盐城市2018届高三年级第三次模拟考试(Word版含答案及详解)

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盐城市2018届高三年级第三次模拟考试(Word版含答案及详解)

盐城市2018届高三年级第三次模拟考试英语试题第一部分: 听力(共两节，满分20分)做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的各案转涂到答题卡上。

第一节(共5小题，每小题1分，满分5分)听下面5段对话。

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

听完每段对话后，你都有10 秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1.What is the woman probably doing?A.W atching a movieB. Reading a newspaper.C. Making an advertisement.2.What are the speakers talking about in general?A.Their best memories of a relaxing holiday.B.Their travelling plans for the summer holiday.C.Their favorite ways of travelling around the world.3.When will the meeting begin?A.At 3:20.B. At 3:40.C. At 4:004.Where are the speakers?A.In a shop.B. In a restaurant.C. In the man’s house.5.What does the woman mean?A.She doesn’t need the man’s help.B.She expects the man to move the desk.C.She wants to remove the books from the desk.第二节(共15 小题; 每小题1分，满分15 分)听下面5段对话或独白。

江苏省盐城市2018届高三第三次模拟考试-英语含答案解析

##市2018届高三年级第三次模拟考试英语试题第一部分听力〔共两节,满分20分〕做题时,先将答案标在试卷上.录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上.第一节〔共5小题；每小题1分,满分5分〕听下面5 段对话.每段对话后有一个小题,从题中所给的A、B、C 三个选项中选出最佳选项.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 is the woman probably doing?A. Watching a movie.B. Reading a newspaper.C. Making an advertisement.2. What are the speakers talking about in general?A. Their best memories of a relaxing holiday.B. Their travelling plans for the summer holiday.C. Their favorite ways of travelling around the world.3. When will the meeting begin?A. At 3:20.B. At 3:40.C. At 4:00.4. Where are the speakers?A. In a shop.B. In a restaurant.C. In the man’s house.5. What does the woman mean?A. She doesn’t need the man’s help.B. She expects the man to move the desk.C. She wants to remove the books from the desk.第二节〔共15小题；每小题1分,满分15分〕听下面 5 段对话或独白.每段对话或独白后有几个小题,从题中所给的A、B、C 三个选项中选出最佳选项.听每段对话或独白前,你将有时间阅读各个小题,每小题5 秒钟；听完后,各小题将给出 5秒钟的作答时间.每段对话或独白读两遍.听第 6 段材料,回答第6、7 题.6. What does the man say about his job?A. The boss is really nice.B. Workmates look friendly.C. Workplace is small and noisy.7. What can we learn from the conversation?A. The boss told the man all the rules.B. Mark will become the boss this year.C. The man will cooperate with Mark.听第7 段材料,回答第8、9 题.8. Why does the woman talk to the man?A. He lost his student visa.B. He often makes trouble.C. He is often absent from class.9. What is the worst result of the man’s behaviour?A. He’ll be sent back home.B. He’ll have to restart his course.C. He’ll stay in the police station.听第8 段材料,回答第10 至12 题.10. What does the man mainly mention about Helen?A. Changes in her character.B. Problems with teachers.C. Relationships with parents.11. What does the man decide to do in the end?A. Have a talk with Helen.B. Talk with Helen’s father.C. Spend more time with Helen.12. What can we learn about Helen?A. Her parents are very busy.B. Classmates dislike her.C. Teachersworry about her.听第9 段材料,回答第13 至16 题.13. What did Nick and Mel do when they were young?A. They learned to act at school.B. They went to live concerts with their father.C. They were encouraged to play music by their mother.14. When was the band Krispy started?A. After Nick and Mel played together.B. After Nick began studying at a music school.C. After two musicians saw Nick and Mel playing.15. What do we know about the band in the first year?A. They recorded twoalbums.B. They joined a music company.C. They were welcome in the concert.16. What has disappointed the man?A. Parents forbade them to join a company.B. His illness delayed a new album recording.C. The first album has sold under a million copies.听第10 段材料,回答第17 至20 题.17. What can be clearly affected if teenagers lack sleep?A. Their spirits.B. Their long-term health.C. Their academic performance.18. What is to blame for teenagers not getting enough sleep according to the speaker?A. Poor diet.B. Lack of exercise.C. Too much entertainment.19. What advice does the speaker give to teenagers who have trouble getting to sleep?A. Listen to music.B. Read a book.C. Drink hot chocolate.20. What does the speaker suggest schools should do?A. Start lessons later.B. Shorten the school day.C. Offer classes in the evenings.第二部分英语知识运用〔共两节,满分35分〕第一节单项填空〔共15小题；每小题1分,满分15分〕请认真阅读下面各题,从题中所给的A、B、C、D四个选项中,选出最佳选项,并在答题卡上将该项涂黑.21. Thomas made ______ his concerns about the changes that had been introduced at work.A. plainB. perfectC. easyD. ambiguous22. Over the last four decades, the Chinese people have ______ enhanced productivity throughhard work with a firm spirit.A. roughlyB. merelyC. significantlyD. equally23. —Can you give me some advice on how to deal with that tough customer?—I ______ to her instead of trying to explain over the phone if in your shoes.A. will writeB. would writeC. were to writeD. would have written24. My printer is five years old, so I’d like to ______ it and purchase a new one online.A. deleteB. retireC. refreshD. declare25. —You should have heard Tom yesterday?—Sure. He’s charged with taking the car without the owner’s ______.A. enquiryB. privilegeC. commandD. consent26. In terms of the urban effects of the technology, our cities are ______ for self-driving cars.A. nowhere near readyB. near nowherereadyC. anywhere near readyD. near anywhere ready27. I sent in my application three weeks ago and they’re probably not going to call me back. I’mstill ______ hope, though.A. holding outB. working outC. figuring outD. setting out28. Many graduates aren’t prepared for what the future has ______ when they come out of school.A. in demandB. at easeC. at lengthD. in store29. ______ to finding ways to make music accessible to all children is Ms. Thompson with a touringcompany of professional musicians.A. DevotingB. DevotedC. Being devotedD. Having been devoted30. By December 2017, the number of Chinese netizens reached 772 million, ______ 753 millionsurf the Internet via mobile phones.A. whichB. whoseC. of whichD.for whom31. Many of life’s failures are experienced by people who did not realize how close they were tosuccess ______ they gave up.A. onceB. whileC. whenD. unless32. Never throughout history has a man who ______ a life of ease left a name worth remembering.A. livesB. livedC. would liveD. had lived33. If we worked on the assumption ______ what is accepted as true really is true, then therewould be little hope for advance.A. whereB. ifC. whetherD. that34. It had been many years since my last visit,but I found the house by ______.A. losing my headB. catching my eyeC. biting my tongueD. following my nose35. —Wall Street English just launched a new WeChat learning platform and we can take freeclasses!—______ ! Let me have a look!A. You don’t sayB. You have got itC. You have me thereD. You can say that again第二节完形填空〔共20小题；每小题1分,满分20分〕请认真阅读下面短文,从短文后各题所给的A、B、C、D四个选项中,选出最佳选项,并在答题卡上将该项涂黑.Stephen Hawking dies at 76 today, the day coinciding with the birthday of another legend of science, Albert Einstein.His 36 are in a state of deep shock and have tweeted that the death of the greatest scientist is 37 for they believed the man to be immortal. There’s one thing that’s certain and it’s death, 38 how and when it comes is something that can’t be in one’s hands. Or can it be? Looking at how Hawking survived over half a century of being 39 the death sentence, the scales are tipping toward a yes.And he indeed just didn’t 40 it; he succeeded. "However difficult life may seem, there is always something you can do and 41 at〞, he said. During the course of his rare disease, his 42 to the field of physics stands out miraculously.Now, what made the man a genius of mind? Was it his willpower or his 43 misfortune that strengthened his mind? He was 44 with a crippling disease at the age of 21, which progressively disables the individual. In the 45 of his genius, however, fortune 46 him since then. His scientific work got 47 and better. The man, whose body was so fragile and seemed to 48 his genius, never gave up.Some experts say perhaps Hawking was 49 diagnosed and this is the reason of his surviving over half a century. Also, doctors say the disease spread 50 than it does in most of the patients, giving him more time. Hawking himself said, "I have been lucky that my condition has 51 more slowly. But it shows that one need not lose 52 .〞Hawking seemed unaffected by the drawbacks life 53 at him. He calls all his accomplishments that followed his disease a 54 . He indeed is a unique example of two mind powers 55 —an astonishing willpower and a strong sense of determination. These two helped him fight against all odds.36. A. ancestorsB. doctorsC. opponentsD. followers37. A. unavoidableB. unreasonableC. unbelievableD. unpredictable38. A. orB. butC. soD. for39. A. handed overB. turned overC. taken overD. run over40. A. shareB. controlC. changeD. survive41. A. succeedB. wonderC. pointD. laugh42. A. explanationB. distributionC. contributionD. attention43. A. mentalB. emotionalC. financialD. physical44. A. infectedB. occupiedC. diagnosedD. concerned45. A. respectB. eventC. faceD. name46. A. trickedB. favoredC. rejectedD. abandoned47. A. thickerB. tougherC. simplerD. lighter48. A. corruptB. failC. spotD. strike49. A. preciselyB. entirelyC. luckilyD. wrongly50. A. easierB. harderC. slowerD. quicker51. A. evolvedB. progressedC. recoveredD. improved52. A. fateB. hopeC. healthD. fortune53. A. screamedB. barkedC. threwD. shot54. A. bonusB. deposit C. rewardD. gain55. A. multipliedB. combinedC. exploitedD. separated第三部分阅读理解〔共15小题；每小题2分,满分30分〕请认真阅读下列短文,从短文后各题所给的A、B、C、D四个选项中,选出最佳选项,并在答题卡上将该项涂黑.AA Yale education emphasizes the importance of learning for public service. Founded in 1886, Dwight Hall at Yale is a student-run, staff-supported public service and social action organization. Below is a list of some active programs, which can be searched by name, cause, or type of service.Logo Overview MissionAIDS Walk New HavenCause<s>:Health & Medicine Types of service:Fundraising and Financial SupportApplication:Join anytime!A 5K off-campus walk to raise money to provide support services to individuals and families infected and affected by HIV/AIDS in the greater New Haven community.Our goals are to raise money, increase public awareness, promote prevention through advocacy.NEW HA VEN56. Which of the following is TRUE in terms of health and medicine causes?A. V olunteers will provide direct support services to AIDS victims.B. AIDS Walk New Haven is targeted at Yale HIV-infected students.C. American Red Cross at Yale encourages Yale students to donate blood.D. Those joining American Red Cross at Yale should be medical experts.57. As a new foreign university freshman, he/she is likely to benefit from ______.A. AIDS Walk New HavenB. American Red Cross at YaleC. Best BuddiesD. Bridges ESLBPersonally, I love writing reviews of any kind mostly because that way I can put all my thoughts about the subject on the paper. Before we go on, I have to mention it is different when you write a review for yourself i.e. your blog or website and for your professor in college.To an untrained eye, reviews may seem pointless. What’s the point of writing about something when other people and your professor have already read the book? Isn’t it enough to talk about it in the class?Just as movie reviews develop your critical thinking, book reviews do the same. It is not enough to read a book and call it a day; you have to establish your opinion, your likes, and dislikes. When a professor gives you this assignment, he/she wants to see your abilities to analyze the book and use vocabulary skills to discuss different parts of the plot.Since we are accustomed to writing book reports at a very young age, it comes as no surprise we don’t think book reviews are different than a book report. Contrary to the popular belief, book reviews and book reports are two different types of writing. Knowing how they differ is essential for writing a high-quality paper that will guarantee a good grade.Book reports usually centre around topical details about the author and the plot of the story. On the other hand, a book review is a more complicated approach to understanding and discussing a book. It doesn’t centre around a summary of each section, but you have to carry out a thorough analysis. As you grow and develop as a student, so does your ability to think critically. You don’t just sum up what you’ve just read but analyze every piece of the puzzle in order to show the ability not only to pay attention to detail but also engage thinking critically. Here, you have to be careful that you aren’t, actually, just retelling the story.While book reviews may contain some elements of book reports e.g. author, characters, plot, the emphasis is to provide a more detailed insight, go deeper and elaborate strengths and weaknesses of the book, and discuss the elements of the story.You know the difference between book reports and book reviews, now what? Now you’re ready to begin the assignment. In order to write a thorough book review, you have to pay attention to everything about the book, which is why writing down the information about the author, genre etc. is strongly advised. That’s why you’ll need a pen and notebook where you can write everything.58. According to the passage, what should be done before starting to write a review?A. Identifying the target reader.B. Consulting with your professor.C. Developing an interest in reviews.D. Listing all your thoughts on the paper.59. The author shows the importance of a book review mainly by ______.A. giving a solid exampleB. making a reasonable assumptionC. drawing a valid comparisonD. providing a detailed description60. Where does the fundamental difference between a report and a review lie?A. Objective summary.B. Critical comments.C. Thorough comprehension.D. Personal abilities.CThe world’s most complex biological