南京市、盐城市2020届高三二模英语考试试题(含解析答案)

京市、盐城市2022届高三年级第二次模拟考试地理一、单项选择题：本大题共22小题，每题2分，共计44分。

小明在自家朝南的外墙壁上自制了一个“墙壁钟表”，架设了与地轴平行的杆子，其日影可用于日常计时，而上还能装饰墙面，下图为其设计的结构图。

据此完成下面小题。

1 该地可能位于我国（）A. 成都B. 兰州C.哈尔滨D. 杭州2. 小明在垂直线上标记了刻度，一年中正午杆子端点的影子会在OP间移动，当其落在OP中点时。

该日北半球（）A. 昼夜等长B. 昼长夜短C. 昼短夜长D. 昼长增加3. 该形式的“墙壁钟表”（）A. 只能用于冬半年计时B. 二分日使用时间最长C. 时间刻度线均匀分布D. 纬度越低计时越方便【1题详解】根据“墙壁钟表”的使用原理，可知主要是通过杆子的影子朝向来确定时间，已知该地地方时13点为正午，可知该地地方时比北京时间慢一个小时，因此可知该地位于120°E的西侧，且慢一个小时，约15°，可知该地经度为105°E左右，成都的经度最为接近，A正确。

兰州经度为100°E左右，B错误。

哈尔滨经度125°E左右，C错误。

杭州经度120°E左右，D错误。

故选A。

【2题详解】由材料可知，影子会在OP之间移动，当在P点是地，应为太阳高度最大的夏至，当其落在O点为冬至日，由于地球公转速度在夏半年慢于冬半年，使夏半年长于冬半年，当影子落在OP中点时，还是北半球的夏半年，因此，此时北半球为昼长夜短，B正确，ACD错误，故选B。

【3题详解】根据墙壁正视图可知，二分日时，太阳从正东升，影子在正西，太阳从正西落下，影子在正东，其反映白天的时长正好符合该钟表设置的白天时长12个小时（7点日出，19点日落），冬半年，太阳从东南升，西南落，时钟无法反映日出和日落时间，夏半年，反映出来白天时间小于12个小时，B正确，A 错误。

“墙壁钟表”的运用原理与地平式日晷一致，晷面必须严格水平，由于日晷的产生是从观察阳光下竖直物体影子的方向变化而开始的，表影随时间变化的实际情况来进行刻划，如果采用均匀刻划的方法会带来相当大的计时误差，因此时间刻度线分布不均匀，C错误。

江苏省南京市、盐城市2024届高三第二次模拟化学试题(满分100分)注意事项：1．答题前，考生先将自己的姓名、考生号、座号填写在相应位置，认真核对条形码上的姓名、考生号和座号，并将条形码粘贴在指定位置上。

2．选择题答案必须使用2B铅笔(按填涂样例)正确填涂；非选择题答案必须使用0.5毫米黑色签字笔书写，字体工整、笔迹清楚。

3．请按照题号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．保持卡面清洁，不折叠、不破损。

可能用到的相对原子质量： H-1 Li-7 C-12 N-14 O-16 Mg-24 S-32 Cl-35.5 K-39 V-51 Fe-56 Ti -48 Co -59一、单项选择题：共13题，每题3分，共39分。

每题只有一个选项最符合题意。

1．“神舟飞船”接力腾飞、“太空之家”遨游苍穹、“福建号”航母下水、国产“C919”大飞机正式交付都彰显了中国力量。

下列成果所涉及的材料为金属材料的是A．“神舟十五”号飞船使用的耐辐照光学窗材料——石英玻璃B．“天宫”空间站使用的太阳能电池板材料——砷化镓C．“福建号”航母使用的高强度甲板材料——合金钢D．“C919”大飞机使用的机身复合材料——碳纤维和环氧树脂2．用NaCN溶液浸取矿粉中金的反应为4Au+2H2O+8NaCN+O2=4Na[Au(CN)2]+4NaOH。

下列说法正确的是A．H2O的空间结构为直线形B．NaCN中含有离子键和共价键C．1mol[Au(CN)2]-中含有2molσ键D．NaOH的电子式为····Na:O:H3．由SiCl4制备高纯SiH4的反应为SiCl4+LiAlH4乙醚SiH4↑+LiCl+AlCl3。

下列说法正确的是A．热稳定性：HCl＞SiH4B．离子半径：r(Li+)＞r(H-)C．第一电离能：I1(Al)＞I1(Cl) D．共价键的极性：Si－Cl＞Al－Cl 4．用如图所示装置制备氨气并验证氨气的还原性，其中不能达到实验目的的是A．用装置甲生成氨气B．用装置乙干燥氨气C．用装置丙验证氨气的还原性D．用装置丁和戊分别收集氨气和氮气5．元素周期表中VIA族元素单质及其化合物有着广泛应用。

