上海市崇明区2021届高三第一学期一模数学试卷 含答案
上海市崇明区2021届高三一模
数学试卷
2020.12
一．填空题（本大题共12题，1-6每题4分，7-12每题5分，共54分〉l 设集合A={1,2，匀，集合B ={3，剑，则AnB=
2.不等式王二.！＿< 0的解集是x+2
3己知复数z 满足（z -2)i = 1 ( i 是虚数单位），则z=
4设函数1·(x ）＝�的反函数为广＇（功，则广1
(2)=x+l
5点（0,0）到直线x+y =2的距离是
1+2+3＋…
6.计算：lim = n →00n (n +2)
14x+6y=l 7若关于x 、γ的方程组｛无解，则实数α＝; E 似－3y=2
8用数字0、1、2、3、4,5组成无重复数字的三位数，其中奇数的个数为（结果用数值表示）
9.若（2α2+ b 3）＂的二项展开式中有一项为ma 4b 12
，则m=. x 2 y 2
I 0.设。
为坐标原点，直线X ＝α与双曲线C.－－；；－－寸＝1Cα＞ 0, b ＞们的两条渐近线a ' b '
分别交于D 、E两点，若A ODE 的面积为1，则双曲线C的焦距的最小值为l l己知函数y=J （吟，对任意xεR，都高f(x+2)·f(x)=k Ck为常数），且当xε（0,2］时，f(x )= x 2 + 1，则／（2021)=
12.己知点D为圆0：对＋y 2 =4的弦MN 的中点，点A的坐标为（1,0），且互M ·五万＝1,
则OA ·O D 的最大值为
二．选择题（本大题共4题，每题5分，共20分〉
13.若α＜O<b ，则下列不等式恒成立的是（
）－70
＞l －G A B.一α＞b C.a 2 >b 2 D.α3 < b 314.正方体上点P 、Q 、R 、S 是其所在棱的中点，则直线PQ 与RS 异面的图形是（
）
s R p ，，Pl : R ’二，J ---I Q P,--1－＿�J ＿＿＿←·· s ，．．．．．．，．．． r……t--I -， ， 二.:.:V Q p
A. ， ’S 'S ， ， Q
B. Q
C.
D.。
2021年上海市崇明县中考数学一模试卷
2021 年上海市崇明县中考数学一模试卷
一.选择题
1．（4 分）已知=，那么的值为（）
A．B．C．D．
2．（4 分）已知Rt△ABC 中，∠C=90°，BC=3，AB=5，那么sinB 的值是（）A．B．C．D．
3．（4 分）将抛物线y=x2先向右平移2 个单位，再向下平移3 个单位，那么得到的新的抛物线的解析式是（）
A．y=（x+2）2+3 B．y=（x+2）2﹣3 C．y=（x﹣2）2+3 D．y=（x﹣2）2﹣3 4．（4 分）如图，在△ABC 中，点D、E 分别在AB、AC 上，∠AED=∠B，那么下列各式中一定正确的是（）
A．AE•AC=AD•AB B．CE•CA=BD•AB C．AC•AD=AE•AB D．AE•EC=AD•DB 5．（4 分）已知两圆的半径分别是3 和5，圆心距是1，那么这两圆的位置关系是（）
A．内切B．外切C．相交D．内含
6．（4 分）如图所示，一张等腰三角形纸片，底边长18cm，底边上的高长18cm，现沿底边依次向下往上裁剪宽度均为3cm 的矩形纸条，已知剪得的纸条中有一张是正方形，则这张正方形纸条是（）。
上海市崇明区2025届高三上学期一模 英语试卷(含答案)
2024学年第一学期高三第一次模拟考试英语（考试时间105分钟，满分115分。
请将答案填涂在答题纸上）I. Grammar and VocabularySection ADirections:After reading the passage below, fill in the blanks to make the passage coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.Report: Harmful Waste Creation Set to IncreaseThe United Nations Environment Programme (UNEP) said in a report that public waste creation will greatly increase by 2050. The rise will cause hundreds of billions of dollars of damage through biodiversity loss, climate change, and deadly pollution, UNEP reports.UNEP’s Global Waste Management Outlook 2024 says worldwide waste creation would greatly increase (1) _____ governments take urgent preventative measures.Damage (2) _____ (cause) by the growing waste would account for about $443 billion of the total cost.The report, called Beyond an Age of Waste: Turning Rubbish into a Resource, (3) _____ (release) during the U.N. Environment Assembly in Kenya early this week. The writers argue that humanity (4) _____ (move) backwards over the past ten years. They say humans are creating more waste, more pollution, and more climate changing gases.Waste prevention measures and improved waste treatment could reduce those costs, the report said. But it notes, there are major barriers (5) _____ such reforms.Negotiators are working toward an agreement (6) _____ (deal) with the especially damaging and dangerous pollution from plastics. They are beginning a fourth round of talks in April. UNEP Executive Director Inger Andersen said she is hopeful they will complete the agreement by the end of this year.Environmentalists and fossil fuel (化石燃料) producers continue to disagree about the terms of the agreement. They especially dispute (7) _____ the deal should center on reducing plastics production or increasing recycling and reuse.“There is an interest, especially among the countries (8) _____ are producing raw polymer (聚合物), but as I keep telling them, this is not an anti-plastic agreement,” Andersen told the reporters, (9) _____ (note) there would still be a need for plastics in vehicles and medical equipment.Andersen said (10) _____ she hopes is that no groups would work to block progress on the agreement, but instead “find a way forward that actually takes into account the fact that we are drowning in plastic.”Section BDirections: Fill in each blank with a proper word chosen from the box. Each word can only be used once. Note that there is one word more than you need.A. appealingB. attainableC. basicallyD.dramaticallyE. freedomF. kicksG. minimum H. prioritizes I. submits J. underlying K. withdrawalFIRE Movement: Financial Independence, Retire EarlyMany Americans are taking early retirement into their own hands by joining the FIRE movement, but it’s not for everyone.“Financial Independence, Retire Early” (FIRE) is a lifestyle movement that 11 extreme saving and investing to be able to retire earlier than traditional methods might allow. The goal of FIRE is to achieve financial 12 so investors can choose how to spend their time.“The concept of the movement is 13 having the financial flexibility to have the ultimate life flexibility,” says Rachael Burns, a financial planner at True Worth Financial Planning, based in Folsom, California.A NerdWallet survey found that of Americans who aren’t retired yet but plan to retire, 25% want to retire before age 50 and 18% want to retire in their 50s. The average age was 57, 10 years earlier than the full retirement age to get Social Security.People who use FIRE to retire early do so by 14 reducing their expenses, looking for ways to increase their income, and investing the money they save in a mix of tax-advantaged accounts as well as regular brokerage (佣金) accounts.Retiring early might sound 15 , but there are risks, and it’s not for everyone. For instance, if you stop working, you’ll have to be responsible for your own medical expenses until Medicare 16 in around age 65, and your investments may not perform as well as you thought. Either one of those situations could have consequences such as having to raise your 17 rate or needing to re-enter the workforce. FIRE also requires cutting down expenses to the bare 18 so you have more income to invest, which not everyone can afford.