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湖北省蕲春县实验中学2014届高三培优补差训练题(四)

培优补差训练题（四）第一部分：阅读理解AIn the kitchen of my mother’s houses there has always been a wooden stand（木架）with a small notepad（记事本）and a hole for a pencil.I’m looking for paper on which to note down the name of a book I am recommending to my mother. Over forty years since my earliest memories of the kitchen pad and pencil, five houses later, the current paper and pencil look the same as they always did. Surely it can’t be the same pencil? The pad is more modern, but the wooden stand is definitely the original one.―I’m just amazed you still have the same stand for holding the pad and pencil after all these year.‖ I say to her, walking ba ck into the living-room with a sheet of paper and the pencil. ―You still use a pencil. Can’t you afford a pen?‖My mother r eplies a little sharply. ―It works perfectly well. I’ve always kept the stand in the kitchen. I never knew when I might want to note down an idea, and I was always in the kitchen in these days.‖Immediately I can picture her, hair wild, blue housecoat covered in flour, a wooden spoon in one hand, the pencil in the other, her mouth moving silently. My mother smiles and says, ―One day I was cooking and watching baby Pauline, and I had a brilliant thought, but the stand was empty. One of the children must have taken the paper. So I just picked up the breadboard and wrote it all down on the back. It turned out to be a real breakthrough for solving the mathematical problem I was working on.‖This story—which happened before I was born—reminds me how extraordinary my mother was, and is also a gifted mathematician. I feel embarrassed that I complain about not having enough child-free time to work. Later, when my mother is in the bathroom, I go into her kitchen and turn over the breadboards. Sure enough, on the back of the smallest one, are some penciled marks I recognize as mathematics. Those symbols have traveled unaffected through fifty years, rooted in the soil of a cheap wooden breadboard, invisible（看不到的）exhibits at every meal. 51．Why has the author’s mother alwa ys kept the notepad and pencil in the kitchen?A．To leave messages. B．To list her everyday tasks.C．To note down maths problems. D．To write down a flash of inspiration.52. What is the author’s original opinion about the wooden stand?A. It has great value for the family.B. It needs to be replaced by a better one.C. It brings her back to her lonely childhood.D .It should be passed on to the next generation.53. The author feels embarrassed for .A. blaming her mother wrongly.B. giving her mother a lot of trouble.C. not making good use of time as her mother did.D. not making any breakthrough in her field.54. What can be inferred from the last paragraph?A .The mother is successful in her career.B. The family members like traveling.C. The author had little time to play when young.D. The marks on the breadboard have disappeared.BUsually, when your teacher asks a question, there is only one correct answer. But there is one question that has millions of current answers. That qu estion is ―What’s your name?‖ Everyone gives a different answer, but everyone is correct.Have you ever wondered about people’s names? Where do they come from? What do they mean?People’s first names, or given names, are chosen by their parents. Sometimes the name of a grandparent or other member of the family is used. Some parents choose the name of a well-known person. A boy could be named George Washington Smith; a girl could be named Helen Keller Jones.Some people give their children names that mean good things. Clara means ―bright‖; Beatrice means ―one who gives happiness‖; Donald means ―world ruler‖; Leonard means ―as brave as a lion‖.The earliest last names, or surnames, were taken from place names. A family with the name Brook or Brooks probably lived near brook(小溪)；someone who was called Longstreet probably lived on a long, paved road. The Greenwood family lived in or near a leafy forest.Other early surnames came from people’s occupations. The most common occupational name is Smith, which means a person who makes things with iron or other metals. In the past, smiths were very important workers in every town and village. Some other occupational names are: Carter— a person who owned or drove a cart; Potter—a person who made pots and pans.The ancestors of the Baker family probably baked bread for their neighbors in their native village. The Carpenter’s great-great-great-grandfather probably built houses and furniture.Sometimes people were known for the color of their hair or skin, or their size, or their special abilities. When there were two men who were named John in the same village, the John with the gray hair probably became John Gray. Or the John was very tall could call himself John Tallman. John Fish was probably an excellent swimmer and John Lightfoot was probably a fast runner or a good dancer.Some family names were made by adding something to the father’s name. English-speaking people added–s or –son. The Johnsons are descendants of John; the Roberts family’s ancestor was Robert. Irish and Scottish people added Mac or Mc or O. Perhaps all of the MacDonnells and the McDonnells and the O’Donnells are descendants of the same Donnell.55. Which of the following aspects do the surnames in the passage NOT cover?A. Places where people lived.B. People’s characters.C. Talents that people possessed.D. People’s occupations.56. According to the passage, the ancestors of the Potter family most probably _______.A. owned or drove a cartB. made things with metalsC. made kitchen tools or containsD. built houses and furniture57. Suppose and English couple whose ancestors lived near a leafy forest wanted their new-bornson to become a world leader, the baby might be named _______.A. Beatrice SmithB. Leonard CarterC. George LongstreetD. Donald Greenwood58. The underlined word ―descendants‖ in the last paragraph means a person’s _____.A. later generationsB. friends and relativesC. colleagues and partnersD. later sponsorsCHaving a husband means an extra seven hours of housework each week for women, according to a new study. For men, getting married saves an hour of housework a week. ―It’s a well-known pattern,‖ said lead researcher Frank Stafford at University of Michigan’s Institute for Social Research. ―Men usually work more outside the home, while women take on more of the housework.‖He points out that differences among households exist. But in general, marriage means more housework for women and less for men. ―And the situatio n gets worse for women when they have children.‖ Stafford said.Overall, times are changing in the American home. In 1976, women busied themselves with 26 weekly hours of sweeping-and-dusting work, compared with 17 hours in 2005. Men are taking on more housework, more than doubling their housework hours from six in 1976 to 13 in 2005.Single women in their 20s and 30s did the least housework, about 12 weekly hours, while married women in their 60s and 70s did the most – about 21 hours a week.Men showed a somewhat different pattern, with older men picking up the broom more often than younger men. Single men worked the hardest around the house, more than that of all other age groups of married men.Having children increases housework even further. With more than three, for example, wives took on more of the extra work, clocking about 28 hours a week compared with husbands’ 10 hours.59. According to the ―well-known pattern‖ in Paragraph 1, a married man ________.A. takes on heavier workB. does more houseworkC. is the main breadwinnerD. is the master of the house60. How many hours of housework did men do every week in the 1970s?A. About 28B. About 26C. About 13D. About 661. What kind of man is doing most housework according to the text?A. An unmarried man.B. An older married man.C. A younger married man.D. A married man with children.62. What can we conclude from Stafford’s research?A. Marriage gives men more freedom.B. Marriage has effects on job choices.C. Housework sharing changes over time.D. Having children means doubled housework.DAsk someone what they have done to help the environment recently and they will almost certainly mention recycling. Recycling in the home is very important of course. However, being forced to recycle often means we already have more material than we need. We are dealing with the results of that over-consumption in the greenest way possible, but it would be far better if we did not need to bring so much material home in the first place.The total amount of packaging increased by 12% between 1999 and 2005. It now makes upa third of a typical household’s waste in the UK. In many supermarkets n owadays food items are packaged twice with plastic and cardboard.Too much packaging is doing serious damage to the environment. The UK, for example, is running out of it for carrying this unnecessary waste. If such packaging is burnt, it gives off greenhouse gases which go on to cause the greenhouse effect. Recycling helps, but the process itself uses energy. The solution is not to produce such items in the first place. Food waste is a serious problem, too. Too many supermarkets encourage customers to buy more than they need. However, a few of them are coming round to the idea that this cannot continue, encouraging customers to reuse their plastic bags, for example.But this is not just about supermarkets. It is about all of us. We have learned to associate packaging with quality. We have learned to think that something unpackaged is of poor quality. This is especially true of food. But is also applies to a wide range of consumer products, which often have far more packaging than necessary.There are signs of hope. As more of us recycle, we are beginning to realize just how much unnecessary material are collecting. We need to face the wastefulness of our consumer culture, but we have a mountain to climb.63. What does the underlined phrase ―over-consumption‖ refer to?A. Using too much packaging.B. Recycling too many wastes.C. Making more products than necessary.D. Having more material than is needed.64. The author uses figures in Paragraph 2 to show _______.A. the tendency of cutting household wasteB. the increase of packaging recyclingC. the rapid growth of super marketsD. the fact of packaging overuse65. What can be inferred from Paragraph 4?A. Unpackaged products are of bad quality.B. Supermarkets care more about packaging.C. It is improper to judge quality by packaging.D. Other products are better packaged than food.66. What can we learn from the last paragraph?A. Fighting wastefulness is difficult.B. Needless material is mostly recycled.C. People like collecting recyclable waste.D. The author is proud of their consumer culture.EHoliday NewsVacancies (空位) now and in the school holidays at a country hotel in Devon. This comfortable, friendly home-from-home lies near the beautiful quiet countryside, but just a drive away from the sea. The food is simple but good. Children and pets are welcome.Reduced prices for low season.The Snowdonia CenterThe Snowdonia Center for young mountain climbers has a mountain 1068.The beginners’ costs are ￡57 for a week, including food and rooms. Equipment is included except walking shoes, which can be hired at a low cost.You must be in good health and prepared to go through a period of body exercises. This could be the beginning of a lifetime of lifetime of mountain climbing adventure.The World Sea Trip of a LifetimeOur World Sea Trip of 2008 will be unlike any holiday you have ever been on before. Instead of one hotel after another, with all its packing and unpacking waiting and traveling, you just go to bed in one country and wake up in another.On board the ship, you will be well taken care of. Every meal will be first-class and every cabin like your home.During the trip, you can rest on deck(甲板), enjoy yourself in the games rooms and in the evening dance to our musical team and watch our wonderful play.You will visit all the places most people only dream about – from Acapulco and Hawaii to Tokyo and Hong Kong.For a few thousand pounds, all you’ve ever hoped for can be yours.67. What can you do if you like to go on holidays with pets?A. Choose the holiday in Devon.B.Go to the Snowdonia CentreC. Join the World Sea Trip of 2008D.Visit Acapulco and Hawaii68. In what way is the Snowdonia Centre different from the other two holidays?A.It provides chances of family gatherings.B. It provides customers with good food.C. It offers a sport lesson.D. It offers comfortable room.69. What is special about the World Sea Trip of 2008?A. You can have free meals on deck every day.B. You can sleep on a ship and tour many places.C. You will have chances to watch and act in a play.D. You have to do your own packing and unpacking.70. At the Snowdonia Centre, the beginners’ costs of ￡57 do not cover .A. foodB. roomsC. body exercisesD. walking shoes第二部分：完成句子71. Yao Ming told the press that he will be focusing his efforts on sports,an area .(familiar)姚明告诉媒体他会致力于体育，那是一个他熟悉的领域。

