2017古美高中高二下册月考试卷含答案

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2017-2018人教版高二语文下学期第三次月考试题附答案解析[最新]

2017-2018人教版高二语文下学期第三次月考试题附答案解析[最新]

第Ⅰ卷阅读题(70分)一、现代文阅读(35分)(一)论述类文本阅读(9分,每小题3分)阅读下面的文字,完成1-3题。

“驼铃古道丝绸路,胡马犹闻唐汉风。

”古代丝绸之路架起了一座交流物产、连通人心的桥梁,对我国各民族交流融合、对东西方经济文化交往都起到了十分重要的作用。

古代丝绸之路大体有草原道、绿洲道、茶马道以及海上道四条。

除了汉族,北方和西北游牧民族也是丝绸之路的重要开拓者,他们的马队和骆驼队踏出了一条横贯欧亚大陆的草原丝路。

他们的迁徒浪潮、相互交往以及游牧经济特点,使其自然而然地成为古代丝绸之路上的重要角色。

继月氏、匈奴之后,鲜卑、吐谷浑、吐蕃、回纥、党项等民族,都曾和丝绸之路结下不解之缘,有的甚至一度控制了草原道和绿洲道,成为经营东西方贸易的主角。

公元439年,鲜卑建立的北魏政权统一了我国北方,使丝绸之路自汉代以来再度繁荣起来。

北魏、西夏占据河西走廊后,吐谷浑控制的“青海道”和吐蕃控制的“青唐道”成为中原和南方通往西域的通道。

因此,“青海道”又称“吐谷浑道”,“青唐道”又称“吐蕃道”。

再看回纥,其与唐朝贸易换回的绸绢,除了供贵族享用,还通过“草原道”大量转输到西方。

“安史之乱”后,吐蕃完全占据了河西走廊及陇右地区,传统的丝绸之路东段受到阻遏,唐朝和西域各国的交往一度绕道回纥居住地。

因此,这一时期的草原丝路有“回纥道”之称。

在肯定我国古代北方和西北游牧民族为丝绸之路的开拓与繁荣做出重要贡献的同时,更要充分认识中原王朝的主体作用。

张骞出使西域之后,汉、唐、元、明各朝代为了经营西域,保障丝绸之路畅通,在丝绸之路沿途设置馆舍以提供食宿,建立都护府、都督府等以加强治理。

这些措施对保障丝绸之路的畅通和安全具有决定性作用。

丝绸之路密切了我国古代民族关系,也密切了东西方关系。

中原、江南以及巴蜀的名茶不仅输入西域、青藏高原与漠北,也输入西方。

在西夏与宋朝的贸易中,“惟茶最为所欲之物”。

同时,西域和中亚、欧洲的物产和文化也传入内地,今天内地随处可见的石榴、葡萄、大蒜、菠菜等,都是汉朝时从西域传入内地的。

上海市古美高级中学高二第二学期月考试卷

上海市古美高级中学高二第二学期月考试卷

2010学年第二学期古美高级中学高二年级月考(1)数 学 试 卷(完卷时间90分钟,满分100分)总分一、填空题(本大题共12小题,每小题3分,共36分)1.若O 是坐标原点,(1,2)A -,(5,6)B -,点P 为线段AB 的中点,则||OP = . 2.若圆22()(1)5x a y ++-=的圆心在直线350x y --=上,则a = . 3.直线210x y ++=的倾斜角α= .(用反三角函数值表示) 4.点(1,1)P -到直线l :10x y -+=的距离d = .5.直线1l:50x ++=与直线2l10y +-=的夹角α= . 6.若直线220ax y ++=与直线320x y --=互相垂直,则a = . 7.经过点(1,0)P -且与直线20x y -=平行的直线的方程是 . 8.圆心为(11),且与直线40x y +-=相切的圆方程是 . 9.若(1,1)A -、(,3)B a -、(3,13)C a -三点共线,则a = . 10.点(2,1)P 关于直线l :20x y +-=的对称点Q 的坐标是 .11.直线l 经过点(1,2)且与直线20x y -+=的夹角为45,则l 的方程是 . 12.原点O 到直线(21)(3)(11)0k x k y k --+--=()k ∈R 的最大距离d = . 二、选择题(本大题共4小题,每小题3分,共12分) 13.关于直线1x =-,下列说法中不正确的是[答]( )(A )向量(2,0)a =是直线1x =-的一个法向量 (B )向量(0,1)b =-是直线1x =-的一个方向向量 (C )直线1x =-的倾斜角不存在 (D )直线1x =-的斜率不存在学校_______________________ 班级__________ 学号_________ 姓名______________ ……………………密○………………………………………封○………………………………………○线……………………………………14.关于直线20x y m -+=和直线4210x y -+=,下列说法正确的是[答]( )(A )可能相交 (B )一定平行 (C(D )夹角为0 15.已知(2,0)A -、(6,0)B ,ABP ∆的面积为16,则点P 的轨迹方程为[答]( )(A )4y = (B )4x = (C )4y =± (D )4x =±16.已知345x y +=,[答]( )(A )1 (B )2 (C )25 (D )45三、解答题(本大题共5小题,共52分.解答应写出文字说明、证明过程或演算步骤) 17.(本题满分8分)过原点向直线250x y -+=作垂线,求垂足H 的坐标.18.(本题满分10分)若动点P 在曲线221y x =+上移动,点(0,1)Q -,线段PQ 的中点为M ,求点M 的轨迹方程.19.(本题满分10分)已知两点(1,2)A -、(3,1)B ,直线l 的方程是2y kx k =+.(1)求线段AB 的垂直平分线的方程;(2)若直线l 与线段AB 有公共点,求k 的取值范围.20.(本题满分12分)已知直线l :1y kx =-和曲线C :21y x =+.(1)若直线l 和曲线C 有公共点,求k 的取值范围;(2)若直线l 截曲线C k 的值.21.(本题满分12分)过点(3,2)P 的直线l 与x 轴的正半轴和y 轴正半轴分别交于A 、B 两点.(1)若P 为线段AB 中点时,求l 的方程; (2)当||||OA OB 最小时,求l 的方程.……………………密○………………………………………封○………………………………………○线…………………………。

17学年下学期高二第二次月考英语试题(附答案)

17学年下学期高二第二次月考英语试题(附答案)

樟树中学2018届高二下学期第二次月考英语试卷考试范围:必修1----选修7 unit3 考试时间: 2017年3月26日命题人:钟世财审题人:殷爱林第一部分听力(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,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. What does the woman like best?A. Water.B. Coffee.C. Tea.2. What are the speakers talking about?A. Market research.B. A job interview.C. An exam paper.3. What was the woman doing?A. Looking for something.B. Admiring a building.C. Selling flowers.4. What will the man probably do?A. Prepare for a test.B. Go to meet friends.C. Rest at home.5. What did the man buy yesterday?A. Shirts.B. Trousers.C. Shoes.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。

