倒数第二次）取出Q ，则须在之前的n 次操作中的某一次取出K ，概率为333+14n n P P ⨯=；……………………………………………………………………12分②当第1n +次（倒数第二次）取出K ，则从第1次起，直到第n 次均须取出Q ，概率为3221111()((2+22+23+1822n n n +⨯⨯=⨯=；…………………………………………14分∴133+1(42n n n P P ++=．……………………………………………………………………15分18．（17分）在直角坐标系xOy 中，已知抛物线2:2(0)C y px p =>的焦点为F ，过F 的直线l 与C 交于M ，N 两点，且当l 的斜率为1时，|8MN =|．（1）求C 的方程；（2）设l 与C 的准线交于点P ，直线PO 与C 交于点Q （异于原点）．记线段MN 的中点为R ，若||3QR ≤，求MNQ △面积的取值范围．解：(1)不妨设l 的方程为2px my =+，11(,)M x y ，22(,)N x y ，联立l 与C 的方程，得2220y mpy p --=，…………………………………………1分∴122y y mp +=，212y y p =-，…………………………………………………………2分则21212||()22(1)MN x x p m y y p p m =++=++=+，…………………………………3分∴由题可知当1m =时，||8MN =，∴2p =，…………………………………………4分∴C 的方程为24y x =．……………………………………………………………………5分(2)由(1)知1222R y y y m +==，将R 的纵坐标2m 代入1x my =+，得2(21,2)R m m +，……………………………6分易知C 的准线方程为1x =-，又l 与C 的准线交于点P ，∴2(1,)P m--，……………7分则直线OP 的方程为2mx y =，………………………………………………………………8分联立OP 与C 的方程，得22y my =，∴2(,2)Q m m ，……………………………………9分∴Q ，R 的纵坐标相等，∴直线QR x ∥轴，……………………………………………11分∴222|||21|1QR m m m =+-=+，…………………………………………………………12分∴MNQ QRM QRN S S S =+△△△121||||2QR y y =-3222(1)2||m QR =+，…………14分∵点Q （异于原点），∴0m ≠，…………………………………………………………15分∵||3QR ≤，∴13||QR <≤，∴3222||QR <≤即MNQ S ∈△．…………………………………………17分19．（17分）若实数集A ，B 对a A ∀∈，b B ∀∈，均有(1)1b a ab +≥+，则称A B →具有Bernoulli 型关系.（1）判断集合{|1}M x x =>，{1,2}N =是否具有Bernoulli 型关系，并说明理由；（2）设集合{|1}S x x =>-，{|}T x x t =>，若S T →具有Bernoulli 型关系，求非负实数t 的取值范围；（3）当*n ∈N时，证明：1158n k k n -=<+∑．解：（1）依题意，M N →是否具有Bernoulli 型关系，等价于判定以下两个不等式对于1x ∀>是否均成立：①1(1)1x x +≥+，②2(1)12x x +≥+，…………………………………2分∵1x ∀>，1(1)1x x +=+，22(1)1212x x x x+=++>+∴M N →具有Bernoulli 型关系．………………………………………………………4分（2）（方法一）令()(1)1b f x x bx =+--，x S ∈，(0,)b ∈+∞，则1()[(1)1]b f x b x -'=+-，…………………………………………………………………5分①当1b =时，显然有(1)1b a ab +=+，∴(1)1b x xb +≥+成立；………………………6分②当1b >时，若10x -<<，则10(1)(1)1b x x -+<+=，即()0f x '<，∴()f x 在区间(1,0)-上单调递减，若0x =，则1(10)10b -+-=，即(0)0f '=，若0x >，则10(1)(1)1b x x -+>+=，即()0f x '>，∴()f x 在区间(0,)+∞上单调递增，∴()f x 的最小值为(0)0f =，∴()(0)0f x f ≥=，∴(1)(1)0b x bx +-+≥，∴(1)1b x xb +≥+成立；………………………………………………………………8分③当01b <<时，若10x -<<，则10(1)(1)1b x x -+>+=，即()0f x '>，∴()f x 在区间(1,0)-上单调递增，若0x =，则1(10)10b -+-=，即(0)0f '=，若0x >，则10(1)(1)1b x x -+<+=，即()0f x '<，∴()f x 在区间(0,)+∞上单调递减，∴()f x 的最大值为(0)0f =，∴()(0)0f x f ≤=，∴(1)(1)0b x bx +-+≤，即(1)1b x bx +≤+，∴当x S ∈，且01b <<时，(1)1b x xb +≥+不能恒成立，…………………………10分综上所述，可知若S T →具有Bernoulli 型关系，则{|1}T x x ⊆≥，∴非负实数t 的取值范围为[1,)+∞．……………………………………………………11分（方法二）当1b =，或01b <<时，与方法一相同；…………………………………8分当1b >时，若10ab +≤，∵(1)01b a ab +>≥+，∴(1)1b a ab +≥+，若10ab +>，则1ab >-，又1b >，∴101b <<，∴由方法一的结论，可知11(1)11b ab ab a b +≤+⋅=+，即1(1)1b ab a +≤+，…………………………………………………………………………9分∵10ab +>，且(1,)a ∈-+∞，∴1[(1)](1)b b b ab a +≤+，即1(1)b ab a +≤+，即(1)1b a ab +≥+；………………………10分∴若集合{|1}S x x =>-，{|}T x x t =>具有Bernoulli 型关系，则{|1}T x x ⊆≥，∴非负实数t 的取值范围为为[1,)+∞．…………………………………………………11分（3）∵1112222211((1)k k k k k k-+==+，…………………………………………12分显然211k >-，且1012k<<，由（2）中的结论：当01b <<时，(1)1b x xb +≤+，可知122231111(1)1+122k k k k k +≤⋅=+，………………………………………………………………………………………13分当2k ≥时，33121(1)111[]24()4(1)(1)4(1)(1)k k k k k k k k k k k k +--≤==---+-+，∴1221111(1)1[4(1)(1)k k k k k k +≤+--+，2k ≥，………………………………………15分当1n =时，1158n k k n -=<+∑显然成立；…………………………………………16分当2n ≥时，11122311[1]24(1)4(1)n n n k k k k k k k k k --====+<++--+∑∑∑211111111515[[24(1)(1)242(1)84(1)8n k n n n n k k k k n n n n ==++-=++⋅-=+-<+-+++∑，综上所述，当*n ∈N时，1158n k k n -=<+∑.……………………………………17分。