金山区2014学年第一学期期末考试试卷和答案
金山区-2014学年第一学期期末考试
金山区20XX-20XX 学年第一学期期末考试高三数学试卷(满分：150分，完卷时间：120分钟)(答案请写在答题纸上)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。
矚慫润厲钐瘗睞枥庑赖賃軔朧。
1+ i1•计算：——=(i为虚数单位)。
3-i3 n 42.若a€ (一 ,2 冗),tan a二-一，贝U sin a二。
2 33•设集合A =迄八a f,集合B=｛a, b｝,若A A B=甘｝则A U B=。
4•不等式：x；1 X <1的解集是。
5•若函数y二f(x)的反函数为y=2x-1-1，贝y f(x)二。
6.若关于x的实系数一元二次方程x2+px+q=0有一个根为3-4i (i是虚数单位)，则实数p与q的乘积pq=。
2 1 7 27•二项式(x2- )7的展开式中含x2的项的系数是。
x8.在等差数列｛a n｝中，c = 3，公差不等于零，且a2、a4、a9恰好是某一个等比数列的前三项，那么该等比数列的公比的值等于。
9•容器中有10个小球，除颜色外，其他性状”完全相同，其中4个是红色球，6个是蓝色球，若从中任意选取3个，则所选的3个小球都是蓝色球的概率是(结果用数值表示)。
聞創沟燴鐺險爱氇谴净祸測樅。
10•从一堆苹果中任取5只，称得它们的质量(单位：克)分别是：125,124,121,123,127,则该堆苹果的总体标准差的点估计值是(结果精确到0.01 )。
残骛楼諍锩瀨濟溆塹籟婭骤東。
11.设数列｛a n｝是公比为q的等比数列，它的前n项和为S n，若n imS n= 2，则此等比数列的首项a1的取值范围是。
12・已知偶函数y二f(x)(x€ R)满足：f(x+2)= f(x)，并且当x € [0,1] 时，f (x) = x，函数y=f (x)( x € R)与函数y= I o gx|的交点个数是。
A 13.如图，已知直线丨:4x-3y+ 6= 0，抛物线C : y2 = 4x图像上的一个动点P 到直线丨与y 轴的距离之和的最小值是。
上海市各区2014学年第一学期期末考试八年级数学试卷合集
2013学年第一学期期末考试八年级数学试卷① （满分100分，考试时间90分钟）一、 选择题：（本大题共5题，每题2分，满分10分）1、下列等式一定成立的是（ ）A =、=、3=± D 、=9 2、下列一元二次方程有两个相等实数根的是（ ）A ．x 2+3=0B ．x 2+2x=0C ．（x+1）2=0D ．（x+3）（x ﹣1）=0 3、下列四组点中，可以在同一个正比例函数图象上的一组点是（ ） A ．（2．-3），（-4，6） B ．（-2，3），（4，6）C ．（-2，-3），（4，-6）D ．（2，3），（-4，6）4、下列函数中，自变量x 的取值范围是x ≥3的是（ ）A ．31-=x y B.31-=x y C. 3-=x y D. 3-=x y5、已知等腰△ABC 中，AD ⊥BC 于点D ，且AD=21BC ，则△ABC 底角的度数为（ ）A ．45oB ．75oC ．15oD ．前述均可二、填空题：（本大题共15题，每题2分，满分30分）DBFECA6、1-b a （0≠a ）的有理化因式可以是____________．7、计算：8214- = .8、已知x=3是方程x 2﹣6x+k=0的一个根，则k= ．9、关于x 的一元二次方程x 2﹣2x+2+m 2=0的根的情况是 .10、在实数范围内分解因式x 2+2x-4 .11、已知矩形的长比宽长2米，要使矩形面积为55.25米2，则宽应为多少米？设宽为x 米，可列方程为 ．12、正比例函数x y 2-=图象上的两上点为（x 1,y 1）,(x 2,y 2)，且x 1<x 2,则y 1 和y 2的大小关系是______________. 13、矩形的长为x ，宽为y ，面积为9，则y 与x 之间的函数关系及定义域是______________. 14、已知正比例函数y=mx 的图象经过（3，4），则它一定经过______________象限.15、函数y ＝1x ＋x 的图象在__________________象限.16如图，在△ABC 中，∠B=47°，三角形的外角∠DAC 和∠ACF 的平分线交 于点E ，则∠ABE=______°.17、若△ABC 的三条边分别为5、12、13，则△ABC 之最大边上的中线长为 ．18、A 、B 为线段AB 的两个端点，则满足PA-PB=AB 的动点P 的轨迹是_____________________________.19、如图是一株美丽的勾股树，其中所有的四边形都是正方形，所有的 三角形都是直角三角形，若正方形A 、B 、C 、D 的面积分别为2，5，1，2．则最大的正方形E 的面积是 ．20、如图，△ABC 中，AB=AC ，∠BAC=56°，∠BAC 的平分线与AB 的 垂直平分线交于点O ，将∠C 沿EF （E 在BC 上，F 在AC 上）折叠，点C 与点O 恰好重合，则∠OEC 为 度．三、（本大题共8题，第21--24题每题6分；第25--27题每题8分．第28题每题12分．满分60分）21、计算：18)21(|322|2+----． 22、解方程：0142=+-x x ．23、已知关于x 的一元二次方程0322=+-m x x 没有实数根，求m 的最小整数值.B24、到三角形三条边距离相等的点，叫做此三角形的内心，由此我们引入 如下定义：到三角形的两条边距离相等的点，叫做此三角形的准内心. 举例：如图若AD 平分∠CAB ，则AD 上的点E 为△ABC 的准内心.应用：（1）如图AD 为等边三角形ABC 的高，准内心P 在高AD 上，且 PD=AB 21，则∠度数为_____________度.（2）如图已知直角△ABC 中斜边AB=5，BC=3，准内心P 在BC 边上，求CP 的长.25、前阶段国际金价大幅波动，在黄金价格涨至每克360元时，大批被戏称为“中国大妈”的非专业人士凭满腔热情纷纷入场买进黄金，但十分遗憾的是国际金价从此下跌，在经历了二轮大幅下跌后，日前黄金价格已跌至每克291.60元，大批 “中国大妈”被套，这件事说明光有热情但不专业也是难办成事的；同学们：你们现在14、15岁，正值学习岁月，务必努力学习。
上海市金山区2014届高三上学期期末考试[2014上海金山一模]
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金山区2013学年第一学期期末考试高三英语试卷（满分：150分考试时间：120分钟）2014.1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. By bus. B. By train. C. By bicycle. D. On foot.2. A. Do some cooking. B. How to become a Shanghainese.C. How to play chess.D. One day in Shanghai.3. A. A pair of stockings. B. Some paper of high quality.C. A large quantity of books.D. A bookshelf.4. A. It is not very difficult. B. He doesn’t believe everyone’s words.C. He finds the maths course too difficult.D. His score is very bad.5. A. He made a lecture to the woman and the man.B. He has just graduated from the university.C. He is too shy to speak in the university.D. He used to be very shy, but now he has overcome it.6. A. He declines the woman’s offer.B. He doesn’t like the food.C. He likes the pudding only.D. He will eat all the food on the table.7. A. The key to the company’s success.B. The changes on the market.C. The management of the woman’s company.D. How the woman’s company goes on.8. A. Frank’s car was accidentally lost.B. Frank was killed in a car accident.C. Frank fell out of a car.D. Frank survived a car accident.9. A. 40 dollars. B. 32 dollars. C. 30 dollars. D. 16 dollars.后期制作10. A. There are too many centers already.B. They aren’t really going to build one.C. She knew about the planned construction.D. She hasn’t been to the other centers.Section BDirections: In Section B, you will hear two short passages, and you will be asked three questionson each of the passages. The passages will be read twice, but the questions will be spoken onlyonce. When you hear a question, read the four possible answers on your paper and decide whichone would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. In 1886. B. In 1856.C. In 1830.D. In 1950.12. A. A doctor. B. A novelist.C. A housekeeper.D. A poet.13. A. She often goes out to get some information.B. She is not a sociable person.C. She only communicated with her seven poems.D. She is not a productive person.Questions 14 through 16 are based on the following passage.14. A. Slow drain. B. Cutting trees.C. Lack of money.D. Heavy rainfall.15. A. Because they lack money.B. Because the bang is used for experiments.C. Because of some technical faults.D. Because people don’t like it.16. A. India. B. South Africa.C. Denmark.D. China.Section CDirections: In Section C, you will hear two longer conversations. The conversations will be readtwice. After you hear each conversation, you are required to fill in the numbered blanks with theinformation you have heard. Write your answers on your answer sheet.Blanks 17 through 20 are based on the following conversation.Blanks 21 through 24 are based on the following conversation.Directions: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 )“Fire! Fire!” What terrible words to hear when one