黄浦区2017学年高考模拟

黄浦区2018年高中学业等级考调研测试物理试卷2018.4.12考生注意：1、答题前，务必在试卷与答题纸上填写学校、姓名、准考证号。

2、试卷满分100分，考试时间60分钟。

3、本考试分设试卷和答题纸。

试卷包括三大部分，第一部分为单项选择题，第二部分为填空题，第三部分为综合题。

4、作答必须涂或写在答题纸上相应的位置，在试卷上作答无效。

一、单项选择题（共40分，1至8题每小题3分，9至12题每小题4分。

每小题只有一个正确选项） 1．下列射线中，属于电磁波的是（ ）（A ）阴极射线 （B ）α射线 （C ）β射线 （D ）γ射线2．白光通过双缝后产生的干涉条纹是彩色的，其原因是不同色光具有不同的（ ） （A ）传播速度 （B ）光强 （C ）振动方向 （D ）波长3．在一个23892U 原子核衰变为一个20682Pb 原子核的过程中，释放出的α粒子个数为（ ） （A ）5 （B ）8 （C ）10 （D ）164．三束单色光①、②、③的频率分别为ν1、ν2、ν3（已知ν1<ν2<ν3）。

某种金属在光束②照射下，能产生光电子。

则该金属（ ）（A ）在光束①照射下，一定能产生光电子 （B ）在光束①照射下，一定不能产生光电子 （C ）在光束③照射下，一定能产生光电子 （D ）在光束③照射下，一定不能产生光电子5．2017年4月“天舟一号”货运飞船与“天宫二号”空间实验室完成了首次交会对接，对接形成的组合体仍沿天宫二号原来的轨道（可视为圆周）运行。

与天宫二号单独运行时相比，组合体运行的（ ） （A ）周期变大 （B ）动能变大 （C ）速率变大 （D ）向心加速度变大6．电场中有a 、c 两点，一正电荷由a 点运动到c 点电势能减少，由此可知（ ） （A ）a 点的场强大于c 点的场强 （B ）a 点的电势高于c 点的电势（C ）电场线的方向由a 沿直线指向c （D ）该正电荷在a 点的动能大于在c 点的动能7．一个质点做简谐运动，其位移随时间变化的s-t 图像如右图。

黄浦区2017年高考模拟考英语试卷（完卷时间：120分钟满分：140分）2017年4月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. 2 litres. B. 13 litres. C. 26 litres. D. 52 litres.2. A. In an English class. B. In a swimming pool. C. On a bus. D. In a sporting goods store.3. A. By bus. B. By underground. C. By taxi. D. By car.4. A. Doctor and patient. B. Teacher and student.C. Employer and Employee.D. Salesman and customer.5. A. Have a lesson. B. Take a test. C. See a film. D. Go to bed.6. A. Difficult. B. Memorable. C. Uninteresting. D. Worthwhile.7. A. She wants a bottle of juice. B. She’d like some alcohol.C. The red wine in this bar is perfect.D. The location of the bar is unknown.8. A. An excellent résumé. B. An entry form.C. A job offer.D. The position of system engineer.9. A. It’s famous. B. It’s professional. C. It’s expensive. D. It’s cheating.10. A. The 26-month-old baby is always busy watching videos.B. TV and videos may hurt a child’s language development.C. Nothing can replace parents in kids’ language development.D. Children usually watch TV too passively to learn something.Section BDirections: In Section B, you will hear one longer conversation and two short passages, and you will be asked several questions on each of the conversation and the passages. The conversation and 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 11 through 13 are based on the following dialogue.11. A. Encouraging. B. Dishonest. C. Interesting. D. Nervous.12. A. How to start his own business. B. How to develop a real interest.C. How to speak to a woman bravely.D. How to balance his study and work.13. A. He has too loose a schedule. B. He loves the feeling with students.C. He is dissatisfied with his current job.D. He wants to determine his future development. Questions 14 through 16 are based on the following passage.14. A. Kids threw litter everywhere. B. The camp director gave rude orders.C. Some mysterious plastic litter was found.D. Kids’ joint efforts led to a clean camp.15. A. By taking pictures of litter he picked up.B. By sharing photos of the terribly dirty planet.C. By keeping a record of crowdsourced cleaning-up.D. By inspiring kids to pick up five pieces of litter every day.16. A. There is strength in numbers. B. Birds can help to pick up litter.C. Litter is artistic and approachable.D. More straws should be used in the café.Questions 17 through 20 are based on the following passage.17. A. To enable students to reject violence. B. To help students face struggles more properly.C. To improve students’ health.D. To eliminate poverty more effectively.18. A. How to calm down by talking to experts. B. How to clear their mind throughout the day.C. How to make their teachers happy.D. How to respond to situations better.19. A. More students dropped out last year. B. There is less bad behavior on campus.C. Students are less responsible for their study.D. More students are willing to be sent to the office.20. A. Its effect remains to be seen. B. Everyone can benefit from it.C. It helps to get rid of poverty to some extent.D. There is enough evidence to show its significance.II. Grammar and VocabularySection ADirections: After reading the passage below, fill in the blanks to make the passage coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.Should Children Ban Their Parents from Social Media?It might be taken for granted - but no previous generation of children will have had the experience of having their entire childhoods intensively and publicly documented in this way. But the very first people to have had some of their childhood pictures __21__ (post) online are not always happy about their formative years being preserved in digital world.Parents may not realize it, but by posting photos and videos of their children online, they are creating an identity for their children __22__ might not be welcomed. Lucy is a good example. She said she had asked her dad to de-tag her from “stuff that doesn’t necessarily represent __23__ I am now. That’s not something I’d want to remember every time I log on to Facebook… It isn’t the best memories, which is the way you’d like to reveal __24__ on social media.”Stories about online privacy are often about children and teenagers being warned of the dangers of publishing too much personal information online. But in this case it’s their parents who are in the spotlight. For some parents, __25__ (safe) option is avoiding social media altogether.Kasia Kurowska from Newcastle is expecting her first child in June and has agreed with her partner Leeto impose a blanket ban __26__ her children are old enough to make their own decisions about social media. But she has two big concerns about her plan. Firstly, it will be difficult __27__ (impose). “When their auntie comes round and takes a picture, we’re going to have to be like paparazzi police, saying, please don’t put these on Facebook. And secondly, the child might dislike __28__ (not own) an online presence, especially if all of their friends do. But I __29__ (keep) a digital record of them. It just won’t have been shared on a platform __30__ the masses.”Section BDirections: Fill in each blank with a proper word chosen from the box. Each word can be used only once.III. Reading ComprehensionSection ADirections: For each blank in the following passages 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.Being Bigger isn’t Necessarily Considered BetterThe firm, which famously started life in 1939, has now declared a new age: that of smaller start-up. By 2014, when Ms Whitman announced HP’s decision to separate its computer and printer business from its corporate hardware and services operations, the company had grown into a clumsy __41__. Its fortunes started to __42__ with a series of expensive and much criticized purchases. By 2012 it had lost its position as theworld’s leading supplier of PCs to Lenovo. The dramatic __43__ was aimed at helping the firm adapt to the new age of mobile and online computing, responding to shareholder demands for more aggresive__44__.“I would go from laser jet printing to our big enterprise services contracts where we were running the back end of IT for many big companies and organizations. These two things are not like each other. So the ability to focus and engage with customers on a(n) __45__ set of objectives and business outcomes... I can already see the difference.” Ms Whitmann, who now heads the new spin-off, Hewlett Packard Enterprise (HPE) selling servers and services, says the change has already __46__ her performance. “One big change is it __47__ each of the divisions to pursue the strategy that is right for them. __48__ , there is ‘no way’ printer and PC company HP Inc’s decision last year to buy Samsung’s printing business for $1bn would have happened when it was part of the larger firm. So it’s that ability to drive your own program, not __49__ by other businesses that don’t have the same characteristics.” Ms Whitman is so convinced her strategy is working that she’s __50__ HPE further, spinning off both its business services division and its software business into separate companies last year.Her assumption that bigger doesn’t always mean better seems __51__. After all, a larger company should find it easier to dominate the market it operates in. But the rapid rise of much smaller start-ups, competing and often overtaking these established powerful companies means the accepted wisdom that __52__ equals success is being challenged. __53__ in 2014, eBay carved PayPal, the electronic payments arm it bought in 2001, off from the main online sale business.Box, a cloud storage company, is another case in point. Founder Aaron Levie says “Whether Uber, Airbnb, those same lessons __54__, which is if you can build something that’s cheaper, faster and more scalable and delivers a far better customer experience than what the traditional sellers were able to do, then you can be extremely __55__.”41. A. appearance B. construction C. giant D. possession42. A. decline B. increase C. stay D. vary43. A. adventure B. combination C. development D. split44. A. behavior B. growth C. markets D. policies45. A. ambitious B. complex C. narrow D. overall46. A. delivered B. improved C. measured D. standardized47. A. allows B. employs C. reminds D. threatens48. A. All in all B. For example C. On the contrary D. What’s more49. A. held back B. kept on C. looked over D. taken down50. A. dissolved B. expanded C. operated D. shrunk51. A. fundamental B. reasonable C. surprising D. widespread52. A. diligence B. discipline C. profit D. size53. A. Comparatively B. Generally C. Similarly D. Unexpectedly54. A. apply B. fail C. hide D. increase55. A. friendly B. miserable C. motivated D. troublesomeSection BDirections: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)Born in 1823 in Wales, Alfred Russel Wallace was a man of modest means, but he had a passion for nature and he chose to follow it. He started out collecting insects as a hobby, but eventually his longing for adventure led him to explore the world.Luckily for Wallace, Victorian Britain was discovering an interest in weird and wonderful insects, so the demand from museums and private collections for these beasts was growing. Wallace was able to make a living doing what he loved: collecting beetles and other insects.But his first trip of exploring the world ended in disaster. Wallace proceeded to the Amazon in South America. Its giant forests promised a wealth of new species, sure to put him on the scientific map. The trip took 6 weeks and involved every mode of transport in existence at the time. After four years Wallace set off for home, but his boat caught fire in the middle of the Atlantic. Everyone survived, but Wallace had to watchin despair as his samples went up in flames – including live animals he was bringing home that were trying to jump free of the flames. But he did not let it stop him.In 1854, Wallace set off on another adventure, this time to the Malay Archipelago. Wallace found himself humbled by the new and exciting things he saw. He later recalled: “As I lie listening to these interesting sounds, I think how many besides myself have longed to see with their own eyes the many wonderful and beautiful things which I am daily encountering.”In 1858, Wallace wrote what became known as the “Ternate essay”: a piece of writing that was to change our understanding of life forever. In his essay, Wallace argued that a species would only turn into another species if it was struggling for existence. Henry W. Bates was one of many scientists delighted by the idea of evolution by natural selection. In a letter to Wallace, he wrote: “The idea is like truth itself, so simple and obvious that those who read and understand it will be struck by its simplicity; and yet it is perfectly original.”56. __________ finally caused Wallace to explore the world.A. His strong affection for natureB. His life-long devotion to beastsC. His deep love for adventureD. Increasing demand for insects57. Which of the following is TRUE about Wallace’s first trip?A. It took him six weeks to explore the Amazon with all kinds of transportation.B. He made a scientific study of a fairly limited number of insects.C. The fire cost him his four years’ collection of animals.D. His passion cooled after the disaster.58. Wallace felt _____ on the Malay Archipelago.A. fearlessB. luckyC. challengedD. risky59. Wallace’s idea on evolution of natural selection __________.A. made no sense at that timeB. built up a new concept of lifeC. was too simple to be trueD. revealed the origin of nature(B)Virtual realityProbably the most exciting tech development of recent times, virtual reality(VR) has arrived, with sufficient options available to the consumer who’ssearching for an extra amount of high-tech fun. The cheapest way to get ahigh-end VR experience comes courtesy of Sony. Its PlayStation VR doesn’trequire a tricked-out PC or expensive phone –it works with the Playstation 4control board and comes with a few great games in its library. There is someequipment you can purchase to enhance the experience, but if you’ve already got a PS4 you can enter the world of VR for just $400. Other high-end offerings like the HTC Vive and Oculus Rift, as well as mobile options like Samsung’s Gear VR, will get your head in the game.Wireless headphonesCombining ease of use with the ability to move wild around your home, gym orworkplace, wireless headphones just make sense. And there are plenty of practicaloptions to suit any budget. The Bose QuietComfort 35 wireless headphones aredefinitely worth a test drive, though. The full-size, around-ear Bluetooth headphoneshighlight active noise cancellation and double as a headset for making phone calls.They’ve even earned the Editor’s Choice award at and can be purchased forless than $400 online.Digital camerasWhile your phone is a worthy assistant, there’s no substitute for a real camera when it comes to taking the perfect picture. And these days you can get qualityspecifications in a package that’s almost as small as your smartphone. The shinydesign of the Fujifilm X70, $699, makes it the perfect companion, or you could go retro with the Olympus PEN-F ($1,200) that offers old school looks alongside cutting edge technology. Domestically, it’s worth checking out Xiaomi’s mirrorless Yi M1 for a more affordable option. With a high-end 20-megapixel（兆像素）sensor and the ability to host multiple lenses, it’s available from just 2,199 yuan.60. Sony can provide high-tech fun at the lowest cost because __________.A. players can play free games onlineB. PS4 owners don’t need any other deviceC. it gives players adequate experienceD. players have purchased expensive PCs61. What is Bose QuietComfort 35 wireless headphones’ selling point promoted in the passage?A. They have various types to meet users’ need s.B. Users can reduce noise manually.C. They work better in the wild.D. Users can make phone calls with the headphones.62. If your friend, who favors everything in the styles of the past, plans to make perfect pictures with a newdevice, you will most probably recommend __________.A. A smart phone.B. Fujifilm X70.C. Olympus PEN-F.D. Yi M1.(C)Naquela Wright’s life took an unexpected turn when she lost her eyesight as a teenager, but even when her world became dark, the New Jersey resident didn’t want to quit social media.Using Facebook was a challenge at first. Diagnosed in 2010 with pseudotumor cerebri, a rare health condition in which pressure increases around the brain and can result in the loss of vision, Wright learned how to use a screen reader to read the site through the touch of the keyboard and sound of a robotic voice. Still, when a friend sends her a photo, Wright often has no clue what the image shows.Now Facebook is trying to solve this problem by exploiting the power of artificial intelligence to create new tools that not only describe items in a photo but allows users to ask what’s in an image.“I can have a basic picture in my mind of what’s going on in the picture and now I can comment on my own,” said Wright, who got to try out the new tools that are still being tested. “Of course, it’s different, but it’s something more than I had.”An estimated 285 million people are visually disabled globally, according to the World Health Organization, and research conducted by Facebook showed that blind users have trouble figuring out what’s in a photo because the description isn’t clear or doesn’t exist.Facebook has made it easier to skim through the content on its website with a screen reader by improving HTML headings, adding alternative text for images, launching keyboard shortcuts, and more. Using artificial intelligence to describe photos is only a part of these ongoing efforts.With 1.5 billion users, Facebook isn’t the only social media company that wants to improve its website for the visually disabled. Along with Facebook and other major tech firms, Twitter and LinkedIn have their own accessibility teams and belong to an initiative called “Teaching Accessibility”.Jeff Wieland, Facebook’s head of accessibility engineering, said the group wants to educate more engineers, especially early in college, about designing products that are compatible with the disabled and others. “We really don’t want accessibility to be the luxury of a handful of companies,” Wieland said. “We want everything around the world to be built with accessibility in mind.”63. What tool helps the visually disabled to read Facebook?A. A screen reader.B. A special keyboard.C. A helpful robot.D. HTML headings.64. What can be inferred from the passage about the new tool created by Facebook?A. It adds a lot of shortcuts on the keyboard.B. It helps users to employ their senses other than sight.C. It meets no competitors with its advanced technology.D. It inspires more engineers to explore artificial intelligence.65. The underlined phrase in the last paragraph “are compatible with” most probably means __________.A. are unaffordable toB. bring harm toC. keep company ofD. well suit66. Which of the following is the best title for this passage?A. Screen reader: tool to access social mediaB. Ongoing efforts: strength to improve websitesC. Artificial intelligence: power to help the blindD. Teaching accessibility: initiative to educate engineersSection CDirections:Read the following passage. Fill in each blank with a proper sentence given in the box. Each sentence can be used only once. Note that there are two more sentences than you need.A.But there’s no need for embarrassment.B.If you want proof, turn on the sports channel.C.As we grow older, we don’t abandon this system – we internalize it.D.This so-called inner speech can improve our performance on various tasks.E.Conducting a dialogue with ourselves might turn out to be one of the keys to human creativity.F.Psychologists refer to this as private speech – language that is spoken out loud but directed at the self.Your Own Best FriendTalking to yourself may seem a little shameful. If you’ve ever been overheard criticizing yourself for a foolish mistake or practicing a tricky speech ahead of time, you’ll have felt the social restriction against communicating with yourself in words. According to the well-known saying, talking to yourself is the first sign of madness.__67__ Talking to ourselves, whether out loud or silently in our heads, is a valuable tool for thought. Far from being a sign of foolishness, self-talk allows us to plan what we are going to do, manage our activities, regulate our emotions and even create a narrative of our experience.Take a trip to any preschool and watch a small child playing with her toys. You are very likely to hear her talking to herself: offering herself directions and giving voice to her frustrations. __68__ We do a lot of it when we are young – perhaps one reason for our shyness about continuing with it as adults.As children, according to the Russian psychologist Lev Vygotsky, we use private speech to regulate our actions in the same way that we use public speech to control the behavior of others. __69__ Psychological experiments have shown that the distancing effect of our words can give us a valuable perspective on our actions. One recent study suggested that self-talk is most effective when we address ourselves in the second person: as “you” rather than “I”.We internalize the private speech we use as children –but we never entirely put away the out-loud version. __70__ You’re sure to see an athlete or two getting themselves ready for a sharp phrase or scolding themselves after a bad shot.Both kinds of self-talk seem to bring a range of benefits to our thinking. Those words to the self, spoken silently or aloud, are so much more than lazy talk.IV．Summary WritingDirections: Read the following passage. Summarize the main idea and the main point(s) of the passage in no more than 60 words. Use your own words as far as possible.Moustache（胡子）for Cash“Movember”, as the annual event is known, sees men in countries including the UK, US and Australia grow out their facial hair while collecting sponsorship money from friends, family and colleagues, with the money going to cancer charities.The month of no shaving began unofficially in 2003, when a pair of men from Australia persuaded their family to join them in growing a moustache in order to encourage men to get themselves checked for cancer,which is seen as distasteful by some males. A year later, the group decided to set up the Movember Foundation, asking friends and colleagues to offer donations of money to support their efforts, and raised a massive A$54,000 which was shared between a number of health projects. With thanks most likely to social media, Movember soon went global and the foundation now operates worldwide, having raised over ￡440 million since 2004. The effects of the fundraising are wide-reaching, which had made a significant discovery in the treatment of cancer.The issue of some men being too self-willed to visit their doctor for a checkup, or perhaps being raised in a culture of “tough it out”, has led some males to neglect their health, which may mean it could be too late if something potentially deadly did develop. However, Movember is helping to break down the shame of male health by making it more accessible, meaning that men are more likely to visit their doctors. They found a way to appeal to men in a way that other campaigns just don’t – with a sort of blokey①jokiness.①blokey: behaving in a way that is supposed to be typical of men , especially men enjoying themselves in a group.V. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.72. 永远不要对你孩子的缺点熟视无睹。

