如皋市2018~2019学年度高三年级第二学期语数英学科模拟(三)(3.5模)高三英语(解析版)
江苏省如皋中学2019届高三上学期第二次月考英语---精校Word版含答案
江苏省如皋中学2018-2019学年度第一学期高三第二次月考试卷高三英语第I卷(选择题共85分)第一部分:听力(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。
第二部分:英语知识运用(共两节,满分35分)第一节:单项填空(共15小题;每小题1分,满分15分)21. His dream of retiring in England ended finally because he hadn’t _________ his poor ability to adapt to the changeable cold weather there.A. stood forB. settled forC. allowed forD. answered for22. The medicine takes effect in the __________ stage of treatment, but the effect declines when it is taken for several months.A. potentialB. terminalC. normalD. initial23. --- Where shall we get together?--- It is at the entrance to the cinema _______ we met last time.A. thatB. thereC. whichD. where24. While digital technology represents a ______ for bridging geographic distance, highly skilled workers are increasingly crowding into cities.A. canalB. channelC. courseD. communication25. Romney isn’t a liar, but he is just to some degree, so it’s vital for us to have trust in him.A. scratching the surfaceB. having butterflies in his stomachC. economical with the truthD. landing on his feet26. _____ having below average intelligence, Forrest earned a scholarship to the University of Alabama.A. DespiteB. On top ofC. BeyondD. By virtue of27. When _______ to danger, men tend to increase blood pressure, _______ nervous or anxious.A. exposed; feelingB. exposed; feltC. exposing; feltD. exposing; feeling28. Music should be taught routinely in schools because of the benefits ______ can have on the development of the brains of young children.A. itB. thatC. whichD. one29. Part of the highway would still be temporarily closed _________the heavy fogcontinue.A. mightB. couldC. wouldD. should30. --How could she have let something so important __________her mind?---She’s fully applied t o work recently and gets burnt out.A. occupyB. crossC. slideD. slip31. In Australia, the weather conditions are perfect for sport, _____, as manyAustralians agree, they owe their love of sport.A. whichB. for whichC. to whichD. where32. According to the report released by the Climate Council in November 2018, periodsof Australia’s warm winter ____ longer, occurring more frequently and becoming more severe.A. have lastedB. are lastingC. lastedD. last33. --- Calgary is the third largest city in Canada.--- But in the 1870s, it was _____ an army camp protecting hunters and merchants.A. not less thanB. little more thanC. really other thanD. quite rather than34. --- He should have been warned not to eat the fish that might give him a stomach upset.--- , but he just couldn’t resist the saltwater fish.A. So he hadB. So had heC. So was heD. So he was35. I have been on many organized trips to Zhejiang, _____.A. the most impressive one being the one to XianjuB. where the most impressive one was one to XianjuC. of them the most impressive one was the one to XianjuD. of which the most impressive one was one to Xianju第二节:完形填空(共20小题;每小题1分,满分20分)Leafing through your family's antique media makes clicking through social media a feast of empty calories __36__ we should throw our computers and phones away, then open every box in every attic and read whatever __37__,such as what I recently found—diaries written by my grandmother when she was 16.I __38__ the diaries would be dark and old-fashioned, but my teenage grandmother had much fun and her genius was so well presented in labeling boys she cared for that I can __39__ keep up with her crushes, wondering who the mysterious “Sunshine”was, the sweetest young man in my grandmother's eyes.Arguments with adults are only referred to but never described in __40__. She doesn't resist her mother's strict __41__,even when she gets a “lovely __42__”for finishing someone else's ice cream. __43__,I recorded every __44__ I suffered in my teenage diary. This, however, further __45__ bitterness. I think my teenage grandmother's superior __46__ was due to her being 16 before the invention of ‘cool’ as a symbol of __47__,or even, for that matter, “teenager” as an identity.I have not __48__ reading the diaries and I do not want to. But my favorite passage so far was the one __49__ on a Monday evening in late summer in 1911. She was sitting on the porch with friends when a neighbor started playing an __50__ tune. The girls “flew” across the street to listen, and when the neighbor started up with “Put Your Arms __51__ Mc, Honey”,something __52__ happened:“We couldn't help dancing __53__ on t he street and felt so sweet and nice.” And then, just when my teenage grandmother thought things couldn't get any __54__,Harvey walked by, like aray of __55__.36. A. because37. A. falls out38. A. concluded39. A. closely40. A. detail41. A. treatment42. A. credit43. A. In addition44. A. injustice45. A. responds to46. A. habit47. A. slogan48. A. finished49. A. mentioned50. A. irresistible51. A. Over52. A. abnormal53. A. right54. A. easier55. A. sunshine B. soB. pulls outB. assumedB. preciselyB. vainB. controlB. screamB. In generalB. defeatB. tends toB. personalityB. fightB. regrettedB. recordedB. antiqueB. BehindB. imaginaryB. straightB. quickerB. heatC. yetC. holds outC. doubtedC. barelyC. defenceC. planningC. treatC. In contrastC. diseaseC. leads toC. effortC. principleC. skippedC. rememberedC. originalC. AroundC. mysteriousC. fastC. crazierC. hopeD. whileD. drops outD. betD. readilyD. effectD. disciplineD. scoldingD. In factD. lossD. corresponds toD. intelligenceD. virtueD. opposedD. celebratedD. oddD. OnD. magicalD. hardD. sweeterD. comfort第三部分:阅读理解(共15小题;每小题2分,满分30分)(A)An exciting opportunity for an enthusiastic and eager individual with excellent people skills, to join our client’s active membership team has become available. They’re one of Europe’s largest wildlife conservation organizations, with over 1.1 million members.If you think that you have the skills and personality to inspire members of the public to support their valuable work through becoming a member and making donations, then this is the job for you.You will be part of a successful and energetic team that’s central to delivering the organizations work that focuses on “Giving Nature a Home”. They achieve this by attending a wide variety ofevents across the country, where they get to engage with members of the public, inform them of their work and encourage to support our client(当事人) with essential financial support.Help to give Nature a Home and make a real difference to the world around us.IMPORTANT: PLEASE READ BEFORE CLICKING THE APPL Y BUTTON FOR MORE INFORMA TION:PLEASE NOTE: You are not applying at this stage, even though you may be asked to attach your CV on the job board.You will then receive an email from CHM Recruit with further details on how to complete your application.Please check your email spam / junk mail folder a nd don’t delay in applying.56. What kind of quality is the most important for the job?A. Being good at the art of gardeningB. Being interested in looking after animalsC. Being best at communicating with othersD. Being generous enough to donate57. Who will most likely be praised for their work once hired?A. Those saving as many animals in natureB. Those attracting most conversation volunteersC. Those daring to speak in public to ask for moneyD. Those raising most financial support for wildlife(B)Subject Art & Design, Craft & Creative, BeautyDelivery methodOnlineStudy level Professional development, Short, AccreditedRef FACE-GUARDPrice £30, was £299, use code: GUARD90Face Painting Academy DiplomaStart a career in Face Painting or simply learn for fun.Do you have a love for entertaining people?Are you artistic and want to impress people with a new skill?Have you ever thought about doing a course in face painting so you can earn fantastic money?If so then with this course you could become a qualified face painter just like hundreds of other people who have taken our courses. For a one-off fee(一次性付款)you can study online and complete the diploma in about 28 hours.The comprehensive syllabus(教学大纲) is supported by 16 instructional videos so you can learn all the designs with ease, and you will learn a wide range of designs including dog, rabbit and spider man. With 14 modules to cover, you can become an accomplished face painter.Your qualification will be recognized and can be checked for validity by all of your future clients too! Take a step in the right direction and get your Face Painting Academy Diploma today.