湖南省衡阳市八中2013届高三第六次月考数学文(附答案)
湖南省衡阳市八中2015届高三上学期第六次月考英语试题及答案
衡阳市八中2015届高三上学期第六次月考试题英语时量：120分钟满分：150分PART ONE LISTENING COMPREHENSIONSECTION A （22.5 marks）Directions: In this section, you will 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. What did the woman do first after work?A．She did some shopping. B．She cooked supper. C．She visited her aunt.2. Why didn‟t the woman think the lecture was useful for her?A．He children have grown up. B．She isn‟t married. C. She has no parents Conversation 23. When will the woman take the written test?A．In five days. B. In two weeks. C. In two months.4. What does the man think of the book?A. Difficult.B. Useful.C. Interesting.Conversation 35. Why is the man there?A．He is giving lessons to interviewees. B．He is giving tips to the interviewees.C．He is having the interview class.6. What does the man think matters most in an interview?A. Appearance.B. Language.C. Honesty.Conversation 47. Why does the man feel lucky?A. He got a well-paid job.B. He can live in a quiet place.C. He can work in a factory.8. How did the man sleep last night?A. He slept soundly.B. He was awake all night.C. He was too excited to sleep well.9. What do we know about the woman?A. She doesn‟t like birds.B. She got up at two this morning.C. She hasn‟t adapted to the silence.Conversation 510. When did the man get two offers below the asking price?A. Today.B. Yesterday.C. The day before yesterday.11. How should the buyers pay, according to the man?A. By credit card.B. By check.C. In cash.12. What is the woman‟s asking price?A. 28,000 dollars.B. 23,000 dollars.C. 18,000 dollars.Conversation 613. When did the man get married?A. Five years ago.B. Ten years ago.C. Fifteen years ago.14. Where is the man living now?A. In Sydney.B. In San Jose.C. In New York.15. Why are the man‟s parents moving in the fall?A. They want to be closer to their family.B. They miss the East Coast very much.C. They are not happy in their present house.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.PART TWO LANGUAGE KNOWLEDGESECTION 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.---What do you think of the furniture on exhibition?--- Well, great! But I don‟t think much of ____ you bought.A. the oneB. itC. thatD. which22. The young man, who by then ___ admission to Peking University, determined to do some part-time jobs to gain more practical skills.A. gainedB. was gainingC. has gainedD. had gained23. ---We need a person very much to think up a creative idea.--- ______ the new manager have a try?A. ShallB. MustC. ShouldD. Need24. The horse covered the distance in six days_____the train's 28 hours.A．comparing with B．being compared with C．to compared with D．compared with25. —Why does the river smell terrible—Because large quantities of water________．A．have polluted B．is being pollutedC．has been polluted D．have been polluted26. The human body consists of organs, each ____a definite job to do.A. haveB. to haveC. hasD. having27. ___, as long as I‟m with nature, I don‟t care.A. However tough the journey isB. Whatever tough the journey isC. However the journey is toughD. Whatever the journey is tough28. News got around quickly_____ a gunman went on a shooting rampage at a movie theatre in Aurora, Colorado.A. which B．what C．that D．when29. It's reported that women with demanding jobs are almost ________ to suffer a heart attack.A. as likely twiceB. likely twice asC. twice as likelyD. twice likely as30. ---I telephoned him twice and I couldn‟t get through to him.--- The line might have been out of order, ____?A．d on‟t you B．was it C．do you D．h adn‟t it31. —Little Daniel is watering the flowers! Why?—Well, I'm not feeling very well today—otherwise, I ________ it myself.A．did B．would do C．had done D．would have done32. As a mother of two young boys, she knows the difficulties of keeping kids ______.A. to entertainB. entertainingC. entertainedD. entertain33. It's much easier to make friends _____you have similar interests.A．unless B．when C．even though D．