江苏省2013年专转本高数真题及答案

2013年江苏省普通高校“专转本”统一考试Part I Reading Comprehension (共20小题，每小题2分，共40分)Directions: There are 4 passages in this part. Each passage is followed by some questions or unfinished statements. For each of them there are 4 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.When Mrs. Joseph Groeger died recently in Vienna, Austria, people asked the obvious question. “W hy did she live to be 107?” Answers were provided by a survey conducted among 148 Viennese men and women who had reached the age of 100. What was surprising was that the majority had lived most of their lives in cities. Although cities are often regarded as unhealthy places, city living provides benefits that country living may lack. One factor seems to be important to the longevity(长寿) of those interviewed.This factor is exercise. In the cities it is often faster to walk short distances than to wait for a bus. Even taking public transportation often requires some walking. Smaller apartment houses have no elevators, and so people must climb stairs. City people can usually walk to local supermarkets. Since parking spaces are hard to find, there is often no choice but to walk.On the other hand, those who live in the country and suburbs do not have to walk every day. In fact, the opposite is often true. To go to school, work, or almost anywhere else, they must ride in cars.1.The Vienna survey may help to explain __________.A.the complaints of people in apartment housesB.the causes of Mrs. Groeger’s deathC.the longevity of people like Mrs. GroegerD.the image of cities in general2.The second paragraph lists some ___________.A.disadvantage of walkingB.occasions for walking in city lifeC.problems of city livingments made by city people3.To reach the third floor of a building, it would probably be most healthful ___________.A.to take the elevatorB.to walk up the stairsC.to ride in a carD.to find an alternative walking4.People who live in the country usually do more driving than walking because ___________.A.they don’t live near business areasB.they don’t need exerciseC.they don’t like walkingD.they can’t afford to take the bus5.We can draw a conclusion from this passage that _________.A.air pollution is not seriousB.anyone can live to be 107C.country people should move to citiesD.walking is a healthful exercisePassage TwoQuestions 6 to 10 are based on the following passage.The Bedouin people, a tribe living in the deserts of Arabia and Northern African, think most highly of people who show loyalty. To them loyalty does not mean that one is devoted to a country, a place, or a leader. Loyalty means being faithful to one’s family and tribe.The Bedouin people are proud of their ancestors(祖先). They do not admire a hero from an ordinary or poor family as much as one who comes from an honored family. They especially respect those who have inherited(继承) a good family name and then have passed it on to their children.A man’s position among his people depends upon his ancestors, relatives, and fellow tribesmen. If they are honored, he is also honored. If they are disagreed, he, too, is disgraced. Therefore, one carefully guards the honor of his family and his tribe.A man can protect his family’s honors by being brave and generous and by giving protection to those who ask for it. He also guards it by carefully watching the women of his family.A Bedouin woman cannot bring honor to her family, but she can bring disgrace. Even if a woman only looks as if she has done something wrong, she may be killed. The honor of her family depends upon her virtue(美德).6. The best title for this passage can be __________.A. Life of the Bedouin PeopleB. Women’s Position in the Bedouin FamilyC. Bedouin People’s Idea of Loyalty and HonorD. How to Guard Honors7. The Bedouin people think most highly of one who is ______.A. devoted to his country and his leaderB. from an ordinary familyC. proud of his ancestor and relativesD. faithful to his family and tribe8. It is learned from the passage that ___________.A. a hero from an honored family is more admiredB. one can not live in the tribe without a good family nameC. all the Bedouin people are kind, brave and generousD. family traditions can never be passed to children9. The word “disgraced” (Para. 3) means “ __________”.A. disappearedB. disappointedC. dishonoredD. discouraged10. It is implied in the passage that the Bedouin women are __________.A. treated as the equals of menB. respected for what they doC. not interested in positionD. not respected as much as menPassage ThreeQuestions 11 to 15 are based on the following passage.Up until about 100 years ago, newspaper in the United States attracted only the most serious readers. They used no illustrations (插图) and the articles were about politics or business.Two men changed that---Joseph Pulitzer of the New York World and William Randolph Hearst of the New York Journal. Pulitzer bought the New York World in 1883. He changed it from a traditional newspaper into a very exciting one overnight. He added lots of illustrations and he told his reporters to write articles on every crime they could find. And they did. One woman reporter even pretended she was mad and was sent to a hospital. She then wrote many articles about the poor treatment of patients in those hospitals where madmen were kept. In 1895, Hearst came to New York from California. He wanted the Journal to be more exciting than the World. He also wanted it to be cheaper, so he lowered the price by a penny. Hearst attracted attention because his newspaper heading were bigger than any one else’s. He often said, “Big print makes big news.”Pulitzer and Hearst did anything they could to sell newspaper. For example, Hearst sent Frederic Remington, the famous illustrator, to draw pictures of the Spanish-American War. When he got there, he told Hearst that no fighting was going on. Hearst answered, “ You provide the pictures. I’ll provide the war.11. About a century ago, American newspaper carried news about __________.A. all kinds of exciting newsB. crimes and mad peopleC. the poor treatment of patientsD. serious matters only12. According to Paragraph 2, which of the following statements is TRUE?A. Pulitzer and a woman news reporter changed the world overnight.B. Pulitzer bought the Journal with the help of his reporter.C. A woman news reporter pretended to be mad.D. A woman reporter became mad and was sent to a hospital.13. Hearst attracted readers’ attention by ______________.A. using bigger headingsB. reporting big war newsC. raising the price of the newspaperD. reporting politics and business only14. __________ was asked to draw war pictures.A. HearstB. RemingtonC. PulitzerD. A woman reporter15. This passage tells us that ________.A. Pulitzer and Hearst were the most famous reporters of the United StatesB. newspaper owners sometimes hired mad men as their news reportersC. the world and the journal were not popular newspaper at that timeD. Pulitzer and Hearst used every possible means to win over each otherPassage FourQuestions 16 to 20 are based on the following passage.It is hard to imagine life without Arabic numbers(数字). No other number system ever invented has a simple way to write a number like 1984. In Roman numeral, it comes out like this MCMLXXXIV. Imagine how difficult it would be for a school pupil to remember it!The Arabs, however, call their numbers “Indian numerals”. This is probably because they got the original idea a long time ago from India. There were no printed newspapers or scientific magazine then, so mathematics traveled along the same routes that silk did, as businessmen sold and bought goods. The Arabs took the new numbers and made improvements that quickly led to advances in technology.The achievement of Arabic technology can be seen from a passage in a 10th-century book about an Arabic city:” The neighboring country showed the results of careful work in agriculture, trade, and industry. Rice and many other plants were grown in fields with water supplied by Arab engineers. The city included 900 public baths, stone-covered streets, fountains, and beautiful buildings.” At this time, London was a small town with dirty streets, and Berlin was a farming village. In fact, cities outside the Arab world did not reach the same level of social development until centuries later.16. This passage tells us that the Arabs were a people _______________.A. inventing Roman numeralsB. having advanced technology long agoC. growing corn and many other plants thenD. supplying other countries with colorful silk17. It takes much more efforts to ___________.A. learn Roman numerals than Arab numeralsB. introduce numerals through printed materialsC. make great use of silk roadsD. teach businessmen numerals18. The Arabs call their numerals “ Indian numerals” because _________.A. the numerals are mainly used in ChinaB. they copied numerals from IndiansC. Arabic numerals were first used by Indian businessmenD. they got some ideas about number from the ancient Indians19. According to the passage, the Arab people ___________.A. did business with the IndiansB. learned some ideas about business from IndiansC. built public baths in London in the 10th centuryD. traveled a long way to learn mathematics in Rome20. From a 10th-century book, we can learn that ___________.A. Berlin was formerly a dirty with sufficient rice and plantsB. London was under the rule of the Romans in the 10th centuryC. the Arab world was more advanced than other countries aroundD. London chiefly developed industry ten centuries agoPart II Vocabulary and Structure （共40小题，每小题1分，共40分）Directions: There are 40 incomplete sentences in this part. For each sentence there are 4 choices marked A, B, C and D. Choose the ONE that best completes the sentence. Then mark the corresponding letter on the Answer Sheet with a single line through the center.21. Jessica ought to stop working; she has a headache because she __________ too long.A. readsB. is readingC. has been readingD. had read22. __________ us harm or good remains to be seen in the near future.A. If it will doB. Whether it will doC. That it will doD. What it will do23. Able-bodied people just don’t realize how difficult it is _________ good jobs.A. of disabled people to getB. of disabled people gettingC. for disabled people to getD. for disabled people getting24. I’ll go camping with you if I ___________ free tomorrow.A. amB. will beC. beingD. was25. After the concert, the rock music star ___________ hundreds of fans outside the theater.A. surrounded byB. was surrounded byC. being surrounded byD. is surrounded by26. That book is really worth _________ if you’re going to take the final examination.A. to readB. reading C .to be read D. being read27. __________ surprised me most was _________ such a little boy of seven could play the violin so well.A. That; whatB. That; whatC. What; thatD. What; what28. Staying in a hotel cost ________ renting a room in an apartment for a week.A. twice many thanB. twice as much asC. twice the same asD. twice as many as29. It is ________ that we’d like to go out for a walk.A. a lovely dayB. too lovely a dayC. so lovely a dayD. such lovely a day30. I can’t find my iPhone 5 in the dormitory. It __________ by somebody.A. must have taken awayB. may have taken awayC. may take awayD. must have been taken away31. When Mr. Smith got there, he found that about ___________ of his friends had been incited.A. twenty-thirdB. two-thirdC. two-thirdsD. twenty-thirds32. In 2012, Mo Yan became the first Chinese writer _________ won the Nobel Prize for Literature.A. thatB. whomC. whoseD. which33. I didn’t see you at the meeting. If you had come, you _________ our sales manager.A. would meetB. would have metC. will have metD. had met34. Ladies and gentlemen! It’s my pleasure __________ Mr. Li, our new Premier to you all.A. introducingB. introduceC. to be introducingD. to introduce35. The audience burst into applause ___________ the singer finished her song.A. after momentB. in a momentC. the momentD. for a moment36. “ Where is the girl we talked about yesterday?”“Look! Here ___________.”A. she comes B . comes she C. she came D. came she37. The number of employees at the factory __________ to a minimum so as to lower production costs.A. is cuttingB. are cuttingC. has been cutD. have been cut38. Many young people nowadays make _________ a rule to buy flower and chocolate for their lovers on Valentine’s Day.A. themselvesB. itC. themD. this39. Dr. Green, together with his wife and two daughters, __________ to arrive in Nanjing tonight.A. will beB. is C .are D. are going40. His father has been suffering from lung cancer for half a year, so the doctor Suggest that he ___________ an operation as soon as possible.A. takesB. would takeC. tookD. should take41. Being out of work, Jane can no longer __________ her friends to dinners as she used to.A. urgeB. treatC. attractD. apply42. It’s a great pity that the girl _________ halfway and dropped out of the competition.A. gave offB. gave outC. gave overD. gave up43. The fact that something is cheap doesn’t _________ mean it is of low quality.A. especiallyB. practicallyC. necessarilyD. specially44. In learning English, many students ___________ an e-dictionary to a paper one.A. likeB. expectC. preferD. respect45. The personnel manager will __________ the applications and pick out the best.A. look upB. look afterC. look outD. look through46. Seventy percent approved ________ the way she handled the situation while thirty percent disapproved.A. withB. aboutC. ofD. up47. The price of houses goes up again, partly _________ the requirement of young people.A. in result ofB. withC. becauseD. due to48. They decided to ___________ the contract because a number of the conditions had not been met.A. attendB. argueC. controlD. cancel49. __________ your English gets better, you’ll begin to feel more comfortable studying in Canada.A. OnceB. As far asC. WhereasD. For50. Although I told him to be careful over and over again, he never took any__________ of what I said.A. attentionB. warningC. noticeD. observation51. Freshmen are usually __________ about everything when they first come to the college.A. troublesomeB. curiousC. differentD. familiar52. Salina came home so thin and weak that her family could hardly ________ her.A. remindB. recognizeC. rememberD. understand53. The manager explained that they didn’t__________ people to bring theirs pets into therestaurant.A. allowB. letC. makeD. have54. ___________ I go, I always seem to bump into him.A. WhateverB. WhereverC. HoweverD. Somehow55. Our working hours are _________, we can go to work in the morning or in the afternoon.A. floatingB. efficientC. flexibleD. personal56. To a great __________, the way you learn at college is different from that when you were in high school.A. amountB. extentC. reasonD. goal57. A skilled tennis player knows how to __________ his compe titor’s weakness so as to get the final victory.A. beatB. take risk ofC. defeatD. take advantage of58. Try this camera, you’ll find it can be ___________ to take pictures in cloudy conditions.A. adjustedB. treatedC. adoptedD. removed59. Being short of food and fuel, we are _________ with a very difficult situation.A. handledB. madeC. facedD. encouraged60. John was late for class again, but he couldn’t _________ his lateness.A. count forB. account forC. count onD. account onPart III Cloze（共20小题，每小题1分，共20分）Directions: There are 20 blanks in the following passage. For each blank there are 4 choices marked A, B, C and D. You should choose the ONE that best fits into the context. Then mark the corresponding letter on the Answer Sheet with a single line through the center.Smoking is considered dangerous to the health. Our tobacco-seller Mr. Johnson, therefore always asking his customers if they are very young, whom the cigarettes are bought __61__. One day, a little girl whom he had never seen before walked boldly __62__ his shop and demanded twenty cigarettes. She had the large __63__ of money in her hand and seemed very ___64___ herself. Mr. Johnson was so __65___ by her confident manner that he __66___ to ask his usual question. ___67___, he asked her what kind of cigarettes she wanted. The girl ___68___ readily and handed him the money. While he was giving her the __69___, Mr. Johson said laughingly that as she was so __70___ she should ___71___ the packet(小盒) in her pocket in __72___ a policeman saw it. However, the little girl did not seem to ___73____ this very funny. Without ___74__ she took the __75___ and walked towards the door. Suddenly she stopped, turned __76___ and looked steadily at Mr. Johnson. __77___ was a moment of silence and the tobacco-seller wondered ___78____ she was going to say. And at once, in a clear, joking ___79__, the girl declared, “___80__ dad is a policeman,” and with that she walked quickly out of the shop.61. A. with B. to C. by D. for62. A. with B. for C. into D. down63. A. amount B. quality C. bits D. part64. A. since B. of C. without D. against65. A. worried B. interested C. surprised D. pleased66. A. forgot B. came C. went D. dreamed67. A. Therefore B. Instead C. Somehow D. Anyhow68. A. Replied B. Screamed C. spoke D. talked69. A. change B. warning C. cheque D. advice70. A. old B. young C. amazed D. excited71. A. hide B. throw C. tip D. take72. A. way B. peace C. case D. end73. A. search B. get C. pass D. find74. A. look B. looked C. smiling D. smiled75. A. packet B. clothes C. knife D. bill76. A. way B. over C. round D. aside77. A. They B. She C. There D. He78. A. where B. what C. when D. if79. A. words B. voice C. sound D. statement80. A. The B. For C. As D. MyPart IV Translation(共35分)Section A (共5小题，每小题4分，共20分)Directions: Translate the following sentences into Chinese. You may refer to the corresponding passages in Part I.81. Although cities are often regarded as unhealthy places, city living provides benefits that country living may lack. (Passage One)82. A man’s position among his people depends upon his ancestors, relatives, and fellow tribesmen. (Passage Two)83. She then wrote many articles about the poor treatment of patients in those hospitals where madmen were kept. (Passage Three)84. The Arabs took the new numbers and made improvements that quickly led to advances in technology. (Passage Four)85. In fact, cities outside the Arab world did not reach the same level of social development until centuries later. (Passage Four)Section B (共5小题，每小题3分，共15分)Directions: Translate the following sentences into English.86. 根据以往的经验，他可能赢得这场比赛。

