2014年广州一模答案

2014广州市一模语文答案广州市2014届高三调研测试语文试题参考答案及评分建议题号考点分值答案说明1 字音识记 3 C chuō/lù，jì/qì，juã/jiáo。

A.yì/sì，xiâ，jiâ/xiâ。

B.ɡuō/tián，kǎ/luò，zhònɡ。

D.cuì，qiān/jiān, tà/tuò。

2 词语使用3 C “不经之谈”，没有根据，不合情理的言论。

用在此处，不合语境。

A．“包罗万象”，指包括一切，形容内容丰富，无所不包，符合文意。

B．“执拗”，意为坚持己见,固执任性，符合文意。

D．“进而”，用于复句的后一分句，表示后一分句承接前一分句，强调在原有基础上更进一点，符合文意。

3 语病辨析 3 D A．搭配不当。

“加强”不能与“设施”搭配。

B．不合逻辑。

“防止”与“不再”否定不当。

C．表意不明。

“部分美声唱法和通俗唱法”可以理解成“部分美声唱法与所有的通俗唱法”或“两种唱法的部分选手”。

4 语言连贯 3 B 文段讨论的话题是“视野的近视”。

第②句中“患上了‘视野近视’病”紧承前文，说明患上“视野的近视”的原因；第③中的“这种视野的近视”紧承第②句，表明“视野近视”给人们带来的影响。

第⑥句“井底之蛙”与第③句中的“井底之蛙”对应，深入讨论话题。

第①句中的“这样的日子”指代第⑥句中的井底之蛙的日子，说明这种日子的特征。

第④句明确说明世界的变化会影响到个人，“迟早”一词与第①句中的“极其微弱”对应。

第⑤句中的关联词“如果”与后文的“那么”构成假设关系，提出预设的结果，收结语段。

5 文言实词 3 A 缘：沿着。

6 文言虚词 3 A 副词，均可译为“然后”“才”。

B．连词，表承接关系/连词，表修饰关系。

C．助词，“的”/代词，代“他”。

D．介词，“用”/连词，“因为”。

2014广州一模参考答案完形填空：1. B 上下文理解型，由下文语境＄25可以得出答案。

2. A 常识型，由常识可知，新娘在商店登记礼物是为了告诉买礼物的人这些是她喜欢的礼物（一般我们买礼物给人家也是买人家可能喜欢的礼物）。

3. C 上下文理解型，由上下文可知，当你买礼物的时候，告诉店家你已买了已登记的商品，以因此新婚夫妇不会收到两次同样的礼物。

解题关键在于twice.4. B 同义复现型，与下一句的句子结构一样，并与appropriate是近义词，因此可以得出答案。

5. D 上下文理解型，由上文的important occasions及常识可以得出答案（一般重要场合都是比较正式的）。

6. A 上下文理解型，由下文the boy’s full name可知，是专门为这小男孩而做的笔。

personalize有一个意思“在（某物上）作物主标志（尤指表示物主的地址或姓名的首字母”在（物品）上标出姓名（记号）; 此题较为容易选Ｃ.design。

Design只是设计的意思，而无“专门为某而做”的意思。

7. C 同义复现型，由下文typical of your homeland可以得出答案。

8. B 近义复现型，由valuable or“有价值的或是罕见的”可以得出答案。

9. D 上下文理解型。

上文讲给美国朋友选礼物的时候，要选一些你国家特有的东西，可能是一本有关你国家的书，不太昂贵的纪念品，或其他代表着你的文化的礼物。

10.C 上下文理解型，一个国家具有代表性的东西，通常代表着一个国家的文化，因此可选出答案。

11.B 上下文理解型，“喜欢收藏的小孩子可能会对你国家的一套硬币或邮票印象非常满意”。

12.C 上下文逻辑关系型，由下文的difficult to find aboard加上前后转折关系可以得出答案。

13.D 上下文理解型，由下文such as a basketball game or a concert可知，这些是娱乐项目，因此可以得出答案。

【一模】2014年广州市普通高中毕业班综合测试和答案第一节完形填空是（共15小题，每小题2分，满分30分）In America, if you are invited to a wedding, baby shower, bar mitzvah(成年礼) or other celebrations, you’re expected to bring a gift. Usually, it should be modest in 1.______, about$25.For a wedding, the bride will often have ―registered‖ a list of gifts at a local department store, indicating the items she 2. ______ . When you buy a registered item, tell the store that you‘re doing this , so the couple doesn‘t receive the 3.______gift twice. For a baby shower, bring a gift 4. ______for a newborn baby. For a bar mitzvah, bring a gift appropriate for a 13-year-old boy. Because they are such important occasions, gifts for bar mitzvahs tend to be more 5. ______ , for example, a gold-plated pen. 6. ______ the pen by carving the boy‘s full name will be appreciated.If you wish to give a gift to American friends, choose something that is 7.______to your country. It needn‘t be valuable or 8.______, just typical of your homeland.9.______ include a book about your country , an inexpensive souvenir , or something else that reflects your 10 . Yong children who like collecting will probably be very 11.______with a set of your country‘s coins or stamps. Items that are 12.______in your country but difficult to find abroad are also good.If staying with an American family, a good way of expressing your thanks is to take them to a form of 13, such as a basketball game or a concert.When giving gifts to a business acquaintance, don‘t give anything too persona l,14.______to a woman. A scarf or a hat is ok, but other types of 15.______ are not. Something appropriate for the office is best.1.A.size B. value C. weight D. appearance2.A. prefers B. owns C. uses D. imagines3.A.first B. best C. same D. similar4.A.general B. suitable C. demanding D. expensive5.A.modest B. cheerful C. normal D. formal6.A.Personalizing B. Replacing C. Designing D. Changing7.A.convenient B. appropriate C. unique D. beneficial8.A.colourful B. rare C. heavy D. nice9.A.Opportunities B. Expectations C. Inventions D. Possibilities10.A.character B. interest C. culture D. progress11.A.annoyed B. impressed C. amused D. puzzled12.A.limited B. banned C. common D. pricelesscation B. discussion C. exercise D. entertainment14.A.directly B. especially C. merely D. deliberately15.A. clothing. B. perfume C. jewelry D.equipment第二节语法填空While thousands of college students headed for warm climates to enjoy sun and fun during their week off from classes, seven local students had other plans.The Northern Essex Community College (NECC) students and one of their teachers spent part of their spring break in New York City, helping repair an area___16___(destroy) by the hurricane.―I want to see for myself what happened,‖ said Terry. ―I couldn‘t imagine ___17__ it is like to lose your home and everything that you know and the __18___ (power) effect the hurricane had on those people, I wanted to do something, to understand their feeling of helplessness.‖The group headed into Brooklyn‘s Red Hook district, which was hit hard by the hurricane. There they met people from other parts of the country, ___19___ had also volunteered to help. Together, those volunteers and the NECC students __20___ (work) to clear rubbish out of a story building. They put on protective suits and gloves __21__ they entered the building.Inside the building, the students saw nothing but broken walls and doors and pieces of the building ___22___ (lie) all over the place.The students returned to school with _23___ sense of achievement, a feeling that __24___ helped people in need. It was remarkable how a community lost so much and was still able to recover, and this left the deepest impression __25___ the students.II阅读（共两节，满分50分）I once met a well-known botanist at a dinner party. I had never talked with a botanist before, and I found him very interesting. I sat there absorbed and listened while he spoke of unusual plants and his experiments (he even told me astonishing facts about the simple potato), I had a small indoor garden of my own – and he was good enough to tell me how to solve some of my problems.As I said, we were at a dinner party. There must have been a dozen other guests, but I broke an important rule of politeness. I ignored everyone else and talked for hours to the botanist.Midnight came. I said good night to everyone and departed. The botanist then turned to our host and said many nice things about ne , Including that I was a ―most interesting conversationalist:.An interesting conversationalist?I had said hardly anything at all. I couldn‘t have said anything if I had wanted to without changing the subject, for I didn‘t know any more about plants than I knew about sharks. But I had done this one thing; I had listened carefully. I listened because I was really interested. And he felt it. Naturally that pleased him. That kind of listening is one of the best ways to show respect to others, and it makes them feel great too. ―Few human beings.‖ Wrote Jack Woodford in Strangers in Love, ―can resist the sweet effect of rapt attention.‖ I went even further that that .I was ―sincere in my admiration and generous in my praise‖. I told him that I had been hugely entertained and instructed. I told him I wished I had his knowledge. I told him that I should love to wander the fields with him. What‘s more, it was all true.And so I had him thinking of me as a good conversationalist when , in reality, I had only been a good listener and had encouraged him to talk.26.From Paragraph 1, we can learn that the writer ______.A. was deeply moved by the botanist‘s talkB. was amazed by what he was hearingC. was not in a comfortable situationD. behaved politely and properly27. Which of the following does the writer describe as a rule of politeness at dinner parties?A. Avoiding discussions about politics and religion.B. Listening carefully to what another guest says.C. Arriving and leaving at the appropriate time.D. Giving attention to all those in attendance.28. The underlined expression ‗rapt attention‖ in Paragraph 4‖ is closet in meaning to ______.A. full understandingB. strong interestC. great uncertaintyD. little curiosity29.According to the writer, which of the following is an important characteristic of a goodconversationalist?A. Listening attentively and encouraging the other side to continue.B. Encouraging the other side by sharing his /her own opinions.C. Promising a future meeting for more communication.D. Expressing respect by nodding his/her head.30.What is the purpose of the passage?A. To prove the writer is an interesting conversationalist.B. To share an interesting experience at a dinner party.C. To explain what makes a good conversationalist.D. To show that botanists can be really talkative.BA British dog-lover has invented a high-tech way of feeding his pet by Twitter( 推特，流行社交网络). Computer expert Nat Morris ,30, has designed a system to give his pet a “tweet treat”by sending him a Twitter message.His dog Toby gets some delicious dog biscuits from a computer-controlled food machine whenever Nat sends a message to ―@ feedtoby‖.Nat often works away from home and isn‘t always able to feed Toby by hand. But his new invention allows Nat to feed his dog from anywhere in the world.Nat said .,‖ Toby absolutely loves it. At first he didn‘t know what was going on. Now he sits underneath the machine, wagging his tail and waiting for the food to drop.‖Nat fills the food machine with small pieces of dog biscuits, but not too many in case four-year- old Toby gets too many messages. And Nat has even equipped his house with an onlinecamera so he can see Toby enjoying he food at his home.But one problem is that friends and family have been so amazed with the ―tweet treat ― machine that they have started sending tweets to Toby too. So Nat ha s had to restrict feeding time to make sure Toby doesn‘t turn into Tubby.“People have been sending him tweets at all hours of the day, so I had to limit it to between 9a.m. and 9 p.m. . I‘m thinking of doing an updated one which can measure his weight be fore he is fed ,just to make sure he‘s not putting on too much puppy fat,‖ explained Nat.How Nat‘s Twitter Feeder works:When a message is sent to @ feedtoby, it is received by a mini –computer that is linked to the feed machine.When the mini-computer receives the message, a bell rings and Toby comes running over and sits in front of the feeding machine. Next , the machine‘s motor pulls open a trap door whi ch releases a serving of food.The doggy biscuits then drop into Toby‘s food bowl. Finally a di gital camera takes a photo of him and sends it back to Nat on Twitter －so he knows Toby has been fed.31.Nat has invented a high-tech way to feed his dog because he ______.A. wants his friends to feed TobyB. has very strong computing skillsC. is often too busy to feed his dogD. doesn‘t like to feed Toby by hand.32.Why has Nat decided to limit the feeding machine‘s operating time?A. He doesn‘t want Toby to get too fat.B. He fears the machine will run out of food.C. He wants his friends to stop feeding Toby.D. He doesn‘t want Toby to be woken up at night .33. It can be learned from the passage that Toby _______.A. sits beneath his feeder all day long.B. is now used to being fed by machineC. doesn‘t know what happens to the feederD. no longer receives tweets from Nat‘s friends34.Which of the following shows the correct order of how the Twitter Feeder works?a. The bell goes off.b. is now used to being fed by machinec. doesn‘t know what happens to the feederd. no longer rece ives tweets from Nat‘s friends.e. The motor starts to work and opens the door to release dog food.A.a,b,v,e, dB.b,c,e,a,dC.b,c,a,e,dD.c,b,a,d,e,35. In which section of the newspaper would you most probably find this passage?A. TechnologyB. Health .C. EnvironmentD. StyleCNo one knows why we dream, but some dreams might be connected to the mental processes that help us learn. In a recent study, scientists found a connection between dreams and better memory in people learning a new skill.So perhaps one way to learn something new is to practice , practice , practice _ and then sleep on it.―I was very surprised by this finding ,‖ said Robert Stickgold , a Harvard University scientist who led the study.In the study ,100 college students each spent an hour on a computer , trying to get through a maze(迷宫). The maze was difficult, and the study participants had to start from a different place each time they tried- making it even more difficult.Then, for the first 90 minutes of a five-hour break, half of the participants were required to stay awake while half were asked to sleep. Participants who stayed awake were asked to describe their thoughts. Participants who slept were asked to describe any dream they had.Stickgold and his colleagues wanted to know about NREM, or non-REM sleep. REM stands for ―rapid eye movement.‖ Which is what happens during REM sleep. This period of sleep often brings strange dreams to a sleeper, although dreams can happen in both kinds of sleep. Stickgold wanted to know what people were dreaming about when their eyes weren‘t moving, during NREM wanted to know what people were dreaming about when their eyes weren‘t moving, during NREM sleep. Other studies have found connection between NREM brain activity and learning ability.Four of the 50 people who slept said their dreams were about the maze. Later, when these four people tried the computer maze again, they were able to complete it faster.Stickgold believes the dream itself d oesn‘t help a person learn-it‘s the other way around. He suspects that such dreams are caused by the brain processes associated with learning All the maze-dreamers had done the task poorly the first time, which makes Stickgold wonder if the NREM dreams show up when a person finds a new task particularly difficult . People who had other dreams ,or people who didn‘t show the same improvement.36. In the first stage of the study, the participants were asked to ____.A. design a maze on computerB. find their way out of a maze.C. decide where to begin a mazeD. remember a location in a maze37. What happened to the participants during the break?A. Half of them were woken up when they started to dream.B. Half of them were asked to dream about the maze.C. All of them were asked to describe their thoughts.D. Half of them were asked to sleep for 90 minutes.38. What can we learn from the passage?A. Everyone will dream about a new skill after learning it.B. Stickgold was the first to study dreams and learning.C. During NREM sleep, people usually don‘t dream.D. Unusual dreams often occur during REM sleep.39.According to the last paragraph , before sleeping the maze –dreamers ___.A. found it difficult to do the maze .B. were greatly interested in the mazeC. were mostly slow and poor thinkersD. completed the maze faster than others40. Which of the following statements best summarizes the study‘s conclusion?A. Dreams have a role in learning.B. Dreams have no basis in reality.C. Dreams are important for health.D. Dreams are the best way to study.D.The recent publication of autobiographies by two of Britain‘s greatest scientists, biologist Richard Dawkins and physicist Stephen Hawking, is a wonderful opportunity to compare and contrast these two remarkable men. Surprisingly, they have rather more in common than we think.Most striking is the similarity in their backgrounds. They were born in the early 1940s to middle class families _ not wealthy but comfortably off , with a strong commitment to academic excellence and public service . Both families were keen to send their boys to Oxford University—and both succeeded, Dawkins studying zoology and Hawking physics.Neither man has a very positive view of his early university life. Hawing describes the attitude at Oxford in the 1950s and 1960s as very anti-work, ―You were supposed to either be brilliant without effort or fail. Hard work was looked down upon by students and we all pretended that nothing was worth making an effort for.”He estimates that he studied for no more than an hour a day as an undergraduate student (本科生)。

