2010年河南专升本高数真题+答案解析

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2010年河南省普通专升本考试《公共英语》真题+答案

2010 年河南省普通专升本考试《公共英语》真题Part ⅠVocabulary and Structure (1 x40)Directions: There are 40 incomplete sentences in this part. For each sentence there are four choices marked A, B, C and D. Choose the ONE that best completes the sentence, and then mark the corresponding letter on the Answer Sheet.1.The fire must have after the shop was closed.A. broken outB. broken downC. broken inD. broken through2.He is join the army.A. too young toB. enough young toC. very young toD. young enough to3.Finally he got time for a glance this report.A. offB. roundC. onD. at4.Your idea seems to be good but it isn’t.A. practicalB. possibleC. plentifulD. precious5.He enjoys pop music while I prefer classical music.A. to listen toB. to listenC. listeningD. listening to6.When the little girl awoke, she found herself by a group of soldiers.A. surroundB. be surroundedC. being surroundedD. being surrounding7.The manager lost his just because his secretary was ten minutes late.A. moodB. temperC. mindD. passion8.There are several characteristics of the textbook attention.A. worthwhileB. worth ofC. worthyD. worthy of9.The new building all the other buildings in the town.A. dwarfsB. distortsC. desertsD. depresses10.I passed the test. I it without your help.A. would not passB. wouldn’t have passedC. didn’t passD. had not passed11.The Internet has brought big changes in the way we work.A. aboutB. outC. backD. up12.The father writes in his will that every son and daughter a share of his property.A. hasB. to haveC. havingD. have13.He hurried to the hospital, only his father had just died.A. to tellB. to be toldC. tellingD. told14.tomorrow, he would be able to see the opening ceremony.A. Would he comeB. If he comesC. Was he comingD. Were he to come15.The speaker could hardly find safe ground his arguments.A. on which to baseB. to base onC. on the baseD. which to base on16.He is a man who is always fault with other people.A. puttingB. seekingC. findingD. looking for17.The factory had to a number of employees because of the economic crisis in the country.A. lay outB. lay offC. lay asideD. lay down18.Would you spare some time to have a chat with me a cup of coffee?A. forB. withC. duringD. over19.Ten days ago the young man his boss his intention to resign.A. informed … ofB. informed … onC. informed … inD. informed … to20.It is necessary that he the task by the end of next week.A. fulfillB. will fulfillC. will have fulfilledD. fulfills21.It is impossible for so workers to do so work in a single day.A. few… muchB. few… manyC. little… muchD. little… many22.No further discussions , the meeting was brought to an end.A. aroseB. arisingC. to ariseD. be arisen23.The other day, Mum and I went to St. James’s Hospital, and they did lots and lots of tests on me, are horrible and frightening.A. most of themB. most of whichC. most of thatD. most of what24.He is a pleasant fellow to .A. workB. work withC. be workingD. be worked25.On his way to the airport, it to him that he had forgotten to take his passport.A. happenedB. occurredC. reflectedD. took place26.Orlando, a city in Florida, for its main attraction, Magic Kingdom.A. which is well knownB. being well knownC. well knownD. is well known27., he couldn’t earn enough to support the family.A. Hard as he workedB. As he worked hardC. As hard he workedD. Hard as did he work28.I used on the left in England, but I soon got used on the right in China.A. to driving… to driveB. to drive… to drivingC. to drive… to driveD. to driving… to driving29.Can machines perform the same tasks ?A. that man doesB. what man doesC. how man doesD. as man does30.that the trade between the two countries reached its highest point.A. During the 1960’sB. It was in the 1960’sC. That it was in the 1960’sD. It was the1960’s31.It’s no use with him since he has made up his mind.A. to argueB. arguingC. to be arguedD. argued32.The more he tried to please her, she seemed to appreciate it.A. lessB. lesserC. the lessD. the lesser33.The information technology has greatly people’s life.A. affectedB. effectC. impactD. infected34.Having a good command of English is an easy thing.A. by all meansB. by any meansC. by every meansD. by no means35.My mobile phone isn’t working. It .A. needs being repairedB. needs repairingC. needs to repairD. needs repaired36.That was so serious a matter that I had no choice but the police.A. called inB. calling inC. call inD. to call in37.He never to his customers in his business except occasionally for some special reasons. This time he cut the price by half, which really shocked me.A. leakedB. drewC. quotedD. yielded38.It is useful to be able to predict the extent which a price change will influence supply and demand.A. fromB. withC. toD. for39.Undergraduate students the rare books in the school library.A. have access forB. keep access inC. keep access onD. have access to40.sat down the phone rang.A. No sooner had he … thanB. No sooner he had … thanC. No sooner had he … whenD. No sooner he had … whenPart Ⅱ(1×20)Directions: There are 20 blanks in the following passage. For each blank there are four choices marked A, B, C and D. You should choose the ONE that best fits into the passage and mark the corresponding letter on the Answer Sheet.What do we mean by a perfect English pronunciation? In one sense there are as many different kinds of (41) as there are speakers of it. No two speakers (42) in exactly the same (43) . We can always hear differences (44) them, and the pronunciation of English (45) a great deal in different geographical (46) . How do we decide what sort of English to use as a (47) ? This is not a question that can be (48) in the same way for all foreign learners of English. (49) you live in a part of the world as (50) , where there is a long (51) of speaking English for general communication purpose, you should select to (52) a good variety of the pronunciation of this area. It would be mistaken in these (53) to use as a model BBC English or (54) of the sort. On the other hand, if you live in a country (55) there is no traditional (56) of English, you must take as your model some forms of (57) English pronunciation. It does not (58) very much which form you choose. The most (59) way is to take as your model the sort of English you can (60) most often.41.A. language42.A. spoke B. linguisticB. spokenC. EnglishC. speaksD. linguistD. speak43. A. way B. form C. sort D. type44. A. of B. among C. between D. from45. A. varies B. changes C. shifts D. alters46. A. spaces B. parts C. countries D. areas47. A. guide B. model C. symbol D. direction48. A. given B. answered C. satisfied D. responded49. A. Because B. When C. Whether D. If50. A. Russia B. Mongolia C. India D. Japan51. A. tradition B. use C. custom D. habit52. A. seize B. acquire C. have D. hold53. A. actions B. decisions C. combinations D. circumstances54. A. everything B. nothing C. things D. anything55. A. which B. that C. where D. wherever56. A. use B. used C. useful D. usefulness57. A. domestic B. practical C. national D. new58. A. matter B. affect C. trouble D. care59. A. ordinary B. sensitive C. effective D. careful60. A. listen B. find C. notice D. hearPart Ⅲ Reading Comprehension (2x20)Directions: There are 4 passages in this part. Each passage is followed by some questions or incomplete sentences. 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.Passage1Thousands of years ago, in the middle of an ocean, miles from the nearest island, an undersea volcano broke out. The hot liquid got higher and higher and spread wider and wider. In this way, an island rose up in the sea.As time went on, hot sun and cool rains made the rock split and break to pieces. Sea waves hit against the rock. Inthis way, soil and sand came into being.Nothing lived on the naked soil. And then the wind and birds brought plant seeds, spiders and other little living things there. Only plants could grow first. Only they, in sunlight, could produce food from the soil, water and air. While many animals landed on the island, they could find no food. A spider made its web uselessly, because there were no insects( 昆虫) for its web to catch. Insects couldn’t stay until there were plants for them to eat. So plants had to be the first life on this new island.61.The passage centers on .A. how an undersea volcano broke outB. how an island rose up in the seaC. how soil was formed on a new islandD. how life began on a volcano produced island62.According to the passage, the island got its first soil from .A. sea wavesB. the sand brought by the windC. its own rockD. cool rains63.The word "naked" (in Para. 3) could be replaced by which of the following?A. redB. newC. oldD. bare64.The order of coming into being on the island is .A. soil, plants and animalsB. soil, little creatures and plantsC. soil, birds and plantsD. soil, human beings and animals65.According to the passage, which of the following is TRUE?A.Spiders were the first life that could live on the island.B.The island is far away from any piece of land.C.Insects could not live on the island without plantsD.Plants were brought to the island by human beingsPassage2Ernest Miller Hemingway was born on July 21, 1899 in Oak Park, Illinois. In the nearly sixty two years of his life that followed, he built a literary fame unsurpassed（无法超越）in the twentieth century.As a boy he was taught by his father to hunt and fish along the shores and in the forests around Lake Michigan. The Hemingways had a summer house in northern Michigan, and the family would spend the summer months there trying to stay cool. Hemingway would either fish the different streams that ran into the lake, or would take the small boat out to do some fishing there. He would also go squirrel hunting in the woods, discovering early in life the peace to be found while alone in the forest or going through a stream. It was something he could always go back to throughout his life, and though he often found himself living in major cities like Chicago, Toronto and Paris early in his life, once he became successful he chose somewhat isolated places to live in.When he wasn’t hunting or fishing his mother taught him the good points of music. She was a skilled singer who once had wished a life on stage, but at last settled down with her husband and spent her time by giving voice and music lessons to local children, including her own. Hemingway was never talented for music and suffered through singing practices and music lessons, however, the musical knowledge he got from his mother helped him share in his first wife Hadley’s interest in the piano.66.Ernest Hemingway died in .A. 1969B. 1979C. 1981D. 196167.Which of the following statements is NOT true according to the passage?A.His father taught him to fish and hunt when he was a boy.B.His family had a summer house in northern Michigan.C.He taught himself music when he was a boy.D.He also went squirrel hunting in the woods.68.After he became successful, Ernest Hemingway .A. preferred to stay in big citiesB. chose to live in somewhat isolated placesC. moved his family to ParisD. killed himself69.Being talented in music, Hemingway’s mother once wanted to .A. be a music teacherB. help Hemingway learn musicC. perform on the stage as a singerD. marry a rich husband70.The passage is most probably from .A. a literary biographyB. a science textbookC. a term paperD. a personal diaryPassage3What will man be like in the future -- in 5000 or even 50000 years from now? We can only make a guess, of course, but we can be sure that he will be different from what he is today. For man is slowly changing all the time.Let us take an obvious example. Man, even five hundred years ago, was shorter than he is today. Now, on average, men are about three inches taller. Five hundred years is a relatively short period of time, so we may assume that man will continue to grow taller.Again, in the modern world we use our brains a great deal. Even so, we still make use of only about 20% of the brains capacity. As time goes on, however, we shall have to use our brains more and more, and eventually we shall need larger ones. This is likely to bring about a physical change --- the head, in particular the forehead, will grow larger.Nowadays our eyes are in constant use. In fact, we use them so much that very often they become weaker and we have to wear glasses. But over very long period of time it is likely that man’s eyes will grow stronger.On the other hand, we tend to make less use of our arms and legs. These, as a result, are likely to grow weaker. At the same time, however, our fingers will grow more sensitive because they are used a great deal in modern life.But what about hair? It will probably disappear from the body altogether in course of time because it does not serve a useful purpose any longer. In the future, then, both sexes are likely to be bald.Perhaps all this gives the impression that future man will not be a very attractive creature to look at. This may well be true. All the same, in spite of all these changes, future man will still have a lot in common with us. He will still be a human being, with thoughts and emotions similar to our own.71.The passage tells us about .A.how man’s life will be in the futureB.how future man will look likeC.the fact that man’s organs will function differently in the futureD.the fact that man is growing uglier as time passes72.There is evidence that man is changing, .A.he has been growing taller over the past 500 yearsB.he has got stronger eyes than he ever hadC.his hair is getting thinner and thinnerD.his limbs are getting weaker because he tends to make less use of them73.Man’s forehead will grow larger because .A.he will make use of only about 20% of the brain’s capacityB.the other 80% of his brain will grow in due timeC.he had rather narrow forehead a few hundred years agoD.he will have to use his brain more and more as time goes on74.Future man will probably .A. have smaller eyesB. have larger eyesC. see betterD. have to wear better glasses75 .The reason for believing that future man will be different is that he .A. will grow strongerB. never stops changingC. hopes for a changeD. will live a different lifePassage4Auctions （拍卖）are public sales of goods, made by an officially approved auctioneer. He asked the crowd assembled in the auction room to make offers, or bids, for the various items on sale. He encouraged buyers to bid higher figures, and finally named the highest bidder as the buyer of the goods. This is called “knocking down the goods, for the bidding ends when the auctione er bangs a small hammer on a table at which he stands”. This is often set on a raised platform called a rostrum.The ancient Romans probably invented sales by auction, and the English word comes from the Latin auction, meaning “increase”. The Romans usually sold in this way the spoils taken in war, these sales were called “sub hash”, meaning “under the spear”, a spear being stuck in the ground as a signal for a crowd to gather. In England in the eighteenth century, goods were often sold “by the candle”: a s hort candle was lit by the auctioneer, and bids could be made while it stayed alight.Practically all goods whose qualities varied are sold by auction. Among these are coffee, hides, skins, wool, tea, cocoa, furs, spices, fruit, vegetables and wines. Auction sales are also usual for land and property, antique furniture, pictures, rare books, old china and similar works of art. The auction rooms at Christie’s and Sotheby’s in London and New York are world famous.An auction is usually advertised beforehand with full particulars of the articles to be sold and where and when they can be viewed by prospective buyers. If the advertisement cannot give full details, catalogues are printed, and each group of goods to be sold together, called a “lot”, is usually giv en a number. The auctioneer need not begin with Lot 1 and continue in numerical order he may wait until he registers the fact that certain dealers are in the room and then produce the lots they are likely to be interested in. The auctioneer’s services are paid for in the form of a percentage of the price the goods are sold for. The auctioneer therefore has a direct interest in pushing up the bidding as high as possible.76.A “bidder” (in Para.1) is a person .A. who sells somethingB. who buys somethingC. who offers a priceD. who borrows something77.Auctioned goods are sold price offered.A. for the highestB. for the fixedC. for the lowestD. for the unexpected78.The end of the bidding is called “knocking down” because .A. the auctioneer knocks the buyer downB. the auctioneer knocks the rostrum downC. the goods are knocked down onto the tableD. the auctioneer bangs the table with a hammer79.The “candle” used in paragraph 2 is .A. because they took place at nightB. as a signal for the crowd to gatherC. to give light to the auctioneerD. to limit the time when offers could be made80.An auction catalogue gives prospective buyers .A. the current market values of the goodsB. details of the goods to be soldC. the order in which goods must be soldD. free admission to the auction salePart ⅣTranslation (30 points)Directions: There are 10 sentences in this part. Please translate sentences 81-85 from Chinese into English, and translate sentences 86-90 from English into Chinese. Write your answer on the Answer Sheet.81.长城是中国的历史文化符号之一。

