吉林省桦甸四中、磐石一中、梅河口五中、蛟河实验中学等2020届高三4月联考理科综合答案

吉林省桦甸四中、梅河五中、蛟河一中、柳河一中、舒兰一中、辉南六中重点高中盟校2024学年高三英语试题质量检测试题（一）英语试题试卷考生须知：1．全卷分选择题和非选择题两部分，全部在答题纸上作答。

选择题必须用2B铅笔填涂；非选择题的答案必须用黑色字迹的钢笔或答字笔写在“答题纸”相应位置上。

2．请用黑色字迹的钢笔或答字笔在“答题纸”上先填写姓名和准考证号。

3．保持卡面清洁，不要折叠，不要弄破、弄皱，在草稿纸、试题卷上答题无效。

第一部分（共20小题，每小题1.5分，满分30分）1．---Excuse me, could you tell me the time, please?---Sorry, I don’t have a watch with me.--- ___________A．Thanks a lot. B．What a pity.C．I’m sorry to hear that D．Thanks all the same2．The high-anxiety focus on reading score may have narrowed student________to knowledge about the world that can improve their reading.A．idea B．aid C．belief D．access3．They did everything in their power to save the building dating back to the Qing Dynasty, but their effortswere .A．in turn B．in timeC．in vain D．in need4．The couple was initially looking forward to having an overseas wedding but had to drop the idea after facing________ from parents.A．recognition B．tolerationC．opposition D．proportion5．— Tony, you seem in high spirits.—________ We won the match 4-1.A．Guess what? B．Oh, come on!C．So what? D．No wonder.6．--- Have you heard that they are working around the clock to compete for the prize?--- Don’t worry. We are ready to the challenge.A．build up B．take upC．stick to D．lead to7．---I went to see The Wandering Earth last night. It was fantastic!---You were so lucky! How I wish I _____ the ticket too.A．get B．gotC．had got D．would get8．He has written two articles for the journal，one to be published in this issue，the other to___________ in the next. A．turn out B．come outC．bring out. D．leave out9．It seems late to say anything. We probably ______ it if we had made an offer sooner.A．would have got B．would getC．had got D．got10．_____ the plant once a week if you are to get fruit in the autumn.A．Watered B．Watering C．To water D．Water11．---Sorry! I’m afraid that I can’t go to the magic show by David Copperfield with you.--- .A．It doesn’t matter.B．Do as you like.C．What a shame.D．Take it easy.12．It is not only blind men who make such stupid mistakes. People who can see sometimes act__________.A．just foolishly B．less foolishly C．as foolishly D．so foolishly13．My summer camp isn’t good fun, but it is real learning experience for me.A．a; / B．the; a C．/; a D．a; the14．A man can fail many times, he is a real failure when he begins to blame someone else.A．but B．or C．and D．so15．The influence of this book might not have been reaching so far, ________for Mr. William, who dared to publish it.A. if had it not been B．is it not being C．had it not been D. its only having been16．How come every kid today is meant to be a champion for something _____ we know every kid can’t be a star? A．in case B．as ifC．when D．unless17．Don’t refer to the diction ary every time you come across a new word as sometimes its meaning may be _______ clearly in a given context.A．picked out B．ruled out C．brought out D．taken out18．Y et _______ in the process of development did they stop to consider the impact of their “progress” on nature.A．in no time B．at no pointC．as likely as not D．more often than not19．The biggest problem for most plants, which ________just get up and run away when threatened, is that animals like to eat them.A．shall not B．can'tC．needn't D．mustn't20．---He was satisfied with the result, wasn't he?---No. It was so difficult that he __________have passed it.A．shouldn't B．mustn'tC．couldn' t D．wouldn't第二部分阅读理解（满分40分）阅读下列短文，从每题所给的A、B、C、D四个选项中，选出最佳选项。

绝密★启用前吉林省四校(桦甸四中、磐石一中、梅河口五中、蛟河实验中学)2020届高三毕业班下学期4月联合质量检测英语试题(解析版)2020年4月注意事项：1.本试卷分为四部分。

2.答题前，考生务必将自己的姓名、准考证号填写在本试卷相应的位置。

3.全部答案写在答题卡上，写在本试卷上无效。

4.本试卷满分150分,测试时间120分钟。

5.考试范围：高考全部内容。

第一部分听力（共两节，满分30分）做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节（共5小题；每小题 1.5分,满分7.5分）听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. What does the man want the woman to do?A. Talk on the phone.B. Turn off the TV.C. Turn down the TV.2. Where does the conversation probably take place?A. In a restaurant.B. In a hotel.C. In an airport.3. What does the man invite the woman to do this afternoon?A See a film. B. Go climbing. C. Go swimming.4. How much will the woman pay altogether?A. $3.50B. $21.50C. $255. What will Peter do this afternoon?A. Drive a car.B. Go skating.C. Play table tennis.第二节（共15小题；每小题1.5分,满分22.5分）听下面5段对话或独白。

动量守恒定律一、选择题(本题共10小题，每小题6分，共60分。

其中1～5题为单选，6～10题为多选)1．两名质量相等的滑冰人甲和乙都静止在光滑的水平冰面上。

现在，其中一人向另一个人抛出一个篮球，另一人接球后再抛回。

如此反复进行几次之后，甲和乙最后的速率关系是( )A ．若甲最先抛球，则一定是v 甲>v 乙B ．若乙最后接球，则一定是v 甲>v 乙C ．只有甲先抛球，乙最后接球，才有v 甲>v 乙D ．无论怎样抛球和接球，都是v 甲>v 乙答案 B解析 两人及篮球组成的系统动量守恒，且总动量为零，由于两人质量相等，故最后球在谁手中，谁的总质量就较大，则速度较小，故B 正确，A 、C 、D 错误。

2. (2020·四川省雅安市模拟)如图所示，子弹水平射入放在光滑水平地面上静止的木块，子弹未穿透木块，此过程木块的动能增加了6 J ，那么此过程产生的内能可能为( )A ．16 JB.2 JC.6 JD.4 J答案 A解析 设子弹的质量为m 0，初速度为v 0，木块的质量为m ，子弹打入木块的过程中，子弹与木块组成的系统动量守恒，即m 0v 0＝(m ＋m 0)v ，此过程产生的内能等于系统损失的动能，即ΔE ＝12m 0v 20－12(m ＋m 0)v 2＝m ＋m 0m 0·12m v 2，而木块获得的动能E k 木＝12m v 2＝6 J ，则ΔE >6 J ，A 正确。

3．(2020·河北衡水中学4月教学质量监测)有一只小船停靠在湖边码头，小船又窄又长(估计重一吨左右)。

一位同学想用一个卷尺粗略测定它的质量，他进行了如下操作：首先将船平行于码头自由停泊，轻轻从船尾上船，走到船头停下，而后轻轻下船。

用卷尺测出船后退的距离d，然后用卷尺测出船长L。

已知他的自身质量为m，水的阻力不计，船的质量为()A.m(L－d)d B.m(L＋d)dC.mLd D.m(L＋d)L答案 A解析设人走动时船的速度大小为v，人的速度大小为v′，人从船尾走到船头所用时间为t。

