考研1996年真题详解
（完整版）1996年考研英语真题超详解
（完整版）1996年考研英语真题超详解1996年年全真试题Part ⅠCloze TestDirections：For each numbered blank in the following passage, there are four choices marked ［A］, ［B］, ［C］ and [D]. Choose the best one and mark your answer on ANSWER SHEET 1 by blackening the corresponding letter in the brackets. (10 points)Vitamins are organic compounds necessary in small amounts in the diet for the normal growth and maintenance of life of animals, including man.They do not provide energy, 1 do they construct or build any part of the body. They are needed for 2 foods into energy and body maintenance. There are thirteen or more of them, and if 3 is missing a deficiency disease becomes 4 .Vitamins are similar because they are made of the same elements—usually carbon, hydrogen, oxygen, and 5 nitrogen. They are different 6 their elements are arranged differently, and each vitamin 7 one or more specific functions in the body.8 enough vitamins is essential to life, although the body has no nutritional use for 9 vitamins. Many people, 10 , believe in being on the “safe side” and thus take extra vitamins. However, a well balanced diet will usually meet all the body’s vitamin needs.1.［A］either ［B］so ［C］nor ［D］never2.［A］shifting ［B］transferring ［C］altering ［D］transforming3.［A］any ［B］some ［C］anything ［D］something4.［A］serious ［B］apparent ［C］severe ［D］fatal5.［A］mostly ［B］partially ［C］sometimes ［D］rarely6.［A］in that ［B］so that ［C］such that ［D］except that7.［A］undertakes ［B］holds ［C］plays ［D］performs8.［A］Supplying ［B］Getting ［C］Providing ［D］Furnishing9.［A］exceptional ［B］exceeding ［C］excess ［D］external10.［A］nevertheless ［B］therefore ［C］moreover ［D］meanwhilePart ⅡReading ComprehensionDirections:Each of the passages below is followed by some questions. For each questions there are four answers marked ［A］, ［B］, ［C］and ［D］. Read the passages carefully and choose the best answer to each of the questions. Then mark your answer on ANSWER SHEET 1 by blackening the corresponding letter in the brackets with a pencil. (40 points)Passage 1Tight lipped elders used to say, “It’s not what you want in this world, but what you get.”Psychology teaches that you do get what you want if you know what you want and want the right things.You can make a mental blueprint of a desire as you would make a blueprint of a house, and each of us is continually making these blueprints in the general routine of everyday living. If we intend to have friends to dinner, we plan the menu, make a shopping list, decide which food to cook first, and such planning is an essential for any type of meal to be served.Likewise, if you want to find a job, take a sheet of paper, and write a brief account of yourself. In making a blueprint for a job,begin with yourself, for when you know exactly what you have to offer, you can intelligently plan where to sell your services.This account of yourself is actually a sketch of your working life and should include education, experience and references. Such an account is valuable. It can be referred to in filling out standard application blanks and is extremely helpful in personal interviews. While talking to you, your could be employer is deciding whether your education, your experience, and other q ualifications will pay him to employ you and your “wares” and abilities must be displayed in an orderly and reasonably connected manner.When you have carefully prepared a blueprint of your abilities and desires, you have something tangible to sell. Then you are ready to hunt for a job. Get all the possible information about your could be job. Make inquiries as to the details regarding the job and the firm. Keep your eyes and ears open, and use your own judgment. Spend a certain amount of time each day seeking the employment you wish for, and keep in mind: Securing a job is your job now.11. What do the elders mean when they say, “It’s not what you want in this world, but what you get.”?［A］You’ll certainly get what you want.［B］It’s no use dreaming.［C］ You should be dissatisfied with what you have.［D］It’s essential to set a goal for yourself.12. A blueprint made before inviting a friend to dinner is used in this passage as .［A］ an illustration of how to write an application for a job ［B］ an indication of how to secure a good job［C］ a guideline for job description［D］ a principle for job evaluation13. According to the passage, one must write an account of himself before starting to finda job because .［A］ that is the first step to please the employer［B］ that is the requirement of the employer［C］ it enables him to know when to sell his services［D］ it forces him to become clearly aware of himself14. When you have carefully prepared a blueprint of your abilities and desires, you have something .［A］ definite to offer ［B］ imaginary to provide［C］ practical to supply ［D］ desirable to presentPassage 2With the start of BBC World Service Television, millions of viewers in Asia and America can now watch the Corporation’s news coverage, as well as listen to it.And of course in Britain listeners and viewers can tune in to two BBC television channels, five BBC national radio services and dozens of local radio stations. They are brought sport, comedy, drama, music, news and current affairs, education, religion, parliamentary coverage, children’s programmes and films for an annual licence fee of ￡83 per household.It is a remarkable record, stretching back over 70 years — yet the BBC’s future is now in doubt. The Corporation will survive as a publicly funded broadcasting organization, at least for the time being, but its role, its size and its programmes are now the subject of a nation wide debate in Britain.The debate was launched by the Government, which invited anyone with an opinion of the BBC — including ordinary listenersand viewers —to say what was good or bad about the Corporation, and even whether they thought it was worth keeping. The reason for its inquiry is that the BBC’s royal charter runs out in 1996 and it must decide whether to keep the organization as it is, or to make changes.Defenders of the Corporation — of whom there are many —are fond of quoting the American slogan “If it ain’t broke, don’t fix it.” The BBC “ain’t broke”, they say, by which they mean it is not broken (as distinct from the word ‘broke’, meaning having no money), so why bother to change it?Yet the BBC will have to change, because the broadcasting world around it is changing. The commercial TV channels ——ITV and Channel 4 ——were required by the Thatcher Government’s Bro adcasting Act to become more commercial, competing with each other for advertisers, and cutting costs and jobs. But it is the arrival of new satellite channels — funded partly by advertising and partly by viewers’subscriptions — which will bring about the biggest changes in the long term.15. The world famous BBC now faces .［A］the problem of news coverage ［B］an uncertain prospect［C］inquiries by the general public ［D］shrinkage of audience16. In the passage, which of the following about the BBC is not mentioned as the key issue?［A］ Extension of its TV service to Far East.［B］ Programmes as the subject of a nation-wide debate.［C］ Potentials for further international co-operations.［D］ Its existence as a broadcasting organization.17. The BBC’s “royal charter” (Line 4, Paragraph 4) standsfor .［A］ the financial support from the royal family.［B］ the privileges granted by the Queen.［C］ a contract with the Queen.［D］ a unique relationship with the royal family.18. The foremost reason why the BBC has to readjust itself is no other than .［A］ the emergence of commercial TV channels.［B］the enforcement of Broadcasting Act by the government.［C］ the urgent necessity to reduce costs and jobs.［D］ the challenge of new satellite channels.Passage 3In the last half of the nineteenth century “capital” and “labour” were enlarging and perfecting their rival organizations on modern lines. Many an old firm was replaced by a limited liability company with a bureaucracy of salaried managers. The change met the technical requirements of the new age by engaging a large professional element and prevented the decline in efficiency that so commonly spoiled the fortunes of family firms in the second and third generation after the energetic founders. It was moreover a step away from individual initiative, towards collectivism and municipal and state-owned business. The railway companies, though still private business managed for the benefit of shareholders, were very unlike old family business. At the same time the great municipalities went into business to supply lighting, trams and other services to the taxpayers.The growth of the limited liability company and municipal business had important consequences. Such large, impersonalmanipulation of capital and industry greatly increased the numbers and importance of shareholders as a class, an element in national life representing irresponsible wealth detached from the land and the duties of the landowners; and almost equally detached from the responsible management of business. All through the nineteenth century, America, Africa, India, Australia and parts of Europe were being developed by British capital, and British shareholders were thus enriched by the world’s movement towards industrialization. Towns like Bournemouth and Eas tbourne sprang up to house large “comfortable” classes who had retired on their incomes, and who had no relation to the rest of the community except that of drawing dividends and occasionally attending a shareholders’ meeting to dictate their orders to the management. On the other hand “shareholding” meant leisure and freedom which was used by many of the later Victorians for the highest purpose of a great civilization.The “shareholders” as such had no knowledge of the lives, thoughts or needs of the workmen employed by the company in which he held shares, and his influence on the relations of capital and labor was not good. The paid manager acting for the company was in more direct relation with the men and their demands, but even he had seldom that familiar personal knowledge of the workmen which the employer had often had under the more patriarchal system of the old family business now passing away. Indeed the mere size of operations and the numbers of workmen involved rendered such personal relations impossible. Fortunately, however, the increasing power and organization of the trade unions, at least in all skilled trades, enabled the workmen to meet on equal terms the managers ofthe companies who employed them. The cruel discipline of the strike and loc kout taught the two parties to respect each other’s strength and understand the value of fair negotiation.19. It’s true of the old family firms that .［A］ they were spoiled by the younger generations［B］ they failed for lack of individual initiative［C］they lacked efficiency compared with modern companies［D］ they could supply adequate services to the taxpayers20. The growth of limited liability companies resulted in .［A］ the separation of capital from management［B］ the ownership of capital by managers［C］ the emergence of capital and labour as two classes［D］ the participation of shareholders in municipal business21. According to the passage, all of the following are true except that .［A］the shareholders were unaware of the needs of the workers［B］ the old firm owners had a better understanding of their workers［C］ the limited liability companies were too large to run smoothly［D］ the trade unions seemed to play a positive role22. The author is most critical of .［A］ family firm owners ［B］ landowners［C］ managers ［D］ shareholdersPassage 4What accounts for the great outburst of major inventions in early America—breakthroughs such as the telegraph, the steamboat and the weaving machine?Among the many shaping factors, I would single out the country’s excellent elementary schools;a labor force that welcomed the new technology; the practice of giving premiums to inventors; and above all the American genius for nonverbal, “spatial” thinking about things technological.Why mention the elementary schools? Because thanks to these schools our early mechanics, especially in the New England and Middle Atlantic states, were generally literate and at home in arithmetic and in some aspects of geometry and trigonometry.Acute foreign observers related American adaptiveness and inventiveness to this educational advantage. As a member of a British commission visiting here in 1853 reported, “With a mind prepared by thorough school discipline, the American boy develops rapidly into the skilled workman.”A furth er stimulus to invention came from the “premium” system, which preceded our patent system and for years ran parallel with it. This approach, originated abroad, offered inventors medals, cash prizes and other incentives.In the United States, multitudes of premiums for new devices were awarded at country fairs and at the industrial fairs in major cities. Americans flocked to these fairs to admire the new machines and thus to renew their faith in the beneficence of technological advance.Given this optimistic approach to technological innovation, the American worker took readily to that special kind of nonverbal thinking required in mechanical technology. As Eugene Ferguson has pointed out, “A technologist thinks about objects that cannot be reduced to unambiguous verbal descriptions; they are dealt with in his mind by a visual, nonverbalprocess … The designer and the inventor … are able to assemble and manipulate in their minds devices that as yet do not exist.”This nonverbal “spatial” thinking can be just a s creative as painting and writing. Robert Fulton once wrote, “The mechanic should sit down among levers, screws, wedges, wheels, etc, like a poet among the letters of the alphabet, considering them as an exhibition of his thoughts, in which a new arrangement transmits a new idea.”When all these shaping forces—schools, open attitudes, the premium system, a genius for spatial thinking —interacted with one another on the rich U.S. mainland, they produced that American characteristic emulation. T oday that word implies mere imitation. But in earlier times it meant a friendly but competitive striving for fame and excellence.23. According to the author, the great outburst of major inventions in early America was ina large part due to .［A］ elementary schools ［B］ enthusiastic workers［C］ the attractive premium system ［D］ a special way of thinking24. It is implied that adaptiveness and inventiveness of the early American mechanics .［A］ benefited a lot from their mathematical knowledge.［B］ shed light on disciplined school management.［C］ was brought about by privileged home training.［D］ owed a lot to the technological development.25. A technologist can be compared to an artist because .［A］ they are both winners of awards.［B］ they are both experts in spatial thinking.［C］ they both abandon verbal description［D］ they both use various instruments26. The best title for this passage might be .［A］ Inventive Mind ［B］ Effective Schooling［C］ Ways of Thinking ［D］ Outpouring of InventionsPassage 5Rumor has it that more than 20 books on creationism/evolution are in the publisher’s pipelines.A few have already appeared. The goal of all will be to try to explain to a confused and often unenlightened citizenry that there are not two equally valid scientific theories for the origin and evolution of universe and life. Cosmology, geology, and biology have provided a consistent, unified, and constantly improving account of what happened. “Scientific” creationism, which is being pushed by some for “equal time” in the classrooms whenever the scientific accounts of evolution are given, is based on religion, not science. Virtually all scientists and the majority of nonfundamentalist religious leaders have come to regard “scientific” creationism as bad science and bad religion.The first four chapters of Kitcher’s book give a very brief introduction to evolution. At appropriate places, he introduces the criticisms of the creationists and provides answers. In the last three chapters, he takes off his gloves and gives the creationists a good beating. He describes their programmes and tactics, and, for those unfamiliar with the ways of creationists, the extent of their deception and distortion may come as an unpleasant surprise. When their basic motivation is religious, one might have expected more Christian behavior.Kitcher is a philosopher, and this may account, in part, for the clarity and effectiveness of his arguments. The non-specialist willbe able to obtain at least a notion of the sorts of data and argument that support evolutionary theory. The final chapters on the creationists will be extremely clear to all. On the dust jacket of this fine book, Stephen Jay Gould says: “This book stands for reason itself.” And so it does - and all would be well were reason the only judge in the creationism/evolution debate.27. “Creationism” in the passage refers to .［A］evolution in its true sense as to the origin of the universe［B］ a notion of the creation of religion［C］ the scientific explanation of the earth formation［D］ the deceptive theory about the origin of the universe28. Kitcher’s book is intended to .［A］ recommend the views of the evolutionists［B］ expose the true features of creationists［C］ curse bitterly at his opponents［D］ launch a surprise attack on creationists29. From the passage we can infer that .［A］ reasoning has played a decisive role in the debate［B］ creationists do not base their argument on reasoning ［C］ evolutionary theory is too difficult for non-specialists ［D］ creationism is supported by scientific findings30. This passage appears to be a digest of .［A］ a book review ［B］ a scientific paper［C］ a magazine feature ［D］ a newspaper editorialPart ⅢEnglish—Chinese TranslationDirections:Read the following text carefully and then translate the underlined segments into Chinese. Your translation should be written clearly on ANSWER SHEET 2. (15 points)The differences in relative growth of various areas of scientific research have several causes. 31）Some of these causes are completely reasonable results of social needs. Others are reasonable consequences of particular advances in science being to some extent self-accelerating. Some, however, are less reasonable processes of different growth in which preconceptions of the form scientific theory ought to take, by persons in authority, act to alter the growth pattern of different areas. This is a new problem probably not yet unavoidable; but it is a frightening trend. 32）This trend began during the Second World War, when several governments came to the conclusion that the specific demands that a government wants to make of its scientific establishment cannot generally be foreseen in detail. It can be predicted, however, that from time to time questions will arise which will require specific scientific answers. It is therefore generally valuable to treat the scientific establishment as a resource or machine to be kept in functional order. 33）This seems mostly effectively done by supporting a certain amount of research not related to immediate goals but of possible consequence in the future.This kind of support, like all government support, requires decisions about the appropriate recipients of funds. Decisions based on utility as opposed to lack of utility are straightforward. But a decision among projects none of which has immediate utility is more difficult. The goal of the supporting agencies is the praisable one of supporting “good” as opposed to “bad” science, but a valid determination is difficult to make. Generally, the idea of good science tends to become confused with the capacity of the field in question to generate an elegant theory. 34）However, the world is so made that elegant systems are inprinciple unable to deal with some of the world’s more fascinating and delightful aspects. 35）New forms of thought as well as new subjects for thought must arise in the future as they have in the past, giving rise to new standards of elegance.Section Ⅳ Writing36. Directions:A. Title: GOOD HEALTHB. Time limit: 40minutesC. Word limit: 120—150 words (not including the given opening sentence)D. Your composition should be based on the “OUTLINE” below and should start with the given opening sentence: “The desire for good health is universal”.E. Your composition must be written clearly on the ANSWER SHEET.Outline:1. Importance of good health.2. Ways to keep fit.3. My own practices.1996年试题答案Part ⅠCloze Test1. C2. D3. A4. B5. C6. A7. D8.B9. C 10. APart ⅡReading ComprehensionPart APassage 1 11. B 12. A 13.D 14. APassage 2 15.B 16.C 17.C 18.DPassage 3 19.C 20.A 21.C 22.DPassage 4 23.D 24.A 25.B 26.APassage 5 27.D 28.B 29.B 30.APart Ⅲ English-Chinese Translation31.在这些原因中，有些纯属社会需求；另一些则是由于科学上某些特定发展在一定程度上自我加速而产生的必然结果。
1996年数学一真题及答案详解