2010“专升本”《高数》试题及答案

《高等数学》试卷一、单项选择题（每题2分，共计60分，在每小题的备选答案中选出一个正确答案，并将其代码写在题干后面的括号内。

不选、错选或多选者，该题无分）1.已知函数)12(-x f 的定义域为]1,0[ ，则)(x f 的定义域为（）A. ]1,21[ B. ]1,1[- C. ]1,0[ D. ]2,1[-解：B x x ⇒≤-≤-⇒≤≤112110.2.)1lg()(2x x x f -+=在),(+∞-∞是（） A ．奇函数 B. 偶函数 C.非奇非偶函数 D. 既奇又偶函数解：01lg )1lg()1lg()()(22==+++-+=-+x x x x x f x f A ⇒. 3. 当0→x 时，x x s i n 2-是x的（） A. 高阶无穷小 B. 低阶无穷小 C. 同阶非等价无穷小 D. 等价无穷小解： 1sin lim20-=-→x x x x ， C ⇒. 4.=+∞→nn n n sin 32lim （）A. ∞B. 2C. 3D. 5 解：B n n n n n n n ⇒=+=+∞→∞→2]sin 32[lim sin 32lim . 5.设函数⎪⎩⎪⎨⎧=+≠-=0,10,1)(2x a x x e x f ax 在0=x 处连续，则 =a （） A. 0 B. 1 C. 2 D. 3 解：B a a a ae x e x f ax x ax x x ⇒=⇒+===-=→→→1122lim 1lim)(lim 20200. 6. 设函数)(x f 在1=x 可导，则=--+→xx f x f x )1()21(lim0 （） A. )1(f ' B. )1(2f ' C. )1(3f ' D. -)1(f '解：x x f f f x f x x f x f x x )1()1()1()21(lim )1()21(lim 00--+-+=--+→→ C f x f x f x f x f x x ⇒'=---+-+=→→)1(3)1()1(lim 2)1()21(lim200 7. 若曲线12+=x y 上点M 处的切线与直线14+=x y 平行，则M 的坐标（）A. （2，5）B. （-2，5）C. （1，2）D.（-1，2）解： A y x x x y ⇒==⇒=⇒='5,5422000.8.设⎪⎩⎪⎨⎧==⎰202cos sin ty du u x t ，则=dx dy （） A. 2t B. t 2 C.-2t D. t 2-解： D t tt t dx dy ⇒-=-=2sin sin 222. 9.已知x x x f n ln )()2(=-，则=)()(x f n （）A.211x+ B. x 1C. x lnD. x x ln 解：B x x f x x f x x x f n n n ⇒=⇒+=⇒=--1)(ln 1)(ln )()()1()2(.10.233222++--=x x x x y 有（）A. 一条垂直渐近线，一条水平渐近线B. 两条垂直渐近线，一条水平渐近线C. 一条垂直渐近线，两条水平渐近线D. 两条垂直渐近线，两条水平渐近线解：A y y y x x x x x x x x y x x x ⇒∞=-==⇒++-+=++--=-→-→∞→2122lim ,4lim ,2lim )2)(1()3)(1(2332 . 11.在下列给定的区间满足罗尔中值定理的是（）A. ]2,0[|,1|-=x yB. ]2,0[,)1(132-=x yC.]2,1[,232+-=x x y D ． ]1,0[,arcsin x x y = 解：由罗尔中值定理条件：连续、可导及端点的函数值相等C ⇒12. 函数x e y -=在区间),(+∞-∞为（）A. 单增且凹B. 单增且凸C. 单减且凹D. 单减且凸解： C e y e y x x ⇒>=''<-='--0,0.13.⎰+=C x F dx x f )()(曲线，则⎰=--dx e f e xx )( （） A.C e F e x x ++--)( B. C e F e x x +---)(C. C e F x +-)(D. C e F x +--)(解：D C e F e d e f dx e f e xx x x x ⇒+-=-=⎰⎰-----)()()()(.14. 设函数x e x f =-')12( ，则 =)(x f （）A. C e x +-1221 B. C e x +-)1(212 C. C e x ++1221 D. C e x ++)1(212解：D C e x f e x f e x f x x x ⇒+=⇒='⇒=-'++)1(21)1(212)()()12(. 15. =⎰b axdx dx darctan （）A.x arctanB. 0C. a b arctan arctan -D. a b arctan arctan + 解：⎰b a xdx arctan 是常数，所以 B xdx dx d ba ⇒=⎰0arctan .16.下列广义积分收敛的为（） A. ⎰+∞1dx e x B. ⎰+∞11dx x C. ⎰+∞+1241dx x D. ⎰+∞1cos xdx 解：C x dx x ⇒-==++∞∞+⎰)21arctan 4(412arctan 4141112π. 17.设区域D 由)(),(,),(,x g y x f y a b b x a x ==>==所围成，则区域D 的面积为（） A. ⎰-b a dx x g x f )]()([ B. ⎰-b a dx x g x f )]()([ C. ⎰-b adx x f x g )]()([ D. ⎰-b adx x g x f |)()(|解：由定积分的几何意义可得D 的面积为 ⎰-badx x g x f |)()(|D ⇒.18. 若直线32311-=+=-z n y x 与平面01343=++-z y x 平行，则常数=n （）A. 2B. 3C. 4D. 5 解： B n n n ⇒=⇒=+-⇒-⊥30943}3,43{}3,,1{.19.设y xy x y x f arcsin)1(),(-+=，则偏导数)1,(x f x '为（） A.2 B.1 C.-1 D.-2 解： B x f x x f x ⇒='⇒=1)1,()1,(. 20. 方程02=-xyz e z 确定函数),(y x f z = ，则x z ∂∂ = （）A. )12(-z x zB. )12(+z x zC. )12(-z x yD. )12(+z x y解：令⇒-='-='⇒-=xy e F yz F xyz e z y x F z z x z 222,),,( A z x zxy xyz yz xy e yz x z z ⇒-=-=-=∂∂⇒)12(222 21.设函数xy y x z +=2，则===11y x dz （）A. dy dx 2+B. dy dx 2-C. dy dx +2D. dy dx -2 解：222x ydx xdy dy x xydx dz -++= A dy dx dx dy dy dx dz y x ⇒+=-++=⇒==2211.22.函数2033222+--=y x xy z 在定义域上（）A.有极大值，无极小值B. 无极大值，有极小值C.有极大值，有极小值D. 无极大值，无极小值解：,6)0,0(),(062,06222-=∂∂⇒=⇒=-=∂∂=-=∂∂x z y x y x y z x y x z⇒=∂∂∂-=∂∂2,6222y x zy z 是极大值A ⇒. 23由012222=+--+y x y x 围成的闭区域D ，则=⎰⎰Ddxdy （）A. πB. 2πC.4πD. 16π解：有二重积分的几何意义知：=⎰⎰Ddxdy 区域D 的面积为π.24累次积分⎰⎰>axa dy y x f dx 0)0(),(交换后为（）A. ⎰⎰a x dx y x f dy 0),( B. ⎰⎰a aydx y x f dy 0),(C. ⎰⎰a a dx y x f dy 0),( D. ⎰⎰a yadx y x f dy 0),(解：积分区域},0|),{(}0,0|),{(a x y a y y x x y a x y x D ≤≤≤≤=≤≤≤≤=B ⇒.25.二重积分⎰⎰20sin 20)sin ,cos (πθθθθrdr r r f d 在直角坐标系下积分区域可表示为（）A. ,222y y x ≤+B. ,222≤+y xC. ,222x y x ≤+D. 220y y x -≤≤ 解：在极坐标下积分区域可表示为：}sin 20,20|),{(θπθθ≤≤≤≤=r r D ，在直角坐标系下边界方程为y y x 222=+，积分区域为右半圆域D ⇒26.设L 为直线1=+y x 坐标从点)0,1(A 到)1,0(B 的有向线段，则⎰-+L dy dx y x )( （） A. 2 B.1 C. -1 D. -2解：L ：,1⎩⎨⎧-==x y xx x 从1变到0 ，⎰⎰⇒-=+=-+012)(D dx dx dy dx y x L . 27.下列级数绝对收敛的是（）A ．∑∞=1sin n n πB ．∑∞=-1sin )1(n n n π C ． ∑∞=-12sin )1(n n n π D ． ∑∞=0cos n n π解： ⇒<22sin n n ππC n n ⇒∑∞=12sin π. 28. 设幂级数n n n n a x a (0∑∞=为常数 ,2,1,0=n ），在 2-=x 处收敛，则∑∞=-0)1(n n na （）A. 绝对收敛B. 条件收敛C. 发散D. 敛散性不确定解：∑∞=0n nn x a 在2-=x 收敛，则在1-=x 绝对收敛，即级数∑∞=-0)1(n n n a 绝对收敛A ⇒.29. 微分方程0sin cos cos sin =+ydx x ydy x 的通解为（） A.C y x =sin cos B. C y x =cos sin C. C y x =sin sin D. C y x =cos cos 解：dx x x dy y y ydx x ydy x sin cos sin cos 0sin cos cos sin -=⇒=+ C C x y x x d y y d ⇒=+⇒-=⇒ln sin ln sin ln sin sin sin sin . 30.微分方程x xe y y y -=-'+''2，特解用特定系数法可设为（） A.x e b ax x y -+=*)( B. x e b ax x y -+=*)(2 C. x e b ax y -+=*)( D. x axe y -=* 解：-1不是微分方程的特征根，x 为一次多项式，可设x e b ax y -+=*)( C ⇒.