so that34. ____ is announced in today‟s newspaper, we have launched another man-made satellite.A.ThatB.WhichC. AsD. What35. Absent for the traffic accident, I have always doubted he announced at the meeting.A. what was it thatB. that it was whatC. what it was thatD. that was it whatSECTION 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 the word or phrase that best fits the context.Recalling my freshman year and first year of drama group, I remember the people I met, and how I soon felt like they all were family. One I greatly36was Christy Hauser, mainly because of her unbelievable ambition to help people.At Christmas I participated in a present-wrapping party for children who would otherwise not have gifts. Christy37this and was so persuasive in motivating other students. She was only 15, but found time to help others, earn outstanding38and manage the make-up department for our drama group.The next year I was shocked to hear through Christy‟s efforts, the drama department was going to try to 39$90, 000 for a former schoolmate named Robin who was suffering from cystic fibrosis(囊肿性纤维化). The most40part of her decision to help Robin was that Christy had41met her! At our first drama meeting Christy gave everyone a piece of paper and asked us to write to talk-show host Rosie to try to gain her42; Christy received over 100 letters, but that wasn’t43. She told us we needed to keepwriting. Christy put all the letters in envelopes, stamped them, and made sure they were sent.Then one cold evening late in October, I gave one of my closest drama friends, Emily, a44to see if she wanted to do something. I was surprised when she answered the phone crying. She45that Christy Hauser had died in a car accident that evening. I had left rehearsal(排练)too early to hear about the46. With the phone still to my ear, I stood still as I remembered all the miraculous47she had organized. The next morning I looked into those eyes around me and was comforted by the fact that despite losing aSECTION C （12 marks）Directions: Complete the following passage by filling in each blank with one word that best fits the context. China has ordered a ban on advertisements during television drama programs. An ad will not be allowed to run in the middle of a program running 45 minutes in length, beginning on January 1st. The authorities say 48.________ ban is important for the growth of culture. Television stations say this will cause 49.________ a loss in revenue(收入). Most audience feel glad 50.________ finally such a policy has been carried out 51.________ some still doubt whether this is an effective policy or not. Simply banning TV stations from broadcasting inserted advertisements is no use.Even if SARFT completely banned domestic TV stations from broadcasting advertisements, they could still implant(植入）advertisements 52.________ TV shows. Why can China not learn some lessons from western TV channels such as NHK and BBC? They are public channels 53.________ any advertisements. But they charge their viewers. In China, TV viewers only need to pay cable television fees, 54.________ are rather low. And TV stations make profits through broadcasting advertisements. If we can learn some lessons from western countries, 55.________ will be rather helpful to the final solution of the conflict between audience and TV stations.PA RT THREE: READING COMPREHENSION (30 points)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.AOne day while shopping in a small town in southern California, it was my misfortune to be approached by a clerk whose personality conflicted with mine. He seemed most unfriendly and not at all concerned about my intended purchase. I bought nothing, and marched angrily out of the store. My hostility toward that clerk increased with each step.On the outside, standing by the road, was a dark-skinned young man in his early twenties. His expressive brown eyes met and held mine, and in the next instant a beautiful, dazzling smile covered his face. I gave way immediately. The magnetic power of that smile dissolved all bitterness within me, and I found the muscles in my own face happily responding.