2017年上海市黄浦区高考数学一模试卷一、填空题（本大题共有12题，满分54分.其中第1～6题每题满分54分，第7～12题每题满分54分）考生应在答题纸相应编号的空格内直接填写结果.[ 1．若集合A={x||x﹣1|＜2，x∈R}，则A∩Z=．2．抛物线y2=2x的准线方程是．3．若复数z满足（i为虚数单位），则z=．4．已知sin（α+）=，α∈（﹣，0），则tanα=．5．以点（2，﹣1）为圆心，且与直线x+y=7相切的圆的方程是．6．若二项式的展开式共有6项，则此展开式中含x4的项的系数是．7．已知向量（x，y∈R），，若x2+y2=1，则的最大值为．8．已知函数y=f（x）是奇函数，且当x≥0时，f（x）=log2（x+1）．若函数y=g （x）是y=f（x）的反函数，则g（﹣3）=．9．在数列{a n}中，若对一切n∈N*都有a n=﹣3a n，且+1=，则a1的值为．10．甲、乙两人从6门课程中各选修3门．则甲、乙所选的课程中至多有1门相同的选法共有．11．已知点O，A，B，F分别为椭圆的中心、左顶点、上顶点、右焦点，过点F作OB的平行线，它与椭圆C在第一象限部分交于点P，若，则实数λ的值为．12．已知为常数），，且当x1，x2∈[1，4]时，总有f（x1）≤g（x2），则实数a的取值范围是．二、选择题（本大题共有4题，满分20分．）每题有且只有一个正确答案，考生应在答题纸的相应编号上，将代表答案的小方格涂黑，选对得5分，否则一律得零分．13．若x ∈R ，则“x ＞1”是“”的（ ）A ．充分非必要条件B ．必要非充分条件C ．充要条件D ．既非充分也非必要条件14．关于直线l ，m 及平面α，β，下列命题中正确的是（ ）A ．若l ∥α，α∩β=m ，则l ∥mB ．若l ∥α，m ∥α，则l ∥mC ．若l ⊥α，m ∥α，则l ⊥mD ．若l ∥α，m ⊥l ，则m ⊥α15．在直角坐标平面内，点A ，B 的坐标分别为（﹣1，0），（1，0），则满足tan ∠PAB•tan ∠PBA=m （m 为非零常数）的点P 的轨迹方程是（ ）A ．B ．C ．D ．16．若函数y=f （x ）在区间I 上是增函数，且函数在区间I 上是减函数，则称函数f （x ）是区间I 上的“H 函数”．对于命题：①函数是（0，1）上的“H 函数”；②函数是（0，1）上的“H 函数”．下列判断正确的是（ ）A ．①和②均为真命题B ．①为真命题，②为假命题C ．①为假命题，②为真命题D ．①和②均为假命题 三、解答题（本大题共有5题，满分76分．）解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤．17．在三棱锥P ﹣ABC 中，底面ABC 是边长为6的正三角形，PA ⊥底面ABC ，且PB 与底面ABC 所成的角为．（1）求三棱锥P ﹣ABC 的体积；（2）若M 是BC 的中点，求异面直线PM 与AB 所成角的大小（结果用反三角函数值表示）．18．已知双曲线C以F1（﹣2，0）、F2（2，0）为焦点，且过点P（7，12）．（1）求双曲线C与其渐近线的方程；（2）若斜率为1的直线l与双曲线C相交于A，B两点，且（O为坐标原点）．求直线l的方程．19．现有半径为R、圆心角（∠AOB）为90°的扇形材料，要裁剪出一个五边形工件OECDF，如图所示．其中E，F分别在OA，OB上，C，D在上，且OE=OF，EC=FD，∠ECD=∠CDF=90°．记∠COD=2θ，五边形OECDF的面积为S．（1）试求S关于θ的函数关系式；（2）求S的最大值．20．已知集合M是满足下列性质的函数f（x）的全体：在定义域内存在实数t，使得f（t+2）=f（t）+f（2）．（1）判断f（x）=3x+2是否属于集合M，并说明理由；（2）若属于集合M，求实数a的取值范围；（3）若f（x）=2x+bx2，求证：对任意实数b，都有f（x）∈M．21．已知数列{a n}，{b n}满足b n=a n﹣a n（n=1，2，3，…）．+1（1）若b n=10﹣n，求a16﹣a5的值；（2）若且a1=1，则数列{a2n+1}中第几项最小？请说明理由；（3）若c n=a n+2a n+1（n=1，2，3，…），求证：“数列{a n}为等差数列”的充分必要（n=1，2，3，…）”．条件是“数列{c n}为等差数列且b n≤b n+12017年上海市黄浦区高考数学一模试卷参考答案与试题解析一、填空题（本大题共有12题，满分54分.其中第1～6题每题满分54分，第7～12题每题满分54分）考生应在答题纸相应编号的空格内直接填写结果.[ 1．若集合A={x||x﹣1|＜2，x∈R}，则A∩Z={0，1，2} ．【考点】交集及其运算．【分析】化简集合A，根据交集的定义写出A∩Z即可．【解答】解：集合A={x||x﹣1|＜2，x∈R}={x|﹣2＜x﹣1＜2，x∈R}={x|﹣1＜x＜3，x∈R}，则A∩Z={0，1，2}．故答案为{0，1，2}．2．抛物线y2=2x的准线方程是．【考点】抛物线的简单性质．【分析】先根据抛物线方程求得p，进而根据抛物线的性质，求得答案．【解答】解：抛物线y2=2x，∴p=1，∴准线方程是x=﹣故答案为：﹣3．若复数z满足（i为虚数单位），则z=1+2i．【考点】复数代数形式的乘除运算．【分析】直接利用复数代数形式的乘除运算化简得答案．【解答】解：由，得z=1+2i．故答案为：1+2i．4．已知sin（α+）=，α∈（﹣，0），则tanα=﹣2．【考点】运用诱导公式化简求值；同角三角函数间的基本关系．【分析】由α∈（﹣，0）sin（α+）=，利用诱导公式可求得cosα，从而可求得sinα与tanα．【解答】解：∵sin（α+）=cosα，sin（α+）=，∴cosα=，又α∈（﹣，0），∴sinα=﹣，∴tanα==﹣2．故答案为：﹣2．5．以点（2，﹣1）为圆心，且与直线x+y=7相切的圆的方程是（x﹣2）2+（y+1）2=18．【考点】圆的切线方程．【分析】由点到直线的距离求出半径，从而得到圆的方程．【解答】解：将直线x+y=7化为x+y﹣7=0，圆的半径r==3，所以圆的方程为（x﹣2）2+（y+1）2=18．故答案为（x﹣2）2+（y+1）2=18．6．若二项式的展开式共有6项，则此展开式中含x4的项的系数是10．【考点】二项式定理的应用．【分析】根据题意求得n=5，再在二项展开式的通项公式中，令x的幂指数等于4，求得r的值，可得展开式中含x4的项的系数．【解答】解：∵二项式的展开式共有6项，故n=5，=•（﹣1）r•x10﹣3r，令10﹣3r=4，∴r=2，则此展开式的通项公式为T r+1中含x4的项的系数=10，故答案为：10．7．已知向量（x，y∈R），，若x2+y2=1，则的最大值为+1．【考点】向量的模．【分析】利用≤+r即可得出．【解答】解：设O（0，0），P（1，2）．=≤+r=+1=+1．∴的最大值为+1．故答案为：．8．已知函数y=f（x）是奇函数，且当x≥0时，f（x）=log2（x+1）．若函数y=g （x）是y=f（x）的反函数，则g（﹣3）=﹣7．【考点】反函数．【分析】根据反函数与原函数的关系，可知反函数的定义域是原函数的值域，即可求解．【解答】解：∵反函数与原函数具有相同的奇偶性．∴g（﹣3）=﹣g（3），∵反函数的定义域是原函数的值域，∴log2（x+1）=3，解得：x=7，即g（3）=7，故得g（﹣3）=﹣7．故答案为：﹣7．9．在数列{a n}中，若对一切n∈N*都有a n=﹣3a n，且+1=，则a1的值为﹣12．【考点】数列的极限．【分析】由题意可得数列{a n}为公比为﹣的等比数列，运用数列极限的运算，解方程即可得到所求．【解答】解：在数列{a n}中，若对一切n∈N*都有a n=﹣3a n+1，可得数列{a n}为公比为﹣的等比数列，=，可得====，可得a1=﹣12．故答案为：﹣12．10．甲、乙两人从6门课程中各选修3门．则甲、乙所选的课程中至多有1门相同的选法共有200．【考点】排列、组合及简单计数问题．【分析】根据题意，甲、乙所选的课程中至多有1门相同，其包含两种情况：①甲乙所选的课程全不相同，②甲乙所选的课程有1门相同；分别计算每种情况下的选法数目，相加可得答案．【解答】解：根据题意，分两种情况讨论：①甲乙所选的课程全不相同，有C63×C33=20种情况，②甲乙所选的课程有1门相同，有C61×C52×C32=180种情况，则甲、乙所选的课程中至多有1门相同的选法共有180+20=200种情况；故答案为：200．11．已知点O，A，B，F分别为椭圆的中心、左顶点、上顶点、右焦点，过点F作OB的平行线，它与椭圆C在第一象限部分交于点P，若，则实数λ的值为．【考点】直线与椭圆的位置关系．【分析】由题意画出图形，求出的坐标，代入，结合隐含条件求得实数λ的值．【解答】解：如图，A（﹣a，0），B（0，b），F（c，0），则P（c，），∴，，由，得，即b=c，∴a2=b2+c2=2b2，．则．故答案为：．12．已知为常数），，且当x1，x2∈[1，4]时，总有f（x1）≤g（x2），则实数a的取值范围是．【考点】函数恒成立问题．【分析】依题意可知，当x1，x2∈[1，4]时，f（x1）max≤g（x2）min，利用对勾函数的单调性质可求g（x2）min=g（1）=3；再对f（x）=2ax2+2x中的二次项系数a分a=0、a＞0、a＜0三类讨论，利用函数的单调性质可求得f（x）在区间[1，4]上的最大值，解f（x）max≤3即可求得实数a的取值范围．【解答】解：依题意知，当x1，x2∈[1，4]时，f（x1）max≤g（x2）min，由“对勾'函数单调性知，=2x+=2（x+）在区间[1，4]上单调递增，∴g（x2）min=g（1）=3；∵=2ax2+2x，当a=0时，f（x）=2x在区间[1，4]上单调递增，∴f（x）max=f（4）=8≤3不成立，故a≠0；∴f（x）=2ax2+2x为二次函数，其对称轴方程为：x=﹣，当a＞0时，f（x）在区间[1，4]上单调递增，f（x）max=f（4）=8≤3不成立，故a＞0不成立；当a＜0时，1°若﹣≤1，即a≤﹣时，f（x）在区间[1，4]上单调递减，f（x）max=f（1）=2a+2≤3恒成立，即a≤﹣时满足题意；2°若1＜﹣＜4，即﹣＜a＜﹣时，f（x）max=f（﹣）=﹣≤3，解得：﹣＜a≤﹣；3°若﹣≥4，即﹣≤a＜0时，f（x）在区间[1，4]上单调递增，f（x）max=f（4）=32a+8≤3，解得a≤﹣∉（﹣，0），故不成立，综合1°2°3°知，实数a的取值范围是：（﹣∞，﹣]．故答案为：．二、选择题（本大题共有4题，满分20分．）每题有且只有一个正确答案，考生应在答题纸的相应编号上，将代表答案的小方格涂黑，选对得5分，否则一律得零分．13．若x∈R，则“x＞1”是“”的（）A．充分非必要条件 B．必要非充分条件C．充要条件D．既非充分也非必要条件【考点】必要条件、充分条件与充要条件的判断．【分析】根据充分必要条件的定义判断即可．【解答】解：由x＞1，一定能得到得到＜1，但当＜1时，不能推出x＞1 （如x=﹣1时），故x＞1是＜1 的充分不必要条件，故选：A．14．关于直线l，m及平面α，β，下列命题中正确的是（）A．若l∥α，α∩β=m，则l∥m B．若l∥α，m∥α，则l∥mC．若l⊥α，m∥α，则l⊥m D．若l∥α，m⊥l，则m⊥α【考点】空间中直线与直线之间的位置关系；空间中直线与平面之间的位置关系．【分析】在A中，l与m平行或异面；在B中，l与m相交、平行或异面；在C 中，由线面垂直的性质定理得l⊥m；在D中，m与α相交、平行或m⊂α．【解答】解：由直线l，m及平面α，β，知：在A中，若l∥α，α∩β=m，则l与m平行或异面，故A错误；在B中，若l∥α，m∥α，则l与m相交、平行或异面，故B错误；在C中，若l⊥α，m∥α，则由线面垂直的性质定理得l⊥m，故C正确；在D中，若l∥α，m⊥l，则m与α相交、平行或m⊂α，故D错误．故选：C．15．在直角坐标平面内，点A，B的坐标分别为（﹣1，0），（1，0），则满足tan ∠PAB•tan∠PBA=m（m为非零常数）的点P的轨迹方程是（）A．B．C．D．【考点】轨迹方程．【分析】设P（x，y），则由题意，（m≠0），化简可得结论．【解答】解：设P（x，y），则由题意，（m≠0），化简可得，故选C．16．若函数y=f（x）在区间I上是增函数，且函数在区间I上是减函数，则称函数f（x）是区间I上的“H函数”．对于命题：①函数是（0，1）上的“H函数”；②函数是（0，1）上的“H函数”．下列判断正确的是（）A．①和②均为真命题B．①为真命题，②为假命题C．①为假命题，②为真命题D．①和②均为假命题【考点】命题的真假判断与应用．【分析】对函数，G（x）=在（0，1）上的单调性进行判断，得命题①是真命题．对函数=，H（x）=在（0，1）上单调性进行判断，得命题②是假命题．【解答】解：对于命题①：令t=，函数=﹣t2+2t，∵t=在（0，1）上是增函数，函数y=﹣t2+2t在（0，1）上是增函数，∴在（0，1）上是增函数；G（x）=在（0，1）上是减函数，∴函数是（0，1）上的“H函数“，故命题①是真命题．对于命题②，函数=是（0，1）上的增函数，H（x）=是（0，1）上的增函数，故命题②是假命题；故选：B．三、解答题（本大题共有5题，满分76分．）解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤．17．在三棱锥P﹣ABC中，底面ABC是边长为6的正三角形，PA⊥底面ABC，且PB与底面ABC所成的角为．（1）求三棱锥P﹣ABC的体积；（2）若M是BC的中点，求异面直线PM与AB所成角的大小（结果用反三角函数值表示）．【考点】棱柱、棱锥、棱台的体积；异面直线及其所成的角．【分析】（1）在Rt△PAB中计算PA，再代入棱锥的体积公式计算；（2）取棱AC的中点N，连接MN，NP，分别求出△PMN的三边长，利用余弦定理计算cos∠PMN即可．【解答】解：（1）∵PA⊥平面ABC，∴∠PBA为PB与平面ABC所成的角，即，∵PA⊥平面ABC，∴PA⊥AB，又AB=6，∴，∴．（2）取棱AC的中点N，连接MN，NP，∵M，N分别是棱BC，AC的中点，∴MN∥BA，∴∠PMN为异面直线PM与AB所成的角．∵PA⊥平面ABC，所以PA⊥AM，PA⊥AN，又，AN=AC=3，BM=BC=3，∴AM==3，，，所以，故异面直线PM与AB所成的角为．18．已知双曲线C以F1（﹣2，0）、F2（2，0）为焦点，且过点P（7，12）．（1）求双曲线C与其渐近线的方程；（2）若斜率为1的直线l与双曲线C相交于A，B两点，且（O为坐标原点）．求直线l的方程．【考点】直线与双曲线的位置关系；双曲线的标准方程．【分析】（1）设出双曲线C方程，利用已知条件求出c，a，解得b，即可求出双曲线方程与渐近线的方程；（2）设直线l的方程为y=x+t，将其代入方程，通过△＞0，求出t的范围，设A（x1，y1），B（x2，y2），利用韦达定理，通过x1x2+y1y2=0，求解t即可得到直线方程．【解答】解：（1）设双曲线C的方程为，半焦距为c，则c=2，，a=1，…所以b2=c2﹣a2=3，故双曲线C的方程为．…双曲线C的渐近线方程为．…（2）设直线l的方程为y=x+t，将其代入方程，可得2x2﹣2tx﹣t2﹣3=0（*）…△=4t2+8（t2+3）=12t2+24＞0，若设A（x1，y1），B（x2，y2），则x1，x2是方程（*）的两个根，所以，又由，可知x1x2+y1y2=0，…即x1x2+（x1+t）（x2+t）=0，可得，故﹣（t2+3）+t2+t2=0，解得，所以直线l方程为．…19．现有半径为R、圆心角（∠AOB）为90°的扇形材料，要裁剪出一个五边形工件OECDF，如图所示．其中E，F分别在OA，OB上，C，D在上，且OE=OF，EC=FD，∠ECD=∠CDF=90°．记∠COD=2θ，五边形OECDF的面积为S．（1）试求S关于θ的函数关系式；（2）求S的最大值．【考点】函数模型的选择与应用．【分析】（1）设M是CD中点，连OM，推出∠COM=∠DOM=，MD=Rsinθ，利用△CEO≌△DFO，转化求解∠DFO=，在△DFO中，利用正弦定理+S ODF+S OCE=S△COD+2S ODF的解析式即可．，求解S=S△COD（2）利用S的解析式，通过三角函数的最值求解即可．【解答】解：（1）设M是CD中点，连OM，由OC=OD，可知OM⊥CD，∠COM=∠DOM=，，MD=Rsinθ，又OE=OF，EC=FD，OC=OD，可得△CEO≌△DFO，故∠EOC=∠DOF，可知，…又DF⊥CD，OM⊥CD，所以MO∥DF，故∠DFO=，在△DFO中，有，可得…所以S=S+S ODF+S OCE=S△COD+2S ODF=△COD=…（2）…=（其中）…当，即时，sin（2θ+φ）取最大值1．又，所以S的最大值为．…20．已知集合M是满足下列性质的函数f（x）的全体：在定义域内存在实数t，使得f（t+2）=f（t）+f（2）．（1）判断f（x）=3x+2是否属于集合M，并说明理由；（2）若属于集合M，求实数a的取值范围；（3）若f（x）=2x+bx2，求证：对任意实数b，都有f（x）∈M．【考点】抽象函数及其应用．【分析】（1）利用f（x）=3x+2，通过f（t+2）=f（t）+f（2）推出方程无解，说明f（x）=3x+2不属于集合M．（2）由属于集合M，推出有实解，即（a﹣6）x2+4ax+6（a﹣2）=0有实解，若a=6时，若a≠6时，利用判断式求解即可．（3）当f（x）=2x+bx2时，方程f（x+2）=f（x）+f（2）⇔3×2x+4bx﹣4=0，令g （x）=3×2x+4bx﹣4，则g（x）在R上的图象是连续的，当b≥0时，当b＜0时，判断函数是否有零点，证明对任意实数b，都有f（x）∈M．【解答】解：（1）当f（x）=3x+2时，方程f（t+2）=f（t）+f（2）⇔3t+8=3t+10…此方程无解，所以不存在实数t，使得f（t+2）=f（t）+f（2），故f（x）=3x+2不属于集合M．…（2）由属于集合M，可得方程有实解⇔a[（x+2）2+2]=6（x2+2）有实解⇔（a ﹣6）x2+4ax+6（a﹣2）=0有实解，…若a=6时，上述方程有实解；若a≠6时，有△=16a2﹣24（a﹣6）（a﹣2）≥0，解得，故所求a的取值范围是．…（3）当f（x）=2x+bx2时，方程f（x+2）=f（x）+f（2）⇔2x+2+b（x+2）2=2x+bx2+4+4b ⇔3×2x+4bx﹣4=0，…令g（x）=3×2x+4bx﹣4，则g（x）在R上的图象是连续的，当b≥0时，g（0）=﹣1＜0，g（1）=2+4b＞0，故g（x）在（0，1）内至少有一个零点；当b＜0时，g（0）=﹣1＜0，，故g（x）在内至少有一个零点；故对任意的实数b，g（x）在R上都有零点，即方程f（x+2）=f（x）+f（2）总有解，所以对任意实数b，都有f（x）∈M．…21．已知数列{a n}，{b n}满足b n=a n+1﹣a n（n=1，2，3，…）．（1）若b n=10﹣n，求a16﹣a5的值；（2）若且a1=1，则数列{a2n+1}中第几项最小？请说明理由；（3）若c n=a n+2a n+1（n=1，2，3，…），求证：“数列{a n}为等差数列”的充分必要条件是“数列{c n}为等差数列且b n≤b n+1（n=1，2，3，…）”．【考点】数列与函数的综合；数列的应用；数列递推式．【分析】（1）判断{b n}是等差数列．然后化简a16﹣a5=（a16﹣a15）+（a15﹣a14）+（a14﹣a13）+…+（a6﹣a5）利用等差数列的性质求和即可．（2）利用a2n+3﹣a2n+1=22n+1﹣231﹣2n，判断a2n+3＜a2n+1，求出n＜7.5，a2n+3＞a2n+1求出n＞7.5，带带数列{a2n+1}中a17最小，即第8项最小．．法二：化简，求出a2n+1=a1+b1+b2+b3+…+b2n=，利用基本不等式求出最小值得到数列{a2n+1}中的第8项最小．（3）若数列{a n}为等差数列，设其公差为d，说明数列{c n}为等差数列．由b n=a n+1﹣a n=d（n=1，2，3，…），推出b n≤b n+1，若数列{c n}为等差数列且b n≤b n+1（n=1，2，3，…），设{c n}的公差为D，转化推出b n+1=b n（n=1，2，3，…），说明数列{a n}为等差数列．得到结果．【解答】解：（1）由b n=10﹣n，可得b n+1﹣b n=（9﹣n）﹣（10﹣n）=﹣1，故{b n}是等差数列．所以a16﹣a5=（a16﹣a15）+（a15﹣a14）+（a14﹣a13）+…+（a6﹣a5）=…（2）a2n+3﹣a2n+1=（a2n+3﹣a2n+2）+（a2n+2﹣a2n+1）=b2n+2+b2n+1=（22n+2+231﹣2n）﹣（22n+1+232﹣2n）=22n+1﹣231﹣2n…由a2n+3＜a2n+1⇔22n+1﹣231﹣2n＜0⇔n＜7.5，a2n+3＞a2n+1⇔22n+1﹣231﹣2n＞0⇔n＞7.5，…故有a3＞a5＞a7＞…＞a15＞a17＜a19＜a20＜…，所以数列{a2n+1}中a17最小，即第8项最小．…法二：由，…可知a2n+1=a1+b1+b2+b3+…+b2n==…（当且仅当22n+1=233﹣2n，即n=8时取等号）所以数列{a2n+1}中的第8项最小．…（3）若数列{a n}为等差数列，设其公差为d，则c n+1﹣c n=（a n+1﹣a n）+2（a n+2﹣a n+1）=d+2d=3d为常数，所以数列{c n}为等差数列．…由b n=a n+1﹣a n=d（n=1，2，3，…），可知b n≤b n+1（n=1，2，3，…）．…若数列{c n}为等差数列且b n≤b n+1（n=1，2，3，…），设{c n}的公差为D，则c n+1﹣c n=（a n+1﹣a n）+2（a n+2﹣a n+1）=b n+2b n+1=D（n=1，2，3，…），…又b n+1+2b n+2=D，故（b n+1﹣b n）+2（b n+2﹣b n+1）=D﹣D=0，又b n+1﹣b n≥0，b n+2﹣b n+1≥0，故b n+1﹣b n=b n+2﹣b n+1=0（n=1，2，3，…），…所以b n+1=b n（n=1，2，3，…），故有b n=b1，所以a n+1﹣a n=b1为常数．故数列{a n}为等差数列．综上可得，“数列{a n}为等差数列”的充分必要条件是“数列{c n}为等差数列且b n ≤b n+1（n=1，2，3，…）”．…2017年2月18日。