2020年吉林省桦甸四中、磐石一中、梅河口五中、蛟河实验中学等联考高考物理
模拟试卷（4月份）
二、选择题：本题共5小题，每小题6分．在每小题给出的四个选项中，第14～18题只有一项符合题目要求，第19～21题有多项符合题目要求．全部选对的得6分，选对但不全的得3分，有选错的得0分．
1．（6分）一个小球从空中由静止释放，与水平地面碰撞反弹后上升到一定的高度，这个过程中小球受到的空气阻力大小恒定，小球运动过程中速度随时间的变化如图所示，已知重力加速度大小为10m/s2，图中t＝1.2s，则图中t2的大小为（）
A．t2＝B．t2＝C．t2＝D．t2＝
2．（6分）如图所示，直杆AB可绕其中心O在竖直面内自由转动，一根细绳的两端分别系于直杆的A、B两端，重物用光滑挂钩吊于细绳上，开始时重物处于静止状态，现将直杆从图示位置绕O点沿顺时针方向缓慢转过90°，则此过程中，细绳上的张力（）
A．先增大后减小B．先减小后增大
C．一直减小D．大小不变
3．（6分）如图所示为某静电场中x轴上电势φ随x变化的情况，且x轴为该静电场中的一根电场线，一个带电粒子在坐标原点O由静止释放，粒子仅在电场力作用下开始沿x轴正向运动，则下列说法正确的是（）
A．粒子一定带负电。