高三英语试卷（2020.05）(满分140分，考试时间120分钟)I Listening ComprehensionSection ADirections: In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A Supermarket. B. Tailors shop. C. Department store. D. Convenience store.2. A. 200 pound. B. 600 pounds. C. 300 pounds. D. 700 pounds.3. A. Call the ticket office later. B. Order the tickets onlineC. Not to buy the ticket on the Internet.D. Order the tickets when it's not busy.4.A. Borrowing money from a business company. B Lending some money to a student.C. Asking for some financial aid. D Reading students’ application.5. A. The bed is to blame for his not falling asleep. B. He can fall asleep if he stops drinking.C. He can drink more to fall asleep easily.D. Drinking is good for sleeping.6. A. He is satisfied with the content. B. He feels sorry for it.C. He thinks it is valueless.D. He thinks it is valuable.7. A Professor Smith spoke Greek when he explained the maths problem.B. The woman still didn’t understand the maths problem.C. Unfortunately, she didn’t hear Professor Smiths explanation.D. Professor Smith didn't explain the problem clearly.8. A. Collect papers for the man. B. Do the typing once again.C. Check the paper for typing errors.D. Read the whole newspaper.9.A. Proceed in his own way. B. Stick to the original plan.C. Negotiate with his colleague.D. Try to change his colleagues mind.10. A. His project proved to be unsuccessful. B. He was unable to get sufficient money.C. Lack of land prevented his success.D. He was successful with his project.Section BDirections: In Section B, you will hear several longer conversation(s)and short passage(s), and you will be asked several questions on each of the conversation(s)and the passage(s). The conversation(s)and the passage (s) will beread twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one is the best answer to the question you have heard.Questions 11 through 13 are based on the following talk.11. A. To guide and help children's play. B. To give children an opportunity to play.C. To make children excited. D To keep children company.12. A. It determines the standard a child can reach.B. It is the happiest period during one's life.C. It is the most important time to shape one’s character.D. It is the best time for children to learn new thing.13. A. The relationship between play and learning.B. The way to help children develop both physically and mentally.C. The importance of children’s play.D. Different stages of children’s development.Questions 14 through 16 are based on the following passage.14. A. It tends to wander towards unpleasant experiences.B. It wanders for almost half of their waking time.C. It has trouble concentrating after a brain injuryD. It tends to be affected by their negative feelings.15. A. To find how happiness relates to daydreaming.B. To observe how one’s mind affects one’s behavior.C. To see why daydreaming impacts what one is doing.D. To study the relation between health and daydreaming.16. A. Participants with clear goals in mind outperformed those without clear goals.B. The difference in performance between the two groups was insignificant.C. Non-daydreamers were more confused on their tasks than daydreamers.D. Daydreamers did better than non-daydreamers in task performance.Questions 17 through 20 are based on the following conversation.17. A. Their average lifespan was less than 50 years.B. It was very common for them to have 12 children.C. They retired from work much earlier than today.D. They were quite optimistic about their future.18. A. Get ready for ecological changes.B. Adapt to the new environment.C. Learn to use new technology.D. Explore ways to stay young.19. A. When all women go out to work.B. When family planning is enforced.C. When a world government is set up.D. When all people become wealthier.20. A. Eliminate poverty and injustice.B. Migrate to other planets.C. Control the environment.D. Find inexhaustible resources.Ⅱ. 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.Plants Scream in the Face of StressFor the first time, researchers appear to have evidence that like animals, those plants deprived of water or (21)__________(force)to endure bodily harm can let out their pain. The study. (22)_________ has yet to be published in a scientific Journal, adds another dimension to scientists(23)________ (grow)understanding of how plants detect and interact with their surroundings.In recent years, it has become very clear that plants are more sensitive than researchers (24)_________ (think). They respond when touched by insects and turn toward sources of light. “Plants are not just robotic stimulus-response devices,” said Frantisek Baluska of the University of Bonn in Germany. “They’re living organisms which have their own problems.”Actually making their suffering hearable, however, is another matter entirely.(25)________(test) that possibility, a team led by Itzhak Khait, a plant scientist at Tel Aviv University in Israel, placed microphones capable of detecting ultrasonic frequencies(超声波频率) four inches from tomato and tobacco plants. The researcher then either stopped watering them or cut their stems.Measuring in the range of 20 to 150 kilohertz (千赫) the researchers found that even happy healthy plants madethe occasional noise. But when cut, tobacco plants emitted (26)_________average of 15 sounds within an hour of being cut, (27)_________tomato plants produced 25 sounds.(28)_______researchers aren’t yet sure how plants produce these sounds, Khait and his colleagues proposed one possibility in their paper (29)____________as water travels through the plants' tubes, air bubbles will form and explode, producing small vibrations.All this “screaming” caused by stress wasn’t in a range detectable by human ears. But organisms that can hear ultrasonic frequencies like mice, bats or perhaps other plants (30)________possibly hear the plants cries from as far away as 15 feet.Section BDirections: Fill in each blank with a proper word chosen from the box. Each word can be used only once. Note that there is one word more than you need.Italians find “Moments of Joy in this Moment of Anxiety”It started with the national anthem. Then came the piano chords, trumpet blasts, violin serenades(小夜曲) and even the clanging of pots and pans--all of it (31)_________from people’s homes, out of windows and from balconies, and resounding across rooftops.Finally, on Saturday afternoon, a nationwide (32)__________of applause broke out for the doctors on the medical front lines fighting the spread of Europe’s worst coronavirus outbreak.Italians remain (33)_________under house arrest as the nation, the European front in the global fight against the coronavirus, has ordered extraordinary restrictions on their movement to prevent infection.But the music and noise erupting over the streets, from people (34)_________in their homes, reflects the spirit, resilience and humor of a nation facing its worst national emergency since the Second World War.To the extent that this is a virus that tries people' s souls, it has also demonstrated the (35)__________of those national characters.In China, patriotic truck drivers risked infection to bring(36)_________needed food to the people of Wuhan, the epicenter of the outbreak. In Iran, videos show doctors in full combat dress and masks dancing to keep (37)_______up. And in Italy, the gestures of gratitude and music ring out above the country’s empty streets, while social media feeds fill with (38)___________, sentimental and humorous web videos.Images of nurses collapsed from exhaustion or their faces bruised(使受瘀伤) from tightly(39)________maskshave also spread across the web in recent days. Parents posted pictures of unicorns and rainbows drawn by young children with the title “It will all be OK.”“We’re Italians, and loving singing is part of our culture,” said Giorgio Albertini, 51, an archaeology professor who clapped from his apartment balcony in the university district of Milan, calling it a way “to feel a community, and to have the (40)_________grief.”Ⅲ 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.Making choices is hard. That would be why researcher Moran Cerf has (41)_________it from his life. As a rule, he always chooses the second menu item at a restaurant.This is (42)___________by his research in neuroeconomics(神经经济学)( a somewhat new, divisive field) at Northwestern University. As Business Insider describes, Cerf has extended his ideas which draw on some controversial ideas in psychology, including ego depletion out into a piece of advice that, to (43)_________happiness, people should "build a life that requires (44)________decisions by surrounding themselves with people who possess traits they prefer.On an instinctive level, Cerf’s idea (45)_____________: Many choices people make are the product of social pressures and the inputs of (46)___________people around them. One example Cerf furnishes is that, (47)________consistently ordering the second menu item. he never picks where to eat. Rather, he (48)________his decision to his dining partner which friend he plans to eat with, probably one he trusts and always lets them pick.While it's (49)__________what, if any, scientific principles underlie those pieces of advice, there is no shortage of research showing that choices can sometimes feel more(50)_______than liberating. An example from Quanta poits (假设): If you have a clear love of Snickers(士力架), choosing that over an Almond Joy(杏仁巧克力) or a Milky Way(牛奶巧克力) should be a(51)________. And, as an experiment conducted by neuroscientist Paul Glimcher at NYU shows, most of the time it is, (52)_________you introduce more choices. When the participants were offered three candy bars (Snickers, Milky Way, and Almond Joy) they had no problem picking their favorite, but when they were given the option of one among 20, including Snickers, they would sometimes drift away from their (53)_____. When the choices were taken away in later trials. the participants would wonder what caused them to make such a bad decision.As Quanta details, according to a model called" 'divisive normalization(分裂归一化), which has gained some popularity, the way the brain encodes choices has a lot to do with how it values all its options. So, if you have twothings that are clearly (54)___________, brain areas involved in decision-making fire in a pattern that makes the decision clear. When the choices are comparable. the brain does its best to focus on the distinctions between the two, but more choices (55)_________ that ability out.41. A. relieved B. released C. eliminated D. liberated42. A. influenced B. inherited C. implemented D. informed43. A. maximize B. balance C. cherish D. seek44. A safer B. fewer C. better D. sounder45. A. stands out B comes into force C makes sense D. play a part46. A. distinguished B. trusted C. authorized D. honored47. A. in addition to B. instead of C. in spite of D. regardless of48. A. conveys B. relates C. submits D. limitsA. evidentB. unclearC. criticalD. inevitable50. A. confusing B. Inspiring C. worrying D. appealing51. A. stressor B. no-brainer C. challenge D. headache52. A. after B. before C. when D. until53. A. preference B. struggle C. status D. directionA. impressiveB. insignificantC. uniqueD. distinct55. A. crowd B. figure C. sort D. putSection 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)One spring day, once the flowers have begun to open, a bee will hover (盘旋)and zip through your yard and dive-bomb your picnic table. While you're thinking about avoiding an attack, that bee is focused on something else entirely: me.A honeybee has about six weeks to live. Today, like most days, her task is to fly as many as three miles from home. stick her long. straw like tongue into a hundred or so flowers. When the bee has had her fill. she’ll fly home. There the bee will deposit what she has got into the mouth of one of her co-workers, who will relay it to another, and so on for about 20 minutes, until the mixture is ready to be placed into the comb. Then she and her 50.000 or so mates will hover in the dark all night every night, flapping their wings to create hot, breezy conditions to remove the waterfrom the mixture. Several sunrises later, they will seal me off in a golden cell of beeswax. In her lifetime, our bee may visit 4,000 flowers, and yet will produce only one-twelfth a tea spoon of me.The average American consumes nearly a pound and a half of me every year, in tea, on toast, and beyond. If I do say so myself, I am a timeless treasure. Literally I never go bad.Unfortunately, my good health is not guaranteed. The problem lies in the growth of industrial agriculture and the use of pest control chemicals, as well as changes in weather patterns, all of which reduce the number of flowers bees have to visit. I’d appreciate your letting your own garden grow just a little wild. My future depends on all of us fostering spring and summers wild flowers, thus helping the bees, who give so much to you, to me without ever asking for anything in return.56. What does “me” refer to in the passage?A. The flower.B. The bee.C. Water. D Honey.57. What is the 2nd paragraph mainly about?A. Bees’ special talent.B. Bees’ hard work.C. Bees’ living environment.D. Bees’ social behavior.58. Which one of the following is true according to the passage?A. A bee will always prioritize attacking picnic lovers.B. Before “me” is sealed off in beeswax, the drying process can take a few nights.C. The lifework of a bee satisfies the average demand of an American consumer annually.D. Bees are more likely to visit those deliberately pest-controlled gardens59. What is the purpose of the passage?A. To appeal for help for honeybees.B. To talk about the history of a treasure.C. To put forward techniques for gardeners.D. To argue against the control of chemicals.(B)Get Your Unlimited Card at Cineworld CinemasEnjoy Unlimited FilmsWatch all the films you want at any Cineworld for just one monthly price.Being an Unlimited card holder gives you access to all the 2D films you canhandle for one monthly price. Watch what you want, when you want as manytimes as you want. Plus, save money when watching films in 3D and others.After you've been with us for more than I year we'll upgrade you to a Premium Card and you' ll get into 3D films completely free too! Start enjoying today by using a temporary pass while you wait for your card to arrive in the post.Save On Snacks And DrinksGet 10% off all in-cinema food and drink. Plus get 10% off atCineworld Starbucks licensed stores. First year card holders get 10%o off,whilst Premium card holders get 25% off Cineworld's in-cinema food anddrinks including, all drinks, popcorn, nachos, hotdogs, ice cream, pick nmix and much more. Plus all card holders get 10% discounts at Cineworld Starbucks licensed stores. All you need to do is show your card at the counter and your discount will be applied.Recommend A FriendUnlimited members can get free months of membership when theyrecommend Unlimited to their friends!Recommend Unlimited to your friends and well give you free monthsof membership to say thank you. For every friend that signs up using yourunique Recommend a Friend code you will both receive one month’s free membership once they have been an active Unlimited member for 90 days. The free month will be automatically added to the end of your current subscription. You can earn a maximum of 12 Free Months with your Recommend a Friend code, so recommend Unlimited to 12 friends and you could get a full year of free Unlimited cinema!60. The cinema names its membership card “Unlimited” because__________.A. one can have the benefits for good upon joining the membership.B. Cineworld members can enjoy as many 2D and 3D films as they like for free.C. it frees a member from any regular payment to the movie tickets at Cineworld.D. card holders can share limitless discounts and offers with friends and relatives.61. Which of the statement is TRUE according to the passage?A. The benefits above are not available until the card is delivered.B. Premium card holders can have 25%o off at a licensed Starbucks.C. Whoever persuades 6 friends into Unlimited can enjoy a half year of free membership.D. A second year of investment is worthwhile if you are a cinema goer.62. This passage is probably written to____________.A. secure the loyalty of potential customers.B. introduce the latest movies and discounts.C promote the popularity of Cineworld cinemas.D. give away movie cards to readers for free.(C)A rare hole has opened up in the ozone layer above the Arctic, in what scientists say is the result of unusually low temperatures in the atmosphere above the north pole.The hole, which has been tracked from space and the ground over the past few days, has reached record dimensions, but is not expected to pose any danger to humans unless it moves further south. If it extends further south overpopulated areas, such as southern Greenland, people would be at increased risk of sunburn. However, on current trends the hole is expected to disappear altogether in a few weeks.Low temperatures in the northern polar regions led to an unusual stable polar vortex(极地漩涡) and the presence of ozone-destroying chemicals such as chlorine(氯)in the atmosphere-from human activities caused the hole to form.“The hole is principally a geophysical curiosity.” said Vincent-Henri Peuch, director of the Copernicus Atmosphere Monitoring Service. “We monitored unusual dynamic(动态的)conditions, which drive the process of chemical depletion of ozone. Those dynamics allowed for lower temperatures and a more stable vortex than usual over the Arctic, which then triggered the formation of polar stratospheric(平流层的)clouds and the catalytic(催化的) destruction of ozone.”The hole is not related to the Covid-19 shutdowns that have dramatically cut air pollution and reduced greenhouse gas emissions. It is also too early to say whether the unusually stable Arctic polar vortex conditions are linked with the climate crisis, or part of normal stratospheric weather variability.Peuch said there were no direct implications for the climate crisis. Temperatures in the region are already increasing, slowing the depletion of ozone, and the hole will start to recover as polar air mixes with ozone rich air from lower latitudes. The last time similar conditions were observed was in spring 2011.While a hole over the Arctic is a rare event, the much larger hole in the ozone layer over the Antarctic has been a major cause for concern for more than four decades. The production of ozone depleting chemicals has been dramatically reduced, under the1987 Montreal Protocol(蒙特利尔协议), but some sources appear still to be functioning in 2018unauthorized emissions were detected from some areas.New sources of ozone depleting chemicals were not a factor in the hole observed in the Arctic, said Peuch. “However, this is a reminder that one should not take the Montreal Protocol measures for granted and that observations from the ground and from satellites are central to avoid a situation where the ozone destroying chemical level in the stratosphere could increase again.”63. What is the possible meaning of the underlined word “depletion”?A. replacement.B. consumption.C. increase.D. production.64. According to the passage, scientists are concerned about the hole because____________.A. it is expected to be a threat to the mankind.B. the new hole is caused by air pollution and greenhouse gas emissions.C. it may encourage further scientific research and environmental awareness.D. it wars us of an oncoming climate crisis.65. What can be learned from the last two paragraphs?A. The hole over the Arctic shares the same causes as the one over the Antarctic.B. Human activities are highly responsible for producing ozone-destroying chemicals.C. The Montreal Protocol has successfully prevented new emissions.D. Some new illegal emissions are to blame for the hole over the Arctic.66. The best title for the passage is probably_________.A. Record-size Hole Opens in Ozone Layer above the Arctic.B Actions Urgently Needed for a New Hole in Ozone Layer.C Environmental Disaster and International Cooperation.D. How a Hole in Ozone Affects our Life on Earth.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.People like to post their selfies(自拍照)on social media. To know more about it, scientists at Syracuse University in New York recently did a research and came up with some surprising findings.People who post selfies and use editing software to make themselves look better show behaviors connected to narcissism, the researchers said. (67)_________Makana Chock, a professor from Syracuse University. said because social media is mostly used by people to share unimportant information about their lives, it is a good place for people to “work towards satisfying their own vanity.” Those “likes” under their Facebook selfies make them feel good.(68)_________Some people feel “peer pressure” to post selfies and some follow the popular belief that if there is no picture of an event or experience. it did not really happen. “Anyway, it shouldn’t be seen as negative. People get sense of satisfaction especially when they get likes. And it does no harm,” Chock said.Other findings from the study include: There are no major differences on how often men and women post selfies and how often they use editing software. (69)_______Chock said posting selfies on social media is not all that different from what people have done for many years. On trips and special events, our parents and grandparents used cameras instead of phones to take photos. They would bring back photos to show friends and family. You had no choice but to look at them. You probably commented about how nice everyone in the photos looked, especially children and the person showing the photos. They were happy to hear your comments. (70)__________On social media, however, people can decide not to look at photos even if they click “like”.TV. 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.It's natural to feel the need to control something when everything around you feels out of control, and you feel helpless. When a friend of mine first heard about the coronavirus outbreak. she got down on her hand. and knees and cleaned her kitchen floor. She told me, “My floor wasn’t even dirty, but doing something constructive made me feel in control and that I was holding on to my power, despite the desperate circumstances.”Your most powerful weapon against uncertainty is your perspective because nobody and no situation can take that from you unless you give it away. Your perspective can victimize or empower you. When you look for the upside in a downside situation and figure out what you can control and what you can't, it's easier to accept whatever is beyond your control.高三英语调研测试（试卷参考答案）I Listening ComprehensionSection A1-5 BDBCB6-10 CBCCDSection B11-13 AAC14-16 BAD17-20 ACDCII. Grammar and V ocabularySection A21. forced22 which23. growing 24. thought25. To test27. while28. Although/While/Thou29. that30 could/canSection B31-40 IBACG FDEKJIII Reading ComprehensionSection A41-55 CDABC BADBA BDADASection B56-59 DBBA60-62CDA63-66 BCBASection C67-70 CBFDTV. Summary WritingFor reference:Strengthening or retrieving a sense of control is a natural demand. First, good perspective is of the greatest help, for evaluating the situation properly brings a clearer picture. Second, kind behaviors during tough times can secure you a sense of control as well. Third, actions and thoughts of positivity also help as they may create hope and optimism.It's essential to feel you can control something in a crisis or emergency. Positive perspective is the most important, since it will ensure you confidence and power to pull through the difficult situation. Then doing small good deeds can free you from worry temporarily by helping others. Lastly. take some positive action to arouse the hope within you.V. Translation72. Never before have we been so eager/keen to go to school as we are now.73. Why not do something you love but don’t have time to do since you can only stay at home?74. In Europe many coffee houses which serve as ideal places for social interaction have to close their stores to respond to the public health crisis.75. It turns out that when our country is facing difficulties, every Chinese, whether at home or abroad is willing to make contributions to their motherland.。

