2018年上海浦东新区高三二模试卷(附问题详解)
2018年浦东新区高三二模语文试卷（时间150分钟，满分150分）2018.4一积累运用（10分）1.按要求填空（5分）（1），幽咽泉流冰下难。
（白居易《》）（2）香远益清，亭亭净植，。
（周敦颐《爱莲说》）（3）苏轼在《江城子》中有“相顾无言，惟有泪千行”的诗句，在柳永的《雨霖铃》中意境与之相似的一句是“，”。
2.按要求选择。
（5分）（1）今年南汇桃花节，小刘去踏青觅胜，欲留影配诗，下列诗句和赏花场景不匹配的一项是（）。
（2分）A.满树和娇烂漫红，万枝丹彩灼春融。
B.桃花一簇开无主，可爱深红爱浅红。
C.花开不并百花丛，独立疏篱趣无穷。
D.一树繁英夺眼红，开时先合占东风。
填入下面语段空白处的句子，最恰当的一项是（）。
（3分）文明是史，未进入文明之前是史前时期，未进入文明的文化是史前文化，未有字，焉有史？文明的标志当然是文字，，中国人大可底气十足地说，中华文明至少肇始于三千年前，其独一无二的持久性正有汉字之功。
A．而文明预示着文字有走向伟大的资本与长寿的禀赋B．而文字预示着文明有走向伟大的资本与长寿的禀赋C．而文明预示着文字有走向长寿与伟大的资本和禀赋D．而文字预示着文明有长寿的资本与走向伟大的禀赋二阅读（70分）（一）阅读下文，完成3-7题。
（16分）导演的限制与自由①导演的地位和作用问题，是近代戏剧史上一个争论不休的话题。
主流派认为，剧本是舞台艺术的基础，导演则是剧本的诠释者和体现者。
导演创作，可以发展或充实刷本，但却不能违背原作的立意与风格。
从俄国的斯坦尼斯拉夫斯基到美国的贝拉斯科、中国的焦菊隐等，都持这种观点。
②也有人认为，导演是现代戏剧的核心，他可以随意篡改或解构剧本，甚至干脆不要据本，正如他有权设计布景，有权摆布演员，有权使用音响灯光一样。
一些先锋派导演或理论家多持这种观点。
如果把这种“导演中心”论限制在演出的范围内，还是有道理的，作为某种创新实验，更是无可厚非，但要推行于全部戏刷活动，恐怕就行不通了。
③导演的职责是排戏。
作为整个创作群体的一员，导演的基本职责是把剧本搬到舞台上去，使文学形象转化为可视可听可感的舞台形象。
而剧本，特别是那些久经考验的经典剧目，往往包含着丰厚的文化内容，所以一个导演必须具备广博的社会、历史、人文知识，方能深入发握原作的思想内涵，充分展现其独特的艺术风貌。
其次，导演是演出的组织者。
为了把各种艺术成分组织起来，融为一体，创造出和谐、统一的舞台形象，导演必须谙悉舞台艺术的方方面面，具有较强的组织领导能力。
第三，导演还承担着培养演员的责任。
演员是舞台艺术的中心，当然也是导演的主要表现手段。
选好演员，往往戏就成功了一半。
好的导演，在排戏当中，能以各种方式启发演员，激起演员的创作欲望，从而大大提高演员的艺术表现能力。
现代剧坛上的许多优秀演员，都是在一些著名导流的培养下，迅速成长起来的。
④导演艺术是以刷本为基出，以完整和谐的舞台艺术为表现形式的二度创作。
导演创作不仅受到剧本的制约，而且必须在舞台艺术和技术条件所能达到的范围内进行。
可以说，导演是戏剧艺术里限制最多，也最富挑战性的工作，是名副其实的戴着脚镣跳舞。
所以，《艺术形态学》的作者莫·卡冈说，“导演是最少独立性的艺术，因为它是其他所有艺术的上层建筑，而它控制其他这些艺术，使它们互相‘迁就”，最终创造出以它们的综合为基础的新的艺术。
”⑤有时候，因为演员、舞台或社会环境的限制，即使是名家执导，也会把一些优秀剧作给弄得支离破碎，面日全非。
欧阳予倩先生曾先后三次执导曹禺的《日出》，每次都不一样，第一次（1937年2月）是个业余剧团，因为找不到扮演翠喜的演员和做效果的人，不得不删去整个第三幕，结果让专程赴上海观摩演出的曹禺很不高兴，认为它没有体现出原作的立意来。
4个月后，欧阳予倩为中国旅行剧团导演《日出》，就保留了第三幕。
为什么呢？因为该团是中国第一个职业话剧团，聚焦了不少人才，可供导演支使，为其二度创作提供了便利条件，演出大获成功。
特别是扮演翠喜的青年演员王荔，获得观众的一致好评，欧阳山尊总管舞美灯光效果，做得也很有层次情调，台口处那面背对观众的空架衣镜和全刷繁复的音响效果，更给观众留下了深刻的印象，曹禺看后非常满意。
演员、布景、效果等舞台艺术条件对导演的制约，由此可见一斑。
⑥然而，从另一方面看，导演又是戏剧艺术里表现手段最多，自由度最大的工作。
对剧本，他有选择和修改的权力，对演员，他有遴选和调教的职责，对舞台艺术的方方面面，他有根据整体构思和排演的需要，随时进行调整的自由。
限制与自由都是相对的。
譬如有些剧本，本身平平，乏善可陈，但是到了优秀导演手里，却可以化腐朽为神奇，排出令观众耳目一新的好戏来。
郭沫若的《蔡文姬》（1959），又是写自己，又是替曹操翻案，主题分裂，合词啰嗦，就连作家本人都说它“很不成热”。
但幸运的是遇到了焦菊隐这样的好导演。
焦弱化了作者对普操文治式功的阿谀奉承，突出了蔡文姬的人生悲剧，刻意追求舞台艺术的民族化，使之成为那个朝代最富创意，也最受欢迎的剧目之一。
⑦剧本是演出的基础但不等于演出，演员、舞关、灯光、音响都很重要，但分开了难以为戏，把他们组合在一起的是导演，导演才是舞台演出的真正作者。
（选自《戏剧艺术十五讲》，有删改）3.第④段中的“戴着脚镣跳舞”形象地描述了导演在戏剧艺术中处于和的矛盾状态。
（2分）4.下列对本文分析理解正确的一项是（）。
（3分）A.剧本是“一剧之本”，剧本的诠释者和体现者是导演。
B.作者反对“剧本中心”论，而主张“导演中心”论。
C.导演在戏剧艺术中受限于剧本、舞台艺术和技术条件。
D.焦菊隐成功改编《蔡文姬》说明剧本对导演限制很小。
5.下列对文中“导演与剧组关系图”各种关系理解正确的一项是（）。
（3分）A.导演是戏剧演出的组织者，直接管控监制。
B.舞美包括布景、道具、灯光和效果四要素。
C.舞美的质量由技术指导负责，与导演无关。
D.舞蹈是由导演做领舞和舞蹈演员共同完成。
6.第③段提到“剧本，特别是那些久经考验的经典剧目，往往包含着丰厚的文化内容”请以高中教材中学过的剧目为例进行分析。
（4分）7.第⑤段论证了演员、布景、效果等舞台艺术条件对导演的制约作用，你认为论证是否充分?请做出判断并说明理由。
（4分）（二）阅读下文，完成8-11题。
（15分）西子湾畔访余光中喻大翔①余光中这辈子善结“海缘”。
○2《高楼对海》里有很多海，西子湾的海，高楼上的海，从窗口和露台望去的海。
诗集“取名《高楼对海》，是纪念这些作品都是在对海的楼窗下写的，波光在望，潮声在耳，所以灵思不绝。
”（《高楼对海·后记》）○3不管多远，我一定要去看看那窗，看看那楼，看看那海。
○42013年七月中上旬，我喜欢的夏天到了。
学校暑假，我随同济大学裴钢校长一行访问台湾的几所大学。
公务告一段落，启程往香港前，获准有两天空闲，可以自行活动。
十日一大早，我毫不犹豫地挎起背包，到台北捷运站买票、登车，到海的另一端——高雄去拜访余光中先生。
○5火车停靠左营，立等片刻，余先生先生就出现在眼前。
车过一条长路，车过一些窄路，车到海边，再上山。
上山时，中大的保安略略弯腰，对他笑了笑，车子就风一样的飙起，那流畅与自由绝不是一般八十五岁老人能够想象的。
○6余教授的办公室在文学院大楼四楼，编号“534”，下面是余光中三字的印刷体。
刚好这时来了一个女学生，她帮余老师和我拍了一张合影。
门的正中是一张菱形的浅底色的抽象画，看过去，画的左下斜边是梵高的自画像，怀疑的眼神和愤怒的黄胡子；余先生就站在他旁边，浅笑着侧耳倾听梵高的声音。
画的左上斜边是红底洒金的一个“福”字，有些像毛笔，又有些像炭笔，现在看来，很像是余先生亲笔书写的。
○7进屋放下背包，余先生第一个召唤就是去看海！不是步行至沙滩，也不是坐船远行，而是到楼西头的露台上眺海。
他办公室的一排窗口朝西南，那里是一座山和一座土红色的楼房，只在远处的右前方，有一线斜斜的绸蓝。
我记得，从他门口到西边的露台，中间只隔着一间办公室。
步上露台，世界大开！海水一望无际，船影艘艘而点点；近海栈桥纵横，浮标漂浪；更近处和左侧，当然是海堤、山坡和住宅，是一部大蓝大绿大红的音色交响曲。
○8余先生指着远方轻轻地说：海峡的对面就是大陆，我已经眺望快三十年了！然后沉默，再然后，还是沉默。
因为这一泓海水，因为六十多年日日夜夜的风波，它将诗人的情思拉得又深又细又长。
○9在诗翁的办公室谈了不少话题：比如永春余光中文学馆的建设，他拿出了一张设计图给我看；比如大陆一些选本的删存与得失；他还谈到了关于他的评论和传记作品……○10余先生的办公桌上堆放着很多书刊，有一册香港的《明报月刊》好像刚刚合上。
沙发旁的一张小几上放着一沓学生的英文作业，余先生在整理物件的当儿，我翻了好几页，每一页都有先生勾画的笔迹与文字。
最后一页原作英文只有三行，其他都是先生的手迹。
在他的允许下，我拍下了这一页，以作意外的纪念。
在给学生“90”分的嘉赏之下，先生花了15行对论文作了评价。
此外，先生还引申到浪漫主义诗人雪莱，谈雪莱创作《古舟子咏》的直觉天赋，让学生触类旁通。
○11离开余先生办公室之前，他签赠了三本新著给我。
○12在海边安静而优雅的大学饭店用了午餐，又在椰叶搭起的凉棚下拍了照片，穿过几幢教学楼长长的走廊，我们进了中山大学图书馆。
下一个活动，就是参访设在该馆的“余光中特藏室”。
○13这间特藏室面积不大，可是，却是中山大学和他的共同心血，它以余先生亲自设计并以收藏他从香港返台后的全部作品、图片、画像、手稿、书法、影像等为特色。
由导览员开了个头、介绍了各种展览形式，并播放了他吟唱的《念奴娇·赤壁怀古》、一位女指挥家指挥合唱的《乡愁四韵》……，其间，他特别拿起人民日报出版社《余光中对话集——凡我在处，就是中国》说：里面有郭虹的一篇，超过一万字，是她把问题写好寄给我，我书面回应的，比较可靠。
○14那时大约下午三点多了，余先生这时也有些疲惫，说我们该返回办公室拿东西，然后到他家里晚餐歇息了。
○15再进办公室，余先生斜躺在沙发上，拿一小瓶眼药水滴眼睛。
片刻，余先生谈兴再起，与我谈起了乡土文学的论争；谈这些年在华人社会的来去对他创作的影响；还有就是心理上的误解、困惑与忧闷……○16在当天最后一班高铁上，我一边啃面包，一边电影似地回忆起从早上开始的经过……（原刊于2018年1月9日《文汇报笔会》，有删改）8.本文叙述采访余光中的经历，为什么先从“海”写起？（4分）9.赏析第○8段划线句的表达效果。
（3分）10.第○13段中对“余光中特藏室”的介绍有何用意？（4分）11.从内容与形式两方面，赏析最后一段。
（4分）（三）阅读下面的词，完成第12-14题。
（8分）青杏儿·风雨替花愁[ 元] 赵秉文风雨替花愁。
上海市浦东新区2018届高三生命科学下学期教学质量检测二模试题含解析
上海市浦东新区2018届高三下学期教学质量检测（二模）生命科学一、选择题1. 图表示某化学键的结构式，则含有该键的化合物可能是OA. ATPB. 抗体C. DNAD. 磷脂【答案】B【解析】据图分析，图示为肽键的结构式，是了解氨基酸的化学键。
ATP的三磷酸腺苷的简称，不含肽键，A错误；抗体的化学本质是蛋白质，含有肽键，B正确；DNA分子中含有磷酸二脂键，但是不含肽键，C错误；磷脂分子中不含肽键，D错误。
2. 大豆营养成分鉴定实验结果见表，据表可知大豆中含有的主要营养成分有试剂碘液班氏试剂双缩脲试剂苏丹川颜色变化棕黄色浅蓝紫色橘红色A. 蛋白质B. 还原性糖C.蛋白质、脂肪D. 还原性糖、脂肪【答案】C【解析】碘液的颜色是棕黄色，淀粉遇碘变蓝，说明大豆的成分中没有淀粉；班氏试剂与还原糖产生砖红色沉淀，而班氏试剂本身的颜色是蓝色的，说明大豆的营养成分中没有还原糖；双缩脲试剂检测蛋白质，出现紫色；苏丹川检测脂肪，出现橘黄色，说明大豆成分中含有蛋白质和脂肪，故选Co3. 如图所示的细胞最可能是A. 颤藻细胞B. 神经细胞C. 浆细胞D. 菠菜叶肉细胞【答案】C【解析】据图分析，图示细胞含有细胞核和多种细胞器，说明该细胞是真核细胞，且该细胞中内质网和高尔基体较多。
颤藻是一种蓝藻，属于原核细胞，A错误；神经细胞具有很多突起，与图示细胞不符，B错误；浆细胞属于真核细胞，能够分泌抗体，因此具有大量的内质网和高尔基体,C正确；菠菜叶肉细胞应该含有叶绿体、不含中心体，D错误。
4.下列不属于RNA功能的是A.生物催化剂B.染色体的组成成分C.翻译的模板D.某些病毒的遗传物质【答案】B【解析】少数酶的化学本质是RNA因此RNA可以作为生物催化剂，A正确；染色体的组成成分为DN A和蛋白质，没有RNA B错误；翻译的模板是mRNAC正确；大多数病毒的遗传物质是DNA少数病毒的遗传物质是RNA D正确。
5. 图是低倍镜下的蚕豆叶下表皮保卫细胞长度测量的局部视野。
【全国区级联考word】上海市浦东新区2018届高三下学期教学质量检测(二模)英语试题(有答案)
上海市浦东新区2018届高三下学期教学质量检测英语试卷I. Listening ComprehensionSection A —10分Directions:In Section A. you will hear fen short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper and decide which one is the best answer to the question you have heard.1. A. Challenges. B. Hobbies. C. Jobs. D. Experiences.2. A. Interesting. B. Boring. C. Difficult. D. Amazing.3. A. Watching TV and videos.B. Replacing videos with TV.C. Parents’ involvement.D. Having baby sitters.4. A. A policeman. B. An accountant. C. A salesman. D. A bank teller.5. A. 7:40. B. 7:15. C. 7:20. D. 7:45.6. A. He will get someone to do it.B. She should do it herself.C. They don’t have to do it.D. He will clean the desk right away.7. A. By bus. B. By subway. C. By taxi. D. By car.8. A. He is not a good mechanic.B. He doesn’t keep his word.C. He spends his spare time doing repairs.D. He is always ready to offer help to others.9. A. She has been having a sad day.B. She needs to take a day off.C. She wants to play basketball, too.D. She has been annoyed by the noise.10. A. The man isn’t sure about the rehearsal.B. It’s better for the woman to wear a costume.C. The