2020届吉林省桦甸四中、磐石一中、梅河口五中、蛟河实验中学等高三4月联考英语试题After a long week, sometimes a short getaway is in order. However, deciding which destination to travel to can be tough when you only have a weekend to spend. To help you choose the right trip, U. S. News considered factors like accessibility, affordability and entertainment options, as well as reader votes, to determine which cities offer the best setting for a weekend away.New York CityVisitors can reach New York City by train, plane or automobile, so weekend getaways are easy. Plus, abundant attractions indoors (like museums and Broadway shows) make unpredictable weather no problem. For history, take the ferry to Ellis Island or the Statue of Liberty.ChicagoChicago’s diverse neighborhoods and abundance of can’t-miss sights make it a great weekend vacation, spot. On warmer days, take a photo of yourself in front of “The Bean” and go aboard an architecture river cruise. Meanwhile, museums like the Art Institute of Chicago and the Field Museum are wonderful places to escape winter s biting temperatures.TorontoCross Lake Ontario to our northern neighbor for an urban escape. It’s hard to see all of Toronto in just a few days, but first- timers can hit the high points, such as the CN Tower, the St. Lawrence Market and the Hockey Hall of Fame. Save an evening to explore the Distillery District.Washington, D．C．Not only is America’s capital full of free attractions, but many of them are concentrated on the National Mall, making it easy to see the Lincoln Memorial, the Washington Monument and other highlig hts in just a few days. While visiting, don’t miss out on the popular weekend brunch scene -just remember to make reservations in advance.1．What make(s) choosing a destination a hard job?A．So many wonderful attractions. B．The limited time.C．Lack of touring experience. D．Having little money.2．Which city suits you most if you are interested in Abrabam Lincoln?A．Washington, D．C．B．Toronto.C．Chicago. D．New York City.3．Why is Toronto different from others?A．You need a few days to tour around it.B．It is situated in a lonely and distant place.C．It has a smaller population.D．It doesn’t belong to the United States.About six years ago I began my job as an IT consultant. I had always been quite good with computers. It was a great move for me, as I quickly learned a ton. Every day was fun. It was one of the best and most challenging jobs I’ve ever had.Flash-forward a few years and things started to taper. I wasn’t learning as much as before. I felt a desire to try something different, to strike out on m y own. I couldn’t get up the courage to make any moves. The job paid very well; I had a lot of freedom, and great benefits.I had grown comfortable and accustomed to an easy life.Last year things came to a head, I was having a particularly stressful month, overloaded with work that was neither challenging nor interesting. I had enough. I reached out to my boss and let him know that I was done. We had a meeting about my decision. He didn’t want to lose me. He addressed my concerns and we agreed that I could change my job and begin working on some new internal organizational systems. I decided to give this new role a chance.Now, a year after my role change, not much is different. Senior staff did not adopt the new systems I built. The work is still unin spiring and unchallenging. I’m comfortable, but comfort to me has become terrible. Change can be very difficult for some and after spending time trying to carry out new solutions, I’ve realized that you simply can’t change other people if they don’t want i t.I’ve again quit my job. It’s both liberating and frightening. I need a greater challenge.I’ve learned a lotfrom my quitting experience. Here are some signs that it might be time to quit your job. 4．What does the underlined word “taper” mean in the pas sage?A．Increase. B．Decline.C．Develop. D．Repeat.5．How did the author feel when considering getting a move?A．Decisive. B．Cautious.C．Contradictory. D．Careless.6．Why was the author upset?A．His leader couldn’t make sense of him.B．He was unsatisfied with his salary.C．He thought his job was boring.D．His boss didn’t allow him to leave.7．What will be continued with the passage?A．Indications that you should give up a job.B．Suggestions on how to live comfortably.C．Instructive stories about some other people.D．Great challenges you should face and accept.For centuries, people have been on the move, to follow a food supply, in search of a better climate, or just because we wanted to know what’s over that next hill. Since our destination is usually a little foggy, we can never be 100% certain; however, we can feel confident that our decision to move is based on clear objectives and not just because we think the grass might be greener.In an attempt to cut overhead costs, companies, like people, are on the move. Company regrouping has resulted in more employees and their families moving to another state or even country. If you find yourself in this position, before you agree to move with your company, take a moment to reflect on where you are in your career and if this move will lead you to that next step.If you’ve just retired and are looking for a sunny, warm place to move to, it’s a good idea to consider all your options. Make sure you have health care services and medical insurance, cost of living and so on, On the other hand, this is a great opportunity to have a more active life, meet new friends and travel. The best way to decide if moving is right for you is to try it for a long-term vacation - three to six months - which will give you a chance to see if a permanent move fits your new lifestyle.lf you made a list of the activities you like to do, whether it’s cycling or dining out at unique restaurants, you may find that the area where you currently live does not allow you topursue these passions. If you would like to live in a place that supports a lifestyle you’ve always wanted, then make your list, take a trip, then make it happen.8．Why are people unsure about moving to a new place?A．They have been accustomed to the current life.B．They have not a clear understanding of it.C．The health care services there are not satisfactory.D．They fear that they will lose what they have.9．What should you do if your company will move to another city?A．You should follow your company without hesitation.B．You should give up your job to take care of your family.C．You should reflect on the overhead cost ahead of time.D．You should first weigh advantages and disadvantages.10．Which is best to determine if a new place agrees with you?A．Your income and the cost of living in a new place.B．Moving to a new place with your good friends.C．Spending some time in a place before you decide.D．Making careful research on your dream lifestyle.11．What does the author want to tell us in the passage?A．How to decide whether you should move or not.B．How to accustom yourself to a new environment.C．How to decide where you will move to.D．How to reduce the cost when you are to move.A special device (设备) worn on the head that sends electrical charges into the brain may improve the memory abilities of older people. The device may even give them the ability to remember things as well as a person 20 to 30 years younger than them. These are the findings of a study published in Nature Neuroscience earlier this month.Researchers working on the study found that the age-related damage to working memory can be repaired. This is possible through stimulating (刺激) two important areas of the brain at a special rhythm. Working memory is the information held temporarily for use in immediate activities such as reasoning and decision-making.42 younger adults aged 20 to 29 and 42 older adults aged 60 to 76 took part in the study.The researchers tested all subjects for their performance in a working memory activity. The findings are still early and only relate to healthy volunteers. But the findings could point to new ways to increase brain operation in older people suffering from diseases that affect these operations, such as dementia and Alzheimer’s patients.Electroencephalography, or EEG, is a method of observing brain activity. Transcranial alternating current stimulation, or tACS, is the means of sending electrical charges into the brain. Using both EEG and tACS, the scientists stimulated the brains of the group of young and old people. They were able to make small changes to the brainwave connections linked to their working memory.Without brain stimulation, the older people were slower and less correct in their memories than the younger ones. This was because the younger people had higher levels of connections of certain brain wave rhythms.Robert Reinhart, a researcher at Boston University, said that when older adults received active brain stimulation, they improved their working-memory test scores to the levels of the younger people. The effect lasted for at least 50 minutes after the stimulation was given. 12．How can the device help the old people?A．It can record what has happened for them.B．It can stimulate the brain and make it active.C．It can turn an old person into a twenty-something.D．It can delay the brain for a short while.13．What did the researchers do using the equipment?A．They repaired some age-related memory damage.B．They found the brain made up of two vital parts.C．They had 200 volunteers join in their research.D．They kept track of two groups of people.14．EEG is different from tACS becauseA．EEG is a method to give motivation to brainB．EEG can change the brainwave connectionsC．tACS can send electrical charges to the brainD．tACS is a means to help see how the brain runs15．Which of the following is the best title for the text?A．Electric Shocks May Help Older People’s MemoryB．As People Age, Their Memory Is DecreasingC．Plenty of Stimulation Can Make a Man Look YoungD．The Young Have a Problem with MemoryBound by manners, you can find it difficult to walk away from a conversation without seeming rude. 16．Next time you feel trapped, follow these tips.17．You can introduce your conversational partner to an individual with a common interest or quality and hope that they hit it off. According to etiquette (礼仪) experts, the simplest way to leave someone who is occupying your time alone at .a party is to pull someone else into the conversation.18．For example, “Bob, have you met Suzy yet?” “Hey Suzy, Bob was justte lling me about his pet dog. I know you also have one.”Tell the person you’ve enjoyed speaking with them.If your acquaintance can’t tell that you’re eager to leave, try summarizing your conversation in the hope that your exchange will come to a natural end. According to etiquette experts, you can politely say, “Beverly, I have so enjoyed speaking with you and if you will excuse me, I have just seen a colleague/ friend/ another person that I need to speak with.” 19．Give the other person a task.If you’re speaking with someone at a party or networking event, you can suggest activities to your conversational partner.20．Or ask, “Have you seen the ballroom downstairs yet? I was down there earlier and it was so wonderful. You should have a look!” “This method of ending a conversation will work best at events.A．Bring someone else into the conversation.B．Speed up your conversation with the other person.C．After a brief introduction, you are able to excuse yourself.D．With the help of etiquette experts, we have some solutions.E.Respecting your partner should come first during a conversation.F.A farewell like this one will make the other person feel respected.G.For example, say, “Hey, you should try the beef in the kitchen!”On the surface, my life seemed great. I had a dream job, and a 21 apartment. But something was off. I couldn’t 22 being chained to my desk in anoffice any longer. So I decided to 23 and travel the world, bringing only my passport, a small 24 , and my great trust fund.My co-workers asked me why I could so 25 throw away everything I fought so hard to 26 . Bu t I don’t expect everyone to. “27 ” me. I set to work packing my bag and throwing out most of my 28 , Whatever didn’t bring me joy went straight in the trash. You don’t need to own a lot of “stuff” to be 29 especially when you can buy whatever you later 30 that you need.Then I 31 a business-class seat, and sent a quick text message to friends 32 them that I was leaving the country forever. As a citizen of the world, I 33 get lonely. Everywhere I go, I meet such diverse groups of people. In hostels(招待所), I’ve 34 beers with friendly British and Australian friends. In hotels, I’ve had fun with some 35 and some locals. We all became lifelong friends, despite the 36 barriers.Once, outside the train station of a small fishing village, I met a man named Greebo, who 37 flowers and various cheap trinkets(小饰品) for a living. Greebo was happy to open up to me about his life, as long as I kept buying roses. Moved by our 38 chatting, some of his friends wandered over to join the conversation. All of our differences soon disappeared.As I left town, I cast one final 39 back at Greebo. I couldn’t help but smile. It felt good to make a 40 in the lives of these simple people.21．A．plain B．novel C．fine D．strange 22．A．imagine B．allow C．consider D．bear 23．A．break B．quit C．try D．handle 24．A．backpack B．tool C．brochure D．invention 25．A．eventually B．firmly C．lately D．casually 26．A．make B．achieve C．live D．meet 27．A．surround B．satisfy C．touch D．get 28．A．possessions B．barrier C．cases D．books 29．A．free B．grateful C．happy D．optimistic 30．A．expect B．realize C．admit D．prove 31．A．chose B．took C．removed D．reserved 32．A．indicating B．promising C．telling D．inviting 33．A．rarely B．merely C．nearly D．regularly34．A．bought B．sought C．shared D．owned 35．A．friends B．workers C．farmers D．strangers 36．A．language B．position C．location D．relation 37．A．fetched B．sold C．raised D．attended’38．A．hard B．familiar C．easy D．meaningful 39．A．glance B．memory C．role D．idea 40．A．decision B．difference C．circle D．contribution阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。