高三英语模拟试卷（二）英语试卷第I卷（选择题，共95分）第一部分：英语知识运用（共两节，满分45分）第一节：单项填空（共15小题；每小题1分，满分15分）从A、B、C、D四个选项中，选出可以填入空白处的最佳选项。

1. —I’d like to take a holiday in August, Mr. Yang.— ___________ We'll be busy then. You'll have to make it another time.A. It's up to you.B. It's out of the question.C. It doesn't matter.D. By all means.2. My computer ___________ start this morning. There must have been something wrong with it.A. shouldn'tB. wouldn'tC. needn'tD. mustn't3. The professor could tell by the _______________ look in Maria's eyes that she didn't understand asingle word of his lecture.A. coldB. emptyC. innocent D・ blank4. It is by no means clear __________ the local government will do with the buildings of poor quality.A. whatB. howC. whetherD. which5. Alice spent an afternoon helping her grandfather to ____________ the stamps which he had collectedfor years.A. make outB. watch outC. sort outD. put out6. Delete the short message at once! Many a man ____________ by such tricks up to now.A. are cheatedB is cheated' C. have been cheatedD. has been cheated7. There is a belief that the robots will take the place of human beings one day.—_______ After all, they don't have their own thinking.A. Don't mention it.B. It depends.C. I don't think so.D. I have no idea.8. The cooling wind swept through our bedroom windows, ___________ air conditioning unnecessary.A. madeB. to makeC. being madeD. making9. Most passengers like sitting in the front of a bus to have a better view. What's more, they believe it's lesslikely that _____________________ will happen.A. car-accidentB. car-sicknessC. car-troubleD. car-crash10. Much disappointed as he is _________ in the job interview, he still keeps his confidence.A. to have failedB. to failC. having failedD. failing11 .---What do you think of the jacket and the hat I wear today?—I don't think this jacket _________ you and that your hat ________ this jacket perfectly.A. suits; fitsB. meets; fitsC. matches; suitsD. fits; matches12. 5G has faster data transmission, lower delay and stronger mobility ______________ the conventional4G.A. in charge ofB. in defense ofC. in contrast toD. in proportion to .13. Although he did not do it ________ , he definitely had some responsibility.A. instantlyB. deliberatelyC. accidentallyD. casually14. Tom is so independent that he never asks his parents' opinion __________ he wants their support.A. sinceB. onceC. unlessD. after15. We have entered into an age _______ dreams have the best chance of coming true.A. whichB. whatC. thatD. when第二节:完形填空（共20小题；每小题1.5分，满分30分）阅读下面短文，掌握其大意，然后从16-35题所给的A、B、C、D四个选项中，选出最佳选项。

