新乡许昌平顶山2014届高三第二次调研考试 答案 _文 数

平顶山许昌新乡2012届高三第二次调研考试语文参考答案及评分标准第Ⅰ卷阅读题必考题一、现代文阅读（9分，每小题3分）1．B（原文第二自然段中“成为‘法’的适用基础”的是“约定俗成”的“俗”。

）2．B（错在“外乡人都会通过文化认同完全融入当地社会”，原文第四自然段中是“也有可能”。

）3．C（错在“应该摒弃具有个体特质的‘俗’文化”，原文第六自然段中是“和而不同”。

）二、古代诗文阅读（36分）（一）文言文阅读（19分）4．D（状：情状，情况。

）5．A（②不能表现有才华，④⑤讲颜之推险被定罪。

）6．B（侯景攻陷郢州时，曾经多次想杀掉颜之推，依靠行台郎中王则说情才将其赦免。

）7．⑴（颜之推）嗜好喝酒，过分任性放纵，不修边幅，当时的舆论因此轻视他。

（“任纵” “以”“少”各1分，句意2分，共5分）⑵天保末年，（颜之推）随从至天池，(显祖)准备任命（颜之推）为中书舍人，派中书郎段孝信带敕书给颜之推看。

（“从”“以为”“出示”各1分，省略句式1分，句意1分，共5分）【参考译文】颜之推，字介，是琅邪临沂人。

父亲名勰，是梁湘东王萧绎的镇西府谘议参军。

颜之推很早继承家传的学业，十二岁时，适逢萧绎亲自讲说《庄子》《老子》，他就参与到门生行列。

他对清谈并不爱好，退学回家，自学《周礼》《左传》。

他广泛地阅读各种书，无所不晓，文词典雅明丽，很受镇西府的人称赞。

萧绎任他为湘东王国的左常侍，加授镇西府的墨曹参军。

他嗜好喝酒，过分任性放纵，不修边幅，当时的舆论因此轻视他。

萧绎派世子方诸出镇郢州，叫颜之推任管记。

正赶上侯景攻陷郢州，多次想要杀死他，依靠他的行台郎中王则说情才得以豁免，被囚禁押送到建业。

侯景被平定后，（颜之推）回到江陵。

此时萧绎已经自立为王，任命颜之推为散骑侍郎，掌管舍人之事。

后来萧绎被周军所击败，周大将军李显庆看重之推，推荐他去弘农，让之推在他的兄长阳平公李远处掌管书翰。

正好黄河水暴涨，颜之推就备办船只带妻子儿女逃奔北齐，中间经历险要的砥柱山，当时人称赞他勇敢有决断。

许昌新乡平顶山三市2014届高三第二次调研考试英语第二部分：阅读理解（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的四个选项（A、B、C和D）中，选出最佳选项，并在答题卡上将该项涂黑。

AJob InterviewsYou have the education and the work experience; now all you need is the job. Most people spend more time getting ready for an evening out than they do preparing for a job interview. There are several things you can do to have a successful job interview. Here are some of them:1. Know about the company where you are trying to get a job. Find out what the company does and who their competitors are.2. Know yourself. Most interviewers will ask a question related to your strengths and weaknesses. Before the interview, examine the job you are applying for and determine the top skills needed for that position. For instance, if you are applying for a job as a telephone operator, it’s more important that you have strong communication skills than that you are tidy.3. Have proof. It’s easy to say you have good organization skills. But if you can tell the interviewer about a time you organized an event, it makes the claim much more solid.4. Prepare questions to ask about the company and the job.An interview is a stressful event. When you are under pressure, it can be difficult to think straight. Preparing before the interview will make it easier to give answers to those tough questions. But you can’t think of every question they might ask. So relax and be confident. Remember, first impressions are lasting.【小题1】This article is about ________.A. how to get a job interviewB. how to communicate at workC. how to prepare for a job interview.D. how to interview people【小题2】A good skill for a telephone operator to have is ________.A. the ability to smileB. tidinessC. good communication skillsD. proof of past experience【小题3】 You should prepare questions about _________ when interviewed for a job.A. the companyB. the salaryC. the interviewer’s strengthsD. the weather【小题4】Which of the following is TRUE?A. The job interview should be conducted over the telephone.B. The first time the interviewer sees you is important.C. You shouldn’t ask questions about the company.D. You shouldn’t give examples of things you have done in the past.【答案】【小题1】C【小题2】C【小题3】A【小题4】B【解析】BGretchen Alexander is sightless. But she refuses to allow her blindness to limit her life activities. She enjoys archery, golf, softball, sailing and water skiing, as well as a number of other activities that those of us who are sighted have yet to learn. She also speaks to groupsabout living life fully. When speaking to a group of high school students, she was once asked if there was anything she wouldn’t try. “I’ve decided to never sky dive,” she answered. “It wo uld scare the heck out of my dog.”Why do some people rise above their problems and live life fully, while others become defeated? Merle Shain explains it this way: “There are only two ways to approach life,” Shain says, “as a victim or as a gallant fighte r. And you must decide if you want to act or to react…"When discouraged, a victim reacts, perhaps in pain or self-pity. But a fighter will act.A fighter will make a decision to change that set of circumstances that left her or him discouraged. Or a fighter will decide to accept those circumstances with grace and move ahead anyway. A fighter will decide to act with courage. A fighter will take responsibility for his or her happiness. No matter how afraid, a fighter will refuse to give in to the most defeating of all human emotions-helplessness. A victim reacts. A fighter acts. It’s your decision. It’s a decision about whether you will live your life fully and with courage, or whether you will be forever defeated by harsh circumstances. Make it well, for it may be one of the most important decisions you ever make.Will you be a victim or a gallant fighter?【小题1】The best title for this article would be _______.A. Gretchen Alexander’s LifeB. Merle Shain’s Attitude to LifeC. Victims and FightersD. The Way of Life【小题2】What can we learn about Gretchen Alexander from this passage?A. She is more athletic than those of us who are sighted.B. She is discouraged when her dog is scared.C. There is no limitation to her life activities.D. She is a brave fighter.【小题3】The underlined word “it”（paragraph 3）is closest in meaning to _________.A. life.B. choice.C. courage.D. circumstance.【小题4】 The third paragraph mainly talks about ________.A. the difference between a victim and a fighter.B. the reactions of helplessness.C. decision-making.D. a fighters responsibility.【答案】【小题1】C【小题2】D【小题3】B【小题4】A【解析】COcean ParkIf you love the sea, Ocean Park is the place for you! Situated on the south side of Hong Kong Island, this 870,000 square metre educational theme park provides many opportunities to learn about marine life.To start with, the park boasts the Atoll Reef, one of the world’s largest aquariums, with about 2,500 fish from nearly 300 different species. What makes this aquarium special, however, is not just its size, but also its design. The Atoll Reef is built with an observation passageway that circles the aquarium on four different levels. This lets visitors view sea life from a variety of depths and angles.Then there’s the Shark Aquarium, a tank w ith more than 200 sharks from more than 30 species.Like the Atoll Reef, this unique aquarium is designed to make sure guests get the most out of their visit. Shaped like an underwater tunnel, guests can watch as sharks swim overhead and dive at them from every side.There’s also the Sea Jelly Spectacular, an aquarium that houses more than 1,000 jellyfish of all shapes, colours and sizes. And at the park’s Dolphin University, visitors can go on educational tours and watch the training of dolphins up close.The park’s most popular attraction is the Ocean Theatre, a huge outdoor pool where dolphins and sea lions entertain the visitors. Sometimes a killer whale even takes part in the performance!Although Ocean Parks focus is on the water, the theme park has plenty of other activities, too. For people seeking excitement, there are rides like the Abyss Turbo Drop, a roller coaster ride that takes passengers on a 20-storey drop straight down. There are also exhibits like the Dinosaur Discovery Trail and Bird Paradise. Finally, no trip to Ocean Park would be complete without visiting the park’s most popular animals--four giant pandas that were given as a gift from China’s central government.【小题1】Why is Hong Kong Ocean Park called an educational theme Park?A. It is specially designed to attract the young who are interested in the sea.B. It provides chances for people to broaden their knowledge of science.C. It offers chances for visitors to enlarge their knowledge of marine life.D. It has plenty of activities for people to have fun.【小题2】What makes the Atoll Reef so special?A. It is one of the symbols of Hong Kong Ocean Park.B. It has thousands of fish from various species.C. It is the largest aquarium in the world.D. It allows visitors to watch sea life from all angles.【小题3】Which of the following activities is NOT mentioned in the passage?A. Interacting with sea life in the huge outdoor pool.B. Visiting exhibitions about dinosaurs and birds.C. Taking a roller coaster ride on a 20-storey drop.D. Enjoying the show of dolphins and sea lions.【答案】【小题1】C【小题2】D【小题3】A【解析】DSunday, 31 August We’ve been in China for a month now. Dad, Mom, Harry and I moved to Tianjin on 25 August. We’re not very far from Beijing. Two days ago, we celebrated my 16th birthday. It was great celebrating in China; the only thing that was strange was the cake--here they’re not as sweet as the ones in New York. On Monday school starts—I wonder what it will be like.Monday, 1 September On my first day I was looking around for a locker to put my books in. However, here all the students keep all of their books at their desks--we stay in the same classroom because apparently we don’t have to go from class to class--teachers come to us!Today we selected teacher assistants for each subject. Their duties are to collect homework, make announcements, and do other stuff for the teacher and the students. It’s kind of a big deal here! Since I am from the US, I was asked to be the English assistant. I felt so proud but quite nervous at the same time because I wasn’t sure what I had to do, but I accepted the jobanyway.Friday, 3 October Boy, what a week! Now we have nine classes every day, including the morning class, a combination of our American schools’ “Homeroom” and “Study Hall”. I think Chinese students work too much! I have to do my homework when I get back home. I don’t even have time to watch TV or surf the Interne t like before. I sometimes miss New York and my school because we didn’t have to study so much. We had more time to hang out with our classmates and neighbors; here, besides their usual classes, students are involved in weekend classes in subjects such as English, Chinese and math.I get a lot of attention, being from another country. Everyone wants to practice English with me! A really cute girl even asked me for my phone number on my second day and sent me a text message! I’m making a lot more friends now. I just need a lot of help to improve my Chinese. Some students want to do a language exchange program with me. Nice!【小题1】 The passage mentions all the following points EXCEPT ________.A. physics study.B. food flavour.C. free time activities.D. language exchange program.【小题2】According to the passage, which of the following is NOT the teacher assistant’s duty?A. Collecting homework.B. Making announcements.C. helping teachers with small errands（差事）D. Teaching classmates.【小题3】Where is this passage most probably from?A. A story book.B. A guide book.C. A diary.D. A magazine.【小题4】The passage is best described by ________.A. culture shock.B. multi-culture.C. unique culture.D. culture background. 【答案】【小题1】A【小题2】D【小题3】C【小题4】A【解析】试题分析：这篇文章是到中国学习的美国学生的日常生活的日记，日记记录了作者的学习生活，也有作者经历的文化冲击。

新乡许昌平顶山2014届高三第二次调研考试语文参考答案 语 文 一、现代文阅读（9分，每小题3分） 1. B（误将未然作已然) 2. C（张冠李戴。

原文是孔子对这种光以较量射中并射穿为目的的“主皮之射”很不以为然。

） 3. B（曲解文意。

原文是“乡射礼的比赛往往非常激烈，但是最后评价一名射手，不仅要看他能否射中靶心，还要看他形体动作是否合适于音乐节奏”，“此外，还要求礼让为先……”） 二、古代诗文阅读（36分） （一）文言文阅读（19分） 4. B（具：全部） 5. C (①勤政为民；⑥ 弃官回家) 6. D（不得已将官吏打发了的是包拯） 7.（1）（5分）县里发洪水，老百姓遭水淹，县令不能拯救他们，胡宿带领公私船只救活了数千人。

（译出大意2分；“被溺”“率”“活”各1分。

） （2）（5分）却暗中袖手旁观，等他离开后便指责他，这难道就是古人分谤的意思吗？（译出大意2分；“阴”“俟”“非”各1分。

） （二）古代诗歌阅读（11分） 8. 通过描绘寂静、昏暗、风雨凄迷的景象渲染了一种悲凉的气氛（2分），为全词奠定了感情基调，为后文表达词人飘零身世和凄凉心境作铺垫。

（3分） 9. ①人在羁旅的寂寞思乡之情，②半生飘零的悲凉之情，③壮志未酬的惆怅之情。

（每点2分） （三）名篇名句默写（6分） 10．（6分）（1）秋天漠漠向昏黑 布衾多年冷似铁（2）侣鱼虾而友麋鹿 举匏樽以相属（3）鼎铛玉石 弃掷逦迤（每答对一空给1分，有错别字、多字、漏字则该空不给分） 三、文学类文本阅读（25分） 11.（1）答A给3分，答E给2分，答D给1分；答B、C不给分。

