【全国百强校】吉林省梅河口市第五中学2018届高三第四次模拟考试英语试题
2018届吉林省梅河口市第五中学高三下学期第一次模拟考试英语试题
梅河口市第五中学2018 届高三第一次模拟考试英语试题第Ⅰ卷（选择题，共100 分）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，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 do we learn from the conversation?A. There will be a math examtomorrow. B. Today is the man’sbirthday.C. The man doesn’t like math exams.2. What colour is the woman’s dress?A. Blue.B. White.C. Black.3. When did the man’s daughter set a new world record?A. In 1999.B. In 2005.C. In 2009.4. What does the man mean?A. He moved the desk alone.B. He had some classmates move the desk.C. His classmates helped him move the desk.5. What time is it now?A: 3:10. B: 3:15. C. 4:10.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。
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，并标在试卷的相应位置。
听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟；听完后，各小题将给出5秒钟的作答时间。
2018年吉林省高考英语试卷(真题)
2018年吉林省高考英语试卷第一部分听力（共两节，满分7.5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．（1.50分）What does John find difficult in learning German？A．Pronunciation．B．Vocabulary．C．Grammar．2．（1.50分）What is the probable relationship between the speakers？A．Colleagues．B．Brother and sister．C．Teather and student．3．（1.50分）Where does the conversation probably take place？A．In a bank．B．At a ticket coffee．C．On the train．4．（1.50分）What are the speakers talking about？A．A restaurant．B．A street．C．A dish．5．（1.50分）How does the woman think of her interview？A．It was tough．B．It was interesting．C It was successful．第二节（共5小题；每小题1.5分，满分22.5分）听下面5段对话或独白。
每段对话或独白后有几个小题，从题中给的A、B、C三个选项中选出最佳选项。
听完每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟；听完后，各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
6．（3.00分）听第6段材料，回答下列各题．（1）When will Judy go to a party？A．On Monday．B．On Tuesday．C．On Wednesday．（2）What will Max do next？A．Fly a kite．B．Read a magazine．C．Do his homework．7．（3.00分）听第7段材料，回答下列各题．（1）What does the man suggest doing at first？A．Going to a concert．B．Watching a movie．C．Playing a computer game．（2）What do the speakers decide to do？A．Visit Mike．B．Go boating．C．Talk a walk．8．（4.50分）听第8段材料，回答下列各题．（1）Which color do cats see better than humans？A．Red．B．Green．C．Blue．（2）Why do cats bring dead birds home？A．To eat them in a safe place．B．To show off their hunting skills．C．To make their owners happy．（3）How does the man sound at the end of the conversation？A．Grateful．B．Humorous．C．Curious．9．（6.00分）听第9段材料，回答下列各题．（1）Who is Macy？A．Ed's mother．B．Ed's teacher．C．Ed's friend．（2）How does Ed usually go to kindergarten？A．By car．B．On foot．C．By bus．（3）What does Ed enjoy doing at the kindergarten？A．Telling stories．B．Singing songs．C．Playing with others．（4）What do the teachers say about Ed？A．He's clever．B．He's quiet．C．He's brave．10．（6.00分）听第10段材料，回答下列各题．（1）At what age did Emily start learning ballet？A．Five．B．Six．C．Nine．（2）Why did Emily move to Toronto？A．To work for a dance school．B．To perform at a dance teacher．C．To learn contemporary dance．（3）Why did Emily quit dancing？A．She was too old to dance．B．She failed to get a scholarship．C．She lost interest in it．（4）How does Emily feel about stopping training？A．She's pleased．B．She's regretful．C．She's upset．第二部分阅读理解（共两节，满分30分）第一节（共4小题；每小题6分，满分30分）阅读下列短文，从每题所给的A、B、C和D四个选项中，选出最佳选项。
吉林省梅河口市第五中学2018届高三政治第四次模拟考试试题201806110388
吉林省梅河口市第五中学2018届高三政治第四次模拟考试试题1. 本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分。
答题前，考生务必在将自己的姓名、考生号填写在答题卡上。
2. 回答第Ⅰ卷时，选出每小题选出答案后，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，再选涂其他答案标号。
写在试卷上无效。
3. 回答第Ⅱ卷时，将答案写在答题卡上，写在试卷上无效。
第1 卷（共140 分）1.小张“五一”期间到厦门旅游。
她前三天住在一家经济型酒店里，免费使用无线网络。
后三天她住在一家五星级酒店里，一天无线上网收费120 元。
小张心想，高端酒店应该提供更多的服务，怎么还收高额的上网费啊？下列解释正确的有：①低端酒店市场竞争更激烈，上网免费可以吸引更多客源②高端酒店顾客对上网的价格相对不敏感，对网络的需求弹性较小③使用无线网络收费可以限制网络资源使用，有效降低经营成本④使用无线网络具有竞争性，免费不会大幅提高经营成本A.①③B.①②C.②③D.③④2. 近日，财政部制定出《关于2018年农业综合开发产业化发展项目申报工作通知》，对今年农业补贴的大方向做出了指导，也对具体补贴细则，比如金额、补贴对象等方面，都做了细致讲解。
对该项财政政策的出台理解正确的是：A. 加大财政对农业的补贴→农业生产规模扩大→提高粮食安全保障水平→促进社会稳定B. 全面实施营改增→增加财政收入→刺激总需求增长→促进经济发展C. 减少税收收入→扩大农业生产队伍→农业投资增加→提高农业生产效率D. 实施积极的财政政策→货币信贷规模扩大→社会消费总量增加→拉动经济发展14．2018 年1 月15 日，国土资源部指出，政府将不再是居住用地唯一提供者，要完善农民闲置宅基地和闲置农房政策，将探索宅基地所有权、资格权、使用权“三权分置”，落实宅基地集体所有权，保障宅基地农户资格权，适度放活宅基地使用权。
此举旨在：①盘活农村集体土地，拓宽土地供应渠道②适当减少农业耕地，提高农村土地利用率③深化农村土地制度改革，保障农民土地所有权④探索农村宅基地入市，增加租赁住房供给A．①②B．①③C．②③D．①④15．2017 年5 月在北京主办“一带一路”国际合作高峰论坛，是各方共商、共建“一带一路”，共享互利合作成果的国际盛会，也是加强国际合作、对接彼此发展战略的重要合作平台。
【全国百强校首发】吉林省梅河口市第五中学2017-2018学年高二下学期期末考试英语试题
梅河口市第五中学2018年高二下学期期末英语试题第I 卷（共90 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，你将有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. Why was the man disappointed?A. The play wasn’t interesting.B.The tickets were unavailable.C. The play was only for daytime.2. What are the speakers going to do?A. To see an exhibition.B. To have a meeting.C. To listen to a lecture.3. What sport does the man like best?A. Swimming.B. Tennis.C. Golf.4. What musical instrument does the man play?A. The piano.B. The violin.C. None.5. What is the man’s opinion?A. He thinks highly of Jim.B. He disagrees with the woman.C. He doesn’t care at all.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。
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，并标在试卷的相应位置。
听每段对话或独白前，你将有时间阅读各个小题，每小题5 秒钟；听完后，各小题将给出5 秒钟的作答时间。
吉林省梅河口一中20182019学年上学期高三期末考试试卷英语Word版含答案
绝密 ★ 启用前 2018-2019学年上学期高三期末考试英 语注意事项：1.答题前，先将自己的姓名、准考证号填写在试题卷和答题卡上，并将准考证号条形码粘贴在答题卡上的指定位置。
用2B 铅笔将答题卡上试卷类型A 后的方框涂黑。
2.选择题的作答：每小题选出答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑，写在试题卷、草稿纸和答题卡上的非答题区域均无效。
3.非选择题的作答：用签字笔直接答在答题卡上对应的答题区域内。
写在试题卷、草稿纸和答题卡上的非答题区域均无效。
4.考试结束后，请将本试题卷和答题卡一并上交。
第Ⅰ卷第一部分 听力（共两节，满分 30 分）(略) 第二部分 阅读理解（共两节，满分40分） 第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的四个选项（A 、B 、C 和D ）中选出最佳选项，并在答题卡上将该项涂黑。
ASports can help us a lot. Taking exercises can make us strong. In collective(集体的) sports like basketball, volleyball or football, we will learn the importance of cooperation(合作). And sports can also help us relax after work or study.However, as the saying goes, “there are two sides of everything.” Sometimes we may hurt other players or ourselves if we are not careful enough when participating(参加) in sports activities. W hat‟s more, too much or hard practice can be bad for our health.Sports can make us healthy both physically and psychologically(心理地). It is also a good way for people to know each other and can improve friendship between people. So long as we are careful enough, sports can do us nothing but good.21. can make us strong. 22. Too much exercise can be for us.A. goodB. enoughC. badD. helpful23. Sports can .A. help people to know each otherB. improve friendship between peopleC. do us nothing but good if we are carefulD. All of the above 24. Which of the following is NOT true? A. Sports can help us relax after work or study. B. Sports can only make us healthy physically.C. Sometimes we may hurt other players or ourselves when participating in sports activities.D. Basketball and volleyball are both collective sports.BMany United States companies have made the search for legal protection from import competition into a major line of work. Since 1980, the United States International Trade Commission(ITC) has receivedabout 280 complaints alleging damage from imports that benefit from subsidies(补贴)by foreign governments. Another 340 charge th at foreign companies “dumped” their products in the United States at “less than fair value”. Even when no unfair practices are claimed, the simple claim that an industry has been injured by imports is sufficient grounds to seek relief(救济).Contrary to the general impression, this request for import relief has hurt more companies than it hashelped. As corporations begin to function globally, they develop a complicated web of marketing,production, and research relationships. The complexity of these relationships makes it