2020年3月18日浙江省学考选考杭州市杭二高2020届高2017级高三第二学期3月月考语文试题参考答案

保密★启用前试卷类型：A杭州第二中学2024年8月高三年级适应性检测英语试题本试卷满分120分。

考试用时100分钟。

注意事项：1. 答卷前，考生务必用黑色字迹的钢笔或签字笔将自己的姓名、考生号、试室号和座位号填写在答题卡上。

用2B 铅笔将试卷类型(B) 填涂在答题卡相应位置上。

并在答题卡相应位置上填涂考生号。

因笔试不考听力，试卷从第二部分开始，试题序号从"21"开始。

2. 作答选择题时，选出每小题答案后，用2B 铅笔把答题卡对应题目选项的答案信息点除黑；如需改动，用橡皮擦干净后，再选涂其他答案。

答案不能答在试卷上。

3.非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内相应位置上，如需改动，先划掉原来的答案，然后再写上新答案；不准使用铅笔和涂改液。

不按以上要求作答无效。

4. 考生必须保持答题卡的整洁。

考试结束后，将试卷和答题卡一并交回。

第二部分阅读理解(共两节，满分50分)第一节(共15小题; 每小题2.5分, 满分37.5分)阅读下列短文，从每题所给的A、B、C和D四个选项中，选出最佳答案。

ABeautiful Guatemala is a land rich in diversity and cultural heritage. Whether you hope to summit a volcano, hike through the jungle or explore ancient ruins, Guatemala will not disappoint.Tikal National ParkTikal National Park is a UNESCO World Heritage Site and one of the most famous places in Guatemala. Tikal covers an estimated 46 miles ( around 74 km) and is one of the most extraordinary archaeological sites in Central America. It is here, centuries ago, that the ancient Mayan civilisation boomed. Chichicastenango“ Chichi,” as it's known by the local Kiche population, hosts the largest market in Central America. The town comes alive on Thursdays and Sundays when vendors(小贩) come from surrounding areas to display their goods. The market bursts with varied colours and a lively atmosphere, stocking everything from vividly- colored cloth to traditional carved wooden masks. Guatemala CityBrightly graffitied(涂鸦) walls line the busy streets in Guatemala City where you'll find trendy stores, relaxing bars, and diverse art galleries. Make sure to check out La Esquina, an indoor food market with some of the best food stalls in the city. The Museo Popol Vu h is a leading museum in the world of Mayan art. Hereyou will be able to spend a couple of hours appreciating the incredible and comprehensive collection of Mayan as well as colonial art.Acatenango VolcanoVolcano Acatenang o towers almost 4,000 metres above the surrounding landscape just outside Antigua. The hike to the summit of this volcano is one of the toughest, yet most popular, in Guatemala. Not only will you get to appreciate the breathtaking views, but you will also have a bird's- eye view of the nearby ( and extremely active) Volcan Fuego.21. What's the best choice for someone interested in Mayan historyA. Tikal National Park.B. Chichicastenango.C. Guatemala City.D. Acatenango Volcano.22. What do we know about Guatemala CityA. It hosts the largest market in Central America.B. It is a perfect destination for art lovers.C. You can buy vivid cloth and masks there.D. Museo Popol Vuh isa great food market.23. What is special about Acatenango VolcanoA. A bird view of Guatemala City.B. The highest tower in Guatemala.C. The hike to the volcano summit.D. It is an extremely active volcano.BA new T- shirt that continuously monitors heart activity and detects abnormalities may help prevent strokes by identifying dangerous heart conditions like atrial fibrillation (AF). This irregular heart rhythm increases the risk of stroke and affects over a million- people in Britain, with an estimated 500,000 more remaining undiagnosed due to a lack of noticeable symptoms. While some individuals experience signs like chest pain, dizziness, or fatigue, many only discover the condition after suffering a stroke. Detecting AF typically involves an electrocardiogram (ECG), performed in a hospital where trained staff attach up to 24 electrodes to different parts of the body. However, since AF often occurs intermittently (间歇地), a short hospital check- up may not detect it. To counter this, doctors sometimes provide patient s with a Holter monitor, a device worn under clothing that records heart activity. However, it's bulky, inconvenient, and requires multiple wires to be attached to the chest, making it uncomfortable for daily wear.The Cardioskin T- shirt offers a more practical alternative. Made from cotton and washable up to35 times, it has15 small electrodes woven into the chest area to monitor heart signals continuously. The T- shirt can be worn 24 hours a day, increasing the likelihoodof detecting abnormal rhythms. Powered by a removable battery, the electrodes feed data to a microchip, which wirelessly transmits the information to an app. The app converts the data into an easy- to- read chart, highlighting any irregular heart activity.The results are then shared with the patient's doctor, allowing for remote monitoring without requiring hospital visits. This innovative design could make diagnosing AF easier and more accessible, potentially reducing the risk of stroke for many.Professor Martin Cowie, a cardiologist at Imperial College London, views Cardioskin as an significant development in heart health monitoring, offering a comfortable and efficient tool for early detection of cardiac problems.24. What is the main idea of Paragraph 2A. The difficulty in detecting atrial fibrillation. B The benefit of wearing a Holter monitor.C. The process of recording heart activity.D. The future of developing a new ECG.25. Why did the author mention a Holter monitor in the textA. To introduce a way to check the patient's heart.B. To explain why doctors like using the device.C. To show the advantage of a Cardiosk in T- shirt over it.D. To encourage people to care about their health.26. What can we learn about the Cardiosk in T- shirtA. It has electrodes placed all over the T- shirt.B. It can be worn a month without being washed.C. It has a battery that can be charged easily and quicklyD. It can make the process of diagnosis timely and convenient.27. What could be the best title for the passageA. New ECC: an important development for doctorsB. Hi- tech T- shirt: a good helper for detecting strokesC. Holter monitor: a device tracking patients' heart problemsD. Cardiosk in T- shirt: a shirt keeping patients from diseasesCThe male western manager(唐纳雀) looks like a little flame, while females are less showy, a dust y yellow. In the spring, they prepare to move thousands of miles to the Mountain West of Central America, flying through grasslands, deserts, and occasionally, suburban yards.To fuel them on their lengthy journey, western tanagers fill up on insects and berries. But as global climate change causes spring to start earlier, birds such as western tanagers are arriving at their destination after wha t's known as“ green- up”, when flowers begin blooming and insects emerge. According to a study published in early March in the journal PNAS, this kind of timing mismatchbetween migrants and their food sources, which is happening across North America, could have serious consequences for migratory birds’ survival.“ In discussing climate change, we often focus on warming,” says Scott Loss, a co- author of the study.“ But the length and timing of seasons——like when winter ends and. spring begins— are some of the most dramatic effects of climate change.”Loss and his colleagues used satellite imagery from 2002 to 2021 to calculate the average start of spring green- up along the typical migration routes of 150 North American bird species, then compared that timing with the current green- up. They found that spring is indeed beginning earlier along birds' migration routes.“ By contrary, previous studies have mainly focused on songbirds in Eastern North America,” says Morgan Tingley, an ornithologist at UCLA,“ but this new investigation shows that bird species in the West and at different levels of the food web might be just as vulnerable(脆弱的).”“ Part of it is knowing which species are vulnerable to various threats,” Loss says.“ This adds to the knowledge about vul nerability of a wide range of bird species.” And he hopes that the information will serve to highlight the urgent need to lower greenhouse- gas emissions as fast as possible.“ It's really important, if we can'taddress climate change immediately, to try to stop habitat loss as much as we can”28. What may pose a direct threat to western tanagers' survivalA. The worsening of global warming.B. The duration of changing seasons.C. Loss of habitats duce to human activities.D. Decreased access to food s during migration.29. What can we learn about the new study on birds like western tanagersA. It covers a wider geographic range.B. It reveals the decline in bird populations.C. It centers on the adaptation of bird species.D. It ensures the existence of a timing mismatch.30. What does Loss suggest we do to safeguard migratory birdsA. Lessen the effects of climate change.B. Preserve ecosystems for bird species.C. Tackle emissions and habitat loss.D. Expand researches on threats to birds.31. Where is the text most likely fromA. A scientific journal.B. A bird- watching guidebook.C. A website about climate change.D. A magazine about botanical conservation.DLord Norman Foster, renowned as one of the world's foremost architects, has devoted decades to redefining the concept of tall buildings.“ What we've done is create a sense of identify draw n from real needs,” Foster said.“ It's not a fashionable idea, but generated from the realities we were in.” His masterpiece includes iconic structures suchas the HSBC building in Hong Kong. Now, he's set his sights on Qatar.Currently nearing completion and set to reach a height of 301 meters, Foster's Lusail Towers are ready to claim the title of Oatar's tallest buildings. Lusail Towers are intended to serve as a center for Oatar's financial institutions, with four distinctive blocks, two standing at 70 stories and two at 50 stories.But the construction itself wasn't all plain sailing. To address the unique challenges presented by Oatar's hot climate, Foster's team had to depart from the materials commonly used in skyscrapers in cooler countries——part of what Foster called a decades- long quest to“ reinvent the tall building.” The design combines advanced shading with ventilation (通风), while the towers' surfaces are coated with“ marine- grade” aluminum(铝) that wraps around the buildings, protecting the glass from strong sunlight whilemaintaining the views and still letting in natural light. Central to the project are special shading fins(鳍), which not only optimize views and natural lighting for occupants but reduce solar radiation by 70% compared to traditional all- glass towers.While the towers are ready to become a recognizable landmark for Lusail, Foster believes that a city's skyline reflects what lies beneath the surface. It was crucial to integrate the towers into a“ low- scale master plan.” They are strategically positioned on top of a subway line and at the end of a commercial avenue linking the waterfront to the nearby football stadium. According to Foster, the ground features of the plaza will play a significant role in the objective of transforming the area into a lively public space for the future32. According to Foster's concept, what should, tall buildings be likeA. Creative.B. Identical.C. Fashionable.D. Practical.33. What can we learn about the Lusail TowersA. They will function as a global financial center. B They are expected to be Qartar's tallest building.C. They are Foster's most renowned iconic masterpiece.D. They have four blocks with the same style and height.34 What was the solution to the challenges caused by Qatar's hot climateA. Replacing the glass with“ marine- grade” aluminum.B. Coating the tower with an unconventional material.C. Reducing the views and natural lighting with shading fins.D. Protecting the glass from natural light and solar radiation.35. What's the goal of“ low- scale master plan”A. To link the waterfront to the football stadium.B. To create a remarkable plaza for the citizens.C. To offer a dynamic public space for the future.D. To connect a subway and a commercial avenue.第二节(共5小题; 每小题2.5分, 满分12.5分)根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。