每段对话或独白读两遍。

听第6段材料,回答6、7题。

6. How is the man probably feeling now?A. Sorry.B. Angry.C. Worried.7. What does the woman want to do?A. Try on some new dresses.B. Wait outside with friends.C. Buy the man something. 听第7段材料,回答第8、9题8. Which service does the man not choose?A. Long-distance call.B. Voicemail.C. Internet service.9. When will the installation be done?A. On Tuesday.B. On Wednesday.C. On Friday.听第8段材料,回答第10至12题。

【英语】全国百校名师联盟2017-2018学年高二下学期月考试题(一)(word版附答案)

【英语】全国百校名师联盟2017-2018学年高二下学期月考试题(一)(word版附答案)

全国百校名师联盟2017-2018学年高二下学期月考(一)英语试题第一部分听力(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 was wrong with the woman?A. She had a toothache.B. She was injured.C. She got a fever.2. What are the speakers probably doing?A. Having dinner.B. Seeing a film.C. Enjoying a concert.3. How long will the woman have to work in her new job per day?A. 6 hours.B. 8 hours.C. 10 hours.4. What are the speakers looking for?A. A hotel.B. A book.C. A person.5. What can we know about the woman?A. She is living in New Zealand.B. She is learning to ride a horse.C. She has friends who have lived on farms.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。

每段对话或独白读两遍。

人教版2017高二下学期月考英语试卷附答案

人教版2017高二下学期月考英语试卷附答案

本试卷分第Ⅰ卷和第Ⅱ卷两部分,共120分,考试时间100分钟。

第Ⅰ卷(选择题)第一部分阅读理解(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。

AMost dictionaries will tell you a number of things about a language. There are three important things. These three things are spelling, pronunciation and meaning.First, a dictionary will tell you the spelling of a word. If you are not sure about the spelling of a word, you can try to find the correct spelling in a dictionary. Words are listed in an alphabetical(字母表的) order--- a, b, c and so on. For example, on a dictionary page the “poor” comes before “poverty” and the word “poverty” comes before the “power”. The words are always given in alphabetical order. The second thing, a dictionary will tell you the pronunciation. Most dictionaries give phonetic (语音的), or sound alphabet. The phonetic alphabet(音标) shows pronunciation. The third thing, a dictionary will tell you the meaning of words. You can look up a word and find out what it means. Many words have more than one meaning, and a good dictionary will tell you all of the word‟s meanings. For example, in English the common word “get” has over 20 diffe rent meanings.21. Many words have_______.A .several meanings B. one meaningC. few meaningsD. no meaning22. A good dictionary will tell you_____A.more of word‟s pronunciationB. more of the word‟s meaningsC. more of grammarD. more of the word‟s spelling.23.. Phonetic alphabets are used to show ________A. handwritingsB. spellingC. meaningsD. pronunciation24.. How many important things will most dictionaries tell you?A. FourB. FiveC. ThreeD. TwoBEverybody has one of those days when everything goes wrong. This is what happened to Harry.He got up one morning very late because he had forgotten to wind up(设定)his alarm clock. He tried to shave quickly and cut himself .when he got blood all over his clean shirt, he had to find another one. The only other shirt that was clean needed ironing(熨烫),so he ironed it . While he was ironing it, there was a knock at the door. It was the man to read the electricity meter(电表).He showed him where the meter was, said goodbye and found that the iron had burnt a hole in his shirt. So he had to wear the one with the blood on it after all. By this time it was very late, so he decided he couldn‟t go to work by bus. He telephoned for a taxi to take him to work. The taxi arrived and Harry got in and began to read the newspaper.In another part of the town, a man had killed a woman with a knife and was seen to run away in a taxi. When Harry‟s taxi stopped outside his office, a policeman happened to be standing there. He saw th e blood on Harry‟s shirt and took him to the police station. He was kept till 3 o‟clock in the afternoon before the policeman was sure that he was not the man they wanted. When he finally arrived at the office at about four, his boss took a look at him and told him to go away and find another job.25. Harry had_____A. a lucky dayB. an unlucky dayC. a busy dayD. a good day26. Why did Harry wear the shirt with blood on it?A.He had only one shirtB. The only other needed ironingC. The iron had burnt a hole on his clean shirt.D. He cut himself and got blood on his shirt.27. His boss told him to go away and find another job because_____.A.he had been kept by the policeB. there was blood on his shirtC. he was late for workD. he had killed a womanCIn today‟s world, almost everyone knows that air pollution and water pollution are harmful to people‟health. However, not all the persons know that noise is also a kind of pollution, and this is harmful to people‟health, too.People who work and live under noisy conditions usually become deaf today. However, scientists believe that 10 percent of workers in Britain are being deafened by the noise where they work. Many of the workers who print newspapers and books, and who weave(编织)cloth become deaf. Quite a few people living near airports also become deaf. Recently it was discovered that many teenagers in America could hear no better than 65-year-old persons, for these young people like to listen to pop music and pop music is a kind of noise, Besides, noise produced by jet planes or machines will make people ill or even drive them mad.It is said a continuous noise of over 85 decibels(分贝)can cause deafness. Now the governments in many countries have made laws to control noise and make it less than 85 decibels.In China, the government is trying to solve not only air and water pollution problems but also noise problems.28. The text is mainly about ______.A. air pollutionB. water pollutionC. noise pollutionD. world pollution29. According to the text, a continuous noise of _____decibels can make people deaf.A. less than 85B. less than 65C. about 65D. more than 8530. Ten percent of the workers in Britain are being deafened because ______.A. they are working in noisy placesB. they often listen to pop musicC. they are busy in listening to other‟s talkD. they live near airports31. The government of China is trying to solve ______.A. only air pollutionB. only air and water pollutionC. only water pollutionD. air, water and noise pollutionDThe Chinese word “Shanzhai” means a small mountain village, but now it becomes an accepted name for fakes (假货), after “Shanzhai Cell-phones” producedby small workshops in southern China became popular in the mainland market over the past two years.Besides “Shanzhai” electronic products, there are “Shanzhai” movies, “Shanzhai” stars and even a “Shanzhai” Spring Festival Gala (联欢晚会), a copy of the25-year-old tr aditional show presented by CCTV on Chinese Lunar New Year‟s Eve.“Shanzhai” has become a culture of its own, meaning anything that imitates(模仿) something famous.In Chongqing, “Shanzhai” version “Bird‟s Nest” and “Water Cube” woven by farmers with bamboo attract wide attention from tourists. Both are copies of the famous Olympic buildings in Beijing.A literature critic (批评家)said that taking the “Shanzhai” Gala as an example, when the traditional CCTV program becomes less and less attractive to the audience, the “Shanzhai” version appears timely to attract people. “Although it is often connected with poor techniques and operation, …Shanzhai‟ culture meets the psychological needs of common people and could be a comfort to their minds,” he said.To the mainstream (主流的) culture, the rise of “Shanzhai” culture is a challenge and a motivation (动力). People believe different kinds of cultures developing together is a perfect situation and it is for the public to choose.32. The Chinese word “Shanzhai” may have started with ______.A. fake cell-phoneB. electronic productsC. Spring Festival GalaD. Olympic buildings33. According to the passage, “Shanzhai” culture refers to ______.A. the action that a person imitatesfamous peopleB. products with poor techniques and qualityC. anything that imitates something famousD. those similar names to famous brands(品牌)34. We can infer that the mainstream culture ______.A. is held back by “Shanzhai” cultureB. is the challenge of “Shanzhai” cultureC. will be replaced by “Shanzhai” cultureD. may develop faster because of the challenge of “Shanzhai” culture35. Which of the following might be the best title of the passage?A. “Shanzhai” culture will surely disappearB. “Shanzhai” culture takes on life of its ownC. “Shanzhai” culture — the mainstream cultureD. “Shanzhai” culture — the mountain village cultureE第二节根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。