unlikely that asystem of import relief laws will meet the strategic needs of all the units under the same parent company. Internationalization increases the danger that foreign companies will use import relief laws against the very companies the laws were designed to protect. Suppose a United States-owned company establishes an overseas plant to manufacture a product while its competitor makes the same product in the United States. If the competitor can prove injury from the imports ——and that the United States company received a subsidy from a foreign government to build its plant abroad ——the United States company‟s products will be uncompetitive in the United States, since they would be subject to duties.Perhaps the most shameful case occurred when the ITC investigated allegations(控诉) that Canadian companies were injuring the United States salt industry by dumping rock salt, used to deice roads. The bizarre aspect of the complaint was that a foreign conglomerate(联合企业) with United States operations 此卷只装订不密封班级 姓名 准考证号 考场号 座位号claiming injury was a unit of a Dutch conglomerate, while the “Canadian” companies included a unit of a Chicago firm that was the second-largest domestic producer of rock salt.25. The passage is chiefly concerned with .A. arguing against the increased internationalization of US corporationsB. recommending a uniform method for handling claims of unfair trade practicesC. warning that the application of laws affecting trade frequently has unintended consequencesD. advocating the use of trade restrictions for “dumped” products but not for other imports26. What can be inferred about the minimal basis for a complaint to the ITC?A. A foreign competitor is selling products in the US at less than fair market value.B. A foreign competitor has greatly increased the volume of products shipped to the US.C. The company requesting import relief has been banned from exporting products.D. The company requesting import relief has been injured by the sale of imports in the US.27. Which of the following is most likely to be true of US trade laws?A. They will eliminate the practice of “dumping” products in the US.B. Those applied to international companies will help to gain more profits.C. They will affect US trade with Canada more negatively than trade with other nations.D. Those helping one unit within a parent company won‟t necessarily help other units.CPalaces are known for their beauty and splendor, but they offer little protection against attacks. It is easy to defend a large building, but usually these buildings are not designed with the comfort of a king in mind. When it comes to structures that are both beautiful and defensive, the European castle is a big success.Castles were originally built in England by the Normans in 1066. They built towers and walls to secure the land they had taken. These castles provided the Normans with a quiet and safe place. They also served as bases of operation for attacks. In this way castles served both defensive and offensive roles. Besides, castles served as offices for governors. Those that were socially beneath the governor would come to report affairs and express their respect. They would address problems, handle business, feast, and enjoy festivities in castles. So castles served as social centers as well.The first castles were made from earth and wood, and they were likely to suffer from attacks by fire. Then wooden castles were gradually replaced by stone, which greatly increased the strength of these towers wooden doors. This led to moving the windows and entrances off of the ground floor and up to the first floor to make them more difficult to access.During the Middle Ages, attacks increased in regularity, so castle defenses were updated. Arrow-slits were added. They were small holes in the castle, which allowed defenders to fire without being hurt. Towers were built from which defenders could provide fire on both sides. The towers were connected to the castle by wooden bridges, so that if one tower fell, the rest of the castle was still easy to defend. A lot of rings of castle walls were constructed, so that even if attackers went past one wall, they would be caught on a killing ground between inner and outer walls. All of these increased the defense of castles.The end of castles can be attributed to gunpowder. During the 15th century, artillery, a kind of large guns, became powerful enough to break through stone walls. This greatly made the role of castles less effective. Though castles no longer serve their original purposes, remaining castles receive millions of visitors each year who wish to experience the situations of ancient times.28. What was the original function of castles according to the passage?A. They served as tourist attractions.B. They were important social centers.C. They marked religious ceremonies.D. They were built for use in emergencies.29. The reason why wooden castles were replaced by stone castles was that .A. stone castles cost less moneyB. stone castles offered better defenseC. wooden castles were uncomfortableD. wooden castles took a long time to build30. Which of the following showed an improvement in castle defenses?A. Castles were totally separated by stones.B. Arrow-slits were made in large quantities.C. Rings of walls were built to defend the towers.D. Windows and entrances were moved to the higher floor.31. What is the best title of the passage?A. Fancy Living: Learning about CastlesB. Normans: Bringing Castles to EnglandD. Defending Castles: Technologies Used to Defend CastlesDA schoolgirl saved her father‟s life by kicking him in the chest after he suffered a serious allergic reaction which stopped his heart.Izzy, nine, restarted father Colm‟s heart by stamping(踩) on his chest after he fell down at home and stopped breathing.Iz zy‟s mother, Debbie, immediately called 999, but Izzy knew doctors would never arrive in time to save her father, so decided to use CPR.However, she quickly discovered her arms weren‟t strong enough, so she stamped on her father‟s chest instead.Debbie then took over with some more conventional chest compressions until the ambulance arrived.Izzy, who has been given a bravery award by her school, said: “I just kicked him really hard. My mum taught me CPR but I knew I wasn‟t strong enough to use hands. I wa s quite scared. The doctor said I might as well be a doctor or a nurse. My mum said that Dad was going to hospital with a big footprint on his chest.”“She‟s a little star,” said Debbie. “I was really upset, but Izzy just took over. I just can‟t believe wh at she did. I really think all children should be taught first aid. Izzy did CPR, then the doctor turned up. Colm had to have more treatment on the way to the hospital and we‟ve got to see an expert.”Truck driver Colm, 35, suffered a mystery allergic reaction on Saturday and was taken to hospital, but was sent home only for it to happen again the next day. The second attack was so serious that his airway swelled, preventing him from breathing, his blood pressure dropped suddenly, and his heart stopped for a moment.He has now made a full recovery from his suffering.32. Izzy kicked her father in the chest _________.A. to express her helplessnessB. to practise CPR on himC. to restart his heartD. to keep him awake33. What‟s the right order o f the events?①Izzy kicked Colm.②Debbie called 999.③Izzy learned CPR.A. ③①②④B. ④②③①C. ④③①②D. ③④②①34. What does Paragraph 8 mainly talk about?A. What caused Colm‟s allergy.B. Colm‟s present condition.C. What Colm suffered.D. Symptoms of Colm‟s allergic reaction.35. Why does the author write the news?A. To report a 9-year-old girl‟s brave act.B. To prove the importance of CPR.C. To describe a serious accident.D. To call people‟s atte ntion to allergic reaction.第二节（共5小题；每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项，选项中有两项为多余选项。