“Beautiful day, isn‟t it?” I remarked in passing. Then, I turned back. “I really owe you a debt of gratitude,” I said softly. His smile deepene d, but he made no attempt to answer. A Mexican woman and two men were standing nearby. The woman stepped forward and eyed me inquiringly. “Sir, but he doesn‟t speak English,” she volunteered. “You want I should tell him something?”At that moment I felt t ransformed. The young man‟s smile had made a big person of me.“Yes,” my reply was enthusiastic and sincere, “tell him I said …Thank you!‟ ”“Thank you?” The woman seeded slightly puzzled. I gave her arm a friendly pat as I turned to leave. “Just tell him that,” I insisted. “He will understand; I am sure!”Oh, what a smile can do! Although I have never seen that young man, I shall never forget the lesson he taught me that morning. From that day on, I became smile-conscious, and I practice the art diligently, anywhere and everywhere, with everybody.56. The author left the store angrily because____.A. his intended purchases were of poor quality.B. there‟s nothing he wanted in the shopC. the clerk didn‟t speak EnglishD. the clerk didn‟t treat him kindly57. The underlined word “hostility” in the first paragraph probably means ____.A. attitudeB. hatredC. ideaD. opinion58. What does the author mean by saying “I found the muscles in my own face happily responding” in Paragraph 2 ?A. He did not want to smile.B. He would thank the young man.C. He smiled back at the young man.D. He was happy to meet the young man.59. The passage tell us that we should____.A. help people in troubleB. smile at othersC. practice smiling every dayD. be generous to strangers60. The author asked the woman to say “Thank you!” to the young man because the young man___.A. taught the author a valuable lessonB. taught the author how to smileC. had offered help to the authorD. was a friendly employee of the shopBIndividuality is the particular character, or aggregate (total) of qualities that distinguishes one person or thing from others. Many artists late in the last century were in search of a means to express their individuality. Modern dance was one of the ways some of these people sought to free their creative spirit. At the beginning there was no exacting technique, no foundation from which to build. In later years, trial, error and genius founded the techniques and the principles of the movement. Eventually, innovators (改革者) even drew from what they considered the dread ballet, but first they had to get rid of all that was academicso that the new could be discovered. The beginnings of modern dance were happening before Isadora Duncan, but she was the first person to bring the new dance to general audiences and see it accepted and acclaimed (称赞).Her search for a natural movement form sent her to nature. She believed movement should be as natural as the swaying of the trees and the rolling waves of the sea, and should be in harmony with the movements of the Earth. Her great contributions are in three areas.First, she began the expansion of the kinds of movement that could be used in dance. Before Duncan danced, ballet was the only type of dance performed in concert. In the ballet the feet and legs were emphasized, with virtuosity (高超技巧) shown by complex, codified positions and movements. Duncan performed dance by using her body in the freest possible way. Her dance stemmed from her soul and spirit. She was one of the pioneers who broke tradition so others might be able to develop the art.Her second contribution lies in dance costume. She rejected ballet shoes and stiff costumes. These were replaced with flowing Grecian (希腊式的) tunes, bare feet, and unbound hair. She believed in the natural body being allowed to move freely, and her dress displayed this ideal.Her third contribution was in the use of music. In her performances she used the symphonies of great masters including Beethoven and Wagner, which was not the usual custom.She was as exciting and eccentric (怪异) in her personal life as in her dance.61．According to the passage, what did nature represent to Isadora Duncan?A．Something to conquer. B．A model for movement.C．A place to find peace. D．A symbol of disorder.62．Compared to those of the balle t, Isadora Duncan‟s costumes were less _________.A．costly B．colorful C．graceful D．restrictive63．