湖南省长沙市一中2011届高三第七次月考(理科数学)

湖南省长沙市一中2011 届高三月考（七）英语试题（考试范围：全部内容）本试卷分为四个部分，包括听力、语言知识运用、阅读和书面表达。

时量120分钟。

满分150分。

PART ONE LISTENING COMPREHENSION（30 marks）SECTION A（22．5 marks）Directions：In this section，you’ll hear six conversations between two speakers．For each conversation，there are several questions and each question is followed by three choices marked A，B and C．Listen carefully and then choose the best answer for each question．You will hear each conversation TWICE．Conversation 11．Why is the man going to read the book？A．Because it is homework．B．Because it is interesting．C．Because it is mysterious．2．What will the man do next？A．Going to the library．B．Going to the bookstore．C．Going to David’s dorm．Conversation 23．What will they do first？A．Buying some food．B．Having supper．C．Reading a newspaper．4．Who will drive the car？A．The man．B．The woman．C．They will drive by turns．Conversation 35．What happened last night？A．There was a big dinner．B．There was a party．C．There was a meeting．6．What is the man going to do？A．Cleaning the apartment．B．Quarrelling with the woman．C．Having a rest．Conversation 47．What probably is Steve？A．A mechanical engineer．B．A shop assistant．C．A hotel chef．8．What does Steve find difficult in his job？A．Working long hours．B．Dealing with customers．C．Keeping everything clean．9．What does Steve hope to do in the future？A．Becoming an expert．B．Becoming a manager．C．Becoming a boss．Conversation 510．What table did the speakers talk about first？A．A table about how children get their pocket money．B．A table about how children spend their pocket money．C．A table about how much pocket money children receive．11．How has the amount of children’s pocket money changed in Britain？A．It has been increasing．B．It has been decreasing．C．It has remained almost the same．12．What can be inferred from the conversation？A．The man thinks pocket money is for school things．B．The man thinks pocket money is for personal things．C．The man thinks pocket money is for Christmas things．Conversation 613．What are the two speakers going to do？A．They are going to sell a farm．B．They are going to buy a farm．C．They are going to rent a farm．14．What do they plan to do when they live on the farm？A．They plan to keep some fish．B．They plan to keep some cows．C．They plan to keep some chickens．15．Why won’t they grow corn on the farm？A．Because the farm is small．B．Because growing corn is tiring．C．Because growing corn isn’t a good way to earn money．SECTION B（7．5 marks）Directions：In this section，you will hear a short passage．Listen carefully and then fill in the numbered blanks with the information you have heard．Fill in each blank with NO MORE THAN THREE WORDS．You’ll hear the short passage TWICE．PART TWO LANGUAGE KNOWLEDGE（45 marks）SECTION A（15 marks）Directions：Beneath each of the following sentences there are four choices marked A，B，C and D．Choose the one answer that best completes the sentence．21．—We have waited for almost 30 minutes．How much longer do we have to wait？—the meeting finishes．I’m afraid there will be another 20 minutes．A．When B．Unless C．Until D．Since22．I’m trying to meet the deadline of my report；on no account me when I’m working．A．you will disturb B．will you disturbC．you must disturb D．must you disturb23．He said we could make it home before 9 p．m．，I personally doubt．A．where B．when C．that D．which24．All along the way to the tourist destination，he would sometimes slow down the car sure if we were going the right way．A．making B．to make C．made D．to be making 25．Sometimes，money is tight and on other occasions time is tight；on occasion，a travel is out of the question．A．all B．any C．either C．every26．It was obvious the moment I got in something unpleasant had happened in the office．A．that B．which C．whether D．so27．When asked which floor Mr．Smith lived，the gatekeeper said there was such person．A．no B．not a C．not any D．never28．Living cost is rising and by the end of last month the prices of most foods ．A．increased B．have increased C．would increase D．had increased29．—I hope you can go over to have dinner with my family．—Sorry，you see I’m in such a hurry，but remember me your parents．A．at B．to C．on D．for30．—How do you plan to spend the weekend？—I plan to go boating in the park．fun it will be boating in the warm autumn sunshine！A．What B．What a C．How D．How a31．I spent several years in Spain，I never learnt to speak Spanish．A．and B．yet C．so D．as32．