2019河南专升本数学真题答案

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2019年专升本公共真题及答案-19年专升本考题

求真务实，助力学子成才！2019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公共英语注意事项：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。

本试卷的试题答案必须答在答题卡上，答在试卷上无效。

Part I Vocabulary and Structure (1x40 points)Directions: There are 40 incomplete sentences in this part. For each sentence there are fourchoices marked A, B, C and D. Choose the ONE that best completes the sentence ，andthen mark the corresponding letter on the Answer Sheet .1. Let’s walk to the shops instead of taking the car, ______?A. shall weB. don’t weC. will weD. won’t we 2. The travel agency is arranging for us ______ at a really nice hotel.A. stayB. of stayingC. to stayD. staying 3. I wouldn’t want ______ of my parents to know I have a boyfriend. A. either B. any C. neither D. none 4. I am beginning to think that ______ people say about him is true. A. which B. this C. what D. \ 5. I ______ the whole of War and Peace by the time I was seven years old. A. was reading B. had read C. had been reading D. had been read 6. Paul is a real introvert in contrast ______ his brother Andrew. A. with B. by C. to D. against 7. If I ______ lottery, I’d give some of the money to each member of my family. A. win B. have won C. will D. won 8. Medicine should not be kept ______ it is approachable to children. A. when B. where C. which D. that 9. Most people are ______ in a dentist’s office than in a doctor’s office. A. obviously nervous B. more obviously nervous C. obviously more nervous D. more nervous obviously10. If we ______ enough rain last year, we could have gained a good harvest.A. haveB. hadC. have hadD. had had 11. We ______ today and I got into trouble because I hadn’t done it. A. had checked our homework B. had our homework checked C. were checked our homework D. have checking our homework 12. Bill asked me why ______ to go shopping with me. A. I hadn’t invited him B. hadn’t invited him C. I hadn’t him invited D. hadn’t him invited 13. After finishing her course, Rachel went on ______ a well -known lawyer. A. to become B. becoming C. to becoming D. become 14. This is the book in ______ Foster described his experience of the war. A. that B. what C. where D. which 15. Never ______ a greater day in the history of mankind. A. did there be B. was there C. there was D. there has been 16. Gymnastics ______ to activities which requires skill such as speed and adaptability. A. is referring B. refers C. referred D. has referred 17. Apple ______ to have put a lot of effort into developing wireless ear -buds. A. said B. it is said C. has said D. is said 18. There are ______ Chinese restaurants in New York City as in Boston. A. twice as many B. twice as much C. as twice many D. as twice much 19. The old temple ______ roof was damaged in a storm is now under repair. A. where B. which C. its D. whose20. The engineers are so busy that they don’t have time for outdoor sports activities, ______ they have the interests. A. wherever B. whenever C. even if D. as if 21. Researchers have ______ to the conclusion that your personality is affected by your genes.A. comeB. gotC. reachedD. arrived 22. She kept practicing ______ she could win the National Speaking Competition.A.toB.in order toC. so thatD. because of 23. Carl wasn't very good at mountain climbing as he's afraid of ______.A. highsB. heightsC. highnessD. heightening 24. Harry and Sam both denied that the fight was their ______.A. blameB. criticismC. causeD. fault 25. The government should do more for ______ people.A. usualB. everydayC. ordinaryD. typical 26. Some drugs produce bad side ______.A. consequencesB. resultsC. productsD. effects 27. It is wise to ______ your property against storm damage.题号 Ⅰ Ⅱ Ⅲ Ⅳ Ⅴ Ⅵ 分值402040201020县（市，区）姓名考生号座位号A. insureB. assureC. reassureD. ensure28. Joe told me a joke today but I didn’t ______ it at all.A. getB. bringC. collectD. fetch29. I think she ______ her fortune in the shipping industry.A. tookB. wonC. madeD. saved30. I really don’t ______ the point of taking the exam when you are not ready for it.A. lookB. seeC. haveD. mind31. The debate will ______ place tonight.A. beB. takeC. haveD. make32. Ivan tells me he really ______ himself at your barbecue last week.A. pleasedB. enjoyedC. playedD. interested33. What's the______ between New York and Chicago.A. lengthB. spaceC. distanceD. gap34. They were accused of interfering in China's internal ______.A. matterB. affairsC. thingD. concern35. Please don’t forget to your essays during Friday' lessonA. handle inB. hand inC. handD. handle36.The teacher asked some questions to make _____ that the student understand the text.A.goodB. trueC. sureD. believe37.She is finding ______with the way I do my hair.A. flawB. mistakeC. faultD. error38.Joe stared at me for a moment and then reluctantly _____ his hand.A. extendedB. increasedC. raisedD. put39.The earth _____ the necessary amount of heat and light from the sun.A.SuppliesB. receivesC. providesD. requires40.Slow down! I can’t _____ with you!A.get onB. put downC. drop outD. keep upPart II Close (1x20 points)Directions:There are 20 blanks in the following passages. For each blank there are four choices marked A, B, C and D. You should choose the one that best fits into the passage andmark the corresponding letter on Answer Sheet.Today it ___41___ have a bad reputation as a cause of addiction and obesity(肥胖), but for a long time sugar was a luxury(奢侈品) and ___42___ the opinion of the medical profession it still should be.During the nineteenth century, however, manufactures discovered___43___ of producing it in vast quantities and it has since become ___44___ of the staple(主要的) articles of diet, particularly ___45___ the lower social classes. It has the advantages of ___46___ comparatively cheap, easily digested, rich ___47___ energy and useful for flavoring.It’s major ___48___are that it lacks nourishing (营养的) quality except that of giving energy, and ___49___ of its attractive flavour it ___50___ to displace other much more valuable foods ___51___ the diet. Most serious of all is its adverse ___52___ on health, since excessive consumption can cause heart ___53___, obesity and dental decay. The latter is widespread among the inhabitant(居民) of western countries. From the very young to the very old. ___54__ anyone escapes. Yet if parents drastically reduced the ___55___ confectionery(甜食) they allow ___56___ children to eat, the extent of dental decay would soon be controlled. And ___57___ they were to ___58___ down their own consumption of sugar, they would suffer much ___59___ from diseases resulting directly or indirectly from their ___60___ overweight.41. A. must B. may C. should D. could42. A. to B. on C. for D. in43. A. problems B. difficulties C. ways D. origins44. A. same B. one C. simple D. well-known45. A. between B. with C. to D. among46. A. be B. having to be C. being D. have to be47. A. in B. with C. to D. for48. A. disadvantages B. qualities C. characteristics D. uses49. A. thanks B. because C. due D. opposite50. A. tends B. owns C. has D. is51. A. out of B. from C.within D. of52. A. strength B. outcome C. result D. effect53. A. broken B. disease C. break D. stop54. A. no way B. hardly C. if D. most unlikely55. A. amount B. variety C. kind D. quality56. A. / B. their C. the D. for57. A. then B. as C. if D. however58. A. shut B. slow C. cut D. turn59. A. least B. less C. / D. more60. A. having B. gaining C. having been D. beingPart III Reading Comprehension (2x20 points)Directions:There are 4 passages in this part. Each passage is followed by some questions or incomplete statements. For each of them there are four choices marked A, B, C and D. You should decide onthe best choice and mark the corresponding letter on Answer Sheet.Passage 1More and more around the world are taking part in dangerous sports and activities. Of course, there have always been people who have looked for adventure--those who have climbed the highest mountains, searched into unknown parts of the world or sailed in small boats across the greatest oceans.I would consider bungee jumping to be a good example of such an activity. You jump froma high place 200 meters above the ground with a rope tied to your feet. You fall at up to 150 kilometers an hour until the rope stops you from hitting the ground.Why do people take part in such activities as these? Some people suggest that it is because life in modern societies has become safe and uninteresting. But in the past, they had to go out and hunt for food, and diseases could not easily be cured.Nowadays, life offers little excitement. They live and work in safe conditions, they buy food in shops and there are doctors and hospitals to look after them if they become ill. The answer for some people is to look for danger in activities such as bungee jumping.61. According to the passage more and more people around the world today_____.A. are taking part in games that are very safeB. are aware of the risk of some activitiesC. are trying dangerous activities for excitementD. are looking for ways of showing their courage62. Bungee jumping is an activity that you_____.A. jump down to the ground without holding to a ropeB. fasten yourself to a rope and fall from a great heightC. jump up as high as you canD. move down a rope to the ground63. Many people take part in sports with the purpose of _____.A. getting healthierB. losing weightC. entertaining themselvesD. seeking adventure64. Nowadays people think that life is _____.A. dullB. interestingC. unsafeD. happy65. What can be the title of the passages?A. The Need for ExcitementB. Interesting Bungee Jumping.C. Dangerous Sports and ActivitiesD. Modern Life and ActivitiesPassage 2The year was 1932. Amelia Earhart was flying alone from North America to England in a small single-engined plane. At midnight, several hours after she had left Newfoundland, she ran into bad weather. To make things worse, her altimeter failed and she didn't know how high she was flying. At night, and in a storm, a pilot is in great difficulty without an altimeter. At times, her plane nearly plunged into the sea.Just before dawn, there was further trouble, Amelia noticed flames coming from the engine. Would she be able to reach land? There was nothing to do except to keep going and to hope.In the end, Amelia Earhart did reach Ireland, and for the courage she had shown, she was warmly welcomed in England and Europe. When she returned to the United States, she was honored by President Hoover at a special dinner in the White House. From that time on. Amelia Earhart was famous. What was so important about her flight? Amelia Earhart was the first woman to fly the Atlantic Ocean alone, and she had set a record of fourteen hours and fifty-six minutes.In the years that followed, Amelia Earhart made several flights across the United States, and on each occasion she set a new record for flying time. Amelia Earhart made these fights to show that women had a place in aviation and that air travel was useful.66. Which of the following statements is NOT the difficulty which Amelia Earhart met in her fight from NorthAmerica to England?A. She was caught in a storm.B. The altimeter went out of order.C. Her engine went wrong.D. She lost her direction.67. When Amelia Earhart saw flames coming from the engine, what did she do?A. She did nothing but pray for herself.B. She changed her direction and landed in Ireland.C. She continued flying.D. She lost hope of reaching land.68. According to the passage, what was Amelia Earhart's reason for making her flights?A. To set a new record for flying time.B. To be the first woman to fly around the world.C. To show that aviation was not just for men.D. To become famous in the world.69. Which of the following statements was NOT mentioned?A. She was the first woman who succeeded in flying across the Atlantic Ocean aloneB. She showed great courage in overcoming the difficulties during the flight.C. She was warmly welcomed in England, Europe and the United States.D. She made plans to fly around the world.70. Which of the following would be the best title for the passage?A. Amelia Earhart---First Across the Atlantic.B. Amelia Earhart---Pioneer in Women’s Aviation.C. A New Record for Flying Time.D. A Dangerous Flight from North America to England.Passage 3The Winter Olympics are also called the White Olympics. At this time, many colorful stamps are published to mark the great Games. The first stamps marking the opening came out on January 25, 1932 in the United States for the 3rd White Olympics. From then on, publishing stamps during the White Olympics became a rule.During the 4th Winter Games a group of stamps were published in Germany in November 1936. The five rings of Olympics were sewn（缝制）on the front of the sportswear. It was the first time that the rings appeared on the stamps of the Whited Olympics.In the 1950s, the stamps of this kind became more colorful. When the White Olympics came, the host countries as well as the non-host countries published stamps to mark those Games. China also published four stamps in February 1980, when the Chinese sportsmen began to take part in the White Olympics.Different kinds of sports were drawn on these small stamps. People can enjoy the beauty of the wonderful movements of some sportsmen.71.The White Olympics and the Winter Olympics are ______.A. the same thingB. different gamesC. not held in winterD. held in summer72. The world made it a rule to publish stamps to mark the Olympics Games ______.A. After the year 1936B. after the 3rd White OlympicsC. Before the 3rd White OlympicsD. before the year 193273. The Winter Olympics are held once every ______ years.A. twoB. threeC. fourD. five74. Which of the following is TRUE?A. Only the host countries can publish stamps to mark those Games.B. Only the non-host countries can publish stamps to mark those Games.C. All the countries can publish stamps to mark those Games.D. Chinese has never published stamps to mark those Games.75. What most probably appear on the stamps of the White Olympic?A. BasketballB. Table tennisC. FootballD. SkatingPassage 4Be aware of those who use the truth to cheat. When someone tells you something that is true, but leaves out important information that should be included, he can create a false impression. For example, someone might say, “I just won a hundred dollars on the lottery. It was great. I took that dollar ticket back to the store and changed it for one hundred dollars!”This guy’s a winner,right Maybe, maybe not. We then discover that he bought twohundred tickets, and only one was a winner. He’s really a big loser!He didn’t say anything that was false, but he deliberately left out some important information. That’s called a half-truth. Half-truths are not technically lies, but they are just as not honest.Untrustworthy candidates in political campaigns often use this strategy. Let’s say that during Governor Smith’s last term, her state lost one million jobs and gained three million jobs. Then she seeks another term. One of her opponents runs an advertisement saying, “During Governor Smith’s term, the state lost one million jobs!” That’s true.However, an honest statement would have been, “During Governor Smith’s term, the state had a net gain of two million jobs.”Advertisers will sometimes use half-truths. It’s against the law to make false claims so they try to mislead you with the truth. An advertisement might boast, “Nine out of ten doctors recommend Yucky Pills to cure nose pimples.” It fails to mention that they only asked ten doctors and nine of them work for the Yucky Corporation.This kind of cheat happens too often. It’s a sad fact of life: Lies are lies, and sometimes the truth can lie as well.76.In which of the following situations is a person telling a half-truth?A. When somebody is telling a white lie.B. When somebody is making up information.C. When somebody is saying something that is wrong.D. When somebody is intentionally leaving out important information.77.It can be inferred from the passage that ______.A.advertisers always make false claimsB.advertisements can be misleading due to partial informationC.Yucky pills are effective in curing nose pimplesD.the Yucky Corporation has a bad reputation for its drug78.The underlined word “deception”（Para. 7）is closest in meaning to “______”.A. dishonestyB. suggestionC. situationD. failure79.The author clearly wants people to ______.A.think carefully about what they read and hearB.be firm supporters of a particular candidate in political campaignsC.never place their trust in lottery winnersD.vote for female candidates in political campaigns80.An appropriate title for this passage would be ______.A.It Pays to Be HonestB. Everyone LiesB.Lying with the Truth D. An Important Tactic in AdvertisingPart IV Translation (2x10 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 Answer81.82.83.84.85.86.87.88.89.90.91. 92. 93. 94. 95. 96. 97. 98. 99.100. The fire quickly spread to the first floor where it destroyed all the equipments in the language lab. D Part -time Jobs . Y ou should2019河南专升本公共英语参考答案Part Ⅰ V ocabulary and Structure (1x40)1-10: ACACB CDBCD11-20: BAADB BDADC 21-30: ACBDC DAACB31-40: BBCBB CCCBD Part Ⅱ Cloze (1x20)41-50: BDCBD CAABA51-60: BDBBA BCCBD Part Ⅲ Reading Comprehension (2x10)61-70: CBDAC DCCDD71-80: ABCCD DBAAC Part Ⅳ Translation (2x10)81. This book is the most interesting book I have ever read.82. He took his students to factory for a visit.83. Given enough time, I can do it well too.84. Thank you for giving us so much help.85. It was in the street that I met him yesterday.86. 众所周知，光速比声速快。