Which of the following is not mentioned as an area of dance that Duncan worked to change?A. The stage set.B. The music.C. Costumes.D. Movements.64. We can infer from the passage that the author _________.A. appreciates modern dance very muchB. dislikes Isadora Duncan‟s danceC. thinks highly of individualityD. knows a lot about modern arts65. Which of the following would be the best title for the passage?A．Artists of the Last Century B．Evolution of Dance in the 20th Century C．Natural Movement in Dance D．A Pioneer in Modern DanceCThree Central Texas men were honored with the Texas Department of Public Safety's Director's Award in a Tuesday morning ceremony for their heroism in saving the victims of a fiery two car accident.The accident occurred on March 25 when a vehicle lost control while traveling on a rain-soaked State Highway 6 near Baylor Camp Road. It ran into an oncoming vehicle, leaving the occupants trapped inside as both vehicles burst into flames.Bonge was the first on the scene and heard children screaming. He broke through a back window and pulled Mallory Smith, 10, and her sister, Megan Smith,9,from the wreckage.The girls' mother, Beckie Smith, was not with them at the time of the wreck, as they were traveling with their baby sitter, Lisa Bow bin.Beckie Smith still remembers the sickening feeling she had up on receiving the call informing her of thewreck and the despair as she drove to the scene.Bozeman and Clemmons arrived shortly after Bonge and helped rescue the other victims and attempted to put out the fires."I was nervous," Bozeman said." I don't feel like I'm a hero. I was just doing what anyone should do in that situation. I hope someone would do the same for me."Everyone at the accident made it out alive, with the victims suffering from nonlife-threatening injuries. Mallory Smith broke both femurs, and Megan had neck and back injuries. Bowbin is still recovering from a broken pelvis, ankle and foot.The rescuers also were taken to the hospital and treated for cuts and smoke breathing, Bonge said.In addition, Bozeman not to meet accident victim Anthony Rus so in the hospital after the accident, where Russo presented him with a glass frame inscribed with" Thank you," Bozeman said. Those involved in that fateful encounter on Highway 6 credited God blessing for bringing them together."Whatever the circumstances, Tuesday's ceremony provided a time to be grateful for those who put their lives on the line for the lives of complete strangers," Beckie Smith said," We're calling it The Miracle on Highway 6."66. Which of the following can be used to describe Bozeman?A. Kind.B. Modest.C. Excited.D. Smart.67. Who saved Megan Smith from the damaged car?A. Clemmons.B. Anthony Russo.C. Bozeman.D. Bonge.68. What's the main idea of the passage?A. Three persons were awarded for rescuing victims in a car accident.B. Three ordinary people were regarded as great heroes.C. Several victims were carried to safety from the burning cars.D. A car accident occurred on a rain-soaked State Highway 6.69. It can be inferred from what Beckie Smith said that ____.A. she regarded the accident as a wonderB. she was frightened by the serious accidentC. she thought highly of the rescuersD. she called on others to learn from the rescuers70. Which of the following is true according to the passage?A. Luckily, no one received too serious injuries in the accident.B. All the victims received slight injuries in the accident.C. The rescuers were taken to the hospital to visit the victims.D. The injured will soon recover from their injuries.PART FOUR: WRITING(45 points)Section A （10 points）Directions: Read the following passage. Complete the diagram / Fill in the numbered blanks by using the information for the passage.Write NO MORE THAN 3 WORDS for each answerWhat is critical thinking? Critical thinking is defined as thinking logically and developing an ability to ask questions or ask for evidence for a particular subject. Developing critical thinking skills in a child throughthe right critical thinking exercises is very important.Critical thinking develops the ability of subjective analysis of a particular fact. Critical thinking is to evaluate the reason behind a particular fact. All possible viewpoints