杭州二中高三 3 月月考化学试题可能用到的相对原子质量：H 1 C 12 N 14 O 16 Na 23 Si 28 S 32 Cl 35.5 Fe 56 Cu 64 Br 80 Ag 108 Ba 137选择题部分一、选择题(本大题共 25 小题，每小题 2 分，共 50 分。

每个小题列出的四个备选项中只有一个是符合题目要求的，不选、多选、错选均不得分)1. 下列物质属于盐的是A. Mg(OH)2B．Al2O3C．Cu2(OH)2CO3D．CH3CH2OH2. 下列仪器名称为“坩埚”的是3. 下列物质属于电解质且能导电的是A．酒精 B．金属铜 C．液态硝酸D．熔融氯化钾4. 下列反应中，非金属单质只作氧化剂的是A．Br2+2NaOH NaBr+NaBrO+H2O B．2Al+2NaOH+2H2O 2NaAlO2+3H2↑C. C+2CuO 2Cu+CO2↑D．4Fe(OH)2+O2+2H2O 4Fe(OH)35. 人类生活离不开化学知识，下列叙述不．正．确．的是A．臭氧是饮用水的理想消毒剂之一，因为它杀菌能力强且不影响水质B．因患“禽流感”而被捕杀的家禽尸体常用生石灰处理C．新冠病毒可以用体积分数为 75%的酒精进行消毒D．食品加工、消毒、防腐常使用福尔马林6. 下列说法不．正．确．的是A. 二氧化硫可用于漂白纸浆B. 热的纯碱溶液可以去除厨房里的油污C. 目前用于饮用水净化的含铝净水剂正逐步被含铁净水剂所取代D．钢铁因含杂质而容易生锈，所以合金一定不耐腐蚀7. 下列表示不正确的是A．16O2-离子结构示意图 B．CF4的球棍模型：C．氯化钠的化学式：NaCl D．乙炔的结构式：CH≡CH 8．下列有关硝酸的说法正确的是A．工业硝酸的质量分数约为 69%，常因溶有少量NO而略显黄色B．硝酸是一种氧化性很强的酸，可以与金、铂等金属反应C．将木炭放入浓硝酸中，可立即观察到有气泡生成D．工业制备硝酸第三步为用水吸收二氧化氮生成硝酸9. 下列说法正确的是A．纸层析法时，亲脂性成分在固定相中分配的多一些，随流动相移动的速度快一些B．溶解度越小，溶液浓度越大，溶剂蒸发的越快，溶液冷却的越快，析出的晶体越小C．吸滤瓶内与液体快到支管口时，应拔掉橡皮管，从支管口将液体倒出D. 可用亚硝酸钠和硝酸银溶液检验氯酸钾中的氯酸根离子10. 化学实验设计和操作中必须十分重视安全和环境保护问题。

2023学年第二学期杭州市高三年级教学质量检测语文试题卷考生须知：1.本试卷分试题卷和答题卡，满分150分，考试时间150分钟。

2.所有答案必须写在答题卡上，写在试题卷上无效。

一、现代文阅读（35分）（一）现代文阅读Ⅰ（本题共5小题，18分）阅读下面的文字，完成下面小题。

材料一：达尔文雀隶属于雀形目燕雀科，一共14种，其中13种分布在加拉帕戈斯群岛上，另一种分布在距加拉帕戈斯群岛600km的可可岛上：达尔文雀羽毛颜色均为暗色，体形相似，体长7-12cm不等，种间最明显的区别是喙部的形状和大小。

据考证，这14种达尔文雀是在过去的100万年至300万年间由同一祖先进化而来。

1835年9月，达尔文乘“贝格尔”号航行到加拉帕戈斯群岛时，在岛上发现一些羽毛颜色暗淡的雀形目鸟类，并采集了标本带回英国。

后来，英国鸟类分类学家G ould在研究达尔文收集的鸟类标本时发现这些雀形目鸟类是一些以前没有描述过的新种。

达尔文也因此受到了启发，在《物种起源》中论述到：“这境。

”这些加拉帕戈斯群岛上的雀形目小鸟促使达尔文产生了生物进化的思想，后人为了纪念达尔文就把这些雀形目小鸟称为达尔文雀。

（摘编自邓文洪、郑光美《达尔文雀与生物进化》）材料二：1938年12月，28岁的英国帅小伙大卫·拉克和他的研究团队登上了圣克里斯托巴尔岛，研究不同种“达尔文雀”的繁殖和觅食行为。

每天上午他外出观察这些小鸟，下午则捕捉个体尝试进行圈养，看不同种之间是否会发生杂交。

正如达尔文曾指出的那样，“达尔文雀”非常温顺，不怕人而易于接近。

这些“很傻很天真”的鸟，是拉克在野外非常难得的理想观察对象。

1939年4月，拉克一行和4种共计40只地雀一起来到美国旧金山的加州科学院。

从4月底至9月初，拉克在加州科学院、加州大学伯克利分校比较动物学博物馆、哈佛大学比较动物学博物馆等地的馆藏做研究，还专程去大英博物馆检视了达尔文当年采集的标本，最后他竟总共测量了近6400号“达尔文雀”标本！根据掌握的翔实资料，拉克很快撰写出了题为《加拉帕戈斯地雀亚科形态变异研究》的专著。

信息技术学业水平考试选择题复习一班级：学号：姓名：（注：请同学认真作答，如遇难点，可参考《学业水平导引》和教科书）1.下图是用音频解霸软件播放歌曲"祝福.mp3"时的界面，从图中可以看出当前歌曲正播放到……（）A. 53秒B. 1分46秒C. 2分53秒D. 4分27秒2.十进制数4与二进制数1101相乘的积是………………………………………………………………（）A.（11010）2B.（10100）2C.（110100）2D.（101100）23.以下文件中能使用Word软件打开编辑的是……………………………………………………………（）A．books.mdb B．秋天不回来.wav C．成绩.xls D．雨巷.doc4.小张想给自己的电脑安装一款杀毒软件，并通过各种途径获得了下面几款软件，请根据软件的相关信息帮助小张做出正确评估，他应选择安装……………………………………………………………………（）A.某网站下载的KV 2008 破解版B.瑞星的官方网站下载的瑞星2008 免费试用版C.朋友处借来正版金山毒霸2008D.某博客上下载的自编超强杀毒软件5.小王用UltraEdit软件观察字符的内码，如第5题图所示，下列说法中正确的是…………………（）A.“同一个梦想”字符在计算机内存储和处理时使用的是ASCII码（中文存储用GB国标码）B.“One dream”字符在计算机内存储时，每个字符占2B （英文数字存储用ASCII码）C.“O”的ASCII码是01001111B O=4F(十六进制)=01001111BD.“世界”内码的十六进制表示是BBB8H CACO BDEF H6.为了获得北京奥运圣火在绍兴传递的火炬手名单，王峰使用全文搜索引擎在网上查找，下列选项中最有效的关键词是………………………………………………………………………………………………（）A.奥运圣火B.圣火传递C.绍兴火炬手D.奥运圣火7.某航空公司的某次航班因“天气原因”临时推迟起飞时间，该航空公司需要及时通知已购票的乘客，下列方式中较合适的有………………………………………………………………………………………（）A.打电话 B.利用传统书信C.上门通知D.利用QQ、MSN等实时交流工具8.如第8题图所示为使用IE保存一个包含文字、图片等内容的网页，单击“保存”按钮后，下列相关描述中正确的是………………………………………………………………………………………………（）第8题图A.该网页中的文字内容将被保存在“huiha.网页”文件中B.该网页中的图片将被保存在“huiha.files”文件夹中C.使用当前保存类型，只能保存网页中的文字信息D.使用当前保存类型，只能保存网页中的图片9.超市收银员用扫描仪扫描顾客选购商品的条形码，然后在计算机中根据识别后的条形码信息在相应数据库中查找其对应商品的名称、价格等信息。

高三化学参考答案 第 1 页 共 2 页2019学年第一学期期中杭州地区(含周边)重点中学高三年级化学学科参考答案一、选择题（本大题共18小题，1-10小题，每题2分；11-18小题，每题3分，共44分）二、非选择题 19．（共8分）（1）第3周期第ⅠA 族（1分） （2）AE （2分）（3）碱性（1分） H 3PO 4（2分）（4）SiCl 4+4H 2O=H 4SiO 4+4HCl （2分，写硅酸也给分） 20．（共9分）（1）SO 3（1分）（2）2K 2S 2O 8=2K 2SO 4+2SO 3↑+O 2↑（2分） （3）BC （2分）（4）8H 2O+2Mn 2++5S 2O 82-=2MnO 4ˉ+10SO 42ˉ+16H +（2分） （5）AD （2分） 21．（共13分）（1）小于（1分） 2.02（1分）COOH*+H*+H 2O*COOH*+2H*+OH*（或H 2O*H*+OH*）（1分）（2）①反应Ⅰ的ΔH ＞0，反应Ⅱ的ΔH ＜0，温度升高使CO 2转化为CO 的平衡转化率上升，使CO 2转化为CH 3OCH 3的平衡转化率下降，且上升幅度超过下降幅度（2分） ②增大压强，使用对反应Ⅱ催化活性更高的催化剂（2分） （3）72.7%或8/11（2分）作图2分（右图）（4）3CO 2+4e −C+2CO 32ˉ（2分） 22．（共14分）（1）①a 、c （2分）②冷凝回流，减少环己醇蒸出，提高原料的利用率（1分）③甲及时将产物环己烯分离出去，有利于平衡正向移动，提高转化率（2分） （2）分液漏斗（1分）增大水层的密度，有利于分层，便于后续分液操作（2分） （3）④②⑤⑦⑥⑧（2分）浙江省学考选考高三化学参考答案 第 2 页 共 2 页（4）822000cv b a⎛⎫−⨯ ⎪⎝⎭（2分） （5）c 、d （2分） 23．（共12分）（1）羰基、羟基（2分） （2）（2分）（3）CD （2分）（4）（K 2CO 3做反应物生成KHCO 3或CO 2也给2分）（5）（2分）（6）（2分）浙江省学考选考。

