全国硕士研究生入学统一考试真题试卷

合集下载

2024 年全国硕士研究生招生考试(教育综合 333)真题及答案

2024年全国硕士研究生招生考试(教育综合333)真题试卷一、单项选择题：第1～30小题，每小题2分，共60分。

下列每小题给出的四个选项中，只有一个选项是最符合题目要求的。

1.在第39个教师带到来之际，习近平总书记专门致信全国优秀教师代表座谈会与会老师，充分肯定与会教师为代表的广大教师的贡献，号召全国广大教师以教育家为榜样，弘扬教育家精神，为强国建设、民族复兴做出新的贡献，教育家精神具有丰富的内涵，其中对教师理想信念的要求是（）。

A.心有大我，至诚报国B.启智润心，因材施教C.勤学笃行，求是创新D.言为士则，行为世范【答案】A【解析】A项“心有大我、至诚报国”中含有对国家的情怀，属于理想信念范畴。

B项“启智润心、因材施教”体现教师的教学方法，是一种教学智慧。

C项“勤学笃行、求是创新”体现教师严谨认真的工作态度。

D项“言为士则、行为世范”强调教师应谨言慎行，为学生做好榜样，是教师应具备的道德情操。

2.20世纪60年代以来，许多国家相继实行教育先行战略，支持这种国家发展战略的理论是（）。

A.社会再生产理论B.人力资本理论C.社会平衡器理论D.文化资本理论【答案】B【解析】人力资本理论认为要重视教育投资对人力资本的作用，教育投资能够转化为人力资本，促进社会经济的发展。

教育先行指教育在本国经济能力可承载的情况下，适度地先于其他行业或经济发展的现有状态而超前、提前发展，体现国家对教育的重视。

3.华虚朋的文纳特卡制在共同要素的教学上，采用了个别化的策略，所遵循的人的身心发展的特点是（）。

A.顺序性B.阶段性C.差异性D.不平衡性【答案】C【解析】人身心发展的差异性要求考虑不同受教育者的个别差异性，做到因材施教、有的放矢，符合华虚朋的个别化策略。

4.教育概念有特定的内涵和外延。

下列故事所述的行为行动中，称得上教育的是（）。

A.曹冲称象B.程门立雪C.岳母刺字D.公车上书【答案】C【解析】教育指有目的地培养人的活动。

2020年全国硕士研究生入学统一考试管理类联考综合能力真题

2020年全国硕士研究生入学统一考试管理类联考综合能力真题(总分：200.00，做题时间：180分钟)一、问题求解(总题数：15，分数：45.00)1.某产品去年涨价 10%，今年涨价 20%，则产品这两年涨价（）。

（分数：3.00）A.15%B.16%C.30%D.32% √E.33%比例之增长率假设产品涨价前（即前年）的价格为 1，两年涨了 p ，则由1(1+p) 1(1+10%)(1+20%)可得 p=0.32，即 32%，故选项 D 正确。

2.，则 A⊂B 的充分必要条件是（）。

（分数：3.00）A.|a-b|≤1√B.|a-b|≥1C.|a-b|＜1D.|a-b|＞1E.|a-b|=1绝对值不等式、集合子集关系3.一项考试的总成绩由甲、乙、丙三项成绩组成，总成绩=甲成绩×30%+乙成绩×20%+丙成绩×50%，考试通过的标准是每部分≥ 50 分，且总成绩≥60 分。

已知某人甲成绩 70 分，乙成绩 75 分，且通过了这项考试，则此人丙成绩的分数至少是（）。

（分数：3.00）A.48B.50 √C.55D.60E.62应用题之不等式最值设丙成绩为 x ，由题意70×30%+75×20%+x.50%≥60, x≥50 ，解得x≥48, x≥50，故 x 至少取 50。

4.从 1 至 10 这 10 个整数中任取 3 个数，恰有 1 个质数的概率是（）。

（分数：3.00）A.B.√C.D.E.古典概型、质数穷举5.若等差数列{an }满足a1=8，且a2+a4=a1，则{an}的前n 项和的最大值为（）。

（分数：3.00）A.16B.17C.18D.19E.20 √等差数列前n 项和最值6.已知实数 x 满足（）。

（分数：3.00）A.12B.15C.18 √D.24E.27公式、整体法7.设实数 x y, 满足|x-2|+|y-2|≤2，则x2+y2的取值范围是（）。

2023 年全国硕士研究生入学统一考试考研英语(一)真题

2023Section I Use of EnglishDirections:Read thefollowing text.Choose the best word(s)for each numbered blank and mark A,B,C or D on the ANSWER SHEET.(10 points)It’s not difficult to set targets for staff. It is much harder, 1 ,to understand their negative consequences.Most work-related behaviors have multiple components. 2 one and the others become distorted.Travel on a London bus and you’ll 3 see how this works with drivers.Watch people get on and show their tickets.Are they carefully inspected? Never. Do people get on without paying? Of course! Are there inspectors to 4 that people have paid? Possibly, but very few. And people who run for the bus? They are 5 . How about jumping lights? Buses do so almost as frequently as cyclists.Why? Because the target is 6 . People complained that buses were late and infrequent. 7 , the number of buses and bus lanes were increased, and drivers were 8 or punished according to the time they took. And drivers hit these targets. But they 9 hit cyclists. If the target was changed to 10 , you would have more inspectors and moresensitive pricing.If the criterion changed to safety,you would get more 11 drivers who obeyed traffic laws. But both these criteria would be at the expense of time.There is another 12 : people became immensely inventive in hitting targets. Have you 13 that you can leave on a flight an hour late but still arrive on time? Tailwinds? Of course not! Airlines have simply changed the time a 14 is meant to take. A one-hour flight is now ballad as a two-hour flight.The 15 of the story is simple. Most jobs are multidimensional, with multiple criteria. Choose one criterion and you may well 16 others.Everything Can be done faster and made cheaper,but there is a 17 Setting targets can and does have unforeseen negative consequences.This is not an argument against target-setting.But it is an argument for exploring consequences first.All good targets should have multiple criteria 18 critical factors such as time, money, quality and customer feedback. The trick is not only to 19 just one or even two dimensions of the objective, but also to understand how to help people better 20 the objective.1. [A] Some [B] Most [C] Few [D] All2 ．[A] put [B] take [C] run [D] come3 ．[A] Since [B] If [C] Though [D] Until4 ．[A] formally [B] relatively [C] gradually [D] literally5 ．[A] back [B] next [C] around [D] away6 ．[A] onto [B] off [C] across [D] alone7 ．[A] unattractive [B] uncrowded [C] unchanged [D] unfamiliar8 ．[A] site9 ．[A] So 10．[A] immediately 11．[A] surprised 12．[A] problem 13．[A] Above all 14．[A] bridge 15．[A] form 16．[A] posts17 ．[A] artificial 18．[A] Finally 19．[A] memories 20．[A] restrict [B] point[B] Yet[B] intentionally[B] annoyed[B]option[B] In contrast[B] avoid[B] through[B] links[B] mysterious[B] Consequently[B] marks[B] adopt[C] way[C] Instead[C]unexpectedly[C] frightened[C] view[C] On average[C] spot[C] beyond[C] shades[C] hidden[C]incidentally[C] notes[C] lead[D] place[D] Besides[D] eventually[D] confused[D] result[D] For example[D] separate[D] under[D] breaks[D] limited[D] Generally[D] belongings[D] exposeSection II Reading ComprehensionPart ADirections:Read thefollowingfour texts.Answer the questions below each text by choosing A,B,C or D.Mark your answers on the ANSWER SHEET.(40points)Text2Communities throughout New England have been attemptingto regulate short-term rentals since sites like Airbnb took off in the2010s. Now with record-high home prices and historically low inventory,there’s an increased urgency in such regulation,particularly among those whoworry that developers will come in and buy up swaths of housing to flip for a fortune on the short-term rental market.In New Hampshire,where the rental vacancy rate has dropped below 1percent,housing advocates fear unchecked short-term rentals will put further pressure on an already strained market. The State Legislature recently voted against a bill that would’ve made it illegal for towns to create legislation restricting short-term rentals.“We are at a crisis level on the supply of rental housing,”said Nick Taylor, executive director of the Workforce Housing Coalition of the Greater Seacoast. Without enough affordable housing in southern New Hampshire towns,“ employers are having a hard time attracting employees, and workers are having a hard time finding a place to live,” Taylor said.However, short-term rentals also provide housing for tourists, pointed out Ryan Castle,CEO of a local association of realter.“A lot of workers are servicing the tourist industry,and the tourism industry is serviced by those people coming in short term,”Castle said,“and so it’s a cyclical effect.”Short-term rentals themselves are not the crux of the issue, said Keren Horn,an expert on affordable housing policy.“I think individuals being able to rent out their second home is a good thing.If it’s their vacation home anyway,and it’s just empty,why can’t you make money off it?”Horn said. Issues arise, however, when developers attempt to create large-scale short-term rental facilities— de facto hotels — to bypass taxesand regulations. “I think the question is, shouldn’t a developer who’s really building a hotel,but disguising it as not a hotel,be treated and taxed and regulated like a hotel?” Horn said.At the end of 2018, governor Charlie Baker of Massachusetts signed a bill to rein in those potential investor-buyers. The bill requires every rental host to register with the state mandates they carry insurance, and opens the potential for local taxes on top of a new state levy. Boston took things even further,requiring renters to register with the city’s Inspectional Services Department.Horn said similar registration requirements could benefit struggling cities and towns, but “if we want to make a change in the housing market, the main one is we have to build a lot more.”26. Which of the following is true of New England?A. Its housing supply is at a very low level.B. Its communities are in need of funding.C. Its rental vacancy rate is going up slowlyD. Its home prices are under strict control.27. The bill mentioned in Paragraph 2 was intended toA. curb short-term rental speculation.B. ensure the supply of cheap housing.C. punish illegal dealings in housing.D. allow a free short-term rental market.28. Compared with Castle, Taylor is more likely to supportA. further investment in local tourism.B. an increase in affordable housing.C. strict management of real estate agents.D. a favorable policy for short-term workers.29. What does Horn emphasize in Paragraph 5?A. The urgency to upgrade short-term rental facilities.B. The efficient operation of the local housing market.C. The necessity to stop developers from evading taxes.D. The proper procedures for renting out spare houses.30. Horn holds that imposing registration requirements isA. an irrational decision.B. an unfeasible proposal.C. an unnecessary measure.D. an inadequate solution.Text3If you’re heading for your nearest branch of Waterstones,the biggest book retailer in the UK, in search of the Duchess of Sussex’s new children’s book The Bench, you might have to be prepared to hunt around a bit, the same may be true of The President’s Daughter, the new thriller by Bill Clinton and James Patterson.Both of these books are published next week by Penguin Random House(PRH), a company currently involved in a stand-offwith Waterstones.The problem began late last year,when PRH confirmed that it had introduced a credit limit with Waterstones “at a very significant level”. The trade magazine The Bookseller reported that Waterstones branch managers were being told to remove PRH books from prominent areas such as tables, display spaces and windows, and were“ quietly retiring them to their relevant sections” .PRH declined to comment on the issue, but a spokesperson for Waterstones told me:“Waterstones are currently operating with reduced credit terms from PRH, the only publisher in the UK to place any limitations on our ability to trade. We are not boycotting PRH titles but we are doing our utmost to ensure that availability for customers remains good despite the lower overall levels of stock.We are hopeful with our shops now open again that normality will return and that we will be allowed to buy appropriately.Certainly,our shops are exceptionally busy.The sales for our May Books of the Month surpassed any month since 2018.”In the meantime,PRH authors have been the losers.Big-name PRH authors may suffer a bit, but it’s those mid-list authors, who normally rely on Waterstones staff’s passion for promoting books by lesser-known writers, who will be praying for an end to the dispute.It comes at a time when authors are already worried about the consequences of the proposed merger between PRH and another big publisher,Simon &Schuster—the reduction in the number of unaligned UK publishers is likely to lead to fewer bidding wars, lower advances, and more conformity in terms of what is published.“This is all part of a wider change towards concentration of power,”says literary agent Andrew Lownie.“The publishing industry talks about diversity in terms of authors and staff but it also needs a plurality of ways of delivering intellectual contact, choice and different voices. After all, many of the most interesting books in recent years have come from small publishers.”We shall see whether that plurality is a casualty of the current need among publishers to be big enough to take on all-comers.31. The author mentions two books in Paragraph 1 to presentA. an ongoing conflict.B. an intellectual concept.C. a prevailing sentiment.D. a literary phenomenon.32. Why did Waterstones shops retire PRH books to their relevant sections?A. To make them easily noticeable.B. To comply with PRH’s requirement.C. To respond to PRH’s business move.D. To arrange them in a systematic way.33. What message does the spokesperson for Waterstones seem to convey?A. Their customers remain loyal.B. The credit limit will be removed.C. Their stock is underestimated.D. The book market is rather slack.34.What can be one consequence of the current dispute?A. Sales of books by mid-list PRH writers fall off considerably.B. Lesser-known PRH writers become the target of criticism.C. Waterstones staff hesitate to promote big-name authors’ books.D. Waterstones branches suffer a severe reduction in revenue.35. Which of the following statements best represents Lownie’s view?A. Small publishers ought to stick together.B. Big publishers will lose their dominance.C. The publishing industry is having a hard time.D. The merger of publishers is a worrying trend.Text4Scientific papers are the recordkeepers of progress in research.Each year researchers publish millions of papers in more than 30,000journals. The scientific community measures the quality of those papers in a number of ways,including the perceived quality of the journal (as reflected by the title’s impact factor) andthe number of citations a specific paper accumulates. The careers of scientists and the reputation of their institutions depend on the number and prestige of the papers they produce,but even more so on the citations attracted by these papers.Citation cartels,where journals,authors,and institutions conspire to inflate citation numbers, have existed for a long time. In 2016, researchers developed an algorithm to recognize suspicious citation patterns,including groups of authors that disproportionately cite one another and groups of journals that cite each other frequently to increase the impact factors of their publications.Recently,another expression of this predatory behavior has emerged:so-called support service consultancies that provide languageand other editorial support to individual authors and to journals sometimes advise contributors to add a number of citations to their articles.The advent of electronic publishing and authors’need to find outlets for their papers resulted in thousands of new journals. The birth of predatory journals wasn’t far behind.These journals can act as milk cows where every single article in an issue may cite a specific paper or a series of papers.In some instances,there is absolutely no relationship between the content of the article and the citations.The peculiar part is that the journal that the editor is supposedly working for is not profiting at all — it is just providing citations to other journals.Such practices can lead an article to accrue more than 150 citations in the same year that it was published.How insidious is this type of citation manipulation?In one example, an individual - acting as author, editor, and consultant - was able to use at least 15 journals as citation providers to articles published by five scientists at three universities. The problem is rampant in Scopus, a citation database, which includes a high number of the new “international” journals. In fact, a listing in Scopus seems to be a criterion to be targeted in this type of citation manipulation.Scopus itself has all the data necessary to detect this malpractice. Red flags include a large number of citations to an article within the first year. And for authors who wish to steer clear of citation cartel activities:whenan editor, a reviewer,or a support service asks you to add inappropriate references, do not oblige and do report the request to the journal.36.According to Paragraph 1,the careers of scientists can be determined by________.[A] how many citations their works contain.[B] how many times their papers are cited.[C] the prestige of the people they work with.[D] the status they have in scientific circles.37. The support service consultancies tend to________.[A] recommend journals to their clients.[B] list citation patterns their clients.[C] ask authors to include extra citations.[D] advise contributors to cite each other.38. The Function of the “milk cow” journals is to________.[A] boost citation counts for certain authors.[B] help scholars publish articles at low cost.[C] instruct First-time contributors in citation.[D] increase the readership of new journals.39. What can be learned about Scopus from the last two paragraphs?[A] It Fosters competition among citation providers.[B] It has the capability to identify suspicious citations.[C] It hinders the growth of “international” journals.[D] It is established to prevent citation manipulation.40. What should an author do to deal with citation manipulators?[A] Take legal action.[B] Demand an apology.[C] Seek professional advice[D] Reveal their misconduct.Part CDirections:Read the following text carefully and then translate the underlined segments into Chinese.Write your answers on the ANSWER SHEET.(10points)46.AI can also be used to identify the lifestyles choices of customers regarding their hobbies, favorite celebrities, and fashions to provide unique content in marketing messages put out through social media.47.Some believe that AI is negatively impacting on the marketer's roleby reducing creativity and removing jobs,but they are aware that it is a way of reducing costs and creating new information.48.Algorithms used to simulate human interactions are creating many of those concerns,especially as no-one is quite sure what the outcomes of using AI to interact with customers will be.49.If customers are not willing to share data,AI will be starved of essential information and will not be able to function effectively or employ machine learning to improve its marketing content and communication.50.The non-intrusive delivery of the marketing message in a way that is sensitive to the needs of target customers is one of the critical challenges to the digital marketer.Section III WritingPart A51.Write a notice to recruit a student for Prof. Smith’s research project on campus sports activities. Specify the duties and requirements of the job.Write your answer about 100 words on the ANSWER SHEET.Do not use your own name in the notice; use “Li Ming”instead.(10 points)Part B52.Directions:Write an essay of 160-200 words based on the picture below. In your essay you should1) describe the picture briefly,2) interpret the implied meaning, and3) give your comments.You should write neatly on the ANSWER SHEET. (20 points)。

