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2009年考研数学三真题及完整解析

2009年全国硕士研究生入学统一考试数学三试题一、选择题：1～8小题，每小题4分，共32分，下列每小题给出的四个选项中，只有一项符合题目要求，把所选项前的字母填在题后的括号内. （1）函数3()sin x xf x xπ-=的可去间断点的个数为：（）()A .1()B . 2 ()C .3()D .无穷多个（2）当0x →时，()sin f x x ax =-与2()ln(1)g x x bx =-是等价无穷小，则（）()A .1a=，16b =-()B . 1a =，16b = ()C .1a =-，16b =-()D .1a =-，16b =（3）使不等式1sin ln xt dt x t>⎰成立的x 的范围是（）()A .(0,1)()B .(1,)2π ()C .(,)2ππ()D .(,)π+∞（4）设函数()y f x =在区间[]1,3-上的图形为：则函数()()0xF x f t dt =⎰的图形为（）()A .()B .()f x 02 3x1 -2-11()f x 0 2 3x1 -2 -11 1()f x -2 0 2 3x-1O()C .()D .（5）设,A B 均为2阶矩阵，*,A B *分别为,A B 的伴随矩阵，若||2,||3A B ==则分块矩阵 00A B⎛⎫⎪⎝⎭的伴随矩阵为（） ()A .**0320B A ⎛⎫⎪⎝⎭()B . **230B A⎛⎫⎪⎝⎭ ()C .**0320A B⎛⎫⎪⎝⎭()D .**0230A B⎛⎫⎪⎝⎭（6）设,A P 均为3阶矩阵，TP 为P 的转置矩阵，且100010002TP AP ⎛⎫ ⎪= ⎪ ⎪⎝⎭，若123122(,,),(,,)P Q ααααααα==+，则TQ A Q 为（） ()A .210110002⎛⎫ ⎪ ⎪ ⎪⎝⎭()B . 110120002⎛⎫⎪⎪ ⎪⎝⎭ ()C .200010002⎛⎫ ⎪ ⎪ ⎪⎝⎭()D .100020002⎛⎫⎪ ⎪ ⎪⎝⎭（7）设事件A 与事件B 互不相容，则（）()A .()0P A B =()B . ()()()P AB P A P B = ()C .()1()P A P B =-()D .()1P A B ⋃=（8）设随机变量X 与Y 相互独立，且X 服从标准正态分布(0,1)N ，Y 的概率分布为1{0}{1}2P Y P Y ====，记()z F Z 为随机变量Z XY =的分布函数，则函数()z F Z 的间断点个数为（）()A .()B . 1 ()C .2()D . 3二、填空题：9-14小题，每小题4分，共24分，请将答案写在答题纸指定位置上.()f x 0 2 3x1 -2-11()f x 02 3x1 -1 1（9）cos 320lim11x x e ex →-=+- .（10）设()y x z x e =+，则(1,0)z x∂=∂（11）幂级数21(1)n nnn e x n∞=--∑的收敛半径为（12）设某产品的需求函数为()Q Q P =,其对应价格P 的弹性0.2p ξ=，则当需求量为10000件时，价格增加1元会使产品收益增加元（13）设(1,1,1)T α=,(1,0,)T k β=，若矩阵T αβ相似于300000000⎛⎫⎪⎪ ⎪⎝⎭，则k = (14)设1X ，2X ,…n X 是来自二项分布总体(,)B n p 的简单随机样本，X 和2S 分别为样本均值和样本方差，记统计量2T X S =-，则E T =三、解答题：15－23小题，共94分.请将解答写在答题纸指定的位置上.解答应写出文字说明、证明过程或演算步骤.（15）（本题满分9分）求二元函数()22(,)2ln f x y x y y y =++的极值。

上海交大历年考研真题

同学，你好！如果你打算考上海交大，那么请你花几分钟看下这份文档，这将改变你的一生！本人为交大在校学生，以下资料都是自己历年来在交大收集的第一手资料，全都出自于历届学长学姐，本人花高价收购而来！虽然市面上还有很多人可提供这些科目资料，但他们很多人本身就不是来自交大，对交考研根本不了解。

当然其中也有一些来自交大，但他们大都离开交大好几年了，交大自07年开始考研就已经改革，而他们提供的一些资料早已不适合现今交大的考研。

考研不同于高考，高考科目全部都是统编，但考研专业课的考试确实一门很大的学问，我这的资料很多都来自官方版权威的资料（网上很多电子其实都是照片，而且真题答案都是请人做的，不少答案都是错的，而我这边的材料都来自往年交大自己办考研班的时候发的资料，绝对权威），还有一部分是从历届学长学姐那收购而来，如果有必要我可以帮你找学长和你交流，最主要的是可以为你提供交大考研专业课的信息，这样就可以使你和交大的学生站在同一起跑线上，不会输在专业课的起跑线上，相对于别的同学外校考研，那你就更有优势了！考研也是一种投资，投资好了，你就会得到相当大的回报，可能你会觉花钱就会觉得不值，但当你以一两分的优势就力压群雄的话，那时你就会发现发这份钱很值了，考研每一分都对你很重要，而我所希望做的就是让你专业课的起点至少和交大本部学生一样高，绝不让考研输在专业课上！现在我已经可以提供以下科目资料，如果这下面有你要报考的专业的话，无疑这对你来说是一个最大的喜讯！如果你需要这些资料，请联系yangweitu@或QQ1449791880或上海交大考研淘宝店/!资料还是实时更新中，如果这里没有你想要的资料，也可来邮询问是否已经收集到你想要的资料！●上海交通大学<经济学>考研专业课资料（代码841）●上海交通大学/上海交大<金融学844>历年考研真题和辅导资料最近和一些同学接触后，让我感觉我很有必要写出如下这段话，如果大家有时间的话，就看看吧！大家都是怀揣梦想的有志者，都希望都过各种渠道获得最多最好的专业课资料，但这个市场鱼龙混杂，资料质量参差不齐，一份好的资料可以助你一臂之力，但一份差的资料也有可能影响你的一生。

上海交通大学《高等代数》《数学分析》历年考研真题汇总(2009-2018真题汇编)

