静安区高三一模(2014-12)

2014年高三一模客观压轴题汇编（学生版）填空题1.（2014年普陀一模文理12）已知全集}8,7,6,5,4,3,2,1{=U ，在U 中任取四个元素组成的集合记为 },,,{4321a a a a A =，余下的四个元素组成的集合记为},,,{4321b b b b A C U =， 若43214321b b b b a a a a +++<+++，则集合A 的取法共有 种2.（2014年普陀一模文理14） 已知函数⎩⎨⎧<+≥-=0),1(0,2)(x x f x a x f x ，若方程0)(=+x x f 有且仅有两个解，则实数a 的取值范围是3.（2014年长宁一模文14）设a 为非零实数，偶函数1||)(2+-+=m x a x x f (x R )在区间(2,3)上存在唯一零点，则实数a 的取值范围是变式：函数2()4f x x x a =-+在(0,3)上有一个零点，则实数a 的取值范围是4.（2014年长宁一模理14）定义：{}123min ,,,,n a a a a 表示123,,,,n a a a a 中的最小值．若定义()f x ={}2min ,5,21x x x x ---，对于任意的n *∈N ，均有(1)(2)(21)(2)()f f f n f n kf n +++-+≤成立，则常数k 的取值范围是5.（2014年嘉定一模理13文14）已知函数⎪⎩⎪⎨⎧<++-≥++=0,,0,12)(22x c bx x x x ax x f 是偶函数，直线t y =与函数 )(x f 的图像自左至右依次交于四个不同点A 、B 、C 、D ，若||||BC AB =，则实数t 的值为6.（2014年嘉定一模理14）某种平面分形图如下图所示，一级分形图是一个边长为1的等边三角形（图（1））；二级分形图是将一级分形图的每条线段三等分，并以中间的那一条线段为一底边向形外作等边三角形， 然后去掉底边（图（2））；将二级分形图的每条线段三等边，重复上述的作图方法，得到三级分形图 （图（3））；…；重复上述作图方法，依次得到四级、五级、…、n 级分形图．则n 级分形图的 周长为7.（2014年静安一模理12）图（1） 图（2） 图（3）……8.（2014年静安一模理14）9.（2014年闸北一模理9）设0,1a a >≠，已知函数()2sin 22,(0)x f x a x x π=+->至少有5个零点，则a 的取值范围为变式练习（2014年闸北区一模文科9）设1,0≠>a a ，函数2|2sin |2)(-+=x a x f xπ（0>x ）有四个零点，则a 的值为10. （2014年闸北一模理10）设曲线22:23()C x y x y ++=+，则曲线C 所围封闭图形的面积为变式练习（2014年闸北区一模文科10）由曲线||||22y x y x +=+所围成的封闭图形的面积为11.（2014年宝山一模理14） 关于函数()1x f x x =-，给出下列四个命题：①当0x >时，()y f x =单调递减且没有最值；②方程()(0)f x kx b k =+≠一定有解；③如果方程()f x k =有解，则解的个数一定是偶数； ④()y f x =是偶函数且有最小值； 则其中真命题是12. （2014年闵行一模理12）设,i j 依次表示平面直角坐标系x 轴、y 轴上的单位向量， 且25a i a j -+-=，则2a i +的取值范围是13.（2014年闵行一模理13）22log (04)()2708(4)33x x f x x x x ⎧<≤⎪=⎨-+>⎪⎩，若,,,a b c d 互不相同， 且()()()()f a f b f c f d ===，则abcd 的取值范围是 变式练习（2014年闵行区一模文科13）已知函数()11f x x =--，若关于x 的方程()f x t =()t R ∈恰有四个 互不相等的实数根1234,,,x x x x （1234x x x x <<<），则1234x x x x ++⋅的取值范围是14.（2014年闵行一模理14）211,1k A x x kt t kt k ⎧⎫==+≤≤⎨⎬⎩⎭，其中2,3,......,2014k =， 则所有k A 的交集为变式练习（2014年闵行区一模文14）已知42421()1x kx f x x x ++=++（k 是实常数），则()f x 的最大值与最小值的乘积 为15.（2014年徐汇一模理12）如图所示，已知点G 是△ABC 的重心，过G 作直线与AB 、AC 两边分别交于M 、N 两点，且,AM x AB AN y AC ==，则xyx y+的值为16.（2014年徐汇一模理13） 一个五位数abcde 满足,,,a b b c d d e <>><且,a d b e >>(如37201,45412)，则称这个五位数符合“正弦规律”.那么，共有 个五位数符合“正弦规律”17.（2014年徐汇区一模理科14）定义区间(),c d 、[),c d 、(],c d 、[],c d 的长度均为()d c d c ->. 已知实数(),a b a b >.则满足111x a x b+≥--的x 构成的区间的长度之和为18.（2014年松江一模理11）对于任意实数x ，x 表示不小于x 的最小整数，如1.22,0.20=-=．定义在R 上的函数()2f x x x =+，若集合{}(),10A y y f x x ==-≤≤，则集合A 中所有元素的和为19.（2014年松江一模理13）已知函数()log 1(0,1)a f x x a a =->≠，若1234x x x x <<<， 且12()()f x f x =34()()f x f x ==，则12341111x x x x +++=20.（2014年松江一模理14） 设集合{1,2,3,,}A n =，若B ≠∅且B A ⊆，记()G B 为B 中元素的最大值与最小值之和，则对所有的B ，()G B 的平均值＝21.（2014年青浦一模理13）已知直角坐标平面上任意两点11(,)P x y 、22(,)Q x y ，定义212121212121(,)x x x x y y d P Q y y x x y y ⎧--≥-⎪=⎨--<-⎪⎩为,P Q 两点的“非常距离”，当平面上动点(,)M x y到定点(,)A a b 的距离满足3MA =时，则(,)d M A 的取值范围是22.（2014年青浦一模理14）若不等式1(1)(1)31n na n +--<++对任意自然数n 恒成立，则实数a 的取值范围是23（2014年金山一模理13）如图，已知直线:4360l x y -+=，抛物线2:4C y x =图像上的一个 动点P 到直线l 与y 轴的距离之和的最小值是24.（2014年金山一模理14）在三棱锥P ABC -中，PA 、PB 、PC 两两垂直，且3,2,1PA PB PC ===.设M 是底面ABC 内一点，定义()(,,)f M m n p =，其中m 、n 、p 分别是三棱锥M PAB -、三棱锥M PBC -、三棱锥M PCA -的体积。

2024届上海市静安区高三一模英语试卷(满分140分,完卷时间120分钟)2023.12 考生注意：1. 完卷时间120分钟,试卷满分140分。

2. 本调研设试卷和答题纸两部分，全卷共12页。

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。

3. 答题前，务必在答题纸上填写准考证号和姓名。

第Ⅰ卷(共100分)I. Listening ComprehensionSection ADirections:In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A. In a gallery. B. At the barber's. C. In a restaurant. D. At the tailor's.2. A. Fellow workers. B. Family members. C. Doctor and patient. D. Driver and passenger.3. A. Choosing psychology. B. Choosing economics.C. Neither is a good choice.D. Choosing a major of interest.4. A. She did not feel sorry for being late for the appointment.B. She did not inform the man of her del ay in advance.C. She wasn't really caught in the traffic jam.D. She wasn't always late for the appointment.5. A. It was lost and won't be found. B. It was transferred to a different city.C. It was delivered to her hotel already.D. It was stolen during her trip.6. A. He has realized he still leaves much to be desired.B. He is angry with not getting the lead role in the play.C. He is confident about getting the lead role next time.D. He feels reluctant to take the new responsibilities.7. A. They told a lot of stories during the meeting.B. There is no need for them to argue so fiercely in public.C. Both perspectives should be considered before judging.D. They should have resolved their issues in private.8. A. She has already been to the café.B. She is not interested in going to the café.C. She knows about the café but hasn't visited it.D. She wants to go to the café right away.9. A. She expects the man to help Brian move to a new house.B. She expects the man to take mum to Brain's new house.C. She expects the man to celebrate mum's birthday together.D. She expects the man to make a phone call to Mum.10. A. The fantastic and high-quality camera. B. The need for better internet connectivity.C. Their favorite photography techniques.D. The pros and cons of a new smartphone.Section BDirections:In Section B, you will hear two short passages and one longer conversation. After each passage or conversation, you will be asked several questions. The passages and the conversation 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 1 1 through 13 are based on the following passage.11. A. Its regular driving practices and poor vehicles. B. Its lack of green spaces and air cleaners.C. Its excessive water pollution and rubbish.D. Its high air pollution and crowded roads.12. A. Establishing a high interest loan scheme. B. Removing outdated black and white taxis.C. Encouraging customers to create new ideas.D. Making advertisements on old vehicles.13. A. Because customers are more friendly and richer.B. Because all new cabs provide air-conditioning.C. Because all new cabs are equipped with meters.D. Because car manufacturers can earn extra money.Questions 14 through 16 are based on the following passage.14. A. Canceling all the gifts. B. Applying a ‘one-gift’ rule.C. Giving children less time to play.D. Encouraging buying second-hand gifts.15. A. Buying a rare and expensive souvenir. B. Buying a hand-made craft product.C. Giving something that won't cost money.D. Giving an experience of something new.16. A. The waste caused by Christmas gifts. B. The importance of buying gifts for children.C. The creative ideas of giving gifts to avoid waste.D. The negative effects of receiving too many gifts.Questions 17 through 20 are based on the following conversation.17. A. By trading physical items. B. By exchanging artistic creativity.C. By hosting art exhibitions.D. By making artistic advertisements.18. A. Painting and writing. B. Graphic design and photography.C. Music and album cover design.D. Video editing and project management.19. A. Members can benefit without efforts.B. Members can make money by providing artistic services.C. Members can get copyrights of other artistic offerings.D. Members can have access to the creative exchange list.20. A. Competitive individualism. B. Artistic cooperation and inspiration.C. Individual fame in the art field.D. Material collaboration and exchange.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.Japan's robot revolution in senior careJapan's artificial intelligence expertise is transforming the elder care industry, with (21) _____(specialize) robotic care accomplishing more than just taking pressure off the critical shortage of caregivers. Senior care facilities across Japan are testing out such new robots (22) _____deliver a collection of social and physical health care and the government-backed initiative has been met with positive reviews by elderly residents.The rapidly graying population (23) _____(eye) by the government as a potential market for medical technology now. Disappointing government predictions show that by 2025, Japan's first baby boomers will have turned 75 and about 7 million people are likely to suffer from some form of dementia (痴呆). The nation won't be able to avoid a dementia crisis (24) _____an additional 3 80.000 senior care workers.The long-standing shortage of professional care workers has encouraged the Japanese government (25) _____(simplify) procedures for foreign caregivers to be trained and certified. The current Technical Intern Training Program between Vietnam, the Philippines, and Indonesia, under (26) _____Economic Partnership Agreement, was extended to include nursing care as well as agriculture, fishery, and construction sectors.(27) _____the government made efforts to increase the numbers of senior care workers, the target number of foreign graduates has still fallen flat, with the national caregiver examination proving a major obstacle to pass. The success rate for foreign students was a merely 106 students last year, (28) _____has slightly improved to 216 students this year. Another depressing reality is that 19 to 38 percent of foreign nurses who pass the exam opt to leave the industry and return home, (29) _____(cite) tough work conditions and long hours. Given the challenges, this is (30) _____ the government believes care robots will be able to step in.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.Sea-level rise predictionsA team of University of Idaho scientists is studying a fast-moving glacier in Alaska in hopes of developing better predictions on how quickly global sea levels will rise.Tim Bartholomaus, a professor in the Department of Geography and Geological Sciences, spent several weeks on Turner Glacier in Alaska's southeastern (31) _____near Disenchantment Bay. The glacier is unique because, unlike other glaciers, it rises greatly every five to eight years.A surging glacier is defined, (32) _____, as one that starts flowing at least 10 times faster than normal. But the how and why of that glacial movement is poorly understood, although recent research suggests that global climate change increases the (33) _____of glacial surging.During Turner's surges, the mass of ice and rock will increase its speed from roughly 3 feet a day to 65 feet per day.All of that is important because glaciers falling into the ocean are a major contributor to sea level rise, and current climate change models don't (34) _____account for these movements. For example, Greenland's glaciers are one of the leading contributors to global sea-level rise. Since the early 2000s, Greenland (35) _____from not having any effect on world sea levels, to increasing sea level by about 1 millimeter per year. Half of that yearly increase is due to warmer average temperatures, which leads to more ice melting. The other half, however, is because glaciers in Greenland are, as a whole, moving faster and running into the ocean more frequently.Glacial movement has something to do with water running underneath the glacier. Glaciers are full of holes, and water runs through those holes. When the water pressure is high underneath a glacier, it starts to move, partly because it's lifting the mass of ice and rock off the ground and partly because it's (36) _____the underside of the glacier.But how exactly does that water move through the glacier, and how does the movement (37) _____the glacier’s speed? Those are the questions the scientists hope to answer.Bartholomaus, some graduate students and researchers from Boise State University, (38) _____onto the ice in August. They set up a base camp at the toe of the glacier and spent their days flying in on helicopters. They placed roughly 30 instruments, burying them deeply into the glacier and (39) _____them on rock outcroppings (露岩) alongside the glacier. This summer the team will return to get the instruments and replace batteries. Those instruments will (40) _____on and around the glacier until the glacier surge stops, providing researchers with before and after data.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.Investors probably expect that following the suggestions of stock analysts would make them better off than doing the exact opposite. (41) _____, recent research by Nicola Gennaioli and his colleagues shows that the best way to gain excess return s would be to invest in the shares least favored by analysts. They compute that, during the last 35 years, investing in the 10 percent of U. S. stocks analysts were most (42) _____about would have yielded on average 3 percent a year. (43) _____, investing in the 10 percent of stocks analysts were most pessimistic aboutwould have yielded a surprising 15 percent a year.Gennaioli and colleagues shed light on this (44) _____with the help of cognitive sciences and, in particular, using Kahneman and Tversky's concept of representativeness. Decision makers, according to this view, (45) _____the representative features of a group or a phenomenon. These are defined as the features that occur more frequently in that group than in a baseline reference group.After observing strong earnings growth—the explanation goes—analysts think that the firm may be the next Google. “Googles” are in fact more frequent among firms experiencing strong growth, which makes them (46) _____. The problem is that “Googles” are very (47) _____in absolute terms. As a result, expectations become too optimistic, and future performance (48) _____. A model of stock prices in which investor beliefs follow this logic can account both qualitatively and quantitatively for the beliefs of analysts and the dynamics(动态变化) of stock returns.In related work, the authors also show that the same model can (49) _____booms and busts in the volume of credit and interest rate spreads.These works are part of a research project aimed at taking insights from cognitive sciences and at (50) _____ them into economic models. Kahneman and Tversky's concept of “representativeness” lies at the heart of this effort. “In a classical example, we (51) _____to think of Irishmen as redheads because red hair is much more frequent among Irishmen than among the rest of the world,”Prof. Gennaioli says. “However, only 10 percent of Irishmen are redheads. In our work, we develop models of belief formation that show this logic and study the (52) _____ of this important psychological force in different fields.”Representativeness helps describe (53) _____and behavior in different fields, not only in financial markets. One such field is the formation of stereotypes about social groups. In a recent experimental paper, Gennaioli and colleagues show that representativeness can explain self-confidence, and in particular the (54) _____of women to compete in traditionally male subjects, such as mathematics. A slight prevalence of (55) _____male math ability in the data is enough to make math ability un-representative for women, driving their under confidence in this particular subject.41. A. Consequently B. Furthermore C. Nevertheless D. Meanwhile42. A. curious B. controversial C. concerned D. optimistic43. A. In brief B. By contrast C. In addition D. Without doubt44. A. engagement B. concentration C. puzzle D. definition45. A. memorize B. prioritize C. modernize D. fertilize46. A. representative B. argumentative C. executive D. sensitive47. A. harsh B. adaptable C. crucial D. rare48. A. cheers B. disappoints C. stabilizes D. improves49. A. account for B. count on C. suffer from D. hold up50. A. pouring B. admitting C. integrating D. tempting51. A. pretend B. afford C. offer D. tend52. A. effects B. delights C. intervals D. codes53. A. companions B. scales C. expectations D. findings54. A. necessity B. involvement C. perseverance D. reluctance55. A. equivalent B. exceptional C. mysterious D. distressingSection 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)Montessori was born in Italy in 1870 with progressive parents, who frequently communicated with the country's leading thinkers and scholars. This enlightened family environment provided Montessori with many advantages over other young girls of the time.Her mother's support was vital for some important decisions, such as her enrolment in a technical school after her elementary education. Her parents' support also proved to be essential for her decision to study medicine, a field that was dominated by men.Soon after graduating, in 1896, Montessori began work as a voluntary assistant in a clinic at the University of Rome, where she cared for children with learning difficulties. The rooms were bare, with just a few pieces of furniture. One day, she found that the children were enthusiastically playing with breadcrumbs(面包屑) that had dropped on the floor. It then occurred to her that the origin of some intellectual disabilities could be related with poverty. With the right learning materials, these and other young minds could be nurtured, Montessori concluded.The observation would lead Montessori to develop a new method of education that focused on providing optimal stimulation during the sensitive periods of childhood.At its centre was the principle that all the learning materials should be child-sized and designed to appeal to all the senses. In addition, each child should also be allowed to move and act freely, and use their creativity and problem- solving skills. Teachers took the role of guides, supporting the children without press or control.Montessori opened her first Children's House in 1907. When the Fascists(法西斯主义者)first came into power in Italy in 1922, they initially embraced her movement. But they soon came to oppose the emphasis on the children's freedom of expression. Montessori's values had always been about human respect, and the rights of children and women, but the Fascists wanted to use her work and her fame.Things reached a breaking point when the Fascist tried to influence the schools' educational content, and in 1934 Montessori and her son decided to leave Italy. She didn't return to her homeland until 1947, and she continued to write about and develop her method until her death in1952, at the age of 81.56. The primary reason for Montessori to develop a new educational method was ______.A. her family's supportive influence on her educationB. her experience as a voluntary assistant in a clinicC. her observation of children playing with breadcrumbs happilyD. her decision to study medicine, a field dominated by men57. What was a central principle of Montessori’s educational method as described in the passage?A. Providing standardized, one-size-fits-all learning materials.B. Encouraging strict discipline and control over children's actions.C. Focusing on rote memorization and competition.D. Creating a free and children-centered learning environment.58. Montessori decided to leave Italy in 1934 because ______.A. she wanted to explore other countries and culturesB. she wanted to avoid the Fascist's influence on her workC. she was offered a better job in a different countryD. she wanted to retire and enjoy a peaceful life in another country59. Which of the following words can best describe Montessori in this passage?A. Observant and innovative.B. Traditional and emotional.C. Progressive and dependent.D. Open-minded and indifferent.(B)Reducing the workweek to four days could have a climate benefit. In addition to improving the well-being of workers, cutting working hours may reduce carbon emissions. But those benefits would depend on a number of factors, experts emphasize, including how people choose to spend nonworking time.Commuting and travelTransportation is the biggest contributor to greenhouse emissions. A November 2021 survey of 2,000 employees and 500 business leaders in the United Kingdom found that if all organizations introduced a four-day week, the reduced trips to work would decrease travel overall by more than 691 million miles a week.But the climate benefits of less commuting could be eliminated, experts said, if people choose to spend their extra time off traveling, particularly if they do so by car or plane.Energy usageShorter working hours could lead to reductions in energy usage, experts said. According to a 2006 paper, if the United States adopted European work standards, the country would consume about 20 percent less energy.Energy could also be conserved if fewer resources are needed to heat and cool large office buildings, reducing demands on electricity. For example, if an entire workplace shuts down on the fifth day, that would help lower consumption—less so if the office stays open to accommodate employees taking different days off.Lifestyle changesIt's possible that fewer working hours may lead some people to have a larger carbon footprint, but experts say research suggests that most people are likely to shift toward more sustainable lifestyles.One theory is that people who work more and have less free time tend to do things in more carbon-intensive ways, such as choosing faster modes of transportation or buying prepared foods. Convenience is often carbon- intensive and people tend to choose convenience when they're time-stressed. Meanwhile, some research suggests that those who work less are more likely to engage in traditionally low-carbon activities, such as spending time with family or sleeping.“When we talk about the four-day workweek and the environment, we focus on the tangible, but actually, in away, the biggest potential benefit here is in the intangible,” experts said.60. What is identified as the leading cause of greenhouse emissions according to the passage?A. The well-being of employees.B. The conservation of energy.C. Commuting and travel.D. The European work standard.61. What can be inferred from the underline d sentence “the biggest potential benefit here is in the intangible” in the last paragraph?A. People will have big potential in achieving intangible benefits while working.B. People are more likely to engage in carbon-intensive activities due to time constraints.C. People may shift toward more sustainable lifestyles and lower carbon footprints.D. People may travel more frequently by car or plane during their extra time off.62. The passage is mainly written to .A. highlight the importance of shortening working time in the context of well-beingB. provide an overview of transportation emissions worldwideC. analyze the impact of reduced working hours on mode of businessD. illustrate factors affecting the climate benefits of a shorter workweek(C)The cultivation of plants by ants is more widespread than previously realized, and has evolved on at least 15 separate occasions.There are more than 200 species of an t in the Americas that farm fungi(真菌) for food, but this trait evolved just once sometime between 45 million and 65 million years ago. Biologists regard the cultivation of fungi by ants as true agriculture appearing earlier than human agriculture because it meets four criteria: the ants plant the fungus, care for it, harvest it and depend on it for food.By contrast, while thousands of ant species are known to have a wide variety of interdependent relationships with plants, none were regarded as true agriculture. But in 2016, Guillaume Chomicki and Susanne Renner at the University of Munich, Germany, discovered that an ant in Fungi cultivates several plants in a way that meets the four criteria for true agriculture.The ants collect the seeds of the plants and place them in cracks in the bar k of trees. As the plants grow, they form hollow structures called domain that the ants nest in. The ants defecate(排便) at designated absorptive places in these domain, providing nutrients for the plant. In return, as well as shelter, the plant provides food in the form of fruit juice.This discovery prompted Chomicki and others to review the literature on ant-plant relationships to see if there are other examples of plant cultivation that have been overlooked. “They have never really been looked at in the framework of agriculture,” says Chomicki, who is now at the University of Sheffield in the UK. “It's definitely widespread.”The team identified 37 examples of tree-living ants that cultivate plants that grow on trees, known asepiphytes(附生植物). By looking at the family trees of the ant species, the team was able to determine on how many occasions plant cultivation evolved and roughly when. Fifteen is a conservative estimate, says Campbell. All the systems evolved relatively recently, around 1million to 3 million years ago, she says.Whether the 37 examples of plant cultivation identified by the team count as true agriculture depends on the definitions used. Not all of the species get food from the plants, but they do rely on them for shelter, which is crucial for ants living in trees, says Campbell. So the team thinks the definition of true agriculture should include shelter as well as food.63. According to biologists, why is ant-fungus cultivation considered as a form of true agriculture?A. Because it occurred earlier than human agriculture.B. Because it fulfills the standards typical of agricultural practices.C. Because it redefines the four criteria for true human agriculture.D. Because it is less common than previously thought.64. What motivated Chomicki and others to review the literature on ant-plant relationships?A. They determined on new family trees of the ant species.B. They overlooked some tree-living ants that provided nutrients for the plants.C. They never studied the ant-plant relationships within the context of agriculture.D. They never identified any an t species that engaged in cultivation of fungi.65. Which of the following statements is supported by the team's findings according to the passage?A. Ants’ cultivation of plants is limited to a few specific species.B. The cultivation of fungi by ants is considered the earliest form of agriculture.C. True agriculture in ants involves only food-related interactions with plants.D. Ants have independently cultivated plants on at least 15 distinct occasions.66. What is the passage mainly about?A. The evolution of ants in the plant kingdom.B. The widespread occurrence of ant-plant cultivation.C. The discovery of a new ant species engaging in agriculture.D. The contrast between ant agriculture and human agriculture.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.did have one, would you want to meet them?Consider how often your facial features are used to identify you. Your passport, ID card and driving license all feature your face. (67) ______You may need your face to unlock your smartphone and possibly even need it to exclude you from being present at a crime scene.The word ‘doppelgänger’ refers to a person who looks the same as you, essentially sharing your features; those that you thought were unique to you and your identity. Not identical twins, as a doppelgänger has no relation to you. The idea originated in German folklore. (68) ______So, let's get real. What are the chances of you having one in the first place? There's said to be a one in 135 chance of an exact match for you existing anywhere in the world, so the chances are pretty low, despite folk wisdom promising you otherwise. And the chances of meeting? The mathematical certainty of finding this particular person is supposedly less than one in a trillion.That said, these statistics may be a good thing. Historically, having a double wasn't always a positive. Back in 1999, an innocent American man, indistinguishable from the real criminal, was sent to prison for robbery, where he stayed for 19 years. (69) ______. In a different case, a woman in New York was accused of trying to poison her doppelgänger with deadly cheesecake so that she could steal her identity!(70) ______The fascination with doppelgängers may be rooted in historical beliefs that facial resemblance meant they were from the same family or had a common ancestor. It leads to the hope that one day you will meet your lookalike, creating the thrill of a potentially strange meeting. However, as these encounters can be both interesting and disturbing, we understand that after such an experience, you might not want to meet your doppelgänger again.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.Competitive CheerleadingOver the years, cheerleading has taken two primary forms: game-time cheerleading and competitive cheerleading. Game-time cheerleaders' main goal is to entertain the crowd and lead them with team cheers, which should not be considered a sport. However, competitive cheerleading is more than a form of entertainment. It is really a competitive sport.Competitive cheerleading includes lots of physical activity. The majority of the teams require a certain level of tumbling (翻腾运动) ability. It's a very common thing for gymnasts, so it's easy for them to go into competitive cheerleading. Usually these cheerleaders integrate lots of their gymnastics experience including their jumps, tumbling, and overall energy. They also perform lifts and throws.Competitive cheerleading is also an activity that is governed by rules under which a winner can be declared. It is awarded points for technique, creativity and sharpness. Usually the more difficult the action is, the better the score is. That's why cheerleaders are trying to experience great difficulty in their performance. Besides, there is also a strict rule of time. The whole performance has to be completed in less than three minutes and fifteen seconds,。