At the entrance of the hall hangs a picture of an elegant and graceful old woman，at everyone passing her．A．seated smiling B．seating smilingC．seating smiled D．seated smiled33．I’m afraid I couldn’t mail you the report until about 8 o’clock in the evening because I on my way to Guangzhou the whole afternoon but my train hasn’t arrived yet．A．had been B．was C．am D．will be34．What is Tom up to？He’s seldom turned in his assignments these days，？A．is he B．isn’t he C．has he D．hasn’t he 35．—We are reading advertisement after advertisement for a suitable decoration company to get my apartment redecorated．—So I before I finally fixed on Oriental，a reliable and experienced company．A．have B．had D．am D．wasSECTION B（18 marks）Directions：For each blank in the following passage there are four words or phrases marked A，B，C and D．Fill in each blank with a word or phrase that best fits the context．Two friends visited the zoo together．The zoo was very large and it was impossible to go 36．They had to decide where and which animal to visit as their time was 37．So both of them agreed not to repeat the journey after choosing a branch road at every fork．A road sign at the first fork indicated one way to the 38quarters and the other to the tiger hill．They decided on the former after a brief discussion because lions were “the king of the grassland”．The second showed a division going 39to the panda and the peacock．They favored panda as it was the nation’s treasure and went its way．Thus they made choices all along the way and each choice meant a rejection that they couldn’t help 40．But they had to 41it，and immediately，for it brooked （容忍）no delay．If they 42they would miss more．Only 43decision could offer more chances for sightseeing and 44possible regret．Life is more or less like this—alternatives often occur that one has to choose，for example，between two 45jobs，two fascinating wooers （追求者）．To obtain one you have to 46the other—you can only get 47of it．If you spend time weighing the pros and cons and calculating gains and losses，you will most likely end up in empty-handedness．Don’t be sad about it．At least you have got half of the desirable things in life—something that is hard to come by．36．A．anywhere B．everywhere C．somewhere D．nowhere37．A．limited B．spared C．lengthened D．permitted 38．A．bear B．monkey C．elephant D．lion 39．A．directly B．immediately C．separately D．deeply 40．A．remembering B．regretting C．refusing D．receiving 41．A．make B．put C．set D．take42．A．left B．went C．hurried D．hesitated 43．A．important B．serious C．careful D．rapid 44．A．reduce B．cause C．increase D．lose 45．A．terrible B．easy C．desirable D．hard46．A．give out B．give in C．give up D．give away 47．A．both B．half C．neither D．allSECTION C（12 marks）Directions：Complete the following passage by filling in each blank with one word that best fits the context．The giving of gifts has been practised in every civilization ever studied．It is 48．basic human phenomenon that has a place in every culture and religion．Gifts are given to mark occasions throughout 49．，such as birth，marriage and retirement．Sociologists view gifts as a marker of the social relationship 50．givers and recipients．When friends exchange gifts，for example，there is an unwritten expectation 51．the gifts will be of roughly the 52．value，showing that the friends have equality of status．However，in hierarchical （等级的）relationships it is a 53．story．If a（n）54．and a boss were exchanging presents，the boss would be expected to give a larger present．In 55．for this generosity，the employee would be expected to both work hard and be respectful．PART THREE READING COMPREHENSION（30 marks）Directions：Read the following three passages．Each passage is followed by several questions or unfinished statements．For each of them there are four choices marked A，B，C and D．Choose the one that fits best according to the information given in the passage．AI went into Harrods in London （a huge department store I am sure you have heard of）to buy some Minton china plates for a wedding anniversary which was to be a group gift from friends to other friends in Geneva．Having only 30 minutes between meetings，I whizzed there in a taxi from the office and battled through the milling people on the 2nd day of the sales on the ground floor up to the 5th to the china department．There was a man standing there who was obviously a sales person who I rushed up to and asked if he had this particular china in stock and if it would take long to wrap，etc．．