江苏省2013年普通高校“专转本”选拔考试高等数学试题卷（二年级）注意事项：1、本试卷分为试题卷和答题卡两部分，试题卷共3页，全卷满分150分，考试时间120分钟．2、必须在答题卡上作答，作答在试题卷上无效。

作答前未必将自己的姓名和准考证号准确清晰地填在试题卷和答题卡上的指定位置。

3、考试结束时，须将试题卷和答题卡一并交回。

一、选择题（本大题共6小题，每小题4分，满分24分。

在下列每小题中，选出一个正确答案，请在答题卡上将所选项的字母标号涂黑）1、当0®x 时，函数()ln(1)f x x x =+-是函数2)(x x g =的( ) A.高阶无穷小 B.低阶无穷小 C.同阶无穷小 D.等价无穷小2、曲线22232x x y x x +=-+的渐近线共有( ) A. 1条B. 2条C. 3条D. 4条3、已知函数sin 20()011x x xf x xx x ì<ïï=í>ï+-î，则点0x =是函数)(x f 的A 、跳跃间断点B 、可去间断点C 、无穷间断点D 、连续点4、设1()y f x =，其中f 具有二阶导数，则22d ydx=A. 231121()()f f x x x x ¢¢¢-+B. 431121()()f f x x x x ¢¢¢+C. 231121()()f f x x xx ¢¢¢-- D. 431121()()f f x x xx¢¢¢-5、下列级数中收敛的是A 、211n n n¥=+åB 、1()1nn n n ¥=+åC 、1!2nn n ¥=åD 、13nn n¥=å6、已知函数)(x f 在点1x =处连续，且21()1lim 12x f x x ®=-，则曲线()y f x =在点(1,(1))f 处的切线方程为A. 1y x =-B. 22y x =-C. 33y x =-D. 44y x =- 二、填空题（本大题共6小题，每小题4分，共24分）分）7、设函数1sin 0()0x x f x xa x ì¹ï=íï=î在点0=x 处连续，则常数a = ▲ ． 8、已知空间三点(1(1,1,1),,1,1),(2,3,4),(3,4,5)A B C ，则ABC D 的面积为的面积为▲ ． 9、设函数)(x y y =由参数方程2311x t y t ì=+ïí=-ïî所确定，则221x d ydx == ▲ ．10、设向量®®b a ,互相垂直，且，，23==®®b a ，则=+®®b a 2 ▲ ． 11、设1lim()xx a x e a x ®+=-，则常数=a ▲ ．12、幂级数12nnn x n¥=å的收敛域为的收敛域为▲ ． 三、计算题（本大题共8小题，每小题8分，共64分）分）13、求极限01lim ln(1)x x e x x ®éù-êú+ëû． 14、设函数(,)z z x y =由方程3331z xy z +-=所确定，求dz 及22z x ¶¶． 15、求不定积分2cos 2x xdx ò．16、计算定积分22024dx x+-ò．17、设函数223(,)x yz f x e+=，其中函数f 具有二阶连续偏导数，求2z y x¶¶¶．18、已知直线10330x y z x y z -+-=ìí--+=î平面P 上，又知直线23132x t y t z t =-ìï=+íï=+î与平面P 平行，求平面P 的方程．方程．19、已知函数()y f x =是一阶微分方程dyy dx=满(0)1y =的特解，的特解，求二阶常系数非齐次线性求二阶常系数非齐次线性微分方程32()y y y f x ¢¢¢-+=的通解．的通解．20、计算二重积分Dxdxdy òò，其中D 是由曲线24(0)y x x =->与三条直线,3,0y x x y ===所围成的平面闭区域．所围成的平面闭区域．四、综合题（本大题共2小题，每小题10分，共20分）分） 21、设平面图形D 由曲线2x y =，y x =-与直线1y =围成，试求：围成，试求：（1）平面图形D 的面积；的面积；（2）平面图形D 绕x 轴旋转一周所形成的旋转体的体积．轴旋转一周所形成的旋转体的体积．22、已知21132()(95)xF x t t dt =-ò是函数()f x 的一个原函数，的一个原函数，求曲线求曲线)(x f y =的凹凸区间与拐点．拐点．五、证明题（本大题共2小题，每小题9分，共18分）分）23、证明：当1x >时，2(1ln )21x x +<-．24、设函数()f x 在[,]a b 上连续，证明：函数2()[()()]a bbaaf x dx f x f a b x dx +=++-òò．江苏省2013年普通高校“专转本”统一考试高等数学（二年级） 试卷答案一、选择题（本大题共6小题，每小题4分，共24分）分） 1、C 2、C 3、B 4、B 5、D 6、A 二、填空题（本大题共6小题，每小题4分，共24分）分）7、0 8、629、34 10、2 11、ln y x x cx =+ 12、11[,)22-三、计算题（本大题共8小题，每小题8分，共64分）分）13、原式=20001ln(1)ln(1)1lim lim lim ln(1)2x xxx x x x e xe xe x xe x x x x x x®®®+--+-++==+ 2013(1)lim 22xxxx e e xe x ®++++== 14、令32(,,)331,3,3,33x y z F x y z z xy z F y F x F z ¢¢¢=+--===-22222233,,33133111yxz z F F z y y z x x y x dz dx dyxF z z y F z z z z ¢¢¶¶=-=-==-=-=\=+¢¢¶--¶----22222222223()(2)()2211(1)(1)(1)z z y y y z yz z y z x x z z x x x z z z ¶¶¶--¶¶¶¶--=====¶¶¶--- 15、22221111cos 2sin 2sin 2sin 2sin 2cos 22222x xdx x d x x x x xdx x x xd x==-=+òòòò 22111111sin 2cos 2cos 2sin 2cos 2sin 2222224x x x x xdx x x x x x C =+-=+-+ò16、令2sin ,2cos ,0,0;2,2x t dx tdt x t x t p======，则原式=222220000222cos 12cos cos 12(1)22cos 1cos 2cos 2cos22t t t dt dt dt dt t tt t p p p p -===-++òòòò 222002011tan 12222cos2t t dt d t pp p p p=-=-=-òò 17、2232323232212223,(22)36x yx yx yx yz z f ef x f e eef yy x++++¶¶¢¢¢¢¢¢=××=×+××+¶¶¶18、直线方向向量12(1,1,1)(1,3,1)(4,2,2),(3,1,2),S S ®®=-´--=-=-平面P 的法向量12(4,2,2)(3,1,2)(6,2,10),n S S ®®®=´=-´-=-在第一条直线上任取一点(1,1,1)，该点也在平面上，所以平面方程为6(1)(2)(1)10(1)0x y z -+--+-=即3570x y z -+-=19、由dy y dx =得111111,,ln ,,x C C C x x x dy dx dy dx y x C y e e e y e e Ce y y +===+===±=òò，由(0)1y =得1C =,所以xy e =，即212,320,1,322xe r y y r r r y -¢¢¢-+==+==，齐次方程的通解为212xxY C e C e =+.令特解为,,xx xy xAe y Ae xAe **¢==+，,x x xy Ae Ae xAe *¢¢=++代入原方程得：,1x xAe e A -==-，所以通解为212xxxy Y C e C exe ==+-20、原式=333cos 4cos 4422127cos cos (8cos )33cos rd r rdrd d p p p q q qq qqq q q==-òòòò241142(27tan 8sin )(27tan 8sin )933443p p pq q =-=-=-. 四、综合题（本大题共2小题，每小题10分，共20分）分） 21、（1）1312320215(2)(2)333S y y dy y y =+=×+=ò（2）22252221010821[1()][1()]()()42802510xxxx Vx dx dx x xp p pp pp p --=--+-=++-=+=òò22、25233()2(95)1810,f x x x x x x =-=-23()3020f x x x ¢=-，13()20200,f x x-¢¢=-=解得1x =，另外0x =为二导不存在的点，为二导不存在的点，通过列表分析得：通过列表分析得：在(,0),(1,)-¥+¥凸，在(0,1)凹， 拐点为(0,0),(1,8)。