2014年上海市静安区中考数学一模试卷一、选择题：（本题共6题，每题4分，满分24分）1．（4分）（2014•青浦区一模）在Rt△ABC中，∠C=90°，如果∠A=α，BC=a，那么AC等于（）A ．a•tanαB．a•cotαC．D．2．（4分）（2014•高邮市模拟）如果抛物线y=mx2+（m﹣3）x﹣m+2经过原点，那么m的值等于（）A ．0 B．1 C．2 D．33．（4分）（2014•青浦区一模）如图，已知平行四边形ABCD中，向量在，方向上的分量分别是（）A ．B．C．、D．、4．（4分）（2014•新泰市一模）抛物线y=﹣（x﹣2）2+1经过平移后与抛物线y=﹣（x+1）2﹣2重合，那么平移的方法可以是（）A ．向左平移3个单位再向下平移3个单位B ．向左平移3个单位再向上平移3个单位C ．向右平移3个单位再向下平移3个单位D ．向右平移3个单位再向上平移3个单位5．（4分）（2014•青浦区一模）在△ABC，点D、E分别在边AB、AC上，如果AD=1，BD=2，那么由下列条件能够判定DE∥BC的是（）A ．B．C．D．6．（4分）（2014•青浦区一模）如图，已知AB、CD分别表示两幢相距30米的大楼，小明在大楼底部点B处观察，当仰角增大到30度时，恰好能通过大楼CD的玻璃幕墙看到大楼AB的顶部点A的像，那么大楼AB的高度为（）A ．B．20米C．30D．60米二、填空题：（本大题共12题，每题4分，满分48分）7．（4分）（2014•青浦区一模）函数y=（x+5）（2﹣x）图象的开口方向是_________．8．（4分）（2014•青浦区一模）在Rt△ABC中，∠C=90°，如果∠A=45°，AB=12，那么BC=_________．9．（4分）（2014•青浦区一模）已知线段a=3cm，b=4cm，那么线段a、b的比例中项等于_________cm．10．（4分）（1999•南京）如果两个相似三角形周长的比是2：3，那么它们面积的比是_________．11．（4分）（2014•青浦区一模）如图，在△ABC于△ADE中，，要使△ABC于△ADE相似，还需要添加一个条件，这个条件是_________．12．（4分）（2014•青浦区一模）已知点G是△ABC的重心，AB=AC=5，BC=8，那么AG=_________．13．（4分）（2014•青浦区一模）已知向量与单位向量方向相反，且，那么=_________（用向量的式子表示）14．（4分）（2014•青浦区一模）如果在平面直角坐标系xOy中，点P的坐标为（3，4），射线OP与x的正半轴所夹的角为α，那么α的余弦值等于_________．15．（4分）（2014•青浦区一模）已知一条斜坡的长度为10米，高为6米，那么坡角的度数约为_________（备用数据：tan31°=cot59°≈0.6，sin37°=cos53°≈0.6）16．（4分）（2014•青浦区一模）如果二次函数y=x2+2kx+k﹣4图象的对称轴为x=3，那么k=_________．17．（4分）（2014•青浦区一模）如图，小李推铅球，如果铅球运行时离地面的高度y（米）关于水平距离x（米）的函数解析式，那么铅球运动过程中最高点离地面的距离为_________米．18．（4分）（2014•青浦区一模）如果将一个三角形绕着它一个角的顶点旋转后使这个角的一边与另一边重叠，再将旋转后的三角形相似缩放，使重叠的两边互相重合，我们称这样的图形为三角形转似，这个角的顶点称为转似中心，所得的三角形称为原三角形的转似三角形．如图，在△ABC中，AB=6，BC=7，AC=5，△A1B1C是△ABC以点C 为转似中心的其中一个转似三角形，那么以点C为转似中心的另一个转似三角形△A2B2C（点A2，B2分别与A、B 对应）的边A2B2的长为_________．三、解答题（本大题共7题，满分78分）19．（10分）（2014•青浦区一模）如图，已知在直角坐标系中，点A在第二象限内，点B和点C在x轴上，原点O 为边BC的中点，BC=4，AO=AB，tan∠AOB=3，求图象经过A、B、C三点的二次函数解析式．20．（10分）（2014•青浦区一模）如图，已知在△ABC中，点D、E分别在边AB、AC上，DE∥BC，，如果，．（1）求（用向量的式子表示）（2）求作向量（不要求写作法，但要指出所作图表中表示结论的向量）21．（10分）（2014•青浦区一模）已知：如图，在平行四边形ABCD中，E、F分别是边BC，CD上的点，且EF∥BD，（1）的值；（2）线段GH的长．22．（10分）（2014•青浦区一模）如图，已知某船向正东方向航行，在点A处测得某岛C在其北偏东60°方向上，前进8海里处到达点B处，测得岛C在其北偏东30°方向上．已知岛C周围6海里内有一暗礁，问：如果该船继续向东航行，有无触礁危险？请说明你的理由．23．（12分）（2014•青浦区一模）已知，如图，在梯形ABCD中，AD∥BC，∠BCD=90°，对角线AC、BD相交于点E，且AC⊥BD．（1）求证：CD2=BC•AD；（2）点F是边BC上一点，联结AF，与BD相交于点G，如果∠BAF=∠DBF，求证：．24．（12分）（2014•青浦区一模）已知在平面直角坐标系xOy中，二次函数y=﹣2x2+bx+c的图象经过点A（﹣3，0）和点B（0，6）．（1）求此二次函数的解析式；（2）将这个二次函数的图象向右平移5个单位后的顶点设为C，直线BC与x轴相交于点D，求∠ABD的正弦值；（3）在第（2）小题的条件下，联结OC，试探究直线AB与OC的位置关系，并说明理由．25．（14分）（2014•青浦区一模）如图，已知在Rt△ABC中，∠ACB=90°，AB=10，tanA=，点D是斜边AB上的动点，联结CD，作DE⊥CD，交射线CB于点E，设AD=x．（1）当点D是边AB的中点时，求线段DE的长；（2）当△BED是等腰三角形时，求x的值；（3）如果y=，求y关于x的函数解析式，并写出它的定义域．2014年上海市静安区中考数学一模试卷参考答案与试题解析一、选择题：（本题共6题，每题4分，满分24分）1．（4分）（2014•青浦区一模）在Rt△ABC中，∠C=90°，如果∠A=α，BC=a，那么AC等于（）A ．a•tanαB．a•cotαC．D．考点：锐角三角函数的定义．分析：画出图形，根据锐角三角函数的定义求出即可．解答：解：cot∠A=，∴AC=BC•cotA=a•cotA，故选B．点评：本题考查了锐角三角函数的定义的应用，主要考查学生的理解能力和计算能力．2．（4分）（2014•高邮市模拟）如果抛物线y=mx2+（m﹣3）x﹣m+2经过原点，那么m的值等于（）A ．0 B．1 C．2 D．3考点：二次函数图象上点的坐标特征．分析：把原点坐标代入函数解析式，计算即可求出m的值．3）x﹣m+2经过原点，∴﹣m+2=0，解得m=2．故选C．点评：本题考查了二次函数图象上点的坐标特征，比较简单，理解函数图象上的点的坐标满足函数关系式是解题的关键．3．（4分）（2014•青浦区一模）如图，已知平行四边形ABCD中，向量在，方向上的分量分别是（）A ．B．C．、D．、考点：*平面向量．分析：由四边形ABCD是平行四边形，根据平行四边形法则求解即可求得答案．解答：解：∵四边形ABCD是平行四边形，∴向量在，方向上的分量分别是：﹣，．故选C．知识．此题难度适中，注意掌握平行四边形法则的应用，注意掌握数形结合思想的应用．4．（4分）（2014•新泰市一模）抛物线y=﹣（x﹣2）2+1经过平移后与抛物线y=﹣（x+1）2﹣2重合，那么平移的方法可以是（）A ．向左平移3个单位再向下平移3个单位B ．向左平移3个单位再向上平移3个单位C ．向右平移3个单位再向下平移3个单位D ．向右平移3个单位再向上平移3个单位考点：二次函数图象与几何变换．分析：根据平移前后的抛物线的顶点坐标确定平移方法即可得解．解答：解：∵抛物线y=﹣（x﹣2）2+1的顶点坐标为（2，1），抛物线y=﹣（x+1）2﹣2的顶点坐标为（﹣1，﹣2），∴顶点由（2，1）到（﹣1，﹣2）需要向左平移3个单位再向下平移3个单位．故选A．目，利用顶点的变化确定抛物线解析式更简便．5．（4分）（2014•青浦区一模）在△ABC，点D、E分别在边AB、AC上，如果AD=1，BD=2，那么由下列条件能够判定DE∥BC的是（）A ．B．C．D．考点：平行线分线段成比例．分析：根据相似三角形的判定得出△ADE∽△ABC即可推出∠ADE=∠B，根据平行线的判定推出即可．解答：解：∵AD=1，BD=2，∴=，只有当=时，DE∥BC，理由是：∵==，∠A=∠A，∴△ADE≌△ABC，∴∠ADE=∠B，∴DE∥BC，而其它选项都不能推出DE∥BC，即选项A、B、C都错误，只有选项D正确；故选D．点评：本题考查了相似三角形的性质和判定，平行线的判定的应用，主要考查学生的推理能力．6．（4分）（2014•青浦区一模）如图，已知AB、CD分别表示两幢相距30米的大楼，小明在大楼底部点B处观察，当仰角增大到30度时，恰好能通过大楼CD的玻璃幕墙看到大楼AB的顶部点A的像，那么大楼AB的高度为（）A ．B．20米C．30D．60米考点：解直角三角形的应用-仰角俯角问题．分析：根据仰角为30°，BD=30米，在Rt△BDE中，可求得ED的长度，根据题意恰好能通过大楼CD的玻璃幕墙看到大楼AB的顶部点A的像，可得AB=2ED．解答：解：在°，BD=30米，∴=tan30°，解得：ED=10（米），∵当仰角增大到30度时，恰好能通过大楼CD的玻璃幕墙看到大楼AB的顶部点A的像，∴AB=2DE=20（米）．故选B．点评：本题考查了解直角三角形的应用，解答本题的关键是根据仰角构造直角三角形，利用三角函数的知识解直角三角形．二、填空题：（本大题共12题，每题4分，满分48分）7．（4分）（2014•青浦区一模）函数y=（x+5）（2﹣x）图象的开口方向是向下．考点：二次函数的性质．分析：首先将二次函数化为一般形式，然后根据二次项系数的符号确定开口方向．解答：解：y=（x+5）（2﹣x）=﹣x2+3x+10，∵a=﹣1＜0，∴开口向下，故答案为：向下．性质，解题的关键是正确的化为一般形式．8．（4分）（2014•青浦区一模）在Rt△ABC中，∠C=90°，如果∠A=45°，AB=12，那么BC=6．考点：等腰直角三角形．分析：由题意可知，此三角形是等腰直角三角形，已知斜边的长，求直角边，可以根据勾股定理求得．解答：解：∵在Rt△ABC中，∠C=90°，∠A=45°，∴Rt△ABC是等腰直角三角形，∴BC=AC，设BC=x，根据勾股定理可得x2+x2=122解得，x=6．故答案为：点评：此题考查等腰直角三角形的判定．在等腰直角三角形中，已知任何一边，根据等腰三角形的性质和勾股定理都可以求出另外两边．9．（4分）（2014•青浦区一模）已知线段a=3cm，b=4cm，那么线段a、b的比例中项等于2cm．比例中项的定义列式计算即可得解．解答：解：∵线段a=3cm，b=4cm，∴线段a、b的比例中项==2cm．故答案为：2．点评：本题考查了比例线段，熟记线段比例中项的求解方法是解题的关键，要注意线段的比例中项是正数．10．（4分）（1999•南京）如果两个相似三角形周长的比是2：3，那么它们面积的比是4：9．考点：相似三角形的性质．分析：相似三角形的周长比等于相似比，而面积比等于相似比的平方，由此得解．解答：解：∵两个相似三角形周长的比是2：3，∴它们的相似比是2：3；∴它们的面积比为4：9．点评：本题重点考查的是相似三角形的性质：相似三角形的周长比等于相似比，方．11．（4分）（2014•青浦区一模）如图，在△ABC于△ADE中，，要使△ABC于△ADE相似，还需要添加一个条件，这个条件是∠B=∠E．考点：相似三角形的判定．专题：开放型．分析：根据两边及其夹角法：两组对应边的比相等且夹角对应相等的两个三角形相似可得添加条件：∠B=∠E．解答：解：添加条件：∠B=∠E；∵，∠B=∠E，∴△ABC∽△AED，故答案为：∠B=∠E．点评：此题主要考查了相似三角形的判定，关键是掌握相似三角形的判定定理．12．（4分）（2014•青浦区一模）已知点G是△ABC的重心，AB=AC=5，BC=8，那么AG=2．考点：三角形的重心．分析：根据题意画出图形，连接AG并延长交形的性质可得出AD⊥BC，再根据勾股定理求出AD的长，由三角形重心的性质即可得出AG的长．解答：解：如图所示：连接AG并延长交BC于点D，∵G是△ABC的重心，AB=AC=5，BC=8，∴AD⊥BC，BD=BC=×8=4，∴AD===3，∴AG=AD=×3=2．故答案为：2．点评：本题考查的是三角形的重心，熟知重心到顶点的距离与重心到对边中点的距离之比为2：1是解答此题的关13．（4分）（2014•青浦区一模）已知向量与单位向量方向相反，且，那么=（用向量的式子表示）考点：*平面向量．分析：由向量与单位向量方向相反，且，根据单位向量与相反向量的知识，即可求得答案．解答：解：∵向量与单位向量方向相反，且，∴=﹣3．故答案为：﹣3．点评：此题考查了平面向量的知识．此题难度不大，注意掌握单位向量与相反向量的定义．14．（4分）（2014•青浦区一模）如果在平面直角坐标系xOy中，点P的坐标为（3，4），射线OP与x的正半轴所夹的角为α，那么α的余弦值等于．考点：锐角三角函数的定义；坐标与图形性质；勾股定理．分析：画出图形，根据勾股定理函数的定义求出即可．解答：解：过P作PA⊥x轴于A，∵P（3，4），∴PA=4，OA=3，由勾股定理得：OP=5，∴α的余弦值是=，过答案为：．点评：本题考查了勾股定理和锐角三角函数的定义的应用，主要考查学生的计算能力．15．（4分）（2014•青浦区一模）已知一条斜坡的长度为10米，高为6米，那么坡角的度数约为37°（备用数据：tan31°=cot59°≈0.6，sin37°=cos53°≈0.6）考点：解直角三角形的应用-坡度坡角问题．分析：做出图形，设坡角为α，根据=sinα，可求得α的度数．得，=sinα，即sinα=0.6，则α=37°．故答案为：37°．点评：本题考查了解直角三角形的应用，解答本题的关键是构造直角三角形，解直角三角形．16．（4分）（2014•青浦区一模）如果二次函数y=x2+2kx+k﹣4图象的对称轴为x=3，那么k=﹣3．考点：二次函数的性质．分析：直接利用对称轴公式求解即可．解答：解：∵二次函数y=x2+2kx+k﹣4图象的对称轴为x=3，∴对称轴为：x=﹣=3，解得：k=﹣3，故答案为：﹣3点评：本题主要考查二次函数的性质，解此题的关键是对二次函数的性质的理解和掌握，知对称轴．17．（4分）（2014•青浦区一模）如图，小李推铅球，如果铅球运行时离地面的高度y（米）关于水平距离x（米）考点：二次函数的应用．分析：直接利用公式法求出函数的最值即可得出最高点离地面的距离．解答：解：∵函数解析式为：，∴y最值===2．故答案为：2．点评：此题主要考查了二次函数的应用，正确记忆最值公式是解题关键．18．（4分）（2014•青浦区一模）如果将一个三角形绕着它一个角的顶点旋转后使这个角的一边与另一边重叠，再将旋转后的三角形相似缩放，使重叠的两边互相重合，我们称这样的图形为三角形转似，这个角的顶点称为转似中心，所得的三角形称为原三角形的转似三角形．如图，在△ABC中，AB=6，BC=7，AC=5，△A1B1C是△ABC以点C 为转似中心的其中一个转似三角形，那么以点C为转似中心的另一个转似三角形△A2B2C（点A2，B2分别与A、B对应）的边A2B2的长为．考点：旋转的性质；相似三角形的判定与性质．专题：新定义．分析：先根据条件证明△ABC∽△A2B2C就可以求出结论．解答：解：∵△ABC∽△A2B2C，∴，∴，∴A2B2=．故答案为：．点评：本题考查了旋转的性质运用，解答时证明三角形相似，运用相似三角形的对应边成比例求解是关键．三、解答题（本大题共7题，满分78分）19．（10分）（2014•青浦区一模）如图，已知在直角坐标系中，点A在第二象限内，点B和点C在x轴上，原点O 为边BC的中点，BC=4，AO=AB，tan∠AOB=3，求图象经过A、B、C三点的二次函数解析式．考点：待定系数法求二次函数解析式；解直角三角形．专题：计算题．分析：先确定B点坐标为（﹣2，0），C点坐标为（2，0），作AH⊥OB于H，根据等腰三角形的性质得到OH=BH=1，再利用三角形函数得到tan∠AOB==3，则AH=3，所以A点坐标为（﹣1，3），设抛物线的交点式y=a（x+2）（x﹣2），然后把A点坐标代入求出a即可．解答：解：∵原点O∴B点坐标为（﹣2，0），C点坐标为（2，0），作AH⊥OB于H，如图，∵AO=AB，∴OH=BH=1，∵tan∠AOB==3，∴AH=3，∴A点坐标为（﹣1，3），设抛物线的解析式为y=a（x+2）（x﹣2），把A（﹣1，3）代入得a×1×（﹣3）=3，解得a=﹣1，∴经过A、B、C三点的二次函数解析式为y=﹣（x+2）（x﹣2）=﹣x2+4．点评：本题考查了待定系数法求二次函数关系式：要根据题目给定的条件，选择恰当的方法设出关系式，从而代入数值求解．一般地，当已知抛般式，用待定系数法列三元一次方程组来求解；当已知抛物线的顶点或对称轴时，常设其解析式为顶点式来求解；当已知抛物线与x轴有两个交点时，可选择设其解析式为交点式来求解．20．（10分）（2014•青浦区一模）如图，已知在△ABC中，点D、E分别在边AB、AC上，DE∥BC，，如果，．（1）求（用向量的式子表示）（2）求作向量（不要求写作法，但要指出所作图表中表示结论的向量）考点：*平面向量．分析：（1）由DE∥BC，，根据平行线分线段成比例定理，可求得AE：AC=2：5，又由，，利用三角形法则，即可求得，案；（2）取点AB的中点M，作=，连接，则即为所求．解答：解：（1）∵DE∥BC，∴=，∵，，∴=+=+，∴=﹣=﹣（+）=﹣﹣；（2）如图，取点AB的中点M，作=，连接，则即为所求．点评：此题考查了平面向量的知识．此题难掌握三角形法则的应用，注意掌握数形结合思想的应用．21．（10分）（2014•青浦区一模）已知：如图，在平行四边形ABCD中，E、F分别是边BC，CD上的点，且EF∥BD，AE、AF分别交BD与点G和点H，BD=12，EF=8．求：（1）的值；（2）线段GH的长．考点：平行线分线段成比例；平行四边形的性质．分析：（1）根据EF∥BD，则=，再利用平行四边形的性质即可得出的值；（2）利用DF∥AB，则==，进而得出==，求出GH即可．解答：解：（1）∵EF∥BD，∴=，∵BD=12，EF=8，∴=，∴=，∵四边形ABCD是平行四边形，∴AB=CD，∴=；（2）∵DF∥AB，∴==，∴=，∵EF∥BD，∴==，∴=，∴GH=6．点评：此题主要考查了平行线分线段成比例定理以及平行四边形的性质，熟练根据平行线分线段成比例定理得出GH的长是解题关键．22．（10分）（2014•青浦区一模）如图，已知某船向正东方向航行，在点A处测得某岛C在其北偏东60°方向上，前进8海里处到达点B处，测得岛C在其北偏东30°方向上．已知岛C周围6海里内有一暗礁，问：如果该船继续向东航行，有无触礁危险？请说明你的理由．考点：解直角三角形的应用-方向角问题．分析：作CD⊥AB于点D，求出C到航线的最近的距离CD的长，与6海里比较大小即可．解答：解：解：作CD⊥AB于点D，由题意可知，∠CAB=30°，∠CBD=60°，∴∠ACB=30°，在Rt△BCD中，∵∠BDC=90°，∠CBD=60°，∴∠BCD=30°，∴∠ACB=∠BCD．∴△CDB∽△ADC．∴=∵AB=CB=8∴BD=4，AD=12．∴=∴CD=4≈6.928＞6．∴船继续向东航行无触礁危险．点评：此题是一道方向角问题，结合航海中的实际问题，将解直角三角形的相关知识有机结合，体现了数际生活的思想．23．（12分）（2014•青浦区一模）已知，如图，在梯形ABCD中，AD∥BC，∠BCD=90°，对角线AC、BD相交于点E，且AC⊥BD．（1）求证：CD2=BC•AD；（2）点F是边BC上一点，联结AF，与BD相交于点G，如果∠BAF=∠DBF，求证：．考点：相似三角形的判定与性质．专题：证明题．分析：（1）首先根据已知得出∠ACD=∠CBD，以及∠ADC=∠BCD=90°，进而求出△ACD∽△DBC，即可得出答案；（2）首先证明△ABG∽△DBA，进而得出=，再利用△ABG∽△DBA，得出=，则AB2=BG•BD，进而得出答案．解答：证明：（1）∵AD∥BC，∠BCD=90°，∴∠ADC=∠又∵AC⊥BD，∴∠ACD+∠ACB=∠CBD +∠ACB=90°，∴∠ACD=∠CBD，∴△ACD∽△DBC，∴=，即CD2=BC×AD ；（2）方法一：∵AD∥BC，∴∠ADB=∠DBF，∵∠BAF=∠DBF，∴∠ADB=∠BAF，∵∠ABG=∠DBA，∴△ABG∽△DBA，∴=，∴=，又∵△ABG∽△DBA，∴=，∴AB2=BG•BD，∴===，方法二：∵AD∥BC，∴∠ADB=∠DBF，∵∠BAF=∠DBF，∴∠ADB=∠BAF，∵∠ABG=∠DBA，∴△ABG∽△DBA，∴=（）2=，而=，∴=．点评：此题主要考查了相似三角形的判定与性质，根据已知得出△ABG∽△DBA是解题关键．24．（12分）（2014•青浦区一模）已知在平面直角坐标系xOy中，二次函数y=﹣2x2+bx+c的图象经过点A（﹣3，0）和点B（0，6）．（1）求此二次函数的解析式；（2）将这个二次函数的图象向右平移5个单位后的顶点设为C，直线BC与x轴相交于点D，求∠ABD的正弦值；（3）在第（2）小题的条件下，联结OC，试探究直线AB与OC的位置关系，并说明理由．考点：二次函数综合题．专题：压轴题．分析：（1）把点A、B的坐标代入函数解析式计算求出b、c的值，即可得（2）先求出抛物线的顶点坐标，再根据向右平移横坐标加，求出点C的坐标，设直线BC的解析式为y=kx+b（k≠0），然后利用待定系数法求出直线BC的解析式，再求出与x轴的交点D 的坐标，过点A作AH⊥BD 于H，先求出OD，再利用勾股定理列式求出BD，然后求出△ADH和△BDO相似，利用相似三角形对应边成比例列式求出AH，再利用勾股定理，然后根据锐角的正弦等于对边比斜边列式计算即可得解；（3）方法一：求出=，然后根据平行线分线段成比例定理解答；方法二：过点C作CP⊥x轴于P，分别求出∠BAO和∠COP的正∠BAO=∠COP，再根据同位角相等，两直线平行解答．解答：解：（1）由题意得，，解得，所以，此二次函数的解析式为y=﹣2x2﹣4x+6；（2）∵y=﹣2x2﹣4x+6=﹣2（x+1）2+8，∴函数y=2x2﹣4x+6的顶点坐标为（﹣1，8），∴向右平移5个单位的后的顶点C（4，8），设直线BC的解析式为y=kx+b（k≠0），则，解得，所以，直线BC的解析式为y=x+6，令y=0，则x+6=0，标为（﹣12，0），过点A作AH⊥BD于H，OD=12，BD===6，AD=﹣3﹣（﹣12）=﹣3+12=9，∵∠ADH=∠BDO，∠AHD=∠B OD=90°，∴△ADH∽△BDO，∴=，即=，解得AH=，∵AB===3，∴sin∠ABD ===；（3）AB∥OC．理由如下：方法一：∵BD=6，BC==2，AD=9，AO=3，∴==3，∴AB∥OC；方法二：过点C作CP⊥x轴于P，由题意得，CP=8，PO=4，AO=3，BO=6，∴tan∠COP===2，tan∠BAO===2，∴tan∠COP=tan∠BAO，∴∠BAO=∠COP，∴AB∥OC．点评：本题是二次函数综合题，主要利用了待定系数法求二次函数解析式，待定系数法求一次函数解析式，锐角三角函数，相似三角形的判定与性质，作辅是解题的关键，作出图形更形象直观．25．（14分）（2014•青浦区一模）如图，已知在Rt△ABC中，∠ACB=90°，AB=10，tanA=，点D是斜边AB上的动点，联结CD，作DE⊥CD，交射线CB于点E，设AD=x．（1）当点D是边AB的中点时，求线段DE的长；（2）当△BED是等腰三角形时，求x的值；（3）如果y=，求y关于x的函数解析式，并写出它的定义域．考点：相似形综合题．专题：综合题．分析：（1）在直角三角形ABC中，由AB与tanA的值，利用锐角三角函数定义及勾股定理求出BC与AC的长，由D为斜边上的中点，利用直角三角形斜边上的中线等于斜边的一半得到CD=AD=BD=5，可得出∠DCB=∠DBC，再由一对直角相等，利用两对对应角相等的三角形相似得到△EDC与△ACB相似，由相似得BC边上时，由△BDE为等腰三角形且∠BED为钝角，得到DE=BE，利用等边对等角得到∠EBD=∠E DB，利用等角的余角相等得到∠CDA=∠A ，利用等角对等边得到CD=AC，作CH垂直于AB，利用三线合一得到AD=2AH，由cosA的值求出AH的长，进而求出AD 的长，即为x 的值；（ii）当E为BC延长线上时，与∠DBE 为钝角得到DB=BE，同理求出x的值；（3）作DM 垂直于BC，得到DM与AC平行，由平行得比例，表示出DM与BM，进而表示出CD与CM，由三角形DEM与三角形CDM相似得比例，表示出DE，由结果，并求出x的范围即可．解答：解：（1）在△ABC中，∵∠ACB=90°，AB=10，tanA=，∴BC=8，AC=6，∵点D为斜边AB的中点，∴CD=AD=BD=5，∴∠DCB=∠DBC，∵∠EDC=∠ACB=90°，∴△EDC∽△ACB，∴=，即=，则DE=；（2）分两种情况情况：（i）当E在BC边长时，∵△BED为等腰三角形，∠BED为钝角，∴EB=ED，∴∠EBD=∠EDB，∵∠EDC=∠ACB=90°，∴∠CDA=∠A，∴=，即AH=，∴AD=，即x=；（ii）当E在CB延长线上时，∵△BED为等腰三角形，∠DBE为钝角，∴BD=BE，∴∠BED=∠BDE，∵∠EDC=90°，∴∠BED+∠BCD=∠BDE +∠BDC=90°，∴∠BCD=∠BDC，∴BD=BC=8，∴AD=x=AB ﹣BD=10﹣8=2；（3）作DM⊥BC，垂足为M，∵DM∥AC，∴==，∴DM=（10﹣x），BM=（10﹣x），=x，CD=，∵△DEM∽△CDM，∴=，即DE==，∴y==，整理得：y=（0＜x＜10）．点评：此题属于相似型综合题，涉及的知识有：相似三角形的判定与性质，锐角三角函数定义，勾股定理，直角三角形斜边上的中线了分类讨论的思想，熟练掌握相似三角形的判定与性质是解本题的关键．参与本试卷答题和审题的老师有：zjx111；星期八；zcx；caicl；sjzx；lf2-9；Linaliu；MMCH；sd2011；ZJX；hdq123；gsls；zhjh；sks（排名不分先后）菁优网2015年1月2日©2010-2015 菁优网。

