2014年同方专转本高数模拟试卷4

江苏省 2014 年一般高校“专转本”一致考试模拟试（一）高等数学注意事项：1.考生务势必密封线内的各项填写清楚。

2.考生一定要钢笔或圆珠笔将答案直接写在试卷上，写在底稿纸上无效。

3.本试卷五大题 24 小题，满分 150 分，考试时间 120 分钟。

一、选择题（本大题共 6 小题，每题 4 分，共 24 分，在每题给出的四个选项中，只有一项为哪一项切合要求的，请把所选项前得字母填在题后的括号内）。

1. 已知 limx 3ax 2 b 存在，则常数 a,b 的值分别为（）x 1x 1A. a 1,b4B. a 1,b 4C. a 1,b 4D. a1,b 42. 函数 f ( x) x 22x的可去中断点是（）x ( x 1)(x 24)A. x 0B.x 1C. x2D. x 23.当 x 0 时，以下无量小中与x 不等价的是（）A. x10x2B. ln(1 x x 2 )C. sin(2sin x x 2 )D. e x2x 2 14.设 f ( x)的一个原函数是ln2x，则 xf ( x 21)dx （）A.ln( x 21) cln 2 (x 21) C. ln( x2 1) c2( x 21) cx 21B.x 2cD. ln15.以下级数绝对收敛的是（ ）A.( 1)n1 1B.( 1)n1 k 2C.( 1)nlnn 1D.( 1)n ln n 1n 1n 2 n 1n 3 n 1nn 1n11 x 26.二重积分dx1 xf ( x, y)dy 互换积分序次后得（）11 y 211 yA.d y 1 yf ( x, y)dxB.dy 1y 2 f ( x, y)dx1 dy1 y 21dy 1 y 2C.1 yf ( x, y)dxD.1 yf (x, y) dx二、填空题（本大题共 6 小题，每题 6 分，共 24 分，请把正确答案的结果添在划线上）。

7、若x lim( x2a) x8， a x a8、设 f ( x) 是连续函数， F ( x) e xf (t )dt ，则 F ( x)2x9、以 A(1,2,0), B(1,3,1),C(2,1,2)为极点的三角形面积=10、设函数zz( x, y) 由方程 e z z x2 y3所确立，则zx11、定积分11x2 )2 dx (x112、幂级数( x2)n的收敛域为n3nn1三、计算题（本大题共8 小题，每题8 分，共 64 分）。