must be thought about before analyzing a fact, and this develops problem-solving skills in a child.Here are some exercises you can use to encourage your child to think critically.Quiz your child. Ask your child about daily activities that he does in the school. Quiz him about certain things that do not have a single correct answer. This will increase his ability to think about the things he‟s just learned in class. Keep on telling your child simple information about nature, like the seasons when flowers bloom. You can either ask him questions or just explain to him in a casual way.Recognize and classify. For critical thinking, your child must learn to recognize important information that is closely connected with the subject. Classification of things on a certain firm and logical basis of information is included in the critical thinking ability. Colorful toys or images can be included in teaching critical thinking. Ask your child to identify the names of the flowers, fruits and animals from a colorful chart.Critical thinking exercises are not intended to make your child intelligent, but it is about making him successful in his decision-making ability and helping him make a successful career. Although thinking habits are cultivated(培养) in early childhood through exercises, they are indeed helpful for a lifetime.Section B （10 points）Directions: Read the following passage. Answer the questions according to the information given in the passage and required words limit. Write your answers on your answer sheet.In the animal kingdom, weakness can bring about aggression in other animal. This sometimes happens with humans also. But I have found that my weakness brings out the kindness in people. I see it every day when people hold doors for me, pour cream into my coffee, or help me to put on my coat. And I have discoveredthat it makes them happy.From my wheelchair experience, I see the best in people, bur sometimes I feel sad because those who appear independent miss the kindness I s ee daily. They don‟t get to see this soft side of others often, we try every way possible to avoid showing our weakness, which includes a lot of pretending. But only when we stop pretending we‟re brave or strong do we allow people to show the kindness that‟s in them.Last month, when I was driving home on a busy highway, I began to feel unwell and drove more slowly than usual. People behind me began to get impatient and angry, with some speeding up alongside me, horning (按喇叭) or even shouting at me. At the moment I decided to do something I had never done in twenty fore years of driving. I put on the car flashlights and drove on at a really low speed.No more angry shouts and no more horns! When I put on my flashlights, I was saying to other drivers, “I have a problem here. I am weak and doing the best I can.” And everyone understood. Several times, I saw drivers who wanted to pass. They couldn‟t get around me because of the stream of passing traffic. But instead of getting impatient and angry, they waited, knowing the driver in front of them was in some way weak.Sometimes situations call for us to act strong and brave even when we don‟t feel that way. But those are and far between. More often, it would be better if we don‟t pretend we feel strong when we fee l weak or pretend that we are brave when we are scared.81. How do people feel when they offer their help?(No more than 2 words) (2 points)____________________________________________________________________82.What reaction did other drivers have when the author drove very slowly at first?(No more than 5 words) (2 points)____________________________________________________________________83.Why did other drivers behave differently when the author put on the car flashlights?(No more than 8 words) (3 points)____________________________________________________________________84. What does the author advise us to do at last? (No more than 8 words) (3 points)____________________________________________________________________Section C（25 points）Directions: Write an English composition according to the instructions given below in Chinese.近年来媒体不断报道一些商家为了获取商业利益生产制造一些对人体有害的食品。
