2017-2018学年曹杨二中高一上期末考试

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上海市曹杨第二中学高一物理期末试卷带解析

上海市曹杨第二中学高一物理期末试卷带解析

上海市曹杨第二中学高一物理期末试卷含解析一、选择题:本题共5小题,每小题3分,共计15分.每小题只有一个选项符合题意1. (多选)物体从静止开始作匀加速直线运动,第3 s内通过的位移是3 m,则( )A.第3 s内的平均速度是3 m/sB.物体的加速度是1.2 m/s2C.前3 s内的位移是6 mD.3 s末的速度是3.6 m/s参考答案:ABD2. (单选)质量为m的物块与转台之间的动摩擦因数为μ,物体与转轴相距R,物块随转台由静止开始转动,直至物块与转台之间产生相对滑动这一过程中,摩擦力对物体做的功为A.0B.2πmgRC.2μmgRD. μmgR/2参考答案:D3. 如图所示,自行车的车轮半径为R,车轮沿直线无滑动地滚动,当气门芯由轮子的正上方第一次运动到轮子的正下方时,气门芯位移的大小为()A.πR B.2 R C.2πR D.参考答案:D试题分析:当气门芯由轮子的正上方第一次运动到轮子的正下方时,轮子向前运动半个周长,气门芯的初位置与末位置如图,由几何知识得,气门芯的位移大小,故选D。

考点:位移4. 举重运动,是我国多次在世界级大赛中摘金夺银的传统强项。

在举重比赛中,运动员举起杠铃时必须使杠铃平衡一段时间才能被裁判视为成功,如图所示。

下列说法中正确的是()A.向上举起的过程中,运动员对杠铃做功B.向上举起的过程中,运动员对杠铃不做功C.举起后保持平衡的一段时间内,运动员对杠铃做功D.举起后保持平衡的一段时间内,运动员对杠铃没有作用力参考答案:A5. (多选题)给滑块一初速度使它沿足够长光滑斜面向上运动,加速度大小为,当滑块速度大小减为时,所用时间可能是A、 B、 C、 D、参考答案:BC二、填空题:本题共8小题,每小题2分,共计16分6. 如图所示,小船用绳牵引,设水对船阻力不变,在小船匀速靠岸过程中,船受绳子的拉力______,船受的浮力______.船受的合力______.(每空均填“变大”“变小”或“不变”)参考答案:增大减小不变7. 如图,轰炸机沿水平方向匀速飞行,到达山坡底端正上方时释放一颗炸弹,并垂直击中山坡上的目标A.已知A点高度为h,山坡倾角为θ,由此求出轰炸机飞行的速度,轰炸机飞行的时间.参考答案:,【考点】平抛运动.【分析】炸弹做平抛运动,垂直击中山坡上的目标A时速度与斜面垂直,因为平抛运动速度与水平方向夹角的正切值是位移与水平方向夹角正切值的2倍,速度方向的夹角得知位移与水平方向夹角的正切值,再通过水平位移求出竖直位移,从而得知轰炸机的飞行高度,炸弹的飞行时间,以及炸弹的初速度.【解答】解:根据几何知识知:A点到斜面底端的水平位移为h?cotθ,即炸弹的水平位移为x=h?cotθ由于炸弹垂直击中目标A,得知速度与竖直方向的夹角为θ,与水平方向的夹角为﹣θ.设炸弹下落的高度为H.根据平抛运动速度与水平方向夹角的正切值是位移与水平方向夹角正切值的2倍,可得:解得:H=h由H=gt2,得:轰炸机飞行的速度为:故答案为:,8. 如图所示,两块木块并排固定在水平面上,一子弹以速度v水平射入,若弹在木块中做匀减速直线运动,穿过两木块时速度刚好减小到零,且穿过每块木块所用的时间相等,则两木块的厚度之比d1:d2=参考答案:3:19. 如图所示为用频闪摄影方法拍摄的研究物体做平抛运动规律的照片.图中A、B、C为三个同时由同一点出发的小球.AA′为A球在光滑水平面上以速度υ运动的轨迹.BB′为B球以速度υ被水平抛出后的运动轨迹.CC′为C球自由下落的运动轨迹.通过分析上述三条轨迹可得出结论.参考答案:平抛运动的物体水平方向做匀速直线运动,竖直方向做自由落体运动.【考点】研究平抛物体的运动.【分析】通ABC三个球运动的比对,分析B球做平抛运动两个方向的分运动的性质.【解答】解:由题,A球沿水平方向做匀速直线运动,B球做平抛运动,C球做自由落体运动,通过对比看出,三个球运动的时间相同,所以说明平抛运动的物体水平方向做匀速直线运动,竖直方向做自由落体运动.故答案为:平抛运动的物体水平方向做匀速直线运动,竖直方向做自由落体运动.10. 如图所示,图给出了电火花计时器在纸带上打出的一些计数点,相邻的两个计数点间的时间间隔为0.1s,相邻的两个计数点间的距离如图中所标。

上海市曹杨二中2018学年高一上学期期末考试数学试题

上海市曹杨二中2018学年高一上学期期末考试数学试题

上海市曹杨二中2018学年度第一学期高一年级期末考试数学试卷命题人:__________校对人:__________审核人:__________试卷共4页1张考生注意:1、答卷前,考生务必将姓名、班级、学号等在指定位置填写清楚。

2、本试卷共有20道试题,满分100分,考试时间90分钟,请考生用水笔或圆珠笔将答案直接写在试卷(或答题卷)上。

一、填空题:(每小题3分,共36分)1.已知集合{}|1=≥A x x ,{}|=≥B x x a ,若⊂A B ,则实数a 的取值范围是__________。

2.若函数()1=f x ()g x ()()+=f x g x __________。

3.函数()2=+f x x ax 为偶函数,则实数a 的值为__________。

4.函数()()21=-≤f x x x 的反函数是()1-=f x __________。

5.在直角坐标系xOy 中,终边在坐标轴上的角α的集合是__________。

6.已知函数()22,3log ,3⎧≤⎪=⎨>⎪⎩xx f x x x ,则()()3=f f __________。

7.若幂函数()()22--=∈mm f x x m Z 在()0,+∞是单调减函数,则m 的取值集合是__________。

8.若不等式1-<x m 成立的充分不必要条件是12<<x ,则实数m 的取值范围是__________。

9.已知等腰三角形的周长为常数l ,底边长为y ,腰长为x ,则函数()=y f x 的定义域为__________.10.已知角α的终边上一点()P m,且sin α=,则tan α的值为__________. 11.已知()=y f x 是定义在R 上的奇函数,且当0≥x 时,()1142=-+x xf x ,则函数()f x 的值域是__________。

12.对于函数()f x ,若存在0∈x R ,使得()00=f x x ,则称0x 是()f x 的一个不动点。

2017-2018学年上海市曹杨二中高一上学期英语期末考试卷

2017-2018学年上海市曹杨二中高一上学期英语期末考试卷

2017-2018学年上海市曹杨二中高一上学期英语期末考试卷一、单项选择1. Sorry, I got up late this morning. I have set the alarm clock before I went to bed last night.A. mightB. shouldC. mustD. would2. Jenny had just walked out of the classroom she heard her desk-mate calling her that she had left her mobile phone on the desk.A. whileB. whenC. asD. since3. with serious employment difficulties, China has to enlarge employment opportunities and spare no effort to ensure economic growth and social stability.A. FacedB. FacingC. To faceD. Having faced4. He was just about to sit down when he felt something near his feet.A. movesB. movingC. to moveD. moved5. How can you expect to be a good speaker you never dare to speak in public?A. in caseB. even ifC. unlessD. if6. doesn’t matter is whether she will join in our activity or not. We have enough members.A. ItB. WhatC. ThatD. Which7. The patient, Mr. Berkey, agreed to an operation without of its seriousness or the risks involved.A. having been informedB. informingC. having informedD. informed8. Whatever we do, we will make it, as long as we take the passion we should have to achieve our goals.A. whenB. whatC. whichD. where9. Is this the best way you thought of people getting into the dangerous areas?A. to stopB. stoppingC. stoppedD. to have stopped10. A free gift will be given to completes the questionnaire before the activity begins.A. whomeverB. whomC. whoeverD. who11. You be hungry already – you had a big cake only half an hour ago.A. wouldn’tB. can’tC. mustn’tD. needn’t12. You can’t move in right now. The library .A. has paintedB. is paintedC. is being paintedD. is painting13. Storm chasing is so dangerous that some chasers are often hurt in accidents by driving in a heavy rain.A. having causedB. causedC. to be causedD. causing14. different kinds of fires, several types of fire extinguisher have been invented.A. ExtinguishedB. To extinguishC. ExtinguishingD. Having extinguished15. I expect is for you to think independently and introduce new ideas.A. WhatB. WhetherC. ThatD. Which16. Taylor’s popularity gave her a lot of power with the movie industry, shegot very high pay for her movies.A. ifB. thoughC. soD. unless17. A new eight-kilometer road is under construction links the port area with motorway system.A. thatB. whereC. itD. as18. In societies where social roles are determined, boys usually copy the behavior oftheir fathers, and girls of their mothers.A. thatB. thoseC. oneD. the ones19. After looking through the art books here, I found Andrew’s paintings to be the closest to I thought good paintings should look.A. howB. whatC. thatD. why20. several times but Mary still has no idea of how to do it properly.A. Being shownB. Having shownC. Having been shownD. I’ve shown her21. He was not here. Since he had already had his papers corrected, he behind.A. needn’t have stayedB. mustn’t have stayedC. didn’t need to stayD. shouldn’t stay22. she seems to put on a smiling face every day, she hasn’t really got over the shadow of exam failure.A. BecauseB. AsC. WhetherD. Though23. No one knows prevented the fire from spreading.A. what was it thatB. what it was thatC. how it was thatD. why it was that24. The patient who had suffered from H1N1 insisted on from others immediately, holding the belief that the disease would infect others.A. having isolatedB. being isolatedC. to be isolatedD. isolating25. The English program normally attracts more than 1,000 students a year, up to half will be from abroad.A. in whichB. for whomC. with whichD. of whom二、语法填空The holiday season is supposed to be the most festive and fun time of the year,but all those plans and expectations of joy can turn tougher and more stressful thanthey sound. This is especially true for (1) of us who struggle with mental illness.The holidays break your routine. Sometimes you (2) (force) to spend timewith family you rarely see and don’t always get along with. Or maybe you’re alone when everyone else is with family. Or you’re at work and can’t be with those you love.Or you are off from work, (3) more time to think troubling thoughts. Or youare thrust into party situations that tempt your demons.When you have a routine, it’s (4)(easy) to manage whatever mental struggles you may face, and when that routine is broken, it (5) trigger thingsyou may not be ready to face. I know it has for me. It was during the holidays (6)_______ I hit a low moment and, with the help of my mother, decided to seek help formy eating disorder.Around the holidays, I often feel like I’m supposed to be everywhere, with everyone—all with the (7) (add) guilt of knowing it’s the season of giving.(8)_______ (fight) this, I’v e developed a mantra(咒语): it’s not selfish to take timefor yourself. Take a walk in nature. Talk to a friend you trust, or a therapist. Sit out (不参加) one of the holiday gatherings in favor of some personal time. Just do (9)_______ helps you calm down and gives you a break from the stress. Download oneof the many meditation apps for your phone. I particularly like Calm and End Anxiety. Trying to spend all of your time (10) (please) everyone else is not only exhausting –it’s impossible. And you know what? If you take a little time for yourself, you will be much better company for those around you.三、选词填空A. outperformedB. comparesC. valueD. benefitsE. accidentallyF. advantagedG. trueH. effectivelyI. influenceJ. mirroredK. involveWhen young people study or take exams, the results are usually about rewarding their individual achievement.But when they get into the workplace they will be told about the importance of social skills and the need to co-operate with other people on solving problems.So, are school systems out of step with what is needed by young people?PISA, which (1) students’ abilities in reading, Maths and science, has now carried out the world’s first global tests on collaborative problem-solving skills.As might have been expected, students who are high achievers in academic tests are also likely to be better at problem solving with other people.They are likely to have the skills in interpreting information and complex reasoning that will help them with any kind of problem solving.The same holds (2) across countries. Top-performing countries in academic tests, such as Japan, South Korea, Singapore, Estonia, Finland and Canada, are also high performers at collaborative problem solving.Working together seems to (3) different types of ability. Five years ago, PISA carried out tests on individual problem-solving skills. These showed that boys tended to do better in most countries. But when the element of collaboration is added to the problem solving, girls (4) boys in every country. In the UK this gender gap is one of the largest.These results are (5) in students’ attitudes. Girls show more positiveattitudes towards relationships, meaning that they tend to be more interested in others’ opinions and want others to succeed. Boys, on the other hand, are more likely to seethe (6) of teamwork and how collaboration can help them work more (7)_______ and efficiently.Disadvantaged students are more likely to see the (8) of teamwork thantheir (9) peers. They tend to report that they prefer working as part of a team to working alone, and that they think teams make better decisions than individuals.The classroom environment seems to (10) how well students work together. When students have a lot of communication-intensive activities –such as taking part in class debates or arguing about science questions – they are more likelyto have positive attitudes towards collaboration.四、完形填空On a cold November afternoon, my mother and I were walking home from a pizza shop. We were dressed (1) and equipped with a rented video we had been (2) _______ to watch. I was feeling a little (3) as I was carrying our shopping,and decided to throw away something. So I started to walk towards a garbage can when I noticed a poor man walking out of the restaurant in front of us. He (4) over to another nearby garbage can and started looking through it.I suddenly felt very guilty because I was about to throw away a new drink just because it was (5) . I walked up to him and handed the drink and some snacksover to him. The man looked up (6) and took what I gave him.A huge smile (7) across his face and this caused me to feel indescribably satisfied. I felt I couldn’t be happier (8)myself, but then he said, “Wow, this ismy son’s lucky day!”With that, he thanked me happily and started off on his bike, I (9) heardhim whistling a song as he rode away.I got a warm (10) inside. I now understand what is meant by the saying “giving is getting”.Although it only (11) a little action and a few words, I gained and learnedmore in those two minutes than I did in the rest of the month. Everyone in the world needs help, everyone can (12) help and everyone will be helped by (13) _______ kindness.The image of that man’s happiness caused by my small gift appears in my mind every (14) I have the chance to do something nice.This is the (15) of charity.1. A. poorly B. coldly C. warmly D. expensively2. A. dying B. exciting C. worrying D. happy3. A. worried B. interested C. bored D. tired4. A. headed B. passed C. crossed D. took5. A. cheap B. heavy C. tasteless D. full6. A. in silence B. in surprise C. in interest D. in a hurry7. A. appeared B. spread C. went D. ran8. A. with B. to C. at D. for9. A. still B. once C. even D. ever10. A. sense B. mind C. thinking D. feeling11. A. held B. took C. called D. asked12. A. offer B. send C. show D. have13. A. showing B. expressing C. lending D. setting14. A. moment B. day C. minute D. time15. A. power B. meaning C. strength D. aim五、阅读理解(A)My six-year-old granddaughter, Caitlynd and I stopped at a Tim Horton’s shop for a blueberry cake. As we were going out the door, a young teenage boy was coming in. This young man had no hair on the sides of his head and a set of blue spiked(竖起的)hair on top of it. One of his nostrils(鼻孔)was pierced, and a ring ran through the hole and a chain went across his face and was attached to a ring he was wearing in hisear. He held a skateboard under one arm and a basketball under the other.Caitlynd, who was walking ahead of me, stopped at once when she saw the teenager. I thought he’d scared her and she’d frozen on the spot.I was wrong.My granddaughter backed up against the door and opened it as wide as it would go. Now I was face to face with the young man. I stepped aside and let him pass. His gracious response was a polite “Thank you very much”.On our way to the car, I praised Caitlynd for her manners in holding open the door for the young man. She didn’t seem to be troubled by his appearance but I wanted to make sure. If a grandmother talked about freedom of self-expression and allowing people their differences was suitable, I wanted to be ready.As it turned out, the person who needed the talk was me.The only thing Caitlynd noticed about the teenager, was the fact that his arms were full. “He would have a hard time in opening the door.”I saw the partially shaved head, the set of spiked hair, the piercings and the chain. She saw a person carrying something under each arm and heading toward a door.In the future, I hope to get down on her level and raise my sights.1. What did the author think of the young man?A. Polite.B. Uncommon.C. Frightening.D. Funny.2. Caitlynd helped the young man because .A. she was scaredB. she didn’t notice his lookC. she wanted to avoid himD. it would be difficult for him to open the door3. The underlined sentence suggests that .A. the author was ashamed of herselfB. the author didn’t know how to give a talk on freedomC. a talk on freedom was useless for the granddaughterD. people should have more freedom to express themselves4. The author intends to tell us that .A. we shouldn’t judge a person by his lookB. we should allow people more freedom to dress differentlyC. we should be more helpful and tolerant to strange dressing of the youngD. we shouldn’t be too particular about people in life(B)Last week Amazon announced Word Runner, a new tool for reading Kindle books. Word Runner is a software update that shows one word at a time, instead of words, sentences and paragraphs on one page.Opinions about Word Runner are mixed. Carmen Blyth, ph.D., and a teacher of English as a Second Language (ESL), says Word Runner is not optimal reading.In the Kindle Forums at Amazon, a commenter called CB Retriever says, “Thanks, but that absol utely will not work for me as I too do lines and blocks of text at a time – I think the last time I read word by word was back when I was learning to read.”Len Edgerly, business journalist and host of The Kindle Chronicles podcast, had a chance to try out Word Runner. In his podcast, Edgerly said that Word Runner is “a dramatically different way to read.”Dynamic PacingWord Runner uses a feature called “Dynamic Pacing” to adjust the speed around complicated words. Dynamic Pacing automatically slows the speed of Word Runner for complicated words, punctuation, and paragraph breaks. It makes changes based on the natural rhythm of your reading.BrakeYou can pause Word Runner by tapping the pause button, called Brake:Brake lets you move words forward and backward. You can go back and re-read words using Brake.AvailabilityAccording to Amazon, Word Runner will be available soon for some English-language books on the Kindle app for Android devices and on Fire tablets.Word Runner will work with select English language Kindle books from Amazon sites in all countries, says an Amazon representative.Amazon has not yet said whether Word Runner will be coming to the Kindle app for iPhone and iPad.1. In CB Retriever’s opinion, .A. Word Runner provides one with necessary information to understand what he is readingB. good reader can read four words at a timeC. Word Runner is not so practical for him as he reads lines and blocks of text at a timeD. Word Runner is “a dramatically different way to read”2. From the passage, we can draw the conclusion that .A. Word Runner will soon be available for iPhone and iPadB. Word Runner will work with all English-language books in all countriesC. Word Runner is a software which updates one word at a timeD. People will soon be able to use Word Runner to read some English-language books on Andriod devices and Fire tablets3. “Dynamic Pacing” is designed to .A. regulate the speed of readingB. quicken the speed of reading something easyC. let you move words forward and backwardD. allow you to see all of the words on a page on your screen(C)In recent years many countries of the world have been faced with the problem of how to make their workers more productive. Some experts claim the answer is to make jobs more varied. But do more varied jobs lead to greater productivity? There is evidence to suggest that while variety certainly makes the worker’s life more enjoyable, it doesn’t actually make him work harder. As far as increasing productivity is concerned, then variety is not an important factor.Other experts feel that giving the worker freedom to do his job in his own way is important and there is no doubt that this is true. The problem is that this kind of freedom cannot easily be given in the modern factory with its complicated machinerywhich must be used in a fixed way. Thus while freedom of choice may be important, there is usually very little that can be done to create it.Another important consideration is how much each worker contributes to the product he is making. In most factories the worker sees only one part of the product. Some car factories are now experimenting with having many small production lines rather than one large one, so that each worker contributes more to the production of the cars on his line. It would seem that not only is degree of the workers’ contribution an important factor, therefore, but it is also one we can do something about.To what extent more money led to greater productivity? The workers themselves certainly think this is important. But perhaps they want more money only because the work they do is so boring. Money just lets them enjoy their spare time more. A similar argument may explain demands for shorter working hours. Perhaps if we succeed in making their jobs more interesting, they will neither want more money, nor will shorter working hours be so important to them.1. Which of these possible factors leading to greater productivity is not true?A. To make jobs more varied.B. To give the worker freedom to do his job in his own way.C. Degree of work contribution.D. Demands for longer working hours.2. Why do workers want more money?A. Because their jobs are too boring.B. In order to enjoy more spare time.C. To make their job more interesting.D. To demand shorter working hours.3. The last sentence in this passage means that if we succeed in making workers’ jobs more interesting, .A. they will want more moneyB. they will demand shorter working hoursC. more money and shorter working hours are important factorsD. more money and shorter working hours will not be so important to them4. In this passage, the author tells us .A. how to make the workers more productiveB. possible factors leading to greater efficiencyC. to a certain extent more money lead to greater productivityD. how to make workers jobs more interestingA. To fight family laziness, set an example.B. Knowing how to fight laziness is important.C. One way to fight laziness is to get enough sleep.D. With strong determination, you will be able to achieve your goal.E. Laziness appears when you no longer feel in charge of your own mind.F. Create a reward system for yourself, just as parents do for a child.Lazy people will never gain anything in life. However, laziness can be defeated once a few changes have been made in your mind.(1) many people lack sleep constantly, since they stay up too late and get up too early to prepare for work. These people have little motivation once they arrive home. Laziness works hand in hand with a lack of motivation and a tendency to put off things. By adjusting your sleep schedule to provide a few more hours of meaningful rest, you can fight laziness throughout the day.Another way to fight laziness is to change your mind from passive to active. Some people treat their lives as if they were pushed from task to task. Others take a more positive approach, viewing each task as a challenge they must overcome alone. (2)__________Some people fight laziness by removing the temptations that surround them. A television in the living room may provide entertainment, but watching too much TV often contributes to laziness. (3)__________complete a few tasks and reward yourself with what you enjoy, such as a good dinner or a film.Laziness can also be a lasting problem at home. Couples and children may all have different energy levels, but laziness can be spread if not dealt with immediately. (4) be the first to collect and wash dishes after a meal. Others in the home may eventually follow your example and perform their own task. It is difficult to practice laziness when you are surrounded by motivated people.Finally, taking exercise regularly can help you fight laziness. Enough exercise and a balanced diet can help you develop a healthy lifestyle, thus enabling you to have more energy and help lift your spirits.1. She was busy (write) out sums upon the blackboard.2. (tell) the story, the old man got more and more excited.3. You should go to your teacher (ask) about it.4. The dictionary (belong) to Jack is missing.5. Time (permit), the students will make a round trip to the nearby mountain village on Sunday afternoon.6. The bridge (build) now will be the longest and the most beautiful one in the country.7. It is the third time you (make) such a stupid mistake.8. We will have a picnic if it (not rain) tomorrow.9. The room is so dirty and it needs (clean).10. The classroom building (complete) by the end of this year.八、翻译1. 你如何解释他的失败呢? (account)2. 科学研究成果往往被运用到现代工业。