(B.对其作用分析过度。

C.没有据为己有的意思。

D.此处主要是“不知所措慌忙掩饰的心理” ) 11.（2）①表现了雨后青草湖清新、明净的自然美景。

②以明净、清新之景烘托李老壮美好的心灵世界，给人以美的感受③为后文人物关系的转机做铺垫，推动了故事情节的发展。

（每点2分） 11.（3）勤劳、朴实、善良（每点2分，其中特点1分，分析1分；若答其他，言之有据，可酌情给分。

平顶山许昌新乡2014届高三第一次调研考试 语文参考答案及评分标准 第Ⅰ卷 阅读题 必考题 一、现代文阅读（9分，每小题3分） 1.C（原文是“重要成分”而非“最主要的内容”。

） 2.D（应该特指“敏锐的、具有艺术气质的文人”。

） 3.C（纸的运用可以使晋人爱用行草字体写书札，但却与其“传世”没有因果关系。

） 二、古代诗文阅读（36分） （一）文言文阅读（19分） 4.D（杜：关闭，堵塞。

） 5.B（⑤⑥是晚年的退隐，不能表现“气侠雄爽”。

） 6.B（“从此打算闭门思过”有误，应为“亲手校雠后，打算闭门读书，度过自己的晚年”。

） 7.⑴许多达官贵人邀请他到家中作客，贺铸或者去或者不去，（遇到）他所不愿意见的人，也始终不说他们的坏话。

（“客致”“所”“贬”各1分，句意2分，共5分） ⑵贺铸家境贫困，经常靠借高利贷维持生活，有亏欠人家的，便拿地契房券等给人家抵押，丝毫不向别人乞讨。

（“贷”“负”“丐”各1分，省略句式1分，句意1分，共5分） （二）古代诗歌阅读（11分） 8.（5分） 这两句对仗工稳，都以拟人手法，动静结合，用“拂”和“侵”两字将句子写得富有动感，有声有色，饶有雅趣。

（3分）竹声“拂琴”写出竹声的美妙和含情，令人神往；竹影“侵棋”写出竹的影子映在棋盘上，使人感到竹似欲与诗人同乐，营造出一个物我为友、物我同趣的意境。

（2分）（若从其他角度分析，言之成理者，亦可酌情给分。

） 9.（6分） 诗歌描写了不争春色、独守严寒、不怕寂寞、保持贞洁、自有情趣“官舍竹”表现了淡泊无争清高C给2分，选D给1分，选A、B不给分。

（A.“无师自通”缺少文本依据。

B.“因我的懒惰” 与文中“多年努力”不符。

D.“更让‘我’感受到了精神上的孤独无助”说法不准确。

） ⑵（6分） ①篇首用一句话开门见山、简明扼要地突出了他的职业和优秀。

②用木匠对木材独具慧眼的敏锐和雕工的神奇来对其手艺的高超进行正面描写。

作为“天才木匠”，他有着超常的天赋和眼光，善于弥补木材本身的不足，常常化腐朽为神奇。

新乡许昌平顶山2014届高三第二次调研考试语文第Ⅰ卷阅读题（共70分）甲必考题一、现代文阅读（9分，每小题3分）阅读下面的文字，完成1～3题。

射箭，在古代是一种重要的生存手段，打仗、狩猎都少不了这一活动，因而古人将“射”列为“六艺”之一。

《周礼·地官·大司徒》就有“三曰六艺：礼、乐、射、御、书、数。

”的语句。

操弓射箭，在今人看来充其量也不过是一项专门技能，但古人却不是这样看的。

从古籍记载可知，“射”在古代曾是选拔人才的标准，而这种标准的确定，主要是因为“射”不光是技艺，还是人的德行的体现。

在儒家看来，“射”与“仁”“礼”是相关联的。

故《礼记·射义》又云：“以立德行者，莫若射，故圣王务焉。

”认为射法是人的德行所由生，又是人的德行所由见。

高超的射术，原本是勇气与技巧相结合的技艺。

春秋时期，诸侯纷争，弓箭成为战争中不可或缺的兵器。

正是在这样崇尚武力的时代，我们的古人有意将弓箭变成礼乐教化的工具，引导社会走向和平，射礼就此诞生。

顾名思义，射礼是一种射箭的礼仪，它融合了比赛、礼乐和宴饮等内容，用于选拔、竞技、宴宾、致礼等活动。

射礼作为周礼之一，也是古代一种民间娱乐活动，讲究谦和、礼让，提倡“发而不中，反求诸己”，重视人的自省，在本质上是一种道德引导方式，也是华夏先民特有的寓教于射的娱乐方式。

古代的箭靶一般用兽皮制作，以较量射中并射穿为目的的比赛，称为“主皮之射”。

孔子对这种光讲究力量的比赛很不以为然，认为它违背了“古之道”。

《论语·八佾》说：“射不主皮，为力不同科，古之道也。

”意思是比赛时射手能否射中、射穿靶子，主要取决于射手的体能，不值得看重；更应当注重的是射手的德行和修养。

因而，儒家巧妙地抓住了射箭与礼乐的结合点，在保留比射的同时，赋予射箭新的灵魂。

射礼按规格分为大射、宾射、燕射、乡射四种。

大射，是天子、诸侯举办盛大祭祀活动之前所行的射礼；宾射，是诸侯朝见天子或诸侯相会时举行的射礼；燕射，是天子、诸侯待客宴会时的射礼；乡射，是举行乡饮酒礼时所行的射礼，用以竞技、选贤等活动。

平顶山新乡许昌2014届高三第一次调研考试文科数学参考答案一.选择题1——5 CAABC 6——10 DACAB 11-----12 AD二.填空题13. 55 14. 3 15. -2 16. 15三.解答题：17.解:1)sin 21(32sin )(2+-+=x x x f++=x x 2cos 32sin 1)32sin(21++=πx ．………………………………………5分 ( I ) 函数)(x f 的最小正周期ππ==22T ．…………………………………… 6分 ( II ) 因为]6,6[ππ-∈x ，所以]32,0[32ππ∈+x ，所以∈+)32sin(πx ]1,0[， ………………………………………10分 所以]3,1[1)32sin(2)(∈++=πx x f ，所以)(x f 的值域为[1，3]．………………………………………12分18.解：（Ⅰ）…………………………………6分（Ⅱ）获一等奖的概率为0.04，所以获一等奖的人数估计为604.0150=⨯（人）. 记这6人为E D C B A A ,,,,,21，其中21,A A 为该班获一等奖的同学. …………………7分 从全校所有一等奖的同学中随机抽取2名同学代表学校参加决赛共有15种情况如下： ()21,A A ，()B A ,1，()C A ,1，()D A ,1，()E A ,1，()B A ,2，()C A ,2，()D A ,2，()E A ,2， ()C B ,，()D B ,，()E B ,，()D C ,，()E C ,，()E D ,. ……………………………9分该班同学参加决赛的人数恰好为1人共有8种情况如下:()B A ,1，()C A ,1，()D A ,1，()E A ,1，()B A ,2，()C A ,2，()D A ,2，()E A ,2. 所以该班同学参加决赛的人数恰好为1人的概率为158=P .……………………………12分 19.（Ⅰ）证：连接DE ，交AF 于点O∵1D D ⊥平面ABCD ，AF ⊂平面ABCD ,∴1D D AF ⊥……………………………2分 ∵点E ，F 分别是BC ，1D C 的中点，∴DF CE =又∵AD DC =，90ADF DCE ∠=∠=∴ADF ∆≌DCE ∆，∴AFD DEC ∠=∠又∵90CDE DEC ∠+∠=,∴90CDE AFD ∠+∠=∴()18090DOF CDE AFD ∠=-∠+∠=，即AF DE ⊥ ………………………………………5分又∵1D D DE D =,∴AF ⊥平面1D DE ,又∵1ED ⊂平面1D DE ,∴1AF ED ⊥………………………………………6分 （Ⅱ）解：∵1D D ⊥平面ABCD ，∴1D D 是三棱锥1D AEF -的高，且1D D a = ∵点E ，F 分别是BC ，1D C 的中点，∴2a DF CF CE BE ====……………7分 ∴AEF ADF FCE ABE ABCD S S S S S ∆∆∆∆=---正方形 2111222a AD DF CF CE AB BE =-⋅⋅-⋅⋅-⋅⋅2222234848a a a a a =---=………………………………………10分∴11E AFD D AEF V V --=113AEF S D D ∆=⋅⋅2313388a a a =⋅⋅= ………………………………………12分20.解：（Ⅰ）()()()2()=3+3131f x x a x a x x a '--3=-+．令()0f x '=得121,x x a ==- ………………………………………1分（i ）当1a -=，即1a =-时，()2()=310f x x '-≥，()f x 在(),-∞+∞单调递增 D 1 D C BA 1 A EF O………………………………………3分（ii ）当1a -<，即1a >-时，当x a <-，或1x >时()0f x '>，()f x 在(),a -∞-、()1+∞,内单调递增当1a x -<<时()0f x '<，()f x 在(),1a -内单调递减. …………………………4分（iii ）当1a ->，即1a <-时，当1,x x a <>-或时()0f x '>，()f x 在()(),1a -∞-+∞和,内单调递增当1x a <<-时()0f x '<，()f x 在()1,a -内单调递减综上，当1a <-时，()f x 在()(),1a -∞-+∞和,内单调递增，()f x 在()1,a -内单调递减；当1a =-时，()f x 在(),-∞+∞单调递增；当1a >-时，()f x 在()(),1a -∞-+∞和,内单调递增，()f x 在(),1a -内单调递减．…………………………6分（Ⅱ）当3a =时，32()391,[,2]f x x x x x m =+-+∈2()3693(3)(1)f x x x x x '=+-=+-，令()0f x '=得121,3x x ==-……………7分 将x ，()f x '，()f x 变化情况列表如下：………………………10分由此表可得 ()(3)28f x f =-=极大，()(1)4f x f ==-极小又(2)328f =< ，故区间[,2]m 内必须含有3-，即m 的取值范围是3]-∞-（，． ………………………12分21. (Ⅰ) 设圆C 和圆D 关于直线20x y +-=对称,由题意知圆D 的直径为12FF ,所以圆心()0,0D ，半径2r c ===，圆心D 与圆心C 关于直线02=-+y x 对称(2,2)C ∴，故圆C 的方程为22(2)(2)4x y -+-=. …… …………………4分(Ⅱ)由(Ⅰ)知2F (2,0), 设直线l 方程为:2,x my m R =+∈，∴圆心C 到直线l 的距离22m 1|2m |m 1|2-22m |=d +=++，由垂径定理和勾股定理得：222224m 4b =4(4-)=1+m 1+m . ………………………………6分设直线与椭圆相交与点1122(,),(,),E x y F x y 由22152,x x y my m R ⎧⎪⎨+==+∈⎪⎩得： 22(5410,m y my ++-=） 由韦达定理可得：12122241,,55m y y y y m m --+==++ 依题意可知：2215m a m +==+ 5158m 14515222222++⋅=+⋅++⋅=∴m m m m ab . (10)分令()0()f x x y f x =≥⇒=在[0,3] 单调递增，在[3,)+∞单调递减，()(3)f x f ≤ ⇒当23m=时，ab 取得最大值，此时直线的方程是 2.x =+ 所以当ab 取得最大值时，直线的方程是 2.x =+…………………………12分22.（Ⅰ）证明：∵AB 切圆于B ，∴2AB AD AE =⋅，又∵AB AC =，∴2AC AD AE =⋅，∴△ACD ∽△AEC ，∴ACD AEC ∠=∠，又∵AEC DGF ∠=∠，∴ACD DGF ∠=∠∴AC //FG………………………………………5分（Ⅱ）证明：连接BD ，BE ，EG由AB AC =，BAD CAD ∠=∠及AD AD =，知△ABD ≅△ACD ，同理有△ABE ≅△ACE ，∴BDE CDE∠=∠，故BE EG =，又BE CE = ∴EC EG = ………………………………………10分 23.解:(Ⅰ)圆C 的普通方程是22(1)1x y -+=，又cos ,sin x y ρθρθ==;所以圆C 的极坐标方程是2cos ρθ=. ………………………………………4分 (Ⅱ)设11(,)ρθ为点P 的极坐标，则有1112cos 3ρθπθ=⎧⎪⎨=⎪⎩解得1113ρπθ=⎧⎪⎨=⎪⎩. ………………………………………6分设22(,)ρθ为点Q的极坐标，则有2222(sin )3ρθθπθ⎧+=⎪⎨=⎪⎩， 解得2233ρπθ=⎧⎪⎨=⎪⎩ ………………………………………8分由于12θθ=，所以122PQ ρρ=-=，所以线段PQ 的长为2.………………10分 24.解:(Ⅰ)211y x -=+==+ 3x >,0y ∴-<,y ∴<----------------------------(5分)。

