上海市虹口区2015届高三二模数学理试题 Word版含答案

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2015届虹口区高三英语二模试卷及答案(官方版)

虹口区2015高三英语二模试卷2015.4考生注意:1. 考试时间120分钟，试卷满分150分。

2. 本考试设试卷和答题纸两部分。

试卷分为第Ⅰ卷（第1—10页）和第Ⅱ卷（第10页），全卷共10页。

第I卷第1-16小题、第41-77小题为选择题，答题必须涂在答题纸上，第I 卷第17-40小题、第78-81小题和第II卷的答案必须写在答题纸上，做在试卷上一律不得分。

3．答题前，务必在答题纸上填写准考证号和姓名，并将核对后的条形码贴在指定位置上，在答题纸反面清楚地填写姓名。

第 I 卷 (共103分)I. Listening ComprehensionSection ADirections：In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A. A policewoman. B. A waitress. C. A shop assistant. D. A worker.2. A. Disappointment. B. Disapproval. C. Sympathy. D. Passion.3. A. At a police station. B. At a car rental house. C. At a post office. D. At a bank.4. A. Go to work. B. Take a break. C. Try another problem. D. Keep doing.5. A. The woman congratulated the wrong person.B. The woman should get another job.C. The woman should be more patient.D. The woman was waiting in the wrong place.6. A. Reading a magazine. B. Writing an article.C. Buying clothes.D. Preparing for a maths test.7. A. The guest has to pay in cash. B. The fee will be added to the hotel bill.C. The guest can pay by check.D. It’s free to watch the hotel movie channel.8. A. The woman will enjoy the trip. B. The woman will be exhausted after the trip.C. The woman had better cancel the trip.D. The woman should go to Los Angeles.9. A. 4 pounds. B. 6 pounds. C. 8 pounds. D. 10 pounds.110. A. Compare notes with his classmates. B. Review the details of all his lessons.C. Focus on the main points of his lectures.D. Talk with her about his learning problems. Section BDirections: In section B, you will hear two short passages, and you will be asked three questions on each of the passages. The passages will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. Sending magazines to friends.B. Sending text messages through mobile phone.C. Sending greeting cards to friends.D. Giving orders to children.12. A. Because it costs much time. B. Because it becomes more popular.C. Because it can be done anywhere.D. Because it makes teachers and parents angry.13. A. Making children clever. B. Saving money and paper.C. Helping students study well.D. Making problems become easy.Questions 14 through 16 are based on the following passage.14. A. they cannot be lost or stolen. B. they are safe and handy.C. they can be used anywhere.D. they can save travelers plenty of money.15. A. The authority that issues you the passport. B. The insurance company.C. The bank where you buy your checks.D. The travel agency that arranges your travel.16. A. People usually get traveler’s checks from foreign banks.B. You are not charged for the safety of your traveler’s checks.C. You cannot get your passport until you get your traveler’s checks.D. Traveler’s checks can be exchanged for th e money of the country you visit.Section CDirections: In section C, you will hear two longer conversations. The conversations will be read twice. After you hear each conversation, you are required to fill in the numbered blanks with the information you have heard. Write your answers on your answer sheet.Blanks 17 through 20 are based on the following conversation.Complete the form. Write ONE WORD for each answer.2Blanks 21 through 24 are based on the following conversation.Complete the form. Write NO MORE THAN THREE WORDS for each answer.II. Grammar and VocabularySection ADirections: After reading the passages below, fill in the blanks to make the passages coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.(A)How I Turned to Be Optimistic (乐观的)I began to grow up that winter night when my parents and I were returning from my aunt’s house, and my mother said that we (25)______(leave) for America soon. We were on the bus then. I was crying, and some people on the bus were turning around to look at me. I remember I could not bear the thought of never hearing again the radio program for school children to (26)______ I listened every morning .I do not remember myself (27)______(cry) for this reason again. In fact I think I cried very little when I was saying goodbye to my friends and relatives. When we were leaving I thought about all the places I was going to see. The country I was leaving never to come back was hardly in my head then.The four years that followed taught me the importance of optimism, but (28)______ idea did not come to me at once. For the first two years in New York I was really lost. I did not quite know what I was or what I should be. Mother remarried, and things became even (29)______(complex) for me. Some time passed before my stepfather and I got used to each other. However, my responsibilities in the family increased a lot since my English was superior (30)______ anyone else’s at home. I translated at interviews with immigration officers, and even discussed telephone bills with company representatives.From my experiences, I believe that my life will turn out all right (31)______ ______ it is not that easy.(B)How Room Designs Affect Our Work and FeelingArchitects have long had the feeling that the place we live in can affect our thoughts, feeling and behaviours. But now scientists are giving this feeling an empirical（实证的）basis. They are discovering how (32)______(design) spaces that promote creativity, keep people focused, and lead to relaxation.3Researches show aspects of the physical environment can influence creativity. In 2012, Joan Meyers-Levy reported that the height of a room’s ceiling affects (33)______ people think. Her research indicates that higher ceilings encourage people to think more freely, (34)______(lead) them to make more abstract connections. Low ceilings, on the other hand, may inspire a more detailed outlook. Besides ceiling height, the view (35)______(afford) by a building may influence an occupant’s ability to concentrate.Using nature to improve focus of attention ought to pay off academically, and (36)______ seems to, according to a study. Students in classrooms with unblocked views of at least 50 feet outside the window had higher scores on tests of vocabulary, language arts and maths than did students (37)______ classrooms primarily overlooked roads and parking lots.Recent study on room lighting design suggests that dim light helps people loosen up.(38)______ that is true generally, keeping the light low during dinner or at parties could increase relaxation.So far public buildings (39)______(focus) on by scientists. “We have a very limited number of studies, so we are almost looking at the problem through a straw(吸管),” architect David says. “How do you take answers to very specific questions and make broad use of them? That is (40)______ we are all struggling wit h.”Section BDirections: Complete the following passage by using the words in the box. Each word can only be used once. Note that there is one word more than you need.Foreign drivers will have a pay on-the-spot fines of up to ￡900 for breaking the traffic law to be carried out next month.If they do not have enough cash or a working credit card, their vehicles will be clamped（扣留）until they pay — and they will face a(n) 41 fee of ￡80 for getting back their vehicles.The law will also be 42 to British citizens. The fines will be described officially as “deposits” when the traffic laws tak e 43 , because the money would be returned if the driver went to court and was found not guilty. In practice, very few foreign drivers are likely to return to Britain to deal with their cases.Foreign drivers are rarely 44 because police cannot take action against them if they fail to appear in court. Instead, officers often 45 give warnings. Foreign vehicles are 30 percent more likely to be in a crash than British-registered vehicles. The number of crashes caused by foreign vehicles rose by 47 percent between 2008 and 2013. There were almost 400 deaths and serious injuries and 3,000 46 injuries from accidents caused by foreign vehicles in 2013.The new law is partly 47 to settle the problem of foreign lorry drivers ignoring limits to weight and hours at the wheel. Foreign lorries are three times more likely to be in a crash than British lorries. Recent spot checks found that three quarters of lorries that failed safety tests were448 overseas.The standard deposit for a careless driving 49 —such as driving too close to the vehicle in front or reading a map at the wheel—will be ￡300.Foreign drivers will not get points as 50 added to their licenses, while British drivers will.III. Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.PlanningIn any planning system, from the simplest budgeting to the most complex corporate planning, there is an annual process. This is partly due to the fact that firms 51 their accounting on a yearly basis, but also because similar 52 often occur in the market.Usually, the larger the firm, the longer the planning takes. But 53 , planning for next year may start nine months or more in advance, with various stages of evaluation leading to 54 of the complete plan three months before the start of the year.Planning continues, however, throughout the year, since managers 55 progress against targets, while looking forward to the next year. What is happening now will 56 the objectives and plans for the future.In today’s business climate, as markets constantly change and become more difficult to 57 , some analysts believe that long-term planning is 58 . In some markets they may be right, as long as companies can build the sort of flexibility into their operations which allows them to 59 to any sudden changes.Most firms, however, need to plan more than one year ahead in order to 60 their long-term goals. This may reflect the time it takes to commission (委任) and build a new production plant, or, in marketing 61 , it may be a question of how long it takes to research and launch a range of new products, and reach a certain 62 in the market. If, for example, it is going to take five years for a particular airline to become the 63 choice amongst business travellers on certain routes, the airline must plan for the various 64 involved.Every one-year plan, therefore, must be 65 in relation to longer-term plans, and it should contain die stages that are necessary to achieve the final goals.51. A. make up B. carry out C. bring about D. put down52. A. patterns B. guides C. designs D. distributions53. A. surprisingly B. contrarily C. equally D. typically54. A. approval B. permission C. admiration D. objection55. A. value B. confirm C. review D. survey56. A. restore B. promote C. influence D. maintain57. A. guess B. advocate C. recognize D. predict58. A. pointless B. meaningful C. realistic D. inevitable59. A. lead B. respond C. refer D. contribute560. A. share B. handle C. develop D. benefit61. A. expressions B. descriptions C. words D. terms62. A. reputation B. position C. situation D. direction63. A. reserved B. selected C. preferred D. supposed64. A. acts B. steps C. means D. points65. A. handed over B. left behind C. made out D. drawn upSection BDirections: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)My dad loved pennies, especially those with the elegant stalk (茎) of wheat curving around each side of the ONE CENT on the back. Those were the pennies he grew up with during the Depression.As a kid, I would go for walks with Dad, spying coins along the way—a penny here, a dime (一角硬币) there. Whenever I picked up a penny, he’d ask, “Is it a wheat?” It always thrilled him when we found one of those special coins produced between 1909 and 1958, the year of my birth.One gray Sunday mor ning in winter, not long after my father’s death in 2002, I was walking down Fifth Avenue, feeling bereft. I found myself in front of the church where Dad once worked. I was warmly shown in and led to a seat. Hearing Dad’s favorite “A Mighty Fortress Is Our God”, I burst into tears. We’d sung that at his funeral.After the service, I shook the pastor’s(牧师) hand and stepped onto the side walk—and there was a penny. I bent to pick it up, turned it over, and sure enough, it was a wheat. A 1944, a year my father was serving on a ship in the South Pacific.That started it. Suddenly wheat pennies began turning up on the sidewalks of New York everywhere. I got most of the important years: his birth year, my mom’s birth year, the year he graduated from college, the year he met my mom, the year they got married, the year my sister was born. But alas, no 1958 wheat penny—my year, the last year they were made.The next Sunday, after the service, I was walking up Fifth Avenue and spotted a penny in the middle of a crossing. Oh, no, it was a busy street；cabs were speeding by—should I risk it? I just had to get it.A wheat! But the penny was worn, and I couldn’t read the date. On arriving home, I took out my glasses and took it to the light. There was my birthday!I found 21 wheat pennies on the streets of Manhattan in the year after my father died, and I don’t think that’s a coincidence.66. The writer’s father loved pennies with wheat because ________.A. when he first saw it, he began to love itB. when he saw the wheat, he thought of his time during the DepressionC. when he was young, he had a lot of pennies with wheatD. when he was a child, he never got a coin with wheat667. The underlined word “bereft” (in Para.3) means ________.A. protestedB. disappointedC. grievedD. offended68. Which of the following statements about the author is NOT true?A. He was born in 1958.B. He knew the church well.C. He once worked in a church.D. He went to church because of his father.69. The best title for the passage would probably be ________.A. Pennies from HeavenB. My father’s life storyC. My father’s hobbyD. Living in New York(B)Do you want to get home from work knowing you have made a real difference in someone’s life? If yes, don’t care about sex or age! Come and join us, then you’ll make it!70. What does the underlined part mean?A. You’ll make others’ lives mo re meaningful with this job.B. You’ll arrive home just in time from this job.C. You’ll earn a good salary from this job.D. You’ll succeed in getting this job.71. The volunteers’ major responsibility is to help people with learning disabilities ________.A. to get some financial supportB. to properly protect themselvesC. to learn some new living skillsD. to realize their own importance772. Which of the following can first be chosen as a volunteer?A. The one who can drive a car.B. The one who has done similar work before.C. The one who has patience to listen to others.D. The one who can use English to communicate.73. The text serves as ________.A. a reminder to social workersB. an advertisement for helpersC. a document on appealing for volunteersD. an introduction about a social care organization(C)There are desert plants which survive the dry season in the form of inactive seeds. There are also desert insects which survive as inactive larvae (幼虫). In addition, difficult as it is to believe, there are desert fish which can survive through years of droughts in the form of inactive eggs. These are the shrimps (小虾) that live in the Mojave Desert, an intensely dry region in the south-west of the United States where shade temperatures of over 50℃ are often recorded.The eggs of the Mojave shrimps are the size and have the appearance of grains of sand. When sufficient spring rain falls to form a lake, once every two to five years, these eggs hatch. Then the water is soon filled with millions of tiny shrimps about a millimetre long which feed on tiny plant and animal organisms which also grow in the temporary desert lake. Within a week, the shrimps grow from their original 1 millimetre to a length of about 1.5 centimetres.Throughout the time that the shrimps are rapidly maturing, the water in the lake equally rapidly evaporates (挥发). Therefore, for the shrimps it is a race against time. By the twelfth day, however, when they are about 3 centimetres long, hundreds of tiny eggs form on the underbodies of the females. Usually by this time, all that remains of the lake is a large, muddy patch of wet soil. On the thirteenth day and the next, during the final hours of their brief lives, the shrimps lay their eggs in the mud. Then, having ensured that their species will survive, the shrimps die as the last of the water evaporates.If sufficient rain falls the next year to form another lake, the eggs hatch, and once again the shrimps pass rapidly through their cycle of growth, adulthood, egg-laying, and death. Some years there is insufficient rain to form a lake: in this case, the eggs will remain dormant for another year, or even longer if necessary. Very, very occasionally, perhaps twice in a hundred years, sufficient rain falls to form a deep lake that lasts a month or more. In this case, the species passes through two cycles of growth, egg-laying, and death. Thus, on such occasions, the species multiplies considerably, which further ensures its survival.74. Which of the following is the most distinctive feature of Mojave shrimps?A. They live a brief and tough life.B. They feed on plant and animal organisms.C. Their eggs can survive years of drought.D. They lay their eggs in the mud.875. The word “dormant” (in Para 4) most probably means ________.A. inactiveB. strongC. alertD. soft76. What can be inferred from the passage?A. appearance and size are important factors for life to survive in the desert.B. a species must be able to multiply quickly in order to survive in the desert.C. for some species one life cycle in a year is enough to survive the desert drought.D. some species develop a unique life pattern to survive in severe conditions.77. The passage mainly deals with ________.A. the life span of the Mojave shrimpsB. the survival of desert shrimpsC. the creatures living in the Mojave desertD. the importance of water to life in the desertSection CDirections: Read the passage carefully. Then answer the questions or complete the statements in the fewest possible words.The greatest recent social changes have been in the lives of women in America, or probably inthe world.During the twentieth century there has been a remarkable shortening of the time of a woman’slife spent in caring for children. A woman marrying at the end of the nineteenth century wouldprobably have been in her middle twenties, and would be likely to have seven or eight children, ofwhom four or five lived till they were five years old. By the time the youngest was fifteen, themother would have been in her early fifties and would expect to live a further twenty years, duringwhich health made it unusual for her to get paid work. Today women marry younger and have fewerchildren. Usually a woman’s youngest child will be fifteen when she is forty-five and can beexpected to live another thirty-five years and is likely to take paid work until retirement at sixty.Even while she has the care of children, her work is lightened by modern living conditions.This important change in women’s life-pattern has only recently begun to have its full effect onwomen’s economic position. Even a few years ago most girls left schools at the first chance, andmost of them took a full-time job. However, when they married, they usually left work at once andnever returned to it. Today the school-leaving age is sixteen, many girls stay at school after that age,and though women usually marry younger, more married women stay at least until shortly beforetheir first child is born. Very many more afterwards return to full or part-time work.Such changes have led to a new relationship in marriage, with the husband accepting a greatershare of the duties and satisfactions of family life, and with the both husband and wife sharing moreequally in providing the money and running the home in terms of the abilities and interests of eachof them.9(Note: Answer the questions or complete the statements in NO MORE THAN EIGHT WORDS.)78. At what age did most women get married in the late nineteenth century?79. A woman today can still take care of her children when doing paid work in their forties becauseof ________.80. Of “such changes” today, one is that many more mothers ________ after their first child is born.81. What are the factors that cause a couple to share economic and family affairs in an equal way?第 II 卷 (共47分)I. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.1. 据我所知，他们学校的面积是我们的两倍。

2015年上海市虹口区高考数学二模试卷(理科)配套款后附答案解析

2015年上海市虹口区高考数学二模试卷（理科）一、填空题（本大题满分56分）本大题共14题，只要求在答题纸相应题号的空格内直接填写结果，每个空格填对得4分，否则一律得零分。

1.计算：201511i i++= .（i 是虚数单位）2.已知函数132,(0)(),(0)x x f x x x ⎧≤⎪=⎨⎪>⎩，则((3))f f -=_______.3. 函数1()ln(1)f x x=+，(0)x >的反函数1()f x -=___________.4.已知正实数x ，y 满足31x y +=，则13xx y+的最小值为_ _____.5.已知复数3sin cos z i θθ=+（i 是虚数单位），且5z =，且当θ为钝角时，tan θ=_______.6.在上海高考改革方案中，要求每位高中生必须在物理、化学、生物、政治、历史、地理6门学科（3门理科学科，3门文科学科）中选择3门学科参加等级考试，小丁同学理科成绩较好，决定至少选择两门理科学科，那么小丁同学的选课方案有____种.7.设数列{}n a 的前n 项的和为n S ，若14a =，且13()n n a S n N *+=∈，则n S =______.8.在极坐标系中，过点(2,)4π且与圆2cos ρθ=相切的直线的方程为____.9.若二项式63()a x x -展开式中含2x 项的系数为52，则2l i m (1...a )nn a a →∞++++=_____.10.若行列式51sin()2cos()214x x πππ++的第1行第2列的元素1的代数余子式为-1，则实数x 的取值集合为____________.11.如图所示，已知1F ，2F 为双曲线22221(0,0)x y a b a b-=>>的两个焦点，且122F F =，若以坐标原点O 为圆心，12F F 为直径的圆与该双曲线的左支相交于A ，B 两点，且2F AB ∆为正三角形，且双曲线的实轴长为________.12.随机变量ζ的分布列如下：ζ -11P abc其中，a 、b 、c 成等差数列，若13E ζ=，则D ζ的值是________.13.已知向量a ，b 满足2a b a b === ，且()()0a c b c --=，则2b c - 的最小值为________.14.若()f x 是定义在R 上的奇函数，且对任意的实数0x ≥，总有正常数T ，使得()()f x T f x T +=+成立，则称()f x 具有“性质p ”，已知函数()g x 具有“性质p ”，且在[]0,T 上，2()g x x =；若当[],4x T T ∈-时，函数()y g x kx =-恰有8个零点，则实数k =__________.二、选择题（本题共4题，满分20分）每题只有一个正确答案，考生在答题纸的相应题号上，将所选答案的代号涂黑，选对得5分，否则一律零分。

2015届高三阶段性诊断考试(二模)数学(文)试题 Word版含答案

高三阶段性诊断考试试题文科数学一、选择题：本大题共10小题，每小题5分，共50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.已知复数z 满足()11z i +=（其中i 为虚数单位），则z 的共轭复数是A. 12i+B. 12i -C. 12i -+D. 12i --2.设{}{}21,,2,xP y y x x R Q y y x R ==-+∈==∈，则A. P Q ⊆B. Q P ⊆C. R C P Q ⊆D. R Q C P ⊆3.设命题21:32,:02x p x x q x --+<0≤-，则p 是q 的 A.充分不必要条件 B.必要不充分条件 C.充要条件 D.既不充分也不必要条件4.某工厂生产的甲、乙、丙三种型号产品的数量之比为2：3：5，现用分层抽样的方法抽取一个容量为n 的样本，其中甲种产品有20件，则n= A.50 B.100 C.150 D.2005.已知不共线向量,,,a b a b a b a b a ---+r r r r r r r r r则与的夹角是A.2πB.3π C.4π D.6π 6. ABC ∆的内角A,B,C 的对边分别为a,b,c ，若a,b,c ，成等比数列，且c=2a ，则cosC=A.4B. 4-C.34D. 34-7.设函数()()()01xx f x a ka a a -=->≠-∞+∞且在，上既是奇函数又是减函数，则()()log a g x x k =+的图象是8.一个几何体的三视图如图所示，其中正视图和侧视图是腰长为1的两个等腰直角三角形，则该几何体外接球的体积为A.B.C.2D. 3π9.已知函数()()f x x R ∈满足()()11,1f f x '=<且，则不等式()2211f g x g x <的解集为A. 10,10⎛⎫⎪⎝⎭B. ()10,10,10⎛⎫⋃+∞ ⎪⎝⎭C. 1,1010⎛⎫⎪⎝⎭D. ()10,+∞10.设双曲线()222210,0x y a b a b-=>>的右焦点为F ，过点F 做与x 轴垂直的直线交两渐近线于A,B 两点，且与双曲线在第一象限的交点为P ，设O 为坐标原点，若()4,,25OP OA OB R λμλμλμ=+=∈uu u r uu r uu u r ，则双曲线的离心率e 是A.B.C.52D.54二、填空题：本大题共5小题，每小题5分，共25分. 11.若x,y都是锐角，且1sin tan ,3x y x y ==+=则_________. 12.在边长为2的正方形ABCD 的内部任取一点M ，则满足90AMB ∠>的概率为___________（结果保留π）. 13.已知0,0a b >>，方程为22420x y x y +-+=的曲线关于直线10ax by --=对称，则2a bab+的最小值为________.14.已知抛物线24y x =上有一条长为6的动弦AB ，则AB 的中点到y 轴的最短距离是_____.15.已知数列{}n a 满足()()11,log 12,n n a a n n n N *==+≥∈.定义：使乘积12k a a a ⋅⋅⋅⋅为正整数的()k k N*∈叫做“易整数”.则在[]1,2015内所有“易整数”的和为________. 三、解答题：本大题共6小题，共75分.16. （本小题满分12分）已知向量()cos ,cos ,3sin cos ,2sin 6m x x n x x x π⎛⎫⎛⎫=+=+ ⎪ ⎪⎝⎭⎝⎭，且满足()f x m n =⋅u r r.（I ）求函数()f x 的的对称轴方程；（II ）将函数()f x 的图象向右平移6π个单位得到()g x 的图象，当[]0,x π∈时，求函数()g x 的单调递增区间.17. （本小题满分12分）如图1，在直角梯形ABCD 中，90,2,3,//A B AD BC EF AB ∠=∠===，且AE=1，M,N 分别是FC,CD 的中点.将梯形ABCD 沿EF 折起，使得1,BM =连接AD,BC,AC 得到（图2）所示几何体.（I ）证明：BC ⊥平面ABFE ；（II ）证明：AF//平面BMN.18. （本小题满分12分）已知函数()()()log 01,,2m n f x x m m a n =>≠且点在函数()f x 的图象上. （I ）若()n n n b a f a m =⋅=，当时，求数列{}n b 的前n 项和n S ；（II ）设2lg n n n c a a =⋅，若数列{}n c 是单调递增数列，求实数m 的取值范围.19. （本小题满分12分）某超市举办促销活动，凡购物满100元的顾客将获得3次模球抽奖机会，抽奖盒中放有除颜色外完全相同的红球、黄球和黑球各1个，顾客每次摸出1个球再放回，规定摸到红球奖励10元，摸到黄球奖励5元，摸到黑球无奖励.（I ）求其前2次摸球所获奖金大于10元的概率；（II ）求其3次摸球获得奖金恰为10元的概率.20. （本小题满分13分）已知椭圆()2222:10x y C a b a b +=>>1，离心率为2.（I ）求椭圆C 的方程；（II ）若过点()2,0M 的直线与椭圆C 交于A,B 两点，设P 为椭圆上一点，且满足OA OB tOP +=uu r uu u r uu u r（O 为坐标原点），当PA PB -<uu r uu r 时，求实数t 的取值范围.21. （本小题满分14分）已知函数()()()2121,ln 23f x x k x kg x x x =+--+=. （I ）若函数()g x 的图象在（1，0）处的切线l 与函数()f x 的图象相切，求实数k 的值；（II ）当0k =时，证明：()()0f x g x +>；（III ）设()()()(),h x f x g x h x '=+若有两个极值点()1212,x x x x ≠，且()()1272h x h x +<，求实数k 的取值范围.。

上海市虹口区2015届高考练习(二模)物理试题(含答案解析)

