成都市“五校联考”高2014级第五学期九月考试题i
四川省成都市五校2013-2014学年高一下学期期中联考英语卷(带解析)
四川省成都市五校2013-2014学年高一下学期期中联考英语卷(带解析)1.–I’m sorry for being late. I should have phoned you earlier.–_____. I’ve just arrived.A. That’s no trouble.B. You are welcome.C. That’s all right.D. You can never tell.【答案】C【解析】试题分析:句意:--很抱歉来晚了。
我应该先打电话给你的。
--_______.我刚刚到。
A项意思是:一点也不麻烦;B项意思是:不用谢;C项意思是:好的,行;没关系;D项意思是:你永远说不清楚。
所填部分是对I’m sorry的应答,根据情景选C。
考点:考查交际用语。
2.The fire was finally brought under control, but extensive damage _________.A. is causedB. was causingC. had causedD. had been caused【答案】D【解析】试题分析:句意:火势被控制住,但是巨大破坏已经造成。
从The fire was finally brought under control,可知造成破坏是在过去的动作之前发生了,是过去的过去的动作,所以用过去完成时,而且damage和cause是被动关系,选D。
考点:考查时态语态3.When and where to build the new factory _____not decided yet .A. isB. areC. hasD. have【答案】A【解析】试题分析:句意:何时何地建新工厂还没有决定呢?When and where to build the new factory是疑问词+不定式作主语,而且何时何地为一个概念,所以谓语动词用单数,工厂是被建造,排除C。
四川省成都市五校2013-2014学年高二下学期期中联考英语卷(带解析)
四川省成都市五校2013-2014学年高二下学期期中联考英语卷(带解析)1.-- I heard that you lost your wallet and your i-Phone.--- Oh,__________ !A.never mind B.that’s all right C.just my luck D.thank you 【答案】C【解析】试题分析:A.never mind没关系,B.that’s all right没关系,C.just my luck真倒霉, D.thank you谢谢,这句话的意思是:--我听说你丢了钱包和i-Phone。
所以第二个人说:--哦,真倒霉。
根据句意选C。
考点:考查交际用语2.I was blessed with a happy childhood, _____ most people would want to have. A. that B. it C. the one D. one【答案】D【解析】试题分析:句意:我幸运地享有一个幸福的童年,一个大多数人想要的童年。
这里需要一个代词泛指a childhood,而one可以泛指前面的可数名词单数,that特指前面的可数名词单数或不可数名词,it是特指前面的同类同物,the one是特指前面的可数名词单数,所以选D。
考点:考查代词3.At long last the child was found on top of _____ we call the museum of science and technology.A. whichB. thatC. whatD. where【答案】C【解析】试题分析:句意:最后,这个孩子在我们称为是科技馆的顶上被找到了。
这句话中的of后面接的是宾语从句,宾语从句中call后面接的是双宾语,而这里显然缺少一个宾语,所以用what连接宾语从句,选C。
考点:考查名词性从句的连接词4.Although her disease her eyesight and forced her to leave the dance floor, she refuses to fall into self-pity.A. has affectedB. affectedC. affectsD. had affected【答案】A【解析】试题分析:句意:虽然他的病已经影响了她的视力并迫使她离开了舞台,但是她拒绝自怨自艾。
四川省成都市五校2014年高一下学期期中联考语文试题
四川省成都市五校2013-2014学年高一下学期期中联考语文试题第Ⅰ卷(单项选择:共27分,每小题3分)一、语言基础1.下列加点字注音完全正确的一项是()(3分)A.敛裾(jù)愆期(qiān)袅娜(nuó)垝垣(yuán)B.羁鸟(jī)梵(fán)语踯躅(zhú)箜篌(kōng)C.窈窕(yáo)嗟(jiē)悼婆娑(suō)徘徊(huái)D. 溘死(kè)伶俜(pīng)房檩(lǐn)茎(jīng)叶2.下列各组中没有错别字的是()A. 气概班驳丰姿萧索B.宛然嬉游缈茫淅沥C.悠闲葱茏形骸殒落D. 落莫葳蕤鸷鸟蜷缩3.下列各句中,加点的成语使用恰当的一项是()A刘邦来了个金蝉脱壳,顺利离开了项羽的军营。
B白居易在地方为官很注意接近民众,不管是乡间农夫还是下里巴人,他都能谈得来,从他们那里得到很多创作素材。
C小王和他哥哥很有默契,不用语言交流,道路以目即可明白彼此的心思。
D.打好基础才能造房子,这个道理很浅显,但是得寸进尺的心态,却往往影响人们对这一真理的认识。
4、下列各句中,没有语病,句意明确的一项是()A出版业当然要讲究装帧艺术,讲究宣传造势和市场营销,但是要想真正赢得读者、赢得市场,还是取决于出版物的内容是否具有吸引力和感染力。
B从上次被老师批评后,我更加小心了,作业的错误率降低了2倍。
C新学期开学后,学校加强了监管的力度,尤其是对学生上网的问题上,更是不遗余力。
D作为一名共产党员,党的领导干部,办事情、想问题都要以党和人民的根本利益为出发点。
二、阅读下面的短文,完成5—7题,每题3分关于“封建”“封建”一词见于《诗经》和《左传》,文献解诂词义为“封邦建国,以藩屏西周王室”。
周朝建立后,为了稳定新征服的地区,实行大规模的“封建”。
所谓“封建”就是“封邦建国”、“封建亲戚”。
当时一共“封建”了71个诸侯国,其中多数是周王室的同姓诸侯。
2014年,5月,九年级,英语联考试题,及答案
九年级英语月考试题一、听力部分(25分)第一节听下面五段小对话和对话后的问题,从每小题A、B、C选项中选出最佳选项。
每段对话仅读一遍。
1.A. Never B. Sometimes C. Every day2.A. Mother’s Day B.Her mother’s birthday C.Her mother doesn’t have3.A.$10 B.$40 C.504.A.By bus B.By bike C.By car5.A.A cook B.A singer C.A nurse第二节听下面四段长对话和一段独白,每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项。
每段对话或独白读两遍。
听下面第一段对话,回答6-7小题。
6. What match was there yesterday evening?A. A tennis matchB. A football matchC. A table tennis match7. What was the woman doing at 8:00 yesterday evening?A. Watching TVB. Playing gamesC. Surfing the Internet.听下面第二段对话,回答8-9小题。
8. What’s the matter with the woman?A. She has a coldB. She has a toothacheC. She has a cough9. What did the doctor ask her to do?A. Stay in hospitalB. Take a restC. Take some medicine听下面第三段对话,回答10-12小题。
10. What will Jane do this weekend?A. Join a club. B. Go to a concert. C. See a movie.11. How much does a ticket cost? A.¥20. B.¥40. C.¥60.12. What will the speakers do to raise money for the concert?A. Sell newspapers.B. Sell flowers.C. Ask their parents for help.听下面第四段对话,回答13-15小题。
2024届四川省成都市“五校联考”高三教学质量检测试题(一模)英语试题含解析
2024届四川省成都市“五校联考”高三教学质量检测试题(一模)英语试题注意事项1.考生要认真填写考场号和座位序号。
2.试题所有答案必须填涂或书写在答题卡上,在试卷上作答无效。
第一部分必须用2B 铅笔作答;第二部分必须用黑色字迹的签字笔作答。
3.考试结束后,考生须将试卷和答题卡放在桌面上,待监考员收回。
第一部分(共20小题,每小题1.5分,满分30分)1.At the end of the historic area,Wilmington displayed its ________ as a working port city:large ware-houses and a few other dated office buildings.A.achievement B.reputationC.character D.standard2.Those successful deaf dancers think that dancing is an activity ________ sight matters more than hearing. A.when B.whose C.which D.where3.I want to tell you is the deep love and respect I have for my parents.A.That B.Which C.Whether D.What4.________about the man wearing sunglasses during night that he was determined to follow him.A.So curious the detective wasB.So curious was the detectiveC.How curious was the detectiveD.How curious the detective was5.Don’t take it seriously,Alice.I wasn’t making _____ fun of you —it’s nothing but _____ joke.A./; the B.the; theC.the; a D./; a6.The writer was so ________ i n her work that she didn’t notice him enter the room.A.abandoned B.focused C.absorbed D.centered7.—Did you go to last night’s concert?— Y es. And the girl playing the violin at the concert _______ all the people present with her excellent ability.A. impressed B.compared C.conveyed D.observed8.A hearty laugh relieves physical tension, _____your muscles relaxed for over half an hour.A.to leave B.left C.leaving D.leave9.In the forest, sound is the best means of communication over distance ________ in comparison with light, it won't be blocked by trees when travelling.A.while B.becauseC.when D.though10."We can not afford limited progress. We need rapid progress," Ban said at the Third World Climate Conference inGeneva, by the World Meteorological Organization (WMO).A.organized B.being organized C.organizing D.was organized11.He felt ________ of cheating in the exam, deciding never to do such things again.A.shame B.ashamedC.sorry D.shameful12.—Next week I will go to a job interview. will you give me some suggestions?—Smiling is a great way to make yourself ________.A.stand out B.turn outC.work out D.pick out13.-----My room gets very cold at night.-----_________________.A.So is mine B.So mine is C.So does mine D.So mine does14.-What if the rainstorm continues?-Come on guys! We have to meet the _______ whatever the weather.A.standard B.demandC.deadline D.satisfaction15.No student ________ go out of school to have lunch without permission of the headteacher.A.might B.mustC.shall D.could16.________ the students from their endless homework the school has decided to take a series of measures. A.Freed B.To freeC.Freeing D.Having freed17.The new playground to be built next year will be ____________ the old one.A.as three times big as B.three times as big asC.as big as three times D.as big three times as18.______ to nuclear radiation, even for a short time, may influence genes in human bodies.A.Having exposed B.Being exposedC.To expose D.Exposed19.They are determined to go into the dark cave, ________ my warning of danger.A.regardless of B.because ofC.apart from D.instead of20.— Can you tell me something about _________ science.— OK. _ Nobel Prize in Chemistry is usually awarded to Americans.A.the; The B./; The C.a; / D.the; /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。
四川省成都市五校2013-2014学年高一下学期期中联考英语试题
四川省成都市五校2013-2014学年高一下学期期中联考英语试题(全卷满分:150分完成时间:120分钟)第一部分听力(共20小题;每小题1.5分,满分30分),从A、B、C三个选项中,选出可以填入空白处的最佳选项。
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,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间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 woman want to do?A. To have an X ray.B. To go to the hospital.C. To help the wounded man.2. Where and when will the meeting be held?A. Room 303, 300 pm.B. Room 303, 200 pm.C. Room 302, 200 pm.3. When would Thomas and Lily like to leave?A. Tomorrow.B. Next Monday or Tuesday.C. This Monday.4. What is the man’s choice?A. He prefers train for trip.B. He doesn’t like traveling.C. Not mentioned.5. According to the woman, what should the man do at first?A. He should ask about the flat on the phone.B. He should read the advertisements for flats in the newspaper.C. He should phone and make an appointment.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。
2014年秋九年级五校联考英语竞赛测试题听力录音稿和答案