1996年全国硕士研究生入学统一考试数学(一)试卷一、填空题(此题共5小题,每题3分，总分值15分把答案填在题中横线上)x 2 a(1)设 lim(—— )x 8,那么 a = ______________ .xx a⑵设一平面经过原点及点(6, 3,2),且与平面4x y 2z 8垂直，那么此平面方程为⑶微分方程y 2y 2y e x 的通解为 _______________________ .⑷函数u In (x ... y 2 z 2)在点A(1,0,1)处沿点 A 指向点B(3, 2,2)方向的方向导数为1 0 2⑸设A 是4 3矩阵,且 A 的秩r(A) 2,而B 02 0 ,那么r(AB)= ______________ . 1 03二、选择题(此题共5小题,每题3分，总分值15分海小题给出的四个选项中，只有一个符合题目 要求，把所选项前的字母填在题后的括号内)(1)么ay)dx2ydy为某函数的全微分，a 那么等于(x y)(A)-1 (C)1(A) f (0)是f(x)的极大值 (B) f (0)是f (x)的极小值 (C) (0, f (0))是曲线y f (x)的拐点(D) f (0)不是f (x)的极值,(0, f (0))也不是曲线y f (x)的拐点⑶设a n 0(n 1,2,L ),且 a n 收敛，常数(0,—),那么级数 (1)n (nta n —)a 2nn 12n 1n(B)条件收敛 (D)散敛性与有关(B)0 (D)2(2)设f( x)具有二阶连续导数，且f (0)0,limx 0f(x) x(A)绝对收敛 (C)发散与X k是同阶无穷小，那k 等于(A)1(B)2 (C)3(D)4a 10 0 b 1(5)四阶行列式a 2b 2 0 的值等于0 a 3 b s 0b 4 00 a 4(A) a£2a 3a 4 bb z b s b q(B) a 1a 2a s a 4 bdb s b q (C) (a^ bib 2 )(a：b s b 4)(D) (a 2a s b 2b s )(a 1a 4 db 4)三、(此题共2小题,每题5分，总分值10分)(1)求心形线r a(1 cos )的全长，其中a 0是常数. ⑵设X i 10,X n 16 X n (n 1,2,L ),试证数列｛X n ｝极限存在，并求此极限五、(此题总分值7分)1求级数 2-的和.n 1(n 2 1)2n六、(此题总分值7分)1 X设对任意 x 0,曲线y f (x)上点(x, f(x))处的切线在 y 轴上的截距等于一 f(t)dt,求x 0f ( X)的一般表达式.七、(此题总分值8分)设f (x)在［0,1］上具有二阶导数，且满足条件 f(x) a, f (x) b,其中a,b 都是非负常数，c⑷设有f(x)连续的导数,f (0)0, f (0) O,F(x)X(x 2 0 \t 2)f (t)dt,且当 x 0 时,F (x)四、(此题共2小题,每题6分，总分值12分) (1)计算曲面积分 (2x z)dydz zdxdy,其中S与z 轴正向的夹角为锐角.S 为有向曲面z2 ，小y (0 x1),其法向量(2)设变换X 2y可把方程6 zx ay2z24 0简化为y2z0,求常数 a.是(0,1)内任意一点•证明八、(此题总分值6分)设A I E弋,其中I是n阶单位矩阵,E是n维非零列向量,f是E的转置•证明(1) A2A的充分条件是f f 1.⑵当f f 1时,A是不可逆矩阵•九、(此题总分值8分)二次型f (x1,x2, x3) 5x： 5x；ex：2x1x2 6x1x3 6x2x3的秩为2,(1) 求参数e及此二次型对应矩阵的特征值•(2) 指出方程f(X1,X2,X3)1表示何种二次曲面•十、填空题(此题共2小题,每题3分，总分值6分•把答案填在题中横线上)(1)设工厂A和工厂B的产品的次品率分别为1%和2%,现从由A和B的产品分别占60%和40% 的一批产品中随机抽取一件，发现是次品，那么该次品属A生产的概率是 _________ •(2)设，是两个相互独立且均服从正态分布N(0,( 1 )2)的随机变量，那么随机变量的数学期望E( )= _____________十一、(此题总分值6分)设，是两个相互独立且服从同一分布的两个随机变量，的分布率为1P( i) -，i 123・3又设X max( , ), Y min(,).(1)(2)求随机变量X的数学期望E(X).1996年全国硕士研究生入学统一考试数学(一)答案详解一、填空题 (1)【答】In 2. 【详解】因为于是 e 3a8 a In 2(2)【答】2x 2y 3z 0【详解】 原点与点(6, 3,2)连线的方向向量为 s (6, 3.2);平面4x y 2z 8的法向量为n 4, 1,2 , 根据题意，所求平面的法向量为j k3 2 2i 2j 3k. 1 2即 2x 2y 3z 0(3)【答】 y C 1e x cosx C 2e x sinx e x【详解】对应齐次方程的特征方程为22 2 0,解得特征根为 1,21 i,由于1不是特征根，可设原方程的特解为y Ae ,代入原方程解得 A 1,故所求通解为 y Ge x cosx C 2e x sinx e x1(4) ［答］丄.2【详解］因为、- y 23axx a x a13a3a3ax aexx 2alim lim xx a x故所求平面方程为2(x0) 2( y 0) 3(z 0) 0,(1,0,1)2!2!CO Sz 2(1,0,1)0,乙22 ,cos 3(1,0,1)1 J3沿AB 方向的单位向量为故u 沿AB 方向的方向导数为uA B(5)【答】2. 【详解】 因为B 10 0,说明矩阵B 可逆，故秩 r(AB) 秩 r(A)2,:■、选择题 (1)【答】应选(D ) 【详解】(x ay)dx ydy(x y)2 为某函数的全微分的充要条件是x (x y)2x ay (x y)2即(a 2)x ay 2y, (a 2)(x y) 0. 当且仅当a 2时上式恒成立，故正确选项( D ). (2)【答】应选(B ) 【详解】由题设x 叫 f (x)1根据极限的性质知，存在 x 0的某领域,在此领域内有 f (x) _x _ 0即 f '(x) 0. 又根据泰勒公式,f(x) f(0) f (0)xt"x 2其中在0与x 之间, 从而 f (x) f (0) 口)x 2f(0)可见f 〔0〕是f 〔x 〕的极小值，故正确选项为〔B 〕〔3〕【答】应选〔A 〕而lim ntan —,所以当n 充分大时，nn ， n ta n a ?n 〔1〕a ?nn又正项级a n 收敛，所以其偶数项数列构成的级数a ?n 也收敛,n 1n 1从而 〔1〕n n tana 2n 绝对收敛， 故正确选项为〔A 〕n 1n〔4〕【答】应选〔C 〕【详解】因为x2x 0f(t)dt.可见应有k 3，故正确选项为〔C 〕 〔5〕【答】应选〔D 〕【详解】按第一行展开，a-i 0 0 b|0 a 2 b 2 0 0b 3 a 3 0b 4 00 a 4【详解】由于〔1〕n ntan — a ?,nntan — a 2n ,n'2 XX 2F (x) x 20f(t)dt0t2f(t)dtX2 22x 0 f (t)dt x 2f (x) x 2f(x)又根据题设F '〔x 〕与x k 是同阶无穷小，且f(0) 0, f '(0) 0,叫Hx一(k叫H X 2_(k叫H Xa 2b 2 00 a 2 t b a 1 th a 30 3 0 b 3 a sa 4b 4F叫ka )a 4b 2b s )(a i a 4隔).n ,都有X n X n 1，即数列X n 为单调减少数列。
1996年考研英语真题超详解
1996年考研英语真题超详解 1996年年全真试题PartⅠClozeTestDirections：Foreachnumberedblankinthefollowingpassage,therearefourchoicesmarked［A］,［B］,［C］and[D].ChoosethebestoneandmarkyouransweronANSWERSHEET1byblackeningthecorresponding letterinthebrackets.(10points)Vitaminsareorganiccompounds necessary insmallamountsinthedietforthenormalgrowthandmaintenance oflifeofanimals,including man.Theydonotprovideenergy,1dotheyconstructorbuildanypartofthebody.Theyareneededfor2foodsintoenergyandbodymaintenance. Therearethirteenormoreofthem,andif3ismissingadeficiencydiseasebecomes4.Vitaminsaresimilarbecausetheyaremadeofthesameelements―usuallyhydrogen,oxygen,and5nitrogen.Theyaredifferent6theirelementsarearrangeddifferently, andeachvitamin7oneormorespecificfunctionsinthe8enoughvitaminsisessential tolife,althoughthebodyhasnonutritional usefor9vitamins. Manypeople,10,believeinbeingonthe“safeside”andthustakeextravitamins.However,awell��balanced dietwillusuallymeetallthebody’svitaminneeds.1.［A］either［B］so［C］nor［D］never2.［A］shifting［B］transferring［C］altering［D］transforming 3.［A］any［B］some［C］anything［D］something4.［A］serious［B］apparent［C］severe［D］fatal5.［A］mostly［B］partially［C］sometimes［D］rarely6.［A］inthat［B］sothat［C］suchthat［D］exceptthat7.［A］undertakes［B］holds［C］plays［D］performs8.［A］Supplying［B］Getting［C］Providing［D］Furnishing9.［A］exceptional［B］exceeding［C］excess［D］external10.［A］nevertheless［B］therefore［C］moreover［D］meanwhilePartⅡReadingComprehensionDirections:Eachofthepassagesbelowisfollowedsomequestions. Foreachquestions therearefouranswersmarked［A］,［B］,［C］and［D］.Readthepassages carefully andchoosethebestanswereachofthequestions.ThenmarkyouransweronANSWERSHEET1byblackeningthecorresponding letterinthebracketswithapencil.(40points)Passage1Tight��lipped eldersusedtosay,“It’snotwhatyouwantinthisworld,butwhatyouget.”Psychologyteachesthatyoudowhatyouwantifyouknowwhatyouwantandwanttherightthings. Youcanmakeamentalblueprint ofadesireaswouldmakeablueprint ofahouse,andeachofusiscontinually makingtheseblueprints inthegeneralroutineofeverydayliving.Ifintend tohavefriends todinner, weplanthemenu,makeashopping list,decide whichfoodtocookfirst, andsuchplanning isessential foranytypeofmealtobeserved. Likewise, ifyouwanttofindajob,takeasheetofpaper,andwritebriefaccount ofyourself. Inmakingablueprint forajob,beginwithyourself, forwhenyouknowexactly whatyouhavetooffer,canintelligently planwheretosellyourservices.Thisaccountofyourselfisactuallyasketchofyourworkinglifeandshouldincludeeducation,experience andreferences. Suchanaccountisvaluable. Itcanbereferredtoinfillingoutstandardapplication blanksandisextremely helpfulinpersonalinterviews.Whiletalkingtoyou,yourcouldbeemployerisdecidingwhetheryoureducation,yourexperience,andotherqualifications willpayhimtoemployyouyour“wares” andabilities mustbedisplayed inanorderlyandreasonably connected manner.Whenyouhavecarefully prepared ablueprint ofyourabilitiesdesires,youhavesomething tangibletosell.Thenyouarereadytohuntforajob.Getallthepossibleinformation aboutyourcouldjob.Makeinquiries astothedetails regarding thejobandthefirm.Keepyoureyesandearsopen,anduseyourownjudgment.acertainamountoftimeeachdayseekingtheemployment youwishfor,andkeepinmind:Securing ajobisyourjobnow.Whatdotheeldersmeanwhentheysay,“It’s notwhatyouwantinthisworld,butwhatyouget.”?［A］You’ll certainly getyouwant.［B］It’snousedreaming.［C］Youshouldbedissatisfied withwhatyouhave.［D］It’sessentialtosetagoalforyourself.12.Ablueprintmadebeforeinvitingafriendtodinnerisusedinthispassageas?????????.［A］anillustration ofhowtowriteanforajob［B］anindication ofhowtosecureagoodjob［C］aguideline forjobdescription［D］aprinciple forjob13.Accordingtothepassage,onemustwriteanaccountofhimselfbeforestartingtofindajobbecause???????? .［A］thatisthesteptopleasetheemployer［B］thatistherequirement oftheemployer［C］itenableshimtoknowwhentosellhisservicesitforceshimtobecomeclearly awareofhimself 14.Whenyouhavecarefully prepared ablueprint ofyourabilities anddesires, youhavesomething????????.［A］definitetooffer［B］imaginarytoprovide［C］practicaltosupply［D］desirabletopresentPassage2WiththestartofBBCWorldServiceTelevision,millionsofviewersinAsiaandAmericacannowwatchtheCorporation’s newscoverage,aswellaslistentoit.AndofcourseBritainlisteners andviewerscantuneintotwoBBCtelevision channels, fiveBBCnational radioservices anddozensoflocalradiostations. Theybroughtsport,comedy,drama,music,newsandcurrentaffairs,education,religion,parliamentary coverage,children’s programmesandfilmsforanannuallicencefeeof￡83household. Itisaremarkable record,stretching backover70years―yettheBBC’sfutureisnowindoubt.TheCorporation willsurviveapublicly��funded broadcastingorganization,atleastforthetimebeing,butitsrole,itssizeanditsprogrammesarenowthesubjectofanation��wide debateinBritain.ThedebatewaslaunchedbytheGovernment,whichinvitedanyonewithanopinionoftheBBC―includingordinarylistenersandviewers―tosaywhatwasgoodorbadabouttheCorporation, andevenwhethertheythoughtitwasworthkeeping.ThereasonforitsinquiryisthattheBBC’sroyalcharterrunsoutin1996anditmustdecidewhethertokeeptheorganization asitis,ortochanges.Defenders oftheCorporation―ofwhomtherearemany―arefondofquotingtheAmericanslogan“Ifitain’tbroke,don’tit.”TheBBC“ain’tbroke”, theysay,bywhichtheymeanitisnotbroken(asdistinct fromtheword‘broke’, meaninghavingnosowhybothertochangeit?YettheBBCwillhavetochange,becausethebroadcasting worldarounditischanging.Thecommercial TV――ITVandChannel4――wererequiredbytheThatcherGovernment’s Broadcasting Acttobecomemorecommercial, competingwitheachotherforadvertisers,cuttingcostsandjobs.Butitisthearrivalofnewsatellitechannels―fundedpartlybyadvertisingandpartlybyviewers’subscriptions―whichbringaboutthebiggestchangesinthelongterm.15.TheworldfamousBBCnowfaces???????? .［A］theproblemofnewscoverage［B］uncertain prospect［C］inquiries bythegeneral public［D］shrinkage ofaudience 16.Inthepassage, whichofthefollowing abouttheBBCismentioned asthekeyissue?［A］Extension ofitsTVservicetoFarEast.［B］Programmes asthesubjectofanation-wide debate.［C］Potentialsforfurtherinternational co-operations.［D］Itsexistenceasabroadcasting organization.17.TheBBC’s“royalcharter”(Line4,Paragraph4)standsfor????????.［A］。
1996年考研数学一真题及答案解析
1 2 1996 年全国硕士研究生入学统一考试理工数学一试题详解及评析一、 填空题⎛ x + 2a ⎫x（1） 设lim ⎪= 8, 则a = .x →∞ ⎝ 【答】 ln 2. x - a ⎭+ x ⎡3axx -a ⎤ x -a 【详解】 因为lim ⎛ x 2a ⎫ = lim ⎢⎛1+ 3a ⎫ 3a ⎥= e 3a , x →∞ x - a ⎪ x →∞ ⎢ x - a⎪ ⎥ ⎝ ⎭于是e 3a = 8 ⇒ a = ln 2⎣⎝ ⎭ ⎦（2）设一平面经过原点及点(6, -3, 2), 且与平面4x - y + 2z = 8 垂直，则此平面方程为.【答】 2x + 2 y - 3z = 0【详解】 原点与点(6, -3, 2) 连线的方向向量为 s = (6, -3, 2) ; 平面4x - y + 2z = 8 的法向量为 n = {4, -1, 2},根据题意，所求平面的法向量为i j ks ⨯ n = 6 4 -3 2 = 2i + 2 j - 3k .-1 2故所求平面方程为 2 ( x - 0) + 2 ( y - 0) - 3( z - 0) = 0,即2x + 2 y - 3z = 0 .(3)微分方程 y '' - 2 y ' + 2 y = e x 的通解为.【答】 y = C e x cos x + C e x sin x + e x【详解】 对应齐次方程的特征方程为λ 2 - 2λ + 2 = 0,解得特征根为 λ1,2 = 1± i ,由于α = 1 不是特征根，可设原方程的特解为 y * = Ae * ，1 2 ⎢ ⎥ 代入原方程解得 A = 1, 故所求通解为（4）函数u = ln (x 1y = C e x cos x + C e x sin x + e x在 A (1, 0,1) 点处沿 A 点指向的方向导数为.【答】 .2【详解】 因为∂u | = 1 |= 1 , ∂x A x (1,0,1) 2∂u | = 1 ⋅ y | = 0∂y A x (1,0,1)∂u | = 1 ⋅ z | = 1 , ∂z A x (1,0,1) 2–––K cos α = 2 , cos β = - 2 , cos γ = 1,3 3 3⎧ 2 2 1 ⎫沿 AB 方向的单位向量为⎨ 3 , - , ⎬,–––K⎩3 3⎭ 故u 沿 AB 方向的方向导数为∂u 1 2 ⎛ 2 ⎫ 1 1 1 –––K = ⋅ + 0 ⋅ - ⎪ + ⋅ = ∂ AB 2 3 ⎝ 3 ⎭ 2 3 2⎡ 1 0 2⎤（5）设 A 是4 ⨯ 3 矩阵，且 A 的秩 r ( A ) = 2, 而 B = ⎢ 0 2 0⎥ , 则 r ( AB ) =.⎢⎣-1 0 3⎥⎦【答】 2.【详解】 因为 B =1 0 22 0 = 10 ≠ 0,-1 0 3说明矩阵 B 可逆，故秩 r ( AB ) = 秩 r ( A ) = 2,二、选择题（1） 已知( x + ay ) dx + ydy ( x + y )2为某函数的全微分，则 a 等于（A ）-1.（B ）0.(C)1.(D)2.【】→ 【答】 应选（D ）.( x + ay ) dx + ydy【详解】( x + y )2为某函数的全微分的充要条件是∂ ⎛ y ⎫ = ∂ ⎛ x + ay ⎫, 2 ⎪ 2 ⎪∂x ( x + y ) ∂y ( x + y ) ⎪即(a - 2) x - ay = -2 y ,⎝ ⎭ ⎝ ⎭(a - 2)( x - y ) = 0.当且仅当 a = 2 时上式恒成立，故正确选项为（D ）. f '' ( x )（2）设 f ( x ) 有二阶连续导数，且 f ' (0) = 0, limx0 = 1, 则（A ） f (0) 是 f ( x ) 的极大值. （B ） f (0) 是 f ( x ) 的极小值.（C ） (0, f (0))是曲线 y = f ( x ) 的拐点.（D ） f (0) 不是 f ( x ) 的极值， (0, f (0))也不是曲线 y = f ( x ) 的拐点【 】【答】 应选（B ）. f '' ( x )【详解】 由题设limx →0在此邻域内有f '' ( x )= 1 根据极限的性质知，存在 x = 0 的某邻域，≥ 0 .即 f '' ( x ) ≥ 0.又根据泰勒公式，f '' (ξ )f ( x ) = f (0) + f ' (0) x +f '' (ξ )x 2 其中ξ 在 0 与 x 之间， 2! 从而 f ( x ) = f (0) +x 2 ≥ 2!f (0)可见 f (0) 是 f ( x ) 的极小值，故正确选项为（B ）∞⎛ π ⎫∞ n ⎛ λ ⎫ （3）设a n > 0 (n = 1, 2,⋯), 且∑a n 收敛，常数λ ∈ 0, 2 ⎪ , 则级数∑(-1) n tan n ⎪ a 2nn =1 ⎝ ⎭ n =1⎝ ⎭（A ）绝对收敛. （B ）条件收敛. (C)发散.（D ）敛散性与λ 有关.【 】【答】 应选（A ）.xxx⎝⎭ ⎝ ⎭ 0= lim∞x n⎛λ ⎫ λ 【详解】 由于 (-1) nn tan n ⎪ a 2n = n tan n ⋅ a 2n ,而lim n tann →∞λ = λ, 所以当n 充分大时，n tan λ⋅ a < (λ +1) an 2n2n∞∞又正项级∑an 收敛，所以其偶数项数列构成的级数∑a2n 也收敛，n =1n =1n⎛λ ⎫ 从而 ∑(-1) n =1n tan n ⎪ a 2n 绝对收敛，故正确选项为（A ）（4）设 f ( x ) 有连续的导数， f (0) = 0, f ' (0) ≠ 0, F ( x ) = ⎰ x(x 2- t 2 )f (t ) dt , 且当 x → 0 时， F ' ( x )是与 xk是同阶无穷小，则 k 等于【答】 应选（C ）. 【详解】 因为' ⎡ 2xx 2【 】⎤'x2 2 F ( x ) = ⎢⎣ x ⎰0xf (t ) dt - ⎰0 t f (t ) d t ⎥⎦= 2x ⎰0 f (t ) dt + x f ( x ) - x f ( x ) = 2x ⎰0 f (t )dt .又根据题设 F ' ( x ) 与xk 是同阶无穷小，且 f (0) = 0, f ' (0) ≠ 0,于是有F ' ( x )2x ⎰ f (t ) dt2 f ( x ) lim x →0 x k = lim 0 x →0 x kx →0 (k -1) x k -2 = 2 lim 1⋅f ( x ) - f (0) x →0 (k -1) x k -3 x - 0= 2 f ' (0)⋅lim 1≠ 0,x →0 (k -1) x k -3可见应有 k = 3 故正确选项为（C ）.（5）四阶行列式 的值等于3 3（A ） a 1a 2 x 3 x 4 - b 1b 2b 3b 4 .（B ） a 1a 2a 3a 4 + b 1b 2b 3b 4 .a 1 0 0b 10 a 2 b 2 0 0 b 4 b 0 a 0 0 a 41 n →∞1 4（C ） (a 1a 2 - b 1b 2 )(a 3a 4 - b 3b 4 ).（D ） (a 2a 3 - b 2b 3 )(a 1a 4 - b 1b 4 ).【 】【答】 应选（D ） 【详解】 按第一行展开，a 1 0 0 0 a 2b 2 b 10 a 2 b 2 ＝ a ⋅ b a 00 0 - b 0 a 2 b 2 b a 0 b 3 a 3 0b 0 0 a1 3 3 0 0 a 4 1 3 3 b 4 0 0 44= aaa 2b 2- b b a 2 b 2 b 3 a 3b 3 a 3故正确选项为（D ）.= (a 2a 3 - b 2b 3 )(a 1a 4 - b 1b 4 ).三、（1）求心形线 r = a (1+ cos θ ) 的全长，其中a > 0 是常数.' 【详解】 因为 r (θ ) = -a sin θ , ds = d θ = 2a cosd θ 利用对称性知，所求心形线的全长s = 2⎰π 2a co sθ d θ = 8a s in θ |π= 8a0 2 2 0（2）设 x 1 = 10, x n +1 =【详解】 由 x 1 = 10, x 2 =n = 1, 2,⋯), 试证数列{x n } 的极限存在，并求此极限.= 4 知， x > x . 2设对某个正整数 k 有 x k > x k +1 则x k +1 =>= x k +2 .故由归纳法知，对一切正整数 n , 都有 x n > x n +1, 即数列{x n } 为单调减少数列.又显然有 x n > 0 (n = 1, 2,⋯),即{x n } 有下界，根据单调有界数列必有极限知，数列{x n } 的极限存在.记lim x n = a , 对 x n +1从而 a 2 - a - 6 = 0两边取极限，得 a解得 a = 3 或 a = -2 （舍去，因为 x n > 0 ）θ21 41 y 0 0 2π 故所求极限值为 a = 3 .四、（1）计算曲面积分⎰⎰(2x + z )dydz + zdxdy , 其中 S 为有向曲面 z = x 2 + y 2(0 ≤ z ≤ 1) ， s其法向量与 z 轴正向的夹角为锐角.【详解 1】 用高斯公式，以 S 表示法向量指向 z 轴负向的有向平面 z = 1(x 2 + y 2 ≤ 1), D 为S 1 在 xOy 平面上的投影区域，则⎰⎰(2x + z ) d yd z + z dxdy = º⎰⎰ (2x + z ) d ydz + zdx dy -⎰⎰ (2x + z ) dyd z + z dxdysS +S 1S 1= -⎰⎰⎰( 2 +1)dV - ⎰⎰ -dxdyΩ D = -3 d θ 1 rdr 2⎡-(-π )⎤dz⎰⎰=- 3π + π2π⎰r 2⎣ ⎦=- .2【详解 2】 用矢量投影法，因为z ' = 2x , z ' = 2 yx于是(2x + z ) dydz + zdxdy =⎡(2x + z )⋅(-z ' ) + z ⎤dxdy⎰⎰⎰⎰ ⎣ x ⎦sS= ⎰⎰(-4x 2 - 2xz + z )dxdyS= ⎰⎰ ⎡⎣-4x 2 - 2x (x 2 + y 2) + x 2 + y 2 ⎤⎦dxdyD= ⎰2πd θ ⎰1(-4r 2 c os 2 θ - 2r 3 c os θ + r 2 )π=- .2【详解 3 】 直接投影法，曲面 S 在 yOz 平面上投影 D yz 对应两个曲面：一是x ≤ z ≤ 1, 其方向指向前侧，因此积分取正号，一是 x =≤ z ≤ 1, 其方向指向后侧，因此积分取负号，再记 D xy 表示 S 在 xOy 平面上的投影区域，则⎩ ∂2 z ) 2 0 五、求级数∑ n - + 2 ⎰⎰(2x + z )dydz + zdxdys= ⎰⎰ (D yz= -4⎰⎰D yz+ z )dyd z + ⎰⎰ (-D yz+ ⎰⎰ (x 2 + y 2 )dxdyD xyz )dy dz + ⎰⎰ (x 2 + y 2 )dxdyD xy112π1 24⎰-1dy ⎰y 2+ ⎰0 d θ ⎰0 r ⋅ rdrπ =- 2 ⎧u = x - 2 y∂2 z∂2 z ∂2 z∂2 z（2）设变换⎨ v = x + ay 可把方程6 ∂x 2 + ∂x ∂y - ∂y 2 = 0 化简为∂u ∂v = 0, 求常数 a .∂z∂z ∂z ∂z ∂z ∂z 【详解】∂x = ∂u + ∂v , ∂y = -2 ∂u + a ∂v ,∂2 z= ∂2 z +∂2 z + ∂2 z∂x 2 ∂2 z ∂u 2 =- 2 ∂u ∂v ∂2 z∂v 2 , ∂2 z ∂2z∂x ∂y 2 ∂u 2 + (a - 2) ∂u ∂v + a ∂v 2 ,∂2 z = ∂y 2 ∂2 z 4 ∂u 2 - 4a ∂2 z ∂u ∂v ∂2 z a ∂v 2. 将上述结果代入原方程，经整理后得(10 + 5a )∂u ∂v+ (6 + a - a 依题意知 a 应满足∂2z = ∂v解之得 6 + a - a 2 = 0, 且10 + 5a ≠ 0,a = 3.∞n =2(n21-1)2n的和.∞n【详解】 令 S ( x ) = ∑ 2n =2 x , 则1=- .2 1x - 1 2 ∞1 ∑ ∞x n1 ⎛ ∞ x n ∞x n ⎫ S ( x ) = ∑ n 2 - = ∑ - ∑ ⎪n =2 1 2 ⎝ n =2 n -1 n =2 n +1 ⎭-1 ∞n +1 = x ∑∞ x n - 1 ∑ x 2 n =2 n -1 2x n =2 n +1= x ∑∞ x n - 1 ⎛ ∑∞n ⎫ x - x .2 n =1 n ⎪ ⎝ n =1 n2 ⎭因为 x n =1 n= - ln (1- x ), 于是有S ( x ) =- x ln (1- x ) + 1 + 1 x + 1ln (1- x )( x < 1, x ≠ 0),2 2 4 2x 1 ⎛ 1 ⎫ ∞ 1 53 令 x = , 得 2S 2⎪ = ∑ 2= - ln 2.⎝ ⎭ n =2 (n -1)2 8 41 x六、设对任意 x > 0, 曲线 y = f ( x ) 的一般表达式.f ( x ) 上点( x , f ( x ))处切线在Y 轴上得截距等于x⎰0f (t )dt , 求【详截】 曲线 y = f ( x ) 上点( x , f ( x ))处切线方程为Y - f ( x ) = f ' ( x )( X - x ) ，令 X = 0 得截距Y = f ( x ) - xf ' ( x )由题意有1⎰ xf (t )dt = f ( x ) - xf ' ( x ),x 0即⎰x f (t )dt = x ⎡ f ( x ) - xf ' ( x )⎤上式对 x 求导，化简得即(xf ' ( x ))'= 0；⎣ ⎦ xf ''' ( x ) + f ' ( x ) = 0积分得因此xf ' ( x ) = C ,f ( x ) = C 1 ln x + C 2 (其中C 1、C 2为任意常数).七、设 f ( x ) 在[0,1]上具有二阶导数，且满足条件常数， c 是(0,1) 内任意一点，证明f ( x ) ≤ a , f '' ( x ) ≤ b 其中 a 、b 都是非负n 2x n1 ⎣⎦ f ' (c ) ≤ 2a + b.2【详解】 对 f ( x ) 在 x = c 处用泰勒公式展开，得f ( x ) = f (c ) + f '(c )( x - c ) + f '' (ξ ) 2!( x - c )2(*)其中ξ = c + θ ( x - c ), 0<θ <1.在(*) 式中令 x = 0, 则有f (0) = f (c ) + f '(c )(0 - c ) +在(*) 式中令 x = 1 ，则有f '' (ξ ) 2!f '' (ξ )(0 - c )22,0<ξ1<c<1,f (1) = f (c ) + f '(c )(1- c ) +(1- c ) 2!,0<ξ2 <1,上述两式相减得f (1) - f (0) = f '(c ) +1 ⎡f ''(ξ)(1 - c )2 - f '' (ξ ) c 2 ⎤2! ⎣于是21⎦f ' (c ) = f (1) - f (0) - 1 ⎡ f '' (ξ )(1 - c )2- f '' (ξ )c 2 ⎤ 2 ⎣ 2 1 ⎦≤ f (1) + f (0) + f '' (ξ ) (1- c )2 + f '' (ξ ) c 2≤ 2a + b ⎡(1- c )2+ c 2 ⎤ .2 又因当c ∈(0,1) 时，有(1- c )2+ c 2 ≤ 1, 故f ' (c ) ≤ 2a + b.2八、设 A = E - ξξ T 其中 E 是 n 阶单位矩阵， ξ 是 n 维非列向量， ξ T 是ξ 的转置，证明：（1） A 2 = A 的充要条件是ξ T ξ = 1;（2） 当ξ T ξ = 1时， A 是不可逆矩阵.【详解】 （1） A 2 = (E - ξξ T )(E - 2ξξ T ) = E - 2ξξ T + ξ (ξ T ξ )ξ T = E - (2 - ξ T ξ )ξξ T ,因此 A 2 = A ⇔ E - (2 - ξ T ξ)ξξ T = E - ξξ T ⇔ (ξ T ξ -1)ξξ T = 01 2 1 22因为ξ≠ 0, 所以ξξT≠ 0故 A2=A 的充要条件为ξTξ= 1;（2）方法一：当ξTξ= 1时，由 A =E -ξξT, 有 Aξ=ξ-ξξTξ=ξ-ξ= 0,因为ξ≠ 0, 故Ax = 0 有非零解，因此A= 0 ，说明A 不可逆.方法二：当ξTξ= 1，由A2=A ⇔A(E -A)= 0, 即E -A 的每一列均为Ax = 0 的解，因为E -A =ξξT≠ 0, 说明Ax = 0 有非零解，故秩(A)<n ，因此A 不可逆.方法三：用反证法.假设 A 可逆，当ξξT= 1, 有 A2=A于是 A-1A2=A-1A, 即 A =E .这与A =E -ξξT≠E 矛盾，故A 是不可逆矩阵.九、已知二次型f (x , x , x )= 5x 2+ 5x 2+cx 2- 2x x+ 6x x- 6x x 的秩为2.1 2 3 1 2 3 1 2 1 3 2 3(1)求参数c及此二次型对应矩阵的特征值；(2)指出方程f (x1, x2 , x3)= 1表示何种二次曲面.【详解】（1）此二次型对应矩阵为⎡5 -1 3 ⎤A =⎢-1 5 -3⎥.⎢⎥⎢⎣3 -3 c⎥⎦因秩(A)= 2, 故A= 0, 由此解得c = 3, 容易验证，此时A 的秩的确为2.又由λ- 5 1 -3λE - A = 1 λ- 5 3 =λ(λ- 4)(λ- 9),-3 3 λ- 3所求特征值为λ1= 0, λ2= 4, λ3= 9.（2）由特征值可知，f (x1, x2 , x3)= 1表示椭球柱面.十、填空题（1）设工厂A 和工厂B 的产品率分别为1％和2％，现从由A 和B 的产品分别占60％和402 2 2π2π +∞=π2 .％的一批产品中随机抽取一件，发现是次品，则该次品属 A 产品的概率是.3 【答】 .7【详解】 设事件 A ={抽取的产品为工厂 A 生产的}， B ＝{抽取的产品为工厂 B 生产的}，C ＝{抽取的是次品},则P ( A ) = 0.6, P ( B ) = 0.4, P (C | A ) = 0.01, P (C | B ) = 0.02,由逆概率公式知P ( A | C ) =P ( AC ) =P ( A )⋅ P (C | A ) P (C ) P ( A ) P (C | A ) + P ( B ) P (C | B )=0.6 ⨯ 0.010.6 ⨯ 0.01+ 0.4 ⨯ 0.02 = 3 . 7⎛ ⎛ ⎫2 ⎫(2)设ξ ,η 是两个相互独立且均服从正态分布 N 0, ⎪ ⎪的随机变量，则随机变量 ξ -η的数学期望 E( ξ -η ) =.⎝ ⎝ ⎭ ⎭【答】.【详解】 因为ξ ,η 是两个相互独立且均服从正态分布 N ⎛ 0,1 ⎫, 2 ⎪ ⎝ ⎭故 Z = ξ -η 也服从正态分布，且 E (Z ) = E ξ - E η = 0, D (Z ) = D ξ + D η = 1 + 1= 1,2 2即 Z ~ N (0,1).于是E ( ξ -η ) = E Z = ⎰ z 1-x22 dz -∞2 +∞ - x 2⎛ z 2 ⎫ = ⎰ e 2 d ⎪0 ⎝ 2 ⎭十一、设 ξ ,η 是两个相互独立且服从同一分布的两个随机变量，已知 ξ 的分布律为P {ξ = i } = 1, i = 1, 2, 3, 又设 X = max (ξ ,η ),Y = min (ξ ,η ).32π（1）写出二维随机变量(X ,Y )的分布律；(2)求随机变量 X 的数学期望 E (X ).P{X <Y}= 0 即【详解】（1 ）由X = max (ξ,η),Y = min (ξ,η). 的定义知，P{X = 1,Y = 2}=P (X = 1,Y = 3)=P (X = 2,Y = 3)= 0,且进已步有P{X = 1,Y = 1}=P{ξ= 1,η= 1}=P{ξ= 1}⋅P{η= 1}=1 ,9P{X = 2,Y = 2}=P{ξ= 2,η= 2}=P{ξ= 2}⋅P{η= 2}=1 ,9P{X = 3,Y = 3}=P{ξ= 3,η= 3}=P{ξ= 3}⋅P{η= 3}=1 ,9P{X = 2,Y = 1}=P{ξ= 1,η= 2}+P{ξ= 2,η= 1}=1 +1 =2 ,9 9 9P{X = 3,Y = 2}=P{ξ= 2,η= 3}+P{ξ= 3,η= 2}=1 +1 =2 ,9 9 9P{X = 3,Y = 1}= 1-7 =2 ;9 9故所求的分布律为（2）X 的边缘分布为故X 的数学期望为E (X )=1 ⨯1+3 ⨯ 2 +5 ⨯ 3 =22 .9 9 9 9。
1996年全国硕士研究生入学统一考试数学一、二、三、四、五试题完整版附答案及评分标准
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1996年考研数学三真题及全面解析