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2009考研数学一真题及参考答案

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2009年全国硕士研究生入学统一考试真题加答案

2009年全国硕士研究生入学统一考试英语试题Section I Use of EnglishDirections:Read the following text. Choose the best word(s) for each numbered blank and mark A, B, C or D on ANSWER SHEET 1. (10 points)Research on animal intelligence always makes me wonder just how smart humans are.1 the fruit-fly experiments described in Carl Zimmer‘s piece in the Science Times on Tuesday. Fruit flies who were taught to be smarter than the average fruit fly 2 to live shorter lives. This suggests that 3 bulbs burn longer, that there is an 4 in not being too terrifically bright.Intelligence, it 5 out, is a high-priced option. It takes more upkeep, burns more fuel and is slow 6 the starting line because it depends on learning — a gradual 7 — instead of instinct. Plenty of other species are able to learn, and one of the things they‘ve apparently learned is when to 8 .Is there an adaptive value to 9 intelligence? That‘s the question behind this new research. I like it. Instead of casting a wistful glance 10 at all the species we‘ve left in the dust I.Q.-wise, it implicitly asks what the real11 of our own intelligence might be. This is 12 the mind of every animal I‘ve ever met.Research on animal intelligence also makes me wonder what experiments animals would 13 on humans if they had the chance. Every cat with an owner, 14 , is running a small-scale study in operant conditioning. we believe that 15 animals ran the labs, they would test us to 16 the limits of our patience, our faithfulness, our memory for terrain. They would try to decide what intelligence in humans is really 17 , not merely how much of it there is. 18 , they would hope to study a 19 question: Are humans actually aware of the world they live in? 20 the results are inconclusive.1. [A] Suppose [B] Consider [C] Observe [D] Imagine2. [A] tended [B] feared [C] happened [D] threatened3. [A] thinner [B] stabler [C] lighter [D] dimmer4. [A] tendency [B] advantage [C] inclination [D] priority5. [A] insists on [B] sums up [C] turns out [D] puts forward6. [A] off [B] behind [C] over [D] along7. [A] incredible [B] spontaneous [C]inevitable [D] gradual8. [A] fight [B] doubt [C] stop [D] think19. [A] invisible [B] limited [C] indefinite [D] different10. [A] upward [B] forward [C] afterward [D] backward11. [A] features [B] influences [C] results [D] costs12. [A] outside [B] on [C] by [D] across13. [A] deliver [B] carry [C] perform [D] apply14. [A] by chance [B] in contrast [C] as usual [D] for instance15. [A] if [B] unless [C] as [D] lest16. [A] moderate [B] overcome [C] determine [D] reach17. [A] at [B] for [C] after [D] with18. [A] Above all [B] After all [C] However [D] Otherwise19. [A] fundamental [B] comprehensive [C] equivalent [D] hostile20. [A] By accident [B] In time [C] So far [D] Better stillSection II Reading ComprehensionPart ADirections:Read the following four texts. Answer the questions below each text by choosing A, B, C or D. Mark your answers on ANSWER SHEET 1. (40 points)Text1Habits are a funny thing. We reach for them mindlessly, setting our brains on auto-pilot and relaxing into the unconscious comfort of familiar routine. ―Not choice, but habit rules the unreflecting herd,‖ William Wordsworth said in the 19th century. In the ever-changing 21st century, even the word ―habit‖ carries a ne gative connotation.So it seems antithetical to talk about habits in the same context as creativity and innovation. But brain researchers have discovered that when we consciously develop new habits, we create parallel synaptic paths, and even entirely new brain cells, that can jump our trains of thought onto new, innovative tracks.But don‘t bother trying to kill off old habits; once those ruts of procedure are worn into the hippocampus, they‘re there to stay. Instead, the new habits we deliberately ingrain into ourselves create parallel pathways that can bypass those old roads.―The first thing needed for innovation is a fascination with wonder,‖ says Dawna Markova, author of ―The Open Mind‖ and an executive change consultant for Professional Thinking Part ners. ―But we are taught instead to ‗decide,‘ just as our president calls himself ‗the Decider.‘‖ She adds, however, that ―to decide is to kill off all possibilities but one. A good innovational thinker is always exploring the many other possibilities.‖A ll of us work through problems in ways of which we‘re unaware, she says. Researchers in the late 1960 covered that humans are born with the capacity to2approach challenges in four primary ways: analytically, procedurally, relationally (or collaboratively) and innovatively. At puberty, however, the brain shuts down half of that capacity, preserving only those modes of thought that have seemed most valuable during the first decade or so of life.The current emphasis on standardized testing highlights analysis and procedure, meaning that few of us inherently use our innovative and collaborative modes of thought. ―This breaks the major rule in the American belief system — that anyone can do anything,‖ explains M. J. Ryan, author of the 2006 book ―This Year I Will...‖ and Ms. Markova‘s business partner. ―That‘s a lie that we have perpetuated, and it fosters commonness. Knowing what you‘re good at and doing even more of it creates excellence.‖ This is where developing new habits comes in.21. The view of Wordsworth habit is claimed by beingA. casualB. familiarC. mechanicalD. changeable.22. The researchers have discovered that the formation of habit can beA. predictedB. regulatedC. tracedD. guided23.‖ ruts‖(in li ne one, paragraph 3) has closest meaning toA. tracksB. seriesC. characteristicsD. connections24. Ms. Markova‘s comments suggest that the practice of standard testing ? A, prevents new habits form being formedB, no longer emphasizes commonnessC, maintains the inherent American thinking modelD, complies with the American belief system25. Ryan most probably agree thatA. ideas are born of a relaxing mindB. innovativeness could be taughtC. decisiveness derives from fantastic ideasD. curiosity activates creative mindsText 2It is a wise father that knows his own child, but today a man can boost his paternal (fatherly) wisdom –or at least confirm that he‘s the kid‘s dad. All he needs to do is shell our $30 for paternity testing kit (PTK) at his local drugstore – and another $120 to get the results.More than 60,000 people have purchased the PTKs since they first become available without prescriptions last years, according to Doug Fog, chief operating officer of Identigene, which makes the over-the-counter kits. More than two dozen companies sell DNA tests Directly to the public , ranging in price from a few hundred dollars to more than $2500.3Among the most popular : paternity and kinship testing , which adopted children can use to find their biological relatives and latest rage a many passionate genealogists-and supports businesses that offer to search for a family‘s geographic roots .Most tests require collecting cells by webbing saliva in the mouth and sending it to the company for testing. All tests require a potential candidate with whom to compare DNA.But some observers are skeptical, ―There is a kind of false precision being hawked by people claiming they are doing ancestry testing,‖ says Trey Duster, a New York University sociologist. He notes that each individual has many ancestors-numbering in the hundreds just a few centuries back. Yet most ancestry testing only considers a single lineage, either the Y chromosome inherited through men in a father‘s line or mito chondrial DNA, which a passed down only from mothers. This DNA can reveal genetic information about only one or two ancestors, even though, for example, just three generations back people also have six other great-grandparents or, four generations back, 14 other great-great-grandparents.Critics also argue that commercial genetic testing is only as good as the reference collections to which a sample is compared. Databases used by some companies don‘t rely on data collected systematically but rather lump together information from different research projects. This means that a DNA database may differ depending on the company that processes the results. In addition, the computer programs a company uses to estimate relationships may be patented and not subject to peer review or outside evaluation.26.In paragraphs 1 and 2 , the text shows PTK‘s ___________.[A]easy availability[B]flexibility in pricing[C] successful promotion[D] popularity with households27. PTK is used to __________.[A]locate one‘s b irth place[B]promote genetic research[C] identify parent-child kinship[D] choose children for adoption28. Skeptical observers believe that ancestry testing fails to__________.[A]trace distant ancestors[B] rebuild reliable bloodlines[C] fully use genetic information[D] achieve the claimed accuracy29. In the last paragraph ,a problem commercial genetic testing faces is __________.4[A]disorganized data collection[B] overlapping database building30. An appropriate title for the text is most likely to be__________.[A]Fors and Againsts of DNA testing[B] DNA testing and It‘s problems[C]DNA testing outside the lab[D] lies behind DNA testingText 3The relationship between formal education and economic growth in poor countries is widely misunderstood by economists and politicians alike progress in both area is undoubtedly necessary for the social, political and intellectual development of these and all other societies; however, the conventional view that education should be one of the very highest priorities for promoting rapid economic development in poor countries is wrong. We are fortunate that is it, because new educational systems there and putting enough people through them to improve economic performance would require two or three generations. The findings of a research institution have consistently shown that workers in all countries can be trained on the job to achieve radical higher productivity and, as a result, radically higher standards of living.Ironically, the first evidence for this idea appeared in the United States. Not long ago, with the country entering a recessing and Japan at its pre-bubble peak. The U.S. workforce was derided as poorly educated and one of primary cause of the poor U.S. economic performance. Japan was, and remains, the global leader in automotive-assembly productivity. Yet the research revealed that the U.S. factories of Honda Nissan, and Toyota achieved about 95 percent of the productivity of their Japanese countere pants a result of the training that U.S. workers received on the job.More recently, while examing housing construction, the researchers discovered that illiterate, non-English- speaking Mexican workers in Houston, Texas, consistently met best-practice labor productivity standards despite the complexity of the building industry‘s work.What is the real relationship between education and economic development? We have to suspect that continuing economic growth promotes the development of education even when governments don‘t force it. After all, that‘s how education got started. When our ancestors were hunters and gatherers 10,000 years ago, they didn‘t have time to wonder much about anything besides finding food. Only when humanity began to get its food in a more productive way was there time for other things.As education improved, humanity‘s productivity potential, they could in turn afford more education. This increasingly high level of education is probably a necessary, but not a sufficient, condition for the complex political systems required by advanced5economic performance. Thus poor countries might not be able to escape their poverty traps without political changes that may be possible only with broader formal education. A lack of formal education, however, doesn‘t const rain the ability of the developing world‘s workforce to substantially improve productivity for the forested future. On the contrary, constraints on improving productivity explain why education isn‘t developing more quickly there than it is.31. The author holds in paragraph 1 that the important of education in poor countries ___________.[A] is subject groundless doubts[B] has fallen victim of bias[C] is conventional downgraded[D] has been overestimated32. It is stated in paragraph 1 that construction of a new education system __________.[A]challenges economists and politicians[B]takes efforts of generations[C] demands priority from the government[D] requires sufficient labor force33.A major difference between the Japanese and U.S workforces is that __________.[A] the Japanese workforce is better disciplined[B] the Japanese workforce is more productive[C]the U.S workforce has a better education[D] ]the U.S workforce is more organize34. The author quotes the example of our ancestors to show that education emerged __________.[A] when people had enough time[B] prior to better ways of finding food[C] when people on longer went hung[D] as a result of pressure on government35. According to the last paragraph , development of education __________.[A] results directly from competitive environments[B] does not depend on economic performance[C] follows improved productivity[D] cannot afford political changesText 4The most thoroughly studied in the history of the new world are the ministers and political leaders of seventeenth-century New England. According to the standard history of American philosophy, nowhere else in colonial America was ―So much6important attached to intellectual pursuits ‖ Accord ing to many books and articles, New England‘s leaders established the basic themes and preoccupations of an unfolding, dominant Puritan tradition in American intellectual life.To take this approach to the New Englanders normally mean to start with the Puritans‘ theological innovations and their distinctive ideas about the church-important subjects that we may not neglect. But in keeping with our examination of southern intellectual life, we may consider the original Puritans as carriers of European culture adjusting to New world circumstances. The New England colonies were the scenes of important episodes in the pursuit of widely understood ideals of civility and virtuosity.The early settlers of Massachusetts Bay included men of impressive education and influence in England. `Besides the ninety or so learned ministers who came to Massachusetts church in the decade after 1629,There were political leaders like John Winthrop, an educated gentleman, lawyer, and official of the Crown before he journeyed to Boston. There men wrote and published extensively, reaching both New World and Old World audiences, and giving New England an atmosphere of intellectual earnestness.We should not forget , however, that most New Englanders were less well educated. While few crafts men or farmers, let alone dependents and servants, left literary compositions to be analyzed, The in thinking often had a traditional superstitions quality. A tailor named John Dane, who emigrated in the late 1630s, left an account of his reasons for leaving England that is filled with signs. sexual confusion, economic frustrations , and religious hope-all name together in a decisive moment when he opened the Bible, told his father the first line he saw would settle his fate, and read th e magical words: ―come out from among them, touch no unclean thing , and I will be your God and you shall be my people.‖ One wonders what Dane thought of the careful sermons explaining the Bible that he heard in puritan churched.Mean while , many se ttles had slighter religious commitments than Dane‘s, as one clergyman learned in confronting folk along the coast who mocked that they had not come to the New world for religion . ―Our main end was to catch fish. ‖36. The author notes that in the seventeenth-century New England___________.[A] Puritan tradition dominated political life.[B] intellectual interests were encouraged.[C] Politics benefited much from intellectual endeavors.[D] intellectual pursuits enjoyed a liberal environment.37. It is suggested in paragraph 2 that New Englanders__________.[A] experienced a comparatively peaceful early history.[B] brought with them the culture of the Old World[C] paid little attention to southern intellectual life[D] were obsessed with religious innovations738. The early ministers and political leaders in Massachusetts Bay__________.[A] were famous in the New World for their writings[B] gained increasing importance in religious affairs[C] abandoned high positions before coming to the New World[D] created a new intellectual atmosphere in New England39. The story of John Dane shows that less well-educated New Englanders were often __________.[A] influenced by superstitions[B] troubled with religious beliefs[C] puzzled by church sermons[D] frustrated with family earnings40. The text suggests that early settlers in New England__________.[A] were mostly engaged in political activities[B] were motivated by an illusory prospect[C] came from different backgrounds.[D] left few formal records for later referencePart BDirections:Directions: In the following text, some sentences have been removed. For Questions (41-45), choose the most suitable one from the list A-G to fit into each of the numbered blank. There are two extra choices, which do not fit in any of the gaps. Mark your answers on ANSWER SHEET 1. (10 points)Coinciding with the groundbreaking theory of biological evolution proposed by British naturalist Charles Darwin in the 1860s, British social philosopher Herbert Spencer put forward his own theory of biological and cultural evolution. Spencer argued that all worldly phenomena, including human societies, changed over time, advancing toward perfection. 41.____________.American social scientist Lewis Henry Morgan introduced another theory of cultural evolution in the late 1800s. Morgan, along with Tylor, was one of the founders of modern anthropology. In his work, he attempted to show how all aspects of culture changed together in the evolution of societies.42._____________.In the early 1900s in North America, German-born American anthropologist Franz Boas developed a new theory of culture known as historical particularism. Historical particularism, which emphasized the uniqueness of all cultures, gave new direction to anthropology. 43._____________ .8Boas felt that the culture of any society must be understood as the result of a unique history and not as one of many cultures belonging to a broader evolutionary stage or type of culture. 44._______________.Historical particularism became a dominant approach to the study of culture in American anthropology, largely through the influence of many students of Boas. But a number of anthropologists in the early 1900s also rejected the particularist theory of culture in favor of diffusionism. Some attributed virtually every important cultural achievement to the inventions of a few, especially gifted peoples that, according to diffusionists, then spread to other cultures. 45.________________.Also in the early 1900s, French sociologist Émile Durkheim developed a theory of culture that would greatly influence anthropology. Durkheim proposed that religious beliefs functioned to reinforce social solidarity. An interest in the relationship between the function of society and culture—known as functionalism—became a major theme in European, and especially British, anthropology.[A] Other anthropologists believed that cultural innovations, such as inventions, had a single origin and passed from society to society. This theory was known as diffusionism.[B] In order to study particular cultures as completely as possible, Boas became skilled in linguistics, the study of languages, and in physical anthropology, the study of human biology and anatomy.[C] He argued that human evolution was characterized by a struggle he called the ―survival of the fittest,‖ in which weaker races and societies must eventu ally be replaced by stronger, more advanced races and societies.[D] They also focused on important rituals that appeared to preserve a people‘s social structure, such as initiation ceremonies that formally signify children‘s entrance into adulthood.[E] Thus, in his view, diverse aspects of culture, such as the structure of families, forms of marriage, categories of kinship, ownership of property, forms of government, technology, and systems of food production, all changed as societies evolved.[F]Supporters of the theory viewed as a collection of integrated parts that work together to keep a society functioning.[G] For example, British anthropologists Grafton Elliot Smith and W. J. Perry incorrectly suggested, on the basis of inadequate information, that farming, pottery9making, and metallurgy all originated in ancient Egypt and diffused throughout the world. In fact, all of these cultural developments occurred separately at different times in many parts of the world.Part CDirections:Read the following text carefully and then translate the underlined segments into Chinese. Your translation should be written carefully on ANSWER SHEET 2. (10 points)There is a marked difference between the education which every one gets from living with others, and the deliberate educating of the young. In the former case the education is incidental; it is natural and important, but it is not the express reason of the association.46It may be said that the measure of the worth of any social institution is its effect in enlarging and improving experience; but this effect is not a part of its original motive. Religious associations began, for example, in the desire to secure the favor of overruling powers and to ward off evil influences; family life in the desire to gratify appetites and secure family perpetuity; systematic labor, for the most part, because of enslavement to others, etc. 47Only gradually was the by-product of the institution noted, and only more gradually still was this effect considered as a directive factor in the conduct of the institution. Even today, in our industrial life, apart from certain values of industriousness and thrift, the intellectual and emotional reaction of the forms of human association under which the world's work is carried on receives little attention as compared with physical output.But in dealing with the young, the fact of association itself as an immediate human fact, gains in importance.48 While it is easy to ignore in our contact with them the effect of our acts upon their disposition, it is not so easy as in dealing with adults. The need of training is too evident; the pressure to accomplish a change in their attitude and habits is too urgent to leave these consequences wholly out of account. 49Since our chief business with them is to enable them to share in a common life we cannot help considering whether or no we are forming the powers which will secure this ability.If humanity has made some headway in realizing that the ultimate value of every institution is its distinctively human effect we may well believe that this lesson has been learned largely through dealings with the young.50 We are thus led to distinguish, within the broad educational process which we have been so far considering, a more formal kind of education -- that of direct tuition or schooling. In undeveloped social groups, we find very little formal teaching and training. These groups mainly rely for instilling needed dispositions into the young upon the same sort of association which keeps the adults loyal to their group.Section & Writing10Part A51. Directions:Restrictions on the use of plastic bags have not been so successful in some regions. ―White pollution ‖is still going on. Write a letter to the editor(s) of your local newspaper to1)give your opinions briefly and2)make two or three suggestionsYou should write about 100 words. Do not sign your own name at the end of the letter. Use "Li Ming" instead. You do not need to write the address.Part B52. Directions:In your essay, you should1) describe the drawing briefly,2) explain its intended meaning, and then3) give your comments.You should write neatly on ANSHWER SHEET 2. (20 points)11。