数学（文科）试题参考答案及评分标准 第 1 页 共 9 页2014年广州市普通高中毕业班综合测试（一）数学（文科）试题参考答案及评分标准说明：1．参考答案与评分标准给出了一种或几种解法供参考，如果考生的解法与参考答案不同，可根据试题主要考查的知识点和能力比照评分标准给以相应的分数．2．对解答题中的计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的得分，但所给分数不得超过该部分正确解答应得分数的一半；如果后继部分的解答有较严重的错误，就不再给分．3．解答右端所注分数，表示考生正确做到这一步应得的累加分数．4．只给整数分数，选择题和填空题不给中间分．一、选择题：本大题考查基本知识和基本运算．共10小题，每小题，满分50分．二、填空题：本大题考查基本知识和基本运算，体现选择性．共5小题，每小题，满分20分．其中14~15题是选做题，考生只能选做一题．三、解答题：本大题共6小题，满分80分．解答须写出文字说明、证明过程和演算步骤． 16．（本小题满分）（本小题主要考查古典概型等基础知识，考查化归与转化的数学思想方法，以及数据处理能力与应用意识）（1）解：记“从6瓶饮料中任意抽取1瓶，抽到没过保质期的饮料”为事件A ，从6瓶饮料中中任意抽取1瓶，共有6种不同的抽法．因为6瓶饮料中有2瓶已过保质期，所以事件A 包含4种情形． 则()4263P A ==． 所以从6瓶饮料中任意抽取1瓶，抽到没过保质期的饮料的概率为23． （2）解法1：记“从6瓶饮料中随机抽取2瓶，抽到已过保质期的饮料”为事件B ，随机抽取2瓶饮料，抽到的饮料分别记为x ，y ，则),(y x 表示第一瓶抽到的是x ，第二瓶抽到的是y ，则),(y x 是一个基本事件．由于是随机抽取，所以抽取到的任何基本事件的概率相等．不妨设没过保质期的饮料为1，2，3，4，数学（文科）试题参考答案及评分标准 第 2 页 共 9 页已过保质期的饮料为a ，b ，则从6瓶饮料中依次随机抽取2瓶的基本事件有：()1,2，()1,3，()1,4，()1,a ，()1,b ，()2,1，()2,3，()2,4，()2,a ，()2,b ， ()3,1，()3,2，()3,4，()3,a ，()3,b ，()4,1，()4,2，()4,3，()4,a ，()4,b ， (),1a ，(),2a ，(),3a ，(),4a ，(),a b ，(),1b ，(),2b ，(),3b ，(),4b ，(),b a ．共30种基本事件．由于2瓶饮料中有1瓶已过保质期就表示抽到已过保质期的饮料，所以事件B 包含的基本事件有：()1,a ，()1,b ，()2,a ，()2,b ，()3,a ，()3,b ，()4,a ，()4,b ，(),1a ，(),2a ， (),3a ，(),4a ，(),a b ，(),1b ，(),2b ，(),3b ，(),4b ，(),b a ．共18种基本事件． 则183()305P B ==． 所以从6瓶饮料中随机抽取2瓶，抽到已过保质期的饮料的概率为35． 解法2：记“从6瓶饮料中随机抽取2瓶，抽到已过保质期的饮料”为事件B ， 随机抽取2瓶饮料，抽到的饮料分别记为x ，y ，则),(y x 是一个基本事件．由于是随机抽取，所以抽取到的任何基本事件的概率相等．不妨设没过保质期的饮料为1，2，3，4， 已过保质期的饮料为a ，b ，则从6瓶饮料中随机抽取2瓶的基本事件有：()1,2，()1,3，()1,4，()1,a ，()1,b ，()2,3，()2,4，()2,a ，()2,b ，()3,4， ()3,a ，()3,b ，()4,a ，()4,b ，(),a b ．共15种基本事件．由于2瓶饮料中有1瓶已过保质期就表示抽到已过保质期的饮料，所以事件B 包含的基本事件有：()1,a ，()1,b ，()2,a ，()2,b ，()3,a ，()3,b ，()4,a ，()4,b ，(),a b ．共9种基本事件． 则93()155P B ==． 所以从6瓶饮料中随机抽取2瓶，抽到已过保质期的饮料的概率为35．数学（文科）试题参考答案及评分标准 第 3 页 共 9 页（本小题主要考查三角函数图象的周期性与单调性、同角三角函数的基本关系、三角函数的化简等知识，考查化归与转化的数学思想方法，以及运算求解能力）解：（1）因为函数()sin cos f x x a x =+的图象经过点π03⎛⎫- ⎪⎝⎭，，所以03f π⎛⎫-= ⎪⎝⎭． 即ππsin cos 033a ⎛⎫⎛⎫-+-= ⎪ ⎪⎝⎭⎝⎭．即02a+=．解得a =（2）由（1）得，()sin f x x x =+12sin 2x x ⎛⎫= ⎪ ⎪⎝⎭2sin cos cos sin 33x x ππ⎛⎫=+ ⎪⎝⎭π2sin 3x ⎛⎫=+ ⎪⎝⎭．所以函数()x f 的最小正周期为2π． 因为函数sin y x =的单调递增区间为2,222k k ππ⎡⎤π-π+⎢⎥⎣⎦()k ∈Z ， 所以当πππ2π2π232k x k -≤+≤+()k ∈Z 时，函数()x f 单调递增， 即5ππ2π2π66k x k -≤≤+()k ∈Z 时，函数()x f 单调递增．所以函数()x f 的单调递增区间为5ππ2π,2π66k k ⎡⎤-+⎢⎥⎣⎦()k ∈Z ．数学（文科）试题参考答案及评分标准 第 4 页 共 9 页（本小题主要考查空间线面关系、几何体的体积等知识，考查数形结合、化归与转化的数学思想方法，以及空间想象能力、推理论证能力和运算求解能力） （1）证明：连结11B D ，BD ，因为四边形1111A B C D 是正方形，所以1111AC B D ⊥． 在正方体1111ABCD A B C D -中，1DD ⊥平面1111A B C D ，11AC ⊂平面1111A B C D ，所以111AC DD ⊥．因为1111B D DD D = ，11B D ，1DD ⊂平面11BB D D ， 所以11AC ⊥平面11BB D D ．因为EF ⊂平面11BB D D ，所以11EF AC ⊥． （2）解：取1C C 的中点H ，连结BH ，则BH AE ．在平面11BB C C 中，过点F 作FG BH ，则FG AE ． 连结EG ，则A ，E ，G ，F 四点共面．因为11122CH C C a ==，11133HG BF C C a ===， 所以1C G 116C C CH HG a =--=．故当1C G 16a =时，A ，E ，G ，F 四点共面．（3）解：因为四边形EFBD 是直角梯形，所以几何体ABFED 为四棱锥A EFBD -．因为()2113222EFBDa a BF DE BD S ⎛⎫+ ⎪+⎝⎭===，点A 到平面EFBD的距离为12h AC ==，所以231153312236A EFBD EFBD V S h a a a -==⨯⨯=． 故几何体ABFED 的体积为3536a ．1D ABCD EF 1A1B1C 1D ABCDEF 1A1B 1CG H数学（文科）试题参考答案及评分标准 第 5 页 共 9 页（本小题主要考查等差数列、分组求和等知识，考查化归与转化的数学思想方法，以及运算求解能力和创新意识）解：（1）因为等差数列{}n a 的首项为10，公差为2，所以()1012n a n =+-⨯， 即28n a n =+． 所以62n n nb a n =-22n n =-． （2）由（1）知()()2228n n b a n n n -=--+()(24822n n n n ⎡⎤⎡⎤=--=+-+⎣⎦⎣⎦，因为526<+，所以当5n ≤时，n n a b >，当5n >时，n n b a >． 所以{}max ,n n n c a b =228,5,2, 5.n n n n n +≤⎧=⎨->⎩当5n ≤时，123n n S c c c c =++++ 123n a a a a =++++ ()10121428n =+++++()10282n n ++=⨯29n n =+．当5n >时，123n n S c c c c =++++()()12567n a a a b b b =+++++++()()()()()222225956267278282n n ⎡⎤=+⨯+-⨯+-⨯+-⨯++-⨯⎣⎦ ()()2222706782678n n ⎡⎤=+++++-++++⎣⎦()()()()22222222265701231234522n n n+-⎡⎤=+++++-++++-⎢⎥⎣⎦()()()()1217055656n n n n n ++⎡⎤=+--+-⎢⎥⎣⎦数学（文科）试题参考答案及评分标准 第 6 页 共 9 页3211545326n n n =--+． 综上可知，n S 2329,5,11545,5.326n n n n n n n ⎧+≤⎪=⎨--+>⎪⎩20．（本小题满分）（本小题主要考查函数的极值、函数的导数、函数的零点与单调性等知识，考查数形结合、化归与转化、分类与讨论的数学思想方法，以及运算求解能力、抽象概括能力与创新意识） 解：（1）因为()32693f x x x x =-+-，所以()23129f x x x '=-+()()313x x =--．令'()0f x =，可得1x =或3x =． 则'(),()f x f x 在R 上的变化情况为：所以当1x =时，函数()f x 有极大值为1，当3x =时，函数()f x 有极小值为3-． （2）假设函数()f x 在()3,+∞上存在“域同区间”[],s t ()3s t <<，由（1）知函数()f x 在()3,+∞上单调递增．所以()(),.f s s f t t =⎧⎪⎨=⎪⎩即3232693,693.s s s s t t t t ⎧-+-=⎪⎨-+-=⎪⎩ 也就是方程32693x x x x -+-=有两个大于3的相异实根． 设32()683g x x x x =-+-()3x >，则2()3128g x x x '=-+． 令()g x '0=，解得123x =<，223x =+>． 当23x x <<时，()g x '0<，当2x x >时，()g x '0>，所以函数()g x 在区间()23,x 上单调递减，在区间()2,x +∞上单调递增．数学（文科）试题参考答案及评分标准 第 7 页 共 9 页因为()3 60g =-<，()()230g x g <<，()5120g =>， 所以函数()g x 在区间()3,+∞上只有一个零点．这与方程32693x x x x -+-=有两个大于3的相异实根相矛盾，所以假设不成立． 所以函数()f x 在()3,+∞上不存在“域同区间”．21．（本小题满分）（本小题主要考查直线的斜率、双曲线的方程、直线与圆锥曲线的位置关系等知识，考查数形结合、化归与转化、函数与方程的数学思想方法，以及推理论证能力和运算求解能力） （1）解：设双曲线E 的半焦距为c ，由题意可得2254.c a c a ⎧=⎪⎨⎪=+⎩解得a =．（2）证明：由（1）可知，直线2533a x ==，点()23,0F ．设点5,3P t ⎛⎫⎪⎝⎭,()00,Q x y ，因为220PF QF = ，所以()0053,3,03t x y ⎛⎫----= ⎪⎝⎭．所以()00433ty x =-． 因为点()00,Q x y 在双曲线E 上，所以2200154x y -=，即()2200455y x =-． 所以20000200005533PQ OQy t y y ty k k x x x x --⋅=⋅=-- ()()2002004453453553x x x x ---==-．所以直线PQ 与直线OQ 的斜率之积是定值45．数学（文科）试题参考答案及评分标准 第 8 页 共 9 页（3）证法1：设点(),H x y ，且过点5,13P ⎛⎫⎪⎝⎭的直线l 与双曲线E 的右支交于不同两点()11,M x y ，()22,N x y ，由（2）知()2211455y x =-，()2222455y x =-． 设PM MH PN HN λ==，则,.PM PN MH HN λλ⎧=⎪⎨=⎪⎩ ． 即()()1122112255,1,1,33,,.x y x y x x y y x x y y λλ⎧⎛⎫⎛⎫--=--⎪⎪ ⎪⎝⎭⎝⎭⎨⎪--=--⎩整理，得()()()1212121251,31,1,1.x x y y x x x y y y λλλλλλλλ⎧-=-⎪⎪⎪-=-⎨⎪+=+⎪+=+⎪⎩①②③④由①×③，②×④得()()22221222221251,31.x x x y y y λλλλ⎧-=-⎪⎨⎪-=-⎩⑤⑥将()2211455y x =-，()2222455y x =-代入⑥，得2221224451x x y λλ-=⨯--． ⑦ 将⑤代入⑦，得443y x =-． 所以点H 恒在定直线43120x y --=上．证法2：依题意，直线l 的斜率k 存在． 设直线l 的方程为513y k x ⎛⎫-=- ⎪⎝⎭，由2251,31.54y k x x y ⎧⎛⎫-=- ⎪⎪⎪⎝⎭⎨⎪-=⎪⎩消去y 得()()()22229453053255690kxk k x k k -+---+=．因为直线l 与双曲线E 的右支交于不同两点()11,M x y ，()22,N x y ，数学（文科）试题参考答案及评分标准 第 9 页 共 9 页则有()()()()()()()22222122212290053900455690,3053,95425569.954k k k k k k k x x k k k x x k ⎧⎪∆=-+--+>⎪⎪-⎪+=⎨-⎪⎪-+⎪=⎪-⎩设点(),H x y ，由PM MH PN HN=，得112125353x x x x x x --=--． 整理得()()1212635100x x x x x x -+++=．将②③代入上式得()()()()()2222150569303553100954954k k x k k x k k -++--+=--． 整理得()354150x k x --+=． ④ 因为点H 在直线l 上，所以513y k x ⎛⎫-=- ⎪⎝⎭． ⑤ 联立④⑤消去k 得43120x y --=． 所以点H 恒在定直线43120x y --=上．（本题（3）只要求证明点H 恒在定直线43120x y --=上，无需求出x 或y 的范围．）①② ③。