英语应试模拟第4套Ⅰ. Phonetics (5 points)Directions: In each of the following groups of words, there are four underlined letters or letter combinations marked A, B, C and D. Compare the underlined parts and identify the one that is different from the others in pronunciation. Mark your answer by blackening the corresponding letter on the Answer Sheet.1. A. MondayB. onionC. wonderfulD. got2. A. baseB. pleaseC. amuseD. because3. A. dreamB. threatC. beatD. heat4. A. visionB. pensionC. leisureD. measure5. A. virusB. T-shirtC. thirdD. virtueⅡ. Vocabulary and Structure (15 points)Directions: There are 15 incomplete sentences in this section. For each sentence there are four choices marked A, B, C and D. Choose one answer that best completes the sentence and blacken the corresponding letter on the Answer Sheet.6. You should buy a computer, you would be able to send emails.A. SoB. ThenC. WellD. Therefore7. The Smiths will move to New York. But we hope to with them.A. keep touchB. keep at touchC. keep on touchD. keep in touch8. "Do you go out on weekends? Yes,we sometimes go out.A. very oftenB. in a timeC. once upon a timeD. once in a while9. My father gave as my birthday gift.A. to me the bike last weekB. to me the bike last nightC. me the bike last nightD. last night the bike to me10. He was have a doctor's appointment today.A. supposedB. supposingC. supposed toD. supposing to11. he was seen to be an aggressive politician, he was a quiet and loving family man at home.A. AlthoughB. DespiteC. In spite ofD. Nevertheless12. I wish you me how to make such a sauce last time.A. teachB. taughtC. have taughtD. had taught13. I don't think my eyes are as good as they used to be. I need to have themA. testedB. cleanedC. servicedD. cared14. The Smiths their breakfast when the morning post came.A. hadB. were havingC. has been havingD. are having15. that she liked curry,he would have brought her to an Indian restaurant.A. If John knowsB. Had John knownC. If John knewD. Dose John know16. My students' sleepless nights became as the examinations approached.A. so frequentlyB. much more frequentlyC. more frequentD. far more frequent than17. After this tour you have whole afternoon free to explore the city.A. aB. anC. /D. the18. My daughter tells me that it is fashionable to wear jeans that are very around the waist.A. looseB. loseC. freeD. restrict19. If you're forbidden to do something, it means you do it,A. haven't got toB. needn'tC. couldn'tD. may not20. He told me that he in Spain the previous year.A. has been workingB. had been workingC. has workedD. had -been workedⅢ. Cioze (30 points)Directions: For each blank in the following passage, there are four choices marked A,B,C andD. Choose the one that is most suitable and mark your answer by blackening the corresponding letter on the Answer Sheet.Most people are born with the natural ability to 21stories, but only a rare few have the determination to become 22authors, and even fewer have the joy of seeing their novels top the23 of bestselling books. Some of the world' s famous crime writers have achieved the24 success of all. Who can 25 the appeal of famous detectives like Sherlock Holmes, Miss Marple or Hercule Poirot? Even if you haven' t read the original books you will have seen 26 in films or on the TV.If you have an ambition to become the 27Agatha Christie what should you do? The best starting 28 is to read lots of examples of crime fiction written by good authors. You will need a notebook to carry around with you or 29 better, some loose 30 of paper that you can 31 notes on and then file into a folder. After all, the most everyday situation -- for example, watching a woman get 32 a train -- may be the 33 for your first bestseller. Like any good recipe you have to know the main 34 of a successful novel. These are: an/a35 story, strong characters and a memorable setting.21. A. sayB. tellC. speakD. announce22. A. amateurB. specialC. professionalD. authoritative23. A. queueB. listC. rowD. line24. A. largestB. greatestC. highestD. prime25. A. denyB. refuseC. insistD. hide26. A. the storyB. the bookC. authorsD. them27. A. nextB. followingC. subsequentD. later28. A. spotB. tipC. pointD. mark29. A. evenB. stillC. veryD. so30. A. stripsB. sheetsC. layersD. pieces31. A. completeB. doC. fillD. make32. A. outB. downC. offD. from33. A. notionB. ideaC. inventionD. inspiration34. A. partsB. ingredientsC. slicesD. sections35. A. originalB. secondhandC. InterestingD. well-knownⅣ. Reading Comprehension (60 points)Directions: There are five reading passages in this part. Each passage is followed by four questions. For each question there are four suggested answers marked A,B, C and D. Choose the best answer and blacken the corresponding letter on the Answer Sheet.Passage OneWhen you imagine the desert,you probably think of a very hot place covered with sand. Although this is a good description for many deserts, Earth' s largest desert is actually a very cold place covered with ice: Antarctica.In order for an area to be considered a desert, it must receive very little rainfall. More specifically, it must receive an average of less than ten inches of precipitation--which can be rain, sleet, hail, or snow--on the ground every year. Antarctica, the coldest place on earth, has an average temperature that usually falls below the freezing point. And because cold air holds less moisture than warm air, the air in Antarctica does not hold much moisture at all. This is evident in the low precipitation statistics recorded for Antarctica. For example, the central part of Antarctica receives an average of less than two inches of snow every year. The coastline of Antarctica receives a little bit more--between seven and eight inches a year. Because Antarctica gets so little precipitation every year, it is considered a desert.When precipitation falls in hot deserts, it quickly evaporates back into the atmosphere. The air over Antarctica is too cold to hold water vapor, so there is very little evaporation. Due to this low rate of evaporation, most of the snow that falls to the ground remains there permanently, eventually building up into thick ice sheets. Any snow that does not freeze into ice sheets becomes caught up in the strong winds that constantly blow over Antarctica. These snow-filled winds can make it look as if it is snowing. Even though snowfall is very rare there, blizzards are actually very common on Antarctica.36. The main purpose of paragraph 1 is toA. accept a conclusionB. introduce an argumentC. provide a brief historyD. deny a common belief37. The best title for this passage would beA. Earth's Many DesertsB. Antarctica: The Coldest Place on EarthC. A Desert of IceD. Unusual Blizzards38. Africa' s Sahara Desert is the second-largest desert on earth. Based on the information in the passage ,what characteristic must the Sahara share with Antarctica?A. Low temperatures.B. High temperatures.C. Frequent blizzards.D. Low precipitation.39. Based on the information in the final paragraph, it can be understood that blizzards in Antarctica are mainly the result ofA. strong windsB. large amounts of snowfallC. low amounts of precipitationD. freezing cold temperaturesPassage TwoStudent Volunteers Needed!On Saturday, December 12th,from 10 A. M. until 4 P. M. , Carverton Middle School will be holding a music festival in the school gymnasium. The special event will feature a variety ofInterested students should speak with Ms. Braxton,the music teacher. Students who would like to help at the festival must have written permission from a parent or guardian.40. What time will the festival begin?A. 10A. M.B. 11A.M.C. 1P.M.D. 2P. M.41. In line 2,the word feature is closest in meaning toA. lookB. keepC. includeD. entertain42. What job will be done the day before the festival begins?A. Making posters.B. Setting up the gym.C. Cleaning up the gym.D. Helping the performers.43. Who is told to talk to Ms. Braxton?A. Parents.B. Students.C. Teachers.D. Performers.Passage ThreeWhen another old cave is discovered in the south of France, it is not usually news. Rather, it is an ordinary event. Such discoveries are so frequent these days that hardly anybody pays heed to them. However, when the Lascaux cave complex was discovered in 1940,the world was amazed. Painted directly on its walls were hundreds of scenes showing how people lived thousands of years ago. The scenes show people hunting animals, such as bison or wild cats. Other images depict birds and, most noticeably, horses, which appear in more than 300 wall images, by far outnumbering all other animals.Early artists drawing these animals accomplished a monumental and difficult task. They did not limit themselves to the easily accessible walls but carried their painting materials to spaces that required climbing steep walls or crawling into narrow passages in the Lascaux complex. Unfortunately, the paintings have been exposed to the destructive action of water and temperature changes, which easily wear the images away. Because the Lascaux caves have many entrances, air movement has also damaged the images inside.Although they are not out in the open air, where natural light would have destroyed them long ago, many of the images have deteriorated and are barely recognizable. To prevent further damage, the site was closed to tourists in 1963,23 years after it was discovered.44. Based on the passage, what is probably true about the south of France?A. It is home to rare animals.B. It has a large number of caves.C. It is known for horse-racing events.D. It has attracted many famous artists.45. According to the passage, which animals appear most often on the cave walls?A. Birds.B. Bison.C. Horses.D. Wild cats.46. Why was painting inside the Lascaux complex a difficult task?A. It was completely dark inside.B. The caves were full of wild animals.C. Painting materials were hard to find.D. Many painting spaces were difficult to reach.47. According to the passage, all of the following have caused damage to the paintings EXCEPTA. temperature changesB. air movementC. waterD. lightPassage FourGold used in jewelry is mixed with harder metals to add strength and durability. The metals added can also be used to change gold' s color, giving it a fashionable rose or white tint, or to lighten or darken the natural yellow tone of pure gold. Mixtures like these, of less costly metals with more valuable ones, are called alloys. Copper and silver are the most common metals mixed with gold to make yellow gold jewelry. White gold is usually made with an alloy of gold andnickel.The measure of gold' s purity is called a karat. The higher the karat rating, the higher the amount of pure gold: 24 karat is pure gold, 15 karat is 75% pure gold, 14 karat is 58.5% pure gold, and 9 karat is 37.5% pure gold. All other things being equal, the higher the percentage of pure gold used in the alloy, the more valuable and expensive the jewelry will be.Gold jewelry pieces are usually stamped with a marking to identify the karat amount. While gold that is 24K is too soft for jewelry, 18K, 14K and 9K gold are all appropriate for jewelry, and they all make pieces that look great and wear beautifully.48. Which of the following statements best captures the main idea of this passage?A. Although gold is very valuable, it is also very expensive.B. Gold jewelry is stamped with its karat weight.C. Gold jewelry is made using alloys.D. Colored gold is more valuable than white gold.49. Based on information in the passage, it can be understood that pure gold isA. not used to make ringsB. stamped with 100KC. an alloy of different metalsD. colorless50. According to the passage, the use of other metals in gold alloysⅠ. can be used to make the gold different colorⅡ. makes jewelry more expensiveⅢ. makes gold more flexibleA. I onlyB. I and ⅡonlyC. Ⅱand ⅢonlyD. I , Ⅱ,and Ⅲ51. Which of the following statements best explains the relationship between gold and its karat rating?A. The lower the karat rating,the purer the gold.B. The higher the karat rating,the more expensive the gold.C. The higher the karat rating,the less valuable the mixture.D. The lower the karat rating,the purer the alloy.Passage FiveIt is bad to have food stuck between your teeth for long periods of time. This is because food attracts germs, germs produce acid ,and acid hurts your teeth and gums. Flossing helps to remove the food that gets stuck between your teeth. This explains why flossing helps to keep your mouth healthy, but some doctors say that flossing can be also good for your heart.It may seem strange that something you do for your teeth can have any effect on your heart. Doctors have come up with a few ideas about how flossing works to keep your heart healthy. One idea is that the germs that hurt your teeth can leave the mouth and travel into your blood. Germs that get into the blood can then attack your heart. Another idea is based on the fact that when there are too many germs in your mouth, the body tries to fight against these germs. For some reason, the way the body fights these mouth germs may end up weakening the heart over time. Not every doctor agrees about these ideas. Some doctors think that the link between goodflossing habits and good heart health is only a coincidence. A coincidence is the occurrence of two or more events at one time apparently by mere chance. The incidence of these events is completely random, as they do not admit of any reliable cause and effect relationship between them. For example, every time I wash my car, it rains. This does not mean that when I wash my car, I somehow change the weather. This is only a coincidence.The theory that flossing your teeth helps to keep your heart healthy might not be true. But every doctor agrees that flossing is a great way to keep your teeth healthy. So even if flossing does not help your heart, it is sure to help your teeth. This is enough of a reason for everyone to floss their teeth every day.52. Flossing effectively helps to keep your mouth healthy by preventingA. germs from producing acidB. food from entering your bodyC. germs from entering into your bloodD. acid from contacting your teeth and gums53. In paragraph 2, the author explains how having too many germs in your mouth can “end up weakening the heart. " Using the passage as a guide, it can be understood that doctors areA. reluctant to hypothesizeB. confident in their estimationsC. extremely knowledgeableD. uncertain but speculative54. Using information in paragraph 3 as a guide, which of the following is the best example of a coincidence?A. Jim wakes up with a sore throat. He eats a piece of bacon for breakfast. By noon, he feels much better. Jim decides that the bacon has cured his sore throat.B. Laura remembers to brush her teeth every day, but she only remembers to floss once a week. She writes a note to herself, reminding herself to floss and sticks it to her bathroom mirror.C. Mario is not very good at baseball. He practices playing every day. After a several months of practice, he is a much better baseball player.D. Jai has a bad heart. Her doctor tells her to eat more vegetables and less junk food. After nearlya year of doing this, the doctor tells Jai that her heart is doing much better.55. Which of the following best states the main idea of the final paragraph?A. Because doctors do not agree that flossing wilt help your heart, it is useless to floss.B. It is a fact that flossing can help your heart as well as your teeth.C. Even if flossing is only good for your teeth, you should still do it every day.D. There is no good reason to believe that flossing will help your heart, but it is still a good idea to do it every day.V. Daily Conversation (15 points)Directions: Pick out five appropriate expressions from the eight choices below and complete the following dialogue by blackening the corresponding letter on the Answer Sheet.A. Thank you for coming alongB. I'm readyC. I was looking for a part-time jobD. When will ! knowE. It is a great jobF. That about covers itG. How long do ! have to waitH. I haveInterviewer: So, you've applied for the Saturday position, right?John: Yes, 56Interviewer: Can you tell me what made you reply to our advertisement?John: Well, 57 to help me through college.Interviewer: Do you know exactly what you would be doing as a shop assistant?John: Well I imagine I would be helping customers, keeping a check on the supplies in the store, and preparing the shop for business.Interviewer: 58 ,you would also be responsible for keeping the front of the store tidy. Have you any previous work experience?John: Yes. I worked part-time at a take-away in the summer holidays.Interviewer: I think I have asked you everything I wanted to.59 to the interview.John: Thank you.60 if ! have been successful?Interviewer: We'll be making our decision next Monday. We'll give you a call.Ⅵ. Writing (25 points)Directions: For this part, you are supposed to write a letter of application in English in 100-120 words based on the following situation. Remember to write it clearly.61．你(Li Yuan)准备参加一个俱乐部(文艺俱乐部、书画俱乐部、旅游俱乐部或者体育俱乐部)。