1996年全国硕士研究生入学统一考试数学三试题一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上.) (1) 设方程yx y =确定y 是x 的函数,则dy =___________. (2) 设()arcsin x f x dx x C =+⎰,则1()dx f x =⎰___________.. (3) 设()00,x y 是抛物线2y ax bx c =++上的一点,若在该点的切线过原点,则系数应满足的关系是___________. (4) 设123222212311111231111n nn n n n n a a a a A a a a a a a a a ----⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎣⎦,123n x x X x x ⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎣⎦,1111B ⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎣⎦, 其中(;,1,2,,)i j a a i j i j n ≠≠=.则线性方程组T A X B =的解是___________.(5) 设由来自正态总体2~(,0.9)X N μ容量为9的简单随机样本,得样本均值5X =,则未知参数μ的置信度为0.95的置信区间为___________.二、选择题(本题共5小题,每小题3分,满分15分.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) 累次积分cos 20(cos ,sin )d f r r rdr πθθθθ⎰⎰可以写成 ( )(A) 10(,)dy f x y dx ⎰(B) 10(,)dy f x y dx ⎰ (C)11(,)dx f x y dy ⎰⎰(D) 10(,)dx f x y dy ⎰(2) 下述各选项正确的是 ( ) (A) 若21nn u∞=∑和21nn v∞=∑都收敛,则21()nn n uv ∞=+∑收敛(B)1n nn u v∞=∑收敛,则21nn u∞=∑与21nn v∞=∑都收敛(C) 若正项级数1nn u∞=∑发散,则1n u n≥(D) 若级数1nn u∞=∑收敛,且(1,2,)n n u v n ≥=,则级数1n n v ∞=∑也收敛(3) 设n 阶矩阵A 非奇异(2n ≥),A *是矩阵A 的伴随矩阵,则 ( ) (A) 1()n A A A -**= (B) 1()n A A A +**= (C) 2()n A AA -**= (D) 2()n A AA +**=(4) 设有任意两个n 维向量组1,,m αα和1,,m ββ,若存在两组不全为零的数1,,m λλ和1,,m k k ,使111111()()()()0m m m m m m k k k k λαλαλβλβ+++++-++-=,则( )(A) 1,,m αα和1,,m ββ都线性相关 (B) 1,,m αα和1,,m ββ都线性无关(C) 1111,,,,,m m m m αβαβαβαβ++--线性无关 (D) 1111,,,,,m m m m αβαβαβαβ++--线性相关(5) 已知0()1P B <<且()1212[]()()P A A B P A B P A B +=+,则下列选项成立的是( ) (A) ()1212[]()()P A A B P A B P A B +=+ (B) ()1212()()P A B A B P A B P A B +=+ (C) ()1212()()P A A P A B P A B +=+ (D) ()()1122()()()P B P A P B A P A P B A =+三、(本题满分6分)设(),0,()0,0,xg x e x f x xx -⎧-≠⎪=⎨⎪=⎩其中()g x 有二阶连续导数,且(0)1,(0)1g g '==-. (1)求()f x '；(2)讨论()f x '在(,)-∞+∞上的连续性.四、(本题满分6分)设函数()z f u =,方程()()xyu u p t dt ϕ=+⎰确定u 是,x y 的函数,其中(),()f u u ϕ可微；()p t ,()u ϕ'连续,且()1u ϕ'≠.求()()z z p y p x x y∂∂+∂∂.五、(本题满分6分)计算2(1)xx xe dx e -+∞-+⎰.六、(本题满分5分)设()f x 在区间[0,1]上可微,且满足条件120(1)2()f xf x dx =⎰.试证:存在(0,1)ξ∈使()()0.f f ξξξ'+=七、(本题满分6分)设某种商品的单价为p 时,售出的商品数量Q 可以表示成aQ c p b=-+,其中a b 、、 c 均为正数,且a bc >.(1) 求p 在何范围变化时,使相应销售额增加或减少.(2) 要使销售额最大,商品单价p 应取何值?最大销售额是多少?八、(本题满分6分)求微分方程dy dx =的通解.九、(本题满分8分)设矩阵010010000010012A y ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦. (1) 已知A 的一个特征值为3,试求y ； (2) 求矩阵P ,使()()TAP AP 为对角矩阵.十、(本题满分8分)设向量12,,,t ααα是齐次线性方程组0AX =的一个基础解系,向量β不是方程组0AX =的解,即0A β≠.试证明:向量组12,,,,t ββαβαβα+++线性无关.十一、(本题满分7分)假设一部机器在一天内发生故障的概率为0.2,机器发生故障时全天停止工作,若一周5个工作日里无故障,可获利润10万元；发生一次故障仍可获得利润5万元；发生两次故障所获利润0元；发生三次或三次以上故障就要亏损2万元.求一周内期望利润是多少?十二、(本题满分6分)考虑一元二次方程20x Bx C ++=,其中B C 、分别是将一枚色子(骰子)接连掷两次先后出现的点数.求该方程有实根的概率p 和有重根的概率q .十三、(本题满分6分)假设12,,,n X X X 是来自总体X 的简单随机样本；已知(1,2,3,4)k k EX a k ==.证明：当n 充分大时,随机变量211n n i i Z X n ==∑近似服从正态分布,并指出其分布参数.1996年全国硕士研究生入学统一考试数学三试题解析一、填空题(本题共5小题,每小题3分,满分15分,把答案填在题中横线上.) (1)【答案】()1ln dxx y +【解析】方法1：方程yx y =两边取对数得ln ln ln yx y y y ==,再两边求微分,()()11ln 1ln 1dx y dy dy dx x x y =+⇒=+()()ln 10x y +≠. 方法2：把yx y =变形得ln y yx e =,然后两边求微分得()()()ln ln 1ln 1ln y y y dx e d y y y y dy x y dy ==+=+,由此可得 ()1.1ln dy dx x y =+(2)【答案】C【解析】由()arcsin x f x dx x C =+⎰,两边求导数有()1()arcsin ()xf x x f x '==⇒=于是有1()dx f x ⎰212==⎰ ()2112x =--C =.(3)【答案】0c a≥(或2ax c =),b 任意 【解析】对2y ax bx c =++两边求导得()0022y ax b,y x ax b,''=+=+ 所以过()00x ,y 的切线方程为()()0002y y ax b x x ,-=+-即()()()200002y ax bx c ax b x x .-++=+-又题设知切线过原点()00,,把0x y ==代入上式,得2200002ax bx c ax bx ,---=--即20ax c.=由于系数0a ≠,所以,系数应满足的关系为0c a≥(或2ax c =),b 任意. (4)【答案】()1000T,,,【解析】因为A 是范德蒙行列式,由i j a a ≠知()0ijA a a =-≠∏.根据解与系数矩阵秩的关系,所以方程组T A X B =有唯一解.根据克莱姆法则,对于2111112122222133332111111111n n n n n nnn x a a a x a a a x a a a x a a a ----⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦, 易见 1230n D A ,D D D .=====所以TA XB =的解为12310n x ,x x x =====,即()1000T,,,,.【相关知识点】克莱姆法则：若线性非齐次方程组11112211211222221122,,.n n n n n n nn n n a x a x a x b a x a x a x b a x a x a x b +++=⎧⎪+++=⎪⎨⎪⎪+++=⎩ 或简记为 112nij ji j a xb ,i ,,,n ===∑其系数行列式1112121222120n n n n nna a a a a a D a a a =≠,则方程组有唯一解12j j D x ,j ,,,n.D==其中j D 是用常数项12n b ,b ,,b 替换D 中第j 列所成的行列式,即1111111121212212111,j ,j n ,j ,j n j n n,j nn,j nna ab a a a a b a a D a a b a a -+-+-+=.(5)【答案】(4.412,5.588) 【解析】可以用两种方法求解：(1)已知方差220.9σ=,对正态总体的数学期望μ进行估计,可根据 因2(,0.9)XN μ,设有n 个样本,样本均值11ni i X X n ==∑,有20.9(,)XN n μ,将其标准化,~(0,1)XN 得：)1,0(~1N nX μ-由正态分布分为点的定义21P uαα⎫⎪<=-⎬⎪⎭可确定临界值2αu ,进而确定相应的置信区间22(x u x u αα-+.(2)本题是在单个正态总体方差已知条件下,求期望值μ的置信区间问题. 由教材上已经求出的置信区间22x u x u αα⎛-+ ⎝,其中21,(0,1)P U u UN αα⎧⎫<=-⎨⎬⎩⎭,可以直接得出答案.方法1：由题设,95.01=-α,可见.05.0=α查标准正态分布表知分位点.96.12=αu 本题9n =, 5X =,因此,根据 95.0}96.11{=<-nX P μ,有 1.96}0.95P <=,即 {4.412 5.588}0.95P μ<<=,故μ的置信度为0.95的置信区间是(4.412,5.588) .方法2：由题设,95.01=-α,22222{}{}2()10.95,()0.975P U u P u U u u u ααααα<=-<<=Φ-=Φ=查得.96.12=αu20.9σ=,9n =, 5X =代入22(x u x u αα-+得置信区间(4.412,5.588).二、选择题(本题共5小题,每小题3分,满分15分.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)【答案】(D)【解析】方法1：由题设知,积分区域在极坐标系cos ,sin x r y r θθ==中是(),|0,0cos ,2D r r πθθθ⎧⎫=≤≤≤≤⎨⎬⎩⎭即是由221124x y ⎛⎫-+= ⎪⎝⎭与x 轴在第一象限所围成的平面图形,如右图.由于D 的最左边点的横坐标是0,最右点的横坐标是1, 下边界方程是0y ,=上边界的方程是y =从而D 的直角坐标表示是(){010D x,y |x ,y ,=≤≤≤≤故(D)正确.方法2：采取逐步淘汰法.由于(A)中二重积分的积分区域的极坐标表示为()1,|0,0sin ,2D r r πθθθ⎧⎫=≤≤≤≤⎨⎬⎩⎭而(B)中的积分区域是单位圆在第一象限的部分, (C)中的积分区域是正方形(){}0101x,y |x ,y ,≤≤≤≤所以,他们都是不正确的.故应选(D).(2)【答案】(A) 【解析】由于级数21nn u∞=∑和21nn v∞=∑都收敛,可见级数()221nn n uv ∞=+∑收敛.由不等式222n n n nu v u v ≤+及比较判别法知级数12n nn u v∞=∑收敛,从而12n nn u v∞=∑收敛.又因为()2222n n nnn n u v u v u v ,+=++即级数()21n n n u v ∞=+∑收敛,故应选(A).设()21112n n u ,v n ,,n ===,可知(B)不正确. 设()21112n u n ,,n n=-=,可知(C)不正确.设()()11112n nn u ,v n ,,nn--==-=,可知(D)不正确.注：在本题中命题(D)“若级数1nn u∞=∑收敛,且(1,2,)n n u v n ≥=,则级数1n n v ∞=∑也收敛.”不正确,这表明：比较判别法适用于正项级数收敛(或级数绝对收敛)的判别,但对任意项级数一般是不适用的.这是任意项级数与正项级数收敛性判别中的一个根本区别. (3)【答案】(C)【解析】伴随矩阵的基本关系式为AA A A A E **==, 现将A *视为关系式中的矩阵A ,则有()A A A E ****=. 方法一：由1n A A-*=及1()AA A*-=,可得 121()().n n A A A A AA A A--****-=== 故应选(C).方法二：由()A A A E ****=,左乘A 得1()()n AA A AA -***=,即1()()n A E A AA -**=.故应选(C). (4)【答案】(D)【解析】本题考查对向量组线性相关、线性无关概念的理解.若向量组12,,,s γγγ线性无关,即若11220s s x x x γγγ+++=,必有120,0,,0s x x x ===.既然1,,m λλ与1,,m k k 不全为零,由此推不出某向量组线性无关,故应排除(B)、(C).一般情况下,对于1122110,s s s s k k k l l αααββ++++++=不能保证必有11220,s s k k k ααα+++=及110,s s l l ββ++=故(A)不正确.由已知条件,有()()()()1111110m m m m m m k k λαβλαβαβαβ+++++-++-=,又1,,m λλ与1,,m k k 不全为零,故1111,,,,,m m m m αβαβαβαβ++--线性相关.故选(D).(5)【答案】(B) 【解析】依题意()()()()()12121212)(,.()()()()()P A A B P A B P A B P A B A B P A B P A B P B P B P B P B P B +⎡⎤++⎣⎦=+=因()0P B >,故有()()1212)(P A B A B P A B P A B +=+.因此应选(B).注：有些考生错误地选择(D).他们认为(D)是全概率公式,对任何事件B 都成立,但是忽略了全概率公式中要求作为条件的事件12,A A 应满足12()0,()0P A P A >>,且12,A A 是对立事件.【相关知识点】条件概率公式：()(|)()P AB P B A P A =.三、(本题满分6分)【解析】(1) 由于()g x 有二阶连续导数,故当0x ≠时,()f x 也具有二阶连续导数,此时,()f x '可直接计算,且()f x '连续；当0x =时,需用导数的定义求(0)f '.当0x ≠时, 22[()]()()()(1)().x x xx g x e g x e xg x g x x e f x x x ---''+-+-++'== 当0x =时,由导数定义及洛必达法则,有2000()()()(0)1(0)lim lim lim 222x x x x x x g x e g x e g x e g f x x ---→→→'''''-+--'==洛洛. 所以 2()()(1),0,()(0)1,0.2xxg x g x x e x x f x g x -'⎧-++≠⎪⎪'=⎨''-⎪=⎪⎩(2) ()f x '在0x =点的连续性要用定义来判定.因为在0x =处,有200()()(1)lim ()lim xx x xg x g x x e f x x -→→'-++'=0()()()(1)lim 2x xx g x xg x g x e x e x --→''''+-+-+= 0()(0)1lim(0)22x x g x e g f -→''''--'===. 而()f x '在0x ≠处是连续函数,所以()f x '在(,)-∞+∞上为连续函数.四、(本题满分6分) 【解析】由()z f u =可得(),()z u z u f u f u x x y y∂∂∂∂''==∂∂∂∂. 在方程()()xyu u p t dt ϕ=+⎰两边分别对,x y 求偏导数,得()(),()().u u u u u p x u p y x x y yϕϕ∂∂∂∂''=+=-∂∂∂∂ 所以()(),1()1()u p x u p y x u y u ϕϕ∂∂-==''∂-∂-. 于是 ()()()()()()()01()1()z z p x p y p x p y p y p x f u x y u u ϕϕ⎡⎤∂∂'+=-=⎢⎥''∂∂--⎣⎦.五、(本题满分6分)【分析】题的被积函数是幂函数与指数函数两类不同的函数相乘,应该用分部积分法. 【解析】方法1:因为21(1)111x x x x x xe x dxdx xd e e e e-----=-++++⎰⎰⎰分部积分 1(1)1111ln(1),1x xx x x x xx x e x dx d e e e e e x e C e---=-=-+++++=-+++⎰⎰所以20lim ln(1)ln 2.(1)1x x x x x x xe xe dx e e e -+∞-→+∞⎡⎤=-++⎢⎥++⎣⎦⎰而 lim ln(1)lim ln (1)11x x x x xxx x x xe xe e e e e e -→+∞→+∞⎡⎤⎧⎫⎡⎤-+=-+⎨⎬⎢⎥⎣⎦++⎣⎦⎩⎭lim ln(1)1x x xx xe x e e -→+∞⎧⎫=--+⎨⎬+⎩⎭lim 001xx xe →+∞-=-=+,故原式ln 2=. 方法2:220001(1)(1)1x x x x x xe xe dx dx xd e e e-+∞+∞+∞-==-+++⎰⎰⎰0000011111(1)ln(1)ln 2.1xxx x xx x xx dx dx e dx e e e e d e e e +∞-+∞+∞+∞-+∞+∞---=-+==++++=-+=-+=+⎰⎰⎰⎰六、(本题满分5分)【分析】由结论可知,若令()()x xf x ϕ=,则()()()x f x xf x ϕ''=+.因此,只需证明()x ϕ在[0,1]内某一区间上满足罗尔定理的条件.【解析】令()()x xf x ϕ=,由积分中值定理可知,存在1(0,)2η∈,使112201()()()2xf x dx x dx ϕϕη==⎰⎰,由已知条件,有1201(1)2()2()(),2f xf x dx ϕηϕη==⋅=⎰于是(1)(1)(),f ϕϕη==且()x ϕ在(,1)η上可导,故由罗尔定理可知,存在(,1)(0,1),ξη∈⊂使得()0,ϕξ'=即()()0.f f ξξξ'+=【相关知识点】1.积分中值定理：如果函数()f x 在积分区间[ ,]a b 上连续,则在[ ,]a b 上至少存在一个点ξ,使下式成立：()()()()baf x dx f b a a b ξξ=-≤≤⎰.这个公式叫做积分中值公式. 2.罗尔定理：如果函数()f x 满足(1)在闭区间[ ,]a b 上连续； (2)在开区间()a,b 内可导；(3)在区间端点处的函数值相等,即()()f a f b =, 那么在()a,b 内至少有一点ξ(a b ξ<<),使得()0f ξ'=.七、(本题满分6分)【分析】利用函数的单调性的判定,如果在x 的某个区间上导函数()0f x '≥,则函数()f x 单调递增,反之递减.【解析】(1)设售出商品的销售额为R ,则()()22(),().ab c p b aR pQ p c R p p b p b -+'==-=++ 令0,R '=得00p b ==>.当0p <<时,0R '>,所以随单价p 的增加,相应销售额R 也将增加.当p >时,有0R '<,所以随单价p 的增加,相应销售额R 将减少. (2)由(1)可知,当p =时,销售额R 取得最大值,最大销售额为2maxR b c ⎡⎤⎫⎥==⎪⎪⎥⎭⎥⎦.八、(本题满分6分) 【解析】令y z x =,则dy dzz x dx dx=+. 当0x >时,原方程化为dzz xz dx +=-,dx x =-,其通解为1ln(ln z x C =-+ 或C z x+=. 代回原变量,得通解(0)y C x =>.当0x <时,原方程的解与0x >时相同,理由如下: 令t x =-,于是0t >,而且dy dy dx dydt dx dt dx =⋅=-===.从而有通解(0)y C t +=>,即(0)y C x =<.综合得,方程的通解为y C =.注：由于未给定自变量x 的取值范围,因而在本题求解过程中,引入新未知函数yz x=后得x =,从而,应当分别对0x >和0x <求解,在类似的问题中,这一点应当牢记.九、(本题满分8分)【分析】本题的(1)是考查特征值的基本概念,而(2)是把实对称矩阵合同于对角矩阵的问题转化成二次型求标准形的问题,用二次型的理论与方法来处理矩阵中的问题. 【解析】(1)因为3λ=是A 的特征值,故31001300313138(2)0,00311311011y E A y y ------==⋅=-=-----所以2y =.(2)由于TA A =,要2()()T T AP AP P A P ==Λ,而21000010000540045A ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦是对称矩阵,故可构造二次型2T x A x ,将其化为标准形Ty y Λ.即有2A 与Λ合同.亦即2T P A P =Λ.方法一：配方法.由于 22222123434558T x A x x x x x x x =++++22222212334444222212344816165()55255495(),55x x x x x x x x x x x x x =+++++-=++++那么,令1122334444,,,,5y x y x y x x y x ===+=即经坐标变换1122334410000100,400150001x y x y x y x y ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥-⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎢⎥⎣⎦有 222221234955Tx A x y y y y =+++. 所以,取 10000100400150001P ⎡⎤⎢⎥⎢⎥=⎢⎥-⎢⎥⎢⎥⎢⎥⎣⎦,有 211()()595T T AP AP P A P ⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦. 方法二：正交变换法.二次型22222123434558T x A x x x x x x x =++++对应的矩阵为21000010000540045A ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦, 其特征多项式23100010(1)(9)005445E A λλλλλλλ---==------.2A 的特征值12341,1,1,9λλλλ====.由21()0E A x λ-=,即12340000000000044000440x x x x ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥--⎢⎥⎢⎥⎢⎥--⎣⎦⎣⎦⎣⎦,和24()0E A x λ-=,即12348000080000044000440x x x x ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥-⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦⎣⎦,分别求得对应1,2,31λ=的线性无关特征向量123(1,0,0,0),(0,1,0,0),(0,0,1,1)T T T ααα===-,和49λ=的特征向量4(0,0,1,1)Tα=.对123,,ααα用施密特正交化方法得123,,βββ,再将4α单位化为4β,其中：1234(1,0,0,0),(0,1,0,0),,T T T Tββββ====. 取正交矩阵[]123410000100000,,,P ββββ⎡⎤⎢⎥⎢⎥⎢==⎢⎢⎢⎢⎣, 则 1221119T P A P P A P -⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎣⎦, 即 211()()19T T AP AP P A P ⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎣⎦.十、(本题满分8分)【解析】证法1： (定义法)若有一组数12,,,,,t k k k k 使得1122()()()0,t t k k k k ββαβαβα+++++++= (1)则因12,,,t ααα是0AX =的解,知0(1,2,,)i A i t α==,用A 左乘上式的两边,有12()0t k k k k A β++++=. (2) 由于0A β≠,故120t k k k k ++++=. 对(1)重新分组为121122()0t t t k k k k k k k βααα++++++++=. (3)把(2)代入(3)得 11220t t k k k ααα+++=.由于12,,,t ααα是基础解系,它们线性无关,故必有120,0,,0t k k k ===.代入(2)式得:0k =. 因此向量组12,,,,t ββαβαβα+++线性无关.证法2： (用秩)经初等变换向量组的秩不变.把第一列的-1倍分别加至其余各列,有()()1212,,,,,,,,.t t ββαβαβαβααα+++→ 因此 ()()1212,,,,,,,,.t t r r ββαβαβαβααα+++=由于12,,,t ααα是基础解系,它们是线性无关的,秩()12,,,t r t ααα=,又β必不能由12,,,t ααα线性表出(否则0A β=),故()12,,,,1t r t αααβ=+.所以 ()12,,,, 1.t r t ββαβαβα+++=+即向量组12,,,,t ββαβαβα+++线性无关.十一、(本题满分7分)【解析】设一周5个工作日内发生故障的天数为X ,则X 服从二项分布即(5,0.2)B . 由二项分布的概率计算公式,有{}500.80.32768,P X ==={}14510.80.20.4096,P X C ==⋅= {}232520.80.20.2048,P X C ==⋅={}{}{}{}310120.05792.P X P X P X P X ≥=-=-=-==设一周内所获利润Y (万元),则Y 是X 的函数,且10,0,5,1,()0,2,2,3.X X Y f X X X =⎧⎪=⎪==⎨=⎪⎪-≥⎩若若若若由离散型随机变量数学期望计算公式,100.3276850.409620.05792 5.20896EY =⨯+⨯-⨯=(万元).【相关知识点】1.二项分布的概率计算公式：若(,)Y B n p ~,则{}(1)kkn kn P Y k C p p -==-, 0,1,,k n =.2.离散型随机变量数学期望计算公式：{}1()nkk k E X xP X x ==⋅=∑.十二、(本题满分6分)【解析】一枚色子(骰子)接连掷两次,其样本空间中样本点总数为36.设事件1A =“方程有实根”,2A =“方程有重根”,则{}221404B A B C C ⎧⎫=-≥=≤⎨⎬⎩⎭.用列举法求有利于i A 的样本点个数(1,2i =),具体做法见下表:有利于的意思就是使不等式24B C ≤尽可能的成立,则需要B 越大越好,C 越小越好.当B 取遍由古典型概率计算公式得到11246619(),3636p P A ++++===2111().3618q P A +===【相关知识点】古典型概率计算公式：().i i A P A =有利于事件的样本点数样本空间的总数十三、(本题满分6分) 【解析】依题意,12,,,n X X X 独立同分布,可见22212,,,n X X X 也独立同分布.由(1,2,3,4)k k EX a k ==及方差计算公式,有224222242222242211,(),111,().ii i i n nn i n ii i EX a DX EX EX a a EZ EX a DZ DX a a n nn ====-=-====-∑∑ 因此,根据中心极限定理n U =的极限分布是标准正态分布,即当n 充分大时,n Z 近似服从参数为2422(,)a a a n-的正态分布.【相关知识点】1.列维-林德伯格中心极限定理,又称独立同分布的中心极限定理：设随机变量12,,,n X X X 独立同分布,方差存在,记μ与2σ()0σ<<+∞分别是它们相同的期望和方差,则对任意实数x ,恒有1lim )(),ni n i P X n x x μ→∞=⎫-≤=Φ⎬⎭∑ 其中()x Φ是标准正态分布函数.2.方差计算公式：22()()()D X E X E X =-.。
1996年考研英语真题超详解
1996年年全真试题Part ⅠCloze TestDirections：For each numbered blank in the following passage, there are four choices marked ［A］, ［B］, ［C］and [D]. Choose the best one and mark your answer on ANSWER SHEET 1 by blackening the corresponding letter in the brackets. (10 points)Vitamins are organic compounds necessary in small amounts in the diet for the normal growth and maintenance of life of animals, including man.They do not provide energy, 1 do they construct or build any part of the body. They are needed for 2 foods into energy and body maintenance. There are thirteen or more of them, and if 3 is missing a deficiency disease becomes 4 .Vitamins are similar because they are made of the same elements—usually carbon, hydrogen, oxygen, and 5 nitrogen. They are different 6 their elements are arranged differently, and each vitamin 7 one or more specific functions in the body.8 enough vitamins is essential to life, although the body has no nutritional use for 9 vitamins. Many people, 10 , believe in being on the “safe side”and thus take extra vitamins. However, a well balanced diet will usually meet all the body’s vitamin needs.1.［A］either ［B］so ［C］nor ［D］never2.［A］shifting ［B］transferring ［C］altering ［D］transforming3.［A］any ［B］some ［C］anything ［D］something4.［A］serious ［B］apparent ［C］severe ［D］fatal5.［A］mostly ［B］partially ［C］sometimes ［D］rarely6.［A］in that ［B］so that ［C］such that ［D］except that7.［A］undertakes ［B］holds ［C］plays ［D］performs8.［A］Supplying ［B］Getting ［C］Providing ［D］Furnishing9.［A］exceptional ［B］exceeding ［C］excess ［D］external10.［A］nevertheless ［B］therefore ［C］moreover ［D］meanwhilePart ⅡReading ComprehensionDirections:Each of the passages below is followed by some questions. For each questions there are four answers marked ［A］, ［B］, ［C］and ［D］. Read the passages carefully and choose the best answer to each of the questions. Then mark your answer on ANSWER SHEET 1 by blackening the corresponding letter in the brackets with a pencil. (40 points)Passage 1Tight lipped elders used to say, “It’s not what you want in this world, but what you get.”Psychology teaches that you do get what you want if you know what you want and want the right things.You can make a mental blueprint of a desire as you would make a blueprint of a house, and each of us is continually making these blueprints in the general routine of everyday living. If we intend to have friends to dinner, we plan the menu, make a shopping list, decide which food to cook first, and such planning is an essential for any type of meal to be served.Likewise, if you want to find a job, take a sheet of paper, and write a brief account of yourself. In making a blueprint for a job, begin with yourself, for when you know exactly what you have to offer, you can intelligentlyplan where to sell your services.This account of yourself is actually a sketch of your working life and should include education, experience and references. Such an account is valuable. It can be referred to in filling out standard application blanks and is extremely helpful in personal interviews. While talking to you, your could be employer is deciding whether your education, your experience, and other qualification s will pay him to employ you and your “wares” and abilities must be displayed in an orderly and reasonably connected manner.When you have carefully prepared a blueprint of your abilities and desires, you have something tangible to sell. Then you are ready to hunt for a job. Get all the possible information about your could be job. Make inquiries as to the details regarding the job and the firm. Keep your eyes and ears open, and use your own judgment. Spend a certain amount of time each day seeking the employment you wish for, and keep in mind: Securing a job is your job now.11. What do the elders mean when they say, “It’s not what you want in this world, but what you get.”?［A］You’ll certainly get what you want.［B］It’s no use dreaming.［C］You should be dissatisfied with what you have.［D］It’s essential to set a goal for yourself.12. A blueprint made before inviting a friend to dinner is used in this passage as .［A］an illustration of how to write an application for a job［B］an indication of how to secure a good job［C］ a guideline for job description［D］ a principle for job evaluation13. According to the passage, one must write an account of himself before starting to find a job because .［A］that is the first step to please the employer［B］that is the requirement of the employer［C］it enables him to know when to sell his services［D］it forces him to become clearly aware of himself14. When you have carefully prepared a blueprint of your abilities and desires, you have something .［A］definite to offer ［B］imaginary to provide［C］practical to supply ［D］desirable to presentPassage 2With the start of BBC World Service Television, millions of viewers in Asia and America can now watch the Corporation’s news coverage, as w ell as listen to it.And of course in Britain listeners and viewers can tune in to two BBC television channels, five BBC national radio services and dozens of local radio stations. They are brought sport, comedy, drama, music, news and current affairs, education, religion, parliamentary coverage, children’s programmes and films for an annual licence fee of ￡83 per household.It is a remarkable record, stretching back over 70 years —yet the BBC’s future is now in doubt. The Corporation will survive as a publicly funded broadcasting organization, at least for the time being, but its role, its size and its programmes are now the subject of a nation wide debate in Britain.The debate was launched by the Government, which invited anyone with an opinion of the BBC — including ordinary listeners and viewers —to say what was good or bad about the Corporation, and even whether theythought it was worth keeping. The reason for its inquiry is that the BBC’s royal charter runs out in 1996 and it must decide whether to keep the organization as it is, or to make changes.Defenders of the Corporation — of whom there are many —are fond of quoting the American slogan “If it ain’t broke, don’t fix it.” The BBC “ain’t broke”, they say, by which they mean it is not broken (as distinct from the word ‘broke’, meaning having no money), so why bother to change it?Yet the BBC will have to change, because the broadcasting world around it is changing. The commercial TV channels —— ITV and Channel 4 —— were required by the Thatcher G overnment’s Broadcasting Act to become more commercial, competing with each other for advertisers, and cutting costs and jobs. But it is the arrival of new satellite channels —funded partly by advertising and partly by viewers’subscriptions — which will bring about the biggest changes in the long term.15. The world famous BBC now faces .［A］the problem of news coverage ［B］an uncertain prospect［C］inquiries by the general public ［D］shrinkage of audience16. In the passage, which of the following about the BBC is not mentioned as the key issue?［A］Extension of its TV service to Far East.［B］Programmes as the subject of a nation-wide debate.［C］Potentials for further international co-operations.［D］Its existence as a broadcasting organization.17. The BBC’s “royal charter” (Line 4, Paragraph 4) stands for.［A］the financial support from the royal family.［B］the privileges granted by the Queen.［C］ a contract with the Queen.［D］ a unique relationship with the royal family.18. The foremost reason why the BBC has to readjust itself is no other than .［A］the emergence of commercial TV channels.［B］the enforcement of Broadcasting Act by the government.［C］the urgent necessity to reduce costs and jobs.［D］the challenge of new satellite channels.Passage 3In the last half of the nineteenth century “capital” and “labour” were enlarging and perfecting their rival organizations on modern lines. Many an old firm was replaced by a limited liability company with a bureaucracy of salaried managers. The change met the technical requirements of the new age by engaging a large professional element and prevented the decline in efficiency that so commonly spoiled the fortunes of family firms in the second and third generation after the energetic founders. It was moreover a step away from individual initiative, towards collectivism and municipal and state-owned business. The railway companies, though still private business managed for the benefit of shareholders, were very unlike old family business. At the same time the great municipalities went into business to supply lighting, trams and other services to the taxpayers.The growth of the limited liability company and municipal business had important consequences. Such large, impersonal manipulation of capital and industry greatly increased the numbers and importance of shareholders as a class, an element in national life representing irresponsible wealth detached from the land and the duties of the landowners; and almost equally detached from the responsible management of business. All through the nineteenth century, America, Africa, India, Australia and parts of Europe were being developed by British capital, and Britishshareholders were thus enriched by the world’s movement towards industri alization. Towns like Bournemouth and Eastbourne sprang up to house large “comfortable” classes who had retired on their incomes, and who had no relation to the rest of the community except that of drawing dividends and occasionally attending a shareholder s’ meeting to dictate their orders to the management. On the other hand “shareholding” meant leisure and freedom which was used by many of the later Victorians for the highest purpose of a great civilization.The “shareholders” as such had no knowledge of the lives, thoughts or needs of the workmen employed by the company in which he held shares, and his influence on the relations of capital and labor was not good. The paid manager acting for the company was in more direct relation with the men and their demands, but even he had seldom that familiar personal knowledge of the workmen which the employer had often had under the more patriarchal system of the old family business now passing away. Indeed the mere size of operations and the numbers of workmen involved rendered such personal relations impossible. Fortunately, however, the increasing power and organization of the trade unions, at least in all skilled trades, enabled the workmen to meet on equal terms the managers of the companies who employed them. The cruel discipline of the strike and lockout taught the two parties to respect each other’s strength and understand the value of fair negotiation.19. It’s true of the old family firms that.［A］they were spoiled by the younger generations［B］they failed for lack of individual initiative［C］they lacked efficiency compared with modern companies［D］they could supply adequate services to the taxpayers20. The growth of limited liability companies resulted in .［A］the separation of capital from management［B］the ownership of capital by managers［C］the emergence of capital and labour as two classes［D］the participation of shareholders in municipal business21. According to the passage, all of the following are true except that .