湖北省蕲春县实验中学2014届高三培优补差训练题(六)

培优补差训练题（六）第一节：多项选择21．Most workers have heard the news that their products, with excellent quality and delicate skill, are enjoying growing ___ among the customers abroad．A．favor B．dignity C．response D．profit22．The best way to undertake big projects is to force them into your ．Spare time doesn't appear from nowhere and you need to create it with efforts．A．presentation B．campaign C．schedule D. procedure23．Don't lend your textbooks to those people; it is difficult to ___the books from them when you want them．A．attain B．recognize C．preserve D．recover24．The city will ____ no more than 10 thousand new car licenses each year in order to set a limit on the number of cars and improve the environment．A．witness B．evaluate C．distribute D．declare25．For an artist who has no lack of passion and great courage to overcome any difficulty, inspirations of his art will never ___ ．A．break out B．run out C．work out D．pull out26．Choking clouds of pollution have virtually ____the northeastern Chinese city of Harbin,forcing schools to call off classes, disturbing traffic and closing the airport．A．left out B．shut down C．blown up D．set aside27．In less developed countries the primary problem is providing ___ employment for the rapidly expanding urban populations．A．sufficient B．essential C．alternative D．convenient28．It is evident that there is a（n）___ increase in the ability of the students to express themselves under your careful instruction these days．A．abundant B．particular C．accessible D．definite29．Although she brushed the old pot ___ , she could not make it completely clean, whichmade her very upset．A．instantly B．roughly C．thoroughly D．occasionally30．He also acknowledges that, in general, state-owned companies do have an advantage ___ the stability, management, and career prospects they offer．A．in terms of B．in need of C．in charge of D．in praise of第二节完形填空I lost my sight when I was four years old by falling off a box car in a yard in Atlantic City and landing on my head．Now I am thirty-two．I can 31 remember the brightness of sunshine and what red color is．It would be wonderful to see again, 32 a disaster can do strange things to people．It occurred to me the other day that I might not have come to 33 life as I do if I hadn't been blind．I believe in life now．I am not so sure that I would have believed in it so deeply, 34 . I don't mean that I would prefer to go without my eyes．I simply mean that the loss of them made me 35 more what I had left．The hardest lesson I had to learn was to believe in myself．That was 36 . If hadn’t been able to do that, I would have 37 and become a chair rocker for the rest of my life．When I say 38 in myself I am not talking about simply the kind of self-confidence that helps me down a(n) 39 staircase alone．That is part of it．But I mean something 40 than that: an assurance that there is a special place where I can make myself fit.It took me years to discover and 41 this assurance．It had to start with the most e4mentary things．Once a man gave me an indoor baseball, I thought he was laughing at me and I was 42 ．"I can't use this．" I said．Take it with you;' he 43 me．t（and roll it around．" The words 44 in my head．"Roll it around!" By rolling the ball I could 45 where it went．This gave me an idea how to achieve a goal I had thought 46 : playing baseball．At Philadelphia's Overbrook School for the Blind I 47 a successful variation of baseball．We called it ground ball．All my life I have set ahead of me a series of 48 and then tried to reach them, one at a time．I had to learn my 49 ．It was no good trying for something I knew at the start was wildly out of reach because that only invited the bitterness of failure．I would fail sometimes anyway but on the average I made 50 .31．A．entirely B．nearly C．vaguely D．simply32．A．and B．but C．so D．for33．A．assess B．fear C．enrich D．love34．A．otherwise B．therefore C．however D．besides35．A．understand B．appreciate C．possess D．accept36．A．enough B．specific C．tough D．basic37．A．survived B．escaped C. collapsed D．sacrificed38．A．hope B．power C．courage D．belief39．A．unfamiliar B．unbelievable C．unexpected D．uncomfortable40．A．harder B．bigger C．warmer D．heavier41．A．weaken B．strengthen C．sharpen D．brighten42．A．upset B．flattered C．ashamed D．hurt43．A．urged B．promised C．convinced D．advised44．A．flashed B．appeared C．stuck D．crowded45．A．notice B．hear C．smell D．touch46．A．impossible B．important C．imaginary D．impressive47．A．produced B．imitated C．invented D．spotted48．A．goals B．efforts C．directions D．barriers49．A．challenges B．strengths C．situations D．limitations50．A．history B．change C．progress D．sense第三节阅读理解Art robbery and art forgery （仿造）are both major themes in crime movies and literature．In the 2012 comedy movie Gambit, British actor Colin Firth plays an art curator who cheats his abusive boss into buying a fake Monet．In reality, art crimes are no less interesting and exciting．According to The New York Times, over the past 15 years, Glafira Rosales fooled two local commercial art galleries into buying 63 false works of art for more than $30 million．She passed off fake paintings as works by 20th century modernist masters such as American artists Mark Rothko and Jackson Pollock．But in fact, these so called "newly discovered works" were all produced by a single man, a Chinese immigrant named Qian Peishen．The art world was shocked by Rosales' deception．But to the public, it was amusing and 6most satisfying to see wealthy people get tricked．So what decides the value of a piece of art? Is it beauty? Is it the artist's talent and craftsmanship? Or is it just because the artist is famous?We should take beauty out．If the buyers were buying paintings only for their beauty, they'll be content displaying good fakes on their walls．They wouldn't be so upset when a forgery is exposed．The art market claims that great artists are inimitable, and that this inimitability justifies the absurd price of their works．We can't deny that most famous artists are good at what they do, but forgers like Qian show that their works are imitable．Otherwise, the difference between the original and the copycats would be obvious and Rosales would not be able to fool anyone．According to an article in the Economist, expensive paintings are what．economists call positional goods．They are valuable because other people can't have them．With other goods, a higher price reduces demand．But art turns down the laws of economics．“When the goods that is really being purchased is evidence that the buyer has paid a lot, price increases cause demand to boom," explained the article．That's why scarcity and authenticity are so important in the art market．Artists sometimes forget this．Demien Hirst, the British pop artist, is famous for his spot paintings．But they dropped in value when it became clear that they had been produced in quantities so vast that nobody knew how many were out there．The art market lost faith in these paintings because no one could be sure which of them were authentic and which were fake．67．The first paragraph is meant to tell the readers that ____ ．A．movies and literature will be popular with art involved in themB．art crimes are as interesting and exciting in reality as in moviesC．Gambit is a good movie with art forgery as the major themeD．real art crime in reality can be adapted into popular films68．Who is amused and satisfied to see the rich buy forged works of art?A．The public．B．Glafira Rosales．C．Qian Peishen．D．Mark Rothlo and Jackson Pollock．69．What kind of art buyers should not be unhappy with its high prices when a forgery is exposed according to the author?A．Those who buy only for its beauty．B．Those who buy for its inimitability．C．Those who 'buy for its authenticity．D．Those who buy for its scarcity．70．What is the law of economics theory behind矾goods according to the economists?A．They are valuable goods．B．High prices reduce the demand of art goods．C．High prices increase the demand of art goods．D．They are produced in quantities to satisfy people．第三节完成句子·1阅读下列各小题，根据括号内的汉语提示，用句末括号内的英语单词完成句子，并将答案写在答题卡上的相应题号后。