2014年秋九年级五校联考英语竞赛测试题听力录音稿和答案Text 1: M: Jay, do you get nervous while listening to English tapes?W: No, never. If you do more listening, you’ll find it easier.Q: What are they talking about?Text 2: M: How was the exam, Kangkang?W: All the students passed the exam except Gina.Q: Who failed the exam?Text 3: W: Be quick, Jack! You’ll be late for school.M: Don’t worry, Mum. It’s only 7:30. I have 20 minutes left.Q: When will the school begin?Text 4: M: That turkey sm ells good! When’s dinner?W: Be patient. Aunt will arrive any minute with salad and pumpkin pies.M: I can hardly wait.Question: Which festival are they talking about?Text 5: M: When are you leaving for your planned holiday in Hawaii?W: It’s out of the question. I’ve got a new job. I won’t be free until my training is over.Q: Why can’t the woman take her holiday?Text 6: M: Linda, do you know if we’ll have a sports meeting?W: Yes, it’ll be held next Saturday.M: Saturday? It would be better if we had it on Sunday.Question: When will they have the sports meeting?Text 7: M: I prefer basketball and volleyball. What about you, Mary?W: Tennis. I started playing it when I was 11. I have been playing it for three years.M: You are great!Question: How old is Mary now?Text 8: M: Look, Mary. Someone is coming. Who can it be? Is it Lisa?W: It must be Carla. She looks like Lisa.Q: Who is the girl?Text 9: W: Did you use to be afraid of the dark, Jack?M: Yes, I did. I used to sleep with the bedroom light on.W: Are you still afraid of the dark now?M: No. But I’m still afraid of snakes.Q: What did Jack use to be afraid of?Text 10: W: Wha t’s your trouble, Jack?M: I felt very sick at school just now.W: I think you got a bad cold. Go back home and stay in bed. Take this medicine.You’ll be better soon.Q: Where is Jack now?Text 11: M: Where would you like to go this winter vacation, Kathy?W. I don’t know, maybe somewhere warm.M: What about Hong Kong? It’s warm all year round.W: Well, I hear it’s very touristy.M: How about Sydney?W: Sydney sounds a good place to visit. It’s not warm but beautiful. It’s also a good place for shopping.Text 12: W: Kangkang, may I ask you some questions?M: Of course. Go ahead.W: How long have you been learning English?M: About three years.W: Do you often speak English with your teacher and classmates?M: Not often. But I listen to the teacher carefully in class.W: How about after class?M: I often try to speak English, but often end with Chinese.W: I think you’d better speak English more after class. Practice makes perfect.M: OK, I will. Thank you.Text 13: M: Hi, Susan. Do you go out on school nights?W: No, I’m not allowed to go out on school nights. I have to stay at home to study.M: How about on weekend nights?W: Sometimes, and my parent must be with me. I can’t go to my friends.M: Then what do you do on weekend nights?W: I can be allowed to watch TV or surf the Internet.M: Wow, you’re luckier.W: Why?M: My parents ask me to do more homework.Text 14: M:Hello,Jane.Have you ever been to the new sports centre?W:No,never,Bob! Where is it?M:On Straight Road.You know,near Long Street,beside the station.W:Oh.Is it good?M:Yes,it’s great! You can do lots of sports there.Last week I went there with my father and we played table tennis and volleyball.W:What about swimming?M:Not yet.They are going to build a swimming pool next year.W:Is it expensive?M:Not really,Jane.It’s $60 a year if you are over 18,and $40 if you are 15 to 18.W:Oh,that’s good because I’m 16.M:And on Saturday and Friday it stays open late-till 10 o’clock.W:Oh,great.How did you get there last week?M:I got the number 14 bus.It’s only ten minutes.Do you want to go next week?W:OK.Any day except Sunday.M:Why don’t we go on Friday? Then we can stay late.W:OK.Let’s meet after sehoo1.Text 15: W: Richard Travel. Good morning, can I help you?M: Yes, please. I want to book a flight.W: Certainly, sir. Where to?M: Madrid, Spain.W: A flight to Madrid. When do you want to travel?M: Um, 15th January, and I want to return to London on 10th March.W: So that’s Madrid on 15th, and you want to come back to London on 10th March.M: Tha t’s right.W: Do you want first or tourist class?M: Tourist, please.W: Thank you, sir. Can I have your name and address?M: Yes, of course. My name is Maggie Shore, and the address is 56 Bridge Street, London.Text 16:People have flown kites in Japan for more than 1,000 years.There are many different kinds of kites there.Some 1ook like bats,others look like birds. Most have pictures on them.There are many interesting stories about kites in Japan.One of the stories tells about a father and a son who were on a small lonely island in the middle of the Japanese sea.There were no boats or ships.They couldn't go back to the mainland. So they made a big kite.His son flew on it back to Japan.There is a K-Day in Japan.The young men in Japan have kite matches.When the kites are flying,the match starts. The competitors try to break each other's kite strings. The last kite left in the sky is the winner.参考答案1~5 CACAB 6~10 CBBAB 11~15 CCBAC 16~20 CCACC 21~25 ABBCB 26~30 BCABC 31~35 BDABC 36~40 CCDAA 41~45 CBADA 46~50 BBACD 51~54 BDCD 55~58 DCBD 59~62 DCDB 63~67 CCBDA68. you’re busy 69. it tastes terrible70. hide bad news 71. everything was fine /OK72. kept/went , possible 73. however, succeed74. 对身处逆境的人说一些鼓励性的话能让他们打起精神,帮助他们渡过难关。
四川省成都市五校协作体2014-2015学年高二上学期期中考试英语试题
四川省成都市五校协作体2014-2015学年高二上学期期中考试英语试题说明: 本试卷共两卷, 第I卷和第II卷。
第I卷的答案请涂在答题卡上, 第II卷的答案请写在答题卡上的规定位置。
交卷时交答题卡。
试卷总分为 150 分, 考试时间为 120 分钟。
第Ⅰ卷(选择题,共90分)第一部分听力测试(共两节,满分20分)第一节(共5小题; 每小题1分,满分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,并标在试卷的相应位置。
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. Whose birthday is it today?A. Mary's.B. Mike's.C. David's.2. What's the probable relationship between the two speakers?A. Husband and wife.B. Customer and assistant.C. Classmates.3. What is the woman going to do?A. Pick up Jack.B. Take her son to school.C. Attend an important meeting.4. What does the man want to do?A. Make friends with the woman.B. Buy a book.C. Send a book to his friend.5. Where does this conversation most likely take place?A. In a supermarket.B. In a restaurant.C. At home.第二节(共15小题,每小题1分,满分15分)听下面 5 段对话或独白。
2014年秋九年级五校联考英语竞赛测试题
2014年秋九年级五校联考英语竞赛测试题第一部分听力(共30分)第一节:请听10段对话及问题,从A、B、C 三个中选出准确的回答。
每段对话听一遍。
(10分)( ) 1. A. Reading English. B. Speaking English. C. Listening English. ( ) 2. A. Gina. B. Kangkang. C. All the students.( ) 3. A. 7:10. B. 7:30. C. 7:50.( ) 4. A. Thanksgiving. B. April Fool’s Day. C. Spring Festival.( ) 5. A. Because she doesn’t have enough money. B. Because she has to take training.C. Because she has to take a train to her work place.( ) 6. A. This Saturday. B. Next Sunday. C. Next Saturday.( ) 7. A. 11 years old. B. 14 years old. C. 3 years old.( ) 8. A. It is Lisa. B. It is Carla. C. It is Mary.( ) 9. A. The dark. B. Snakes. C. Dogs.( ) 10. A. At school. B. At a doctor’s. C. At home.第二节:请听五段长对话,从A、B、C三个选项中选出问题的准确回答。
每段对话听两遍。
(16分)听第11段对话,完成第11、12 小题。
( ) 11. How does Kathy think of Hong Kong?A. Warm.B. Touristy.C. Warm but touristy. ( ) 12. Where do you think Kathy might go this winter vacation?A. Hong Kong.B. Singapore.C. Sydney.听第12段对话,完成第13-15 小题。
四川省成都市五校2013-2014学年高二下学期期中联考英语试题