二、填空题（每题2分，共30分） 31.设 ,1||,01||,1)(⎩⎨⎧>≤=x x x f ，则=)(sin x f _________ 解：1)(sin 1}sin |=⇒≤x f x .32.若=--+→x x x x 231lim 22=_____________ 解：=++=++--=--+→→→)31(1lim )31)(2()2(lim 231lim 2222x x x x x x x x x x x x 123341==. 33.已知x y 2arctan =，则=dy __________ 解：dx xdy 2412+= . 34.函数 bx x a x x f ++=23)(，在1-=x 处取得极值-2，则_______,==b a . 解：b a b a b ax x x f -+-=-=+-⇒++='12,02323)(2.5,4==⇒b a .35.曲线12323-+-=x x x y 的拐点为 __________解：)1,1(),(0662632-=⇒=-=''⇒+-='y x x y x x y .36.设)(),(x g x f 是可微函数，且为某函数的原函数，有1)1(,3)1(==g f 则=-)()(x g x f _________解：2)1()1()()(=-=⇒=-g f C C x g x f 2)()(=-⇒x g x f .37.⎰-=+ππ)sin (32x x _________解：3202sin )sin (023232ππππππππ=+=+=+⎰⎰⎰⎰---x xdx dx x x x . 38.设⎪⎩⎪⎨⎧<≥=0,0,)(2x x x e x f x ，则 ⎰=-20)1(dx x f __________解：⎰⎰⎰⎰--=--=+==-201110012132)()1(e dx e dx x dt t f dx x f x t x .39. 已知 }1,1,2{},2,1,1{-==b a，则向量a 与b 的夹角为=__________解：3,21663||||,cos π>=⇒<==⋅>=<b a b a b a b a.40.空间曲线⎩⎨⎧==022z xy 绕x 轴旋转所得到的曲面方程为 _________.解：把x y 22=中的2y 换成22y z +即得所求曲面方程x y z 222=+.41. 函数y x x z sin 22+=，则 =∂∂∂yx z2_________解： ⇒+=∂∂y x x x z sin 22y x yx z cos 22==∂∂∂ . 42.设区域}11,10|),{(≤≤-≤≤=y x y x D ，则___)(2⎰⎰=-Ddxdy xy . 解：⎰⎰⎰⎰⎰-=-=-=--Ddx x dy x y dx dxdy x y 102101122322)()( .43. 函数2)(x e x f -=在0=x 处的展开成幂级数为________________解： ∑∞=⇒=0!n n xn x e ∑∑∞=∞=-+∞-∞∈-=-==0022),(,!1)1(!)()(2n n n n n x x x n n x e x f .44.幂级数∑∞=+++-0112)1()1(n n n nn x 的和函数为 _________ 解：∑∑∑∞=∞=-+∞=+++=-=+-=+-0111011)21ln()2()1(1)2()1(2)1()1(n n nn n n n n n nx n x n x n x .45.通解为x x e C e C y 321+=-的二阶线性齐次常系数微分方程为_________解：x x e C e C y 321+=-0323,1221=--⇒=-=⇒λλλλ032=-'-''⇒y y y .三、计算题（每小题5分，共40分）46． x x e x xx 2sin 1lim 3202-→-- 解：20300420320161lim 3222lim 81lim 2sin 1lim2222x e x xe x x ex xx e x x x x x x x x x -=+-=--=---→-→-→-→ 161lim 161322lim220000-=-=-=-→-→x x x x e x xe . 47.设x x x y 2sin 2)3(+=，求dxdy解：取对数得：)3ln(2sin ln 2x x x y +=，两边对x 求导得：xx x x x x x y y 3322sin )3ln(2cos 2122++++='所以]3322sin )3ln(2cos 2[)3(222sin 2xx x x x x x x x y x +++++=' xx x x x x x x x x x 2sin )32()3()3ln(2cos )3(212sin 222sin 2+++++=-.48.求 ⎰-dx x x 224解：⎰⎰⎰⎰-===-=dt t tdt tdt t tdx x x tx )2cos 1(2sin 4cos 2cos 2sin 4422sin 222C x x x C t t x C t t +--=+-=+-=242arcsin 2cos sin 22arcsin 22sin 2249.求⎰--+102)2()1ln(dx x x解：⎰⎰⎰+---+=-+=-+101010102)1)(2(12)1ln(21)1ln()2()1ln(dx x x x x x d x dx x x⎰=-=+-+=++--=10102ln 312ln 322ln 12ln 312ln )1121(312ln x x dx x x ..50.设),()2(xy x g y x f z ++= ，其中),(),(v u g t f 是可微函数，求 yzx z ∂∂∂∂,解：xv v g x u u g x y x y x f x z ∂∂∂∂+∂∂∂∂+∂+∂+'=∂∂)2()2( ),(),()2(2xy x g y xy x g y x f v u'+'++'==∂∂∂∂+∂∂∂∂+∂+∂+'=∂∂y vv g y u u g y y x y x f y z )2()2(),()2(xy x g x y x f v '++'. 51．计算积分⎰⎰=Dydxdy x I 2 ，其中:D 由直线1,2,===x x y x y 所围成的闭区域.解：积分区域如图所示，可表示为：x y x x 2,10≤≤≤≤.所以 ⎰⎰⎰⎰==1222xx Dydy x dx ydxdy x I10310323)2(10510421022====⎰⎰x dx x y dx x xx52．求幂级数nn nx ∑∞=--+0)1()3(11的收敛区间（不考虑端点）. 解：令t x =-1，级数化为 n n nt ∑∞=-+0)3(11，这是不缺项的标准的幂级数. 因为 313)3(11)3(1lim )3(1)3(1lim lim 11=--+-=-+-+==∞→+∞→+∞→nnn n n n n n n a a ρ，故级数nn nt ∑∞=-+0)3(11的收敛半径31==ρR ，即级数收敛区间为（-3，3）. 对级数nn nx ∑∞=--+0)1()3(11有313<-<-x ，即42<<-x . 故所求级数的收敛区间为），（42-.53．求微分方程 0)12(2=+-+dy x xy dy x 通解.解：微分方程0)12(2=+-+dx x xy dy x 可化为 212xxy x y -=+'，这是一阶线性微分方程，它对应的齐次线性微分方程02=+'y x y 通解为2xCy =.设非齐次线性微分方程的通解为2)(x x C y =，则3)(2)(xx C x C x y -'='，代入方程得C x x x C x x C +-=⇒-='2)(1)(2.故所求方程的通解为2211xCx y +-=.四、应用题（每题7分，共计14分）54.某公司甲乙两厂生产一种产品，甲乙两厂月产量分别为y x ,千件；甲厂月产量成本为5221+-=x x C ，乙厂月产量成本为3222++=y y C ；要使月产量为8千件，且总成本最小，求甲乙两厂最优产量和最低成本？解：由题意可知：总成本8222221++-+=+=y x y x C C C ，约束条件为8=+y x .问题转化为在8=+y x 条件下求总成本C 的最小值 . 由8=+y x 得x y -=8，代入得目标函数为0(882022>+-=x x x C 的整数）.则204-='x C ，令0='C 得唯一驻点为5=x ，此时有04>=''C . 故5=x 使C 得到极小唯一极值点，即最小值点.此时有38,3==C y . 所以甲乙两厂最优产量分别为5千件和3千件，最低成本为38成本单位. 55.求曲线)2)(1(--=x x y 和x 轴所围成图形绕y 轴旋转一周所得的体积. 解：平面图形如下图所示：此立体可看作x 区域绕y利用体积公式⎰=ba y dx x f x V |)(|2π.显然，抛物线与x 两交点分别为（1，0）；（2平面图形在x 轴的下方.故⎰⎰---==21)2)(1(2|)(|2x x x dx x f x V ba y ππ2)4(2)23(2212342123πππ=+--=+--=⎰x x x dx x x x .xx五、证明题（6分）56设)(x f 在],[a a -上连续，且>a ，求证⎰⎰--+=aaadx x f x f dx x f 0)]()([)(.并计算⎰--+441cos ππdx e xx .证明：因为⎰⎰⎰--+=aaaadx x f dx x f dx x f 0)()()(，而⎰⎰⎰⎰-=-=--=-=-0)()()()()(aaa tx a dx x f dt t f t d t f dx x f ，故⎰⎰⎰⎰⎰-+=+=--aaa aa adx x f dx x f dx x f dx x f dx x f 0)()()()()( 即有⎰⎰--+=aaadx x f x f dx x f 0)]()([)(.利用上述公式有dx e e e x dx e x e x dx e x x x x x x x ⎰⎰⎰⎥⎦⎤⎢⎣⎡+++=+-++=+---404044111cos ]1)cos(1cos [1cos ππππ 22sin cos 4040===⎰ππx dx x .说明：由于时间紧，个别题目语言叙述与试卷有点不近相同，没有进行认真检查，考生仅作参考.河南省“专升本”考试《高等数学》辅导专家葛云飞提供.。