“FIRE is a long-term strategy, and you can’t be too reactive to short-term economic events,” Burns said. “You may need to adjust spending or saving in certain years based on market events, but the 19strategy should stay the same.”If you don’t earn enough to cover your basic needs and save greatly for early retirement at the same time, FIRE may not be for you, Burns says. Lacking an emergency fund or owing high-interest debt may be other reasons that FIRE may not be 20 .II. Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.Administrators of the Mogao Caves in Dunhuang, Gansu Province, are trying to harmonize tourists’ exploration of the site with the need to safeguard the murals(壁画), through innovative measures.Sandstorms, rainfall and tourist visits constitute the most severe 21 to the UNESCO World Heritage Site, said Wang Xiaowei, director of the Dunhuang Grottoes Monitoring Center at the Dunhuang Academy. Since the Mogao Caves opened to the public in 1979, the number of visitors has been 22 at an average annual rate of around 20 percent, reaching 2.15 million in 2019. “If you enter the caves during the 23 tourism months of July, August and September, you’ll find it hard to breathe,” Wang said. The carbon dioxide and moisture (潮气) breathed out by visitors increase inside the caves and cause damage to the murals, Wang said.To preserve the caves, the duration of visits is 24 and sometimes stopped during rain or dust storms. To ensure visitors aren’t 25 when restrictions are in place, the center provides a digital exhibition, he said. Currently, the center is being 26 to accommodate an additional 3,000 visitors on top of the existing capacity of 6,000.The Dunhuang Academy began 27 recording and storing images of murals and painted sculptures over 30 years ago. The digitization project has successfully 28 over 200 caves, with a dedicated team of 110 experts currently undertaking the work. “Digital technology not only serves cultural tourism but also 29 a historical record for future generations,” said Ding Xiaosheng, deputy director of the Institute of Cultural Heritage Digitization at the academy. Digitization also brings the wonders of the Mogao Caves to a 30 audience, according to Su Bomin, head of the Dunhuang Academy. “The Mogao Caves are 31 , and transporting them is impossible,” Su said. “However, with digitization, we can perfectly copy Dunhuang art exactly and show it worldwide, introducing Eastern culture to the world.”In 2016, the Digital Dunhuang went live, sharing high-definition images and 32 tours of the most beautiful 30 caves globally. Currently, visitors from 78 countries have 33 the murals, totaling over 16.8 million visits.Su said Dunhuang can 34 diverse cultural exchanges through its cultural relics (遗迹). “By digitizing these relics, we enable people worldwide to understand Dunhuang’s culture, thereby gaining a deeper appreciation for China’s historical 35 to diverse cultural exchanges — that is, an idea of inclusivity, mutual learning and a shared future,” he said.21. A. shortages B. restrictions C. contributions D. threats22. A. doubling B. growing C. continuing D. varying23. A. cultural B. previous C. peak D. commercial24. A. limited B. extended C. publicized D. concealed25. A. confused B. amazed C. scared D. disappointed26. A. expanded B. constructed C. decorated D. repaired27. A. exclusively B. digitally C. subjectively D. autonomously28. A. clarified B. highlighted C. covered D. strategized29. A. comes across B. turns over C. leaves behind D. lets alone30. A. global B. professional C. technological D. different31. A. complicated B. irreproducible C. controversial D. immovable32. A. virtual B. temporary C. conventional D. steady33. A. imitated B. accessed C. praised D. purchased34. A. reject B. provide C. adjust D. classify35. A. adaptation B. attention C. admission D. commitmentSection BDirections:Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.（A）The Pulitzers are American awards given within the United States for outstanding achievement in journalism as well as books, drama and music. Under these headings, there are 22 categories including online journalism, newspaper reporting, fiction, history, music, drama, poetry, photography and more. Recipients can be a news organization, an individual or a group of people.Each of the first 21 winners receives a $15,000 cash prize and a certificate. The twenty-second prize, however, is the Public Service Prize, which is always given to a news agency. Instead of cash, the news organization receives the Pulitzer Gold Medal. The name of the year’s winner is on one side of the coin and the year is on the other. All winners are honored and awarded their prizes at an annual lunch party. The event takes place at Columbia University, which was originally tasked with administering the award, and usually occurs in May.If it hadn’t been for a Hungarian-American named Joseph Pulitzer, the awards would never have come into existence. Pulitzer was born into a wealthy family in Hungary in 1847. He made his way to America and built a career in journalism. He developed a reputation as an ambitious and energetic young journalist. By 1872 Pulitzer had become a publisher, and six