四川省成都市五校2013-2014学年高二下学期期中联考英语试题(全卷满分:150分完成时间:120分钟)第I卷(选择题共90分)第一部分英语知识运用(共两节,共40分)第一节单项填空(共10小题;每小题1分,共10分)1. -- I heard that you lost your wallet and your i-Phone.--- Oh,__________ !A.never mind B.that’s all right C.just my luck D.thank you2. I was blessed with a happy childhood, _____ most people would want to have.A. thatB. itC. the oneD. one3. At long last the child was found on top of _____ we call the museum of science and technology.A. whichB. thatC. whatD. where4. Although her disease her eyesight and forced her to leave the dance floor, she refuses to fall into self-pity.A. has affectedB. affectedC. affectsD. had affected5. ----He thinks that girls are smarter than boys when it comes to learning English.----But two years ago he thought_______.A. soB. otherwiseC. oppositeD. different6. Will was stupefied with exhaustion, and he _____ his head on the grass under one of the trees and slept, but he saw a cat acting strangely.A. should have liedB. ought to have layC. might have laidD. must have laid7. The new policy allows a couple to have a second birth _____either is an only child.A. thoughB. ifC. unlessD. until8. With our country entering an aging society, ______ the retirement age is probably unavoidable.A. having delayedB. delayingC. being delayedD. to be delayed9. This restaurant has an inviting, homelike atmosphere ___ many others are short of.A. whereB. whenC. thatD. what10. “I hope the dialogue,” said the spokesman, “______ between the two presidents next week will give us some active signals.A. makingB. to makeC. to be madeD. made第二节完形填空(共20小题,每小题1.5分,共30分)Music to My EarsRobby was 10 for his first piano lesson in my class. Much as he tried, he __11__ even the basic rhythm. However, he dutifully reviewed the pieces that I required.Over the months he tried and tried while I __12__ and encouraged him. At the end of each lesson he'd always say, "My mom's going to hear me play some day." __13__ it seemed hopeless.I only knew his mother from a __14__as she waited in her aged car to pick him up. Then one day Robby stopped coming. I was secretly __15__ that he stopped because of his lack of ability.Weeks later I informed the students, including Robby, of the coming recital(独奏). To my __16__, Robby asked me if he could be included. I told him he really did not qualify because he had __17__ out. He said his mom had been sick and unable to take him to lessons but he was still __18__ .“I've just got to play!" he __19__. Somet hing inside me let me allow him to.Then came the recital night. The gym was __20__ with parents. I put Robby up __21__, thinking that I could save his poor performance through my “curtain closer(谢幕).”The recital went off smoothly. Then Robby came up on stage. His clothes were wrinkled andhis hair was __22__. "Why didn't his mother at least make him comb his hair for this special night?" I thought.Robby pulled out the piano bench and began. I was not __23__ for what I heard next. His fingers were __24__ on the keys. Never had I heard Mozart played so well by people of his age. After he ended, everyone was __25__ their feet in wild applause.In __26__ I ran up on stage and put my arms around Robby. "I've never heard you play like that, Robby! How did you __27__ it?"Robby explained, "Well, Miss Hondorf…remember I told you my mom was sick? …__28__ she had cancer and passed away yesterday. She was born deaf, so tonight she could hear me play in heaven. I wanted to make it special."There wasn’t a __29__ e ye in the house. That night I felt he was the teacher and I was the pupil, for it was he who taught me the meaning of perseverance and __30__.11. A. had B. lacked C. showed D. got12. A. learned B. checked C. listened D. played13. A. And B. But C. So D. Or14. A. conversation B. performance C. distance D. picture15. A. guilty B. anxious C. glad D. sad16. A. surprise B. relief C. pleasure D. satisfaction17. A. stepped B. worn C. run D. dropped18. A. acting B. practicing C. performing D. recording19. A. insisted B. suggested C. complained D. threatened20. A. packed B. lined C. piled D. covered21. A. least B. most C. first D. last22. A. cool B. messy C. neat D. dull23. A. eager B. concerned C. prepared D. grateful24. A. hesitating B. crawling C. touching D. dancing25. A. over B. under C. in D. on26. A. chaos B. return C. silence D. tears27. A. find B. feel C. make D. like28. A. Gradually B. Suddenly C. Frequently D. Actually29. A. bright B. curious C. dry D. wet30. A. love B. talent C. regret D. courage第二部分阅读理解(共两节,共50分)第一节阅读理解(共20小题;每小题2分,共40分)ADreams can be familiar and strange, fantastical or boring, but some dreams might be connected to the mental processes that help us learn. In a recent study, scientists found a connection between nap-time dreams and better memory in people who were learning a new skill.In the study,99 college students between the ages of 18 and 30 each spent an hour on a compute, trying to get through a virtual maze(迷宫).The maze was different place each time they tired—making it even more difficult. They were also told to find a particular picture of a tree and remember where it was.For the first 90 minutes of a five-hour break, half of the participants stayed awake and an half were told to take a short nap .Participants who stayed awake were asked to describe their thoughts. Participants who took a nap were asked about their dreams before sleep and after sleep—and they were awakened within a minute of sleep to describe their dreams.Stickgold, a neuroscientist(神经科学家),wanted to know what people were dreaming about when their eyes weren't moving during sleep.Four of the 50 people who slept said their dreams were connected to the maze. Some dreamedabout the music that had been playing when they were working ; others said they dreamed about seeing people in the maze. When these four people tried the computer maze again, they were able to find the tree faster than before their naps.Stickgold suggests the dream itself doesn‘t help a person learn—it's the other way around.He suspects that the dream was caused by the brain processes associated with learning.All four of the people who dreamed about the task had done poorly the first time, which makes Stickgold wonder if the dreams show up when a person finds a new task particularly difficult. People who had other dreams, or people who didn't take a nap, didn't show the same improvement.31. Before having a short nap, participants of the experiment were asked to .A. stay in a different place in the mazeB. design a virtual maze which is difficult to get throughC. experience the experiment and try to remember somethingD. get through a virtual maze on a computer from the same place32. What can we learn from the text?A. Participants who took a nap were required to express their thoughts.B. Some dreams may encourage people to invent something new.C. Participants who dreamed about films could finish the task more easily.D. Participants whose dreams had something to do with the maze could findthe tree faster.33. According to Stickgold, .A. every person may dream about what they learnedB. people's brain processes may still be connected with their learningin their dreamsC. once people's eyes stop moving, they are sure to dream about somethingD. no matter how fantastical or boring, dreams are connected with people's life34.What is the best title for this text?A. Dreams Are StrangeB. Not All Dreams Are TrueC. Dreaming Makes PerfectD. Stickgold, a Dream ExpertBFarmers, especially in developing countries, are often criticized for cutting down forests. But a new study suggests that many farmers recognize the value of keeping trees.Researchers using satellite images found at least ten percent tree cover on more than one billion hectares of farmland. That is almost half the farmland in the world. Earlier estimates were much lower but incomplete. The authors of the new study say it may still underestimate the true extent worldwide.The study found the most tree cover in South America. Next comes Africa south of the Sahara, followed by Southeast Asia. North Africa and West Asia have the least.The study found that climate conditions alone could not explain the amount of tree cover in different areas. Nor could the size of nearby populations, meaning people and trees can live together. There are areas with few trees but also few people, and areas with many trees and many people. The findings suggest that things like land rights, markets or government policies can influence tree planting and protection.Dennis Garrity, who heads the World Agroforestry Center, says farmers are acting on their own to protect and plant trees. The problem, he says, is that policy makers and planners have been slow to recognize this and to support such efforts.The satellite images may not show what the farmers are using the trees for, but trees provide nuts, fruit, wood and other products. They also help prevent soil loss and protect water supplies. Even under drought(干旱)conditions, trees can often provide food and a way to earn money until the next growing season.Some trees act as natural fertilizers. They take nitrogen(氮气)out of the air and put it in the soil. Scientists at the Center say the use of fertilizer trees can re-duce the need for chemical nitrogen by up to three-fourths. Trees