2009年全国硕士研究生入学统一考试数学一真题及答案

在本题中，
当时，
当时，
因此函数仅在处间断，故选（B）.
二、填空题：9-14小题，每小题4分，共24分，请将答案写在答题纸指定位置上.
（9）设函数具有二阶连续偏导数，，则 .
【答案】
【考点】多元函数的偏导数
【难易度】★★
【详解】本题涉及到的主要知识点：
利用复合函数的链式求导法则求多元函数的偏导数的方法。
在本题中，
，
（10）若二阶常系数线性齐次微分方程的通解为，则非齐次方程满足条件的解为 .
【答案】
【考点】简单的二阶常系数非齐次线性微分方程
【难易度】★★
【详解】本题涉及到的主要知识点：
线性微分方程的解的性质即叠加原理，线性微分方程通解的结构为齐次方程的通解加上特解。
在本题中，
由通解表达式该二阶线性常系数齐次方程的特征值为，于是特征方程为
，
而在上，有连续的一阶偏导数且，于是
（在：上用高斯公式）
（20）（本题满分11分）
设， .
（Ⅰ）求满足的 . 的所有向量 , .
（Ⅱ）对（Ⅰ）中的任意向量 , 证明 , , 线性无关.
【考点】向量组的线性无关，非齐次线性方程组的通解
【难易度】★★★
【详解】本题涉及到的主要知识点：
非齐次线性微分方程的解的性质即叠加原理，非齐次线性微分方程通解的结构为齐次方程的通解加上特解。
收敛级数的和的概念，称为无穷级数的前n项的部分和。若部分和数列的极限存在，即，则称级数收敛。当级数收敛时，其和。
在本题中，
（Ⅰ）先求 .易求得与的交点为，，于是曲线与所围成区域的面积为
（Ⅱ）按定义求
（Ⅲ）求 .

考研数学一真题解析 2009

2009年全国硕士研究生入学统一考试数学(一)试卷一、选择题(1-8小题,每小题4分,共32分,下列每小题给出的四个选项中,只有一项符合题目要求,把所选项前的字母填在题后的括号内.)(1)当时,与等价无穷小,则(A) (B)(C)(D) 【考点分析】：等价无穷小，洛必达法则，泰勒公式【求解过程】：⏹ 方法一：利用洛必达法则和等价无穷小0x →时，ln(1)~bx bx --2320000()sin sin 1cos limlim lim lim 1()ln(1)3x x x x f x x ax x ax a axJ g x x bx bx bx→→→→---=====--- 1a ⇒=否则，J =∞⇒2220011cos 12lim lim 1336x x x x J bx bx b→→-====---16b ⇒=-。

选A ⏹ 方法二：利用泰勒公式或者三角函数的幂级数展开式由三角函数的幂级数展开式：357111sin 3!5!7x x x x x =-+-+ 所以，3331sin ()(0)6ax ax a x o x x =-+→ 由泰勒公式：3331sin ()(0)6ax ax a x o x x =-+→332301(1)()sin 6lim 1ln(1)x a x x o x x ax J x bx bx →-++-⇒===-- 1a ⇒=，否则J =∞⇒116J b ==-16b ⇒=-。