江苏省2014年普通高校对口单招文化统考数学模拟试卷（四）一、单项选择题（本大题共10小题，每小题4分，共40分．在下列每小题中，只有一个正确选项，请在答题卡上将所选的字母标号涂黑）1．若全集{}{}0,1,2,3,42,3U U C A ==且，则集合A 的真子集共有 （ ▲ ）A ．3个B ．5个C ．7个D ．8个 2．函数2log 2-=x y 的定义域是 （ ▲ ）A ．),3(+∞B ．),3[+∞C ．),4(+∞D ．),4[+∞3．函数1)4(cos 22--=πx y 是 （ ▲ ）A ．最小正周期为π的奇函数 B. 最小正周期为π的偶函数 C. 最小正周期为2π的奇函数 D. 最小正周期为2π的偶函数4．设向量)2(，x a -=，)31(，-=b ，且b a -与b 共线，则=x （ ▲ ） A ．31B ．32 C ．31- D ．32- 5．已知i 是虚数单位，复数z 的共轭复数是z ，若4)1(2=+z i ，则=z （ ▲ ）A ．2B ．i 2C ．2-D ．i 2- 6．若d c b a >>,，则下面不等式中成立的一个是 （ ▲ ）A ．c b d a +>+B ．bd ac >C ．dbc a > D ．b c ad -<- 7．曲线的参数方程为⎩⎨⎧+=-=2sin 31cos 3t y t x (t 是参数)，则曲线是 （ ▲ ）A ．线段B ．双曲线的一支C ．圆D ．射线8．关于直线m 、n 与平面α、β，下列四个命题正确的是 （ ▲ ）A ．βα//,//n m 且βα//，则n m //；B ．βα⊥⊥n m ,且βα⊥，则n m ⊥；C ．βα//,n m ⊥且βα//，则n m //；D ．βα⊥n m ,//且βα⊥，则n m //. 9．由点M(5,3)向圆222690x y x y +-++=所引切线长是 （ ▲ ）A ．B. C. 51 D . 110．设直线的方程是0=+By Ax ，从2，3，4，5，6这五个数中每次取两个不同的数作为A 、B 的值，则所得不同直线的条数是（ ▲ ）A ．20B ．19C ．18D ．16二、填空题（本大题共5小题，每小题4分，共20分） 11．若αtan =3，则αα2cos 2sin ． 12．5()a x +展开式中2x 的系数为－10， 则实数a 的值为 .13．已知变量x 、y 满足条件⎪⎩⎪⎨⎧≤-+≤-≥09201y x y x x 则z x y =+的最大值是 ．14．过点)1,4(-A 和双曲线116922=-y x 右焦点的直线方程为 ． 15．已知定义域为R 的偶函数)(x f 在区间[)+∞,0上单调增加，则满足)31()12(f x f <-的 x 取值范围是 ．三、解答题（本大题共10小题．其中17—20四小题中，考生可任选其中两题解答，每题9分，多解不给分．共计90分） 16．（8分）解不等式：22531649x x --⎛⎫<⎪⎝⎭．第17—20题是选做题，每题9分，考生可任选其中两题解答，多解不给分． 17．（9分）在一次面试中，有A ，B ，C 三位考官，当至少有两位考官认为应试者面试合格，才能认定应试者面试合格．①写出逻辑关系；②化简逻辑关系式．)0()1(2<+x x18．（9分）已知函数=y 8 )0(=x ，请画出程序框图，要求输入自变量x 的值， )1()1(2>-x x输出函数值y ．19．（9分）某旅游公司第一季度接待国内某旅游景点的游客双飞价格如下表（单位：游客①试用数组表示每月的旅游收入；②试用数组运算求第一季度旅行社接待到四个城市旅游的月平均收入．20．（9分）某项工程的网络图如图所示：第20题（1）写出所有不同的路径；（2）指出关键路径及总工期．21（11分）在锐角三角形ABC中，a，b，c分别为内角A，B，C所对的边，且满足-Aa．(1)求角B的大小；(2)若b＝7，c＝2，求ABCb3=sin2∆的面积．22．（12分）某地区有小学21所，中学14所，大学7所，现采取分层抽样的方法从这些学校中抽取6所学校对学生进行视力调查．（1）求应从小学、中学、大学中分别抽取的学校数目．（2）若从抽取的6所学校中随机抽取2所学校做进一步数据分析，求抽取的2所学校均为小学的概率．23．（13分）在数列{}n a 中，1a =1，c a a n n +=+1(c 为常数，+∈N n )，且1a ，2a ，5a 成等比数列．(1) 求数列{}n a 的通项公式； (2) 设12+=n a n b ，数列{}n b 的前n 项和为n T ，①求证：数列{}n b 是等比数列；②求5T ．24．（14分）一辆新汽车使用一段时间后，就值不到原来的价钱了．假若一辆新车价值18万元，按下列方式贬值：从第二年起，每年的车价是上一年车价的32．如果从购买日起t 个月后汽车价贬值量为w 万元．（1）求出汽车贬值量w 万元关于使用时间t 个月的函数关系式（贬值量＝原价－汽车现在价值）；（2）求18个月后此车价值？（45.26≈）25．（14分）已知椭圆的离心率e =12,F F ，定点P (，点2F 在线段1PF 的中垂线上．（1）求椭圆C 的方程；（2）设直线:l y kx m =+与椭圆C 交于M 、N 两点，直线22,F M F N 的倾斜角分别为,,αβαβπ+=且，求证：直线l 过定点，并求该定点的坐标．江苏省2014年普通高校对口单招文化统考数学模拟试卷（四）答案及评分参考11．6 12．－1 13．6 14．05=--y x 15．⎥⎦⎤⎢⎣⎡3231，三、解答题（本大题共10小题，共计90分．其中17—20四小题中，考生可任选其中两题解答，每题9分，多解不给分．） 16．（本小题8分）{}分或不等式解集为分或解得分分上是单调减函数，在分）解：由题意得：（8..............................................................316.....................................................................................311......................................................................................2521.................................................)()43(1.......................................................)43(4322522>-<∴>-<->--∴∞+-∞=<---x x x x x x x y x x x 第17—20题是选做题，每题9分，考生可任选其中两题解答，每题10分，多解不给分． 17．（本小题9分）解：（1）分4..................................................ABC C B A C AB BC A Y +++=（2）分9.......................................................................BC AC AB Y ++=解： 19．（本小题9分）（1）解： 1月份的旅游收入数组为：)9017180145(1，，，=a ，2月份旅游收入数组为： )5.1622251215.227(2，，，=a ，2月份旅游收入数组为：)5.7396605.94(3，，，=a …………………………… 4分 （2）解：第一季度旅行社四个城市的月平均收入[)5.1622251215.227()9017180145(31)(31321，，，，，，+=++=a a a b ]分，，，9.....................................)5.7396605.94(+ 20．（本小题9分）解：（1）路径有：EJ G A EC I G A E CD A EF B H A →→→→→→→→→→→→→→④③②① ……………………………4分（2）关键路径：E F B H A →→→→………………………………………9分分，是锐角三角形分，解：4 (3)23sin 2........................sin 2sin 323sin 0sin 23)1(π=∴=∴∆==∴=-B B ABC BAb a A A b a分分分是锐角三角形，11 (2)33sin 218 (14)213sin cos cos sin )sin(sin 6 (7)72cos 721sin sin sin sin )2(==∴=+=+=∴=∴∆==∴=∆A bc S C B C B C B A C ABC B b c C b B c C ABC22．（本小题12分）解：(1)从小学、中学、大学中分别抽取的学校数目为3，2，1. …………………………5分(2)（法一）①在抽取到的6所学校中，3所小学分别记为A 1，A 2，A 3，2所中学分别记为A 4，A 5，大学记为A 6，则抽取2所学校的所有可能结果为{A 1，A 2}，{A 1，A 3}，{A 1，A 4}，{A 1，A 5}，{A 1，A 6}，{A 2，A 3}，{A 2，A 4}，{A 2，A 5}，{A 2，A 6}，{A 3，A 4}，{A 3，A 5}，{A 3，A 6}，{A 4，A 5}，{A 4，A 6}，{A 5，A 6}，共15种．…………………………………………9分②从6所学校中抽取的2所学校均为小学(记为事件B )的所有可能结果为{A 1，A 2}，{A 1，A 3}，{A 2，A 3}，共3种．所以P (B )＝315＝15．……………………………………………12分（法二）从6所学校中抽取的2所学校均为小学记为事件B …………………6分则31153)(2623===C C B P ……………………………………………12分23．（本小题13分）{}分分舍去或成等差数列，，分，的等差数列是公差为解：6......................................................................................122)1(15........................................................................................).........(0241)1(2. (4111))1(25215211-=⨯-+=∴==∴+=+∴+=+=∴=∴+=+n n a c c cc a a a c a c a a c a c a a n n n n{}分，②分的等比数列是公比为，分①13 (13644)-1)4-1(41)1(410......................................................................447...........................................................................................422)2(55151121==--==∴=∴===++q q b T b b b b b n nn n n a n n 24．（本小题14分）解：（1）建立汽车的现价Q 与使用时间t 个月后的函数关系)(t f Q = 当0=t 时，即刚买来，显然)0(f ＝180000当12=t 时，即买了一年，)12(f ＝180000×32＝120000 当买了两年后，)24(f ＝180000×2)32(＝80000 一般地，)12(n f ⨯＝180000×n)32(设n t 12=，则)(t f ＝180000×12)32(t…………………………………………………6分则-=18w )(t f ，)()32(18000018000012N t w t∈⨯-=∴………………………9分分个月后汽车价值：13 (980003)632180000)32(180000)32(18000018)2(231218=⨯⨯=⨯=⨯=Q 25．（本小题14分）分得：由分，、，设分椭圆的方程为：分，分舍去，，解得：分的中垂线上，在，，，设解：8......................................0224)12(126..................................................................).........()()2(5. (12)4 (22)21.3....................).........(3712)3()2(1....................................................2)0()0()1(2222222112221222121221=-+++⎪⎩⎪⎨⎧+==+=+∴=∴===-===+-∴==∴-m km x x k m kx y y x y x N y x M y x a a a c e c c c c c F F P F PF F c F c F分，9 (1)2221242221221+-=+-=+∴k m x x k km x x 直线22,F M F N 的倾斜角分别为,,αβαβπ+=且分直线过定点直线方程可化为：，分化得：，分14..............................................................................).........0,2()2(202)124()(1222212...................................................02))((2111110..........................................tan )tan(tan 22221212211221122∴-=∴-=∴=-+-⨯-++-⨯∴=-+-+-+-=-+∴--=-∴-=-=-==∴x k y km m k kmk m k m k m x x k m x kx x m kx x m kx x y x y k k N F M F ββπα。

