浙江大学2006-2007学年秋冬学期_微积分I期末试卷

浙江大学2006–2007学年秋冬学期《无机及分析化学》课程期末考试试卷(A)开课学院：理学院考试形式：半开卷，允许带一页A4纸资料和计算器入场考试时间：2007年1月22日上午 8：00 –10:00 所需时间：120分钟考生姓名： _____学号：专业： ________一、选择题（单选题，把正确答案序号填入括号内，共27分）1.某金属离子生成的两种八面体配合物的磁距分别为μ＝4.90B.M和μ＝0B.M，则该金属离子可能是（）A.Cr3+B. Mn2+C.Mn3+D.Fe2+2. 在Cr(H2O)4Cl3溶液中，加入过量AgNO3溶液，只有1/3的Cl－被沉淀，说明（）A.反应进行的不完全B. Cr(H2O)4Cl3的量不足C.反应速度快D. 其中两个Cl－与Cr3+形成了配位键3. 用K2Cr2O7法测定Fe2+加入H3PO4的主要目的是（）A. 提高酸度B.防止Fe2+的水解C. 减少E （Fe3+/Fe2+）的数值, 减少终点误差D. 同Fe3+形成稳定的无色的配合物，减少黄色对终点的干扰4.下列物质中，其分子具有V形几何构型的是（）A. NO2+B. CO 2C. CH 4D. O 35. 下列分子中偶极距不为零的是 ( )A.BeCl2B.SO2C.CO2D.CH46. 已知lg K f (CuY) = 18.8, lg K f (ZnY) =16.5, 用EDTA滴定Cu2+,Zn2+混合溶液中的Cu2。

为消除Zn2+的干扰，应采用的方法是（）A.控制酸度法B.配位掩蔽法C.氧化还原掩蔽法D.沉淀掩蔽法7.0.010mol·L-1的一元弱碱(K b ＝1.0×10－8)溶液与等体积水混合后，溶液的pH值为( )A. 8.7B. 8.85C. 9.0D. 10.58.将50.0 mL 0.100 mol·L-1 (NH4)2SO4 溶液，加入到50.0 mL 0.200 mol·L-1 NH3·H2O（K(NH3·H2O) = 1.8×10-5 ）溶液中，得到的缓冲溶液pH 值是. （）A. 8.70B. 9.56C. 9.26D. 9.009. 已知298 K 时，MnO2 (s) →MnO (s) + 1/2 O2 (g) 的∆r H m (1) = 134.8 kJ·mol-1MnO2(s) + Mn (s) →2 MnO (s) 的∆r H m(2) = -250.4 kJ·mol-1，则∆f H m (MnO2 , s)为. （）A. -385.2 kJ·mol-1B. 385.2 kJ·mol-1C. -520.0 kJ·mol-1D. 520.0 kJ·mol-110. 分光光度法测定中，若显色剂或其它所加试剂在测定波长处略有吸收，而试液本身无吸收，应采用的参比溶液是（）A. 水B.试剂空白C.试样空白D.纯溶剂空白11. 已知K sp(CaF2)=2.07⨯10-11，K sp(AgCl) =1.8 ⨯ 10-10。

诚信应考 考出水平 考出风格浙江大学城市学院2006—2007学年第二学期考试试卷《 微积分（B ）II 》开课单位：计算分院 ；考试形式：闭卷；考试日期：07年7月10日；时间：120分钟一． 微分方程问题(本大题共 3 题，每题 5分，共15 分)1． 求解微分方程 20(4)2,1x y x xy y ='-==.解：222(4)2, (4)dy dy x x xy dx dx y x -==-； ()()122222121,　4(4)(4)ln ln 4, 4,　c dy x dy dxd x y x y x y x C ye x ==---=-+=±-⎰⎰⎰⎰记1cC e =±，得通解：()24y C x =-， 由01x y==，得14C =-，所以微分方程特解为()2144y x =-- 点评：此题考可分离变量微分方程掌握情况。

可分离变量微分方程的关键是将方程通过因式分解，使,x y “分家”，变成：()()f y dy g x dx =形式，然后积分。

本题还要注意1cC e =±的变化。

2.求解微分方程 22x y xy xe -'-=.解: 2()2,()x p x x q x xe -=-=,()222222222222(2)(2)22221241144x dx x dxx x x x x x x x x x x xy e xe e dx C e xe e dx C e xe dx C e e d x C e e C e Ce ----------⎡⎤⎰⎰⎡⎤=+=+⎢⎥⎣⎦⎣⎦⎡⎤⎡⎤=+=--+⎢⎥⎣⎦⎣⎦⎛⎫=-+=-+ ⎪⎝⎭⎰⎰⎰⎰点评： 本题为典型的一阶线性微分方程()()y p x y q x '+=，这类方程只要记住公式：()()()p x dx p x dxy e q x e dx C -⎡⎤⎰⎰=+⎢⎥⎣⎦⎰注意公式中三个不定积分计算后不需要另再加积分常数，因为本公式中已经有C 了。

浙江大学级微积分（上）期终考试试卷系班级学号姓名考试教室一、选择题：（每小题分，共分）在每题的四个选项中，只有一个是正确的，请把正确那项的代号填入空格中.设()()()()()f x x a x b x c x d=----，其中a，b，c，d互不相等，且'()()()()f k k a k b k c=---，则k的值等于（）.（）.a（）.b（）.c（）.d.曲线y=x→-∞时，它有斜渐进线（）.（）.1y x=+（）.1y x=-+（）.1y x=--（）.1y x=-.下面的四个论述中正确的是（）.（）.“函数()f x在[],a b上有界”是“()f x在[],a b上可积”的必要条件；（）.函数()f x在区间(),a b内可导，(),x a b∈，那末'()0f x=是()f x在x处取到极值的充分条件；（）.“函数()f x在点x处可导”对于“函数()f x在点x处可微”而言既非充分也非必要；（）.“函数()f x在区间E上连续”是“()f x在区间E上原函数存在”的充要条件..下面四个论述中正确的是（）.（）.若0nx≥(1,2,)n=，且{}n x单调递减，设lim nnx a→+∞=，则0a>；（）. 若0nx>(1,2,)n=，且limnnx→+∞极限存在，设limnnx a→+∞=，则0a>；（）. 若lim0nnx a→+∞=>，则0nx≥(1,2,)n=；（）. 若lim0nnx a→+∞=>，则存在正整数N，当n N>时，都有2nax>.二、填空题：（每空格分，共分）只填答案. 2lim (1)tgxx x π→-；2lim (1)tgxx x π→--..函数()f u 可导，(sin )y f x x =，则dy dx.. cos sin x xxe e dx e ⎰. . 50sin tdt π⎰；50cos tdt π⎰.三、求极限：（每小题分，共分）.数列{}n x通项21n x n =++++，求lim n n x →+∞..求300sin lim sin xx t dt t x x→-⎰.四、求导数：（每小题分，共分）. 2sin 1xx y x x =+，求dydx.. 2,sin ,x t y t ⎧=⎨=⎩求dy dx ，22d ydx ..函数()y y x =由sin x y y +=确定，求221,;x y dydxππ=-=22221,.x y d y dx ππ=-=五、求积分：（每小题分，共分） .求21(1)x dx x x ++⎰..求0sin cos x x dx π-⎰..求0⎰(0)a >..计算2cos x e xdx π+∞-⎰.六、（分）下面两题做一题，其中学过常微分方程的专业做第题，未学常微分方程的专业做第题..求解常微分方程：22(),(1) 1.x dy xy x dx y ⎧=-⎨=⎩.有一半径为M 的半球形水池注满了水，现要把水全部抽到距水池水面高M 的水箱内，问至少要做多少功？七、（分）在xoy 平面上将连结原点(0,0)O 与点(1,0)A 的线段OA （即区间[]0,1）作n 等分，分点(,0)k n记作k P ，对1,2,,1k n =-，过k P 作抛物线2y x =的切线，切点为k Q ..设k k P Q A ∆的面积为k S ，求k S ；.求极限111lim n k n k S n -→+∞=∑.八、证明题（分）设()f x 在(),-∞+∞上连续，且()0f x >，0()()xG x tf x t dt =-⎰.证明：对任意,(,)a b ∈-∞+∞，且a b ≠，必有()()'()()0G b G a G a b a --->.浙江大学级微积分（下）期终考试试卷系班级学号姓名考试教室一、填空题：（每小题分，共分）只填答案.设一平面经过原点及点()6,3,2-，且与平面428x y z-+=垂直，则此平面的方程是。