2河南省平顶山、许昌、新乡2008—2009学年高三第二次调研考试文科数学(必修+选修I )、选择题： M U ｛a 1, a 2｝= ｛a 1, a 2, a 3｝，则这样的集合 M 共有 A . 1个 等差数列｛a n ｝的前 A . 8 B . 2个 n 项的和为Si ，且公比 B . 10 C. 3个 q = 2, S2= 3,C . 13D . 4个则S 4的值为D . 151的实轴的长是B . 4C. 2 3—1函数y = f (x)是y =f (x)的反函数，若 过点 B. (0, 1)——1f (0) = 1，则函数y = f (1-x)的图像一定经A . (1 , 0)1(x + —)n 的展开式的二项式的系数和为xA . 31 个 B. 32 个a 、b 是两条异面直线，则"a 丄b ”是A .充要条件 C.必要而不充分条件uuu uuu凸四边形ABCD 中，AB 丄BC ,C . (0, 0)D . (1, 1)32,则集合｛a 1, a 2,…，a n ｝的真子集的个数为 D. 64 个C . 63 个"存在经过 a 且与b 垂直的平面”的B .充分而不必要条件 D .既不充分也不必要条件uuuuuu CD 丄 DA ,I AB | =、. 3 , | BC|= 1, | BD | 则/ BAD 的大小为 A . 45 ° B. 75°点P(a , 3)到直线4x — 3y + l = 0的距离等于4，且在2x + y — 3<0表示的平面区域内，则 a的值为 A . 3B . 7有以下四种变换方式：C .105 ° D . 135 ° C . —3D .①向左平行移动 一个单位长度，再将每个点的横坐标缩短为原来的4②向右平行移动个单位长度，再将每个点的横坐标缩短为原来的81 ； ;2;③每个点的横坐标缩短为原来的 ④每个点的横坐标缩短为原来的1，再向右平行移动个单位长度;2 81 一一 、、、 ，再向左平行移动 个单位长度.2 8其中能将函数 y = sinx 的图像变为函数 y = sin(2x +)的图像的是4A .①和③ B.①和④ C.②和④ D .②和③矩形 ABCD 中，AB = 4, BC = 3,沿AC 把厶ABC 折起使二面角 B - CA - D 的大小为 120°,则点D 到平面ABC 的距离为先从数字0、1、2、3、4、5中每次取出3个成等差数列的不同数字，再把每一组的 3个数字组成三位数，则所有的三位数共有 A . 22 个 B . 24 个 C . 42 个 D. 32 个uuur uuu OC | 2+| AB | 2填空题：本大题共 4小题。

第一部分 一 9一、选择题1．(文)(2014·东北三省三校联考)等差数列{a n }的前n 项和为S n ，若a 2＋a 4＋a 6 ＝12，则S 7的值是( )A ．21B ．24C ．28D ．7[答案] C[解析] ∵a 2＋a 4＋a 6＝3a 4＝12，∴a 4＝4， ∴2a 4＝a 1＋a 7＝8，∴S 7＝7(a 1＋a 7)2＝7×82＝28.[方法点拨] 1.熟记等差、等比数列的求和公式． 2．形如a n ＋1＝a n ＋f (n )的递推关系用累加法可求出通项； 3．形如a n ＋1＝a n f (n )的递推关系可考虑用累乘法求通项a n ；4．形如a n ＋1＝ka n ＋b (k 、b 为常数)可通过变形，设b n ＝a n ＋bk －1构造等比数列求通项a n .(理)在等比数列{a n }中，a 1＝a ，前n 项和为S n ，若数列{a n ＋1}成等差数列，则S n 等于( ) A ．a n ＋1－a B ．n (a ＋1) C ．na D ．(a ＋1)n －1[答案] C[解析] 利用常数列a ，a ，a ，…判断，则存在等差数列a ＋1，a ＋1，a ＋1，…或通过下列运算得到：2(aq ＋1)＝(a ＋1)＋(aq 2＋1)，∴q ＝1，S n ＝na .2．(文)已知S n 为等差数列{a n }的前n 项和，若S 1＝1，S 4S 2＝4，则S 6S 4的值为( )A.94 B.32 C.53 D ．4[答案] A[解析] 由等差数列的性质可知S 2，S 4－S 2，S 6－S 4成等差数列，由S 4S 2＝4得S 4－S 2S 2＝3，则S 6－S 4＝5S 2，所以S 4＝4S 2，S 6＝9S 2，S 6S 4＝94.(理)(2014·全国大纲文，8)设等比数列{a n }的前n 项和为S n .若S 2＝3，S 4＝15，则S 6＝( )A ．31B ．32C ．63D ．64[答案] C[解析] 解法1：由条件知：a n >0，且⎩⎪⎨⎪⎧a 1＋a 2＝3，a 1＋a 2＋a 3＋a 4＝15， ∴⎩⎪⎨⎪⎧a 1(1＋q )＝3，a 1(1＋q ＋q 2＋q 3)＝15，∴q ＝2. ∴a 1＝1，∴S 6＝1－261－2＝63.解法2：由题意知，S 2，S 4－S 2，S 6－S 4成等比数列，即(S 4－S 2)2＝S 2(S 6－S 4)，即122＝3(S 6－15)，∴S 6＝63.[方法点拨] 下标成等差的等差、等比数列的项或前n 项和的问题，常考虑应用等差、等比数列的性质求解．3．(2015·浙江理，3)已知{a n }是等差数列，公差d 不为零，前n 项和是S n .若a 3，a 4，a 8成等比数列，则( )A ．a 1d >0，dS 4>0B ．a 1d <0，dS 4<0C ．a 1d >0，dS 4<0D ．a 1d <0，dS 4>0 [答案] B[解析] 考查等差数列的通项公式及其前n 项和；等比数列的概念． ∵{a n }为等差数列，且a 3，a 4，a 8成等比数列， ∴(a 1＋3d )2＝(a 1＋2d )(a 1＋7d )⇒ a 1＝－53d ，∴S 4＝2(a 1＋a 4)＝2(a 1＋a 1＋3d )＝－23d ，∴a 1d ＝－53d 2＜0，dS 4＝－23d 2＜0，故选B.4．等比数列{a n }的前n 项和为S n ，已知S 3＝a 2＋10a 1，a 5＝9，则a 1＝( ) A.13 B ．－13C.19 D ．－19[答案] C[解析] ∵S 3＝a 2＋10a 1，∴a 1＋a 2＋a 3＝a 2＋10a 1，a 3＝9a 1＝a 1q 2，∴q 2＝9， 又∵a 5＝9，∴9＝a 3·q 2＝9a 3，∴a 3＝1， 又a 3＝9a 1，故a 1＝19.[方法点拨] 求基本量的问题，熟记等差、等比数列的定义、通项及前n 项和公式，利用公式、结合条件，建立方程求解．5．(2015·江西省质检)已知数列{a n }满足a 1＝1，a 2＝3，a n ＋2＝3a n (n ∈N *)，则数列{a n }的前2015项的和S 2015等于( )A ．31008－2B ．31008－3C ．32015－2D ．32015－3[答案] A[解析] 因为a 1＝1，a 2＝3，a n ＋2a n＝3， 所以S 2015＝(a 1＋a 3＋…＋a 2015)＋(a 2＋a 4＋…＋a 2014)＝1－310081－3＋3(1－31007)1－3＝31008－2.6．(文)(2014·新乡、许昌、平顶山调研)设{a n }是等比数列，S n 是{a n }的前n 项和，对任意正整数n ，有a n ＋2a n ＋1＋a n ＋2＝0，又a 1＝2，则S 101的值为( )A ．2B ．200C ．－2D ．0[答案] A[解析] 设公比为q ，∵a n ＋2a n ＋1＋a n ＋2＝0，∴a 1＋2a 2＋a 3＝0，∴a 1＋2a 1q ＋a 1q 2＝0，∴q 2＋2q ＋1＝0，∴q ＝－1，又∵a 1＝2，∴S 101＝a 1(1－q 101)1－q ＝2[1－(－1)101]1＋1＝2.(理)(2014·哈三中二模)等比数列{a n }，满足a 1＋a 2＋a 3＋a 4＋a 5＝3，a 21＋a 22＋a 23＋a 24＋a 25＝15，则a 1－a 2＋a 3－a 4＋a 5的值是( )A ．3 B. 5 C ．－ 5 D ．5[答案] D[解析] 由条件知⎩⎪⎨⎪⎧a 1(1－q 5)1－q＝3a 21(1－q10)1－q2＝15，∴a 1(1＋q 5)1＋q＝5，∴a 1－a 2＋a 3－a 4＋a 5＝a 1[1－(－q )5]1－(－q )＝a 1(1＋q 5)1＋q＝5.7．(文)在等差数列{a n }中，a 1＋a 2＋a 3＝3，a 18＋a 19＋a 20＝87，则此数列前20项的和等于( )A ．290B ．300C ．580D ．600[答案] B[解析] 由a 1＋a 2＋a 3＝3，a 18＋a 19＋a 20＝87得， a 1＋a 20＝30，∴S 20＝20×(a 1＋a 20)2＝300.(理)已知等比数列{a n }中，各项都是正数，且a 1，12a 3,2a 2成等差数列，则a 8＋a 9a 6＋a 7＝( )A ．1＋ 2B ．1－ 2C ．3＋2 2D ．3－2 2 [答案] C[解析] 由条件知a 3＝a 1＋2a 2， ∴a 1q 2＝a 1＋2a 1q ， ∵a 1≠0，∴q 2－2q －1＝0， ∵q >0，∴q ＝1＋2， ∴a 8＋a 9a 6＋a 7＝q 2＝3＋2 2. 8．(2015·福建理，8)若a ，b 是函数f (x )＝x 2－px ＋q (p ＞0，q ＞0)的两个不同的零点，且a ，b ，－2这三个数可适当排序后成等差数列，也可适当排序后成等比数列，则p ＋q 的值等于( )A ．6B ．7C ．8D ．9[答案] D[解析] 由韦达定理得a ＋b ＝p ，a ·b ＝q ，因为p >0，q >0，则a ＞0，b ＞0，当a ，b ，－2适当排序后成等比数列时，－2必为等比中项，故a ·b ＝(－2)2＝4，故q ＝4，b ＝4a .当适当排序后成等差数列时，－2必不是等差中项，当a 是等差中项时，2a ＝4a －2，解得a ＝1，b ＝4，；当b 是等差中项时，8a ＝a －2，解得a ＝4，b ＝1，综上所述，a ＋b ＝p ＝5，所以p ＋q ＝9，选D.9．已知数列{a n }，{b n }满足a 1＝b 1＝1，a n ＋1－a n ＝b n ＋1b n＝2，n ∈N ＋，则数列{ba n }的前10项的和为( )A.43(49－1) B.43(410－1) C.13(49－1) D.13(410－1) [答案] D[解析] 由a 1＝1，a n ＋1－a n ＝2得，a n ＝2n －1， 由b n ＋1b n＝2，b 1＝1得b n ＝2n －1， ∴ba n ＝2a n －1＝22(n －1)＝4n －1，∴数列{ba n }前10项和为1×(410－1)4－1＝13(410－1)．10．(文)若数列{a n }为等比数列，且a 1＝1，q ＝2，则T n ＝1a 1a 2＋1a 2a 3＋…＋1a n a n ＋1等于( )A ．1－14nB.23(1－14n ) C ．1－12nD.23(1－12n ) [答案] B[解析] 因为a n ＝1×2n －1＝2n －1，所以a n ·a n ＋1＝2n －1·2n ＝2×4n －1， 所以1a n a n ＋1＝12×(14)n －1，所以{1a n a n ＋1}也是等比数列，所以T n ＝1a 1a 2＋1a 2a 3＋…＋1a n a n ＋1＝12×1×(1－14n )1－14＝23(1－14n )，故选B.(理)(2014·唐山市一模)已知等比数列{a n }的前n 项和为S n ，且a 1＋a 3＝52，a 2＋a 4＝54，则S na n( )A ．