【数学】吉林省梅河口市第五中学2018届高三第四次模拟考试数学(理)试题
数学试卷（理工类）一．选择题(每小题 5 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. 已知复数z53i，则下列说法正确的是（）1 iA．z的虚部为 4i B. z的共轭复数为1 4iC．z 5 D. z在复平面内对应的点在第二象限已知 m, n R ，集合 A 2, log7 m ，集合 B m, n ，若 A B 0 ，则 m n( ) A．1 B．2 C．4 D．83. (2x 1)8的二项展开式中，各项系数和为（）xA． 28B． 28C．1 D. 1 4.下列命题中正确命题的个数是（）（1） cos 0 是 2k(k Z ) 的充分必要条件2（2）f(x) sin x cos x 则 f (x)最小正周期是（3）若将一组样本数据中的每个数据都加上同一个常数后， 则样本的方差不变（4）设随机变量 X 服从正态分布 N (0,1) ,若 P ( X 1) p ，则 P ( 1 X 0)1 p 2A.4B.3C.2D.15. 如果函数 y 2 sin(2x ) 的图像关于点43 ，0 中心对称，那么 || 的最小值为A ．B ．C ．D ． 643 26．执行如图程序框图其输出结果是 （ ）A ． 29B ． 31C ． 33D ． 35某四棱锥的三视图如图所示，其中正(主)视图是等腰直角三角形,侧(左)视图是等腰三角形,俯视图是正方形,则该四棱锥的体积是 ( )A. 4B.8C. 4D. 2 33318. 3 log2x, x，则不等式 f (x)5的解集为( )已知函数 f (x)2x2 3x, xA. 1,1B. , 1 0,1C. 1,4D., 1 0,47.若函数 y cos 2x 与函数 y sin(x )在[0, 2]上的单调性相同，则 的一个值为( )A．B．C．D．643212.从 P 点出发的三条射线 PA, PB, PC 两两所成角均为60，且分别与球O切于点A,B, C,若球 O 的体积为4，则 OP 两点间的距离为（）33A. 2B. 3 C . D .2211. 直线 l 与抛物线 C : y 22x 交于 A , B 两点，O 为坐标原点，若直线 OA , OB 的斜率 k 1 ，k 2 满足 k 1k 2 2 ，则 l 一定过点( )3A. ( 3,0)B. (3,0)C.( 1,3)D. ( 2,0)已知函数 f (x ) x ln x k ，在区间[1e , e ] 上任取三个数 a , b , c 均存在以f (a ) ，f (b ) ，f (c ) 为边长的三角形，则 k 的取值范围是（ ）A ． ( 1， )B. ( , 1)C. ( , e 3)D. (e 3， )二、填空题（共 4 小题，每小题 5 分，共 20 分，将答案填在答题卡相应的位置上．）如图,在边长为 1 的正方形中随机撒 1000 粒豆子,有 380 粒落到阴影部分,据此估计阴影部分的面积为．y 514.若实数 x、 y 满足不等式组 2x y 3 0.则 z =2 y | x |的最大值是x 3 y 1 0下列命题：①已知 m, n 表示两条不同的直线， , 表示两个不同的平面，并且m , n ，则“ ”是“ m // n ”的必要不充分条件；②不存在x (0,1) ，使2不等式成立 log2x log3x；③“若am2 bm2，则a b”的逆命题为真命题；④ R ，函数 f (x) sin(2x )都不是偶函数.正确的命题序号是．在 ABC 中，角 A ， B ， C 所对边的长分别为 a ， b ， c ， M 为 AB 边上一点，CA CB c CM MP( R)且 MP ，又已知CM ，2a2 b22 ab ，则角 C CA cos A CB cos B2．三、解答题（本大题共 6 小题，共 70 分，解答应写出文字说明，证明过程或演算步骤．）17.（本小题满分 12 分）数列{a n } 满足a1 2 ，且na n 1 (n 1)a n n(n 1) ．(Ⅰ)求数列 a n 的通项公式；(Ⅱ) 已知b (n 1)2，求证：1115na1 b1a2 b2a n b n1218.（本小题满分12分）集成电路 E 由 3 个不同的电子元件组成，现由于元件老化，三个电子元件能正常工作的概率分别降为12 , 12 , 23，且每个电子元件能否正常工作相互独立．若三个电子元件中至少有 2 个正常工作，则 E 能正常工作，否则就需要维修，且维修集成电路 E 所需费用为 100 元．(I)求集成电路 E 需要维修的概率；(II)若某电子设备共由 2 个集成电路 E 组成，设 X 为该电子设备需要维修集成电路所需的费用，求 X 的分布列和期望．19.（本小题满分12分）如图，在四棱锥 P-ABCD 中，底面 ABCD 为梯形， ABC= BAD=90 ，AP=AD=AB=2 , BC t , PAB= PAD=(I)当t 3 2 时，试在棱 PA 上确定一个点 E，使得 PC∥平面 BDE，并求出此时EP AE 的值；(II)当 =60 时，若平面 PAB 平面 PCD，求此时棱 BC 的长．320.（本小题满分 12 分）已知椭圆E : x 2 y 2 1 的左，右顶点分别为 A , B ，圆 x 2 y 2 4 上有一4动点 P ,点 P 在 x 轴的上方， C 1,0 ，直线 PA 交椭圆 E 于点 D ，连接 DC , PB .(1)若 ADC 90,求△ ADC 的面积 S ;(2)设直线 PB , DC 的斜率存在且分别为 k 1 , k 2 ,若 k 1 k 2 ,求 的取值范围.21.（本小题满分 12 分）已知函数 f (x ) ln x1 ax 2x , a R ．． 2（Ⅰ）若 f (1) 0 ，求函数 f (x ) 的最大值；（Ⅱ）令 g (x ) f (x ) (ax 1) ，讨论函数 g ( x ) 的单调区间；1（Ⅲ）若 a 2 ，正实数 x , x满足 f (x ) f (x ) x x 20 ，证明 x x2 5 1 212 1 1 2．请考生在第 22、23、24 三题中任选一题作答，如果多做，则按所做的第一题记分.22.（本小题满分 10 分）选修 4-1：几何证明选讲如图，E 是圆内两弦 AB 和 CD 的交点，F 为 AD 延长线上一点，FG 切圆于 G ，且FE=FG ．(I)证明：FE∥BC；AF(II)若 AB⊥CD，∠DEF=30°，求 FG ．423（本小题满分10分）选修4-4:坐标系与参数方程x2cos已知曲线 C1（ 为参数），以坐标原点 O 为极点，x 轴的正半的参数方程为y3sin轴为极轴建立极坐标系，曲线 C2的极坐标方程为 =2.(1)分别写出C1的普通方程，C2的直角坐标方程．(2)已知 M，N 分别为曲线C1的上、下顶点，点 P 为曲线C2上任意一点，求PM PN的最大值．24（本小题满分10分）选修4—5：不等式选讲设函数 f(x)=|x 一 a|，a∈R．(I)若a=1，解不等式f(x) ≥12 (x+l)；( II)记函数g(x)=f (x) x 2 的值域为 A，若 A 1,3 ，求a的取值范围．5一．选择题BACCC BACDB AD二．填空题19504 13. 14 . 10 15. ①16.三．解答题（阅卷时发现其他正确解答，请教师参阅此评分标准酌情给分）17.解：(1)由na n 1 (n 1)a n n(n 1)得, a n1a n1, …………………..2 分n1 n2.a11 2 ,所以 an 是以 2 位首项，1 为公差的等差数列…………………..3 分 na n n1 na n n(n1) …………………..5 分（Ⅱ）115．…………………..6 分a b6121 1n≥2时，由（Ⅰ）知 a n b n (n 1)(2n 1) 2(n1)n ．·····································9分故11…11111…1a1 b1a2 b2a n622334n (n1)b n111111…1162 2 334nn1111111562 2 n 16412综上，原不等式成立．·················································································· 12 分解：（Ⅰ）三个电子元件能正常工作分别记为事件 A,B,C ，则p( A) 1, p(B)1, p(C)2.232依题意，集成电路 E 需要维修有两种情形：①3 个元件都不能正常工作，概率为p ) ) ) ) 1 1 1 1 ；p (ABCp ( Ap ( Bp ( C…………2 分1 2 2 3 126②3 个元件中的 2 个不能正常工作，概率为p 2 p ( ABC ABC ABC ) p ( ABC ) p ( ABC ) p ( ABC )1 1 1 1 1 1 1 12 4 1 ……………5 分2 22 2 2 123 2 3 3 3 1 15所以,集成电路 E 需要维修的概率为 pp． ……………6 分 1 2 123 12（Ⅱ）设 Y 为维修集成电路的个数，则 Y B (2, 5) ，而 X 100Y ， 12 57P ( X 100k ) P (Y k ) C k ()k()2 k, k 0,1, 2. ... ... ... (9)122 12分的分布列为：………………10 分EX 0 14449 100 3572 200 14425 2503或 EX 100EY 100 25250. …………12 分12 319 解：(1)（方法一）连接 AC, BD 交于点 F ,在平面 PCA 中作 EF // BC 交 PA于 E ,因为 PC 平面 BDE ， EF 平面 BDEPC ∥平面 BDE ，---------------2分因为AD ∥ BC,所以FC AF BC AD 13,因为 E F∥ P C , 所以EP AE FC AF = 13 . -------------4 分（方法二）在棱 PA 上取点 E ，使得AE 1,---------------2分EP3因为AD ∥ BC,所以AF AD 1 ,FC BC3所以， E F∥ P C7因为 PC 平面 BDE ， EF 平面 BDE所以 PC ∥平面 BDE------------- 4 分(2)取BC上一点G使得BG2, 连结DG ,则ABGD为正方形.P 作 PO ⊥平面 ABCD ,垂足为 O.连结 OA, OB, OD, OG .AP AD AB, PAB PAD 600，Array所以 PAB 和 PAD 都是等边三角形，因此 PA PBPD ,所以 OA OB OD ,即点 O 为正方形 ABGD 对角线的交点,---------------7 分因为 OG , OB , OP 两两垂直以 O 坐标原点，分别以 OG ,OB , OP 的方向为 x 轴， y 轴, z 轴的正方向建立如图所示的空间直角坐标系 O xyz .O （ 0，0，0）， P （ 0，0，1）， A （ 1，0，0）， B （ 0，1，0）， D （ 0，1，0） G （1，0，0）设棱 BC 的长为 t ，则 C ( 22t ,1 22t , 0) ,tt2 2 --------------9PA ( 1, 0, 1), PB (0,1, 1), PC, 1), PD (0,1,( 2 ,1 2 1)分设平面PAB 的法向量m (x 1 , y 1 , z 1 ),x z 0m PA, 则,即m PB 0y z不妨令x 1, 可得m ( 1,1,1)为平面PAB 的一个法向量.-----------10 分 设平面PCD 的法向量n ( x 2 , y 2 , z 2 ),2 2n PC 0tx (1t ) y z 02 2则 ,即n PDy z2 2不妨令y 1, 可得n (1 ,1, 1)为平面PCD 的一个法向量.t-----------11 分m n 0, 解得 t=2即棱BC 的长为2----------------12 分2 2.820.解（1）依题意，A( 2,0).设D(x, y)x2y 21.由 ADC 90 得kk1,，则1AD CD 1 1411 x2y y y212, 4,解得 x, x 2(舍去）11 1x2x1x 2x 1 x x2 312 1 1 11 1 1 1 1 12212y1 , S2 3.2----------------5 分3 23(2)设D x2 , y2 ,动点 P 在圆 x 2 y 2 4上, k PB k PA 1.又 k k , 1y2, 即x22 x21=x22 x211 2y2x21 y22x221x2 24x2 2 x21x2112,2 , 且== 4 =4 1. 又由题意可知x21x222 x2 24 4 x2x21,----------------10分1则问题可转化为求函数 f x4 1x 2,2 ,且x1 的值域.x2函数 f x 在其定义域内为减函数,函数 f x 的值域为,0 0,3从而 的取值范围为 ,00,3----------------12 分21.解:（Ⅰ）因为f(1) 1 a2 0，所以a 2，此时f(x) ln x x2x,x 0，f1 2x 12 x2 x1(x0) ,x(x) xf (x) 0，得 x 1，所以 f (x)在(0,1)上单调递增，在(1, )上单调递减，Y当 x 1时函数有极大值，也是最大值，所以 f (x)的最大值为则(1) 0 ．………………4 分（Ⅱ） g(x) f (x)-(ax 1) ln x 1 ax2 (1 a)x 1，9所以 1ax (1 a)ax2(1 a)x1．xg (x) x当 a ≤0时，因为 x 0，所以 g (x) 0．所以 g(x)在(0, )上是递增函数，1当 a 0ax2 (1 a)x1a(x)(x 1) ，时，ag (x)x x令 g (x) 0，得x 1．所以当 x(0,1) 时，g(x) 0 ；当x(1, )时，aa ag (x) 0，因此函数 g(x)在 x (0,1a)是增函数，在 x (1a, )是减函数．综上，当 a ≤0时，函数 g(x)的递增区间是(0,)，无递减区间；当 a 0时，函数 g(x)的递增区间是(0, 1) ，递减区间是a(1, ) ．………………8 分a（Ⅲ）当 a 2时， f (x) ln x x 2 x, x 0．f (x1) f (x2) x1 x2 0，即ln x1 x12 x1 ln x2 x22x2 x1 x2 0．从而 (x1 x2 )2 (x1 x2 ) x1 x2 ln(x1 x2 ) ．令 t x1 x2，则由 (t) t ln t 得， (t) t t1．可知， (t) 在区间 (0,1) 上单调递减，在区间 (1, ) 上单调递增．所以16. (t ) ≥ (1) 1，所以 (x x )2(x x )≥1 ，因为 x0, x 0 , 1 21 2121成立．……………………………………………… 12 分 因此 x x ≥ 51 2 210…….10 分23.解：（1）曲线 C 的普通方程为x 2 y 21，……………………2 分143曲线 C 2 的普通方程为 x 2y 24 . (4)分（2）法一：由曲线 C 2 ： x 2y 24 x 2cos ，所以 P 点，可得其参数方程为y 2sin坐标为 (2 cos , 2sin ) ，由题意可知 M(0, 3), N (0,3) .因此 PM + PN (2cos )2 (2sin 3)2 (2cos )2 (2sin3)27 4 3 sin 7 4 3 sin (6)分( PM + PN )2 14 249 48sin2 .所以当 sin 0 时， ( PM + PN )2有最大值 28，……………………8分因此 PM + PN 的最大值为27 . ……………………10 分法二：设 P 点坐标为( x, y)，则 x2 y2 4，由题意可知 M (0, 3), N (0, 3) .因此 PM + PN x2 ( y 3)2 x2 ( y 3)27 23y 7 23y……………………6分( PM + PN )2 14 249 12 y2 .11所以当 y 0时，( PM + PN )2有最大值28，……………………8分因此 PM + PN 的最大值为27 . ……………………10 分学生用几何法求最大值，只要论证严密，酌情给分122018年高考考前猜题卷理科数学 第Ⅰ卷（共60分）一、选择题：本大题共12个小题，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．设复数z 满足iii z 2|2|++=，则=||z （ ） A ．3 B ．10 C ．9 D ．102．已知全集R U =，集合}012|{2≥--=x x x M ，}1|{x y x N -==，则=N M C U )(（ ）A ．}1|{≤x xB ．}121|{≤<-x xC ．}121|{<<-x x D ．}211|{<<-x x3．已知蚂蚁在边长为4的正三角形区域内随机爬行，则它在离三个顶点的距离都大于2的区域内的概率P 为（ ） A ．631π-B ．43C ．63π D ．414．已知双曲线)0,0(12222>>=-b a by a x ，过双曲线左焦点1F 且斜率为1的直线与其右支交于点M ，且以1MF 为直径的圆过右焦点2F ，则双曲线的离心率是（ ） A ．12+ B ．2 C ．3 D ．13+5．一个算法的程序框图如图所示，如果输出y 的值是1，那么输入x 的值是（ ）A ．2-或2B ．2-或2C ．2-或2D ．2-或2 6．