浙江⼤学远程教育学院试题卷浙江⼤学远程教育学院试题卷2006—2007学年秋冬学期期中作业课程名称英语（3）学习中⼼专业年级学号姓名VocabularySection A: Choose the best answer to complete the following sentences:1). When I said some people are lazy, I didn’t ________ to you.A. applyB. meanC. mentionD. refer2) The speaker ________ the whole audience with his enthusiasm.A. infectedB. effectedC. defectedD. defended3) I am ________ of watching TV all the time; let’s go out for a walk.A. wornB. boredC. tiredD. fed up4) After leaving the cinema, he ________ straight for home.A. setB. headedC. wentD. proceeded5) I’ve got a ________ in the tax office who I think might help us out of trouble.A. contrastB. contactC. contractD. concrete6) Jack ________ in enthusiasm what he lacked in experience.A. made forB. made outC. made up ofD. make up for7) He always ________ himself hard to get his work finished on time.A. drivesB. programsC. depressesD. rules8) Janet felt ashamed that she hadn’t been able to ________ her fear.A. commitB. commandC. conquerD. conserve9) She ________ the tube hard until the last bit of the toothpaste came out.A. squeezedB. extendedC. minimizedD. oppressed10) His work as a simultaneous interpreter requires strong powers of ________.A. considerationB. concentrationC. intensityD. attention11) These research results are ________ over a wide range of circumstances.A. applicableB. changeableC. enjoyableD. bearable12) Silence ________ on the room. Everybody was worried about the child’s safety.A. aroseB. occupiedC. appearedD. fell13. My sister Lucy ________ her boss with her capabilities and efficiency. Figured B. impressed C. reached D. volunteered14. He is _________ two foreign languages.A. essential toB. involved withC. familiar withD. anxious about15. She ________ the way she is treated in school.A. rewritesB. reducesC. resentsD. rents16. My father had a very ________ nature; we had to do whatever she ordered us to and otherwise she would get anger.A. physicalB. dominantC. financialD. sole17. The teacher pointed out that all of the students need, in short, to learn to read ________.A. exclusivelyB. activelyC. suddenlyD. socially18. Mother often asked us not to judge people by ________ .A. separationB. involvementC. adherenceD. appearances19. Tom declared that he didn’t want to have anything to do ________ the political party.A. atB. intoC. offD. with20. Roy must have gone out early this morning, for he did not ________ for breakfast.A. bring alongB. sign upC. show upD. set upStructure1. There was a park near our house and we ________ play football with our classmates in it when we were young.A. were used toB. use toC. were using toD. used to2. The president’s suggestion that the manager of the personnel department ________ the candidate again was accepted at the meeting.A. interviewsB. interviewedC. interviewD. were to interview3. All the arrangements should have been completed ________ the departure.A. prior toB. next toC. apart fromD. other from4. Catherine didn’t run ________ catch the bus to the Space Museum.A. enough fast toB. fast enough toC. enough fast andD. fast enough and5. Ten ponds ________ not enough if you want to have a good dinner in a good restaurant.A. isB. areC. hasD. have6. ________ I like the color of the hat, I don’t like its shape.A. UnlessB. AsC. WhileD. Whether7. After ________ seemed to be an endless wait, the chairman finally showed up.A. whatB. whichC. thatD. whatever8. More American people fail to pay medical insurance fee than ________ .A. they are believedB. generally believedC. believedD. is generally believed9. “I hope the children won’t go near the water. ”“I warned them ________.”A. notB. not toC. not goD. wouldn’t10. I t is worthwhile for parents to spend an hour or two ________ with their children every week.A. talkB. to talkC. talkingD. talked toCloze(1) The horse and carriage is a thing of the past, but love and marriage are still with us and still closely interrelated. Most American marriages, particularly first marriages 1 young couples, are