woman would regret it if she wore a costume.D. It wouldn’t make any difference if the woman did it.Section B—15 分Directions: In Section B, you will hear two short passages and one longer conversation, and you will be asked several questions on each of the passages and the conversation. The passages and the conversation will be read twice, but the questions will be spoken only once. When you hear q question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. He qualified as a teacher.B. He became a student.C. He became a government researcher.D. He conducted a research on Zimbabwe.12. A. Children’s minds are not used to the full.B. It is a great drain on children’s time and energy.C. It highlights the flexibility of children’s minds.D. It prevents children from seeking answers by themselves.13. A. To teach people to understand the worldB. To instruct people how to raise good questions.C. To encourage people to study as they get older.D. To inform people of problems in foreign countries.Questions 14 through 16 are based on the following passage.14. A. To serve as a time killer.B. To cultivate people’s reading killsC. To promote the sales of some books.D. To encourage people to take public transportation15. A. The stories are the short edition of some website articles.B. Users can choose the length and type of the stories.C. The stories are obtained by simply pressing a button.D. Users don’t need to pay for the short stories.16. A. From the boring travel experience.B. From the love for short stories.C. From the positive feedbackD. From the snack vending machine.Qusions17 through 20 are based on the following conversation.17. A. 5. B. 7. C. 8. D. 10.18. A. Because his friends don’t get off work till 5 p.m.B. Because there will be more friends to go to the cinema on Friday.C. Because the film will be more popular than the Wednesday’s.D. Because there are not enough tickets left for the 9 p.m. showing.19. A. Paying a deposit.B. E-ordering in advance.C. Paying right away.D. Collecting tickets one day ahead.20. A. The film. B. The date C. The seating. D. The viewers.II. Grammar and vocabularySection A—10分Directions:After reading the passage below, fill in the blanks to make the passage 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.Pumas are large, cat-like animals which are found in America. When reports came into the London Zoo that a wild puma 21(spot) forty miles south of London, they were not taken seriously. However, as the evidence began to accumulate, experts decided to investigate.The hunt 22 the puma began in a small village where a woman 23 (pick) blackbe rries saw “a large cat” only five yards away from her. It immediately ran away when she saw it, and experts confirmed that a puma will not attack a human being 24 it is cornered. The search proved difficult, for the puma was often observed at one place in the morning and at 25 place twenty miles away in the evening. 26 it went, it left behind it a trail of dead deer and small animals like rabbits. Several people complained of cat-like noises at night and a businessman on a 27(fish) trip saw the puma up a tree.The experts were now fully convinced that the animal was a puma, 28 where had it come from? As no pumashad been reported missing from any zoo in the country, this one 29 have been in the possession of a private collector and somehow managed to escape. The hunt went on for several weeks, but the puma was not caught. It is disturbing 30 (think) a dangerous wild animal is still at large in the quiet countryside.Section B—10分Directions: Fill in each blank with a proper word chosen from the box. Each word can be used only once. Note that there is one word more than you need.A. networkB. specifyC. traditionallyD. ingredientE. uneasyF. additionalG. culturallyH. blockI. determine J. requirement K. criticalA multicultural person is someone who is deeply convinced that all cultures are equally good, enjoys learning the rich variety of cultures in the world, and most likely has been exposed to more than one culture in his or her lifetime.You cannot motivate anyone, especially someone of another culture, until that person has accepted you. A multilingual salesperson can explain the advantages of a product in other languages, but a multicultural salesperson can motivate foreigners to buy it. That’s a(an) 31difference.No one likes foreigners who are arrogant(自大的) about their own culture. The trouble is most people are arrogantly monocultural without being aware of it and even those who are can’t hide it. Foreigners sense monocultural arrogance at once and set up their own cultural barriers, which may effectively 32 any attempt by the monocultural person to motivate them.Multiculturalism is a(an) 33 that has been neglected too often in hiring managers for international positions. Even if your company is not a multinational one, chances are you’re in to uch with foreign customers or manufacturers Do you have the right employee to buildup the 34?For 20-odd years, I’ve run an executive-search firm from Brussels. When clients ask us to find the right person for a new pan-European sales or management position, I start by asking them to 35 the qualifications their ideal candidate would have. Most often they list the same qualities they would want for a domestic position, but with the 36 requirement that the new manager be fluent enough in English, German and French to cope with faxes and email. It sometimes takes me hours to persuade clients that the linguistic (语言的)abilities they see as crucial are not enough.Of course, it’s far more difficult to 37 candidates multiculturalism than it is to check their language skills—but it’s also a far more important 38 to success. I remember a company that asked me to check out a salesman they wereplanning to send to Mexico. He’d studied Spanish, and had grown up in New York City—the most 39 diverse place in America. But when I interviewed him, he turned out to have no concept of the great pride Mexicans took in their culture, and moreover he was 40 about Mexican restaurants and markets being dirty and unsafe. I rejected him just as Mexican buyers would have if he’d been s elected for the job.III. Reading ComprehensionSection A—15分Directions: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.Hailing from Sw eden, “plogging” is a fitness craze that sees participants pick up plastic litter while jogging adding a virtuous, environmentally driven element to the sport. Plogging appears to have started around 2016, but is now going global, due