［A］the shareholders were unaware of the needs of the workers［B］the old firm owners had a better understanding of their workers［C］the limited liability companies were too large to run smoothly［D］the trade unions seemed to play a positive role22. The author is most critical of .［A］family firm owners ［B］landowners［C］managers ［D］shareholdersPassage 4What accounts for the great outburst of major inventions in early America—breakthroughs such as the telegraph, the steamboat and the weaving machine?Among the many shaping factors, I would single out the country’s excellent elementary schools; a labor force that welcomed the new technology; the practice of giving premiums to inventors; and above all the American genius for nonver bal, “spatial” thinking about things technological.Why mention the elementary schools? Because thanks to these schools our early mechanics, especially in the New England and Middle Atlantic states, were generally literate and at home in arithmetic and in some aspects of geometry and trigonometry.Acute foreign observers related American adaptiveness and inventiveness to this educational advantage. As amember of a British commission visiting here in 1853 reported, “With a mind prepared by thorough school d iscipline, the American boy develops rapidly into the skilled workman.”A further stimulus to invention came from the “premium” system, which preceded our patent system and for years ran parallel with it. This approach, originated abroad, offered inventors medals, cash prizes and other incentives.In the United States, multitudes of premiums for new devices were awarded at country fairs and at the industrial fairs in major cities. Americans flocked to these fairs to admire the new machines and thus to renew their faith in the beneficence of technological advance.Given this optimistic approach to technological innovation, the American worker took readily to that special kind of nonverbal thinking required in mechanical technology. As Eugene Ferguson has poin ted out, “A technologist thinks about objects that cannot be reduced to unambiguous verbal descriptions; they are dealt with in his mind by a visual, nonverbal process … The designer and the inventor … are able to assemble and manipulate in their minds dev ices that as yet do not exist.”This nonverbal “spatial” thinking can be just as creative as painting and writing. Robert Fulton once wrote, “The mechanic should sit down among levers, screws, wedges, wheels, etc, like a poet among the letters of the alpha bet, considering them as an exhibition of his thoughts, in which a new arrangement transmits a new idea.”When all these shaping forces—schools, open attitudes, the premium system, a genius for spatial thinking —interacted with one another on the rich U.S. mainland, they produced that American characteristic emulation. Today that word implies mere imitation. But in earlier times it meant a friendly but competitive striving for fame and excellence.23. According to the author, the great outburst of major inventions in early America was in a large part due to .［A］elementary schools ［B］enthusiastic workers［C］the attractive premium system ［D］ a special way of thinking24. It is implied that adaptiveness and inventiveness of the early American mechanics .［A］benefited a lot from their mathematical knowledge.［B］shed light on disciplined school management.［C］was brought about by privileged home training.［D］owed a lot to the technological development.25. A technologist can be compared to an artist because .［A］they are both winners of awards.［B］they are both experts in spatial thinking.［C］they both abandon verbal description［D］they both use various instruments26. The best title for this passage might be .［A］Inventive Mind ［B］Effective Schooling［C］Ways of Thinking ［D］Outpouring of InventionsPassage 5Rumor has it that more than 20 books on creationism/evolution are in the publisher’s pipelines. A few have already appeared. The goal of all will be to try to explain to a confused and often unenlightened citizenry that there are not two equally valid scientific theories for the origin and evolution of universe and life. Cosmology, geology, and biology have provided a consistent, unified, and constantly improv ing account of what happened. “Scientific”creationism, which is being pushed by some for “equal time” in the classrooms whenever the scientific accounts of evolution are given, is based on religion, not science. Virtually all scientists and the majority of nonfundamentalist religious leaders have come to regard “scientific” creationism as bad science and bad religion.The first four chapters of Kitcher’s book give a very brief introduction to evolution. At appropriate places, he introduces the criticisms of the creationists and provides answers. In the last three chapters, he takes off his gloves and gives the creationists a good beating. He describes their programmes and tactics, and, for those unfamiliar with the ways of creationists, the extent of their deception and distortion may come as an unpleasant surprise. When their basic motivation is religious, one might have expected more Christian behavior.Kitcher is a philosopher, and this may account, in part, for the clarity and effectiveness of his arguments. The non-specialist will be able to obtain at least a notion of the sorts of data and argument that support evolutionary theory. The final chapters on the creationists will be extremely clear to all. On the dust jacket of this fine book, Stephen Jay Gould says: “This book stands for reason itself.”And so it does - and all would be well were reason the only judge in the creationism/evolution debate.27. “Creationism” in the passage refers to .［A］evolution in its true sense as to the origin of the universe［B］ a notion of the creation of religion［C］the scientific explanation of the earth formation［D］the deceptive theory about the origin of the universe28. Kitcher’s book is intended to.［A］recommend the views of the evolutionists［B］expose the true features of creationists［C］curse bitterly at his opponents［D］launch a surprise attack on creationists29. From the passage we can infer that .［A］reasoning has played a decisive role in the debate［B］creationists do not base their argument on reasoning［C］evolutionary theory is too difficult for non-specialists［D］creationism is supported by scientific findings30. This passage appears to be a digest of .［A］ a book review ［B］ a scientific paper［C］ a magazine feature ［D］ a newspaper editorialPart ⅢEnglish—Chinese TranslationDirections:Read the following text carefully and then translate the underlined segments into Chinese. Your translation should be written clearly on ANSWER SHEET 2. (15 points)The differences in relative growth of various areas of scientific research have several causes. 31）Some of these causes are completely reasonable results of social needs. Others are reasonable consequences of particular advances in science being to some extent self-accelerating. Some, however, are less reasonable processes of different growth in which preconceptions of the form scientific theory ought to take, by persons in authority, act to alter the growth pattern of different areas. This is a new problem probably not yet unavoidable; but it is a frightening trend. 32）This trend began during the Second World War, when several governments came to the conclusion that the specific demands that a government wants to make of its scientific establishment cannotgenerally be foreseen in detail. It can be predicted, however, that from time to time questions will arise which will require specific scientific answers. It is therefore generally valuable to treat the scientific establishment as a resource or machine to be kept in functional order. 33）This seems mostly effectively done by supporting a certain amount of research not related to immediate goals but of possible consequence in the future.This kind of support, like all government support, requires decisions about the appropriate recipients of funds. Decisions based on utility as opposed to lack of utility are straightforward. But a decision among projects none of which has immediate utility is more difficult. The goal of the supporting agencies is the praisable one of supp orting “good” as opposed to “bad” science, but a valid determination is difficult to make. Generally, the idea of good science tends to become confused with the capacity of the field in question to generate an elegant theory. 34）However, the world is so made that elegant systems are in principle unable to deal with some of the world’s more fascinating and delightful aspects. 35）New forms of thought as well as new subjects for thought must arise in the future as they have in the past, giving rise to new standards of elegance.Section ⅣWriting36. Directions:A. Title: GOOD HEALTHB. Time limit: 40minutesC. Word limit: 120—150 words (not including the given opening sentence)D. Your composition should be based on the “OUTLINE” below and should start with th e given opening sentence: “The desire for good health is universal”.E. Your composition must be written clearly on the ANSWER SHEET.Outline:1. Importance of good health.2. Ways to keep fit.3. My own practices.1996Part ⅠCloze Test1. C2. D3. A4. B5. C6. A7. D8.B9. C 10. APart ⅡReading ComprehensionPart APassage 1 11. B 12. A 13.D 14. APassage 2 15.B 16.C 17.C 18.DPassage 3 19.C 20.A 21.C 22.DPassage 4 23.D 24.A 25.B 26.APassage 5 27.D 28.B 29.B 30.APart ⅢEnglish-Chinese Translation31.在这些原因中，有些纯属社会需求；另一些则是由于科学上某些特定发展在一定程度上自我加速而产生的必然结果。
1996年数学一真题及答案详解
1996年全国硕士研究生入学统一考试数学(一)试卷一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上)(1)设2lim()8,xx x a x a→∞+=-则a =_____________.(2)设一平面经过原点及点(6,3,2),-且与平面428x y z -+=垂直,则此平面方程为_____________. (3)微分方程22e xy y y '''-+=的通解为_____________. (4)函数ln(u x =+在点(1,0,1)A 处沿点A 指向点(3,2,2)B -方向的方向导数为_____________.(5)设A 是43⨯矩阵,且A 的秩()2,r =A 而102020,103⎡⎤⎢⎥=⎢⎥⎢⎥-⎣⎦B 则()r AB =_____________.二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)已知2()()x ay dx ydyx y +++为某函数的全微分,a 则等于(A)-1 (B)0 (C)1(D)2(2)设()f x 具有二阶连续导数,且0()(0)0,lim1,x f x f x→'''==则 (A)(0)f 是()f x 的极大值 (B)(0)f 是()f x 的极小值(C)(0,(0))f 是曲线()y f x =的拐点(D)(0)f 不是()f x 的极值,(0,(0))f 也不是曲线()y f x =的拐点 (3)设0(1,2,),n a n >=且1n n a ∞=∑收敛,常数(0,),2πλ∈则级数21(1)(tan )n n n n a n λ∞=-∑(A)绝对收敛 (B)条件收敛 (C)发散(D)散敛性与λ有关(4)设有()f x 连续的导数220,(0)0,(0)0,()()(),xf f F x x t f t dt '=≠=-⎰且当0x →时,()F x '与kx 是同阶无穷小,则k 等于(A)1(B)2 (C)3(D)4(5)四阶行列式112233440000000a b a b a b b a 的值等于(A)12341234a a a a b b b b -(B)12341234a a a a b b b b + (C)12123434()()a a b b a a b b --(D)23231414()()a a b b a a b b --三、(本题共2小题,每小题5分,满分10分) (1)求心形线(1cos )r a θ=+的全长,其中0a >是常数. (2)设1110,1,2,),n x x n +==试证数列{}n x 极限存在,并求此极限.四、(本题共2小题,每小题6分,满分12分)(1)计算曲面积分(2),Sx z dydz zdxdy ++⎰⎰其中S 为有向曲面22(01),z xy x =+≤≤其法向量与z 轴正向的夹角为锐角.(2)设变换 2u x y v x ay =-=+可把方程2222260z z zx x y y∂∂∂+-=∂∂∂∂简化为20,z u v ∂=∂∂求常数.a五、(本题满分7分) 求级数211(1)2nn n ∞=-∑的和.六、(本题满分7分)设对任意0,x >曲线()y f x =上点(,())x f x 处的切线在y 轴上的截距等于01(),xf t dt x⎰求()f x 的一般表达式.七、（本题满分8分）设()f x 在[0,1]上具有二阶导数,且满足条件(),(),f x a f x b ''≤≤其中,a b 都是非负常数,c 是(0,1)内任意一点.证明()2.2b fc a '≤+八、（本题满分6分）设,TA =-I ξξ其中I 是n 阶单位矩阵,ξ是n 维非零列向量,Tξ是ξ的转置.证明(1)2=A A 的充分条件是 1.T=ξξ(2)当1T=ξξ时,A 是不可逆矩阵. 九、（本题满分8分）已知二次型222123123121323(,,)55266f x x x x x cx x x x x x x =++-+-的秩为2,(1)求参数c 及此二次型对应矩阵的特征值. (2)指出方程123(,,)1f x x x =表示何种二次曲面.十、填空题(本题共2小题,每小题3分,满分6分.把答案填在题中横线上)(1)设工厂A 和工厂B 的产品的次品率分别为1%和2%,现从由A 和B 的产品分别占60%和40%的一批产品中随机抽取一件,发现是次品,则该次品属A 生产的概率是____________.(2)设,ξη是两个相互独立且均服从正态分布2)N 的随机变量,则随机变量ξη-的数学期望()E ξη-=____________.十一、（本题满分6分）设,ξη是两个相互独立且服从同一分布的两个随机变量,已知ξ的分布率为1(),1,2,3.3P i i ξ=== 又设max(,),min(,).X Y ξηξη==(1)(2)求随机变量X 的数学期望().E X1996年全国硕士研究生入学统一考试数学(一)答案详解一、填空题 （1）【答】 .2ln 【详解】因为,31lim 2lim 333a ax ax aa x x x x e a x a a x a x =⎥⎥⎦⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛-+=⎪⎭⎫ ⎝⎛-+--∞→∞→于是 2ln 83=⇒=a e a （2）【答】0322=-+z y x【详解】 原点与点)2,3,6(-连线的方向向量为);2.3,6(-=s 平面824=+-z y x 的法向量为{},2,1,4-=n 根据题意，所求平面的法向量为.322214236k j i kj in s -+=--=⨯ 故所求平面方程为,0)0(3)0(2)0(2=---+-z y x 即 0322=-+z y x（3）【答】 xxxe x e C x e C y ++=sin cos 21【详解】 对应齐次方程的特征方程为 ,0222=+-λλ 解得特征根为 ,12,1i ±=λ由于1=α不是特征根，可设原方程的特解为,**=Ae y代入原方程解得,1=A 故所求通解为 xxxe x e C x e C y ++=sin cos 21 （4）【答】.21【详解】因为,211)1,0,1(22=++=∂∂z y x x u A,01)1,0,1(2222=+⋅++=∂∂zy yzy x y u A,211)1,0,1(2222=+⋅++=∂∂z y z zy x z u A ,31cos ,32cos ,32cos =-==γβα 沿AB 方向的单位向量为,31,32,32⎭⎬⎫⎩⎨⎧-故u 沿AB 方向的方向导数为 2131213203221=⋅+⎪⎭⎫ ⎝⎛-⋅÷+⋅=∂∂AB u （5）【答】 .2【详解】 因为,010301020201≠=-=B说明矩阵B 可逆，故秩=)(AB r 秩,2)(=A r 二、选择题（1）【答】应选（D ） 【详解】2)()(y x ydydx ay x +++为某函数的全微分的充要条件是,)()(22⎪⎪⎭⎫ ⎝⎛++∂∂=⎪⎪⎭⎫ ⎝⎛+∂∂y x ay x y y x y x 即,2)2(y ay x a -=-- .0))(2(=--y x a 当且仅当2=a 时上式恒成立，故正确选项（D ）.（2）【答】应选（B ）【详解】由题设1)(lim '0=→x x f x 根据极限的性质知，存在0=x 的某领域，在此领域内有 .0)(.0)(''≥≥x f xx f 即又根据泰勒公式，2"'!2)()0()0()(x f x f f x f ξ++=其中ξ在0与x 之间， 从而)0(!2)()0()(2"f x f f x f ≥+=ξ 可见)0(f 是)(x f 的极小值，故正确选项为（B ）.（3）【答】应选（A ） 【详解】由于,tan tan)1(22n n n a nn a n n ⋅=⎪⎭⎫⎝⎛-λλ 而,tan lim λλ=∞→nn n 所以当n 充分大时，n n a a nn 22)1(tan+<⋅λλ又正项级∑∞=1n na收敛，所以其偶数项数列构成的级数∑∞=12n na也收敛，从而n n n a n n 21tan )1(⎪⎭⎫ ⎝⎛-∑∞=λ绝对收敛， 故正确选项为（A ）. （4）【答】应选（C ）【详解】因为)()()(2)()()(220'0022'x f x x f x dt t f x dt t f t dt t f x x F x x x -+=⎥⎦⎤⎢⎣⎡-=⎰⎰⎰⎰=xdt t f x.)(2又根据题设)('x F 与k x 是同阶无穷小，且,0)0(,0)0('≠=f f于是有 2000'0)1()(2lim)(2lim )(lim-→→→-==⎰k x kxx k x x k x f x dtt f x xx F)0()()1(1lim230--⋅-=-→x f x f x k k x,0)1(1lim)0(230'≠-⋅=-→k x x k f可见应有3=k ，故正确选项为（C ）. （5）【答】应选（D ）【详解】 按第一行展开，00000000000043322143322144332211b a b b a b a a b b a a a b a b b a b a -⋅= 332241332241a b b a b b a b b a a a -=).)((41413232b b a a b b a a --= 故正确选项为（D ）. 三、（1）【详解】因为θθθθθθθd a d r r ds a r 2cos 2)()(,sin )(2'2'=+=-=利用对称性知，所求心形线的全长 a a d a s 82sin 82cos 220===⎰ππθθθ（2）【详解】由46.10121=+==x x x 知，.21x x >设对某个正整数k 有1+>k k x x 则 .66211+++=+>+=k k k k x x x x故由归纳法知，对一切正整数n ，都有1+>n n x x ，即数列{}n x 为单调减少数列。
1996真题及解析
1996年全国硕士研究生入学统一考试数学三试题一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上.)⑴ 设方程x = y y确定y是x的函数,则dy = _____________ .1⑵ 设『xf (x)dx = arcsinx+C,则］------ dx = ___________ ..f(x)⑶ 设x o,y o是抛物线y = ax2 bx c上的一点，若在该点的切线过原点，则系数应满足的关系是 . ⑷设-1 1 1 1 H 1f xj 1a1 a2 a3 1 H a n X2 1A = 2a1+2a2■a；I■H a2q,X = X3++,B = 1+n A.a1An A.a24a异Iq\\ an：+1 其中a i Ha j(i式j;i, j =12111, n).则线性方程组A J X = B的解是 ____________ . (5)设由来自正态总体X~N(^0.92)容量为9的简单随机样本，得样本均值X -5,则未知参数卩的置信度为0.95的置信区间为___________________________ .二、选择题(本题共5小题，每小题3分,满分15分.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.)cos •二(1)累次积分.02小，［f(rco^,rsinRrdr可以写成()1 j y-y2(A) 0dy 0 f(x,y)dx1 1(C) 0dx0f(x,y)dy(2)下述各选项正确的是O0 QO QO(A)若7 u：和7 v2都收敛，则7 (u n v n)2收敛nF nF n T1 j1T(B) pdy。
f (x, y)dx1 x」2(D) dx f (x, y)dyL0 %()(C)oO若正项级数v u nn d1发散,则u n丄-n(B) Z u n v n收敛，则瓦u；与瓦v；都收敛n m n T n TOQQ Q(D)若级数7 u n 收敛，且U n _Vn ( n =12|)l),则级数v V n 也收敛nJ nJ⑶ 设n 阶矩阵A 非奇异(n_ 2), A”是矩阵A 的伴随矩阵,则()n 1n(A) (A)二 A A (B) (A )二 A A (C) (A)二 A n，A(D)(A ) =A 「2A⑷ 设有任意两个n 维向量组：elllCm 和若存在两组不全为零的数’l」l(,'m和 k i ,||( ,k m ，使(「kjr(m • k m)> m(1 — kJ 「( m - KJ F = 0 ,则()(A)〉1」l(,〉m 和：1」l(, ：m 都线性相关(B)〉1」l(,〉m 和-1^1, ：m 都线性无关(C)M ' -iJH ,■ ：m/^ - -iJH/^ - ：m 线性无关(D)M ■ -lJH, ：^ ■ -m/^ - -iJH/^ - -m 线性相关⑸ 已知0 ::： P(B) ::： 1且P[ A • A B]二P(A B) • P(A B),则下列选项成立的是()(A) P[ A A 2 ：B] =P(A B) P(A 2 B) (B) P(AB+"B )=P(AB) +卩(砂) (C) P (A +A2)=P (A|B )+ P (4|B ) (D) p B 二 P A P(B A) P(A 2)P(B A 2)x 式0其中g(x)有二阶连续导数，且g(0) =1,g(0) --1 . x = 0,(1)求 f (x)；(2)讨论f (x)在(」：，=)上的连续性(本题满分6分)[g(x)-e*设 f(x)二 x0,四、(本题满分6分)x设函数z 二f(u),方程U 二「(u) • p(t)dt 确定u 是x,y 的函数,其中f(u)」(u)可Jy微；p(t),「(U)连续,且：(U)".求 p(y)三 p(x)三. ex dy五、(本题满分6分)丄 -x 严 xe(1 e")2六、(本题满分5分)1设f (x)在区间［0,1］上可微，且满足条件f(1) = 2 jxf (x)dx .试证：存在-(0,1)使f( ) f ( )=0.七、(本题满分6分)设某种商品的单价为 p 时，售出的商品数量 Q 可以表示成Q=—^-c ,其中a b 、p + bc 均为正数，且a bc .(1) 求p 在何范围变化时，使相应销售额增加或减少•(2) 要使销售额最大，商品单价p 应取何值？最大销售额是多少？ 八、(本题满分6分)九、(本题满分8分)(1)已知A 的一个特征值为3,试求y ；⑵ 求矩阵P,使(AP )T(AP )为对角矩阵 十、(本题满分8分)设向量〉1」2」l(「t 是齐次线性方程组 AX =0的一个基础解系，向量［不是方程组AX =0的解，即A lt^O.试证明：向量组 打」心1」■■:上“线性无关• 十一、（本题满分7分）假设一部机器在一天内发生故障的概率为0.2,机器发生故障时全天停止工作，若一周5计算 dx .求微分方程 也」j 的通解.dxx-0 1 0设矩阵A = 1 0 0 0 0 y0 0 1011 2个工作日里无故障，可获利润10万元；发生一次故障仍可获得利润5万元；发生两次故障所获利润0元；发生三次或三次以上故障就要亏损2万元.求一周内期望利润是多少？十二、（本题满分6分）考虑一元二次方程x2 Bx 0 ,其中B、C分别是将一枚色子（骰子）接连掷两次先后出现的点数•求该方程有实根的概率p和有重根的概率q.十三、（本题满分6分）假设X i,X2,|l（,X n是来自总体X的简单随机样本；已知EX k二a k（k =123,4）•1n2证明：当n充分大时，随机变量Z n=丄^ X：近似服从正态分布，并指出其分布参数.n y1996年全国硕士研究生入学统一考试数学三试题解析(本题共5小题,每小题3分,满分15分,把答案填在题中横线上.)dx = e yln y d y ln y = y y 1 In y dy = x 1 In y dy ,【解析】对y 二ax 2bx c 两边求导得y = 2ax b,y ' x 0二2ax 0 b,所以过x °,y 。
1996考研数学一真题及答案解析
1996年全国硕士研究生入学统一考试数学一试题一、填空题(本题共5个小题,每小题3分,满分15分,把答案填在题中横线上.)(1) 设,则___________.2lim(8xx x a x a→∞+=-a =(2) 设一平面经过原点及点(6,-3,2),且与平面垂直,则此平面方程为428x y z -+= ___________.(3) 微分方程的通解为___________.22xy y y e '''-+=(4) 函数在点处沿点指向点方向的方向导数为ln(u x =+(1,0,1)A A (3,2,2)B -___________.(5) 设是矩阵,且的秩,而,则___________.A 43⨯A ()2r A =102020103B ⎛⎫ ⎪= ⎪ ⎪-⎝⎭()r AB =二、选择题(本题共5个小题,每小题3分,满分15分.在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) 已知为某函数的全微分,则等于 ( )2()()x ay dx ydyx y +++a (A) -1 (B) 0 (C) 1 (D) 2(2) 设有二阶连续导数,且,,则 ( )()f x (0)0f '=0()lim 1||x f x x →''==(A) 是的极大值 (0)f ()f x (B) 是的极小值(0)f ()f x (C) 是曲线的拐点(0,(0))f ()y f x = (D) 不是的极值,也不是曲线的拐点(0)f ()f x (0,(0))f ()y f x =(3) 设,且收敛,常数,则级数0(1,2,)n a n >= 1n n a ∞=∑(0,2πλ∈21(1)(tan )n n n n a n λ∞=-∑ ( )(A) 绝对收敛 (B) 条件收敛 (C) 发散 (D) 收敛性与有关λ(4) 设有连续的导数,,,,且当()f x (0)0f =(0)0f '≠220()()()xF x x t f t dt =-⎰x →时,与是同阶无穷小,则等于 ( )()F x 'kx k (A) 1 (B) 2 (C) 3 (D) 4(5) 四阶行列式的值等于 ( )112233440000000a b a b b a b a (A) (B)12341234a a a a b b b b -12341234a a a a b b b b +(C) (D) 12123434()()a a b b a a b b --23231414()()a a b b a a b b --三、(本题共2小题,每小题5分,满分10分.)(1) 求心形线的全长,其中是常数.(1cos )r a θ=+0a >(2) 设,,试证数列极限存在,并求此极限.110x=11,2,)n x n +== {}n x 四、(本题共2小题,每小题6分,满分12分.)(1) 计算曲面积分,其中为有向曲面,其法(2)Sx z dydz zdxdy ++⎰⎰S 22(01)z x y z =+≤≤向量与轴正向的夹角为锐角.z (2) 设变换可把方程化简为,求常数,其中2,u x y u x ay =-⎧⎨=+⎩2222260z z z x x y y ∂∂∂+-=∂∂∂∂20zu v ∂=∂∂a 有二阶连续的偏导数.(,)z z x y =五、(本题满分7分)求级数的和.221(1)2nn n ∞=-∑六、(本题满分7分)设对任意,曲线上点处的切线在轴上的截距等于0x >()y f x =(,())x f x y ,求的一般表达式.01()xf t dt x⎰()f x 七、(本题满分8分)设在上具有二阶导数,且满足条件,,其中都是非()f x [0,1]|()|f x a ≤|()|f x b ''≤,a b负常数,是(0,1)内任一点,证明.c |()|22b fc a '≤+八、(本题满分6分)设,其中是阶单位矩阵,是维非零列向量,是的转置,证明：T A E ξξ=-E n ξn Tξξ(1) 的充要条件是；(2) 当时,是不可逆矩阵.2A A =1Tξξ=1T ξξ=A 九、(本题满分8分)已知二次型的秩为2.222123123121323(,,)55266f x x x x x cx x x x x x x =++-+-(1) 求参数及此二次型对应矩阵的特征值；c (2) 指出方程表示何种二次曲面.123(,,)1f x x x =十、填空题(本题共2小题,每小题3分,满分6分.)(1) 设工厂和工厂的产品的次品率分别为1%和 2%,现从由和的产品分别占60%和A B A B 40%的一批产品中随机抽取一件,发现是次品,则该次品属生产的概率是__________.A(2) 设、是两个相互独立且均服从正态分布的随机变量,则随机变量ξη2)N 的数学期望__________.ξη-()E ξη-=十一、(本题满分6分.)设、是相互独立且服从同一分布的两个随机变量,已知的分布律为,ξηξ{}13P i ξ===1,2,3,又设,.i max(,)X ξη=min(,)Y ξη=(1) 写出二维随机变量的分布律：(,)X YX Y123123(2) 求随机变量的数学期望.X ()E X1996年全国硕士研究生入学统一考试数学一试题解析一、填空题(本题共5个小题,每小题3分,满分15分,把答案填在题中横线上.)(1)【答案】ln 2【解析】这是型未定式求极限.1∞方法一： ,3323lim()lim(1)x a axx a xax x x a a x a x a-⋅-→∞→∞+=+--令,则当时,,3at x a=-x →∞0t →则 ,1303lim(1)lim(1)x aa t x t a t e x a -→∞→+=+=-即 .33lim lim 312lim()x x ax ax a x a x x a e e e x a→∞→∞-→∞+===-由题设有,得.38ae=1ln 8ln 23a ==方法二：,2223()2221lim 112lim lim lim 11lim 1x xa xax a x a x x a x x x a a x a a a x a e x x x e a x a e a a x x x ⋅→∞-→∞→∞→∞-⋅-→∞⎛⎫⎛⎫⎛⎫+++ ⎪ ⎪ ⎪+⎛⎫⎝⎭⎝⎭===== ⎪ ⎪-⎝⎭⎛⎫ ⎪-⎛⎫- ⎪-⎝⎭ ⎪⎝⎭⎝⎭由题设有,得.38ae=1ln 8ln 23a ==(2)【答案】2230x y z +-=【解析】方法一：所求平面过原点与,其法向量；O 0(6,3,2)M -{}06,3,2n OM ⊥=-平面垂直于已知平面,它们的法向量也互相垂直：；428x y z -+={}04,1,2n n ⊥=-由此, .00//632446412i j kn OM n i j k ⨯=-=--+-取,则所求的平面方程为.223n i j k =+-2230x y z +-=方法二：所求平面即为过原点,与两个不共线的向量(一个是从原点到点的向量0(6,3,2)M -,另一是平面的法向量)平行的平面,{}06,3,2OM =- 428x y z -+={}04,1,2n =-即 ,即 .6320412x y z-=-2230x y z +-=(3)【答案】12(cos sin 1)xe c x c x ++【解析】微分方程所对应的齐次微分方程的特征方程为22xy y y e '''-+=,解之得.故对应齐次微分方程的解为.2220r r -+=1,21r i =±12(cos sin )x y e C x C x =+由于非齐次项不是特征根,设所给非齐次方程的特解为,代入,1xe αα=*()xy x ae =得(也不难直接看出),故所求通解为22x y y y e '''-+=1a =*()x y x e =.1212(cos sin )(cos sin 1)x x x y e C x C x e e C x C x =++=++【相关知识点】① 二阶线性非齐次方程解的结构：设是二阶线性非齐次方程*()y x 的一个特解.是与之对应的齐次方程()()()y P x y Q x y f x '''++=()Y x的通解,则是非齐次方程的通解.()()0y P x y Q x y '''++=*()()y Y x y x =+② 二阶常系数线性齐次方程通解的求解方法：对于求解二阶常系数线性齐次方程的通解,可用特征方程法求解：即中的、均是常数,方程()Y x ()()0y P x y Q x y '''++=()P x ()Q x 变为.其特征方程写为,在复数域内解出两个特征根；0y py qy '''++=20r pr q ++=12,r r 分三种情况：(1) 两个不相等的实数根,则通解为12,r r 1212;rx r x y C eC e =+(2) 两个相等的实数根,则通解为12r r =()112;rxy C C x e =+(3) 一对共轭复根,则通解为其中1,2r i αβ=±()12cos sin .xy e C x C x αββ=+12,C C 为常数.③ 对于求解二阶线性非齐次方程的一个特解,可用待定()()()y P x y Q x y f x '''++=*()y x 系数法,有结论如下：如果则二阶常系数线性非齐次方程具有形如()(),xm f x P x e λ=*()()k xm y x x Q x eλ=的特解,其中是与相同次数的多项式,而按不是特征方程的根、是特征方()m Q x ()m P x k λ程的单根或是特征方程的重根依次取0、1或2.如果,则二阶常系数非齐次线性微分方程()[()cos ()sin ]xl n f x e P x x P x x λωω=+的特解可设为()()()y p x y q x y f x '''++=,*(1)(2)[()cos ()sin ]k x m m y x e R x x R x x λωω=+其中与是次多项式,,而按(或)不是特征(1)()m R x (2)()m R x m {}max ,m l n =k i λω+i λω-方程的根、或是特征方程的单根依次取为或.01(4)【答案】12【分析】先求方向的方向余弦和,然后按方向导数的计算公式l ,,u u ux y z∂∂∂∂∂∂求出方向导数.cos cos cos u u u u l x y zαβγ∂∂∂∂=++∂∂∂∂【解析】因为与同向,为求的方向余弦,将l AB l {}{}31,20,212,2,1AB =----=-单位化，即得 .{}{}12,2,1cos ,cos ,cos 3||AB l AB αβγ==-=将函数分别对求偏导数得ln(u x =,,x y z,12Au x ∂==∂,0Au y ∂==∂,12Au z∂==∂所以cos cos cos AA AA u u u ulx y z αβγ∂∂∂∂=++∂∂∂∂ .1221110()233232=⨯+⨯-+⨯=(5)【答案】2【解析】因为,所以矩阵可逆,故.102020100103B ==≠-B ()()2r AB r A ==【相关知识点】.若可逆,则()min((),())r AB r A r B ≤A .1()()()[()]()r AB r B r EB r A AB r AB -≤==≤从而,即可逆矩阵与矩阵相乘不改变矩阵的秩.