-2018年上海市南模中学高三上十月月考
2018届上海实验学校高三10月考英语试卷
2018届高三第一学期英语十月月考试卷（考试时间：120 分钟）第I 卷(共90分)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. He often works extra hours.B. He'll help Tina prepare for the meeting.C. He's afraid the meeting won't end on time.D. He's disappointed that he can't attend the meeting.2. A. The woman will show the man how to use the camera.B. The man will take the camera to be repaired.C. The woman will take a picture of the man.D. The man will lend a camera to the woman.3. A. He'd like to apply for a replacement card.B. He needed to see a doctor two weeks ago.C. He's pleased that the woman found the card.D. He's glad he was finally able to get an appointment.4. A. She has to leave in a little while.B. She doesn't understand what the man meant.C. She has no time to repeat the explanation now.D. She doesn't mind answering the man's questions.5. A. The woman doesn't want to accept the man's apology.B. The woman wasn't annoyed with the man's being late.C. The man didn't realize that the woman was waiting.D. The man waited a long time for the bus to come.6. A. She bought the coat last winter.B. The coat is the only warm coat she owns.C. The coat isn't warm enough to wear in cold weather.D. She needs to have the coat cleaned before next winter.7. A. She won't be a candidate next year. B. The news won't disappoint Mary.C. She doesn't believe the news.D. The news doesn't upset her.8. A. Give the woman a ride to the bookstore. B. Try to find the woman's roommate.C. Buy tickets for the film festival.D. Get a schedule for the woman.9. A. Annie might spend a lot of money on classical music.B. He doesn't know many composers of classical music.C. He has known Annie's neighbor for many years.D. Annie should try to avoid getting sick.10. A. She'll probably postpone her appointment with the dentist.B. She won't take the bus with the man to the museum.C. She forgot that she was going to the museum with the man.D. She'll meet the man in front of the library.Section BDirections: In Section B, you will hear several short passages and longer conversations, and you will be asked several questions on each of the passages and the conversations. The passages and the conversations will be read 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 would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. Visiting the UK. B. Gardening.C. Doing some exercise.D. Going clubbing.12. A. Going to work. B. Taking dance classes.C. Sleeping.D. Running.13. A. What brings them happiness. B. How they enjoy in their work.C. What places they like visiting.D. How they keep fit and healthy.Questions 14 through 16 are based on the following passage.14. A. He forbade anyone but the sheep raiser to talk to them.B. He left them with sheep living in the wild.C. He isolated them from anybody else.D. He brought up them by himself.15. A. It's unbelievable. B. It's well-established.C. It's unprecedented.D. It's challenging.16. A. To confirm the way children learned to talk.B. To discover the oldest ever language in the world.C. To find out whether children could talk if left alone.D. To prove that children living with animals could speak.Questions 17 through 18 are based on the following conversation.17. A. Looking for a new place to live. B. Leaving Salford for Manton.C. Doing shopping for their bathroom.D. Walking along a big river.18. A. Its size. B. Its view. C. Its garden. D. Its location.Questions 19 through 20 are based on the following conversation.19. A. Making a TV programme.B. Looking for the world's first woman pilot.C. Reading an article about Harriet Quimby.D. Finding more information about the Titanic.20. A. She used to be a famous film star. B. She left England for the US in 1912.C. She got her pilot's licence at 37.D. She flew across the English Channel.II. Grammar and VocabularySection A 10%Directions: 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 ofthe given word; for the other blanks, use one word that best fits each blank.Sizing Up Carbon FootprintsKelsey Schroed er was “born green”, according to her mother, and she takes that environmental enthusiasm to class with her at the Oak Knoll School of the Holy Child in Summit, N. J.. The 12-year-old 21)_____ (be) a driving force in greening her school since she was a fourth-grader. But 22)______really motivates kids — especially the sort of achievers who attend an exemplary private school like Oak Knoll — is a little competition. So when Schroeder and her classmates found out about a website 23) _____ (launch) last year that sets teams from around the country against one another in a contest to see who could be 24)______(green), they jumped on board. Her seventh-grade Royal Acorns team is ’s the most recent champion, 25)_____ (save) 11.21 tons of climate-changing CO2 to date.26)_____Americans grow more green-minded, more of them want to approach environmentalism in concrete terms. Thanks to websites like Carbonrally, one increasingly popular way to do so is by measuring and measurably reducing our carbon footprints —the greenhouse gases we’re responsible for 27)_____(emit). The more dependent we are 28)_____ fossil fuels, the bigger our carbon footprints; unsurprisingly, Americans, who are responsible for more than 20 tons of CO2 per capita annually, have some of the biggest feet in the world. How big? A recent study by a class at the Massachusetts Institute of Technology found that even a homeless American 29)_____ have a carbon footprint of 8.5 tons —twice 30)_______global average. “We have contributed m ore than our fair share to this problem,” says Katherine Wroth, a senior editor at the green website . “It seems logical that we would want to contribute to the solution.”Section B 10%Directions: Complete the following passage by using the words in the box. Each word can only beYou may still remember the “Eight Minutes of Tokyo” in the closing ceremony of the Rio Olympics last August. Even if the Tokyo Games wasn’t going to be here for another four years, the performance of the new host successfully 31 the show with its famous animated characters –Doraemon, Hello Kitty and Super Mario.It was a wise choice since there is probably nothing that shouts “Japan” more loudly than the country’s animation, known as “Japanimation”. And this year32 the 100th anniversary of the very first Japanese cartoon, made in 1917.In the past century, the wild imagination of Japanese animators continued to feed our 33 . Monsters, fairies, robots and magic feature often in their work. It has also been inspiring film industries in other parts of the world. The 34 story of Disney’s The Lion King (1994), for example, actually comes from Japan’s Kimba the White Lion. And the 1999 Hollywood 35 film, The Matrix, was also 36 influenced by the 1989 Japanese manga Ghost in the Shell .“I love his films. I study his films. I watch his films when I’m looking for 37 ,” John Lasseter, director of Pixar’s Toy Story, once said about famous Japanese animator Hayao Miyazaki. And our 38 for this imaginary world is only growing.At the end of last year, for example, the story of Japanese cartoon Pokemon (was brought to life with the help of augmented reality technology. People in many countries are often seen searching for Pokemon in real life locations through the screens of their phones. They play it on their way toschool, to work, and during holiday outings. Although Pokemon began as a video game a year before the cartoon came out, people should give the animation a “Thanks” for bringing it to a wider audience.Besides being a source of entertainment, Japanese animation is also a 39 force of our daily lives. For example, wearing glasses used to be considered as uncool and geeky (书呆子气的), but after the 1981 TV animation Arale (《阿拉蕾》), in which there is a heavily-nearsighted girl with wings and magic powers, glasses soon became fashionable. And the language we use –the word meng (萌), to name one – is also 40 from Japanese animation.But interestingly, with all the imagination that is so admired by modern fans, it would still be hard for Japanese animators in 1917 to believe that the two-dimensional worlds that they were creating at the time would have such a big influence in today’s three-dimensional world.III. Reading ComprehensionSection A 15%Directions: 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 of phrase that best fits the context.Shoppers on Black Friday, the traditional start of the holiday shopping season in America are notoriously(臭名昭著的)__41__. Some even start queuing outside stores before dawn to be the first to lay their hands on heavily discounted goods. Despite the madness at many stores, __42__, the global economic recession appears to have accelerated the pace at which shoppers are ___43___ bricks and mortar(灰浆) in favor of online retailers---e-retailers,E-commerce holds particular __44__ in poor times as it enables people to compare prices across retailers quickly and easily. Buyers can sometimes avoid local sales taxes online, and shipping is often free. No wonder, then，that online shopping continues to grow even as the offline sort __45__. // The __46__ in spending to the Internet is good news for companies like P&G that lack retail shops of their own．But it is a big __47___ for physical retailers, whose prices are often higher than those of e-retailers, since they must __48___ the extra expense of running stores.The most obvious response to the growth of e-retailing is for ___49___ retailers to redouble their own efforts online. The online arms of big retailers are performing well, ___50__ _.The concept of “multi-channel” shopping，where people can purchase the same items from the same retailer in several different ways online, is gaining popularity, and retailers are trying to __51__ users of one channel to try another.Retailers are also trying to make online shopping seem ___52___ and exciting to act against the low economy．One common tactic is to set up “pop—up” stores，which appear for a short time before disappearing again, to develop a sense of novelty and __53__.Stores are also trying to attract customer by offering services that are not ___54___ online. Best Buy, a consumer-electronics retailer, has started selling music lessons along with its musical instruments. Lululemon athletica, which sells sports clothes, offers free yoga classes: The idea is to bring people back to its shops ___55___, increasing the likelihood that they will develop the habit of shopping there.41. A.positive B. productive C.aggressive petitive42. A.moreover B.otherwise C. meanwhile D.however43. A.abandoning B.applying C.foreseeing D. predicting44. A.opinion B.intention C.interest D.appeal45. A.soars B.shrinks C. contracts D. squeeze46. A.shift B.demand C.impact D. pattern47. A.concern B.care C. influence D. contrast48. A.carry B.cease C. bear D. record49. A.additional. B.mysterious C.relevant D. conventional50. A.ontheirown B. on the whole C. on the contrary D. on the other hand .51. A.encourage B.represent C.stimulate D.allow52. A.solution B.irony C.fun D. vision53. A. emergency B. urgency C. humor D. promotion54. A. available B.expected C.apparent D.incredible55. A.extremely B.regularly C.especially D.properlySection B ( 22%)Directions: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)Zelda Fitzgerald, as is revealed by numerous personal books and letters, wore many labels in her life. She was “the original flapper” girl and “the spirit of the Jazz Age.” Married to the celebrated writer F. Scott Fitzgerald (author of The Great Gatsby), she was by turns his muse and the woman who ruined his life. In her later years she was “Crazy Zelda.”Accurate as all these descriptions may be, they do not tell the whole story. Born in Montgomery, Alabama, she was noted for her beauty and high spirits in dancing. In July 1918, at a country club dance, Fitzgerald was hooked immediately by the beautiful and charming 18-year-old Zelda who outshined other beauties with her distinguished ballet. A light affection evolved into a lengthy long-distance pursuit of weekly letters, with Fitzgerald aware of her uncommitted dating of other men. He courted her feverishly, reading her his stories and parts of his unfinished novel. He proposed after his discharge from the Army in February 1919, but Zelda had doubts. Her fiancéwasn’t rich and there was no guarantee he’d ever be famous. His short stories didn’t sell. His attempt was a dump. Zelda gave back the ring. Hoping to fix the “no money” part of his problem, Fitzgerald quit the job and started to rewrite novels for success and money so that he could win back his girl. Finally, he made it！