years later he owned a newspaper company. In 1883 he purchased yet another newspaper. He became known as someone who was not afraid to take a public stand against corruption (腐败).After a successful career in journalism, seven years before his death, Pulitzer wrote a will, leaving $2,000,000 to Columbia University. The money was to establish a school of journalism at Columbia and a set of awards. Pulitzer’s desire was to raise the standards of journalism. He hoped the prize would act as a stimulus for journalists to work for excellence — then and into the future.Faithfully following Pulitzer’s instructions, Columbia University awarded the very first Pulitzer Prizes in 1917. Today these awards are considered some of the most distinguished prizes in America.36. What can be known about the Pulitzers?A.Their recipients all get cash prizes.B. There are 21 winners each year.C. They are awarded worldwide.D. They go beyond journalism.37. It can be inferred from the passage that Joseph Pulitzer _____.A. was humorous and courageousB. was raised up in a journalist’s familyC. had a strong sense of justiceD. sought a goal of producing wealth38. Pulitzer established the prizes mainly to _____.A. facilitate better journalismB. diversify ways of access to journalismC. support Columbia UniversityD. stimulate people to become journalists39. Which of the following is the best title for the passage?A. Pulitzer Prizes: Joseph Pulitzer’s Lifetime AmbitionB. Pulitzer Prizes: Some of America’s Greatest HonorsC. Pulitzer Prizes: Awards for Distinguished JournalistsD. Pulitzer Prizes: Annual Events at Columbia University（B）Winter storms create a higher risk of car accidents, hypothermia(体温过低), frostbite, and carbon monoxide (CO) poisoning. Winter storms can bring extreme cold, freezing rain, snow, ice and high winds.IF YOU ARE UNDER A WINTER STORM WARNING, FIND SHELTER RIGHT AWAYWinter Storm WarningIssued when hazardous winter weather in the form of heavy snow or heavy freezing rain is upcoming or occurring. Winter Storm Warnings are usually issued 12 to 24 hours before the event is expected to begin.Winter Weather AdvisoryIssued for accumulations (积聚) of snow and freezing rain which will cause significant inconveniences and, if measures are not taken, could lead to life-threatening situations.Preparing for Winter WeatherPrepare your home to keep out the cold. Learn how to keep pipes from freezing. Install and test smoke alarms and carbon monoxide detectors with battery backups. Gather supplies in case you need to stay home for several days without power. Keep in mind each person’s specific needs, including medication. Remember the needs of your pets. Have extra batteries for radios and flashlights.In Case of EmergencyBe prepared for winter weather at home, at work and inyour car. Create an emergency supply kit for your car. Include sand, a flashlight, warm clothes, blankets, bottled water and snacks. Keep a full tank of gas.Avoid carbon monoxide poisoning. Only use generators and grills(烤架) outdoors andaway from windows. Never heat your home with gas appliances.∙Stay off roads if at all possible. If trapped in your car, then stay inside.∙Limit your time outside. If you need to go outside, then wear layers of warm clothing. Watch for signs of frostbite and hypothermia.40. Which of the following pictures when clicked will most probably lead you to the above webpage?A. B.C. D.41. To get ready for winter weather, you should _____.A. send your pets to animal sheltersB. warm your home with gas appliancesC. remove pipes to prevent them freezingD. get necessities in case of life without power42. _____ is a good way to stay safe during winter weather.A. Trying to stay insideB. Driving a car when going outC. Wearing water-proof clothesD. Using generators to keep warm（C）A recent groundbreaking study by the Potsdam Institute for Climate Impact Research (PIK) forecasts massive financial damages due to climate change, projecting annual damages of about $38 trillion ($38 million million) by the year 2050. This figure highlights the severe economic challenges that lie ahead on a global scale, with the greatest impacts expected to burden the countries least responsible for greenhouse emissions (排放).The economic damage, estimated between $19 trillion to $59 trillion by mid-century, primarily stems from increased temperatures and their subsequent impact on agriculture, labor productivity, and infrastructure (基础设施). The study also notes potential increases in costs due to more frequent and severe weather events, which are expected to become more common as global temperatures rise.The findings reveal a disconcerting disparity: nations within the tropical (热带的) regions, which have contributed least to historical carbon emissions, are predicted to suffer income losses 60% greater than those in higher-income, higher-emission countries. This highlights a significant climate justiceissue, as these least-developed nations possess fewer resources to adapt to increasing climate impacts.Anders Levermann, head of complexity science at PIK and co-author of the study, emphasized the inequity and urgency of the situation. “Countries in the tropics will suffer the most because they are already warmer. Further temperature increases will therefore be most harmful there,” he explained. Levermann argues for a rapid structural shift towards renewable energy to ease these impacts and stabilize global temperatures.When