to increasing awareness and 41 over plastic levels in the ocean.The appeal of plogging is its 42—all you need is running gear and a bin bag, and the feeling of getting fit while supporting a good cause. By adding regular squats(蹲) to pick up junk and carrying 43 to jogging. we can assume the health benefits are increased.Running and good causes have always gone 44 —just think of all the fundraising marathon runners do. But there couldn’t be a more on-trend way of keeping fit than plogging.Anything that’s getting people o ut in nature and connecting 45 with their I environment is a good thing, says Lizzie Carr, an environmentalist who helped set up Plastic Patrol, a nationwide campaign to 46 our inland waterways of plastic pollution. There’s been a real 47 in the public mindset around plastics, helped by things like Blue Planet highlighting how disastrous the crisis is,” she says.We need to keep momentum high and the pressure up, and empower people through 48 like plogging and Plastic Patrol.The plastic Patrol app allows users to 49 plastic anywhere in the world by collecting discarded items, photographing them and 50 to the app, giving us a better knowledge of what sorts of plastic and which brands are being thrown out. “I’d urge all ploggers to get involved,” adds Carr.P logging isn’t the first fitness trend to combine running with a good cause, Here are some of our favourites: Good GymIts idea is simple: go for a run, visit an elderly person, have a chat and some tea, and run back.51 among the elderly is a growing problem in the UK. With over 10,000 runs so far, 52, Good Gym is finding a solution.Guide RunningGuide runners volunteer their time to helping blind people get 53. By linking themselves together, the 54 —impaired individual can feel safe while both work of a sweat.55 for the HomelessStart-up Stuart Delivery and the Church Housing Trust collaborated last year in bringing clothing and healthy food to the homeless. Deliveries are mostly made by bike, so those who deliver keep fit while helping rough sleepers(无家可归者).41. A. satisfaction B. hesitation C. fear D. control42. A. complexity B. simplicity C. instrument D. expense43. A. substance B. responsibility C. value D. weight44. A. one on one B. head to toe C. hand in hand D. on and off45. A. positively B. neutrally C. objectively D. fairly46. A. accuse B. rid C. assure D. rob47. A. shift B. interest C. aid D. delight48. A. motives B. performances C. exercises D. initiatives49. A. eliminate B. map C. seek D. degrade50. A. leading B. devoting C. ending D. uploading51. A. Disappointment B. Tiredness C. Sickness D. Loneliness52. A. therefore B. moreover C. however D. instead53. A. excited B. ready C. active D. smart54. A. visually B. audibly C. visibly D. sensibly55. A. Running B. Plogging C. Driving D. CyclingSection B—22 分Directions: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)In 1982, I had responsibility for Stephen Hawking’s third academic book for the Press,Superspace and Supergravity. This was a messy collection of papers from a technical workshop on how to devise a new theory of gravity. While that book was in production, I suggested he try something easier: a popular book about the nature ofthe Universe, suitable for the general market.Stephen hesitated over my suggestion. He already had an international reputation as a brilliant theoretical physicist working on rotating black holes and theories of gravity. And he had concerns about financial matters: importantly, it was impossible for him to obtain any form of life insurance to protect his family in the event of his death or becoming totally dependent on nursing care. So, he took precious time out from his research to prepare the rough draft of a book.At the time, several bestselling physics authors had already published non-technical books on the early Universe and black holes. Stephen decided to write a more personal approach, by explaining his own research in cosmology and quantum theory.One afternoon, in the 1980s, he invited me to take a look at the first draft, but first he wanted to discuss cash. He told me he had spent considerable time away from his research, and that he expected advances and royalties (定金和版税) to be large. When I pressed him on the market that he foresaw, he insisted that it be on sale, up front, at all airport bookshops in the UK and the US. I told that was a tough call for a university press. Then I thumbed the typescript. To my dismay, the text was far too technical for a general reader.A few weeks later he showed me a revision, much improved. Eventually, he decided to place it with a mass market publisher rather than a university press. Bantam published A Brief History of Time in March 1988. Sales took off like a rocket, and it ranked as a bestseller for at least five years. The book’s impact on the popularization of science has been incalculable.56. What suggestion did the writer give to Stephen Hawking?A. Simplifying Superspace and Supergravity.B. Formulating a new theory of gravity.C. Writing a popular book on the nature of the universe.D. Revising a book based on a new theory.57. Which of the following was Stephen Hawking most concerned about?A. Financial returns.B. Other competitors.C. Publishing houses.D. His family’s life insurance.58. The underlined word thumbed is closest in meaning to .A. praisedB. typedC. confirmedD. browsed59. The greatest contribution of the book A Brief History of Time lies in .A. bringing him overnight fame in the scientific worldB. keeping up the living standard of his familyC. making popular science available to the general publicD creating the rocketing sales of a technical book(B)Conventional wisdom may tell you that a master’s degree from Harvard Business School in the US is the key to a Fortune 500 job, while the same degree from the Wharton School of the University of Pennsylvania, US, means a possible career on Wall Street.It seems that the graduate school you go to somewhat decides your future. And a recent New York Times article reveals the correlation between MBA (Master of Business Administration) graduates at certain US schools and career prospects.To work at AmazonRoss School of Business (University of Michigan)Amazon regularly hires more MBAs from top 10 business schools than big Wall Streetfirms. And a large chunk of Americans employees are from Ross. Graduate Peter Faricy, vice president of Amazon Marketplace, says the reason behind this is that Ross’ curriculum-related offerings, a problem-solvingcourse for instance, are particularly