He was amazing．He got the plates in seconds，wrapped them up and asked me if I wanted a store card，to which I replied no，because I lived in Switzerland，to which he replied asking if，as I lived abroad，would like a tax rebate form （退税表）．After that he showed me what to do，produced a map of the store and told me where I should go for the formalities （程序）．That was fabulous！I thanked him and said what wonderful service he had given me．I even asked if he gave this to everyone，to which he answered：“I’m just doing what is required at Harrods”．With that，a tall man in a grey suit approached me and offered his hand to shake mine saying，“Can I introduce myself？I am the Chief Executive of Harrods and what an interesting conversation I have just heard．．．” He had been wandering through the store （as youshould do as a hands-on CEO）and had overheard me thanking this salesman，whose face，I can hardly describe，was frozen in a mixture of delight awe and astonishment！Can you imagine the salesman going home to his family and friends recounting what the CEO spoke to him after overhearing him being praised by a customer？56．The writer went to Harrods to ．A．kill time B．buy some giftsC．go sightseeing D．shop for her wedding57．What does the underlined word “whiz” in Paragraph 2 most probably mean？A．Drive．B．Arrive．C．Rush．D．Leave．58．W hich of the following sentences can best replace “I’m just doing what is required at Harrods” in the second paragraph？A．“I’m an honest man”．B．“I must obey the rules．”C．“Harrods is strict with its employees．”D．“Anybody will receive our good service”．59．As soon as she bought what she wanted，the writer ．A．went to a meeting B．flew back to GenevaC．visited one of her friends D．attended a wedding ceremony60．The writer wrote the above passage because she thought what she had experienced was so ．A．pleasant B．strange C．exciting D．amusingBYou never see him，but they’re with you every time you fly．They record where you are going，how fast you’re traveling and whether everything on your airplane is functioning normally．Their ability to withstand （经受住）almost any disaster makes them seem like something out of a comic book．They’re known as the black box．When planes fall from the sky，as a Yemeni airliner did on its way to Comoros Islands in the India ocean June 30，2009，the black box is the best bet for identifying what went wrong．So when a French submarine （潜水艇）detected the device’s homing signal five days later，the discovery marked a huge step toward determining the cause of a tragedy in which 152 passengers were killed．In 1958，Australian scientist David Warren developed a flight-memory recorder that would track basic information like altitude and direction．That was the first mode for a black box，which became a requirement on all U．S．commercial flights by 1960．Early models often failed to withstand crashes，however，so in 1965 the device was completely redesigned and moved to the rear （后部）of the plane—the area least subject to impact—from its original position in the landing wells （起落架舱）．The same year，the Federal Aviation Authority required that the boxes，which were never actually black，be painted orange or yellow to aid visibility．Modern airplanes have two black boxes： a voice recorder，which tracks pilots’conversations，and a flight-data recorder，which monitors fuel levels，engine noises and other operating functions that help investigators reconstruct the aircraft’s final moments．Placed in an insulated （隔绝的）case and surrounded by a quarter-inch-thick panels of stainless steel，theboxes can withstand massive force and temperatures up to 2,000℉．When submerged （淹没），they’re also able to emit signals from depths of 20,000 ft．Experts believe the boxes from Air France Flight 447，which crashed near Brazil on June 1,2009，are in water nearly that deep，but statistics say they’re still likely to turn up．In the approximately 20 deep-sea crashes over the past 30 years，only one plane’s black boxes were never recovered．61．Which of the following is the magic feature about a black box，according to the text？A．It can record clearly．B．It can be seen clearly．C．It can survive a serious disaster．D．It can tell altitude and direction．62．Which of the following statements about the air disaster happening to Yemeni airliner is TRUE？A．Its black box hasn’t been found yet．B．Its black box was found under the sea．C．