£30, was £299, use code: GUARD90Module 1 Your Introduction to Becoming a Face PainterModule 2 The Equipment and Materials You Will Need for Face PaintingModule 3 Health & Safety and Risk AssessmentsModule 4 Starting / Running Your Own BusinessModule 5 Pricing and CostsModule 6 Marketing Your Business & Social MediaModule 7 The Do’s and Don’ts and What to Do If Your Business Doesn’t Go WellModule 8 How to do a Dog / Cat Face Paint DesignModule 9 How to do a Butterfly / Dolphin Face Paint DesignModule 10 How to do a Monkey / Frog Face Paint DesignModule 11 How to do a Rabbit / Swan Face Paint DesignModule 12 How to do a Tiger / Dinosaur Face Paint DesignModule 13 How to do a Spiderman / Batman Face Paint DesignModule 14 How to do a Minnie Mouse / Princess Face Paint Design58. The course is intended mainly for those _____.A. keen on showing off new skillsB. eager to get an academy diplomaC. interested in learning face paintingD. equipped with a unique taste for art59. Which of the following statements is TRUE about the course?A. It is presented both online and offline.B. It provides not only lessons on business.C. The diploma can be obtained in one day.D. Some clients will be invited to examine your qualification.60. In which module are you likely to learn how to advertise your business?A. Module 4.B. Module 6.C. Module 7. . Module 10.(C)Although it might have happened anywhere, my encounter with the green banana started on a steep mountain road in the interior of Brazil. My ancient jeep was straining up through spectacular countryside when the radiator(散热器) began to leak ten miles from the nearest mechanic. The over-heated engine forced me to stop at the next village, which consisted of a small store and scattering of houses. People gathered to look. Three fine streams of hot water spouted from holes in the jacket of the radiator. “That's easy to fix,” a man said. He sent a boy running for some green bananas. He patted me on the shoulder, assuring me everything would work out. “Green bananas,” he smiled. Everyone agreed.We exchanged pleasantries while I thought over the effects of the green banana. Asking questions would betray my ignorance, so I remarked on the beauty of the place. Huge rock formations, like Sugar Loaf in Rio, rose up all around us.“Do you see that tall one right over there?”asked my benefactor, pointing to a particular tall, slender pinnacle of dark rock. “That rock marks the center of the world.”I looked to see if he was teasing me, but his face was serious. He in turn inspected me carefully to be sure I grasped the significance of his statement. The occasion demanded some show of recognition on my part. “The center of the world?” I repeated trying to convey interest if not complete acceptance. He nodded.“The absolute center. Everyone around here knows it.”At that moment the boy returned with my green bananas. The man sliced one in half and pressed the cut end against the radiator jacket. The banana melted into a glue against the hot metal, plugging the leaks instantly. Everyone laughed at my astonishment. They refilled my radiator and gave me extra bananas to take along. An hour later, after one more application of green banana, my radiator and I reached our destination. The local mechanic smiled, “Who taught you about the green banana?”I named the village.“Did they show you the rock marking the center of the world?” he asked. I assured him they had.“My grandfather came from there,” he said.“The exact center. Everyone around here has always known about it.”__①__ As a product of American higher education, I had never paid the slightest attention to the green banana, except to regard it as a fruit whose time had not yet come. __②__ But as I reflected on it further, I realized that the green banana had been there all along. __③__ Itstime reached back to the very origins of the banana. __④__ The people in that village had known about it for years. My own time had come in relation to it. This chance encounter showed me the special genius of those people, and the special potential of the green banana. I had been wondering for some time about those episodes of clarity which educators like to call “learning moments ” and knew I had just experienced two of them at once.The importance of the rock marking the center of the world took a while to filter through. I had initially doubted their claim, knowing for a fact that the center was located somewhere in New England. After all, my grandfather had come from there. But gradually I realized they had a valid belief, a universal concept, and I agreed with them. We tend to define the center as that special place where we are known, where we know others, where things mean much to us, and where we ourselves have both identity and meaning; family, school, town, and local region.The lesson which gradually filtered through was the simple concept that every place has special meanings for the people in it; every place represents the center of the world. The number of such centers is incalculable, and no one student or traveler can experience all of them, but once a conscious breakthrough to a second center is made, a life-long perspective and collection can begin.61. What is the best title for the passage?A. A Car AccidentB. An Identity IssueC. The Unforgettable MomentD. The Green Banana62. What can we infer from Paragraph 3?A. The author was open-minded enough to respect their wisdom and beliefs.B. The author was polite trying not to show disagreement with the helper.C. It occurred to the author that the center of the world would be the tall slender rock.D. The author came to realize that every place has special meanings for the people in it.63. Where could the following “Suddenly on that mountain road, its time and my need hadmet.” be best added in Paragraph 5?A. ①B. ②C. ③D. ④64. What is the author's purpose of writing the passage?A. To inspire people to rethink and redefine the center of the world in their eyes.B. To illustrate that ignorance can sometimes be a blessing in disguise.C. To encourage people to discover something with special value and meaning.D. To point out that traveling is a good way for people to search for their identity.DCraig Smallwood, a disabled American war veteran, spent more than 20,000 hours over five years playing an online role-playing game called "Lineage II’. When NCsoft, the South Korean firm behind the game, accused him of breaking the game’s rules and banned him, he was plunged into depression. After he spent three weeks in hospital, he accused NCsoft of fraud and negligence (疏忽),demanding over $9,000,000 in damages and claiming that the company acted negligently by failing to warn him of the danger that he would become “addicted” to the game.But does it make sense to talk of addiction to online activity? Mental-health specialists say someonline behaviors can become problematic for many people, such as video games and messaging via e-mail and social networks. But there is far less agreement about whether any of this should be called “Internet addiction”一or how to treat it.Skeptics say there is nothing uniquely addictive about the Internet. Back in 2000 Joseph Walther, a communications professor at Michigan State University, co-wrote an article in which he suggested,professors were “addicted” to academia. He argued that other factors, such as depression, are the real problem. He stands by that