上海交通大学研究生入学考试双控考研试题2009
上海交通大学2009年法学基础考研试题
上海交通大学2009年法学基础考研试题法理综合
一、简述法学研究的方法。
（20分）
二、简述正当程序的特征及其意义。
（30分）
宪法综合
一、全国人民代表大会常务委员会的职权有哪些？（15分）
二、如何理解“公民在法律面前一律平等”？（15分）
三、我国法院审判工作的基本原则是什么？（20分）
民法综合
一、法人机关的构成。
（10分）
二、我国物权法规定的物权类型。
（10分）
三、债的原因。
（10分）
四、被继承人遗产清偿债务的原则。
（10分）
刑法综合
一、名词解释：危害结果、假释。
（10分）
二、简述继续犯和连续犯的区别。
（15分）
三、被判处缓刑和被假释的人应当遵守哪些共同规定？（10分）。
2009年全国硕士研究生入学考试数学二真题及答案
y y x 0 。求 y(x)的表达式。
（21）（本题满分 11 分）（I）证明拉格朗日中值定理：若函数 f (x) 在[a,b]上连续，在（a,b）
可导，则存在 (a,b) ，使得 f (b) f (a) f ( )(b a) 。（II）证明：若函数 f (x) 在 x=0
阵
0 B
A 0
的伴随矩阵为（）
（A）
0 2 A
3B
0
（B）
0 3 A
2B
0
பைடு நூலகம்（C）
0 2B
3A
0
（D）
0 3B
2 A
0
100
（8）设
A，P
均为
3
阶矩阵，
PT
为
P
的转置矩阵，且
1
x
1
y
A
2
dx
4x f x, y dy
1
1
C
2
dy
4y f x, y dx
1
1
B
2
dx
4x f x, y dy
1
x
D
2
1
2
dyy
f
x,
y dx
【答案】 C
2
2
2
2
【解析】 dx f (x, y)dy dy f (x, y)dx 的积分区域为两部分：
sin x
A 1 B 2 C 3 D 无穷多个
【答案】 C
【解析】由于 f x x x3 ,则当 x 取任何整数时, f x 均无意义.
上海交通大学《高等代数》《数学分析》历年考研真题汇总(2009-2018真题汇编)
(x − 1)n | (f (x) + 1), (x + 1)n | (f (x) − 1).
Ê! V •ê• F þ n ‘‚5˜m, A • V þ ‚5C†÷v A 3 − 2A 2 − A = −2id, Ù¥ id • V þð C†.
(1) A ´ÄŒé z, e´, žy². (2) - V1 = {(A − 2id)v | v ∈ V }, V2 = {(A 2 − id)v | v ∈ V }. y²: V = V1 ⊕ V2.
8
5 þ° ÏŒÆ 2015 ca¬ïÄ)\Æ•ÁÁK£828 p “ê¤
9
6 þ° ÏŒÆ 2018 ca¬ïÄ)\Æ•ÁÁK£828 p “ê¤
10
7 þ° ÏŒÆ 2010 ca¬ïÄ)\Æ•ÁÁK( 614 êÆ©Û)
11
8 þ° ÏŒÆ 2011 ca¬ïÄ)\Æ•ÁÁK( 614 êÆ©Û)
16
3
1. 2010年þ° ÏŒÆ828《高等代数》a¬ïÄ)\Æ•ÁÁK
˜! ( 20 ©) OŽ1 ª
an1
an2
(1) Dn+1 =
...
an1 −1b1 · · ·
an2 −1b2 · · · ...
ann+1 ann−+11bn+1 · · ·
1 + a1 + b1 a1 + b2
a1bn1 −1
›˜! A ´ n ‘m¥ f˜m.
C†, V1 ´ V A − ØCf˜m. y²: V1
Ö•´ V A − ØC
› ! A, B þ• n ¢é¡ , y²: AB A ŠÑŒu".
4
1996-2005上海交大自控真题答案
(
) (
)
则求得 s =
− 3.414 − 0.586
T +2 4 s+ = 0 T T
2）特征方程： Ts (s + 1) + 2 s + 4 = 0 即 s 2 +
∴ ∴
2 ωn =
4 T
2ξωn =
1
T +2 T
1
1 1 − ∴ξ = T 2 + T 2 4 2
dξ 1 1 1 1 1 −3 = ⋅ T 2 + − T 2 = 0 dT 4 2 2 2
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
88
89
90
91
92
93
94
95
96
97
上海交通大学 ２００３ 自控原理研究生考试试题题解
一．设一反馈控制系统的结构如图所示，
1）试求取系统输出对输入的传递函数 C (s ) 2）试求取系统输出对扰动的传递函数 C (s )
R( s ) F (s)
； 。
（10 分） （10 分）
解：1）设 F ( s ) = 0
G2 G1 1 − G H C (s ) G1G2 2 2 = = R(s ) 1 − G2 H 2 + G1G2 H 3 G2 1 + G1 H 3 1− G H 2 2
2009年全国硕士研究生入学统一考试真题加答案
2009年全国硕士研究生入学统一考试英语试题Section I Use of EnglishDirections:Read the following text. Choose the best word(s) for each numbered blank and mark A, B, C or D on ANSWER SHEET 1. (10 points)Research on animal intelligence always makes me wonder just how smart humans are.1 the fruit-fly experiments described in Carl Zimmer‘s piece in the Science Times on Tuesday. Fruit flies who were taught to be smarter than the average fruit fly 2 to live shorter lives. This suggests that 3 bulbs burn longer, that there is an 4 in not being too terrifically bright.Intelligence, it 5 out, is a high-priced option. It takes more upkeep, burns more fuel and is slow 6 the starting line because it depends on learning — a gradual 7 — instead of instinct. Plenty of other species are able to learn, and one of the things they‘ve apparently learned is when to 8 .Is there an adaptive value to 9 intelligence? That‘s the question behind this new research. I like it. Instead of casting a wistful glance 10 at all the species we‘ve left in the dust I.Q.-wise, it implicitly asks what the real11 of our own intelligence might be. This is 12 the mind of every animal I‘ve ever met.Research on animal intelligence also makes me wonder what experiments animals would 13 on humans if they had the chance. Every cat with an owner, 14 , is running a small-scale study in operant conditioning. we believe that 15 animals ran the labs, they would test us to 16 the limits of our patience, our faithfulness, our memory for terrain. They would try to decide what intelligence in humans is really 17 , not merely how much of it there is. 18 , they would hope to study a 19 question: Are humans actually aware of the world they live in? 20 the results are inconclusive.1. [A] Suppose [B] Consider [C] Observe [D] Imagine2. [A] tended [B] feared [C] happened [D] threatened3. [A] thinner [B] stabler [C] lighter [D] dimmer4. [A] tendency [B] advantage [C] inclination [D] priority5. [A] insists on [B] sums up [C] turns out [D] puts forward6. [A] off [B] behind [C] over [D] along7. [A] incredible [B] spontaneous [C]inevitable [D] gradual8. [A] fight [B] doubt [C] stop [D] think19. [A] invisible [B] limited [C] indefinite [D] different10. [A] upward [B] forward [C] afterward [D] backward11. [A] features [B] influences [C] results [D] costs12. [A] outside [B] on [C] by [D] across13. [A] deliver [B] carry [C] perform [D] apply14. [A] by chance [B] in contrast [C] as usual [D] for instance15. [A] if [B] unless [C] as [D] lest16. [A] moderate [B] overcome [C] determine [D] reach17. [A] at [B] for [C] after [D] with18. [A] Above all [B] After all [C] However [D] Otherwise19. [A] fundamental [B] comprehensive [C] equivalent [D] hostile20. [A] By accident [B] In time [C] So far [D] Better stillSection II Reading ComprehensionPart ADirections:Read the following four texts. Answer the questions below each text by choosing A, B, C or D. Mark your answers on ANSWER SHEET 1. (40 points)Text1Habits are a funny thing. We reach for them mindlessly, setting our brains on auto-pilot and relaxing into the unconscious comfort of familiar routine. ―Not choice, but habit rules the unreflecting herd,‖ William Wordsworth said in the 19th century. In the ever-changing 21st century, even the word ―habit‖ carries a ne gative connotation.So it seems antithetical to talk about habits in the same context as creativity and innovation. But brain researchers have discovered that when we consciously develop new habits, we create parallel synaptic paths, and even entirely new brain cells, that can jump our trains of thought onto new, innovative tracks.But don‘t bother trying to kill off old habits; once those ruts of procedure are worn into the hippocampus, they‘re there to stay. Instead, the new habits we deliberately ingrain into ourselves create parallel pathways that can bypass those old roads.―The first thing needed for innovation is a fascination with wonder,‖ says Dawna Markova, author of ―The Open Mind‖ and an executive change consultant for Professional Thinking Part ners. ―But we are taught instead to ‗decide,‘ just as our president calls himself ‗the Decider.‘‖ She adds, however, that ―to decide is to kill off all possibilities but one. A good innovational thinker is always exploring the many other possibilities.‖A ll of us work through problems in ways of which we‘re unaware, she says. Researchers in the late 1960 covered that humans are born with the capacity to2approach challenges in four primary ways: analytically, procedurally, relationally (or collaboratively) and innovatively. At puberty, however, the brain shuts down half of that capacity, preserving only those modes of thought that have seemed most valuable during the first decade or so of life.The current emphasis on standardized testing highlights analysis and procedure, meaning that few of us inherently use our innovative and collaborative modes of thought. ―This breaks the major rule in the American belief system — that anyone can do anything,‖ explains M. J. Ryan, author of the 2006 book ―This Year I Will...‖ and Ms. Markova‘s business partner. ―That‘s a lie that we have perpetuated, and it fosters commonness. Knowing what you‘re good at and doing even more of it creates excellence.‖ This is where developing new habits comes in.21. The view of Wordsworth habit is claimed by beingA. casualB. familiarC. mechanicalD. changeable.22. The researchers have discovered that the formation of habit can beA. predictedB. regulatedC. tracedD. guided23.‖ ruts‖(in li ne one, paragraph 3) has closest meaning toA. tracksB. seriesC. characteristicsD. connections24. Ms. Markova‘s comments suggest that the practice of standard testing ? A, prevents new habits form being formedB, no longer emphasizes commonnessC, maintains the inherent American thinking modelD, complies with the American belief system25. Ryan most probably agree thatA. ideas are born of a relaxing mindB. innovativeness could be taughtC. decisiveness derives from fantastic ideasD. curiosity activates creative mindsText 2It is a wise father that knows his own child, but today a man can boost his paternal (fatherly) wisdom –or at least confirm that he‘s the kid‘s dad. All he needs to do is shell our $30 for paternity testing kit (PTK) at his local drugstore – and another $120 to get the results.More than 60,000 people have purchased the PTKs since they first become available without prescriptions last years, according to Doug Fog, chief operating officer of Identigene, which makes the over-the-counter kits. More than two dozen companies sell DNA tests Directly to the public , ranging in price from a few hundred dollars to more than $2500.3Among the most popular : paternity and kinship testing , which adopted children can use to find their biological relatives and latest rage a many passionate genealogists-and supports businesses that offer to search for a family‘s geographic roots .Most tests require collecting cells by webbing