河南省2019年对口升学数学真题答案及解析

第 1 页共 8 页河南省2019年普通高等学校对口招收中等职业学校毕业生考试数学试题及其解析一、选择题（每小题3分，共30分。

每小题中只有一个选项是正确的，请将正确选项写在括号内上）00.000.000.000..0,0,0.12222222222≠≠≠+>+≠≠≠≠≠+≠+≠≠===+b a b a D b a b a C b a b a B b a b a A b a b a 且，则如果，则或如果或，则如果，则或如果）题的逆否命题（下列哪个命题是前述命则已知【考点】：命题【解析】选择A.命题：已知022=+b a 则0=a ，0=b 逆命题：如果0=a ，0=b ，则022=+b a 否命题：如果022≠+b a ，则 0≠a 或0≠b 逆否命题：如果0≠a 或0≠b ，则022≠+b a注意：0=a ，0=b 即0=a 且0=b ，它的否定形式为0=a 或0=b . 可参考德摩根率：在命题逻辑中存在着下面这些关系：非（p 且q ）=（非p ）或（非q ）非（p 或q ）=（非p ）且（非q ） q p q p q p q p ∧=∨∨=∧,2222..11..,0,,.2bab a D ba ab C b a B bc ac A b a R c b a >><<><<∈）则下列式子正确的是（，且已知【考点】：不等式性质【解析】选择D.（1）用特值法：设2-=a ，1-=b ，.0=c显然A 选项不成立，1112121-=->-=-，B 选项不成立，2122121=--<=--，C 选项不成立，2)1()2(,4)2(2=-⨯-=-，124,1)1(2>>=-，故选择D.（2）因为a ，b ，R c ∈，且0<<b a ，所以0<-b a ，因为0)(2>-=-b a a ab a ，所以ab a >2，因为0)(2>-=>b a b b ab ，所以2b ab >，根据不等式性质的传递性得.22b ab a >>故选择D. ()[]()）的定义域为（12，则函数4,2的定义域为1已知函数.3+-+x f x f第 2 页共 8 页[][][]2,1.9,3.3,3.23,23.---⎥⎦⎤⎢⎣⎡-D C B A【考点】：函数的定义域【解析】选择D.函数)1(+x f 的定义域为[]4,2-，则其中的42≤≤-x ，所以511≤+≤-x ，函数)(x f y =的定义域为[].5,1-由5121≤+≤-x 得21,422≤≤-≤≤-x x ，则函数)12(+x f 的定义域为[].2,1-）同一函数的是（下列各组函数中，表示.4①()()x x x g x x f 223-=-=和②()()2x x g x x f ==和③()()42x x g x x f ==和④()()121222+-=+-=t t t g x x x f 和.A ①② .B ①③ .C ③④ .D ①④【考点】：同一函数【解析】选择C.定义域与对应法则都相同的函数是同一个函数①两个函数32)(x x f -=和x x x g 2)(-=的定义域均为).0,(-∞因为x x =2，所以.22223x x xx x -=-=-x x x f 2)(-=，x x ≠，因此).()(x g x f ≠②两个函数x x f =)(和2)(x x g =的定义域均为).,(+∞-∞因为 x x =2，x x ≠，因此).()(x g x f ≠③两个函数2)(x x f =和4)(x x g =的定义域均为).,(+∞-∞因为()22224x x x x ===，所以).()(x g x f =④两个函数12)(2+-=x x x f 和12)(2+-=t t t g 的定义域均为).,(+∞-∞有相同的对应法则，与表示函数所选用的字母无关，因此，正确的选项为C.{}{}3.2.1.21.）的值为（的公差，数列123，若项和为的前已知等差数列.523D C B A da S S S n a n n n -=-【考点】：等差数列的前n 项和公式【解析】选择C.d a S d a S 33,21212+=+=，由13223=-S S ,得12233311=+-+da d a ，化简得2,12,1211===⎪⎭⎫ ⎝⎛+-+d d d a d a ，故选C.第 3 页共 8 页()()()3.3.4.4.43,31,12.6D C B A AC AB C B A --=•-）（，则，，，已知【考点】：1.有向线段的坐标表示。