南京市、盐城市2022届高三年级第二次模拟考试数 学 2022.03一、填空题1．设集合A ＝{x |－2＜x ＜0}，B ＝{x |－1＜x ＜1}，那么A ∪B ＝▲________．2．假设复数z ＝(1＋m i)(2－i)(i 是虚数单位)是纯虚数，那么实数m 的值为 ▲ ． 3．将一骰子连续抛掷两次，至少有一次向上的点数为1的概率是 ▲ ．4．如下图，一家面包销售店根据以往某种面包的销售记录，绘制了日销售量的频率分布直方图．假设一个月以30天计算，估计这家面包店一个月内日销售量不少于150个的天数为▲________．5．执行如下图的流程图，那么输出的k 的值为 ▲ ．6．设公差不为0的等差数列{a n }的前n 项和为S n ．假设S 3＝a 22，且S 1，S 2，S 4成等比数列，那么a 10等于 ▲ ．7．如图，正三棱柱ABC —A 1B 1C 1中，AB ＝4，AA 1＝6．假设E ，F 分别是棱BB 1，CC 1上的点，那么三棱锥A —A 1EF 的体积是▲________．〔第5题图〕〔第4题图〕〔第7题图〕ABCA 1B 1FC 1EANBPMC8．函数f (x )＝2sin(ωx ＋φ)(ω＞0，|φ|＜π2)的最小正周期为π，且它的图象过点(－π12，－2)，那么φ的值为▲________．9．函数f (x )＝⎩⎪⎨⎪⎧12x ＋1，x ≤0，－(x －1)2，x ＞0，那么不等式f (x )≥－1的解集是▲________．10．在平面直角坐标系xOy 中，抛物线y 2＝2px (p ＞0)的焦点为F ，双曲线x 2a 2－y 2b2＝1(a ＞0，b ＞0)的两条渐近线分别与抛物线交于A ，B 两点(A ，B 异于坐标原点O )．假设直线AB 恰好过点F ，那么双曲线的渐近线方程是▲________.11．在△ABC 中，A ＝120°，AB ＝4．假设点D 在边BC 上，且BD →＝2DC →，AD ＝273，那么AC的长为▲________．12．圆O ：x 2＋y 2＝1，圆M ：(x －a )2＋(y －a ＋4)2＝1．假设圆M 上存在点P ，过点P 作圆O 的两条切线，切点为A ，B ，使得∠APB ＝60°，那么实数a 的取值范围为▲________． 13．函数f (x )＝ax 2＋x －b (a ，b 均为正数)，不等式f (x )＞0的解集记为P ，集合Q ＝{x |－2－t ＜x ＜－2＋t }．假设对于任意正数t ，P ∩Q ≠∅，那么1a －1b 的最大值是▲________．14．假设存在两个正实数x 、y ，使得等式x ＋a (y －2e x )(ln y －ln x )＝0成立，其中e 为自然对数的底数，那么实数a 的取值范围为▲________． 二、解答题15．(本小题总分值14分)α为锐角，cos (α＋π4)＝55．（1）求tan(α＋π4)的值； 〔2〕求sin(2α＋π3)的值．16．(本小题总分值14分)如图，在三棱锥P —ABC 中，平面P AB ⊥平面ABC ，P A ⊥PB ，M ，N 分别为AB ，P A 的中点．（1）求证：PB ∥平面MNC ；〔2〕假设AC ＝BC ，求证：P A ⊥平面MNC .17．(本小题总分值14分)如图，某城市有一块半径为1〔单位：百米〕的圆形景观，圆心为C ，有两条与圆形景观相切且互相垂直的道路．最初规划在拐角处〔图中阴影局部〕只有一块绿化地，后来有众多市民建议在绿化地上建一条小路，便于市民快捷地往返两条道路．规划部门采纳了此建议，决定在绿化地中增建一条与圆C 相切的小道AB ．问：A ，B 两点应选在何处可使得小道AB 最短？18． (本小题总分值16分)在平面直角坐标系xOy 中，点C 在椭圆M ：x 2a 2＋y 2b 2＝1(a ＞b ＞0)上．假设点A (－a ，0)，B (0，a3)，且AB →＝32BC →．〔1〕求椭圆M 的离心率；〔2〕设椭圆M 的焦距为4，P ，Q 是椭圆M 上不同的两点，线段PQ 的垂直平分线为直线l ，且直线l 不与y 轴重合．①假设点P (－3，0)，直线l 过点(0，－67)，求直线l 的方程；②假设直线l 过点(0，－1) ，且与x 轴的交点为D ，求D 点横坐标的取值范围．〔第16题图〕〔第17题图〕19．(本小题总分值16分)对于函数f (x )，在给定区间[a ，b ]内任取n ＋1(n ≥2，n ∈N *)个数x 0，x 1，x 2，…，x n ，使得a ＝x 0＜x 1＜x 2＜…＜x n －1＜x n ＝b ，记S ＝n -1∑i =0|f (x i ＋1)－f (x i )|．假设存在与n 及x i (i ≤n ，i ∈N )均无关的正数A ，使得S ≤A 恒成立，那么称f (x )在区间[a ，b ]上具有性质V ． 〔1〕假设函数f (x )＝－2x ＋1，给定区间为[－1，1]，求S 的值；〔2〕假设函数f (x )＝xex ，给定区间为[0，2]，求S 的最大值；〔3〕对于给定的实数k ，求证：函数f (x )＝k ln x －12x 2 在区间[1，e ]上具有性质V ．20．(本小题总分值16分)数列{a n }的前n 项和为S n ，且对任意正整数n 都有a n ＝(－1)n S n ＋p n (p 为常数，p ≠0)． 〔1〕求p 的值；〔2〕求数列{a n }的通项公式；〔3〕设集合A n ＝{a 2n －1，a 2n }，且b n ，c n ∈A n ，记数列{nb n }，{nc n }的前n 项和分别为P n ，Q n ． 假设b 1≠c 1，求证：对任意n ∈N *，P n ≠Q n ．南京市、盐城市2022届高三年级第二次模拟考试数学附加题 2022.0321．【选做题】在A 、B 、C 、D 四小题中只能选做2题，每题10分，共计20分．请在答．卷纸指定．．．．区域内．．．作答．解容许写出文字说明、证明过程或演算步骤． A ．选修4—1：几何证明选讲如图，在Rt △ABC 中，AB ＝BC ．以AB 为直径的⊙O 交AC 于点D ，过D 作DE ⊥BC ，垂足为E ，连接AE 交⊙O 于点F ．求证：BE ⋅CE ＝EF ⋅EA ．B ．选修4—2：矩阵与变换 a ，b 是实数，假如矩阵A ＝⎣⎢⎡⎦⎥⎤3 a b －2 所对应的变换T 把点(2，3)变成点(3，4)．〔1〕求a ，b 的值．〔2〕假设矩阵A 的逆矩阵为B ，求B 2．C ．选修4—4：坐标系与参数方程在平面直角坐标系xOy 中，以坐标原点为极点，x 轴正半轴为极轴建立极坐标系．直线l 的极坐标方程为ρsin(π3－θ)=32，椭圆C 的参数方程为⎩⎨⎧x ＝2cos t ，y ＝3sin t(t 为参数) ．〔1〕求直线l 的直角坐标方程与椭圆C 的普通方程；〔2〕假设直线l 与椭圆C 交于A ，B 两点，求线段AB 的长．D ．选修4—5：不等式选讲解不等式：|x －2|＋x |x ＋2|＞2A【必做题】第22题、第23题，每题10分，共计20分．请在答．卷卡指定区域内．．．．．．．作答．解容许写出 文字说明、证明过程或演算步骤． 22．〔本小题总分值10分〕甲、乙两人投篮命中的概率分别为23与12，各自互相独立．现两人做投篮游戏，共比赛3局，每局每人各投一球．〔1〕求比赛完毕后甲的进球数比乙的进球数多1个的概率；〔2〕设ξ表示比赛完毕后甲、乙两人进球数的差的绝对值，求ξ的概率分布和数学期望E (ξ)．23．〔本小题总分值10分〕设(1－x )n ＝a 0＋a 1x ＋a 2x 2＋…＋a n x n ，n ∈N *，n ≥2．〔1〕设n ＝11，求|a 6|＋|a 7|＋|a 8|＋|a 9|＋|a 10|＋|a 11|的值；〔2〕设b k ＝k ＋1n －k a k ＋1(k ∈N ，k ≤n －1)，S m ＝b 0＋b 1＋b 2＋…＋b m (m ∈N ，m ≤n －1)，求|S mC m n －1 |的值．南京市、盐城市2022届高三年级第二次模拟考试数学参考答案一、填空题〔本大题共14小题，每题5分，计70分. 不需写出解答过程，请把答案写在答题纸的指定位置上〕1． {x |－2＜x ＜1} 2．－2 3．1136 4． 9 5． 5 6． 19 7． 8 38．－π12 9． [－4，2] 10．y ＝±2x 11．3 12． [2－22，2＋22]13． 12 14．a ＜0或a ≥1e二、解答题〔本大题共6小题，计90分.解容许写出必要的文字说明，证明过程或演算步骤，请把答案写在答题纸的指定区域内〕 15．(本小题总分值14分)解：〔1〕因为α∈(0，π2〕，所以α＋π4∈(π4，3π4)，所以sin (α＋π4)＝1－cos 2(α＋π4)＝255，……………………………………………………………3分所以tan(α＋π4)＝sin(α＋π4)cos(α＋π4)＝2． (6)分〔2〕因为sin(2α＋π2)＝sin[2(α＋π4)]＝2 sin (α＋π4) cos (α＋π4)＝45， (9)分cos(2α＋π2)＝cos[2(α＋π4)]＝2 cos 2(α＋π4)－1＝－35， (12)分所以sin(2α＋π3)＝sin[(2α＋π2)－π6]＝sin(2α＋π2)cos π6－cos(2α＋π2)sin π6＝43＋310． (14)分ANBPMC16．(本小题总分值14分)证：〔1〕因为M ，N 分别为AB ，P A 的中点，所以MN ∥PB ． …………………………………2分 因为MN ⊂平面MNC ，PB ⊄平面MNC ，所以PB ∥平面MNC . ……………………………………4分 〔2〕因为P A ⊥PB ，MN ∥PB ，所以P A ⊥MN . ……………6分因为AC ＝BC ，AM ＝BM ，所以CM ⊥AB . ……………8分 因为平面P AB ⊥平面ABC ，CM ⊂平面ABC ，平面P AB ∩平面ABC ＝AB ， 所以CM ⊥平面P AB ． …………………………………12分 因为P A ⊂平面P AB ，所以CM ⊥P A ．因为P A ⊥MN ，MN ⊂平面MNC ，CM ⊂平面MNC ，MN ∩CM ＝M ，所以P A ⊥平面MNC. ……………………………………………………………………14分 17．(本小题总分值14分)解法一：如图，分别由两条道路所在直线建立直角坐标系xOy ． 设A (a ，0)，B (0，b )(0＜a ＜1，0＜b ＜1)， 那么直线AB 方程为x a ＋yb ＝1，即bx ＋ay －ab ＝0．因为AB 与圆C 相切，所以|b ＋a －ab |b 2＋a 2＝1．……………4分化简得 ab －2(a ＋b )＋2＝0，即ab ＝2(a ＋b )－2．……………6分因此AB ＝ a 2＋b 2＝ (a ＋b )2－2ab ＝ (a ＋b )2－4(a ＋b )＋4＝ (a ＋b －2)2．………………8分因为0＜a ＜1，0＜b ＜1，所以0＜a ＋b ＜2， 于是AB ＝2－(a ＋b )． 又ab ＝2(a ＋b )－2≤(a +b 2)2，解得0＜a ＋b ≤4－22，或a ＋b ≥4＋22．因为0＜a ＋b ＜2，所以0＜a ＋b ≤4－22，………………………………………12分所以AB ＝2－(a ＋b ) ≥2－(4－22)＝22－2，当且仅当a ＝b ＝2-2时取等号，所以AB 最小值为22－2，此时a ＝b ＝2-2．答：当A ，B 两点离道路的交点都为2－2(百米)时，小道AB 最短．……………14分解法二：如图，连接CE ，CA ，CD ，CB ，CF ． 设∠DCE ＝θ，θ∈(0，π2)，那么∠DCF ＝π2－θ．在直角三角形CDA 中，AD ＝tan θ2．………………4分在直角三角形CDB 中，BD ＝tan(π4－θ2)，………6分所以AB ＝AD ＋BD ＝tan θ2＋tan(π4－θ2)＝tan θ2＋1－tanθ2 1＋tanθ2．………………………8分令t ＝tan θ2，0＜t ＜1，那么AB ＝f (t )＝t ＋1－t 1＋t ＝＝t ＋1＋21＋t－2≥22－2，当且仅当t ＝2－1时取等号．………………………12分所以AB 最小值为22－2，此时A ，B 两点离两条道路交点的间隔 是1－(2－1)＝2－2．答：当A ，B 两点离道路的的交点都为2－2(百米)时，小道AB 最短．……………14分18．(本小题总分值16分)解：〔1〕设C (x 0，y 0)，那么AB →＝(a ，a 3)，BC →＝(x 0，y 0－a 3)．因为AB →＝32BC →，所以(a ，a 3)＝32(x 0，y 0－a 3)＝(32x 0，32y 0－a 2)，得⎩⎨⎧x 0＝23a ，y 0＝59a ，………………………………………………………2分 代入椭圆方程得a 2＝95b 2．因为a 2－b 2＝c 2，所以e ＝c a ＝23．………………………………………4分〔2〕①因为c ＝2，所以a 2＝9，b 2＝5，所以椭圆的方程为x 29＋y 25＝1，设Q (x 0，y 0)，那么x 029＋y 025＝1．……① ………………………………………………6分因为点P (－3，0)，所以PQ 中点为(x 0－32，y 02)，因为直线l 过点(0，－67)，直线l 不与y 轴重合，所以x 0≠3，所以y 02＋67x 0－32·y 0x 0＋3＝－1， ………………………………………………8分化简得x 02＝9－y 02－127y 0．……②将②代入①化简得y 02－157y 0＝0，解得y 0＝0〔舍〕，或y 0＝157．将y 0＝157代入①得x 0＝±67，所以Q 为(±67，157)，所以PQ 斜率为1或59，直线l 的斜率为－1或－95，所以直线l 的方程为y ＝－x ＋67或y ＝－95x ＋67．