2010成人高考专升本高数一真题及答案解析

☆基本分配制度：
内容：以按劳分配为主体、多种分配方式并存；
意义：有利于发挥激励作用，调动人们的积极性、创造性，吸引人才，发挥人才的作用，促进国家和社会的发展。
☆基本经济制度、分配制度与人民的职业及收入来源的关系：
是因果关系，由于我国实行这种充满生机和活力的基本经济制度、分配制度，极大地调动人民的积极性、创造性，
非公有制经济：个体、私营、外资
☆基本经济制度的意义：
1.是由我国基本国情决定的；
2.公有制是社会主义经济制度的基础，占主体地位，是维护广大人民根本利益和实现共同富裕的保证；
3.非公有制经济是我国经济的重要组成部分；
4.有利于促进经济发展、社会进步，解放、发展生产力；
5.有利于提高人民生活水平，为公民施展才干提供极大的活动舞台和发展空间。
质文明的发展提供政治保证和法律保障；精神文明为物质文明的发展提供思想保证、精神动力和智
力支持；生态文明有利于促进我国经济、社会的可持续发展。
基本经济制度：以公有制为主体，多种所有制经济共同发展。
国有经济
公有制经济集体经济
经济混合所有制经济中的国有成分和集体成分
不要嫌多，这可是初中全部的精点，热点，希望可以帮助你社会主义初级阶段
历史原因：由于中国脱胎于半殖民地半封建社会，是在一穷二白的基础上起步建设的；
.. 1.生产力水平还比较低（根本原因）；
原因现实原因（表现）2.地区发展不平衡；
是否有利于发展社会主义社会的生产力；
“三个有利于”是否有利于增强社会主义国家的综合国力；
是否有利于提高人民的生活水平。
☆我国发展的有利条件：
建立社会主义公有制，能集中力量办大事，有较雄厚的工业基础和经济基础；科学技术提高很大，农业和农

2010成人高考专升本高数二真题及答案解析

统计分析软件应用学期报告具体要求：1、课程学期报告要求：1）报告：请各位同学选择感兴趣的主题，并收集相应数据，尝试利用常用统计分析方法、工具（如：1、描述性统计分析；2、相关分析；3、回归分析等）对所选主题进行分析，并依照附件论文的格式完成课程报告，主要结果以及程序代码以附件形式给出；2）格式要求：论文包含学期报告封面、标题、作者、摘要、正文、参考资料、附件等部分，具体可以参考附件样稿，正文使用五号字体，小节标题小四字体，行距设为1.5倍，每幅图控制在1/6~1/3篇幅，正文部分不得少于6页。

论文中引用他人工作请采用上标序号标出，同时在参考文献中给出相应文献。

（如：关于非线性随机效应模型[4]我们可知，……相应参考文献：[4] 赵为华，冯予. 非线性随机效应模型基于统计曲率的统计诊断[J]. 南京理工大学学报，2005，243-247.）2、关于主题选择及参考资料的说明1）主题可以任选，但要注意数据的可获得性，但不可以直接抄袭相关数据结论。

2）统计年鉴数据可登陆中国统计局年鉴和各个省份统计局网页查询：历年中国年鉴：/tjsj/；地方年鉴：/tjlj/。

3）软件与数据及相关资料网络资源推荐：3、特别提示：学期论文重在参与，主要目的是让每个同学通过亲自实践操作分析，掌握基本的统计分析方法，因此要求每个人务必自己做，可以互相帮忙参考，但绝不可以抄袭，雷同者一律为零分（抄袭与被抄袭的均如此）！4、请各位同学认真按要求完成报告，不要急于提交，5月29日上机时间安排答辩，答辩完后由学委统一收取报告电子稿、打印稿各一份。

（电子稿发送到：Email：ahualian@ ，文档以“学号+姓名+报告题目.doc”命名，如“2010030005 张珊珊农林牧渔总产值及其指数情况的分析.doc）课程学期报告2012-2013 学年第1学期课程名称统计分析软件应用报告题目学生姓名学号院系专业任课教师朱连华二Ｏ一二年五月二十八日农林牧渔总产值及其指数情况的分析张珊珊2010030005 2010统计（1）班摘要：本文采用了……的数据，研究了………的问题，得出……的结论。

2010河南专升本高等数学真题及答案详解

2010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答题前，考生务必将自己的姓名、考场号、座位号、考生号填写在答题卡上。