实验班培优3语文参考答案

实验班培优3语文参考答案一、基础题（76分，4分一个）1. C2. A3. C4. B5. B6. D7．C（A项bân／bēn bì／pì cān shânɡ／chãnɡ; B项bāo／bōài／yì bào／pù b ó; C项bì／bài ān／yīn páo／fú nào／chuò; D项chāi／châ jiǎo jī／yī liáo／liǎo）8．A（纵横捭阖：指在政治、外交上运用手段进行联合或分化。

异曲同工：不同的曲调演得同样好，比喻不同的人的辞章或言论同样精彩，或者不同的做法收到同样好的效果。

振振有词：形容理由似乎充分，说个不休。

八面玲珑：原指窗户宽敞明亮，后用来形容人处事圆滑，不得罪任何一方）9.答案A.【A.zhã/zhâ、gù/gào、cháng/ chǎng ；B. hã/ hâ、juàn、tiáo / diào；C. ｃｈìch à/zhà，bì；D. huòmù/mò、zǎn/cuán10、答案A .【解析】A．登堂入室：比喻学问或技能从浅到深，循序渐进，达到了很高的水平。

易误用为“进入”。

B．大肆：谓无顾忌地进行某种活动，常用于贬义。

此处符合语境。

C.痛心疾首：疾首：头痛。

心痛，头也痛。

比喻痛恨到极点或悲伤到了极点。

D.不可思议：原有神秘奥妙的意思。

现多指无法想象，难以理解。

11、答案C.（⑤是对史铁生总的评价，后面的几句是对这一评价的解释。

②是对史铁生品格的赞扬，④是通过比较进一步突出史铁生的特点；⑥①③三句联系较为明显，⑥是说史铁生的创作，①是说创作的艰辛，有“文字”“一个个字”相照应，③是对史铁生以生命创作出的“文字”的赞扬。

2021年高三实验班物理培优模拟考试题(一)

2021年高三实验班物理培优模拟考试题〔一〕2021年3月09日一、选择题〔5*4=20分〕1.如图7所示，体重为510N 的人，用滑轮组拉重500N 的物体A 沿程度方向以0.02m/s 的速度匀速运动。

运动中物体A 受到地面的摩擦阻力为200N 。

动滑轮重为20N 〔不计绳重和摩擦，地面上的定滑轮与物体A 相连的绳子沿程度方向，地面上的定滑轮与动滑轮相连的绳子沿竖直方向，人对绳子的拉力与对地面的压力始终竖直向下且在同一直线上，〕。

那么以下计算结果中正确的选项是〔〕A ．绳子自由端受到的拉力大小是100NB ．人对地面的压力为400NC ．人对地面的压力为250ND ．绳子自由端运动速度是0.06m/s 2.家用电吹风由电动机和电热丝等组成。

为了保证电吹风的平安使用，要求：电动机不工作时，电热丝不能发热；电热丝不发热时，电动机仍能工作。

以下电路中符合要求的是：( )3.在如下图的电路中，当电键S 闭合后，电压表有示数，调节可变电阻R 的阻值，电压表的示数增大了△U 。

那么〔〕A ．可变电阻R 被调到较小的阻值B ．电阻R 2两端的电压减小，减小量等于△UC ．通过电阻R 2的电流减小，减小量小于△U/ R 2D ．通过电阻R 2的电流减小，减小量等于△U/ R 14．在光滑的程度桌面上有两个质量均为m 的小球，由长度为2l 的拉紧细线相连。

以一恒力作用于细线中点，恒力的大小为F ，方向平行于桌面。

两球开场运动时，细线与恒力方向垂直。

在两球碰撞前瞬间，两球的速度在垂直于恒力方向的分量为〔〕A ．m Fl 2B ．m FlC ．m Fl 2D ．mFl 2 5.如下图的皮带轮传动装置中，A 为主动轮，B 为被动轮，L 为扁平的传动皮带，A 轮与B 轮的轮轴程度放置且互相平行，那么能传递较大功率的情况是( )A ．A 轮逆时针转且皮带L 较宽B ．A 轮逆时针转且皮带L 较窄C ．A 轮顺时针转且皮带L 较宽D ．A 轮顺时针转且皮带L 较窄 A二．计算题〔12+15+15+18+20=80分〕6.〔12分〕一个用电阻丝绕成的线圈，浸没在量热器所盛的油中，油的温度为0℃，当线圈两端加上一定电压后，油温渐渐上升，0℃时温度升高的速率为5.0 K·min−1，持续一段时间后，油温上升到30℃，此时温度升高的速率为4.5 K·min−1，这是因为线圈的电阻与温度有关。

实验班培优题库

实验班培优题库
实验班培优题库是一套针对学生提高学习成绩和能力的练习题集。

该题库涵盖了各个学科，包括语文、数学、英语、物理、化学等，旨在通过练习题的形式来帮助学生巩固所学知识，提高解题技巧和能力。

实验班培优题库的题目难度较高，适合学习成绩较好或者对学科知识有更高要求的学生使用。

通过练习这些题目，学生可以提高自己的思维能力、解决问题的能力以及应试能力等。

此外，实验班培优题库还有一定的权威性，因为这些题目都是由经验丰富的教师和专家编写而成，具有一定的参考价值。

同时，该题库还会根据学生的学习情况和考试要求进行不断更新和调整，以保证题目的针对性和有效性。

总之，实验班培优题库是一套非常有用的练习题集，能够帮助学生提高学习成绩和能力，增强自信心和应试能力。

高三重点班培优7

ab cd高三重点班培优7一选择题（1-5题为单选题，6-8题为多选题）1（10海南）在水平的足够长的固定木板上，一小物块以某一初速度开始滑动，经一段时间t 后停止．现将该木板改置成倾角为45°的斜面，让小物块以相同的初速度沿木板上滑．若小物块与木板之间的动摩擦因数为μ．则小物块上滑到最高位置所需时间与t 之比为（）A ．1μ+BCD 2（01理综)在抗洪抢险中，战士驾驶摩托艇救人。

假设江岸是平直的，洪水沿江向下游流去，水流速度为v 1，摩托艇在静水中的航速为v 2，战士救人的地点A 离岸边最近处O 的距离为d 。

如战士想在最短时间内将人送上岸，则摩托艇登陆的地点离O 点的距离为（）A ．21222v v dv - B ．0 C ．21v dv D ．12v dv3（04年全国理综）如图所示，ad 、bd 、cd 是竖直面内三根固定的光滑细杆，a 、b 、c 、d 位于同一圆周上，a 点为圆周的最高点，d 点为最低点。

每根杆上都套着一个小滑环(图中未画出)，三个滑环分别从a 、b 、c 处释放(初速为0)，用t 1、t 2、t 3依次表示滑环到达d 所用的时间，则（） A ．t 1 < t 2 < t 3 B ．t 1 > t 2 > t 3 C ．t 3 > t 1 > t 2 D ．t 1 = t 2 = t 3 4如图所示，岸上的人通过定滑轮用绳子拖动小船靠岸，则当人匀速运动时，船的运动情况是( ) A ．加速运动 B ．减速运动C ．匀速运动D ．条件不足，不能判定 5如图所示，在光滑的水平地面上有两个质量相等的物体，中间用劲度系数为k 的轻质弹簧相连，在外力F 1、F 2的作用下运动．已知F 1＞F 2，当运动达到稳定时，弹簧的伸长量为( )A ．F 1－F 2kB ．F 1－F 22kC ．F 1＋F 22kD ．F 1＋F 2k6如图所示，小船用绳牵引靠岸，设水的阻力不变，在小船匀速靠岸的过程中，有( )A ．绳子的拉力不断增大B ．绳子的拉力不变C ．船受的浮力减小D ．船受的浮力不变7在光滑的水平面上，用F=6N 的恒力，水平作用在质量为2kg 的质点上，使其由静止开始运动．试比较经过5s 的时间或经过5m 的位移时，突然撤去拉力，得到的正确结论是 ( ) A ．撤力时，前种情况的末速度大于后种情况的末速度 B ．撤力时，前种情况的末速度小于后种情况的末速度 C ．撤力前，前种情况比后种情况滑行的路程长 D ．撤力前，前种情况比后种情况滑行的时间长8如图所示，一名消防队员在模拟演习训练中，沿着长为12m 的竖立在地面上的钢管向下滑。