江苏省20XX 年普通高校“专转本”统一考试模拟试（一）高等数学注意事项：1.考生务必将密封线内的各项填写清楚。

2.考生必须要钢笔或圆珠笔将答案直接写在试卷上，写在草稿纸上无效。

3.本试卷五大题24小题，满分150分，考试时间120分钟。

一、选择题（本大题共6小题，每小题4分，共24分，在每小题给出的四个选项中，只有一项是符合要求的，请把所选项前得字母填在题后的括号内）。

1. 已知312lim1x x ax b x →+-=-存在，则常数,a b 的值分别为（ ） A. 1,4a b ==- B. 1,4a b == C. 1,4a b =-=- D. 1,4a b =-=2. 函数222()(1)(4)x xf x x x x -=--的可去间断点是（ ）A. 0x =B. 1x =C. 2x =-D. 2x =3.当0x →时，下列无穷小中与x 不等价的是（ ）A. 210x x - B. 2ln(1)x x+ C. 2sin(2sin )x x + D. 221x e x -- 4.设()f x 的一个原函数是2ln x ，则2(1)xf x dx '+⎰（ ） A.22ln(1)1x c x +++ B. 222ln (1)1x c x +++ C. 2ln(1)x c ++ D. 22ln (1)x c ++5.下列级数绝对收敛的是（ ）A.1(1)nn ∞=-∑B. 1(1)nn ∞=-∑ C. 11(1)ln nn n n ∞=+-∑D.1(1)lnn n ∞=-∑6.二重积分11(,)xdx f x y dy -⎰交换积分次序后得（ ）A.11(,)ydy f x y dx -⎰B.110(,)y dy f x y dx -⎰C.11(,)ydy f x y dx +⎰D.11(,)ydy f x y dx -⎰二、填空题（本大题共6小题，每小题6分，共24分，请把正确答案的结果添在划线上）。