本试卷的试题答案必须答在答题卡上，答在试卷上无效。

一、选择题（每小题2分，共60分）在每小题的四个备选答案中选出一个正确答案，用铅笔把答题卡上对应题目的答案标号涂黑。

如需改动，用橡皮擦干净后，再选涂其他答案标号。

1．设函数)(x f 的定义域为区间(1,1]-，则函数(1)e f x -的定义域为A ．[2,2]-B ．(1, 1]-C ．(2, 0]-D ．(0, 2]2．若()f x ()x R ∈为奇函数，则下列函数为偶函数的是A ．()y x =，[1, 1]x ∈-B ．3()tan y xf x x =+，(π, π)x ∈-C ．3sin ()y x x f x =-，[1, 1]x ∈-D ．25()e sin x y f x x =，[π, π]x ∈- 3．当0→x 时，2e1x-是sin 3x 的A ．低阶无穷小B ．高阶无穷小C ．等价无穷小D ．同阶非等价无穷小4．设函数2511sin , 0()e , 0xx x x f x x ⎧>⎪=⎨⎪<⎩，则0x =是)(x f 的 A ．可去间断点 B ．跳跃间断点 C ．连续点D ．第二类间断点5．下列方程在区间(0, 1)内至少有一个实根的为 A ．220x +=B ．sin 1πx =-C ．32520x x +-=D ．21arctan 0x x ++=6．函数)(x f 在点0x x =处可导，且1)(0-='x f ，则000()(3)lim2h f x f x h h→-+=A ．23B ．23-C ．32-D ．327．曲线x x y ln =的平行于直线01=+-y x 的切线方程是 A ．1-=x y B ．)1(+-=x y C ．1y x =-+D ．)1)(1(ln -+=x x y8．设函数π2sin 5y =，则='y A．π2cos 5-B．CD．2πcos 55-9．若函数()f x 满足2d ()2sin d f x x x x =-，则()f x = A ．2cos xB ．2cos x C +C ．2sin x C +D ．2cos x C -+10．d e sin(12)d d b xax x x --=⎰ A ．e sin(12)x x -- B ．e sin(12)d x x x -- C ．e sin(12)x x C --+D ．011．若()()f x f x -=，在区间(0, )+∞内，()0f x '>，()0f x ''>，则()f x 在区间(, 0)-∞内A ．()0f x '<，()0f x ''<B ．()0f x '>，()0f x ''>C ．()0f x '>，()0f x ''<D ．()0f x '<，()0f x ''>12．若函数()f x 在区间(, )a b 内连续，在点0x 处不可导，0(, )x a b ∈，则 A ．0x 是()f x 的极大值点 B ．0x 是()f x 的极小值点 C ．0x 不是()f x 的极值点 D ．0x 可能是()f x 的极值点13．曲线e xy x -=的拐点为 A ．1x =B ．2x =C ．222,e ⎛⎫ ⎪⎝⎭D ．11,e ⎛⎫ ⎪⎝⎭14．曲线2arctan 35xy x=+ A ．仅有水平渐近线 B ．仅有垂直渐近线C ．既有水平渐近线，又有垂直渐近线D ．既无水平渐近线，又无垂直渐近线 15．若x cos 是)(x f 的一个原函数，则=⎰)(d x fA ．sin x C -+B ．sin xC + C ．cos x C -+D ．cos x C +16．设曲线()y f x =过点(0, 1)，且在该曲线上任意一点(, )x y 处切线的斜率为e x x +，则=)(x fA ．2e 2x x -B ．2e 2x x +C ．2e x x +D ．2e x x -17．2 π4πsin d 1x xx x -=+⎰A ．2B ．0C ．1D ．1-18．设)(x f 是连续函数，则2()d x af t t ⎰是A ．)(x f 的一个原函数B ．)(x f 的全体原函数C ．)(22x xf 的一个原函数D ．)(22x xf 的全体原函数19．下列广义积分收敛的是 A．1x +∞⎰ B ．2 e ln d xx x +∞⎰C ．2e1d ln x x x+∞⎰D ．21d 1xx x+∞+⎰20．微分方程0)(224=-'+''y x y y x 的阶数是 A ．1B ．2C ．3D ．421．已知向量{5, , 2}a x =-和{, 6, 4}b y = 平行，则x 和y 的值分别为A ．4-，5B ．3-，10-C ．4-，10-D ．10-，3-22．平面1x y z ++=与平面2=-+z y x 的位置关系是 A ．重合 B ．平行C ．垂直D ．相交但不垂直23．下列方程在空间直角坐标系中表示的曲面为柱面的是 A ．221y z += B ．22z x y =+ C ．222z x y =+D ．22z x y =-24．关于函数222222,0(,)0,0xy x y x y f x y x y ⎧+≠⎪+=⎨⎪+=⎩下列表述错误的是A ．(, )f x y 在点(0, 0)处连续B ．(0, 0)0x f =C ．(0, 0)0y f =D ．(, )f x y 在点(0, 0)处不可微25．设函数)ln(y x y x z -=，则=∂∂yzA ．)(y x y x -B ．2ln()x x y y --C ．ln()()x y xy y x y -+- D ．2ln()()x x y xy y x y ---- 26．累次积分2d (, )d x f x y y ⎰⎰写成另一种次序的积分是A ．1d (, )d yyy f x y x -⎰⎰B．2d (, )d y f x y x ⎰⎰C．11d (,)d y f x y x -⎰⎰D．11 11d (, )d y f x y x -⎰⎰27．设{(, )|D x y x =≤2, y ≤2}，则⎰⎰=Dy x d dA ．2B ．16C ．12D ．428．若幂级数∑∞=0n nnx a的收敛半径为R ，则幂级数∑∞=-02)2(n n n x a 的收敛区间为A．( B ．(2, 2)R R -+ C ．(, )R R -D．(2 229．下列级数绝对收敛的是 A ．∑∞=-11)1(n nnB ．∑∞=-1223)1(n n nnC ．∑∞=-+-1121)1(n n n nD ．∑∞=--1212)1(n nn n30．若幂级数(3)nn n a x ∞=-∑在点1x =处发散，在点5x =处收敛，则在点0x =，2x =，4x =，6x =中使该级数发散的点的个数有A ．0个B ．1个C ．2个D ．3个二、填空题（每空2分，共20分）31．设(32)f x -的定义域为(3, 4]-，则)(x f 的定义域为________． 32．极限limx =________．33．设函数()(1)(2)(3)(4)f x x x x x =++--，则(4)()f x =________．34．设参数方程22 1 31x t y t =+⎧⎨=-⎩所确定的函数为()y y x =，则22d d yx =________． 35．(ln 1)d x x +=⎰________．36．点(3, 2, 1)-到平面10x y z ++-=的距离是________． 37．函数(1)x z y =+在点(1, 1)处的全微分d z =________．38．设L 为三个顶点分别为(0, 0)，(1, 0)和(0, 1)的三角形边界，L 的方向为逆时针方向，则2322()d (3)d Lxyy x x y xy y -+-=⎰ ________．39．已知微分方程x ay y e =+'的一个特解为x x y e =，则a =________．40．级数03!nn n ∞=∑的和为________．三、计算题（每小题5分，共45分）41．求极限2040sin d (e 1)sin lim 1cos x x x t t x x x →⎛⎫- ⎪- ⎪- ⎪⎝⎭⎰． 42．设由方程22e e y xy -=确定的函数为)(x y y =，求d d x yx =． 43．求不定积分2xx ．44．求定积分( 2d x x ⎰．45．求过点(1, 2, 5)-且与直线213 3 x y z x y -+=⎧⎨-=⎩平行的直线方程．46．求函数x xy y x y x f 823),(22+-+=的极值． 47．将23()21xf x x x =+-展开成x 的幂级数． 48．计算二重积分Dσ⎰⎰，其中D 是由圆223x y +=所围成的闭区域．49．求微分方程069=+'-''y y y 的通解．四、应用题（每小题8分，共16分）50．要做一个容积为V 的圆柱形带盖容器，问它的高与底面半径的比值是多少时用料最省？ 51．平面图形D 由曲线2x y =，直线x y -=2及x 轴所围成．求：（1）D 的面积；（2）D 绕x 轴旋转形成的旋转体的体积．五、证明题（9分）52．设函数)(x f 在闭区间]1,0[上连续，在开区间)1,0(内可导，且(0)0f =，(1)2f =．证明：在)1,0(内至少存在一点ξ，使得()21f ξξ'=+成立．2010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试题参考答案及评分标准一、选择题（每小题2分，共60分）二、填空题（每小题2分，共20分）31．[5, 9)- 32．5233．24 34．3235．ln x x C + 3637．2ln 2d d x y + 38．0 39．1- 40．3e三、计算题（每小题5分，共45分）41．3242．222002d d 24e d d e 0x x y y y xx-======- 43．322(e 1)3x C +-44．π22+ 45．125315x y z --+==- 46．函数在(6, 2)--处有极小值(6, 2)24f --=- 47．00111()(1)2[(1)2], , 22nnnnn n nn n n f x x x x x ∞∞∞===⎛⎫=--=--∈- ⎪⎝⎭∑∑∑48．49．1312()e x y C C x =+（1C ，2C 是任意常数）四、应用题（每小题8分，共16分）50．3232ππ2πππV h V V V r r r r V===⋅=⋅= 51．（1） 1201d 112A x x =+⋅⋅⎰ 13015326x =+= （2） 14201πd π113x V x x =+⋅⋅⎰ 150π8ππ5315x =+=第51题图五、证明题（9分）52．证明：构造函数2()()F x f x x =-，因)(x f 在闭区间]1,0[上连续，在开区间)1,0(内可导，所以函数)(x F 在闭区间]1,0[上连续，在开区间)1,0(内可导，且()()2F x f x x ''=-．于是)(x F 在]1,0[上满足拉格朗日中值定理的条件，故在开区间)1,0(内至少存在一点ξ，使得(1)(0)()10F F F ξ-'=-，将(0)0f =，(1)2f =代入上式，得(1)(0)()[(1)1][(0)0]110F F F f f ξ-'==---=-，即()21f ξξ'-=，于是()21f ξξ'=+．。

2010成人高考专升本高数一真题及答案解析

2010成人高考专升本高数一真题及答案解析2010成人高考专升本高数一真题及答案解析——2010年成人高等学校招生全国统一考试高等数学（一）答案必须答在答题卡上指定的位置，答在试卷上无效。

一、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，只有一项是符合题目要求的，将近选项前的字母填涂在答题卡相应题号的信息点上。