2019年河南成人高考高起点数学(理)真题及答案

1 x2 2019年河南成人高考高起点数学(理)真题及答案本试卷分第 I 卷(选择题)和第Ⅱ卷(非选择题)两部分。

满分 150 分。

考试时间 120 分钟。

第Ⅰ卷(选择题，共 85 分)一、选择题(本大题共 17 小题，每小题 5 分，共 85 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.设全集 U={1,2,3,4}集合 M={3,4}，则C U M =【】A.{2,3}B.{2，4}C.{1，2}D.{1，4} 2. 函数 y=cos4x 的最小正周期为【】A. B. C. D. 2 2 4 3.设甲：b=0；乙：函数 y=kx+b 的图像经过坐标原点，则【】A. 甲是乙的充分条件但不是必要条件B. 甲是乙的充要条件C. 甲是乙的必要条件但不是充分条件D. 甲既不是乙的充分条件也不是乙的必要条件4. 已知tan1.则tan(【】2 A.-3 B.1 34C.3D. 1 35. 函数 y 的定义域是【】A. x x 1B. x x 1C. x 1 x 1D. x x 16. 设 0<x<1，则【】A. log 2 x0 B. 0 2x 1C. log 1 x0 2D.1 2x 27. 不等式 x 1 1 的解集为【】2 2 A. x x 0或x 1C. x x 1B. x 1 x 0D. x x 0)3 y8. 甲、乙、丙、丁 4 人排成一行，其中甲、乙必须排在两端，则不同的排法共有【】 A.4 种 B.2 种 C.8 种 D.24 种9.若向量 a =(1，1)，b =(1，一 1)，则 1 a 3b 【】2 2A.(1.2)B.(-1.2)C.(1，-2)D.(-1，-2)110. log 1162 (2)0 【】A.2B.4C.3D.511. 函数 y x 2 4x 5 的图像与 x 轴交于 A ，B 两点，则|AB|=A.3B.4C.6D.512.下列函数中，为奇函数的是【】A. y 2x13.双曲线 x 9 B.y=-2x+3 C. y x 232- 1的焦点坐标是【】16 D.y=3cosxA.(0，- )，(0， )B.(- ，0)，( ，0)C.(0，-5)，(0，5)D.(-5，0)，(5，0)14.若直线mx y 1 0 与直线4x 2 y 1 0 平行，则 m=【】A.-1B .0C.2D.115.在等比数列a n 中，若a 4a 5 6, 则a 2a 3a 6a 7 【】A.12B.36C.24D.7216.已知函数 f x 的定义域为 R ，且 f (2x ) 4x 1, 则 f (1) 【】A.9B.5C.7D.3 17. 甲、乙各自独立地射击一次，已知甲射中 10 环的概率为 0.9，乙射中 10 环的概率为 0.5，则甲、乙都射中 10 环的概率为【】 A.0.2 B.0.45 C.0.25 D.0.75第Ⅱ卷(非选择题，共 65 分) 二、填空题(本大题共 4 小题，每小题 4 分，共 16 分)18.椭圆 x 4 + y 21的离心率为。

河南省2019年专升本考试《高等数学》试题+答案

河南省2019年普通高等学校专科毕业生进入本科阶段学习考试《高等数学》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。

本卷的试题答案必须答在答题卡上，答在卷上无效。

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。

1.函数)1ln()(2x x x f -+=在定义域是（）A.不确定B.偶函数C.非奇非偶函数D.奇函数[解析]D由()()f x f x -=-得，为奇函数2.已知()f x 的定义域为[]1e ，，则()xf e 的定义域为（）A.(]1,0B.[]0,1 C.()1,0 D.[)10，[解析]B由101xe e x ≤≤⇒≤≤；3.曲线32116132y x x x =+++在点(0,1)处的切线与x 轴的交点坐标为（）A.1(,0)6-B.()10，C.⎪⎭⎫ ⎝⎛0,61 D.()0,1-[解析]A由200=6661x x k y x x y x =='=++=⇒=+，与x 轴的交点即当0y =时得交点坐标为1(,0)6-；4.当0x →1与212x -等价，则=a （）A.32-B.32-C.21-D.32[解析]A由当0x →2113ax -→，所以22113=322ax x a -⇒=-；5.极限22324lim 354x n n n n →∞+-=-+（）A.1B.43 C.52-D.34-[解析]D由抓大头口诀：相同即为系数比，可得223244lim 3543x n n n n →∞+-=--+；6.极限0sin 4lim =5x xx→（）A.45B.51C.54 D.1[解析]C00sin 444lim=lim 555x x x x x x →→=；7.当0x →时，221x e -是2x 的_______无穷小（）A.高阶B.低阶C.等价D.同阶非等价[解析]D 当0x →时，22212x ex -→，故是2x 的同阶非等价；8.已知函数()ln 21a xf x ax +⎧=⎨-⎩在1x =处连续，则a =（）A.1B.1- C.0D.3题号一二三四五总分分值602050146150班级：姓名：准考证号：[解析]A9.设1,1()=cos ,12x x f x x x π-≥<⎧⎪⎨⎪⎩则1x =是____间断点（）A.连续点B.可去C.跳跃D.第二类[解析]A 10.函数()f x 在x a =处可导，则()()limf a x f a x xx +--→（）A.()2f a 'B.0C.()a f ' D.()a f '21[解析]A11.已知()12x f x x=+，求1(1)f -=（）A.1- B.1C.13-D.13[解析]反解12y x y=-，交换,x y 得反函数12x y x =-，则1(1)1f -=-。

河南省2019年对口升学数学真题答案及解析

)
. 3
【考点】
：等差数列的前 n 项和公式
【解析】选择 C.

2
2
3
2 = 21 + , 3 = 31 + 3，由 3 −

化简得1 + − (1 + ) = 1,
2

2
31 +3
= 1,得
= 1,
3
−
21 +
= 2，
2
= 1，
故选 C.
⃗⃗⃗⃗⃗ •
⃗⃗⃗⃗⃗ = ( )
因为2 − = ( − ) > 0，所以2 > ，
因为 > 2 = ( − ) > 0，所以 > 2 ，
根据不等式性质的传递性得2 > > 2 .故选择 D.
3.已知函数( + 1)的定义域为[-2,4]，则函数(2 + 1)的定义域为( )
3 3
B.如果2 +2 ≠0，则 ≠ 0或 ≠ 0
C.如果 ≠ 0或 ≠ 0，则2 +2 >0
D.如果2 +2 ≠0，则 ≠ 0且 ≠ 0
【考点】
：命题
【解析】选择 A.
命题：已知2 + 2 = 0则 = 0， = 0
逆命题：如果 = 0， = 0，则2 + 2 = 0
显然 A 选项不成立，
1
−2
1
1
2
−1
=− >
= −1，
−1
B 选项不成立，
−2
1
−2
2
−1
= <
= 2，
C 选项不成立，(−2)2 = 4, (−2) × (−1) = 2，(−1)2 = 1, 4 > 2 > 1，故选择 D.

河南省专升本高等数学真题(带答案详细讲解)

2009年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注意事项：答题前，考生务必将自己的、座位号、考生号涂写在答题卡上。