2020届吉林省通化市梅河口重点中学高三4月高考模拟数学（理）试题一､选择题:本题共12小题,每小题5分,共60分｡在每小题给出的四个选项中,只有一项是符合题目要求的｡ 1. (5分)已知集合A={x ∈Z|-2<x,1}, B ⊆A,则集合B 中的元素个数最多是( ) A.1B.2C.3D.42. (5分)若复数z=1-2i,则41izz =-() A. –iB. iC.-1D.13.(5分)已知函数||,0(),0x lnx x f x e x >⎧=⎨≥⎩则f(f(-1))=( )A.-1B.0C.1D. e4. (5分)某公司从A ､B 两个部门中各选出6名员工参加本季度的笔试考核,他们取得的成绩(满分100 分)的茎叶图如图,其中A 部门员工成绩的中位数是83,B 部门员工成绩的平均数是85,则x+y 的值为()A.2B.3C.4D.55.(5分)在△ABC 中,延长BC 至点M 使得BC=2CM,连接AM,点N 为AM 上一点且1,3AN AM =若AN=λAB + μAC,则λ+ μ=( )1.3A1.2B1.2C -1.3D -6. (5分)如图,在某观测塔塔顶A 处测得信号站B,C 的俯角分别为57°和45°,已知观测塔的高度AO=100m,则信号站B,C 间的距离约为(结果保留整数｡参考数据:sin57°≈0.84,cos57° ≈0.54)A.30mB.32mC.34mD.36m7. (5分)阳马和鳖臑(bie nao)是《九章算术商功》里对两种锥体的称谓｡如图所示,取一个长方体,按如图斜割一分为二,得两个模一样的三棱柱,称为堑堵(如图)｡再沿其中一个堑堵的一个顶点与相对的棱剖开,得四棱锥和三棱锥各一个,有一棱与底面垂直的四棱锥称为阳马(四棱锥S- ABCD),余下三棱锥称为鳖臑(三棱锥S- ECD),若将某长方体沿上述切割方法得到一个阳马一个鳖臑,且该阳马的正视图和鳖臑的侧视图如图所示,则可求出该阳马和鳖臑的表面积之和为( ).121335A ++.111335B ++.123135C ++.113135D ++8. (5分)如果执行如图的程序框图,那么输出的S=( )A.402B.440C.441D.4839.(5分)如图,已知双曲线C 2222:1(0,0)x y a b a b-=>>的左､右焦点分别为12,F F ，过2F 直线l 与双曲线C左,右两支交于点B, A,若1ABF ∆为正三角形,则双曲线C 的渐近线方程为( ).2A y x =± .3B y x =± 3.3C y x =±.6D y x =±10. (5分)421(1)x x+-展开式常数项为() A.11B. -11C.8D. -711.(5分)如图,在△ABC中,a,b,c分别为∠A,∠B,∠C所对的边,且222sin sin sin ()sin sin ,B C B C B C +=++b=c,点D 为△ABC 外一点, ∠ADC=θ,AD= CD=1,则四边形ABCD面积的最大值为( )853.A +1.2B2.C23.D +12.(5分)已知四棱锥S-ABCD 的底面边长均为1,其顶点S 在底面的射影恰好为四边形ABCD 对角线的交点,且四条侧棱与底面所成的角都相等,异面直线SA 与CD 所成角的正弦值为25,则四棱锥S-ABCD 外接球的半径为()53.A3.B3.C3.D 二､填空题:本题共4小题,每小题5分,共20分｡ 13. (5分)若0<x+β<π,tan(α+ β)=1,3tan ,4α=则tanβ=__ 14.(5分)A,B,C 三人中只有一人去游玩过七彩云南欢乐世界,当他们被问到谁去过时,A 说:C 去过了; B 说:A 去过了; C 说:我没有去过｡事实证明:三人中,只有一人说的是假话,那么游玩过七彩云南欢乐世界的人是_____15. (5分)已知定义在R 上的函数f(x)周期为2,且∀x ∈R , f(x)- f(-x)=0恒成立,当x ∈[-1, 0]时,2(),f x x =若2020()()log g x f x x =-在(0, m]上恰有2019个零点,则整数m 的最小值为___16. (5分)椭圆22:14x C y +=的左,右焦点分别为12,,F F 点M 为其上的动点,当12F MF ∠为钝角时,点M 的纵坐标的取值范围是____.三､共70分｡解答应写出文字说明､证明过程或演算步骤｡第17~21题为必考题,每个试题考生都必须作答｡第22､23题为选考题,考生根据要求作答｡(一)必考题:60分｡17. (12分)已知{}n a 是公比为q(q> 0)的等比数列,{}n b 是公差为2的等差数列,满足113133,.a b a b ===(1)求数列{}n a 和{}n b 的通项公式;(2)若数列,n n n c a b =求数列{}n c 的前n 项和为.n S18. (12分)如图,在四棱锥S - ABCD 中,底面ABCD 为直角梯形,//,,2AD BC ABC π∠=侧面SAB 为正三角形且平面SAB ⊥底面ABCD,1,2AB BC AD ==F,F 分别为SD,SB 的中点｡(1)证明:EC//平面SAB;(2)求EC 与平面FCD 所成角θ的正弦值｡19. (12分)已知抛物线2:2(0)C y px p =>的焦点到准线的距离为1,过抛物线C 上点P(2,00)(0)y y >作两条弦PA,PB 交抛物线于A, B,设其斜率分别为12,,k k 且12(k k m m =常数,m≠0)｡(1)求抛物线C 的方程; (2)求证:直线AB 恒过定点｡20. (12 分)近年来,国家相关政策大力鼓励创新创业种植业户小李便是受益者之一,自从2017年毕业以来,其通过自主创业而种植的某种农产品广受市场青睐,他的种植基地也相应地新增加了一个平时小李便带着部分员工往返于新旧基地之间进行科学管理和经验交流,新旧基地之间开车单程所需时间为i,由于不同时间段车流量的影响,现对50名员工往返新旧基地之间的用时情况进行统计,结果如表:(1)若有50名员工参与调查7人,再从这7人中随机抽取3人进行座谈,用X 表示抽取的3人中时间在40分钟的人数,求X 的分布列和数学期望:(2)某天,小李需要从旧基地驾车赶往新基地召开一个20分钟的紧急会议,结束后立即返回旧基地.(以50名员工往返新旧基地之间的用时的频率作为用时发生的概率)①求小李从离开旧基地到返回旧基地共用时间不超过110分钟的概率;｡②若用随机抽样的方法从旧基地抽取8名骨干员工陪同小李前往新基地参加此次会议,其中有Y 名员工从离开旧基地到返回旧基地共用时间不超过110 分钟,求随机变量Y 的方差｡21. (12分)已知函数22()ln ,()(1)1,f x x ax g x a x ax a a R =+=--+-∈.(1)讨论函数f(x)的单调性;(2)设F(x)= f(x)+g(x), 若对∀x ∈[1, +∞), F(x)≥0恒成立,求实数a 的取值范围.(二)选考题: 10分.请考生在第22, 23题中任选一题作答,如果多做,则按所做的第一题计分.作答时请写清题号. [选修4-4;坐标系与参数方程] (10 分)22. (10 分)在平面直角坐标系xOy 中,曲线C 的参数方程为cos (4sin x a a b y b ϕϕ=⎧>>⎨=⎩0, φ为参数),且曲线C 上的点M 对应的参数,3πϕ=直线l：2x t y ⎧=-⎪⎪⎨⎪=⎪⎩(t 为参数).(1)求曲线C 的普通方程;(2)若点A 是曲线C 上的一动点,求点A 到直线l 距离的最小值.[选修4-5:不等式选讲] (10 分) 23.已知函数f(x)=|2x+1|+|1-x|. (1)解不等式f(x)≥-x+2; (2) ∀x ∈R,不等式21()22f x t t ≥-+恒成立,求实数t 的取值范围.- 21 -。