1978-2019年全国硕士研究生入学统一考试(数学一)真题及部分答案

历年考研数学一真题1987-20191987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)当x =_____________时,函数2x y x =⋅取得极小值. (2)由曲线ln y x=与两直线e 1y x=+-及y =所围成的平面图形的面积是_____________.1x =(3)与两直线 1y t =-+2z t =+及121111x y z +++==都平行且过原点的平面方程为_____________.(4)设L 为取正向的圆周229,x y +=则曲线积分2(22)(4)L xy y dx x x dy -+-⎰= _____________. (5)已知三维向量空间的基底为123(1,1,0),(1,0,1),(0,1,1),===ααα则向量(2,0,0)=β在此基底下的坐标是_____________.二、(本题满分8分)求正的常数a 与,b 使等式201lim 1sin x x bx x →=-⎰成立.三、(本题满分7分) (1)设f 、g 为连续可微函数,(,),(),u f x xy v g x xy ==+求,.u v x x∂∂∂∂ (2)设矩阵A 和B 满足关系式2,+AB =A B 其中301110,014⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦A 求矩阵.B四、(本题满分8分)求微分方程26(9)1y y a y ''''''+++=的通解,其中常数0.a >五、选择题(本题共4小题,每小题3分,满分12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)设2()()lim1,()x af x f a x a →-=--则在x a =处 (A)()f x 的导数存在,且()0f a '≠ (B)()f x 取得极大值 (C)()f x 取得极小值 (D)()f x 的导数不存在 (2)设()f x 为已知连续函数0,(),s t I t f tx dx =⎰其中0,0,t s >>则I 的值(A)依赖于s 和t (B)依赖于s 、t 和x (C)依赖于t 、x ,不依赖于s (D)依赖于s ,不依赖于t (3)设常数0,k >则级数21(1)n n k n n∞=+-∑(A)发散 (B)绝对收敛 (C)条件收敛 (D)散敛性与k 的取值有关 (4)设A 为n 阶方阵，且A 的行列式||0,a =≠A 而*A 是A 的伴随矩阵，则*||A 等于(A)a (B)1a(C)1n a - (D)na六、（本题满分10分）求幂级数1112n nn x n ∞-=∑的收敛域，并求其和函数.七、（本题满分10分）求曲面积分2(81)2(1)4,I x y dydz y dzdx yzdxdy ∑=++--⎰⎰其中∑是由曲线13()0z y f x x ⎧=≤≤⎪=⎨=⎪⎩绕y 轴旋转一周而成的曲面,其法向量与y 轴正向的夹角恒大于.2π八、（本题满分10分）设函数()f x 在闭区间[0,1]上可微,对于[0,1]上的每一个,x 函数()f x 的值都在开区间(0,1)内,且()f x '≠1,证明在(0,1)内有且仅有一个,x 使得().f x x =九、（本题满分8分）问,a b 为何值时,现线性方程组123423423412340221(3)2321x x x x x x x x a x x b x x x ax +++=++=-+--=+++=-有唯一解,无解,有无穷多解?并求出有无穷多解时的通解.十、填空题(本题共3小题,每小题2分,满分6分.把答案填在题中横线上)(1)设在一次实验中,事件A 发生的概率为,p 现进行n 次独立试验,则A 至少发生一次的概率为____________;而事件A 至多发生一次的概率为____________.(2)有两个箱子,第1个箱子有3个白球,2个红球, 第2个箱子有4个白球,4个红球.现从第1个箱子中随机地取1个球放到第2个箱子里,再从第2个箱子中取出1个球,此球是白球的概率为____________.已知上述从第2个箱子中取出的球是白球,则从第一个箱子中取出的球是白球的概率为____________. (3)已知连续随机变量X 的概率密度函数为221(),xx f x-+-=则X 的数学期望为____________,X 的方差为____________.十一、（本题满分6分）设随机变量,X Y 相互独立,其概率密度函数分别为()X f x = 1001x ≤≤其它,()Y f y = e 0y - 00y y >≤, 求2Z X Y =+的概率密度函数.1988年全国硕士研究生入学统一考试数学(一)试卷一、(本题共3小题,每小题5分,满分15分)(1)求幂级数1(3)3nnn x n ∞=-∑的收敛域. (2)设2()e ,[()]1x f x f x x ϕ==-且()0x ϕ≥,求()x ϕ及其定义域.(3)设∑为曲面2221x y z ++=的外侧,计算曲面积分333.I x dydz y dzdx z dxdy ∑=++⎰⎰二、填空题(本题共4小题,每小题3分,满分12分.把答案填在题中横线上)(1)若21()lim (1),tx x f t t x→∞=+则()f t '= _____________.(2)设()f x 连续且31(),x f t dt x -=⎰则(7)f =_____________. (3)设周期为2的周期函数,它在区间(1,1]-上定义为()f x =22x1001x x -<≤<≤,则的傅里叶()Fourier 级数在1x =处收敛于_____________.(4)设4阶矩阵234234[,,,],[,,,],==A αγγγB βγγγ其中234,,,,αβγγγ均为4维列向量,且已知行列式4,1,==A B 则行列式+A B = _____________.三、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)设()f x 可导且01(),2f x '=则0x ∆→时,()f x 在0x 处的微分dy 是(A)与x ∆等价的无穷小 (B)与x∆同阶的无穷小(C)比x ∆低阶的无穷小 (D)比x ∆高阶的无穷小 (2)设()y f x =是方程240y y y '''-+=的一个解且00()0,()0,f x f x '>=则函数()f x 在点0x 处(A)取得极大值 (B)取得极小值(C)某邻域内单调增加 (D)某邻域内单调减少 (3)设空间区域2222222212:,0,:,0,0,0,x y z R z x y z R x y z Ω++≤≥Ω++≤≥≥≥则(A)124xdv dv ΩΩ=⎰⎰⎰⎰⎰⎰(B)124ydv ydv ΩΩ=⎰⎰⎰⎰⎰⎰(C)124zdv zdv ΩΩ=⎰⎰⎰⎰⎰⎰(D)124xyzdv xyzdv ΩΩ=⎰⎰⎰⎰⎰⎰(4)设幂级数1(1)n n n a x ∞=-∑在1x =-处收敛,则此级数在2x =处(A)条件收敛 (B)绝对收敛(C)发散 (D)收敛性不能确定(5)n 维向量组12,,,(3)s s n ≤≤ααα线性无关的充要条件是(A)存在一组不全为零的数12,,,,s k k k 使11220s s k k k +++≠ααα(B)12,,,s ααα中任意两个向量均线性无关(C)12,,,s ααα中存在一个向量不能用其余向量线性表示(D)12,,,s ααα中存在一个向量都不能用其余向量线性表示四、(本题满分6分)设()(),x y u yf xg yx=+其中函数f 、g 具有二阶连续导数,求222.u u x y x x y∂∂+∂∂∂五、(本题满分8分)设函数()y y x =满足微分方程322e ,x y y y '''-+=其图形在点(0,1)处的切线与曲线21y x x =--在该点处的切线重合,求函数().y y x =六、（本题满分9分）设位于点(0,1)的质点A 对质点M 的引力大小为2(0kk r>为常数,r 为A 质点与M 之间的距离),质点M 沿直线y =(2,0)B 运动到(0,0),O 求在此运动过程中质点A 对质点M 的引力所作的功.七、（本题满分6分）已知,=AP BP 其中100100000,210,001211⎡⎤⎡⎤⎢⎥⎢⎥==-⎢⎥⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦B P 求5,.A A八、（本题满分8分）已知矩阵20000101x ⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦A 与20000001y ⎡⎤⎢⎥=⎢⎥⎢⎥-⎣⎦B 相似. (1)求x 与.y (2)求一个满足1-=PAP B的可逆阵.P九、（本题满分9分）设函数()f x 在区间[,]a b 上连续,且在(,)a b 内有()0,f x '>证明:在(,)a b 内存在唯一的,ξ使曲线()y f x =与两直线(),y f x a ξ==所围平面图形面积1S 是曲线()y f x =与两直线(),y f x b ξ==所围平面图形面积2S 的3倍.十、填空题(本题共3小题,每小题2分,满分6分.把答案填在题中横线上)(1)设在三次独立试验中,事件A 出现的概率相等,若已知A 至少出现一次的概率等于19,27则事件A 在一次试验中出现的概率是____________.(2)若在区间(0,1)内任取两个数,则事件”两数之和小于65”的概率为____________.(3)设随机变量X 服从均值为10,均方差为0.02的正态分布,已知22(),(2.5)0.9938,u xx du φφ-==⎰则X 落在区间(9.95,10.05)内的概率为____________.十一、（本题满分6分）设随机变量X 的概率密度函数为21(),(1)X f x x π=-求随机变量1Y =的概率密度函数().Y f y1989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)已知(3)2,f '=则0(3)(3)lim2h f h f h→--= _____________. (2)设()f x 是连续函数,且10()2(),f x x f t dt =+⎰则()f x =_____________.(3)设平面曲线L为下半圆周y =则曲线积分22()Lxy ds +⎰=_____________.(4)向量场div u在点(1,1,0)P 处的散度div u =_____________.(5)设矩阵300100140,010,003001⎡⎤⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦A I 则矩阵1(2)--A I =_____________.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)当0x >时,曲线1sin y x x=(A)有且仅有水平渐近线 (B)有且仅有铅直渐近线(C)既有水平渐近线,又有铅直渐近线 (D)既无水平渐近线,又无铅直渐近线(2)已知曲面224z x y =--上点P 处的切平面平行于平面2210,x y z ++-=则点的坐标是(A)(1,1,2)- (B)(1,1,2)-(C)(1,1,2) (D)(1,1,2)-- (3)设线性无关的函数都是二阶非齐次线性方程的解是任意常数,则该非齐次方程的通解是(A)11223c y c y y ++ (B)1122123()c y c y c c y +-+(C)1122123(1)c y c y c c y +--- (D)1122123(1)c y c y c c y ++--(4)设函数2(),01,f x x x =≤<而1()sin ,,n n S x b n x x π∞==-∞<<+∞∑其中12()sin ,1,2,3,,n b f x n xdx n π==⎰则1()2S -等于(A)12- (B)14-(C)14(D)12(5)设A 是n 阶矩阵,且A 的行列式0,=A 则A 中 (A)必有一列元素全为0 (B)必有两列元素对应成比例(C)必有一列向量是其余列向量的线性组合 (D)任一列向量是其余列向量的线性组合三、(本题共3小题,每小题5分,满分15分) (1)设(2)(,),z f x y g x xy =-+其中函数()f t 二阶可导,(,)g u v 具有连续二阶偏导数,求2.zx y∂∂∂ (2)设曲线积分2()c xy dx y x dy ϕ+⎰与路径无关,其中()x ϕ具有连续的导数,且(0)0,ϕ=计算(1,1)2(0,0)()xy dx y x dy ϕ+⎰的值.(3)计算三重积分(),x z dv Ω+⎰⎰⎰其中Ω是由曲面z =与z =所围成的区域.四、(本题满分6分)将函数1()arctan 1x f x x+=-展为x 的幂级数.五、(本题满分7分)设0()sin ()(),xf x x x t f t dt =--⎰其中f 为连续函数,求().f x六、（本题满分7分）证明方程0ln exx π=-⎰在区间(0,)+∞内有且仅有两个不同实根.七、（本题满分6分）问λ为何值时,线性方程组13x x λ+= 123422x x x λ++=+ 1236423x x x λ++=+有解,并求出解的一般形式. 八、（本题满分8分）假设λ为n 阶可逆矩阵A 的一个特征值,证明 (1)1λ为1-A 的特征值.(2)λA为A 的伴随矩阵*A 的特征值.九、（本题满分9分）设半径为R 的球面∑的球心在定球面2222(0)x y z a a ++=>上,问当R 为何值时,球面∑在定球面内部的那部分的面积最大?十、填空题(本题共3小题,每小题2分,满分6分.把答案填在题中横线上)(1)已知随机事件A 的概率()0.5,P A =随机事件B 的概率()0.6P B =及条件概率(|)0.8,P B A =则和事件AB的概率()P AB =____________.(2)甲、乙两人独立地对同一目标射击一次,其命中率分别为0.6和0.5,现已知目标被命中,则它是甲射中的概率为____________.(3)若随机变量ξ在(1,6)上服从均匀分布,则方程210x x ξ++=有实根的概率是____________.十一、（本题满分6分）设随机变量X 与Y 独立,且X 服从均值为1、标准差(均方差)的正态分布,而Y 服从标准正态分布.试求随机变量23Z X Y =-+的概率密度函数.1990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)2x t =-+(1)过点(1,21)M -且与直线 34y t =-垂直的平面方程是_____________.1z t =-(2)设a 为非零常数,则lim()x x x a x a→∞+-=_____________.(3)设函数()f x =111x x ≤>,则[()]f f x =_____________.(4)积分2220e y x dx dy -⎰⎰的值等于_____________. (5)已知向量组1234(1,2,3,4),(2,3,4,5),(3,4,5,6),(4,5,6,7),====αααα则该向量组的秩是_____________.