本试卷的试卷答案在答题卡上，答试卷上无效。

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

如需改动，用橡皮擦干净后，再涂其他答案标号.1.下列函数相等的是（）A.2x y x=，y x = B. y =y x =C.x y =，2y = D. y x =，y =【答案】D.解：注意函数的定义围、解读式，应选D.2.下列函数中为奇函数的是（）A.e e ()2x xf x -+= B. ()tan f x x x =C. ()ln(f x x =D. ()1x f x x=- 【答案】C.解:()ln(f x x -=-，()()ln(ln(ln10f x f x x x +-=-+==()()f x f x -=-，选C.3．极限11lim1x x x →--的值是( ) A.1B.1-C.0 D.不存在【答案】D. 解：11lim 11x x x +→-=-，11lim 11x x x -→-=--，应选D.4.当0x →时，下列无穷小量中与x 等价是（）A.22x x - C. ln(1)x + D.2sin x【答案】C.解:由等价无穷小量公式，应选C.5.设e 1()x f x x-=，则0=x 是()f x 的（）A.连续点B.可去间断点C.跳跃间断点D.无穷间断点【答案】B.解:00e 1lim ()lim1x x x f x x→→-==⇒0=x 是)(x f 的可去间断点,应选B. 6. 已知函数()f x 可导，且0(1)(1)lim12x f f x x→--=-，则(1)f '= （）A. 2B. -1C.1D.-2 【答案】D. 解：0(1)(1)1lim(1)1(1)222x f f x f f x →--''==-⇒=-，应选D.7.设()f x 具有四阶导数且()f x ''=(4)()f x = （）AB ．1 D ．3214x --【答案】D. 解：1(3)21()2fx x -=，(4)()f x =3214x --，应选D.8.曲线sin 2cos y t x t=⎧⎨=⎩在π4t =对应点处的法线方程（）A.2x =B.1y =C.1y x =+D.1y x =- 【答案】A.解：0d 2cos 20d sin y t k x x x t =⇒=⇒==切，应选A. 9.已知d e ()e d x xf x x -⎡⎤=⎣⎦，且(0)0f =，则()f x =（）A ．2e e x x + B. 2e e x x - C. 2e e x x -+ D. 2e e x x -- 【答案】B.解:由d e ()e d x x f x x -⎡⎤=⎣⎦得2d e ()d(e )e ()e ()e e x x x x x xf x f x C f x C --⎡⎤=⇒=+⇒=+⎣⎦，把(0)0f =代入得1C =-,所以2()e e x x f x =-,应选B. 10.函数在某点处连续是其在该点处可导的（）A. 必要条件B. 充分条件C. 充分必要条件D. 无关条件【答案】A.解:根据可导与连续的关系知，应选A.11.曲线42246y x x x =-+的凸区间为（） A.(2,2)- B.(,0)-∞ C.(0,)+∞ D. (,)-∞+∞ 【答案】A.解:34486y x x '=-+，212480(2,2)y x x ''=-<⇒∈-，应选A.12.设e xy x=（）A.仅有水平渐近线B.既有水平又有垂直渐近线C.仅有垂直渐近线D.既无水平又无垂直渐近线【答案】B.解:e lim0x x x →-∞=，0e lim xx x→=∞，应选B. 13.下列说确的是（） A. 函数的极值点一定是函数的驻点 B. 函数的驻点一定是函数的极值点C. 二阶导数非零的驻点一定是极值点D. 以上说法都不对【答案】D.解:根据极值点与驻点的关系和第二充分条件，应选D.14. 设函数()f x 在[,]a b 连续，且不是常数函数,若()()f a f b =，则在(,)a b （）A. 必有最大值或最小值B.既有最大值又有最小值C.既有极大值又有极小值D.至少存在一点ξ，使()0f ξ'= 【答案】A.解:根据连续函数在闭区间上的性质及()()f a f b =的条件,在对应的开区间至少有一个最值,应选A.15.若()f x 的一个原函数为ln x ，则()f x '=（）A.1x B.21x- C.ln x D.ln x x 【答案】B.解:()1()ln f x x x '==⇒21()f x x'=-,应选B.16.若2()f x dx x C =+⎰，则2(1)xf x dx -=⎰（） A. 222(1)x C --+ B. 222(1)x C -+C. 221(1)2x C --+D. 221(1)2x C -+【答案】C. 解:2221(1)(1)(1)2xf x dx f x d x -=---⎰⎰=221(1)2x C --+,应选C. 17.下列不等式不成立的是（）A. 22211ln (ln )xdx x dx >⎰⎰ B.220sin xdx xdx ππ<⎰⎰C.22ln(1)x dx xdx +<⎰⎰ D.22(1)x e dx x dx <+⎰⎰【答案】D.解:根据定积分的保序性定理，应有22(1)x e dx x dx ≥+⎰⎰，应选D.18.1ln eex dx ⎰= （）A.111ln ln e exdx xdx +⎰⎰ B.111ln ln eexdx xdx -⎰⎰C. 111ln ln e exdx xdx -+⎰⎰ D.111ln ln eexdx xdx --⎰⎰【答案】C.解:因1ln ,1|ln |ln ,1x x x ex x e⎧-≤≤⎪=⎨⎪≤≤⎩,考察积分的可加性有 1111ln ln ln eeeexdx xdx xdx =-+⎰⎰⎰，应选C.19．下列广义积分收敛的是（）A.lnex dx x +∞⎰B.1ln e dx x x+∞⎰ C.21(ln )e dx x x +∞⎰ D.e +∞⎰ 【答案】C.解:由广义积分性质和结论可知:21(ln )edx x x +∞⎰是2p =的积分,收敛的,应选C.20.方程220x y z +-=在空间直角坐标系中表示的曲面是（） A.球面 B.圆锥面C. 旋转抛物面D.圆柱面【答案】C.解:根据方程的特点是抛物面,又因两个平方项的系数相等,从而方程220x y z +-=在空间直角坐标系中表示的曲面是旋转抛物面,应选C.21. 设{}1,1,2a =-r ，{}2,0,1b =r，则a r 与b r 的夹角为（）A ．0B ．6πC ．4πD ．2π 【答案】D.解:0(,)2a b a b a b π=⇒⊥⇒=r r r r r r g ,应选D.22.直线34273x y z++==--与平面4223x y z --=的位置关系是 ( ) A.平行但直线不在平面 B.直线在平面 C. 垂直 D.相交但不垂直【答案】A.解:因{}2,7,3s =--r ,{}4,2,20n s n s n =--⇒⋅=⇒⊥⇒r r r r直线在平面或平行但直线不在平面.又直线上点(3,4,0)--不在平面.故直线与平面的位置关系是平行但直线不在平面,应选A.23.设(,)f x y 在点(,)a b 处有偏导数，则0(,)(,)limh f a h b f a h b h→+--=（ )A.0B.2(,)x f a b 'C.(,)x f a b 'D.(,)y f a b ' 【答案】B. 解:原式00(,)(,)(,)(,)limlimh h f a h b f a b f a h b f a b h h→→+---=- 00(,)(,)(,)(,)limlim 2(,)x h h f a h b f a b f a h b f a b f a b h h→-→+---'=+=- 应选B. 24．函数x yz x y+=-的全微dz =（） A ．22()()xdx ydy x y -- B ．22()()ydy xdx x y -- C ．22()()ydx xdy x y --D ．22()()xdy ydx x y --【答案】D 解：22()()()()2()()()x y x y d x y x y d x y xdy ydx z dz x y x y x y +-+-+--=⇒==---，应选D25．0(,)ady f x y dx ⎰化为极坐标形式为（）A ．20(cos ,sin )ad f r r rdr πθθθ⎰⎰B ．2cos 0(cos ,sin )d f r r rdr πθθθθ⎰⎰C ．sin 20(cos ,sin )a d f r r rdr πθθθθ⎰⎰D ．20(cos ,sin )ad f r r rdr πθθθ⎰⎰【答案】D.解:积分区域{(,)|0,0(,)|0,02x y y a x r r a πθθ⎧⎫≤≤≤≤=≤≤≤≤⎨⎬⎩⎭有(,)ady f x y dx ⎰2(cos ,sin )ad f r r rdr πθθθ=⎰⎰,应选D.26.设L 是以A(-1,0),B(-3,2),C(3,0)为顶点的三角形区域的边界，方向为ABCA,则(3)(2)Lx y dx x y dy -+-=⎰ÑA.-8B.0 C 8 D.20【答案】A.解:由格林公式知,(3)(2)228LDx y dx x y dy d S σ∆-+-=-=-=-⎰⎰⎰Ñ，应选A.27.下列微分方程中，可分离变量的是（） A ．tan dy y ydx x x=+ B ．22()20x y dx xydy +-= C ．220x y x dx e dy y ++=D ． 2x dy y e dx+= 【答案】C.解:根据可分离变量微分的特点,220x y xdx e dy y++=可化为 22y x ye dy xe dx -=-知,应选C.28.若级数1n n u ∞=∑收敛，则下列级数收敛的是（）A ．110nn u ∞=∑ B ．1(10)n n u ∞=+∑C ．110n nu ∞=∑D ．1(10)nn u∞=-∑【答案】A.解:由级数收敛的性质知,110nn u ∞=∑收敛,其他三个一定发散,应选A. 29.函数()ln(1)f x x =-的幂级数展开为（）A ．23,1123x x x x +++-<≤LB ．23,1123x x x x -+--<≤LC ．23,1123x x x x -----≤<LD ． 23,1123x x x x -+-+-≤<L【答案】C.解:根据23ln(1),1123x x x x x +=-+--<≤L 可知,23ln(1),1123x x x x x -=-----≤<L ,应选C.30.级数1(1)n n n a x ∞=-∑在1x =-处收敛，则此级数在2x =处（）A ．条件收敛B ．绝对收敛C ．发散D ．无法确定【答案】B.解:令1x t -=,级数1(1)nn n a x ∞=-∑化为1n n n a t ∞=∑,问题转化为:2t =-处收敛,确定1t =处是否收敛.由阿贝尔定理知是绝对收敛的,故应选B.二、填空题（每小题2分，共30分）31.已知()1xf x x=-，则[()]______f f x =. 解：()1[()](1,)1()122f x x f f x x x f x x ==≠≠--.32.当0x →时，()f x 与1cos x -等价，则0()lim_______sin x f x x x→=. 解：2211cos ()1cos 2220sin 00()1cos 12lim lim lim sin 2x x f x x x x x x x x f x x x x x x --→→→-==============:::. 33.若2lim 8xx x a x a →∞+⎛⎫= ⎪-⎝⎭，则_______a =. 解：因2223()221lim 12lim lim 1lim 1x xa axa x a x x a x x a a x a a x a e x x e x a e a a x x ⋅→∞-→∞→∞⋅--→∞⎛⎫⎛⎫++ ⎪ ⎪+⎛⎫⎝⎭⎝⎭==== ⎪-⎝⎭⎛⎫⎛⎫- ⎪- ⎪⎝⎭⎝⎭，所以有 38a e =ln 2a ⇒=.34.设函数sin ,0(),0xx f x x a x ⎧≠⎪=⎨⎪=⎩在(,)-∞+∞处处连续，则_______a =.解：函数在(,)-∞+∞处处连续，当然在0x =处一定连续，又因为0sin lim ()lim1;(0)x x x f x f a x→→===，所以0lim ()(0)1x f x f a →=⇒=.35.曲线31xy x=+在（2，2）点处的切线方程为___________. 解：因2231340(1)3x y k y x y x =''=⇒==⇒-+=+. 36.函数2()2f x x x =--在区间[0，2]上使用拉格朗日中值定理结论中____ξ=. 解：(2)(0)()2121120f f f x x ξξ-'=-⇒-=⇒=-.37.函数()f x x =-的单调减少区间是 _________.解：1()100,4f x x ⎛⎫'=<⇒∈ ⎪⎝⎭,应填10,4⎛⎫ ⎪⎝⎭或10,4⎡⎤⎢⎥⎣⎦或10,4⎡⎫⎪⎢⎣⎭或10,4⎛⎤⎥⎝⎦. 38.已知(0)2,(2)3,(2)4,f f f '===则20()______xf x dx ''=⎰. 解：222200()()()()2(2)(2)(0)7xf x dx xdf x xf x f x dx f f f ''''''==-=-+=⎰⎰⎰.39.设向量b r 与}{1,2,3a =-r共线，且56a b ⋅=r r ,则b =r _________. 解：因向量b r 与a r共线,b r 可设为{},2,3k k k -，5649564a b k k k k ⋅=⇒++=⇒=rr ,所以{}4,8,12b =-r . 40.设22x y z e +=，则22zx∂=∂_______.解：22222222222(12)x y x y x y z z z exe x e x x+++∂∂=⇒=⇒=+∂∂. 41．函数22(,)22f x y x xy y =+-的驻点为________.解：40(,)(0,0)40fx y xx y f x y y∂⎧=+=⎪∂⎪⇒=⎨∂⎪=-=∂⎪⎩.42．区域D 为229x y +≤，则2______Dx yd σ=⎰⎰.解：利用对称性知其值为0或232420cos sin 0Dx yd d r dr πσθθθ==⎰⎰⎰⎰.43.交换积分次序后,10(,)_____________xdx f x y dy =⎰.解：积分区域{{}2(,)|01,(,)|01,D x y x x y x y y y x y =≤≤≤≤=≤≤≤≤,则有21100(,)(,)yxydx f x y dy dy f x y dx =⎰⎰⎰.44.14x y xe -=-是23x y y y e -'''--=的特解，则该方程的通解为_________.解：230y y y '''--=的通解为312x x y C e C e -=+，根据方程解的结构,原方程的通解为31214x x x y C e C e xe --=+-.45.已知级数1n n u ∞=∑的部分和3n S n =，则当2n ≥时，_______n u =.解：当2n ≥时,3321(1)331n n n u S S n n n n -=-=--=-+.三、计算题（每小题5分，共40分）46．求011lim 1x x x e →⎛⎫- ⎪-⎝⎭.解：20001111lim lim lim 1(1)x x x x x x x e x e x x e x e x →→→----⎛⎫-== ⎪--⎝⎭0011lim lim 222x x x e x x x →→-===. 47.设()y y x =是由方程ln sin 2xy e y x x +=确定的隐函数，求dxdy . 解：方程两边对x 求导得()ln 2cos 2xy ye xy y x x x''++= 即()ln 2cos 2xy e x y xy y y x x x x ''+++=2(ln )2cos 2xy xy x e x x y x x e xy y '+=--所以dydx=22cos 2ln xy xy x x e xy y y x e x x --'=+.48.已知2()x xf x dx e C -=+⎰，求1()dx f x ⎰. 解：方程2()x xf x dx e C -=+⎰两边对x 求导得2()2xxf x e-=-，即22()xe f x x--=，所以211()2x xe f x =-. 故22111()24x x dx xe dx xde f x =-=-⎰⎰⎰ 222211114448x x x x xe e dx xe e C =-+=-++⎰.49.求定积分44|(1)|x x dx --⎰.解：4014441|(1)||(1)||(1)||(1)|x x dx x x dx x x dx x x dx ---=-+-+-⎰⎰⎰⎰01441(1)(1)(1)x x dx x x dx x x dx -=-+-+-⎰⎰⎰14322332401322332x x x x x x -⎛⎫⎛⎫⎛⎫=-+-+- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭ 641164118843323332=++-+--+=. 50.已知22x xy y z e +-= 求全微分dz .解：因222222()(2)x xy y x xy y x ze x xy y e x y x+-+-∂'=+-=+∂, 222222()(2)x xy y x xy y y ze x xy y e x y y+-+-∂'=+-=-∂, 且它们在定义域都连续，从而函数22xxy y z e +-=可微，并有z zdz dx dy x y∂∂=+∂∂22[(2)(2)]x xy y e x y dx x y dy +-=++-. 51.求(2)Dx y d σ+⎰⎰，其中区域D 由直线,2,2y x y x y ===围成.x x y =→=2yx =2解：积分区域D 如图所示：把D 看作Y 型区域，且有(,)|02,2y D x y y x y ⎧⎫=≤≤≤≤⎨⎬⎩⎭故有22(2)(2)yy Dx y d dy x y dx σ+=+⎰⎰⎰⎰2222025()4yy x xy dy y dy =+=⎰⎰230510123y ==. 52.求微分方程22x y xy xe -'-=的通解. 解：这是一阶线性非齐次微分方程，它对应的齐次微分方程20y xy '-=的通解为2x y Ce =，设原方程的解为2()x y C x e =代入方程得22()x x C x e xe -'=, 即有22()x C x xe -'=，所以222222211()(2)44x x x C x xe dx e d x e C ---==--=-+⎰⎰, 故原方程的通解为2214x x y e Ce -=-+.53.求幂级数212nn n n x ∞=∑的收敛区间（考虑区间端点）. 解：这是规缺项的幂级数，考察正项级数212nn n n x ∞=∑，因221112limlim 22n n n n n nu n x l x u n ++→∞→∞+==⨯=, 当212x l =<，即||x <212n n n nx ∞=∑是绝对收敛的；当212x l =>，即||x >212n n n nx ∞=∑是发散的；当212x l ==，即x =212nn n n x ∞=∑化为1n n ∞=∑，显然是发散的。