选A(2)如图，正方形{(,)|1,1}x y x y ≤≤被其对角线划分为四个区域(1,2,3,4)k D k =，cos kk D I y xdxdy =⎰⎰，则{}14max k k I ≤≤=(A)(B)(C)(D)0x →()sin f x x ax =-()()2ln 1g x x bx =-11,6a b ==-11,6a b ==11,6a b =-=-11,6a b =-=1I 2I 3I 4I【考点分析】：利用对称性化简二重积分，二重积分的估值【求解过程】：1234111222331444(,)cos ,cos ,(,)0,0cos ,(,)0,cos ,(,)0,0cos ,(,)0,A D D D D f x y y x I y xdxdy D f x y I I y xdxdy D x f x y y I I y xdxdy D f x y I I y xdxdy D x f x y y I ==≥≥===≤≤==⎰⎰⎰⎰⎰⎰⎰⎰记在上则，关于轴对称，且关于为奇函数，则在上则，关于轴对称，且关于为奇函数，则所以选择。

2009考研数学三真题及答案解析

2009年全国硕士研究生入学统一考试数学三试题一、选择题：1～8小题，每小题4分，共32分，下列每小题给出的四个选项中，只有一个选项是符合题目要求的，请把所选项前的字母填在答题纸指定位置上.（1）函数的可去间断点的个数为3()sin x x f x xπ-=(A)1.(B)2.(C)3.(D)无穷多个.（2）当时，与是等价无穷小，则0x →()sin f x x ax =-2()ln(1)g x x bx =-(A)，. （B ），. 1a =16b =-1a =16b =(C)，. （D ），.1a =-16b =-1a =-16b =（3）使不等式成立的的范围是1sin ln x tdt x t>⎰x (A).(B). (C).(D).(0,1)(1,2π(,)2ππ(,)π+∞（4）设函数在区间上的图形为()y f x =[]1,3-则函数的图形为()()0xF x f t dt =⎰(A)(B)(C)(D)（5）设均为2阶矩阵，分别为的伴随矩阵，若，则分块矩,A B *,A B *,A B ||2,||3A B ==阵的伴随矩阵为O A B O ⎛⎫⎪⎝⎭(A).(B). **32O B A O ⎛⎫ ⎪⎝⎭**23OB A O ⎛⎫⎪⎝⎭(C).(D).**32O A B O ⎛⎫⎪⎝⎭**23O A BO ⎛⎫⎪⎝⎭（6）设均为3阶矩阵，为的转置矩阵，且，,A P T P P 100010002TP AP ⎛⎫ ⎪= ⎪ ⎪⎝⎭若，则为1231223(,,),(,,)P Q ααααααα==+TQ AQ (A).(B).210110002⎛⎫⎪ ⎪ ⎪⎝⎭110120002⎛⎫⎪ ⎪ ⎪⎝⎭(C). (D).200010002⎛⎫ ⎪ ⎪ ⎪⎝⎭100020002⎛⎫ ⎪ ⎪ ⎪⎝⎭（7）设事件与事件B 互不相容，则A (A). (B). ()0P AB =()()()P AB P A P B =(C).(D).()1()P A P B =-()1P A B ⋃=（8）设随机变量与相互独立，且服从标准正态分布，的概率分布为X Y X (0,1)N Y ，记为随机变量的分布函数，则函数1{0}{1}2P Y P Y ====()z F Z Z XY =()z F Z的间断点个数为(A)0.(B)1. (C)2.(D)3.二、填空题：9~14小题，每小题4分，共24分，请将答案写在答题纸指定位置上.（9）.0x →=（10）设，则.()y xz x e =+(1,0)zx ∂=∂（11）幂级数的收敛半径为 .21(1)n n nn e x n ∞=--∑（12）设某产品的需求函数为,其对应价格的弹性，则当需求量为()Q Q P =P 0.2p ξ=10000件时，价格增加1元会使产品收益增加元.（13）设,，若矩阵相似于，则.(1,1,1)T α=(1,0,)T k β=Tαβ300000000⎛⎫⎪ ⎪ ⎪⎝⎭k = (14)设，,…,为来自二项分布总体的简单随机样本，和分别为样1X 2X n X (,)B n p X 2S 本均值和样本方差，记统计量，则.2T X S =-ET =三、解答题：15～23小题，共94分.请将解答写在答题纸指定的位置上.解答应写出文字说明、证明过程或演算步骤.（15）（本题满分9分）求二元函数的极值.()22(,)2ln f x y xy y y =++（16）（本题满分10 分）计算不定积分 .ln(1dx +⎰(0)x >（17）（本题满分10 分）计算二重积分，其中.()Dx y dxdy -⎰⎰22{(,)(1)(1)2,}D x y x y y x =-+-≤≥（18）（本题满分11 分）（Ⅰ）证明拉格朗日中值定理，若函数在上连续，在上可导，则()f x [],a b (),a b ，得证.(),a b ξ∈()'()()()f b f a f b a ξ-=-（Ⅱ）证明：若函数在处连续，在内可导，且()f x 0x =()0,,(0)σσ>，则存在，且.'0lim ()x f x A +→='(0)f +'(0)f A +=（19）（本题满分10 分）设曲线，其中是可导函数，且.已知曲线与直线()y f x =()f x ()0f x >()y f x =及所围成的曲边梯形绕轴旋转一周所得的立体体积值是该曲边梯0,1y x ==(1)x t t =>x 形面积值的倍，求该曲线的方程.t π（20）（本题满分11 分）设,.111A=111042--⎛⎫ ⎪- ⎪ ⎪--⎝⎭1112ξ-⎛⎫⎪= ⎪⎪-⎝⎭（Ⅰ）求满足，的所有向量，.21A ξξ=231Aξξ=2ξ3ξ（Ⅱ）对（Ⅰ）中的任意向量,，证明,,线性无关.2ξ3ξ1ξ2ξ3ξ（21）（本题满分11 分）设二次型.2221231231323(,,)(1)22f x x x ax ax a x x x x x =++-+-（Ⅰ）求二次型的矩阵的所有特征值.f （Ⅱ）若二次型的规范形为，求的值.f 2211y y +a （22）（本题满分11 分）设二维随机变量的概率密度为(,)X Y 0(,)0xe y xf x y -⎧<<=⎨⎩其他（Ⅰ）求条件概率密度；()Y X f y x （Ⅱ）求条件概率.11P X Y =⎡≤≤⎤⎣⎦（23）（本题满分11分）袋中有一个红球，两个黑球，三个白球，现在放回的从袋中取两次，每次取一个，求以、X 、分别表示两次取球所取得的红、黑与白球的个数.Y Z （Ⅰ）求；10P X Z ⎡==⎤⎣⎦（Ⅱ）求二维随机变量的概率分布.(,)X Y 2009年全国硕士研究生入学统一考试数学三试题解析一、选择题：1～8小题，每小题4分，共32分，下列每小题给出的四个选项中，只有一个选项是符合题目要求的，请把所选项前的字母填在答题纸指定位置上.（1）函数的可去间断点的个数为3()sin x x f x xπ-=(A)1. (B)2. (C)3.(D)无穷多个.【答案】C. 【解析】()3sin x x f x xπ-=则当取任何整数时，均无意义x ()f x 故的间断点有无穷多个，但可去间断点为极限存在的点，故应是的解()f x 30x x -=1,2,30,1x =±320032113211131lim lim sin cos 132lim lim sin cos 132lim lim sin cos x x x x x x x x x x x x x x x x x x x x x ππππππππππππ→→→→→-→---==--==--==故可去间断点为3个，即0,1±（2）当时，与是等价无穷小，则0x →()sin f x x ax =-2()ln(1)g x x bx =-(A)，. （B ），. 1a =16b =-1a =16b =(C)，.（D ），.1a =-16b =-1a =-16b =【答案】A.【解析】为等价无穷小，则2()sin ,()(1)f x x ax g x x ln bx =-=-222200000()sin sin 1cos sin lim lim lim lim lim ()ln(1)()36x x x x x f x x ax x ax a ax a ax g x x bx x bx bx bx→→→→→---==-⋅---洛洛故排除(B)、(C).230sin lim 166x a ax a b b axa→==-=-⋅36a b ∴=-另外存在，蕴含了故排除(D).201cos lim3x a axbx →--1cos 0a ax -→()0x → 1.a =所以本题选(A).（3）使不等式成立的的范围是1sin ln xtdt x t>⎰x (A).(B). (C).(D).(0,1)(1,2π(,)2ππ(,)π+∞【答案】A.【解析】原问题可转化为求成立时的111sin sin 1()ln xx x tt f x dt x dt dt t t t =-=-⎰⎰⎰11sin 11sin 0x x t t dt dt t t--==>⎰⎰x 取值范围，由，时，知当时，.故应选(A).1sin 0tt->()0,1t ∈()0,1x ∈()0f x >（4）设函数在区间上的图形为()y f x =[]1,3-则函数的图形为()()0xF x f t dt =⎰(A)(B)(C)(D)【答案】D.【解析】此题为定积分的应用知识考核，由的图形可见，其图像与轴及轴、()y f x =x y 所围的图形的代数面积为所求函数，从而可得出几个方面的特征：0x x =()F x ①时，，且单调递减.[]0,1x ∈()0F x ≤②时，单调递增.[]1,2x ∈()F x ③时，为常函数.[]2,3x ∈()F x ④时，为线性函数，单调递增.[]1,0x ∈-()0F x ≤⑤由于F(x)为连续函数结合这些特点，可见正确选项为(D).（5）设均为2阶矩阵，分别为的伴随矩阵，若，则分块矩,A B *,A B *,A B ||2,||3A B ==阵的伴随矩阵为O A B O ⎛⎫⎪⎝⎭(A).(B). **32O B A O ⎛⎫ ⎪⎝⎭**23OB A O ⎛⎫⎪⎝⎭(C).(D).**32O A B O ⎛⎫⎪⎝⎭**23O A BO ⎛⎫⎪⎝⎭【答案】B.【解析】根据，若CC C E *=111,C C C CC C*--*==分块矩阵的行列式，即分块矩阵可逆O A B O ⎛⎫ ⎪⎝⎭221236O A A B B O ⨯=-=⨯=（）1111661O B BO A O A O A O B B O B O B O AO A O A**---*⎛⎫ ⎪⎛⎫⎛⎫⎛⎫ ⎪=== ⎪ ⎪ ⎪⎪⎝⎭⎝⎭⎝⎭⎪⎝⎭1236132O B O B AO A O ****⎛⎫⎪⎛⎫== ⎪ ⎪ ⎪⎝⎭⎪⎝⎭故答案为(B).（6）设均为3阶矩阵，为的转置矩阵，且，,A P T P P 100010002TP AP ⎛⎫ ⎪= ⎪ ⎪⎝⎭若，则为1231223(,,),(,,)P Q ααααααα==+TQ AQ (A).(B).210110002⎛⎫⎪ ⎪ ⎪⎝⎭110120002⎛⎫⎪ ⎪ ⎪⎝⎭(C). (D).200010002⎛⎫ ⎪ ⎪ ⎪⎝⎭100020002⎛⎫ ⎪ ⎪ ⎪⎝⎭【答案】A.【解析】，即：122312312312100(,,)(,,)110(,,)(1)001Q E αααααααααα⎡⎤⎢⎥=+==⎢⎥⎢⎥⎣⎦12121212122112(1)[(1)][(1)](1)[](1)100(1)010(1)002110100100210010010110110001002001002T T TT Q PE Q AQ PE A PE E P AP E E E ===⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥==⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦⎣⎦（7）设事件与事件B 互不相容，则A (A).(B). ()0P AB =()()()P AB P A P B =(C).(D).()1()P A P B =-()1P A B ⋃=【答案】D.【解析】因为互不相容，所以,A B ()0P AB =(A)，因为不一定等于1，所以(A)不正确.()()1()P AB P A B P A B ==- ()P A B (B)当不为0时，(B)不成立，故排除.(),()P A P B (C)只有当互为对立事件的时候才成立，故排除.,A B(D)，故(D)正确.()()1()1P A B P AB P AB ==-= （8）设随机变量与相互独立，且服从标准正态分布，的概率分布为X Y X (0,1)N Y ，记为随机变量的分布函数，则函数1{0}{1}2P Y P Y ====()z F Z Z XY =()z F Z 的间断点个数为（）(A)0.(B)1. (C)2.(D)3.【答案】 B.【解析】()()(0)(0)(1)(1)Z F z P XY z P XY z Y P Y P XY z Y P Y =≤=≤==+≤==1[(0)(1)]21[(00)(1)]2P XY z Y P XY z Y P X z Y P X z Y =≤=+≤==⋅≤=+≤=独立,X Y 1()[(0)()]2Z F z P x z P x z ∴=⋅≤+≤（1）若，则0z <1()()2Z F z z =Φ（2）当，则0z ≥1()(1())2Z F z z =+Φ为间断点，故选(B).0z ∴=二、填空题：9~14小题，每小题4分，共24分，请将答案写在答题纸指定位置上.