第二部分应试模拟高等数学(－)应试模拟第1套－、选择题：1～10小题，每小题4分，共40分．在每小题给出的四个选项中，只有－项是符合题目要求的．1．当x→0时，3x是x的（）．A．高阶无穷小量B．等价无穷小量C．同阶无穷小量，但不是等价无穷小量D．低阶无穷小量2．设函数f(x)在区间(0，1)内可导f'(x)>0，则在(0，1)内f(x)（）．A．单调增加B．单调减少C．为常量D．既非单调，也非常量3．A．3B．2C．1D．4．设y＝sin(x-2)，则dy＝（）．A．－cosxdxB．cosxdXC．－cos(x－2)dxD．cos(x－2)dx5．A．B．C．D．6．A．B．C．D．7．A．sin(x－1)＋CB．－sin(x－1)＋CC．sinx＋CD．－sinx＋C8．A．B．C．D．9．A．B．C．D．10．A．为所给方程的解，但不是通解B．为所给方程的解，但不－定是通解C．为所给方程的通解D．不为所给方程的解二、填空题：11～20小题，每小题4分，共40分．11．设y＝sin(2＋x)，则dy＝．12．13．14．15．16．17．18．设曲线y＝f(x)在点(1，f(1))处的切线平行于x轴，则该切线方程为．19．20．21．（本题满分8分）22．(本题满分8分)设y=y(x)由方程x2＋2y3＋2xy＋3y－x＝1确定，求y’23．(本题满分8分)24．(本题满分8分)25．(本题满分8分)26．(本题满分10分)求由曲线y＝3－x2与y＝2x，y轴所围成的平面图形的面积及该封闭图形绕x轴旋转－周所成旋转体的体积．27．(本题满分10分)28．(本题满分10分)将f(x)＝ln(1＋x2)展开为x的幂级数．高等数学(－)应试模拟第1套参考答案与解析－、选择题1．【答案】C．【解析】本题考查的知识点为无穷小量阶的比较．应依定义考察由此可知，当x→0时，3x是x的同阶无穷小量，但不是等价无穷小量，故知应选C．本题应明确的是：考察当x→x0时无穷小量β与无穷小量α的阶的关系时，要判定极限这里是以α为“基本量”，考生要特别注意此点，才能避免错误．2．【答案】A．【解析】本题考查的知识点为利用导数符号判定函数的单调性．由于f(x)在(0，1)内有f'(x)>0，可知f(x)在(0，1)内单调增加，故应选A．3．【答案】B．【解析】本题考查的知识点为导数的运算．可知应选B．4．【答案】D．【解析】本题考查的知识点为微分运算．可知应选D．5．【答案】C．【解析】本题考查的知识点为复合函数导数的运算．由复合函数的导数链式法则知可知应选C．6．【答案】B．【解析】本题考查的知识点为定积分运算．因此选B．7．【答案】A．【解析】本题考查的知识点为不定积分运算．可知应选A．8．【答案】D．【解析】本题考查的知识点为偏导数的计算．9．【答案】B．【解析】本题考查的知识点为级数收敛性的定义．10．【答案】B．【解析】本题考查的知识点为线性常系数微分方程解的结构．二、填空题11．【参考答案】cos(2＋x)dx【解析】这类问题通常有两种解法．解法1因此dy＝cos(2＋x)dx．解法2利用微分运算公式dy＝d(sin(2＋x))＝cos(2＋x)·d(2＋x)＝cos(2＋x)dx．12．【参考答案】【解析】本题考查的知识点为初等函数的求导运算．本题需利用导数的四则运算法则求解．本题中常见的错误有这是由于误将sin 2认作sinx，事实上sin 2为－个常数，而常数的导数为0，即请考生注意，不论以什么函数形式出现，只要是常数，它的导数必定为0．13．【参考答案】0．【解析】本题考查的知识点为连续函数在闭区间上的最小值问题．通常求解的思路为：14．【参考答案】【解析】本题考查的知识点为定积分计算．可以利用变量替换，令u＝2x，则du＝2dx，当x＝0时，u＝0；当x＝1时，u＝2．因此15．【参考答案】【解析】本题考查的知识点为二元函数的偏导数计算．16．【参考答案】【解析】本题考查的知识点为二阶线性常系数齐次微分方程的求解．二阶线性常系数齐次微分方程求解的－般步骤为：先写出特征方程，求出特征根，再写出方程的通解．17．【参考答案】3(x－1)－(y＋2)＋z＝0(或3x-y＋z＝5)．【解析】本题考查的知识点为平面与直线的方程．由题设条件可知应该利用点法式方程来确定所求平面方程．所给直线z的方向向量s＝(3，－1，1)．若所求平面π垂直于直线1，则平面π的法向量n ∥s，不妨取n＝s＝(3，－1，1)．则由平面的点法式方程可知3(x－1)－[y－(－2)]＋(z－0)＝0，即3(x－1)－(y＋2)＋z＝0为所求平面方程．或写为3x-y＋z－5＝0．上述两个结果都正确，前者3(x－1)－(y＋2)＋z＝0称为平面的点法式方程，而后者3x-y＋z －5＝0称为平面的－般式方程．18．【参考答案】y＝f(1)．【解析】本题考查的知识点有两个：－是导数的几何意义，二是求切线方程．设切点为(x0，f(x0))，则曲线y＝f(x)过该点的切线方程为y－f(x0)＝f'(x0)(x－x0)．由题意可知x0＝1，且在(1，f(1))处曲线y＝f(x)的切线平行于x轴，因此应有f'(x0)＝0，故所求切线方程为y—f(1)＝0．本题中考生最常见的错误为：将曲线y＝f(x)在点(x0，f(x0))处的切线方程写为y－f(x0)＝f'(x)(x－x0)而导致错误．本例中错误地写为y－f(1)＝f'(x)(x－1)．本例中由于f(x)为抽象函数，－些考生不习惯于写f(1)，有些人误写切线方程为y－1＝0．19．【参考答案】1．【解析】本题考查的知识点为反常积分，应依反常积分定义求解．20．【参考答案】【解析】本题考查的知识点为计算二重积分．其积分区域如图1—1阴影区域所示．可利用二重积分的几何意义或将二重积分化为二次积分解之．解法1解法2化为先对y积分，后对x积分的二次积分．作平行于y轴的直线与区域D相交，沿Y轴正向看，人口曲线为y＝x，作为积分下限；出口曲线为y＝1，作为积分上限，因此x≤y≤1．区域D在x轴上的投影最小值为x＝0，最大值为x＝1，因此0≤x≤1．可得知解法3化为先对x积分，后对y积分的二次积分．作平行于x轴的直线与区域D相交，沿x轴正向看，入口曲线为x＝0，作为积分下限；出口曲线为x＝y，作为积分上限，因此0≤x≤y．区域D在y轴上投影的最小值为y＝0，最大值为y＝1，因此0≤y≤1．可得知三、解答题21．【解析】解法1解法222．【解析】本题考查的知识点为隐函数求导法．解法1将所给方程两端关于x求导，可得解法2【解题指导】y＝y(x)由方程F(x，y)＝0确定，求y'通常有两种方法：－是将F(x，y)＝0两端关于x求导，认定y为中间变量，得到含有y'的方程，从中解出y'．对于－些特殊情形，可以从F(x，y)＝0中较易地解出y＝y(x)时，也可以先求出y＝y(x)，再直接求导．23．【解析】本题考查的知识点为定积分的计算．24．【解析】本题考查的知识点为曲线的切线方程．25．【解析】本题考查的知识点为求解－阶线性微分方程．所给方程为－阶线性微分方程26．【解析】本题考查的知识点有两个：利用定积分求平面图形的面积；用定积分求绕坐标轴旋转所得旋转体的体积．所给曲线围成的平面图形如图1－2所示．解法1利用定积分求平面图形的面积。

浙江省 2014 年选拔优秀高职高专毕业生进入本科学习统一考试高等数学请考生按规定用笔将所有试题的答案涂、写在答题纸上。

选择题部分注意事项:1.答题前,考生务必将自己的姓名、 准考证号用黑色字迹的签字笔或钢笔填写在答题纸规定的位置上。

2.每小题选出答案后，用2B 铅笔把答题纸上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其它答案标号。

不能答在试题卷上。

一、选择题: 本大题共5小题,每小题4分,共 20分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

1.当x →x0时，f(x)的极限存在，g(x)的极限不存在，那么下面说法正确的是A ．当x →x0时，f(x)⋅g(x)必定极限存在B ．当x →x0时，f(x)⋅g(x)必定极限不存在C ．当x →x0时，若f(x)⋅g(x)极限若存在，极限必定为零D ．当x →x0时，f(x)⋅g(x)极限可能存在，也可能不存在2.函数x x x f 3)(3-=上切线方程平行x 轴的点是A.（0，0）B.（1，2）C.（-1，2）D.（1，3）3.函数f(x)=不可导点的个数是||)2(32x x x x ---是A.3B.1C.0D.24.若f(x)=dt x t x dx d )sin(0-⎰,则f(x)A.-sinxB.-1+cosxC.sinxD.05. 微分方程)1(1'2+=+x x xy y 的通解为 A.C x +1arctan B. )(arctan 1C x x + C.C x x +arctan 1 D.C x x ++arctan 1非选择题部分注意事项:1.用黑色字迹的签字笔或钢笔将答案写在答题纸上，不能答在试题卷上。

2.在答题纸上作图，可先使用2B 铅笔，确定后必须使用黑色字迹的签字笔或钢笔描黑。

二、 填空题: 本大题共10小题，每小题 4分，共40分。

6. )2sin (3sin lim 则,3)2(f 上连续，在)(f 设0xx f x x R x x →= 7. )]([f 求,0x x 0x 1x )(f 设x f x ⎩⎨⎧>=<+= 8. 的渐近线是)0)(1ln(函数>+=x x e x y 9. 的拐点)0(11求2>+=x xy 10.)0('则,11ln y xx y +-= 11. 由曲线x y =，y=x 所围成的平面图形的面积是12. 的值c ·]）c b （×）b a （[，求1c ·）b ×a 已知向量（→→→→→→→→++= 13. 的通解是0)1()1(微分方程=-++xdx y ydy x 14.=+++=++by ay by ay '"y ，则e c e c 的通解为0'"y 已知x 22x 11满足 y(0)=2,y’(0)=-1的解是 15. 的幂级数x 展开成x sin )(f将函数2=x三、计算题：本题共有8小题，其中16-19 小题每小题7分，20-23 小题每小题8分，共 60分。