2019年专升本高等数学考试题及答案1.doc

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2019年河南省普通专升本高等数学真题及答案

2019年河南省普通高等学校选拔专科优秀毕业生进入本科学校学习考试高等数学试卷一、单项选择题(每小题2分，共60分)在每个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.如需更改，用橡皮擦干净后，再选涂其他答案标号.1．函数)1ln()(2x x x f -+=在定义域内是()A ．不确定奇偶性B ．偶函数C ．非奇非偶函数D ．奇函数【答案】D【解析】因为函数()x f 定义域为R ，且()()x f x x xx x x x x x f -=⎪⎭⎫ ⎝⎛-+-=-+-+=⎪⎭⎫ ⎝⎛++=-222221ln 11ln 1ln ，所以()x f 为奇函数．2．函数)(x f 的定义域为],1[e ，则)(x e f 的定义域是()A ．](1,0B ．[]1,0C ．()1,0D ．)[1,0【答案】B【解析】由题可知()x e f 中有e e x ≤≤1，解得10≤≤x ，所以()x e f 的定义域为[]1,0．3．曲线16213123+++=x x x y 在)1,0(处的切线与x 轴的交点坐标为()A ．)0,61(-B ．()1,0C ．)0,61(D ．)(0,1-【答案】A【解析】切线斜率()662=++'===x x x xy k ，则切线方程为(),061-=-x y 即16+=x y ，令0=y 得61-=x ，故切线与x 轴的交点坐标为)0,61(-．4．当0→x 时，1132-+ax 与221x -等价，则=a ()A ．23-B ．32-C ．21-D ．32【答案】A【解析】因为当0→x 时，23231~11ax ax -+，则1322131lim 2111lim 2202320=-=-=--+→→a x axx ax x x ，所以23-=a ．5．极限=+--+∞→453423lim 22n n n n n ()A ．1B ．43C ．52-D ．34-【答案】D【解析】34453423lim 453423lim 2222-=+--+=+--+∞→∞→nn n n n n n n n n ．6．极限=→xxx 54sin lim0()A ．45B ．51C ．54D ．1【答案】C 【解析】5454lim 54sin lim00==→→x x x x x x ．7．当0→x 时，122-x e 是2x 的()无穷小A ．高阶B ．低阶C ．等价D ．同阶非等【答案】D【解析】当0→x 时，222~12x e x-，则122lim 1lim 2202202≠==-→→x x xe x x x ，故当0→x 时，122-x e 是2x 的同阶非等价无穷小．8．已知函数⎩⎨⎧<-≥+=1,121,ln )(x ax x x a x f 在1=x 处连续，则=a ()A ．1B ．1-C ．0D ．3【答案】A【解析】因为函数()x f 在1=x 处连续，且()()()()()1212lim lim ,1ln lim lim 1111-=-===+=--++→→→→a ax x f f a x a x f x x x x ，所以a a =-12，故1=a ．9．设⎪⎩⎪⎨⎧<≥-=1,2cos 1,1)(x x x x x f π，则1=x 是其()A ．连续点B ．可去间断点C ．跳跃间断点D ．第二类间断点【答案】A【解析】因为函数()x f 在1=x 处有定义，且()()()()02coslim lim 101lim lim 1111=====-=--++→→→→x x f f x x f x x x x π，所以1=x 是函数的连续点．10．函数)(x f 在a x =处可导，则=--+→xx a f x a f x )()(lim0()A ．()a f '2B ．0C ．()a f 'D ．()a f '21【答案】A 【解析】()()()()()()()()()()().2lim lim )()(limlim0000a f a f a f x a f x a f x a f x a f xa f x a f a f x a f x x a f x a f x x x x '='+'=---+-+=+---+=--+→→→→11．已知xxx f 21)(+=，则=-)1(1f ()A ．1-B ．1C ．31-D ．31【答案】A【解析】由()x x x f 21+=，可得到其反函数xx x f 21)(1-=-，故()121111-=-=-f ，故应选A ．12．已知x xe y =，则=dy ()A ．dxxe x B ．dxe x C ．()dxe x x+1D ．()dxx e x+【答案】C【解析】()x x x e x xe e y +=+='1，故()dx e x dy x +=1．13．xx y +=12的垂直渐近线为()A ．1=xB ．1-=x C ．1=y D ．1-=y 【答案】B【解析】由题意可知，令01=+x ，可得1-=x 为其无定义点，故由定义可知∞=+-→xx x 1lim 21，所以垂直渐近线是1-=x ，故选B ．14．方程)(0sin 23+∞<<-∞=-x x x 的实根个数为()A ．0B ．1C ．2D ．无数个【答案】B【解析】设()x x x f sin 23-=，则()x x f cos 23-='，由于,1cos 1≤≤-x 5cos 231≤-≤x ，故()0>'x f ，()x f 在()+∞∞-,内单调递增，又因为()00=f ，所以函数()x f 只有一个零点，即方程0sin 23=-x x 只有一个实根．15．12213123++-=x x x y 的拐点为()A ．0=xB ．)1,1(C ．)0,0(D ．)1,0(【答案】D【解析】函数123++=x x y 的定义域为R ，x y x y 12,162=''+='，令0=''y 得0=x ，且0>x 时，0>''y ；0<x 时0<''y ，所以函数的拐点为()1,0．16．可导函数()x f 和()x g 满足)()(x f x g '='，则下列选项正确的是()A ．()()g x f x =B ．(())(())g x dx f x dx ''=⎰⎰C ．()()g x f x C =-D ．()()g x dx f x dx=⎰⎰【答案】C【解析】由()()x f x g '='两边同取积分得()()C x f x g -=，再积分得()()()()()Cx dx x f Cdx dx x f dx C x f dx x g -=-=-=⎰⎰⎰⎰⎰，两边求导得()()()()()()C dx x f Cx dx x f dx x g -'='-='⎰⎰，故选C ．17．计算不定积分=-⎰dx x211()A ．1ln 122x C-+B ．1ln(12)2x C-+C ．1ln 122x C--+D ．1ln(12)2x C--+【答案】C 【解析】()C x x d x dx x +--=---=-⎰⎰21ln 212121121211．18．cos bad tdt dx =⎰()A ．a b cos cos -B ．0C ．a b sin sin -D ．ba sin sin -【答案】B【解析】定积分的结果是一个确定的常数，常数求导是0，故选B ．19．当k 为何值时，dx e kx ⎰∞--0收敛()A ．0>kB ．0≥k C ．0<k D ．0≤k 【答案】C【解析】因为()()()⎪⎪⎩⎪⎪⎨⎧⎪⎩⎪⎨⎧>∞<-=-=∞==∞--∞-∞--⎰⎰,0,,0,11,0,1000k k k e k k dx dx e kx kx发散收敛发散所以当0≥k 时，dx ekx⎰∞--0发散；当0<k 时dx e kx ⎰∞--0收敛，故选C ．20．若)(x f 在)5,1(上可积，⎰-=111)(dx x f ，⎰-=512)(dx x f ，则⎰=15)(3dx x f ()A ．2-B ．2C ．3-D ．3【答案】C【解析】由定积分的性质，可知()()()⎰⎰⎰--+=115151dx x f dx x f dx x f ，故1)(51⎰=dx x f ，即313)(3)(35115-=⋅-=-=⎰⎰dx x f dx x f ．21．平面0172=++-z y x 和平面055=+--z y x 的位置关系为()A ．重合B ．垂直C ．平行D ．相交但不垂直【答案】B【解析】平面0172=++-z y x 的法向量()7,2,11-=n，055=+--z y x 的法向量()1,1,52--=n ，因为021=⋅n n，所以两平面垂直．22．若)2,2,(),4,,6(--=-=y b x a ，已知b a //，则y x ,的值分别是()A ．3,4-B ．4,3--C ．4,3-D ．3,4-【答案】D【解析】因为→→b a //，所以22426=--=-=x y ，故3,4=-=y x ．23．已知)ln(y x x z +=，则=∂∂∂yx z2()A ．()2y x x +B ．()2y x y +C ．()22y x y x ++D ．()22y x y x ++【答案】B【解析】()()()2221,ln y x y y x x y x y x z y x x y x x z +=+-+=∂∂∂+++=∂∂．24．一元函数在某点处极限存在是在该点可导的()条件A ．必要B ．充分C ．充要D ．无关【答案】A【解析】一元函数在某点处极限存在但在该点不一定可导；反之一元函数在某点处可导则在该点一定连续，进而在该点极限一定存在．25．级数nn n x n ∑∞=+122的收敛区间为()A ．⎪⎭⎫⎝⎛-41,41B ．⎪⎭⎫ ⎝⎛-21,21C ．()1,1-D ．()2,2-【答案】B【解析】因为()()()2322lim 2322lim lim 11=++=⋅++⋅==∞→+∞→+∞→n n n n a a n n n n n n n ρ，所以级数n n n x n ∑∞=+122的收敛半径211==ρR ，故收敛区间为⎪⎭⎫ ⎝⎛-21,21．26．已知L 为0=+y x 上从点)2,2(-到点)2,2(-上的一段弧，则=⎰Lydx cos ()A ．2sin 2-B ．2sin 2C ．2cos 2-D ．2cos 【答案】A【解析】22:,:-→-=x x y L ，则原式()2sin 2sin cos cos 2222-==-==--⎰⎰xdx x ydx L．