……………………………………………10分②设PQ ：y ＝kx +m ，那么直线l 的方程为：y ＝－1kx －1，所以x D ＝－k ．将直线PQ 的方程代入椭圆的方程，消去y 得(5＋9k 2)x 2＋18kmx ＋9m 2－45＝0．…………①， 设P (x 1，y 1)，Q (x 2，y 2)，中点为N ，x N ＝x 1＋x 22＝－9km 5+9k 2，代入直线PQ 的方程得y N ＝5m 5+9k 2，……………………………………12分 代入直线l 的方程得9k 2＝4m －5． ……② 又因为△＝(18km )2－4(5＋9k 2) (9m 2－45)＞0，化得m 2－9k 2－5＜0． ………………………………………………14分 将②代入上式得m 2－4m ＜0，解得0＜m ＜4，所以－113＜k ＜113，且k ≠0，所以x D ＝－k ∈(－113，0)∪(0，113)．综上所述，点D 横坐标的取值范围为(－113，0)∪(0，113)．………………………………16分19．(本小题总分值16分)〔1〕解：因为函数f (x )＝－2x ＋1在区间[－1，1]为减函数， 所以f (x i ＋1)＜f (x i )，所以|f (x i ＋1)－f (x i )|＝ f (x i )－f (x i ＋1)．S ＝n -1∑i =0|f (x i ＋1)－f (x i )|＝[ f (x 0)－f (x 1)]+[ f (x 1)－f (x 2)]+…+[ f (x n -1)－f (x n )]＝f (x 0)－f (x n )＝f (－1)－f (1)＝4． …………………………………………2分 (2) 解：由f ′(x )＝1－xex ＝0，得x ＝1．当x ＜1时，f ′(x )＞0，所以f (x )在(－∞，1)为增函数； 当x ＞1时，f ′(x )＜0，所以f (x )在(1，＋∞)为减函数；所以f (x )在x ＝1时取极大值1e ． ……………………………………4分设x m ≤1＜x m ＋1，m ∈N ，m ≤n －1，那么S ＝n -1∑i =0|f (x i ＋1)－f (x i )|＝|f (x 1)－f (0)|＋…＋|f (x m )－f (x m －1)|＋|f (x m ＋1)－f (x m )|＋|f (x m ＋2)－f (x m ＋1)|＋…＋|f (2)－f (x n －1)| ＝[f (x 1)－f (0)]＋…＋[f (x m )－f (x m －1)]＋|f (x m ＋1)－f (x m )|＋[f (x m ＋1)－f (x m ＋2)]＋…＋[f (x n －1)－f (2)] ＝[f (x m )－f (0)]＋|f (x m ＋1)－f (x m )|＋[f (x m ＋1)－f (2)]． …………………………………………6分 因为|f (x m ＋1)－f (x m )|≤[f (1)－f (x m )]＋[f (1)－f (x m ＋1)]，当x m ＝1时取等号， 所以S ≤f (x m )－f (0)＋f (1)－f (x m )＋f (1)－f (x m ＋1)＋f (x m ＋1)－f (2) ＝2 f (1)－f (0)－f (2)＝2(e －1)e 2.所以S 的最大值为2(e －1)e 2． …………………………………………8分〔3〕证明：f ′(x )＝kx －x ＝k －x 2x，x ∈[1，e]．①当k ≥e 2时，k －x 2≥0恒成立，即f ′(x )≥0恒成立，所以f (x )在[1，e]上为增函数，所以S ＝n -1∑i =0|f (x i ＋1)－f (x i )|＝[ f (x 1)－f (x 0)]+[ f (x 2)－f (x 1)]+…+[ f (x n )－f (x n -1)]＝f (x n )－f (x 0)＝f (e)－f (1)＝k +12－12e 2．因此，存在正数A ＝k +12－12e 2，都有S ≤A ，因此f (x )在[1，e]上具有性质V ． (10)分②当k ≤1时，k －x 2≤0恒成立，即f ′(x )≤0恒成立，所以f (x )在[1，e]上为减函数，所以S ＝n -1∑i =0|f (x i ＋1)－f (x i )|＝[ f (x 0)－f (x 1)]+[ f (x 1)－f (x 2)]+…+[ f (x n -1)－f (x n )]＝f (x 0)－f (x n )＝ f (1)－f (e)＝ 12e 2－k －12．因此，存在正数A ＝12e 2－k －12，都有S ≤A ，因此f (x )在[1，e]上具有性质V ． (12)分③当1＜k ＜e 2时，由f ′(x )＝0，得x ＝k ；当f ′(x )＞0，得1≤x ＜k ；当f ′(x )＜0，得k ＜x ≤e ，因此f (x )在[1，k )上为增函数，在(k ，e]上为减函数． 设x m ≤k ＜x m +1，m ∈N ，m ≤n －1那么S ＝n －1∑i ＝1|f (x i ＋1)－f (x i )|＝|f (x 1)－f (x 0)|+…+|f (x m )－f (x m －1)|+ |f (x m +1)－f (x m )|+ |f (x m +2)－f (x m +1)|+…+|f (x n )－f (x n －1)| ＝f (x 1)－f (x 0)+…+f (x m )－f (x m －1) + |f (x m +1)－f (x m )|+ f (x m +1)－f (x m +2) +…+f (x n －1)－f (x n ) ＝f (x m )－f (x 0) + |f (x m +1)－f (x m )| + f (x m +1)－f (x n )≤f (x m )－f (x 0) + f (x m +1)－f (x n )+ f (k )－f (x m +1)+ f (k )－f (x m )＝2 f (k )－f (x 0)－f (x n )＝k ln k －k －[－12+k －12e 2]＝k ln k －2k ＋12+12e 2．因此，存在正数A ＝k ln k －2k ＋12+12e 2，都有S ≤A ，因此f (x )在[1，e]上具有性质V ．综上，对于给定的实数k ，函数f (x )＝k ln x －12x 2 在区间[1，e]上具有性质V ．……………16分20．(本小题总分值16分)解：〔1〕由a 1＝－S 1＋p ，得a 1＝p2．………………………………………………………2分由a 2＝S 2＋p 2，得a 1＝－p 2，所以p2＝－p 2．又p ≠0，所以p ＝－12． …………………………………………………………3分〔2〕由a n ＝(－1)n S n ＋(－12)n ，得⎩⎨⎧a n ＝(－1)n S n ＋(－12)n ， ……①a n ＋1＝－(－1)nS n ＋1＋(－12)n ＋1， ……②①＋②得a n ＋a n ＋1＝(－1)n (－a n ＋1)＋12×(－12)n ． …………………………………………5分当n 为奇数时，a n ＋a n ＋1＝a n ＋1－12×(12)n ，所以a n ＝－(12)n ＋1． ………………………………………………………………7分当n 为偶数时，a n ＋a n ＋1＝－a n ＋1＋12×(12)n ，所以a n ＝－2a n ＋1＋12×(12)n ＝2×(12)n ＋2＋12×(12)n ＝(12)n ，所以a n＝⎩⎨⎧－12n ＋1，n 为奇数， n ∈N *，12n， n 为偶数，n ∈N *．………………………………………………9分〔3〕A n ＝{－14n ，14n }，由于b 1≠c 1，那么b 1 与c 1一正一负，不妨设b 1＞0，那么b 1＝14，c 1＝－14．那么P n ＝b 1＋2b 2＋3b 3＋…＋nb n ≥14－(242＋343+…＋n4n )．……………………………………………12分设S ＝242＋343+…＋n 4n ，那么14S ＝243＋…＋n －14n +n 4n ＋1，两式相减得34S ＝242＋143+…＋14n －n 4n ＋1＝116＋116×1－(14)n －11－14－n 4n ＋1＝748－112×14n －1－n 4n ＋1＜748．所以S ＜748×43＝736，所以P n ≥14－(242＋143+…＋14n )＞14－736＝118＞0．………………………14分因为Q n = c 1＋2 c 2＋3 c 3＋…＋n c n ≤－14＋S ＜－14＋736 ＝－118＜0，所以P n ≠Q n ． ………………………………………………………………16分南京市、盐城市2022届高三年级第二次模拟考试 数学附加题参考答案及评分标准 2022.0321．【选做题】A ．选修4—1：几何证明选讲证明：连接BD ．因为AB 为直径，所以BD ⊥AC ． 因为AB ＝BC ，所以AD ＝DC ．……………………4分 因为DE ⊥BC ，AB ⊥BC ，所以DE ∥AB ，…………6分 所以CE ＝EB ．………………………………………8分 因为AB 是直径，AB ⊥BC ，所以BC 是圆O 的切线，所以BE 2＝EF ⋅EA ，即BE ⋅CE ＝EF ⋅EA ．…………………………………………………………10分 B ．选修4—2：矩阵与变换解：〔1〕由题意，得⎣⎢⎡⎦⎥⎤ 3 a b －2 ⎣⎡⎦⎤23＝⎣⎡⎦⎤34，得6＋3a ＝3，2b －6＝4，…………………………4分所以a ＝－1，b ＝5．………………………………………………………………………………6分〔2〕由〔1〕，得A ＝⎣⎢⎡⎦⎥⎤ 3 －1 5 －2．由矩阵的逆矩阵公式得B ＝⎣⎢⎡⎦⎥⎤2 －1 5 －3．…………………8分所以B 2＝⎣⎢⎡⎦⎥⎤－1 1 －5 4． …………………………………………………………10分 C ．选修4—4：坐标系与参数方程解：〔1〕由ρsin(π3－θ)=32 ，得ρ(32cos θ－12sin θ)=32，即32x －12y=32，化简得y=3x －3，所以直线l 的直角坐标方程是y=3x －3．………………………………2分由(x 2)2+(y 3)2=cos 2t +sin 2t =1，得椭圆C 的普通方程为x 24+y 23=1．……………………………4分〔2〕联立直线方程与椭圆方程，得⎩⎪⎨⎪⎧y=3x －3， x 24+y 23=1，消去y ，得x 24+(x －1)2=1，化简得5x 2－8x =0，解得x 1=0，x 2=85， ………………………………8分所以A (0，－3)，B (85，353)，那么AB =(0－85)2+(－3－353)2=165． ………………………………10分D ．选修4—5：不等式选讲A解：当x ≤－2时，不等式化为(2－x )＋x (－x －2)＞2，解得－3＜x ≤－2； ………………………………………………3分 当－2＜x ＜2时，不等式化为(2－x )＋x (x ＋2)＞2，解得－2＜x ＜－1或0＜x ＜2； …………………………………………………6分 当x ≥2时，不等式化为(x －2)＋x (x ＋2)＞2,解得x ≥2； ………………………………………………………9分 所以原不等式的解集为{x |－3＜x ＜－1或x ＞0}． …………………………………………10分 【必做题】第22题、第23题，每题10分，共计20分． 22．〔本小题总分值10分〕解：〔1〕比赛完毕后甲的进球数比乙的进球数多1个有以下几种情况： 甲进1球，乙进0球；甲进2球，乙进1球；甲进3球，乙进2球． 所以比赛完毕后甲的进球数比乙的进球数多1个的概率P ＝C 1323(13)2(12)3＋C 23(23)2(13)C 13(12)3＋C 33(23)3C 23(12)3＝1136．……………………………………………4分〔2〕ξ的取值为0，1，2，3，所以 ξ的概率分布列为……………………………………………………………………………………8分所以数学期望E (ξ)＝0×724＋1×1124＋2×524＋3×124＝1．………………………………………10分23．〔本小题总分值10分〕解：〔1〕因为a k ＝(－1)k C kn ，当n ＝11时，|a 6|＋|a 7|＋|a 8|＋|a 9|＋|a 10|＋|a 11|＝C 611＋C 711＋C 811＋C 911＋C 1011＋C 1111＝12( C 011＋C 111＋…＋C 1011＋C 1111)＝210＝1024．………………………………………………3分〔2〕b k ＝k ＋1n －k a k ＋1＝(－1)k ＋1 k ＋1n －kC k ＋1n ＝(－1)k ＋1 C kn ，……………………………………5分当1≤k ≤n －1时，b k ＝(－1)k ＋1 C k n ＝ (－1)k ＋1 (C k n －1＋C k －1n －1)＝(－1)k ＋1 C k －1n －1＋(－1)k ＋1 C k n －1＝(－1)k －1 C k －1n －1－(－1)kC k n －1． ……………………………………7分当m ＝0时，|S m C m n －1 |＝|b 0C 0n －1|＝1． ……………………………………8分当1≤m ≤n －1时，S m ＝－1＋k =1∑m[(－1)k －1 C k －1n －1－(－1)k C k n －1]＝－1＋1－(－1)m C m n －1＝－(－1)mC m n －1，S m C m n－1|＝1．综上，|S mC m n－1|＝1．………………………10分所以|。