placed alongside other major climate impact forecasts, the PIK study stands out for its comprehensive data analysis and harsh projections. Previous studies have similarly highlighted the economic damages of climate change but lacked the extensive regional and specific insights this study provides.The study strongly advocates for great and immediate reductions in greenhouse gas emissions to avoid the worst of these economic damages. It suggests that slowing climate change is not only a moral and environmental must but also economically advantageous, as the cost of inaction is far greater than the expenses associated with reducing global warming to manageable levels.The socioeconomic consequences of such extensive economic losses could lead to heightened global inequality and potentially, social unrest. This projection stresses the importance of international cooperation in climate action plans and the development of fair policies that recognize and make up for the inequities faced by the less developed nations most affected by climate change.43. Climate change will cause massive economic damage worldwide by 2050 mainly because _____.A. the average world temperature will increase much faster than expectedB. rising temperatures will damage farming, infrastructure and productivityC. big countries will experience economic crises due to more severe weather eventsD. tropical regions are unprepared to adapt to the damages caused by climate change44. By “a disconcerting disparity” in paragraph 4, the author means _____.A. poor countries least responsible for emission will suffer most in economyB. it is projected that people in tropical countries will earn 60% less by 2050C. higher-income countries are reluctant to invest in slowing global warmingD. higher-emission countries are trying to justify their role in climate impacts45. The PIK study arouses the public attention due to _____.A. its rigid climate impact forecastsB. its overall statistics and the severity of its findingsC. its different approach to analyzing dataD. its similar regional insights on economic damages46. What does the PIK study imply?A. Fair policies should be established to avoid further rising of temperatures.B. Reducing gas emissions will benefit the environment but harm the economy.C. Inequality should be reduced through cooperation in fighting climate change.D. Wealthy countries should bear the whole cost of coping with global warming.Section CDirections: Read the following passage. Fill in each blank with a proper sentence given in the box. Each sentence can be used only once. Note that there are two more sentences than you need.A.One 7-minute block had people sitting in silence.B.In this case, the brain might be tuning into the music.C.The authors suggested it was evidence of music being more than a distraction from anunpleasant experience.D.And the most effective pain relievers were found to be sad songs detailing bittersweet andemotional experiences.E.Nevertheless, while our bodies still feel the pain, the messages to make our conscious mindperceive the pain may not be relayed.F.People who listened to bittersweet songs also reported more thrills (震颤感) and shivers (哆嗦)on the skin from listening to pleasurable music.Certain Types of Music Could Help You Feel Less Pain, New Study Says There is no doubt that music can calm the soul for some, and it turns out that it could also be a temporary reliever for physical pain.Listening to favorite songs could reduce people’s perception of pain, according to a new study.47The small study invited 63 young adults to bring two of their favorite songs, and the only requirement was that they needed to be at least 3 minutes and 20 seconds long. One selection represented their favorite music of all time. The other was the song they would bring with them on a desert island. The researchers also had the young adults pick one of seven songs that the team considered relaxing and were unfamiliar to the study participants.Each person underwent 7-minute blocks where they were instructed to stare at a monitor screen while listening to their favorite music, one of the seven relaxing instrumental songs (each of which lasted for 6 minutes and 40 seconds), or a scrambled (杂乱的) version of both songs and the relaxing song chosen. 48 All the while, the researchers stuck a hot object — similar to the pain of a boiling hot teacup on your skin — to the participants’ left inner forearms.When rating their experiences, people were more likely to report feeling less pain when listening to their favorite songs compared with hearing the unfamiliar relaxing song or silence. The scrambled songs did not reduce pain either. 