高中上海市曹杨中学高一上学期期末复习卷一数学试题

高中上海市曹杨中学高一上学期期末复习卷一数学试题

上海市曹杨中学【精品】高一上学期期末复习卷一数学试题学校:___________姓名:___________班级:___________考号:___________一、填空题1.满足{}{},,,,a b M a b c d ⋃=的所有集合M 的个数是________.2.关于x 的不等式()()221110m x m x ----<的解集为R ,则实数m 的取值范围为______ .3.已知幂函数()()223m m f x x m --=∈Z 为偶函数,且在()0,∞+上是减函数,则()f x 的解析式为________.4.已知{}{}0,1,P M x x P ==⊆,则P 与M 的关系为________.5.设220,0,12b a b a ≥≥+=,则的最大值为6.已知α,β满足1122αβ-<<<,则2αβ-的取值范围是________. 7.若0a >,0b > ,则关于x 的不等式1b a x-<<的解集为________.8.若x ,y ∈R +,且2x +8y -xy =0,则x +y 的最小值为________. 9.函数()2320y x x x=+>的最小值为________. 10.函数()22f x x x =--在[],a b 上的值域是[]3,1-,则+a b 的取值范围是________. 11.()f x 是定义在R 上的函数,(1)若存在1212,,x x R x x ∈<,使()()12f x f x <,则函数()f x 在R 上单调递增; (2)若存在1212,,x x R x x ∈<,使()()12f x f x ≥,则函数()f x 在R 上不可能单调递增;(3)对任意1212,,x x R x x ∈<,使()()12f x f x ≥,则函数()f x 在R 上单调递增; (4)函数()f x 对任意实数x 都有()()1f x f x <+,那么()f x 在R 上是增函数. 以上命题正确的序号是________. 12.设2()lg()1f x a x=+-是奇函数,则使()0f x <的x 的取值范围是_____________ 13.设函数(),()f x g x 的定义域分别为,f g D D ,且fg D D ⊂≠.若对于任意f x D ∈,都有()()g x f x =,则称函数()g x 为()f x 在g D 上的一个延拓函数.设2()2f x x x =+,(],0x ∈-∞,()g x 为()f x 在R 上的一个延拓函数,且()g x 是偶函数,则()g x =____________ .二、单选题14.设()y f x =和()y g x =是两个不同幂函数,集合()()(){},M x y f x g x ==,则集合M 中元素个数为( ) A .1或2或0B .1或2或3C .1或2或3或4D .0或1或2或315.若()22f x x ax =-+与()1ag x x =+,在区间[]1,2是减函数,则a 的取值范围是( ) A .()()1,00,1- B .()(]1,00,1- C .()1,0- D .(]0,116.设定义域为R 的函数()()()lg 1101x x f x x ⎧-≠⎪=⎨=⎪⎩,则关于x 的方程()()20f x bf x c ++=,有7个不同实数根的充要条件是( )A .0b <且0c >B .0b <且0c <C .0b <且0cD .0b ≥且0c三、解答题17.已知关于x 的不等式250ax x a-<-的解集为M . (1)当4a =时,求集合M ;(2)若3M ∈且5M ∉,求实数a 的取值范围. 18.求下列函数的值域(1)5121x y =-+ (2)22211x y x -=+ (3)2123y x x =+- (4)y (5)y x =- (6)y x =+(7)222231x x y x x ++=++ (8)23y x x =++- (9)y = (10) 141,02xxy x -⎛⎫=++≥ ⎪⎝⎭(11)()212log 32y x x =+-19.已知函数()224422f x x ax a a =-+-+在区间[0,2]上的最小值为3,求a 的值.20.已知函数22x x a y x++=.(1)当4a =时,求函数()f x 在区间1,44⎡⎤⎢⎥⎣⎦的值域;(2)求函数()f x 在区间1,44⎡⎤⎢⎥⎣⎦的最小值.21.已知函数()12x m f x x +-=-,0m >且()11f =-.(1)求实数m 的值;(2)判断函数()y f x =在区间(],1m -∞-上的单调性,并用函数单调性的定义证明; (3)求实数k 的取值范围,使得关于x 的方程()f x kx =分别为: ①有且仅有一个实数解;②有两个不同的实数解;③有三个不同的实数解. 22. 函数2()1ax b f x x +=+是定义在(),-∞+∞上的奇函数,且12()25f =. (1)求实数a ,b ,并确定函数()f x 的解析式;(2)判断()f x 在(-1,1)上的单调性,并用定义证明你的结论;(3)写出()f x 的单调减区间,并判断()f x 有无最大值或最小值?如有,写出最大值或最小值.(本小问不需要说明理由)23.在区间1,22⎡⎤⎢⎥⎣⎦上,函数()2f x x px q =++与()212g x x x =+在同一点取得相同的最小值,那么()f x 在1,22⎡⎤⎢⎥⎣⎦上的最大值是多少?24.已知函数()()2232log ,log f x x g x x =-=.(1)当[]1,4x ∈时,求函数()()()1h x f x g x =+⋅⎡⎤⎣⎦的值域;(2)如果对任意的[]1,4x ∈,不等式()()2f x f kg x ⋅>⋅恒成立,求实数k 的取值范围.25.已知函数()221xf x x =+,用定义判断:(1)()f x 的奇偶性;(2)()f x 的单调性、并求出最值. 26.设函数()a f x x x=+,()222g x x x a =-+-,其中0a >. (1)若1x =是关于x 的不等式()()f x g x >的解,求a 的取值范围; (2)求函数()af x x x=+在(]0,2x ∈上的最小值; (3)若对任意的(]12,0,2x x ∈,不等式()()12f x g x >恒成立,求a 的取值范围; (4)当32a =时,令()()()(),0,h x f x g x x =+∈+∞,试研究函数()h x 的单调性,求()h x 在该区间上的最小值.27.已知定义域为R 的函数()122x x b f x a+-+=+是奇函数.(1)求b 的值;(2)判断并证明函数()f x 的单调性;(3)若对任意t R ∈,不等式()()22220f t t f t k -+-<恒成立,求k 的取值范围.参考答案1.4 【解析】 【分析】先从集合等式中到,c M d M ∈∈,而,a b 可在M 中或不在M 中,从而可得M 的个数. 【详解】因为{}{},,,,a b M a b c d ⋃=,故,c M d M ∈∈,故,a b 可在M 中或不在M 中, 所以M 的个数为{},a b 的子集的个数即224=. 故答案为4. 【点睛】本题考虑集合子集个数的计算,一般地,如果集合中元素的个数为n ,则其子集的个数为2n ,此类问题为基础题. 2.3,15⎛⎤- ⎥⎝⎦【分析】分210m -=以及210m -≠两种情况讨论. 【详解】当210m -=时,1m =±,若1m =,原不等式变为:10-<,满足;若1m =-,原式变为:210x -<,此时解集不为R ,不满足;当210m -≠时,因为解集为R ,所以()()()2214110m m ∆=-----<⎡⎤⎣⎦,解得: 315m -<<; 综上:3,15m ⎛⎤∈- ⎥⎝⎦.【点睛】形如20ax bx c ++<在实数集上恒成立的问题,首先需要考虑的是a 是否为零,也就是说()2f x ax bx c =++是二次函数还是一次函数,这一定要分析清楚,其次才是分析恒成立.3.()4f x x -=【分析】根据函数的单调性可得m 满足的不等式,再根据其为整数可得具体的值,代入检验可得()f x 的解析式.【详解】因为()f x 在()0,∞+上为减函数,故2230m m --<,所以13m -<<. 因为m 为整数,故0,1,2m =.当0m =时,()3f x x -=,其为奇函数,舍去;当1m =时,()4f x x -=,其为偶函数,符合;当2m =时,()3f x x -=,其为奇函数,舍去.故答案为()4f x x -=.【点睛】幂函数在()0,∞+上的单调性和奇偶性是由幂指数决定的,解题中注意根据给定的性质确定幂指数的性质,此类问题为基础题. 4.P M ∈ 【分析】M 中的元素为P 的子集,从而可得P 与M 的关系.【详解】{}{}{}{}{},0,1,0,1M x x P =⊆=∅,所以P M ∈.故答案为:P M ∈. 【点睛】一般地,元素与集合之间的关系用,∈∉,集合与集合之间的关系用,⊆⊄,但集合可以作为另一个集合的元素,因此关系判断的关键是弄清楚集合中元素的属性. 5.1 【详解】 令,则,而由可得,所以,令,由,可得,,所以时,的最大值为1,所以t 最大值也为1.6.31,22⎛⎫-⎪⎝⎭ 【分析】利用不等式的性质可得2αβ-的范围. 【详解】因为1111,2222αβ-<<-<<,αβ<,故10αβ-<-<, 所以3122αβα-<-+<即31222αβ-<-<,故答案为:31,22⎛⎫- ⎪⎝⎭ 【点睛】本题考查不等式的性质,注意不可算出2α再求2αβ-的范围,因为有αβ<这样的限制,此类题属于基础题. 7.11,,b a ⎛⎫⎛⎫-∞-+∞ ⎪ ⎪⎝⎭⎝⎭【分析】可将该不等式转化为一元二次不等式组,从而可求原不等式的解集. 【详解】不等式1b a x -<<等价于1010b xa x⎧+>⎪⎪⎨⎪-<⎪⎩即1010bx x ax x +⎧>⎪⎪⎨-⎪<⎪⎩,故()()1010x bx x ax ⎧+>⎪⎨->⎪⎩,整理得到1010x x bx x a ⎧-⎪⎪⎨⎪⎪⎩或或,该不等式组的解为1x b <-或1x a >.故原不等式的解集为11,,b a ⎛⎫⎛⎫-∞-+∞ ⎪⎪⎝⎭⎝⎭.故答案为:11,,b a ⎛⎫⎛⎫-∞-+∞ ⎪ ⎪⎝⎭⎝⎭.【点睛】本题考查分式不等式解的求法,一般地,分式不等式可以转化为一元二次不等式来求解,注意转化时分母不为零. 8.18 【解析】 【分析】转化已知280x y xy +-=为右边是1的式子,然后去乘以x y +,再利用基本不等式求得最小值. 【详解】由280x y xy +-=得28x y xy +=,281x y xy+=,即281y x +=,所以()2828101010818x y x y x y y x y x ⎛⎫+=++=++≥+=+= ⎪⎝⎭,当且仅当28x yy x=,即12,6x y ==时等号成立.故最小值为18. 【点睛】本小题主要考查利用基本不等式求和式的最小值,主要的方法是“1”代换的方法,属于基础题.9.23362【分析】通过单调性的定义可判断函数的单调性,再利用单调性可求函数的最小值. 【详解】 令()()2320f x x x x=+>,设120x x <<,则()()221212123322f x f x x x x x -=+--()()1212121223x x x x x x x x -+-⎡⎤⎣⎦=,若任意的1202x x <<≤,则120x x -<,120x x >, ()312122623434308x x x x x +-<⨯-≤⨯-=,故()()120f x f x ->即()()12f x f x >,所以()f x在0,2⎛ ⎝⎦上为减函数,若任意的122x x ≤<,则120x x -<,120x x >, ()312121623434308x x x x x +->⨯-≥⨯-=,故()()120f x f x -<即()()12f x f x <,所以()f x在,+2⎫∞⎪⎪⎣⎭上为增函数. 所以()f x 在()0,∞+上的最小值为223326222f ⎛⎛== ⎝⎭⎝⎭故答案为23362.【点睛】函数的最值,一般要依据函数的单调性来求,如果函数不是基本初等函数,那么单调性的判断可以依据复合函数的单调性或依据定义来判断,后者需作差后利用“逼近”的方法来寻找单调性的分界点,如本题中,为了确定()121223x x x x +-的符号,可令12,x x 近似相等,从而得到代数式3143x -变号的分界点为12x =,从而得到两个区间⎛ ⎝⎦及⎫∞⎪⎪⎣⎭,在这两个区间上讨论函数的单调性即可. 10.[]4,0- 【分析】在平面直角坐标系画出()22f x x x =--的图像,结合图像可得,a b 满足的条件,从而得到+a b 的取值范围.【详解】函数()22f x x x =--的图像如图所示,作出直线1y =,它和()f x 的图像相切于顶点()1,1C -,作出直线3y =-,令223x x --=-,解得3x =-或1x =,故()()3,3,1,3A B ---. 因为()f x 的值域为[]3,1-,故3a =-或1b =, 若3a =-,则11b -≤≤,此时42a b -≤+≤-, 若1b =,则31a -≤≤-,此时40a b -≤+≤, 故40a b -≤+≤. 故答案为:[]4,0-. 【点睛】本题考查二次函数的图像和性质,注意根据值域可初步确定定义域的某个端点,从而得到定义域区间两个端点之间的关系,本题属于中档题. 11.(2) 【分析】依据单调性的定义及反例可得正确的选项. 【详解】对于(1)(3),根据单调性的定义,只有对任意的1212,,x x R x x ∈<,总有()()12f x f x <, 函数()f x 才在R 上单调递增,故(1)(3)错误;对于(2),如果函数()f x 在R 上单调递增,则必有()()12f x f x <,与()()12f x f x ≥矛盾,故函数()f x 在R 上不可能单调递增,故(2)正确;对于(4),取函数()3,021,0x x f x x x ⎧+≤⎪=⎨⎪+>⎩,因171366f f ⎛⎫⎛⎫-== ⎪ ⎪⎝⎭⎝⎭, 所以()f x 在R 不是单调递增函数,但()5,1122,1x x f x x x ⎧+≤-⎪+=⎨⎪+>-⎩, 当1x ≤-时,()()()()531,,122f x x f x x f x f x +=+=++>成立, 当10x -<≤时,()()()()312,,12f x x f x x f x f x +=+=++>成立,当0x ≥时,()()()()12,1,1f x x f x x f x f x +=+=++>也成立, 故(4)错. 故答案为(2) 【点睛】本题考查函数单调性的定义的理解,注意单调性定义的关键词“任意的12,x x ”,本题属于基础题. 12.()1,0- 【分析】先根据奇函数性质求参数a ,再解对数不等式得结果. 【详解】由f(x)是奇函数可得a =-1, ∴f(x)=lg,定义域为(-1,1).由f(x)<0,可得0<<1,∴-1<x<0.【点睛】利用奇偶性求值的类型及方法(1)求函数值:利用奇偶性将待求值转化到已知区间上的函数值,进而得解.(2)求参数值:在定义域关于原点对称的前提下,根据奇函数满足f(-x)=-f(x)或偶函数满足f(-x)=f(x)列等式,根据等式两侧对应相等确定参数的值.特别要注意的是:若能够确定奇函数的定义域中包含0,可以根据f(0)=0列式求解,若不能确定则不可用此法.13.22x x - 【分析】设函数()(),f x g x 的定义域分别为,f g D D ,且fg D D ⊂≠.若对于任意f x D ∈,都有()()g x f x =,则称函数()g x 为()f x 在g D 上的一个延拓函数.设()22f x x x =+,(],0x ∈-∞,()g x 为()f x 在R 上的一个延拓函数,且()g x 是偶函数,则()g x =【详解】因为()22f x x x =+,(],0x ∈-∞,()g x 为()f x 在R 上的一个延拓函数,且()g x 是偶函数,当0x ≤时,()()22g x f x x x ==+,0x >时,0x -<,()()()222g x x x x x g x 2-=--=-=,所以()22g x x x =-,故答案为22x x -.【点睛】本题主要考查函数的奇偶性以及新定义问题,属于中档题. 本题题型可归纳为“已知当0x >时,函数()y f x =,则当0x <时,求函数的解析式”.有如下结论:若函数()f x 为偶函数,则当0x <时,函数的解析式为()y f x =-;若()f x 为奇函数,则函数的解析式为()y f x =--.14.B 【分析】考虑不同幂函数构成的方程,解方程后可得图像的交点及交点的个数,从而得到正确的选项. 【详解】取()()133,f x x g x x ==,由133x x =可得0x =或1x =或1x =-,故()()()(){}133,=0,0,1,1,1,1M x y x x ⎧⎫⎪⎪==--⎨⎬⎪⎪⎩⎭;取()()132,f x x g x x ==,由132x x =可得0x =或1x =,故()()(){}132,=0,0,1,1M x y x x ⎧⎫⎪⎪==⎨⎬⎪⎪⎩⎭,取()()23,f x x g x x -==,由23x x -=可得1x =,故(){}(){}23,=1,1M x y xx -==,注意,任意幂函数的图像必过()1,1点,故()1,1M ∈,任意两个幂函数的图像不可能有4个交点,故M 中元素个数为1或2或3, 故选B. 【点睛】本题考查幂函数的图像和性质,解答本题的关键是熟悉三类幂函数(即幂指数小于0、大于等于0小于1及大于等于1)在第一象限内的图像和性质,此类问题属于中档题. 15.D 【分析】根据两个函数的单调性得到a 的不等式组,其解即为a 的取值范围. 【详解】因为()22f x x ax =-+、()1ag x x =+在[]1,2是减函数,故10a a ≤⎧⎨>⎩,所以01a <≤, 故选D. 【点睛】本题考查二次函数的单调性及分式函数的单调性,前者取决于对称轴的问题,后者取决于平移前反比例函数的比例系数的正负,此类问题属于基础题. 16.C 【分析】画出()f x 的图像,根据方程有7个不同的实数根可得方程20t bt c ++=有一个零根和正根,从而得到,b c 满足的条件. 【详解】令()t f x =,考虑方程20t bt c ++=的根,该方程必有解,设解为12,t t t t ==,由题设方程()1t f x =和方程()2t f x =的解即为方程()()20fx bf x c ++=的解,因为方程()()20fx bf x c ++=的解有7个不同的解,根据()f x 的图像(如图所示)可得,直线1y t =与()y f x =的图像有3个不同公共点, 直线2y t =与()y f x =的图像有4个不同公共点, 故10t =,20t >, 所以0c ,20t b =->即0b <,故选C.【点睛】复合方程()g f x m =⎡⎤⎣⎦的解的个数问题,其实质就是方程组()()g t m t f x ⎧=⎪⎨=⎪⎩的解的个数问题,可先利用图像变换等工具刻画()f x 的图像特征,结合原来方程解的个数得到t 的限制条件,再利用常见函数的性质(如二次函数等)刻画()g t 的图像特征从而得到参数的取值范围. 17.(1)()5,2,24⎛⎫-∞- ⎪⎝⎭;(2)(]51,9,253⎡⎫⎪⎢⎣⎭.【分析】(1)代入4a =后将分式不等式转化为高次不等式,求解后可得M . (2)根据3M ∈且5M ∉可得关于a 的不等式组,其解为实数a 的取值范围. 【详解】(1)因为4a =,故24504x x -<-即()()()45220x x x --+<, 所以2x <-或524x <<,故M 为()5,2,24⎛⎫-∞- ⎪⎝⎭.(2)因为3M ∈且5M ∉,故350955025a aa a -⎧<⎪⎪-⎨-⎪≥⎪-⎩或250a -=,故()()()()35901250a a a a ⎧-->⎪⎨--≤⎪⎩,解得513a ≤<或925a <≤,故a 的取值范围为(]51,9,253⎡⎫⎪⎢⎣⎭.【点睛】一般地,()()0f x g x >等价于()()0f x g x >,而()()0f x g x ≥则等价于()()()00f x g x g x ⎧≥⎪⎨≠⎪⎩,注意分式不等式转化为整式不等式时分母不为零.解本题时还应注意5M ∉对应的a 满足的条件中容易遗漏250a -=这个情况.18.(1)()4,1-;(2)[)1,2-;(3)()1,0,4⎛⎤-∞-+∞ ⎥⎝⎦;(4)50,2⎡⎤⎢⎥⎣⎦;(5)(],1-∞;(6)(],2-∞;(7)102,3⎛⎤⎥⎝⎦;(8)[)5,+∞;(9)[)0,3;(10)(]1,3;(11)[)2,-+∞. 【分析】根据函数的特点,可利用换元法、基本初等函数的性质(如单调性等)、反表示、分离常数法等可求题设中的11个函数的值域. 【详解】(1)函数的定义域为R ,当x ∈R 时,211x +>, 故50521x <<+,所以541121x-<-<+,故函数的值域为()4,1-. (2)函数的定义域为R ,由22211x y x -=+可以得到2221yx y x +=-,整理得到212y yx +=-. 