2023年河南省普通高等学校选拔优秀专科毕业生进入本科阶段学习考试高等数学 解析及解析一、选择题（每小题2分,共60分）1．解析:A【解析】:2901310x x x ⎧-≥⇒<≤⎨->⎩,应选A.2．解析:C 【解析】:2211(2)(2)2()44f x x x f x x x =-⇒=-,应选C.3．解析:B【解析】:()()()()g x f x f x g x -=--=-,所以()g x 是奇函数,应选B.4．解析:A【解析】:222lim(2)0lim(4)04401x x x ax a a →→-=⇒+=⇒+=⇒=-,应选A.5．解析:B【解析】:因221(1)(1)2(1)(2)x x x y x x x x --+==--+-,所以1x =-是函数2212x y x x -=--地可去间断点,应选B.6．解析:D【解析】:211cos 2x x - ,33arctan x x ,所以比与1cos x -高价地无穷小是3arctan x ,应选D.7．解析:B【解析】:222200()()1()()limlim 22h h f x h f x f x h f x h h→→+-+-=()()2211()ln 22f x x ''==ln x x =,应选B.8．解析:B 【解析】:πππ222d cos =0d 2sin t t t t t y y t k x x t==='==='-切,π2t =对应点为（0,1）,所以切线方程为1y =,应选B.9．解析:C【解析】:函数()f x 在[0,1],[1,2],[2,3],[3,4]四个区间上均满足罗尔中值定理,至少存在4个实数使得()0f x '=成立,而方程()0f x '=是4次多项式方程,最多有4个实根.故方程()0f x '=实根地个数为4,应选C.10．解析:B【解析】:d d d d (1)d ()d xxy x y y x e x x y y e x =++⇒-=+,所以d 2d 11x y y e y xy x x x+-==--,应选B.11．解析:C【解析】:()f x 在区间[0,](0)a a >上是增函数,有()(0)0f x f >>,从而120()d (0)d (0)a as f x x f x af s =>==⎰⎰,应选C.12．解析:B【解析】:60y x ''==,只有一个拐点（0,1）,应选B.13. 解析:D【解析】:因为1lim lim02x x y x →±∞→±∞==-;221lim lim 2x x y x →→==∞-所以渐近线方程为2,0x y ==,应选D.14. 解析:B 【解析】:()d ()d ()xx x x x e f e x f e e F e C -----=-=-+⎰⎰,应选B.15. 解析:C【解析】:根据定积分几何意义可知,围成平面图形面积为|()|d b af x x ⎰,应选C.16．解析:B 【解析】:令11()d f x x a -=⎰,则21sin ()1xf x a x +=-+,所以11112211111sin ()d d d d 11x f x x x x a x x x ----=+-++⎰⎰⎰⎰,即有π22a a =-,故π6a =,从而1211sin πlim ()lim lim ()d 16x x x x f x f x x a x -→∞→∞→∞+=-=-=-+⎰,应选B.17．解析:D【解析】:()(1)sin f x x x '=-,应选D.18．解析:C 【解析】:21d 1x x -⎰是12q =地q 广义积分,是收敛地,应选C.19．解析:C【解析】:方程化为2222d d 0d()0x x y y x y x y C +=⇒+=⇒+=,应选C.20．解析:D【解析】:xxe 中多项式函数是一次函数,指数函数中x 系数1是二重特征根,特解应设)(2B Ax e x y x+=*,应选D.21．解析:B【解析】:0a b a b ⋅=⇒⊥, 0//b c b c ⨯=⇒,应选B.22．解析:D【解析】:因{3,2,5}//{6,4,10}--,所以直线与平面垂直,应选D.23．解析:D【解析】:2221x y -=在平面内表示双曲线,从而在空间直角坐标内表示双曲柱面,应选D.24．解析:B【解析】:0000002(11)2limlim 2lim(11)411x x x y y y xy xy xyxy xy xy →→→→→→++==++=+-,应选B.25．解析:A 【解析】:因0,0z zy x x y∂∂====∂∂,所以点（0,0）函数z xy =地驻点,应选A.26．解析:A【解析】:根据二重积分地对称性有()d d 0Dxy y x y +=⎰⎰,应选A.27. 解析:C【解析】:积分区域为{(,)|01,0}{(,)|12,02}x y x y x x y x y x ≤≤≤≤⋃≤≤≤≤-,画出图形,也可表示为{(,)|01,2}x y y y x y ≤≤≤≤-,应选C.28. 解析:A【解析】:从(0,0)到(1,0)曲线可表示为0x xy =⎧⎨=⎩x 从0 变到1,有12d d 0L x y y x +=⎰,从(1,0)到(1,1)曲线可表示为1x y y=⎧⎨=⎩y 从0 变到1,2120d d d 1L x y y x y +==⎰⎰,故有2d d 1Lx y y x +=⎰,应选A.29. 解析:D 【解析】:显然级数∑∞=-11)1(n nn是收敛地,而级数11n n∞=∑是发散地,应选D.30．解析:C【解析】:21111114122121n n n n n ∞∞==⎛⎫=- ⎪--+⎝⎭∑∑,所以111221n S n ⎛⎫=- ⎪+⎝⎭,111lim lim 12212n n n S S n →∞→∞⎛⎫==-= ⎪+⎝⎭,应选C.二、填空题（每小题2分,共20分）31．解析:x1.【解析】:因为111x f x x x-⎛⎫=⎪-⎝⎭,所以1()f x x =.32．解析:98.【解析】:设20()d f x x a =⎰,则2()f x x a =-,所以222008()d ()d 23a f x x x a x a ==-=-⎰⎰,从而有89a =,即208()d 9f x x =⎰.33．解析:1=a .【解析】:因11lim ()lim ln 0x x f x x ++→→==,11lim ()lim()1x x f x x a a --→→=-=-,所以10a -=,即1a =.34．解析:12--x x .【解析】:因()3312(1)1f x x '+=+-,所以()21f x x '=-,即有()2f x x x C =-+,把(0)1f =-代入得1C =-,故()21f x x x =--.35．解析:C x +2sin 21.【解析】:11cos 2d cos 2d(2)sin 222x x x x x C ==+⎰⎰.36．解析:2.【解析】:因011{1,1,1}101i j k a b ⨯==-,所以()1111102a b c ⨯⋅=⨯+⨯-⨯= .37．解析:()xex C C 221+.【解析】:微分方程地特征方程为2440r r -+=,特征根为122r r ==,故微分方程地通解为212()()xy x C C x e =+.38．解析:0.【解析】:因2(,0)ln f x x =,所以2ln (,0)x xf x x'=,故(1,0)0x f '=.39．解析:32.【解析】:方向导数地最大值就是梯度地模,梯度为{}(1,1,1)grad (1,1,1)2,2,2{2,2,2}f x y z ==,|grad (1,1,1)|23f =,故方向导数地最大值为23.40．解析:⎪⎭⎫⎝⎛<<-∑∞=2121,20x x n n n .【解析】:00111()(2)2,1222nn n n n f x x x x x ∞∞==⎛⎫===-<< ⎪-⎝⎭∑∑.三、计算题（每小题5分,共50分）41．求极限20(1)lim 1tan 1x x x e x x→-+-+.【解析】:2300(1)(1tan 1)lim limtan 1tan 1x x x x e x x x x xx x →→-+++=-+-+300lim(1tan 1)lim tan x x x x x x x →→=+++⨯-22220032lim 6lim 6sec 1tan x x x x x x→→===-.42．设n a 为曲线ny x =与1n y x+=(1,2,3,4,)n =所围成地面积,判定级数1n n na ∞=∑地敛散性.【解析】:因两曲线n y x =、1n y x+=交点为（0,0）,（1,1）,所以110111()d 12(1)(2)n n n a x x x n n n n +=-=-=++++⎰.级数11(1)(2)n n n nna n n ∞∞===++∑∑,又因为232(1)(2)limlim 1(1)(2)n n n n n n n n n→∞→∞++==++,而级数3121n n∞=∑是收敛地,根据比较判别法地极限形式知,级数1(1)(2)n nn n ∞=++∑收敛.所以 级数1n n na ∞=∑收敛.43．求不定积分2d 1x x x -⎰．【解析】:22211d d(1)211x x x x x =---⎰⎰122221(1)d(1)12x x x C -=--=-+⎰.44．计算定积分4|2|d x x -⎰.【解析】:4242422|2|d |2|d |2|d (2)d (2)d x x x x x x x x x x-=-+-=-+-⎰⎰⎰⎰⎰ 242202112222422x x x x ⎛⎫⎛⎫=-+-=+= ⎪ ⎪⎝⎭⎝⎭.45．解方程3xy y x '-=地通解．【解析】:方程化为21y y x x'-=,这是一阶线性非齐次微分方程,它对应地齐次方程10y y x'-=地通解为y Cx =.设()y C x x =是原方程地解,代入方程得2()C x x x '=所以()C x x '=,即21()2C x x C =+,故 原方程通解为312y Cx x =+.46．已知函数(,)z f x y =由方程20xyz e z e --+=所确定,求d z ．【解析】:方程两边微分得 [d d ]2d d 0xy z ey x x y z e z --+-+=,即 (2)d [d d ]zxye z ey x x y --=+,所以 d d d 22xy xy zz e y e xz x y e e --=+--.47．已知点(4,1,2),(1,2,2),(2,0,1)A B C --,求ABC ∆地面积.【解析】:因{3,3,4},{2,1,1}AB AC =--=--,所以334{1,5,3}211i j kAB AC ⨯=--=--,故ABC ∆地面积为11351259222S AB AC =⨯=++= .48．计算二重积分22ln d d Dx y x y +⎰⎰,其中22{(,)|14}D x y x y =≤+≤.【解析】:积分区域在极坐标下表示为(){},02π,12D r r θθ=≤≤≤≤,所以2π2221ln d d d ln d Dx y x y r r r θ+=⎰⎰⎰⎰221πln d r r =⎰()222113πln d π4ln22r r r r ⎛⎫=-=- ⎪⎝⎭⎰.49．计算曲线积分22(1)d (1)d Ly x x x y y ++-⎰,其中L 是圆周221x y +=（逆时针方向）.【解析】:令2(,)(1)P x y y x =+,2(,)(1)Q x y x y =-,则有21P x y ∂=+∂,21Qy x∂=-∂.又L 为封闭曲线且取正方向,故由格林公式可得:2222(1)d (1)d d d ()d d L D DQ P y x x x y y x y x y x y x y ⎛⎫∂∂++-=-=-+ ⎪∂∂⎝⎭⎰⎰⎰⎰⎰ 2π13001d d π2r r θ=-=-⎰⎰.50．试确定级数01nn x n ∞=+∑地收敛域并求出和函数．【解析】:级数01nn x n ∞=+∑是标准不缺项地幂级数,收敛半径为112limlim 111n n n n a n R a n →∞→∞++==⨯=+,当1x =时,级数化为011n n ∞=+∑,是调和级数,发散地；当1x =-时,级数化为0(1)1nn n ∞=-+∑,是交错级数,收敛地；故所求级数地收敛域为[1,1)-.设和函数为()S x ,即0()1nn x S x n ∞==+∑,当(1,1)x ∈-且0x ≠时,10000001()d d d 11n x x x nn n n n x xS x t t t t t n t +∞∞∞=======+-∑∑∑⎰⎰⎰ln(1)x =--,所以ln(1)()x S x x-=-；当0x =时,00ln(1)1(0)lim lim 11x x x S x x →→-=-==-,当1x =-时,ln(1)()x S x x-=-有意义,故所求和函数为ln(1),[1,0)(0,1)()1,0x x S x xx -⎧-∈-⋃⎪=⎨⎪=⎩.四、应用题（每小题7分,共14分）51．欲围一个面积为150平方米地矩形场地.所用材料地造价其正面是每平方米6元,其余三面是每平方米3元.问场地地长、宽各为多少时,才能使造价最低？【解析】:设场地地长、宽各为,x y ,高为h ,造价为z ,则有63(2)z xh x y h =++,且150xy =,即9009(0)z xh h x x=+>,h 为常数,令290090x z h h x'=-=得定义域内唯一驻点10x =,此时15y =；在10x =时,有318000x z h x''=>,所以10x =是极小值点即最小值点,故场地地长、宽各为10米、15米时,才能使造价最低.52．已知D 是抛物线2:2L y x =和直线12x =所围成平面区域.试求:(1) 区域D 地面积；（2）区域D 绕Ox 轴旋转所形成空间旋转体地体积.【解析】:平面图形如下图所示取x 为积分变量,10,2x ⎡⎤∈⎢⎥⎣⎦,(1)根据抛物线地对称性,区域D 地面积是x 轴上方图形面积地2倍. 112202()d 22d s D f x x x x==⎰⎰1222y x=xyo13220222233x ==；（2）区域D 绕Ox 轴旋转所形成空间旋转体地体积为 1122220π()d πd D V f x x y x ==⎰⎰112220ππ2d π4x x x===⎰.五、证明题（6分）53．设2e a b e <<<,证明 2224ln ln ()b a b a e ->-.【证明】:设2()ln f x x =,显然它在(0,)+∞内可导,从而()f x 在区间[,]a b 上满足拉格朗日中值定理,即存在(,)a b ξ∈,使得2ln ()()()f b f a b a ξξ-=-成立,所以有()2222ln ln ln (),b a b a e a b e ξξξ-=-<<<<,又因为函数ln ()x g x x=在区间2[,]e e 上是减函数,所以有2()()g g e ξ>,即2ln 2eξξ>,故 22ln 4()()b a b a eξξ->-所以 2224ln ln ()b a b a e->-.。