27．已知∑∞=-+1212)(n n n u u 收敛，则()A ．∑∞=0n n u 收敛B ．0lim =∞→n n uC ．∑∞=0n n u 敛散性不确定D ．∑∞=0n n u 发散【答案】C【解析】收敛级数加括号后所成的级数仍收敛于原级数的和.如果加括号后所成的级数收敛，则不能断定去括号后原来的级数也收敛.例如，级数()()() +-++-+-111111收敛于零，但级数() +-+-1111却是发散的．28．x Ce y =是0=-''y y 的()A ．解B ．通解C ．特解D ．所有解【答案】A【解析】微分方程0y y ''-=的特征方程为012=-r ，特征根为1,121-==r r ，则微分方程的通解为12x x y C e C e -=+．又因为x y Ce =，x Ce y ='，x Ce y =''，所以把y y '',代入方程可知，x Ce y =满足微分方程0=-''y y ，即x Ce y =是微分方程的解，但x Ce y =只有一个任意常数，则x Ce y =不是通解，不是特解，也不是所有解，故选A ．29．若122++-=x x e y x ，则=)520(y ()A ．520xe B ．2e xC ．5202xe D ．0【答案】B【解析】221x y e x '=-+，22x y e ''=-，2x y e '''=，，(520)2x y e =，故选B ．30．122=-y x 表示的二次曲面是()A ．椭圆柱面B ．抛物面C ．双曲柱面D ．单叶双曲面【答案】C【解析】由方程特点可知，221x y -=表示双曲柱面．二、填空题(每小题2分，共20分)31．极限3lim 13xx x →∞⎛⎫+= ⎪+⎝⎭________．【答案】3e 【解析】33333lim3333lim 1lim 133x x xx xxxx x ee x x →∞+⋅⋅++→∞→∞⎛⎫⎛⎫+=+== ⎪ ⎪++⎝⎭⎝⎭．32．微分方程1090y y y '''-+=的通解是________．【答案】912x x y C e C e =+，12,C C 为任意常数【解析】特征方程为21090r r -+=，特征根11r =，29r =，故通解为912x x y C e C e =+，其中12,C C 为任意常数．33．设(1)23f x x +=+，则(()3)f f x -=________．【答案】43x -【解析】(1)232(1)1f x x x +=+=++，()21f x x =+，故(()3)(22)43f f x f x x -=-=-．34．若⎰-=xdt t x f 0)1()(，则()f x 的单调增区间是________．【答案】(1,)+∞【解析】1])1([)(0-='-='⎰x dt t x f x，令()0f x '>，解得1x >，故()f x 的单调增区间是(1,)+∞．35．不定积分=________．C【解析】12221(1)(1)2x d x C -=++=⎰．36．62(sin )x x x dx ππ-⋅+=⎰________．【答案】323π【解析】因为x x sin 6⋅在区间],[ππ-上为奇函数，2x 在区间],[ππ-上为偶函数，所以由偶倍奇零，可得3622312(sin )2233x x x dx x dx x πππππ-⋅+==⋅=⎰⎰．37．交换积分次序21(,)x dx f x y dy =⎰⎰________．【答案】110(,)dy f x y dx⎰【解析】积分区域{}{}1,10),(0,10),(2≤≤≤≤=≤≤≤≤x y y y x x y x y x ，则交换积分次序21(,)x dx f x y dy =⎰⎰110(,)dy f x y dx ⎰．38．22z x y =+的全微分dz =________．【答案】22xdx ydy +【解析】22z zdz dx dy xdx ydy x y∂∂=+=+∂∂．39．将1()2f x x =-展开成2x +的幂级数为________．【答案】101(2)4n n n x ∞+=+∑，(6,2)x ∈-【解析】,)2(41)42(41421141)2(4121010∑∑∞=+∞=+=+=+-⋅=+-=-n n nnn x x x x x ，其中1421<+<-x ，即(6,2)x ∈-．40．参数方程331cos 21sin 2x t y t ⎧=⎪⎪⎨⎪=⎪⎩的导数dy dx =________．【答案】tan t-【解析】t t t tt dt dx dt dy dx dy tan )sin (cos 23cos sin 2322-=-⋅⋅==．三、计算题(每小题5分，共50分)41．求不定积分cos x xdx ⎰．【答案】sin cos x x x C++【解析】cos sin sin sin sin cos x xdx xd x x x xdx x x x C ==-=++⎰⎰⎰．42．求极限21lim ln 1x x x x →∞⎡⎤⎛⎫-+ ⎪⎢⎥⎝⎭⎣⎦．【答案】12【解析】令t x=1，则22200001111ln(1)ln(1)111lim ln 1lim lim lim lim 22(1)2x t t t t t t t t x x x t t t t t →∞→→→→-⎡⎤+-+⎛⎫⎡⎤+-+=-==== ⎪⎢⎥⎢⎥+⎝⎭⎣⎦⎣⎦．43．设方程y x xyz xy 42-=+所确定的隐函数),(y x z z =，其中0≠xy ，求zx∂∂，z y ∂∂．【答案】2z y yz x xy ∂+-=-∂，4z x xz y xy∂++=-∂【解析】令(,,)24F x y z xy xyz x y =+-+，则2x F y yz =+-，4y F x xz =++，z F xy =，由于0≠xy ，故2x z F z y yz x F xy ∂+-=-=-∂，4y z F z x xz y F xy∂++=-=-∂．44．已知2,0()0x x f x x ≤⎧⎪=>，求31(2)f x dx --⎰．【答案】325-【解析】令2x t -=，则2+=t x ，dx dt =，当1x =-时，3t =-，当3x =时，1t =，故原式325322)()()(10230320313113-=+=+=+==----⎰⎰⎰⎰⎰t t dtt tdt dt t f dt t f dt t f ．45．求过点(9,8,5)且与直线3210210x y y z ++=⎧⎨+-=⎩平行的直线方程．【答案】985236x y z ---==-【解析】设所求直线的方向向量为s ，由题意知320236(2,3,6)021==-+=-i j ks i j k ，又由于直线过点(9,8,5)，故所求直线的方程为985236x y z ---==-．46．计算Dxdxdy ⎰⎰，其中D 是由1y =，y x =，2x =所围成的闭区域．【答案】56【解析】由题意可知，积分区域{}x y x y x D ≤≤≤≤=1,21),(，故2223221111115()326x Dxdxdy dx xdy x x dx x x ⎛⎫==-=-=⎪⎝⎭⎰⎰⎰⎰⎰．47．求幂级数15(1)nn n x n ∞=+∑的收敛区间（不考虑端点）．【答案】(5,5)-【解析】因为115(1)1limlim 5(2)5n n n n n n a n a n ρ++→∞→∞+===+，所以收敛半径15R ρ==，故收敛区间(5,5)-．48．求方程cos (0)xy y x x '+=>的通解．【答案】1(sin )y x C x=+，C 为任意常数【解析】方程可化简为1cos x y y x x'+=，由公式可得故()11cos 11cos (sin )dxx x x y e e dx C xdx C x C x xx-⎛⎫⎰⎰=⋅+=+=+ ⎪⎝⎭⎰⎰，C 为任意常数．49．求321312323y x x x =-+-的极值．【答案】在1x =时，取得极大值12，在2x =时取得极小值13【解析】函数的定义域为R ，232(1)(2)y x x x x '=-+=--，令0y '=，得11x =，22x =．列表，定义域被分为3个区间x (,1)-∞1(1,2)2(2,)+∞y '+0-0+y极大值极小值综上，函数在1x =时，取得极大值12，在2x =时取得极小值13．50．求椭球面222239x y z ++=在点(2,1,1)处的切平面方程．【答案】22390x y z ++-=【解析】令222(,,)239F x y z x y z =++-，则2x F x =，4y F y =，6z F z =，所以切平面的法向量(4,4,6)2(2,2,3)==n ，又由于切平面过点(2,1,1)，故切平面的方程为2(2)2(1)3(1)0x y z -+-+-=，即22390x y z ++-=．四、应用题(每小题7分，共14分)51．曲线2y x =，2x =，0y =所围图形D 绕x 轴旋转一周所得旋转体的体积．【答案】325π【解析】联立方程22y x x ⎧=⎨=⎩，解得交点(2,4)，由题意可知，222520032()55x V x dx x πππ===⎰．52．已知血液浓度C 关于时间t 的函数为32004.004.003.0)(t t t t C -+=，求时间t 为多0.0885≈）【答案】02.7【解析】由题意可知，20012.008.003.0)(t t t C -+=')0(>t ，令0012.008.003.0)(2=-+='t t t C ，得35.01≈t (舍)，02.72≈t ．故02.7≈t 为唯一的驻点．又由于002.7012.008.0)02.7(<⨯-=''C ，故在02.7≈t 处，取得极大值，由实际意义可知，在02.7≈t 处，可取得最大值，即在02.7≈t 处，血液浓度最大．五、证明题(6分)53．函数()f x 在[],a b 上连续，在(,)a b 内可导，()f a a =，()f b b =且()0f x ≠，证明：在(,)a b 内至少存在一点ξ，使得()()f f ξξξ'=⋅．【解析】令()()xF x f x =，则()F x 在[],a b 上连续，在(,)a b 内可导，且2)]([)()()(x f x f x x f x F '-='，又因为()1()a F a f a ==，()1()bF b f b ==，由罗尔定理知，至少存在一个点(,)a b ξ∈，使得()0F ξ'=，又0)(≠ξf ，所以0)()(='-ξξξf f ，即()()f f ξξξ'=⋅．。