the result of 2 attraction and affection 3 than practical considerations.In the United States, parents do not arrange marriages for their children. Teen-agers begin 4 in high school and usually find mates through their own academic and social 5 .Though young people feel 6 to choose their friends from 7 groups, most choose a mate of similar background. This is 8 in part to parentalguidance. Parents cannot select spouses for their children, but they can usually 9 choices by 10 disapproval of someone they consider unsuitable.11 , marriages between members of different groups (interclass, interfaith, and interracial ) occur because of the greater 12 of today's youth and thefact that hey are restricted by 13 prejudices than their parents. Many young people leave their hometowns to attend college,14 in the armed forces.15 pursue a career in a big city.Once away from home and family, they are more 16 to date and marry outside their own social group. In mobile American society, interclass marriages are neither 17 nor shocking. Interfaith marriages are 18 the rise particularly between Protestants and Catholics. On the other hand, interracial marriage is still very uncommon. I t can be difficult for interracial couples to find a place to live, maintain friendships, and 19 a family. Marriages between people of different national 20 (but the same race and religion) have been commonplace here since colonial times.1. A. involving B. linking C. connecting D. correlating2. A. personal B. emotional C. mutual D. magnetic3. A. more B. less C. other D. rather4. A. dating B. appointment C. engaging D. matching5. A. position B. association C. contract D. contacts6. A. certain B. embarrassed C. hesitated D. free7. A. similar B. identical C. diverse D. differential8. A. for B. likely C. due D. because9. A. give B. influence C. make D. offer10. A sounding B. avoiding C. expecting D. voicing11. A. However B. Moreover C. Therefore D. Furthermore12. A. mobility B. motive C. moral D. mission13. A. scarcity B. satisfactory C. abundant D. fewer14. A. work B. serve C. stay D. remain15. A. but B. otherwise C. or D. likewise16. A. certainly B. likely C. reluctant D. readily17. A. lack B. rare C. scared D. relieved18. A. in B. at C. on D. for19. A. raise B. obtain C. grow D. unite20. A source B. origin C. resource D. baseReading Comprehension 1Directions: There are 4 passages in this part. Each passage is followed by some questions or unfinished statements. For each of them there are four choices marked A), B), C), and D). You should decide on the best choice and mark the corresponding letter on the Answer Sheet with a single line through the center.Passage OneQuestions 1 to 5 are based on the following passage:Washington Irving was America's first man of letters to be known internationally. His works were received enthusiastically both in England and in the United States. He was, in fact, one of the most successful writers of his time in either country, delighting a large general public and at the same time winning the admiration of fellow writers like Scott in Britain and Poe and Hawthorne in the United States. The respect in which he was held was partly owing to the man himself, with his warm friendliness, his good sense, his urbanity, his gay spirits, has artistic integrity, his love of both the Old World and the new. Thackeray described Irving as "a gentleman, who, though himself born in no very high sphere, was most finished, polished, witty; socially the equal of the most refined Europeans." In England he was granted an honorary degree from Oxford—an unusual honor for a citizen of a young, uncultured nation---and he received the medal of the Royal Society of Literature; America made him ambassador to Spain.Irving's background provides little to explain his literary achievements. A gift but deliberate child, he had little schooling, He