well suited to Amazon.To work at McKinsey& CompanyKellogg School of Management (Northwestern)For an MBA, landing a job at McKinsey is like trying to get into a competitive business school all over again. However, Kellogg graduates perform well in the fierce competition. The school’s MBAs are in dem and at elite consulting firms, which hired 35 percent of Kellogg graduates last year, a higher percentage than at Harvard (23 percent) and Stanford (16 percent).To work at AppleFuqua School of Business (Duke)Silicon alley hasn’t always welcomed MBAs. However, two of Apple’s top 10 executives come from Fuqua. Apple has hired 32 Fuqua graduates over the past five years, and provided 42 internships for Duke students.To start your own companyHarvard Business SchoolThe extensive resources Harvard has devoted to its entrepreneurial offerings in recent years are starting to show real results. By many accounts, it has surpassed Stanford as the top entrepreneurial hot-bed in the US.60. Which university offers students a course on various approaches to difficulties at work?A. Kellogg School of Management.B. Ross School of Business.C. Harvard Business School.D. Fuqua School of Business.61. According to the passage, which of the following is true?A. Consulting companies favor MBA students from Kellogg.B. Stanford produces the greatest number of business leaders.C. To work at Apple, MBA graduates have an advantage.D. Wall Street employs more MBAs from top 10 than Amazon.62. If you want to work in the area of hi-tech electronic products, you may choose to study in .A. Wharton SchoolB. Kellogg School of ManagementC. Ross School of BusinessD. Fuqua School of Business(C)“Two centuries ago, Lewis and Clark left St. Louis to explore the new lands acquired in the Louisiana Purchase,” George W. Bush said, announcing his desire for a program to send men and women to Mars. They made that journey in the spirit of discovery. America has ventured forth into space for the same reasons.”Yet there are vital differences between Lewis and Clark’s expediti on and a Mars mission. First, they wereheaded to a place where hundreds of thousands of people were already living. Second, they were certain to discover places and things of immediate value to the new nation. Third, their venture cost next to nothing by today’s standards.A Mars mission may be the single most expensive non-wartime undertaking in U.S. history.Appealing as the thought of travel to Mars is, it does not mean the journey makes sense, even considering the human calling to explore. And Mars as a destination for people makes absolutely no sense with current technology.Present systems for getting from Earth’s surface to low-Earth orbit are so fantastically expensive that merely launching the 1,000 tons or so of spacecraft and equipment a Mars mission would require could be accomplished only by cutting health-care benefits, education spending, or other important programs—or by raising taxes. Absent some remarkable discovery, astronauts, geologists, and biologists once on Mars could do little more than analyze rocks and feel awestruck(敬畏的) staring into the sky of another world. Yet rocks can be analyzed by automated probes without risk to human life, and at a tiny fraction of the cost of sending people.It is interesting to note that when President Bush unveiled his proposal, he listed these recent major achievements of space exploration pictures of evidence of water on Mars, discovery of more than 100 planets outside our solar system, and study of the soil of Mars. All these accomplishments came from automated probes or automated space telescopes. Bush’s proposal, which calls for reprogramming some of NASA’s present budget into the Mars effort, might actually lead to a reduction in such unmanned science—the one aspect of space exploration that’s worki ng really well.Rather than spend hundreds of billions of dollars to hurl tons toward Mars using current technology, why not take a decade or two or however much time is required researching new launch systems and advanced propulsion (推进力)? lf new launch systems could put weight into orbit affordably, and advanced propulsion could speed up that long, slow transit to Mars, the dream of stepping onto the red planet might become reality. Mars will still be there when the technology is ready.63. What do Lewi s and Clark’s expedition and a Mars mission have in common?A. Instant value.B. Human inhabitance.C. Venture cost.D. Exploring spirit.64. Bush’s proposal is challenged for the following reasons except that.A. its expenditure is too huge for the government to afford.B. American people’s well-being will suffer a lot if it is implementedC. great achievements have already been made in Mars exploration in AmericaD. unmanned Mars exploration sounds more practical and economical for the moment65. Which cannot be concluded from the passage?A. Going to Mars using current technology is quite unrealistic.B. A Mars mission will in turn promote the development of unmanned program.C. Bush’s proposal is based on three recent great achievements o f space explorationD. The achievements in space exploration show how well unmanned science has developed.66. What is the main idea of the passage?A. Risky as it is, a Mars mission helps to retain Americas position as a technological leader.B. A Mars mission is so costly that it may lead to an economic disaster in America.C. Someday people may go to Mars but not until it makes technological sense.D. A Mars mission is unnecessary since the scientists once there won’t make great discoveries.Section C—8分Directions: Read the passage carefully. Fill in each blank with a proper sentence given in the box Each sentence can be used only once. Note that there are two more sentences than you need.A. Being simple might be another reason.B. It was the only affordable way to play them.C. We should have admiration for this old technology.D. The current trend for old games shows no sign of slowing.E. Newer consoles and their games are incredibly expensive.F. So it seems like its not ‘game over’ for ol d-school technologyRetro GamingThere’s no doubt that in today’s digital world, computer games are extremely sophisticated and capable of creating virtual reality experiences that were unimaginable only a few years ago. So I am interested to see that the simplistic games that I grew up with, are making a revival. But Why?In the 1970s, the original place to play a computer game was at an arcade. Here, you and your mates could try out the new big names in games such as Space Invaders and Pacman. 