------------------------2014年文亮“专升本”《高等数学》模拟试卷------------------------ 2014年文亮“专升本”《高等数学》模拟试卷（一） 题 号 一 二 三 四 总 分 得 分 考试说明： 1．考试时间为150分钟； 2．满分为150分； 3．答案请写在试卷纸上，用蓝色或黑色墨水的钢笔、圆珠笔答卷，否则无效； 4．密封线左边各项要求填写清楚完整。

一. 选择题（每个小题给出的选项中，只有一项符合要求：本题共有5个小题，每小题4分，共20分） 1. 函数11()1x f x e k ⎧⎪=⎨+⎪⎩在点0x =处左连续，则k =（ ） A. 0 B. 12 C. 1 D. 2 2. 函数0()x t f x e dt =⎰在(),-∞+∞内是（ ） A. 单调增加，曲线为凹的 B. 单调增加，曲线为凸的 C. 单调减少，曲线为凹的 D. 单调减少，曲线为凸的 3. 若1()1x f x x =+，则10()f x dx =⎰（ ）. A. 0 B. 1 C. 1ln 2- D. ln 2 4. 已知1,2,a b →→==且向量a →与b →的夹角为4π，则a b →→+=（ ） A. 1 B. 5 C. 2 D. 12+ 5. 对于正项级数1n n u ∞=∑，下列命题中错误的是（ ） 得分 阅卷人 姓名：教室： ------------------------------------------------------------------------------------------密封线---------------------------------------------------------------------------------------------------A. 如果 1lim 1n n nu u ρ+→∞=< ，则1n n u ∞=∑收敛 B. 如果 1lim 1n n nu u ρ+→∞=> ，则1n n u ∞=∑发散 C ．如果 11n n u u +< ，则1n n u ∞=∑收敛 D. 如果 11n n u u +> ，则1n n u ∞=∑发散 二.填空题 (只须在横线上直接写出答案,不必写出计算过程,每小题4分,共40分) 1．已知函数()f x 的定义域为[]0,1，则2(1)f x -的定义域为2. 曲线arctan y x =在横坐标为1的点处的切线方程为3. 设函数()f x 在2x =的某邻域内可导，且()()f x f x e '=，(2)1f =，则(3)(2)f =4. 设函数1sin x y x =，则dy =5. 已知2,x e dx a π+∞--∞=⎰为实数，0b >，则()2x a b e dx --+∞-∞=⎰6. 422222sin (31)[cos ]1x x x x dx x ππ-+++=+⎰ 7. 不定积分sin cos x x xdx =⎰8. 微分方程0y y ''+=的通解为9. 平行平面xoy 且垂直于向量()3,4,7-的单位向量为 得分 阅卷人10. 设有级数12nn n x a ∞=⎛⎫ ⎪⎝⎭∑，若11lim 3n n n a a →∞+=，则该级数的收敛半径为 三.计算题 ( 计算题必须写出必要的计算过程，只写答案的不给分，共60分)1. 求极限2101tan lim 1sin x x x x x →+⎛⎫ ⎪+⎝⎭（7分）2. 已知函数()y y x =由方程(1)2x xy x y e xe e -+=所确定，求(0)y '（7分）3. 已知22ln(1),0()1sin ,0x x f x x x x ⎧-≤⎪=⎨>⎪⎩求()f x '。