静安区2014学年第二学期高三年级教学质量检测数学试卷（文科） 2015.04.（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，考生应在答题纸相应编号的空格内直接填写结果，每个空格填对得4分，否则一律得零分．1．已知抛物线22y px =的准线方程是2x =-，则p = ．2．已知扇形的圆心角是1弧度，半径为5cm ，则此扇形的弧长为 cm ． 3．复数34ii-（i 为虚数单位）的模为 ． 4．函数2y x =的值域为 ．5．若2021310x y -⎛⎫⎛⎫⎛⎫= ⎪⎪ ⎪-⎝⎭⎝⎭⎝⎭，则x y += ．6．在921x x ⎛⎫- ⎪⎝⎭的展开式中，31x 的系数是 ．7cos x x =的解集为 ．8．已知{}1,0,1m ∈-，{}1,1n ∈-，若随机选取,m n ，则直线10mx ny ++=不经过第二象限的概率是 .9．圆22420x y x y +-+=的圆心到直线3430x y ++=的距离为 ．10．已知M 、N 是不等式组⎪⎩⎪⎨⎧≤+≥+-≥≥6011,1y x y x y x 所表示的平面区域内的不同两点，则M 、N 两点之间距离||MN 的最大值是 .11．把一个大金属球表面涂漆，共需油漆2.4公斤．若把这个大金属球熔化制成64个大小都相同的小金属球，不计损耗，将这些小金属球表面都涂漆，需要用漆 公斤．12．设12,e e 是平面内两个不共线的向量，12(1)AB a e e =-+ ，122AC be e =- ，0,0a b >>.若,,A B C三点共线，则12a b+的最小值是 ． 13．设等差数列{}n a 的前n 项和为n A ，等比数列{}n b 的前n 项和为n B ，若33a b =，44a b =，且53427A A B B -=-，则数列{}n b 的公比q = ．14．已知：当0x >时，不等式11kx b x ≥++恒成立，当且仅当13x =时取等号，则k = ． 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，考生应在答案纸的相应编号上，填上正确的答案，选对得5分，否则一律得零分. 15.如图，ABCDEF 是正六边形，下列等式成立的是（ ）（A ）0AE FC ⋅= （B ）0AE DF ⋅>（C ）FC FD FB =+ （D ）0FD FB ⋅<16.已知偶函数)(x f 的定义域为R ，则下列函数中为奇函数的是（ ） （A ）)](sin[x f （B ）)(sin x f x ⋅（C ）)(sin )(x f x f ⋅（D ）2)](sin [x f 17. 如图所示是一个循环结构的算法，下列说法不正确的是（ ） （A ）①是循环变量初始化，循环就要开始 （B ）②为循环体（C ）③是判断是否继续循环的终止条件（D ）输出的S 值为2，4，6，8，10，12，14，16，18.18.定义：最高次项的系数为1的多项式1110n n n p (x)x a x a x a --=++鬃?+（*∈n N ）的其余系数(0,1,,1)=⋅⋅⋅-i a i n 均是整数，则方程()0=p x 的根叫代数整数．下列各数不是代数整数的是（ ）F三、解答题（本大题满分74分）本大题共5题，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.19. (本题满分12分)本题共有2个小题，第1小题满分4分，第2小题满分8分． 如图，在正三棱柱111ABC A B C -中，已知16AA =， 三棱柱111ABC A B C -的体积为．(1)求正三棱柱111ABC A B C -的表面积； (2)求异面直线1BC 与1AA 所成角的大小.20. (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． 已知函数)(),(x g x f 满足关系)()()(α+⋅=x f x f x g ，其中α是常数. （1）若x x x f sin cos )(+=，2πα=，求)(x g 的解析式，并写出)(x g 的递增区间；（2）设()f x x =，若()1g x ≥在1[,)2x ∈+∞上恒成立，求常数α的取值范围．21. (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．某公园有个池塘，其形状为直角ABC ∆，090C ∠=，AB 的长为2百米，BC 的长为1百米． （1）若准备养一批供游客观赏的鱼，分别在AB 、BC 、CA 上取点D E F 、、，如图(1)，使得EF//AB ，EF ED ⊥，在DEF ∆内喂食，求当DEF ∆的面积取最大值时EF 的长；（2）若准备建造一个荷塘，分别在AB 、BC 、CA 上取点D E F 、、，如图(2)，建造DEF ∆连廊（不考虑宽度）供游客休憩，且使DEF ∆为正三角形，记FEC α∠=，求DEF ∆边长的最小值及此时α的值．(精确到1米和0.1度)A C BC A F EFE B B 1C 122. （本题满分16分）本题共有3个小题，第1小题4分，第2小题5分，第3小题7分．在平面直角坐标系xoy 中，已知椭圆C 的方程为2218x y +=，设AB 是过椭圆C 中心O 的任意弦，l 是线段AB 的垂直平分线，M 是l 上与O 不 重合的点． （1）求以椭圆的焦点为顶点，顶点为焦点的双曲线方程；（2）若2MO OA =，当点A 在椭圆C 上运动时，求点M 的轨迹方程；（3） 记M 是l 与椭圆C 的交点，若直线AB 的方程为(0)y kx k =>，当△AMB求直线AB 的方程．23. （本题满分18分）本题共有3个小题，第1小题4分，第2小题6分，第3小题8分．设{}n a 是公比为(q 1)q ≠的等比数列，若{}n a 中任意两项之积仍是该数列中的项，那么称{}n a 是封闭数列. （1）若123a q ==，，判断{}n a 是否为封闭数列，并说明理由；（2）证明{}n a 为封闭数列的充要条件是：存在整数1m ≥-，使1m a q =；（3）记n ∏是数列{}n a 的前n 项之积，2log nn b =∏，若首项为正整数，公比2q=，试问：是否存在这样的封闭数列{}n a ，使1211111lim 9n n b b b →∞⎛⎫++⋅⋅⋅+= ⎪⎝⎭，若存在，求{}n a 的通项公式；若不存在，说明理由．。