高中数学复习提升高三实验班练习题

高三实验班练习题一、填空题：1．已知圆1)sin 2()cos 2(:221=-+-θθy x C 与圆1:222=+y x C ，在下列说法中： ①对于任意的θ，圆1C 与圆2C 始终相切； ②对于任意的θ，圆1C 与圆2C 始终有四条公切线；时，圆1C 被直线④Q P ,分别为圆1C 与圆2C 上的动点，则||PQ 的最大值为4．其中正确命题的序号为______．2．在直角坐标系内，点实施变换后，对应点为，给出以下命题：①圆上任意一点实施变换后，对应点的轨迹仍是圆；②若直线上每一点实施变换后，对应点的轨迹方程仍是则；③椭圆上每一点实施变换后，对应点的轨迹仍是离心率不变的椭圆；④曲线：上每一点实施变换后，对应点的轨迹是曲线，是曲线上的任意一点，是曲线上的任意一点，则的最小值为。

以上正确命题的序号是（写出全部正确命题的序号）.二、解答题：1．设平面直角坐标系xOy 中，曲线G x ∈R ）．（1）若a ≠0，曲线G 的图象与两坐标轴有三个交点，求经过这三个交点的圆C 的一般方程；（2）在（1）的条件下，求圆心C 所在曲线的轨迹方程；（3）若a=0，已知点M （0，3），在y 轴上存在定点N （异于点M ）满足：对于圆C 上任一点P 为一常数，试求所有满足条件的点N 的坐标及该常数．),(y x A f ),(1x y A )0(222≠=+r r y x f 222r y x =+b kx y +=f ,b kx y +=1-=k )0(12222>>=+b a by a x f C )0(ln >-=x x x y f 1C M C N 1C MN )2ln 1(2+2.已知过原点的动直线l与圆C1：x2+y2﹣6x+5=0相交于不同的两点A，B．（1）求圆C1的圆心坐标；（2）求线段AB 的中点M的轨迹C的方程；（3）是否存在实数k，使得直线L：y=k（x﹣4）与曲线C只有一个交点？若存在，求出k的取值范围；若不存在，说明理由．3.数列{a n}满足：a1+2a2+…na n=4﹣，n∈N+．（1）求a3的值；（2）求数列{a n}的前n项和T n；（3）令b1=a1，b n=+（1+++…+）a n（n≥2），证明：数列{b n}的前n项和S n满足S n＜2+2lnn．参考答案1．（1）x2+y2+ax+（a2﹣2）y﹣2a2=0（2）y=1﹣2x2（x≠0）（3）存在定点N（0【解析】试题分析：（1）利用待定系数法，求经过这三个交点的圆C的一般方程；（2）由（1）可知C（x，y）a得到圆心C所在曲线的轨迹方程；（3）利用勾股定理，计算，即可得出结论试题解析：（1）令x=0，得曲线与y轴的交点是（0，﹣a2），令y=0x=﹣2a或x=a，∴曲线与x轴的交点是（﹣2a，0），（a，0）．设圆的一般方程为x2+y2+Dx+Ey+F=0，则42220 420 a Ea Fa Da Fa Da F⎧-+=⎪++=⎨⎪-+=⎩，解得D=a，E=a2﹣2，F=﹣2a2，∴圆的一般方程为x2+y2+ax+（a2﹣2）y﹣2a2=0；（2）由（1）可得C设C（x，y），则x=a，得到y=1﹣2x2，∵a≠0，∴x≠0，∴圆心C所在曲线的轨迹方程为y=1﹣2x2（x≠0）；（3）若a=0，圆C的方程为x2+（y﹣1）2=1，令x=0，得到圆C与y轴交于点（0，0），（0，2）由题意设y轴上的点N（0，t）（t≠3），当P点为（0，2当P点为（0，0t=3舍去）下面证明点N （0，对于圆C 上任一点P设P （x ，y ），则x 2+（y ﹣1）2=1，∴在y 轴上存在定点N （0，满足：对于圆C 上任一点P考点：圆的方程，轨迹方程及直线与圆的位置关系 2．①③④ 【解析】试题分析：对于①，我们知道两个圆相切等价于两个圆的圆心距刚好等于两个圆的半径之和，有题意，有：圆1C 的半径为：1，圆心为：()2cos ,2sin θθ；圆2C 的半径为：1，圆心为：()0,0，又因为，两圆的半径之和为：112+=圆心距，所以对于任意θ，圆1C 和圆2C 始终相切;对于②，从①有，两圆相切，所以两圆只有三条公切线，所以②错误;对于③，我们有圆1C ，故有圆1C 的圆心为，设其被l 所截弦为CD ，过圆心1C 做1C P 垂直于CD ，则由圆的性质，有P 是弦CD 的中点，所以圆心到直线l 的距离为又因为圆1C 的半径为1，所以有其所截弦CD 的长为;对于④，由①有，两圆相切，所以两圆上的点的最大距离就是两圆的直径之和，因为1C 的直径为2，2C 的直径也为2，也就是说224+=.考点：直线与圆及圆与圆的位置关系.方法点睛：本题通过命题的形式考查了直线与圆及圆与圆的位置关系，属于基础题.但受限于题型和运算量大，考生往往得分率不高.圆与圆的位置关系离不开圆心距与半径的和、差的关系,本题中利用两点间的距离公式和三角函数知识即可得到圆心距为定值2，恰好等于半径的和，得到两个圆为外切关系，公切线有3条；关于圆的弦长通常求出弦心距利用勾股定理即可求得弦长；两动点间的距离根据图形转化为两定点间的距离来解决就容易多了. 3．①③④ 【解析】试题分析：由题意点),(y x A 实施变换f 后，对应点为),(x y A '，对应曲线来说，就是求曲线关于直线x y =的对应曲线，对于①，因为圆)0(222≠=+r r y x 的圆心在直线x y =上，所以圆)0(222≠=+r r y x 上任意一点实施变换f 后，对应点的轨迹仍是圆)0(222≠=+r r y x ，所以①正确；对于②，直线b kx y +=关于直线x y =对称的曲线方程为而直线b kx y +=上每一点实施变换f 后，解得⎩⎨⎧==01b k ，所以②不正确；上的每一点实施f 后，对应的轨迹方程为离心率不变，故③正确；对于④，令)0(ln 2)(ln )(>-=--=x x x x x x x g ，易求得时，)(x g 为减函数，时，)(x g 为增函数，所以2ln1)(min +=x g ，由对称性可知，曲线x x y -=ln 上的点与其关于直线x y =的对称曲线上的点的最小值为故答案为①③④.考点：命题的真假判断与应用.。

2016年高三实验班物理培优模拟考试题(三)

（2）求从地球表面发射同步轨道卫星时的速度至少为多少。
8．（15分）如图所示，长为2L的板面光滑且不导电的平板小车C放在光滑水平面上，车的右端有块挡板，车的质量，绝缘小物块B的质量。若B以一定速度沿平板向右与C车的挡板相碰，磁后小车的速度总等于碰前物块B速度的一半。今在静止的平板车的左端放一个带电量、质量为的小物块A，将物块B放在平板车的中央，在整个空间加上一个水平方向的匀强电场时，金属块A由静止开始向右运动，当A以速度与B发生碰撞，碰后A以的速率反弹回来，B向右运动，求
题号
1
2
3
4
5
答案
命题陈新生
2016年1月29日
一、选择题（5*4=20分）
二、计算题（12+15+15+18+20=80分）
6.（12分）
7.（15分）
8.（15分）
9.（18分）
10.（20分）
(1)若再在MN左侧空间加一个宽度也为L的匀强磁场，使得荧光屏上的亮点恰好位于原点O处，求这个磁场的磁感应强度B的大小和方向；
（2）如果磁场的磁感应强度B的大小保持不变，但把方向变为与电场方向相同，则荧光屏上的亮点位于图乙中的A点，已知A点的纵坐标y= L，求A点横坐标的数值(最后结果用L和其他常数表示)。
用代入解得 B碰后做匀速运动，碰到挡板的时间
A的加速度 A在段时间的位移为
因，故A第二次与B相碰必在B与C相碰之后
（3）B与C相碰，由动量守恒定律可得
A从第一次相碰到第二次与B相碰的位移为L，因此电场力做的功
9.(18分)解析：（1）粒子若直线前进，应加一竖直向上的匀强磁场
由
（2）如果加一个垂直纸面向里、大小为的匀强磁场，粒子在垂直于磁场的平面内的分运动是匀速圆周运动（见图），在荧光屏上