saliva in the mouth and sending it to the company for testing. All tests require a potential candidate with whom to compare DNA.But some observers are skeptical, ―There is a kind of false precision being hawked by people claiming they are doing ancestry testing,‖ says Trey Duster, a New York University sociologist. He notes that each individual has many ancestors-numbering in the hundreds just a few centuries back. Yet most ancestry testing only considers a single lineage, either the Y chromosome inherited through men in a father‘s line or mito chondrial DNA, which a passed down only from mothers. This DNA can reveal genetic information about only one or two ancestors, even though, for example, just three generations back people also have six other great-grandparents or, four generations back, 14 other great-great-grandparents.Critics also argue that commercial genetic testing is only as good as the reference collections to which a sample is compared. Databases used by some companies don‘t rely on data collected systematically but rather lump together information from different research projects. This means that a DNA database may differ depending on the company that processes the results. In addition, the computer programs a company uses to estimate relationships may be patented and not subject to peer review or outside evaluation.26.In paragraphs 1 and 2 , the text shows PTK‘s ___________.[A]easy availability[B]flexibility in pricing[C] successful promotion[D] popularity with households27. PTK is used to __________.[A]locate one‘s b irth place[B]promote genetic research[C] identify parent-child kinship[D] choose children for adoption28. Skeptical observers believe that ancestry testing fails to__________.[A]trace distant ancestors[B] rebuild reliable bloodlines[C] fully use genetic information[D] achieve the claimed accuracy29. In the last paragraph ,a problem commercial genetic testing faces is __________.4[A]disorganized data collection[B] overlapping database building30. An appropriate title for the text is most likely to be__________.[A]Fors and Againsts of DNA testing[B] DNA testing and It‘s problems[C]DNA testing outside the lab[D] lies behind DNA testingText 3The relationship between formal education and economic growth in poor countries is widely misunderstood by economists and politicians alike progress in both area is undoubtedly necessary for the social, political and intellectual development of these and all other societies; however, the conventional view that education should be one of the very highest priorities for promoting rapid economic development in poor countries is wrong. We are fortunate that is it, because new educational systems there and putting enough people through them to improve economic performance would require two or three generations. The findings of a research institution have consistently shown that workers in all countries can be trained on the job to achieve radical higher productivity and, as a result, radically higher standards of living.Ironically, the first evidence for this idea appeared in the United States. Not long ago, with the country entering a recessing and Japan at its pre-bubble peak. The U.S. workforce was derided as poorly educated and one of primary cause of the poor U.S. economic performance. Japan was, and remains, the global leader in automotive-assembly productivity. Yet the research revealed that the U.S. factories of Honda Nissan, and Toyota achieved about 95 percent of the productivity of their Japanese countere pants a result of the training that U.S. workers received on the job.More recently, while examing housing construction, the researchers discovered that illiterate, non-English- speaking Mexican workers in Houston, Texas, consistently met best-practice labor productivity standards despite the complexity of the building industry‘s work.What is the real relationship between education and economic development? We have to suspect that continuing economic growth promotes the development of education even when governments don‘t force it. After all, that‘s how education got started. When our ancestors were hunters and gatherers 10,000 years ago, they didn‘t have time to wonder much about anything besides finding food. Only when humanity began to get its food in a more productive way was there time for other things.As education improved, humanity‘s productivity potential, they could in turn afford more education. This increasingly high level of education is probably a necessary, but not a sufficient, condition for the complex political systems required by advanced5economic performance. Thus poor countries might not be able to escape their poverty traps without political changes that may be possible only with broader formal education. A lack of formal education, however, doesn‘t const rain the ability of the developing world‘s workforce to substantially improve productivity for the forested future. On the contrary, constraints on improving productivity explain why education isn‘t developing more quickly there than it is.31. The author holds in paragraph 1 that the important of education in poor countries ___________.[A] is subject groundless doubts[B] has fallen victim of bias[C] is conventional downgraded[D] has been overestimated32. It is stated in paragraph 1 that construction of a new education system __________.[A]challenges economists and politicians[B]takes efforts of generations[C] demands priority from the government[D] requires sufficient labor force33.A major difference between the Japanese and U.S workforces is that __________.[A] the Japanese workforce is better disciplined[B] the Japanese workforce is more productive[C]the U.S workforce has a better education[D] ]the U.S workforce is more organize34. The author quotes the example of our ancestors to show that education emerged __________.[A] when people had enough time[B] prior to better ways of finding food[C] when people on longer went hung[D] as a result of pressure on government35. According to the last paragraph , development of education __________.[A] results directly from competitive environments[B] does not depend on economic performance[C] follows improved productivity[D] cannot afford political changesText 4The most thoroughly studied in the history of the new world are the ministers and political leaders of seventeenth-century New England. According to the standard history of American philosophy, nowhere else in colonial America was ―So much6important attached to intellectual pursuits ‖ Accord ing to many books and articles, New England‘s leaders established the basic themes and preoccupations of an unfolding, dominant Puritan tradition in American intellectual life.To take this approach to the New Englanders normally mean to start with the Puritans‘ theological innovations and their distinctive ideas about the church-important subjects that we may not neglect. But in keeping with our examination of southern intellectual life, we may consider the original Puritans as carriers of European culture adjusting to New world circumstances. The New England colonies were the scenes of important episodes in the pursuit of widely understood ideals of civility and virtuosity.The early settlers of Massachusetts Bay included men of impressive education and influence in England. `Besides the ninety or so learned ministers who came to Massachusetts church in the decade after 1629,There were political leaders like John Winthrop, an educated gentleman, lawyer, and official of the Crown before he journeyed to Boston. There men wrote and published extensively, reaching both New World and Old World audiences, and giving New England an atmosphere of intellectual earnestness.We should not forget , however, that most New Englanders were less well educated. While few crafts men or farmers, let alone dependents and servants, left literary compositions to be analyzed, The in thinking often had a traditional superstitions quality. A tailor named John Dane, who emigrated in the late 1630s, left an account of his reasons for leaving England that is filled with signs. sexual confusion, economic frustrations , and religious hope-all name together in a decisive moment when he opened the Bible, told his father the first line he saw would settle his fate, and read th e magical words: ―come out from among them, touch no unclean thing , and I will be your God and you shall be my people.‖ One wonders what Dane thought of the careful sermons explaining the Bible that he heard in puritan churched.Mean while , many se ttles had slighter religious commitments than Dane‘s, as one clergyman learned in confronting folk along the coast who mocked that they had not come to the New world for religion . ―Our main end was to catch fish. ‖36. The author notes that in the seventeenth-century New England___________.[A] Puritan tradition dominated political life.[B] intellectual interests were encouraged.[C] Politics benefited much from intellectual endeavors.[D] intellectual pursuits enjoyed a liberal environment.37. It is suggested in paragraph 2 that New Englanders__________.