67 And because of the technology involved, the gaming machines were too big to fit into your house.But in the 1980s and 90s, gaming arrived in our homes and people like me were addicted. The sound of beeping became a familiar sound emanating from bedrooms across the land! Names such as Tetris, Sonic and Street Fighterbecame popular language in the playground—and now they are being talked about—and played again. One of the reasons is the low cost. The BBC spoke to gamer, Gemma Wood, who says that: 68 I understand that a lot of hard work has gone into the design etc... but how can anyone justify £50 to £60 for a game that you might not even enjoy?69 The graphics on old games may not compare with the detail and definition of modern games but they are fun and easy to use by children and adults alike. And of course, nostalgia plays its part. Some people want to relive their childhood while for others, it is a chance to show their children the computer games they grew up with.Technology journalist, KG Orphanides, says “it’s important to recognize how well-designed many of those classic games are…the developers had so little space to work with-your average Sega Mega Drive or SNES cartridge had a maximum capacity of just 4mb-and limited graphics and sound capabilities. This compares to an average capacity of 40G in today’s games. 70This craze for using retro hardware and grabbing an old joystick is certainly catching on. And to persuade those of us who are not sure about downgrading the gaming experience, manufacturers such as Nintendo, are bringing back some of their older consoles in new style casing.IV. 71. Summary Writing—10分Directions: Read the following passage. Summarize the main idea and the main point(s) of the passage in no more than 60wonds. Use your own words as, far as possible.Every time there is a mass shooting, the debate surrounding guns tends to flare up in America. The abuse of guns has been a serious problem in the US all along, but why doesn’t the US government just dismiss owning guns privately?The right to own a gun and defend oneself is central to American society. As early as the1600s, when the first Europeans set foot on the continent of North America, they had to face a lot of dangers. They could only rely on themselves. Therefore, guns played a significant role in self-defense. Guns were also important in American’s Independent War and the Civil War.Secondly, the American founding fathers believed that gun ownership was necessary for a truly free country. If the government distrusts the people and disarms them, then that government no longer represents the people. The Second Amendment to the US Constitution specifies that the American people cannot be deprived of the “right to keep and bear arms.” So the sale and purchase of firearms are legal in the United States according to law.The importance of guns is also derived from the role of hunting in American culture. In the nation’s early years, hunting was essential for food and shelter. Today, guns are a vital part of hunting, which remains very popular asboth a sport and a way of life in many parts of the country. People spend time with friends, sharing the pleasure that the sport brings.For those reasons, when critics say guns mean violence, they miss a large part of the picture, and they misrepresent the complex nature of America’s diverse gun culture. Most people who own guns privately, are actually part of the gun culture. They have rational and thoughtful reasons to own and use guns.V. Translation—15分Directions: Translate the following sentences into English, using the words given in the brackets.72. 我们常常忍不住秒回刚收到的信息。
2018届浦东新区高考数学二模(附答案)
2018届浦东新区高考数学二模(附答案)D18. 在ABC ∆中，边a 、b 、c 分别为角A 、B 、C 所对应的边. （1）若2(2)sin 0(2)sin 1sin (2)sin c a b Ab a BC a b A-=-+-，求角C 的大小；（2）若4sin 5A =，23C π=，3c =ABC ∆的面积.19. 已知双曲线22:1C x y -=.（1）求以右焦点为圆心，与双曲线C 的渐近线相切的圆的方程；（2）若经过点(0,1)P -的直线与双曲线C 的右支交于不同两点M 、N ，求线段MN 的中垂线l 在y 轴上截距t 的取值范围.20. 已知函数()y f x =定义域为R ，对于任意x ∈R 恒有(2)2()f x f x =-.（1）若(1)3f =-，求(16)f 的值；（2）若(1,2]x ∈时，2()22f x x x =-+，求函数()y f x =，(1,8]x ∈的解析式及值域；（3）若(1,2]x ∈时，3()||2f x x =--，求()y f x =在区间(1,2]n，*n N ∈上的最大值与最小值.21. 已知数列{}n a 中11a =，前n 项和为nS ，若对任意的*n N ∈，均有n n k S a k +=-（k 是常数，且*k N ∈）成立，则称数列{}na 为“()H k 数列”.（1）若数列{}na 为“(1)H 数列”，求数列{}na 的前n 项和nS ；（2）若数列{}na 为“(2)H 数列”，且2a 为整数，试问：是否存在数列{}na ，使得211||40nn n a a a -+-≤对一切2n ≥，*n N ∈恒成立？如果存在，求出这样数列{}n a 的2a 的所有可能值，如果不存在，请说明理由； （3）若数列{}na 为“()H k 数列”，且121ka a a==⋅⋅⋅==，证明：211(1)2n kn kk a -+-≥+.参考答案2018.04一. 填空题1. 22. ()0,13.114.35.846.()1,07.1638. ,36k k ππππ⎡⎤-+⎢⎥⎣⎦，k ∈Z 9.6 10.1311.[]1,0- 12.6二. 选择题 13-16. ABAD 三. 解答题 17.（1）圆锥的底面积214S r ππ== ……………3分 圆锥的侧面积2410S rl ππ==……………3分 圆锥的全面积124(110)S S S π=+=+……………1分 （2）2BOC π∠= OC OB ∴⊥ 且OC OA ⊥，OC ⊥平面AOB ……………2分CDO∴∠是直线CD 与平面AOB 所成角 ……………1分 在Rt CDO 中，2OC =,10OD , ……………1分10tan 5CDO ∠=,10arctan 5CDO ∴∠= ……………2分 所以，直线CD 与平面AOB 所成角的为101分 18.（1）由题意，()()2sin 2sin 2sin c C a b A b a B =-+-；……………2分 由正弦定理得()()2222c a b a b a b=-+-，∴222c a b ab=+-，……………2分∴2221cos 22a b c C ab +-==，∴3C π=；……………2分 （2）由4sin 5A =，3c =sin sin a c A C =，∴85a =；…………2分 由23a c A C π<⇒<=，∴3cos 5A =,…………2分∴()334sin sin sin cos cos sin B A C A C A C -=+=+=；…………2分∴11883sin 225ABCSca B ∆-==…………2分19.（1）2(2,0)F …………1分 渐近线x y ±=………1分1R = (2)分22(2)1x y +=………………2分（2）设经过点B 的直线方程为1y kx =-,交点为1122(,),(,)M x y N x y ………………1分22221(1)2201x y k x kx y kx ⎧-=⇒-+-=⎨=-⎩…1分 则212121,00120k x x k x x ⎧≠∆>⎪+>⇒<<⎨⎪>⎩ (2)分MN的中点为221(,)11k k k ----，…1分 得中垂线2211:()11kl y x k k k+=-+--…1分令0x =得截距2222211t kk -==>--………………2分即线段MN 的中垂线l 在y 轴上截距t 的取值范围是(2,)+∞. 20.（1）(1)3f =-且(2)2()f x f x =- (2)3(2)f ∴=-⋅-……………1分 22(2)3(2)f ∴=-⋅-……………1分 33(2)3(2)f ∴=-⋅-………1分44(16)(2)3(2)48f f ∴==-⋅-=-……1分（2）(2)2()()2()2xf x f x f x f =-⇒=-，(1,2]x ∈时，22()22(1)1f x x x x =-+=-+，()(1,2]f x ∈……………1分(2,4]x ∈时，221()2()2[(1)1](2)2222x x f x f x =-=--+=---，……………1分 ()[4,2)f x ∈--……………1分(4,8]x ∈时，2211()2()2[(2)2](4)42224x xf x f x =-=----=-+， (1)分()(4,8]f x ∈……………1分得：222(1)1,(1,2]1()(2)2,(2,4]21(4)4,(4,8]4x x f x x x x x ⎧⎪-+∈⎪⎪=---∈⎨⎪⎪-+∈⎪⎩，值域为[4,2)12](4,8]--（， (1)分（3）(2)2()()2()2xf x f x f x f =-⇒=-当(1,2]x ∈时，3()2f x x =--得：当2(2,2]x ∈时，()2()32xf x f x =-=-……1分当1(2,2]n n x -∈时，1(1,2]2n x-∈，21122113()2()(2)()(2)()(2)(1)3222222n n n n n n x xxxf x f f f x -----=-=-=-=---=--⋅……………2分当1(2,2]n nx -∈，n 为奇数时，22()32[,0]4nn f x x -=--⋅∈-当1(2,2]n nx -∈，n 为偶数时，22()32[0,]4nn f x x -=-⋅∈综上：1n =时，()f x 在(1,2]上最大值为0，最小值为12-……………1分2n ≥，n 为偶数时，()f x 在(1,2]n上最大值为24n ，最小值为28n -……………1分3n ≥，n 为奇数时，()f x 在(1,2]n上最大值为28n ，最小值为24n -……………1分21.