computer, made from a group of engineered cells, could one day be implanted into the body to detect diseases and deliver treatments.In an early research in 2012, Martin Fussenegger at ETH Zurish in Switzerland and his colleagues engineered two kidney cells to become a biological circuit capable of simple mathematics. One of the cells was able to calculate addition: the presence or absence of each of two chemicals would switch on a reaction inside the cell that would make it shine different colours. The other cell worked in the same way but could subtract amounts. This kind of biological circuit resembles a simple logic circuit in a computer. In theory, it could be used to indicate the presence of an infectious substance while in fact it failed.Most biological reactions in the body aren’t that simple, though. They rarely rely on "one input and one output〞– instead, multiple inputs lead to different outputs. For instance, a high level of calcium in the body in the presence of a specific hormone may suggest one disease, but a high level of calcium along with another hormone might indicate a completely different condition.To be more practical, biological computers need to be able to perform more complex mathematics. However, it is hard to pack multiple calculations into a single cell. To get around this, Fussenegger and his team have engineered a multicellular system, in which different cells each perform a separate calculation and pass onthe results to each other.The system has nine cells, each containing a biochemical reaction that responds to three chemical inputs – similar to an AND, NOT and OR system in a traditional electronic circuit. These cells coordinate their activities by releasing chemicals that pass from one cell to the other. Together, they form a fully biological circuit that can respond to multiple inputs."Although it is not at a stage yet where we can test on animals, we believe it is the most complex biological computer ever assembled,〞says Fussenegger. "This work addresses one of the major limitations in synthetic biology <合成生物学> – a lack of programmable devices,〞says Ángel Goni-Moreno, a synthetic biologist at Newcastle University, UK. He says that Fussenegger’s multicellular approach enables you to programme the circuit and achieve different calculations just by connecting the nine cells in different configurations <设置>.In the future, a biological computer like this could be used to monitor more complex medical conditions. For example, it could respond to a rise in calcium, a drop in a hormone and an increase in a biomarker, which together would signal the presence of a specific type of cancer, help diagnose it and alert the user to seek appropriate treatment.61. The underlined word "subtract〞in Paragraph 2 is closest in meaning to ______.A. add upB. take awayC. split upD. give away62. What was the progress made in Fussenegger’s early research?A. A biological circuit was implanted in one of kidney cells.B. The indication of infectious substances became a reality.C. Engineered kidney cells could switch on biological reactions.D. Certain cells were made capable of performing mathematics.63. What has made Fussenegger’s current multicellular system so special?A. It has all the functions of a traditional electronic circuit.B. It is programmable and able to perform different mathematics.C. It has successfully packed multiple calculations into a single cell.D. It has been tested through a series of experiments on animals.64. What is the best title for the passage?A. Smart cells indicating various cancersB. Electronic circuit made from multi-cellsC. Programmable cells implanted in human bodiesD. Biological computer made from human cellsDAccording to Guglielmo Cavallo and Roger Chartier, reading aloud was a common practice in the ancient world, the Middle Ages, and as late as the sixteenth and seventeenth centuries. Readers were "listeners attentive to a reading voice,〞and "the text addressed to the ear as much as to the eye.〞The significance of reading aloud continued well into the nineteenth century.Using Charles Dickens’s nineteenth century as a point of departure, it would be useful to look at the familial and social uses of reading aloud and reflect on the functional change of the practice. Dickens habitually read his work to a domestic audience or friends. In his later years he also read to a broader public crowd. Chapters of reading aloud also abound in Dickens’s own literary works. More importantly, he took into consideration the Victorian practice when composing his prose, so much so that his writing is meant to be heard, not only read on the page.Performing a literary text orally in a Victorian family is well documented. Apart from promoting a pleasant family relationship, reading aloud was also a means of protecting young people from the danger of solitary <孤独的> reading. Reading aloud was a tool for parental guidance. By means of reading aloud, parents could also introduce literature to their children, and as such the practice combined leisure and more serious purposes such as religious cultivation in the youths. Within the family, it was commonplace for the father to read aloud. Dickens read to his children: one of his surviving and often-reprinted photographs features him posing on a chair, reading to his two daughters.Reading aloud in the nineteenth century was as much a class phenomenon as a family affair, which points to a widespread belief that Victorian readership primarily meant a middle-class readership. Those who fell outside this group tended to be overlooked by Victorian publishers. Despite this, Dickens, with his publishers Chapman and Hall, managed to distribute literary reading materials to people from different social classes by reducing the price of novels. This was also made possible with the technological and mechanical advances in printing and the spread of railway networks at the time.Since the literacy level of this section of the population was still low before school attendance was made compulsory in 1870 by the Education Act, a considerable number of people from lower classes would listen to recitals of texts. Dickens’s readers, who were from such social backgrounds, might have heard Dickens in this manner. Several biographers of Dickens also draw attention to the fact that it was typical for his texts to be read aloud in Victorian England, and thus illiteracy was not an obstacle for reading Dickens. Reading was no longer a chiefly closeted form of entertainment practiced by the middle class at home.A working-class home was in many ways not convenient for reading: there were too many distractions, the lighting was bad, and the home was also often half a workhouse. As a result, the Victorians from the non-middle classes tended to find relaxation outside the home such as in parks and squares, which were ideal places for the public to go while away their limited leisure time. Reading aloud, in particular publicreading, to some extent blurred the distinctions between classes. The Victorian middle class defined its identity through differences with other classes. Dickens’s popularity among readers from the non-middle classes contributed to the creation of a new class of readers who read through listening.Different readers of Dickens were not reading solitarily and "jealously,〞to use Walter Benjamin’s term. Instead, they often enjoyed a more communal experience, an experience that is generally lacking in today’s world. Modern audiobooks can be considered a contemporary version of the practice. However, while the twentieth-and twentieth-first-century trend for individuals to listen to audiobooks keeps some characteristics of traditional reading aloud—such as "listeners attentive to a reading voice〞and the ear being the focus—it is a far more solitary activity.65. What does the author want to convey in Paragraph 1?A. The history of reading aloud.B. The significance of reading aloud.C. The development of reading practice.D. The roles of readers in reading practice.66. How did the practice of reading aloud influence Dickens’s works?A. He started to write for a broader public crowd.B. He included more readable contents in his novels.C. Scenes of reading aloud became common in his works.D. His works were intended to be both heard and read.67. How many benefits did reading aloud bring to a Victorian family?A. 2.B. 3.C. 4.D. 5.68. Where could a London steel worker possibly have gone to for reading?A. Working place.B. His/her own house.C. Nearby bookstores.D. Trafalgar Square.69. What change did reading aloud bring to Victorian society?A. Different classes started to appreciate and read literary works together.B. People from lower social classes became accepted as middle-class.C. The differences between classes grew less significant than before.D. A non-class society in which everyone could read started to form.70. What is likely to be discussed after the last paragraph?A. New reading trends for individuals.B. The harm of modern audiobooks.C. The material for modern reading.D. Reading aloud in contemporary societies.第四部分任务型阅读<共10小题；每小题1分,满分10分>请认真阅读下列短文,并根据所读内容在文章后表格中的空格里填入一个．．最恰当的单词.注意：请将答案写在答题卡上相应题号的横线上.每个空格只填1个单词.Artificial intelligence <AI> is rushing into business. Firms of all types are using AI to forecast demand, hire workers and deal with customers. The McKinsey Global Institute, a think-tank within a consultancy, forecasts that just applying AI to marketing, sales and supply chains could create economic value of $2.7trn over the next 20 years.Such grand forecasts fuel anxiety as well as hope. Less familiar, but just as important, is how AI will transform the workplace.Start with the benefits. AI ought to improve productivity. Humanyze,a people analytics software provider, combines data from its badges <工牌> with employees’ calendars and e-mails to work out, say, whether office layouts favour teamwork. Slack, a workplace messaging app, helps managers assess how quickly employees accomplish tasks. Companies will see when workers are not just dozing off but also misbehaving.Employees will gain, too. Thanks to advance in computer vision, AI can check that workers are wearing safety equipment and that no one has been harmed on the factory floor. Some will appreciate more feedback on their work and welcome a sense of how to do better.Machines can help ensure that pay rises and promotions go to those who deserve them. That starts with hiring. People often have biases but algorithms <算法>, if designed correctly, can be more unprejudiced. Software can flag patterns that people might miss.Yet AI’s benefits will come with many potential drawbacks. Algorithms may not be free of the biases of their programmers, which canhave unintended consequences. The length of a travel may predict whether an employee will quit a job, but this focus may harm poorer applicants. Older staff might work more slowly than younger ones and could risk losing their positions if all AI looks for is productivity. And surveillance <监控> may feel Orwellian—a sensitive matter now that people have begun to question how much Facebook and other tech giants know about their private lives.As regulators and employers weigh the pros and cons of AI in the workplace, three principles ought toguide its spread. First, data should be anonymised where possible. Microsoft, for example, has a product that shows individuals how they manage their time in the office, but gives managers information only in aggregated <整合> form. Second, the use of AI ought to be transparent. Employees should be told what technologies are being used in their workplaces and which data are being gathered. As a matter of routine, algorithms used by firms to hire, fire and promote should be tested for bias and unintended consequences. Last, countries should let individuals request their own data, whether they are ex-workers wishing to contest a dismissal or jobseekers hoping to demonstrate their ability to prospective employers.The march of AI into the workplace calls for trade-offs between privacy and performance. A fairer, more productive workforce is a prize worth having, but not if it chains employees. Striking a balance will require thought, a willingness for both employers and employees to adapt, and a strong dose of humanity.第五部分书面表达〔满分25分〕81. 请阅读下面短文,并按要求用英语写一篇150词左右的文章.Culture is the crystal of a national character and soul. It is widely acknowledged that the Chinese culture is an important aspect of the country’s opening up and maintaining a close relationship with the rest of the world.Over the past five years, many international cultural events were held, such as the China Shanghai International Arts Festival, the Beijing Music Festival, the International Festival of Intangible Cultural Heritage Chengdu, the China Xinjiang International Dance Festival, the China International Chorus Festival, the China Wuqiao International Circus Festival, and the Silk Road International Arts Festival. These activities have not only enhanced the relations between Chinese and international artists, but also provided them with good opportunities to learn from each other. Through international platforms like UNESCO, China has also strengthened its participation in multilateral cultural communication and cooperation.A series of policies for promoting Chinese culture overseas have reflected culture’s increasing importance, which also make clear our mission to improve the nation’s soft power by delivering a favorable impression of China to the world and promoting communication between people.[写作内容]1．用约30个单词概述上面信息的主要内容；2．用约120个单词就Delivering Chinese culture to the world这一话题发表你的看法.〔1〕分析推广中国文化的意义；〔2〕提出推广中国文化的建议〔至少两点〕.[写作要求]1．写作过程中不能直接引用原文语句；2．作文中不能出现真实##和学校名称；3．不必写标题.[评分标准]内容完整,语言规X,语篇连贯,词数适当.##市2018届高三年级第三次模拟考试英语参考答案第一部分听力〔共两节,满分20分〕1. B2. C3. C4. B5. A6. B7. C8. C9. A 10. A。