高三数学实验班培优训练卷一试题

姜堰中学高三数学实验班培优训练卷一制卷人：打自企；成别使；而都那。

审核人：众闪壹；春壹阑；各厅…… 日期：2022年二月八日。

1．某人的密码箱上的密码是一种五位数字号码，每位上的数字可在0到9这10个数字中选取，该人记得箱子的密码1，3，5位均为0，而忘记了2，4位上的数字，只要随意按下2，4位上的数字，那么他按对2，4位上的数字的概率是〔〕 A ．52 B ．51 C ．101 D ．1001 2．设命题P ：函数f (x )=ax x+(a >0)在区间(1, 2)上单调递增；命题Q ：不等式|x －1|－|x +2|<4a 对任意x ∈R 都成立。

假设“P 或者Q 〞是真命题，“P 且Q 〞是假命题，那么实数a 的取值范围是〔〕 A ．43<a ≤1 B 。

43≤a <1 C ．0<a ≤43或者a >1 D 。

0<a <43或者a ≥13．由等式223144322314)1()1()1(+++++=++++x b x b x a x a x a x a x413)1(b x b +++定义),,,(),,,(43214321b b b b a a a a f =，那么),1,2,3,4(f 等于〔〕〔A 〕)4,3,2,1( 〔B 〕)0,4,3,0( 〔C 〕)2,2,0,1(-- 〔D 〕)1,4,3,0(--4．点F 1、F 2分别是双曲线22a x －22b y=1的左、右焦点，过F 1且垂直于x 轴的直线与双曲线交于A 、B 两点，假设△ABF 2为锐角三角形，那么该双曲线的离心率e 的取值范围是．5．有两个向量1(1,0)e =，2(0,1)e =，今有动点P ，从0(1,2)P -开场沿着与向量12e e +一样的方向作匀速直线运动，速度为12||e e +；另一动点Q ，从0(2,1)Q --开场沿着与向量1232e e +一样的方向作匀速直线运动，速度为12|32|e e +．设P 、Q 在时刻0t =秒时分别在0P 、0Q 处，那么当00PQ P Q ⊥时，t = 秒．6．：OF =〔c ，0〕〔c>0〕,))(,(R n n n OG ∈=,)0>>=c a ②OF PE λ=其中),0)(,(2R t t ca OE ∈≠=λ③动点P 的轨迹C 过点B(0,-1).(1) 求c 的值; (2) 求曲线C 的方程;(3) 过点M(0,2)的直线l 与曲线C 的轨迹交于A,B 两点,求MB MA ⋅的取值范围.7．对数列{}n a ，规定{}n a ∆为数列{}n a 的一阶差分数列，其中)(*1N n a a a n n n ∈-=∆+。

江苏省赣榆县清华园学校2019届高三培优班考前测验试题(一)(数学)