2024届浙江省杭州地区高三第二次调研英语试卷考生须知：1．全卷分选择题和非选择题两部分，全部在答题纸上作答。

选择题必须用2B铅笔填涂；非选择题的答案必须用黑色字迹的钢笔或答字笔写在“答题纸”相应位置上。

2．请用黑色字迹的钢笔或答字笔在“答题纸”上先填写姓名和准考证号。

3．保持卡面清洁，不要折叠，不要弄破、弄皱，在草稿纸、试题卷上答题无效。

第一部分（共20小题，每小题1.5分，满分30分）1．---Where is my Chinese book? I remember I put it here yesterday.---You _________ it in the wrong place.A．must put B．should have putC．might have put D．might put2．Recently, a programmer criticized the “996” work schedule ______ employees work from 9 am to 9 pm, 6 days a week, with the prospect of ending up in an intensive care unit.A．where B．when C．which D．whose3．-- -____that he manage to get the information?---Oh，a friend of his helped him.A．Where was it B．What was itC．How was it D．Why was it4．An exhibition with 885 cultural relics to be displayed ________ at the Meridian Gate Gallery.A．was held B．would hold C．has hold D．will be held5．Anyone with an eye on the employment situation knew the assessment about economic recovery _______ just around the corner was correct.A．being B．to be C．was D．having been6．When they first came to the city, my parents often went to neighbors for a talk, just as they ________ in the countryside.A．will do B．had doneC．have done D．were doing7．—I am putting on weight again! Maybe I should start doing yoga.—You _______ that the whole morning!A．are saying B．have said C．have been saying D．were saying8．Many netizens are impressed with the excuse given by a teacher for quitting her job ______ she owes the world a visit. A．because B．that C．where D．why9．If you want to go further in the new sport, the best way is to ______ and practise more frequently.A．dive in B．drop outC．catch on D．spring up10．The manager is trying to find a man to recommend how the job .A．is done B．be done C．should done D．to do11．The two girls are getting on very well and share _______ with each other.A．little B．much C．some D．none12．—Hi, Tom! I got a chance to be an exchange student in Harvard University.—_________! I had been expecting to study there.A．Lucky you B．Have funC．Take it easy D．Forget it13．The professor _____about how to protect the endangered animal in the conference at this time tomorrow. A．talked B．talks C．has been talking D．will be talking14．—I have something important to tell John. But I can’t find him.—His cell phone is here, so he ________ have gone too far.A．mustn’t B．needn’tC．wouldn’t D．can’t15．––The small restaurant is always crowded in every part.––That’s ______ it has a unique dining environment and quite a few wonderfu l dishes.A．why B．becauseC．where D．when16．As is often the case, there are always some obstacles in the way，something ________ before we realize the real goal of education.A．to be got through B．got throughC．getting through D．having been got through17．Everything is amazing. Thank you all. We without your help.A．can’t make it B．mustn’t have madeC．won’t make D．couldn’ t have made it18．Creating an atmosphere ______ employees feel part of a team is a big challenge.A．where B．whoseC．that D．which19．The possibility that Frank was lying ______ through my mind.A．swallowed B．masked C．flashed20．The old woman who ________ in the deserted house alone for ten years has been settled in a nursing home now. A．lived B．has livedC．had lived D．has been living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。

2024届浙江省杭二中高三第二次诊断性检测生物试卷注意事项1．考试结束后，请将本试卷和答题卡一并交回．2．答题前，请务必将自己的姓名、准考证号用0．5毫米黑色墨水的签字笔填写在试卷及答题卡的规定位置．3．请认真核对监考员在答题卡上所粘贴的条形码上的姓名、准考证号与本人是否相符．4．作答选择题，必须用2B铅笔将答题卡上对应选项的方框涂满、涂黑；如需改动，请用橡皮擦干净后，再选涂其他答案．作答非选择题，必须用05毫米黑色墨水的签字笔在答题卡上的指定位置作答，在其他位置作答一律无效．5．如需作图，须用2B铅笔绘、写清楚，线条、符号等须加黑、加粗．一、选择题：（共6小题，每小题6分，共36分。

每小题只有一个选项符合题目要求）1．细胞衰老和干细胞耗竭是机体衰老的重要标志，转录激活因子YAP是发育和细胞命运决定中发挥重要作用的蛋白质。

研究人员利用CRISPR/Cas9基因编辑技术和定向分化技术来产生YAP特异性缺失的人胚胎干细胞，YAP缺失的干细胞会表现出严重的加速衰老。

下列叙述错误的是（）A．推测YAP在维持人体干细胞年轻态中发挥着关键作用B．转录激活因子YAP的合成需要在核糖体上进行C．YAP缺失的干细胞的形态结构和功能会发生改变D．CRISPR/Cas9基因编辑过程中，基因中的高能磷酸键会断裂2．2020年全球爆发了新冠肺炎疫情。

新冠肺炎由新型冠状病毒感染所致，如图表示新型冠状病毒的结构模式图，该病毒的遗传物质为单链RNA，其通过刺突糖蛋白（S）与宿主细胞表面受体血管紧张素转化酶II（ACE2）专一性结合进入宿主细胞，在宿主细胞内完成增殖。

下列相关叙述错误的是（）A．新型冠状病毒的组成有RNA和蛋白质，但其不属于生命系统的结构层次B．该病毒容易发生变异是因为RNA中A与U碱基配对时形成的氢键太少C．具有ACE2受体的细胞可将该病毒作为自己需要的成分吞入细胞内部D．抑制该病毒表面的刺突糖蛋白或细胞ACE2受体活性可能会阻止该病毒的入侵3．某家系中有甲、乙两种单基因遗传病（如下图），其中一种是伴性遗传病。

浙江省杭州第二中学2024年高三考前热身物理试卷注意事项：1．答题前，考生先将自己的姓名、准考证号填写清楚，将条形码准确粘贴在考生信息条形码粘贴区。

2．选择题必须使用2B铅笔填涂；非选择题必须使用0．5毫米黑色字迹的签字笔书写，字体工整、笔迹清楚。

3．请按照题号顺序在各题目的答题区域内作答，超出答题区域书写的答案无效；在草稿纸、试题卷上答题无效。

4．保持卡面清洁，不要折叠，不要弄破、弄皱，不准使用涂改液、修正带、刮纸刀。

一、单项选择题：本题共6小题，每小题4分，共24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、如图所示，理想变压器原副线圈匝数之比为10︰1，原线圈两端连接正弦交流电源u=311sin314t(V)，副线圈接电阻R，同时接有理想电压表和理想电流表。

下列判断正确的是（）A．电压表读数约为31.1VB．若仅将副线圈匝数增加到原来的2倍，则电流表的读数增大到原来的2倍C．若仅将R的阻值增加到原来的2倍，则输入功率也增加到原来的2倍D．若R的阻值和副线圈匝数同时增加到原来的2倍，则输出功率增加到原来的4倍2、一个单摆在海平面上的振动周期是T0，把它拿到海拔高度很高的山顶上，该单摆的振动周期变为T，关于T与T0的大小关系，下列说法中正确的是（）A．T=T0B．T>T0C．T<T0D．无法比较T与T0的大小关系3、1960年10月第十一届国际计量大会确定了国际通用的国际单位制，简称SI制。

国际单位制共有七个基本单位，其中力学单位制中的3个基本单位是（）①kg ②③N ④m ⑤s ⑥⑦m/s2⑧A．①④⑤B．①③④C．②③⑦D．⑤⑥⑧4、一正三角形导线框ABC（高度为a）从图示位置沿x轴正向匀速穿过两匀强磁场区域。

两磁场区域磁感应强度大小均为B、方向相反、垂直于平面、宽度均为a。

下图反映感应电流Ⅰ与线框移动距离x的关系，以逆时针方向为电流的正方向。

图像正确的是（）A ．B ．C ．D ．5、如图所示，一磁感应强度为B 的圆形匀强磁场区域，圆心为O ，半径为r ，MN 是直径，一粒子发射装置S 置于M 端，可从M 端向圆平面内任意方向发射速率相等的同种带电粒子，某个粒子从N 端离开磁场，在磁场中运动的时间为2kB π，其中k 为带电粒子的比荷，下列说法正确的是（ ）A ．该粒子的速率为krB ，发射方向垂直于MNB 2krB ，发射方向与MN 的夹角为45°C ．该粒子在磁场中运动的时间最短D ．若该粒子沿直径MN 方向射入磁场，其运动的时间为3kB π6、下列关于原子核的说法正确的是（ ）A ．质子由于带正电，质子间的核力表现为斥力B ．原子核衰变放出的三种射线中，α粒子的穿透能力最强C ．铀核发生链式反应后能自动延续下去，维持该反应不需要其他条件D ．比结合能小的原子核结合成比结合能大的原子核时一定放出核能二、多项选择题：本题共4小题，每小题5分，共20分。

2017 学年第二学期杭州市高三年级教学质量检测英语试题卷考生须知：本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）。