2024届吉林省桦甸市第四中学物理高三上期中联考试题 注意事项：1．答卷前，考生务必将自己的姓名、准考证号填写在答题卡上。

2．回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其它答案标号。

回答非选择题时，将答案写在答题卡上，写在本试卷上无效。

3．考试结束后，将本试卷和答题卡一并交回。

一、单项选择题：本题共6小题，每小题4分，共24分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

1、如图所示,导线框中电流为I ,导线框垂直于磁场放置,磁感应强度为B ,AB 与CD 相距为d ,则棒MN 所受安培力大小( )A ．F BId =B ．F BIdsin = θC ．F BIdcos = θD ．sin BId F θ= 2、如图所示，平行边界MN 、PQ 间有垂直纸面向里的匀强磁场，磁场的磁感应强度为B ，两边界间距为d ，MN 上有一粒子源A ，可在纸面内沿各个方向向磁场中射入质量均为m ，电量均为q 的带正电的粒子，粒子射入磁场的速度23qBd v m =，不计粒子的重力，则粒子能从PQ 边界射出的区域长度为（ ）A ．dB ．23dC ．233dD ．32d 3、一辆汽车在平直公路上做刹车实验，0 时刻起运动过程的位移 x 与速度 v 的关系式为 x =(10－0.1v 2)(各物理量均采用国际单位制单位)，下列分析正确的是( ) A ．上述过程的加速度大小为 10m/s 2B ．刹车过程持续的时间为 5sC ．0 时刻的初速度为 l0m/sD ．刹车过程的位移为 5m4、蹦床是一项运动员利用从蹦床反弹中表现杂技技巧的竞技运动，一质量为50kg 的运动员从离蹦床1.8m 处自由下落，若从运动员接触蹦床到运动员陷至最低点经历了0.2s ，（取g=10m/s 2，不计空气阻力）则这段时间内，下列说法错误．．．．的是（ ） A ．运动员受到的合力冲量大小300N•s B ．重力的冲量大小100N•sC ．蹦床对运动员的冲量大小200N•sD ．运动员动量变化量大小300 N•s5、小红在探究弹簧的弹力与弹簧伸长量的关系时，将刻度尺悬挂在弹簧下端，刻度尺的零刻线在下方。