wakes up in a strange house in the middle of the night! It was a large, old, wooden house and my room was on the top floor. I jumped out of bed, opened the door and stepped outside the house. There was full of thick smoke.I began to run, but as I was still only half-awake, ____25___ ________ going towards the stairs，I went in the opposite direction. The smoke grew___26____ (thick) and I could see fire all around. The floor became hot under my bare feet. I found an open door and ran into a room to get to the window. But ____27___I could reach it, one of my feet caught in something soft and I fell down. The thing I had fallen over felt like a bundle of clothes, and I picked it up ___28___ (protect) my face from the smoke and heat. Just then the floor gave way under me and I crashed to the floor below with pieces of ____29__ (burn)wood all around me.As I reached the cold air outside, my bundle of clothes gave a thin cry. I nearly dropped ___30___in my surprise. Then I was in a crowd gathered in the street. A woman in ___31__ night dress and a borrowed man’s coat scr eamed as she saw me and _____32______(come) running madly.She was the Mayor’s wife, and I had saved her baby.( B )Today, roller skating is easy and fun．But a long time ago, it wasn’t easy at all. Before 1750, the idea of skating didn’t exist. That ch anged because of a man___33_____(name)Joseph Merlin.One day Merlin received an invitation to attend a fancy dress ball. He was very pleased and a little excited. ___34___ the day of the party came near, Merlin began to think ___35____ to make a grand entrance at the party. He had an idea. He thought he____36____(get) a lot of attention ifhe could skate into the room.Merlin tried different ways to make himself ___37___(roll). Finally, he decided to put two wheels under each shoe. These were the first roller skates. Merlin was very proud of his invention and dreamed of arriving at the party___38____ wheels while playing the violin.On the night of the party Merlin rolled into the room playing his violin. Everyone___39____ (amaze) to see him. There was just one problem. Merlin had no way to stop his roller skates. He rolled on and on. Suddenly, he ran into a huge mirror___40___ was hanging on the wall. Down fell the mirror, breaking to pieces. Nobody forgot Merlin’s grand entrance for a long time! Section BDirections: Complete the following passage by using the words in the box. Each word can onlylittle environmental influence. But with ___41___ numbers of people wanting to escape into the wilderness, it is becoming more and more important to camp unobtrusively (不引人注目地) and leave no mark.Wild camping is not permitted in many places, ___42___ in crowded lowland Britain. Wherever you are, find out about organizations ___43___ for managing wild spaces, and ___44___ them to find out their policy on camping and shelter building. For example, it is fine to camp wild in remote parts of Scotland, but in England you must ask the landowner’s ___45___, except in national parks.Camping is about having relaxation, sleeping outdoors, ___46___ bad weather, and making food without modern conveniences. A busy, fully-equipped campsite seems to go against this, so ___47___ out smaller, more remote places with easy ___48___ to open spaces and perhaps beaches. Better still, find a campsite with no road access: walking in makes a real adventure.Finding the right spot to camp is the first step to ___49___ a good night’s sleep. Choose a campsite with privacy and minimum influence on others and the environment. Try to use an area where people have ___50___ camped before rather than creating a new spot. Always consider what influence you might have on the natural world. Avoid damaging plants. A good campsite is found, not made—changing it should be unnecessary.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.Recent years have seen considerable growth in the number of children learning a second or foreign language, as the importance of being able to use a language other than one’s first language has become recognized in an increasingly globalized world. In Asia and Europe ___51___, there has been a tendency to ___52___ the age at which school children begin to learn a foreign language, since it is believed that the earlier a child starts to learn a foreign language, the greater the ultimate achievement will be.In many countries, the language of education is not the same as the language of the homefor__53__ children. Furthermore, in many countries, young language learners comprise the most rapidly growing segment of the elementary (primary) school population.___54__ in some schools there is no extra support to help young language learners acquire the language of instruction, in most countries where there are large numbers of young learners, there is a ___55___ awareness of their special needs. There is ___56___ a need to identify the needs of young language learners, to ___57___ what level, if any, of proficiency they have in the target language to diagnose their strengths and areas in need of improvement. Language ___58___, whether this is informal, classroom- based, or large-scale, thus has a __59___ role to play in gathering the information needed for these purposes.Unfortunately, the vast majority of teachers who work with young language learners have had little or no ___60___ training or education in language assessment. Teachers are involved in assessment on a daily basis, as they ___61___ their pupils’ classroom performance, and as they develop formal classroom assessments. Assessment should