满分150分，考试用时120分。

第Ⅰ卷第一部分：听力略第二部分：阅读理解（共两节，满分35分）第一节（共10个小题，每小题2.5分，满分25分）AIt is halfway through October. The sun arrives early and I stare out my window, feeling shame for invading (侵犯）the privacy of the homeless woman who has settled across the street. I am not sure when she first appeared, maybe a month ago. Each morning I watch as she gets out from the narrow space between the two large buildings across from mine.She wears the same dirty, badly torn dress every day. I can see she fights to pull herself up into a standing position. She fails in the first two or three tries, but determination always wins out. She remains leaning against the wall for several long minutes, appearing to gather the necessary strength needed to pick up her possession--all packed into five plastic bags. With tears in my eyes, I watch as she places her feet pointing outward to avoid falling. Her steps are measured and cautious as she slowly makes her way to the corner and disappears.When I first noticed this woman, I went across the street and offered her 20 dollars, but she turned her back to me and pretended not to see me. The expression on her face was one of shame, causing me to feel guilt for invading her privacy. The next night I bought a hotdog and a Coke, and left them in the spot when she was not there. Feeling like a spy, I watched as she ate the hotdog like a starving cat that had not eaten for days.I too often find myself complaining about life, but when I look out my window and see the old woman, I feel the need to apologize for my foolish pride in not realizing how fortunate I really am. There are many poor people in my town. I have become familiar with many of their faces, and then suddenly they disappear. I know that one day the old woman will be gone and I will have no idea where she is. I fear that day.21.What is the main idea of paragraph 2 ?A. Life is tough for the homeless woman.B.Poverty causes the woman’s poor health.C.The woman doesn’t care about her appearance.D.Possessions mean much to the woman.22.Why did the woman refuse the author’s 20 dollars?A. The money didn’t help much.B. She didn’t like the author.C. Accepting money was guilty.D. She felt he dignity was hurt.23.What lesson does the author learn from the homeless woman?A. Complaints can sometimes make life worse.B. He should appreciate his present life more.C. More is to be done to make life meaningful.D. We should all apologize for ignoring the poor.24.Which of the following words best describe the author of the text?A. OptimisticB. SympatheticC. DeterminedD. Easy-goingBA laptop, a hair straightener, Christmas lights, an e-reader, a kettle, two bags, a pair of jeans, a remote-control helicopter, a spoon, and a dining-room chair. All broken.It sounds like a pile of things that you’s stuck in boxes and take to the dustbin. In fact, it’s a list of things mended in a single afternoon by British volunteers determined to get people to stop throwing things away. This is the Reading Repair Cafe, part of an international network aimed at dealing with a world of rubbish.The helicopter belongs to William, who cheerfully describes himself as “ mechanically imcompetent”. He has owned it for 8 years, but 3 years ago it stopped working and it has been sitting unused in his cupboard ever since.He sits down at the table of Colin Haycock, an IT professional who volunteers at the repair cafe, which has been running for about four years and is a place where people can bring all manner of household items to be fixed for free. In less than five minutes, Haycock has fixed the helicopter and it works well.William looks ashamed;Haycock looks pleased.“I wish they were all that simple,” he says. Today, the repairs will stop 24 kg of waste from going to landfill (垃圾填埋场）and save 284kg of CO2. Some items can’t be fixed on the spot but very little needs to be thrown away.Gabrille Stanley, who used to run a clothing business, says she was drawn to volunteering at the repair cafe to fight the “throwaway culture” she sees. About 300,000 tons of clothing was sent to landfill in UK in 2016 and a report from WRAP puts the average lifespan (寿命）for a piece of clothing in the UK at 3.3 years.25.What is the author’s purpose in writing paragraph 1?A. To present the main idea.B.To support the author’s opinion.C.To raise reader’s interest.D.To establish an argument.26.Why doesn’t William repair the helicopter himself?A.He is both lazy and busy.B. He is ashamed of repair work.C. He doesn’t treasure the helicopter.D. He lakes the necessary skills.27.Which of the following argument will Cabrille Stanley probably agree with?A. People should send old clothing to landfill.B. People should hold on to their clothes longer.C. Clothing factories are to blame for the waste.D. Running a clothing business is no easy task.CThe descriptions are scary. Some people felt fine in the morning and then were dead by night. Patients coughed up blood. Their faces turned blue. That was what happened in 1918. A deadly flu killed ten of millions of people around the world. It is 100 years later. Scientists are thinking about how to keep us safe from another super-flu.Dr. Anthony Fauci says we need one vaccine （疫苗）that can protect against most or all types of the flu. Scientists want a shot that people could get every 5 or 10 years to get rid of normal flu shots. Maybe someday we will need just one shot. It could last our whole lives. It will not be easy. Scientists have been learning about flu for 100 years. Still, the flu often beats our best plans tostop it. That is because it is always changing.The immune system (免疫系统）protects the body. It keeps us healthy by fighting things that might harm the body. To get around our immune system, the flu changes itself every year. Scientists have a new plan, though. They are learning how the flu tricks the immune system. They are finding which part of the flu stays the same each year. The 1918 flu outbreak shows why this is so important. Back then, there was no flu vaccine. There was no way to stop it. By winter 1919. the flu had infected one-third of the world’s population, killing more than 50 million people in a short time.In 2009, scientists made a great discovery. They learned something new. Sometimes, people’s immune system make a small number of special antibodies. Antibodies are proteins （蛋白质）in the blood.These proteins stop us from getting sick. These special antibodies can stop the flu. Scientists are trying different tricks to get people’s bodies to make more of those antibodies. They still need to learn more, though. In the meantime, Fauci says it is silly to think about what the next flu might bring. “We just need to be prepared,” he said.28.The flu tricks the immune system by__________________.A. changing itself every yearB. remaining totally unchangedC.producing special antibodiesD. harming proteins in the body.29.What can we learn from the text?A. The flu causes more deaths than it did 100 years ago.B. A flu vaccine shot with long-term effects is easy to make.C. The super flu will take about 5 or 10 years to get rid of.D. People don’t have to be too concerned about the flu.30.What is the main idea of the text?A. The deadly flu killed ten of millions of people in such a short time.B. Scientist are trying to figure out the best protecting against the flu.C. The immune system plays a vital part in keeping us from being sick.D. People around the world are scared of the sudden outbreak of the flu.第二节（共5小题;每小题2分，满分10分）Why I Got Rid of My Cell PhoneIn 2017 I decided to take a break from my cell phone. Instead of shutting it off completely, I decided to turn it off a month to see how it goes. By the end of the month I found that I could live without a cell phone. It has been seven months since I have lived without a cell phone. I would like to share with you the reasons why I decided to get rid of my cell phone.1. 31 Checking my phone was the first thing I did when I woke up and the last thing I did before going to sleep. Throughout the day I would check my phone countless times. It became a bad habit of mine in which the only way to break it was to be far away from it.2.When I owned a phone, I was constantly connected to it. I wasn’t taking enough time to see what was right in front of me. There was no end of the day when it came ti work and there was not nearly enough time to relax. Instead, I was constantly going. 323. 33 If I was having a conversation with someone, I would think about what else I could be doing or texting other people instead of being fully present in the conversation. Now thatI don’t have a cell phone, I am more present in my conversations and to my surroundings.4. I was completely dependent on my cell phone. 34 A cell phone was my quick and easy problem solver. Whatever problems I met during the day, I used my cell phone for the answers. Without my cell phone I have become more self-dependent and self-problem solver/5. I also found that my sleep wasn’t as good as it had been in the past, Screens stimulate（刺激）your mind and can affect your natural sleep patterns 35 However, I did not have the self-control to do this so I would go to bed with a stimulated mind causing me to have a poor night sleep.A. It made me less social.B. I found that I was addicted to my phone.C. Having a cell phone split my attention in half.D. I know that is not good for my health.E. The answer to this is to turn off your phone a few hours before bed.F. These phone-free travels were some of the most liberating moments of my life.G. It was my watch, my alarm clock, my email, my maps. my wallet, and my way to talkwith friends.第三部分；语言运用第一节完形填空（共20题小题，每小题1.5分，满分30分）Mariana was just leaving a doctor’s appointment lase Wednesday. All of a sudden, she found herself in the middle of what felt like a 36 in a movie. Mariana had been returning to he office in downtown Ottawa when she heard a woman down the street yell that a man had 37 her wallet.Sure enough, a man started running away from the scene of 38 -but not before Mariana could 39 after him. After running for two blocks, Mariana rounded the corner into an alleyway and found that 40 had stopped running and had started to 41 instead.“He came out from behind the dustbin and said in a very 42 tone, ‘ I’m sorry, I won’t do this anymore. Here is the 43 . Just take it, take it ,’” she told the local media. “So I took the wallet, and the woman 44 soon after. I gave it back to her and he stayed there. 45 a lot. ” Recognizing that the man was extremely 46 , Mariana decided to take him to a 47 down the street and bought him a large black coffee.The man 48 that he had come to the city with his friends from Toronto, but they had 49 him in Ottawa without any money. Mariana did her best to 50 that men before showing him how to get to the public library where he could seek help from the 51 workers on duty. She says that she 52 that the man gets the help that he needs, and she has no 53 about her actions from the day.“You kill more flies with 54 than you do with vinegar,” she said, “ I wanted to show him some 55 . It is a more powerful remedy （治疗方法）to forgive and forget past crimes.”36.A. song B. scene C. disaster D. party37.A. check B. returned C. stolen D. spotted38.A. crime B. accident C. match D. quarrel39.A. line up B. get on C. stand up D. take off40.A. victim B. waiter C. burglar D. policeman41.A. struggle B. cry C. flee D. pray42.A. guilty B. aggressive C. friendly D. innocent43.A. coffee B. ticket C. menu D. wallet44.A. left B. woke C. caught up D. broke down45.A. complaining B. apologizing C. suffering D. charging46.A. upset B. calm C. stubborn D. gentle47.shelter B. clinic C. station D. cafe48.A. insisted B. announced C. explained D. intended49.A. reminded B. deserted C. tease D. arranged50.A. comfort B. connect C. approach D. thank51.A. retired B. skilled C. medical D. social52.A. suspects B. admits C. expects D. understand53.A. debates B. regret C. confidence D. worry54.A. honey B. wine C. poison D. glue55.A. courage B. appreciation C. reflection D. sympathy第Ⅱ卷第二节（共10个小题，每小题1.5分，满分15分）The first bicycle appeared in France in the 1790s. It was 56 little wooded horse with a fixed front wheel. Because the wheel was fixed, it could not be turned right or left. The only way it could be controlled was the rider 57 ( push ) against the ground with his or her feet. In 1817, a German replaced the fixed wheel with one 58 could be turned. Now the wooden horse could be directed right or left. The rider still needed to push it with his or her feet on the ground. In 1839, a Scot 59 ( name ) MacMillan designed the first bicycle-like machine with foot board and rode it for 40 miles in only 5 hours. The bicycle got more 60 ( comfort ) in 1869 when rubber tires were introduced. Around the same time, the front wheel began to grow larger while the back wheel got smaller. During the 1880s, 61 ( bicycle )enjoyed a sudden 62 ( grow )in popularity. However, they weren’t very safe. Sitting high up towards the front of the bicycle and traveling very fast, the rider could be 63 ( easy ) thrown over the front wheel if the bicycle hit a small stone. Fortunately, the “ safety bicycle ” 64 ( invent ) in 1884. The safety bicycle had equal-sized wheels and tires with air in them. And more improvements quickly followed. No doubt there will be 65 ( far ) improvement in design and materials in the future.第四部分；写作（共两节，满分40分）第一节应用文写作（满分15分）假定你是李华，注意到你们学校学生很少进行英语文学阅读。