2017年上海市黄浦区高考语文一模试卷一、（10分）1．（6分）识记和理解，依据题干要求，填写正确内容。

（1）黄庭坚《登快阁》一诗的尾联是“万里归船弄长笛，”。

（2）李商隐《夜雨寄北》中，最温暖的句子是“何当共剪西窗烛，”。

（3）韩愈在《师说》中指出教师职责的句子是“师者，”。

（4）周邦彦《苏幕遮》，描绘鸟儿生活的句子是“，”。

（5）苏轼《水调歌头》中有“此事古难全”一句，句中的“此事”指的是“，。

”（6）孔子在《论语•里仁》里，表明对“贤”“不贤”态度的句子是“，。

”2．（2分）理解和应用。

新年第一天，小李向同学发送微信，最适合表达新春祝福的一项是（）A．乡心新岁切，天畔独潸然B．遥知兄弟登高处，遍插茱萸少一人C．爆竹声中一岁除，春风送暖入屠苏D．故乡今夜思千里，霜鬓明朝又一年E．遥知兄弟登高处，遍插茱萸少一人F．爆竹声中一岁除，春风送暖入屠苏3．（2分）根据文意推断，填入画直线上的句子，最恰当的一项是（）本报讯11月7日下午2点30，“她眼中的都市之美”﹣﹣第五届上海女记者摄影展开幕仪式将在东方明珠4.2米环廊举办。

此届摄影展为徐汇专场，展出的100幅作品选自今年举办的上海女记者摄影大赛的入围及获奖作品。

摄影大赛的参赛者来自上海市各家平面、广播、电视和网络媒体，大赛一共收到1235件作品。

女新闻工作者们独具“汇”眼，将镜头对准了徐汇这一历史底蕴深厚，文化资源丰富的中西文化交汇地。

她们以女性的视角、发现的眼光，聚焦霓虹闪耀的楼宇商厦，活力四射的创业基地，深藏历史的经典建筑，气象一新的滨江大道，多姿多彩的街巷民宅，安静优美的社区绿地，以及怡然舒心的养老院，欢笑不断的幼儿园。

本届摄影展由上海新闻工作者协会女记者工作委员会与中共徐汇区委宣传部联合举办，上海东方明珠广播电视塔有限公司协办。

A．用她们的镜头勾勒出美丽徐汇、创新徐汇、幸福徐汇、文化徐汇的不同影像B．用她们的镜头勾勒出创新徐汇、文化徐汇、美丽徐汇、幸福徐汇的不同影像C．用她们的镜头勾勒出文化徐汇、幸福徐汇、创新徐汇、美丽徐汇的不同影像D．用她们的镜头勾勒出幸福徐汇、美丽徐汇、文化徐汇、创新徐汇的不同影像二、阅读70分4．（16分）阅读下文，完成下列各题。

黄浦区2017年高考模拟考数学试卷（完卷时间：120分钟满分：150分）一、填空题（本大题共有12题，满分54分. 其中第1~6题每题满分4分，第7~12题每题满分5分）考生应在答题纸相应编号的空格内直接填写结果.[1. 函数的定义域是________．【答案】；【解析】试题分析：考点：函数的定义域的求法.2. 若关于的方程组有无数多组解，则实数_________．【答案】；【解析】当时，，不合题意；当时，，得，综上：.3. 若“”是“”的必要不充分条件，则的最大值为_________．【答案】；【解析】由得：或；若“”是“”的必要不充分条件，则，所以的最大值为.【点睛】从集合的角度看充要条件，若对应集合，对应集合，如果，则是的充分条件；如果，则是的充分不必要条件；如果，则是的必要条件；如果，则是的必要不充分条件；如果，则是的充要条件，如果无上述包含关系，则是的既不充分也不必要条件；4. 已知复数，(其中i为虚数单位)，且是实数，则实数t等于________．【答案】；【解析】为实数，则.5. 若函数 (a>0,且a≠1)是R上的减函数，则a的取值范围是________．【答案】；【解析】当时，在上为减函数，而在上为减函数，要使函数在R上为减函数，则a满足，解得.6. 设变量满足约束条件，则目标函数的最小值为___________【答案】；【解析】先画出二元一次不等式组所表示的平面区域，目标函数为截距型目标函数，令，作直线，由于，表示直线的截距，平移直线得最优解为，的最小值为.【点睛】线性规划问题要搞清目标函数的几何意义，常见的目标函数线有截距型、距离型（两点间的距离、点到直线的距离）、斜率型等，主要考查最值或范围.另外有时考查线性规划的逆向思维问题，难度稍大一点. 线性规划问题为高考高频考点，属于必得分题.7. 已知圆和两点，若圆上至少存在一点，使得，则的取值范围是________．【答案】；【解析】由于两点在以原点为圆心，为半径的圆上，若圆上至少存在一点，使得，则两圆有公共点，设圆心距为，，则，则，则的取值范围是.8. 已知向量，，如果∥，那么的值为________．【答案】；【解析】，则，.【点睛】有关三角函数计算问题，“异名化同名，异角化同角”，注意弦切互化，最关键问题是寻找角与角之间的关系，角与角之间是否存在和、差、倍关系，再借助诱导公式，同角三角函数关系，和、差公式，二倍角公式等求值.9. 若从正八边形的8个顶点中随机选取3个顶点，则以它们作为顶点的三角形是直角三角形的概率是________．【答案】；【解析】正八边形的八个顶点，无三点在同一直线上，任取3点可连成一个三角形，共可作个三角形，其中4条对角线为其外接圆的直径，根据直径所对的圆周角为直角，每条直径可连接6个直角三角形，共计可作个直角三角形，概率为.10. 若将函数的图像向左平移个单位后，所得图像对应的函数为偶函数，则的最小值是________．【答案】；【点睛】11. 三棱锥满足：，，，，则该三棱锥的体积V的取值范围是________．【答案】；【解析】由于平面，，在中，，要使面积最大，只需，的最大值为，的最大值为，该三棱锥的体积V的取值范围是.【答案】（或，或）．【解析】数列满足，，，当，时，，，若时，，，当时，，，解得，填写 .继续讨论可求出其他的解（略）.二、选择题（本大题共有4题，满分20分．）每题有且只有一个正确答案，考生应在答题纸的相应编号上，将代表答案的小方格涂黑，选对得5分，否则一律得零分．13. 下列函数中，周期为π，且在上为减函数的是（）A. y = sin(2x+B. y = cos(2x+C. y = sin(x+D. y = cos(x+【答案】A【解析】根据正、余函数周期公式可知，排除C、D. 对于，，，则在上为减函数，选.14. 如图是一个几何体的三视图，根据图中数据，可得该几何体的表面积是（）A. B.C. D.【答案】D【解析】试题分析：由三视图可知，该几何体是下部为圆柱体，上部是半径为1的球，直接求表面积即可。