【KS5U解析】上海市虹口区2015届高考练习（二模）物理试题（含答案解析）高考真题高考模拟高中联考期中试卷期末考试月考试卷学业水平同步练习【KS5U解析】上海市虹口区2015届高考练习（二模）物理试题（含答案解析）1 关于科学家的贡献，下列叙述中正确的是（）A．牛顿创立了微积分B．法拉第最早发现了电流的磁效应C．库仑最先准确测出了电子的电量D．亚里士多德首先提出了惯性的概念【答案解析】 A：解：A、牛顿创立了微积分，故A正确；B、奥斯特最早发现了电流的磁效应，故B错误；C、密立根首次比较准确测出电子的电量，故C错误；D、牛顿首先提出了惯性的概念，故D错误；故选：A【点评】：本题考查物理学史，是常识性问题，对于物理学上重大发现、发明、著名理论要加强记忆，这也是考试内容之一．2 关于原子核的变化，下列说法中正确的是（）A．温度越高，放射性元素的半衰期越长B．天然放射现象说明原子是可以再分的C．重核在裂变过程中向外界放出巨大的核能D．α粒子散射实验说明了原子核是可以再分的【答案解析】 C：解：A、半衰期与外界因素无关，故A错误；B、天然放射现象说明原子核内部是有结构的，并不是原子可以再分的，故B错误；C、重核的裂变和轻核的聚变过程都有质量亏损，都向外界放出核能，故C正确；D、卢瑟福通过α粒子散射实验提出了原子的核式结构，故D错误；故选：C．【点评】：本题考查了原子核的知识和物理学史，象半衰期和裂变、聚变的知识都是考查的重点，要重点掌握．3 有关宇宙的理解，下列说法中正确的是（）A．质量越大的恒星寿命越长B．太阳发出的光和热来自于碳、氧等物质的燃烧C．在天空中呈现暗红色的恒星的温度比呈现白色的恒星的温度高D．由于光速有限，因此观察遥远的天体就相当于观察宇宙的过去【答案解析】 D：解：A、质量越大的恒星寿命越短，故A错误；B、太阳发出的光和热来自于在太内部阳进行着大规模的核聚变释放的能量，故B错误B、恒星的颜色是由温度决定的，温度越低，颜色越偏红，温度越高，颜色越偏蓝．故在天空中呈现暗红色的恒星的温度比呈现白色的恒星的温度低．故C错误D、由于光速有限，遥远的天体发出的光线到达我们时，我们看到的是过去的宇宙射线；故因此观察遥远的天体就等于在观察宇宙的过去，故D正确故选：D【点评】：本题考查学生对恒星的理解，本部分属于记忆型的内容，要求在平时的学习中要对积累相关的知识，做到处处留心．4 用光照射某种金属，有光电子从金属表面逸出．如果换一种频率更大、强度较弱的光照射该金属，则（）A．单位时间内逸出的光电子数减少，光电子的最大初动能减小B．单位时间内逸出的光电子数增大，光电子的最大初动能减小C．单位时间内逸出的光电子数减少，光电子的最大初动能增大D．单位时间内逸出的光电子数增大，光电子的最大初动能增大【答案解析】 C：解：根据光电效应方程Ekm=hγ﹣W0得，光强度不影响光电子的最大初动能，光电子的最大初动能与入射光的频率有关；光电效应的条件是入射光的频率大于极限频率，与光的强度无关；换一种频率更大的光，则光电子的最大初动能增大；入射光的强度影响单位时间内发出光电子的数目，光的强度减弱，单位时间内发出光电子数目减少．故ABD错误，C正确．故选：C．【点评】：解决本题的关键掌握光电效应的条件，以及知道光的强度不影响能否发生光电效应，只影响单位时间内发出光电子的数目．5 下列说法中，正确的是（）A．只有热传递才可以改变物体的内能B．气体温度越高，每个分子运动的速率一定越大C．布朗运动是指在显微镜下观察到的液体分子的无规则运动D．热量不可能由低温物体传给高温物体而不发生其他变化【答案解析】 D：解：A、改变物体内能的方式有做功和热传递，故A错误；B、温度是分子平均动能的标志，而不是每个分子动能的标志，高温物体也有速率小的分子，故B错误；C、布朗运动的现象是固体微粒的无规则运动，反映的是液体分子的无规则运动，故C错误；D、热量不能自发的从低温物体传给高温物体，但在引起其它变化的情况下可以由低温物体传给高温物体，故D正确；故选：D．【点评】：该题的关键是掌握布朗运动的现象和实质，此考点为考察热点，要注意掌握牢固．6 伽利略研究变速运动规律时做了著名的“斜面实验”：他测量了铜球在较小倾角斜面上运动的位移和时间，发现位移与时间的平方成正比，增大斜面倾角，该规律仍然成立．于是，他外推到倾角为90°的情况，得出结论（）A．自由落体运动是一种匀变速直线运动B．力是使物体产生加速度的原因C．力不是维持物体运动的原因D．物体具有保持原来运动状态的惯性【答案解析】 A：解：铜球在较小倾角斜面上的运动情况，发现铜球做的是匀变速直线运动，且铜球加速度随斜面倾角的增大而增大，倾角最大的情况就是90°时，这时物体做自由落体运动，由此得出的结论是自由落体运动是一种匀变速直线运动．故选：A【点评】：该题属于实验推论题，要求同学们正确理解科学家的基本观点和佐证实验，该题难度不大，属于基础题．7 科学家在研究某两个重离子结合成超重元素的反应时，发现生成超重元素的原子核经过6次α衰变后的产物是．则元素的原子序数和质量数分别为（）A． 112、265 B． 112、277 C． 124、259 D． 124、265【答案解析】 B：解：每经过一次α衰变质量数少4，质子数少2，经过6次α衰变质量数减少24，质子数减少12，超重元素的原子序数和质量数分别，100+12=112，253+24=277，故B正确，ACD错误．故选：B．【点评】：本题考查了原子核的组成以及质量数和电荷数守恒在衰变方程中的应用，对于这些基础知识，注意平时加强理解与练习．8 （多选题）一物体在四个共点力作用下做匀速直线运动．若突然撤去一个沿运动方向的力，其余三个力保持不变，则物体做（）A．匀速圆周运动 B．匀加速直线运动C．类平抛运动 D．匀减速直线运动【答案解析】 BCD：解：A、其余三个力的合力恒定，而匀速圆周运动合力一直指向圆心，是变力，所以物体不可能做匀速圆周运动．故A错误；B、C、D、有一个作匀速直线运动的物体受到四个力的作用，这四个力一定是平衡力，如果其中的一个力突然消失，剩余的三个力的合力与撤去的力等值、反向、共线，这个合力恒定不变，故做匀变速运动．若物体的速度方向与此合力方向相同，则物体将匀加速直线运动；若物体的速度方向与此合力方向相反，则物体将匀减速直线运动；若物体的速度方向与此合力方向垂直，则物体将类平抛运动，故B正确，C正确，D正确．故选：BCD【点评】：本题考查了曲线运动的条件以及共点力平衡的知识，关键根据平衡得到其余三个力的合力恒定，然后结合曲线运动的条件分析．9 电视台体育频道讲解棋局的节目中通常有一个竖直放置的棋盘．该棋盘具有磁性，每个棋子都可视为能被棋盘吸引的小磁体．对于静止在棋盘上的棋子，下列说法中正确的是（） A．棋盘对棋子施加三个力的作用B．磁力越大，棋子所受的摩擦力也越大C．棋盘对棋子总的作用力比棋子的重力大D．只要磁力足够大，即使棋盘光滑，棋子也能静止在棋盘上【答案解析】 A：解：A、小棋子受到重力G、棋盘面的吸引力F、弹力N和静摩擦力f，那么棋盘对棋子施加三个力的作用，故A正确．B、棋盘对棋子的吸引力与棋盘面对棋子的弹力平衡；而静摩擦力与棋子的重力平衡，与磁力大小无关，故B错误．C、由A选项分析可知，棋盘对棋子总的作用力与棋子的重力相平衡，故C错误．D、当G＞fm=μN=μF时，棋子将下滑，故D错误．故选：A．【点评】：掌握平衡力时应注意其关键点：二力是作用在同一个物体上的，明确这一点即可与作用力与反作用力进行区分．10 将一只苹果水平抛出，苹果在空中依次飞过三个完全相同的窗户1、2、3，图中曲线为苹果在空中运行的轨迹，不计空气阻力．下列说法中正确的是（）A．苹果通过第1个窗户所用的时间最短B．苹果通过第1个窗户的过程中，重力做功最多C．苹果通过第3个窗户的过程中，重力的平均功率最大D．苹果通过第3个窗户的过程中，竖直方向的平均速度最小【答案解析】 C：解：A、平抛运动在竖直方向上做自由落体运动，速度越来越快，可知通过相同竖直位移所用的时间越来越短，所以通过第3个窗户时间最短，故A错误．B、通过三个窗户下降的高度相同，则重力做功相同，故B错误．C、通过第3个窗户时间最短，根据P=知，重力的平均功率最大，故C正确．D、苹果通过第3个窗户的过程中，由于时间最少，则竖直方向的平均速度最大，故D错误．故选：C．【点评】：解决本题的关键知道重力做功与路径无关，与首末位置的高度差有关，知道平均功率等于重力做功与时间的比值，基础题．11 用于火灾报警的离子烟雾传感器如图所示．在网罩Ⅰ内有电极Ⅱ和Ⅲ，a、b两端接电源，Ⅳ是一小块放射性同位素镅241，它能放射出一种很容易使空气电离的粒子．正常情况下镅241放射出的粒子使两个极板间的空气电离，在a、b间形成较强的电流；发生火灾时，大量烟雾进入网罩Ⅰ内，烟尘颗粒吸收其中的带电粒子，导致电流发生变化，从而报警．下列说法中正确的是（）A．镅241射出的是α粒子，有烟雾时电流增强B．镅241射出的是α粒子，有烟雾时电流减弱C．镅241射出的是β粒子，有烟雾时电流增强D．镅241射出的是β粒子，有烟雾时电流减弱【答案解析】 B：解：放射性同位素镅241，它能放射出一种很容易使气体电离的粒子，知该粒子是α粒子．发生火灾时，烟雾进入网罩内，烟尘颗粒吸收空气中的离子和镅发出的粒子，使得电流减弱．故B正确，A、C、D错误．故选：B．【点评】：在三种射线中，α射线的电离能力最强，烟雾进入网罩内，烟尘颗粒吸收空气中的离子和镅发出的粒子，使得烟雾电流减弱．12 如图所示，虚线AB和CD分别为椭圆的长轴和短轴，相交于O点，两个等量异号点电荷分别位于椭圆的两个焦点M、N上．下列说法中正确的是（）A． O点的电场强度为零B． A、B两点的电场强度相同C． C点的电势高于D点的电势D．将电荷+q沿C、D连线从C移到D的过程中，电势能先减少后增加【答案解析】 B：解：A、根据等量异种电荷电场线的特点可知，O点场强的方向向右，不是0．故A错误B、根据等量异种电荷电场线的特点可知，A点场强的方向向左，B点场强的方向向左，两点场强的大小相等．故B正确．C、根据等量异种电荷电场线、等势面分布特点知，OCD在同一条等势线上，所以D点电势等于C点电势．故C错误．D、根据等量异种电荷电场线、等势面分布特点知，OCD在同一条等势线上，所以D点电势等于C点电势，试探电荷从O处移到C处电场力不做功．电势能始终不变．故D错误故选：B【点评】：这类问题要巧妙利用电场线、等势面分布对称性的特点，再根据电场线方向判断电势高低，电场线的疏密判断场强的大小13 一列简谐横波沿x轴正方向传播，图（甲）是t=3s时的波形图，图（乙）是波中某质点P的振动图象，下列判断中正确的是（）A．质点P在t=3s 时沿y轴负方向振动B．质点P在t=3s时的速度为零C．质点P的平衡位置坐标可能是x=4cmD．该简谐波的波速为1m/s【答案解析】 C：解：A、由图乙知，t=3s时图象切线的斜率为正值，说明质点P此时正沿y轴正方向振动，故A错误．B、由图乙知，t=3s时质点P的位移为0，正通过平衡位置，速度最大，故B错误．C、由图甲知，简谐横波沿x轴正方向传播，x=4cm处的质点正通过平衡位置向上，与质点P 在t=3s时的状态相同，所以质点P的平衡位置坐标可能是x=4cm，故C正确．D、由甲读出波长λ=4cm，由图乙读出周期T=4s，则该波的波速v==1cm/s．故D错误．故选：C．【点评】：本题考查基本的读图能力，由波动图象读出波长，由波的传播方向判断质点的振动方向，由振动图象读出周期，判断质点的振动方向等等都是基本功，要加强训练，熟练掌握．14 如图所示，一圆形闭合小铜环从高处由静止开始下落，穿过一根竖直悬挂的、质量为m 的条形磁铁，铜环的中心轴线与条形磁铁的中心轴线始终保持重合．则细绳中弹力F随时间t的变化关系图象可能是（）A BCD 【答案解析】 B：解：铜环闭合，铜环在下落过程中，穿过铜环的磁通量不断变化，铜环中产生感应电流；由楞次定律可知，感应电流总是阻碍磁体间的相对运动，当铜环在磁铁上方时，感应电流阻碍铜环靠近磁铁，给铜环一个向上的安培力，因此拉力大于重力；当铜环位于磁铁下方时，铜环要远离磁铁，感应电流阻碍铜环的远离对铜环施加一个向上的安培力，则拉力大于重力；当铜环处于磁铁中央时，磁通量不变，则没有感应电流，没有安培阻力，因此拉力等于重力，故ACD错误，B正确．故选：B．【点评】：该题考查了楞次定律的应用，应全面、正确理解楞次定律中“阻碍”的含义．该题也可以由楞次定律的规范化的步骤来解答，比较麻烦．15 如图所示，直线A为某电源的U﹣I图线，曲线B为标识不清的小灯泡L1的U﹣I图线，将L1与该电源组成闭合电路时，L1恰好能正常发光．若将相同材料制成的标有“3V，20W”的灯泡L2与该电源组成闭合电路，下列说法中正确的是（）A．电源的内阻为ΩB．把灯泡L1换成L2，L2可能正常发光C．把灯泡L1换成L2，电源的输出功率可能相等D．把灯泡L1换成L2，电源的输出功率一定变小【答案解析】 C：解：A、由图读出电源的电动势为 E=4V，图线A的斜率大小表示电源的内阻，则 r=Ω=0.5Ω，故A错误；BCD、灯泡与电源连接时，A、B两图线的交点表示灯泡的工作状态，则知其电压U=3V，I=2A，则灯泡L1的额定电压为3V，功率为=UI=6W．把灯泡L1换成“3V，20W”的灯泡L2，不能正常发光，而由P=知：灯泡L2的正常工作时的电阻为 R2===0.45Ω灯泡L1的电阻为R1==Ω=1.5Ω，则知灯泡L2的电阻更接近电源的内阻，电源的输出功率可能相等，且电源的输出功率将变大，故BD错误，C正确；故选：C．【点评】：解决这类问题的关键在于从数学角度理解图象的物理意义，抓住图象的斜率、面积、截距、交点等方面进行分析，更加全面地读出图象的物理内涵．16 将一小球以20m/s的初速度竖直上抛，经3.5s落回原处．已知空气阻力的大小与速率成正比，则小球落回原处的速度大小为（）A． 5m/s B． 10m/s C． 12.5m/s D． 15m/s【答案解析】 D：解：在上升过程中物体的加速度为a1=，设上升过程所用时间为t在时间他内上升的高度为h=，下降过程中加速度为，下降过程所需时间为3.5﹣t下降的高度h=，落地时的速度为v=，联立解得v=15m/s．故ABC错误，D正确．故选：D．【点评】：本题主要考查了利用微元法表示位移和速度，抓住题目中给的信息即可顺利解决．17 （多选题）英国物理学家托马斯•杨巧妙地解决了相干光源问题，第一次在实验室观察到了光的干涉现象．图为实验装置简图，M为竖直线状光源，N和O均为有狭缝的遮光屏，P 为像屏．现有四种刻有不同狭缝的遮光屏，实验时正确的选择是（）A． N应选用遮光屏1 B． N应选用遮光屏3C． O应选用遮光屏2 D． O应选用遮光屏4【答案解析】 AC：解：单缝衍射条纹宽度是中央亮纹最宽；双缝干涉的图线是明暗相间的条纹，条纹间距等宽；是双缝干涉；因此要先通过单缝后找到双缝上，形成相干光源，则N应选用遮光屏1，O应选用遮光屏2；故AC正确，BD错误故选：AC．【点评】：本题关键明确双缝干涉条纹和单缝衍射条纹的区别，注意确定遮光屏与条纹的方向关系是解题的关键．18 （多选题）将横截面积为S的圆柱形气缸固定在铁架台上，内有可自由移动的轻质活塞，活塞通过轻杆与重物m相连，将一团燃烧的轻质酒精棉球经阀门K放置于活塞上，棉球熄灭时立即关闭阀门K，此时活塞距离气缸底部为L．之后，缸内气体冷却至环境温度时，重物上升高度为．已知环境温度恒为27℃，外界大气压为p0，缸内气体可以看作是理想气体，则（）A．重物离开地面稳定后，气体压强可能大于p0B．重物离开地面稳定后，气体压强一定小于p0C．酒精棉球熄灭的瞬间，缸内气体的温度t可能等于120℃D．酒精棉球熄灭的瞬间，缸内气体的温度t可能等于140℃【答案解析】 BD：解：酒精棉球熄灭时，活塞受到封闭气体向下的压力，大气压向上的支持力，由平衡得：P1S=P0S解得：P1=P0此时体积为：V1=LS，温度为：T1=273+t重物被吸起稳定后，活塞受绳子得拉力，封闭气体向下得压力和大气压向上得支持力，由平衡得：P2S+mg=P0S解得：P2=P0﹣＜P0此时体积为：V2=0.75LS，温度为T2=273+27K=300K有理想气体状态方程得：t=140℃故选：BD【点评】：本题得关键是以活塞为研究对象，受力分析利用平衡求出初末状态压强，然后利用理想气体状态方程列式即可求解19 （多选题）如图所示，一位同学玩飞镖游戏．圆盘最上端有一P点，飞镖抛出时与P 等高，且距离P点为L．当飞镖以初速度v0垂直盘面瞄准P点抛出的同时，圆盘以经过盘心O点的水平轴在竖直平面内匀速转动．忽略空气阻力，重力加速度为g，若飞镖恰好击中P点，则（）A．飞镖击中P点所需的时间为B．圆盘的半径可能为C．圆盘转动角速度的最小值为D． P点随圆盘转动的线速度可能为【答案解析】 AD：解：A、飞镖水平抛出做平抛运动，在水平方向做匀速直线运动，因此t=，故A正确．B、飞镖击中P点时，P恰好在最下方，则2r=，解得圆盘的半径 r=，故B错误．C、飞镖击中P点，则P点转过的角度满足θ=ωt=π+2kπ（k=0，1，2…）故ω==，则圆盘转动角速度的最小值为．故C错误．D、P点随圆盘转动的线速度为v=ωr=•=当k=2时，v=．故D正确．故选：AD．【点评】：本题关键知道恰好击中P点，说明P点正好在最低点，利用匀速圆周运动的周期性和平抛运动规律联立求解．20 (多选题)如图（甲）所示，相距为2L的光滑平行金属导轨水平放置，右侧接有定值电阻R，导轨电阻忽略不计，OO′的左侧存在垂直于导轨平面向下、磁感应强度为B的匀强磁场．在OO′左侧L处垂直导轨放置一质量为m、电阻为0.5R的金属杆ab，ab在恒力F的作用下由静止开始向右运动3L的距离，其速度与位移的变化关系如图（乙）所示．下列判断中正确的是（）A． ab即将离开磁场时，安培力的大小为B．整个运动的过程中，通过电阻R上的电量为C． ab即将离开磁场时，加速度的大小为D．整个过程中，电阻R上产生的焦耳热为m（v22﹣3v12）【答案解析】 BCD：解：A、ab即将离开时，速度为v1，电动势E=2BLv1，电流I=；安培力F=2BIL=；故A错误；B、整个过程中，磁通量的变化量为△Φ=2BL2；产生的电量q===；故B正确；C、ab杆在离开磁场前瞬间，水平方向上受安培力F安和外力F作用，设加速度为a，则 F安=BILI=a=联立解得：a=﹣；故C正确；D、ab杆在位移L到3L的过程中，由动能定理得：F（3L﹣L）=ab杆在磁场中发生L位移过程中，恒力F做的功等于ab杆增加的动能和回路产生的电能（即电阻R上产生的电热Q1），由能量守恒定律得：FL=联立解得：Q1=；故D正确；故选：BCD．【点评】：本题考查导体切割磁感线中的能量转化规律；要能够把法拉第电磁感应定律与电路知识结合运用．电磁感应中动力学问题离不开受力分析和运动过程分析．关于电磁感应中能量问题我们要从功能关系角度出发研究．21 如图所示，R为一含有的放射源，它能放出α、β、γ三种射线，变为．LL′为一张厚纸板，MN为涂有荧光物质的光屏，虚线框内存在平行于边界ab的匀强电场．若射线正对光屏的中心O点射出，在光屏上只观察到O、P两个亮点，则打在O点的是射线，虚线框内匀强电场的方向（选填“由a指向b”或“由b指向a”）．【答案解析】γ；由b指向a．：解：α、β、γ三种射线中γ射线不带电，在电场中不偏转，所以打在O点的是γ射线．一张纸就能挡住α射线，所以打在P点的是β射线，而β粒子带负电，要使β射线向下偏转，虚线框内匀强电场的方向应由b指向a．故答案为：γ；由b指向a．【点评】：本题考查了三种射线的特点，知道α、β和γ电离本领依次减弱，贯穿本领依次增强，γ射线不带电，特别要记住一张纸就能挡住的是α射线．22 如图所示，气球吊着A、B两个重物以速度v匀速上升，已知A与气球的总质量为m1，B的质量为m2，且m1＞m2．某时刻A、B间细线断裂，当气球的速度增大为2v时，B的速度大小为，方向．（不计空气阻力）【答案解析】，竖直向下：解：规定向上为正方向，根据动量守恒定律得：（m1+m2）v=m1•2v+m2v′，解得：，因为m1＞m2．所以v′为负值，可知方向竖直向下，大小为：．故答案为：，竖直向下【点评】：解决本题的关键知道气球、A、B组成的系统竖直方向上合力为零，动量守恒，运用动量守恒定律解题时注意公式的矢量性．23 北斗一号卫星系统的三颗卫星均定位在距离地面36000km的地球同步轨道上，而GPS系统中的24颗卫星距离地面的高度均为20000km，已知地球半径为6400km．则北斗一号卫星线速度的大小 GPS卫星线速度的大小（选填“大于”、“小于”或“等于”）； GPS卫星加速度的大小约为北斗一号卫星的倍（取2位有效数字）．【答案解析】小于，2.6：解：根据万有引力提供向心力得：，解得：v=，a=，因为北斗一号卫星的半径大于GPS卫星的半径，所以北斗一号卫星线速度的大小小于GPS卫星线速度的大小，．故答案为：小于，2.6【点评】：解决本题的关键掌握万有引力提供向心力公式：，特别注意卫星的轨道半径等于卫星离地面的高度加上地球的半径，难度不大，属于基础题．24 图示为一列沿x轴负方向传播的机械波，实线和虚线分别为t时刻和t+△t时刻的波形，B和C是横坐标分别为d和3d的两个质点．则t时刻质点B的振动方向为y轴正方向，该波的波速为（其中k=1，2，3…）．【答案解析】 y轴正方向，，：解：波沿x轴负方向传播，运用波形平移法分析知，t时刻质点B的振动方向为y轴正方向．由图知，该波的波长为λ=3d（其中k=1，2，3…）．根据波的周期性，可知波在△t时间传播的距离为△x=kλ+d=3kd+，则波速为 v==，（其中k=1，2，3…）．故答案为：y轴正方向，，（其中k=1，2，3…）．【点评】：本题是多解问题，关键要理解波的周期性及重复性，是会通过波形微平移法得到波的传播距离的通项，最后求解传播速度通项．25 如图所示，在竖直平面内有两根质量相等的均匀细杆A和C，长度分别为60cm和40cm，它们的底端相抵于地面上的B点，另一端分别搁置于竖直墙面上，墙面间距为80cm，不计一切摩擦．系统平衡时两杆与地面的夹角分别为α和β，两侧墙面所受压力的大小分别为FA和FC，则FAFC（选填“大于”、“小于”或“等于”），夹角β=．【答案解析】等于；37°：解：对整体分析，整体处于平衡状态，整体在水平方向上受到两侧墙壁的弹力，可知FA=FB，对AB分析，A受到重力、墙壁的弹力、地面对它的支持力以及BC对AB的作用力，同样度BC分析，受重力、墙壁的弹力、AB对BC的作用力，和地面的支持力，因为两杆重力相等，墙壁的作用力相等，根据平衡知，α=β，根据xABcosα+xBCcosβ=d，可知（60+40）cosβ=80，解得β=37°．故答案为：等于；37°【点评】：解决本题的关键能够正确地受力分析，运用共点力平衡进行求解，掌握整体法和。

上海市虹口区2015届高三二模英语试题Word版含答案

虹口区2015年英语学科高考练习题2015.4第I卷(共103分)I. Listening（略）II. Grammar and VocabularySection ADirections: After reading the passages below, fill in the blanks to make the passages coherent arid grammatically con-ect- For the blanks with a given word, fill in each blank with the proper formofthe given word; for the other blanks, use one ward that best fits each blank.(A)How I Turned to Be Optimistic(乐观的)I began to grow up that winter night when my parents and I were returning from my aunt's House, and my mother said that we (25) (leave) for America soon. We were on the bus then. Iwas crying, and some people on the bus were turning around to look at me. I remember I could not bear the thought of never hearing again the radio program for school children to (26) I listened every morning.I do not remember myself (27) (cry) for this reason again. In fact I think I cried very little when I was saying goodbye to my friends and relatives. When we were leaving I thought aboutall the places I was going to see. The country I was leaving never to come back was hardly in my head then.The four years that followed taught me the importance of optimism, but (28) idea did not come to me at once. For the first two years in New York I was really lost. I did not quiteknow what I was or what I should be. Mother remarried, and things became even (29) (complex) for me. Some time passed before my stepfather and I got used to each orher. However,。

2015年静安(青浦、宝山)区高考数学二模试卷含答案

2015年静安区高考数学二模含答案数学试卷（理科） 2015.04.（满分150分，考试时间120分钟）一．填空题（本大题满分56分）本大题共有14题，考生应在答题纸相应编号的空格内直接填写结果，每个空格填对得4分，否则一律得零分．1．已知抛物线22y px =的准线方程是2x =-，则p = ．2．已知扇形的圆心角是1弧度，半径为5cm ，则此扇形的弧长为 cm ． 3．复数34ii-（i 为虚数单位）的模为． 4．函数221y x x =+-的值域为． 5．若2021310x y -⎛⎫⎛⎫⎛⎫=⎪⎪ ⎪-⎝⎭⎝⎭⎝⎭，则x y += ．6．在921x x ⎛⎫- ⎪⎝⎭的展开式中，31x 的系数是．7．方程)cos (lg )sin 3(lg x x -=的解集为．8．射击比赛每人射2次，约定全部不中得0分，只中一弹得10分，中两弹得15分，某人每次射击的命中率均为45，则他得分的数学期望是分． 9．过圆0422=+-+my x y x 上一点)1,1(P 的切线方程为． 10．在极坐标系中，点P (2，6π11)到直线πsin 16ρθ⎛⎫-= ⎪⎝⎭的距离等于． 11．把一个大金属球表面涂漆，共需油漆2.4公斤．若把这个大金属球熔化制成64个大小都相同的小金属球，不计损耗，将这些小金属球表面都涂漆，需要用漆公斤．12．设12,e e 是平面内两个不共线的向量，12(1)AB a e e =-+，122AC be e =-，0,0a b >>.若,,A B C 三点共线，则12a b+的最小值是．13．设等差数列{}n a 的前n 项和为n A ，等比数列{}n b 的前n 项和为n B ，若33a b =，44a b =，且53427A A B B -=-，则5353a ab b +=+． 14．已知：当0x >时，不等式11kx b x≥++恒成立，当且仅当13x =时取等号，则k = ．二、选择题（本大题满分20分）本大题共有4题，每题有且只有一个正确答案，考生应在答案纸的相应编号上，填上正确的答案，选对得5分，否则一律得零分.15.如图，ABCDEF 是正六边形，下列等式成立的是（）（A ）0AE FC ⋅= （B ）0AE DF ⋅> （C ）FC FD FB =+ （D ）0FD FB ⋅<16.已知偶函数)(x f 的定义域为R ，则下列函数中为奇函数的是（）（A ）)](sin[x f （B ）)(sin x f x ⋅（C ）)(sin )(x f x f ⋅（D ）2)](sin [x f 17. 如图所示是一个循环结构的算法，下列说法不正确的是（）（A ）①是循环变量初始化，循环就要开始（B ）②为循环体（C ）③是判断是否继续循环的终止条件（D ）输出的S 值为2，4，6，8，10，12，14，16，18.18.定义：最高次项的系数为1的多项式1110n n n p (x)x a x a x a --=++鬃?+（*∈n N ）的其余系数(0,1,,1)=⋅⋅⋅-i a i n 均是整数，则方程()0=p x 的根叫代数整数．下列各数不是代数整数的是（）（A ）22 （B ）3 （C ）152+ （D ）1322i-+三、解答题（本大题满分74分）本大题共5题，解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤.19. (本题满分12分)本题共有2个小题，第1小题满分4分，第2小题满分8分．FA BCE D如图，在直三棱柱111C B A ABC -中，已知21===AB BC AA ，AB ⊥BC ．（1）求四棱锥111A BCC B -错误！未指定书签。