view today. No scientific evidence has emerged to suggest that Internet use is a cause rather than a consequence of some other sort of issue,” he says. “Focusing on and treating people for Internet addiction, rather than looking for underlying (潜在的)clinical issues, is unwise.”Others disagree. “That would be wrong,” says Kimberly Young,a researcher and therapist who has worked on Internet addiction since 1994. She insists that the Internet, with its powerfully immersive environments, creates new problems that people must learn to tackle.No one disputes that online habits can turn toxic (有毒的).Take South Korea, where widespread broadband means that the average high-school student can play video games for 23 hours each week. In 2007 the government estimated that around 210,000 children needed treatment for Internet addiction. And several South Korean men have died from exhaustion after marathon, multi-day gaming sessions. The South Korea government has recently asked game developers to adopt a gaming curfew (宵禁) for children, to prevent them playing between midnight and 8 am. It has also opened more than 100 clinics for I nternet addiction and sponsored an “Internet rescue camp” for serious cases.Treatment centres have popped up around the world. In 2006 Amsterdam’s Smith & Jones facility claimed to be “ the first and, currently,the only residential video-game treatment program in the world’. In America the reSTART Internet Addiction Recovery Program claims to treat Internet addiction and gaming addiction. In China, meanwhile, military-style “boot camps” are the preferred way to treat internet problems.But compulsive behavior is not limited to gamers. When something can be summoned in an instant via broadband, whether it is a game world or an e-mail inbox, it is harder to resist. Getting through a business lunch in which no one pulls out a phone to check their messages now counts as a minor miracle in many quarters. When online auction sites first became popular, talk of “eBay addiction” soon followed. Dr Young says women complain to her now about addiction to Facebook---or even to “FarmVille”,a game playable only within Facebook.Yet many people like feeling permanently connected. As Arikia Millikan, an American blogger, once put it, “If I could be jacked in at every waking hour of the day, I would, and I think a lot of my peers would do the same.” Bob LaRose, an Internet sp ecialist at Michigan State University,doesn’t believe her. In his research on college students, he found that most sense when they are Agoing overboard and restore self-control”. For most people,Internet use “is just a habit—and one that brings us pleasure”,he adds.65.The author cites the example of Craig Smallwood to _________ .A.criticize him for his ridiculous accusationB.warn against the online game companiesC.introduce the topic of “Internet addiction”D.show how serious “Internet addiction” is66.The underlined part in Paragraph 3 indicates that Joseph Walther was _________ .A. sincereB. jokingC. seriousD. criticizing67..According to Joseph Walther, ________ .A.addiction to the Internet must be specially treatedB.Internet addiction is a serious social phenomenonC.the Internet is a cause of many sorts of medical issuesD.clinical issues behind Internet addiction should be found68.Which of the following is true about the world’s efforts to fight Internet addiction?A.The South Korea government aids organizations to treat Internet addiction.B.Amsterdam once started the world first video-game treatment program.C.Treatment centers in America have cured Internet and gaming addiction.itary-style camps in China prove to be the best to treat Internet problems.69.What can we infer from Paragragh 7?A.People communicate more in the virtual world than in real life.B.E-mail or web-use behaviors can also show signs of addiction.C.Websites such as eBay and Facebook are likely to cause complaints.D.Women rather than men tend to be easily addicted to the social network.70.What is the author’s tone in writing this article?A. Objective.B. Subjective.C. Doubtful.D. Approving.第Ⅱ卷(非选择题共35分)第四部分:任务型阅读(共10小题;每小题1分,满分10分)认真阅读下面短文,并根据所读内容在文章后表格中的空格里填入最恰当的单词。
江苏省如皋市2019届高三下学期三模考试语文试题 Word版含解析
如皋市2018~2019学年高三下学期三模语文试题一、语言文字运用1.在下面一段话的空缺处依次填入词语,最恰当的一组是传统手工艺是民众的重要物质与精神诉求,在社会历史长河中具有________的地位。
然而,民众生活方式的变革使传统手工艺正面临着________的生存危机。
由于传统手工艺根植于民众的日常生活,与其他文化要素________,在此背景下,传统手工艺的文化生态保护应运而生。
A. 举重若轻史无前例唇齿相依B. 举重若轻前所未有相得益彰C. 举足轻重史无前例相得益彰D. 举足轻重前所未有唇齿相依【答案】D【解析】【详解】试题分析:本题考查正确使用词语(包括熟语)的能力,此类题要在理解句意的基础上,结合具体语境及词语的意思从三个方面综合考虑分析,即成语的基本义。
感情色彩和语境。
需要注意色彩不明、断词取义,对象误用、谦敬错位,功能混乱,不合语境、望文生义等错误使用类型。
选段摘编自王明月《传统手工艺的文化生态保护与手艺人的身份实践》。
①举重若轻:举重的东西就像举轻东西那样,形容做繁难的事或处理棘手的问题轻松而不费力。
举足轻重:所处地位重要,一举一动都关系到全局。
语境中说的是地位重要,此处应选“举足轻重”;②史无前例:指的是历史上从来没有过的事,即前所未有。
前所未有:是从来没有发生过的。
“前所未有”比“史无前例”语气轻,适用范围较广,根据语境此处应选“前所未有”;③唇齿相依:像嘴唇和牙齿一样互相依存,形容关系非常密切。
相得益彰:互相帮助,互相补充,更能显出各自的好处。
语境中说的是两者关系密切,应选“唇齿相依”。
故选D。
【点睛】正确运用成语,要弄清楚一下几种类型的错误:一、看成语含义与前后文的修饰限制成分是否协调;二、看成语意思与所处的语境是否吻合,是否造成大词小用或小词大用;三、看成语的褒贬感情色彩是否适合所在的语境;四、看成语适用的对象、范围和场合是否造成张冠李戴;五、看成语运用是否因望文生义而误用;六、看成语是否因画蛇添足而造成前后内容重复。
2019届如皋市高三年级第三次数学模拟试卷
2018~2019学年度高三年级第二学期语数英学科模拟(三)数学试题Ⅰ(考试时间:120分钟 总分:160分)一、填空题:本大题共14小题,每小题5分,共计70分.请把答案填写在答题卡相应位置.......上..1. 已知集合 =1,2=2,3A B ,,则=A B ▲ .2. 若复数1i z (i 为虚数单位),则1z z▲ . 3. 某批产品共100件,将它们随机编号为001,002,…,100,计划用系统抽样方法随机抽取20件产品进行检测,若抽取的第一个产品编号为003,则第三件产品的编号为 ▲ .4. 已知双曲线过点23P (,)20y ,则其标准方程为 ▲ .5. 若Z ,条件:1p ,条件:q 函数22()f x x在+ (0,)上是单调递减函数,则条件p 是条件q 成立的 ▲ 条件.6. 口袋内有大小、形状完全相同的红球、白球各两个,现从中随机摸出两个球,则摸出的两球颜色恰好相同的概率为 ▲ . 7. 一个算法的伪代码如右图所示,最后输出的值为 ▲ .8. 若1sin(62,则sin(2+)6▲ .9. 已知边长为2的正方形纸片ABCD ,现将其沿着对角线AC 翻折,使得二面角B AC D 的大小等于45 ,则四面体ABCD 的体积为 ▲ .10.过原点作函数32()22f x x x 图像的切线,设切点为00,P x y (),则200y x ▲ .11. 已知等差数列 n a 的公差0d , 22251011,0a a a ,则15=S ▲ . 12. 如图所示,已知等腰直角三角形ABC 中=2AB AC ,半径为2的圆O 在三角形外与斜边BC 相切,P 为圆上任意一点,且满足AP xAB yAC,则x y 的最大值为 ▲ .(第12题图)C13. 已知函数 12log (1),121,1x x f x x x,若函数 g x f x kx 有且只有三个零点,则实数k 的取值范围为 ▲ .14. 已知单位圆的一内接ABC 的内角,,A B C 所对的边分别为,,a b c ,若sinsin()sin()2B AC A B C A C B (),则abc 的值为 ▲ .二、解答题:本大题共6小题,共计90分.请在答题卡指定区域.......内作答.解答时应写出文字说明、证明过程或演算步骤. 15.(本小题满分14分)已知四棱锥P ABCD 中,底面ABCD 是菱形,3BAD,面PAD 面ABCD .(1)若Q 为AD 中点,证明:BQ 面PAD ;(2)若M 为PA 中点,面BCM PD N ,证明:N 为PD 中点.16.(本小题满分14分)已知函数π()2sin()(02f x x ,的图像的一部分如图所示,5(,0)2C 是图像与x 轴的交点,,A B 分别是图像的最高点与最低点且5AB .(1)求函数()y f x 的解析式;(2)求函数31()()(,0,22g x f x f x x的最大值.(第16题图)(第15题图)17. (本小题满分14分)为了纪念五四青年节,学校决定举办班级黑板报主题设计大赛,高三(1)班李明同学将班级长AB=4米、宽BC =2米的黑板做如图所示的区域划分:取AB 中点F ,连接CF ,以AB 为对称轴,过A 、C 两点作一抛物线弧,在抛物线弧上取一点P ,作PE AB 垂足为E ,作//PG AB 交CF 于点G .在四边形PEFG 内设计主题LOGO ,其余区域用于文字排版.(1)设PE x ,求PG 的长度()f x ;(2)求四边形PEFG 面积的最大值.18.(本小题满分16分)已知椭圆2214x y ,直线1:2l y kx 与椭圆交于,A B 两点,P 为椭圆右顶点.(1)若1k ,求PAB 的面积;(2)设PAB 的外接圆与x 轴另有一个交点0(,0)Q x ,求0x 的取值范围.(第17题图)xGP ECBDFA19.(本小题满分16分)已知2e ()e ln ,()e xxf x xg x x.(1)求函数()y g x 的极小值;(2)求函数()y f x 的单调区间;(3)证明:()()1f x g x .20.(本小题满分16分) 已知如下数阵: 1,2,3 1,2,3,41,2,3,1,2,3,4,5 ……其排列规则为:第一行为 1,2,3;第 n (n 2) 行从左至右依次排列着第 1 行、第 2 行、……第 n 1行的各项且保留它们间的原有顺序不变,最后一项为n 2 .(1)设第n 行的项数为n a ,求5a ;(2)设该数阵第n 行所有项之和为n S ,求n S ; (3)设数阵的第i 行、第j 项为i j b ,,求2019i b ,.2018~2019学年度高三年级第二学期语数英学科模拟(三)数学Ⅱ(附加题)21. 已知矩阵2111A ,且12AX,求X .22.在极坐标系中,已知曲线1:sin 4C与曲线2:=2cos C 相交于,A B 两点,求OA OB 的值(O 为极点).23. 已知抛物线24y x ,过点(1,2)P 作两直线12,l l 与抛物线另交于,A B 两点,设直线12,l l 的斜率分别为12,k k ,且121k k .证明:直线AB 过定点.24.n 个同学12,,,n A A A 在操场上围成一圈进行传球游戏,规则如下:从1A 开始传球,每人只能将球随机传给自己左边或右边相邻的同学,球从一名同学传到另一名同学后记为一次传球,若第(1)k k n 次传球后球又回到1A ,则游戏结束;否则第n 次传球后,游戏也必须结束.(1)若3n ,求3次传球后,球恰好又回到1A 的概率;(2)若7n ,记游戏结束时传球的次数为随机变量 ,求 的分布列与 E .。
江苏省如皋市2020届高三下学期语数英学科模拟(三) 语文试题 Word版含答案 (2)
2019~2020学年度高三年级第二学期语数英学科模拟(三)语文I试题一、语言文字运用(12分)阅读下面的文字,完成1~3题。
在当代中国文化建设不断取得进步的同时,对外文化交流也在不断开拓。
但也出现了将一些“非主流”思潮的文艺作品视作当代中国文化主流,热捧那些格调低下、▲的文艺人士的现象。
( ▲ )要知道,各种文化艺术间的借鉴交流、各种文化艺术的发展有其基本规律。
对此,我们固然要注意正确引导国际舆论,但也要重视各种艺术形式的全面发展,不能▲。
从文化大国到文化强国的升级不可能▲,我们任重道远。
1.依次填入文中横线上的词语,全都恰当的一项是(3分)A.夸大其词顾此失彼一蹴而就B.不攒一词厚此薄彼一马平川C.夸大其词厚此薄彼一马平川D.不攒一词顾此失彼一蹴而就2.下列填入文中括号内的语句,衔接最恰当的一项是(3分)A.国外对当代中国主要文化艺术的认知匮乏、当代中国文化艺术作品与产品欠缺国际竞争力是根本原因。