studied law, but without zeal, and never did practice seriously. He was immune to his strict Presbyterian (基督教长⽼会的) home environment, frequenting both social gatherings and the theatre.1. The main point of the first paragraph is that Washington Irving was ______.A) America's first man of lettersB) a great writer who was successful in his own country andother parts of the world as wellC) a man who won the respect of other writers because of hishigh social statusD) a man who was able to move from literature to politics2. What is implied by the comment about Scott, Poe and Hawthorne?A) Irving's great popularity resulted in the admiration of Scott, Poe andHawthorne.B) More Americans than Britains admired Irving.C) Irving's work was not only popular, but also of high literary quality.D) Irving's success was attributed to his family background.3. What can be said about Irving's law career?A) He only began to practice law late in life.B) He spent very little time working as a lawyer.C) He never practiced law although he studied it.D) He worked as a lawyer with great enthusiasm.4. Why did Thackeray think that Irving's social grace was unusual?A) Because Irving's degree was honorable and unusual.B) Because his parents were not aristocratic.C) Because he had good sense and gay spirits.D) Because he often exhibited warm friendliness.5. Which of the following best describes the effect of Irving's Presbyterian background on his life?A) It had almost no effect on his life.B) It promoted his interest in law.C) It fostered his love for literature.D) It enabled him to become a successful writer.Passage TwoQuestions 6 to 10 are based on the following passage:Psychologists now believe that noise has a considerable effect on people's attitudes and behavior. Experiments have proved that in noisy situations (even temporary ones), people be have more irritably and less cooperatively; in more permanent noisy situations, many people cannot work hard, and they suffer from severe anxiety as well as other psychological problems. However, psychologists distinguish between "sound" and "noise". "Sound" is measured physically in decibels. "Noise" cannot be measured in the same way because it refers to the psychological effect of sound and its level of "intensity" depends on the situation. Thus, for passengers at an airport who expect to hear airplanes taking off and landing, there may be a lot of sound, but not much noise (that is, they are not bothered by the noise). By contrast, if you are at a concert and two people behind you are whispering, you feel they are talking noisily even if there is not much sound. You notice the noise because it affects you psychologically. Both sound and noise can have negative effects, but what is most important is if the person has control over the sound. People walking down the street with earphones, listening to music that they enjoy, are receiving a lot of decibels of sound, but they are probably happy hearing sounds which they control. On the other hand, people in the street without earphones must tolerate a lot of noise which they have no control over. It is noise pollution that we need to control in order to help people live more happily.6. According to the passage, people _____.A) can not work better in a noisy situationB) will suffer from complete deafness because of noise pollutionC) can be psychologically affected by working in very noise factoriesD) may cooperate well in a noisy surrounding7. “Sound”, as defined by the psychologist, _____.A) can be measured in the same way that "noise" is measuredB) may be extremely harmful to healthC) is not at all different from "noise"D) can be measured by machines8. People waiting at an airport _____.A) enjoy hearing airplanes taking off and landingB) are usually not troubled by the noiseC) can easily tell sound from noiseD) are often physically affected by