已知函数)2||,0)(3sin()(πϕωπω<>+=x x f 的图象中相邻两条对称轴之间的距离为2π，将函数)(x f y =的图象向左平移3π个单位后，得到的图象关于y 轴对称，那么)(x f y =的图象（ ）A ．关于点)0,12(π对称 B ．关于点)0,12(π-对称C ．关于直线12π=x 对称 D ．关于直线12π-=x 对称7．如下图，网格纸上小正方形的边长为1，图中实线画的是某几何体的三视图，则该几何体最长的棱的长度为（ ）A.32 B.43C. 2D. 411 8．已知等差数列}{n a 的第6项是6)2(xx -展开式中的常数项，则=+102a a （ ）A ．160B ．160-C ．350D ．320- 9．已知函数)0(212)(<-=x x f x 与)(log )(2a x x g +=的图象上存在关于y 轴对称的点，则a 的取值范围是（ ）A ．)2,(--∞B ．)2,(-∞C ．)22,(--∞D ．)22,22(- 10．已知正四棱台1111D C B A ABCD -的上、下底面边长分别为22,2，高为2，则其外接球的表面积为（ ）A ．π16B ．π20C ．π65D ．π465 11．平行四边形ABCD 中，2,3==AD AB ，0120=∠BAD ，P 是平行四边形ABCD 内一点，且1=AP ，若y x +=，则y x 23+的最大值为（ ） A ．1 B ．2 C ．3 D ．412．设n n n C B A ∆的三边长分别为n n n c b a ,,，n n n C B A ∆的面积为,3,2,1,=n S n …，若n n a a a c b ==++1111,2，2,211nn n n n n a b c a c b +=+=++，则（ ） A ．}{n S 为递减数列 B ．}{n S 为递增数列C ．}{12-n S 为递增数列，}{2n S 为递减数列D ．}{12-n S 为递减数列，}{2n S 为递增数列二、填空题（每题4分，满分20分，将答案填在答题纸上）13．函数x a x a x x f )3()1()(24-+--=的导函数)('x f 是奇函数，则实数=a .14．已知y x ,满足约束条件⎪⎩⎪⎨⎧≥+≤-≥+-002043y x x y x （R y x ∈,），则22y x +的最大值为 .15．已知F 为抛物线x y C 4:2=的焦点，过点F 作两条互相垂直的直线21,l l ，直线1l 与C 交于B A ,两点，直线2l 与C 交于E D ,两点，则||||DE AB +的最小值为 . 16．在锐角三角形ABC 中，角C B A ,,的对边分别为c b a ,,，且满足ac a b =-22，则BA tan 1tan 1-的取值范围为 . 三、解答题 （本大题共6题，共70分．解答应写出文字说明、证明过程或演算步骤．）17．已知等比数列}{n a 的前n 项和为n S ，且满足)(221R m m S n n ∈+=+. （1）求数列}{n a 的通项公式；（2）若数列}{n b 满足)(log )12(112+⋅+=n n n a a n b ，求数列}{n b 的前n 项和n T .18．小张举办了一次抽奖活动.顾客花费3元钱可获得一次抽奖机会.每次抽奖时，顾客从装有1个黑球，3个红球和6个白球（除颜色外其他都相同）的不透明的袋子中依次不放回地摸出3个球，根据摸出的球的颜色情况进行兑奖.顾客中一等奖，二等奖，三等奖，四等奖时分别可领取的奖金为a 元，10元，5元，1元.若经营者小张将顾客摸出的3个球的颜色分成以下五种情况：1:A 个黑球2个红球；3:B 个红球；:c 恰有1个白球；:D 恰有2个白球；3:E 个白球，且小张计划将五种情况按发生的机会从小到大的顺序分别对应中一等奖，中二等奖，中三等奖，中四等奖，不中奖.（1）通过计算写出中一至四等奖分别对应的情况（写出字母即可）； （2）已知顾客摸出的第一个球是红球，求他获得二等奖的概率；（3）设顾客抽一次奖小张获利X 元，求变量X 的分布列；若小张不打算在活动中亏本，求a 的最大值.19．如图，三棱柱111C B A ABC -中，侧面C C BB 11为菱形，0160=∠CBB ，1AC AB =.（1）证明：平面⊥C AB 1平面C C BB 11；（2）若C B AB 1⊥，直线AB 与平面C C BB 11所成的角为030，求直线1AB 与平面C B A 11所成角的正弦值.20．如图，圆),(),0,2(),0,2(,4:0022y x D B A y x O -=+为圆O 上任意一点，过D 作圆O 的切线，分别交直线2=x 和2-=x 于F E ,两点，连接BE AF ,，相交于点G ，若点G 的轨迹为曲线C .（1）记直线)0(:≠+=m m x y l 与曲线C 有两个不同的交点Q P ,，与直线2=x 交于点S ，与直线1-=y 交于点T ，求OPQ ∆的面积与OST ∆的面积的比值λ的最大值及取得最大值时m 的值.（注：222r y x =+在点),(00y x D 处的切线方程为200r yy xx =+）21．已知函数x a x g x x f ln )(,21)(2==. （1）若曲线)()(x g x f y -=在2=x 处的切线与直线073=-+y x 垂直，求实数a 的值；（2）设)()()(x g x f x h +=，若对任意两个不等的正数21,x x ，2)()(2121>--x x x h x h 恒成立，求实数a 的取值范围；（3）若在],1[e 上存在一点0x ，使得)(')()('1)('0000x g x g x f x f -<+成立，求实数a 的取值范围.请考生在22、23二题中任选一题作答，如果都做，则按所做的第一题记分.22．选修4-4：坐标系与参数方程在直角坐标系xOy 中，曲线1C 的参数方程为⎪⎩⎪⎨⎧==21t a y t x （其中t 为参数，0>a ），以坐标原点O 为极点，x 轴的正半轴为极轴建立的极坐标系中，直线l ：0sin cos =+-b θρθρ与2C ：θρcos 4-=相交于B A ,两点，且090=∠AOB .（1）求b 的值；（2）直线l 与曲线1C 相交于N M ,两点，证明：||||22N C M C ⋅（2C 为圆心）为定值. 23．选修4-5：不等式选讲已知函数|1||42|)(++-=x x x f . （1）解不等式9)(≤x f ；（2）若不等式a x x f +<2)(的解集为A ，}03|{2<-=x x x B ，且满足A B ⊆，求实数a 的取值范围.参考答案一、选择题：本大题共12小题，每小题5分，共60分．在每个小题给出的四个选项中，只有一项是符合题目要求的．二、填空题：本大题共4小题，每小题5分，共20分．13．3 14．8 15．16 16．)332,1( 三、解答题：本大题共6小题，满分70分．解答须写出文字说明、证明过程和演算步骤．17．解：（1）由)(221R m m S n n ∈+=+得⎪⎪⎪⎩⎪⎪⎪⎨⎧+=+=+=282422321m S m S m S ，)(R m ∈，从而有4,2233122=-==-=S S a S S a ， 所以等比数列}{n a 的公比223==a a q ，首项11=a ，因此数列}{n a 的通项公式为)(2*1N n a n n ∈=-.（2）由（1）可得12)22(log )(log 1212-=⋅=⋅-+n a a n n n n ， ∴)121121(21)12)(12(1+--⨯=-+=n n n n b n ∴)1211215131311(2121+--++-+-⨯=+++=n n b b b T n n 12+=n n. 18．解：（1）4011203)(31023===C C A P ；12011)(310==C B P ，10312036)(3102416===C C C C P ，2112060)(3101426===C C C D P ，6112020)(31036===C C E P∵)()()()()(D P C P E P A P B P <<<<， ∴中一至四等奖分别对应的情况是C E A B ,,,.（2）记事件F 为顾客摸出的第一个球是红球，事件G 为顾客获得二等奖，则181)|(2912==C C F G P .（3）X 的取值为3,2,2,7,3---a ，则分布列为由题意得，若要不亏本，则03212103)2(61)7(401)3(1201≥⨯+⨯+-⨯+-⨯+-⨯a ， 解得194≤a ，即a 的最大值为194.19．解：（1）证明：连接1BC ，交C B 1于O ，连接AO ， ∵侧面C C BB 11为菱形，∴11BC C B ⊥ ∵为1BC 的中点，∴1BC AO ⊥ 又O AO C B = 1，∴⊥1BC 平面C AB 1又⊂1BC 平面C C BB 11，∴平面⊥C AB 1平面C C BB 11.（2）由B BO AB C B BO C B AB =⊥⊥ ,,11，得⊥C B 1平面ABO 又⊂AO 平面ABO ，∴C B AO 1⊥，从而1,,OB OB OA 两两互相垂直，以O 为坐标原点，的方向为x 轴正方向，建立如图所示的空间直角坐标系xyz O -∵直线AB 与平面C C BB 11所成角为030，∴030=∠ABO设1=AO ，则3=BO ，∵0160=∠CBB ，∴1CBB ∆是边长为2的等边三角形∴)0,1,0(),0,1,0(),0,0,3(),1,0,0(1-C B B A ，则)1,0,3(),0,2,0(),1,1,0(1111-==-=-=AB B A C B AB 设),,(z y x =是平面C B A 11的法向量，则⎪⎩⎪⎨⎧=⋅=⋅00111C B n B A n 即⎩⎨⎧=-=-0203y z x ，令1=x ，则)3,0,1(=n设直线1AB 与平面C B A 11所成的角为θ， 则46||||||,cos |sin ==><=n AB θ. 20．解：（1）易知过点),(00y x D 的切线方程为400=+y y x x ，其中42020=+y x ，则)24,2(),2,2(000y x F y x E +--， ∴4116416416424424220020000021-=-=--=-⋅-+=y y y x y x y x k k 设),(y x G ，则144122412221=+⇒-=+⋅-⇒-=y x x y x y k k （0≠y ） 故曲线C 的方程为1422=+y x （0≠y ） （2）联立⎩⎨⎧=++=4422y x mx y 消去y ，得0448522=-++m mx x ，设),(),,(2211y x Q y x P ，则544,5822121-=-=+m x x m x x ，由0)44(206422>--=∆m m 得55<<-m 且2,0±≠≠m m ∴22221221255245444)58(24)(11||m m m x x x x PQ -=-⨯--⨯=-++=，易得)1,1(),2,2(---+m T m S ， ∴)3(2)3()3(||22m m m ST +=+++=，∴22)3(554||||m m ST PQ S S OSTOPQ +-===∆∆λ，令)53,53(,3+-∈=+t t m 且5,3,1≠t ，则45)431(4544654222+--⨯=-+-=t t t t λ， 当431=t ，即43=t 时，λ取得最大值552，此时35-=m . 21．解：（1）xax y x a x x g x f y -=-=-=',ln 21)()(2 由题意得322=-a，解得2-=a （2）)()()(x g x f x h +=x a x ln 212+=对任意两个不等的正数21,x x ，2)()(2121>--x x x h x h 恒成立，令21x x >，则)(2)()(2121x x x h x h ->-，即2211)(2)(x x h x x h ->-恒成立 则问题等价于x x a x x F 2ln 21)(2-+=在),0(+∞上为增函数 2)('-+=xax x F ，则问题转化为0)('≥x F 在),0(+∞上恒成立，即22x x a -≥在),0(+∞上恒成立，所以1)2(max 2=-≥x x a ，即实数a 的取值范围是),1[+∞. （3）不等式)(')()('1)('0000x g x g x f x f -<+等价于0000ln 1x ax a x x -<+， 整理得01ln 000<++-x ax a x ，构造函数x a x a x x m ++-=1ln )(， 由题意知，在],1[e 上存在一点0x ，使得0)(0<x m2222)1)(1()1(11)('x x a x x a ax x x a x a x m +--=+--=+--=因为0>x ，所以01>+x ，令0)('=x m ，得a x +=1①当11≤+a ，即0≤a 时，)(x m 在],1[e 上单调递增，只需02)1(<+=a m ，解得2-<a ； ②当e a ≤+<11，即10-≤<e a 时，)(x m 在a x +=1处取得最小值.令01)1ln(1)1(<++-+=+a a a a m ，即)1ln(11+<++a a a ，可得)1ln(11+<++a aa （*）令1+=a t ，则e t ≤<1，不等式（*）可化为t t t ln 11<-+ 因为e t ≤<1，所以不等式左端大于1，右端小于或等于1，所以不等式不能成立. ③当e a >+1，即1->e a 时，)(x m 在],1[e 上单调递减，只需01)(<++-=eaa e e m 解得112-+>e e a .综上所述，实数a 的取值范围是),11()2,(2+∞-+--∞e e . 22．解：（1）由题意可得直线l 和圆2C 的直角坐标方程分别为0=+-b y x ，4)2(22=++y x∵090=∠AOB ，∴直线l 过圆2C 的圆心)0,2(2-C ，∴2=b . （2）证明：曲线1C 的普通方程为)0(2>=a ay x ，直线l 的参数方程为⎪⎪⎩⎪⎪⎨⎧=+-=ty t x 22222（t 为参数），代入曲线1C 的方程得04)2222(212=++-t a t ， 04212>+=∆a a 恒成立，设N M ,两点对应的参数分别为21,t t ，则821=t t ， ∴8||||22=N C M C ， ∴||||22N C M C 为定值8.23．解：（1）由9)(≤x f 可得9|1||42|≤++-x x ，即⎩⎨⎧≤->9332x x 或⎩⎨⎧≤-≤≤-9521x x 或⎩⎨⎧≤+--<9331x x解得42≤<x 或21≤≤-x 或12-<≤-x ， 故不等式9)(≤x f 的解集为]4,2[-.（2）易知)3,0(=B ，由题意可得a x x x +<++-2|1||42|在)3,0(上恒成立⇒1|42|-+<-a x x 在)3,0(上恒成立1421-+<-<+-⇒a x x a x 在)3,0(上恒成立 3->⇒x a 且53+->x a 在)3,0(上恒成立⎩⎨⎧≥≥⇒50a a 5≥⇒a .。
【全国百强校首发】吉林省梅河口市第五中学2017-2018学年高二下学期期末考试英语试题
梅河口市第五中学2018年高二下学期期末英语试题第I 卷（共90 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，你将有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. Why was the man disappointed?A. The play wasn’t interesting.B.The tickets were unavailable.C. The play was only for daytime.2. What are the speakers going to do?A. To see an exhibition.B. To have a meeting.C. To listen to a lecture.3. What sport does the man like best?A. Swimming.B. Tennis.C. Golf.4. What musical instrument does the man play?A. The piano.B. The violin.C. None.5. What is the man’s opinion?A. He thinks highly of Jim.B. He disagrees with the woman.C. He doesn’t care at all.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。
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，并标在试卷的相应位置。
听每段对话或独白前，你将有时间阅读各个小题，每小题5 秒钟；听完后，各小题将给出5 秒钟的作答时间。
吉林省梅河口市第五中学2018届高三第四次模拟考试数学理试题 含答案 精品