19专升本真题高数

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x,

2019年河南临床专升本真题

2019年河南临床专升本真题1金属和类金属毒物主要经何途径排泄 [单选题] *A 呼吸道B 消化道C 毛发D 汗腺E 肾脏(正确答案)2与职业病发生有关的细菌是 [单选题] *A 脑膜炎双球菌B 肺炎双球菌C 葡萄球菌D 痢疾杆菌E 炭疽杆菌(正确答案)3属于我国法定职业病的有： [单选题] *A 教师的咽喉炎和气管炎B 接触噪声引起的高血压C 高温作业工人的消化不良D 矽尘作业工人的肺结核E 苯作业工人的白血病(正确答案)4就业前健康检查的目的主要是 [单选题] *A 劳动能力鉴定B 建立健康档案C 发现职业禁忌症(正确答案)D 发现职业病E 筛选传染病5下列哪项不属于职业病范围 [单选题] *A 伐木工人的森林脑炎B 凿岩工人的矽肺C 石墨工人的炭黑尘肺D 铅作业工人的消化性溃疡(正确答案)E 水银作业工人的汞中毒6.治疗铅中毒首选药物为 [单选题] *A 二巯基丙醇B 依地酸二钠钙(正确答案)C 二巯基丙磺酸钠D 青霉胺E 二巯基丁二酸钠7患者从事蓄电池生产3年，诉头昏、无力、肌肉关节酸痛、记忆力减退、时有便秘、腹绞痛，体检发现舌、手指及眼睑均有轻度震颤，皮肤划痕阳性，于门齿、犬齿、牙龈的内外侧边缘处可见蓝黑色线带，化验血色素9g%，你认为何种诊断可能性大 [单选题] *A.急性汞中毒B.急性铅中毒C.急性铅中毒D.慢性铅中毒(正确答案)E.慢性苯中毒8.一次高浓度吸入苯蒸气，主要是引起的损害 [单选题] *A 化学性肺炎B 中枢神经麻痹(正确答案)C 造血机能极度障碍D 植物神经功能紊乱E 肝功能受损9.下列哪些物质不属于刺激性气体 [单选题] *A HClB CL2C FD SO3E CO(正确答案)10.田间喷洒有机磷农药时发生中毒的主要原因是 [单选题] *A.呼吸道吸入(正确答案)B.消化道摄入C.皮肤吸收D.体质虚弱E.以上都不是11.我国尘肺病例中，占比重最大的是 [单选题] *A 石棉肺B 矽肺(正确答案)C 煤工尘肺D 农民肺E 水泥尘肺12.矽肺的并发症中，最常见和危害最大的是 [单选题] *A 肺心病B 肺结核(正确答案)C 肺炎D 支气管炎E 自发性气胸13.热射病的临床特点不包括下列哪项 [单选题] *A 起病急，体温可达40℃以上B 早期无汗，继之大汗(正确答案)C 皮肤干热D 不同程度意识障碍E 病死率较高14.局部振动病典型的临床症状主要是 [单选题] *A.手痛B.手震颤C.白指(正确答案)D.多汗E.手冷15.下列粉尘何者为无机粉尘 [单选题] *A 农药粉尘B 橡胶粉尘C 水泥粉尘(正确答案)D 木尘E铅尘16.高气湿是指环境相对湿度大于( ) [单选题] * A．50%B．60%C．70%D．80%(正确答案)17.热辐射能量大小主要取决于 ( ) [单选题] * A．环境的温度B．环境的湿度C．辐射源的温度(正确答案)D．辐射源的湿度18.高温作业指工作地点有生产性热源，其气温等于或高于本地区夏季室外通风设计计算温度（）的作业 [单选题] *A．2℃及以上(正确答案)B．0.2℃及以上C．1℃及以上D．1℃及以上19.中暑按发病机理分为（） [单选题] *A．热射病，热痉挛和热衰竭(正确答案)B．轻症中暑，重症中暑C．热适应，热射病和热衰竭D．热适应，热痉挛和热衰竭20.生产性噪声按其产生的来源来分，可分为（） [单选题] *A．机械性噪声、流体动力性噪声和电磁性噪声(正确答案)B．稳态噪声、非稳态噪声、脉冲噪声C．交通噪声、生活噪声、工业噪声D．低频噪声、中频噪声、高频噪声21.永久性听阈位移包括（） [单选题] *A．听觉适应、听觉疲劳B．听觉疲劳、听觉损伤C．听觉适应、噪声性耳聋D．听力损伤、噪声性耳聋(正确答案)22.职业性噪声聋的诊断原则 ( ) [单选题] *A．有明确的职业噪声接触史B．排除其他致聋原因C．纯音测定为感音性耳聋D．以上都是(正确答案)23.除了听觉系统外，噪声对其他系统的影响还包括 ( ) [单选题] * A．神经系统B．消化系统C．心血管系统D．以上都是(正确答案)24.影响噪声对机体作用的因素有 ( ) [单选题] *A．噪声强度、接触时间B．噪声的频谱、类型、接触方式C．机体的健康状况和敏感性D．上述A、B和C三项都是(正确答案)25.以下哪种不属于非电离辐射（） [单选题] *A．激光B．X射线(正确答案)C．紫外线D．红外线26.高温作业的主要类型有（） [单选题] *A．高温强辐射作业B．高温高湿作业C．夏季露天作业D．以上都是(正确答案)27.某纺织女工，45岁，经检查听阈提高20dB，10-20小时后恢复听力，此听力损害属于（） [单选题] *A．噪声性耳聋B．听觉适应C．听觉疲劳(正确答案)D．爆震性耳聋34.炼钢、陶瓷、铸造工种属于何种作业（） [单选题] *A．高温低辐射作业B．高气湿作业C．高温高湿作业D．干热作业(正确答案)28.一般用（）作为衡量高温作业劳动者劳动强度和受热程度的综合指标。