()()r AB r B =二、选择题(本题共5个小题,每小题3分,满分15分.在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)【答案】(D)【解析】由于存在函数,使得 ,(,)u x y 22()()()x ay dx ydydu x y x y +=+++由可微与可偏导的关系,知,,2()u x ay x x y ∂+=∂+2()u yy x y ∂=∂+分别对求偏导数,得,y x ,2243()()2()(2)()()u a x y x ay x y a x ayx y x y x y ∂+-+⋅+--==∂∂++.232()u yy x x y ∂-=∂∂+由于与连续,所以,即2u y x ∂∂∂2u x y ∂∂∂22u u y x x y∂∂=∂∂∂∂,33(2)2()()a x ay yx y x y ---=++2a ⇒=故应选(D).(2)【答案】(B)【解析】因为有二阶连续导数,且所以由函数极限的局部保号()f x 0()lim 10,||x f x x →''=>性可知,在的空心领域内有,即,所以为单调递增.0x =()0||f x x ''>()0f x ''>()f x '又由,在由负变正,由极值的第一充分条件,是的极(0)0f '=()f x '0x =0x =()f x 小值点,即是的极小值.应选(B).(0)f ()f x 【相关知识点】极限的局部保号性：设若(或)当lim ().x x f x A →=0A >0A <⇒0,δ∃>时,(或).00x x δ<-<()0f x >()0f x <(3)【答案】(A)【解析】若正项级数收敛,则也收敛,且当时,有1nn a∞=∑21nn a∞=∑n →+∞.tanlim (tan limn n n n n nλλλλλ→+∞→+∞=⋅=用比较判别法的极限形式,有.22tanlim0nn nn a na λλ→+∞=>因为收敛,所以也收敛,所以原级数绝对收敛,应选(A).21n n a ∞=∑2lim tann x n a nλ→+∞【相关知识点】正项级数比较判别法的极限形式：设和都是正项级数,且则1n n u ∞=∑1n n v ∞=∑lim,nn nv A u →∞=(1) 当时,和同时收敛或同时发散；0A <<+∞1nn u∞=∑1nn v∞=∑(2) 当时,若收敛,则收敛；若发散,则发散；0A =1nn u∞=∑1nn v∞=∑1nn v∞=∑1nn u∞=∑(3) 当时,若收敛,则收敛；若发散,则发散.A =+∞1nn v∞=∑1nn u∞=∑1nn u∞=∑1nn v∞=∑(4)【答案】(C)【解析】用洛必达法则.由题可知 ,220()()()xxF x xf t dt t f t dt =-⎰⎰对该积分上限函数求导数,得,220()2()()()2()x xF x x f t dt x f x x f x x f t dt '=+-=⎰⎰所以 0010002()2()()lim lim lim x xk k k x x x x f t dt f t dt F x x x x-→→→'==⎰⎰ .23002()2()limlim(1)(1)(2)k k x x f x f x k x k k x --→→'---洛洛因为与是同阶无穷小,且,所以为常数,即时()F x 'kx (0)0f '≠302()lim(1)(2)k x f x k k x -→'--3k =有 ,300()2()limlim (0)0(1)(2)k k x x F x f x f x k k x-→→'''==≠--故应选(C).【相关知识点】设在同一个极限过程中,为无穷小且存在极限 ,(),()x x αβ()lim()x l x αβ=(1) 若称在该极限过程中为同阶无穷小；0,l ≠(),()x x αβ(2) 若称在该极限过程中为等价无穷小,记为；1,l =(),()x x αβ()()x x αβ(3) 若称在该极限过程中是的高阶无穷小,记为.0,l =()x α()x β()()()x o x αβ=若不存在(不为),称不可比较.()lim()x x αβ∞(),()x x αβ(5)【答案】(D)【解析】可直接展开计算,22221331334400000000a b a b D a b a b b a a b =-,22221414232314143333()()a b a b a a b b a a b b a a b b b a b a =-=--所以选(D).三、(本题共2小题,每小题5分,满分10分.)(1)【解析】由极坐标系下的弧微分公式得ds a θθ==.2cos2a a d θθθ==由于以为周期,因而的范围是.()(1cos )r r a θθ==+2πθ[0,2]θπ∈又由于,心形线关于极轴对称.由对称性,()()r r θθ=-.24cos 8sin 822s ds a d a a πππθθθ⎡⎤====⎢⎥⎣⎦⎰⎰(2)【解析】用单调有界准则.由题设显然有,数列有下界.0n x >{}n x 证明单调减：用归纳法.；设,则nx 214x x ===<1n n x x -<.1n n x x +=<=由此,单调减.由单调有界准则,存在.n x lim n n x →+∞设,在恒等式两边取极限,即lim ,(0)n n x a a →+∞=≥1n x +=,1lim lim n n n x a +→+∞=⇒=解之得(舍去).3a =2a =-【相关知识点】1.单调有界准则：单调有界数列必有极限.2. 收敛数列的保号性推论：如果数列从某项起有(或),且,那{}n x 0n x ≥0n x ≤lim n n x a →∞=么(或).0a ≥0a ≤四、(本题共2小题,每小题6分,满分12分.)(1)【分析一】见下图所示,在平面与平面上的投影均易求出,分别为S xOy yOz ；22:1xy D x y +≤,或.2:11,1yz D y y z-≤≤≤≤01,z y ≤≤≤≤图1求,自然投影到平面上.求时,若投影到平面上,被积函Szdxdy ⎰⎰xOy (2)Sx z dydz +⎰⎰xOy 数较简单且可利用对称性.【分析二】令,则.(,,)2,(,,)0,(,,)P x y z x z Q x y z R x y z z =+==SI Pdydz Rdxdy =+⎰⎰这里,,若用高斯公式求曲面积分,则较简单.因不是封闭曲213P Q R x y z∂∂∂++=+=∂∂∂I S 面,故要添加辅助曲面.【解析】方法一：均投影到平面上,则xOy ,22(2)[(2)(()]xySD zI x z dydz zdxdy x z x y dxdy x∂=++=+-++∂⎰⎰⎰⎰其中,.22z x y =+22:1xy D x y +≤把代入,得2zx x∂=∂,2222242()()xyxyxyD D D I x dxdy x x y dxdy x y dxdy =--+++⎰⎰⎰⎰⎰⎰由对称性得,,222()0xyD x x y dxdy +=⎰⎰22242()xyxyD D x dxdy x y dxdy =+⎰⎰⎰⎰所以 .22()xyD I x y dxdy =-+⎰⎰利用极坐标变换有.121340001242I d r dr r ππθπ⎡⎤=-=-=-⎢⎥⎣⎦⎰⎰方法二：分别投影到平面与平面.yOz xOy 投影到平面时要分为前半部分yOz S 1:S x=2:S x =(见图1),则.12(2)(2)S S SI x z dydz x z dydz zdxdy =++++⎰⎰⎰⎰⎰⎰由题设,对法向量与轴成钝角,而对法向量与轴成锐角.将化成二重积分得1S x 2S xI 2222)()()4().yzyzxyyzxy D D D D D I z dydz z dydz x y dxdyx y dxdy =-++-+++=-++⎰⎰⎰⎰⎰⎰⎰⎰2213111221131242200sin 2()344(1)cos 3343,34224yzz yD z y y t dy z y dyy dy tdt πππ=--====-=-=⋅⋅=⋅⎰⎰⎰⎰⎰⎰⎰或21101.24yzD dz dz ππ===⎰⎰⎰⎰(这里的圆面积的一半.)(同方法一).22()2xyD x y dxdy π+=⎰⎰因此, 4.422I πππ=-⋅+=-方法三：添加辅助面,法方向朝下,则221:1(1)S z x y =+≤,11(2)1S S Dx z dydz zdxdy dxdy dxdy π++==-=-⎰⎰⎰⎰⎰⎰其中是在平面的投影区域：.D 1S xy 221x y +≤与即与围成区域,与的法向量指向内部,所以在上S 1S 22z x y =+1z =ΩS 1S ΩΩ满足高斯公式的条件,所以1(2)3S S x z dydz zdxdy dVΩ++=-⎰⎰⎰⎰⎰ ,11()3332D z dz dxdy zdz ππ=-=-=-⎰⎰⎰⎰其中,是圆域：,面积为.()D z 22x y z +≤z π因此,.133(2)()222S I x z dydz zdxdy ππππ=--++=---=-⎰⎰(2)【解析】由多元复合函数求导法则,得,z z u z v z zx u x v x u v ∂∂∂∂∂∂∂=+=+∂∂∂∂∂∂∂,2z z u z v z z a y u y v y u v∂∂∂∂∂∂∂=+=-+∂∂∂∂∂∂∂所以 22222222((z z z z u z v z v z ux x u x v u x u v x v x v u x ∂∂∂∂∂∂∂∂∂∂∂∂∂=+=⋅+⋅+⋅+⋅∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂ ,222222z z zu u v v∂∂∂=++∂∂∂∂2222222()(z z z z u z v z v z u x y y u y v u y u v y v y v u y∂∂∂∂∂∂∂∂∂∂∂∂∂=+=⋅+⋅+⋅+⋅∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂,222222(2)z z za a u u v v ∂∂∂=-+-+∂∂∂∂222222222222222((2()()44.z z z a y y u y vz u z v z v z ua u y u v y v y v u yz z z a a u u v v ∂∂∂∂∂=-+∂∂∂∂∂∂∂∂∂∂∂∂∂=-⋅+⋅+⋅+⋅∂∂∂∂∂∂∂∂∂∂∂∂∂=-+∂∂∂∂代入,并整理得2222260z z zx x y y∂∂∂+-=∂∂∂∂.2222222226(105)(6)0z z z z z a a a x x y y u v v∂∂∂∂∂+-=+++-=∂∂∂∂∂∂∂于是,令得或.260a a +-=3a =2a =-时,,故舍去,时,,因此仅当时化简为.2a =-1050a +=3a =1050a +≠3a =20zu v∂=∂∂【相关知识点】多元复合函数求导法则：若和在点处偏导数存(,)u u x y =(,)v v x y =(,)x y 在,函数在对应点具有连续偏导数,则复合函数在(,)z f u v =(,)u v [(,),(,)]z f u x y v x y =点处的偏导数存在,且(,)x y .,z f u f v z f u f vx u x v x y u y v y∂∂∂∂∂∂∂∂∂∂=+=+∂∂∂∂∂∂∂∂∂∂五、(本题满分7分)【解析】先将级数分解,212211222131111)(1)2211111111.212122n n n n n n n n n n n n A n n n n n n n ∞∞+==∞∞∞∞+++======---+=⋅-⋅=--+⋅⋅∑∑∑∑∑∑令 ,122131122n nn n A A nn∞∞+====⋅⋅∑∑则 .12A A A =-由熟知幂级数展开式,即,得ln(1)x +11(1)ln(1)(11)n nn x x x n -∞=-+=-<≤∑,1121111(1)1111()ln(1ln 2242424n n n n n A n n -∞∞+==-==--=--=⋅∑∑12331211(1)1(22(1)11111115(()ln(1ln 2,22222288n nn n n n n n A n n n -∞∞==-∞=-==--⋅-=-----=----=-∑∑∑因此, .1253ln 284A A A =-=-六、(本题满分7分)【解析】曲线上点处的切线方程为()y f x =(,())x f x.()()()Y f x f x X x '-=-令得轴上的截距.由题意,0X =y ()()Y f x f x x '=-.01()()()xf t dt f x f x x x' =-⎰为消去积分,两边乘以,得 , (*)x 20()()()xf t dt xf x f x x ' =-⎰将恒等式两边对求导,得x ,2()()()2()()f x f x xf x xf x x f x ''''=+--即 .()()0xf x f x '''+=在(*)式中令得自然成立.故不必再加附加条件.就是说是微分方程0x =00=()f x 的通解.下面求解微分方程.0xy y '''+=0xy y '''+=方法一：,()100xy y xy xy C ''''''+=⇒=⇒=因为,所以,0x >1C y x'=两边积分得 .12()ln y f x C x C ==+方法二：令,则,解得.()y P x '=y P '''=0xP P '+=1C y P x'==再积分得.12()ln y f x C x C ==+七、(本题满分8分)【解析】由于问题涉及到与的关系,自然应当利用泰勒公式,而且应在点展开：,f f 'f ''c ,在与之间.2()()()()()()2!f f x f c f x x c x c ξ'''=+-+-ξc x 分别取得0,1x =,在与之间,20()(0)()()(0)(0)2!f f f c f c c c ξ'''=+-+-0ξc 0 ,在与之间,21()(1)()()(1)(1)2!f f f c f c c c ξ'''=+-+-1ξc 1两式相减得 ,22101(1)(0)()[()(1)()]2!f f f c f c f c ξξ'''''-=+--于是 .22101()(1)(0)[()(1)()]2!f c f f f c f c ξξ'''''=----由此 221011()(1)(0)()(1)()2!2!f c f f f c f c ξξ'''''≤++-+ .2212[(1)]222b a bc c a ≤+-+<+八、(本题满分6分)【解析】(1)因为,为数,为阶矩阵,所以TA E ξξ=-TξξT ξξn ,2()()2()(2)T T T T T T T A E E E E ξξξξξξξξξξξξξξ=--=-+=--因此, 2(2)(1)0TT T T T A A E E ξξξξξξξξξξ=⇔--=-⇔-=因为是非零列向量,所以,故即.ξ0Tξξ≠210,TA A ξξ=⇔-=1Tξξ=(2)反证法.当时,由(1)知,若可逆,则.1Tξξ=2A A =A 121A A A A A E --===与已知矛盾,故是不可逆矩阵.T A E E ξξ=-≠A 九、(本题满分8分)【解析】(1)此二次型对应的矩阵为.51315333A c -⎛⎫⎪=-- ⎪ ⎪-⎝⎭因为二次型秩 ,由()()2r f r A ==513440400153153163333336A c c c -⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪=--→--→-- ⎪ ⎪ ⎪⎪ ⎪ ⎪---⎝⎭⎝⎭⎝⎭可得.再由的特征多项式3c =A 513||153(4)(9)333E A λλλλλλλ---=-=----求得二次型矩阵的特征值为.0,4,9(2)因为二次型经正交变换可化为,故222349y y +,即.123(,,)1f x x x =2223491y y +=表示椭圆柱面.【相关知识点】主轴定理：对于任一个元二次型n,12(,,,)T n f x x x x Ax = 存在正交变换(为阶正交矩阵),使得x Qy =Q n ,2221122()T T T n nx Ax y Q AQ y y y y λλλ==+++ 其中是实对称矩阵的个特征值,的个列向量是对应于12,,,n λλλ A n Q n 12,,,n ααα A 特征值的标准正交特征向量.12,,,n λλλ 十、填空题(本题共2小题,每小题3分,满分6分.)(1)【答案】37【解析】设事件“抽取的产品是次品”,事件“抽取的产品是工厂生产的”,C =D =A 则事件表示“抽取的产品是工厂生产的”,依题意有D B .()0.60,()0.40,(|)0.01,(|)0.02P D P D P C D P C D ====应用贝叶斯公式可以求得条件概率：(|)P D C .()(|)0.60.013(|)0.60.010.40.027()(|)()(|)P D P C D P D C P D P C D P D P C D ⨯===⨯+⨯+【相关知识点】贝叶斯公式：设试验的样本空间为.为的事件,为E S A E 12,,,n B B B S 的一个划分,且,则()0,()0(1,2,,)i P A P B i n >>= (*)1()(|)(|),1,2,,.()(|)i i i njjj P B P A B P B A i n P B P A B ===∑ (*)式称为贝叶斯公式.(2)【解析】由于与相互独立且均服从正态分布,因此它们的线性函数ξη2)N 服从正态分布,且U ξη=-()0,EU E E E ξηξη=-=-=,()11122DU D D D ξηξη=-=+=+=所以有 .(0,1)U N 代入正态分布的概率密度公式,有.22()u f u du --∞=⎰应用随机变量函数的期望公式有22(||)(||)||u E E U u du ξη+∞--∞-= =⎰222u udu+∞-=⎰由凑微分法,有.222(||)2(2u uE d ξη--=--⎰22u +∞-==【相关知识点】对于随机变量与均服从正态分布,则与的线性组合亦服从正态分X Y X Y 布.若与相互独立,由数学期望和方差的性质,有X Y ,()()()E aX bY c aE X bE Y c ++=++,22()()()D aX bY c a D X b D Y ++=+其中为常数.,,a b c 十一、(本题满分6分.)【解析】易见的可能取值为(1,1),(2,1),(2,2),(3,1),(3,2),(3,3).依题意(,)X Y ,故,即{}X Y <=∅{}0P X Y <=,{}{}{}1,21,32,30P X Y P X Y P X Y ========={}{}1,1max(,)1,min(,)1P X Y P ξηξη=====.{}{}{}11,1119P P P ξηξη=======类似地可以计算出所有的值列于下表中,得到随机变量的联合分布律：ij p (,)X YX Y123119002291903292919(2)将表中各行元素相加求出的边缘分布X ,123135999X ⎡⎤⎢⎥⎢⎥⎣⎦由离散型随机变量数学期望计算公式可得.135221239999EX =⋅+⋅+⋅=【相关知识点】1.离散型随机变量的边缘分布计算公式：二维离散型随机变量关于与的边缘概率分布或边缘分布律分别定义为：(,)X Y X Y {}{},,1,2,i i i j ij jjp P X x P X x Y y p i ⋅=======∑∑{}{},,1,2,j j i j ij iip P Y y P X x Y y p j ⋅=======∑∑它们分别为联合分布律表格中第行与第列诸元素之和.i j 2. 离散型随机变量数学期望计算公式：.{}1()nkk k E X xP X x ==⋅=∑。
考研1996年真题详解
本题考核的知识点是 ( 习惯搭配 ? 动词词义辨析 ! " 精准定位#空格所在句子的含义是 需要维生素将食物 ++ 为能量 " 以维持身体的健康 . !空 并符合文意 ! 强调的是 事物大的变革或质 格填入的动词需和" & # -搭配 " # ( $ & ’ 1 ( % 常与" & # -搭配 " 的改变. !在此从1 ) 食物* 到) ) 能量* 的转变是一种质的改变 " 因此 " 0 & ) ( G 符合句意 ! 其他项都 2 + 不与" " 状态 / 观点态度的改变 " & # -搭配 " ’ / " 1 #与# ( $ & ’ 1 ) (常用 + 1 ( %4 4 4 # 4 4 4 ’ / " 1 #常指位置 / # ( $ & ’ ; 程度较轻 ! 1 ) (多用于位置的改变’ $ , # ) (强调部分或少量的变动 " " 空格设置# 且有较多的近义词汇可以考查 ! 例 # ( $ & ’ 1 ( % 是此处说明维生素的作用的关键词 " 句如 ( # / )6 ( * ) ’ ’9 / " * /* $ # ) ( " , , $ ( ’ $ ( ) # ( $ & ’ 1 ( % ) 0 " & # -9 . # # ) ( 1 , " ) ’毛虫变为蝴蝶的过程 ! +3 6 " 干扰设置#其他项都构成对# 风 ( $ & ’ 1 ( % 的近义干扰 !例句如 ( 5 / )3 " & 0’ / " 1 # ) 0 # # / )’ . # / 4 转向南吹 !M 老板把他调到了另一个办公室 !5 " ’ ) % , ) ( # ( $ & ’ 1 ) ( ( ) 0/ " %# -$ & # / ) (1 1 " * ) 4 / ) ’ ) 6 + ’ 这些衣服太大 " 得修改 ! * , # / ) ’ $ ( ) # , $ ( ) # / ) . ’ # 9 ) $ , # ) ( ) 0 4 2 +% 知识点补充 ( 为前缀 " 与动词连用 " 表示 横过 " 越过 . 或转变 " 转移 . " 例如 ( ) 办理" # ( $ & ’ ; # ( $ & ’ $ * # 交易 "谈判"处理* " ) 横渡大西洋的 * " ) 跨越国界的 * " ) 移植 * " # ( $ & ’ $ # , $ & # " * # ( $ & ’ ; & $ # " & $ , # ( $ & ’ , $ & # 6 ) 运输* ! # ( $ & ’ ( # 6 任何一个 < 4 # E$ $ & +任何 " # $ > $ & # / " & + 2任何事物 本题考核的知识点是 ( 不定代词的用法 ! # $ F ’ % )一些 # G$ ’ % ) # / " & 2某事物
1996年考研数学三真题及全面解析
1996年全国硕士研究生入学统一考试数学三试题一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上.)(1) (1) 设方程设方程yx y =确定y 是x 的函数的函数,,则dy =___________. (2) (2) 设设()arcsin x f x dx x C =+ò,则1()dx f x =ò___________.. (3) (3) 设设()00,x y 是抛物线2y ax bx c =++上的一点上的一点,,若在该点的切线过原点若在该点的切线过原点,,则系数应满足的关系是的关系是___________. ___________. (4) (4) 设设123222212311111231111n n n n n n n a a a a A a a a a a a a a ----éùêúêúêú=êúêúêúëû,123n x x X x x éùêúêúêú=êúêúêúëû,1111B éùêúêúêú=êúêúêúëû,其中(;,1,2,,)i j a a i j i j n ¹¹=.则线性方程组TA XB =的解是的解是___________. ___________.(5) (5) 设由来自正态总体设由来自正态总体2~(,0.9)X N m 容量为9的简单随机样本的简单随机样本,,得样本均值5X =,则未知参数m 的置信度为0.95的置信区间为的置信区间为___________. ___________.二、选择题(本题共5小题,每小题3分,满分15分.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) (1) 累次积分累次积分cos20(cos ,sin )d f r r rdr pqq q q òò可以写成可以写成 ( ) ( )(A) 210(,)y y dy f x y dx -òò(B) 21100(,)y dy f x y dx -òò (C)1100(,)dx f x y dy òò(D) 21(,)x x dx f x y dy -òò(2) (2) 下述各选项正确的是下述各选项正确的是下述各选项正确的是 ( ) ( ) (A) (A) 若若21nn u ¥=å和21nn v ¥=å都收敛都收敛,,则21()n n n u v ¥=+å收敛收敛(B)1n n n u v ¥=å收敛收敛,,则21nn u ¥=å与21nn v ¥=å都收敛都收敛 (C) (C) 若正项级数若正项级数1n n u ¥=å发散发散,,则1n u n³(D) (D) 若级数若级数1nn u¥=å收敛收敛,,且(1,2,)n n u v n ³=,则级数1n n v ¥=å也收敛也收敛(3) (3) 设设n 阶矩阵A 非奇异非奇异((2n ³),A *是矩阵A 的伴随矩阵的伴随矩阵,,则 ( )(A) 1()nA AA -**= (B) 1()nA AA +**= (C) 2()nA AA -**= (D) 2()nA AA +**=(4) (4) 设有任意两个设有任意两个n 维向量组1,,m a a 和1,,m b b ,若存在两组不全为零的数1,,m l l和1,,m k k ,使111111()()()()0m m m m m m k k k k l a l a l b l b +++++-++-=,则( )(A) 1,,m a a 和1,,m b b 都线性相关都线性相关(B) 1,,m a a 和1,,m b b 都线性无关都线性无关(C) 1111,,,,,m m m m a b a b a b a b ++--线性无关线性无关(D) 1111,,,,,m m m m a b a b a b a b ++--线性相关线性相关(5) (5) 已知已知0()1P B <<且()1212[]()()P A A B P A B P A B +=+,则下列选项成立的是则下列选项成立的是( ) ( ) (A) ()1212[]()()P A A B P A B P A B +=+(B) ()1212()()P A B A B P A B P A B +=+ (C) ()1212()()P A A P A B P A B +=+ (D) ()()1122()()()P B P A P B A P A P B A =+三、(本题满分6分)设(),0,()0,0,xg x e x f x x x -ì-¹ï=íï=î其中()g x 有二阶连续导数有二阶连续导数,,且(0)1,(0)1g g ¢==-. (1)(1)求求()f x ¢；(2)(2)讨论讨论()f x ¢在(,)-¥+¥上的连续性上的连续性. .四、(本题满分6分)设函数()z f u =,方程()()xyu u p t dt j =+ò确定u 是,x y 的函数的函数,,其中(),()f u u j 可微；()p t ,()u j ¢连续连续,,且()1u j ¢¹.求()()z z p y p x x y¶¶+¶¶.五、(本题满分6分)计算2(1)xx xe dx e -+¥-+ò.六、(本题满分5分)设()f x 在区间[0,1]上可微上可微,,且满足条件120(1)2()f xf x dx =ò.试证试证::存在(0,1)x Î使()()0.f f x x x ¢+=七、(本题满分6分)设某种商品的单价为p 时,售出的商品数量Q 可以表示成a Q c p b=-+,其中a b 、、c 均为正数均为正数,,且a bc >.(1) (1) 求求p 在何范围变化时在何范围变化时,,使相应销售额增加或减少使相应销售额增加或减少. .(2) (2) 要使销售额最大要使销售额最大要使销售额最大,,商品单价p 应取何值应取何值??最大销售额是多少最大销售额是多少? ?八、(本题满分6分)求微分方程22yx y dy dx x-+=的通解的通解. .九、(本题满分8分)设矩阵010010000010012A y éùêúêú=êúêúëû.(1) (1) 已知已知A 的一个特征值为3,3,试求试求y ； (2) (2) 求矩阵求矩阵P ,使()()TAP AP 为对角矩阵为对角矩阵. .十、(本题满分8分)设向量12,,,t a a a 是齐次线性方程组0AX =的一个基础解系的一个基础解系,,向量b 不是方程组不是方程组0AX =的解的解,,即0A b ¹.试证明试证明::向量组12,,,,t b b a b a b a +++线性无关线性无关. .十一、(本题满分7分)假设一部机器在一天内发生故障的概率为0.2,0.2,机器发生故障时全天停止工作机器发生故障时全天停止工作机器发生故障时全天停止工作,,若一周5个工作日里无故障个工作日里无故障,,可获利润10万元；发生一次故障仍可获得利润5万元；发生两次故障所获利润0元；发生三次或三次以上故障就要亏损2万元万元..求一周内期望利润是多少求一周内期望利润是多少? ?十二、(本题满分6分)考虑一元二次方程20x Bx C ++=,其中B C 、分别是将一枚色子分别是将一枚色子((骰子骰子))接连掷两次先后出现的点数先后出现的点数..求该方程有实根的概率p 和有重根的概率q .十三、(本题满分6分)假设12,,,n X X X 是来自总体X 的简单随机样本；已知(1,2,3,4)kk EX a k ==.证明：当n 充分大时充分大时,,随机变量211nn i i Z X n==å近似服从正态分布近似服从正态分布,,并指出其分布参数并指出其分布参数. .1996年全国硕士研究生入学统一考试数学三试题解析一、填空题(本题共5小题,每小题3分,满分15分,把答案填在题中横线上.) (1)(1)【答案】【答案】()1ln dx x y +【解析】方法1：方程y x y =两边取对数得ln ln ln yx y y y ==,再两边求微分再两边求微分, ,()()11ln 1ln 1dx y dy dy dx x x y =+Þ=+()()ln 10x y +¹. 方法2：把yx y =变形得ln y yx e=,然后两边求微分得然后两边求微分得()()()ln ln 1ln 1ln y y y dx e d y y y y dy x y dy ==+=+,由此可得由此可得()1.1ln dydx x y =+(2)(2)【答案】【答案】()32113x C --+【解析】由()arcsin x f x dx x C =+ò,两边求导数有两边求导数有()2211()arcsin 1()1xf x x x x f x x ¢==Þ=--, 于是有于是有1()dx f x ò2221112x x dx x dx =-=-òò()221112x d x =---ò()32113x C =--+. (3)(3)【答案】【答案】0c a³(或20ax c =),b 任意任意【解析】对22y ax bx c =++两边求导得()0022y ax b,y x ax b,¢¢=+=+ 所以过()00x ,y 的切线方程为()()0002y y ax b x x ,-=+-即()()()200002y ax bx c ax b x x .-++=+-又题设知切线过原点()00,,把0x y ==代入上式代入上式,,得2200002ax bx c ax bx ,---=--即20ax c.=由于系数0a ¹,所以所以,,系数应满足的关系为0c a³(或20ax c =),b 任意任意.. (4)(4)【答案】【答案】()1000T,,,【解析】因为A 是范德蒙行列式是范德蒙行列式,,由i j a a ¹知()0ijA a a =-¹Õ.根据解与系数矩阵秩的关系秩的关系,,所以方程组T A X B =有唯一解有唯一解. .根据克莱姆法则根据克莱姆法则,,对于对于2111112122222133332111111111n n n n n nnn x a a a x a a a x a a a x a a a ----éùéùéùêúêúêúêúêúêúêúêúêú=êúêúêúêúêúêúêúêúêúëûëûëû, 易见易见 1230n D A ,D D D .=====所以TA XB =的解为12310n x ,x x x =====,即()1000T,,,,.【相关知识点】克莱姆法则：若线性非齐次方程组【相关知识点】克莱姆法则：若线性非齐次方程组11112211211222221122,,.n n n n n n nn n n a x a x a x b a x a x a x b a x a x a x b +++=ìï+++=ïíïï+++=î或简记为或简记为 112nij ji j a xb ,i ,,,n ===å其系数行列式其系数行列式1112121222120nn n n nna a a a a a D a a a =¹, 则方程组有唯一解则方程组有唯一解12j j D x ,j ,,,n.D==其中j D 是用常数项12n b ,b ,,b 替换D 中第j 列所成的行列式列所成的行列式,,即1111111121212212111,j ,j n ,j ,j nj n n ,j nn ,j nna ab a a a a b a a D a a b a a -+-+-+=.(5)(5)【答案】【答案】(4.412,5.588) 【解析】可以用两种方法求解：【解析】可以用两种方法求解：(1)(1)已知方差已知方差220.9s =,对正态总体的数学期望m 进行估计进行估计,,可根据可根据 因2(,0.9)XN m ,设有n 个样本个样本,,样本均值11ni i X X n ==å,有20.9(,)XN nm ,将其标准化将其标准化,,由公式()~(0,1)()X E X N D X n-得：得：)1,0(~1N nX m- 由正态分布分为点的定义211X P u n a ma ìü-ïï<=-íýïïîþ可确定临界值2a u ,进而确定相应的置信区间22(,)x u x u nnaass-+.(2)(2)本题是在单个正态总体方差已知条件下本题是在单个正态总体方差已知条件下本题是在单个正态总体方差已知条件下,,求期望值m 的置信区间问题的置信区间问题. .由教材上已经求出的置信区间22,x u x u n n a as s æö-+ç÷èø, 其中21,(0,1)P U u U N aa ìü<=-íýîþ,可以直接得出答案可以直接得出答案. .方法1：由题设由题设,,95.01=-a ,可见.05.0=a 查标准正态分布表知分位点.96.12=a u 本题9n =, 5X =, , 因此因此因此,,根据根据 95.0}96.11{=<-nX P m,有5{ 1.96}0.9519P m -<=,即 {4.412 5.588}0.95P m <<=,1 xyO1212故m 的置信度为0.95的置信区间是(4.412,5.588) .方法2：由题设由题设,,95.01=-a ,22222{}{}2()10.95,()0.975P U u P u U u u u a a a a a <=-<<=F -=F =查得.96.12=a u20.9s =,9n =, 5X =代入22(,)x u x u nnaass-+得置信区间(4.412,5.588).二、选择题(本题共5小题,每小题3分,满分15分.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)(1)【答案】【答案】【答案】(D) (D) 【解析】方法1：由题设知：由题设知,,积分区域在极坐标系cos ,sin x r y r q q ==中是中是(),|0,0cos ,2Dr rp q q qìü=££££íýîþ即是由221124x y æö-+=ç÷èø与x轴在第一象限所围成的轴在第一象限所围成的 平面图形平面图形,,如右图如右图. .由于D 的最左边点的横坐标是0,最右点的横坐标是1, 下边界方程是0y ,=上边界的方程是2y x x =-,从而D 的直角坐标表示是的直角坐标表示是(){}2010D x,y |x ,y x x,=££££-故(D)(D)正确正确正确. .方法2：采取逐步淘汰法：采取逐步淘汰法..由于由于(A)(A)(A)中二重积分的积分区域的极坐标表示为中二重积分的积分区域的极坐标表示为中二重积分的积分区域的极坐标表示为()1,|0,0sin ,2D r r pq q q ìü=££££íýîþ而(B)(B)中的积分区域是单位圆在第一象限的部分中的积分区域是单位圆在第一象限的部分中的积分区域是单位圆在第一象限的部分, , (C)(C)中的积分区域是正方形中的积分区域是正方形(){}0101x,y |x ,y ,££££所以所以,,他们都是不正确的他们都是不正确的..故应选故应选(D). (D).(2)(2)【答案】【答案】【答案】(A) (A) 【解析】由于级数21nn u¥=å和21nn v¥=å都收敛都收敛,,可见级数()221nn n uv ¥=+å收敛收敛..由不等式由不等式 222n n n n u v u v £+及比较判别法知级数12n nn u v¥=å收敛收敛,,从而12n nn u v¥=å收敛收敛. .又因为()2222n n nnn n u v u v u v ,+=++即级数()21n n n u v ¥=+å收敛收敛,,故应选故应选(A). (A). 设()21112n n u ,v n ,,n ===,可知可知(B)(B)(B)不正确不正确不正确. . 设()21112n u n ,,n n =-=,可知可知(C)(C)(C)不正确不正确不正确. .设()()11112n n nu ,v n ,,n n --==-=,可知可知(D)(D)(D)不正确不正确不正确. . 注：在本题中命题在本题中命题(D)(D)(D)“若级数“若级数1nn u¥=å收敛收敛,,且(1,2,)n n u v n ³=,则级数1n n v ¥=å也收敛也收敛..”不正确不正确,,这表明：比较判别法适用于正项级数收敛比较判别法适用于正项级数收敛((或级数绝对收敛或级数绝对收敛))的判别的判别,,但对任意项级数一般是不适用的一般是不适用的..这是任意项级数与正项级数收敛性判别中的一个根本区别这是任意项级数与正项级数收敛性判别中的一个根本区别. . (3)(3)【答案】【答案】【答案】(C) (C)【解析】伴随矩阵的基本关系式为AA A A A E **==, 现将A *视为关系式中的矩阵A ,则有()A A A E ****=.方法一：由1n A A-*=及1()A A A*-=,可得可得 121()().n nA A A A AAA A--****-===故应选故应选(C). (C).方法二：由()A A A E ****=,左乘A 得1()()n AA A AA -***=,即1()()n A E A AA -**=.故应选故应选(C). (C). (4)(4)【答案】【答案】【答案】(D) (D)【解析】本题考查对向量组线性相关、线性无关概念的理解【解析】本题考查对向量组线性相关、线性无关概念的理解..若向量组12,,,s g g g 线性无关无关,,即若11220s s x x x g g g +++=,必有120,0,,0sx x x ===.既然1,,m l l 与1,,m k k 不全为零不全为零,,由此推不出某向量组线性无关由此推不出某向量组线性无关,,故应排除故应排除(B)(B)(B)、、(C). 一般情况下一般情况下一般情况下,,对于对于1122110,s s s s k k k l l a a a b b ++++++=不能保证必有11220,s s k k k a a a +++=及110,s s l l b b ++=故(A)(A)不正确不正确不正确..由已知条件由已知条件,,有()()()()1111110m m m m m m k k l a b l a b a b a b +++++-++-=,又1,,m l l 与1,,m k k 不全为零不全为零,,故1111,,,,,m m m m a b a b a b a b ++--线性相关线性相关..故选故选(D). (D).(5)(5)【答案】【答案】【答案】(B) (B) 【解析】依题意【解析】依题意 ()()()()()12121212)(,.()()()()()P A A B P A B P A B P A B A B P A B P A B P B P B P B P B P B +éù++ëû=+=因()0P B >,故有()()1212)(P A B A B P A B P A B +=+.因此应选因此应选(B).(B). 注：有些考生错误地选择：有些考生错误地选择(D).(D).(D).他们认为他们认为他们认为(D)(D)(D)是全概率公式是全概率公式是全概率公式,,对任何事件B 都成立都成立,,但是忽略了全概率公式中要求作为条件的事件12,A A 应满足12()0,()0P A P A >>,且12,A A 是对立事件.【相关知识点】条件概率公式：()(|)()P AB P B A P A =.三、(本题满分6分)【解析】【解析】(1) (1) (1) 由于由于()g x 有二阶连续导数有二阶连续导数,,故当0x ¹时,()f x 也具有二阶连续导数也具有二阶连续导数,,此时,()f x ¢可直接计算可直接计算,,且()f x ¢连续；当0x =时,需用导数的定义求(0)f ¢.当0x ¹时, 22[()]()()()(1)().x xxx g x e g x exg x g x x ef x x x ---¢¢+-+-++¢==当0x =时,由导数定义及洛必达法则由导数定义及洛必达法则,,有2000()()()(0)1(0)lim lim lim 222x x x x x x g x e g x e g x e g f x x ---®®®¢¢¢¢¢-+--¢==洛洛. 所以所以 2()()(1),0,()(0)1,0.2xxg x g x x e x xf xg x -¢ì-++¹ïï¢=í¢¢-ï=ïî(2) ()f x ¢在0x =点的连续性要用定义来判定点的连续性要用定义来判定..因为在0x =处,有200()()(1)lim ()lim xx x xg x g x x e f x x -®®¢-++¢=()()()(1)lim 2x xxg x xg x g x e x e x --®¢¢¢¢+-+-+=()(0)1lim(0)22x x g x e g f -®¢¢¢¢--¢===.而()f x ¢在0x ¹处是连续函数处是连续函数,,所以()f x ¢在(,)-¥+¥上为连续函数上为连续函数. .四、(本题满分6分)【解析】由()z f u =可得(),()zu z u f u f u x x y y¶¶¶¶¢¢==¶¶¶¶. 在方程()()xyu u p t dt j =+ò两边分别对,x y 求偏导数求偏导数,,得()(),()().u u u uu p x u p y x x y y j j ¶¶¶¶¢¢=+=-¶¶¶¶所以所以()(),1()1()u p x u p y x u y u j j ¶¶-==¢¢¶-¶-. 于是于是 ()()()()()()()01()1()z z p x p y p x p y p y p x f u xy u u j j éù¶¶¢+=-=êú¢¢¶¶--ëû.五、(本题满分6分)【分析】题的被积函数是幂函数与指数函数两类不同的函数相乘【分析】题的被积函数是幂函数与指数函数两类不同的函数相乘,,应该用分部积分法应该用分部积分法. . 【解析】方法1:因为因为21(1)111xxx x xxe x dx dx xd ee e e -----=-++++òòò分部积分 1(1)1111ln(1),1x x x x x x x xx e x dx d e e e e e x e C e ---=-=-+++++=-+++òò 所以所以 2lim ln(1)ln 2.(1)1xx x x x x xe xe dx e e e -+¥-®+¥éù=-++êú++ëûò而 lim ln(1)lim ln (1)11x x x x x x xx x xe xe e e ee e -®+¥®+¥éùìüéù-+=-+íýêúëû++ëûîþlim ln(1)1x x x x xe x e e-®+¥ìü=--+íý+îþlim001xx x e®+¥-=-=+,故原式ln 2=. 方法2:221(1)(1)1x x xxxxe xe dx dx xde e e-+¥+¥+¥-==-+++òòò00011111(1)ln(1)ln 2.1xxx x x xxxx dx dx e dx ee e e d e e e +¥-+¥+¥+¥-+¥+¥---=-+==++++=-+=-+=+òòòò六、(本题满分5分)【分析】由结论可知【分析】由结论可知,,若令()()x xf x j =,则()()()x f x xf x j ¢¢=+.因此因此,,只需证明()x j 在[0,1]内某一区间上满足罗尔定理的条件内某一区间上满足罗尔定理的条件. .【解析】令()()x xf x j =,由积分中值定理可知由积分中值定理可知,,存在1(0,)2h Î,使112201()()()2xf x dx x dx j j h ==òò,由已知条件由已知条件,,有1201(1)2()2()(),2f xf x dx j h j h ==×=ò于是于是(1)(1)(),f j j h ==且()x j 在(,1)h 上可导上可导,,故由罗尔定理可知故由罗尔定理可知,,存在(,1)(0,1),x h ÎÌ使得使得()0,j x ¢=即()()0.f f x x x ¢+=【相关知识点】【相关知识点】1.1.1.积分中值定理：如果函数积分中值定理：如果函数()f x 在积分区间[ ,]a b 上连续上连续,,则在[ ,]a b 上至少存在一个点x ,使下式成立：使下式成立：()()()()baf x dx f b a a b x x =-££ò.这个公式叫做积分中值公式这个公式叫做积分中值公式. . 2.2.罗尔定理：如果函数罗尔定理：如果函数()f x 满足满足(1)(1)在闭区间在闭区间[ ,]a b 上连续；上连续；(2)(2)在开区间在开区间()a,b 内可导；内可导；(3)(3)在区间端点处的函数值相等在区间端点处的函数值相等在区间端点处的函数值相等,,即()()f a f b =, 那么在()a,b 内至少有一点x (a b x <<),),使得使得()0f x ¢=.七、(本题满分6分)【分析】利用函数的单调性的判定【分析】利用函数的单调性的判定,,如果在x 的某个区间上导函数()0f x ¢³,则函数()f x 单调递增单调递增,,反之递减反之递减. .【解析】【解析】(1)(1)(1)设售出商品的销售额为设售出商品的销售额为R ,则()()222(),().ab c p b aR pQ p c R p p b p b -+¢==-=++ 令0,R ¢=得 0()0ab b p b a bc cc=-=->.当0()b p a bc c<<-时,0R ¢>,所以随单价p 的增加的增加,,相应销售额R 也将增加也将增加. .当()b p a bc c>-时,有0R ¢<,所以随单价p 的增加的增加,,相应销售额R 将减少将减少. .(2)(2)由由(1)(1)可知可知可知,,当()bp a bc c =-时,销售额R 取得最大值取得最大值,,最大销售额为最大销售额为2max()ab a R b c a bc c ab céùæöêú=--=-ç÷ç÷êúèøêúëû.八、(本题满分6分)【解析】令yz x =,则dy dz z x dx dx=+.当0x >时,原方程化为21dz z x z z dx +=-+,即21dz dx x z=-+,其通解为其通解为 21ln(1)ln z z x C ++=-+ 或 2C1z z x++=. 代回原变量代回原变量,,得通解22(0)y x y C x ++=>.当0x <时,原方程的解与0x >时相同时相同,,理由如下理由如下: : 令t x =-,于是0t >,而且而且222222yx y y x y y t y dy dy dx dy dt dx dt dx x xt -+-+-+=×=-=-==-.从而有通解22(0)y t y C t ++=>,即22(0)y x y C x ++=<.综合得综合得,,方程的通解为22y x y C ++=.注：由于未给定自变量x 的取值范围的取值范围,,因而在本题求解过程中因而在本题求解过程中,,引入新未知函数yz x=后得后得 2221x y x z +=+,从而从而,,应当分别对0x >和0x <求解求解,,在类似的问题中在类似的问题中,,这一点应当牢记这一点应当牢记. .