上海市虹口区高考数学二模试题文

（第11题图）虹口区2015年高考练习题数学(文科)试卷一、填空题（本大题满分56分）本大题共14题，只要求在答题纸相应题号的空格内直接填写结果，每个空格填对得4分，否则一律得零分.1．计算：201511i i +=+____________.（i 是虚数单位）2. 已知函数132,(0)(),((3))_______.,(0)x x f x f f x x ⎧≤⎪=-=⎨⎪>⎩则3．函数1()ln(1)(0)f x x x =+>的反函数1()f x -=___________.4．已知正实数,x y 满足31,x y +=则xy 的最大值为___________.5．已知复数3sin cos z i θθ=+（i 是虚数单位），且z 则当θ为钝角时，tan _____.θ=6. 在上海高考改革方案中，要求每位高中生必须在物理、化学、生物、政治、历史、地理6门学科(3门理科学科、3门文科学科)中选择3门学科参加等级考试. 小丁同学理科成绩较好，决定至少选择两门理科学科，那么小丁同学的选科方案有__________种（结果用数值表示）.7.设数列{}n a 前n 项的和为,n S 若114,3(),n n a a S n N *+==∈且则_______.n S =8. 已知抛物线22(0)y px p =>的焦点在圆22(1)4x y -+=上，则p =________.9.6225(lim(1)2n n x x a a a →∞++++=若二项式展开式中含项的系数为，则________.10 .若行列式51in()0cos()24s x x ππ++的第1行第2列的元素1的代数余子式为1-，则实数x 的取值集合为_________.11.如图所示，已知2212221(0,0)x y F F a b a b -=>>、为双曲线的两个焦点，且122,F F =若以坐标原点O 为圆心，12F F 为直径的圆与该双曲线的左支相交于A B 、两点，且2F AB ∆为正三角形，则双曲线的实轴长为_________.12. 设二元一次不等式组2190802140x y x y x y +-≥⎧⎪-+≥⎨⎪+-≤⎩所表示的平面区域为,M 若函数xy a =(0,1)a a >≠且的图像经过区域,M 则实数a 的取值范围为_________.13. 已知直线1:125150l x y -+=和2:2,l x =-28P y x =点为抛物线上的动点，则1P l 点到直线2l 和直线的距离之和的最小值为_________.抛物线的准线即为x=-2,所以根据抛物线的定义，点P 到x=-2的距离即为点P 到焦点（2,0）的距离，所以1P l 点到直线2l 和直线的距离之的和为1P l 点到直线和到焦点的距离之和，这个距离的最小值是 3.d ==14.已知向量,a b 满足2,a b a b ==⋅=且()()0,a c b c -⋅-=则2b c-的最小值为________.根据2,a b a b ==⋅=cos ,3a b π<⋅>=不防设(2,0),(1,3),(,),a b c xy ===所以22()()(2,)(1,320,a cbc x y x y x x y -⋅-=-⋅-=-++=即223()(1,2x y -+-=则2b c-的最小值为点到点3(,22距离减去1，即1-二、选择题（本大题共4题，满分20分）每题有且只有一个正确答案，考生应在答题纸的相应题号上，将所选答案的代号涂黑，选对得 5分，否则一律零分.{}23,0,12,2x U R A x B x x x ⎧+⎫==>=-<⎨⎬-⎩⎭15.设全集已知()()U A B ⋂=则ð（A ）3,12⎛⎫-- ⎪⎝⎭ （B ）(]1,2- （C ）(]2,3 （D ）[)2,3 16.设,a R ∈则1a =-“”是“函数()(2)(0,)f x ax x =-+∞在上单调递增”的 ( )（第17题图）俯视图左视图主视图（A ）充要条件（B ）既不充分也不必要条件（C ）充分不必要条件（D 17. 一个四棱锥的三视图如图所示，则此四棱锥的（）（A ） 24 （B ）16 （C ） 12 （D ）818．设函数()()()(),()(2),f x x R f x f x f x f x ∈-==-满足且当[]20,1().x f x x ∈=时，又函数()sin(),g x x π=则函数[]()()()1,3h x g x f x =--在区间上零点的个数为（）（A ） 6 （B ）7 （C ） 8 （D ）9由函数()()()(),()f x x R f x f x f x ∈-=满足则为偶函数，()(2)=(2),f x f x f x =--函数的周期为2，所以在同一坐标系下画出两个函数的图象，[]()()1,3y g x y f x ==-和在区间的交点有6个，[]()()()1,3h x g x f x =--在区间上零点的个数为6个三、解答题（本大题共5题，满分74分）解答下列各题必须在答题纸的规定区域内写出必要的步骤.19．(本题满分12分) 本题共2个小题，第1小题5分，第2小题7分. 已知函数()log (01)a f x b x a a =+>≠且的图像经过点（8，2）和(1,1).-(1) 求函数()f x 的解析式；(2) 令()2(1)(),()g x f x f x g x =+-求的最小值及取最小值时x 的值.20．(本题满分14分) 本题共2个小题，第1小题6分，第2小题8分.（第21题图）θBCPNMA (第20题图）D 1C 1B 1BCDA 1A在如图所示的直四棱柱1111ABCD A BC D -中，底面ABCD是边长为2的菱形，且60,BAD ∠=︒1 4.AA =(1) 求直四棱柱1111ABCD A BC D -的体积；（2）求异面直线11AD BA 与所成角的大小．21．(本题满分14分) 本题共2个小题，第1小题6分，第2小题8分.如图，经过村庄A 有两条夹角为60︒的公路AB AC 、，根据规划拟在两条公路之间的区域内建一工厂P ，分别在两条公路边上建两个仓库M N 、（异于村庄A ），要求2PM PN MN === （单位：千米）. 记.AMN θ∠=（1）将AN AM 、θ用含的关系式表示出来；（2）如何设计（即AN AM 、为多长时），使得工厂产生的噪声对居民的影响最小（即工厂与村庄的距离AP 最大）？22. (本题满分16分) 本题共3个小题，第1小题5分，第2小题5分，第3小题6分.已知圆1F ：22(1)8x y ++=，点2F (1, 0)，点Q 在圆1F 上运动，2QF 的垂直平分线交1QF 于点P .(1) 求动点P 的轨迹C 的方程；(2) 设M N 、分别是曲线C 上的两个不同点，且点M 在第一象限，点N 在第三象限，若122OM ON OF +=, O 为坐标原点，求直线MN 的斜率；(3)过点1(0,)3S -的动直线l 交曲线C 于A B 、两点，求证：以AB 为直径的圆恒过定点(0,1).T23. (本题满分18分) 本题共3个小题，每小题6分.(第20题图）D 1C 1B 1BCDA 1A设各项均为正数的数列{}n a 的前n 项和为,nS且满足：211,(1)().n n a S a n N *==+∈4（1）求数列{}n a 的通项公式；（2）设1121(),lim(2n n n n n n n a ab n N b b b n a a *+→∞+=+∈+++-试求)的值；（3）是否存在大于2的正整数,m k 、使得12300?m m m m k a a a a +++++++=若存在，求出所有符合条件的m k 、；若不存在，请说明理由.虹口区2015年高考练习题数学(文科)参考答案与评分标准 2015年4月一、填空题（本大题共14题,每题4分,满分56分）1．i - 2．12 3．1(0)1xx e >- 4． 1125． 1- 6. 10 7. 4n8． 69. 23 10．(21),()k k Z π+∈ 111 12. []2,913． 3 14．1二、选择题（本大题共4题，每题5分，满分20分）15. B 16. C 17. D 18. A 三、解答题（本大题共5题，满分74分）19．(本题满分12分) 本题共2个小题，第1小题5分，第2小题7分.解：（1）由已知，得log 82,log 11a a b b +=⎧⎨+=-⎩解得2.1a b =⎧⎨=-⎩ ……3分故2()log 1.f x x =- ……5分（2）由于[]22()2(1)()2log (1)1(log 1)g x f x f x x x =+-=+---222(1)1log 1log (2)1(0)x x x x x +=-=++-> ……8分故221()log (2)1log (22)1 1.g x x x =++-≥+-= ……10分于是，当1x =时，()g x 取得最小值1. ……12分20．(本题满分14分) 本题共2个小题，第1小题6分，第2小题8（第21题图）θBCPNMA分.解：(1) 因菱形ABCD的面积为2sin60AB ⋅︒= ……2分故直四棱柱1111ABCD A BC D -的体积为：14ABCD S AA ⋅==底面 ……6分(2) 连接111BC AC 、，易知11//BC AD ，故11ABC ∠等于异面直线11AD BA 与所成角. ……8分由已知，可得1111A B BC AC=== ……10分则在11A BC ∆中，由余弦定理，得222111111117cos .210A B BC AC A BC A B BC +-∠==⋅ ……12分故异面直线11AD BA 与所成角的大小为7cos.10arc ……14分21．(本题满分14分) 本题共2个小题，第1小题6分，第2小题8分. 解：（1）在AMN ∆中，由正弦定理，得sin sin(120)sin 60AN AM MN θθ===︒-︒ ……2分于是4,60)(0120).3AN AM θθθ==+︒︒<<︒ (6)分（2）在ANP ∆中，由余弦定理，得2222222cos 44222cos(180)331616820sin 4cos 2cos 2)33331620sin(230)(0120).1133AP AN NP AN NP ANPθθθθθθθθθθ=+-⋅∠⎛⎫=+-⋅⋅⋅︒- ⎪⎝⎭=++⋅=-+=-︒+︒<<︒⋯⋯分故当2max 2309060()12.AP θθ-︒=︒=︒=，即时，此时 2.AN AM == 于是，设计2()AN AM ==千米时，工厂与村庄的距离AP最大，为；工厂产生的噪声对居民的影响最小. ……14分22. (本题满分16分) 本题共3个小题，第1小题5分，第2小题5分，第3小题6分. 解：(1) 因为2QF 的垂直平分线交1QF 于点P . 所以2PF PQ =，从而1211122,PF PF PF PQ FQ F F +=+==>=所以，动点P 的轨迹C 是以点12F F 、为焦点的椭圆. ……3分设椭圆的方程为12222=+b y a x ，则22,222==c a ，1222=-=c a b ，故动点P 的轨迹C 的方程为 2212x y += ……5分(2) 设1122(,),(,)M a b N a b 1122(0,0,0,0)a b a b >><<，则2222112222,22a b a b +=+= ①因为122OM ON OF+=，则121222,20a a b b +=-+= ② 由①、② 解得112215,,24a b a b ===-= ……8分所以直线MN 的斜率MNk 212114b b a a -==- . ……10分(3)设直线l 的方程为1,3y kx =-则由221312y kx x y ⎧=-⎪⎪⎨⎪+=⎪⎩，得229(21)12160,k x kx +--= 由题意知，点1(0,)3S -在椭圆C 的内部，所以直线l 与椭圆C 必有两个交点，设11(,)A x y 、22(,)B x y ，则121222416,.3(21)9(21)k x x x x k k +==-++ ……12分假设在y 轴上存在定点(0,)T m 满足题设，则1122(,),(,),TA x y m TB x y m =-=-因为以AB 为直径的圆恒过点T , 所以1122(,)(,)0,T A T B x y m x y m ⋅=-⋅-=即1212()()0()x x y m y m +--=* ……14分因为112211,,33y kx y kx =-=-故()*可化为 2121212221212()121(1)()()339x x y y m y y m k x x k m x x m m +-++=+-+++++2222222216(1)1421()9(21)33(21)3918(1)3(325)9(21)k k k m m m k k m k m m k +=--+⋅+++++-++-=+由于对于任意的R k ∈,0,TA TB ⋅=恒成立，故2210,3250m m m ⎧-=⎪⎨+-=⎪⎩ 解得 1m =.因此，在y 轴上存在满足条件的定点T ，点T 的坐标为(0,1). …… 16分23. (本题满分18分) 本题共3个小题，每小题6分.解：（1）由2(1),n n S a =+4及211(1),n n S a ++=+4 两式相减，得 222211111(1)(1)22,n n n n n n n n n a S S a a a a a a +++++=-=+-+=-+-44411()(2)0.n n n n a a a a ++⇒+--= ……3分由于{}n a 各项均为正数，故由上式，可得 12().n n a a n N *+-=∈于是数列{}n a 是以11a =为首项，2为公差的等差数列，其通项公式为：21().n a n n N *=-∈……6分（2）因为1121211122(),21212121n n n n n a a n n b a a n n n n +++-=+=+=+--+-+ ……8分故121111111122(1)()()()22(1)33557212121n b b b n n n n n ⎡⎤+++=+-+-+-++-=+-⎢⎥-++⎣⎦……10分于是121lim(2lim 2(1) 2.21n n n b b b n n →∞→∞⎡⎤+++--=⎢⎥+⎣⎦)= ……12分（3）假设存在大于2的正整数,m k 、使得12300.m m m m k a a a a +++++++=由（1），可得12(21)(1),m m m m k a a a a m k k +++++++=+-+从而 (21)(1)300.m k k +-+= ……14分由于正整数m k 、均大于2，知2114,211m k k m k k +->+≥+-+且与的奇偶性相同. ……16分故由22300235,=⨯⨯得212312559,21235=23=112125k k k k m k m m m k +=⨯+=⨯==⎧⎧⎧⎧⇒⎨⎨⎨⎨+-=⨯⨯+-=⨯⎩⎩⎩⎩或，或.因此，存在大于2的正整数:m k 、59,=23=11k k m m ==⎧⎧⎨⎨⎩⎩或使得12300.m m m m k a a a a +++++++=……18分。

上海市虹口区2015届高三二模英语试题含答案

上海市虹口区2015届高三二模英语试题含答案2015.4第I卷(共103分)I. Listening（略）II. Grammar and VocabularySection ADirections: After reading the passages below, fill in the blanks to make the passages coherent arid grammatically con-ect- For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one ward that best fits each blank.(A)How I Turned to Be Optimistic(乐观的)I began to grow up that winter night when my parents and I were returning from my aunt's House, and my mother said that we (25) (leave) for America soon. We were on the bus then. Iwas crying, and some people on the bus were turning around to look at me. I remember I could notbear the thought of never hearing again the radio program for school children to (26) I listened every morning.I do not remember myself (27) (cry) for this reason again. In fact I think I cried very little when I was saying goodbye to my friends and relatives. When we were leaving I thought aboutall the places I was going to see. The country I was leaving never to come back was hardly in my head then.The four years that followed taught me the importance of optimism, but (28) idea did not come to me at once. For the first two years in New York I was really lost. I did not quite know what I was or what I should be. Mother remarried, and things became even (29) (complex) for me. Some time passed before my stepfather and I got used to each orher. However, my responsibilities in the family increased a lot since my English vas superior (30) anyone else's at home. I translated at interviews with immigration officers,and even discussed telephone bills with company representatives.From my experiences, I believe that my life will turn out all right (31) it is not that easy.(B)How Room Designs Affect Our Work and FeelingArchitects have long had the feeling that the place we live in can affect our thoughts, feelingand behaviours. But now scientists are giving this feeling an empirical(实证的)basis. They are discovering how(32) (design) spaces that promote creativity, keep people focused, and lead to relaxation.Researches show aspects of the physical environment can influence creativity. In 2012, Joan Meyers-Levy reported that the height of a room's ceiling affects (33) people think. Her research indicates that higher ceilings encourage people to think more freely, (34) (lead) them to make more abstract connections. Low ceilings, on the other hand, may inspire a more detailed outlook. Besides ceiling height, the view (35) (afford) by a building may influence anoccupant's ability to concentrate.Using nature to improve focus of attention ought to pay off academically, and(36) seems to, according to a study. Students in classrooms with unblocked views of at least 50 feet outside the window had higher scores on tests of vocabulary, language arts and maths than did students (37) classrooms primarily overlooked roads and parking lots.Recent study on room lighting design suggests that dim light helps people loosen up. (38) that is true generally, keeping the light low during dinner or at parties could increase relaxation.So far public buildings (39) (focus) on by scientists. "We have a very limited number ofstudies, so we are almost looking at the problem through a straw(吸管),” architect David says. “How do you take answers to very specific questions and make broad use of them? That is (40) we are all struggling with.”Section BDirections: Complete the following passage by using the words in the box. Each word can only beForeign drivers will have a pay on-the-spot fines of up to £900 for breaking the traffic law to be carried out next month.If they do not have enough cash or a working credit card, their vehicles will be clamped(扣留)until they pay--and they will face a(n) 41 fee of £80 for getting back their vehicles.The law will also be 42 to British citizens. The fines will be described officially as "deposits" when the traffic laws take 43 , because the money would be returned if the driverwent to court and was found not guilty. In practice, very few foreign drivers are likely to return to Britain to deal with their cases.Foreign drivers are rarely 44 because police cannot take action against them if they fail to appear in court. Instead, officers often 45 give warnings. Foreign vehicles are 30 percentMore likely to be in a crash than British-registered vehicles. The number of crashes caused by foreign vehicles rose by 47 percent between 2008 and 2013, There were almost 400 deaths and serious injuries and 3,000 46 injuries from accidents caused by foreign vehicles in 2013.The new law is partly 47 to settle the problem of foreign lorry drivers ignoring limits to weight and hours at the wheel. Foreign lorries are three times more likely to be in a crash than British lorries. Recent spot checks found that three quarters of lorries that failed safety tests were48 overseas.The standard deposit for a careless driving 49 --such as driving too close to the vehiclefront or reading a map at the wheel--will be £300. Foreign drivers will not get points as 50 added to their licenses, while British drivers will.III. Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.In any planning system, from the simplest budgeting to the most complex corporate planning,there is an annual process. This is partly due to the fact that firms _ 51__ their accounting on a yearly basis, but also because similar _ 52__ often occur in the market.Usually, the larger the firm, the longer the planning takes. But 53 , planning for next yearmay start nine months or more in advance, with various stages of evaluation leading to 54 ofthe complete plan three months before the start of the year.Planning continues, however, throughout the year, since managers 55 progress againsttargets, while looking forward to the next year. What is happening now will 56 the objectives and plans for the future.In today's business climate, as markets constantly change and become more difficult to 57 ,some analysts believe that long-term planning is 58 . In some markets they may be right, as long as companies can build the sort of flexibility into their operations which allows them to59 to any sudden changes.Most firms, however, need to plan more than one year ahead in order to 60 .their long-termgoals. This may reflect the time it takes to commission and build a new production plant, or, in marketing 61 , it may be a question of how long it takes to research and launch a range of new products, and reach a certain 62 in the market. If, for example, it is going to take five years for a particular airline to become the 63 choice amongst business travelers on certainroutes, the airline must plan for the various 64 involved.Every one-year plan, therefore, must be 65 in relation to longer-term plans,and it shouldcontain die stages that are necessary to achieve the final goals.51.A make up B carry out C bring about D put down52.A patterns B guides C designs D distributions53.A surprisingly B contrarily C equally D typically54.A approval B permission C admiration D objection55.A value B confirm C review D survey56.A restore B promote C influence D maintain57.A guess B advocate C recognize D predict58.A pointless B meaningful C realistic D inevitable59.A lead B respond C refer D contribute60.A share B handle C develop D benefit61.A expressions B descriptions C words D terms62.A reputation B position C situation D direction63.A reserved B selected C preferred D supposed64.A acts B steps C means D points65.A handed over B left behind C made out D drawn upSection BDirections: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)My dad loved pennies, especially those with the elegant stalk(茎) of wheat curving around each side of the ONE CENT on the back. Those were the pennies he grew up with during the Depression.As a kid, I would go for walks with Dad, spying coins along the way-a penny here, a dime(一角硬币) there. Whenever I picked up a penny, he'd ask, "Is it a wheat?" It always thrilled him when we found one of those special coins produced between 1909 and 1958, the year of my birth.One gray Sunday morning in winter, not long after my father's death in 2002, I was walking down Fifth Avenue, feeling bereft. I found myself in front of the church where Dad once worked.I was warmly shown in and led to a seat. Hearing Dad's favorite "A Mighty Fortress Is Our God", I burst into tears. We'd sung that at his funeral.After the service, I shook the pastor's(牧师) hand and stepped onto the sidewalk-and there was apenny. I bent to pick it up, turned it over, and sure enough, it was a wheat. A 1944, a year my father was serving on a ship in the South Pacific.That started it. Suddenly wheat pennies began turning up on the sidewalks of New York everywhere. I got most of the important years: his birth year, my mom's birth year, the year he graduated from college, the year he met my mom, the year they got married, the year my sister was born. But alas, no 1958 wheat penny-my year, the last year they were made.The next Sunday, after the service, I was walking up Fifth Avenue and spotted a penny in the middle of a crossing. Oh, no, it was a busy street；cabs were speeding by-should I risk it? I just had to get it.A wheat! But the penny was worn, and I couldn't read the date. On arriving home, I took out myglasses and took it to the light. There was my birthday!I found 21 wheat pennies on the streets of Manhattan in the year after my father died, and I don't think that's a coincidence.66. The writer's father loved pennies with wheat because ________.A. when he first saw it, he began to love itB. when he saw the wheat, he thought of his time during the DepressionC. when he was young, he had a lot of pennies with wheatD. when he was a child, he never got a coin with wheat67. The underlined word "bereft"(in Para.3) means ________.A. protestedB. disappointedC. grievedD. offended68. Which of the following statements about the author is NOT true?A. He was born in 1958.B. He went to church because of his father.C. He once worked in a church.D. He knew the church well.69. The best title for the passage would probably be ________.A. Pennies from HeavenB. My father's life storyC. My father's hobbyD. Living in New York(B)Do you want to get home from work knowing you have made a real difference in someone's life? If yes. don't care about sex or age! Come and join us, then you'll make it!Position: Volunteer Social Care Assistant(No Pay with Free Meals)Place: ManchesterHours: Part TimeWe are now looking for volunteers to support people with learning disabilities to live active lives! Only 4 days left. Don't miss the chance of lending your warm hands to help others!Role:You will provide people with learning disabilities with all aspects of their daily lives. You will help them to develop new skills. You will help them to protect their rights and their safety. But your primary concern is to let them know they are valued.Skills and Experience Required:You will have the right values and great listening skills. You will be honest and patient. You will have the ability to drive a car and to communicate in fluent written and spoken English since you'll have to help those people with different learning disabilities. Previous care-related experience will be a great advantage for you.70. What docs the underlined part mean?A. You'll make others' lives more meaningful with this job.B. You'll arrive home just in time from this job.C. You'll earn a good salary from this job.D. You'll succeed in getting this job.71. The volunteers' major responsibility is to help people with learning disabilities ______ .A. to get some financial supportB. to properly protect themselvesC. to learn some new living skillsD. to realize their own importance72. Which of the following can first be chosen as a volunteer?A. The one who can drive a car.B. The one who has done similar work before.C. The one who has patience to listen to others.D. The one who can use English to communicate.73. The text serves as a _______ .A. a reminder to social workersB. an advertisement for helpersC. a document on appealing for volunteersD. an introduction about a social care organization(C)There are desert plants which survive the dry season in the form of inactive seeds. There are also desert insects which survive as inactive larvae (幼虫). In addition, difficult as it is to believe, there are desert fish which can survive through years of drought (干旱) in the form of inactive eggs. These are the shrimps (虾) that live in the Mojave Desert, an extremely dry place in the southwest of the United States where shade temperatures of over 50℃ are often recorded.The eggs of the Mojave shrimps look like grains of sand both in size and appearance. When enough spring rain falls to form a lake, once every two to five years, these eggs hatch. Then the water is soon filled with millions of tiny shrimps about a millimetre long which feed on tiny plant and animal organisms (有机物）which also grow in the temporary desert lake. Within a week, the shrimpsgrow from their original 1 millimetre to a length of about 1.5 centimetres.Throughout the time that the shrimps are rapidly maturing, the water in the lake equally rapidly evaporates(蒸发). Therefore, for the shrimps it is a race against time. By the twelfth day, however, when they are about 3 centimetre long, hundreds of tiny eggs form on the underbodies of the females. Usually by this time, all that remains of the lake is a large, muddy patch of wet soil. On the 13th day, the shrimps lay their eggs in the mud. Then, having ensured that their species will survive, the shrimps die as the last of the water evaporates.If enough rain falls the next year to form another lake, the eggs hatch, and once again the shrimps pass rapidly through their cycle of growth, adulthood, egg-laying, and death. Some years there is not enough rain to form a lake: in this case, the eggs will remain dormant for another years, or even longer if necessary.74. Which of the following is the MOST unique feature of Mojave shrimps?A. Their lives are brief.B. They feed on plant and animal organisms.C. Their eggs can survive years of drought.D. They lay their eggs in the mud.75. What doe the underlined word “dormant” in the last paragraph most probably mean?A. Inactive.B. Sleeping.C. Safe.D. Dead.76. What can be inferred from the passage?A. Appearance and size are the most important for life to survive in the desert.B. A species must be able to grow up quickly in order to survive.C. Shrimps are the only species with a life cycle of 13 days.D. Some species develop a unique life pattern to survive in extremely hard condition.77. What is the passage mainly about?A.The life span of Mojave shrimpsB. The survival of desert shrimpsC. The creatures living in the Mojave desertD. The importance of water to life in the desertSection CDirections: Read the passage carefully. Then answer the questions or complete the statements in the fewest possible words.The greatest recent social changes have been in the lives of women in America, or probably in the world.During the twentieth century there has been a remarkable shortening of the proportion of a woman's life spent in caring for the children. A woman marrying at the end of the nineteenth century would probably have been in her middle twenties? And would be likely to have seven or eight children, of whom four or five lived till they were five years old. By the time the youngest was fifteen, the mother would have been in her early fifties and would expect to live a further twenty years, during which custom, opportunity and health made it unusual for her to get paid work. Today women marry younger and have fewer children. Usually a woman's youngest child will be fifteen when she is forty-five years and is likely to take paid work until retirement at sixty. Even while she has the care of children, her work is lightened by modern living conditions.This important change in women's life-pattern has only recently begun to have its full effect on women's economic position. Even a few years ago most girls left school at the first opportunity, and most of them took a full-time job. However, when they married, they usually left work at once and never returned to it. Today the school leaving age is sixteen, many girls stay at school after that age, and though women tend to marry younger, more married women stay at work at least until shortly before their first child is born. Very many more afterwards return to full-or-part-time work.Such changes have led to a new relationship in marriage, with the husband accepting a greater share of the duties and satisfactions of family life, and with both husband and wife sharing more equally in providing the money, and running the home, according to the abilities and interests of each of them.(Note: Answer the questions or complete the statements in NO MORE THAN TEN WORDS.)78. At what age did most women get married in the late nineteenth century?______________________________________________.79. A women today can still take care of her children when doing paid work in their forties because of ______________________________________.80. Of “such changes” today, one is that many more mothers _________________________ after their first child is born.81. What are the factors that cause a couple to share economic and family affairs in an equalway?______________________________________________.第II卷(共47分)I. TranslationDirections: Translate the following sentences into English, using the word or phrase given in the brackets.82、据我所知，他们学校的面积是我们的两倍。