On March 20, 1920, his novel This Side of Paradise got published and Zelda agreed to marry him.However, their marriage was troubled by wild drinking, fighting, infidelity（不忠）and bitter recriminations(互相指责). Earnest Hemingway, whom Zelda disliked, blamed her for Scott’s declining literary output, though she has also been portrayed as the victim of an overbearing husband. Actually, Zelda was also creative, pursuing both dancing and writing. Some scholars have portrayed Zelda as a creative talent ignored by the patriarchal（男权的）society of the day. Her inspiration was even drawn by her husband in literary creation. Scott used their relationship as material in his novels, even borrowing ep isodes from Zelda’s diary and applying them into his fictional writings. She detested her husband’s practice: “Mr. Fitzgerald—I believe that is how he spells his name—seems to believe that plagiarism begins at home.” To seek an artistic identity of her own, Zelda wrote independently to declare her own value, as she put it “I wish I could write a beautiful book to break those hearts that are soon to cease to exist.”Nevertheless her unique personality was starting to seem more unbalanced than charming. The couple —like the rest of the nation—was living on borrowed time. In October 1929 the stock market crashed, triggering the Great Depression. Six months later, Zelda suffered her first nervousbreakdown. After being diagnosed with schizophrenia（精神分裂）, she was increasingly confined to specialist clinics, and since then has departed with her husband. Zelda died later in a fire at her hospital in Asheville, North Carolina, putting an end to her flamboyant life.56.Where will you most probably find this article?A.In a newspaper.B. In a literary magazine.C. In a prepared speech.D. In a research report.57.Which phrase can best summarize the relationship between F. Scott Fitzgerald and Zelda?A. Ideal partnership.B. Unbalanced love relationship.C. Love-hate relationship.D. Mutually-jealous relationship.58.Which of the following is TRUE according to the article?A.Fitzgerald successfully won Zelda’s heart by reading her his novels and writing her weeklyletters.B.Hemingway disliked Zelda because of her female identity and talent that outshined herhusband.C.Zelda was glad to be her husband’s muse and provided him with literary materials.D.The “Crazy Zelda” died without Fitzgerald’s companion after severe schizophrenia.（B）Hillary vs. The DonaldFor years, general and politicians used sports metaphors(比喻) to explain some of the most consequential issues of our time. So just this once, let’s flip the script. Forget for a moment that the future of America—not to mention civilization—hangs in the balance. Instead, imagine the debate as a sporting event...Think Trump as Mike Tyson. Announcer: “He’s still looking to land that one big punch. And he’s had it with the fact—checking low blows and policy-question clinches. Hold on! Did he just bite the top of her ear off?”Or Clinton as the late, usually mid-mannered Arizona Cardinals coach Dennis Green, firing back at critics who suggested she didn’t take her opponent seriously enough: “He is who we thought he was! That’s why I took the stage! Now if you want to crown him, yo u can crown him! But he is who we thought he was!”It’s so tempting that a number of political analysts are using sports-writing tools to break down the matchup: offense vs. defense, strengths and weaknesses, how the coaching staffs and straining facilities compete with each other, even how the practice sessions have been going.No one has posted a point spread yet, so feel free to add your own, mercifully.The best comparisons to the Clinton-Trump battle from different sports:FOOTBALLClinton is serious and prepared. She uses her practice time purposefully. She has experience, a multitude of game plans and knows how to manage the clock. If she was a player, she’d be Peyton Manning. If she was a coach—let’s see: wonky, secretive, willing to bend rules, eve n got caught recently fudging an injury report—she’d be Bill Bellicose. Definitely Bellicose.As a plays and never saw a throw he wouldn’t make. If he was a coach, he’d be Chip Kelly in his Oregon days, breaking the mold with his house-on-fire offensive attack. Or maybe Steve Hurried at Florida, less worried about what his opponents might do than whether he could sneak out of a film session and squeeze in nine holes before sundown.59. The writer’s purpose of writing this passage is to ____________.A. educate the public on the wide use of sports metaphorsB. entertain the public by using sports metaphors in politicsC. compare the candidates and issue an appeal for votes for HillaryD. indicate the subtle relationship between sporting events and election campaign60. The underlined phrase “break down” in Paragraph 4 is closest in meaning to __________.A. stop working because of a faultB. change something into a different formC. change people’s fixed attitude towards somethingD. divide something into parts to understand it better61. According to the passage, which of the following statement is TRUE?A. Bill Belichick is Hillary’s favorite football coach.B. If Trump went in for boxing, he would bite his opponent’s ear.C. Hillary met with criticism for underestimating her opponent.D. Issues of great importance can be solved in an easy-going way.(C)Open data-sharers are still in the minority in many fields. Although many researchers broadly agree that public access to raw data would accelerate science—because other scientists might be able to make advances not foreseen by the data's producers—most are reluctant to post the results of their own labours online. When Wolkovich, for instance, went hunting for the data from the 50 studies in her meta-analysis, only 8 data sets were available online, and many of the researchers whom she e-mailed refused to share their work. Forced to extract data from tables or figures in publications, Wolkovich's team could conduct only limited analyses.Some communities have agreed to share online—geneticists, for example, post DNA sequences at the GenBank repository(库), and astronomers are accustomed to accessing images of galaxies and stars from, say, the Sloan Digital Sky Survey, a telescope that has observed some 500 million objects –but these remain the exception, not the rule. Historically, scientists have objected to sharing for many reasons: it is a lot of work; until recently, good databases did not exist; grant funders were not pushing for sharing; it has been difficult to agree on standards for formatting data and the contextual information called metadata; and there is no agreed way to assign credit for data.But the barriers are disappearing in part because journals and funding agencies worldwide are encouraging scientists to make their data public. Last year, the Royal Society in London said in its report that scientists need to ‘shift away from a research culture where data is viewed as private preserve’. Funding agencies note that data paid for with public money should be public information, and the scientific community is recognizing that data can now be shared digitally in ways that were not possible before. To match the growing demand, services are springing up to make it easier to publish research products online and enable other researchers to discover and cite them.Although calls to share data often concentrate on the moral advantages of sharing, the practice is not purely altruistic(利他的). Researchers who share get plenty of personal benefits, including more connections with colleagues, improved visibility and increased citations. The most successful sharers –those whose data are downloaded and cited the most often – get noticed, and their work gets used. For example, one of the most popular data sets on multidisciplinary repository Dryad is about wood density around the world; it has been downloaded 5,700 times. Co-author Amy Zanne, a biologist at George Washington University in Washington DC, thinks that users probably range from climate-change researchers wanting to estimate how much carbon is stored in biomass, to foresters looking for information on different grades of timber. ‘I would much prefer to have my data used bythe maximum number of people to ask their own questions,’ she says. ‘It's important to allow readers and reviewers to see exactly how you arrive at your results. Publishing data and code allows your science to be reproducible.Even people whose data are less popular can benefit. By making the effort to organize and label files so others can understand them, scientists become more organized and better disciplined themselves, thus avoiding confusion later on.62. What do many researchers generally accept?A. Open data sharing is conducive to scientific advancement.B. Open data sharing is most important to medical science.C. Repositories are essential to scientific research.D. It is imperative to protect scientists’ patents.63. What is the attitude of most researchers towards making their own data public?A. Ambiguous.B. Neutral.C. Liberal.D. Opposed.64. According to the passage, what might hinder open data sharing?A. The fear of massive copying.B. The lack of a research culture.C. The belief that research data is private intellectual property.D. The concern that certain agencies may make a profit out of it.65. What helps lift some of the barriers to open data sharing?A. The ever-growing demand for big data.B. The advancement of digital technology.C. The changing attitude of journals and funders.D. The trend of social and economic development.66. Dryad serves as an example to show how open data sharing ________.A. benefits sharers and users alikeB. saves both money and laborC. makes researchers successfulD. is becoming increasingly popularSection C ( 8%)Directions: 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 world of difference.Scientists at Columbia University and the University of Pittsburgh found that when making choices between a right and wrong answer, people’s accuracy in making the right decision increased dramatically when they gave themselves a small amount of time. __67___It’s all about giving our attention enough time to sort out the relevant information from the distracters, says one of the study's co-authors, Vincent Ferrera, an associate professor of neurology at Columbia University Medical Center. “The little extra time to sort out the irr elevant information makes the decision-making more efficient," he says.___68___. They were told which set was their target, and then asked to indicate which direction the target dots were moving when the distracting set was introduced ---the distracting set either moved in the same direction or the opposite direction of the target dots. T he participants performed this task under different conditions: in some experiments they aimed for accuracy, in others, for speed.Ferrera and his colleagues found that when the participants had a little more 1 as 50 milliseconds to make their decision, their accuracy improved. The reason? ___69___. "If you take attention offline, you go from almost perfect accuracy to almost chance accuracy," he says.That doesn’t m ean taking longer to decide necessarily leads to a better choice. With more time, there's more chance for distracting and even misleading information to filter into the process, confusing the situation. Ferrera's findings suggest that simply accumulating more and more information isn't always helpful. "What we're saying is that before you start gathering evidence, take a short moment to determine whether that evidence is really relevant to the decision you are making,” he says."___70___"第II 卷(共50分)I. Read the following passage and write a summary of less than 60 words. (10%)If you suffer from shyness, you are not alone, for shyness is a universal phenomenon. It is not surprising that social scientists are learning more about its causes.The first environmental cause of shyness many be a child’s home and family life. Today’s children are growing up in smaller and smaller families, with fewer and fewer relatives living nearby. Growing up in single-parent homes or in homes in which both parents work full time, children may not have the socializing experience of frequent visits by neighbours and friends. Because of their lack of social skills, they may begin to feel socially inhibited, or shy, when they start school.A second environmental cause of shyness in an individual may be one’s culture. In a study conducted in Japan, 57 percent of participants rated themselves as shy. Researchers Henderson and Zimbardo say, “One explanation is that in Japan, an individual’s performance success is credited externally to parents, grandparents, teachers, coaches, and others, while failure is entirely blamed on the person.” Therefore, Japanese learn not to take risks in public and rely instead on group-shared decisions.Technology may play a role as well. In the United States, the number of young people who report being shy has risen from 40 percent to 50 percent in recent years. Due to huge advances in technology, watching television, playing video games, and surfing the Web have displaced recreational activities that involve social interaction for many young people. Adults, too, are becoming more isolated as a result of technology. Face-to-face interactions with bank tellers, gas station attendants, and store clerks are no longer necessary because people can use machines to do their banking, fill their gas tanks, and order merchandise. In short, they become shy.It appears that most people have experienced shyness at some time in their lives. Therefore, if you are shy, you have lots of company.。