49After interviewing the participants about the song they brought and their rating of pain, the researchers found people who listened to bittersweet and moving songs felt less pain than when they listened to songs with calming or cheerful themes.50 This was associated with lower ratings of unpleasantness produced by the burning pain they felt in the experiment.III. Summary WritingDirections: Read the following passage. Summarize the main idea and the main point(s) of the passage in no more than 60 words. Use your own words as far as possible.51. Body Dysmorphic Disorder (BDD)Most humans have some physical characteristics that they wish they could change, such as a facial feature or extra fat. Many people can comfortably live with these characteristics, but for some, their negative body image significantly affects their daily lives. The latter group are said to have body dysmorphic disorder (BDD). They find some common aspects, including the face or stomach, impossible to accept.People with BDD often try to alter these features by exercising, changing their hairstyle, using skin care products or even getting plastic surgery. Or they may try to hide the features under clothing or makeup. They may look at mirrors constantly or avoid them altogether. They may ask others for reassurance (安慰) that they look OK but not believe that reassurance when it comes. People with BDD sometimes pick at their skin or constantly comb (梳) their hair; at its worst, BDD can lead people to consider or attempt suicide.BDD is often not diagnosed until 10 to 15 years after it develops, largely because it takes time for people to realize their feelings are not normal. But early detection makes a significant difference in quality of life as it allows people to take measures to prevent BDD from getting worse. The primary treatment for BDD is therapy, which often involves teaching the patient to challenge negative thoughts as they come. Some doctors may also prescribe antidepressants (开抗抑郁药). Plastic surgery is not effective at making people with BDD feel better. In fact, the changes in appearance brought about by surgery often compound the problem.If someone you know has BDD, assuring them that they look good will probably not help. Instead, listen patiently and encourage them to seek treatment. And if you think you have BDD, consult a doctor or a trusted friend. The earlier the problem is addressed, the better.IV. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.52. 这位老人虽无儿无女，但侄女一直在精心照料着他的起居。
2023年上海崇明区高三数学高考一模试卷含答案
2022学年第一学期高三第一次模拟考试数学考生注意：1.本试卷共4页，21道试题，满分150分，考试时间120分钟.2.本试卷分设试卷和答题纸.试卷包括试题与答题要求.作答必须涂(选择题)或写(非选择题)在答题纸上，在试卷上作答一律不得分.3.答卷前，务必用钢笔或圆珠笔在答题纸正面清楚地填写姓名、准考证号码等相关信息.一、填空题(本大题共有12题，满分54分，其中1~6题每题4分，7~12题每题5分)【考生应在答题纸相应编号的空格内直接填写结果.】1.已知集合{}{}|04.1,2,3,4,5A x x B =<≤=-,则A B ⋂=2.不等式2102x x +<-的解集为3.已知复数2,3,z ai z i =+=+₁₂若z z ₁₂是纯虚数,则实数a =.4.已知对数函数(0,1)a y log x a a =>≠的图像经过点()4,2,则实数a =.5.设等比数列{}n a 满足12131,3a a a a +=--=-,则4a =.6.已知方程组2168x my mx y +=⎧⎨+=⎩无解,则实数m 的值等于.7.已知角α的终边与单位圆221x y +=交于点1,,2P y ⎛⎫⎪⎝⎭则sin 2πα⎛⎫+= ⎪⎝⎭8.将半径为2的半圆形纸片卷成一个无盖的圆锥筒，则该圆锥筒的高为.9.已知函数()2f x x =,则曲线()y f x =在点()1,1P 处的切线方程是.10.设函数()sin (0),6f x x k πωω⎛⎫=-+> ⎪⎝⎭若()3f x f π⎛⎫≤ ⎪⎝⎭对任意的实数x 都成立，则ω的最小取值等于.11.在边长为2的正六边形ABCDEF 中,点P 为其内部或边界上一点，则AD BP ⋅的取值范围为.12.已知椭圆1Γ与双曲线2Γ的离心率互为倒数,且它们有共同的焦点12,F F P 、是1Γ与2Γ在第一象限的交点，当126F PF π∠=时,双曲线1Γ的离心率等于.二、选择题(本大题共有4题，满分18分，其中13、14题每题4分，15、16题每题5分)【每题有且只有一个正确答案，考生应在答题纸的相应编号上，将代表答案的小方格涂黑，选对得满分，否则一律得零分.】13.下列函数中，既是奇函数又在区间()0,1上是严格增函数的是().A y x= B.3y x =-.lg C y x= D.y sinx=14.设x R ∈,则1“2x x+>”是“1x ≠”的()A.充分不必要条件B.必要不充分条件C.充分必要条件D.既不充分也不必要条件15.设函数()sin ,6f x x π⎛⎫=-⎪⎝⎭若对于任意5,,62ππα⎡⎤∈--⎢⎥⎣⎦在区间[]0,m 上总存在唯一确定的β,使得()()0f f αβ+=,则m 的最小值为().6A π.2B π7.6C π D.π16.已知曲线C:()3222216x yx y +=,命题p ：曲线C 仅过一个横坐标与纵坐标都是整数的点；命题q ：曲线C 上的点到原点的最大距离是2.则下列说法正确的是()A.p q 、都是真命题B.p 是真命题，q 是假命题C.p 是假命题，q 是真命题D.p q 、都是假命题三、解答题(本大题共有5题，满分78分)【解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤.】17.(本题满分14分，本题共有2个小题，第(1)小题满分7分，第(2)小题满分7分)如图，长方体1111ABCD A B C D -中，12,AB BC A C ==与底面ABCD 所成的角为45°(1)求四棱锥1A ABCD -的体积;(2)求异面直线1A B 与11B D 所成角的大小.18.(本期满分15分，本题具有2个小题，第(1)小题满分7分，第(2)小题满分8分)已知函数()21,2f x sinx cosx sin x =⋅-+(1)求()f x 的单调递增区间:(2)在ABC 中,a b c 、、为角A B C 、、的对边,且满足2bcos A bcosA asinB =-,且0,2A π<<求()f B 的取值范围.19、(本题满分15分，本题共有3个小题，第(1)小题满分3分，第(2)小题满分5分，第(3)小题满分7分)某公园有一块如图所示的区域OACB ，该场地由线段OA OB AC 、、及曲线段BC 围成.经测量，90AOB ∠=︒,100OA OB ==米,曲线BC 是以OB 为对称轴的抛物线的一部分,点C 到OA OB 、的距离都是50米，现拟在该区域建设一个矩形游乐场OEDF ，其中点D 在线段AC 或曲线段BC 上，点E 、F 分别在线段OA 、OB 上，且该游乐场最短边长不低于30米.x =米，游乐场的面积为S 平方米.(1)试建立平面直角坐标系，求曲线段BC 的方程；(2)求面积S 关于x 的函数解析式()S f x =;(3)试确定点D 的位置，使得游乐场的面积S 最大.(结果精确到0.1米)20、(本题满分16分，本题共有3个小题，第(1)小题满分4分，第(2)小题满分5分，第(3)小题满分7分)已知椭圆2221(1)x y a a+=>的右焦点为F ，左右顶点分别为A B 、，直线l 过点B 且与x 轴垂直，点P 是椭圆上异于A B 、的点，直线AP 交直线l 于点D.(1)若E 是椭圆的上顶点，且AEF 是直角三角形，求椭圆的标准方程；(2)若2,45,a PAB =∠=︒求PAF 的面积;(3)判断以BD 为直径的圆与直线PF 的位置关系，并加以证明.21.(本题满分18分，本题共有3个小题，第(1)小题满分4分，第(2)小题满分6分，第(3)小题满分8分)已知数列{}n a 满足()1121,2,,2.i i i i a a a a i n +++-≤-=- (1)若数列{}n a 的前4项分别为4,2,,1,a 求3a 的取值范围;(2)已知数列{}n a 中各项互不相同.令()11,2,,1,m m m b a a m n +=-=- 求证：数列{}n a 是等差数列的充要条件是数列{}m b 是常数列；(3)已知数列{}n a 是(m m N ∈且3)m ≥个连续正整数1,2,,m ⋯的一个排列.1112,m k k k a a m -=+∑-=+若求m 的所有取值.崇明区2022学年第一学期高三第一次模拟考试参考答案及评分标准一、填空题1.{}2,3,4;12.,22⎛⎫- ⎪⎝⎭; 3.6; 4.2; 5.8-; 6.4-;17.;29.21y x =-;10.2;11[].4,12;-12.2+二、选择题13.D;14.A;15.B;16.A.三、解答题17.解(1)因为A A ₁⊥平面ABCD ,所以A CA ∠₁是AC 与底面ABCD 所成的角所以45ACA ∠=︒……………………………………2分所以2A A =₁………………………………………4分所以1433A ABCD V Sh -==………………………………7分(2)联结BD ,则1//,BD B D ₁所以A BD ∠₁就是异面直线AB 与B D ₁₁所成的角………………3分A BD ₁中,112A B A D BD ===所以2221116cos 26AB BD A D A BD A B BD +-∠==⋅………………6分所以异面直线A B ₁与B D ₁₁所成角的大小为6arccos 6………………7分18.解(1)由题意()()111sin21cos2222f x x x =--+112sin2cos22,2224x x x π⎛⎫=+=+ ⎪⎝⎭…………………………5分由()222,242k x k k Z πππππ-≤+≤+∈解得3,88k x k ππππ-≤≤+所以()f x 单调递增区间为()3,88k k k Z ππππ⎡⎤-+∈⎢⎥⎣⎦………………………7分(2)由正弦定理，得2sinBcos A sinBcosA sinAsinB =-,因为在三角形中0sinB ≠,所以2cos A cosA sinA =-,即()()10cosA sinA cosA sinA -+-=,……………………2分当 cosA sin A =时,,4A π=当1cosA sinA +=时,²²21,0cos A sin A sinAcosA sinA ++=≠,cos 0,,2A A π∴==由于0,2A π<<所以,4A π=………………………5分故30,4B π<<又72,444B πππ<+<所以1sin 21,4B π⎛⎫-≤+≤ ⎪⎝⎭由()sin 2,24f B B π⎛⎫=+ ⎪⎝⎭所以()f B 的取值范围是22,22⎡-⎢⎣⎦…………………………8分19.