苏教版2017高二(下册)语文月考考试卷(附答案)

苏教版2017高二(下册)语文月考考试卷(附答案)

苏教版2017高二(下册)语文月考考试卷考生注意:1.本试卷分第一部分(阅读题)和第二部分(表达题)两部分,共150分。

考试时间150分钟.2.请将各题答案填在试卷后面的答题卡上。

3.本试卷主要考试内容:高考全部范围。

一、语言表达运用(15分)1.依次填入下列各句横线处的词语恰当..的一组是()(3分)⑴《唐诗三百首》是一部很广、风行海内、老幼皆宜、雅俗共赏的唐诗选集。

⑵墨西哥的“老虎大巡游”是当地居民为了来年风调雨顺而举行的大型活动。

⑶在2015年度江苏省青少年校园足球比赛中,我校初中部男子足球队_______勇夺冠军。

A.流传乞求不孚众望B.留传乞求不负众望C.留传祈求不孚众望D.流传祈求不负众望2. 对下面古诗词中的词语,解读不正确的一项是() (3分) A.熊咆龙吟殷岩泉,栗.深林兮惊层巅栗:使……战栗B.烹羊宰牛且.为乐,会须一饮三百杯且:将要C. 惊风乱飐.芙蓉水,密雨斜侵薜荔墙。

飐:吹动D.谁道闲情..抛掷久,每到春来,惆怅还依旧。

闲情:闲愁3.对下面加点诗句所用的修辞理解有误..的一项是()(3分)A.高堂明镜悲白发,朝如青丝暮成雪.......——夸张B.明月松间照.....,清泉石上流——比拟C.牙璋辞凤阙.....,铁骑绕龙城——借代D.飘飘何所似,天地一沙鸥...........——比喻4. 下面四项文学文化常识的表述,不恰当的一项是() (3分) A.初唐后期,随着“四杰”(王勃、杨炯、卢照邻、骆宾王)登上诗坛,突破了宫廷诗风的格局。

他们的诗歌情思浓郁,气势壮大。

B.在群星璀璨的盛唐诗坛上,李白无疑是最为耀眼的一颗巨星。

韩愈赞美他“笔落惊风雨,诗成泣鬼神”,他自己却爱好“清水出芙蓉,天然去雕饰”。

C. 中国古代有一副对联:“闭月羞花之貌,沉鱼落雁之容”。

“落雁”,说的就是昭君出塞的故事。

杜甫的《咏怀古迹(其三)》被人们誉为“咏昭君之绝唱”D.号称“吴中四子”的张若虚诗风近于齐梁体。

高二下学期第二次月考试题_00001

2017—2018学年度下学期第二次月考高二语文试卷时间:150分钟满分:150分第Ⅰ卷阅读题一、现代文阅读(35分)(一)论述类文本阅读(共3小题, 9分)“文化全球化”是一个伪命题随着市场化、信息化在世界范围持续发展,跨时空的全球性交流互动不断由经济、科技领域走向政治、文化领域。

在这个过程中,全球文化一体化、世界文化趋同化、全球文化同质化等论调甚嚣尘上。

这种观点认为,经济全球化决定政治、文化全球化,世界市场使得消费主义走向全球各个角落,而通信、交通和网络的超地域性加速了不同民族和国家的文化融合,让不同民族和国家的文化最终走向趋同。

事实上,文化既有时代性又有民族性和地域性,其发展离不开自身所处时代和固有文化传统,所谓“文化全球化”是一个具有欺骗性的伪命题。

经济全球化是在不同民族和国家融入世界市场过程中发展起来的。

不同民族和国家经济发展、国家治理、民众生活的实际情况千差万别,在经济全球化进程中的地位和作用不尽相同,因而其参与经济全球化的利益诉求也各不相同。

在经济全球化进程中,为维护自身利益,不同民族和国家根据自身实际情况进行决策,并相应实行不同的国家治理模式。

在这种不同民族和国家基于维护与发展自身利益而形成的世界格局中,连经济都很难趋同,就更谈不上所谓的文化趋同。

文化是一个民族、一个国家的灵魂。

民族文化是长期发展和积累起来的,是一个民族的根脉。

由于人口种族、地理环境和社会生产方式等存在差异,不同民族和国家在历史发展中形成了不同的思维方式、价值取向、风俗习惯,造就了多元文化,而且每一种文化都具有无可替代性和不可复制性。