试卷类型：A2014年广州市普通高中毕业班综合测试（一）数学（理科）2014.3本试卷共4页，21小题， 满分150分．考试用时120分钟 注意事项：1．答卷前，考生务必用2B 铅笔在“考生号”处填涂考生号。

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考试结束后，将试卷和答题卡一并交回。

参考公式：锥体的体积公式Sh V 31=，其中S 是锥体的底面积，h 是锥体的高． ()()22221211236n n n n ++++++=()*n ∈N ． 一、选择题：本大题共8小题，每小题5分，满分40分．在每小题给出的四个选项中，只有一项是符合题目要求的． 1．已知i 是虚数单位，若()2i 34i m +=-，则实数m 的值为A ．2-B ．2±C ．D ．22．在△ABC 中，角A ，B ，C 所对的边分别为a ，b ，c ，若2C B =，则c b为 A ．2sin C B ．2cos B C ．2sin B D ．2cos C 3．圆()()22121x y -+-=关于直线y x =对称的圆的方程为A ．()()22211x y -+-= B ．()()22121x y ++-= C ．()()22211x y ++-= D ．()()22121x y -++=4．若函数()f x =R ，则实数a 的取值范围为A ．()2,2-B ．()(),22,-∞-+∞C ．(][),22,-∞-+∞D ．[]2,2-5．某中学从某次考试成绩中抽取若干名学生的分数，并绘制２成如图1的频率分布直方图．样本数据分组为[)50,60，[)60,70，[)70,80，[)80,90，[]90,100．若用分层抽样的方法从样本中抽取分数在[]80,100范围内的数据16个， 则其中分数在[]90,100范围内的样本数据有A ．5个B ．6个C ．8个D ．10个 6．已知集合32A x x x ⎧⎫=∈∈⎨⎬-⎩⎭Z Z 且，则集合A 中的元素个数为 A ．2 B ．3 C ．4D ．57．设a ，b 是两个非零向量，则使a b =a b 成立的一个必要非充分条件是A ．=a bB ．⊥a bC ．λ=a b ()0λ>D ．ab8．设a ，b ，m 为整数（0m >），若a 和b 被m 除得的余数相同，则称a 和b 对模m 同余，记为()mod a b m ≡．若0122202020202020C C 2C 2C 2a =+⋅+⋅++⋅，()mod10a b ≡，则b 的值可以是A ．2011B ．2012C ．2013D ．2014 二、填空题：本大题共7小题，考生作答6小题，每小题5分，满分30分． （一）必做题（9～13题）9．若不等式1x a -<的解集为{}13x x <<，则实数a 的值为 ． 10．执行如图2的程序框图，若输出7S =，则输入k ()*k ∈N 的值为 ． 11．一个四棱锥的底面为菱形，其三视图如图3所示，则这个四棱锥的体积是 ．12．设α为锐角，若cos 65α⎛⎫+= ⎪⎝⎭，则sin 12απ⎛⎫-= ⎪⎝⎭． 侧（左）视图图3俯视图爱迪教育 D 爱迪个性化教育发展中心D Idea Personalized Education Development C ３13．在数列{}n a 中，已知11a =，111n n a a +=-+，记n S 为数列{}n a 的前n 项和，则2014S = ．（二）选做题（14～15题，考生只能从中选做一题） 14．（坐标系与参数方程选做题）在极坐标系中，直线()sin cos a ρθθ-=与曲线2cos 4sin ρθθ=-相交于A ，B 两点，若AB =3a 的值为 ．15．（几何证明选讲选做题）如图4，PC 是圆O 的切线，切点为C ，直线PA 与圆O 交于A ，B 两点，APC ∠的平分线分别交弦CA ，CB 于D ，E两点，已知3PC =，2PB =，则PEPD的值为 ．三、解答题：本大题共6小题，满分80分．解答须写出文字说明、证明过程和演算步骤． 16．（本小题满分12分）已知函数()sin cos f x x a x =+的图象经过点π03⎛⎫- ⎪⎝⎭，． （1）求实数a 的值；（2）设[]2()()2g x f x =-，求函数()g x 的最小正周期与单调递增区间．17．（本小题满分12分）甲，乙，丙三人参加某次招聘会，假设甲能被聘用的概率是25，甲，丙两人同时不能被聘用的概率是625，乙，丙两人同时能被聘用的概率是310，且三人各自能否被聘用相互独立． （1）求乙，丙两人各自能被聘用的概率；（2）设ξ表示甲，乙，丙三人中能被聘用的人数与不能被聘用的人数之差的绝对值，求ξ的分布列与均值（数学期望）．18．（本小题满分14分）如图5，在棱长为a 的正方体1111ABCD A B C D -中，点E 是棱1D D 的PEABCD 图4O 1C 1D DE1A 1B４中点，点F 在棱1B B 上，且满足12B F FB =． （1）求证：11EF A C ⊥；（2）在棱1C C 上确定一点G ， 使A ，E ，G ，F 四点共面，并求此时1C G 的长；（3）求平面AEF 与平面ABCD 所成二面角的余弦值． 19．（本小题满分14分）已知等差数列{}n a 的首项为10，公差为2，等比数列{}n b 的首项为1，公比为2，*n ∈N ．（1）求数列{}n a 与{}n b 的通项公式；（2）设第n 个正方形的边长为{}min ,n n n c a b =，求前n 个正方形的面积之和n S ． （注：{}min ,a b 表示a 与b 的最小值．） 20．（本小题满分14分）已知双曲线E ：()222104x y a a -=>的中心为原点O ，左，右焦点分别为1F ，2F ，离心率为35，点P 是直线23a x =上任意一点，点Q 在双曲线E 上，且满足220PF QF =．（1）求实数a 的值；（2）证明：直线PQ 与直线OQ 的斜率之积是定值；（3）若点P 的纵坐标为1，过点P 作动直线l 与双曲线右支交于不同两点M ，N ，在线段MN上取异于点M ，N 的点H ，满足PM MHPN HN=，证明点H 恒在一条定直线上． 21．（本小题满分14分）已知函数()()221e x f x x x =-+（其中e 为自然对数的底数）． （1）求函数()f x 的单调区间；（2）定义：若函数()h x 在区间[],s t ()s t <上的取值范围为[],s t ，则称区间[],s t 为函数()h x 的“域同区间”．试问函数()f x 在()1,+∞上是否存在“域同区间”？若存在，求出所有符合条件的“域同区间”；若不存在，请说明理由．2014年广州市普通高中毕业班综合测试（一）数学（理科）试题参考答案及评分标准说明：1．参考答案与评分标准给出了一种或几种解法供参考，如果考生的解法与参考答案不同，可C爱迪教育 D 爱迪个性化教育发展中心D Idea Personalized Education Development C ５根据试题主要考查的知识点和能力比照评分标准给以相应的分数．2．对解答题中的计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的得分，但所给分数不得超过该部分正确解答应得分数的一半；如果后继部分的解答有较严重的错误，就不再给分．3．解答右端所注分数，表示考生正确做到这一步应得的累加分数．4．只给整数分数，选择题和填空题不给中间分．一、选择题：本大题考查基本知识和基本运算．共8小题，每小题，满分40分．题号 1 23 4 5 6 7 8答案 A B A D B C D A二、填空题：本大题考查基本知识和基本运算，体现选择性．共7小题，每小题，满分30分．其中14~15题是选做题，考生只能选做一题．题号 9 10 11 12131415答案23421020112-1-或5- 23三、解答题：本大题共6小题，满分80分． 16．（本小题满分1）（本小题主要考查三角函数图象的周期性、单调性、同角三角函数的基本关系和三角函数倍角公式等等知识，考查化归与转化的数学思想方法，以及运算求解能力）解：（1）因为函数()sin cos f x x a x =+的图象经过点π03⎛⎫- ⎪⎝⎭，，所以03f π⎛⎫-= ⎪⎝⎭． 即ππsin cos 033a ⎛⎫⎛⎫-+-= ⎪ ⎪⎝⎭⎝⎭． 即302a+=． 解得3a =（2）方法1：由（1）得()sin 3f x x x =．所以2()[()]2g x f x =-()2sin 32x x=+-22sin 23cos 3cos 2x x x x =++-６2cos 2x x =+122cos 22x x ⎫=+⎪⎪⎝⎭ 2sin 2cos cos 2sin 66x x ππ⎛⎫=+ ⎪⎝⎭π2sin 26x ⎛⎫=+ ⎪⎝⎭．所以()g x 的最小正周期为22π=π． 因为函数sin y x =的单调递增区间为2,222k k ππ⎡⎤π-π+⎢⎥⎣⎦()k ∈Z ， 所以当πππ2π22π262k x k -≤+≤+()k ∈Z 时，函数()g x 单调递增， 即ππππ36k x k -≤≤+()k ∈Z 时，函数()g x 单调递增．所以函数()g x 的单调递增区间为πππ,π36k k ⎡⎤-+⎢⎥⎣⎦()k ∈Z ． 方法2：由（1）得()sin f x x x =+2sin cos cos sin 33x x ππ⎛⎫=+ ⎪⎝⎭π2sin 3x ⎛⎫=+ ⎪⎝⎭．所以2()[()]2g x f x =-2π2sin 23x ⎡⎤⎛⎫=+- ⎪⎢⎥⎝⎭⎣⎦2π4sin 23x ⎛⎫=+- ⎪⎝⎭2π2cos 23x ⎛⎫=-+ ⎪⎝⎭分所以函数()g x 的最小正周期为22π=π分 因为函数cos y x =的单调递减区间为[]2,2k k ππ+π()k ∈Z ， 所以当22223k x k ππ≤+≤π+π()k ∈Z 时，函数()g x 单调递增．爱迪教育 D 爱迪个性化教育发展中心D Idea Personalized Education Development C ７即ππππ36k x k -≤≤+（k ∈Z ）时，函数()g x 单调递增．所以函数()g x 的单调递增区间为πππ,π36k k ⎡⎤-+⎢⎥⎣⎦()k ∈Z ．17．（本小题满分1）（本小题主要考查相互独立事件、解方程、随机变量的分布列与均值（数学期望）等知识，考查或然与必然的数学思想方法，以及数据处理能力、运算求解能力和应用意识） 解：（1）记甲，乙，丙各自能被聘用的事件分别为1A ，2A ，3A ，由已知1A ，2A ，3A 相互独立，且满足()()()()()113232,5611,253.10P A P A P A P A P A ⎧=⎪⎪⎪--=⎡⎤⎡⎤⎨⎣⎦⎣⎦⎪⎪=⎪⎩解得()212P A =，()335P A =． 所以乙，丙各自能被聘用的概率分别为12，35． （2）ξ的可能取值为1，3．因为()()()1231233P P A A A P A A A ξ==+()()()()()()123123111P A P A P A P A P A P A =+---⎡⎤⎡⎤⎡⎤⎣⎦⎣⎦⎣⎦213312525525=⨯⨯+⨯⨯625=． 所以()()113P P ξξ==-=61912525=-=． 所以ξ的分布列为所以19613252525E ξ=⨯+⨯=．ξ 1 3P1925625８18．（本小题满分1）（本小题主要考查空间线面关系、四点共面、二面角的平面角、空间向量及坐标运算等知识，考查数形结合、化归与转化的数学思想方法，以及空间想象能力、推理论证能力和运算求解能力）推理论证法：（1）证明：连结11B D ，BD ，因为四边形1111A B C D 是正方形，所以1111A C B D ⊥． 在正方体1111ABCD A B C D -中，1DD ⊥平面1111A B C D ，11A C ⊂平面1111A B C D ，所以111A C DD ⊥．因为1111B D DD D =，11B D ，1DD ⊂平面11BB D D ，所以11A C ⊥平面11BB D D ．因为EF ⊂平面11BB D D ，所以11EF A C ⊥． （2）解：取1C C 的中点H ，连结BH ，则BHAE ．在平面11BB C C 中，过点F 作FGBH ，则FGAE ．连结EG ，则A ，E ，G ，F 四点共面．因为11122CH C C a ==，11133HG BF C C a ===， 所以1C G 116C C CH HG a =--=．故当1C G 16a =时，A ，E ，G ，F 四点共面．（3）延长EF ，DB ，设EFDB M =，连结AM ，则AM 是平面AEF 与平面ABCD 的交线．过点B 作BN AM ⊥，垂足为N ，连结FN ， 因为FB AM ⊥，FB BN B =， 所以AM ⊥平面BNF ．因为FN ⊂平面BNF ，所以AM ⊥FN ． 所以FNB ∠为平面AEF 与平面ABCD 所成二面角的平面角．因为123132aMB BF MD DE a ===，23=，1D ABCD EF 1A1B1C MN1D ABCD EF 1A1B1C 1DABCDE F 1A1B 1C G H爱迪教育 D 爱迪个性化教育发展中心D Idea Personalized Education Development C ９所以22MB a =．