------------------------2014年文亮“专升本”《高等数学》模拟试卷------------------------2014年文亮“专升本”《高等数学》模拟试卷（四）题 号 一 二 三 四 总 分 得 分考试说明：1．考试时间为150分钟； 2．满分为150分；3．答案请写在试卷纸上，用蓝色或黑色墨水的钢笔、圆珠笔答卷，否则无效； 4．密封线左边各项要求填写清楚完整。

一. 选择题（每个小题给出的选项中，只有一项符合要求：本题共有5个小题，每小题4分，共20分）1. 设()f x 在点x 的某邻域内连续，且0()lim1.2x x f x x x →'=-，则（ ） A.0()f x 是()f x 的极小值； B. 0()f x 是()f x 的极大值；C. 00(,())x f x 是曲线()y f x =的拐点；D. 0()f x 不是()f x 的极值，00(,())x f x 也不是曲线()y f x =的拐点。

2. 设函数()f x 是连续函数，且0()s t I tf tx dx =⎰，其中0t ≠则I （ ）A. 依赖于s 与t ；B. 依赖于t ，不依赖于s ；C. 依赖于s ，不依赖于t ；D. 不依赖于s 与t ； 3. 下列反常积分发散的是（ ）. A.22ln dxx x+∞⎰B.1211dx x--⎰C.131sin xdx x-⎰ D.11sin dx x -⎰4. 若()f x 为(,)-∞+∞的奇函数，在(,0)-∞内()0f x '>，且()0f x ''<，则在(0,)+∞内有（ ）A. ()0,()0f x f x '''><B. ()0,()0f x f x '''>>得分 阅卷人姓名： 班级：------------------------------------------------------------------------------------------密封线---------------------------------------------------------------------------------------------------C. ()0,()0f x f x '''<<D. ()0,()0f x f x '''<>5. 数项级数2(1)1n n n n ∞=--∑为（ ）A.绝对收敛B. 条件收敛 C ．发散 D. 无法判断 二.填空题 (只须在横线上直接写出答案,不必写出计算过程,每小题4分,共40分)1．极限ln(1)lim31n n e n →∞+=+ 2.设函数 2sin()y x x =+，则()y x ''=3. 曲线ln y x =上与直线1x y +=垂直的切线方程为4. 设2x y xe =，则(11)(0)y = 5.设函数21y x x =-+的渐近线为6. 设函数()y y x =由方程cos 1xy ye x x +=确定，则0x dy == 7点(1,2,1)P -到直线112:212x y z l -+-==-的距离为 8. 设22,2x y x y ==和1y =在第一象限所围面积为9. 幂级数211(1)31n n n x n -∞=-+∑的收敛区间为10. 微分方程22sin xy y y x e x -'''++=+的特解形式应设为 三.计算题 ( 计算题必须写出必要的计算过程，只写答案的不给分，共60分)1. 设1(cos )xy x =，求dy dx。

江苏省2014年普通高校专转本选拔考试高等数学 试题卷答案一、选择题（本大题共6小题，每小题4分，共24分）1、C2、B3、B4、A5、D6、D二、填空题（本大题共6小题，每小题4分，共24分）7、2y e -= 8、5 9、2π10、2222y x dz dx dy x y x y =-+++ 11、3π 12、[0,2) 三、计算题（本大题共8小题，每小题8分，共64分）13、原式=230000arcsin arcsin lim lim arcsin x x x x x x x x x x x→→→→--===20116x x →-==- 14、2(32)y t dy y dy dt e t dx dx e t dt-+==+,013t dy dx e==-. 15、2222222221111ln ln ln ln ln ln 2222x xdx xdx x x x d x x x x xdx ==-=-⎰⎰⎰⎰222222222211111111ln ln ln ln ln ln ln 22222222x x xdx x x x x x d x x x x x xdx =-=-+=-+⎰⎰⎰2222111ln ln 224x x x x x C =-++ 16、令t x =-12，则原式=222222220002444(1)22arctan 2044422t t t dt dt dt t t t π+-==-=-=-+++⎰⎰⎰ 17、平面∏的法向量(1,2,3)(1,0,0)(0,3,2)n MN i →→=⨯=⨯=-，直线方程：0(1)3(1)2(1)0x y z -+---=.即3210y z --=.18、12cos 2z xf xf x∂''=+∂212221222cos (2)2(2)2cos 4z xf y xf y y xf xyf x y∂''''''''=⋅-+⋅-=--∂∂ 19、2101001()()26y D y x y dxdy dy x y dx dy -+=+==⎰⎰⎰⎰⎰ 20、特征方程：220r r -=，120,2r r ==，齐次方程的通解为212x Y C C e =+.令特解为2()x y x Ax B e *=+，则22(222)x y Ax Bx Ax B e *'=+++，22(44824)x y Ax Bx Ax A B e *''=++++代入原方程得：22(422)x x Ax A B e xe ++=， 有待定系数法得：41220A A B =⎧⎨+=⎩，解得1414A B ⎧=⎪⎪⎨⎪=-⎪⎩，所以通解为221211()44x x y C C e x x e =++-. 四、证明题（本大题共2小题，每小题9分，共18分）21、令()ln 3f x x x =-,显然在区间(2,3)上连续，且38(2)2ln 23ln ln10,f e =-=<< (3)3ln333(ln31)0,f =-=->根据零点定理，(2,3),()0f ξξ∃∈=成立.又()ln 10f x x '=->，(2,3)x ∈，)(x f '单调递增，唯一性得证.22、令21()1ln(1)2x f x e x x =---+，则1()1x f x e x x '=--+，21()1(1)x f x e x ''=-++， 在0x >时，()f x ''单调递增，()(0)10f x f ''''>=>，所以()f x '单调递增，()(0)0f x f ''>=，所以()f x 单调递增，()(0)0f x f >=，得证.五、综合题（本大题共2小题，每小题10分，共20分）23、（1）2k y x '==-切，切线：,02(1)y x -=--，即2(1)y x =--，D 面积1201[2(1)(1)]3x x dx ----=⎰. (2) 21200211(1)(1)2326y y V d y y d y πππππ=---=-=⎰⎰ 24、已知0()1()xt t dt x ϕϕ=-⎰两边同时对x 求导得：()()x x x ϕϕ'=-，22()x x Ce ϕ-=，令0x =代入0()1()xt t dt x ϕϕ=-⎰得(0)1ϕ=，所以求得221,()x C x e ϕ-==.（2）因为2222232222(),(),()(1),()(3)x x x x x e x xe x x e x x x e ϕϕϕϕ----''''''==-=-=-(0)1ϕ=，(0)0ϕ'=，(0)1,(0)0ϕϕ'''''=-=. 20000()1()()(0)1lim ()lim lim lim (0)2222x x x x x x x f x f x x ϕϕϕϕ→→→→'''''-=====-=. 所以()f x 在0=x 处的连续.223000()11()(0)2()22lim lim lim 2x x x x f x f x x x x x x ϕϕ→→→-+--+== 20002()2()()11lim lim lim 6666x x x x x x x x x x ϕϕϕ→→→''''''+++====. 所以()f x 在0=x 处可导，1(0)6f '=.。