杭州二中2022学年第二学期高三年级3月考试数学试卷第I 卷（选择题）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.已知集合{}(){}214,ln 4A x x B x y x =-≤≤==-，则A B ⋃=（）A.(,1][2,)-∞-+∞B.[1,2)- C.[1,4]- D.(2,4]-2.已知复数()2iR 1ib z b +=∈-的实部为1-，则b 的值为（）A.2B.4C.2-D.4-3.已知圆锥的侧面展开图是一个半径为4，弧长为4π的扇形，则该圆锥的表面积为（）A.4πB.8πC.12πD.20π4.2022年10月22日，中国共产党第二十次全国代表大会胜利闭幕．某班举行了以“礼赞二十大、奋进新征程”为主题的联欢晚会，原定的5个学生节目已排成节目单，开演前又临时增加了两个教师节目，如果将这两个教师节目插入到原节目单中，则这两个教师节目相邻的概率为（）A.16 B.17C.13D.275.已知OAB ，1OA =，2OB =，1OA OB ⋅=-，过点O 作OD 垂直AB 于点D ，点E满足12OE ED = ，则EO EA ⋅的值为（）A.328-B.121-C.29-D.221-6.已知1132,5,(2)e a b c e ===+，则,,a b c 的大小关系为（）A .b<c<aB.c b a <<C.b a c<< D.c<a<b7.已知1F ，2F 是椭圆和双曲线的公共焦点，P 是它们的一个公共点，且12π4F PF ∠=，则椭圆和双曲线的离心率乘积的最小值为（）A.B.22C. D.28.已知在矩形ABCD 中，2AB =，4=AD ，E ，F 分别在边AD ，BC 上，且1AE =，3BF =，如图所示，沿EF 将四边形AEFB 翻折成A EFB ''，设二面角B EF D '--的大小为α，在翻折过程中，当二面角B CD E '--取得最大角，此时sin α的值为（）A.35B.45C.223D.13二、选择题：本题共4小题，每小题5分，共20分.在每小题给出的选项中，有多项符合题目要求.全部选对的得5分，部分选对的得2分，有选错的得0分.9.下列说法正确的是（）A.用简单随机抽样从含有50个个体的总体中抽取一个容量为10的样本，个体m 被抽到的概率是0.2B.已知一组数据1，2，m ，6，7的平均数为4，则这组数据的方差是5C.数据27，12，14，30，15，17，19，23的50%分位数是17D.若样本数据1x ，2x ，…，10x 的标准差为8，则数据121x -，221x -，…，1021x -的标准差为1610.已知函数()()()sin cos cos sin f x x x =+，下列关于该函数结论正确的是（）A.()f x 的图象关于直线π2x =对称 B.()f x 的一个周期是2πC.()f x 的最大值为sin11+ D.()f x 是区间3ππ,2⎛⎫⎪⎝⎭上的减函数11.已知正四棱锥P ABCD -的所有棱长均为E ，F 分别是PC ，AB 的中点，M 为棱PB 上异于P ，B 的一动点，则以下结论正确的是（）A.异面直线EF 、PD 所成角的大小为3π B.直线EF 与平面ABCD 所成角的正弦值为66C.EMF +D.存在点M 使得PB ⊥平面MEF12.已知定义域为R 的函数()f x 在(]1,0-上单调递增，()()11f x f x +=-，且图像关于()2,0对称，则()f x （）A.()()02f f =-B.周期2T =C.在()2,3单调递减D.满足()()()202120222023f f f >>第II 卷（非选择题）三、填空题：本题共4小题，每题5分，共20分.13.已知抛物线E ：()220x py p =>的焦点为F ，过点F 的直线l 与抛物线交于,A B 两点，与准线交于C 点，F 为AC 的中点，且3AF =，则p =__________.14.在6()x a +的展开式中的3x 系数为160，则=a _______．15.已知正实数,a b 满足()3386311a a b b +≤+++，则23a b +的最小值是___________.16.函数2()2e x f x a bx =++，其中,a b 为实数，且(0,1)a ∈.已知对任意23e b >，函数()f x 有两个不同零点，a 的取值范围为____________.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.已知,,a b c 分别为ABC 内角,,A B C 的对边，若ABC 同时满足下列四个条件中的三个：①a =2b =；③sin sin sin ++=-B C a c A b c ；④21cos sin sin 24-⎛⎫-= ⎪⎝⎭B C B C ．（1）满足有解三角形的序号组合有哪些？（2）请在（1）所有组合中任选一组，求对应ABC 的面积．18.已知数列{}n a 满足22113,2221++==+-++n n n a a a n n ．（1）求证：22⎧⎫-⎨⎩⎭n na n 是等差数列；（2）令2⎡⎤=⎢⎥⎣⎦nn n a b （[]x 表示不超过x 的最大整数．提示：当a ∈Z 时，[][]a x a x +=+），求使得12100n b b b ++≤+L 成立的最大正整数n 的值．19.如图，四棱锥P -ABCD 的底面为梯形，PD⊥底面ABCD ，90BAD CDA ∠=∠=︒，1AD AB ==，2CD =，E 为PA 的中点．（1）证明：平面PBD ⊥平面BCE ；（2）若二面角P -BC -E 的余弦值为265，求三棱锥P -BCE 的体积．20.法国数学家庞加莱是个喜欢吃面包的人，他每天都会到同一家面包店购买一个面包.该面包店的面包师声称自己所出售的面包的平均质量是1000g ，上下浮动不超过50g .这句话用数学语言来表达就是：每个面包的质量服从期望为1000g ，标准差为50g 的正态分布.（1）已知如下结论：若()2,X Nμσ ，从X 的取值中随机抽取()*,2k k N k ∈≥个数据，记这k 个数据的平均值为Y ，则随机变量2,Y N k σμ⎛⎫~ ⎪⎝⎭.利用该结论解决下面问题.（i ）假设面包师的说法是真实的，随机购买25个面包，记随机购买25个面包的平均值为Y ，求()980P Y ≤；（ii ）庞加莱每天都会将买来的面包称重并记录，25天后，得到的数据都落在()950,1050上，并经计算25个面包质量的平均值为978.72g .庞加莱通过分析举报了该面包师，从概率角度说明庞加莱举报该面包师的理由；（2）假设有两箱面包（面包除颜色外，其他都一样），已知第一箱中共装有6个面包，其中黑色面包有2个；第二箱中共装有8个面包，其中黑色面包有3个.现随机挑选一箱，然后从该箱中随机取出2个面包.求取出黑色面包个数的分布列及数学期望.附：①随机变量η服从正态分布()2,N μσ，则()0.6827P μσημσ-≤≤+=，()()220.9545,330.9973P P μσημσμσημσ-≤≤+=-≤≤+=；②通常把发生概率小于0.05的事件称为小概率事件，小概率事件基本不会发生.21.已知抛物线21:C y x =，开口向上的抛物线2C 与1C 有一个公共点(2,4)T ，且在该点处有相同的切线，（1）求所有抛物线2C 的方程；（2）设点P 是抛物线2C 上的动点，且与点T 不重合，过点P 且斜率为k 的直线l 交抛物线1C 于,A B 两点，其中PA PB ≥，问是否存在实常数k ，使得PA PB为定值？若存在，求出实常数k ；若不存在，说明理由.22.已知221ln ,0(),0x x x x f x e x --⎧->=⎨≤⎩.（1）当(0,)x ∈+∞时，求()f x 的最大值；（2）若存在[0,)a ∈+∞使，得关于x 的方程2()0f x ax bx ++=有三个不相同的实数根，求实数b 的取值范围.杭州二中2022学年第二学期高三年级3月考试数学试卷第I 卷（选择题）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.已知集合{}(){}214,ln 4A x x B x y x =-≤≤==-，则A B ⋃=（）A.(,1][2,)-∞-+∞B.[1,2)- C.[1,4]- D.(2,4]-【答案】D 【解析】【分析】根据对数函数的定义域，先求出集合B ，然后利用并集的运算即可求解.【详解】因为集合22{|ln(4)}{|40}{|22}B x y x x x x x ==-=->=-<<，又因为集合{|14}A x x =-≤≤，由并集的概念可知，{|24}(2,4]A B x x =-<≤=- ，故选：D .2.已知复数()2iR 1ib z b +=∈-的实部为1-，则b 的值为（）A.2B.4C.2- D.4-【答案】B 【解析】【分析】先利用复数的四则运算得出(2)(2)i2b b z -++=，然后根据题意即可求解.【详解】复数2i (2i)(1i)(2)(2)i1i (1i)(1i)2b b b b z +++-++===--+，因为复数()2iR 1ib z b +=∈-的实部为1-，所以22b -=-，则4b =，故选：B .3.已知圆锥的侧面展开图是一个半径为4，弧长为4π的扇形，则该圆锥的表面积为（）A.4πB.8πC.12πD.20π【答案】C 【解析】【分析】圆锥的侧面展开图是一个半径为4，弧长为4π的扇形，可知底面圆的半径，再求的底面圆的面积和圆锥的侧面积，即可求得该圆锥的表面积.【详解】由于圆锥的侧面展开图是一个半径为4，弧长为4π的扇形，则圆锥底面圆的半径为4π22πr ==，底面圆的面积为22ππ24πr =⨯=，圆锥的表面积为14π44π12π2⨯⨯+=.故选：C.4.2022年10月22日，中国共产党第二十次全国代表大会胜利闭幕．某班举行了以“礼赞二十大、奋进新征程”为主题的联欢晚会，原定的5个学生节目已排成节目单，开演前又临时增加了两个教师节目，如果将这两个教师节目插入到原节目单中，则这两个教师节目相邻的概率为（）A.16 B.17C.13D.27【答案】D【解析】【分析】先插入第一个节目，再插入第二个节目，再按照分步乘法计数原理分别计算插入的情况数量及这两个教师节目恰好相邻的情况数量,再应用古典概率公式求概率即可.【详解】由题意可知，先将第一个教师节目插入到原节目单中，有6种插入法，再将第二个教师节目插入到这6个节目中，有7种插入法，故将这两个教师节目插入到原节目单中，共有6742⨯=（种）情况，其中这两个教师节目恰好相邻的情况有2612⨯=（种），所以所求概率为122427=.故选:D.5.已知OAB ，1OA =，2OB =，1OA OB ⋅=-，过点O 作OD 垂直AB 于点D ，点E满足12OE ED = ，则EO EA ⋅的值为（）A.328-B.121-C.29-D.221-【答案】D 【解析】【分析】作出图形，由平面向量数量积的定义及余弦定理可得OD =，再由平面向量数量积的运算律即可得解.【详解】由题意，作出图形，如图，1OA = ，2OB =，1OA OB ⋅=-12cos 2cos 1OA OB AOB AOB ∴⋅=⨯∠=∠=- ，1cos 2AOB ∴∠=-，由()0,AOB π∠∈可得23AOB π∠=，AB ∴==又113sin 222AOB S OA OB AOB OD AB =⋅⋅⋅∠=⋅⋅=△，则OD =()222232299721EO EA OE ED DA OE OD ∴⋅=-⋅+=-=-⋅=-⨯=- .故选：D ．