（9）.0x →=【答案】.32e 【解析】.00x x →→=02(1cos )lim13x e x x→-=20212lim 13x e x x →⋅=32e =（10）设，则.()y xz x e =+(1,0)zx ∂=∂【答案】.2ln 21+【解析】由，故()xy z x e=+()(),01xz x x =+()''ln(1)ln(1)1ln(1)1x x x x x dz x x e e x dx x ++⎡⎤⎡⎤⎡⎤=+==++⎣⎦⎢⎥⎣⎦+⎣⎦代入得，.1x =()ln 21,01ln 22ln 212z e x∂⎛⎫=+=+ ⎪∂⎝⎭（11）幂级数的收敛半径为 .21(1)n n nn e x n ∞=--∑【答案】.1e【解析】由题意知，()210nn n e a n--=>()()()()111122122111()11111n n n n n nn n nn e e ea n n e n a n e n e e +++++⎡⎤⎛⎫--⎢⎥ ⎪⎝⎭--⎢⎥⎣⎦=⋅=⋅→→∞⎡⎤+--+⎛⎫--⎢⎥⎪⎝⎭⎢⎥⎣⎦所以，该幂级数的收敛半径为1e（12）设某产品的需求函数为,其对应价格的弹性，则当需求量为()Q Q P =P 0.2p ξ=10000件时，价格增加1元会使产品收益增加元.【答案】8000.【解析】所求即为()QP Q P Q ''=+因为，所以0.2p Q PQξ'==-0.2Q P Q '=-所以()0.20.8QP Q Q Q '=-+=将代入有.10000Q =()8000QP '=（13）设,，若矩阵相似于，则.(1,1,1)T α=(1,0,)T k β=Tαβ300000000⎛⎫ ⎪ ⎪ ⎪⎝⎭k =【答案】2.【解析】相似于，根据相似矩阵有相同的特征值，得到的特征值为T αβ300000000⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦Tαβ3，0，0.而为矩阵的对角元素之和，，.TαβT αβ1300k ∴+=++2k ∴= (14)设，,…,为来自二项分布总体的简单随机样本，和分别为样1X 2X n X (,)B n p X 2S 本均值和样本方差，记统计量，则 .2T X S =-ET =【答案】2np 【解析】由.222()(1)ET E X S E X ES np np p np =-=-=--=三、解答题：15～23小题，共94分.请将解答写在答题纸指定的位置上.解答应写出文字说明、证明过程或演算步骤.（15）（本题满分9分）求二元函数的极值.()22(,)2ln f x y xy y y =++【解析】，，故.2(,)2(2)0x f x y x y '=+=2(,)2ln 10y f x y x y y '=++=10,x y e= =.2212(2),2,4xxyy xyf y f x f xy y''''''=+ =+ =则，，.12(0,)12(2xxef e ''=+1(0,)0xyef ''=1(0,)yyef e ''=而0xxf ''> 2()0xy xx yy f f f ''''''-<二元函数存在极小值.∴11(0,)f e e=-（16）（本题满分10 分）计算不定积分 .ln(1dx+⎰(0)x >得t =22212,1(1)tdtx dx t t -= =--2221ln(1ln(1)1ln(1)11111dx t d t t dt t t t +=+-+=---+⎰⎰⎰而22111112(11411(1)111ln(1)ln(1)2441dt dtt t t t t t t C t =---+-++--++++⎰⎰所以2ln(1)111ln(1ln 1412(1)1ln(1.2t t dx C t t t x C +++=+-+--+=+++⎰（17）（本题满分10 分）计算二重积分，其中.()Dx y dxdy -⎰⎰22{(,)(1)(1)2,}D x y x y y x =-+-≤≥【解析】由得，22(1)(1)2x y -+-≤2(sin cos )r θθ≤+32(sin cos )4()(cos sin )04Dx y dxdy d r r rdr πθθθθθπ+∴-=-⎰⎰⎰⎰332(sin cos )14(cos sin )034r d πθθθθθπ⎡+⎤=-⋅⎢⎥⎣⎦⎰2384(cos sin )(sin cos )(sin cos )34d πθθθθθθθπ=-⋅+⋅+⎰3384(cos sin )(sin cos )34d πθθθθθπ=-⋅+⎰.3344438814(sin cos )(sin cos )(sin cos )3344d πππθθθθθθπ=++=⨯+⎰83=-（18）（本题满分11 分）（Ⅰ）证明拉格朗日中值定理，若函数在上连续，在上可导，则()f x [],a b (),a b ，得证.(),a b ξ∈()'()()()f b f a f b a ξ-=-（Ⅱ）证明：若函数在处连续，在内可导，且()f x 0x =()0,,(0)σσ>，则存在，且.'0lim ()x f x A +→='(0)f +'(0)f A +=【解析】（Ⅰ）作辅助函数，易验证满足：()()()()()()f b f a x f x f a x a b aϕ-=----()x ϕ；在闭区间上连续，在开区间内可导，且()()a b ϕϕ=()x ϕ[],a b (),a b .''()()()()f b f a x f x b aϕ-=--根据罗尔定理，可得在内至少有一点，使，即(),a b ξ'()0ϕξ='()f ξ'()()0,()()()()f b f a f b f a f b a b aξ--=∴-=--（Ⅱ）任取，则函数满足：在闭区间上连续，开区间内可导，0(0,)x δ∈()f x []00,x ()00,x 从而有拉格朗日中值定理可得：存在，使得()()000,0,x x ξδ∈⊂……()0'00()(0)x f x f f x ξ-=-()*又由于，对上式（*式）两边取时的极限可得：()'lim x f x A +→=00x +→()()000000'''0000()00lim lim ()lim ()0x x x x x f x f f f f A x ξξξ++++→→→-====-故存在，且.'(0)f +'(0)f A +=（19）（本题满分10 分）设曲线，其中是可导函数，且.已知曲线与直线()y f x =()f x ()0f x >()y f x =及所围成的曲边梯形绕轴旋转一周所得的立体体积值是该曲边梯0,1y x ==(1)x t t =>x 形面积值的倍，求该曲线的方程.t π【解析】旋转体的体积为22()()11x x t t V f dx f dxππ==⎰⎰曲边梯形的面积为：，则由题可知()1x ts f dx =⎰22()()()()1111x x x x t t t tV ts f dx t f dx f dx t f dxπππ=⇒=⇒=⎰⎰⎰⎰两边对t 求导可得 22()()()()()()11t x t t t x t t f f dx tf f tf f dx =+⇒-=⎰⎰继续求导可得，化简可得''2()()()()()f t f t f t tf t f t --=，解之得'1(2())()2()12dt f t t f t f t t dy y-=⇒+=1223t c y y-=⋅+在式中令，则，代入得 1t =2(1)(1)0,()0,(1)1f f f t f -=>∴= 1223t cyy -=+.11,2)33c t y =∴=所以该曲线方程为：.230y x +=（20）（本题满分11 分）设,.111A=111042--⎛⎫ ⎪- ⎪ ⎪--⎝⎭1112ξ-⎛⎫⎪= ⎪⎪-⎝⎭（Ⅰ）求满足，的所有向量，.21A ξξ=231Aξξ=2ξ3ξ（Ⅱ）对（Ⅰ）中的任意向量,，证明,,线性无关.2ξ3ξ1ξ2ξ3ξ【解析】（Ⅰ）解方程21A ξξ=()1111111111111,111100000211042202110000A ξ---------⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪=-→→ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪---⎝⎭⎝⎭⎝⎭故有一个自由变量，令，由解得，()2r A =32x =0Ax =211,1x x =-= 求特解，令，得120x x ==31x = 故，其中为任意常数21101021k ξ⎛⎫⎛⎫ ⎪ ⎪=-+ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭1k解方程231Aξξ=2220220440A ⎛⎫ ⎪=-- ⎪ ⎪⎝⎭()21111022012,2201000044020000A ξ-⎛⎫ ⎪-⎛⎫ ⎪ ⎪=--→⎪ ⎪ ⎪ ⎪-⎝⎭ ⎪⎝⎭故有两个自由变量，令，由得231,0x x =-=20A x =11x =令，由得230,1x x ==-20A x =10x =求得特解21200η⎛⎫- ⎪ ⎪= ⎪ ⎪ ⎪⎝⎭故，其中为任意常数3231102100010k k ξ⎛⎫-⎪⎛⎫⎛⎫⎪ ⎪ ⎪=-++ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭ ⎪⎝⎭23,k k （Ⅱ）证明：由于12121212122111121112(21)()2((21)22221k k k k k k k k k k k k k -+--=+++-+-+-+102=≠故线性无关. 123,,ξξξ（21）（本题满分11 分）设二次型.2221231231323(,,)(1)22f x x x ax ax a x x x x x =++-+-（Ⅰ）求二次型的矩阵的所有特征值.f （Ⅱ）若二次型的规范形为，求的值.f 2211y y +a【解析】（Ⅰ） 0101111a A a a ⎛⎫ ⎪=- ⎪⎪--⎝⎭0110||01()1111111aa aE A a a a a λλλλλλλλ-----=-=---+---+222()[()(1)1][0()]()[()(1)2]()[22]19(){[(12)]}24()(2)(1)a a a a a a a a a a a a a a a a a λλλλλλλλλλλλλλλλ=---+--+-=---+-=--++--=-+--=--+--.123,2,1a a a λλλ∴==-=+（Ⅱ）若规范形为，说明有两个特征值为正，一个为0.则2212y y +1)若，则，，不符题意10a λ==220λ=-<31λ=2)若，即，则，，符合20λ=2a =120λ=>330λ=>3)若，即，则，，不符题意30λ=1a =-110λ=-<230λ=-<综上所述，故2a =（22）（本题满分11 分）设二维随机变量的概率密度为(,)X Y 0(,)0x e y xf x y -⎧<<=⎨⎩其他（Ⅰ）求条件概率密度()Y X f y x （Ⅱ）求条件概率11P X Y =⎡≤≤⎤⎣⎦【解析】（Ⅰ）由得其边缘密度函数0(,)0x y xe f x y -<<⎧= ⎨⎩其它()0xx x x f x e dy xe x --== >⎰故 |(,)1(|)0()y x x f x y f y x y x f x x== <<即|1(|)0y x y xf y x x ⎧ 0<<⎪=⎨⎪⎩其它（Ⅱ）[1,1][1|1][1]P X Y P X Y P Y ≤≤≤≤=≤而11111[1,1](,)12xxx x y P X Y f x y dxdy dx e dy xe dx e ---≤≤≤≤====-⎰⎰⎰⎰⎰()|,0x x yY yf y e dx e e y y+∞---+∞==-= >⎰11101[1]|110y y P Y e dy e e e ----∴ ≤==-=-+=-⎰.11122[1|1]11e e P X Y e e ----∴ ≤≤==--（23）（本题满分11分）袋中有一个红球，两个黑球，三个白球，现在放回的从袋中取两次，每次取一个，求以、X 、分别表示两次取球所取得的红、黑与白球的个数.Y Z ①求.10P X Z ⎡==⎤⎣⎦②求二维随机变量的概率分布.(,)X Y 【解析】（Ⅰ）在没有取白球的情况下取了一次红球，利用压缩样本空间则相当于只有1个红球，2个黑球放回摸两次，其中摸了一个红球.12113324(10)9C P X Z C C ⨯∴====⋅（Ⅱ）X ，Y 取值范围为0，1，2，故()()()()()()()()()1111332311116666111223111166661122116611221166110,0,1,0461112,0,0,136311,1,2,10910,291,20,2,20C C C C P X Y P X Y C C C C C C C P X Y P X Y C C C C C C P X Y P X Y C C C C P X Y C C P X Y P X Y ⋅⋅========⋅⋅⋅⋅========⋅⋅⋅=======⋅⋅====⋅======012 XY01/41/61/36 11/31/9021/900。