湖南省衡阳八中2014-2015学年高二10月月考(六科联赛)数学(文)试题 Word版含答案
高二10月月考（六科联赛）数学（文）试题时量 120分钟 满分 100分一、选择题 (本大题共10小题，每小题3分，共30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．命题“若p 则q ”的逆命题是 A ．若q 则p B ．若¬p 则¬q C ．若¬q 则¬p D ．若p 则¬q 2．若p 是真命题，q 是假命题，则A ．p ∧q 是真命题B ．p ∨q 是假命题C ．¬p 是真命题D ．¬q 是真命题 3.“1a =”是“(1)(2)0a a --=”成立的A . 充分非必要条件 .B 必要非充分条件.C 充要条件 .D 既不充分也不必要条件4．一个命题与它的逆命题、否命题、逆否命题这四个命题中A ．真命题与假命题的个数相同B ．真命题的个数一定是奇数C ．真命题的个数一定是偶数D ．真命题的个数可能是奇数，也可能是偶数 5．命题“∈∃x R,0123=+-x x ”的否定是A ．∈∃x R,0123≠+-x xB ．不存在∈x R, 0123≠+-x xC ．∈∀x R,0123=+-x xD ．∈∀x R, 0123≠+-x x6. 椭圆的中心在坐标原点，对称轴是坐标轴，且经过点Ａ（－３，０），Ｂ（０，22）， 则椭圆的标准方程是A.18922=+y xB.19822=+y xC.122322=+y xD. 122322=+x y7.设双曲线)0,0(12222>>=-b a b y a x 的虚轴长为6，焦距为10，则双曲线的实轴长为A. 8B. 6C. 4D. 2 8.下列曲线中离心率为62的是A.22124x y -=B.22142x y -=C.22146x y -=D.221410x y -= 9.已知点P 是椭圆5922y x +=1上的动点，过P 作椭圆长轴的垂线，垂足为M ，则PM 中点的轨迹方程为 A 、159422=+y x B 、154922=+y x C 、120922=+y x D 、53622y x +=110.已知椭圆2222:1(0)x y E a b a b+=>>的右焦点为(3,0)F ,过点F 的直线交椭圆于,A B 两点.若AB 的中点坐标为(1,1)-,则E 的方程为 A ．2214536x y += B ．2213627x y += C ．2212718x y += D ．221189x y += 二.填空题(每小题3分,共15分)11.命题“若a ≥b ，则a 3≥b 3”的否命题是 ．12.椭圆22192x y +=的焦点为12,F F ，点P 在椭圆上，若1||4PF =，则2||PF = . 13.若双曲线2x 4-22y b=1(b>0)的渐近线方程式为y=1x 2±，则ｂ等于 ;14. 命题“∈∀x R,x 2+2x +m >0”是真命题,则实数m 的取值范围 ．15.已知椭圆22221(0)x y a b a b+=>>的左、右焦点分别为12(,0),(,0)F c F c -，若椭圆上存在一点P 使1221sin sin a cPF F PF F =，则该椭圆的离心率的取值范围为 .三、解答题(本大题共6小题，共55分.解答应写出文字说明、证明过程或演算步骤。
湖南省衡阳八中2013-2014学年高一下学期期末试题 数学 Word版含答案
衡阳市八中2014年下期高一期末考试试题数 学 命题人: 彭学军（本卷共21道小题,考试时间120分钟,满分100分）注意事项：答题前,考生务必将自己的班级、姓名、考试号写在答题卡的密封线内．答题时,答案写在答题卡上对应题目的空格内,答案写在试卷上无效．．．．．．．．．．考试结束后,上交答题卡．一、选择题（每小题3分，共10小题，满分30分）1.cos9cos36sin 36sin 9︒︒-︒︒的值为（ B ）A ．12BCD ．12.若R,a b c a b ∈>、、，则下列不等式成立的是( D ）A ．11a b< B ．22a b > C ．2b aa b +≥ D ．22(1)(1)a c b c +>+3.在ABC ∆中，内角,,A B C 的对边分别为,,a b c ，若135A =︒,30B =︒,2=a ,则b 等于( A )A. 1B.2C. 3D. 24.已知数列{}n a 满足11a =，*12,3n n a a n N +=∈，其前n 项和为n S ，则（ D ）. A.21n n S a =- B.32n n S a =- C.43n n S a =- D.32n n S a =-5.在ABC ∆中，若||||BA BC AC +=，则ABC ∆一定是（ C ） A ．钝角三角形 B ．锐角三角形 C ．直角三角形 D ．不能确定6. 同时具有性质“①最小正周期是π；②图象关于直线3π=x 对称；③在]3,6[ππ-上是增函数”的一个函数是（ C ） A ．)62sin(π+=x y B ．)32cos(π+=x y C ．)62sin(π-=x y D ．)62cos(π-=x y7.在四边形ABCD 中，(2,4)AC =u u u r ，(6,3)BD =-uu u r,则该四边形的面积为 （ D ）.A. B.52 C.5 D.158.已知点(,)x y 在如图所示的平面区域（阴影部分）内运动， 则22z x y =+的最大值是（ D ）A ．1B ．3C ．5D ．139.已知等差数列{}n a 的前n 项和n S 满足65S S <且876S S S >=，则下列结论错误．．的是（ D ）A ．6S 和7S 均为n S 的最大值B ．07=aC ．公差0d <D ．59S S >10. 已知,OA OB 是两个单位向量，且OA OB ⋅=0．若点C 在AOB ∠内，且30AOC ∠=︒， 则,(,)OC mOA nOB m n R =+∈, 则mn等于（ C ） A ．13 BCD ．3二、填空题（每小题4分,共5小题，满分20分）11．已知集合{|320}M x R x =∈+>,{|(1)(3)0}N x R x x =∈+-≤，则M N =2(,3]3-12．已知等比数列}{n a 的公比为正数，且1a =2，23954a a a ⋅=，则2a = 1 .13. 在ABC ∆中，内角,,A B C 的对边分别为,,a b c ，若35a b =，且si n A 是sin B 与sin C 的等差中项，则角C =___120︒______. ．已知正实数15. 已知数列{}n a 通项为cos(),*,2n n a n n N π=∈，则123201a a a a +++⋅⋅⋅+= -1008 .三、解答题（共6小题，满分50分）16. (本题满分6分) 已知关于x 的不等式2320ax x -+≤的解集为{|1}x x b ≤≤. （1）求实数,a b 的值；（2）解关于x 的不等式：0x cax b->-（c 为常数）. 解：（1）由题知b ,1为关于x 的方程0232=+-x ax 的两根，即⎪⎪⎩⎪⎪⎨⎧=+=a b ab 312 ∴2,1==b a . ………………3分（2）不等式等价于0)2)((>--x c x ， 所以：当2>c 时解集为{}2|<>x c x x 或；当2=c 时解集为{}R x x x ∈≠,2|；当2<c 时解集为{}c x x x <>或2|. ……………6分17.(本题满分8分) 某同学在一次研究性学习中发现，以下五个式子的值都等于同一个常数. 22sin 45cos 75+sin 45cos75,+ 22sin 36cos 66+sin 36cos66,+ 22sin 15cos 45+sin15cos45,+ 22sin (15)cos 15+sin(15)cos15,-+- 22sin (45)cos (15)+sin(45)cos(15),-+--- （1）试从上述五个式子中选择一个，求出这个常数；（2）根据（1）的计算结果，将该同学的发现推广为三角恒等式，并证明你的结论.解：（1）22113sin 15cos 15+sin(15)cos151sin 301244︒︒︒︒-︒+-=-=-=（）. …………3分（2）22sin cos ()sin cos()66ππαααα++++=43. (5)分 111cos21[cos(2)cos2]22342211111[cos22cos2]2cos22244131.44πααααααααα-=++-+-⋅=+--+-+=-=左边……………8分18.(本题满分8分) 等比数列{}n a 的前n 项和为n S ，公比0q >，已知3614,126S S ==. （1）求数列{}n a 的通项公式；（2）若35,a a 分别为等差数列{}n b 的第4项和第16项，试求数列{}n b 的通项公式及前n 项和n T .解：（1）易知1q ≠,由已知得3161(1)61(1)541a q q a q q⎧-=⎪-⎪⎨-⎪=⎪-⎩，解得12a q ==.所以2n n a =. …4分（2）由（1）得38a =，532a =，则48b =，1632b =,设{}n b 的公差为d ，则有1138,1532,b d b d +=⎧⎨+=⎩解得12,2.b d =⎧⎨=⎩ ……………………6分1(1)2(1)22.n b b n d n n ∴=+-=+-⨯=且数列{}n b 的前n 项和1(1)2n n n T na d -=+2(1)22.2n n n n n -=+⨯=+ ………8分19.(本题满分8分) 已知()3s i n ,c o s ,2c o s,c o s a x x b x x ⎛⎫==-⎪⎝⎭,函数1(),.2f x a b x R =⋅-∈ （1）求函数()f x的最小值和最小正周期；（2）设ABC ∆的内角,,A B C 的对边分别为,,a b c ，且c =，()0f C =，若sin 2sin B A =，求ABC ∆的面积．