盐城市2018届高三年级第三次模拟考试(含答案)

开始 k ←0 S ←0S ＜20k ←k ＋2 S ←S ＋2kYN 输出S 结束第6题图盐城市2018届高三年级第三次模拟考试数学试题(总分160分，考试时间120分钟)注意事项：1．本试卷考试时间为120分钟，试卷满分160分，考试形式闭卷． 2．本试卷中所有试题必须作答在答题卡上规定的位置，否则不给分．3．答题前，务必将自己的姓名、准考证号用0.5毫米黑色墨水签字笔填写在试卷及答题卡上．参考公式：锥体体积公式：13V Sh =，其中S 为底面积,h 为高. 圆锥侧面积公式：S rl π=，其中r 为底面半径,l 为母线长.样本数据12,,,n x x x ⋅⋅⋅的方差2211()n i i s x x n ==-∑，其中11n i i x x n ==∑.一、填空题（本大题共14小题，每小题5分，计70分. 不需写出解答过程，请把答案写在答题纸的指定位置上）1．已知(,]A m =-∞，(1,2]B =，若B A ⊆，则实数m 的取值范围为 ▲ ．2．设复数1a iz i+=+（i 为虚数单位）为纯虚数，则实数a 的值为 ▲ ． 3．设数据12345,,,,a a a a a 的方差为1，则数据123452,2,2,2,2a a a a a 的方差为 ▲ ．4．一个袋子中装有2个红球和2个白球（除颜色外其余均相同），现从中随机摸出2个球，则摸出的2个球中至少有1个是红球的概率为 ▲ ．5．“2,6x k k Z ππ=+∈”是“1sin 2x =”成立的 ▲条件（选填“充分不必要”、“必要不充分”、“充要”、“既不充分又不必要”）．6．运行如图所示的算法流程图，则输出S 的值为 ▲ ． 7．若双曲线22221(0,0)x y a b a b -=>>的两条渐近线与抛物线24y x =交于,,O P Q 三点，且直线PQ 经过抛物线的焦点，则该双曲线的离心率为 ▲ ．8．函数()ln(13)f x x =-的定义域为 ▲ ．9．若一圆锥的底面半径为1，其侧面积是底面积的3倍，则该圆锥的体积为 ▲ ．10．已知函数()3sin()cos()(0,0)f x x x πωϕωϕωϕ+-+><<为偶函数，且其图象的两条相邻对称轴间的距离为2π，则(8f π-的值为 ▲ ．11．设数列{}n a 的前n 项和为n S ，若*2()n n S a n n N =+∈，则数列{}n a 的通项公式为n a = ▲ ．A12．如图，在18AB B ∆中，已知183B AB π∠=，16AB =，84AB =，点234567,,,,,B B B B B B 分别为边18B B 的7等分点，则当9(18)i j i +=≤≤时，i j AB AB ⋅的最大值为 ▲ ．13．定义：点00(,)M x y 到直线:0l ax by c ++=0022a b+．已知点(1,0)A -,(1,0)B ，直线m 过点(3,0)P ，若圆22(18)81x y +-=上存在一点C ，使得,,A B C 三点到直线m 的有向距离之和为0，则直线l 的斜率的取值范围为 ▲ ．14．设ABC ∆的面积为2，若,,A B C 所对的边分别为,,a b c ，则22223a b c ++的最小值为 ▲ ．二、解答题（本大题共6小题，计90分. 解答应写出必要的文字说明，证明过程或演算步骤，请把答案写在答题纸的指定区域内） 15．(本小题满分14分)在直四棱柱1111ABCD A BC D -中，已知底面ABCD 是菱形，,M N 分别是棱11,A D 11D C 的中点．（1）求证：AC ∥平面DMN ；（2）求证：平面DMN ⊥平面11BB D D ．16．(本小题满分14分)在ABC ∆中，角,,A B C 的对边分别为,,a b c ，AD 为边BC 上的中线．（1）若4a =，2b =，1AD =，求边c 的长；（2）若2AB AD c ⋅=，求角B 的大小．A B CD D 1A 1B 1C 1M N第15题图17．(本小题满分14分)如图，是一个扇形花园，已知该扇形的半径长为400米，2AOB π∠=，且半径OC 平分AOB ∠．现拟在OC 上选取一点P ，修建三条路PO ,PA ,PB 供游人行走观赏，设PAO α∠=．（1）将三条路PO ,PA ,PB 的总长表示为α的函数()l α，并写出此函数的定义域；（2）试确定α的值，使得()l α最小．18．(本小题满分16分)如图，已知12,F F 分别是椭圆2222:1(0)x y C a b a b+=>>的左、右焦点，点(2,3)P -是椭圆C 上一点，且1PF x ⊥轴．（1）求椭圆C 的方程；（2）设圆222:()(0)M x m y r r -+=>．①设圆M 与线段2PF 交于两点,A B ，若2MA MB MP MF +=+，且2AB =，求r 的值；②设2m =-，过点P 作圆M 的两条切线分别交椭圆C 于,G H 两点（异于点P ）．试问：是否存在这样的正数r ，使得,G H 两点恰好关于坐标原点O 对称？若存在，求出r 的值；若不存在，请说明理由．19．(本小题满分16分)若对任意实数,k b 都有函数()y f x kx b =++的图象与直线y kx b =+相切，则称函数()f x 为“恒切函数”．设函数()xg x ae x pa =--，,a p R ∈．（1）讨论函数()g x 的单调性；（2）已知函数()g x 为“恒切函数”．①求实数p 的取值范围；②当p 取最大值时，若函数()()xh x g x e m =-也为“恒切函数”，求证：3016m ≤<． AO BCPα第17题图O P F 1 F 2 yx 第18题图（参考数据：320e ≈）20．(本小题满分16分)在数列{}n a 中，已知121,a a λ==，满足111221222,,,,n n n n a a a a ---++⋅⋅⋅是等差数列（其中2,n n N ≥∈），且当n 为奇数时，公差为d ；当n 为偶数时，公差为d -．（1）当1λ=，1d =时，求8a 的值；（2）当0d ≠时，求证：数列{}2*22||()n n a a n N +-∈是等比数列；（3）当1λ≠时，记满足2m a a =的所有m 构成的一个单调递增数列为{}n b ，试求数列{}n b 的通项公式．盐城市2018届高三年级第三次模拟考试数学附加题部分（本部分满分40分，考试时间30分钟）21．[选做题]（在A 、B 、C 、D 四小题中只能选做2题，每小题10分，计20分．请把答案写在答题纸的指定区域内）A .（选修4-1：几何证明选讲）如图，已知半圆O 的半径为5，AB 为半圆O 的直径，P 是BA 延长线上一点，过点P 作半圆O 的切线PC ，切点为C ，CD AB ⊥于D ．若2PC PA =，求CD 的长．B .（选修4-2：矩阵与变换）已知矩阵 2 0 a b ⎡⎤=⎢⎥⎣⎦M 的属于特征值1的一个特征向量为11⎡⎤⎢⎥⎣⎦，求矩阵M 的另一个特征值和对应的一个特征向量．C ．（选修4-4：坐标系与参数方程）在平面直角坐标系中，直线l 的参数方程为21222x y ⎧=-⎪⎪⎨⎪=⎪⎩（t 为参数）．以坐标原点O 为极点，x 轴的正半轴为极轴建立极坐标系（单位长度相同），设曲线C 的极坐标方程为2ρ=，求直线l 被曲线C 截得的弦长．D ．(选修4-5：不等式选讲）已知正数,,x y z 满足232x y z ++=，求222x y z ++的最小值．[必做题]（第22、23题，每小题10分，计20分．请把答案写在答题纸的指定区域内） 22．（本小题满分10分）某公司的一次招聘中，应聘者都要经过三个独立项目,,A B C 的测试，如果通过两个或三个项目的测试即可被录用．若甲、乙、丙三人通过,,A B C 每个项目测试的概率都是12．（1）求甲恰好通过两个项目测试的概率；（2）设甲、乙、丙三人中被录用的人数为X ，求X 的概率分布和数学期望． A BCDO· 第21（A ）图23．（本小题满分10分）（1）已知*0,0()i i a b i N >>∈，比较221212b b a a +与21212()b b a a ++的大小，试将其推广至一般性结论并证明；（2）求证：3*01213521(1)()2n nn n nn n n n N C C C C ++++++≥∈．盐城市2018届高三年级第三次模拟考试数学参考答案一、填空题：本大题共14小题，每小题5分，计70分.1．2m ≥ 2．1- 3．4 4．565．充分不必要 6．21 758．(2,3] 9．23102 11．12n - 12．1327 13．3(,]4-∞- 14．二、解答题：本大题共90小题.15．（1）证明：连接11AC ，在四棱柱1111ABCD A BC D -中，因为11//AA BB ，11//BB CC ，所以11//AA CC ，所以11A ACC 为平行四边形，所以11//AC AC ． ……2分又,M N 分别是棱11,A D 11D C 的中点，所以11//MN AC ，所以//AC MN ． ……4分又AC ⊄平面DMN ，MN ⊂平面DMN ，所以AC ∥平面DMN ． ……6分（2）证明：因为四棱柱1111ABCD A BC D -是直四棱柱，所以1DD ⊥平面1111A B C D ，而MN ⊂平面1111A B C D ，所以1MN DD ⊥． ……8分又因为棱柱的底面ABCD 是菱形，所以底面1111A B C D 也是菱形，所以1111AC B D ⊥，而11//MN AC ，所以11MN B D ⊥．……10分又1MN DD ⊥，111,DD B D ⊂平面1111A B C D ，且1111DD B D D =，所以MN ⊥平面1111A B C D ． ……12分而MN ⊂平面DMN ，所以平面DMN ⊥平面11BB D D ． ……14分 16．解：（1）在ADC ∆中，因为11,2,22AD AC DC BC ====，所以由余弦定理，得2222222217cos 22228AC DC AD C AC DC +-+-===⋅⨯⨯． ……3分故在ABC ∆中，由余弦定理，得2222272cos 4224268c a b ab C =+-=+-⨯⨯⨯=，所以6c = ……6分（2）因为AD 为边BC 上的中线，所以1()2AD AB AC =+，所以21()2c AB AD AB AB AC =⋅=⋅+221111cos 2222AB AB AC c cb A =+⋅=+，得cos c b A =． ……10分则2222b c a c b bc+-=⋅，得222b c a =+，所以90B =︒． ……14分17．解：（1）在APO ∆中，由正弦定理，得sin sin sin AP OP AOAOP PAO APO==∠∠∠，即400sin sin sin()44AP OP ππαα==+，从而2002sin()4AP α=+，400sin sin()4OP απα=+． ……4分所以()l α＝400sin 200222sin()sin()44OP PA PB OP PA αππαα++=+=+⨯++， ABCDDABCM N故所求函数为2sin )()sin()4l ααπα=+，3(0,)8πα∈． ……6分（2）记2sin 22sin 3()(0,)8sin()4f ααπααα++==∈+，因为22(sin cos )(22)(cos sin )()(sin cos )f ααααααααα+-+-'=+2)24(sin cos )πααα-+=+， ……10分由()0f α'=，得1sin()42πα-=-，又3(0,)8πα∈，所以12πα=． ……12分列表如下：α(0,)12π12π 3(,)128ππ()f α' － 0 ＋ ()f α递减极小递增所以，当12πα=时，()l α取得最小值．答：当12πα=时，()l α最小． ……14分18．解：（1）因点(2,3)P -是椭圆C 上一点，且1PF x ⊥轴，所以椭圆的半焦距2c =，由22221c y a b+=，得2b y a =±，所以2243b a a a -==， ……2分化简得2340a a --=，解得4a =，所以212b =，所以椭圆C 的方程为2211612x y +=． ……4分（2）①因2MA MB MP MF +=+，所以2MA MP MF MB -=-，即2PA BF =，所以线段2PF 与线段AB 的中点重合（记为点Q ），由（1）知3(0,)2Q ， ……6分因圆M 与线段2PF 交于两点,A B ，所以21MQ AB MQ PF k k k k ⋅=⋅=-，所以303021m --⋅=-，解得98m =-， ……8分所以229315(0)(0)828MQ =--+-=，故221517()188r =+=. ……10分② 由,G H 两点恰好关于原点对称，设00(,)G x y ，则00(,)H x y --，不妨设00x <，因(2,3)P -，2m =-，所以两条切线的斜率均存在，设过点P 与圆M 相切的直线斜率为k ，则切线方程为3(2)y k x -=+，即230kx y k -++=，由该直线与圆M 相切，得21r k =+，即229r k r -=±，……12分所以两条切线的斜率互为相反数，即PG PH k k =-，所以00003322y y x x ---=-+-+，化简得006x y =-，即006y x -=，代入220011612x y +=，化简得420016480x x -+=，解得02x =-（舍），023x =-03y = ……14分所以(23,3)G -，(23,3)H -，所以333223PG k -==-+，所以26731()2r ==+ 故存在满足条件的r ，且67r =……16分 19．