also capture carbon dioxide, a gas linked to climate change.35. Through the study, the researchers found that .A. there are more trees on farmlands than expectedB. fewer trees are being cut in developing countriesC. most farmers still don’t realize the value of treesD. trees play a key role in preventing climate change36. Which of the following has the least tree cover?A. Southeast AsiaB. West Asia.C. South America.D. Africa south of the Sahara.37. In Dennis Garrity’s opinion, .A. most farmers care about nothing but their own interestsB. there are usually few people living in areas with few treesC. government plays a small role in tree planting and protectionD. government should support farmers in planting and protecting trees38. The sixth paragraph mainly tells about .A. how farmers plant treesB. what products trees can bringC. the importance of trees to farmersD. the environmental value of tree coverCNuclear power’s danger to health, safety, and even life itself can be summed up in one word radiation.Nuclear radiation has a certain mystery about it, partly because it cannot be detected by human senses. It can’t be seen or heard, or touched or tasted, even though it may be all around us. There are other things like that. For example, radio waves are all around us but we can’t detect them, sense them, without a radio receiver. Similarly, we can’t sense rad ioactivity without a radiation detector. But unlike common radio waves, nuclear radiation is not harmless to human beings and other living things.At very high levels, radiation can kill an animal or human being immediately by killing masses of cells in vital organs. But even the lowest levels can do serious damage. There is no levels of radiation that is completely safe. If the radiation does not hit anything important, the damage may not be serious. This is the case when only a few cells are hit, and if they are killed immediately, your body will replace the dead cells with healthy ones. But if the few cells are only damaged, and if they reproduce themselves, you may be in trouble. They can grow into cancer. Sometimes this does not show up for many years.This is another reason for some of the mystery about nuclear radiation. Serious damage can be done without the victim being aware at the time that damage has occurred. A person can be irradiated(辐射)and feel fine, then die for cancer five, ten, or twenty years later as a result. Or a child can be born weak as a result of radiation absorbed by its grandparents.Radiation can hurt us. We must know the truth.39. What is the main idea of the passage?A. How to detect nuclear radiation.B. How radiation kill a man.C. The mystery about nuclear radiation.D. Serious damage caused by nuclear radiation.40. Which of the following statements is true?A. Nuclear radiation can cause cancer to human beings.B. Nuclear radiation can be safe to human beings if its level is low.C. Nuclear radiation can be detected by human senses.D. Nuclear radiation is just like common radio waves.41. What is not the reason why nuclear radiation has a certain mystery?A. The hurt cells can stay in the body many years and then grow into cancer.B. It can do harm to a person while the victim isn't aware the damagehas occurred.C. Nuclear radiation can kill a person very easily.D. Radiation can seldom kill a person immediately.42. If a human being is hit by nuclear radiation, he may _______.A. die of cancer after many yearsB. die immediatelyC. have a child who may be born weakD. all of the aboveDFrom good reading we can derive pleasure, companionship, experience, and instruction. A good book may absorb our attention so completely that for the time being we forget our surroundings and even our identity. Reading good books is one of the greatest pleasures in life. It increases our contentment when we are cheerful, and lessens our troubles when we are sad. Whatever may be our main purpose in reading, our contact with good books should never fail to give us enjoyment and satisfaction.With a good book in our hands we need never be lonely. Whether the characters portrayed are taken from real life or are purely imaginary, they may become our companions and friends. In the pages of books we can walk with the wise and the good of all lands and all times. The people we meet in books may delight us either because they resemble human friends whom we hold dear or because they present unfamiliar types whom we are glad to welcome as new acquaintances. Our human friends sometimes may bore us, but the friends we make in books need never weary us with their company. By turning the page we can dismiss them without any fear of hurting their feelings. When human friends desert us, good books are always ready to give us friendship, sympathy, and encouragement.One of the most valuable gifts bestowed by books is experience. Few of us can travel far from home or have a wide range of experiences, but all of us can lead varied lives through the pages of books. Whether we wish to escape from the seemingly dull realities of everyday life or whether we long to visit some far-off place, a book will help us when nothing else can. To travel by book we need no bank account to pay our way; no airship or ocean liner or stream-lined train to transport us; no passport to enter the land of our heart's desire. Through books we may get the thrill of hazardous adventure without danger. We can climb lofty mountains, brave the perils of an Antarctic winter, or cross the scorching sands of the desert, all without hardship. In books we may visit the studios of Hollywood; we may mingle with the gay throngs of the Paris boulevards; we may join the picturesque peasants in an Alpine village or the kindly natives on a South Sea island. Indeed, through books the whole world is ours for the asking. The possibilities of our literary experiences are almost unlimited. The beauties of nature, the enjoyment of music, the treasures of art, the triumphs of architecture, the marvels of engineering, are all open to the wonder and enjoyment of those who read.43. Why is it that we sometimes forget our surroundings and even our identity while reading?A. No one has come to disturb you.B. The book you are reading is so interesting and attractive.C. Everything is so quiet and calm around you.D. Your book is overdue; you are finishing it at a very fast speed.44. How would you account for the fact that people like their acquaintances in books even more?A. They resemble human friends exactly.B. They are unfamiliar types we like.C. They never desert us.D. They never hurt our feelings.45. Which of the following is true?A. Your wish to visit some far-off place can be realized through the pages of the books.B. To escape from the dull realities of everyday life you should take up reading.C. You may obtain valuable experience from reading good books.D. Books can always help you to live a colorful life.46. The word “weary” means ______.A. “to attract someone’s attention”B. “to distract someone’s attention”C. “to make someone interested”D. “to make someone very tired”EHave you ever picked a job based on the fact that you were good at it but later found it made you feel very uncomfortable over time? When you select your career, there's a whole lot more to it than assessing your skills and matching them with a particular position. If you ignore your personality, it will hurt you long-term regardless of your skills or the job's pay. There are several areas of your personality that you need to consider to help you find a good job. Here are a few of those main areas;1) Do you prefer working alone or with other people?There are isolating(使孤立)jobs that will drive an outgoing person crazy and also interactive jobs that will make a shy person uneasy. Most people are not extremes in either direction but do have a tendency that they prefer. There are also positions that are sometimes a combination of the two, which may be best for someone in the middle who adapts easily to either situation.2) How do you handle change?Most jobs these days have some elements of change to them, but some are more than others. If you need stability in your life, you may need a job where the changes don't happen so often. Other people would