[A] experienced a comparatively peaceful early history.[B] brought with them the culture of the Old World[C] paid little attention to southern intellectual life[D] were obsessed with religious innovations738. The early ministers and political leaders in Massachusetts Bay__________.[A] were famous in the New World for their writings[B] gained increasing importance in religious affairs[C] abandoned high positions before coming to the New World[D] created a new intellectual atmosphere in New England39. The story of John Dane shows that less well-educated New Englanders were often __________.[A] influenced by superstitions[B] troubled with religious beliefs[C] puzzled by church sermons[D] frustrated with family earnings40. The text suggests that early settlers in New England__________.[A] were mostly engaged in political activities[B] were motivated by an illusory prospect[C] came from different backgrounds.[D] left few formal records for later referencePart BDirections:Directions: In the following text, some sentences have been removed. For Questions (41-45), choose the most suitable one from the list A-G to fit into each of the numbered blank. There are two extra choices, which do not fit in any of the gaps. Mark your answers on ANSWER SHEET 1. (10 points)Coinciding with the groundbreaking theory of biological evolution proposed by British naturalist Charles Darwin in the 1860s, British social philosopher Herbert Spencer put forward his own theory of biological and cultural evolution. Spencer argued that all worldly phenomena, including human societies, changed over time, advancing toward perfection. 41.____________.American social scientist Lewis Henry Morgan introduced another theory of cultural evolution in the late 1800s. Morgan, along with Tylor, was one of the founders of modern anthropology. In his work, he attempted to show how all aspects of culture changed together in the evolution of societies.42._____________.In the early 1900s in North America, German-born American anthropologist Franz Boas developed a new theory of culture known as historical particularism. Historical particularism, which emphasized the uniqueness of all cultures, gave new direction to anthropology. 43._____________ .8Boas felt that the culture of any society must be understood as the result of a unique history and not as one of many cultures belonging to a broader evolutionary stage or type of culture. 44._______________.Historical particularism became a dominant approach to the study of culture in American anthropology, largely through the influence of many students of Boas. But a number of anthropologists in the early 1900s also rejected the particularist theory of culture in favor of diffusionism. Some attributed virtually every important cultural achievement to the inventions of a few, especially gifted peoples that, according to diffusionists, then spread to other cultures. 45.________________.Also in the early 1900s, French sociologist Émile Durkheim developed a theory of culture that would greatly influence anthropology. Durkheim proposed that religious beliefs functioned to reinforce social solidarity. An interest in the relationship between the function of society and culture—known as functionalism—became a major theme in European, and especially British, anthropology.[A] Other anthropologists believed that cultural innovations, such as inventions, had a single origin and passed from society to society. This theory was known as diffusionism.[B] In order to study particular cultures as completely as possible, Boas became skilled in linguistics, the study of languages, and in physical anthropology, the study of human biology and anatomy.[C] He argued that human evolution was characterized by a struggle he called the ―survival of the fittest,‖ in which weaker races and societies must eventu ally be replaced by stronger, more advanced races and societies.[D] They also focused on important rituals that appeared to preserve a people‘s social structure, such as initiation ceremonies that formally signify children‘s entrance into adulthood.[E] Thus, in his view, diverse aspects of culture, such as the structure of families, forms of marriage, categories of kinship, ownership of property, forms of government, technology, and systems of food production, all changed as societies evolved.[F]Supporters of the theory viewed as a collection of integrated parts that work together to keep a society functioning.[G] For example, British anthropologists Grafton Elliot Smith and W. J. Perry incorrectly suggested, on the basis of inadequate information, that farming, pottery9making, and metallurgy all originated in ancient Egypt and diffused throughout the world. In fact, all of these cultural developments occurred separately at different times in many parts of the world.Part CDirections:Read the following text carefully and then translate the underlined segments into Chinese. Your translation should be written carefully on ANSWER SHEET 2. (10 points)There is a marked difference between the education which every one gets from living with others, and the deliberate educating of the young. In the former case the education is incidental; it is natural and important, but it is not the express reason of the association.46It may be said that the measure of the worth of any social institution is its effect in enlarging and improving experience; but this effect is not a part of its original motive. Religious associations began, for example, in the desire to secure the favor of overruling powers and to ward off evil influences; family life in the desire to gratify appetites and secure family perpetuity; systematic labor, for the most part, because of enslavement to others, etc. 47Only gradually was the by-product of the institution noted, and only more gradually still was this effect considered as a directive factor in the conduct of the institution. Even today, in our industrial life, apart from certain values of industriousness and thrift, the intellectual and emotional reaction of the forms of human association under which the world's work is carried on receives little attention as compared with physical output.But in dealing with the young, the fact of association itself as an immediate human fact, gains in importance.48 While it is easy to ignore in our contact with them the effect of our acts upon their disposition, it is not so easy as in dealing with adults. The need of training is too evident; the pressure to accomplish a change in their attitude and habits is too urgent to leave these consequences wholly out of account. 49Since our chief business with them is to enable them to share in a common life we cannot help considering whether or no we are forming the powers which will secure this ability.If humanity has made some headway in realizing that the ultimate value of every institution is its distinctively human effect we may well believe that this lesson has been learned largely through dealings with the young.50 We are thus led to distinguish, within the broad educational process which we have been so far considering, a more formal kind of education -- that of direct tuition or schooling. In undeveloped social groups, we find very little formal teaching and training. These groups mainly rely for instilling needed dispositions into the young upon the same sort of association which keeps the adults loyal to their group.Section & Writing10Part A51. Directions:Restrictions on the use of plastic bags have not been so successful in some regions. ―White pollution ‖is still going on. Write a letter to the editor(s) of your local newspaper to1)give your opinions briefly and2)make two or three suggestionsYou should write about 100 words. Do not sign your own name at the end of the letter. Use "Li Ming" instead. You do not need to write the address.Part B52. Directions:In your essay, you should1) describe the drawing briefly,2) explain its intended meaning, and then3) give your comments.You should write neatly on ANSHWER SHEET 2. (20 points)11。
2009年全国硕士研究生入学统一考试数学一真题及答案
当 时，
当 时，
因此函数 仅在 处间断，故选（B）.
二、填空题：9-14小题，每小题4分，共24分，请将答案写在答题纸指定位置上.
（9）设函数 具有二阶连续偏导数， ，则 .
【答案】
【考点】多元函数的偏导数
【难易度】★★
【详解】本题涉及到的主要知识点：
利用复合函数的链式求导法则求多元函数的偏导数的方法。
在本题中，
，
（10）若二阶常系数线性齐次微分方程 的通解为 ，则非齐次方程 满足条件 的解为 .
【答案】
【考点】简单的二阶常系数非齐次线性微分方程
【难易度】★★
【详解】本题涉及到的主要知识点：
线性微分方程的解的性质即叠加原理，线性微分方程通解的结构为齐次方程的通解加上特解。
在本题中，
由通解表达式 该二阶线性常系数齐次方程的特征值为 ，于是特征方程为
，
而在 上， 有连续的一阶偏导数且 ，于是
（在 ： 上用高斯公式）
（20）（本题满分11分）
设 ， .
（Ⅰ）求满足 的 . 的所有向量 , .
（Ⅱ）对（Ⅰ）中的任意向量 , 证明 , , 线性无关.
【考点】向量组的线性无关，非齐次线性方程组的通解
【难易度】★★★
【详解】本题涉及到的主要知识点：
非齐次线性微分方程的解的性质即叠加原理，非齐次线性微分方程通解的结构为齐次方程的通解加上特解。
收敛级数的和的概念， 称为无穷级数 的前n项的部分和。若部分和数列 的极限存在，即 ，则称级数 收敛。当级数收敛时，其和 。
在本题中，
（Ⅰ）先求 .易求得 与 的交点为 ， ，于是曲线 与 所围成区域的面积为
（Ⅱ）按定义求
（Ⅲ）求 .
2009-数二真题、标准答案及解析
（6）设函数 y = f ( x) 在区间−1,3 上的图形为：
则函数 F ( x) = x f (t ) dt 的图形为 0
( A)
(B)
(C)
(D)
【答案】 D
【解析】此题为定积分的应用知识考核，由 y = f (x) 的图形可见，其图像与 x 轴及 y 轴、
x = x0 所围的图形的代数面积为所求函数 F (x) ，从而可得出几个方面的特征：
a
a3 = −6b ,故排除 B,C .
另外,
lim
x→0
1− a cos ax −3bx2
存在,蕴含了1−
a
cos
ax
→
0
(
x
→
0)
,故
a
=
1.
排除
D
.
所以本题选 A .