解：（1）()1x g x ae '=-， ……2分当0a ≤时，()0g x '<恒成立，函数()g x 在R 上单调递减；当0a >时，由()0g x '=得ln x a =-，由()0g x '>得ln x a >-，由()0g x '<得ln x a <-，得函数()g x 在(,ln )a -∞-上单调递，在(ln ,)a -+∞上单调递增． ……4分（2）①若函数()f x 为“恒切函数”，则函数()y f x kx b =++的图像与直线y kx b =+相切，设切点为00(,)x y ，则0()f x k k '+=且000()f x kx b kx b ++=+，即0()0f x '=，0()0f x =. 因为函数()g x 为“恒切函数”，所以存在0x ，使得0()0g x '=，0()0g x =，即0000 10xx ae x pa ae ⎧--=⎪⎨-=⎪⎩，得00x a e -=>，00(1)x p e x =-，设()(1)x m x e x =-， ……6分则()xm x xe '=-，()0m x '<，得0x >，()0m x '>，得0x <，故()m x 在(,0)-∞上单调递增，在(0,)+∞上单调递减，从而[]max ()(0)1m x m ==，故实数p 的取值范围为(,1]-∞． ……8分②当p 取最大值时，1p =，00x =，01x a e-==，()(1)x x h x e x e m =---，()(22)x x h x e x e '=--，因为函数()h x 也为“恒切函数”，故存在0x ，使得0()0h x '=，0()0h x =，由0()0h x '=得000(22)0x x e x e --=，00220x e x --=，设()22xn x e x =--， ……10分则()21xn x e '=-，()0n x '>得ln 2x >-，()0n x '<得ln 2x <-，故()n x 在(,ln 2)-∞-上单调递减，在(ln 2,)-+∞上单调递增，1°在单调递增区间(ln 2,)-+∞上，(0)0n =，故00x =，由0()0h x =，得0m =；…12分 2°在单调递减区间(,ln 2)-∞-上，2(2)20n e--=>，31223111()22(20)022225n e ---=-≈⨯-=<，又()n x 的图像在(,ln 2)-∞-上不间断，故在区间3(2,)2--上存在唯一的0x ，使得00220xe x --=，故0022x x e +=，此时由0()0h x =，得00000022(1)(1)22x xx x m e x e x ++=--=-- 001(2)4x x =-+2011(1)44x =-++，函数211()(1)44r x x =-++在3(2,)2--上递增，(2)0r -=，33()216r -=，故3016m <<．综上1°2°所述，3016m ≤<． ……16分20．解：（1）由1λ=，1d =，所以21a =，234,,a a a 为等差数列且公差为1-，所以4221a a =-=-，又458,,a a a 为等差数列且公差为1，所以8443a a =+=． ……2分（2）当21n k =+时，22221221222,,,,k k k k a a a a +++⋅⋅⋅是等差数列且公差为d ，所以2122222k k k a a d +=+，同理可得22121222k k k a a d --=-， ……4分两式相加，得212121222k k k a a d +---=；当2n k =时，同理可得2222222k k k a a d +-=-， ……6分所以222||2n n na a d +-=．又因为0d ≠，所以21122122||22(2)||2n n n n n n a a n a a ++---==≥-，所以数列{}2*22||()n n a a n N +-∈是以2为公比的等比数列． ……8分（3）因为2a λ=，所以4222a a d d λ=-=-，由（2）知212121222k k k a a d +--=+，所以212123212321222222k k k k k k a a d a d d +-----=+=++，依次下推，得211132*********k k k a a d d d d +--=+++++，所以21222(21)3k ka d λ+=+-， ……10分当212222k k n ++≤≤时，212321222(2)()33k k k n a a n d n d λ+++=--=+--，由2m a a =，得232233k m +=-，所以23212233k k b ++=-，所以22233n n b +=-（n 为奇数）； ……12分由（2）知222222222222222k k k k k k a a d a d d +--=-=--，依次下推，得22224222222222k k k a a d d d d +-=-----，所以22224(21)23k k a d d λ+-=--， ……14分当222322k k n ++≤≤时，222422222(2)()33k k k n a a n d n d λ+++=+-=+--，由2m a a =，得242233k m +=+，所以24222233k k b ++=+．所以22233n n b +=+（n 为偶数）．综上所述，2222(3322(33n n n n b n ++⎧+⎪⎪=⎨⎪-⎪⎩为偶数）为奇数）． ……16分方法二：由题意知，23121231222222n n n n n b b b b b +++=<<<<<⋅⋅⋅<<<<<<⋅⋅⋅， ……10分当n 为奇数时，1221222,,,,n n n n a a a a +++⋅⋅⋅的公差为d -，1112221222,,,,n n n n a a a a ++++++⋅⋅⋅的公差为d ，所以112(2)()n n n b n a a b d ++=---，11112(2)n n n b n a a b d ++++=+-，则由12n n b b a a a +==，得111(2)()(2)n n n n b d b d +++---=-，即212n n n b b +++=．同理，当n 为偶数时，也有212n n n b b +++=．故恒有2*12()n n n b b n N +++=∈． ……12分①当n 为奇数时，由3212n n n b b ++++=，212n n n b b +++=，相减，得222n n n b b ++-=，所以532311()()(222)2n n n n b b b b b b -=-+⋅⋅⋅+-+=+⋅⋅⋅+++13222(14)2221433n n -+-=+=--．……14分②当n 为偶数时，同理可得22233n n b +=+．综上所述，2222(3322(33n n n n b n ++⎧+⎪⎪=⎨⎪-⎪⎩为偶数）为奇数）． ……16分附加题答案21．（A ）解：连,AC BC ，因PC 为半圆O 的切线，所以PCA B ∠=∠．又P P ∠=∠，所以PCA ∆∽PBC ∆，所以12PA AC PC BC ==，即2AC BC =． ……5分因AB 为半圆O 的直径，所以22225AB AC BC AC =+=，因半圆O 的半径为5，所以21005AC =，所以25,45AC BC ==由射影定理，得2AC AD AB =⋅，解得2AD =，所以224CD AC AD =-=． ……10分（B ）解：由题意得 2 110 11a b ⎡⎤⎡⎤⎡⎤=⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦，解得11a b =-⎧⎨=⎩，所以 2 10 1-⎡⎤=⎢⎥⎣⎦M ． ……2分矩阵M 的特征多项式为 2 1()(2)(1)0 1f λλλλλ-==---，由()0f λ=，得2,1λλ==，所以矩阵M 的另一个特征值为2． ……6分此时0 1()0 1f λ=，对应方程组为010010x y x y ⋅+⋅=⎧⎨⋅+⋅=⎩，所以0y =，所以另一个特征值2对应的一个特征向量为10⎡⎤⎢⎥⎣⎦． ……10分（C ）解：直线的普通方程为10x y +-=；由2ρ=，得曲线C 的普通方程为224x y +=， ………………………5分所以1222d -==l 被曲线C 截得的弦长为22222()142-= ……10分（D ）解：根据柯西不等式，有2222222(23)(123)()x y z x y z ++≤++++，因232x y z ++=，所以222222421237x y z ++≥=++， ……5分当且仅当123x y z ==时等号成立，解得123,,777x y z ===，即当123,,777x y z ===时，222x y z ++取最小值27． ……10分A BCDO·22．解：（1）甲恰好通过两个项目测试的概率为223113()(1)228C -=． ……4分（2）因为每人可被录用的概率为22331111()(1)()2222C -+=，所以311(0)(1)28P X ==-=，1123113(1)()(1)228P X C ==-=，2213113(2)()(1)228P X C ==-=，311(3)()28P X ===．故X 的概率分布表为：X0 1 2 3 P18 38 38 18…………8分所以，X 的数学期望13313()012388882E X =⨯+⨯+⨯+⨯=． ……10分23．解：（1）22222212211212121212()()()b b a b a b a a b b a a a a ++=+++，因为0i a >，0i b >，所以222112120,0a b a b a a >>，则22222112211212121222a b a b a b a b b b a a a a +≥⨯=，所以22222121212121212()()2()b b a a b b b b b b a a ++≥++=+，即22121212()()b b a a a a ++212()b b ≥+．所以221212b b a a +≥21212()b b a a ++，当且仅当22211212a b a b a a =，即2112a b a b =时等号成立． ……2分推广：已知0i a >,0i b >(,1i N i n *∈≤≤)，则2221212n n b b b a a a +++21212()n nb b b a a a +++≥+++． ……………………………4分证明：①当1n =时命题显然成立；当2n =时，由上述过程可知命题成立； ②假设(2)n k k =≥时命题成立，即已知0i a >,0i b >(,1i N i k *∈≤≤)时，有2221212k k b b b a a a +++21212()k kb b b a a a +++≥+++成立，则1n k =+时，222222112112121121()()k k k k k k k k b b b b b b b b a a a a a a a a +++++++++++≥++++，由221212b b a a +≥21212()b b a a ++，可知222121*********()()k k k k k k k k b b b b b b b b a a a a a a a a ++++++++++++≥+++++++，故2222112121k k k k b b b b a a a a ++++++2121121()k k k k b b b b a a a a ++++++≥++++，故1n k =+时命题也成立．综合①②，由数学归纳法原理可知，命题对一切n N ∈*恒成立． ……6分（注：推广命题中未包含1n =的不扣分）（2）证明：由（1）中所得的推广命题知01213521nn n n nn C C C C +++++ 2222012135(21)35(21)nn n nn n C C C n C +=+++++[]212135(21)35(21)nn nn nn C C C n C +++++≥+++++ ①， …8分记01235(21)nn n n n nS C C C n C =+++++，则1(21)(21)n n n n n nS n C n C C -=++-++，两式相加，得0122(22)(22)(22)(22)nn n n n nS n C n C n C n C =++++++++， 012(22)()(22)2nn n n n n n C C C C n =+++++=+⨯，故(1)2n n S n =+⨯ ②，又[]2241(21)135(21)(1)(1)2n n n n ++⎡⎤+++++=⨯+=+⎢⎥⎣⎦③，将②③代入①，得222243012135(21)(1)(1)35(21)(1)22n n nn n n n n n n C C C n C n +++++++≥=++，所以，301213521(1)2n nn n nn n n C C C C ++++++≥，证毕． ……10分。