4n －1B ．4n －1 C ．2n －1 D ．2n －1[答案] C[解析] 设公比为q ，则a 1(1＋q 2)＝52，a 2(1＋q 2)＝54，∴q ＝12，∴a 1＋14a 1＝52，∴a 1＝2.∴a n ＝a 1q n －1＝2×(12)n －1，S n ＝2[1－(12)n ]1－12＝4[1－(12)n ]，∴S n a n ＝4[1－(12)n ]2×(12)n －1＝2(2n －1－12)＝2n －1.[点评] 用一般解法解出a 1、q ，计算量大，若注意到等比数列的性质及求S na n，可简明解答如下：∵a 2＋a 4＝q (a 1＋a 3)，∴q ＝12，∴S na n ＝a 1(1－q n )1－q a 1q n －1＝1－q n (1－q )·qn －1＝1－12n 12·12n －1＝2n －1. 11．给出数列11，12，21，13，22，31，…，1k ，2k －1，…，k1，…，在这个数列中，第50个值等于1的项的序号．．是( ) A ．4900 B ．4901 C ．5000 D ．5001[答案] B[解析] 根据条件找规律，第1个1是分子、分母的和为2，第2个1是分子、分母的和为4，第3个1是分子、分母的和为6，…，第50个1是分子、分母的和为100，而分子、分母的和为2的有1项，分子、分母的和为3的有2项，分子、分母的和为4的有3项，…，分子、分母的和为99的有98项，分子、分母的和为100的项依次是：199，298，397，…，5050，5149，…，991，第50个1是其中第50项，在数列中的序号为1＋2＋3＋…＋98＋50＝98(1＋98)2＋50＝4901.[点评] 本题考查归纳能力，由已知项找到规律，“1”所在项的特点以及项数与分子、分母的和之间的关系，再利用等差数列求和公式即可．二、填空题12．(文)(2015·广东理，10)在等差数列{a n }中，若a 3＋a 4＋a 5＋a 6＋a 7＝25，则a 2＋a 8＝________.[答案] 10[解析] 本题考查等差数列的性质及简单运算，属于容易题．因为{a n }是等差数列，所以a 3＋a 7＝a 4＋a 6＝a 2＋a 8＝2a 5，a 3＋a 4＋a 5＋a 6＋a 7＝5a 5＝25 即a 5＝5，a 2＋a 8＝2a 5＝10.(理)(2015·湖南理，14)设S n 为等比数列{a n }的前n 项和．若a 1＝1，且3S 1,2S 2，S 3成等差数列，则a n ＝________.[答案] 3n －1[解析] 考查等差数列与等比数列的性质．∵3S 1,2S 2，S 3成等差数列，∴4S 2＝3S 1＋S 3，∴4(a 1＋a 2)＝3a 1＋a 1＋a 2＋a 3⇒a 3＝3a 2⇒q ＝3.又∵{a n }为等比数列，∴a n ＝a 1q n －1＝3n －1.[方法点拨] 条件或结论中涉及等差或等比数列中的两项或多项的关系时，先观察分析下标之间的关系，再考虑能否应用性质解决，要特别注意等差、等比数列性质的区别．13．(文)(2015·安徽理，14)已知数列{a n }是递增的等比数列，a 1＋a 4＝9，a 2a 3＝8，则数列{a n }的前n 项和等于________．[答案] 2n －1[解析] 考查1.等比数列的性质；2.等比数列的前n 项和公式．由题意，⎩⎪⎨⎪⎧ a 1＋a 4＝9，a 2·a 3＝8.∴⎩⎪⎨⎪⎧a 1＋a 4＝9，a 1·a 4＝8，解得a 1＝1，a 4＝8或者a 1＝8，a 4＝1，而数列{a n }是递增的等比数列，所以a 1＝1，a 4＝8，即q 3＝a 4a 1＝8，所以q ＝2，因而数列{a n }的前n 项和S n ＝a 1(1－q n )1－q ＝1－2n 1－2＝2n －1.(理)(2015·江苏，11)设数列{a n }满足a 1＝1，且a n ＋1－a n ＝n ＋1(n ∈N *)，则数列⎩⎨⎧⎭⎬⎫1a n 前10项的和为________．[答案]2011[解析] 考查数列通项，裂项求和．由题意得：a n ＝(a n －a n －1)＋(a n －1－a n －2)＋…＋(a 2－a 1)＋a 1＝n ＋(n －1)＋…＋2＋1＝n (n ＋1)2，所以1a n ＝2(1n －1n ＋1)，S n ＝2(1－12)＋2(12－13)＋…＋2(1n －1n ＋1)＝2(1－1n ＋1)＝2nn ＋1，S 10＝2011.三、解答题14．(文)设数列{a n }的前n 项和为S n ，且S n ＝4a n －p (n ∈N *)，其中p 是不为零的常数． (1)证明：数列{a n }是等比数列；(2)当p ＝3时，若数列{b n }满足b n ＋1＝a n ＋b n (n ∈N *)，b 1＝2，求数列{b n }的通项公式． [解析] (1)证明：因为S n ＝4a n －p (n ∈N *)， 则S n －1＝4a n －1－p (n ∈N *，n ≥2)，所以当n ≥2时，a n ＝S n －S n －1＝4a n －4a n －1， 整理得a n ＝43a n －1.由S n ＝4a n －p ，令n ＝1，得a 1＝4a 1－p ，解得a 1＝p3.所以{a n }是首项为p 3，公比为43的等比数列．(2)因为a 1＝1，则a n ＝(43)n －1，由b n ＋1＝a n ＋b n (n ＝1,2，…)，得b n ＋1－b n ＝(43)n －1，当n ≥2时，由累加法得b n ＝b 1＋(b 2－b 1)＋(b 3－b 2)＋…＋(b n －b n －1) ＝2＋1－(43)n －11－43＝3(43)n －1－1，当n ＝1时，上式也成立．∴b n ＝3·(43)n －1－1.[方法点拨] 证明数列是等差(等比)数列时，应用定义分析条件，结合性质进行等价转化． (理)(2015·河南高考适应性测试)已知数列{a n }的各项均为正数，且a 1＝2，a n ＝a 2n ＋1＋4a n ＋1＋2.(1)令b n ＝log 2(a n ＋2)，证明：数列{b n }是等比数列． (2)设c n ＝nb n ，求数列{c n }的前n 项和S n .[解析] (1)由a n ＝a 2n ＋1＋4a n ＋1＋2，得a n ＋2＝a 2n ＋1＋4a n ＋1＋4＝(a n ＋1＋2)2.因为a n >0，所以a n ＋2＝a n ＋1＋2. 因为b n ＋1b n ＝log 2(a n ＋1＋2)log 2(a n ＋2)＝log 2a n ＋2log 2(a n ＋2)＝12，又b 1＝log 2(a 1＋2)＝2，所以数列{b n }是首项为2，公比为12的等比数列．(2)由(1)知，b n ＝2·⎝⎛⎭⎫12n －1，则c n ＝2n ⎝⎛⎭⎫12n －1. S n ＝2×⎝⎛⎭⎫120＋4×⎝⎛⎭⎫121＋…＋2(n －1)⎝⎛⎭⎫12n －2＋2n ⎝⎛⎭⎫12n －1，① 12S n ＝2×⎝⎛⎭⎫121＋4×⎝⎛⎭⎫122＋…＋2(n －1)⎝⎛⎭⎫12n －1＋2n ⎝⎛⎭⎫12n .② ①－②得：12S n ＝2×⎝⎛⎭⎫120＋2×⎝⎛⎭⎫121＋2×⎝⎛⎭⎫122＋…＋2×⎝⎛⎭⎫12n －1－2n ·⎝⎛⎭⎫12n ＝21－⎝⎛⎭⎫12n1－12－2n ·⎝⎛⎭⎫12n ＝4－(4＋2n )⎝⎛⎭⎫12n . 所以S n ＝8－(n ＋2)⎝⎛⎭⎫12n －2.15．(2015·南昌市一模)已知等差数列{a n }的前n 项和为S n ，a 1＝1，S 3＝6，正项数列{b n }满足b 1·b 2·b 3·…·b n ＝2S n .(1)求数列{a n }，{b n }的通项公式；(2)若λb n >a n 对n ∈N *均成立，求实数λ的取值范围．[解析] (1)等差数列{a n }，a 1＝1，S 3＝6，∴d ＝1，故a n ＝n⎩⎪⎨⎪⎧b 1·b 2·b 3·…·b n ＝2S n (1)b 1·b 2·b 3·…·b n －1＝2S n －1 (2)，(1)÷(2)得b n ＝2S n －S n －1＝2a n ＝2n (n ≥2)， b 1＝2S 1＝21＝2，满足通项公式，故b n ＝2n(2) 设λb n >a n 恒成立⇒λ>n 2n 恒成立，设c n ＝n 2n ⇒c n ＋1c n ＝n ＋12n当n ≥2时，c n <1，{c n }单调递减， ∴(c n )max ＝c 1＝12，故λ>12.16．(文)(2014·湖北理，18)已知等差数列{a n }满足：a 1＝2，且a 1，a 2，a 5成等比数列． (1)求数列{a n }的通项公式；(2)记S n 为数列{a n }的前n 项和，是否存在正整数n ，使得S n >60n ＋800？若存在，求n 的最小值；若不存在，说明理由．[分析] (1)设数列{a n }的公差为d ，利用等比数列的性质得到a 22＝a 1·a 5，并用a 1、d 表示a 2、a 5，列等式求解公差d ，进而求出通项，注意对公差d 分类讨论；(2)利用(1)的结论，对数列{a n }的通项分类讨论，分别利用通项公式及等差数列的前n 项和公式求解S n ，然后根据S n >60n ＋800列不等式求解．[解析] (1)设数列{a n }的公差为d ，依题意，2,2＋d,2＋4d 成等比数列，故有(2＋d )2＝2(2＋4d )．化简得d 2－4d ＝0，解得d ＝0或d ＝4. 当d ＝0时，a n ＝2；当d ＝4时，a n ＝2＋(n －1)·4＝4n －2，从而得数列{a n }的通项公式为a n ＝2或a n ＝4n －2. (2)当a n ＝2时，S n ＝2n ，显然2n <60n ＋800， 此时不存在正整数n ，使得S n >60n ＋800成立， 当a n ＝4n －2时，S n ＝n [2＋(4n －2)]2＝2n 2，令2n 2>60n ＋800，即n 2－30n －400>0， 解得n >40或n <－10(舍去)．此时存在正整数n ，使得S n >60n ＋800成立，n 的最小值为41. 综上，当a n ＝2时，不存在满足题意的n ；当a n ＝4n －2时，存在满足题意的n ，其最小值为41.[方法点拨] 存在型探索性问题解答时先假设存在，依据相关知识(概念、定理、公式、法则、性质等)，结合所给条件进行推理或运算，直到得出结果或一个明显成立或错误的结论，从而断定存在与否．(理)(2014·新课标Ⅰ理，17)已知数列{a n}的前n项和为S n，a1＝1，a n≠0，a n a n＋1＝λS n－1，其中λ为常数．(1)证明：a n＋2－a n＝λ；(2)是否存在λ，使得{a n}为等差数列？并说明理由．[分析](1)利用a n＋1＝S n＋1－S n用配凑法可获证；(2)假设存在λ，则a1，a2，a3应成等差数列求出λ的值，然后依据a n＋2－a n＝λ推证{a n}为等差数列．[解析](1)由题设：a n a n＋1＝λS n－1，a n＋1a n＋2＝λS n＋1－1，两式相减得a n＋1(a n＋2－a n)＝λa n＋1.由于a n＋1≠0，所以a n＋2－a n＝λ.(2)由题设，a1＝1，a1a2＝λS1－1，可得a2＝λ－1.由(1)知，a3＝λ＋1，令2a2＝a1＋a3，解得λ＝4.故a n＋2－a n＝4，由此可得{a2n－1}是首项为1，公差为4的等差数列，a2n－1＝4n－3；{a2n}是首项为3，公差为4的等差数列，a2n＝4n－1.所以a n＝2n－1，a n＋1－a n＝2.因此存在λ＝4，使得数列{a n}为等差数列.。