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)设()f x 是连续函数,且e ()(),xx F x f t dt -=⎰则()F x '等于(A)e (e )()x x f f x ---- (B)e (e )()x x f f x ---+(C)e (e )()x x f f x ---(D)e (e )()x x f f x --+ (2)已知函数()f x 具有任意阶导数,且2()[()],f x f x '=则当n 为大于2的正整数时,()f x 的n 阶导数()()n f x 是(A)1![()]n n f x + (B)1[()]n n f x +(C)2[()]n f x(D)2![()]n n f x(3)设a 为常数,则级数21sin()[n na n∞=∑ (A)绝对收敛 (B)条件收敛(C)发散 (D)收敛性与a 的取值有关 (4)已知()f x 在0x =的某个邻域内连续,且0()(0)0,lim2,1cos x f x f x→==-则在点0x =处()f x (A)不可导 (B)可导,且(0)0f '≠(C)取得极大值 (D)取得极小值(5)已知1β、2β是非齐次线性方程组=AX b 的两个不同的解1,α、2α是对应其次线性方程组=AX 0的基础解析1,k 、2k 为任意常数,则方程组=AX b 的通解(一般解)必是(A)1211212()2k k -+++ββααα(B)1211212()2k k ++-+ββααα (C)1211212()2k k -+++ββαββ(D)1211212()2k k ++-+ββαββ三、(本题共3小题,每小题5分,满分15分)(1)求120ln(1).(2)x dx x +-⎰(2)设(2,sin ),z f x y y x =-其中(,)f u v 具有连续的二阶偏导数,求2.zx y∂∂∂(3)求微分方程244e x y y y -'''++=的通解(一般解).四、(本题满分6分)求幂级数0(21)n n n x ∞=+∑的收敛域,并求其和函数.五、(本题满分8分) 求曲面积分2SI yzdzdx dxdy =+⎰⎰其中S 是球面2224x y z ++=外侧在0z ≥的部分.六、（本题满分7分）设不恒为常数的函数()f x 在闭区间[,]a b 上连续,在开区间(,)a b 内可导,且()().f a f b =证明在(,)a b 内至少存在一点,ξ使得()0.f ξ'>七、（本题满分6分）设四阶矩阵1100213401100213,0011002100010002-⎡⎤⎡⎤⎢⎥⎢⎥-⎢⎥⎢⎥==⎢⎥⎢⎥-⎢⎥⎢⎥⎣⎦⎣⎦B C 且矩阵A 满足关系式1()-''-=A E C B C E其中E 为四阶单位矩阵1,-C 表示C 的逆矩阵,'C 表示C 的转置矩阵.将上述关系式化简并求矩阵.A八、（本题满分8分）求一个正交变换化二次型22212312132344448f x x x x x x x x x =++-+-成标准型.九、（本题满分8分）质点P 沿着以AB 为直径的半圆周,从点(1,2)A 运动到点(3,4)B 的过程中受变力F 作用(见图).F 的大小等于点P 与原点O 之间的距离,其方向垂直于线段OP 且与y 轴正向的夹角小于.2π求变力F 对质点P 所作的功.十、填空题(本题共3小题,每小题2分,满分6分.把答案填在题中横线上)(1)已知随机变量X 的概率密度函数1()e ,2xf x x -=-∞<<+∞ 则X 的概率分布函数()F x =____________.(2)设随机事件A 、B 及其和事件的概率分别是0.4、0.3和0.6,若B 表示B 的对立事件,那么积事件AB 的概率()P AB =____________.(3)已知离散型随机变量X 服从参数为2的泊松()Poisson 分布,即22e {},0,1,2,,!k P X k k k -===则随机变量32Z X =-的数学期望()E Z =____________.十一、（本题满分6分）设二维随机变量(,)X Y 在区域:01,D x y x <<<内服从均匀分布,求关于X 的边缘概率密度函数及随机变量21Z X =+的方差().D Z1991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 21cos x t y t=+=,则22d ydx =_____________.(2)由方程xyz +=所确定的函数(,)z z x y =在点(1,0,1)-处的全微分dz =_____________.(3)已知两条直线的方程是1212321:;:.101211x y z x y zl l ---+-====-则过1l 且平行于2l 的平面方程是_____________.(4)已知当0x →时123,(1)1ax +-与cos 1x -是等价无穷小,则常数a =_____________.(5)设4阶方阵52002100,00120011⎡⎤⎢⎥⎢⎥=⎢⎥-⎢⎥⎣⎦A 则A的逆阵1-A =_____________.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)曲线221e 1ex x y --+=-(A)没有渐近线 (B)仅有水平渐近线(C)仅有铅直渐近线 (D)既有水平渐近线又有铅直渐近线 (2)若连续函数()f x 满足关系式20()()ln 2,2tf x f dt π=+⎰则()f x 等于(A)e ln 2x(B)2e ln 2x(C)eln 2x+(D)2eln 2x+(3)已知级数12111(1)2,5,n n n n n a a ∞∞--==-==∑∑则级数1n n a ∞=∑等于(A)3 (B)7 (C)8 (D)9 (4)设D 是平面xoy 上以(1,1)、(1,1)-和(1,1)--为顶点的三角形区域1,D 是D 在第一象限的部分,则(cos sin )Dxy x y dxdy +⎰⎰等于(A)12cos sin D x ydxdy ⎰⎰ (B)12D xydxdy ⎰⎰(C)14(cos sin )D xy x y dxdy +⎰⎰ (D)0(5)设n 阶方阵A 、B 、C 满足关系式,=ABC E 其中E 是n 阶单位阵,则必有(A)=ACB E (B)=CBA E(C)=BAC E (D)=BCA E三、(本题共3小题,每小题5分,满分15分) (1)求20lim ).x π+→(2)设n 是曲面222236x y z ++=在点(1,1,1)P 处的指向外侧的法向量,求函数u =在点P 处沿方向n 的方向导数.(3)22(),x y z dv Ω++⎰⎰⎰其中Ω是由曲线 220yz x ==绕z 轴旋转一周而成的曲面与平面4z =所围城的立体.四、(本题满分6分)过点(0,0)O 和(,0)A π的曲线族sin (0)y a x a =>中,求一条曲线,L 使沿该曲线O 从到A 的积分3(1)(2)Ly dx x y dy +++⎰的值最小.五、(本题满分8分)将函数()2(11)f x x x =+-≤≤展开成以2为周期的傅里叶级数,并由此求级数211n n∞=∑的和.六、（本题满分7分）设函数()f x 在[0,1]上连续,(0,1)内可导,且1233()(0),f x dx f =⎰证明在(0,1)内存在一点,c 使()0.f c '=七、（本题满分8分）已知1234(1,0,2,3),(1,1,3,5),(1,1,2,1),(1,2,4,8)a a ===-+=+αααα及(1,1,3,5).b =+β(1)a 、b 为何值时,β不能表示成1234,,,αααα的线性组合?(2)a 、b 为何值时,β有1234,,,αααα的唯一的线性表示式?写出该表示式.八、（本题满分6分）设A 是n 阶正定阵,E 是n 阶单位阵,证明+A E 的行列式大于1.九、（本题满分8分）在上半平面求一条向上凹的曲线,其上任一点(,)P x y 处的曲率等于此曲线在该点的法线段PQ 长度的倒数(Q是法线与x 轴的交点),且曲线在点(1,1)处的切线与x 轴平行.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)若随机变量X 服从均值为2、方差为2σ的正态分布,且{24}0.3,P X <<=则{0}P X <=____________.(2)随机地向半圆0y a <<为正常数)内掷一点,点落在半圆内任何区域的概率与区域的面积成正比,则原点和该点的连线与x 轴的夹角小于4π的概率为____________.十一、（本题满分6分）设二维随机变量(,)X Y 的密度函数为(,)f x y =(2)2e 0,00 x y x y -+>>其它求随机变量2Z X Y =+的分布函数.1992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)设函数()y y x =由方程e cos()0x yxy ++=确定,则dydx=_____________.(2)函数222ln()u x y z =++在点(1,2,2)M -处的梯度grad Mu=_____________.(3)设()f x =211x-+ 00x x ππ-<≤<≤,则其以2π为周期的傅里叶级数在点x π处收敛于_____________. (4)微分方程tan cos y y x x'+=的通解为y=_____________.(5)设111212121212,n n n n n n a b a b a b a b a b a b a b a b a b ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦A 其中0,0,(1,2,,).i i a b i n ≠≠=则矩阵A的秩()r A =_____________.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)当1x →时,函数1211e 1x x x ---的极限(A)等于2 (B)等于0 (C)为∞ (D)不存在但不为∞(2)级数1(1)(1cos )(n n a n∞=--∑常数0)a >(A)发散 (B)条件收敛(C)绝对收敛 (D)收敛性与a 有关(3)在曲线23,,x t y t z t ==-=的所有切线中,与平面24x y z ++=平行的切线(A)只有1条 (B)只有2条(C)至少有3条 (D)不存在(4)设32()3,f x x x x =+则使()(0)n f 存在的最高阶数n 为 (A)0 (B)1(C)2 (D)3(5)要使12100,121⎛⎫⎛⎫ ⎪ ⎪== ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭ξξ都是线性方程组=AX 0的解,只要系数矩阵A 为(A)[]212- (B)201011-⎡⎤⎢⎥⎣⎦(C)102011-⎡⎤⎢⎥-⎣⎦(D)011422011-⎡⎤⎢⎥--⎢⎥⎢⎥⎣⎦三、(本题共3小题,每小题5分,满分15分) (1)求x x →(2)设22(e sin ,),x z f y x y =+其中f 具有二阶连续偏导数,求2.zx y∂∂∂ (3)设()f x = 21exx -+ 00x x ≤>,求31(2).f x dx -⎰四、(本题满分6分)求微分方程323e x y y y -'''+-=的通解.五、(本题满分8分) 计算曲面积分323232()()(),xaz dydz y ax dzdx z ay dxdy ∑+++++⎰⎰其中∑为上半球面z =.六、（本题满分7分）设()0,(0)0,f x f ''<=证明对任何120,0,x x >>有1212()()().f x x f x f x +<+七、（本题满分8分）在变力F yzizxj xyk=++的作用下,质点由原点沿直线运动到椭球面2222221x y z a b c++=上第一卦限的点(,,),M ξηζ问当ξ、η、ζ取何值时,力F 所做的功W 最大?并求出W 的最大值.八、（本题满分7分）设向量组123,,ααα线性相关,向量组234,,ααα线性无关,问:(1)1α能否由23,αα线性表出?证明你的结论. (2)4α能否由123,,ααα线性表出?证明你的结论.九、（本题满分7分）设3阶矩阵A 的特征值为1231,2,3,λλλ===对应的特征向量依次为1231111,2,3,149⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪=== ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭ξξξ又向量12.3⎛⎫⎪= ⎪ ⎪⎝⎭β (1)将β用123,,ξξξ线性表出. (2)求(n n A β为自然数).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)已知11()()(),()0,()(),46P A P B P C P AB P AC P BC ======则事件A、B 、C 全不发生的概率为____________.(2)设随机变量X 服从参数为1的指数分布,则数学期望2{e }X E X -+=____________.十一、（本题满分6分）设随机变量X 与Y 独立,X 服从正态分布2(,),N Y μσ服从[,]ππ-上的均匀分布,试求Z X Y =+的概率分布密度(计算结果用标准正态分布函数Φ表示,其中22()e)t xx dt --∞Φ=.1993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)函数1()(2(0)x F x dt x =->⎰的单调减少区间为_____________.(2)2232120x y z +==绕y 轴旋转一周得到的旋转面在点处的指向外侧的单位法向量为_____________.(3)设函数2()()f x x x x πππ=+-<<的傅里叶级数展开式为1(cos sin ),2n n n a a nx b nx ∞=++∑则其中系数3b 的值为_____________. (4)设数量场u =则div(grad )u =_____________.(5)设n 阶矩阵A 的各行元素之和均为零,且A 的秩为1,n -则线性方程组=AX 0的通解为_____________.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,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)设sin 2340()sin(),(),xf x t dtg x x x ==+⎰则当0x →时,()f x 是()g x 的(A)等价无穷小 (B)同价但非等价的无穷小(C)高阶无穷小 (D)低价无穷小(2)双纽线22222()x y x y +=-所围成的区域面积可用定积分表示为(A)402cos 2d πθθ⎰ (B)404cos 2d πθθ⎰(C)2θ(D)2401(cos 2)2d πθθ⎰(3)设有直线1158:121x y z l --+==-与2:l 623x y y z -=+=则1l 与2l 的夹角为(A)6π(B)4π(C)3π(D)2π(4)设曲线积分[()e ]sin ()cos xL f t ydx f x ydy --⎰与路径无关,其中()f x 具有一阶连续导数,且(0)0,f =则()f x 等于(A)e e 2x x--(B)e e 2x x--(C)e e 12x x-+-(D)e e 12x x-+-(5)已知12324,369t ⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦Q P 为三阶非零矩阵,且满足0,=PQ 则(A)6t =时P 的秩必为1 (B)6t =时P的秩必为2(C)6t ≠时P 的秩必为1 (D)6t ≠时P的秩必为2三、(本题共3小题,每小题5分,满分15分)(1)求21lim(sin cos ).x x x x→∞+(2)求.x(3)求微分方程22,x y xy y '+=满足初始条件11x y ==的特解.