上海交通大学一九九九年硕士研究生入学考试试题电子技术基础

上海交通大学一九九九年硕士研究生入学考试试题试题编号：１６试题名称：电子技术基础（含数字电路与模拟电路）一、放大电路为图一所示，三极管参数为β１和β２，R be1,R be2,U BE=0.7V,设电容为足够大。

１、指出Ｔ１，Ｔ２各起什么作用。

２、估算静态时的电流ＩＣＱ１。

３、写出中频时电压放大倍数ＡＵ，输入电阻Ｒi和Ｒo的表达式。

（８分）一、判别下图所示电路中，级间交流反馈的组织和组态，若是负反馈则计算在浓度负反馈系数和闭环电压增益A uf=U0/U I,设各三极管的参数为β，R be为已知，电容足够大。

（8分）二、在下图的放大电路中，已知三极管的β＝１００，ＵＢＥ＝0.7V Ｒbe1=R be2=43.4KΩ,稳压管Ｄ2＝6.7V，Ａ为理想运放。

试求１、电压放大系数Ａu=U0/U i1-U i2=?２、运算放大器Ａ的共模输入电压ＵＩＣ＝？共模输出电压ＵＯＣ＝？三、一种增益可调的差动数大电路如下图所示，试推导出其输入与输出的关系式，设Ａ为理想运放。

（８分）ＵＵ四、由理想运放组成的电路如下图所示，设ＵI=V M cosωt,β>>1,试求输入信号电流i0EE六、下图是用热敏电阻Ｒt作检测元件的测温电路，该Ｒt每度变化１Ω，在0°C时R t=1KΩ.三极管的β＞＞1，U BE＝0.7V。

１、表头每伏对应几度？２、设A1和A2的CMRR为无穷大，A3的CMRR为100dB。

试估算由此引进的测量误差是多少度？七、下图所示电路中,A1,A2为理想运放，二极管D的正向压降为0.6V,动态电阻很小，反向电流为零。

１、画出U01，U02的波形，要求时间坐标对应，标明电压幅值。

２、求振幅频率f与输入电压Ui的近似函数关系式。

３、设Ui为一缓慢变化的锯齿波，试定性画出U02的波形。

(10分)一、下图所示电路是一个多功能函数发生器，其中，C1，C2，C0均为控制信号，x,y为数据输入：1、试列表说明当C2C1C0为不同取值组合时，输出端L的逻辑功能（L（ x,y））的表达式。

2009年上海交通大学841经济学(Ⅰ)考研真题及详解

2009年上海交通大学841经济学（Ⅰ）考研真题及详解跨考网独家整理最全经济学考研真题资料库，您可以在这里查阅历年经济学考研真题，经济学考研资料，经济学参考书等内容，更有跨考考研历年辅导的经济学学哥学姐的经济学考研经验，从前辈中获得的经验对初学者来说是宝贵的财富，这或许能帮你少走弯路，躲开一些陷阱。

以下内容为跨考网独家整理，如您还需更多考研资料，可选择经济学一对一在线咨询进行解答。

一、（10分）消费者的效用函数为(){}min 33U x y x y y x =++，，，两种商品的价格分别为x p 和y p ，且消费者收入为m 。

求该消费者的最优选择。

解：根据消费者的效用函数(){}min 33U x y x y y x =++，，可知：33x y y x +=+消费者的约束条件为：x y xp yp m +=联立解得：x y m x p p =+，x ymy p p =+ 此即为消费者的最优选择。

二、（15分）写出斯勒茨基方程（slutsky equation ）的表达式，并运用显示偏好原理证明斯勒茨基替代效应是非正的，同时说明什么条件下其为零。

答：（1）斯勒茨基方程表达式为：111s n x x x ∆=∆+∆，即：()()()()()()111111111111,,,,,,x p m x p m x p m x p m x p m x p m ''''''-=-+-⎡⎤⎡⎤⎣⎦⎣⎦该方程表明需求的总变动等于替代效应和收入效应的和。