Its black box saved all the passengers’ lives．D．Its black box was found immediately after the crash．63．When the first mode of black box appeared，．A．it wasn’t black B．it was painted yellowC．it was at the rear of the plane D．it wasn’t able to track direction 64．According to the passage，modern black boxes ．A．are placed in insulated plastic boxesB．have seldom been found in deep-sea crashesC．are painted bright colors and can emit signalsD．will melt only when the temperature rises to 1,000℉65．Several changes to the black box are mentioned in the passage except the change in its ．A．position B．number C．strength D．sizeCThe U．S．Food and Drug Administration （FDA）is considering to put stricter limits over tanning salons （晒黑廊）and wants to ban anyone younger than 18 years of age from using a tanning bed，an advisory panel （专家团）announced last week．The panel is calling for tighter controls on the industry such as requiring teenagers to get the approval from their parents before using tanning beds or limiting the use of artificial tanning to a certain age．“Given the absence of any demonstrated benefit，I think it is an obligation for us to ban artificial tanning for those under 18，” said panelist Dr．Michael Olding．Along with a possible ban for teenagers，the panel also recommended that visible warning labels should be placed either on the tanning machines or in the salons in order to caution tanners of the possible dangers．In addition，the committee decided that stricter regulations and classifications were critical to make the machines safer．At this time the machines are categorized as FDA Class 1 devices，the ones that are least likely to cause harm．In case the FDA decided to change their classification from Class 1 to Class 2，as advised by the panel，the FDA could limit the levels of radiation the machines emit．Class 2 devices include X-ray machines and powered wheelchairs．Getting a tan，whether from a tanning bed or the sun，increases the risk of developing skin cancer．Last year，the World Health Organization’s International Agency for Research on Cancer （IARC）declared tanning beds as “carcinogenic （致癌的）to humans”．It wasdiscovered that young individuals in their teens and 20s who use tanning beds on a regular basis have a 75 per cent higher risk of suffering from melanoma （黑素瘤），the deadliest form of skin cancer．According to the American Cancer Society，melanoma accounted for nearly 69,000 cases of skin cancer in 2009 and will account for most （about 8,650）of the 11,590 mortality cases due to skin cancer each year．66．According to the passage，what measures will U．S．FDA most probably take？A．Banning tanning salons．B．Posing heavier tax over tanning salons．C．Having tighter controls over tanning salons．D．Limiting the number of tanning salons in every state．67．Which of the following suggestions for making tanning salons safer is NOT mentioned？A．Visible Caution．B．Setting age limit．C．Professional personnel．D．Parental approval for teenagers．68．Which of the following is TRUE according to the passage？A．X-ray machines are less likely to cause harm than tanning machines．B．X-ray machines are more likely to cause harm than tanning machines．C．Powered wheelchairs are as likely to cause harm as tanning machines．D．Powered wheelchairs are less likely to cause harm than tanning machines．69．What does the writer want to express in the last paragraph？A．Tanning in one’s youth may mean death．B．Tanning in the sun is safer than on the tanning bed．C．People should get tanned without getting melanoma．D．Getting tanned is only a good idea for those above thirty years old．70．What will most probably happen，if the advisory panel’s suggestions are adopted and put into practice？A．Fewer people will suffer from skin cancer．B．Tanning salons will have more customers．C．Getting a tan in a tanning salon will cost less．D．Parents will be more anxious about their tanning children．PART FOUR WRITING（45 marks）SECTION A（10 marks）Directions：Read the following passage．Fill in the numbered blanks by using the information for the passage．Write NO MORE THAN THREE WORDS for each answer．According to the surprising findings of a new study by U．S．investigators，chewing sugarless gum during class and while doing homework can have a positive effect on academic performance in teenagers．Study leader Craig A．Johnston of the Children’s Nutrition Research Center at Baylor College of Medicine in Houston，Texas，and his colleagues studied more than 100 eighth-grade students，52 girls and 56 boys，aged between 13 and 16 years，in four math classes．