2014年07月21日NIUXS的高中数学组卷2014年07月21日niuxs的高中数学组卷一．解答题（共24小题）1．（2014•上海模拟）已知，其中b是常数．（1）若y=f（x）是奇函数，求b的值；（2）求证：y=f（x）的图象上不存在两点A、B，使得直线AB平行于x轴．2．（2014•静安区一模）已知函数是奇函数，（其中a＞1）（1）求实数m的值；（2）讨论函数f（x）的增减性；（3）当x时，f（x）的值域是（1，+∞），求n与a的值．3．（2014•崇明县一模）解方程：．4．（2014•长宁区一模）已知函数为奇函数．（1）求常数a的值；（2）判断函数的单调性，并说明理由；（3）函数g（x）的图象由函数f（x）的图象先向右平移2个单位，再向上平移2个单位得到，写出g（x）的一个对称中心，若g（b）=1，求g（4﹣b）的值．5．（2013•静安区一模）函数y=f（x），x∈D，其中D≠∅．若对任意x∈D，f（|x|）=|f（x）|，则称y=f（x）在D内为对等函数．（1）指出函数，y=x3，y=2x在其定义域内哪些为对等函数；（2）试研究对数函数y=log a x（a＞0且a≠1）在其定义域内是否是对等函数？若是，请说明理由；若不是，试给出其定义域的一个非空子集，使y=log a x在所给集合内成为对等函数；（3）若{0}⊆D，y=f（x）在D内为对等函数，试研究y=f（x）（x∈D）的奇偶性．6．（2012•虹口区一模）已知函数（a＞0，a≠1）．（1）若m=﹣1时，判断函数f（x）在上的单调性，并说明理由；（2）若对于定义域内一切x，f（1+x）+f（1﹣x）=0恒成立，求实数m的值；（3）在（2）的条件下，当时，f（x）的取值恰为，求实数a，b的值．7．（2011•黄浦区二模）已知函数是奇函数，定义域为区间D（使表达式有意义的实数x 的集合）．（1）求实数m的值，并写出区间D；（2）若底数a＞1，试判断函数y=f（x）在定义域D内的单调性，并说明理由；（3）当x∈A=[a，b）（A⊆D，a是底数）时，函数值组成的集合为[1，+∞），求实数a、b的值．8．（2009•静安区一模）已知函数f（x）=．（1）求函数f（x）的定义域；（2）若函数f（x）的定义域关于坐标原点对称，试讨论它的奇偶性和单调性；（3）在（2）的条件下，记f﹣1（x）为f（x）的反函数，若关于x的方程f﹣1（x）=5k•2x﹣5k有解，求k的取值范围．9．已知函数f（x）=log a（2x+2），g（x）=log a（2x﹣2）（a＞0，且a≠1）．（1）求函数h（x）=f（x）﹣g（x）的定义域；（2）判断函数h（x）=f（x）﹣g（x）在x∈（1，+∞）内的单调性，并用定义给予证明；（3）当a=2时，若对[3，5]上的任意x都有h（x）＜2x+m成立，求m的取值范围．10．若函数f（x）满足下列条件：在定义域内存在x0，使得f（x0+1）=f（x0）+f（1）成立，则称函数f（x）具有性质M；反之，若x0不存在，则称函数f（x）不具有性质M．（Ⅰ）证明：函数f（x）=2x具有性质M，并求出对应的x0的值；（Ⅱ）已知函数h（x）=具有性质M，求a的取值范围；（Ⅲ）试探究形如①y=kx+b（k≠0）、②y=ax2+bx+c（a≠0）、③y=（k≠0）、④y=ax（a＞0且a≠1）、⑤y=log a x（a＞0且a≠1）的函数，指出哪些函数一定具有性质M？并加以证明．11．（2008•闸北区二模）设f（x）=a x+b同时满足条件f（0）=2和对任意x∈R都有f（x+1）=2f（x）﹣1成立．（1）求f（x）的解析式；（2）设函数g（x）的定义域为[﹣2，2]，且在定义域内g（x）=f（x），且函数h（x）的图象与g（x）的图象关于直线y=x对称，求h（x）；（3）求函数y=g（x）+h（x）的值域．12．（2008•长宁区二模）设函数f（x）=2x﹣1的反函数为f﹣1（x），g（x）=log4（3x+1）．（1）若f﹣1（x）≤g（x），求x的取值范围D；（2）设，当x∈D（D为（1）中所求）时函数H（x）的图象与直线y=a有公共点，求实数a的取值范围．13．（2004•宝山区一模）已知f（x）=log4（4x+1）+kx（k∈R）是偶函数．（1）求k的值；（2）证明：对任意实数b，函数y=f（x）的图象与直线最多只有一个交点；（3）设，若函数f（x）与g（x）的图象有且只有一个公共点，求实数a的取值范围．14．定义在R上的函数f（x）满足对任意x，y∈R都有f（x+y）=f（x）+f（y）．当x＞0，f（x）＞0，（1）求证：f（x）为奇函数；（2）判断f（x）的单调性并证明；（3）解不等式：．15．已知函数f（x）=log2+log2（x﹣1）+log2（p﹣x）．（1）当p=7时，求函数f（x）的定义域与值域；（2）求函数f（x）的定义域与值域．16．已知函数f（x）=log a（x+1）（a＞1），若函数y=g（x）的图象上任意一点P关于原点的对称点Q的轨迹恰好是函数f（x）的图象．（1）写出函数g（x）的解析式；（2）当x∈[0，1）时，总有f（x）+g（x）≥m成立，求m的取值范围．17．设f（x）=（10﹣3x）．（1）求使f（x）≥1的x的取值范围；（2）若对于区间[2，3]上的每一个x的值，不等式f（x）＞+m恒成立，求实数m的取值范围．18．已知函数是偶函数．（1）求k的值；（2）若f（2t2+1）＜f（t2﹣2t+1），求t的取值范围；（3）设函数，其中a＞0，若函数f（x）与g（x）的图象有且只有一个公共点，求实数a的取值范围．19．已知函数y=f（x）存在反函数，定义：若对给定的实数a（a≠0），函数y=f（x+a）与y=f﹣1（x+a）互为反函数，则称y=f（x）满足“a和性质”．（1）判断函数g（x）=x2+1（x＞0）是否满足“1和性质”，说明理由；（2）求所有满足“2和性质”的一次函数．20．“学习曲线”可以用来描述学习达到某一水平所需的学习时间．假设“学习曲线”符合函数N（单位：字）表示某一英文词汇量水平，t（单位：天）表示达到这一英文词汇量所需要的学习时间．（1）已知某人练习达到40个词汇量时需要10天，求该人的学习曲线解析式；（2）如果他学习几天能掌握160个词汇量？（3）如果他学习时间大于30天，他的词汇量情况如何？21．对于．（1）函数的“定义域为R”和“值域为R”是否是一回事？分别求出实数a的取值范围；（2）结合“实数a的取何值时f（x）在[﹣1，+∞）上有意义”与“实数a的取何值时函数的定义域为（﹣∞，1）∪（3，+∞）”说明求“有意义”问题与求“定义域”问题的区别．22．（2007•崇文区二模）已知f（x）=2（x﹣1）2+2，g（x）=x2﹣1，求函数f[g（x）]的单调递增区间．23．已知函数f（x）=λ•2x﹣4x，定义域为[1，3]．（1）若λ=6求函数f（x）的值域；（2）若函数f（x）在区间[1，3]上是增函数，求实数λ的取值范围．24．已知函数，（1）若函数定义域为（﹣∞，﹣1）∪（3，+∞），求a的值；（2）若函数值域为（﹣∞，﹣1]，求a的值；（3）若f（x）在（﹣∞，1]单调递增，求a的取值范围．2014年07月21日niuxs的高中数学组卷参考答案与试题解析一．解答题（共24小题）1．（2014•上海模拟）已知，其中b是常数．（1）若y=f（x）是奇函数，求b的值；（2）求证：y=f（x）的图象上不存在两点A、B，使得直线AB平行于x轴．，设）∵=0，为奇函数．，设∵，，∴时，∵∴2．（2014•静安区一模）已知函数是奇函数，（其中a＞1）（1）求实数m的值；（2）讨论函数f（x）的增减性；（3）当x时，f（x）的值域是（1，+∞），求n与a的值．a=a a，∴，＞a，任取a a•a∴21+2）上为减函数，值域为（a=﹣+33．（2014•崇明县一模）解方程：．由原方程可化简得解：由原方程化简得∴4．（2014•长宁区一模）已知函数为奇函数．（1）求常数a的值；（2）判断函数的单调性，并说明理由；（3）函数g（x）的图象由函数f（x）的图象先向右平移2个单位，再向上平移2个单位得到，写出g（x）的一个对称中心，若g（b）=1，求g（4﹣b）的值．）根据函数的定义域关于原点对称，以及根据，单调递增，）因为函数为奇函数，所以定义域关于原点对称，由，满足.利用已有函数的单调性加以说明．∵在（﹣5．（2013•静安区一模）函数y=f（x），x∈D，其中D≠∅．若对任意x∈D，f（|x|）=|f（x）|，则称y=f（x）在D内为对等函数．（1）指出函数，y=x3，y=2x在其定义域内哪些为对等函数；（2）试研究对数函数y=log a x（a＞0且a≠1）在其定义域内是否是对等函数？若是，请说明理由；若不是，试给出其定义域的一个非空子集，使y=log a x在所给集合内成为对等函数；（3）若{0}⊆D，y=f（x）在D内为对等函数，试研究y=f（x）（x∈D）的奇偶性．）根据对等函数的定义，我们判断）6．（2012•虹口区一模）已知函数（a＞0，a≠1）．（1）若m=﹣1时，判断函数f（x）在上的单调性，并说明理由；（2）若对于定义域内一切x，f（1+x）+f（1﹣x）=0恒成立，求实数m的值；（3）在（2）的条件下，当时，f（x）的取值恰为，求实数a，b的值．由于在）∪2）的取值恰为），记∴单调递减，）在）∪，故值域即为，得7．（2011•黄浦区二模）已知函数是奇函数，定义域为区间D（使表达式有意义的实数x 的集合）．（1）求实数m的值，并写出区间D；（2）若底数a＞1，试判断函数y=f（x）在定义域D内的单调性，并说明理由；（3）当x∈A=[a，b）（A⊆D，a是底数）时，函数值组成的集合为[1，+∞），求实数a、b的值．）令，即，解得∴．时，函数上是单调减函数．理由：令在时，函数上是单调减函数．时，函数上是增函数，，解得上的函数值组成的集合为的值是8．（2009•静安区一模）已知函数f（x）=．（1）求函数f（x）的定义域；（2）若函数f（x）的定义域关于坐标原点对称，试讨论它的奇偶性和单调性；（3）在（2）的条件下，记f﹣1（x）为f（x）的反函数，若关于x的方程f﹣1（x）=5k•2x﹣5k有解，求k的取值范围．）2=log2=9．已知函数f（x）=log a（2x+2），g（x）=log a（2x﹣2）（a＞0，且a≠1）．（1）求函数h（x）=f（x）﹣g（x）的定义域；（2）判断函数h（x）=f（x）﹣g（x）在x∈（1，+∞）内的单调性，并用定义给予证明；（3）当a=2时，若对[3，5]上的任意x都有h（x）＜2x+m成立，求m的取值范围．）的解析式可得===﹣，＞10．若函数f（x）满足下列条件：在定义域内存在x0，使得f（x0+1）=f（x0）+f（1）成立，则称函数f（x）具有性质M；反之，若x0不存在，则称函数f（x）不具有性质M．（Ⅰ）证明：函数f（x）=2x具有性质M，并求出对应的x0的值；（Ⅱ）已知函数h（x）=具有性质M，求a的取值范围；（Ⅲ）试探究形如①y=kx+b（k≠0）、②y=ax2+bx+c（a≠0）、③y=（k≠0）、④y=ax（a＞0且a≠1）、⑤y=log a x（a＞0且a≠1）的函数，指出哪些函数一定具有性质M？并加以证明．y=lg=（+a+2ax，满足题意+2ax3+3+，]．（（（（11．（2008•闸北区二模）设f（x）=a x+b同时满足条件f（0）=2和对任意x∈R都有f（x+1）=2f（x）﹣1成立．（1）求f（x）的解析式；（2）设函数g（x）的定义域为[﹣2，2]，且在定义域内g（x）=f（x），且函数h（x）的图象与g（x）的图象关于直线y=x对称，求h（x）；（3）求函数y=g（x）+h（x）的值域．12．（2008•长宁区二模）设函数f（x）=2x﹣1的反函数为f﹣1（x），g（x）=log4（3x+1）．（1）若f﹣1（x）≤g（x），求x的取值范围D；（2）设，当x∈D（D为（1）中所求）时函数H（x）的图象与直线y=a有公共点，求实数a的取值范围．）先化简得到：时，，∴，∴时，∴因此当13．（2004•宝山区一模）已知f（x）=log4（4x+1）+kx（k∈R）是偶函数．（1）求k的值；（2）证明：对任意实数b，函数y=f（x）的图象与直线最多只有一个交点；（3）设，若函数f（x）与g（x）的图象有且只有一个公共点，求实数a的取值范围．有且只有一个实根，令有且k=x）的图象与直线）的图象与直线）的图象与直线有且只有一个实根有且只有一个正根①②14．定义在R上的函数f（x）满足对任意x，y∈R都有f（x+y）=f（x）+f（y）．当x＞0，f（x）＞0，（1）求证：f（x）为奇函数；（2）判断f（x）的单调性并证明；（3）解不等式：．原不等式等价于或解得解集为15．已知函数f（x）=log2+log2（x﹣1）+log2（p﹣x）．（1）当p=7时，求函数f（x）的定义域与值域；（2）求函数f（x）的定义域与值域．）由题意，可得2）由题意，可得2x+p=当当（16．已知函数f（x）=log a（x+1）（a＞1），若函数y=g（x）的图象上任意一点P关于原点的对称点Q的轨迹恰好是函数f（x）的图象．（1）写出函数g（x）的解析式；（2）当x∈[0，1）时，总有f（x）+g（x）≥m成立，求m的取值范围．）∴即17．设f（x）=（10﹣3x）．（1）求使f（x）≥1的x的取值范围；（2）若对于区间[2，3]上的每一个x的值，不等式f（x）＞+m恒成立，求实数m的取值范围．）由已知得：，由此求得）由题意可得（，设由已知得：（≥，，∴＜[）＞，∴）（（∵∴∴，即实数，﹣18．已知函数是偶函数．（1）求k的值；（2）若f（2t2+1）＜f（t2﹣2t+1），求t的取值范围；（3）设函数，其中a＞0，若函数f（x）与g（x）的图象有且只有一个公共点，求实数a的取值范围．2）的图象有且只有一个交点，即方程=在区间（2的定义域为（=在区间（a 在区间（，＞t，，不合题意；﹣t=，，＞）＜，即（）﹣19．已知函数y=f（x）存在反函数，定义：若对给定的实数a（a≠0），函数y=f（x+a）与y=f﹣1（x+a）互为反函数，则称y=f（x）满足“a和性质”．（1）判断函数g（x）=x2+1（x＞0）是否满足“1和性质”，说明理由；（2）求所有满足“2和性质”的一次函数．的定义可先根据求反函数的步骤求出x+1=x=y=①∵∴∴∴∴20．“学习曲线”可以用来描述学习达到某一水平所需的学习时间．假设“学习曲线”符合函数N（单位：字）表示某一英文词汇量水平，t（单位：天）表示达到这一英文词汇量所需要的学习时间．（1）已知某人练习达到40个词汇量时需要10天，求该人的学习曲线解析式；（2）如果他学习几天能掌握160个词汇量？（3）如果他学习时间大于30天，他的词汇量情况如何？得：，∴）当时，∴，解得21．对于．（1）函数的“定义域为R”和“值域为R”是否是一回事？分别求出实数a的取值范围；（2）结合“实数a的取何值时f（x）在[﹣1，+∞）上有意义”与“实数a的取何值时函数的定义域为（﹣∞，1）∪（3，+∞）”说明求“有意义”问题与求“定义域”问题的区别．：，则的取值范围为或22．（2007•崇文区二模）已知f（x）=2（x﹣1）2+2，g（x）=x2﹣1，求函数f[g（x）]的单调递增区间．，或的单调递增区间为23．已知函数f（x）=λ•2x﹣4x，定义域为[1，3]．（1）若λ=6求函数f（x）的值域；（2）若函数f（x）在区间[1，3]上是增函数，求实数λ的取值范围．24．已知函数，（1）若函数定义域为（﹣∞，﹣1）∪（3，+∞），求a的值；（2）若函数值域为（﹣∞，﹣1]，求a的值；（3）若f（x）在（﹣∞，1]单调递增，求a的取值范围．则函数，则函数。

静安区2013-2014学年度第一学期期末质量抽查初三英语试卷（满分150分，考试时间100分钟）2014.01考生注意：本卷有7大题，共94小题。

试题均采用连续编号，所有答案务必按照规定在答题卡上完成，做在试卷上不给分。

Part 1 Listening （第一部分听力）I. Listening comprehension (听力理解) (共30 分)A.Listen and choose the right picture (根据你听到的内容，选出相应的图片) (6 分)1. _________2. _________3._______4._______.5._______6._______B. Listen to the dialogue and choose the best answer to the question you hear (根据你听到的对话和问题，选出最恰当的答案)：(8分)7. A）15 B）18 C) 21 D) 338. A) Surprised B) Happy C) Worried D) Excited9. A）Kitty B) Gary C) Mike D) Mary10. A) A student B) A manager C) A worker D) A math teacher11. A) Shanghai B) Beijing C) Korea D) Thailand12. A) Because it is hotter B) Because it is sweeterC) Because it tastes better D) Because it looks nicer13. A) Before the meeting B) After the meetingC) In the middle of the meeting D) At the end of the meeting14. A) He‟s been away from home for a long time B) He has lost some moneyC) He works hard day and night D) He works with his wifeC．Listen to the passage and tell whether the following statements are true or false (判断下列句子是否符合你听到的短文内容，符合的用“T”表示，不符合的用“F”表示): (6分)15. Tom attended classes every evening to improve himself.16. After Tom was offered a good job, he spent more time with his family.17. With his hard work, Tom managed to get the position of manager.18. Tom decided to hire s servant to help his wife with the housework.19. Tom lived with his family in the beautiful house for the rest of his life.20. Tom was so tired that he got up very late the next day.D. Listen to the passage and fill in the blanks (听短文填空，完成下列内容。