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2013年普通高等学校夏季招生全国统一考试数学(江苏卷)数学Ⅰ试题参考公式：样本数据x 1，x 2，…，x n 的方差2211()n i i s x x n ==-∑，其中11n i i x x n ==∑。

棱锥的体积公式：13V Sh =，其中S 是锥体的底面积，h 为高．棱柱的体积公式：V ＝Sh ，其中S 是柱体的底面积,h 为高．一、填空题:本大题共14小题,每小题5分，共计70分．请把答案填写在答题卡相应位置．．．．．．．上．．．1．（2013江苏，1）函数π3sin 24y x ⎛⎫=+ ⎪⎝⎭的最小正周期为__________．答案：π解析：函数π3sin 24y x ⎛⎫=+ ⎪⎝⎭的最小正周期2ππ2T ==。

2．（2013江苏，2)设z ＝（2－i ）2（i 为虚数单位)，则复数z 的模为__________．答案:5解析：|z ｜＝｜（2－i)2|＝｜4－4i ＋i 2|＝｜3－4i5=＝5。

3．（2013江苏,3）双曲线22=1169x y -的两条渐近线的方程为__________．答案：34y x =±解析：由题意可知所求双曲线的渐近线方程为34y x =±。

4．（2013江苏，4）集合｛－1,0,1｝共有__________个子集．答案：8解析：由于集合{－1,0,1｝有3个元素，故其子集个数为23＝8. 5．（2013江苏，5)下图是一个算法的流程图,则输出的n 的值是__________．答案：3解析：第一次循环后：a ←8，n ←2; 第二次循环后:a ←26，n ←3； 由于26＞20，跳出循环， 输出n ＝3。