解：（1）21()2cos sin(2)126f x x x x π=--=--，()f x 的最小值为2-，最小正周期为.π……………3分 （2）()sin(2)106f C C π=--=，则sin(2)16C π-=．∵0C π<<，∴112666C πππ-<-<，因此26C π-＝2π，∴3C π=．……………5分∵sin 2sin B A =及正弦定理，得2b a =．①由余弦定理，得2222cos3c a b ab π=+-，且c =∴223a b ab +-=. ②由①②联立，得1a =,2b =． ……………7分1sin 22ABC S ab C ∆∴== ……………8分20.(本题满分10分)如图，山顶有一座石塔BC ，已知石塔的高度为a . （1）若以,B C 为观测点，在塔顶B 处测得地面上一点A 的俯角为α，在塔底C 处测得A 处的俯角为β，用,,a αβ表示山的高度h ；（2）若将观测点选在地面的直线AD 上，其中D 是塔顶B 在地面上的射影. 已知石塔高度20a =,当观测点E 在AD 上满足DE =BC 的视角(即BEC ∠)最大，求山的高度h .21.(本题满分10分) 设函数2()(31)n f x x n x =--（其中*n N ∈），区间{|()0}n n I x f x =>.（Ⅰ）定义区间(,)αβ的长度为βα-，求区间n I 的长度； （Ⅱ）把区间n I 的长度记作数列{}n a ，令1=n n n b a a +⋅， （1）求数列{}n b 的前n 项和n T ；（2）是否存在正整数m ，n （1m n <<）,使得1T ，m T ，n T 成等比数列？若存在，求出所有的m ，n 的值；若不存在，请说明理由.解：（Ⅰ）由()0n f x >，得2(31)0x n x -->，解得1031x n <<-， 即1(0,)31n I n =-，所以区间n I 的长度为1103131n n -=--； …………3分 （Ⅱ）由（Ⅰ）知 131n a n =-.（1）∵111111()(31)[3(1)1](31)(32)33132n n n b a a n n n n n n +====--+--+-+ ∴12n n T b b b =+++ 111111111()()()32535833132n n =-+-++--+111()3232n =-+2(32)n n =+ …………6分 （2）由（1）知，1110T =，2(32)m m T m =+，2(32)n n T n =+假设存在正整数m 、n (1)m n <<，使得1T 、m T 、n T 成等比数列，则 21m n T T T =⋅，即 21[]2(32)102(32)m n m n =⨯++, 经化简得22(32)5(32)m n m n =++. ∴222(32)1510m n m n m +=+ ∴22(362)5m m n m -++= （*） 当2m =时，（*）式可化为 220n =，所以10n =. 当3m ≥时，223623(1)570m m m -++=--+≤-<.又∵250m >，∴（*）式可化为 2250362m n m m =<-++，所以此时n 无正整数解. 综上可知，存在满足条件的正整数m 、n ，此时2m =，10n =. …………10分。
2014-2015湖南省衡阳市八中高三上学期第六次月考试题 生物
衡阳市八中2015届高三上学期第六次月考试题理科综合考生注意：1.时量：150分钟，总分：300分，所有答案必须写到答题卡上并按答题卡要求作答，否则不计分；2.可能用到的相对原子质量：H 1 ；O 16 ；N 14 ；S 32；Mg 24；Al 27；Fe 56；Cu 64；Ba 137第I 卷（共126分）一，选择题：本题共13小题，每小题6分，在给出的四个选项中，只有一个符合要求的。
1.埃博拉病毒传染性强，2014年非洲埃博拉病毒出血热的感染及死亡人数都达到历史最高。
下列有关埃博拉病毒的叙述中，正确的有几项？①遗传物质是DNA和RNA②结构简单，只有核糖体一种细胞器③在实验室中可用牛肉膏蛋白胨培养基培养病毒用于科学研究A．0项B．1项C．2项D．3项2.甲、乙、丙、丁分别为绿色植物新陈代谢过程中有关变化的示意图，下列叙述错误是A．图甲是胡萝卜在不同的含氧情况下从硝酸钾溶液中吸收K＋和NO3-的曲线。
影响A、B两点吸收量不同的因素是载体数量的多少B．图乙表示野外松树光合作用强度与光照强度的关系，当光照强度为b时，光合作用强度达到最大C. 图丙表示大气中氧的浓度对植物组织内CO2释放的影响，为了有利于贮藏蔬菜和水果，贮藏室内的氧气通常调节到图中B点所对应的浓度D．图丁表示豌豆种子萌发时吸水量随时间的变化关系，科学家研究发现，在第Ⅱ阶段种子中O2的吸收量大大低于CO2的释放量，故第Ⅱ阶段细胞呼吸强度最弱3.如图表示DNA复制的过程，结合图示判断，下列有关叙述不正确的是A．DNA复制过程中首先需要解旋酶破坏DNA双链之间的氢键，解开双链B．DNA分子的复制具有双向复制的特点，生成的两条子链的方向相反C．从图示可知，DNA分子具有多起点复制的特点，缩短了复制所需的时间D．DNA分子的复制需要DNA聚合酶将单个脱氧核苷酸连接成为DNA片段4.γ- 氨基丁酸和某种局部麻醉药在神经兴奋传递过程中的作用机理如下图所示。
【解析】湖南省衡阳市八中2014届高三第六次月考试题 历史
【解析】湖南省衡阳市八中2014届高三第六次月考试题历史一．单项选择题：每题2分，共计60分1．春秋战国时期，我国社会生产力水平有很大提高，社会经济得到较快的发展，其中推动社会经济发展的主要因素有①铁农具的使用②牛耕的使用③水利灌溉事业的发展④封建制度的确立A．①②③④ B．①②④ C．①②③ D．②③④【答案】A考点：本题考查学生分析归纳能力。
春秋战国时期出现都江堰、郑国渠等水利工程铁农具和牛耕推广使用大大推动农业生产的发展封建土地私有制确立、地租剥削形式出现有利于解放农业生产力推动农业发展故①②③④均为推动因素。
2. 《吕氏春秋·上农》载：“是故丈夫不织而衣，妇人不耕而食，男女贸功以长生，此圣人之制也。
故敬时爱日，非老不休，非疾不息，非死不舍。
”小农经济的这种形态，主要取决于A．个体分工 B．社会制度 C．生产力水平 D．劳动力短缺【答案】C考点：小农经济的特点。
“丈夫不织而衣，妇人不耕而食”反映出小农经济下男耕女织，自给自足的状态，“非老不休，非疾不息，非死不舍”反映出生产落后，由此选择C。
3．“新皇帝……废除了所有的封建国家和王国，将广阔的国土划分为若干行政区，每一行政区都配备一批由中央政府任命、并向中央政府负责的官员”(摘自《全球通史》)，这些官员可以①制定和颁布法律②征收辖区内的赋税③管理辖区行政事务④世袭相应的特殊地位A．②③ B．②③④ C．①②③ D．①②③④【答案】A考点：秦朝政治体制。
据材料可以分析出，分封制被废除，封建国家和王国瓦解，国土划分为若干行政区，由中央任命的官员进行管理，官员向中央政府负责。
第④项官员可以世袭官位说法错误。
4.柳宗元《封建论》指出：秦始皇建立帝国，以郡县取代封建，固然出自“一己之私”，却成就了“天下之公”。
黄宗羲《明夷待访录》批评皇帝是“以我之大私为天下之公”，以满足君主“一己之私”。
以下对柳、黄二人的观点，理解正确的是①柳宗元的目的是肯定帝制的合理性②黄宗羲对帝制的批判符合当时的社会现实③二人所处时代不同，批判的内涵不同，都有其合理性④两种观点恰好相反，所以其中应该有一个是错误的A.①②③④B.①②③C.②③④D.②③【答案】D考点：郡县制。
2013年全国大纲高考数学文科试卷带详解
2013年普通高等学校招生全国统一考试数学（文科）一、选择题：本大题共12小题，每小题5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．设集合{}{}1,2,3,4,5,1,2,U U A A ===集合则ð ( )A.{}1,2B.{}3,4,5C.{}1,2,3,4,5D.∅ 【测量目标】集合的补集.【考查方式】直接给出集合，用列举法求集合补集. 【参考答案】B【试题解析】依据补集的定义计算. {}1,2,3,4,5U =，{}1,2A =，∴ U A =ð{3,4,5}. 2．已知α是第二象限角，5sin ,cos 13αα==则 （ ） A.1213- B.513- C.513 D.1213【测量目标】同角三角函数基本关系.【考查方式】直接给出角的象限和正弦值，求余弦值. 【参考答案】A【试题解析】利用同角三角函数基本关系式中的平方关系计算.因为α为第二象限角，所以12cos .13α==-3．已知向量()()()()1,1,2,2,,=λλλ=+=++⊥-若则m n m n m n （ ）A.-4B.-3C.-2D.1- 【测量目标】平面向量的坐标运算与两向量垂直的坐标公式等.【考查方式】给出两向量的坐标表示，两向量坐标运算的垂直关系，求未知数.λ 【参考答案】B【试题解析】利用坐标运算得出+-与m n m n 的坐标，再由两向量垂直的坐标公式求λ， 因为()()23,3,1,1,λ+=+-=--m n m n 由()(),+⊥-m n m n 可得()()()()23,31,1260,λλ+-=+--=--= m n m n (步骤1)解得 3.λ=- (步骤2)4．不等式222x -<的解集是 （ ）A.()1,1-B.()2,2-C.()()1,00,1-D.()()2,00,2- 【测量目标】含绝对值的一元二次不等式的解.【考查方式】给出绝对值不等式，求出满足不等式的解集. 【参考答案】D【试题解析】将绝对值不等式转化为一元二次不等式求解.