四、(本题满分6分)计算22,xzdydz yzdzdx z dxdy ∑+-⎰⎰其中∑是由曲面z =与z =所围立体的表面外侧.五、(本题满分7分)求级数20(1)(1)2n nn n n ∞=--+∑的和.六、(本题共2小题,每小题5分,满分10分) (1)设在[0,)+∞上函数()f x 有连续导数,且()0,(0)0,f x k f '≥><证明()f x 在(0,)+∞内有且仅有一个零点.(2)设,b a e >>证明.ba ab >七、（本题满分8分）已知二次型22212312323(,,)2332(0)f x x x x x x ax x a =+++>通过正交变换化成标准形22212325,f y y y =++求参数a 及所用的正交变换矩阵.八、（本题满分6分）设A 是n m ⨯矩阵,B 是m n ⨯矩阵,其中,n m <I 是n 阶单位矩阵,若,=AB I 证明B 的列向量组线性无关.九、（本题满分6分）设物体A 从点(0,1)出发,以速度大小为常数v 沿y 轴正向运动.物体B 从点(1,0)-与A 同时出发,其速度大小为2,v 方向始终指向,A 试建立物体B 的运动轨迹所满足的微分方程,并写出初始条件.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)一批产品共有10个正品和2个次品,任意抽取两次,每次抽一个,抽出后不再放回,则第二次抽出的是次品的概率为____________.(2)设随机变量X 服从(0,2)上的均匀分布,则随机变量2Y X =在(0,4)内的概率分布密度()Y f y =____________.十一、（本题满分6分）设随机变量X的概率分布密度为1()e ,.2xf x x -=-∞<<+∞ (1)求X 的数学期望EX 和方差.DX(2)求X 与X 的协方差,并问X 与X 是否不相关? (3)问X 与X 是否相互独立?为什么?1994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)011lim cot ()sin x x xπ→-= _____________.(2)曲面e 23x z xy -+=在点(1,2,0)处的切平面方程为_____________. (3)设e sin ,xxu y-=则2u x y ∂∂∂在点1(2,)π处的值为_____________.(4)设区域D为222,x y R +≤则2222()Dx y dxdy a b +⎰⎰=_____________. (5)已知11[1,2,3],[1,,],23==αβ设,'=A αβ其中'α是α的转置,则nA =_____________.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)设4342342222222sin cos ,(sin cos ),(sin cos ),1x M xdx N x x dx P x x x dx x ππππππ---==+=-+⎰⎰⎰则有(A)N P M << (B)MP N<<(C)N MP <<(D)P MN<<(2)二元函数(,)f x y 在点00(,)x y 处两个偏导数00(,)x f x y '、00(,)y f x y '存在是(,)f x y 在该点连续的(A)充分条件而非必要条件 (B)必要条件而非充分条件(C)充分必要条件 (D)既非充分条件又非必要条件(3)设常数0,λ>且级数21n n a ∞=∑收敛,则级数1(1)nn ∞=-∑(A)发散 (B)条件收敛(C)绝对收敛 (D)收敛性与λ有关 (4)2tan (1cos )lim2,ln(12)(1)x x a x b x c x d e-→+-=-+-其中220,a c +≠则必有(A)4b d = (B)4b d =- (C)4a c = (D)4a c =- (5)已知向量组1234,,,αααα线性无关,则向量组1994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)011lim cot ()sin x x xπ→-= _____________.(2)曲面e 23x z xy -+=在点(1,2,0)处的切平面方程为_____________. (3)设e sin ,xxu y-=则2u x y ∂∂∂在点1(2,)π处的值为_____________. (4)设区域D为222,x y R +≤则2222()Dx y dxdy a b +⎰⎰=_____________. (5)已知11[1,2,3],[1,,],23==αβ设,'=A αβ其中'α是α的转置,则nA =_____________.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)设4342342222222sin cos ,(sin cos ),(sin cos ),1x M xdx N x x dx P x x x dx x ππππππ---==+=-+⎰⎰⎰则有(A)N P M << (B)MP N<<(C)N MP <<(D)P MN<<(2)二元函数(,)f x y 在点00(,)x y 处两个偏导数00(,)x f x y '、00(,)y f x y '存在是(,)f x y 在该点连续的(A)充分条件而非必要条件 (B)必要条件而非充分条件(C)充分必要条件 (D)既非充分条件又非必要条件 (3)设常数0,λ>且级数21n n a ∞=∑收敛,则级数1(1)nn ∞=-∑(A)发散 (B)条件收敛(C)绝对收敛 (D)收敛性与λ有关 (4)2tan (1cos )lim2,ln(12)(1)x x a x b x c x d e -→+-=-+-其中220,a c +≠则必有(A)4b d = (B)4b d =-(C)4a c = (D)4a c =- (5)已知向量组1234,,,αααα线性无关,则向量组 (A)12233441,,,++++αααααααα线性无关 (B)12233441,,,----αααααααα线性无关 (C)12233441,,,+++-αααααααα线性无关 (D)12233441,,,++--αααααααα线性无关三、(本题共3小题,每小题5分,满分15分)(1)设 2221cos()cos()tx t y t t udu ==-⎰,求dy dx 、22d y dx在t =的值.(2)将函数111()ln arctan 412x f x x x x +=+--展开成x 的幂级数.(3)求.sin(2)2sin dxx x+⎰四、(本题满分6分)计算曲面积分2222,Sxdydz z dxdyx y z +++⎰⎰其中S是由曲面222x y R +=及,(0)z R z R R ==->两平面所围成立体表面的外侧.五、(本题满分9分) 设()f x 具有二阶连续函数,(0)0,(0)1,f f '==且2[()()][()]0xy x y f x y dx f x x y dy '+-++=为一全微分方程,求()f x 及此全微分方程的通解.六、(本题满分8分)设()f x 在点0x =的某一邻域内具有二阶连续导数,且()lim0,x f x x →=证明级数11()n f n ∞=∑绝对收敛.七、（本题满分6分）已知点A 与B 的直角坐标分别为(1,0,0)与(0,1,1).线段AB绕x 轴旋转一周所成的旋转曲面为.S 求由S 及两平面0,1z z ==所围成的立体体积.八、（本题满分8分）设四元线性齐次方程组(Ⅰ)为122400x x x x +=-=,又已知某线性齐次方程组(Ⅱ)的通解为12(0,1,1,0)(1,2,2,1).k k +-(1)求线性方程组(Ⅰ)的基础解析.(2)问线性方程组(Ⅰ)和(Ⅱ)是否有非零公共解?若有,则求出所有的非零公共解.若没有,则说明理由.九、（本题满分6分）设A 为n 阶非零方阵*,A 是A 的伴随矩阵,'A 是A 的转置矩阵,当*'=AA 时,证明0.≠A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)已知A 、B 两个事件满足条件()(),P AB P AB =且(),P A p =则()P B =____________.(2)设相互独立的两个随机变量,X Y 具有同一分布率,且X 的分布率为则随机变量max{,}Z X Y =的分布率为____________.十一、（本题满分6分）设随机变量X和Y 分别服从正态分布2(1,3)N 和2(0,4),N 且X 与Y 的相关系数1,2xy ρ=-设,32X Y Z =+(1)求Z 的数学期望EZ 和DZ 方差. (2)求X 与Z 的相关系数.xz ρ (3)问X 与Y 是否相互独立?为什么?1995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)2sin 0lim(13)xx x →+=_____________.(2)202cos x d x t dt dx ⎰= _____________.(3)设()2,⨯=a b c 则[()()]()+⨯++a b b c c a =_____________.(4)幂级数2112(3)n n nn nx ∞-=+-∑的收敛半径R=_____________.(5)设三阶方阵,A B 满足关系式16,-=+A BA A BA 且100310,41007⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦A 则B =_____________.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)设有直线:L321021030x y z x y z +++=--+=,及平面:4220,x y z π-+-=则直线L(A)平行于π (B)在π上(C)垂直于π (D)与π斜交(2)设在[0,1]上()0,f x ''>则(0),(1),(1)(0)f f f f ''-或(0)(1)f f -的大小顺序是(A)(1)(0)(1)(0)f f f f ''>>-(B)(1)(1)(0)(0)f f f f ''>->(C)(1)(0)(1)(0)f f f f ''->>(D)(1)(0)(1)(0)f f f f ''>->(3)设()f x 可导,()()(1sin ),F x f x x =+则(0)0f =是()F x 在0x =处可导的(A)充分必要条件 (B)充分条件但非必要条件(C)必要条件但非充分条件 (D)既非充分条件又非必要条件 (4)设(1)ln(1n n u =-则级数 (A)1n n u ∞=∑与21nn u ∞=∑都收敛 (B)1n n u ∞=∑与21nn u ∞=∑都发散(C)1n n u ∞=∑收敛,而21nn u ∞=∑发散 (D)1n n u ∞=∑收敛,而21nn u ∞=∑发散(5)设11121311121321222321222312313233313233010100,,100,010,001101a a a a a a a a a a a a a a a a a a ⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥====⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦⎣⎦A B P P 则必有(A)12AP P =B (B)21AP P =B (C)12P P A =B (D)21P P A =B三、(本题共2小题,每小题5分,满分10分) (1)设2(,,),(,e ,)0,sin ,y u f x y z x z y x ϕ===其中,f ϕ都具有一阶连续偏导数,且0.zϕ∂≠∂求.du dx(2)设函数()f x 在区间[0,1]上连续,并设1(),f x dx A =⎰求110()().x dx f x f y dy ⎰⎰四、(本题共2小题,每小题6分,满分12分) (1)计算曲面积分,zdS ∑⎰⎰其中∑为锥面z =在柱体222x y x +≤内的部分.(2)将函数()1(02)f x x x =-≤≤展开成周期为4的余弦函数.五、(本题满分7分)设曲线L 位于平面xOy 的第一象限内,L 上任一点M 处的切线与y 轴总相交,交点记为.A 已知,MA OA =且L 过点33(,),22求L 的方程.六、(本题满分8分)设函数(,)Q x y 在平面xOy 上具有一阶连续偏导数,曲线积分2(,)L xydx Q x y dy +⎰与路径无关,并且对任意t 恒有(,1)(1,)(0,0)(0,0)2(,)2(,),t t xydx Q x y dy xydx Q x y dy +=+⎰⎰求(,).Q x y七、（本题满分8分）假设函数()f x 和()g x 在[,]a b 上存在二阶导数,并且()0,()()()()0,g x f a f b g a g b ''≠====试证:(1)在开区间(,)a b 内()0.g x ≠(2)在开区间(,)a b 内至少存在一点,ξ使()().()()f fg g ξξξξ''=''八、（本题满分7分）设三阶实对称矩阵A 的特征值为1231,1,λλλ=-==对应于1λ的特征向量为101,1⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦ξ求.A九、（本题满分6分）设A 为n 阶矩阵,满足('=AA I I 是n 阶单位矩阵,'A 是A 的转置矩阵),0,<A 求.+A I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X 表示10次独立重复射击命中目标的次数,每次射中目标的概率为0.4,则2X 的数学期望2()E X =____________.(2)设X 和Y 为两个随机变量,且34{0,0},{0}{0},77P X Y P X P Y ≥≥=≥=≥= 则{max(,)0}P X Y ≥=____________.十一、（本题满分6分）设随机变量X 的概率密度为()X f x = e 0x- 00x x ≥<,求随机变量e XY =的概率密度().Y f y1996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,x x 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 x y 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,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 nλ∞=-∑(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k x 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 x y 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,zu 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(),x f t dt x⎰求()f x 的一般表达式.。