数学试卷（理工类）一．选择题(每小题 5 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. 已知复数z53i，则下列说法正确的是（）1 iA．z的虚部为 4i B. z的共轭复数为1 4iC．z 5 D. z在复平面内对应的点在第二象限已知 m, n R ，集合 A 2, log7 m ，集合 B m, n ，若 A B 0 ，则 m n( ) A．1 B．2 C．4 D．83. (2x 1)8的二项展开式中，各项系数和为（）xA． 28B． 28C．1 D. 1 4.下列命题中正确命题的个数是（）（1） cos 0 是 2k(k Z ) 的充分必要条件2（2）f(x) sin x cos x 则 f (x)最小正周期是（3）若将一组样本数据中的每个数据都加上同一个常数后，则样本的方差不变（4）设随机变量X服从正态分布N(0,1) ,若P( X 1) p ，则 P( 1 X 0)1p2A.4B.3C.2D.15.如果函数 y 2 sin(2x )的图像关于点 43 ，0 中心对称，那么||的最小值为A．B．C．D．64326．执行如图程序框图其输出结果是（）A． 29B． 31 C． 33 D． 35某四棱锥的三视图如图所示，其中正(主)视图是等腰直角三角形,侧(左)视图是等腰三角形,俯视图是正方形,则该四棱锥的体积是( )A. 4B.8C. 4D.2 33318. 3 log2x, x，则不等式 f (x)5的解集为( )已知函数 f (x)2x2 3x, xA. 1,1B. , 1 0,1C. 1,4D., 1 0,47.若函数 y cos 2x 与函数 y sin(x )在[0, 2]上的单调性相同，则 的一个值为( )A．B．C．D．643212.从 P 点出发的三条射线 PA, PB, PC 两两所成角均为60，且分别与球O切于点A,B, C,若球 O 的体积为4，则 OP 两点间的距离为（）3 3A. 2B. 3 C .D .2211. 直线 l 与抛物线 C : y 22x 交于 A , B 两点，O 为坐标原点，若直线 OA , OB 的斜率 k 1 ，k 2 满足 k 1k 2 2 ，则 l 一定过点( )3A. ( 3,0)B. (3,0)C.( 1,3)D. ( 2,0)已知函数 f (x ) x ln x k ，在区间[1e , e ] 上任取三个数 a , b , c 均存在以f (a ) ，f (b ) ，f (c ) 为边长的三角形，则 k 的取值范围是（ ）A ． ( 1， )B. ( , 1)C. ( , e 3)D. (e 3， )二、填空题（共 4 小题，每小题 5 分，共 20 分，将答案填在答题卡相应的位置上．）如图,在边长为 1 的正方形中随机撒 1000 粒豆子,有 380 粒落到阴影部分,据此估计阴影部分的面积为．y 514.若实数 x、 y 满足不等式组 2x y 3 0.则 z =2 y | x |的最大值是x 3 y 1 0下列命题：①已知 m, n 表示两条不同的直线， , 表示两个不同的平面，并且m , n ，则“ ”是“ m // n ”的必要不充分条件；②不存在x (0,1) ，使2不等式成立 log2x log3x；③“若am2 bm2，则a b”的逆命题为真命题；④ R ，函数 f (x) sin(2x )都不是偶函数.正确的命题序号是．在 ABC 中，角 A ， B ， C 所对边的长分别为 a ， b ， c ， M 为 AB 边上一点，CA CB cCM MP( R)且 MP ，又已知CM ，2 a2 b2 ab ，则角 C CA cos A CB cos B22．三、解答题（本大题共 6 小题，共 70 分，解答应写出文字说明，证明过程或演算步骤．）17.（本小题满分 12 分）数列{a n } 满足a1 2 ，且na n 1 (n 1)a n n(n 1) ．(Ⅰ)求数列 a n 的通项公式；(Ⅱ) 已知b (n 1)2，求证：1115na1 b1a2 b2a n b n1218.（本小题满分12分）集成电路 E 由 3 个不同的电子元件组成，现由于元件老化，三个电子元件能正常工作的概率分别降为1 , 1 , 23，且每个电子元件能否正常工作相互独立．若三个电子元件中至少有 2 个正常工作，则 E 能正常工作，否则就需要维修，且维修集成电路 E 所需费用为 100 元．(I)求集成电路 E 需要维修的概率；(II)若某电子设备共由 2 个集成电路 E 组成，设 X 为该电子设备需要维修集成电路所需的费用，求 X 的分布列和期望．19.（本小题满分12分）如图，在四棱锥 P-ABCD 中，底面 ABCD 为梯形， ABC= BAD=90 ，AP=AD=AB=2 , BC t , PAB= PAD=(I)当t 3 2 时，试在棱 PA 上确定一个点 E，使得 PC∥平面 BDE，并求出此时EP AE 的值；(II)当 =60 时，若平面 PAB 平面 PCD，求此时棱 BC 的长．320.（本小题满分 12 分）已知椭圆E : x 2 y 2 1 的左，右顶点分别为 A , B ，圆 x 2 y 2 4 上有一4动点 P ,点 P 在 x 轴的上方， C 1,0 ，直线 PA 交椭圆 E 于点 D ，连接 DC , PB .(1)若 ADC 90,求△ ADC 的面积 S ;(2)设直线 PB , DC 的斜率存在且分别为 k 1 , k 2 ,若 k 1 k 2 ,求 的取值范围.21.（本小题满分 12 分）已知函数 f (x ) ln x1 ax 2x , a R ．． 2（Ⅰ）若 f (1) 0 ，求函数 f (x ) 的最大值；（Ⅱ）令 g (x ) f (x ) (ax 1) ，讨论函数 g ( x ) 的单调区间；1（Ⅲ）若 a 2 ，正实数 x , x满足 f (x ) f (x ) x x 20 ，证明 x x2 5 1 212 1 1 2．请考生在第 22、23、24 三题中任选一题作答，如果多做，则按所做的第一题记分.22.（本小题满分 10 分）选修 4-1：几何证明选讲如图，E 是圆内两弦 AB 和 CD 的交点，F 为 AD 延长线上一点，FG 切圆于 G ，且FE=FG ．(I)证明：FE∥BC；AF(II)若 AB⊥CD，∠DEF=30°，求 FG ．423（本小题满分10分）选修4-4:坐标系与参数方程x2cos已知曲线 C1（ 为参数），以坐标原点 O 为极点，x 轴的正半的参数方程为y3sin轴为极轴建立极坐标系，曲线 C2的极坐标方程为 =2.(1)分别写出C1的普通方程，C2的直角坐标方程．(2)已知 M，N 分别为曲线C1的上、下顶点，点 P 为曲线C2上任意一点，求PM PN的最大值．24（本小题满分10分）选修4—5：不等式选讲设函数 f(x)=|x 一 a|，a∈R．(I)若a=1，解不等式f(x) ≥12 (x+l)；( II)记函数g(x)=f (x) x 2 的值域为 A，若 A 1,3 ，求a的取值范围．5一．选择题BACCC BACDB AD二．填空题19504 13. 14 . 10 15. ①16. 三．解答题（阅卷时发现其他正确解答，请教师参阅此评分标准酌情给分）17.解：(1)由na n 1 (n 1)a n n(n 1)得, a n1a n1, …………………..2 分n1 n2.a11 2 ,所以 an 是以 2 位首项，1 为公差的等差数列…………………..3 分 na n n1 na n n(n1) …………………..5 分（Ⅱ）115．…………………..6 分a b6121 1n≥2时，由（Ⅰ）知 a n b n (n 1)(2n 1) 2(n1)n ．·····································9分故 1 1 ...1 1 1 1 1 (1)a1 b1a2 b2a n622334n (n1)b n111111…1162 2 334nn1111111562 2 n 16412综上，原不等式成立．·················································································· 12 分解：（Ⅰ）三个电子元件能正常工作分别记为事件 A,B,C ，则p( A) 1, p(B)1, p(C)2.232依题意，集成电路 E 需要维修有两种情形：①3 个元件都不能正常工作，概率为p p()p()p()p()1 1 1 1 ；ABC A B C (2)分 1 2 2 3 126②3 个元件中的 2 个不能正常工作，概率为p 2 p ( ABC ABC ABC ) p ( ABC ) p ( ABC ) p ( ABC )1 1 1 1 1 1 1 12 4 1 ……………5 分2 22 2 2 123 2 3 3 3 1 15所以,集成电路 E 需要维修的概率为 pp． ……………6 分 1 2 123 12（Ⅱ）设 Y 为维修集成电路的个数，则 Y B (2, 5) ，而 X 100Y ， 12 57P ( X 100k ) P (Y k ) C k ()k()2 k, k 0,1, 2. ... ... ... (9)122 12分的分布列为：………………10 分EX 0 14449 100 3572 200 14425 2503或 EX 100EY 100 25250. …………12 分12 319 解：(1)（方法一）连接 AC, BD 交于点 F ,在平面 PCA 中作 EF // BC 交 PA于 E ,因为 PC 平面 BDE ， EF 平面 BDEPC ∥平面 BDE ，---------------2分因为AD ∥ BC,所以FC AF BC AD 13,因为 E F∥ P C , 所以EP AE FC AF = 13 . -------------4 分（方法二）在棱 PA 上取点 E ，使得AE 1,---------------2分EP3因为AD ∥ BC,所以AF AD 1 ,FC BC3所以， E F∥ P C7因为 PC 平面 BDE ， EF 平面 BDE所以 PC ∥平面 BDE------------- 4 分(2)取BC上一点G使得BG, 连结DG ,则ABGD为正方形.2P 作 PO ⊥平面 ABCD ,垂足为 O.连结 OA, OB, OD, OG .AP AD AB, PAB PAD 600，Array所以 PAB 和 PAD 都是等边三角形，因此 PA PBPD ,所以 OA OB OD ,即点 O 为正方形 ABGD 对角线的交点,---------------7 分因为 OG , OB , OP 两两垂直以 O 坐标原点，分别以 OG ,OB , OP 的方向为 x 轴， y 轴, z 轴的正方向建立如图所示的空间直角坐标系 O xyz .O （ 0，0，0）， P （ 0，0，1）， A （ 1，0，0）， B （ 0，1，0）， D （ 0，1，0） G （1，0，0）设棱 BC 的长为 t ，则 C ( 22t ,1 22t , 0) ,tt2 2 --------------9PA ( 1, 0, 1), PB (0,1, 1), PC, 1), PD (0,1,( 2 ,1 2 1)分设平面PAB 的法向量m (x 1 , y 1 , z 1 ),x z 0m PA, 则,即m PB 0y z 0不妨令x 1, 可得m ( 1,1,1)为平面PAB 的一个法向量.-----------10 分 设平面PCD 的法向量n ( x 2 , y 2 , z 2 ),2 2n PC 0tx (1t ) y z 02 2则 ,即n PDy z2 2不妨令y 1, 可得n (1 ,1, 1)为平面PCD 的一个法向量.t-----------11 分m n 0, 解得 t=2即棱BC 的长为2----------------12 分2 2.820.解（1）依题意，A ( 2,0) .设 D (x , x 2 y 2 .由 ADC 90得y)，则1 1kAD kCD1,1 1411 x2y y y212, 4,解得 x, x 2(舍去）11 1x2x1x 2x 1 x x2 312 1 1 11 1 1 1 1 12212y1 , S2 3.2----------------5 分3 23(2)设D x2 , y2 ,动点 P 在圆 x 2 y 2 4上, k PB k PA 1.又 k k , 1y2, 即x22 x21=x22 x211 2y2x21 y22x221x2 24x 2 2 x 21 x2 1 12,2 , 且 == 4 = 4 1. 又 由 题 意 可 知x 21x 2 22 x 2 24 4 x 2x 2 1,----------------10分1则问题可转化为求函数 f x 4 1 x 2,2 , 且x 1 的值域. x 2函数 f x 在其定义域内为减函数,函数 f x 的值域为,0 0,3从而 的取值范围为 ,0 0,3----------------12 分21. 解:（Ⅰ）因为 f (1) 1 a 2 0 ，所以 a 2 ， 此时 f (x ) ln x x 2x , x 0 ，f12x 12 x 2x1(x 0) ,x (x ) xf (x) 0，得 x 1，所以 f (x)在(0,1)上单调递增，在(1, )上单调递减，Y当 x 1时函数有极大值，也是最大值，所以 f (x)的最大值为则(1) 0 ．………………4 分（Ⅱ） g(x) f (x)-(ax 1) ln x 12 ax2 (1 a)x 1，9所以 1ax (1 a)ax2(1 a)x1．xg (x) x当 a ≤0时，因为 x 0，所以 g (x) 0．所以 g(x)在(0, )上是递增函数，1当 a 0ax2 (1 a)x1a(x)(x 1) ，时，ag (x)x x令 g (x) 0，得1．所以当 x 1) 时，g (x) 0 ；当x1, )x (0, (a 时，a ag (x) 0，因此函数 g(x)在 x (0,1a)是增函数，在 x (1a, )是减函数．综上，当 a ≤0时，函数 g(x)的递增区间是(0,)，无递减区间；当 a 0时，函数 g(x)的递增区间是(0, 1) ，递减区间是a(1, ) ．………………8 分a（Ⅲ）当 a 2时， f (x) ln x x 2 x, x 0．f (x1) f (x2) x1 x2 0，即ln x1 x12 x1 ln x2 x22x2 x1 x2 0．从而 (x1 x2 )2 (x1 x2 ) x1 x2 ln(x1 x2 ) ．令 t x1 x2，则由 (t) t ln t 得， (t) t t1．可知， (t) 在区间 (0,1) 上单调递减，在区间 (1, ) 上单调递增．所以16.(t) ≥ (1) 1，所以 (x x )2(x x )≥1 ，因为 x0, x 0 , 1 21 2121成立．……………………………………………… 12 分 因此 x x ≥ 51 2 210…….10 分23.解：（1）曲线 C 的普通方程为x 2 y 21，……………………2 分143曲线 C 2 的普通方程为 x 2y 24 . (4)分（2）法一：由曲线 C 2 ： x 2y 24 x 2cos ，所以 P 点，可得其参数方程为y 2sin坐标为 (2 cos , 2sin ) ，由题意可知 M(0, 3), N (0,3) .因此 PM + PN (2cos )2(2sin3)2(2cos )2(2sin3)27 4 3 sin 7 4 3 sin (6)分( PM + PN )2 14 249 48sin2 .所以当 sin 0 时， ( PM + PN )2有最大值 28，……………………8分因此 PM + PN 的最大值为27 . ……………………10 分法二：设 P 点坐标为( x, y)，则 x2 y2 4，由题意可知 M (0, 3), N (0, 3) .因此 PM + PN x2 ( y 3)2 x2 ( y 3)27 23y 7 23y……………………6分( PM + PN )2 14 249 12 y2 .11所以当 y 0时，( PM + PN )2有最大值28，……………………8分因此 PM + PN 的最大值为27 . ……………………10 分学生用几何法求最大值，只要论证严密，酌情给分12。
2017-2018学年吉林省梅河口市第五中学高一4月考英语试题(PDF版)
梅河口市第五中学2018年高一4月月考英语试题第I卷(选择题，共85分)第一部分：听力理解（共两节，20小题，每题1.5分，满分30分）第一节听下面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.Where does the conversation take place probably?A.At home.B.In a shop.C.In a swimming pool.2.On which days doesn’t the man’s son work?A.Monday,Wednesday and Friday.B.Tuesday,Thursday and Sunday.C.Monday,Friday and Saturday.3.Who is going after Jack?A.Mr.Green.B.Mrs.Green.C.His brother.4.What does the man mean?A.The car doesn’t need cleaning.B.He cleaned the car last time.C.Mark should clean the car this time.5.When should the delivery be made to the man?A.On Sunday.B.On Saturday.C.On Thursday.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。
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。
并标在试卷的相应位置。
听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟；听完后，各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
请听第6段材料，回答第6、7题。
吉林省梅河口一中20182019学年上学期高三期末考试试卷 英语 Word版含答案
绝密 ★ 启用前2018-2019学年上学期高三期末考试英 语注意事项：1.答题前，先将自己的姓名、准考证号填写在试题卷和答题卡上，并将准考证号条形码粘贴在答题卡上的指定位置。