2014年河南省许昌、新乡、平顶山三市高考数学二模试卷（理科） 参考答案与试题解析 一、选择题：1．集合()(){}1231A x x x =--≤，312B x x ⎧⎫=-<<⎨⎬⎩⎭，则A B 为（ ）A ．1322x x ⎧⎫<⎨⎬⎩⎭≤B ．312x x ⎧⎫<⎨⎬⎩⎭≤C ．1322x x ⎧⎫⎨⎬⎩⎭≤≤ D ．1322x x ⎧⎫<⎨⎬⎩⎭≤ 答案：D【考点】交集及其运算． 【专题】集合．【分析】求出A 中不等式的解集确定出A ，找出A 与B 的交集即可．【解答】解：由A 中的不等式变形得：22520x x -+≤，即()()2120x x --≤，解得：122x ≤≤，即122A x ⎧⎫=⎨⎬⎩⎭≤≤； 312B x x ⎧⎫∴=-<<⎨⎬⎩⎭， 1322A B x x ⎧⎫∴=<<⎨⎬⎩⎭ ．故选：D ．【点评】此题考查了交集及其运算，熟练掌握交集的定义是解本题的关键． 2．在样本频率分布直方图中，共有五个小长方形，这五个小长方形的面积由小到大成等差数列{}n a ．已知212a a =，且样本容量为300，则小长方形面积最大的一组的频数为（ ） A ．100 B ．120 C ．150 D ．200 答案：A【考点】频率分布直方图． 【专题】概率与统计．【分析】根据直方图中的各个矩形的面积代表了频率，各个矩形面积之和为1，求出小长方形面积最大的一组的频率，再根据频数=频率⨯样本容量，求出频数即可．【解答】解： 直方图中的各个矩形的面积代表了频率，这5个小方形的面积由小到大构成等差数列{}n a ，212a a =，1d a ∴=，313a a =，414a a =，515a a =根据各个矩形面积之和为1，则123451151a a a a a a ++++==1115a ∴=，小长方形面积最大的一组的频率为5115153a =⨯= 根据频率=频数样本容量可求出频数13001003=⨯=故选：A ．【点评】本题考查了频率、频数的应用问题，各小组频数之和等于样本容量，各小组频率之和等于1． 3．复数1z 、2z 满足()214i z m m =+-，()()22cos 3sin i ,,z m θλθλθ=++∈R ，并且12z z =，则λ的取值范围是（ ）A ．[]1,1-B ．9,116⎡⎤-⎢⎥⎣⎦C ．9,716⎡⎤-⎢⎥⎣⎦D ．9,116⎡⎤⎢⎥⎣⎦答案：C【考点】复数代数形式的混合运算． 【专题】数系的扩充和复数．【分析】利用12z z =，可得22cos 43sin m m θλθ=⎧⎨-=+⎩，化为2394sin 816λθ⎛⎫=-- ⎪⎝⎭，利用1sin 1θ-≤≤和二次函数的单调性即可得出．【解答】解：12z z = ，22cos 43sin m m θλθ=⎧∴⎨-=+⎩， 化为24sin 3sin θλθ=+，2394sin 816λθ⎛⎫∴=-- ⎪⎝⎭，1sin 1θ- ≤≤，∴当3sin 8θ=时，λ取得最小值916-；当sin 1θ=-时，λ取得最大值7．9716λ∴-≤≤．∴λ的取值范围是9,716⎡⎤-⎢⎥⎣⎦．故选：C ．【点评】本题考查了复数相等、正弦函数的单调性、二次函数的单调性，属于基础题．4．已知α是三角形的最大内角，且1cos22α=，则曲线221cos sin x y αα+=的离心率为（ ）ABCD答案：D【考点】双曲线的简单性质；二倍角的余弦． 【专题】圆锥曲线的定义、性质与方程．【分析】由已知条件推导出150α=︒，曲线221cos sin x y αα+=等价转化为22112y =，由此能求出结果． 【解答】解：α 是三角形的最大内角，且1cos22α=，2300α∴=︒，150α∴=︒，cos cos150cos30α∴=︒=-︒=，1sin sin150sin302α=︒=︒=，∵曲线221cos sin x y αα+=，2112y 2∴=，a ∴=c =e=c a ∴==． 故选：D ．【点评】本题考查双曲线的求法，是中档题，解题时要熟练掌握三角函数的性质．5．已知实数x ，y 满足不等式组315033505x y x y y -+⎧⎪+-⎨⎪⎩≥≤≥，则z x y =+的最大值为（ ）A ．15B ．17C ．20D ．30 答案：B【考点】简单线性规划． 【专题】数形结合．【分析】由线性约束条件作出可行域，求出最优解，则目标函数的最大值可求．【解答】解：由不等式组315033505x y x y y -+⎧⎪+-⎨⎪⎩≥≤≥作可行域如图，联立31503350x y x y -+=⎧⎨+-=⎩，解得98x y =⎧⎨=⎩．()9,8B ∴．由图可知，使z x y =+取得最大值的最优解为()9,8B ． z x y ∴=+的最大值为9817+=．故选：B ．【点评】本题只是直接考查线性规划问题，近年来线性规划问题高考数学考试的热点，数形结合法是重要的数学思想方法，是连接代数和几何的重要方法．随着要求数学知识从书本到实际生活的呼声不断升高，线性规划这一类新型数学应用问题要引起重视．是中档题．6．已知i为执行如图所示的程序框图输出的结果，则二项式6⎛ ⎝的展开式中含2x -的系数是（ ）A ．192B ．32C ．42-D ．192- 答案：C【考点】程序框图；二项式定理的应用． 【专题】算法和程序框图．【分析】根据框图的流程依次计算运行的结果，直到不满足条件100S ≤，求得输出i 的值，再利用二项展开式定理的通项公式求得2x -的系数．【解答】解：由程序框图知：程序第一次运行i=1，11021S -=+=； 第二次运行i=1+1=2，21123S -=+=； 第三次运行i=2+1=3，21227S =++=； 第四次运行i=3+1=4，37215S =+=； 第五次运行i=4+1=5，415231S =+=； 第六次运行i=5+1=6，531263S =+=； 第七次运行i=6+1=7，6632127S =+=． 不满足条件100S ≤，输出i=7，6⎛∴ ⎝的通项()662216C 71r rr r r r T x x ---+=⋅⋅-⋅，令6222r r --=-得5r =，2x-∴的系数为()5561C 742-⋅⋅=-．故选：C ．【点评】本题考查了循环结构的程序框图，考查了二项展开式定理，根据框图的流程依次计算运行的结果是解答此类问题的常用方法．7．若双曲线()2210,0x y a b a b -=>>和椭圆()2210x y m n m n+=>>有共同的焦点1F ，2F ，P 是两条曲线的一个交点，则12PF PF ⋅=（ ） A ．22m a - B．()12m a - D ．()m a - 答案：D【考点】双曲线的标准方程．【专题】圆锥曲线的定义、性质与方程．【分析】在同一直角坐标系中作出双曲线()2210,0x y a b a b -=>>和椭圆()2210x y m n m n+=>>的图形，利用双曲线与椭圆的定义得到1PF 与2PF 的关系式，从而可求得12PF PF ⋅的值．【解答】解：依题意，作图如下：不妨设点P 为第一象限的交点则12PF PF +=，①12PF PF -=②22①-②得：()1244PF PF m a ⋅=-，12PF PF m a ∴⋅=-，故选：D ．【点评】本题考查双曲线与椭圆的定义及其标准方程，考查作图与运算求解能力，属于中档题． 8．已知函数()e x f x =，如果1x ，2x ∈R ，且12x x ≠，下列关于()f x 的性质： ①()()()12120x x f x f x -->⎡⎤⎣⎦； ②()y f x =不存在反函数；③()()121222x x f x f x f +⎛⎫+< ⎪⎝⎭；④方程()2f x x =在()0,+∞上没有实数根，其中正确的是（ ）A ．①②B ．①④C ．①③D ．③④答案：B【考点】命题的真假判断与应用． 【专题】函数的性质及应用．【分析】利用函数的单调性判断①的正误；通过函数具有反函数的性质判断②的正误；利用函数的凹凸性判断③的正误；函数的零点判断④的正误．【解答】解：函数()e x f x =，函数是单调增函数，如果1x ，2x ∈R ，且12x x ≠， ①()()()12120x x f x f x -->⎡⎤⎣⎦；说明函数是增函数，满足题意，∴①正确； ②()y f x =不存在反函数；函数有反函数函数必须是单调函数，∴②不正确；③具有性质()()121222x x f x f x f +⎛⎫+< ⎪⎝⎭的函数是凸函数，而()e x f x =是凹函数；∴③不正确； ④方程()2f x x =，即2e x x =，函数()e x f x =，()2g x x =．在()0,+∞上没有交点，就是说分没有实数根，∴④正确．综上正确的结果为：①④． 故选：B ．【点评】本题考查函数的基本性质的应用，函数的单调性、反函数函数的凹凸性以及函数的零点，基本知识考查．9．设{}n a 是等比数列，n S 是{}n a 的前n 项和，对任意正整数n ，有1220n n n a a a ++++=，又12a =，则101S =（ ） A ．200 B ．2 C ．2- D ．0答案：B【考点】等比数列的性质；等比数列的前n 项和． 【专题】计算题．【分析】设出等比数列的公比为q ，利用等比数列的性质化简已知的等式，根据0n a ≠，等式左右两边同时除以n a ，得到关于q 的方程，求出方程的解得到公比q 的值，由1a 及q 的值，利用等比数列的前n 项和公式即可求出101S 的值．【解答】解析：设等比数列{}n a 的公比为q ，对任意正整数n ，有1220n n n a a a ++++=， 220n n n a a q a q ∴++=，又0n a ≠，可得：2120q q ++=， 解得： 1q =-，又12a =， 则()101211211S ⨯+==+．故选B【点评】此题考查了等比数列的性质，以及等比数列的前n 项和公式，熟练掌握性质及公式是解本题的关键．10．在三棱椎P ABC -中，PA ⊥平面ABC ，AC BC ⊥，D 为侧棱PC 上的一点，它的正视图和侧视图如图所示，则下列命题正确的是（ ）CDAP正视图侧视图A ．AD ⊥平面PBC 且三棱椎D ABC -的体积为83B ．BD ⊥平面PAC 且三棱椎D ABC -的体积为83C ．AD ⊥平面PBC 且三棱椎D ABC -的体积为163D ．BD ⊥平面PAC 且三棱椎D ABC -的体积为163答案：C【考点】直线与平面垂直的判定；命题的真假判断与应用；简单空间图形的三视图． 【专题】空间位置关系与距离．【分析】通过证明直线与平面内的两条相交直线垂直即可证明直线与平面垂直，求出几何体的体积即可．【解答】解：PA ⊥ 平面ABC ，PA BC ∴⊥，又AC BC ⊥，PA AC A = ， BC ∴⊥平面PAC ， BC AD ∴⊥，又由三视图可得在PAC △中，4PA AC ==，D 为PC 的中点， AD PC ∴⊥，AD ∴⊥平面PBC ．又4BC =，90ADC ∠=︒，BC ⊥平面PAC ．故11164323D ABC B ADC V V --==⨯⨯=．故选：C ．【点评】本题考查直线与平面垂直的判断，几何体的体积的求法，考查命题的真假的判断与应用．11．已知函数()2cos sin f x x x =，下列结论中错误的是（ ）A ． ()f x 既是偶函数又是周期函数B ．()f x 最大值是1C ． ()f x 的图象关于点π,02⎛⎫⎪⎝⎭对称 D ．()f x 的图象关于直线πx =对称答案：B【考点】三角函数中的恒等变换应用；三角函数的周期性及其求法． 【专题】三角函数的图像与性质．【分析】利用函数的周期性、奇偶性、对称性的概念对A 、B 、C 、D 四个选项逐一分析即可． 【解答】解：A ，()2cos sin f x x x = ，()()()()22cos sin cos sin f x x x x x f x ∴-=--==， ()f x ∴是偶函数；又()()()()222πcos 2πsin 2πcos sin f x x x x x f x +=+=+==， ()f x 是周期函数；()f x ∴既是偶函数又是周期函数，即A 正确；B ，cos 1x ≤，2sin 1x ≤，二者不能同时取到等号，∴无论x 取什么值，()2cos sin f x x x =均取不到值1，故B 错误；C ，()()()()2222πcos sin cos πsin πcos sin cos sin 0f x f x x x x x x x x x +-=+--=-= ， ()f x ∴的图象关于点π,02⎛⎫⎪⎝⎭对称，即C 正确；D ，()()()()222πcos 2πsin 2πcos sin f x x x x x f x -=--== ， ()f x ∴的图象关于直线πx =对称，即D 正确．综上所述，结论中错误的是：B ．故选：B ．【点评】本题考查三角函数的性质，着重考查函数的周期性、奇偶性、对称性及最值，考查分析问题、解决问题的能力，属于中档题．12．自平面上一点O 引两条射线OA ，OB ，点P 在OA 上运劝，点Q 在OB 上运动且保持PQ为定值a（点P ，Q 不与点O 重合），已知60AOB ∠=︒，a =PQ PO QP QOPO QO ⋅⋅+ 的取值范围为（ ）答案：BA ．1,2⎡⎢⎣ B ．,⎝ C ．1,2⎛- ⎝ D ．7⎛⎤ ⎥ ⎝⎦答案：B【考点】平面向量数量积的运算． 【专题】平面向量及应用．【分析】作图，记向量PQ 与PO 的夹角为α，0120α︒<<︒可得向量QP 与QO的夹角为120α︒-，可得()cos cos 120PQ PO QP QO PQ QP PO QOαα⋅⋅+=+︒-，由三角函数的公式化简结合角的范围可得所求．【解答】解：（如图）记向量PQ 与PO的夹角为α，0120α︒<<︒可得向量QP 与QO的夹角为()18060120αα︒-︒+=︒-， ()cos cos 120PQ PO QP QO PQ QP PO QO αα⋅⋅∴+=+︒-()1120cos cos 2ααααα⎫=+︒-=-+⎪⎪⎭()1cos302ααα⎫==+︒⎪⎪⎭0120α∴︒<<︒，3030150α∴︒<+︒<︒()1sin301α∴<+︒≤()30α<+︒≤.PQ PO QP QOPO QO⋅∴+的取值范围为,⎝故选：BA120°-ααOQB【点评】本题考查平面向量的数量积的运算，涉及三角函数的化简及应用，属中档题．二、填空题：13．过圆22240x y x y++-=的圆心，且与直线230x y+=垂直的直线方程为．答案：3270x y-+=【考点】圆的一般方程．【专题】直线与圆．【分析】求出圆的圆心，以及直线的斜率，利用点斜式方程即可得到直线的方程．【解答】解： 圆的标准方程为()()22125x y++-=，∴圆心坐标为()1,2-，直线230x y+=的斜率23k=-，则与直线230x y+=垂直的直线斜率32k=，∴所求的直线方程为()3212y x-=+，即3270x y-+=，故答案为：3270x y-+=【点评】本题主要考查直线方程的求法，求出圆心坐标以及直线斜率是解决本题的关键，比较基础．14．四棱锥P ABCD-中，底面ABCD是矩形，PA⊥底面ABCD，则这个五面体的五个面中两两互相垂直的共有对．答案：5【考点】平面与平面垂直的判定；棱锥的结构特征．【专题】证明题；空间位置关系与距离．【分析】因为PA⊥平面ABCD，得到2组互相垂直的平面．再利用四边形ABCD为正方形得到其他互相垂直的平面即可．【解答】解：因为PA⊥平面ABCD，所以平面PDA⊥平面ABCD，平面PAB⊥平面ABCD，又因为四边形ABCD为正方形，所以AB⊥平面PAD⇒平面ABP⊥平面PAD，同理可得平面PBC⊥平面PAB．平面PAD⊥平面PAB．故图中互相垂直的平面共有5组．故答案为：5．CBDAP【点评】本题考查面面垂直的判定．在证明面面垂直时，其常用方法是在其中一个平面内找两条相交直线和另一平面内的某一条直线垂直．15．已知()24g x x =--，()f x 为二次函数，满足()()()()0f x g x f x g x ++-+-=，且()f x 在[]1,2-上的最大值为7，则()f x = ．答案：2142x x -+或224x x -+【考点】二次函数的性质． 【专题】函数的性质及应用．【分析】设出函数的解析式，由()()()()0f x g x f x g x ++-+-=，可得二次项系数和常数项，结合二次函数的图象和性质分类讨论()f x 在[]1,2-上的最大值为7时，一次项系数的取值，最后综合讨论结果，可得答案．【解答】解：()f x 为二次函数，∴设()()20f x ax bx c a =++≠，则()()()()()()()()()222224422280f xg x f x g x ax bx c x ax bx c x a x c ++-+-=+++--+-++--=-+-=即220280a c -=⎧⎨-=⎩解得：14a c =⎧⎨=⎩()24f x x bx ∴=++，()f x 的图象是开口朝上且以直线2bx =-为对称轴的抛物线故当122b -≤，即1b -≥时，()f x 在[]1,2-上的最大值为()2287f b =+=，解得12b =-故当122b -≥，即1b -≤时，()f x 在[]1,2-上的最大值为()157f b -=-+=，解得2b =-，()2142f x x x ∴=-+或()224f x x x =-+，故答案为：2142x x -+或224x x -+．【点评】本题考查的知识点是二次函数的图象和性质，待定系数法求函数的解析式，熟练掌握选定系数法的步骤和二次函数的图象和性质是解答的关键． 