6.已知1132,5,(2)e a b c e ===+，则,,a b c 的大小关系为（）A.b<c<aB.c b a <<C .b a c<< D.c<a<b【答案】A 【解析】【分析】化简由题意，可得11132(22),(23),(2)ea b c e =+=+=+，构造()()1ln 2f x x x=⋅+，得到则()()2ln 22xx x f x x-+'+=，再令()()ln 22x g x x x =-++，求得函数的单调性，结合单调性，即可求解.【详解】由题意，可得11132(22),(23),(2)ea b c e =+=+=+，所以令()()1ln 2,(0)f x x x x=⋅+>，则()()2ln 22xx x f x x -+'+=，令()()ln 2,(0)2xg x x x x =-+>+，则()20(2)x g x x +'-=<，所以()g x 在()0,∞+上单调递减，()()00g x g <=，所以()0f x '<恒成立，所以()f x 在()0,∞+上单调递减，因为23e <<，所以()()()23f f e f >>，即()()()111ln 22ln 2ln 2323e e +>+>+，所以11132ln(22)ln(2)ln(23)ee +>+>+，所以111324(2)5ee >+>，即b<c<a .故选：A.7.已知1F ，2F 是椭圆和双曲线的公共焦点，P 是它们的一个公共点，且12π4F PF ∠=，则椭圆和双曲线的离心率乘积的最小值为（）A.B.22C. D.2【答案】B 【解析】【分析】根据双曲线以及椭圆的定义可得112||PF a a =+，212||PF a a =-，进而在焦点三角形中运用余弦定理即可得2212224e e +=，结合均值不等式即可求解.【详解】如图，设椭圆的长半轴长为1a ，双曲线的半实轴长为2a ，则根据椭圆及双曲线的定义：121||||2PF PF a +=，122||||2PF PF a -=，112||PF a a ∴=+，212||PF a a =-，设12||2F F c =，12π4F PF ∠=，则：在△12PF F 中由余弦定理得，22212121212π4()()2()()cos4c a a a a a a a a =++--+-，化简得：22212(2(24a a c ++=，即2212224e e +=，又221212222212·e e e e ++≥，∴121e e ≤12·2e e ≥，即椭圆和双曲线的离心率乘积的最小值为22．故选：B8.已知在矩形ABCD 中，2AB =，4=AD ，E ，F 分别在边AD ，BC 上，且1AE =，3BF =，如图所示，沿EF 将四边形AEFB 翻折成A EFB ''，设二面角B EF D '--的大小为α，在翻折过程中，当二面角B CD E '--取得最大角，此时sin α的值为（）A.35B.45C.23D.13【答案】B 【解析】【分析】过B 作EF 的垂线交EF 与O ，交AD 于M ，CD 于G ，然后利用定义法可得B KH '∠为二面角B CD E '--的平面角，设B OH α'∠=，可得2B H α'=，53cos 22HK α=-，从而sin tan 3253cos B H B KH HK αα''∠==-，然后求函数最大值时的sin α值即可.【详解】过B 作EF 的垂线交EF 与O ，交AD 于M ，CD 于G ，设B '在平面AC 内的投影为H ，则H 在直线BM 上，过H 作CD 的垂线，垂足为K ，则B KH '∠为二面角B CD E '--的平面角，设B OH α'∠=，由题意2B O BO '==sin 2B H B O αα''==，则cos cos )2BH BO B O αα'=++，由45GBC ∠=︒，42BG =，得42cos )2HG BG BH α=-=+，所以3534(1cos )cos 2222HK αα==-+=-，所以sin tan 3253cos B H B KH HK αα''∠==-，令sin 53cos t αα=-，可得2sin 3cos 519t t t αα+=≤+，则14t ≤，所以，当14t =即sin 153cos 4αα=-，也即4sin 5α=时，tan B KH ∠'取到最大值324，此时B KH '∠最大，即二面角B CD E '--取得最大角.故选：B二、选择题：本题共4小题，每小题5分，共20分.在每小题给出的选项中，有多项符合题目要求.全部选对的得5分，部分选对的得2分，有选错的得0分.9.下列说法正确的是（）A.用简单随机抽样从含有50个个体的总体中抽取一个容量为10的样本，个体m 被抽到的概率是0.2B.已知一组数据1，2，m ，6，7的平均数为4，则这组数据的方差是5C.数据27，12，14，30，15，17，19，23的50%分位数是17D.若样本数据1x ，2x ，…，10x 的标准差为8，则数据121x -，221x -，…，1021x -的标准差为16【答案】AD 【解析】【分析】利用概率对于即可判断A ；根据平均数求得m 的值，然后利用方差公式求解即可判断B ；根据百分位数的求法即可判断C ；利用方差公式求解即可判断D.【详解】对于A ，一个总体含有50个个体，某个个体被抽到的概率为150，以简单随机抽样方式从该总体中抽取一个容量为10的样本，则指定的某个个体被抽到的概率为11100.2055⨯==，故A 正确；对于B ， 数据1，2，m ，6，7的平均数是4，4512674m =⨯----=，这组数据的方差是()()()()()222222114244464745s ⎡⎤=-+-+-+-+-⎣⎦=265，故B 错误；对于C ，8个数据50百分为850%4⨯=，第50百分位数为1719=182+，故C 错误；对于D ，依题意，()28D x =，则()()2221216D x D x -=⨯=，所以数据121021,21,,21x x x --⋯-的标准差为16，D 正确；10.已知函数()()()sin cos cos sin f x x x =+，下列关于该函数结论正确的是（）A.()f x 的图象关于直线π2x =对称 B.()f x 的一个周期是2πC.()f x 的最大值为sin11+ D.()f x 是区间3ππ,2⎛⎫⎪⎝⎭上的减函数【答案】BC 【解析】【分析】利用诱导公式判断()f x 与()πf x -是否相等判断A ，判断()f x 与()2πf x +是否相等判断B ，利用三角函数及复合函数的单调性判断CD.【详解】由()()()sin cos cos sin f x x x =+，对于A ，()()()()()()()()πsin cos πcos sin πsin cos cos sin f x x x x x f x -=-+-=-+≠，故A不正确；对于B ，()()()()()()()()2πsin cos 2πcos sin 2πsin cos cos sin f x x x x x f x +=+++=+=，故B 正确；对于C ，因为1cos 1x -≤≤，所以()sin cos y x =的最大值为sin1，当cos 1x =时，()cos sin cos 01y x ===，取得最大值，所以()f x 的最大值为sin11+，故C 正确；对于D ，()3ππ3ππsin1cos111110244f f ⎛⎫-=+-=+->-=⎪⎝⎭（），又函数连续，故D 错误；故选：BC11.已知正四棱锥P ABCD -的所有棱长均为E ，F 分别是PC ，AB 的中点，M 为棱PB 上异于P ，B 的一动点，则以下结论正确的是（）A.异面直线EF 、PD 所成角的大小为3πB.直线EF 与平面ABCD 所成角的正弦值为66C.EMF +D.存在点M 使得PB ⊥平面MEF【解析】【分析】根据空间中异面直线所成角，直线与平面所成角的定义，空间中折叠问题以及垂直关系的判定与性质，逐个选项运算求解即可．【详解】如图1，取PD 的中点Q ，连接EQ ，AQ ，因为E ，F 分别是PC ，AB 的中点，所以EQ DC AF ，且EQ AF =，所以四边形AFEQ 为平行四边形，则EF AQ ，又正四棱锥P ABCD -的所有棱长均为，则AQ PD ⊥，所以异面直线EF ，PD 所成角为π2，故A 错误；设正方形ABCD 的中心为O ，连接OC ，PO ，则PO ⊥平面ABCD ，2OC OP ==，设OC 的中点为H ，连接EH ，FH ，则EH OP ，且EH ⊥平面ABCD ，所以EFH ∠为直线EF 与平面ABCD 所成角，所以112EH PO ==，OFH 中，1OH =，OF =，135FOC ︒∠=，所以由余弦定理可得FH =EF ==，所以6sin6EH EFH EF ∠==，故B 正确；将正PAB 和PBC 沿PB 翻折到一个平面内，如图2，当E ，M ，F 三点共线时，ME MF +取得最小值，此时，点M 为PB 的中点，ME MF BC +==，所以EMF V +C 正确；若PB ⊥平面MEF ，则PB ME ⊥，此时点M 为PB 上靠近点P 的四等分点，而此时，PB 与FM 显然不垂直，故D 错误；12.已知定义域为R 的函数()f x 在(]1,0-上单调递增，()()11f x f x +=-，且图像关于()2,0对称，则()f x （）A.()()02f f =-B.周期2T =C.在()2,3单调递减D.满足()()()202120222023f f f >>【答案】AC 【解析】【分析】根据题意化简得到()()4f x f x =+，得到()f x 的周期为4T =,结合()()22f f -=，求得()()02f f =-，得到A 正确，B 错误；再由()f x 的对称性和单调性，得出()f x 在()2,3单调递减，可判定C 正确；根据()f x 的周期求得()()20211f f =，()()20222f f =，()()20233f f =，结合特殊函数()f x 的图象，可判定D 不正确.【详解】由()()11f x f x +=-，可得()f x 的对称轴为1x =，所以()()02,f f =又由()()11f x f x +=-知：()()2f x f x +=-，因为函数()f x 图像关于()2,0对称，即()()22f x f x +=--，故()()4f x f x +=--，所以()()24f x f x -+=+，即()()2f x f x -=+，所以()()4f x f x =+，所以()f x 的周期为4,所以()()22f f -=，所以()()02f f =-，故A 正确，B 错误；因为()f x 在(]1,0-上单调递增，且4T =，所以()f x 在(]3,4上单调递增，又图像关于()2,0对称，所以()f x 在(]0,1上单调递增，因为关于1x =对称，所以()f x 在(]1,2上单调递减，又因为关于()2,0对称，可得函数()f x 在()2,3单调递减，故C 正确；根据()f x 的周期为4T =，可得()()()()()()20211,20222,20233f f f f f f ===，因为关于1x =对称，所以()()20f f =且()()31f f =-，即()()()()()()20211,20220,20231f f f f f f ===-，由函数()f x 在(]1,2上单调递减，且关于1x =对称，可得()f x 在(]0,1上单调递增，如图所示的函数()f x 中，此时()()()()10,01f f f f -<>，所以()()()202120222023f f f >>不正确.