the noise9. People enjoy listening to music,_____.A) though they are receiving a lot of decibels of sound in factB) because it does not have any negative effectC) because they do not have to tolerate the noise around themD) even though it is sometimes unpleasant hearing strange sounds10. We can conclude from the passage that we need to control noise pollution if _____.A) we want to stay both psychologically and physically healthyB) we don't want to be physically dentC) we want to cooperate wellD) we don't want to be anxiousPassage ThreeQuestions 11 to 15 are based on the following passage:Is there a "success personality"? Some winning combination of qualities that leads almost inevitably to achievement? If so, exactly what is that secret success formula, and can anyone develop it?At the Gallop Organization we recently focused in depth on success, probing the attitudes of 1500 prominent people selected at random from Who's Who in America. Our research finds out a number of qualities that occur regularly among top achievers. Here is one of the most important, that is common sense.Common sense is the most prevailing quality possessed by our respondents. Seventy-nine percent award themselves a top score in this quality. And 61 percent say that common sense was very important in contributing to their success.To most, common sense means the ability to present sound, practical judgments on everyday affairs. To do this, one has to sweep aside extra ideas and get right to the core of what matters. A Texas oil and gas businessman puts it this way: "The key ability for success is simplifying. In conduction of meeting and dealing with industry reducing a complex problem to the simplest term is highly important."Is common sense a quality a person is born with, or can you do something to increase it? The oil man's answer is that common sense can definitely be developed. He attributes his to learning how to debate in school. Another way to increase your store of common sense is to observe it in others, learning from their and your own mistakes.Besides common sense, there are many other factors that influence success: knowing your field, self-reliance, intelligence, the ability to get things done, leadership, creativity, relationships with others, and of course, luck. But common sense standsout. If you develop these qualities, you'll succeed. And you might even find yourself listed in Who's Who someday.11. It can be known from the passage that Who's Who _____.A) is a very useful book telling us how to succeedB) is a book providing us with the information about the family life of some famouspeopleC) is a book providing us with the names and brief biographies of the top successfulpeopleD) is a book from which we can find out the names of different peoples in the world12. According to the author, common sense _____.A) is something that common people like bestB) is a popular quality a person is born withC) is something that enables one to form correct opinionsD) is a quality that is possessed by common people13. It can be inferred from the passage that a successful businessman _____.A) tries to get experience through practiceB) pays attention to the essence of a problem when he tries to solve itC) keeps on learning in order to be successfulD) has strong willpower, extensive interest and intelligence14. The passage is mainly concerned with _____.A) organizational ability and good work habitsB) the way to obtain big profits and achieve fame and successC) knowledge and interest which are primary to successD) what successful people have in common15. According to the author, how can one develop one's common sense?A) To become a businessman.B) To learn how to debate and learn from mistakes.C) To become famous.D) To be simplifying.Passage FourQuestions 1 to 5 are based on the following passage:Brazil has become one of the developing