(3) 设函数 z = f ( x, y) 的全微分为 dz = xdx + ydy ,则点 (0, 0)
( A) 不是 f ( x, y) 的连续点 ( B) 不是 f ( x, y) 的极值点
【解析】1 =
+ ek x dx = 2 + ekxdx = 2 lim 1 ekx b
−
0
k b→+
0
【答案】 −2
因为极限存在所以 k 0 1=0− 2
k k = −2
（11） lim 1e−x sin nxdx = n→ 0
【答案】0
【解析】令 In = e−x sin nxdx = −e−x sin nx + n e−x cos nxdx
y = t2 ln(2 − t2 )
【答案】 y = 2x
【解析】
2009考研数学三真题及答案解析
2009年全国硕士研究生入学统一考试数学三试题一、选择题：1～8小题，每小题4分，共32分，下列每小题给出的四个选项中，只有一个选项是符合题目要求的，请把所选项前的字母填在答题纸指定位置上.（1）函数的可去间断点的个数为3()sin x x f x xπ-=(A)1.(B)2.(C)3.(D)无穷多个.（2）当时，与是等价无穷小，则0x →()sin f x x ax =-2()ln(1)g x x bx =-(A)，. （B ），. 1a =16b =-1a =16b =(C)，. （D ），.1a =-16b =-1a =-16b =（3）使不等式成立的的范围是1sin ln x tdt x t>⎰x (A).(B). (C).(D).(0,1)(1,2π(,)2ππ(,)π+∞（4）设函数在区间上的图形为()y f x =[]1,3-则函数的图形为()()0xF x f t dt =⎰(A)(B)(C)(D)（5）设均为2阶矩阵，分别为的伴随矩阵，若，则分块矩,A B *,A B *,A B ||2,||3A B ==阵的伴随矩阵为O A B O ⎛⎫⎪⎝⎭(A).(B). **32O B A O ⎛⎫ ⎪⎝⎭**23OB A O ⎛⎫⎪⎝⎭(C).(D).**32O A B O ⎛⎫⎪⎝⎭**23O A BO ⎛⎫⎪⎝⎭（6）设均为3阶矩阵，为的转置矩阵，且，,A P T P P 100010002TP AP ⎛⎫ ⎪= ⎪ ⎪⎝⎭若，则为1231223(,,),(,,)P Q ααααααα==+TQ AQ (A).(B).210110002⎛⎫⎪ ⎪ ⎪⎝⎭110120002⎛⎫⎪ ⎪ ⎪⎝⎭(C). (D).200010002⎛⎫ ⎪ ⎪ ⎪⎝⎭100020002⎛⎫ ⎪ ⎪ ⎪⎝⎭（7）设事件与事件B 互不相容，则A (A). (B). ()0P AB =()()()P AB P A P B =(C).(D).()1()P A P B =-()1P A B ⋃=（8）设随机变量与相互独立，且服从标准正态分布，的概率分布为X Y X (0,1)N Y ，记为随机变量的分布函数，则函数1{0}{1}2P Y P Y ====()z F Z Z XY =()z F Z的间断点个数为(A)0.(B)1. (C)2.(D)3.二、填空题：9~14小题，每小题4分，共24分，请将答案写在答题纸指定位置上.（9）.0x →=（10）设，则.()y xz x e =+(1,0)zx ∂=∂（11）幂级数的收敛半径为 .21(1)n n nn e x n ∞=--∑（12）设某产品的需求函数为,其对应价格的弹性，则当需求量为()Q Q P =P 0.2p ξ=10000件时，价格增加1元会使产品收益增加元.（13）设,，若矩阵相似于，则.(1,1,1)T α=(1,0,)T k β=Tαβ300000000⎛⎫⎪ ⎪ ⎪⎝⎭k = (14)设，,…,为来自二项分布总体的简单随机样本，和分别为样1X 2X n X (,)B n p X 2S 本均值和样本方差，记统计量，则.2T X S =-ET =三、解答题：15～23小题，共94分.请将解答写在答题纸指定的位置上.解答应写出文字说明、证明过程或演算步骤.（15）（本题满分9分）求二元函数的极值.()22(,)2ln f x y xy y y =++（16）（本题满分10 分）计算不定积分 .ln(1dx +⎰(0)x >（17）（本题满分10 分）计算二重积分，其中.()Dx y dxdy -⎰⎰22{(,)(1)(1)2,}D x y x y y x =-+-≤≥（18）（本题满分11 分）（Ⅰ）证明拉格朗日中值定理，若函数在上连续，在上可导，则()f x [],a b (),a b ，得证.(),a b ξ∈()'()()()f b f a f b a ξ-=-（Ⅱ）证明：若函数在处连续，在内可导，且()f x 0x =()0,,(0)σσ>，则存在，且.'0lim ()x f x A +→='(0)f +'(0)f A +=（19）（本题满分10 分）设曲线，其中是可导函数，且.已知曲线与直线()y f x =()f x ()0f x >()y f x =及所围成的曲边梯形绕轴旋转一周所得的立体体积值是该曲边梯0,1y x ==(1)x t t =>x 形面积值的倍，求该曲线的方程.t π（20）（本题满分11 分）设,.111A=111042--⎛⎫ ⎪- ⎪ ⎪--⎝⎭1112ξ-⎛⎫⎪= ⎪⎪-⎝⎭（Ⅰ）求满足，的所有向量，.21A ξξ=231Aξξ=2ξ3ξ（Ⅱ）对（Ⅰ）中的任意向量,，证明,,线性无关.2ξ3ξ1ξ2ξ3ξ（21）（本题满分11 分）设二次型.2221231231323(,,)(1)22f x x x ax ax a x x x x x =++-+-（Ⅰ）求二次型的矩阵的所有特征值.f （Ⅱ）若二次型的规范形为，求的值.f 2211y y +a （22）（本题满分11 分）设二维随机变量的概率密度为(,)X Y 0(,)0xe y xf x y -⎧<<=⎨⎩其他（Ⅰ）求条件概率密度；()Y X f y x （Ⅱ）求条件概率.11P X Y =⎡≤≤⎤⎣⎦（23）（本题满分11分）袋中有一个红球，两个黑球，三个白球，现在放回的从袋中取两次，每次取一个，求以、X 、分别表示两次取球所取得的红、黑与白球的个数.Y Z （Ⅰ）求；10P X Z ⎡==⎤⎣⎦（Ⅱ）求二维随机变量的概率分布.(,)X Y 2009年全国硕士研究生入学统一考试数学三试题解析一、选择题：1～8小题，每小题4分，共32分，下列每小题给出的四个选项中，只有一个选项是符合题目要求的，请把所选项前的字母填在答题纸指定位置上.（1）函数的可去间断点的个数为3()sin x x f x xπ-=(A)1. (B)2. (C)3.(D)无穷多个.【答案】C. 【解析】()3sin x x f x xπ-=则当取任何整数时，均无意义x ()f x 故的间断点有无穷多个，但可去间断点为极限存在的点，故应是的解()f x 30x x -=1,2,30,1x =±320032113211131lim lim sin cos 132lim lim sin cos 132lim lim sin cos x x x x x x x x x x x x x x x x x x x x x ππππππππππππ→→→→→-→---==--==--==故可去间断点为3个，即0,1±（2）当时，与是等价无穷小，则0x →()sin f x x ax =-2()ln(1)g x x bx =-(A)，. （B ），. 1a =16b =-1a =16b =(C)，.（D ），.1a =-16b =-1a =-16b =【答案】A.【解析】为等价无穷小，则2()sin ,()(1)f x x ax g x x ln bx =-=-222200000()sin sin 1cos sin lim lim lim lim lim ()ln(1)()36x x x x x f x x ax x ax a ax a ax g x x bx x bx bx bx→→→→→---==-⋅---洛洛 故排除(B)、(C).230sin lim 166x a ax a b b axa→==-=-⋅36a b ∴=-另外存在，蕴含了故排除(D).201cos lim3x a axbx →--1cos 0a ax -→()0x → 1.a =所以本题选(A).（3）使不等式成立的的范围是1sin ln xtdt x t>⎰x (A).(B). (C).(D).(0,1)(1,2π(,)2ππ(,)π+∞【答案】A.【解析】原问题可转化为求成立时的111sin sin 1()ln xx x tt f x dt x dt dt t t t =-=-⎰⎰⎰11sin 11sin 0x x t t dt dt t t--==>⎰⎰x 取值范围，由，时，知当时，.故应选(A).1sin 0tt->()0,1t ∈()0,1x ∈()0f x >（4）设函数在区间上的图形为()y f x =[]1,3-则函数的图形为()()0xF x f t dt =⎰(A)(B)(C)(D)【答案】D.【解析】此题为定积分的应用知识考核，由的图形可见，其图像与轴及轴、()y f x =x y 所围的图形的代数面积为所求函数，从而可得出几个方面的特征：0x x =()F x ①时，，且单调递减.[]0,1x ∈()0F x ≤②时，单调递增.[]1,2x ∈()F x ③时，为常函数.[]2,3x ∈()F x ④时，为线性函数，单调递增.[]1,0x ∈-()0F x ≤⑤由于F(x)为连续函数结合这些特点，可见正确选项为(D).（5）设均为2阶矩阵，分别为的伴随矩阵，若，则分块矩,A B *,A B *,A B ||2,||3A B ==阵的伴随矩阵为O A B O ⎛⎫⎪⎝⎭(A).(B). **32O B A O ⎛⎫ ⎪⎝⎭**23OB A O ⎛⎫⎪⎝⎭(C).(D).**32O A B O ⎛⎫⎪⎝⎭**23O A BO ⎛⎫⎪⎝⎭【答案】B.