在△ABM 中，AB a =，135ABM ∠=， 所以2222cos135AM AB MB AB MB =+-⨯⨯⨯ ()222222222a aa a ⎛=+-⨯⨯⨯- ⎝⎭213a =． 即13AM a =． 因为11sin13522AM BN AB MB ⨯=⨯⨯， 所以222sin13521321313a a AB MB BN a AMa⨯⨯⨯===．所以2222121371331339FN BF BN a a ⎛⎫⎛⎫=+=+= ⎪ ⎪ ⎪⎝⎭⎝⎭． 所以6cos 7BN FNB FN ∠==．故平面AEF 与平面ABCD 所成二面角的余弦值为67．空间向量法：（1）证明：以点D 为坐标原点，DA ，DC ，1DD 所在的直线分别为x 轴，y 轴，z 轴，建立如图的空间直角坐标系， 则(),0,0A a ，()1,0,A a a ，()10,,C a a ，10,0,2E a ⎛⎫ ⎪⎝⎭，1,,3F a a a ⎛⎫ ⎪⎝⎭，所以()11,,0AC a a =-，1,,6EF a a a ⎛⎫=- ⎪⎝⎭． 因为221100AC EF a a =-++=， 所以11AC EF ⊥．1D ABC D EF 1A1B1C xyz１０所以11EF A C ⊥．（2）解：设()0,,G a h ，因为平面11ADD A 平面11BCC B ，平面11ADD A 平面AEGF AE =，平面11BCC B 平面AEGF FG =，所以FGAE ．所以存在实数λ，使得FG AE λ=． 因为1,0,2AE a a ⎛⎫=- ⎪⎝⎭，1,0,3FG a h a ⎛⎫=-- ⎪⎝⎭， 所以11,0,,0,32a h a a a λ⎛⎫⎛⎫--=- ⎪ ⎪⎝⎭⎝⎭． 所以1λ=，56h a =． 所以1C G 15166CC CG a a a =-=-=． 故当1C G 16a =时，A ，E ，G ，F 四点共面． （3）解：由（1）知1,0,2AE a a ⎛⎫=- ⎪⎝⎭，10,,3AF a a ⎛⎫= ⎪⎝⎭． 设(),,x y z =n 是平面AEF 的法向量，则0,0.AE AF ⎧=⎪⎨=⎪⎩n n 即10,210.3ax az ay az ⎧-+=⎪⎪⎨⎪+=⎪⎩取6z =，则3x =，2y =-．所以()3,2,6=-n 是平面AEF 的一个法向量． 而()10,0,DD a =是平面ABCD 的一个法向量， 设平面AEF 与平面ABCD 所成的二面角为θ， 则11cos DD DD θ=n n (1)爱迪教育 D 爱迪个性化教育发展中心D Idea Personalized Education Development C １１()()2220302667326a a⨯+⨯-+⨯==+-+⨯． 故平面AEF 与平面ABCD 所成二面角的余弦值为67．第（1）、（2）问用推理论证法，第（3）问用空间向量法： （1）、（2）给分同推理论证法． （3）解：以点D 为坐标原点，DA ，DC ，1DD 所在的直线分别为x 轴，y 轴，z 轴，建立如图的空间直角坐标系， 则(),0,0A a ，10,0,2E a ⎛⎫ ⎪⎝⎭，1,,3F a a a ⎛⎫ ⎪⎝⎭，则1,0,2AE a a ⎛⎫=- ⎪⎝⎭，10,,3AF a a ⎛⎫= ⎪⎝⎭．设(),,x y z =n 是平面AEF 的法向量，则0,0.AE AF ⎧=⎪⎨=⎪⎩nn即10,210.3ax az ay az ⎧-+=⎪⎪⎨⎪+=⎪⎩取6z =，则3x =，2y =-．所以()3,2,6=-n 是平面AEF 的一个法向量． 而()10,0,DD a =是平面ABCD 的一个法向量， 设平面AEF 与平面ABCD 所成的二面角为θ， 则11cos DD DD θ=n n (1)()()2220302667326a a⨯+⨯-+⨯==+-+⨯． 故平面AEF 与平面ABCD 所成二面角的余弦值为67． 19．（本小题满分1）（本小题主要考查等差数列、等比数列、分组求和等知识，考查化归与转化的数学思想方法，以及运算求解能力和创新意识）1D ABC DEF 1A1B1C xyz１２解：（1）因为等差数列{}n a 的首项为10，公差为2，所以()1012n a n =+-⨯， 即28n a n =+．因为等比数列{}n b 的首项为1，公比为2，所以112n n b -=⨯， 即12n n b -=．（2）因为110a =，212a =，314a =，416a =，518a =，620a =，11b =，22b =，34b =，48b =，516b =，632b =．易知当5n ≤时，n n a b >．下面证明当6n ≥时，不等式n n b a >成立．方法1：①当6n =时，616232b -==620268a >=⨯+=，不等式显然成立．②假设当n k =()6k ≥时，不等式成立，即1228k k ->+．则有()()()()122222821826218kk k k k k -=⨯>+=++++>++．这说明当1n k =+时，不等式也成立．综合①②可知，不等式对6n ≥的所有整数都成立． 所以当6n ≥时，n n b a >． 方法2：因为当6n ≥时()()()112281128n n n n b a n n ---=-+=+-+()()01211111C C C C 28n n n n n n -----=++++-+()()012321111111C C C C C C 28n n n n n n n n n n ---------≥+++++-+ ()()0121112C C C 28n n n n ---=++-+()()236460n n n n n =--=-+->，所以当6n ≥时，n n b a >．所以{}min ,n n n c a b =12,5,28,5.n n n n -⎧≤=⎨+>⎩爱迪教育 D 爱迪个性化教育发展中心D Idea Personalized Education Development C １３则()22222,5,44, 5.n n n c n n -⎧≤⎪=⎨+>⎪⎩当5n ≤时，2222123n n S c c c c =++++ 2222123n b b b b =++++024222222n -=++++1414n -=-()1413n=-．当5n >时，2222123n n S c c c c =++++()()22222212567n b b b a a a =+++++++()51413=-()()()222464744n ⎡⎤+++++++⎣⎦()()()222341467867165n n n ⎡⎤=+++++++++-⎣⎦ ()()()()2222223414121253267645n n n ⎡⎤=++++-++++++++-⎣⎦()()()()()121653414553264562n n n n n n +++-⎡⎤=+-+⨯+-⎢⎥⎣⎦3242421867933n n n =++-． 综上可知，n S ()32141,5,3424218679, 5.33nn n n n n ⎧-≤⎪⎪=⎨⎪++->⎪⎩20．（本小题满分1）（本小题主要考查直线的斜率、双曲线的方程、直线与圆锥曲线的位置关系等知识，考查数形结合、化归与转化、函数与方程的数学思想方法，以及推理论证能力和运算求解能力） （1）解：设双曲线E 的半焦距为c ，由题意可得22354.c a c a ⎧=⎪⎨⎪=+⎩解得5a =．１４（2）证明：由（1）可知，直线2533a x ==，点()23,0F ．设点5,3P t ⎛⎫⎪⎝⎭,()00,Q x y ， 因为220PF QF =，所以()0053,3,03t x y ⎛⎫----= ⎪⎝⎭． 所以()00433ty x =-． 因为点()00,Q x y 在双曲线E 上，所以2200154x y -=，即()2200455y x =-． 所以20000200005533PQ OQy t y y ty k k x x x x --⋅=⋅=--()()2002004453453553x x x x ---==-．所以直线PQ 与直线OQ 的斜率之积是定值45．（3）证法1：设点(),H x y ，且过点5,13P ⎛⎫⎪⎝⎭的直线l 与双曲线E 的右支交于不同两点()11,M x y ，()22,N x y ，则22114520x y -=，22224520x y -=，即()2211455y x =-，()2222455y x =-． 设PM MH PN HN λ==，则,.PM PN MH HN λλ⎧=⎪⎨=⎪⎩． 即()()1122112255,1,1,33,,.x y x y x x y y x x y y λλ⎧⎛⎫⎛⎫--=--⎪⎪ ⎪⎝⎭⎝⎭⎨⎪--=--⎩整理，得()()()1212121251,31,1,1.x x y y x x x y y y λλλλλλλλ⎧-=-⎪⎪⎪-=-⎨⎪+=+⎪+=+⎪⎩①②③④由①×③，②×④得()()22221222221251,31.x x x y y y λλλλ⎧-=-⎪⎨⎪-=-⎩⑤⑥将()2211455y x =-，()2222455y x =-代入⑥，爱迪教育 D 爱迪个性化教育发展中心D Idea Personalized Education Development C １５得2221224451x x y λλ-=⨯--． ⑦ 将⑤代入⑦，得443y x =-． 所以点H 恒在定直线43120x y --=上．证法2：依题意，直线l 的斜率k 存在． 设直线l 的方程为513y k x ⎛⎫-=- ⎪⎝⎭，由2251,31.54y k x x y ⎧⎛⎫-=- ⎪⎪⎪⎝⎭⎨⎪-=⎪⎩消去y 得()()()22229453053255690k x k k x k k -+---+=． 因为直线l 与双曲线E 的右支交于不同两点()11,M x y ，()22,N x y ，则有()()()()()()()22222122212290053900455690,3053,95425569.954k k k k k k k x x k k k x x k ⎧⎪∆=-+--+>⎪⎪-⎪+=⎨-⎪⎪-+⎪=⎪-⎩设点(),H x y ，由PM MH PN HN =，得112125353x x x x x x --=--． 整理得()()1212635100x x x x x x -+++=．1 将②③代入上式得()()()()()2222150569303553100954954k k x k k x k k -++--+=--．整理得()354150x k x --+=． ④①② ③１６因为点H 在直线l 上，所以513y k x ⎛⎫-=- ⎪⎝⎭． ⑤ 联立④⑤消去k 得43120x y --=． 所以点H 恒在定直线43120x y --=上．（本题（3）只要求证明点H 恒在定直线43120x y --=上，无需求出x 或y 的范围．） 21．（本小题满分1）（本小题主要考查函数的单调性、函数的导数、函数的零点等知识，考查数形结合、化归与转化、分类与讨论的数学思想方法，以及运算求解能力、抽象概括能力与创新意识） 解：（1）因为()()221e x f x x x =-+，所以2()(22)e (21)e x x f x x x x '=-+-+()21e xx =-(1)(1)e x x x =+-． 当1x <-或1x >时，()0f x '>，即函数()f x 的单调递增区间为(),1-∞-和()1,+∞． 当11x -<<时，()0f x '<，即函数()f x 的单调递减区间为()1,1-．所以函数()f x 的单调递增区间为(),1-∞-和()1,+∞，单调递减区间为()1,1-． （2）假设函数()f x 在()1,+∞上存在“域同区间”[,](1)s t s t <<，由（1）知函数()f x 在()1,+∞上是增函数，所以(),().f s s f t t =⎧⎨=⎩ 即22(1)e ,(1)e .s ts s t t ⎧-⋅=⎨-⋅=⎩也就是方程2(1)e xx x -=有两个大于1的相异实根． 设2()(1)e (1)xg x x x x =-->，则2()(1)e 1xg x x '=--． 设()h x =2()(1)e 1xg x x '=--，则()()221e x h x x x '=+-．因为在(1,)+∞上有()0h x '>，所以()h x 在()1,+∞上单调递增． 因为()110h =-<，()223e 10h =->，即存在唯一的()01,2x ∈，使得()00h x =．当()01,x x ∈时，()()0h x g x '=<，即函数()g x 在()01,x 上是减函数； 当()0,x x ∈+∞时，()()0h x g x '=>，即函数()g x 在()0,x +∞上是增函数．因为()110g =-<，0()(1)0g x g <<，2(2)e 20g =->，爱迪教育 D 爱迪个性化教育发展中心D Idea Personalized Education Development C １７所以函数()g x 在区间()1,+∞上只有一个零点．这与方程2(1)e xx x -=有两个大于1的相异实根相矛盾，所以假设不成立． 所以函数()f x 在()1,+∞上不存在“域同区间”．。