The experts randomly assigned teenagers into two groups：one group was asked to chew Wrigley’s sugar-free gum during class，while doing homework，and also while performing a standardized test．They chewed at least one stick of gum 86 per cent of the time they were in math class and 36 per cent of the time they were doing homework．The participants of the other group did not．Johnston and his team found that 14 weeks later，the gum chewers had a 3 percent increasein their math scores on the Texas Assessment of Knowledge and Skills achievement test，a small but statistically significant change，according to experts．There was no difference found in math scores between the participants in the two groups in another test called the Woodcock Johnston III Tests of Achievement．However，the experiment revealed that gum-chewers had better final grades in the class compared to their non-chewing peers．According to Johnston，chewing gum reduces stress and anxiety as well as it increases arousal （兴奋）．Researchers say that the studies and research are focused on investigating the effect of chewing gum on focus，alertness，concentration，situational stress，weight control and oral health．According to them，the study is really meaningful and should raise interest in parents “when related to small steps that can lead to better academic performance．”The new study is being built on previous research that was conducted in a laboratory setting and showed that gum chewing can help reduce stress，improve alertness and relieve anxiety．The current findings，for the first time，provide a possible role for chewing gum in helping to improve academic performance in a “real life” classroom setting．A surprising finding—chewing gum results in 71．Ⅰ．Purpose of the study：To investigate the effect of chewing gum on focus，alertness，concentration，situational stress，weight control and 72．．Ⅱ．The research：◆Researchers：Craig A．Johnston and 73．◆Participants：74．eighth-grade students aged between 13 and 16 years◆Contents：◆Results：● Gum­chewers’math scores increasing by 78．on the Texas Assessment of Knowledge and Skills achievement test● 79．in the Woodcock Johnston III Tests of Achievement between the two groups ● Gum­chewers having bett er final gradesⅢ．Researchers’ Analysis：◆Chewing gum reduces 80．．◆Chewing gum increases arousal．SECTION B （10 marks）Directions：Read the following passage．Answer the questions according to the information given in the passage．The period of engagement is the time between the marriage proposal and the wedding ceremony．Two people agree to marry when they decide to spend their lives together．The manusually gives the woman a diamond engagement ring．That tradition is said to have started when an Austrian man gave a diamond ring to the woman he wanted to marry．The diamond represented beauty．He placed it on the third finger of her left hand．He chose that finger because it was thought that a blood vessel （血管）in that finger went directly to the heart．Today，we know that this is not true．Yet the tradition continues．Americans generally are engaged for a period of about one year if they are planning a wedding ceremony and party．During the time，friends of the bride may hold a party at which women friends and family members give the bride gifts that she will need as a wife．These could include cooking equipment or new clothing．Friends of the man who is getting married may have a bachelor party for him．This usually takes place the night before the wedding．Only men are invited to the bachelor party．During the marriage ceremony，the bride and her would-be husband usually exchange gold rings that represent the idea that their union will continue forever．The wife often wears both the wedding ring and engagement ring on the same finger．The husband wears his ring on the third finger of his left hand．Many people say the purpose of the engagement period is to permit enough time to plan the wedding．But the main purpose is to let enough time pass so the two people are sure they want to marry each other．Either person may decide to break the engagement．If this happens，the woman usually returns the ring to the man；they also return any wedding gifts they have received．81．According to the passage，on which finger does a woman wear her engagement ring today？（No more than 12 words）（3 marks）82．Why will a bride receive such gifts as cooking equipment？（No more than 8 words）（2 marks）83．When does a bachelor party usually take place？（No more than 10 words）（2 marks）84．What is the most important reason for there to be the engagement period？（No more than 18 words）（3 marks）SECTION C（25 marks）Directions：Write an English composition according to the instructions given below in Chinese．happyfamily．com是一个致力于消除家长和孩子之间沟通障碍的网站。