因210,02y yx +≥∴≥-,即12y -≤<,故函数的值域为[)1,2-.(3)函数2123y x x =+-的定义域为()(),31,-∞-⋃+∞,当()(),31,x ∈-∞-+∞,[)()2234,00,x x +-∈-+∞,故()211,0,234x x ⎛⎤∈-∞-+∞ ⎥+-⎝⎦, 所以函数的值域为()1,0,4⎛⎤-∞-+∞ ⎥⎝⎦.(4)函数的定义域为[]1,4-,令223253424t x x x ⎛⎫=-++=--+ ⎪⎝⎭,当[]1,4x ∈-时,2504t ≤≤,故502y ≤≤,所以函数的值域为50,2⎡⎤⎢⎥⎣⎦. (5)函数的定义域为(],1-∞,因为y x =为(],1-∞的增函数,y =(],1-∞上的减函数,故y x =-(],1-∞上的增函数,当1x =时,函数的函数值1,故函数的值域为(],1-∞.(6)函数的定义域为(],1-∞,令t =,则21x t =-,0t ≥, 所以()221212y t t t =-+=--+,因为0t ≥,故2y ≤,故函数的值域为(],2-∞. (7)函数的定义域为R ,又2222231211x x y x x x x ++==+++++,而221331244x x x ⎛⎫++=++≥ ⎪⎝⎭, 所以214013x x <≤++,故1023y <≤,故函数的值域为102,3⎛⎤⎥⎝⎦. (8)函数的定义域为R , 当2x -≤时,125y x =-≥ ; 当23x <<时,235y x x =++-=, 当3x ≥时,215y x =-≥,综上,函数的值域为[)5,+∞. (9)函数的定义域为(],2-∞,当(],2x ∈-∞时,039x <≤,故0939x ≤-<,所以03y ≤<, 所以函数的值域为[)0,3.(10)函数可变形为2111,022x xy x ⎛⎫⎛⎫=++≥ ⎪ ⎪⎝⎭⎝⎭, 令12xt ⎛⎫= ⎪⎝⎭,则01t <≤且21y t t =++,所以13y <≤,故函数的值域为(]1,3.(11)函数的定义域为()1,3-,令()223421t x x x =--=-++,因为()1,3x ∈-,故04t <≤,故1122log log 42t ≥=-,故函数的值域为[)2,-+∞. 【点睛】函数值域的求法,应根据函数的特点选取合适的方法来求,主要的方法有:(1)换元法:当函数是简单函数的复合时(如指对数函数与分式函数、二次函数、幂函数的复合),可用此法把值域归结为简单函数的值域问题;(2)单调性法:如果函数在给定的区间上是单调的,则可直接求出函数的值域; (3)反表示法:如果可以用y 来表示x ,则可以根据x 的范围求出y 的范围(就是函数的值域);(4)分离常数法:如果函数是分式的形式,则可以分离常数,把函数的值域归结为一个简单的函数的值域.19.1a =5a =. 【分析】将f (x )转化为顶点式,求得对称轴,讨论区间和对称轴的关系,结合函数单调性,得最小值所对应方程,解方程可得a 的值 【详解】函数()f x 的表达式可化为()()24222a f x x a ⎛⎫=-+- ⎪⎝⎭. ① 当022a<<,即04a <<时,()f x 有最小值22a -,依题意应有223a -=,解得12a =-,这个值与04a ≤≤相矛盾.②当2a 0≤,即a 0≤时,()2022f a a =-+是最小值,依题意应有2223a a -+=,解得1a =±a 0≤,∴1a = ③当2a 2≥ ,即a 4≥时,()2216822f a a a =-+-+是最小值, 依题意应有2168223a a a -+-+=,解得5a =a 4≥,∴5a =+综上所述,1a =-5a =. 【点睛】本题考查了二次函数求最值,解题中要注意对称轴和区间的关系,考查分类讨论的思想方法和运算能力.20.(1)[]6,7;(2)()min9144161216166164a a f x a aa ⎧+≤⎪⎪⎪=<<⎨⎪⎪+≥⎪⎩,,,.【分析】(1)讨论函数()42f x x x =++在1,44⎡⎤⎢⎥⎣⎦的单调性后可得()f x 的值域. (2)就116a ≤、11616a <<、16a ≥三种情况分别讨论函数的单调性后可得函数的最小值. 【详解】(1)由题设有()42f x x x=++ 设12144x x ≤<≤,()()12121244f x f x x x x x -=+--()1212124x x x x x x -=-,当任意的12124x x ≤<≤时,120x x -<,121240,0x x x x -<>, 故()()120f x f x ->即()()12f x f x >,故()f x 在1,24⎡⎤⎢⎥⎣⎦为减函数.当任意的1224x x ≤<≤时,120x x -<,121240,0x x x x ->>, 故()()120f x f x -<即()()12f x f x <,故()f x 在[]2,4为增函数.故()()min 26f x f ==,因()125,4744f f ⎛⎫== ⎪⎝⎭,故()max 7f x =, 故()f x 的值域为[]6,7. .(2)设任意的12144x x ≤<≤,则()()()()12121212x x x x a f x f x x x ---=, 若116a ≤,则对任意的12144x x ≤<≤, 总有120x x -<,120x x a ->,120x x >,所以()()120f x f x -<即()()12f x f x <,所以()f x 为1,44⎡⎤⎢⎥⎣⎦上的增函数,故()f x 的最小值为19444f a ⎛⎫=+ ⎪⎝⎭. 若16a ≥,则对任意的12144x x ≤<≤, 总有120x x -<,120x x a -<,120x x >,所以()()120f x f x ->即()()12f x f x >,所以()f x 为1,44⎡⎤⎢⎥⎣⎦上的减函数,故()f x 的最小值为()464a f =+. 若11616a <<,则对任意的1214x x ≤<≤总有120x x -<,120x x a -<,120x x >,所以()()120f x f x ->即()()12f x f x >,所以()f x为14⎡⎢⎣上的减函数,124x x ≤<≤,总有120x x -<,120x x a ->,120x x >,所以()()120f x f x -<即()()12f x f x <,所以()f x为4⎤⎦上的增函数,故()f x的最小值为2f=.综上,()min9144161216166164a a f x a aa ⎧+≤⎪⎪⎪=<<⎨⎪⎪+≥⎪⎩,,,【点睛】 函数()()0af x x a x=+>常称为“双勾函数”,它在(,()上是减函数,在)+∞,(,-∞上是增函数,注意()()0af x x a x=+<不是双勾函数,该函数在()0,∞+上是增函数,在(),0-∞上是减函数.注意在高中数学的初始阶段,函数的单调性的证明只能依据定义,并且在运用该函数时需要证明其单调性,不可直接使用. 21.(1)1m =;(2)函数()f x 在区间(],0-∞上是单调递增函数,证明见解析; (3)答案不唯一,见解析 【分析】(1)将已知条件()11f =-,解得1m =,再结合m 是正数,可得1m =; (2)将(1)的结论代入得(](],1,0m -∞-=-∞,根据函数单调性的定义,可设(]12,,0x x ∈-∞,且12x x <,通过作差化简整理,最后得到()()120f x f x -<,说明函数在区间(],1m -∞-上是增函数;(3)首先,方程()f x kx =有一个解0x =,然后分0x >和0x <加以讨论:当0x >且2x ≠时,方程转化为2x kx x =-,解得12x k=+,解不等式得12k <-或0k >,当0x <时,则2x kx x -=-,解得12x k=-,解不等式得102k <<;最后综合可得方程()f x kx =解集的情况. 【详解】(1)由()11f =-,得11m=--,1m =,∵0m >,∴1m =. (2)由(1),1m =,从而()2xf x x =-,只需研究()f x 在(],0-∞上的单调性. 当(],0x ∈-∞时,()2xf x x -=-. 设(]12,,0x x ∈-∞,且12x x <,则()()12121222x x f x f x x x ---=---()()()1212222x x x x -=--, ∵120x x <≤,∴120x x -<,120x -<,220x -<, ∴()()120f x f x -<,即()()12f x f x <. ∴函数()f x 在区间(],0-∞上是单调递增函数.(3)原方程即为2xkx x =- ……① 0x =恒为方程①的一个解.若0x <时方程①有解,则2x kx x -=-,解得12x k=-, 由120k-<,得102k <<;若0x >且2x ≠时方程①有解,则2x kx x =-,解得12x k=+,由120k +>且122k+≠,得12k <-或0k >.综上可得,当1,02k ⎡⎤∈-⎢⎥⎣⎦时,方程()f x kx =有且仅有一个解; 当11,,22k ⎛⎫⎡⎫∈-∞-+∞ ⎪⎪⎢⎝⎭⎣⎭时,方程()f x kx =有两个不同解; 当10,2k ⎛⎫∈ ⎪⎝⎭时,方程()f x kx =有三个不同解. 【点睛】本题考查了函数零点的分布与单调性等知识点,属于难题. 22.(1)2()1x f x x ∴=+(2)见解析(3)单调减区间为(][),1,1,-∞-+∞x=-1时,min12y =-,当x=1时,min 12y =. 【解析】试题分析:(1)先根据函数为奇函数()求出值,再利用12()25f =求出值,即可其解析式;(2)利用函数的单调性定义进行判定与证明;(3)结合(2)问容易得到单调递减区间,进而写出最值.解题思路:(1)求解析式的一种主要方法是待定系数法;(2)利用函数单调性的定义证明函数的单调性的一般步骤为:设值代值、作差变形、判定符号、下结论. 试题解析:(1)()f x 是奇函数,()()f x f x ∴-=-.即2211ax b ax bx x -++=-++,ax b ax b -+=--,0b ∴=2()1ax f x x ∴=+,又12()25f =,1221514a∴=+,1a =,2()1xf x x ∴=+ (2)任取12,(1,1)x x ∈-,且12x x <,1212121222221212()(1)()()11(1)(1)x x x x x x f x f x x x x x ---=-=++++ 12121211,11,0x x x x x x -<<<∴-<<-<,1210x x ->2110x +>,2210x +>,12()()0f x f x ∴-<,12()()f x f x <,()f x ∴在(-1,1)上是增函数.(3)单调减区间为(][),1,1,-∞-+∞ 当x=-1时,min 12y =-,当x=1时,.考点:1.函数的奇偶性;2.函数的解析式;3.函数的单调性与最值. 23.4 【分析】先考虑函数()g x 在何处取何最小值,从而得到()f x 在何处取何最小值,求出,p q 后可求()f x 的最大值.【详解】设12122x x ≤<≤,()()121222121122g x g x x x x x -=+-- ()()2221121222221212121122x x x x x x x x x x x x ⎛⎫-=-+=--- ⎪⎝⎭,当任意的12112x x ≤<≤时,120x x -<,221212111,1x x x x >>,故2212121120x x x x --<, 故()()120g x g x ->即()()12g x g x >,故()g x 在1,12⎡⎤⎢⎥⎣⎦为减函数.当任意的1212x x ≤<≤时,120x x -<,221212111,1x x x x <<,故2212121120x x x x -->, 故()()120g x g x -<即()()12g x g x <,故()g x 在[]1,2为增函数. 所以()g x 在1x =取最小值且最小值为()13g =. 故()f x 在1x =取最小值且最小值为3.所以21234pp q ⎧-=⎪⎪⎨⎪-=⎪⎩,解得24p q =-⎧⎨=⎩,故()224f x x x =-+,因为1,22x ⎡∈⎤⎢⎥⎣⎦,故()()max 24f x f ==.【点睛】对于函数()y g x =的单调性的讨论,需要对()()12g x g x -因式分解后才能找到决定()()12g x g x -正负的核心代数式(如221212112x x x x --),为了找到该代数式变号的分界点,可令12,x x 近似相等,从而得到代数式3122x -变号的分界点为11x =,从而得到两个区间1,12⎡⎤⎢⎥⎣⎦及[]1,2,在这两个区间上讨论函数的单调性即可. 24.(1)[]0,2;(2) (),3-∞-. 【分析】(1)利用配方法化简函数,根据函数的定义域,换元得到t =2log x ∈[0,2],由二次函数的性质,即可求出函数的值域;(2)先利用对数运算化简不等式,换元,再通过分离参数法,转化为最值问题,利用基本不等式求出最值,即可求出实数k 的取值范围. 【详解】(1)h (x )=(4-22log x )·2log x =-2(2log x -1)2+2,因为x ∈[1,4],所以t =2log x ∈[0,2],2()2(1)2h x t =--+,故函数h (x )的值域为[0,2]. (2)由f (x 2)·f>k ·g (x ),得(3-42log x )(3-2log x )>k ·2log x ,令2log t x =,因为x ∈[1,4],所以t =2log x ∈[0,2], 所以(3-4t )(3-t )>k ·t 对一切t ∈[0,2]恒成立, ①当t =0时,k ∈R ; ②当t ∈(0,2]时,()()343t t k t--<恒成立,即9415k t t<+-, 因为9412t t +,当且仅当94t t=,即32t =时取等号,所以9415t t+-的最小值为-3.所以k <-3.综上,实数k 的取值范围为(-∞,-3). 【点睛】本题主要考查含有对数式的二次函数的值域的求法,利用分离参数法解决不等式恒成立问题,以及利用基本不等式求最值.意在考查学生的转化与化归思想和数学运算能力.25.(1)奇函数;(2)在(][),1,1,-∞-+∞单调递减,在()1,1-单调递增;min 1y =-;max 1y =. 【分析】(1)依据定义可判断该函数为奇函数.(2)先考虑函数在[)0,+∞上的单调性,该单调性可依据定义来判断,再根据函数为奇函数得到函数在(],0-∞上的单调性,根据单调性可求函数的最小值. 【详解】(1)函数的定义域为R ,因为()()221xf x f x x -=-=-+,故()f x 为R 上的奇函数. (2)设任意的1201x x ≤<≤,()()()()()()12121222122111x x x x f x f x x x ---=++,因为1201x x ≤<≤,故120x x -<,1210x x -<,()()2212110x x ++>,所以()()120f x f x -<,所以()f x 在[]0,1为增函数, 同理可证:()f x 在[)1,+∞上为减函数,因为()f x 为R 上的奇函数,故()f x 在(),1-∞-为减函数,在[]1,0-上为增函数, 所以()f x 在[]1,1-上为增函数.所以当1x ≤-时,()()110f f x -=-≤<, 而当1x ≥时,()0f x >且()()011f x f <≤=, 而当11x -≤≤时,()()()1111f f x f -=-≤≤=,故当x ∈R 时,min 1y =-,max 1y =. 【点睛】函数的最值问题,往往需要讨论函数的单调性,后者应利用定义来讨论,注意讨论函数最值时,要观察函数的图像是否具有渐近线(如本题中当1x ≤-时,()0f x <总成立,x 轴为图像的渐近线).26.(1)1a >;(2)min04()242a f x a a ⎧<<⎪=⎨+≥⎪⎩;(3)4a > ;(4)在()0,2单调递减,在()2,+∞单调递增;最小值为6-, 【分析】(1)在不等式()()f x g x >中令1x =,则可以得到关于a 的不等式,其解即为a 的取值范围.(2)就是4a ≥、04a <<分类讨论函数的单调性后可求()f x 在(]0,2上的最小值. (3)由()()min max f x g x >可得实数a 的取值范围.(4)设任意120x x <<,考虑()()12h x h x -的符号后可得()h x 的单调性,从而可求()h x 的最小值. 【详解】(1)由题设有()()11f g >,故13a a +>-,故1a >. (2)若4a ≥,设任意的1202x x <<≤,则()()()()12121212x x x x a f x f x x x ---=,因为1202x x <<≤,故120x x a -<,120x x -<,所以()()120f x f x ->即()()12f x f x >,所以()f x 为(]0,2上的减函数, 故()f x 的最小值为()222a f =+. 若04a <<,则设任意的1202x x <<≤,则()()()()12121212x x x x a f x f x x x ---=,因为120x x <<≤120x x a -<,120x x -<,所以()()120f x f x ->即()()12f x f x >,所以()f x为(上的减函数, 同理可证:()f x为2⎤⎦上的增函数.所以()f x的最小值为f=,故()min42,42a f x a a ⎧<<⎪=⎨+≥⎪⎩. (3)因为对任意的(]12,0,2x x ∈,不等式()()12f x g x >恒成立, 故()()()min max 28f x g x g a >==-.由(2)可知:当04a <<时,由()min f x =4a ≥时,由()min 22af x =+,所以048a a <<⎧⎪⎨>-⎪⎩或4282a aa ≥⎧⎪⎨+>-⎪⎩即044a a <<⎧⎨>⎩(无解)或4a >, 故4a >.(4)若32a =,则()220323h x xx =+-, 设任意的1202x x <<≤,则()()()()1212121212216x x x x x x h x h x x x -+-⎡⎤⎣⎦-=,因为1202x x <<≤,故()1212160x x x x +-<,120x x -<,所以()()120h x h x ->即()()12h x h x >,所以()h x 为(]0,2上的减函数, 同理可证()h x 为[)2,+∞上的增函数, 所以()h x 在()0,∞+上的最小值为()26h =-. 【点睛】 函数()()0af x x a x=+>常称为“双勾函数”,它在(,()上是减函数,在)+∞,(,-∞上是增函数,注意()()0af x x ax=+<不是双勾函数,该函数在()0,∞+上是增函数,在(),0-∞上是减函数.注意在高中数学的初始阶段,函数的单调性的证明只能依据定义.27.(1)1b=;(2)单调递减,证明略;(3)13k<-【分析】(1)根据()()f x f x-=恒成立可求得1b=.(2)()f x为减函数,利用单调性的定义可证明该结论.(3)函数不等式可以转化为2320t t k-->在R上恒成立,从而可求实数k的取值范围. 【详解】(1)()()112122,2222x x xx x xb b bf x f xa a a--++-+-+⨯-+-==-=-++⨯+,因为()f x为奇函数,故1122222x xx xb ba a+-+⨯-+=-+⨯+,化简得到()()1222240x xab a b a b+⨯-+-+-⨯=恒成立,所以22a bab=⎧⎨=⎩,解得21ab=⎧⎨=⎩或21ab=-⎧⎨=-⎩,当21ab=-⎧⎨=-⎩,()1121212222x xx xf x++--+==---,此时()f x的定义域不为R,当21ab=⎧⎨=⎩,()12122xxf x+-+=+,满足定义域为R,故1b=.(2)()f x为减函数,证明如下:设任意的12x x<,()()()()()21121212121111322121221212121x xx xx x x xf x f x++++⨯----=-=++++,因为12x x<,故21220x x->,而()()121121210x x++++>,故()()12f x f x->即()()12f x f x>,所以()f x为R上的减函数.(3)因为()f x为奇函数,故不等式()()22220f t t f t k-+-<等价于()()2222f t t f t k-<-+,而()f x 为R 上的减函数,2222t t t k ->-+即2320t t k -->对任意t R ∈. 所以4120k ∆=+<,故13k <-. 【点睛】含参数的偶函数(或奇函数),可通过取自变量的特殊值来求参数的大小,注意最后检验必不可少,也可以利用()()f x f x =-(或()()f x f x -=-)恒成立来求参数的大小.解函数不等式要利用函数的单调性、奇偶性去掉对应法则f .。