盐城市2018届高三三模英语试卷及答案

2018届高三年级第三次模拟考试(十六)英语第一部分听力(共两节，满分20分)第一节(共5小题；每小题1分，满分5分)听下面5段对话。

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

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

() 1. What is the woman probably doing?A. Watching a movie.B. Reading a newspaper.C. Making an advertisement.() 2. What are the speakers talking about in general?A. Their best memories of a relaxing holiday.B. Their travelling plans for the summer holiday.C. Their favorite ways of travelling around the world.() 3. When will the meeting begin?A. At 3：20.B. At 3：40.C. At 4：00.() 4. Where are the speakers?A. In a shop.B. In a restaurant.C. In the man's house.() 5. What dose the woman mean?A. She doesn't need the man's help.B. She expects the man to move the desk.C. She wants to remove the book from the desk.第二节(共15小题；每小题1分，满分15分)听下面5段对话或独白。

盐城市2018届高三年级第三次模拟考试

第6题图盐城市2018届高三年级第三次模拟考试数学试题(总分160分，考试时间120分钟)注意事项：1．本试卷考试时间为120分钟，试卷满分160分，考试形式闭卷． 2．本试卷中所有试题必须作答在答题卡上规定的位置，否则不给分．3．答题前，务必将自己的姓名、准考证号用0.5毫米黑色墨水签字笔填写在试卷及答题卡上．参考公式：锥体体积公式：13V Sh =，其中S 为底面积,h 为高.圆锥侧面积公式：S rl π=，其中r 为底面半径,l 为母线长.样本数据12,,,n x x x ⋅⋅⋅的方差2211()n i i s x x n ==-∑，其中11n i i x x n ==∑.一、填空题（本大题共14小题，每小题5分，计70分. 不需写出解答过程，请把答案写在答题纸的指定位置上）1．已知(,]A m =-∞，(1,2]B =，若B A ⊆，则实数m 的取值范围为 ▲ ． 2．设复数1a iz i+=+（i 为虚数单位）为纯虚数，则实数a 的值为 ▲ ． 3．设数据12345,,,,a a a a a 的方差为1，则数据123452,2,2,2,2a a a a a 的方差为 ▲ ． 4．一个袋子中装有2个红球和2个白球（除颜色外其余均相同），现从中随机摸出2个球，则摸出的2个球中至少有1个是红球的概率为 ▲ ． 5．“2,6x k k Z ππ=+∈”是“1sin 2x =”成立的 ▲ 条件（选填“充分不必要”、“必要不充分”、“充要”、“既不充分又不必要”）．6．运行如图所示的算法流程图，则输出S 的值为 ▲ ．7．若双曲线22221(0,0)x ya b a b -=>>的两条渐近线与抛物线24y x =交于,,O P Q 三点，且直线PQ 经过抛物线的焦点，则该双曲线的离心率为 ▲． 8．函数()ln(13)f x x =-的定义域为 ▲ ．9．若一圆锥的底面半径为1，其侧面积是底面积的3倍，则该圆锥的体积为 ▲ ． 10．已知函数())cos()(0,0)f x x x πωϕωϕωϕ=+-+><<为偶函数，且其图象的两条相邻对称轴间的距离为2π，则()8f π-的值为 ▲ ．11．设数列{}n a 的前n 项和为n S ，若*2()n n S a n n N =+∈，则数列{}n a 的通项公式为n a = ▲ ． 12．如图，在18AB B ∆中，已知183B AB π∠=，16AB =，第12题图AB 1 B 2 B 3 B 4 B 5 B 6 B 7 B 884AB =，点234567,,,,,B B B B B B 分别为边18B B 的7等分点，则当9(18)i j i +=≤≤时，i j AB AB ⋅的最大值为 ▲ ．13．定义：点00(,)M x y 到直线:0l ax by c ++=的有向距离为．已知点(1,0)A -,(1,0)B ，直线m 过点(3,0)P ，若圆22(18)81x y +-=上存在一点C ，使得,,A B C 三点到直线m 的有向距离之和为0，则直线l 的斜率的取值范围为 ▲ ． 14．设ABC ∆的面积为2，若,,A B C 所对的边分别为,,a b c ，则22223a b c ++的最小值为 ▲ ．二、解答题（本大题共6小题，计90分. 解答应写出必要的文字说明，证明过程或演算步骤，请把答案写在答题纸的指定区域内） 15．(本小题满分14分)在直四棱柱1111ABCD A B C D -中，已知底面ABCD 是菱形，,M N 分别是棱11,A D 11D C 的中点．（1）求证：AC ∥平面DMN ；（2）求证：平面DMN ⊥平面11BB D D ．16．(本小题满分14分)在ABC ∆中，角,,A B C 的对边分别为,,a b c ，AD 为边BC 上的中线．（1）若4a =，2b =，1AD =，求边c 的长；（2）若2AB AD c ⋅=，求角B 的大小．ABCD D 1A 1B 1C 1MN 第15题图17．(本小题满分14分)如图，是一个扇形花园，已知该扇形的半径长为400米，2AOB π∠=，且半径OC 平分AOB ∠．现拟在OC 上选取一点P ，修建三条路PO ,PA ,PB 供游人行走观赏，设PAO α∠=．（1）将三条路PO ,PA ,PB 的总长表示为α的函数()l α，并写出此函数的定义域；（2）试确定α的值，使得()l α最小．18．(本小题满分16分)如图，已知12,F F 分别是椭圆2222:1(0)x y C a b a b+=>>的左、右焦点，点(2,3)P -是椭圆C上一点，且1PF x ⊥轴．（1）求椭圆C 的方程；（2）设圆222:()(0)M x m y r r -+=>．①设圆M 与线段2PF 交于两点,A B ，若2MA MB MP MF +=+，且2AB =，求r 的值；②设2m =-，过点P 作圆M 的两条切线分别交椭圆C 于,G H 两点（异于点P ）．试问：是否存在这样的正数r ，使得G 出r 的值；若不存在，请说明理由．AO BCPα第17题图19．(本小题满分16分)若对任意实数,k b 都有函数()y f x kx b =++的图象与直线y kx b =+相切，则称函数()f x 为“恒切函数”．设函数()xg x ae x pa =--，,a p R ∈．（1）讨论函数()g x 的单调性；（2）已知函数()g x 为“恒切函数”．①求实数p 的取值范围；②当p 取最大值时，若函数()()xh x g x e m =-也为“恒切函数”，求证：3016m ≤<．（参考数据：320e ≈）20．(本小题满分16分)在数列{}n a 中，已知121,a a λ==，满足111221222,,,,n n n n a a a a ---++⋅⋅⋅是等差数列（其中2,n n N ≥∈），且当n 为奇数时，公差为d ；当n 为偶数时，公差为d -．（1）当1λ=，1d =时，求8a 的值；（2）当0d ≠时，求证：数列{}2*22||()n n a a n N +-∈是等比数列；（3）当1λ≠时，记满足2m a a =的所有m 构成的一个单调递增数列为{}n b ，试求数列{}n b 的通项公式．盐城市2018届高三年级第三次模拟考试数学附加题部分（本部分满分40分，考试时间30分钟）21．[选做题]（在A 、B 、C 、D 四小题中只能选做2题，每小题10分，计20分．请把答案写在答题纸的指定区域内） A .（选修4-1：几何证明选讲）如图，已知半圆O 的半径为5，AB 为半圆O 的直径，P 是BA 延长线上一点，过点P 作半圆O 的切线PC ，切点为C ，CD AB ⊥于D ．若2PC PA =，求CD 的长．B .（选修4-2：矩阵与变换）已知矩阵 2 0 a b ⎡⎤=⎢⎥⎣⎦M 的属于特征值1的一个特征向量为11⎡⎤⎢⎥⎣⎦，求矩阵M 的另一个特征值和对应的一个特征向量．C ．（选修4-4：坐标系与参数方程）在平面直角坐标系中，直线l的参数方程为122x t y ⎧=-⎪⎪⎨⎪=⎪⎩（t 为参数）．以坐标原点O 为极点，x 轴的正半轴为极轴建立极坐标系（单位长度相同），设曲线C 的极坐标方程为2ρ=，求直线l 被曲线C 截得的弦长．D ．(选修4-5：不等式选讲）已知正数,,x y z 满足232x y z ++=，求222x y z ++的最小值．[必做题]（第22、23题，每小题10分，计20分．请把答案写在答题纸的指定区域内） 22．（本小题满分10分）A BCD O ·第21（A ）图某公司的一次招聘中，应聘者都要经过三个独立项目,,A B C 的测试，如果通过两个或三个项目的测试即可被录用．若甲、乙、丙三人通过,,A B C 每个项目测试的概率都是12．（1）求甲恰好通过两个项目测试的概率；（2）设甲、乙、丙三人中被录用的人数为X ，求X 的概率分布和数学期望．23．（本小题满分10分）（1）已知*0,0()i i a b i N >>∈，比较221212b b a a +与21212()b b a a ++的大小，试将其推广至一般性结论并证明；（2）求证：3*01213521(1)()2n nn n nn n n n N C C C C ++++++≥∈．盐城市2018届高三年级第三次模拟考试数学参考答案一、填空题：本大题共14小题，每小题5分，计70分.1．2m ≥ 2．1- 3．4 4．565．充分不必要 6．21 78．(2,3] 9．3 1011．12n - 12．1327 13．3(,]4-∞- 14．二、解答题：本大题共90小题.15．（1）证明：连接11A C ，在四棱柱1111ABCD A B C D -中，因为11//AA BB ，11//BB CC ，所以11//AA CC ，所以11A ACC 为平行四边形，所以11//A C AC ． ……2分又,M N 分别是棱11,A D 11D C 的中点，所以11//MN AC ，所以//AC MN ． ……4分又AC ⊄平面DMN ，MN ⊂平面DMN ，所以AC ∥平面DMN ． ……6分（2）证明：因为四棱柱1111ABCD A B C D -是直四棱柱，所以1DD ⊥平面1111A B C D ，而MN ⊂平面1111A B C D ，所以1MN DD ⊥． ……8分又因为棱柱的底面ABCD 是菱形，所以底面1111A B C D 也是菱形，所以1111A C B D ⊥，而11//MN AC ，所以11MN B D ⊥．……10分又1MN DD ⊥，111,DD B D ⊂平面1111A B C D ，且1111DD B D D =，所以MN ⊥平面1111A B C D ． ……12分而MN ⊂平面DMN ，所以平面DMN ⊥平面11BB D D ． ……14分 16．解：（1）在ADC ∆中，因为11,2,22AD AC DC BC ====，所以由余弦定理，得2222222217cos 22228AC DC AD C AC DC +-+-===⋅⨯⨯． ……3分故在ABC ∆中，由余弦定理，得2222272cos 4224268c a b ab C =+-=+-⨯⨯⨯=，所以6c =……6分（2）因为AD 为边BC 上的中线，所以1()2AD AB AC =+，所以21()2c AB AD AB AB AC =⋅=⋅+221111cos 2222AB AB AC c cb A =+⋅=+，得cos c b A =． ……10分则2222b c a c b bc+-=⋅，得222b c a =+，所以90B =︒． ……14分17．解：（1）在APO ∆中，由正弦定理，得sin sin sin AP OP AOAOP PAO APO==∠∠∠，A BCDDABCM N即400sin sin sin()44APOP ππαα==+，从而sin()4AP πα=+，400sin sin()4OP απα=+． ……4分所以()l α＝400sin 22sin()sin()44OP PA PB OP PA αππαα++=+=+⨯++，故所求函数为sin )()sin()4l ααπα=+，3(0,)8πα∈． ……6分（2）记sin 23(),(0,)sin cos 8sin()4f ααπααπααα+==∈++，因为2(sin cos )(2)(cos sin )()(sin cos )f ααααααααα+--'=+2)4(sin cos )πααα-+=+， ……10分由()0f α'=，得1sin()42πα-=-，又3(0,)8πα∈，所以12πα=． ……12分列表如下：所以，当12α=时，()l α取得最小值．答：当12πα=时，()l α最小． ……14分18．