江苏省普通高校“专转本”统一考试高等数学专转本高数试卷结构知识分类与历年真题●函数、极限和连续●一元函数微分学●一元函数积分学●向量代数与空间解析几何●多元函数微积分●无穷级数●常微分方程时间排序与参考答案◆2004年高等数学真题参考答案◆2005年高等数学真题参考答案◆2006年高等数学真题参考答案◆2007年高等数学真题参考答案◆2008年高等数学真题参考答案◆2009年高等数学真题参考答案◆2010年高等数学真题参考答案◆2011年高等数学真题参考答案◆2012年高等数学真题参考答案◆2013年高等数学真题参考答案江苏省普通高校“专转本”统一考试高等数学试卷结构全卷满分150分一、单选题（本大题共6小题，每小题4分，满分24分） 二、填空题（本大题共6小题，每小题4分，满分24分） 三、解答题（本大题共8小题，每小题8分，满分64分） 四、综合题（本大题共2小题，每小题10分，满分20分） 五、证明题（本大题共2小题，每小题9分，满分18分）知识分类与历年真题一、函数、极限和连续（一）函数（0401）[](]333,0()0,2x x f x x x ⎧∈-⎪=⎨-∈⎪⎩是（ ） A.有界函数 B.奇函数 C.偶函数 D.周期函数 （0801）设函数)(x f 在),(+∞-∞上有定义，下列函数中必为奇函数的是（ ）A.()y f x =-B.)(43x f x y = C.()y f x =-- D.)()(x f x f y -+= （二）极限（0402）当0→x 时，x x sin 2-是关于x 的（ ）A.高阶无穷小B.同阶无穷小C.低阶无穷小D.等价无穷小（0407）设xx x x f ⎪⎭⎫⎝⎛++=32)(，则=∞→)(lim x f x ．（0601）若012lim 2x x f x →⎛⎫ ⎪⎝⎭=，则0lim 3x xx f →=⎛⎫ ⎪⎝⎭（ ） A.21 B.2C.3D.31 （0607）已知0→x 时，(1cos )a x ⋅-与x x sin 是等价无穷小，则=a ．（0613)计算x →． （0701）若0(2)lim2x f x x→=，则1lim 2x xf x →∞⎛⎫= ⎪⎝⎭（ ） A.41B.21 C.2D.4（0702）已知当0→x 时，)1ln(22x x +是x n sin 的高阶无穷小，而x nsin 又是x cos 1-的高阶无穷小，则正整数=n （ ） A.1B.2C.3D.4（0813）求极限：32lim xx x x →∞-⎛⎫⎪⎝⎭． （0901）已知22lim32x x ax bx →++=-，则常数b a ,的取值分别为（ ） A.2,1-=-=b a B.0,2=-=b aC.0,1=-=b aD.1,2-=-=b a（0907）已知lim 2xx x x C →∞⎛⎫= ⎪-⎝⎭，则常数=C ． （1001）设当0x →时，()sin f x x x =-与()ng x ax =是等价无穷小，则常数,a n 的值为 ( ) A.1,36a n == B.1,33a n == C.1,412a n == D.1,46a n == （1007） 1lim 1xx x x →∞+⎛⎫= ⎪-⎝⎭． （1101）当0→x 时，函数1)(--=x e x f x是函数2)(x x g =的( )A.高阶无穷小B.低阶无穷小C.同阶无穷小D.等价无穷小（1107）已知22lim kxx x e x →∞-⎛⎫= ⎪⎝⎭，则=k _________． （1201）极限1sin 3lim 2sinx x x x x →∞⎛⎫+= ⎪⎝⎭（ ） A.0 B.2 C.3D.5（1301）当0x →时，函数()ln(1)f x x x =+-是函数2()g x x =的（ ） A.高阶无穷小 B.低阶无穷小C.同阶无穷小D.等价无穷小（1310）设10lim xx a x e a x →+⎛⎫=⎪-⎝⎭，则常数a = ． （三）连续（0413）求函数xxx f sin )(=的间断点，并判断其类型． （0501）0=x 是xx x f 1sin )(=的（ ） A.可去间断点B.跳跃间断点C.第二类间断点D.连续点（0513）设()2sin 0()0f x xx F x xa x +⎧≠⎪=⎨⎪=⎩在R 内连续，并满足0)0(=f ，(0)6f '=，求a ． （0602）函数21sin 0()00x x f x xx ⎧≠⎪=⎨⎪=⎩在0x =处（ ） A.连续但不可导B.连续且可导C.不连续也不可导D.可导但不连续（0608）若A x f x x =→)(lim 0，且)(x f 在0x x =处有定义，则当=A 时，)(x f 在0x x =处连续．（0707）设函数1(1)0()20x kx x f x x ⎧⎪+≠=⎨⎪=⎩，在点0=x 处连续，则常数=k ．（0807）设函数21()(1)x f x x x -=-，则其第一类间断点为 ．（0808）设函数0()tan 30a x x f x x x x+≥⎧⎪=⎨<⎪⎩在点0=x 处连续，则a ＝ ．（0902）已知函数423)(22-+-=x x x x f ，则2=x 为)(x f 的（ ）A.跳跃间断点B.可去间断点C.无穷间断点D.震荡间断点（1123）设210arctan ()1010sin 2ax axe x ax x x xf x x e x x ⎧---<⎪⎪⎪==⎨⎪-⎪>⎪⎩，问常数为何值时：（1）0=x 是函数)(x f 的连续点？ （2）0=x 是函数)(x f 的可去间断点？ （3）0=x 是函数)(x f 的跳跃间断点？ （1202）设()2(2)sin ()4x xf x x x -⋅=⋅-，则函数)(x f 的第一类间断点的个数为（ ） A.0 B.1C.2D.3（1207）要使函数()1()12xf x x =-在点0=x 处连续，则需补充定义(0)f =_________．（1303）设sin 20()0xx x f x x ⎧<⎪⎪=⎨>，这点0x =是函数()f x 的（ ）A.跳跃间断点B.可去间断点C.无穷间断点D.连续点（1307）设1sin0()0x x f x xa x ⎧≠⎪=⎨⎪=⎩在点0x =处连续，则常数a = ． 二、一元函数微分学（一） 导数与微分（0403）直线L 与x 轴平行且与曲线xe x y -=相切，则切点的坐标是（ ） A.()1,1B.()1,1-C.()0,1-D.()0,1（0409）设()(1)(2)()f x x x x x n =+++，N n ∈，则=)0('f ．（0415）设函数)(x y y =由方程1=-yxe y 所确定，求22d d x yx=的值．（0502）若2=x 是函数1ln 2y x ax ⎛⎫=-+ ⎪⎝⎭的可导极值点，则常数=a （ ） A.1-B.21C.21- D.1 （0514）设函数)(x y y =由方程cos sin cos x t y t t t =⎧⎨=-⎩所确定，求d d y x 、22d d yx ．（0614）若函数)(x y y =是由参数方程2ln (1)arctan x t y t t⎧=+⎨=-⎩所确定，求d d y x 、22d d yx ．（0708）若直线m x y +=5是曲线232++=x x y 的一条切线，则常数=m ．（0714）设函数)(x y y =由方程xy e e yx=-确定，求d d x yx=、22d d x y x =．（0802）设函数)(x f 可导，则下列式子中正确的是（ ） A.0(0)()lim(0)x f f x f x →-'=- B.000(2)()lim ()x f x x f x f x x→+-'=C.0000()()lim ()x f x x f x x f x x ∆→+∆--∆'=∆D.0000()()lim 2()x f x x f x x f x x∆→-∆-+∆'=∆ （0814）设函数)(x y y =由参数方程sin 1cos x t t y t =-⎧⎨=-⎩（2t n π≠，n Z ∈）所决定，求d d y x 、22d d y x ．（0903）设函数00()1sin 0x f x x x x α≤⎧⎪=⎨>⎪⎩在点0=x 处可导，则常数α的取值范围为（ ） A.10<<αB.10≤<αC.1>αD.1≥α（0914）设函数)(x y y =由参数方程2ln (1)23x t y t t =+⎧⎨=+-⎩所确定，d d y x 、22d d yx ． （0923）已知函数0()10x e x f x x x -⎧<=⎨+≥⎩，证明函数)(x f 在点0=x 处连续但不可导．（1008）.若(0)1f '=，则0()()limx f x f x x→--= ．（1014）设函数()y y x =由方程2x yy ex ++=所确定，求d d y x 、22d d yx ．（1022）设()0()1x x f x xx ϕ⎧≠⎪=⎨⎪=⎩，其中函数()x ϕ在0x =处具有二阶连续导数，且(0)0ϕ=，(0)1ϕ'=，证明：函数()f x 在0x =处连续且可导．（1102）设函数)(x f 在点0x 处可导，且4)()(lim 000=+--→hh x f h x f h ，则=')(0x f ( )A.4-B.2-C.2D.4（1110）设函数x y arctan=，则1d x y==_____________．（1114）设函数)(x y y =由参数方程⎪⎩⎪⎨⎧=++=22ty e tt x y 所确定，求d d y x ．（1208）设函数()22221x y x x x e =⋅+++，则=)0()7(y________．（1209）设xy x =（0x >），则函数y 的微分=dy ___________．（1214）设函数)(x y y =由参数方程⎪⎩⎪⎨⎧+=-=tt y tt x ln 212所确定，求d d y x 、22d d y x ． （1304）设1y f x ⎛⎫= ⎪⎝⎭，其中f 具有二阶导数，则22d d y x =（ ）A.231121f f x x x x ⎛⎫⎛⎫'''-+ ⎪ ⎪⎝⎭⎝⎭ B.231121f f x x x x ⎛⎫⎛⎫'''+ ⎪ ⎪⎝⎭⎝⎭ C.231121f f x x x x ⎛⎫⎛⎫'''--⎪ ⎪⎝⎭⎝⎭D.231121f f x x x x ⎛⎫⎛⎫'''-⎪ ⎪⎝⎭⎝⎭（1306）已知函数()f x 在点1x =处连续，且21()1lim 12x f x x →=-，则曲线()f x 在点()1,()f x 处切线方程为（ ） A.1y x =-B.22y x =-C.33y x =-D.44y x =-（1309）设函数由参数方程2211x t y t ⎧=+⎨=-⎩所确定，则221d d t yx == ．（二）中值定理及导数的应用（0423）甲、乙二城位于一直线形河流的同一侧，甲城位于岸边，乙城离河岸40公里，乙城在河岸的垂足与甲城相距50公里，两城计划在河岸上合建一个污水处理厂，已知从污水处理厂到甲乙二城铺设排污管道的费用分别为每公里500、700元．问污水处理厂建在何处，才能使铺设排污管道的费用最省？（0507）02limsin x x x e e xx x-→--=- ． （0508）函数x x f ln )(=在区间[]1,e 上满足拉格郎日中值定理的=ξ ． （0521）证明方程：0133=+-x x 在[]1,1-上有且仅有一根．（0603）下列函数在[]1,1-上满足罗尔定理条件的是（ ） A.xe y =B.1y x =+C.21x y -=D.xy 11-= （0621）证明：当2x ≤时，332x x -≤．（0703）设函数()(1)(2)(3)f x x x x x =---，则方程()0f x '=的实根个数为（ ） A.1B.2C.3D.4（0713）求极限01lim tan x x e x x x→--．（0722）设函数9)(23-++=cx bx ax x f 具有如下性质：（1）在点1-=x 的左侧临近单调减少； （2）在点1-=x 的右侧临近单调增加； （3）其图形在点(1,2)的两侧凹凸性发生改变． 试确定a ，b ，c 的值．（0724）求证：当0>x 时，22(1)ln (1)x x x -⋅≥-．（0809）已知曲线543223++-=x x x y ，则其拐点为 ． （0821）求曲线1y x=（0x >）的切线，使其在两坐标轴上的截距之和最小，并求此最小值． （0823）设函数)(x f 在闭区间[]0,2a （0a >）上连续，且)()2()0(a f a f f ≠=，证明：在开区间(0,)a 上至少存在一点ξ，使得()()f f a ξξ=+． （0824）对任意实数x ，证明不等式：(1)1xx e -⋅≤． （0904）曲线221(1)x y x +=-的渐近线的条数为（ ）A.1B.2C.3D.4（0913）求极限30lim sin x x x x→-．（0921）已知函数13)(3+-=x x x f ，试求： （1）函数)(x f 的单调区间与极值； （2）曲线)(x f y =的凹凸区间与拐点；（3）函数)(x f 在闭区间[2,3]-上的最大值与最小值．（0924）证明：当12x <<时，24ln 23x x x x >+-．（1002）曲线223456x x y x x -+=-+的渐近线共有 ( )A.1条B.2条C.3条D.4条 （1006）设3()3f x x x =-，则在区间(0,1)内 ( ) A.函数()f x 单调增加且其图形是凹的 B.函数()f x 单调增加且其图形是凸的 C.函数()f x 单调减少且其图形是凹的 D.函数()f x 单调减少且其图形是凸的（1013）求极限2|011lim tan x x x x →⎛⎫-⎪⎝⎭．（1021）证明：当1x >时，121122x e x ->+． （1103）若点(1,2)-是曲线23bx ax y -=的拐点，则（ ） A.3,1==b aB.1,3-=-=b aC.3,1-=-=b aD.6,4==b a（1113）求极限()()22limln 1xx x eex -→-+．（1121）证明：方程()2ln 12x x ⋅+=有且仅有一个小于2的正实根． （1122）证明：当0>x 时，x x201120102011≥+．（1203）设232152)(x x x f -=，则函数)(x f ( ) A.只有一个最大值 B.只有一个极小值 C.既有极大值又有极小值D.没有极值（1213）求极限()2302cos 2lim ln 1x x x x x →+-+． （1223）证明：当10<<x 时，361arcsin x x x +>． （1302）曲线22232x xy x x +=-+的渐近线共有（ ）A.1条B.2条C.3条D.4条（1313）求极限01lim ln (1)x x e x x →⎡⎤-⎢⎥+⎣⎦．（1323）证明：当1x >时，2(1ln )21x x +<-．三、一元函数积分学（一）不定积分（0410）求不定积分3x = ．（0416）设)(x f 的一个原函数为xe x，计算(2)d x f x x '⎰．（0503）若()d ()f x x F x C =+⎰，则sin (cos )d x f x x =⎰（ ）A.C x F +)(sinB.C x F +-)(sinC.C F +(cos)D.C x F +-)(cos（0515）计算3tan sec d x x x ⎰．（0522）设函数)(x f y =的图形上有一拐点(2,4)P ，在拐点处的切线斜率为3-，又知该函数的二阶导数6y x a ''=+，求)(x f ．（0604）已知2()d x f x x e C =+⎰，则()d f x x '-=⎰（ ）A.C ex+-22B.C e x +-221 C.C e x +--22 D.C e x +--221（0615）计算x ． （0622）已知曲线)(x f y =过原点且在点),(y x 处的切线斜率等于y x +2，求此曲线方程． （0704）设函数)(x f 的一个原函数为x 2sin ，则(2)d f x x '=⎰（ ）A.C x +4cosB.C x +4cos 21C.C x +4cos 2D.C x +4sin（0715）求不定积分2d x x e x -⎰．（0810）设函数)(x f 的导数为x cos ，且21)0(=f ，则不定积分()d f x x =⎰ ． （0815）求不定积分3d 1x x x +⎰． （0905）设()ln (31)F x x =+是函数)(x f 的一个原函数，则(21)d f x x '+=⎰（ ）A.C x ++461B.C x ++463C.C x ++8121D.C x ++8123（0915）求不定积分x ⎰．（1015）求不定积分arctan d x x x ⎰．（1115）设)(x f 的一个原函数为x x sin 2，求不定积分()d f x x x⎰． （1215）求不定积分sin 2d x x x ⎰． （1315）求不定积分sin 2d x x x ⎰．（二）定积分（0404）2228R y x =+设所围的面积为S ，则0x ⎰的值为（ ）A.SB.4S C.2S D.S 2（0421）证明：0(sin )d (sin )d 2x f x x f x x πππ=⎰⎰，并利用此式求20sin d 1cos xxx xπ+⎰．（0509）1211d 1x x x π-+=+⎰．（0516）计算10arctan d x x ⎰．（0609）设)(x f 在[]0,1上有连续的导数且(1)2f =，10()d 3f x x =⎰，则1()d x f x x '=⎰ ．（0616）计算22cos d x x x π⎰．（0709）定积分)231cos d x x x -+⎰的值为 ．（0716）计算定积分x ． （0811）定积分1212sin d 1xx x -++⎰的值为 ．（0816）求定积分10d x ⎰．（0916）求定积分：210⎰．（1009）定积分31211d 1x x x -++⎰的值为 ． （1016）计算定积分40x ⎰． （1111）定积分()32221sin d xx x ππ-+⋅⎰的值为____________．（1116）计算定积分3⎰　． （1216）计算定积分21⎰．（1316）计算定积分20⎰（1324）设函数()f x 在[,]a b 上连续，证明：[]2()d ()()d a b b aaf x x f x f a b x x +=++-⎰⎰．（三）变限积分与广义积分（0417）计算广义积分2+∞⎰（0422）设函数)(x f 可导，且满足方程20()d 1()x t f t t x f x =++⎰，求)(x f ．（0705）设221()sin d x f x t t =⎰，则()f x '=（ ）A.4sin x B.2sin 2x xC.2cos 2x xD.4sin 2x x（0803）设函数)(x f 122sin d xt t t =⎰，则()f x '等于（ ）A.x x 2sin 42B.x x 2sin 82C.x x 2sin 42-D.x x 2sin 82-（0908）设函数20()d x t x te t ϕ=⎰，则()x ϕ'＝ ．（1003）设函数22()cos d t xx e t t Φ=⎰，则函数()x Φ的导数()x 'Φ等于 ( )A.222cos x xe x B.222cos x xe x - C.2cos xxe x - D.22cos x e x - （1108）设函数2()ln (1)d x x t t Φ=+⎰　，则=Φ'')1(____________．（1211）设反常积分1d 2x ae x +∞-=⎰，则常数=a ______． （1222）已知定义在(),-∞+∞上的可导函数)(x f 满足方程31()4()d 3xx f x f t t x -=-⎰，试求：（1）函数()f x 的表达式； （2）函数)(x f 的单调区间与极值； （3）曲线()y f x =的凹凸区间与拐点．（1224）设0()d 0()(0)0x g t t x f x g x ⎧≠⎪=⎨⎪=⎩⎰，其中函数)(x g 在(,)-∞+∞上连续，且3cos 1)(lim 0=-→xx g x ．证明：函数)(x f 在0=x 处可导，且1(0)2f '=． （1322）已知251320()95d x F x t t t ⎛⎫=- ⎪⎝⎭⎰是()f x 的一个原函数，求曲线()y f x =的凹凸区间、拐点． （四）定积分的几何应用（0523）已知曲边三角形由x y 22=、0=x 、1=y 所围成，求：（1）曲边三角形的面积；（2）曲边三角形绕x 轴旋转一周的旋转体体积．（0623）已知一平面图形由抛物线2x y =、82+-=x y 围成．（1）求此平面图形的面积；（2）求此平面图形绕y 轴旋转一周所得的旋转体的体积．（0721）设平面图形由曲线21x y -=（0≥x ）及两坐标轴围成．（1）求该平面图形绕x 轴旋转所形成的旋转体的体积；（2）求常数a 的值，使直线a y =将该平面图形分成面积相等的两部分．（0822）设平面图形由曲线2x y =，22x y =与直线1=x 所围成．（1）求该平面图形绕x 轴旋转一周所得的旋转体的体积；（2）求常数a ，使直线a x =将该平面图形分成面积相等的两部分．（0922）设1D 是由抛物线22x y =和直线x a =，0y =所围成的平面封闭区域，2D 是由抛物线22x y =和直线x a =，2x =及0=y 所围成的平面封闭区域，其中20<<a ．试求：（1）1D 绕y 轴旋转所成的旋转体的体积1V ，以及2D 绕x 轴旋转所成的旋转体的体积2V ； （2）求常数a 的值，使得1D 的面积与2D 的面积相等．（1023）设由抛物线2y x =（0x ≥），直线2y a =（01a <<）与y 轴所围成的平面图形绕x 轴旋转一周所形成的旋转体的体积记为1()V a ，由抛物线2y x =（0x ≥），直线2y a =（01a <<）与直线1x =所围成的平面图形绕x 轴旋转一周所形成的旋转体的体积记为2()V a ，另12()()()V a V a V a =+，试求常数a 的值，使()V a 取得最小值．（1024）设函数()f x 满足方程()()2xf x f x e '+=，且(0)2f =，记由曲线'()()f x y f x =与直线1y =，x t =（0t >）及y 轴所围平面图形的面积为()A t ，试求lim ()t A t →+∞．（1124）设函数)(x f 满足微分方程()2()(1)x f x f x a x '-=-+（其中a 为正常数），且1)1(=f ，由曲线()y f x =（1x ≤）与直线1x =，0y =所围成的平面图形记为D ．已知D 的面积为32． （1）求函数)(x f 的表达式；（2）求平面图形D 绕x 轴旋转一周所形成的旋转体的体积x V ； （3）求平面图形D 绕y 轴旋转一周所形成的旋转体的体积y V ．（1221）在抛物线2y x =（0x >）上求一点P ，使该抛物线与其在点P 处的切线及x 轴所围成的平面图形的面积为32，并求该平面图形绕x 轴旋转一周所形成的旋转体的体积．（1321）设平面图形D 是由曲线x =y =1y =所围成，试求：（1）平面图形D 的面积；（2）平面图形D 绕x 轴旋转一周所形成的旋转体的体积．四、向量代数与空间解析几何（一）向量代数（0510）设向量{}3,4,2=-a 、{}2,1,k =b ；a 、b 互相垂直，则=k ． （0610）设1=a ，⊥a b ，则()⋅+=a a b ． （0710）已知a 、b 均为单位向量，且12⋅=a b ，则以a 、b 为邻边的平行四边形面积为 ． （0804）设向量(1,2,3)=a ，(3,2,4)=b ，则⨯a b 等于（ ）A.(2,5,4)B.(2,5,4)--C.(2,5,4)-D.(2,5,4)--（0909）已知向量{}1,0,1=-a ，{}1,2,1=-b ，则+a b 与a 的夹角为 ． （1010）设{}1,2,3=a ，{}2,5,k=b ，若a 与b 垂直，则常数k = ．（1109）若1=a ，4=b ，2⋅=a b ，则⨯=a b ____________．（1210）设向量a 、b 互相垂直，且3=a ，2=b ，则2+=a b ________．（1308）已知空间三点(1,1,1)A ，(2,3,4)B ，(3,4,5)C ，则ABC ∆的面积为 ．（二）平面与直线（0518）求过点(3,1,2)A -且通过直线L ：43521x y z-+==的平面方程． （0619）求过点(3,1,2)M -且与二平面07=-+-z y x 、0634=-+-z y x 都平行的直线方程．