上海曹杨二中数学高一上期末经典练习题(专题培优)

上海曹杨二中数学高一上期末经典练习题(专题培优)

一、选择题1.(0分)[ID :12119]已知()f x 在R 上是奇函数,且2(4)(),(0,2)()2,(7)f x f x x f x x f +=∈==当时,则A .-2B .2C .-98D .98 2.(0分)[ID :12118]已知a =21.3,b =40.7,c =log 38,则a ,b ,c 的大小关系为( ) A .a c b <<B .b c a <<C .c a b <<D .c b a <<3.(0分)[ID :12113]已知()f x 是偶函数,它在[)0,+∞上是增函数.若()()lg 1f x f <-,则x 的取值范围是( )A .1,110⎛⎫⎪⎝⎭B .10,10,10C .1,1010⎛⎫⎪⎝⎭D .()()0,110,⋃+∞4.(0分)[ID :12085]已知0.11.1x =, 1.10.9y =,234log 3z =,则x ,y ,z 的大小关系是( ) A .x y z >>B .y x z >>C .y z x >>D .x z y >>5.(0分)[ID :12121]若函数f(x)=a |2x -4|(a>0,a≠1)满足f(1)=19,则f(x)的单调递减区间是( ) A .(-∞,2] B .[2,+∞) C .[-2,+∞)D .(-∞,-2]6.(0分)[ID :12105]已知131log 4a =,154b=,136c =,则( ) A .a b c >> B .a c b >>C .c a b >>D .b c a >>7.(0分)[ID :12081]设函数()f x 的定义域为R ,满足(1) 2 ()f x f x +=,且当(0,1]x ∈时,()(1)f x x x =-.若对任意(,]x m ∈-∞,都有8()9f x ≥-,则m 的取值范围是 A .9,4⎛⎤-∞ ⎥⎝⎦B .7,3⎛⎤-∞ ⎥⎝⎦C .5,2⎛⎤-∞ ⎥⎝⎦D .8,3⎛⎤-∞ ⎥⎝⎦8.(0分)[ID :12058]已知函数()2log 14x f x x ⎧+=⎨+⎩ 00x x >≤,则()()3y f f x =-的零点个数为( ) A .3B .4C .5D .69.(0分)[ID :12052]根据有关资料,围棋状态空间复杂度的上限M 约为3361,而可观测宇宙中普通物质的原子总数N 约为1080.则下列各数中与MN最接近的是 (参考数据:lg3≈0.48) A .1033 B .1053 C .1073D .109310.(0分)[ID :12033]若二次函数()24f x ax x =-+对任意的()12,1,x x ∈-+∞,且12x x ≠,都有()()12120f x f x x x -<-,则实数a 的取值范围为( )A .1,02⎡⎫-⎪⎢⎣⎭B .1,2⎡⎫-+∞⎪⎢⎣⎭C .1,02⎛⎫-⎪⎝⎭D .1,2⎛⎫-+∞ ⎪⎝⎭11.(0分)[ID :12070]定义在[]7,7-上的奇函数()f x ,当07x <≤时,()26x f x x =+-,则不等式()0f x >的解集为A .(]2,7B .()(]2,02,7- C .()()2,02,-+∞D .[)(]7,22,7--12.(0分)[ID :12061]若0.33a =,log 3b π=,0.3log c e =,则( )A .a b c >>B .b a c >>C .c a b >>D .b c a >>13.(0分)[ID :12046]已知[]x 表示不超过实数x 的最大整数,()[]g x x =为取整函数,0x 是函数()2ln f x x x=-的零点,则()0g x 等于( ) A .1 B .2C .3D .414.(0分)[ID :12079]已知全集U={1,2,3,4,5,6},集合P={1,3,5},Q={1,2,4},则()UP Q ⋃=A .{1}B .{3,5}C .{1,2,4,6}D .{1,2,3,4,5}15.(0分)[ID :12050]已知定义在R 上的函数()f x 在(),2-∞-上是减函数,若()()2g x f x =-是奇函数,且()20g =,则不等式()0xf x ≤的解集是( )A .][(),22,-∞-⋃+∞B .][)4,20,⎡--⋃+∞⎣C .][(),42,-∞-⋃-+∞D .][(),40,-∞-⋃+∞二、填空题16.(0分)[ID :12225]若155325a b c ===,则111a b c+-=__________. 17.(0分)[ID :12223]若函数()1f x mx x =--有两个不同的零点,则实数m 的取值范围是______.18.(0分)[ID :12214]如果函数()22279919mm y m m x--=-+是幂函数,且图像不经过原点,则实数m =___________.19.(0分)[ID :12208]已知()y f x =是定义在R 上的奇函数,且当0x 时,11()42x xf x =-+,则此函数的值域为__________. 20.(0分)[ID :12201]已知函数2()log f x x =,定义()(1)()f x f x f x ∆=+-,则函数()()(1)F x f x f x =∆++的值域为___________.21.(0分)[ID :12177]已知偶函数()f x 的图象过点()2,0P ,且在区间[)0,+∞上单调递减,则不等式()0xf x >的解集为______.22.(0分)[ID :12175]若函数()()()()22,0,0x x x f x g x x ⎧+≥⎪=⎨<⎪⎩为奇函数,则()()1f g -=________.23.(0分)[ID :12163]对于函数()y f x =,若存在定义域D 内某个区间[a ,b ],使得()y f x =在[a ,b ]上的值域也为[a ,b ],则称函数()y f x =在定义域D 上封闭,如果函数4()1xf x x=-+在R 上封闭,则b a -=____. 24.(0分)[ID :12149]若存在实数(),m n m n <,使得[],x m n ∈时,函数()()2log x a f x a t =+的值域也为[],m n ,其中0a >且1a ≠,则实数t 的取值范围是______.25.(0分)[ID :12137]已知函数()()212log 22f x mx m x m ⎡⎤=+-+-⎣⎦,若()f x 有最大值或最小值,则m 的取值范围为______.三、解答题26.(0分)[ID :12308]已知函数2()(8)f x ax b x a ab =+--- 的零点是-3和2 (1)求函数()f x 的解析式.(2)当函数()f x 的定义域是0,1时求函数()f x 的值域.27.(0分)[ID :12269]已知函数2()log (421)x xf x a a =+⋅++,x ∈R .(Ⅰ)若1a =,求方程()3f x =的解集;(Ⅱ)若方程()f x x =有两个不同的实数根,求实数a 的取值范围.28.(0分)[ID :12262]已知函数()f x 是二次函数,(1)0f -=,(3)(1)4f f -==. (1)求()f x 的解析式;(2)函数()()ln(||1)h x f x x =-+在R 上连续不断,试探究,是否存在()n n ∈Z ,函数()h x 在区间(,1)n n +内存在零点,若存在,求出一个符合题意的n ,若不存在,请说明由. 29.(0分)[ID :12256]某镇在政府“精准扶贫”的政策指引下,充分利用自身资源,大力发展养殖业,以增加收入.政府计划共投入72万元,全部用于甲、乙两个合作社,每个合作社至少要投入15万元,其中甲合作社养鱼,乙合作社养鸡,在对市场进行调研分析发现养鱼的收益M 、养鸡的收益N 与投入a(单位:万元)满足25,1536,49,3657,a M a ⎧⎪=⎨<⎪⎩1202N a =+.设甲合作社的投入为x (单位:万元),两个合作社的总收益为()f x (单位:万元). (1)若两个合作社的投入相等,求总收益;(2)试问如何安排甲、乙两个合作社的投入,才能使总收益最大?30.(0分)[ID :12230]设全集为R ,集合A ={x |3≤x <7},B ={x |2<x <6},求∁R (A ∪B ),∁R (A ∩B ),(∁R A )∩B ,A ∪(∁R B ).【参考答案】2016-2017年度第*次考试试卷 参考答案**科目模拟测试一、选择题 1.A 2.C 3.C 4.A 5.B 6.C 7.B 8.C 9.D 10.A 11.B 12.A 13.B15.C二、填空题16.1【解析】故答案为17.【解析】【分析】令可得从而将问题转化为和的图象有两个不同交点作出图形可求出答案【详解】由题意令则则和的图象有两个不同交点作出的图象如下图是过点的直线当直线斜率时和的图象有两个交点故答案为:【点睛】本18.3【解析】【分析】根据幂函数的概念列式解得或然后代入解析式看指数的符号负号就符合正号就不符合【详解】因为函数是幂函数所以即所以所以或当时其图象不过原点符合题意;当时其图象经过原点不合题意综上所述:故19.【解析】【分析】可求出时函数值的取值范围再由奇函数性质得出时的范围合并后可得值域【详解】设当时所以所以故当时因为是定义在上的奇函数所以当时故函数的值域是故答案为:【点睛】本题考查指数函数的性质考查函20.【解析】【分析】根据题意以及对数的运算性质得出进而可由基本不等式可得出从而可得出函数的值域【详解】由题意即由题意知由基本不等式得(当且仅当时取等号)所以(当且仅当时取等号)即所以的值域为故答案为:【21.【解析】【分析】根据函数奇偶性和单调性的性质作出的图象利用数形结合进行求解即可【详解】偶函数的图象过点且在区间上单调递减函数的图象过点且在区间上单调递增作出函数的图象大致如图:则不等式等价为或即或即22.【解析】根据题意当时为奇函数则故答案为23.6【解析】【分析】利用定义证明函数的奇偶性以及单调性结合题设条件列出方程组求解即可【详解】则函数在R上为奇函数设即结合奇函数的性质得函数在R上为减函数并且由题意可知:由于函数在R上封闭故有解得:所以24.【解析】【分析】由已知可构造有两不同实数根利用二次方程解出的范围即可【详解】为增函数且时函数的值域也为相当于方程有两不同实数根有两不同实根即有两解整理得:令有两个不同的正数根只需即可解得故答案为:【25.或【解析】【分析】分类讨论的范围利用对数函数二次函数的性质进一步求出的范围【详解】解:∵函数若有最大值或最小值则函数有最大值或最小值且取最值时当时由于没有最值故也没有最值不满足题意当时函数有最小值没三、解答题26.28. 29. 30.2016-2017年度第*次考试试卷 参考解析【参考解析】**科目模拟测试一、选择题 1.A 解析:A 【解析】∵f(x+4)=f(x),∴f(x)是以4为周期的周期函数,∴f(2 019)=f(504×4+3)=f(3)=f(-1).又f(x)为奇函数,∴f(-1)=-f(1)=-2×12=-2,即f(2 019)=-2. 故选A2.C解析:C 【解析】 【分析】利用指数函数2xy =与对数函数3log y x =的性质即可比较a ,b ,c 的大小. 【详解】1.30.7 1.4382242c log a b =<<===<,c a b ∴<<. 故选:C . 【点睛】本题考查了指数函数与对数函数的单调性,考查了推理能力与计算能力,属于基础题.3.C解析:C【分析】利用偶函数的性质将不等式()()lg 1f x f <-变形为()()lg 1f x f <,再由函数()y f x =在[)0,+∞上的单调性得出lg 1x <,利用绝对值不等式的解法和对数函数的单调性即可求出结果. 【详解】由于函数()y f x =是偶函数,由()()lg 1f x f <-得()()lg 1f x f <, 又函数()y f x =在[)0,+∞上是增函数,则lg 1x <,即1lg 1x -<<,解得11010x <<. 故选:C. 【点睛】本题考查利用函数的单调性和奇偶性解不等式,同时也涉及了对数函数单调性的应用,考查分析问题和解决问题的能力,属于中等题.4.A解析:A 【解析】 【分析】利用指数函数、对数函数的单调性直接比较. 【详解】 解:0.1x 1.11.11=>=, 1.100y 0.90.91<=<=,22334z log log 103=<<,x ∴,y ,z 的大小关系为x y z >>. 故选A . 【点睛】本题考查三个数的大小的比较,利用指数函数、对数函数的单调性等基础知识,考查运算求解能力,是基础题.5.B解析:B 【解析】 由f(1)=得a 2=, ∴a=或a=-(舍), 即f(x)=(.由于y=|2x-4|在(-∞,2]上单调递减,在[2,+∞)上单调递增,所以f(x)在(-∞,2]上单调递增,在[2,+∞)上单调递减,故选B.6.C解析:C【分析】首先将b 表示为对数的形式,判断出0b <,然后利用中间值以及对数、指数函数的单调性比较32与,a c 的大小,即可得到,,a b c 的大小关系. 【详解】因为154b=,所以551log log 104b =<=,又因为(133331log log 4log 3,log 4a ==∈,所以31,2a ⎛⎫∈ ⎪⎝⎭, 又因为131133336,82c ⎛⎫⎛⎫⎛⎫ ⎪=∈ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭ ⎪⎝⎭,所以3,22c ⎛⎫∈ ⎪⎝⎭, 所以c a b >>. 故选:C. 【点睛】本题考查利用指、对数函数的单调性比较大小,难度一般.利用指、对数函数的单调性比较大小时,注意数值的正负,对于同为正或者负的情况可利用中间值进行比较.7.B解析:B 【解析】 【分析】本题为选择压轴题,考查函数平移伸缩,恒成立问题,需准确求出函数每一段解析式,分析出临界点位置,精准运算得到解决. 【详解】(0,1]x ∈时,()=(1)f x x x -,(+1)= ()f x 2f x ,()2(1)f x f x ∴=-,即()f x 右移1个单位,图像变为原来的2倍.如图所示:当23x <≤时,()=4(2)=4(2)(3)f x f x x x ---,令84(2)(3)9x x --=-,整理得:2945560x x -+=,1278(37)(38)0,,33x x x x ∴--=∴==(舍),(,]x m ∴∈-∞时,8()9f x ≥-成立,即73m ≤,7,3m ⎛⎤∴∈-∞ ⎥⎝⎦,故选B .【点睛】易错警示:图像解析式求解过程容易求反,画错示意图,画成向左侧扩大到2倍,导致题目出错,需加深对抽象函数表达式的理解,平时应加强这方面练习,提高抽象概括、数学建模能力.8.C解析:C 【解析】 【分析】 由题意,函数()()3y ff x =-的零点个数,即方程()()3f f x =的实数根个数,设()t f x =,则()3f t =,作出()f x 的图象,结合图象可知,方程()3f t =有三个实根,进而可得答案. 【详解】 由题意,函数()()3y ff x =-的零点个数,即方程()()3f f x =的实数根个数,设()t f x =,则()3f t =,作出()f x 的图象,如图所示,结合图象可知,方程()3f t =有三个实根11t =-,214t =,34t =, 则()1f x =- 有一个解,()14f x =有一个解,()4f x =有三个解, 故方程()()3ff x =有5个解.【点睛】本题主要考查了函数与方程的综合应用,其中解答中合理利用换元法,结合图象,求得方程()3f t =的根,进而求得方程的零点个数是解答的关键,着重考查了分析问题和解答问题的能力,以及数形结合思想的应用.9.D解析:D 【解析】试题分析:设36180310M x N == ,两边取对数,36136180803lg lg lg3lg10361lg38093.2810x ==-=⨯-=,所以93.2810x =,即M N 最接近9310,故选D.【名师点睛】本题考查了转化与化归能力,本题以实际问题的形式给出,但本质就是对数的运算关系,以及指数与对数运算的关系,难点是令36180310x =,并想到两边同时取对数进行求解,对数运算公式包含log log log a a a M N MN +=,log log log a a aM M N N-=,log log n a a M n M =.10.A解析:A 【解析】 【分析】由已知可知,()f x 在()1,-+∞上单调递减,结合二次函数的开口方向及对称轴的位置即可求解. 【详解】∵二次函数()24f x ax x =-+对任意的()12,1,x x ∈-+∞,且12x x ≠,都有()()12120f x f x x x -<-,∴()f x 在()1,-+∞上单调递减, ∵对称轴12x a=, ∴0112a a<⎧⎪⎨≤-⎪⎩,解可得102a -≤<,故选A . 【点睛】本题主要考查了二次函数的性质及函数单调性的定义的简单应用,解题中要注意已知不等式与单调性相互关系的转化,属于中档题.11.B解析:B 【解析】 【分析】当07x <≤时,()f x 为单调增函数,且(2)0f =,则()0f x >的解集为(]2,7,再结合()f x 为奇函数,所以不等式()0f x >的解集为(2,0)(2,7]-⋃.