2018级高三10月月考
2018级高三10月月考本试卷分第Ⅰ卷和第Ⅱ卷两部分，共12页。
第Ⅰ卷为选择题, 共100分；第Ⅱ卷为非选择题，共50分。
全卷共150分，考试时间为120分钟。
第I卷第一部分听力（共两节，满分30分）第一节听下面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. What will the man probably do?A. Visit Bill.B. Going swimming.C. Play tennis.2. What time is Alex supposed to arrive?A. At 8:00.B. At 7:30.C. At 8:15.3. Where will the speakers probably spend their vacation?A. In .B. In .C. In .4. Which color MP4 player does the man want?A. Blue.B. Brown.C. Green.5. What’s the man doing?A. Making an invitation.B. Making an apology.C. Offering help.第二节（共15小题；每小题1.5分，满分22.5分）听下而5段对活或独白。
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。
听第6段材料，回答第6、7题。
6. What’s the probable relationship between the speakers?A. Mother and son.B. Husband and wife.C. Brother and sister.7. What will the woman most probably do?A. Buy some flowers.B. Clean the house.C. Make a card.听第7段材料，回答第8、9题。
2017-2018学年上海市徐汇区南模中学高三(上)10月月考数学试卷(解析版)
2017-2018学年上海市徐汇区南模中学高三（上）10月月考数学试卷一、填空题1．若α是第四象限角，则所在是象限是第象限．2．已知集合A={x|k，k∈Z}，B={x|4﹣x2≥0}，则A∩B=．3．函数y=的值域为．4．已知tanα=2，则sinαcos2α=．5．函数f （x）=的单调递增区间为．6．f（x）=sin2ωx+1（ω＞0）在区间[﹣，]上为增函数，则ω的最大值为．7．有一个解三角形的题因纸张破损有一个条件不清，具体如下：“在△ABC中，角A、B、C所对的边分别为a、b、c，已知a=，B=45°，，求角A：“经推断破损处的条件为三角形一边的长度，且答案提示A=60°，试将条件补充完整．8．函数f（x）=，f（x）max=M，f（x）min=m，则M+m=．9．已知函数f（x）=sin2x﹣cos2x+1，若f（x）≥log2t对x∈R恒成立，则t 的取值范围为．10．设a1、a2∈R，且，则|10π﹣a1﹣a2|的最小值等于．11．设a，b∈R，c∈[0，2π），若对于任意实数x都有2sin（3x﹣）=asin（bx+c），则满足条件的有序实数组（a，b，c）的组数为．12．关于x的方程=|sin|在[﹣2016，2016]上解的个数为．二、选择题13．在△ABC中，sin2A≤sin2B+sin2C﹣sinBsinC，则A的取值范围是（）A．（0，]B．[，π）C．（0，]D．[，π）14．把y=（cos2x﹣sin2x）的图象平移后得到y=sin2x的图象（）A．向右平移个单位 B．向左平移个单位C．向右平移个单位 D．向左平移个单位15．函数y=﹣的图象按向量=（1，0）平移之后得到的函数图象与函数y=2sinπx （﹣2≤x≤4）的图象所有交点的橫坐标之和等于（）A．2 B．4 C．6 D．816．存在函数f（x）满足，对任意x∈R都有（）A．f（sin2x）=sinx B．f（sin2x）=x2+x C．f（x2+1）=|x+1| D．f（x2+2x）=|x+1|三、解答题17．已知函数f（x）=cos2x﹣sin2x+，x∈（0，π）．（1）求f（x）的单调递增区间；（2）设△ABC为锐角三角形，角A所对边a=，角B所对边b=5，若f（A）=0，求△ABC的面积．18．设函数f（x）=cos（2x+）+sin2x．（1）求函数f（x）的最小周期；（2）设函数g（x）对任意x∈R，有g（x+）=g（x），且当x∈[0，]时，g（x）=﹣f（x）．求函数g（x）在[﹣π，0]上的解析式．19．如图所示：湖面上甲、乙、丙三艘船沿着同一条直线航行，某一时刻，甲船在最前面的A点处，乙船在中间B点处，丙船在最后面的C点处，且BC：AB=3：1．一架无人机在空中的P点处对它们进行数据测量，在同一时刻测得∠APB=30°，∠BPC=90°．（船只与无人机的大小及其它因素忽略不计）（1）求此时无人机到甲、丙两船的距离之比；（2）若此时甲、乙两船相距100米，求无人机到丙船的距离．（精确到1米）20．如图，某园林单位准备绿化一块直径为BC的半圆形空地，△ABC的外面种草，△ABC的内接正方形PQRS为一水池，其余的地方种花，若BC=a，∠ABC=θ，设△ABC的面积为S1，正方形的面积为S2．（1）用a，θ表示S1和S2；（2）当a固定，θ变化时，求取最小值时的角．21．已知函数y=f（x），x∈D，y∈A；g（x）=x2﹣（4）x+1，（1）当f（x）=sin（x+φ）为偶函数时，求φ的值．（2）当f（x）=sin（2x+）+sin（2x+）时，g（x）在A上是单调递增函数，求θ的取值范围．（3）当f（x）=a1sin（ωx+φ1）+a2sin（ωx+φ2）+…+a n sin（ωx+φn）时，（其中a i∈R，i=1，2，3…n，ω＞0），若f2（0）+f2（）≠0，且函数f（x）的图象关于点（，0）对称，在x=π处取得最小值，试探讨ω应该满足的条件．2017-2018学年上海市徐汇区南模中学高三（上）10月月考数学试卷参考答案与试题解析一、填空题1．若α是第四象限角，则所在是象限是第一或三象限．【分析】用不等式表示第四象限角α，再利用不等式的性质求出满足的不等式，从而确定角的终边在的象限．解：∵α是第四象限角，∴+2kπ＜α＜2π+2kπ，k∈Z，∴+kπ＜＜π+kπ，k∈Z，∴+kπ＜﹣＜+kπ，k∈Z，当k为偶数时，属于第一象限，当k为奇数时，属于第三象限，故答案为：一或三【点评】本题考查象限角的表示方法，不等式性质的应用，通过角满足的不等式，判断角的终边所在的象限．2．已知集合A={x|k，k∈Z}，B={x|4﹣x2≥0}，则A∩B={x|﹣2≤x＜﹣或≤x＜} ．【分析】化简集合B，根据交集的定义写出A∩B．解：集合A={x|k，k∈Z}，B={x|4﹣x2≥0}={x|﹣2≤x≤2}，则A∩B={x|﹣2≤x＜﹣或≤x＜}．故答案为：{x|﹣2≤x＜﹣或≤x＜}．【点评】本题考查了集合的化简与运算问题，是基础题．3．函数y=的值域为[，2] ．【分析】把已知函数解析式变形，分离常数，可得y=，由cosx的范围得答案．解：y==．∵﹣1≤cosx≤1，∴2≤cosx+3≤4，∴[]，则∈[，2]．故答案为：[，2]．【点评】本题考查函数值域的求法，训练了利用分离参数法求解函数值域，是中档题．4．已知tanα=2，则sinαcos2α=±．【分析】根据同角三角函数的基本关系、以及三角函数在各个象限中的符号，分类讨论求得要求式子的值．解：∵tanα==2，∴α为第一象限角或α为第三象限角，再根据sin2α+cos2α=1，若α为第一象限角，则sinα=，cosα=，则sinαcos2α=•=．若α为第三象限角，则sinα=﹣，cosα=﹣，则sinαcos2α=﹣•=﹣，故答案为：±．【点评】本题主要考查同角三角函数的基本关系、以及三角函数在各个象限中的符号，属于基础题．5．函数 f （x）=的单调递增区间为，k∈Z．【分析】利用复合函数的单调性的规律：同增异减将原函数的单调性转化为t的单调性，利用三角函数的单调性的处理方法：整体数学求出单调区间．解：∵y=log0.5t为减函数，所以函数f （x）=的单调递增区间为即为单调减区间且令解得故答案为（k∈Z）【点评】本题考查复合函数的单调性的规律、三角函数的单调区间的求法．6．f（x）=sin2ωx+1（ω＞0）在区间[﹣，]上为增函数，则ω的最大值为．【分析】由题意可得可得﹣•2ω≥2kπ﹣，且•2ω≤2kπ+，k∈z，求得ω的最大值．解：∵f（x）=sin2ωx+1（ω＞0）在区间[﹣，]上为增函数，可得﹣•2ω≥2kπ﹣，且•2ω≤2kπ+，k∈z，求得ω≤，故ω的最大值为，故答案为：．【点评】本题主要考查求正弦函数的单调性，属于基础题．7．有一个解三角形的题因纸张破损有一个条件不清，具体如下：“在△ABC中，角A、B、C所对的边分别为a、b、c，已知a=，B=45°，c=，求角A：“经推断破损处的条件为三角形一边的长度，且答案提示A=60°，试将条件补充完整．【分析】先把A=60°当做已知条件根据正弦定理计算出b，c，然后把b，c当做已知条件利用正弦定理解出A进行验证．解：∵A=60°，B=45°，∴C=75°．由正弦定理得，即，解得b=，c=．若条件为b=，则由正弦定理得，解得sinA=，∴A=60°或A=120°，答案不唯一，不符合题意．故答案为：c=．【点评】本题考查了正弦定理，解三角形，属于中档题．8．函数f（x）=，f（x）max=M，f（x）min=m，则M+m=2．【分析】利用分离法化简f（x），构造函数g（x），讨论g（x）的奇偶性，求解最值，即可求解M+m的值．解：函数f（x）==1+，令函数g（x）=的最大值为N，最小值为n，那么f（x）max=1+N=M，f（x）min=1+n=m，∵g（﹣x）=﹣=﹣g（x），则函数g（x）是奇函数，∴n+N=0．则M+m=2．故答案为：2．【点评】本题主要考查函数最值的求解，根据分离后构造新函数，求解最值是解决本题的关键．9．已知函数f（x）=sin2x﹣cos2x+1，若f（x）≥log2t对x∈R恒成立，则t 的取值范围为（0，1] ．【分析】先化简函数解析式，f（x）≥log2t恒成立，只需求出f（x）的最小值大于log2t，求出t的范围即可．解：f（x）=sin2x﹣cos2x+1=sin（2x﹣）+1，函数f（x）=sin（2x﹣）+1的最小值为：0，若f（x）≥log2t恒成立，只需0≥log2t恒成立，所以t∈（0，1]．所以t的取值范围：（0，1]．故答案为：（0，1]．【点评】本题是基础题，考查三角函数的基本性质，三角函数的化简，恒成立问题的应用，考查计算能力，逻辑推理能力，属于常考题型、基本知识的考查．10．设a1、a2∈R，且，则|10π﹣a1﹣a2|的最小值等于．【分析】由题意，要使+=2，可得sinα1=﹣1，sin2α2=﹣1．求出α1和α2，即可求出|10π﹣α1﹣α2|的最小值解：根据三角函数的性质，可知sinα1，sin2α2的范围在[﹣1，1]，要使+=2，∴sinα1=﹣1，sin2α2=﹣1．则：，k1∈Z．，即，k2∈Z．那么：α1+α2=（2k1+k2）π，k1、k2∈Z．∴|10π﹣α1﹣α2|=|10π﹣（2k1+k2）π|的最小值为．故答案为：．【点评】本题主要考察三角函数性质，有界限的范围的灵活应用，属于基本知识的考查．11．设a，b∈R，c∈[0，2π），若对于任意实数x都有2sin（3x﹣）=asin（bx+c），则满足条件的有序实数组（a，b，c）的组数为4．【分析】根据三角函数恒成立，则对应的图象完全相同．解：∵对于任意实数x都有2sin（3x﹣）=asin（bx+c），∴必有|a|=2，若a=2，则方程等价为sin（3x﹣）=sin（bx+c），则函数的周期相同，若b=3，此时C=，若b=﹣3，则C=，若a=﹣2，则方程等价为sin（3x﹣）=﹣sin（bx+c）=sin（﹣bx﹣c），若b=﹣3，则C=，若b=3，则C=，综上满足条件的有序实数组（a，b，c）为（2，3，），（2，﹣3，），（﹣2，﹣3，），（﹣2，3，），共有4组，故答案为：4．【点评】本题主要考查三角函数的图象和性质，结合三角函数恒成立，利用三角函数的性质，结合三角函数的诱导公式进行转化是解决本题的关键．12．关于x的方程=|sin|在[﹣2016，2016]上解的个数为4031．【分析】根据函数与方程的关系转化为两个函数的交点个数问题，作出两个函数的图象，利用数形结合进行求解即可得到结论．解：y==，作函数y=与y=|sinπx|在[﹣2016，2016]上的图象如下，由图象知函数y=|sin|的周期是2，两个函数都关于x=1对称，当x≤0时，两个函数在每个周期内都有两个交点，此时在[﹣2016，0]内有1008×2=2016个交点，在[0，2]内两个函数只有一个交点，当x≥2时，两个函数在每个周期内都有两个交点，此时在[2，2016]内有1007×2=2014个交点，则在[﹣2016，2016]上解的个数为2016+1+2014=4031，故答案为：4031【点评】本题主要考查方程根式的个数的求解，利用函数与方程的关系转化为两个函数的交点个数，利用数形结合进行求解是解决本题的关键．综合性较强，有一定的难度．二、选择题13．在△ABC中，sin2A≤sin2B+sin2C﹣sinBsinC，则A的取值范围是（）A．（0，]B．[，π）C．（0，]D．[，π）【分析】先利用正弦定理把不等式中正弦的值转化成边，进而代入到余弦定理公式中求得cosA的范围，进而求得A的范围．解：由正弦定理可知a=2RsinA，b=2RsinB，c=2RsinC，∵sin2A≤sin2B+sin2C﹣sinBsinC，∴a2≤b2+c2﹣bc，∴bc≤b2+c2﹣a2∴cosA=≥∴A≤∵A＞0∴A的取值范围是（0，]故选：C．【点评】本题主要考查了正弦定理和余弦定理的应用．作为解三角形中常用的两个定理，考生应能熟练记忆．14．把y=（cos2x﹣sin2x）的图象平移后得到y=sin2x的图象（）A．向右平移个单位 B．向左平移个单位C．向右平移个单位 D．向左平移个单位【分析】化函数y为正弦型函数，再根据函数图象平移法则得出答案．解：函数y=（cos2x﹣sin2x）=cos2x﹣sin2x=cos（2x+）=sin（2x++）=sin（2x+）=sin2（x+），把函数y=（cos2x﹣sin2x）的图象向右平移个单位后得到y=sin2x的图象．故选：A．【点评】本题考查了三角函数的图象平移问题，也考查了三角恒等变换问题，是中档题．15．函数y=﹣的图象按向量=（1，0）平移之后得到的函数图象与函数y=2sinπx （﹣2≤x≤4）的图象所有交点的橫坐标之和等于（）A．2 B．4 C．6 D．8【分析】y1=的图象由奇函数y=﹣的图象向右平移1个单位而得，所以它的图象关于点（1，0）中心对称，再由正弦函数的对称中心公式，可得函数y2=2sinπx 的图象的一个对称中心也是点（1，0），故交点个数为偶数，且每一对对称点的横坐标之和为2．由此不难得到正确答案．解：函数y=﹣的图象按向量=（1，0）平移之后得到函数y1=，y2=2sinπx 的图象有公共的对称中心（1，0），作出两个函数的图象如图：当1＜x≤4时，y1＜0，而函数y2在（1，4）上出现1.5个周期的图象，在（1，）和（，）上是减函数；在（，）和（，4）上是增函数．∴函数y1在（1，4）上函数值为负数，且与y2的图象有四个交点E、F、G、H，相应地，y1在（﹣2，1）上函数值为正数，且与y2的图象有四个交点A、B、C、D，且：x A+x H=x B+x G═x C+x F=x D+x E=2，故所求的横坐标之和为8，故选：D．【点评】发现两个图象公共的对称中心是解决本题的入口，讨论函数y2=2sinπx 的单调性找出区间（1，4）上的交点个数是本题的难点所在．16．存在函数f（x）满足，对任意x∈R都有（）A．f（sin2x）=sinx B．f（sin2x）=x2+x C．f（x2+1）=|x+1| D．f（x2+2x）=|x+1|【分析】利用x取特殊值，通过函数的定义判断正误即可．解：A．取x=0，则sin2x=0，∴f（0）=0；取x=，则sin2x=0，∴f（0）=1；∴f（0）=0，和1，不符合函数的定义；∴不存在函数f（x），对任意x∈R都有f（sin2x）=sinx；B．取x=0，则f（0）=0；取x=π，则f（0）=π2+π；∴f（0）有两个值，不符合函数的定义；∴该选项错误；C．取x=1，则f（2）=2，取x=﹣1，则f（2）=0；这样f（2）有两个值，不符合函数的定义；∴该选项错误；D．令x+1=t，则f（x2+2x）=|x+1|，化为f（t2﹣1）=|t|；令t2﹣1=x，则t=±；∴；即存在函数f（x）=，对任意x∈R，都有f（x2+2x）=|x+1|；∴该选项正确．故选：D．【点评】本题考查函数的定义的应用，基本知识的考查，但是思考问题解决问题的方法比较难．三、解答题17．已知函数f（x）=cos2x﹣sin2x+，x∈（0，π）．（1）求f（x）的单调递增区间；（2）设△ABC为锐角三角形，角A所对边a=，角B所对边b=5，若f（A）=0，求△ABC的面积．【分析】（1）由二倍角的余弦公式和余弦函数的递增区间，解不等式可得所求增区间；（2）由f（A）=0，解得A，再由余弦定理解方程可得c，再由三角形的面积公式，计算即可得到所求值．解：（1）函数f（x）=cos2x﹣sin2x+=cos2x+，x∈（0，π），由2kπ﹣π≤2x≤2kπ，解得kπ﹣π≤x≤kπ，k∈Z，k=1时，π≤x≤π，可得f（x）的增区间为[，π）；（2）设△ABC为锐角三角形，角A所对边a=，角B所对边b=5，若f（A）=0，即有cos2A+=0，解得2A=π，即A=π，由余弦定理可得a2=b2+c2﹣2bccosA，化为c2﹣5c+6=0，解得c=2或3，若c=2，则cosB=＜0，即有B为钝角，c=2不成立，则c=3，△ABC的面积为S=bcsinA=×5×3×=．【点评】本题考查二倍角公式和余弦函数的图象和性质，考查解三角形的余弦定理和面积公式的运用，考查运算能力，属于中档题．18．设函数f（x）=cos（2x+）+sin2x．（1）求函数f（x）的最小周期；（2）设函数g（x）对任意x∈R，有g（x+）=g（x），且当x∈[0，]时，g（x）=﹣f（x）．求函数g（x）在[﹣π，0]上的解析式．【分析】（1）利用二倍角和两角和与差以及辅助角公式基本公式将函数化为y=Asin（ωx+φ）的形式，再利用周期公式求函数的最小正周期．（2）当x∈[0，]时，g（x）=﹣f（x）．在讨论x∈[，0]时，g（x）的解析式，在函数g（x）在[﹣π，0]上的解析式．解：（1）函数f（x）=cos（2x+）+sin2x化简可得：f（x）=（cos2xcos﹣sin2xsin+sin2x）=cos2x﹣sin2x cos2x=﹣sin2x∴函数f（x）的最小周期T=；（2）当x∈[0，]时，g（x）=﹣f（x）．即g（x）=（sin2x）=sin2x．当x∈[﹣，0]时，由于g（x+）=g（x），则（x+）∈[0，]那么：g（x）=sin2（x+）=sin2x．当x∈[﹣π，﹣]时，则（x+π）∈[0，]可得：g（x）=sin2（x+π）=sin2x．∴函数g（x）在[﹣π，0]上的解析式为f（x）=【点评】本题主要考查对三角函数的化简能力和三角函数的图象和性质的运用，以及分段函数的解析式的求法．利用三角函数公式将函数进行化简是解决本题的关键．属于中档题．19．如图所示：湖面上甲、乙、丙三艘船沿着同一条直线航行，某一时刻，甲船在最前面的A点处，乙船在中间B点处，丙船在最后面的C点处，且BC：AB=3：1．一架无人机在空中的P点处对它们进行数据测量，在同一时刻测得∠APB=30°，∠BPC=90°．（船只与无人机的大小及其它因素忽略不计）（1）求此时无人机到甲、丙两船的距离之比；（2）若此时甲、乙两船相距100米，求无人机到丙船的距离．（精确到1米）【分析】（1）利用正弦定理，即可求此时无人机到甲、丙两船的距离之比；（2）若此时甲、乙两船相距100米，由余弦定理求无人机到丙船的距离．解：（1）在△APB中，由正弦定理，得，，在△BPC中，由正弦定理，得，又，sin∠ABP=sin∠CBP，故．即无人机到甲、丙两船的距离之比为．（2）由BC：AB=3：1得AC=400，且∠APC=120°，由（1），可设AP=2x，则CP=3x，在△APC中，由余弦定理，得160000=（2x）2+（3x）2﹣2（2x）（3x）cos120°，解得，即无人机到丙船的距离为≈275米．【点评】本题考查利用数学知识解决实际问题，考查正弦定理、余弦定理的运用，属于中档题．20．如图，某园林单位准备绿化一块直径为BC的半圆形空地，△ABC的外面种草，△ABC的内接正方形PQRS为一水池，其余的地方种花，若BC=a，∠ABC=θ，设△ABC的面积为S1，正方形的面积为S2．（1）用a，θ表示S1和S2；（2）当a固定，θ变化时，求取最小值时的角．【分析】（1）据题知三角形ABC为直角三角形，根据三角函数分别求出AC和AB，求出三角形ABC的面积S1；设正方形PQRS的边长为x，利用三角函数分别表示出BQ和RC，利用BQ+QR+RC=a列出方程求出x，算出S2；（2）由比值称为“规划合理度”，可设t=sin2θ来化简求出S1与S2的比值，利用三角函数的增减性求出比值的最小值即可求出此时的θ．解：（1）在Rt△ABC中，AB=acosθ，AC=asinθ，设正方形的边长为x则，由BP+AP=AB，得，故所以（2），令t=sin2θ，因为，所以0＜2θ＜π，则t=sin2θ∈（0，1]所以，，所以函数g（t）在（0，1]上递减，因此当t=1时g（t）有最小值，此时所以当时，“规划合理度”最小，最小值为．【点评】考查学生会根据实际问题选择合适的函数关系的能力，以及在实际问题中建立三角函数模型的能力．21．已知函数y=f（x），x∈D，y∈A；g（x）=x2﹣（4）x+1，（1）当f（x）=sin（x+φ）为偶函数时，求φ的值．（2）当f（x）=sin（2x+）+sin（2x+）时，g（x）在A上是单调递增函数，求θ的取值范围．（3）当f（x）=a1sin（ωx+φ1）+a2sin（ωx+φ2）+…+a n sin（ωx+φn）时，（其中a i∈R，i=1，2，3…n，ω＞0），若f2（0）+f2（）≠0，且函数f（x）的图象关于点（，0）对称，在x=π处取得最小值，试探讨ω应该满足的条件．【分析】（1）根据函数f（x）=sin（x+φ）为偶函数，可得sin（x+φ）=sin（﹣x+φ），化简为cosφ=0，可得φ的值．（2）利用三角恒等变换化简函数f（x）的解析式为sin（2x+α）∈[﹣，]，可得A，再根据g（x）的解析式结合题意可得tanθ≤﹣，由此可得θ的取值范围．（3）由于f（x）的解析式以及f2（0）+f2（）≠0，可得f（x）=msinωx+ncosωx=sin（ωx+φ），且m2+n2≠0．由函数f（x）的图象关于点（，0）对称，可得ω+φ=kπ，k∈z ①．又函数f（x）在x=π处取得最小值，可得ωπ+φ=2k′π+，k′∈z ②．由①②可得ω 满足的条件．解：（1）因为函数f（x）=sin（x+φ）为偶函数，所以，sin（x+φ）=sin（﹣x+φ），化简为2sinxcosφ=0，∴cosφ=0，所以φ=kπ+，k∈z．（2）∵函数f（x）=sin（2x+）+sin（2x+）=sin2x+2cos2x=sin（2x+α）∈[﹣，]，其中，sinα=，cosα=，所以A=[﹣，]．g（x）=x2﹣（4tanθ）x+1=+1﹣28tan2θ，由题意可知：2tanθ≤﹣，tanθ≤﹣，∴kπ﹣≤θ≤kπ﹣arctan，k∈z，即θ的取值范围是[kπ﹣，kπ﹣arctan]，k∈z．（3）由于f（x）=a1sin（ωx+φ1）+a2sin（ωx+φ2）+…+a n sin（ωx+φn）=a1（sinωxcosφ1 +cosωxsinφ1）+a2（sinωxcosφ2 +cosωxsinφ2）+…+a n（sinωxcosφn+cosωxsinφn）=sinωx （a1•cosφ1+a2•cosφ2+…+a n•cosφn）+cosωx（a1•sinφ1+a2•sinφ2+…+a n•sinφn）．∵f2（0）+f2（）≠0，∴a1•cosφ1+a2•cosφ2+…+a n•cosφn =0与a1•sinφ1+a2•sinφ2+…+a n•sinφn =0 不能同时成立．不妨设a1•cosφ1+a2•cosφ2+…+a n•cosφn =m，a1•sinφ1+a2•sinφ2+…+a n•sinφn =n，则f（x）=msinωx+ncosωx==sin（ωx+φ），且m2+n2≠0．由函数f（x）的图象关于点（，0）对称，可得sin（ω+φ）=0，故ω+φ=kπ，k∈z ①．又函数f（x）在x=π处取得最小值，∴sin（ωπ+φ）=﹣1，∴ωπ+φ=2k′π+，k′∈z ②．由①②可得，ω=（4k′﹣2k）+3．由于k和k′都是整数，故4k′﹣2k为偶数，∴ω=2h+1，h∈N．【点评】本题主要考查三角函数的恒等变换和化简求值，复合三角函数的单调性和对称性，属于中档题．。
2018届高三10月月考英语试题 含答案
七宝中学高三月考（十月）英语试卷第I卷（共95分）I. Listening Comprehension (30)Section 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. 