（0719）求过点(1,2,3)且垂直于直线20210x y z x y z +++=⎧⎨-++=⎩的平面方程．（0817）设平面∏经过点(2,0,0)A ，(0,3,0)B ，(0,0,5)C ，求经过点(1,2,1)P 且与平面∏垂直的直线方程． （0917）求通过直线12213-=-=z y x 且垂直于平面02=+++z y x 的平面方程． （1017）求通过点(1,1,1)，且与直线23253x ty t z t =+⎧⎪=+⎨⎪=+⎩垂直，又与平面250x z --=平行的直线的方程．（1117）求通过x 轴与直线132zy x ==的平面方程． （1217）已知平面∏通过(1,2,3)M 与x 轴，求通过(1,1,1)N 且与平面∏平行，又与x 轴垂直的直线方程．（1318）已知直线10330x y z x y z -+-=⎧⎨--+=⎩在平面∏上，又知直线23132x ty t z t=-⎧⎪=+⎨⎪=+⎩与平面∏平行，求平面∏的方程．五、多元函数微积分（一）多元函数微分学（0418）设(,)z f x y xy =-，且具有二阶连续的偏导数，求x z ∂∂、yx z∂∂∂2．（0505）设yxy x u arctan),(=，(,)v x y =，则下列等式成立的是（ ）A.yv x u ∂∂=∂∂ B.xvx u ∂∂=∂∂ C.x v y u ∂∂=∂∂ D.y v y u ∂∂=∂∂ （0517）已知函数2(sin ,)z f x y =，其中),(v u f 有二阶连续偏导数，求x z ∂∂、yx z∂∂∂2．（0611）设x e u xysin =，=∂∂xu． （0620）设2(,)z x f x xy =⋅其中(,)f u v 的二阶偏导数存在，求y z ∂∂、xy z∂∂∂2．（0711）设yxz =，则全微分d z = ． （0717）设(23,)z f x y xy =+其中f 具有二阶连续偏导数，求yx z∂∂∂2．（0805）函数xyz ln =在点(2,2)处的全微分d z 为（ ）A.11d d 22x y -+B.11d d 22x y +C.11d d 22x y -D.11d d 22x y --（0818）设函数,y z f x y x ⎛⎫=+ ⎪⎝⎭，其中)(x f 具有二阶连续偏导数，求y x z ∂∂∂2．（0910）设函数(,)z z x y =由方程12=+yz xz 所确定，则xz∂∂＝ ． （0919）设函数(sin ,)z f x xy =，其中)(x f 具有二阶连续偏导数，求yx z∂∂∂2．（1011）设函数z =，则10d x y z=== ．（1018）设()2,xz y f xy e =⋅，其中函数f 具有二阶连续偏导数，求2zx y∂∂∂．（1104）设),(y x f z =为由方程8333=+-x yz z 所确定的函数，则=∂∂==00y x yz （ ）A.21-B.21C.2-D.2（1118）设)(y xyxf z ，=，其中函数f 具有二阶连续偏导数，求y x z ∂∂∂2．（1204）设3ln 2z x y=+在点()1,1处的全微分为 ( )A.d 3d x y -B.d 3d x y +C.1d 3d 2x y +D.1d 3d 2x y -（1218）设函数22(,)()z f x xy x y ϕ=++，其中函数f 具有二阶连续偏导数，函数()x ϕ具有二阶连续导数，求yx z∂∂∂2．（1314）设函数(,)z z x y =由方程3331z xy z +-=所确定，求d z 及22zx∂∂．（1317）设()223,x yz fx e+=，其中函数f 具有二阶连续偏导数，求2zy x∂∂∂．（二）二重积分（0411）交换二次积分的次序2120d (,)d x x x f x y y -=⎰⎰．（0419）计算二重积分sin d d Dyx y y ⎰⎰，其中D 由曲线x y =及x y =2所围成． （0504）设区域D 是xoy 平面上以点(1,1)A 、(1,1)B -、(1,1)C --为顶点的三角形区域，区域1D 是D 在第一象限的部分，则(cos sin )d d Dxy x y x y +=⎰⎰（ ）A.⎰⎰1)sin (cos 2D dxdy y xB.⎰⎰12D xydxdyC.⎰⎰+1)sin cos (4D dxdy y x xyD. 0（0511）交换二次积分的次序11d (,)d x x f x y y -+=⎰；（0524）设)(x f 为连续函数，且1)2(=f ，1()d ()d uuyF u y f x x =⎰⎰（1u >）． （1）交换)(u F 的积分次序； （2）求(2)F '．（0606）设对一切x 有(,)(,)f x y f x y -=-，22{(,)|1,0}D x y x y y =+≤≥，=1D 22{(,)|1,0,0}x y x y x y +≤≥≥，则(,)d d Df x y x y =⎰⎰（ ）A. 0B.1(,)d d D f x y x y ⎰⎰C.21(,)d d D f x y x y ⎰⎰D.41(,)d d D f x y x y ⎰⎰（0612）D 为以点(0,0)O 、(1,0)A 、(0,2)B 为顶点的三角形区域，d d Dx y =⎰⎰ ．（0624）设⎪⎩⎪⎨⎧=≠=⎰⎰00)(1)(t a t dxdy x f t t g tD ，其中t D 是由t x =、t y =以及坐标轴围成的正方形区域，函数)(x f 连续．（1）求a 的值使得)(t g连续；（2）求)('t g ．（0720）计算二重积分d Dx y ，其中{}22(,)|2,0D x y x y x y =+≤≥．（0723）设0>>a b ，证明：()232d ()d ()d b b b x y xx a ayay f x e x ee f x x ++⋅=-⎰⎰⎰．（0819）计算二重积分2d d Dx x y ⎰⎰，其中D 是由曲线xy 1=，直线y x =，2x =及0=y 所围成的平面区域． （0918）计算二重积分d Dy σ⎰⎰，其中22{(,)02,2,2}D x y x x y x y =≤≤≤≤+≥． （1005）二次积分111d (,)d y y f x y x +⎰⎰交换积分次序后得 ( )A.1101d (,)d x x f x y y +⎰⎰B.2110d (,)d x x f x y y -⎰⎰C.2111d (,)d x x f x y y -⎰⎰D.2111(,)d x dx f x y y -⎰⎰（1019）计算d d Dx x y ⎰⎰，其中D 是由曲线x =y x =及x 轴所围成的闭区域．（1105）若(,)d d Df x y x y ⎰⎰可转化为二次积分1201d (,)d y y f x y x +⎰⎰ ，则积分域D 可表示为（ ） A.{}(,)01,11x y x x y ≤≤-≤≤ B.{}(,)12,11x y x x y ≤≤-≤≤C.{}(,)01,10x y x x y ≤≤-≤≤D.{}(,)12,01x y x y x ≤≤≤≤-（1119）计算二重积分d d Dy x y ⎰⎰，其中D 是由曲线y =直线x y -=及y 轴所围成的平面闭区域． （1205）二次积分dx y x f dy y),(11⎰⎰ 在极坐标系下可化为（ ）A.sec 40d (cos ,sin )d f πθθρθρθρ⎰⎰ B.sec 40d (cos ,sin )d f πθθρθρθρρ⎰⎰C.sec 24d (cos ,sin )d f πθπθρθρθρ⎰⎰D .sec 24d (cos ,sin )d f πθπθρθρθρρ⎰⎰ （1220）计算二重积分d d Dy x y ⎰⎰，其中D 是由曲线y =2xy =及x 轴所围成的平面闭区域．（1320）计算二重积分d d Dx x y ⎰⎰，其中D 是由曲线y =0x >）与三条直线y x =，3x =，0y =所围成的平面闭区域．六、无穷级数（一）数项级数（0506）正项级数（1）∑∞=1n n u 、（2）∑∞=13n n u ，则下列说法正确的是（ ） A.若（1）发散、则（2）必发散 B.若（2）收敛、则（1）必收敛C.若（1）发散、则（2）不确定D.（1）、（2）敛散性相同（0605）设∑∞=1n nu为正项级数，如下说法正确的是（ ）A.若0lim 0=→n n u ，则∑∞=1n nu必收敛 B.若l u u nn n =+∞→1lim )0(∞≤≤l ，则∑∞=1n n u 必收敛C.若∑∞=1n nu收敛，则∑∞=12n nu必定收敛D.若∑∞=-1)1(n n nu 收敛，则∑∞=1n n u 必定收敛（0706）下列级数收敛的是（ ）A.∑∞=122n nnB.∑∞=+11n n n C.∑∞=-+1)1(1n nnD.∑∞=-1)1(n nn（0906）设α为非零常数，则数项级数∑∞=+12n n n α（ ）A.条件收敛B.绝对收敛C.发散D.敛散性与α有关（1004）下列级数收敛的是（ ）A.11n nn ∞=+∑B.2121n n n n ∞=++∑C.nn ∞= D.212n n n ∞=∑（1206）下列级数中条件收敛的是（ ）A.1(1)21nn nn ∞=-+∑B.13(1)2nnn ∞=⎛⎫- ⎪⎝⎭∑C.21(1)nn n ∞=-∑D.1nn ∞=（1305）下列级数中收敛的是（ ）A.211n n n∞=+∑ B.11nn n n ∞=⎛⎫ ⎪+⎝⎭∑C.1!2n n n ∞=∑D.1n ∞= （二）幂级数（0412）幂级数∑∞=-12)1(n nnx 的收敛区间为 ．（0420）把函数21)(+=x x f 展开为2-x 的幂级数，并写出它的收敛区间． （0512）幂级数1(21)nn n x∞=-∑的收敛区间为 ．（0519）把函数222)(x x x x f --=展开为x 的幂级数，并写出它的收敛区间．（0618）将函数()ln (1)f x x x =+展开为x 的幂函数（要求指出收敛区间）．（0812）幂函数12n nn x n ∞=⋅∑的收敛域为 ． （0911）若幂函数21n nn a x n∞=∑（0a >）的收敛半径为21，则常数=a ．（1012）幂级数0(1)n nn x n ∞=-∑的收敛域为 ．（1106）若x x f +=21)(的幂级数展开式为0()nn n f x a x ∞==∑（22x -<<），则系数=n a ( )A.n 21B.121+n C.(1)2nn -D.1(1)2nn +-（1112）幂级数0nn ∞=的收敛域为_ _ _________． （1212）幂级数1(1)(3)3nn nn x n ∞=--⋅∑的收敛域为____________． （1312）幂级数1n nn ∞=的收敛域为 ． 七、常微分方程（一）一阶微分方程（0520）求微分方程0'=-+xe y xy 满足1x ye ==的特解．（0617）求微分方程22x y xy y '=-的通解． （0718）求微分方程22007xy y x '-=满足初始条件12008x y==的特解．（0820）求微分方程22xy y x '=+的通解．（0912）微分方程2(1)d (2)d 0x y x y x y +--=的通解为 ． （1311）微分方程d d y x y x x+=的通解为 ． （二）二阶线性微分方程（0406）微分方程232xy y y xe '''-+=的特解*y 的形式应为（ ）A.xAxe 2B.xe B Ax 2)(+C.xeAx 22D.xeB Ax x 2)(+（0712）设x xe C eC y 3221+=为某二阶常系数齐次线性微分方程的通解，则该微分方程为 ．（0806）微分方程321y y y '''++=的通解为（ ）A.1221++=--x xe c e c yB.21221++=--x xe c ec yC.1221++=-xxec e c yD.21221++=-xxec e c y （0920）求微分方程y y x ''-=的通解． （1020）已知函数xy e =和2xy e-=是二阶常系数齐次线性微分方程0y py qy '''++=的两个解，试确定常数p 、q 的值，并求微分方程xy py qy e '''++=的通解．（1120）已知函数(1)xy x e =+⋅是一阶线性微分方程2()y y f x '+=的解，求二阶常系数线性微分方程)(23x f y y y =+'+''的通解．（1219）已知函数)(x f 的一个原函数为xxe ，求微分方程)(44x f y y y =+'+''的通解． （1319）已知函数()y f x =是一阶微分方程d d yy x=满足初始条件(0)1y =的特解，求二阶常系数非齐次线性微分方程32()y y y f x '''-+=的通解．时间排序与参考答案2004年高等数学真题参考答案1、A ．2、B ．3、C ．4、B ．5、A ．6、D ．7、1-e ． 8、32241-+==-z y x ． 9、!n ． 10、C x +4arcsin 41． 11、12201d (,)d d (,)d y y f x y x y f x y x -+⎰⎰⎰．12、()3,1-．13、解：间断点为πk x =（Z k ∈），当0=x 时，1sin lim)(lim 00==→→xxx f x x ，为可去间断点；当πk x =（0≠k ，Z k ∈）时，∞=→xxx sin lim0，为第二类间断点．14、解：原式0430(tan sin )d tan sin limlim312xx x t t tx xx x →→--==⎰233001tan (1cos )12lim lim 121224x x x x x x x x →→⋅-===． 15、解：0=x 代入原方程得1)0(=y ，对原方程求导得0''=--y xe e y yy，对上式求导并将0=x 、1=y 代入，解得：22''e y =．16、解：因为)(x f 的一个原函数为x e x，所以2')1()(x e x x e x f xx -=⎪⎪⎭⎫ ⎝⎛=， 原式11(2)d(2)d (2)22xf x x x f x '==⎰⎰11(2)(2)d 22x f x f x x =-⎰ 222211(21)1(2)(2)d(2)24884x x xx x e e x x f x f x x C e C x x x--=-=-+=+⎰． 17211122d d 22arctan (1)12t tt tt t t π+∞∞+∞+===++⎰．18、解：12zf f y x∂''=+⋅∂； []21112221221112222(1)(1)()zf f x f y f f x f x y f xy f f x y∂''''''''''''''''=⋅-+⋅++⋅-+⋅=-+-⋅+⋅+∂∂．19、解：原式21100sin sin d d d d (1)sin d y y Dyy x y y x y y y y y ===-⎰⎰⎰⎰⎰ 1100(1)cos cos d 1sin1y y y y =--=-⎰．20、解：01111(2)()(1)24244414n n nn x f x x x ∞=-==⋅=--+-+∑)62(<<-x ． 21、证：00(sin )d ()[sin ()]d ()(sin )d t xx f x xt f t t t f t I t πππππππ=-=---=-⎰⎰⎰(sin )d (sin )d (sin )d f x x x f x x f x x I πππππ=-=-⎰⎰⎰解得： 0(sin )d (sin )d 2f x x f x x I x πππ==⎰⎰， 原命题证毕．222000sin sin d d arctan (cos )1cos 21cos 24x x x x x x x x ππππππ⋅==-=++⎰⎰． 22、解：等式两边求导得()2()x f x x f x '=+，即()()2f x x f x x '-=-，且(0)1f =-，x p -=，x q 2-=，而2()d 2x x xe e --⎰=，由公式求得通解：222222()2d 2x x x f x e xq x C C e -⎡⎤⎛⎫=-+=+⎢⎥ ⎪ ⎪⎢⎥⎝⎭⎣⎦⎰， 将初始条件(0)1f =-代入通解，解得：3-=C ，故22()23x f x e =-．23、解：设污水厂建在河岸离甲城x 公里处，则()500M x x =+500≤≤x ），由150070002M '=+⨯=解得：650050-=x （公里），唯一驻点，即为所求．2005年高等数学真题参考答案1、A ．2、C ．3、D ．4、A ．5、A ．6、C ．7、2． 8、1-e ． 9、2π． 10、5． 11、11d (,)d y y f x y x -⎰⎰．12、(1,1)-．13、解：因为)(x F 在0=x 处连续，所以)0()(lim 0F x F x =→，'00()2sin ()(0)lim ()limlim 2(0)28x x x f x x f x f F x f x x→→→+-==+=+=， 解得：a F =)0(，故8=a ．14、解：d d cos cos sin d d d sin d yy t t t t t t x x t t-+===--，22d ()csc d (cos )y t t x t '-=='．15、解：原式22tan tan sec d (sec1)d(sec )x x x xx x =⋅-⎰⎰积进去231sec d(sec )d(sec )sec sec 3x x x x x C =-=-+⎰⎰．16、解：原式211120002d 1d(1)arctan 1421x x x x x x x π+=--++⎰⎰积进去 ()12011ln 1ln 24242x ππ⎡⎤=-+=-⎣⎦．17、解：1cos zx f x∂'=⋅∂，()21212cos 22cos z x f y y x f x y ∂''''=⋅⋅=⋅∂∂． 18、解：直线L 的方向向量{}5,2,1=s ，过点()4,3,0B -，{}1,4,2AB =-；所求平面的法向量{}5218,9,22142AB =⨯==---ij kn s ，点法式为8(3)9(1)22(2)0x y z ----+=，即592298=--z y x ．19、解：2222101111(1)()13216313212n nn n x x x x f x x x x x x ∞+=⎡⎤-⎛⎫=+=⋅+⋅=+⋅ ⎪⎢⎥+--⎝⎭⎣⎦+∑， 收敛域为：11<<-x ．20、解：1x e y y x x '+⋅=，即1p x=，x e q x =，而1d 1x x e x -⎰=；故通解为1d xx e e C y x x C x x x ⎛⎫+=+= ⎪⎝⎭⎰．把初始条件1x y e ==解得：0=C ；故所求特解为：xe y x=．21、证：令13)(3+-=x x x f ，[]1,1x ∈-，且(1)30f -=>，(1)10f =-<，(1)(1)0f f -⋅<；由连续函数零点定理知：)(x f 在(1,1)-内至少有一实根；对于()1,1x ∈-恒有()22()33310f x x x '=-=-<，即)(x f 在(1,1)-内单调递减， 故方程0133=+-x x 在[]1,1-上有且仅有一根； 原命题获证．22、解：设所求函数为)(x f y =，则有4)2(=f ，(2)3f '=-，(2)0f ''=；由()6f x x a ''=+和(2)0f ''=解得：12-=a ，即()612f x x ''=-，故21()312f x x x C '=-+，由(2)3f '=-解得：91=C ，故22396C x x x y ++-=，由(2)4f =解得：22=C ； 所求函数为：29623++-=x x x y ．23、解：（1）112300111d 266S y y y ===⎰；（如图1所示） （2）()()112222012d 4x V x x x xπππ=-=-=⎰．24、解：积分区域D 为：u y ≤≤1，u x y ≤≤；（1）111()()d d ()d (1)()d u xuDF u f x x f x y x f x x σ===-⎰⎰⎰⎰⎰；（2）()(1)()F u u f u '=-，(2)(21)(2)(2)1F f f '=-==．2006年高等数学真题参考答案1、C ．2、B ．3、C ．4、C ．5、C ．6、A ．7、2． 8、)(0x f ． 9、1-． 10、1． 11、(sin cos )xye y x x +． 12、1．13、解：原式322131lim 21341==--→x xx ．图114、解：2211d 12d 21t t y y t t t x x t-'+==='+，2222d 1d d 122d 41ty x y t t x x t t '⎛⎫ ⎪+⎝⎭==='+． 15、解：原式322ln )(1ln )3x x C =+=++．16、解：原式()2222220d(sin )sin 2sin d x x x xx x πππ=-⎰⎰积进去222220sin 2sin d 2d(cos )4x xx x xx x ππππ-+⎰⎰积进去导出来2222002cos 2cos d 244x x x x ππππ=+-=-⎰．17、解：方程变形为2y y y x x ⎛⎫'=- ⎪⎝⎭，即得到了形如d d y y f x x ⎛⎫= ⎪⎝⎭齐次方程； 令y u x =，则d d d d y u u x x x =+，代入得：2d d u x u x =-，分离变量得：211d d u x u x-=； 两边积分，得：211d d u x u x -=⎰⎰，1ln x C u=+，故ln x y x C =+． 18、解：令()ln (1)g x x =+，则(0)0g =；由于01()(1)1n n n g x x x ∞='==-+∑（(]1,1x ∈-）， 所以01(1)((1))d x n n n g x n x g t t ∞+='=+=-∑⎰（(]1,1x ∈-），故20(1)()1n n n f x x n ∞+=-=+∑，收敛域为：11x -<≤．19、解：由题意知：{}11,1,1=-n ，{}24,3,1=-n ；{}12311232,3,1431=⨯=-=++=-i j ks n n i j k ，故所求直线方程的对称式方程为：123123+=-=-z y x ．20、解：22z x f x∂'=∂，2'2'''''3''2''22122221222(2)22z x f x f x f y x f x f x y f y x ∂=+⋅+⋅=++∂∂． 21、证：令33)(x x x f -=，[]2,2x ∈-，由2()330f x x '=-=解得驻点：1±=x ，比较以下函数值的大小：(1)2f -=-，(1)2f =，(2)2f =-，(2)2f -=； 所以2min -=f ，2m ax =f ，故2)(2≤≤-x f ，即332x x -≤，原命题获证．22、解：0)0(=y ，2y x y '=+，通解为：xCe x y +--=)22(；将0)0(=y 代入通解解得：2=C ，故所求特解为：xe x y 222+--=．23、解：（1）()2222648d 3S x x x -=--=⎰； （2）224804d d 16y V y y πππ=+=⎰⎰．24、解：()d d d ()d ()d tt tt D f x x y x f x y t f x x ==⎰⎰⎰⎰⎰，0()d 0()0t f x x t g t a t ⎧≠⎪=⎨⎪=⎩⎰；（1）00lim ()lim()d 0t t t g t f x x →→==⎰，由)(t g 的连续性可知：0)(lim )0(0===→t g g a t ；（2）当0≠t 时，()()g t f t '=，当0=t 时，0000()d ()(0)(0)limlim lim ()(0)hh h h f x x g h g g f h f h h→→→-'====⎰； 综上，()()g t f t '=．2007年高等数学真题参考答案1、B ．2、C ．3、C ．4、A ．5、D ．6、D ．7、2ln ． 8、1． 9、π2． 10、23． 11、21d d xx y y y-． 12、06'5''=+-y y y ． 13、解：212lim 21lim 1lim tan 1lim00200==-=--=--→→→→x x x x x x x x e x e x x e x x x e ． 14、解：当0=x 时，0=y ；。