【解析】根据，若CC C E *=111,C C C CC C*--*==分块矩阵的行列式，即分块矩阵可逆O A B O ⎛⎫ ⎪⎝⎭221236O A A B B O ⨯=-=⨯=（）1111661O B BO A O A O A O B B O B O B O AO A O A**---*⎛⎫ ⎪⎛⎫⎛⎫⎛⎫ ⎪=== ⎪ ⎪ ⎪⎪⎝⎭⎝⎭⎝⎭⎪⎝⎭1236132O B O B AO A O ****⎛⎫⎪⎛⎫== ⎪ ⎪ ⎪⎝⎭⎪⎝⎭故答案为(B).（6）设均为3阶矩阵，为的转置矩阵，且，,A P T P P 100010002TP AP ⎛⎫ ⎪= ⎪ ⎪⎝⎭若，则为1231223(,,),(,,)P Q ααααααα==+TQ AQ (A).(B).210110002⎛⎫⎪ ⎪ ⎪⎝⎭110120002⎛⎫⎪ ⎪ ⎪⎝⎭(C). (D).200010002⎛⎫ ⎪ ⎪ ⎪⎝⎭100020002⎛⎫ ⎪ ⎪ ⎪⎝⎭【答案】A.【解析】，即：122312312312100(,,)(,,)110(,,)(1)001Q E αααααααααα⎡⎤⎢⎥=+==⎢⎥⎢⎥⎣⎦12121212122112(1)[(1)][(1)](1)[](1)100(1)010(1)002110100100210010010110110001002001002T T TT Q PE Q AQ PE A PE E P AP E E E ===⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥==⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦⎣⎦（7）设事件与事件B 互不相容，则A (A).(B). ()0P AB =()()()P AB P A P B =(C).(D).()1()P A P B =-()1P A B ⋃=【答案】D.【解析】因为互不相容，所以,A B ()0P AB =(A)，因为不一定等于1，所以(A)不正确.()()1()P AB P A B P A B ==- ()P A B (B)当不为0时，(B)不成立，故排除.(),()P A P B (C)只有当互为对立事件的时候才成立，故排除.,A B(D)，故(D)正确.()()1()1P A B P AB P AB ==-= （8）设随机变量与相互独立，且服从标准正态分布，的概率分布为X Y X (0,1)N Y ，记为随机变量的分布函数，则函数1{0}{1}2P Y P Y ====()z F Z Z XY =()z F Z 的间断点个数为（ ）(A)0.(B)1. (C)2.(D)3.【答案】 B.【解析】()()(0)(0)(1)(1)Z F z P XY z P XY z Y P Y P XY z Y P Y =≤=≤==+≤==1[(0)(1)]21[(00)(1)]2P XY z Y P XY z Y P X z Y P X z Y =≤=+≤==⋅≤=+≤=独立,X Y 1()[(0)()]2Z F z P x z P x z ∴=⋅≤+≤（1）若，则0z <1()()2Z F z z =Φ（2）当，则0z ≥1()(1())2Z F z z =+Φ为间断点，故选(B).0z ∴=二、填空题：9~14小题，每小题4分，共24分，请将答案写在答题纸指定位置上.（9）.0x →=【答案】.32e 【解析】.00x x →→=02(1cos )lim13x e x x→-=20212lim 13x e x x →⋅=32e =（10）设，则.()y xz x e =+(1,0)zx ∂=∂【答案】.2ln 21+【解析】由，故()xy z x e=+()(),01xz x x =+()''ln(1)ln(1)1ln(1)1x x x x x dz x x e e x dx x ++⎡⎤⎡⎤⎡⎤=+==++⎣⎦⎢⎥⎣⎦+⎣⎦代入得，.1x =()ln 21,01ln 22ln 212z e x∂⎛⎫=+=+ ⎪∂⎝⎭（11）幂级数的收敛半径为 .21(1)n n nn e x n ∞=--∑【答案】.1e【解析】由题意知，()210nn n e a n--=>()()()()111122122111()11111n n n n n nn n nn e e ea n n e n a n e n e e +++++⎡⎤⎛⎫--⎢⎥ ⎪⎝⎭--⎢⎥⎣⎦=⋅=⋅→→∞⎡⎤+--+⎛⎫--⎢⎥⎪⎝⎭⎢⎥⎣⎦所以，该幂级数的收敛半径为1e（12）设某产品的需求函数为,其对应价格的弹性，则当需求量为()Q Q P =P 0.2p ξ=10000件时，价格增加1元会使产品收益增加 元.【答案】8000.【解析】所求即为()QP Q P Q ''=+因为，所以0.2p Q PQξ'==-0.2Q P Q '=-所以()0.20.8QP Q Q Q '=-+=将代入有.10000Q =()8000QP '=（13）设,，若矩阵相似于，则.(1,1,1)T α=(1,0,)T k β=Tαβ300000000⎛⎫ ⎪ ⎪ ⎪⎝⎭k =【答案】2.【解析】相似于，根据相似矩阵有相同的特征值，得到的特征值为T αβ300000000⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦Tαβ3，0，0.而为矩阵的对角元素之和，，.TαβT αβ1300k ∴+=++2k ∴= (14)设，,…,为来自二项分布总体的简单随机样本，和分别为样1X 2X n X (,)B n p X 2S 本均值和样本方差，记统计量，则 .2T X S =-ET =【答案】2np 【解析】由.222()(1)ET E X S E X ES np np p np =-=-=--=三、解答题：15～23小题，共94分.请将解答写在答题纸指定的位置上.解答应写出文字说明、证明过程或演算步骤.（15）（本题满分9分）求二元函数的极值.()22(,)2ln f x y xy y y =++【解析】，，故.2(,)2(2)0x f x y x y '=+=2(,)2ln 10y f x y x y y '=++=10,x y e= =.2212(2),2,4xxyy xyf y f x f xy y''''''=+ =+ =则，，.12(0,)12(2xxef e ''=+1(0,)0xyef ''=1(0,)yyef e ''=而0xxf ''> 2()0xy xx yy f f f ''''''-<二元函数存在极小值.∴11(0,)f e e=-（16）（本题满分10 分）计算不定积分 .ln(1dx+⎰(0)x >得t =22212,1(1)tdtx dx t t -= =--2221ln(1ln(1)1ln(1)11111dx t d t t dt t t t +=+-+=---+⎰⎰⎰而22111112(11411(1)111ln(1)ln(1)2441dt dtt t t t t t t C t =---+-++--++++⎰⎰所以2ln(1)111ln(1ln 1412(1)1ln(1.2t t dx C t t t x C +++=+-+--+=+++⎰（17）（本题满分10 分）计算二重积分，其中.()Dx y dxdy -⎰⎰22{(,)(1)(1)2,}D x y x y y x =-+-≤≥【解析】由得，22(1)(1)2x y -+-≤2(sin cos )r θθ≤+32(sin cos )4()(cos sin )04Dx y dxdy d r r rdr πθθθθθπ+∴-=-⎰⎰⎰⎰332(sin cos )14(cos sin )034r d πθθθθθπ⎡+⎤=-⋅⎢⎥⎣⎦⎰2384(cos sin )(sin cos )(sin cos )34d πθθθθθθθπ=-⋅+⋅+⎰3384(cos sin )(sin cos )34d πθθθθθπ=-⋅+⎰.3344438814(sin cos )(sin cos )(sin cos )3344d πππθθθθθθπ=++=⨯+⎰83=-（18）（本题满分11 分）（Ⅰ）证明拉格朗日中值定理，若函数在上连续，在上可导，则()f x [],a b (),a b ，得证.(),a b ξ∈()'()()()f b f a f b a ξ-=-（Ⅱ）证明：若函数在处连续，在内可导，且()f x 0x =()0,,(0)σσ>，则存在，且.'0lim ()x f x A +→='(0)f +'(0)f A +=【解析】（Ⅰ）作辅助函数，易验证满足：()()()()()()f b f a x f x f a x a b aϕ-=----()x ϕ；在闭区间上连续，在开区间内可导，且()()a b ϕϕ=()x ϕ[],a b (),a b .''()()()()f b f a x f x b aϕ-=--根据罗尔定理，可得在内至少有一点，使，即(),a b ξ'()0ϕξ='()f ξ'()()0,()()()()f b f a f b f a f b a b aξ--=∴-=--（Ⅱ）任取，则函数满足：在闭区间上连续，开区间内可导，0(0,)x δ∈()f x []00,x ()00,x 从而有拉格朗日中值定理可得：存在，使得()()000,0,x x ξδ∈⊂……()0'00()(0)x f x f f x ξ-=-()*又由于，对上式（*式）两边取时的极限可得：()'lim x f x A +→=00x +→()()000000'''0000()00lim lim ()lim ()0x x x x x f x f f f f A x ξξξ++++→→→-====-故存在，且.'(0)f +'(0)f A +=（19）（本题满分10 分）设曲线，其中是可导函数，且.