九、(本题满分8分) 【分析】本题的【分析】本题的(1)(1)(1)是考查特征值的基本概念是考查特征值的基本概念是考查特征值的基本概念,,而(2)(2)是把实对称矩阵合同于对角矩阵的问题是把实对称矩阵合同于对角矩阵的问题转化成二次型求标准形的问题转化成二次型求标准形的问题,,用二次型的理论与方法来处理矩阵中的问题用二次型的理论与方法来处理矩阵中的问题. . 【解析】【解析】(1)(1)(1)因为因为3l =是A 的特征值的特征值,,故31001300313138(2)0,003113110011y E A y y ------==×=-=-----所以2y =.(2)(2)由于由于T A A =,要2()()T T AP AP P A P ==L ,而21000010000540045A éùêúêú=êúêúëû是对称矩阵是对称矩阵,,故可构造二次型2Tx A x ,将其化为标准形Ty y L .即有2A 与L 合同合同..亦即2T P A P =L .方法一：配方法配方法. .由于由于 22222123434558T x A x x x x x x x =++++22222212334444222212344816165()55255495(),55x x x x x x x x x x x x x =+++++-=++++那么那么,,令1122334444,,,,5y x y x y x x y x ===+=即经坐标变换即经坐标变换1122334410000100,400150001x y x y x y x y éùéùéùêúêúêúêúêúêú=êúêú-êúêúêúêúëûëûêúëû有 222221234955T x A x y y y y =+++. 所以所以,,取 10000100400150001P éùêúêú=êú-êúêúêúëû,有 211()()595T TAP AP P A P éùêúêú==êúêúêúêúëû.方法二：正交变换法正交变换法. .二次型22222123434558T x A x x x x x x x =++++对应的矩阵为对应的矩阵为21000010000540045A éùêúêú=êúêúëû,其特征多项式其特征多项式2310000100(1)(9)0054045E A l l l l l l l ---==------.2A 的特征值12341,1,1,9l l l l ====.由21()0E A x l -=,即123400000000000044000440x x x x éùéùéùêúêúêúêúêúêú=êúêúêú--êúêúêú--ëûëûëû, 和224()0E A x l -=,即123480000080000044000440x x x x éùéùéùêúêúêúêúêúêú=êúêúêú-êúêúêú-ëûëûëû, 分别求得对应1,2,31l =的线性无关特征向量的线性无关特征向量1231(1,0,0,0),0,0,0),(0,1(0,1,0,0),0,0),(0,0,1(0,0,1,,1)TTT a a a ===-, 和49l =的特征向量4(0,0,1(0,0,1,1,1,1))T a =.对123,,a a a 用施密特正交化方法得123,,b b b ,再将4a 单位化为4b ,其中：其中：12341111(1,0,0,0),(0,1,0,0),(0,0,,),(0,0,,)2222T T T T b b b b ===-=.取正交矩阵取正交矩阵[]123411100001000000,,,221122P b b b b éùêúêúêú==êúêúêúêû-úë, 则 1221119T P A P P A P -éùêúêú==êúêúëû, 即 211()()19T T AP AP P A P éùêú==êúêúëû.十、(本题满分8分)【解析】证法1： ( (定义法定义法定义法))若有一组数12,,,,,tk k k k 使得使得1122()()()0,t t k k k k b b a b a b a +++++++= (1)则因12,,,t a a a 是0AX =的解的解,,知0(1,2,,)i A i t a ==,用A 左乘上式的两边左乘上式的两边,,有12()0t k k k k A b ++++=. (2)由于0A b ¹,故120t k k k k ++++=.对(1)(1)重新分组为重新分组为121122()0t t t k k k k k k k b a a a ++++++++=. (3)把(2)(2)代入代入代入(3)(3)(3)得得 11220t t k k k a a a +++=.由于12,,,t a a a 是基础解系是基础解系,,它们线性无关它们线性无关,,故必有120,0,,0tk k k ===.代入代入(2)(2)(2)式得式得式得::0k =. 因此向量组12,,,,t b b a b a b a +++线性无关线性无关. .证法2： ( (用秩用秩用秩))经初等变换向量组的秩不变经初等变换向量组的秩不变..把第一列的把第一列的-1-1倍分别加至其余各列倍分别加至其余各列,,有()()1212,,,,,,,,.t t b b a b a b a b a a a +++®因此因此 ()()1212,,,,,,,,.t t r r b b a b a b a b a a a +++= 由于12,,,t a a a 是基础解系是基础解系,,它们是线性无关的它们是线性无关的,,秩()12,,,t r t a a a =,又b 必不能由12,,,t a a a 线性表出线性表出((否则0A b =),),故故()12,,,,1tr t a a a b =+.所以所以 ()12,,,, 1.t r t b b a b a b a +++=+即向量组12,,,,t b b a b a b a +++线性无关线性无关. .十一、(本题满分7分)【解析】设一周5个工作日内发生故障的天数为X ,则X 服从二项分布即(5,0.2)B . 由二项分布的概率计算公式由二项分布的概率计算公式,,有{}500.80.32768,P X ==={}14510.80.20.4096,P X C ==×= {}232520.80.20.2048,P X C ==×={}{}{}{}310120.05792.P X P X P X P X ³=-=-=-==设一周内所获利润Y (万元万元),),),则则Y 是X 的函数的函数,,且10,0,5,1,()0,2,2, 3.X XY f X X X =ìï=ï==í=ï-³î若若若若由离散型随机变量数学期望计算公式由离散型随机变量数学期望计算公式, ,100.3276850.409620.05792 5.20896EY =´+´-´=(万元万元). ).【相关知识点】【相关知识点】1.1.1.二项分布的概率计算公式：二项分布的概率计算公式：二项分布的概率计算公式：若(,)Y B n p ~,则{}(1)k kn knP Y k C p p-==-, 0,1,,k n =.2.2.离散型随机变量数学期望计算公式：离散型随机变量数学期望计算公式：{}1()nkk k E X xP X x ==×=å.ý1,2,3,4,5,6时,统计C 可能出现的点数有多少种可能出现的点数有多少种. .B1 2 3 4 5 6有利于1A 的样本点数的样本点数 0 1 2 4 6 6 有利于2A 的样本点数的样本点数 0 1 0 1 0 0,3636==.3618=【相关知识点】古典型概率计算公式：.A =有利于事件的样本点数样本空间的总数,,n X 独立同分布独立同分布,2,,n X 也独立同分布也独立同分布.2111(nn n ===åå2242()n Z a a a n-=-242,)a a n-布.【相关知识点】【相关知识点】1.1.1.列维,,n X 独立同分布独立同分布,相同的期望和方差相同的期望和方差,,则对任意实数x ,恒有恒有11lim ()(),ni n i P X n x x n m s ®¥=ìü-£=F íýîþå 其中()x F 是标准正态分布函数是标准正态分布函数. .2.2.方差计算公式：方差计算公式：22()()()D X E X E X =-.。
1996考研数三真题及解析
1996年全国硕士研究生入学统一考试数学三试题一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上.) (1) 设方程yx y =确定y 是x 的函数,则dy =___________. (2) 设()arcsin x f x dx x C =+⎰,则1()dx f x =⎰___________.. (3) 设()00,x y 是抛物线2y ax bx c =++上的一点,若在该点的切线过原点,则系数应满足的关系是___________. (4) 设123222212311111231111n n n n n n n a a a a A a a a a a a a a ----⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎣⎦L L L M M M M L,123n x x X x x ⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎣⎦M ,1111B ⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎣⎦M ,其中(;,1,2,,)i j a a i j i j n ≠≠=L .则线性方程组T A X B =的解是___________. (5) 设由来自正态总体2~(,0.9)X N μ容量为9的简单随机样本,得样本均值5X =,则未知参数μ的置信度为0.95的置信区间为___________.二、选择题(本题共5小题,每小题3分,满分15分.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) 累次积分cos 20(cos ,sin )d f r r rdr πθθθθ⎰⎰可以写成 ( )(A) 10(,)dy f x y dx ⎰(B) 10(,)dy f x y dx ⎰ (C)11(,)dx f x y dy ⎰⎰(D) 10(,)dx f x y dy ⎰(2) 下述各选项正确的是 ( ) (A) 若21nn u∞=∑和21nn v∞=∑都收敛,则21()nn n uv ∞=+∑收敛(B)1n nn u v∞=∑收敛,则21nn u∞=∑与21nn v∞=∑都收敛(C) 若正项级数1nn u∞=∑发散,则1n u n≥(D) 若级数1nn u∞=∑收敛,且(1,2,)n n u v n ≥=L ,则级数1nn v∞=∑也收敛(3) 设n 阶矩阵A 非奇异(2n ≥),A *是矩阵A 的伴随矩阵,则 ( ) (A) 1()n A A A -**= (B) 1()n A A A +**= (C) 2()n A AA -**= (D) 2()n A AA +**=(4) 设有任意两个n 维向量组1,,m ααL 和1,,m ββL ,若存在两组不全为零的数1,,m λλL和1,,m k k L ,使111111()()()()0m m m m m m k k k k λαλαλβλβ+++++-++-=L L ,则( )(A) 1,,m ααL 和1,,m ββL 都线性相关 (B) 1,,m ααL 和1,,m ββL 都线性无关(C) 1111,,,,,m m m m αβαβαβαβ++--L L 线性无关 (D) 1111,,,,,m m m m αβαβαβαβ++--L L 线性相关(5) 已知0()1P B <<且()1212[]()()P A A B P A B P A B +=+,则下列选项成立的是( ) (A) ()1212[]()()P A A B P A B P A B +=+ (B) ()1212()()P A B A B P A B P A B +=+ (C) ()1212()()P A A P A B P A B +=+ (D) ()()1122()()()P B P A P B A P A P B A =+三、(本题满分6分)设(),0,()0,0,xg x e x f x xx -⎧-≠⎪=⎨⎪=⎩其中()g x 有二阶连续导数,且(0)1,(0)1g g '==-. (1)求()f x '；(2)讨论()f x '在(,)-∞+∞上的连续性.四、(本题满分6分)设函数()z f u =,方程()()xyu u p t dt ϕ=+⎰确定u 是,x y 的函数,其中(),()f u u ϕ可微；()p t ,()u ϕ'连续,且()1u ϕ'≠.求()()z z p y p x x y∂∂+∂∂.五、(本题满分6分)计算2(1)xx xe dx e -+∞-+⎰.六、(本题满分5分)设()f x 在区间[0,1]上可微,且满足条件120(1)2()f xf x dx =⎰.试证:存在(0,1)ξ∈使()()0.f f ξξξ'+=七、(本题满分6分)设某种商品的单价为p 时,售出的商品数量Q 可以表示成aQ c p b=-+,其中a b 、、 c 均为正数,且a bc >.(1) 求p 在何范围变化时,使相应销售额增加或减少.(2) 要使销售额最大,商品单价p 应取何值?最大销售额是多少?八、(本题满分6分)求微分方程dy dx =的通解.九、(本题满分8分)设矩阵01010000010012A y ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦. (1) 已知A 的一个特征值为3,试求y ； (2) 求矩阵P ,使()()TAP AP 为对角矩阵.十、(本题满分8分)设向量12,,,t αααL 是齐次线性方程组0AX =的一个基础解系,向量β不是方程组0AX =的解,即0A β≠.试证明:向量组12,,,,t ββαβαβα+++L 线性无关.十一、(本题满分7分)假设一部机器在一天内发生故障的概率为0.2,机器发生故障时全天停止工作,若一周5个工作日里无故障,可获利润10万元；发生一次故障仍可获得利润5万元；发生两次故障所获利润0元；发生三次或三次以上故障就要亏损2万元.求一周内期望利润是多少?十二、(本题满分6分)考虑一元二次方程20x Bx C ++=,其中B C 、分别是将一枚色子(骰子)接连掷两次先后出现的点数.求该方程有实根的概率p 和有重根的概率q .十三、(本题满分6分)假设12,,,n X X X L 是来自总体X 的简单随机样本；已知(1,2,3,4)kk EX a k ==.证明：当n 充分大时,随机变量211n n i i Z X n ==∑近似服从正态分布,并指出其分布参数.1996年全国硕士研究生入学统一考试数学三试题解析一、填空题(本题共5小题,每小题3分,满分15分,把答案填在题中横线上.) (1)【答案】()1ln dxx y +【解析】方法1：方程yx y =两边取对数得ln ln ln yx y y y ==,再两边求微分,()()11ln 1ln 1dx y dy dy dx x x y =+⇒=+()()ln 10x y +≠. 方法2：把yx y =变形得ln y yx e =,然后两边求微分得()()()ln ln 1ln 1ln y y y dx e d y y y y dy x y dy ==+=+,由此可得 ()1.1ln dy dx x y =+(2)【答案】C【解析】由()arcsin x f x dx x C =+⎰,两边求导数有()1()arcsin ()xf x x f x '==⇒=于是有1()dx f x ⎰212==⎰ ()2112x =--C =.(3)【答案】0c a≥(或2ax c =),b 任意 【解析】对2y ax bx c =++两边求导得()0022y ax b,y x ax b,''=+=+ 所以过()00x ,y 的切线方程为()()0002y y ax b x x ,-=+-即()()()200002y ax bx c ax b x x .-++=+-又题设知切线过原点()00,,把0x y ==代入上式,得2200002ax bx c ax bx ,---=--即20ax c.=由于系数0a ≠,所以,系数应满足的关系为0c a≥(或2ax c =),b 任意. (4)【答案】()1000T,,,L【解析】因为A 是范德蒙行列式,由i j a a ≠知()0ijA a a =-≠∏.根据解与系数矩阵秩的关系,所以方程组T A X B =有唯一解.根据克莱姆法则,对于2111112122222133332111111111n n n n n nnn x a a a x a a a x a a a x a a a ----⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦L L L M M M M M M L, 易见 1230n D A ,D D D .=====L所以TA XB =的解为12310n x ,x x x =====L ,即()1000T,,,,L .【相关知识点】克莱姆法则：若线性非齐次方程组11112211211222221122,,.n n n n n n nn n n a x a x a x b a x a x a x b a x a x a x b +++=⎧⎪+++=⎪⎨⎪⎪+++=⎩L L L L L L L L L L L L L 或简记为 112nij ji j a xb ,i ,,,n ===∑L其系数行列式1112121222120n n n n nna a a a a a D a a a =≠L L M M M L,则方程组有唯一解12j j D x ,j ,,,n.D==L其中j D 是用常数项12n b ,b ,,b L 替换D 中第j 列所成的行列式,即1111111121212212111,j ,j n ,j ,j nj n n,j nn,j nna ab a a a a b a a D a a b a a -+-+-+=L L L L M M M M M LL. (5)【答案】(4.412,5.588) 【解析】可以用两种方法求解：(1)已知方差220.9σ=,对正态总体的数学期望μ进行估计,可根据因2(,0.9)X N μ:,设有n 个样本,样本均值11ni i X X n ==∑,有20.9(,)X N n μ:,将其标准化,~(0,1)X N 得： )1,0(~1N nX μ-由正态分布分为点的定义21P u αα⎫⎪<=-⎬⎪⎭可确定临界值2αu ,进而确定相应的置信区间22(x u x u αα-+.(2)本题是在单个正态总体方差已知条件下,求期望值μ的置信区间问题.由教材上已经求出的置信区间22x u x u αα⎛-+ ⎝,其中21,(0,1)P U u U N αα⎧⎫<=-⎨⎬⎩⎭:,可以直接得出答案.方法1：由题设,95.01=-α,可见.05.0=α查标准正态分布表知分位点.96.12=αu 本题9n =, 5X =, 因此,根据 95.0}96.11{=<-nX P μ,有1.96}0.95P <=,即 {4.412 5.588}0.95P μ<<=,故μ的置信度为0.95的置信区间是(4.412,5.588) .方法2：由题设,95.01=-α,22222{}{}2()10.95,()0.975P U u P u U u u u ααααα<=-<<=Φ-=Φ=查得.96.12=αu20.9σ=,9n =, 5X =代入22(x u x u αα-+得置信区间(4.412,5.588).二、选择题(本题共5小题,每小题3分,满分15分.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)【答案】(D)【解析】方法1：由题设知,积分区域在极坐标系cos ,sin x r y r θθ==中是(),|0,0cos ,2D r r πθθθ⎧⎫=≤≤≤≤⎨⎬⎩⎭即是由221124x y ⎛⎫-+= ⎪⎝⎭与x 轴在第一象限所围成的平面图形,如右图.由于D 的最左边点的横坐标是0,最右点的横坐标是1, 下边界方程是0y ,=上边界的方程是y =从而D 的直角坐标表示是(){010D x,y |x ,y ,=≤≤≤≤故(D)正确.方法2：采取逐步淘汰法.由于(A)中二重积分的积分区域的极坐标表示为()1,|0,0sin ,2D r r πθθθ⎧⎫=≤≤≤≤⎨⎬⎩⎭而(B)中的积分区域是单位圆在第一象限的部分, (C)中的积分区域是正方形(){}0101x,y |x ,y ,≤≤≤≤所以,他们都是不正确的.故应选(D).(2)【答案】(A) 【解析】由于级数21nn u∞=∑和21nn v∞=∑都收敛,可见级数()221nn n uv ∞=+∑收敛.由不等式222n n n nu v u v ≤+及比较判别法知级数12n nn u v∞=∑收敛,从而12n nn u v∞=∑收敛.又因为()2222n n nnn n u v u v u v ,+=++即级数()21n n n u v ∞=+∑收敛,故应选(A).设()21112n n u ,v n ,,n ===L ,可知(B)不正确. 设()21112n u n ,,n n=-=L ,可知(C)不正确.设()()11112n nn u ,v n ,,nn--==-=L ,可知(D)不正确. 注：在本题中命题(D)“若级数1nn u∞=∑收敛,且(1,2,)n n u v n ≥=L ,则级数1nn v∞=∑也收敛.”不正确,这表明：比较判别法适用于正项级数收敛(或级数绝对收敛)的判别,但对任意项级数一般是不适用的.这是任意项级数与正项级数收敛性判别中的一个根本区别. (3)【答案】(C)【解析】伴随矩阵的基本关系式为AA A A A E **==, 现将A *视为关系式中的矩阵A ,则有()A A A E ****=. 方法一：由1n A A-*=及1()AA A*-=,可得 121()().n n A A A A AA A A--****-=== 故应选(C).方法二：由()A A A E ****=,左乘A 得1()()n AA A AA -***=,即1()()n A E A AA -**=.故应选(C). (4)【答案】(D)【解析】本题考查对向量组线性相关、线性无关概念的理解.若向量组12,,,s γγγL 线性无关,即若11220s s x x x γγγ+++=L ,必有120,0,,0s x x x ===L .既然1,,m λλL 与1,,m k k L 不全为零,由此推不出某向量组线性无关,故应排除(B)、(C). 一般情况下,对于1122110,s s s s k k k l l αααββ++++++=L L不能保证必有11220,s s k k k ααα+++=L 及110,s s l l ββ++=L 故(A)不正确.由已知条件,有()()()()1111110m m m m m m k k λαβλαβαβαβ+++++-++-=L L ,又1,,m λλL 与1,,m k k L 不全为零,故1111,,,,,m m m m αβαβαβαβ++--L L 线性相关. 故选(D).(5)【答案】(B) 【解析】依题意()()()()()12121212)(,.()()()()()P A A B P A B P A B P A B A B P A B P A B P B P B P B P B P B +⎡⎤++⎣⎦=+=因()0P B >,故有()()1212)(P A B A B P A B P A B +=+.因此应选(B).注：有些考生错误地选择(D).他们认为(D)是全概率公式,对任何事件B 都成立,但是忽略了全概率公式中要求作为条件的事件12,A A 应满足12()0,()0P A P A >>,且12,A A 是对立事件.【相关知识点】条件概率公式：()(|)()P AB P B A P A =.三、(本题满分6分)【解析】(1) 由于()g x 有二阶连续导数,故当0x ≠时,()f x 也具有二阶连续导数,此时,()f x '可直接计算,且()f x '连续；当0x =时,需用导数的定义求(0)f '.当0x ≠时, 22[()]()()()(1)().x x xx g x e g x e xg x g x x e f x x x ---''+-+-++'== 当0x =时,由导数定义及洛必达法则,有2000()()()(0)1(0)lim lim lim 222x x x x x x g x e g x e g x e g f x x ---→→→'''''-+--'==洛洛. 所以 2()()(1),0,()(0)1,0.2xxg x g x x e x x f x g x -'⎧-++≠⎪⎪'=⎨''-⎪=⎪⎩(2) ()f x '在0x =点的连续性要用定义来判定.因为在0x =处,有200()()(1)lim ()lim xx x xg x g x x e f x x -→→'-++'=0()()()(1)lim 2x xx g x xg x g x e x e x --→''''+-+-+= 0()(0)1lim(0)22x x g x e g f -→''''--'===. 而()f x '在0x ≠处是连续函数,所以()f x '在(,)-∞+∞上为连续函数.四、(本题满分6分) 【解析】由()z f u =可得(),()z u z u f u f u x x y y∂∂∂∂''==∂∂∂∂. 在方程()()xyu u p t dt ϕ=+⎰两边分别对,x y 求偏导数,得()(),()().u u u u u p x u p y x x y yϕϕ∂∂∂∂''=+=-∂∂∂∂ 所以()(),1()1()u p x u p y x u y u ϕϕ∂∂-==''∂-∂-. 于是 ()()()()()()()01()1()z z p x p y p x p y p y p x f u x y u u ϕϕ⎡⎤∂∂'+=-=⎢⎥''∂∂--⎣⎦.五、(本题满分6分)【分析】题的被积函数是幂函数与指数函数两类不同的函数相乘,应该用分部积分法. 【解析】方法1:因为21(1)111x x x x x xe x dxdx xd e e e e-----=-++++⎰⎰⎰分部积分 1(1)1111ln(1),1x xx x x x xx x e x dx d e e e e e x e C e---=-=-+++++=-+++⎰⎰所以20lim ln(1)ln 2.(1)1x x x x x x xe xe dx e e e -+∞-→+∞⎡⎤=-++⎢⎥++⎣⎦⎰而 lim ln(1)lim ln (1)11x x x x xxx x x xe xe e e e e e -→+∞→+∞⎡⎤⎧⎫⎡⎤-+=-+⎨⎬⎢⎥⎣⎦++⎣⎦⎩⎭lim ln(1)1x x xx xe x e e -→+∞⎧⎫=--+⎨⎬+⎩⎭lim 001xx xe →+∞-=-=+,故原式ln 2=. 方法2:220001(1)(1)1x x x x x xe xe dx dx xd e e e-+∞+∞+∞-==-+++⎰⎰⎰0000011111(1)ln(1)ln 2.1xxx x xx x xx dx dx e dx e e e e d e e e +∞-+∞+∞+∞-+∞+∞---=-+==++++=-+=-+=+⎰⎰⎰⎰六、(本题满分5分)【分析】由结论可知,若令()()x xf x ϕ=,则()()()x f x xf x ϕ''=+.因此,只需证明()x ϕ在[0,1]内某一区间上满足罗尔定理的条件.【解析】令()()x xf x ϕ=,由积分中值定理可知,存在1(0,)2η∈,使112201()()()2xf x dx x dx ϕϕη==⎰⎰,由已知条件,有1201(1)2()2()(),2f xf x dx ϕηϕη==⋅=⎰于是(1)(1)(),f ϕϕη==且()x ϕ在(,1)η上可导,故由罗尔定理可知,存在(,1)(0,1),ξη∈⊂使得()0,ϕξ'=即()()0.f f ξξξ'+=【相关知识点】1.积分中值定理：如果函数()f x 在积分区间[ ,]a b 上连续,则在[ ,]a b 上至少存在一个点ξ,使下式成立：()()()()baf x dx f b a a b ξξ=-≤≤⎰.这个公式叫做积分中值公式. 2.罗尔定理：如果函数()f x 满足(1)在闭区间[ ,]a b 上连续； (2)在开区间()a,b 内可导；(3)在区间端点处的函数值相等,即()()f a f b =, 那么在()a,b 内至少有一点ξ(a b ξ<<),使得()0f ξ'=.七、(本题满分6分)【分析】利用函数的单调性的判定,如果在x 的某个区间上导函数()0f x '≥,则函数()f x 单调递增,反之递减.【解析】(1)设售出商品的销售额为R ,则()()22(),().ab c p b aR pQ p c R p p b p b -+'==-=++ 令0,R '=得00p b ==>.当0p <<时,0R '>,所以随单价p 的增加,相应销售额R 也将增加.当p >时,有0R '<,所以随单价p 的增加,相应销售额R 将减少. (2)由(1)可知,当p =时,销售额R 取得最大值,最大销售额为2maxR b c ⎡⎤⎫⎥==⎪⎪⎥⎭⎥⎦.八、(本题满分6分) 【解析】令y z x =,则dy dzz x dx dx=+. 当0x >时,原方程化为dzz xz dx +=-,dx x =-,其通解为1ln(ln z x C =-+ 或C z x=. 代回原变量,得通解(0)y C x =>.当0x <时,原方程的解与0x >时相同,理由如下: 令t x =-,于是0t >,而且dy dy dx dydt dx dt dx =⋅=-===.从而有通解(0)y C t +=>,即(0)y C x =<.综合得,方程的通解为y C =.注：由于未给定自变量x 的取值范围,因而在本题求解过程中,引入新未知函数yz x=后得=从而,应当分别对0x >和0x <求解,在类似的问题中,这一点应当牢记.九、(本题满分8分)【分析】本题的(1)是考查特征值的基本概念,而(2)是把实对称矩阵合同于对角矩阵的问题转化成二次型求标准形的问题,用二次型的理论与方法来处理矩阵中的问题. 【解析】(1)因为3λ=是A 的特征值,故31001300313138(2)0,00311311011y E A y y ------==⋅=-=-----所以2y =.(2)由于TA A =,要2()()T T AP AP P A P ==Λ,而21000010000540045A ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦是对称矩阵,故可构造二次型2T x A x ,将其化为标准形Ty y Λ.即有2A 与Λ合同.亦即2T P A P =Λ.方法一：配方法.由于 22222123434558T x A x x x x x x x =++++22222212334444222212344816165()55255495(),55x x x x x x x x x x x x x =+++++-=++++那么,令1122334444,,,,5y x y x y x x y x ===+=即经坐标变换最新整理1122334410000100,400150001x y x y x y x y ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥-⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎢⎥⎣⎦有 222221234955Tx A x y y y y =+++. 所以,取 10000100400150001P ⎡⎤⎢⎥⎢⎥=⎢⎥-⎢⎥⎢⎥⎢⎥⎣⎦,有 211()()595T T AP AP P A P ⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦. 方法二：正交变换法.二次型22222123434558T x A x x x x x x x =++++对应的矩阵为21000010000540045A ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦, 其特征多项式231000010(1)(9)005445E A λλλλλλλ---==------.2A 的特征值12341,1,1,9λλλλ====.由21()0E A x λ-=,即12340000000000044000440x x x x ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥--⎢⎥⎢⎥⎢⎥--⎣⎦⎣⎦⎣⎦,和24()0E A x λ-=,即12348000080000044000440x x x x ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥-⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦⎣⎦,分别求得对应1,2,31λ=的线性无关特征向量123(1,0,0,0),(0,1,0,0),(0,0,1,1)T T T ααα===-,和49λ=的特征向量4(0,0,1,1)Tα=.对123,,ααα用施密特正交化方法得123,,βββ,再将4α单位化为4β,其中：1234(1,0,0,0),(0,1,0,0),,T T T Tββββ====. 取正交矩阵[]123410000100000,,,P ββββ⎡⎤⎢⎥⎢⎥⎢==⎢⎢⎢⎢⎣, 则 1221119T P A P P A P -⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎣⎦, 即 211()()19T T AP AP P A P ⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎣⎦.十、(本题满分8分)【解析】证法1： (定义法)若有一组数12,,,,,t k k k k L 使得1122()()()0,t t k k k k ββαβαβα+++++++=L (1)则因12,,,t αααL 是0AX =的解,知0(1,2,,)i A i t α==L ,用A 左乘上式的两边,有12()0t k k k k A β++++=L . (2)由于0A β≠,故120t k k k k ++++=L .对(1)重新分组为121122()0t t t k k k k k k k βααα++++++++=L L . (3) 把(2)代入(3)得 11220t t k k k ααα+++=L .由于12,,,t αααL 是基础解系,它们线性无关,故必有120,0,,0t k k k ===L .代入(2)式得:0k =.因此向量组12,,,,t ββαβαβα+++L 线性无关.证法2： (用秩)经初等变换向量组的秩不变.把第一列的-1倍分别加至其余各列,有()()1212,,,,,,,,.t t ββαβαβαβααα+++→L L因此 ()()1212,,,,,,,,.t t r r ββαβαβαβααα+++=L L由于12,,,t αααL 是基础解系,它们是线性无关的,秩()12,,,t r t ααα=L ,又β必不能由12,,,t αααL 线性表出(否则0A β=),故()12,,,,1t r t αααβ=+L . 所以 ()12,,,, 1.t r t ββαβαβα+++=+L 即向量组12,,,,t ββαβαβα+++L 线性无关.十一、(本题满分7分)【解析】设一周5个工作日内发生故障的天数为X ,则X 服从二项分布即(5,0.2)B . 由二项分布的概率计算公式,有{}500.80.32768,P X ==={}14510.80.20.4096,P X C ==⋅= {}232520.80.20.2048,P X C ==⋅={}{}{}{}310120.05792.P X P X P X P X ≥=-=-=-==设一周内所获利润Y (万元),则Y 是X 的函数,且10,0,5,1,()0,2,2,3.X X Y f X X X =⎧⎪=⎪==⎨=⎪⎪-≥⎩若若若若由离散型随机变量数学期望计算公式,100.3276850.409620.05792 5.20896EY =⨯+⨯-⨯=(万元).【相关知识点】1.二项分布的概率计算公式：若(,)Y B n p ~,则{}(1)kkn kn P Y k C p p -==-, 0,1,,k n =L .2.离散型随机变量数学期望计算公式：{}1()nkk k E X xP X x ==⋅=∑.十二、(本题满分6分)【解析】一枚色子(骰子)接连掷两次,其样本空间中样本点总数为36.设事件1A =“方程有实根”,2A =“方程有重根”,则{}221404B A B C C ⎧⎫=-≥=≤⎨⎬⎩⎭.用列举法求有利于i A 的样本点个数(1,2i =),具体做法见下表:有利于的意思就是使不等式24B C ≤尽可能的成立,则需要B 越大越好,C 越小越好.当B 取遍由古典型概率计算公式得到11246619(),3636p P A ++++===2111().3618q P A +===【相关知识点】古典型概率计算公式：().i i A P A =有利于事件的样本点数样本空间的总数十三、(本题满分6分)【解析】依题意,12,,,n X X X L 独立同分布,可见22212,,,n X X X L 也独立同分布.由(1,2,3,4)k k EX a k ==及方差计算公式,有224222242222242211,(),111,().i i i i n nn i n ii i EX a DX EX EX a a EZ EX a DZ DX a a n nn ====-=-====-∑∑ 因此,根据中心极限定理n U =的极限分布是标准正态分布,即当n 充分大时,n Z 近似服从参数为2422(,)a a a n-的正态分布.【相关知识点】1.列维-林德伯格中心极限定理,又称独立同分布的中心极限定理：设随机变量12,,,n X X X L 独立同分布,方差存在,记μ与2σ()0σ<<+∞分别是它们相同的期望和方差,则对任意实数x ,恒有1lim )(),ni n i P X n x x μ→∞=⎫-≤=Φ⎬⎭∑ 其中()x Φ是标准正态分布函数.2.方差计算公式：22()()()D X E X E X =-.。
1996年考研真题解析(英一)
1995年试题答案与解析SectionⅠUse of English一、文章结构总体分析睡眠分为浅睡阶段和较长时间的深睡阶段。
虽然人们对两个阶段的睡眠都不是十分了解，但是人们推测浅睡对大脑起修复作用。
深睡的作用更令人难以理解。
在最近睡眠研究协会的会议上，专家们首次描述的新实验，对非快眼动睡眠的作用进行了吸引人的阐释。
因剥夺睡眠而死亡的老鼠的尸体检查发现，其免疫系统似乎已崩溃了。
二、试题解析1.[答案]B[解析]本题考核的知识点是：不定代词的用法。
空格上文提到，睡眠分为浅睡阶段和深睡阶段。
因此空格处____kind of sleep指的就是这两种睡眠。
四个备选项中，首先排除[D]项。
any指“任何一个”，如：You can buy sugar at any big store.你可以在任何一家大商店里买到糖。
余下的选项中，each指两个或多个中的每一个，例如：Each of them thinks different thoughts.他们中的每一个人都有不同的想法。
either表示两者之一，表示肯定，如：You can park on either side of the street.在街道的哪边停车都可以。
neither表否定含义，表示“两者都不”。
如：Neither of us could understand German.我们两个谁也不懂德语。
空格所在的分句里出现的关键词at all一般与否定词连用，即not…at all，意为“一点也不，根本不”，因而这里应填表否定含义的词，[B]正确，意为“人们对两类睡眠都没有完全了解”。
2.[答案]C[解析]本题考核的知识点是：动词词义辨析。
intend意为“想要，打算，企图”，intend to do sth.(=mean to do sth.)意为“打算做某事”，主语一般是有生命的事物。
如：I hear they intend to marry.我听说他们要结婚了。
1996考研数学一真题及答案解析
1996年全国硕士研究生入学统一考试数学一试题一、填空题(本题共5个小题,每小题3分,满分15分,把答案填在题中横线上.) (1) 设2lim()8xx x a x a→∞+=-,则a =___________. (2) 设一平面经过原点及点(6,-3,2),且与平面428x y z -+=垂直,则此平面方程为___________.(3) 微分方程22xy y y e '''-+=的通解为___________.(4) 函数ln(u x =+在(1,0,1)A 点处沿A 点指向(3,2,2)B -点方向的方向导数为___________.(5) 设A 是43⨯矩阵,且A 的秩()2r A =,而102020103B ⎛⎫ ⎪= ⎪ ⎪-⎝⎭,则()r AB =___________.二、选择题(本题共5个小题,每小题3分,满分15分.在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) 已知2()()x ay dx ydyx y +++为某函数的全微分,则a 等于 ( ) (A) -1 (B) 0 (C) 1 (D) 2 (2) 设()f x 有二阶连续导数,且(0)0f '=,0()lim 1||x f x x →''==,则 ( ) (A) (0)f 是()f x 的极大值 (B) (0)f 是()f x 的极小值(C) (0,(0))f 是曲线()y f x =的拐点(D) (0)f 不是()f x 的极值,(0,(0))f 也不是曲线()y f x =的拐点(3) 设0(1,2,)n a n >=,且1n n a ∞=∑收敛,常数(0,)2πλ∈,则级数21(1)(tan )n n n n a n λ∞=-∑( )(A) 绝对收敛 (B) 条件收敛 (C) 发散 (D) 收敛性与λ有关 (4) 设()f x 有连续的导数,(0)0f =,(0)0f '≠,220()()()xF x x t f t dt =-⎰,且当0x →时,()F x '与kx 是同阶无穷小,则k 等于 ( )(A) 1 (B) 2 (C) 3 (D) 4(5) 四阶行列式112233440000000a b a b b a b a 的值等于 ( ) (A) 12341234a a a a b b b b - (B) 12341234a a a a b b b b +(C) 12123434()()a a b b a a b b -- (D) 23231414()()a a b b a a b b -- 三、(本题共2小题,每小题5分,满分10分.)(1) 求心形线(1cos )r a θ=+的全长,其中0a >是常数. (2) 设110x =,11,2,)n x n +==,试证数列{}n x 极限存在,并求此极限.四、(本题共2小题,每小题6分,满分12分.) (1) 计算曲面积分(2)Sx z dydz zdxdy ++⎰⎰,其中S 为有向曲面22(01)z xy z =+≤≤,其法向量与z 轴正向的夹角为锐角.(2) 设变换2,u x y u x ay=-⎧⎨=+⎩可把方程2222260z z z x x y y ∂∂∂+-=∂∂∂∂化简为20zu v ∂=∂∂,求常数a ,其中(,)z z x y =有二阶连续的偏导数. 五、(本题满分7分)求级数221(1)2nn n ∞=-∑的和. 六、(本题满分7分)设对任意0x >,曲线()y f x =上点(,())x f x 处的切线在y 轴上的截距等于01()xf t dt x⎰,求()f x 的一般表达式. 七、(本题满分8分)设()f x 在[0,1]上具有二阶导数,且满足条件|()|f x a ≤,|()|f x b ''≤,其中,a b 都是非负常数,c 是(0,1)内任一点,证明|()|22b fc a '≤+. 八、(本题满分6分)设TA E ξξ=-,其中E 是n 阶单位矩阵,ξ是n 维非零列向量,Tξ是ξ的转置,证明：(1) 2A A =的充要条件是1Tξξ=；(2) 当1Tξξ=时,A 是不可逆矩阵.九、(本题满分8分)已知二次型222123123121323(,,)55266f x x x x x cx x x x x x x =++-+-的秩为2.(1) 求参数c 及此二次型对应矩阵的特征值； (2) 指出方程123(,,)1f x x x =表示何种二次曲面. 十、填空题(本题共2小题,每小题3分,满分6分.)(1) 设工厂A 和工厂B 的产品的次品率分别为1%和 2%,现从由A 和B 的产品分别占60%和40%的一批产品中随机抽取一件,发现是次品,则该次品属A 生产的概率是__________. (2) 设ξ、η是两个相互独立且均服从正态分布2)N 的随机变量,则随机变量 ξη-的数学期望()E ξη-=__________.十一、(本题满分6分.)设ξ、η是相互独立且服从同一分布的两个随机变量,已知ξ的分布律为{}13P i ξ==, i =1,2,3,又设max(,)X ξη=,min(,)Y ξη=.(1) 写出二维随机变量(,)X Y 的分布律：(2) 求随机变量X 的数学期望()E X .1996年全国硕士研究生入学统一考试数学一试题解析一、填空题(本题共5个小题,每小题3分,满分15分,把答案填在题中横线上.) (1)【答案】ln 2【解析】这是1∞型未定式求极限.方法一： 3323lim()lim(1)x a axx a xax x x a a x a x a-⋅-→∞→∞+=+-- ,令3at x a=-,则当x →∞时,0t →,则 1303lim(1)lim(1)x aa t x t a t e x a -→∞→+=+=-, 即 33lim lim 312lim()x x ax ax a x a x x a e e e x a→∞→∞-→∞+===-. 由题设有38ae=,得1ln8ln 23a ==.方法二：2223()2221lim 112lim lim lim 11lim 1x xa xaxa x a x x a x x x a a x a a a x a e x x x e a x a e a a x x x ⋅→∞-→∞→∞→∞-⋅-→∞⎛⎫⎛⎫⎛⎫+++ ⎪ ⎪ ⎪+⎛⎫⎝⎭⎝⎭===== ⎪ ⎪-⎝⎭⎛⎫ ⎪-⎛⎫- ⎪-⎝⎭ ⎪⎝⎭⎝⎭, 由题设有38ae=,得1ln8ln 23a ==.(2)【答案】2230x y z +-=【解析】方法一：所求平面过原点O 与0(6,3,2)M -,其法向量{}06,3,2n OM ⊥=-；平面垂直于已知平面428x y z -+=,它们的法向量也互相垂直：{}04,1,2n n ⊥=-；由此, 00//632446412ij kn OM n i j k ⨯=-=--+-.取223n i j k =+-,则所求的平面方程为2230x y z +-=.方法二：所求平面即为过原点,与两个不共线的向量(一个是从原点到点0(6,3,2)M -的向量{}06,3,2OM =-,另一是平面428x y z -+=的法向量{}04,1,2n =-)平行的平面,即 6320412xy z-=-,即 2230x y z +-=.(3)【答案】12(cos sin 1)xe c x c x ++【解析】微分方程22xy y y e '''-+=所对应的齐次微分方程的特征方程为2220r r -+=,解之得1,21r i =±.故对应齐次微分方程的解为12(cos sin )x y e C x C x =+.由于非齐次项,1xe αα=不是特征根,设所给非齐次方程的特解为*()xy x ae =,代入22x y y y e '''-+=得1a =(也不难直接看出*()x y x e =),故所求通解为1212(cos sin )(cos sin 1)x x x y e C x C x e e C x C x =++=++.【相关知识点】① 二阶线性非齐次方程解的结构：设*()y x 是二阶线性非齐次方程()()()y P x y Q x y f x '''++=的一个特解.()Y x 是与之对应的齐次方程 ()()0y P x y Q x y '''++=的通解,则*()()y Y x y x =+是非齐次方程的通解.② 二阶常系数线性齐次方程通解的求解方法：对于求解二阶常系数线性齐次方程的通解()Y x ,可用特征方程法求解：即()()0y P x y Q x y '''++=中的()P x 、()Q x 均是常数,方程变为0y py qy '''++=.