承认文化差异、实现文化共存,是各个民族和国家实现生存发展、开展国际合作的基础。

否认这种差异,盲目推动趋同,不但会导致人们自我身份认同的弱化甚至消失,而且将导致民族文化衰落和国家衰亡。

西方人鼓吹“文化全球化”,实质是向全世界兜售以美国为代表的西方文化。

一个人如果认同西方文化,就会更加乐于消费其商品、接受其制度规则。

宁夏石嘴山市高二数学下学期第二次月考试卷 文(含解析)-人教版高二全册数学试题

2016-2017学年某某某某高二(下)第二次月考数学试卷(文科)一、选择题(共12小题,每小题5分,满分60分)1.已知集合A={x|x>0},集合B={x|2≤x≤3},则A∩B=()A.[3,+∞)B.[2,3] C.(0,2]∪[3,+∞)D.(0,2]2.已知全集U=R,N={x|x(x+3)<0},M={x|x<﹣1},则图中阴影部分表示的集合是()A.{x|﹣3<x<﹣1} B.{x|﹣3<x<0} C.{x|﹣1≤x<0} D.{x|x<﹣3}3.满足条件M∪{1}={1,2,3}的集合M的个数是()A.1 B.2 C.3 D.44.下列各组函数f(x)与g(x)相同的是()A.f(x)=1,g(x)=x0B.f(x)=x2,g(x)=(x+1)2C.f(x)=x,g(x)=e lnx D.f(x)=|x|,g(x)=5.若函数y=(x+1)(x﹣a)为偶函数,则a=()A.﹣2 B.﹣1 C.1 D.26.已知a,b∈R,则“b≠0”是“复数a+bi是纯虚数”的()A.充分而不必要条件 B.必要而不充分条件C.充分必要条件 D.既不充分也不必要条件7.已知复数z满足(z﹣5)(1﹣i)=1+i,则复数z的共轭复数为()A.5+i B.5﹣i C.﹣5+i D.﹣5﹣i8.下列函数中与函数y=﹣3|x|奇偶性相同且在(﹣∞,0)上单调性也相同的是()A.y=﹣B.y=log2|x| C.y=1﹣x2D.y=x3﹣19.设U=R,A={x|mx2+8mx+21>0},∁U A=∅,则m的取值X围是()A.[0,)B.{0}∪(,+∞)C.(﹣∞,0] D.(﹣∞,0]∪(,+∞)10.A、B为两个非空集合,定义集合A﹣B={x|x∈A且x∉B},若A={﹣2,﹣1,0,1,2},B={x|(x﹣1)(x+2)<0},则A﹣B=()A.{2} B.{1,2} C.{﹣2,1,2} D.{﹣2,﹣1,0}11.曲线C的参数方程为(α为参数),M是曲线C上的动点,若曲线T极坐标方程2ρsinθ+ρcosθ=20,则点M到T的距离的最大值()A.B.C.D.12.已知函数f(x)是定义在R上的偶函数,且对任意的x∈R,都有f(x+2)=f(x).当0≤x≤1时,f(x)=x2.若直线y=x+a与函数y=f(x)的图象在[0,2]内恰有两个不同的公共点,则实数a的值是()A.0 B.0或C.或D.0或二、填空题13.函数f(x)=x2+2(a﹣1)x+2在区间(﹣∞,4]上递减,则实数a的取值X围是.14.已知函数f(x)=若f(1)+f(a)=2,则a的值为.15.已知定义在R上的奇函数f(x),当x<0时,f(x)=2x﹣3.若f(a)=7,实数a的值是.16.给出下列四个命题:①“若x+y≠5,则x≠2或y≠3”是假命题;②已知在△ABC中,“A<B”是“sinA<sinB”成立的充要条件;③若函数,对任意的x1≠x2都有<0,则实数a的取值X围是;④若实数x,y ∈[﹣1,1],则满足x2+y2≥1的概率为.其中正确的命题的序号是(请把正确命题的序号填在横线上).三.解答题17.已知集合A={x|3≤x<7},B={x|2<x<10},求:A∪B,(∁R A)∩B.18.设函数f(x)=ln(2x﹣m)的定义域为集合A,函数g(x)=﹣的定义域为集合B.(Ⅰ)若B⊆A,某某数m的取值X围;(Ⅱ)若A∩B=∅,某某数m的取值X围.19.已知m∈R,命题p:对任意x∈[0,1],不等式2x﹣2≥m2﹣3m 恒成立;命题q:存在x∈[﹣1,1],使得m≤ax 成立.(1)若p为真命题,求m 的取值X围;(2)当a=1 时,若p且q为假,p或q为真,求m的取值X围.20.已知函数f(x)=|x+3|+|2x﹣4|.(1)当x∈[﹣3,3]时,解关于x的不等式f(x)<6;(2)求证:∀t∈R,f(x)≥4﹣2t﹣t2.21.在平面直角坐标系中,以坐标原点为极点,x轴的正半轴为极轴建立极坐标系,已知曲线C的极坐标方程为ρsin2θ=mcosθ(m>0),过点P(﹣2,﹣4)且倾斜角为的直线l 与曲线C相交于A,B两点.(1)写出曲线C的直角坐标方程和直线l的普通方程;(2)若|AP|•|BP|=|BA|2,求m的值.22.已知f(x)是定义在[﹣2,2]上的奇函数,且f(2)=3.若对任意的m,n∈[﹣2,2],m+n≠0,都有>0.(1)判断函数f(x)的单调性,并说明理由;(2)若f(2a﹣1)<f(a2﹣2a+2),某某数a的取值X围;(3)若不等式f(x)≤(5﹣2a)t+1对任意x∈[﹣2,2]和a∈[﹣1,2]都恒成立,某某数t的取值X围.2016-2017学年某某某某三中高二(下)第二次月考数学试卷(文科)参考答案与试题解析一、选择题(共12小题,每小题5分,满分60分)1.已知集合A={x|x>0},集合B={x|2≤x≤3},则A∩B=()A.[3,+∞)B.[2,3] C.(0,2]∪[3,+∞)D.(0,2]【考点】1E:交集及其运算.【分析】利用交集定义直接求解.【解答】解:∵集合A={x|x>0},集合B={x|2≤x≤3},∴A∩B=[2,3].故选:B.2.已知全集U=R,N={x|x(x+3)<0},M={x|x<﹣1},则图中阴影部分表示的集合是()A.{x|﹣3<x<﹣1} B.{x|﹣3<x<0} C.{x|﹣1≤x<0} D.{x|x<﹣3}【考点】1J:Venn图表达集合的关系及运算.【分析】首先化简集合N,然后由Venn图可知阴影部分表示N∩(C U M),即可得出答案.【解答】解:N={x|x(x+3)<0}={x|﹣3<x<0}由图象知,图中阴影部分所表示的集合是N∩(C U M),又M={x|x<﹣1},∴C U M={x|x≥﹣1}∴N∩(C U M)=[﹣1,0)故选:C.3.满足条件M∪{1}={1,2,3}的集合M的个数是()A.1 B.2 C.3 D.4【考点】1D:并集及其运算.【分析】先由M∪{1}={1,2,3}可知集合M必含2和3,是否含1,不确定,则得出两种可能集合,得出答案.【解答】解:满足条件M∪﹛1﹜=﹛1,2,3﹜的集合M,M必须包含元素2,3,所以不同的M集合,其中的区别就是否包含元素1.那么M可能的集合有{2,3}和{1,2,3},故选:B.4.下列各组函数f(x)与g(x)相同的是()A.f(x)=1,g(x)=x0B.f(x)=x2,g(x)=(x+1)2C.f(x)=x,g(x)=e lnx D.f(x)=|x|,g(x)=【考点】32:判断两个函数是否为同一函数.【分析】分别判断两个函数的定义域和对应法则是否一致,否则不是同一函数.【解答】解:A.f(x)的定义域为R,而g(x)的定义域为(﹣∞,0)∪(0,+∞),所以定义域不同,所以函数f(x)与g(x)不相同.B.