其中，()()1111,,x p m x p m ''-⎡⎤⎣⎦表示替代效应，()()1111,,x p m x p m '''-⎡⎤⎣⎦表示收入效应。

它是一个恒等式：对于1p 、1p '、m 和m '的一切值，它都成立。

（2）斯勒茨基替代效应是非正的证明斯勒茨基替代效应是指当价格变动但购买力保持不变时，消费者如何用一种商品“替代”其他商品的情况。

考研数学2009真题及分析

2009年全国硕士研究生入学统一考试数学一试题一、选择题：1~8小题，每小题8分，共32分，下列每小题给出的四个选项中，只有一项符合题目要求，把所选项前的字母填在题后的括号内。

（1）当0x 时，()sin fxxax 与2()ln(1)gxxbx 等价无穷小，则（）（A ）11,6ab （B ）11,6ab （C ）11,6ab （D ）11,6ab 【解析与点评】考点：无穷小量比阶的概念与极限运算法则。

参见水木艾迪考研数学春季基础班教材《考研数学通用辅导讲义》（秦华大学出版社）例 4.67，强化班教材《大学数学强化 299》16、17 等例题。

【答案】A22220000sinsin1cossin limlimlimlim ln(1)()36xxxx xaxxaxaxaax xbxxbxbxbx230sin lim166.x aaxa b b axa 36ab 意味选项B ，C 错误。

再由201cos lim 3x aax bx存在，故有1cos0(0)aaxx ，故a=1，D 错误，所以选A 。

（2）如图，正方形{(,)|||1,||1}xyxy 被其对角线划分为四个区域，(1,2,3,4),cos KKKD DkIyxdxdy，则14max{}KK I =（）【解析与点评】本题利用二重积分区域的对称性及被积函数的奇偶性。

对称性与轮换对称性在几分钟的应用是水木艾迪考研数学重点打造的技巧之一。

参见水木艾迪考研数学春季班教材《考研数学通用辅导讲义----微积分》例 12.3、12.14、12.16、12.17，强化班教材《大学数学同步强化 299》117 题，以及《考研数学三十六技》例 18-4。

24,DD 关于x 轴对称，而cos yx 即被积函数是关于y 的奇函数，所以2413;,IIDD 两区域关于y 轴对称，cos()cos yxyx即被积函数是关于x 的偶函数，由积分的保号性，13{(,)|,01}{(,)|,01}2cos0,2cos0xyyxxxyyxx IyxdxdyIyxdxdy，所以正确答案为A 。

上海交大法学考研真题

1999年硕士生入学考试试题名称解释1．作品2．著作人身权3．软件开发者4．进口权5．证明商标二、简答题1．比较著作权与物权的异同2.合作作者有共同创作的行为3．我国专利权的获得方式和条件4．美国和我国相比较，在发明和实用新型专利保护范围界定原则方面有什么不同？各有什么特点？5．如何理解商标申请的异议和注册商标的争议论述题：论专利权的相对性2000年硕士生入学考试试题名称解释1．巴黎公约的临时保护原则2专利独家实施许可3．版权独立性原则4．商业秘密5．平行进口简答题：1．根据《世界知识产权组织公约》有关规定，简述“知识产权”这个概念包含的主要权利。

2。

外观设计与注册商标的主要区别。

3．哪些外国作品不受我国《著作权法》保护.4根据《反不正当竞争法》，例举我国法律禁止的不正当竞争行为。

论述题1．论述如何进一步完善我国对计算机软件的保护。

.2。

结合TRIPS协议，论述如何进一步完善我国知识产权法律制度。

2001年硕士生入学考试试题名词解释1．专利间接侵权2．KNOW-HOW3．专利分许可合同.4。

商标权继受取得简答题1．简述TRIPS协议最惠国待遇及其修正与限制。

.2。

根据《版权法》《专利法》规定，简述对职务作品，职务作品发明创造专利权归属的认定。

3．简述商标法的基本原则.4。

我国专利法进行了那些最新修改。

5。

简述商业秘密罪的犯罪构成论述题简要论述驰名商标或域名权的冲突与解决。

2002年硕士生入学考试试题名词解释1知识产权法2平行进口3专利独占许可4商标代理5商业秘密简答题1简述知识产权的特点。

2四种商标专用权制度。

3TRIPS确定了哪些保护知识产权的基本原则。

4简述著作邻接权。

5简述我国对地理标志的保护论述题1．论述知识产权为私权原则。

2009年试题简答论述我国刑法管辖权的规定属人管辖权的问题丢失枪支不报罪好像也考到了：过于自信的过失与间接故意走私罪虚报注册资本骗取公司登记罪与虚假出资、抽逃出资罪组织、领导和参加黑社会性质的组织罪关于刑罚方面的题目应该也有考到案例：关于共同犯罪，有身份者与无身份者共同犯罪的问题，案件牵涉到贪污罪与职务侵占罪一、2009年的题目实在是记不情了，但大体是和2007年的题目格式布局、考察风格是一样的。

2009_上海交大工硕-试卷-解答_数学

要考工硕啊，努力吧！！！上海交通大学工程硕士研究生入学考试数学（高等数学，线性代数）模拟试卷 2009.4考试时间为180分钟；试卷总分为100分准考证号码_____________________ 报考领域____________ 姓名_________一．单项选择题（共18分，每小题3分）1. 求极限 22301l i m s i n xx x e x x-→--= ( ) A ． 1； B ．0.5； C ．1-； D ．0.5- 2. 若函数()f x 与()h x 在实数轴上均可导，且()()f x g x <，则必有（） A ．()()f x h x ->-； B ．'()'()f x h x <； C ．0lim ()lim ()x x x x f x h x →→<； D ．()d ()d xxf t t h t t <⎰⎰。

3. 设(1,1),(6,3),(2,7)A B C 是xOy 平面上的三点，则三角形ABC 的面积为（） A ．6； B ．14； C ．28； D ．32。

4. 设,αβ是非齐次线性方程组()I A x b λ-=的两个不同的解，其中A 为n 阶矩阵，则下列选项中一定是A 对应的特征值λ的特征向量的为 ( )A ． αβ+；B ．αβ-；C ．α；D ．β。

5. n 维向量12,,,(3)s s n ααα≤≤ 线性无关的充要条件是 ( ) A ．存在不全为零的数12,,,s c c c ,使11220s s c c c ααα+++≠ ； B ．12,,,(3)s s n ααα≤≤ 中任意两个向量都线性无关；C ．12,,,(3)s s n ααα≤≤ 中任意一个向量都不能用其余向量线性表示；D ．12,,,(3)s s n ααα≤≤ 中存在一个向量，它不能用其余向量线性表示。

6. 设函数()f x 在0x 满足000'()''()0,'''()0f x f x f x ==>，则（） A ．0'()f x 是'()f x 的极大值； B ．0()f x 是()f x 的极大值；C ．0()f x 是()f x 的极小值；D ．00(,())x f x 是曲线()y f x =的拐点。

2009年上海交通大学化工原理考研真题【圣才出品】

2009年上海交通大学化工原理考研真题一、选择题（每题3分，共30分）1．流体在管内流动时，如要测取管截面上的流速分布，应选用（）流量计测量。

A．皮托管B．孔板流量计C．文丘里流量计D．转子流量计2．离心泵原来输送水时的流量为q v，现改用输送密度为水的1.2倍的水溶液，其它物理性质可视为与水相同，管路状况不变，则泵的输送流量（）。

A．增大B．减小C．不变D．无法确定3．某板框过滤机恒压操作过滤某悬浮液，滤框充满滤饼所需过滤时间为，若s=0，则压差提高一倍后的过滤时间'与原来过滤时间的关系为：（）。

A．'=0.B．'=C．'=D ．无法确定4．对于由三层平壁组成的多层平壁稳定热传导而言，若三层的传热推动力123t t t ∆>∆>∆，则三层平壁的传热阻力123R R R 、、之间的关系为：（）。

A ．123R R R >>B ．123R R R <<C ．132R R R >>D ．213R R R >>5．已知当温度为T 时，耐火砖的辐射能力大于铝板的辐射能力，则铝的黑度（）耐火砖的黑度。

A ．大于B ．等C ．不能确定D ．小于6．在吸收操作中，下列各项数值的变化不影响吸收传质系数的是：（）。

A ．传质单元数的改变B ．传质单元高度的改变C ．吸收塔结构尺寸的改变D ．吸收塔填料类型及尺寸的改变7．若某精馏塔正常、稳定地生产，现想增加进料量，且要求产品质量维持不变，宜采取的措施为：（）。

A．加大塔顶回流量B．加大塔釜加热蒸汽量C．加大塔釜加热蒸汽量以及冷凝器冷却水量D．加大冷凝器冷却水量8．采用多级逆流萃取和单级萃取比较，如果溶剂比、萃取浓度相同，则多级逆流萃取可使萃余分数：（）。

A．增加B．减少C．不变D．不确定9．在溶剂S与稀释剂B完全不互溶的多级逆流萃取塔操作中，原用纯溶剂，现改用再生溶剂（z A>0），其他条件不变，则对操作的影响是：（）。