黄浦(卢湾)区高考 数学(文科)二模卷考生注意：1.每位考生应同时收到试卷和答题卷两份材料，解答必须在答题卷上进行，写在试卷上的解答一律无效；2.答卷前，考生务必将姓名、准考证号等相关信息在答题卷上填写清楚；3.本试卷共23道试题，满分150分；考试时间120分钟. 一、填空题(本大题满分56分) 本大题共有14题，考生应在答题卷的相应编号的空格内直接填写结果，每题填对得4分，否则一律得零分.1.函数xxy -+=11log 2的定义域是 . 2.函数x x y 22sin cos -=的最小正周期=T . 3.已知全集U =R ，集合{}|0,A x x a x =+∈R …，{}||1|3,B x x x =-∈R ….若()[2,4]U A B =- ð，则实数a 的取值范围是 .4.已知等差数列{}()n a n ∈*N 的公差为3，11-=a ，前n 项和为n S ，则n nn S na ∞→lim的数值是 . 5.函数)1,0(|log |)(≠>=a a x x f a 且的单调递增区间是 .6.函数2()(0)f x x x =-…的反函数是)(1x f -，则反函数的解析式是=-)(1x f .7.方程1)34(log 2+=-x x 的解=x .8.在△ABC 中，角C B A 、、所对的边的长度分别为c b a 、、，且ab c b a 3222=-+，则=∠C .9.已知i (i 11-=x 是虚数单位，以下同)是关于x 的实系数一元二次方程02=++b ax x 的一个根，则实数=a ，=b . 10.若用一个平面去截球体，所得截面圆的面积为16π，球心到该截面的距离是3，则这个球的表面积是 .11.已知直线05301221=+-=-+y x l y x l ：，：，则直线21l l 与的夹角的大小是 .(结果用反三角函数值表示)12.已知实数y x 、满足线性约束条件30,40,350.x y x y x y -⎧⎪+-⎨⎪-+⎩………则目标函数1--=y x z 的最大值是 .13.某个不透明的袋中装有除颜色外其它特征完全相同的7个乒乓球(袋中仅有白色和黄色两种颜色的球)，若从袋中随机摸一个乒乓球，得到的球是白色乒乓球的概率是72，则从袋中一次随机摸两个球，得到一个白色乒乓球和一个黄色乒乓球的概率是 . 14.已知函数)(x f y =是定义域为R 的偶函数. 当0x …时，161,02()2log 2xx f x x x ⎧⎛⎫<⎪ ⎪=⎨⎝⎭⎪⎩．……若关于x 的方程2[()]()0f x a f x b +⋅+=()a b ∈R 、有且只有7个不同实数根，则b a +的值是 .二、选择题(本大题满分20分) 本大题共有4题，每题有且只有一个正确答案，考生应在答题卷的相应编号上，将代表答案的小方格涂黑，选对得5分，否则一律得零分.15.已知a b ∈R 、，且0ab ≠，则下列结论恒成立的是 ( )A.a b +…B.2ab ba+… C.||2a bb a+… D.222a b ab +>16.已知空间直线l 不在平面α内，则“直线l 上有两个点到平面α的距离相等”是“l α ”的 ( )A.充分非必要条件B.必要非充分条件C.充要条件D.非充分非必要条件17.已知22,0a b a b ∈+≠R 、，则直线0=+by ax l ：与圆：022=+++by ax y x 的位置关系是( )A.相交B.相切C.相离D.不能确定18.四棱锥S ABCD -的底面是矩形，锥顶点在底面的射影是矩形对角线的交点，四棱锥及其三视图如下(AB 平行于主视图投影平面)则四棱锥S ABCD -的体积= ( ) A.24 B.18D.8 三、解答题(本大题满分74分)本大题共有5题，解答下列各题必须在答题卷的相应编号规定区域内写出必要的步骤.19.(本题满分12分)本题共有2个小题，第1小题满分6分，第2小题满分6分.已知矩形11ABB A 是圆柱体的轴截面，1O O 、分别是下底面圆和上底面圆的圆心，母线长与底面圆的直径长之比为2:1，且该圆柱体的体积为32π，如图所示. (1)求圆柱体的侧面积S 侧的值；(2)若1C 是半圆弧11A B 的中点，点C 在半径OA 上，且12OC OA =，异面直线1CC 与1BB 所成的角为θ，求sin θ的值.20.(本题满分14分)本题共有2个小题，第1小题满分7分，第2小题满分7分.已知复数12cos i,1isin ,R z x z x x =+=-∈.(1)求||21z z -的最小值；(2)设21z z z ⋅=，记z z x f (Im Im )(=表示复数z 的虚部). 将函数)(x f 的图像上所有点的横坐标伸长到原来的2倍(纵坐标不变)，再把所得的图像向右平移π2个单位长度，得到函数)(x g 的图像. 试求函数)(x g 的解析式.21.(本题满分12分)本题共有2个小题，第1小题满分6分，第2小题满分6分.某通讯公司需要在三角形地带OAC 区域内建造甲、乙两种通信信号加强中转站，甲中转站建在区域BOC 内，乙中转站建在区域AOB 内.分界线OB 固定，且OB =(1百米，边界线AC 始终过点B ，边界线OC OA 、满足75,30,45AOC AOB BOC ∠=∠=∠= .设OA x =(36x 剟)百米，OC y =百米.(1)试将y 表示成x 的函数，并求出函数y 的解析式；(2)当x 取何值时？整个中转站的占地面积OAC S ∆最小，并求出其面积的最小值.第21题图22.(本题满分18分)本题共有3个小题，第1小题满分6分，第2小题满分6分，第3小题满分6分.已知数列{}n a 满足n n n n n n a a a a a 3,)1(,12121221+=-+==+-(*n ∈N ). (1)求753a a a 、、的值；(2)求12-n a (用含n 的式子表示)；(3)记n n n a a b 212+=-，数列{}n b *()n ∈N 的前n 项和为n S ，求n S (用含n 的式子表示).23.(本题满分18分)本题共有3个小题，第1小题满分6分，第2小题满分6分，第3小题满分6分.已知点D 在双曲线22221(0,0)x y C a b a b-=>>：上，且双曲线的一条渐近线的方程是03=+y x .(1)求双曲线C 的方程；(2)若过点)1,0(且斜率为k 的直线l 与双曲线C 有两个不同交点，求实数k 的取值范围；(3)设(2)中直线l 与双曲线C 交于B A 、两个不同点，若以线段AB为直径的圆经过坐标原点，求实数k 的值.黄埔(卢湾)区高考数学(文科)二模卷一、填空题1.()1,1-【解析】(探究性理解水平/对数函数的定义域)由对数函数的定义域，()()2101101,1 1.110xx x x x xx +⎧>⎪⇒+->⇒<∴-<<-⎨⎪-≠⎩则21log 1xy x +=-的定义域为()1,1.-2.π【解析】(探究性理解水平/二倍角公式、余弦函数周期性)22cos sin cos2.y x x x =-=2π=π.2T ∴=3.4a <-【解析】(探究性理解水平、解释性理解水平/集合的交集、补集、含有绝对值的不等式的解法)由题意，{}{}|,,|24,.A x x a x B x x x =-∈=-∈R R 厔?()[2,4].U A B =- ð.U B A ∴⊆ð4 4.a a ∴->⇒<- 4. 2 【解析】(探究性理解水平/等差数列、数列的极限)对于等差数列，()()111,3,11313 4.n a d a a n d n n =-=∴=+-=-+-=-()21135.222n n n S na d n n -=+=-2234lim lim 3522n n n n na n n S n n →∞→∞-∴=-43lim3522n n n→∞-=-2=. 5.[)1,+∞【解析】(探究性理解水平/对数函数的性质与图像)由对数函数的图像，得()|log |a f x x =的图像为：a >1 0<a <1第5题图()|log |(0,1)a f x x a a ∴=>≠且的单调递增区间为[)1,+∞.6.())10f x x -=…【解析】(探究性理解水平/反函数)()()20.f x x x =- …2.x y ∴=-()10,0).x x f x x-∴== 剟7.2log 3x =【解析】(探究性理解水平/对数) 2log (43)1x x -=+，∴1122log (43)log 2432.x x x x ++-=⇒-=令2(0)x μμ=>，则2230μμ--=，解得3μ=或1μ=-(舍)，2log 3.x ∴=8.π6【解析】(探究性理解水平/余弦定理)由余弦定理，222cos 2a b c C ab ++===π.6C ∴=9.2,2a b =-=【解析】(探究性理解水平/复数的四则运算) 11i x =-是20x ax b ++=的一个根，将11i x =-代入方程，并化简得：()2i=+.a a b +202, 2.0a ab a b +=⎧∴⇒=-=⎨+=⎩10.100π【解析】(探究性理解水平/球的表面积) 圆的面积为16π，∴半径为4，又球心到截面距离为3，∴由勾股定理，球的半径为5.∴表面积为2=4π=100π.S R11.arccosarctan7)10或【解析】(探究性理解水平/直线的倾斜角与斜率、两角差的正切)设12,l l 的倾斜角分别为,αβ，则其夹角为()12tan tan 3tan 711tan tan 1(2)3αβαβαβ----===-++-⨯因为直线的夹角小于等于90 ,所以其夹角的正切值为7，即arctan7)或. 12.32-【解析】(探究性理解水平/二元一次不等式表示的平面区域)约束区域如图阴影部分所示:第12题图由图知,1z x y =--在B 点取得最大值，即40350x y x y +-=⎧⎨-+=⎩解得7494x y ⎧=⎪⎪⎨⎪=⎪⎩,所以max 7931442z =--=-.13.1021【解析】(探究性理解水平/等可能事件的概率)记所求事件为A ,分析题意知112527C C 10()C 21P A ==.14.1-【解析】(探究性理解水平/一元二次不等式的解法、函数的基本性质)由题意知，方程有解的条件:212402a b a x x ⎧⎪∆=-⎪-⎪=⎨⎪⎪=⎪⎩… 又因为该方程有7个不同的实根，即121141x x ⎧<<⎪⎨⎪=⎩, 解得1a b +=-.第14题图二、选择题15. C 【解析】(解释性理解水平／不等式的基本性质及其证明)A 、B 项要满足0,0a b >>，故不成立，D 项 1,1a b ==时，不成立，故选C. 16. B 【解析】(解释性理解水平、探究式理解水平／充分必要条件，直线与平面垂直、相交关系)由“l α ”可以得到“直线l 上有两个点到平面α的距离相等”，但由“直线l 上有两个点到平面α的距离相等”得到l 与平面α相交或平行，所以“直线l 上有两个点到平面α的距离相等”是“l α ”的必要非充分条件.17. B 【解析】(探究式理解水平／直线与圆的位置关系)由题可得圆的方程为2222()()224a b a b x y ++++=，圆心坐标为(,22a b --)，直线l 到圆心距离为d ==，所以直线到圆心距离等于圆的半径，所以直线l 与圆相切，故选B.18. D 【解析】(探究式理解水平／解释性理解水平锥体，三视图)由四棱锥的三视图知四棱锥为底面长为4，宽为2的矩形，而四棱锥的高为3，所以四棱锥的体积为142383v =⨯⨯⨯=,故选D. 三、解答题19. (本题满分12分)本题共有2个小题，第1小题满分６分，第2小题满分６分.【解】(探究性理解水平/圆柱的表面积与体积公式，异面直线所成角)(1)设圆柱的底面圆的半径为R ，依据题意，有21124,32AA AB R R AA ==π⋅=π， ２分∴2R =.3分 ∴1=2π32πS R AA ⋅=侧.6分(2)设D 是线段11AO 的中点，联结111DC DC OC 、、，则11111,C O A B CD BB ⊥ . ８分因此，1C CD ∠就是异面直线1CC 与1BB 所成的角，即1C CD θ∠=. 9分 又2R =，11190CDC C O D ∠=∠=，∴11DC CC ==.11分∴sin θ==.12分小题满分7分.【解】(记忆水平、探究性理解水平/函数sin()y A x ωϕ=+的图像变化，复数的模，复数的四则运算,两角差的正弦，二倍角)(1)∵12cos i,1isin ,z x z x x =+=-∈R ，∴12||z z -=分=.4分∴当πsin()14x -=-，即2()4x k k π=π-∈Z 时，6分12min ||1z z -==.7分(2)∵12z z z =⋅，∴12sin cos (1sin cos )i z z z x x x x =⋅=++-.9分 ∴1()1sin cos 1sin 2()2f x x x x x =-=-∈R .10分将函数)(x f 的图像上所有点的横坐标伸长到原来的2倍(纵坐标不变)后，得到的图像所对应的函数是111sin 2y x =-. 11分把函数11sin 2y x =-的图像向右平移π2个单位长度，得到的图像对应的函数是211sin()22y x π=--.12分 ∴11()1sin()1cos ()222g x x x x π=--=+∈R .14分小题满分6分.【解】(探究性理解水平/基本不等式，函数模型的建立) (1)结合图形可知，BOC AOB AOCS S S +=△△△.1分于是，111(130(145sin 75222x y xy ++= ，4分解得(36)2y x x =-剟.6分(2)由(1)知，(36)2y xx =-剟，因此，21sin 7522AOCx S xy x ==-△ 8分42)4]2x x =-++-9分2+…(当且仅当422x x -=-，即4x =时，等号成立). 10分答：当400x =米时，整个中转站的占地面积OAC S △最小，最小面积是4(210+⨯平方米.12分22. (本题满分18分)本题共有3个小题，第1小题满分6分，第2小题满分6分，第3小题满分6分.【解】(探究性理解水平/数列的通项及求和) (1) n n n n n n a a a a a 3,)1(,12121221+=-+==+-(*n ∈N )，∴1211324325465376(1)0,33,14,313,112,339.a a a a a a a a a a a a =+-==+==+==+==-==+=6分 (2)由题知，有*21213(1)()n n n n a a n +--=+-∈N .7分∴112123222325225311313(1)3(1)3(1)3(1)n n n n n n n n a a a a a a a a --------⎫-=+-⎪-=+-⎪⎪⎬⎪-=+-⎪⎪-=+-⎭211n a a -⇒-121121(333)[(1)(1)(1)]n n --=++++-+-++- .10分 ∴*213(1)1()2n nn a n ---=-∈N .12分(3)由(2)可知，2213(1)(1)12n nnn n a a -+-=+-=-，n ∈*N .13分 ∴21232()n n n n b a a n -=+=-∈*N .14分 ∴123n n S b b b b =++++23(32)(32)(32)(32)n =-+-+-++-16分1*3(13)13232(N )1322n n n n n +-=-=⋅--∈-.18分23. (本题满分18分)本题共有3个小题，第1小题满分6分，第2小题满分6分，第3小题满分6分.【解】(探索性理解水平/双曲线的标准方程和几何性质，直线与双曲线的位置关系)(1)由题知，有22121,a b b a⎧-=⎪⎪⎨⎪=⎪⎩ 2分 解得221,31.a b ⎧=⎪⎨⎪=⎩4分因此，所求双曲线C的方程是221113x y -= 6分(2)∵直线l 过点)1,0(且斜率为k ，∴直线l ：1y kx =+. 7分 联立方程组2231,1x y y kx ⎧-=⎨=+⎩得22(3)220k x kx ---=.又直线l 与双曲线C 有两个不同交点，∴22230,(2)4(3)(2)0.k k k ⎧-≠⎪⎨∆=---->⎪⎩ 10分 解得((k ∈ .12 分(3)设交点为1122(,)(,)A x y B x y 、，由(2)可得1221222,32.3k x x k x x k ⎧+=⎪⎪-⎨-⎪=⎪-⎩14分又以线段AB 为直径的圆经过坐标原点，因此，(OA OB O ⊥为坐标原点).15分于是，0,OA OB ⋅=即12120x x y y +=，21212(1)()10k x x k x x ++++=，22222(1)21033k k k k-+++=--， 解得1k =±.17分又1k =±满足230k -≠，且0∆>，所以，所求实数1k =±.。