其特征方程写为20r pr q ++=,在复数域内解出两个特征根12,r r ； 分三种情况：(1) 两个不相等的实数根12,r r ,则通解为1212;rx r x y C eC e =+(2) 两个相等的实数根12r r =,则通解为()112;rxy C C x e =+(3) 一对共轭复根1,2r i αβ=±,则通解为()12cos sin .xy e C x C x αββ=+其中12,C C 为常数.③ 对于求解二阶线性非齐次方程()()()y P x y Q x y f x '''++=的一个特解*()y x ,可用待定系数法,有结论如下：如果()(),x m f x P x e λ=则二阶常系数线性非齐次方程具有形如*()()k xm y x x Q x e λ=的特解,其中()m Q x 是与()m P x 相同次数的多项式,而k 按λ不是特征方程的根、是特征方程的单根或是特征方程的重根依次取0、1或2.如果()[()cos ()sin ]xl n f x e P x x P x x λωω=+,则二阶常系数非齐次线性微分方程()()()y p x y q x y f x '''++=的特解可设为*(1)(2)[()cos ()sin ]k x m m y x e R x x R x x λωω=+,其中(1)()m R x 与(2)()m R x 是m 次多项式,{}max ,m l n =,而k 按i λω+(或i λω-)不是特征方程的根、或是特征方程的单根依次取为0或1. (4)【答案】12【分析】先求方向l 的方向余弦和,,u u ux y z∂∂∂∂∂∂,然后按方向导数的计算公式cos cos cos u u u u l x y zαβγ∂∂∂∂=++∂∂∂∂求出方向导数. 【解析】因为l 与AB 同向,为求l 的方向余弦,将{}{}31,20,212,2,1AB =----=-单位化，即得 {}{}12,2,1cos,cos ,cos 3||AB l AB αβγ==-=. 将函数ln(u x =+分别对,,x y z 求偏导数得12Au x ∂==∂,0Au y∂==∂,12Au z∂==∂, 所以cos cos cos AA A A u u u ulx y z αβγ∂∂∂∂=++∂∂∂∂ 1221110()233232=⨯+⨯-+⨯=. (5)【答案】2【解析】因为10220100103B ==≠-,所以矩阵B 可逆,故()()2r AB r A ==.【相关知识点】()min((),())r AB r A r B ≤.若A 可逆,则1()()()[()]()r AB r B r EB r A AB r AB -≤==≤.从而()()r AB r B =,即可逆矩阵与矩阵相乘不改变矩阵的秩.二、选择题(本题共5个小题,每小题3分,满分15分.在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)【答案】(D)【解析】由于存在函数(,)u x y ,使得 22()()()x ay dx ydydu x y x y +=+++, 由可微与可偏导的关系,知2()u x ay x x y ∂+=∂+,2()u y y x y ∂=∂+, 分别对,y x 求偏导数,得2243()()2()(2)()()u a x y x ay x y a x ayx y x y x y ∂+-+⋅+--==∂∂++, 232()u yy x x y ∂-=∂∂+. 由于2u y x ∂∂∂与2u x y ∂∂∂连续,所以22u uy x x y∂∂=∂∂∂∂,即33(2)2()()a x ay yx y x y ---=++2a ⇒=, 故应选(D).(2)【答案】(B)【解析】因为()f x 有二阶连续导数,且0()lim10,||x f x x →''=>所以由函数极限的局部保号性可知,在0x =的空心领域内有()0||f x x ''>,即()0f x ''>,所以()f x '为单调递增. 又由(0)0f '=,()f x '在0x =由负变正,由极值的第一充分条件,0x =是()f x 的极小值点,即(0)f 是()f x 的极小值.应选(B).【相关知识点】极限的局部保号性：设0lim ().x x f x A →=若0A >(或0A <)⇒0,δ∃>当00x x δ<-<时,()0f x >(或()0f x <).(3)【答案】(A) 【解析】若正项级数1nn a∞=∑收敛,则21nn a∞=∑也收敛,且当n →+∞时,有tanlim (tan )limn n n n n nλλλλλ→+∞→+∞=⋅=. 用比较判别法的极限形式,有22tanlim0nn nn a na λλ→+∞=>.因为21n n a ∞=∑收敛,所以2lim tann x n a nλ→+∞也收敛,所以原级数绝对收敛,应选(A).【相关知识点】正项级数比较判别法的极限形式：设1n n u ∞=∑和1n n v ∞=∑都是正项级数,且lim,nn nv A u →∞=则(1) 当0A <<+∞时,1nn u∞=∑和1nn v∞=∑同时收敛或同时发散；(2) 当0A =时,若1nn u∞=∑收敛,则1nn v∞=∑收敛；若1nn v∞=∑发散,则1nn u∞=∑发散；(3) 当A =+∞时,若1nn v∞=∑收敛,则1nn u∞=∑收敛；若1nn u∞=∑发散,则1nn v∞=∑发散.(4)【答案】(C)【解析】用洛必达法则.由题可知 220()()()xxF x xf t dt t f t dt =-⎰⎰,对该积分上限函数求导数,得220()2()()()2()x xF x x f t dt x f x x f x x f t dt '=+-=⎰⎰,所以 01002()2()()limlim limxxk kk x x x x f t dtf t dtF x xxx-→→→'==⎰⎰23002()2()limlim (1)(1)(2)k k x x f x f x k x k k x --→→'---洛洛.因为()F x '与kx 是同阶无穷小,且(0)0f '≠,所以302()lim(1)(2)k x f x k k x -→'--为常数,即3k =时有 300()2()limlim (0)0(1)(2)k k x x F x f x f x k k x-→→'''==≠--, 故应选(C).【相关知识点】设在同一个极限过程中,(),()x x αβ为无穷小且存在极限 ()lim()x l x αβ=,(1) 若0,l ≠称(),()x x αβ在该极限过程中为同阶无穷小； (2) 若1,l =称(),()x x αβ在该极限过程中为等价无穷小,记为()()x x αβ；(3) 若0,l =称在该极限过程中()x α是()x β的高阶无穷小,记为()()()x o x αβ=. 若()lim()x x αβ不存在(不为∞),称(),()x x αβ不可比较. (5)【答案】(D)【解析】可直接展开计算,22221414232314143333()()a b a b a a b b a a b b a a b b b a b a =-=--,所以选(D).三、(本题共2小题,每小题5分,满分10分.) (1)【解析】由极坐标系下的弧微分公式得2cos2a a d θθθ==.由于()(1cos )r r a θθ==+以2π为周期,因而θ的范围是[0,2]θπ∈. 又由于()()r r θθ=-,心形线关于极轴对称.由对称性,24cos 8sin 822s ds a d a a πππθθθ⎡⎤====⎢⎥⎣⎦⎰⎰.(2)【解析】用单调有界准则.由题设显然有0n x >,数列{}n x 有下界.证明n x 单调减：用归纳法.214x x ==<；设1n n x x -<,则1n n x x +<=.由此,n x 单调减.由单调有界准则,lim n n x →+∞存在.设lim ,(0)n n x a a →+∞=≥,在恒等式1n x +=两边取极限,即1lim lim n n n x a +→+∞=⇒=解之得3a =(2a =-舍去).【相关知识点】1.单调有界准则：单调有界数列必有极限.2. 收敛数列的保号性推论：如果数列{}n x 从某项起有0n x ≥(或0n x ≤),且lim n n x a →∞=,那么0a ≥(或0a ≤).四、(本题共2小题,每小题6分,满分12分.)(1)【分析一】见下图所示,S 在xOy 平面与yOz 平面上的投影均易求出,分别为22:1xy D x y +≤；2:11,1yz D y y z -≤≤≤≤,或01,z y ≤≤≤≤ 求S⎰⎰被积函,z =. 这里,213P Q Rx y z∂∂∂++=+=∂∂∂,若用高斯公式求曲面积分I ,则较简单.因S 不是封闭曲面,故要添加辅助曲面.【解析】方法一：均投影到平面xOy 上,则22(2)[(2)()()]xySD zI x z dydz zdxdy x z x y dxdy x∂=++=+-++∂⎰⎰⎰⎰, 其中22z x y =+,22:1xy D x y +≤.把2zx x∂=∂代入,得 2222242()()xyxyxyD D D I x dxdy x x y dxdy x y dxdy =--+++⎰⎰⎰⎰⎰⎰,由对称性得222()0xyD x x y dxdy +=⎰⎰,22242()xyxyD D x dxdy x y dxdy =+⎰⎰⎰⎰, 所以 22()xyD I x y dxdy =-+⎰⎰. 利用极坐标变换有121340001242I d r dr r ππθπ⎡⎤=-=-=-⎢⎥⎣⎦⎰⎰.方法二：分别投影到yOz 平面与xOy 平面.投影到yOz 平面时S要分为前半部分1:S x =2:S x =(见图1),则12(2)(2)S S SI x z dydz x z dydz zdxdy =++++⎰⎰⎰⎰⎰⎰.由题设,对1S 法向量与x 轴成钝角,而对2S 法向量与x 轴成锐角.将I 化成二重积分得 或21101.24yzD dz dz ππ===⎰⎰⎰⎰(这里的圆面积的一半.)22()2xyD x y dxdy π+=⎰⎰(同方法一).因此, 4.422I πππ=-⋅+=-方法三：添加辅助面221:1(1)S z x y =+≤,法方向朝下,则11(2)1S S Dx z dydz zdxdy dxdy dxdy π++==-=-⎰⎰⎰⎰⎰⎰,其中D 是1S 在平面xy 的投影区域：221x y +≤.S 与1S 即22z x y =+与1z =围成区域Ω,S 与1S 的法向量指向Ω内部,所以在Ω上满足高斯公式的条件,所以11()3332D z dz dxdy zdz ππ=-=-=-⎰⎰⎰⎰, 其中,()D z 是圆域：22x y z +≤,面积为z π. 因此,133(2)()222S I x z dydz zdxdy ππππ=--++=---=-⎰⎰. (2)【解析】由多元复合函数求导法则,得z z u z v z zx u x v x u v∂∂∂∂∂∂∂=+=+∂∂∂∂∂∂∂, 2z z u z v z z a y u y v y u v∂∂∂∂∂∂∂=+=-+∂∂∂∂∂∂∂,所以 22222222()()z z z z u z v z v z ux x u x v u x u v x v x v u x ∂∂∂∂∂∂∂∂∂∂∂∂∂=+=⋅+⋅+⋅+⋅∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂ 222222z z zu u v v ∂∂∂=++∂∂∂∂, 222222(2)z z z a a u u v v∂∂∂=-+-+∂∂∂∂,代入2222260z z zx x y y∂∂∂+-=∂∂∂∂,并整理得 2222222226(105)(6)0z z z z z a a a x x y y u v v∂∂∂∂∂+-=+++-=∂∂∂∂∂∂∂. 于是,令260a a +-=得3a =或2a =-.2a =-时,1050a +=,故舍去,3a =时,1050a +≠,因此仅当3a =时化简为20zu v∂=∂∂. 【相关知识点】多元复合函数求导法则：若(,)u u x y =和(,)v v x y =在点(,)x y 处偏导数存在,函数(,)z f u v =在对应点(,)u v 具有连续偏导数,则复合函数[(,),(,)]z f u x y v x y =在点(,)x y 处的偏导数存在,且,z f u f v z f u f v x u x v x y u y v y∂∂∂∂∂∂∂∂∂∂=+=+∂∂∂∂∂∂∂∂∂∂. 五、(本题满分7分) 【解析】先将级数分解, 令 1221311,22n nn n A A nn ∞∞+====⋅⋅∑∑, 则 12A A A =-.由熟知ln(1)x +幂级数展开式,即11(1)ln(1)(11)n nn x x x n -∞=-+=-<≤∑,得 1121111(1)1111()ln(1)ln 2242424n n n n n A n n -∞∞+==-==--=--=⋅∑∑,因此, 1253ln 284A A A =-=-.六、(本题满分7分)【解析】曲线()y f x =上点(,())x f x 处的切线方程为()()()Y f x f x X x '-=-.令0X =得y 轴上的截距()()Y f x f x x '=-.由题意,01()()()xf t dt f x f x x x' =-⎰. 为消去积分,两边乘以x ,得 20()()()xf t dt xf x f x x ' =-⎰, (*)将恒等式两边对x 求导,得2()()()2()()f x f x xf x xf x x f x ''''=+--,即 ()()0xf x f x '''+=.在(*)式中令0x =得00=自然成立.故不必再加附加条件.就是说()f x 是微分方程0xy y '''+=的通解.下面求解微分方程0xy y '''+=.方法一：()100xy y xy xy C ''''''+=⇒=⇒=, 因为0x >,所以1C y x'=, 两边积分得 12()ln y f x C x C ==+.方法二：令()y P x '=,则y P '''=,解0xP P '+=得1C y P x'==. 再积分得12()ln y f x C x C ==+. 七、(本题满分8分)【解析】由于问题涉及到,f f '与f ''的关系,自然应当利用泰勒公式,而且应在点c 展开：2()()()()()()2!f f x f c f x x c x c ξ'''=+-+-,ξ在c 与x 之间. 分别取0,1x =得20()(0)()()(0)(0)2!f f f c f c c c ξ'''=+-+-,0ξ在c 与0之间, 21()(1)()()(1)(1)2!f f f c f c c c ξ'''=+-+-,1ξ在c 与1之间, 两式相减得 22101(1)(0)()[()(1)()]2!f f f c f c f c ξξ'''''-=+--,于是 22101()(1)(0)[()(1)()]2!f c f f f c f c ξξ'''''=----. 由此 221011()(1)(0)()(1)()2!2!f c f f f c f c ξξ'''''≤++-+2212[(1)]222b a bc c a ≤+-+<+.八、(本题满分6分)【解析】(1)因为TA E ξξ=-,Tξξ为数,Tξξ为n 阶矩阵,所以2()()2()(2)T T T T T T T A E E E E ξξξξξξξξξξξξξξ=--=-+=--,因此, 2(2)(1)0TTTTTA A E E ξξξξξξξξξξ=⇔--=-⇔-= 因为ξ是非零列向量,所以0Tξξ≠,故210,TA A ξξ=⇔-=即1Tξξ=.(2)反证法.当1Tξξ=时,由(1)知2A A =,若A 可逆,则121A A A A A E --===.与已知TA E E ξξ=-≠矛盾,故A 是不可逆矩阵. 九、(本题满分8分)【解析】(1)此二次型对应的矩阵为51315333A c -⎛⎫ ⎪=-- ⎪ ⎪-⎝⎭.因为二次型秩 ()()2r f r A ==,由 可得3c =.再由A 的特征多项式 求得二次型矩阵的特征值为0,4,9.(2)因为二次型经正交变换可化为222349y y +,故123(,,)1f x x x =,即2223491y y +=.表示椭圆柱面.【相关知识点】主轴定理：对于任一个n 元二次型12(,,,)T n f x x x x Ax =,存在正交变换x Qy =(Q 为n 阶正交矩阵),使得2221122()T T T n n x Ax y Q AQ y y y y λλλ==+++,其中12,,,n λλλ是实对称矩阵A 的n 个特征值,Q 的n 个列向量12,,,n ααα是A 对应于特征值12,,,n λλλ的标准正交特征向量.十、填空题(本题共2小题,每小题3分,满分6分.) (1)【答案】37【解析】设事件C =“抽取的产品是次品”,事件D =“抽取的产品是工厂A 生产的”,则事件D 表示“抽取的产品是工厂B 生产的”,依题意有()0.60,()0.40,(|)0.01,(|)0.02P D P D P C D P C D ====.应用贝叶斯公式可以求得条件概率(|)P D C ：()(|)0.60.013(|)0.60.010.40.027()(|)()(|)P D P C D P D C P D P C D P D P C D ⨯===⨯+⨯+.【相关知识点】贝叶斯公式：设试验E 的样本空间为S .A 为E 的事件,12,,,n B B B 为S 的一个划分,且()0,()0(1,2,,)i P A P B i n >>=,则1()(|)(|),1,2,,.()(|)i i i njjj P B P A B P B A i n P B P A B ===∑ (*)(*)式称为贝叶斯公式. (2)【解析】由于ξ与η相互独立且均服从正态分布2)N ,因此它们的线性函数U ξη=-服从正态分布,且()11122DU D D D ξηξη=-=+=+=, 所以有 (0,1)UN .代入正态分布的概率密度公式,有22()u f u du +∞--∞=⎰.应用随机变量函数的期望公式有 由凑微分法,有222(||)2()2u uE dξη+∞--=--⎰22u+∞-==【相关知识点】对于随机变量X与Y均服从正态分布,则X与Y的线性组合亦服从正态分布.若X与Y相互独立,由数学期望和方差的性质,有()()()E aX bY c aE X bE Y c++=++,22()()()D aX bY c a D X b D Y++=+,其中,,a b c为常数.十一、(本题满分6分.)【解析】易见(,)X Y的可能取值为(1,1),(2,1),(2,2),(3,1),(3,2),(3,3).依题意{}X Y<=∅,故{}0P X Y<=,即{}{}{}1,21,32,30P X Y P X Y P X Y=========,{}{}{}11,1119P P Pξηξη=======.类似地可以计算出所有ijp的值列于下表中,得到随机变量(,)X Y的联合分布律：(2)将表中各行元素相加求出的边缘分布123135999X⎡⎤⎢⎥⎢⎥⎣⎦,由离散型随机变量数学期望计算公式可得135221239999EX=⋅+⋅+⋅=.【相关知识点】1.离散型随机变量的边缘分布计算公式：二维离散型随机变量(,)X Y 关于X 与Y 的边缘概率分布或边缘分布律分别定义为： 它们分别为联合分布律表格中第i 行与第j 列诸元素之和. 2. 离散型随机变量数学期望计算公式：{}1()nkk k E X xP X x ==⋅=∑.。
1996年考研英语真题超详解
1996年考研英语真题超详解 1996年年全真试题PartⅠClozeTestDirections：Foreachnumberedblankinthefollowingpassage,therearefourchoicesmarked［A］,［B］,［C］and[D].ChoosethebestoneandmarkyouransweronANSWERSHEET1byblackeningthecorresponding letterinthebrackets.(10points)Vitaminsareorganiccompounds necessary insmallamountsinthedietforthenormalgrowthandmaintenance oflifeofanimals,including man.Theydonotprovideenergy,1dotheyconstructorbuildanypartofthebody.Theyareneededfor2foodsintoenergyandbodymaintenance. Therearethirteenormoreofthem,andif3ismissingadeficiencydiseasebecomes4.Vitaminsaresimilarbecausetheyaremadeofthesameelements―usuallyhydrogen,oxygen,and5nitrogen.Theyaredifferent6theirelementsarearrangeddifferently, andeachvitamin7oneormorespecificfunctionsinthe8enoughvitaminsisessential tolife,althoughthebodyhasnonutritional usefor9vitamins. Manypeople,10,believeinbeingonthe“safeside”andthustakeextravitamins.However,awell��balanced dietwillusuallymeetallthebody’svitaminneeds.1.［A］either［B］so［C］nor［D］never2.［A］shifting［B］transferring［C］altering［D］transforming 3.［A］any［B］some［C］anything［D］something4.［A］serious［B］apparent［C］severe［D］fatal5.［A］mostly［B］partially［C］sometimes［D］rarely6.［A］inthat［B］sothat［C］suchthat［D］exceptthat7.［A］undertakes［B］holds［C］plays［D］performs8.［A］Supplying［B］Getting［C］Providing［D］Furnishing9.［A］exceptional［B］exceeding［C］excess［D］external10.［A］nevertheless［B］therefore［C］moreover［D］meanwhilePartⅡReadingComprehensionDirections:Eachofthepassagesbelowisfollowedsomequestions. Foreachquestions therearefouranswersmarked［A］,［B］,［C］and［D］.Readthepassages carefully andchoosethebestanswereachofthequestions.ThenmarkyouransweronANSWERSHEET1byblackeningthecorresponding letterinthebracketswithapencil.(40points)Passage1Tight��lipped eldersusedtosay,“It’snotwhatyouwantinthisworld,butwhatyouget.”Psychologyteachesthatyoudowhatyouwantifyouknowwhatyouwantandwanttherightthings. Youcanmakeamentalblueprint ofadesireaswouldmakeablueprint ofahouse,andeachofusiscontinually makingtheseblueprints inthegeneralroutineofeverydayliving.Ifintend tohavefriends todinner, weplanthemenu,makeashopping list,decide whichfoodtocookfirst, andsuchplanning isessential foranytypeofmealtobeserved. Likewise, ifyouwanttofindajob,takeasheetofpaper,andwritebriefaccount ofyourself. Inmakingablueprint forajob,beginwithyourself, forwhenyouknowexactly whatyouhavetooffer,canintelligently planwheretosellyourservices.Thisaccountofyourselfisactuallyasketchofyourworkinglifeandshouldincludeeducation,experience andreferences. Suchanaccountisvaluable. Itcanbereferredtoinfillingoutstandardapplication blanksandisextremely helpfulinpersonalinterviews.Whiletalkingtoyou,yourcouldbeemployerisdecidingwhetheryoureducation,yourexperience,andotherqualifications willpayhimtoemployyouyour“wares” andabilities mustbedisplayed inanorderlyandreasonably connected manner.Whenyouhavecarefully prepared ablueprint ofyourabilitiesdesires,youhavesomething tangibletosell.Thenyouarereadytohuntforajob.Getallthepossibleinformation aboutyourcouldjob.Makeinquiries astothedetails regarding thejobandthefirm.Keepyoureyesandearsopen,anduseyourownjudgment.acertainamountoftimeeachdayseekingtheemployment youwishfor,andkeepinmind:Securing ajobisyourjobnow.Whatdotheeldersmeanwhentheysay,“It’s notwhatyouwantinthisworld,butwhatyouget.”?［A］You’ll certainly getyouwant.［B］It’snousedreaming.［C］Youshouldbedissatisfied withwhatyouhave.［D］It’sessentialtosetagoalforyourself.12.Ablueprintmadebeforeinvitingafriendtodinnerisusedinthispassageas?????????.［A］anillustration ofhowtowriteanforajob［B］anindication ofhowtosecureagoodjob［C］aguideline forjobdescription［D］aprinciple forjob13.Accordingtothepassage,onemustwriteanaccountofhimselfbeforestartingtofindajobbecause???????? .［A］thatisthesteptopleasetheemployer［B］thatistherequirement oftheemployer［C］itenableshimtoknowwhentosellhisservicesitforceshimtobecomeclearly awareofhimself 14.Whenyouhavecarefully prepared ablueprint ofyourabilities anddesires, youhavesomething????????.［A］definitetooffer［B］imaginarytoprovide［C］practicaltosupply［D］desirabletopresentPassage2WiththestartofBBCWorldServiceTelevision,millionsofviewersinAsiaandAmericacannowwatchtheCorporation’s newscoverage,aswellaslistentoit.AndofcourseBritainlisteners andviewerscantuneintotwoBBCtelevision channels, fiveBBCnational radioservices anddozensoflocalradiostations. Theybroughtsport,comedy,drama,music,newsandcurrentaffairs,education,religion,parliamentary coverage,children’s programmesandfilmsforanannuallicencefeeof￡83household. Itisaremarkable record,stretching backover70years―yettheBBC’sfutureisnowindoubt.TheCorporation willsurviveapublicly��funded broadcastingorganization,atleastforthetimebeing,butitsrole,itssizeanditsprogrammesarenowthesubjectofanation��wide debateinBritain.ThedebatewaslaunchedbytheGovernment,whichinvitedanyonewithanopinionoftheBBC―includingordinarylistenersandviewers―tosaywhatwasgoodorbadabouttheCorporation, andevenwhethertheythoughtitwasworthkeeping.ThereasonforitsinquiryisthattheBBC’sroyalcharterrunsoutin1996anditmustdecidewhethertokeeptheorganization asitis,ortochanges.Defenders oftheCorporation―ofwhomtherearemany―arefondofquotingtheAmericanslogan“Ifitain’tbroke,don’tit.”TheBBC“ain’tbroke”, theysay,bywhichtheymeanitisnotbroken(asdistinct fromtheword‘broke’, meaninghavingnosowhybothertochangeit?YettheBBCwillhavetochange,becausethebroadcasting worldarounditischanging.Thecommercial TV――ITVandChannel4――wererequiredbytheThatcherGovernment’s Broadcasting Acttobecomemorecommercial, competingwitheachotherforadvertisers,cuttingcostsandjobs.Butitisthearrivalofnewsatellitechannels―fundedpartlybyadvertisingandpartlybyviewers’subscriptions―whichbringaboutthebiggestchangesinthelongterm.15.TheworldfamousBBCnowfaces???????? .［A］theproblemofnewscoverage［B］uncertain prospect［C］inquiries bythegeneral public［D］shrinkage ofaudience 16.Inthepassage, whichofthefollowing abouttheBBCismentioned asthekeyissue?［A］Extension ofitsTVservicetoFarEast.［B］Programmes asthesubjectofanation-wide debate.［C］Potentialsforfurtherinternational co-operations.［D］Itsexistenceasabroadcasting organization.17.TheBBC’s“royalcharter”(Line4,Paragraph4)standsfor????????.［A］。
1996年考研数学三真题及全面解析
1996年全国硕士研究生入学统一考试数学三试题一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上.) (1) 设方程yx y =确定y 是x 的函数,则dy =___________. (2) 设()arcsin x f x dx x C =+⎰,则1()dx f x =⎰___________.. (3) 设()00,x y 是抛物线2y ax bx c =++上的一点,若在该点的切线过原点,则系数应满足的关系是___________. (4) 设123222212311111231111n nn n n n n a a a a A a a a a a a a a ----⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎣⎦,123n x x X x x ⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎣⎦,1111B ⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎣⎦, 其中(;,1,2,,)i j a a i j i j n ≠≠=.则线性方程组T A X B =的解是___________.(5) 设由来自正态总体2~(,0.9)X N μ容量为9的简单随机样本,得样本均值5X =,则未知参数μ的置信度为0.95的置信区间为___________.二、选择题(本题共5小题,每小题3分,满分15分.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) 累次积分cos 20(cos ,sin )d f r r rdr πθθθθ⎰⎰可以写成 ( )(A) 10(,)dy f x y dx ⎰(B) 10(,)dy f x y dx ⎰ (C)11(,)dx f x y dy ⎰⎰(D) 10(,)dx f x y dy ⎰(2) 下述各选项正确的是 ( ) (A) 若21nn u∞=∑和21nn v∞=∑都收敛,则21()nn n uv ∞=+∑收敛(B)1n nn u v∞=∑收敛,则21nn u∞=∑与21nn v∞=∑都收敛(C) 若正项级数1nn u∞=∑发散,则1n u n≥(D) 若级数1nn u∞=∑收敛,且(1,2,)n n u v n ≥=,则级数1n n v ∞=∑也收敛(3) 设n 阶矩阵A 非奇异(2n ≥),A *是矩阵A 的伴随矩阵,则 ( ) (A) 1()n A A A -**= (B) 1()n A A A +**= (C) 2()n A AA -**= (D) 2()n A AA +**=(4) 设有任意两个n 维向量组1,,m αα和1,,m ββ,若存在两组不全为零的数1,,m λλ和1,,m k k ,使111111()()()()0m m m m m m k k k k λαλαλβλβ+++++-++-=,则( )(A) 1,,m αα和1,,m ββ都线性相关 (B) 1,,m αα和1,,m ββ都线性无关(C) 1111,,,,,m m m m αβαβαβαβ++--线性无关 (D) 1111,,,,,m m m m αβαβαβαβ++--线性相关(5) 已知0()1P B <<且()1212[]()()P A A B P A B P A B +=+,则下列选项成立的是( ) (A) ()1212[]()()P A A B P A B P A B +=+ (B) ()1212()()P A B A B P A B P A B +=+ (C) ()1212()()P A A P A B P A B +=+ (D) ()()1122()()()P B P A P B A P A P B A =+三、(本题满分6分)设(),0,()0,0,xg x e x f x xx -⎧-≠⎪=⎨⎪=⎩其中()g x 有二阶连续导数,且(0)1,(0)1g g '==-. (1)求()f x '；(2)讨论()f x '在(,)-∞+∞上的连续性.四、(本题满分6分)设函数()z f u =,方程()()xyu u p t dt ϕ=+⎰确定u 是,x y 的函数,其中(),()f u u ϕ可微；()p t ,()u ϕ'连续,且()1u ϕ'≠.求()()z z p y p x x y∂∂+∂∂.五、(本题满分6分)计算2(1)xx xe dx e -+∞-+⎰.六、(本题满分5分)设()f x 在区间[0,1]上可微,且满足条件120(1)2()f xf x dx =⎰.试证:存在(0,1)ξ∈使()()0.f f ξξξ'+=七、(本题满分6分)设某种商品的单价为p 时,售出的商品数量Q 可以表示成aQ c p b=-+,其中a b 、、 c 均为正数,且a bc >.(1) 求p 在何范围变化时,使相应销售额增加或减少.(2) 要使销售额最大,商品单价p 应取何值?最大销售额是多少?八、(本题满分6分)求微分方程dy dx =的通解.九、(本题满分8分)设矩阵010010000010012A y ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦. (1) 已知A 的一个特征值为3,试求y ； (2) 求矩阵P ,使()()TAP AP 为对角矩阵.十、(本题满分8分)设向量12,,,t ααα是齐次线性方程组0AX =的一个基础解系,向量β不是方程组0AX =的解,即0A β≠.试证明:向量组12,,,,t ββαβαβα+++线性无关.十一、(本题满分7分)假设一部机器在一天内发生故障的概率为0.2,机器发生故障时全天停止工作,若一周5个工作日里无故障,可获利润10万元；发生一次故障仍可获得利润5万元；发生两次故障所获利润0元；发生三次或三次以上故障就要亏损2万元.求一周内期望利润是多少?十二、(本题满分6分)考虑一元二次方程20x Bx C ++=,其中B C 、分别是将一枚色子(骰子)接连掷两次先后出现的点数.求该方程有实根的概率p 和有重根的概率q .十三、(本题满分6分)假设12,,,n X X X 是来自总体X 的简单随机样本；已知(1,2,3,4)k k EX a k ==.证明：当n 充分大时,随机变量211n n i i Z X n ==∑近似服从正态分布,并指出其分布参数.1996年全国硕士研究生入学统一考试数学三试题解析一、填空题(本题共5小题,每小题3分,满分15分,把答案填在题中横线上.) (1)【答案】()1ln dxx y +【解析】方法1：方程yx y =两边取对数得ln ln ln yx y y y ==,再两边求微分,()()11ln 1ln 1dx y dy dy dx x x y =+⇒=+()()ln 10x y +≠. 方法2：把yx y =变形得ln y yx e =,然后两边求微分得()()()ln ln 1ln 1ln y y y dx e d y y y y dy x y dy ==+=+,由此可得 ()1.1ln dy dx x y =+(2)【答案】C【解析】由()arcsin x f x dx x C =+⎰,两边求导数有()1()arcsin ()xf x x f x '==⇒=于是有1()dx f x ⎰212==⎰ ()2112x =--C =.(3)【答案】0c a≥(或2ax c =),b 任意 【解析】对2y ax bx c =++两边求导得()0022y ax b,y x ax b,''=+=+ 所以过()00x ,y 的切线方程为()()0002y y ax b x x ,-=+-即()()()200002y ax bx c ax b x x .-++=+-又题设知切线过原点()00,,把0x y ==代入上式,得2200002ax bx c ax bx ,---=--即20ax c.=由于系数0a ≠,所以,系数应满足的关系为0c a≥(或2ax c =),b 任意. (4)【答案】()1000T,,,【解析】因为A 是范德蒙行列式,由i j a a ≠知()0ijA a a =-≠∏.根据解与系数矩阵秩的关系,所以方程组T A X B =有唯一解.根据克莱姆法则,对于2111112122222133332111111111n n n n n nnn x a a a x a a a x a a a x a a a ----⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦, 易见 1230n D A ,D D D .=====所以TA XB =的解为12310n x ,x x x =====,即()1000T,,,,.【相关知识点】克莱姆法则：若线性非齐次方程组11112211211222221122,,.n n n n n n nn n n a x a x a x b a x a x a x b a x a x a x b +++=⎧⎪+++=⎪⎨⎪⎪+++=⎩ 或简记为 112nij ji j a xb ,i ,,,n ===∑其系数行列式1112121222120n n n n nna a a a a a D a a a =≠,则方程组有唯一解12j j D x ,j ,,,n.D==其中j D 是用常数项12n b ,b ,,b 替换D 中第j 列所成的行列式,即1111111121212212111,j ,j n ,j ,j n j n n,j nn,j nna ab a a a a b a a D a a b a a -+-+-+=.(5)【答案】(4.412,5.588) 【解析】可以用两种方法求解：(1)已知方差220.9σ=,对正态总体的数学期望μ进行估计,可根据 因2(,0.9)XN μ,设有n 个样本,样本均值11ni i X X n ==∑,有20.9(,)XN n μ,将其标准化,~(0,1)XN 得：)1,0(~1N nX μ-由正态分布分为点的定义21P uαα⎫⎪<=-⎬⎪⎭可确定临界值2αu ,进而确定相应的置信区间22(x u x u αα-+.(2)本题是在单个正态总体方差已知条件下,求期望值μ的置信区间问题. 由教材上已经求出的置信区间22x u x u αα⎛-+ ⎝,其中21,(0,1)P U u UN αα⎧⎫<=-⎨⎬⎩⎭,可以直接得出答案.方法1：由题设,95.01=-α,可见.05.0=α查标准正态分布表知分位点.96.12=αu 本题9n =, 5X =,因此,根据 95.0}96.11{=<-nX P μ,有 1.96}0.95P <=,即 {4.412 5.588}0.95P μ<<=,故μ的置信度为0.95的置信区间是(4.412,5.588) .方法2：由题设,95.01=-α,22222{}{}2()10.95,()0.975P U u P u U u u u ααααα<=-<<=Φ-=Φ=查得.96.12=αu20.9σ=,9n =, 5X =代入22(x u x u αα-+得置信区间(4.412,5.588).二、选择题(本题共5小题,每小题3分,满分15分.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)【答案】(D)【解析】方法1：由题设知,积分区域在极坐标系cos ,sin x r y r θθ==中是(),|0,0cos ,2D r r πθθθ⎧⎫=≤≤≤≤⎨⎬⎩⎭即是由221124x y ⎛⎫-+= ⎪⎝⎭与x 轴在第一象限所围成的平面图形,如右图.由于D 的最左边点的横坐标是0,最右点的横坐标是1, 下边界方程是0y ,=上边界的方程是y =从而D 的直角坐标表示是(){010D x,y |x ,y ,=≤≤≤≤故(D)正确.方法2：采取逐步淘汰法.由于(A)中二重积分的积分区域的极坐标表示为()1,|0,0sin ,2D r r πθθθ⎧⎫=≤≤≤≤⎨⎬⎩⎭而(B)中的积分区域是单位圆在第一象限的部分, (C)中的积分区域是正方形(){}0101x,y |x ,y ,≤≤≤≤所以,他们都是不正确的.故应选(D).(2)【答案】(A) 【解析】由于级数21nn u∞=∑和21nn v∞=∑都收敛,可见级数()221nn n uv ∞=+∑收敛.由不等式222n n n nu v u v ≤+及比较判别法知级数12n nn u v∞=∑收敛,从而12n nn u v∞=∑收敛.又因为()2222n n nnn n u v u v u v ,+=++即级数()21n n n u v ∞=+∑收敛,故应选(A).设()21112n n u ,v n ,,n ===,可知(B)不正确. 设()21112n u n ,,n n=-=,可知(C)不正确.设()()11112n nn u ,v n ,,nn--==-=,可知(D)不正确.注：在本题中命题(D)“若级数1nn u∞=∑收敛,且(1,2,)n n u v n ≥=,则级数1n n v ∞=∑也收敛.”不正确,这表明：比较判别法适用于正项级数收敛(或级数绝对收敛)的判别,但对任意项级数一般是不适用的.这是任意项级数与正项级数收敛性判别中的一个根本区别. (3)【答案】(C)【解析】伴随矩阵的基本关系式为AA A A A E **==, 现将A *视为关系式中的矩阵A ,则有()A A A E ****=. 方法一：由1n A A-*=及1()AA A*-=,可得 121()().n n A A A A AA A A--****-=== 故应选(C).方法二：由()A A A E ****=,左乘A 得1()()n AA A AA -***=,即1()()n A E A AA -**=.故应选(C). (4)【答案】(D)【解析】本题考查对向量组线性相关、线性无关概念的理解.若向量组12,,,s γγγ线性无关,即若11220s s x x x γγγ+++=,必有120,0,,0s x x x ===.既然1,,m λλ与1,,m k k 不全为零,由此推不出某向量组线性无关,故应排除(B)、(C).一般情况下,对于1122110,s s s s k k k l l αααββ++++++=不能保证必有11220,s s k k k ααα+++=及110,s s l l ββ++=故(A)不正确.由已知条件,有()()()()1111110m m m m m m k k λαβλαβαβαβ+++++-++-=,又1,,m λλ与1,,m k k 不全为零,故1111,,,,,m m m m αβαβαβαβ++--线性相关.故选(D).(5)【答案】(B) 【解析】依题意()()()()()12121212)(,.()()()()()P A A B P A B P A B P A B A B P A B P A B P B P B P B P B P B +⎡⎤++⎣⎦=+=因()0P B >,故有()()1212)(P A B A B P A B P A B +=+.因此应选(B).注：有些考生错误地选择(D).他们认为(D)是全概率公式,对任何事件B 都成立,但是忽略了全概率公式中要求作为条件的事件12,A A 应满足12()0,()0P A P A >>,且12,A A 是对立事件.【相关知识点】条件概率公式：()(|)()P AB P B A P A =.