therefore, wherever possible, be familiar and involve familiar adults, rather than ___62_____. The environment should be safe for the learner. Teachers responding___63___ to the child’s efforts is ideal for young learners. Such feedback maintains attention and ___64___. As children grow, they are able to work more ____65___ and for long spans of time without ongoing feedback.51. A. in particular B. as a result C. for example D. in other words52. A. shorten B. enlarge C. lower D. increase53. A. the majority of B. the amount of C. the quantity of D. the number of54. A. Before B. While C. As D. If55. A. reducing B. watching C. growing D. slipping56. A. however B. moreover C. instead D. therefore57. A. discuss B. determine C. teach D. train58. A. draft B. performance C.assessment D. arrangement59. A. unnecessary B. uninteresting C. concrete D. critical60. A. personal B. valuable C. professional D. approval61. A. monitor B. master C. inspect D. control62. A. students B. children C. strangers D. neighbors63. A. quickly B. kindly C. rudely D. friendly64. A. friendship B. relationship C. quality D. confidence65. A. independently B. dependently C. roughly D. carelesslySection 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.Studies show that 30 minutes of mild exercise a day will increase your life span. This doesn’t mean you have to run three miles or start jumping. Simple activities such as walking, gardening or taking exercise classes will work.2. Enjoy peopleContacts with family and friends help fight life-sapping depression and stress. Studies show that people socialize regularly live longer than loners. Interact daily with family members and friends. If you are isolated, make a point of joining social clubs or church groups so you can laugh and share life’s joys.3. Stay mentally activeThis is very important—use it or lose it! People who allow their mental faculties to decline run the risk of shortening their lives through falls and other injuries, and not being able to take care of themselves. Read the newspaper, visit the public library, balance your checkbook withouta calculator, and exercise your brain by doing crossword puzzle.4. DietThe key words are “ high-fiber, low-fat”. Cut back on red meat, salt, white flour, white sugar, alcohol and coffee. Start eating a balanced diet including lots of fruit, grains, raw vegetables and nuts.5. Positive attitudeBy approaching life with a positive outlook you increase your chances of living longer. Stay optimistic—always search for the silver lining. Studies show people who see life as an enjoyable challenge, rather than a constant trial, cope better and prolong their life spans.6. medication mixAs we age, we are more likely to take medications. Sometimes this leads to over-medication, which can be disabling and even deadly. Ask your doctor if the drugs you take are really necessary. Make sure there is no danger of a bad drug interaction from your medication. Used correctly, medicines can help you live longer and more comfortably.7. VolunteerHelping others increases your self-esteem and makes you feel like a valuable contributor.66. Which of the following does NOT increase your life span?A. Staying happy about lifeB. Exercising as much as possibleC. Exercising your brain frequentlyD. Enjoying your social life67. People who do not stay mentally active are more likely to _________.A. become ignorantB. lose their mindsC. shorten their livesD. become slow in movement68. The underlined phrase silver lining can be replaced with __________.A. something made of silverB. new informationC. good aspectsD. long life span69. Which of the following is correct about medication ?A. Appropriate medication is necessary.B. Medication is always helpful to you.C. Doctors suggest taking expensive medicines.D. Drugs are unnecessary because of side-effects.( B )“W e are going to have to get rid of Bay this year,” Meg’s dad said. “That horse can’t work another winter.”It was October, and snow was falling lightly over their part of Montana. Meg knew the ranch needed strong, steady horses to bring the sheep back from pasture. Still, she could hardly take in the enormity of her father’s words.“I know Bay’s old,” Meg told her dad, “but he is my horse. Won’t you let me keep him? I’d feed him. He won’t be a problem.”“We use horses for labor,” her dad said, “We don’t keep them as pets.”Meg understood. She thought about Casey, a dog she had once loved. He became lost when he chased a wandering sheep and never came home. Meg had been sad for weeks. She knew that on a ranch, animals come and go. You couldn’t grow attached to them. Bay was different, though. The horse had a personality all his own.She went to the stable and stroked the old brown horse’s head. As Bay nuzzled her hand, Meg tried to imagine what the ranch would be like without her favorite horse.The snow was falling faster, and dark gray clouds were settling over the peaks. Meg had an idea. She saddled Bay, put her foot in the stirrup, and swung up. “Let’s go !”she cried, and nudged him with heels.They galloped over the meadow, onto a steep and narrow trail, and up a wooded slope. “We’ll round up the sheep now, before the snow gets too thick,” Meg said. “We will show Dad how well you work.”