B.当代中国文化艺术作品与产品欠缺国际竞争力、国外对当代中国文化艺术的认知匮乏是大家的共识。
C.大家认为国外对当代中国文化艺术的认知匮乏,当代中国文化艺术作品与产品欠缺国际竞争力。
D.这主要是因为国外对当代中国文化艺术的认知匮乏,当代中国文化艺术作品与产品欠缺国际竞争力。
3.文中画横线的句子有语病,下列修改最恰当的一项是(3分)A.各种文化艺术的繁荣发展有其基本规律,各种文化艺术间的借鉴交流有其基本原则。
B.各种文化艺术间的借鉴交流有其基本原则,各种文化艺术的繁荣发展有其基本规律。
C.各种文化艺术间的交流借鉴有其基本原则,各种文化艺术的发展繁荣有其基本规律。
D.各种文化艺术的发展繁荣有其基本规律,各种文化艺术间的交流借鉴有其基本原则。
4.下列各句中,与右图漫画的内容最吻合的一项是A.点额不成龙,归来伴凡鱼。
B.海阔凭鱼跃,天高任鸟飞。
C.吾闻池中鱼,不识海水深。
D.游鱼潜渌水,翔鸟薄天飞。
二、文言文阅读(20分)阅读下面的文言文,完成5~8题。
如皋高三数学2.5模定稿
2018~2019学年度高三年级第二学期语数英学科模拟(二)数学试题Ⅰ(考试时间:120分钟 总分:160分)参考公式:锥体的体积公式13V Sh =锥体,其中S 为锥体的底面积,h 为高.一、填空题:本大题共14小题,每小题5分,共计70分.请把答案填写在答题卡相应位置.......上.. 1. 已知集合{}220A x x x =-≤,{}024B =,,,C A B =I ,则集合C 的子集共有 ▲ 个. 2. 已知复数z 满足43i iz+=(i 为虚数单位),则z 的共轭复数z = ▲ .3. 已知双曲线()2210x y m m-=>的一条渐近线方程为x +3y =0,则m = ▲ .4. 随机抽取100名年龄在[10,20),[20,30),…,[50,60)年龄段的市民进行问卷调查,由此得到样本的频率分布直方图如图所示,从不小于40岁的人中按年龄段分层抽样的方法随机抽取8人,则在[50,60)年龄段抽取的人数为 ▲ . 5. 为强化环保意识,环保局每周从当地的5所化工厂(甲,乙,丙,丁,戊)中随机抽取3所进行污水合格检测,则在一周抽检中,甲,乙化工厂都被抽测的概率是 ▲ 6. 如图,若输入的x 值为π3,则相应输出的值y 为 ▲ .7. 已知一个圆锥的底面半径为 3 cm ,侧面积为6π cm 2,则该圆锥的体积是 ▲ cm 3.8. 已知实数x ,y 满足03040x y x y x y ->⎧⎪-⎨⎪+-⎩,≤,≤,则1y z x +=的取值范围是 ▲ .(第4题)m(第6题图)开始 输入xsin cos x x >cos y x←sin y x ←输出y 结束YN9.在△ABC 中,角A ,B ,C 的对边分别为a ,b ,c ,若a =2,b =3,C =2A ,则cos C 的值为 ▲ .10.已知F 1,F 2分别为椭圆E :()222210x y a b a b+=>>的左,右焦点,点A ,B 分别是椭圆E的右顶点和上顶点,若直线AB 上存在点P ,使得PF 1⊥PF 2,则椭圆C 的离心率e 的取值范围是 ▲ .11.已知数列{}n a 的首项118a =,数列{}nb 是等比数列,且52b =,若1n n n a b a +=则10a = ▲ .12.在平面四边形OABC 中,已知OA =u u u r OA ⊥OC ,AB ⊥BC ,∠ACB =60°.若6OB AC ⋅=u u u r u u u r,则OC =u u u r▲ .13.已知正数x ,y 满足121332x y x y +++=,则1x y-的最小值为 ▲ . 14.定义{}min a a b a b b a b ⎧=⎨>⎩,≤,,,. 已知函数()1e xf x m =-,()()()2121g x x mx m m =-+--,若()h x ()(){}min f x g x =,恰有3个零点,则实数m 的取值范围是 ▲ . 二、解答题:本大题共6小题,共计90分.请在答题卡指定区域.......内作答.解答时应写出文字说明、证明过程或演算步骤. 15.(本小题满分14分)如图,已知四棱锥P -ABCD 中,CD ⊥平面P AD ,AP =AD ,AB ∥CD ,CD =2AB ,M 是PD 的中点.(1)求证:AM ∥平面PBC ; (2)求证:平面PBC ⊥平面PCD .APBCDM (第15题图)16.(本小题满分14分)如图,在平面直角坐标系xOy 中,点P ,Q 是以AB 为直径的上半圆弧上两点(点P 在Q 的右侧),点O 为半圆的圆心,已知AB =2,∠BOP =θ,∠POQ =α. (1)若点P 的横坐标为45,点Q 的纵坐标为12,求cos α的值; (2)若PQ =1,求AQ BP ⋅u u u r u u u r的取值范围.17.(本小题满分14分)在平面直角坐标系xOy 中,点A ,F 分别是椭圆C :()222210x y a b a b+=>>左顶点,右焦点,椭圆C 的右准线与x 轴相交于点Q ,已知右焦点F 恰为AQ 的中点,且椭圆C 的焦距为2.(1)求椭圆C 的标准方程;(2)过右焦点F 的直线l 与椭圆C 相交于M ,N .记直线AM ,AN 的斜率分别为k 1,k 2,若k 1+k 2=-1,求直线l 的方程.18.(本小题满分16分)如图,矩形ABCD 是某生态农庄的一块植物栽培基地的平面图,现欲修一条笔直的小路MN (宽度不计)经过该矩形区域,其中MN 都在矩形ABCD 的边界上.已知AB =8,AD =6(单位:百米),小路MN 将矩形ABCD 分成面积分别为S 1,S 2(单位:平方百米)的两部分,其中S 1≤S 2,且点A 在面积为S 1的区域内,记小路MN 的长为l 百米. (1)若l =4,求S 1的最大值; (2)若S 2=2S 1,求l 的取值范围.19.(本小题满分16分)已知函数()e x f x mx =-,x ∈R ,其导函数为()'f x . (1)讨论函数()f x 的单调性;(2)若0x >,关于x 的不等式()21f x x +≥恒成立,求实数m 的取值范围; (3)若函数()f x 有两个零点1x ,2x ,求证:()()12'+'0f x f x >.20.(本小题满分16分)已知数列{}n a 的前n 项和S n 满足2S n -na n =n . (1)求证:数列{}n a 是等差数列; (2)若数列{}n a 的公差0d >,设1n n S b n+=,求证:存在唯一的正整数n ,使得12n n n a b a ++<≤;(3)若a 2=2,设1n n na c a +=,求证:数列{}n c 中的任意一项都可以表示成其他两项的乘积.D ABC(第18题图)。
江苏省如皋市2020届高三下学期语数英学科模拟(三) 语文试题 Word版含答案
2019~2020学年度高三年级第二学期语数英学科模拟(三)语文I试题一、语言文字运用(12分)阅读下面的文字,完成1~3题。
在当代中国文化建设不断取得进步的同时,对外文化交流也在不断开拓。
但也出现了将一些“非主流”思潮的文艺作品视作当代中国文化主流,热捧那些格调低下、▲的文艺人士的现象。
( ▲ )要知道,各种文化艺术间的借鉴交流、各种文化艺术的发展有其基本规律。
对此,我们固然要注意正确引导国际舆论,但也要重视各种艺术形式的全面发展,不能▲。
从文化大国到文化强国的升级不可能▲,我们任重道远。
1.依次填入文中横线上的词语,全都恰当的一项是(3分)A.夸大其词顾此失彼一蹴而就B.不攒一词厚此薄彼一马平川C.夸大其词厚此薄彼一马平川D.不攒一词顾此失彼一蹴而就2.下列填入文中括号内的语句,衔接最恰当的一项是(3分)A.国外对当代中国主要文化艺术的认知匮乏、当代中国文化艺术作品与产品欠缺国际竞争力是根本原因。
B.当代中国文化艺术作品与产品欠缺国际竞争力、国外对当代中国文化艺术的认知匮乏是大家的共识。
C.大家认为国外对当代中国文化艺术的认知匮乏,当代中国文化艺术作品与产品欠缺国际竞争力。
D.这主要是因为国外对当代中国文化艺术的认知匮乏,当代中国文化艺术作品与产品欠缺国际竞争力。
3.文中画横线的句子有语病,下列修改最恰当的一项是(3分)A.各种文化艺术的繁荣发展有其基本规律,各种文化艺术间的借鉴交流有其基本原则。
B.各种文化艺术间的借鉴交流有其基本原则,各种文化艺术的繁荣发展有其基本规律。
C.各种文化艺术间的交流借鉴有其基本原则,各种文化艺术的发展繁荣有其基本规律。
D.各种文化艺术的发展繁荣有其基本规律,各种文化艺术间的交流借鉴有其基本原则。
4.下列各句中,与右图漫画的内容最吻合的一项是A.点额不成龙,归来伴凡鱼。
B.海阔凭鱼跃,天高任鸟飞。
C.吾闻池中鱼,不识海水深。
D.游鱼潜渌水,翔鸟薄天飞。
二、文言文阅读(20分)阅读下面的文言文,完成5~8题。
精品解析:【市级联考】江苏省如皋市2019届高三第二学期语数英学科模拟语文试题(解析版)
江苏省如皋市2018-2019学年度高三年级第二学期语数英学科模拟一语文一、语言文字运用1.在下面一段话的空缺处依次填入词语,最恰当的一组是反对“一把手”领导事必躬亲,并不是说就可以当________,美其名曰“无为而治”。
有的________,上项目怕被人怀疑,搞改革怕惹是非,该管的事情不管,该负的责任不负,当起了________的“公堂木偶”。
不管怎么“伪装”和“铺陈”,都丝毫掩盖不了这种行为敷衍塞责的实质。
当“班长”就要敢担当,怕担当就不要当“班长”。
A. 后台老板滥竽充数明哲保身B. 甩手掌柜滥竽充数独善其身C. 后台老板尸位素餐独善其身D. 甩手掌柜尸位素餐明哲保身【答案】D【解析】【详解】试题分析:题干是“在下面一段话的空缺处依次填入词语,最恰当的一组是”。
本题考查学生结合语境辨析近义词语的能力。
辨析词语运用是否恰当,可从多方面分析。
解答此类题目,必须了解所提供词语的词义。
词性、程度轻重、范围大小、适用对象、感情色彩等,必须抓住同义词或近义词的不同语素的意义进行区别,必须结合具体的语言环境进行辨别筛选。
本题,甩手掌柜:指光指挥别人,自己什么事也不干的人。
也指只挂名,不负责,也不做事的主管人员。
后台老板:指背后操纵、支持的人或集团。
语境说的是反对“一把手”领导事必躬亲,并不是说“一把手”领导自己就可以什么不干。
所以此处选用“甩手掌柜”。
尸位素餐:空占着职位而不做事,白吃饭。
滥竽充数:比喻没有真才实学的人混在内行人之中,以次充好。
语境“上项目怕被人怀疑,搞改革怕惹是非,该管的事情不管,该负的责任不负”说的是“一把手”领导空占着领带的位置,不真正做事。
所以选用“尸位素餐”。
明哲保身:明智的人善于保全自己,现指因怕连累自己而回避斗争的处世态度。
独善其身:原意是做不上官就修养好自身。
现指只顾自己,不管别人。
鉴于上述的分析,这些领导之所以这样做,是怕连累自己的消极工作态度,所以此处选用“明哲保身”。
江苏省盐城市如皋中学2018-2019学年高三英语期末试卷含解析
江苏省盐城市如皋中学2018-2019学年高三英语期末试卷含解析一、选择题1. I have learned a lot about Asian customs, ______ in the small village for three years in the early 1990s.A. livedB. to liveC. having livedD. to h ave lived参考答案:C略2. A recent survey found that 52% of Americans questioned said they supported restricting guns or them illegal.A. makesB. madeC. makingD. to make参考答案:C3. The famous actor keeps fit by________ for half an hour every morning.A.working out B.acting out C.giving out D.bring out参考答案:A4. _____ you keep on trying, I don’t really mind whether you can come out top in your class.A. As soon asB. Evenif C. The moment D. So long as参考答案:D略5. He had a burning ________ to win back the teacher's confidence in him.A.desire B.feeling C.emotion D.impression参考答案:A6. The CNSA released the first picture of the moon captured by Chang’e –l onNov.26,_________the full success of the lunar probe projectA. having markedB.marking C. to mark D. marked参考答案:B略7. Experts at the conference agreed that such a product with environmentally friendly technology is ______.A. worth being promotedB. worthy promotingC. worthy of promotingD. worthy of being promoted参考答案:D略8. His contribution to the country was never officially ____ , which made all of us feel discouraged.A. acknowledgedB. appealedC. recommendedD. evaluated参考答案:A略9. As a teacher you have to ______ your methods to suit the needs of slower children.A. devoteB. offerC. makeD. adjust参考答案:D10. A number of protective measures have been _______ in order to reduce the possible damage caused by the deadly disease.A. in needB. in sightC. inorder D. in place参考答案:D11. _______ in a friendly way, their quarrel came to an end.A. Being settledB. SettledC. SettlingD. Having settled参考答案:B12. Gillard’s ride to top of Australian politicswas rough one.A. 不填;aB. the;aC. a;theD. a;不填参考答案:B略13. ---How could she have let something so important __________her mind?---She’s fully applied to work recently and gets burnt out.A. occupyB. crossC. slideD. slip参考答案:D考查动词辨析。
如皋市2018~2019学年度高三年级第二学期语数英学科模拟(三)(3.5模)高三英语
B. He drove too fast.
C. He crashed on the freeway.
12. What had the speakers been doing?
A. Shopping.
B. Looking at cars.
C. Attending a wedding.
C. controversialD. confidential
15.—What do you think of your preparations for the final exams?
A. is climbingB. is being climbed
C. has been climbingD. has been climbed
8.It’s strongly advised that smokers not be allowed to smoke in any room ________ babies currently occupy.
第一节单项填空(共15小题;每小题1分,满分15分)
请认真阅读下面各题,从题中所给的A、B、C、D四个选项中,选出最佳选项,并在答题卡上将该项涂黑。
1.As is often the case, there are always some obstacles in the way,something ________ before we realize the real goal of education.
10. Why is the man impressed by the woman?
A. She is very athletic.
B. She likes loud music.
2019届江苏省如皋市高三下学期语数英学科模拟二英语试题(PDF版)
was still fully ____48____ from neck down and I couldn’t even lift a finger at that time. I started
my physical ____49____ from home. It took me one year to take my first few ____50____ with the
-2-
A. you cry for the moon
B. pigs fly
C. all good things come to an end
D. you get a new lease on life
【答案】B
第二节 完形填空 (共 20 小题;每小题 1 分,满分 20 分)
请认真阅读下面短文,从短文后各题所给的 A、B、C、D 四个选项中,选出最佳选项,并在答题卡上将该
visitors were amazed by the complex architectural space and abundant building types.