2015年上海市虹口区高考一模数学试卷【解析版】

2015年上海市虹口区高考数学一模试卷一、填空题（本大题满分56分）本大题共14题，只要求在答题纸相应题号的空格内直接填写结果，每个空格填对得4分，否则一律得零分.1．（4分）椭圆：的焦距是．2．（4分）在的展开式中，各项系数之和为．3．（4分）若复数z满足＝2﹣i（i为虚数单位），则复数z＝．4．（4分）若正实数a，b满足ab＝32，则2a+b的最小值为．5．（4分）行列式的最小值为．6．（4分）在△ABC中，角A、B、C所对的边分别为a、b、c，若A＝75°，B ＝60°，b＝，则c＝．7．（4分）若f（x）＝则方程f（x）＝1的解的个数为：．8．（4分）若数列{a n}为等差数列，且a1＝1，a2+a3+a4＝21，则＝．9．（4分）设等比数列{a n}的公比为q，前n项和为S n，若S n﹣1，S n，S n+1成等差数列，则q＝．10．（4分）已知l1，l2是分别经过A（2，1），B（0，2）两点的两条平行直线，当l1，l2之间的距离最大时，直线l1的方程是．11．（4分）若抛物线y2＝4x上的两点A、B到焦点的距离之和为6，则线段AB 的中点到y轴的距离为．12．（4分）10件产品中有8件正品，2件次品，从中任取3件，则恰好有一件次品的概率为．（结果用最简分数表示）13．（4分）如图是正四面体的平面展开图，M、N、G分别为DE、BE、FE的中点，则在这个正四面体中，MN与CG所成角的大小为．14．（4分）右图为函数f（x）＝A sin（ωx+φ）（A＞0，ω＞0，0＜φ＜）的部分图象，M、N是它与x轴的两个交点，D、C分别为它的最高点和最低点，E（0，1）是线段MD的中点，且，则函数f（x）的解析式为．二、选择题（本大题共4题，满分20分）每题有且仅有一个正确答案，考生应在答题纸的相应题号上，将所选答案的代号涂黑，选对得5分，否则一律零分.15．（5分）设全集U＝R，A＝{x|y＝ln（1﹣x）}，B＝{x||x﹣1|＜1}，则（∁U A）∩B＝（）A．（﹣2，1）B．（﹣2，1]C．[1，2）D．（1，2）16．（5分）设、都是非零向量，下列四个条件中，使成立的充分条件是（）A．B．C．D．且17．（5分）关于曲线C：x4+y2＝1，给出下列四个命题：①曲线C关于原点对称；②曲线C关于直线y＝x对称③曲线C围成的面积大于π④曲线C围成的面积小于π上述命题中，真命题的序号为（）A．①②③B．①②④C．①④D．①③18．（5分）若直线y＝kx+1与曲线有四个公共点，则k的取值集合是（）A．B．C．D．三、解答题（本大题共5题，满分74分）解答下列各题必须在答题纸的规定区域内写出必要步骤.19．（12分）已知，求的值．20．（14分）一个透明的球形装饰品内放置了两个公共底面的圆锥，且这两个圆锥的顶点和底面圆周都在这个球面上，如图，已知圆锥底面面积是这个球面面积的，设球的半径为R，圆锥底面半径为r．（1）试确定R与r的关系，并求出较大圆锥与较小圆锥的体积之比；（2）求出两个圆锥的体积之和与球的体积之比．21．（14分）已知函数f（x）和g（x）的图象关于原点对称，且f（x）＝x2+2x．（Ⅰ）求函数g（x）的解析式；（Ⅱ）若h（x）＝g（x）﹣mf（x）在[﹣1，1]上是增函数，求实数m的取值范围．22．（16分）已知各项均不为零的数列{a n}的前n项和为S n，且4S n＝a n•a n+1+1（n∈N*），其中a1＝1．（1）求证：a1，a3，a5成等差数列；（2）求证：数列{a n}是等差数列；（3）设数列{b n}满足＝1+，且T n为其前n项和，求证：对任意正整数n，不等式2T n＞log2a n+1恒成立．23．（18分）已知F1、F2为为双曲线C：＝1的两个焦点，焦距|F1F2|＝6，过左焦点F1垂直于x轴的直线，与双曲线C相交于A，B两点，且△ABF2为等边三角形．（1）求双曲线C的方程；（2）设T为直线x＝1上任意一点，过右焦点F2作TF2的垂线交双曲线C与P，Q两点，求证：直线OT平分线段PQ（其中O为坐标原点）；（3）是否存在过右焦点F2的直线l，它与双曲线C的两条渐近线分别相交于R，S两点，且使得△F1RS的面积为6？若存在，求出直线l的方程；若不存在，请说明理由．2015年上海市虹口区高考数学一模试卷参考答案与试题解析一、填空题（本大题满分56分）本大题共14题，只要求在答题纸相应题号的空格内直接填写结果，每个空格填对得4分，否则一律得零分.1．（4分）椭圆：的焦距是2．【解答】解：因为椭圆：，所以a2＝2，b2＝1，所以c2＝1，所以2c＝2．所以椭圆的焦距为2．故答案为：2．2．（4分）在的展开式中，各项系数之和为1．【解答】解：在的展开式中，令x＝1，可得各项系数之和为19＝1，故答案为：1．3．（4分）若复数z满足＝2﹣i（i为虚数单位），则复数z＝﹣5i．【解答】解：∵复数z满足＝2﹣i，∴z＝＝＝﹣5i，故答案为：﹣5i．4．（4分）若正实数a，b满足ab＝32，则2a+b的最小值为16．【解答】解：∵正实数a，b满足ab＝32，∴2a+b＝16，当且仅当2a＝b＝8时取等号．∴2a+b的最小值为16．故答案为：16．5．（4分）行列式的最小值为﹣5．．【解答】解：∵原式＝3sin x tan（+x）﹣4cos x tan（π﹣x）＝4sin x﹣3cos x＝5sin （x﹣φ），（其中，tanφ＝﹣），∴根据正弦函数的图象和性质可知行列式的最小值为﹣5，故答案为：﹣5．6．（4分）在△ABC中，角A、B、C所对的边分别为a、b、c，若A＝75°，B＝60°，b＝，则c＝．【解答】解：∵A＝75°，B＝60°，∴C＝45°，由正弦定理＝得：c＝＝＝，故答案为：7．（4分）若f（x）＝则方程f（x）＝1的解的个数为：3．【解答】解：f（x）＝当x＜0时，f（x）＝1，即x＝﹣1，1个解．当0≤x≤π时，f（x）＝1，x2＝2sin x，2个解．∴方程f（x）＝1的所有解的个数3个故答案为：38．（4分）若数列{a n}为等差数列，且a1＝1，a2+a3+a4＝21，则＝．【解答】解：设等差数列{a n}的公差为d，由a1＝1，a2+a3+a4＝21，得3a1+6d＝21，即3+6d＝21，d＝3．∴＝＝＝＝．故答案为：．9．（4分）设等比数列{a n}的公比为q，前n项和为S n，若S n﹣1，S n，S n+1成等差数列，则q＝1．，S n，S n+1成等差数列，【解答】解：因为S n﹣1所以S n﹣S n＝S n+1﹣S n，即a n＝a n+1，所以a n+1＝a n•q，﹣1解得q＝1，故答案为：1．10．（4分）已知l1，l2是分别经过A（2，1），B（0，2）两点的两条平行直线，当l1，l2之间的距离最大时，直线l1的方程是2x﹣y﹣3＝0．【解答】解：由题意可得，l1，l2间的距离最大时，AB和这两条直线都垂直．由于AB的斜率为＝，故直线l1的斜率为：2，故它的方程是y﹣1＝2（x﹣2），化简为2x﹣y﹣3＝0，故答案为：2x﹣y﹣3＝0．11．（4分）若抛物线y2＝4x上的两点A、B到焦点的距离之和为6，则线段AB 的中点到y轴的距离为2．【解答】解：∵F是抛物线y2＝4x的焦点∴F（1，0），准线方程x＝﹣1设A（x1，y1），B（x2，y2）∴|AF|+|BF|＝x1+1+x2+1＝6∴x1+x2＝4，∴线段AB的中点横坐标为2，∴线段AB的中点到y轴的距离为2，故答案为：212．（4分）10件产品中有8件正品，2件次品，从中任取3件，则恰好有一件次品的概率为．（结果用最简分数表示）【解答】解：所有的取法共有种，而满足条件的取法有•种，故恰好有一件次品的概率为＝，故答案为：．13．（4分）如图是正四面体的平面展开图，M、N、G分别为DE、BE、FE的中点，则在这个正四面体中，MN与CG所成角的大小为arccos．【解答】解：把这个正四面体的平面展开图还原得到如图所示的正四面体P﹣DEF，其中A，B，C三点重合为点P，设正四面体P﹣DEF的棱长为2，∵M，N，G分别为DE，PE，EF的中点，∴MN∥PD，∴MN与CG所成角为∠DPG，连结DG，则DG＝PG＝，PD＝2，∴cos∠DPG＝＝．∴∠DPG＝arccos．故答案为：arccos．14．（4分）右图为函数f（x）＝A sin（ωx+φ）（A＞0，ω＞0，0＜φ＜）的部分图象，M、N是它与x轴的两个交点，D、C分别为它的最高点和最低点，E（0，1）是线段MD的中点，且，则函数f（x）的解析式为f（x）＝2sin（2x+）．【解答】解：由已知点E（0，1）是线段MD的中点知A＝2，根据，可得||•||•cos∠DMN＝•＝＝，求得ω＝2．∴函数f（x）＝2sin（2x+φ），又由E（0，1）是线段MD的中点，则D的纵坐标为2，且点M、D的横坐标互为相反数．又由ω＝2，则周期T＝＝π．设点D的横坐标为a，则点M的横坐标为﹣a，2a＝•T＝，∴a＝，故M的坐标为（﹣，0），D的坐标为（，2）．根据五点法作图可得2•（﹣）+φ＝0，可得φ＝，∴f（x）＝2sin（2x+），故答案为：f（x）＝2sin（2x+）．二、选择题（本大题共4题，满分20分）每题有且仅有一个正确答案，考生应在答题纸的相应题号上，将所选答案的代号涂黑，选对得5分，否则一律零分.15．（5分）设全集U＝R，A＝{x|y＝ln（1﹣x）}，B＝{x||x﹣1|＜1}，则（∁U A）∩B＝（）A．（﹣2，1）B．（﹣2，1]C．[1，2）D．（1，2）【解答】解：A＝{x|y＝ln（1﹣x）}＝（﹣∞，1），B＝{x||x﹣1|＜1}＝（0，2），故（∁U A）∩B＝[1，2）；故选：C．16．（5分）设、都是非零向量，下列四个条件中，使成立的充分条件是（）A．B．C．D．且【解答】解：⇔⇔与共线且同向⇔且λ＞0，故选：C．17．（5分）关于曲线C：x4+y2＝1，给出下列四个命题：①曲线C关于原点对称；②曲线C关于直线y＝x对称③曲线C围成的面积大于π④曲线C围成的面积小于π上述命题中，真命题的序号为（）A．①②③B．①②④C．①④D．①③【解答】解：对于①，将方程中的x换成﹣x，y换成﹣y方程不变，所以曲线C 关于x轴、y轴、原点对称，故①对对于②，将方程中的x换为y，y换为x方程变为y4+x2＝1与原方程不同，故②错对于③，在曲线C上任取一点M（x0，y0），x04+y02＝1，∵|x0|≤1，∴x04≤x02，∴x02+y02≥x04+y02＝1，即点M在圆x2+y2＝1外，故③对，④错．故选：D．18．（5分）若直线y＝kx+1与曲线有四个公共点，则k的取值集合是（）A．B．C．D．【解答】解：t＝x﹣＝＝，①若x≤﹣1，t≤0，y＝|x+|﹣|x﹣|＝（﹣x﹣）﹣（﹣x）＝﹣；②若﹣1＜x＜0，t＞0，y＝|x+|﹣|x﹣|＝（﹣x﹣）﹣（x﹣）＝﹣2x；③若0＜x＜1，t＜0，则y＝|x+|﹣|x﹣|＝（x+）﹣（﹣x）＝2x；④若x≥1，t≥0，则曲线y＝|x+|﹣|x﹣|＝（x+）﹣（x﹣）＝．∴y＝，作图如右：由于直线y＝kx+1经过定点A（0，1），当过A点的直线m与曲线y＝﹣相切时，直线m与曲线y＝|x+|﹣|x﹣|有四个公共点，设切点坐标为：（x0，y0），则k＝（﹣）′|x＝x0＝，∴y0＝﹣＝kx0+1＝•x0+1，解得，x0＝﹣4，∴k＝＝；同理，可得当直线n与曲线y＝相切时，直线n与曲线y＝|x+|﹣|x﹣|有四个公共点，可求得直线n的斜率为k′＝﹣；当过A点的直线l∥x轴，即其斜率为0时，直线l与曲线y＝|x+|﹣|x﹣|有四个公共点；综上所述，实数k的取值范围是{，0，﹣}．故选：A．三、解答题（本大题共5题，满分74分）解答下列各题必须在答题纸的规定区域内写出必要步骤.19．（12分）已知，求的值．【解答】解：由于，则x﹣∈（，），即有sin（x﹣）＝＝＝；sin x＝sin[（x）+]＝sin（x﹣）cos+cos（x﹣）sin＝（）＝；cos2x＝sin（﹣2x）＝﹣2sin（x﹣）cos（x﹣）＝﹣2×＝﹣．20．（14分）一个透明的球形装饰品内放置了两个公共底面的圆锥，且这两个圆锥的顶点和底面圆周都在这个球面上，如图，已知圆锥底面面积是这个球面面积的，设球的半径为R，圆锥底面半径为r．（1）试确定R与r的关系，并求出较大圆锥与较小圆锥的体积之比；（2）求出两个圆锥的体积之和与球的体积之比．【解答】解：（1）不妨设球的半径为：4；则球的表面积为：64π，圆锥的底面积为：12π，∴圆锥的底面半径为：2；由几何体的特征知球心到圆锥底面的距离，球的半径以及圆锥底面的半径三者可以构成一个直角三角形由此可以求得球心到圆锥底面的距离是＝2，所以圆锥体积较小者的高为：4﹣2＝2，同理可得圆锥体积较大者的高为：4+2＝6；又由这两个圆锥的底面相同，∴较大圆锥与较小圆锥的体积之比等于它们高之比，即3：1（2）由（1）可得两个圆锥的体积和为：＝32π，球的体积为：＝，故两个圆锥的体积之和与球的体积之比为：32π：＝3：821．（14分）已知函数f（x）和g（x）的图象关于原点对称，且f（x）＝x2+2x．（Ⅰ）求函数g（x）的解析式；（Ⅱ）若h（x）＝g（x）﹣mf（x）在[﹣1，1]上是增函数，求实数m的取值范围．【解答】解：（Ⅰ）设g（x）的图象上任意一点（x，y），则关于原点对称点的坐标为（﹣x，﹣y），代入f（x）＝x2+2x，可得﹣y＝x2﹣2x，∴y＝﹣x2+2x，∴g（x）＝﹣x2+2x…（6分）（Ⅱ）h（x）＝g（x）﹣mf（x）＝﹣x2+2x﹣m（x2+2x）＝﹣（1+m）x2+2（1﹣m）x求导函数可得h′（x）＝﹣2（1+m）x+2（1﹣m）…（9分）依题设知：h（x）在[﹣1，1]上是增函数且非常函数，则在[﹣1，1]上h′（x）≥0恒成立．∴，解得：m≤0…（12分），22．（16分）已知各项均不为零的数列{a n}的前n项和为S n，且4S n＝a n•a n+1+1（n∈N*），其中a1＝1．（1）求证：a1，a3，a5成等差数列；（2）求证：数列{a n}是等差数列；（3）设数列{b n}满足＝1+，且T n为其前n项和，求证：对任意正整数n，不等式2T n＞log2a n+1恒成立．【解答】（1）证明：各项均不为零的数列{a n}的前n项和为S n，且4S n＝a n•a n+1+1（n∈N*），其中a1＝1．则：当n＝1时，解得：a2＝3当n＝2时，解得：a3＝5当n＝3时，解得：a4＝7当n＝4时，解得：a5＝9由于：2a3＝a1+a5所以：a1，a3，a5成等差数列．（2）证明：由于4S n＝a n a n+1+1①＝a n a n﹣1+1②所以：4S n﹣1①﹣②得：a n+1﹣a n﹣1＝4则数列的相邻项成等差数列．③当数列是奇数项时，a1＝1公差为4则：数列a n＝1+4（n﹣1）＝4n﹣3④当数列是偶数项时，a2＝3则：数列a n＝3+4（n﹣1）＝4n﹣1则相邻项的差值为2，所以数列{a n}为等差数列．（3）解：由（2）得到：a n＝1+2（n﹣1）＝2n﹣1所以：整理得：T n＝b1+b2+…+b n＝则：要使不等式2T n＞log2a n+1恒成立只需满足恒成立即可．即：恒成立用数学归纳法证明：①当n＝1时，2＞恒成立．②当n＝k时，恒成立则：当n＝k+1时，＝＝＝所以无论n取任意正整数上述不等式恒成立．23．（18分）已知F1、F2为为双曲线C：＝1的两个焦点，焦距|F1F2|＝6，过左焦点F1垂直于x轴的直线，与双曲线C相交于A，B两点，且△ABF2为等边三角形．（1）求双曲线C的方程；（2）设T为直线x＝1上任意一点，过右焦点F2作TF2的垂线交双曲线C与P，Q两点，求证：直线OT平分线段PQ（其中O为坐标原点）；（3）是否存在过右焦点F2的直线l，它与双曲线C的两条渐近线分别相交于R，S两点，且使得△F1RS的面积为6？若存在，求出直线l的方程；若不存在，请说明理由．【解答】（1）解：∵F1、F2为为双曲线C：＝1的两个焦点，焦距|F1F2|＝6，∴2c＝6，即c＝3，设|AF 2|＝2x，则36+x2＝4x2，解得，，∴2a＝||AF2|﹣|AF1||＝2．∴a＝，∴b2＝9﹣3＝6，∴双曲线C的方程为．（2）证明：设P（x1，y1），Q（x2，y2），中点T′（x0，y0），则x1+x2＝2x0，y1+y2＝2y0，把P（x1，y1），Q（x2，y2）分别代入双曲线C的方程，得，两式相减，得2（x1+x2）（x1﹣x2）﹣（y1+y2）（y1﹣y2）＝0，∴4x0（x1﹣x2）﹣2y0（y1﹣y2）＝0，∴k PQ＝＝＝﹣＝，∵T为直线x＝1上任意一点，过右焦点F2作TF2的垂线交双曲线C与P，Q两点，∴＝＝＝k OT，∴直线OT′与OT重合，∴OT过PQ的中点T'，∴直线OT平分线段PQ．（3）解：假设存在这样的直线，设直线l：x＝my+3，R（x R，y R），S（x S，y S），联立，得y R＝，联立，得y S＝，＝＝6，∴||＝2，解得m＝，∴直线l为：x＝y+3．。

【推荐】高三数学(理)优题精练：复数与极限 Word版含答案[ 高考]

上海市2016届高三数学理优题精练复数与极限一、复数1、（2015年上海高考）若复数z 满足3z+=1+i ，其中i 是虚数单位，则z= ．2、（2014年上海高考）若复数12z i =+，其中i 是虚数单位，则1z z z ⎛⎫+⋅= ⎪⎝⎭. 3、（2013年上海高考）设m R ∈，222(1)i m m m +-+-是纯虚数，其中i 是虚数单位，则________m =4、（静安、青浦、宝山区2015届高三二模）复数34ii-（i 为虚数单位）的模为． 5、（闵行区2015届高三二模）若复数z 满足(1i )2z ⋅+=（其中i 为虚数单位），则1z += ．6、（浦东新区2015届高三二模）设i 是虚数单位，复数)1)(3(i i a -+是实数，则实数a = 3 .7、（普陀区2015届高三二模）若1m ii i+=+（i 为虚数单位），则实数m 1- . 8、（徐汇、松江、金山区2015届高三二模）若复数i i z (21-=为虚数单位），则=+⋅z z z 9、（长宁、嘉定区2015届高三二模）若bi i ai -=+2)1(，其中a 、b R ∈，i 是虚数单位，则=+||bi a ________10、（松江区2015届高三上期末）若复数z 满足014=-zz ,则z 的值为 ▲11、（徐汇区2015届高三上期末）设i 是虚数单位，复数z 满足(2)5i z +⋅=，则z =112、（杨浦区2015届高三上期末）已知122ω=-+，集合{}2*1,n A z z n N ωωω==++++∈，集合1212{|,}B x x z z z z A ==⋅∈、（1z 可以等于2z )，则集合B 的子集个数为__________ 13、（闸北区2015届高三上期末）若复数i21i2+-a （i 是虚数单位）是纯虚数，则实数a = 14、（嘉定区2015届高三上期末）设i 是虚数单位，则=-+ii i 123_________15、（金山区2015届高三上期末）如果复数z =i1i2--b （b ∈R ）的实部与虚部相等，则z 的共轭复数z = ▲二、极限1、（2015年上海高考）设 P n （x n ，y n ）是直线2x ﹣y=（n ∈N *）与圆x 2+y 2=2在第一象限的交点，则极限=（）A ．﹣1B ． ﹣C ． 1D ．22、（2014年上海高考）设无穷等比数列{}n a 的公比为q ，若()134lim n n a a a a →∞=+++，则q = .3、（2013年上海高考）计算：20lim______313n n n →∞+=+4、（闵行区2015届高三二模）已知等比数列{}n a 满足232,1a a ==，则1223l i m ()n n n a a a a a a +→+∞+++= .5、（徐汇、松江、金山区2015届高三二模）矩阵1211222232332123i n i n i n n ninn a a a a a a a a a n a a a ⎛⎫ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎝⎭中每一行都构成公比为2的等比数列，第i 列各元素之和为i S ，则2lim2nnn S n →∞=⋅ ．6、（长宁、嘉定区2015届高三二模）已知对任意*N ∈n ，向量⎪⎪⎭⎫ ⎝⎛-=++n n n n n a a a a d 211,41都是直线x y =的方向向量，设数列}{n a 的前n 项和为n S ，若11=a ，则=∞→n n S lim ____________7、（宝山区2015届高三上期末）计算lim n →∞2123n n ++++＝8、（长宁区2015届高三上期末）已知()214732lim6752n a n n n →∞++++-⎡⎤⎣⎦=--，则a =9、（虹口区2015届高三上期末）若数列{}n a 为等差数列，且12341,21a a a a =++=，则122limnn a a a n →∞+++=10、（浦东区2015届高三上期末）若0lim =∞→nn x ，则实数x 的取值范围是 .11、（普陀区2015届高三上期末）若1lim=+∞→an ann ，则常数=a .12、（青浦区2015届高三上期末）已知1cos 22n n n a π=，则无穷数列{}n a 前n 项和的极限为13、（徐汇区2015届高三上期末）已知函数222111()1()()(1)2222015n n n f x x n =+++++++，其中*n N ∈．当1 2 3 n =，，，时，()n f x 的零点依次记作123 x x x ，，，，则lim n n x →∞=14、（上海市八校2015届高三3月联考）23111lim ()123n n n n n n →∞---=+++ ； 15、（奉贤区2015届高三4月调研测试（二模））已知3lim =∞→n n a ，31lim =∞→n n b ，则n nn n a b a 23lim-∞→=____________．参考答案一、复数1、解：设z=a+bi ，则=a ﹣bi （a ，b ∈R ），又3z+=1+i ，∴3（a+bi ）+（a ﹣bi ）=1+i ，化为4a+2bi=1+i ，∴4a=1，2b=1，解得a=，b=． ∴z=．故答案为：．2、：原式＝211516z z z ⋅+=+=+=3、【解答】2220210m m m m ⎧+-=⇒=-⎨-≠⎩． 4、5 5、136、37、－18、62i -9、510、i 2± 1112、16 13、4 14、-1 15、1-i二、极限1、解：当n →+∞时，直线2x ﹣y=趋近于2x ﹣y=1，与圆x 2+y 2=2在第一象限的交点无限靠近（1，1），而可看作点 P n （x n ，y n ）与（1，1）连线的斜率，其值会无限接近圆x 2+y 2=2在点（1，1）处的切线的斜率，其斜率为﹣1． ∴=﹣1．故选：A ．2、：223111011a a q a q q q q q ==⇒+-=⇒=--，∵01q <<，∴q = 3、【解答】根据极限运算法则，201lim 3133n n n →∞+=+．4、3235、146、27、128、289、1.5 10、)1,1(-11、1 12、15- 13、3- 14、6 15、31。