.The sheep stood in a high pasture, bleating at the storm clouds. Meg heard a lone bleat from above. She looked for the lost sheep but couldn’t see it. “It must be stuck in the bush above the rock wall,” she thought.She got off her horse and began to climb up the steep wall. She was almost at the top when the heel of her foot slipped into a crack. She lost her balance and fell onto a rock. She tried to stand, but her foot hurt too much.Then Bay got into position under the rock. He was telling Meg to crawl onto his back!Painfully, Meg got her feet into the stirrups. Slowly and carefully, Bay carried her down the snow trail.On the way down, they met Dad riding his gray horse. “What happened ?”he asked anxiously.“Bay saved me.” Meg told him what happened.Dad’s voice shook when he said, “ I wouldn’t get rid of that horse for anything.”70. Meg’s dad intended to ______________.A. trade Bay for sheepB. let Meg keep Bay as a pet.C. sell or kill BayD. keep Bay working for another winter.71. Meg hoped to keep Bay because______________.A. Bay was a good sheep-keeper.B. Bay was like a good friend.C. Bay was strong and beautiful.D. Bay was the only animal she loved.72. Meg rode Bay to the pasture to ___________.A. look for a lost sheep.B. round up the sheep.C. climb over the meadow.D. meet her father.73. Why does Dad change his mind about Bay?A. Bay found the lost sheep and Meg.B. Bay was such a good friend to Meg.C. Bay proved his importance by rescuing Meg.D. Bay proved that he had a personality all his own.( C )One picture in the Wonder Book of knowledge I had as a little boy showed a man reading a book while floating in the Dead Sea. What a miracle! How would it feel to lie back in water so thick with salt that it was impossible to sink?Fed by the Jordan River and smaller streams, the Dead Sea is the lowest point on the earth’s surface, and its water is ten times saltier than the Mediterranean. With evaporation its only outlet, salt and other minerals become super-concentrated.Earlier this year, I drove down the long, steep hill to realize my dream. The shoreline was a broad area of bare salt-mud, but the water edge was far out of sight. Had somebody pulled the Dead Sea’s plug? I wondered. Eli Dior, an Israeli official, explained the problem: “ The Dead Sea is drying up. Every year, the surface drops about one meter, and as the water level falls, shadow areas are left high and dry.”Over the last half-century, the five neighboring countries have collectively diverted nearly all the water flowing into the Dead Sea to meet human and agriculture needs. Result: the Dead Sea is being emptied.With population in the region set to double at least in the next 50 years, there is little hope of restoring the water being diverted for human consumption. No country has a drop to spare for the Dead Sea, where they know it will just evaporate. To dream of opening the dams and restoring natural balance is plainly unrealistic.Yet one ambitious high-tech dream may turn out to be not only the salvation of the Dead Sea but also a ticket to peace around its shores. The “ Red-Dead” is a proposed $5 billion project to bring sea water some 240 kilometers by pipeline and canal from the Red Sea to the Dead Sea. The Red-Dead may be the only solution, but even if the project is carried out successfully, the Dead Sea will be 10 to 20 meters lower than now and two thirds of its current size.Whatever the future holds, the Dead Sea’s magical mix of sun, mud, sea and salt will surely survive. Many might complain that the Dead Sea is half empty—but for me the Dead sea will always be half full.74. What’s the passage mainly about ?A. Dead Sea—miracle of the world.B. Save the environment of the Dead Sea.C. Slow shrinking of the Dead Sea.D. Why is the Dead Sea so salty.75. The shrinking of the Dead Sea is mainly caused by _________ according to the passage.A. a severe reduction of the water flowing into the sea.B. rapid evaporation of the water in the Dead Sea area.C. the increasing quantity of water drawn from the sea.D. very low annual rainfall in the Dead Sea Area.76. Which of the following is right according to the passage?A. With no outlet to any ocean, the Dead Sea has become by evaporation the saltiest on earth.B. Though burdened with the growing population, the neighboring countries haven’t cut offthe sources of the Dead Sea.C. All the countries in the area will consider diverting less water from the Jordan River.D. The Red-Dead Project has not only brought water to the Dead Sea, but peace to the area aswell.77. Which of the following statements will the author approve of ?A. If the Dead Sea dried up, great natural disasters would happen in the region.B. The Dead Sea will not survive no matter what people do to save it.C. The five neighboring countries should stop diverting water from the Jordan River.D. Though the Dead Sea is shrinking gradually, it will not die.Section CDirections: Read the passage carefully. Then answer the questions or complete the statements in the fewest possible words.However important we may regard school life to be, there is no denying the fact that children spend more time at home than in the classroom. Therefore, the great influence of parents cannot be ignored or discounted by the teacher. They can become strong allies of the school personnel or they can consciously or unconsciously hinder and frustrate curricular objectives.Administrators have been aware of the need to keep parents informed of the newer methods used in schools. Many principles have conducted workshops explaining such matters as the reading readiness program, manuscript writing and developmental mathematics. Moreover, the classroom teacher, with the permission of the supervisors, can also play an important role in enlightening parents. The informal tea and the many interviews carried on during the year, as well as new ways of