全国硕士研究生入学统一考试管理类综合能力真题及答案

全国硕士研究生入学统一考试真题（管理类综合能力）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 、C 、C 、D 、E 五个选项中，只有一项是符合试题要求的。

1、学科竞赛设一、二、三等奖，比例1：3：8获奖率30%，已知10人已获一等奖，则参赛人数（）A.300B.400C.500D.550E.602、为了解某公司员工年龄结构，按男女人数比例进行随机抽样，结果如下：男员工年龄（岁）23 26 28 30 32 3436 38 41女员工年龄（岁）23 25 27 27 29 31据表中数据统计，该公司男员工的平均年龄与全体员工平均年龄分别是（）A.32，30B.32，29.5C.32，27D.30，27E.29.5，273、某单位分段收费收网站流量（单位：）费：每日20（含）以内免，20到30（含）每收1元，30到40（含）每 3元，40以上每 5元，小王本月用45该交费（）A.45B.65C.75D.85E.1354、圆O 是△内切圆△面积与周长比1：2，则图O 面积（）A. B.π2πC. D.3π4πE.5π5、实数满足，，则（）,a b 3326a b -=2a b -=22=a b +A.30 B.22 C.15 D.13 E.106、6张不同卡片两张一组分别装入甲乙丙3个袋中，指定两张要在同一组，不同装法有（）种。

A.12B.18C.24D.30E.367、四边形A 、B 、C 、D 是平行四边形，是四边的中点2222A B C D 1111A B C D 是四边中点依次下去，得到四边形序列(123)n n n n A B C D n =、、… 3333A B C D 2222A B C D ，设面积为且则（）n n n n A B C D n S 112S =123S S S +++⋅⋅⋅⋅= A.16 B.20 C.24D.28E.308、甲乙比赛围棋，约定先胜2局者胜，已知每局甲胜概率0.6，乙为0.4，若第一局乙胜，则甲赢得比赛概率为（）A.0.144B.0.288C.0.36D.0.4E.0.69、圆，若圆在点（1，2）处的切线与y 轴及点为22:()C x y a b +-=C （0，3）则（）ab =2 1 C.0 D.1 E.210、96顾客至少购甲、乙、丙3种商品中一种，经调查同时购甲、乙两种的有8位，同时购甲丙的有12位，同购乙、丙的有6位，同购3种的有2位，则仅购一种的有（）A.70B.72C.74D.76E.8211、函数的最小值为（）{}22()max 8f x x x =-+，A.8 B.7 C.6 D.5 E.412、某单位为检查3个印前工作，由这3个部门主任和外聘3名人员组成检查组，每组1名外聘，规定本部门主任不能检查本部门，则不同的安排方式有（）A.6种B.8种C.12种D.18种E.36种13、从标号1到10中的10张卡片中随机抽2张，而它们的标号2种能被5整除的概率（）A. B. C. D. E. 15192921571514、圆柱体底面半径2，高3，垂直于底面的平面截圆柱体所得截面为矩形ABCD ，若弦AB 所对圆心角是，则截去3π部分（较小那部分）体积（）B.3π-6π-D. E.2π-π15、羽毛球队4名男运动员3女足动员，从中选出2对参加混双比赛，不同选派方式（）A.19B.18C.24D.36E.72二、条件充分性判断：第16~25小题，每小题3分，共30分。