函数的连续性与导数、微分的概念的练习题答案一、单项选择题（每小题4分，共24 分）1．若()f x 为是连续函数，且()()01,10f f ==， 则1lim sin x f x x →∞⎛⎫= ⎪⎝⎭( )A ． -1B ．0C ．1D ． 不存在解： 原式1sin 1lim sin lim1x x f x f x f x x →∞→∞⎡⎤⎢⎥⎡⎤=⎢⎥⎢⎥⎣⎦⎢⎥⎣⎦连续()10f ==，选B2． 要使()()ln 1mx f x kx =+在点0x =处连续，应给()0f 补充定义的数值是( )A ． kmB ． kmC ． ln kmD ． km e解：()00lim ln lim(1)mx x x f x kx →→⎡⎤=+⎢⎥⎣⎦0lim ln ln x m kx km x e e km →⋅===()0f km ∴= 选A3．若lim ()x a f x A →=，则下列正确的是 （ ）A ． ()lim x a f x A →=B ．x a →=C ． ()lim x a f x A →=-D ． lim ()x a f x A →=解：x →=选B4．设()()(),00,0f x x F x x f x ⎧≠⎪=⎨⎪=⎩且()f x 在0x =处可导，()00,f '≠()00f =,则0x =是()F x 的 ( )A ． 可去间断点B ． 跳跃间断点C ． 无穷间断点D ． 连续点解：()()()()000lim lim 0,0x x f x f F x f x →→-'==- ()()00f f '≠()()()000lim 0x F f F →∴=≠，故0x =是()F x 的第一类可去间断点。