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第Ⅰ卷第一部分 听力（共两节，满分 30 分）(略)第二部分 阅读理解（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的四个选项（A 、B 、C 和D ）中选出最佳选项，并在答题卡上将该项涂黑。
ASports can help us a lot. Taking exercises can make us strong. In collective(集体的) sports likebasketball, volleyball or football, we will learn the importance of cooperation(合作). And sports can alsohelp us relax after work or study.However, as the saying goes, “there are two sides of everything.” Sometimes we may hurt otherplayers or ourselves if we are not careful enough when participating(参加) in sports activities. What’s more, too much or hard practice can be bad for our health.Sports can make us healthy both physically and psychologically(心理地). It is also a good way forpeople to know each other and can improve friendship between people. So long as we are careful enough,sports can do us nothing but good. 21. can make us strong.A. SleepingB. SingingC. Making friendsD. Taking exercise22. Too much exercise can be for us. A. good B. enough C. bad D. helpful 23. Sports can . A. help people to know each other B. improve friendship between people C. do us nothing but good if we are careful D. All of the above 24. Which of the following is NOT true? A. Sports can help us relax after work or study. B. Sports can only make us healthy physically. C. Sometimes we may hurt other players or ourselves when participating in sports activities. D. Basketball and volleyball are both collective sports. B Many United States companies have made the search for legal protection from import competition into a major line of work. Since 1980, the United States International Trade Commission(ITC) has received about 280 complaints alleging damage from imports that benefit from subsidies(补贴)by foreign governments. Another 340 charge that foreign companies “dumped” their products in the United States at “less than fair value”. Even when no unfair practices are claimed, the simple claim that an industry has been injured by imports is sufficient grounds to seek relief(救济). Contrary to the general impression, this request for import relief has hurt more companies than it has helped. As corporations begin to function globally, they develop a complicated web of marketing, production, and research relationships. The complexity of these relationships makes it unlikely that a system of import relief laws will meet the strategic needs of all the units under the same parent company. Internationalization increases the danger that foreign companies will use import relief laws against the very companies the laws were designed to protect. Suppose a United States-owned company establishes an overseas plant to manufacture a product while its competitor makes the same product in the United States. If the competitor can prove injury from the imports ——and that the United States company received a subsidy from a foreign government to build its plant abroad ——the United States company’s products will be uncompetitive in the United States, since they would be subject to duties. Perhaps the most shameful case occurred when the ITC investigated allegations(控诉) that Canadian companies were injuring the United States salt industry by dumping rock salt, used to deice roads. The bizarre aspect of the complaint was that a foreign conglomerate(联合企业) with United States operations was crying for help against a United States company with foreign operations. The “United States” company claiming injury was a unit of a Dutch conglomerate, while the “Canadian” companies included a unit of a此卷只装订不密封班级姓名准考证号考场号座位号Chicago firm that was the second-largest domestic producer of rock salt.25. The passage is chiefly concerned with .A. arguing against the increased internationalization of US corporationsB. recommending a uniform method for handling claims of unfair trade practicesC. warning that the application of laws affecting trade frequently has unintended consequencesD. advocating the use of trade restrictions for “dumped” products but not for other imports26. What can be inferred about the minimal basis for a complaint to the ITC?A. A foreign competitor is selling products in the US at less than fair market value.B. A foreign competitor has greatly increased the volume of products shipped to the US.C. The company requesting import relief has been banned from exporting products.D. The company requesting import relief has been injured by the sale of imports in the US.27. Which of the following is most likely to be true of US trade laws?A. They will eliminate the practice of “dumping” products in the US.B. Those applied to international companies will help to gain more profits.C. They will affect US trade with Canada more negatively than trade with other nations.D. Those helping one unit within a parent company won’t necessarily help other units.CPalaces are known for their beauty and splendor, but they offer little protection against attacks. It is easy to defend a large building, but usually these buildings are not designed with the comfort of a king in mind. When it comes to structures that are both beautiful and defensive, the European castle is a big success.Castles were originally built in England by the Normans in 1066. They built towers and walls to secure the land they had taken. These castles provided the Normans with a quiet and safe place. They also served as bases of operation for attacks. In this way castles served both defensive and offensive roles. Besides, castles served as offices for governors. Those that were socially beneath the governor would come to report affairs and express their respect. They would address problems, handle business, feast, and enjoy festivities in castles. So castles served as social centers as well.The first castles were made from earth and wood, and they were likely to suffer from attacks by fire. Then wooden castles were gradually replaced by stone, which greatly increased the strength of these towers and walls. However, attackers could throw flaming objects into castles through the windows or burn the wooden doors. This led to moving the windows and entrances off of the ground floor and up to the first floor to make them more difficult to access.During the Middle Ages, attacks increased in regularity, so castle defenses were updated. Arrow-slits were added. They were small holes in the castle, which allowed defenders to fire without being hurt. Towers were built from which defenders could provide fire on both sides. The towers were connected to the castle by wooden bridges, so that if one tower fell, the rest of the castle was still easy to defend. A lot of rings of castle walls were constructed, so that even if attackers went past one wall, they would be caught on a killing ground between inner and outer walls. All of these increased the defense of castles.The end of castles can be attributed to gunpowder. During the 15th century, artillery, a kind of large guns, became powerful enough to break through stone walls. This greatly made the role of castles less effective. Though castles no longer serve their original purposes, remaining castles receive millions of visitors each year who wish to experience the situations of ancient times.28. What was the original function of castles according to the passage?A. They served as tourist attractions.B. They were important social centers.C. They marked religious ceremonies.D. They were built for use in emergencies.29. The reason why wooden castles were replaced by stone castles was that .A. stone castles cost less moneyB. stone castles offered better defenseC. wooden castles were uncomfortableD. wooden castles took a long time to build30. Which of the following showed an improvement in castle defenses?A. Castles