【详解】当07x <≤时,()26xf x x =+-,所以()f x 在(0,7]上单调递增,因为2(2)2260f =+-=,所以当07x <≤时,()0f x >等价于()(2)f x f >,即27x <≤,因为()f x 是定义在[7,7]-上的奇函数,所以70x -≤< 时,()f x 在[7,0)-上单调递增,且(2)(2)0f f -=-=,所以()0f x >等价于()(2)f x f >-,即20x -<<,所以不等式()0f x >的解集为(2,0)(2,7]-⋃ 【点睛】本题考查函数的奇偶性,单调性及不等式的解法,属基础题.应注意奇函数在其对称的区间上单调性相同,偶函数在其对称的区间上单调性相反.12.A解析:A 【解析】因为00.31,1e <,所以0.3log 0c e =<,由于0.30.3031,130log 31a b ππ>⇒=><<⇒<=<,所以a b c >>,应选答案A .13.B解析:B 【解析】 【分析】根据零点存在定理判断023x <<,从而可得结果. 【详解】 因为()2ln f x x x=-在定义域内递增, 且()2ln 210f =-<,()23ln 303f =->, 由零点存在性定理可得023x <<,根据[]x 表示不超过实数x 的最大整数可知()02g x =, 故选:B. 【点睛】本题主要考查零点存在定理的应用,属于简单题.应用零点存在定理解题时,要注意两点:(1)函数是否为单调函数;(2)函数是否连续.14.C解析:C 【解析】试题分析:根据补集的运算得{}{}{}{}2,4,6,()2,4,61,2,41,2,4,6UP UP Q =∴⋃=⋃=.故选C.【考点】补集的运算.【易错点睛】解本题时要看清楚是求“⋂”还是求“⋃”,否则很容易出现错误;一定要注意集合中元素的互异性,防止出现错误.15.C解析:C 【解析】 【分析】由()()2g x f x =-是奇函数,可得()f x 的图像关于()2,0-中心对称,再由已知可得函数()f x 的三个零点为-4,-2,0,画出()f x 的大致形状,数形结合得出答案. 【详解】由()()2g x f x =-是把函数()f x 向右平移2个单位得到的,且()()200g g ==,()()()4220f g g -=-=-=,()()200f g -==,画出()f x 的大致形状结合函数的图像可知,当4x ≤-或2x ≥-时,()0xf x ≤,故选C. 【点睛】本题主要考查了函数性质的应用,作出函数简图,考查了学生数形结合的能力,属于中档题.二、填空题16.1【解析】故答案为 解析:1 【解析】155325a b c ===因为,1553log 25,log 25,log 25a b c ∴===,252525111log 15log 5log 3a b c∴+-=+-25log 251==,故答案为1. 17.【解析】【分析】令可得从而将问题转化为和的图象有两个不同交点作出图形可求出答案【详解】由题意令则则和的图象有两个不同交点作出的图象如下图是过点的直线当直线斜率时和的图象有两个交点故答案为:【点睛】本 解析:0,1【解析】 【分析】 令0f x,可得1mx x =-,从而将问题转化为y mx =和1y x =-的图象有两个不同交点,作出图形,可求出答案. 【详解】由题意,令()10f x mx x =--=,则1mx x =-, 则y mx =和1y x =-的图象有两个不同交点, 作出1y x =-的图象,如下图,y mx =是过点()0,0O 的直线,当直线斜率()0,1m ∈时,y mx =和1y x =-的图象有两个交点. 故答案为:0,1.【点睛】本题考查函数零点问题,考查函数图象的应用,考查学生的计算求解能力,属于中档题.18.3【解析】【分析】根据幂函数的概念列式解得或然后代入解析式看指数的符号负号就符合正号就不符合【详解】因为函数是幂函数所以即所以所以或当时其图象不过原点符合题意;当时其图象经过原点不合题意综上所述:故解析:3 【解析】 【分析】根据幂函数的概念列式解得3m =,或6m =,然后代入解析式,看指数的符号,负号就符合,正号就不符合. 【详解】因为函数()22279919mm y m m x--=-+是幂函数,所以29191m m -+=,即29180m m -+=,所以(3)(6)0m m --=, 所以3m =或6m =-, 当3m =时,12()f x x-=,其图象不过原点,符合题意;当5m =时,21()f x x =,其图象经过原点,不合题意. 综上所述:3m =. 故答案为:3 【点睛】本题考查了幂函数的概念和性质,属于基础题.19.【解析】【分析】可求出时函数值的取值范围再由奇函数性质得出时的范围合并后可得值域【详解】设当时所以所以故当时因为是定义在上的奇函数所以当时故函数的值域是故答案为:【点睛】本题考查指数函数的性质考查函解析:11,44⎡⎤-⎢⎥⎣⎦【解析】 【分析】可求出0x ≥时函数值的取值范围,再由奇函数性质得出0x ≤时的范围,合并后可得值域. 【详解】设12x t =,当0x ≥时,21x ≥,所以01t <≤,221124y t t t ⎛⎫=-+=--+ ⎪⎝⎭, 所以104y ≤≤,故当0x ≥时,()10,4f x ⎡⎤∈⎢⎥⎣⎦. 因为()y f x =是定义在R 上的奇函数,所以当0x <时,()1,04f x ⎡⎫∈-⎪⎢⎣⎭,故函数()f x 的值域是11,44⎡⎤-⎢⎥⎣⎦.故答案为:11,44⎡⎤-⎢⎥⎣⎦.【点睛】本题考查指数函数的性质,考查函数的奇偶性,求奇函数的值域,可只求出0x ≥时的函数值范围,再由对称性得出0x ≤时的范围,然后求并集即可.20.【解析】【分析】根据题意以及对数的运算性质得出进而可由基本不等式可得出从而可得出函数的值域【详解】由题意即由题意知由基本不等式得(当且仅当时取等号)所以(当且仅当时取等号)即所以的值域为故答案为:【 解析:[)2,+∞【解析】【分析】根据题意以及对数的运算性质得出()21log 2F x x x ⎛⎫=++ ⎪⎝⎭,进而可由基本不等式可得出124x x ++≥,从而可得出函数()F x 的值域. 【详解】由题意,()()()()22212log 1log F x f x f x x x =+-=+-,即()222211log log 2x x F x x x x ++⎛⎫==++⎪⎝⎭,由题意知,0x >,由基本不等式得12x x +≥=(当且仅当1x =时取等号), 所以124x x ++≥(当且仅当1x =时取等号),即221log 2log 42x x ⎛⎫++≥= ⎪⎝⎭,所以()F x 的值域为[)2,+∞. 故答案为:[)2,+∞. 【点睛】本题考查了函数值域的定义及求法,对数的运算性质,基本不等式的运用,考查了计算能力,属于基础题.21.【解析】【分析】根据函数奇偶性和单调性的性质作出的图象利用数形结合进行求解即可【详解】偶函数的图象过点且在区间上单调递减函数的图象过点且在区间上单调递增作出函数的图象大致如图:则不等式等价为或即或即 解析:()(),20,2-∞-⋃【解析】 【分析】根据函数奇偶性和单调性的性质作出()f x 的图象,利用数形结合进行求解即可. 【详解】偶函数()f x 的图象过点()2,0P ,且在区间[)0,+∞上单调递减,∴函数()f x 的图象过点()2,0-,且在区间(),0-∞上单调递增,作出函数()f x 的图象大致如图:则不等式()0xf x >等价为()00x f x >⎧>⎨⎩或()00x f x <⎧<⎨⎩,即02x <<或2x <-,即不等式的解集为()(),20,2-∞-⋃, 故答案为()(),20,2-∞-⋃ 【点睛】本题主要考查不等式的解集的计算,根据函数奇偶性和单调性的性质作出()f x 的图象是解决本题的关键.22.【解析】根据题意当时为奇函数则故答案为 解析:15-【解析】根据题意,当0x <时,()()(),f x g x f x =为奇函数,()()()()()()()()()211113(323)15f g f f f f f f f -=-=-=-=-=-+⨯=-,则故答案为15-.23.6【解析】【分析】利用定义证明函数的奇偶性以及单调性结合题设条件列出方程组求解即可【详解】则函数在R 上为奇函数设即结合奇函数的性质得函数在R 上为减函数并且由题意可知:由于函数在R 上封闭故有解得:所以解析:6 【解析】 【分析】利用定义证明函数()y f x =的奇偶性以及单调性,结合题设条件,列出方程组,求解即可. 【详解】44()()11x xf x f x x x--=-==-+-+,则函数()f x 在R 上为奇函数设120x x ≤<,4()1xf x x=-+()()()2112121212444()()01111x x x x f x f x x x x x --=-+=>++++,即12()()f x f x > 结合奇函数的性质得函数()f x 在R 上为减函数,并且(0)0f = 由题意可知:0,0a b <>由于函数()f x 在R 上封闭,故有4141()()a bab f a b f b aa b-=-⎧⎪=⎧⎪⇒⎨⎨=⎩-=+⎪⎪⎩ ,解得:3,3a b =-=所以6b a -= 故答案为:6 【点睛】本题主要考查了利用定义证明函数的奇偶性以及单调性,属于中档题.24.【解析】【分析】由已知可构造有两不同实数根利用二次方程解出的范围即可【详解】为增函数且时函数的值域也为相当于方程有两不同实数根有两不同实根即有两解整理得:令有两个不同的正数根只需即可解得故答案为:【解析:10,4⎛⎫⎪⎝⎭【解析】 【分析】由已知可构造()2log xa a t x +=有两不同实数根,利用二次方程解出t 的范围即可.【详解】()2()log x a f x a t =+为增函数,且[],x m n ∈时,函数()()2log xa f x at =+的值域也为[],m n ,(),()f m m f n n ∴==,∴相当于方程()f x x =有两不同实数根,()2log x a a t x ∴+=有两不同实根,即2x x a a t =+有两解, 整理得:20x x a a t -+=, 令,0xm a m => ,20m m t ∴-+=有两个不同的正数根, ∴只需1400t t ∆=->⎧⎨>⎩即可,解得104t <<,故答案为:10,4⎛⎫ ⎪⎝⎭【点睛】本题主要考查了对数函数的单调性,对数方程,一元二次方程有两正根,属于中档题.25.或【解析】【分析】分类讨论的范围利用对数函数二次函数的性质进一步求出的范围【详解】解:∵函数若有最大值或最小值则函数有最大值或最小值且取最值时当时由于没有最值故也没有最值不满足题意当时函数有最小值没解析:{|2m m >或2}3m <- 【解析】 【分析】分类讨论m 的范围,利用对数函数、二次函数的性质,进一步求出m 的范围. 【详解】解:∵函数()()212log 22f x mx m x m ⎡⎤=+-+-⎣⎦,若()f x 有最大值或最小值,则函数2(2)2y mx m x m =+-+-有最大值或最小值,且y 取最值时,0y >.当0m =时,22y x =--,由于y 没有最值,故()f x 也没有最值,不满足题意. 当0m >时,函数y 有最小值,没有最大值,()f x 有最大值,没有最小值.故y 的最小值为24(2)(2)4m m m m ---,且 24(2)(2)04m m m m--->,求得 2m >;当0m <时,函数y 有最大值,没有最小值,()f x 有最小值,没有最大值.故y 的最大值为24(2)(2)4m m m m ---,且 24(2)(2)04m m m m--->,求得23m <-. 综上,m 的取值范围为{|2m m >或2}3m <-. 故答案为:{|2m m >或2}3m <-. 【点睛】本题主要考查复合函数的单调性,二次函数、对数函数的性质,二次函数的最值,属于中档题.三、解答题 26.(1)2()3318f x x x =--+(2)[12,18]【解析】 【分析】 【详解】 (1)832,323,5b a aba b a a----+=--⨯=∴=-= ,()23318f x x x =--+ (2)因为()23318f x x x =--+开口向下,对称轴12x =- ,在[]0,1单调递减,所以()()max min 0,18,1,12x f x x f x ====当当 所以函数()f x 的值域为[12,18] 【点睛】本题将函数的零点、解析式、最大小值等有关知识与性质有机整合在一起,旨在考查函数的表示、零点、最大小值等基础知识及综合运用.求解时先依据函数零点与方程的根之间的关系,求出函数解析式中的参数的值;解答第二问时,借助二次函数的图像和性质,运用数形结合的数学思想求出最大小值从而使得问题获解.27.(Ⅰ){}1(Ⅱ)13a -<<-【解析】 【分析】(Ⅰ)将1a =代入直接求解即可;(Ⅱ)设2x t =,得到()()2110t a t a +-++=在()0,+∞有两个不同的解,利用二次函数的性质列不等式组求解即可. 【详解】(Ⅰ)当1a =时,()()2log 4223xxf x =++=,所以34222x x ++=, 所以4260x x +-=,因此()()23220xx+-=,得22x = 解得1x =, 所以解集为{}1.(Ⅱ)因为方程()2log 421x xa a x +⋅++=有两个不同的实数根, 即4212x x x a a +⋅++=,设2x t =,()()2110t a t a +-++=在()0,+∞有两个不同的解,令()()()211f t t a t a =+-++,由已知可得()()()2001021410f a a a ⎧>⎪-⎪->⎨⎪⎪=--+>⎩解得13a -<<- 【点睛】本题主要考查了对数函数与指数函数的复合函数的处理方式,考查了函数与方程的思想,属于中档题.28.(1)2()(1)f x x =+;(2)存在,1-. 【解析】 【分析】(1)由(3)(1)f f -=,知此二次函数图象的对称轴为1x =-, 由(1)0f -=可设出抛物线的解析式为2()(1)f x a x =+,再利用(1)4f =求得a 的值;(2)利用零点存在定理,证明(0)(1)0h h ⋅<即可得到n 的值. 【详解】(1)由(3)(1)f f -=,知此二次函数图象的对称轴为1x =-, 又因为(1)0f -=,所以(1,0)-是()f x 的顶点, 所以设2()(1)f x a x =+,因为(1)4f =,即2(11)4a +=,所以设1a = 所以2()(1)f x x =+(2)由(1)知2()(1)ln(||1)h x x x =+-+因为2(1)(11)ln(|1|1)ln(2)0h -=-+--+=-<2(0)(01)ln(|0|1)10h =+-+=>即(0)(1)0h h ⋅<因为函数()()ln(||1)h x f x x =-+在R 上连续不断, 由零点存在性定理,所以函数()h x 在(1,0)-上存在零点. 所以存在1n =-使得函数()h x 在区间(,1)n n +内存在零点. 【点睛】本题考查一元二次函数的解析式、零点存在定理,考查函数与方程思想考查逻辑推理能力和运算求解能力.29.(1)87万元;(2)甲合作社投入16万元,乙合作社投入56万元 【解析】 【分析】(1)先求出36x =,再求总收益;(2)(2)设甲合作社投入x 万元(1557)x ≤≤,乙合作社投入72x -万元,再对x 分类讨论利用函数求出如何安排甲、乙两个合作社的投入,才能使总收益最大.【详解】(1)两个合作社的投入相等,则36x =, 1(36)436253620872f =++⨯+=(万元) (2)设甲合作社投入x 万元(1557)x ≤≤,乙合作社投入72x -万元.当1536x ≤≤时,11()425(72)2048122f x x x x x =++-+=-++, 令t x =,得156t ≤≤,则总收益2211()481(4)8922g t t t t =-++=--+, 当4t =即16x =时,总收益取最大值为89;当3657x <≤时,11()49(72)2010522f x x x =+-+=-+, ()f x 在(36,57]上单调递减,所以()(36)87f x f <=.因为8987>,所以在甲合作社投入16万元,乙合作社投入56万元时,总收益最大,最大总收益为89万元.【点睛】本题主要考查函数的应用和最值的求法,意在考查学生对这些知识的理解掌握水平和应用能力.30.见解析【解析】【分析】根据题意,在数轴上表示出集合,A B ,再根据集合的运算,即可得到求解.【详解】解:如图所示.∴A ∪B ={x |2<x <7},A ∩B ={x |3≤x <6}.∴∁R (A ∪B )={x |x ≤2或x ≥7},∁R (A ∩B )={x |x ≥6或x <3}.又∵∁R A ={x |x <3或x ≥7},∴(∁R A )∩B ={x |2<x <3}.又∵∁R B ={x |x ≤2或x ≥6},∴A∪(∁R B)={x|x≤2或x≥3}.【点睛】本题主要考查了集合的交集、并集与补集的混合运算问题,其中解答中正确在数轴上作出集合,A B,再根据集合的交集、并集和补集的基本运算求解是解答的关键,同时在数轴上画出集合时,要注意集合的端点的虚实,着重考查了数形结合思想的应用,以及推理与运算能力.。