2017年上海市黄浦区高考数学一模试卷一、填空题（本大题共有12题，满分54分.其中第1～6题每题满分54分，第7～12题每题满分54分）考生应在答题纸相应编号的空格内直接填写结果.[ 1．（4分）若集合A＝{x||x﹣1|＜2，x∈R}，则A∩Z＝．2．（4分）抛物线y2＝2x的准线方程是．3．（4分）若复数z满足（i为虚数单位），则z＝．4．（4分）已知sin（α+）＝，α∈（﹣，0），则tanα＝．5．（4分）以点（2，﹣1）为圆心，且与直线x+y＝7相切的圆的方程是．6．（4分）若二项式的展开式共有6项，则此展开式中含x4的项的系数是．7．（5分）已知向量（x，y∈R），，若x2+y2＝1，则的最大值为．8．（5分）已知函数y＝f（x）是奇函数，且当x≥0时，f（x）＝log2（x+1）．若函数y＝g（x）是y＝f（x）的反函数，则g（﹣3）＝．9．（5分）在数列{a n}中，若对一切n∈N*都有a n＝﹣3a n+1，且＝，则a的值为．10．（5分）甲、乙两人从6门课程中各选修3门．则甲、乙所选的课程中至多有1门相同的选法共有．11．（5分）已知点O，A，B，F分别为椭圆的中心、左顶点、上顶点、右焦点，过点F作OB的平行线，它与椭圆C在第一象限部分交于点P，若，则实数λ的值为．12．（5分）已知（a为常数），，且当x1，x2∈[1，4]时，总有f（x1）≤g（x2），则实数a的取值范围是．二、选择题（本大题共有4题，满分20分．）每题有且只有一个正确答案，考生应在答题纸的相应编号上，将代表答案的小方格涂黑，选对得5分，否则一律得零分．13．（5分）若x∈R，则“x＞1”是“”的（）A．充分非必要条件B．必要非充分条件C．充要条件D．既非充分也非必要条件14．（5分）关于直线l，m及平面α，β，下列命题中正确的是（）A．若l∥α，α∩β＝m，则l∥m B．若l∥α，m∥α，则l∥mC．若l⊥α，m∥α，则l⊥m D．若l∥α，m⊥l，则m⊥α15．（5分）在直角坐标平面内，点A，B的坐标分别为（﹣1，0），（1，0），则满足tan∠P AB•tan∠PBA＝m（m为非零常数）的点P的轨迹方程是（）A．B．C．D．16．（5分）若函数y＝f（x）在区间I上是增函数，且函数在区间I上是减函数，则称函数f（x）是区间I上的“H函数”．对于命题：①函数是（0，1）上的“H函数”；②函数是（0，1）上的“H函数”．下列判断正确的是（）A．①和②均为真命题B．①为真命题，②为假命题C．①为假命题，②为真命题D．①和②均为假命题三、解答题（本大题共有5题，满分76分．）解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤．17．（14分）在三棱锥P﹣ABC中，底面ABC是边长为6的正三角形，P A⊥底面ABC，且PB与底面ABC所成的角为．（1）求三棱锥P﹣ABC的体积；（2）若M是BC的中点，求异面直线PM与AB所成角的大小（结果用反三角函数值表示）．18．（14分）已知双曲线C以F1（﹣2，0）、F2（2，0）为焦点，且过点P（7，12）．（1）求双曲线C与其渐近线的方程；（2）若斜率为1的直线l与双曲线C相交于A，B两点，且（O为坐标原点）．求直线l的方程．19．（14分）现有半径为R、圆心角（∠AOB）为90°的扇形材料，要裁剪出一个五边形工件OECDF，如图所示．其中E，F分别在OA，OB上，C，D在上，且OE＝OF，EC＝FD，∠ECD＝∠CDF＝90°．记∠COD＝2θ，五边形OECDF的面积为S．（1）试求S关于θ的函数关系式；（2）求S的最大值．20．（16分）已知集合M是满足下列性质的函数f（x）的全体：在定义域内存在实数t，使得f（t+2）＝f（t）+f（2）．（1）判断f（x）＝3x+2是否属于集合M，并说明理由；（2）若属于集合M，求实数a的取值范围；（3）若f（x）＝2x+bx2，求证：对任意实数b，都有f（x）∈M．21．（18分）已知数列{a n}，{b n}满足b n＝a n+1﹣a n（n＝1，2，3，…）．（1）若b n＝10﹣n，求a16﹣a5的值；（2）若且a1＝1，则数列{a2n+1}中第几项最小？请说明理由；（3）若c n＝a n+2a n+1（n＝1，2，3，…），求证：“数列{a n}为等差数列”的充分必要条件是“数列{c n}为等差数列且b n≤b n+1（n＝1，2，3，…）”．2017年上海市黄浦区高考数学一模试卷参考答案与试题解析一、填空题（本大题共有12题，满分54分.其中第1～6题每题满分54分，第7～12题每题满分54分）考生应在答题纸相应编号的空格内直接填写结果.[ 1．（4分）若集合A＝{x||x﹣1|＜2，x∈R}，则A∩Z＝{0，1，2}．【解答】解：集合A＝{x||x﹣1|＜2，x∈R}＝{x|﹣2＜x﹣1＜2，x∈R}＝{x|﹣1＜x＜3，x∈R}，则A∩Z＝{0，1，2}．故答案为{0，1，2}．2．（4分）抛物线y2＝2x的准线方程是．【解答】解：抛物线y2＝2x，∴p＝1，∴准线方程是x＝﹣故答案为：x＝﹣．3．（4分）若复数z满足（i为虚数单位），则z＝1+2i．【解答】解：由，得z＝1+2i．故答案为：1+2i．4．（4分）已知sin（α+）＝，α∈（﹣，0），则tanα＝﹣2．【解答】解：∵sin（α+）＝cosα，sin（α+）＝，∴cosα＝，又α∈（﹣，0），∴sinα＝﹣，∴tanα＝＝﹣2．故答案为：﹣2．5．（4分）以点（2，﹣1）为圆心，且与直线x+y＝7相切的圆的方程是（x﹣2）2+（y+1）2＝18．【解答】解：将直线x+y＝7化为x+y﹣7＝0，圆的半径r＝＝3，所以圆的方程为（x﹣2）2+（y+1）2＝18．故答案为（x﹣2）2+（y+1）2＝18．6．（4分）若二项式的展开式共有6项，则此展开式中含x4的项的系数是10．【解答】解：∵二项式的展开式共有6项，故n＝5，则此展开式的通项公式为T r+1＝•（﹣1）r•x10﹣3r，令10﹣3r＝4，∴r＝2，中含x4的项的系数＝10，故答案为：10．7．（5分）已知向量（x，y∈R），，若x2+y2＝1，则的最大值为+1．【解答】解：设O（0，0），P（1，2）．＝≤+r＝+1＝+1．∴的最大值为+1．故答案为：．8．（5分）已知函数y＝f（x）是奇函数，且当x≥0时，f（x）＝log2（x+1）．若函数y＝g（x）是y＝f（x）的反函数，则g（﹣3）＝﹣7．【解答】解：∵反函数与原函数具有相同的奇偶性．∴g（﹣3）＝﹣g（3），∵反函数的定义域是原函数的值域，∴log2（x+1）＝3，解得：x＝7，即g（3）＝7，故得g（﹣3）＝﹣7．故答案为：﹣7．9．（5分）在数列{a n}中，若对一切n∈N*都有a n＝﹣3a n+1，且的值为﹣12．＝，则a【解答】解：在数列{a n}中，若对一切n∈N*都有a n＝﹣3a n+1，可得数列{a n}为公比为﹣的等比数列，＝，可得＝＝＝＝，可得a1＝﹣12．故答案为：﹣12．10．（5分）甲、乙两人从6门课程中各选修3门．则甲、乙所选的课程中至多有1门相同的选法共有200．【解答】解：根据题意，分两种情况讨论：①甲乙所选的课程全不相同，有C63×C33＝20种情况，②甲乙所选的课程有1门相同，有C61×C52×C32＝180种情况，则甲、乙所选的课程中至多有1门相同的选法共有180+20＝200种情况；故答案为：200．11．（5分）已知点O，A，B，F分别为椭圆的中心、左顶点、上顶点、右焦点，过点F作OB的平行线，它与椭圆C在第一象限部分交于点P，若，则实数λ的值为．【解答】解：如图，A（﹣a，0），B（0，b），F（c，0），则P（c，），∴，，由，得，即b＝c，∴a2＝b2+c2＝2b2，．则．故答案为：．12．（5分）已知（a为常数），，且当x1，x2∈[1，4]时，总有f（x1）≤g（x2），则实数a的取值范围是．【解答】解：法1°：依题意知，当x1，x2∈[1，4]时，f（x1）max≤g（x2）min，由“对勾“函数单调性知，＝2x+＝2（x+）在区间[1，4]上单调递增，∴g（x2）min＝g（1）＝3；∵＝2ax2+2x，当a＝0时，f（x）＝2x在区间[1，4]上单调递增，∴f（x）max＝f（4）＝8≤3不成立，故a≠0；∴f（x）＝2ax2+2x为二次函数，其对称轴方程为：x＝﹣，当a＞0时，f（x）在区间[1，4]上单调递增，f（x）max＝f（4）＝8≤3不成立，故a＞0不成立；当a＜0时，1°若﹣≤1，即a≤﹣时，f（x）在区间[1，4]上单调递减，f（x）max＝f（1）＝2a+2≤3恒成立，即a≤﹣时满足题意；2°若1＜﹣＜4，即﹣＜a＜﹣时，f（x）max＝f（﹣）＝﹣≤3，解得：﹣＜a≤﹣；3°若﹣≥4，即﹣≤a＜0时，f（x）在区间[1，4]上单调递增，f（x）max＝f（4）＝32a+8≤3，解得a≤﹣∉（﹣，0），故不成立，综合1°2°3°知，实数a的取值范围是：（﹣∞，﹣]．法2°：由法1°知g（x2）min＝g（1）＝3，∵＝2ax2+2x，∴当x1∈[1，4]时，f（x1）＝2ax2+2x≤3恒成立，∴a≤＝（﹣）2﹣，∴当＝，即x＝3时，＝﹣，∴实数a的取值范围是：（﹣∞，﹣]．故答案为：．二、选择题（本大题共有4题，满分20分．）每题有且只有一个正确答案，考生应在答题纸的相应编号上，将代表答案的小方格涂黑，选对得5分，否则一律得零分．13．（5分）若x∈R，则“x＞1”是“”的（）A．充分非必要条件B．必要非充分条件C．充要条件D．既非充分也非必要条件【解答】解：由x＞1，一定能得到得到＜1，但当＜1时，不能推出x＞1 （如x＝﹣1时），故x＞1是＜1 的充分不必要条件，故选：A．14．（5分）关于直线l，m及平面α，β，下列命题中正确的是（）A．若l∥α，α∩β＝m，则l∥m B．若l∥α，m∥α，则l∥mC．若l⊥α，m∥α，则l⊥m D．若l∥α，m⊥l，则m⊥α【解答】解：由直线l，m及平面α，β，知：在A中，若l∥α，α∩β＝m，则l与m平行或异面，故A错误；在B中，若l∥α，m∥α，则l与m相交、平行或异面，故B错误；在C中，若l⊥α，m∥α，则由线面垂直的性质定理得l⊥m，故C正确；在D中，若l∥α，m⊥l，则m与α相交、平行或m⊂α，故D错误．故选：C．15．（5分）在直角坐标平面内，点A，B的坐标分别为（﹣1，0），（1，0），则满足tan∠P AB•tan∠PBA＝m（m为非零常数）的点P的轨迹方程是（）A．B．C．D．【解答】解：设P（x，y），则由题意，（m≠0），化简可得，故选：C．16．（5分）若函数y＝f（x）在区间I上是增函数，且函数在区间I上是减函数，则称函数f（x）是区间I上的“H函数”．对于命题：①函数是（0，1）上的“H函数”；②函数是（0，1）上的“H函数”．下列判断正确的是（）A．①和②均为真命题B．①为真命题，②为假命题C．①为假命题，②为真命题D．①和②均为假命题【解答】解：对于命题①：令t＝，函数＝﹣t2+2t，∵t＝在（0，1）上是增函数，函数y＝﹣t2+2t在（0，1）上是增函数，∴在（0，1）上是增函数；G（x）＝在（0，1）上是减函数，∴函数是（0，1）上的“H函数“，故命题①是真命题．对于命题②，函数＝是（0，1）上的增函数，H（x）＝是（0，1）上的增函数，故命题②是假命题；故选：B．三、解答题（本大题共有5题，满分76分．）解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤．17．（14分）在三棱锥P﹣ABC中，底面ABC是边长为6的正三角形，P A⊥底面ABC，且PB与底面ABC所成的角为．（1）求三棱锥P﹣ABC的体积；（2）若M是BC的中点，求异面直线PM与AB所成角的大小（结果用反三角函数值表示）．【解答】解：（1）∵P A⊥平面ABC，∴∠PBA为PB与平面ABC所成的角，即，∵P A⊥平面ABC，∴P A⊥AB，又AB＝6，∴，∴．（2）取棱AC的中点N，连接MN，NP，∵M，N分别是棱BC，AC的中点，∴MN∥BA，∴∠PMN为异面直线PM与AB所成的角．∵P A⊥平面ABC，所以P A⊥AM，P A⊥AN，又，AN＝AC＝3，BM＝BC＝3，∴AM＝＝3，，，所以，故异面直线PM与AB所成的角为．18．（14分）已知双曲线C以F1（﹣2，0）、F2（2，0）为焦点，且过点P（7，12）．（1）求双曲线C与其渐近线的方程；（2）若斜率为1的直线l与双曲线C相交于A，B两点，且（O为坐标原点）．求直线l的方程．【解答】解：（1）设双曲线C的方程为，半焦距为c，则c＝2，，a＝1，…（2分）所以b2＝c2﹣a2＝3，故双曲线C的方程为．…（4分）双曲线C的渐近线方程为．…（6分）（2）设直线l的方程为y＝x+t，将其代入方程，可得2x2﹣2tx﹣t2﹣3＝0（*）…（8分）△＝4t2+8（t2+3）＝12t2+24＞0，若设A（x1，y1），B（x2，y2），则x1，x2是方程（*）的两个根，所以，又由，可知x1x2+y1y2＝0，…（11分）即x1x2+（x1+t）（x2+t）＝0，可得，故﹣（t2+3）+t2+t2＝0，解得，所以直线l方程为．…（14分）19．（14分）现有半径为R、圆心角（∠AOB）为90°的扇形材料，要裁剪出一个五边形工件OECDF，如图所示．其中E，F分别在OA，OB上，C，D在上，且OE＝OF，EC＝FD，∠ECD＝∠CDF＝90°．记∠COD＝2θ，五边形OECDF的面积为S．（1）试求S关于θ的函数关系式；（2）求S的最大值．【解答】解：（1）设M是CD中点，连OM，由OC＝OD，可知OM⊥CD，∠COM＝∠DOM＝，，MD＝R sinθ，又OE＝OF，EC＝FD，OC＝OD，可得△CEO≌△DFO，故∠EOC＝∠DOF，可知，…（2分）又DF⊥CD，OM⊥CD，所以MO∥DF，故∠DFO＝，在△DFO中，有，可得…（5分）所以S＝S+S ODF+S OCE＝S△COD+2S ODF＝△COD＝…（8分）（2）…（10分）＝（其中）…（12分）当，即时，sin（2θ+φ）取最大值1．又，所以S的最大值为．…（14分）20．（16分）已知集合M是满足下列性质的函数f（x）的全体：在定义域内存在实数t，使得f（t+2）＝f（t）+f（2）．（1）判断f（x）＝3x+2是否属于集合M，并说明理由；（2）若属于集合M，求实数a的取值范围；（3）若f（x）＝2x+bx2，求证：对任意实数b，都有f（x）∈M．【解答】解：（1）当f（x）＝3x+2时，方程f（t+2）＝f（t）+f（2）⇔3t+8＝3t+10…（2分）此方程无解，所以不存在实数t，使得f（t+2）＝f（t）+f（2），故f（x）＝3x+2不属于集合M．…（4分）（2）由属于集合M，可得方程有实解⇔a[（x+2）2+2]＝6（x2+2）有实解⇔（a ﹣6）x2+4ax+6（a﹣2）＝0有实解，…（7分）若a＝6时，上述方程有实解；若a≠6时，有△＝16a2﹣24（a﹣6）（a﹣2）≥0，解得，故所求a的取值范围是．…（10分）（3）当f（x）＝2x+bx2时，方程f（x+2）＝f（x）+f（2）⇔2x+2+b（x+2）2＝2x+bx2+4+4b⇔3×2x+4bx﹣4＝0，…（12分）令g（x）＝3×2x+4bx﹣4，则g（x）在R上的图象是连续的，当b≥0时，g（0）＝﹣1＜0，g（1）＝2+4b＞0，故g（x）在（0，1）内至少有一个零点；当b＜0时，g（0）＝﹣1＜0，，故g（x）在内至少有一个零点；故对任意的实数b，g（x）在R上都有零点，即方程f（x+2）＝f（x）+f（2）总有解，所以对任意实数b，都有f（x）∈M．…（16分）21．（18分）已知数列{a n}，{b n}满足b n＝a n+1﹣a n（n＝1，2，3，…）．（1）若b n＝10﹣n，求a16﹣a5的值；（2）若且a1＝1，则数列{a2n+1}中第几项最小？请说明理由；（3）若c n＝a n+2a n+1（n＝1，2，3，…），求证：“数列{a n}为等差数列”的充分必要条件是“数列{c n}为等差数列且b n≤b n+1（n＝1，2，3，…）”．【解答】解：（1）由b n＝10﹣n，可得b n+1﹣b n＝（9﹣n）﹣（10﹣n）＝﹣1，故{b n}是等差数列．所以a16﹣a5＝（a16﹣a15）+（a15﹣a14）+（a14﹣a13）+…+（a6﹣a5）＝…（4分）（2）a2n+3﹣a2n+1＝（a2n+3﹣a2n+2）+（a2n+2﹣a2n+1）＝b2n+2+b2n+1＝（22n+2+231﹣2n）﹣（22n+1+232﹣2n）＝22n+1﹣231﹣2n…（6分）由a2n+3＜a2n+1⇔22n+1﹣231﹣2n＜0⇔n＜7.5，a2n+3＞a2n+1⇔22n+1﹣231﹣2n＞0⇔n＞7.5，…（8分）故有a3＞a5＞a7＞…＞a15＞a17＜a19＜a20＜…，所以数列{a2n+1}中a17最小，即第8项最小．…（10分）法二：由，…（5分）可知a2n+1＝a1+b1+b2+b3+…+b2n＝＝…（8分）（当且仅当22n+1＝233﹣2n，即n＝8时取等号）所以数列{a2n+1}中的第8项最小．…（10分）（3）若数列{a n}为等差数列，设其公差为d，则c n+1﹣c n＝（a n+1﹣a n）+2（a n+2﹣a n+1）＝d+2d＝3d为常数，所以数列{c n}为等差数列．…（12分）由b n＝a n+1﹣a n＝d（n＝1，2，3，…），可知b n≤b n+1（n＝1，2，3，…）．…（13分）若数列{c n}为等差数列且b n≤b n+1（n＝1，2，3，…），设{c n}的公差为D，则c n+1﹣c n＝（a n+1﹣a n）+2（a n+2﹣a n+1）＝b n+2b n+1＝D（n＝1，2，3，…），…（15分）又b n+1+2b n+2＝D，故（b n+1﹣b n）+2（b n+2﹣b n+1）＝D﹣D＝0，又b n+1﹣b n≥0，b n+2﹣b n+1≥0，故b n+1﹣b n＝b n+2﹣b n+1＝0（n＝1，2，3，…），…（17分）所以b n+1＝b n（n＝1，2，3，…），故有b n＝b1，所以a n+1﹣a n＝b1为常数．故数列{a n}为等差数列．综上可得，“数列{a n}为等差数列”的充分必要条件是“数列{c n}为等差数列且b n≤b n+1（n＝1，2，3，…）”．…（18分）。