100 dollars. B. 200 dollars. C. 300 dollars. D. 600 dollars.2. A. In the hospital. B. At a nursery. C. At a police station. D. In a library.3. A. April. B. May. C. June. D. July.4. A. Go to visit the writer. B. Buy her new book.C. Ignore her new book.D. Write a book review.5. A. Jogging. B. Basketball. C. Swimming. D. Throwing.6. A. The lady’s room is far.B. She has to sign up for using the lady’s room.C. She is not able to use the lady’s room right now.D. He will lead her to the lady’s room.7. A. He will read it on Saturday. B. He will lend it to Sandy first.C. He will lend it to Jane first.D. He will keep the paper until Sunday.8. A. He probably just has got a headache.B. There’s no air-conditioner in the room.C. She thinks he is not seriously sick.D. She thinks he should go to see the doctor.9. A. They couldn’t change the plan.B. They should change their plan.C. She doesn’t believe the weather forecast.D. She thinks the basketball game won’t last.10. A. He can come for next party. B. He can bring his wife along to the party.C. He can stay at home for his wife.D. She will change the time of the party.Section BDirections:In Section B, you will hear two short passages, and you will be asked three questions on each of the passages. The passages will be read 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 would be the best answer to the question you have heard.Questions 11through 13 are based on the following passage.11. A. They believed that he would realize his dream.B. They offered him help to achieve his dream.C. They didn’t believe that his dream would come true.D. They made no response to his announcement.12. A. That he was encouraged by the teacher.B. That he was encouraged by the local paper editor.C. That his first poem was published.D. That he was encouraged by a professional writer.13. A. It reminds him of his school life.B. It reminds him that dreams will come true through efforts.C. It reminds him how poor he used to be.D. It reminds him of the local paper.Questions 14 through 16 are based on the following passage.14. A. Gardening. B. How to take care of birds.C. How to feed birds.D. How to get birds to the garden.15. A. Fruit, water and insects. B. Plants, sleeping place and food.C. Variety of plants, food and water.D. Different types of plants, seed and insects.16. A. Change the water for them. B. Play with them.C. Not to approach them.D. Play the music for them.Section CListen to a conversation between a student and a clerk in the bookstore.17. Why can the man NOT find the book he needs?A. The bookstore is sold out of the book.B. The bookstore he is in does not carry the book.C. His professor did not order enough copies of the book.D. The book is not being used for any course offered at the university.18. What does the woman offer to do for the student?A. Save a copy of the book for him as soon as it comes inB. Order more copies of the bookC. Call the computer store across the streetD. Find a store that sells the book if he cannot find it19. How does the man react to the information the woman gives him about where computerscience books are sold?A. He is embarrassed that he did not think of it himself.B. He suggests that the information be posted in the store.C. He apologizes for bothering the woman.D. He is annoyed that the woman did not tell him sooner.20. Why does the woman say this [ You are not buying it in advance for next year oranything.]A. To determine how urgent the student’s need is.B. To figure out why the book is not listed on the computer.C. To find out what level computer science course the man is taking.D. To explain why the book might be sold out.II. Grammar and VocabularySection ADirections: After reading the passages below, fill in the blanks to make the passages 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.Mystery of the White Gardenia（Every year on my birthday, from the time I turned 12, a white gardenia栀子花was delivered to my house. No card or note came with it. ）…I don’t remember ever slamming my door ( 21 )________ anger at her and shouting,” you just don’t understand!”, ( 22 )____________ she did understand.One month before my high-school graduation, my father died of a heart attack. My feelings ranged from grief to abandonment, fear and ( 23 )_________( overwhelm ) anger that my dad was missing some of the most important events in my life. I became completely uninterested in my upcoming graduation, the senior-class play and the prom. But my mother, in the midst of her own grief, ( 24 )_______not hear of my skipping any of those things.The day before my father died, my mother and I ( 25 ) ____________ ( go ) shopping for a prom dress. We’d found a spectacular one, with yards and yards of dotted Swi ss in red, white and blue, ( 26 ) _______ made me feel like Scarlett O’ Hara, ( 27 )________it was the wrong size. When my father died, I forgot about the dress.My mother didn’t. The day before the prom, I found that dress---in the right size---draped （织物）majestically over the living-room sofa. It wasn’t just delivered, still in the box. It was presented to me---beautifully, artistically, lovely. I didn’t care if I had a new dress or not. But my mother did.She wanted her children to feel （28 ）_______ ( love ) and lovable, creative and imaginative, imbued with a sense （29 ）_______ there was magic in the world and beauty even in the face of adversity. In truth, my mother wanted her children to see （30 ）________ much like the gardenia---lovely, strong and perfect---with an aura(气氛，氛围)of magic and perhaps a bit of mystery.My mother died ten days after I married. I was 22 years old. That was the year the gardenias stopped coming.Section BDirections: Complete the following passage by using the words in the box. Each word can only be used once. Note that there is one word more than you need.Poetry is a kind of writing in which the sound and meaning of groups of words express ideas or emotion in addition to the experiences or strong feelings the writer ____31____. Unlike most other forms of writing, poetry is often written in lines, rather than paragraphs. Poetry also sounds different from other forms of writing, often using rhythm and rhyme to create an interesting sound when read aloud. Poetry catches the attention of a reader because it ____32____to both emotions and senses.Sound is ____33____ the single most important aspect of any poem. The sound that any given word makes, or the sounds that come from specific groups of words used together, are what make poetry so unique as a form of writing. A typical story or report does not focus on the sounds that each _____34____ word makes when read. But poems generally contain few words, so it is important that each word plays a role in making an impact on the reader. Rhythm is the flow of sounds created by successive words in a poem. When you read a poem you can often hear this ____35____ pattern, or “beat,” in the sounds. This is called meter.Some of the oldest and best-known poetry in the world came from Ancient Greece. As far back as 700 BCE, poets there recited their work at public _____36____ and religious ceremonies. The great epic poems The Iliad and The Odyssey by Homer came from Greece. The Greeks eventually ____37____ Roman poets, such as Virgil, who wrote the Aeneid around 200r 30 BCE. In medieval times, poems such as Beowulf, The Divine Comedy by Dante, and The Canterbury Tales by Chaucer were written. Religion and romance became the ____38___ of choice for many poets at that time.Poetry _____39____ even more during the Renaissance period of history, an era of many great cultural achievements. This was the period during which Shakespeare, the most well-known poet, was making his mark! Needless to say, a trend had started. Poetry has continued to grow and change as a form of ____40____ expression in modern times.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.When a human infant is born into any other community in any part of the world it has two things in common with any infant, ____41____neither of them has been ____42____in any way either before or during birth.Firstly, and most obviously, new born children are completely ____43____. Apart from a powerful capacity to draw attention on their helplessness by using sound, there is nothing the new born child can do to ____44____ his own survival. Without ____45____ from some other human being or beings, be it mother, grandmother, or human group, a child is very unlikely to survive. This helplessness of human infants is in marked contrast with the ____46____ of many new born animals to get to their ____47____ within minutes of birth and run with the herd within a few hours. Although young animals are certainly in danger, sometimes for weeks or even months after birth, compared with the human infant, they very quickly develop the capacity to ____48____ for themselves. It is during this very long period in which the human infant is totally ____49____ on the others that it reveals second ____50____ which it shares with all other undamaged human infants, a capacity to learn ____51____.For this reason, biologists now suggest that language is “species specific特有的；特定的” to the human race, that is to say, they consider the hum an infant to be genetically ____52____ in such a way that it can ____53____ language. This suggestion implies that just as human beings are designed to see three-dimensionally and in color and just as they are designed to ____54____ upright rather than to move on all fours, so they are designed to learn and use language as part of their normal ____55____ as well-formed human beings.41. A. provided B. assume C. promised D. predicted42. A. unprotected B. hurt C. damaged D. unhealthy43. A. ignorant B. unknown C. inexperienced D. helpless44. A. ensure B. assure C. emphasize D. solidify45. A. love B. affection C. care D. attention46. A. possibility B. capacity C. try D. attempt47. A. arms B. body C. feet D. limbs48. A. feed B. defend C. protect D. prevent49. A. dependent B. based C. focused D. operated50. A. ability B. feature C. aspect D. specialty51. A. walking B. feeding C. language D. racing52. A. programmed B. set C. arranged D. born53. A. get B. learn C. speak D. acquire54. A. sit B. walk C. stand D. move55. A. abilities B. development C. performance D. behaviorSection 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.AMen of HonorA knight was a mounted warrior of medieval Europe who served a king or other feudal superior, usually in return for land. Knighthood was taken quite seriously and had to be earned.At about the age of eight, a boy would begin training in preparation for knighthood. This young trainee, known as a page(男侍者), would train with mentors to learn about horses, armor, and weapons. Pages practiced fighting with a sword against a wooden stake and learned to skillfully use a bow and arrow. The lady of the castle taught a young page about manners and social graces, as well as how to sing, play instruments, and dance. A priest might give a page religious training and teach him to read and write.By the age of fourteen, the page would become a squire（护卫）. A squire was responsible for dressing a knight for battles and tournaments and taking care of the knight's armor and weapons. He would even follow his master on the battlefield to protect him if the knight fell.A squire had to gain skill in using a lance, spear, or sword, so he would practice against a wooden dummy called a quintain（枪靶）. The quintain and a shield were hung on a wooden pole, and when hit, the whole structure would spin. The squire would learn to ride up and hit the shield's center, but then quickly move out of the way without getting hit and knocked off his horse by the quintain.At about age twenty, a squire was finally prepared to be called a knight, which involved an extended ceremony. On the evening before becoming a knight, the squire confessed his sins to a priest, was given a symbolic bath, and then fasted in order to cleanse his soul. The squire would dress all in white and stay in a chapel all night praying and watching over his weapons and armor.In the morning, the squire would dress in symbolically-colored clothing: red for his blood, white for purity, and brown for his return to the earth after death. At his induction ([in'dʌkʃən] 入会仪式), the knight swore a code of chivalry, which required him always to be brave, loyal, courteous, and to protect the defenseless. Knighthood was granted by the overlord with an accolade, during which the new knight was tapped on the shoulders or neck with the flat side of the sword.If this new knight ever broke his vows or acted dishonorably, he would be stripped of his knighthood in another ceremony, in which he was "buried." In the Middle Ages, a knight without honor was considered as good as dead.56.What were the responsibilities of a squire?A.Practicing fighting with a sword and using a bow and arrow skillfully.B.Looking after his own weapons and learning manners from the Lady of the castle.C.Confessing his sin and praying for his mentors.D.Dressing a knight for battles or competitions and protecting him.57.What does the underlined word “ chivalry” mean ?A.the noble spirits a knight possessesB.the tough task a knight has to finishC.the high goal a knight must achieveD.the military discipline a knight should obey58.If a knight were to betray the king, what do you think might happen?A. He would be sentenced to death.B. He would be robbed of his title.C. He would be forced to leave Court.D. His land would be returned to the King.59.Which of the following statements is right according to the passage?A. A knight had to be highly born.B. A knight had gone through different stages of training to become a KNIGHT.C. A knight was militarily skillful but not necessarily literate.D. Knighthood started in the Medieval Ages and existed only in England.BReading Your MindModern technology allows scientists to look inside a living human brain to see what is happening. These procedures are safe and painless. By understanding the normal brain activity, doctors and scientists are better able to assess the brain’s behavior during times of injury, disease, and mental illness.CT or CAT scans: Computed tomography (CT) or computerized axial tomography (CAT) shows images of the brain by passing multiple X-ray beams through the brain tissue. CT or CAT scans show a cross-section of the brain. These scans can be used to find brain tumors.MRI scans: Magnetic resonance imaging (MRI) uses powerful magnet to cause the atoms of the brain to shake. MRI sensors pick up the signals emitted ([i'mit] 发出) from the brain’s atoms and a computer interprets them as a picture. MRIs show more detail than CT or CAT scans can. They are especially useful in finding brain tumors that grow on the back of the brain, between the ears.PET scans:Positron emission tomography (PET) is different from other scans because it shows how the brain functions. After a person’s bloodstream is injected with a small dose of glucose (['glu:kəus] 【生化】葡萄糖), which is what gives the brain energy, scanners around the head detect where the glucose moves. The PET scan shows whichpart of the brain use a lot of glucose, which are the more active parts. PET scans are helpful for diagnosing strokes, studying mental illness, and learning how the brain process language.60.How can doctors and scientists understand the brain’s reaction to injury, disease, ormental illness?A.By contrasting/comparing it with normal brain activities.B.By passing the X-Ray beams through the brain tissue.C.By picking up the signals emitted from the brain’s atoms.D.By being injected with glucose and detecting where it moves.61.By Which two scans mentioned are helpful for diagnosing brain tumors?A. CT / CA T and PETB. MRI and PETC. CT/CA T and MRID. CT and CA T62. If a person suffers from defect in speaking, which scan will a doctor be mostlikely to suggest?A. CT / CATB. MRIC. PETD. Any one of themCThe psychology of innovationWhy are so few companies truly innovative?Innovation is key to business survival, and companies put substantial resources into inspiring employees to develop new ideas. There are, nevertheless, people working in luxurious, state-of-the-art centers designed to stimulate innovation who find that their environment doesn’t make them feel at all creative. And there are those who don’t have a budget, or much space, but who innovate successfully.For Robert B. Cialdini, Professor of Psychology at Arizona State University, one reason that companies don’t succeed as often as they should is that innovation starts with recruitment. Research shows that the fit between an employee’s values and a company’s values makes a difference to what contribution they make and whether, two years after they join, they’re still at the company.One of the most famous photographs in the story of rock’n’roll emphasizes Cialdini’s views. The 1956 picture of singers Elvis Presley, Carl Perkins, Johnny Cash and Jerry Lee Lewis jamming at a piano in Sun Studios in Memphis tells a hidden story. Sun’s ‘million-dollar quartet’ could have been a quintet. Missing from the picture is Roy Orbison, a greater natural singer than Lewis, Perkins or Cash. Sam Phillips, who owned Sun, wanted to revolutionize popular music with songs that fused black and white music, and country and blues. Presley, Cash, Perkins and Lewis instinctively understood Phillips’s ambition and believed in it. Orbison wasn’t inspired by the goal, and only ever achieved one hit with the Sun label.Managing innovation is a delicate art. It’s easy for a company to be pulled in conflictingdirections as the marketing, product development, and finance departments each get different feedback from different sets of people. And without a system which ensures collaborative exchanges within the company, it’s also easy for small ‘pockets of innovation’ to disappear. Innovation is a contact sport. You can’t brief people just by saying, ‘We’re going in this direction and I’m going to take you with me.’Cialdini believes that this ‘follow-the-leader syndrome’ is dangerous, not least because it encourages bosses to go it alone. ‘It’s been scientifically proven that three people will be better than one at solving problems, even if that one person is the smartest person in the field.’To prove his point, Cialdini cites an interview with molecular biologist James Watson. Watson, together with Francis Crick, discovered the structure of DNA, the genetic information carrier of all living organisms. ‘When asked how they had cracked the code ahead of an array of highly accomplished rival investigators, he said something that stunned me. He said he and Crick had succeeded because they were aware that they weren’t the most intelligent of the scientists pursuing the answer. The smartest scientist was called Rosalind Franklin who, Watson said, “was so intelligent she rarely sought advice”.’Writing, visualizing and prototyping can stimulate the flow of new ideas. Cialdini cites scores of research papers and historical events that prove that even something as simple as writing deepens every individual’s engagement in the project. It is, he says, the reason why all those competitions on breakfast cereal packets encouraged us to write in saying, in no more than 10 words: ‘I like Kellogg’s Corn Flakes because… .’ The very act of writing makes us more likely to believe it.Authority doesn’t have to inhibit innovation but it often does. Many theorist believe the ideal boss should lead from behind, taking pride in collective accomplishment and giving credit where it is due. Cialdini says: ‘Leaders should encourage everyone to contribute and simultaneously assure all concerned that every recommendation is important to making the right decision and will be given full attention.’ The frustrating thing about innovation is that there are many approaches, but no magic formula. However, a manager who wants to create a truly innovative culture can make their job a lot easier by recognizing these psychological realities.63. The example of the ‘million-dollar quartet’ underlines the writer’s point aboutA. recognizing talent.B. working as a team.C. having a shared objective.D. being an effective leader.64. James Watson suggests that he and Francis Crick won the race to discover theDNA code because theyA. were conscious of their own limitations.B. brought complementary skills to their partnership.C. were determined to outperform their brighter rivals.D. encouraged each other to realize their joint ambition.65. The writer mentions competitions on breakfast cereal packets as an example ofhow to______________________________________________________.A. inspire creative thinking.B. generate concise writing.C. promote loyalty to a group.D. strengthen commitment to an idea.66. In the last paragraph, the writer suggests that it is important for employees toA. be aware of their company’s goals.B. feel that their contributions are valued.C. have respect for their co-workers’ achievements.D. understand why certain management decisions are made.Section CDirections: Complete the following passage by using the sentences given below. Each sentence can be used only once. Note that there are two more sentences than you need.Many people go through life's usual phases of ups and downs, which are common and normal stages in life, but when your self-esteem is low, you may need to know how to boost your self-esteem, since it can lead to problems you may not have thought of.Self-esteem is the measurement or the value of how a person knows his worth and its effects are life-changing and dramatic, since it make s up a person’s attitude and outlook towards life. (67) ______________________ But when self-esteem is low, you can either sulk and be sad or be afraid to try out new things or take risks and chances with your love-life, career, relationships and self-improvement.Self-esteem plays a key role in the maturity of a person, especially when trying to get away from a dangerous situation, going through a series of trials in life. Our normal responses to these circumstances and situations are governed by how we value ourselves and how our decisions are dictated by how we value ourselves and how our decisions are dictated by these conditions.Faced with tough decisions in life, the more self-esteem one has, the better. It is for thatperson to make sound decisions, even in the face of peer pressure or stress at work and at home. Let us try to look into some of the common and best practices which have been tried and tested to help boost self-esteem.Always compliment yourself daily, especially by trying to look for specific tasks you did well for that day and congratulate yourself for it. (68) ____________________ List down all things you are good at doing and achieving, be it a talent, skill, sport or building up other people.You can add more focus to these good points and fuel our passion to do better and make you not only understand yourself more, but also give you the true meaning and measurement of self-worth and this is how you see yourself as important.Appreciation of one’s physical appearance and bearing can also be your source of self-esteem, be it the size and shape of your body, your overall physical structure or unique features. Your body can be your source of pride and will help you understand how you would like others to see you, or work on your physical appearance to boost self-morale and satisfaction.Sometimes when you tend to see things in a different light or perspective from others, don’t focus too much on making sure that what you think will cause things to change. (69) _____________________________When you have good self-esteem you will realize that what you did was right and was made under your own food judgment, sound principles and concepts based on your personal outlook and attitude towards life.Do not let negative feedback affect you. Of course, one cannot help but feel bad about negative comments or reactions, but you have to consider that these are tests against your character and personality. (70) ________________________________ So try to look at yourself and see, and if you feel less important or are not satisfied with how you see and look at things, then think about ways on how to boost your self-esteem. You’ll thank yourself for it.第II 卷I．SummaryDirections: Read the following passage. Summarize in no more than 60 words the mainidea of the passage and how it is illustrated. Use your own words as far as possible.1. One day, a poor boy who was trying to pay his way through school by selling goods door to door found that he only had one dime left. He was hungry so he decided to beg for a meal at the next house.2. However, he lost his nerve when a lovely young woman opened the door. Instead of a meal he asked for a drink of water. She thought he looked hungry so she brought him a large glass of milk. He drank it slowly, and then asked, “How much do I owe you？”3. “You don't owe me anything,” she replied. “Mother has taught me never to accept pay for a kindness.” He said, “Then I thank you from the bottom of my heart.” As Howard Kelly left that house, he not only felt stronger physically, but it also increased his faith in God and the human race. He was about to give up and quit before this point.4. Years later the young woman became critically ill. The local doctors were baffled. They finally sent her to the big city, where specialists can be called in to study her rare disease. Dr. Howard Kelly, now famous, was called in for the consultation. When he heard the name of the town she came from, a strange light filled his eyes. Immediately, he rose and went down through the hospital hall into her room.5. Dressed in his doctor's gown he went in to see her. He recognized her at once. He went back to the consultation room and determined to do his best to save her life. From that day on, he gave special attention to her case.6. After a long struggle, the battle was won. Dr. Kelly requested the business office to pass the final bill to him for approval. He looked at it and then wrote something on the side. The bill was sent to her room. She was afraid to open it because she was positive that it would take the rest of her life to pay it off. Finally she looked, and the note on the side of the bill caught her attention. She read these words...“Paid in full with a glass of milk.（Signed）Dr. Howard KellyTears of joy flooded her eyes as she prayed silently:”Thank You, God. Your love has spread through human hearts and hands.”。
上海市南洋模范中学2018届高三上学期数学周练试题(第三周) pdf版缺答案
南模中学高三周练卷（三）2017.9一、填空题1.将函数2log (31)y x =-的图像向左平移2个单位，得到函数_____________的图像2.已知函数()f x 在[0,)+∞上是减函数，()()g x f x =-，若(lg )(1)g x g <，则x 的取值范围是______________3.已知关于x 的二次方程22210x mx m +++=有两根，其中一根在区间(1,0)-内，另一根在区间(1,2)内，则m 的范围为_____________4.函数()f x 与()x g x =图像关于直线0x y -=对称，则2(4)f x -的单调增区间是___________5.函数()y f x =是定义域为R 的奇函数，当0x <时，13()21x f x x -=+-，则函数的解析式()f x =___________________（结果用分段函数表示）6.已知实数0a ≠，函数2,1()2,1x a x f x x a x +<⎧=⎨--≥⎩，若(1)(1)f a f a -=+，则a 的值为____________7.函数12log 1y x =-的单调递增区间是_____________8.设1()fx -是函数1()2()3x x f x x =-+的反函数，则1()1f x ->成立的x 的取值范围是______________9.已知1()log ()2a f x ax =-（其中0a >且1a ≠）在区间[1,2]上是减函数，则实数a 的取值范围_______________10.已知定义在R 上的奇函数()f x 和偶函数()g x 满足()()2x xf xg x a a -+=-+（0a >且1a ≠），若(2)g a =，则(2)f =_____________11.设()y f x =的反函数为1()y f x -=，又(2)y f x =+与1(1)y f x -=-互为反函数，则11(2)(1)f f ---的值为____________12.已知关于x 的方程2(21)4312x xx a a -+=++有奇数个解，则a 的值为___________二、选择题13.“(0)0f =”是“函数()f x 是奇函数”的（ ）条件A.充分非必要B. 必要非充分C.充要D.非充分非必要14.如图所示，()(1,2,3,4)i f x i =是定义在[0,1]上的四个函数，其中满足性质：“对[0,1]中的任意的1x 和2x ，任意[0,1]λ∈，1212((1))()(1)()f x x f x f x λλλλ+-<+-恒成立”的只有（ ）A B C D 15.若1()2x f x x+=-，对于n N +∈，定义1()()f x f x =，1()[()]n n f x f f x +=，如果1331()()f x f x =，那么16()f x 的解析式为（ ）A.161()2x f x x +=- B.161()x f x x -= C.16116()216x f x x +=- D.16116()xf x x+= 16.若1()1(1)f x f x +=+，当[0,1x ∈时，()f x x =，若在区间(1,1]-内()()g x f x mx m =--有两个零点，则实数m 的取值范围是（ ）A.[0,2)B.1[,)2+∞C.1[0,)3D.1(0,]2三、解答题17.已知函数2()21f x ax x a =-+-（a 为实常数） （1）若1a =，作函数()f x 的图像； （2）设()()f x h x x=，若函数()h x 在区间[1,2]上是增函数，求实数a 的取值范围18.已知指数函数()y g x =满足：(2)4g =，定义域为R 的函数()()2()g x nf xg x m-+=+是奇函数，（1）确定()y g x =的解析式；（2）求m ，n 的值；（3）若对任意的t R ∈，不等式22(2)(2)0f t t f t k -+-<恒成立，求实数k 的取值范围。