2014学年度第一学期静安区高三数学文理科期末检测试卷参考答案及评分标准 2014.12说明1．本解答列出试题一种或几种解法，如果考生的解法与所列解法不同，可参照解答中评分标准的精神进行评分．2．评阅试卷，应坚持每题评阅到底，不要因为考生的解答中出现错误而中断对该题的评阅．当考生的解答在某一步出现错误，影响了后续部分，但该步以后的解答未改变这一题的内容和难度时，可视影响程度决定后面部分的给分，但是原则上不应超出后面部分应给分数之半，如果有较严重的概念性错误，就不给分．3．第19题至第23题中右端所注的分数，表示考生正确做到这一步应得的该题分数． 4．给分或扣分均以1分为单位．一．填空题（本大题满分56分）本大题共有14题，考生应在答题纸相应编号的空格内直接填写结果，每个空格填对得4分，否则一律得零分． 1．理：)2,0(；文：121； 2. 理：25628=；文：)2,0( 3. 理：)4,21(；文：n n +22； 4. 理：21；文：45 5. 理：)212(4nn-；文：)4,21(； 6. 理：-2，文： 25628= 7. 理：221log 2+=x ；文：π3； 8. 55arccos ；文：54-9. 理：31-；文：-2； 10. 32π-；文：10103arccos （或31arctan ）11. 理：03213=++-y x 或03213=-+--y x ； 文：31-12．理：]2,2[-； 文：03213=++-y x 或03213=-+--y x13. 理：12S <<；文：]2,2[-； 14. 理：228+=+n C nk ．7或者14；文：12S <<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，考生应在答案纸的相应编号上，填上正确的答案，选对得5分，否则一律得零分. 15．D ； 16．B ； 17． D ；18．C三、解答题（本大题满分74分）本大题共5题，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 .19．(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. （1）根据正弦定理BbA a sin sin =，得b B b a A sin 23sin ==，所以23sin =B ，………（4分） 又由角B 为锐角，得3π=B ；…………………………（6分）（2）B ac S ABC sin 21=∆，又ABC S ∆=3=ac ，…………………………（8分） 根据余弦定理B ac c a b cos 2222-+=，得1037cos 2222=+=+=+B ac b c a ，…………………………（12分）所以ac c a c a 2)(222++=+=16，从而a c +=4．…………………………（14分）20．(本题满分14分) 本题共有2个小题，第1小题满分4分，第2小题满分10分. （1）他应付出租车费26元；……………………………（ 4分）（2）　, )10( 107c )013( 3b )30( ,⎪⎩⎪⎨⎧>-++≤<-+≤<=x c b a x x b a x x a y 文21．(本题满分14分) 本题共有2个小题，第1小题满分8分，第2小题满分6分. （1）因为点P 为面11A ADD 的对角线1AD 的中点.⊥PM 平面ABCD ，所以PM 为△1ADD 的中位线，得1=PM ， 又BD MN ⊥，所以2222===MD ND MN ………………（ 2分） 因为在底面ABCD 中，BD AC B M ⊥⊥,D N ，所以AC MN //，又AC C A //11，∠PNM 为异面直线PN 与11C A 所成角的平面角，………………（ 6分）在△PMN 中，∠PMN 为直角，2tan =∠PNM ，所以2arctan =∠PNM 。

【考点训练】三角形的形状判断-2(扫描二维码可查看试题解析)一、选择题（共20小题）1．（2014•静安区校级模拟）若，则△ABC为（）A ．等腰三角形B．直角三角形C．锐角三角形D．不能判断2．（2014秋•郑州期末）若△ABC 的三个内角A、B、C满足6sinA=4sinB=3sinC，则△ABC（）A ．一定是锐角三角形B ．一定是直角三角形C ．一定是钝角三角形D ．可能是锐角三角形，也可能是钝角三角形3．（2014秋•祁县校级期末）A为三角形ABC的一个内角，若sinA+cosA=，则这个三角形的形状为（）A ．锐角三角形B．钝角三角形C ．等腰直角三角形D．等腰三角形4．（2014•天津学业考试）在△ABC中，sinA•sinB＜cosA•cosB，则这个三角形的形状是（）A ．锐角三角形B．钝角三角形C．直角三角形D．等腰三角形5．（2014春•禅城区期末）已知：在△ABC中，，则此三角形为（）A ．直角三角形B．等腰直角三角形C ．等腰三角形D．等腰或直角三角形6．（2014•南康市校级模拟）已知△ABC满足，则△ABC是（）A ．等边三角形B．锐角三角形C．直角三角形D．钝角三角形7．（2014•马鞍山二模）已知非零向量与满足且=．则△ABC为（）A ．等边三角形B．直角三角形C ．等腰非等边三角形D．三边均不相等的三角形8．（2014•蓟县校级二模）在△ABC中，a，b，c分别是角A，B，C所对的边，且2c2=2a2+2b2+ab，则△ABC是（）A ．钝角三角形B．直角三角形C．锐角三角形D．等边三角形9．（2014•黄冈模拟）已知在△ABC中，向量与满足（+）•=0，且•=，则△ABC为（）A ．三边均不相等的三角形B．直角三角形C ．等腰非等边三角形D．等边三角形10．（2014•奉贤区二模）三角形ABC中，设=，=，若•（+）＜0，则三角形ABC的形状是（）A ．锐角三角形B．钝角三角形C．直角三角形D．无法确定11．（2015•温江区校级模拟）已知向量，则△ABC的形状为（）A ．直角三角形B．等腰三角形C．锐角三角形D．钝角三角形12．（2014秋•景洪市校级期末）在△ABC中，内角A、B、C的对边分别为a、b、c，且，则△ABC的形状为（）A ．等边三角形B．等腰直角三角形C ．等腰或直角三角形D．直角三角形13．（2014•咸阳三模）△ABC的三个内角A、B、C成等差数列，，则△ABC一定是（）A ．直角三角形B．等边三角形C ．非等边锐角三角形D．钝角三角形14．（2014•奎文区校级模拟）在△ABC中，P是BC边中点，角A、B、C的对边分别是a、b、c，若，则△ABC的形状是（）A．等边三角形B．钝角三角形C．直角三角形D ．等腰三角形但不是等边三角形15．（2014秋•正定县校级期末）在△ABC中，tanA•sin2B=tanB•sin2A，那么△ABC一定是（）A锐角三角形B直角三角形．．C ．等腰三角形D．等腰三角形或直角三角形16．（2014•漳州四模）在△ABC中的内角A、B、C所对的边分别为a，b，c，若b=2ccosA，c=2bcosA则△ABC的形状为（）A ．直角三角形B．锐角三角形C ．等边三角形D．等腰直角三角形17．（2014•云南模拟）在△ABC中，若tanAtanB＞1，则△ABC是（）A ．锐角三角形B．直角三角形C．钝角三角形D．无法确定18．（2013秋•金台区校级期末）双曲线=1和椭圆=1（a＞0，m＞b＞0）的离心率互为倒数，那么以a，b，m为边长的三角形是（）A ．锐角三角形B．钝角三角形C．直角三角形D．等腰三角形19．（2014•红桥区二模）在△ABC中，“”是“△ABC为钝角三角形”的（）A ．充分不必要条件B．必要不充分条件C ．充要条件D．既不充分又不必要条件20．（2014秋•德州期末）在△ABC中，若acosA=bcosB，则△ABC的形状是（）A ．等腰三角形B．直角三角形C ．等腰直角三角形D．等腰或直角三角形二、填空题（共10小题）（除非特别说明，请填准确值）21．（2014春•沭阳县期中）在△ABC中，已知sinA=2sinBcosc，则△ABC的形状为．22．（2014秋•思明区校级期中）在△ABC中，若a=9，b=10，c=12，则△ABC的形状是．23．（2013•文峰区校级一模）已知△ABC中，AB=，BC=1，tanC=，则AC等于．24．（2013春•广陵区校级期中）在△ABC中，若2cosBsinA=sinC，则△ABC的形状一定是三角形．25．（2014秋•潞西市校级期末）在△ABC中，已知c=2acosB，则△ABC的形状为．26．（2014春•常熟市校级期中）在△ABC中，若，则△ABC的形状是．27．（2014春•石家庄期末）在△ABC中，若sin2A+sin2B＜sin2C，则该△ABC是三角形（请你确定其是锐角三角形、直角三角形还是钝角三角形）．28．（2013春•遵义期中）△ABC中，b=a，B=2A，则△ABC为三角形．29．（2013秋•沧浪区校级期末）若△ABC的三个内角满足sinA：sinB：sinC=5：11：13，则△ABC为（填锐角三角形、直角三角形、钝角三角形．）30．（2014春•宜昌期中）在△ABC中，sinA=2cosBsinC，则三角形为三角形．【考点训练】三角形的形状判断-2参考答案与试题解析一、选择题（共20小题）1．（2014•静安区校级模拟）若，则△ABC为（）A ．等腰三角形B．直角三角形C．锐角三角形D．不能判断考点：三角形的形状判断．专题：计算题．分析：利用平方差公式，由，推出AB=AC，即可得出△ABC为等腰三角形．解答：解：由，得：，∴故AB=AC，△ABC为等腰三角形，故选A．点评：本小题主要考查向量的数量积、向量的模、向量在几何中的应用等基础知识，考查运算求解能力，考查数形结合思想、化归与转化思想．属于基础题．2．（2014秋•郑州期末）若△ABC 的三个内角A、B、C满足6sinA=4sinB=3sinC，则△ABC （）A ．一定是锐角三角形B ．一定是直角三角形C ．一定是钝角三角形D ．可能是锐角三角形，也可能是钝角三角形考点：三角形的形状判断．专题：计算题；解三角形．分析：根据题意，结合正弦定理可得a：b：c=4：6：8，再由余弦定理算出最大角C的余弦等于﹣，从而得到△ABC是钝角三角形，得到本题答案．解答：解：∵角A、B、C满足6sinA=4sinB=3sinC，∴根据正弦定理，得6a=4b=3c，整理得a：b：c=4：6：8设a=4x，b=6x，c=8x，由余弦定理得：cosC===﹣∵C是三角形内角，得C∈（0，π），∴由cosC=﹣＜0，得C为钝角因此，△ABC是钝角三角形故选：C点评：本题给出三角形个角正弦的比值，判断三角形的形状，着重考查了利用正、余弦定理解三角形的知识，属于基础题．3．（2014秋•祁县校级期末）A为三角形ABC的一个内角，若sinA+cosA=，则这个三角形的形状为（）A ．锐角三角形B．钝角三角形C ．等腰直角三角形D．等腰三角形考点：三角形的形状判断．专题：计算题；解三角形．分析：将已知式平方并利用sin2A+cos2A=1，算出sinAcosA=﹣＜0，结合A∈（0，π）得到A为钝角，由此可得△ABC是钝角三角形．解答：解：∵sinA+cosA=，∴两边平方得（sinA+cosA）2=，即sin2A+2sinAcosA+cos2A=，∵sin2A+cos2A=1，∴1+2sinAcosA=，解得sinAcosA=（﹣1）=﹣＜0，∵A∈（0，π）且sinAcosA＜0，∴A∈（，π），可得△ABC是钝角三角形故选：B点评：本题给出三角形的内角A的正弦、余弦的和，判断三角形的形状．着重考查了同角三角函数的基本关系、三角形的形状判断等知识，属于基础题．4．（2014•天津学业考试）在△ABC中，sinA•sinB＜cosA•cosB，则这个三角形的形状是（）A ．锐角三角形B．钝角三角形C．直角三角形D．等腰三角形考点：三角形的形状判断；两角和与差的余弦函数．专题：计算题．分析：对不等式变形，利用两角和的余弦函数，求出A+B的范围，即可判断三角形的形状．解答：解：因为在△ABC中，sinA•sinB＜cosA•cosB，所以cos（A+B）＞0，所以A+B∈（0，），C＞，所以三角形是钝角三角形．故选B．点评：本题考查三角形的形状的判定，两角和的余弦函数的应用，注意角的范围是解题的关键．5．（2014春•禅城区期末）已知：在△ABC中，，则此三角形为（）A ．直角三角形B．等腰直角三角形C ．等腰三角形D．等腰或直角三角形考点：三角形的形状判断．专题：计算题．分析：由条件可得sinCcosB=cosCsinB，故sin（C﹣B）=0，再由﹣π＜C﹣B＜π，可得C﹣B=0，从而得到此三角形为等腰三角形．解答：解：在△ABC中，，则ccosB=bcosC，由正弦定理可得sinCcosB=cosCsinB，∴sin（C﹣B）=0，又﹣π＜C﹣B＜π，∴C﹣B=0，故此三角形为等腰三角形，故选C．点评：本题考查正弦定理，两角差的正弦公式，得到sin（C﹣B）=0及﹣π＜C﹣B＜π，是解题的关键．6．（2014•南康市校级模拟）已知△ABC满足，则△ABC 是（）A ．等边三角形B．锐角三角形C．直角三角形D．钝角三角形考点：三角形的形状判断．专题：计算题；平面向量及应用．分析：根据向量的加减运算法则，将已知化简得=+•，得•=0．结合向量数量积的运算性质，可得CA⊥CB，得△ABC是直角三角形．解答：解：∵△ABC中，，∴=（﹣）+•=•+•即=+•，得•=0∴⊥即CA⊥CB，可得△ABC是直角三角形故选：C点评：本题给出三角形ABC中的向量等式，判断三角形的形状，着重考查了向量的加减法则、数量积的定义与运算性质等知识，属于基础题．7．（2014•马鞍山二模）已知非零向量与满足且=．则△ABC为（）A ．等边三角形B．直角三角形C ．等腰非等边三角形D．三边均不相等的三角形考点：三角形的形状判断．专题：计算题．分析：通过向量的数量积为0，判断三角形是等腰三角形，通过=求出等腰三角形的顶角，然后判断三角形的形状．解答：解：因为，所以∠BAC的平分线与BC垂直，三角形是等腰三角形．又因为，所以∠BAC=60°，所以三角形是正三角形．故选A．点评：本题考查向量的数量积的应用，考查三角形的判断，注意单位向量的应用，考查计算能力．8．（2014•蓟县校级二模）在△ABC中，a，b，c分别是角A，B，C所对的边，且2c2=2a2+2b2+ab，则△ABC是（）A ．钝角三角形B．直角三角形C．锐角三角形D．等边三角形考点：三角形的形状判断．专题：计算题．分析：整理题设等式，代入余弦定理中求得cosC的值，小于0判断出C为钝角，进而可推断出三角形为钝角三角形．解答：解：∵2c2=2a2+2b2+ab，∴a2+b2﹣c2=﹣ab，∴cosC==﹣＜0．则△ABC是钝角三角形．故选A点评：本题主要考查了三角形形状的判断，余弦定理的应用．一般是通过已知条件，通过求角的正弦值或余弦值求得问题的答案．9．（2014•黄冈模拟）已知在△ABC中，向量与满足（+）•=0，且•=，则△ABC为（）A ．三边均不相等的三角形B．直角三角形C ．等腰非等边三角形D．等边三角形考点：三角形的形状判断．专题：计算题．分析：设，由=0，可得AD⊥BC，再根据边形AEDF是菱形推出∠EAD=∠DAC，再由第二个条件可得∠BAC=60°，由△ABH≌△AHC，得到AB=AC，得到△ABC是等边三角形．解答：解：设，则原式化为=0，即=0，∴AD⊥BC．∵四边形AEDF是菱形，|•=||•||•cos∠BAC=，∴cos∠BAC=，∴∠BAC=60°，∴∠BAD=∠DAC=30°，∴△ABH≌△AHC，∴AB=AC．∴△ABC是等边三角形．点评：本题考查两个向量的加减法的法则，以及其几何意义，三角形形状的判断，属于中档题．10．（2014•奉贤区二模）三角形ABC中，设=，=，若•（+）＜0，则三角形ABC的形状是（）A ．锐角三角形B．钝角三角形C．直角三角形D．无法确定考点：三角形的形状判断．专题：计算题；解三角形．分析：依题意，可知+=；利用向量的数量积即可判断三角形ABC的形状．解答：解：∵=，=，∴+=+=；∵•（+）＜0，∴•＜0，即||•||•cos∠BAC＜0，∵||•||＞0，∴cos∠BAC＜0，即∠BAC＞90°．∴三角形三角形．故选B．点评：本题考查三角形的形状判断，+=的应用是关键，考查转化思想与运算能力，属于中档题．11．（2015•温江区校级模拟）已知向量，则△ABC的形状为（）A ．直角三角形B．等腰三角形C．锐角三角形D．钝角三角形考点：三角形的形状判断；数量积表示两个向量的夹角．专题：平面向量及应用．分析：由数量积的坐标运算可得＞0，而向量的夹角=π﹣B，进而可得B为钝角，可得答案．解答：解：由题意可得：=（cos120°，（cos30°，sin45°）=（，）•（，）==＞0，又向量的夹角=π﹣B，故cos（π﹣B）＞0，即cosB＜0，故B为钝角，故△ABC为钝角三角形故选D点评：本题为三角形性质的判断，由向量的数量积说明角的范围是解决问题的关键，属中档题．12．（2014秋•景洪市校级期末）在△ABC中，内角A、B、C的对边分别为a、b、c，且，则△ABC的形状为（）A ．等边三角形B．等腰直角三角形C ．等腰或直角三角形D．直角三角形考点：三角形的形状判断．专题：计算题．分析：利用二倍角的余弦函数公式化简已知等式的左边，整理后表示出cosA，再利用余弦定理表示出cosA，两者相等，整理后得到a2+b2=c2，根据勾股定理的逆定理即可判断出此三角形为直角三角形．解答：解：∵cos2=，∴=，∴cosA=，又根据余弦定理得：cosA=，∴=，∴b2+c2﹣a2=2b2，即a2+b2=c2，∴△ABC为直角三角形．故选D．点评：此题考查了三角形形状的判断，考查二倍角的余弦函数公式，余弦定理，以及勾股定理的逆定理；熟练掌握公式及定理是解本题的关键．13．（2014•咸阳三模）△ABC的三个内角A、B、C成等差数列，，则△ABC一定是（）A ．直角三角形B．等边三角形C ．非等边锐角三角形D．钝角三角形考点：三角形的形状判断．专题：计算题；解三角形．分析：由，结合等腰三角形三线合一的性质，我们易判断△ABC为等腰三角形，又由△ABC的三个内角A、B、C成等差数列，我们易求出B=60°，综合两个结论，即可得到答案．解答：解：∵△ABC的三个内角A、B、C成等差数列，∴2B=A+C．又∵A+B+C=180°，∴B=60°．设D为AC边上的中点，则+=2．又∵，∴．∴即△ABC为等腰三角形，AB=BC，又∵B=60°，故△ABC为等边三角形．故选：B．点评：本题考查的知识点是平面向量的数量积运算和等差数列的性质，其中根据平面向量的数量积运算，判断△ABC为等腰三角形是解答本题的关键．14．（2014•奎文区校级模拟）在△ABC中，P是BC边中点，角A、B、C的对边分别是a、b、c，若，则△ABC的形状是（）A．等边三角形B．钝角三角形C．直角三角形D ．等腰三角形但不是等边三角形考点：三角形的形状判断．专题：计算题；解三角形．分析：将c+a+b=转化为以与为基底的关系，即可得到答案．解答：解：∵=﹣，=﹣，∴c+a+b=c﹣a+b（﹣）=即c+b﹣（a+b）=，∵P是BC边中点，∴=（+），∴c+b﹣（a+b）（+）=，∴c﹣（a+b）=0且b﹣（a+b）=0，∴a=b=c．故选A．点评：本题考查三角形的形状判断，突出考查向量的运算，考查化归思想与分析能力，属于中档题．15．（2014秋•正定县校级期末）在△ABC中，tanA•sin2B=tanB•sin2A，那么△ABC一定是（）A ．锐角三角形B．直角三角形C ．等腰三角形D．等腰三角形或直角三角形考点：三角形的形状判断．专题：综合题．分析：把原式利用同角三角函数间的基本关系变形后，得到sin2A=sin2B，由A和B为三角形的内角，得到2A与2B相等或互补，从而得到A与B相等或互余，即三角形为等腰三角形或直角三角形．解答：解：原式tanA•sin2B=tanB•sin2A，变形为：=，化简得：sinBcosB=sinAcosA，即sin2B=sin2A，即sin2A=sin2B，∵A和B都为三角形的内角，∴2A=2B或2A+2B=π，即A=B或A+B=，则△ABC为等腰三角形或直角三角形．故选D．点评：此题考查了三角形形状的判断，熟练掌握三角函数的恒等变换把原式化为sin2A=sin2B是解本题的关键．16．（2014•漳州四模）在△ABC中的内角A、B、C所对的边分别为a，b，c，若b=2ccosA，c=2bcosA则△ABC的形状为（）A ．直角三角形B．锐角三角形C ．等边三角形D．等腰直角三角形考点：三角形的形状判断．专题：计算题．分析：通过两个等式推出b=c，然后求出A的大小，即可判断三角形的形状．解答：解：因为在△ABC中的内角A、B、C所对的边分别为a，b，c，若b=2ccosA，c=2bcosA所以，所以b=c，2bcosA=c，所以cosA=，A=60°，是正三角形．故选C．点评：本题考查三角形的形状的判断，三角函数值的求法，考查计算能力．17．（2014•云南模拟）在△ABC中，若tanAtanB＞1，则△ABC是（）A ．锐角三角形B．直角三角形C．钝角三角形D．无法确定考点：三角形的形状判断．专题：综合题．分析：利用两角和的正切函数公式表示出tan（A+B），根据A与B的范围以及tanAtanB＞1，得到tanA和tanB都大于0，即可得到A与B都为锐角，然后判断出tan（A+B）小于0，得到A+B为钝角即C为锐角，所以得到此三角形为锐角三角形．解答：解：因为A和B都为三角形中的内角，由tanAtanB＞1，得到1﹣tanAtanB＜0，＞0，tanB＞0，即A，B为锐角，所以tan（A+B）=＜0，则A+B∈（，π），即C都为锐角，所以△ABC是锐角三角形．故答案为：锐角三角形点评：此题考查了三角形的形状判断，用的知识有两角和与差的正切函数公式．解本题的思路是：根据tanAtanB＞1和A与B都为三角形的内角得到tanA和tanB都大于0，即A和B都为锐角，进而根据两角和与差的正切函数公式得到tan（A+B）的值为负数，进而得到A+B的范围，判断出C也为锐角．18．（2013秋•金台区校级期末）双曲线=1和椭圆=1（a＞0，m＞b＞0）的离心率互为倒数，那么以a，b，m为边长的三角形是（）A ．锐角三角形B．钝角三角形C．直角三角形D．等腰三角形考点：三角形的形状判断；椭圆的简单性质；双曲线的简单性质．专题：计算题．分析：求出椭圆与双曲线的离心率，利用离心率互为倒数，推出a，b，m的关系，判断三角形的形状．解答：解：双曲线=1和椭圆=1（a＞0，m＞b＞0）的离心率互为倒数，所以，所以b2m2﹣a2b2﹣b4=0即m2=a2+b2，所以以a，b，m为边长的三角形是直角三角形．故选C．点评：本题是中档题，考查椭圆与双曲线基本性质的应用，三角形形状的判断方法，考查计算能力．19．（2014•红桥区二模）在△ABC中，“”是“△ABC为钝角三角形”的（）A ．充分不必要条件B．必要不充分条件C ．充要条件D．既不充分又不必要条件考点：三角形的形状判断．专题：计算题．分析：利用平面向量的数量积运算法则化简已知的不等式，得到两向量的夹角为锐角，从而得到三角形的内角为钝角，即可得到三角形为钝角三角形；反过来，三角形ABC若为钝角三角形，可得B不一定为钝角，故原不等式不一定成立，可得前者是后者的充分不必要条件．解答：解：∵，即||•||cosθ＞0，∴cosθ＞0，且θ∈（0，π），所以两个向量的夹角θ为锐角，又两个向量的夹角θ为三角形的内角B的补角，所以B为钝角，所以△ABC为钝角三角形，反过来，△ABC为钝角三角形，不一定B为钝角，则“”是“△ABC为钝角三角形”的充分条件不必要条件．故选A点评：此题考查了三角形形状的判断，涉及的知识有平面向量的数量积运算，以及充分必要条件的证明，熟练掌握平面向量的数则是解本题的关键．20．（2014秋•德州期末）在△ABC中，若acosA=bcosB，则△ABC的形状是（）A ．等腰三角形B．直角三角形C ．等腰直角三角形D．等腰或直角三角形考点：三角形的形状判断．专题：计算题．分析：利用正弦定理化简已知的等式，再根据二倍角的正弦函数公式变形后，得到sin2A=sin2B，由A和B都为三角形的内角，可得A=B或A+B=90°，从而得到三角形ABC为等腰三角形或直角三角形．解答：解：由正弦定理asinA=bsinB化简已知的等式得：sinAcosA=sinBcosB，∴sin2A=sin2B，∴sin2A=sin2B，又A和B都为三角形的内角，2A+2B=π，即A=B或A+B=，则△ABC为等腰或直角三角形．故选D点评：此题考查了三角形形状的判断，涉及的知识有正弦定理，二倍角的正弦函数公式，以及正弦函数的图象与性质，其中正弦定理很好得解决了三角形的边角关系，利用正弦定理化简已知的等式是本题的突破点．二、填空题（共10小题）（除非特别说明，请填准确值）21．（2014春•沭阳县期中）在△ABC中，已知sinA=2sinBcosc，则△ABC的形状为等腰三角形．考点：三角形的形状判断．专题：计算题．分析：通过三角形的内角和，以及两角和的正弦函数，化简方程，求出角的关系，即可判断三角形的形状．解答：解：因为sinA=2sinBco（B+C）=2sinBcosC，所以sinBcosC﹣sinCcosB=0，即sin（B﹣C）=0，因为A，B，C是三角形内角，所以B=C．三角形的等腰三角形．故答案为：等腰三角形．点评：本题考查两角和的正弦函数的应用，三角形的判断，考查计算能力．22．（2014秋•思明区校级期中）在△ABC中，若a=9，b=10，c=12，则△ABC的形状是锐角三角形．考点：三角形的形状判断．专题：计算题；解三角形．分析：因为c是最大边，所以C是最大角．根据余弦定理算出cosC是正数，得到角C是锐角，所以其它两角均为锐角，由此得到此三角形为锐角三角形．解答：解：∵c=12是最大边，∴角根据余弦定理，得cosC==＞0∵C∈（0，π），∴角C是锐角，由此可得A、B也是锐角，所以△ABC是锐角三角形故答案为：锐角三角形点评：本题给出三角形的三条边长，判断三角形的形状，着重考查了用余弦定理解三角形和知识，属于基础题．23．（2013•文峰区校级一模）已知△ABC中，AB=，BC=1，tanC=，则AC等于2．考点：三角形的形状判断．专题：解三角形．分析：画出图形，利用已知条件直接求出AC的距离即可．解答：解：由题意AB=，BC=1，知C=60°，B=90°，三角形ABC是直角三角形，所以AC==2．故答案为：2．点评：本题考查三角形形状的判断，勾股定理的应用，考查计算能力．24．（2013春•广陵区校级期中）在△ABC中，若2cosBsinA=sinC，则△ABC的形状一定是等腰三角形．考点：三角形的形状判断．专题：计算题．分析：等式即2cosBsinA=sin（A+B），展开化简可得sin（A﹣B）=0，由﹣π＜A﹣B＜π，得A﹣B=0，故三角形ABC是等腰三角形．解答：解：在△ABC中，若2cosBsinA=sinC，即2cosBsinA=sin（A+B）=sinAcosB+cosAsinB，∴sinAcosB﹣cosAsinB=0，即sin（A﹣B）=0，∵﹣π＜A﹣B＜π，∴A﹣B=0，故△ABC 为等腰三角形，故答案为：等腰．点评：本题考查两角和正弦公式，诱导公式，根据三角函数的值求角，得到sin（A﹣B）=0，是解题的关键．25．（2014秋•潞西市校级期末）在△ABC中，已知c=2acosB，则△ABC的形状为等腰三角形．考点：三角形的形状判断．专题：计算题．分析：由正弦定理可得sin（A+B）=2sinAcosB，由两角和的正弦公式可求得sin（A﹣B）=0，根据﹣π＜A﹣B＜π，故A﹣B=0，从而得到△ABC的形状为等腰三角形．解答：解：由正弦定理可得sin（A+B）=2sinAcosB，由两角和的正弦公式可得sinAcosB+cosAsinB=2sinAcosB，∴sin（A﹣B）=0，又﹣π＜A﹣B＜π，∴A﹣B=0，故△ABC的形状为等腰三角形，故答案为等腰三角形．点评：本题考查正弦定理的应用，已知三角函数值求角的大小，得到sin（A﹣B）=0，是解题的关键．26．（2014春•常熟市校级期中）在△ABC中，若，则△ABC的形状是等腰或直角三角形．考点：三角形的形状判断．专题：计算题；解三角形．分析：在△ABC中，利用正弦定理将中等号右端的边化为其所对角的正弦，再由二倍角公式即可求得答案．解答：解：在△ABC中，由正弦定理得：=，∴=，∴⇔=，∴sin2A=sin2B，又A，B为三角形的内角，∴2A=2B或2A+2B=π，∴A=B或A+B=．∴△ABC为等腰三角形或直角三角形．故答案为：等腰或直角三角形．点评：本题考查三角形的形状判断，着重考查正弦定理与二倍角公式的应用，属于中档题．27．（2014春•石家庄期末）在△ABC中，若sin2A+sin2B＜sin2C，则该△ABC是钝角三角形（请你确定其是锐角三角形、直角三角形还是钝角三角形）．考点：三角形的形状判断．专题：解三角形．分析：由正弦定理可得a2+b2＜c2，则再由余弦定理可得cosC＜0，故C为钝角，从而得出结论．解答：解：在△ABC中，若sin2A+sin2B＜sin2C，由正弦定理可得a2+b2＜c2，再由余弦定理可得cosC=＜0，故C为钝角，故△ABC是钝角三角形，故答案为钝角．点评：本题主要考查正弦定理、余弦定理的应用，求出cosC＜0，是解题的关键，属于中档题．28．（2013春•遵义期中）△ABC中，b=a，B=2A，则△ABC为等腰直角三角形．考点：三角形的形状判断．专题：计算题；解三角形．分析：利用正弦定理以及二倍角的正弦函数，求出A，然后求出B即可判断三角形的形状．解答：解：因为△ABC中，b=a，B=2A，所以由正弦定理可知：sinB=sinA，即sin2A=sinA，∴cosA=，∵A是三角形内角，∴A=，则B=，C=，∴△ABC为等腰直角三角形．故答案为：等腰直角．点评：本题主要考查了解三角形的应用和三角形形状的判断．解题的关键是利用正弦定理这一桥梁完成了问题的转化．29．（2013秋•沧浪区校级期末）若△ABC的三个内角满足sinA：sinB：sinC=5：11：13，则△ABC为钝角三角形（填锐角三角形、直角三角形、钝角三角形．）考点：三角形的形状判断．专题：计算题．分析：由正弦定理可得，△ABC的三边之比a：b：c=5：11：13，设a=5k，则b=11k，c=13k，由余弦定理可得cosC＜0，故角C为钝角，故△ABC为钝角三角形．解答：解：由正弦定理可得，△ABC的三边之比a：b：c=5：11：13，设a=5k，则b=11k，c=13k，由余弦定理可得cosC==﹣＜0，故角C为钝角，故△ABC形，故答案为：钝角三角形．点评：本题考查正弦定理、余弦定理的应用，求出cosC＜0，是解题的关键．30．（2014春•宜昌期中）在△ABC中，sinA=2cosBsinC，则三角形为等腰三角形．考点：三角形的形状判断．专题：计算题．分析：由三角形的内角和及诱导公式得到sinA=sin（B+C），右边利用两角和与差的正弦函数公式化简，再根据已知的等式，合并化简后，再利用两角和与差的正弦函数公式得到sin（B﹣C）=0，由B与C都为三角形的内角，可得B=C，进而得到三角形为等腰三角形．解答：解：∵A+B+C=π，即A=π﹣（B+C），∴sinA=sin（B+C）osBsinC，又sinA=2cosBsinC，∴sinBcosC+cosBsinC=2cosBsinC，变形得：sinBcosC﹣cosBsinC=0，即sin（B﹣C）=0，又B和C都为三角形内角，∴B=C，则三角形为等腰三角形．故答案为：等腰三角形点评：此题考查了三角形形状的判断，涉及的知识有诱导公式，两角和与差的正弦函数公式，以及特殊角的三角函数值，熟练掌握公式是解本题的关键，同时注意三角形内角和定理及三角形内角的范围的运用．。