2020届吉林省梅河口五中（实验班）等联谊校高三上学期期中数学（理）试题一、单选题1．已知集合{}2|50A x x x =->，则C R A =（） A ．{|05}x x ≤≤ B ．{|0}x x < C ．{|5}x x >D ．{|50}x x -≤≤【答案】A【解析】求出集合A 后，根据补集定义求得结果. 【详解】{}{2500A x x x x x =-=<或}5x > {}05R C A x x ∴=≤≤本题正确选项：A 【点睛】本题考查集合运算中的补集运算，属于基础题. 2．设i 是虚数单位,如果复数i2ia ++的实部与虚部是互为相反数,那么实数a 的值为 ( ) A ．13B ．13-C ．3D ．3-【答案】D【解析】分析：由复数代数形式的乘除运算化简复数，再由已知条件列出方程，求解即可得答案．详解：2a i i ++=()()()()()()2212225a i i a a i i i +-++-=+-=21255a ai +-+， ∵复数2a ii ++的实部与虚部是互为相反数， ∴212055a a+-+=，即a=3-． 故选：D ．点睛：本题考查了复数代数形式的乘除运算，考查了复数的实部与虚部的概念，属于基础题．3．若向量(0,2)m =-u r ，n =r ，则与2m n +u r r共线的向量可以是（ ）A ．(3,1)-B ．(1,3)-C ．(3,1)--D ．(1,3)--【答案】B【解析】先利用向量坐标运算求出向量2m n +v v,然后利用向量平行的条件判断即可. 【详解】()()0,2,3,1m n =-=v v Q()23,3m n ∴+=-v v()()31,33,3-=--故选B 【点睛】本题考查向量的坐标运算和向量平行的判定,属于基础题,在解题中要注意横坐标与横坐标对应,纵坐标与纵坐标对应,切不可错位.4．设a ，b∈R，那么“＞1”是“a＞b ＞0”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 【答案】B【解析】试题分析：a ＞b ＞0，可推出，而当，时，例如取a=﹣2，b=﹣1，显然不能推出a ＞b ＞0，由充要条件的定义可得答案． 解：由不等式的性质，a ＞b ＞0，可推出，而当，时，例如取a=﹣2，b=﹣1，显然不能推出a ＞b ＞0． 故是a ＞b ＞0的必要不充分条件．故选B ．【考点】必要条件、充分条件与充要条件的判断．5．某四棱锥的三视图如图所示，则该四棱锥的最长的长度为（ ）．A ．23B ．32C ．22D ．2【答案】A【解析】先由三视图得出该几何体的直观图，结合题意求解即可. 【详解】由三视图可知其直观图，该几何体为四棱锥P-ABCD ，最长的棱为PA ，则最长的棱长为2223PA PB PC =+=故选A ． 【点睛】本题主要考查几何体的三视图，属于基础题型. 6．等比数列{}n a 的首项为32，公比为12-，前n 项和为n S ，则当*n N ∈时，1nn S S -的最小值与最大值的比值为（ ） A ．512-B ．710-C ．910D ．512【答案】B【解析】先计算得到11()2n n S =--，111()121()21n n n n S S ---=---，构造函数1()11f t t t=---，证明函数单调递减，得到最大值和最小值.【详解】等比数列{}n a 的首项为32，公比为12-，前n 项和为11()3121()12212n n n S --=⨯=--+ 111()121()21n n n n S S ---=---设1()2n t -=，则max min 11,42t t ==-数列对应函数为：1()11f t t t=---易知：1t -在11,24⎡⎤-⎢⎥⎣⎦上递减，11t --在11,24⎡⎤-⎢⎥⎣⎦上递减故1()11f t t t =---在11,24⎡⎤-⎢⎥⎣⎦上递减 max min 1517()(),()()26412f t f f t f =-===-1n n S S -的最小值与最大值的比值为710- 故选：B 【点睛】本题考查了数列的最大最小值，构造数列1()11f t t t=---是解题的关键，可以简化运算.7．.某汽车公司的A,B 两个装配厂可装配甲、乙两种不同型号的汽车,若A 厂每小时可装配1辆甲型车和2辆乙型车,B 厂每小时可装配3辆甲型车和1辆乙型车.现要装配40辆甲型车和40辆乙型车,若要使所费的总工作时数最少,则这两个装配厂的工作时数分别为( ) A ．16,8 B ．15,9C ．17,7D ．14,10【答案】A【解析】根据条件列可行域与目标函数，结合图象确定最小值取法，即得结果. 【详解】设A 厂工作x 小时， B 厂工作y 小时，总工作时数为z ，则目标函数为z x y =+，约束条件为340,240,0,0.x y x y x y +≥⎧⎪+≥⎪⎨≥⎪⎪≥⎩作出可行域如图所示，由图知当直线y x z =-+经过Q 点时，z取得最小值，由340,240,x y x y +=⎧⎨+=⎩可得()16,8Q ，故A 厂工作16小时，B 厂工作8小时，可使所费的总工作时数最少.选A.【点睛】线性规划的实质是把代数问题几何化，即数形结合的思想.需要注意的是：一，准确无误地作出可行域；二，画目标函数所对应的直线时，要注意与约束条件中的直线的斜率进行比较，避免出错；三，一般情况下，目标函数的最大或最小值会在可行域的端点或边界上取得.8．已知正数x 、y 满足1x y +=，则141x y++的最小值为（ ） A ．2 B ．92 C ．143D ．5【答案】B【解析】由1x y +=得(1)2x y ++=，再将代数式(1)x y ++与141x y++相乘，利用基本不等式可求出141x y++的最小值． 【详解】1x y +=Q ，所以，(1)2x y ++=，则141441412()[(1)]()5591111x y x yx y x y x y y x y x+++=+++=++=++++g …，所以，14912x y ++…， 当且仅当4111x y y x x y +⎧=⎪+⎨⎪+=⎩，即当2313x y ⎧=⎪⎪⎨⎪=⎪⎩时，等号成立，因此，141x y ++的最小值为92， 故选：B ． 【点睛】本题考查利用基本不等式求最值，对代数式进行合理配凑，是解决本题的关键，属于中等题．9．已知函数()cos f x x x =+，把函数()f x 的图象向右平移3π个单位，再把图象的横坐标缩小到原来的一半，得到函数()g x 的图象，当0,2x π⎡⎤∈⎢⎥⎣⎦时，方程()0g x k -=有两个不同的实根，则实数k 的取值范围为（ ）A．⎡⎣B．)2C ．[]1,2D ．[)1,2【答案】D【解析】化简函数为()26f x sin x π⎛⎫=+ ⎪⎝⎭，由平移变换与伸缩变换得到()226g x sin x π⎛⎫=- ⎪⎝⎭，然后数形结合可得实数k 的取值范围.【详解】函数()cos 26f x x x sin x π⎛⎫=+=+ ⎪⎝⎭，把函数()f x 的图象向右平移3π个单位，再把图象的横坐标缩小到原来的一半， 得到函数()226g x sin x π⎛⎫=-⎪⎝⎭当0,2x π⎡⎤∈⎢⎥⎣⎦时，方程()0g x k -=有两个不同的实根等价于函数()226g x sin x π⎛⎫=- ⎪⎝⎭与y k =有两个不同交点，令t 52666x πππ⎡⎤=-∈-⎢⎥⎣⎦，，即y 2sint =与y k =有两个不同交点， 结合图象可知：12k ≤< 故选D 【点睛】函数零点的求解与判断(1)直接求零点：令()0f x =，如果能求出解，则有几个解就有几个零点； (2)零点存在性定理：利用定理不仅要函数在区间[],a b 上是连续不断的曲线，且()()0f a f b ⋅<，还必须结合函数的图象与性质(如单调性、奇偶性)才能确定函数有多少个零点；(3)利用图象交点的个数：将函数变形为两个函数的差，画两个函数的图象，看其交点的横坐标有几个不同的值，就有几个不同的零点． 10．执行如图所示的程序框图，输出的S 值为（ ）A 20171B 20181C ．20191D 20201【答案】D【解析】程序框图表示的是数列1n a n n =++2019项和，利用裂项相消法得到答案. 【详解】 设数列1n a n n =++2019项和11n a n n n n ==+++即2019122019...2132 (2020201920201)S S a a a ==+++== 故选：D【点睛】本题考查了程序框图，确定程序框图表示的是数列n a =2019项和是解题的关键.11．甲、乙、丙三人中，一人是教师、一人是记者、一人是医生.已知：丙的年龄比医生大；甲的年龄和记者不同；记者的年龄比乙小.根据以上情况，下列判断正确的是（ ） A ．甲是教师，乙是医生，丙是记者 B ．甲是医生，乙是记者，丙是教师 C ．甲是医生，乙是教师，丙是记者 D ．甲是记者，乙是医生，丙是教师 【答案】C【解析】由甲的年龄和记者不同和记者的年龄比乙小可以推得丙是记者，再由丙的年龄比医生大，可知甲是医生，故乙是教师. 故选C12．已知定义在R 上的连续奇函数()f x 的导函数为()f x '，当0x >时，()()0f x f x x'+>，则使得()()()2213310xf x x f x +-->成立的x 的取值范围是（ ）A ．()1,+∞B ．()11,1,5⎛⎫-+∞ ⎪⎝⎭U C ．1,15⎛⎫ ⎪⎝⎭D ．(),1-∞【答案】C【解析】根据0x >时()()0f x f x x'+>可得：()()0xf x f x '+>；令()()g x xf x =可得函数在()0,∞+上单调递增；利用奇偶性的定义可证得()g x 为偶函数，则()g x 在(),0-∞上单调递减；将已知不等式变为()()231g x g x >-，根据单调性可得自变量的大小关系，解不等式求得结果. 【详解】当0x >时，()()0f x f x x'+> ()()0xf x f x '∴+>令()()g x xf x =，则()g x 在()0,∞+上单调递增()f x Q 为奇函数 ()()()()g x xf x xf x g x ∴-=--== ()g x ∴为偶函数则()g x 在(),0-∞上单调递减()()()2213310xf x x f x ∴+-->等价于()()231g x g x >-可得：231x x >-，解得：115x << 本题正确选项：C 【点睛】本题考查函数奇偶性和单调性的综合应用问题，关键是能够构造函数，根据导函数的符号确定所构造函数的单调性，并且根据奇偶性的定义得到所构造函数的奇偶性，从而将函数值的大小关系转变为自变量之间的比较.二、填空题13．已知函数2()(1)f x ax ab x b =+--，如果不等式()0f x >的解集为()1,3-，那么不等式()20f x -<的解集为________________. 【答案】31{|}22x x x <->或 【解析】先得到不等式()0f x <的解集为(,1)(3,)-∞-+∞U ，再确定()20f x -<的解为21x -<- 或23x ->，解得答案. 【详解】不等式()0f x >的解集为()1,3-，则不等式()0f x <的解集为(,1)(3,)-∞-+∞U()20f x -<的解为：21x -<- 或23x ->解得答案：31{|}22x x x <->或 故答案为：31{|}22x x x <->或【点睛】本题考查了解不等式，将2x -看成整体可以简化运算，是解题的关键.14．观察下列式子：3211=，332123+=，33321236++=，33332123410+++=，…，根据以上式子可猜想：3333123n ++++=L ________________．【答案】()212n n +⎡⎤⎢⎥⎣⎦.【解析】因为33321=11+2=12+，（）,33321+2+3=123（）++,333321+2+3+4=1234+++（）, 所以()233331+2+3+123n n =++++L L =()2214n n +15．若函数的图象如图所示,则图中的阴影部分的面积为 ；【答案】【解析】试题分析：由图像，得,即，即；令，得；由定积分的几何意义，得所求阴影部分的面积为.【考点】1.三角函数的图像与性质；2.定积分的几何意义. 16．底面半径为1cm 的圆柱形容器里放有四个半径为12cm 的实心铁球，四个球两两相切，其中底层两球与容器底面相切. 现往容器里注水，使水面恰好浸没所有铁球，则需要注水________________. 