2017年上海市黄浦区高考数学一模试卷学校:___________姓名：___________班级：___________考号：___________一、填空题(本大题共12小题，共54.0分)1.若集合A={x||x-1|＜2，x∈R}，则A∩Z= ______ ．【答案】{0，1，2}【解析】解：集合A={x||x-1|＜2，x∈R}={x|-2＜x-1＜2，x∈R}={x|-1＜x＜3，x∈R}，则A∩Z={0，1，2}．故答案为{0，1，2}．化简集合A，根据交集的定义写出A∩Z即可．本题考查了集合的化简与运算问题，是基础题目．2.抛物线y2=2x的准线方程是______ ．【答案】【解析】解：抛物线y2=2x，∴p=1，∴准线方程是x=-故答案为：-先根据抛物线方程求得p，进而根据抛物线的性质，求得答案．本题主要考查了抛物线的性质．属基础题．3.若复数z满足（i为虚数单位），则z= ______ ．【答案】1+2i【解析】解：由，得z=1+2i．故答案为：1+2i．直接利用复数代数形式的乘除运算化简得答案．本题考查了复数代数形式的乘除运算，是基础题．4.已知sin（α+）=，α∈（-，0），则tanα= ______ ．【解析】解：∵sin（α+）=cosα，sin（α+）=，∴cosα=，又α∈（-，0），∴sinα=-，∴tanα==-2．故答案为：-2．由α∈（-，0）sin（α+）=，利用诱导公式可求得cosα，从而可求得sinα与tanα．本题考查运用诱导公式化简求值，考查同角三角函数间的基本关系，属于中档题．5.以点（2，-1）为圆心，且与直线x+y=7相切的圆的方程是______ ．【答案】（x-2）2+（y+1）2=18【解析】解：将直线x+y=7化为x+y-7=0，圆的半径r==3，所以圆的方程为（x-2）2+（y+1）2=18．故答案为（x-2）2+（y+1）2=18．由点到直线的距离求出半径，从而得到圆的方程．本题考查直线与圆相切的性质，圆的标准方程等知识的综合应用，属于基础题．6.若二项式的展开式共有6项，则此展开式中含x4的项的系数是______ ．【答案】10【解析】解：∵二项式的展开式共有6项，故n=5，则此展开式的通项公式为T r+1=•（-1）r•x10-3r，令10-3r=4，∴r=2，中含x4的项的系数=10，故答案为：10．根据题意求得n=5，再在二项展开式的通项公式中，令x的幂指数等于4，求得r的值，可得展开式中含x4的项的系数．本题主要考查二项式定理的应用，二项展开式的通项公式，求展开式中某项的系数，二项式系数的性质，属于基础题．7.已知向量，（x，y∈R），，，若x2+y2=1，则的最大值为______ ．【答案】+1=≤+r=+1=+1．∴的最大值为+1．故答案为：．利用≤+r即可得出．本题考查了向量的模的计算公式、点与圆的位置关系，考查了推理能力与计算能力，属于基础题．8.已知函数y=f（x）是奇函数，且当x≥0时，f（x）=log2（x+1）．若函数y=g（x）是y=f（x）的反函数，则g（-3）= ______ ．【答案】-7【解析】解：∵反函数与原函数具有相同的奇偶性．∴g（-3）=-g（3），∵反函数的定义域是原函数的值域，∴log2（x+1）=3，解得：x=7，即g（3）=7，故得g（-3）=-7．故答案为：-7．根据反函数与原函数的关系，可知反函数的定义域是原函数的值域，即可求解．本题考查了反函数与原函数的性质关系．属于基础题．9.在数列{a n}中，若对一切n∈N*都有a n=-3a n+1，且=，则a1的值为______ ．【答案】-12【解析】解：在数列{a n}中，若对一切n∈N*都有a n=-3a n+1，可得数列{a n}为公比为-的等比数列，=，可得====，可得a1=-12．故答案为：-12．由题意可得数列{a n}为公比为-的等比数列，运用数列极限的运算，解方程即可得到所求．本题考查等比数列的通项和求和公式，以及数列极限的运算，属于中档题．10.甲、乙两人从6门课程中各选修3门．则甲、乙所选的课程中至多有1门相同的选法共有______ ．【答案】解：根据题意，分两种情况讨论：①甲乙所选的课程全不相同，有C63×C33=20种情况，②甲乙所选的课程有1门相同，有C61×C52×C32=180种情况，则甲、乙所选的课程中至多有1门相同的选法共有180+20=200种情况；故答案为：200．根据题意，甲、乙所选的课程中至多有1门相同，其包含两种情况：①甲乙所选的课程全不相同，②甲乙所选的课程有1门相同；分别计算每种情况下的选法数目，相加可得答案．本题考查排列组合的运用，涉及分类计数问题，注意“甲、乙所选的课程中至多有1门相同的选法”的理解．11.已知点O，A，B，F分别为椭圆：＞＞的中心、左顶点、上顶点、右焦点，过点F作OB的平行线，它与椭圆C在第一象限部分交于点P，若，则实数λ的值为______ ．【答案】【解析】解：如图，A（-a，0），B（0，b），F（c，0），则P（c，），∴，，，，由，得，即b=c，∴a2=b2+c2=2b2，．则．故答案为：．由题意画出图形，求出、的坐标，代入，结合隐含条件求得实数λ的值．本题考查椭圆的简单性质，考查了直线与椭圆位置关系的应用，训练了平面向量在求解圆锥曲线问题中的应用，是中档题．12.已知为常数），，且当x1，x2∈[1，4]时，总有f（x1），【解析】解：依题意知，当x1，x2∈[1，4]时，f（x1）max≤g（x2）min，由“对勾'函数单调性知，=2x+=2（x+）在区间[1，4]上单调递增，∴g（x2）min=g（1）=3；∵=2ax2+2x，当a=0时，f（x）=2x在区间[1，4]上单调递增，∴f（x）max=f（4）=8≤3不成立，故a≠0；∴f（x）=2ax2+2x为二次函数，其对称轴方程为：x=-，当a＞0时，f（x）在区间[1，4]上单调递增，f（x）max=f（4）=8≤3不成立，故a＞0不成立；当a＜0时，1°若-≤1，即a≤-时，f（x）在区间[1，4]上单调递减，f（x）max=f（1）=2a+2≤3恒成立，即a≤-时满足题意；2°若1＜-＜4，即-＜a＜-时，f（x）max=f（-）=-≤3，解得：-＜a≤-；3°若-≥4，即-≤a＜0时，f（x）在区间[1，4]上单调递增，f（x）max=f（4）=32a+8≤3，解得a≤-∉（-，0），故不成立，综合1°2°3°知，实数a的取值范围是：（- ，-]．故答案为： ，．依题意可知，当x1，x2∈[1，4]时，f（x1）max≤g（x2）min，利用对勾函数的单调性质可求g（x2）min=g（1）=3；再对f（x）=2ax2+2x中的二次项系数a分a=0、a＞0、a ＜0三类讨论，利用函数的单调性质可求得f（x）在区间[1，4]上的最大值，解f（x）≤3即可求得实数a的取值范围．max本题考查函数恒成立问题，由题意分析出当x1，x2∈[1，4]时，f（x1）max≤g（x2）min 是关键，考查等价转化思想与分类讨论思想的综合运用，考查运算能力，属于难题．三、解答题(本大题共5小题，共76.0分)17.在三棱锥P-ABC中，底面ABC是边长为6的正三角形，PA⊥底面ABC，且PB与底面ABC所成的角为．（1）求三棱锥P-ABC的体积；（2）若M是BC的中点，求异面直线PM与AB所成角的大小（结果用反三角函数值表示）．【答案】解：（1）∵PA⊥平面ABC，∴∠PBA为PB与平面ABC所成的角，即∠，∵PA⊥平面ABC，∴PA⊥AB，又AB=6，∴，∴．（2）取棱AC的中点N，连接MN，NP，∵M，N分别是棱BC，AC的中点，∴MN∥BA，∴∠PMN为异面直线PM与AB所成的角．∵PA⊥平面ABC，所以PA⊥AM，PA⊥AN，又，AN=AC=3，BM=BC=3，∴AM==3，，，所以∠，故异面直线PM与AB所成的角为．【解析】（1）在R t△PAB中计算PA，再代入棱锥的体积公式计算；（2）取棱AC的中点N，连接MN，NP，分别求出△PMN的三边长，利用余弦定理计算cos∠PMN即可．本题考查了棱锥的体积计算，空间角的计算，属于中档题．18.已知双曲线C以F1（-2，0）、F2（2，0）为焦点，且过点P（7，12）．（1）求双曲线C与其渐近线的方程；（2）若斜率为1的直线l与双曲线C相交于A，B两点，且（O为坐标原点）．求直线l的方程．【答案】解：（1）设双曲线C的方程为＞，＞，半焦距为c，则c=2，，a=1，…（2分）所以b2=c2-a2=3，故双曲线C的方程为．…（4分）双曲线C的渐近线方程为．…（6分）可得2x2-2tx-t2-3=0（*）…（8分）△=4t2+8（t2+3）=12t2+24＞0，若设A（x1，y1），B（x2，y2），则x1，x2是方程（*）的两个根，所以，，又由，可知x1x2+y1y2=0，…（11分）即x1x2+（x1+t）（x2+t）=0，可得，故-（t2+3）+t2+t2=0，解得，所以直线l方程为．…（14分）【解析】（1）设出双曲线C方程，利用已知条件求出c，a，解得b，即可求出双曲线方程与渐近线的方程；（2）设直线l的方程为y=x+t，将其代入方程，通过△＞0，求出t的范围，设A（x1，y1），B（x2，y2），利用韦达定理，通过x1x2+y1y2=0，求解t即可得到直线方程．本题考查双曲线的方程的求法，双曲线的简单性质的应用，直线与双曲线的位置关系的综合应用，考查计算能力．19.现有半径为R、圆心角（∠AOB）为90°的扇形材料，要裁剪出一个五边形工件OECDF，如图所示．其中E，F分别在OA，OB上，C，D在上，且OE=OF，EC=FD，∠ECD=∠CDF=90°．记∠COD=2θ，五边形OECDF的面积为S．（1）试求S关于θ的函数关系式；（2）求S的最大值．【答案】解：（1）设M是CD中点，连OM，由OC=OD，可知OM⊥CD，∠COM=∠DOM=，∠，MD=R sinθ，又OE=OF，EC=FD，OC=OD，可得△CEO≌△DFO，故∠EOC=∠DOF，可知∠∠∠，…（2分）又DF⊥CD，OM⊥CD，所以MO∥DF，故∠DFO=，在△DFO中，有，∠∠可得…（5分）=＜＜…（8分）（2）…（10分）=（其中）…（12分）当，即时，sin（2θ+φ）取最大值1．又，，所以S的最大值为．…（14分）【解析】（1）设M是CD中点，连OM，推出∠COM=∠DOM=∠，MD=R sinθ，利，用△CEO≌△DFO，转化求解∠DFO=，在△DFO中，利用正弦定理∠∠求解S=S△COD+S ODF+S OCE=S△COD+2S ODF的解析式即可．（2）利用S的解析式，通过三角函数的最值求解即可．本题考查函数与方程的实际应用，三角函数的最值的求法，考查转化思想以及计算能力．20.已知集合M是满足下列性质的函数f（x）的全体：在定义域内存在实数t，使得f（t+2）=f（t）+f（2）．（1）判断f（x）=3x+2是否属于集合M，并说明理由；（2）若属于集合M，求实数a的取值范围；（3）若f（x）=2x+bx2，求证：对任意实数b，都有f（x）∈M．【答案】解：（1）当f（x）=3x+2时，方程f（t+2）=f（t）+f（2）⇔3t+8=3t+10…（2分）此方程无解，所以不存在实数t，使得f（t+2）=f（t）+f（2），故f（x）=3x+2不属于集合M．…（4分）（2）由属于集合M，可得方程有实解⇔a[（x+2）2+2]=6（x2+2）有实解⇔（a-6）x2+4ax+6（a-2）=0有实解，…（7分）若a=6时，上述方程有实解；若a≠6时，有△=16a2-24（a-6）（a-2）≥0，解得，2=2x+bx2+4+4b⇔3×2x+4bx-4=0，…（12分）令g（x）=3×2x+4bx-4，则g（x）在R上的图象是连续的，当b≥0时，g（0）=-1＜0，g（1）=2+4b＞0，故g（x）在（0，1）内至少有一个零点；当b＜0时，g（0）=-1＜0，＞，故g（x）在，内至少有一个零点；故对任意的实数b，g（x）在R上都有零点，即方程f（x+2）=f（x）+f（2）总有解，所以对任意实数b，都有f（x）∈M．…（16分）【解析】（1）利用f（x）=3x+2，通过f（t+2）=f（t）+f（2）推出方程无解，说明f（x）=3x+2不属于集合M．（2）由属于集合M，推出有实解，即（a-6）x2+4ax+6（a-2）=0有实解，若a=6时，若a≠6时，利用判断式求解即可．（3）当f（x）=2x+bx2时，方程f（x+2）=f（x）+f（2）⇔3×2x+4bx-4=0，令g（x）=3×2x+4bx-4，则g（x）在R上的图象是连续的，当b≥0时，当b＜0时，判断函数是否有零点，证明对任意实数b，都有f（x）∈M．本题考查抽象函数的应用，函数的零点以及方程根的关系，考查转化思想以及计算能力．21.已知数列{a n}，{b n}满足b n=a n+1-a n（n=1，2，3，…）．（1）若b n=10-n，求a16-a5的值；（2）若且a1=1，则数列{a2n+1}中第几项最小？请说明理由；（3）若c n=a n+2a n+1（n=1，2，3，…），求证：“数列{a n}为等差数列”的充分必要条件是“数列{c n}为等差数列且b n≤b n+1（n=1，2，3，…）”．【答案】解：（1）由b n=10-n，可得b n+1-b n=（9-n）-（10-n）=-1，故{b n}是等差数列．所以a16-a5=（a16-a15）+（a15-a14）+（a14-a13）+…+（a6-a5）=…（4分）（2）a2n+3-a2n+1=（a2n+3-a2n+2）+（a2n+2-a2n+1）=b2n+2+b2n+1=（22n+2+231-2n）-（22n+1+232-2n）=22n+1-231-2n…（6分）由a2n+3＜a2n+1⇔22n+1-231-2n＜0⇔n＜7.5，a2n+3＞a2n+1⇔22n+1-231-2n＞0⇔n＞7.5，…（8分）故有a3＞a5＞a7＞…＞a15＞a17＜a19＜a20＜…，所以数列{a2n+1}中a17最小，即第8项最小．…（10分）法二：由，…（5分）可知a2n+1=a1+b1+b2+b3+…+b2n==取等号）所以数列{a2n+1}中的第8项最小．…（10分）（3）若数列{a n}为等差数列，设其公差为d，则c n+1-c n=（a n+1-a n）+2（a n+2-a n+1）=d+2d=3d为常数，所以数列{c n}为等差数列．…（12分）由b n=a n+1-a n=d（n=1，2，3，…），可知b n≤b n+1（n=1，2，3，…）．…（13分）若数列{c n}为等差数列且b n≤b n+1（n=1，2，3，…），设{c n}的公差为D，则c n+1-c n=（a n+1-a n）+2（a n+2-a n+1）=b n+2b n+1=D（n=1，2，3，…），…（15分）又b n+1+2b n+2=D，故（b n+1-b n）+2（b n+2-b n+1）=D-D=0，又b n+1-b n≥0，b n+2-b n+1≥0，故b n+1-b n=b n+2-b n+1=0（n=1，2，3，…），…（17分）所以b n+1=b n（n=1，2，3，…），故有b n=b1，所以a n+1-a n=b1为常数．故数列{a n}为等差数列．综上可得，“数列{a n}为等差数列”的充分必要条件是“数列{c n}为等差数列且b n≤b n+1（n=1，2，3，…）”．…（18分）【解析】（1）判断{b n}是等差数列．然后化简a16-a5=（a16-a15）+（a15-a14）+（a14-a13）+…+（a6-a5）利用等差数列的性质求和即可．（2）利用a2n+3-a2n+1=22n+1-231-2n，判断a2n+3＜a2n+1，求出n＜7.5，a2n+3＞a2n+1求出n＞7.5，带带数列{a2n+1}中a17最小，即第8项最小．．法二：化简，求出a2n+1=a1+b1+b2+b3+…+b2n=，利用基本不等式求出最小值得到数列{a2n+1}中的第8项最小．（3）若数列{a n}为等差数列，设其公差为d，说明数列{c n}为等差数列．由b n=a n+1-a n=d （n=1，2，3，…），推出b n≤b n+1，若数列{c n}为等差数列且b n≤b n+1（n=1，2，3，…），设{c n}的公差为D，转化推出b n+1=b n（n=1，2，3，…），说明数列{a n}为等差数列．得到结果．本题考查数列的综合应用，等差数列的性质等比数列的判断，数列求和，转化思想的应用，考查分析问题解决问题的能力．二、选择题(本大题共4小题，共20.0分)13.若x∈R，则“x＞1”是“＜”的（）A.充分非必要条件B.必要非充分条件C.充要条件D.既非充分也非必要条件【答案】A【解析】解：由x＞1，一定能得到得到＜1，故x＞1是＜1的充分不必要条件，故选：A．根据充分必要条件的定义判断即可．本题考查充分条件、必要条件的定义，通过给变量取特殊值，举反例来说明某个命题不正确，是一种简单有效的方法．14.关于直线l，m及平面α，β，下列命题中正确的是（）A.若l∥α，α∩β=m，则l∥mB.若l∥α，m∥α，则l∥mC.若l⊥α，m∥α，则l⊥mD.若l∥α，m⊥l，则m⊥α【答案】C【解析】解：由直线l，m及平面α，β，知：在A中，若l∥α，α∩β=m，则l与m平行或异面，故A错误；在B中，若l∥α，m∥α，则l与m相交、平行或异面，故B错误；在C中，若l⊥α，m∥α，则由线面垂直的性质定理得l⊥m，故C正确；在D中，若l∥α，m⊥l，则m与α相交、平行或m⊂α，故D错误．故选：C．在A中，l与m平行或异面；在B中，l与m相交、平行或异面；在C中，由线面垂直的性质定理得l⊥m；在D中，m与α相交、平行或m⊂α．本题考查命题真假的判断，是中档题，解题时要认真审题，注意空间中线线、线面、面面的位置关系的合理运用．15.在直角坐标平面内，点A，B的坐标分别为（-1，0），（1，0），则满足tan∠PAB•tan∠PBA=m（m为非零常数）的点P的轨迹方程是（）A. B. C. D.【答案】C【解析】解：设P（x，y），则由题意，（m≠0），化简可得，故选C．设P（x，y），则由题意，（m≠0），化简可得结论．本题考查直接法求轨迹方程，考查斜率公式的运用，属于中档题．16.若函数y=f（x）在区间I上是增函数，且函数在区间I上是减函数，则称函数f（x）是区间I上的“H函数”．对于命题：①函数是（0，1）上的“H函数”；②函数是（0，1）上的“H函数”．下列判断正确的是（）A.①和②均为真命题B.①为真命题，②为假命题C.①为假命题，②为真命题D.①和②均为假命题【答案】B【解析】解：对于命题①：令t=，函数=-t2+2t，∵t=在（0，1）上是增函数，函数y=-t2+2t在（0，1）上是增函数，∴在（0，1）上是增函数；G（x）=在（0，1）上是减函数，∴函数是（0，1）上的“H函数“，故命题①是真命题．对于命题②，函数=是（0，1）上的增函数，H（x）=是（0，1）上的增函数，故命题②是假命题；故选：B．对函数，G（x）=在（0，1）上的单调性进行判断，得命题①是真命题．对函数=，H（x）=在（0，1）上单调性进行判断，得命题②是假命题．本题考查了命题真假的判定，涉及到了函数的单调性，属于中档题．。

黄浦区2017年高考模拟考语文试卷2017年4月（完卷时间：150分钟满分：150分）考生注意：1．本考试设试卷和答题纸两部分，试卷包括试题与答题要求，所有答题必须涂（选择题）或写（非选择题）在答题纸上，做在试卷上一律不得分。

2．答卷前，务必在答题纸正面清楚地填写姓名、准考证号。

3．答题纸与试卷在试题编号上是一一对应的，答题时应特别注意，不能错位。

用2B铅笔作答选择题，非选择题部分必须使用0.5毫米的黑色墨水笔书写。

4．完卷时间150分钟；试卷满分150分。

一积累应用 10分1．按要求填空。

（5分）（1）念去去、千里烟波，________________。

（柳永《雨霖铃》）（2）落花水香茅舍晚，_________________。

（马致远《[________]寿阳曲·远浦帆归》）（3）同样是杜甫的诗，《江南逢李龟年》中“正是江南好风景，落花时节又逢君”与《登楼》“___________________，____________________”一联所用手法相似。

2．按要求选择。

（5分）（1）小王不幸遭遇车祸，情绪低落，以下句子适．合用来．．．鼓．励．他战胜困难的一项是（）。

（2分）A．沉舟侧畔千帆过，病树前头万木春。

B．删繁就简三秋树，领异标新二月花。

C．九层之台，起于累土。

D．玉不琢，不成器。

（2）把下列画线句放回原文序号处，最．通顺．．的一项是（）。

（3分）2017年3月3日，《科学》杂志发表了题为《中国许昌发现的更新世晚期古老型人类头骨》的论文。

该文称在中国许昌发现了生活在10.5万年至12.5万年之前的“许昌人”，①这是一种新的古老类型人类，②中国古人类学家在近十年中开展了大量野外调查、发掘和化石研究工作，先后在湖北郧西黄龙洞、湖南道县福岩洞、安徽东至华龙洞等地发现了珍贵的古人类化石，③从而对中国更新世中、晚期人类演化的认识进一步深入。

④发现“许昌人”的灵井旧石器时代晚期遗址，在2007年和2014年两次出土人类头骨化石，这些发现被评为当年的十大考古发现之一，遗址已被国务院公布为国家级文物保护单位。

上海市黄浦区2017届高考模拟考试（二模）英语试题考生注意：1. 考试时间120分钟, 试卷满分150分。

2. 本试卷设试卷和答题卷两部分。

试卷分为第I卷和第II卷。

所有答案必须涂（选择题）或写（非选择题）在答题卷上，做在试卷上一律不得分。

3. 答题前，务必在答题卷纸上填写准考证号和姓名，并将核对后的条形码贴在指定位置上。

第I卷I. Listening ComprehensionSection ADirections: In Section A, you will hear ten short conversationsbetween two speakers. At the end of each conversation, a questionwill be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and thequestion 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. August 6. B. August 7. C. August 13.D. August 19.2. A. A coat store. B. A chemist’s. C. A dry cleaner.D. A watch shop.3. A. Policeman and driver. B. Teacher and student.C. Boss and clerk.D. Doctor and nurse.4. A. A brand of jeans. B. A new hair style. C. Popular。