解(1)以O 为坐标原点，OA OB 、所在直线分别为x 轴、y 轴建立平面直角坐标系，如图所示,则()()()100,0,50,50,0,60,A CB 设曲线段BC 所在抛物线的方程为²(0),y ax b a =+<由题意可知,点()0,100B 和()50,50C 在此抛物线上,故0.02,100a b =-=所以曲线段BC 的方程为：()0.02²100050y x x =-+≤≤……………………4分(2)由题意，线段AC 的方程为：()10050100y x x =-+≤≤当点D 在曲线段BC 上时，()()0.02²1003050S x x x =-+≤≤当点D 在线段AC 上时，()()10050 70S x x x =-+≤≤所以()()()20.02100,3050,100,5070.x x x f x x x x ⎧-+≤≤⎪=⎨-+<≤⎪⎩………………………4分(3)当3050x ≤≤时,()0.06²100,f x x =-+'令0.06²1000x -+=,得1506,3x =25063x =-(舍去)当30,3x ⎡⎫∈⎪⎢⎪⎣⎭时,()0;f x '>当,50]3x ⎛⎫∈ ⎪ ⎪⎝⎭时,()0.f x '<因此当5063x =时,5061000639S f ⎛⎫== ⎪ ⎪⎝⎭是极大值,也是最大值…………………………4分当5070x <≤时()()2,502500f x x =--+当50x =时,()502500S f ==是最大值………………6分因为10000625009>所以5063x =时,S 取得最大值，此时506200,33D ⎛⎫ ⎪ ⎪⎝⎭所以当点D 在曲线段BC 上且其到66.7米时，游乐场的面积S 最大………7分20.解(1)由题意()()(),,0,,0(0),0,1A a F c c E ->由题意， 90AEF ∠=︒,故0,EA EF ⋅=所以1ac =又²²1a c =+,所以215,2a =221152y =………………………4分(2)当2a =时,椭圆方程为221,4x y +=由对称性，不妨设点P 在x 轴上方，则直线AP 的方程为2y x =+，代入椭圆方程，得2516120x x ++=,解得2x =-₁(舍去),26,5x =-所以64,55P ⎛⎫- ⎪⎝⎭………………………3分所以142325PAF p S AF y ∆+=⋅=………………………5分(3)设()00,,P x y 则220021x y a+=直线AP 的方程为()00,y y x a x a =++所以002,,ay D a x a ⎛⎫ ⎪+⎝⎭BD 中点00,ay M a x a ⎛⎫ ⎪+⎝⎭直线PF 方程为()()000..y x c x c y ---= 3分点M 到直线PF 的距离d =20000000|a cx x a ay y y MBc x a x a a -+====+-所以以BD 为直径的圆与直线PF 相切…………………………………………………7分21.解(1)由题意,33221a a ≤-≤-,解得34a ≥………………………4分(2)必要性：若数列{}n a 是等差数列，设公差为d ，则1,m m m b a a d +=-=所以数列{}m b 是常数列.……………………………2分充分性：若数列{}m b 是常数列，则()11,2,,2,m m b b m n +==- 即()1121,2,,2.m m m m a a a a m n +++-=-=- 所以112m m m m a a a a +++-=-或()112.m m m m a a a a +++-=--因为数列{}n a 的各项互不相同，所以112.m m m m a a a a +++-=-所以数列{}n a 是等差数列.…………………………6分(3)当3m =时,因为()121,2,i i a a i +-≤=所以12235a a a a -+-<,不符合题意；当4m =时,数列为3,2,4,1.此时122334 6a a a a a a -+-+-=,符合题意；当5m =时,数列为2,3,4,5,1.此时122334457a a a a a a a a -+-+-+-=,符合题意;…………………………3分下证当6m ≥时，不存在m 满足题意.令()11,2,,1,k k k b a a k m +=-=- 则1211,m b b b -≤≤≤≤ 且112,m k k b m -=∑=+所以k b 有以下三种可能：①()()1,1,2,,24,1k k m b k m ⎧=-⎪=⎨=-⎪⎩ ②()()()1,1,2,,32,2,3,1k k m b k m k m ⎧=-⎪==-⎨⎪=-⎩③()()1,1,2,,4.2,3,2,1k k m b k m m m ⎧=-⎪=⎨=---⎪⎩ 当()()1,1,2,,24,1k k m b k m ⎧=-⎪=⎨=-⎪⎩ 时，因为122,m b b b -=== 由(2)知:121,,,m a a a - 是公差为1(或1-)的等差数列。
上海市崇明区2021届高考数学二模试卷(含答案解析)
上海市崇明区2021届高考数学二模试卷一、单选题（本大题共4小题，共20.0分） 1.在同一平面直角坐标系中经过伸缩变换{x′=5xy′=3y 后，曲线C 变为曲线2x′2+8y′2=0，则曲线C的方程为( )．A. 25x 2+36y 2=0B. 9x 2+100y 2=0C. 25x 2+36y 2=1D. 225x 2+89y 2=12.已知函数y =f(x)是R 上的偶函数，且在x ≤0上是减函数，若f(2x )>f(12)，则实数x 的取值范围是( )A. x <−1B. x >−1C. x ≤−1D. x ≥−13.设x ，y ∈R ，则“x ≠1或y ≠1”是“xy ≠1”的( )A. 充分不必要条件B. 必要而不充分条件C. 充分必要条件D. 既不充分也必要条件4.下列命题：①5>4或4>5；②9≥3；③命题“若a >b ，则a +c >b +c ”的否命题；④命题“矩形的两条对角线相等”的逆命题．其中假命题的个数为( )A. 0B. 1C. 2D. 3二、单空题（本大题共12小题，共54.0分） 5. 若集合则M ∩N = ．6.已知复数z =(2−i)(1+3i)，其中i 是虚数单位，则复数z 在复平面上对应的点位于第______ 象限．7. 过圆锥的轴的截面是顶角为120°的等腰三角形，若圆锥的体积为π，则圆锥的母线长为______．8. 已知向量a ⃗ =(6,2)，b ⃗ =(−4,12)，过点A(3,−1)且与向量a ⃗ +2b ⃗ 平行的直线l 的方程为______ ． 9.曲线y =xe x +2x +1在点(0,1)处的切线方程为_______________．10. 已知x ，y 满足：{x ≥0x +y ≤2x −y ≤0，若目标函数z =ax +y 取最大值时的最优解有无数多个，则实数a的值是______．11.若函数为定义在R上的偶函数，且在内是增函数，又，则不等式的解集为，满足的x的取值范围为______________12.二项式(x2−√5x3)5的展开式中的常数项为______ ．13.气象意义上从春季进入夏季的标志为：“连续5天的日平均温度均不低于22℃”.现有甲、乙、丙、丁四地连续5天的日平均温度的记录数据(记录数据都是正整数)：①甲地：5个数据的中位数为24，众数为22；②乙地：5个数据的中位数为27，总体均值为24；③丙地：5个数据的总体均值为24，且极差小于或等于4；④丁地：5个数据中有一个数据是32，总体均值为26，总体方差为10.8．则肯定进入夏季的地区有______ (写出所有正确编号)14.A，B，C，D，E，F六人并排站成一排，A，B必须站在一起，且C，D不能相邻，那么不同的排法共有______种(结果用数字表示)．15.已知函数y=x3，则此函数的反函数是______ ．16.若直线y=kx−1与圆x2+y2+kx+my−4=0的交点M，N关于直线2x−y−1=0对称，则m=______ ．三、解答题（本大题共5小题，共76.0分）17.如图ABC−A1B1C1是直三棱柱，底面△ABC是等腰直角三角形，且AB=AC=4，直三棱柱的高等于4，线段B1C1的中点为D，线段BC的中点为E，线段CC1的中点为F．(1)求异面直线AD、EF所成角的大小；(2)求三棱锥D−AEF的体积．18.已知函数f(x)=2sinx(sinx+cosx)．(1)求f(x)的最小正周期和单调递减区间；(2)画出函数y=f(x)在区间[−π2,π2]上的图象．19. 某公司有价值a 万元的一条流水线，要提高该流水线的生产能力，就要对其进行技术改造，从而提高产的附加值．改造需要投入，假设附加值y(万元)与技术改造投入x(万元)之间的关系满足：①y 与(a −x)和x 2的乘积成正比；②当x =a2时，y =a 3；③0≤x2(a−x)≤t ，其中常数t ∈(0,2]． (1)设y =f(x)，求函数f(x)的解析式与定义域； (2)求出附加值y 的最大值，并求此时的技术改造投入x ．20. 已知双曲线x 24−y 23=1的左、右顶点分别为A 1和A 2，M(x 1,−y 1)和N(x 1,y 1)是双曲线上两个不同的动点．(1)求直线A 1M 与A 2N 交点Q 的轨迹C 的方程；(2)过点P(l,0)作斜率为k(k ≠0)的直线l 交轨迹C 于A 、B 两点， ①求OA ⃗⃗⃗⃗⃗ ⋅OB⃗⃗⃗⃗⃗⃗ 的取值范围； ②若AP ⃗⃗⃗⃗⃗ =λPB ⃗⃗⃗⃗⃗ ，问在x 轴上是否存在定点E ，使得OP ⃗⃗⃗⃗⃗ ⊥EA ⃗⃗⃗⃗⃗ −λEB ⃗⃗⃗⃗⃗ ？若存在，求出E 点的坐标；若不存在，说明理由．21. 已知{a n }为等差数列，{b n }为等比数列，{b n }的前n 项和为S n ，且a 1=b 1=1，a 2=a 3−b 3，a 3=S 3+b 2．(1)求数列{a n }，{b n }的通项公式；(2)设c n =a n ⋅b n ，T n 为数列{c n }的前n 项和，求T n ．【答案与解析】1.答案：A解析：解：把{x′=5xy′=3y 代入曲线2x′2+8y′2=0，可得2(5x)2+8(3y)2=0，化为25x 2+36y 2=0，即为曲线C 的方程． 故选：A ．把{x′=5x y′=3y 代入曲线2x′2+8y′2=0，即可得出． 本题考查了曲线的变换公式的应用，属于基础题．2.答案：B解析：解：因为f(x)为偶函数且在x ≤0上上是减函数， 所以f(x)在(0,+∞)上是增函数， 则f(2x )>f(12)⇔2x >12，解得x >−1， 所以实数x 的取值范围为x >−1． 故选B ．利用f(x)的奇偶性及在(−∞,0)上的单调性可判断其在(0,+∞)上的单调性，由f(x)的性质可把f(2x )>f(12)，转化为具体不等式，解出即可．本题考查函数奇偶性、单调性的综合运用，解决本题的关键是利用函数的基本性质化抽象不等式为具体不等式，体现转化思想．3.答案：B解析：解：若“x ≠1或y ≠1”，则“xy ≠1， 其逆否命题为：若xy =1，则x =1且y =1．由x =1且y =1⇒xy =1，反之不成立，例如取x =2，y =12． ∴xy =1是x =1且y =1的必要不充分条件． ∴“x ≠1或y ≠1”是“xy ≠1”的必要不充分条件． 故选：B ．若“x ≠1或y ≠1”，则“xy ≠1，其逆否命题为：若xy =1，则x =1且y =1.即可判断出关系． 本题考查了命题之间的关系、简易逻辑的判定方法，考查了推理能力与计算能力，属于基础题．4.答案：B解析：解：①是p 或q 形式的复合命题，p 真q 假，根据真值表，故p 或q 为真，①为真命题；②是真命题；③否命题是“若a≤b，则a+c≤b+c”，根据不等式的性质，③是真命题；④逆命题是“两条对角线相等的四边形是矩形”，是假命题，例如等腰梯形的对角线也相等，但不是矩形．故选B5.答案：[0,1]解析：解得：N=[0,2]，所以得：M∩N=[0,1]6.答案：一解析：解：复数z=(2−i)(1+3i)=5+5i，复数z在复平面上对应的点(5,5)位于第一象限．故答案为：一．利用复数的运算法则、几何意义即可得出．本题考查了复数的运算法则、几何意义，属于基础题．7.答案：2解析：解：由题意可知，如图圆锥的轴截面的顶角∠ASB=120°，所以在直角三角形中，∠OSB=12∠ASB=60°，圆锥的底面半径为r=SB×sin60°=SB×√32=√32SB，高ℎ=SB×cos60°=12SB，所以该圆锥的体积为：V=13×πr2×ℎ=13×π×(√32SB)2×12SB=π．解得SB=2．∴圆锥的母线长为2．故答案为：2．根据题意，求出圆锥的底面半径和高，代入公式即可．本题考查圆锥的体积，求出圆锥的底面半径和高是解决问题的关键，考查空间中线线、线面、面面间的位置关系等基础知识，考查运算求解能力，是中档题．