（1）数列{}na 为“()1H 数列”，则11nn Sa +=-，故121n n Sa ++=-，两式相减得：212n n a a ++=， …………………1分又1n =时，121a a =-，所以2122a a ==，………………1分 故12n na a +=对任意的N*n ∈恒成立，即12n na a +=（常数），故数列{}na 为等比数列，其通项公式为12,*n na n N -=∈；………………1分21,*n n S n N =-∈………………1分（2）2132321132()2N*nn n n n n n n n n Sa a a a a a a n Sa +++++++++=-⎧⇒=-⇒=+∈⎨=-⎩21(2,)N*n n n a a a n n ++⇒=+≥∈………………1分 当*2,n n N ≥∈时，()222121111()n n n n n n n n n n n aa a a a a a a a a a ++++++-=-+=--因为*11,(3,)n n n a a a n n N +--=≥∈，则22*1211,(3,)n n n n n n aa a a a a n n N ++-+-=-≥∈；则22*1211,(3,)n n n n n n aa a a a a n n N ++-+-=-≥∈………………2分则22*11324(3,)nn n aa a a a a n n N -+-=-≥∈，因为432aa a =+则222*113232(3,)n n n a a a a a a a n n N -+-=--≥∈………………1分因为13132,13Sa a a =-=⇒=，则2229340aa --≤，且2n =时，22340a-≤，解得：20,1,2,3,4,5,6a=±±±±±-………………2分（3）*1*11(2,)(2,)n k n n k n k n n k n a S k a a a n n N a S k n n N +++--+-=+⎧⎪⇒=+≥∈⎨=+≥∈⎪⎩…………1分110k a S k +=+>，由归纳知，20,,0k na a +>⇒>，…………1分1211,1kk a a a a k +=====+，由归纳知，*1,()n n a a n N +≤∀∈， (2)分 则*11112(2,)n kn k n n k n k n k aa a a a a n n N ++-+-+-+-=+≤+=≥∈*12(2,)n k n k a a n n N ++-≤≥∈…………1分*122121111,()222n k n k n k n k k a a a a n N ++++++--⇒≥≥≥≥∈…………1分于是*2212111(1),()2n kn k n k n k k a a a a n N ++-++--=+≥+∈于是1*2211(1),()2n n kk k aa n N -+-≥+∈…………1分22k k a S k k=+=，∴112111111(1)2(1),(2(1))222n n k kn kk k k ak k ----+---≥+⋅>+>+…1分结论显然成立.。
2018年上海市浦东新区高考数学二模试卷
2018年上海市浦东新区高考数学二模试卷一.填空题（本大题共12题，1-6每题4分，7-12每题5分，共54分）1. limn→∞2n+1n−1=________【答案】2【考点】极限及其运算【解析】变形得到limn→∞2n+1n−1=limn→∞2+1n1−1n，而limn→∞1n=0，从而求出该极限的值．【解答】lim n→∞2n+1n−1=limn→∞2+1n1−1n=2．2. 不等式xx−1<0的解集为________．【答案】(0, 1)【考点】其他不等式的解法【解析】由不等式xx−1<0可得x(x−1)<0，由此解得不等式的解集．【解答】由不等式xx−1<0可得x(x−1)<0，解得0<x<1，3. 已知{a n}是等比数列，它的前n项和为S n，且a3=4，a4=−8，则S5=________ 【答案】11【考点】等比数列的前n项和【解析】根据等比数列的定义求出公比，结合数列前n项和公式的定义进行求解即可．【解答】∵a3=4，a4=−8，∴公比q=a4a3=−84=−2，则a2=−2，a1=1，a5=16，则S5=1−2+4−8+16=11，4. 已知f−1(x)是函数f(x)=log2(x+1)的反函数，则f−1(2)=________ 【答案】3反函数 【解析】令f(x)=log 2(x +1)=2，解得x 值，进而可得答案． 【解答】∵ f −1(x)是函数f(x)=log 2(x +1)的反函数， 令f(x)=log 2(x +1)=2， 解得：x =3， 故f −1(2)=3，5. (√x +1x )9二项展开式中的常数项为________ 【答案】 84【考点】二项式定理的应用 二项式系数的性质 【解析】写出二项展开式的通项，由x 的指数为0求得r 值，则答案可求． 【解答】(√x +1x )9的展开式的通项为T r+1=C 9r ∗(√x)9−r ∗(1x)r =C 9r ∗x 9−3r2．取9−3r 2=0，得r =3．∴ (√x +1x )9二项展开式中的常数项为C 93=84．6. 椭圆{x =2cosθy =√3sinθ （θ为参数）的右焦点坐标为________【答案】 (1, 0) 【考点】椭圆的参数方程 【解析】根据题意，将椭圆的参数方程变形为标准方程，分析可得a 、b 的值，计算可得c 的值，即可得椭圆的右焦点坐标，即可得答案． 【解答】根据题意，椭圆{x =2cosθy =√3sinθ （θ为参数）的普通方程为x 24+y 23=1， 其中a =2，b =√3，则c =1；故椭圆的右焦点坐标为(1, 0)；7. 满足约束条件{x +2y ≤42x +y ≤3x ≥0y ≥0 的目标函数f =3x +2y 的最大值为________163【考点】 简单线性规划 【解析】由约束条件作出可行域，化目标函数为直线方程的斜截式，数形结合得到最优解，联立方程组求得最优解的坐标，代入目标函数得答案． 【解答】由约束条件{x +2y ≤42x +y ≤3x ≥0y ≥0 作出可行域如图， 联立{x +2y =42x +y =3，解得A(23, 53)． 化目标函数f =3x +2y 为y =−32x +f2，由图可知， 当直线y =−32x +f2过A 时，直线在y 轴上的截距最大，f 有最大值为163．8. 函数f(x)=cos 2x +√32sin2x ，x ∈R 的单调递增区间为________【答案】[kπ−π3, kπ+π6]，k ∈Z ．【考点】三角函数中的恒等变换应用 【解析】利用二倍角和辅助角化简，结合三角函数性质即可求单调性； 【解答】函数f(x)=cos 2x +√32sin2x=12cos2x +√32sin2x +12=sin(2x +π6)+12，令2kπ−π2≤2x +π6≤π2+2kπ，k ∈Z ． 可得：kπ−π3≤x ≤kπ+π6，∴ 单调递增区间为[kπ−π3, kπ+π6]，k ∈Z ．9. 已知抛物线型拱桥的顶点距水面2米时，量得水面宽为8米，当水面下降1米后，水面的宽为________米 【答案】 4√6抛物线的求解【解析】根据题意，求出抛物线的方程，进而利用当水面下降1米后，y=−3，可求水面宽度．【解答】由题意，设y=ax2，代入(4, −2)，∴a=−18，∴−3=−18x2，解得x=2√6∴水面的宽为4√6，10. 一个四面体的顶点在空间直角坐标系O−xyz中的坐标分别是(0, 0, 0)、(1, 0, 1)、(0, 1, 1)、(1, 1, 0)，则该四面体的体积为________．【答案】13【考点】柱体、锥体、台体的体积计算【解析】如图所示，满足条件的四面体为正方体的内接正四面体O−ABC．利用正方体的体积与三棱锥的体积计算公式即可得出．【解答】如图所示，满足条件的四面体为正方体的内接正四面体O−ABC．∴该四面体的体积V=13−4×13×12×12×1=13．11. 已知f(x)是定义在R上的偶函数，且f(x)在[0, +∞)上是增函数，如果对于任意x∈[1, 2]，f(ax+1)≤f(x−3)恒成立，则实数a的取值范围是________．【答案】[−1, 0]【考点】奇偶性与单调性的综合【解析】由题意可得|ax+1|≤|x−3|在x∈[1, 2]恒成立，即x−3≤ax+1≤3−x，即1−4 x ≤a≤2x−1在x∈[1, 2]恒成立，运用函数的单调性求得最值，即可得到a的范围．【解答】f(x)是定义在R上的偶函数，且f(x)在[0, +∞)上是增函数，如果对于任意x∈[1, 2]，f(ax+1)≤f(x−3)恒成立，可得|ax+1|≤|x−3|在x∈[1, 2]恒成立，即有|ax+1|≤3−x，即x−3≤ax+1≤3−x，即1−4x ≤a ≤2x −1在x ∈[1, 2]恒成立，由y =1−4x 在x ∈[1, 2]递增，可得y 的最大值为1−2=−1； y =2x −1在x ∈[1, 2]递减，可得y 的最小值为1−1=0，则−1≤a ≤0，12. 已知函数f(x)=x 2−5x +7，若对于任意的正整数n ，在区间[1, n +5n ]上存在m +1个实数a 0、a 1、a 2、…a m ，使得f(a 0)>f(a 1)+f(a 2)+...+f(a m )成立，则m 的最大值为________ 【答案】 6【考点】二次函数的性质 二次函数的图象 【解析】求出n +5n 的最小值，得出f(x)在此区间上的最值，根据最值的倍数关系得出m 的值． 【解答】∵ n 为正整数，∴ n +5n ≥92， ∴ f(x)在区间[1, 92]上最大值为f(92)=194，最小值为f(52)=34， ∵ 194=34×6+14，∴ m 的最大值为6． 故最大值为6．二.选择题（本大题共4题，每题5分，共20分）已知方程x 2−px +1=0的两虚根为x 1、x 2，若|x 1−x 2|=1，则实数p 的值为（ ） A.±√3 B.±√5 C.√3，√5 D.±√3，±√5 【答案】 A【考点】 复数的运算 【解析】根据方程x 2−px +1=0有两虚根x 1、x 2， 知△<0，写出方程x 2−px +1=0的两虚根， 由|x 1−x 2|=1求得实数p 的值． 【解答】方程x 2−px +1=0的两虚根为x 1、x 2， ∴ △=p 2−4<0， 解得−2<p <2，即x 1=−p+i√4−p 22，x 2=−p−i√4−p 22，∴ |x 1−x 2|=√4−p 2=1， 解得p =±√3．在复数运算中下列三个式子是正确的：(1)|z 1+z 2|≤|z 1|+|z 2|；(2)|z 1⋅z 2|=|z 1|⋅|z 2|；(3)(z 1⋅z 2)⋅z 3=z 1⋅(z 2⋅z 3)，相应的在向量运算中，下列式子：(1)|a →+b →|≤|a →|+|b →|；(2)|a →∗b →|=|a →|⋅|b →|；(3)(a →∗b →)∗c →=a →∗(b →∗c →)，正确的个数是（ ） A.0B.1C.2D.3【答案】 B【考点】复数的运算 【解析】根据在复数运算性质、向量运算的性质即可判断出结论． 【解答】根据在复数运算中下列三个式子是正确的：(1)|z 1+z 2|≤|z 1|+|z 2|；(2)|z 1⋅z 2|=|z 1|⋅|z 2|；(3)(z 1⋅z 2)⋅z 3=z 1⋅(z 2⋅z 3)，相应的在向量运算中，下列式子：(1)|a →+b →|≤|a →|+|b →|，正确；（2）而|a →∗b →|=|a →|⋅|b →|cos <a →,b →>，因此不正确；（3）由于c →与a →不一定共线，因此(a →∗b →)∗c →=a →∗(b →∗c →)不正确．因此正确的个数是1． 故选：B ．唐代诗人杜牧的七绝唐诗中有两句诗为：“今来海上升高望，不到蓬莱不成仙．”其中后一句中“成仙”是“到蓬莱”的（ ） A.充分条件 B.必要条件 C.充要条件D.既非充分又非必要条件 【答案】 A【考点】必要条件、充分条件与充要条件的判断 【解析】根据充分必要条件的定义判断即可． 【解答】解：根据充分必要条件的定义可知： ∵ 不到蓬莱就不能成仙， ∴ 成仙就必须到蓬莱，即成仙必须到蓬莱，而到蓬莱不一定成仙，故“成仙”是“到蓬莱”的充分不必要条件.设P、Q是R上的两个非空子集，如果存在一个从P到Q的函数y=f(x)满足：（1）Q= {f(x)|x∈P}；（2）对任意x1，x2∈P，当x1<x2时，恒有f(x1)<f(x2)，那么称这两个集合构成“P→Q恒等态射”，以下集合可以构成“P→Q恒等态射”的是（）A.R→ZB.Z→QC.[1, 2]→(0, 1)D.(1, 2)→R【答案】D【考点】映射【解析】利用题目给出的“P→Q恒等态射”的概念，对每一个选项中给出的两个集合，利用所学知识，找出能够使两个集合满足题目所给出的条件的函数，即Q是函数的值域，且函数为定义域上的增函数，即可得到要选择的答案．【解答】根据题意，函数f(x)的定义域为P，单调递增，值域为Q，由此判断，对于A，定义域为R，值域为整数集，且为递增函数，找不出这样的函数；对于B，定义域为Z，值域为Q，且为递增函数，找不出这样的函数；对于C，定义域为[1, 2]，值域为(0, 1)，且为递增函数，找不出这样的函数；)，对于D，可取f(x)=tan(πx−3π2且f(x)在(1, 2)递增，可得值域为R，满足题意．三.解答题（本大题共5题，共14+14+14+16+18=76分）已知圆锥AO的底面半径为2，母线长为2√10，点C为圆锥底面圆周上的一点，O为圆．心，D是AB的中点，且∠BOC=π2（1）求圆锥的全面积；（2）求直线CD与平面AOB所成角的大小．（结果用反三角函数值表示）【答案】∵圆锥AO的底面半径为r=2，母线长为l=2√10，∴圆锥的全面积S=πrl+πr2=π×2×2√10+π×22=(4√10+4)π．∵圆锥AO的底面半径为2，母线长为2√10，点C为圆锥底面圆周上的一点，O为圆心，D是AB的中点，且∠BOC=π．OA =√(2√10)2−22=6，C(2, 0, 0)，A(0, 0, 6)，B(0, 2, 0)，D(0, 1, 3)， DC →=(2, −1, −3)，平面ABO 的法向量n →=(1, 0, 0)， 设直线CD 与平面AOB 所成角为θ， 则sinθ=|CD →∗n →||CD →|∗|n →|=√14=√147． ∴ θ=arcsin √147．∴ 直线CD 与平面AOB 所成角为arcsin √147．【考点】直线与平面所成的角 【解析】（1）由圆锥AO 的底面半径为r =2，母线长为l =2√10能求出圆锥的全面积．（2）以O 为圆心，OC 为x 轴，OB 为y 轴，OA 为z 轴，建立空间直角坐标系，利用向量法能求出直线CD 与平面AOB 所成角． 【解答】∵ 圆锥AO 的底面半径为r =2，母线长为l =2√10， ∴ 圆锥的全面积S =πrl +πr 2 =π×2×2√10+π×22 =(4√10+4)π．∵ 圆锥AO 的底面半径为2，母线长为2√10，点C 为圆锥底面圆周上的一点，O 为圆心， D 是AB 的中点，且∠BOC =π2．