2018 上海市南模中学高三月考数学试卷
2018 上海市南模中学高三月考数学试卷2017.10一. 填空题1. 已知集合2{|0}5x A x x -=<+，2{|230,}B x x x x R =--≥∈，则A B = 2. 已知函数()arcsin(21)f x x =+，则1()6f π-=3. 在(0,2)π内，使sin cos x x ≤成立的x 的取值范围为4. 函数212()log (32)f x x x =-+-的单调递增区间为5. 函数||y x x =的反函数为6. 若对任意x R ∈，不等式2sin 22sin 0x x m +-<恒成立，则m 的取值范围为7. 设数列{}n a 的前n 项和为n S ，若11a =，114n n S a +=（*n N ∈），则n S = 8. 数列{}n x 中，10x =，111n n n x x n n++=+（*n N ∈），则数列的通项公式n x = 9. 设()f x 是定义域为R 的奇函数，()g x 是定义域为R 的偶函数，若函数()()f x g x +的值 域为[1,3)，则函数()()f x g x -的值域为10. 已知()sin()f x A x A ωϕ=++（0ω>），若两个不相等的实数12,{|()}2Ax x x f x ∈=， 且12min ||x x π-=，则()f x 的最小正周期为 11. 函数2()|2|4sin4xf x x x π=-+-（16x ≤≤）的值域为12. 函数()2|}f x x =-，其中min{,}a a ba b b a b ≤⎧=⎨>⎩，若动直线y m =与函数()y f x =的图像有三个不同的交点，它们的横坐标分别为1x 、2x 、3x ，则1x 、2x 、3x 是否存在最大值？若存在，在横线处填写其最大值；若不存在，直接填写“不存在”二. 选择题13. 已知x R ∈，则“1x ≠”是“2430x x -+≠”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14. 若||a c h -<，||b c h -<，则下列不等式一定成立的是（ ）A. ||2a b h -<B. ||2a b h ->C. ||a b h -<D. ||a b h -> 15. 已知ABC ∆内接于单位圆，则长为sin A 、sin B 、sin C 的三条线段（ ） A. 能构成一个三角形，其面积大于ABC ∆面积的一半 B. 能构成一个三角形，其面积等于ABC ∆面积的一半 C. 能构成一个三角形，其面积小于ABC ∆面积的一半 D. 不一定能构成一个三角形16. 已知函数()21x x e mf x +=+，若对任意123,,x x x R ∈，总有1()f x 、2()f x 、3()f x 为某一个三角形的边长，则实数m 的取值范围是（ ）A. 1[,1]2B. [0,1]C. [1,2]D. 1[,2]2三. 解答题17. 已知()2f x kx =+，不等式|()|3f x <的解集为(1,5)-，不等式1()xf x ≥的解集为A ； （1）求集合A ；（2）设22()log (22)g x ax x =-+的定义域为B ，若B A ≠∅，求实数a 的取值范围；18. 右图为函数()sin()f x A x ωϕ=+（0A >，0ω>，02πϕ<<）的部分图像，M 、N是它与x 轴的两个交点，D 、C 分别为它的最高点和最低点，(0,1)E 是线段MD 的中点， 且OME ∆为等腰直角三角形； （1）求()f x 的解析式；（2）将函数()f x 图像上的每个点的横坐标缩短为原来的一半，再向左平移12的单位长度 得到()y g x =的图像，求()g x 的解析式及单调增区间，对称中心；19. 在ABC ∆中，a 、b 、c 分别为内角A 、B 、C 所对的边，且满足(2)cos cos b A C =；（1）求A 的大小；（2）现给出三个条件：（1）2a =；（2）4B π=；（3）c =，试从中选出两个可以确定ABC ∆的条件，写出你的选择，并以此为依据求ABC ∆的面积；（需写出所有可行的方案）20. 对于函数1()f x 、2()f x 、()h x ，如果存在实数a 、b 使得12()()()h x a f x b f x =⋅+⋅， 那么称()h x 为1()f x 、2()f x 的生成函数； （1）若1()sin f x x =，2()cos f x x =，()sin()3h x x π=+，则()h x 是否分别为1()f x 、2()f x的生成函数？并说明理由；（2）设12()log f x x =，212()log f x x =，2a =，1b =，生成函数()h x ，若不等式(4)(2)0h x th x +<在[2,4]x ∈上有解，求实数t 的取值范围；（3）设1()f x x =（0x >），21()f x x=（0x >）取0a >，0b >，生成函数()h x 图象的 最低点坐标为(2,8)，若对于任意正实数1x 、2x 且121x x +=，试问是否存在最大的常数m ，使12()()h x h x m ≥恒成立？如果存在，求出这个m 的值；如果不存在，请说明理由；21. 对于给定的正整数k ，若数列{}n a 满足11(21)n k n k n k n k n a a a a k a --++-+++⋅⋅⋅++=+对任 意正整数n ()n k >恒成立，则称数列{}n a 是()AP k 数列，若正数项数列{}n b ，满足2111()k n k n k n k n k n b b b b b +--++-+⋅⋅⋅=对任意正整数)(k n n >恒成立，则称{}n b 是()GP k 数列；（1）已知正数项数列{}n a 是(3)GP 数列，且前五项分别为4、2a 、2、4a 、1，求4a 的值； （2）若2sin n a n n ω=+（0ω>）为常数，且{}n a 是(2)AP 数列，求ω的最小值； （3）对于下列两种情形，只要选作一种，满分分别是 ① 7分，② 8分，若选择了多于一种情形，则按照序号较小的解答积分；① 证明：数列{}n a 是等差数列的充要条件为“{}n a 既是(2)AP 数列，又是(3)AP 数列”； ② 证明：正数项数列{}n b 是等比数列的充要条件为“数列{}n b 既是(2)GP 数列，又是(3)AP 数列”；参考答案一. 填空题1. (,2)[3,)-∞+∞2. 14-3. 5(0,][,2)44πππ4. 3(,2)25. 10()x f x x -⎧≥⎪=⎨<⎪⎩ 6. 1,)+∞ 7. 15n -（*n N ∈）8. 1n -（*n N ∈） 9. (3,1]-- 10. 3π 11. [3,29] 12. 1二. 选择题13. B 14. A 15. C 16. D三. 解答题17.（1）[1,2)A =；（2）(0,)a ∈+∞； 18.（1）()2sin()44f x x ππ=+；（2）()2cos 2g x x π=；增区间：[42,4]k k -（*n N ∈）； 对称中心：(21,0)k +（k Z ∈）； 19.（1）6A π=；（2）选（1）（3），得S =1）（2），得1S =）；20.（1）是，12a =，b =；（2）4(,)3-∞-；（3）289；21.（1）2±；（2）2ωπ=；（3）略。
上海市延安中学2017-2018学年高三上学期10月月考数学试题
○…………外○…………内绝密★启用前上海市延安中学2017-2018学年高三上学期10月月考数学试题试卷副标题注意事项：1．答题前填写好自己的姓名、班级、考号等信息 2．请将答案正确填写在答题卡上第I 卷（选择题)请点击修改第I 卷的文字说明 一、单选题1．已知,a b 都是实数，那么“22a b >”是“a 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2．若数列{}n a 的通项公式1,1,211,3,3n nn n a n n N *⎧=⎪⎪+=⎨⎪≥∈⎪⎩前n 项和为n S ，则下列结论中正确的是（ ） A ．lim n n a →∞不存在 B ．8lim 9n n S →∞=C ．lim 0n n a →∞=或1lim 3n n a →∞=D ．1lim 18n n S →∞=3．有一塔形几何体由若干个正方体构成，构成方式如图所示，上层正方体下底面的四个顶点是下层正方体上底面各边的中点．已知最底层正方体的棱长为2，且该塔形的表面积(含最底层正方体的底面面积)超过39，则该塔形中正方体的个数至少是○…………外○…………内4．对任意一个复数z ，定义集合{}21,n z M z n N αα-==∈，设12ω=-+（i 为虚数单位），则集合M ω与2M ω的关系是（ ） A ．{}21,M M ωωω=I B ．2M M ωω⊆ C ．2M M ωω= D ．M ω和2M ω没有关系第II 卷（非选择题)请点击修改第II 卷的文字说明 二、填空题5．已知集合{}{}1,3,,3,4A m B ==，且B A ⊆，则实数m 的值是__________. 6．函数()f x =的定义域是_________. 7．如果圆锥的底面积为π，母线长为2，那么该圆锥的高为___________. 8．在()1521x +的二项展开式中，含5x 的项是二项展开式的第__________项.9．已知复数03z i =+（i 为虚数单位），复数z 满足003z z z z ⋅=+，则z =__________. 10．等差数列{a n }的前10项和为30，则14710a a a a +++=________11．已知双曲线与椭圆221166x y +=有相同的焦点，且双曲线的渐进线方程为12y x =±，则此双曲线方程为_________12．某班从4位男生和3位女生志愿者选出4人参加校运会的点名签到工作，则选出的志愿者中既有男生又有女生的概率的是__________.（结果用最简分数表示） 13．如图，将四个边长为1的小正方形拼成一个大正方形，,A B 是原来小正方形的其中两个顶点，()1,2,7i P i =L 是小正方形的其余顶点，在所有()1,2,7i AB AP i ⋅=u u u r u u u rL 中，不同的数值有__________个.……○…………订________班级：___________考……○…………订14．函数()()()1,01,0x x x f x x x x ⎧-≥⎪=⎨-+<⎪⎩，()()21g x f x =-+，若()3g a =，则()4g a -=__________.15．已知实数数列{}n a 满足()2123n n n a a a n N *+=-+∈，n S 是数列{}n a 的前n 项和.若()32k a k N *=∈，则2017k S +=__________. 16．已知a b vv、是平面内两个互相垂直的单位向量，且此平面内另一向量c v在满足()()340a c b c +-=v v v v，均能使c b k v v -≤成立 ，则k 的最小值是_________.三、解答题17．如图，在直三棱柱111ABC A B C -中，12AA AB BC ===，且AB BC ⊥.求：（1）四棱锥11C ABB A -的体积； （2）AC 与平面11A B C 所成角的大小.18．设函数()4sin ,0,3f x x x x π⎡⎤=+∈⎢⎥⎣⎦.（1）求函数()f x 的单调递减区间；（2）ABC ∆中，1,2AB AC ==,且()f A =ABC ∆为直角三角形. 19．某公司进行共享单车的投放与损耗统计，到去年2016年底单车的市场保有量（已投入市场且能正常使用的单车数量）为10000辆，预计今后每年新增单车1000辆，随着单车的频繁使用，估计每年将有200辆车的损耗，并且今后若干年内，年平均损耗在上一年损耗基础上增加10％.（1）预计2019年底单车的市场保有量是多少？（2）到哪一年底，市场的单车保有量达到最多？该年的单车保有量是多少辆（最后结…………线……………………线…………20．如图，过抛物线24y x=焦点F的直线与抛物线交于,A B（其中A点在x轴的上方）两点.（1）若线段AF的长为3，求O到直线AB的距离；（2）证明：OAB∆为钝角三角形；（3）已知AF FBλ=uu u r uu r且[]2,3λ∈，求三角形OAB的面积S的取值范围.21．A是定义在[]1,2上且满足如下条件的函数()xϕ组成的集合：①对任意的[]1,2x∈，都有()()1,2xϕ∈；②存在常数()01L L<<，使得对任意的[]12,1,2x x∈，都有()()1212x x L x xϕϕ-≤-.（1）设()[]211,1,25x x xϕ=+∈，问()xϕ是否属于A？说明你的判断理由；（2）若()x Aϕ∈，如果存在()1,2x∈，使得()00x xϕ=，证明这样的x是唯一的；（3）设,k b为正实数，是否存在函数()[]1,2x xϕ=∈，使()x Aϕ∈？作出你的判断，并说明理由.参考答案1．D 【解析】 【分析】举反例说明既不充分也不必要. 【详解】当2,1a b =-=时，满足22a b >，但a b >不成立；所以不充分； 当1,2a b ==-时，满足a b >，但22a b >不成立；所以不必要； 故选：D 【点睛】本题考查充要关系的判定，考查基本分析判断能力，属基础题. 2．B 【解析】 【分析】先利用等比数列求和公式求和，再求极限得结果. 【详解】1lim lim03n nn n a →∞→∞==3234211(1)11111551133+++(1)1233336618313n n n n S ---=++=+=+--L因此2511518lim lim[(1)]61836189n n n n S -→∞→∞=+-=+=故选：B 【点睛】本题考查数列解析以及等比数列求和公式，考查基本分析求解能力，属中档题. 3．C 【解析】 【分析】根据相邻正方体的关系得出个正方体的棱长为等比数列，求出塔形表面积n S 的通项公式，令39n S >，即可得出n 的范围． 【详解】设从最底层开始的第n 层的正方体棱长为n a ，则{}n a 是以2为首项，以2为公比的等比数列. ∴{}2na 是以4为首项，以12为公比的等比数列 ∴塔形的表面积为222222222123123114(1)32264444444248401212n n n n n S a a a a a a a a a -=+++⋅⋅⋅=+++⋅⋅⋅+=⨯+=--. 令3240392n ->，解得5n >. ∴塔形正方体最少为6个. 故选C. 【点睛】此题考查了立体图形的表面积问题以及等比数列求和公式的应用．解决本题的关键是得到上下正方体的棱长之间的关系，从而即可得出依次排列的正方体的一个面的面积，这里还要注意把最下面的正方体看做是6个面之外，上面的正方体都是露出了4个面． 4．C 【解析】 【分析】根据ω的性质化简集合M ω与2M ω，再进行判定选择. 【详解】因为12ω=-，所以231,12ωω=-={}{}212,,1,n M n N ωααωωω-==∈=，{}{}2422,,1,n M n N ωααωωω-==∈=所以2M M ωω=， 故选：C 【点睛】本题考查ω的性质以及集合相等的判定，考查基本分析化简判断能力，属基础题. 5．4 【解析】 【分析】根据集合包含关系确定方程，解得结果. 【详解】因为B A ⊆，所以{3,4}{1,3,}4m m ⊆∴= 故答案为：4 【点睛】本题考查根据集合包含关系求参数，考查基本分析求解能力，属基础题. 6．()[),02,-∞+∞U 【解析】 【分析】 由210x -≥，化为20x x-≥,解分式不等式可得结果. 【详解】要使函数()f x =有意义， 则210x -≥， 即20x x-≥,解得0x <或2x ≥,即函数()f x =的定义域是()[),02,-∞+∞U ， 故答案为()[),02,-∞+∞U . 【点睛】本题主要考查抽象函数的定义域、不等式的解法，属于中档题.定义域的三种类型及求法：(1)已知函数的解析式，则构造使解析式有意义的不等式(组)求解；(2) 对实际问题：由实际意义及使解析式有意义构成的不等式(组)求解；(3) 若已知函数()f x 的定义域为[],a b ，则函数()()f g x 的定义域由不等式()a g x b ≤≤求出.7【解析】 【分析】由底面积求出底面半径，利用勾股定理可得结果. 【详解】设圆锥底面半径为r , 因为圆锥的底面积为π， 所以21,r r ππ=⇒=又因为母线长为2=【点睛】本题主要考查圆锥的性质，意在考查对基础知识的掌握情况，考查了空间想象能力，属于基础题. 8．11 【解析】 【分析】根据二项展开式通项公式即可确定结果. 【详解】15115(2)15510rr r T C x r r -+=∴-=∴=Q ,即含5x 的项是二项展开式的第11项.故答案为：11 【点睛】本题考查二项展开式通项公式，考查基本分析求解能力，属基础题. 9【解析】 【分析】变形003z z z z ⋅=+可得0033z iz z i+==-，分子分母同乘以i ，可得13i z =-，利用复数模的公式可得结果.Q 复数03z i =+，复数z 满足003z z z z ⋅=+，()020333i iz i z z i i ++∴===-13131ii -+==--，z ==.【点睛】复数是高考中的必考知识，主要考查复数的概念及复数的运算．要注意对实部、虚部的理解，掌握纯虚数、共轭复数、复数的模这些重要概念，复数的运算主要考查除法运算，通过分母实数化转化为复数的乘法，运算时特别要注意多项式相乘后的化简，防止简单问题出错，造成不必要的失分. 10．12 【解析】 【分析】利用等差数列的前n 项和公式即可得到a 1+a 10=6．由等差数列的性质可得a 1+a 10=a 4+a 7，进而可得答案． 【详解】∵等差数列{a n }的前10项和为30，∴()11010302a a +=，解得a 1+a 10=6．由等差数列的性质可得a 1+a 10=a 4+a 7， ∴a 1+a 4+a 7+a 10=2（a 1+a 10）=2×6=12． ∴a 1+a 4+a 7+a 10=12． 故答案为12． 【点睛】熟练掌握等差数列的前n 项和公式、等差数列的性质是解题的关键．11．22182x y -=【分析】根据双曲线的渐进线方程为12y x =±，设双曲线222214x y b b-=，计算椭圆焦点为()，根据双曲线焦点公式得到答案. 【详解】221166x y +=的焦点为：()双曲线的渐进线方程为12y x =±，则设双曲线方程为：222214x y b b-=，焦点为()故2224102b b b +=∴= ，双曲线方程为22182x y -=故答案为：22182x y -=【点睛】本题考查了求双曲线方程，根据渐近线设双曲线为222214x y b b-=是解题的关键.12．3435【解析】 【分析】“4位男生和3位女生志愿者选出4人参加校运会的点名签到工作，若这4人中必须既有男生又有女生”的对立事件是“只有男生”，利用组合知识求出总事件数，根据古典概型概率公式以及对立事件的概率公式可得结果. 【详解】“4位男生和3位女生志愿者选出4人参加校运会的点名签到工作，若这4人中必须既有男生又有女生”的对立事件是“只有男生”，事件“只有男生”只包含一个基本事件，而总的基本事件数是4735C =，故事件“只有男生”的概率是135， 事件“4位男生和3位女生志愿者选出4人参加校运会的点名签到工作，若这4人中必须既有男生又有女生”的概率是13413535-=，故答案为3435. 【点睛】本题主要考查对立事件的概率公式以及古典古典概型概率公式的应用，属于基础题. 在求解有关古典概型概率的问题时，首先求出样本空间中基本事件的总数n ，其次求出概率事件中含有多少个基本事件m ，然后根据公式mP n=求得概率. 13．5 【解析】 【分析】建立直角坐标系，根据向量数量积坐标计算，再统计不同结果个数. 【详解】建立如图所示直角坐标系，则123(2,1)(2,0)4,(2,1)(0,1)1,(2,1)(1,1)3,AB AP AB AP AB AP ⋅=⋅=⋅=⋅=⋅=⋅=u u u r u u u r u u u r u u u u r u u u r u u u r 4567(2,1)(1,0)2,(2,1)(0,2)2,(2,1)(1,2)4,(2,1)(2,2)6,AB AP AB AP AB AP AB AP ⋅=⋅=⋅=⋅=⋅=⋅=⋅=⋅=u u u r u u u u r u u u r u u u r u u u r u u u u r u u u r u u u u r所以有5个不同的数值， 故答案为：5 【点睛】本题考查向量数量积坐标计算，考查基本分析求解能力，属基础题. 14．1- 【解析】 【分析】先根据分段函数求a 的值，再代入求()2f a -的值，即得结果. 【详解】因为()3g a =，所以()()()21322g a f a f a =-+=∴-=因此20(2)(3)2a a a -≥⎧⎨--=⎩或20(2)(1)2a a a -<⎧⎨---=⎩解得4a =或a ∈∅从而()()()2(2)24211f a f g a f a -=-=-∴-=-+=- 故答案为：1- 【点睛】本题考查求分段函数值，考查基本分析求解能力，属基础题. 15．32【解析】 【分析】根据递推关系求各项，再求和. 【详解】2211323232()02n n n k k k k k a a a a a a a a ++=-+∴=-+=--=Q当1n k ≥+时，0n a =若2k ≥，则由2123n n n a a a +=-+得2211111323232k k k k k k a a a a a a -----=-+∴-+=∴∈∅，因此=1k ，从而20173,133+02220,2n k n a S n +⎧=⎪=∴==⎨⎪≥⎩， 故答案为：32【点睛】本题考查根据递推关系求通项，考查基本分析求解能力，属中档题. 16．52+ 【解析】根据题意，()()()1,0,0,1,,a b c x y v v v===，利用()()340a c b c +⋅-=r r r r ，求得,x y 的关系，利用圆的几何性质，再求出c b -vv 的最大值，从而求出k 的最小值.【详解】因为a b v v 、是平面内两个互相垂直的单位向量，所以可设 ()()()1,0,0,1,,a b c x y v v v===， ()33,a c x y ∴+=+r r， ()4,4b c x y -=--r r，又()()340a c b c +⋅-=r r r r，()()340x x y y ∴-++-=， 即()22325224x y ⎛⎫++-= ⎪⎝⎭， 它表示的圆心在3,22M ⎛⎫-⎪⎝⎭，半径为52的圆，c b -vv 表示圆上的点到(0,1)B 的距离，圆心M 到点(0,1)B 的距离为d =c b ∴-r r的最大值为52=，要使c b k -≤r r 恒成立，52k +≥即k 的最小值是52+，故答案为52. 【点睛】本题主要考查向量模的几何意义、轨迹方程的应用以及圆的几何意义，考查了转化思想的应用，属于难题. 转化是数学解题的灵魂，合理的转化不仅仅使问题得到了解决，还可以使解决问题的难度大大降低，本题将不等式恒成立问题转化为圆上动点到定点距离的最值问题是17．（1）83；（2）6π【解析】 【分析】（1）先证线面垂直得高，再根据锥体体积公式求结果；（2）先补成正方体，再根据线面垂直确定线面角，最后解三角形得结果. 【详解】（1）因为直三棱柱111ABC A B C -，所以1AA ⊥平面ABC ∴1AA BC ⊥1,AB BC AB AA A BC ⊥=∴⊥Q I 平面11ABB A因此四棱锥11C ABB A -的体积为11118=222333ABB A BC S ⋅⋅⨯⨯⨯=矩形； （2）先补成正方体1111ABCD A B C D -,则AC 与平面11A B C 所成角为AC 与平面11A B CD 所成角，因为1AD ⊥平面11A B CD ,设11A O AD D =I ,则ACO ∠为AC 与平面11A B C 所成角, 因为1sin 26AO ACO ACO AC π∠==∴∠=,因此AC 与平面11A B C 所成角为6π.【点睛】本题考查锥体体积公式、线面角以及线面垂直判定定理，考查基本分析求解能力，属中档题.18．（1）766ππ⎡⎤⎢⎥⎣⎦，（2）详见解析【解析】 【分析】（1）先根据辅助角公式化简函数，再根据正弦函数性质求单调区间；（2）先根据()f A A ，再根据余弦定理求BC ，最后根据勾股定理证结论. 【详解】（1）()sin 2sin(),3f x x x x π==+由322,()232k x k k Z πππππ+≤+≤+∈得722,()66k x k k Z ππππ+≤≤+∈ 因为40,3x π⎡⎤∈⎢⎥⎣⎦，所以7,66x ππ⎡⎤∈⎢⎥⎣⎦，即单调递减区间为766ππ⎡⎤⎢⎥⎣⎦，；（2）因为()f A =2sin()sin()33A A ππ+=+因为(0,)A π∈，所以2333A A πππ+=∴= 222211221232BC AC AB ∴=+-⨯⨯⨯==- 因此ABC ∆为直角三角形. 【点睛】本题考查辅助角公式、正弦函数性质以及余弦定理，考查基本分析求解能力，属基础题. 19．（1）12338（2）2033年底；最多18891辆 【解析】 【分析】（1）根据等差数列与等比数列进行列式计算；（2）根据题意先列市场的单车保有量函数关系式，再根据数列单调性确定单车保有量最大值. 【详解】（1）2019年底单车的市场保有量是210000+10003200200 1.1200 1.112338⨯--⨯-⨯=（2）到2016n +年底，市场的单车保有量为2110000+1000200200 1.1200 1.1200 1.1n n a n -=⨯--⨯-⨯--⨯L1 1.110000+100020010200+10002000 1.11 1.1n n n n -=-⨯=-⨯-11000200 1.1200(5 1.1)n n n n a a +-=-⨯=-Q1116,;17,;n n n n n a a n a a ++∴≤>≥<即17,n n a =取最大值18891，此时为2033年底；即到2033年底，市场的单车保有量达到最多，为18891辆. 【点睛】本题考查数列单调性以及等比数列再实际问题中的应用，考查综合分析求解能力，属中档题.20．（1（2）详见解析；（3）⎣⎦ 【解析】 【分析】（1）先根据抛物线定义求出A 点坐标，再根据点斜式求直线AB 的方程，最后根据点到直线距离公式求结果；（2）先设直线方程，与抛物线方程联立，结合韦达定理化简OA OB ⋅u u u r u u u r ,根据OA OB ⋅u u u r u u u r为负证明结果；（3）先设直线方程，与抛物线方程联立，结合韦达定理以及面积公式表示三角形OAB 的面积S ，再根据对勾函数单调性求值域. 【详解】（1）设111(,),(0)A x y y >，因为3AF =，所以111132x x y +=∴=∴=因此:(1),021AB y x y =----从而O 到直线AB3=； （2）设直线AB 的方程为1x my =+，11122(,),(0)(,),A x y y B x y ，>由214x my y x=+⎧⎨=⎩得211124404,4y my y y m y y --=∴+==- 从而22121212123016y y OA OBx x y y y y u u u r u u u r ?+=+=-<，因此OAB ∆为钝角三角形；（3）因为AF FB λ=uu u r uu r，所以12y y λ=-，由（2）得124y y =-，所以2224|y y λ=，12211||1(1)||22S y y y λ=-⨯=+==因为[]2,3λ∈，而1y tt =+在上单调递增，所以S ∈= 【点睛】本题考查抛物线定义、点到直线距离公式、抛物线中三角形面积以及直线与抛物线位置关系，考查综合分析求解与论证能力，属中档题.21．（1）是，详见解析（2）详见解析（3）详见解析 【解析】 【分析】（1）根据定义逐一验证，即求函数在[]1,2上值域，再判断是否为()1,2子集；根据不等式寻找满足条件的常数()01L L <<；（2）利用反证法，假设存在两个，根据条件得到1L ≥，即假设不成立，原命题成立； （3）先根据条件①解不等式确定13k <<，再根据条件②利用恒成立转化为对应函数最值，再解不等式确定12k <<.b 的条件由k 确定. 【详解】（1）因为()2115x x ϕ=+在[]1,2上单调递增，所以()69[,](1,2)55x ϕ∈⊆；()()121212121455x x x x x x x x ϕϕ-=-+≤-Q ， 所以存在常数45，使得对任意的[]12,1,2x x ∈，都有()()121245x x x x ϕϕ-≤-， 综上()x ϕ属于A ；（2）设存在()12121,,2,x x x x ≠∈，满足()()1122,x x x x ϕϕ==， 因为()x A ϕ∈，所以存在常数()01L L <<，使得()()1212x x L x x ϕϕ-≤-， 即12121201x x L x x x x L -≤-->∴≥Q ，与01L <<矛盾，因此满足条件的0x 是唯一的；（3）假设存在，则因为()x A ϕ∈，且()x ϕ=[]1,2上单调递增，所以12<121210222b bb k k b b ∴+<<+∴>+>+∴<<，2411243k b k k k k ∴-<<-∴->-∴<，因此13k <<；存在常数()01L L <<，使得对任意的[]12,1,2x x ∈，12L x x ≤-，12L x x ≤-，所以k ≤<，≥>=因此2416022k k k k <+-<∴--<<从而12k <<即当12241,0k k b k b <<-<<->，时存在函数()[]1,2x x ϕ=∈，使()x A ϕ∈；否则不存在.【点睛】本题考查函数新定义、反证证以及恒成立问题，考查综合分析求解与论证能力，属难题.。