8.答案：3x+2y−7=0解析：解：∵向量a⃗=(6,2)，b⃗ =(−4,12)，∴a⃗+2b⃗ =(6−8,2+1)=(−2,3)；∴过点A(3,−1)且与向量a⃗+2b⃗ 平行的直线l的斜率为k=−32，∴直线l的方程为y−(−1)=−32(x−3)，化简为3x+2y−7=0．故答案为：3x+2y−7=0．根据向量a⃗+2b⃗ 与直线l平行，求出直线的斜率k，利用点斜式求出直线l的方程．本题考查了平面向量的应用问题，也考查了直线方程的应用问题，是基础题目．9.答案：y=3x+1解析：y=x·e x+2x+1，则y′=e x+x·e x+2．∴y′|x=0=3．故在点(0,1)处切线方程为y−1=3x，即y=3x+1．10.答案：1解析：解：由约束条件{x≥0x+y≤2x−y≤0作出可行域如图，化目标函数z=ax+y为y=−ax+z，若a≤0，则−a≥0，由图可知使目标函数取得最大值的最优解唯一，为(0,2)，不合题意；若a>0，则−a<0，要使目标函数z=ax+y取最大值时的最优解有无数多个，则直线y=−ax+z 与直线x+y=2重合，此时a=1．故答案为：1．由约束条件作出可行域，化目标函数为直线方程的斜截式，对a分类可知，若a≤0，则−a≥0，由图可知使目标函数取得最大值的最优解唯一，为(0,2)，不合题意；若a>0，则−a<0，要使目标函数z=ax+y取最大值时的最优解有无数多个，则直线y=−ax+z与直线x+y=2重合，由此求得a值．本题考查简单的线性规划，考查了数形结合的解题思想方法，是中档题．11.答案：，．解析：本题考查的知识点是函数的奇偶性和单调性，由题意，f(x)>f(2)，故|x|>2，解得：x>2或x<−2，故的解集为；同理由可得：|3x|<1，解得：．12.答案：2解析：解：二项展开式的第r+1项为T r+1=C5r(x2)5−r√5−3)r=√5)r C5r x10−5r，令10−5r=0，求得r=2，可得得常数项为√5)2⋅C52=2，故答案为：2．在二项展开式的通项公式中，令x的幂指数等于0，求出r的值，即可求得展开式中的常数项．本题主要考查二项式定理的应用，二项展开式的通项公式，二项式系数的性质，属于基础题．13.答案：①④解析：解：①甲地：5个数据的中位数为24，众数为22，根据数据得出：甲地连续5天的日平均温度的记录数据可能为：22，22，24，25，26.其连续5天的日平均温度均不低于22.②乙地：5个数据的中位数为27，总体均值为24.当5个数据为19，20，27，27，27可知其连续5天的日平均温度有低于22，故不确定．③丙地：5个数据的总体均值为24，且极差小于或等于4，当5个数据，21，24，25，25，25，可知其连续5天的日平均温度有低于22，故不确定．④丁地：5个数据中有一个数据是32，总体均值为26，若有低于22，则取21，此时方差就超出了10.8，可知其连续5天的日平均温度均不低于22．则肯定进入夏季的地区有甲、丁．根据数据的特点进行估计出甲、乙、丙、丁四地连续5天的日平均温度的记录数据，分析数据的可能性进行解答即可得出答案．本题考查中位数、众数、平均数、方差的数据特征，简单的合情推理，解答此题应结合题意，根据平均数的计算方法进行解答、取特值即可．14.答案：144解析：解：根据题意，分2步进行分析：①，将AB两人看成一个元素，与EF2人进行全排列，有A22A33=12种排法，排好后有4个空位，②，在4个空位中任选2个，安排C、D，有A42=12种情况，则有12×12=144种不同的排法，故答案为：144根据题意，分2步进行分析：①，将AB两人看成一个元素，与EF2人进行全排列，易得排好后有4个空位，②，在4个空位中任选2个，安排C、D，由分步计数原理计算可得答案．本题考查排列、组合的应用，注意常见问题的处理方法，属于基础题．15.答案：f−1(x)=√x3(x∈R)解析：解：函数f(x)=x3，反函数为f−1(x)=√x3(x∈R)．故答案为：f−1(x)=√x3(x∈R)．直接利用原函数的定义求出反函数．本题考查的知识要点：原函数的反函数的求法，主要考查学生的运算能力和数学思维能力，属于基础题．16.答案：1解析：解：由题意可得，圆心(−k2−m2)在直线2x−y−1=0上，故有−k+m2−1=0，∴m−2k−2=0．再根据直线y=kx−1与直线2x−y−1=0垂直，可得k=−12，∴m=1，故答案为：1．由题意可得圆心(−k2−m2)在直线2x−y−1=0上，由此求得m−2k−2=0.再根据直线y=kx−1与直线2x−y−1=0垂直，求得k的值，由此求得m的值．本题主要考查直线和圆相交的性质，两条直线垂直的性质，属于基础题．17.答案：解：(1)以A 为坐标原点，AB 、AC 、AA 1分别为x 轴，y 轴，z轴建立空间直角坐标系．依题意有D(2,2，4)，A(0,0，0)，E(2,2，0)，F(0,4，2)， 所以AD ⃗⃗⃗⃗⃗⃗ =(2,2,4),EF⃗⃗⃗⃗⃗⃗⃗ =(−2,2,2)． 设异面直线AD 、EF 所成角为α，则cosα=|AD⃗⃗⃗⃗⃗⃗ ⋅EF ⃗⃗⃗⃗⃗⃗⃗ ||AD ⃗⃗⃗⃗⃗⃗ |⋅|EF ⃗⃗⃗⃗⃗ |=√4+4+16⋅√4+4+4=√23， 所以α=arccos √23，即异面直线AD 、EF 所成角的大小为arccos √23．(2)∵AB =AC =4，AB ⊥AC ，∴BC =4√2，AE =2√2，DE =AA 1=4， ∴S △DEF =12×4×2√2=4√2，由E 为线段BC 的中点，且AB =AC ， ∴AE ⊥BC ，又BB 1⊥面ABC ，∴AE ⊥BB 1， ∴AE ⊥面BB 1C 1C ，∴V D−AEF =V A−DEF =13S △DEF ⋅AE =13⋅4√2⋅2√2=163，∴三棱锥D −AEF 的体积为163．解析：(1)以A 为原点建立空间坐标系，求出AD ⃗⃗⃗⃗⃗⃗ ，EF ⃗⃗⃗⃗⃗ 的坐标，利用向量的夹角公式得出AD ，EF 的夹角；(2)证明AE ⊥平面DEF ，求出AE 和S △DEF ，代入体积公式计算． 本题考查了异面直线所成的角，棱锥的体积计算，属于中档题．18.答案：解：(1)f(x)=2sin 2x +2sinxcosx =1−cos2x +sin2x =1+√2sin(2x −π4)，∴f(x)的最小正周期T =π； f(x)的递减区间为[kπ+3π8,kπ+7π8]，k ∈z ；(2)(2)由(1)列表得： x −3π8−π8 π83π85π8y1 1−√2 11+√2 1故函数y=f(x)在区间[−π2,π2]上的图象是：．解析：(1)利用二倍角的正弦、余弦函数公式化简后，利用两角差的正弦函数公式的逆运算及特殊角的三角函数值化简为一个角的正弦函数，利用周期的计算公式求出函数的周期，根据正弦函数的递减区间求出所找区间；(2)由(1)的解析式列出表格，在平面坐标系中描出五个点，然后用平滑的曲线作出函数的图象即可．本小题主要考查三角函数的基本性质和恒等变换的基本技能，考查画图的技能．19.答案：解：(1)设y=k(a−x)⋅x2，∵当x=a2时，y=a3；∴k=8，∴y=8(a−x)⋅x2，∵y>0，∴0<x<a，又由0≤x2(a−x)≤t，得x≤2at1+2t=a⋅2t1+2t<a．所以定义域为(0,2at1+2t]；(2)y=8(a−x)⋅x2=−8x3+8ax2，y′=−16x2+16ax.=−16x(x−a)在(0,2at1+2t]上，y′>0，f(x)单调递增，当x=2at1+2t 时，y的最大值为f(2at1+2t)=32a3t2(1+2t)3解析：(1)设y=k(a−x)⋅x2，利用当x=a2时，y=a3；求出k=8，从而可得函数表达式，即可求得y=f(x)的定义域；(2)利用函数单调性与导数关系，进行求最值．20.答案：解：(Ⅰ)设直线A1M与A2N的交点为Q(x,y)，∵A 1，A 2是双曲线x 24−y 23=1的左、右顶点，∴A 1(−2,0)，A 2(2,0)，∵M(x 1,−y 1)和N(x 1,y 1)是双曲线上两个不同的动点．∴直线A 1M ：y =−y 1x 1+2(x +2)，直线A 2N ：y =y 1x 1−2(x −2)， 两式相乘，得：y 2=−y 12x 12−4(x 2−4)， 而M(x 1,−y 1)在双曲线x 24−y 23=1上， ∴x 124−y 123=1，即x 12−4=43y 12，∴x 24+y 23=1∴直线A 1M 与A 2N 交点Q 的轨迹C 的方程是x 24+y 23=1．(2)①设直线AB 的方程为y =k(x −1)，A(x 3,y 3)，B(x 4,y 4)，由{y =k(x −1)x 24+y 23=1，得(4k 2+3)x 2−8k 2x +4k 2−12=0， ∴x 3+x 4=8k 23+4k 2，x 3⋅x 4=4k 2−123+4k 2，∴y 3y 4=k(x 3−1)⋅k(x 4−1)=k 2x 3x 4−k 2(x 3+x 4)+k 2，从而OA ⃗⃗⃗⃗⃗ ⋅OB ⃗⃗⃗⃗⃗⃗ =x 3x 4+y 3y 4=(k 2+1)x 3x 4−k 2(x 3+x 4)+k 2=−5k2+124k 2+3， 令t =4k 2+3，t ≥3，则OA ⃗⃗⃗⃗⃗ ⋅OB ⃗⃗⃗⃗⃗⃗ =−(54+334t)， ∵t ≥3，∴0<1t ≤13，54<54+334t <4，∴−4≤−(54+334t )<−54， 故OA ⃗⃗⃗⃗⃗ ⋅OB ⃗⃗⃗⃗⃗⃗ 的取值范围是[−4,−54). ②设在x 轴上存在点E(x 0,0)，则AP ⃗⃗⃗⃗⃗ =(1−x 3,−y 3)，PB ⃗⃗⃗⃗⃗ =(x 4−1,y 4)，∵AP ⃗⃗⃗⃗⃗ =λPB ⃗⃗⃗⃗⃗ ，∴−y 3=λy 4，∵y 4≠0，∴λ=−y3y 4， ∵OP ⃗⃗⃗⃗⃗ =(1,0),EA ⃗⃗⃗⃗⃗ =(x 3−x 0,y 3)，EB ⃗⃗⃗⃗⃗ =(x 4−x 0,y 4)，∴EA ⃗⃗⃗⃗⃗ −λEB ⃗⃗⃗⃗⃗ =(x 3−x 0−λx 4+λx 0,y 3−λy 4)，又∵OP ⃗⃗⃗⃗⃗ ⊥(EA ⃗⃗⃗⃗⃗ −λEB ⃗⃗⃗⃗⃗ )，∴OP ⃗⃗⃗⃗⃗ ⋅(EA ⃗⃗⃗⃗⃗ −λEB ⃗⃗⃗⃗⃗ )=0，∴x 3−x 0−λx 4+λx 0=0，将x 3=1k y 3+1,x 4=1k y 4+1,λ=−y3y 4代入上式，并整理，得： 2k y 3y 4+y 3+y 4=(y 3+y 4)x 0，当y 3+y 4≠0时，x 0=2y 3y 4k(y 3+y 4)+1=2(−9k 2)k(−6k)+1=4，当y 3+y 4=0时，k =0不合题意，∴在x 轴上存在定点E ，使得OP ⃗⃗⃗⃗⃗ ⊥EA ⃗⃗⃗⃗⃗ −λEB ⃗⃗⃗⃗⃗ ，E 点坐标为E(4,0)．解析：(Ⅰ)设直线A 1M 与A 2N 的交点为Q(x,y)，由已知得y 2=−y 12x 12−4(x 2−4)，由此能求出直线A 1M 与A 2N 交点Q 的轨迹C 的方程是x 24+y 23=1．(2)①设直线AB 的方程为y =k(x −1)，A(x 3,y 3)，B(x 4,y 4)，由{y =k(x −1)x 24+y 23=1，得(4k 2+3)x 2−8k 2x +4k 2−12=0，由此利用韦达定理结合已知条件能推导出OA⃗⃗⃗⃗⃗ ⋅OB ⃗⃗⃗⃗⃗⃗ 的取值范围． ②设在x 轴上存在点E(x 0,0)，则AP ⃗⃗⃗⃗⃗ =(1−x 3,−y 3)，PB ⃗⃗⃗⃗⃗ =(x 4−1,y 4)，由AP ⃗⃗⃗⃗⃗ =λPB ⃗⃗⃗⃗⃗ ，得λ=−y3y 4，由OP ⃗⃗⃗⃗⃗ ⊥(EA ⃗⃗⃗⃗⃗ −λEB ⃗⃗⃗⃗⃗ )，得：2ky 3y 4+y 3+y 4=(y 3+y 4)x 0，由此推导出在x 轴上存在定点E ，使得OP ⃗⃗⃗⃗⃗ ⊥EA ⃗⃗⃗⃗⃗ −λEB ⃗⃗⃗⃗⃗ ，E 点坐标为E(4,0)．本题考查点的轨迹方程的求法，考查向量积的取值范围的求法，考查满足条件的定点坐标是否存在的判断与求法，解题时要认真审题，注意函数与方程思想的合理运用． 21.答案：解：(1)设等差数列{a n }的公差为d ，等比数列{b n }的公比为q ，由a 1=b 1=1，a 2=a 3−b 3，a 3=S 3+b 2，即为{1+d =1+2d −q 21+2d =1+q +q 2+q， 解得{d =4q =2或{d =0q =0(舍去)， 则a n =1+4(n −1)=4n −3，b n =2n−1；(2)c n =a n ⋅b n =(4n −3)⋅2n−1，T n =1⋅20+5⋅21+9⋅22+⋯+(4n −3)⋅2n−1，2T n =1⋅2+5⋅22+9⋅23+⋯+(4n −3)⋅2n ，两式相减可得，−T n =1+4(21+22+⋯+2n−1)−(4n −3)⋅2n=1+4⋅2(1−2n−1)1−2−(4n −3)⋅2n ，化简可得，T n =7+(4n −7)⋅2n ．解析：(1)设等差数列{a n }的公差为d ，等比数列{b n }的公比为q ，由等差数列和等比数列的通项公式，解方程可得公差和公比，即可得到所求；(2)求得c n=a n⋅b n=(4n−3)⋅2n−1，由数列的错位相减法求和，结合等比数列的求和公式，计算可得所求和．本题考查等差数列和等比数列的通项公式和求和公式的运用，以及数列的错位相减法求和，考查方程思想和运算求解能力，属于中档题．。