reporting pupils’progress, can significantly aid in achieving a harmonious interplay between school and home.To illustrate, suppose that a father has been drilling Junior in arithmetic processes night after night. In a friendly interview, the teacher can help the parent sublimate (升华) his natural paternal interest into productive channels. He might be persuaded to let Junior participate in discussing the family budget, buying the food, using a standard or measuring cup at home, setting the clock, calculating mileage on a trip and engaging in scores of other activities that have a mathematical basis.If the father follows the advice, it is reasonable to assume that he will soon realize his son is making satisfactory progress in mathematics, and at the same time, enjoying the work. Too often, however, teachers’conferences with parents are devoted to petty(不重要的) accounts of children’s misdeeds, complaints about laziness and poor work habits, and suggestion for penalties and rewards at home.What is needed is a more creative approach in which the teacher, as a professional adviser, plants ideas in parents’ minds for the best utilization of the many hours that the child spends out ofthe classroom.In this way, the school and the home join forces in fostering the fullest development of youngsters’capacities.(Note: Answer the questions or complete the statements with NO MORE THAN 14 WORDS)78. Why do parents also have great influence on children?_____________________________________________________________________________79. Through which ways can the teacher play an important role in enlightening parents?_____________________________________________________________________________80. According to the teacher, that parent should let the boy _____________________________ ifhe wants to sublimate his natural paternal interest into productive channels in teaching his sonarithmetic.81. A more creative approach is needed for _____________________of children out of classroom.第Ⅱ卷(共47分)I. TranslationDirections:Translate the following sentences into English, using the words given in the brackets.1．我不认为我今天所说的话能改变别人对我的看法。
上海市金山区2013-2014学年度初三第一学期期末质量抽查英语试卷及答案(word版)
上海市金山区2013-2014学年度初三第一学期期末质量抽查英语试卷（满分150分，完卷时间100分钟）2014.01考生注意：本卷有7大题，共94小题。
试题均采用连续编号，所有答案务必按照规定在答题卡上完成，做在试卷上不给分。
Part 1 Listening （第一部分听力）I. Listening comprehension (听力理解) (共30 分)B. Listen to the dialogue and choose the best answer to the question you hear(根据你听到的对话和问题，选出最恰当的答案)：(8分)7. A) 2:15..B) 2:30.C) 2:50.D) 3:05.8. A) Near the moon.B) In the dream.C) In China.D) On the moon.9. A) Mum’s.B) Peter’s.C) Jane’s.D) Theirs.10. A) By bus.B) On foot.C)By taxi.D)By bike.11. A) Jack.B) Steven.B) Frank.D) The two boys.12. A) Yellow and white.B) White.C) Yellow.D)Blue.13. A) Shop owner and shop assistant.B)Mother and son.C) Teacher and student.D) Manager and secretary.14. A) Taiwan Island.B) Italy.C) The summer holidays.D) The tower of Pisa.C．Listen to the passage and tell whether the following statements are true or false (判断下列句子是否符合你听到的短文内容，符合的用“T”表示，不符合的用“F”表示): (6分) ( )15. The love for dogs makes Gail Mirabella become a dog trainer in a circus.( )16. Gail did hqr first dog show when she was 14 years old.( )17. Her dogs started performing in the small shows when they were eight months old. ( )18. In order to train dogs better, Gail tried to make friends with her dogs in the circus. ( )19. How long it takes Gail to train a dog well depends on the dog and herself.( )20. Gail thinks her life in the circus is wonderful because she can stay with her dogs.D. Listen to the passage and fill in the blanks (听短文填空，完成下列内容。
金山区高三数学上学期期末试卷附答案
2014-2015金山区高三数学上学期期末试卷（附答案）2014-2015金山区高三数学上学期期末试卷（附答案）(满分：150分，完卷时间：120分钟)（答题请写在答题纸上）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．若集合M={，xR}，N={，x≥–2}，则M∩N=▲．2．计算：=▲．3．不等式：的解是▲．4．如果复数z=（bR）的实部与虚部相等，则z 的共轭复数=▲．5．方程：sinx+cosx=1在[0，π]上的解是▲．6．等差数列中，a2=8，S10=185，则数列的通项公式an=▲(nN*)．7．当a0，b0且a+b=2时，行列式的值的最大值是▲．8．若的二项展开式中的常数项为m，则m=▲．9．从一堆苹果中任取5只，称得它们的质量分别是：(单位：克)125，124，121，123，127，则该样本的标准差是▲克．10．三棱锥O–ABC中，OA=OB=OC=2，且∠BOC=45，则三棱锥O–ABC体积的最大值是▲．11．从集合{1,2,3,4,5,6,7,8,9,10}中任取两个数，欲使取到的一个数大于k，另一个数小于k(其中k{5,6,7,8,9})的概率是，则k=▲．12．已知点A(–3,–2)和圆C：(x–4)2+(y–8)2=9，一束光线从点A发出，射到直线l：y=x–1后反射(入射点为B)，反射光线经过圆周C上一点P，则折线ABP的最短长度是▲．13．如图所示，在长方体ABCD–EFGH中，AD=2，AB=AE=1，M为矩形AEHD内的一点，如果∠MGF=∠MGH，MG和平面EFG所成角的正切值为，那么点M到平面EFGH 的距离是▲．14．已知点P(x0,y0)在椭圆C：(ab0)上，如果经过点P的直线与椭圆只有一个公共点时，称直线为椭圆的切线，此时点P称为切点，这条切线方程可以表示为：．根据以上性质，解决以下问题：已知椭圆L：，若Q(u，v)是椭圆L外一点(其中u，v为定值)，经过Q点作椭圆L的两条切线，切点分别为A、B，则直线AB的方程是▲．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，考生应在答题纸的相应编号上，将代表答案的小方格涂黑，选对得5分，否则一律得零分.15．复数z1＝a+bi(a、bR，i为虚数单位)，z2＝–b+i，且|z1||z2|，则a的取值范围是(▲)．(A)a＞1(B)a＞0(C)–l＜a＜1(D)a＜–1或a＞116．用1，2，3，4，5组成没有重复数字的五位数，其中偶数有(▲)．(A)60个(B)48个(C)36个(D)24个17．设k1，f(x)=k(x–1)(xR)，在平面直角坐标系xOy中，函数y=f(x)的图像与x轴交于A点，它的反函数y=f–1(x)的图像与y轴交于B点，并且这两个函数的图像相交于P点.已知四边形OAPB的面积是3，则实数k等于(▲)．(A)3(B)(C)(D)18．若集合A1、A2满足A1∪A2=A，则称(A1，A2)为集合A的一个分拆，并规定：当且仅当A1=A2时，(A1，A2)与(A2，A1)为集合A的同一种分拆，则集合A={a1，a2，a3}的不同分拆种数是(▲)．(A)8(B)9(C)26(D)27三、解答题（本大题满分74分）本大题共有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)a、b、c分别是锐角△ABC的内角A、B、C的对边，向量=(2–2sinA，cosA+sinA)，=(sinA–cosA，1+sinA)，且∥．已知a=，△ABC面积为，求b、c的大小．20．(本题满分14分)本题共有2个小题，第1小题满分8分，第2小题满分6分．如图，在四棱锥P–ABCD的底面梯形ABCD中，AD∥BC，AB⊥BC，AB=2，AD=3，∠ADC=45.已知PA⊥平面ABCD，PA=1．求：(1)异面直线PD与AC所成角的大小(结果用反三角函数值表示)；(2)三棱锥C–APD的体积．21．(本题满分14分)本题共有2个小题，第1小题满分7分，第2小题满分7分.已知a0且a1，数列是首项与公比均为a的等比数列，数列满足bn=anlgan(nN*)．(1)若a=3，求数列的前n项和Sn；(2)若对于nN*，总有bnbn+1，求a的取值范围．22．（本题满分16分）本题共有3个小题，第1小题满分4分，第2小题满分6分，第3小题满分6分．动点与点的距离和它到直线的距离相等，记点的轨迹为曲线．