world's great successes at reducing population growth but more by accident than design. While countries such as India have made joint efforts to reduce birth rates, Brazil has had better result without really trying, says George Martine at Harvard.Brazil's population growth rate has dropped from 2.99 a year between 1951 and 1960 to 1.93 a year between 1981 and 1990, and Brazilian women now have only 2.7 children on average. Martine says this figure may have fallen still further since 1990, an achievement that makes it the envy of many other Third World countries.Martine puts it down to, among other things, soap operas and installment plans introduced in the 1970s. Both played animportant, although indirect, role in lowering the birth rate. Brazil is one of the world's biggest producers of soap operas. Global, Brazil's most popular television network, shows three hours of soaps six nights a week, while three others show at least one hour a night. Most soaps are based on wealthy characters living the high life in big cities."Although they have never really tried to work in a massage towards the problems of reproduction, they describe middle and upper class values not man children, different attitudes towards sex, women working," says Martine. "They sent his image to all parts of Brazil and made people conscious of other patterns of behavior and other values, which were put into a very attractive package."Meanwhile, the installment plans tried to encourage the poor to become consumers. "This led to an enormous change in consumption patterns and consumption was in compatible with unlimited reproduction." says Martine.16. According to the passage, Brazil has cut back its population growth _______.A) by educating its citizensB) by careful family panningC) by developing TV programmersD) by chance17. According to the passage, many Third World countries _______.A) haven't attached much importance to birth controlB) would soon join Brazil in controlling their birth rateC) haven't yet found an effective measure to control their populationD) neglected the role of TV plays in family planning18.The phrase “puts it down to” (Line 1, Para. 3) is closest in meaning to “_______”.A) attributes it toB) sums it up asC) finds it a reason forD) compares it to19. Soap operas have helped in lowering Brazil's birth rate because _______.A) they keep people sitting long hours watching TVB) they have gradually changed people's way of lifeC) people are drawn to their attractive packageD) they popularize birth control measures20. What is Martine's conclusion about Brazil's population growth?A) The increase in birth rate will promote consumption.B) The desire for consumption helps to reduce birth rate.C) Consumption patterns and reproduction patterns are contradictory.D) A country's production is limited by its population growth.浙江⼤学远程教育学院试题卷2006—2007学年秋冬学期期中作业课程名称英语（3）学习中⼼专业年级学号姓名Section A: Choose the best answer to complete the following sentences: 1. 2 . 3. 4. 5.6. 7. 8. 9. 10.11. 12. 13. 14. 15.16. 17. 18. 19. 20Structure1. 2. 3. 4. 5. 6. 7. 8. 9. 10.Cloze1. 2. 3. 4. 5. 6. 7. 8. 9. 10.11. 12. 13. 14. 15. 16. 17. 18. 19. 20.Reading Comprehension 1Passage One1. 2. 3. 4. 5.Passage Two6. 7. 8. 9. 10.Passage Three11. 12. 13. 14. 15.Passage Four16. 17. 18. 19. 20.浙江⼤学远程教育学院试题卷（期中）2006-2007学年冬季学期课程名称⽣物化学学习中⼼专业（层次）学号姓名⼀、选择题（20分）1．有关蛋⽩质三级结构描述，错误的是A．具有三级结构的多肽链都有⽣物学活性B．亲⽔基团多位于三级结构的表⾯C．三级结构的稳定性由次级键维系D．三级结构是单体蛋⽩质或亚基的空间结构E．三级结构是各个单键旋转⾃由度收到各种限制的结果2. 蛋⽩质变性是由于A.蛋⽩质⼀级结构的改变B. 蛋⽩质亚基的解聚C.蛋⽩质空间构象的破坏E. 蛋⽩质⽔解3. DNA的解链温度指的是A．A260nm达到最⼤值时的温度B．A260nm达到最⼤值的50％时的温度C．DNA开始解链时所需要的温度D．DNA完全解链时所需要的温度F．A280nm达到最⼤值的50％时的温度4. 酶原激活作⽤在酶分⼦结构上的变化是：A.⼀级结构的改变B.空间构象的改变C.酶蛋⽩亚基的聚合D.酶蛋⽩与特异的辅基结合形成全酶E.⼀级结构和空间构象同时发⽣改变5.．在下列关于三羟酸循环及氧化磷酸化中能产⽣ATP最多的步骤是A.苹果酸－草酰⼄酸B.琥珀酸－苹果酸C.柠檬酸－异柠檬酸D.异柠檬酸－α－酮戊⼆酸E.α－酮戊⼆酸－琥珀酸6．糖原的1个葡萄糖经糖酵解可⽣成⼏个ATPA.1B.2C.3D.4E.57.．糖酵解途径最重要的调控位点是A.磷酸化酶B.乳酸脱氢酶C.葡萄糖激酶D.丙酮酸激酶E.磷酸果糖激酶-18. 在胞液中，乳酸脱氢⽣成的NADHA.可直接进⼊呼吸链氧化B.在线粒体内膜外侧使琥珀酸－苹果酸酸⽢油转变成磷酸⼆羟丙酮后进⼊线粒体C.仅仅需要内膜外侧的磷酸⽢油脱氢酶的催化后即可直接进⼊呼吸链D.经α－磷酸⽢油穿梭作⽤后可进⼊琥珀酸氧化呼吸链E.上述各条都不能使胞液中的NADH进⼊呼吸链氧化9. 磷酸戊糖途径A.是体内产⽣CO2的主要来源B. 可⽣成NADPH供合成代谢需要C．是体内⽣成糖醛酸的途径D．饥饿时葡萄糖经此途径代谢增加E．可⽣成NADPH，后者经电⼦传递链可⽣成ATP10. ⽣物体内能量转换的基本⽅式是A.三羧酸循环B.底物⽔平磷酸化C.氧化磷酸化D.ATP-ADP循环E.磷酸肌酸脱磷酸⼆、填空（20分）1.DNA双螺旋结构稳定的维系横向维系，纵向则靠维持。