be bored of the same daily routine.3) Do you enjoy working with computers?I do see this as a kind of personality characteristic. There are people who are happy to spend more than 40 hours a week on a computer, while there are others who need a lot of human interaction throughout the day. Again, these are extremes and you'll likely find a lot of positions somewhere in the middle as well.4) What type of work environment do you enjoy?This can range from being in a large building with a lot of people you won't know immediately to a smaller setting where you'll get to know almost all the people there fairly quickly.5) How do you like to get paid?Some people are motivated by the pay they get, while others feel too stressed to be like that. The variety of payment designs in the sales industry is a typical example for this.Anyway, these are a great starting point for you. I've seen it over and over again with people that they make more money over time when they do something they love. It may take you a little longer, but making a move to do what you have a passion for can change the course of your life for the better.47. Which of the following is TRUE according to the passage?A. Isolating jobs usually drive people mad.B. Almost everyone has a tendency in jobs.C. Interactive jobs make people shy easily.D. Extreme people tend to work with others.48. What does the underlined sentence in paragraph one mean?A. Before you select your job, you should assess your skills and match them withyour positionB . You should ignore your skills when you select job.C. Nothing is important than assessing skills and match them with the positionwhen you select job.D. There are more important things than assessing skills and match them with theposition when you select job.49. What is the missing word about a job search in the following chart?A. Design.B. Changes.C. Cooperation.D. Hobbies.A. Lifestyles and Job PayB. Jobs and EnvironmentC. Personalities and JobsD. Job Skills and Abilities第二节根据对话内容,从对话后的选项中选出能填入空白处的最佳选项,并在答题卡上将该项涂黑。
四川省成都市五校2013-2014学年高二下学期期中联考数学(文)试题
四川省成都市五校2013-2014学年高二下学期期中联考数学(文)试题)(全卷满分:150分完成时间:120分钟)一选择题(本题共10个小题,每小题5分)1.“”是“方程表示焦点在y轴上的椭圆”的()。
(A)充分而不必要条件(B)必要而不充分条件(C)充要条件 (D) 既不充分也不必要条件2.双曲线的实轴长是 ( ) 。
(A)2 (B) (C) 4 (D) 43.已知命题:,则()。
A. B.C. D.4.下列命题中的假命题是( )。
A.∀x∈R,2x-1>0 B.∀x∈N*,(x-1)2>0C.∃x∈R,lg x<1 D.∃x∈R,tan x=25.设曲线在点处的切线斜率为,则点的坐标为()。
A、B、C、D、6.下列说法中正确的是( )。
A.一个命题的逆命题为真,则它的逆否命题一定为真B.“a>b”与“a+c>b+c”不等价C.“a2+b2=0,则a,b全为0”的逆否命题是“若a,b全不为0,则a2+b2≠0”D.一个命题的否命题为真,则它的逆命题一定为真7. 抛物线y=4x2的准线方程是()。
(A)x=-1 (B)y=-1 (C)x=-(D)y=-8. 离心率e=是双曲线的两条渐近线互相垂直的()。
(A)充分条件(B)必要条件(C)充要条件(D)不充分不必要条件9. 抛物线y2=-8x中,以(-1, 1)为中点的弦所在的直线方程是()。
(A)x-4y-3=0 (B)x+4y+3=0 (C)4x+y-3=0 (D)4x+y+3=010. 设A(-2, ),椭圆3x2+4y2=48的右焦点是F,点P在椭圆上移动,当|AP|+2|PF| 取最小值时P点的坐标是()。
(A)(0, 2) (B)(0, -2) (C)(2, ) (D)(-2, )二.填空题(本题共5个小题,每小题5分)11.若,则的值为;12. 已知F1,F2是椭圆的两焦点,过点F2的直线交椭圆于A,B两点,在中,若有两边之和是10,则第三边的长度为 _____13.设双曲线的渐近线方程为,则的值为__ ____14.若双曲线的离心率是2,则的最小值为_ __ 15.下列四个命题中,真命题的序号有__ __ __(写出所有真命题的序号).①将函数y=的图象按向量v=(-1,0)平移,得到的图象的函数表达式为y=②圆x2+y2+4x+2y+1=0与直线y=相交,所得弦长为2③若sin(+)= ,sin(-)=,则tan cot=5④如图,已知正方体ABCD- A1B1C1D1,P为底面ABCD内一动点,P到平面AA1D1D的距离与到直线CC1的距离相等,则P点的轨迹是抛物线的一部分.三.简答题(本题共6个小题,共75分)16.(本题12分)双曲线与椭圆有相同焦点,且经过点,求双曲线方程。
成都市2014年高中阶段统一招生考试英语真题(含答案)
成都市2014年高中阶段教育学校统一招生考试(含成都市毕业初三会考)A 卷(选择题:共100分)第一部分听力测试(共25小题,25分)一、听句子,根据所听到的内容选择正确答语。
每小题念两遍。
(共6小题,每小题1 分;计6 分)1 A.Thank you. B.You are welcome. C.Do you want more?2 A.He’s in the library.B.He’s at home.C.He’s from Russian.3 A.She is tall. B.She is outgoing. C.She likes art.4 A.Sorry to hear that. B.Yes, I’d love to.C.Yes, you are right.5 A.Do some sports. B.Clean up your room. C.Take this medicine.6 A.They are American. B.They are beautiful. C.They have a sweet voice.二、听句子,选出与所听句子相符的图片,并将代表图片的字母填在答题卡的相应位置,每小题念两遍。
(共4小题,每小题1分;计4分)7.___________ 8.___________9.___________10.___________三、听对话,根据对话内容选择正确的答案。
每小题念两遍。
(共10 小题,每小题1 分;计10 分)11 A.It’s cloudy.B.It’s raining.C.It’s sunny.12 A.The bank. B.The bookstore. C.The post office.13 A.In a college. B.In a school. C.In a hospital.14 A.Excited. B.Worried C.Relaxed.15 A.Small. B.Medium. C.Large.16 A.Nothing. B.A little. C.A lot.17 A.No, she doesn’t.B.She isn’t.C.Yes, she does.18 A.Mary B.Linda. C.Jane.19 A.After-school activities B.Looking after the old C.Study plans.20 A.At 7:00. B.At 7:50. C.At 8:20.四、听短文,根据短文内容选择正确的答案。
成都市高2014级五校联考语文试题与答案
成都市高2014级五校联考语文试题与答案成都市高2014级五校联考语文试题第Ⅰ卷阅读题甲必考题一、现代文阅读(9分,每小题3分)阅读下面的文字,完成1—3题。
骚扰电话该谁管?最近,网上流行这样一句话:“每天叫醒我的,不是闹钟,也不是梦想,而是骚扰电话。
”虽有戏谑与夸张成分,但未尝不是人们深受骚扰电话之害的真实写照:理财推销、发票开具、中奖兑换、房屋租售、辅导培训,骚扰电话五花八门,商业推广与电信诈骗混为一体,让人感觉“接不完,不胜烦”。
骚扰电话到底有多少?今年3月,百度发布的《中国互联网安全白皮书》显示,2015年,全国骚扰电话总量为948亿条,较2014年上涨57%。
这么大的数量,即使设套诈骗的概率有限,最后被骗用户的绝对数也不少。
最不可思议的是,很多人手机上显示,一些被安全软件标注了上千次甚至上万次的骚扰电话还能继续拨打,岂非怪事?客观地看,骚扰电话的界定确实不容易。
以常见的骚扰号码400+电话为例,除非含有色情、暴力、淫秽等明显违法内容,否则很难认定哪些电话是骚扰、哪些内容属诈骗。
更有人认为,电话通信是公众权利,在不能判定某个号码涉嫌违法的情况下,运营商没有权力擅自停止服务。
这种似是而非的“付费就能使用论”,好像有一定道理,但听起来不免让人心塞。
对每年数以几百亿计的骚扰电话,难道只能听之任之?据知情人透露,目前骚扰电话分工精细,已经形成了完整的产业链条。
一些所谓高科技软件公司开发出的系统,既能批量外呼,还能语音自动群呼,更能随意更改主叫号码,为骚扰电话推波助澜;有的地方电信运营商为了追求利润,滥用透传技术,出售通道帮不法分子建立呼叫中心,提供交换机等设备支持,更助长了骚扰电话的气焰。
有媒体报道,在这条黑色利益链背后,每通话1分钟,运营商能获得五六分钱的收益。
打击骚扰电话,离不开电信运营商的主动作为。
对个别地方运营商参与骚扰电话产业链牟利问题,各大运营商总部应严格行业规范,定期开展网络安全普查,清除内部害群之马。
【物理】四川省成都市五校协作体联考2014-2015学年高二上学期期中试卷
四川省成都市五校协作体联考2014-2015学年高二上学期期中物理试卷参考答案与试题解析一、选择题(本题包括8个小题,每小题3分,共24分,每小题只有一个选项符合题意)1.(3分)下列说法正确的是()A.只有体积很小的带电体才能看作点电荷B.电场强度是矢量,方向就是电荷在该点的受力方向C.电容器的电容跟它两极所加电压成反比,跟它所带电量成正比D.外电路断开时,电动势数值上等于电源两极的电压考点:电场强度;电容.分析:带电体看作点电荷的条件,当一个带电体的形状及大小对它们间相互作用力的影响可忽略时,这个带电体可看作点电荷,是由研究问题的性质决定.电场强度方向与正电荷所受的电场力方向相同.电容与电容器板间电压和电量无关.外电路断开时,电动势数值上等于电源两极的电压.解答:解:A、带电体能否看作点电荷是由研究问题的性质决定,与自身大小形状无具体关系,故A错误;B、电场强度是矢量,方向就是正电荷在该点的受力方向,而与负电荷在该点的受力方向相反,故B错误.C、电容表征电容器容纳电荷本领大小,与两极所加电压、带电量无关,故C错误.D、外电路断开时,电源的内电压为零,根据闭合电路欧姆定律可知电动势数值上等于电源两极的电压,故D正确.故选:D.点评:本题考查的知识点较多,关键要理解点电荷的条件、电容的物理意义和电动势与电压的关系.2.(3分)两个分别带有电荷量﹣Q和+3Q的相同金属小球(均可视为点电荷),固定在相距为r的两处,它们间库仑力的大小为F.两小球相互接触后将其固定距离变为,则两球间库仑力的大小为()A.B.C.D.12F考点:库仑定律;电荷守恒定律.专题:计算题.分析:清楚两小球相互接触后,其所带电量先中和后均分.根据库仑定律的内容,根据变化量和不变量求出问题.解答:解:接触前两个点电荷之间的库仑力大小为F=k,两个相同的金属球各自带电,接触后再分开,其所带电量先中和后均分,所以两球分开后各自带点为+Q,距离又变为原来的,库仑力为F′=k,所以两球间库仑力的大小为.故选C.点评:本题考查库仑定律及带电题电量的转移问题.3.(3分)三个点电荷电场的电场线分布如图所示,图中a、b两点处的场强大小分别为E a、E b,电势分别为φa、φb,则()A.E a>E b,φa<φb B.E a<E b,φa<φb C.E a>E b,φa>φb D.E a<E b,φa>φb考点:电场线.分析:电场线密的地方电场强度大,电场线稀的地方电场强度小,沿电场线电势降低,据此可正确解答本题.解答:解:电场线密的地方电场强度大,电场线稀的地方电场强度小,故E a>E b;根据电场线与等势线垂直,在b点所在电场线上找到与a点电势相等的,依据沿电场线电势降低,可知φa<φb,故BCD错误,A正确.故选:A.点评:本题考查了如何根据电场线的分布判断电场强度大小以及电势高低,对于不在同一电场线上的两点,可以根据电场线和等势线垂直,将它们转移到同一电场线上进行判断.4.(3分)如图所示,在A板附近有一电子由静止开始向B板运动,则关于电子到达了B 板时的速率,下列解释正确的是()A.两板间距越大,加速的时间就越长,则获得的速率越大B.两板间距越小,加速度就越大,则获得的速率越大C.与两板间的距离无关,仅与加速电压U有关D.以上解释都不正确考点:匀强电场中电势差和电场强度的关系.专题:电场力与电势的性质专题.分析:根据动能定理求出电子获得的速度,判断与什么因素有关,根据牛顿第二定律和运动学公式求出加速度和运动时间,判断与什么因素有关.解答:解:根据动能定理知,qU=,解得:v=,获得的速率与加速电压有关,与间距无关.加速度a=,知间距越小,加速度越大.根据d=,解得:t=,间距越大,时间越长,故A、B、D错误,C正确.故选:C.点评:根据电子的运动的规律,列出方程来分析电子的加速度、运动的时间和速度分别与哪些物理量有关,根据关系式判断即可.5.(3分)平行板间加如图所示周期变化的电压,重力不计的带电粒子静止在平行板中央,从t=0时刻开始将其释放,运动过程无碰板情况.图中,能定性描述粒子运动的速度图象正确的是()A.B.C.D.考点:带电粒子在匀强电场中的运动;牛顿第二定律.专题:带电粒子在电场中的运动专题.分析:不计重力的带电粒子在周期变化的电场中,在电场力作用下运动.速度随着时间变化的关系由加速度来确定,而加速度是由电场力来确定,而电场力却由电势差来确定.解答:解:开始粒子在匀强电场中从静止运动,前半个周期是匀加速运动,后半个周期是匀减速运动,在下一个周期中仍是这样:继续向前匀加速运动,再匀减速运动,这样一直向前运动下去.速度的方向不变,而大小先增大后减小,再增大,再减小.故选:A点评:带电粒子正好是从零时刻由静止开始运动,加速度、速度具有周期性与对称性.6.(3分)热敏电阻是传感电路中常用的电子元件,其电阻R随温度t变化的图线如图甲所示.如图乙所示电路中,热敏电阻R t与其它电阻构成的闭合电路中,当R t所在处温度升高时,两电表读数的变化情况是()A.A变大,V变大B.A变小,V变小C.A变小,V变大D.A变大,V变小考点:闭合电路的欧姆定律.专题:恒定电流专题.分析:当R t所在处温度升高时,R t减小,并联部分电阻减小,总电阻减小,根据闭合电路欧姆定律分析路端电压和总电流的变化,再分析R2电压的变化,确定通过R2的电流变化,根据总电流和通过R2的变化,确定电流表示数的变化.解答:解:当R t所在处温度升高时,R t减小,并联部分电阻减小,外电路的总电阻减小,根据闭合电路欧姆定律可知:干路电流I总增大,路端电压U变小,则V变小.R t和R2并联的电压U并=E﹣I总(R1+r),I总增大,E、R1、r均不变,则U并减小,通过R2的电流I2减小,则A变小.故B正确.故选:B.点评:本题首先要读懂图象,明确热敏电阻与温度的关系,其次按“分部→整体→分部”的顺序进行动态分析.7.(3分)电流表的内阻足R g=100Ω,满刻度电流值是I g=1mA,现欲把这电流表改装成量程为3V的电压表,正确的方法是()A.应串联一个2900Ω的电阻B.应并联一个0.1Ω的电阻C.应串联一个0.1Ω的电阻D.应并联一个2900Ω的电阻考点:把电流表改装成电压表.专题:实验题;恒定电流专题.分析:把电流表改装成电压表需要串联分压电阻,应用串联电路特点与欧姆定律求出电阻阻值.解答:解:把电流表改装成电压表需要串联一个分压电阻,串联电阻阻值:R=﹣R g=﹣100=2900Ω;故选:A.点评:本题考查了电压表的改装,知道电压表的改装原理,应用串并联电路特点即可正确解题.8.(3分)如图所示,实线表示电场线,虚线表示只受电场力作用的带电粒子的运动轨迹.粒子先经过M点,再经过N点,以下正确的是()A.无法判断带电粒子的电性B.粒子在M点的动能小于在N点的动能C.粒子在M点受到的电场力大于在N点受到的电场力D.粒子在M点的电势能小于在N点的电势能考点:电场线;电势能.分析:带电粒子的轨迹向下弯曲,则带电粒子所受的电场力沿电场线切线向下,则知带电粒子带正电,由电场线的疏密可判断场强的大小,再判断电场力的大小.由带电粒子的轨迹可判定电场力的方向,确定电场力做功情况,分析电势能和动能的变化,再分析速度的变化.