选A5．()1sin ,00,0x f x x x x ⎧⎪=≠⎨⎪=⎩在0x =处 ( )A ． 极限不存在B ．极限存在但不连续C ．连续但不可导D ．可导但不连续解：()001lim lim sin 0x x f x x x→→=⋅= ，且()00f = ()f x ∴在0x =连续，又()0f '01sin 0lim 0x x x x →-==-不存在，()f x ∴在0x =不可导 选C 6．设()21,1,1x x f x ax b x ⎧+≤=⎨+>⎩在1x =可导，则,a b 为 （ ） A ． 2,2a b =-= B ． 0,2a b ==C ． 2,0a b ==D ． 1,1a b ==解：（1）()f x 在1x =连续，()()211lim 12,lim x x x ax b a b -+→→∴+=+=+ 故()21a b +=⋯（2）()()2111lim 2,11x x f f x --+→-''==-()()11112lim lim 11x x a x ax b a x x ++→→-+-==--2a ∴=，代入()1得0b =，选C二、 填空题（每小题4分，共24分）7．设()f x 为连续奇函数，则()0f =解：（1）()f x 为奇函数，()()f x f x ∴-=-（2）()()00lim lim x x f x f x →→-=-⎡⎤⎣⎦又()f x 在0x =连续()()00f f ∴=- 故()00f =8．若()f x 为可导的偶函数，则()0f '=解：（1）()f x 为偶函数，()()f x f x ∴-=(2)()f x 可导，()()f x f x ''∴--= 故()()00f f ''-=()200f '= 即()00f '=9．设6y x k =+是曲线23613y x x =-+的一条切线，则k =解： （1）6,66,666,2y y x x x ''==-∴-==（2）62346213,12121213,k k ⨯+=⨯-⨯+∴+=-+故1k =10． 若()y f x =满足：()()0f x f =x +()x +α，且()0lim 0x x x →α=则()0f '=解：()()()000lim 0x f x f f x →-'=-()0lim 101x x x x α→-==+=11． 设()f x 在2x =连续，且(2)f =4， 则2214lim ()24x f x x x →⎛⎫-= ⎪--⎝⎭解： 原式=2224(2)lim 4x x f x →+--2114lim 4124x x →==⋅=+12．()5sin 1()x x f x x x ⋅-=-的间断点个数为解： 令()()()520,1110x x x x x x -=-++=0,1,1x x x ==-=为间断点，故()f x 有三个间断点三 、计算题（每小题8分，共64分）13． 已知2sin 21,0(),0ax x e x f x xa x ⎧+-≠⎪=⎨⎪=⎩在(),-∞+∞上连续，求a 的值解：()f x 在0x =连续()200sin 21lim lim axx x x e f x x →→+-∴=200sin 21lim lim 22ax x x x e a x x →→-=+=+且()0,22f a a a =∴+=故2a =-14． 讨论1,0()0,01ln ,11x e x f x x xx x ⎧⎪<⎪=≤≤⎨⎪⎪>-⎩在0,1x x ==连续性解：（1）在0x =处，10lim 0,xx e -→= 0lim 00x +→=且()00f =()f x ∴在0x =处连续（2）在1x =处，1lim 00,x -→= ()10ln 1ln 1lim lim 11x x t x x t x t++→→+-===- ()f x ∴在1x =不连续15． 设()f x 有连续的导函数，且()()00,0f f b '==若()()sin ,0,0f x a x x F x xA x +⎧≠⎪=⎨⎪=⎩在0x =连续，求常数A 。