两个函数的对应法则不相同,所以函数f(x)与g(x)不相同.C.f(x)的定义域为R,而g(x)的定义域为(0,+∞),所以定义域不同,所以C函数f (x)与g(x)不相同.D.f(x)=,两个函数的定义域和对应法则相同,所以函数f(x)与g(x)相同.故选D.5.若函数y=(x+1)(x﹣a)为偶函数,则a=()A.﹣2 B.﹣1 C.1 D.2【考点】3J:偶函数.【分析】本小题主要考查函数的奇偶性的定义:f(x)的定义域为I,∀x∈I都有,f(﹣x)=f(x).根据定义列出方程,即可求解.【解答】解:f(1)=2(1﹣a),f(﹣1)=0∵f(x)是偶函数∴2(1﹣a)=0,∴a=1,故选C.6.已知a,b∈R,则“b≠0”是“复数a+bi是纯虚数”的()A.充分而不必要条件 B.必要而不充分条件C.充分必要条件 D.既不充分也不必要条件【考点】2L:必要条件、充分条件与充要条件的判断.【分析】a,b∈R,复数a+bi是纯虚数⇔,即可判断出结论.【解答】解:a,b∈R,复数a+bi是纯虚数⇔,∴“b≠0”是“复数a+bii是纯虚数”的必要不充分条件.故选:B.7.已知复数z满足(z﹣5)(1﹣i)=1+i,则复数z的共轭复数为()A.5+i B.5﹣i C.﹣5+i D.﹣5﹣i【考点】A5:复数代数形式的乘除运算.【分析】把已知等式变形,利用复数代数形式的乘除运算化简求得z,再由共轭复数的概念得答案.【解答】解:由(z﹣5)(1﹣i)=1+i,得z﹣5=,∴z=5+i,则,故选:B.8.下列函数中与函数y=﹣3|x|奇偶性相同且在(﹣∞,0)上单调性也相同的是()A.y=﹣B.y=log2|x| C.y=1﹣x2D.y=x3﹣1【考点】3E:函数单调性的判断与证明;3K:函数奇偶性的判断.【分析】先判定函数y=﹣3|x|的奇偶性以及在(﹣∞,0)上的单调性,再对选项中的函数进行判断,找出符合条件的函数.【解答】解:∵函数y=﹣3|x|是偶函数,且在(﹣∞,0)上是增函数,∴对于A,y=﹣是奇函数,不满足条件;对于B,y=log2|x|是偶函数,在(﹣∞,0)上是减函数,∴不满足条件;对于C,y=1﹣x2是偶函数,且在(﹣∞,0)上是增函数,∴满足条件;对于D,y=x3﹣1是非奇非偶的函数,∴不满足条件.故选:C.9.设U=R,A={x|mx2+8mx+21>0},∁U A=∅,则m的取值X围是()A.[0,)B.{0}∪(,+∞)C.(﹣∞,0] D.(﹣∞,0]∪(,+∞)【考点】1F:补集及其运算.【分析】由补集的定义可得A=R,即不等式mx2+8mx+21>0恒成立,讨论m=0,m>0,m<0,结合二次函数的图象和性质,解不等式即可得到所求X围.【解答】解:设U=R,A={x|mx2+8mx+21>0},∁U A=∅,可得A=R,即不等式mx2+8mx+21>0恒成立,当m=0时,21>0成立;当m>0,△<0,即64m2﹣84m<0,解得0<m<;当m<0时,不等式不恒成立.综上可得,0≤m<.故选:A.10.A、B为两个非空集合,定义集合A﹣B={x|x∈A且x∉B},若A={﹣2,﹣1,0,1,2},B={x|(x﹣1)(x+2)<0},则A﹣B=()A.{2} B.{1,2} C.{﹣2,1,2} D.{﹣2,﹣1,0}【考点】1H:交、并、补集的混合运算.【分析】先分别求出集合A、B,由此能求出A﹣B.【解答】解:∵A、B为两个非空集合,定义集合A﹣B={x|x∈A且x∉B},A={﹣2,﹣1,0,1,2},B={x|(x﹣1)(x+2)<0}={x|﹣2<x<1},∴A﹣B={﹣2,1,2}.故选:C.11.曲线C的参数方程为(α为参数),M是曲线C上的动点,若曲线T极坐标方程2ρsinθ+ρcosθ=20,则点M到T的距离的最大值()A.B.C.D.【考点】QH:参数方程化成普通方程.【分析】先求出曲线C的普通方程,使用参数坐标求出点M到曲线T的距离,得到关于α的三角函数,利用三角函数的性质求出距离的最值.【解答】解:曲线T的普通方程是:x+2y﹣20=0.点M到曲线T的距离为=,∴sin(α+θ)=﹣1时,点M到T的距离的最大值为2+4,故选B.12.已知函数f(x)是定义在R上的偶函数,且对任意的x∈R,都有f(x+2)=f(x).当0≤x≤1时,f(x)=x2.若直线y=x+a与函数y=f(x)的图象在[0,2]内恰有两个不同的公共点,则实数a的值是()A.0 B.0或C.或D.0或【考点】3P:抽象函数及其应用.【分析】先作出函数f(x)在[0,2]上的图象,再分类讨论,通过数形结合与方程思想的应用即可解决问题.【解答】解:∵f(x)是定义在R上的偶函数,当0≤x≤1时,f(x)=x2,∴当﹣1≤x≤0时,0≤﹣x≤1,f(﹣x)=(﹣x)2=x2=f(x),又f(x+2)=f(x),∴f(x)是周期为2的函数,又直线y=x+a与函数y=f(x)的图象在[0,2]内恰有两个不同的公共点,其图象如下:当a=0时,直线y=x+a变为直线l1,其方程为:y=x,显然,l1与函数y=f(x)的图象在[0,2]内恰有两个不同的公共点;当a≠0时,直线y=x+a与函数y=f(x)的图象在[0,2]内恰有两个不同的公共点,由图可知,直线y=x+a与函数y=f(x)相切,切点的横坐标x0∈[0,1].由得:x2﹣x﹣a=0,由△=1+4a=0得a=﹣,此时,x0=x=∈[0,1].综上所述,a=﹣或0故选D.二、填空题13.函数f(x)=x2+2(a﹣1)x+2在区间(﹣∞,4]上递减,则实数a的取值X围是(﹣∞,﹣3].【考点】3W:二次函数的性质.【分析】f(x)是二次函数,所以对称轴为x=1﹣a,所以要使f(x)在区间(﹣∞,4]上递减,a应满足:4≤1﹣a,解不等式即得a的取值X围.【解答】解:函数f(x)的对称轴为x=1﹣a;∵f(x)在区间(﹣∞,4]上递减;∴4≤1﹣a,a≤﹣3;∴实数a的取值X围是(﹣∞,﹣3].故答案为:(﹣∞,﹣3].14.已知函数f(x)=若f(1)+f(a)=2,则a的值为 4 .【考点】3T:函数的值.【分析】根据函数的表达式先求出f(1),从而求出f(a)的值,求出a即可.【解答】解:f(1)=log21=0,即由f(1)+f(a)=2得f(a)=2﹣f(1)=2﹣0=2,若a>0,则由f(a)=log2a=2,得a=4,若a≤0,则由f(a)=2a=2,得a=1,不成立,综上a=4,故答案为:4.15.已知定义在R上的奇函数f(x),当x<0时,f(x)=2x﹣3.若f(a)=7,实数a的值是 2 .【考点】3L:函数奇偶性的性质.【分析】先求出x>0时的解析式,再利用条件,即可求出a的值.【解答】解:设x>0,则﹣x<0,∴f(x)=﹣f(﹣x)=﹣(﹣2x﹣3)=2x+3,∴a<0,2a﹣3=7,a=5(舍去);a>0,2a+3=7,∴a=2.故答案为:2.16.给出下列四个命题:①“若x+y≠5,则x≠2或y≠3”是假命题;②已知在△ABC中,“A<B”是“sinA<sinB”成立的充要条件;③若函数,对任意的x1≠x2都有<0,则实数a的取值X围是;④若实数x,y ∈[﹣1,1],则满足x2+y2≥1的概率为.其中正确的命题的序号是②④(请把正确命题的序号填在横线上).【考点】2K:命题的真假判断与应用;21:四种命题.【分析】①根据逆否命题的等价性进行转化证明即可.