湖南省长沙市一中2011届高三月考试卷(七)数 学(理科)长沙市一中高三理科数学备课组组稿 命题人：李湘斌 审题人：赵意扬 (考试范围：高考理科内容(不含选修系列4))本试题卷包括选择题、填空题和解答题三部分，共6页。

时量120分钟。

满分150分。

一、选择题：本大题共8小题，每小题5分，共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.直线x ＋y －1＝0的倾斜角是( )A.－π4B.π4C.3π4D.π22.“p ∧q 是真命题”是“p ∨q 是真命题”的( ) A.充分不必要条件 B.必要不充分条件C.充要条件D.既不充分也不必要条件3.右图是一个几何体的三视图，根据图中数据，可得该几何体的表面积是( )A.9πB.10πC.11πD.12π4.在等差数列{a n }中，已知a 1＝2，a 2＋a 3＝13，则a 4＋a 5＋a 6等于( ) A.40 B.42 C.43 D.455.设m >0，则直线x ＋3y ＋1＋m ＝0与圆x 2＋y 2＝m 的位置关系是( ) A.相切 B.相交 C.相切或相离 D.相交或相切6.下列函数中，图象的一部分如右图所示的是( )A.y ＝sin(2x ＋π6)B.y ＝sin(2x －π6)C.y ＝cos(2x ＋π3)D.y ＝cos(2x －π6)7.函数y ＝lg x －9x的零点所在的大致区间是( )A.(6,7)B.(7,8)C.(8,9)D.(9,10)8.设m ∈N *，F (m )表示log 2m 的整数部分，则F (210＋1)＋F (210＋2)＋F (210＋3)＋…＋F (211)的值为( )A.10×210B.10×210＋1C.10×210＋2D.10×210－1二、填空题：本大题共7小题，每小题5分，共35分，把答案填在答题卡中对应题号后的横线上.9.∫10x 2d x ＝ .10.某工厂生产A 、B 、C 三种不同型号的产品.产品数量之比依次为2∶3∶5，现用分层抽样方法抽出一个容量为n 的样本，已知A 种型号产品共抽取了16件，那么此样本的容量n ＝ .11.如右图所示的算法流程图中，输出S 的值为 .12.设A 、B 为x 轴上两点，点P 的横坐标为2，且|PA|＝|PB|，若直线PA 的方程为x －2y ＋1＝0，则直线PB 的方程是 .13.已知双曲线x 2a 2－y 2b2＝1(a ＞0，b ＞0)的左、右焦点分别为F 1、F 2，点P 在双曲线的右支上，且|PF 1|＝3|PF 2|，则此双曲线的离心率e 的最大值为 .14.在△ABC 中，P 为中线AM 上的一个动点，若|AM |＝2，则PA ·(PB ＋PC )的最小值为 .15.如图，将平面直角坐标系的格点(横、纵坐标均为整数的点)按如下规则标上数字标签：原点处标数字0，点(1,0)处标数字1，点(1，－1)处标数字2，点(0，－1)处标数字3，点(－1，－1)处标数字4，点(－1,0)处标数字5，点(－1,1)处标数字6，点(0，1)处标数字7，…以此类推，①标数字50的格点的坐标为 .②记格点坐标为(m ，n)的点(m 、n 均为正整数)处所标的数字为f(m ，n)，若n>m ，则f(m ，n)＝ .三、解答题：本大题共6小题，共75分.解答应写出文字说明，证明过程或演算步骤. 16.(本小题满分12分)已知sin x 2－2cos x2＝0.(1)求tan x 的值；(2)求cos 2x2cos (π4＋x)·sin x的值.17.(本小题满分12分)某旅游公司为3个旅游团提供甲、乙、丙、丁4条旅游线路，每个旅游团只能任选其中一条线路，不同的旅游团可选相同的旅游线路.(1)求3个旅游团选择3条不同的线路的概率； (2)求选择甲线路旅游团的团数的分布列和期望.如右图，简单组合体ABCDPE ，其底面ABCD 为正方形，PD ⊥平面ABCD ，EC ∥PD ，且PD ＝2EC.(1)若N 为线段PB 的中点，求证：EN ⊥平面PDB ；(2)若PDAD ＝2，求平面PBE 与平面ABCD 所成的锐二面角的大小.19.(本小题满分13分)已知函数f(x)＝12ax 2＋(1－a)x －1－ln x ，a ∈R .(1)若函数在区间(2,4)上存在．．单调递增区间，求a 的取值范围； (2)求函数的单调增区间.20.(本小题满分13分)某公园的大型中心花园的边界为椭圆，花园内种植各种花草，为增强观赏性，在椭圆内以其中心为直角顶点且关于中心对称的两个直角三角形内种植名贵花草(如图)，并以该直角三角形斜边开辟观赏小道(不计小道的宽度)，某园林公司承接了该中心花园的施工建设，在施工时发现，椭圆边界上任意一点到椭圆两焦点距离和为4(单位：百米)，且椭圆上点到焦点的最近距离为1(单位：百米).(1)试以椭圆中心为原点建立适当的坐标系，求出该椭圆的标准方程； (2)请计算观赏小道的长度(不计小道宽度)的最大值.顶点在坐标原点，开口向上的抛物线经过A 0(1,1)，过A 0作抛物线的切线交x 轴于B 1，过B 1点作x 轴的垂线交抛物线于A 1，过A 1作抛物线的切线交x 轴于B 2，…，过A n (x n ，y n )作抛物线的切线交x 轴于B n ＋1(x n ＋1,0)(1)求{x n }，{y n }的通项公式；(2)设a n ＝11＋x n ＋11－x n ＋1，数列{a n }的前n 项和为T n .求证：T n >2n －12.(3)设b n ＝1－log 2y n ，若对任意正整数n ，不等式(1＋1b 1)(1＋1b 2)…(1＋1b n)≥a 2n ＋3成立，求正数a的取值范围.炎德·英才大联考长沙市一中2011届高三月考试卷(七)数 学(理科) 教师用卷长沙市一中高三理科数学备课组组稿 命题人：李湘斌 审题人：赵意扬 (考试范围：高考理科内容(不含选修系列4))本试题卷包括选择题、填空题和解答题三部分，共6页。

湖南省长沙市一中-高三第七次月考理科数学命题：卿 科 审卷：卿 科本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分．满分150分，考试时间120分钟．参考公式： 正棱锥、圆锥的侧面积公式如果事件A 、B 互斥，那么 cl S 21=锥侧 P （A+B ）=P （A ）+P （B ）如果事件A 、B 相互，那么 其中，c 表示底面周长、l 表示斜高或 P （A ·B ）=P （A ）·P （B ） 母线长如果事件A 在1次实验中发生的概率是 球的体积公式 P ，那么n 次重复实验中恰好发生k 334R V π=球 次的概率kn k k n n P P C k P --=)1()( 其中R 表示球的半径第I 卷(共50分)一．选择题：本大题共10小题，每小题5分，共50分，在每小题给出的四个选项中，只有一项是符合要求的． １．复数2(1)z i i =-⋅等于Ａ．2- Ｂ．2 Ｃ．2i Ｄ．2i - ２．设全集Ｉ是实数集R. 22{|4}{|1}1M x x N x x =>=≥-与都是Ｉ的子集（如图所示， 则阴影部分所表示的集合为 Ａ．{}2x x <Ｂ．{}21x x -≤<Ｃ．{}22x x -≤≤ Ｄ．{}12x x <≤３．函数x x x f 44cos sin )(+=的最小正周期是Ａ．4π Ｂ．2πＣ．3πＤ．π ４．设等差数列{}n a 的公差为2，前n 项和为n S ，则下列结论正确的是Ａ．()12--=n n na S n nＢ．()12-+=n n na S n n Ｃ．()1--=n n na S n nＤ．()1-+=n n na S n n５．抛物线x y 82-=的准线与双曲线12822=-y x 的两条渐近线所围成的三角形的面积为 Ａ．8 Ｂ．6 Ｃ．4 Ｄ．2 ６．已知b a <<0，且a+b=1，则下列不等式中，正确的是Ａ．0log 2>aＢ．212<-ba Ｃ．2log log 22-<+b aＤ．212<+abb a ７．在空间给出下列四个命题：①如果平面α内的一条直线a 垂直于平面β内的任意一条直线，则α⊥β； ②如果直线a 与平面β内的一条直线平行，则a ∥β； ③如果直线a 与平面β内的两条直线都垂直，则a ⊥β；④如果平面α内的两条直线都平行于平面β，则a ∥β．其中正确的个数是 Ａ．1 Ｂ．2 Ｃ．3 Ｄ．4８．已知点(3,3)A ，O 是坐标原点，点(,)P x y 的坐标满足303200x y x y -≤+≥⎨⎪≥⎪⎩，设z 为OA在OP 上的投影，则z 的取值范围是Ａ．[3,3]- Ｂ．[3,3]- Ｃ．[33]Ｄ．[3,3]-９．把半径都为1的四个小球装入一个大球内，则此大球的半径的最小值为 Ａ．261+Ｂ．231+ Ｃ．62+ Ｄ．3321+ 10．设点P 是函数),0(,sin )(π∈=x x x f 图象上的任意一点．点A 的坐标为)0,(π，O 为坐标原点，则使得OAP ∆为直角三角形的点P 的个数是Ａ．0 Ｂ．2 Ｃ．4 Ｄ．6第II 卷二．填空题：本大题共５小题，每小题5分（第14、15题第一空２分，第二空3分），共25分．把答案填在答题卡．．．中对应题号后的横线上．11．二项式92()2x x-展开式中1x的系数为252-． 12．若311x b x →=-，则=+b a 2．