江苏省2013年普通高校“专转本”选拔考试高等数学 试题卷（二年级）注意事项：1、本试卷分为试题卷和答题卡两部分，试题卷共3页，全卷满分150分，考试时间120分钟．2、必须在答题卡上作答，作答在试题卷上无效。

作答前未必将自己的姓名和准考证号准确清晰地填在试题卷和答题卡上的指定位置。

3、考试结束时，须将试题卷和答题卡一并交回。

一、选择题（本大题共6小题，每小题4分，满分24分。

在下列每小题中，选出一个正确答案，请在答题卡上将所选项的字母标号涂黑）1、当0→x 时，函数()ln(1)f x x x =+-是函数2)(x x g =的( ) A.高阶无穷小 B.低阶无穷小 C.同阶无穷小 D.等价无穷小2、曲线22232x xy x x +=-+的渐近线共有( )A. 1条B. 2条C. 3条D. 4条3、已知函数sin 20()0xx xf x x ⎧<⎪⎪=⎨> ，则点0x =是函数)(x f 的A 、跳跃间断点B 、可去间断点C 、无穷间断点D 、连续点4、设1()y f x=，其中f 具有二阶导数，则22d y dx =A. 231121()()f f x x x x '''-+ B. 431121()()f f x x x x '''+ C. 231121()()f f x x x x'''-- D.431121()()f f x x x x'''- 5、下列级数中收敛的是A 、211n n n∞=+∑B 、1()1nn n n ∞=+∑ C 、1!2n n n ∞=∑D、13n n ∞=∑6、已知函数)(x f 在点1x =处连续，且21()1lim 12x f x x →=-，则曲线()y f x =在点(1,(1))f 处的切线方程为A. 1y x =-B. 22y x =-C. 33y x =-D. 44y x =- 二、填空题（本大题共6小题，每小题4分，共24分）7、设函数1sin 0()0x x f x xa x ⎧≠⎪=⎨⎪=⎩ 在点0=x 处连续，则常数a = ▲ ． 8、已知空间三点(1,1,1),(2,3,4),(3,4,5)A B C ，则ABC ∆的面积为 ▲ ．9、设函数)(x y y =由参数方程2311x t y t ⎧=+⎪⎨=-⎪⎩所确定，则221x d y dx == ▲ ．10、设向量→→b a ,互相垂直，且，，23==→→b a ，则=+→→b a 2 ▲ ．11、设10lim()x x a x e a x→+=-，则常数=a ▲ ． 12、幂级数1n nn ∞=∑的收敛域为 ▲ ． 三、计算题（本大题共8小题，每小题8分，共64分）13、求极限01lim ln(1)x x e x x →⎡⎤-⎢⎥+⎣⎦．14、设函数(,)z z x y =由方程3331z xy z +-=所确定，求dz 及22zx∂∂．15、求不定积分2cos 2x xdx ⎰． 16、计算定积分20⎰　．17、设函数223(,)x yz f x e+=，其中函数f 具有二阶连续偏导数，求2zy x∂∂∂．18、已知直线10330x y z x y z -+-=⎧⎨--+=⎩平面∏上，又知直线23132x ty t z t=-⎧⎪=+⎨⎪=+⎩与平面∏平行，求平面∏的方程．19、已知函数()y f x =是一阶微分方程dyy dx=满(0)1y =的特解，求二阶常系数非齐次线性微分方程32()y y y f x '''-+=的通解．20、计算二重积分Dxdxdy ⎰⎰，其中D 是由曲线0)y x =>与三条直线,3,0y x x y ===所围成的平面闭区域．四、综合题（本大题共2小题，每小题10分，共20分） 21、设平面图形D由曲线x =y =1y =围成，试求：（1）平面图形D 的面积；（2）平面图形D 绕x 轴旋转一周所形成的旋转体的体积． 22、已知21132()(95)x F x t t dt =-⎰是函数()f x 的一个原函数，求曲线)(x f y =的凹凸区间与拐点．五、证明题（本大题共2小题，每小题9分，共18分） 23、证明：当1x >时，2(1ln )21x x +<-． 24、设函数()f x 在[,]a b 上连续，证明：函数2()[()()]a b baaf x dx f x f a b x dx +=++-⎰⎰．江苏省2013年普通高校“专转本”统一考试高等数学（二年级） 试卷答案一、选择题（本大题共6小题，每小题4分，共24分）1、C2、C3、B4、B5、D6、A 二、填空题（本大题共6小题，每小题4分，共24分）7、0 89、34 10、2 11、ln y x x cx =+ 12、11[,)22- 三、计算题（本大题共8小题，每小题8分，共64分）13、原式=20001ln(1)ln(1)1lim lim lim ln(1)2x x x x x x x e xe xe x xe x x x x xx→→→+--+-++==+213(1)lim22x x x x e e xe x →++++==14、令32(,,)331,3,3,33x y z F x y z z xy z F y F x F z '''=+--===-22222233,,33133111y x z z F F z y y z x x y xdz dx dy x F z z y F z z z z ''∂∂=-=-==-=-=∴=+''∂--∂----22222222223()(2)()2211(1)(1)(1)z z y y y z yz z y z x x z z x x x z z z ∂∂∂--∂∂∂∂--=====∂∂∂--- 15、22221111cos 2sin 2sin 2sin 2sin 2cos 22222x xdx x d x x x x xdx x x xd x ==-=+⎰⎰⎰⎰22111111sin 2cos 2cos 2sin 2cos 2sin 2222224x x x x xdx x x x x x C =+-=+-+⎰ 16、令2sin ,2cos ,0,0;2,2x t dx tdt x t x t π======，则原式=22222000222cos 12cos cos 12(1)22cos 1cos 2cos 2cos 22ttt dt dt dt dt t ttt ππππ-===-++⎰⎰⎰⎰　 2222011tan 12222cos 2t t dt dtπππππ=-=-=-⎰⎰　17、2232323232212223,(22)36x yx y x y x y z z f e f x f e e e f y y x++++∂∂''''''=⋅⋅=⋅+⋅⋅+∂∂∂ 18、直线方向向量12(1,1,1)(1,3,1)(4,2,2),(3,1,2),S S →→=-⨯--=-=-平面∏的法向量12(4,2,2)(3,1,2)(6,2,10),n S S →→→=⨯=-⨯-=-在第一条直线上任取一点(1,1,1)，该点也在平面上，所以平面方程为6(1)(2)(1)10(1)0x y z -+--+-=即3570x y z -+-=19、由dyy dx=得111111,,ln ,,x C C C x x x dy dx dy dx y x C y e e e y e e Ce y y +===+===±=⎰⎰，由(0)1y =得1C =,所以xy e =，即212,320,1,322x e r y y r r r y -'''-+==+==， 齐次方程的通解为212x xY C e C e =+.令特解为,,x x x y xAe y Ae xAe **'==+，,x x x y Ae Ae xAe *''=++代入原方程得：,1x x Ae e A -==-，所以通解为212x x xy Y C e C e xe ==+-20、原式=333cos 4cos 442002127cos cos (8cos )33cos r d r rdr d d πππθθθθθθθθθ==-⎰⎰⎰⎰ 24011(27tan 8sin )(27tan 8sin )933443πππθθ=-=-=-. 四、综合题（本大题共2小题，每小题10分，共20分）21、（1）131232215)(2)333 S y dy y y==⋅+=⎰（2）0222502221010821[1][1()]()()42802510 xx x xV dx dx x xπππππππ--=-+-=++-=+=⎰⎰22、25233()2(95)1810,f x x x x x x=-=-23()3020f x x x'=-，13()20200,f x x-''=-=解得1x=，另外0x=为二导不存在的点，通过列表分析得：在(,0),(1,)-∞+∞凸，在(0,1)凹，拐点为(0,0),(1,8)。

2013年专升本（高等数学一）真题试卷(题后含答案及解析) 题型有：1. 选择题 2. 填空题 3. 解答题选择题1．( )A．eB．1C．eD．-e正确答案：C2．设y=3+x2，则y’=( )A．2xB．3+2xC．3D．x2正确答案：A3．设y=2x3，则dy=( )A．2x2dxB．6x2dxC．3x2dxD．x2dx正确答案：B4．设y=-2ex，则y’=( )A．exB．2exC．-exD．-2ex正确答案：D5．设y=3+sinx，则y’=( )A．-cosxB．cosxC．1-cosxD．1+cosx正确答案：B6．( )A．x2B．2x2C．xD．2x正确答案：A7．( )A．B．-3ln|x|+CC．D．3ln|x|+C正确答案：D8．( )A．B．0C．D．1正确答案：B9．设z=3x2+5y，则( )A．5yB．3xC．6xD．6x+5正确答案：C10．微分方程(y’)2=x的阶数为( ) A．1B．2C．3D．4正确答案：A填空题11．=________。

正确答案：2e12．设y=(x+3)2，则y’=________。

正确答案：2(x+3)13．设y=2ex-1，则y”=________。

正确答案：2ex-114．设y=5+lnx，则dy=________。

正确答案：15．∫cos(x+2)dx=________。

正确答案：sin(x+2)+C16．∫012exdx=________。

正确答案：2(e-1)17．过坐标原点且与平面2x-y+z+1=0平行的平面方程为________。

正确答案：2x-y+z=018．设z=xy，则dz=________。

正确答案：ydx+xdy19．幂级数的收敛半径R=________。

正确答案：120．设区域D={(x，y)|x2+y2≤4}，则=________。

正确答案：π解答题21．设函数f(x)=在x=1处连续，求a。

正确答案：(x2-2x+3)=2。