江苏省赣榆县清华园学校2019届高三培优班考前测验试题（一）数学试题一、填空题：本大题共14小题，每小题5分，共70分．请把答案填写在答题卡相应的位置上．．．．．．．．．． 1．已知集合M ={－1，1}，{|124}x N x =≤≤，则MN =▲ ．2．已知射手甲射击一次，命中9环以上（含9环）的概率为0.5，命中8环的概率为0.2，命中7环的概率为0.1，则甲射击一次，命中6环以下（含6环）的概率为 ▲ ．3．设(12i)34i z +=-（i 为虚数单位），则||z = ▲ ． 4．根据右图的算法，输出的结果是 ▲ ．5．某校对全校1200名男女学生进行健康调查，选用分层抽样法抽取一个容量为200的样本．已知女生抽了85人，则该校的男生数应是 ▲ 人． 6．若“2230x x -->”是 “x a <”的必要不充分条件，则a 的最大值为 ▲ ． 7．设a ，b 为空间的两条直线，α，β为空间的两个平面，给出下列命题：（1）若a ∥α，a ∥β，则α∥β；（2）若a ⊥α，a ⊥β，则α∥β；（3）若a ∥α，b ∥α，则a ∥b ；（4）若a ⊥α，b ⊥α，则a ∥b ．For from 1 to 10 End for Print EndS I S S I S ←←+（第4题）上述命题中，所有真命题的序号是 ▲ ．8．已知S 是△ABC 所在平面外一点，D 是SC 的中点，若BD ＝xAB yAC zAS ++，则x ＋y ＋z ＝ 9．函数()()sin f x x x x ωω=∈R ，又()2f α=-，()0f β=，且αβ-的最小值等于π2，则正数ω的值为 ▲ ．10．若圆C ：22()(1)1x h y -+-=在不等式10x y ++≥所表示的平面区域内，则h 的最小值为 ▲ ．11．在平面直角坐标系xOy 中，已知A （0，－1），B （－3，－4）两点，若点C 在AOB ∠的平分线上，且10OC=C 的坐标是 ▲ ．12．已知函数3221()(21)13f x x x a x a a =++-+-+，若()0f x '=在（1，3]上有解，则实数a 的取值范围为 ▲ ．13．已知21(),()()2x f x x g x m ==-，若对[]11,3x ∀∈-，[]20,2x ∃∈，12()()f x g x ≥，则实数m 的取值范围是 ▲ ．14．已知等腰三角形腰上的中线长为▲ ．二、解答题：15.（本小题满分14分）在ABC △中，已知角,,A B C 的对边分别为,,a b c 且()()3a b c b c a bc +++-=. ⑴求A ;ABC C 1B 1A 1 FDE（第16题） O M⑵若90,4B C c -=︒=，求b .（结果用根式表示）16.如图，直三棱柱111ABC A B C -中，D 、E 分别是棱BC 、AB 的中点，点F 在棱1CC 上，已知AB AC =，13AA =，2BC CF ==．（1）求证：1//C E 平面ADF ；（2）设点M 在棱1BB 上，当BM 为何值时，平面CAM ⊥平面ADF ？17. （本小题满分14分）有一隧道既是交通拥挤地段，又是事故多发地段.为了保证安全，交通部门规定，隧道内的车距()d m 正比于车速()/v km h 的平方与车身长()l m 的积，且车距不得小于一个车身长l （假设所有车身长均为l ）.而当车速为()60/km h 时，车距为1.44个车身长. ⑴求通过隧道的最低车速；⑵在交通繁忙时，应规定怎样的车速，可以使隧道在单位时段内通过的汽车数量Q 最多？18..如图，在平面直角坐标系xOy 中，椭圆C ：22221x y a b +=（0a b >>）的左焦点为F ，右顶点为A ，动点M 为右准线上一点（异于右准线与x 轴的交点），设线段FM 交椭圆C 于点P ，已知椭圆C 的离心率为23，点M 的横坐标为92．（1）求椭圆C 的标准方程；（2）设直线P A 的斜率为1k ，直线MA 的斜率为2k ，求12k k ⋅的取值范围． 19．（本小题16分）已知数列{a n }的通项公式为a n = 2⨯3n + 23n – 1(n ∈N *)．⑴求数列{a n }的最大项；⑵设b n = a n + pa n– 2，求实常数p ，使得{b n }为等比数列；⑶设*,,,N m n p m n p ∈<<，问：数列{a n }中是否存在三项m a ，n a ，p a ，使数列m a ，n a ，p a 是等差数列？如果存在，求出这三项；如果不存在，说明理由． 20．（本小题16分）已知二次函数g （x ）对任意实数x 都满足()()21121g x g x x x -+-=--，且()11g =-．令()219()23ln (0,0)24f x gx mx m x m x =++-+>>．（1）求 g (x )的表达式；（2）若函数()f x 在[1,)x ∈+∞上的最小值为0，求m 的值；（3）记函数22()[()1][(1)1]H x x x a x a x a =--⋅-+-+-，若函数()y H x =有5个不同的零点，求实数a 的取值范围．数学详解详析 1.【答案】{1} 2．【答案】0.2【解析】p=1-(0.5+0.2+0.1)=0.23．设(12i)34i z +=-（i 为虚数单位），则||z = ▲ ．【答案】z=【解析】()()()()23412343108121212125i i i i iz i z i i i ----+====--++-4．【答案】55【解析】由For 语句可得如下具体10次循环I= 1 2 3 4 5 6 7 8 9 10S=1 3 6 10 15 21 28 36 45 55 所以S=55. 5．某校对全校1200名男女学生进行健康调查，选用分层抽样法抽取一个容量为200的样本．已知女生抽了85人，则该校的男生数应是 ▲ 人．【答案】690【解析】设女生和男生总人数分别为n ，m200851156901200m n m===.6．若“2230x x -->”是 “x a <”的必要不充分条件，则a 的最大值为 ▲ ．【答案】1-【解析】()()31031x x x x -+>><-或由题可知集合{}31A x x =><-或真包含集合{}B x a =<由上图可知：1a ≤-，则a 的最大值为-1.For from 1 to 10 End for Print EndS I S S I S ←←+（第4题）7．【答案】（2），（4）【解析】⑴错误.因为α与β可能相交；⑶错误.因为直线a 与b 可能异面、相交和平行. 8．解：因为BD ＝11()()221122++=++-=-++xAB y AC z AS BA AS AC AB AB AC AS故x+y+z=0. 9．【答案】1【解析】()22sin 3f x x T ππϖϖ⎛⎫=+=⎪⎝⎭122124T T πππϖϖ=∴==∴= 10．2【解析】当圆C 与直线10x y ++=相切时，h 的取值最小122h h ==-=-.11．在平面直角坐标系xOy 中，已知A （0，－1），B （－3，－4）两点，若点C 在AOB ∠的平分线上，且10OC =C 的坐标是 ▲ ．【答案】()1,3C --【解析】设直线BC 的倾斜角为α，则∠AOB 的角平分线所在直线的倾斜角为22πα-，21tan312tan cot tan3tan242232tan 2απααααα-⎛⎫-====-=⎪⎝⎭或,角平分线方程为()30y x x =≤于是点()1,3C --. 12．【答案】71a -<≤ 函数21122y xx =--+的对称轴为1x =-，于是函数21122y x x =--+在区间(]1,3单调减，[)7,1a ∈-- .13．已知21(),()()2x f x x g x m ==-，若对[]11,3x ∀∈-，[]20,2x ∃∈，12()()f x g x ≥，则实数m 的取值范围是 ▲ ．【答案】14m ≥【解析】()()[][]1212min min 21,3,0,211024f x g x x x m m ≥∈-∈⎡⎤⎡⎤⎣⎦⎣⎦⎛⎫≥-∴≥⎪⎝⎭14．【答案】2【解析】设等腰三角形的顶角为A,腰长为2a ，则22224353cos 224a a a A a a a +--==⨯⨯2422421930922sin 4sin 24a a S a a A a A a -+-⎛⎫=⨯⨯⨯⨯==< ⎪⎝⎭⎝令2133u a u ⎛⎫=<< ⎪⎝⎭，则2293094u u S -+-=，当53u =时max 2S =.15.【解析】（1）由条件，得()22b c a bc +-=. 所以2221cos .22b c a A bc +-== 因为A 是三角形内角，所以60.A ︒= （2）由12090B C B C ︒︒⎧+=⎨-=⎩得105,15.B C ︒︒==由正弦定理得44,sin1054tan 75.sin105sin15sin15b b ︒︒︒︒︒==⨯= 因为()1tan 30tan 75tan 453021tan 30︒︒︒︒︒+=+==-所以8b =+16.解：（1）连接CE 交AD 于O ，连接OF ．因为CE ，AD 为△ABC 中线，所以O 为△ABC 的重心，123CF CO CC CE ==．从而OF//C 1E ．………………………………………………………………………………3分OF ⊂面ADF ，1C E ⊄平面ADF ，所以1//C E 平面A．……………………………………………………………………6分（2）当BM=1时，平面CAM ⊥平面ADF ．在直三棱柱111ABC A B C -中，由于1B B ⊥平面ABC ，BB 1⊂平面B 1BCC 1，所以平面B 1BCC 1⊥平面ABC ．由于AB =AC ，D 是BC 中点，所以AD BC ⊥．又平面B 1BCC 1∩平面ABC =BC ，所以AD ⊥平面B 1BCC 1．而CM ⊂平面B 1BCC 1，于是AD ⊥CM ．…………………………………………………9分因为BM =CD =1，BC = CF =2，所以Rt CBM ∆≌Rt FCD ∆，所以CM ⊥DF ． ………11分DF 与AD 相交，所以CM ⊥平面ADF ． CM ⊂平面CAM ，所以平面CAM ⊥平面ADF ． (13)分当BM=1时，平面CAM ⊥平面ADF ． (14)分17.【解析】（1）依题意，设2d kv l =，其中k 是待定系数，因为当60v =时， 1.44d l = 所以21.4460l k l =⨯，0.0004k =，所以20.0004.d v l =因为d l ≥，所以20.0004v l l ≥，50.v ≥ 所以最低车速为50/.km h(2)因为两车间距为d ，则两辆车头间的距离为.l d + 一小时内通过汽车的数量为21000100010.00040.0004v Q l v l l v v ==+⎛⎫+ ⎪⎝⎭，因为10.00040.04,v v+≥=所以25000.Q l ≤ 所以当10.0004v v=即50/v km h =时，单位时段内通过的汽车数量最多.分18．（本题满分15分）（1）由已知，(4,0),(4,0),(2,0)A B F -，直线8l x =的方程为．设N （8，t ）（t >0），因为AM =MN ，所以M （2，2t ）．由M 在椭圆上，得t =6．故所求的点M 的坐标为M （2，3）．………………………4分所以(6,3),(2,3)MA MB =--=-，1293MA MB ⋅=-+=-．cos ||||36MA MB AMB MA MB ⋅∠===．……………………………………7分（2）设圆的方程为220x y Dx Ey F ++++=，将A ，F ，N 三点坐标代入，得22,1640,72420,,6480,8.D D F D FE t t t D EtF F =⎧⎧-+=⎪⎪⎪++=⇒=--⎨⎨⎪⎪++++=⎩⎪=-⎩ ∵ 圆方程为22722()80x y x t y t++-+-=，令0x =，得272()80y t y t-+-=．…11分设12(0,),(0,)P y Q y ，则12y =、．由线段PQ 的中点坐标为（0，9），得1218y y +=，7218t t+=．此时所求圆的方程为2221880x y x y ++--=． (15)分（2）（法二）由圆过点A 、F 得圆心横坐标为－1，由圆与y 轴交点的纵坐标为（0，9），得圆心的纵坐标为9，故圆心坐标为（－1，9）．…………………………………… 11分易求得圆的半径为13分所以，所求圆的方程为221)(9)90x y ++-=（． (15)分18.解：（1）由已知，得22,39,2ca a c ⎧=⎪⎪⎨⎪=⎪⎩ …………………2分解得3,2.a c =⎧⎨=⎩ ∴ 229,5.a b ⎧=⎪⎨=⎪⎩ ………………………………4分 ∴椭圆C 的标准方程为22195xy +=．………………………………6分（2）设点11(,)P x y （123x -<<），点M 29(,)2y ，∵点F 、P 、M 三点共线，12x ≠-， ∴1211322y yx =+，121132(2)y y x =+， ∴点M 11139(,)22(2)y x +． (8)分∵1113y k x =-，121133(2)y k x =+， ∴12k k ⋅=11111333(2)y y x x ⨯-+=2111133(2)(3)y x x +-． ………10分∵点P 在椭圆C 上， ∴2211195x y +=， ∴22115(9)9y x =--． ∴12k k ⋅=2111513()(9)93(2)(3)x x x ⨯--+-=11365272x x +-⨯+=1651(1)272x -⨯++．……………12分∵123x -<<， ∴12269k k ⋅<-．∴12k k ⋅的取值范围是26(,)9-∞-． ……………………………………14分19.解：⑴由题意a n = 2 + 43n – 1,随着n 的增大而减小,所以{a n }中的最大项为a 1 = 4．⑵b n = 2 + 43n – 1 + p 4n = (2 + p )(3n – 1) + 44 = (2 + p )3n + (2 – p )4，若{b n }为等比数列，则b 2n +1 – b n b n +2= 0(n ∈N * )，所以 [(2 + p )3n +1 + ( 2 – p )]2 – [{2 + p )3n + (2 – p )][(2 + p )3n +2 + (2 – p )] = 0(n ∈N *)，化简得(4 – p 2)(2·3n +1 – 3n +2 – 3n ) = 0即– (4 – p 2)·3n ·4 = 0,解得p = ±2．反之,当p = 2时,b n = 3n ,{b n }是等比数列;当p = – 2时,b n = 1,{b n }也是等比数列．所以,当且仅当p = ±2时{b n }为等比数列． ⑶因为4231m m a =+-，4231n n a =+-，4231p pa =+-，若存在三项m a ，n a ，p a ，使数列m a ，n a ，p a 是等差数列，则2n m p a a a =+，所以42(2)31n +-=4231m +-4231p ++-，化简得3(2331)1323n p n p m p m n m ----⨯--=+-⨯（*），因为*,,,N m n p m n p ∈<<，所以1p m p n -≥-+，1p m n m -≥-+，所以13333p m p n p n --+-≥=⨯，13333p m n m n m --+-≥=⨯，（*）的左边3(23331)3(31)0n p n p n n p n ---≤⨯-⨯-=--<，右边13323130n m n m n m ---≥+⨯-⨯=+>，所以（*）式不可能成立，故数列{a n }中不存在三项m a ，n a ，p a ，使数列m a ，n a ，p a 是等差数列．20解：（1）设()2g x ax bx c =++，于是()()()()2211212212g x g x a x c x -+-=-+=--，所以121.a c ⎧=⎪⎨⎪=-⎩，又()11g =-，则12b =-．所以()211122g x x x =--．（2）()22219()23ln 3ln f x gx mx m x x mx m x =++-+=+-则222323(23)()()2m x mx m x m x m f x x m x x x +-+-'=+-==．令()0f x '=，得32mx =-（舍），x m =．①当m >1时，∴当x m =时, 2223ln ()min m x m f m -=．令2223ln 0mm m -=，得23m =e．②当01m <≤时，()f x '≥0在[1,)x ∈+∞上恒成立，()f x 在[1,)x ∈+∞上为增函数，当1x =时, min ()1f x m =+．令10m +=，得1m =-（舍）．综上所述，所求m 为23e m =．（3）记21()()h x x x a =-，22()(1)h x x a x a =-+-+，则据题意有1()10h x -=有3个不同的实根， 2()10h x -=有2个不同的实根，且这5个实根两两不相等．（ⅰ）2()10h x -=有2个不同的实根，只需满足1()1132a g a a ->⇒><-或；（ⅱ）1()10h x -=有3个不同的实根，因221()34(3)()h x x ax a x a x a '=-+=--，令1()0h x '=，得x a =或3a ，1当3aa >即0a <时，1()h x 在x a =处取得极大值，而1()0h a =，不符合题意，舍； 2当3aa =即0a =时，不符合题意，舍；3当3a a <即0a >时，1()h x 在3ax =处取得极大值，1()13a h a >⇒>所以a >因为（ⅰ）（ⅱ）要同时满足，故a > 下证：这5个实根两两不相等，即证：不存在0x 使得10()10h x -=和20()10h x -=同时成立；若存在0x 使得1020()()1h x h x ==，由1020()()h x h x =，即220000(1)x x a x a x a -=-+-+（），得20000(1)0x a x ax x --++=（），当0x a =时，00()()0f x g x ==，不符合，舍去；当0x a ≠时，既有200010x ax x -++= ①；又由0()1g x =，即200(1)1x a x a -+-+= ②；联立①②式，可得0a =；而当0a =时，32()(1)(1)0H x x x x =----=没有5个不同的零点，故舍去，所以这5个实根两两不相等．综上，当a >()y H x =有5个不同的零点。