故选：AC.【点睛】规律探求：对于函数的基本性质综合应用问题解答时，涉及到函数的周期性有时需要通过函数的对称性得到，函数的对称性体现的是一种对称关系，而函数的单调性体现的时函数值随自变量变化而变化的规律，因此在解题时，往往西药借助函数的对称性、奇偶性和周期性来确定另一区间上的单调性，即实现区间的转换，再利用单调性解决相关问题.第II 卷（非选择题）三、填空题：本题共4小题，每题5分，共20分.13.已知抛物线E ：()220x py p =>的焦点为F ，过点F 的直线l 与抛物线交于,A B 两点，与准线交于C 点，F 为AC 的中点，且3AF =，则p =__________.【答案】32##1.5【解析】【分析】利用抛物线的定义结合三角形中位线定理求解即可.【详解】设y 轴交准线于N ，过A 作准线的垂线，垂足为Q ，因为F 为AC 的中点，且3AF =，则由抛物线的定义可得3AQ =，在Rt ACQ 中，1322FN AQ ==，所以32p =，故答案为：3214.在6()x a +的展开式中的3x 系数为160，则=a _______．【答案】2【解析】【分析】首先求出6()x a +的展开项中3x 的系数，然后根据3x 系数为160即可求出a 的取值.【详解】由题知616r rr r T C xa -+=，当3r =时有333333466160160T C x a x C a ==⇒=，解得2a =.故答案为：2.【点睛】本题主要考查了二项式展开项的系数，属于简单题.15.已知正实数,a b 满足()3386311a a b b +≤+++，则23a b +的最小值是___________.【答案】3-【解析】【分析】根据不等式特征可通过构造函数()33,0f x x x x =+>，利用函数单调性解不等式可得21a b ≥+，再根据基本不等式即可求得23a b +的最小值是3-.【详解】由题意可得将不等式变形成33223311a a b b ⎛⎫⎛⎫+⨯≤+ ⎪ ⎪++⎝⎭⎝⎭；又因为,a b 都是正数，所以20,01a b +>>；可构造函数()33,0f x x x x =+>，易知函数为增函数，由()3386311a a b b +≤+++可得33223311a ab b ⎛⎫⎛⎫+⨯≤+ ⎪ ⎪++⎝⎭⎝⎭，即()21f f a b ⎛⎫≤⎪+⎝⎭，根据函数单调性可得21a b ≥+，则()233313443311b b a b b b ++=++-≥=++-≥，当且仅当()3124,11a b b b +=++=，即2313a b ==-取等号，因此23a b +的最小值是3-.故答案为：316.函数2()2e x f x a bx =++，其中,a b 为实数，且(0,1)a ∈.已知对任意23e b >，函数()f x有两个不同零点，a 的取值范围为____________.【答案】)6,1-⎡⎣e 【解析】【分析】由题意可得2ln 22e e 2e 0x x a a bx bx ++=++=有两个不相等的实根，利用换元法，分离参数，令ln t x a =，则22ln t b a t +-=e e ，再利用导函数求2e e t t+的最小值即可.【详解】因为()f x 有两个不同零点()0f x ⇔=有两个不相等的实根，即2ln 22e e 2e 0x x a a bx bx ++=++=有两个不相等的实根，令ln t x a =，则220ln tbt a ++=e e ，t 显然不为零，所以22ln t b a t+-=e e ，因为()0,1a ∈，23e b >，所以20ln ba->，所以0t >，令()()2e e 0t g t t t +=>，则()()22t t t g t t-+'=e e e ，令()()()2e e e0tth t t t =-+>，则()0tttth t t t '=+-=>e e e e ，所以()h t 在()0,∞+上单调递增，又()20h =，所以当()0,2t ∈时，()0h t <；当()2,t ∈+∞时，()0h t >，所以当()0,2t ∈时，()0g t '<；当()2,t ∈+∞时，()0g t '>，故()g t 在()0,2上单调递减，在()2,+∞上单调递增，所以()()2min 2g t g ==e ，所以22ln ba-≥e ，又23e b >，所以23b >e ，所以ln 32a -≤即ln 6a ≥-，6a -≥e ，又()0,1a ∈，所以)6,1a -⎡∈⎣e ，故答案为：)6,1-⎡⎣e 【点睛】利用换元法，令ln t x a =，根据t 不为零，分离参数得22ln t b a t+-=e e ，构造函数，通过求解函数的最值，即可得出a 的取值范围.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.已知,,a b c 分别为ABC 内角,,A B C 的对边，若ABC 同时满足下列四个条件中的三个：①a =2b =；③sin sin sin ++=-B C a c A b c ；④21cos sin sin 24-⎛⎫-= ⎪⎝⎭B C B C ．（1）满足有解三角形的序号组合有哪些？（2）请在（1）所有组合中任选一组，求对应ABC 的面积．【答案】（1）序号组合为①②③，①②④（2）答案不唯一，具体见解析【解析】【分析】（1）判断出③，④不能同时存在，由此确定正确答案.（2）选①②③，则利用余弦定理求得c ，进而求得三角形ABC 的面积；选①②④，则利用余弦定理求得c ，进而求得三角形ABC 的面积.【小问1详解】对于③，()22212π,0,223b c a c a c b B B a b c ac π+++-=⇒=-∈∴=-；对于④，()()1cos 11sin sin cos 2sin sin 242B C B C B C B C +--=⇒--=-，即()1cos 2B C +=-，且π,0,,πA B C A B C ++=<<，则π3A =，故③，④不能同时存在，则满足有解三角形的序号组合为①②③，①②④．【小问2详解】选①②③：2π2,3a b B ===时，由余弦定理：22221cos 22a c b B ac +-=⇒-=整理得：210c -=且0c >，则2c =，ABC ∴的面积为31sin 28ABCSac B == ．选①②④：π2,3a b A ===时，由余弦定理：2222143cos 224b c a c A bc c+-+-=⇒=，整理得：2210c c -+=，则1c =，ABC ∴ 的面积13sin 22ABC S bc A ==．18.已知数列{}n a 满足22113,2221++==+-++n n n a a a n n ．（1）求证：22⎧⎫-⎨⎩⎭n na n 是等差数列；（2）令2⎡⎤=⎢⎥⎣⎦nn n a b （[]x 表示不超过x 的最大整数．提示：当a ∈Z 时，[][]a x a x +=+），求使得12100n b b b ++≤+L 成立的最大正整数n 的值．【答案】（1）证明见解析（2）9【解析】【分析】（1）根据递推关系，结合等差数列定义证明即可；（2）结合（1）得()2221nn a n n =-+，故2212n n n b n ⎡⎤=-+⎢⎥⎣⎦，再根据函数()ln xf x x =的单调性得当5x ≥时，22x x <，进而解5n时，2123100n b b b n +++=+≤ 即可得答案.【小问1详解】证明：因为2212221n n n a a n n ++=+-++，所以222222111(1)2221(1)2222n n n n n n n n na n a n a n n n a n ++++-+-+-++-+--=-()2221222222n n n n a n a n +++---==，所以数列22⎧⎫-⎨⎬⎩⎭n na n 是以1112a -=为首项，2d =为公差的等差数列．【小问2详解】解：由（1）知，2212n na n n -=-，即()2221n n a n n =-+，所以()()22222121212222n n n n nn n n n a n n b n n ⎡⎤-+⎡⎤⎡⎤⎡⎤===-+=-+⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦⎣⎦．令函数()ln x f x x =，所以()21ln xf x x -'=，当()0,e x ∈时，()()0,f x f x '>单调递增；当()e,x ∈+∞时，()()0,f x f x '<单调递减．注意到：2552<，两边同时取对数25ln5ln2<，即ln5ln252<，所以当5x ≥时，ln ln5ln252x x ≤<，即22x x <，特别地，1n =时，21022n n ⎡⎤⎡⎤==⎢⎥⎢⎥⎣⎦⎣⎦；当2n =时，24124n n ⎡⎤⎡⎤==⎢⎥⎢⎥⎣⎦⎣⎦；当3n =时，29128n n ⎡⎤⎡⎤==⎢⎥⎢⎥⎣⎦⎣⎦；当4n =时，2161216n n ⎡⎤⎡⎤==⎢⎥⎢⎥⎣⎦⎣⎦；当5n ≥时，22nn <，则202n n ⎡⎤=⎢⎥⎣⎦．显然使得12100n b b b ++≤+L 成立的最大正整数n 的值大于5，则5n时，()2121352133100n b b b n n +++=++++-+=+ ，所以满足条件的n 的最大值为9．19.如图，四棱锥P -ABCD 的底面为梯形，PD⊥底面ABCD ，90BAD CDA ∠=∠=︒，1AD AB ==，2CD =，E 为PA 的中点．（1）证明：平面PBD ⊥平面BCE ；（2）若二面角P -BC -E 的余弦值为265，求三棱锥P -BCE 的体积．【答案】（1）证明见解析；（2）312.【解析】【分析】（1）线面垂直的性质可得PD BC ⊥，若F 为CD 中点，连接BF ，由正方形的性质及勾股定理可得BD BC ⊥，再由线面垂直的性质有BC ⊥面PBD ，最后根据面面垂直的判定证结论.（2）构建空间直角坐标系，设PD m =求相关点坐标，再求面PBC 、面EBC 的法向量，应用空间向量夹角的坐标表示，结合二面角的余弦值求参数m ，最后求PBC S 、向量法求E 到面PBC 的距离，再由体积公式求棱锥的体积.【小问1详解】因为PD⊥底面ABCD ，BC ⊂面ABCD ，则PD BC ⊥，由90BAD ∠=︒，1AD AB ==，则BD =，又90CDA ∠=︒，则//AB DC，若F 为CD 中点，连接BF ，易知：ABFD 为正方形，则1BF =，又2CD =，即1FC =，所以BC =综上，222BC BD CD +=，即BD BC ⊥，又BD PD D = ，则BC ⊥面PBD ，又BC ⊂面BCE ，所以平面PBD ⊥平面BCE .【小问2详解】由题设，可构建如下图示的空间直角坐标系，若PD m =，则(0,0,0)D ，(1,1,0)B ，(0,2,0)C ，1(,0,)22mE ，(0,0,)P m，所以(1,1,)PB m =- ，1(,1,)22mEB =- ，(1,1,0)BC =- ，若(,,)x y z α= 为面PBC 的一个法向量，则0BC x y PB x y zm αα⎧⋅=-+=⎪⎨⋅=+-=⎪⎩，令1x =，则2(1,1,)mα= ，若(,,)a b c β= 为面EBC 的一个法向量，则0022BC a b a cmEB b ββ⎧⋅=-+=⎪⎨⋅=+-=⎪⎩，令1a =，则3(1,1,mβ=，所以262|cos ,|||5||||m αβαβαβ+⋅<>==，整理得429610m m-+=，所以m =，即PD =，易得：2,PA PC ==由PD⊥底面ABCD ，AB ⊂面ABCD ，则PD ⊥AB ，又90BAD ∠=︒，即AD ⊥AB ，由=PD AD D ⋂，则AB ⊥面PAD ，PA ⊂面PAD ，即AB ⊥PA ，所以在直角△PAB中，PB ，在△PBC中，PB =PC =、BC =222PB BC PC +=，则PB BC ⊥，所以11022PBC S ==.