A、3B、2C、1D、0正确答案：C【安通名师解析】根据函数的连续性立即得出结果【安通名师点评】这是计算极限最常见的重要题型。

在教学中一直被高度重视。

在上课时多次强调的重点，必须记住。

正确答案：B【安通名师解析】根据基本初等函数求导公式复合函数求导法则或直接用微分计算【安通名师点评】这样的题目已经在安通学校保过班讲义中练习过多次，属于特别重要内容。

【安通名师解析】基本积分公式,直接积分法。

【安通名师点评】这是每年都有的题目。

考的就是公式是否记住了。

课堂上讲过练过多次，要求学生对基本积分公式背熟。

正确答案：C【安通名师解析】使用基本初等函数求导公式【安通名师点评】这是本试卷中第二个直接使用基本初等函数求导公式的计算题。

考的就是公式是否掌握了。

我们在平时教学中一再要求学生对基本公式背熟。

否则寸步难行。

正确答案：D【安通名师解析】用洛必达法则求解【安通名师点评】这类问题在以往的考试中经常出现，重要但并不难。

是一种典型的题目。

也始终是讲课的重点。

正确答案：A【安通名师解析】把y看作常数，对x求导。

【安通名师点评】本题仍然属于基本题目，是年年考试都有的内容正确答案：A【安通名师解析】因为是选择题，只要验证点的坐标满足方程就可以了。

【安通名师点评】本题如果是填空或解答题，难度将大为增加。

现在是选择题，理解概念就行。

正确答案：B【安通名师解析】直接使用公式【安通名师点评】这是计算收敛半径最常见的题型。

比较简单比较重要。

在教学中一直被高度重视。

二、11－20小题，每小题4分，共40分，把答案写在答题卡相应题号后。

2010年河南专升本高数真题+答案解析

2010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一、选择题（每小题2 分，共60 分）1．设函数()f x 的定义域为区间(1,1]-，则函数(1)f x e -的定义域为（）A ．[]2,2-B ．(1,1]-C ．(2,0]-D ．(0,2]【答案】D【解析】由题意得，()f x 的定义域为(1,1]-，则在(1)f x e -中，1(1,1]x -∈-，即02x <≤，故选D ．2．若()()f x x R ∈为奇函数，则下列函数为偶函数的是（） A ．[]331(),1,1y x x x -∈- B ．3()tan ,(,)y xf x x x ππ=+∈-C ．[]3sin (),1,1y x x f x x =-∈-D ．[]25()sin ,,x y f x e x x ππ=∈-【答案】D【解析】()f x 为奇函数，对于选项D ，22()55()sin ()()sin x x f x e x f x e x ---=，故选D ．3．当0x →时，21x e -是sin3x 的（） A ．低阶无穷小 B ．高阶无穷小C ．等价无穷小D ．同阶非等价无穷小【答案】D【解析】200122lim lim sin333x x x e x x x →→-==，从而21x e -是sin3x 的同阶非等价无穷小，故选D ．4．设函数2511sin ,0(),0xx x x f x e x ⎧>⎪=⎨⎪<⎩，则0x =是()f x 的（）A ．可去间断点B ．跳跃间断点C ．连续点D ．第二类间断点【解析】2501lim sin 0x x x+→=，10lim 0x x e -→=，00lim ()lim ()x x f x f x +-→→=，从而0x =是()f x 的可去间断点，故选A ．5．下列方程在区间(0,1)内至少有一个实根的为（） A ．20x += B ．sin 1x π=-C ．32520x x +-=D ．21arctan 0x x ++=【答案】C【解析】对于选项C ，构造函数32()52f x x x =+-，(0)20f =-<，(1)40f =>，由零点定理得，()0f x =在(0,1)上至少存在一个实根，故选C ．6．函数()f x 在点0x x =处可导，且0()1f x '=-，则000()(3)lim2x f x f x h h→-+=（）A ．23 B ．23-C ．32-D ．32【答案】D 【解析】0000000()(3)(3)()333limlim ()23222x x f x f x h f x h f x f x h h →→-++-⎛⎫'=⋅-=-= ⎪⎝⎭，故选D ．7．曲线ln y x x =平行于直线10x y -+=的切线方程是（） A ．1y x =- B ．(1)y x =-+C ．1y x =-+D ．(ln 1)(1)y x x =+-【答案】A【解析】ln 1y x '=+，又直线10x y -+=的斜率1k =，令1y '=得1x =，0y =，从而与直线平行的切线方程为01y x -=-，即1y x =-，故选A ．8．设函数212sin 5y x π=-，则y '=（）A ．22cos51x π-- B ．21x-C 21x-D ．22cos 551x π-【解析】(2212sin 51y x xπ''⎛⎫'=--= ⎪⎝⎭-B ．9．若函数()f x 满足2()2sin df x x x dx =-，则()f x =（）A ．2cos xB ．2cos xC +C ．2sin x C +D ．2cos x C -+【答案】B【解析】2()2sin df x x x dx =-，则2222()(2sin )sin cos f x x x dx x dx x C =-=-=+⎰⎰，故选B ． 10．sin(12)b xa d e x dx dx--=⎰（）A ．sin(12)x e x --B ．sin(12)x e x dx --C ．sin(12)x e x C --+D ．0【答案】D【解析】sin(12)bx a e x dx --⎰为一常数，从而sin(12)0b xa d e x dx dx--=⎰，故选D ．11．若()()f x f x -=，在区间(0,)+∞内，()0f x '>，()0f x ''>，则()f x 在区间(,0)-∞内（） A ．()0,()0f x f x '''<< B ．()0,()0f x f x '''>>C ．()0,()0f x f x '''><D ．()0,()0f x f x '''<>【答案】D【解析】()()f x f x -=，则()f x 为偶函数，又在(0,)+∞上，()0f x '>，()0f x ''>，所以在(,0)-∞上()0f x '<，()0f x ''>，故选D ．12．若函数()f x 在区间(,)a b 内连续，在点0x x =处不可导，0(,)x a b ∈，则（） A ．0x 是()f x 的极大值点 B ．0x 是()f x 的极小值点C ．0x 不是()f x 的极值点D ．0x 可能是()f x 的极值点【答案】D【解析】由判断极值的方法知，0x 可能是()f x 的极值点，故选D ．13．曲线x y xe -=的拐点为（）A ．1x =B ．2x =C ．222,e ⎛⎫ ⎪⎝⎭D ．11,e ⎛⎫ ⎪⎝⎭【答案】C【解析】(1)x y x e -'=-，(2)x y x e -''=-，令0y ''=，得2x =，22y e=．当2x >时，0y ''>，2x <，0y ''<，所以曲线的拐点为222,e ⎛⎫⎪⎝⎭，故选C ．14．曲线2arctan 5xy x=（） A ．仅有水平渐近线 B ．仅有垂直渐近线C ．既有水平渐近线，又有垂直渐近线D ．既无水平渐近线，又无垂直渐近线【答案】A 【解析】002arctan 22limlim 555x x x x x x →→==，所以曲线没有垂直渐近线；2arctan lim 05x xx→∞=，所以0y =为曲线的水平渐近线，故选A ．15．若cos x 是()f x 的一个原函数，则()df x =⎰（）A ．sin x C -+B ．sin xC +C ．cos x C -+D ．cos x C +【答案】A【解析】令()cos F x x =，则()()sin f x F x x '==-，所以()(sin )sin df x d x x C =-=-+⎰⎰，故选A ．16．设曲线()y f x =过点(0,1)，且在该曲线上任意一点(,)x y 处切线的斜率为x x e +，则()f x =（）A ．22x x e -B ．22x x e +C ．2x x e +D ．2x x e -【答案】B【解析】由题意得xy x e '=+，则2()2xx x y x e dx e C =+=++⎰，又因为曲线过点(0,1)，有0C =，从而2()2x x y f x e ==+，故选B ．17． 24sin 1x xdx x ππ-=+⎰（）A ．2B ．0C ．1D ．1-【答案】B【解析】24sin 1x xx +为奇函数，积分区间关于原点对称，从而24sin 01x x dx xππ-=+⎰．18．设()f x 是连续函数，则20()x f t dt ⎰是（）A ．()f x 的一个原函数B ．()f x 的全体原函数C ．22()xf x 的一个原函数D ．22()xf x 的全体原函数【答案】C【解析】220()2()x f t dt xf x '⎛⎫= ⎪⎝⎭⎰，由原函数的定义可知，它是22()xf x 的一个原函数，故选C ．19．下列广义积分收敛的是（）A ．1x+∞⎰B ．2ln exdx x+∞⎰C ．21ln edx x x+∞⎰D ．21exdx x +∞+⎰【答案】C 【解析】22111ln 011ln ln ln eee dx d x x x x x+∞+∞+∞==-=+=⎰⎰，故选C ．20．微分方程422()0x y y x y '''+-=的阶数是（）A ．1B ．2C ．3D ．4【答案】B【解析】由微分方程的概念知，阶数为方程中的最高阶导数的阶数，故选B ．21．已知向量{}5,,2x =-a 和{},6,4y =b 平行，则x 和y 的值分别为（）A ．4,5-B ．3,10--C ．4,10--D ．10,3--【答案】B【解析】向量a 与b 平行，所以5264x y -==，得3x =-，10y =-，故选B ．22．平面1x y z ++=与平面2x y z +-=的位置关系是（）A ．重合B ．平行C ．垂直D ．相交但不垂直【答案】D【解析】两平面的法向量分别为1(1,1,1)=n ，2(1,1,1)=-n ，而111111=≠-，从而两平面不平行，又121⋅=n n ，从而两平面不垂直但相交，故选D ．23．下列方程在空间直角坐标系中表示的曲面为柱面的是（）A ．221y z +=B ．22z x y =+C ．222z x y =+D ．22z x y =-【答案】A【解析】由柱面方程的特点可知，221y z +=表示圆柱面，故选A ．24．关于函数222222,0(,)0,0xyx y x y f x y x y ⎧+≠⎪+=⎨⎪+=⎩，下列表述错误的是（）A ．(,)f x y 在点(0,0)处连续B ．(0,0)0f =C ．(0,0)0y f '=D ．(,)f x y 在点(0,0)处不可微【答案】A【解析】令y kx =，则222222000lim lim (1)1x x y kx xy kx kx y k x k →→=→==+++．