16．如图所示，将正整数排成三角形数阵，每排的数称为一个群，从上到下顺次为第一群，第二群， ，第n 群， ，第n 群恰好n 个数，则第n 群中n 个数的和是 ．111828404832914202416710128564321答案：3223nn ⋅-- 【考点】归纳推理．【专题】规律型；等差数列与等比数列．【分析】观察图例，我们可以得到每一行的数第一个构成一个以1为首项，以2为公比的等比数列，每一行的从右边的第k 个数都构成一个以2k 为公差的等差数列，进而可分析出第n 群中n 个数的和的表达式．【解答】解：观察图例，我们可以得到每一行的数第一个构成一个以1为首项，以2为公比的等比数列，每一行的从右边的第k 个数都构成一个以2k 为公差的等差数列， 故第n 群的第一个数为：12n -，第n 群的第二个数为：2122232n n n ---+=⋅， 第n 群的第三个数为：22322252n n n ---+⨯=⋅， …第n 群的第1n -个数为：()()2222232n n +-⨯=-⋅， 第n 群的第n 个数为：()11221n n +-⨯=-，故第n 群中n 个数的和()()1232325223221n n n n S n n ---=+⋅+⋅++-⋅+- ，…① 故()()122223252232212n n n n S n n --=+⋅+⋅++-⋅+-⋅ ，…② ②-①得：()()122222222213223n n n n n S n n --=+++++--=⋅-- ，故答案为： 3223n n ⋅--【点评】归纳推理的一般步骤是：（1）通过观察个别情况发现某些相同性质；（2）从已知的相同性质中推出一个明确表达的一般性命题（猜想）．三、解答题：解答应写出文字说明、证明过程或演算步骤．17．在ABC △中，角A 、B 、C 所对的边分别为a 、b 、c ，己知()πcos ,A A =，()2cos ,2cos n A A =-，π1n ⋅=- ．（Ⅰ）若a =2c =，求ABC △的面积；（Ⅱ）求()2cos 60b ca C -︒+的值．【考点】正弦定理；平面向量数量积的运算． 【专题】三角函数的求值． 【分析】（Ⅰ）由两向量的坐标及两向量数量积为1-，利用平面向量数量积运算法则计算列出关系式，再利用两角和与差的正弦函数公式化为一个角的正弦函数，确定出A 的度数，由a 与c 的值，利用正弦定理求出sin C 的值，即可确定出ABC △的面积；（Ⅱ）原式利用正弦定理化简后，根据A 的度数，得到B C +的度数，用C 表示出B ，代入关系式整理后约分即可得到结果．【解答】解：（Ⅰ）()πcos ,A A = ，()2cos ,2cos n A A =-，π1n ⋅=- ．222cos cos cos 211A A A A A ∴-=+=-，即2212cos 22A A ⎫--=-⎪⎪⎝⎭， πsin 216A ⎛⎫∴-= ⎪⎝⎭，A 为三角形内角，ππ262A ∴-=，即π3A =，a = 2c =，∴由正弦定理sin sin a cA C=，得：2sin 1sin 2c A C a ===， C 为三角形内角，π6C ∴=，π2B ∴=，则122ABC S =⨯⨯△；（Ⅱ）2sin sin sin a b cR A B C=== ，即2sin a R A =，2sin b R B =，2sin c R C =，∴原式()1sin 2sin sin 1202sin 60sin 2sin 2sin cos 60C C C C C C B C A C +-︒--︒+-======︒+【点评】此题考查了正弦定理，平面向量的数量积运算，两角和与差的正弦函数公式，熟练掌握定理及公式是解本题的关键．18．甲、乙、丙、丁、戊5名学生进行劳动技术比赛，决出第一名至第五名的名次．比赛之后甲乙两位参赛者去询问成绩，回答者对甲说“根遗憾，你和乙都投有得到冠军”，对乙说“你当然不会是最差的”． （Ⅰ）从上述回答分析，5人的名次排列可能有多少种不同的情况；（Ⅱ）比赛组委会规定，第一名获奖金1000元，第二名获奖金800元，第三名获奖金600元，第四及第五名没有奖金，求丙获奖金数的期望．【考点】离散型随机变量的期望与方差；排列、组合的实际应用． 【专题】概率与统计． 【分析】（Ⅰ）由已知条件，先求出冠军有几种可能，再求乙的名次有几种可能，上述位置确定后，求出甲连同其余二人可任意排列，有几种可能，按乘法原理计算名次排列的可能情况的种数．（Ⅱ）丙可能获得第一名、第二名、第三名、第四名或第五名，并分别求出相应的概率，能得到随机变量丙获得奖金数X 的可能取值为1000，800，600，0，由此能求出结果． 【解答】解：（Ⅰ） 甲、乙都没有得冠军， ∴冠军是其余3人中的一个，有13A 种可能， 乙不是第五名，∴乙是第二、第三或第四名中的一名，有13A 种可能，上述位置确定后，甲连同其余二人可任意排列，有33A 种可能， ∴名次排列的可能情况的种数有：113333A A A 54⋅⋅=种可能．（Ⅱ）丙可能获得第一名、第二名、第三名、第四名或第五名，P （丙获第一名）13=，P （丙获第二名）111222C C C 45427==， P （丙获第三名）P =（丙获第四名）427=，P （丙获第五名）29=，∴随机变量丙获得奖金数X 的可能取值为1000，800，600，0，()110003P X p ==，()480027P X ==， ()460027P X ==， ()4210027927P X ==+=， 1441460010008006003272727EX =⨯+⨯+⨯=（元）． 【点评】本题考查离散型随机变量的分布列和数学期望，是中档题，在历年高考中都是必考题．解题时要注意排列组合的合理运用．19．已知四棱锥P ABCD -中，PC ⊥底面ABCD ，2PC =，且底面ABCD 是边长为1的正方形．E 是最短的侧棱PC 上的动点．（Ⅰ）求证：P 、A 、B 、C 、D 五点在同一个球面上，并求该球的体积；（Ⅱ）如果点F 在线段BD 上，3DF BF =，EF ∥平面PAB ，求PEEC 的值．DAFBCEP【考点】与二面角有关的立体几何综合题；球的体积和表面积． 【专题】综合题；空间位置关系与距离． 【分析】（Ⅰ）设PA 的中点为M ，证明CM PM AM BM DM ====，即可得出结论； （Ⅱ）连接CF 并延长交AB 于K ，连接PK ，则利用线面平行的性质，可得EF PK ∥，利用3DF BF =，AB CD ∥，即可得出结论． 【解答】（Ⅰ）证明：设PA 的中点为M ，则 PAC △为直角三角形，CM PM AM ∴===．设正方形ABCD 的中心为点O ，则OM PC ∥，1OM =且PC ⊥底面ABCD ， OM ∴⊥底面ABCD ， O 为BD 的中点，BM DM ∴==，CM PM AM BM DM ∴====，P ∴、A 、B 、C 、D 五点在以M 为球心，球的体积为34π3⋅=⎝⎭； （Ⅱ）解：连接CF 并延长交AB 于K ，连接PK ，则EF ∥平面PAB ，EF ⊂面PCK ，面PCK 平面PAB PK =， EF PK ∴∥，3DF BF = ，AB CD ∥，3CF KF ∴=， EF PK ∥，3CE PE ∴=， 13PE EC ∴=．EP【点评】本题考查线面平行的性质，考查线面垂直，考查学生分析解决问题的能力，属于中档题．20．已知椭圆()2222:10x y E a b ab+=>>，过其右焦点2F 作与x 轴垂直的直线l 与该椭圆交于A 、B 两点，与抛物线24y x =交于C 、D 两点，且AB = ． （Ⅰ）求椭圆E 的方程；（Ⅱ）若过点()2,0M 的直线与椭圆E 相交于G 、H 两点，设P 为椭圆E 上一点，且满足OG OH tOP +=（O 为坐标原点），当OG OH -< 时，求实数t 的取值范围．【考点】直线与圆锥曲线的综合问题．【专题】圆锥曲线中的最值与范围问题．【分析】（Ⅰ）由题设条件推导出2222c a baa b c ⎧=⎪⎪⎪=⎨⎪⎪=+⎪⎩，由此能求出椭圆E 的方程．（Ⅱ）设直线GH 的方程为2x my =+，联立22213216x my x y =+⎧⎪⎨+=⎪⎩，得()2224280m y my ++-=，由此入手能求出实数t 的取值范围． 【解答】解：（Ⅰ） 直线l 过右焦点2F 且于x 轴垂直，22bAB a∴=，CD =又 椭圆E，且AB =，2222c ab a a bc ⎧=⎪⎪⎪∴=⎨⎪⎪=+⎪⎩，解得223216a b ⎧=⎪⎨=⎪⎩， ∴椭圆E 的方程为：2213216x y +=．（Ⅱ）由题意知直线GH 的斜率不为0，设直线GH 的方程为2x my =+，联立22213216x my x y =+⎧⎪⎨+=⎪⎩，消去x 得()2224280m y my ++-=，设(),P x y ，()11,G x y ，()22,H x y ，12242m y y m ∴+=-+，122282y y m =-+， ()12122842x x m y y m ∴+=++=+， OG OH tOP += ，1221228242tx x x m m ty y y m ⎧=+=⎪⎪+∴⎨⎪=+=-⎪+⎩，()()2284,22m P t m t m ⎛⎫ ⎪∴- ⎪++⎝⎭， P 点在椭圆上，∴将P 点代入椭圆方程，得2212t m =+，OG OH -()()222121GH m y y ∴=+-()()22121214m y y y y ⎡⎤=++-⎣⎦()22224428122m m m m ⎡⎤-⨯⎛⎫=++⎢⎥ ⎪++⎝⎭⎢⎥⎣⎦ ()()()222232147641192m m m++⨯=<+, 421411250m m +-<,201m ∴<≤，22111,232t m ⎛⎫∴=∈ ⎪+⎝⎭，,t ⎡∴∈⎢⎣⎭⎝⎦． ∴实数t的取值范围是,⎡⎢⎣⎭⎝⎦． 【点评】本题考查椭圆方程的求法，考查实数的取值范围的求法，综合性强，难度大，解题时要综合运用直线与圆锥曲线的位置关系，合理地进行等价转化．21．已知函数()()()32ln 2123x f x ax x ax a =++--∈R ，（Ⅰ）若()y f x =在[)3,+∞上为增函数，求实数a 的取值范围；（Ⅱ）当12a =-时，方程()()3113x b f x x --=+有实根，求实数b 的最大值． 【考点】导数在最大值、最小值问题中的应用．有 【专题】综合题；导数的综合应用．【分析】（Ⅰ）()y f x =在[)3,+∞上为增函数，等价于()'f x ()()2221442021x ax a x a ax ⎡⎤+--+⎣⎦=+≥在[)3,+∞上恒成立，分类讨论，当0a ≠时，由函数()f x 的定义域可知，必须有210ax +>对3x ≥恒成立，故只能0a >，所以()()22214420ax a x a +--+≥在[)3,+∞上恒成立，构造函数()()()2221442g x ax a x a =+--+，要使()0g x ≥在[)3,+∞上恒成立，只要()30g ≥即可，从而可求实数a 的取值范围；（Ⅱ）当12a =-时，方程()()3113x b f x x --=+有实根，等价于23ln b x x x x =+-在()0,+∞上有解，即求()23ln g x x x x x =+-的值域．构造()()2ln 0h x x x x x =+->，证明()h x 在()0,1上为增函数，在()1,+∞上为减函数，即可得出结论．【解答】解：（I ）因为函数()y f x =在[)3,+∞上为增函数， 所以()()()2221442'021x ax a x a f x ax ⎡⎤+--+⎣⎦=+≥在[)3,+∞上恒成立当0a =时，()()'20f x x x =-≥在[)3,+∞上恒成立，所以()y f x =在[)3,+∞上为增函数，故0a =符合题意当0a ≠时，由函数()f x 的定义域可知，必须有210ax +>对3x ≥恒成立，故只能0a >， 所以()()22214420ax a x a +--+≥在[)3,+∞上恒成立 令函数()()()2221442g x ax a x a =+--+，其对称轴为114x a=-， 因为0a >，所以1114a-<， 要使()0g x ≥在[)3,+∞上恒成立，只要()30g ≥即可， 即()234610g a a =-++≥，a ≤因为0a >，所以0a <≤综上所述，a 的取值范围为0,⎡⎢⎣⎦；（Ⅱ）当12a =-时，方程()()3113x b f x x --=+有实根，等价于23ln b x x x x =+-在()0,+∞上有解， 即求()23ln g x x x x x =+-的值域．令()()2ln 0h x x x x x =+->，则()()()211'x x h x x+-=，01x ∴<<时，()'0h x >，从而()h x 在()0,1上为增函数，当1x >时()'0h x <，从而()h x 在()1,+∞上为减函数， ()()10h x h ∴=≤， 0x > ，()0b xh x ∴=≤， 1x ∴=时，b 取得最大值0．【点评】本题考查导数知识的综合运用，考查函数的单调性，考查函数的最值，构建函数是关键，也是难点．四、请考生在第22、23、24三题中任选一题作答，如果多做，则按所做的第一题记分．22．如图所示，ABC △是圆O 的内接三角形，AC BC =，D 为弧AB 上任一点，延长DA 至点E ，使CE CD =．（Ⅰ）求证：BD AE =；（Ⅱ）若AC BC ⊥，求证：AD BD +=．【考点】与圆有关的比例线段． 【专题】直线与圆． 【分析】（Ⅰ）由题意知CAD E ECA CAB BAD ∠=∠+∠=∠+∠，由此能够证明ECAQD DCB △△，从而得到BD AE =．（Ⅱ）由已知条件推导出90ECA ACD ∠+∠=︒，DE=，由此能够证明AD CD +． 【解答】（Ⅰ）证明：由题意知CAD E ECA CAB BAD ∠=∠+∠=∠+∠， AC BC = ，CAB DCB ∴∠=∠，ECA DCB ∴∠=∠， ECAQD DCB ∴△△，BD AE ∴=．（Ⅱ）证明：AC BC ⊥ ，90ACB DAB ACD ∴∠=︒=∠+∠， 90ECA ACD ∴∠+∠=︒，CECD = ，DE ∴=， BD AE = ，AD BD DE +=，AD CD ∴+=．【点评】本题考查线段长相等的证明，是中档题，解题时要认真审题，注意圆的简单性质的灵活运用． 五、坐标系与参数方程23．己知直线112:x t l y ⎧=+⎪⎪⎨⎪=⎪⎩．曲线1cos :sin x C y θθ=⎧⎨=⎩，（θ为参数）．（I ）设l 与1C 相交于A ，B 两点，求AB ；（Ⅱ）若把曲线1C 上各点的横坐标压缩为原来的12倍，得到曲线2C ，设点P 是曲线2C 上的一个动点，求它到直线l 的距离的最小值． 【考点】简单曲线的极坐标方程． 【专题】坐标系和参数方程． 【分析】（I ）把参数方程化为普通方程，联立方程组求得点A 、B 的坐标，可得AB 的值．（Ⅱ）由题意求得曲线2C 的参数方程，设点1cos ,2P θθ⎛⎫ ⎪ ⎪⎝⎭，求得点P到直线l的距离π24d θ⎤⎛⎫=-+ ⎪⎥⎝⎭⎦，再根据正弦函数的值域，求得d 的最小值． 【解答】解：（I）直线112:x t l y ⎧=+⎪⎪⎨⎪=⎪⎩的普通方程为)1y x -；曲线1cos :sin x C y θθ=⎧⎨=⎩，（θ为参数）的直角坐标方程为221x y +=．由)2211y x x y ⎧=-⎪⎨+=⎪⎩，求得11x y =⎧⎨=⎩，或12x y ⎧=⎪⎪⎨⎪=⎪⎩，()1,0A ∴、1,2B ⎛ ⎝⎭．1AB ∴==． （Ⅱ）由题意可得曲线2C的参数方程为1cos 2x y θθ⎧=⎪⎪⎨⎪=⎪⎩（θ为参数），设点1cos ,2P θθ⎛⎫ ⎪ ⎪⎝⎭，则点P 到直线l 的距离π24d θ⎤⎛⎫=-+ ⎪⎥⎝⎭⎦， 故当πsin 14θ⎛⎫-=- ⎪⎝⎭时，d)1． 【点评】本题主要考查把参数方程化为普通方程的方法，点到直线的距离公式的应用，直线和圆的位置关系，属于基础题 六、不等式选讲24．已知函数()1f x x x a =-+-．（Ⅰ）若2a =，解不等式()2f x ≥；（Ⅱ）若1a >，x ∀∈R ，()11f x x +-≥，求实数a 的取值范围． 【考点】绝对值不等式的解法．【专题】计算题；不等式的解法及应用．【分析】（Ⅰ）当2a =时，()23,1121,1223,2x x f x x x x x x -+<⎧⎪=-+-=⎨⎪->⎩≤≤，解不等式()2f x ≥即可求得答案；（Ⅱ）令()()1F x f x x =+-，则()32,12,132,x a x F x x a x a x a x a -++<⎧⎪=-+<⎨⎪--⎩≤≥函数先单调递减，再单调增，从而可得实数a 的取值范围．【解答】解：（Ⅰ）当2a =时，()23,1121,1223,2x x f x x x x x x -+<⎧⎪=-+-=⎨⎪->⎩≤≤，而()2f x ≥，解得12x ≤或52x ≥．（Ⅱ）令()()1F x f x x =+-，则()32,12,132,x a x F x x a x a x a x a -++<⎧⎪=-+<⎨⎪--⎩≤≥()y F x = 在(),1-∞上单调递减，在[)[)1,,a a +∞ 上单调递增，∴当1x =时，()F x 有最小值()11F a =-，11a ∴-≥，解得2a ≥，∴实数a 的取值范围为[)2,+∞．【点评】本题考查绝对值不等式的解法，分类讨论去掉绝对值符号是关键，考查运算求解能力，属于中档题．。