2009考研数学真题及答案详解

2009年全国硕士研究生入学统一考试数学一试题答案解析一、选择题：1～8小题，每小题4分，共32分，下列每小题给出的四个选项中，只有一项符合题目要求，把所选项前的字母填在题后的括号内. （1）当0x ®时，()sin f x x ax =-与()()2ln 1g x x bx =-等价无穷小，则()A 11,6a b ==-.()B 11,6a b ==.()C 11,6a b =-=-.()D 11,6a b =-=.【答案】 A【解析】2()sin ,()ln(1)f x x ax g x x bx =-=-为等价无穷小，则222200000()sin sin 1cos sinlim lim lim lim lim ()ln(1)()36x x x x x f x x ax x ax a ax a ax g x x bx x bx bx bx ®®®®®---==-×---洛洛230sin lim 166x aax a b b ax a®==-=-× 36a b \=- 故排除,B C 。

另外201cos lim 3x a axbx ®--存在，蕴含了1cos 0a ax -®()0x ®故 1.a =排D 。

所以本题选A 。

（2）如图，正方形(){},1,1x y x y ££被其对角线划分为四个区域()1,2,3,4k D k =，cos kk D I y xdxdy =òò，则{}14max k k I ££=()A 1I .()B 2I . ()C 3I .()D 4I .【答案】A【解析】本题利用二重积分区域的对称性及被积函数的奇偶性。

24,D D 两区域关于x 轴对称，而(,)cos (,)f x y y x f x y -=-=-，即被积函数是关于y 的奇函数，所以240I I ==;13,D D 两区域关于y 轴对称，而(,)cos()cos (,)f x y y x y x f x y -=-==，即被积函数是-1 -1 1 1 xy 1D 2D3D4D关于x 的偶函数，所以{}1(,),012cos 0x y y x x I y xdxdy ³££=>òò；{}3(,),012cos 0x y y x x I y xdxdy £-££=<òò.所以正确答案为A. （3）设函数()y f x =在区间[]1,3-上的图形为：则函数()()0x F x f t dt =ò的图形为()A ()B()C ()D【答案】D【解析】此题为定积分的应用知识考核，由()y f x =的图形可见，其图像与x 轴及y 轴、0x x =所围的图形的代数面积为所求函数()F x ，从而可得出几个方面的特征：①[]0,1x Î时，()0F x £，且单调递减。

2009年全国硕士研究生入学考试数学一真题(2009考研数一真题答案解析)

2009年全国硕士研究生入学统一考试部分数学一试题答案解析一、选择题：1～8小题，每小题4分，共32分，下列每小题给出的四个选项中，只有一项符合题目要求，把所选项前的字母填在题后的括号内.（1）当0x →时，()sin f x x ax =-与()()2ln 1g x x bx =-等价无穷小，则()A 11,6a b ==-. ()B 11,6a b ==. ()C 11,6a b =-=-. ()D 11,6a b =-=. 【答案】 A【解析】2()sin ,()ln(1)f x x ax g x x bx =-=-为等价无穷小，则222200000()sin sin 1cos sin lim lim lim lim lim ()ln(1)()36x x x x x f x x ax x ax a ax a ax g x x bx x bx bx bx→→→→→---==-⋅---洛洛230sin lim 166x a ax a b b axa→==-=-⋅ 36a b ∴=- 故排除,B C 。

另外201cos lim3x a axbx→--存在，蕴含了1cos 0a ax -→()0x →故 1.a =排除D 。

所以本题选A 。

（2）如图，正方形(){},1,1x y x y ≤≤四个区域()1,2,3,4k D k =，cos kk D I y xdxdy =⎰⎰，则{}14max k k I ≤≤=()A 1I .()B 2I . ()C 3I .()D 4I .【答案】A【解析】本题利用二重积分区域的对称性及被积函数的奇偶性。

24,D D 两区域关于x 轴对称，而(,)cos (,)f x y y x f x y -=-=-，即被积函数是关于y 的奇函数，所以240I I ==;13,D D 两区域关于y 轴对称，而(,)cos()cos (,)f x y y x y x f x y -=-==，即被积函数是x关于x 的偶函数，所以{}1(,),012cos 0x y y x x I y xdxdy ≥≤≤=>⎰⎰；{}3(,),012cos 0x y y x x I y xdxdy ≤-≤≤=<⎰⎰.所以正确答案为A.（3）设函数()y f x =在区间[]1,3-上的图形为：则函数()()0xF x f t dt =⎰的图形为()A ()B()C ()D【答案】D【解析】此题为定积分的应用知识考核，由()y f x =的图形可见，其图像与x 轴及y 轴、0x x =所围的图形的代数面积为所求函数()F x ，从而可得出几个方面的特征：①[]0,1x ∈时，()0F x ≤，且单调递减。

2009年上海交通大学机械原理与设计考研试题

2009年上海交通大学机械原理与设计考研试题选择填空18个选择1个1分5~6个填空吧每题2分具体题有下来难度大了点~然后是问答每题5分记得的有1 描述凸轮反转法的原理2 螺栓组受力的设计考虑3 求5个构件的所速度瞬心4 蜗杆传动和齿轮传动强度设计的比较5 判断斜齿轮传动的转向和旋向然后画出轴向力计算题1 求自由度两小题第二题很多杆件。

很多虚约束。

好像每题8分吧2 证明偏置曲柄滑块机构的行程大于2倍曲柄长度比较简单画出极位然后任意两边之差小于第三边10分3 轮系先算轮系自由度然后求某一轮的转速是两个周转轮系公用一个转臂的情形15分或20分4 斜齿轮和锥齿轮组合25分由主动件转矩转速求下面一系列齿轮的转矩转速然后求轴向力判断旋向最后要你分析该传动不合理的地方5 轴承反向安装求当量荷载无非是压紧放松的问题每年必考吧。

然后求极限转速20分6 机械速度波动调节已知阻力矩的图形求恒定的驱动力矩然后给了J 求不均匀系数和最大最小角速度好像15分很多题没图没真相不过例题一抓一把了解下就好上海交通大学2010年机械设计原理考研真题（回忆版）一．选择题（20个 20分）1 两构件之间以线接触的平面运动副是（高副）2 平底垂直于导路的直动平底从动件凸轮机构的压力角3 渐开线齿轮齿廓上任意一点的法线与齿轮的哪个圆相切、4 非液体摩擦滑动轴承中，限制比压P的主要目的是5 零件失效形式的强度问题刚度问题的划分6 链条节数采用偶数7 渐开线斜齿轮的当量齿轮计算8 蜗轮涡杆的传动效率计算9 斜齿轮螺旋角取的越大，对传动平稳性和轴向力的影响10 单个万向连轴器的主要缺点是………………….其他的想不起来了，反正大部分或者说全部的都是原题，非常简单，只要是能有以前考试过的真题，并且自己好好做过，一分钟就能选好了，题目基本上没有变动，包括选项什么的都是一样的。

二填空题（30个空 30分）填空题主要还是考基本的知识点，没有偏题，比如考直齿轮斜齿轮锥齿轮蜗轮蜗杆的正确啮合条件，考了齿轮啮合基本定律这个地方就考了五六分，考了飞轮安装位置（高速轴），型号相同的轴承90%的寿命是相同的，这里考了“90%”；其他的具体的题目我想不起来了，反正填空基本上都是知识点，没有涉及计算的题目，只是写字花时间，不用怎么想，专业课要考查的知识点一定要看全，并且反复的复习，熟练了以后，做这个题目也很快并且不会出现什么大的错误。

上海交通大学2009年法学基础考研试题

上海交通大学2009年法学基础考研试题法理综合
一、简述法学研究的方法。

（20分）
二、简述正当程序的特征及其意义。

（30分）
宪法综合
一、全国人民代表大会常务委员会的职权有哪些？（15分）
二、如何理解“公民在法律面前一律平等”？（15分）
三、我国法院审判工作的基本原则是什么？（20分）
民法综合
一、法人机关的构成。

（10分）
二、我国物权法规定的物权类型。

（10分）
三、债的原因。

（10分）
四、被继承人遗产清偿债务的原则。

（10分）
刑法综合
一、名词解释：危害结果、假释。

（10分）
二、简述继续犯和连续犯的区别。

（15分）
三、被判处缓刑和被假释的人应当遵守哪些共同规定？（10分）。

2009年考研数学三答案解析

2009年全国硕士研究生入学统一考试数学三试题及解析一、选择题：1～8小题，每小题4分，共32分，下列每小题给出的四个选项中，只有一项符合题目要求，把所选项前的字母填在题后的括号内.（1）函数3()sin x x f x xπ-=的可去间断点的个数为：（）()A .1()B . 2 ()C .3()D .无穷多个【答案】C 【解析】()3s i n x x f x xπ-=则当x 取任何整数时，()f x 均无意义故()f x 的间断点有无穷多个，但可去间断点为极限存在的点，故应是30x x -=的解1,2,30,1x =±320032113211131lim lim sin cos 132lim lim sin cos 132lim lim sin cos x x x x x x x x x x x x x x x x x x x x x ππππππππππππ→→→→→-→---==--==--== 故可去间断点为3个，即0,1±（2）当0x →时，()sin f x x ax =-与2()ln(1)g x x bx =-是等价无穷小，则（）()A .1a =，16b =-()B . 1a =，16b = ()C .1a =-，16b =- ()D .1a =-，16b =【答案】 A【解析】2()sin ,()(1)f x x ax g x x ln bx =-=-为等价无穷小，则222200000()sin sin 1cos sin lim lim lim lim lim ()ln(1)()36x x x x x f x x ax x ax a ax a ax g x x bx x bx bx bx→→→→→---==-⋅---洛洛230sin lim 166x a ax a b b axa→==-=-⋅ 36a b ∴=- 故排除,B C 。

另外201cos lim3x a axbx→--存在，蕴含了1cos 0a ax -→()0x →故 1.a =排D 。