静安区2013-2014学年度第一学期期末质量抽查初三英语试卷初三英语试卷（满分150分，考试时间100分钟）2014.01 考生注意：本卷有7大题，共94小题。

试题均采用连续编号，所有答案务必按照规定在答题卡上完成，做在试卷上不给分。

做在试卷上不给分。

Part 1 Listening （第一部分听力）（第一部分 听力）I. Listening comprehension (听力理解) (共30 分)A.Listen and choose the right picture (根据你听到的内容，选出相应的图片根据你听到的内容，选出相应的图片) (6 分) 1. _________ 2. _________ 3._______ 4._______. 5._______ 6._______ B. Listen to the dialogue and choose the best answer to the question you hear (根据你听到的对话和问题，选出最恰当的答案)：(8分) 7. A）15 B）18 C) 21 D) 33 8. A) Surprised B) Happy C) W orried D) Excited 9. A）Kitty B) Gary C) Mike D) Mary 10. A) A student B) A manager C) A worker D) A math teacher 11. A) Shanghai B) Beijing C) Korea D) Thailand 12. A) Because it is hotter B) Because it is sweeter C) Because it tastes better D) Because it looks nicer 13. A) Before the meeting B) After the meeting C) In the middle of the meeting D) At the end of the meeting 14. A) He‟s been away from home for a long time B) He has lost some money C) He works hard day and night D) He works with his wife 你听到的短文内容，符合的用“T”表示，不符合的用“F”表示): (6分) 15. Tom attended classes every evening to improve himself. 16. After Tom was offered a good job, he spent more time with his family. 17. With his hard work, Tom managed to get the position of manager. 18. Tom decided to hire s servant to help his wife with the housework. 19. Tom lived with his family in the beautiful house for the rest of his life. 20. Tom was so tired that he got up very late the next day. D. Listen to the passage and fill in the blanks (听短文填空，完成下列内容。

静安区学年第一学期质量检测试卷本试卷共页，满分分．考试时间为分钟。

全卷包括六大题，第一、二大题为单项选择题，第三大题为多项选择题，第四大题为填空题，第五大题为实验题，第六大题为计算题，考生注意：、答卷前，务必用钢笔或圆珠笔在答题纸上清楚填写学校、班级、姓名、考号，并用铅笔在答题卡上正确涂写考号．、第一、第二和第三大题的作答必须用铅笔涂在答题纸上相应的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。

第四、第五和第六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）。

、第、、、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（共分，每小题分。

每小题只有一个正确选项。

）．刘翔能够获得雅典奥运会米跨栏冠军，是由于他在这米中（）某时刻的瞬时速度大（）撞线时的瞬时速度大（）平均速度大（）起跑时的加速度大．如图，是某逻辑电路，当和分别输入和时，则和输出信号是（）和（）和（）和（）和．一物块静止在粗糙的水平桌面上。

从某时刻开始，物块受到一方向不变的水平拉力作用。

假设物块与桌面间的最大静摩擦力等于滑动摩擦力。

以表示物块的加速度大小，表示水平拉力的大小。

能正确描述与之间关系的图像是．如图所示，一均匀木棒可绕过点的水平轴自由转动，现有一方向不变的水平力作用于棒的点，使棒从竖直位置缓慢转到偏角θ<的某一位置，设为力对转轴的力矩，则在此过程中（）不断变大，不断变小（）不断变大，不断变大（）不断变小，不断变小（）不断变小，不断变大．关于点电荷和电场线，下列说法中正确的是（）点电荷和电场线可以等效替代它们各自描述的对象（）点电荷是物理模型，而电场线不是理想模型（）电场线上任一点的切线方向与点电荷在该点所受电场力的方向相同（）点电荷和电场线都不是真实存在的．足球的容积为．足球内已有的气体与外部大气的温度相同，压强等于大气压强，现再从球外取体积为△的空气充入球内，使足球内的压强增大到，设足球容积保持不变，充气过程气体温度不变，则△为（）（）（）（）．如图所示，电源内电阻为，定值电阻的阻值等于，滑动变阻器的最大阻值是。

静安区2013学年第二学期教学质量检测高三年级地理试卷（考试时间：120分钟，满分：150分） 2014.41、答题前，考生务必将自己的学校、班级、姓名、准考证号填写清楚，并认真核对准考证号与填涂数字的一致。

2、选择题部分必须使用2B铅笔填涂，非选择题部分使用0、5毫米以上蓝、黑色字迹的钢笔、圆珠笔或签字笔书写，字体工整、笔迹清楚。

3、请按照题号顺序在各题目的答题区内作答，超出答题区域书写的答案无效；在草稿纸、试题卷上答题无效。

4、保持答题纸清洁，不折叠、不破损。

5、满分150分。

考试时间120分钟。

一、选择题（共60分，每小题2分。

每小题只有一个正确答案）（一）中国南极第四个考察站——泰山站于2014年2月建成。

读“我国南极考察站分布图”。

1．四个考察站中，地理坐标最接近62°S，59°W的是A．长城站B．中山站 C．昆仑站 D. 泰山站2．结合所学知识分析下列叙述正确的是A．中山站常年吹东北风 B．昆仑站气压最高C．泰山站光照时间最长 D．长城站降水量最大（二）下图为我国某地地形发育示意图。

3．图中甲地反映的地形主要分布在我国的A．黄土高原 B．云贵高原 C．内蒙古高原 D．准噶尔盆地4．图中地形的发育主要受到的外力影响是A．流水侵蚀 B．风力侵蚀 C．冰川侵蚀 D．流水沉积5．在甲地建设铁路面临的主要困难最可能是A．冻土和冰川 B．地质基础不稳固 C．沼泽、软土 D．流沙和水土流失（三）读①②③三地气温与降水统计表。

6A．甲—③；乙—②；丙—①B．甲—②；乙—①；丙—③C．甲—②；乙—③；丙—①D．甲—③；乙—①；丙—②7．右图中甲乙丙三地气候的叙述，正确的是A．三地7月平均气温均高于4月B．三地4月降水量均高于7月C．甲、乙两地气候差异的主导因素是地形D．甲、丙两地气候差异的主导因素是洋流（四）俄罗斯南部城市索契是2014年冬奥会举办城市。

8．索契是世界上纬度最高的亚热带气候分布区，其主要原因是①北部有高大山脉阻挡，冬季削弱了南下的冷空气②临近水域，冬季对沿岸有增温作用③终年受西风带控制，冬不冷夏不热④地处大陆内部，冬季太阳辐射强A．①② B．②③ C．③④ D．①④9．索契有利于冬奥会室外项目比赛的主要条件是A．依山面海，交通发达 B．冬季降水量大，山地积雪多C．位于大洲分界线，地理位置优越 D．沿岸有寒流经过，降温增湿（五）下图为我国某地区某月14日6时的气压形势，L为低压。

静安区2014学年第一学期高三年级高考数学模拟理卷（试卷满分150分 考试时间120分钟） 2014.12一、填空题（本大题满分56分）本大题共有14题，考生应在答题纸相应编号的空格内直接填写结果，每个空格填对得4分，否则一律得零分．1．已知集合{}0,2>==x x y y M ，{})2lg(2x x y x N -==，则=N M . 答案：)2,0(考点：集合的描述法备考建议：强调，对集合描述法要区分集合的代表元。

2．设8877108)1(x a x a x a a x ++++=- ，则=++++8710a a a a . 答案：25628=考点：二项式定理解法：将1x =-代入式子中备考建议：让学生理解，二项式题型中的赋值法，并补充一些通过某一项系数判断二项式次数的题型。

3．不等式01271<--x 的解集是 . 答案：)4,21(考点：分式不等式的解法备考建议：分式不等式建议通分后再解不等式，易错点是：不等式性质中，若要两边同乘除，要注意所乘所除数的正负性。

4．如图，在四棱锥ABCD P -中，已知⊥PA 底面ABCD ，1=PA ，底面ABCD 是正方形，PC 与底面ABCD 所成角的大小为6π，则该四棱锥的体积是 . 答案：12考点：锥体体积的求法备考建议：让学生熟练掌握各简单几何体面积与体积的公式。

5．已知数列{}n a 的通项公式1222+-+=n n n a （其中*N n ∈），则该数列的前n 项和=n S .答案：)212(4n n-考点：数列分组求和，等比数列求和。

备考建议：此类题型要让学生观察数列通项公式的结构，从而选择正确的求和方法。

同时，也可带领回忆一下倒序相加、错位相减、裂项相消的常用求和方法及其适用情况。

AB CDP6．已知两个向量a ，b 的夹角为303=，b 为单位向量，t t )1(-+=, 若c ⋅=0，则t = . 答案： 2考点：向量的数量积：解法：由于b 与c 、a 、b 的数量积都有联系，故等式两边同乘上一个b 。