2020届南京盐城二模英语试卷第一部分听力(共两节，满分20分)第一节(共5小题；每小题1分，满分5分)听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

() 1. Why is Mike so unhappy?A. He failed one of his exams.B. He is upset about others.C. He is worried about physics.() 2. How does the man probably feel?A. Happy.B. Nervous.C. Disappointed.() 3. Who is the woman speaking to?A. A bank clerk.B. A tour guide.C. A customs officer.() 4. What can we learn from the conversation?A. Ann left here two years ago.B. The woman has covered a long way.C. The man has lived here for two years.() 5. What will the man probably do tomorrow?A. Stay at home.B. Do some shopping.C. Take an interview.第二节(共15小题；每小题1分，满分15分)听下面5段对话或独白。

每段对话或独白后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟；听完后，各小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

南京市、盐城市2021届高三年级第二次模拟考试数学注意事项：1．本试卷考试时间为120分钟，试卷满分150分，考试形式闭卷．2．本试卷中所有试题必须作答在答题卡上规定的位置，否则不给分．3．答题前，务必将自己的姓名、准考证号用0．53米黑色墨水签字笔填写在试卷及答题卡上．第I 卷(选择题共60分)一、单项选择题(本大题共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．设复数z1，z2在复平面内的对应点关于实轴对称，z1＝3＋4i，则z1z2＝A．25 B．－25 C．7－24i D．－7－24i 2．设集合A，B是全集U的两个子集，则“A∩B＝ ”是“A✶ U B”的A．充分不必要条件B．必要不充分条件C．充要条件D．既不充分也不必要条件3．已知a，b是相互垂直的单位向量，与a，b共面的向量c满足a⋅c＝b⋅c＝2，则c的模为A．1 B． 2 C．2 D．224．在流行病学中，基本传染数是指每名感染者平均可传染的人数．当基本传染数高于1时，每个感染者平均会感染一个以上的人，从而导致感染这种疾病的人数量指数级增长．当基本传染数持续低于1时，疫情才可能逐渐消散．广泛接种疫苗可以减少疾病的基本传染数．假设某种传染病的基本传染数为R，1个感染者在每个传染期会接触到N个新人，这N人中有V个人接种过疫苗(VN称为接种率)，那么1个感染者新的传染人数为()RN VN-．已知新冠病毒在某地的基本传染数R＝2.5，为了使1个感染者传染人数不超过1，该地疫苗的接种率至少为A．40% B．50% C．60% D．70%5．计算2cos10sin20cos20︒-︒︒所得的结果为A．1 B． 2 C． 3 D．26．密位制是度量角的一种方法．把一周角等分为6000份，每一份叫做1密位的角．以密位作为角的度量单位，这种度量角的单位制，叫做角的密位制．在角的密位制中，采用四个数码表示角的大小，单位名称密位二字可以省去不写．密位的写法是在百位数与十位数字之间画一条短线，如7密位写成“0－07”，478密位写成“4－78．1周角等于6000密位，记作1周角＝60－00，1直角＝15－00．如果一个半径为2的扇形，它的面积为76π，则其圆心角用密位制表示为A ．12－50B ．17－50C ．21－00D ．35－007．已知双曲线()2222100x y C a b a b-=>>：，的左、右焦点分别为F 1，F 2，过点F 2作倾斜角为θ的直线l 交双曲线C 的右支于A ，B 两点，其中点A 在第一象限，且cos θ＝14.若|AB |＝|AF 1|，则双曲线C 的离心率为A ．4B ．15C ．32 D ．28．已知f (x )是定义在R 上的奇函数，其导函数为f ′(x )，且当x ＞0时，()()ln 0f x f x x x'⋅+>，则不等式(x 2－1)f (x )＜0的解集为A ．(－1，1)B ．(－∞，－1)∪(0，1)C ．(－∞，－1)∪(1，＋∞)D ．(－1，0)∪(1，＋∞)二、多项选择题(本大题共4小题，每小题5分，共20分在每小题给出的四个选项中，有多项符合题目要求的．全部选对的得5分，部分选对的得2分，有选错的得0分) 9．对于两条不同直线m ，n 和两个不同平面α，β，下列选项中正确的为A ．若m ⊥α，n ⊥β，α⊥β，则m ⊥nB ．若m //α，n //β，α⊥β，则m ⊥n 或m //nC ．若m //α，α//β，则m //β或m ⊂βD ．若m ⊥α，m ⊥n ，则n //α或n ⊂α 10．已知a ＞b ＞0，下列选项中正确的为A ．若a －b ＝1，则a －b ＜1B ．若a 2－b 2＝1，则a －b ＜1C ．若2a －2b ＝1，则a －b ＜1D ．若22log log 1a b -=，则a －b ＜1 11．已知函数f (x )＝|sin x |＋|cos x |,则A ．f (x )是周期函数B ．f (x )的图象必有对称轴C ．f (x )的增区间为2k k k Z πππ⎡⎤+∈⎢⎥⎣⎦，， D ．f (x )的值域为⎡⎣ 12．已知*n N ∈，n ≥2，p ＋q ＝1，设()22k n kn f k C q-=，其中k ∈N ，k ≤2n ，则 A ．()201nk f k ==∑ B ．()202nk kf k npq ==∑C ．若np ＝4，则f (k )≤f (8)D ．()()0112212nnk k f k f k ==<<-∑∑第II 卷 (非选择题 共90分)三，填空题(本大题共4小题，每小题5分，共20分)13．某班4名同学去参加3个社团，每人只参加1个社团，每个社团都有人参加，则满足上述要求的不同方案共有 ▲ 种．(用数字填写答案)14．已知椭圆22143x y +=的右顶点为A ，右焦点为F ，以A 为圆心，R 为半径的圆与椭圆相交于B ，C 两点，若直线BC 过点F ，则R 的值为 ▲ ．15．在四棱锥P －ABCD 中，P A ⊥面ABCD ，四边形ABCD 是边长为2的正方形，且P A ＝2．若点E 、F 分别为AB ，AD 的中点，则直线EF 被四棱锥P －ABCD 的外接球所截得的线段长为 ▲ ．16．牛顿选代法又称牛顿—拉夫逊方法，它是牛顿在17世纪提出的一种在实数集上近似求解方程根的一种方法．具体步骤如下：设r 是函数y ＝f (x )的一个零点，任意选取x 0作为r 的初始近似值，过点()()00x f x ，作曲线y ＝f (x )的切线l 1，设l 1与x 轴交点的横坐标为x 1，并称x 1为r 的1次近似值；过点()()11x f x ，作曲线y ＝f (x )的切线l 2，设l 2与x 轴交点的横坐标为x 2，称x 2为r 的2次近似值．一般的，过点(x n ，f (x n ))(n ∈N )作曲线y ＝f (x )的切线l n+1， 记l n+1与x 轴交点的横坐标为x n+1，并称x n+1为r 的的n ＋1次近似值．设()31f x x x =+-(x ≥0)的零点为r ，取x 0＝0，则r 的2次近似值为 ▲ ；设33321n n n n x x a x +=+，n ∈N *，数列{}n a 的前n 项积为T n ．若任意n ∈N *，T n ＜λ恒成立，则整数λ的最小值为 ▲ ．四、解答题(本大题共6小题，共70分．解答时应写出文字说明、证明过程或演算步骤) 17．(本小题满分10分)在①b ＝3a ；②a ＝3cos B ；③a sin C ＝1这三个条件中任选一个，补充在下面问题中．若问 题中的三角形存在，求该三角形面积的值；若问题中的三角形不存在，说明理由． 问题：是否存在△ABC ，它的内角A ，B ，C 的对边分别为a ，b ，c ，且()sin sin B A C C --=，c ＝3， ▲ ？18．(本小题满分12分)已知等比数列{a n }的前n 项和S n ＝2n ＋r ，其中r 为常数．(1)求r 的值；(2)设()221log n n b a =+，若数列{b n }中去掉数列{a n }的项后余下的项按原来的顺序组成数列{c n }，求123100c c c c ++++的值．某公司对项目A 进行生产投资，所获得的利润有如下统计数据表：(1)请用线性回归模型拟合y 与x 的关系，并用相关系数加以说明；(2)该公司计划用7百万元对A ，B 两个项目进行投资．若公司对项目B 投资x (1≤x ≤6)百万 元所获得的利润y 近似满足：y ＝0.16x －0.49x ＋1＋0.49，求A ，B两个项目投资金额分别为多少时，获得的总利润最大?附：①对于一组数据(x 1，y 1)，(x 2，y 2)，……，(x n ，y n )，其回归直线方程ˆˆˆybx a =+的斜率和截距的最小二乘法估计公式分别为：1221ˆni ii nii x y nx ybxnx ==-⋅=-∑∑，ˆˆay bx =-． ②线性相关系数ni ix y nx yr -⋅=∑一般地，相关系数r 的绝对值在0.95以上(含0.95)认为线性相关性较强；否则，线性相关性较弱． 参考数据：对项目A 投资的统计数据表中111ni ii x y==∑，212.24ni i y ==∑， 4.4≈2.1．如图，三棱柱ABC －A 1B 1C 1的所有棱长都为2，B 1C ＝6，AB ⊥B 1C. (1)求证：平面ABB 1A 1⊥平面ABC ；(2)若点P 在棱BB 1上且直线CP 与平面ACC 1A 1所成角的正弦值为45，求BP 的长21．(本小题满分12分)已知直线l ：y ＝x ＋m 交抛物线C ：24y x =于A ，B 两点． (1)设直线l 与x 轴的交点为T ．若→AT ＝2→TB ，求实数m 的值；(2)若点M ，N 在抛物线C 上，且关于直线l 对称，求证：A ，B ，M ，N 四点共圆．22．(本小题满分12分)已知函数f (x )＝e x －ax sin x －x －1，x ∈[]0π，，a ∈R ． (1)当a ＝12时，求证：f (x )≥0；(2)若函数f (x )有两个零点，求a 的取值范围．南京市、盐城市 2021 届高三年级第二次模拟考试数 学 试 题(总分 150 分，考试时间 120 分钟)注意事项：1．本试卷考试时间为 120 分钟，试卷满分 150 分，考试形式闭卷． 2．本试卷中所有试题必须作答在答题卡上规定的位置，否则不给分．3．答题前，务必将自己的姓名、准考证号用 0．5 毫米黑色墨水签字笔填写在试卷及答题卡上．第I 卷（选择题 共 60 分）一、单项选择题(本大题共 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．设复数 z 1，z 2 在复平面内的对应点关于实轴对称，z 1＝3＋4i ，则 z 1z 2＝A ．25B ．－25C ．7－24iD ．－7－24i 【答案】A【解析】＋4i)( 3－4i)＝32＋42＝25，故选择A. 2．设集合 A ，B 是全集 U 的两个子集，则“A ∩B ＝∅”是“A ⊆∁U B ”的 A ．充分不必要条件 B ．必要不充分条件C ．充要条件D ．既不充分也不必要条件【答案】C【解析】由韦恩图，A ∩B ＝∅，而显然可得 A ⊆∁U B ，又 A ⊆∁U B ，可得 A ∩B ＝∅，所以“A ∩B ＝∅”是“A ⊆∁U B ”的充要条件，故选择 C.3．已知 a ，b 是相互垂直的单位向量，与 a ， b 共面的向量 c 满足 a ·c ＝b ·c ＝2，则 c 的模为A ．1 【答案】DB ． 2C ．2D ．2 2【解析】不妨设 a ，b 分别为平面直角坐标系中 x 轴，y 轴上的单位向量，则 a ＝(1,0)，b ＝(0,1)，设 c ＝(x ,y )，则 a ·c ＝x ＝2，b ·c ＝y ＝2，所以 c ＝(2,2)，所以|c |＝ 22＋22＝2 2，故选择 D.4．在流行病学中，基本传染数是指每名感染者平均可传染的人数．当基本传染数高于 1 时， 每个感染者平均会感染一个以上的人，从而导致感染这种疾病的人数呈指数级增长．当基本传染数持续低于 1 时，疫情才可能逐渐消散．广泛接种疫苗可以减少疾病的基本传染数．假 设某种传染病的基本传染数为 R 0，1 个感染者在每个传染期会接触到 N 个新人，这 N 人中 有 V 个人接种过疫苗(V 称为接种率)，那么 1 个感染者新的传染人数为 N R 0(N －V )．已知新冠 N 病毒在某地的基本传染数 R 0＝2.5，为了使 1 个感染者传染人数不超过 1，该地疫苗的接种 率至少为()A ．40% 【答案】CB ．50%C ．60%D ．70%R 0 V【解析】为使 1 个感染者传染人数不超过 1，即 (N －V )≤1，即 R 0 (1－ )≤1，由题 R 0＝N N 2.5，所以 2.5(1－V)≤1 V 60%，即接种率至少为 60%，故选择 C. ，所以可解得N ≥N 2cos10º－sin20º 5．计算所得的结果为 cos20ºA ．1B ． 2C ． 3D ．2【答案】C【解析】cos10° ＝ c os(30° － 20°) ＝ c os30°cos20° ＋ sin30°sin20°＋ 1sin20°. 故 22cos10°－sin20°3cos20° ＝＝ 3，故选择C. cos20°6．密位制是度量角的一种方法．把一周角等分为 6000 份，每一份叫做 1 密位的角．以密位 作为角的度量单位，这种度量角的单位制，叫做角的密位制．在角的密位制中，采用四个数 码表示角的大小，单位名称密位二字可以省去不写．密位的写法是在百位数与十位数字之间 画一条短线，如 7 密位写成“0-07”，478 密位写成“4-78”．1 周角等于 6000 密位，记作 71 周角＝60-00，1 直角＝15-00．如果一个半径为2 的扇形，它的面积为 π ，则其圆心角用密6 位制表示为 A ．12-50 B ．17-50C ．21-00D ．35-00【答案】B7π 6 7πS 7 【解析】面积 6 ，半径为 2 的扇形所对的圆心角弧度大小为 θ＝2π·πr 2＝2π·4π＝12π，由题 7 π12意，其密位大小为 6000× 2π ＝1750，故用密位制表示为 17-50.故选择B.x 2 y 27 ．已知双曲线 C ：a 2－b 2＝1(a ＞0，b ＞0)的左、右焦点分别为 F 1，F 2，过点 F 2 作倾斜角 1为 θ 的直线 l 交双曲线 C 的右支于 A ，B 两点，其中点 A 在第一象限，且 cos θ ＝4．若|AB |＝|AF 1|，则双曲线 C 的离心率为3 A ．4 B ． 15C ．2D ．2【答案】D1【解析】由双曲线的性质，|AF 1|－|AF 2|＝2a 即|AB |－|AF 2|＝|BF 2|＝2a ，由 cos θ＝ 知 B 点的4a 215 (c －2) () 21 a横坐－ ＝1， a 2 b 2c结合 c 2＝a 2＋b 2 消去 b 2 即离心率为 2.故选择 D.，可得a ＝f (x ) 8．已知 f (x )是定义在 R 上的奇函数，其导函数为 f ′(x )，且当 x ＞0 时， f ′(x ) ·ln x 0，＋ ＞x 则不等式(x 2－1)f (x )＜0 的解集为 A ．(－1， 1)C ． (－∞，－1)∪(1，＋∞) B ．(－∞，－1)∪(0，1)D ．(－1，0)∪(1，＋∞)【答案】B【解析】设 g (x )＝f (x )·ln x ，则 g'(x )＝f'(x )·ln x ＋f (x )·1(x ＞0)，则由题意 g (x )在(0,＋∞)单调递 x ， 增，且由 g (1)＝0 知，当 x ∈(0,1)时 g (x )＜0，当 x ∈(1,＋∞)时 g (x )＞0，又由 g (x )＝f (x )·ln x ， 故有 x ∈(0,1)或(1,＋∞)时 f(x)＞0.因为 f (x )为奇函数，所以 x ∈(－∞,－1)或(－1,0)时 f (x )＜0. 综上(x 2－1) f (x )＜0 的解集为(－∞,－1)∪(0,1).故选择 B.二、多项选择题(本大题共 4 小题，每小题 5 分，共 20 分．在每小题给出的四个选项中，有多项符合题目要求的．全部选对的得 5 分，部分选对的得 2 分，有选错的得 0 分) 9. 对于两条不同直线 m ，n 和两个不同平面 α，β，下列选项中正确的为 A ．若 m ⊥α，n ⊥β，α⊥β，则 m ⊥n B ．若 m //α，n //β，α⊥β，则 m ⊥n 或 m //nC. 若 m //α，α//β，则 m //β 或 m ⊂βD. 若 m ⊥α，m ⊥n ，则 n //α 或 n ⊂α【答案】ACD 【解析】略10．已知 a ＞b ＞0，下列选项中正确的为A ．若 a － b ＝1，则 a －b ＜1B ．若 a 2－b 2＝1，则 a －b ＜1C ．若 2a －2b ＝1，则 a －b ＜1D ．若 log 2a －log 2b ＝1，则 a －b ＜1 【答案】BCa 2－b 2 1【解析】a －b ＝( a － b )( a ＋ b )＝ a ＋ b ＞ a － b ＝1，A 错误；a －b ＝ a ＋b ＝a ＋b 1 ＜ ，a －b ＜1，B 正确；2a －2b ＝1＝2b (2a －b －1)＞2a －b －1，a －b ＜1，C 正确；log 2a a －b －log 2b ＝1＝log a，a ＝2b ，a －b 无法判断，D 错误；故选择BC.2b 11．已知函数 f (x )＝ |sin x |＋ |cos x |，则A ．f (x )是周期函数B ．f (x )的图象必有对称轴π，k ⊥Z D ．f (x )的值域为[1，4 8]C ．f (x )的增区间为[k π，k π ＋2] 【答案】ABD【解析】A 显然正确；注意到 f (－x )＝ |sin(－x )|＋ |cos(－x )|＝ |sin x |＋ |cos x |＝f (x )， π＝1， π＝4 8，C 错误；f (x )＝ |sin x | 故 y 轴为 f (x )的一条对称轴，B 正确；注意到 f (0)＝f (2) f (4) k π π(k ∈Z )时，取“＝”，又 f (x )＝＋ |cos x |≤(1＋1)(sin x ＋cos x )≤ 4 8，当且仅当 x ＝ ＋24|sin x |＋ |cos x |≥ |sin x |2＋ |cos x |2＝|sin x |＋|cos x |≥1，当且仅当 x ＝k π(k ∈Z )时，取2 “＝”，D 正确；故选择ABD.k * k 2n － k12．