∴ 以O 为圆心，OC 为x 轴，OB 为y 轴，OA 为z 轴，建立空间直角坐标系， OA =√(2√10)2−22=6，C(2, 0, 0)，A(0, 0, 6)，B(0, 2, 0)，D(0, 1, 3)， DC →=(2, −1, −3)，平面ABO 的法向量n →=(1, 0, 0)， 设直线CD 与平面AOB 所成角为θ， 则sinθ=|CD →∗n →||CD →|∗|n →|=√14=√147． √14∴ 直线CD 与平面AOB 所成角为arcsin √147．在△ABC 中，边a 、b 、c 分别为角A 、B 、C 所对应的边． （1）若|2c (2a −b)sinA1+(2b−a)sinB (2a−b)sinA sinC |=0，求角C 的大小；（2）若sinA =45，C =2π3，c =√3，求△ABC 的面积．【答案】由题意，2csinC =(2a −b)sinA ⋅(1+(2b−a)sinB(2a−b)sinA )， 即2csinC =(2a −b)sinA +(2b −a)sinB 由正弦定理得2c 2=(2a −b)a +(2b −a)b ． ∴ c 2=a 2+b 2−ab ． ∴ cosC =a 2+b 2−c 22ab=12．∵ 0<C <π． ∴ C =π3 由sinA =45，C =2π3，c =√3，根据正弦定理：asinA =csinC ， 可得：a =85 由a <c 即A <C ， ∴ cosA =35那么：sinB =sin(A +C)=sinAcosC +sinCcosA =3√3−410故得△ABC 的面积S =12acsinB =18−8√325． 【考点】 三角形求面积（1）根据矩阵的计算法则，可得2csinC =(2a −b)sinA ⋅(1+(2b−a)sinB(2a−b)sinA )，利用公式化简可得角C 的大小．（2）根据正弦定理求解a ，由余弦定理求解b ，即可求解△ABC 的面积． 【解答】由题意，2csinC =(2a −b)sinA ⋅(1+(2b−a)sinB(2a−b)sinA )， 即2csinC =(2a −b)sinA +(2b −a)sinB 由正弦定理得2c 2=(2a −b)a +(2b −a)b ． ∴ c 2=a 2+b 2−ab ． ∴ cosC =a 2+b 2−c 22ab=12．∵ 0<C <π． ∴ C =π3 由sinA =45，C =2π3，c =√3，根据正弦定理：asinA =csinC ， 可得：a =85 由a <c 即A <C ， ∴ cosA =35那么：sinB =sin(A +C)=sinAcosC +sinCcosA =3√3−410故得△ABC 的面积S =12acsinB =18−8√325．已知双曲线C:x 2−y 2=1．（1）求以右焦点为圆心，与双曲线C 的渐近线相切的圆的方程；（2）若经过点P(0, −1)的直线与双曲线C 的右支交于不同两点M 、N ，求线段MN 的中垂线l 在y 轴上截距t 的取值范围． 【答案】双曲线的右焦点为F 2(√2, 0)，渐近线方程为：x ±y =0． ∴ F 2到渐近线的距离为√2√2=1，∴ 圆的方程为(x −√2)2+y 2=1．设经过点P 的直线方程为y =kx −1，M(x 1, y 1)，N(x 2, y 2)， 联立方程组{x 2−y 2=1y =kx −1 ，消去y 得：(1−k 2)x 2+2kx −2=0， ∴ {1−k 2≠0−2k1−k 2>0−21−k 2>0 ，解得1<k <√2．∴ 线段MN 的中垂线方程为：y +11−k2=−1k(x +k 1−k 2)，令x =0得截距t =−21−k 2=2k 2−1>2．即线段MN 的中垂线l 在y 轴上截距t 的取值范围是(2, +∞)． 【考点】双曲线的离心率 【解析】（1）求出右焦点到渐近线的距离，得出圆的方程；（2）设直线MN 的方程为y =kx −1，联立方程组消元，根据方程在(1, +∞)上有两解求出k 的范围，得出线段MN 的中垂线方程，从而得出截距t 关于k 的函数，得出t 的范围． 【解答】双曲线的右焦点为F 2(√2, 0)，渐近线方程为：x ±y =0． ∴ F 2到渐近线的距离为√2√2=1，∴ 圆的方程为(x −√2)2+y 2=1．设经过点P 的直线方程为y =kx −1，M(x 1, y 1)，N(x 2, y 2)， 联立方程组{x 2−y 2=1y =kx −1 ，消去y 得：(1−k 2)x 2+2kx −2=0， ∴ {1−k 2≠0−2k1−k 2>0−21−k 2>0 ，解得1<k <√2．∴ MN 的中点为(−k1−k 2, −11−k 2)， ∴ 线段MN 的中垂线方程为：y +11−k 2=−1k(x +k 1−k 2)，令x =0得截距t =−21−k =2k −1>2．即线段MN 的中垂线l 在y 轴上截距t 的取值范围是(2, +∞)．已知函数y =f(x)定义域为R ，对于任意x ∈R 恒有f(2x)=−2f(x)． （1）若f(1)=−3，求f(16)的值；（2）若x ∈(1, 2]时，f(x)=x 2−2x +2，求函数y =f(x)，x ∈(1, 8]的解析式及值域；（3）若x ∈(1, 2]时，f(x)=−|x −32|，求y =f(x)在区间(1, 2n ]，n ∈N ∗上的最大值与最小值． 【答案】=−3，f(2x)=−2f(x)． 那么f =−2f=−3×(−2)∴ f(1)=f(22)=−2f(2)=−3×(−2)2 ∴ f(23)=−3×(−2)3∴ f(3)=f(24)=−3×(−2)4=−48 （4）由f(2x)=−2f(x)．可得f(x)=−2f(x2) 当x ∈(1, 2]时，f(x)=x 2−2x +2，那么：x ∈(2, 4]时，f(x)=−2f(x2)=−2[(x2)2−2(x2)+2)]=−12(x −2)2−2 那么：x ∈(4, 8]时，f(x)=−2f(x2)=−2[−12(x2−2)2−2]=14(x −4)2+4 故得x ∈(1, 8]的解析式为f(x)={ x 2−2x +2,(1,2brack −12(x −2)2−2,(2,4brack 14(x −4)2+4,(4,8brack根据二次函数的性质，可得值域为[−4, −2)∪(1, 2]∪(4, 8]． （5）（6）由f(2x)=−2f(x)．可得f(x)=−2f(x2) 当x ∈(1, 2]时，f(x)=−|x −32|，得当x ∈(2, 22]时，f(x)=−2f(x 2)=|x −3|； 当x ∈(2n−1, 2n ]时，x2∈(1, 2]，f(x)=−2f(x2)=(−2)n−1f(x2n−1)=(−1)n |x −3⋅2n−2|； 当x ∈(2n−1, 2n ]时，n 为奇数时，f(x)=|x −3⋅2n−2|∈[−2n 4, 0]当x ∈(2n−1, 2n ]时，n 为偶数时，f(x)=−|x −3⋅2n−2|∈[0, 2n4]综上：n =1时，f(x)在(1, 2]上最大值为0，最小值为−12 n ≥2，n 为偶数时，f(x)在(1, 2n ]上最大值为2n4，最小值为−2n 8n ≥3，n 为奇数时，f(x)在(1, 2n ]上最小值为−2n 4，最大值为2n 8．【考点】函数的最值及其几何意义 【解析】（1）根据f(1)=−3，f(2x)=−2f(x)．即可求解f(2)，f(4)依此类推，即可求解求f(16)的值；（2）根据x ∈(1, 2]时，f(x)=x 2−2x +2，结合f(2x)=−2f(x)．即可递推出x ∈(1, 8]的解析式及值域；（3）根据x ∈(1, 2]时，f(x)=−|x −32|，根据规律，即可求y =f(x)在区间(1, 2n ]，n ∈N ∗上的最大值与最小值． 【解答】=−3，f(2x)=−2f(x)． 那么f =−2f=−3×(−2)∴ f(1)=f(22)=−2f(2)=−3×(−2)2 ∴ f(23)=−3×(−2)3∴ f(3)=f(24)=−3×(−2)4=−48 （4）由f(2x)=−2f(x)．可得f(x)=−2f(x2) 当x ∈(1, 2]时，f(x)=x 2−2x +2，那么：x ∈(2, 4]时，f(x)=−2f(x2)=−2[(x2)2−2(x2)+2)]=−12(x −2)2−2 那么：x ∈(4, 8]时，f(x)=−2f(x2)=−2[−12(x2−2)2−2]=14(x −4)2+4 故得x ∈(1, 8]的解析式为f(x)={ x 2−2x +2,(1,2brack −12(x −2)2−2,(2,4brack 14(x −4)2+4,(4,8brack根据二次函数的性质，可得值域为[−4, −2)∪(1, 2]∪(4, 8]． （5）（6）由f(2x)=−2f(x)．可得f(x)=−2f(x2) 当x ∈(1, 2]时，f(x)=−|x −32|，得当x ∈(2, 22]时，f(x)=−2f(x 2)=|x −3|； 当x ∈(2n−1, 2n ]时，x2n−1∈(1, 2]， f(x)=−2f(x2)=(−2)n−1f(x 2)=(−1)n |x −3⋅2n−2|；当x ∈(2n−1, 2n ]时，n 为奇数时，f(x)=|x −3⋅2n−2|∈[−2n 4, 0]当x ∈(2n−1, 2n ]时，n 为偶数时，f(x)=−|x −3⋅2n−2|∈[0, 2n4]综上：n =1时，f(x)在(1, 2]上最大值为0，最小值为−12 n ≥2，n 为偶数时，f(x)在(1, 2n ]上最大值为2n4，最小值为−2n 8n ≥3，n 为奇数时，f(x)在(1, 2n]上最小值为−2n 4，最大值为2n 8．已知数列{a n }中a 1=1，前n 项和为S n ，若对任意的n ∈N ∗，均有S n =a n+k −k(k 是常数，且k ∈N ∗)成立，则称数列{a n }为“H(k)数列”． （1）若数列{a n }为“H(1)数列”，求数列{a n }的前n 项和S n ；（2）若数列{a n }为“H(2)数列”，且a 2为整数，试问：是否存在数列{a n }，使得|a n 2−a n−1a n+1|≤40对一切n ≥2，n ∈N ∗恒成立？如果存在，求出这样数列{a n }的a 2的所有可能值，如果不存在，请说明理由；（3）若数列{a n }为“H(k)数列”，且a 1=a 2=...=a k =1，证明：a n+2k ≥(1+12k−1)n−k ．【答案】数列{a n }为“H(1)数列”，则S n =a n+1−1，可得：S n+1=a n+2−1， 两式相减得：a n+2=2a n+1，又n =1时，a 1=a 2−1，∴ a 2=2=2a 1． 故a n+1=2a n ，对任意的n ∈N ∗恒成立，故数列{a n }为等比数列，其通项公式为a n =2n−1，n ∈N ∗． ∴ S n =2n −1．S n =a n+2−2，S n+1=a n+3−2，相减可得：a n+1=a n+3−a n+2，a n+1+a n+2=a n+3，n ≥2时，a n+2=a n+1+a n (n ≥2)，∴ n ≥3时，a n+12−a n a n+2=a n+12−a n (a n+1+a n )=a n+1(a n+1−a n )−a n 2=a n+1a n−1−a n 2．则|a n+12−a n a n+2|=|a n 2−a n−1a n+1|，则|a n 2−a n−1a n+1|=|a 32−a 2a 4|(n ≥3)，∵ a 4=a 3+a 2．∴ |a n 2−a n−1a n+1|=|a 32−a 2a 3−a 22|，∵ S 1=a 3−2，a 1=1，可得：a 3=3，∴ |9−3a 2−a 22|≤40，且|a 22−3|≤40．解得：a 2=0，±1，±2，±3，±4，5，−6．证明：a n+k =S n +k ，a n−1+k =S n−1+k(n ≥2)，可得：a n+k =a n+k−1+a n ，a k+1=S 1+k >0，由归纳知，a k+2>0，……，a n >0， a 1=a 2=……=a k =1，a k+1=k +1， 由归纳知，a n ≤a n+1．则a n+k =a n+k−1+a n ≤a n+k−1+a n+k−1=2a n+k−1，n ≥2， a n+k ≤2a n+k−1，n ≥2，∴ a n+k ≥12a n+k+1≥122a n+k+2≥……≥12k−1a n+2k−1(n ∈N ∗)， 于是：a n+2k =a n+2k−1+a n+k ≥(1+12k−1)a n+2k−1(n ∈N ∗)， 于是：a n+2k ≥(1+12k−1)n−1a 2k ． a 2k =S k +k =2k ，∴ a n+2k ≥(1+12k−1)n−1⋅2k >(1+12k−1)n−k−1(2k >(1+12k−1)−k )． ∴ a n+2k ≥(1+12k−1)n−k ．【考点】数列与不等式的综合 【解析】（1）数列{a n }为“H(1)数列”，可得S n =a n+1−1，S n+1=a n+2−1，两式相减得：a n+2=2a n+1，又n =1时，a 1=a 2−1，可得a 2=2=2a 1．利用等比数列的通项公式可得a n ，即可得出S n ．（2）S n =a n+2−2，S n+1=a n+3−2，相减可得：a n+1=a n+3−a n+2，a n+1+a n+2=a n+3，n ≥2时，a n+2=a n+1+a n (n ≥2)，n ≥3时，a n+12−a n a n+2=a n+12−a n (a n+1+a n )=a n+1(a n+1−a n )−a n 2，可得：|a n+12−a n a n+2|=|a n 2−a n−1a n+1|，|a n 2−a n−1a n+1|=|a 32−a 2a 4|(n ≥3)，根据a 4=a 3+a 2．可得|a n 2−a n−1a n+1|=|a 32−a 2a 3−a 22|，由S 1=a 3−2，a 1=1，可得：a 3=3，可得|9−3a 2−a 22|≤40，且|a 22−3|≤40．解得：a 2．（3）a n+k=S n+k，a n−1+k=S n−1+k(n≥2)，可得：a n+k=a n+k−1+a n，a k+1=S1+k>0，由归纳知，a k+2>0，……，a n>0，a1=a2=……=a k=1，a k+1=k+ 1，由归纳知，a n≤a n+1．则a n+k=a n+k−1+a n≤a n+k−1+a n+k−1=2a n+k−1，n≥2，a n+k≤2a n+k−1，n≥2，可得a n+k≥12a n+k+1≥122a n+k+2≥……≥12k−1a n+2k−1(n∈N∗)，于是：a n+2k≥(1+1 2k−1)n−1a2k．a2k=S k+k=2k，进而得到：a n+2k≥(1+12k−1)n−k．【解答】数列{a n}为“H(1)数列”，则S n=a n+1−1，可得：S n+1=a n+2−1，两式相减得：a n+2=2a n+1，又n=1时，a1=a2−1，∴a2=2=2a1．故a n+1=2a n，对任意的n∈N∗恒成立，故数列{a n}为等比数列，其通项公式为a n=2n−1，n∈N∗．∴S n=2n−1．S n=a n+2−2，S n+1=a n+3−2，相减可得：a n+1=a n+3−a n+2，a n+1+a n+2=a n+3，n≥2时，a n+2=a n+1+a n(n≥2)，∴n≥3时，a n+12−a n a n+2=an+12−a n(a n+1+a n)=a n+1(a n+1−a n)−a n2=a n+1a n−1−a n2．则|a n+12−a n a n+2|=|a n2−a n−1a n+1|，则|a n2−a n−1a n+1|=|a32−a2a4|(n≥3)，∵a4=a3+a2．∴|a n2−a n−1a n+1|=|a32−a2a3−a22|，∵S1=a3−2，a1=1，可得：a3=3，∴|9−3a2−a22|≤40，且|a22−3|≤40．解得：a2=0，±1，±2，±3，±4，5，−6．证明：a n+k=S n+k，a n−1+k=S n−1+k(n≥2)，可得：a n+k=a n+k−1+a n，a k+1=S1+k>0，由归纳知，a k+2>0，……，a n>0，a1=a2=……=a k=1，a k+1=k+1，由归纳知，a n≤a n+1．则a n+k=a n+k−1+a n≤a n+k−1+a n+k−1=2a n+k−1，n≥2，a n+k≤2a n+k−1，n≥2，∴a n+k≥12a n+k+1≥12a n+k+2≥……≥12a n+2k−1(n∈N∗)，于是：a n+2k=a n+2k−1+a n+k≥(1+12k−1)a n+2k−1(n∈N∗)，于是：a n+2k≥(1+12k−1)n−1a2k．a2k=S k+k=2k，∴a n+2k≥(1+12k−1)n−1⋅2k>(1+12k−1)n−k−1(2k>(1+12k−1)−k)．∴a n+2k≥(1+12k−1)n−k．。