许昌新乡平顶山三市2014届高三第二次调研考试英语第二部分：阅读理解（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的四个选项（A、B、C和D）中，选出最佳选项，并在答题卡上将该项涂黑。

AJob InterviewsYou have the education and the work experience; now all you need is the job. Most people spend more time getting ready for an evening out than they do preparing for a job interview. There are several things you can do to have a successful job interview. Here are some of them:1. Know about the company where you are trying to get a job. Find out what the company does and who their competitors are.2. Know yourself. Most interviewers will ask a question related to your strengths and weaknesses. Before the interview, examine the job you are applying for and determine the top skills needed for that position. For instance, if you are applying for a job as a telephone operator, it’s more important that you have strong communication skills than that you are tidy.3. Have proof. It’s easy to say you have good organization skills. But if you can tell the interviewer about a time you organized an event, it makes the claim much more solid.4. Prepare questions to ask about the company and the job.An interview is a stressful event. When you are under pressure, it can be difficult to think straight. Preparing before the interview will make it easier to give answers to those tough questions. But you can’t think of every question they might ask. So relax and be confident. Remember, first impressions are lasting.【小题1】This article is about ________.A. how to get a job interviewB. how to communicate at workC. how to prepare for a job interview.D. how to interview people【小题2】A good skill for a telephone operator to have is ________.A. the ability to smileB. tidinessC. good communication skillsD. proof of past experience【小题3】 You should prepare questions about _________ when interviewed for a job.A. the companyB. the salaryC. the interviewer’s strengthsD. the weather【小题4】Which of the following is TRUE?A. The job interview should be conducted over the telephone.B. The first time the interviewer sees you is important.C. You shouldn’t ask questions about the company.D. You shouldn’t give examples of things you have done in the past.【答案】【小题1】C【小题2】C【小题3】A【小题4】B【解析】BGretchen Alexander is sightless. But she refuses to allow her blindness to limit her life activities. She enjoys archery, golf, softball, sailing and water skiing, as well as a number of other activities that those of us who are sighted have yet to learn. She also speaks to groupsabout living life fully. When speaking to a group of high school students, she was once asked if there was anything she wouldn’t try. “I’ve decided to never sky dive,” she answered. “It wo uld scare the heck out of my dog.”Why do some people rise above their problems and live life fully, while others become defeated? Merle Shain explains it this way: “There are only two ways to approach life,” Shain says, “as a victim or as a gallant fighte r. And you must decide if you want to act or to react…"When discouraged, a victim reacts, perhaps in pain or self-pity. But a fighter will act.A fighter will make a decision to change that set of circumstances that left her or him discouraged. Or a fighter will decide to accept those circumstances with grace and move ahead anyway. A fighter will decide to act with courage. A fighter will take responsibility for his or her happiness. No matter how afraid, a fighter will refuse to give in to the most defeating of all human emotions-helplessness. A victim reacts. A fighter acts. It’s your decision. It’s a decision about whether you will live your life fully and with courage, or whether you will be forever defeated by harsh circumstances. Make it well, for it may be one of the most important decisions you ever make.Will you be a victim or a gallant fighter?【小题1】The best title for this article would be _______.A. Gretchen Alexander’s LifeB. Merle Shain’s Attitude to LifeC. Victims and FightersD. The Way of Life【小题2】What can we learn about Gretchen Alexander from this passage?A. She is more athletic than those of us who are sighted.B. She is discouraged when her dog is scared.C. There is no limitation to her life activities.D. She is a brave fighter.【小题3】The underlined word “it”（paragraph 3）is closest in meaning to _________.A. life.B. choice.C. courage.D. circumstance.【小题4】 The third paragraph mainly talks about ________.A. the difference between a victim and a fighter.B. the reactions of helplessness.C. decision-making.D. a fighters responsibility.【答案】【小题1】C【小题2】D【小题3】B【小题4】A【解析】COcean ParkIf you love the sea, Ocean Park is the place for you! Situated on the south side of Hong Kong Island, this 870,000 square metre educational theme park provides many opportunities to learn about marine life.To start with, the park boasts the Atoll Reef, one of the world’s largest aquariums, with about 2,500 fish from nearly 300 different species. What makes this aquarium special, however, is not just its size, but also its design. The Atoll Reef is built with an observation passageway that circles the aquarium on four different levels. This lets visitors view sea life from a variety of depths and angles.Then there’s the Shark Aquarium, a tank w ith more than 200 sharks from more than 30 species. Like the Atoll Reef, this unique aquarium is designed to make sure guests get the most out oftheir visit. Shaped like an underwater tunnel, guests can watch as sharks swim overhead and dive at them from every side.There’s also the Sea Jelly Spectacular, an aquarium that houses more than 1,000 jellyfish of all shapes, colours and sizes. And at the park’s Dolphin University, visitors can go on educational tours and watch the training of dolphins up close.The park’s most popular attraction is the Ocean Theatre, a huge outdoor pool where dolphins and sea lions entertain the visitors. Sometimes a killer whale even takes part in the performance!Although Ocean Parks focus is on the water, the theme park has plenty of other activities, too. For people seeking excitement, there are rides like the Abyss Turbo Drop, a roller coaster ride that takes passengers on a 20-storey drop straight down. There are also exhibits like the Dinosaur Discovery Trail and Bird Paradise. Finally, no trip to Ocean Park would be complete without visiting the park’s most popular animals--four giant pandas that were given as a gift from China’s central government.【小题1】Why is Hong Kong Ocean Park called an educational theme Park?A. It is specially designed to attract the young who are interested in the sea.B. It provides chances for people to broaden their knowledge of science.C. It offers chances for visitors to enlarge their knowledge of marine life.D. It has plenty of activities for people to have fun.【小题2】What makes the Atoll Reef so special?A. It is one of the symbols of Hong Kong Ocean Park.B. It has thousands of fish from various species.C. It is the largest aquarium in the world.D. It allows visitors to watch sea life from all angles.【小题3】Which of the following activities is NOT mentioned in the passage?A. Interacting with sea life in the huge outdoor pool.B. Visiting exhibitions about dinosaurs and birds.C. Taking a roller coaster ride on a 20-storey drop.D. Enjoying the show of dolphins and sea lions.【答案】【小题1】C【小题2】D【小题3】A【解析】DSunday, 31 August We’ve been in China for a month now. Dad, Mom, Harry and I moved to Tianjin on 25 August. We’re not very far from Beijing. Two days ago, we celebrated my 16th birthday. It was great celebrating in China; the only thing that was strange was the cake--here they’re not as sweet as the ones in New York. On Monday school starts—I wonder what it will be like.Monday, 1 September On my first day I was looking around for a locker to put my books in. However, here all the students keep all of their books at their desks--we stay in the same classroom because apparently we don’t have to go from class to class--teachers come to us!Today we selected teacher assistants for each subject. Their duties are to collect homework, make announcements, and do other stuff for the teacher and the students. It’s kind of a big deal here! Since I am from the US, I was asked to be the English assistant. I felt so proud but quite nervous at the same time because I wasn’t sure what I had to do, but I accepted the job anyway.Friday, 3 October Boy, what a week! Now we have nine classes every day, including the morning class, a combination of our American schools’ “Homeroom” and “Study Hall”. I think Chinese students work too much! I have to do my homework when I get back home. I don’t even have time to watch TV or surf the Interne t like before. I sometimes miss New York and my school because we didn’t have to study so much. We had more time to hang out with our classmates and neighbors; here, besides their usual classes, students are involved in weekend classes in subjects such as English, Chinese and math.I get a lot of attention, being from another country. Everyone wants to practice English with me! A really cute girl even asked me for my phone number on my second day and sent me a text message! I’m making a lot more friends now. I just need a lot of help to improve my Chinese. Some students want to do a language exchange program with me. Nice!【小题1】 The passage mentions all the following points EXCEPT ________.A. physics study.B. food flavour.C. free time activities.D. language exchange program.【小题2】According to the passage, which of the following is NOT the teacher assistant’s duty?A. Collecting homework.B. Making announcements.C. helping teachers with small errands（差事）D. Teaching classmates.【小题3】Where is this passage most probably from?A. A story book.B. A guide book.C. A diary.D. A magazine.【小题4】The passage is best described by ________.A. culture shock.B. multi-culture.C. unique culture.D. culture background. 【答案】【小题1】A【小题2】D【小题3】C【小题4】A【解析】试题分析：这篇文章是到中国学习的美国学生的日常生活的日记，日记记录了作者的学习生活，也有作者经历的文化冲击。