上海市南洋模范中学2018-2019学年上学期高三期中数学模拟题
上海市南洋模范中学2018-2019学年上学期高三期中数学模拟题 班级__________ 座号_____ 姓名__________ 分数__________一、选择题（本大题共12小题，每小题5分，共60分.每小题给出的四个选项中，只有一项是符合题目要求的.）1． 设集合,,则( )A BCD2． 已知函数⎩⎨⎧≤>=)0(||)0(log )(2x x x x x f ，函数)(x g 满足以下三点条件：①定义域为R ；②对任意R x ∈，有1()(2)2g x g x =+；③当]1,1[-∈x 时，()g x 则函数)()(x g x f y -=在区间]4,4[-上零点的个数为（ ）A ．7B ．6C ．5D ．4【命题意图】本题考查利用函数图象来解决零点问题，突出了对分段函数的转化及数形结合思想的考查，本题综合性强，难度大.3． 记集合{}22(,)1A x y x y =+?和集合{}(,)1,0,0B x y x y x y =+３?表示的平面区域分别为Ω1，Ω2，若在区域Ω1内任取一点M （x ，y ），则点M 落在区域Ω2内的概率为（ ） A ．12p B ．1p C ．2pD ．13p【命题意图】本题考查线性规划、古典概型等基础知识，意在考查数形结合思想和基本运算能力． 4． 12,e e 是平面内不共线的两向量，已知12AB e ke =-，123CD e e =-，若,,A B D 三点共线，则的值是（ ）A ．1B ．2C ．-1D ．-25． 执行如图所示的程序框图，输出的s 值为( )。
A-3BC D26． 在ABC ∆中，60A =，1b =sin sin sin a b cA B C++++等于（ ）A ．B ．3C ．3D ．27． 某几何体的三视图如图所示，则该几何体的体积为（ ） A ．16163π-B ．32163π-C ．1683π-D ．3283π-【命题意图】本题考查三视图、圆柱与棱锥的体积计算，意在考查识图能力、转化能力、空间想象能力．8． 在复平面内，复数1zi+所对应的点为(2,1)-，i 是虚数单位，则z =（ ） A ．3i -- B ．3i -+ C ．3i - D ．3i +9． 函数()()f x x R Î是周期为4的奇函数，且在02[,]上的解析式为(1),01()sin ,12x x x f x x x ì-＃ï=íp <?ïî，则1741()()46f f +=（ ） A ．716 B ．916 C ．1116 D ．1316【命题意图】本题考查函数的奇偶性和周期性、分段函数等基础知识，意在考查转化和化归思想和基本运算能力．10．一个几何体的三视图如图所示，则该几何体的体积是（ ）A ．64B ．72C ．80D ．112【命题意图】本题考查三视图与空间几何体的体积等基础知识，意在考查空间想象能力与运算求解能力.11．“3<-b a ”是“圆056222=++-+a y x y x 关于直线b x y 2+=成轴对称图形”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【命题意图】本题考查圆的一般方程、圆的几何性质、常用逻辑等知识，有一定的综合性，突出化归能力的考查，属于中等难度．12．以下四个命题中，真命题的是（ ） A ．(0,)x π∃∈，sin tan x x =B ．“对任意的x R ∈，210x x ++>”的否定是“存在0x R ∈，20010x x ++<C ．R θ∀∈，函数()sin(2)f x x θ=+都不是偶函数D ．ABC ∆中，“sin sin cos cos A B A B +=+”是“2C π=”的充要条件【命题意图】本题考查量词、充要条件等基础知识，意在考查逻辑推理能力．二、填空题（本大题共4小题，每小题5分，共20分.把答案填写在横线上）13．若x ，y 满足约束条件⎩⎪⎨⎪⎧x ＋y －5≤02x －y －1≥0x －2y ＋1≤0，若z ＝2x ＋by （b ＞0）的最小值为3，则b ＝________．14．阅读如图所示的程序框图，则输出结果S 的值为 .【命题意图】本题考查程序框图功能的识别，并且与数列的前n 项和相互联系，突出对逻辑判断及基本运算能力的综合考查，难度中等.15．要使关于x 的不等式2064x ax ≤++≤恰好只有一个解，则a =_________. 【命题意图】本题考查一元二次不等式等基础知识，意在考查运算求解能力. 16．在等差数列}{n a 中，20161-=a ，其前n 项和为n S ，若2810810=-S S ，则2016S 的值等于 . 【命题意图】本题考查等差数列的通项公式、前n 项和公式，对等差数列性质也有较高要求，属于中等难度.三、解答题（本大共6小题，共70分。
最新-上海市上南中学2018学年第一学期高三年级水平测
上南中学2018学年第一学期高三年级水平测试物理 试卷一．（20分）填空题。
本大题共5小题，每小题4分。
答案写在题中横线上的空白处或指定位置，不要求写出演算过程。
1、 用a ，b 两根细绳悬挂小球，如图一所示。
现 保持小球位置不变，b 绳由水平方向逆时针缓慢转动到竖直方向，在此过程中， a 绳拉力的变化 图一是 ，b 绳拉力的变化是 。
2、总质量为M 的热气球由于故障在高空以速度v 0匀速竖直下降，为了阻止继续下降，在t ＝0时刻，从吊篮中释放一个质量为m 的沙袋。
不计空气阻力，当t ＝ 时，热气球停止下降，这时沙袋的速度为 （此时沙袋未着地）。
3、 根均匀直棒AB ，BC 长度相等，AB 重为2G ，BC 重为G 。
现在B ，C两处分别施加竖直向上的力F B 和F C ，使AB ，BC 均在水平位置平衡，则此两力的大小分别为 图二 F B ＝ ，F C ＝ 。
4、一定质量的气体经过如图三所示的变化过程，在0℃时，气体的压强P 0＝2×118Pa ，体积V 0＝100ml ，那么气体在状态A 的压强为Pa ，在状态B 的体积为 ml 。
图三5、如图所示，以速度V 0＝12m/s 沿光滑地面滑行的木块上升到顶部水平的跳板后由跳板飞出，当跳板的高度h ＝ m 时，木块飞行的水平距离s 最大，最大距离s ＝ m （g ＝10m/s 2）。
图四二．（40分）选择题。
本大题共8小题，每小题5分。
每小题给出的四个答案中，至少有一个是正确的。
把正确答案全选出来，并将正确答案前面的字母填写在题后的方括号内。
每一小题全选对的得5分；选对但不全，得部分分；有选错或不答的，得0分。
填写在方括号外的字母，不作为选出的答案。
6、一木板B 放在水平地面上，木板A 放在B 的上面，A 的右端通过轻的弹簧秤与墙壁连接。
用水平拉力F 拉动木板B ，使它以速度v 匀速向左运动，此时弹簧秤示数为T 。
下面说法中正确的是t(℃)-273546[ ]A ．木块B 受到的滑动摩擦力的大小等于T 。
上海市南洋模范中学2020届高三数学10月检测试题(三)(含解析)
上海市南洋模范中学2020届高三数学10月检测试题（三）（含解析）一、填空题（本大题满分56分，每小题4分）1. 若实数a、b满足a2+b2=1，则ab的取值范围是______________．【答案】．【解析】因为实数满足，解得的取值范围是，故答案为.2. 设是一元二次方程的两个实根，则的最小值为______________．【答案】8．【解析】根据题意得，即，或，，当时，，当时，，的最小值，故答案为.3. 设f(x)为定义在R上的奇函数，当x≥0时，f(x)=2x+2x+b（b为常数），则f(-1)=______________．【答案】-3．【解析】因为为定义在上的奇函数，所以，解得，所以当时，，又因为为定义在上的奇函数，所以，故答案为.4. 已知集合A={(x,y) |-2<y<1,x∈Z,y∈Z}，，则A⋂B的真子集的个数为______________．【答案】15．【解析】或，或，，所以集合的真子集的个数为，故答案为.5. 函数的单调递增区间是______________．【答案】．【解析】由，解得，令，则外函数为为减函数，求函数的单调递增区间，即求的减区间，函数在上为减函数，则原函数的增区间为，故答案为.【方法点睛】本题主要考查二次函数的性质、复合函数的单调性，属于中档题.复合函数的单调性的判断可以综合考查两个函数的单调性，因此也是命题的热点，判断复合函数单调性要注意把握两点：一是要同时考虑两个函数的的定义域；二是同时考虑两个函数的单调性，正确理解“同增异减”的含义（增增增，减减增，增减减，减增减）.6. 不等式的解集为______________．【答案】 )【解析】因为且，所以原不等式的解集是，故答案为.7. 已知二次函数的值域为[0, )，则的最小值为__________．【答案】4．【解析】因为二次函数的值域为，，，当且仅当时取等号，而，故答案为.8. 若三角方程有解，则实数m的取值范围是______________．【答案】．【解析】令，则，因为三角方程有解，所以直线与正弦曲线有公共点，，故答案为.【点睛】已知函数有零点(方程有根)求参数取值范围的三种常用的方法：(1)直接法：直接根据题设条件构建关于参数的不等式，再通过解不等式确定参数范围；(2)分离参数法：先将参数分离，转化成求函数值域问题加以解决；(3)数形结合法：先对解析式变形，在同一平面直角坐标系中，画出函数的图象，然后数形结合求解．9. 若y=f(2x-1)是周期为t的周期函数，则函数y=f(x)的一个周期是______________．【答案】．【解析】若是周期为的周期函数，则，则，故的一个周期是，故答案为.【方法点晴】本题主要考查抽象函数的周期性，属于难题.对函数周期性的考查主要命题方向由两个，一是三角函数，可以用公式求出周期；二是抽象函数，往往需要根据条件判断出周期，抽象函数给出条件判断周期的常见形式为:(1)；（2）；（3）.10. 已知若，则=______________．【答案】．【解析】因为，所以，所以，因为，所以，所以，故答案为.11. 若正数a,b满足2+log2a=3+log3b=log6(a+b)，则的值为______________．【答案】108．【解析】因为正数满足，，所以设，则，，故答案为 .12. 设集合中的最大元素与最小元素分别为M,m，则M-m的值为______．【答案】．【解析】由题意得，，当且仅当时，等号成立，，，故答案为.13. 若函数f(x)=x2+a|x-1|在[0,+∞)上单调递增，则实数a的取值范围是______________．【答案】．【解析】,要使在上单调递增，则，得，所以实数的取值范围是，故答案为...................【答案】．【解析】当时，，解得，当时，，解得，的阶周期点的个数是，当时，解得，当时，解得，当时，解得，当时，解得，的阶周期点的个数是…由依次类推，有个不同的解析式，f n(x) x的点有个，的阶周期点的个数是，故答案为.二、选择题（本大题满分20分，每小题5分）15. 把下列命题中的“=”改为“>”，结论仍然成立的是（）A. 如果，，那么B. 如果，那么C. 如果，，那么D. 如果，，那么【答案】D【解析】把下列命题中的“=”改为“>”, 对于选项，如果，那么，若时，不成立，对于选项，如果，那么，取时，不成立，对于选项，如果，取不成立，对于选项，如果，那么根据不等式的性质可知正确，故选D.16. 设p,q是两个命题，，，则p是q（）A. 充分非必要条件B. 必要非充分条件C. 充要条件D. 既非充分又非必要条件【答案】B【解析】可化为，可得，显然后者可以推出前者，前者不能推出后者，所以是必要非充分条件，故选B.17. 定义在上的函数，当时，，且对任意的满足（常数），则函数f(x)在区间的最小值是（）A. B. C. D.【答案】D【解析】当，所以；当，所以；当，所以；所以当时，，故选D.18. 如图放置的边长为1的正方形沿轴滚动（向右为顺时针，向左为逆时针）．设顶点的轨迹方程是，则关于的最小正周期及在其两个相邻零点间的图像与x轴所围区域的面积S的正确结论是（）A. B.C. D.【答案】A【解析】从某一个顶点（比如）落在轴上的时候开始计算，到下一次点落在轴上，这个过程中四个顶点依次落在了轴上，而每两个顶点间距离为正方形的边长,因此该函数的周期为.下面考查点的运动轨迹,不妨考查正方形向右滚动，点从轴上开始运动的时候，首先是围绕点运动个圆，该圆半径为，然后以点为中心，滚动到点落地，其间是以为半径旋转，再以为圆心，旋转,这时候以为半径,因此最终构成图象如下：所以两个相邻零点间的图象与轴所围成区域的面积，故选A.三、解答题（本大题满分74分）19. 已知函数，且．(1) 求实数c的值；(2) 解不等式．【答案】(1) (2)【解析】试题分析：（1）函数求值时首先确定自变量的值对应的取值范围，进而代入相应的函数解析式；（2）解不等式时需分与两种情况分别代入函数式求解试题解析：（1）因为，所以；由，即，．（2）由（1）得由得，当时，解得，当时，解得，所以的解集为．考点：分段函数求值及解不等式20. 已知函数．(1) 若，求x的取值范围；(2) 若是以2为周期的偶函数，且当时，有，求函数的反函数．【答案】(1) (2) ，【解析】试题分析：（1）考虑对数函数的定义域，结合对数运算法则。
数学---上海市黄浦区格致中学2018届高三(上)10月月考试卷(解析版)
上海市黄浦区格致中学2018届高三（上）10月月考数学试卷一.填空题1．（3分）集合A={1，2，3，4，5}，B={x||x﹣3|≤1}，则A∩B=．2．（3分）直线l过点M（2，1）与直线3x+y﹣1=0垂直，则直线l的方程为．3．（3分）已知tanα=2，则=．4．（3分）等差数列{a n}的公差为d，前n项和为S n，若d＞0，且S10=0，则使得a n＜0的n 的最大值等于．5．（3分）已知实数x满足|log2x+4i|≥5（i是虚数单位），则x的取值范围是．6．（3分）设集合M={（x，y）|x﹣y+1≥0且x+y﹣2≤0且y≥0}，P∈M，A点坐标为（0，﹣1），则|AP|最大值与最小值之差等于．7．（3分）与椭圆有共同焦点，且焦点到渐近线距离等于1的双曲线方程为．8．（3分）从由0，1，2，3，4，5这六个数字组成的没有重复数字的四位数中任取一个，所取到的数大于3400的概率等于（结果用最简分数作答）9．（3分）设正整数m满足：二项式的展开式含有x的一次项．若将满足条件的正整数m由从小到大排成一个数列{a n}，则此数列的第2017项a2017=．10．（3分）定义在D上的函数f（x），若存在x0∈D，对于任意的x∈D都有f（x）≤f（x0）成立，则称x0为函数f（x）的极大值点，若函数f（x）=sinωx（ω＞0）在区间（0，π）上恰好有三个极大值点，则ω的取值范围是．11．（3分）正方体ABCD﹣A1B1C1D1的棱长为1，P是正方体（包括表面）中的动点，且满足，则点P所形成的几何体的体积等于．12．（3分）设[x]表示不超过x的最大整数，如[1.5]=1，[﹣1.5]=﹣2．若函数（a ＞0，a≠1），则g（x）=[f（x）﹣]+[f（﹣x）﹣]的值域为．二.选择题13．（3分）给出下列命题：（1）若奇函数存在反函数，则其反函数也是奇函数；（2）函数f（x）在区间[a，b]上存在反函数的充要条件是f（x）在区间[a，b]上是单调函数；（3）函数f（x）在定义域D上的反函数为f﹣1（x），则对于任意的x0∈D都有成立；其中正确的命题为（）A．（1）B．（1）（2）C．（1）（3）D．（1）（2）（3）14．（3分）已知a1、a2、b1、b2、c1、c2都是非零实数，集合A={x|a1x2+b1x+c1=0，x∈R}，B={x|a2x2+b2x+c2=0，x∈R}，则“A=B”是“”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15．（3分）在空间坐标系中，高为1，底面边长也为1的正三棱柱ABC﹣A1B1C1的侧棱AA1与z轴平行，设此三棱柱的左视图的面积为S，则S的最大值与最小值之差为（）A．B．C．D．16．（3分）已知，B={（x，y）|（x﹣1）2+（y﹣1）2≤1}，则A∩B所表示的平面区域的面积等于（）A．B．C．D．三.解答题17．如图，直三棱柱ABC﹣A1B1C1中，∠BAC=90°，AB=AC=λAA1，M、N分别是A1B与B1C1的中点；（1）求证：MN∥平面ACC1A1；（2）是否存在λ的值，使得MN与AC所成角为？若存在，求出λ的值，若不存在，说明理由．18．已知向量，，设函数的图象关于直线x=π对称，其中ω、λ为常数，且；（1）求函数f（x）的最小正周期；（2）若y=f（x）的图象经过点，求函数f（x）在区间上的取值范围．19．函数y=f（x）的定义域为D，若存在x0∈D，使得f（x0）=x0成立，则称x0为函数y= f（x）的“不动点”．（1）若f（x）=ax2+bx+c（a≠0）有两个不动点﹣1、3，求bc的最小值；（2）若a＞0，且f（x）=ax2+bx+c有两个不动点m、n满足：，求证：当x∈（0，m）时，f（x）＜m．20．已知圆M：x2+（y﹣2）2=1，Q是x轴上的动点，QA、QB分别切圆M于A、B两点；（1）若Q（1，0），求直线QA、QB的方程；（2）设|MQ|=t，用t表示∠AQB的余弦值，并求的最小值；（3）若，试求直线MQ的方程．21．已知点P1（a1，b1）、P2（a2，b2）、…、P n（a n，b n）（n∈N*），都在函数f（x）=p x（p ＞0，p≠1）的图象上；（1）若数列{a n}是等差数列，求证：数列{b n}是等比数列；（2）设p＞1，函数f（x）的反函数为f﹣1（x），若函数y=f（x）与函数f﹣1（x）的图象有公共点M，求证：M在直线y=x上；（3）设，a n=n（n∈N*），过点P n、P n+1的直线l n与两坐标轴围成的三角形面积为c n，问：数列{c n}是否存在最大项？若存在，求出最大项的值，若不存在，请说明理由．【参考答案】一.填空题1．{2，3，4}【解析】∵集合A={1，2，3，4，5}，B={x||x﹣3|≤1}={x|2≤x≤4}，∴A∩B={2，3，4}．故答案为：{2，3，4}．2．x﹣3y+1=0【解析】∵直线3x+y﹣1=0的斜率为﹣3，直线l与直线3x+y﹣1=0垂直，∴直线l的斜率为，又直线l过点M（2，1），∴直线l的方程为：y﹣1=（x﹣2），即x﹣3y+1=0．故答案为：x﹣3y+1=0．3．1【解析】tanα=2，则===1．故答案为：1．4．5【解析】等差数列{a n}的公差为d，前n项和为S n，若d＞0，且S10=0，可得10a1+45d=0，即a1=﹣d，则a n=a1+（n﹣1）d=（n﹣）d，d＞0，a n＜0，可得n＜，则n的最大值为5．故答案为：5．5．【解析】由|log2x+4i|≥5，得，即，∴log2x≤﹣3①或log2x≥3②，解①得：0，解②得x≥8．∴x的取值范围是：．故答案为：．6．【解析】∵集合M={（x，y）|x﹣y+1≥0且x+y﹣2≤0且y≥0}，∴集合M表示的平面区域是如图所示的△BDC，∵A点坐标为（0，﹣1），P∈M，|AP|min=PO=1，联立，得C（），|AP|max=|AC|=，∴|AP|最大值与最小值之差为：．故答案为：．7．【解析】根据题意，椭圆的焦点为（±2，0），则要求双曲线的焦点在x轴上，且c=2，设要求双曲线的方程为：﹣=1，则有a2+b2=c2=4，其渐近线方程为：y=±x，即ay±bx=0，又由焦点到渐近线距离等于1，则有=1，即b=1；则a2=c2﹣b2=3；则双曲线的标准方程为：；故答案为：．8．【解析】由0，1，2，3，4，5这六个数字组成的没有重复数字的四位数，能组成：=300个，其中大于3400的有：+=144，从由0，1，2，3，4，5这六个数字组成的没有重复数字的四位数中任取一个，基本事件总数n=300，所取到的数大于3400包含的基本事件个数m=144，∴所取到的数大于3400的概率p==．故答案为：．9．6050【解析】二项式的展开式的通项公式公式为C m r x2m﹣3r，令2m﹣3r=1，则r=，则当m=2时，r=1满足，当m=5，则r=3，当m=8，则r=5，故满足条件的正整数m由从小到大排成一个数列{a n}为2，5，8，11，…，故其通项公式为a n=2+3（n﹣1）=3n﹣1，则a2017=3×2017﹣1=6050，故答案为：605010．【解析】由题意，恰好有三个极大值点，可得，且，解得：，∵T=，∴＜ω，故答案为：．11．【解析】如图所示，B（0，0，0），A（0，1，0），A1（0，1，1）．设P（x，y，z），（x，y，z∈[0，1]）．则•=1﹣y．∵，∴，解得，x，z∈[0，1]．∴点P所形成的几何体的体积=V正方体=．故答案为：．12．{0，﹣1}【解析】=∈（0，1）∴f（x）﹣∈（﹣，）[f（x）﹣]=0 或﹣1∵f（﹣x）=∈（0，1）∴f（﹣x）﹣∈（，）则[f（﹣x）﹣]=﹣1或0∴g（x）=[f（x）﹣]+[f（﹣x）﹣]的值域为{0，﹣1}故答案为：{0，﹣1}二.选择题13．A【解析】（1）设奇函数f（x）的反函数为f﹣1（x），∵f（x）是奇函数，∴f（x）的值域关于原点对称，即f﹣1（x）的定义域关于原点对称．假设f（x）=y，则f（﹣x）=﹣y．∴f﹣1（y）=x，f﹣1（﹣y）=﹣x．∴f﹣1（﹣y）=﹣f﹣1（y），即f﹣1（﹣x）=﹣f﹣1（x）∴f﹣1（x）是奇函数故（1）正确；（2）函数f（x）在区间[a，b]上存在反函数不一定f（x）在区间[a，b]上是单调函数，比如f（x）=存在反函数，但f（x）在R上不单调，故（2）不正确；（3）函数f（x）在定义域D上的反函数为f﹣1（x），则对于任意的x0∈D都有不一定成立，其中x0不一定属于f（x）的值域，即f﹣1（x0）不一定存在，故（3）不正确．故选A．14．B【解析】A=B=∅时，“”不一定成立，不是充分条件，“”时，A=B，是必要条件，故选：B．15．C【解析】空间坐标系中，高为1，底面边长也为1的正三棱柱ABC﹣A1B1C1的侧棱AA1与z 轴平行，设此三棱柱的左视图的面积为S，当AB与x轴平行时，此时左视图的面积最大，最大为1×1=1，当底面三角形的高与x轴成45°时，此时左视图的面积最小，最大为××1=．故S的最大值与最小值之差为，故选：C16．B【解析】由B={（x，y）|（x﹣1）2+（y﹣1）2≤1}，可知集合B表示的图形是以（1，1）为圆心，1为半径的圆面．由（y﹣x）（y﹣）≥0，得，或．∴集合A∩B所表示的平面图形如图所示的阴影部分：由于圆和函数y=的对称性可知：圆面的阴影部分的面积和剩下的部分面积相等，故S阴影=π．故选：B三.解答题17．证明：（1）∵直三棱柱ABC﹣A1B1C1中，∠BAC=90°，AB=AC=λAA1，M、N分别是A1B 与B1C1的中点，∴以A为原点，AB为x轴，AC为y轴，AA1为z轴，建立空间直角坐标系，设AB=AC=λAA1=λ，则M（，0，），N（，，1），=（0，，），平面ACC1A1的法向量=（1，0，0），∵•=0，MN⊄平面ACC1A1，∴MN∥平面ACC1A1．（2）假设存在λ的值，使得MN与AC所成角为．A（0，0，0），C（0，λ，0），=（0，λ，0），∵MN与AC所成角为，∴cos==，由λ＞0，解得λ=1．∴存在λ=1，使得MN与AC所成角为．18．解：（1）由函数，可得：f（x）=sin2ωx﹣cos2ωx+2sinωx cosωx+λ=sin2ωx﹣cos2ωx=2sin（2ωx）+λ．图象关于直线x=π对称，即x=π时，f（x）取得最值，∴2πω=，k∈Z．∵；∴当k=0，可得ω=，∴f（x）=2sin（x）+λ．那么：周期T==3π．（2）由（1）可知f（x）=2sin（x）+λ．图象经过点，∴2sin（×）+λ=0可得λ=0∵x∈上，∴（x）∈[，]，∴≤sin（x）≤1故得函数f（x）在区间上的取值范围是[﹣1，2]．19．解：（1）∵f（x）=ax2+bx+c（a≠0）有两个不动点﹣1、3，∴，消去a可得：3b﹣2c=3，即c=，∴bc==（b﹣）2﹣，∴当b=时，bc取得最小值﹣．（2）∵f（x）=x有两个解m，n，ax2+（b﹣1）x+c=0的两解为m，n，∴m+n==﹣，∵0＜m，∴m+n＜m+，即﹣＜m+，∴﹣＜，又a＞0，即f（x）的图象开口向上，∴当x∈（0，m）时，f（x）＜f（m）=m．20．解：（1）设过点Q的圆M的切线方程为x=my+1，则圆心M到切线的距离为1，∴=1，∴|2m+1|=，平方得4m2+4m+1=1+m2，即3m2+4m=0，得m=﹣或0，∴QA，QB的方程分别为x=1，；即3x+4y﹣3=0和x=1．（2）设∠AQB=2θ，则|QA|=|QB|==，sinθ==，则=||||cosθ=•（1﹣2sin2θ）=（t2﹣1）（1﹣）=t2+﹣3≥2﹣3=2﹣3，当且仅当t2=，即t2=时取等号，即的最小值是2﹣3；（3）设AB与MQ交于P，则MP⊥AB，MB⊥BQ，∴|MP|====．在Rt△MBQ中，|MB|2=|MP||MQ|，即1=|MQ|，∴|MQ|=3，∴x2+（y﹣2）2=9．设Q（x，0），则x2+22=9，则x2=5，∴x=±，∴Q（±，0），∴MQ的方程为2x+y﹣2=0或2x﹣y+2=0．21．（1）证明：由题意可得：b n=＞0，a n+1﹣a n=d常数．∴===p d＞0，∴数列{b n}是等比数列，首项为，公比为p d．（2）证明：f（x）=p x（p＞1），函数f（x）的反函数为f﹣1（x）=log p x，设函数y=f（x）与函数f﹣1（x）的图象有公共点M（a，b），则b=p a，b=log p a即a=p b．∴=p a﹣b，a＜b或a＞b时都不成立，∴必有a=b．因此M在直线y=x上．（3），a n=n（n∈N*），b n=．===﹣．直线l n：y﹣=（x﹣n），与坐标轴的交点A（，0），B（0，n+1）．c n=×（n+1）=．c n+1﹣c n=（2﹣n2）．∴c1＜c2＞c3＞c4＞…．∴（c n）max=c2=．。
上海市南洋模范中学2024届高三上学期10月月考数学试题
上海市南洋模范中学2024届高三上学期10月月考数学试题学校:___________姓名：___________班级：___________考号：___________二、多选题13．有一组样本数据1x ，2x ，…，6x ，其中1x 是最小值，6x 是最大值，则（ ）A ．2x ，3x ，4x ，5x 的平均数等于1x ，2x ，…，6x 的平均数B ．2x ，3x ，4x ，5x 的中位数不等于1x ，2x ，…，6x 的中位数C ．2x ，3x ，4x ，5x 的标准差不小于1x ，2x ，…，6x 的标准差D ．2x ，3x ，4x ，5x 的极差不大于1x ，2x ，…，6x 的极差假设存在常数0M >，使得n a M <恒成立，取[]1m M =+，其中[]1M M M -<£，且[]Z M Î，因为11n n a a +>+，所以[][]213211,1,,1M M a a a a a a +>+>+>+L ，上式相加得，[][]1191M aa M M M +>+>+->，则[]1m M a a M +=>，与n a M <恒成立矛盾，故D 错误.故选：B.【点睛】关键点睛：本题解决的关键是根据首项给出与通项性质相关的相应的命题，再根据所得命题结合放缩法得到通项所满足的不等式关系，从而可判断数列的上界或下界是否成立.17．(1)证明见解析(2)30°【分析】（1）以点D 为坐标原点，DA 为x 轴，DC 为y 轴，1DD 为z 轴建立空间直角坐标系，求出MN uuuu r和平面11BCC B 的法向量，利用空间向量证明即可，（2）求出平面11A B CD 的法向量，利用空间向量求解即可.【详解】（1）如图，以点D 为坐标原点，DA 为x 轴，DC 为y 轴，1DD 为z 轴建立空间直角坐标系．。
上海市南洋模范中学高一上学期10月月考数学试题(解析版)