上海市各区2014届高三数学（理科）一模试题分类汇编函数 2014.01.23（浦东新区2014届高三1月一模，理）6.已知函数的反函数为，则11()24x x f x -=1()f x -___________.1(12)f -=（ 6. 2log 3（杨浦区2014届高三1月一模，理）6．若函数的反函数为，则 ．()23-=x x f ()x f 1-()=-11f 6. 1 ； （（嘉定区2014届高三1月一模，理）1．函数的定义域是_____________．)2(log 2-=x y 1． ),2(∞+（徐汇区2014届高三1月一模，理）7. 若函数()f x 的图像经过(0,1)点，则函数()3f x +的反函数的图像必经过点.长宁区2014届高三1月一模，理）1、设是上的奇函数，当时，，则 ()x f R 0≤x ()x x x f -=22()=1f 1、 3-（浦东新区2014届高三1月一模，理）17.已知函数则,1)(22+=x x x f （ ）()()()111112(2013)20142320132014f f f f f f f f ⎛⎫⎛⎫⎛⎫⎛⎫+++++++++= ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭K L (A) 2010 (B) 2011 (C) 2012 (D) 2013 21212121 17. D （普陀区2014届高三1月一模，理）6. 函数)1(log )(2-=x x f )21(≤<x 的反函数 .=-)(1x f 6. （不标明定义域不给分）； =-)(1x f )0(21≤+x x（嘉定区2014届高三1月一模，理）13．已知函数是偶函数，直线⎪⎧≥++=,0,12)(2x x ax x f 与函数的图像自左t y =)(x f至右依次交于四个不同点、、、，若，则实数的值为________．A B C D ||||BC AB =t 13． 47（嘉定区2014届高三1月一模，理）3．已知函数存在反函数，若函数的)(x f y =)(1x f y -=)1(-=x f y 图像经过点，)1,3(则的值是___________．)1(1-f 3． 2（杨浦区2014届高三1月一模，理）8. 已知函数，若，则 ()lg f x x =()1f ab =22()()f a f b +=_________.8. 2；（浦东新区2014届高三1月一模，理）14. 已知函数，对任意都有**(),,y f x x y =∈∈N N *n ∈N ，且是增函数，则 [()]3f f n n =()f x (3)f =14.6（长宁区2014届高三1月一模，理）3、已知函数的图像关于直线对称，则5()2x f x x m -=+y x =m =3、 1-（普陀区2014届高三1月一模，理）14.已知函数，若方程有且仅有两⎩⎨⎧<+≥-=0),1(0,2)(x x f x a x f x 0)(=+x x f 个解，则实数的取值范围是 .a 14.；2<a （徐汇区2014届高三1月一模，理）14. 定义区间(),c d 、[),c d 、(],c d 、[],c d 的长度均为()d c d c ->.已知实数(),a b a b >.则满足111x a x b +≥--的x 构成的区间的长度之和为 .14. 2（杨浦区2014届高三1月一模，理）18．定义一种新运算：，已知函数,(),()b a b a b a a b ≥⎧⊗=⎨<⎩，若函数24()(1log f x x x =+⊗ 恰有两个零点，则的取值范围为 ………（ ）.()()g x f x k =-k ． ． ． ． )(A (]1,2)(B (1,2))(C (0,2))(D (0,1)18.理B ；（嘉定区2014届高三1月一模，理）18．设函数的定义域为，若存在闭区间，使得函数)(x f D D b a ⊆],[满足：①)(x f )(x f 在上是单调函数；②在上的值域是，则称区间是函],[b a )(x f ],[b a ]2,2[b a ],[b a 数的“和谐区间”．下列结论错误的是………………………………………（ ）)(x f A ．函数（）存在“和谐区间”2)(x x f =0≥x B ．函数（）不存在“和谐区间”x e x f =)(R ∈x C ．函数）存在“和谐区间”14)(2+=x x x f (0≥x D ．函数（，）不存在“和谐区间”⎪⎭⎫ ⎝⎛-=81log )(x a a x f 0>a 1≠a 18．D （长宁区2014届高三1月一模，理）18、函数的定义域为，值域为，变动时，方程表示的图形可2x y =[,]a b [1,16]a ()b g a =以是 （）A ．B ．C ．D ．18、B （普陀区2014届高三1月一模，理）23.（本题满分18分） 本大题共有3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.定义在上的函数，如果对任意，恒有（，）成()0,+∞()f x ()0,x ∈+∞()()f kx kf x =2k ≥*k N ∈立，则称为阶缩放函数.()f x k （1）已知函数为二阶缩放函数，且当时，，求的值；()f x (]1,2x ∈()121log f x x=+(f （2）已知函数为二阶缩放函数，且当时，()f x (]1,2x ∈()f x =在上无零点；()y f x x =-()1,+∞（3）已知函数为阶缩放函数，且当时，的取值范围是，求在()f x k (]1,x k ∈()f x [)0,1()f x （）上的取值范围.(10,n k +⎤⎦n N ∈23. （本题满分18分） 本大题共有3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.解：（1）由得，………………2分]2,1(2∈212log 1)2(21=+=f 由题中条件得……………………4分1212)2(2)22(=⨯==f f （2）当（）时，，依题意可得：]2,2(1+∈i i x i N ∈(]1,22i x ∈分()222222222i i x x x f x f f f ⎛⎫⎛⎫⎛⎫====== ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭ 方程或，与均不属于……8分0)(=-x x f ⇔x =⇔0x =2i x =0i 2]2,2(1+i i 当（）时，方程无实数解。

1 上海市各区2014届高三数学（理科）一模试题分类汇编：概率和统计 Word 版含答案（长宁区2014届高三1月一模，理）8、不透明的袋子中装有除颜色不同其它完全一样的黑、白小球共10只，从中任意摸出一只小球得到是黑球的概率为25．则从中任意摸出2只小球，至少得到一只白球的概率为 ．8、1513 （长宁区2014届高三1月一模，理）10、已知总体的各个体的值由小到大依次为2，3，3，7，a ，b ,12，13.7，18.3，20，且总体的中位数为10.5，若要使该总体的方差最小，则._______=ab 10、25.10（浦东新区2014届高三1月一模，理）5.甲校有3600名学生，乙校有5400名学生，丙校有1800名学生.为统计三校学生某方面的情况，计划采用分层抽样法，抽取一个样本容量为90人的样本，则应在甲校抽取的学生数是___________.5. 30（徐汇区2014届高三1月一模，理）8. 某小组有10人，其中血型为A 型有3人，B 型4人，AB 型3人，现任选2人，则此2人是同一血型的概率为 .（结论用数值表示）（浦东新区2014届高三1月一模，理）11. 某学校要从5名男生和2名女生中选出2人作为社区志愿者，若用随机变量ξ表示选出的志愿者中女生的人数，则随机变量ξ的数学期望E ξ=_____（结果用最简分数表示）.11. （理）4712. 1＜a ＜4 （虹口区2014届高三1月一模，理）4、从长度分别为1、2、3、4的四条线段中任意取三条，则以这三条线段为边可以构成三角形的概率是 ．8、1513 （徐汇区2014届高三1月一模，理）13. 一个五位数abcde 满足,,,a b b c d d e <>><且,a d b e >>(如37201,45412)，则称这个五位数符合“正弦规律”.那么，共有 个五位数符合“正弦规律”.。

上海静安区2023-2024学年第一学期期末教学质量调研高三数学试卷本试卷满分150分，考试时间120分钟．一、填空题（本大题共12小题，满分54分）第1小题至第6小题每个空格填对得4分，第7小题至第12小题每个空格填对得5分，考生应在答题纸的相应编号后填写答案，否则一律得零分．1.准线方程为10x +=的抛物线标准方程为______．2.32x x ⎛⎫+ ⎪⎝⎭的二项展开式中x 的系数为______.3.若一个圆柱的底面半径和母线长都是1，则这个圆柱的体积是______.4.已知R a ∈，i 是虚数单位，1i a -的虚部为______.5.计算123ii +∞=⎛⎫=⎪⎝⎭∑_____________．6.某果园种植了222棵苹果树，现从中随机抽取了20棵苹果树，算得这20棵苹果树平均每棵产量为28kg ，则预估该果园的苹果产量为______kg ．7.下列幂函数在区间()0,∞+上是严格增函数，且图象关于原点成中心对称的是______（请填入全部正确的序号）．①12y x =；②13y x =；③23y x =；④13y x-=．8.若不等式35x x a-+-≥对所有实数x 恒成立，则实数a 的取值范围是______．9.如图，在四棱锥P ABCD -中，PA ⊥平面ABCD ，底面ABCD 是矩形，||||2AP AB ==，||4AD =，E 是BC 上的点，直线PB 与平面PDE 所成的角是3arcsin6，则BE 的长为______.10.不等式2log 42x x +<的解集为______.11.在国家开发西部的号召下，某西部企业得到了一笔400万元的无息贷款用做设备更新．据预测，该企业设备更新后，第1个月收入为20万元，在接下来的5个月中，每月收入都比上个月增长20%，从第7个月开始，每个月的收入都比前一个月增加2万元．则从新设备使用开始计算，该企业用所得收入偿还400万无息贷款只需______个月．（结果取整）12.记22()ln 2f x x x kx k =+-+，若存在实数a b 、，满足122a b ≤<≤，使得函数()y f x =在区间[],a b 上是严格增函数，则实数k 的取值范围是______.二、选择题（本大题共4小题，满分18分）第13题、14题各4分，第15题、16题各5分．每题有且仅有一个正确答案，考生应在答题纸的相应编号上，将代表答案的小方格涂黑．13.已知α：1x >，β：11x <，则α是β的（）A.必要非充分条件B.充分非必要条件C.充要条件D.既非充分又非必要条件14.设α是第一象限的角，则2α所在的象限为（）A.第一象限B.第三象限C.第一象限或第三象限D.第二象限或第四象限15.教材在推导向量的数量积的坐标表示公式“1212a b x x y y ⋅=+（其中1122(,),(,)x y x y ==a b ）”的过程中，运用了以下哪些结论作为推理的依据（）①向量坐标的定义；②向量数量积的定义；③向量数量积的交换律；④向量数量积对数乘的结合律；⑤向量数量积对加法的分配律．A.①③④ B.②④⑤C.①②③⑤D.①②③④⑤16.记点P 到图形C 上每一个点的距离的最小值称为点P 到图形C 的距离，那么平面内到定圆C 的距离与到定点A 的距离相等的点的轨迹不可能是（）A.圆B.椭圆C.双曲线的一支D.直线三、解答题（本大题共5题，满分78分）解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤．17.记22()sin cos cos ()f x x x x x x λ=-++∈R ，其中λ为实常数．（1）求函数()y f x =的最小正周期；（2）若函数()y f x =的图像经过点π,02⎛⎫⎪⎝⎭，求该函数在区间20,π3⎡⎤⎢⎥⎣⎦上的最大值和最小值．18.甲、乙两人每下一盘棋，甲获胜的概率是0.4，甲不输的概率为0.9．（1）若甲、乙两人下一盘棋，求他们下成和棋的概率；（2）若甲、乙两人连下两盘棋，假设两盘棋之间的胜负互不影响，求甲至少获胜一盘的概率．19.已知双曲线C ：2212x y -=，点M 的坐标为()0,1．（1）设直线l 过点M ，斜率为12，它与双曲线C 交于A 、B 两点，求线段AB 的长；（2）设点P 在双曲线C 上，Q 是点P 关于y 轴的对称点．记k MP MQ =⋅，求k 的取值范围．20.如下图，某公园东北角处有一座小山，山顶有一根垂直于水平地平面的钢制笔直旗杆AB ，公园内的小山下是一个水平广场（虚线部分）．某高三班级数学老师留给同学们的周末作业是：进入该公园，提出与测量有关的问题，在广场上实施测量，并运用数学知识解决问题．老师提供给同学们的条件是：已知10AB =米，规定使用的测量工具只有一只小小的手持激光测距仪（如下图，该测距仪能准确测量它到它发出的激光投射在物体表面上的光点之间的距离）．（1）甲同学来到通往山脚下的笔直小路l 上，他提出的问题是：如何测量小山的高度？于是，他站在点C 处，独立的实施了测量，并运用数学知识解决了问题．请写出甲同学的解决问题方案，并用假设的测量数据（字母表示）表示出小山的高度H ；（2）乙同学是在一阵大风过后进入公园的，广场上的人纷纷议论：旗杆AB 似乎是由于在根部A 处松动产生了倾斜．她提出的问题是：如何检验旗杆AB 是否还垂直于地面？并且设计了一个不用计算就能解决问题的独立测量方案．请你写出她的方案，并说明理由；（3）已知（1）中的小路l 是东西方向，且与点A 所确定的平面垂直于地平面．又已知在（2）中的乙同学已经断定旗杆AB 大致向广场方向倾斜．如果你是该班级的同学，你会提出怎样的有实际意义的问题？请写出实施测量与解决问题的方案，并说明理由（如果需要，可通过假设的测量数据或运算结果列式说明，不必计算）．21.如果函数()y f x =满足以下两个条件，我们就称()y f x =为L 型函数．①对任意的()0,1x ∈，总有()0f x >；②当12120,0,1x x x x >>+<时，总有1212()()()f x x f x f x +<+成立．（1）记21()2g x x =+，求证：()y g x =为L 型函数；（2）设R b ∈，记()ln()p x x b =+，若()y p x =是L 型函数，求b 的取值范围；（3）是否存在L 型函数()y r x =满足：对于任意的()0,4m ∈，都存在()00,1x ∈，使得等式0()r x m =成立？请说明理由．参考答案一．填空题：1、24y x =；2、6；3、π；4、211a +；5、2；6、6216；7、②；8、(,2]-∞；9、2；10、()0,4；11、10；12、9,4⎛⎫-∞ ⎪⎝⎭；二．选择题：13、B ；14、C ；15、D ；16、D ；三．解答题：17、（1）()cos 22f x x x =-+π2sin 26x ⎛⎫=- ⎪⎝⎭λ+．∴函数()y f x =的最小正周期为π．（2） π102f λ⎛⎫=+=⎪⎝⎭，∴1λ=-，则π()2sin 216f x x ⎛⎫=-- ⎪⎝⎭．令2π6x t -=，则π7π,66t ⎡⎤∈-⎢⎥⎣⎦．当ππ266x -=-或7π6，即0x =或2π3时，()min 2f x =-．当ππ262x -=，即π3x =时，max ()1f x =．18、设事件A 表示甲获胜，事件B 表示和棋，事件C 表示甲不输．则C A B = ．因为和棋与获胜是互斥的，由概率的可加性，得()()()()P C P A B P A P B ==+ ．因为()0.9,()0.4P C P A ==，所以()0.90.40.5.P B =-=（2）设事件A 表示甲获胜，则A 表示甲未获胜.设下两次棋至少有一次获胜的事件为E ，则()()()E A A A A A A =⋂⋃⋂⋃⋂，因为两盘棋之间的胜负互不影响，且至少有一次获胜包括的三种情况是互斥的.所以()0.40.4(10.4)0.40.4(10.4)0.64P E =⨯+-⨯+⨯-=19、（1）直线l 的方程为112y x =+．由方程组2211,21,2y x x y ⎧=+⎪⎪⎨⎪-=⎪⎩得2480x x --=．设()()1122,,,A x y B x y ,则12124,8x x x x +==-，AB ===．（2）设点(),P x y ，则点Q 的坐标为(),x y -．(),1MP x y =- ，(),1MQ x y =--，∴()221k x y =-+-222221y y y =--+-+2221(1)y y y =---=-+．因为R y ∈，所以(],0k ∞∈-．20、（1）解一：(1)如图1，设点A 在水平面的投影点为O ．用测距仪测得CA m =，CB n =．在ABC 中，22100cos 20m n BAC m +-∠=，在AOC 中，22100cos 20m n OAC m +-∠=-，所以22100cos 20n m H m OAC --=∠=．解二：如图2，在平面ABC 上，以点C 为原点，向量CO为x 轴，建立平面直角坐标系xCy ，设点(),A x H ，则(),10B x H +，用测距仪测得CA m =，CB n =，则()22222210x H mx H n⎧+=⎪⎨++=⎪⎩，解得22100.20n m H --=（2）如图，用电子尺测得CA m =，CB n =，在广场上从点C 移动至点D ，使得DB n =，再移至点E ，使得EA n =，此时再测量DA EA 、，若CA DA EA ==，则可知旗杆AB 垂直于地面，否则就是倾斜了．理由如下：已知CB DB =，CA DA =，设点M 是CD 的中点，则在等腰CBD △中，BM CD ⊥．同理AM CD ⊥，又,AM BM ⊂平面ABM ，所以AM ⊥平面ABM ；又因为AB ⊂平面ABM ，故AB CD ⊥．同理可证AB DE ⊥．综上所述，旗杆AB 垂直于地面．（3）提问：旗杆AB 向哪个方向倾斜多少角度？说明：用AB 在地平面上的投影来刻画AB 的倾斜方向是合理的，也可以采用在广场上确定一个位于在地平面上投影上的点来刻画，用AB 与小路l 的夹角刻画扣1分．关于如何刻画AB 倾斜多少角度的问题，既可以用AB 与垂直于地面的直线所成角的大小，也可以用AB 与地平面所成角的大小来刻画．解答方案1：如图，在地面画出离点A 距离相等的点的轨迹圆O ，再在圆O 上找到离点B 距离最近的点D ，作BH 垂直于地面，垂足为H ，则ABH ∠的大小就是旗杆AB 倾斜角度．理由如下：先证明OH 与圆O 的交点既是点D ．只需证明：对于圆O 上任意一点M ，MB DB >．因为在MHD 中，ODM OMD ∠>∠，所以MH DH >，故MB DB >．如图5，从图4中的点D 向点A 的方向走到点P ，放置一个物体，测得PD 、PA 、DA 的长，利用余弦定理可得ADO ∠的大小．同理可得BDO ∠的大小．因此，可以求得图4中的BH 、AO 、DH 、DO 的长．在COD △中，三边已知，利用余弦定理可求得COD ∠，即旗杆AB 向西偏南COD ∠的方向倾斜．又由于DH 、DO 已求得，故AB 倾斜角度为arccos10DO DH-．测量倾斜角的大小方案2：如图5，从点D 向点A 的方向走到点P ，测得PD 、PA 、DA 的长，利用余弦定理可得ADO ∠的大小，从而求得A 点的高度1h ．同理可求得B 点的高度2h ．如图，1210h h +-即是由于旗杆倾斜旗杆顶点所下降的高度1B G．所以21AG h h =-，在Rt ABG △中，21arccos 10h h BAG -∠=即为所求，测量倾斜角的大小方案3：在图5中，以点O 为原点，以OA 为y 轴建立平面直角坐标系xOy ，则容易求出点A 与点B 的坐标(),A A x y 与(),B B x y ，故AB 的倾斜角为arctanB AB Ay y x x --．21、（1）当()0,1x ∈时，1()02g x >>，当1>0x ，20x >，121x x +<时，()()2121212g x x x x +=++，()()2212121g x g x x x +=++，则()()()()2221212121212111222g x g x g x x x x x x x x +-+=++-+-=-12142x x -=，121x x >+≥∴12140x x ->，∴()()()1212g x g x g x x +>+，∴21()2g x x =+为L 型函数.（2）当()0,1x ∈时，由()()ln 00p x b >+≥得1b ≥，当1>0x ，20x >，121x x +<时，()()1212ln p x x x x b +=++，()()()()1212ln ln p x p x x b x b +=+++，由()()()1212p x x p x p x +<+，得()()()1212ln ln ln x x b x b x b ++<+++，即()()1212x x b x b x b ++<++，即()2121212x x b x x b x x b ++<+++，即()()212121210b b x x x x x x ++-+-+>，令()()()21212121h b b b x x x x x x =++-+-+，则对称轴()12110,22x x b -+⎛⎫=∈ ⎪⎝⎭，所以()h b 在[)1,+∞上的最小值为()1h ，只要()10h >，则()0h b >，因为()()()2121212111h x x x x x x =++-+-+120x x =>，所以[)1,b ∈+∞．（3）存在，举例1：()r x =理由如下：当()0,1x ∈时，()()04r x ∈,符合()0r x >；当1>0x ，20x >，121x x +<时，()12r x x +=()()12r x r x +=，212x x =++，21212x x x x =+<++，故22<，∴<()()()1212r x x r x r x +<+，即()y r x =是L 型函数，且对任意的()0,4m ∈，存在()00,1x ∈，使得等式0()r x m =成立；举例2：()()1r x x =+；理由如下：当()0,1x ∈时，()()04r x ∈,，符合()0r x >，当1>0x ，20x >，121x x +<时，()()12121r x x x x +=++，()()()()121211r x r x x x +=+++，()()121212121111x x x x x x x x ++=+++>++ ，∴()()()1212111x x x x ++<+++，即()()()1212r x x r x r x +<+，即()y r x =是L 型函数，且对任意的()0,4m ∈，都存在()00,1x ∈，使得等式0()r x m =成立.由此可知存在L 型函数()y r x =满足：对于任意的()0,4m ∈，都存在()00,1x ∈，使得等式0()r x m =成立.。