2019河南专升本高等数学真题及其答案

2019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。

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，再选涂其它选项。

1.函数f (x )=ln(√1+x 2−x)在定义域内是（） A.不确定 B.偶函数 C.非奇非偶函数 D.奇函数2.已知f (x )的定义域是[1,e ]，则f (e x )的定义域为（）A.(0,1]B.[0,1]C.[0,1)D.(0,1) 3.曲线y =13x 3+12x 2+6x +1在点(0,1)处的切线与x 轴的交点坐标为（）A.(−16,0)B.(−1,0)C.(16,0)D.(1,0) 4.当x →0时，√1+ax 23−1与−12x 2等价，则a = （）A.−32 B.32C.12D.−125.极限lim x→∞3+2n−4n 23n 2−5n+4=（）A.1B.−1C.−43D.436.极限lim x→0sin 4x 5x= （）A.−45B.45C.−54D.547.当x →0时，e 2x 2−1是x 2的（）A.低阶无穷小B.高阶无穷小x u a ny iC.等价无穷小D.同阶非等价无穷小8.已知函数f (x )={a +ln x ,x ≥12ax −1, x <1，在x =1处连续，则a =（）A.1B.−1C.12D.−129.设f (x )={1−x, x <−1cos π2x ,x ≥−1,则x =−1是f (x )的（）A.连续点B.可去间断点C.跳跃间断点D.第二类间断点10.函数f (x )在点x =a 处可导，则：lim x→0f (a+x )−f (a−x )x=（）A.2f ′(a )B.−2f ′(a )C.f ′(a )D.−f ′(a )11.设f (x )=x 1+2x，那么f −1(1)= （）A.−1B.1C.−13D.1312.已知y =xe x ，求 dy = （）A.(x −1)e 2x dxB.(x −1)e x dxC.(x +1)e x dxD.xe x dx13.曲线y =x 21+x 的垂直渐近线为（）A.x =1B.x =−1C.y =1D.y =−114.方程3x −2sin x =0 (−∞<x <+∞)的实根的个数为（） A.0B.1C.2D.无数个15.曲线y =2x 3+x +1的拐点为（） A.x =0B.(1,1)C.(0,0)D.(0,1)16.若在区间[a,b ]内，g ′(x )=f ′(x )，则下列等式正确的是（）A .g (x )=f(x)B.(∫g(x)dx )′=(∫f(x)dx )′C.g (x )=f (x )−CD.∫g(x)dx =∫f(x)dx17.计算不定积分∫11−2xdx =（）A.12ln |1−2x |+CB.12ln (1−2x )+CC.−12ln |1−2x |+C D.−12ln (1−2x )+C18.ddx ∫cos t dt ba = （）A.sinxB.cosxC.0D.xsinx19.当k 为何值时，广义积分∫e −kx dx 0−∞收敛（）A.k >0B.k ≥0C.k <0D.k ≤0x u a ny i20.已知函数f (x )在(1,5)上可积，∫f (x )dx =1,∫f (x )dx =2,5−11−1求∫3f (x )dx 15 （）A.−2B.2C.−3D.321.平面x −2y +7z +1=0与平面5x −y −z +5=0的位置关系是（）A.重合B.垂直C.平行D.相交但不垂直22.已知向量a ⃗=(6,x,−4),b ⃗⃗={y,−2,−2}，已知a ⃗与b⃗⃗平行，则x,y =（）A.4,−3B.−3,−4C.−4,3D.3,−423.二元函数z =x ln (x +y )，则ð2zðxðy = （）A.x(x+y )2B.−x(x+y )2C.y(x+y )2D.−y(x+y )224.一元函数在某点处极限存在是在该点可导的_____条件 A.必要B.充分C.充要D.无关25.已知D ={(x,y )|x 2+y 2≤9}，则二重积分∬D√9−x 2−y 2dxdy =（）A.18πB.36πC.9πD.6π26.设L 是直线x +y =0上从(2,−2)到(−2,2)的一段弧，则曲线积分∫Lcos y dx =（）A.−2sin 2B.2sin 2C.−2cos 2D.2cos 227.已知∑∞n=1(u 2n−1+u 2n )收敛，则（）A.∑∞n=1u n 收敛B.lim n→∞u n =0C.∑∞n=1u n 不确定D.∑∞n=1u n 发散28.y =Ce x 是y ′′−y =0的（） A.解B.通解C.特解D.所有解29.已知y =2e x −x 2+x +1，则y (520)= （） A.520e x B.2e xC.2e 520xD.030.x 2−y 2=1表示的二次曲面是（）A.锥面B.抛物面C.双曲柱面D.单叶双曲面二、填空题（每小题2分，共20分）31.极限lim x→∞(1+33+x)x =_____。

2019河南专升本高数真题

2019年河南省专升本高等数学真题一、单项选择题（每小题2分，共60分）1.函数()()x x x f -+=21ln 在定义域是（D ）A.不确定B.偶函数C.非奇非偶函数D.奇函数解析：由()()x f x f -=-得，()()x x x f -+=21ln 为奇函数。

C.C x +--21ln 21D.()C x +--21ln 21解析：()x d dx 2-12x -11212x -11⎰⎰-==C x +--21ln 2118.⎰b atdt dx d cos =（）解析：定积分⎰ba tdt cos 就是一个常数，常数求导等于0.19.当k 为何值时，dx e kx ⎰∞--0收敛。

A.0>k B.0≥k C.0<k D.0≤k 解析：⎪⎩⎪⎨⎧><∞--==-∞-∞--⎰⎰0,0,01,0100k k e k k dx dx e kx kx 发散收敛，发散，，故选C 20.()x f 在()5,1上可积，()()2,15111==⎰⎰--dx x f dx x f ，求()=⎰dx x f 153（）A.x e 520B.xe 2 C.x e 5202 D.0解析:()x n e y 2=，故()x e y 2520=，选B30.122=-y x 表示的二次曲面是（）解析：由曲面方程可知122=-y x 表示的二次曲面是双曲柱面。

二、填空题（每小题2分。

共20分）31.极限x x x ⎪⎭⎫ ⎝⎛++∞→331lim =解析：333lim 3333331lim 331lim e e x x x xx x x x x x x ==⎪⎭⎫ ⎝⎛++=⎪⎭⎫ ⎝⎛++++⋅+∞→∞→→∞32.微分方程0910'''=+-y y y 的通解是解析：9,10910212==⇒=+-r r r r ，通解为21921,,C C e C e C yx x +=为任意常数解析：令t x t x 1,1==，则原式=()()()212111lim 1ln lim 1ln 1lim 02020=+-=⎦⎤⎢⎣⎡+-=⎥⎦⎤⎢⎣⎡+-→→→t t t t t t t t t t t 43.0,42≠-=+xy y x xyz xy ，求y z x z ∂∂∂∂,解析：令()=z y x F ,,yx xyz xy 42+-+xyxz x y z xy yz y x z xy F xz x F yz y F z y x 4,2,,4,2'''++-=∂∂-+-=∂∂=++=-+=44.已知()⎩⎨⎧>≤=0,0,2x x x x x f ，求()dx x f ⎰-312解析：令2,2+==-t x t x ，则()()()()()dx x f dx x f dx x f dt t f dx x f ⎰⎰⎰⎰⎰+===----10011111312()x f 在[]b a ,上连续，()b a ,内可导，()()()0,≠==x f b b f a a f ，，证明：在()b a ,内至少存在一点ξ，使得()()ξξξ'f f ⋅=。

2019河南专升本高数真题及答案解析

2019河南专升本高数真题及答案解析基本信息：[矩阵文本题] *1. lim x→0sin3xx= [单选题]A.-1B.1C.2D.3(正确答案)2. 当x→0时，In (1+x) 与x 比较是（） [单选题]A.高阶的无穷小B.等价的无穷小(正确答案)C.非等价的同阶无穷小D.低阶的无穷小3. 设函数f(x)=2x−1的断点为x=（） [单选题]A.1(正确答案)B.0C.-1D.24. 设函数y=cosx+1,则 dy=( ) [单选题]A.(sinx+1)dxB.(cosx+1)dxC.-sinxdx(正确答案)D.sin xdx5. 设函数f(x)在区间[a,b]l 连续，在（a,b)可导，f'(x)>0,若f(a).f(b)<0,则y=f(x)在（a,b)内（） [单选题]A.不存在零点B.存在唯一零点(正确答案)C.存在极大值点D.存在极小值点6. 曲线 y=x3的拐点坐标是（） [单选题]A.(-1,-1)B.(0,0)(正确答案)C.(1,1)D.(2,8)7. 函数f(x)=ln(x2 +2x+2)的单调递减区间是（） [单选题]A.（-∞，-1）(正确答案)B. （-1，0）C.（0，1）D.（1，+∞）8. 设函数 y=ln sinx ,则 dy=_______________. [单选题]A.2 cos 2x dx(正确答案)B.cos 2x dxC.--2cos2xdxD.-cos 2x dx9. 设函数y=2+sinx ,则y'= ( ) [单选题]A. cosx(正确答案)B.-cosxC.2+cosxD.2-cosx10. d(sin2x)=( ) [单选题]A.2 cos 2xdx(正确答案)B.cos 2x dxC.-2cos2xdxD.-cos 2x dx11. ∫-11x2 sinxdx=( ) [单选题]A.-1B.0(正确答案)C.1D.212. ∫01 (5x4 +2)dx [单选题]A.1B.3(正确答案)C.5D.713. 设函数f(x)=∫0x costdt ,则F'(x)= [单选题]A.cosx(正确答案)B.sinxC.costD.sint14. 设A,B是两随机事件，则事件A ----表示（） [单选题]A.事件A,B都是随机发生B.事件B发生而事件A不发生C.事件A发生而事件B不发生(正确答案)D.事件A，B,都不发生15. 甲乙两人同时向一个目标射击，甲击中的概率是0.9，乙击中的概率是0.8，求两个人都击中目标的概率（ ) [单选题]A.0.8B.0.9C.0.72(正确答案)D.0.1816. 设函数z=x2 +y,则dz= ( )[单选题]A.x(正确答案)B.2x+y+z=1C.x+2y+z=1D.x+y+2z=117. 当x →0时，sin3x 是2x 的（） [单选题]A.低级无穷小量B.等价无穷小量C.同阶但不等价无穷小量(正确答案)D.高阶无穷小量18. 设函数f(x)=ln (3x),则f'(2)=( ) [单选题]A.6B.ln6C.1/2(正确答案)D.1/619. 已知函数f(x)的导函数f'(x)=3x2 -x-1,则曲线y=f(x)在x=2 处切线的斜率是（） [单选题]A.3B.5C.9(正确答案)D.1120. 函数f(x)在[0，2]上连续，且在（0，2）内f'(x)>0则下列不等式成立的是（） [单选题]A.f(0)>f(1)>f(2)B.f(0)<f(1)<f(2)(正确答案)C.f(0)<f(2)<f(1)D.f(0)>f(2)>f(1)。

2019河南专升本数学真题答案

2019河南专升本数学真题答案
河南2019年统招专升本考试结束了，很多同学都在问小编有没有真题试卷，好老师教育河南专升本教研团队整理了本次《高等数学》的真题试卷并进行了整体分析。

整体情况：本次考试中，难度系数适中，所考题型不变，无超题目，重点仍然集中在微积分学，考察题型与知识点为集训班常讲内容，高等数学总分150分，原题命中25题；选择题共30题命中11题，命中分值22分；填空题共10题命中6题，命中分值12分；计算题共10题命中7题，命中分值35分；证明题命中1题，命中分值7分，且相似题占比100%，知识点覆盖率100%。

从协议集训班学员反馈情况来看，此次考试题型在集训时均有专项版块训练，部分学员分数预估可以达到140分左右，甚至满分。

一、2019年河南专升本整体报考情况分析
全省报考127500人，仅郑州地区报考“专升本”考生65262人，比去年增加了11619人，增长21.7%。

2019年河南专升本报名人数居于第一位的是护理学，高达10000人，成为河南专升本历史上第一个上万人专业！
第二名会计学8710人，
第三名临床医学7194人，
临床医学招生院校：河南大学民生学院、河南科技大学、新乡医学院三全学院！总共招生850人。

第四名财务管理6956人，
第五名小学教育6863人!
2019河南专升本报名人数12.75万，2018年10.9961万，2017年6.4205万！
综合对比，所以今年河南专升本考试竞争将会空前激烈！。