【答案】12(3π+cm 3 【解析】设四个实心铁球的球心为1234,,,O O O O ，其中12,O O 为下层两球的球心，四个球心连线组成棱长为1 的正四面体，,,,A B C D 分别为四个球心在底面的射影，则ABCD是一个边长为2的正方形，所以注水高为正四面体相对棱的距离与球半径的二倍的和，即为12+， 故应注水的体积等于以注入水的高度为高的圆柱的体积减去四个球的体积，341(14232ππ⎛⎫+-⨯ ⎪⎝⎭＝1(3π+，故答案为313cm π⎛+ ⎝⎭.三、解答题17．设数列{}n a 的前n 项和为n S .已知24S =，121n n a S +=+，*n N ∈. （1）求通项公式n a .（2）求数列{}2n a n --的前n 项和.【答案】（1）13-=n n a ，*n N ∈；（2）n S 23152n n n---=【解析】（1）根据11n n n a S S ++=-即可化简得13n n a a +=，可证明数列为等比数列，即可求出通项公式（2）采用分组求和的方法，利用等差数列、等比数列的求和公式，即可求解. 【详解】（1）由题意得12214,21,a a a a +=⎧⎨=+⎩则121,3.a a =⎧⎨=⎩又当n 2≥时，由()()1121212n n n n n a a S S a +--=+-+=，得13n n a a +=， 所以数列{}n a 是以1为首项，公比为3的等比数列， 所以13-=n n a ，*n N ∈.（2）记()()()()1232122232n n n S a a a a --=--+--+--++L L()12[345(2)]n a a a n =+++-+++++L L L L L2213(32)315315132222n n n n n n n n n-++-+---=-=-=-. 【点睛】本题主要考查了等比数列的证明、通项公式，求和公式，等差数列的求和公式，分组求和，属于中档题.18．某工厂利用辐射对食品进行灭菌消毒，现准备在该厂附近建一职工宿舍，并对宿舍进行防辐射处理，建房防辐射材料的选用与宿舍到工厂距离有关.若建造宿舍的所有费用p (万元)和宿舍与工厂的距离x (km )的关系为(08)35kp x x =≤≤+,若距离为1km 时，测算宿舍建造费用为100万元.为了交通方便，工厂与宿舍之间还要修一条道路，已知购置修路设备需5万元，铺设路面每公里成本为6万元，设f (x )为建造宿舍与修路费用之和. (1)求f (x )的表达式(2)宿舍应建在离工厂多远处，可使总费用f (x )最小并求最小值. 【答案】（1）800()56,0835f x x x x =++≤≤+ （2）宿舍应建在离厂5km 处可使总费用()f x 最小为75万元. 【解析】(1)先代入数据计算800k =，再把两部分费用相加得到答案. (2)先变形800()2(35)535f x x x =++-+,再利用均值不等式得到答案. 【详解】（1）根据题意，距离为1km 时，测算宿舍建造费用为100万元100800315kk =∴=⨯+Q800()56,0835f x x x x ∴=++≤≤+（2）800()2(35)58057535f x x x =++-≥-=+ 当且仅当8002(35)35x x =++即5x = 时min ()75f x = 【点睛】本题考查了函数的应用，均值不等式，意在考查学生的应用能力和解决问题的能力. 19．如图，在四边形ABCD 中，7,2,AC CD AD ==2.3ADC π∠=（1）求CAD ∠的正弦值；（2）若2BAC CAD ∠=∠，且△ABC 的面积是△ACD 面积的4倍，求AB 的长. 【答案】（1）217（27【解析】（1）ACD ∆中，设(0)AD x x =>，利用余弦定理得到1x =，再利用正弦定理得到答案.（2）利用面积关系得到sin 4sin .AB BAC AD CAD ⋅∠=⋅∠化简得到cos 2.AB CAD AD ⋅∠=根据（1）中sin 7CAD ∠=解得答案. 【详解】（1）在ACD ∆中，设(0)AD x x =>， 由余弦定理得2227=422cos 3x x x x +-⨯⋅π 整理得277x =，解得1x =. 所以1, 2.AD CD ==由正弦定理得2sin sin 3DC ACDAC =∠π，解得sin DAC ∠= （2）由已知得4ABC ACD S S ∆∆=， 所以11sin 4sin 22AB AC BAC AD AC CAD ⋅⋅∠=⨯⋅⋅∠， 化简得sin 4sin .AB BAC AD CAD ⋅∠=⋅∠所以2sin cos 4sin ,AB CAD CAD AD CAD ⋅∠⋅∠=⋅∠ 于是cos 2.AB CAD AD ⋅∠=因为sin 7CAD ∠=，且CAD ∠为锐角，所以cos CAD ∠==代入计算217AB ⨯=⨯因此AB = 【点睛】本题考查了正弦定理，余弦定理，面积公式，意在考查学生利用正余弦定理解决问题的能力.20．各项均为正数的等比数列{}n a 中，已知152,512,n a a T ==是数列{}2log n a 的前n 项和.（1）求数列{}n a 的通项公式； （2）求n T ； （3）求满足231111011(1)(1)(1)2013n T T T --->L L 的最大正整数n 的值. 【答案】（1）212n na -=，（2）2n T n =，（3）223【解析】（1）直接利用等比数列公式计算得到答案. （2）先计算得到22og 1l n a n =-，前N 项和2n T n = （3）化简231111(1)(1)(1)2n n T T T n +---=L L 再解不等式1101122013n n +>得到答案.【详解】（1）41512,512(0),4a a a q q q ===>=，故121242n n n a --=⨯=（2）22212o 1l g log 2n n a n -==-，2(121)2n n nT n +-== （3）231111111111(1)(1)(1)(1)(1)(1)(1)...(1)(1)22332n n T T T n n n+---=-+-+-+=L L 即1101122013n n +>解得6713n < 故最大正整数223n = 【点睛】本题考查了等比数列通项公式，等差数列前N 项和，数列不等式，意在考查学生对于数列公式方法的灵活运用.21．已知函数()ln 3f x a x ax =-- (0)a ≠． (1)讨论()f x 的单调性；(2)若()(1)40f x a x e +++-≤对任意2[,]x e e ∈恒成立，求实数a 的取值范围（e 为自然常数）； (3)求证：22221111ln(1)ln(1)ln(1)...ln(1)1234n++++++++<*(2,)n n ≥∈N ． 【答案】（1）当0a >时，()f x 的单调增区间为(0,1]，单调减区间为[1,)+∞；当0a <时，()f x 的单调增区间为[1,)+∞，单调减区间为(0,1]；（2）212e e a --≤（3）证明见解析【解析】(1)求导得到'(1)()a x f x x-=，讨论0a >和0a <两种情况得到答案. (2) 令()()(1)4ln 1F x f x a x e a x x e =+++-=++-，讨论()F x 的单调性，计算()F x 的最值得到答案.(3) 令1a =-，()ln 3f x x x =-+-在[1,)+∞上单调递增，得到ln 1x x <-对一切(1,)x ∈+∞成立，故2211111ln(1)(1)1n n n n n n+<<=---代入计算得到到答案. 【详解】（1）函数的定义域为()0,∞+，'(1)()a x f x x-=当0a >时，()f x 的单调增区间为(0,1]，单调减区间为[1,)+∞； 当0a <时，()f x 的单调增区间为[1,)+∞，单调减区间为(0,1]； （2）令()ln 3(1)4ln 1F x a x ax a x e a x x e =--+++-=++-, 则'()a x F x x +=，令'()0a x F x x+==，则x a =-, （a ）若a e -≤,即a e ≥- 则()F x 在2[,]e e 是增函数,22max()()210F x F e a e e ==++-≤ , 212e e a --≤ 无解.（b ）若2a e -≥即2a e ≤-，则()F x 在2[,]e e 是减函数,max ()()10F x F e a ==+≤ 1a ≤- 所以2a e ≤-,（c ）若2e a e <-<,即2e a e -<<-，()F x 在[,]e a -是减函数, 在2[,]a e -是增函数,最大值22()210F e a e e =++-≤可得212e e a --≤,()10F e a =+≤可得1a ≤-所以2212e e e a ---≤≤ ,综上所述212e e a --≤ ,（3）令1a =-，此时()ln 3f x x x =-+-，所以(1)2f =-，由（1）知()ln 3f x x x =-+-在[1,)+∞上单调递增，∴当(1,)x ∈+∞时，()(1)f x f >即ln 10x x -+->，∴ln 1x x <-对一切(1,)x ∈+∞成立， ∵*2,n n N ≥∈，则有2211111ln(1)(1)1n n n n n n+<<=---， 所以 22221111ln(1)ln(1)ln(1)...ln(1)234n ++++++++ 1111111(1)()()...()223341n n <-+-+-+--111n=-<【点睛】本题考查了函数的单调性，恒成立问题，不等式的证明，其中放缩2211111ln(1)(1)1n n n n n n+<<=---并用裂项相消法是解题的关键. 22．在直角坐标系xOy 中，圆C 的参数方程1cos sin x y ϕϕ=+⎧⎨=⎩（ϕ为参数），以O 为极点，x 轴的非负半轴为极轴建立极坐标系.（1）求圆C 的极坐标方程；（2）直线l的极坐标方程是2sin 3πρθ⎛⎫+= ⎪⎝⎭:3OM πθ=与圆C 的交点为O 、P ，与直线l 的交点为Q ，求线段PQ 的长. 【答案】（1）2cos ρθ=；（2）2【解析】（1）首先利用221cos sin ϕϕ+=对圆C 的参数方程1{x cos y sin ϕϕ=+=（φ为参数）进行消参数运算，化为普通方程，再根据普通方程化极坐标方程的公式得到圆C 的极坐标方程．（2）设11P ρθ（，），联立直线与圆的极坐标方程，解得11ρθ，；设22Q ρθ（，），联立直线与直线的极坐标方程，解得22ρθ，，可得PQ ． 【详解】（1）圆C 的普通方程为()2211x y -+=，又cos x ρθ=，sin y ρθ= 所以圆C 的极坐标方程为2cos ρθ=.（2）设()11,ρθP ，则由2{3cos ρθπθ==解得11ρ=，13πθ=，得1,3P π⎛⎫⎪⎝⎭；设()22Q ,ρθ，则由2sin 3{3πρθπθ⎛⎫+= ⎪⎝⎭=解得23ρ=，23πθ=，得3,3Q π⎛⎫⎪⎝⎭； 所以Q 2P = 【点睛】本题考查圆的参数方程与普通方程的互化，考查圆的极坐标方程，考查极坐标方程的求解运算，考查了学生的计算能力以及转化能力，属于基础题. 23．已知函数()|1|f x x =- （1）解不等式()(4)8f x f x ++≥；（2）若||1,||1,0a b a <<≠，求证：()||()bf ab a f a>. 【答案】（1）{}|53x x x ≤-≥或 （2）证明见解析【解析】（1）得到分段函数()22,3(4)4,3122,1x x f x f x x x x --<-⎧⎪++-≤≤⎨⎪+>⎩＝，分别计算不等式得到答案.（2）不等式等价于1||||ab a b >--，证明22|1|||0ab a b --->得到答案. 【详解】（1）()22,3+(4)134,3122,1x x f x f x x x x x x --<-⎧⎪+-++-≤≤⎨⎪+>⎩＝＝当3x <-时，由228x --≥，解得5x ≤-； 当31x -≤≤时，()8f x ≥不成立； 当1x >时，由228x +≥，解得3x ≥．综上所述：不等式()4f x ≤ 的解集为5{}3|x x x ≤-≥或． （2）()||()b f ab a f a>，即1||||ab a b >-- ．11a b <<Q ，，()()()()22222222|1|||212110ab a b a b ab a ab b a b ∴---=-+--+=--> ，所以1||||ab a b >--．故所证不等式成立．【点睛】本题考查了解绝对值不等式，不等式的证明，将绝对值不等式转化为分段函数是常用的技巧，需要灵活掌握.。