A. that
B. one
C. the one
D. those
【答案】A
25.Maybe some of you are curious about what my life was like on the streets because I’ve never
B. distinction
C. division
D. distribution
【答案】B
22.It is said that the only survivor in the car crash was badly injured. However, somehow the
江苏省如皋中学2018—2019学年度第二学期高三数学三模模拟试卷
江苏省如皋中学2018—2019学年度第二学期高三数学三模模拟试卷第I 卷(必做题 共160分)一、填空题:本大题共14小题,每小题5分,共70分.把答案填在题中横线上。
1.已知集合[)3,9A =,[),B a =+∞.若B A ⊆,则实数a 的取值范围是 .(],3-∞ 2. 设复数z 满足z +i =3-i ,则z =_______3+2i3. 已知袋子中装有大小相同的6个小球,其中有2个红球、4个白球.现从中随机摸出3个小球,则至少有2个白球的概率为_________454. 设S n 是等比数列{a n }的前n 项和,a n >0,若S 6-2S 3=5,则S 9-S 6的最小值为_______205. 在△ABC 中,若(a ﹣b )(sinA +sinB)=(c ﹣b )sinC ,a b 2+c 2的取值范围是_____ (5,6]6. 在长方体ABCD —A 1B 1C 1D 1中,AB =BC =2,AC 1与平面BB 1C 1C 所成的角为30°,则该长方体的体积为______7. 设12014S =++S 的最大整数[]S 是__. 20148. 已知函数33,(){,ax x a x af x ax x a x a-+<=+-≥,若对于任意的实数a ,总存在0x ,使得0()0f x ≠,则0x 的取值范围是____[1,0)(0,1]-⋃9.已知函数3()3||(),[2,2]f x x x a a R x =+-∈∈-,若函数()y f x =的最大值为11,则实数a 的取值为____1,310.函数 函数2()l o g ,x g x =则方程()()f x g x =的实根个数是 . 311.已知圆12,O O 均与x 相切,且两圆心与原点共线,两圆心的横坐标之积为8.则两圆交点到直线2100x y --=距离的最小值为_______12. 在△ABC 中,角A ,B ,C 的对边分别为a ,b ,c ,若a 2cos A =b 3cos B =c 6cos C,则cos cos cos A B C = . 11013. 在平面直角坐标系xOy 中已知圆C 满足:圆心在x 轴上,且与圆22(2)1x y +-=相外切.设圆C 与x 轴的交点为M,N ,若圆心C 在x 轴上运动时,在y 轴正半轴上总存在定点P ,使得MPN ∠为定值,则点P 的纵坐标...为 .3 14.已知正实数,x y 满足3320x y x y +-+=,若存在正实数,x y 使得22112x ay +≥成立,则实数a 的最小值为__1二、解答题:本大题共6小题,共90分.解答应写出文字说明、证明过程或演算步骤.15. (本小题满分14分)如图,在四棱柱中,已知平面平面且,. (1)求证:(2)若为棱的中点,求证:平面.解:⑴在四边形中,因为,,所以,又平面平面,且平面平面,平面,所以平面,又因为平面,所以.⑵在三角形中,因为,且为中点,所以, 又因为在四边形中,,, 所以,,所以,所以,因为平面,平面,所以平面.16. (本小题满分14分)已知向量()1sin 2A =,m与()3sin A A =,n 共线,其中A 是△ABC 的内角.(1)求角A 的大小; (2)若BC =2,求△ABC 面积S 的最大值,并判断S 取得最大值时△ABC 的形状.(1)因为m //n ,所以3sin (sin 3cos )02A A A ⋅+-=. ……………2分所以1cos232022A A --=,12cos212A A -=,即()πs i n 21A -=. …4分因为(0,π)A ∈ , 所以()ππ11π2666A -∈-,. …5分故ππ2A -=,πA =.……7分 (2)由余弦定理,得 224b c bc =+-.…8分 又1sin 2ABC S bc A ∆==,…9分 而222424b c bc bc bc bc +⇒+⇒≥≥≤,(当且仅当b c =时等号成立) ……11分所以1sin 42ABC S bc A ∆==. ……12分当△ABC 的面积取最大值时,b c =.又π3A =,故此时△ABC 为等边三角形. …14分1111D C B A ABCD -⊥C C AA 11,ABCD 3===CA BC AB 1==CD AD ;1AA BD ⊥E BC //AE 11D DCC ABCD BA BC =DA DC =BD AC ⊥11AAC C ⊥ABCD 11AA C CABCD AC =BD ⊂ABCD BD ⊥11AA C C 1AA ⊂11AA C C 1BD AA ⊥ABC AB AC =E BC BC AE ⊥ABCD AB BC CA ==1DA DC ==60ACB ∠=︒30ACD ∠=︒BC DC ⊥AE DC DC ⊂11D DCC AE ⊄11D DCC AE11D DCC 1AE CDBA1D1B1C第15题17. (本小题满分14分)如图,GH 是东西方向的公路北侧的边缘线,某公司准备在GH 上的一点B 的正北方向的A 处建一仓库,设AB = y km ,并在公路同侧建造边长为x km 的正方形无顶中转站CDEF (其中边EF 在GH 上),现从仓库A 向GH 和中转站分别修两条道路AB ,AC ,已知AB = AC + 1,且∠ABC = 60o.(1)求y 关于x 的函数解析式;(2)如果中转站四周围墙造价为1万元/km ,两条道路造价为3万元/km ,问:x 取何值时,该公司建中转站围墙和两条道路总造价M 最低? 【解析】试题分析:(1)利用题意结合余弦定理可得函数的解析式,其定义域是.(2)结合(1)的结论求得利润函数,由均值不等式的结论即可求得当km时,公司建中转站围墙和两条道路最低总造价为490万元.(1)(x > 1);(2)时,该公司建中转站围墙和道路总造价M 最低.18. (本小题满分16)在平面直角坐标系中,椭圆的离心率,且点在椭圆上. (1)求椭圆的方程; (2)若点都在椭圆上,且中点在线段(不包括端点)上. ①求直线的斜率; ②求面积的最大值.xOy )0(12222>>=+b a by a x 22=e )1,2(P C C B A ,C AB M OP AB AOB ∆(2)①法一、设,直线的斜率为则,∴, ∴.............................................6分 又直线在线段上,所以,所以.............................................................8分()()()112200,,,,,A x y B x y M x y AB k 22112222163163x y x y ⎧+=⎪⎪⎨⎪+=⎪⎩22221212063x x y y --+=0022063x y k +=1:,2OP y x M =OP 0012y x =1k =-法三、设,直线的方程为,则,∴,由题意, 所以,..............................................6分∴ 又直线在线段上,所以,在直线上,∴解得:,...............................................8分 ②设直线的方程为,则,∴,()()()112200,,,,,A x y B x y M x y AB y kx m =+22163y kx m x y =+⎧⎪⎨+=⎪⎩()222124260k x kmx m +++-=0∆>122412kmx x k +=-+()02212kmx i k =-+1:,2OP y x M =OP ()0012y x ii =M AB ()00y kx m iii =+()()()i ii iii 1k =-AB (),0,3y x m m =-+∈22163y x mx y =-+⎧⎪⎨+=⎪⎩2234260x mx m -+-=19. (本小题满分16)数列{a n }满足a n =2a n -1+2n +1(n ∈N *,n ≥2),a 3=27.(1)求a 1,a 2的值;(2)是否存在一个实数t ,使得b n =12n (a n +t )(n ∈N *),且数列{b n }为等差数列?若存在,求出实数t ;若不存在,请说明理由;(3)求数列{a n }的前n 项和S n .解 (1)由a 3=27,得27=2a 2+23+1,∴a 2=9, ∵9=2a 1+22+1,∴a 1=2.(2)假设存在实数t ,使得{b n }为等差数列, 则2b n =b n -1+b n +1(n ≥2且n ∈N *), ∴2×12n (a n +t )=12n -1(a n -1+t )+12n +1(a n +1+t ), ∴4a n =4a n -1+a n +1+t ,∴4a n =4×a n -2n -12+2a n +2n +1+1+t ,∴t =1. 即存在实数t =1,使得{b n }为等差数列. (3)由(1),(2)得b 1=32,b 2=52,∴b n =n +12, ∴a n =⎝ ⎛⎭⎪⎫n +12·2n -1=(2n +1)2n -1-1,S n =(3×20-1)+(5×21-1)+(7×22-1)+…+[(2n +1)×2n -1-1] =3+5×2+7×22+…+(2n +1)×2n -1-n ,① ∴2S n =3×2+5×22+7×23+…+(2n +1)×2n -2n ,② 由①-②得-S n =3+2×2+2×22+2×23+…+2×2n -1-(2n +1)×2n+n =1+2×1-2n1-2-(2n +1)×2n +n =(1-2n )×2n +n -1,∴S n =(2n -1)×2n -n +1.20. (本小题满分16)已知函数()e (1)x f x a x =-+,其中e 为自然对数的底数,a ∈R .(1)讨论函数()f x 的单调性,并写出相应的单调区间;(2)已知0a >,b ∈R ,若()≥f x b 对任意x ∈R 都成立,求ab 的最大值; (3)设()(e)g x a x =+,若存在0x ∈R ,使得00()()f x g x =成立,求a 的取值范围. 解:(1)由()e (1)x f x a x =-+,知()e x f x a '=-.若0a ≤,则()0f x '>恒成立,所以()f x 在()-∞+∞,上单调递增; 若0a >,令()0f x '=,得ln x a =,当ln x a <时,()0f x '<,当ln x a >时,()0f x '>, 所以()f x 在(ln )a -∞,上单调递减;在(ln )a +∞,上单调递增. (2)由(1)知,当0a >时,min ()(ln )ln f x f a a a ==-.因为()f x b ≥对任意x ∈R 都成立,所以ln b a a -≤,所以2ln ab a a -≤. 设2()ln t a a a =-,(0a >),由21()(2ln )(2ln 1)t a a a a a a a '=-+⋅=-+,令()0t a '=,得1e a -=,当10e a -<<时,()0t a '>,所以()t a 在()120e-,上单调递增;当12e a ->时,()0t a '<,所以()t a 在()12e -∞,+上单调递减,所以()t a 在12e a -=处取最大值,且最大值为12e.所以21ln 2e ab a a -≤≤,当且仅当12e a -=,121e 2b -=时,ab 取得最大值为12e .(3)设()()()F x f x g x =-,即()e e 2x F x x ax a =---,题设等价于函数()F x 有零点时的a 的取值范围.① 当0a ≥时,由(1)30F a =-≤,1(1)e e 0F a --=++>,所以()F x 有零点. ② 当e 02a -<≤时,若0x ≤,由e 20a +≥,得()e (e 2)0x F x a x a =-+->;若0x >,由(1)知,()(21)0F x a x =-+>,所以()F x 无零点.③ 当e 2a <-时,(0)10F a =->,又存在010e 2a x a -=<+,00()1(e 2)0F x a x a <-+-=,所以()F x 有零点.综上,a 的取值范围是e 2a <-或0a ≥.。
江苏省如皋市2018~2019学年度高三年级第二学期语数英学科模拟(二)语文试题答案