13．在1,2,3,4,5五个数字组成没有重复数字的三位数中，各位数字之和为奇数的共有24． 14．已知⊙1:22=+y x O 及直线052:=+-y x l ．点),(00y x P 是直线l 上的任意一点．过),(00y x P 作⊙O 的两条切线PB PA ,，B A ,为切点．（i ）当20=x 时，则直线AB 的方程为0192=-+y x ；（ii ）OB OA ⋅的最大值为53-． 15．已知函数|37||37|)(-++=x x x f ．（i ）函数)(x f 的对称中心为)0,0(；（）若函数a x x a x a x x g 42|)3||1)(|()(+--+-++=的图象有对称中心，则=a 32-．三．解答题：本大题共6小题，共75分．解答应写出文字说明，证明过程或演算步骤． 16．（本小题满分12分）一袋中装有分别标记着1、2、3、4数字的4个球, 从这只袋中每次取出1个球, 取出后放回, 连续取三次, 设三次取出的球中数字最大的数为ξ. （１）求3ξ=时 的概率；（２）求ξ的概率分布列及数学期望. 解：（１）3ξ=表示取出的三个球中数字最大者为3．①三次取球均出现最大数字为3的概率311()4P =②三取取球中有2次出现最大数字3的概率2223126()()4464P C ==③三次取球中仅有1次出现最大数字3的概率12331212()()4464P C == ∴12319(3)64P P P P ξ==++=． …………………………………………………6分 （２）在k ξ=时, 利用（１）的原理可知：2322123311111331()()()()()()4444464k k k k P k C C ξ---+==++=,(k =1,2,3,4)ξ的概率分布为：ξ １ ２ ３ ４P641647 6419 6437E =1×164＋2×764＋3×1964＋4×3764 = 5516．………………………………………………12分17．（本小题满分12分）如图, 在正方体ABCD —1111D C B A 中, E 为AB 的中点．（１）证明：平面D EB 1⊥平面CD B 1；（２）求CE 与平面DE B 1所成角的大小的正弦值． 解：（１）取C B 1的中点,F D B 1的中点,G 连结.GF ,EG ,BF⊥CD 平面11B BCC , B F DC ⊥∴.又C B BF 1⊥, ,C C B DC 1=⋂⊥∴B F 平面CD B 1．……………………………3分 GFBE ,CD 21,CD 21B E∴GF ,∴四边形B FGE 是平行四边形, .//EG BF ∴⊥∴EG 平面.CD B 1又⊂EG 平面D EB 1, ∴平面⊥D EB 1平面.CD B 1 ………………………………6分 （２）过C 作D B CH 1⊥于H ，连结EH ．由（１）中的平面D EB 1⊥平面CD B 1知⊥CH 面D EB 1，所以CE 在面DE B 1上的射影为HE ，所以CEH ∠就是所求的角． …………………………………………9分 令正方体的棱长为1，所以36,25==CH CE ，所以15302sin =∠CEH ． 即CE 与平面DE B 1所成角的大小的正弦值为15302． …………………………12分 18．（本小题满分12分）已知函数xe xf =)(，过该函数图象上任意一点b kx xg x f x +=)())(,(00的切线为（１）证明：)(x f y =图象上的点总在)(x g y =图象的上方； （２）若R x ax e x∈≥在上恒成立，求实数a 的取值范围． 解：（１）000000)(,)(x x xxe ex x e x g e x f +-==，设))()()(0000x x x x e e x x e e x f x f x h +--=-=)(,0)(,)(00x h x h x x e e x h x x >>-=∴时，当为增，当0)()()(,0)(000==<<x h x h x x x h x h x x 取最小值时，为减，当时，)()(,0)()()(,0)()(0x g x f x g x f x h x h x h ≥≥-==≥，所以)(x f y =图象上的点总在)(x g y =图象的上方． …………………………6分（２）当2')1()(,)(0xx e x F x e x F x x x -==≠时，令． x （－∞，0） （0，1）1 （1，+∞）F ‘（x ） － － 0 + F （x ） 减 减 e增①当x ＞0时，F （x ）在x=1时有最小值e ，e a a ax e a xe x x≤≥≥∴的范围是恒成立的即,．②当x ＜0时，F （x ）为减函数，0),0,()(,0)()(0<-∞∈∴→-∞→→xe x F x F x F x x，，，0,≥≥≤∴a a ax e a xe x x的范围是恒成立的即．③当x =0时，a ∈R ．由①②③，ax e x≥恒成立的a 的范围是],0[e ． ……………………………………12分 19．（本小题满分13分）如图，一船在海上由西向东航行，在A 处测得某岛M 的方位角为北偏东α角，前进4km 后在B 处测得该岛的方位角为北偏东β角，已知该岛周围5.3km 范围内有暗礁，现该船继续东行． （１）若0602==βα，问该船有无触礁危险？如果没有，请说明理由；如果有，那么该船自B 处向东航行多少距离会有触礁危险？ （２）当α与β满足什么条件时，该船没有触礁危险？解：（１）作AB MC ⊥，垂足为C ，由已知060=α，030=β，所以0120=∠ABM ，030=∠AMB 所以4==AB BM ，060=∠MBC ，所以5.33260sin 0<=⋅=BM MC ，所以该船有触礁的危险． 设该船自B 向东航行至点D 有触礁危险，则5.3=MD ， 在△MBC 中，4=BM ，2=BC ，32=MC ，5.0)32(5.322=-=CD ，所以，5.1=BD （km ）．所以，该船自B 向东航行5.1km 会有触礁危险． ……6分 （２）设x CM =，在△MAB 中，由正弦定理得，MABBMAMB AB ∠=∠sin sin ，β 北 MABCα β北 MAB Cα D即αβαcos )sin(4BM =-，)sin(cos 4βαα-=BM ， 而)sin(cos cos 4cos sin βαβαβ-=⋅=∠⋅=BM MBC BM x ，所以，当5.3>x ，即27)sin(cos cos 4>-βαβα，即87)sin(cos cos >-βαβα时，该船没有触礁危险． …………13分20．（本小题满分13分）在直角坐标平面中，ABC ∆的两个顶点B A ,的坐标分别为)0,1(-A ，)0,1(B ，平面内两点M G ,同时满足下列条件：①0=++GC GB GA MC MB MA ==GM ∥AB ． （１）求ABC ∆的顶点C 的轨迹方程；（２）过点)0,3(P 的直线l 与（１）中轨迹交于不同的两点F E ,，求OEF ∆面积的最大值．解：（１）设).,( , ),( , ),(00M M y x M y x G y x C MB MA = ，M ∴点在线段AB 的中垂线上．由已知(1,0) , (1,0) ,0M A B x -∴=． …………1分又GM ∥AB ，0y y M =∴．又0=++GC GB GA ，()()()()0,0,,1,1000000=--+--+---∴y y x x y x y x ，33 , 300y y y y x x M =∴==∴． ………………………………………………3分MC MB = ，()()2222300310⎪⎭⎫⎝⎛-+-=⎪⎭⎫⎝⎛-+-∴y y x y ， 1322=+∴y x ()0≠y ，∴顶点C 的轨迹方程为1322=+y x ()0≠y ． …5分（２）设直线l 方程为：)0)(3(≠-=k x k y ，),(11y x E ，),(22y x F ，由⎪⎩⎪⎨⎧=+-=13)3(22yx x k y 消去y 得：()039632222=-+-+k x k x k ①362221+=+∴k k x x ， 3392221+-=k k x x ． ………………………………………7分由方程①知()()()393462222-+-=∆k k k ＞0，2k ∴＜83，0≠k ，0∴＜2k ＜83． …………………………………………9分而2122121214)(2||3||||23||321x x x x k x x k y y S ABC -+=-⋅⨯=-⨯⨯=∆9624939636)3(2||3244222++-=-+=k k k k k k k ．………………………………………11分 令t k =2，则)83,0(∈t ，=∆ABC S 96249322++-t t t t ．记)830()3(249)(22<<+-=t t t t t f ， 求导易得当173=t 时有OEF ∆面积的最大值23． ……………………13分21．（本小题满分13分）已知数列}{n a 满足：))()(1(*12N n a a n a n n n ∈++=++，且1,021==a a ．求证：（１）数列})1({1n n a n a +-+为等比数列；（２）2)1(1!98nn n a -++⨯<．解：（１）由))(1(12n n n a a n a ++=++得1112)2())(1()2(+++++-++=+-n n n n n a n a a n a n a])1([)1(11n n n n a n a a n a +--=++-=++．而01212≠=-a a ，所以11)1()1(++-=+-n n n a n a ，所以数列})1({1n n a n a +-+为等比数列． …………………………………………4分（２）由（１）有)2(!)1()!1(!1≥-=---n n n a n a nn n ． ……………………………………6分 所以)!12(1)!22()!12(2212--=-----n n a n a n n ，)!22(1)!32()!22(3222-=-----n n a n a n n ，……，)!2(1)!1()!2(12=-a a ，累和得!21!31)!22(1)!12(1)!12(12+-++-+--=-- n n n a n )!12(22!54!32))!12(1)!22(1()!51!41()!31!21(--+++=---++-+-=n n n n ． …8分因为1223)1(!-≥⨯⨯⨯-⨯=n n n n ，………………………………………………9分所以)1)((21232221)!12(*325312也成立检验=∈-++++≤---n N n n n a n n ．记325321232221--++++=n n S ，用错位相减法得 1212134])41(1[98---⨯--=n n n S ，所以98<S ．所以2)1(1)!12(98)!12(981212---++-⨯=-⨯<n n n n a ． 即当n 为奇数时命题成立．……………………………………………………………11分 又)!2(198!21!31)!22(1)!12(1)!2(1)!2(2n n n n n a n +<+-++-+--+= ， 所以2)1(1)!2(981)!2(9822nnn n a -++⨯=+⨯<．即当n 为偶数时命题成立． 综合以上得2)1(1!98nn n a -++⨯<．………………………………………………13分。