2024年全国硕士研究生招生考试英语一真题考研英语一

2024年全国硕士研究生入学统一考试英语(一)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 the ANSWER SHEET.(10 points)There's nothing more welcoming than a door opening for you. 1 the need to be touched to open or close, automatic doors are essential in 2disabled access to buildings and helping provide general3to commercial buildings.Self-sliding doors began to emerge as a commercial product in 1960 after being invented six years4by Americans Dee Horton and Lew Hewitl. They5as a novelty feature, but as their use has grown, their 6 have extended within our technologically advanced world.Particularly7in busy locations or during times of emergency, the doors8crowdmanagement by reducing the obstacles put in people's way.9 making access both in and out of buildings easier for people, the difference in the way many of these doors open helps reduce the total area10 by them. Automatic doors often open to the side, with the panels sliding across one another. Replacing swing doors, these 11 smaller buildings to maximise the usable space inside without having to12 the way for a large, sticking-out door. There are many different types of automatic door, with each13 specific signals to tell them when to open.14 these methods differ, the main15 remain the same.Each automatic door system 16 the light, sound weight or movement in their vicinity as a signal to open. Sensor types are chosen to 17 the different environments they are needed in.18,a busy street might not19 a motion-sensored door, as it would constantly be opening for passers-by. A pressure sensitive mat would be more20 to limit the surveyed area.1.[A]Through [B]Despite [C]Besides [D]Without2.[A]revealing [B]demanding [C]improving[D]tracing3.[A]experience [B]convenience[C]guidance [D]reference4.[A]previously[B]temporarily [C]successively [D]eventually5.[A]held on [B]started out[C]settled down [D]went by6.[A]relations [B]volumes [C]benefits[D]sources7.[A]useful[B]simple [C]flexible [D]stable8.[A]call for [B]yield to [C]insist on [D]act as9.[A]As well as[B]In terms of [C]Thanks to [D]Rather than10.[A]connected[B]shared [C]represented [D]occupied11.[A]allow[B]expect [C]require [D]direct12.[A]adopt [B]lead [C]clear[D]change13.[A]adapting te [B]deriving from [C]relying on[D]pointing at14.[A]Once [B]Since [C]Unless [D]Although15.[A]records [B]positions [C]principles[D]reasons16.[A]controls [B]analyses[C]produces [D]mixes17.[A]decorate [B]compare [C]protect [D]complement18.[A]In conclusion [B]By contrast [C]For example[D]Above all19.[A]identify [B]suit[C]secure [D]include20.[A]appropriate[B]obvious [C]impressive [D]delicateSection 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 the ANSWER SHEET.(40 points)Text 1Nearly 2000 years ago, as the Romans began to pull out of Scotland, they left behind a curious treasure:10 tons of nails, nearly a million of the things. The nail hoard was discovered in 1960 in a four-metre-deep pit covered by two metres of gravel.Why had the Romans bunied a million nails? The likely explanation is that the withdrawal was rushed, and they didn't want the local Caledonians getting their hands on 10 tons of weapon-grade iron. The Romans buried the nails so deep that they would not be discovered for almost two millennia.Later civilisations would value the skilled blacksmith's labour in a nail even more than the raw material. As Roma Agrawal explains in her new delightful book Nuts and Bolts, early 17th-century Virginians would sometimes bum down their homes if they were planning to relocate. This was an attempt to recover the valuable nails, which could be reused after sifting the ashes. The idea that one might bum down an entire house just to reclaim the nails underlines how scarce, costly and valuable the simple-seeming technology was.The price of nails fell by 90%between the late 1700s and mid-1900s,as economist Daniel Sichel points out in a research paper. According to Sichel, although the falling price of nails was driven partly by cheaper iron and cheaper energy, most of the credit goes to nail manufactures who simply found more efficient ways to turn steel into nails.Nails themselves have changed over the years, but Sichel studied them because they haven't changed much. Roman lamps and Roman chariots are very different from LED strips and sports cars,but Roman nails are still clearly nails. It would be absurd to try to track the changing price of sports cars since 1695,but to ask the same question of nails makes perfect sense.I make no apology for being obsessed by a particular feature of these objects: their price. I am an economist, after all. After writing two books about the history of inventions, one thing Ive leamt is that while it is the enchantingly sophisticated technologies that get all the hype, it's thecheap technologies that change the world.The Gutenberg printing press transformed civilisation not by changing the nature of writing but by changing its cost-and it would have achieved little without a parallel collapse in the price of surfaces to write on, thanks to an often-overlooked technology called paper. Solar panels had few niche uses until they became cheap; now they are transforming the global energy system.21.The Romans buried the nails probably for the sake of[A]saving them for future use[B]keeping them from rusting[C]letting them grow in value[D]hiding them from the locals22.The example of early 17th-century Virginians is used to[A]highlight the thriftiness of early American colonists[B]illustrate the high status of blacksmiths in that period[C]contrast the attitudes of different civilisations toward nails[D]show the preciousness of nail-making technology at that time23.What played the major role in lowering the price of nails after the late 1700s?[A]Increased productivity.[B]Wider use of new energies.[C]Fiercer market competition.[D]Reduced cost of raw materials.24.It can be leamed from Paragraph 5 that nails[A]have undergone many technological improvements[B]have remained basically the same since Roman times[C]are less studied than other everyday products[D]are one of the world's most significant inventions25.Which of the following best summarises the last two paragraphs?[A]Cheap technologies bring about revolutionary change.[B]Technological innovation is integral to economic success.[C]Technology defines people's understanding of the world.[D]Sophisticated technologies develop from small inventions.Text 2Parenting tips obtained from hunter-gatherers in Africa may be the key to bringing up more contented children, researchers have suggested. The idea is based on studies of communities such as the Kung of Botswana, where each child is cared for by many adults. Kung children as young as four will help to look after younger ones and baby-wearing", in which infants are carried in slings, is considered the norm.According to Dr Nikhil Chaudhary, an evolutionary anthropologist at Cambridge University, these practices, Known as all oparenting, could lead to less anxiety for children and parents.Dr Annie Swanepoel, a child psychiatrist, believes that there are ways to incorporate them into western life. In Germany, one scheme has paired an old people's home with a nursery. The residents help to look after the children, an arrangement akin to alloparenting. Another measure could be encouraging friendships between children indifferent school years to miror the unsupervised mixed-age playgroups in hunter-gatherer communities.In a paper published in the Journal of Child Psychology and Psychiatry, researchers said that the western nuclear family was a recent invention which family broke with evolutionary history. This abrupt shift to an "intensive mothering narrative", which suggests that mothers should manage childcare alone, was likely to have been harmful. "Such narratives can lead to maternal exhaustion and have dangerous consequences, "they wrote.By contrast, in hunter-gatherer societies adults other than the parents can provide almost half of a child's care. One previous study looked at the Efe people of the Democratic Republic of Congo. It found that infants had an average of 14 alloparents a day by the time they were 18 weeks old and were passed between caregivers eight times an hour.Chaudhary said that parents now had less childcare support from family and social networks than during most of humans'evolutionary history,but introducing additional caregivers could reduce stress and maternal depression,which could have a“knock-on"benefit to a child's wellbeing.An infant bom to a hunter-gatherer society could have more than ten caregivers-this contrasts starkly to nursery settings in the UK where regulations call for a ratio of one carer to four children aged two to three.While hunter-gatherer children leamt from observation and imitation in mixed-age playgroups,researchers said that western"instructive teaching",where pupils are asked to sit still,may contribute to conditions such as attention deficit hyperactivity disorder.Chaudhary said that Britain should explore the possibility that older siblings helping their parents“might also enhance their own social development”26.According to the first two paragraph,alloparenting refers to the practice of[A]sharing child care among community members[B]assigning babies to specific adult caregivers[C]teaching parenting details to older children[D]carrying infants around by their parent27.The scheme in Germany is mentioned to illustrate[A]an attempt to facilitate intergenerational communication[B]an approach to integrating alloparenting into western culture[C]the conventional parenting style in western culture[D]the differences between western African ways ofliving28.According to Paragraph 4,the"intensive mothering narrative"_[A]alleviate parenting pressure[B]considerate family relationships[C]results in the child-centered family[D]departs from the course of evolution29.According to paragraph 6,what can we lean about nursery in the UK?[A]They tend to fall short of official requirements.[B]They have difficulty finding enough caregivers.[C]They ought to improve their carer-to-child ratio.[D]They should try to prevent parental depression.30.Which of the following would be the best title?[A]Instructive teaching:a dilemma for anxious parents[B]For a happier family,learn from the hunter-gatherers[C]Mix-aged playgroup,a better choice for lonely children[D]Tracing the history of parenting:from Africa to EuropeText 3Rutkowski is a Polish digital artist who uses classical painting styles to create dreamy fantasy landscapes.He has made illustrations for games such as Sony's Horizon Forbidden West,Ubisoft's Anno,Dungeons&Dragons,and Magic:The Gathering.And he's become a sudden hit in the new world of text-to-image AI generation.His distinctive style is now one of the most commonly used prompts in the new open-source AI art generator Stable Diffusion,which was launched late last month.The tool,along with other popular image-generation AI models,allows anyone to create impressive images based on text prompts.For example,type in"Wizard with sword and a glowing orb of magic fire fights a fierce dragon GregRutkowski,"and the system will produce something that looks not a milion miles away from works in Rutkowski's style.But these open-source programs are built by scraping images from the Internet,often without permission and proper attribution to artists.As a result,they are raising tricky questions about ethics and copyright.And artists like Rutkowski have had enough.According to the website Lexica,which tracks over 10 million images and prompts generated by Stable Diffusion,Rutkowski's name has been used as a prompt around 93,000 times.Some of the world's most famous artists,such as Michelangelo,Pablo Picasso,and Leonardo da Vinci,brought up around 2,000 prompts each or less.Rutkowski's name also features as a prompt thousands of times in the Discord of another text-to-image generator,Midjourney.Rutkowski was initially surprised but thought it might be a good way to reach new audiences.Then he tried searching for his name to see if a piece he had worked on had been published.The online search brought back work that had his name attached to it but wasn't his.“It's been just a month.What about in a year?I probably won't be able to find my work out there because[the internet]will be flooded with AI art,"Rutkowski says."That's concerning."“There is a coalition growing within artist industries to figure out how to tackle or mitigate this,"says Ortiz.The group is in its early days of mobilization,which could involve pushing for new policies or regulation.One suggestion is that AI models could be trained on images in the public domain,and AI companies could forge partnerships with museums and artists,Ortiz says.31.What can be leamed about Rutkowski from the first two paragraphs?[A]He is enthusiastic about AI generation painting.[B]He is popular with the users of an Al art generator.[C]He attracts admiration from other illustrators.[D]He specializes in classical painting digitalization.32.The problem with open-source AI art generators is that they[A]lack flexibility in responding to prompts[B]produce artworks in unpredictable styles[C]make unauthorized use of online images[D]collect user information without consent33.After searching online,Rutkowski found[A]a unique way to reach audiences[B]a new method to identify Al images[C]AI-generated work bearing his name[D]heated disputes regarding his copyright34.According to Ortiz,AI companies are advised to[A]campaign for new policies or regulations[B]offer their services to public institutions[C]strengthen their relationships with AI users[D]adopt a different strategy for Al model training35.What is the text mainly about?[A]Artists'responses to Al art generation.[B]AI's expanded role in artistic creation.[C]Privacy issues in the application of Al.[D]Opposing views on AI development.Text 4The miracle of the Chesapeake Bay lies not in its depths,but in the complexity of its natural construction,the interaction of fresh and saline water and the mix of land and water.The shallows provide homes for hundreds of species while storing floodwaters,filtering pollutants from water,and protecting nearby communities from potentially destructive storm surges.All this was put at great risk late last month,when the US Supreme Court issued a ruling in an Idaho case that provides the EPA far less authority to regulate wetlands and waterways.Specifically,a 5-4 majority decided that wetlands protected by the EPA under it Clean Water Act authority must have a“continuous surface connection”to bodies of water.This narrowing of the regulatory scope was a victory for builders,mining operators and other commercial interests often at odds with environmental rules.And it carries"significant repercussions for water quality and flood control throughout the US,"as Justice Brett Kavanaugh observed.In Maryland,the good news is that there are many state laws in place that provide wetlands protections.But that's a very shortsighted view,particularly when it comes to the Chesapeake Bay.The reality is that water and the pollutants that so often come with it,don't respect state boundaries.The Chesapeake draws from a 64000-square-mile watershed that extends to Virginia,Pennsylvania,New York,West Virginia,the District of Columbia and Delaware.Will thosejurisdictions extend the same protections now denied under Sackett V.EPA?Perhaps some,but all?That seems unlikely.It is too easy,and misleading,to see such court rulings as merely standing up for the rights of land owners when the consequences can be so dire for their neighbors.And it's reminder that they EPA's involvement in the Chesapeake Bay program has long been crucial as the means to transcend the influence of deep-pocketed special interests in neighboring states.Pennsylvania farmers,to use one telling example,aren't thinking about next year's blue crab harvest in Maryland when they decide whether to spread animal waste on their fields,yet the runoff into nearby creeks can have enormous impacts downstream.And so we would also call on state lawmakers from Richmond to Albany to consider reviewing their own wetlands protections and see for themselves the enormous stakes involved.We can't offer them a trip to the Chesapeake Bay model.It's been gone since the 1980s but perhaps a visit to Blackwater National Wildlife Refuge in Dorchester County where American bald eagles fly over tidal marshes so shallow you could not paddle a boat across them but teaming with aquatic life.It's worth the scenic drive.36.The Chesapeake Bay is described in Paragraph 1 as[A]a value natural environment[B]a controversial conservation area[C]a place with commercial potential[D]a headache for nearby communities37.The U.S.Supreme Court's ruling in the Idaho case[A]reinforces water pollution control[B]weakens the EPA's regulatory power[C]will end conflicts among local residents[D]may face opposition from mining operators38.How does the author feel about the future of the Chesapeake Bay?[A]Worried.[B]Puzzled.[C]Relieved[D]Encouraged.39.What can be inferred about the EPA's involvement in the Chesapeake BayProgram?[A]It has restored the balance among neighboring jurisdictions.[B]It has triggered a radical reform in commercial fisheries.[C]It has set a fine example of respecting state authorities.[D]It has ensured the coordination of protection efforts.40.The author holds that the state lawmakers should[A]be cautious about the influence of landowners[B]attach due importance to wetlands protections[C]recognize the need to expand wildlife refugesPart BDirections:Read the following comments on a report about American museums returning artifacts to their countries of origin and a list of statements summarizing the comments.Choose the best statement from the list A-G for each numbered name(41—45).There are two extra choices which you do not need to use.Mark your answers on the ANSWER SHEET.(10 points)(41)HannabSimply,there are people in Nigeria who cannot travel to the Smithsonian Institution to see that part of their history and culture represented by the Benin Bronzes.These should be available to them as part of their cultural heritage and history and as a source of national pride.There is no good reason that these artifacts should be beyond the ordinary reach of the educational objectives or inspiration of the generations to which they were left.They serve no purpose in a museum in the United States or elsewhere except as curious objects.They cannot be compared to works of art produced for sale which can be passed from hand to hand and place to place by purchase.(42)BuckWe know very exact reproductions of artwork can be and are regularly produced.Perhaps museums and governments might explore some role for the use of nearly exact reproductions as a means of resolving issues relating to returning works of art and antiquities.The context of any exhibit is more important to me than whether the object being displayed is 2,000 years old or 2 months old.In many cases the experts have a hard time agreeing on what is the real object and what is a forgery.Again,the story an exhibit is trying to tell is what matters.The monetary value of the objects on display is a distant second place in importance.(43)SaraWhen visiting the Baltimore Museum of Art,I came across a magnificentlSth-century Chinese sculpture.It inspired me to learm more about the culture that it represented.Artifacts in museums have the power to inspire,and perhaps spark that need to learn and understand the nature of their creators.Having said that,I do feel that whatever artifacts find their way to public museums should,in fact,be sanctioned as having been obtained on loan,legally purchased,or obtained by treaty.Stealing artifacts from other peoples'cultures is obscene;it robs not only the physical objects,but the dignity and spirit of their creators.(44)VictorAncient art that is displaced in foreign countries should be returned.…(缺失)(45)JuliaTo those of you in the comments section,by all means,who are having strong feeling about artifacts being removed from cities in the US and Britain,I would ask you to consider…(缺失)[A]It is clear that countries of origin have never been compensated for stolen artifacts.[B]It is a flawed line of reasoning to argue against returning artifacts to their countries of origin.[C]Museum visitor can still learn as much from artifacts copies after the originals are returmed.[D]Reproductions,even if perfectly made,cannot take the place of the authentic objects.[E]The real value of artifacts can only be recognized in their countries of origin rather than anywhere else[F]Ways to get artifacts from other countries must be decent and lawful.[G]Concern over security is no excuse for refusing to return artifacts to their countries of origin. 参考答案：41.E 42.C 43.F 44.G 45.BPart CDirections:Read the following text carefully and then translate the underlined segments into Chinese.Your translation should be written clearly on the ANSWER SHEET.(10 points)“Elephants never forget”—or so they say—and that piece of folklore seems to have some foundation.The African savanna elephant also known as the African bush elephant,is distributed across 37 African countries.(46)They sometimes travel more than sixty miles to find food or water,anc are very good at working out where other elephants are—even when they are out of ing tracking devices,researchers have shown that they have"remarkable spatial acuity",when finding their way to waterholes,they headed off in exactly the right direction,on one occasion from a distance of roughly thirty miles.What is more,they almost always seem to choose the nearest water hole.(47)The researchers are convinced that the elephants always know precisely where they are in relation to all the resources they need,and can therefore take shortcuts,as well as following familiar routes.Although the cues used by African elephants for long-distance navigation are not yet understood,smell may well play a part.Elephants are very choosy eaters,but until recently litle was known about how they selected their food.(48)One possibility was that they merely used their eyes and tried out the plants they found,but that would probably result in a lot of wasted time and energy,not least because their eyesight is actually not very good.(49)The volatile chemicals produced by plants can be carried a long way,and they are very characteristic:Each plant or tree has its own particular odor signature.What is more,they can be detected even when they are not actually visible.New research suggests that smell is a crucial factor in guiding elephants—and probably other herbivores—to the best food resources.The researchers first established what kinds of plant the elephants preferred either to eat or avoid when foraging freely.They then set up a“food station”experiment,in which they gave elephants a series of choices based only on smell.(50)The experiment showed that elephants may well use smell to identify patches of trees that are good to eat,and secondly to assess the quality ofthe trees within each patch.Free-ranging elephants presumably also use this information to locate their preferred food.参考译文：(46)它们有时跋涉六十多英里寻找食物或水，并且非常善于寻找其他大象的位置——即使它们不在视线范围内。