2024学年吉林省桦甸四中、梅河五中、蛟河一中、柳河一中、舒兰一中、辉南六中重点高中盟校高三暑期调研考试物理试题试卷注意事项：1． 答题前，考生先将自己的姓名、准考证号填写清楚，将条形码准确粘贴在考生信息条形码粘贴区。

2．选择题必须使用2B 铅笔填涂；非选择题必须使用0．5毫米黑色字迹的签字笔书写，字体工整、笔迹清楚。

3．请按照题号顺序在各题目的答题区域内作答，超出答题区域书写的答案无效；在草稿纸、试题卷上答题无效。

4．保持卡面清洁，不要折叠，不要弄破、弄皱，不准使用涂改液、修正带、刮纸刀。

一、单项选择题：本题共6小题，每小题4分，共24分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

1、两质点A 、B 同时、同地、同向出发，做直线运动。

v t -图像如图所示。

直线A 与四分之一椭圆B 分别表示A 、B 的运动情况，图中横、纵截距分别为椭圆的半长轴与半短轴（椭圆面积公式为S ab π=，a 为半长轴，b 为半短轴）。

则下面说法正确的是（ ）A ．当2s t =时，a b 1.5m/s v v ==B ．当a 3m/s v =，两者间距最小C ．A 的加速度为23m/s 2D ．当B 的速度减小为零之后，A 才追上B 2、带电粒子仅在电场力作用下，从电场中a 点以初速度进入电场并沿虚线所示的轨迹运动到b 点，如图所示，实线是电场线，关于粒子，下列说法正确的是A ．在a 点的加速度大于在b 点的加速度B ．在a 点的电势能小于在b 点的电势能C ．在a 点的速度小于在B 点的速度D ．电场中a 点的电势一定比b 点的电势高3、一物体静止在升降机的地板上，在升降机加速上升的过程中，地板对物体的支持力所做的功等于（ ）A．物体势能的增加量B．物体动能的增加量C．物体动能的增加量加上物体势能的增加量D．物体动能的增加量减去物体势能的增加量4、2019年7月9日，在沈阳进行的全国田径锦标赛上，来自上海的王雪毅以1米86的成绩获得女子跳高冠军。