已知曲线与直线()y f x =()f x ()0f x >()y f x =及所围成的曲边梯形绕轴旋转一周所得的立体体积值是该曲边梯0,1y x ==(1)x t t =>x 形面积值的倍，求该曲线的方程.t π【解析】旋转体的体积为22()()11x x t t V f dx f dxππ==⎰⎰曲边梯形的面积为：，则由题可知()1x ts f dx =⎰22()()()()1111x x x x t t t tV ts f dx t f dx f dx t f dxπππ=⇒=⇒=⎰⎰⎰⎰两边对t 求导可得 22()()()()()()11t x t t t x t t f f dx tf f tf f dx =+⇒-=⎰⎰继续求导可得，化简可得''2()()()()()f t f t f t tf t f t --=，解之得'1(2())()2()12dt f t t f t f t t dy y-=⇒+=1223t c y y-=⋅+在式中令，则，代入得 1t =2(1)(1)0,()0,(1)1f f f t f -=>∴= 1223t cyy -=+.11,2)33c t y =∴=所以该曲线方程为：.230y x +=（20）（本题满分11 分）设,.111A=111042--⎛⎫ ⎪- ⎪ ⎪--⎝⎭1112ξ-⎛⎫⎪= ⎪⎪-⎝⎭（Ⅰ）求满足，的所有向量，.21A ξξ=231Aξξ=2ξ3ξ（Ⅱ）对（Ⅰ）中的任意向量,，证明,,线性无关.2ξ3ξ1ξ2ξ3ξ【解析】（Ⅰ）解方程21A ξξ=()1111111111111,111100000211042202110000A ξ---------⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪=-→→ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪---⎝⎭⎝⎭⎝⎭故有一个自由变量，令，由解得，()2r A =32x =0Ax =211,1x x =-= 求特解，令，得120x x ==31x = 故 ，其中为任意常数21101021k ξ⎛⎫⎛⎫ ⎪ ⎪=-+ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭1k解方程231Aξξ=2220220440A ⎛⎫ ⎪=-- ⎪ ⎪⎝⎭()21111022012,2201000044020000A ξ-⎛⎫ ⎪-⎛⎫ ⎪ ⎪=--→⎪ ⎪ ⎪ ⎪-⎝⎭ ⎪⎝⎭故有两个自由变量，令，由得231,0x x =-=20A x =11x =令，由得230,1x x ==-20A x =10x =求得特解21200η⎛⎫- ⎪ ⎪= ⎪ ⎪ ⎪⎝⎭故 ，其中为任意常数3231102100010k k ξ⎛⎫-⎪⎛⎫⎛⎫⎪ ⎪ ⎪=-++ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭ ⎪⎝⎭23,k k （Ⅱ）证明：由于12121212122111121112(21)()2((21)22221k k k k k k k k k k k k k -+--=+++-+-+-+102=≠故 线性无关. 123,,ξξξ（21）（本题满分11 分）设二次型.2221231231323(,,)(1)22f x x x ax ax a x x x x x =++-+-（Ⅰ）求二次型的矩阵的所有特征值.f （Ⅱ）若二次型的规范形为，求的值.f 2211y y +a【解析】（Ⅰ） 0101111a A a a ⎛⎫ ⎪=- ⎪⎪--⎝⎭0110||01()1111111aa aE A a a a a λλλλλλλλ-----=-=---+---+222()[()(1)1][0()]()[()(1)2]()[22]19(){[(12)]}24()(2)(1)a a a a a a a a a a a a a a a a a λλλλλλλλλλλλλλλλ=---+--+-=---+-=--++--=-+--=--+--.123,2,1a a a λλλ∴==-=+（Ⅱ） 若规范形为，说明有两个特征值为正，一个为0.则2212y y +1)若，则， ，不符题意10a λ==220λ=-<31λ=2)若 ，即，则，，符合20λ=2a =120λ=>330λ=>3)若 ，即，则 ，，不符题意30λ=1a =-110λ=-<230λ=-<综上所述，故2a =（22）（本题满分11 分）设二维随机变量的概率密度为(,)X Y 0(,)0x e y xf x y -⎧<<=⎨⎩其他（Ⅰ）求条件概率密度()Y X f y x （Ⅱ）求条件概率11P X Y =⎡≤≤⎤⎣⎦【解析】（Ⅰ）由得其边缘密度函数0(,)0x y xe f x y -<<⎧= ⎨⎩其它()0xx x x f x e dy xe x --== >⎰故 |(,)1(|)0()y x x f x y f y x y x f x x== <<即|1(|)0y x y xf y x x ⎧ 0<<⎪=⎨⎪⎩其它（Ⅱ）[1,1][1|1][1]P X Y P X Y P Y ≤≤≤≤=≤而11111[1,1](,)12xxx x y P X Y f x y dxdy dx e dy xe dx e ---≤≤≤≤====-⎰⎰⎰⎰⎰()|,0x x yY yf y e dx e e y y+∞---+∞==-= >⎰11101[1]|110y y P Y e dy e e e ----∴ ≤==-=-+=-⎰.11122[1|1]11e e P X Y e e ----∴ ≤≤==--（23）（本题满分11分）袋中有一个红球，两个黑球，三个白球，现在放回的从袋中取两次，每次取一个，求以、X 、分别表示两次取球所取得的红、黑与白球的个数.Y Z ①求.10P X Z ⎡==⎤⎣⎦②求二维随机变量的概率分布.(,)X Y 【解析】（Ⅰ）在没有取白球的情况下取了一次红球，利用压缩样本空间则相当于只有1个红球，2个黑球放回摸两次，其中摸了一个红球.12113324(10)9C P X Z C C ⨯∴====⋅（Ⅱ）X ，Y 取值范围为0，1，2，故()()()()()()()()()1111332311116666111223111166661122116611221166110,0,1,0461112,0,0,136311,1,2,10910,291,20,2,20C C C C P X Y P X Y C C C C C C C P X Y P X Y C C C C C C P X Y P X Y C C C C P X Y C C P X Y P X Y ⋅⋅========⋅⋅⋅⋅========⋅⋅⋅=======⋅⋅====⋅======012 XY01/41/61/36 11/31/9021/900。
2009_上海交大工硕-试卷-解答_数学
要考工硕啊，努力吧！！！上海交通大学工程硕士研究生入学考试 数学（高等数学，线性代数）模拟试卷 2009.4考试时间为180分钟；试卷总分为100分准考证号码_____________________ 报考领域____________ 姓名_________一．单项选择题（共18分，每小题3分）1. 求极限 22301l i m s i n xx x e x x-→--= ( ) A ． 1； B ．0.5； C ．1-； D ．0.5- 2. 若函数()f x 与()h x 在实数轴上均可导，且()()f x g x <，则必有 （ ） A ．()()f x h x ->-； B ．'()'()f x h x <； C ．0lim ()lim ()x x x x f x h x →→<； D ．()d ()d xxf t t h t t <⎰⎰。
3. 设(1,1),(6,3),(2,7)A B C 是xOy 平面上的三点，则三角形ABC 的面积为（ ） A ．6； B ．14； C ．28； D ．32。
4. 设,αβ是非齐次线性方程组()I A x b λ-=的两个不同的解，其中A 为n 阶矩阵，则下列选项中一定是A 对应的特征值λ的特征向量的为 ( )A ． αβ+；B ．αβ-；C ．α；D ．β。
5. n 维向量12,,,(3)s s n ααα≤≤ 线性无关的充要条件是 ( ) A ．存在不全为零的数12,,,s c c c ,使11220s s c c c ααα+++≠ ； B ．12,,,(3)s s n ααα≤≤ 中任意两个向量都线性无关；C ．12,,,(3)s s n ααα≤≤ 中任意一个向量都不能用其余向量线性表示；D ．12,,,(3)s s n ααα≤≤ 中存在一个向量，它不能用其余向量线性表示。
6. 设函数()f x 在0x 满足000'()''()0,'''()0f x f x f x ==>，则（ ） A ．0'()f x 是'()f x 的极大值； B ．0()f x 是()f x 的极大值；C ．0()f x 是()f x 的极小值；D ．00(,())x f x 是曲线()y f x =的拐点。
2009年全国硕士研究生入学考试数学二真题及答案
,
) 的光滑曲线，当
22
x 0 时，曲线上任一点处的发现都过原点，当 0 x 时，函数 y(x)满足
y y x 0 。求 y(x)的表达式。
（21）（本题满分 11 分）（I）证明拉格朗日中值定理：若函数 f (x) 在[a,b]上连续，在（a,b）
可导，则存在 (a,b) ，使得 f (b) f (a) f ( )(b a) 。（II）证明：若函数 f (x) 在 x=0
（A）
0 2 A
3B
0
（B）
0 3 A
2B
0
（C）
0 2B
3A
0
（D）
0 3B
2 A
0
100
（8）设
A，P
均为
3
阶矩阵，
PT
为
P
的转置矩阵，且
PT
AＰ＝
010
，若
002
P (1,2,3),Q (1 2,2,3) ，则 QT AQ 为（）
由考研云助手整理( 专注免费考研资料 微信公众号提供更多资讯)
tan
x)]
（16）（本题满分 10 分）计算不定积分 ln(1 1 x )dx(x 0)
x
（17）（本题满分 10 分）设 z f (x y, x y, xy) ，其中 f 具有 2 阶连续偏导数，求 dz 与
2z xy
（18）（本题满分 10 分）设非负函数 y=y(x)(x 0)，满足微分方程 xy y 2 0 ，当曲线
（23）（本题满分 11 分）设二次型 f (x1, x2, x3) ax12 ax22 (a 1)x32 2x1x3 2x2x3
（I）求二次型 f 的矩阵的所有特征值；（II）若二次型 f 的规范形为 y12 y22 ，求 a 的值。