解：（1）因点(2,3)P -是椭圆C 上一点，且1PF x ⊥轴，所以椭圆的半焦距2c =，由22221c y a b+=，得2b y a =±，所以2243b a a a -==， ……2分化简得2340a a --=，解得4a =，所以212b =，所以椭圆C 的方程为2211612x y +=． ……4分（2）①因2MA MB MP MF +=+，所以2MA MP MF MB -=-，即2PA BF =，所以线段2PF 与线段AB 的中点重合（记为点Q ），由（1）知3(0,)2Q ，……6分因圆M 与线段2PF 交于两点,A B ，所以21MQ AB MQ PF k k k k ⋅=⋅=-，所以30302122m --⋅=---，解得98m =-， ……8分所以158MQ ==，故178r ==. ……10分② 由,G H 两点恰好关于原点对称，设00(,)G x y ，则00(,)H x y --，不妨设00x <，因(2,3)P -，2m =-，所以两条切线的斜率均存在，设过点P 与圆M 相切的直线斜率为k ，则切线方程为3(2)y k x -=+，即230kx y k -++=，由该直线与圆M相切，得r =，即k =12分所以两条切线的斜率互为相反数，即PG PH k k =-，所以00003322y y x x ---=-+-+，化简得006x y =-，即006y x -=，代入220011612x y +=，化简得420016480x x -+=，解得02x =-（舍），0x =-0y = (14)分所以(G -，H，所以2223PG k ==-+，所以r ==. 故存在满足条件的r ，且7r =． ……16分 19．解：（1）()1xg x ae '=-， ……2分当0a ≤时，()0g x '<恒成立，函数()g x 在R 上单调递减；当0a >时，由()0g x '=得ln x a =-，由()0g x '>得ln x a >-，由()0g x '<得ln x a <-，得函数()g x 在(,ln )a -∞-上单调递，在(ln ,)a -+∞上单调递增． ……4分（2）①若函数()f x 为“恒切函数”，则函数()y f x kx b =++的图像与直线y kx b =+相切，设切点为00(,)x y ，则0()f x k k '+=且000()f x kx b kx b ++=+，即0()0f x '=，0()0f x =.因为函数()g x 为“恒切函数”，所以存在0x ，使得0()0g x '=，0()0g x =，即0000 10xx ae x pa ae ⎧--=⎪⎨-=⎪⎩，得00x a e -=>，00(1)x p e x =-，设()(1)x m x e x =-， (6)分则()xm x xe '=-，()0m x '<，得0x >，()0m x '>，得0x <，故()m x 在(,0)-∞上单调递增，在(0,)+∞上单调递减，从而[]max ()(0)1m x m ==，故实数p 的取值范围为(,1]-∞． ……8分②当p 取最大值时，1p =，00x =，01x a e-==，()(1)x x h x e x e m =---，()(22)x x h x e x e '=--，因为函数()h x 也为“恒切函数”，故存在0x ，使得0()0h x '=，0()0h x =，由0()0h x '=得000(22)0xx e x e --=，00220x e x --=，设()22x n x e x =--， (10)分则()21xn x e '=-，()0n x '>得ln 2x >-，()0n x '<得ln 2x <-，故()n x 在(,ln 2)-∞-上单调递减，在(ln 2,)-+∞上单调递增，1°在单调递增区间(ln 2,)-+∞上，(0)0n =，故00x =，由0()0h x =，得0m =；…12分2°在单调递减区间(,ln 2)-∞-上，2(2)20n e--=>，31223111()22(20)02222n e ---=-≈⨯-=-<，又()n x 的图像在(,ln 2)-∞-上不间断，故在区间3(2,)2--上存在唯一的0x ，使得00220xe x --=，故0022x x e+=，此时由0()0h x =，得0000022(1)(1)22x x x x m e x e x ++=--=-- 001(2)4x x =-+2011(1)44x =-++，函数211()(1)44r x x =-++在3(2,)2--上递增，(2)0r -=，33()216r -=，故3016m <<．综上1°2°所述，3016m ≤<． ……16分20．解：（1）由1λ=，1d =，所以21a =，234,,a a a 为等差数列且公差为1-，所以4221a a =-=-，又458,,a a a 为等差数列且公差为1，所以8443a a =+=． (2)分（2）当21n k =+时，22221221222,,,,k k k k a a a a +++⋅⋅⋅是等差数列且公差为d ，所以2122222k k ka a d +=+，同理可得22121222k k k a a d --=-， ……4分两式相加，得212121222k k k a a d +---=；当2n k =时，同理可得2222222k k ka a d +-=-， ……6分所以222||2n n na a d +-=．又因为0d ≠，所以21122122||22(2)||2n n n n nn a a n a a ++---==≥-，所以数列{}2*22||()n n a a n N +-∈是以2为公比的等比数列． ……8分（3）因为2a λ=，所以4222a a d d λ=-=-，由（2）知212121222k k k a a d +--=+，所以212123212321222222k k k k k k a a d a d d +-----=+=++，依次下推，得211132*********k k k a a d d d d +--=+++++，所以21222(21)3k ka d λ+=+-， ……10分当212222k k n ++≤≤时，212321222(2)()33k k k n a a n d n d λ+++=--=+--，由2m a a =，得232233k m +=-，所以23212233k k b ++=-，所以22233n n b +=-（n 为奇数）； ……12分由（2）知222222222222222k k k k k k a a d a d d +--=-=--，依次下推，得22224222222222k k k a a d d d d +-=-----，所以22224(21)23k k a d d λ+-=--， ……14分当222322k k n ++≤≤时，222422222(2)()33k k k n a a n d n d λ+++=+-=+--，由2m a a =，得242233k m +=+，所以24222233k k b ++=+．所以22233n n b +=+（n 为偶数）．综上所述，2222(3322(33n n n n b n ++⎧+⎪⎪=⎨⎪-⎪⎩为偶数）为奇数）． ……16分方法二：由题意知，23121231222222n n n n n b b b b b +++=<<<<<⋅⋅⋅<<<<<<⋅⋅⋅， (10)分当n 为奇数时，1221222,,,,n n n n a a a a +++⋅⋅⋅的公差为d -，1112221222,,,,n n n n a a a a ++++++⋅⋅⋅的公差为d ，所以112(2)()n n n b n a a b d ++=---，11112(2)n n n b n a a b d ++++=+-，则由12n n b b a a a +==，得111(2)()(2)n n n n b d b d +++---=-，即212n n n b b +++=．同理，当n 为偶数时，也有212n n n b b +++=．故恒有2*12()n n n b b n N +++=∈． ……12分①当n 为奇数时，由3212n n n b b ++++=，212n n n b b +++=，相减，得222n n n b b ++-=，所以532311()()(222)2n n n n b b b b b b -=-+⋅⋅⋅+-+=+⋅⋅⋅+++13222(14)2221433n n -+-=+=--．……14分②当n 为偶数时，同理可得22233n n b +=+．综上所述，2222(3322(33n n n n b n ++⎧+⎪⎪=⎨⎪-⎪⎩为偶数）为奇数）． ……16分附加题答案21．（A ）解：连,AC BC ，因PC 为半圆O 的切线，所以PCA B ∠=∠．又P P ∠=∠，所以PCA ∆∽PBC ∆，所以12PA AC PC BC ==，即2AC BC =． ……5分因AB 为半圆O 的直径，所以22225AB AC BC AC =+=，因半圆O 的半径为5，所以21005AC =，所以AC BC ==由射影定理，得2AC AD AB =⋅，解得2AD =，所以4CD ==． (10)分（B ）解：由题意得 2 110 11a b ⎡⎤⎡⎤⎡⎤=⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦，解得11a b =-⎧⎨=⎩，所以 2 10 1-⎡⎤=⎢⎥⎣⎦M ． ……2分矩阵M 的特征多项式为 2 1()(2)(1)0 1f λλλλλ-==---，由()0f λ=，得2,1λλ==，所以矩阵M 的另一个特征值为2． ……6分此时0 1()0 1f λ=，对应方程组为010010x y x y ⋅+⋅=⎧⎨⋅+⋅=⎩，所以0y =，所以另一个特征值2对应的一个特征向量为10⎡⎤⎢⎥⎣⎦． ……10分（C ）解：直线的普通方程为10x y +-=；由2ρ=，得曲线C 的普通方程为224x y +=， ………………………5分所以2d ==，所以直线l 被曲线C截得的弦长为=． (10)A BCDO·分（D ）解：根据柯西不等式，有2222222(23)(123)()x y z x y z ++≤++++，因232x y z ++=，所以222222421237x y z ++≥=++， ……5分当且仅当123x y z ==时等号成立，解得123,,777x y z ===，即当123,,777x y z ===时，222x y z ++取最小值27． ……10分22．解：（1）甲恰好通过两个项目测试的概率为223113()(1)228C -=． ……4分（2）因为每人可被录用的概率为22331111()(1)()2222C -+=，所以311(0)(1)28P X ==-=， 1123113(1)()(1)228P X C ==-=，2213113(2)()(1)228P X C ==-=，311(3)()28P X ===．故X 的概率分布表为： (8)分所以，X 的数学期望13313()012388882E X =⨯+⨯+⨯+⨯=． ……10分23．解：（1）22222212211212121212()()()b b a b a b a a b b a a a a ++=+++，因为0i a >，0i b >，所以222112120,0a b a b a a >>，则22211212122a b a b b b a a +≥=，所以22222121212121212()()2()b b a a b b b b b b a a ++≥++=+，即22121212()()b b a a a a ++212()b b ≥+．所以221212b b a a +≥21212()b b a a ++，当且仅当22211212a b a b a a =，即2112a b a b =时等号成立． ……2分推广：已知0i a >,0i b >(,1i N i n *∈≤≤)，则2221212n n b b b a a a +++21212()n nb b b a a a +++≥+++． ……………………………4分证明：①当1n =时命题显然成立；当2n =时，由上述过程可知命题成立； ②假设(2)n k k =≥时命题成立，即已知0i a >,0i b >(,1i N i k *∈≤≤)时，有2221212k k b b b a a a +++21212()k kb b b a a a +++≥+++成立，则1n k =+时，222222112112121121()()k k k k k k k k b b b b b b b b a a a a a a a a +++++++++++≥++++，由221212b b a a +≥21212()b b a a ++，可知222121*********()()k k k k k k k k b b b b b b b b a a a a a a a a ++++++++++++≥+++++++，故2222112121kk k k b bb b a a a a ++++++2121121()k k k k b b b b a a a a ++++++≥++++，故1n k =+时命题也成立．综合①②，由数学归纳法原理可知，命题对一切n N ∈*恒成立． ……6分（注：推广命题中未包含1n =的不扣分）（2）证明：由（1）中所得的推广命题知01213521nn n nnn C C C C +++++2222012135(21)35(21)nn n nnn C C C n C +=+++++[]2012135(21)35(21)n n n n n n C C C n C +++++≥+++++ ①， …8分记01235(21)nn n n n n S C C C n C =+++++，则1(21)(21)n n n n n n S n C n C C -=++-++，两式相加，得0122(22)(22)(22)(22)nn n n n n S n C n C n C n C =++++++++，012(22)()(22)2nn n n n n n C C C C n =+++++=+⨯，故(1)2n n S n =+⨯ ②，又[]2241(21)135(21)(1)(1)2n n n n ++⎡⎤+++++=⨯+=+⎢⎥⎣⎦③，将②③代入①，得222243012135(21)(1)(1)35(21)(1)22n n nn n n n n n n C C C n C n +++++++≥=++，所以，301213521(1)2nnnn nn n n C C C C ++++++≥，证毕． ……10分。