(1)求曲线的方程；(2)设点2,动点在曲线上运动时，的最短距离为，求的值以及取到最小值时点的坐标；(3)设为曲线的任意两点，满足(为原点)，试问直线是否恒过一个定点？如果是，求出定点坐标；如果不是，说明理由．23．(本小题满分18分)本题共有3个小题，第1小题满分4分，第2小题满分6分，第3小题满分8分.设函数f(x)=2kax+(k–3)a–x(a0且a1)是定义域为R的奇函数．(1)求k值；(2)若f(2)0，试判断函数f(x)的单调性，并求使不等式f(x2–x)+f(tx+4)0恒成立的t的取值范围；(3)若f(2)=3，且g(x)=2x+2–x–2mf(x)在2,+∞上的最小值为–2，求m的值．上海市金山区2014—2015学年第一学期期末考试评分标准一、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．[0,5]；2．；3．0x1；4．1–i；5．或0；6．3n+2；7．08．7920；9．2；10．；11．7；12．10；13．；14．二、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，考生应在答题纸的相应编号上，将代表答案的小方格涂黑，选对得5分，否则一律得零分.15．C；16．B；17．B；18．D三、解答题（本大题满分74分）本大题共有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)解：，，又‖(2–2sinA)(1+sinA)–(cosA+sinA)(sinA–cosA)=0，即：又为锐角，则，所以∠A=60 (6)分因为△ABC面积为，所以bcsinA=，即bc=6，又a=，所以7=b2+c2–2bccosA，b2+c2=13，解之得：或………………………………………………………………12分20．(本题满分14分)本题共有2个小题，第1小题满分8分，第2小题满分6分．解：(1)过点C作CF∥AB交AD于点F，延长BC至E，使得CE=AD，连接DE，则AC∥DE，所以∠PDE就是异面直线PD与AC所成的角或其补角，………………2分因为∠ADC=45，所以FD=2，从而BC=AF=1，且DE=AC=，AE=，PE=，PD=，在△中，，所以，异面直线与所成角的大小为………………………………………………………………8分(2)因为VC–APD=VP–ACD，S△ACD=CFAD=3PA⊥底面ABCD，三棱锥P–ACD的高为PA=1，VP–ACD=S△ACDPA=1，所以，三棱锥C–APD的体积为1.………………………………………………………14分21．(本题满分14分)本题共有2个小题，第1小题满分7分，第2小题满分7分.(1)由已知有，，，所以，.………………………………………………………7分(2)即.由且，得，所以或即或对任意nN*成立，且，所以或……………………………………………14分22．（本题满分16分）本题共有3个小题，第1小题满分4分，第2小题满分6分，第3小题满分6分．(1)根据抛物线的定义可知,动点的轨迹是抛物线所以曲线C的方程为x2=4y；……………………………………………………………4分(2)设点T(x0,y0),x02=4y0(y0≥0)，|AT|==，a–20，则当y0=a–2时，|AT|取得最小值为2，2=a–1,a2–6a+5=0，a=5或a=1(舍去)，所以y0=a–2=3，x0=2，所以T坐标为(2,3)；……………………………10分(3)显然直线OP1、OP2的斜率都必须存在，记为k，，，解之得P1(,)，同理P2(–4k,4k2)，直线P1P2的斜率为，直线P1P2方程为：整理得：k(y–4)+(k2–1)x=0，所以直线P1P2恒过点(0,4)………………………………16分23．(本小题满分18分)本题共有3个小题，第1小题满分4分，第2小题满分6分，第3小题满分8分.解(1)因为f(x)是定义域为R的奇函数，所以f(0)＝0，所以2k+(k–3)=0，即k=1，检验知，符合条件………………………………………4分(2)f(x)=2(ax–a–x)(a0且a1)因为f(2)0，0，又a0且a1，所以0a1因为y=ax单调递减，y=a–x单调递增，故f(x)在R上单调递减。
上海市金山区2014年中考一模(即期末)数学试题及答案(word版本)
2013学年第一学期期末质量检测初三数学试卷（测试时间：100分钟，满分：150分） 2014.01考生注意： 1．本试卷含三个大题，共25题．答题时，考生务必按答题要求在答题纸规定的位置上作答，在草稿纸、本试卷上答题一律无效．2．除第一、二大题外，其余各题如无特别说明，都必须在答题纸的相应位置上写出证明或计算的主要步骤． 一、选择题：（本大题共6题，每题4分，满分24分）1．两个相似三角形的面积比为1∶4，那么这两个三角形的周长比为（ ） （A ）1∶2； （B ）1∶4；（C ）1∶8；（D ）1∶16．2．如果向量a 与单位向量e方向相反，且长度为12，那么向量a 用单位向量e表示为（ ） （A ）12a e = ； （B ）2a e =；（C ）12a e =- ； （D ）2a e =-．3．将抛物线2y x =向右平移1个单位，所得新抛物线的函数解析式是（ ） （A ）2(1)y x =+； （B ）2(1)y x =-； （C ）21y x =+； （D ）21y x =-．4．在Rt △ABC 中，∠A =90°，如果把这个直角三角形的各边长都扩大2倍，那么所得到的直角三角形中，∠B 的正切值（ ） （A ）扩大2倍； （B ）缩小2倍； （C ）扩大4倍； （D ）大小不变 ． 5．已知在Rt △ABC 中，∠C =90°，∠A =a ，BC =m ，那么AB 的长为（ ） （A ）sin m α；（B ）cos m α； （C ）sin mα； （D ）cos mα． 6．在平面直角坐标系中，抛物线()221y x =--+的顶点是点P ，对称轴与x 轴相交于点Q ，以点P 为圆心，PQ 长为半径画⊙P ，那么下列判断正确的是（ ） （A ）x 轴与⊙P 相离； （B ）x 轴与⊙P 相切； （C ）y 轴与⊙P 与相切； （D ）y 轴与⊙P 相交． 二、填空题：（本大题共12题，每题4分，满分48分） 7．如果23x y =，那么22x yx y+-= ▲ ． 8．已知在△ABC 中，点D 、E 分别在边AB 、AC 上，DE //BC ，35DE BC =，那么CE AE的值等于 ▲ ．9．计算：()223a b b +-=▲ ．10．抛物线22y x x =+的对称轴是 ▲ ．11．二次函数22y x t =+的图像向下平移2个单位后经过点（1，3），那么t = ▲ ． 12．已知在△ABC 中，∠C =90°，AB =12，点G 为△ABC 的重心，那么CG = ▲ ． 13．已知在Rt △ABC 中，∠C =90°，BC，那么∠A = ▲ 度． 14．已知在Rt △ABC 中，∠C =90°，1cot 3B =，BC =3，那么AC = ▲ ．15．已知内切两圆的圆心距为6，其中一个圆的半径为4，那么另一个圆的半径为 ▲ ． 16．如果正n 边形的每一个内角都等于144°，那么n = ▲ ．17．正六边形的边长为a ，面积为S ，那么S 关于a 的函数关系式是 ▲ ． 18．在Rt △ABC 中，∠C =90°，3cos 5B =， 把这个直角三角形绕顶点C 旋转后得到 Rt △A'B'C ，其中点B' 正好落在AB 上， A'B'与AC 相交于点D ，那么B DCD'= ▲ ．三、解答题：（本大题共7题，满分78分） 19．（本题满分10分）计算：222sin 60cos 45tan 60cos30tan 30cot 45---20．（本题满分10分, 其中第（1）小题6分，第（2）小题4分）已知一个二次函数2y x b x c =++的图像经过点（4，1）和（1-，6）． （1）求这个二次函数的解析式；（2）求这个二次函数图像的顶点坐标和对称轴． 21．（本题满分10分）如图，已知AB 是⊙O 的弦，点C 在线段AB 上，OC =AC =4，CB =8． 求⊙O 的半径． 22．（本题满分10分）第18题图如图，某超市从底楼到二楼有一自动扶梯，右图是侧面示意图。
上海市金山区中考化学一模(即期末)试题
上海市金山区2014年中考化学一模（即期末）试题（理化完卷时间100分钟，满分150分） 2014年1月相对原子质量：H-1，O-16，C-12, Ca-40六、选择题（每小题只有一个正确选项，共20分） 27.下列属于化学变化的是（ ）A. 氯化钠溶解于水B. 用石灰水检验二氧化碳C. 过滤D. 酒精挥发28.下列物质属于纯净物的是（ ）A. 空气B. 牛奶C. 糖水D. 碳酸钙29.下列空气的组成物质中属于氧化物的是（ ） A. 氮气B. 氧气C. 二氧化碳D. 稀有气体30.下列各组物质中，属于同素异形体的是（ ）A. 氧气和臭氧B.CO 和CO 2C. H 2O 和H 2O 2D.水和冰31.新版100元人民币采用含MgF 2光变色防伪油墨印刷，MgF 2中F 元素的化合价是（ ）A. +2B. +1C. -2D. -132.下列物质的化学式，书写正确的是（ ）A. 硫酸铁（FeSO 4）B. 氯化铝（AlCl 2）C. 氧化钠（NaO ）D. 硫化钠（Na 2S ）33.下列化学方程式书写正确的是（ ） A. 4Fe+3O 2 2Fe 2O 3B. 2H 2O 22H 2O+O 2C. CaO+H 2OCa(OH)2D. CuSO 4+H 2OCuSO 4·5H 2O34.以下饮料和食品中，属于溶液的是（ ） A. 豆浆B. 矿泉水C. 牛奶D. 果酱35.下图表示的是身边一些物质在常温时的近似pH 。
下列叙述不正确．．．的是（ ）A. 苏打水呈碱性B. 鸡蛋清呈碱性C. 橘子汁的酸性比萝卜汁的酸性弱D. 柠檬汁能使紫色石蕊溶液变红 36.下列有关物质的用途描述不正确．．．的是（ ） A. 自来水生产中用明矾作消毒剂B. 戴含有活性炭的口罩，可减少雾霾对人体的影响C. 金刚石可用来切割玻璃D. 干冰可用来保藏食品和进行人工降雨柠檬汁橘子汁萝卜汁 鸡蛋清苏打水点燃MnO 237.下列各组混合物按溶解、过滤、蒸发的操作顺序进行分离的是（ ）A .水和花生油B. 蔗糖和泥沙C. 面粉和泥沙D. 蔗糖和食盐38.有关2CuO + C −−→−高温2Cu + CO 2↑的反应，说法正确的是（ ） A. 氧化铜发生了氧化反应 B. 碳发生了还原反应 C. 反应前后固体质量不变D. 碳在反应中作还原剂39.电解水的实验装置如右图所示。
2014-2015年上海市金山区高一上学期数学期末试卷和解析
2014-2015学年上海市金山区高一（上）期末数学试卷一、填空题（本大题满分36分）本大题共有12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3分，否则一律得零分）1．（3.00分）已知全集U=R，A={x|x≥2}，则∁U A=．2．（3.00分）函数y=lg的定义域是．3．（3.00分）函数y=x+（x＞0）的最小值为．4．（3.00分）若集合A={﹣1，0，1}，集合B={x|x=t2，t∈A}，用列举法表示B=．5．（3.00分）若4x﹣2x+1=0，则x=．6．（3.00分）已知关于x的不等式x2﹣（a﹣1）x+（a﹣1）＞0的解集是R，则实数a取值范围是．7．（3.00分）已知函数y=a x﹣1+1（a＞0，a≠1）的图象经过一个定点，则顶点坐标是．8．（3.00分）已知y=f（x）是定义在R上的偶函数，且在[0，+∞）上单调递增，则满足f（m）＜f（1）的实数m的范围是．9．（3.00分）用二分法求函数f（x）=3x﹣x﹣4的一个零点，其参考数据如下：据此数据，可得方程3x﹣x﹣4=0的一个近似解（精确到0.01）是．10．（3.00分）方程|x2+4x+3|﹣a=0有2解，则实数a的取值范围是．11．（3.00分）已知y=f（x）是定义在R上的奇函数，且当x≥0时，，则此函数的值域为．12．（3.00分）设a+b=3，b＞0，则当a=时，取得最小值．二、选择题（本大题满分18分）本大题共6题，每题有且只有一个正确答案，考生应在答题纸的相应编号上，将代表答案的小方格涂黑，选对得3分，否则一律得零分.13．（3.00分）下列命题中，与命题“如果x2+3x﹣4=0，那么x=﹣4或x=1”等价的命题是（）A．如果x2+3x﹣4≠0，那么x≠﹣4或x≠1B．如果x≠﹣4或x≠1，那么x2+3x﹣4≠0C．如果x≠﹣4且x≠1，那么x2+3x﹣4≠0D．如果x=﹣4或x=1，那么x2+3x﹣4=014．（3.00分）已知实数a，b满足ab＞0，则“＜成立”是“a＞b成立”的（）A．充分非必要条件 B．必要非充分条件C．充要条件D．既非充分又非必要条件15．（3.00分）若a，b∈R，且ab＞0，则下列不等式中，恒成立的是（）A．a2+b2＞2ab B．C． D．16．（3.00分）如图所示曲线是幂函数y=x a在第一象限内的图象，其中a=±，a=±2，则曲线C1，C2，C3，C4对应a的值依次是（）A．