由上有：1(,1,)22EB =- 且面PBC的一个法向量α= ，则1152|cos ,||20EB α<>==，故E 到面PBC的距离|||cos ,|2020d EB EB α=<>==，所以11301033320212P BCE PBC V d S -=⋅⋅=⨯⋅=.20.法国数学家庞加莱是个喜欢吃面包的人，他每天都会到同一家面包店购买一个面包.该面包店的面包师声称自己所出售的面包的平均质量是1000g ，上下浮动不超过50g .这句话用数学语言来表达就是：每个面包的质量服从期望为1000g ，标准差为50g 的正态分布.（1）已知如下结论：若()2,X Nμσ ，从X 的取值中随机抽取()*,2k k Nk ∈≥个数据，记这k 个数据的平均值为Y ，则随机变量2,Y N k σμ⎛⎫~ ⎪⎝⎭.利用该结论解决下面问题.（i ）假设面包师的说法是真实的，随机购买25个面包，记随机购买25个面包的平均值为Y ，求()980P Y ≤；（ii ）庞加莱每天都会将买来的面包称重并记录，25天后，得到的数据都落在()950,1050上，并经计算25个面包质量的平均值为978.72g .庞加莱通过分析举报了该面包师，从概率角度说明庞加莱举报该面包师的理由；（2）假设有两箱面包（面包除颜色外，其他都一样），已知第一箱中共装有6个面包，其中黑色面包有2个；第二箱中共装有8个面包，其中黑色面包有3个.现随机挑选一箱，然后从该箱中随机取出2个面包.求取出黑色面包个数的分布列及数学期望.附：①随机变量η服从正态分布()2,N μσ，则()0.6827P μσημσ-≤≤+=，()()220.9545,330.9973P P μσημσμσημσ-≤≤+=-≤≤+=；②通常把发生概率小于0.05的事件称为小概率事件，小概率事件基本不会发生.【答案】（1）（i ）0.02275；（ii ）理由见解析.（2）ξ012p5314044984073840()1724E ξ=【解析】【分析】（1）（i ）由正太分布的对称性及3σ原则进行求解；（ii ）结合第一问求解的概率及小概率事件进行说明；（2）设取出黑色面包个数为随机变量ξ，则ξ的可能取值为0,1,2,求出相应的概率，进而求出分布列及数学期望.【小问1详解】（i ）因为25010025=，所以()21000,10Y N ，因为()220.9545P μσημσ-≤≤+=，所以()10.954520.022752P ημσ-≤-==，因为9801000210=-⨯，所以()()98020.02275P Y P Y μσ≤=≤-=；（ii ）由第一问知()()98020.02275P Y P Y μσ≤=≤-=，庞加莱计算25个面包质量的平均值为978.72g ，978.72980<，而0.022750.05<，为小概率事件，小概率事件基本不会发生，这就是庞加莱举报该面包师的理由；【小问2详解】设取出黑色面包个数为随机变量ξ，则ξ的可能取值为0,1,2,则()143154530265287140p ξ==⨯⨯+⨯⨯=；()124135449122265287840p ξ==⨯⨯⨯+⨯=，()121132732265287840p ξ==⨯⨯+⨯=，故分布列为：ξ012p5314044984073840其中数学期望()53449731701214084084024E ξ=⨯+⨯+⨯=21.已知抛物线21:C y x =，开口向上的抛物线2C 与1C 有一个公共点(2,4)T ，且在该点处有相同的切线，（1）求所有抛物线2C 的方程；（2）设点P 是抛物线2C 上的动点，且与点T 不重合，过点P 且斜率为k 的直线l 交抛物线1C 于,A B 两点，其中PA PB ≥，问是否存在实常数k ，使得PAPB为定值？若存在，求出实常数k ；若不存在，说明理由.【答案】（1）2(2)4(1)y a x x =-+-（0a >且1)a ≠（2）存在，4k =.【解析】【分析】（1）设22:C y ax bx c =++，根据题意结合导数的几何意义，得到44a b +=，再由2C 过点T ，求得44c a =-，即可求得抛物线2C 的方程；（2）根据题意得到l 即为公共点T 处的切线，得出4k =，设2(,(2)4(1))P t a t t -+-，求得切线方程为()()()24241y x t a t t =-+-+-，联立方程组，得到12PA x t PBx t-=-，令m x t =-，得到12PA mPB m =，并代入整理得222(24)(2)4(1)0m t m t a t t +-+----=，根据根与系数的关系，化简求得22212212(22)(88)882(1)(1)(44)441m m a t a t a a m m a t a t a a++-++++==----+-为定值，分1a >和01a <<，两种情况讨论，结合21y y <，得到,A B 在点P 的两侧和同侧，进而得到答案.【小问1详解】解：设22:,(0)C y ax bx c a =++>，可得2y ax b '=+，抛物线21:C y x =，可得2y x '=，因为抛物线2C 与1C 有一个公共点(2,4)T ，且在该点处有相同的切线，可得44a b +=，即44b a =-，所以22:(44)C y ax a x c =+-+，因为抛物线2C 过点(2,4)T ，代入可得44c a =-，即满足条件的22:(44)(44)C y ax a x a =+-+-即抛物线2C 的方程为2(2)4(1),(0y a x x a =-+->且1)a ≠.【小问2详解】解：当0PB →时，若PA PB为常数，则0PA →，此时l 即为公共点T 处的切线，故若存在，则4k =.下面证明：4k =时，PAPB为常数，设2(,(2)4(1))P t a t t -+-，则切线方程为()()()24241y x t a t t =-+-+-，联立方程组()()()224241y xy x t a t t ⎧=⎪⎨=-+-+-⎪⎩，整理得224()(2)4(1)0x x t a t t ------=，设1122(,),(,)A x y B x y ，则12PA x t PBx t -=-，令m x t =-，可得x m t =+，所以12PA m PBm =，代入上式得22()4(2)4(1)0m t m a t t +-----=，即222(24)(2)4(1)0m t m t a t t +-+----=，可得()()12221242241m m t m m t a t t +=-⎧⎪⎨=----⎪⎩，所以222121224(2)m m m m t ++=-，则2222222124(2)22(2)8(1)22(2)8(1)m m t t a t t t a t t +=--+-+-=+---，所以22212212(22)(88)882(1)(1)(44)441m m a t a t a a m m a t a t a a++-++++==----+-为定值，且2221(1)4(1)4(1)(1)(2)y y a x a x a a x -=-+-+-=--，①当1a >时，由21y y >，可得,A B 在点P 的两侧，所以11221PA x t mPB x t m -==->-，令12m t m =，可得12(1)1a t t a ++=-，即2(1)2(1)10a t a t a --++-=，解得121a t a+±=-，因为1t <-，所以121a t a+-=-为定值；②当01a <<时，由21y y <，可得,A B 在点P 的同侧，所以11221PA x t mPB x t m -==>-，因为1t >，所以11a t a++=-为定值，综上可得，存在4k =时，使得PAPB为定值.【点睛】方法技巧：解答圆锥曲线的定点、定值问题的常见策略：1、参数法：参数解决定点问题的思路：①引进动点的坐标或动直线中的参数表示变化量，即确定题目中核心变量（通常为变量k ）；②利用条件找到k 过定点的曲线0(),F x y =之间的关系，得到关于k 与,x y 的等式，再研究变化量与参数何时没有关系，得出定点的坐标；2、由特殊到一般发：由特殊到一般法求解定点问题时，常根据动点或动直线的特殊情况探索出定点，再证明该定点与变量无关.22.已知221ln ,0(),0x x x x f x e x --⎧->=⎨≤⎩.（1）当(0,)x ∈+∞时，求()f x 的最大值；（2）若存在[0,)a ∈+∞使，得关于x 的方程2()0f x ax bx ++=有三个不相同的实数根，求实数b 的取值范围.【答案】（1）112e +；（2）1(,,e ⎡⎫-∞-⋃+∞⎪⎢⎣⎭.【解析】【分析】（1）利用导数判断出函数的单调性，再根据函数的单调性即可求出最值.（2）验证0x =不是方程的根，将原方程的根等价于()f x ax b x=--的根，记0A a =-≤，B b =-，令() t x Ax B =+，令2()g()(0)x f x e x x x x--==<，讨论B 的取值，利用导数求出函数()g x 的最值，通过比较即可确定答案.【详解】（1）当(0,)x ∈+∞时，2()1ln f x x x =-，即()2ln (2ln 1)f x x x x x x '=--=-+当x <时，()0f x ¢>，()f x 单调递增；当x >时，()0f x '<，()f x 单调递减，所以max 1()12f x f e==+（2）()20f x ax bx ++=，经验证0x =不是方程的根，所以原方程的根等价于()f x ax b x=--的根，记0A a =-≤，B b =-，令() t x Ax B =+，0A ≤，单调递减，令2()g()(0)x f x e x x x x --==<，即22(1)()x x e g x x---+'=，令()01g x x '=⇒=-为极大值点，其在(),1-∞-上单调递增，在()1,0-上单调递减，当B >，1()(1)()(0)t x B g g x x e>>-=-≥<，所以()()g x t x =在0x <无实数根当0x >时，21()()()ln B g x t x h x x A x x=⇔=--=……①2323212()B x Bx h x x x x x -+'=--+=-()h x 有两个极值点12,x x，且121220x x B x x ⎧+=>⎪⎨⋅=>⎪⎩即120x x <<，22B x =故222213()ln ln422B Bx h x x x B Bx++=--=--3ln0422<-⨯-=-<，所以()20h x<，存在A使①有三个实根所以B>.当B ()h x'的分子中2=80B∆-≤，()0h x'≤，显然()0,0x h x+→>，所以①仅有一个正根，要使2xe Ax Bx--=+有两个负根，则max1()(0)B g x xe≤=-<﹐综上所()1,Be⎛⎤∈-∞-⋃+∞⎥⎝⎦﹐即1(,,be⎡⎫∈-∞-⋃+∞⎪⎢⎣⎭.【点睛】本题考查了利用导数研究方程的根、利用导数求函数的最值，考查了分类讨论的思想，属于难题.。