were totally separated by stones.B. Arrow-slits were made in large quantities.C. Rings of walls were built to defend the towers.D. Windows and entrances were moved to the higher floor.31. What is the best title of the passage?A. Fancy Living: Learning about CastlesB. Normans: Bringing Castles to EnglandC. A History of Castles: The Rise and Fall of CastlesD. Defending Castles: Technologies Used to Defend CastlesDA schoolgirl saved her father’s life by kicking him in the chest after he suffered a serious allergicreaction which stopped his heart.Izzy, nine, restarted father Colm’s heart by stamping(踩) on his chest after he fell down at home and stopped breathing.Izzy’s mother, Debbie, immediately called 999, but Izzy knew doctors would never arrive in time to save her father, so decided to use CPR.However, she quickly discovered her arms weren’t strong enough, so she stamped on her father’s chest instead.Debbie then took over with some more conventional chest compressions until the ambulance arrived.Izzy, who has been given a bravery award by her school, said: “I just kicked him really hard. My mum taught me CPR but I knew I wasn’t strong enough to use hands. I was quite scared. The doctor said I might as well be a doctor or a nurse. My mum said that Dad was going to hospital with a big footprint on his chest.”“She’s a little star,” said Debbie. “I was really upset, but Izzy just took over. I just can’t believe what she did. I really think all children should be taught first aid. Izzy did CPR, then the doctor turned up. Colm had to have more treatment on the way to the hospital and we’ve got to see an expert.”Truck driver Colm, 35, suffered a mystery allergic reaction on Saturday and was taken to hospital, but was sent home only for it to happen again the next day. The second attack was so serious that his airway swelled, preventing him from breathing, his blood pressure dropped suddenly, and his heart stopped for a moment.He has now made a full recovery from his suffering.32. Izzy kicked her father in the chest _________.A. to express her helplessnessB. to practise CPR on himC. to restart his heartD. to keep him awake33. What’s the right order of the events?①Izzy kicked Colm.②Debbie called 999.③Izzy learned CPR.④Colm’s heart stopped.A. ③①②④B. ④②③①C. ④③①②D. ③④②①34. What does Paragraph 8 mainly talk about?A. What caused Colm’s allergy.B. Colm’s present condition.C. What Colm suffered.D. Symptoms of Colm’s allergic reaction.35. Why does the author write the news?A. To report a 9-year-old girl’s brave act.B. To prove the importance of CPR.C. To describe a serious accident.D. To call people’s attention to allergic reaction.第二节（共5小题；每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项，选项中有两项为多余选项。
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英语本试卷分第 I 卷（选择题）和第 II 卷（非选择题），第 I 卷 1 至 10 页，第 II 卷 11 至 12 页，共 12 页。
满分150 分，考试时间 120 分钟。
考生作答时，须将答案答在答题卡上，在本试题卷、草稿纸上答题无效。
考试结束后，将本试题卷和答题卡一并交回。
注意事项：19.必须使用 3B 铅笔在答题卡上将所选答案对应标号涂黑。
3.第 I 卷共两部分，共计 100 分。
第Ⅰ卷（选择题，共 100 分）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。
每段对话后有一道小题，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.W hat will the woman do tonight?A. Be a volunteerB. Be a guest.C. Buy a ticket.2.W hat did the speakers just buy?A. A microwave.B. Stay at home for a while.C. A fridge3.W hat is the man thinking about?A. Food for breakfast.B. The life of a farmer.C. The noise from his neighbors.4.H ow much are the chicken sandwiches?A. $6.B. $12.C. $18.5.W hat will the man probably do?A. Quite the band today.B. Join the chess club later.C. Forget about the chess club.第二节 (共 15 小题；每小题 1.5 分，满分 33.5 分)听下面 5 段对话或独白。
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，并标在试卷的相应位置。
听完每段对话或独白前，你将有时间阅读各个小题，每小题 5 秒钟；听完后，各小题将给出 5 秒钟的做答时间。
每段对话或独白读两遍。
听第 6 段材料，回答第 6、7 题。
6.Why does the woman feel tired?A.She recently gave birth.B.She has to work all the time.C.Her baby is awake all night.7.What did the doctor tell the woman?A. Her baby was healthy.B. She should hire a babysitter.C. It’s normal for babies to cry.听第 7 段材料，回答第 8、9 题。
8. Where does the man want to go?A. The 23rd floor.B. The 15th floor.C. The 5th floor.9. Who does the woman blame for their situation?A. The man.B.The HELP button.C. The woman with the groceries.听第 8 段材料，回答第 10 至 12 题。
What does the woman think she knows the man?She has seen him somewhere in her opinion.She’s seen a movie with him.She used to exercise together with him.How does the man most likely feel?A. Grateful.B. Confused.C. Angry.12. What is the last place that the woman mentions?A. A movie theather.B. A fitness center.C. A coffee shop.听第 9 段材料，回答第 13 至 16 题。
13. Where will Janet Jackson’s tour begin?A. In New York.B. In Los Angeles.C. In Vancouver.What happened last Monday?Jackson’s new song came out.Jackson attended an award show.Jackson’s new jewelry line came out.What can we learn about Jackson’s necklace?It is made of white gold.It is shaped like a star.It is her favourite piece of jewelry.Who might the man be?A. The woman’s husband.B. A host.C. A jewelry designer.听第 10 段材料，回答第 17 至 20 题。
17. What does the speaker do?A. He is a student.B. He’s a coach.C. He’s a high school principal.18. What is happening this weekend?A. A graduation ceremony.B. A school get-together.C. A football game.What is the speaker’s advice?Train as much as you can.Take advantage of all opportunities to learn.C. Don’t take college too seriously.20. What’s the big educational process?A. Learning in life.B. Reading textbooks.C. Listening to lectures.第二部分阅读理解（共两节，满分 40 分）第一节（共 15 小题：每小题 2 分，满分 30 分）阅读下列短文，从每题所给的四个选项（A、B、C 和 D）中，选出最佳选项，并在答题卡上将该项涂黑。
AI have been watching my daughters play sports since they were four, and have been shocked at how some parents in the crowd behave.At soccer games, when my girls were very young, parents would be screaming at their children to get their heads into the game, to charge, to “get” an opposing player, or to “kill it.” I understand parents want to encourage their children and their team—but at some point it becomes too much.My daughter Grace is a high school and AAU (Association of American Universities) basketball player and a very good one at that. During a recent game, a parent sitting near me was shouting loudly to her daughter, standing up, expressing displeasure about the way her team was playing and, during halftime, speaking loudly about her personal life. After having her in my ear for more than half of the game, I was so put off that I had to move.Admittedly, I am the quiet parent at games. I clap for the schoolgirls when they make wonderful plays. For the most part, however, I remain silent. Shouting out makes them nervous.I asked Grace to weigh in on this, and she tells me: “Some parents become extremely emotionally invested in sports games, which can become a problem for the players. The worst thing is when people point out that ‘crazy parent,’ and they don’t know that’s your parent.”I am all for parent spectators cheering on their children and the rest of the team. But I hope you can chee r your children on in a positive way. Shouting things like, “Where’s your head?” “Go after her!” is not, in my view, the way to cheer on high school students or younger children.If you have nothing positive to say, hold your tongue. If you find yourself struggling to keep it together, get some fresh air during halftime.21.Unable to stand the mother nearby, the author.A. told her to shut upB. found another seatC. asked Grace for helpD. refused to talk to her22.In Grace’s opinion, c razy parents in the audience will make players feel.A. helplessB. impatientC. scaredD. ashamed23.The author is against parent spectators.A. keeping silent all the timeB. saying unpleasant wordsC. cheering on the playersD. leaving during halftime24.Who are the intended readers of the text?A. Parents.B. Teenage players.C. Soccer spectators.D. Coaches.BRiding School:You can start horse-riding at any age. Choose private or group lessons any weekday. There are 10 kilometers of tracks and paths for leisurely rides across farmland and open country. You will need a riding hat.Opening Hours: Monday through Friday: 9:00 a.m.