带电粒子的动能和电势能总和守恒.解答:解:A、带电粒子的轨迹向下弯曲,则带电粒子所受的电场力沿电场线切线向下,则知带电粒子带正电,故A错误.B、粒子先经过M点,再经过N点,电场力做正功,动能增大,速度也增大,故带电粒子在N点的速度大于在M点的速度,粒子在N点的动能大于在M点的动能.故B正确.C、电场线的疏密表示场强大小,由图知粒子在M点的场强小于N点的场强,在M点的加速度小于N点的加速度.故C错误.D、粒子先经过M点,再经过N点,电场力做正功,电势能减小,粒子在M点的电势能大于在N点的电势能.故D错误.故选:B点评:此类轨迹问题,由轨迹的弯曲方向可判定电场力的方向,并判断电场力做功正负情况.二、不定项选择题(本题包括5个小题,每小题4分,共20分,每小题给出一个或多个选项符合题意.全部选对得4分,选对且不全的得2分,有选错的得0分.)9.(4分)如图所示,两个带等量正电荷的小球A、B(可视为点电荷),被固定在光滑绝缘的水平面上.P、N是小球A、B的连线的水平中垂线,且PO=ON.现将一个电荷量很小的带负电的小球C(可视为质点),由P点静止释放,在小球C向N点运动的过程中,下列关于小球C的速度图象中,可能正确的是()A.B.C.D.考点:电场的叠加.专题:电场力与电势的性质专题.分析:A、B为两个等量的正点电荷,其连线中垂线上电场强度方向O→P,负点电荷q 从P点到O点运动的过程中,电场力方向P→O,速度越来越大.但电场线的疏密情况不确定,电场强度大小变化情况不确定,则电荷所受电场力大小变化情况不确定,加速度变化情况不确定.越过O点后,负电荷q做减速运动,点电荷运动到O点时加速度为零,速度达最大值,加速度变化情况同样不确定.解答:解:在AB的垂直平分线上,从无穷远处到O点电场强度先变大后变小,到O点变为零,负电荷受力沿垂直平分线运动,电荷的加速度先变大后变小,速度不断增大,在O 点加速度变为零,速度达到最大,由速度与时间的图象的斜率先变大后变小,由O点到无穷远,速度变化情况与另一侧速度的变化情况具有对称性.如果PN足够远,则B正确,如果PN很近,则A正确,CD错误.故选:AB.点评:本题考查对等量异种电荷电场线的分布情况及特点的理解和掌握程度,要抓住电场线的对称性.10.(4分)某同学设计了一种静电除尘装置,如图甲所示,其中有一长为L、宽为b、高为d的矩形通道,其前、后面板为绝缘材料,上、下面板为金属材料.图乙是装置的截面图,上、下两板与电压恒定为U的高压直流电源相连.带负电的尘埃被吸入矩形通道的水平速度为v0,当碰到下板后其所带电荷被中和,同时被收集.将被收集尘埃的数量与进入矩形通道尘埃的数量的比值,称为除尘率.不计尘埃的重力及尘埃之间的相互作用.要增大除尘率,则下列措施可行的是()A.只增大电压UB.只增大高度dC.只增大长度LD.只增大尘埃被吸入水平速度v0考点:带电粒子在匀强电场中的运动.专题:带电粒子在电场中的运动专题.分析:带电尘埃在矩形通道内做类平抛运动,在沿电场的方向上的位移为y=,增大y便可增大除尘率.解答:解:增加除尘率即是让离下极板较远的粒子落到下极板上,带电尘埃在矩形通道内做类平抛运动,在沿电场的方向上的位移为y=即增加y即可.A、只增加电压U可以增加y,故A满足条件;B、只增大高度d,由题意知d增加则位移y减小,故不满足条件;C、只增加长度L,可以增加y,故C满足条件;D、只增加水平速度v0,y减小,故不足条件.故选:AC.点评:此题为结合生活背景的题目,考查频率较高,注意构建物理情景类平抛运动,应用运动的分解知识求解11.(4分)M、N是一对水平放置的平行板电容器,将它与一电动势为E,内阻为r的电源组成如图所示的电路,R是并联在电容器上的滑动变阻器,G是灵敏电流计,在电容器的两极板间有一带电的油滴处于悬浮状态,如图所示,现保持开关S闭合,将滑动变阻器的滑片向上滑动,则()A.在滑片滑动时,灵敏电流计中有从b向a的电流B.在滑片滑动时,灵敏电流计中有从a向b的电流C.带电油滴将向上运动D.带电油滴将向下运动考点:闭合电路的欧姆定律.专题:恒定电流专题.分析:电容器两极板的电压等于变阻器两端的电压.将滑动变阻器的滑片向上滑动时,根据欧姆定律分析电容器板间电压如何变化,判断电容器是充电还是放电,分析通过电流计的电流方向.由E=分析板间场强如何变化,判断油滴如何运动.解答:解:AB、电容器两极板间的电压U=,当将滑动变阻器的滑片向上滑动时,R增大,U增大,电容器的电量增加,处于充电状态,灵敏电流计中有电流,由于电容器上板带正电,则灵敏电流计中有从a向b的电流.故A错误,B正确.CD、U增大,由E=分析得知,板间场强增大,则带电油滴将向上运动.故C正确,D错误.故选:BC点评:本题是电容的动态分析问题,关键确定电容器的电压,电路稳定时,电容器的电压等于所并联电路两端的电压.12.(4分)如图所示的电路中,电源内阻不可忽略,若调整可变电阻R的阻值,可使电压表V的示数减小△U(电压表为理想电表),在这个过程中()A.通过R1的电流减小,减少量一定等于B.R2两端的电压增加,增加量一定等于△UC.路端电压减小,减少量一定等于△UD.通过R2的电流增加,但增加量一定小于考点:闭合电路的欧姆定律.专题:恒定电流专题.分析:由图可知,R1与R并联后与R2串联,电压表测并联部分的电压;由欧姆定律可知,通过R1的电流的变化量;由闭合电路欧姆定律可得出电路中电流的变化量,则可得出路端电压的变化量及R2两端电压的关系.解答:解:A、因电压表示数减小,而R1为定值电阻,故电流的减小量一定等于,故A正确;B、R1两端的电压减小,则说明R2及内阻两端的电压均增大,故R2两端的电压增加量一定小于△U,故B错误;C、因内电压增大,故路端电压减小,由B的分析可知,路端电压的减小量一定小于△U,故C错误;D、由B的分析可知,R2两端的电压增加量一定小于△U,故电流的增加量一定小于,故D正确;故选AD.点评:本题考查欧姆定律中的动态分析,因本题要求分析电流及电压的变化量,故难度稍大,要求学生能灵活选择闭合电路的欧姆定律的表达形式.13.(4分)如图所示,粗糙程度均匀的绝缘斜面下方O点处有一正点电荷,带负电的小物体以初速度v1从M点沿斜面上滑,到达N点时速度为零,然后下滑回到M点,此时速度为v2(v2<v l).若小物体电荷量保持不变,OM=ON,则()A.从N到M的过程中,小物体的电势能逐渐减小B.从M到N的过程中,电场力对小物体先做负功后做正功C.从N到M的过程中,小物体受到的摩擦力和电场力均是先增大后减小D.小物体上升的最大高度为考点:电势能;电场强度.分析:小物块带负电,正点电荷对它有吸引力,根据距离变化即可判断电场力做功和电势能的变化.分析小球受力情况,根据斜面对小球支持力的变化,分析摩擦力的变化.根据动能定理分别上升和下滑两个过程,求出物体上升的最大高度.解答:解:A、由题分析可知,正点电荷对物体有吸引力,从N到M的过程中,物体与正点电荷间的距离先减小后增大,则电场力先做正功后做负功,物体的电势能先减小后增加.故A错误.B、从M到N的过程中,物体与正点电荷间的距离先减小后增大,则电场力先做正功后做负功.故B错误.C、从N到M的过程中,小物体受到的电场力先增大后减小,电场力在垂直于斜面的分力也先增大后减小,根据物体在垂直于斜面方向力平衡得知,斜面对物体的支持力也先增大后减小,则物体受到的摩擦力先增大后减小.故C正确.D、设物体上升过程中,摩擦力做功为W,上升的最大高度为h.由于OM=ON,M、N两点的电势相等,上升和下滑过程中电场力做功都为0,则根据动能定理得,上升过程:W﹣mgh=0﹣mv12下滑过程:W+mgh=mv22﹣0联立解得,h=.故D正确.故选:CD点评:本题根据物体与正电荷间距离的变化,来判断电场力做功的正负.求物体上升的最大高度时,要抓住OM=ON这个条件,得到M、N的电势相等,则知上升和下降电场力做功为0.三、实验题(本题包括2个小题,共16分)14.(4分)用控制变量法,可以研究影响平行板电容器的因素(如图).设两极板正对面积为S,极板间的距离为d,静电计指针偏角为θ.实验中,极板所带电荷量不变,若()A.保持S不变,增大d,则θ变大B.保持S不变,增大d,则θ变小C.保持d不变,减小S,则θ变小D.保持d不变,减小S,则θ不变考点:影响平行板电容器电容的因素.分析:静电计指针偏角θ表示电容器两端电压的大小,根据电容的定义式C=,判断电容的变化,再根据C=,判断电压的变化,从而得知静电计指针偏角的变化.解答:解:根据电容的定义式C=,保持S不变,增大d,电容C减小,再根据U=,知U增大,所以θ变大.故A正确,B错误.保持d不变,减小S,电容减小,再根据C=,知U增大,所以θ变大.故CD错误.故选:A.点评:解决电容器的动态分析问题关键抓住不变量.若电容器与电源断开,电量保持不变;若电容器始终与电源相连,电容器两端间的电势差保持不变.15.(12分)在“测定金属的电阻率”实验中,所用测量仪器均已校准.待测金属丝接入电路部分的长度约为50cm.(1)用螺旋测微器测量金属丝的直径,其中某一次测量结果如图1所示,其读数应为0.397mm(该值接近多次测量的平均值).(2)用伏安法测金属丝的电阻R x.实验所用器材为:电池组(电动势3V,内阻约1Ω)、电流表(内阻约0.1Ω)、电压表(内阻约3kΩ)、滑动变阻器R(0~20Ω,额定电流2A)、开关、导线若干.某小组同学利用以上器材正确连接好电路,进行实验测量,记录数据如下:由以上实验数据可知,他们测量R x是采用图2中的甲(选填“甲”或“乙”).(3)如图3是测量R x的实验器材实物图,图中已连接了部分导线,滑动变阻器的滑片P置于变阻器的一端.请根据(2)所选的电路图,补充完成下图中实物间的连线,并使闭合开关的瞬间,电压表或电流表不至于被烧坏.(4)这个小组的同学在坐标纸上建立U、I坐标,如图4所示,图中已标出了与测量数据对应的4个坐标点.请在图4中标出第2、4、6次测量数据的坐标点,并描绘出U﹣I图线.由图线得到金属丝的阻值R x=4.5Ω(保留两位有效数字).(5)根据以上数据可以估算出金属丝电阻率约为C(填选项前的符号).A.1×10﹣2Ω•m B.1×10﹣3Ω•mC.1×10﹣6Ω•m D.1×10﹣8Ω•m.考点:测定金属的电阻率.专题:实验题.分析:(1)关于螺旋测微器的读数,要先读出固定刻度,再读出可动刻度,然后相加即可得出结果.(2)根据数据比较电压表、电流表和被测电阻的阻值关系,确定可采取的电路.(3)按照电路原理图进行实物图的连接,注意导线不能交叉和滑动变阻器的连接方式.(4)根据图上所标的点,做出U﹣﹣I图线,从而可得出电阻值(5)把以上数据代入电阻定律,可得出结果解答:解:(1)固定刻度读数为0,可动刻度读数为39.7,所测长度为0+39.7×0.01=0.397mm (0.395~0.399)(2)由记录数据根据欧姆定律可知金属丝的电阻R x约5Ω.则有R x<,属于小电阻,用外接法测量误差小,由(3)知是用伏安特性曲线来测量电阻的,就要求电压电流从接近0开始调节,所以应该采用分压接法故选甲.(3)注意连图时连线起点和终点在接线柱上并且不能交叉,结合(2)可知应该连接成外接分压接法(甲)那么在连线时断开开关且使R x两端的电压为0.先连外接电路部分,再连分压电路部分,此时滑片P必须置于变阻器的左端.实物图如右图所示,(4)描绘出第2、4、6三个点后可见第6次测量数据的坐标点误差太大舍去,然后出U﹣I 图线.如右图所示;其中第4次测量数据的坐标点在描绘出的U﹣I图线上,有:R x==4.5Ω(5)根据电阻定律R=ρ,得ρ=R,代入数据可计算出ρ=1×10﹣6Ω•m,故选C.故答案为:(1)0.397 (2)甲(3)如图(4)4.5 (5)C点评:该题是综合性较强的题,解答时注意一下几方面:1、对于长度的测量注意高中所要求的游标卡尺和螺旋测微器的使用方法,读书时是固定刻度的值与可动刻度的值得和.2、会根据电压表、电流表及被测电阻的阻值关系,确定电流表是内接还是外接.3、实物连接时,注意导线不能相交叉,并且要注意闭合电建时,分压电路的输出端电压要为零.4、会用电阻定律来求解导线的电阻率四、计算题16.(8分)在真空中的O点放一点电荷Q=1.0×10﹣9C,直线MN过O点,OM=30cm,M 点放有一点电荷q=﹣2×10﹣10C,如图所示.求:(1)M点的场强大小;(2)若M点的电势比N点的电势高15V,则电荷q从M点移到N点,电势能变化了多少?(k=9.0×109N•m2/C2)考点:电势能;电场强度.专题:电场力与电势的性质专题.分析:(1)知道点电荷的电荷量及M点到点电荷的距离,由点电荷的场强公式E=k可以直接求得结果.(2)根据电场力做功的公式可以直接求得电场力做的功的大小,从而可以知道电势能的变化.解答:解:(1)由点电荷的场强公式E=k可知:M点的场强大小E=k=9.0×109×N/C=100N/C,所以电荷Q在M点的电场强度大小是100N/C.(2)电荷q从M点移到N点,电场力做的功为:W MN=qU MN=q(φM﹣φN)=﹣2×10﹣10×15J=﹣3×10﹣9J,所以电势能增加了3×10﹣9J.答:(1)M点的场强大小100N/C;(2)电势能增加了3×10﹣9J.点评:本题是对点电荷的场强公式和电场力做的功与电势能的转化之间关系的考查,掌握住基本内容就可以解决这道题.17.(10分)质量都是m的两个完全相同、带等量异种电荷的小球A、B分别用长L的绝缘细线悬挂在同一水平面上相距为2L的M、N两点,平衡时小球A、B的位置如图甲所示,线与竖直方向夹角α=37°,当外加水平向左的匀强电场时,两小球平衡位置如图乙所示,线与竖直方向夹角也为α=37°,(sin37°=,cos37°=)求:(1)A、B小球电性及所带电荷量Q;(2)外加匀强电场的场强E.考点:库仑定律;共点力平衡的条件及其应用.分析:(1)根据小球在匀强电场中受电场力的方向可得出小球所带电性;因两球均受重力、拉力及库仑力而处于平衡,由共点力的平衡规律可得出小球的电性及电荷量;(2)外加匀强电场后,小球仍然处于平衡状态,则由共点力的平衡条件可得出电场强度的大小.解答:解:(1)A球带正电,B球带负电,两小球相距d=2l﹣2lsin37°=l由A球受力平衡可得mgtanα=k解得Q=(2)外加匀强电场时两球相距d′=2l+2lsin37°=根据A球受力平衡可得QE﹣k=mgtanα解得:E=答:(1)A、B小球电性及所带电荷量Q为;(2)外加匀强电场的场强E为.点评:小球在电场中的受力平衡和力学中物体的平衡处理方法类似,只要正确做好受力分析,可根据题意选用共点力的平衡或牛顿第二定律求解.18.(10分)如图所示的电路中,两平行金属板A、B水平放置,极板长L=80cm,两板间的距离d=40cm.电源电动势E=40V,内电阻r=1Ω,电阻R=15Ω,闭合开关S,待电路稳定后,将一带负电的小球从B板左端且非常靠近B板的位置以初速度v0=4m/s水平向右射入两板间,该小球可视为质点.若小球带电量q=1×10﹣2C,质量为m=2×10﹣2kg,不考虑空气阻力,电路中电压表、电流表均是理想电表.若小球恰好从A板右边缘射出(g取10m/s2).求:(1)滑动变阻器接入电路的阻值为多少?(2)此时电流表、电压表的示数分别为多少?(3)此时电源的输出功率是多少?考点:带电粒子在匀强电场中的运动;闭合电路的欧姆定律.专题:带电粒子在电场中的运动专题.分析:(1)小球进入电场中做类平抛运动,小球恰好从A板右边缘射出时,水平位移为L,竖直位移为d,根据运动学和牛顿第二定律结合可求出板间电压,再根据串联电路分压特点,求解滑动变阻器接入电路的阻值.(2)根据闭合电路欧姆定律求解电路中电流,由欧姆定律求解路端电压,即可求得两电表的读数.(3)电源的输出功率P=UI,U是路端电压,I是总电流.。
四川省成都市成华区成都列五中学2024-2025学年高三上学期9月月考英语试题(含答案)
高三英语第I卷(共80分)第一部分:听力(共两节,满分30分)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,从试题所给的A、B、C 三个选项中选出最佳选项,并标在试卷的相应位置。
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。
第一节(共5小题;每小题1.5分,共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,并标在试卷的相应位置。
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 are the speakers talking aboutA.The stars. B.A rainbow. C.The fireworks.2.What do you know about the womanA.She doesn’t like her own clothes.B.She wants to be a dress designer.C.She is a famous dress designer.3.What does the man like to play nowA.Tennis. B.Baseball. C.Basketball.4.How did the man feel about the lectureA.Boring. B.Excellent. C.Too long.5.What does the man meanA.The lady’s room is a long way from here.B.The woman has to sign up for using the lady’s room.C.The woman is not able to use the lady’s room right now.第二节(共15小题;每题1.5分,满分22.5分)听下面5段对话或独白。
2024-2025学年四川省成都列五中学高三上学期9月月考数学试题及答案