试卷类型：A 2014届广州市普通高中毕业班综合测试（一）文科综合2014.3文科综合试题A参考答案及评分标准文科综合·政治试题参考答案及评分标准一、选择题二、非选择题评分标准：若考生答案言之有理，持之有据，可酌情给分。

36．（27分）（1）①联系的观点，发展的观点。

（4分）②二选一个角度作答。

○选“联系的观点”：世界是普遍联系的，要求我们用联系的观点看问题。

我们要正确认识保护环境和经济增长、居民就业收入等之间的固有联系，善于分析和把握影响当前环境保护问题的各种条件；正确认识整体与部分的辩证关系，立足于人民的根本利益，促进经济社会的全面协调发展，同时需要调动各方力量，共同致力于解决环境保护问题。

（11分）○选“发展的观点”：世界是永恒发展的，要求我们用发展的观点看问题。

我们既要坚持发展是硬道理，在发展中以创新精神推动环境保护问题的解决，又要做好充分的思想准备，认识到各种利益冲突对我国经济社会发展带来的复杂性和曲折性；既需要各方积极做好量的积累，落实应有责任，又需要我们立足于当前经济社会发展转型的关键时期，完善环境保护法，着重解决环境保护中的突出问题。

（11分）（2）①根据经济社会和环境保护的新要求，全国人大适时启动修法，通过实地调研、广泛听取有关部门和专家意见，提高了立法的科学性。

（4分）②全国人大自觉贯彻民主集中制原则，发挥人大代表在修法中的主体作用，向社会公布法律草案公开征求意见，不断拓宽公民参与立法的途径，提高了立法的民主性。

（4分）③全国人大自觉贯彻依法治国基本方略，严格按照法律程序修订法律，坚持法制化和规范化的工作机制，保证了立法的科学性和民主性。

（4分）或①全国人大自觉贯彻民主集中制原则，通过实地调研、广泛听取有关部门和专家意见网上征求意见等方式，不断拓宽公民参与立法的途径，坚持了科学立法和民主立法。

（4分）②全国人大代表自觉履行职责，依法参加行使国家权力，通过提出和审议反映客观实际与人民诉求的议案，提高了立法的科学性和民主性。

广州市2014届普通高中毕业班综合测试（一）数学（理科）本试卷共4页，21小题， 满分150分．考试用时120分钟 注意事项：1．答卷前，考生务必用2B 铅笔在“考生号”处填涂考生号。

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用2B 铅笔将试卷类型（A ）填涂在答题卡相应位置上。

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3．非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内的相应位置上；如需改动，先划掉原来的答案，然后再写上新的答案；不准使用铅笔和涂改液。

不按以上要求作答的答案无效。

4．作答选做题时，请先用2B 铅笔填涂选做题题号对应的信息点，再作答。

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考试结束后，将试卷和答题卡一并交回。

参考公式：锥体的体积公式Sh V 31=，其中S 是锥体的底面积，h 是锥体的高． ()()22221211236n n n n ++++++=()*n ∈N ． 一、选择题：本大题共8小题，每小题5分，满分40分．在每小题给出的四个选项中，只有一项是符合题目要求的． 1．已知i 是虚数单位，若()2i 34i m +=-，则实数m 的值为 A ．2- B ．2± C ．2±D ．22．在△ABC 中，角A ，B ，C 所对的边分别为a ，b ，c ，若2C B =，则c b为 A ．2sin C B ．2cos B C ．2sin B D ．2cos C3．圆()()22121x y -+-=关于直线y x =对称的圆的方程为A ．()()22211x y -+-= B ．()()22121x y ++-= C ．()()22211x y ++-= D ．()()22121x y -++= 4．若函数()21f x x ax =++的定义域为实数集R ，则实数a 的取值范围为A ．()2,2-B ．()(),22,-∞-+∞C ．(][),22,-∞-+∞ D ．[]2,2-5．某中学从某次考试成绩中抽取若干名学生的分数，并绘制成如图1的频率分布直方图．样本数据分组为[)50,60，[)60,70，[)70,80，[)80,90，[]90,100．若用分层抽样的方法从样本中抽取分数在[]80,100范围内的数据16个， 则其中分数在[]90,100范围内的样本数据有A ．5个B ．6个C ．8个D ．10个 6．已知集合32A x x x ⎧⎫=∈∈⎨⎬-⎩⎭Z Z 且，则集合A 中的元素个数为 A ．2 B ．3 C ．4D ．57．设a ，b 是两个非零向量，则使a b =a b 成立的一个必要非充分条件是 A ．=a b B ．⊥a b C ．λ=a b()0λ> D ．ab8．设a ，b ，m 为整数（0m >），若a 和b 被m 除得的余数相同，则称a 和b 对模m 同余，记为()mod a b m ≡．若0122202020202020C C 2C 2C 2a =+⋅+⋅++⋅，()mod10a b ≡，则b 的值可以是A ．2011B ．2012C ．2013D ．2014二、填空题：本大题共7小题，考生作答6小题，每小题5分，满分30分． （一）必做题（9～13题）9．若不等式1x a -<的解集为{}13x x <<，则实数a 的值为 ．10．执行如图2的程序框图，若输出7S =，则输入k ()*k ∈N 的值为 ．11．一个四棱锥的底面为菱形，其三视图如图3所示，则这个四棱锥的体积是 ．11 正（主）视图 侧（左）视图图3俯视图452 2图2开始 结束输入k否 是输出S 1n n =+?n k < 0,0n S ==log y x =12n S S -=+图1分数频率/组距50 60 70 80 90 100 0.0100.015 0.020 0.025 0.030 012．设α为锐角，若3cos 65απ⎛⎫+= ⎪⎝⎭，则sin 12απ⎛⎫-= ⎪⎝⎭ ．13．在数列{}n a 中，已知11a =，111n n a a +=-+，记n S 为数列{}n a 的前n 项和，则2014S = ．（二）选做题（14～15题，考生只能从中选做一题） 14．（坐标系与参数方程选做题）在极坐标系中，直线()sin cos a ρθθ-=与曲线2cos 4sinρθθ=-相交于A ，B 两点，若AB =23，则实数a 的值为 ．15．（几何证明选讲选做题）如图4，PC 是圆O 的切线，切点为C ，直线PA 与圆O 交于A ，B 两点，APC ∠的平分线分别交弦CA ，CB 于D ，E两点，已知3PC =，2PB =，则PEPD的值为 ．题号 1 23 4 5 67 8答案 A B A D B C D A二、填空题：本大题考查基本知识和基本运算，体现选择性．共7小题，每小题，满分30分．其中14~15题是选做题，考生只能选做一题．题号 9 10 11 12131415答案23421020112-1-或5-23PEABC D 图4O。

20 1 4年广州市普通高中毕业班综合测试（一）语文参考答案及评分标准9．（10分）（1）【文言翻译】（7分）①（郭舒）因为犯了擅自释放司马彪的罪过，被廷尉拘囚，当时的人大多认为他为人仗义。

[4分。

“坐”“系”“义”各1分，大意1分。

]②郭舒年轻时与杜曾交情深厚，杜曾曾经征召郭舒，郭舒不肯前往，杜曾对他（这件事）怀恨在心。

[3分。

“厚”“衔”各1分，大意1分。

]（2）【信息筛选】（3分）①郭舒直言劝告王澄不要终日饮酒，而要以政务为重。

（澄终日酣饮，不以众务在意，舒常切谏之。

）②郭舒在天下大乱时劝王澄要修养德行，树立威望，保全州境。

（及天下大乱，又劝澄修德养威，保完州境。

）③郭舒制止王澄手下棒打士人宗庾廞，甘受王澄的责罚。

（舒厉色谓左右曰：“使君过醉，汝辈何敢妄动！”“澄遣掐其鼻，灸其眉头，舒跪而受之”。

）④郭舒说服王敦让手下退还了强夺菜农的土地。

（“公为胜尧、舜邪？乃逆折舒，使不得言。

何与古人相远！”“缪坦可谓小人……舒等不敢不言”。

）［3分。

答对一点2分，两点3分。

找出的事例能体现郭舒的“忠亮”，即可给分。

］【参考译文】郭舒，字稚行。

他年幼的时候请求母亲让他拜师求学，学了一年多就回家了，粗略懂得了为人处事的要义。

乡里人、宗族人都称他会是后起之秀，最终成为国家的栋梁之材。

郭舒开始做官担任领军校尉，因为犯了擅自释放司马彪的罪过，被廷尉拘囚，当时的人大多认为他为人仗义。

王澄听说了郭舒的声名，征召他担任别驾。

王澄整天痛饮，不把政务放在心上。

郭舒经常直言劝谏他。

等到天下大乱的时候，郭舒又劝王澄修养德行，树立威望，保全州境。

王澄虽然没有听从郭舒的劝告，但是敬重他的忠诚坚贞。

荆州当地的读书人宗庾廞曾经因为饮酒得罪了王澄，王澄发怒，命令手下人棒打宗庾廞。

郭舒神色严厉地对王澄的手下说：‚使君喝得太多了，你们这些人怎么敢胡乱行动！‛王澄发怒说：‚别驾发狂了吗？诓骗说我醉了！‛于是让人掐他的鼻子，烫他的眉头，郭舒跪着承受。

2014 广州一模试题及答案解析一．单选题13．用某色光照射金属表面时，有光电子从金属表面飞出，如果改用频率更大的光照射该金属表面，则A．金属的逸出功增大B．金属的逸出功减小C．光电子的最大初动能增大D．光电子的最大初动能不变答案：C，解析：根据k hγ =W + E ，材料的逸出功W 不变，判断出选项C 正确。

14．如图是氢原子从n=3、4、5、6 能级跃迁到n=2 能级时辐射的四条光谱线，其中频率最大的是A．HαB．HβC．HγD．Hδ答案：D，解析：能量大的频率大，选项D 正确。

15．如图是荷质比相同的a、b 两粒子从O 点垂直匀强磁场进入正方形区域的运动轨迹，则A．a 的质量比易的质量大B．a 带正电荷、b 带负电荷C．a 在磁场中的运动速率比b 的大D．a 在磁场中的运动时间比b 的长答案：C，解析：根据a 的半径大，a 的速率大，所以选项C 正确。