三、(本题满分6分)【解析】(1) 由于()g x 有二阶连续导数,故当0x ≠时,()f x 也具有二阶连续导数,此时,()f x '可直接计算,且()f x '连续；当0x =时,需用导数的定义求(0)f '.当0x ≠时, 22[()]()()()(1)().x x xx g x e g x e xg x g x x e f x x x ---''+-+-++'== 当0x =时,由导数定义及洛必达法则,有2000()()()(0)1(0)lim lim lim 222x x x x x x g x e g x e g x e g f x x ---→→→'''''-+--'==洛洛. 所以 2()()(1),0,()(0)1,0.2xxg x g x x e x x f x g x -'⎧-++≠⎪⎪'=⎨''-⎪=⎪⎩(2) ()f x '在0x =点的连续性要用定义来判定.因为在0x =处,有200()()(1)lim ()lim xx x xg x g x x e f x x -→→'-++'=0()()()(1)lim 2x xx g x xg x g x e x e x --→''''+-+-+= 0()(0)1lim(0)22x x g x e g f -→''''--'===. 而()f x '在0x ≠处是连续函数,所以()f x '在(,)-∞+∞上为连续函数.四、(本题满分6分) 【解析】由()z f u =可得(),()z u z u f u f u x x y y∂∂∂∂''==∂∂∂∂. 在方程()()xyu u p t dt ϕ=+⎰两边分别对,x y 求偏导数,得()(),()().u u u u u p x u p y x x y yϕϕ∂∂∂∂''=+=-∂∂∂∂ 所以()(),1()1()u p x u p y x u y u ϕϕ∂∂-==''∂-∂-. 于是 ()()()()()()()01()1()z z p x p y p x p y p y p x f u x y u u ϕϕ⎡⎤∂∂'+=-=⎢⎥''∂∂--⎣⎦.五、(本题满分6分)【分析】题的被积函数是幂函数与指数函数两类不同的函数相乘,应该用分部积分法. 【解析】方法1:因为21(1)111x x x x x xe x dxdx xd e e e e-----=-++++⎰⎰⎰分部积分 1(1)1111ln(1),1x xx x x x xx x e x dx d e e e e e x e C e---=-=-+++++=-+++⎰⎰所以20lim ln(1)ln 2.(1)1x x x x x x xe xe dx e e e -+∞-→+∞⎡⎤=-++⎢⎥++⎣⎦⎰而 lim ln(1)lim ln (1)11x x x x xxx x x xe xe e e e e e -→+∞→+∞⎡⎤⎧⎫⎡⎤-+=-+⎨⎬⎢⎥⎣⎦++⎣⎦⎩⎭lim ln(1)1x x xx xe x e e -→+∞⎧⎫=--+⎨⎬+⎩⎭lim 001xx xe →+∞-=-=+,故原式ln 2=. 方法2:220001(1)(1)1x x x x x xe xe dx dx xd e e e-+∞+∞+∞-==-+++⎰⎰⎰0000011111(1)ln(1)ln 2.1xxx x xx x xx dx dx e dx e e e e d e e e +∞-+∞+∞+∞-+∞+∞---=-+==++++=-+=-+=+⎰⎰⎰⎰六、(本题满分5分)【分析】由结论可知,若令()()x xf x ϕ=,则()()()x f x xf x ϕ''=+.因此,只需证明()x ϕ在[0,1]内某一区间上满足罗尔定理的条件.【解析】令()()x xf x ϕ=,由积分中值定理可知,存在1(0,)2η∈,使112201()()()2xf x dx x dx ϕϕη==⎰⎰,由已知条件,有1201(1)2()2()(),2f xf x dx ϕηϕη==⋅=⎰于是(1)(1)(),f ϕϕη==且()x ϕ在(,1)η上可导,故由罗尔定理可知,存在(,1)(0,1),ξη∈⊂使得()0,ϕξ'=即()()0.f f ξξξ'+=【相关知识点】1.积分中值定理：如果函数()f x 在积分区间[ ,]a b 上连续,则在[ ,]a b 上至少存在一个点ξ,使下式成立：()()()()baf x dx f b a a b ξξ=-≤≤⎰.这个公式叫做积分中值公式. 2.罗尔定理：如果函数()f x 满足(1)在闭区间[ ,]a b 上连续； (2)在开区间()a,b 内可导；(3)在区间端点处的函数值相等,即()()f a f b =, 那么在()a,b 内至少有一点ξ(a b ξ<<),使得()0f ξ'=.七、(本题满分6分)【分析】利用函数的单调性的判定,如果在x 的某个区间上导函数()0f x '≥,则函数()f x 单调递增,反之递减.【解析】(1)设售出商品的销售额为R ,则()()22(),().ab c p b aR pQ p c R p p b p b -+'==-=++ 令0,R '=得00p b ==>.当0p <<时,0R '>,所以随单价p 的增加,相应销售额R 也将增加.当p >时,有0R '<,所以随单价p 的增加,相应销售额R 将减少. (2)由(1)可知,当p =时,销售额R 取得最大值,最大销售额为2maxR b c ⎡⎤⎫⎥==⎪⎪⎥⎭⎥⎦.八、(本题满分6分) 【解析】令y z x =,则dy dzz x dx dx=+. 当0x >时,原方程化为dzz xz dx +=-,dx x =-,其通解为1ln(ln z x C =-+ 或C z x+=. 代回原变量,得通解(0)y C x =>.当0x <时,原方程的解与0x >时相同,理由如下: 令t x =-,于是0t >,而且dy dy dx dydt dx dt dx =⋅=-===.从而有通解(0)y C t +=>,即(0)y C x =<.综合得,方程的通解为y C =.注：由于未给定自变量x 的取值范围,因而在本题求解过程中,引入新未知函数yz x=后得x =,从而,应当分别对0x >和0x <求解,在类似的问题中,这一点应当牢记.九、(本题满分8分)【分析】本题的(1)是考查特征值的基本概念,而(2)是把实对称矩阵合同于对角矩阵的问题转化成二次型求标准形的问题,用二次型的理论与方法来处理矩阵中的问题. 【解析】(1)因为3λ=是A 的特征值,故31001300313138(2)0,00311311011y E A y y ------==⋅=-=-----所以2y =.(2)由于TA A =,要2()()T T AP AP P A P ==Λ,而21000010000540045A ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦是对称矩阵,故可构造二次型2T x A x ,将其化为标准形Ty y Λ.即有2A 与Λ合同.亦即2T P A P =Λ.方法一：配方法.由于 22222123434558T x A x x x x x x x =++++22222212334444222212344816165()55255495(),55x x x x x x x x x x x x x =+++++-=++++那么,令1122334444,,,,5y x y x y x x y x ===+=即经坐标变换1122334410000100,400150001x y x y x y x y ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥-⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎢⎥⎣⎦有 222221234955Tx A x y y y y =+++. 所以,取 10000100400150001P ⎡⎤⎢⎥⎢⎥=⎢⎥-⎢⎥⎢⎥⎢⎥⎣⎦,有 211()()595T T AP AP P A P ⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦. 方法二：正交变换法.二次型22222123434558T x A x x x x x x x =++++对应的矩阵为21000010000540045A ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦, 其特征多项式23100010(1)(9)005445E A λλλλλλλ---==------.2A 的特征值12341,1,1,9λλλλ====.由21()0E A x λ-=,即12340000000000044000440x x x x ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥--⎢⎥⎢⎥⎢⎥--⎣⎦⎣⎦⎣⎦,和24()0E A x λ-=,即12348000080000044000440x x x x ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥-⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦⎣⎦,分别求得对应1,2,31λ=的线性无关特征向量123(1,0,0,0),(0,1,0,0),(0,0,1,1)T T T ααα===-,和49λ=的特征向量4(0,0,1,1)Tα=.对123,,ααα用施密特正交化方法得123,,βββ,再将4α单位化为4β,其中：1234(1,0,0,0),(0,1,0,0),,T T T Tββββ====. 取正交矩阵[]123410000100000,,,P ββββ⎡⎤⎢⎥⎢⎥⎢==⎢⎢⎢⎢⎣, 则 1221119T P A P P A P -⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎣⎦, 即 211()()19T T AP AP P A P ⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎣⎦.十、(本题满分8分)【解析】证法1： (定义法)若有一组数12,,,,,t k k k k 使得1122()()()0,t t k k k k ββαβαβα+++++++= (1)则因12,,,t ααα是0AX =的解,知0(1,2,,)i A i t α==,用A 左乘上式的两边,有12()0t k k k k A β++++=. (2) 由于0A β≠,故120t k k k k ++++=. 对(1)重新分组为121122()0t t t k k k k k k k βααα++++++++=. (3)把(2)代入(3)得 11220t t k k k ααα+++=.由于12,,,t ααα是基础解系,它们线性无关,故必有120,0,,0t k k k ===.代入(2)式得:0k =. 因此向量组12,,,,t ββαβαβα+++线性无关.证法2： (用秩)经初等变换向量组的秩不变.把第一列的-1倍分别加至其余各列,有()()1212,,,,,,,,.t t ββαβαβαβααα+++→ 因此 ()()1212,,,,,,,,.t t r r ββαβαβαβααα+++=由于12,,,t ααα是基础解系,它们是线性无关的,秩()12,,,t r t ααα=,又β必不能由12,,,t ααα线性表出(否则0A β=),故()12,,,,1t r t αααβ=+.所以 ()12,,,, 1.t r t ββαβαβα+++=+即向量组12,,,,t ββαβαβα+++线性无关.十一、(本题满分7分)【解析】设一周5个工作日内发生故障的天数为X ,则X 服从二项分布即(5,0.2)B . 由二项分布的概率计算公式,有{}500.80.32768,P X ==={}14510.80.20.4096,P X C ==⋅= {}232520.80.20.2048,P X C ==⋅={}{}{}{}310120.05792.P X P X P X P X ≥=-=-=-==设一周内所获利润Y (万元),则Y 是X 的函数,且10,0,5,1,()0,2,2,3.X X Y f X X X =⎧⎪=⎪==⎨=⎪⎪-≥⎩若若若若由离散型随机变量数学期望计算公式,100.3276850.409620.05792 5.20896EY =⨯+⨯-⨯=(万元).【相关知识点】1.二项分布的概率计算公式：若(,)Y B n p ~,则{}(1)kkn kn P Y k C p p -==-, 0,1,,k n =.2.离散型随机变量数学期望计算公式：{}1()nkk k E X xP X x ==⋅=∑.十二、(本题满分6分)【解析】一枚色子(骰子)接连掷两次,其样本空间中样本点总数为36.设事件1A =“方程有实根”,2A =“方程有重根”,则{}221404B A B C C ⎧⎫=-≥=≤⎨⎬⎩⎭.用列举法求有利于i A 的样本点个数(1,2i =),具体做法见下表:有利于的意思就是使不等式24B C ≤尽可能的成立,则需要B 越大越好,C 越小越好.当B 取遍由古典型概率计算公式得到11246619(),3636p P A ++++===2111().3618q P A +===【相关知识点】古典型概率计算公式：().i i A P A =有利于事件的样本点数样本空间的总数十三、(本题满分6分) 【解析】依题意,12,,,n X X X 独立同分布,可见22212,,,n X X X 也独立同分布.由(1,2,3,4)k k EX a k ==及方差计算公式,有224222242222242211,(),111,().i i i i n nn i n ii i EX a DX EX EX a a EZ EX a DZ DX a a n nn ====-=-====-∑∑ 因此,根据中心极限定理n U =的极限分布是标准正态分布,即当n 充分大时,n Z 近似服从参数为2422(,)a a a n-的正态分布.【相关知识点】1.列维-林德伯格中心极限定理,又称独立同分布的中心极限定理：设随机变量12,,,n X X X 独立同分布,方差存在,记μ与2σ()0σ<<+∞分别是它们相同的期望和方差,则对任意实数x ,恒有1lim )(),ni n i P X n x x μ→∞=⎫-≤=Φ⎬⎭∑ 其中()x Φ是标准正态分布函数.2.方差计算公式：22()()()D X E X E X =-.物业管理执行方案（赠送文档）一、 物业管理人员要具备服务意识1、 物业管理人员形象。
考研真题96年
考研真题96年考研真题一直是考生备考的重要资料之一，对于了解考试难度、把握备考方向有着很大的帮助。
本文将深入分析和解读1996年的考研真题，帮助考生更好地了解历年考试内容和趋势。
一、多项选择题1. （） Use of atomic power is unique in being the only way of generating electricity that does not cause pollution in the process of power generation.A. One can say thatB. It can be said thatC. That one can sayD. That can be said在1996年的考研英语真题中，我们可以看到这道多项选择题的类型为判断。
答案是B。
然而，我深刻认识到遗漏了审题中提示信息的重要性。
二、阅读理解Dreams are often a series of images and emotions that occur during a long period of sleep. For centuries, people have wondered why we dream and what purpose dreams serve. In understanding dreams and their significance, the field of psychology has come up with several theories.One of the most famous dream theories is Sigmund Freud’s theory of dream symbolism. According to Freud, dreams are symbols of our unconscious desires and needs. They are a way for us to understand and fulfill our deepest wishes. For example, if you dream about flying, Freud might say that you have a desire for freedom or a need to escape from something in your life.Another theory about dreams is that they are our mind’s way of making sense of the events and emotions that we experience during the day. This theory, known as the information processing theory, suggests that dreams are a way for our brain to process, sort, and organize the information that we encounter when we are awake. For example, if you have a stressful day at work, you might have a dream about being chased by a monster. This dream could be your mind’s way of dealing with the stress and fear that you experienced during the day.Overall, dreams are still a mystery to scientists and psychologists. While there are many theories about why we dream, no one theory can fully explain the purpose and meaning of our dreams. However, studying dreams can provide insight into the human mind and the mysteries of our subconscious.1996年考研英语阅读理解题中包含了对梦的理论和含义的阐述。
1996考研数学三真题和详解
1996年全国硕士研究生入学统一考试数学三试题一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上.) (1) 设方程y x y =确定y 是x 的函数,则dy =___________. (2) 设()arcsin x f x dx x C =+⎰,则1()dx f x =⎰___________.. (3) 设()00,x y 是抛物线2y ax bx c =++上的一点,若在该点的切线过原点,则系数应满足的关系是___________. (4) 设123222212311111231111n n n n n n n a a a a A a a a a a a a a ----⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎣⎦,123n x x X x x ⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎣⎦,1111B ⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎣⎦,其中(;,1,2,,)i j a a i j i j n ≠≠=.则线性方程组T A X B =的解是___________.(5) 设由来自正态总体2~(,0.9)X N μ容量为9的简单随机样本,得样本均值5X =,则未知参数μ的置信度为0.95的置信区间为___________.二、选择题(本题共5小题,每小题3分,满分15分.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) 累次积分cos 20(cos ,sin )d f r r rdr πθθθθ⎰⎰可以写成 ( )(A) 10(,)dy f x y dx ⎰(B) 10(,)dy f x y dx ⎰(C)1100(,)dx f x y dy ⎰⎰(D) 10(,)dx f x y dy ⎰(2) 下述各选项正确的是 ( ) (A) 若21nn u∞=∑和21nn v∞=∑都收敛,则21()nn n uv ∞=+∑收敛(B)1n nn u v∞=∑收敛,则21nn u∞=∑与21nn v∞=∑都收敛(C) 若正项级数1nn u∞=∑发散,则1n u n≥(D) 若级数1nn u∞=∑收敛,且(1,2,)n n u v n ≥=,则级数1n n v ∞=∑也收敛(3) 设n 阶矩阵A 非奇异(2n ≥),A *是矩阵A 的伴随矩阵,则 ( )(A) 1()n A AA -**= (B) 1()n A A A +**= (C) 2()n A AA -**= (D) 2()n A AA +**=(4) 设有任意两个n 维向量组1,,m αα和1,,m ββ,若存在两组不全为零的数1,,m λλ和1,,m k k ,使111111()()()()0m m m m m m k k k k λαλαλβλβ+++++-++-=,则( )(A) 1,,m αα和1,,m ββ都线性相关 (B) 1,,m αα和1,,m ββ都线性无关(C) 1111,,,,,m m m m αβαβαβαβ++--线性无关 (D) 1111,,,,,m m m m αβαβαβαβ++--线性相关(5) 已知0()1P B <<且()1212[]()()P A A B P A B P A B +=+,则下列选项成立的是( ) (A) ()1212[]()()P A A B P A B P A B +=+ (B) ()1212()()P A B A B P A B P A B +=+ (C) ()1212()()P A A P A B P A B +=+ (D) ()()1122()()()P B P A P B A P A P B A =+三、(本题满分6分)设(),0,()0,0,xg x e x f x xx -⎧-≠⎪=⎨⎪=⎩其中()g x 有二阶连续导数,且(0)1,(0)1g g '==-. (1)求()f x '；(2)讨论()f x '在(,)-∞+∞上的连续性.四、(本题满分6分)设函数()z f u =,方程()()xyu u p t dt ϕ=+⎰确定u 是,x y 的函数,其中(),()f u u ϕ可微；()p t ,()u ϕ'连续,且()1u ϕ'≠.求()()z z p y p x x y∂∂+∂∂.五、(本题满分6分)计算2(1)xx xe dx e -+∞-+⎰.六、(本题满分5分)设()f x 在区间[0,1]上可微,且满足条件120(1)2()f xf x dx =⎰.试证:存在(0,1)ξ∈使()()0.f f ξξξ'+=七、(本题满分6分)设某种商品的单价为p 时,售出的商品数量Q 可以表示成aQ c p b=-+,其中a b 、、 c 均为正数,且a bc >.(1) 求p 在何范围变化时,使相应销售额增加或减少.(2) 要使销售额最大,商品单价p 应取何值?最大销售额是多少?八、(本题满分6分)求微分方程dy dx =的通解.九、(本题满分8分)设矩阵010010000010012A y ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦. (1) 已知A 的一个特征值为3,试求y ； (2) 求矩阵P ,使()()TAP AP 为对角矩阵.十、(本题满分8分)设向量12,,,t ααα是齐次线性方程组0AX =的一个基础解系,向量β不是方程组0AX =的解,即0A β≠.试证明:向量组12,,,,t ββαβαβα+++线性无关.十一、(本题满分7分)假设一部机器在一天内发生故障的概率为0.2,机器发生故障时全天停止工作,若一周5个工作日里无故障,可获利润10万元；发生一次故障仍可获得利润5万元；发生两次故障所获利润0元；发生三次或三次以上故障就要亏损2万元.求一周内期望利润是多少?十二、(本题满分6分)考虑一元二次方程20x Bx C ++=,其中B C 、分别是将一枚色子(骰子)接连掷两次先后出现的点数.求该方程有实根的概率p 和有重根的概率q .十三、(本题满分6分)假设12,,,n X X X 是来自总体X 的简单随机样本；已知(1,2,3,4)k k EX a k ==.证明：当n 充分大时,随机变量211n n i i Z X n ==∑近似服从正态分布,并指出其分布参数.1996年全国硕士研究生入学统一考试数学三试题解析一、填空题(本题共5小题,每小题3分,满分15分,把答案填在题中横线上.) (1)【答案】()1ln dxx y +【解析】方法1：方程y x y =两边取对数得ln ln ln y x y y y ==,再两边求微分,()()11ln 1ln 1dx y dy dy dx x x y =+⇒=+()()ln 10x y +≠. 方法2：把y x y =变形得ln y yx e=,然后两边求微分得()()()ln ln 1ln 1ln y y y dx e d y y y y dy x y dy ==+=+,由此可得 ()1.1ln dy dx x y =+(2)【答案】C【解析】由()arcsin x f x dx x C =+⎰,两边求导数有()1()arcsin ()xf x x f x '==⇒=于是有1()dx f x ⎰212==⎰()2112x =--⎰C =.(3)【答案】0c a≥(或20ax c =),b 任意 【解析】对2y ax bx c =++两边求导得()0022y ax b,y x ax b,''=+=+ 所以过()00x ,y 的切线方程为()()0002y y ax b x x ,-=+-即()()()200002y ax bx c ax b x x .-++=+-又题设知切线过原点()00,,把0x y ==代入上式,得2200002ax bx c ax bx ,---=--即20ax c.=由于系数0a ≠,所以,系数应满足的关系为0c a≥(或20ax c =),b 任意. (4)【答案】()1000T,,,【解析】因为A 是范德蒙行列式,由i j a a ≠知()0ijA a a =-≠∏.根据解与系数矩阵秩的关系,所以方程组TA XB =有唯一解.根据克莱姆法则,对于2111112122222133332111111111n n n n n nnn x a a a x a a a x a a a x a a a ----⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦, 易见 1230n D A ,D D D .=====所以TA XB =的解为12310n x ,x x x =====,即()1000T,,,,.【相关知识点】克莱姆法则：若线性非齐次方程组11112211211222221122,,.n n n n n n nn n n a x a x a x b a x a x a x b a x a x a x b +++=⎧⎪+++=⎪⎨⎪⎪+++=⎩或简记为 112nij ji j a xb ,i ,,,n ===∑其系数行列式1112121222120n n n n nna a a a a a D a a a =≠,则方程组有唯一解12j j D x ,j ,,,n.D==其中j D 是用常数项12n b ,b ,,b 替换D 中第j 列所成的行列式,即1111111121212212111,j ,j n ,j ,j n j n n,j nn,j nna ab a a a a b a a D a a b a a -+-+-+=.(5)【答案】(4.412,5.588) 【解析】可以用两种方法求解：(1)已知方差220.9σ=,对正态总体的数学期望μ进行估计,可根据 因2(,0.9)XN μ,设有n 个样本,样本均值11ni i X X n ==∑,有20.9(,)XN n μ,将其标准化,~(0,1)XN 得：)1,0(~1N nX μ-由正态分布分为点的定义21P uαα⎫⎪<=-⎬⎪⎭可确定临界值2αu ,进而确定相应的置信区间22(x u x u αα-+.(2)本题是在单个正态总体方差已知条件下,求期望值μ的置信区间问题. 由教材上已经求出的置信区间22x u x u αα⎛-+ ⎝,其中21,(0,1)P U u UN αα⎧⎫<=-⎨⎬⎩⎭,可以直接得出答案.方法1：由题设,95.01=-α,可见.05.0=α查标准正态分布表知分位点.96.12=αu 本题9n =, 5X =, 因此,根据 95.0}96.11{=<-nX P μ,有 1.96}0.95P <=,即 {4.412 5.588}0.95P μ<<=,故μ的置信度为0.95的置信区间是(4.412,5.588) .方法2：由题设,95.01=-α,22222{}{}2()10.95,()0.975P U u P u U u u u ααααα<=-<<=Φ-=Φ=查得.96.12=αu20.9σ=,9n =, 5X =代入22(x u x u αα-+得置信区间(4.412,5.588).二、选择题(本题共5小题,每小题3分,满分15分.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)【答案】(D)【解析】方法1：由题设知,积分区域在极坐标系cos ,sin x r y r θθ==中是(),|0,0cos ,2D r r πθθθ⎧⎫=≤≤≤≤⎨⎬⎩⎭即是由221124x y ⎛⎫-+= ⎪⎝⎭与x 轴在第一象限所围成的平面图形,如右图.由于D 的最左边点的横坐标是0,最右点的横坐标是1, 下边界方程是0y ,=上边界的方程是y =从而D的直角坐标表示是(){010D x,y |x ,y ,=≤≤≤≤故(D)正确.方法2：采取逐步淘汰法.由于(A)中二重积分的积分区域的极坐标表示为()1,|0,0sin ,2D r r πθθθ⎧⎫=≤≤≤≤⎨⎬⎩⎭而(B)中的积分区域是单位圆在第一象限的部分, (C)中的积分区域是正方形(){}0101x,y |x ,y ,≤≤≤≤所以,他们都是不正确的.故应选(D).(2)【答案】(A) 【解析】由于级数21nn u∞=∑和21nn v∞=∑都收敛,可见级数()221nn n uv ∞=+∑收敛.由不等式222n n n nu v u v ≤+及比较判别法知级数12n nn u v∞=∑收敛,从而12n nn u v∞=∑收敛.又因为()2222n n nnn n u v u v u v ,+=++即级数()21n n n u v ∞=+∑收敛,故应选(A).设()21112n n u ,v n ,,n ===,可知(B)不正确. 设()21112n u n ,,n n=-=,可知(C)不正确.设()()11112n nn u ,v n ,,nn--==-=,可知(D)不正确.注：在本题中命题(D)“若级数1nn u∞=∑收敛,且(1,2,)n n u v n ≥=,则级数1n n v ∞=∑也收敛.”不正确,这表明：比较判别法适用于正项级数收敛(或级数绝对收敛)的判别,但对任意项级数一般是不适用的.这是任意项级数与正项级数收敛性判别中的一个根本区别. (3)【答案】(C)【解析】伴随矩阵的基本关系式为AA A A A E **==,现将A *视为关系式中的矩阵A ,则有()A A A E ****=.方法一：由1n A A-*=及1()AA A*-=,可得 121()().n n A A A A AA A A--****-=== 故应选(C).方法二：由()A A A E ****=,左乘A 得1()()n AA A AA -***=,即1()()n A E A AA -**=.故应选(C). (4)【答案】(D)【解析】本题考查对向量组线性相关、线性无关概念的理解.若向量组12,,,s γγγ线性无关,即若11220s s x x x γγγ+++=,必有120,0,,0s x x x ===.既然1,,m λλ与1,,m k k 不全为零,由此推不出某向量组线性无关,故应排除(B)、(C). 一般情况下,对于1122110,s s s s k k k l l αααββ++++++=不能保证必有11220,s s k k k ααα+++=及110,s s l l ββ++=故(A)不正确.由已知条件,有()()()()1111110m m m m m m k k λαβλαβαβαβ+++++-++-=,又1,,m λλ与1,,m k k 不全为零,故1111,,,,,m m m m αβαβαβαβ++--线性相关.故选(D).(5)【答案】(B) 【解析】依题意()()()()()12121212)(,.()()()()()P A A B P A B P A B P A B A B P A B P A B P B P B P B P B P B +⎡⎤++⎣⎦=+=因()0P B >,故有()()1212)(P AB A B P AB P A B +=+.因此应选(B).注：有些考生错误地选择(D).他们认为(D)是全概率公式,对任何事件B 都成立,但是忽略了全概率公式中要求作为条件的事件12,A A 应满足12()0,()0P A P A >>,且12,A A 是对立事件.【相关知识点】条件概率公式：()(|)()P AB P B A P A =.三、(本题满分6分)【解析】(1) 由于()g x 有二阶连续导数,故当0x ≠时,()f x 也具有二阶连续导数,此时,()f x '可直接计算,且()f x '连续；当0x =时,需用导数的定义求(0)f '.当0x ≠时, 22[()]()()()(1)().x x xx g x e g x e xg x g x x e f x x x ---''+-+-++'== 当0x =时,由导数定义及洛必达法则,有2000()()()(0)1(0)lim lim lim 222x x x x x x g x e g x e g x e g f x x ---→→→'''''-+--'==洛洛. 所以 2()()(1),0,()(0)1,0.2xxg x g x x e x x f x g x -'⎧-++≠⎪⎪'=⎨''-⎪=⎪⎩(2) ()f x '在0x =点的连续性要用定义来判定.因为在0x =处,有200()()(1)lim ()lim xx x xg x g x x e f x x -→→'-++'=0()()()(1)lim 2x xx g x xg x g x e x e x --→''''+-+-+= 0()(0)1lim(0)22x x g x e g f -→''''--'===. 而()f x '在0x ≠处是连续函数,所以()f x '在(,)-∞+∞上为连续函数.四、(本题满分6分) 【解析】由()z f u =可得(),()z u z u f u f u x x y y∂∂∂∂''==∂∂∂∂. 在方程()()xyu u p t dt ϕ=+⎰两边分别对,x y 求偏导数,得()(),()().u u u u u p x u p y x x y yϕϕ∂∂∂∂''=+=-∂∂∂∂ 所以()(),1()1()u p x u p y x u y u ϕϕ∂∂-==''∂-∂-. 于是 ()()()()()()()01()1()z z p x p y p x p y p y p x f u x y u u ϕϕ⎡⎤∂∂'+=-=⎢⎥''∂∂--⎣⎦.五、(本题满分6分)【分析】题的被积函数是幂函数与指数函数两类不同的函数相乘,应该用分部积分法. 【解析】方法1:因为21(1)111x x x x x xe x dxdx xd e e e e-----=-++++⎰⎰⎰分部积分 1(1)1111ln(1),1x xx x x x xx x e x dx d e e e e e x e C e---=-=-+++++=-+++⎰⎰所以20lim ln(1)ln 2.(1)1x x x x x x xe xe dx e e e -+∞-→+∞⎡⎤=-++⎢⎥++⎣⎦⎰而 lim ln(1)lim ln (1)11x x x x xxx x x xe xe e e e e e -→+∞→+∞⎡⎤⎧⎫⎡⎤-+=-+⎨⎬⎢⎥⎣⎦++⎣⎦⎩⎭lim ln(1)1x x xx xe x e e -→+∞⎧⎫=--+⎨⎬+⎩⎭lim 001xx xe →+∞-=-=+,故原式ln 2=. 方法2:220001(1)(1)1x x x x x xe xe dx dx xd e e e-+∞+∞+∞-==-+++⎰⎰⎰0000011111(1)ln(1)ln 2.1xxx x x x x xx dx dx e dx e e e e d e e e +∞-+∞+∞+∞-+∞+∞---=-+==++++=-+=-+=+⎰⎰⎰⎰六、(本题满分5分)【分析】由结论可知,若令()()x xf x ϕ=,则()()()x f x xf x ϕ''=+.因此,只需证明()x ϕ在[0,1]内某一区间上满足罗尔定理的条件.【解析】令()()x xf x ϕ=,由积分中值定理可知,存在1(0,)2η∈,使112201()()()2xf x dx x dx ϕϕη==⎰⎰,由已知条件,有1201(1)2()2()(),2f xf x dx ϕηϕη==⋅=⎰于是(1)(1)(),f ϕϕη==且()x ϕ在(,1)η上可导,故由罗尔定理可知,存在(,1)(0,1),ξη∈⊂使得()0,ϕξ'=即()()0.f f ξξξ'+=【相关知识点】1.积分中值定理：如果函数()f x 在积分区间[ ,]a b 上连续,则在[ ,]a b 上至少存在一个点ξ,使下式成立：()()()()baf x dx f b a a b ξξ=-≤≤⎰.这个公式叫做积分中值公式. 2.罗尔定理：如果函数()f x 满足(1)在闭区间[ ,]a b 上连续； (2)在开区间()a,b 内可导；(3)在区间端点处的函数值相等,即()()f a f b =, 那么在()a,b 内至少有一点ξ(a b ξ<<),使得()0f ξ'=.七、(本题满分6分)【分析】利用函数的单调性的判定,如果在x 的某个区间上导函数()0f x '≥,则函数()f x 单调递增,反之递减.【解析】(1)设售出商品的销售额为R ,则()()22(),().ab c p b aR pQ p c R p p b p b -+'==-=++ 令0,R '=得00p b ==>.当0p <<时,0R '>,所以随单价p 的增加,相应销售额R 也将增加.当p >时,有0R '<,所以随单价p 的增加,相应销售额R 将减少. (2)由(1)可知,当p =时,销售额R 取得最大值,最大销售额为2maxR b c ⎡⎤⎫⎥=-=⎪⎪⎥⎭⎥⎦.八、(本题满分6分) 【解析】令y z x =,则dy dz z x dx dx=+. 当0x >时,原方程化为dz z xz dx +=dxx =-,其通解为1ln(ln z x C +=-+ 或Cz x=. 代回原变量,得通解(0)y C x +=>.当0x <时,原方程的解与0x >时相同,理由如下: 令t x =-,于是0t >,而且dy dy dx dydt dx dt dx =⋅=-===从而有通解(0)y C t +=>,即(0)y C x =<.综合得,方程的通解为y C +=.注：由于未给定自变量x 的取值范围,因而在本题求解过程中,引入新未知函数yz x=后得=从而,应当分别对0x >和0x <求解,在类似的问题中,这一点应当牢记.九、(本题满分8分)【分析】本题的(1)是考查特征值的基本概念,而(2)是把实对称矩阵合同于对角矩阵的问题转化成二次型求标准形的问题,用二次型的理论与方法来处理矩阵中的问题. 【解析】(1)因为3λ=是A 的特征值,故31001300313138(2)0,003113110011y E A y y ------==⋅=-=-----所以2y =.(2)由于TA A =,要2()()T T AP AP P A P ==Λ,而21000010000540045A ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦是对称矩阵,故可构造二次型2T x A x ,将其化为标准形Ty y Λ.即有2A 与Λ合同.亦即2T P A P =Λ.方法一：配方法.由于 22222123434558T x A x x x x x x x =++++22222212334444222212344816165()55255495(),55x x x x x x x x x x x x x =+++++-=++++那么,令1122334444,,,,5y x y x y x x y x ===+=即经坐标变换1122334410000100,400150001x y x y x y x y ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥-⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎢⎥⎣⎦有 222221234955Tx A x y y y y =+++. 所以,取 10000100400150001P ⎡⎤⎢⎥⎢⎥=⎢⎥-⎢⎥⎢⎥⎢⎥⎣⎦,有 211()()595T T AP AP P A P ⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦. 方法二：正交变换法.二次型22222123434558T x A x x x x x x x =++++对应的矩阵为21000010000540045A ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦, 其特征多项式2310000100(1)(9)00540045E A λλλλλλλ---==------.2A 的特征值12341,1,1,9λλλλ====.由21()0E A x λ-=,即12340000000000044000440x x x x ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥--⎢⎥⎢⎥⎢⎥--⎣⎦⎣⎦⎣⎦,和24()0E A x λ-=,即12348000080000044000440x x x x ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥-⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦⎣⎦,分别求得对应1,2,31λ=的线性无关特征向量123(1,0,0,0),(0,1,0,0),(0,0,1,1)T T T ααα===-,和49λ=的特征向量4(0,0,1,1)T α=.对123,,ααα用施密特正交化方法得123,,βββ,再将4α单位化为4β,其中：1234(1,0,0,0),(0,1,0,0),,T T T Tββββ====. 取正交矩阵[]123410000100000,,,P ββββ⎡⎤⎢⎥⎢⎥⎢⎥==⎢⎢⎢⎢⎣, 则 1221119T P A P P A P -⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎣⎦, 即 211()()19T T AP AP P A P ⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎣⎦.十、(本题满分8分)【解析】证法1： (定义法)若有一组数12,,,,,t k k k k 使得1122()()()0,t t k k k k ββαβαβα+++++++= (1)则因12,,,t ααα是0AX =的解,知0(1,2,,)i A i t α==,用A 左乘上式的两边,有12()0t k k k k A β++++=. (2) 由于0A β≠,故120t k k k k ++++=. 对(1)重新分组为121122()0t t t k k k k k k k βααα++++++++=. (3)把(2)代入(3)得 11220t t k k k ααα+++=.由于12,,,t ααα是基础解系,它们线性无关,故必有120,0,,0t k k k ===.代入(2)式得:0k =. 因此向量组12,,,,t ββαβαβα+++线性无关.证法2： (用秩)经初等变换向量组的秩不变.把第一列的-1倍分别加至其余各列,有()()1212,,,,,,,,.t t ββαβαβαβααα+++→ 因此 ()()1212,,,,,,,,.t t r r ββαβαβαβααα+++=由于12,,,t ααα是基础解系,它们是线性无关的,秩()12,,,t r t ααα=,又β必不能由12,,,t ααα线性表出(否则0A β=),故()12,,,,1t r t αααβ=+.所以 ()12,,,, 1.t r t ββαβαβα+++=+即向量组12,,,,t ββαβαβα+++线性无关.十一、(本题满分7分)【解析】设一周5个工作日内发生故障的天数为X ,则X 服从二项分布即(5,0.2)B . 由二项分布的概率计算公式,有{}500.80.32768,P X ==={}14510.80.20.4096,P X C ==⋅= {}232520.80.20.2048,P X C ==⋅={}{}{}{}310120.05792.P X P X P X P X ≥=-=-=-==设一周内所获利润Y (万元),则Y 是X 的函数,且10,0,5,1,()0,2,2,3.XX Y f X X X =⎧⎪=⎪==⎨=⎪⎪-≥⎩若若若若由离散型随机变量数学期望计算公式,100.3276850.409620.05792 5.20896EY =⨯+⨯-⨯=(万元).【相关知识点】1.二项分布的概率计算公式：若(,)Y B n p ~,则{}(1)kkn kn P Y k C p p -==-, 0,1,,k n =.2.离散型随机变量数学期望计算公式：{}1()nkk k E X xP X x ==⋅=∑.十二、(本题满分6分)【解析】一枚色子(骰子)接连掷两次,其样本空间中样本点总数为36.设事件1A =“方程有实根”,2A =“方程有重根”,则{}221404B A B C C ⎧⎫=-≥=≤⎨⎬⎩⎭.用列举法求有利于i A 的样本点个数(1,2i =),具体做法见下表:有利于的意思就是使不等式24B C ≤尽可能的成立,则需要B 越大越好,C 越小越好.当B 取遍1,2,3,4,5,6时,统计C 可能出现的点数有多少种.由古典型概率计算公式得到11246619(),3636p P A ++++===2111().3618q P A +===【相关知识点】古典型概率计算公式：().i i A P A =有利于事件的样本点数样本空间的总数十三、(本题满分6分) 【解析】依题意,12,,,n X X X 独立同分布,可见22212,,,nX X X 也独立同分布.由 (1,2,3,4)k k EX a k ==及方差计算公式,有224222242222242211,(),111,().i i i i n n ni n i i i EX a DX EX EX a a EZ EX a DZ DX a a n n n====-=-====-∑∑ 因此,根据中心极限定理n U =的极限分布是标准正态分布,即当n 充分大时,n Z 近似服从参数为2422(,)a a a n-的正态分布.【相关知识点】1.列维-林德伯格中心极限定理,又称独立同分布的中心极限定理：设随机变量12,,,n X X X 独立同分布,方差存在,记μ与2σ()0σ<<+∞分别是它们相同的期望和方差,则对任意实数x ,恒有1lim )(),ni n i P X n x x μ→∞=⎫-≤=Φ⎬⎭∑ 其中()x Φ是标准正态分布函数.2.方差计算公式：22()()()D X E X E X =-.。
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