高三语文试卷参考答案及评分建议语文Ⅰ一、语言文字运用(12分)1.(3分)D(“南辕北辙”比喻行动和目的相反,“背道而驰”意思是比喻方向、目标完全相反,结合语境“背道而驰”准确;“如数家珍”好像在数自己家里的珍宝一样,通常用来形容某人对所讲的事情十分熟悉,“耳熟能详”指听得多了,能够说得很清楚、很详细,结合语境“如数家珍”准确;“淡薄”意思是印象因淡忘而模糊,“淡泊”指对于名利冷淡不看重,“淡泊”准确)2.(3分)A(“璧还”敬辞,原璧退还。
用于归还原物或推辞谢绝赠品)3.(3分)A(⑤①因果关系,④③内容紧密,②和后面的内容相承接)4.(3分)D(语段强调的是人的思想对字体风格的影响)二、文言文阅读(20分)5.(3分)D(良:的确)6. (3分)B(“在乡里备受推崇”,不符合原文)7.(10分)(1)(5分)恰巧天下动荡不安,读书人大多前往依附他,他们互相推许奖誉,因此(司空图)名声更盛。
评分建议:“属”版荡”“推奖”“藉”,语句通顺,各1分。
(2)(5分)河中节度使王重荣请司空图撰写碑文,(司空图)得到数千匹绢,司空图将绢送到虞乡的闹市中,任凭乡人获取,(数千匹绢)一日之内就被拿完。
评分建议:“撰碑”“得绢数千匹”“致”“恣”,语句通顺,各1分。
8.(4分)知恩图报,不求功名,不贪财物,坚守气节(坚守忠义)评分建议:任选4点,每点1分,意思对即可。
三、古诗词鉴赏(10分)9.(5分)以动衬静,以忽至的幽禽反衬出环境的静谧;作者以静谧的环境烘托出内心的宁静以及孤独;并照应开头的“空斋寂寂”。
评分建议:前两点2分,第三点1分,意思对即可。
10.(6分)劝勉、宽慰友人莫负风景;作者身处困境的豁达乐观;也暗含远离政治中心,只能消磨时光的一丝无奈。
评分建议:每点2分,意思对即可。
四、名句名篇默写(8分)11.(8分)(1)知止不殆;(2)惟草木之零落兮;(3)砯崖转石万壑雷;(4)而无车马喧;(5)后人哀之而不鉴之;(6)山肴野蔌;(7)一尊还酹江月;(8)种德者必养其心。
如皋市2018~2019学年度高三年级第二学期语数英学科模拟英语试题
如皋市2018~2019学年度高三年级第二学期语数英学科模拟英语试题(总12页)-CAL-FENGHAI.-(YICAI)-Company One1-CAL-本页仅作为文档封面,使用请直接删除2018~2019学年度高三年级第二学期语数英学科模拟(二)英语试题第一部分听力 (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。
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 kind of movie will the speakers watchA. An action movie.B. A comedy.C. A thriller.2. What will the man do next?A. Pour the milk in the sink.B. Buy some milk.C. Eat breakfast.3. How many fish did the man catch at the beginning?A. Two.B. Three.C. Six.4. What is the woman trying to doA. Solve a crime.B. Decorate her bedroom.C. Study a language.5. What aspect of the jeans are the speakers discussingA. The style.B. The color.C. The quality.第二节听下面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。
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。
2018年如皋高三2.5模数学试卷(定稿)精品资料
2018年如皋高三2.5模数学试卷(定稿)(第5题图)2017~2018学年度高三年级第二学期语数英学科模拟(二)数 学 试 题(考试时间:120分钟 总分:160分)一.填空题:本大题共14小题,每题5分,共70分.请把答案填写在答题卡相....应位置...上. 1. 已知集合A ={1,2m },B ={0,2}.若A ∪B ={0,1,2,8},则实数m 的值为 ▲ .2. 设复数z 满足i z =1+2i (i 为虚数单位),则复数z 的模为 ▲ . 3. 高三(1)班共有56人,学号依次为1,2,3,…,56,现用系统抽样的办法抽取一个容量为4的样本,已知学号为6,34,48的同学在样本中,那么还有一个同学的学号应该为▲ .4. 设直线l 1:x -my +m -2=0,l 2:mx +(m -2)y -1=0,则“m =-2”是直线“l 1∥l 2”的条件.(从“充要”,“ 充分不必要”,“ 必要▲不充分”及“既不充分也不必要”中选择一个填空) 5. 根据下图所示的算法,输出的结果为 ▲ .(第12题6. 若将一个圆锥的侧面沿一条母线剪开,其展开图是半径为3,圆心角为23π的扇形,则该圆锥的体积为 ▲ .7. 已知F 1、F 2是双曲线的两个焦点,以线段F 1F 2为边作正方形MF 1F 2N ,若M ,N 都在该双曲线上,则该双曲线的离心率为 ▲ .8. 在正项等比数列{a n }中,S n 为其前n 项和,已知2a 6=3S 4+1,a 7=3S 5+1,则该数列的公比q 为 ▲ .9. 函数f (x )=A sin(ωx +φ)(A >0,ω>0,0≤φ<2π)在R 上的部分图象如图所示,则f (2 018)的值为 ▲ .10.已知变量x ,y 满足约束条件00201422x y x y x y --⎧⎪+-⎨⎪+-⎩≤,≤,,≥则1y x z x +=+的最大值为 ▲ .11. 在△ABC 中,角A ,B ,C 的对边分别为a ,b ,c 是△ABC的面积,若22213b c a +=,则角A 的值是 ▲ .12.如图,在平面四边形ABCD 中,2AB =,△BCD 是等边三 角形,若1AC BD ⋅=,则AD 的长为 ▲ .13.在平面直角坐标系xOy 中,已知点()0A m ,,()40B m +,,若圆C :()2238x y m +-=上存在点P ,使得45APB ∠=︒,则实数m 的取值范围是▲ .14.已知函数()e 0102x x f x x x ⎧⎪=⎨->⎪⎩,≤,,,若关于x 的方程()1f f x m -=⎡⎤⎣⎦有两个不同的根1x ,2x ,则12x x +的取值范围是 ▲ .二、解答题:本大题共6小题,共90分.请在答题卡指定区域内........作答,解答时应写出文字说明、证明过程或演算步骤.15.在△ABC 中,角A ,B ,C 所对的边分别为a ,b ,c ,已知cos 2cos a Ab c C=-. (1)求角A 的值;(2)求2sin B -sin C 的取值范围.16.如图,在四棱锥P ─ABCD 中,底面ABCD 是菱形,PA ⊥平面ABCD .(1)证明:平面PBD ⊥平面P AC ;(2)设E 为线段PC 上一点,若AC ⊥BE ,求证:P A ∥平面BED .17.如图,OA ,OB 是两条互相垂直的笔直公路,半径2OA =km 的扇形AOB是某地的一名胜古迹区域.当地政府为了缓解该古迹周围的交通压力,欲在圆弧AB 上新增一个入口P (点P 不与A ,B 重合),并新建两条都与圆弧AB 相切的笔直公路MB ,MN ,切点分别是B ,P .当新建的两条公路总长最小时,投资费用最低.设POA θ∠=,公路MB ,MN 的总长为()f θ. (1)求()f θ关于θ的函数关系式,并写出函数的定义域;(2)当θ为何值时,投资费用最低?并求出()f θ的最小值.O A BMNPθPABCDE(第16题图)18.在平面直角坐标系xOy 中,设椭圆C :()222210x y a b a b+=>>的下顶点为A ,右焦点为F.已知点P 是椭圆上一点,当直线AP 经过点F 时,原点O 到直线AP的距离为. (1)求椭圆C 的方程;(2)设直线AP 与圆O :222x y b +=相交于点M (异于点A ),设点M 关于原点O 的对称点为N ,直线AN 与椭圆相交于点Q (异于点A ).①若2AP AM =,求△APQ 的面积;②设直线MN 的斜率为k 1,直线PQ 的斜率为k 2,求证:12k k 是定值.19.已知函数()2114ln 22f x x ax a x a =-+++(1)当1a =时,求函数()f x 在1x =处的切线方程;(2)记函数()f x 的导函数是()'f x ,若不等式()()'f x xf x <对任意的实数()1x ∈+∞,恒成立,求实数a 的取值范围;(3)设函数()()2g x f x a =+,()'g x 是函数()g x 的导函数,若函数()g x 存在两个极值点1x ,2x ,且()()()1212'g x g x g x x +≥,求实数a 的取值范围.(第18题图)20.已知正项数列{}n a 的前n 项和为n S,且满足1n a =,数列{}n b 是首项12b =,公比为()1q q ≠的等比数列.(1)求数列{}n a 的通项公式;(2)设三个互不相等的正整数k ,t ,r ()k t r <<满足2t k r =+,若k t t r r k a b a b a b +=+=+,求实数q 的最大值;(3)将数列{}n a 与{}n b 的项相间排列成新数列{}n c :1a ,1b ,2a ,2b ,3a ,3b ,…,设新数列{}n c 的前n 项和为n T ,当3q =时,是否存在正整数m ,使得22-1mm T T 恰好是数列{}n c 中的项?若存在,求出m 的值,若不存在,说明理由.。
江苏省如皋市2018~2019学年度高三年级第二学期语数英学科模拟数学试题
.........6 的终边经过点 P(﹣1, -2 2 ),则 sin α =江苏省如皋市 2018~2019 学年度高三年级第二学期语数英学科模拟(一)数学试题Ⅰ(考试时间:120 分钟总分:160 分)一、填空题(本大题共 14 小题,每小题 5 分,共计 70 分.不需要写出解答过程,请将答案填写在答题卡相应的位置上.)1.已知全集 U ={1,2,3},A ={2},则 U A =.2.已知复数 z = m - i( m ∈ R ,i 是虚数单位)是纯虚数,则实数 m 的值为.1 + i3.某学校高一、高二、高三年级的学生人数之比为 5:5:4,现用分层抽样的方法抽取若干人,若抽取的高三年级为 12 人,则抽取的样本容量为 人. 4.一个算法的伪代码如图所示,执行此算法,最后输出的 T 的值为 .5.在平面直角坐标系 xOy 中,双曲线 x 2 y 2 - a 2 b 2= 1 (a >0,b >0)的一条渐近线经过点(1,2),则双曲线的离心率为 .6.将一颗质地均匀的骰子(一种各个面上分别标有 1,2,3,4,5,6 个点的正方体玩具)先后抛掷 2 次,则出现向上的点数之和大于 9 的概率为 .7.已知变量 x ,y 满足约束条件 2 x + y - 2 ≤ 1 , x ≥ 0 , y ≥ 0 ,则 x - 2 y + 1 的最大值为.8.已知角 α +π.9.如图,直三棱柱 ABC —A 1B 1C 1 中,∠CAB =90°,AC =AB =2,CC 1=2,P 是 BC 1 的中点,则三棱锥 C —A 1C 1P 的体积为 .10.已知数列{a n }的前 n 项和为 S n , a = 1 ,且满足 S = a 1 nn +1,则数列{Sn}的前 10 项的和为.11.已知函数 f ( x) = ⎨ 1 ,若函数 h( x) = f ( x) + x - a 恰有 3 个不同的⎪ x ,x ≥ 0 二、解答题(本大题共 6 小题,共计 90 分.请在答题纸指定区域内作答,解答应写出文字⎧2 x 2 + 4 x + 1,x < 0⎪1 2 ⎩ e零点,则实数 a 的取值集合为.△12.若等边 ABC 的边长为 2,其所在平面内的两个动点 P ,M 满足 AP = 1 ,PM = MB ,则 CM ⋅ CB 的最大值为.13.已知正数 a ,b ,c ,d 满足 1 2 2 3+ = 1, + = 2 ,则 a + bcd 的最小值为 .a b c d914.在平面直角坐标系 xOy 中,已知点 A 是圆 C : ( x - 4)2 + ( y - 1)2 = 上一动点,点 B2OB是直线 x - y + 2 = 0 上一动点,若∠AOB =90°,则 的最小值为 .OA.......说明,证明过程或演算步骤.)15.(本题满分 14 分)在△ABC 中,角 A ,B ,C 所对的边分别是 a ,b ,c ,且 3cos(B + C) + 2sin 2 A = 0 .(1)求角 A 的大小;(2)若 B =π4,a = 2 3 ,求边长 c .16.