2023考研管综真题试卷及答案解析

绝密★启用前2023年全国硕士研究生入学统一考试管理类专业硕士学位联考综合试卷（科目代码：199）考试时间：08:30-11:30一、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，只有一项是符合试题要求的。

1、油价上涨5%后，加一箱油比原来多花20元。

一个月后原油价下降了4%，则加一箱油需要花（）钱？A.384元B.401元C.402.8元D.403.2元E.404元2、已知甲、乙两公司的利润之比为3：4，甲、丙两公司的利润之比为1:2。

若乙公司的利润为3000万元，则丙公司的利润为（）？A.5000万元B.4500万元C.4000万元D.3500万元E.2500万元3、一个分数的分子与分母之和为38，其分子分母都减去15，约分后得到13，则这个分数的分母与分子之差为（）A.1B.2C.3D.4E.54、5+26−3=A.2B.3C.6D.22E.235、某公司财务部有2名男员工，3名女员工，销售部有4名男员工，1名女员工，现要从中选2名男员工，1名女员工组成工作小组，并要求每部门至少有1名男员工入选，则工作小组的构成方式有（）种A.24 B.36 C.50 D.51 E.686、甲、乙两人从同一地点出发，甲先出发10分钟，若乙跑步追赶甲，则10分钟可追上；若乙骑马追赶甲，每分钟比跑步多行100米，则5分钟可追上，那么甲每分钟走的距离为（）？A.50m B.75m C.100m D.125m E.150m7、如图，已知点A（-1,2），点B（3,4），若点P（m，0）使得PB −PA 最大，则A.m=-5B.m=-3C.m=-1D.m=1E.m=38、由于疫情防控，电影院要求不同家庭之间至少间隔1个座位，同一家庭的成员座位要相连。

两个家庭看电影，一家3人，一家2人，现有一排7个相连的座位，则符合要求的坐法有（）种？A.36 B.48 C.72 D.144 E.2169、方程x 2−3x −2−4=0的所有实根之和为（）A.-4 B.-3 C.-2 D.-1 E.010、如图，从一个边长为6的正方体中裁去两个相同的正三棱锥，若正三棱锥的底面边长AB=42，则剩余的几何体的表面积为（）A.168B.168+163C.168+323D.112+323E.124+32311、如图，在三角形ABC中，∠BAC=60°，BD平分∠ABC，交AC于D，CE平分∠ACB 交AB于E，BD和CE交于F，则∠EFB=（）A.45°B.52.5°C.60°D.67°E.75°12、跳水比赛中，裁判给某选手的一个动作打分，其平均值为8.6，方差为1.1，若去掉一个最高分9.7和一个最低分7.3，则剩余得分的：A.平均值变小，方差变大B.平均值变小，方差变小C.平均值变小，方差不变D.平均值变大，方差变大E.平均值变大，方差变小13.设x为正实数，则�8�3+5�+2的最大值为？A．115B.111C.19D.16E.1514.如图，在矩形ABCD中AD=2AB，E，F分别为AD，BC为中点，A,B,C,D,E,F中任选3个点，则这3个点为原点可组成直角三角形的概率为A．12B.1124C.35D.1320E.71015.快递员收到3个同城快递任务，取送地点各不相同，取送件可穿插进行，不同的送件方式有A.6B.27C.36D.90E.360二、条件充分性判断：第16-25小题，每小题3分，共30分。