广东省普宁二中高三数学实验班精选综合训练(共两套)

广东省普宁二中高三实验班数学精选综合训练（一）1. 在ABC △中，已知角A 为锐角，22sin 3A =，（Ⅰ）求22tansin 22B C A++的值；（Ⅱ）若22=a ，2ABC S =△b 的值2.一个盒子里装有标号为1，2，3，L ，n 的n (3,n ≥且*n N ∈)张标签，今随机地从盒子里无放回地抽取两张标签，记ξ为这两张标签上的数字之和，若ξ=3的概率为110。

（1）求n 的值；（2）求ξ的分布列；（3）求ξ的期望。

3．如图所示，ABC —A 1B 1C 1是各条棱长均为a 的正三棱柱，D 是侧棱CC 1的中点. （1）求证：平面AB 1D ⊥平面ABB 1A 1；（2）求点C 到平面AB 1D 的距离；（3）求平面AB 1D 与平面ABC 所成二面角（锐角）的大小.4．已知椭圆的中心在原点,离心率为12,一个焦点是(,0)F m -(m 为大于0的常数). （1）求椭圆的方程;（2）设Q 是椭圆上一点,且过点,F Q 的直线l 与y 轴交于点M ,若||2||MQ QF =u u u u r u u u r ,求直线l 的斜率.5设函数32()(,,,,0)f x a x b x cx d a b c d R a =+++∈>, 其中(0)3f =,'()f x 是()f x 的导函数.(Ⅰ)若'(1)'(3)36,'(5)0f f f -==-=,求函数()f x 的解析式;(Ⅱ)若6c =-,函数()f x 的两个极值点为12,x x 满足12112x x -<<<<. 设226210a b a b λ+=+-+, 试求实数λ的取值范围.6．设数列{}n a 是首项为6，公差为1的等差数列；n S 为数列{}n b 的前n 项和，且22n S n n =+（1）求{}n a 及{}n b 的通项公式n a 和n b ；（2）若对任意的正整数n ，不等式1201112(1)(1)(1)nna n ab b b -≤-++++…恒成立，求正数a 的取值范围。

2020届高三物理精准培优专练1：直线运动的图象(附解析)

2020届高三物理精准培优专练1：直线运动的图象（附解析）一、考点分析1．物理图象能形象地表达物理规律、直观地描述物理过程、鲜明地表示物理量之间的相互关系，是分析物理问题的有效手段之一，也是当今高考命题的热点。

近几年高考，主要是在选择题中考查两种运动学图象的理解和应用。

2．图象问题的解题技巧：(1)图象反映了两个物理量之间的函数关系，因此首先要由运动学公式推导出两个物理量间的关系式，再分析图象及斜率、截距、面积等几何元素的物理意义。

(2)注意把处理常规图象问题的思想方法加以迁移应用，必要时可将该图象所反映的物理过程转换为常见的x－t或v－t图象。

二、考题再现典例1.( 2019浙江4月选考∙9)甲、乙两物体零时刻开始从同一地点向同一方向做直线运动，位移－时间图象如图所示，则在0～t4时间内( )A．甲的速度总比乙大 B．甲、乙位移相同C．甲经过的路程比乙小 D．甲、乙均做加速运动典例2.( 2018全国II卷∙19)甲、乙两汽车在同一条平直公路上同向运动，其速度－时间图象分别如图中甲、乙两条曲线所示。

已知两车在t2时刻并排行驶。

下列说法正确的是( ) A．两车在t1时刻也并排行驶B．在t1时刻甲车在后，乙车在前C．甲车的加速度大小先增大后减小D．乙车的加速度大小先减小后增大三、对点速练1．如图所示为某质点做直线运动的ν－t图象。

已知t0时刻的速度为v0，2t0时刻的速度为2v0，图中OA与AB是关于A点中心对称的曲线，由图可得( )A ．0～t 0时间内的位移为12v 0t 0B ．0～2t 0时间内的位移为2v 0t 0C ．t 0时刻的加速度为00v t D ．2t0时刻的加速度为00v t2．(多选)如图所示为一个质点运动的位移x 随时间t 变化的图象，由此可知质点在0～4 s 内( )A ．先沿x 轴正方向运动，后沿x 轴负方向运动B ．一直做匀变速运动C ．t ＝2 s 时速度一定最大D ．速率为5 m/s 的时刻有两个3．宇航员的训练、竞技体育的指导、汽车的设计等多种工作都会用到急动度的概念。