-8:30 p.m. (3:30 p.m. on Saturdays)Phone:(412)396-6754 Fax:(412)396-6752Sailing Club:Our Young Sailor’s Course leads to the Stage 1 Sailing qualification. You will learn how to sail safely and the course also covers sailing theory and first aid. Have fun with other course members, afterwards in the clubroom. There are 10 weekly two-hour lessons.Opening Hours: Tuesdays: 6:00 p.m.-8:00 p.m.Phone:(412)396-6644Fax:(412)396-6644Diving Centre:Our experienced instructors offer one-month courses in deep-sea diving for beginners. There are two evening lessons a week, in which you learn to breathe underwater and use the equipment safely. You only need swimming costume and a towel. Reduced rates for couples.Opening Hours: Monday and Friday: 6:30 p.m.-8:30 p.m.Phone:(412)396-6312Fax:(412)396-6706Medical Center:The staff of the Medical Center aim to provide convenient and comprehensive medical care to students and staff of the university. The center is well equipped and the staff here are trained to deal with a broad range of medical problems. Both female and male doctors as well as nursing staff are available for consultation. Also, all kinds of medicines are sold here and are cheaper for students than other drugstores.Opening Hours:24 hours from Monday to SundayPhone:(412)396-6649Fax:(412)396-6648Watersports Club:We use a two-kilometer length of river for speedboat racing, and water-skiing. A beginners’ course consists of ten 20-minute lessons. You will learn to handle boats safely and confidently, but must be able to swim. The club is in a convenient central position and is open daily from 9 a.m. to 4 p.m., with lessons all through the day.Opening Hours: Monday through Friday: 9:00 a.m.-4:00 p.m.Phone:(412)396-6899Fax:(412)396-6890If you want to swim and enjoy activities which are fast and a bit dangerous, you shouldjoin________.A. Diving CentreB. Riding SchoolC. Watersports ClubD. Sailing Club26. If you want to experience a new activity in the countryside in the mornings, you may fax____.A. (412)396-6752B. (412)396-6648C. (412)396-6890D. (412)396-6706If you are planning to explore the ocean depths, you should attend your lessons at________.Monday through Friday: 9:00 a.m.-8:30 p.m.24 hours from Monday to SundayC.Tuesdays:6:00 p.m.-8:00 p.m.D. Monday and Friday: 6:30 p.m.-8:30 p.m.28. Which is NOT the convenience that the Medical Center provides?A. Well trained staff members.B. Nursery for newly-born babiesC. Good equipment.D. Various less expensive medicines.CTo help make up for sleep lost during marathon night flights, migrating( 迁徙 ) birds take hundreds of power nap(小睡) during the day, each lasting a few seconds.Every autumn, Swainson’s thrushes(画眉) fly up to 3,000 miles from their breeding grounds in northern Canada and Alaska to winter in Central and South America. Come spring, the birds make the long trip back.The birds fly mostly at night and often for long hours at a time, leaving little time for sleep. So to get through these tiresome periods, thrushes, during autumn and spring when the birds are normally migrating, change their sleep patterns, staying awake at night and resting during day. But instead of sleeping for a long period at a time, the birds took several naps a day, each one lasting only 9 seconds on average.The thrushes also mixed up their shut-eye sessions with two other forms of sleep. In one, called unilateral eye closure, or UEC, the birds rested one eye and one half of their brains while their other eye and half brain remained open and active, protecting them from danger.The birds also occasionally slipped into another state, one that any college student who has ever been stuck in a boring lecture can relate to. Called drownsiness( 睡意), this state is characterized by a partial shutting of both eyes that still allows for some visual processing.“In terms of quality, drowsiness may be less useful than normal sleep, but it may also be safer.” said Fuchs.What’s interesting is that e ven animals should make up for sleep loss. That a need for sleep cannot be lost even in these birds means the importance of sleep for many, if not all, animals, not to mention human.What would be the best title for the text?.Migrating Birds Travel for Long Hours.Migrating Need HelpC. How Birds Migrate during AutumnD. Migrating Birds Take Many Daily Power NapsAccording to the text, when having drowsiness, the thrushes______.will rest one eye and the other activewill often sleep during daywill still be careful with the possible dangerswill not see anything in front of themWhat can we learn from the text?Sleep is important for human being.Animals need sleep to rest.Migrating birds need sleep only during day.College students often have drowsiness in fall.DThere is a tread in travelling that is sweeping the nation—it’s being called voluntourism. Foreign destinations are attracting American citizens who want to sightsee, while at the same time engage in community service.Photos on a wall are of recent visits to an orphanage(孤儿院) in India. It’s a trip Jim Inners first made after looking a more meaningful way to see the world. Through the Austin-based Miracle Foundation, Inners has traveled to India three times as part of a growing trend called voluntourism, which combines foreign travel with charity work.“You’re living. You’re not just sending a check to help somebody but you have a little girl, a little boy. You’re their life line to a new life.” said InnersSilicon Labs co-founder Nav Sooch recently raised money to build a village for orphans in India and has traveled there several times. He says it’s possible to sightsee and serve the needy.“I think it would be a shame to go to India and not see the Taj Mahal. It’s a won derful place and I think you can have that as well as the real interaction with real people.” said Sooch.“We’ve had all ages. We’ve had from 21/2 with a family all the way up to an 82-year-old and that’s the beauty of these trips.” says Barbara Joubert. She plans the 9-day journeys on behalf of the Miracle Foundation. She says it appeals to those who want to give back. “So the most basic work or what you’re doing there is loving the kids and it’s not hard.” said Joubert.“Spending time with all these kids has been wonderful.” says Inners. Voluntourism will take Jim Inners back to “Pinky”, the girl who calls him Papa Jim.中According to the text, voluntourism_________.is kind of travel organized by travel companiesis a way of earning money by travelingis a certain travel while offering community serviceis kind of travel to enjoy the local cultures中According to the text, the Miracle Fountain_________.A. is intended to help the poor areasB. mainly helps plan voluntourismC. belongs to a large travel companyD. is organized to offer service in IndiaWhat does Inners mean by saying “You’re their life line to a new life.” in paragraph 3?.You’ll have to look after their future life..You need to send money to help the poor..You will allow them to be your family members.D. You have a great effect on their future life.What’s the purpose of the text?A. To make voluntourism more popular.B. To persuade readers to help the poor.C. To give you advice on voluntourism.D. To introduce readers voluntourism.第二节（共 5 小题：每小题 2 分，满分 10 分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。