已知 n ⊥N ，n ≥2，p ，q ＞0，p ＋q ＝1．设 f (k )＝C p q，其中 k ⊥N ，k ≤2n ，则2n 2nA ． ∑ f (k )＝1k =02nB ． ∑ kf (k )＝2npqk =0n1 nC ．若 np =4，则 f (k )≤f (8)D ． ∑ f (2k ) f (2k －1)＜ ＜∑ 2 k =0k =1 【答案】AC2n2n2n 2n －1k k － k k 2n k － 1 k 2n k －p k q 2n －1－k ＝ 【解析】A 显然正确； ∑ kf (k )＝ ∑ kC p q ＝ ∑ 2nC p q ＝2np ∑ C 2n 2n －12n －1 k =0 k =0 k =1 k =0k k 2n k－f (k ) C p qp (2n ＋1－k ) f (k ＋1) p (2n －k ) p (2n －k ) 2n 2np ，B 错误； ＝ ＝ ， ＝ ， ≤1≤ k － qkf (k ) f (k －1) 1 k — ＋ －1 2n 1 k q (k ＋1) q (k ＋1) C p q 2n p (2n ＋1－k ) 1n ，2np －p ≤k ≤2np ＋q ，8－p ≤k ≤8＋q ，k ＝8，C 正确；当 p ＝q ＝2时，∑f (2k )qk k =01 n＝ ＝∑f (2k －1)，D 错误；故选 AC. 2 k =1三、填空题(本大题共 4 小题，每小题 5 分，共 20 分)13．某班 4 名同学去参加 3 个社团，每人只参加 1 个社团，每个社团都有人参加，则满足上述要求的不同方案共有【答案】36▲ 种．（用数字填写答案） 【解析】依题意，四名同学可分为(1，1，2)，有 C 2A 3＝6×6＝36 种. 4 3 x 2 y 2 14 ．已知椭圆4 ＋ 3 ＝1 的右顶点为 A ，右焦点为 F ，以 A 为圆心，R 为半径的圆与椭圆相交 于 B ，C 两点．若直线 B C 过点 F ，则 R 的值为 ⊥ ．13【答案】2【解析】A (2,0), F (1,0), B ，C 两点关于 x 轴对称，即横坐标为 1，代入椭圆方程，得 B ,C 坐 33 2＝ .标为(1， ±2）,R ＝ (2－1)2＋(0 －2) 15．在四棱锥 P －ABCD 中，P A ⊥面 ABCD ，四边形 ABCD 是边长为 2 的正方形，且 P A ＝ 2．若点 E ，F 分别为 AB ，AD 的中点，则直线 EF 被四棱锥 P －ABCD 的外接球所截得的线段长为▲ ． 【答案】 6【解析】注意到⊥P AC ，⊥PBC ，⊥PDC 均为以 PC 为斜边的直角三角形，故外接球球心O为 PC 中点，R ＝2PC ＝ 3，取 EF 中点 G ，又AC ＝OC ＝故 GO ⊥PC ，d ＝GO ＝ 1P C GC 6l ＝2 R 2－d 2＝ 6.16．牛顿迭代法又称牛顿－拉夫逊方法，它是牛顿在 17 世纪提出的一种在实数集上近似求解方程根的一种方法．具体步骤如下：设 r 是函数 y ＝f (x )的一个零点，任意选取 x 0 作为 r 的初始近似值，过点(x 0，f (x 0))作曲线 y ＝f (x )的切线 l 1，设 l 1 与 x 轴交点的横坐标为 x 1，并称 x 1 为 r 的 1 次近似值；过点(x 1，f (x 1))作曲线 y ＝f (x )的切线 l 2，设 l 2 与 x 轴交点的横坐标为 x 2，并称 x 2 为 r 的 2 次近似值．一般的，过点(x n ，f (x n ))(n ⊥N )作曲线 y ＝f (x )的切线 l n ＋1，记 l n ＋1 与 x 轴交点的横坐标为 x n ＋1，并称 x n ＋1 为 r 的 n ＋1 次近似值．设 f (x )＝x 3＋x －1(x 3x 3＋x n n，n ⊥N *，数列{a n }≥0)的零点为 r ，取 x 0＝0，则 r 的 2 次近似值为 ▲ ；设 a n ＝ 2x 3＋1n 的前 n 项积为 T n ．若任意 n ⊥N *，T n ＜λ 恒成立，则整数 λ 的最小值为 ▲ ．3【答案】4，2【解析】(1) f '(x )＝3x 2＋1，取 x 0＝0，f (0)＝－1，f '(0)＝1，即过点(0，－1)作曲线 y ＝f (x )的切线 l 1 斜率为 1，l 1 方程为 y ＝x －1,交 x 轴点横坐标为 1，即 x 1＝1，f (1)＝1，f '(1)＝4，过点(1，1)作曲线 y ＝f (x )的切线 l 2 斜率为 4，l 2 方程为 y ＝4x －3 交 x 轴点横坐标为3(2)f (x 0)＝； 42 x 3＋1 0x 3＋x －1，f '(x )＝3x 2＋1，切线方程为 y ＝(3x 2＋1)(x －x )＋x 3＋x －1,即 x ＝,可得出0 0 0 0 0 0 0 0 1 3 x 2＋1 03 2 32x ＋1 n －1 1 3x ＋1 x n －1 n －1 3x ＋x n －1x n －1 n －1 ,即 a ＝ ，所以 n ⊥N * {x }的递推关系式为 x ＝, ＝ , ＝ n n n －1 3x ＋1 x n 2x ＋1 2 3 x n 3x n 2x ＋1n －1 n －1 n －1 x 11 3 1 ，因为 f '(x )＞0，且 f ( )＝－ ，f (1)＝1，所以 f (x )的有唯一零点 x '∈( ，1)，所以 时 T n ＝2 8 2 x n ＋11x 1 当 n ≥1 时，x ⊥(x '，x ) (2， 1)，所以 T ＝ ∈(1，2).故 λ 的最小值为 2. n ＋1 1 n x n ＋1四、解答题(本大题共 6 小题，共 70 分．解答时应写出文字说明、证明过程或演算步骤) 17．(本小题满分 10 分)在①b ＝ 3a ；②a ＝3cos B ；③a sin C ＝1 这三个条件中任选一个，补充在下面问题中．若问题 中的三角形存在，求该三角形面积的值；若问题中的三角形不存在，说明理由．问题：是否存在⊥ABC ，它的内角 A ，B ，C 的对边分别为 a ，b ，c ，且 sin B －sin(A －C )＝ 3sin C ，c ＝3，?解：因为 A ＋B ＋C ＝π，所以 sin B ＝sin(A ＋C )，所以 sin B －sin(A －C )＝(sin A cos C ＋cos A sin C ) －(sin A cos C －cos A sin C )＝2cos A sin C ＝ 3sin C ，因为 C ∈(0，π)，所以 sin C ≠0，所以 cos A ＝π又 A ∈(0，π)，所以 A ＝6.若选①，由正弦定理，sin B ＝ 3sin A π 2π所以 B ＝3或 3 ，ππ 若 B ＝3，则 C ＝π－A －B ＝2，所以 b ＝c cos A ＝1 1 S ⊥ABC ＝2bc sin A ＝3×2＝2π π若 B ＝ 3 ，则 C ＝π－A －B ＝6，所以 a ＝c ＝3，1 1 S ⊥ABC ＝2ac sin B ＝2×3×3×若选②，因为 c ＝3，由正弦定理，sin A ＝sin C cos B ，又因为 A ＋B ＋C ＝π， 所以 sin A ＝sin(C ＋B )＝sin C cos B ＋cos C sin B ， 所以 cos C sin B ＝0，又 B ∈(0，π)，所以 sin B ≠0，π所以 cos C ＝0，C ＝2，所以 b ＝c cos A ＝1 1 S ⊥ABC ＝2bc sin A ＝3×2＝1若选③，由正弦定理 c sin A ＝a sin C ＝1，由 c ＝3，sin A ＝2，矛盾，所以这样的三角形不存在 . 18．(本小题满分 12 分)已知等比数列{a n }的前 n 项和 S n ＝2n ＋r ，其中 r 为常数． (1)求 r 的值；(2)设 b n ＝2(1＋log 2a n )，若数列{b n }中去掉数列{a n }的项后余下的项按原来的顺序组成数列 {c n }，求 c 1＋c 2＋c 3＋···＋c 100 的值． 解:(1)n ＝1 时，a 1＝S 1＝2＋r ，－1n ≥2 时，a n ＝S n －S n －1＝2n ，所以 a 2＝2，a 3＝4，a 22＝1，即 2＋r ＝1，所以 r ＝－1，因为{a n }为等比数列，所以 a 1＝ a 3n此时，对任意 n ⊥N ，a ＝2 ，所以 n ≥2 时，a * n 1－ ≠0， ＝2，故{a }为等比数列，所 n n －1 na n －1以 r ＝－1．(2)b n ＝2(1＋log 2a n )＝2n ，b n ＋1－b n ＝2，所以{b n }是首项为 2，公差为 2 的等差数列．数列{b n }前 100 项为 2，4，6，8，…，200，其中 2，4，8，16，32，64，128 为数列{a n } 中的项，所以{c n }前 100 项为{b n }中前 107 项去除 2，4，8，16，32，64，128 后按原来顺 序构成的数列．故 c 1＋c 2＋c 3＋···＋c 100＝(b 1＋b 2＋…＋b 107)－(a 2＋a 3＋…＋a 8) 107(2＋214) ＝ －2(2 －1)＝11556－256＋2＝11302． 7 2 19．(本小题满分 12 分)某公司对项目 A 进行生产投资，所获得的利润有如下统计数据表：(1)请用线性回归模型拟合 y 与 x 的关系，并用相关系数加以说明；(2)该公司计划用 7 百万元对 A ，B 两个项目进行投资．若公司对项目 B 投资 x (1≤x ≤6)百 万元所获得的利润 y 近似满足：y ＝0.16x －0.49＋0.49，求对 A ，B 两个项目投资金额分别x ＋1 为多少时，获得的总利润最大？附：①对于一组数据(x 1，y 1)，(x 2，y 2)，···，(x n ，y n )，其回归直线方程^y ＝b ^x ＋a^的斜率和 项目 A 投资金额 x（单位：x 百万元）12345所获利润 y（单位：y 百万元）0.30.30.50.91n∑ x i y i －nx · y— －i=截距的最小二乘法估计公式分别为：b ^＝n ∑ i i =1n∑ x i y i －nx · y i =1 — －②线性相关系数 r ＝．一般地，相关系数 r 的绝对值在 0.95 以n ( n∑ x i －nx ) ( ∑ y i －ny 2 －2 2－2 )i =1i =1上(含 0.95)认为线性相关性较强；否则，线性相关性较弱．n n参考数据：对项目 A 投资的统计数据表中∑ x y ＝11， ∑ y ＝2.24， 4.4≈2.1．2i i i i =1 i =1解(y ＝(0.3＋0.3＋0.5＋0.9＋1)÷5＝0.6， 5∑ i ＝1 5∑ 22i =1 5 ∑2 2i =1 5∑ x i y i －5 x · y — －i =1 则b ^ ＝^－ ^ ^ － ＝0.2，a ＝ y －bx ＝0.6－0.2×3＝0，则有y ＝0.2x ， 5 ∑ i i =15∑ x i y i －5 x · y— －2 2＝ ＝ ≈0.9524＞0.95， i ＝1 r ＝2.1 5 5 10×0.44 ∑ x i －5 x ) ( ∑ y i －5 y 2 －2 2－2 ( )i ＝1i ＝1答：线性回归方程为：^y ＝0.2x ；y 与 x 线性相关性较强．(2)由于对项目 B 投资 x (1≤x ≤6)百万元，则对项目 A 投资(7－x )百万元，则总利润为：y ＝0.16x －0.49＋0.49＋0.2(7－x )，(1≤x ≤6)x ＋1 y ＝1.89－0.04x －0.49 ＝1.93－[0.04(x ＋1)＋0.49] x ＋1 ≤1.93－0.28＝1.65x ＋1当且仅当 x ＋1＝3.5，即 x ＝2.5 时，取到最大值 1.65 百万元，答：投资 A 项目 4.5 百万元，B 项目 2.5 百万元，利润最大值为 1.65 百万元． 20．(本小题满分 12 分)如图，三棱柱 ABC －A 1B 1C 1 的所有棱长都为 2，B 1C ＝ 6，且 AB ⊥B 1C ． （1）求证：平面 ABB 1A 1⊥平面 ABC ；4（2）若点 P 在棱 BB 1 上且直线 CP 与平面 ACC 1A 1 所成角的正弦值为 ，求 BP 的长．5z C 1C 1B 1B 1A 1A 1PxCCBOAy （第 20 题图）A （第 20 题图）解(1)证明：取 AB 中点 O ，连结 B 1O ，CO ，在正三角形 ABC 中，CO ⊥AB ，且 CO ＝ 3，因为 AB ⊥B 1C ，CO ∩B 1C ＝C ，所以 AB ⊥平面 B 1CO ，所以 AB ⊥B 1O ，因为 BO ＝1，BB 1＝2，所以 B 1O ＝ 3，因为 B 1O 2＋CO 2＝6＝B 1C 2，所以 B 1O ⊥CO ， 因为 CO ∩AB ＝O ，所以 B 1O 垂直平面 ABC ，又 B 1O ⊆平面 ABB 1A 1，所以平面 ABB 1A 1⊥平 面 ABC ；(2)由(1)，OC ，OA ，OB 1 两两垂直，故可分别以 OC ，OA ，OB 1 方向为 x ，y ，z 轴建立如图 所示的空间直角坐标系，所以 A (0，1，0)，C( 3，0，0)，B (0，－1，0)，B 1(0，0， 3)，→ → - - 所以AC ＝( 3，－1，0)，CB ＝(－ 3，－1，0)，AA 1＝BB 1＝(0，1， 3)，设BP ＝λBB 1＝(0，- →→ λ， 3λ) ，则CP ＝ C B ＋ BP ＝ (－ 3，λ－1， 3λ).设平面 ABB 1A 1 的一个法向量为 n ＝(x ，y ，z )，⎧⎪→ ⎧y ＝ 3 则⎨ AC ·n ＝ 3x －y ＝0，取 x ＝1，得⎨ ， ⎪ → ⎩z ＝－1 ⎩ AA 1·n ＝y ＋ 3z ＝0所以 n ＝(1， 3，－1)，设直线 CP 与平面 ACC 1A 1 所成角的大小为 θ， →则 sin θ＝|cos<n ， C P >| ＝(1， 3，－1)·(－ 3，λ－1， 3λ)||12＋( 3)2＋(－1)2× (－ 3)2＋(λ－1)2＋( 3λ)2＝ 2 3 1 1 4 ＝ ，得 4λ －2λ＋ ＝0，解得 λ＝ ， 2 4 4 55× 4λ2－2λ＋41 1所以 BP ＝ BB 1＝ .4 221.已知直线 l :y ＝x ＋m 交抛物线 C :y 2＝4x 于 A ，B 两点． -(1)设直线 l 与 x 轴的交点为 T ，若AT ＝2 TB ，求实数 m 的值；(2)若点 M ，N 在抛物线 C 上，且关于直线 l 对称，求证:A ，B ，M ，N 四点共圆． 解:(1)在 y ＝x ＋m 中令 y ＝0，可得 T (－m ，0)， 设 A (x 1，y 1)，B (x 2，y 2)，- - → → 因为AT ＝2 TB ，所以OA ＝3 OT －2OB ，即(x 1，y 1)＝(－3m －2x 2，－2y 2)，所以 y 1＝－2y 2， 将 y ＝x ＋m 代入 y 2＝4x 可得 y 2－4y ＋4m ＝0， 所以 y 1＋y 2＝4，y 1y 2＝4m ， 所以 y 1＝8，y 2＝－4，m ＝－8， 所以实数 m 的值为－8．(2)证法 1：设 M ，N 两点的坐标分别为(x 1，y 1)，(x 2，y 2)， 因为点 M ，N 在抛物线 C 上，且关于直线 l 对称，所以可设直线 MN :x ＋y ＋n ＝0，代入 y 2＝4x 得 y 2＋4y ＋4n ＝0， 所以 y 3＋y 4＝－4，y 3y 4＝4n ， x ＋x 3 4所以 MN 中点为 ( ，－2)，2y 2＋y 2 x 3＋x 4 3 4＝ (y 3＋y 4)2－2y 3y 4 因为 ＝ ＝2－n ，2 8 8所以 MN 中点为(2－n ，－2)， 所以－2＝2－n ＋m ，即 m －n ＝－4，y 3－y 4 4(y 3－y 4) 4因为 k MN ＝ ＝ ＝ ， y 2－y 2x 3－x 4 3 4y 3＋y 4 4 16 所以 k AM ·k BM ＝ 4· ＝ ， y 2＋(y 1＋y 2)y 3＋y 1y 2y 3＋y 1 y 3＋y 2 3因为 y 1＋y 2＝4，y 1y 2＝4m ，16 4 所以 k AM ·k BM ＝ 16＝ ＝ ＝－1，y 2＋4y 3＋4m 4x 3＋4y 3＋4m m －n 3 所以⊥AMB ＝90º，同理⊥ANB ＝90º， 所以 A ，B ，M ，N 都在以 AB 为直径的圆上， 所以 A ，B ，M ，N 四点共圆．证法 2：因为点 M ，N 在抛物线 C 上，且关于直线 l 对称， 所以可设直线 MN :x ＋y ＋n ＝0，所以 A ，B ，M ，N 满足方程(x －y ＋m )(x ＋y ＋n )＋2(y 2－4x )＝0， 即 x 2＋y 2＋(m ＋n －8)x ＋(m －n )y ＋mn ＝0， 所以 A ，B ，M ，N 四点共圆．注：圆锥曲线上四点共圆的充要条件是两条对棱斜率相反或斜率均不存在，参考我拙作《高 中数学－解析几何系统解析》. 22．(本小题满分 12 分)已知函数 f (x )＝e x －ax sin x －x －1，x ⊥[0，π]，a ⊥R ． 1 (1)当 a ＝2 时，求证：f (x )≥0；(2)若函数 f (x )有两个零点，求 a 的取值范围． 1 1解：(1)当 a f (x )=e x －2x sin x －x －1， ＝2时， 1f'(x )＝e x －2(sin x ＋x cos x )－1，1 1 f'(x )＝e x －2(cos x ＋cos x －x sin x )＝(e x －1)＋(1－cos x ) ＋2x sin x ≥0(因为 x ∈[0，π])， 所以 f'(x )在区间[0，π]为单调递增函数，所以 f'(x )≥f ’(0)＝0， 所以 f (x )在区间[0，π]为单调递增函数，所以 f (x )≥f (0)＝0．1 1≤2时，f (x )≥e x －2x sin x (2)由(1)知，当 a －x －1≥0，当且仅当 x ＝0 时取等号， 此时函数 f (x )仅有 1 个零点．1当a＞2时，因为f(x)＝e x－ax sin x－x－1，所以f′(x)＝e x－a(x cos x＋sin x)－1，f′′(x)＝e x＋a(x sin x－2cos x)．当x∈ π[2，π]时，f′′(x)＞0，所以f′(x)单调递增．π时，f′′′(x)＝e x＋a(3sin x＋x cos x)．当x∈[0，2]因为e x＞0，a(3sin x＋x cos x)≥0，所以f′′′(x)＞0，所以f′′(x)单调递增．πππ又f′′(0)＝1－2a＜0，f′′(2)＝e2＋2a＞0，ππ因此f′′(x)在[0，]上存在唯一的零点x0，且x0⊥(0，)．2当x⊥(0，x0)时，f′′(x)＜0，所以f′(x)单调递减；2π当x⊥(x0，)时，f′′(x)＞0，所以f′(x)单调递增．2又f′(0)＝0，f′(x0)＜f′(0)＝0，f′(π)＝eπ＋aπ－1＞0，因此f′(x)在[0，π]上存在唯一的零点x1，且x1⊥(x0，π)．当x⊥(0，x1)时，f′(x)＜0，所以f(x)单调递减；当x⊥(x1，π)时，f′(x)＞0，所以f (x)单调递增．又f (0)＝0，f (x1)＜f (0)＝0，f(π)＝eπ－π－1＞0，所以f(x)在(x1，π)上存在唯一零点，因此f(x)在[0，π]上有两个零点．综上，a 的取值范围是1(2，＋∞)．18。