(本题满分 14 分)如图,四棱锥 P —ABCD 中,底面为直角梯形,AD ∥BC ,AD =2BC ,且∠BAD =∠ BPA =90°,平面 APB ⊥底面 ABCD ,点 M 为 PD 的中点.(1)求证:CM ∥平面 PAB ; (2)求证:PB ⊥PD .两点,直线MB2与直线NB1交于点T.①若直线l的斜率为,求点T的坐标;②试问点现需要设计一个仓库,它由上下两部分组成,上部的形状是圆锥,下部的形状是圆柱(如图所示),并要求圆柱的高是圆锥的高的2倍.(1)若圆柱的底面圆的半径为3m,仓库的侧面积为63πm2,则仓库的容积是多少?(2)若圆锥的母线长为6m,则当PO1为多少时,仓库的容积最大.18.(本题满分16分)如图,在平面直角坐标系xOy中,椭圆间的距离为83.3(1)求椭圆的方程;x2y2+a2b2=1(a>b>0)过点P(2,0),且两准线(2)已知B2,B1分别是椭圆的上、下顶点,过点E(0,12)的直线l与椭圆交于M,N12T是否在某定直线上?若在定直线上,求出定直线方程;若不在定直线上,请说明理由.{ }- a2 (n ∈ N * ),且等差数列 {a }的公差为 ,存在3已知函数 f ( x ) = x 2 + (a + 2) x + ae x(a ∈ R) , g ( x ) = e x f ( x ) .(1)若 A = x g ( x ) ≤ 9, x ∈[a, + ∞) ≠ ∅ ,求实数 a 的取值范围;(2)设 f ( x ) 的极大值为 M ,极小值为 N ,求M N的取值范围.20.(本题满分 16 分)已知数列 {a n}是公差不为零的等差数列,数列{b }满足 b n n= a ⋅ ann +1⋅ an +2(n ∈ N * ).(1)若数列 {a n}满足 a 10= -2 , a , a , a 成等比数列.①求数列 {a }的通项公4 14 9 n式;②数列{b }的前 n 项和为 S ,当 n 多大时, S 取最小值.n nn(2)若数列 {c }满足 c nn= a a n +1n +2n n1正整数 p ,q ,使得 a + c 是整数,求 a 的最小值.p q11-26611.⎨1,+ln2⎬12.413.13+4314.()=3在∆ABC中,由正弦定理得:c所以c数学试题(Ⅰ卷)答案一、填空题:本大题共14小题,每小题5分,共70分.1.{1,3}2.13.424.155.56.1527.8.9.10.1023 623⎧11⎫⎩22⎭14二、解答题:本大题共6小题,计90分.15.⑴在∆ABC中,由A+B+C=π,sin2A+cos2A=1及3cos(B+C)+2sin2A=0得:3cos(π-A)+21-cos2A=0………………………………………………2分所以2cos2A+3cos A-2=0,所以(2cos A-1)(cos A+2)=0,因为cos A∈(-1,1),所以cos A=12,因为A∈(0,π),所以A=π3………………………………………………6分⑵sin C=sin(π-A-B)=sin(A+B)=sin A c os B+cos A s in B2126+2⨯+⨯=………………………………………………10分22224a=,sin C sin A23=6+23,所以c=6+2.………………………………14分4216.证明:⑴取AP的中点H,连接BH,HM,因为H,M分别为AP,DP的中点,所以HM=12AD且HM//AD………2分因为AD//BC且AD=2BC,所以HM=BC且HM//BC,所以四边形BCMH为平行四边形,所以CM//BH………………………4分因为CM⊄平面PAB,BH⊂平面PAB,所以CM//平面PAB…………………………………………………………6分⑵因为∠BAD=900,所以BA⊥AD.因为平面APB⊥平面ABCD,AD⊂平面ABCD,平面APB I平面ABCD=AB, V = ⨯ π ⨯ r 2 ⨯ x + π ⨯ r 2 ⨯ 2x = π r 2 x = π (- x 3+ 36x ), x ∈ (0,6 ) 所以 AD ⊥ 平面 APB ………………………………………………………………9 分 因为 PB ⊂ 平面 PAB ,所以 PB ⊥ AD ,因为 ∠BPA = 900 ,所以 PB ⊥ P A ,因为 P AIPD = P , P A PD ⊂ 平面 P AD ,所以 PB ⊥ 平面 PAD ………………………………………………………………12 分 因为 PD ⊂ 平面 PAD ,所以 PB ⊥ PD . …………………………………………14 分17. ⑴ 解:设圆锥的高为 h m ,因为圆柱的高是圆锥的高的 2 倍,所以圆柱的高为 2h m .仓库的侧面积 S = 1⨯ 2π ⨯ 3 9 + h 2 + 2π ⨯ 3 ⨯ 2h = 63π2所以 9 + h 2 = 21 - 4h ,所以 9 + h 2 = (21 - 4h )2 ,所以 5h 2 - 56h + 144 = (h - 4)(5h - 36) = 0 ,………………………2 分所以 h = 4 或 h = 36 5,当 h = 36 5时,21 - 4h < 0 ,所以h = 4 m ………………………………………4 分1所以仓库的容积为 π ⨯ 32 ⨯ 4 + π ⨯ 32 ⨯ 8 = 84π m 2 3答:仓库的容积是 84π m 2⑵ 设 PO 为 x m ,圆柱的底面圆的半径为 r m .1……………………………6 分……………………………7 分仓库的容积1 7 7 3 3 3设 f (x ) = - x 3 + 36x, x ∈ (0,6 ) ……………………………………………………9 分令 f ' (x ) = -3x 2 + 36 = 0 得: x = 2 3 ,xf ' (x )f (x )(0,2 3 )+Z2 3极大值(0,2 3 )-]所以 x = 2 3 m 时,仓库的容积V 取得极大值,也是最大值………………13 分答:当 PO 为 2 3 m 时,仓库的容积最大……………………………………14 分118.⑴ 设椭圆的半焦距为 c .1 1 ,所以直线 l 的方程为 y = x + ,⎪⎪ 所以 x = -1 - 7 2 2 y = 1 x + 1x ⎛ y - 1 y + 1 ⎫ ⎪ x = 2 1 - 2 由 ⎨ 得: ⎪ ⎝ x2 ⎭⎪ y = 2 y + 1 x ⎪⎩所以 x = 2 x 1x 2 = x 2 + ⎪- x 1 - ⎪y = ( ) ()( )⎪⎪所以 x + x = - 4k1 + 4k2 1 + 4k 2因为椭圆过点 P (2,0 ) ,且两准线间的距离为 83 ,3所以 a = 2, 2 ⨯ a 2 8= 3 , 所以 a = 2, c = 3, b = a 2 - c 2 = 1 ,c 3所以椭圆的方程为 x 2 4+ y 2= 1 ………………………………………………3 分⑵ ① 设 M (x , y ), N (x , y 1122)因为直线 l 的斜率为 12 2 2⎧ x 2+ y 2 = 1 由 ⎨ 4得: 2 x 2 + 2 x - 3 = 0 ,⎪ y = 1 x + 1 ⎪⎩ 2 2-1 + 7 , x = ……………………………………………5 分1 2⎧y - 1 ⎪ 1 x - 1 1 x2x ( y + 1)- x ( y - 1) 1 2 2 12 x x1 2⎛ x 3 ⎫ ⎛ x 1 ⎫ 1⎝ 2 2 ⎭ 2 ⎝ 2 2 ⎭= 4 x 1x2 3x + x12= 2 7 - 4 ………………………………………………………7 分y - 1 x - 1 1 2 7 - 4 + 1 = 1 2 7 - 4 + 1 = 2 .x 2 x11点 T 的坐标为 (2 7 - 4, 2 )………………………………………………10 分⎧ x 2+ y 2 = 1 ② 由 ⎨ 4得: 1 + 4k 2 x 2 + 4kx - 3 = 0 ,⎪ y = kx + 1 ⎪⎩ 23 , x x = - …………………………………12 分1 2 1 2y = 1 x + 1x 由 ⎨得: ⎪ ⎪ y = 2x - 1 ⎪⎩所以 y = ⎡⎣ x ( y -1)+x (y+1)⎤⎦ ( y + 1)- x ( y - 1)⎤⎦ x y - x 2 y 1 + x 2 + x 1 , = 1 2 x kx + ⎪ + x kx + ⎪ - x + x x kx + ⎪ - x kx + ⎪ + x + x 1 ⎝ 2 ⎭ 2 ⎭ 2 ⎝ 1 ⎫ 2 ⎝ 1 1 ⎫ 4k -3 1 + 4k 1 + 4k { }10 当 a ≥ - 时,函数 g (x )的对称轴为 - 20 a < - 时,函数 g (x )的对称轴为 - 综上:实数 a 的取值范围为 -∞, ⎦⎧y - 1 ⎪ 1 y + 1 x2⎡⎣ x 1 ( y 2 + 1)- x 2 ( y 1 -1)⎤⎦ y = ⎡⎣ x 2 ( y 1 -1)+ x 1 (y 2 + 1)⎤⎦x y + x y - x + x 2 1 1 2 ⎡⎣ x 1 2 2 1 1 2 2 1 2 1⎛ ⎛ 1 ⎫ 1 ⎝ 2 2 ⎭ 2 ⎭ 2 14kx x + 3x - x =1 2 1 2 ⎛ ⎛ 1 ⎫ 3x + x 1 2 2 1 2 1= 4kx 1x 2 - 3 (x + x )+ 6 x 1 2 1 3x + x1 2+ 2 x2 = -4k- 3 + 6 x + 2 x 2 2 1 2 3x + x 1 2= 2所以点 T 是否在直线 y = 2 上……………………………………………………16 分19.⑴ 因为 A = x g (x) ≤ 9, x ∈ [ a, +∞ ) ≠ ∅ ,所以函数 g (x ) = x 2 + (a + 2) x + a 的最小值小于等于 9 .2 a + 23 2≤ a ,所以 g (x ) min= g (a ) = 2a 2 + 3a ≤ 9 ,所以 -3 ≤ a ≤ 3 2,因为 a ≥ - 2 2 3,所以 - ≤ a ≤ ……………………………………………………3 分3 3 22 a + 23 2> a ,所以 g (x ) min -a 2 - 4 2= ≤ 9 恒成立,所以 a < - ………………………………5 分4 3⑵ f ' (x ) =- x 2 - ax + 2ex⎛⎝3 ⎤2 ⎥ ………………………………………………6 分设 h (x ) = - x 2 - ax + 2 ,因为 ∆ = a 2 + 8 > 0 ,M f (x ) x 2 + (a + 2) x + a 2 x + a + 2 所以 == 2 2e x 1 - x 2 = e x 1 - x 2 (*) ………10 分 e x 1 - x 2 = e t ,设Q (t ) = e t ,t ≤- 2 2 ………………13 分 (t + 2)2 < 0 ,所以函数 Q (t ) 在 ( -∞, 2 2 ⎤⎦ 上为单调减函数,( )( )的取值范围为 ⎡- 3 + 2 2 e -2 2 ,0 ………………………………16 分所以函数 h (x ) 有两个不同的零点,不妨设 x , x 且 x < x ,1 212x + x = -a, x x = -2……………………………………………………………8 分1 21 2当 x ∈ (-∞, x ) 时, h (x ) > 0 ,函数 f (x )为单调增函数, 1当 x ∈ (x , x 12) 时, h (x ) < 0 ,函数 f (x )为单调减函数,当 x ∈ (x , +∞ )时, h (x ) > 0 ,函数 f (x )为单调增函数,2所以当 x = x 时,函数 f (x )取得极小值,当 x = x 时,函数 f (x )取得极大值,1 2N f(x ) x 2 + (a + 2) x + a 2x + a + 21111将 x + x = -a 代入 (*) 得:1 2x - x + 2 2 1 x - x + 2 12e x 1 - x 2 ,设 t = x - x = - 1 2 (x 1 - x 2 )2 = - a 2 + 8 ≤ -2 2 ,所以 x - x + 2 2 - t 2 - t 2 1x - x + 2 t + 2 t + 21 2Q ' (t ) = -t 2e t- 3 + 2 2 e -22≤ Q (t ) < 0 ,综上: M N⎣ )20.