2015届上海市高考数学·二模汇编解析几何

2015届高中数学·二模汇编（专题：解析几何）2015届高中数学·二模汇编解析几何一、填空题1.（2015崇明二模文6理6）设直线0132=++y x 和圆22230x y x +--=相交于点A 、B ，则弦AB 的垂直平分线方程是．2.（2015崇明二模文12理11）已知双曲线2212y x -=的焦点为1F 、2F ，点M 在双曲线上且120MF MF ⋅=，则点M 到x 轴的距离等于．3. （2015奉贤二模文6理6）以抛物线x y 42=的焦点F 为圆心，与抛物线的准线相切的圆的标准方程为__________．4. （2015奉贤二模理11）关于x 的实系数一元二次方程2240x px -+=的两个虚根1z 、2z ，若1z 、2z 在复平面上对应的点是经过原点的椭圆的两个焦点，则该椭圆的长轴长为__________．5. （2015奉贤二模文13）设12,F F 是曲线()0,012222>>=+n m ny m x 的两个焦点，曲线上一点与12,F F 构成的三角形的周长是16，曲线上的点到1F 的最小距离为2，则=n ____________．6. （2015虹口二模文8）已知抛物线22(0)y px p =>的焦点在圆22(1)4x y -+=上，则p =________.7. （2015虹口二模理11文11）如图所示，已知12,F F 为双曲线()222210,0x y a b a b-=>>的两个焦点，且122F F =，若以坐标原点O 为圆心，12F F 为直径的圆与该双曲线的左支相交于,A B 两点，且2F AB ∆为正三角形，则双曲线的实轴长为__________.8.（2015虹口二模文13）已知直线1:125150l x y -+=和2:2,l x =-28P y x =点为抛物线上的动点，则1P l 点到直线2l 和直线的距离之和的最小值为_________.9．（2015黄浦二模文8理8）已知点(2,3)(1,4)A B --、，则直线AB 的点法向式方程是．10．（2015黄浦二模文9理9）已知抛物线216y x =的焦点与双曲线2221(0)12x y a a -=>的一个焦点重合，则双曲线的渐近线方程是．11．（2015静安二模文9）圆22420x y x y +-+=的圆心到直线3430x y ++=的距离为． 12.（2015静安二模理9）过圆0422=+-+my x y x 上一点)1,1(P 的切线方程为．xy2F 1F A BO13.（2015闵行二模理11文11）斜率为22的直线与焦点在x 轴上的椭圆2221(0)y x b b +=>交于不同的两点P 、Q .若点P 、Q 在x 轴上的投影恰好为椭圆的两焦点，则该椭圆的焦距为 .14.（2015闵行二模理13）如图，已知点(2,0)P ，且正方形ABCD 内接于O ：221x y +=， M 、N 分别为边AB 、BC 的中点.当正方形ABCD 绕圆心O 旋转时， PM ON ⋅的取值范围为．15．（2015浦东二模理6文6）已知直线0243=++y x 与圆()2221r y x =+-相切，则该圆的半径大小为．16.（2015普陀二模理6文6）如图，若,66π∠=⋅=-OFB OF FB ，则以OA 为长半轴，OB 为短半轴，F 为左焦点的椭圆的标准方程为．17．（2015徐汇二模理3文3）已知直线l 的一个法向量是()1,3n =-，则此直线的倾斜角的大小为． 18．（2015徐汇二模理14文14）对于曲线C 所在平面上的定点0P ，若存在以点0P 为顶点的角α，使得0AP B α≥∠对于曲线C 上的任意两个不同的点B A ,恒成立，则称角α为曲线C 相对于点0P 的“界角”，并称其中最小的“界角”为曲线C 相对于点0P 的“确界角”．曲线⎪⎩⎪⎨⎧<--≥+=)0(12)0(1:22x x x x y C 相对于坐标原点O 的“确界角”的大小是．19.（2015闸北二模文9理8）从双曲线()222210,0x y a b a b-=>>的左焦点F 引圆222x y a +=的切线，切点为T ，延长FT 交双曲线右支于点P ，若M 是线段FP 的中点，O 为原点，则MO MT -的值是____________．20．（2015长宁二模文2理2）抛物线28x y =的焦点到准线的距离是______________．二、选择题1. （2015虹口二模理17）如图所示，PAB ∆所在平面α和四边形ABCD 所在的平面β互相垂直，且AD α⊥， BC α⊥，4AD =，8BC =，6AB =，若tan 2tan 1ADP BCP ∠-∠=，则动点P 在平面α内的轨迹是（）A.线段B.椭圆的一部分C.抛物线D.双曲线的一部分2. （2015虹口二模理18）已知F 为抛物线24y x =的焦点，,,A B C 为抛物线上的三点，O 为坐标原点，F 若为ABC ∆的重心，,,OFA OFB OFC ∆∆∆面积分别记为123,,S S S ，则222123S S S ++的值为（）A.3B.4C.6D.9βαP BA DCABDy xCP NMO3．（2015浦东二模理17文17）若直线30ax by +-=与圆223x y +=没有公共点，设点P 的坐标(,)a b ，则过点P的一条直线与椭圆22143x y +=的公共点的个数为 ( ) )(A 0 )(B 1)(C 2 )(D 1或24．（2015长宁二模文17）设双曲线12222=-by a x （0>a ，0>b ）的虚轴长为2，焦距为32，则双曲线的渐近线方程为（）A ．x y 2±=B ．x y 2±=C ．x y 22±= D ．x y 21±=三、解答题1.（2015崇明二模理22文22）已知椭圆的中心在坐标原点O ，焦点在x 轴上，短轴长为2，且两个焦点和短轴的两个端点恰为一个正方形的顶点．过右焦点F 与x 轴不垂直的直线交椭圆于,P Q 两点．（1）求椭圆的方程；（2）当直线l 的斜率为1时，求POQ ∆的面积；（3）在线段OF 上是否存在点(,0)M m ，使得以,MP MQ 为邻边的平行四边形是菱形？若存在，求出m 的取值范围；若不存在，请说明理由．2.（2015奉贤二模理21文21）平面直角坐标系中，点()0,2-A 、()0,2B ，平面内任意一点P 满足：直线PA 的斜率1k ，直线PB 的斜率2k ，4321-=k k ，点P 的轨迹为曲线1C ．双曲线2C 以曲线1C 的上下两顶点N M ,为顶点，Q 是双曲线2C 上不同于顶点的任意一点，直线QM 的斜率3k ，直线QN 的斜率4k ．（1）求曲线1C 的方程；(5分)（2）如果04321≥+k k k k ，分别求双曲线2C 的两条渐近线倾斜角的取值范围．(9分)(第22题图）F 2F1y xPQ O 3.（205虹口二模文22理22）已知圆()221:18F x y ++=，点()21,0F ，点Q 在圆1F 上运动，2QF 的垂直平分线交1QF 于点P .（1）求动点P 的轨迹的方程C ；（2）设,M N 分别是曲线C 上的两个不同点，且点M 在第一象限，点N 在第三象限，若122OM ON OF +=， O 为坐标原点，求直线MN 的斜率；（3）过点10,3S ⎛⎫- ⎪⎝⎭的动直线l 交曲线C 于,A B 两点，在y 轴上是否存在定点T ，使以AB 为直径的圆恒过这个点？若存在，求出点T 的坐标，若不存在，请说明理由.4.（2015黄浦二模理23）已知点()12,0F -、()22,0F ，平面直角坐标系上的一个动点(),P x y 满足124PF PF +=，设动点P 的轨迹为曲线C . （1）求曲线C 的轨迹方程；（2）点M 是曲线C 上的任意一点，GH 为圆()22:31N x y -+=的任意一条直径，求MG MH ⋅的取值范围；（3）已知点,A B 是曲线C 上的两个动点，若OA OB ⊥（O 是坐标原点），试证明：直线AB 与某个定圆恒相切，并写出定圆的方程．5.（2015黄浦二模文23）已知点12(2,0)(2,0)F F -、，平面直角坐标系上的一个动点(,)P x y 满足12||+||=4PF PF ．设动点P 的轨迹为曲线C ．（1）求曲线C 的轨迹方程；（2）点M 是曲线C 上的任意一点，GH 为圆22:(3)1N x y -+=的任意一条直径，求MG MH ⋅的取值范围；（3）已知点A B 、是曲线C 上的两个动点，若OA OB ⊥(O 是坐标原点)，试证明：原点O 到直线AB 的距离是定值．6.（2015静安二模理22）在平面直角坐标系xoy 中，已知椭圆C 的方程为2218x y +=，设AB 是过椭圆C 中心O 的任意弦，l 是线段AB 的垂直平分线，M 是l 上与O 不重合的点．（1）求以椭圆的焦点为顶点，顶点为焦点的双曲线方程；（2）若2MO OA =，当点A 在椭圆C 上运动时，求点M 的轨迹方程；（3）记M 是l 与椭圆C 的交点，若直线AB 的方程为(0)y kx k =>，当△AMB 面积取最小值时，求直线AB 的方程．7.（2015静安二模文22）在平面直角坐标系xoy 中，已知椭圆C 的方程为2218x y +=，设AB 是过椭圆C 中心O 的任意弦，l 是线段AB 的垂直平分线，M 是l 上与O 不重合的点．（1）求以椭圆的焦点为顶点，顶点为焦点的双曲线方程；（2）若2MO OA =，当点A 在椭圆C 上运动时，求点M 的轨迹方程；（3）记M 是l 与椭圆C 的交点，若直线AB 的方程为(0)y kx k =>，当△AMB 面积为4147时，求直线AB 的方程．8.（2015闵行二模理22）已知两动圆2221:(3)F x y r ++=和2222:(3)(4)F x y r -+=-（04r <<），把它们的公共点的轨迹记为曲线C ,若曲线C 与y 轴的正半轴的交点为M ，且曲线C 上的相异两点A B 、满足0MA MB ⋅=．（1）求曲线C 的方程；（2）证明直线AB 恒经过一定点，并求此定点的坐标；（3）求ABM △面积S 的最大值．9.（2015闵行二模文22）已知两动圆2221:(3)F x y r ++=和2222:(3)(4)F x y r -+=-（04r <<），把它们的公共点的轨迹记为曲线C ，若曲线C 与y 轴的正半轴的交点为M ，且曲线C 上的相异两点A B 、满足：0MA MB ⋅=．（1）求曲线C 的方程；（2）若A 的坐标为(2,0)-，求直线AB 和y 轴的交点N 的坐标；（3）证明直线AB 恒经过一定点，并求此定点的坐标．10．（2015浦东二模理22）已知直线l 与圆锥曲线C 相交于两点,A B ，与x 轴，y 轴分别交于D E 、两点，且满足1EA AD λ=、2EB BD λ=．（1）已知直线l 的方程为24y x =-，抛物线C 的方程为24y x =，求12λλ+的值；（2）已知直线():11l x my m =+>，椭圆22:12x C y +=，求1211λλ+的取值范围；（3）已知双曲线()222122222:10,0,x y a C a b a b bλλ-=>>+=，试问D 是否为定点？若是，求点D 的坐标；若不是，说明理由．11．（2015浦东二模文22）已知直线l 与圆锥曲线C 相交于两点,A B ，与x 轴，y 轴分别交于D E 、两点，且满足1EA AD λ=、2EB BD λ=．（1）已知直线l 的方程为24y x =-，抛物线C 的方程为24y x =，求12λλ+的值；（2）已知直线():11l x my m =+>，椭圆22:12x C y +=，求1211λλ+的取值范围；（3）已知双曲线C ：1322=-y x ，621=+λλ，求点D 的坐标．11.（2015普陀二模理22文22）如图，射线OA OB 、所在的直线的方向向量分别是()()()121,1,0==->d k d k k 、，点P 在∠AOB 内，⊥PM OA 于M ，⊥PN OB 于N ．（1）若311,,22k P ⎛⎫= ⎪⎝⎭，求OM 的值；（2）若()2,1,∆P OMP 的面积为65，求k 的值；（3）已知k 为常数，M N 、的中点为T ，且1∆=MON S k，当P 变化时，求动点T 的轨迹方程．22465NMPyxAOBS RPQDC BAO12.（2015年徐汇二模文21理21）用细钢管焊接而成的花坛围栏构件如右图所示，它的外框是一个等腰梯形PQRS ，内部是一段抛物线和一根横梁．抛物线的顶点与梯形上底中点是焊接点O ，梯形的腰紧靠在抛物线上，两条腰的中点是梯形的腰、抛物线以及横梁的焊接点,A B ，抛物线与梯形下底的两个焊接点为,C D ．已知梯形的高是40厘米，C D 、两点间的距离为40厘米．（1）求横梁AB 的长度;（2）求梯形外框的用料长度．（注:细钢管的粗细等因素忽略不计，计算结果精确到1厘米．）13.（2015年杨浦文23理23）已知抛物线x y C 4:2=的焦点F ，线段PQ 为抛物线C 的一条弦. （1）若弦PQ 过焦点F ，求证：11FP FQ+为定值；（2）求证:x 轴的正半轴上存在定点M ，对过点M 的任意弦PQ ，都有2211MP MQ +为定值；（3）对于（2）中的点M 及弦PQ ，设PM MQ λ=，点N 在x 轴的负半轴上，且满足()NM NP NQ λ⊥-，求N 点坐标.14.（2015年闸北二模文17理16）已知圆()221:18C x y ++=，点()21,0C ，点Q 在圆1C 上运动，2QC 的垂直平分线交1QC 于点P ．（1）求动点P 的轨迹W 方程；（2）过点10,3S ⎛⎫- ⎪⎝⎭且斜率为k 的动直线l 交曲线W 于,A B 两点，在y 轴上是否存在定点D ，使以AB 为直径的圆恒过这个点？若存在，请求出点D 的坐标；若不存在，请说明理由．15.（2015长宁二模文22）已知椭圆1:2222=+by a x C （0>>b a ）的焦距为2，且椭圆C 的短轴的一个端点与左、右焦点1F 、2F构成等边三角形．（1）求椭圆C 的标准方程；（2）设M 为椭圆上C 上任意一点，求21MF MF ⋅的最大值与最小值；（3）试问在x 轴上是否存在一点B ，使得对于椭圆上任意一点P ，P 到B 的距离与P 到直线4=x 的距离之比为定值．若存在，求出点B 的坐标，若不存在，请说明理由．16.（2015长宁二模理22）已知椭圆1:2222=+by a x C （0>>b a ）的左、右焦点分别为1F 、2F ，点B ),0(b ，过点B 且与2BF垂直的直线交x 轴负半轴于点D ，且→=+02221D F F F ．（1）求证：△21F BF 是等边三角形；（2）若过B 、D 、2F 三点的圆恰好与直线l ：033=--y x 相切，求椭圆C 的方程；（3）设过（2）中椭圆C 的右焦点2F 且不与坐标轴垂直的直线l 与C 交于P 、Q 两点，M 是点P 关于x 轴的对称点．在x 轴上是否存在一个定点N ，使得M 、Q 、N 三点共线，若存在，求出点N 的坐标；若不存在，请说明理由．。

2015年虹口区高考数学汇编一模数学试卷(含答案)

虹口区2014学年第一学期高三期终教学质量监测试卷（文理合卷）2015.1.8一、填空题（本大题满分56分）本大题共14题，只要求在答题纸相应题号的空格内直接填写结果，每个空格填对得4分，否则一律得零分.1、椭圆2214x y +=的焦距为 .2、在91x ⎛⎫ ⎪⎝⎭的展开式中，各项系数之和为 .3、若复数z 满足22zii i=-+（i 为虚数单位），则复数z = . 4、若正实数a b ，满足ab =32，则2a b +的最小值为 .5、行列式()3sin tan 4cos tan()2x x x x ππ-+的最小值为 .6、在ABC ∆中，角A B C 、、所对的边分别为a b c 、、，若75,60,A B b =︒=︒=，则c = .7、若()22sin 00x x f x x x π≤≤⎧=⎨<⎩，，，，则方程()1f x =的所有解之和等于 .8、若数列{}n a 为等差数列，且12341,21a a a a =++=，则122limnn a a a n →∞+++= .9、设等比数列{}n a 的公比为q ，前n 项和为n S ，若12,,n n n S S S ++成等差数列，则q = .10、已知12,l l 是分别经过()()2102A B ,，,两点的两条平行直线，当12,l l 之间的距离最大时，直线1l 的方程是 .11、若抛物线24y x =上的两点A 、B 到焦点的距离之和为6，则线段AB 的中点到y 轴的距离为 .12、10件产品中有8件正品，2件次品，从中任取3件，则恰好有一件次品的概率为 .（结果用最简分数表示）13、右图是正四面体的平面展开图，M N G 、、分别为DE BE FE 、、的中点，则在这个正四面体中，MN 与CG 所成角的大小为 .14、右图为函数()()=sin (0,0,0)2f x A x A πωϕωϕ+>><<的部分图像，M N 、是它与x 轴的两个交点，D C 、分别为它的最高点和最低点，()0,1E 是线段MD 的中点，且28MD MN π⋅=，则函数()f x 的解析式为 .二、选择题（本大题共4题，满分20分）每题有且仅有一个正确答案，考生应在答题纸的相应题号上，将所选答案的代号涂黑，选对得5分，否则一律零分. 15、设全集(){}{},ln 1,11U R A x y x B x x ===-=-<，则()U C A B = （）.A.()2,1-B.(]2,1-C.[)1,2D.()1,216、设,a b 均为非零向量，下列四个条件中，使a b ab=成立的必要条件是（）.A.a b =-B.//a bC.2a b =D.//a b 且a b =17、关于曲线42:1C x y +=，给出下列四个命题：①曲线C 关于原点对称； ②曲线C 关于直线y x =对称 ③曲线C 围成的面积大于π ④曲线C 围成的面积小于π上述命题中，真命题的序号为（）A.①②③B.①②④C.①④D.①③18、若直线1y kx =+与曲线11y x x x x=+--有四个不同交点，则实数k 的取值范围是（）.A.11,0,88⎧⎫-⎨⎬⎩⎭ B.11,88⎧⎫-⎨⎬⎩⎭ C.11,88⎡⎤-⎢⎥⎣⎦ D.11,88⎛⎫- ⎪⎝⎭三、解答题（本大题共5题，满分74分）解答下列各题必须在答题纸的规定区域内写出必要步骤.19、（本题满分12分）已知3cos ,424x x πππ⎛⎫⎛⎫-=∈ ⎪ ⎪⎝⎭⎝⎭，求sin ,sin ,cos 24x x x π⎛⎫- ⎪⎝⎭的值20、（本题满分14分）本题共2个小题，每小题7分一个透明的球形装饰品内放置了两个公共底面的圆锥，且这两个圆锥的顶点和底面圆周都在这个球面上，如图，已知圆锥底面面积是这个球面面积的316，设球的半径为R ，圆锥底面半径为r .（1）试确定R 与r（2）求出两个圆锥的体积之和与球的体积之比.21、（本题满分14分）本题共2小题，第1小题6分，第2小题8分已知函数()f x 和()g x 的图像关于原点对称，且2()f x x x =+ （1）求函数()y g x =的解析式；（2）若()()()3h x g x m f x =-⋅+在[]1,1-上是增函数，求实数m 的取值范围.22、（本题满分16分）本题共3小题，第1小题5分，第2小题5分，第3小题6分. 已知各项均不为零的数列{}n a 的前n 项和为n S ，且()141n n n S a a n N *+=⋅+∈，其中11a =. （1）求证：135,,a a a 成等差数列；（2）求证：数列{}n a 是等差数列；（3）设数列{}n b 满足()121n b nn N a *=+∈，且n T 为其前n 项和，求证：对任意正整数n ，不等式212log n n T a +>恒成立.23、（本题满分18分）本题共3个小题，第1小题5分，第2小题7分，第3小题6分.已知12F F 、为为双曲线22221x y C a b-=：的两个焦点，焦距12=6F F ，过左焦点1F 垂直于x 轴的直线，与双曲线C 相交于,A B 两点，且2ABF ∆为等边三角形. （1）求双曲线C 的方程；（2）设T 为直线1x =上任意一点，过右焦点2F 作2TF 的垂线交双曲线C 与,P Q 两点，求证：直线OT 平分线段PQ （其中O 为坐标原点）；（3）是否存在过右焦点2F 的直线l ，它与双曲线C 的两条渐近线分别相交于,R S 两点，且使得1F RS ∆的面积为l 的方程；若不存在，请说明理由.2015年虹口区高三一模数学试卷理科（参考答案）一．填空题1. 2. 1； 3. 5i -； 4. 16； 5. 5-；； 7. 1π-； 8. 1.5； 9. 2-； 10. 230x y --=； 11. 3； 12.715；13. arccos 3； 14. 2sin(2)4y x π=+；二．选择题15. C ； 16. B ； 17. D ； 18. A ；三．解答题19. 解：(,)442x πππ-∈，在第一象限，∴sin()410x π-==； 4sin sin()sin()cos cos()sin 4444445x x x x ππππππ=-+=-+-=； 27cos 212sin 25x x =-=-；20. （1）解：223416r R ππ=⨯，2r R =；::3:1V V h h ==大小大小；（2）解：22232321143():()::3338h r V V V r h r h R r h R R R πππ+=+==⋅=小大小球大小小；21. （1）解：2()g x x x =-+；（2）解：2()(1)(1)3h x m x m x =--+-+，当10m -->，即1m <-时，对称轴112(1)mx m -=≤-+，∴31m -≤<-；当10m --=，即1m =-时，()23h x x =+，符合题意，∴1m =-；当10m --<，即1m >-时，对称轴112(1)m x m -=≥+，∴113m -<≤-；综上，133m -≤≤-； 22. （1）解：141n n n S a a +=+ ①；1141n n n S a a --=+ ②；①－②得114n n a a +--=，得证；（2）解：由11a =，得23a =，结合第（1）问结论，即可得{}n a 是等差数列；（3）解：根据题意，22log 21n n b n =-，22462log 13521n nT n =⨯⨯⨯⨯-…；要证2122log log (21)n n T a n +>=+，即证246213521nn ⨯⨯⨯⨯>-…；当1n =时，2>假设当n k =时，246213521kk ⨯⨯⨯⨯>-… 当1n k =+时，24622222135212121k k k k k k ++⨯⨯⨯⨯⨯>-++…=>2(22)(21)(23)k k k +>++，展开后显然成立，所以对任意正整数n ，不等式212log n n T a +>恒成立；23. （1）3c =，∵等边三角形，∴2AF =，1AF =a =∴22136x y -=；（2）解：设11(,)P x y ，22(,)Q x y ，中点为00(,)T x y '，然后点差法，即得2121212122()1312()PQ PF T Tx x y y k y y x x k y y +--===-==+-， ∴001TOT OT y y k k x '===，即点T '与点T 重合，所以T 为PQ 中点，得证；（3）解：假设存在这样的直线，设直线:3l x my =+，(,)R R R x y ，(,)S S S x y联立3y x my ⎧=⎪⎨=+⎪⎩得R y =3y x my ⎧=⎪⎨=+⎪⎩得S y =116()2F RSR S Sy y =⨯⨯-=()R S y y -==l .。

(优辅资源)上海市虹口区高考数学二模试卷文(含解析)