江苏省南京市、盐城市高三第二次模拟考试地理试题本试卷分为选择题和综合题两部分。

本次考试满分为120分，考试时间为100分钟。

注意事项：答题前，务必在试题卷、答题卡规定的地方填写自己的姓名、座位号，答案按要求填涂在答题卡上；非选择题答案写在答题卡上对应题目的答案空格内，答案写在试卷上无效。

考试结束后，交汇答题卡。

一、选择题（共60分）（一）单项选择题：本大题共18小题，每小题2分，共36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。

2019年伊始，一条关于“美联航890航班穿越了”的新闻在网络上火了。

该航班联系上海和旧金山（西八区）两城市。

2019年起飞，2016年到达，飞行时间11个小时。

据此回答1-2题。

1.飞机起飞时，太平洋上的日照情况最可能是下图中的（）2.关于该次航班的叙述，正确的是（）A.飞机自东向西穿越日界线B.飞行过程中穿过晨线C.飞机先向东南方向飞，再向东北方向飞D.飞行过程中太阳处于偏北的方向“七下八上”是指每年7月下旬到8月上旬我国气象部门高度关注的一个多雨时期。

读中国雨带进程示意图（图1），回答3-4题。

3.与“七下八上”多雨期密切相关的天气系统是（）A.蒙古-西伯利亚高压B.西太平洋副热带高压C.阿留申低压D.江淮准静止锋4.“七上八下”期间，我国各地与其面临的气象灾害组合正确的是（）A. 长江中下游—梅雨、洪涝B. 华北平原—洪涝、暴雨C. 东北地区—冰雹、滑坡D. 华南沿海—伏旱、台风当海面有空气平流运动时，海面温度和气温之间产生温度差异，空气和海面之间发生热量交换，空气达到饱和状态而形成雾，海雾的发生和洋流的运动密切相关。

图2为图3中AB航线附近海雾时空分布图（图中数值越大，出现海雾的频率越高），据此回答5—6题。

5. 关于AB航线附近海域海雾时空分布，描述正确的是（）A. 海面和大气温差较大季节海雾发生频率更大B. 东侧海雾发生的频率比西侧高C. 55°W地区海雾发生的季节变化最大D. 西侧和东侧海雾发生频率多的季节是相同的6. 关于B附近海域海雾的状况，理解正确的是（）A. 寒暖流交汇是当地海雾形成的重要原因B. 海雾导致该地区阴雨天气较多C. 海雾的形成与流经该地区的暖流相关D. 海雾对当地环境产生比较严重的大气污染图4是我国贵州某地地质地形图（从①--⑤，岩层年龄由老到新），该地大部分地区已开辟为梯田。

南京市、盐城市2020届高三年级第一次模拟考试英语参考答案第一部分听力（共两节，满分20分）1.A2.C3.B4.A5.C6.B7.A8.C9.B10.A11.B12.C13.A14.A15.B16.C17.A18.C19.B20.A第二部分英语知识运用（共两节，满分35分）第一节单项填空（共15小题；每小题1分，满分15分）21.C22.B23.B24.C25.C26.C27.D28.A29.B30.A31.D32.B33.D34.C35.D第二节完形填空（共20小题；每小题1分，满分20分）36.A37.B38.A39.D40.C41.B42.C43.A44.D45.B46.B47.A48.C49.D50.C51.D52.B53.C54.A55.D第三部分阅读理解（共15小题；每小题2分，满分30分）56.A57.B58.D59.B60.C61.C62.B63.D64.A65.D66.B67.C68.D69.A70.C第四部分任务型阅读(共10小题；每小题1分，满分10分)71.adults72.including73.aging/ageing74.subjective75.Trying76.tips/suggestions77.requires/demands/needs78.effective79.absorb80.extends第五部分书面表达（满分25分）81.One possible versionAccording to a new regulation issued by China’s Ministry of Education,teachers have the right to discipline students who break school rules,which,however,has generated a lot of discussion.From my perspective,the significance of school discipline cannot be stressed too much.For one thing,the regulation will help students better understand the consequences of misbehaving,which is of great benefit to their healthy growth and development.For another,the regulation can delegate more power to teachers,who shoulder pressure in dealing with misbehaving students.In order for the regulation to be conducted effectively,many joint efforts need to be made.First of all,authorities and schools should set up specific and reasonable rules for school discipline.Secondly,teachers ought to understand the rules thoroughly to employ them properly instead of abusing them.More importantly,parents are supposed to cooperate with teachers to reach a consensus on the level of punishments.书面表达评分建议一、评分原则1．本题总分为25分，按5个档次给分。