2014年上海静安区高三英语一模(附答案)高三年级英语试卷（考试时间：120分钟，满分：150分）2014.01I. 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 conversation 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. In a sports center. B. At a birthday party.C. At a gift shop.D. In a department store.2. A. Turn the alarm off. B. Live near his work place.C. Go to bed earlier.D. Move his alarm clock far away.3. A. The man could watch the ballet with her.B. She happened to have bought two tickets.C. She can get a ticket for the man.D. Her sister can go to watch the ballet herself.4. A. She doesn’t know anything about it.B. A good name hasn’t been found for it.C. They decided to postpone building it.D. It hasn’t been designed yet.5. A. At 5:35. B. At 5:25. C. At 4:40. D. At 5:50.6. A. Chicken is tasty. B. Roast beef is tasty.C. Not very satisfactory.D. Very satisfactory.7. A. He will continue his work on vacation.B. Papers piled while he was on vacation.C. He has too much work to do.D. He has made his vacation plans.8. A. Dr Smith usually sees patients at once.B. Dr Smith is very busy on Mondays.C. Dr Smith didn’t put the man on his schedule.D. Dr Smith is hard to see.9. A. Colleagues. B. Husband and wife.C. Employer and employee.D. Mother and son.10. A. The man should change his plan.B. The man can go camping tomorrow.C. Weather forecasts are not available.D. The man won’t have to go camping.Section BDirections: In Section B, you will hear two short passages, and you will be asked three questions on each of the passages. The passages will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. The poor places are getting richer. B. The rich places are getting richer.C. The poor places are getting poorer.D. Both B and C.12. A. The poor are unemployed. B. All the poor have no land.C. The poor have no houses in big cities.D. There is no hope for the poor in the village.13. A. Rural unemployment. B. Urban unemployment.C. No housing in the villages.D. No foreign aid in the villages.Questions 14 through 16 are based on the following passage.14. A. She could never overcome difficulties.B. She had problems with other children.C. She had trouble communicating with others.D. She suffered from severe learning disability.15.A. A person whose experience can inspire others,B. A person with a remarkable memory,C. A person who has a better understanding,D. A person with special education,16.A. Always listen to doctor’s advice. B. Never give up in face of difficulties.C. Always get encouragement from others.D. Never compare yourself with others.Section CDirections: In Section C, you will hear two longer conversations. The conversations will be read twice. After you hear each conversation, you are required to fill in the numbered blanks with information you have heard. Write your answers on your answer sheet.Blanks 17 through 20 are based on the following conversation.Complete the message. Write ONE WORD for each answer.The man’s problem Doesn’t know how to__17__ the numbers on.The woman’s suggestion First, press Shift F- __18__.Second, press __19__.Thirdly, print.Explanation Numbers don’t show up on the screen but they willbe on the printed __20__.Blanks 21 through 24 are based on the following conversation.Complete the form. Write NO MORE THAN THREE WORDS for each answer.What does the man complain about?The quality of ____21____.What is the problem?The screen always ___22___.Where is the repair center?On the 7th Floor, ____23____.What will the company do if there is anyTo let the customer ____24____.problem concerning the quality?II. Grammar and VocabularySection ADirections: After reading the passages below, fill in the blanks to make the passages coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.(A)Mother Teresa was born in Yugoslavia, on August 27, 1910. She attended the government school near her home until she was eighteen. At that time, some doctors and nurses from Yugoslavia were working in India, and they often (25)_______(write) to the school about their work. She decided to join them one day.When she left school, she first went to Britain. Then a year later she went to India, where she began (26) _______ (train) to be a teacher. After training, she was sent to Calcutta, (27)_______she taught geography at a school and soon after became headmistress.However, (28) _______she loved teaching; in 1946 Mother Teresa left the school and went to work in the poor parts of Calcutta. Later she was trained to become a nurse in Patna, and then began her work helping the poor and comforting the dying in the streets of the city. Slowly, (29) _______ came to help her, and her work spread to other parts of India.Mother Teresa is now a well-known person. Many photos (30) _______ (take) of her, (31)_______ she travels around the world to open new schools and hospitals in poor countries. In 1979, she was given the Nobel Prize for the lifetime of love and service she has given to the poor.(B)On any collecting trip, obtaining the animals is, as a rule, the simplest part of the job. As soon as the local people discover that you are willing to buy live wild creatures, the stuff comes (32) _______ (pour) in; ninety percent is, of course, the more common types, but they do bring (33) _______occasional rarity. If you want the really rare stuff, you generally have to go out and find ityourself.The chief difficulty you have when you have got a newly (34)______ (catch) animal is not so much the shock it might be suffering, but the fact (35)_______being caught forces it to exist close to a creature it regards as an enemy of the (36)_______ (bad) possible sort: yourself. On many occasions an animal may take beautifully to being in a cage but (37) _______ (get) used to the idea of living with people is another matter. This is the difficulty you (38) _______only deal with by patience and kindness. For month after month an animal may try to bite you every time you approach its cage, (39 )_______you despair of ever making a favorable impression on it. Then, one day, sometimes without any preliminary warning, it will trot forward and take food from your hand, or allow you to tickle it behind the ears. (40)_______ such moments you feel that all the waiting in the world was worthwhile.Section BDirections: Complete the following passage by using the words in the boxEach word can only be used once. Note that there is one word more than you need.A. changeB. repeatedlyC. dependentD. limitedE. flexibleF. properG. respect H. concepts I. explained J. freely K. figureIn giving advice, you must learn to understand the person’s level of judgment. There will be some people who come to you with unconnected knowledge, talking about, say, Vitamin B12 and other modern__41__. When they ask advice, begin at that level. Unless you start at their own level, they will not be able to understand. You must explain to them in scientific terms the effects of the food they eat, and how they need to __42__it.You have to train yourself to be very__43__. Staying at one level is not being a free man. If we stay at a very high level all the time, that is not practical. A limitless person goes __44__from one level of thinking to another, according to his circumstances. To do this we must get rid of our unwillingness to change our way of thinking or behavior, become friends with everyone, and have the same loving feelings for everyone. Then we can give advice to all kinds of people. If there is someone or something you dislike, you are still__45__, and your ability to give advice is reduced. For anyone, the same. A free person acts like that.You cannot stay with a sick person all the time. You must __46__ a person’s freedom as much as you can. If people really want to die, let them—it is their freedom. The point is never to become an authority__47__; remain a friend or advisor. People should not come back __48__for consultation; if they do, your advice has been incomplete—you did not know how to give the __49__advice about freedom. If they do not understand that, sick people become slaves; they are still afraid inside, and are__50__. That is no way to build a healthy world and help people become happy and free.III. Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.You’ve now heard it so many times; you can probably repeat it in your sleep. President Obama will no doubt __51__the point publicly when he gets to Beijing: the Chinese need to __52__more; they need—believe it or not—to become more like Americans, for the sake of the global economy.And it’s all true. __53__the other side of that equation is that the U.S. needs to save more. For the moment, American households actually are doing so. After the personal-savings rate __54__to zero in 2005, the shock of the economic __55__last year prompted people to snap __56__their wallets. In China, the household-savings rate exceeds 20%. It is partly for policy__57__. As we’ve seen, wage earners are expected to __58__not only their children but their aging parents. And there is, to date, only the flimsiest (脆弱的) of publicly-funded health care and pension systems, which increases incentives for individuals to save __59__they are working. But China is a society that has __60__esteemed personal financial prudence (谨慎). There is no __61__that will change anytime soon, even if the government creates a better social safety net and successfully encourages greater consumer spending.Why does the U.S. need to learn a little frugality (节俭)？Because healthy savings rates are one of the surest indicators of a country’s long-term financial health. High savings lead, over time, to increased investment, which in turn generates productivity gains, __62__and job growth. __63__, savings are the seed corn of a good economic harvest.The U.S. government thus needs to act as well. By running __64__deficits, it is dis-saving, even as households save more. Peter Orszag, Obama’s Budget Director, __65__called the U.S. budget deficits unsustainable and he’s right. To date, the U.S. has seemed unable to see the consequences of spending so much more than is taken in. That needs to change.51. A. play B. take C. make D. give52. A. concern B. process C. promote D. consume53. A. But B. Therefore C. However D. Furthermore54. A. drained B. dipped C. discounted D. dissolved55. A. issues B. crisis C. troubles D. questions56. A. cut B. put C. shut D. get57. A. reasons B. situations C. areas D. zones58. A. take off B. break out C. make up D. care for59. A. unless B. before C. after D. while60. A. long B. short C. good D. bad61. A. doubt B. wonder C. chance D. problem62. A. condition B. action C. innovation D. location63. A. In general B. In short C. In addition D. In a sense64. A. significant B. constant C. conscious D. stable65. A. occasionally B. consequently C. recently D. accidentallySection 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)When people think of improving their diet, they often talk about eating more fruits and vegetables. Others want to eat more fish and less red meat, in addition to reducing the amount of food they eat. But, they can improve their diets even more with just a simple addition. American researchers have found that a diet rich in spices can help reduce the harmful effects of eating highfat meals.Pennsylvania State University Associate Professor Shiela West led an investigation of the health effects of a spice-rich diet. Her team knew that a high-fat meal produces high levels of triglycerides（甘油三脂）, a kind of fat, in the blood. She said,“If this happens too frequently, or if triglyceride levels are raised too much, your risk of heart disease is increased.”As part of the study, her team prepared meals on two separate days for six men between the ages of 30 and 65. The men were overweight, but healthy. The researchers added about 30 milliliters of spices to each serving of the test meal, which included chicken curry, Italian herb bread and a cinnamon(肉桂树皮)biscuit. The meal for the control group was the same, but it did not include any spices.During the experiment, the researchers removed blood from the men every 30 minutes for three hours. They found that antioxidant activity(抗氧化活性) in the blood of the men who ate the spicy meal was 13 percent higher than it was for the men who did not. In addition, insulin(胰岛素) activity dropped by about 20 percent in the men who ate the spicy food.Shiela West says many scientists think that oxidative stress leads to heart disease. And what exactly is oxidative stress? Think of an apple that has been cut in half and set aside for half an hour or so. The cut side of the apple turns brown. That is a simple explanation of what happens when oxidative stress comes in contact with the inside and outside of our bodies.Professor West says,“Antioxidants, like spices, may be important in reducing oxidative stress and thus reducing the risk of chronic disease.” She adds that the level of spices used in the study provided the same amount of antioxidants found in 150 milliliters of red wine or about 38 grams of dark chocolate.66. What does the author advise people to do in their diets?A. Eating large amount of food.B. Eating less fruits and vegetables.C. Eating more vegetables and fish.D. Eating small amount of food with spices.67. What is the function of spices according to the passage?A. To help people lose weight.B. To cure chronic disease.C. To reduce the risk of heart disease.D. To cause oxidative stress.68. What happened to the men who ate the spicy meal according to the experiment?A. The antioxidant activity in their blood became increased.B. The insulin activity in their body became increased.C. The level of triglyceride in their blood was increased.D. The oxidative stress in their body was strengthened.69. What does Professor West show by citing the example of a half apple?A. The whole thing can be divided into two parts.B. It implies oxidative stress is harmful to our health.C. An apple is the only food that contains antioxidants.D. We can keep diseases away if we have an apple a day.( B )Which tablet computer should YOU be buying: They are this year's must have... and there's a style to suit everyone？Best for young childrenLeapPad Explorer 2, £68Aimed at children betweenthree and nine (though a nine-year-old might find it a littlesimple), it comes in pink or blueand with five built-in educationgames (you can buy more). Besides, the LeapPaddoes not allow access to the internet — so it isimpossible for your child to stumble acrossanything inappropriate.Pros: The education games are well-designed, the built-in video camera is a fun wayto play at being a film director.Cons:Some of the games are shockinglyexpensive. And the power adaptor is notincluded.Best for teenagersiPad 4th generation, £399-£659The iPad is still themarket leader, and for goodreason. If the teenager inyour house enjoys playingcomputer games, the latestoffering from Apple is the one to choose.Pros: No other tablet can compete with thenear one million ‘apps’ (the name Applecreated for specially-designed downloadableprograms) available for the iPad. Simple to use,even for those who usually struggle withtechnology.Cons: Considerably more expensive thanmost competitors.Best for working parentsMicrosoft Surface, £399-£559Tablets are brilliantfor leisure — but what ifyou want to do a bit ofwork? No tablet can yet compete with a full-sizelaptop computer, but this is the only tabletthat allows you to use Microsoft Word, Excel andPowerpoint (they are all pre-installed and included in the price) and you can buy a pretty lovely mini- keyboard for typing letters and emails, which also doubles up as the cover.Pros: The Surface is good for watching movies — a bonus when stuck in the airport on a business trip — and surfing the internet.Con: The keyboard is an expensive add-on — costing up to £109. It might be cheaper to buy a laptop (though a tablet is much smaller and lighter).Best for bookworms Amazon Kindle Paperwhite, £109Nearly all tablets let you download books. It's a great way to take a mountainous pile of hardbacks on holiday without stuffing your suitcase. But most tablets have a shiny screen — which can be very distracting when you're trying to read. The Paperwhite is different: its matt screen and crisp black lettering imitate thelook of words on paper brilliantly. And yet you can still read the words in the dark.Pros: Easy on the eye, excellent batterylife, 180,000 free books (if you subscribe to the Amazon Prime customer loyalty service) plus hundreds of thousands more to buy.Cons: No TV, films, games, internet orcamera.70. The underlined phrase ‘stumble across ’ most probably means ‘___________’.A. meet withB. quarrel withC. compare withD. compete with71. Which of the following about Surface is NOT TRUE?A. The keyboard will add to the cost.B. The keyboard can serve as a cover.C. You have to pay extra to install Microsoft Word.D. You can watch movies or surf the Internet with it.72. If you are a game lover, which tablet is least likely to be your choice?A. LeapPad Explorer 2.B. iPad 4th generation.C. Microsoft Surface.D. Amazon Kindle Paper73. If you want to add something to your prepared PPT for a presentation at a meeting, whichtablet is most helpful?A. LeapPad Explorer 2.B. iPad 4th generation.C. Microsoft Surface.D. Amazon Kindle Paper.(C )We are not who we think we are.The American self-image is suffused with the golden glow of opportunity. We think of the United States as a land of unlimited possibility, not so much a classless society but as a place where class is mutable—a place where brains, energy and ambition are what counts, not the circumstances of one's birth.The Economic Mobility Project, an ambitious research initiative led by Pew Charitable Trusts, looked at the economic fortunes of a large group of families over time, comparing the income of parents in the late 1960s with the income of their children in the late 1990s and early 2000s. Here is the finding: "The 'rags to riches' story is much more common in Hollywood than on Main Street. Only 6 percent of children born to parents with family income at the very bottom move to the top.That is right, just 6 percent of children born to parents who ranked in the bottom fifth of the study sample, in terms of income, were able to bootstrap their way into the top fifth. Meanwhile, an incredible 42 percent of children born into that lowest quintile are still stuck at the bottom, having been unable to climb a single rung of the income ladder.It is noted that even in Britain-a nation we think of as burdened with a hidebound class system-children who are born poor have a better chance of moving up. When the three studies were released, most reporters focused on the finding that African-Americans born to middle-class or upper middle-class families are earning slightly less, in inflation-adjusted dollars, than did their parents.One of the studies indicates, in fact, that most of the financial gains white families have made in the past three decades can be attributed to the entry of white women into the labor force. This is much less true for African-Americans.The picture that emerges from all the quintiles, correlations and percentages is of a nation in which, overall, "the current generation of adults is better off than the previous one", as one of the studies notes.The median income of the families in the sample group was $55,600 in the late 1960s; their children's median family income was measured at $71,900. However, this rising tide has not lifted all boats equally. The rich have seen far greater income gains than have the poor.Even more troubling is that our notion of America as the land of opportunity gets little support from the data. Americans move fairly easily up and down the middle rungs of the ladder, but there is "stickiness at the ends" —four out of ten children who are born poor will remain poor, and four out often who are born rich will stay rich.74. What did the Economic Mobility Project find in its research?A. Children from low-income families are unable to bootstrap their way to the top.B. Hollywood actors and actresses are upwardly mobile from rags to riches.C. The rags to riches story is more fiction than reality.D. The rags to riches story is only true for a small minority of whites.75. It can be inferred from the undertone of the writer that America, as a classless society, should________.A. perfect its self-image as a land of opportunityB. have a higher level of upward mobility than BritainC. enable African-Americans to have exclusive access to well-paid employmentD. encourage the current generation to work as hard as the previous generation76. Which of the following statements is TRUE according to the passage?A. The US is a land where brains, energy and ambition are what counts.B. Inequality persists between whites and blacks in financial gains.C. Middle-class families earn slightly less with inflation considered.D. Children in lowest-income families manage to climb a single rung of the ladder.77. What might be the best title for this passage?A. Social Upward Mobility.B. Incredible Income Gains.C. Inequality in Wealth.D. America Not Land of Opportunity.Section CDirections: Read the passage carefully. Then answer the questions or complete the statements in the fewest possible words.It is 2035. You have a job, a family and you are about 40 years old! Welcome to your future life.Getting ready for work, you pause in front of the mirror. “Turn red,” you say. Your shirt changes from sky blue to deep red. Tiny preprogrammed electronics are rearranged in your shirt to change its color. Looking into the mirror, you find it hard to believe you’re 40. You look much younger. With amazing progress in medicine, people in your generation may live to be 150 years old. You are not even middle-aged!As you go into the kitchen and prepare to pour your breakfast cereal into a bowl, you hear, “To lose weight, you shouldn’t eat that,” from your shoes. They read the tiny electronic code on the cereal box to find out the nutrition details. You decide to listen to your shoes. “Kitchen, what can I have for breakfast?” A list of possible foods appears on the counter as the kitchen checks its food supplies.“Ready for your trip to space?” you ask your son and daughter. In 2005 only specially trained astronauts went into space—and very few of them. Today anyone can go to space for day trips or longer vacations. Your best friend even works in space. Handing your children three strawberries each, you add, “The doctor said you need these for space travel.” Thanks to medical progress, vaccination shots(防疫针) are a thing of the past. Ordinary foods contain the vaccines. With the strawberries in their mouths, the kids head for the front door.It’s time for you to go to work. Your car checks your fingerprints and unlocks the doors. “My office, Autopilot,” you order. Your car drives itself down the road and moves smoothly into traffic on the highway. You sit back and unroll your e-newspaper. The latest news downloads and fills the viewer. Looking through the pages, you watch the news as video film rather than read it. (Notes: Answer the questions or complete the statements in NO MORE THAN TWELVE WORDS.)78. What changes the color of your shirt?79. The shoes know that you shouldn’t eat the breakfast cereal by__________.80. What do the strawberries the children eat serve as?81. In the future, when you look through the pages in the e-newspapers, ___________.第II卷（共47分）I. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.1. 这份工作这么难, 没几个人能胜任。