2024年上海市曹杨第二中学高一上学期期末考试数学试卷含答案

2024年上海市曹杨第二中学高一上学期期末考试数学试卷含答案

上海市曹杨二中2023学年度第一学期高一年级期终考试数学试卷一、填空题(本大题共有12题,满分54分,第1~6题每题4分,第7~12题每题5分)1.已知全集{}1,2,3,4,5U =,集合{}13,5A =,,则A =____________.2.函数()f x =的定义域为_________.3.函数21xy =-的反函数为____________.4.已知扇形的弧长为4cm ,面积为24cm ,则该扇形的圆心角的大小为___________.5.已知正数a 、b 满足a +b =1,则a ·b 的最大值为_____.6.已知πsin sin8x =,且π3π,22x ⎛⎫∈ ⎪⎝⎭,则x =___________.7.已知lg 2a =,103b=,则24log 5可以用a 、b 表示为_________.8.已知a ∈R ,()y f x =是定义在R 上的偶函数,且当0x <时,()3axf x =.若()3log 24f =,则=a _________.9.已知a ∈R ,若函数()3312,1,1a x a x y x x ⎧-+>=⎨≤⎩的值域为R ,则a 的取值范围是________.10.对于实数x ,用[]x 表示不超过x 的最大整数,并记{}[]x x x =-,例如{}10=,{}1.230.23=.则关于x 的方程{}10x x ⋅=在区间[]0,2024上解的个数为_________.11.已知函数()f x 是定义在R 上的奇函数,且(1)0f -=,若对任意的()12,,0x x ∈-∞,当12x x ≠时,都有112212()()x f x x f x x x ⋅-⋅<-成立,则不等式()0f x <的解集为_____.12.已知b ∈R ,设函数()2log 2f x x x b=++在区间[](),10t t t +>上的最大值为()t M b .若(){}2tb M b ≥=R ,则正实数t 的最大值为_________.二、选择题(本大题共有4题,满分18分,第13~14题每题4分,第15~16题每题5分.13.若0a b <<,则下列不等式中不成立的是()A.11a b >; B.11a b a>-;C.a b >;D.22a b >.14.已知0a >且1a ≠,则“2a >”是“函数()2log a y a x =-是严格增函数”的().A.充分非必要条件B.必要非充分条件C .充要条件D.既非充分条件又非必要条件15.设方程2|lg |x x -=的两个根为12,x x ,则A.120x x < B.121=x x C.121x x > D.1201x x <<16.已知函数()y f x =满足()()111f x f x +=+,且当[]0,1x ∈时,()f x x =.若在区间(]1,1-上关于x 的方程()0f x mx m --=有且仅有一解,则实数m 的取值范围是().A.1,2∞⎛⎫+ ⎪⎝⎭B.1,2⎡⎫+∞⎪⎢⎣⎭C.10,2⎡⎤⎢⎥⎣⎦D.{}10,2∞⎛⎫⋃+ ⎪⎝⎭三、解答题(本大题共有5题,满分78分)17.已知1a ≤,设集合111A x x ⎧⎫=>⎨⎬-⎩⎭,集合()(){}210B x x a x a =--->.(1)分别求集合A 和B ;(2)若A B A = ,求a 的取值范围.18.(1)已知m ∈R ,若sin α、cos α是关于x 的一元二次方程()210x mx m -++=的两实根,求m 的值;(2)已知()0,πα∈,且1sin cos 3αα-=,求sin cos αα及()11πcos 2πcos 2αα++⎛⎫- ⎪⎝⎭的值.19.某机构为了研究某种药物对某种疾病的治疗效果,准备利用小白鼠进行试验,研究发现,药物在血液内的浓度与时间的关系因使用方式的不同而不同:若使用注射方式给药,则在注射后的4小时内,药物在白鼠血液内的浓度1y (单位:毫克/升)与时间t (单位:小时)满足关系式16y at =-(0a >,a 为常数);若使用口服方式给药,则药物在白鼠血液内的浓度2y (单位:毫克/升)与时间t (单位:小时)满足关系式22014614t t y t t ≤<⎧⎪=⎨-≤≤⎪⎩,,,现对小白鼠同时进行注射和口服该种药物,同时使用两种方式给药后,小白鼠血液中药物的浓度等于单独使用每种方式给药的浓度之和.(1)若1a =,求4小时内,该小白鼠何时血液中药物的浓度最高,并求出最大值;(2)若小白鼠在用药后4小时内血液中的药物浓度都不低于6毫克/升,求正数a 的取值范围.20.已知k ∈R ,设()()14122x xf x k k k =⋅+-⋅++.(1)若0k =,求函数()y f x =的值域;(2)已知0k <,若函数()y f x =的最大值为12,求k 的值;(3)已知01k <<,若存在两个不同的正实数m 、n ,使得当函数()y f x =的定义域为[],m n 时,其值域为1122m n ++⎡⎤⎣⎦,,求k 的取值范围.21.已知函数()y f x =的定义域为D .若存在实数a ,使得对于任意1x D ∈,都存在2x D ∈,使得()12x f x a +=,则称函数()y f x =具有性质()P a .(1)分别判断:2x y =及21y x =+是否具有性质()0P ;(结论不需要证明)(2)若函数()y f x =的定义域为D ,且具有性质()1P ,证明:“1D ∈”是“函数()y f x =存在零点”的充分非必要条件;(3)已知t ∈R ,设()22g x tx x =+,若存在唯一的实数a ,使得函数()y g x =,[]0,2x ∈具有性质()P a ,求t 的值.上海市曹杨二中2023学年度第一学期高一年级期终考试数学试卷一、填空题(本大题共有12题,满分54分,第1~6题每题4分,第7~12题每题5分)【1题答案】{}2,4【2题答案】【3题答案】()()2log 11y x x =+>-【4题答案】2【5题答案】14【6题答案】7π8【7题答案】13a a b-+【8题答案】2-【9题答案】1235a <≤【10题答案】2014【11题答案】()()101-∞-⋃,,;【12题答案】13二、选择题(本大题共有4题,满分18分,第13~14题每题4分,第15~16题每题5分.【13题答案】B 【14题答案】A 【15题答案】D 【16题答案】D三、解答题(本大题共有5题,满分78分)【17题答案】(1){}01A x x =<<,{1B x x a =>+或}2x a <(2)(]1,1,12⎡⎤-∞-⎢⎥⎣⎦【18题答案】(1)1m =-;(2)49;4【19题答案】(1)当2t =时血液中药物的浓度最高,最大值为8(2)50,4⎛⎤ ⎥⎝⎦【20题答案】(1)1,2⎛⎫+∞⎪⎝⎭(2)1k =-(3)13,23⎛⎫⎪⎪⎝⎭【21题答案】(1)2x y =不具有性质()0P ,21y x =+具有性质()0P (3)12-或354--。

2018-2019学年上海市曹杨二中高一上学期期末数学试题(解析版)

2018-2019学年上海市曹杨二中高一上学期期末数学试题(解析版)

2018-2019学年上海市曹杨二中高一上学期期末数学试题一、单选题1.如果,a b c d >>,则下列不等式成立的是( ) A.a c b d ->- B.a c b d +>+C.a b d c> D.ac bd >【答案】B【解析】根据不等式的性质,分别将各个选项分析求解即可。

【详解】A 项,当54,31a b c d =>==>=时,2,3a c b d -=-=,则a c b d -<-,故A 项不一定成立;因为,a b c d >>,两式相加得a c b d +>+,故B 项一定成立; 当21,11a b c d =>==>=-时,2,1a bd c =-=,则a b d c<,故C 项不一定成立; D 项,当12,34a b c d =->=-=->=-时,3,8ac bd ==,则ac bd <,故D 项不一定成立; 故选:B 【点睛】本题主要考查不等式的性质,此题比较简单,需掌握不等式的性质,注意排除法在解选择题中的应用。

2.唐代诗人杜牧的七绝唐诗中的两句诗为“今来海上升高望,不到蓬莱不成仙。

”其中后一句“成仙”是“到蓬莱”的( ) A.充分非必要条件 B.必要不充分条件C.充要条件D.既不充分又不必要条件 【答案】A【解析】根据命题的“真、假”,条件与结论的关系即可得出选项。

【详解】不到蓬莱⇒不成仙,∴成仙⇒到蓬莱,“成仙”是到“到蓬莱”的充分条件,但“到蓬莱”是否“成仙”不确定,因此“成仙”是“到蓬莱”的充分非必要条件。

故选:A 【点睛】充分、必要条件有三种判断方法:1、定义法:直接判断“若p 则q ”和“若q 则p ”的真假。

2、等假法:利用原命题与逆否命题的关系判断。

3、若A B ⊆,则A 是B 的充分条件或B 是A 的必要条件;若A B =,则A 是B 的充要条件。

高中上海市曹杨二中高一上学期期末数学试题

高中上海市曹杨二中高一上学期期末数学试题

上海市曹杨二中【精品】高一上学期期末数学试题学校:___________姓名:___________班级:___________考号:___________一、填空题1.已知集合{|1}A x x =≥,{|}B x x a =≥,若A B ⊆,则实数a 的取值范围是________. 2.若函数()1f x =+()g x =,则()()f x g x +=________. 3.函数()2||f x x ax =+为偶函数,则实数a 的值为________.4.函数()()21f x x x =≤-的反函数是______.5.在直角坐标系xOy 中,终边在坐标轴上的角α的集合是________.6.已知函数22,3()log ,3x x f x x x ⎧≤=⎨>⎩,则((3))f f =________.7.若幂函数22()()mm f x x m --=∈Z 在(0,)+∞是单调减函数,则m 的取值集合是________.8.若不等式||1x m -<成立的充分不必要条件是12x <<,则实数m 的取值范围是________.9.已知等腰三角形的周长为常数l ,底边长为y ,腰长为x ,则函数()y f x =的定义域为________.10.已知角α的终边上一点()P m,且sin α=,则tan α的值为________. 11.已知()y f x =是定义在R 上的奇函数,且当0x 时,11()42x xf x =-+,则此函数的值域为__________. 12.对于函数f (x ),若存在x 0∈R ,使f (x 0)=x 0,则称x 0是f (x )的一个不动点,已知f (x )=x 2+ax +4在[1,3]恒有两个不同的不动点,则实数a 的取值范围______.二、单选题13.若0a <,0b >,则下列不等式恒成立的是( )A .22a b < B<C .11a b < D .2abb a +≥14.函数ln ||y x =与y = )A .B .C .D .15.已知函数1()|lg |2x f x x ⎛⎫=- ⎪⎝⎭有两个零点1x ,2x ,则有( ) A .120x x < B .121=x x C .121x x > D .1201x x << 16.对于函数()f x ,若存在区间[,]A m n =,使得{|(),}y y f x x A A =∈=,则称函数()f x 为“可等域函数”,区间A 为函数的一个“可等域区间”.给出下列四个函数:①()||f x x =;②2()21f x x =-;③()|12|x f x =-;④2()log (22)f x x =-.其中存在唯一“可等域区间”的“可等域函数”的个数是( )A .1B .2C .3D .4三、解答题17.已知一个扇形的周长为定值a ,求其面积的最大值,并求此时圆心角α的大小. 18.若方程2(3)0x m x m +-+=,m ∈R ,在x ∈R 上有两个不相等的实数根,求m 的取值范围.19.设函数()|1|||f x x x a =-+-.(1)若1a =-,解不等式()3f x ≥;(2)若不等式()3f x ≥对一切x ∈R 恒成立,求实数a 的取值范围.20.已知集合M 是具有下列性质的函数()f x 的全体:存在实数对(,)a b ,使得()()f a x f a x b +⋅-=对定义域内任意实数x 都成立.(1)判断函数1()f x x =,2()3x f x =是否属于集合M ;(2)若函数1()1tx f x x-=+具有反函数1()f x -,是否存在相同的实数对(,)a b ,使得()f x 与1()f x -同时属于集合M ?若存在,求出相应的,,a b t ;若不存在,说明理由;(3)若定义域为R 的函数()f x 属于集合M ,且存在满足有序实数对(0,1)和(1,4);当[0,1]x ∈时,()f x 的值域为[1,2],求当[2016,2016]x ∈-时函数()f x 的值域.参考答案1.(,1]-∞【解析】【分析】根据子集的定义和不等式的性质,即可求得答案.【详解】集合{|1}A x x =≥,{|}B x x a =≥,A B ⊆,∴1a ≤.∴实数a 的取值范围是(,1]-∞.故答案为:(,1]-∞.【点睛】本题考查了根据集合的包含关系求解参数,在集合运算比较复杂时,可以使用数轴来辅助分析问题..2.1+01x ≤≤【分析】因为()1f x =()g x =-,故1()()f x g x =++,此时()()f x g x +的定义域,是()f x 和()g x 定义域的交集,即可求得答案.【详解】函数()1f x =()g x =-∴()()(11f x g x +=++-=+此时()()f x g x +的定义域,是()f x 和()g x 定义域的交集∴100x x -≥⎧⎨≥⎩,即01x ≤≤,∴ 1()()f x g x =+,01x ≤≤故答案为:1+01x ≤≤.【点睛】本题考查求解函数解析式,掌握函数定义域的求法是解题关键,考查了计算能力,属于基础题.3.0【分析】根据偶函数的定义,建立方程关系进行求解,即可求得答案.【详解】()2||f x x ax =+为偶函数,∴()()f x f x -=,即2||2||x ax x ax --=+,则0a =,故答案为:0.【点睛】本题主要考查了根据奇偶性求解函数解析式,掌握偶函数定义是解本题关键,考查了计算能力,属于基础题.4.())11f x x -=≥【分析】根据反函数的求法,求得原函数的反函数的解析式并求出定义域.【详解】令()21,1y x x y =≤-≥,解得)1x y =≥,交换,x y 的位置得)1y x =≥,所以函数()()21f x xx =≤-的反函数是())11f x x -=≥.故填:())11f x x -=≥. 【点睛】本小题主要考查反函数的求法,属于基础题.5.|,2n n παα⎧⎫=∈⎨⎬⎩⎭Z 【分析】分别写出终边在x 轴上的角的集合、终边在y 轴上的角的集合,进而可得到终边在坐标轴上的角的集合.【详解】终边在x 轴上的角的集合为{|,}k k αα=π∈Z ,终边在y 轴上的角的集合为|,2k k πααπ⎧⎫=+∈⎨⎬⎩⎭Z , 故终边在坐标轴上的角α的集合是:|,2n n παα⎧⎫=∈⎨⎬⎩⎭Z 故答案为:|,2n n παα⎧⎫=∈⎨⎬⎩⎭Z 【点睛】 本题考查终边相同的角的表示方法,掌握终边相同角的集合写法是解题关键,属于基础题. 6.3【分析】由已知得3(3)28f ==,从而((3))(8)f f f =,由此能求出结果.【详解】函数22,3()log ,3x x f x x x ⎧≤=⎨>⎩, ∴3(3)28f ==,2((3))(8)log 83f f f ===.故答案为:3.【点睛】本题考查了分段函数的求值以及分类讨论思想.求分段函数的函数值时,注意判断自变量的范围,自变量在哪一段的范围内,就选择哪一段的解析式求值,如果自变量不确定在哪一段的范围内,就必须要分类讨论.7.{}0,1【分析】由幂函数()f x 为(0,)+∞上递减,推知220m m --<,解得12m -<<,结合m 为整数,即可求得答案.【详解】幂函数22()()m m f x x m --=∈Z 在区间(0,)+∞上是减函数,∴220m m --<,解得12m -<<,m 为整数,∴0,1m =∴满足条件的m 的值的集合是{0,1},故答案为:{0,1}.【点睛】本题考查根据幂函数的单调性求解析式,掌握幂函数的基础知识是解题关键,考查了分析能力和计算能力,属于基础题.8.[1,2]【分析】根据不等式的性质,以及充分条件和必要条件的定义即可得到结论.【详解】由||1x m -<得11m x m -≤≤+,12x <<是不等式||1x m -<成立的充分不必要条件,∴满足1112m m -≤⎧⎨+≥⎩,且等号不能同时取得, 即21m m ≤⎧⎨≥⎩,解得12m ≤≤, 故答案为:[1,2].【点睛】本题主要考查充分条件和必要条件的应用,根据不等式之间的关系是解决本题的关键. 9.,42l l ⎛⎫ ⎪⎝⎭根据周长得出x 、y 、l 三者的关系,再根据三角形的三边大小关系及不等式的性质即可得出.【详解】由题意得:2y x l +=,20x y >>,解得:42l l x <<, 故答案为:,42l l ⎛⎫⎪⎝⎭. 【点睛】 熟练不等式的基本性质和三角形的三边大小关系是解题的关键,考查了计算能力,属于基础题.10.3±或0 【分析】利用正弦函数的定义求出m ,利用正切函数的定义求出tan α的值.【详解】角α的终边上一点()P m根据正弦函数的定义得:sin 4m α==解得0m =或m =当0m =时,tan 0α=;当m =, tan α=当m =, tan α=则tan α的值为:3±或0故答案为: 3±或0.本题考查三角函数的定义,掌握三角函数的定义是解本题关键,考查学生的计算能力,是基础题.11.11,44⎡⎤-⎢⎥⎣⎦ 【分析】可求出0x ≥时函数值的取值范围,再由奇函数性质得出0x ≤时的范围,合并后可得值域.【详解】 设12x t =,当0x ≥时,21x ≥,所以01t <≤,221124y t t t ⎛⎫=-+=--+ ⎪⎝⎭, 所以104y ≤≤,故当0x ≥时,()10,4f x ⎡⎤∈⎢⎥⎣⎦. 因为()y f x =是定义在R 上的奇函数,所以当0x <时,()1,04f x ⎡⎫∈-⎪⎢⎣⎭,故函数()f x 的值域是11,44⎡⎤-⎢⎥⎣⎦. 故答案为:11,44⎡⎤-⎢⎥⎣⎦. 【点睛】本题考查指数函数的性质,考查函数的奇偶性,求奇函数的值域,可只求出0x ≥时的函数值范围,再由对称性得出0x ≤时的范围,然后求并集即可.12.10,33⎡⎫--⎪⎢⎣⎭【分析】不动点实际上就是方程f (x 0)=x 0的实数根,二次函数f (x )=x 2+ax +4有不动点,是指方程x =x 2+ax +4有实根,即方程x =x 2+ax +4有两个不同实根,然后根据根列出不等式解答即可.【详解】解:根据题意,f (x )=x 2+ax +4在[1,3]恒有两个不同的不动点,得x =x 2+ax +4在[1,3]有两个实数根,即x 2+(a ﹣1)x +4=0在[1,3]有两个不同实数根,令g (x )=x 2+(a ﹣1)x +4在[1,3]有两个不同交点,∴2(1)0(3)01132(1)160g g a a ≥⎧⎪≥⎪⎪⎨-<<⎪⎪-->⎪⎩,即24031001132(1)160a a a a +≥⎧⎪+≥⎪⎪⎨-<<⎪⎪-->⎪⎩, 解得:a ∈10,33⎡⎫--⎪⎢⎣⎭; 故答案为:10,33⎡⎫--⎪⎢⎣⎭. 【点睛】本题考查了二次函数图象上点的坐标特征、函数与方程的综合运用,属于中档题. 13.C【解析】【分析】已知0a <,0b >,根据不等式的基本性质,逐项检验,即可求得答案.【详解】对于A ,因为0a <,0b >,可取3a =-,1b =,则22a b >.故A 错误;对于B , 因为0a <,0b >,可取9a =-,1b =,>故B 错误;对于C ,若0a <,则10a <,而0b >,则10b >,故11a b<,故C 正确; 对于D ,若0a <,0b >,故0a b <,0b a <,则有0a b b a +<,故D 错误; 故选C.【点睛】本题考查不等式的性质,关键是熟悉不等式的性质,对于不成立的不等式,可以举出反例,进行判断.14.C【解析】【分析】根据函数ln ||y x =是偶函数,且在(0,)+∞上单调递增,排除A 、B ;再根据y =示一个半圆(圆位于x 轴下方的部分),可得结论.【详解】由于函数ln ||y x =是偶函数,且在(0,)+∞上单调递增,故排除A 、B ;由于y =即221(0)y x y +=<,表示一个半圆(圆位于x 轴下方的部分), 故选:C.【点睛】本题主要考查函数的图像特征,掌握函数基础知识是解题关键,属于基础题.15.D【解析】【分析】 先将1()|lg |2x f x x ⎛⎫=- ⎪⎝⎭有两个零点转化为|lg |y x =与2x y -=有两个交点,然后在同一坐标系中画出两函数的图像得到零点在(0,1)和(1,)+∞内,即可得到112lg x x --=和222lg x x -=,然后两式相加即可求得12x x 的范围.【详解】1()|lg |2xf x x ⎛⎫=- ⎪⎝⎭有两个零点1x ,2x ,即|lg |y x =与2x y -=有两个交点 由题意0x >,分别画2x y -=和|lg |y x =的图像∴ 发现在(0,1)和(1,)+∞有两个交点不妨设1x 在(0,1)内,2x 在(1,)+∞内,∴ 在(0,1)上有112lg x x -=-,即112lg x x --=——①在(1,)+∞有222lg x x -=——② ①②相加有211222lg x x x x ---= 21x x >,∴2122x x --<即21220x x ---<∴12lg 0x x <∴1201x x <<故选:D .【点睛】本题主要考查确定函数零点所在区间的方法,转化为两个函数的交点问题.函数的零点等价于函数与x 轴的交点的横坐标,等价于对应方程的根.16.B【解析】【分析】根据存在区间[,]A m n =,使得{|(),}y y f x x A A =∈=,则称函数()f x 为“可等域函数”,区间A 为函数的一个“可等域区间”,对四个函数逐一判断,即可得到答案.【详解】在①中,如在区间(0,)+∞、(1,2)都是()||f x x =的可等域区间,故①不合题意;在②中,2()211f x x =-≥-,且()f x 在0x ≤时递减,在0x ≥时递增,若0[,]m n ∈,则1[,]m n -∈,于是1m =-,又()11f -=,(0)1f =-,而(1)1f =,故1n =,[1,1]-是一个可等域区间;若0n ≤,则222121n m m n ⎧-=⎨-=⎩,解得m =,0n =>,不合题意, 若0m ≥,则221x x -=有两个非负解,但此方程的两解为1和12-,也不合题意, 故函数2()21f x x =-只有一个等可域区间[1,1]-,故②成立;在③中,函数()|12|x f x =-的值域是[0,)+∞,所以0m ≥,函数()|12|xf x =-在[0,)+∞上是增函数,考察方程21x x -=,由于函数2x y =与1y x =+只有两个交点(0,1),(1,2),即方程21x x -=只有两个解0和1,因此此函数只有一个等可域区间[0,1],故③成立;在④中,函数2()log (22)f x x =-在定义域(1,)+∞上是增函数,若函数有2()log (22)f x x =-等可域区间[,]m n ,则()f m m =,()f n n =,但方程2log (22)x x -=无解(方程2log x x =无解),故此函数无可等域区间,故④不成立. 综上只有②③正确.故选:B.【点睛】本题考查了函数的新定义.解题关键是理解所给的函数新定义:“可等域区间”的“可等域函数”,考查了分析能力和计算能力,属于中等题. 17.2α=时,扇形面积最大为2a 16. 【分析】设扇形面积为S ,半径为r ,圆心角为α,则扇形弧长为2a r -,,1(2)2S a r r =-,结合二次函数的图像与性质求解最值即可.【详解】设扇形面积为S ,半径为r ,圆心角为α,则扇形弧长为2a r -, 所以221(2)2416a a S a r r r ⎛⎫=-=--+ ⎪⎝⎭. 故当4a r =且2α=时,扇形面积最大为2a 16. 【点睛】本题重点考查了扇形的面积公式、弧长公式、二次函数的最值等知识,属于基础题. 18.1m <,或9m >.【分析】根据二次函数的性质求出m 的范围,即可求得答案.【详解】若方程2(3)0x m x m +-+=,m ∈R ,在x ∈R 上有两个不相等的实数根,则2(3)40m m ∆=-->,解得:1m <,或9m >.【点睛】本题考查了二次函数的性质,根据判别式求出m 的范围即可.19.(1)35,,22⎛⎤⎡⎫-∞-⋃+∞ ⎪⎥⎢⎝⎦⎣⎭(2)(,2][4,)-∞-⋃+∞ 【分析】(1)利用1a =-,化简不等式,通过分类讨论取得绝对值求解即可.(2)利用函数恒成立,转化求解即可.【详解】(1)当1a =-时,不等式()3f x ≥,即|1||1|3x x -++≥,①当1x ≥时,不等式即115x x -++≥,解得52x ≥; ②当11x -<<时,不等式即115x x ---≥,无解;③当1x ≤-时,不等式即113x x ---≥,解得32x ≤-;综上,不等式()5f x ≥的解集为35,,22⎛⎤⎡⎫-∞-⋃+∞ ⎪⎥⎢⎝⎦⎣⎭. (2)()|1||||(1)()||1|f x x x a x x a a =-+-≥---=-,∴min ()|1|f x a =-.()3f x ≥对任意x ∈R 恒成立,∴|1|3a -≥,解得2a ≤-或4a ≥,即实数a 的取值范围为(,2][4,)-∞-⋃+∞.【点睛】本题考查函数恒成立绝对值不等式的解法,考查分类讨论思想以及转化思想的应用,考查计算能力.20.(1)()x 23f x M =∈(2)不存在实数对(,)a b ,使得()f x 与1()f x -同时属于集合M .见解析(3)201620162,2-⎡⎤⎣⎦【分析】 (1)根据已知中集合M 的定义,分别判断两个函数是否满足条件,即可求得答案;(2)假定1()1tx f x x-=+,求出相应的,,a b t 值,得到矛盾,即可求得答案; (3)利用题中的新定义,列出两个等式恒成立;将x 用2x +代替,两等式结合得到函数值的递推关系;用不完全归纳的方法求出值域.【详解】(1)当()f x x =时,22()()()()f a x f a x a x a x a x +⋅-=+⋅-=-,其值不为常数,故1()f x x M =∉,当()3x f x =时,2()()333a x a x a f a x f a x +-+⋅-=⋅=,当0a =时,1b =,故存在实数对(0,1),使得(0)(0)1f x f x +⋅-=对定义域内任意实数x 都成立, 故()x 23f x M =∈;(2)若函数1()1tx f x x -=+具有反函数1()f x -,且1()1tx f x M x-=∈+, 则222221()1()(1)()()1()1()(1)t a x t a x ta t x f a x f a x b a x a x a x-+----+⋅-=⋅==+++-+-, 则21b t b t b ⎧=⎪=-⎨⎪=⎩,解得:011a b t =⎧⎪=⎨⎪=-⎩,此时,不存在反函数,故不存在实数对(,)a b ,使得()f x 与1()f x -同时属于()1(1)f x x =≠-集合M .(3)函数()f x M ∈,且存在满足条件的有序实数对(0,1)和(1,4),于是()()1f x f x ⋅-=,(1)(1)4f x f x +⋅-=,用1x -替换(1)(1)4f x f x +⋅-=中x 得:()(2)4f x f x -=,当[1,2]x ∈时,2[0,1]x -∈,4()[2,4](2)f x f x =∈-, ∴[0,2]x ∈时,()[1,4]f x ∈.又由()()1f x f x ⋅-=得:1()()f x f x =-, 故14()(2)f x f x =--,即4()(2)f x f x -=-, 可得:(2)4()f x f x +=.∴ [2,4]x ∈时,()[4,16]f x ∈,[4,8]x ∈时,()[16,64]f x ∈,……依此类推可知[2,22]x k k ∈+时,222()2,2k k f x +⎡⎤∈⎣⎦,故[2014,2016]x ∈时,20142016()2,2f x ⎡⎤∈⎣⎦,综上所述,[0,2016]x ∈时,2016()1,2f x ⎡⎤∈⎣⎦,[2016,0]x ∈-时,20161()2,1()f x f x -⎡⎤=∈⎣⎦-, 综上所述,当[2016,2016]x ∈-时函数()f x 的值域为201620162,2-⎡⎤⎣⎦.【点睛】本题考查理解题中的新定义,解题关键是判断函数是否具有特殊函数的条件,利用新定义得到恒等式和通过仿写的方法得到函数的递推关系,考查利用归纳的方法得结论.。