浙江大学2007-2008学年春季学期 《微积分Ⅱ》课程期末考试试卷一 、填空题（每小题5分.共25分.把答案填在题中横线上） 1.点M （1,－1, 2）到平面2210x y z -+-=的距离d = . 2.已知2a =,3b =,3a b ⋅=,则a b += . 3.设(,)f u v 可微.(,)yxz f x y =,则dz = .4.设()f x 在[0.1]上连续.且()f x ＞0, a 与b 为常数.()}{,01,01D x y x y =≤≤≤≤,则()()()()Daf x bf y d f x f y σ++⎰⎰= .5.设(,)f x y 为连续函数.交换二次积分次序2220(,)x x dx f x y dy -=⎰⎰.二 、选择题（每小题5分.共20分.在每小题给出的四个选项中只有一个是符合题 目要求的.把所选字母填入题后的括号内）6.直线l 1:155121x y z --+==-与直线l 2:623x y y z -=⎧⎨+=⎩的夹角为 （A ）2π . （B ）3π . （C ）4π . （D ）6π. [ ] 7.设(,)f x y 为连续函数.极坐标系中的二次积分cos 2d (cos ,sin )d f r r r r πθθθθ⎰⎰可以写成直角坐标中的二次积分为（A）100(,)dy f x y dx ⎰⎰ （B）100(,)dy f x y dx ⎰⎰（C）10(,)dx f x y dy ⎰⎰（D）10(,)dx f x y dy ⎰⎰[ ]8.设1, 02()122, 12x x f x x x ⎧≤≤⎪⎪=⎨⎪-≤⎪⎩ ()S x 为()f x 的以2为周期的余弦级数.则5()2S -=（A ）12. （B ）12-. （C ）34. （D ）34-. [ ] ＜9.设,)(0,0),(,)0, (,)(0,0),x y f x y x y ≠==⎩则(,)f x y 在点O (0,0)处（A ）偏导数存在.函数不连续 （B ）偏导数不存在.函数连续（C ）偏导数存在.函数连续 （D ）偏导数不存在.函数不连续 [ ] 三、解答题10.（本题满分10分）求曲线L ：2222222393x y z z x y⎧++=⎪⎨=+⎪⎩在其上点M （1.－1.2）处的切线方程与法平面方程.11.（本题满分10分）设F 可微.z 是由F (x y -,,)0y z z x --=确定的可微函数.并设23F F ''≠.求z zx y∂∂+∂∂. 12.（本题满分10分）设D 是由曲线3y x =与直线y x =围成的两块有界闭区域的并集.求2[e sin()]d xDx y σ++⎰⎰. 13.（本题满分10分）求空间曲线L ：222920335x y z x y z ⎧+-=⎨++=⎩上的点到xOy 平面的距离最大值与最小值.14.（本题满分10分）设平面区域D ={}(,)01,01x y x y ≤≤≤≤.计算二重积分22 1 d Dx y σ+-⎰⎰.15.（本题满分5分）设当y ＞0时(,)u x y 可微.且已知222222(,)()(2)y x du x y xy dx x y y dy x y x y=++-++++. 求(,)u x y .浙江大学2007－2008学年春季学期《微积分II 》课程期末考试试卷答案一、填空题（每小题5分.共25分） 1．231421=-++=d .2．22()()2496a b a b a b a b a b +=+⋅+=++⋅=++=3．()()dy xy f x x f dx y y f yx f dz x y x y 121211ln ln --'+⋅'+'+⋅'=4．()()()()()()()()⎰⎰⎰⎰++=++=D Dd x f y f x bf y af d y f x f y bf x af I σσ. ()()⎰⎰+=+=+=∴Db a I b a d b a I 21,2σ.5．()()2220111,,x x dx f x y dy dy f x y dx --=⎰⎰⎰⎰或 ()0111,dy f x y dx -⎰⎰或 ()1101,dy f x y dx -⎰⎰.二、选择题（每小题5分.共20分） 6．选（B ）.l 1的方向向量{}1,2,1-.l 2的方向向量{}2,1,1--.{}{}3,2163662,1,11,2,1cos πθθ===--⋅-=.7．选（D ）. 积分区域(){}0,,22≥≤+=y x y x y x D .化成直角坐标后故知选（D ）.8．选（C ）. 511111113()()()((0)(0))(1)222222224S S S f f -=-==-++=+=.9．选（A ）. ()()0000,0lim0,0,00x y x f f x→-''===.偏导数存在. 取kx y =.()4411lim,lim kk kk kx x f x x +=+=→→随k 而异.所以不连续．三、解答题（10～14每题10分.15题5分.共55分） 10．由L .视x 为自变量.有⎪⎩⎪⎨⎧=-+=++.0226,0264dx dz z dx dy y x dx dz z dx dy y x 以()()2,1,1,,-=z y x 代入并解出dxdzdx dy ,.得 87,45==dx dz dx dy . 所以切线方程为87245111-=+=-z y x .法平面方程为()()()57112048x y z -+++-=.即0127108=-++z y x .11．133212232332,,1y x z z F F F F F F F F z z z z x F F F y F F F x y F F ''''''''--+∂∂∂∂=-=-=-=-+==''''''''∂-+∂-+∂∂-.12．D 在第一象限中的一块记为D 1.D 在第三象限中的一块记为D 2.()()()()⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰+++++=++2122122sin sin sin D D DD x D x x d y x d y x d e d e d y x eσσσσσ.32222312101xx x x x xxxD D e d e d dx e dy dx e dy σσ-+=+⎰⎰⎰⎰⎰⎰⎰⎰ ()()()()222210103333011x x x x x x e dx xx e dx x x e dx xx e dx -=-+-=-+-⎰⎰⎰⎰()2111130021()112x u u u u x x e dx e du ue du e ue e e e =-=-=---=--=-⎰⎰⎰()()()()3312101sin sin sin sin x x xxD D x y d x y d dx x y dy dx x y dy σσ-+++=+++⎰⎰⎰⎰⎰⎰⎰⎰()()()()103301cos cos cos cos x x x x dx x x x x dx -⎡⎤⎡⎤=-+-+-+-+⎣⎦⎣⎦⎰⎰ ()()()()13301cos cos cos cos 0x x x x dx x x x x dx ⎡⎤⎡⎤=-+-+++-+=⎣⎦⎣⎦⎰⎰ 所以.原式2-=e .13．L 上的点到平面xoy 的距离为z .它的最大值点.最小值点与2z 的一致.用拉格朗日乘数法.设()()()53329,,,,2222-+++-++=z y x zy x z z y x F μλμλ.求偏导数.并令其为零有：20F x x λμ∂=+=∂.1830F y x λμ∂=+=∂. 2430F z z z λμ∂=-+=∂.22920Fx y z x∂=+-=∂ . 3350Fx y z μ∂=++-=∂ ． 解之得两组解()()1215,,(1,,1);,,(5,,5)33x y z x y z ==--. 所以当31,1==y x 时.1=z 最小；当35,5-=-=y x 时.5=z 最大.14．将分成如图的两块.41的圆记为D 1.