高三数学一、单项选择题:本题共8小题,每小题5分,共40分,在每小题给出的四个选项中,只有一项是符合要求的.1. 集合{}260A x x x =--<,集合{}2B x x =<,则A B = ( )A.()2,3- B. (),3-∞ C. ()2,2- D. ()0,22. 函数()f x 是定义在R 上的奇函数,当0x >时,()2log f x x =,则()4f -=( )A.12B. 2C. 12-D. 2-3. ()102x -展开式中第6项二项式系数是( )A. 610C B. ()6610C 2⋅- C. 510C D. ()5510C 2⋅-4. 已知函数()f x 是定义在[0,)+∞上的增函数,则满足1(21)3f x f ⎛⎫-< ⎪⎝⎭的x 的取值范围是( )A 12,33⎡⎤⎢⎥⎣⎦ B. 12,33⎡⎫⎪⎢⎣⎭ C. 12,23⎛⎫ ⎪⎝⎭ D. 12,23⎡⎫⎪⎢⎣⎭5. 体育课上甲、乙两名同学进行投篮比赛(甲、乙各投篮一次),甲投中的概率为0.7,乙投中的概率为0.8,则甲、乙两人恰好有一人投中的概率为( )A. 0.38 B. 0.24C. 0.14D. 0.56. 函数213x ax y -+=在区间()1,2上单调递增,则实数a 的取值范围是( )A. 2a ≤ B. 4a ≤ C. 2a > D. 4a >7. 函数()log 31(0,1)a y x a a =+->≠的图像恒过定点A ,若点A 在直线10mx ny ++=上,其中0mn >,则12m n+的最小值为( )A. 4B.C. D. 88. 已知函数()y f x =是R 上的偶函数,对于R x ∈都有()()()63f x f x f +=+成立,且()42f -=-,当[]12,0,3x x ∈,且12x x ≠时,都有1212()()0f x f x x x ->-.则给出下列命题:的.①()20082=-f ;②函数()y f x =图象的一条对称轴为6x =-;③函数()y f x =在[]9,6--上为严格减函数;④方程()0f x =在[]9,9-上有4个根;其中正确的命题个数为( )A. 1B. 2C. 3D. 4二、多项选择题:本题共3小题,每小题6分,共18分.在每小题给出的选项中,有多项符合题目要求.全选对的得6分,部分选对的得部分分,有选错的得0分.9. 在下列函数中,既是偶函数又在()0,1上单调递增的函数有( )A cos y x= B. sin y x =C. 2xy = D. 3y x =10. 已知函数()ln f x x x =,则( )A. ()f x 在()1,+∞单调递增B. ()f x 有两个零点C. ()f x 的最小值为1e-D. ()y f x =在()1,0点处切线为1y x =-11. 在2024年巴黎奥运会艺术体操项目集体全能决赛中,中国队以69.800分的成绩夺得金牌,这是中国艺术体操队在奥运会上获得的第一枚金牌.艺术体操的绳操和带操可以舞出类似四角花瓣的图案,它可看作由抛物线2:2(0)C y px p =>绕其顶点分别逆时针旋转90180270 、、后所得三条曲线与C 围成的(如图阴影区域),,A B 为C 与其中两条曲线的交点,若1p =,则( )A. 开口向上的抛物线的方程为212y x =B. |AB |=4.C. 直线x y t +=截第一象限花瓣的弦长最大值为34D. 阴影区域的面积大于4三、填空题:本大题共3小题,每小题5分,共计15分.12. 12011lg125lg 1)864-⎛⎫-+-+= ⎪⎝⎭__________.13. 已知二次函数()2f x x bx c =++满足()()11f x f x +=-,则()1f -与()4f 大小关系是______.14. 在棱长为2正方体1111ABCD A B C D -中,点,E F 分别为棱1,AD BB 的中点. 点P 为正方体表面上的动点,满足1A P EF ⊥. 给出下列四个结论:①线段1A P长度的最大值为;②存在点P ,使得//DP EF ;③存在点P ,使得1B P DP =;④EPF 是等腰三角形.其中,所有正确结论的序号是________.四、解答题:共77分.解答应写出文字说明,证明过程或演算步骤.15. 已知数列{}n a 为公差不为零的等差数列,其前n 项和为n S ,749=S ,且2a ,5a ,14a 成等比数列.(1)求{}n a 的通项公式;(2)若数列{}n n a b +是公比为3的等比数列,且322b =,求{}n b 的前n 项和n T .16. 如图,在四棱锥S ABCD -中,底面ABCD 为正方形、SA ⊥平面ABCD M N ,,分别为棱SB SC ,的中点的的(1)证明://MN 平面SAD ;(2)若SA AD =,求直线SD 与平面ADNM 所成角的正弦值17. 为了调查学生喜欢跑步是否与性别有关,高三年级特选取了200名学生进行了问卷调查,得到如下的22⨯列联表:已知在这200名学生中随机抽取1人抽到喜欢跑步的概率为0.6.(1)判断:是否有90%的把握认为喜欢跑步与性别有关?(2)从上述不喜欢跑步的学生中用分层抽样的方法抽取8名学生,再在这8人中抽取3人调查其喜欢的运动,用X 表示3人中女生的人数,求X 的分布及数学期望.附:()()()()22()n ad bc a b c d a c b d χ-=++++,其中n a b c d =+++.α0.100.050.005x α2.7063.8417.87918. 已知函数()ln f x a x x =-.(1)讨论()f x 的单调性;(2)证明:当0a >时,()1e aa f x ⎛⎫≤- ⎪⎝⎭.19. 已知双曲线()2222:10,0x y E a b a b -=>>的实轴长为2(1)求双曲线E 的标准方程;(2)若直线l 与E 的右支及渐近线的交点自上而下依次为C A B D 、、、,证明:AC BD =;(3)求二元二次方程2231x y -=的正整数解()()*,,,n n n n n Q x y x y n ∈N,可先找到初始解()11,x y ,其中1x 为所有解n x 中的最小值,因为(22122231=+-=-⨯,所以()12,1Q ;因为(22221(2(277734=+-=+-=-⨯,所以()27,4Q ;重复上述过程,因为(2n+与(2n -的部分互为相反数,故可设()()221(2(23n n n nn n n n x xx y ==-=-⨯,所以(),n n n Q x y .若方程E 的正整数解为(),n n n Q x y ,且初始解()13,2Q ,则1n n OQ Q +△的面积是否为定值?若是,请求出该定值,并说明理由.高三数学一、单项选择题:本题共8小题,每小题5分,共40分,在每小题给出的四个选项中,只有一项是符合要求的.1. 集合{}260A x x x =--<,集合{}2B x x =<,则A B = ( )A.()2,3- B. (),3-∞ C. ()2,2- D. ()0,2【答案】C 【解析】【分析】先解不等式求出集合A ,B ,再根据交集的定义求解即可.【详解】由{}{}26023A x x x x x =--<=-<<,{}{}222B x x x x =<=-<<,则{}{}{}232222A B x x x x x x ⋂=-<<⋂-<<=-<<,即()2,2A B ⋂=-,故选:C.2. 函数()f x 是定义在R 上奇函数,当0x >时,()2log f x x =,则()4f -=( )A.12B. 2C. 12-D. 2-【答案】D 【解析】【分析】根据奇函数的性质求解即可.【详解】解:因为()f x 是定义在R 上奇函数,当0x >时,()2log f x x =,所以()24(4)log 42f f -=-=-=-.故选:D.3. ()102x -展开式中第6项的二项式系数是( )A. 610C B. ()6610C 2⋅- C. 510C D. ()5510C 2⋅-【答案】C 【解析】的的【分析】根据第6项的二项式系数即可求解.【详解】()102x -展开式中第6项的二项式系数是510C ,故选:C.4. 已知函数()f x 是定义在[0,)+∞上的增函数,则满足1(21)3f x f ⎛⎫-< ⎪⎝⎭的x 的取值范围是( )A. 12,33⎡⎤⎢⎥⎣⎦B. 12,33⎡⎫⎪⎢⎣⎭C. 12,23⎛⎫ ⎪⎝⎭D. 12,23⎡⎫⎪⎢⎣⎭【答案】D 【解析】【分析】利用函数的定义域及单调性计算即可.【详解】由题意可知2101213x x -≥⎧⎪⎨-<⎪⎩,解不等式得12,23x ⎡⎫∈⎪⎢⎣⎭.故选:D5. 体育课上甲、乙两名同学进行投篮比赛(甲、乙各投篮一次),甲投中的概率为0.7,乙投中的概率为0.8,则甲、乙两人恰好有一人投中的概率为( )A. 0.38 B. 0.24 C. 0.14 D. 0.5【答案】A 【解析】【分析】根据相互独立事件的概率乘法公式即可求解.【详解】甲、乙两人恰好有一人投中的概率为()()0.710.80.810.70.38⨯-+⨯-=,故选:A 6. 函数213x ax y -+=在区间()1,2上单调递增,则实数a 的取值范围是( )A. 2a ≤B. 4a ≤ C. 2a > D. 4a >【答案】A 【解析】【分析】根据复合函数单调性的性质,结合指数函数和二次函数的单调性进行求解即可.【详解】因为函数3x y =是实数集上的增函数,213x ax y -+=在区间()1,2上单调递增,所以函数21y x ax =-+在区间()1,2上单调递增,因为二次函数21y x ax =-+的对称轴为2a x =,所以有12a≤,即2a ≤,故选:A7. 函数()log 31(0,1)a y x a a =+->≠的图像恒过定点A ,若点A 在直线10mx ny ++=上,其中0mn >,则12m n+的最小值为( )A. 4B.C. D. 8【答案】D 【解析】【分析】先求出对数函数的定点(2,1)--,再结合点在线上得出1,最后应用基本不等式常值代换计算求解.【详解】因为当2x =-时,log 111,a y =-=-所以函数()log 31(0,1)a y x a a =+->≠的图像恒过定点(2,1)--,即(2,1)A --,因为点A 在直线10mx ny ++=上,所以210,m n --+=即 21,m n +=因为0,mn > 所以0,0,m n >>122424448,m n m n n m m n m n m n +++=+=++≥+=当且仅当4,n mm n =即11,42m n ==时取等号.故选:D.8. 已知函数()y f x =是R 上的偶函数,对于R x ∈都有()()()63f x f x f +=+成立,且()42f -=-,当[]12,0,3x x ∈,且12x x ≠时,都有1212()()0f x f x x x ->-.则给出下列命题:①()20082=-f ;②函数()y f x =图象的一条对称轴为6x =-;③函数()y f x =在[]9,6--上为严格减函数;④方程()0f x =在[]9,9-上有4个根;其中正确的命题个数为( )A. 1 B. 2C. 3D. 4【答案】D【解析】【分析】对于①,令3x =-代入已知等式可求出()30f -=,再结合其为偶函数可得f (3)=0,从而可求出函数的周期为6,利用周期可求得结果;对于②,由()f x 为偶函数,结合周期为6分析判断;对于③,由当[]12,0,3x x ∈,且12x x ≠时,都有1212()()0f x f x x x ->-,可得y =f (x )在[]0,3上为严格增函数,再结合其为偶函数及周期为6分析判断;对于④,由f (3)=0,()f x 的周期为6,及函数的单调性分析判断.【详解】①:对于任意R x ∈,都有()()()63f x f x f +=+成立,令3x =-,则()()()3633f f f -+=-+,解得()30f -=,又因为()f x 是R 上的偶函数,所以f (3)=0,所以()()6f x f x +=,所以函数()f x 的周期为6,所以()()()200844f f f ==-,又由()42f -=-,故()20082f =-;故①正确;②:由(1)知()f x 的周期为6,又因为()f x 是R 上的偶函数,所以()()6f x f x +=-,而()f x 的周期为6,所以()()66f x f x +=-+,()()6f x f x -=--,所以:()()66f x f x --=-+,所以直线6x =-是函数y =f (x )的图象的一条对称轴.故②正确;③:当[]12,0,3x x ∈,且12x x ≠时,都有1212()()0f x f x x x ->-.所以函数y =f (x )在[]0,3上为严格增函数,因为()f x 是R 上的偶函数,所以函数y =f (x )在[]3,0-上为严格减函数,而()f x 的周期为6,所以函数y =f (x )在[]9,6--上为严格减函数.故③正确;④:f (3)=0,()f x 的周期为6,所以()()()()93390f f f f -=-===,又()f x 在[]3,3-先严格递减后严格递增,所以()f x 在[]3,3-上除端点外不存在其他零点,所以()f x 在[9,3)--和(3,9]上各有一个零点,所以函数y =f (x )在[]9,9-上有四个零点.故④正确;故选:D .【点睛】关键点点睛:此题考查抽象函数的奇偶性,对称性,单调性和周期性,解题的关键是利用赋值法求出f (3)=0,从而可得()()6f x f x +=,得到周期为6,然后结合周期性和奇偶性分析判断,考查分析问题的能力,属于较难题.二、多项选择题:本题共3小题,每小题6分,共18分.在每小题给出的选项中,有多项符合题目要求.全选对的得6分,部分选对的得部分分,有选错的得0分.9. 在下列函数中,既是偶函数又在()0,1上单调递增的函数有( )A. cos y x = B. sin y x =C. 2xy = D. 3y x =【答案】BC 【解析】【分析】对所给的函数注意判断即可.【详解】对A :cos y x =是偶函数,在()0,1上递减,排除A ;对B :sin y x =为偶函数,在()0,1上递增,故B 正确;对C :2xy =为偶函数,在()0,1上递增,故C 正确;对D :3y x =为奇函数,排除D.故选:BC10. 已知函数()ln f x x x =,则( )A. ()f x 在()1,+∞单调递增B. ()f x 有两个零点C. ()f x 的最小值为1e-D. ()y f x =在()1,0点处切线为1y x =-【答案】ACD【解析】【分析】首先求函数的导数,并判断函数的单调性,即可判断AB C ,再根据导数的几何意义求切线方程,判断D.【详解】()ln 1f x x ='+,0x >,对于A ,当1x >时,()ln 110f x x +>'=>,所以()f x 在(1,+∞)单调递增,故A 正确;()0f x '=,得1ex =,当10,e x ⎛⎫∈ ⎪⎝⎭,f ′(x )<0,()f x 单调递减,当1,ex ∞⎛⎫∈+ ⎪⎝⎭,f ′(x )>0,()f x 单调递增,所以当1e x =时,()f x 取得最小值1e-,C 正确,当01x <<时,()0f x <,当1x >时,()0f x >,所以函数()f x 只有1个零点,故B 错误,对于D ,f (1)=0,()11f '=,所以曲线y =f (x )在点(1,f (1))处的切线方程为1y x =-,故D 正确;故选:ACD.11. 在2024年巴黎奥运会艺术体操项目集体全能决赛中,中国队以69.800分的成绩夺得金牌,这是中国艺术体操队在奥运会上获得的第一枚金牌.艺术体操的绳操和带操可以舞出类似四角花瓣的图案,它可看作由抛物线2:2(0)C y px p =>绕其顶点分别逆时针旋转90180270 、、后所得三条曲线与C 围成的(如图阴影区域),,A B 为C 与其中两条曲线的交点,若1p =,则( )A. 开口向上的抛物线的方程为212y x =B. |AB |=4C. 直线x y t +=截第一象限花瓣的弦长最大值为34D. 阴影区域的面积大于4【答案】ABD 【解析】【分析】对于A ,利用旋转前后抛物线焦点和对称轴变化,即可确定抛物线方程;对于B ,联立抛物线方程,求出点,A B 的坐标,即得;对于C ,将直点线与抛物线方程联立求出,M N 的坐标,由两点间距离公式求得弦长,利用换元和函数的图象即可求得弦长最大值;对于D ,利用以直线近似取代曲线的思想求出三角形面积,即可对阴影部分面积大小进行判断.【详解】由题意,开口向右的抛物线方程为2:2C y x =,顶点在原点,焦点为11(,0)2F ,将其逆时针旋转90 后得到的抛物线开口向上,焦点为21(0,)2F ,则其方程为22x y =,即212y x =,故A 正确;对于B ,根据A 项分析,由2222y xx y⎧=⎨=⎩可解得,0x =或2x =,即2A x =,代入可得2A y =,由图象对称性,可得(2,2),(2,2)A B -,故4AB =,即B 正确;对于C ,如图,设直线x y t +=与第一象限花瓣分别交于点,M N ,由22y x t y x =-+⎧⎨=⎩解得11M M x t y ⎧=+⎪⎨=⎪⎩22y x t x y =-+⎧⎨=⎩解得,11N N x y t ⎧=-⎪⎨=+-⎪⎩,即得(11),1,1M t N t +---+-,则弦长为:|||2|MN t ==+-,由图知,直线x y t +=经过点A 时t 取最大值4,经过点O 时t 取最小值0,即在第一象限部分满足04t <<,不妨设u =13u <<,且212u t -=,代入得,221|||22||(2)1|2u MN u u -=+-=--,(13u <<)由此函数图象知,当2u =时,||MN,即C 错误;对于D ,根据对称性,每个象限的花瓣形状大小相同,故可以先求18部分面积的近似值.如图,在抛物线21,(0)2y x x =≥上取一点P ,使过点P 的切线与直线OA 平行,由1y x '==可得切点坐标为1(1,2P ,因:0OA l x y -=,则点P 到直线OA的距离为d ==,于是1122OPA S == ,由图知,半个花瓣的面积必大于12,故原图中的阴影部分面积必大于1842⨯=,故D 正确.故选:ABD.【点睛】思路点睛:本题主要考查曲线与方程的联系的应用问题,属于难题.解题思路是,理解题意,结合图形对称性特征,通过曲线方程联立,计算判断,并运用函数的图象单调性情况,有时还需要以直代曲的思想进行估算、判断求解.三、填空题:本大题共3小题,每小题5分,共计15分.12. 12011lg125lg 1)864-⎛⎫-+-+= ⎪⎝⎭__________.【答案】12【解析】【分析】根据对数和指数运算法则计算可得结果.的【详解】易知1122201111lg125lg 1)lg 125186848--⨯⎛⎫⎛⎫⎛⎫-++=÷++ ⎪⎪ ⎪⎝⎭⎝⎭⎝⎭1312lg118100081-⎛⎫++ =+⎪+⎭=⎝=.故答案为:1213. 已知二次函数()2f x x bx c =++满足()()11f x f x +=-,则()1f -与()4f 的大小关系是______.【答案】()()14f f -<【解析】【分析】由()()11f x f x +=-得到函数的对称轴,从而得到方程,求出2b =-,再根据对称性和单调性比较出大小.【详解】以为()()11f x f x +=-,所以1x =为二次函数图象的对称轴,所以12b-=,解得2b =-.根据对称性知,()()13f f -=,又函数在[)1,+∞上单调递增,所以()()34f f <,即()()14f f -<.故答案为:()()14f f -<14. 在棱长为2的正方体1111ABCD A B C D -中,点,E F 分别为棱1,AD BB 的中点. 点P 为正方体表面上的动点,满足1A P EF ⊥. 给出下列四个结论:①线段1A P 长度的最大值为;②存在点P ,使得//DP EF ;③存在点P ,使得1B P DP =;④EPF 是等腰三角形.其中,所有正确结论的序号是________.