2009年全国硕士研究生入学考试数学一真题(2009考研数一真题答案解析)
2009年全国硕士研究生入学统一考试部分数学一试题答案解析一、选择题：1～8小题，每小题4分，共32分，下列每小题给出的四个选项中，只有一项符合题目要求，把所选项前的字母填在题后的括号内.（1）当0x →时，()sin f x x ax =-与()()2ln 1g x x bx =-等价无穷小，则()A 11,6a b ==-. ()B 11,6a b ==. ()C 11,6a b =-=-. ()D 11,6a b =-=. 【答案】 A【解析】2()sin ,()ln(1)f x x ax g x x bx =-=-为等价无穷小，则222200000()sin sin 1cos sin lim lim lim lim lim ()ln(1)()36x x x x x f x x ax x ax a ax a ax g x x bx x bx bx bx→→→→→---==-⋅---洛洛230sin lim 166x a ax a b b axa→==-=-⋅ 36a b ∴=- 故排除,B C 。
另外201cos lim3x a axbx→--存在，蕴含了1cos 0a ax -→()0x →故 1.a =排除D 。
所以本题选A 。
（2）如图，正方形(){},1,1x y x y ≤≤四个区域()1,2,3,4k D k =，cos kk D I y xdxdy =⎰⎰，则{}14max k k I ≤≤=()A 1I .()B 2I . ()C 3I .()D 4I .【答案】A【解析】本题利用二重积分区域的对称性及被积函数的奇偶性。
24,D D 两区域关于x 轴对称，而(,)cos (,)f x y y x f x y -=-=-，即被积函数是关于y 的奇函数，所以240I I ==;13,D D 两区域关于y 轴对称，而(,)cos()cos (,)f x y y x y x f x y -=-==，即被积函数是x关于x 的偶函数，所以{}1(,),012cos 0x y y x x I y xdxdy ≥≤≤=>⎰⎰；{}3(,),012cos 0x y y x x I y xdxdy ≤-≤≤=<⎰⎰.所以正确答案为A.（3）设函数()y f x =在区间[]1,3-上的图形为：则函数()()0xF x f t dt =⎰的图形为()A ()B()C ()D【答案】D【解析】此题为定积分的应用知识考核，由()y f x =的图形可见，其图像与x 轴及y 轴、0x x =所围的图形的代数面积为所求函数()F x ，从而可得出几个方面的特征：①[]0,1x ∈时，()0F x ≤，且单调递减。
上海交通大学历年考研真题2002-2012(1)
上海交通大学历年考研试题2002年行政学一、名词解释（每题4分，10题，共40分）1、行政制度2、行政行为3、行政主体4、编制5、行为科学6、行政协调7、行政文化8、行政过程9、预算 10、组织行政二、简述题（每题10分，4题，共40分）1、行政激励的方法。
2、非正式组织及其特点。
3、公正性的标准。
4、我国社会主义民主政治下的行政监督。
三、论述题（1题，共20分）略论新形势下提高我国政府机构行政效率的方法与途径。
2002硕士研究生入学考试试题-管理学一、简述题（每小题5分，共40分）1、如何理解管理的自然属性？2、简述三种控制类型的含义。
3、简述谈判的基本方法。
4、简述矩阵结构组织形式的优缺点。
5、组织设计需要考虑哪几个基本原则？6、管理的法律方法及作用是什么？7、有效沟通的障碍有哪些？8、常用的盈利比是什么？二、判断题（每小题2分，共20分）1、高层管理者的主要工作是决策。
2、领到这之所以对部下有影响力，全靠手中的权力，拥有权力才能有影响力，权力越大，影响力越大。
3、股份有限公司的最高权力机构是股东会。
4、对于管理人员的选聘来说，从组织外部招聘一般优于从组织内部提升。
5、企业的规章制度一旦制定出来，就必须不折不扣地加以执行。
6、知识越来越引起人们的高度重视，原因在于知识是一种特殊的资源。
其特殊性表现在快速贬值、价值难以评估等方面。
7、管理学是有国界的。
8、由高层管理者制定而且时间跨度很长的计划就成为战略计划。
9、依据系统管理理论的思想，企业的竞争优势取决于策略组合的优势而不是单个策略的优势。
10、创新是管理的基本职能，而维持则已过时。
三、选择题（每题2分，共20分）1、在一定规模条件下，管理幅度越大，其管理层次就会：[A]越少越多[C]不变[D]时多时少2、下列非正式组织的作用中，哪一种是对组织管理工作最不利的？[A]不同正式组织的成员集中于同一非正式组织中在非正式组织中传播着小道消息[C]非正式组织间有明显的竞争关系[D]非正式组织中的核心人物具有相同或者大于政治组织领导的影响力和号召力3、“科学管理理论”的创始人是：[A]法约尔[B]泰罗[C]梅奥[D]福特4、当人们认为自己的报酬与劳动之比与他们的报酬与劳动之比相等时，就会有较大的激励作用，这种理论称为：[A]双因素理论效用理论[C]公平理论[D]强化理论5、以下哪种情况不是由于过分集权引起的？[A]降低决策质量降低企业员工的工作热情[C]增加企业各部门之间的摩擦[D]削弱了企业的应变能力6、根据某地统计年鉴，该地区去年的电话占有率为每百人34.5部。
上海交通大学研究生试题.doc
上海交通大学研究生试题(2007至2008学年第二学期)科目________________ 班级代号_________ 学号__________ 姓名_________ 成绩______^o.oi5 =2.17,Zo.o3 =1.88,仏(9) = 1.8331,仏(15) = 1.7531,仏(4) = 2.1318加。
5 ⑷=9.48&加。
5 ⑸=11・°71,加95 ⑷二0.71 l，Zo.95(5) = 1-145 , 九05(9」6) = 2.54, F005 (4,7) = 3.51,尬(16,7) = 2.81, F005 (3,16) = 3.24.填空：(6X5分)1.设X\,X“…，XJidNWQ?),〃未知，则假设检验H°:Q2 vs H l:a<2的水平为Q的拒绝域为____________________________________________________ .2.设X〜F(n,m),贝0丄.的分布为 _________________________________ .X3・设X],X…，X9〃4V(3,9),则EX= _____________________ , DX = ___________________ .4.设某保险丝熔化时间X〜N取一样本，容量为16,均值与方差分别、为15,0.36,则“的95%的单侧置信区间下限为 ______________________________ .5.在一为零假设，为对立假设，显著水平为Q的假设检验中，问：P(第一类错误)二 ___________ o6.设X], X?,…，XQidN(0,9),与乙,匕,YjidN(0,16)独立，贝I](X +••• + X )2z= -------------------- 为__________________ 分布。
(需写岀自由度)(乙+…+绻).(10分)理论上压缩机的冷却用水的温度升高值T〜N0Q冷,升高的平均值不多于5度，现随机抽取5台压缩机，测得温度升高为：6.4, 4.3, 5.7, 4.9, 5.4,问：是否可认为这批数据与理论一致？(0 = 5%)。
2009年10月全国硕士研究生入学统一考试管理类专业学位联考数学考试真题
2009 年10 月全国硕士研究生入学统一考试管理类专业学位联考数学试题一、问题求解：第1～15小题，每小题3分，共45分。
下列每题给出的A、B、C、D、E五个选项中，只有一个选项符合试题要求。
请在答题卡上将所选项的字母涂黑。
．．．1. 已知某车间的男工人数比女工人数多80%，若在该车间一次技术考核中全体工人的平均成绩为75分，而女工平均成绩比男工平均成绩高20%，则女工的平均成绩为（）分.（A）88（B）86（C）84（D）82（E）802. 某人在市场上买猪肉，小贩称得肉重为 4 斤. 但此人不放心，拿出一个自备的 100 克重的砝码，将肉和砝码放在一起让小贩用原称复称，结果重量为 4.25 斤. 由此可知顾客应要求小贩补猪肉（）两.（A）3（B）6（C）4（D）7（E）83. 甲、乙两商店某种商品的进货价格都是200元，甲店以高于进货价格20%的价格出售，乙店以高于进货价格15%的价格出售，结果乙店的售出件数是甲店的 2倍. 扣除营业税后乙店的利润比甲店多5400 元. 若设营业税率是营业额的5%，那么甲、乙两店售出该商品各为（）件.（A）450 ，900（C）550，1100（B）500，1000（D）600 ，1200（E）650，13004. 甲、乙两人在环形跑道上跑步，他们同时从起点出发，当方向相反时每隔48 秒相遇一次，当方向相同时每隔 10 分钟相遇一次. 若甲每分钟比乙快40 米，则甲、乙两人的跑步速度分别是（）米/分.（A）470 ，430 （B）380，340 （C）370，330 （D） 280 ，240 （E）270，2305. 一艘小轮船上午8: 00起航逆流而上（设船速和水流速度一定），中途船上一块木板落入水中，直到8: 50船员才发现这块重要的木板丢失，立即调转船头去追，最终于9: 20追上木板. 由上述数据可以算出木板落水的时间是（）.（A）8:35（B）8:30（C）8:25（D）8:20（E）8:156. 若，是有理数，且满足(1 + 2√3)+ (1 −√3)−2 + 5√3= 0，则，的值分别为（）.第1页（A）1，3 （B）−1 ，2 （C）−1 ，3 （D）1，2 （E）以上结论都不正确7. 设与之和的倒数的2007 次方等于1，的相反数与之和的倒数的2009 次方也等于1，则2007 +2009=（）.第1页第2页（A ） −1 8. 设 = |是（（A ） 10− | + | ）.