上海市南洋模范中学高一上学期10月月考数学试题一、单选题1．不等式1a b a b +≤+成立的充要条件是（ ）A ．0ab ≠B ．220a b +≠C ．0ab >D ．0ab < 【答案】B【解析】由于1a b a b +≤+，可得出0a b +≠，进而可得出220a b +≠，由此可得出+≤+a b a b ，在所得不等式两边平方化简后得出ab ab ≤，进而可得出结论.【详解】 由于1a b a b +≤+，则0a b +≠，即a 、b 不同时为零，即220a b +≠，则0a b +>. 由1a b a b +≤+可得+≤+a b a b ，不等式两边平方可得222222a ab b a ab b ++≤++， 即ab ab ≤，显然ab ab ≤恒成立， 因此，不等式1a b a b +≤+成立的充要条件是220a b +≠.故选：B.【点睛】 本题考查充要条件的寻找，考查分析问题和解决问题的能力，属于中等题.2．x 为实数，且|5||3|x x m -+-<有解，则m 的取值范围是（ ）A ．1mB ．m 1≥C ．2m >D ．2m ≥【答案】C【解析】求出|x ﹣5|+|x ﹣3|的最小值，只需m 大于最小值即可满足题意．【详解】 53x x m -+-<有解，只需m 大于53x x -+-的最小值，532x x -+-≥，所以2m >，53x x m -+-<有解．故选C ．【点睛】本题考查绝对值不等式的解法，考查计算能力，是基础题．3．已知关于x 的不等式0ax b +>的解集是(1,)+∞，则关于x 的不等式02ax b x ->-的解集是（ ） A ．{|1x x <-或2}x >B ．{|12}x x -<<C ．{|12}x x <<D ．{|2}x x > 【答案】A【解析】由题意可得0a >，且1b a -=，进而可得=-b a ，代入不等式解分式不等式即可求解. 【详解】因为不等式0ax b +>的解集是(1,)+∞，可得0a >，且1b a-=， 所以=-b a ， ()()0012022ax b ax a a x x x x -+>⇒>⇒+->-- ()()1202x x x ⇒+->⇒>或1x <-，所以不等式的解集为{|1x x <-或2}x >.故选：A【点睛】本题考查了分式不等式的解法，考查了基本运算求解能力，属于基础题.4．不等式组03232x x x x x >⎧⎪--⎨>⎪++⎩的解集是（ ） A ．{}02x x <<B ．{}0 2.5x x << C．{0x x <<D ．{}03x x << 【答案】C【解析】原不等式组等价于0303323323x x xx x x x x x⎧⎪>⎪-⎪>⎨+⎪---⎪<<⎪+++⎩，解出该不等式组可得出解集.【详解】 原不等式组等价于0303323323x x x x x x x x x ⎧⎪>⎪-⎪>⎨+⎪---⎪<<⎪+++⎩. 解不等式303x x->+，即()()330x x -+<，解得33x -<<，03x ∴<<； 解不等式2323x x x x--<++，即()()2023x x x >++，即()()230x x x ++>， 解得32x -<<-或0x >，此时，0x >； 解不等式3232x x x x --<++，即()()()226023x x x -<++，即()()()26230x x x -++<，解得3x -<<2x -<<0x <<综上所述，原不等式组的解集为{0x x <<，故选C. 【点睛】 本题考查分段不等式的解集，同时也考查了利用穿根法求解高次不等式，考查运算求解能力，属于中等题.二、填空题5．已知集合{|(4)0}M x x x =-<，{|(1)(6)0,}N x x x x =--<∈Z ，则MN =________【答案】{5}【解析】利用一元二次不等式的解法求出集合,M N ，再根据集合的交运算即可求解.【详解】 {{|(4)0}4M x x x x x =-<=>或}0x <，{}{}{|(1)(6)0,}16,2,3,4,5N x x x x Z x x x Z =--<∈=<<∈=，所以{}5M N =.故答案为：{5}【点睛】本题考查了集合的基本运算、一元二次不等式的解法，考查了基本运算求解能力，属于基础题.6．不等式112x <的解集是____________． 【答案】【解析】【详解】 解：111120,0,(2)0,0222x x x x x x x-<∴-<∴<∴->∴<，或2x > 7．不等式514x x -≥+的解集为________. 【答案】14,2⎛⎤- ⎥⎝⎦ 【解析】利用分式不等式的解法，求得原不等式的解集.【详解】 55121100444x x x x x x ---≥⇒-≥⇒≥+++ ()()124040x x x ⎧-+≥⇒⎨+≠⎩142x ⇒-<≤， 所以原不等式的解集为14,2⎛⎤- ⎥⎝⎦. 故答案为：14,2⎛⎤- ⎥⎝⎦【点睛】本小题主要考查分式不等式的解法，属于基础题.8．不等式()()()()2321120x x x x ++--≤的解集为________ 【答案】(]{}[],211,2-∞--【解析】根据题意作出数轴，将各个因式等于零的值标记在数轴上，然后采用“穿针引线法”求解出不等式的解集.【详解】如下图所示：根据图象可知：当2x -≤或1x =-或12x ≤≤时，()()()()2321120x x x x ++--≤， 所以不等式的解集为：(]{}[],211,2-∞--， 故答案为：(]{}[],211,2-∞--.【点睛】 本题考查高次不等式的解法，难度一般.利用“穿针引线法”求解高次不等式的解集时，注意从数轴的右上方开始，每经过一个因式对应的数轴上点，要判断该因式是奇次还是偶次，如果是奇次，则穿过该点，如果是偶次，则选择穿而不过.9．若不等式ax 2-bx +c ＜0的解集是{|23}x x -<<，则不等式bx 2+ax +c ＜0的解集是______【答案】（-3，2）【解析】由题分析得b ＞0，且a b =1，c b=-6，再解一元二次不等式得解. 【详解】∵不等式ax 2-bx +c ＜0的解集是（-2，3），∴a ＞0，且对应方程ax 2-bx +c =0的实数根是-2和3， 由根与系数的关系，得2323c a b a⎧=-⨯⎪⎪⎨⎪=-+⎪⎩， 即c a =-6，b a=1, ∴b ＞0，且a b =1，c b =-6， ∴不等式bx 2+ax +c ＜0可化为x 2+x -6＜0，解得-3＜x ＜2；∴该不等式的解集为（-3，2）．故答案为（-3，2）.【点睛】本题主要考查一元二次不等式的解的求法和应用，意在考查学生对这些知识的理解掌握水平.10．已知{||23|}A x x a =-<，{|||10}B x x =≤，且A B ，则实数a 的取值范围________【答案】(,17]-∞【解析】求出集合A 、B ，由A B ，讨论A =∅或A ≠∅，再由集合的包含关系即可求解.【详解】{||23|}A x x a =-<，{}{|10}1010B x x x x =≤=-≤≤，由A B ，当0a ≤时，A =∅满足题意；当0a >时，332322a a x a x -+-<⇒<<， 因为A B ， 所以310231001720a a a a -⎧≥-⎪⎪+⎪≤⇒<≤⎨⎪>⎪⎪⎩， 综上所述，实数a 的取值范围为(,17]-∞.【点睛】本题考查了集合的包含关系求参数的取值范围、绝对值不等式的解法，属于基础题.11．关于x 的方程2(3)3m x m x -+=的解为不大于2的实数，则m 的取值范围________ 【答案】3(,](0,1)(1,)2-∞-+∞【解析】讨论m 的取值，当0m =或1m =或0m ≠且1m ≠时，根据题意可得()()31321m x m m m--==-≤-，解分式不等式即可求解. 【详解】由2(3)3m x m x -+=可得：()233m m x m -=-+，若0m =，不成立；1m =，解得x ∈R ，不成立；若0m ≠且1m ≠时，则()()31321m x m m m--==-≤-， 即230m m +≥，可化为()2300m m m ⎧+≥⎨≠⎩， 解得0m >或32m ≤-， 综上，m 的取值范围为3(,](0,1)(1,)2-∞-+∞. 故答案为：3(,](0,1)(1,)2-∞-+∞【点睛】 本题考查了分式不等式的解法，考查了分类讨论的思想以及基本运算求解能力，属于基础题.12．若不等式()2211x m x ->-对满足22m -≤≤的所有m 都成立，则x 的取值范围是_________.【答案】11,22⎛-++ ⎝⎭【解析】将不等式()2211x m x ->-化为含参数x 的m 的一次不等式()21(21)0m x x ---<，再令()2()1(21)f m m x x =---，只要(2)0,(2)0f f -<<即可． 【详解】不等式化为：()21(21)0m x x ---<，令()2()1(21)f m m x x =---，则22m -≤≤时，()0f m <恒成立， 所以只需(2)0(2)0f f -<⎧⎨<⎩，即()()2221(21)021(21)0x x x x ⎧----<⎪⎨---<⎪⎩， 所以x的范围是1122x ⎛⎫-+∈ ⎪ ⎪⎝⎭，故答案为：⎝⎭. 【点睛】本题主要考查将一元二次不等式转化为一元一次不等式进行求解的问题，是基础题.13．已知集合2{|540}A x x x =-+≤，2{|220}B x x ax a =-++≤，若B A ⊆，则a 的取值范围________ 【答案】18(1,]7- 【解析】求出集合A 、B ，再由B A ⊆，讨论B =∅或B ≠∅，根据集合的包含关系即可求解.【详解】 由集合{}2{|540}14A x x x x x =-+≤=≤≤，2{|220}B x x ax a =-++≤， 若B A ⊆，若B =∅，可得()()222424480a a a a ∆=--+=--<，解得1a 2-<<，若B ≠∅，()()222424480a a a a ∆=--+=--≥，可得2a ≥或1a ≤-，{B x a x a =≤+，则42a a ⎧+⎪⎨-≥⎪⎩①②， 解不等式①可得187a ≤， 解不等式②可得13a ≤≤，取交集得1817a ≤≤， 又0∆≥，可得2a ≥或1a ≤-， 可得1827a ≤≤， 经验证，当187a =符合题意； 当2a =符合题意；1827a ∴≤≤， 综上所述，1817a -<≤， 故答案为：18(1,]7-. 【点睛】 本题考查了由集合的包含关系求参数的取值范围，考查了基本运算求解能力，属于基础题.14．已知不等式222xy ax y ≤+对于[1,2]x ∈，[2,3]y ∈恒成立，则a 的取值范围________【答案】[1,)-+∞ 【解析】分离参数可得22y y a x x ⎛⎫≥- ⎪⎝⎭，令y t x =，则13t ≤≤，再利用二次函数配方求最值，只需2max2y y a x x ⎡⎤⎛⎫≥-⎢⎥ ⎪⎝⎭⎢⎥⎣⎦即可. 【详解】由题意可知：不等式222xy ax y ≤+对于[1,2]x ∈，[2,3]y ∈恒成立， 即22y y a x x ⎛⎫≥- ⎪⎝⎭对于[1,2]x ∈，[2,3]y ∈恒成立， 令y t x=，则13t ≤≤， 22a t t ∴≥-在[]1,3上恒成立，2112248y t t t ⎛⎫=-+=--+ ⎪⎝⎭， max 1y ∴=-，1a ∴≥-，故答案为：[1,)-+∞【点睛】本题考查了不等式恒成立问题以及二次函数的性质，考查了分离参数法，属于基础题.三、解答题15．已知()()2366f x x a a x =-+-+. (1)解关于a 的不等式()10f >；(2)若不等式()f x b >的解集为()1,3-，求实数,a b 的值．【答案】（1）{|33a a -<+；（2）33a b ⎧=⎪⎨=-⎪⎩【解析】(1)由f (1)＝－3＋a (6－a )＋6＝－a 2＋6a ＋3，得a 2－6a －3<0，求解即可；（2）f (x )>b 的解集为(－1,3)等价于方程－3x 2＋a (6－a )x ＋6－b ＝0的两根为－1,3，由根与系数的关系求解即可.【详解】(1)∵f (x )＝－3x 2＋a (6－a )x ＋6，∴f (1)＝－3＋a (6－a )＋6＝－a 2＋6a ＋3，∴原不等式可化为a 2－6a －3<0，解得3－a <3＋∴原不等式的解集为{a |3－<a <3＋(2)f (x )>b 的解集为(－1,3)等价于方程－3x 2＋a (6－a )x ＋6－b ＝0的两根为－1,3， 等价于()61+3=3613=3a a b ⎧--⎪⎪⎨-⎪-⨯-⎪⎩解得33a b ⎧=±⎪⎨=-⎪⎩16．已知a ∈R ，解关于x 的不等式1(1)x a x x-≥-. 【答案】当1a =，(,0)[1,)-∞⋃+∞；当1a <，1[,0)[1,)1a +∞-；当12a <≤，1(,0)[1,]1a -∞-；当2a >，1(,0)[,1]1a -∞-. 【解析】将不等式化为()()1110x a x x --+⎡⎤⎣⎦≥，讨论a 的取值范围，当1a =或1a <或12a <≤或2a >，解分式不等式即可求解.【详解】()()()2111111(1)00x a x a x ax x a x x x x--+⎡⎤-+-⎣⎦-≥-⇒≥⇒≥，（*） （1）当1a =时，（*）式为10x x-≥， 解得0x <或1≥x ， （2）当1a ≠时，（*）式为()()11110a x x a x⎛⎫--+ ⎪-⎝⎭≥， ①若1a <，则10a -<，101a <-， 解得101x a ≤<-或1≥x ； ②若12a <≤，则10a -<，111a ≥-， 解得0x <，或111x a ≤≤-， ③若2a >，则11a ->，1011a <<-， 解得0x <，或111x a ≤≤-，， 综上所述，当1a =，(,0)[1,)-∞⋃+∞；当1a <，1[,0)[1,)1a +∞-；当12a <≤，1(,0)[1,]1a -∞-；当2a >，1(,0)[,1]1a -∞-. 【点睛】 本题考查了分式不等式的解法，注意高次不等式中“穿针引线”法的应用，考查了分类讨论的思想，属于基础题.17．已知a 是实数, 关于x 的方程22230ax x a +--=在区间[]1,1-上有实根, 求a 的取值范围.【答案】[)37,1,2⎛⎫---∞+∞ ⎪ ⎪⎝⎭．【解析】【详解】 （1）当0a =时,()23f x x =-, 令230x -=得[]31,12x =∉-, ()f x ∴在[]1,1-上无零点, 故0a ≠.（2）当0a >时,()2223f x ax x a =+--的对称轴为12x a=-. ① 当112a-≤-,即102a <≤时,须使()()10{10f f -≤≥,即5{,1a a a ≤∴≥的解集为∅. ②当1102a -<-<,即12a >时,须使,即130{21a a a ---≤≥,解得1a ≥, a ∴的取值范围是[)1,+∞. （3）当0a <时, ① 当1012a <-≤,即12a ≤-时,须有,即5{1302a a a≤---≥, 解得37a --≤375a -+≤≤,又1.2a a ≤-∴的取值范围是37,2⎡---∞⎢⎣⎭. ②当112a ->时,即102a -<<时,须有()()10{10f f -≤≥,即5{1a a ≤≥,a ∴的解集为∅. 综上所述 ,a 的取值范围是[)371,⎛---∞+∞ ⎝⎭.18．已知1S 、2S 、3S 为非空整数集合，对于1、2、3的任意一个排列i 、j 、k ，若i x S ∈，j y S ∈，则k x y S -∈.（1）证明：三个集合中至少有两个相等；（2）三个集合中是否可能有两个集合无公共元素？说明理由.【答案】（1）证明见解析；（2）可能，如12{S S ==奇数}，3{S =偶数}【解析】（1）由题意三个集合中的元素都为零时，成立；不妨设1a S ∈，b 为2S 、3S 中最小的非负元素，若0b >，可得0b a -≥的取法矛盾，即证.（2）举特例比如12{S S ==奇数}，3{S =偶数}即可证出.【详解】（1）若i x S ∈，j y S ∈，则k x y S -∈，所以每个集合中均有非负元素，当三个集合中的元素都为零时， 命题显然成立，否则，设1S 、2S 、3S 中的最小正元素为a ， 不妨设1a S ∈，设b 为2S 、3S 中最小的非负元素，不妨设2b S ∈，则3b a S -∈，若0b >，则0b a -≥的取法矛盾，所以0b =，任取1x S ∈，因20S ∈，故30x x S -=∈，所以1S 包含3S ，同理3S 包含1S ，所以12S S .（2）可能，比如12{S S ==奇数}，3{S =偶数}，这时1S 与3S ，2S 与3S 都无公共元素.【点睛】本题考查了元素与集合的关系，考查了考生的分析能力，属于中档题.。
2018届上海市位育中学高三上学期10月月考数学试题(word版)
2018届上海市位育中学高三上学期10月月考数学试题（word 版）2017.10一. 填空题1. 设集合2{|}M x x x ==，{|lg 0}N x x =≤，则MN = 2. 已知1tan 3α=-，则sin 2α=3. 函数y =的定义域是4. 函数()cos (sin )222xx x f x =-的最小正周期是 5. 设{|||2}A x x a =-<，21{|1}2x B x x -=<+，若A B ⊆，则a 的取值范围是 6. 函数213x y -=（10x -≤≤）的反函数是7. 已知0x >，0y >，1211x y +=+，则x y +的最小值为 8. 若212log (42)0ax x a -+-<对任意x R ∈恒成立，则实数a 的取值范围是9. 设函数()min{||,||}f x x x t =+的图像关于直线3x =-对称，其中min{,}a b 表示a 、b 中的最小值，则实数t =10. 已知函数10()42x x x x f x x --⎧+>=⎨-≤⎩，若函数(32)y f x a =--恰有三个不同的零点，则实数a 的取值范围是 11. 在锐角ABC ∆中，2A B ∠=∠，B ∠、C ∠的对边长分别是b 、c ，则b bc +的取值范围是 12. 在三角形ABC 中，若220AB BC AB ⋅+=，且2b =，一个内角为30°，则ABC ∆的面积为二. 选择题13. 函数41()2x xf x +=的图像（ ） A. 关于原点对称 B. 关于直线y x =对称C. 关于x 轴对称D. 关于y 轴对称14. 若0x 是方程131()2x x =的解，则0x 属于区间（ ） A. 2(,1)3 B. 12(,)23 C. 11(,)32 D. 1(0,)315. 已知函数()3sin()6f x x πω=-（0ω>）和()3cos(2)g x x ϕ=+的图像的对称中心完 全相同，若[0,]2x π∈，则()f x 的取值范围是（ ） A. 3[,3]2- B. 3[,3]2 C. 3[,1]2- D. [2,3]-16. 已知函数21()log ()2a f x ax x =-+（0a >且1a ≠）在[1,3]上恒正，则实数a 的取值范围为（ ） A. 183(,)(,)292+∞ B. 3(,)2+∞ C. 18(,)29 D. 13(,)22三. 解答题 17. 已知函数()||f x x a =-，2()21g x x ax =++（a 为正常数），且函数()f x 与()g x 的图像在y 轴上的截距相等；（1）求a 的值；（2）若()()h x f x =+b 为常数），试讨论函数()h x 的奇偶性；18. 已知ABC ∆中，三个内角A 、B 、C 满足下列等式：sin sin sin (cos cos )B C A B C +=+；（1）求A 的度数；（2）若ABC ∆面积为4，求ABC ∆的周长的最小值；19.（1）a 取何值时，方程22sin 2sin cos cos x x x x a +-=（[0,]x π∈）无解？有一解？有两解？有三解？（2）函数的性质通常指函数的定义域、值域、周期性、单调性、奇偶性等，请选择适当的探究顺序，研究函数()f x =的性质，并在此基础上，作出其在[,]ππ-的草图；20. 已知1()log 1amx f x x -=-是奇函数（其中0a >，1a ≠）； （1）求m 的值；（2）讨论()f x 的单调性；（3）当()f x 的定义域区间为(1,2)a -时，()f x 的值域为(1,)+∞，求a 的值；21. 已知函数421()421x x x x k f x +⋅+=++； （1）若对任意的x R ∈，()0f x >恒成立，求实数k 的取值范围；（2）若()f x 的最小值为3-，求实数k 的值；（3）若对任意实数1x 、2x 、3x ，均存在以1()f x 、2()f x 、3()f x 为三边边长的三角形，求实数k 的取值范围；参考答案一. 填空题1. [0,1]2. 35- 3. 7(2,2)()66k k k Z ππππ-++∈ 4. 2π5. [0,1]6. 11()[,1]3f x x -=∈7. 2+4a >9. 6 10. (2,3] 11. 11(,)3212. 1二. 选择题13. D 14. C 15. A 16. B三. 简答题17.（1）1a =；（2）1b =-时，()h x 为奇函数；1b =时，()h x 为偶函数；1b ≠±时，()h x 为非奇非偶函数；18.（1）2A π=；（2）4+19.（1）(,(2,)a ∈-∞+∞时，无解；a =(1)(1,2)a ∈--时，有两解；1a =-时，有三解；（2）定义域为R ，值域为，周期为π，在[,,]2k k πππ-+为增函数，在[,]2k k πππ+ 上为减函数，偶函数；20.（1）1m =-；（2）当01a <<时，()f x 在(,1)-∞-或(1,)+∞上为增函数；当1a >时，()f x 在(,1)-∞-或(1,)+∞上为减函数；（3）2a =；21.（1）2k >-；（2）11k =-；（3）1[,4]2k ∈-；。
上海市南洋模范中学2018-2019学年高三上学期第三次月考试卷数学含答案
上海市南洋模范中学2018-2019学年高三上学期第三次月考试卷数学含答案 班级__________ 座号_____ 姓名__________ 分数__________一、选择题（本大题共12小题，每小题5分，共60分.每小题给出的四个选项中，只有一项是符合题目要求的.）1． 已知e 为自然对数的底数，若对任意的1[,1]x e∈，总存在唯一的[1,1]y ∈-，使得2ln 1y x x a y e -++= 成立，则实数a 的取值范围是（ ）A.1[,]e eB.2(,]e eC.2(,)e +∞D.21(,)e e e+【命题意图】本题考查导数与函数的单调性，函数的最值的关系，函数与方程的关系等基础知识，意在考查运用转化与化归思想、综合分析问题与解决问题的能力．2． 在数列{}n a 中，115a =，*1332()n n a a n N +=-∈，则该数列中相邻两项的乘积为负数的项是 （ ）A ．21a 和22aB ．22a 和23aC ．23a 和24aD ．24a 和25a 3． 沿一个正方体三个面的对角线截得几何体如图所示，则该几何体的侧视图为（ ）A ．B ．C ．D ．4． 函数f （x ）＝sin （ωx ＋φ）（ω＞0，－π2≤φ≤π2）的部分图象如图所示，则φω的值为（ ）A.18 B ．14C.12D ．15． 已知圆M 过定点)1,0(且圆心M 在抛物线y x 22=上运动，若x 轴截圆M 所得的弦为||PQ ，则弦长||PQ 等于（ ）A ．2B ．3C ．4D ．与点位置有关的值【命题意图】本题考查了抛物线的标准方程、圆的几何性质，对数形结合能力与逻辑推理运算能力要求较高，难度较大.6． 已知抛物线C ：24y x =的焦点为F ，定点(0,2)A ，若射线FA 与抛物线C 交于点M ，与抛 物线C 的准线交于点N ，则||:||MN FN 的值是（ ）A ．B ．C ．1:D (1 7． 如图是某几何体的三视图，正视图是等腰梯形，俯视图中的曲线是两个同心的半圆组成的半圆环，侧视图是直角梯形．则该几何体表面积等于（ ）A ．12+B ．12+23πC ．12+24πD ．12+π8． 设函数()''y f x =是()'y f x =的导数.某同学经过探究发现，任意一个三次函数()()320f x ax bx cx d a =+++≠都有对称中心()()00,x f x ，其中0x 满足()0''0f x =.已知函数()3211533212f x x x x =-+-，则1232016...2017201720172017f f f f ⎛⎫⎛⎫⎛⎫⎛⎫++++= ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭（ ）A ．2013B ．2014 C ．2015 D ．20161111] 9． 集合{}{}2|ln 0,|9A x x B x x =≥=<,则AB =（ ）A ．()1,3B ．[)1,3C ．[]1,+∞D ．[],3e 10．“24x ππ-<≤”是“tan 1x ≤”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【命题意图】本题主要考查充分必要条件的概念与判定方法，正切函数的性质和图象，重点是单调性. 11．棱长为2的正方体被一个平面截去一部分后所得的几何体的三视图如图所示，则该几何体的表面积为（ ）A ．B ．18C ．D ．12．由直线与曲线所围成的封闭图形的面积为（ ）A B1C D二、填空题（本大题共4小题，每小题5分，共20分.把答案填写在横线上）13．函数()2log f x x =在点()1,2A 处切线的斜率为 ▲ ．14．1F ，2F 分别为双曲线22221x y a b-=（a ，0b >）的左、右焦点，点P 在双曲线上，满足120PF PF ⋅=，若12PF F ∆______________.【命题意图】本题考查双曲线的几何性质，直角三角形内切圆半径与外接圆半径的计算等基础知识，意在考查基本运算能力及推理能力．15．圆上的点（2，1）关于直线x+y=0的对称点仍在圆上，且圆与直线x ﹣y+1=0相交所得的弦长为，则圆的方程为 ．16．在（1+2x ）10的展开式中，x 2项的系数为 （结果用数值表示）．三、解答题（本大共6小题，共70分。
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2017-2018年上海市南模中学高三上十月月
考
-CAL-FENGHAI-(2020YEAR-YICAI)_JINGBIAN
2017学年南模中学高三年级十月份月考试卷
2017.10
一、填空题 1. 若α是第四象限角，则
2
2
α
π
-
所在象限是第________象限
2.
已知集合2|,,{|40}32A x k x k k Z B x x ππππ⎧⎫
=+≤<+∈=-≥⎨⎬⎩⎭
，则
A B =____________.
3. 函数3cos 3cos x
y x -=+的值域为____________.
4. 已知tan 2α=，则2sin cos αα=____________.
5.
函数12
()log cos 34x f x π⎛⎫
=+ ⎪⎝⎭的单调递增区间为____________.
6.
若()21(0)f x x ωω=+>在区间3,22ππ⎡⎤
-
⎢⎥⎣⎦
上为增函数，则ω的最大值是____________. 7.
有一个解三角形的题因纸张破损有一个条件不清，具体如下“在ABC 中，角,,A B C
所对的边分别为,,a b c
，已知4
a B π
==，__________. 求角A ”。
经推断破损处的条件为三
角形的一边的长度，且答案提示3
A π
=，试将条件补充完整
8.
函数2max min 2(1)sin (),(),()1x x
f x f x M f x m x ++=
==+，则M m +=____________. 9.
已知函数1()sin 2212f x x x =+，若2()log f x t ≥对x R ∈恒成立，则t 的取值范围为____________. 10.
设12,R αα∈，且
1211
22sin 2sin(2)
αα+=++，则12|10|παα--的最小值等于
____________. 11.
设,,[0,2)a b R c π∈∈，若对任意实数x 都有2sin 3sin()3x a bx c π⎛
⎫-=+ ⎪⎝
⎭，则满足条件
的有序实数组(,,)a b c 的组数为____________.
12. 关于x 的方程
11
|sin ||1|12
x x π=--在[2016,2016]-上解的个数是____________.
二、选择题 13. 在ABC 中，222sin sin sin sin sin A B C B C ≤+-，则角A 的取值范围是（ ）
A. (0,]6
π
B. [,)6
ππ
C. (0,]3
π
D.
[,)3
π
π
14. 把(cos 2sin 2)2
y x x =
-的图像作适当的移动得sin 2y x =的图像，这样的移动可以是（ ）
A. 向右平移
38π
个单位 B. 向左平移
38π
个单位 C. 向右平移34
π
个单位 D. 向左平移
34
π
个单位 15.
将函数1
y x
=-的图像按向量(1,0)a =平移，得到的函数图像与函数
2sin (24)y x x π=-≤≤的图像的所有交点的横坐标之和等于（ ） A. 2
B. 4
C. 6
D. 8
16. 存在函数()f x 满足：对任意x R ∈都有（ ） A. (sin 2)sin f x x = B. 2(sin 2)f x x x =+ C. 2(1)|1|f x x +=+
D. 2(2)|1|f x x x +=+
三、解答题 17.
已知函数221
()cos sin ,(0,)2
f x x x x π=-+∈
（1）求()f x 的单调递增区间；
（2）设ABC 为锐角三角形，角A 所对边a =，角B 所对边5b =，若()0f A =，求
ABC 的面积
18.
设函数2()cos 2sin 24f x x x π⎛
⎫=
++ ⎪⎝
⎭ （1）求函数()f x 的最小正周期；
（2）设函数()g x 对任意x R ∈，有()2g x g x π⎛⎫+= ⎪⎝⎭，且当0,2x π⎡⎤
∈⎢⎥⎣⎦时，1()()2g x f x =-，求
函数()g x 在[,0]π-上的解析式
19.
如图所示，湖面上甲、乙、丙三艘船沿着同一条直线航行，某一时刻，甲船在最前面
的A 点处，乙船在中间的B 点处，丙船在最后的C 点处，且:3:1BC AB =，一架无人机在空中的P 点处对它们进行数据测量，在同一时刻测得30,90APB BPC ∠=︒∠=︒（船只与无人机的大小及其他因素忽略不计）
（1）求此时无人机到甲、丙两船的距离之比；
（2）若此时甲、乙两船相距100米，求无人机到丙船的距离（精确到1米）
20.
如图，某园林单位准备绿化一块直径为BC 的半圆形空地，ABC 外的地方种草，
ABC 的内接正方体PQRS 为一水池，其余的地方种花，若,BC a ABC θ=∠=，设ABC 的面积为1S ，正方形的面积为2S （1）用,a θ表示1S 和2S ； （2）当a 固定，θ
变化时，求
1
2
S S 取最小值时的角θ
21.
已知函数2(),,;()(47)1y f x x D y A g x x x θ=∈∈=-+
（1）当()sin()f x x ϕ=+为偶函数时，求ϕ的值；
（2）当()sin 2263f x x x ππ⎛⎫⎛
⎫=++ ⎪ ⎪⎝⎭⎝⎭时，()g x 在A 上是单调递减函数，求θ的取值范
围；
（3）当1122()sin()sin()sin()n n f x a x a x a x ωϕωϕωϕ=++++++时，
（,1,2,3,
,,0i a R i n ω∈=>），若22(0)02f f πω⎛⎫+≠ ⎪⎝⎭
，且函数()f x 的图像关于点,02π⎛⎫ ⎪⎝⎭对
称，在x π=处取得最小值，试探讨论ω应该满足的条件
参考答案
一、填空题 1. 一、三
2. [2,)[,)232
πππ
--
3. 1,22⎡⎤
⎢⎥⎣⎦
4.
225
5. 36,6,4k k k Z πππ⎛
⎫+∈ ⎪⎝⎭
6. 14
7.
8. 2 9. [0,1]
10. 1,13⎡⎤⎢⎥⎣⎦
11. 4
12. 4031
二、选择题 13-16 CADD
三、解答题
17. （1）,2ππ⎛⎫
⎪⎝⎭
； （2
18. （1）π； （2）1sin 2,,22()1sin(2),(,0]22
x x g x x x πππ⎧⎡⎤∈--⎪⎢⎥⎪⎣⎦=⎨
⎪-∈-⎪⎩ 19. （1）2
3
； （2）275
20. （1）2221221sin 2sin 2,44sin 24sin 2a S a S θθθθ==++； （2）9
4
21. （1）,2k k Z π
ϕπ=+
∈； （2）1,arctan ,22k k k Z πθππ⎡
⎤∈--∈⎢⎥⎣
⎦； （3）*21,k k N ω=+∈。