上海市曹杨二中2018学年高一上学期期末考试物理试题

上海市曹杨二中2018学年高一上学期期末考试物理试题

上海市曹杨二中2015学年度第一学期高一年级期末考试物理试卷命题人:__________ 校对人:__________ 审核人:__________试卷共4页1张考生注意:1、答卷前,考生务必将姓名、班级、学号等在指定位置填写清楚。

2、本试卷共有24道试题,满分100分,考试时间60分钟。

请考生用水笔或圆珠笔将答案直接写在试卷(或答题卷)上。

3、本试卷重力加速度g 取210m/s 。

sin 370.6︒=,cos370.8︒=。

一、单选题(每题3分,共30分)1.汽车以10m/s 的速度在水平路面上匀速前进,紧急制动时以22m/s -的加速度在粗糙水平面上滑行,则在6s 内汽车通过的路程为( ) A .24mB .30mC .25mD .以上答案都不对2.1F 、2F 的合力为F ,已知120N =F ,28N =F ,那么2F 的取值可能是( ) A .40NB .70NC .100ND .6N3.静止在水平桌面上的物体,对水平桌面的压力( ) A .就是物体的重力B .大小等于物体的重力C .是由于地球的吸引而产生的D .是由于桌面的形变而产生的4.如图,在倾角为45︒的光滑斜面上有一圆球,在球前放一光滑竖直挡板使球保持静止,此时球对斜面的正压力为1N ;若去掉挡板,球对斜面的正压力为2N ,则下列判断正确的是( )A .2112=N NB .21=N NC .212=N ND .21=N5.如图所示,用一水平力F 将两铁块A 和B 紧压在竖直墙上而静止。

对此,下列说法中正确的是( )A .铁块A 对铁块B 的摩擦力方向可能向上,也可能向下 B .铁块A 对铁块B 摩擦力一定向上C .墙对铁块B 的摩擦力方向可能向上,也可能向下D .墙对铁块B 的摩擦力一定向上6.下列关于惯性的说法,正确的是( ) A .物体只有静止或做匀速直线运动时才具有惯性B .惯性是保持物体运动状态的力,起到阻碍物体运动状态变化的作用C .一切物体都有惯性,速度越大惯性越大D .两个物体质量相等,它们的惯性大小一定相等 7.关于力和运动的下列说法中,正确的是( ) A .物体运动的方向总跟物段所受合外力的方向一致B .作用在物体上的合外力不为零时,物体的瞬时速度可能为零C .当物体所受合外力减小,物体的速度也随之减小D .当物体所受合外力不变时,物体的速度也一定不变8.一个在水平粗糙地面上做直线运动的物体,在水平方向除摩擦力外只受向右推力作用,下面叙述的四种情况中,不可能出现的是( ) A .物体向右运动,加速度为零 B .物体向左运动,加速度为零 C .物体加速度的方向向右D .物体加速度的方向向左9.雨滴从空中由静止落下,若雨滴下落时空气对其阻力随雨滴下落速度的增大而增大,下面图象能反映雨滴下落运动情况的是( )10.如图所示,小车固定着三角硬杆,杆的端点固定着一个质量为m 的小球。

上海市曹杨二中2018-2019学年高三第一学期数学期末考试

上海市曹杨二中2018-2019学年高三第一学期数学期末考试

曹杨二中2018学年第一学期期终考试数学试卷一、填空题1、函数()sin cos f x x x =的最小正周期为_________2、2lim 12n P n →∞++=_________3、函数()()()3log 212x f x x =-≥的反函数()1f x -_________ 4、在62x x ⎛⎫- ⎪⎝⎭的二项式展开式中,常数项为___________ 5、一直一组数据为2,11,9,8,10,则这组数据的方差为_________6、双曲线221x y -=的一条渐近线被圆()2224x y -+=截得线段长为________7、已知数列{}n a 的首项12a =,且满足()*22n n n a a n N +=∈,则20a =________8、已知函数()f x 是奇函数,()112f =,且()()()()22f x f x f x R +=+∈,则()5f =________ 9、将一颗均匀的骰子掷两次,第一次得到的点数记为a ,第一次得到的点数记为b ,则方程组322ax by x y +=⎧⎨+=⎩有唯一解的概率是___________ 10、已知等差数列{}n a 的前n 项和为n S ,若1313,615a S ≤≤≤≤,则21a a 的取值范围是__________ 11、设函数()3,1,1x a x f x x a x ⎧-<=⎨-≥⎩,若()f x 有且仅有1个零点,则实数a 的取值范围是___________12、定义全集U 的子集M 的特征函数()10M U x M f x x C M ∈⎧=⎨∈⎩,对于两个集合,M N ,定义集合()(){}*1M N M N x f x f x =+=,已知集合{}{}2,4,6,8,10,1,2,4,8,16A B ==,并用S 表示有限集S 的元素个数,则对于任意有限集,**M M A M B +的最小值为________二、选择题13、若1+i 为虚数单位)是关于x 的实系数一元二次方程20z bz c ++=的一个复数根,则()A. 2,3b c ==B. 2,1b c ==-C. 2,3b c =-=D. 2,1b c =-=-14、已知,,x y z 为正实数,且230x y -+=,则2y xz的最小值为() A.1 B.2 C.3 D.615、设平面α和平面β相交于直线m ,直线a 在平面α上,直线b 在平面β上,且b m ⊥,则“αβ⊥”是“a b ⊥”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16、在ABC ∆中,若623AC AB AB BC BC CA ⋅=⋅=⋅,则A ∠=()A.45°B.60°C.120°D.135°二、解答题17、在ABC ∆中,角,,A B C 的对边分别为,,a b c ,已知2,6a C π==,且ABC ∆(1)求c ;(2)若F 为边AC 上一点,且CF =,求sin BFC ∠18、如图,某甜品创作一种冰淇淋,其上半部分呈半球形,下半部分呈圆锥形,现把半径为10cm 的圆形蛋皮等分成5个扇形,用一个扇形蛋皮固成圆锥的侧面(蛋皮厚度忽略不计)。

上海市曹杨第二中学高一物理上学期期末试卷含解析

上海市曹杨第二中学高一物理上学期期末试卷含解析

上海市曹杨第二中学高一物理上学期期末试卷含解析一、选择题:本题共5小题,每小题3分,共计15分.每小题只有一个选项符合题意1. (单选)汽车以大小为20m/s的速度做匀速直线运动,刹车后,获得的加速度的大小为5m/s2,那么刹车后2s内与刹车后6s内汽车通过的路程之比为()A.1:1 B.3:1C.4:3 D.3:4参考答案:D2. (单选)如图所示,要在客厅里挂一幅质量m=1.2kg的画(含画框),已知画框背面有两个相距l=0.8m、位置固定的挂钩。

现将轻质细绳的两端分别固定在两个挂钩上,把画对称地挂在插入竖直墙壁的光滑钉子上,挂好后整条细绳呈绷紧状态。

若细绳能够承受的最大拉力为Fmax=10N,g取10m/s2,要使细绳不被拉断,求细绳的最小长度应为()A.0.5m B.1m C.2m D.2.5m参考答案:B3. (单选)下列说法符合史实的是A.牛顿发现了行星的运动规律B.胡克发现了万有引力定律C.卡文迪许测出了引力常量G,被称为“称量地球重量的人”D.伽利略用“月—地检验”证实了万有引力定律的正确性参考答案:C4. (单选)从匀速直线行驶的火车窗口释放一石子,不计风对石子的影响,站在路边的人看到石子做( )A.自由落体运动B.平抛运动C.匀速直线运动D.匀变速直线运动参考答案:B5. (单选)如图所示,汽车以速度通过一弧形的拱桥顶端,且汽车对桥面有压力。

关于汽车受力的说法中正确的是()A.汽车受重力、支持力、牵引力、摩擦力和向心力的作用B.汽车的向心力是它所受的重力与压力的合力C.汽车的向心力是它所受的重力与支持力的合力D.汽车的向心力是它所受的重力、支持力与摩擦力的合力参考答案:C二、填空题:本题共8小题,每小题2分,共计16分6. 如图所示, 一颗水平飞行的子弹(长度不计)穿过紧挨着的固定在水平桌面上的三块同样的木块之后, 速度恰好为零, 该子弹射穿木块1、木块2、木块3所用时间之比t1∶t2∶t3=__________.(设子弹在木块内做匀减速运动)参考答案:(-):(-1):17. 已知地球自转周期为T,地球半径为R,同步卫星离地表的高度为h,万有引力恒量为G,则同步卫星绕地球运动的线速度为______ __,地球的质量为___ _____。

上海市曹杨二中2018学年度第一学期高一年级期末考试物理卷.pdf

上海市曹杨二中2018学年度第一学期高一年级期末考试物理卷.pdf

上海市曹杨二中2018学年度第一学期高一年级期末考试物理卷一、单选题1、下列不属于国际单位制中基本单位的是()A.牛顿B.米C.千克D.秒2、用比值法定义物理量是物理学中一种很重要的思想方法,下列物理量的表达式不属于比值法定义的是( )A.密度V m=ρ B.速度t sv = C.加速度m Fa = D.电阻IUR =3、甲、乙两个物体沿同一直线向同一方向运动时,取物体的初速度方向为正方向,甲的加速度恒为2/2s m ,乙的加速度恒为2/3-s m ,则下列说法中正确的是( )A.两物体都做匀加速直线运动,乙的速度变化快B.甲做匀加速直线运动,它的速度变化快C.乙做匀减速直线运动,它的速度变化率大D.甲的加速度比乙的加速度大4、游乐园中,游客乘坐能匀速、加速或减速运动的升降机,可以体会超重与失重的感觉。

下列描述正确的是( )A.当升降机匀速上升时,游客是处在超重状态B.当升降机加速上升时,游客是处在失重状态C.当升降机减速上升时,游客是处在失重状态D.当升降机加速下降时,游客是处在超重状态5、下面说法中正确的是( )A.当物体的运动状态发生变化时,它一定受到外力作用B.静止或做匀速直线运动的物体,一定不受外力的作用C.当物体的速度等于零时,它一定处于平衡状态D.物体的运动方向一定是它所受的合外力的方向6、汽车拉着拖车在水平道路上沿直线加速行驶,下列说法中正确的是( )A.汽车拉拖车的力大于拖车拉汽车的力B.汽车拉拖车的力与拖车拉汽车的力大小相等C.汽车拉拖车的力大于拖车受到的阻力D.汽车所受的牵引力大于其所受的阻力7、如图,物体静止在光滑的水平面上,水平力F作用于O点,现要使物体在水平面上沿OO’方向做加速运动,必须在F和OO’所决定的水平面内再加一个力,那么F’的最小值应为( )A.FcosAC.Ftan AD.FcotA8、如图,在水平地面上运动的小车车厢底部有一质量为m1的木块,木块和车厢通过一根轻弹簧相连接。

2017-2018学年上海市普陀区曹杨二中高一(上)期中物理试卷和答案

2017-2018学年上海市普陀区曹杨二中高一(上)期中物理试卷和答案

第1页(共18页)2017-2018学年上海市普陀区曹杨二中高一(上)期中物理试卷一、单选题(每题只有一个正确答案,1-6题每题2分,7-10题每题3分)1.(2分)下列几种运动中的物体,可以看作质点的是()A .从广州飞往北京的飞机B .绕地轴做自转的地球C .绕太阳公转的地球D .在平直公路上行驶的汽车2.(2分)在匀变速直线运动中,下列说法中正确的是()A .相同时间内位移的变化相同B .相同时间内速度的变化相同C .相同时间内加速度的变化相同D .相同路程内速度的变化相同3.(2分)几个做匀变速直线运动的物体,在时间t 内位移一定最大的是()A .加速度最大的物体B .初速度最大的物体C .末速度最大的物体D .平均速度最大的物体4.(2分)一个物体静止在水平桌面上,下列说法正确的是()A .物体受的重力和桌面对它的支持力是一对作用力和反作用力B .物体对桌面的压力就是重力C .桌面对物体的支持力大小等于重力,这两个力是一对平衡力D .物体受到的重力的反作用力是物体对桌面的压力5.(2分)关于物体的重心,下列说法正确的是()A .重心就是物体内最重的点B .重心是重力的作用点,所以重心一定在物体上C .任何有规则形状的物体,它的几何中心必然与重心重合D .物体的重心在物体中的位置不因物体升高、降低、倾斜而发生改变6.(2分)分)如图所示,如图所示,细绳竖直拉紧,细绳竖直拉紧,小球和光滑斜面上接触,小球和光滑斜面上接触,小球和光滑斜面上接触,并处于静止状态,并处于静止状态,则小球受到的力是()A.重力、绳的拉力 B.重力、绳的拉力、斜面的弹力C.重力、斜面的弹力.重力、斜面的弹力 D.绳的拉力、斜面的弹力7.(3分)某物体沿一直线运动,其v﹣t图象如图所示,则下列说法中不正确的是( )是(A.第2s内和第3s内速度方向相反B.第2s内和第3s内的加速度方向相反C.第3s内速度方向与加速度方向相反D.第5s内速度方向与加速度方向相反8.(3分)为了测定某辆轿车在平直公路上启动时的加速度(轿车启动时的运动可近似看做匀加速直线运动),某人拍摄了一张在同一底片上多次曝光的照片,如图所示。