2015年上海市虹口区高考数学二模试卷（文科）一、填空题（本大题满分56分）本大题共14题，只要求在答题纸相应题号的空格内直接填写结果，每个空格填对得4分，否则一律得零分.1 = ．（i是虚数单位）2．已知函数f（x）f（f（﹣3））= ．3．函数f（x）=ln）（x＞0）的反函数f﹣1（x）= ．4．已知正实数x，y满足x+3y=1，则xy的最大值为．5．已知复数z=3sinθ+icosθ（i是虚数单位），且θ为钝角时，tanθ= ．6．在上海高考改革方案中，要求每位高中生必须在物理、化学、生物、政治、历史、地理6门学科（3门理科学科，3门文科学科）中选择3门学科参加等级考试，小丁同学理科成绩较好，决定至少选择两门理科学科，那么小丁同学的选科方案有种．7．设数列{a n}前n项的和为S n，若a1=4，且a n+1=3S n（n∈N*），则S n= ．8．已知抛物线y2=2px（p＞0）的焦点在圆（x﹣1）2+y2=4上，则p= ．9x2= ．101行第2列的元素1的代数余子式为﹣1，则实数x的取值集合为．11．如图所示，已知F1，F2且|F1F2|=2，若以坐标原点O为圆心，|F1F2|为直径的圆与该双曲线的左支相交于A，B两点，且△F2AB为正三角形，则双曲线的实轴长为．12．M，若函数y=a x（a＞0，且a≠1）的图象经过区域M，则实数a的取值范围为．13．已知直线l1：12x﹣5y+15=0和l2：x=﹣2，点P为抛物线y2=8x上的动点，则点P到直线l1和直线l2的距离之和的最小值为．14的最小值为．二、选择题（本大题共4题，满分20分）每题有且只有一个正确答案，考生应在答题纸的相应题号上，将所选答案的代号涂黑，选对得5分，否则一律零分.15．设全集U=R，已知0}，B={x||x﹣1|＜2}，则（∁U A）∩B=（）A1）B．（﹣1，﹣2] C．（2，3] D．[2，3）16．设a∈R，则“a=﹣1”是“f（x）=|（ax﹣2）x|在（0，+∞）上单调递增”的（）A．充要条件 B．既不充分也不必要条件C．充分不必要条件D．必要不充分条件17．一个四棱锥的三视图如图所示，则此四棱锥的体积为（）A．24 B．16 C．12 D．818．设函数f（x）（x∈R）满足f（﹣x）=f（x），f（x）=f（2﹣x），且当x∈[0，1]时，f（x）=x2．又函数g（x）=|sin（πx）|，则函数h（x）=g（x）﹣f（x）在区间[﹣1，3]上零点的个数为（）A．6 B．7 C．8 D．9三、解答题（本大题共5题，满分74分）解答下列各题必须在答题纸的规定区域内写出必要的步骤.19．函数f（x）=m+log a x（a＞0且a≠1）的图象过点（8，2）和（1，﹣1）．（Ⅰ）求函数f（x）的解析式；（Ⅱ）令g（x）=2f（x）﹣f（x﹣1），求g（x）的最小值及取得最小值时x的值．20．在如图所示的直四棱柱ABCD﹣A1B1C1D1中，底面ABCD是边长为2的菱形，且∠BAD=60°，AA1=4．（1）求直四棱柱ABCD﹣A1B1C1D1的体积；（2）求异面直线AD1与BA1所成角的大小．21．如图，经过村庄A有两条夹角60°为的公路AB，AC，根据规划拟在两条公路之间的区域内建一工厂P，分别在两条公路边上建两个仓库M，N（异于村庄A），要求PM=PN=MN=2（单位：千米）．记∠AMN=θ．（1）将AN，AM用含θ的关系式表示出来；（2）如何设计（即AN，AM为多长时），使得工厂产生的噪声对居民的影响最小（即工厂与村庄的距离AP最大）？22．已知圆F1：（x+1）2+y2=8，点F2（1，0），点Q在圆F1上运动，QF2的垂直平分线交QF1于点P．（1）求动点P的轨迹C的方程；（2）设M、N分别是曲线C上的两个不同点，且点M在第一象限，点N在第三象限，若O为坐标原点，求直线MN的斜率；（3l交曲线C于A、B两点，求证：以AB为直径的圆恒过定点T（0，1）．23．设各项均为正数的数列{a n}的前n项和为S n，且满足：a1=1，4S n=（a n+1）2（n∈N*）．（1）求数列{a n}的通项公式；（2）设b n N*b1+b2+…+b n﹣2n）的值；（3）是否存在大于2的正整数m、k，使得a m+a m+1+a m+2+…+a m+k=300？若存在，求出所有符合条件的m、k；若不存在，请说明理由．2015年上海市虹口区高考数学二模试卷（文科）参考答案与试题解析一、填空题（本大题满分56分）本大题共14题，只要求在答题纸相应题号的空格内直接填写结果，每个空格填对得4分，否则一律得零分.1 = ﹣i ．（i是虚数单位）【考点】复数代数形式的乘除运算；虚数单位i及其性质．【专题】数系的扩充和复数．【分析】由虚数单位i的运算性质化简，然后利用复数代数形式的乘除运算化简求值．【解答】故答案为：﹣i．【点评】本题考查了复数代数形式的乘除运算，是基础的计算题．2．已知函数f（x）f（f（﹣3））【考点】函数的值．【专题】计算题；函数的性质及应用．【分析】由分段函数f（x）f（﹣3），再求f（f（﹣3））即可．【解答】解：∵函数f（x）∴f（﹣3）=2﹣3f（f（﹣3））=f【点评】本题考查了分段函数的简单应用，属于基础题．3．函数f（x）=ln）（x＞0）的反函数f﹣1（x）【考点】反函数．【专题】函数的性质及应用．【分析】直接利用反函数的求法求解即可．【解答】解：函数f（x）=ln）（x＞0），f（x）∈（0，+∞）．y，解得函数f（x）=ln）（x＞0）的反函数f﹣1（x）x∈（0，+∞）．x∈（0，+∞）．【点评】本题考查反函数与原函数的关系，考查计算能力．注意函数的定义域．4．已知正实数x，y满足x+3y=1，则xy【考点】基本不等式．【专题】不等式的解法及应用．【分析】运用基本不等式得出【解答】解；∵正实数x，y满足x+3y=1，化简得出xy【点评】本题考查了运用基本不等式求解二元式子的最值问题，关键是判断，变形得出不等式的条件，属于容易题．5．已知复数z=3sinθ+icosθ（i是虚数单位），且θ为钝角时，tanθ= ﹣1 ．【考点】复数求模．【专题】三角函数的求值；数系的扩充和复数．【分析】直接利用复数的模，得到θ的三角方程，然后求解即可．【解答】解：复数z=3sinθ+icosθ（i是虚数单位），且可得9sin2θ+cos2θ=5，可得sin2θθ为钝角时，sinθθ=135°，∴tanθ=﹣1．故答案为：﹣1．【点评】本题考查复数的模以及三角函数的化简求值，考查计算能力．6．在上海高考改革方案中，要求每位高中生必须在物理、化学、生物、政治、历史、地理6门学科（3门理科学科，3门文科学科）中选择3门学科参加等级考试，小丁同学理科成绩较好，决定至少选择两门理科学科，那么小丁同学的选科方案有10 种．【考点】计数原理的应用．【专题】应用题；排列组合．【分析】分类讨论：选择两门理科学科，一门文科学科；选择三门理科学科，即可得出结论．【解答】种；选择三门理科学科，有1种，故共有10种．故答案为：10．【点评】本题考查计数原理的应用，考查学生的计算能力，比较基础．7．设数列{a n}前n项的和为S n，若a1=4，且a n+1=3S n（n∈N*），则S n= 4n．【考点】数列的求和；数列递推式．【专题】等差数列与等比数列．【分析】a n+1=3S n（n∈N*），变形为S n+1﹣S n=3S n，S n+1=4S n，再利用等比数列的通项公式即可得出．【解答】解：∵a n+1=3S n（n∈N*），∴S n+1﹣S n=3S n，化为S n+1=4S n，∴数列{S n}是等比数列，首项为4，公比为4．∴S n=4n．故答案为：4n．【点评】本题考查了递推式的应用、等比数列的通项公式，考查了推理能力与计算能力，属于中档题．8．已知抛物线y2=2px（p＞0）的焦点在圆（x﹣1）2+y2=4上，则p= 6 ．【考点】抛物线的简单性质．【专题】计算题；圆锥曲线的定义、性质与方程．【分析】0），把它代入圆的方程求出p的值．【解答】解：抛物线y2=2px（p＞00），代入圆（x﹣1）2+y2=41）2=4，∴p=6，故答案为：6．【点评】本题考查由抛物线的方程求焦点坐标，以及点在圆上的性质．9x2=【考点】极限及其运算；二项式系数的性质．【专题】计算题；二项式定理．【分析】x2项的系数，得出a 的值；【解答】T r+16﹣r（﹣a）r令6，解得r=3；∴展开式中含x2项的系数为（﹣a）3解得a=【点评】本题考查了二项式定理的应用问题，也考查了数列求和的应用问题以及极限的计算问题，是基础题目．101行第2列的元素1的代数余子式为﹣1，则实数x的取值集合为{x|x=π+2kπ，k∈Z} ．【考点】三阶矩阵．【专题】三角函数的求值；矩阵和变换．【分析】本题直接根据行列式的代数余子式的定义进行计算，即可得到本题结论．【解答】解：1行第2列的元素1的代数余子式为﹣1，，∴﹣sinx cosx﹣sinx）=1，即cosx=﹣1，∴x=π+2kπ（k∈Z），故答案为：{x|x=π+2kπ，k∈Z}．【点评】本题考查了行列式的代数余子式，三角函数的计算，记住常用常见角的三角函数值是解决本题的关键，注意解题方法的积累，属于中档题．11．如图所示，已知F1，F2且|F1F2|=2，若以坐标原点O为圆心，|F1F2|为直径的圆与该双曲线的左支相交于A，B两点，且△F2AB【考点】双曲线的简单性质．【专题】计算题；圆锥曲线的定义、性质与方程．【分析】根据△F2AB是等边三角形，判断出∠AF2F1=30°，进而在RT△AF1F2中求得|AF1|，|AF2|，进而根据双曲线的简单性质求得a可得．【解答】解：∵△F2AB是等边三角形，∴∠AF2F1=30°，∵|F1F2|=2，∴|AF1|=1，|AF21．1．【点评】本题主要考查了双曲线的简单性质．考查了学生综合分析问题和数形结合的思想的运用．属基础题．12．M，若函数y=a x（a＞0，且a≠1）的图象经过区域M，则实数a的取值范围为[2，9] ．【考点】简单线性规划的应用．【专题】不等式的解法及应用．【分析】画出其表示的平面区域，再利用函数y=a x（a＞0，a≠1）的图象特征，结合区域的角上的点即可解决问题【解答】解：平面区域M如如图所示．求得A（2，10），C（3，8），B（1，9）．由图可知，欲满足条件必有a＞1且图象在过B、C两点的图象之间．当图象过B点时，a1=9，∴a=9．当图象过C点时，a3=8，∴a=2．故a的取值范围为[2，9]．【点评】本题主要考查了用平面区域二元一次不等式组、指数函数的图象与性质，以及简单的转化思想和数形结合的思想，属中档题．巧妙识别目标函数的几何意义是我们研究规划问题的基础．13．已知直线l1：12x﹣5y+15=0和l2：x=﹣2，点P为抛物线y2=8x上的动点，则点P到直线l1和直线l2的距离之和的最小值为 3 ．【考点】直线与圆锥曲线的关系．【专题】圆锥曲线的定义、性质与方程．【分析】由抛物线方程求出其焦点坐标和准线方程，把抛物线y2=8x上的点P到两直线l1：x=﹣2，l2：12x﹣5y+15=0的距离之和的最小值转化为焦点到l2：12x﹣5y+15=0的距离，由点到直线的距离公式求解．【解答】解：如图，由抛物线y2=8x，得其焦点F（2，0），准线方程为x=﹣2．∴l1：x=﹣2为抛物线的准线，P到两直线l1：x=﹣2，l2：12x﹣5y+15=0的距离之和，即为P到F和l2：12x﹣5y+15=0的距离之和．最小值为F到l2：12x﹣5y+15=0故答案为：3．【点评】本题考查了直线与圆锥曲线的关系，考查了数形结合的解题思想方法和数学转化思想方法，是中档题．14【考点】平面向量数量积的运算．【专题】平面向量及应用．【分析】为等边三角形，且边长为2，而C点在以AB为直径的圆上，延长OB到D，使|OB|=|BD|，而D和圆心E，设C点是与圆的交点，从而|CD|而由余弦定理可求出|DE|，而圆半径为1，从而能得出|CD|的值．【解答】解：由已知条件知cos∴C点在以AB为直径的圆上，如下图所示：延长OB到D，使|OB|=|BD|，连接CD；设圆心为E，连接D点和圆心，设与圆交点为C，则|CD|便是的最小值；由上面知△AOB为等边三角形，边长为2；∴|BE|=1，|BD|=2，∠EBD=120°；∴在△BED中由余弦定理得【点评】考查数量积的计算公式，向量夹角的范围，两向量垂直的充要条件，直径所对圆周角为直角，以及余弦定理，圆外一点到圆的最近距离．二、选择题（本大题共4题，满分20分）每题有且只有一个正确答案，考生应在答题纸的相应题号上，将所选答案的代号涂黑，选对得5分，否则一律零分.15．设全集U=R，已知0}，B={x||x﹣1|＜2}，则（∁U A）∩B=（）A1）B．（﹣1，﹣2] C．（2，3] D．[2，3）【考点】交、并、补集的混合运算．【专题】集合．【分析】求出A与B中不等式的解集确定出A与B，找出A补集与B的交集即可．【解答】解：由A中不等式变形得：（2x+3）（x﹣2）＞0，解得：x x＞2，即A=2，+∞），∴∁U A=[2]，由B中不等式变形得：﹣2＜x﹣1＜2，解得：﹣1＜x＜3，即B=（﹣1，3），∴（∁U A）∩B=（﹣1，2]，故选：B．【点评】此题考查了交、并、补集的混合运算，熟练掌握各自的定义是解本题的关键．16．设a∈R，则“a=﹣1”是“f（x）=|（ax﹣2）x|在（0，+∞）上单调递增”的（）A．充要条件 B．既不充分也不必要条件C．充分不必要条件D．必要不充分条件【考点】必要条件、充分条件与充要条件的判断．【专题】简易逻辑．【分析】根据二次函数的性质结合充分必要条件的定义进行判断即可．【解答】解：①若a=﹣1，则f（x）=|（﹣x﹣2）x|=|（x+2）x|，x∈（0，+∞）如图示：f（x）在（0，+∞）单调递增，∴“a=﹣1”是“f（x）=|（ax﹣2）x|在（0，+∞）上单调递增”的充分条件；②若f（x）=|（ax﹣2）x|在（0，+∞）上单调递增，a＞0时，f（x）在（0a≤0时，f（x）在（0，+∞）单调递增，∴f（x）=|（ax﹣2）x|在（0，+∞）上单调递增推不出a=﹣1，不是必要条件，故选：C．【点评】本题考查了充分必要条件，考查二次函数的性质，是一道中档题．17．一个四棱锥的三视图如图所示，则此四棱锥的体积为（）A．24 B．16 C．12 D．8【考点】由三视图求面积、体积．【专题】空间位置关系与距离．【分析】画出图形，利用三视图的数据，求解棱锥的体积即可．【解答】解：由题意可知几何体为如图所示的四棱锥：棱锥的底面是边长为：2，3的矩形，棱锥的高为4，四棱锥的体积为：．故选：D．【点评】本题考查三视图与几何体是直观图的关系，几何体的体积的求法，考查计算能力．18．设函数f（x）（x∈R）满足f（﹣x）=f（x），f（x）=f（2﹣x），且当x∈[0，1]时，f（x）=x2．又函数g（x）=|sin（πx）|，则函数h（x）=g（x）﹣f（x）在区间[﹣1，3]上零点的个数为（）A．6 B．7 C．8 D．9【考点】根的存在性及根的个数判断．【专题】函数的性质及应用．【分析】根据条件判断函数f（x）的周期性，令h（x）=0，得g（x）=f（x），分别作出函数f（x）和g（x）的图象，利用图象判断两个函数的交点个数即可得到结论．【解答】解：∵f（﹣x）=f（x），f（x）=f（2﹣x），∴f（x）=f（2﹣x）=f（x﹣2），即函数是偶函数，且函数是周期为2的周期数列，设x∈[﹣1，0]，则﹣x∈[0，1]，则f（x）=f（﹣x）=（﹣x）2=x2，即f（x）=x2．x∈[﹣1，1]，由h（x）=g（x）﹣f（x）=0，则f（x）=g（x），∵g（x）=|sin（πx）|，∴在坐标系中作出函数f（x），g（x）的图象如图：由图象可知，两个图象的交点个数为6个，故函数h（x）=g（x）﹣f（x）在区间[﹣1，3]上零点的个数为6个，故选：A【点评】本题主要考查函数零点个数的判断，利用数形结合转化为两个函数的图象交点个数是解决本题的关键．三、解答题（本大题共5题，满分74分）解答下列各题必须在答题纸的规定区域内写出必要的步骤.19．函数f（x）=m+log a x（a＞0且a≠1）的图象过点（8，2）和（1，﹣1）．（Ⅰ）求函数f（x）的解析式；（Ⅱ）令g（x）=2f（x）﹣f（x﹣1），求g（x）的最小值及取得最小值时x的值．【考点】函数解析式的求解及常用方法；基本不等式．【专题】综合题．【分析】（1）根据题意，将点的坐标代入即可；（2）先求出g（x）的表达式，观察到函数是复合函数，故应该先研究真数的范围再利用对数函数的单调性求出最值．【解答】解得m=﹣1，a=2，故函数解析式为f（x）=﹣1+log2x，（Ⅱ）g（x）=2f（x）﹣f（x﹣1）=2（﹣1+log2x）﹣[﹣1+log2（x﹣1）其中x＞1，因为x=2时，“=”成立，而函数y=log2x﹣1在（0故当x=2时，函数g（x）取得最小值1．【点评】该题目第一问是送分的，第二问比较有难度，解题时应该注意复合函数的最值拆分开来求：本题先分离常数利用基本不等式求真数的范围，利用对数函数的单调性求出最值．20．在如图所示的直四棱柱ABCD﹣A1B1C1D1中，底面ABCD是边长为2的菱形，且∠BAD=60°，AA1=4．（1）求直四棱柱ABCD﹣A1B1C1D1的体积；（2）求异面直线AD1与BA1所成角的大小．【考点】棱柱、棱锥、棱台的体积；异面直线及其所成的角．【专题】空间角；立体几何．【分析】（1）根据体积公式得出：菱形ABCD的面积×h即可，关键求面积，高．（2）根据性质得出：∠A1BC1等于异面直线AD1与BA1所成角．在△A1BC1中，由余弦定理可求解．【解答】解：（1）因菱形ABCD的面积为AB2故直四棱柱ABCD﹣A1B1C1D1的体积为：S底面ABCD•AA1（2）连接BC1，A1C1，易知BC1∥AD1，故∠A1BC1等于异面直线AD1与BA1所成角．由已知，可得A1B=BC1A1C1则在△A1BC1中，由余弦定理，得cos∠A1BC1故异面直线AD1与BA所成角的大小为【点评】本题考查了空间几何体的性质，运用求解体积，空间想象能力，思维能力的运用，属于中档题．21．如图，经过村庄A有两条夹角60°为的公路AB，AC，根据规划拟在两条公路之间的区域内建一工厂P，分别在两条公路边上建两个仓库M，N（异于村庄A），要求PM=PN=MN=2（单位：千米）．记∠AMN=θ．（1）将AN，AM用含θ的关系式表示出来；（2）如何设计（即AN，AM为多长时），使得工厂产生的噪声对居民的影响最小（即工厂与村庄的距离AP最大）？【考点】在实际问题中建立三角函数模型．【专题】解三角形．【分析】（1）根据正弦定理，即可θ表示出AN，AM；（2）设AP2=f（θ），根据三角函数的公式，以及辅助角公式即可化简f（θ）；根据三角函数的图象和性质，即可求出函数的最值．【解答】解：：（1）∠AMN=θ，在△AMN所以（2）AP2=AM2+MP2﹣2AM•MP•cos∠AMP2（θ+60°）+4θ+60°）cos（θ+60°）﹣cos（2θ（2θ+120°）+4（2θ+120°）+cos（2θ+120°）（2θ+150°），θ∈（0°，120°）（其中利用诱导公式可知sin（120°﹣θ）=sin（θ+60°））当且仅当2θ+150°=270°，即θ=60°时，工厂产生的噪声对居民的影响最小，此时AN=AM=2．故答案为：（1）（2）AN=AM=2时，工厂产生的噪声对居民的影响最小．【点评】本题主要考查与三角函数有关的应用问题，利用正弦定理以及三角函数的三角公式是解决本题的关键．22．已知圆F1：（x+1）2+y2=8，点F2（1，0），点Q在圆F1上运动，QF2的垂直平分线交QF1于点P．（1）求动点P的轨迹C的方程；（2）设M、N分别是曲线C上的两个不同点，且点M在第一象限，点N在第三象限，若O为坐标原点，求直线MN的斜率；（3l交曲线C于A、B两点，求证：以AB为直径的圆恒过定点T（0，1）．【考点】直线与圆锥曲线的综合问题；轨迹方程．【专题】圆锥曲线中的最值与范围问题．【分析】（1）如图所示，|PF1|+|PF2|=|QF1|F1F2|=2，可知：动点P的轨迹为椭a＞b＞0），可得c=1，b2=a2﹣c2．即可得出椭圆C的方程．（2）设M（x1，y1），N（x2，y2），F1（﹣1，0x1+2x2=﹣2，y1+2y2=0k MN（3）假设在y轴上存在定点T（0，t），使以AB为直径的圆恒过这个点．设直线AB的方程为y=kx A（x1，y1），B（x2，y2）．联立直线与椭圆方程化为（1+2k2）x2，，解出即可．【解答】解：（1）如图所示，∵|PF1|+|PF2|=|QF1|F1F2|=2，∴动点P a＞b＞0），c=1，b2=1．∴方程C．（2）设M（x1，y1），N（x2，y2），F1（﹣1，0∴x1+2x2=﹣2，y1+2y2=0．∴x1=﹣2﹣2x2，y1=﹣2y2∴k MN（3）假设在y轴上存在定点T（0，t），使以AB为直径的圆恒过这个点．设直线AB的方程为y=kx A（x1，y1），B（x2，y2）．（x1，y1﹣t）•（x2，y2﹣t）=x1x2+（y1﹣t）（y2﹣t）=x1x22=（1+k2）x1x2k+kt）（x1+x2）（x1+x2）2=0，1+2k2）x2，△＞0恒成立．∴x1+x2x1x2=代入上式可得：2=0，化为（18t2﹣18）k2+（9t2+6t ﹣15）=0，t=1．满足△＞0．∴在y轴上存在定点T（0，1），使以AB为直径的圆恒过这个点T．【点评】本题考查了椭圆的定义及其标准方程、直线与椭圆相交问题转化为方程联立可得根与系数的关系、圆的性质、向量坐标运算，考查了推理能力与计算能力，属于难题．23．设各项均为正数的数列{a n}的前n项和为S n，且满足：a1=1，4S n=（a n+1）2（n∈N*）．（1）求数列{a n}的通项公式；（2）设b n N*b1+b2+…+b n﹣2n）的值；（3）是否存在大于2的正整数m、k，使得a m+a m+1+a m+2+…+a m+k=300？若存在，求出所有符合条件的m、k；若不存在，请说明理由．【考点】数列的极限；数列的求和．【专题】点列、递归数列与数学归纳法．【分析】（1）通过4a n+1=4S n+1﹣4S n得（a n+1+a n）（a n+1﹣a n﹣2）=0，进而可得结论；（2）通过分离b n的分母可得b n=2+2（3）假设存在大于2的正整数m、k使得a m+a m+1+…+a m+k=300，通过（1）可得300=（2m+k ﹣1）（k+1），利用2m+k﹣1＞k+1≥4，且2m+k﹣1与k+1的奇偶性相同，即得结论．【解答】解：（1）∵4S n=（a n+1）2，∴4S n+1=（a n+1+1）2，两式相减，得4a n+1=4S n+1﹣4S n=（a n+1）2﹣（a n+1+1）2n+1﹣2a n，化简得（a n+1+a n）（a n+1﹣a n﹣2）=0，又∵数列{a n}各项均为正数，∴a n+1﹣a n=2 （n∈N*），∴数列{a n}是以1为首项，2为公差的等差数列，∴a n=2n﹣1 （n∈N*）．（2）因为b n故b1+b2+…+b n=2n+2[（1+]=2n+2（1b1+b2+…+b n﹣2n）（1；（3）结论：存在大于2的正整数m、k使得a m+a m+1+…+a m+k=300．理由如下：假设存在大于2的正整数m、k使得a m+a m+1+…+a m+k=300，由（1），可得a m+a m+1+…+a m+k=（2m+k﹣1）（k+1），从而（2m+k﹣1）（k+1）=300，由于正整数m、k均大于2，知2m+k﹣1＞k+1≥4，且2m+k﹣1与k+1的奇偶性相同，故由300=22×3×52，得因此，存在大于2的正整数m、k a m+a m+1+…+a m+k=300．【点评】本题考查求数列的通项，涉及到极限等知识，注意解题方法的积累，属于中档题．。

上海市虹口区2015届高三第三次模拟考试数学理试题（无答案）

上海市虹⼝区2015届⾼三第三次模拟考试数学理试题（⽆答案）2015年虹⼝区第三次模拟考试（理科试卷）满分150分，时间120分钟2015.05.19⼀、填空题（本⼤题满分56分）本⼤题共14题，只要求在答题纸相应题号的空格内直接填写结果，每个空格填对得4分，否则⼀律得零分. 1.若{}{}2,,0,2,4A a aB ==，且A B ?，则实数a 的值为_________.2.不等式220x x-≤的解集为_________.3.已知20151232z i i =+（i 为虚数单位），则复数z 的值为_______.4.6x ?展开式的常数项为___________.5.已知数列121,,,9a a 是等差数列，数列1231,,,,9b b b 是等⽐数列，则122a ab +的值为_______. 6.⼀个算法的程序框图如右，则其输出结果为_________.7.随机投掷两枚质地均匀的骰⼦，它们向上的点数之积为6的倍数的概率为_________. 8.设函数()()()log 0,1a f x x b a a =+>≠的图象过点()2,1，其反函数的图象过点()2,8，则a b +等于_________.9.已知向量()()()1,cos ,sin ,10m x n x ωωω==->，函数()f x m n =?，且()f x 图象上⼀个最⾼点的坐标为38π?，与之相邻的⼀个最低点的坐标为8π?- ?，，则()f x 的解析式为__________.10.如果⼀个球的外切圆锥的⾼是这个球半径的3倍，那么圆锥侧⾯积和球⾯积的⽐为___________.11.在极坐标系中，若直线sin 4a πρθ??+= ??被圆2ρ=截得的弦长为，则实数a =__________.12.已知,a b 为平⾯内两个互相垂直的单位向量，若向量c 满⾜()()c a c b R λλ+=+∈，则c 的最⼩值为_________.13.关于x 的实系数⼀元⼆次⽅程230x px ++=的两个虚根为1z 、2z ，若1z 、2z 在复平⾯上对应的点是经过原点的椭圆的两个焦点，则该椭圆的长轴长为_______. 14.关于曲线2211:1C x y +=，有如下结论：①曲线C 关于原点对称；②曲线C 关于直线0x y ±=对称；③曲线C 是封闭图形，且封闭图形的⾯积⼤于2π；④曲线C 不是封闭图形，且它与圆222x y +=⽆公共点；⑤曲线C与曲线:D x y +=4个交点，这4点构成正⽅形.其中所有正确结论的序号为__________.⼆、选择题（本题共4题，满分20分）每题只有⼀个正确答案，考⽣在答题纸的相应题号上，将所选答案的代号涂⿊，选对得5分，否则⼀律零分.15.对⼀个容量为N 的总体抽取容量为n 的样本，当选取简单随机抽样、系统抽样和分层抽样三种不同⽅法抽取样本时，总体中每个个体被抽中的概率分别是1p 、2p 、3p ，则（） A.123p p p =<B. 321p p p =<C. 132p p p =<D.123p p p ==16.对于函数()f x ，若存在0a ≠，使得x 取定义域内的每⼀个值，都有()()2f x f a x =-，则称()f x 为准偶函数，下列函数是准偶函数的是（） A. ()()cos 1f x x =+B. ()f x x =C. ()tan f x x =D. ()3f x x =17.点,,,A B C D 是半径为1的球⾯上的四个点，线段,,AB AC AD 两两垂直，若⽤123,,S S S 分别表⽰,,ABC ABD ACD 的⾯积，则123S S S ++的最⼤值为（） A.12B.2C.4D.8ABCDO18.已知12,F F 是双曲线()222:10y C x b b-=>的两个焦点，P 是双曲线C12PF PF ⊥，2PF 与两渐近线相交于,M N 两点（如图），若点N 恰好平分线段2PF ，则双曲线C 的焦距为（） A. B.C.D.4三、解答题（本⼤题共5题，满分74分）解答下列各题必须在答题纸的规定区域内写出必要的步骤.19.（本题满分12分）本题共2⼩题，第1⼩题6分，第2⼩题6分. 如图，在所有棱长为2的正三棱柱111ABC A B C -中，D 为11AC 的中点，求（1）直线AD 与平⾯1B DC 所成⾓的⼤⼩；（2）点A 到平⾯1B DC 的距离.20.（本题满分14分）本题共2⼩题，第1⼩题7分，第2⼩题7分.已知函数()22sin 24f x x x π??=+ ，,42x ππ??∈.（1）求函数()f x 的最⼤值与最⼩值；（2）若不等式()2f x m -<在,42x ππ??∈恒成⽴，求实数m 的取值范围.21.（本题满分14分）本题共2⼩题，第1⼩题6分，第2⼩题8分. 已知抛物线2:4C y x =的焦点为F .（1）点,A F 满⾜2AP AF =，当点A 在抛物线上运动时，求动点P 的轨迹⽅程；（2）在x 轴上是否存在点Q ，使得点Q 关于直线2y x =的对称点在抛物线C 上？如果存在，求出所有满⾜条件的点Q 的坐标，若不存在，请说明理由.ABC1B 1C 1A D22.（本题满分16分）本题共3⼩题，第1⼩题4分，第2⼩题6分，第3⼩题6分. 定义⾮零向量(),OM a b =的“相伴函数”为()sin cos f x a x b x =+（R x ∈），向量(),OM a b =称为函数()sin cos f x a x b x =+的“相伴向量”（其中O 为坐标原点），记平⾯内所有向量的“相伴函数”构成的集合为S .（1）已知点(),M a b 满⾜250a b +-=，求向量OM 的模的最⼩值；（2）设()()c o s 2c o s 6h x x x a π?=+-+（R a ∈），求证：()h x S ∈，并求函数()h x 的“相伴向量”模的取值范围；（3）已知点(),M a b （0b ≠）满⾜(()2211a b -+-=，向量OM 的“相伴函数”()f x 在0x x =处取得最⼤值，当M 点运动时，求0tan 2x 的取值范围.23.（本题满分18分）本题共3⼩题，第1⼩题5分，第2⼩题6分，第3⼩题7分. 某情报部门获取⼀些信息密码需要破译，密码可以看成⼀个数列，记为{}n a ，并且满⾜11a >，()()()*612N n n n S a a n =++∈，其中n S 表⽰该数列的前n 项和.（1）求出1a 的值，对任意的*N n ∈，写出1n a +与n a 的关系的各种表达式；（2）有⼀条密码可由数列{}n a 的前100项组成，试问共可组成多少条不同的密码，为什么？（3）若⼀条密码可由数列的前10项组成，且10100S =，试问组成的密码是否唯⼀确定？并说明理由.。