②根据大角对大边以及正弦定理进行证明.③根据分段函数单调性的性质进行证明.④根据几何概型的概率公式进行证明.【解答】解:①“若x+y≠5,则x≠2或y≠3”的等价命题为x=2且y=3时,x+y=5,则等价命题为真命题,则原命题为真命题,故①错误,②已知在△ABC中,“A<B”等价为a<b,根据正弦定理得“sinA<sinB”成立,即,“A <B”是“sinA<sinB”成立的充要条件;故②正确,③若对任意的x1≠x2都有<0,则函数f(x)为减函数,则满足,即,得≤a<,故③错误,④由题意可得,的区域为边长为2的正方形,面积为4,∵x2+y2≥1的区域是圆的外面的阴影区域,其面积S=4﹣π,∴在区间[﹣1,1]上任取两个实数x,y,则满足x2+y2≥1的概率为=.故④正确.故正确的答案是②④,故答案为:②④三.解答题17.已知集合A={x|3≤x<7},B={x|2<x<10},求:A∪B,(∁R A)∩B.【考点】1F:补集及其运算;1D:并集及其运算;1E:交集及其运算.【分析】根据并集的定义,由集合A={x|3≤x<7},B={x|2<x<10},求出A与B的并集即可;先根据全集R和集合A求出集合A的补集,然后求出A补集与B的交集即可.【解答】解:由集合A={x|3≤x<7},B={x|2<x<10},把两集合表示在数轴上如图所示:得到A∪B={x|2<x<10};根据全集为R,得到C R A={x|x<3或x≥7};则(C R A)∩B={x|2<x<3或7≤x<10}.18.设函数f(x)=ln(2x﹣m)的定义域为集合A,函数g(x)=﹣的定义域为集合B.(Ⅰ)若B⊆A,某某数m的取值X围;(Ⅱ)若A∩B=∅,某某数m的取值X围.【考点】33:函数的定义域及其求法;1E:交集及其运算.【分析】(Ⅰ)分别求出集合A、B,根据B⊆A,求出m的X围即可;(Ⅱ)根据A∩B=∅,得到关于m的不等式,求出m的X围即可.【解答】解:由题意得:A={x|x>},B={x|1<x≤3},(Ⅰ)若B⊆A,则≤1,即m≤2,故实数m的X围是(﹣∞,2];(Ⅱ)若A∩B=∅,则≥3,故实数m的X围是[6,+∞).19.已知m∈R,命题p:对任意x∈[0,1],不等式2x﹣2≥m2﹣3m 恒成立;命题q:存在x∈[﹣1,1],使得m≤ax 成立.(1)若p为真命题,求m 的取值X围;(2)当a=1 时,若p且q为假,p或q为真,求m的取值X围.【考点】2E:复合命题的真假.【分析】(1)对任意x∈[0,1],不等式2x﹣2≥m2﹣3m 恒成立,可得﹣2≥m2﹣3m,解得mX围.(2)a=1时,存在x∈[﹣1,1],使得m≤ax 成立.可得m≤1.由p且q为假,p或q为真,可得p与q必然一真一假,即可得出.【解答】解:(1)对任意x∈[0,1],不等式2x﹣2≥m2﹣3m 恒成立,∴﹣2≥m2﹣3m,解得1≤m≤2.(2)a=1时,存在x∈[﹣1,1],使得m≤ax 成立.∴m≤1.∵p且q为假,p或q为真,∴p与q必然一真一假,∴或,解得1<m≤2或m<1.∴m的取值X围是(﹣∞,1)∪(1,2].20.已知函数f(x)=|x+3|+|2x﹣4|.(1)当x∈[﹣3,3]时,解关于x的不等式f(x)<6;(2)求证:∀t∈R,f(x)≥4﹣2t﹣t2.【考点】R5:绝对值不等式的解法;R4:绝对值三角不等式.【分析】(1)通过讨论a的X围,求出不等式的解集即可;(2)求出f(x)的分段函数的形式,求出f(x)的最小值,得到关于t的不等式,证出即可.【解答】解:(1)当﹣3≤x≤2时,f(x)=x+3﹣(2x﹣4)=﹣x+7,故原不等式可化为﹣x+7<6,解得:x>1,故1<x≤2;当2<x≤3时,f(x)=x+3+(2x﹣4)=3x﹣1,故原不等式可化为3x﹣1<6,解得;综上,可得原不等式的解集为.(2)证明:,由图象,可知f(x)≥5,又因为4﹣2t﹣t2=﹣(t+1)2+5≤5,所以f(x)≥4﹣2t﹣t2.21.在平面直角坐标系中,以坐标原点为极点,x轴的正半轴为极轴建立极坐标系,已知曲线C的极坐标方程为ρsin2θ=mcosθ(m>0),过点P(﹣2,﹣4)且倾斜角为的直线l 与曲线C相交于A,B两点.(1)写出曲线C的直角坐标方程和直线l的普通方程;(2)若|AP|•|BP|=|BA|2,求m的值.【考点】Q4:简单曲线的极坐标方程.【分析】(1)曲线C的极坐标方程为ρsin2θ=mcosθ(m>0),即ρ2sin2θ=mρcosθ(m >0),利用互化公式可得直角坐标方程.过点P(﹣2,﹣4)且倾斜角为的直线l参数方程为:(t为参数).相减消去参数化为普通方程.(2)把直线l的方程代入曲线C的方程为:t2﹣(m+8)t+4(m+8)=0.由于|AP|•|BP|=|BA|2,可得|t1•t2|=,化为:5t1•t2=,利用根与系数的关系即可得出.【解答】解:(1)曲线C的极坐标方程为ρsin2θ=mcosθ(m>0),即ρ2sin2θ=mρcosθ(m>0),可得直角坐标方程:y2=mx(m>0).过点P(﹣2,﹣4)且倾斜角为的直线l参数方程为:(t为参数).消去参数化为普通方程:y=x﹣2.(2)把直线l的方程代入曲线C的方程为:t2﹣(m+8)t+4(m+8)=0.则t1+t2=(m+8),t1•t2=4(m+8).∵|AP|•|BP|=|BA|2,∴|t1•t2|=,化为:5t1•t2=,∴20(m+8)=2(m+8)2,m>0,解得m=2.22.已知f(x)是定义在[﹣2,2]上的奇函数,且f(2)=3.若对任意的m,n∈[﹣2,2],m+n≠0,都有>0.(1)判断函数f(x)的单调性,并说明理由;(2)若f(2a﹣1)<f(a2﹣2a+2),某某数a的取值X围;(3)若不等式f(x)≤(5﹣2a)t+1对任意x∈[﹣2,2]和a∈[﹣1,2]都恒成立,某某数t的取值X围.【考点】3R:函数恒成立问题;3E:函数单调性的判断与证明;3F:函数单调性的性质.【分析】(1)设任意x1,x2,满足﹣2≤x1<x2≤2,利用函数单调性的定义证明;(2)由(1)知,f(2a﹣1)<f(a2﹣2a+2)可化为﹣2≤2a﹣1)<a2﹣2a+2≤2,从而解得.(3)不等式f(x)≤(5﹣2a)t+1对任意x∈[﹣2,2]和a∈[﹣1,2]都恒成立,f max(x)≤(5﹣2a)t+1对任意的a∈[﹣1,2]都恒成立,令g(a)=2ta﹣5t+2,a∈[﹣1,2],从而求t.【解答】解:(1)设任意x1,x2,满足﹣2≤x1<x2≤2,由题意可得f(x1)﹣f(x2)=f(x1)+f(﹣x2)=(x1﹣x2)<0,即f(x1)<f(x2),∴f(x)在定义域[﹣2,2]上是增函数.(2)由(1)知,f(2a﹣1)<f(a2﹣2a+2)可化为﹣2≤2a﹣1)<a2﹣2a+2≤2,解得0≤a<1,∴a的取值X围为[0,1).(3)由(1)知,不等式f(x)≤(5﹣2a)t+1对任意x∈[﹣2,2]和a∈[﹣1,2]都恒成立,f max(x)≤(5﹣2a)t+1对任意的a∈[﹣1,2]都恒成立,∴3≤(5﹣2a)t+1恒成立,即2ta﹣5t+2≤0对任意的a∈[﹣1,2]都恒成立,令g(a)=2ta﹣5t+2,a∈[﹣1,2],则只需,解得t≥2,∴t的取值X围是[2,+∞).。