黄浦区2016-2017学年度第一学期高三年级期终调研测试英语试卷（完卷时间：120分钟满分：150分）2016年12月9日上午II. Grammar and VocabularySection ADirections: After reading the passage 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.Infant Day Care, Good or Bad?The British psychoanalyst John Bowlby maintains that separation from the parents during th e sensitive “attachment” period from birth to three may influence a child’s personality and lead to emotional problems in later life. Some people have drawn the conclusion from Bowlby’s work (21) _________ children should not be sent to day care before the age of three because of the parental separation (22) _________ involves, and many people do believe this. But there are also arguments (23) _________ such a strong conclusion.Firstly, experts point out that the isolated love affair between children and parents (24) _________ (find) in modern societies does not usually exist in traditional societies. For example, in some tribal societies, such as the Ngoni, the father and mother of a child did not raise their infant alone –far from it. Secondly, common sense tells us that day care would not be so widespread today (25) _________ parents and care-takers found children had problems with it. Statistical studies of this kind have not yet been carried out, and they have regularly reported that day care had a sli ghtly positive effect on children’s development. But tests (26) ________ have been used to measure this development are not widely enough accepted to settle the issue.But Bowlby’s analysis raises the possibility that early day care has delayed eff ects. The possibility that such care might lead to, say, more mental illness or crime 15 or 20 years later can only be explored by the use of statistics. Whatever the long-term effects, parents sometimes find the immediate effects difficult (27) _________ (deal) with. Children under three are likely to protest at (28) _________ (leave) their parents and show unhappiness. At the age of three or three and a half almost all children find the change to nursery easy, and this is undoubtedly (29) _________ more and more parents make use of child care at this time. The matter, then, is far from clear-cut, though experience and available evidence (30) _________ (indicate) early care is reasonable for infants.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.First Aid: Difference between Death and LifeFirst aid is emergency care for a victim of sudden illness or injury until more skillful medical treatment is available. It may save a life or improve certain ___31___ signs including pulse, temperature, and breathing. First aid must be ___32___ as quickly as possible. In the case of the critically injured, a few minutes can make the difference between complete recovery and loss of life.First-aid ___33___ depend upon a victim’s needs and the provider’s level of knowledge and skill. Knowing what not to do in an emergency is as important as knowing what to do. For example, ___34___ moving a person with a neck injury can lead to permanent health problems.Despite the variety of injuries possible, several ___35___ of first aid apply to all emergencies. The first step is to call for professional medical help. The victim, if conscious, should be reassured that medical aid has been requested, and asked for permission to provide any first aid. Next, ___36___ the scene, asking other people or the injured person’s family or friends about details of the injury or illness, any care that may have already been given, and ___37___ conditions such as heart trouble. Unless the accident scene becomes unsafe or the victim may suffer further injury, do not move the victim.First aid requires rapid assessment of victims to determine whether ___38___ conditions exist. One method for ___39___ a victim’s condition is known by the acronym ABC, which stands for:A – Airway: is it open and clear?B – Breathing: is the person breathing? Look, listen and feel for breathing.C –Circulation: is there a pulse? Is the person bleeding ___40___? Check skin color and temperature foradditional indications of circulation problems.III. Reading ComprehensionSection ADirections:For each blank in the following passages 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.Animal RightsEvery conscious being has interests that should be respected. No being who is conscious of being alive should be devalued to thinghood, dominated, and used as a resource or ___41___. The key point of the idea known as animal rights is a movement to extend moral consideration to all ___42___ beings. Nobody should have to demonstrate a specific level of intelligence or be judged beautiful to be given moral consideration. No being should have to be useful to humanity or capable of accepting “duties” in order to be extended moral considera tion. ___43___, what other animals need from us is being free from duties to us.Animal rights is about letting animals live on their own terms. It can be written into our laws, but is not an actual list or bill of rights as we have for human society. It begins with ourpromises not to act like ___44___ of others. Animal rights is about justice ─ treating animals fairly.Why is animal rights ___45 ___? It is because we humans often act as though we are the only beings on the planet. Although we depend on other animals for our very survival, humans are the only animals that have upset the balance of nature. There are lots of ways that humans ___46___ animals. We domesticate them and use them for food, even though our nutritional needs can be completely supplied by a(n) ___47___ diet. Although other materials are available, we use animal’s skin and other body parts for clothing, furs, hats, boots, jewellery and even pet toys. Humans can talk about it but animals cannot. All animals wish to experience life in its fullness. Unlike many animals who have to kill to survive, humans do not. Why should humans cause ___48___ to other beings when it’s not necessary?As we do, animals protect their children; they feel fear; they warn each other of dangers; they play. We might differ from other animals in some ways, but that doesn’t give us the right to ___49___ them down, take their lands, pollute their waters, or use them for our conveniences. Animals also experience pain and it’s not difficult to observe ___50___ of pain in the way a conscious being reacts to it. We take advantage, cause distress, and act ___51___ when we use animals for amusement. Lots of pets are ___52___ on the streets when their owners no longer find it convenient or affordable to keep or care for them.Whether we admit it or not, it’s a prejudice to think we are ___53___ to animals and that it is our right to control them, which can only make people act mean, hateful or neglectful. However, each of us has within us the power to ___54___. We can adopt a different attitude, one that reshape our destiny. This will have wonderful effects on the planet’s other communities, for life is ___55___ avoiding suffering. It is interacting, singing, pursuing joy. We humans can learn to live responsibly, with respect, kindness and love.41. A. companies B. goods C. insects D. providers42. A. active B. conscious C. intelligent D. strange43. A. Indeed B. Moreover C. Nevertheless D. Otherwise44. A. followers B. friends C. masters D. tutors45. A. necessary B. neglected C. respected D. revolutionary46. A. distinguish B. eliminate C. exploit D. raise47. A. animal-free B. eco-friendly C. low-salt D. well-balanced48. A. conflict B. confusion C. isolation D. misery49. A. calm B. chase C. pull D. tear50. A. signs B. symbols C. symptoms D. performances51. A. differently B. enthusiastically C. gently D. unfairly52. A. abandoned B. chosen C. oppressed D. spoiled53. A. accessible B. appealing C. reasonable D. superior54. A. change B. dominate C. persist D. proceed55. A. contrary to B. more than C. owing to D. rather thanSection 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 read.(A)①Did English football finally find a new star? At the age of 19, Theo Walcott came onto the scene by scoring a hat-trick for England in a 4-1 victory over Croatia in 2010 World Cup.②Walcott’s lightning speed and accurate shooting turned the teenager into an overnight star. Many thought he was a new dawn for the England team. He was building his fame for his fast pace, with former Barcelona manager Pep declaring that “you would need a gun to stop him.” FIFA World Player of the Year winner Lionel Messi once also described Walcott as “one of the most dangerous players I have ever played against.” In addition to his speed, Walcott also possessed good balance, movement and technique.③It was symbolic that Walcott’s goals came from the right-wing. The position had been played by “golden boy” David Beckham for more than 10 years. No longer were the cheers for Beckham. The fans’ hopes now rested on the shoulders of Walcott.④Walcott was born in London to a black British Jamaican father and a white English mother. He grew up as a Liverpool fan due to his father’s support of Liverpool. When Chelsea asked him to be a ball boy, he used the opportunity to meet his Liverpool idols.⑤The teenage r’s rise to fame was not completely out of blue. He was part of England’s World Cup team in 2006, but he did not get to play a match. He also spent over two years at Arsenal, where he was fast becoming a key player.⑥But that year, few were expecting the wonderful performance between England and Croatia. The teenager was the first England player to score three goals in a game since Michael Owen did so seven years before.⑦Although England was full of superstars, they had a poor record in major tournaments. But things were beginning to change. The win against Croatia was sweet revenge. Croatia was the team which knocked England out of Euro 2008.⑧Walcott’s wonderful performance lighted England fans’ hope for World Cup victory in South Africa in 2010, since England had not lifted the cup since 1966.⑨But before England fans got too carried away, our reflection on the past history told us that placing a country’s hopes on one teenager was dangerous.56. Which of the following CANNOT account for Walcott’s increa sing fame?A. Fast speed.B. Masterly skills.C. Positional sense.D. Unusual family.57. Why did the author mention David Beckham in the 3rd paragraph?A. To show that England football once had a glorious history.B. To illustrate that Walcot t could be entitled “golden boy”.C. To indicate that England fans were difficult to please.D. To imply that people had high expectation on Walcott.58. In the 5th paragraph, the underlined phrase “out of blue” most probably means “________”.A. impoliteB. unexpectedC. impossibleD. unintentional59. What is the author most likely to agree with?A. Walcott might not live up to fans’ expectation.B. Walcott might transfer from Arsenal to Liverpool.C. Croatia might change the history of the World Cup.D. England might be defeated by the opponent in the next round.(B)✓OverviewExplore Stewart Island and the surrounding bays in our modern mini-buses. Our guides enjoy sharing their local knowledge of the history and environment of Stewart Island. Highlights include Lee Bay, the gateway to Rakiura National Park, beautiful Horseshoe Bay and amazing views of✧More information♦Departure location: Oban Visitor Centre.♦What to bring: Comfortable walking shoes or boots, waterproof jacket, warm sweater or fleece jacket, sunscreen or sunglasses, insect repellent and camera.♦Car parking: Vehicle parking is available at Oban (extra cost—reservations recommended).♦Wheelchair access: Available.♦Children ticket: Children under ten go free for travel as long as they are accompanied by an adult.✧Reviews♦“There was so much to see and learn that it was hard to take everything in. The bays we stopped at were beautiful with golden sandy beaches, the forests were overpoweringand we expected dinosaurs to appear at any time, the views from lookout point weresplendid and the anchor point with Bluff brought a smile. Thank you to Chris and theexperienced team for such an informative tour.”Ron P♦“Any visitor to Stewart Island could do no better than take one of the guided tours from the Oban Visitor Centre—especially if you only have limited time available. We hadthe delightful and extremely informative Kylie conduct a small number on one of thevillage tours. This is a beautiful place—a few fascinating shops and restaurants,wonderful walks and warm and friendly people.”Michael Mason“I love findin g out about places and the guide was full of information and stories as we visited every interesting place and view in Oban (it didn’t take too long...). A great wayto start a visit as it helps you know where everything is.”Kiwieric60. If a traveler plans to leave a car at Oban, he had better ________.A. refer to the guides firstB. use wheelchair accessC. make a reservationD. walk to the center in advance61. Herry, a six-year-old boy, wanted to have a sightseeing of the Stewart Island with his parents.How much should they pay for the mini-bus tour?A. $135.B. $90.C. $ 45.D. Free.62. If a traveler takes the guided tour, he can experience all the following EXCEPT ________.A. breath-taking sceneryB. charming walksC. dinosaur samplesD. detailed tour guide(C)①What does it say about the future of meat when the country’s largest processor of chicken, pork, and beef buys a stake(股份) in a start-up that aims to “perfectly replace animal protein with plant protein”?②Tyson Foods announced this week that it purchased a 5 percent stake in Beyond Meat, the Southern California-based food-tech start-up that made headlines earlier this year with its veggie burger that reportedly cooks and tastes like real beef.③To be sure, Beyond Meat’s meatless creations have yet to take the country by storm. Although the 100 percent plant-based burgers have achieved plenty of positive press since they appeared for the first time in May, so far they’re only available at Whole Foods stores in seven states. Even though the company’s “chicken” strips, “beef” pies, and meatless frozen dinners are available nationwide, Beyond Meat is hardly a household name.④That may be what makes the news of Tyson’s investment all the more noteworthy. While the two companies declined to give details about the deal, it’s doubtful that Tyson’s 5 percent stake made much of dent(凹陷) in the meat giant’s coffers(金库). The company posted $41.4 billion in sales last year; prior to the deal with Tyson, Beyond Meat had reportedly raised $64 million in project capital funding—about what Tyson earns before lunch on any given day.⑤Tyson is doing pretty great. The company reported record third-quarter earnings per share in August and says that it expects overall meat production to increase 2 to 3 percent during the next financial year. But like a big oil company shelling out cash to invest in wind power, Tyson’s toe-in-the-water move to team up with a start-up devoted to bringing more plant-based protein to American dinner tables seems to suggest the meat industry is starting to see which way the winds are blowing.⑥Sales of plant-based protein, which totaled an estimated $5 billion last year, continue to pale compared with the market for meat in America—but vegetarian alternatives to meat are booming, with sales growing at more than double the rate for food products overall. The steadydrumbeat of news about the negative health impacts, environmental problems, and animal welfare concerns associated with meat consumption appears to be sinking in. According to a survey released in April, more than half of Americans surveyed said they plan to eat more plant-based foods in the coming year.63. Beyond Meat’s veggie burger made headlines probably because __________.A. it makes perfect use of animal proteinB. it uses high tech in the making processC. it tastes as good as a genuine beef burgerD. it represents the diet trend in South California64. Which of the following statements is TRUE regarding the state of Beyond Meat?A. It is the creator of the country’s first 100 percent plant-based burgers.B. It has been well received as its products are available nationwide.C. It is far from being a match to real food processing giants like Tyson.D. It provides high-quality dining experience in selected Whole Foods stores.65. What can we infer from paragraph 4?A. The purchase of the stake barely costs a thing for Tyson.B. The 5 percent stake in Beyond Meat means a lot to Tyson.C. Tyson’s investment hasn’t caught the attention of the media as expected.D. Tyson is relying on this investment to raise more project capital funding.66. What does the passage mainly talk about?A. Meat will still take over the market in spite of other alternatives.B. A major American meat company is betting on plant-based protein.C. Tyson and Beyond Meat work together to build a global meat giant.D. Plants have been found to contain protein that does more good to human beings.Section CDirections: Read the following passage. Fill in each blank with a proper sentence given in the box. Each sentence can be used only once. Note that there are two more sentences than you need.Would You B ully(欺负) a Driverless Car or Show It Respect?Say you’re driving down a two-way street and there’s a truck unloading a delivery in theopposite lane. The oncoming traffic needs to pull out into your lane to overtake.What do you do?___67___ Eventually one of us feels charitable and slows down to allow the oncoming car to overtake and give permission with a quick flash of headlights or a wave of the hand.But what if the car waiting patiently behind the parked truck is a driverless or autonomous vehicle (AV)? Will this robot car be able to understand what you mean when you flash your lights or wave your hands?Its sensors could decide that it’s only safe to overtake when there’s no oncoming traffic at all. On a busy road at school home time, this may be never, leading to increasingly angry drivers queuing behind. ___68___ This is one of the conclusions to be drawn from research carried out by Dr Chris Tennant of the psychological and behavioural science department at the London School of Economics.His Europe-wide survey finds that nearly two-thirds of drivers think machines won’t have enough common sense to interact with human drivers, and more than two-fifths think a robot car would remain stuck behind our assumed parked truck for a long time.Driving isn’t just about technology and engineering, it’s about human interactions and psychology. The road is a social space. ___69___ “If you view the road as a social space, you will consciously negotiate your journey with other drivers. People who like that negotiation process appear to feel less comfortable engaging with AVs than with human drivers,” says Mr Tennant in his report.___70___ A statistic often trotted out(动不动就搬出) is that human error is responsible for more than 90% of accidents, with our tendency to road anger, tiredness and lack of concentration.IV. Summary WritingDirections: Read the following passage. Summarize the main idea and the main point(s) of the passage in no more than 60 words. Use your own words as far as possible.Super Size MeFast food, otherwise known as junk food, is a huge passion for a large number of people across the Western world. But what would happen if you ate lots of junk food, every day? Would it seriously damage your health? These were the questions which led Morgan Spurlock, an independent film-maker, to do an experiment, which he made into a documentary film entitled Super Size Me.The main basis of his experiment was that Spurlock promised to eat three McDonald’s meals a day, every day, for a month. He could only eat food from McDonald’s and every time an employee asked if he would like to “super size” the meal, he had to agree. “Super sizing” refers to the fact that with this type of meal you get a considerable larger portion of everything.Spurlock knew that by ea ting three McDonald’s meals a day, he would consume a lot of fat and a great deal of salt and sugar in each meal—much more than he needed. Although Spurlock knew he would put on a bit of weight, and that this diet was unhealthy, he wasn’t quite prepared for just how unhealthy it turned out to be. The changes in his body were horrifying in the first week, he put on 4.5 kilos and by the end of the thirty days he had gained nearly 14 kilos, bringing his total weight to a massive 98kg.Spurlock says “I’d love people to walk out of the movie and say, ’Next time I’m not going to “super size”. Maybe I’m not going have any junk food at all. I’m going to sit down and eat dinner with my kids, with the TV off, so that we can eat healthy food, talk about what we’re eat ing and have a relationship with each other.’” Food for thought indeed.第II卷（共40分）V. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.1.这款手表不防水。

黄浦区2016学年度第一学期高三年级期终调研测试物理试卷2017年1月12日（本试卷共4页, 满分100分, 考试时间60分钟。

）考生注意:1.答卷前, 务必用钢笔或圆珠笔在答题纸正面清楚地填写姓名、准考证号, 并将核对后的条形码贴在指定位置上, 在答题纸反面清楚地填写姓名。

1．2、第一大题的作答必须用2B铅笔涂在答题纸上相应区域内与试卷题号对应的位置, 需要更改时, 必须将原选项用橡皮擦去, 重新选择。

第二、第三大题的作答必须用黑色的钢笔或圆珠笔写在答题纸上与试卷题号对应的位置（作图可用铅笔）。

2．3.第19、20题要求写出必要的文字说明、方程式和重要的演算步骤。

只写出最后答案, 而未写出主要演算过程的, 不能得分。

有关物理量的数值计算问题, 答案中必须明确写出数值和单位。

一. 单项选择题（共40分, 1-8题每小题3分, 9-12题每小题4分, 每小题只有一个正确选项。

）在国际单位制（SI）中, 下列属于基本单位的是（）（A）千克（B）牛顿（C）库仑（D）焦耳3．奥斯特首先通过实验（）（A）提出了单摆的周期公式（B）测出了万有引力恒量G（C）发现了电流周围存在磁场（D）发现了电磁感应现象质量为2kg的质点仅受两个力作用, 两个力的大小分别为3N和5N。

则该质点的加速度的值可能为（）（A）0.5m/s2（B）0.75m/s2（C）3.5m/s2（D）4.5 m/s24．下列事例中属于利用静电现象的是（）（A）油罐车上连接地线（B）复印机复印文件资料（C）屋顶安装避雷针（D）印染厂车间保持湿度三段材质完全相同且不可伸长的细绳OA.OB.OC, 它们共同悬挂一重物, 如图所示, 其中OB水平, A端、B端固定。

若逐渐增加C端所挂重物的质量, 则最先断的绳（）（A）必定是OA（B）必定是OB（C）必定是OC （D）可能是OB, 也可能是OC根据分子动理论可知, 在使两个分子间的距离由很远（r＞10-9m）变到很难再靠近的过程中, 分子间的作用力的大小将（）（A）先减小后增大（B）先增大后减小（C）先增大后减小再增大（D）先减小后增大再减小如图所示, P 为固定的点电荷, 周围实线是其电场的电场线。

2017届上海市黄浦区高三下学期二模考试语文试题及答案上海市黄浦区高考模拟（二模）语文试卷（时间：150分钟满分：150分）考生注意：1、答卷前，考生务必在答题卷上将自己的姓名等相关信息及准考证号填写清楚。

2、本考试设试卷和答题卷两部分，所有试题的答案及作文必须全部写在答题卷上，写在试卷上一律不给分；答题时应注意试题题号和答题卷题号一一对应，不能错位。

3、本试卷共6页，试卷总分150分，考试时间150分钟。

一、阅读80分（一）阅读下文，完成第1-6题。

（17分）收藏是一种物证马未都①文物兼精神与物质两方面属性。

②文物收藏之所以诱人，藏品增值是最直接的原因，再加之中国文物浩如烟海，种类繁多，以个人智力财力精力与之挑战，败多胜少，刺激无穷，这．就使具备一定条件的现代人执着于此，且乐此不疲。