河南省新乡、许昌、平顶山三市2014届高三语文第二次调研考试试题（扫描版）新人教版新乡许昌平顶山2014届高三第二次调研考试语文参考答案语文一、现代文阅读（9分，每小题3分）1. B（误将未然作已然)2. C（张冠李戴。

原文是孔子对这种光以较量射中并射穿为目的的“主皮之射”很不以为然。

）3. B（曲解文意。

原文是“乡射礼的比赛往往非常激烈，但是最后评价一名射手，不仅要看他能否射中靶心，还要看他形体动作是否合适于音乐节奏”，“此外，还要求礼让为先……”）二、古代诗文阅读（36分）（一）文言文阅读（19分）4. B（具：全部）5. C (①勤政为民；⑥弃官回家)6. D（不得已将官吏打发了的是包拯）7.（1）（5分）县里发洪水，老百姓遭水淹，县令不能拯救他们，胡宿带领公私船只救活了数千人。

（译出大意2分；“被溺”“率”“活”各1分。

）（2）（5分）却暗中袖手旁观，等他离开后便指责他，这难道就是古人分谤的意思吗？（译出大意2分；“阴”“俟”“非”各1分。

）（二）古代诗歌阅读（11分）8. 通过描绘寂静、昏暗、风雨凄迷的景象渲染了一种悲凉的气氛（2分），为全词奠定了感情基调，为后文表达词人飘零身世和凄凉心境作铺垫。

（3分）9. ①人在羁旅的寂寞思乡之情，②半生飘零的悲凉之情，③壮志未酬的惆怅之情。

（每点2分）（三）名篇名句默写（6分）10．（6分）（1）秋天漠漠向昏黑布衾多年冷似铁（2）侣鱼虾而友麋鹿举匏樽以相属（3）鼎铛玉石弃掷逦迤（每答对一空给1分，有错别字、多字、漏字则该空不给分）三、文学类文本阅读（25分）11.（1）答A给3分，答E给2分，答D给1分；答B、C不给分。

(B.对其作用分析过度。

C.没有据为己有的意思。

D. 此处主要是“不知所措慌忙掩饰的心理” )11.（2）①表现了雨后青草湖清新、明净的自然美景。

②以明净、清新之景烘托李老壮美好的心灵世界，给人以美的感受③为后文人物关系的转机做铺垫，推动了故事情节的发展。

许昌新乡平顶山三市2014届高三第二次调研考试英语第二部分：阅读理解（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的四个选项（A、B、C和D）中，选出最佳选项，并在答题卡上将该项涂黑。

AJob InterviewsYou have the education and the work experience; now all you need is the job. Most people spend more time getting ready for an evening out than they do preparing for a job interview. There are several things you can do to have a successful job interview. Here are some of them:1. Know about the company where you are trying to get a job. Find out what the company does and who their competitors are.2. Know yourself. Most interviewers will ask a question related to your strengths and weaknesses. Before the interview, examine the job you are applying for and determine the top skills needed for that position. For instance, if you are applying for a job as a telephone operator, it’s more important that you have strong communication skills than that you are tidy.3. Have proof. It’s easy to say you have good organization skills. But if you can tell the interviewer about a time you organized an event, it makes the claim much more solid.4. Prepare questions to ask about the company and the job.An interview is a stressful event. When you are under pressure, it can be difficult to think straight. Preparing before the interv iew will make it easier to give answers to those tough questions. But you can’t think of every question they might ask. So relax and be confident. Remember, first impressions are lasting.【小题1】This article is about ________.A. how to get a job interviewB. how to communicate at workC. how to prepare for a job interview.D. how to interview people【小题2】A good skill for a telephone operator to have is ________.A. the ability to smileB. tidinessC. good communication skillsD. proof of past experience【小题3】You should prepare questions about _________ when interviewed for a job.A. the companyB. the salaryC. the interviewer’s strengthsD. the weather【小题4】Which of the following is TRUE?A. The job interview should be conducted over the telephone.B. The first time the interviewer sees you is important.C. You shouldn’t ask questions about the company.D. You shouldn’t give examples of things you have done in the past.【答案】【小题1】C【小题2】C【小题3】A【小题4】B【解析】BGretchen Alexander is sightless. But she refuses to allow her blindness to limit her life activities. She enjoys archery, golf, softball, sailing and water skiing, as well as a number of other activities that those of us who are sighted have yet to learn. She also speaks to groups about living life fully. When speaking to a group of high school students, she was once asked if there was anything she wouldn’t try. “I’ve decided to never sky dive,” she an swered. “It would scare the heck out of my dog.”Why do some people rise above their problems and live life fully, while others become defeated? Merle Shain explains it this way: “There are only two ways to approach life,” Shain says, “as a victim or as a gallant fighter. And you must decide if you want to act or to react…"When discouraged, a victim reacts, perhaps in pain or self-pity. But a fighter will act. A fighter will make a decision to change that set of circumstances that left her or him discouraged. Or a fighter will decide to accept those circumstances with grace and move ahead anyway. A fighter will decide to act with courage. A fighter will take responsibility for his or her happiness. No matter how afraid, a fighter will refuse to give in to the most defeating of all human emotions-helplessness. A victim reacts. A fighter acts. It’s your decision. It’s a decision about whether you will live your life fully and with courage, or whether you will be forever defeated by harsh circumstances. Make it well, for it may be one of the most important decisions you ever make.Will you be a victim or a gallant fighter?【小题1】The best title for this article would be _______.A. Gretchen Alexander’s LifeB. Merle Shain’s Attitude to LifeC. Victims and FightersD. The Way of Life【小题2】What can we learn about Gretchen Alexander from this passage?A. She is more athletic than those of us who are sighted.B. She is discouraged when her dog is scared.C. There is no limitation to her life activities.D. She is a brave fighter.【小题3】The underlined word “it”（paragraph 3）is closest in meaning to _________.A. life.B. choice.C. courage.D. circumstance.【小题4】The third paragraph mainly talks about ________.A. the difference between a victim and a fighter.B. the reactions of helplessness.C. decision-making.D. a fighters responsibility.【答案】【小题1】C【小题2】D【小题3】B【小题4】A【解析】COcean ParkIf you love the sea, Ocean Park is the place for you! Situated on the south side of Hong Kong Island, this 870,000 square metre educational theme park provides many opportunities to learn about marine life.To start with, the park boasts the Atoll Reef, one of the world’s largest aq uariums, with about 2,500 fish from nearly 300 different species. What makes this aquarium special, however, is not just its size, but also its design. The Atoll Reef is built with an observation passageway that circles the aquarium on four different levels. This lets visitors view sea life from a variety of depths and angles.Then there’s the Shark Aquarium, a tank with more than 200 sharks from more than 30 species. Like the Atoll Reef, this unique aquarium is designed to make sure guests get the most out of their visit. Shaped like an underwater tunnel, guests can watch as sharks swim overhead and dive at them from every side.There’s also the Sea Jelly Spectacular, an aquarium that houses more than 1,000 jellyfish of all shapes, colours and sizes. And at the park’s Dolphin University, visitors can go on educational tours and watch the training of dolphins up close.The park’s most popular attraction is the Ocean Theatre, a huge outdoor pool where dolphins and sea lions entertain the visitors. Sometimes a killer whale even takes part in the performance!Although Ocean Parks focus is on the water, the theme park has plenty of other activities, too. For people seeking excitement, there are rides like the Abyss Turbo Drop, a roller coaster ride that takes passengers on a 20-storey drop straight down. There are also exhibits like the Dinosaur Discovery Trail and Bird Paradise. Finally, no trip to Ocean Park would be complete without visiting the park’s most popular animals--four giant pandas that were given as a gift from China’s central government.【小题1】Why is Hong Kong Ocean Park called an educational theme Park?A. It is specially designed to attract the young who are interested in the sea.B. It provides chances for people to broaden their knowledge of science.C. It offers chances for visitors to enlarge their knowledge of marine life.D. It has plenty of activities for people to have fun.【小题2】What makes the Atoll Reef so special?A. It is one of the symbols of Hong Kong Ocean Park.B. It has thousands of fish from various species.C. It is the largest aquarium in the world.D. It allows visitors to watch sea life from all angles.【小题3】Which of the following activities is NOT mentioned in the passage?A. Interacting with sea life in the huge outdoor pool.B. Visiting exhibitions about dinosaurs and birds.C. Taking a roller coaster ride on a 20-storey drop.D. Enjoying the show of dolphins and sea lions.【答案】【小题1】C【小题2】D【小题3】A【解析】DSunday, 31 August We’ve been in China for a month now. Dad, Mom, Harry and I moved to Tianjin on 25 August. We’re not very far from Beijing. Two days ago, we celebrated my 16th birthday. It was great celebrating in China; the only thing that was strange was the cake--here they’re not as sweet as the ones in New York. On Monday school starts—I wonder what it will be like.Monday, 1 September On my first day I was looking around for a locker to put my books in. However, here all the students keep all of their books at their desks--we stay in the same classroom because apparently we don’t have to go from class to class--teachers come to us!Today we selected teacher assistants for each subject. Their duties are to collect homework, make announcements, and do other stuff for the teacher and the students. It’s kind of a big deal here! Since I am from the US, I was asked to be the English assistant. I felt so proud but quite nervous at the same time because I wasn’t sure what I had to do, but I accepted the job anyway.Friday, 3 October Boy, what a week! Now we have nine classes every day, including the morning class, a combination of our American schools’ “Homeroom” and “Study Hall”. I think Chinese students work too much! I have to do myhomework when I get back home. I don’t even have time to watch TV or surf the Internet like before. I sometimes miss New York and my school because we didn’t have to study so much. We had more time to hang out with our classmates and neighbors; here, besides their usual classes, students are involved in weekend classes in subjects such as English, Chinese and math.I get a lot of attention, being from another country. Everyone wants to practice English with me! A really cute girl even asked me for my phone number on my second day and sent me a text message! I’m making a lot more friends now. I just need a lot of help to improve my Chinese. Some students want to do a language exchange program with me. Nice!【小题1】The passage mentions all the following points EXCEPT ________.A. physics study.B. food flavour.C. free time activities.D. language exchange program.【小题2】According to the passage, which of the f ollowing is NOT the teacher assistant’s duty?A. Collecting homework.B. Making announcements.C. helping teachers with small errands（差事）D. Teaching classmates.【小题3】Where is this passage most probably from?A. A story book.B. A guide book.C. A diary.D. A magazine.【小题4】The passage is best described by ________.A. culture shock.B. multi-culture.C. unique culture.D. culture background.【答案】【小题1】A【小题2】D【小题3】C【小题4】A【解析】试题分析：这篇文章是到中国学习的美国学生的日常生活的日记，日记记录了作者的学习生活，也有作者经历的文化冲击。