16．如图是悬绳对称且长度可调的自制降落伞．用该伞挂上重为G 的物体进行两次落体实验，悬绳的长度l1<l2，匀速下降时每根悬绳的拉力大小分别为F1、F2，则A．F l<F2 B．F1>F2 C．F1=F2<G D．F1=F2>G答案：B二．双选题17．如图为某压力锅的结构简图．将压力阀套在出气孔上，给压力锅加热，在气体把压力阀顶起之前，锅内气体A．压强增大B．内能不变C．对外界做正功D．分子平均动能增大答案：AD，解析：根据PV=nRT，选项A 正确。

分子平均动能是温度T 的函数，选项D 正确。

18．如图，在绕地运行的天宫一号实验舱中，宇航员王亚平将支架固定在桌面上，摆轴末端用细绳连接一小球．拉直细绳并给小球一个垂直细绳的初速度，它做圆周运动．在a、b 两点时，设小球动能分别为E ka、E kb，细绳拉力大小分别为T a、T b，阻力不计，则A．E ka>E kb B．E ka=E kb C．T a>T b D．T a=T b答案：BD，解析：在完全失重的条件下，小球是漂在空中的，只有绳拉力提供向心力，选项BD 正确。

2014年广州市普通高中毕业班综合测试（一）数学（文科）参考公式：()()22221211236n n n n ++++++=()*n ∈N ． 一、选择题：1．函数()()ln 1f x x =+的定义域为（ ）A ．(),1-∞-B ．(),1-∞C ．()1,-+∞D ．()1,+∞ 2．已知i 是虚数单位，若()2i 34i m +=-，则实数m 的值为（ ） A ．2- B ．2±C ．D ．23．在△ABC 中，角A ，B ，C 所对的边分别为a ，b ，c ，若2C B =，则cb为（ ） A ．2sin C B ．2cos B C ．2sin B D ．2cos C4．圆()()22121x y -+-=关于直线y x =对称的圆的方程为（ ）A ．()()22211x y -+-= B ．()()22121x y ++-= C ．()()22211x y ++-= D ．()()22121x y -++= 5．已知1x >-，则函数11y x x =++的最小值为（ ） A ．1- B ．0 C ．1 D ．2 6．函数()21xf x x =+的图象大致是（ ）7．已知非空集合M 和N ，规定{}M N x x M x N -=∈∉且，那么()M M N --等于（ ） A ．MN B ．M N C ．M D ．N8．任取实数a ，b ∈[]1,1-，则a ，b 满足22a b -≤的概率为（ ） A ．18 B ．14 C ．34 D ．789．设a ，b 是两个非零向量，则使a b =a b 成立的一个必要非充分条件是（ ） A ．=a b B ．ab C ．⊥a b D ．λ=a b ()0λ>10．在数列{}n a 中，已知11a =，()11sin2n n n a a ++π-=，记n S 为数列{}n a 的前n 项和，则2014S =（ ）（一）必做题（11～13题）11．执行如图1的程序框图，若输入=3k ，则输出S 的值为 ．12．一个四棱锥的底面为菱形，其三视图如图2所示，则这个四棱锥的体积是 ．13．由空间向量()1,2,3=a ，()1,1,1=-b 构成的向量集合{},A k k ==+∈Z x x a b ，则向量x 的模x 的最小 值为 ．（二）选做题（14～15题，考生只能从中选做一题）14．（坐标系与参数方程选做题）在极坐标系中，直线()sin cos a ρθθ-=与曲线2cos 4sin ρθθ=-相交于A ，B 两点，若AB =a 的值为 ．15．（几何证明选讲选做题）如图3，PC 是圆O 的切线，切点为C ，直线PA与圆O 交于A ，B 两点，APC ∠的平分线分别交弦CA ，CB 于D ，E两点，已知3PC =，2PB =，则PEPD的值为 ． 三、解答题：本大题共6小题，满分80分．解答须写出文字说明、证明过程和演算步骤． 16．（本小题满分12分）已知某种同型号的6瓶饮料中有2瓶已过保质期． （1）从6瓶饮料中任意抽取1瓶，求抽到没过保质期的饮料的概率； （2）从6瓶饮料中随机抽取2瓶，求抽到已过保质期的饮料的概率．侧（左）视图图2俯视图P图317．（本小题满分12分）已知函数()sin cos f x x a x =+的图象经过点π03⎛⎫- ⎪⎝⎭，． （1）求实数a 的值；（2）求函数()x f 的最小正周期与单调递增区间．18．（本小题满分14分）如图4，在棱长为a 的正方体1111ABCD A B C D -中，点E 是棱1D D 的中点，点F 在棱1B B 上，且满足12B F FB =．（1）求证：11EF AC ⊥；（2）在棱1C C 上确定一点G ，使A ，E ，G ，F 四点共面，并求此时1C G 的长；（3）求几何体ABFED 的体积． 1D ABCDEF1A1B1C 图419．（本小题满分14分） 已知等差数列{}n a 的首项为10，公差为2，数列{}n b 满足62n n nb a n =-，*n ∈N ． （1）求数列{}n a 与{}n b 的通项公式；（2）记{}max ,n n n c a b =，求数列{}n c 的前n 项和n S ．（注：{}max ,a b 表示a 与b 的最大值．）20．（本小题满分14分） 已知函数()32693f x x x x =-+-．（1）求函数()f x 的极值；（2）定义：若函数()h x 在区间[],s t ()s t <上的取值范围为[],s t ，则称区间[],s t 为函数()h x 的“域同区间”．试问函数()f x 在()3,+∞上是否存在“域同区间”？若存在，求出所有符合条件的“域同区间”；若不存在，请说明理由．21．（本小题满分14分）已知双曲线E ：()222104x y a a -=>的中心为原点O ，左，右焦点分别为1F ，2F ，离，点P 是直线23a x =上任意一点，点Q 在双曲线E 上，且满足220PF QF =．（1）求实数a 的值；（2）证明：直线PQ 与直线OQ 的斜率之积是定值；（3）若点P 的纵坐标为1，过点P 作动直线l 与双曲线右支交于不同两点M ，N ，在线段MN 上取异于点M ，N 的点H ，满足PM MHPN HN=，证明点H 恒在一条定直线上．2014年广州市普通高中毕业班综合测试（一）数学（文科）试题参考答案及评分标准一、选择题：CABAC ABDBC二、填空题：11. 7 12. 4 13.14. 1-或5- 15.2316．解：记“从6瓶饮料中任意抽取1瓶，抽到没过保质期的饮料”为事件A ，从6瓶饮料中中任意抽取1瓶，共有6种不同的抽法．因为6瓶饮料中有2瓶已过保质期，所以事件A 包含4种情形．则()4263P A ==． 所以从6瓶饮料中任意抽取1瓶，抽到没过保质期的饮料的概率为23． （2）解法1：记“从6瓶饮料中随机抽取2瓶，抽到已过保质期的饮料”为事件B ，随机抽取2瓶饮料，抽到的饮料分别记为x ，y ，则),(y x 表示第一瓶抽到的是x ，第二瓶抽到的是y ，则),(y x 是一个基本事件．由于是随机抽取，所以抽取到的任何基本事件的概率相等．不妨设没过保质期的饮料为1，2，3，4， 已过保质期的饮料为a ，b ，则从6瓶饮料中依次随机抽取2瓶的基本事件有：()1,2，()1,3，()1,4，()1,a ，()1,b ，()2,1，()2,3，()2,4，()2,a ，()2,b ，()3,1，()3,2，()3,4，()3,a ，()3,b ，()4,1，()4,2，()4,3，()4,a ，()4,b ，(),1a ，(),2a ，(),3a ，(),4a ，(),a b ，(),1b ，(),2b ，(),3b ，(),4b ，(),b a ．共30种基本事件．由于2瓶饮料中有1瓶已过保质期就表示抽到已过保质期的饮料，所以事件B 包含的基本事件有：()1,a ，()1,b ，()2,a ，()2,b ，()3,a ，()3,b ，()4,a ，()4,b ，(),1a ，(),2a ，(),3a ，(),4a ，(),a b ，(),1b ，(),2b ，(),3b ，(),4b ，(),b a ．共18种基本事件．则183()305P B ==．所以从6瓶饮料中随机抽取2瓶，抽到已过保质期的饮料的概率为35． 解法2：记“从6瓶饮料中随机抽取2瓶，抽到已过保质期的饮料”为事件B ，随机抽取2瓶饮料，抽到的饮料分别记为x ，y ，则),(y x 是一个基本事件．由于是随机抽取，所以抽取到的任何基本事件的概率相等．不妨设没过保质期的饮料为1，2，3，4， 已过保质期的饮料为a ，b ，则从6瓶饮料中随机抽取2瓶的基本事件有：()1,2，()1,3，()1,4，()1,a ，()1,b ，()2,3，()2,4，()2,a ，()2,b ，()3,4，()3,a ，()3,b ，()4,a ，()4,b ，(),a b ．共15种基本事件．由于2瓶饮料中有1瓶已过保质期就表示抽到已过保质期的饮料，所以事件B 包含的基本事件有：()1,a ，()1,b ，()2,a ，()2,b ，()3,a ，()3,b ，()4,a ，()4,b ，(),a b ．共9种基本事件．则93()155P B ==． 所以从6瓶饮料中随机抽取2瓶，抽到已过保质期的饮料的概率为35． 17．解：（1）因为函数()sin cos f x x a x =+的图象经过点π03⎛⎫- ⎪⎝⎭，，所以03f π⎛⎫-= ⎪⎝⎭．即ππsin cos 033a ⎛⎫⎛⎫-+-= ⎪ ⎪⎝⎭⎝⎭．即022a +=．解得a =（2）由（1）得，()sin f x x x =12sin 2x x ⎛⎫= ⎪⎪⎝⎭2sin cos cos sin 33x x ππ⎛⎫=+ ⎪⎝⎭π2sin 3x ⎛⎫=+ ⎪⎝⎭． 所以函数()x f 的最小正周期为2π．因为函数sin y x =的单调递增区间为2,222k k ππ⎡⎤π-π+⎢⎥⎣⎦()k ∈Z ， 所以当πππ2π2π232k x k -≤+≤+()k ∈Z 时，函数()x f 单调递增，即5ππ2π2π66k x k -≤≤+()k ∈Z 时，函数()x f 单调递增．所以函数()x f 的单调递增区间为5ππ2π,2π66k k ⎡⎤-+⎢⎥⎣⎦()k ∈Z ． 18．（1）证明：连结11B D ，BD ，因为四边形1111A B C D 是正方形，所以1111AC B D ⊥． 在正方体1111ABCD A B C D -中，1DD ⊥平面1111A B C D ，11AC ⊂平面1111A B C D ， 所以111AC DD ⊥．因为1111B D DD D =，11B D ，1DD ⊂平面11BB D D ，所以11AC ⊥平面11BB D D ．因为EF ⊂平面11BB D D ，所以11EF AC ⊥．（2）解：取1C C 的中点H ，连结BH ，则BH AE ．在平面11BB C C 中，过点F 作FG BH ，则FGAE ．连结EG ，则A ，E ，G ，F 四点共面． 因为11122CH C C a ==，11133HG BF C C a ===，所以 1C G 116C C CH HG a =--=．故当1C G 16a =时，A ，E ，G ，F 四点共面． （3）解：因为四边形EFBD 是直角梯形，所以几何体ABFED 为四棱锥A EFBD -．因为()211322212EFBDa a BF DE BD S a ⎛⎫+ ⎪+⎝⎭===，点A 到平面EFBD的距离为122h AC a ==，所以231153336A EFBD EFBD V S h a -===．故几何体ABFED 的体积为3536a ．19．解：（1）因为等差数列{}n a 的首项为10，公差为2，所以()1012n a n =+-⨯，即28n a n =+． 所以62n n nb a n =-22n n =-． （2）由（1）知()()2228n n b a n n n -=--+()(24822n n n n ⎡⎤⎡⎤=--=+-+⎣⎦⎣⎦，因为526<+<，所以当5n ≤时，n n a b >，当5n >时，n n b a >．所以{}max ,n n n c a b =228,5,2, 5.n n n n n +≤⎧=⎨->⎩当5n ≤时，123n n S c c c c =++++123n a a a a =++++()10121428n =+++++()10282n n ++=⨯29n n =+．1DABCDEF 1A1B1C 1DABCD EF 1A1B 1C G H当5n >时，123n n S c c c c =++++()()12567n a a a b b b =+++++++()()()()()222225956267278282n n ⎡⎤=+⨯+-⨯+-⨯+-⨯++-⨯⎣⎦()()2222706782678n n ⎡⎤=+++++-++++⎣⎦()()()()22222222265701231234522n n n +-⎡⎤=+++++-++++-⎢⎥⎣⎦()()()()1217055656n n n n n ++⎡⎤=+--+-⎢⎥⎣⎦3211545326n n n =--+．综上可知，n S 2329,5,11545,5.326n n n n n n n ⎧+≤⎪=⎨--+>⎪⎩20．解：（1）因为()32693f x x x x =-+-，所以()23129f x x x '=-+()()313x x =--． 令'()0f x =，可得1x =或3x =．则'(),()f x f x 在R 上的变化情况为：所以当1x =时，函数()f x 有极大值为1，当3x =时，函数()f x 有极小值为3-．（2）假设函数()f x 在()3,+∞上存在“域同区间”[],s t ()3s t <<，由（1）知函数()f x 在()3,+∞上单调递增．所以()(),.f s s f t t =⎧⎪⎨=⎪⎩即3232693,693.s s s s t t t t ⎧-+-=⎪⎨-+-=⎪⎩也就是方程32693x x x x -+-=有两个大于3的相异实根．设32()683g x x x x =-+-()3x >，则2()3128g x x x '=-+．令()g x '0=，解得123x =-，223x =>．当23x x <<时，()g x '0<，当2x x >时，()g x '0>，所以函数()g x 在区间()23,x 上单调递减，在区间()2,x+∞上单调递增．因为()3 60g =-<，()()230g x g <<，()5120g =>，所以函数()g x 在区间()3,+∞上只有一个零点．这与方程32693x x x x -+-=有两个大于3的相异实根相矛盾，所以假设不成立．所以函数()f x 在()3,+∞上不存在“域同区间”．21．（1）解：设双曲线E 的半焦距为c ，由题意可得22 4.c a c a ⎧=⎪⎨⎪=+⎩解得a =． （2）证明：由（1）可知，直线2533a x ==，点()23,0F ．设点5,3P t ⎛⎫⎪⎝⎭,()00,Q x y ，220PF QF =，5⎛⎫422x y 4∴2000020*******PQ OQy t y y ty k k x x x x --⋅=⋅=--()()2002004453453553x x x x ---==-．∴直线PQ 与直线OQ 斜率之积是定值45．（3）证法1：设点(),H x y ，且过点5,13P ⎛⎫⎪⎝⎭的直线l 与双曲线E 的右支交于不同两点()11,M x y ，()22,N x y ，由（2）知()2211455y x =-，()2222455y x =-．设PM MH PN HN λ==，则,.PM PN MH HN λλ⎧=⎪⎨=⎪⎩．即()()1122112255,1,1,33,,.x y x y x x y y x x y y λλ⎧⎛⎫⎛⎫--=--⎪⎪ ⎪⎝⎭⎝⎭⎨⎪--=--⎩整理，得()()()1212121251,31,1,1.x x y y x x x y y y λλλλλλλλ⎧-=-⎪⎪⎪-=-⎨⎪+=+⎪+=+⎪⎩①②③④由①×③，②×④得()()22221222221251,31.x x x y y y λλλλ⎧-=-⎪⎨⎪-=-⎩⑤⑥将()2211455y x =-，()2222455y x =-代入⑥，得2221224451x x y λλ-=⨯--． ⑦，将⑤代入⑦，得443y x =-．所以点H 恒在定直线43120x y --=上． 证法2：依题意，直线l 的斜率k 存在．设直线l 的方程为513y k x ⎛⎫-=- ⎪⎝⎭，由2251,31.54y k x x y ⎧⎛⎫-=- ⎪⎪⎪⎝⎭⎨⎪-=⎪⎩ 消去y 得()()()22229453053255690k x k k x k k -+---+=．因为直线l 与双曲线E 的右支交于不同两点()11,M x y ，()22,N x y ，则有()()()()()()()22222122212290053900455690,3053,95425569.954k k k k k k k x x k k k x x k ⎧⎪∆=-+--+>⎪⎪-⎪+=⎨-⎪⎪-+⎪=⎪-⎩ 设点(),H x y ，由PM MH PN HN =，得112125353x x x x x x --=--．整理得()()1212635100x x x x x x -+++=．将②③代入上式得()()()()()2222150569303553100954954k k x k k x k k -++--+=--．整理得()354150x k x --+=． ④，因为点H 在直线5⎛⎫ ①② ③。