⑴① 设数列{a n}的公差为 d ,因为 a , a , a 成等比数列,所以 (-2 + 4d )2 = (-2 - 6d )(-2 - d ),4149所以 d 2 - 3d = 0 ,因为 d ≠ 0 , 所以 d = 3 ,所以 a = a + (n - 10)d = 3n - 32……………………………………………3 分n 10② 当1 ≤ n ≤ 10 时, a < 0 ,当 n ≥ 11时, a > 0 ,n n因为 b = a ⋅ annn +1⋅ a n +2 ,所以当1 ≤ n ≤ 8 时, b < 0 ,当 n ≥ 11时, b > 0 ,n n10 ………………………………………………………6 分- a 2 = a + ⎪ a + ⎪ - a 2 = a +3 ⎭⎝ n 3 ⎭9 ⎝ n 则 a + c = a + ( p - 1)⨯ + a + (q - 1)⨯ +1 2 3 3 9 3 9 18b > 0, b < 0 ,所以 S > S > L > S < S > S < S < L9 1012891011所以 S 的最小值为 S 或 S n 8因为 S - S = b + b = a a 10891010 11(a 9+ a 12),又因为 a < 0, a > 0, a + a = -1 < 0 ,所以 S - S > 010 11912108所以当 n = 8 时, S 取最小值………………………………………………………9 分n⑵ c = ann +1 an +2n n n⎛ 1 ⎫⎛ 2 ⎫ 2 ……………………10 分若存在正整数 p , q ,使得 a + c 是整数,p q1 p + q -2 2= 2a + + ∈ Z ,p q 1 1 1设 m = 2a + 1 p + q - 2 2+ , m ∈ Z ,3 9所以18a = 3 (3m - p - q + 1)+ 1是一个整数,1所以 18a ≥ 1 ,从而 a ≥ 1 1 1 18…………………………………………………14 分又当 a = 1 1 18时,有 a + c = 1∈ Z .1 31综上: a 的最小值为 …………………………………………………………16 分1所以 ⎢ = ⎢ ⎥ , 1 3⎥⎦ ⎢⎣ y ⎥⎦ ⎣ y ⎦ ⎣ 得: ⎨ ⎧( ) ( ) ( )将直线 l 与曲线 C 联立方程组 ⎨ x 2 y 2(2,0 ), ⎛ 10 , - 4 7 7 ⎪⎭ - 2 ⎪ + - 7 ⎪⎭所以直线 l 被曲线 C 截得的线段长为 ⎪⎝ ⎭数学Ⅱ附加题21. 解:设直线 l 上任意一点 (x , y0 0⎡2 0⎤ ⎡ x ⎤ ⎡ x ⎤ 0 0)在矩阵 M 变换作用下变为 (x, y ),⎧ 2 x = x⎩ x 0 + 3 y 0 = y因为 ax + by - 2 = 0 , 所以 (2a + b ) x + 3by - 2 = 0 (*) ………………………6 分0 0(x , y )为直线 l 上任意一点,所以 (*) 与 2 x- 2 y - 2 = 0 为同一方程,所以 ⎨2a + b = 2 ⎩ 3b = -24 2 , 所以 a = , b = - ………………………………………10 分3 322.⑴ 因为曲线 C 的极坐标方程是 ρ 2 =4cos 2θ + 3sin 2 θ,所以 ρ 2 cos2θ + 3sin 2 θ = ρ 2 cos 2 θ - sin 2 θ + 3sin 2 θ = ρ 2 cos 2 θ + 2sin 2 θ ,因为 x = ρ cos θ , y = ρ sin θ ,所以 x 2 + 2 y 2 = 4 ,所以曲线 C 的直角坐标方程为 x 2 y 2+ = 1 ………………………………………4 分4 2⑵ 因为直线 l 过点 (2,0 ),且倾斜角为 600,所以直线 l 的直角坐标方程为 y =3x - 2 3 ……………………………………6 分⎧ y = 3x - 2 3 ⎪ ⎪ + = 1 ⎩ 4 2得: 7 x 2 - 24x + 20 = (7 x -10)(x - 2 ) = 0 , 所以 x = 2 或 x =107,所以直线 l 与曲线 C 的交点为⎝3 ⎫⎛ 10 ⎫2 ⎛ 4 3 ⎫⎝2=8 7……………10 分23.⑴ 因为抛物线 C : y 2 = 2 px ( p > 0)的焦点是 F (1,0 ),所以 p 2= 1,即 p = 2 ,抛物线 C 的方程为 y 2 = 4 x …………………………………………………………2 分11⨯ AF ⨯ DF ⨯ sin ∠AFD + ⨯ BF ⨯ DF ⨯ sin ∠BFD所以 1 = 2 = = 4 …4 分S DF ⨯ AF ⨯ C F ⨯ sin ∠AFC + ⨯ BF ⨯ CF ⨯ sin ∠BFC CF 2 2 ⎪ 2),所以 ⎨ y 2 = 4 x F ⎩ ⎩ y 2 = 4 x因此,共有 4 C 2 + C 2 + L + C 2 ⎪ + C 2 = 4C 3 + C 2 = ⎝-1 ⎭因此,共有 4 C 2 + C 2 + L + C 2 ⎪ - C 2 = 4C 3 - C 2 =⎝ ⎭⑵ 设 ∆ABD 的面积为 S , ∆ABC 的面积为 S12因为 ∠AFD + ∠BFD = 1800 , ∠AFC + ∠BFC = 1800 ,1 12S1 1 2u uur u uurF D = 4 C , 设 C (x , y ), D (x , y 1122⎧ x - 1 = 4 (1 - x )1⎪ y = -4 y 2 1⎪ 1 1 ⎪ y 22 = 4 x 2⎧4 y 2 = 5 - 4 x 得: ⎨ 1 1 , 所以 5 - 4 x = 16 x , 所以 x = 1 1 1 1 11 4, y = ±1 ,1所以直线 l 的方程为 4 x + 3 y - 4 = 0 或 4 x - 3 y - 4 = 0………………………10 分224.⑴ 因为1 + 4 = 2 + 3,1 + 5 = 2 + 4,1 + 6 = 2 + 5,1 + 6 = 3 + 4,2 + 5 = 3 + 4,2 + 6 = 3 + 5 ,3 + 6 =4 +5 , 所以 f (6) = 7 ;同理: f (7) = 13 ……………………………2 分⑵ 10 当 n ≥ 4 的偶数时,和 a + c = b + d = s 可以取以下值: 5,6,L , n + 1,L ,2 n - 3 ,在 s 取定后,相应的两个最小的加数取值分别有:C 2 , C 2 , C 2 , C 2 ,L , C 2 , C 2 , C 2 , C 2 , C 2 ,L , C 2 , C 2 种取法,2233n n n n n 222 -1 2 -122 -12 -1⎛ ⎫2 3 n n n n 2 2 2 2n (n - 2)(2n - 5)24 种取法……………………………5 分20 当 n ≥ 4 的奇数时,和 a + c = b + d = t 可以取以下值: 5,6,L , n + 1,L ,2 n - 3 ,在 s 取定后,相应的两个最小的加数取值分别有:C 2 , C 2 , C 2 , C 2 ,L , C 2 , C 2 , C 2 ,L , C 2 , C 2 种取法,2233n -1n -1 n -1 22222⎛ ⎫2 3 n -1 n -1 n +1 n -12 2 2 2(2n - 1)(n - 1)(n - 3) 24种取法 ………………………………………………………………………………8 分12⎪⎪ 24⎧ n (n - 2)(2n - 5), n = 2k + 2 ,综上所述: f (n ) = ⎨(k ∈ N *) ……………………10 分 ⎪ (2n - 1)(n - 1)(n - 3), n = 2k + 3 ⎪⎩ 2413。
江苏省如皋中学19届高三英语上学期第二次月考试卷.doc
江苏省如皋中学2018-2019学年度第一学期高三第二次月考试卷高三英语第I卷(选择题共85分)第一部分:听力(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. What's wrong with the man?A. He is sick.B. He is thirsty.C. He is tired.2. What will the speakers discuss?A. A paper.B. A new computer.C. A new viewpoint.3. Why does the man make the phone call?A. To book a room.B. To apply for a job.C. To put an advertisement.4.How much time did the man spend on the exam?A. One hour and 20 minutes.B. One hour and 40 minutes.C. Two hours and 20 minutes.5.Where does the conversation probably take place?A. In a bookstore.B. In a museum.C. In a library.第二节(共15小题;每小题1分,满分15分)听下面5段对话。
每段对话后有几个小题,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。
听每段对话前,你将有时间阅读各个小题,每小题5秒钟;听完后,每小题将给出5秒钟的作答时间。
每段对话读两遍。
听第6段材料,回答第6、7题。
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如皋市2018~2019学年度高三年级第二学期语数英学科模拟(三)
英语试题
第一部分听力(共两节,满分20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. What is the man going to do today?
A. Go surfing.
B. Jog on the beach.
C. Get some work done early.
2. How long is the man’s speech allowed to be?
A. Ten minutes long.
B. Twenty minutes long.
C. Thirty minutes long.
3. What are the speakers mainly talking about?
A. The ocean.
B. Sea lions.
C. Dogs.
4. Where will the woman probably check for her phone?
A. In the kitchen.
B. In the bathroom.
C. In the dentist’s office.
5. How does the woman probably feel in the end?
A. Tired.
B. Sick.
C. Annoyed.
第二节
听下面5段对话或独白。
每段对话或独白后有2至4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5秒钟的时间阅读各个小题;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。
听第6段材料,回答第6、7题。
6. What does the woman do for a living?
A. She’s a cook.
B. She’s a guide.
C. She takes care of animals.
7. What does the man recommend?
A. A hotel.
B. A bakery.
C. A jewelry store. 听第7段材料,回答第8至10题。
8. Where are the speakers?
A. In a park.
B. At a concert.
C. In a gym.
9. When will the speakers meet?
A. In an hour.
B. In 45 minutes.
C. In 10 minutes.
10. Why is the man impressed by the woman?
A. She is very athletic.
B. She likes loud music.
C. She rides her bike to work every day.
听第8段材料,回答第11至13题。
11. What did the Ferrari driver do?
A. He hit someone.
B. He drove too fast.
C. He crashed on the freeway.
12. What had the speakers been doing?
A. Shopping.
B. Looking at cars.
C. Attending a wedding.
13. What is special about the vase?
A. It was on sale.
B. It’s a gift for the woman’s sister.
C. It’s for the speakers’ house.
听第9段材料,回答第14至17题。