【精选高中试题】上海市虹口区高考数学二模试卷文(含解析)

2015年上海市虹口区高考数学二模试卷（文科）一、填空题（本大题满分56分）本大题共14题，只要求在答题纸相应题号的空格内直接填写结果，每个空格填对得4分，否则一律得零分.1．计算： = ．（i是虚数单位）2．已知函数f（x）=，则f（f（﹣3））= ．3．函数f（x）=ln（+1）（x＞0）的反函数f﹣1（x）= ．4．已知正实数x，y满足x+3y=1，则xy的最大值为．5．已知复数z=3sinθ+icosθ（i是虚数单位），且|z|=，则当θ为钝角时，tanθ= ．6．在上海高考改革方案中，要求每位高中生必须在物理、化学、生物、政治、历史、地理6门学科（3门理科学科，3门文科学科）中选择3门学科参加等级考试，小丁同学理科成绩较好，决定至少选择两门理科学科，那么小丁同学的选科方案有种．7．设数列{a n}前n项的和为S n，若a1=4，且a n+1=3S n（n∈N*），则S n= ．8．已知抛物线y2=2px（p＞0）的焦点在圆（x﹣1）2+y2=4上，则p= ．9．若二项式展开式中含x2项的系数为，则= ．10．若行列式的第1行第2列的元素1的代数余子式为﹣1，则实数x的取值集合为．11．如图所示，已知F1，F2为双曲线的两个焦点，且|F1F2|=2，若以坐标原点O为圆心，|F1F2|为直径的圆与该双曲线的左支相交于A，B两点，且△F2AB为正三角形，则双曲线的实轴长为．12．设二元一次不等式组所表示的平面区域为M，若函数y=a x（a＞0，且a≠1）的图象经过区域M，则实数a的取值范围为．13．已知直线l1：12x﹣5y+15=0和l2：x=﹣2，点P为抛物线y2=8x上的动点，则点P到直线l1和直线l2的距离之和的最小值为．14．已知向量，满足，且，则|2﹣|的最小值为．二、选择题（本大题共4题，满分20分）每题有且只有一个正确答案，考生应在答题纸的相应题号上，将所选答案的代号涂黑，选对得5分，否则一律零分.15．设全集U=R，已知A={x|＞0}，B={x||x﹣1|＜2}，则（∁U A）∩B=（）A．（﹣，﹣1）B．（﹣1，﹣2] C．（2，3] D．[2，3）16．设a∈R，则“a=﹣1”是“f（x）=|（ax﹣2）x|在（0，+∞）上单调递增”的（）A．充要条件 B．既不充分也不必要条件C．充分不必要条件D．必要不充分条件17．一个四棱锥的三视图如图所示，则此四棱锥的体积为（）A．24 B．16 C．12 D．818．设函数f（x）（x∈R）满足f（﹣x）=f（x），f（x）=f（2﹣x），且当x∈[0，1]时，f（x）=x2．又函数g（x）=|sin（πx）|，则函数h（x）=g（x）﹣f（x）在区间[﹣1，3]上零点的个数为（）A．6 B．7 C．8 D．9三、解答题（本大题共5题，满分74分）解答下列各题必须在答题纸的规定区域内写出必要的步骤. 19．函数f（x）=m+log a x（a＞0且a≠1）的图象过点（8，2）和（1，﹣1）．（Ⅰ）求函数f（x）的解析式；（Ⅱ）令g（x）=2f（x）﹣f（x﹣1），求g（x）的最小值及取得最小值时x的值．20．在如图所示的直四棱柱ABCD﹣A1B1C1D1中，底面ABCD是边长为2的菱形，且∠BAD=60°，AA1=4．（1）求直四棱柱ABCD﹣A1B1C1D1的体积；（2）求异面直线AD1与BA1所成角的大小．21．如图，经过村庄A有两条夹角60°为的公路AB，AC，根据规划拟在两条公路之间的区域内建一工厂P，分别在两条公路边上建两个仓库M，N（异于村庄A），要求PM=PN=MN=2（单位：千米）．记∠AMN=θ．（1）将AN，AM用含θ的关系式表示出来；（2）如何设计（即AN，AM为多长时），使得工厂产生的噪声对居民的影响最小（即工厂与村庄的距离AP最大）？22．已知圆F1：（x+1）2+y2=8，点F2（1，0），点Q在圆F1上运动，QF2的垂直平分线交QF1于点P．（1）求动点P的轨迹C的方程；（2）设M、N分别是曲线C上的两个不同点，且点M在第一象限，点N在第三象限，若，O为坐标原点，求直线MN的斜率；（3）过点的动直线l交曲线C于A、B两点，求证：以AB为直径的圆恒过定点T（0，1）．23．设各项均为正数的数列{a n}的前n项和为S n，且满足：a1=1，4S n=（a n+1）2（n∈N*）．（1）求数列{a n}的通项公式；（2）设b n=+（∈N*），试求（b1+b2+…+b n﹣2n）的值；（3）是否存在大于2的正整数m、k，使得a m+a m+1+a m+2+…+a m+k=300？若存在，求出所有符合条件的m、k；若不存在，请说明理由．2015年上海市虹口区高考数学二模试卷（文科）参考答案与试题解析一、填空题（本大题满分56分）本大题共14题，只要求在答题纸相应题号的空格内直接填写结果，每个空格填对得4分，否则一律得零分.1．计算： = ﹣i ．（i是虚数单位）【考点】复数代数形式的乘除运算；虚数单位i及其性质．【专题】数系的扩充和复数．【分析】由虚数单位i的运算性质化简，然后利用复数代数形式的乘除运算化简求值．【解答】解： =．故答案为：﹣i．【点评】本题考查了复数代数形式的乘除运算，是基础的计算题．2．已知函数f（x）=，则f（f（﹣3））= ．【考点】函数的值．【专题】计算题；函数的性质及应用．【分析】由分段函数f（x）=，先求f（﹣3），再求f（f（﹣3））即可．【解答】解：∵函数f（x）=，∴f（﹣3）=2﹣3=，f（f（﹣3））=f（）==，故答案为：．【点评】本题考查了分段函数的简单应用，属于基础题．3．函数f（x）=ln（+1）（x＞0）的反函数f﹣1（x）= ，x∈（0，+∞）．【考点】反函数．【专题】函数的性质及应用．【分析】直接利用反函数的求法求解即可．【解答】解：函数f（x）=ln（+1）（x＞0），f（x）∈（0，+∞）．+1=e y，解得x=，函数f（x）=ln（+1）（x＞0）的反函数f﹣1（x）=，x∈（0，+∞）．故答案为：，x∈（0，+∞）．【点评】本题考查反函数与原函数的关系，考查计算能力．注意函数的定义域．4．已知正实数x，y满足x+3y=1，则xy的最大值为．【考点】基本不等式．【专题】不等式的解法及应用．【分析】运用基本不等式得出x+3y=1，化简求解xy即可．【解答】解；∵正实数x，y满足x+3y=1，∴x+3y=1，化简得出xy（x=3y=等号成立）xy的最大值为（=，y=等号成立）故答案为；【点评】本题考查了运用基本不等式求解二元式子的最值问题，关键是判断，变形得出不等式的条件，属于容易题．5．已知复数z=3sinθ+icosθ（i是虚数单位），且|z|=，则当θ为钝角时，tanθ= ﹣1 ．【考点】复数求模．【专题】三角函数的求值；数系的扩充和复数．【分析】直接利用复数的模，得到θ的三角方程，然后求解即可．【解答】解：复数z=3sinθ+icosθ（i是虚数单位），且|z|=，可得9sin2θ+cos2θ=5，可得sin2θ=，当θ为钝角时，sinθ=，θ=135°，∴tanθ=﹣1．故答案为：﹣1．【点评】本题考查复数的模以及三角函数的化简求值，考查计算能力．6．在上海高考改革方案中，要求每位高中生必须在物理、化学、生物、政治、历史、地理6门学科（3门理科学科，3门文科学科）中选择3门学科参加等级考试，小丁同学理科成绩较好，决定至少选择两门理科学科，那么小丁同学的选科方案有10 种．【考点】计数原理的应用．【专题】应用题；排列组合．【分析】分类讨论：选择两门理科学科，一门文科学科；选择三门理科学科，即可得出结论．【解答】解：选择两门理科学科，一门文科学科，有=9种；选择三门理科学科，有1种，故共有10种．故答案为：10．【点评】本题考查计数原理的应用，考查学生的计算能力，比较基础．7．设数列{a n}前n项的和为S n，若a1=4，且a n+1=3S n（n∈N*），则S n= 4n．【考点】数列的求和；数列递推式．【专题】等差数列与等比数列．【分析】a n+1=3S n（n∈N*），变形为S n+1﹣S n=3S n，S n+1=4S n，再利用等比数列的通项公式即可得出．【解答】解：∵a n+1=3S n（n∈N*），∴S n+1﹣S n=3S n，化为S n+1=4S n，∴数列{S n}是等比数列，首项为4，公比为4．∴S n=4n．故答案为：4n．【点评】本题考查了递推式的应用、等比数列的通项公式，考查了推理能力与计算能力，属于中档题．8．已知抛物线y2=2px（p＞0）的焦点在圆（x﹣1）2+y2=4上，则p= 6 ．【考点】抛物线的简单性质．【专题】计算题；圆锥曲线的定义、性质与方程．【分析】求出抛物线的焦点（，0），把它代入圆的方程求出p的值．【解答】解：抛物线y2=2px（p＞0）的焦点为（，0），代入圆（x﹣1）2+y2=4得（﹣1）2=4，∴p=6，故答案为：6．【点评】本题考查由抛物线的方程求焦点坐标，以及点在圆上的性质．9．若二项式展开式中含x2项的系数为，则= ．【考点】极限及其运算；二项式系数的性质．【专题】计算题；二项式定理．【分析】根据二项式展开式的通项公式求出展开式中含x2项的系数，得出a的值；再计算的值．【解答】解：∵二项式展开式的通项公式为T r+1=•x6﹣r•=（﹣a）r••，令6﹣r=2，解得r=3；∴展开式中含x2项的系数为（﹣a）3•=，解得a=﹣；∴===．故答案为：．【点评】本题考查了二项式定理的应用问题，也考查了数列求和的应用问题以及极限的计算问题，是基础题目．10．若行列式的第1行第2列的元素1的代数余子式为﹣1，则实数x的取值集合为{x|x=π+2kπ，k∈Z} ．【考点】三阶矩阵．【专题】三角函数的求值；矩阵和变换．【分析】本题直接根据行列式的代数余子式的定义进行计算，即可得到本题结论．【解答】解：∵行列式的第1行第2列的元素1的代数余子式为﹣1，∴﹣=﹣1，∴sin（π+x）﹣=1，∴﹣sinx﹣×（cosx﹣sinx）=1，即cosx=﹣1，∴x=π+2kπ（k∈Z），故答案为：{x|x=π+2kπ，k∈Z}．【点评】本题考查了行列式的代数余子式，三角函数的计算，记住常用常见角的三角函数值是解决本题的关键，注意解题方法的积累，属于中档题．11．如图所示，已知F1，F2为双曲线的两个焦点，且|F1F2|=2，若以坐标原点O为圆心，|F1F2|为直径的圆与该双曲线的左支相交于A，B两点，且△F2AB为正三角形，则双曲线的实轴长为﹣1 ．【考点】双曲线的简单性质．【专题】计算题；圆锥曲线的定义、性质与方程．【分析】根据△F2AB是等边三角形，判断出∠AF2F1=30°，进而在RT△AF1F2中求得|AF1|，|AF2|，进而根据双曲线的简单性质求得a可得．【解答】解：∵△F2AB是等边三角形，∴∠AF2F1=30°，∵|F1F2|=2，∴|AF1|=1，|AF2|=，∴a=，∴2a=﹣1．故答案为：﹣1．【点评】本题主要考查了双曲线的简单性质．考查了学生综合分析问题和数形结合的思想的运用．属基础题．12．设二元一次不等式组所表示的平面区域为M，若函数y=a x（a＞0，且a≠1）的图象经过区域M，则实数a的取值范围为[2，9] ．【考点】简单线性规划的应用．【专题】不等式的解法及应用．【分析】先依据不等式组，结合二元一次不等式（组）与平面区域的关系画出其表示的平面区域，再利用函数y=a x（a＞0，a≠1）的图象特征，结合区域的角上的点即可解决问题【解答】解：平面区域M如如图所示．求得A（2，10），C（3，8），B（1，9）．由图可知，欲满足条件必有a＞1且图象在过B、C两点的图象之间．当图象过B点时，a1=9，∴a=9．当图象过C点时，a3=8，∴a=2．故a的取值范围为[2，9]．【点评】本题主要考查了用平面区域二元一次不等式组、指数函数的图象与性质，以及简单的转化思想和数形结合的思想，属中档题．巧妙识别目标函数的几何意义是我们研究规划问题的基础．13．已知直线l1：12x﹣5y+15=0和l2：x=﹣2，点P为抛物线y2=8x上的动点，则点P到直线l1和直线l2的距离之和的最小值为 3 ．【考点】直线与圆锥曲线的关系．【专题】圆锥曲线的定义、性质与方程．【分析】由抛物线方程求出其焦点坐标和准线方程，把抛物线y2=8x上的点P到两直线l1：x=﹣2，l2：12x﹣5y+15=0的距离之和的最小值转化为焦点到l2：12x﹣5y+15=0的距离，由点到直线的距离公式求解．【解答】解：如图，由抛物线y2=8x，得其焦点F（2，0），准线方程为x=﹣2．∴l1：x=﹣2为抛物线的准线，P到两直线l1：x=﹣2，l2：12x﹣5y+15=0的距离之和，即为P到F和l2：12x﹣5y+15=0的距离之和．最小值为F到l2：12x﹣5y+15=0的距离．故答案为：3．【点评】本题考查了直线与圆锥曲线的关系，考查了数形结合的解题思想方法和数学转化思想方法，是中档题．14．已知向量，满足，且，则|2﹣|的最小值为﹣1 ．【考点】平面向量数量积的运算．【专题】平面向量及应用．【分析】可设，根据已知条件容易判断出△AOB为等边三角形，且边长为2，而C点在以AB为直径的圆上，延长OB到D，使|OB|=|BD|，这样即可得到．而，连接D和圆心E，设C点是与圆的交点，从而|CD|便是的最小值，而由余弦定理可求出|DE|，而圆半径为1，从而能得出|CD|的值．【解答】解：由已知条件知cos＜＞=；∴；设，∵；∴；∴；∴C点在以AB为直径的圆上，如下图所示：延长OB到D，使|OB|=|BD|，连接CD；则，；设圆心为E，连接D点和圆心，设与圆交点为C，则|CD|便是|2|的最小值；由上面知△AOB为等边三角形，边长为2；∴|BE|=1，|BD|=2，∠EBD=120°；∴在△BED中由余弦定理得|ED|=；∴的最小值为．故答案为：．【点评】考查数量积的计算公式，向量夹角的范围，两向量垂直的充要条件，直径所对圆周角为直角，以及余弦定理，圆外一点到圆的最近距离．二、选择题（本大题共4题，满分20分）每题有且只有一个正确答案，考生应在答题纸的相应题号上，将所选答案的代号涂黑，选对得5分，否则一律零分.15．设全集U=R，已知A={x|＞0}，B={x||x﹣1|＜2}，则（∁U A）∩B=（）A．（﹣，﹣1）B．（﹣1，﹣2] C．（2，3] D．[2，3）【考点】交、并、补集的混合运算．【专题】集合．【分析】求出A与B中不等式的解集确定出A与B，找出A补集与B的交集即可．【解答】解：由A中不等式变形得：（2x+3）（x﹣2）＞0，解得：x＜﹣或x＞2，即A=（﹣∞，﹣）∪（2，+∞），∴∁U A=[﹣，2]，由B中不等式变形得：﹣2＜x﹣1＜2，解得：﹣1＜x＜3，即B=（﹣1，3），∴（∁U A）∩B=（﹣1，2]，故选：B．【点评】此题考查了交、并、补集的混合运算，熟练掌握各自的定义是解本题的关键．16．设a∈R，则“a=﹣1”是“f（x）=|（ax﹣2）x|在（0，+∞）上单调递增”的（）A．充要条件 B．既不充分也不必要条件C．充分不必要条件D．必要不充分条件【考点】必要条件、充分条件与充要条件的判断．【专题】简易逻辑．【分析】根据二次函数的性质结合充分必要条件的定义进行判断即可．【解答】解：①若a=﹣1，则f（x）=|（﹣x﹣2）x|=|（x+2）x|，x∈（0，+∞）如图示：，f（x）在（0，+∞）单调递增，∴“a=﹣1”是“f（x）=|（ax﹣2）x|在（0，+∞）上单调递增”的充分条件；②若f（x）=|（ax﹣2）x|在（0，+∞）上单调递增，a＞0时，f（x）在（0，）递增，在（，）递减，在（，+∞）递增，a≤0时，f（x）在（0，+∞）单调递增，∴f（x）=|（ax﹣2）x|在（0，+∞）上单调递增推不出a=﹣1，不是必要条件，故选：C．【点评】本题考查了充分必要条件，考查二次函数的性质，是一道中档题．17．一个四棱锥的三视图如图所示，则此四棱锥的体积为（）A．24 B．16 C．12 D．8【考点】由三视图求面积、体积．【专题】空间位置关系与距离．【分析】画出图形，利用三视图的数据，求解棱锥的体积即可．【解答】解：由题意可知几何体为如图所示的四棱锥：棱锥的底面是边长为：2，3的矩形，棱锥的高为4，四棱锥的体积为：=8．故选：D．【点评】本题考查三视图与几何体是直观图的关系，几何体的体积的求法，考查计算能力．18．设函数f（x）（x∈R）满足f（﹣x）=f（x），f（x）=f（2﹣x），且当x∈[0，1]时，f（x）=x2．又函数g（x）=|sin（πx）|，则函数h（x）=g（x）﹣f（x）在区间[﹣1，3]上零点的个数为（）A．6 B．7 C．8 D．9【考点】根的存在性及根的个数判断．【专题】函数的性质及应用．【分析】根据条件判断函数f（x）的周期性，令h（x）=0，得g（x）=f（x），分别作出函数f（x）和g（x）的图象，利用图象判断两个函数的交点个数即可得到结论．【解答】解：∵f（﹣x）=f（x），f（x）=f（2﹣x），∴f（x）=f（2﹣x）=f（x﹣2），即函数是偶函数，且函数是周期为2的周期数列，设x∈[﹣1，0]，则﹣x∈[0，1]，则f（x）=f（﹣x）=（﹣x）2=x2，即f（x）=x2．x∈[﹣1，1]，由h（x）=g（x）﹣f（x）=0，则f（x）=g（x），∵g（x）=|sin（πx）|，∴在坐标系中作出函数f（x），g（x）的图象如图：由图象可知，两个图象的交点个数为6个，故函数h（x）=g（x）﹣f（x）在区间[﹣1，3]上零点的个数为6个，故选：A【点评】本题主要考查函数零点个数的判断，利用数形结合转化为两个函数的图象交点个数是解决本题的关键．三、解答题（本大题共5题，满分74分）解答下列各题必须在答题纸的规定区域内写出必要的步骤. 19．函数f（x）=m+log a x（a＞0且a≠1）的图象过点（8，2）和（1，﹣1）．（Ⅰ）求函数f（x）的解析式；（Ⅱ）令g（x）=2f（x）﹣f（x﹣1），求g（x）的最小值及取得最小值时x的值．【考点】函数解析式的求解及常用方法；基本不等式．【专题】综合题．【分析】（1）根据题意，将点的坐标代入即可；（2）先求出g（x）的表达式，观察到函数是复合函数，故应该先研究真数的范围再利用对数函数的单调性求出最值．【解答】解：（Ⅰ）由得，解得m=﹣1，a=2，故函数解析式为f（x）=﹣1+log2x，（Ⅱ）g（x）=2f（x）﹣f（x﹣1）=2（﹣1+log2x）﹣[﹣1+log2（x﹣1）]=，其中x ＞1，因为当且仅当即x=2时，“=”成立，而函数y=log2x﹣1在（0，+∞）上单调递增，则，故当x=2时，函数g（x）取得最小值1．【点评】该题目第一问是送分的，第二问比较有难度，解题时应该注意复合函数的最值拆分开来求：本题先分离常数利用基本不等式求真数的范围，利用对数函数的单调性求出最值．20．在如图所示的直四棱柱ABCD﹣A1B1C1D1中，底面ABCD是边长为2的菱形，且∠BAD=60°，AA1=4．（1）求直四棱柱ABCD﹣A1B1C1D1的体积；（2）求异面直线AD1与BA1所成角的大小．【考点】棱柱、棱锥、棱台的体积；异面直线及其所成的角．【专题】空间角；立体几何．【分析】（1）根据体积公式得出：菱形ABCD的面积×h即可，关键求面积，高．（2）根据性质得出：∠A1BC1等于异面直线AD1与BA1所成角．在△A1BC1中，由余弦定理可求解．【解答】解：（1）因菱形ABCD的面积为AB2•sin60°=故直四棱柱ABCD﹣A1B1C1D1的体积为：S底面ABCD•AA1=（2）连接BC1，A1C1，易知BC1∥AD1，故∠A1BC1等于异面直线AD1与BA1所成角．由已知，可得A1B=BC1=，A1C1=则在△A1BC1中，由余弦定理，得cos∠A1BC1==故异面直线AD1与BA所成角的大小为arcos【点评】本题考查了空间几何体的性质，运用求解体积，空间想象能力，思维能力的运用，属于中档题．21．如图，经过村庄A有两条夹角60°为的公路AB，AC，根据规划拟在两条公路之间的区域内建一工厂P，分别在两条公路边上建两个仓库M，N（异于村庄A），要求PM=PN=MN=2（单位：千米）．记∠AMN=θ．（1）将AN，AM用含θ的关系式表示出来；（2）如何设计（即AN，AM为多长时），使得工厂产生的噪声对居民的影响最小（即工厂与村庄的距离AP最大）？【考点】在实际问题中建立三角函数模型．【专题】解三角形．【分析】（1）根据正弦定理，即可θ表示出AN，AM；（2）设AP2=f（θ），根据三角函数的公式，以及辅助角公式即可化简f（θ）；根据三角函数的图象和性质，即可求出函数的最值．【解答】解：：（1）∠AMN=θ，在△AMN中，由正弦定理得： ==所以AN=，AM=（2）AP2=AM2+MP2﹣2AM•MP•cos∠AMP=sin2（θ+60°）+4﹣sin（θ+60°）cos（θ+60°）= [1﹣cos（2θ+120°）]﹣sin（2θ+120°）+4= [sin（2θ+120°）+cos（2θ+120°）]+=﹣sin（2θ+150°），θ∈（0°，120°）（其中利用诱导公式可知sin（120°﹣θ）=sin （θ+60°））当且仅当2θ+150°=270°，即θ=60°时，工厂产生的噪声对居民的影响最小，此时AN=AM=2．故答案为：（1）AN=，AM=（2）AN=AM=2时，工厂产生的噪声对居民的影响最小．【点评】本题主要考查与三角函数有关的应用问题，利用正弦定理以及三角函数的三角公式是解决本题的关键．22．已知圆F1：（x+1）2+y2=8，点F2（1，0），点Q在圆F1上运动，QF2的垂直平分线交QF1于点P．（1）求动点P的轨迹C的方程；（2）设M、N分别是曲线C上的两个不同点，且点M在第一象限，点N在第三象限，若，O为坐标原点，求直线MN的斜率；（3）过点的动直线l交曲线C于A、B两点，求证：以AB为直径的圆恒过定点T（0，1）．【考点】直线与圆锥曲线的综合问题；轨迹方程．【专题】圆锥曲线中的最值与范围问题．【分析】（1）如图所示，|PF1|+|PF2|=|QF1|=R=2＞|F1F2|=2，可知：动点P的轨迹为椭圆，设标准方程为（a＞b＞0），可得a=，c=1，b2=a2﹣c2．即可得出椭圆C的方程．（2）设M（x1，y1），N（x2，y2），F1（﹣1，0）．由于，可得x1+2x2=﹣2，y1+2y2=0．代入椭圆方程可得=1，又，联立解出即可得出k MN=．（3）假设在y轴上存在定点T（0，t），使以AB为直径的圆恒过这个点．设直线AB的方程为y=kx﹣，A（x1，y1），B（x2，y2）．联立直线与椭圆方程化为（1+2k2）x2﹣kx﹣=0，把根与系数的关系代入=0，解出即可．【解答】解：（1）如图所示，∵|PF1|+|PF2|=|QF1|=R=2＞|F1F2|=2，∴动点P的轨迹为椭圆，设标准方程为（a＞b＞0），a=，c=1，b2=1．∴方程C为=1．（2）设M（x1，y1），N（x2，y2），F1（﹣1，0）．∵，∴x1+2x2=﹣2，y1+2y2=0．∴x1=﹣2﹣2x2，y1=﹣2y2，代入椭圆方程可得=1，又，联立解得，∴．∴k MN==．（3）假设在y轴上存在定点T（0，t），使以AB为直径的圆恒过这个点．设直线AB的方程为y=kx﹣，A（x1，y1），B（x2，y2）．则=（x1，y1﹣t）•（x2，y2﹣t）=x1x2+（y1﹣t）（y2﹣t）=x1x2+﹣t+t2=（1+k2）x1x2﹣（k+kt）（x1+x2）（x1+x2）+++t2=0，联立，化为（1+2k2）x2﹣kx﹣=0，△＞0恒成立．∴x1+x2=，x1x2=﹣．代入上式可得：﹣﹣+++t2=0，化为（18t2﹣18）k2+（9t2+6t﹣15）=0，∴，解得t=1．满足△＞0．∴在y轴上存在定点T（0，1），使以AB为直径的圆恒过这个点T．【点评】本题考查了椭圆的定义及其标准方程、直线与椭圆相交问题转化为方程联立可得根与系数的关系、圆的性质、向量坐标运算，考查了推理能力与计算能力，属于难题．23．设各项均为正数的数列{a n}的前n项和为S n，且满足：a1=1，4S n=（a n+1）2（n∈N*）．（1）求数列{a n}的通项公式；（2）设b n=+（∈N*），试求（b1+b2+…+b n﹣2n）的值；（3）是否存在大于2的正整数m、k，使得a m+a m+1+a m+2+…+a m+k=300？若存在，求出所有符合条件的m、k；若不存在，请说明理由．【考点】数列的极限；数列的求和．【专题】点列、递归数列与数学归纳法．【分析】（1）通过4a n+1=4S n+1﹣4S n得（a n+1+a n）（a n+1﹣a n﹣2）=0，进而可得结论；（2）通过分离b n的分母可得b n=2+2（﹣），累加后取极限即可；（3）假设存在大于2的正整数m、k使得a m+a m+1+…+a m+k=300，通过（1）可得300=（2m+k﹣1）（k+1），利用2m+k﹣1＞k+1≥4，且2m+k﹣1与k+1的奇偶性相同，即得结论．【解答】解：（1）∵4S n=（a n+1）2，∴4S n+1=（a n+1+1）2，两式相减，得4a n+1=4S n+1﹣4S n=（a n+1）2﹣（a n+1+1）2=﹣+2a n+1﹣2a n，化简得（a n+1+a n）（a n+1﹣a n﹣2）=0，又∵数列{a n}各项均为正数，∴a n+1﹣a n=2 （n∈N*），∴数列{a n}是以1为首项，2为公差的等差数列，∴a n=2n﹣1 （n∈N*）．（2）因为b n=+=+=2+2（﹣），故b1+b2+…+b n=2n+2[（1﹣）+（﹣）+…+（﹣）]=2n+2（1﹣），于是（b1+b2+…+b n﹣2n）= [2（1﹣）]=2；（3）结论：存在大于2的正整数m、k使得a m+a m+1+…+a m+k=300．理由如下：假设存在大于2的正整数m、k使得a m+a m+1+…+a m+k=300，由（1），可得a m+a m+1+…+a m+k=（2m+k﹣1）（k+1），从而（2m+k﹣1）（k+1）=300，由于正整数m、k均大于2，知2m+k﹣1＞k+1≥4，且2m+k﹣1与k+1的奇偶性相同，故由300=22×3×52，得或，解得或，因此，存在大于2的正整数m、k：或，使得a m+a m+1+…+a m+k=300．【点评】本题考查求数列的通项，涉及到极限等知识，注意解题方法的积累，属于中档题．。