、2、﹣2、﹣B．2、、﹣、﹣2 C．﹣、﹣2、2、D．2、、﹣2、﹣17．（3.00分）下列函数中，在其定义域内既是奇函数又是减函数的是（）A．y=﹣|x|（x∈R）B．y=﹣x3﹣x（x∈R）C．D．18．（3.00分）对于函数f（x），若在定义域内存在实数x，满足f（﹣x）=﹣f（x），称f（x）为“局部奇函数”，若f（x）=4x﹣m2x+1+m2﹣3为定义域R上的“局部奇函数”，则实数的取值范围是（）A．1﹣≤m≤1+B．1﹣≤m≤2C．﹣2≤m≤2D．﹣2≤m≤1﹣三、解答题（本大题满分46分）本大题共有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．（6.00分）本题共有2题，第1小题满分4分，第2小题满分2分已知集合A={x||x﹣1|≤1}，B={x|x≥a}．（1）当a=1时，求集合A∩B；（2）若A⊆B，求实数a的取值范围．20．（8.00分）已知a≠0，试讨论函数f（x）=在区间（0，1）上单调性，并加以证明．21．（8.00分）某商场对顾客实行购物优惠活动，规定一次购物总额：（1）如果不超过500元，那么不予优惠；（2）如果超过500元但不超过1000元，那么按标价给予8折优惠；（3）如果超过1000元，那么其中1000元给予8折优惠，超过1000元部分按5折优惠．设一次购物总额为x元，优惠后实际付款额为y元．（1）试写出用x（元）表示y（元）的函数关系式；（2）某顾客实际付款1600元，在这次优惠活动中他实际付款比购物总额少支出多少元？22．（12.00分）已知函数f（x）=3x+k（k为常数），A（﹣2k，2）是函数y=f1（x）图象上的点．（1）求实数k的值及函数y=f1（x）的解析式：（2）将y=f1（x）的图象向右平移3个单位，得到函数y=g（x）的图象，若2f1（x+﹣3）﹣g（x）≥1对任意的x＞0恒成立，试求实数m的取值范围．23．（12.00分）已知集合H是满足下列条件的函数f（x）的全体：在定义域内存在实数x0，使得f（x0+1）=f（x0）+f（1）成立．（1）幂函数f（x）=x﹣1是否属于集合H？请说明理由；（2）若函数g（x）=lg∈H，求实数a的取值范围；（3）证明：函数h（x）=2x+x2∈H．2014-2015学年上海市金山区高一（上）期末数学试卷参考答案与试题解析一、填空题（本大题满分36分）本大题共有12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3分，否则一律得零分）1．（3.00分）已知全集U=R，A={x|x≥2}，则∁U A={x|x＜2} ．【解答】解：∵全集U=R，A={x|x≥2}，∴∁U A={x|x＜2}，故答案为：{x|x＜2}2．（3.00分）函数y=lg的定义域是（﹣∞，﹣1）∪（1，+∞）．【解答】解：∵函数y=lg，∴x应满足：；解得0＜x＜1，或x＞1，∴函数y的定义域是（﹣∞，﹣1）∪（1，+∞）．故答案为：（﹣∞，﹣1）∪（1，+∞）．3．（3.00分）函数y=x+（x＞0）的最小值为2．【解答】解：∵x＞0，∴≥2，当且仅当x=时取等号，此时x=，即函数的最小值是2，故答案为：2．4．（3.00分）若集合A={﹣1，0，1}，集合B={x|x=t2，t∈A}，用列举法表示B= {0，1} ．【解答】解：当t=±1时，x=1，当t=0时，x=0，∴B={0，1}，故答案为：{0，1}．5．（3.00分）若4x﹣2x+1=0，则x=1．【解答】解：∵4x﹣2x+1=0，∴2x（2x﹣2）=0，∴2x﹣2=0，解得x=1．故答案为：16．（3.00分）已知关于x的不等式x2﹣（a﹣1）x+（a﹣1）＞0的解集是R，则实数a取值范围是（1，5）．【解答】解：∵关于x的不等式x2﹣（a﹣1）x+（a﹣1）＞0的解集是R，∴△＜0，即（a﹣1）2﹣4（a﹣1）＜0；整理得（a﹣1）（a﹣5）＜0，解得1＜a＜5；∴实数a取值范围是（1，5）．故答案为：（1，5）．7．（3.00分）已知函数y=a x﹣1+1（a＞0，a≠1）的图象经过一个定点，则顶点坐标是（1，2）．【解答】解：当x=1时，f（1）=a1﹣1+1=a0+1=2，∴函数f（x）=a x﹣1+1的图象一定经过定点（1，2）．故答案为：（1，2）．8．（3.00分）已知y=f（x）是定义在R上的偶函数，且在[0，+∞）上单调递增，则满足f（m）＜f（1）的实数m的范围是﹣1＜m＜1．【解答】解：∵函数f（x）是定义在R上的偶函数，且在区间[0，+∞）上单调递增．∴不等式f（m）＜f（1）等价为f（|m|）＜f（1），即|m|＜1，∴﹣1＜m＜1，即实数m的取值范围是﹣1＜m＜1，故答案为：﹣1＜m＜1．9．（3.00分）用二分法求函数f（x）=3x﹣x﹣4的一个零点，其参考数据如下：据此数据，可得方程3x﹣x﹣4=0的一个近似解（精确到0.01）是 1.56．【解答】解：由表格作数轴如下，故f（1.5625）f（1.5563）＜0；故方程3x﹣x﹣4=0的一个近似解在（1.5563，1.5625）之间，故可取（1.5563+1.5625）=1.5594≈1.56作为近似解．故答案为：1.56．10．（3.00分）方程|x2+4x+3|﹣a=0有2解，则实数a的取值范围是a=0或a ＞1．【解答】解：方程|x2+4x+3|﹣a=0有2解可化为y=|x2+4x+3|与y=a有两个交点，作函数y=|x2+4x+3|的图象如右图，故当a=0或a＞1时，有两个交点；故答案为：a=0或a＞1．11．（3.00分）已知y=f（x）是定义在R上的奇函数，且当x≥0时，，则此函数的值域为．【解答】解：设t=，当x≥0时，2x≥1，∴0＜t≤1，f（t）=﹣t2+t=﹣+，∴0≤f（t）≤，故当x≥0时，f（x）∈[0，]；∵y=f（x）是定义在R上的奇函数，∴当x≤0时，f（x）∈[﹣，0]；故函数的值域时[﹣，]．12．（3.00分）设a+b=3，b＞0，则当a=﹣时，取得最小值．【解答】解：∵a+b=3，b＞0，∴b=3﹣a＞0，即a＜3，当0＜a＜3时，=+=++≥+=+=，当且仅当a=取等号，故当a=时，取得最小值；当a＜0时，=﹣﹣=﹣﹣﹣≥﹣+2=﹣+=，当且仅当a=﹣取等号，故当a=﹣时，取得最小值；综上所述a的值为﹣时，取得最小值．故答案为：﹣．二、选择题（本大题满分18分）本大题共6题，每题有且只有一个正确答案，考生应在答题纸的相应编号上，将代表答案的小方格涂黑，选对得3分，否则一律得零分.13．（3.00分）下列命题中，与命题“如果x2+3x﹣4=0，那么x=﹣4或x=1”等价的命题是（）A．如果x2+3x﹣4≠0，那么x≠﹣4或x≠1B．如果x≠﹣4或x≠1，那么x2+3x﹣4≠0C．如果x≠﹣4且x≠1，那么x2+3x﹣4≠0D．如果x=﹣4或x=1，那么x2+3x﹣4=0【解答】解：原命题与其逆否命题等价，故命题“如果x2+3x﹣4=0，那么x=﹣4或x=1”等价的命题是：如果x≠﹣4且x≠1，那么x2+3x﹣4≠0，故选：C．14．（3.00分）已知实数a，b满足ab＞0，则“＜成立”是“a＞b成立”的（）A．充分非必要条件 B．必要非充分条件C．充要条件D．既非充分又非必要条件【解答】解：ab＞0，＜⇔⇔b＜a．∴实数a，b满足ab＞0，则“＜成立”是“a＞b成立”的充要条件．故选：C．15．（3.00分）若a，b∈R，且ab＞0，则下列不等式中，恒成立的是（）A．a2+b2＞2ab B．C． D．【解答】解：对于A；a2+b2≥2ab所以A错对于B，C，虽然ab＞0，只能说明a，b同号，若a，b都小于0时，所以B，C 错∵ab＞0∴故选：D．16．（3.00分）如图所示曲线是幂函数y=x a在第一象限内的图象，其中a=±，a=±2，则曲线C1，C2，C3，C4对应a的值依次是（）A．、2、﹣2、﹣B．2、、﹣、﹣2 C．﹣、﹣2、2、D．2、、﹣2、﹣【解答】解：根据幂函数y=x a在第一象限内的图象，知；当a=2时，幂函数y=x2在第一象限内是增函数，图象向上靠近y轴，符合C1特征；当a=时，幂函数y=在第一象限内是增函数，图象向右靠近x轴，符合C2特征；当a=﹣时，幂函数y=在第一象限内是减函数，图象向右靠近x轴，符合C3特征；当a=﹣2时，幂函数y=x﹣2在第一象限内是减函数，图象向右更靠近x轴，符合C4特征．综上，曲线C1，C2，C3，C4对应a的值依次是2、、﹣、﹣2．故选：B．17．（3.00分）下列函数中，在其定义域内既是奇函数又是减函数的是（）A．y=﹣|x|（x∈R）B．y=﹣x3﹣x（x∈R）C．D．【解答】解：A选项不正确，因为y=﹣|x|（x∈R）是一个偶函数，且在定义域内不是减函数；B选项正确，y=﹣x3﹣x（x∈R）是一个奇函数也是一个减函数；C选项不正确，是一个减函数，但不是一个奇函数；D选项不正确，是一个奇函数，但在定义域上不是减函数．综上，B选项正确故选：B．18．（3.00分）对于函数f（x），若在定义域内存在实数x，满足f（﹣x）=﹣f（x），称f（x）为“局部奇函数”，若f（x）=4x﹣m2x+1+m2﹣3为定义域R上的“局部奇函数”，则实数的取值范围是（）A．1﹣≤m≤1+B．1﹣≤m≤2C．﹣2≤m≤2D．﹣2≤m≤1﹣【解答】解：根据“局部奇函数”的定义可知，函数f（﹣x）=﹣f（x）有解即可，即f（﹣x）=4﹣x﹣m2﹣x+1+m2﹣3=﹣（4x﹣m2x+1+m2﹣3），∴4x+4﹣x﹣2m（2x+2﹣x）+2m2﹣6=0，即（2x+2﹣x）2﹣2m⋅（2x+2﹣x）+2m2﹣8=0有解即可．设t=2x+2﹣x，则t=2x+2﹣x≥2，∴方程等价为t2﹣2m⋅t+2m2﹣8=0在t≥2时有解，设g（t）=t2﹣2m⋅t+2m2﹣8，对称轴x=，①若m≥2，则△=4m2﹣4（2m2﹣8）≥0，即m2≤8，∴﹣2，此时2，②若m＜2，要使t2﹣2m⋅t+2m2﹣8=0在t≥2时有解，则，即，解得1﹣，综上：1﹣．故选：B．三、解答题（本大题满分46分）本大题共有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．（6.00分）本题共有2题，第1小题满分4分，第2小题满分2分已知集合A={x||x﹣1|≤1}，B={x|x≥a}．（1）当a=1时，求集合A∩B；（2）若A⊆B，求实数a的取值范围．【解答】解：由题意，A={x||x﹣1|≤1}=[0，2]，（1）B={x|x≥1}，故A∩B=[1，2]．（2）∵A⊆B，∴a≤0．20．（8.00分）已知a≠0，试讨论函数f（x）=在区间（0，1）上单调性，并加以证明．【解答】解：a＜0时，f（x）在（0，1）上是减函数，a＞0时，f（x）在（0，1）上是增函数；证明如下：任取x1，x2∈（0，1），且x1＜x2；∴f（x1）﹣f（x2）=﹣=；∵0＜x1＜x2＜1，∴x1+x2＞0，x1﹣x2＜0，（1﹣）（1﹣）＞0；∴当a＜0时，f（x1）﹣f（x2）＞0，f（x）在（0，1）上是减函数；当a＞0时，f（x1）﹣f（x2）＜0，f（x）在（0，1）上是增函数．综上，a＜0时，f（x）在（0，1）上是减函数，a＞0时，f（x）在（0，1）上是增函数．21．（8.00分）某商场对顾客实行购物优惠活动，规定一次购物总额：（1）如果不超过500元，那么不予优惠；（2）如果超过500元但不超过1000元，那么按标价给予8折优惠；（3）如果超过1000元，那么其中1000元给予8折优惠，超过1000元部分按5折优惠．设一次购物总额为x元，优惠后实际付款额为y元．（1）试写出用x（元）表示y（元）的函数关系式；（2）某顾客实际付款1600元，在这次优惠活动中他实际付款比购物总额少支出多少元？【解答】解：（1）由题可知：y=．（6分）（2）∵y=1600＞900，∴x＞1000，∴500+400+0.5（x﹣1000）=1600，解得，x=2400，2400﹣1600=800，故此人在这次优惠活动中他实际付款比购物总额少支出800元．…（12分）22．（12.00分）已知函数f（x）=3x+k（k为常数），A（﹣2k，2）是函数y=f1（x）图象上的点．（1）求实数k的值及函数y=f1（x）的解析式：（2）将y=f1（x）的图象向右平移3个单位，得到函数y=g（x）的图象，若2f1（x+﹣3）﹣g（x）≥1对任意的x＞0恒成立，试求实数m的取值范围．【解答】解：（1）∵函数f（x）=3x+k（k为常数），且A（﹣2k，2）是函数y=f1（x）图象上的点；∴32+k=﹣2k，解得k=﹣3；∴f（x）=3x﹣3，∴函数y=f1（x）=log3（x+3）；（2）将y=f1（x）=log3（x+3）的图象向右平移3个单位，得到函数y=g（x）的图象，∴y=g（x）=log3x；∵2f1（x+﹣3）﹣g（x）≥1，即2log3（x+﹣3+3）﹣log3x≥1，∴log3≥1；即≥3对任意的x＞0恒成立，∴x+2+≥3，即x++2≥3对任意的x＞0恒成立；∵x＞0，∴x+≥2，当且仅当x=时取“=”，∴函数h（m）=x++2≥2+2=4，令4≥3，解得m≥，∴实数m的取值范围[，+∞）．