盐城市2018届高三年级第三次模拟考试

第6题图盐城市2018届高三年级第三次模拟考试数学试题(总分160分，考试时间120分钟)注意事项：1．本试卷考试时间为120分钟，试卷满分160分，考试形式闭卷． 2．本试卷中所有试题必须作答在答题卡上规定的位置，否则不给分．3．答题前，务必将自己的姓名、准考证号用0.5毫米黑色墨水签字笔填写在试卷及答题卡上．参考公式：锥体体积公式：13V Sh =，其中S 为底面积,h 为高. 圆锥侧面积公式：S rl π=，其中r 为底面半径,l 为母线长.样本数据12,,,n x x x ⋅⋅⋅的方差2211()n i i s x x n ==-∑，其中11n i i x x n ==∑.一、填空题（本大题共14小题，每小题5分，计70分. 不需写出解答过程，请把答案写在答题纸的指定位置上）1．已知(,]A m =-∞，(1,2]B =，若B A ⊆，则实数m 的取值范围为 ▲ ．2．设复数1a iz i+=+（i 为虚数单位）为纯虚数，则实数a 的值为 ▲ ． 3．设数据12345,,,,a a a a a 的方差为1，则数据123452,2,2,2,2a a a a a 的方差为 ▲ ．4．一个袋子中装有2个红球和2个白球（除颜色外其余均相同），现从中随机摸出2个球，则摸出的2个球中至少有1个是红球的概率为 ▲ ．5．“2,6x k k Z ππ=+∈”是“1sin 2x =”成立的 ▲条件（选填“充分不必要”、“必要不充分”、“充要”、“既不充分又不必要”）．6．运行如图所示的算法流程图，则输出S 的值为 ▲ ． 7．若双曲线22221(0,0)x y a b a b -=>>的两条渐近线与抛物线24y x =交于,,O P Q 三点，且直线PQ 经过抛物线的焦点，则该双曲线的离心率为 ▲ ．8．函数()ln(1f x =的定义域为 ▲ ．9．若一圆锥的底面半径为1，其侧面积是底面积的3倍，则该圆锥的体积为 ▲ ．10．已知函数())cos()(0,0)f x x x πωϕωϕωϕ=+-+><<为偶函数，且其图象的两条相邻对称轴间的距离为2π，则(8f π-的值为 ▲ ．11．设数列{}n a 的前n 项和为n S ，若*2()n n S a n n N =+∈，则数列{}n a 的通项公式为n a = ▲ ． 12．如图，在18AB B ∆中，已知183B AB π∠=，16AB =，84AB =，点234567,,,,,B B B B B B 分别为边18B B 的7等分点，则当9(18)i j i +=≤≤时，i j AB AB ⋅的最大值为▲ ．13．定义：点00(,)M x y 到直线:0l ax by c ++=的有向距离为．已知点(1,0)A -,(1,0)B ，直线m 过点(3,0)P ，若圆22(18)81x y +-=上存在一点C ，使得,,A B C 三点到直线m 的有向距离之和为0，则直线l 的斜率的取值范围为 ▲ ．14．设ABC ∆的面积为2，若,,A B C 所对的边分别为,,a b c ，则22223a b c ++的最小值为 ▲ ．二、解答题（本大题共6小题，计90分. 解答应写出必要的文字说明，证明过程或演算步骤，请把答案写在答题纸的指定区域内） 15．(本小题满分14分)在直四棱柱1111ABCD A B C D -中，已知底面ABCD 是菱形，,M N 分别是棱11,A D 11D C 的中点．（1）求证：AC ∥平面DMN ；（2）求证：平面DMN ⊥平面11BB D D ．16．(本小题满分14分)在ABC ∆中，角,,A B C 的对边分别为,,a b c ，AD 为边BC 上的中线．（1）若4a =，2b =，1AD =，求边c 的长；（2）若2AB AD c ⋅=，求角B 的大小．A B CD D 1A 1B 1C 1M N第15题图第12题图AB 1 B 2 B 3 B 4 B 5 B 6 B 7817．(本小题满分14分)如图，是一个扇形花园，已知该扇形的半径长为400米，2AOB π∠=，且半径OC 平分AOB ∠．现拟在OC 上选取一点P ，修建三条路PO ,PA ,PB 供游人行走观赏，设PAO α∠=．（1）将三条路PO ,PA ,PB 的总长表示为α的函数()l α，并写出此函数的定义域；（2）试确定α的值，使得()l α最小．18．(本小题满分16分)如图，已知12,F F 分别是椭圆2222:1(0)x y C a b a b+=>>的左、右焦点，点(2,3)P -是椭圆C 上一点，且1PF x ⊥轴．（1）求椭圆C 的方程；（2）设圆222:()(0)M x m y r r -+=>．①设圆M 与线段2PF 交于两点,A B ，若2MA MB MP MF +=+，且2AB =，求r 的值； ②设2m =-，过点P 作圆M 的两条切线分别交椭圆C 于,G H 两点（异于点P ）．试问：是否存在这样的正数r ，使得,G H 两点恰好关于坐标原点O 对称？若存在，求出r 的值；若不存在，请说明理由．19．(本小题满分16分)若对任意实数,k b 都有函数()y f x kx b =++的图象与直线y kx b =+相切，则称函数()f x 为“恒切函数”．设函数()xg x ae x pa =--，,a p R ∈．（1）讨论函数()g x 的单调性；（2）已知函数()g x 为“恒切函数”．①求实数p 的取值范围；②当p 取最大值时，若函数()()xh x g x e m =-也为“恒切函数”，求证：3016m ≤<．（参考数据：320e ≈）20．(本小题满分16分)在数列{}n a 中，已知121,a a λ==，满足111221222,,,,n n n n a a a a ---++⋅⋅⋅是等差数列（其中2,n n N ≥∈），且当n 为奇数时，公差为d ；当n 为偶数时，公差为d -．（1）当1λ=，1d =时，求8a 的值；（2）当0d ≠时，求证：数列{}2*22||()n n a a n N +-∈是等比数列；（3）当1λ≠时，记满足2m a a =的所有m 构成的一个单调递增数列为{}n b ，试求数列{}n b 的通项公式．AO BCPα第17题图盐城市2018届高三年级第三次模拟考试数学附加题部分（本部分满分40分，考试时间30分钟）21．[选做题]（在A 、B 、C 、D 四小题中只能选做2题，每小题10分，计20分．请把答案写在答题纸的指定区域内）A .（选修4-1：几何证明选讲）如图，已知半圆O 的半径为5，AB 为半圆O 的直径，P 是BA 延长线上一点，过点P 作半圆O 的切线PC ，切点为C ，CD AB ⊥于D ．若2PC PA =，求CD 的长．B .（选修4-2：矩阵与变换）已知矩阵 2 0 a b ⎡⎤=⎢⎥⎣⎦M 的属于特征值1的一个特征向量为11⎡⎤⎢⎥⎣⎦，求矩阵M 的另一个特征值和对应的一个特征向量．C ．（选修4-4：坐标系与参数方程）在平面直角坐标系中，直线l的参数方程为122x y ⎧=-⎪⎪⎨⎪=⎪⎩（t 为参数）．以坐标原点O 为极点，x轴的正半轴为极轴建立极坐标系（单位长度相同），设曲线C 的极坐标方程为2ρ=，求直线l 被曲线C 截得的弦长．D ．(选修4-5：不等式选讲）已知正数,,x y z 满足232x y z ++=，求222x y z ++的最小值．[必做题]（第22、23题，每小题10分，计20分．请把答案写在答题纸的指定区域内） 22．（本小题满分10分）某公司的一次招聘中，应聘者都要经过三个独立项目,,A B C 的测试，如果通过两个或三个项目的测试即可被录用．若甲、乙、丙三人通过,,A B C 每个项目测试的概率都是12．（1）求甲恰好通过两个项目测试的概率；（2）设甲、乙、丙三人中被录用的人数为X ，求X 的概率分布和数学期望． 23．（本小题满分10分）（1）已知*0,0()i i a b i N >>∈，比较221212b b a a +与21212()b b a a ++的大小，试将其推广至一般性结论并证明；（2）求证：3*01213521(1)()2n nn n nn n n n N C C C C ++++++≥∈． A BCDO· 第21（A ）图盐城市2018届高三年级第三次模拟考试数学参考答案一、填空题：本大题共14小题，每小题5分，计70分.1．2m ≥ 2．1- 3．4 4．565．充分不必要 6．21 78．(2,3] 9．3 1011．12n - 12．1327 13．3(,]4-∞- 14．二、解答题：本大题共90小题.15．（1）证明：连接11A C ，在四棱柱1111ABCD A B C D -中，因为11//AA BB ，11//BB CC ，所以11//AA CC ，所以11A ACC 为平行四边形，所以11//A C AC ． ……2分又,M N 分别是棱11,A D 11D C 的中点，所以11//MN AC ，所以//AC MN ． ……4分又AC ⊄平面DMN ，MN ⊂平面DMN ，所以AC ∥平面DMN ． ……6分（2）证明：因为四棱柱1111ABCD A B C D -是直四棱柱，所以1DD ⊥平面1111A B C D ，而MN ⊂平面1111A B C D ，所以1MN DD ⊥． ……8分又因为棱柱的底面ABCD 是菱形，所以底面1111A B C D 也是菱形，所以1111A C B D ⊥，而11//MN AC ，所以11MN B D ⊥．……10分又1MN DD ⊥，111,DD B D ⊂平面1111A B C D ，且1111DD B D D =，所以MN ⊥平面1111A B C D ． ……12分而MN ⊂平面DMN ，所以平面DMN ⊥平面11BB D D ． ……14分16．解：（1）在ADC ∆中，因为11,2,22AD AC DC BC ====，所以由余弦定理，得2222222217cos 22228AC DC AD C AC DC +-+-===⋅⨯⨯． ……3分故在ABC ∆中，由余弦定理，得2222272cos 4224268c a b ab C =+-=+-⨯⨯⨯=，所以c = ……6分（2）因为AD 为边BC 上的中线，所以1()2AD AB AC =+，所以21()2c AB AD AB AB AC =⋅=⋅+221111cos 2222AB AB AC c cb A =+⋅=+，得cos c b A =． ……10分则2222b c a c b bc+-=⋅，得222b c a =+，所以90B =︒． ……14分17．解：（1）在APO ∆中，由正弦定理，得sin sin sin AP OP AOAOP PAO APO==∠∠∠，即400sin sin sin()44AP OP ππαα==+，从而sin()4AP πα=+，400sin sin()4OP απα=+． ……4分所以()l α＝400sin 22sin()sin()44OP PA PB OP PA απαα++=+=+++，故所求函数为sin )()sin()4l ααπα=+，3(0,)8πα∈． ……6分（2）记3()(0,)8sin()4f πααα==∈+，因为2(sin cos )(2)(cos sin )()(sin cos )f ααααααααα+--'=+2)4(sin cos )πααα-=+， ……10分由()0f α'=，得1sin()42πα-=-，又3(0,)8πα∈，所以12πα=． ……12分列表如下：所以，当12πα=时，()l α取得最小值．答：当12πα=时，()l α最小． ……14分18．解：（1）因点(2,3)P -是椭圆C 上一点，且1PF x ⊥轴，所以椭圆的半焦距2c =，由22221c y a b+=，得2b y a =±，所以2243b a a a -==， ……2分化简得2340a a --=，解得4a =，所以212b =，所以椭圆C 的方程为2211612x y +=． ……4分（2）①因2MA MB MP MF +=+，所以2MA MP MF MB -=-，即2PA BF =，ABCDDABCM N所以线段2PF 与线段AB 的中点重合（记为点Q ），由（1）知3(0,)2Q ， ……6分因圆M 与线段2PF 交于两点,A B ，所以21MQ AB MQ PF k k k k ⋅=⋅=-，所以303021m --⋅=-，解得98m =-，……8分所以158MQ ==，故178r ==. ……10分② 由,G H 两点恰好关于原点对称，设00(,)G x y ，则00(,)H x y --，不妨设00x <，因(2,3)P -，2m =-，所以两条切线的斜率均存在，设过点P 与圆M 相切的直线斜率为k ，则切线方程为3(2)y k x -=+，即230kx y k -++=，由该直线与圆M 相切，得r =，即k =12分所以两条切线的斜率互为相反数，即PG PH k k =-，所以00003322y y x x ---=-+-+，化简得006xy =-，即006y x -=，代入220011612x y +=，化简得420016480x x -+=，解得02x =-（舍），0x =-0y ……14分所以(G -，H，所以332PG k -==，所以r ==. 故存在满足条件的r ，且7r =． ……16分 19．解：（1）()1xg x ae '=-， ……2分当0a ≤时，()0g x '<恒成立，函数()g x 在R 上单调递减；当0a >时，由()0g x '=得ln x a =-，由()0g x '>得ln x a >-，由()0g x '<得ln x a <-，得函数()g x 在(,ln )a -∞-上单调递，在(ln ,)a -+∞上单调递增． ……4分（2）①若函数()f x 为“恒切函数”，则函数()y f x kx b =++的图像与直线y kx b =+相切，设切点为00(,)x y ，则0()f x k k '+=且000()f x kx b kx b ++=+，即0()0f x '=，0()0f x =. 因为函数()g x 为“恒切函数”，所以存在0x ，使得0()0g x '=，0()0g x =，即0000 10xx ae x pa ae ⎧--=⎪⎨-=⎪⎩，得00x a e -=>，00(1)x p e x =-，设()(1)x m x e x =-， ……6分则()xm x xe '=-，()0m x '<，得0x >，()0m x '>，得0x <，故()m x 在(,0)-∞上单调递增，在(0,)+∞上单调递减，从而[]max ()(0)1m x m ==，故实数p 的取值范围为(,1]-∞． ……8分 ②当p 取最大值时，1p =，00x =，01x a e-==，()(1)x x h x e x e m =---，()(22)x x h x e x e '=--，因为函数()h x 也为“恒切函数”，故存在0x ，使得0()0h x '=，0()0h x =，由0()0h x '=得000(22)0xx e x e--=，00220x e x --=，设()22x n x e x =--， ……10分则()21xn x e '=-，()0n x '>得ln 2x >-，()0n x '<得ln 2x <-，故()n x 在(,ln 2)-∞-上单调递减，在(ln 2,)-+∞上单调递增，1°在单调递增区间(ln 2,)-+∞上，(0)0n =，故00x =，由0()0h x =，得0m =；…12分 2°在单调递减区间(,ln 2)-∞-上，2(2)20n e--=>，31223111()22(20)02222n e ---=-≈⨯-=-<，又()n x 的图像在(,ln 2)-∞-上不间断，故在区间3(2,)2--上存在唯一的0x ，使得00220x e x --=，故0022xx e +=，此时由0()0h x =，得00000022(1)(1)22x xx x m e x e x ++=--=-- 001(2)4x x =-+2011(1)44x =-++，函数211()(1)44r x x =-++在3(2,)2--上递增，(2)0r -=，33()216r -=，故3016m <<．综上1°2°所述，3016m ≤<． ……16分20．解：（1）由1λ=，1d =，所以21a =，234,,a a a 为等差数列且公差为1-，所以4221a a =-=-，又458,,a a a 为等差数列且公差为1，所以8443a a =+=． ……2分（2）当21n k =+时，22221221222,,,,k k k k a a a a +++⋅⋅⋅是等差数列且公差为d ，所以2122222k k k a a d +=+，同理可得22121222k k k a a d --=-， ……4分两式相加，得212121222k k k a a d +---=；当2n k =时，同理可得2222222k k ka a d +-=-， ……6分所以222||2n n na a d +-=．又因为0d ≠，所以21122122||22(2)||2n n n n n n a a n a a ++---==≥-，所以数列{}2*22||()n n a a n N +-∈是以2为公比的等比数列． ……8分（3）因为2a λ=，所以4222a a d d λ=-=-，由（2）知212121222k k k a a d +--=+，所以212123212321222222k k k k k k a a d a d d +-----=+=++，依次下推，得211132321222222k k k a a d d d d +--=+++++，所以21222(21)3k ka d λ+=+-， ……10分当212222k k n ++≤≤时，212321222(2)()33k k k n a a n d n d λ+++=--=+--，由2m a a =，得232233k m +=-，所以23212233k k b ++=-，所以22233n n b +=-（n 为奇数）； ……12分由（2）知222222222222222k k k k k k a a d a d d +--=-=--，依次下推，得22224222222222k k ka a d d d d +-=-----，所以22224(21)23k ka d d λ+-=--， ……14分当222322k k n ++≤≤时，222422222(2)()33k k k n a a n d n d λ+++=+-=+--，由2m a a =，得242233k m +=+，所以24222233k k b ++=+．所以22233n n b +=+（n 为偶数）．综上所述，2222(3322(33n n n n b n ++⎧+⎪⎪=⎨⎪-⎪⎩为偶数）为奇数）． ……16分方法二：由题意知，23121231222222n n n n n b b b b b +++=<<<<<⋅⋅⋅<<<<<<⋅⋅⋅， ……10分当n 为奇数时，1221222,,,,n n n n a a a a +++⋅⋅⋅的公差为d -，1112221222,,,,n n n n a a a a ++++++⋅⋅⋅的公差为d ，所以112(2)()n n n b n a a b d ++=---，11112(2)n n n b n a a b d ++++=+-，则由12n n b b a a a +==，得111(2)()(2)n n n n b d b d +++---=-，即212n n n b b +++=．同理，当n 为偶数时，也有212n n n b b +++=．故恒有2*12()n n n b b n N +++=∈． ……12分 ①当n 为奇数时，由3212n n n b b ++++=，212n n n b b +++=，相减，得222n n n b b ++-=，所以532311()()(222)2n n n n b b b b b b -=-+⋅⋅⋅+-+=+⋅⋅⋅+++13222(14)2221433n n -+-=+=--．……14分②当n 为偶数时，同理可得22233n n b +=+．综上所述，2222(3322(33n n n n b n ++⎧+⎪⎪=⎨⎪-⎪⎩为偶数）为奇数）． ……16分附加题答案21．（A ）解：连,AC BC ，因PC 为半圆O 的切线，所以PCA B ∠=∠．又P P ∠=∠，所以PCA ∆∽PBC ∆，所以12PA AC PC BC ==，即2AC BC =． ……5分因AB 为半圆O 的直径，所以22225AB AC BC AC =+=，因半圆O 的半径为5，所以21005AC =，所以AC BC ==由射影定理，得2AC AD AB =⋅，解得2AD =，所以4CD ==． ……10分（B ）解：由题意得 2 110 11a b ⎡⎤⎡⎤⎡⎤=⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦，解得11a b =-⎧⎨=⎩，所以 2 10 1-⎡⎤=⎢⎥⎣⎦M ． ……2分矩阵M 的特征多项式为 2 1()(2)(1)0 1f λλλλλ-==---，由()0f λ=，得2,1λλ==，所以矩阵M 的另一个特征值为2． ……6分此时0 1()0 1f λ=，对应方程组为010010x y x y ⋅+⋅=⎧⎨⋅+⋅=⎩，所以0y =，所以另一个特征值2对应的一个特征向量为10⎡⎤⎢⎥⎣⎦． ……10分（C ）解：直线的普通方程为10x y +-=；由2ρ=，得曲线C 的普通方程为224x y +=， ………………………5分所以2d ==，所以直线l 被曲线C截得的弦长为=． ……10分（D ）解：根据柯西不等式，有2222222(23)(123)()x y z x y z ++≤++++，因232x y z ++=，所以222222421237x y z ++≥=++， ……5分当且仅当123x y z ==时等号成立，解得123,,777x y z ===，即当123,,777x y z ===时，222x y z ++取最小值27． ……10分22．解：（1）甲恰好通过两个项目测试的概率为223113()(1)228C -=． ……4分A BCDO·（2）因为每人可被录用的概率为22331111()(1)()2222C -+=，所以311(0)(1)28P X ==-=， 1123113(1)()(1)228P X C ==-=，2213113(2)()(1)228P X C ==-=，311(3)()28P X ===．故X 的概率分布表为：…………8分所以，X 的数学期望13313()012388882E X =⨯+⨯+⨯+⨯=． ……10分23．解：（1）22222212211212121212()()()b b a b a b a a b b a a a a ++=+++，因为0i a >，0i b >，所以222112120,0a b a b a a >>，则22211212122a b a b b b a a +≥=，所以22222121212121212()()2()b b a a b b b b b b a a ++≥++=+，即22121212()()b b a a a a ++212()b b ≥+．所以221212b b a a +≥21212()b b a a ++，当且仅当22211212a b a ba a =，即2112a b a b =时等号成立． ……2分推广：已知0i a >,0i b >(,1i N i n *∈≤≤)，则2221212n n b b b a a a +++21212()n nb b b a a a +++≥+++． ……………………………4分证明：①当1n =时命题显然成立；当2n =时，由上述过程可知命题成立； ②假设(2)n k k =≥时命题成立，即已知0i a >,0i b >(,1i N i k *∈≤≤)时，有2221212k k b b b a a a +++21212()k kb b b a a a +++≥+++成立，则1n k =+时，222222112112121121()()k k k k k k k k b b b b b b b b a a a a a a a a +++++++++++≥++++，由221212b b a a +≥21212()b b a a ++，可知222121*********()()k k k k k k k k b b b b b b b b a a a a a a a a ++++++++++++≥+++++++，故2222112121k k k k b b b b a a a a ++++++2121121()k k k k b b b b a a a a ++++++≥++++，故1n k =+时命题也成立．综合①②，由数学归纳法原理可知，命题对一切n N ∈*恒成立． ……6分（注：推广命题中未包含1n =的不扣分）（2）证明：由（1）中所得的推广命题知01213521nn n nnn C C C C +++++2222012135(21)35(21)nn n n n n C C C n C +=+++++[]2012135(21)35(21)n n n n nn C C C n C +++++≥+++++ ①， …8分记01235(21)nn n n n n S C C C n C =+++++，则10(21)(21)n n n n n n S n C n C C -=++-++，两式相加，得0122(22)(22)(22)(22)nn n n n n S n C n C n C n C =++++++++，012(22)()(22)2n n n n n n n C C C C n =+++++=+⨯，故(1)2n n S n =+⨯ ②，又[]2241(21)135(21)(1)(1)2n n n n ++⎡⎤+++++=⨯+=+⎢⎥⎣⎦③，将②③代入①，得222243012135(21)(1)(1)35(21)(1)22nn nn n n n n n n C C C n C n +++++++≥=++，所以，301213521(1)2n nn n nn n n C C C C ++++++≥，证毕． ……10分。