【答案】①③④【解析】【分析】建立空间直角坐标系,利用坐标验证垂直判断①,找出平行直线再由坐标判断是否垂直可判断B ,设点的坐标根据条件列出方程组②,探求是否存在符合条件的解判断③④【详解】如图,建立空间直角坐标系,则()()()()()()112,0,2,1,0,0,2,2,1,0,2,0,0,0,0,2,2,2A E F C D B ,对①,由正方体性质知当P 在C 时,线段1A P长度的最大值为此时()()12,2,2,1,2,1A P EF =--= ,12420A P EF ⋅=-+-=,所以1A P EF ⊥,即满足1A P EF ⊥,故①正确;对②,取正方形11BB C C 的中心M ,连接,DM MF ,易知//,MF DE MF DE =,所以四边形DMFE 为平行四边形,所以//DM EF ,故P 运动到M 处时,//DP EF ,此时()1,2,1P ,()11,2,1A P =-- ,114120A P EF ⋅=-+-=≠,即不满足1A P EF ⊥,综上不存在点P ,使得//DP EF ,故②错误;对③,设(),,P x y z ,则()12,,2A P x y z =-- ,()1,2,1EF =,若存在,由1B P DP =,1A P EF ⊥可得方程组2220x y z -++-=⎧=化简可得243x y z x y z ++=⎧⎨++=⎩,解得 2,1x z y +==,显然当0,2,1x z y ===时满足题意,即存在点P ,使得1B P DP =,故③正确;对④,设(),,P x y z ,若PE PF =,=24x y z ++=,由③知1A P EF ⊥时可得24x y z ++=,所以不妨取0,1,2x y z ===,此时()0,1,2P 在正方体表面上,满足题意,故④正确.故答案为:①③④【点睛】关键点点睛:本题的关键之处在于建立空间直角坐标系,利用坐标运算建立方程,探求是否存在满足条件的点,运算比较复杂,属于难题.四、解答题:共77分.解答应写出文字说明,证明过程或演算步骤.15. 已知数列{}n a 为公差不为零的等差数列,其前n 项和为n S ,749=S ,且2a ,5a ,14a 成等比数列.(1)求{}n a 的通项公式;(2)若数列{}n n a b +是公比为3的等比数列,且322b =,求{}n b 的前n 项和n T .【答案】(1)()*21Nn a n n =-∈(2)2n T n=【解析】【分析】(1)设公差为d ,根据等差数列的前n 项和公式与等比中项公式列出关于1a 和d 的方程,求解即可得{a n }的通项公式;(2)由(1)可得等比数列{}n n a b +的第三项33a b +,进而得11a b +,从而得到{b n }的通项公式,利用等差和等比数列前n 项和公式分组求和即可求出n T .【小问1详解】因为{a n }为等差数列,设公差为d ,由749=S ,得()17477492a a a+⨯==,47a ⇒=即137a d +=,由2a ,5a ,14a 成等比数列得22514a a a =⋅,()()()2772710d d d ⇒+=-+,化简()()()2772710d d d +=-+得220d d -=,因为0d ≠,所以2=d .所以()()*4421N n a a n d n n =+-=-∈.综上()*21Nn a n n =-∈.【小问2详解】由21n a n =-知11a =,35a =,又{}n n a b +为公比是3的等比数列,322b =,所以()3311952227a b a b +=+⨯=+=,即11113a b b +=+=, 所以1333n n n n a b -+=⨯=,()321n n b n =--,()*N n ∈所以()123123333313521nn n T b b b b n ⎡⎤=+++⋅⋅⋅+=+++⋅⋅⋅+-+++⋅⋅⋅+-⎣⎦()()12313121331322nn n n n +⨯-+--=-=--.综上12332n n T n +-=-.16. 如图,在四棱锥S ABCD -中,底面ABCD 为正方形、SA ⊥平面ABCD M N ,,分别为棱SB SC ,的中点(1)证明://MN 平面SAD ;(2)若SA AD =,求直线SD 与平面ADNM 所成角的正弦值【答案】(1)证明见解析; (2)12.【解析】【分析】(1)由题意易知//MN BC ,根据线面平行的判定定理证明即可;(2)由题意,,,AB AD AS 两两垂直,所以建立空间直角坐标系,求出直线SD 的方向向量与平面ADNM 的法向量,再通过空间角的向量求解即可.【小问1详解】M N 、分别为,SBSC 的中点//MN BC ABCD ∴ 为正方形//BC AD ∴//MN AD MN ∴ ⊄平面,SAD AD ⊂平面SAD//MN ∴平面SAD .【小问2详解】由题知SA ⊥平面,ABCD AB AD ⊥建立如图所示的空间直角坚标系,2SA AD ==设,则()()()()()0,0,2,0,0,0,0,2,0,2,0,0,2,2,0S A D B C ,()()1,0,1,1,1,1M N ∴,()0,2,2SD ∴=- ,()0,2,0AD = ,()1,0,1AM =设平面ADNM 的一个法向量为n =(x,y,z )则20n AD y n AM x z ⎧⋅==⎪⎨⋅=+=⎪⎩,令1,x =则0,1y z ==-,()1,0,1n ∴=-设直线SD 与平面ADNM 所或的角为θ,1sin cos ,2n SD n SD n SDθ⋅∴====⋅,所以直线SD 与平面ADNM 所成角的正弦值为12.17. 为了调查学生喜欢跑步是否与性别有关,高三年级特选取了200名学生进行了问卷调查,得到如下的22⨯列联表:喜欢跑步不喜欢跑步合计男生80女生20合计已知在这200名学生中随机抽取1人抽到喜欢跑步概率为0.6.(1)判断:是否有90%的把握认为喜欢跑步与性别有关?(2)从上述不喜欢跑步的学生中用分层抽样的方法抽取8名学生,再在这8人中抽取3人调查其喜欢的运动,用X 表示3人中女生的人数,求X 的分布及数学期望.附:()()()()22()n ad bc a b c d a c b d χ-=++++,其中n a b c d =+++.α0.100.050.005x α2.7063.8417.879【答案】(1)没有 (2)分布列见解析,34【解析】【分析】(1)根据卡方计算公式求解卡方,即可与临界值比较求解,(2)根据分层抽样比求解抽取人数,即可利用超几何分布概率公式求解概率,进而得分布列求解.【小问1详解】由题可知,从200名学生中随机抽取1人抽到喜欢跑步的概率为0.6,故喜欢跑步的人有2000.6120⨯=(人),不喜欢跑步的人有20012080-=(人).喜欢跑步不喜欢跑步合计男生8060140女生402060合计12080200∴80a =,60b =,40c =,20d =,()2220080204060 1.587 2.7061208014060χ⨯⨯-⨯=≈<⨯⨯⨯,故无90%把握认为喜欢跑步与性别有关.【小问2详解】的的按分层抽样,设女生x 名,男生y 名,8802060x y ==,解得2x =,6y =,∴从不喜欢跑步的学生中抽取女生2名,男生6名,故X 0=,1,2.()032638C C 50C 14P X ⋅===,()122638C C 151C 28P X ⋅===,()212638C C 32C 28P X ⋅===, 故X 的分布为:X012P5141528328∴()5153213012142828284E X =⨯+⨯+⨯==.18. 已知函数()ln f x a x x =-.(1)讨论()f x 的单调性;(2)证明:当0a >时,()1e aa f x ⎛⎫≤- ⎪⎝⎭.【答案】(1)答案见解析 (2)证明见解析【解析】【分析】(1)先明确函数定义域和求导,根据导数结构特征对a 进行0a ≤和0a >的分类讨论导数正负即可得单调性.(2)证()1e a a f x ⎛⎫≤-⇔ ⎪⎝⎭()max1e a a f x ⎛⎫≤- ⎪⎝⎭,故问题转化成证()0ln 1e aa a a a a ⎛⎫- ⎪-⎭≤⎝>10ln e e aaa a ⎛⎫⎛⎫⇔ ⎪ ⎪⎝⎭-+⎭≤⎝,接着构造函数()()ln 10g x x x x =-+>研究其单调性和最值即可得证.【小问1详解】由题函数定义域为()0,∞+,()1a a x f x x x-'=-=,故当0a ≤时,()0f x '<恒成立,所以函数()f x 在()0,∞+上单调递减;当0a >时,()f x '在()0,∞+上单调递减,令()0f x x a '=⇒=,则()0,x a ∈时,()0f x '>;(),x a ∈+∞时,()0f x '<,所以函数()f x 在()0,a 上单调递增,在(),a +∞上单调递减,综上,当0a ≤时,函数()f x 在()0,∞+上单调递减;当0a >时,函数()f x 在()0,a 上单调递增,在(),a +∞上单调递减.【小问2详解】由(1)当0a >时,函数()f x 在()0,a 上单调递增,在(),a +∞上单调递减,故()()ln f f a a a x a ≤=-在()0,∞+上恒成立,故证()()10e a a f x a ⎛⎫≤->⇔ ⎪⎝⎭证()0ln 1e a a a a a a ⎛⎫- ⎪-⎭≤⎝>,即()0ln 1ln 10e e e e a a a a a a a a a ⎛⎫⎛⎫⎛⎫⎛⎫⇔-⇔ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎝⎭≤-+⎭≤>,令()()ln 10g x x x x =-+>,则()()1110x g x x x x-'=-=>,故当()0,1x ∈时,()0g x '>;()1,x ∈+∞时,()0g x '<,所以()g x 在()0,1上单调递增,在()1,+∞上单调递减,所以()()10g x g ≤=在()0,∞+上恒成立,故0ln 1e e a aa a ⎛⎫⎛⎫ ⎪ ⎪⎝⎭-+⎭≤⎝,所以当0a >时,()1e aa f x ⎛⎫≤- ⎪⎝⎭.【点睛】思路点睛:证明含参函数不等式问题通常转化成研究函数最值问题,第(2)问证当0a >时,()1e a a f x ⎛⎫≤- ⎪⎝⎭可将问题转化成证()max 1e aa f x ⎛⎫≤- ⎪⎝⎭,接着根据其结构特征进行变形转化和构造函数,利用导数确定所构造的函数单调性和最值即可得证.19. 已知双曲线()2222:10,0x y E a b a b -=>>的实轴长为2(1)求双曲线E 的标准方程;(2)若直线l 与E 的右支及渐近线的交点自上而下依次为C A B D 、、、,证明:AC BD =;(3)求二元二次方程2231x y -=的正整数解()()*,,,n n n n n Q x y x y n ∈N ,可先找到初始解()11,x y ,其中1x 为所有解n x中的最小值,因为(22122231=+-=-⨯,所以()12,1Q;因为(22221(2(277734=+-=+-=-⨯,所以()27,4Q;重复上述过程,因为(2n +与(2n -的部分互为相反数,故可设()()221(2(23n n n n n n n n x x x y ==-=-⨯,所以(),n n n Q x y .若方程E 的正整数解为(),n n n Q x y ,且初始解()13,2Q ,则1n n OQ Q +△的面积是否为定值?若是,请求出该定值,并说明理由.【答案】(1)22112y x -=; (2)证明见解析;(3)为定值1,理由见解析.【解析】【分析】(1)根据实轴长和顶点到渐近线的距离求解即可;(2,AB CD 的中点重合,结合韦达定理求解即可;(3)知识迁移,类比二元二次方程2231x y -=的正整数解(),n n n Q x y ,求方程E 的正整数解,然后将1n n OQ Q +△的面积表示出来即可.【小问1详解】由题意22222a c a b ==⎨⎪=+⎪⎪⎩,解得22112a b ⎧=⎪⎨=⎪⎩,所以双曲线E 的标准方程为22112y x -=;【小问2详解】由题意直线l 的斜率不为0,设直线:l x my t =+,因为直线l 与E 的右支交于两点,所以0t >,联立2221x my t x y =+⎧⎨-=⎩得()2222210m y mty t -++-=,所以222A B mt y y m +=--,且()22Δ4220m t =+->,即2222m t >-,联立2220x my t x y =+⎧⎨-=⎩得()222220m y mty t -++=,所以222C D mt y y m -+=-,所以222C D A B mt y y y y m +=-=+-,即线段,AB CD 的中点重合,所以AC BD =.【小问3详解】由题意得方程2221x y -=的初始解为()3,2,则根据循环构造原理得(3,(3n n n n n n x x +=+=-,从而1(3(3,(3(32n n n n n n x y ⎡⎤⎤=++-=+--⎣⎦⎦,记(),n n n OQ x y = ,则()111,n n n OQ x y +++= ,设1,n n OQ OQ + 的夹角为α,则1n n OQ Q +△的面积111sin 2n n OQ Q n n S OQ OQ α++=⋅===1112n n n n x y x y ++==-,令(3,(3,1n n a b ab =+=-=,则)((()((}13333n n OQ Q S a b a b a b a b +⎡⎤⎡⎤=++----++-⎣⎦⎣⎦1==,于是1n n OQ Q +△的面积为定值1.【点睛】思路点睛:结合题目给的数学情景,运用到新的数学问题中,将学习过的知识方法迁移到新的问题中.。
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数学(文)
命题人: 审题人: (全卷满分: 150 分 完成时间: 120 分钟)
一、选择题(本大题共 12 小题,共 60 分)
1.已知集合 A x | 1 x 2 , B x | 0 x 3 ,则 A B ( )
A. ( 1,3) B . ( 1,0) C . (0,2) D . (2,3)
x2 C . y 1 D . y x x x
3 4.已知 cos(
2
3 ) ,且
5
,则 tan 为( ) 2
4
A.
3
4
D.
4
5.下列说法中,正确的是(
)
A.命题“若 a b ,则 am2 bm2 ”的否命题是假命题
B.设 , 为两不同平面,直线 l
,则“ l
”是 “
” 成立的充分不必要条件
B
. { x | x 3或0 x 3}
C. { x | x 3或 x 3}
D
. { x | 3 x 0或0 x 3}
2x y 10 10. 设实数 x, y 满足 x 2 y 14 ,则 xy 的最大值为( )
xy6
25
A.
B
2
11.已知 f ( x)
49
.
C
2
ln( x2 1), g( x)
. 12
19.如图,在四棱锥 P-ABCD中,底面 ABCD是正方形,侧棱 PD 底面 ABCD, PD=DC=,2 E 是 PC的中
点.
(Ⅰ)证明 PA// 平面 EDB;
(Ⅱ)求三棱锥 A-BDP的体积.
20.已知 P 为圆 A : ( x 1)2 y 2 8上的动点,点 B 1,0 ,线段 PB
的垂直平分线与半径 PA 相交于点 M ,记点 M 的 轨迹为 . ( 1)求曲线 的方程;
2 1
倍,再向右平移
2
单位
6
单位
12
C.先把各点的横坐标伸长到原来的
2 倍,再向左平移 单位
6
D.先把各点的横坐标伸长到原来的
2 倍,,再向左平移 单位
12
9.函数 f (x) 是奇函数, 且在 (0, ) 内是增函数, f ( 3) 0 ,则不等式 x f (x) 0 的解集为( )
A. { x | 3 x 0或x 3}
32
32
3
D .3 2
7.已知命题 p1 :函数 y 2x 2 x 在 R 上为增函数 , p2 :函数 y 2x 2 x 在 R 上为减函数,则在
命题 q1 : p1 p2; q2 : p1 p2; q3 : ( p1) p2 和 q4 : p1 ( p2) 中,真命题是(
)
A. q1,q 3
B . q2 , q3
2.已知函数 f (x) (cos 2 xcos x sin 2xsin x)sin x, x R ,则 f ( x) 是( )
A.最小正周期为 的奇函数
B
.最小正周期为 的偶函数
C.最小正周期为 的奇函数
2
D
.最小正周期为 的偶函数
2
3.下列函数中,既是奇函数又是增函数的为(
)
A. y ln x3 B . y
CU (A B)
.
14.若 cos(
) sin
3 3 ,则 sin(
5 )=
.
6
5
6
15.数列 {a n} 满足 an+1 =3an 1,且 a1 1,则数列 {a n} 的通项公式 an =
.
16 . 已 知 曲 线 y
x ln x 在 点 (1,1) 处 的 切 线 与 曲 线 y
2
ax
a 2 x 1相 切, 则
( 1)x 2
则实数 m的取值范围是( )
D . 14
m ,若对 ? x1∈ [0 ,3] ,? x2 ∈ [1 ,2] ,使得 f (x1)
g( x2) ,
1
1
1
A.[ ,+∞) B .(-∞, ] C .[ ,+∞)
4
4
2
1
D .(-∞,- ]
2
12.已知函数 F x ex 满足 F x g x h x ,且 g x ,h x 分别是 R 上的偶函数和奇函数,
C . q1 ,q4
D . q2, q4
8 .已知 f (x) A sin( x )(A 0,
0,
, xR 在一) 个周期内的图像如图所示,则
2
y f (x) 的图像可由函数 y cos x 的图像(纵坐标不变) ( )得到.
A.先把各点的横坐标缩短到原来的 B.先把各点的横坐标缩短到原来的
1
倍,再向左平移
C.命题“存在 x R, x 2 x 0 ”的否定是“对任意 x R, x 2 x 0 ”
D.已知 x R ,则“ x 1 ”是“ x 2 ”的充分不必要条件
6.在等比数列 { an} 中, a7a11 6 , a4 a14 5, 则 a20 等于(
)
a10
A. 2 或 3 B . 1 或 1 C . 2
( 2)当点 P 在第一象限,且 cos BAP
22 时,求点 M 的坐标.
3
若 x 0,2 使得不等式 g 2x ah x 0 恒成立,则实数 a 的取值范围是( )
A. , 2 2 B . , 2 2 C . 0, 2 2
二、填空题(本大题共 4 小题,共 20 分)
D . 2 2,
w w w .x k b 1.c o m
13.若 U n n 是小于 9 的正整数 } , A { n U n 是奇数 } , B={ n U n 是 3 的倍数 } ,则
a
.
三、解答题(本大题共 6 小题,共 70 分)
17.在 ABC 中,角 A, B, C 的对边分别为 a,b, c ,且 2b
3c cosC
.
3a cosA
( 1)求角 A 的值; [来源 : 学*科*网 ]
( 2)若 B , BC 边上中线 AM 6
7 ,求 ABC 的面积.
18.某车间将 10 名 技工平均分为甲,乙两组加工某种零件,在单位时间内每个技工加工零件若干,
其中合格零件的个数如下表:
( 1)分别求出甲,乙两组技工在单位时间内完成合格零件的平均数及方差,并由此分析两组技工的 技术水平;
( 2)质检部门从该车间甲,乙两组中各随机抽取
1 名技工,对其加工的零件进行检测,若两人完成
合格零件个数之和超过 1 2 件,则称该车间“质量合格”,求该车间“质量合格”的概率.