1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。

全国硕士研究生入学统一考试真题试卷《数学三》试题一、选择题:1—8小题．每小题4分，共32分．1．若函数10(),0x f x ax b x ⎧->⎪=⎨⎪≤⎩在0x =处连续，则（A ）12ab = （B ）12ab =- （C ）0ab = （D ）2ab = 2．二元函数(3)z xy x y =--的极值点是（）（A ）(0,0) （B ）03(,) （C ）30(,) （D ）11(,)3．设函数()f x 是可导函数，且满足()()0f x f x '>，则（A ）(1)(1)f f >- （B ）11()()f f <-（C ）11()()f f >- （D ）11()()f f <-4．若级数211sin ln(1)n k n n ∞=⎡⎤--⎢⎥⎣⎦∑收敛，则k =（）（A ）1 （B ）2 （C ）1- （D ）2-5．设α为n 单位列向量，E 为n 阶单位矩阵，则（A ）T E αα-不可逆（B ）T E αα+不可逆（C ）2T E αα+不可逆（D ）2T E αα-不可逆6．已知矩阵200021001A ⎛⎫ ⎪= ⎪ ⎪⎝⎭，210020001B ⎛⎫ ⎪= ⎪ ⎪⎝⎭，100020002C ⎛⎫ ⎪= ⎪ ⎪⎝⎭，则（A ）,A C 相似，,B C 相似（B ）,A C 相似，,B C 不相似（C ）,A C 不相似，,B C 相似（D ）,A C 不相似，,B C 不相似7．设,A B ，C 是三个随机事件，且,A C 相互独立，,B C 相互独立，则A B 与C 相互独立的充分必要条件是（）（A ）,A B 相互独立（B ）,A B 互不相容（C ）,AB C 相互独立（D ）,AB C 互不相容8．设12,,,(2)n X X X n ≥为来自正态总体(,1)N μ的简单随机样本，若11ni i X X n ==∑，则下列结论中不正确的是（）（A ）21()n i i X μ=-∑服从2χ分布（B ）()212n X X -服从2χ分布（C ）21()n i i X X =-∑服从2χ分布（D ）2()n X μ-服从2χ分布二、填空题（本题共6小题，每小题4分，满分24分. 把答案填在题中横线上）9．3(sin x dx ππ-=⎰ ．10．差分方程122t t t y y +-=的通解为．11．设生产某产品的平均成本()1Q C Q e -=+，其中产量为Q ，则边际成本为 .12．设函数(,)f x y 具有一阶连续的偏导数，且已知(,)(1)y y df x y ye dx x y e dy =++，(0,0)0f =，则(,)f x y =13．设矩阵101112011A ⎛⎫ ⎪= ⎪ ⎪⎝⎭，123,,ααα为线性无关的三维列向量，则向量组123,,A A A ααα的秩为．14．设随机变量X 的概率分布为{}122P X =-=，{}1P X a ==，{}3P X b ==，若0EX =，则DX = ．三、解答题15．（本题满分10分）求极限0lim t x dt +→16．（本题满分10分）计算积分3242(1)Dy dxdy x y ++⎰⎰，其中D是第一象限中以曲线y =x 轴为边界的无界区域．17．（本题满分10分）求21lim ln 1nn k k k n n →∞=⎛⎫+ ⎪⎝⎭∑18．（本题满分10分）已知方程11ln(1)k x x -=+在区间(0,1)内有实根，确定常数k 的取值范围．19．（本题满分10分）设011111,0,()(1,2,3),1n n n a a a na a n n +-===+=+，()S x 为幂级数0n n n a x ∞=∑的和函数（1）证明0n n n a x ∞=∑的收敛半径不小于1．（2）证明(1)()()0((1,1))x S x xS x x '--=∈-，并求出和函数的表达式．20．（本题满分11分）设三阶矩阵()123,,A ααα=有三个不同的特征值，且3122.ααα=+（1）证明：()2r A =；（2）若123,βααα=+，求方程组Ax β=的通解．21．（本题满分11分）设二次型222123123121323(,,)2282f x x x x x ax x x x x x x =-++-+在正交变换x Qy =下的标准形为221122y y λλ+，求a 的值及一个正交矩阵Q ．22．（本题满分11分）设随机变量,X Y 相互独立，且X 的概率分布为{}10{2}2P X P X ====，Y 的概率密度为2,01()0,y y f y <<⎧=⎨⎩其他．（1）求概率P Y EY ≤（）；（2）求Z X Y =+的概率密度．23．（本题满分11分）某工程师为了解一台天平的精度，用该天平对一物体的质量做了n 次测量，该物体的质量μ是已知的，设n 次测量结果12,,,n X X X 相互独立且均服从正态分布2(,).N μσ该工程师记录的是n 次测量的绝对误差,(1,2,,)i i Z X i n μ=-=，利用12,,,n Z Z Z 估计参数σ．（1）求Z的概率密度；i（2）利用一阶矩求σ的矩估计量；（3）求参数σ最大似然估计量．2017年全国硕士研究生入学统一考试真题试卷《数学三》试题答案一、选择题:1—8小题．每小题4分，共32分．1．解：000112lim ()lim lim 2x x x xf x ax a+++→→→===，0lim ()(0)x f x b f -→==，要使函数在0x =处连续，必须满足1122b ab a =⇒=．所以应该选（A ）2．解：2(3)32zy x y xy y xy y x∂=---=--∂，232z x x xy y ∂=--∂，2222222,2,32z z z zy x x x y x y y x∂∂∂∂=-=-==-∂∂∂∂∂∂ 解方程组22320320zy xy y xzx x xy y∂⎧=--=⎪∂⎪⎨∂⎪=--=∂⎪⎩，得四个驻点．对每个驻点验证2AC B -，发现只有在点11(,)处满足230AC B -=>，且20A C ==-<，所以11(,)为函数的极大值点，所以应该选（D ）3．解：设2()(())g x f x =，则()2()()0g x f x f x ''=>，也就是()2()f x 是单调增加函数．也就得到()()22(1)(1)(1)(1)f f f f >-⇒>-，所以应该选（C ） 4．解：iv n →∞时22221111111111sin ln(1)(1)22k k k o k o n n n n n n n n n ⎛⎫⎛⎫⎛⎫⎛⎫--=---+=++ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭显然当且仅当(1)0k +=，也就是1k =-时，级数的一般项是关于1n的二阶无穷小，级数收敛，从而选择（C ）．5．解：矩阵T αα的特征值为1和1n -个0，从而,,2,2T T T T E E E E αααααααα-+-+的特征值分别为0,1,1,1；2,1,1,,1；1,1,1,,1-；3,1,1,,1．显然只有T E αα-存在零特征值，所以不可逆，应该选（A ）．6．解：矩阵,A B 的特征值都是1232,1λλλ===．是否可对解化，只需要关心2λ=的情况．对于矩阵A ，0002001001E A ⎛⎫⎪-=- ⎪ ⎪⎝⎭，秩等于1 ，也就是矩阵A 属于特征值2λ=存在两个线性无关的特征向量，也就是可以对角化，也就是~A C ．对于矩阵B ，010*******E B -⎛⎫⎪-= ⎪ ⎪⎝⎭，秩等于2 ，也就是矩阵A 属于特征值2λ=只有一个线性无关的特征向量，也就是不可以对角化，当然,B C 不相似故选择（B ）．7．解：(())()()()()()()()()()P A B C P AC AB P AC P BC P ABC P A P C P B P C P ABC =+=+-=+- ()()(()()())()()()()()()()P A B P C P A P B P AB P C P A P C P B P C P AB P C =+-=+-显然，A B 与C 相互独立的充分必要条件是()()()P ABC P AB P C =，所以选择（C ）．8．解：（1）显然22()~(0,1)()~(1),1,2,i i X N X i n μμχ-⇒-=且相互独立，所以21()nii Xμ=-∑服从2()n χ分布，也就是（A ）结论是正确的；（2）222221(1)()(1)~(1)ni i n S X X n S n χσ=--=-=-∑，所以（C ）结论也是正确的；（3）注意221~(,)()~(0,1)()~(1)X N X N n X nμμμχ⇒-⇒-，所以（D ）结论也是正确的；（4）对于选项（B ）：22111()~(0,2)~(0,1)()~(1)2n n X X N N X X χ-⇒⇒-，所以（B ）结论是错误的，应该选择（B ）二、填空题（本题共6小题，每小题4分，满分24分. 把答案填在题中横线上）9．解：由对称性知330(sin 22x dx ππππ-==⎰⎰．10．解：齐次差分方程120t t y y +-=的通解为2x y C =；设122t t t y y +-=的特解为2t t y at =，代入方程，得12a =；所以差分方程122t t t y y +-=的通解为12 2.2t ty C t =+11．解：答案为1(1)Q Q e -+-．平均成本()1Q C Q e -=+，则总成本为()()Q C Q QC Q Q Qe -==+，从而边际成本为()1(1).Q C Q Q e -'=+-12．解：(,)(1)()y y y df x y ye dx x y e dy d xye =++=，所以(,)y f x y xye C =+，由(0,0)0f =，得0C =，所以(,)y f x y xye =．13．解：对矩阵进行初等变换101101101112011011011011000A ⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪=→→ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭，知矩阵A 的秩为2，由于123,,ααα为线性无关，所以向量组123,,A A A ααα的秩为2． 14．解：显然由概率分布的性质，知112a b ++=12133102EX a b a b =-⨯+⨯+⨯=+-=，解得11,44a b ==29292EX a b =++=，229()2DX EX E X =-=．三、解答题15．（本题满分10分）解：令x t u-=，则,t x u dt du=-=-，00t x udt du-=⎰⎰00002 lim lim lim lim33t x u u x x x x xdt e du du++++---→→→→==== 16．（本题满分10分）解：3324224200242420022(1)(1)1(1)4(1)111141128Dy ydxdy dx dyx y x yx ydxx ydxx xπ+∞+∞+∞=++++++=++⎛⎛⎫=-=⎪++⎝⎭⎝⎭⎰⎰⎰⎰⎰17．（本题满分10分）解：由定积分的定义12011121lim ln1lim ln1ln(1)11ln(1)24n nn nk kk k k kx x dxn n n n nx dx→∞→∞==⎛⎫⎛⎫+=+=+⎪ ⎪⎝⎭⎝⎭=+=∑∑⎰⎰18．（本题满分10分）解：设11(),(0,1)ln(1)f x xx x=-∈+，则22222211(1)ln(1)()(1)ln(1)(1)ln(1)x x xf xx x x x x x++-'=-+=++++令22()(1)ln(1)g x x x x=++-，则2(0)0,(1)2ln21g g==-2()ln(1)2ln(1)2,(0)0g x x x x g''=+-+-=2(ln(1))()0,(0,1)1x xg x xx+-''=<∈+，所以()g x'在(0,1)上单调减少，由于(0)0g '=，所以当(0,1)x ∈时，()0)0g x g ''<=，也就是()g x ()g x '在(0,1)上单调减少，当(0,1)x ∈时，()(0)0g x g <=，进一步得到当(0,1)x ∈时，()0f x '<，也就是()f x 在(0,1)上单调减少．00011ln(1)1lim ()lim lim ln(1)ln(1)2x x x x x f x x x x x +++→→→⎛⎫-+=-== ⎪++⎝⎭，1(1)1ln 2f =-，也就是得到111ln 22k -<<． 19．（本题满分10分）解：（1）由条件11111()(1)1n n n n n n a na a n a na a n +-+-=+⇒+=++ 也就得到11(1)()()n n n n n a a a a +-+-=--，也就得到111,1,2,1n n n n a a n a a n +--=-=-+1112110112101(1)(1)!n n n n n n n n n n n a a aa a a a a a a a a a a a a n ++--------=⨯⨯⨯=-----+ 也就得到111(1),1,2,(1)!n n n a a n n ++-=-=+111121121()()()(1)!nk n n n n n k a a a a a a a a k +++-==-+-++-+=-∑ 1lim 1!n n n n ρ=≤++≤=，所以收敛半径1R ≥ （2）所以对于幂级数0nn n a x ∞=∑，由和函数的性质，可得11()n n n S x na x ∞-='=∑，所以11111101111111(1)()(1)(1)((1))()n n nn n n n n n nnn n n n nn n n nn n n n n n n n x S x x na xna xna x n a x na x a n a na x a x a xx a x xS x ∞∞∞--===∞∞+==∞+=∞∞∞+-==='-=-=-=+-=++-====∑∑∑∑∑∑∑∑∑也就是有(1)()()0((1,1))x S x xS x x '--=∈-．解微分方程(1)()()0x S x xS x '--=，得()1xCe S x x-=-，由于0(0)1S a ==，得1C =所以()1xe S x x-=-．20．（本题满分11分）解：（1）证明：因为矩阵有三个不同的特征值，所以A 是非零矩阵，也就是()1r A ≥．假若()1r A =时，则0r =是矩阵的二重特征值，与条件不符合，所以有()2r A ≥，又因为31220ααα-+=，也就是123,,ααα线性相关，()3r A <，也就只有()2r A =．（2）因为()2r A =，所以0Ax =的基础解系中只有一个线性无关的解向量．由于31220ααα-+=，所以基础解系为121x ⎛⎫⎪= ⎪ ⎪-⎝⎭；又由123,βααα=+，得非齐次方程组Ax β=的特解可取为111⎛⎫ ⎪ ⎪ ⎪⎝⎭；方程组Ax β=的通解为112111x k ⎛⎫⎛⎫⎪ ⎪=+ ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭，其中k 为任意常数． 21．（本题满分11分）解：二次型矩阵21411141A a -⎛⎫⎪=- ⎪ ⎪-⎝⎭因为二次型的标准形为221122y y λλ+．也就说明矩阵A 有零特征值，所以0A =，故 2.a =114111(3)(6)412E A λλλλλλλ---=+=+---令0E A λ-=得矩阵的特征值为1233,6,0λλλ=-==．通过分别解方程组()0i E A x λ-=得矩阵的属于特征值13λ=-的特征向量1111ξ⎛⎫⎪=-⎪⎪⎭，属于特征值特征值26λ=的特征向量2101ξ-⎛⎫⎪=⎪⎪⎭，30λ=的特征向量3121ξ⎛⎫⎪=⎪⎪⎭，所以()123,,0Q ξξξ⎛ == ⎝为所求正交矩阵． 22．（本题满分11分）解：（1）1202()2.3Y EY yf y dy y dy +∞-∞===⎰⎰所以{}230242.39P Y EY P Y ydy ⎧⎫≤=≤==⎨⎬⎩⎭⎰（2）Z X Y =+的分布函数为{}{}{}{}{}{}{}[](),0,20,2,211{}2221()(2)2Z Y Y F z P Z z P X Y z P X Y z X P X Y z X P X Y z P X Y z P Y z P Y z F z F z =≤=+≤=+≤=++≤===≤+=≤-=≤+≤-=+-故Z X Y =+的概率密度为[]1()()()(2)2,012,230,Z Z f z F z f z f z z z z z '==+-≤≤⎧⎪=-≤<⎨⎪⎩其他 23．（本题满分11分）解：（1）先求i Z 的分布函数为{}{}()i Z i i X z F z P Z z P X z P μμσσ⎧-⎫=≤=-≤=≤⎨⎬⎩⎭当0z <时，显然()0Z F z =；当0z ≥时，{}{}()21i Z i i X z zF z P Z z P X z P μμσσσ⎧-⎫⎛⎫=≤=-≤=≤=Φ-⎨⎬ ⎪⎝⎭⎩⎭；所以i Z的概率密度为222,0()()0,0z Z Z z f z F z z σ-⎧≥'==<⎩．（2）数学期望22200()z i EZ z f z dz ze dz σ-+∞+∞===⎰⎰ 令11n i i EZ Z Z n ===∑，解得σ的矩估计量1ni i Z σ===．（3）设12,,,n Z Z Z 的观测值为12,,,n z z z ．当0,1,2,i z i n >=时似然函数为221121()(,)ni i n nz i i L f z σσσ=-=∑==∏，取对数得：2211ln ()ln 2ln(2)ln 22nii n L n n zσπσσ==---∑令231ln ()10n i i d L n z d σσσσ==-+=∑，得参数σ最大似然估计量为σ=。