另一块记为D 2()⎰⎰⎰⎰--=-+DD d y x d y x 1222211σσ+()⎰⎰-+2122D d y x σ ()()()σσσd y x d y x d y xD DD ⎰⎰⎰⎰⎰⎰-+--++--=11111222222()()()()1222211122220211211211()43343D Dx y d x y d d r rdr dy xy dx πσσθππ=--++-=-++-=+-+=-⎰⎰⎰⎰⎰⎰⎰⎰15．由()222222,()(2)y x du x y xy dx x y y dy x y x y =++-++++.有222xy y x y x u ++=∂∂.从而知()()y y x y x y x u ϕ++=2221arctan,.又由y y x yx x y u 2222+++-=∂∂.推知 ()22222221()xx y x y y x y y x x y y ϕ-'++=-++++. ()()22,y y y y C ϕϕ'==+所以.()2221,arctan2x u x y x y y C y =+++. 注：若用凑的办法亦可：222222()(2)y x xy dx x y y dy x y x y++-++++()()22222211221()ydx xdy ydx xdy xy ydx xdy ydy d xy dy x x y y y--=+++=++++ ()221(arctan)2x d xy y y =++ 所以.()C y y x y x y x u +++=22221arctan,． ()()u f u F ='．浙江大学2006–2007学年春季学期 《 微积分Ⅱ 》课程期末考试试卷开课学院： 理学院 考试形式：闭卷 考试时间： 年 月 日 所需时间：120 分钟 考生姓名： _____学号： 专业： ________一、 填空题（每小题5分.满分30分） 1． 直线63321-==+z y x 在平面0522=--+z y x 上的投影直线方程为.2． 数量场2),,(zye z y x g x +=在)0,3,1(P 点的梯度为 .=u函数)ln(),,(22z y x z y x f ++=在P 点沿u的方向导数为 .3． 设ϕϕ,),2,3(),,(f y x x u u x f z+== 具有二阶连续偏导数.则=∂∂∂yx z 2.4． 设}1,11|),{(3≤≤≤≤-=y x x y x D.则=+⎰⎰+Dy xy x e y x x d d )(222.5． 已知曲面1=z y x 与椭球面193222=++z y x 在第一卦限内相切.则切点坐标为 .公共切平面方程为.6． 设函数⎪⎩⎪⎨⎧<≤<≤=121,210,)(2x x x x x f .∑∞=+=10cos 2)(n n x n a a x S π.其中,2,1,0,d cos )(210==⎰n x x n x f a n π.则.)27(=S二、 （满分10分）求直线 ⎩⎨⎧=-++=-+-022012z y x z y x 绕x 轴旋转一周所得的旋转曲面方程.1002 22dd x yex y.三、（满分10分）计算⎰⎰-四、 (满分15分)已知),(y x z z =由方程013=++zxe z y 确定.试求1022==∂∂y x x z.五、 （满分15分）设平面),,(,1:z y x d y x =+π为曲线⎪⎩⎪⎨⎧=++=++014222z y x z y x 上的点),,(z y x 到平面π的距离.求),,(z y x d 的最大.最小值 .六、 (满分15分)如图是一块密度为ρ（常数）的薄板的平面图形（在一个半径为R 的半圆直 径上拼上一个矩形.矩形的另一边为h ）,已知平面图形的形心位于原点(0, 0). 试求：1. 长度 h ；2.薄板绕x 轴旋转的转动惯量.七、 (满分5分) 求证:当0,1≥≥s t 时.成立不等式 s e t t t ts +-≤ln .参考解答：一．1．⎩⎨⎧=--+=+-0522043z y x z y x ； 2. 21},0,,3{e e ；3. )3(2))(3(2222122222122212ϕϕϕϕϕϕ''+''⋅'+'+'⋅'⋅''+'''f f f ； 4.;32 5. ;03313,3,1,31=-++⎪⎭⎫⎝⎛z y x 6. 83.二．直线：t z t y t x -=-==1,1,曲面上点→),,(z y x P 直线上点00000001,1),,,(x z x y z y x -=-=22222020220)1()1(,,x x z y z y z y x x -+-=+⇒+=+=则旋转曲面方程：222)1(2x z y -=+三．⎰⎰10222d d xy ex y -⎰⎰⎰-==--212212220142)d 41(d d y y e x e y 2y yy2120202020221d d d d 212212212212212------=-+=+=⎰⎰⎰⎰e y e ey y e e y y e yy y y y四．,1)1,0(-=z ,032=∂∂++∂∂⋅x z xe e x z z y z z ex z y x 3110-=∂∂∴== ,02632222222=∂∂+⎪⎪⎭⎫ ⎝⎛∂∂+∂∂+⎪⎪⎭⎫ ⎝⎛∂∂⋅+∂∂⋅x z xe x z xe x z e x z z y x z z y z z z 2102294ex zy x =∂∂∴== 五．|1|21),,(-+=y x z y x d )14()()1(2222-++++++-+=z y x z y x y x L μλ⎪⎪⎪⎪⎩⎪⎪⎪⎪⎨⎧=-++='=±===++='==+='-==⇒≠=++-+='=⇒==++-+='014,01302,002)1(20,002)1(22223231221z y x L z y x z y x L x z L xz x y y y x L x y x L z y xμλμλμμλλμμλ，无解最小距离：2236),,(323131-=-d .最大距离：2236),,(323131+=--d六．形心：01,0=⇒==⎰⎰⎰⎰DDxdxdy xdxdyx y σ即0d cos d d d 220=⋅+⎰⎰⎰⎰---ππθθRhRRr r r y x xR h R h R 320312)21(232=⇒=⋅+-⋅ ⎰⎰=Dxdxdy y I 2302202)832(d θsin d d d 22R R h r r r y y x RhRR πθππ+=⋅+=⎰⎰⎰⎰--- 七．设0)0,1(,ln ),(=-+-=F ts e t t t s t F s.ln ,0),(t s e t t e s t F s s s ==⇒=-=' 且对固定的1>t . 当,0),(,ln 0<'<<s t F t s s 当,0),(,ln >'>s t F t ss所以.t s ln =取得最小值且为0.则 0),(≤s t F .即s e t tt ts +-≤ln1、已知22(,)yf x y x y x +=-,则=),(y x f _____________.2、已知,则=⎰∞+--dx e x x21___________.π=⎰∞+∞--dx e x 23、函数22(,)1f x y x xy y y =++-+在__________点取得极值. 4、已知y y x x y x f arctan )arctan (),(++=,则=')0,1(x f ________.5、以x e x C C y 321)(+=(21,C C 为任意常数)为通解的微分方程是____________________. 6 知dxexp ⎰∞+- 0)1(与⎰-ep x x dx11ln 均收敛,则常数p 的取值范围是( c ).(A) 1p > (B) 1p < (C) 12p << (D) 2p >7 数⎪⎩⎪⎨⎧=+≠++=0 ,0 0,4),(222222y x y x y x x y x f 在原点间断,是因为该函数( b ).(A) 在原点无定义 (B) 在原点二重极限不存在 (C) 在原点有二重极限,但无定义(D) 在原点二重极限存在,但不等于函数值 8、若2211x y I +≤=⎰⎰,22212x y I ≤+≤=⎰⎰,22324x y I ≤+≤=⎰⎰,则下列关系式成立的是( a).(A)123I I I >> (B)213I I I >> (C)123I I I << (D)213I I I <<9、方程xe x y y y 3)1(596+=+'-''具有特解( d ). (A) b ax y += (B) x e b ax y 3)(+= (C) x e bx ax y 32)(+= (D) x e bx ax y 323)(+=10、设∑∞=12n na收敛.则∑∞=-1)1(n nna ( d ).(A) 绝对收敛 (B) 条件收敛 (C) 发散 (D) 不定 一、填空题(每小题3分,共15分)1、2(1)1x y y -+. 2、、)32,31(-. 4、1. 5、"6'0y y y -+=. 11、求由23x y =,4=x ,0=y 所围图形绕y 轴旋转的旋转体的体积.解：32y x =的函数为23,0x y y =>。