江苏省盐城市2018届高三第三次模拟考试英语含答案解析

答案：A 解析：根据上下文语境，此处需要一个表示“解释”的动词，而A选项 “explain”恰好符合语境，表示“解释”。
04 解析：根据上下文语境，此处需要一个表示“解释”的动词，而A选项 “explain”恰好符合语境，表示“解释”。
语法填空答案解析
答案：本题考查的是动词时态语态，根据语境可知，此处应用一般过去时态。答案：本题考查的是非谓语动词，此处应用动词-ing形式作伴随状语。答案：本题考查的是名词性从句，此处应用what引导宾语从句。答案：本题考查的是定语从句，此处应用that引导定语从句修饰先行词。
江苏省盐城市2018届高三第三次模拟考试英语
目录
单击此处添加文本考试概况试卷分析答案解析备考建议
考试科目：英语
考试时间
考试时间：2018年5月19日（星期六）上午9:00-11:30
总分：150分
考试时长：1小时30分钟
考试科目：英语
考试形式
考试时间：2018年5月4日
考试形式：笔试
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多做模拟试题和真题，熟悉考试题和语法规则多读英文文章，提高语言表达能力练习写作时注意时态、语态、主谓一致等细节问题多写多练，不断修改和优化自己的作文
阅读理解答案解析
答案：C 解析：根据文章第一段可知，作者认为人们应该更加关注自己的健康，而不是过度关注自己的外貌。因此，正确答案是C，即人们应该更加关注自己的健康。解析：根据文章第一段可知，作者认为人们应该更加关注自己的健康，而不是过度关注自己的外貌。因此，正确答案是C，即人们应该更加关注自己的健康。

江苏省盐城市2018届高三第三次模拟考试英语试卷含答案

盐城市2018届高三年级第三次模拟考试英语试题第一部分听力（共两节，满分20分）做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节（共5小题；每小题1分，满分5分）听下面 5 段对话。

每段对话后有一个小题，从题中所给的 A、B、C 三个选项中选出最佳选项。

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 is the woman probably doing?A. Watching a movie.B. Reading a newspaper.C. Making an advertisement.2. What are the speakers talking about in general?A. Their best memories of a relaxing holiday.B. Their travelling plans for the summer holiday.C. Their favorite ways of travelling around the world.3. When will the meeting begin?A. At 3:20.B. At 3:40.C. At 4:00.4. Where are the speakers?A. In a shop.B. In a restaurant.C. In the man’s house.5. What does the woman mean?A. She doesn’t need the man’s help.B. She expects the man to move the desk.C. She wants to remove the books from the desk.第二节（共15小题；每小题1分，满分15分）听下面 5 段对话或独白。

盐城市2018届高三年级第三次模拟考试英语试题含答案

盐城市2018届高三年级第三次模拟考试英语试题第一部分听力（共两节，满分20分）做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节（共5小题；每小题1分，满分5分）听下面5段对话。

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

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. Who will go to London with the man?A. Linda.B. Bruce.C. Mike.2. What does the man do?A. A doctor.B. A coach.C. A teacher.3. What does the man mean?A. She had better not wear sunglasses.B. She is a humorous and lovely woman.C. She can choose the sunglasses on the left.4. Why did the man give up his job?A. He will study in another city.B. The wage is far from satisfactory.C. He dislikes the location of the company.5. What time do you think it is now?A. 9:15.B. 10:30.C. 11:15.第二节（共15小题；每小题1分，满分15分）听下面5段对话或独白。

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

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

江苏省盐城市2018届高三第三次模拟考试英语试卷含答案

盐城市2018届高三年级第三次模拟考试英语试题第一部分听力（共两节，满分20分）做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节（共5小题；每小题1分，满分5分）听下面 5 段对话。

每段对话后有一个小题，从题中所给的 A、B、C 三个选项中选出最佳选项。

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。

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盐城市2018届高三年级第三次模拟考试英语试题第一部分: 听力(共两节，满分20分)做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的各案转涂到答题卡上。

第一节(共5小题，每小题1分，满分5分)听下面5段对话。

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

听完每段对话后，你都有10 秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

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

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

每段对话或独白读两遍。

听第6段材料，回答第6至7题。

6. What does the man say about his job?A.The boss is really nice.B. Workmates look friendly.C. Workplace is small and noisy.7.What can we learn from the conversation?A. The boss told the man all the rules.B. Mark will become the boss this year.C. The man will cooperate with Mark.听第7段材料，回答第8 至9题。

8.Why does the woman talk to the man?A.He lost his student visa.B. He often makes trouble.C. He is often absent from class.9.What is the worst result of the man’s behavior?A.He’ll be sent back home.B.He’ll have to restart his course.C.He’ll stay in the police station.听第8段材料，回答第10 至12题。

10.What does the man mainly mention about Helen?A.Changes in her character.B. Problems with teachers.C. Relationships with parents.11.What does the man decide to do in the end?A.Have a talk with Helen.B. Talk with Helen’s father.C. Spend more time with Helen.12.What can we learn about Helen?A.Her parents are very busy.B. Classmates dislike her.C. Teachers worry about her.听第9段材料，回答第13 至16题。

13.What did Nick and Mel do when they were young?A.They learned to act at school.B.They went to live concerts with their father.C.They were encouraged to play music by their mother14.When was the band Krispy started?A.After Nick and Mel played together.B.After Nick began studying at a music school.C.After two musicians saw Nick and Mel playing.15.What do we know about the band in the first year?A.They recorded two albums.B.They joined a music company.C.They were welcome in the concert16.What has disappointed the man?A.Parents forbade them to join a company.B.His illness delayed a new album recording.C.The first album has sold under a million copies.听第10段材料，回答第17 至20题。

17.What can be clearly affected if teenagers lack sleep?A.Their spirits.B. Their long-term health.C. Their academic performance.18.What is to blame for teenagers not getting enough sleep according to the speaker?A.Poor diet.B. Lack of exerciseC. Too much entertainment19.What advice does the speaker give to teenagers who have trouble getting to sleep?A.Listen to music.B. Read a book.C. Drink hot chocolate.20.What does the speaker suggest schools should do?A.Start lessons later.B. Shorten the school day.C. Offer classes in the evenings.第二部分: 英语知识运用(共两节，满分35分)第一节: 单项填空(共15 小题: 每小题1分，满分15 分)请认真阅读下面各题，从题中所给的A、B、C、D 四个选项中，选出最佳选项，并在答题卡上将该项涂黑。

21. Thomas made _____ his concerns about the changes that had been introduced at work.A. plainB. perfectC. easyD. ambiguous22. Over the last four decades, the Chinese people have _______ enhanced productivity through hard work with a firm spirit.A. roughlyB. merely C significantly D. equally23. ----Can you give me some advice on how to deal with that tough customer?---- I ______ to her instead of trying to explain over the phone if in your shoes.A. will writeB. would writeC. were to writeD. would have written24. My printer is five years old, so I’d like to ________ it and purchase a new one online.A. deleteB. retireC. refreshD. declare25. ----You should have heard Tom yesterday?---- Sure. He’s charged with taking the car without the owner’s _______.A. enquiryB. privilegeC. commandD. consent26. In terms of the urban effects of the technology, our cities are _______ for self-driving cars.A. nowhere near readyB. near nowhere readyC. anywhere near readyD. near anywhere ready27. I sent in my application three weeks ago and they’re probably not going to call me back. I’m still ______ hope, though.A. holding outB. working outC. figuring outD. setting out28. Many graduates aren’t prepared for what the future has _______ when they come out of school.A. in demandB. at easeC. at lengthD. in store29. _______ to finding ways to make music accessible to all children is Ms. Thompson with a touring company of professional musicians.A. DevotingB. DevotedC. Being devotedD. Having been devoted30. By December 2017, the number of Chinese netizens reached 772 million, ______ 753 million surf the Internet via mobile phones.A. whichB. whoseC. of whichD. for whom31. Many of life’s failures are experienced by people who did not realize how close they were to success _______ they gave up.A. onceB. whileC. whenD. unless32. Never throughout history has a man who _______ a life of ease left a name worth remembering.A. livesB. livedC. would liveD. had lived33. If we worked on the assumption _______ what is accepted as true really is true, then there would be little hope for advance.A. whereB. ifC. whetherD. that34. It had been many years since my last visit, but I found the house by _____.A. losing my headB. catching my eyeC. biting my tongueD. following my nose35. –Wall Street English just launched a new WeChat learning platform and we can take free classes!---______! Let me have a look!A. You don’t sayB. You have got itC. You have me thereD. You can say that again.第二节：完形填空(共20个小题，每小题1分，满分20分)请认真阅读下面短文，从短文后各题所给的A.、B、C 和D 四个选项中，选出可以填入空白处的最佳选项，并在答题卡上将该项涂黑。

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