2014年广州市一模试题参考答案30．（16分）（1）C8H7Br (2分) (2) (2分)(3) (6分，每个化学方程式3分。

错、漏条件均只得2分，条件不重复扣分)(4) （4分，每个2分）(5) （2分）31．（16分）（1）NO ；产生红棕色气体 （4分，每空2分） （2）1:2 （3分）（3）c(Ce4+)•[c(OH —)]4 （3分） ；8 （2分）（4）②其它条件相同时，S 随c(HNO3)减小而增大 （2分）③c(HNO3)越小，温度对S 的影响越大或c(HNO3)越大，温度对S 的影响越小 （2分） 32．（16分） （1）Al2O3+6H+===2Al3++3H2O ； Fe2O3+6H+===2Fe3++3H2O （4分，每个2分） （2）盐酸的浓度、反应温度、煤矸石颗粒大小、是否充分搅拌、反应时间（任写两个） （4分，每个2分）（3）CO2 ； Al(OH)3 +OH —=AlO2—+2H2O （4分，每个2分）（4）加入CaCO3调节pH 到3.2，过滤除去Fe(OH) 3后，再加入CaCO3调节pH 到5.4，过滤得到Al(OH) 3 （2分）（5）AlCl3饱和溶液中存在溶解平衡：AlCl3·6H2O(s)Al3+(aq) +3Cl —(aq) +6H2O(l)，通入HCl 气体使溶液中c(Cl —)增大，平衡向析出固体的方向移动从而析出AlCl3晶体。

（2分）33．（16分）（1）①品红、溴水或KMnO4溶液 ； 溶液颜色很快褪色或指出相应试剂的正确颜色（红色、黄色等）很快褪色也可得分。

（2分，每个1分）②增大SO2的接触面积、控制SO2的流速、适当升高温度（答其中二条即可） （4分，每个2分）（3）6000a/214V 或3000a/107V (其它正确答案也给分) （2分）。

广州市2014届普通高中毕业班综合测试（英语）Ⅰ语言知识及应用（共两节，满分45分）第一节完形填空（共15小题；每小题2分，满分30分）In America, if you are invited to a wedding, baby shower, bar mitzvah(成年礼)or other celebrations, you're expected to bring a gift. Usually, it should be modest in 1 , about$25.For a wedding, the bride will often have "registered" a list of gifts at a local department store, indicating the items she 2 ．When you buy a registered item, tell the store that you're doing this, so the couple doesn't receive the 3 gift twice. For a baby shower, bring a gift 4 for a newborn baby. For a bar mitzvah, bring a gift appropriate for a 13-year-old boy. Because they are such important occasions, gifts for bar mitzvahs tend to be more 5 , for example, a gold-plated pen. 6 the pen by carving the boy's full name will be appreciated.If you wish to give a gift to American friends, choose something that is 7 to your country. It needn't be valuable or 8 , just typical of your homeland. 9 include a book about your country, an inexpensive souvenir, or something else that reflects your 10 .Young children who like collecting will probably be very 11 with a set of your country's coins or stamps. Items that are 12 in your country but difficult to find abroad are also good.If staying with an American family, a good way of expressing your thanks is to take them to a form of 13 , such as a basketball game or a concert.When giving gifts to a business acquaintance, don't give anything too personal, 14 to a woman. A scarf or a hat is ok, but other types of 15 are not. Something appropriate for the office is best.1. A. size B. value C. weight D. appearance2. A. prefers B. owns C. uses D. imagines3. A. first B. best C. same D. similar4. A. general B. suitable C. demanding D. expensive5. A. modest B. cheerful C. normal D. formal6. A. Personalizing B. Replacing C. Designing D. Changing7. A. convenient B. appropriate C. unique D. beneficial8. A. colorful B. rare C. heavy D. nice9. A. Opportunities B. Expectations C. Inventions D. Possibilities10. A. character B. interest C. culture D. progress11. A. annoyed B. impressed C. amused D. puzzled12. A. limited B. banned C. common D. priceless13. A. education B. discussion C. exercise D. entertainment14. A. directly B. especially C. merely D. deliberately15. A. clothing B. perfume C. jewelry D. equipment笫二节语法填空（共10小题；每小题1.5分，满分15分）While thousands of college students headed for warm climates to enjoy sun and fun during their week off from classes, seven local students had other plans.The Northern Essex Community College( NECC) students and one of their teachers spent part of their spring break in New York City, helping repair an area 16. (destroy) by the hurricane.“I wanted to see for myself what happened,” said Terry. “I couldn't imagine 17. it is like to lose your home and everything that you know and the 18. (power) effect the hurricane had on those people. I wanted to do something, to understand their feeling ofhelplessness.” The group headed into Brooklyn's Red Hook district, which was hit hard by the hurricane. There they met people from other parts of the country, 19. had also volunteered to help. Together, those volunteers and the NECC students 20. (work) to clear rubbish out of a three-story building. They put on protective suits and gloves 21. they entered the building.Inside the building, the students saw nothing but broken walls and doors and pieces of the building 22. (lie) all over the place.The students returned to school with 23. sense of achievement, a feeling that 24.______ helped people in need. It was remarkable how a community lost so much and was still able to recover, and this left the deepest impression 25. the students.Ⅱ阋读(共两节,满分50分)第一节阅读理解（共20小题；每小题2分，满分40分）AI once met a well-known botanist at a dinner party. I had never talked with a botanist before, and I found him very interesting. I sat there absorbed and listened while he spoke of unusual plants and his experiments (he even told me astonishing facts about the simple potato). I had a small indoor garden of my own -- and he was good enough to tell me how to solve some of my problems.As I said, we were at a dinner party. There must have been a dozen other guests, but I broke an important rule of politeness. I ignored everyone else and talked for hours to the botanist.Midnight came. I said good night to everyone and departed. The botanist then turned to our host and said many nice things about me, including that I was a “most interesting conversationalist ”.An interesting conversationalist? I had said hardly anything at all. I couldn't have said anything if I had wanted to without changing the subject, for I didn't know any more about plants than I knew about sharks. But I had done this one thing: I had listened carefully. I listened because I was really interested. And he felt it. Naturally that pleased him. That kind of listening is one of the best ways to show respect to others, and it makes them feel great too. “Few human beings," wrote Jack Woodford in Strangers in Love "can resist the sweet effect of rapt attention.” I went even further than that. I was “sincere in my admiration and generous in my praise”.I told him that I had been hugely entertained and instructed. I told him I wished I had his knowledge. I told him that I should love to wander the fields with him. What's more, it was all true.And so I had him thinking of me as a good conversationalist when, in reality, I had only been a good listener and had encouraged him to talk.26. From Paragraph l, we can learn that the writer_________A. was deeply moved by the botanist's talkB. was amazed by what he was hearingC. was not in a comfortable situationD. behaved politely and properly27. Which of the following does the writer describe as a rule of politeness at dinner parties?A. Avoiding discussions about politics and religion.B. Listening carefully to what another guest says.C. Arriving and leaving at the appropriate time.D. Giving attention to all those in attendance.28. The underlined expression "rapt attention" in Paragraph 4 is closest in meaning to__________A. full understandingB. strong interestC. great uncertaintyD. little curiosity29. According to the writer, which of the following is an important characteristic of a good conversationalist?A. Listening attentively and encouraging the other side to continue.B. Encouraging the other side by sharing his/her own opinions.C. Promising a future meeting for more communication.D. Expressing respect by nodding his/her head.30. What is the purpose of the passage?A. To prove the writer is an interesting conversationalist.B. To share an interesting experience at a dinner party.C. To explain what makes a good conversationalist.D. To show that botanists can be really talkative.BA British dog-lover has invented a high-tech way of feeding his pet by Twitter(推特，流行社交网络)．Computer expert Nat Morris, 30, has designed a system to give his pet a "tweet treat" by sending him a Twitter message.His dog Toby: gets some delicious dog biscuits from a computer-controlled food machine whenever Nat sends a message to “@ feedtoby “.Nat often works away from home and isn't always able to feed Toby by hand. But his new invention allows Nat to feed his dog from anywhere in the world.Nat said, "Toby absolutely loves it. At first he didn't know what was going on. Now he sits underneath the machine, wagging his tail and waiting for the food to drop. "Nat fills the food machine with small pieces of dog biscuits, but not too many in case four-year-old Toby gets too many messages. And Nat has even equipped his house with an online camera so he can see Toby enjoying the food at his home.But one problem is that friends and family have been so amazed with the “tweet treat" machine that they have started sending tweets to Toby too. So Nat has had to restrict feeding time to make sure Toby doesn't turn into Tubby.“People have been sending him tweets at all hours of the day, so I had to limit it to between 9 a. m. and 9 p. m. I'm thinking of doing an updated one which can measure his weight before he is fed, just to make sure he's not putting on too much puppy fat, “explained Nat.How Nat's Twitter Feeder works:When a message is sent to @ feedtoby, it is received by a mini-computer that is linked to the food machine.When the mini-computer receives the message, a bell rings and Toby comes running over and sits in front of the feeding machine. Next, the machine's motor pulls open a trap door which releases a serving of food.The doggy biscuits then drop into Toby's food bowl. Finally a digital camera takes a photo of him and sends it back to Nat on Twitter - so he knows Toby has been fed.31. Nat has invented a high-tech way to feed his dog because he________.A. wants his friends to feed TobyB. has very strong computing skillsC. is often too busy to feed his dogD. doesn't like to feed Toby by hand32. Why has Nat decided to limit the feeding machine's operating time?A. He doesn't want Toby to get too fat.B. He fears the machine will run out of food.C. He wants his friends to stop feeding Toby.D. He doesn't want Toby to be woken up at night.33. It can be learned from the passage that Toby_________A. sits beneath his feeder all day longB. is now used to being fed by machineC. doesn't know what happens to the feederD. no longer receives tweets from Nat's friends34. Which of the following shows the correct order of how the Twitter Feeder works?a. The bell goes off.b. A message is sent to @ feedtoby.c. The mini-computer gets the message.d. The digital camera takes a photo of Toby and sends it to Nat.e. The motor starts to work and opens the door to release dog food.A. a, b, c, e, d.B. b, c, e, a, d.C. b, c, a, e, d.D. c, b, a, d, e35. In which section of the newspaper would you most probably find this passage?A. Technology.B. Health.C. Environment.D. Style.CNo one knows why we dream, but some dreams might be connected to the mental processes that help us learn. In a recent study, scientists found a connection between dreams and better memory in people learning a new skill.So perhaps one way to learn something new is to practice, practice, practice -- and then sleep on it."I was very surprised by this finding,” said Robert Stickgold, a Harvard University scientist who led the study.In the study, 100 college students each spent an hour on a computer, trying to get through amaze (谜宫). The maze was difficult, and the study participants had to start from a different place each time they tried -- making it even more difficult.Then, for the first 90 minutes of a five-hour break, half of the participants were required to stay awake while half were asked to sleep. Participants who stayed awake were asked to describe their thoughts. Participants who slept were asked to describe any dream they had.Stickgold and his colleagues wanted to know about NREM, or non-REM sleep. REM stands for “rapid eye movement, “which is what happens during REM sleep. This period of sleep often brings strange dreams to a sleeper, although dreams can happen in both kinds of sleep. Stickgold wanted to know what people were dreaming about when their eyes weren't moving, during NREM sleep. Other studies have found a connection between NREM brain activity and learning ability.Four of the 50 people who slept said their dreams were about the maze. Later, when these four people tried the computer maze again, they were able to complete it faster.Stickgold believes the dream itself doesn't help a person learn -- it's the other way around. He suspects that such dreams are caused by the brain processes associated with learning.All the maze-dreamers had done the task poorly the first time, which makes Stickgold wonder if the NREM dreams show up when a person finds a new task particularly difficult. People who had other dreams, or people who didn't sleep, didn't show the same improvement.36. In the first stage of the study, the participants were asked to________A. design a maze on computerB. find their way out of a mazeC. decide where to begin a mazeD. remember a location in a maze37. What happened to the participants during the break?A. Half of them were woken up when they started to dream.B. Half of them were asked to dream about the maze.C. All of them were asked to describe their thoughts.D. Half of them were asked to sleep for 90 minutes.38. What can we learn from the passage?A. Everyone will dream about a new skill after learning it.B. Stickgold was the first to study dreams and learning.C. During NREM sleep, people usually don't dream.D. Unusual dreams often occur during REM sleep.39. According to the last paragraph, before sleeping the maze-dreamers________A. found it difficult to do the mazeB. were greatly interested in the mazeC. were mostly slow and poor thinkersD. completed the maze faster than others40. Which of the following statements best summarizes the study's conclusion?A. Dreams have a role in learning.B. Dreams have no basis in reality.C. Dreams are important for health.D. Dreams are the best way to study.DThe recent publication of autobiographies by two of Britain's greatest scientists, biologist Richard Dawkins and physicist Stephen Hawking is a wonderful opportunity to compare and contrast these two remarkable men. Surprisingly, they have rather more in common than we think.Most striking is the similarity in their backgrounds. They were born in the early 1940s to middle class families -- not wealthy but comfortably off, with a strong commitment to academic excellence and public service. Both families were keen to send their boys to Oxford University --and both succeeded, Dawkins studying zoology and Hawking physics.Neither man has a very positive view of his early university life. Hawking describes the attitude at Oxford in the 1950s and 1960s as very anti-work, "You were supposed to either be brilliant without effort or fail. Hard work was looked down upon by students and we all pretended that nothing was worth making an effort for. “He estimates that he studied for no more than an hour a day as an undergraduate student (本科生).Undergraduate life was somewhat more rewarding for Dawkins. Like Hawking, he wasn't particularly hard-working and never attended his lectures. But he found Oxford's system of weekly essay-based lessons with an academic tutor useful, "It was really only the tutorial system that educated me.”For both men, scientific life really got going as postgraduates after 1962. Dawkins, who remained at Oxford, describes brilliantly the academic competition among the postgraduate students, which he believed helped push him to develop the ideas that formed the basis of his most famous book, The Selfish Gene. This volume transformed scientific thinking about Darwinian evolution.Hawking, on the other hand, moved to Cambridge University after graduation, where his research into the universe would eventually make him the most famous physicist since Albert Einstein. He writes movingly about the disease which progressively crippled his entire body, leaving him unable to move and only able to communicate using a computer controlled by his eyes.Although communication is slow - he can write only 3 words a minute using the machine - his illness has not affected his mind or his research on space-time and the origins of the universe.Each book is recommended individually as a personal introduction to an important scientific thinker. Read together, they provide a superb background to the academic and social climate of postwar British research.41. Which of the following describes a similarity in Hawking's and Dawkins' backgrounds?A. They were both from wealthy families.B. They studied the same subject in university.C. They graduated from the same secondary school.D. They both came from families that valued good education.42. Why did Hawking study very little as an undergraduate student?A. He preferred doing his own research and experiments.B. Students considered it inappropriate to study too much.C. The materials discussed in lectures were very easy for him.D. He was more interested in making friends with his classmates.43. According to Dawkins, what helped him develop his most important ideas?A. His hard work as an undergraduate.B. The support he received from his family.C. The excellent tutors at Oxford University.D. The competition from other postgraduate students.44. What can we reasonably infer about the two scientists from the passage?A. Dawkins worked much harder than Hawking as an undergraduate.B. Hawking is more respected by the scientific community.C. They knew each other during their studies at Oxford.D. Hawking has experienced more physical difficulties.45. What is the function of the last paragraph?A. To state which book the writer prefers.B. To recommend the reviewed books to readers.C. To summarize the achievements of the two scientists.D. To suggest the order in which the books should be read.笫二节信息匹配（共5小题；每小题2分，满分10分）阅读下列应用文及相关信息，并按照要求匹配信息。