③中国历史上的收藏大都形成于物阜民丰之际。

我们熟知的赵宋皇帝宋徽宗，就是一位身体力行的收藏爱好者。

时隔千年，我们仍能读到成书于北宋宣和年间的有关收藏的各类书籍。

徽宗酷爱艺术，不谙政事，做了国家画院院长，却丢了国家。

这让后人对这一段幽怨的历史唏嘘不已，百思不得其解。

这就是收藏的魅力。

我们今天仍对赵宋江山所遗留的书画、陶瓷等艺术珍品怀着崇高的敬意。

对宋徽宗——赵佶本人的艺术造诣也怀着同样的敬意。

他的《写生珍禽图》以2300万的高价拍卖成交。

④2002年春季，中国嘉德拍卖公司在北京成功地拍卖出宋徽宗的《写生珍禽图》，每一个在拍卖现场的人当时都激动不已。

十数人不断勇敢地举起了手中的号牌，，争先恐后，轮番冲刺。

我在现场，享受着那振奋人心的一刻，所有人对每一次的价格攀升发出惊呼。

一位老者，尽自身全部财力，倾终生对艺术的向往，在1300万高价位勇敢地伸出了自己那只苍老却苍劲的手。

我知道，那是老先生全部的财产。

他用行动阐释着收藏的魅力。

事后，他对我说，那东西(《写生珍禽图》)属于我了一秒钟。

1300万与一秒钟，是老先生酷爱艺术的写照，我甚至觉得他的1300万与成交价2300万相比，前者更让人感动，虽败犹荣。

15.在直角坐标平面内，点A 、B 的坐标分别为(-1,0)、(1,0)，则满足上海市黄浦区2017届高三一模数学试卷2017.1一.填空题(本大题共 12题，1-6每题4分，7-12每题5分，共54分) 1•若集合 A =｛x||x-1| ：：：2,x ・ R ｝，则 A 「|Z 二 ___ 2. 抛物线y =2x 的准线方程是 ____________i 13. 若复数z 满足(i 为虚数单位)，则z =z_1 2----------H 1兀4. 已知sin( )，圧三(,0)，则tan 的值为 ___________2 325. 以点(2, -1)为圆心，且与直线 x • y=7相切的圆的方程是 _____________16•若二项式(x 2-—)n 的展开式共有6项，则此展开式中含 x 4的项的系数是 _______________x2 27.已知向量 ^(x,y) (x, y R)，b =(1,2)，若 x y ^1，则 |a - b| 的最大值为 ____________________ 8•已知函数y 二f(x)是奇函数，且当 x _0时，f (x) = Iog 2(x • 1)，若函数y = g(x)是y 二f(x)的反函数，贝U g(-3)二 __________*99. 在数列｛a n ｝中，若对一切 n • N 都有 a *二-3a n 1，且 lim( a 2a 4- a^ ^2 )则a 1的值为 _________10. 若甲、乙两人从6门课程中各选修 3门，则甲、乙所选修的课程中至多有 1门相同的选法种数为 _________2 211. 已知点o 、A 、B 、F 分别为椭圆c ：务•召 =1 (a b 0)的中心、左顶点、上a b顶点、右焦点，过点F 作OB 平行线，它与椭圆C 在第一象限部分交于点P ，若A^ OP ,则实数■的值为f(xj^g(x 2)，则实数a 的取值范围是 _____________ 二.选择题(本大题共 4题，每题5分，共20分)113. 若x R ，则“ x 1 ”是“1 ”的()条件 xA.充分非必要B.必要非充分C.充要D.既不充分也不必要14. 关于直线I 、m 及平面-：：、-，下列命题中正确的是( )ax x12.已知 f(x) =_2 2x ( a 为常数)，g(x)亠且当x 1、 X 2 [1,4]时，总有A.若I = m，则I // mB.若I //〉，m /,则I // mC.若I —：, m / -■,贝U I - mD.若I //〉，m — I，则m —〉tan. PAB tan • PBA二m （m为非零常数）的点P的轨迹方程是（A.22 y x =1(y-0) B.22 y x=1m m 22C .2 yx=1(y-0) D. 2 yx=1 m m16.若函数y = f （x）在区间I上是增函数，且函数y = 1凶在区间I上是减函数，则称函x数f （x）是区间I上的“ H函数”，对于命题：① 函数f （x）二-x - 2.浪是（0,1）上的2x “ H函数”；②函数g x 2是（0,1）上的“ H函数”；下列判断正确的是（）4 一xA.①和②均为真命题B.①为真命题，②为假命题C.①为假命题，②为真命题D.①和②均为假命题三.解答题（本大题共5题，共14+14+14+16+18=76分）17.在三棱锥P- ABC中，底面ABC是边长为6的正三角形，PA _底面ABC，且PB与底面ABC所成的角为一；6（1 ）求三棱锥P - ABC的体积；（2）若M是BC的中点，求异面直线PM与AB所成角的大小；（结果用反三角函数值表示）18.已知双曲线C以斤（-2,0）、F2（2,0）为焦点，且过点P（7,12）；（1 ）求双曲线C与其渐近线的方程；（2）若斜率为1的直线|与双曲线C相交于A、B两点，且OA_OB（O为坐标原点），求直线I的方程；19.如图，现有半径为R，圆心角（• AOB）为90的扇形材料，要裁剪出一个五边形工件OECDF，其中E、F 分别在OA、OB 上，C、D 在AB 上，且OE = OF，EC 二FD，20. 已知集合M是满足下列性质的函数f(x)的全体：在定义域内存在实数t，使得f(t 2) = f(t) f(2)；(1)判断f(x) =3x 2是否属于集合M，并说明理由；(2)若f(x) =©器属于集合M，求实数a的取值范围；x2十2(3)若f(x)=2x bx2，求证：对任意实数b，都有f(x)・M ；21.已知数列｛a n｝、｛b n｝满足b n =a ni-a n( n =1,2,3,…)；(1)若b n =10—■ n，求a-i^ —a5 的值；(2)若b n =(-1)n(2n・233』)，a^1，则数列｛a2n,｝中第几项最小？请说明理由；(3)若C n二a n • 2a n 1 (n =1,2,3,)，求证：“数列｛a.｝为等差数列”的充分必要条件是“数列｛C n｝为等差数列且b^ib n 1 (n =1,2,3,) ”；参考答案一.填空题1 .{0,1,2}12. X 二23. 1 2i4.-2.25 .(x_2)2(y 1)2 =18 6. 107..518. -79 .-1210. 20011. 212. a < -16二.选择题13. A 14. C15. C16. B - 解答题17 .(1) V =18 ; (2)3癒arccos—218. (1) x2- y 1 ; (2) y = x=、、3 ;2632 2 5 1 219. (1) S =(sin" —sin ^)R ； (2) R ;220. (1)不属于；(2) [1^6.3,12 6.3] ; (3)略;21. (1) 0 ; (2) 8 ; (3)略;。

二、阅读70分4．（16分）阅读下文，完成下列各题。

①历史学家詹姆逊在《政治无意识》一书里写道：“历史并不是一个文本，因为从本质上说它是非叙事的、非再现性的。

然而，还必须附加一个条件，历史只有以文本的形式才能接近我们，换言之，我们只有通过预先的（再）文本化才能接近历史。

”历史有没有“本质化存在”？答案是“有”，否则便陷入了历史虚无主义和历史不可知论。

那么，谁掌握历史的“本质化存在”？基督教徒认为是“上帝之眼”，中国老百姓则认为是“老天爷”。

“上帝之眼”也好，“老天爷”也好，这两种说法都提供了一个共同的信息，即在已经逝去的时空里，存在着一个可能遥远而难以企及但确实存在的客观历史。

历史一旦走过“彼在”的客观时空，我们对历史的认知就只能通过各种转述和记忆进行信息比对，而历史的完全信息在一次又一次的不同的转述和记忆中不断地“沙漏”，同时也不断地“捡漏”。

因此，对于历史的认知，如果不是别有用心的话，后世之人的所有努力都是在尽可能地靠拢“彼在”，接近历史的真相。

我的一个研究希腊艺术的朋友说，希腊语中“历史”一词的词根“historia”原义，就是“调查”“寻找”“研究”，它说明人类在早期对于历史真相及本原的认识就是清醒的。

②历史存在于文本叙述之中，这是新历史主义的一种观点。

从逻辑上讲，不仅后世之人借由前世遗留的各种叙述文本进入已逝的历史，即便是当其世者，接触和把握的信息也是局部的和零碎的，也需要借由他人的叙述来了解现世的信息。

那么，谁在叙述历史？谁在建构历史的不同文本？③目前认为主要有两类叙述者：一类是建立在历史自觉建构意识之上的撰史修志者，他们通过拥有的信息资源，撰写各种年鉴、年志、备忘录，我们通常所谓的“正史”便是指此一类；一类是以历史以及当下现实的事件、人物为原型素材进行文艺创作的文艺家。

对于后者来说，叙述是一种主观创造性行为，叙述风格因为叙述主体的不同而差异性很大，不同的叙述风格形成不同的历史文本。

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（第11题图）黄浦区2017年高考模拟考数 学 试 卷 2017年4月（完卷时间：120分钟 满分：150分）一、填空题（本大题共有12题，满分54分. 其中第1~6题每题满分4分，第7~12题每题满分5分）考生应在答题纸相应编号的空格内直接填写结果.[1．函数y 的定义域是 ．2．若关于,x y 的方程组10420ax y x ay +-=⎧⎨+-=⎩，有无数多组解，则实数a =_________．3．若“2230x x -->”是“x a <”的必要不充分条件，则a 的最大值为 ． 4．已知复数134i z =+，2i z t =+(其中i 为虚数单位)，且12z z ⋅是实数，则实数t 等于 ．5．若函数3 (0),() 1 (0)x x a x f x a x -+<⎧=⎨+≥⎩(a >0,且a ≠1)是R 上的减函数，则a 的取值范围是 ．6．设变量,x y 满足约束条件212x y x y y +≥⎧⎪-≤⎨⎪≤⎩，，， 则目标函数2z x y =-+的最小值为 ．7. 已知圆22:(4)(3)4C x y -+-=和两点 (, 0), (, 0)(0)A m B m m ->，若圆C 上至少存在一点P ，使得90APB ∠=︒，则m 的取值范围是 ．8. 已知向量π(cos(), 1)3a α=+ ，(1,4) b = ，如果a ∥b ，那么πcos(2)3α-的值为 ． 9．若从正八边形的8个顶点中随机选取3个顶点，则以它们作为顶点的三角形是直角三角形的概率是 ．10．若将函数()f x =π|sin()|(0)8x ωω->的图像向左平移π12个单位后，所得 图像对应的函数为偶函数，则ω的最小值是 ．11．三棱锥P ABC -满足：AB AC ⊥，AB AP ⊥，2AB =，4AP AC +=，则该三棱锥的体积V 的取值范围是 ．12．对于数列{}n a ，若存在正整数T ，对于任意正整数n 都有n T n a a +=成立，则称数列{}n a 是以T 为周期的周期数列．设1(01)b m m =<<，对任意正整数n 都有 111)1(01) (n n n n nb b b b b +->⎧⎪=⎨<⎪⎩≤，，若数列{}n b是以5为周期的周期数列，则m 的值可以是 ．(只要求填写满足条件的一个m 值即可)二、选择题（本大题共有4题，满分20分．）每题有且只有一个正确答案，考生应在答题13（ ）A ．y = sin(2x +)2B ．y = cos(2x +)2C ．y = sin(x +π)2D ．y = cos(x +π)214．如图是一个几何体的三视图，根据图中数据，可得该几何体的表面积是 （ ）A ．9πB ．10πC ．11πD ．12π15．已知双曲线22221(0,0)x y a b a b-=>>的右焦点到左顶点的距离等于它到渐近线距离的2倍，则其渐近线方程为（ ）A ．20x y ±=B ．20x y ±=C ．430x y ±=D ．340x y ±=16．如图所示，2π3BAC ∠=，圆M 与,AB AC 分别相切于点,D E ，AD 1=，点P 是圆M 及其内部任意一点，且AP xAD yAE =+(,)x y ∈R ，则x y +的取值范围是（ ）A ．[1,4+B ．[4-+C ．[1,2+D ．[2三、解答题（本大题共有5题，满分76分．）解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤．17．（本题满分14分） 本题共有2个小题，第1小题满分6分，第2小题满分8分． 如图，在直棱柱111ABC A B C -中，12AA AB AC ===，AB AC ⊥，D E F ,,分别是111,,A B CC BC 的中点．（1）求证：AE D F ⊥；（2）求AE 与平面DEF 所成角的大小及点A 到平面DEF 的距离．18．（本题满分14分）本题共有2小题，第小题满分6分，第小题满分8分．在ABC ∆中，角,,A B C 的对边分别为,,a b c ，且cos ,cos ,cos b C a A c B 成等差数列． （1）求角A 的大小；（2）若a =6b c +=，求AB AC +的值．19．（本题满分14分）本题共有2个小题，第1小题6分，第2小题8分．如果一条信息有n 1,)n n >∈N （种可能的情形（各种情形之间互不相容），且这些情形发生的概率分别为12,,,n p p p ，则称H =12()()()n f p f p f p ++ （其中()f x =log ,a x x -(0,1)x ∈）为该条信息的信息熵．已知11()22f =．（1）若某班共有32名学生，通过随机抽签的方式选一名学生参加某项活动，试求“谁被选中”的信息熵的大小；（2）某次比赛共有n 位选手（分别记为12,,,n A A A ）参加，若当1,2,k =,1n - 时，选手k A 获得冠军的概率为2k -，求“谁获得冠军”的信息熵H 关于n 的表达式．20．（本题满分16分）本题共有3个小题，第1小题满分4分，第2小题满分6分，第3小题满分6分．设椭圆M :22221(0)x y a b a b +=>>的左顶点为A 、中心为O ，若椭圆M 过点11(,)22P -，且AP PO ⊥．（1）求椭圆M 的方程；（2）若△APQ 的顶点Q 也在椭圆M 上，试求△APQ 面积的最大值；（3）过点A 作两条斜率分别为12,k k 的直线交椭圆M 于,D E 两点，且121k k =，求证：直线DE 恒过一个定点．xy21．（本题满分18分）本题共有3个小题，第1小题满分4分，第2小题满分6分，第3小题满分8分．若函数()f x 满足:对于任意正数,s t ，都有()0,()0f s f t >>，且()()()f s f t f s t +<+，则称函数()f x 为“L 函数”．（1）试判断函数21()f x x =与2()f x =)31(31)x x x a -=-+-为“L 函数”，求实数a 的取值范围；()x 为“L 函数”，且(1)1f =，求证：对任意1(2,2)(*)N k k x k -∈∈，都有高三数学参考答案与评分标准一、填空题：（1~6题每题4分；7~12题每题5分）1. [0 2]，； 2. 2； 3.1-； 4.34；5.2[ 1)3，；6. 4-；7. [3 7]，； 8. 78； 9.37； 10. 32； 11. 4(0,]3； 122，1）．二、选择题：（每题5分）13.A 14.D 15. C 16. B 三、解答题：（共76分）17．解：（1）以A 为坐标原点、AB 为x 轴、AC 为y 轴、1AA 为z 轴建立如图的空间直角坐标系．由题意可知(0,0,0),(0,1,2),(2,0,1),(1,1,0)A D E F --， 故(2,0,1),(1,0,2)AE DF =-=--，…………………4分由2(1)1(2)0AE DF ⋅=-⨯-+⨯-=，可知AE DF ⊥，即AE D F ⊥． …………………6分（2）设(,,1)n x y =是平面DEF 的一个法向量， 又(1,0,2) (1,1,1)DF EF =--=- ，， 故由20,10,n DF x n EF x y ⎧⋅=--=⎪⎨⋅=+-=⎪⎩解得2,3,x y =-⎧⎨=⎩ 故(2,3,1)n =- ． …………9分 设AE 与平面DEF 所成角为θ，则||sin ||||n AE n AE θ⋅===⋅，…………12分 xyzO所以AE 与平面DEF 所成角为点A 到平面DEF 的距离为sin AE θ⋅ …………………14分18．解：（1）由cos ,cos ,cos b C a A c B 成等差数列，可得cos cos 2cos b C c B a A =+， …………………2分 故sin cos sin cos 2sin cos B C C B A A =+，所以sin()2sin cos B C A A =+， ………4分 又A B C π++=，所以sin()sin B C A +=，故sin 2sin cos A A A =， 又由(0,π)A ∈，可知sin 0A ≠，故1cos 2A =，所以π3A =． …………………6分 （另法：利用cos cos b C cB a =+求解）（2）在△ABC 中，由余弦定理得2222cos3b c bc π+-=， …………………8分即2218b c bc +-=，故2()318b c bc +-=，又6b c +=，故6bc =，………………10分所以2222()2AB AC AB AC AB AC AB AC ==++⋅++22||||2||||cos AB AC AB AC A =++⋅…………………12分22c b bc =++2()30b c bc =+-=，故AB AC =+…………………14分19．解：（1）由11()22f =，可得111log 222a -=，解之得2a =. …………………2分由32种情形等可能，故1(1,2,,32)32kP k == ， ……………………4分 所以21132(log )53232H =⨯-=，答：“谁被选中”的信息熵为5． ……………………6分（2）n A 获得冠军的概率为111111111+)1(1)24222n n n ----++=--=(，……………8分 当1,2,k =,1n - 时，2()2log 22k k k k k f p --=-=，又11()2n n n f p --=，故111231124822n n n n H ----=+++++ ， ……………………11分1112211 +248222n n n n n n H ----=++++ ， 以上两式相减，可得11111111+1224822n n H --=+++=- ，故422n H =-，答：“谁获得冠军”的信息熵为422n -． ……………………14分20．解：（1）由 AP OP ⊥，可知1AP OP k k ⋅=-，又A 点坐标为(,0),a -故1122111+22a ⋅=---，可得1a =， ……………………………2分因为椭圆M 过P 点，故211+144b =，可得213b =， 所以椭圆M 的方程为22113yx +=． ……………………………4分 （2）AP 的方程为01110122y x -+=--+，即10x y -+=， 由于Q 是椭圆M上的点，故可设(cos )Q θθ， ……………………………6分所以12APQ S ∆=……………………………8分)16πθ++ 当2()6k k θπ+=∈Z ，即2()6k k πθπ=-∈Z 时，APQ S ∆取最大值．故APQ S ∆14． ……………………………10分 法二：由图形可知，若APQ S ∆取得最大值，则椭圆在点Q 处的切线l 必平行于AP ，且在直线AP 的下方． …………………………6分设l 方程为(0)y x t t =+<，代入椭圆M 方程可得2246310x tx t ++-=， 由0∆=，可得t =0t <，故t = …………………………8分 所以APQ S ∆的最大值1124==． ……………………………10分 （3）直线AD 方程为1(1)y k x =+，代入2231x y +=，可得2222111(31)6310k x k x k +++-=，21213131A D k x x k -⋅=+，又1A x =-，故21211313D k x k -=+，21112211132(1)1313D k k y k k k -=+=++， ………………12分 同理可得22221313E k x k -=+，222213E k y k =+，又121k k =且12k k ≠，可得211k k =且11k ≠±， 所以212133E k x k -=+，12123E k y k =+，112211122211122112231323133(1)313E D DE E D k k y y k k k k x x k k k k k --++===---+-++，直线DE 的方程为21112221112213()133(1)13k k k y x k k k --=-+++， ………………14分 令0y =，可得22112211133(1)21313k k x k k -+=-=-++． 故直线DE 过定点(2,0)-． ………………16分（法二）若DE 垂直于y 轴，则,E D E D x x y y =-=，此时221222111133D E D D D E D D y y y y k k x x x y =⋅===++-与题设矛盾．若DE 不垂直于y 轴，可设DE 的方程为+x ty s =，将其代入2231x y +=，可得222(3)210t y tsy s +++-=，可得22221,33D E D E ts s y y y y t t --+=⋅=++，………12分 又12111(1)(1)D E D ED E D E y y y y k k x x ty s ty s =⋅==++++++， 可得22(1)(1)()(1)0D E D E t y y t s y y s -+++++=， ………………14分故2222212(1)(1)(1)033s tst t s s t t ---++++=++， 可得2s =-或1-，又DE 不过A 点，即1s ≠-，故2s =-．所以DE 的方程为2x ty =-，故直线DE 过定点(2,0)-． ………………16分 21．解：（1）对于函数21()f x x =，当0,0t s >>时，2211()0,()0f t t f s s =>=>， 又222111()()()()20f t f s f t s t s t s ts +-+=+-+=-<，所以111()()()f s f t f s t +<+， 故21()f x x =是“L 函数”. ………………2分对于函数2()f x =1t s ==时，222()()2()f t f s f t s +=+，故2()f x =不是“L 函数”. ………………4分 （2）当0,0t s >>时，由()31(31)x x g x a -=-+-是“L 函数”，可知()31(31)0t t g t a -=-+->，即(31)(3)0t t a -->对一切正数t 恒成立， 又310t ->，可得3t a <对一切正数t 恒成立，所以1a ≤． ………………6分由()()()g t g s g t s +<+，可得+3331(3331)0s t s t s t s t a ------++--+>， 故(31)(31)(3)0+s t s t a +-->，又(31)(31)0t s -->，故30+s t a +>，由30+s t a +>对一切正数,s t 恒成立，可得10a +≥，即1a ≥-． ………………9分综上可知，a 的取值范围是[11] -，． ………………………10分 （3）由函数()f x 为“L 函数”， 可知对于任意正数,s t ，都有()0,()0f s f t >>，且()()()f s f t f s t +<+，令s t =，可知(2)2()f s f s >，即(2)2()f s f s >， ………………………12分 故对于正整数k 与正数s ，都有112(2)(2)(2)(2)2k k k k k k f s f s f s f s ---=⋅⋅⋅> ， ………………………………14分 16分 18分。