（B ）2 − 2 | + | 0（B ） 15（C ） 1− − 2 |，其中 < 0（C ） 20（E ）22007< 2 ，则对于满足 ≤ ≤ 2 的 值， 的最小值 0 0（D ）259. 若关于 的二次方程 2 − ( − 1) + − 5 = 有两个实根 、 ， 且满足−1 < < 和 < <1，则 的取值范围是（ ）. 0 0 0 （A ）3 < < 4 （B ）4 < < 5 （C ） 5 < < 6 （D ） > 6或 < 5 （E ） > 5或 < 410. 一个球从 100 米高处自由落下，每次着地后又跳回前一次高度的一半再落下 . 当它第 10 次着地时，共经过的路程是（ ）米（精确到 1 米且不计任何阻力） .（A ）300 （B ）250 （C ） 200 （D ） 150 （E ） 100 11. 曲线 2 − 2 +2 =上的点到直线3 + 4 − 12 = 的最短距离是（ ）.（A ）35B ）4（C ） 1 0（D ）4 3（E ） √212. 曲线| + 1 = | | + | 所围成的图形的面积为（ ）.（A ）| 2 （C ） 1 （D ）2 （E ）413. 如图所示，向放在水槽底部的口杯注水（流量一定） ，注满口杯后继续注水，直到注满水槽，水槽中水平面上升高度ℎ与注水时间 之间的函数关系大致是（ ）.（B ）11 |4 5 （E ）30（D ）0（A）（B）（C）（D）（E）以上图形均不正确14. 若将 10 只相同的球随机放入编号为 1、2、3、4 的四个盒子中，则每个盒子不空的投放方法有（）.第2页（A）72 （B）84 （C）96 （D）108 （E）12015. 若以连续两次掷骰子得到的点数和作为点的坐标，则点( ，落在直线+ = 6与两坐标轴围成的三角形内的概率为（）.（A）1 6（B）736（C） 29（D）4（E）518二、条件充分性判断：解题说明：本大题要求判断所给出的条件能否充分支持题干中陈述的结论。
2009年全国硕士研究生入学统一考试数学二试题及答案
⎝e ⎠
e
−2
2 −
最小值为 y = e e 。【答案】 e e 。
⎜⎛ 2 0 0⎟⎞ （14）设α , β 为 3 维列向量，β T 为 β 的转置向量，若αβ T 相似于 ⎜ 0 0 0⎟ ，则 β Tα ＝
2009 年全国硕士研究生入学统一考试 数学二试题
一、选择题：1~8 小题，每小题 4 分，共 32 分，下列每小题给出的四个选项中，只有一项 符合题目要求，把所选项前的字母填在题后的括号内。
（1）函数 f (x) = x − x3 的可去间断点的个数，则（ ） sin πx
（A）1
（B）2
（C） 3
（5）若 f ′′(x) 不变号，且曲线 y = f (x) 在点 (1,1) 上的曲率圆为 x2 + y 2 ＝2，则 f (x) 在区
间（1，2）内（ ） （A）有极值点，无零点（B）无极值点，有零点 （C）有极值点，有零点（D）无极值点，无零点
【解析与点评】在点（1，1）处的领域内 f (x) 凸性不变（上凸），即 f ′′( x) < 0 ，由曲率圆
z = 1 ( x 2 + y 2 ) ，有最小值 0，立即有结果 D。这是水木艾迪一再强调的凑微分方法。 2
（方法 2）由 dz = xdx + ydy 可得 ∂z = x, ∂z = y
∂x
∂y
A = ∂ 2 z = 1, B = ∂ 2 z = ∂ 2 z = 0, C = ∂ 2 z = 1
e −1
+
n
n2
) +1
=
0
（12）设 y = y(x) 是方程 xy + e y = x + 1 确定的隐函数，则 d 2 y
上海交通大学控制理论专业课考研真题
上海交通大学2007年硕士研究生入学考试试题试题序号：805试题名称： 控制理论基础（答案必须写在答题纸上，写在试题纸上的一律不给分）一、（10分）什么事非线性系统的极限环振荡它与临界稳定的线性系统的等幅振荡主要有哪些区别二、（15分）一个简单的水位调节系统有水箱、浮球、杠杆机构、及进出水阀门等组成，如图（a ）所示。
显然，如果阀门2的开度不断地发生变化，改变出水流量，会引起水箱水位发生波动；而水箱水位的变化会通过浮球和杠杆带动阀门1，从而使进水流量发生变化，假定进水流量与液位（分别设为1Q 和h ）的关系如图（b ）所示。
试问：当阀门2的开度保持不变时，系统是否可能有平衡状态如有的话，请说明1） 应该如何确定平衡状态下的水位和流量2） 实际水位偏离平衡水位时，系统如何工作，最后实际水位是否能回到平衡水位三、（20分）已知电路如图，设1R =9k Ω，2R =1k Ω，2C =10μF 。
试求出其传递函数，并通过其波德图说明该电路有何种相应调节作用，最大相位调节量是多大更多资源：四、（20分）以下两个系统方块图中，大写字母均表示传递函数，小写字母均表示输入或输出信号。
试将图（a ）所示的系统等价变换为图（b ）所示的系统，求出相应的传递函数)(s P 、)(s Q ；另外，请以d 为输入e 为输出，求系统的传递函数)(/)(s D s E 。
五、（30分）设单位反馈系统开环波德图的渐进折线如下，已知这是一个最小相位系统，其闭环增益稳定裕量为20dB ，其他条件如图所示1） 试求系统开环传递函数，并请计算（或估算）其相应的相位稳定裕量 2） 若要求系统的相位稳定裕量为45，应如何调整开环增益（用符号K 表示）3） 为保持系统闭环稳定，K 应在什么范围内取值六、（20分）假定有一复合控制系统如图所示，其输入信号为t t r )(，且被控制对象的参数0K 和0T 为已知。
1） 为使系统的稳态误差严格为零，前馈控制环节的参数f K 应如何设定2） 这个方法在实际实现时可能会有什么问题吗如有的话，还应该怎么办七、（20分）一阶单位反馈系统如图（a ）所示，现欲通过实验确定参数T 。
2009年考研数学二真题答案解析
2009年全国硕士研究生入学统一考试 数学二试题一、选择题：1～8小题，每小题4分，共32分，下列每小题给出的四个选项中，只有一项符合题目要求，把所选项前的字母填在题后的括号内.（1）函数()3sin x x f x nx -=的可去间断点的个数，则（ ） ()A 1.()B 2. ()C 3.()D 无穷多个.【答案】C【解析】()3s i n x x f x x π-=则当x 取任何整数时，()f x 均无意义故()f x 的间断点有无穷多个，但可去间断点为极限存在的点，故应是30x x -=的解1,2,30,1x =±320032113211131lim lim sin cos 132lim lim sin cos 132lim lim sin cos x x x x x x x x x x x x x x x x x x x x x ππππππππππππ→→→→→-→---==--==--==故可去间断点为3个，即0,1±（2）当0x →时，()sin f x x ax=-与()()2ln 1g x x bx =-是等价无穷小，则（ ）()A 11,6a b ==-.()B 11,6a b ==. ()C 11,6a b =-=-. ()D 11,6a b =-=.【答案】 A【解析】2()sin ,()(1)f x x ax g x x ln bx =-=-为等价无穷小，则 222200000()sin sin 1cos sin lim lim lim lim lim ()ln(1)()36x x x x x f x x ax x ax a ax a axg x x bx x bx bx bx→→→→→---==-⋅---洛洛230sin lim 166x a ax a b b ax a →==-=-⋅ 36a b ∴=- 故排除,B C 。
另外201cos lim3x a axbx →--存在，蕴含了1cos 0a ax -→()0x →故 1.a =排除D 。
2009年上海交通大学机械原理与设计考研试题
2009年上海交通大学机械原理与设计考研试题选择填空18个选择1个1分5~6个填空吧每题2分具体题有下来难度大了点~然后是问答每题5分记得的有1 描述凸轮反转法的原理2 螺栓组受力的设计考虑3 求5个构件的所速度瞬心4 蜗杆传动和齿轮传动强度设计的比较5 判断斜齿轮传动的转向和旋向然后画出轴向力计算题1 求自由度两小题第二题很多杆件。
很多虚约束。
好像每题8分吧2 证明偏置曲柄滑块机构的行程大于2倍曲柄长度比较简单画出极位然后任意两边之差小于第三边10分3 轮系先算轮系自由度然后求某一轮的转速是两个周转轮系公用一个转臂的情形15分或20分4 斜齿轮和锥齿轮组合25分由主动件转矩转速求下面一系列齿轮的转矩转速然后求轴向力判断旋向最后要你分析该传动不合理的地方5 轴承反向安装求当量荷载无非是压紧放松的问题每年必考吧。
然后求极限转速20分6 机械速度波动调节已知阻力矩的图形求恒定的驱动力矩然后给了J 求不均匀系数和最大最小角速度好像15分很多题没图没真相不过例题一抓一把了解下就好上海交通大学2010年机械设计原理考研真题（回忆版）一．选择题（20个 20分）1 两构件之间以线接触的平面运动副是（高副）2 平底垂直于导路的直动平底从动件凸轮机构的压力角3 渐开线齿轮齿廓上任意一点的法线与齿轮的哪个圆相切、4 非液体摩擦滑动轴承中，限制比压P的主要目的是5 零件失效形式的强度问题刚度问题的划分6 链条节数采用偶数7 渐开线斜齿轮的当量齿轮计算8 蜗轮涡杆的传动效率计算9 斜齿轮螺旋角取的越大，对传动平稳性和轴向力的影响10 单个万向连轴器的主要缺点是………………….其他的想不起来了，反正大部分或者说全部的都是原题，非常简单，只要是能有以前考试过的真题，并且自己好好做过，一分钟就能选好了，题目基本上没有变动，包括选项什么的都是一样的。
二填空题（30个空 30分）填空题主要还是考基本的知识点，没有偏题，比如考直齿轮斜齿轮锥齿轮蜗轮蜗杆的正确啮合条件，考了齿轮啮合基本定律这个地方就考了五六分，考了飞轮安装位置（高速轴），型号相同的轴承90%的寿命是相同的，这里考了“90%”；其他的具体的题目我想不起来了，反正填空基本上都是知识点，没有涉及计算的题目，只是写字花时间，不用怎么想，专业课要考查的知识点一定要看全，并且反复的复习，熟练了以后，做这个题目也很快并且不会出现什么大的错误。