当k 取不同值时，极限值不同，因此2200limx y xyx y →→+不存在，所以在点(0,0)处不连续，故选A ．25．设函数ln()x z x y y =-，则zy∂=∂（） A ．()x y x y -B ．2ln()x x y y --C ．ln()()x y xy y x y -+- D ．2ln()()x x y xy y x y ---- 【答案】D 【解析】221ln()ln()(1)()z x x x x y xx y y y y x y y y x y ∂-=--+⋅⋅-=--∂--．26．累次积分222202(,)x x x x dx f x y dy --⎰写成另一种次序的积分是（）A ．10(,)yydy f x y dx -⎰⎰B ．222202(,)y y y y dy f x y dx ---⎰C ．221111(,)y y dy f x y dx ----⎰D ．22111111(,)y y dy f x y dx +----⎰⎰【答案】D【解析】由题意知，02x ≤≤，2222x x y x x -≤≤-11y -≤≤，221111y x y -≤-，所以交换积分次序后为22111111(,)y y dy f x y dx +----⎰⎰．27．设{}(,)2,2D x y x y =≤≤，则Ddxdy =⎰⎰（）A ．2B ．16C ．12D ．4【答案】B【解析】222216Ddxdy dx dy --==⎰⎰⎰⎰，故选B ．28．若幂级数0nn n a x ∞=∑的收敛半径为R ，则幂级数20(2)n n n a x ∞=-∑的收敛区间为（）A ．(,)R RB ．(2,2)R R -+C ．(,)R R -D ．(2,2)R R【答案】D【解析】令2(2)t x =-，则0n n n a t ∞=∑的收敛半径为R ，即R t R -<<，则2(2)x R -<，即22R x R <<D ．29．下列级数绝对收敛的是（）A ．1(1)nn n∞=-∑B ．213(1)2nnn n ∞=-∑C ．11(1)21nn n n ∞=+--∑D ．21(1)21nn n ∞=--∑【答案】B【解析】对选项B ，21133(1)24nn nn n n ∞∞==⎛⎫-= ⎪⎝⎭∑∑，级数收敛，从而原级数绝对收敛，故选B ．30．若幂级数0(3)n n n a x ∞=-∑在点1x =处发散，在点5x =收敛，则在点0x =，2x =，4x =，6x =中使该级数发散的点的个数有（）A ．0 个B ．1个C ．2个D ．3个【答案】C【解析】由幂级数发散、收敛性质及收敛区间的讨论可得，在这4个点中发散点的个数有两个，即0x =，6x =，故选C ．二、填空题（每空 2分，共 20分）31．设(32)f x -的定义域为(3,4]-，则()f x 的定义域为________．【答案】[5,9)-【解析】(32)f x -的定义域为(3,4]-，即34x -<≤，所以5329x -≤-<，即()f x 的定义域为[5,9)-．32．极限lim (23)x x x x +-=________．【答案】52【解析】55lim (23)limlim2232311x x x x x x x x x x x+-===++-++-．33．设函数()(1)(2)(3)(4)f x x x x x =++--，则(4)()f x =________．【答案】24【解析】(4)()4!24f x ==．34．设参数方程22131x t y t =+⎧⎨=-⎩所确定的函数为()y y x =，则22d ydx =________．【答案】32【解析】632dydy t dt t dx dx dt===，22(3)322d dy d y t dt dx dx dx dt ⎛⎫ ⎪'⎝⎭===．35．(ln 1)x dx +=⎰________．【答案】ln x x C +【解析】1(ln 1)ln ln ln x dx xdx dx x x x dx x x x C x+=+=-⋅+=+⎰⎰⎰⎰．36．点(3,2,1)-到平面10x y z ++-=的距离是________． 3【解析】321131113d +--===++．37．函数(1)x z y =+在点(1,1)处的全微分dz =________．【答案】2ln 2dx dy + 【解析】(1)ln(1)x zy y x∂=++∂，1(1)x z x y y -∂=+∂，(1,1)(1,1)2ln 2z z dz dx dy dx dy xy ⎛⎫∂∂=+=+ ⎪∂∂⎝⎭．‘38．设L 为三个顶点分别为(0,0)，(1,0)和(0,1)的三角形边界，L 的方向为逆时针方向，则2322()(3)Lxy y dx x y xy dy -+-=⎰________．【答案】0 【解析】223P xy y y ∂=-∂，223Qxy y x∂=-∂，P Q y x ∂∂=∂∂，由格林公式得，该曲线积分为0．39．已知微分方程x y ay e '+=的一个特解为x y xe =，则a =________．【答案】1-【解析】将x y xe =代入微分方程得x x x x e xe axe e ++=，即1a =-．40．级数03!nn n ∞=∑的和为________．【答案】3e【解析】23012!3!!!n n xn x x x x e x n n ∞==++++++=∑，故303!nn e n ∞==∑．三、计算题（每小题5 分，共45 分）41．求极限2040sin (1)sin lim 1cos x x x tdt e x x x →⎡⎤-⎢⎥-⎢⎥-⎢⎥⎣⎦⎰．【答案】32【解析】220044000sin sin (1)sin (1)sin lim lim lim 1cos 1cos x x x x x x x tdt tdt e x e x x x x x →→→⎡⎤--⎢⎥-=-⎢⎥--⎢⎥⎣⎦⎰⎰ 230022sin 13lim lim 214222x x x x x x x x→→⋅=-=-=．42．设由方程22y e xy e -=确定的函数为()y y x =，求0x dy dx=．【答案】24e -【解析】方程两边同时关于x 求导，得220y e y y xy y ''⋅--⋅=，当0x =时，2y =，代入得 204x dy e dx-==．43．求不定积分21x xe +．32(1)213x x e e C ++ 【解析】令1x t e =+21x e t =-，2ln(1)x t =-，则221tdx dt t =-，于是 2222332(1)222(22)2(1)211331xx x x t t dt t dt t t C e e C t t e -=⋅=-=-+=++-+⎰⎰．44．求定积分220(2)x x x dx +-⎰．【答案】22π+【解析】22222000(2)221(1)(1)x x x dx xdx x x dx x d x -=+-=----⎰⎰⎰⎰令1t x =-，则122220111(1)(1)11122x d x t dt t dt ππ-----=-=--=-⋅⋅=-⎰⎰⎰，故220(2)22x x x dx π-=+⎰．45．求过点(1,2,5)-且与直线2133x y z x y -+=⎧⎨-=⎩平行的直线方程．【答案】125315x y z --+==- 【解析】由题意得，两平面的法向量分别为1(2,1,1)=-n ，2(1,3,0)=-n ，所以该直线的方向向量为12211(3,1,5)130=⨯=-=--i j ks n n ，又直线过点(1,2,5)-，故该直线的方程为125315x y z --+==-．46．求函数22(,)328f x y x y xy x =+-+的极值．【答案】24-【解析】228x f x y =-+，62y f y x =-，令00x y f f =⎧⎪⎨=⎪⎩，得驻点为62x y =-⎧⎨=-⎩，又2xx f =，2xy f =-，6yy f =，对于驻点(6,2)--，280B AC -=-<，20A =>，故函数在点(6,2)--处取得极小值(6,2)24f --=-．47．将23()21xf x x x =+-展开成x 的幂级数．【答案】011()(1)222n n n n f x x x ∞=⎛⎫⎡⎤=-+-<< ⎪⎣⎦⎝⎭∑ 【解析】2311()21112x f x x x x x ==-+-+-，其中01(1)(11)1n n n x x x ∞==--<<+∑，00111(2)21222n n nn n x x x x ∞∞==⎛⎫==-<< ⎪-⎝⎭∑∑，故00011()(1)2(1)222nnnnn n n n n n f x x x x x ∞∞∞===⎛⎫⎡⎤=-+=-+-<< ⎪⎣⎦⎝⎭∑∑∑．48．计算二重积分22Dx y d σ+，其中D 是由圆223x y +=所围成的闭区域．【答案】3π【解析】用极坐标计算，{}(,)03,02D r r θθπ=≤≤≤≤，于是232220323Dx y d d rdr d ππσθθπ+=⋅==⎰．49．求微分方程960y y y '''-+=的通解．【答案】1312()x y C C x e =+（12,C C 是任意常数）【解析】对应的特征方程为29610r r -+=，特征根为1213r r ==，因此所给方程的通解为1312()x y C C x e =+（12,C C 是任意常数）．四、应用题（每小题8 分，共 16 分）50．要做一个容积为V 的圆柱形带盖容器，问它的高与底面半径的比值是多少时用料最省？【答案】当2hr=时，用料最省【解析】设该容器的高为h ，底面半径为r ，则该容器的容积2V r h π=，即2Vh r π=，该带盖容器的用料222222V S r rh r r πππ=+=+，则224V S r rπ'=-，令0S '=，解得唯一驻点32V r π=，故当32Vr πS 取值最小，此时 323322V h V V V r r r r ππππ===⋅=．51．平面图形D 由曲线2y x =直线2y x =-及x 轴所围成．求：（1）D 的面积；（2）D 绕x 轴旋转形成的旋转体的体积．【答案】（1）56 （2）815π 【解析】（1）由题意可得，此平面区域D 如图所示，则1312200125(2)2236S y y dy y y y ⎡⎤⎡=-=--=⎢⎥⎣⎣⎦⎰．（2）平面D 绕x 轴旋转形成的旋转体的体积为124251322101118(2)245315x V x dx x dx x x x x πππππ⎛⎫=+-=+-+=⎪⎝⎭⎰⎰．五、证明题（9 分）52．设函数()f x 在闭区间[]0,1上连续，在开区间(0,1)内可导，且(0)0f =，(1)2f =．证明：在(0,1)内至少存在一点ξ，使得()21f ξξ'=+．【解析】构造函数2()()F x f x x =-，由题意可知()F x 在[]0,1上满足拉格朗日中值定理的条件，故在(0,1)内至少存在一点ξ，使得(1)(0)()10F F F ξ-'=-，代入得，()()21F f ξξξ''=-=，即()21f ξξ'=+．。