2014年广州市普通高中毕业班综合测试(一)理科综合本试卷共12页，36小题，满分300分。

考试用时150分钟。

注意事项:1．答卷前，考生务必用2B铅笔在“考生号”处填涂考生号,用黑色字迹钢笔或签字笔将自己所在的市、县／区、学校,以及自己的姓名和考生号、试室号、座位号填写在答题忙上。

用2B铅笔将试卷类型(A）填涂在答题卡相应位置上.2．选择题每小题选出答案后，用2B铅笔把答题卡上对应题目选项的答案信息点涂黑，如需改动，用橡皮擦干净后，再选涂其他答案，答案不能答在试卷上。

3．非选择题必须用黑色字迹钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内相应位置上；如需改动,先划掉原来的答案，然后再写上新的答案；不准使用铅笔和涂改液。

不按以上要求作答的答案无效。

4．考生必须保持答题卡的整洁.考试结束后，将试卷和答题卡一并交回.5. 本卷所用相对原子质量:H－1、C－12、N－14、O－16、K－39、I－127一、单项选择题:本题包括16小题，每小题4分,共64分。

在每小题给出的四个选项中，只有一个选项符合题目要求，选对的得4分,选错或不答的得0分.1．下列有关细胞中化合物的叙述,正确的是A．叶肉细胞中含量最多的化合物是蛋白质B．肝细胞中含有葡萄糖和蔗糖C．脂肪是细胞中良好的储能物质D．细胞中的无机盐都以离子形式存在2．研究发现,一些癌细胞能进入“平衡”阶段，此阶段的癌细胞不易被免疫系统识别,不会恶性增殖,也不会导致机体发病.下列有关分析不．合理的是A．这种癌细胞是细胞正常分化的结果B．免疫系统不易清除这种癌细胞C．可以通过癌基因检测的手段来监测细胞是否癌变D．该发现有助于寻找治疗癌症的新方法3．在摩洛哥有一种被称为“奶树"的古老树种,成年树分泌的“乳液”有利于由其根部细胞发育而来的幼树成长。

下列有关推论合理的是A．“奶树”是通过细胞工程培育而成的B．成年树与其根部细胞发育而来的幼树的基因型不同C．分泌“乳液”的植株是雌性植株D．成年树分泌“乳液”是表现型4．右图为突触结构和功能的模式图,下列有关叙述不．恰当的是A．瞬间增大轴突末端细胞膜对Ca2+的通透性会加速神经递质的释放B．过程①体现了细胞膜具有流动性C．过程②表示神经递质进入突触后膜所在的神经元D．过程⑧可避免突触后膜持续兴奋5.下列相关实验组合不．正确的是A。

2014广州市普通高中毕业班综合测试（一）参考答案完形填空1-5 BACBD 6-10 ACBDC 11-15 BCDBA语法填空16. destroyed 17. what 18. powerful 19. who 20. worked21. before 22. lying 23. a 24. they 25. on阅读理解26-30 BDBAC 31-35 CABCA 36-40 BDDAA 41-45 DBDDB信息匹配46-50 BACFE基础写作（满分15分）White Tea, which gets its name from its silver-colour, is regarded as one of the six major Chinese tea types. Dating back to the Beisong Dynasty, White Tea has a long history and it is mainly grown in Fujian and Taiwan. White Tea is famous for its beautiful silver-needle-like shape, its sweat taste and its medicinal effects. The longer it is preserved, the better effect it has. White Tea is good for health if drunk frequently and can be used in preventing diseases like high blood pressure and helping relieve the pain of toothache and fever.读写任务（满分25分）概括要点：1. family never liked to waste money;2. …was deeply in debt;3. being thrifty is being smart with your money… find ways to do everything for less.第二部分：1. 对节俭的理解；2. 对节俭的重要性的论述；3. 就生活中如何做到节俭提出建议。

2014年广州市普通高中毕业班综合测试（一）语文本试卷共8页，24小题，满分为150分。

考试用时150分钟。

注意事项：1．答卷前，考生务必用2B铅笔在“考生号”处填涂考生号。

用黑色字迹的钢笔或签字笔将自己所在的市、县／区、学校以及自己的姓名和考生号、试室号、座位号填写在答题卡上。

用2B铅笔将试卷类型（B）填涂在答题卡相应位置上。

2．选择题每小题选出答案后，用2B铅笔把答题卡上对应题目选项的答案信息点涂黑；如需改动，用橡皮擦干净后，再选涂其他答案，答案不能答在试卷上。

3．非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内的相应位置上；如需改动，先划掉原来的答案，然后再写上新的答案；不准使用铅笔和涂改液。

不按以上要求作答的答案无效。

4．作答选做题时，请先用2B铅笔填涂选做题的题组号对应的信息点，再作答。

漏涂、错涂、多涂的，答案无效。

5．考生必须保持答题卡的整洁。

考试结束后，将试卷和答题卡一并交回。

一、本大题4小题，每题3分，共12分。

1．下列词语中加点的字，每对读音都不相同的一组是A．箴言／斟酌国粹／仓猝复辟／开天辟地B．撰写／编纂贬谪／嫡系冠名／冠冕堂皇C．对峙／嗜好竣工／疏浚提防／提心吊胆D．清澈／掣肘粗犷／旷达识别／博闻强识2.下面语段中画线的词语，使用不恰当的一项是近几年，国内许多风景名胜区实行“一票制”，将景区内多个景点门票捆绑搭售。

这种做法引起了人们的置疑和不满，许多游客认为这是变相涨价。

一个知名景区要可持续发展，首先必须赢得游客的口碑，如果过分依赖“门票经济”做“一锤子买卖”，对游客的意见充耳不闻，一意孤行，一旦引起游客的反感乃至抵触，就可能造成难以挽回的损失。

A．置疑B．一锤子买卖C．充耳不闻D．乃至3.下列句子中，没有语病的一句是A．著名作家村上春树连续五年排在诺贝尔文学奖获奖预测名单榜首，却年年与该奖无缘，可以堪称诺贝尔文学奖史上“最悲壮的入围者”。

B．广州恒大足球队首次参加“世俱杯”比赛，与非洲、欧洲和南美洲的冠军同场竞技，在收获自信的同时也看到了与世界强队的差距。