上海曹杨二中2017-2018学年高一上学期期末考试化学试题 Word版缺答案

上海曹杨二中2017-2018学年高一上学期期末考试化学试题 Word版缺答案

上海市曹杨二中2017-2018学年度第一学期高一年级期末考试化学试卷一、选择题(每小题只有1个正确选项)(本大题共40分)1.碘缺乏病是目前已知的导致人类智力障碍的主要原因. 为解决这一全国性问题,我国很久以前已经开始实施“智力工程”,最经济可行的措施是() A .食盐加碘盐B .面包加碘盐C .大量食用海带D .注射含碘药剂 2.下列属于电解质的是()A .硫酸溶液B .铜C .NaClD .2CO 3.下列有关化学用语正确的是()A .氮气分子的结构式N N ≡B .钾离子的电子式:KC .硫原子的结构示意图D .二氧化硅的分子式:2SiO4.下列不能使淀粉KI 试纸变蓝的是()A .次氯酸钠溶液B .食盐溶液C .溴水D .碘水 5.下列电离方程式书写正确的是()A .22424H SO H SO +-→+B .324NH H O NH OH +-⋅+C .()22Ca OH Ca 20H +-+D .233NaHCO Na H CO ++-→++6.下列叙述正确的是()A .有氧化还原反应中,肯定有一种元素被氧化,另一种元素被还原B .有单质参加或生成的反应一定属于氧化还原反应C .失电子难的原子,获得电子的能力一定强D .元素由化合态变成游离态时,它可能被氧化,也可能被还原7.在烧瓶中充满干燥的气体,胶头滴管及烧杯内分别盛放液体,下列组合中不可形成喷泉的是()A .氯气和B .氯化氢和水C .二氧化碳和烧碱溶液D .氯气和烧碱溶液 8.实验室里保存下列试剂的方法,有错误的是()A .新制氯水盛放在棕色试剂瓶中,存放于低温避光的地方B .液溴易挥发,盛放在用水液封、用橡皮胶塞密封的棕色试剂瓶中C .碘易升华,盛放在棕色试剂瓶中D .浓盐酸易挥发,盛装在无色密封的细口玻璃试剂瓶中 9.下列物质中即含有共价键又含有离子键的是()A .24H SOB .2HC .KClD .NaOH 10.下列各组关于强电解质、弱电解质、非电解质的归类正确的是()11.下列叙述中不正确的是()A .卤素单质的颜色按222Cl Br I 、、的顺序逐渐变深B .222Cl Br I 、、的氧化性逐渐增强C .氯、溴、碘的原子半径或离子半径随电子层数的增多而增大D .氯气易液化,溴单质易挥发,碘单质易升华12.将碘研碎后与锌粉混合于蒸发皿中,然后滴加少量水,可观察到紫红色蒸气产生,此实验被称为“滴水生烟”. 关于该实验的下列说法中错误的是()A .Zn 与2I 的反应是放热反应B .水在该反应中作催化剂C .生成的碘化锌为紫红色D .紫红色的蒸气是碘升华所致13.如图所示的五种尾气吸收装置中,适合吸收极易溶于水的气体,且能防止产生倒吸现象(溶液进入气体发生装置中)的是()A .②④⑤B .①②④C .①②③D .①②④⑤14.分类方法在化学学科的发展中起着非常重要的作用. 下列分类标准合理的是()A .根据物质能否导电将物质分为电解质和非电解质B .根据反应中的热效应将化学反应分为放热反应和吸热反应C .根据溶液导电性强弱将电解质分为强电解质和弱非电解质D .根据元素的化合价的高低将化学反应分为氧化还原反应和非氧化还原反应 15.下列有关化学健与晶体结构说法正确的是()A .两种元素组成的分子中一定只有极性键B .非金属元素组成的化合物一定是共价化合物C .含有阴离子的化合物一定有阳离子D .离子化合物的熔点一定比共价化合物的高16.氯化碘()ICl 的化学性质跟氯气相似,预计它跟水反应的最初生成物是()A .HI 和HClOB .HCl 和HIOC .3HClO 和HIOD .HClO 和HIO17.某澄清溶液可能由32KI NaNO CaCl 、、和23Na CO 中的一种或几种混合而成,为鉴定其组成,依次进行如下实验;⑴上述溶液中滴加3AgNO 溶液有浅黄色沉淀产生;⑵将沉淀滤出加3HNO 时,沉淀部分溶解,有气体放出;⑶向一份溶液中加淀粉溶液,再加适量氯水,溶液显蓝色. 根据以上实验可判断()A .一定有2CaCl 和23Na CO ,一定没有KI ,可能有3NaNOB .一定有23KI Na CO 、,可能有32NaNO CaCl 、C .一定有23KI Na CO 、,可能有3NaNO ,一定没有2CaClD .只有23Na CO ,可能有3NaNO ,一定没有2KI CaCl 、18.将碘水滴入2Na S 溶液中,溶液浑浊:将溴水滴入KI 溶液中,溶液由无色变为褐色,通过以上两个实验可得到的正确结论是()A .离子的氧化性由强到弱的顺序为:2Br I S --->>B .离子的还原性由强到弱的顺序为:2Br I S --->>C .元素的非金属性由强到弱的顺序为:2Br I S --->>D .离子的还原性由强到弱的顺序为:2Br I S ---<<19.硫酸镁和硫酸铝溶液等体积混合后,铝离子浓度为0.1mol/L ,硫酸根离子的浓度为0.3mol/L ,则混合液中镁离子的浓度为()A .0.15mol/LB .0.3mol/LC .0.45mol/LD .0.2mol/L20.标准状况下22.24LH 和2Cl 组成的混合气体,经光照后缓缓通入100ml lmol/LNaOH 溶液中充分反应,测得最终溶液中NaClO 的浓度为0.1mol/L (假设溶液体积不变),则原混合气体中2Cl 的体积分数为()A .55%B .60%C .65%D .70% 二、选择题:(每小题有1~2个正确选项)(共15分) 21.A N 表示阿伏加德罗常数,下列判断正确的是()A .在18218g O 中含有A N 个氧原子.B .标准状况下,22.4L 空气含有A N 个单质分子C .2lmolCl 参加反应转移电子数一定为A 2ND .含A N 个Na +的2Na O 溶解于水中形成1L 溶液,Na +物质的量深度为1mol/L 22.氯水中存在多种分子和离子,可通过实验的方法加以确定。

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曹杨二中2017学年度第一学期高一英语期末考试II. Grammar and VocabularySection AThe holiday season is supposed to be the most festive and fun time of the year, but all those plans and expectations of joy can turn tougher and more stressful than they sound. This is especially true for (21)______ of us who struggle with mental illness.The holidays break your routine. Sometimes you (22)______ (force) to spend time with family you rarely see and don’t always get along with. Or maybe you’re alone when everyone else is with family. Or you’re at work and can’t be with those you love. Or you are off from work, (23)______ more time to think troubling thoughts. Or you are thrust into party situations that tempt your demons.When you have a routine, it’s (24)______ (easy) to manage whatever mental struggles you may face, and when that routine is broken, it (25)______ trigger tings you may not be ready to face. I know it has for me. It was during the holidays (26)______ I hit a low moment and, with the help of my mother, decided to seek help for my eating disorder.Around the holidays, I often feel like I’m supposed to be everywhere, with everyone—all with the (27)______ (add) guilt of knowing it’s the season of giving. (28)______ (fight) this, I’ve developed a mantra(咒语): it’s not selfish to take time for yourself. Take a walk in nature. Talk to a friend you trust, or a therapist. Sit out one of the holiday gatherings in favor of some personal time. Just do (29)______ helps you calm down and gives you a break from the stress. Download one of the many meditation apps for your phone. I particularly like Calm and End Anxiety.Trying to spend all of your time (30)______ (please) everyone else is not only exhausting – it’s impossible. And you know what? If you take a little time for yourself, you will be much better company for those around you.21. some 22. are forced 23. 24. easier 25. can 26. that 27. added 28. To fight 29. what 30. pleasingWhen young people study or take exams the results are usually about rewarding their individual achievement.But when they get into the workplace they will be told about the importance of social skills and the need to co-operate with other people on solving problems.So, are school systems out of step with what is needed by young people?PISA, which (31)______ students’ abilities in reading, maths and science, has now carried out the world’s first global tests on collaborative problem-solving skills.As might have been expected, students who are high achievers in academic tests are also likely to be better at problem solving with other people.They are likely to have the skills in interpreting information and complex reasoning that will help them with any kind of problem solving.The same holds (32)______ across countries. Top-performing countries in academic tests, suchas Japan, South Korea, Singapore, Estonia, Finland and Canada, are also high performers at collaborative problem solving.Working together seems to (33)______ different types of ability. Five years ago, PISA carried out tests on individual problem-solving skills. These showed that boys tended to do better in most countries. But when the element of collaboration is added to the problem solving, girls (34)______ boys in every country. In the UK this gender gap is one of the largest.These results are (35)______ in students’ attitudes. Girls show more positive attitudes towards relationships, meaning that they tend to be more interested in others’ opinions and want others to succeed. Boys, on the other hand, are more likely to see the (36)______ of teamwork and how collaboration can help them work more (37)______ and efficiently.Disadvantaged students are more likely to see the (38)______ of teamwork than their (39)______ peers. They tend to report that they prefer working as part of a team to working alone, and that they think teams make better decisions than individuals.The classroom environment seems to (40)______ how well students work together. When students have a lot of communication-intensive activities – such as taking part in class debates or arguing about science questions – they are more likely to have positive attitudes towards collaboration.III. Reading ComprehensionSection AOn a cold November afternoon, my mother and I were walking home from a pizza shop. We were dressed __41__ and equipped with a rented video we had been __42__ to watch. I was feeling a little __43__ as I was carrying our shopping, and decided to throw away something. So I started to walk towards a garbage can when I noticed a poor man walking out of the restaurant in front of us. He __44__ over to another nearby garbage can and started looking through it.I suddenly felt very guilty because I was about to throw away a new drink just because it was __45__. I walked up to him and handed the drink and some snacks over to him. The man looked up __46__ and took what I gave him.A huge smile __47__ across his face and this caused me to feel indescribably satisfied. I felt I couldn’t be happier __48__ myself, but then he said, “Wow, this is my son’s lucky day!”With that, he thanked me happily and started off on his bike, I __49__ heard him whistling a song as he rode away.I got a warm __50__ inside. I now understand what is meant by the saying “giving is getting”.Although it only __51__ a little action and a few words, I gained and learned more in those two minutes than I did in the rest of the month. Everyone in the world needs help, everyone can __52__ help and everyone will be helped by __53__ kindness.The image of that man’s happiness caused by my small gift appears in my mind every __54__ I have the chance to do something nice.This is the __55__ of charity.41. A. poorly B. coldly C. warmly D. expensively42. A. dying B. exciting C. worrying D. happy43. A. worried B. interested C. bored D. tired44. A. headed B. passed C. crossed D. took45. A. cheap B. heavy C. tasteless D. full46. A. in silence B. in surprise C. in interest D. in a hurry47. A. appeared B. spread C. went D. ran48. A. with B. to C. at D. for49. A. still B. once C. even D. ever50. A. sense B. mind C. thinking D. feeling51. A. held B. took C. called D. asked52. A. offer B. send C. show D. have53. A. showing B. expressing C. lending D. setting54. A. moment B. day C. minute D. time55. A. power B. meaning C. strength D. aimSection B(A)My six-year-old granddaughter, Caitlynd and I stopped at a Tim Horton’s shop for a blueberry cake. As we were going out the door, a young teenage boy was coming in.This young man had no hair on the sides of his head and a set of blue spiked(竖起的)hair on top of it. One of his nostrils(鼻孔)was pierced, and a ring ran through the hole and a chain went across his face and was attached to a ring he was wearing in his ear. He held a skateboard under one arm and a basketball under the other.Caitlynd, who was walking ahead of me, stopped at once when she saw the teenager. I thought he’d scared her and she’d frozen on the spot.I was wrong.My granddaughter backed up against the door and opened it as wide as it would go. Now I was face to face with the young man. I stepped aside and let him pass. His gracious response was a polite “Thank you very much”.On our way to the car, I praised Caitlynd for her manners in holding open the door for the young man. She didn’t seem to be troubled by his appearance but I wanted to make sure. If a grandmother talked about freedom of self-expression and allowing people their differences was suitable, I wanted to be ready.As it turned out, the person who needed the talk was me.The only thing Caitlynd noticed about the teenager, was the fact that his arms were full. “He would have a hard time in opening the door.”I saw the partially shaved head, the set of spiked hair, the piercings and the chain. She saw a person carrying something under each arm and heading toward a door.In the future, I hope to get down on her level and raise my sights.56.What did the author think of the young man?A. Polite.B. Uncommon.C. Frightening.D. Funny.57.Caitlynd helped the young man because ______.A. she was scaredB. she didn’t notice his lookC. she wanted to avoid himD. it would be difficult for him to open the door58.The underlined sentence suggests that ______.A. the author was ashamed of herselfB. the author didn’t know how to give a talk on freedomC. a talk on freedom was useless for the granddaughterD. people should have more freedom to express themselves59.The author intends to tell us that ______.A. we shouldn’t judge a person by his lookB. we should allow people more freedom to dress differentlyC. we should be more helpful and tolerant to strange dressing of the youngD. we shouldn’t be too particular about people in life(B)Last week Amazon announced Word Runner, a new tool for reading Kindle books. Word Runner is a software update that shows one word at a time, instead of words, sentences and paragraphs on one page.Opinions about Word Runner are mixed. Carmen Blyth, ph.D., and a teacher of English as a Second Language (ESL), says Word Runner is not optimal reading.In the Kindle Forums at Amazon, a commenter called CB Retriever says, “Thanks, but that absolutely will not work for me as I too do lines and blocks of text at a time – I think the last time I read word by word was back when I was learning to read.”Len Edgerly, business journalist and host of The Kindle Chronicles podcast, had a chance to try out Word Runner. In his podcast, Edgerly said that Word Runner is “a dramatically different way to read.”Dynamic PacingWord Runner uses a feature called “Dynamic Pacing” to adjust the speed around complicated words. Dynamic Pacing automatically slows the speed of Word Runner for complicated words, punctuation, and paragraph breaks. It makes changes based on the natural rhythm of your reading.BrakeYou can pause Word Runner by tapping the pause button, called Brake:Brake lets you move words forward and backward. You can go back and re-read words using Brake.AvailabilityAccording to Amazon, Word Runner will be available soon for some English-language books on the Kindle app for Android devices and on Fire tablets.Word Runner will work with select English language Kindle books from Amazon sites in all countries, says an Amazon representative.Amazon has not yet said whether Word Runner will be coming to the Kindle app for iPhone and iPad.60.In CB Retriever’s opinion, ______.A. Word Runner provides one with necessary information to understand what he is readingB. good reader can read four words at a timeC. Word Runner is not so practical for him as he reads lines and blocks of text at a timeD. Word Runner is “a dramatically different way to read”61.From the passage, we can draw the conclusion that ______.A. Word Runner will soon be available for iPhone and iPadB. Word Runner will work with all English-language books in all countriesC. Word Runner is a software which updates one word at a timeD. People will soon be able to use Word Runner to read some English-language books on Andriod devices and Fire tablets62.“Dynamic Pacing” is designed to ______.A. regulate the speed of readingB. quicken the speed of reading something easyC. let you move words forward and backwardD. allow you to see all of the words on a page on your screen(C)In recent years many countries of the world have been faced with the problem of how to make their workers more productive. Some experts claim the answer is to make jobs more varied. But do more varied jobs lead to greater productivity? There is evidence to suggest that while variety certainly makes the worker’s life more enjoyable, it doesn’t actually make him work harder. As far as increasing productivity is concerned, then variety is not an important factor.Other experts feel that giving the worker freedom to do his job in his own way is important and there is no doubt that this is true. The problem is that this kind of freedom cannot easily be given in the modern factory with its complicated machinery which must be used in a fixed way. Thus while freedom of choice may be important, there is usually very little that can be done to create it.Another important consideration is how much each worker contributes to the product he is making. In most factories the worker sees only one part of the product. Some car factories are now experimenting with having many small production lines rather than one large one, so that each worker contributes more to the production of the cars on his line. It would seem that not only is degree of the workers’ contribution an important factor, therefore, but it is also one we can do something about.To what extent more money led to greater productivity? The workers themselves certainly think this is important. But perhaps they want more money only because the work they do is so boring. Money just lets them enjoy their spare time more. A similar argument may explain demands for shorter working hours. Perhaps if we succeed in making their jobs more interesting, they will neither want more money, nor will shorter working hours be so important to them.63.Which of these possible factors leading to greater productivity is not true?A. To make jobs more varied.B. To give the worker freedom to do his job in his own way.C. Degree of work contribution.D. Demands for longer working hours.64.Why do workers want more money?A. Because their jobs are too boring.B. In order to enjoy more spare time.C. To make their job more interesting.D. To demand shorter working hours.65.The last sentence in this passage means that if we succeed in making workers’ jobs moreinteresting, ______.A. they will want more moneyB. they will demand shorter working hoursC. more money and shorter working hours are important factorsD. more money and shorter working hours will not be so important to them66.In this passage, the author tells us ______.A. how to make the workers more productiveB. possible factors leading to greater efficiencyC. to a certain extent more money lead to greater productivityD. how to make workers jobs more interestingLazy people will never gain anything in life. However, laziness can be defeated once a few changes have been made in your mind.__67__ Many people lack sleep constantly, since they stay up too late and get up too early to prepare for work. These people have little motivation once they arrive home. Laziness works hand in hand with a lack of motivation and a tendency to put off things. By adjusting your sleep schedule to provide a few more hours of meaningful rest, you can fight laziness throughout the day.Another way to fight laziness is to change your mind from passive to active. Some people treat their lives as if they were pushed from task to task. Others take a more positive approach, viewing each task as a challenge they must overcome alone. __68__Some people fight laziness by removing the temptations that surround them. A television in the living room may provide entertainment, but watching too much TV often contributes to laziness. __69__ Complete a few tasks and reward yourself with what you enjoy, such as a good dinner or a film.Laziness can also be a lasting problem at home. Couples and children may all have different energy levels, but laziness can be spread if not dealt with immediately. __70__ Be the first to collect and wash dishes after a meal. Others in the home may eventually follow your example and perform their own task. It is difficult to practice laziness when you are surrounded by motivated people.Finally, taking exercise regularly can help you fight laziness. Enough exercise and a balanced diet can help you develop a healthy lifestyle, thus enabling you to have more energy and help lift your spirits.第II卷IV. Translation71.你如何解释他的失败呢?(account)72.科学研究成果往往被运用到现代工业。

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