江苏省姜堰中学高三数学实验班培优训练卷二苏教版

江苏省姜堰中学高三数学实验班培优训练卷二1．对于R 上可导的任意函数f (x )，若满足()()10x f x '-≥则必有（）（A ）()()()02＜21f f f + （B ）()()()0221f f f +≤ （C ）()()()0221f f f +≥（D ）()()()02＞21f f f +2．已知函数y=31x 3+x 2+x 的图像C 上存在一定点P 满足：若过点p 的直线l 与曲线C 交于不同于P 的两点M(x 1，y 1)，N(x 2，y 2)，就恒有y 1+y 2为定值y 0，则y 0的值为 ( ) A.-31 B.-32 C.-34D.-2 3．函数)(x f 在R 上可导，且)1(2)(2f x x x f '⋅+=，则)1()1(f f 与-的大小关系是（） A ．)1()1(f f =- B ．)1()1(f f <-C ．)1()1(f f >-D ．不能确定4．已知实数x 、y 满足⎪⎩⎪⎨⎧≤--≥-+≥+-0520402y x y x y x ，则|42|-+=y x z 的最大值为（）A 、21B 、20C 、19D 、185．已知P 在直线1+=x y 上的一点，M 、N 分别为圆C 1：4)1()4(22=-+-y x 与 C 2：1)2(22=-+y x 上的点，则||||PN PM -的最大值为（） A 、4 B 、5 C 、6 D 、76．定义在R 上的增函数f(x)满足f[f(x)]=x,则｜f(x)|-|f(x)-1|的范围是 ( ) A. (],1-∞- B.[-1,1] C. (]1,2 D. [)2,+∞7．设)(x f 定义域为D ，若满足：（1）)(x f 在D 内是单调函数；（2）存在D b a ⊆],[，使)(x f 在],[b a x ∈时值域也为],[b a ，则称)(x f 为D 上的闭函数,当42)(++=x k x f 时，k 的取值范围是。

河南省示范性高中罗山高中高三物理培优考试题(实验班)

河南省示范性高中罗山高中2015届高三实验班培优考试题物理一、选择题：本大题共8小题，每小题6分，15、16、21题为多选题，全部选对的得6分，选对但不全的得3分，有选错的得0分；其余为单选题。

14．如图所示，两个质量为m1的小球套在竖直放置的光滑支架上，支架的夹角为120°，用轻绳将两球与质量为m2的小球连接，绳与杆构成一个菱形，则m1∶m2为（） A ．1∶1 B ．1∶2 C ． D ．15．如图所示，长方体物块C 置于水平地面上，物块A 、B 用不可伸长的轻质细绳通过滑轮连接(不计滑轮与绳之间的摩擦)，A 物块与C 物块光滑接触，整个系统中的A 、B 、C 三物块在水平恒定推力F 作用下从静止开始以相同的加速度一起向左运动,下列说法正确的是 ( )A ．B 与C 之间的接触面可能是光滑的B ．若推力F 增大，则绳子对B 的拉力必定增大C ．若推力F 增大，则定滑轮所受压力必定增大D ．若推力F 增大，则C 物块对A 物块的弹力必定增大16．如图所示，a 、b 、c 、d 是某匀强电场中的四个点，它们正好是一个矩形的四个顶点，ab = cd = L ，ad = bc = 2L ，电场线与矩形所在平面平行．已知a 点电势为20V ，b 点电势为24V ，d 点电势为12V ．一个质子从b 点以v0的速度射入此电场，入射方向与bc 成450角，一段时间后经过c 点．不计质子的重力．下列判断正确的是 ( ) A ．c 点电势低于a 点电势B ．场强的方向由b 指向dC ．质子从b 运动到c 所用的时间为02v LD ．质子从b 运动到c ，电场力做功为4eV17. 如图所示的甲、乙、丙图中，MN 、PQ 是固定在同一水平面内足够长的平行金属导轨。

导体棒ab 垂直放在导轨上，导轨都处于垂直水平面向下的匀强磁场中。

导体棒和导轨间接触良好且摩擦不计，导体棒、导轨和直流电源的电阻均可忽略，甲图中的电容器C 原来不带电。

清泉州阳光实验学校高三数学培优班资料(文七)

清泉州阳光实验学校一中高三培优班资料〔文数七〕一、选择题〔本大题一一一共8小题，每一小题5分，一一共40分〕 1．设集合}2|||{},0|{2<=<-=x x P x x x M，那么A ．φ=⋂P MB ．M P M=⋂C ．M P M =⋃D ．R P M =⋃2．假设向量,2,2,()满足a b a b a b a ==-⊥，那么向量与a b 的夹角等于A ．4πB ．6π C ．43π D ．65π 3．数列}{n a 是等比数列，且4,34231=-=+a a a a ，那么公比q 的值是A ．2B ．-2C ．2±D ．2±4．10<<<<a y x ，那么 A ．1)(log 0<<xy aB ．2)(log 1<<xy aC ．2)(log >xy aD ．0)(log <xy a5．如图是函数)2,0)(sin(2πϕωϕω<>+=x y与的图象，那么A ．6,2πϕω-==B ．6,2πϕω==C ．6,1110πϕω==D ．6,1110πϕω-==6．以下函数中既是奇函数，又在区间),0(+∞上单调递增的是A ．x y 2sin =B ．xey -=C ．1ln+=x xy D ．⎩⎨⎧≥+<--=0)2(30)2(3x x x x x x y 7．假设),2(,ππβα∈，且βαcot tan <，那么必有A ．πβα23>+B ．πβα23<+ C ．βα>D ．βα<8．在区间),0(+∞上函数)(x f 是减函数，且当b a x f x <<>>0,0)(,0若时，那么A ．)()(b af a bf <B ．)()(a bf b af <C ．)()(b f a af <D ．)()(a f b bf <第二卷〔非选择题，一一共80分〕二、填空题：〔本大题一一一共6小题，每一小题5分，一一共30分。

高考物理实验班培优训练13教研

高二物理实验班培优训练一1、如图所示，一端可绕O 点自由转动的长木板上方向一个物块，手持木板的另一端，使木板从水平位置沿顺时针方向缓慢旋转，则在物体相对于木板滑动前（） A.物块对木板的压力变小 B.物块的机械能变小C.物块对木板的作用力不变D.物块受到的静摩擦力增大2、半圆柱体P 放在粗糙的水平地面上，其右端有一固定放置的竖直挡板MN ．在半圆柱体P 和MN 之间放有一个光滑均匀的小圆柱体Q ，整个装置处于平衡状态，如图所示是这个装置的截面图．现使MN 保持竖直并且缓慢地向右平移，在Q 滑落到地面之前，发现P 始终保持静止．则在此过程中，下列说法中正确的是A ．P 对Q 的弹力逐渐增大B ．MN 对Q 的弹力逐渐减小C ．Q 所受的合力逐渐增大D ．地面对P 的摩擦力逐渐增大 3、如图所示，一水平方向足够长的传送带以恒定的速度v 1沿顺时针方向运动，一物体以水平速度v 2从右端滑上传送带后，经过一段时间又返回光滑水平面，此时速率为v 2'，则下列说法正确的是： A ．若v 1< v 2，则v 2'＝v 1 B ．若v 1> v 2，则v 2'＝v 2 C ．v 1＝v 2时，有v 2'＝v 2 D ．不管v 2多大，总有v 2'＝v 24、如图所示，一斜面固定在水平地面上，质量不相等的物体，A 、B 叠放后．一起沿斜面下滑，已知物体B 的上表面水平，则下列判断正确的是（） A ．若A 、B 一起匀速下滑，增加A 的质量，A 、B 仍一起匀速下滑 B ．若A 、B 一起匀速下滑，给A 施加一个竖直向下的力F ，A 、B 将加速下滑C ．若A 、B 一起加速下滑，增加A 的质量，A 、B 仍保持原来的加速度一起加速下滑D ．若A 、B 一起加速下滑．给A 施加一个竖直向下的力F ，A 、B 仍保持原来的加速度一起加速下滑5．如图所示，一质量为m 的木块靠在粗糙竖直的墙上，且受到水平力F 的作用，下列说法正确的是（）A ．若木块静止，则受到的静摩擦力大小等于mg ，方向竖直向上B ．若木块静止，当F 增大时，木块受到的静摩擦力随之增大C ．若木块与墙壁间的动摩擦因数为μ，则撤去F 后，木块受到的滑动摩擦力大小等于mg μD ．若撤去F ，木块沿墙壁下滑时，木块不受滑动摩擦力作用6．小球在离地面高为h 处以初速度v 水平抛出，球从抛出到着地，速度变化量的大小和方向为（）A ．gh v 22+，竖直向下 B ．gh 2，竖直向下 C ．gh v 22+，斜向下 D ．v gh v -+22，斜向下v 1 v 2Fo7．如图所示，电梯质量为M ，地板上放置一质量为m 的物体，钢索拉着电梯由静止开始向上做加速运动，当上升高度为H 时，速度达到v ，则A．地板对物体的支持力做的功等于221mv B ．地板对物体的支持力做的功等于mgHC ．钢索的拉力做的功等于MgH Mv +221D ．合力对电梯M 做的功等于221Mv8．一个质量为m 的小球，在光滑曲轨道上的1位置由静止释放，经过时间t 后，沿轨道运行了d 的路程到达了2位置，如图所示。