—17学年下学期高二第二次月考语文试题(附答案)

第二次月考语文试卷命题人:杨辉命题人:梅学斌满分150分,考试时间150分钟本试卷分第Ⅰ卷(阅读读题)和第Ⅱ卷(表达题)两部分。

满分150分,考试时间150分钟。

第Ⅰ卷阅读题一、现代文阅读(35分)(一)论述类文本阅读(9分,每小题3分)阅读下面文字,完成1—3题宋朝是一个重视传统文化的朝代,每一个节日都被宋人发挥到极致。

清明节是当时一个非常重要的节日。

人们扫墓、颁新火、踏青、荡秋千、蹴鞠、斗鸡、放风筝,各种民俗活动内容丰富、形式多样,寄托了人们美好的愿望。

宋朝的清明节的最大亮点应该是蹴鞠。

《水浒传》中写高俅球技高超,因陪侍宋徽宗踢球,被提拔当了殿前都指挥使。

诗圣杜甫《清明》诗中说到,“十年蹴鞠将雏远,万里秋千习俗同”,诗人陆游《感旧末章盖思有以自广》诗中有“路入梁州似掌平,秋千蹴鞠趁清明”的诗句。

这说明从唐朝到宋朝清明节都有踢球娱乐的习俗。

蹴鞠在宋朝获得了极大的发展。

上层踢球已经成为时尚,一幅《宋太祖蹴鞠图》,描绘的就是当时皇帝和大臣踢球的情景。

宋代社会上还有专门靠踢球技艺维持生活的足球艺人。

宋代的足球有用球门的间接比赛和不用球门的“白打”两种方式,但书上讲的大多都是白打踢法。

所谓“脚头十万踢,解数百千般”,就是指踢球花样动作和由几个花样组成的成套动作,指用头、肩、背、胸、膝、腿、脚等一套完整的踢技,使“球终日不坠”。

由此看来,宋朝的足球,已由射门比准向灵巧和控制球技术方面发展。

为了维护自身利益和发扬互助精神,宋朝的踢球艺人还组织了自己的团体,叫做“齐云社”,又称“圆社”。

《水浒传》中写到宋徽宗也是“齐云社”的成员。

这是专门的蹴鞠组织,专事负责蹴鞠比赛的组织和宣传推广。

宋朝清明节还有一个习俗就是市民携带炊饼出游踏青。

宋朝民间,习惯把无馅的食品称为饼,用火烤得叫烧饼,蒸的叫蒸饼,面条叫汤饼,油炸的叫油饼。

宋仁宗赵祯做皇帝之后,因为宋仁宗的名字叫赵祯,而蒸饼的“蒸”字和赵祯的“祯”字发音相似,那时说话写字都讲究避皇帝或长辈的名讳,所以,蒸饼就改称为“炊饼”。

河北曲周县2016-2017学年高二语文下学期第一次月考试题(扫描版)