炎德㊃英才大联考理科数学(一中版)-1 炎德·英才大联考长沙市一中2011届高三月考试卷(一)(一中版)理科数学参考答案一㊁选择题题 号12345678答 案A C D A D B C B 二㊁填空题9.-8 10.(-1,1) 11.x 2=2.764或x 2=3.236(只要写对其中一个给全分) 12.4 13.(-∞,5]14.102 15.1 三㊁解答题16.解:p :∵f (x )=2|x -a |在区间(4,+∞)上递增,∴u =|x -a |在(4,+∞)上递增,故a ≤4.(3分)………………………………………………………………q :由l o g a 2<1=l o g a a ⇒0<a <1或a >2.(6分)………………………………………………………………如果 ⌝p ”为真命题,则p 为假命题,即a >4.(8分)…………………………………………………………又因为p 或q 为真,则q 为真即0<a <1或a >2.由0<a <1或a >2a >{4可得实数a 的取值范围是a >4.(12分)…………………………………………………17.解:(1)f (x )=x 2-x -3,因为x 0为不动点,所以f (x 0)=x 20-x 0-3=x 0,解得x 0=-1或x 0=3,-1和3是函数的两个不动点.(4分)………………………………………………(2)因为函数f (x )恒有两个相异的不动点,所以方程f (x )=a x 2+(b +1)x +(b -1)=x ,也就是a x 2+b x +(b -1)=0对任何实数b 恒有两个不相等的实数根,即b 2-4a (b -1)>0对任意的b ∈R 恒成立.(8分)……………………………………………………………这个不等式可化为b 2-4a b +4a >0,所以(4a )2-16a <0,解得0<a <1.(12分)……………………………………………………………………18.解:(1)圆锥曲线x =2c o s θy =3s i n {θ化为普通方程是x 24+y 23=1,所以F 1(-1,0),F 2(1,0),(2分)…………………………………………………………………………………则直线A F 2的斜率k =0-31-0=-3,于是经过点F 1垂直于直线A F 2的直线l 的斜率k '=33,直线l 的倾斜角是30°,(5分)…………………………………………………………………………………………………所以直线l 的参数方程是x =-1+t c o s 30°y =0+t {s i n 30°(t 为参数),即x =32t -1y =12ìîíïïïït (t 为参数).(7分)…………………………………………………………………………………炎德㊃英才大联考理科数学(一中版)-2(2)解法一:直线A F 2的斜率k =0-31-0=-3,倾斜角是120°,(8分)………………………………………设P (ρ,θ)是直线A F 2上任意一点,则ρs i n 120°=1s i n (120°-θ),即ρs i n (120°-θ)=s i n 120°,即ρs i n θ+3ρc o s θ=3.(12分)………………………………………………………………………………………………………解法二:直线A F 2的直角坐标方程是y =-3(x -1),(9分)…………………………………………………将x =ρc o s θy =ρs i n {θ代入得直线A F 2的极坐标方程是ρs i n θ=-3ρc o s θ+3,即ρs i n θ+3ρc o s θ=3.(12分)…………………………………………………………………………………19.解:(1)∵四边形A B C D 是梯形,A D ∥B C ,∴△AMD ∽△C M B ,∴S △A M D S △C M B =(A D B C )2=14.∵种满△AMD 地带花费160元,∴S △A M D =1608=20(m 2),(4分)……………………………………………∴S △C M B =80m 2,∴种满△B M C 地带的花费为80×8=640(元).(6分)……………………………………(2)设△AMD ,△B M C 的高分别为h 1,h 2,梯形A B C D 的高为h .∵S △A M D =12×10h 1=20,∴h 1=4(m ).又∵h 1h 2=12,∴h 2=8(m ),h =h 1+h 2=12(m ),(9分)…………………………………………………………∴S 梯形A B C D =12(A D +B C )h =12×30×12=180(m 2),∴S △A M B +S △D M C =180-20-80=80(m 2).又∵160+640+80×10=1600(元),∴应选择种植茉莉花可刚好用完所筹集的资金.(13分)………………………………………………………20.解:(1)∵f (x )的定义域为{x |x >0}不关于原点对称,∴f (x )为非奇非偶函数,(2分)…………………………………………………………………………………而g t (x )的定义域为R ,且g t (-x )=(1+t )(-x )-e t ≠±g t (x ),∴g t (x )也为非奇非偶函数.(4分)………………………………………………………………………………(2)函数y =f (x )-g 2(x )=x l n x -3x +e 2的定义域为(0,+∞),y'=l n x -2.由y '>0得x >e2,由y '<0得0<x <e 2,故y =f (x )-g 2(x )的单调递增区间为(e 2,+∞);单调递减区间为(0,e 2).(8分)…………………………(3)解法一:令h (x )=f (x )-g t (x )=x l n x -(1+t )x +e t ,(10分)……………………………………………则h '(x )=l n x -t .由h '(x )=0,得x =e t ,当x >e t 时,h '(x )>0;当0<x <e t 时,h '(x )<0,∴h (x )在(0,e t )上单调递减,在(e t ,+∞)上单调递增,∴h (x )在(0,+∞)上有唯一极小值h (e t ),也是它的最小值,而h (x )在(0,+∞)上的最小值h (e t )=0,∴h (x )≥0,即f (x )≥g t (x ).(13分)……………………………………………………………………………解法二:对任意x >0,令h (t )=f (x )-g t (x )=x l n x -(1+t )x +e t ,则h '(t )=-x +e t .由h '(t )=0,得t =l n x ,当t >l n x 时,e t >e l n x =x ,e t -x >0,∴h '(t )>0;当t <l n x 时,h '(t )<0,∴l n x 为h (t )的唯一极小值点,h (t )≥h (l n x )=x l n x -x -x l n x +e l n x =0,炎德㊃英才大联考理科数学(一中版)-3 ∴x l n x ≥(1+t )x -e t ,即f (x )≥g t (x ).(13分)………………………………………………………………21.解:(1)∵f (x )=l n x +92(x +1)(x >0),∴f '(x )=1x -92(x +1)2=2x 2-5x +22x (x +1)2,(2分)…………………………………………………………………故当12<x <2时,f '(x )<0,即f (x )单调递减,从而x ∈[1,2)时,f (x )单调递减.当0<x ≤12或x ≥2时,f '(x )≥0,即f (x )单调递增,从而x ∈[2,e ]时,f (x )单调递增.(4分)……………故f m i n (x )=f (2)=l n 2+32.又f (1)=94>f (e )=1+92(e +1),故f m a x (x )=f (1)=94.(6分)……………………………………………………………………………………(2)由g (x 2)-g (x 1)x 2-x 1<-1可知g (x 2)+x 2-[g (x 1)+x 1]x 2-x 1<0,所以可设ω(x )=g (x )+x =|l n x |+x +a x +1(a >0,x ∈(0,2]),(8分)………………………………………故由题设可知ω(x )在x ∈(0,2]上为减函数.∵ω'(x )=1x +1-a (x +1)2,1≤x ≤2-1x +1-a (x +1)2,0<x <ìîíïïïï1,(10分)………………………………………………………………而由1x +1-a (x +1)2<0(1≤x ≤2)可得a >x 2+3x +3+1x (1≤x ≤2),而y =x 2+3x +3+1x 在x ∈[1,2]上是增函数,∴a >272.显然当a >272且0<x <1时,-1x +1-a (x +1)2<0,所以a 的取值范围是(272,+∞).(13分)………………………………………………………………………。