2015届虹口区高三英语二模试卷及答案(官方版)

虹口区2015高三英语二模试卷2015.4考生注意:1. 考试时间120分钟，试卷满分150分。

2. 本考试设试卷和答题纸两部分。

试卷分为第Ⅰ卷（第1—10页）和第Ⅱ卷（第10页），全卷共10页。

第I卷第1-16小题、第41-77小题为选择题，答题必须涂在答题纸上，第I卷第17-40小题、第78-81小题和第II卷的答案必须写在答题纸上，做在试卷上一律不得分。

3．答题前，务必在答题纸上填写准考证号和姓名，并将核对后的条形码贴在指定位置上，在答题纸反面清楚地填写姓名。

第 I 卷 (共103分)I. Listening ComprehensionSection ADirections： In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A. A policewoman. B. A waitress. C. A shop assistant.D. A worker.2. A. Disappointment. B. Disapproval. C. Sympathy.D. Passion.3. A. At a police station. B. At a car rental house. C. At a post office. D. At a bank.4. A. Go to work. B. Take a break. C. Try another problem. D. Keep doing.5. A. The woman congratulated the wrong person.B. The woman should get another job.C. The woman should be more patient.D. The woman was waiting in the wrong place.6. A. Reading a magazine. B. Writing an article.C. Buying clothes.D. Preparing for a maths test.7. A. The guest has to pay in cash. B. The fee will be added to the hotel bill.C. The guest can pay by check.D. It’s free to watch the hotel movie channel.8. A. The woman will enjoy the trip. B. The woman will be exhausted after the trip.C. The woman had better cancel the trip.D. The woman should go to Los Angeles.9. A. 4 pounds. B. 6 pounds. C. 8 pounds.D. 10 pounds.10. A. Compare notes with his classmates. B. Review the details of all his lessons.C. Focus on the main points of his lectures.D. Talk with her about his learning problems.Section BDirections:In section B, you will hear two short passages, and you will be asked three questions on each of the passages. The passages will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. Sending magazines to friends.B. Sending text messages through mobile phone.C. Sending greeting cards to friends.D. Giving orders to children.12. A. Because it costs much time. B. Because it becomes more popular.C. Because it can be done anywhere.D. Because it makes teachers and parents angry.13. A. Making children clever. B. Saving money and paper.C. Helping students study well.D. Making problems become easy.Questions 14 through 16 are based on the following passage.14. A. they cannot be lost or stolen. B. they are safe and handy.C. they can be used anywhere.D. they can save travelers plenty of money.15. A. The authority that issues you the passport. B. The insurance company.C. The bank where you buy your checks.D. The travel agency that arranges your travel.16. A. People usually get traveler’s checks from foreign banks.B. You are not charged for the safety of your traveler’s checks.C. You cannot get your passport until you get your traveler’s checks.D. Traveler’s checks can be exchanged for the money of the country you visit.Section CDirections: In section C, you will hear two longer conversations. The conversations will be read twice. After you hear each conversation, you are required to fill in the numbered blanks with the information you have heard. Write your answers on your answer sheet.Blanks 17 through 20 are based on the following conversation.Complete the form. Write ONE WORD for each answer.Blanks 21 through 24 are based on the following conversation. Complete the form. Write NO MORE THAN THREE WORDS for each answer.II. Grammar and VocabularySection ADirections: After reading the passages below, fill in the blanks to make the passages coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.(A)How I Turned to Be Optimistic (乐观的)I began to grow up that winter night when my parents and I were returning from my aunt’s house, and my mother said that we (25)______(leave) for America soon. We were on the bus then.I was crying, and some people on the bus were turning around to look at me. I remember I could not bear the thought of never hearing again the radio program for school children to (26)______ I listened every morning .I do not remember myself (27)______(cry) for this reason again. In fact I think I cried very little when I was saying goodbye to my friends and relatives. When we were leaving I thought about all the places I was going to see. The country I was leaving never to come back was hardly in my head then.The four years that followed taught me the importance of optimism, but (28)______ idea did not come to me at once. For the first two years in New York I was really lost. I did not quite know what I was or what I should be. Mother remarried, and things became even (29)______(complex) for me. Some time passed before my stepfather and I got used to each other. However, my responsibilities in the family increased a lot since my English was superior (30)______ anyone else’s at home. I translated at interviews with immigration officers, and even discussed telephone bills with company representatives.From my experiences, I believe that my life will turn out all right (31)______ ______ it is not that easy.(B)How Room Designs Affect Our Work and Feeling Architects have long had the feeling that the place we live in can affect our thoughts, feeling and behaviours. But now scientists are giving this feeling an empirical（实证的）basis. They are discovering how (32)______(design) spaces that promote creativity, keep people focused, and lead to relaxation.Researches show aspects of the physical environment caninfluence creativity. In 2012, Joan Meyers-Levy reported that the height of a room’s ceiling affects (33)______ people think. Her research indicates that higher ceilings encourage people to think more freely, (34)______(lead) them to make more abstract connections. Low ceilings, on the other hand, may inspire a more detailed outlook. Besides ceiling height, the view (35)______(afford) by a building may influence an occupant’s ability to concentrate.Using nature to improve focus of attention ought to pay off academically, and (36)______ seems to, according to a study. Students in classrooms with unblocked views of at least 50 feet outside the window had higher scores on tests of vocabulary, language arts and maths than did students (37)______ classrooms primarily overlooked roads and parking lots.Recent study on room lighting design suggests that dim light helps people loosen up. (38)______ that is true generally, keeping the light low during dinner or at parties could increase relaxation.So far public buildings (39)______(focus) on by scientists. “We have a very limited number of studies, so we are almost looking at the problem through a straw(吸管),” architect David says. “How do you take answers to very specific questions and make broad use of them That is (40)______ we are all struggling with.”Section BDirections: Complete the following passage by using the words in the box. Each word can only be used once. Note that there is one word more than you need.￡900 for breaking the traffic law to be carried out next month.If they do not have enough cash or a working credit card,their vehicles will be clamped（扣留）until they pay —and they will face a(n) 41 fee of ￡80 for getting back their vehicles.The law will also be 42 to British citizens. The fines will be described officially as “deposits” when the traffic laws take 43 , because the money would be returned if the driver went to court and was found not guilty. In practice, very few foreign drivers are likely to return to Britain to deal with their cases.Foreign drivers are rarely 44 because police cannot take action against them if they fail to appear in court. Instead, officers often 45 give warnings. Foreign vehicles are 30 percent more likely to be in a crash than British-registered vehicles. The number of crashes caused by foreign vehicles rose by 47 percent between 2008 and 2013. There were almost 400 deaths and serious injuries and 3,000 46 injuries from accidents caused by foreign vehicles in 2013.The new law is partly 47 to settle the problem of foreign lorry drivers ignoring limits to weight and hours at the wheel. Foreign lorries are three times more likely to be in a crash than British lorries. Recent spot checks found that three quarters o f l o r r i e s t h a t f a i l e d s a f e t y t e s t s w e r e 48 overseas.The standard deposit for a careless driving 49 —such as driving too close to the vehicle in front or reading a map at the wheel—will be ￡300.Foreign drivers will not get pointsa s50 added to their licenses, while British drivers will. III. Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.PlanningIn any planning system, from the simplest budgeting to the most complex corporate planning, there is an annual process. This is partly due to the fact that firms 51 their accounting on a yearly basis, but also because similar 52 often occur in the market.Usually, the larger the firm, the longer the planning takes. But 53 , planning for next year may start nine months or morein advance, with various stages of evaluation leading to 54 of the complete plan three months before the start of the year.Planning continues, however, throughout the year, since managers 55 progress against targets, while looking forward to the next year. What is happening now will 56 the objectives and plans for the future.In today’s business climate, as markets constantly change and become more difficult to 57 , some analysts believe that long-term planning is 58 . In some markets they may be right, as long as companies can build the sort of flexibility into theiro p e r a t i o n s w h i c h a l l o w s t h e m t o 59 to any sudden changes.Most firms, however, need to plan more than one year aheadin order to 60 their long-term goals. This may reflect the time it takes to commission (委任) and build a new production plant, or, in marketing 61 , it may be a question of how longit takes to research and launch a range of new products, and reacha certain 62 in the market. If, for example, it is going to take five years for a particular airline to become the 63 choice amongst business travellers on certain routes, the airline must plan for the various 64 involved.Every one-year plan, therefore, must be 65 in relation to longer-term plans, and it should contain die stages that are necessary to achieve the final goals.51. A. make up B. carry out C. bring aboutD. put down52. A. patterns B. guides C. designsD. distributions53. A. surprisingly B. contrarily C. equallyD. typically54. A. approval B. permission C. admirationD. objection55. A. value B. confirm C. reviewD. survey56. A. restore B. promote C. influenceD. maintain57. A. guess B. advocate C. recognizeD. predict58. A. pointless B. meaningful C. realisticD. inevitable59. A. lead B. respond C. refer D. contribute60. A. share B. handle C. developD. benefit61. A. expressions B. descriptions C. wordsD. terms62. A. reputation B. position C. situationD. direction63. A. reserved B. selected C. preferredD. supposed64. A. acts B. steps C. means D. points65. A. handed over B. left behind C. made outD. drawn upSection BDirections: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)My dad loved pennies, especially those with the elegant stalk (茎) of wheat curving around each side of the ONE CENT on the back. Those were the pennies he grew up with during the Depression.As a kid, I would go for walks with Dad, spying coins along the way—a penny here, a dime (一角硬币) there. Whenever I picked up a penny, he’d ask, “Is it a wheat” It always thrilled him when we found one of those special coins produced between 1909 and 1958, the year of my birth.One gray Sunday morning in winter, not long after my father’s death in 2002, I was walking down Fifth Avenue, feeling bereft. I found myself in front of the church where Dad once worked. I was warmly shown in and led to a seat. Hearing Dad’s favorite “A Mighty Fortress Is Our God”, I burst into tears. We’d sung that at his funeral.After the service, I shook the pastor’s(牧师) hand and stepped onto the side walk—and there was a penny. I bent to pick it up, turned it over, and sure enough, it was a wheat. A 1944, a year my father was serving on a ship in the South Pacific.That started it. Suddenly wheat pennies began turning up on the sidewalks of New York everywhere. I got most of the important years: his birth year, my mom’s birth year, the year he gradua ted from college, the year he met my mom, the year they got married, the year my sister was born. But alas, no 1958 wheat penny—my year, the last year they were made.The next Sunday, after the service, I was walking up Fifth Avenue and spotted a penny in the middle of a crossing. Oh, no, it was a busy street；cabs were speeding by—should I risk it I just had to get it.A wheat! But the penny was worn, and I couldn’t read the date. On arriving home, I took out my glasses and took it to the light. There was my birthday!I found 21 wheat pennies on the streets of Manhattan in the year after my father died, and I don’t think that’s a coincidence.66. The writer’s father loved pennies with wheat because ________.A. when he first saw it, he began to love itB. when he saw the wheat, he thought of his time during the DepressionC. when he was young, he had a lot of pennies with wheatD. when he was a child, he never got a coin with wheat67. The underlined word “bereft” (in Para.3) means ________.A. protestedB. disappointedC. grievedD. offended68. Which of the following statements about the author is NOTtrueA. He was born in 1958.B. He knew the church well.C. He once worked in a church.D. He went to church because of his father.69. The best title for the passage would probably be ________.A. Pennies from HeavenB. My father’s life storyC. My father’s hobbyD. Living in New York(B)Do you want to get home from work knowing you have made a real difference in someone’s life If yes, don’t care about sex or age! Come and join us, then you’ll make it!70. What does the underlined part meanA. You’ll make others’ lives more meaningful with this job.B. You’ll arrive home just in time from this job.C. You’ll earn a good salary from this job.D. You’ll succeed in getting this job.71. The volunteers’ major responsibility is to help people withlearning disabilities ________.A. to get some financial supportB. to properlyprotect themselvesC. to learn some new living skillsD. to realizetheir own importance72. Which of the following can first be chosen as a volunteerA. The one who can drive a car.B. The one who has done similar work before.C. The one who has patience to listen to others.D. The one who can use English to communicate.73. The text serves as ________.A. a reminder to social workersB. an advertisementfor helpersC. a document on appealing for volunteersD. anintroduction about a social care organization(C)There are desert plants which survive the dry season in the form of inactive seeds. There are also desert insects which survive as inactive larvae (幼虫). In addition, difficult as it is to believe, there are desert fish which can survive through years of droughts in the form of inactive eggs. These are the shrimps (小虾) that live in the Mojave Desert, an intensely dry region in the south-west of the United States where shade temperatures of over 50℃ are often recorded.The eggs of the Mojave shrimps are the size and have the appearance of grains of sand. When sufficient spring rain falls to form a lake, once every two to five years, these eggs hatch. Then the water is soon filled with millions of tiny shrimps about a millimetre long which feed on tiny plant and animal organisms which also grow in the temporary desert lake. Within a week, the shrimps grow from their original 1 millimetre to a length of about 1.5 centimetres.Throughout the time that the shrimps are rapidly maturing,the water in the lake equally rapidly evaporates (挥发). Therefore, for the shrimps it is a race against time. By the twelfth day, however, when they are about 3 centimetres long, hundreds of tiny eggs form on the underbodies of the females. Usually by this time, all that remains of the lake is a large, muddy patch of wet soil. On the thirteenth day and the next, during the final hours of their brief lives, the shrimps lay their eggs in the mud. Then, having ensured that their species will survive, the shrimps die as the last of the water evaporates.If sufficient rain falls the next year to form another lake, the eggs hatch, and once again the shrimps pass rapidly through their cycle of growth, adulthood, egg-laying, and death. Some years there is insufficient rain to form a lake: in this case, the eggs will remain dormant for another year, or even longer if necessary. Very, very occasionally, perhaps twice in a hundred years, sufficient rain falls to form a deep lake that lasts a month or more. In this case, the species passes through two cycles of growth, egg-laying, and death. Thus, on such occasions, the species multiplies considerably, which further ensures its survival.74. Which of the following is the most distinctive feature of Mojave shrimpsA. They live a brief and tough life.B. They feed on plant and animal organisms.C. Their eggs can survive years of drought.D. They lay their eggs in the mud.75. The word “dormant” (in Para 4) most probably means ________.A. inactiveB. strongC. alertD. soft76. What can be inferred from the passageA. appearance and size are important factors for life to survive in the desert.B. a species must be able to multiply quickly in order to survive in the desert.C. for some species one life cycle in a year is enough to survive the desert drought.D. some species develop a unique life pattern to survive in severe conditions.77. The passage mainly deals with ________.A. the life span of the Mojave shrimpsB. the survival of desert shrimpsC. the creatures living in the Mojave desertD. the importance of water to life in the desert Section CDirections:Read the passage carefully. Then answer thequestions or complete the statements in the fewest possible words.The greatest recent social changes have been in the lives of women in America, or probably in the world.During the twentieth century there has been a remarkable shortening of the time of a woman’s life spent in caring for children. A woman marrying at the end of the nineteenth century would probably have been in her middle twenties, and would be likely to have seven or eight children, of whom four or five lived till they were five years old. By the time the youngest was fifteen, the mother would have been in her early fifties and would expect to live a further twenty years, during which health made it unusual for her to get paid work. Today women marry younger and have fewer children. Usually a woman’s youngest child will be fifteen when she is forty-five and can be expected to live another thirty-five years and is likely to take paid work until retirement at sixty. Even while she has the care of children, her work is lightened by modern living conditions.This important change in women’s life-pattern has only recently begun to have its full effect on women’s economic position. Even a few years ago most girls left schools at the first chance, and most of them took a full-time job. However, when they married, they usually left work at once and never returned to it. Today the school-leaving age is sixteen, many girls stay at school after that age, and though women usually marry younger, more married women stay at least until shortly before their first child is born. Very many more afterwards return to full or part-time work.Such changes have led to a new relationship in marriage, with the husband accepting a greater share of the duties and satisfactions of family life, and with the both husband and wife sharing more equally in providing the money and running the home in terms of the abilities and interests of each of them. (Note: Answer the questions or complete the statements in NO MORETHAN EIGHT WORDS.)78. At what age did most women get married in the late nineteenth century79. A woman today can still take care of her children when doingpaid work in their forties because of ________.80. Of “such changes” today, one is that many more m others ________ after their first child is born.81. What are the factors that cause a couple to share economic and family affairs in an equal way第 II 卷 (共47分)I. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.1. 据我所知，他们学校的面积是我们的两倍。

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虹口区2015年数学学科（理科）高考练习卷
时间120分钟，满分150分 2015.4.21
一、填空题（本大题满分56分）本大题共14题，只要求在答题纸相应题号的空格内直接填写结果，每个空格填对得4分，否则一律得零分.
1、计算：2015
1+1i i
=+_________.（i 是虚数单位）
2、已知函数()()()132,0,0x x f x x x ⎧≤⎪
=⎨⎪>⎩，则()()3f f -=_________.
3、函数()()1ln 10f x x x ⎛⎫
=+> ⎪⎝⎭
的反函数()1f x -=_______.
4、已知正实数,x y 满足31x y +=，则
13x x y
+的最小值为___________. 5、已知复数3sin cos z i θθ=+（i 是虚数单位），
且z 且当θ为钝角时，tan θ=_______. 6、在上海高考改革方案中，要求每位高中生必须在物理、化学、生物、政治、历史、地理6门学科（3门理科学科，3门文科学科）中选择3门学科参加等级考试，小丁同学理科成绩较好，决定至少选择两门理科学科，那么小丁同学的选科方案有_________种.
7、设数列{}n a 前n 项的和为n S ，若14a =，且()
*13N n n a S n +=∈，则n S =_________.
8
、在极坐标系中，过点4π⎫⎪⎭且与圆2cos ρθ=相切的直线的方程为_______________.
9
、若二项式6
x ⎛ ⎝
展开式中含2x 项的系数为5
2，则()2lim 1n n a a a →∞+++
+=__________.
10、若行列式(
)
51sin 0
cos 24x x ππ+⎛⎫
+ ⎪⎝⎭
的第1行第2列的元素1的代数余子式为1-，则实数x 的取值集合为___________.
11、如图所示，已知12,F F 为双曲线
()22
2210,0x y
a b a b -=>>
的两个焦点，且122F F =，若以坐标原点O 为圆心，12F F
为直径的圆与该双曲线的左支相交于,A B 两点，且2F AB ∆ 为正三角形，则双曲线的实轴长为
__________.
12、随机变量ξ的分布列为
其中,,a b c 成等差数列，若1
3
E ξ=
，则D ξ=_________. 13、已知向量,a b ，满足2a b a b ==⋅=，且()()
0a c b c -⋅-=，则2b c -的最小值为_______. 14、若()f x 是定义在R 上的奇函数，且对任意的实数0x ≥，总有正常数T ，使得
()()f x T f x T +=+成立，则称()f x 具有“性质p ”，已知函数()g x 具有“性质p ”，且在[]0,T 上，()2g x x =；若当[],4x T T ∈-时，函数()y g x kx =-恰有8个零点，则实数k =__________.
二、选择题（本题共4题，满分20分）
15.设全集R U =，已知2302x A x x ⎧+⎫=>⎨⎬-⎩⎭
，{}
12B x x =-<，则()U A B =ð（）
A. 3,12⎛⎫- ⎪⎝⎭
B. (]1,2-
C. (]2,3
D. [)2,3
16.设R a ∈，则“1a =-”是“()()2f x ax x =-在()0,+∞上单调递增”的（） A.充要条件 B.既不充分也不必要条件 C.充分不必要条件
D.必要不充分条件
17.如图所示，PAB ∆所在平面α和四边形ABCD 所在的平面β互相垂直，且AD α⊥，BC α⊥，4AD =，
8BC =，6AB =，若t a n 2t a n 1A D P B C P ∠-∠=，则动点P 在平面α内的轨迹是（）
A.线段
B.椭圆的一部分
C.抛物线
D.双曲线的一部分 18.已知F 为抛物线24y x =的焦点，,,A B C 为抛物线上的三点，
O 为坐标原点，F 若为ABC ∆的重心，,,OFA OFB OFC ∆∆∆
面积分别记为123,,S S S ，则2
2
2
1
23S S S ++的值为（） A.3 B.4 C.6 D.9
三、解答题（本大题共5题，满分74分）
19.（本题满分12分）本题共2小题，第1小题5分，第2小题7分. 已知函数()log a f x b x =+（0a >且1a ≠）的图像经过点()8,2和()1,1-. （1）求函数()f x 的解析式；
（2）令()()()21g x f x f x =+-，求()g x 的最小值及取最小值时x 的值.
β
αP B
A D
C
20.（本题满分14分）本题共2小题，第1小题6分，第2小题8分.
在如图所示的几何体中，四边形CDPQ 为矩形，四边形ABCD 为直角梯形，且90BAD ADC ∠=∠=，平面CDPQ ⊥平面ABCD ，1
12
AB AD CD ===
，PD （1）若M 为PA 的中点，求证：AC //平面DMQ ；（2）求平面PAD 与平面PBC 所成的锐二面角的大小.
21.（本题满分14分）本题共2小题，第1小题6分，第2小题8分.
如图，经过村庄A 有两条夹角60为的公路,AB AC ，根据规划拟在两条公路之间的区域内建一工厂P ，分别在两条公路边上建两个仓库,M N （异于村庄A ），要求2PM PN MN ===（单位：千米）.记AMN θ∠=.
（1）将,AN AM 用含θ的关系式表示出来；
（2）如何设计（即,AN AM 为多长时），使得工厂产生的噪声对居民的影响最小（即工厂与村庄的距离AP 最大）？
A
B
C
Q
P
D
M
A M
B
P
N
C
22.（本题满分16分）本题共3小题，第1小题5分，第2小题5分，第2小题6分.
已知圆()2
21:18F x y ++=，点()21
,0F ，点Q 在圆1F 上运动，2QF 的垂直平分线交1QF 于点P . （1）求动点P 的轨迹的方程C ；
（2）设,M N 分别是曲线C 上的两个不同点，且点M 在第一象限，点N 在第三象限，若
122OM ON OF +=，O 为坐标原点，求直线MN 的斜率；
（3）过点10,3S ⎛
⎫- ⎪⎝
⎭的动直线l 交曲线C 于,A B 两点，在y 轴上是否存在定点T ，使以AB 为
直径的圆恒过这个点？若存在，求出点T 的坐标，若不存在，请说明理由.
23.（本题满分18分）本题共3小题，第1小题6分，第2小题6分，第2小题6分. 已知数列{}n a 满足：121a a ==，且()
*22N n n n a a n +-=∈，设3n n b a =. （1）求数列{}n a 的通项公式；
（2）在数列{}n b 中，是否存在连续的三项构成等差数列？若存在，求出所有符合条件的项；若不存在，请说明理由；
（3）试证明：在数列{}n b 中，一定存在正整数(),1k l k l <<，使得1,,k l b b b 构成等比数列；并求出,k l 之间的关系.
虹口区2015年数学学科（理科）高考练习卷答案
（仅供参考）
一．填空题 1. i -； 2. 12； 3. 1
1()(0)1
x
f x x e -=>-； 4. 7； 5. 1-； 6. 10； 7. 4(41)3
n
-； 8. 1y =； 9. 23； 10. {|2,}x x k k Z ππ=+∈
11.
1； 12.
5
9
；
13. 1；
14. 6；
二．选择题
15. B ； 16. C ； 17. D ； 18. A ；三．解答题
19.（1）2()1log f x x =-+；（2）1x =，()1g x =； 20.（1）联结PC ，证明略；（2）
3
π
； 21.（1
）3AN θ=
；)3
AM θ=︒-；
（2）2AM AN ==
，AP =22.（1）2212x y +=；（2
；（3）(0,1)； 23.（1）1(21)3
1(21)3
n
n n n a n ⎧+⎪⎪=⎨⎪-⎪⎩，为奇数，为偶数；（2）23b =，39b =，415b =成等差数列；（3）略。

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