2010年河南省专升本数学

2010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题号一二三四五总分分值602045169150注意事项：答题前，考生务必将自己的姓名、考场号、座位号、考生号填写在答题卡上。

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1．设函数)(x f 的定义域为区间(1,1]-，则函数(1)e f x -的定义域为A ．[2,2]-B ．(1, 1]-C ．(2, 0]-D ．(0, 2]2．若()f x ()x R ∈为奇函数，则下列函数为偶函数的是A ．331()y x f x =-，[1, 1]x ∈-B ．3()tan y xf x x =+，(π, π)x ∈- C ．3sin ()y x x f x =-，[1, 1]x ∈- D ．25()e sin x y f x x =，[π, π]x ∈- 3．当0→x 时，2e1x-是sin 3x 的A ．低阶无穷小B ．高阶无穷小C ．等价无穷小D ．同阶非等价无穷小4．设函数2511sin , 0()e , 0xx x x f x x ⎧>⎪=⎨⎪<⎩，则0x =是)(x f 的 A ．可去间断点 B ．跳跃间断点 C ．连续点D ．第二类间断点5．下列方程在区间(0, 1)内至少有一个实根的为 A ．220x += B ．sin 1πx =- C ．32520x x +-=D ．21arctan 0x x ++=6．函数)(x f 在点0x x =处可导，且1)(0-='x f ，则000()(3)lim2h f x f x h h →-+=A ．23B ．23-C ．32-D ．327．曲线x x y ln =的平行于直线01=+-y x 的切线方程是 A ．1-=x y B ．)1(+-=x y C ．1y x =-+D ．)1)(1(ln -+=x x y8．设函数2π12sin 5y x =--，则='y A ．2π2cos 51x x ---B ．21x x--C ．221x x -D ．222πcos 551xx ---9．若函数()f x 满足2d ()2sin d f x x x x =-，则()f x = A ．2cos x B ．2cos x C +C ．2sin x C +D ．2cos x C -+10．d e sin(12)d d b xa x x x--=⎰ A ．e sin(12)x x -- B ．e sin(12)d x x x -- C ．e sin(12)xx C --+D ．011．若()()f x f x -=，在区间(0, )+∞内，()0f x '>，()0f x ''>，则()f x 在区间(, 0)-∞内A ．()0f x '<，()0f x ''<B ．()0f x '>，()0f x ''>C ．()0f x '>，()0f x ''<D ．()0f x '<，()0f x ''>12．若函数()f x 在区间(, )a b 内连续，在点0x 处不可导，0(, )x a b ∈，则 A ．0x 是()f x 的极大值点 B ．0x 是()f x 的极小值点 C ．0x 不是()f x 的极值点 D ．0x 可能是()f x 的极值点13．曲线e xy x -=的拐点为 A ．1x =B ．2x =C ．222,e ⎛⎫ ⎪⎝⎭D ．11, e ⎛⎫ ⎪⎝⎭14．曲线2arctan 35xy x=+ A ．仅有水平渐近线 B ．仅有垂直渐近线C ．既有水平渐近线，又有垂直渐近线D ．既无水平渐近线，又无垂直渐近线 15．若x cos 是)(x f 的一个原函数，则=⎰)(d x f A ．sin x C -+ B ．sin x C +C ．cos x C -+D ．cos x C +16．设曲线()y f x =过点(0, 1)，且在该曲线上任意一点(, )x y 处切线的斜率为e x x +，则=)(x fA ．2e 2x x - B ．2e 2x x + C ．2e xx +D ．2e xx -17．2 π4πsin d 1x xx x -=+⎰ A ．2B ．0C ．1D ．1-18．设)(x f 是连续函数，则2()d x af t t ⎰是A ．)(x f 的一个原函数B ．)(x f 的全体原函数C ．)(22x xf 的一个原函数D ．)(22x xf 的全体原函数19．下列广义积分收敛的是 A ．11d x x+∞⎰ B ．2e ln d x x x +∞⎰ C ．2e1d ln x x x+∞⎰D ．21d 1xx x+∞+⎰20．微分方程0)(224=-'+''y x y y x 的阶数是 A ．1B ．2C ．3D ．421．已知向量{5, , 2}a x =-和{, 6, 4}b y =平行，则x 和y 的值分别为 A ．4-，5B ．3-，10-C ．4-，10-D ．10-，3-22．平面1x y z ++=与平面2=-+z y x 的位置关系是 A ．重合 B ．平行C ．垂直D ．相交但不垂直23．下列方程在空间直角坐标系中表示的曲面为柱面的是 A ．221y z += B ．22z x y =+ C ．222z x y =+ D ．22z x y =-24．关于函数222222,0(,)0,0xy x y x y f x y x y ⎧+≠⎪+=⎨⎪+=⎩下列表述错误的是A ．(, )f x y 在点(0, 0)处连续B ．(0, 0)0x f =C ．(0, 0)0y f =D ．(, )f x y 在点(0, 0)处不可微 25．设函数)ln(y x y x z -=，则=∂∂yzA ．)(y x y x -B ．2ln()x x y y --C ．ln()()x y xy y x y -+- D ．2ln()()x x y xy y x y ---- 26．累次积分2222 02d (, )d x x x x x f x y y ---⎰⎰写成另一种次序的积分是A ．1d (, )d yyy f x y x -⎰⎰B ．2222 02d (, )d y y y y y f x y x ---⎰⎰C ．2211 11d (,)d y y y f x y x ----⎰⎰D ．22111 111d (, )d y y y f x y x +----⎰⎰27．设{(, )|D x y x =≤2, y ≤2}，则⎰⎰=Dy x d dA ．2B ．16C ．12D ．428．若幂级数∑∞=0n nn x a的收敛半径为R ，则幂级数∑∞=-02)2(n nnx a 的收敛区间为 A ．(, )R R - B ．(2, 2)R R -+ C ．(, )R R -D ．(2, 2)R R -+29．下列级数绝对收敛的是 A ．∑∞=-11)1(n nn B ．∑∞=-1223)1(n n nnC ．∑∞=-+-1121)1(n n n nD ．∑∞=--1212)1(n nn n30．若幂级数(3)nn n a x ∞=-∑在点1x =处发散，在点5x =处收敛，则在点0x =，2x =，4x =，6x =中使该级数发散的点的个数有A ．0个B ．1个C ．2个D ．3个31．设(32)f x -的定义域为(3, 4]-，则)(x f 的定义域为________． 32．极限lim(23)x x x x →+∞+--=________．33．设函数()(1)(2)(3)(4)f x x x x x =++--，则(4)()f x =________．34．设参数方程22 1 31x t y t =+⎧⎨=-⎩所确定的函数为()y y x =，则22d d yx =________． 35．(ln 1)d x x +=⎰________．36．点(3, 2, 1)-到平面10x y z ++-=的距离是________． 37．函数(1)x z y =+在点(1, 1)处的全微分d z =________．38．设L 为三个顶点分别为(0, 0)，(1, 0)和(0, 1)的三角形边界，L 的方向为逆时针方向，则2322()d (3)d Lxy y x x y xy y -+-=⎰________．39．已知微分方程x ay y e =+'的一个特解为x x y e =，则a =________．40．级数03!nn n ∞=∑的和为________．三、计算题（每小题5分，共45分）41．求极限2040sin d (e 1)sin lim 1cos x x x t t x x x →⎛⎫- ⎪- ⎪- ⎪⎝⎭⎰． 42．设由方程22e e y xy -=确定的函数为)(x y y =，求d d x yx=．43．求不定积分2e d e 1xxx +⎰． 44．求定积分()22 02d x x x x +-⎰．45．求过点(1, 2, 5)-且与直线213 3x y z x y -+=⎧⎨-=⎩平行的直线方程．46．求函数x xy y x y x f 823),(22+-+=的极值． 47．将23()21xf x x x =+-展开成x 的幂级数．48．计算二重积分22d Dx y σ+⎰⎰，其中D 是由圆223x y +=所围成的闭区域．49．求微分方程069=+'-''y y y 的通解．50．要做一个容积为V 的圆柱形带盖容器，问它的高与底面半径的比值是多少时用料最省？ 51．平面图形D 由曲线2x y =，直线x y -=2及x 轴所围成．求：（1）D 的面积；（2）D 绕x 轴旋转形成的旋转体的体积．五、证明题（9分）52．设函数)(x f 在闭区间]1,0[上连续，在开区间)1,0(内可导，且(0)0f =，(1)2f =．证明：在)1,0(内至少存在一点ξ，使得()21f ξξ'=+成立．。

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