上海市崇明区2020-2021学年高一上学期期中考试数学试题 2020.11含答案
三、解答题
17．（1）计算： ；
（2）已知 ， ，化简： ．
18．解下列不等式：
（1） ；（2） ．
19．已如全集为 ，集合 ， ．
（1）当 时，求 ；（2）若 ，求实数 的范围．
20．（1）已知 பைடு நூலகம்实数，集合 ， ．求证：“ ”是“ ”
的充要条件．
（2）设 ．用反证法证明命题“若 ，则 或 ．”
∴②正确；对于③：由韦达定理，知 ，∴③错误；
对于④：由韦达定理，知 ，
则 ，解得 ，∴④正确；
综上，真命题的个数是3，选C．
三、解答题
17．（1）2；（2） ．
18．（1） ；（2） ．
19．（1） ， ， ；
（2） ，∵ ，∴ 且 ，解得， ．
20．（1）先证充分性（即证 ）．当 时， ．又因为 ，所以 ．
再证必要性（即证 ）．当 时，由 ，得 ，
因此 ．
综上所述，“ ”是“ ”的充要条件．
（2）假设 且 ，则 ，这与已知条件 矛盾．
所以假设不成立，即 或 ．
21．（1）由题意， ，∵ ，∴ ，即 ，
两边平方，得 ；
（2）即证 ，即证 ，
∵ ，∴ ，即证 ，即证 （*），
∵ ，∴ ，
∴ ，（*）成立，即 比 更远离 ；
（3）∵ ，∴ ，
从而 ，
① 时，
，
即 ；
② 时，
，
即 ；
综上， ，即 比 更远离 ．
9．已知 都是正实数，且 ，则当 时， 取得最小值．
10．不等式 的解集为 ，且 ，则实数 的取值范围是．
11．已知 ， 且 ．式子 的最小值是．
12．从集合 的子集中选出两个非空集合 ，满足以下两个条件：① ， ；②若 ，则 ．共有种不同的选择．
上海市崇明区2025届高三第一次模拟考试语文试卷(含答案)
上海市崇明区2025届高三第一次模拟考试语文试卷（考试时间150分钟，满分150分。
请将答案填写在答题纸上）一、积累应用10分1.按要求填空。
(5分)(1)自见者不明，______________________。
（《老子》）(2) ______________________,愿乞终养。
(《陈情表》)(3)李白《梦游天姥吟留别》中“_________________，_________________”两句直抒胸臆，表现了作者蔑视权贵、傲岸不屈的精神。
2.按要求选择。
(5分)(1)下列内容是小明围绕“词语的语体色彩”准备写作语言札记收集的例句，请为他选择语体色彩使用最恰当的一项( )。
(2分)A.全校领导会议开场白：尊敬的各位同事，大家好！我谨以个人的名义邀请大家一起来聊聊咱们学校未来的五年规划。
B.公司季度业绩报告：本季度业绩下滑，主要是因为竞争激烈，加上原材料价格疯长，真是“屋漏偏逢连夜雨”啊。
C.广播电台儿童广播剧：从前有一只小兔子，它生活在一片广袤无垠的森林之中，每天都在奋力地探索未知的世界。
D.给专家发送的邀请函：尊敬的赵老师，我们诚挚地邀请您参加我校举办的读书节活动并为广大师生做主旨演讲。
(2)将下列编号的语句依次填入语段空白处，语意连贯的一项是( )。
(3分)以深度学习为代表的人工智能技术，可有效地表示或逼近高维空间所形成的函数，这种解题模式，___________——___________，___________。
___________，___________，通过生成模型和判别模型间的互相博弈来产生更优输出，在极短时间内遍历更大的未知空间，为科学的发展带来新的可能。
①需通过大算力来破解多维参数间的逻辑关系②人工智能具有的深度探索优势可以开启远超科学家能力范围的潜在知识域③直接绕开传统分析方法的困境④更为值得一提的是⑤直接获知多维参数及数据中所隐藏的模式A.①③②④⑤B.⑤④③②①C.③①⑤④②D.④⑤①②③二、阅读70分(一)阅读《事实与虚构——论边界》一书的书评，完成第3-7题。
2021届-上海市崇明区高三一模-物理试卷+参考答案+评分标准
高三物理 共5页 第1页崇明区2020学年第一学期等级考第一次模拟考试试卷高 三 物 理（考试时间60分钟，满分100分．请将答案填写在答题纸上）一、选择题（40分，1-8题每题3分，9-12题每题4分）1．首先发现小磁针在通电导线周围会发生偏转现象的物理学家是 (A) 法拉第 (B) 奥斯特(C) 科拉顿(D) 麦克斯韦2．楞次定律是下列哪个定律在电磁感应现象中的具体体现？(A) 电阻定律(B) 库仑定律(C) 欧姆定律(D) 能量守恒定律3．湖面上的水波可以绕过障碍物是由于波在传播过程中发生了(A) 反射(B) 折射(C) 干涉(D) 衍射4．根据牛顿第一定律，我们可以认识到物体 (A) 只有在不受力时才具有惯性 (B) 维持运动状态的原因是受力的作用 (C) 运动状态改变的原因是受力的作用(D) 只有不受力时牛顿第一定律才适用5．2020年12月1日，我国嫦娥5号月球探测器成功登陆月球．在着陆月球之前，首先绕月球做绕月飞行．设月球的质量为M ，嫦娥5号的总质量为m ，绕月做圆周运动的半径为R ，引力常量为G ，则嫦娥5号绕月飞行速度(A)v(B)=v(C)=v(D)v 6．某物体以20 m/s 的初速度竖直上抛，不计空气阻力，g 取10m/s 2．则物体3s 内的 (A) 路程为15 m(B) 位移大小为20m(C) 速度改变量大小为10 m/s(D) 平均速度大小为5 m/s7．真空中某点电荷的等势面示意如图，图中相邻等势面间电势差相等．则 (A) 该点电荷一定为正电荷 (B) P 点的场强一定比Q 点的场强大 (C) P 点电势一定比Q 点电势低(D) 正检验电荷在P 点比在Q 点的电势能大高三物理 共5页 第2页8．动车组是由几节自带动力的车厢（动车）和几节不带动力的车厢（拖车）编成的组．设动车组运行过程中的阻力与质量成正比，每节动车与拖车的质量相等，每节动车的额定功率都相等。
若开一节动车带三节拖车时，最大速率为120 km/h ，那么当开五节动车带三节拖车时，最大速率为(A) 60 km/h(B) 240 km/h(C) 300 km/h(D) 600 km/h9．图示为同一位置的甲乙两个单摆的振动图像，根据 图像可以知道两个单摆的 (A) 甲的摆长大于乙的摆长 (B) 甲摆球质量大于乙摆球质量 (C) 甲摆球机械能大于乙摆球机械能(D) 摆球甲的最大偏角大于乙的最大偏角10．以下各图中的p 表示质子，e 表示电子，距离D ＞d ，其中O 点电场强度最大的粒子排布方式为11．如图，两端开口的弯管，左管插入水银槽中，管内外水银面高度差为h 1，右侧管有一段水银柱，两端液面高度差为h 2，中间封有一段空气．若 (A) 温度升高，则h 1增大，h 2增大 (B) 大气压升高，则h 1增大，h 2增大 (C) 弯管下移少许距离，则h 1增大，h 2不变(D) 右管中滴入少许水银，则h 1不变，h 2增大12．如图所示电路中，R 1、R 2为定值电阻，电源内阻为r ．闭合电键S ，电压表显示有读数，调节可变电阻R 的阻值，电压表示数增大量为ΔU ，则在此过程中 (A) 路端电压一定增大，变化量大于ΔU (B) 电阻R 2两端的电压减小，变化量等于ΔU (C) 可变电阻R 阻值增大，流过它的电流增大(D) 通过电阻R 2的电流减小，变化量小于2URt/sx/cm 0甲乙高三物理 共5页 第3页二、填空题（20分，每题4分）13．如图所示，一只质量为m 的小虫子沿弧形树枝缓慢向上爬行，A 、B 两点中在 点容易滑落；弧形树枝B 点切线的倾角为θ，则虫子经该位置时对树枝的作用力大小为 ．14．如图，一列简谐横波平行于x 轴正方向传播，经过0.1t =s 时间，从图中的实线波形变为虚线波形。
上海市崇明区2023届高三上学期高考一模数学试卷带讲解
（3）在不同情况计算最大值，然后比较两个最大值就可以得到面积最大值，然后确定 的位置.
【小问1详解】
以 为坐标原点， 、 所在直线分别为 轴、 轴建立平面直角坐标系，
如图所示，则 ， ， ，
设曲线段 所在抛物线的方程为 ，
由题意可知，点 和 在此抛物线上，
【小问1详解】
由题意， ， ， ，
由题意， ， ， ，故 ，所以 ，
又 ，所以 ，
所以椭圆的标准方程为 .
【小问2详解】
当 时，椭圆方程为 ，则 ， ，
由对称性，不妨设点P在x轴上方，则直线AP的方程为 ，代入椭圆方程，得
，解得 （舍去）， ，所以 ，
所以 .
【小问3详解】
设 ，则 ，
直线AP 方程为 ，
【详解】设椭圆 标准方程为 ，椭圆离心率为 ，
设双曲线 标准方程为 ，双曲线离心率为 ，
由题可知： ．
设 ， ，
则 ，
由①②得， ， ，
代入③整理得， ，
两边同时除以 得， ，
即 ，
即 ，
解得 ，即 ．
故答案为：
【点睛】本题综合考查椭圆和双曲线的几何性质，解题关键是熟练应用椭圆和双曲线的定义，结合焦点三角形中的余弦定理，列出方程组即可求解.
【详解】将点 代入 得 ，解得
故答案为：2.
5.设等比数列 满足a1+a2= –1,a1–a3= –3，则a4=___________．
【答案】-8
【详解】设等比数列 的公比为 ，很明显 ，结合等比数列的通项公式和题意可得方程组：
，由 可得： ，代入①可得 ，
由等比数列的通项公式可得 .
上海市崇明县2021届高考数学达标测试试题
2019-2020学年高考数学模拟试卷一、选择题：本题共12小题，每小题5分，共60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．若0a b <<，则下列不等式不能成立的是（ ） A ．11a b> B ．11a b a>- C ．|a|>|b|D ．22a b >2．已知底面为正方形的四棱锥，其一条侧棱垂直于底面，那么该四棱锥的三视图可能是下列各图中的（ ）A ．B ．C ．D ．3．复数21iz i=-（i 为虚数单位），则z 等于（ ） A ．3 B ．2 C ．2D 24．5()(2)x y x y +-的展开式中33x y 的系数为( ) A ．－30B ．－40C ．40D ．505．已知双曲线2221x y a -=的一条渐近线方程是33y x =，则双曲线的离心率为（ ）A 3B 6C 3D 236．已知(2)f x +是偶函数，()f x 在(]2-∞，上单调递减，(0)0f =，则(23)0f x ->的解集是 A ．2()(2)3-∞+∞，，B ．2(2)3， C ．22()33-，D ．22()()33-∞-+∞，， 7．已知双曲线2222:1(0,0)x y C a b a b-=>>的一个焦点为F ，点,A B 是C 的一条渐近线上关于原点对称的两点，以AB 为直径的圆过F 且交C 的左支于,M N 两点，若|MN|=2，ABF ∆的面积为8，则C 的渐近线方程为（ ） A ．3y x =± B ．33y x =± C ．2y x =±D ．12y x =±8．在直角梯形ABCD 中，0AB AD ⋅=，30B ∠=︒，23AB =，2BC =，点E 为BC 上一点，且AE xAB y AD =+，当xy 的值最大时，||AE =( )A ．5B ．2C ．302D ．239．若424log 3,log 7,0.7a b c ===，则实数,,a b c 的大小关系为（ ） A ．a b c >>B ．c a b >>C ．b a c >>D ．c b a >>10．已知函数()()sin 0,2f x x πωϕωϕ⎛⎫=+>< ⎪⎝⎭，1,03A ⎛⎫ ⎪⎝⎭为()f x 图象的对称中心，若图象上相邻两个极值点1x ，2x 满足121x x -=，则下列区间中存在极值点的是（ ） A ．,06π⎛⎫-⎪⎝⎭B ．10,2⎛⎫ ⎪⎝⎭C ．1,3π⎛⎫⎪⎝⎭D ．,32ππ⎛⎫ ⎪⎝⎭11．某几何体的三视图如图所示，图中圆的半径为1，等腰三角形的腰长为3，则该几何体表面积为（ ）A ．7πB ．6πC ．5πD ．4π12．已知()f x 是定义是R 上的奇函数，满足3322f x f x ⎛⎫⎛⎫-+=+ ⎪ ⎪⎝⎭⎝⎭，当30,2x ⎛⎫∈ ⎪⎝⎭时， ()()2ln 1f x x x =-+，则函数()f x 在区间[]0,6上的零点个数是（ ）A ．3B ．5C ．7D ．9二、填空题：本题共4小题，每小题5分，共20分。