23．（12.00分）已知集合H是满足下列条件的函数f（x）的全体：在定义域内存在实数x0，使得f（x0+1）=f（x0）+f（1）成立．（1）幂函数f（x）=x﹣1是否属于集合H？请说明理由；（2）若函数g（x）=lg∈H，求实数a的取值范围；（3）证明：函数h（x）=2x+x2∈H．【解答】（1）解：若f（x）=x﹣1∈H，则有，即，而此方程无实数根，所以f（x）=x﹣1∉H．（4分）（2）解：由题意有实数解即，也即有实数解．当a=2时，有实数解．当a≠2时，应有．综上得，a的取值范围为．（3）证明：∵，∴令m（x）=2x+2x﹣2，∵m（x）在R上连续不断，且m（0）=﹣1＜0，m（1）=2＞0，∴存在x0∈（0，1），使得m（x0）=0成立．∴存在x0∈（0，1），使得h（x0+1）=h（x0）+h（1）成立．∴h（x）∈H．赠送初中数学几何模型【模型二】半角型：图形特征：AB正方形ABCD 中，∠EAF =45° ∠1=12∠BAD 推导说明：1.1在正方形ABCD 中，点E 、F 分别在BC 、CD 上，且∠FAE ＝45°，求证：EF ＝BE +DF45°DEa +b-a45°A1.2在正方形ABCD 中，点E 、F 分别在BC、CD 上，且EF ＝BE +DF ，求证：∠FAE ＝45°E-aaBE挖掘图形特征：x-aa-a运用举例：1．正方形ABCD的边长为3，E、F分别是AB、BC边上的点,且∠EDF=45°.将△DAE绕点D逆时针旋转90°，得到△DCM.（1）求证：EF=FM（2）当AE=1时，求EF的长．DE2.如图，△ABC是边长为3的等边三角形，△BDC是等腰三角形，且∠BDC=120°．以D为顶点3．如图，梯形ABCD 中，AD ∥BC ，∠C ＝90°，BC ＝CD ＝2AD ＝4，E 为线段CD 上一点，∠ABE ＝45°.（1）求线段AB 的长；（2）动点P 从B 出发，沿射线．．BE 运动，速度为1单位/秒，设运动时间为t ，则t 为何值时，△ABP 为等腰三角形； （3）求AE －CE 的值.变式及结论：4.在正方形ABCD中，点E，F分别在边BC，CD上，且∠EAF=∠CEF=45°．（1）将△ADF绕着点A顺时针旋转90°，得到△ABG（如图1），求证：△AEG≌△AEF；（2）若直线EF与AB，AD的延长线分别交于点M，N（如图2），求证：EF2=ME2+NF2；（3）将正方形改为长与宽不相等的矩形，若其余条件不变（如图3），请你直接写出线段EF，BE，DF之间的数量关系．F。
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金山区2014学年第一学期期末考试高三化学试卷(完卷时间:120分钟满分:150分)考生注意：1．本试卷满分150分，考试时间120分钟。
2．本卷使用答题纸，所有答案均填写在答题纸上，填写在试卷上一律不给分。
相对原子质量：H－1；C－12；N－14；O－16；Mg－24；Al－27；Na－23；S－32；Cl－35.5；Fe－56；Ba－137一、选择题（本题共10分，每小题2分，每题只有一个正确选项）1．有关化学资源的合成、利用与开发的叙述合理的是A. 大量使用化肥和农药，能不断提高农作物产量B. 通过有机合成，可以制造出比钢铁更强韧的新型材料C. 安装煤炭燃烧过程的“固硫”装置，主要是为了提高煤的利用率D. 开发利用可燃冰（固态甲烷水合物），有助于海洋生态环境的治理2．对于排布在2s轨道上的电子，不能确定的是A. 电子所在的电子层B．电子的自旋方向C．电子云的形状D．电子云的伸展方向3．能确定为丙烯的化学用语是4．下列有关共价键的说法正确的是A. 分子晶体中共价键越强，熔沸点越高B. 只含共价键的物质，一定是共价化合物C. 两种元素组成的分子中一定只含有极性共价键D. 分子晶体中，可能不存在共价键，但一定存在分子间作用力5．下列各组物质既不是同系物又不是同分异构体的是A. 甲酸甲酯和乙酸B. 对甲基苯酚和苯甲醇C. 油酸甘油酯和乙酸乙酯D. 软脂酸甘油酯和硬脂酸甘油酯二、选择题（本题共12小题，每小题3分，每题只有一个正确选项）6．有关物质的性质可以用元素周期律解释的是A．酸性：HCl>H2S>H2O B．密度：Na>K>LiC．沸点：NH3>AsH3>PH3D．稳定性：HF>HCl>HBr7．NH5属于离子晶体。
与水反应的化学方程式为：NH5+H2O→NH3·H2O+H2↑，它也能跟乙醇发生类似的反应，并都产生氢气。
有关NH5叙述正确的是A. 与乙醇反应时，NH5被氧化B．NH5中N元素的化合价为+5价C ．1 mol NH 5中含有5 molN —H 键D ．1 mol NH 5与H 2O 完全反应，转移电子2 mol8．某一化学反应在不同条件下的能量变化曲线如右图所示。
下列说法正确的是A. 化学催化比酶催化的效果好B. 使用不同催化剂可以改变反应的能耗C. 反应物的总能量低于生成物的总能量D. 使用不同催化剂可以改变反应的热效应9. 关于Na 2O 2的叙述正确的是（N A 表示阿伏伽德罗常数）A ．7.8g Na 2O 2含有的共价键数为0.2N AB ．7.8 g Na 2S 与Na 2O 2的混合物，含离子总数为0.3 N AC ．7.8g Na 2O 2与足量的CO 2充分反应，转移的电数为0.2N AD ．0.2 mol Na 被完全氧化生成7.8g Na 2O 2，转移电子的数目为0.4N AA ．利用化学平衡常数判断化学反应进行的快慢B ．利用溶解度数据判断氧化还原反应发生的可能性C ．利用沸点数据推测将一些液体混合物分离的可能性D ．利用物质的摩尔质量判断相同状态下不同物质密度的大小12．根据右图装置分析，有关说法错误的是A. 虚线框中接直流电源，铁可能会被腐蚀B. 虚线框中接灵敏电流计或接直流电源，锌都是负极C. 虚线框中接灵敏电流计，该装置可将化学能转化为电能D. 若将电解液改成硫酸锌溶液并接上直流电源，该装置可用于铁皮上镀锌13．在化学反应中，有时存在“一种物质过量，另一种物质仍不能完全反应”的情况，下列反应不属于这种情况的是A. 过量的铜与浓硝酸反应B. 过量的铜与浓硫酸在加热条件下反应C. 过量二氧化锰与浓盐酸在加热条件下反应D. 过量的氢气与氮气在工业生产条件下反应14．25℃时将10mLpH = 11的氨水加水稀释至100mL ，下列判断正确的是A. 稀释后溶液的pH = 7B. 氨水的电离度增大，溶液中所有离子的浓度均减小C. 稀释过程中 增大D. pH = 11氨水的浓度为0.001mol /L 15．对已达化学平衡的反应：2X(g)+Y(g)2Z(g)，减小压强后，对反应产生的影响是A ．逆反应速率增大，正反应速率减小，平衡向逆反应方向移动c (NH 4+) c (NH 3∙H 2O)稀H 2SO 4B．逆反应速率减小，正反应速率增大，平衡向正反应方向移动C．正反应速率先减小后增大，逆反应速率减小，平衡向逆反应方向移动D．逆反应速率先减小后增大，正反应速率减小，平衡向逆反应方向移动16．关于实验室中四个装置的说明符合实验要求的是A B C DA.若需制备较多量的乙炔可用此装置B.可用此装置来制备硝基苯，但产物中可能会混有苯磺酸C.可用此装置来分离含碘的四氯化碳液体，最终在锥型瓶中可获得碘D.可用此装置来制备乙酸乙酯并在烧瓶中获得产物17．向100 mL 0.1 mol·L-1硫酸铝铵[NH4Al(SO4)2]溶液中逐滴滴入0.1 mol·L-1 Ba(OH)2溶液。
随着Ba(OH)2溶液体积V的变化，沉淀总物质的量n的变化如右图所示。
下列说法正确的A．a点的溶液呈中性B．a点沉淀的质量比c点沉淀的质量大C．b点加入Ba(OH)2溶液的体积为250 mLD．至c点完成反应，反应的离子方程式可表示为：Al3++2SO42-+NH4++2Ba2++5OH—→ AlO2-+2BaSO4↓+NH3·H2O+2H2O三、选择题（本题共5小题，每小题4分，每小题有一个或两个正确选项。
只有一个正确选项的，多选不给分；有两个正确选项的，选对一个给2分，选错一个该小题不给分。
）18．与实际化工生产功能相符合的化学方程式是A. 工业合成氨：N2 +3H2 2NH3B. 工业合成盐酸：H2 + Cl2 2HClC. 工业获取氯化钠：2Na + Cl2 2NaClD. 工业制取氯气：2NaCl + 2H2O 2NaOH + H2↑+ Cl2↑19.已知酸性：；现要将转变为，可行的方法是A．向该溶液中加入足量的稀硫酸，加热B．将该物质与稀硫酸共热后，再加入足量的NaOH溶液C．将该物质与足量的NaOH溶液共热，再通入足量CO2气体D．将该物质与稀硫酸共热后，再加入足量的NaHCO3溶液20. 常温下，向20mL 0.2mol/L H2A溶液中滴加0.2 mol/L NaOH溶液。
有关微粒的物质的量变化如右图，下列说法正确的是A．滴加过程中当溶液呈中性时，V(NaOH)≥20mLB．当V(NaOH)=30mL时，则有：2c(Na+) = 3c(A2-) + 3c(HA-)C．H 2A在水中的电离方程式是： H 2A H+ + HA－； HA—H+ + A2－D．当V(NaOH)=20mL时，则有：c(Na+) > c(HA-)> c(H+) > c(A2-) > c(OH-)21．室温时将100mLH2S和O2的混合气体，点燃完全反应后恢复到原来状况，体积变为40mL。
下列叙述中正确的是A. 剩余40mL气体一定是O2B. 剩余40mL气体是H2S或SO2C. 混合气体中含H2S 40mL或80mLD. 剩余40mL气体是SO2和O2混合气体22. 6克含杂质的Na2SO3样品与足量盐酸反应，可生成1.12升气体（S、T、P），气体质量为3克，该样品的组成可能是A. Na2SO3，Na2CO3B. Na2SO3，Na2SO4，NaHCO3C. Na2SO3，NaHCO3D.Na2SO3，NaHCO3，Na2CO3四、（本题共12分）工业上制取冰晶石(Na3AlF6)的化学方程式如下：2Al(OH)3+ 12HF+ 3Na2CO3→ 2Na3AlF6+ 3CO2↑+ 9H2O完成下列填空：23. 反应中四种元素对应的简单离子核外电子排布相同，请按离子半径从大到小的顺序排列这这四种离子，其中原子序数最大的元素原子核外有种能量不同的电子，其最外层电子的电子云有种不同的伸展方向。
24. 反应中有两种元素在元素周期表中位置相邻，能比较它们的金属性或非金属性强弱的事实是（选填编号）。
a．气态氢化物的稳定性b．最高价氧化物对应水化物的酸性c．单质与氢气反应的难易d．单质与同浓度酸发生反应的快慢25. 反应中两种金属元素，它们的最高价氧化物对应的水化物之间发生反应的离子方程式为。
26. 冰晶石在工业上可做电解氧化铝的助熔剂，此反应中若有0.6mol电子转移，则在极可得金属铝的质量为克。
27. 工业上不采用电解氯化铝的方法而是采用电解氧化铝的方法获得铝单质的原因是：。
五、（本题共12分）以下是处于研究阶段的“人工固氮”的新方法。
N 2在催化剂表面与水发生反应：2N 2(g)+6H 2O(l)4NH 3(g)+3O 2(g) ―1530.4 k J完成下列填空： 28．该反应平衡常数K 的表达式 。
29．上述反应达到平衡后，保持其他条件不变，升高温度，重新达到平衡，则 。
a ．平衡常数K 增大b ．H 2O 的浓度减小c ．容器内的压强增大d ．v 逆(O 2)减小30．研究小组分别在四个容积为2升的密闭容器中，充入N 2 1mol 、H 2O 3mol ，在催化剂条件下进行反应3小时。
实验数据见下表：第四组实验中以NH 3表示反应的速率是 ，与前三组相比，NH 3 生成量最小的原因可能是 。
31．氨水是实验室常用的弱碱。
（1）往CaCl 2溶液中通入CO 2至饱和，无明显现象。
再通入一定量的NH 3后产生白色沉淀，此时溶液中一定有的溶质是_____。
请用电离平衡理论解释上述实验现象______。
（2）向盐酸中滴加氨水至过量，该过程中离子浓度大小关系可能正确的是 。
a ．c (C1－)=c (NH 4+)>c (H +)=c (OH －) b ．c (C1－)>c (NH 4+)= c (OH －) >c (H +) c ．c (NH 4+)>c (OH －)>c (C1－)>c (H +) d ．c (OH －)>c (NH 4+)>c (H +)>c (C1－) 六、（本题共12分）利用下图装置探究温度对氨气还原Fe 2O 3的影响(固定装置略)。
完成下列填空：32．实验时A 中有大量紫红色的烟气，则NH 4I 的分解产物为 （至少填三种），碱石灰的作用是 ；33．装置B 中的反应化学方程式 ，D 装置的作用是 ；按上图装置进行对比实验，甲组用酒精灯、乙组用酒精喷灯对装置C 加热，反应产物均催化剂34．乙组得到的黑色粉末是 ；35．步骤1甲组中反应的离子方程式为 ；36．步骤3乙组中溶液变红的原因为 ；37．若装置C 中反应后的固体是Fe 2O 3和FeO 组成的混合物。