河北省曲周县2016-2017学年高二语文下学期第一次月考试题(扫描版)高二月考语文参考答案1、D(A、“化解了龌龊与清洁的冲突”错。

B、“全部内容”绝对化。

C、“还原宇宙的本来面目”无中生有,“我们”应是“画家”)2、D(“反衬”错,原文是“在静寒氛围中展现生命的跃迁。

以静观动,动静相宜,可以说是中国艺术的通则,它一般是在静寒中表现生趣,静寒为盎然的生机跃动提供了一个背景”)3、A(B、目的是论证“以静观动,动静相宜,可以说是中国艺术的通则,它一般是在静寒中表现生趣,静寒为盎然的生机跃动提供了一个背景”。

C、应是融合成宁静空茫的境界,“听觉艺术”属无中生有。

D、原文是“不介入社会的复杂文化活动”,而非“隔绝俗世”)4.C E(A项守旧并不合理;B项急切粗鲁是欣喜的缘故,自私粗俗无教养不合情理;D项“小说语言冷峻、客观”不准确,本文语言细腻雅致,有情感热度。

)5.①作为线索,统摄全篇。

整篇小说围绕“家”展开,以租房的所见所闻所为来推动情节发展。

②交代主要内容。

小说以夫妻俩回忆过去的“家”,观察现在的“家”,谈论自己心中的“家”为主要内容。

③揭示小说主旨。

小说围绕“家”行文,到结尾笔锋一转,点出“这不是你的家”,更显对家和归宿的渴望之强烈。

(共计4分,答对两条即可给4分,言之成理可酌情给分。

)6.能找到。

①从主题上看,有爱的地方就有“家”。

“家”是避风的港湾,夫妻俩真诚爱护彼此,简陋的旧房子也因为这爱而温暖,双方互相依偎相濡以沫,对家有着强烈的渴求,有这样的情感基础,两人相依为命的居所就是“家”。

②从人物来看,两人顽强坚韧乐观。

虽然贫困漂泊,居无定所,但言谈举止纯真美好,比明眼人更懂得生活的情趣;且渴望归宿,并顽强坚韧、辛勤努力去奔波寻觅,一定会有自己的“家”。

③从情节上看,二人融入新生活迅速,适应性强。

进新屋时妻子称离开的租处为“现在的房子”,还不时抱怨和旧房子水乳交融,格外亲切;可是刚进新房间不久,两人就欣喜于陈设的完备,兴奋自在地弹奏起钢琴来,在床边并肩倚坐弹跳,吹口哨,稳健迈步行动自如,适应很快,他们定能找到自己的“家”。

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2017七宝高三3月测试试卷 II. Grammar and Vocabulary Section A Directions After reading the passages below fill in the blanks to make the passages coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word for. The other blanks, use one word that best fits each blank. In the countries of south and Southeast Asia the elephant has been an intimate part of the culture, economy and religion, and nowhere more so than in Thailand. Unlike its African cousin, the Asian elephant is easily domesticated (驯化). The rare so-called white elephants have actually lent the authority of kingship to its rulers and 21_______ the 1920s the national flag was a white elephant on a red background.To the early Western visitors the country's romantic name was ―Land of the White Elephant‖. Today, however, the story is very different. Out of work and out of land, the Thai elephant struggles 22______ survival in a nation that no longer needs it. The elephant has found itself more or less abandoned by previous owners who have moved on to a different economic world and a westernized society. And 23________ the elephant's problems began many years ago, now it rates a very low national priority. How24________ this reversal from national icon (圣像) to neglected animal came about is a tale of worsening environmental and the changing lives of the Thais 25_______. According to Richard Lair, Thailand’s experts on the Asian elephant and author of the report Gone Astray, at the turn of the century there 26________ well have beenas many as 100,000 domestic elephants in the country. In the north of Thailand alone it was estimated that more than 20,000 elephants were employed in transport, 1,000 of them alone on the road between the cities of Chiang Mai and Chiang Saen. This was at a time when 90 per cent of Thailand was still forest—a habitat (栖息地) 27________ not only supported the animals but also made them necessary to carry goods and people. Nothing ploughs through dense forest 28______(well) than a massive but sure-footed elephant. By 1950 the elephant population 29________(drop) to a still substantial 13,397, but today there are probably nomore than 3,800, with another 1,350 30________(wander) free in the national parks. But now, Thailand’s forest coversonly 20 per cent of the land. This deforestation (采伐森林) is the central point of the elephant's difficult situation, for it has effectively put the animals out of work. This century, as the road network grew, so the elephant's role as a beast of burden declined.

Section B Directions Complete the following passage by using the words in the box. Each word can only be used once. Note that there is one word more than you need.

Do you take part in office gossip? I don't like to think of myself as a gossip, but I have to admit I often do it. In my turbulent industry, I 31________________ my behavior--perhaps wrongly--by reasoning that gossip helps me get information and figure out what is going on. Amid a rise in office gossip, researchers are disagreeing over whether it is 32_____________ good or bad. Some defend it as a way of building 33_________ among people and sharing essential information. But others hold that office gossip can be savage and 34_____________, as the New York Times reports. At one company, PrintingForLess.com, which has a strict no-gossip policy, gossiping about colleagues can become a 35___________offense. In one case analyzed in a ___36_______ journal, middle school teachers' gossip about their principal became so poisonous that the principal retaliated, many teachers fled the school and students' test scores declined. In this case, gossip ___________ to "a form of warfare that brought everyone down." On the other hand, less malignant gossip that stops short of repeating lies or breaching confidences can serve as a source of understanding. Gossip helps us 38_____________ the motivations of other people, and enables those low on the food chain, in particular, to understand how power is used in their organizations, says this New York Times article. It is relaxing, it brings people together, and as a 39_______________ it beats gambling, drinking or doing drugs, this reasoning holds.

A.destructive B.transfers C.equaled D.justify AB.bonds AC.scholarly AD.pastime BC.amounted BD.analyze CD.firing ABC.fundamentally

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