上海市浦东新区2018届高三下学期质量调研(二模)数学试(含详细解答)
上海市浦东新区 2018 届高三二模数学试卷2018.04一 . 填空题（本大题共 12 题， 1-6 每题 4 分， 7-12 每题 5 分，共 54 分）1. lim2n 1nn 12. 不等式xx 0 的解集为13. 已知 { a n } 是等比数列，它的前 n 项和为 S n ，且 a 3 4， a 48，则 S 54. 已知 f 1( x) 是函数 f ( x) log 2 ( x 1) 的反函数，则 f 1 (2)5. ( x1)9二项睁开式中的常数项为x6. 椭圆x 2cos （ 为参数）的右焦点坐标为y3sinx 2 y 47. 2x y3的目标函数f3x 2 y 的最大值为知足拘束条件xy 08. 函数 f ( x) cos 2 x3sin2x ， x R 的单一递加区间为29. 已知抛物线型拱桥的极点距水面 2 米时，量得水面宽为 8 米，当水面降落1 米后，水面的宽为米10. 一个四周体的极点在空间直角坐标系O xyz 中的坐标分别是 (0,0,0)、 、、，(1,0,1) (0,1,1) (1,1,0)则该四周体的体积为11. 已知 f (x) 是定义在 R 上的偶函数，且 f ( x) 在 [0, ) 上是增函数，假如关于随意x [1,2] ， f (ax 1)f (x 3) 恒建立，则实数 a 的取值范围是12. 已知函数 f (x)x 2 5x 7 ，若关于随意的正整数n ，在区间 [1,n5] 上存在 m 1个n实数 a 0 、 a 1 、 a 2 、、 a m ，使得 f (a 0 )f (a 1 ) f (a 2 )f ( a m ) 建立，则 m 的最大值为二 .选择题（本大题共 4 题，每题5 分，共20 分）13. 已知方程x 2px10 的两虚根为 x 1 、x 2 ，若 | x 1x 2 | 1 ，则实数p 的值为（）A.3B.5C.3，5D. 3 ，514.在复数运算中以下三个式子是正确的：（1 ）| z1z2| | z1|| z2 |；（2）| z1z2 | | z1 | | z2 |；（3）( z1z2 )z3 z1 ( z2 z3 ) ，相应的在向量运算中，以下式子：（1）| a b | | a || b | ；（2）| a b || a | | b | ；（3） ( a b) c a (b c) ，正确的个数是（）A. 0B. 1C. 2D. 315.唐朝诗人杜牧的七绝唐诗中有两句诗为：“今来海上涨高望，不到蓬莱不可仙。
2018年上海浦东新区高三二模语文试卷(附答案)
2018年浦东新区高三二模语文试卷（时间150分钟，满分150分）2018.4一积累运用（10分）1.按要求填空（5分）（1），幽咽泉流冰下难。
（白居易《》）（2）香远益清，亭亭净植，。
（周敦颐《爱莲说》）（3）苏轼在《江城子》中有“相顾无言，惟有泪千行”的诗句，在柳永的《雨霖铃》中意境与之相似的一句是“，”。
2.按要求选择。
（5分）（1）今年南汇桃花节，小刘去踏青觅胜，欲留影配诗，下列诗句和赏花场景不匹配的一项是（）。
（2分）A.满树和娇烂漫红，万枝丹彩灼春融。
B.桃花一簇开无主，可爱深红爱浅红。
C.花开不并百花丛，独立疏篱趣无穷。
D.一树繁英夺眼红，开时先合占东风。
填入下面语段空白处的句子，最恰当的一项是（）。
（3分）文明是史，未进入文明之前是史前时期，未进入文明的文化是史前文化，未有字，焉有史？文明的标志当然是文字，，中国人大可底气十足地说，中华文明至少肇始于三千年前，其独一无二的持久性正有汉字之功。
A．而文明预示着文字有走向伟大的资本与长寿的禀赋B．而文字预示着文明有走向伟大的资本与长寿的禀赋C．而文明预示着文字有走向长寿与伟大的资本和禀赋D．而文字预示着文明有长寿的资本与走向伟大的禀赋二阅读（70分）（一）阅读下文，完成3-7题。
（16分）导演的限制与自由①导演的地位和作用问题，是近代戏剧史上一个争论不休的话题。
主流派认为，剧本是舞台艺术的基础，导演则是剧本的诠释者和体现者。
导演创作，可以发展或充实刷本，但却不能违背原作的立意与风格。
从俄国的斯坦尼斯拉夫斯基到美国的贝拉斯科、中国的焦菊隐等，都持这种观点。
②也有人认为，导演是现代戏剧的核心，他可以随意篡改或解构剧本，甚至干脆不要据本，正如他有权设计布景，有权摆布演员，有权使用音响灯光一样。
一些先锋派导演或理论家多持这种观点。
如果把这种“导演中心”论限制在演出的范围内，还是有道理的，作为某种创新实验，更是无可厚非，但要推行于全部戏刷活动，恐怕就行不通了。
2018年浦东新区高考数学二模含答案
2018年浦东新区⾼考数学⼆模含答案2018年浦东新区⾼考数学⼆模含答案 2018．4注意：1．答卷前，考⽣务必在试卷上指定位置将学校、班级、姓名、考号填写清楚．2．本试卷共有21道试题，满分150分，考试时间120分钟．⼀、填空题（本⼤题共有12⼩题，满分54分）只要求直接填写结果，1-6题每个空格填对得4分，7-12题每个空格填对得5分，否则⼀律得零分．21lim 1n n n →+∞+=- .2 2.不等式01xx <-的解集为________.(0,1)3.已知{}n a 是等⽐数列，它的前n 项和为n S ，且34,a =48a =-，则5S = ________.114.已知1()f x -是函数2()log (1)f x x =+的反函数，则1(2)f -=________.35.91)x⼆项展开式中的常数项为________.846.椭圆2cos ,x y θθ=（θ为参数）的右焦点为________.(1,0)7.满⾜约束条件2423x y x y x y +≤??+≤?≥≥的⽬标函数32f x y =+的最⼤值为________.1638.函数2()cos 2,R f x x x x =+∈的单调递增区间为____________.,,36Z k k k ππππ?-+∈9.已知抛物线型拱桥的顶点距⽔⾯2⽶时，量得⽔⾯宽为8⽶。
当⽔⾯下降1⽶后，⽔⾯的宽为_____⽶。
10.—个四⾯体的顶点在空间直⾓坐标系xyz O -中的坐标分别是(0,0,0)，(1,0,1)，(0,1,1)，(1,1，0)，则该四⾯体的体积为________.111.已知()f x 是定义在R 上的偶函数，且()f x 在[)0,+∞上是增函数，如果对于任意[1,2]x ∈，(1)(3)f ax f x +≤-恒成⽴，则实数a 的取值范围是________.[1,0]-12.已知函数2()57f x x x =-+.若对于任意的正整数n ，在区间51,n n ??+上存在1m +个实数012,,,,m a a a a 使得012()()()()m f a f a f a f a >+++成⽴，则m 的最⼤值为________.6⼆、选择题(本⼤题共有4⼩题，满分20分) 每⼩题都给出四个选项，其中有且只有⼀个选项是正确的，选对得 5分，否则⼀律得零分．13.已知⽅程210x px -+=的两虚根为12,x x ，若121x x -=，则实数p 的值为（）A A ． 3± B ．5± C. 3,5 D ． 3,5±± 14.在复数运算中下列三个式⼦是正确的：（1）1212z z z z +≤+，（2）1212z z z z ?=?，（3）123123()()z z z z z z ??=??；相应的在向量运算中，下列式⼦：（1）a b a b +≤+，（2）a b a b ?=?，（3）()()a b c a b c ??=??；正确的个数是（）BA ． 0B ．1 C. 2 D ．315.唐代诗⼈杜牧的七绝唐诗中两句诗为“今来海上升⾼望，不到蓬莱不成仙。
上海市浦东新区2018届高三下学期教学质量检测(二模)数学试卷
后一句中“成仙”是“到蓬莱”的（
）
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f (am ) 成立，则 m的最大
值为
二. 选择题（本大题共 4 题，每题 5 分，共 20 分）
.
2
（1）求圆锥的全面积；
（2）求直线 CD 与平面 AOB 所成角的大小 .
（结果用反三角函数值表示）
18. 在 ABC 中，边 a 、 b 、 c 分别为角 A 、 B 、 C 所对应的边 .
2c （1）若 (2b a)sin B
1 (2a b)sin A
(2a b)sin A sin C
0 ，求角 C 的大小；
米
10. 一个四面体的顶点在空间直角坐标系 O xyz 中的坐标分别是 (0,0,0) 、(1,0,1) 、(0,1,1) 、
(1,1,0) ，则该四面体的体积为
11. 已知 f ( x) 是定义在 R 上的偶函数，且 f ( x) 在 [0, ) 上是增函数，如果对于任意
x [1,2] ， f (ax 1) f ( x 3) 恒成立，则实数 a 的取值范围是
【全国大联考】【上海】上海市浦东新区2018届高三下学期质量检测(二模)化学试题(解析版)
【试卷整体分析】考试范围：高考范围试题难度：一般【题型考点分析】上海市浦东新区2018届高三下学期质量检测（二模）化学试题第I卷（选择题）1．合金在生产生活中具有广泛的用途。
不属于合金的是A．明矾B．硬铝C．生铁D．青铜【答案】A【解析】A、明矾是十二水硫酸铝钾，属于盐，属于纯净物不是合金；B、硬铝是铝合金；C、生铁是铁碳合金；D、青铜是铜锡合金；答案选A。
2．12C和13C原子之间，相等的是A．原子质量B．中子数C．质量数D．电子数【答案】D【解析】12C和13C原子之间，质量数分别为12和13，不相等选项C错误；故原子质量不相等，选项A错误；中子数分别为6和7，不相等，选项B错误；质子数均为6，原子核内质子数等于核外电子数，也均为6，选项D正确。
答案选D。
3．含有共价键的离子晶体是A．H2SO4B．KOH C．MgCl2D．Na2O【答案】B点睛：本题考查化学键的判断。
一般活泼的金属和活泼的非金属容易形成离子键，非金属元素的原子间容易形成共价键。
氢氧根离子、酸根离子中的非金属原子之间形成共价键。
4．硫化氢能与二氧化硫反应，说明硫化氢具有A．还原性B．氧化性C．酸性D．不稳定性【答案】A【解析】硫化氢能与二氧化硫反应生成硫和水，硫化氢中硫元素化合价由-2价变为0价，被氧化，体现硫化氢的还原性，答案选A。
5．化学反应中的能量变化符合如图所示的是A．甲烷燃烧B．碳酸钙高温分解C．电石与水反应D．酸碱中和【答案】B【解析】A．燃烧反应是放热反应，选项A错误；B．碳酸钙高温分解是吸热反应，选项B正确；C．电石与水反应是放热反应，选项C错误；D．酸碱中和反应是放热反应，选项D错误。
答案选B。
点睛：本题考查化学反应的热量变化，学生应注重归纳中学化学中常见的吸热或放热的反应，对于特殊过程中的热量变化的要熟练记忆来解答此类习题。
生成物具有的总能量高于反应物具有的总能量，该反应为吸热反应。
①放热反应：有热量放出的化学反应，因为反应物具有的总能量高于生成物具有的总能量．常见放热反应：燃烧与缓慢氧化，中和反应；金属与酸反应制取氢气，生石灰和水反应等；②吸热反应：有热量吸收的化学反应，因为反应物具有的总能量低于生成物具有的总能量．常见的吸热反应：C（s）+H2O（g）→CO（g）+H2O；C+CO2→CO的反应，以及KClO3、KMnO4、CaCO3的分解等。
上海市浦东新区2018届高三下学期质量调研(二模)数学试(含详细解答)
上海市浦东新区2018届高三二模数学试卷2018.04一. 填空题（本大题共12题，1-6每题4分，7-12每题5分，共54分）1. 21lim1n n n →+∞+=-2. 不等式01xx <-的解集为3. 已知{}n a 是等比数列，它的前n 项和为n S ，且34a =，48a =-，则5S =4. 已知1()f x -是函数2()log (1)f x x =+的反函数，则1(2)f -=5. 91()x x+二项展开式中的常数项为6. 椭圆2cos 3sin x y θθ=⎧⎪⎨=⎪⎩（θ为参数）的右焦点坐标为7. 满足约束条件242300x y x y x y +≤⎧⎪+≤⎪⎨≥⎪⎪≥⎩的目标函数32f x y =+的最大值为8. 函数23()cos sin 22f x x x =+，x ∈R 的单调递增区间为 9. 已知抛物线型拱桥的顶点距水面2米时，量得水面宽为8米，当水面下降1米后，水 面的宽为 米10. 一个四面体的顶点在空间直角坐标系O xyz -中的坐标分别是(0,0,0)、(1,0,1)、(0,1,1)、(1,1,0)，则该四面体的体积为11. 已知()f x 是定义在R 上的偶函数，且()f x 在[0,)+∞上是增函数，如果对于任意[1,2]x ∈，(1)(3)f ax f x +≤-恒成立，则实数a 的取值范围是12. 已知函数2()57f x x x =-+，若对于任意的正整数n ，在区间5[1,]n n+上存在1m +个 实数0a 、1a 、2a 、⋅⋅⋅、m a ，使得012()()()()m f a f a f a f a >++⋅⋅⋅+成立，则m 的最大 值为二. 选择题（本大题共4题，每题5分，共20分）13. 已知方程210x px -+=的两虚根为1x 、2x ，若12||1x x -=，则实数p 的值为（ ） A. 3± B. 5± C. 3，5 D. 3±，5±14. 在复数运算中下列三个式子是正确的：（1）1212||||||z z z z +≤+；（2）1212||||||z z z z ⋅=⋅；（3）123123()()z z z z z z ⋅⋅=⋅⋅，相应的在向量运算中，下列式子：（1）||||||a b a b +≤+；（2）||||||a b a b ⋅=⋅；（3）()()a b c a b c ⋅⋅=⋅⋅，正确的个数是（ ）A. 0B. 1C. 2D. 315. 唐代诗人杜牧的七绝唐诗中有两句诗为：“今来海上升高望，不到蓬莱不成仙。