由222,x -<得2222,x -<-<即204,x <<(步骤1)所以20x -<<或02,x <<故解集为()()2,00,2.- (步骤2)5．()862x x +的展开式中的系数是 （ ）A.28B.56C.112D.224 【测量目标】二项式定理.【考查方式】由二项式展开式，求满足条件的项的系数. 【参考答案】C【试题解析】写出二项展开式的通项，从而确定6x 的系数.该二项展开式的通项为88188C 22C ,r r r r r r r T x x --+==(步骤1)令2,r =得2266382C 112,T x x ==所以6x 的系数是112. (步骤2)6．函数()()21log 10f x x x ⎛⎫=+> ⎪⎝⎭的反函数1()f x -= ( ) A.()1021x x >- B.()1021xx ≠- C.()21x x -∈R D.()210x x -> 【测量目标】反函数的求解方法，函数的值域求法. 【考查方式】给出函数的解析式，求它的反函数.. 【参考答案】A【试题解析】由已知函数解出,x 并由x 的范围确定原函数的值域，按照习惯把,x y 互换，得出反函数. 由21log 1y x ⎛⎫=+⎪⎝⎭得112,yx ⎛⎫+= ⎪⎝⎭故1.21yx =-(步骤1)把x 和y 互换，即得()11.21x f x -=-(步骤2) 由0,x >得111,x+>可得0.y > 故所求反函数为()11(0).21xf x x -=>-(步骤3) 7．已知数列{}n a 满足{}12430,,103n n n a a a a ++==-则的前项和等于 ( )A.()10613---B.()101139-- C.()10313-- D.()1031+3-【测量目标】等比数列的定义及等比数列前n 项和.【考查方式】给出一个数列{n a }、它的前后项的关系，判断是否为特殊数列，从而求出它的前n 项和. 【参考答案】C【试题解析】先根据等比数列的定义判断数列{}n a 是等比数列，得到首项与公比，再代入等比数列前n 项和公式计算. 由130,n n a a ++=得11,3n n a a +=-故数列{}n a 是公比13q =-的等比数列. (步骤1)又24,3a =-可得1 4.a =(步骤2)所以()1010101413313.113S -⎡⎤⎛⎫--⎢⎥⎪⎝⎭⎢⎥⎣⎦==-⎛⎫-- ⎪⎝⎭(步骤3)8．()()1221,0,1,0,F F C F x -已知是椭圆的两个焦点过且垂直于轴的直线交于A B 、两点，且3AB =，则C 的方程为 ( )A.2212x y += B.22132x y += C.22143x y += D.22154x y += 【测量目标】椭圆的标准方程及简单几何性质.【考查方式】给出椭圆焦点，由椭圆与直线的位置关系，利用待定系数法求椭圆的标准方程. 【参考答案】C【试题解析】设出椭圆的方程，依据题目条件用待定系数法求参数.由题意知椭圆焦点在x 轴上，且1,c =可设C 的方程为()22221,1x y a a a +>-(步骤1)由过2F 且垂直于x 轴的直线被C 截得的弦长3,AB =知点21,3⎛⎫ ⎪⎝⎭必在椭圆上，(步骤2)代入椭圆方程化简得4241740,a a -+=所以24a =或214a =（舍去）. (步骤3) 故椭圆C 的方程为221.43x y +=(步骤4) 9．若函数()()sin 0=y x ωϕωω=+>的部分图像如图，则 ( ) A.5 B.4 C.3 D.2第9题图【测量目标】根据函数的部分图象确定函数解析式.【考查方式】给出正弦函数的未知解析式及正弦函数的部分图象.根据图象求出T ，确定ω的值.【参考答案】B【试题解析】根据图象确定函数的最小正周期，再利用2πT ω=求.ω设函数的最小正周期为T ，由函数图象可知0ππ=,244T x x ⎛⎫+-= ⎪⎝⎭所以π.2T =(步骤1)又因为2π,T ω=可解得 4.ω=(步骤2)10．已知曲线()421128=y x ax a a =++-+在点，处切线的斜率为， ( )A.9B.6C.9-D.6- 【测量目标】导数的几何意义及求导公式等知识.【考查方式】已知曲线在未知点处的切线斜率，利用导数的几何意义求未知数a . 【参考答案】D【试题解析】先对函数求导，利用导数的几何意义得出点()1,2a -+处的切线斜率，解方程所得.342,y x ax '=+由导数的几何意义知在点(1,2)a -+处的切线斜率1|428,x k y a =-'==--=解得 6.a =-11．已知正四棱柱1111112,ABCD A B C D AA AB CD BDC -=中，则与平面所成角的正弦值等于 ( )A.23 D.13 【测量目标】直线与平面所成角和线面垂直的判定.【考查方式】已知正四棱柱，利用其性质和几何体中的垂直关系求线面角的正弦值. 【参考答案】A【试题解析】利用正四棱柱的性质，通过几何体中的垂直关系，判断点C 在平面1BDC 上的射影位置，确定线平面角，并划归到直角三角形中求解.如图，连接AC ，交BD 于点O ，由正四棱柱的性质，有.AC BD ⊥ 因为1CC ⊥平面ABCD ，所以 BD ⊥（步骤1）又1,CC AC C = 所以BD ⊥平面 O （步骤2） 在平面1CC O 内作1,CH C O ⊥垂足为H ，则.BD CH ⊥又1,BD C O O = 所以CH ⊥平面1,BDC （步骤3） 第11题图 连接DH ，则DH 为CD 在平面1BDC 上的射影，所以CDH ∠为CD 与1BDC 所成的角.（步骤4）设12 2.AA AB ==在1Rt COC △中，由等面积变换易求得2,3CH =在Rt CDH △中，2sin .3CH CDH CD ∠==（步骤5） 12．已知抛物线2:8C y x =与点()2,2M -，过C 的焦点且斜率为k 的直线与C 交于,A B 两点，若0MA MB =，则k = ( )A ．12 D.2 【测量目标】直线与抛物线的位置关系，平面向量的坐标运算等知识.【考查方式】已知抛物线标准方程,利用抛物线性质及直线与抛物线的位置关系求解过焦点的直线的斜率. 【参考答案】D【试题解析】联立直线与抛物线的方程，消元得一元二次方程并得两根之间的关系，由0MA MB =进行坐标运算解未知量k .抛物线C 的焦点为()2,0,F 则直线方程为()2,y k x =-与抛物线方程联立，消去y 化简得()22224840.k x k x k -++=(步骤1)设点()()1122,,,,A x y B x y 则1212284, 4.x x x x k +=+=所以()121284,y y k x x k k+=+-=()21212122416.y y k x x x x =-++=-⎡⎤⎣⎦(步骤2) ()()()()()()112212122,22,22222MA MB x y x y x x y y =+-+-=+++--()()121212122280,x x x x y y y y =+++-++=(步骤3)将上面各个量代入，化简得2440,k k -+=所以 2.k =(步骤4)二、填空题：本大题共4小题，每小题5分.13．设()[)()21,3=f x x f x ∈是以为周期的函数，且当时， . 【测量目标】函数周期的应用及根据函数解析式求值.【考查方式】给出函数()f x 的周期及取值范围，代入解析式求函数值.【参考答案】1-【试题解析】利用周期将自变量转化到已知解析式中x 的范围内，代入解析式计算 . 由于()f x 的周期为2，且当[)1,3x ∈时，()2,f x x =-(步骤1)()2,f x x =-()()()112112 1.f f f -=-+==-=-(步骤2)14．从进入决赛的6名选手中决出1名一等奖，2名二等奖，3名三等奖，则可能的决赛结果共有 种.（用数字作答）【测量目标】简单的排列组合知识的应用. 【考查方式】直接利用排列组合知识列式求解. 【参考答案】60【试题解析】利用排列组合知识列式求解. 由题意知，所有可能的决赛结果有12365354C C C 61602⨯=⨯⨯=（种）.15．若x y 、满足约束条件0,34,34,x x y x y ⎧⎪+⎨⎪+⎩………则z x y =-+的最小值为 .【测量目标】二元线性规划求目标函数最值.【考查方式】直接给出函数的约束条件，利用线性规划性质及借助数形结合思想求z 的最小值.【参考答案】0【试题解析】作出定义域，借助数形结合寻找最优解.由不等式组作出可行域，如图阴影部分所示()包括边界，且()()41,1040,.3A B C ⎛⎫⎪⎝⎭，，，，由数形结合知，直线y x z =+过点()1,1A 时，min 110.z =-+= 16．已知圆O 和圆K 是球O 的大圆和小圆，其公共弦长等于球O 的半径，3602OK O K = ，且圆与圆所在的平面所成角为，则球O 的表面积等于 .【测量目标】球的大圆、小圆及球的截面性质，二面角的平面角，球的表面积公式等知识. 【考查方式】已知二面角的平面角，根据球的截面性质，直角三角形的性质，求出球的半径，并由球的表面积公式求球的表面积. 【参考答案】16π 【试题解析】根据球的截面性质以及二面角的平面角的定义确定平面角，把球的半径转化到三角形中计算，进而求得球的表面积.如图所示，公共弦为AB ，设球的半径为R ，则,AB R =取AB 为中点M ，连接OM 、,KM由圆的性质知,,OM AB KM AB ⊥⊥ 所以KMO ∠为圆O 与圆K 所在平面所成的一个二面角的平面角，则60.KOM ∠=(步骤1)Rt KOM △中，3,2OK =所以sin 60OK OM == (步骤2) 在Rt OMA △中，因为222,OA OM AM =+所以2213,4R R =+解得24,R =(步骤3)所以球O 的表面积为24π16π.R =(步骤4)三、解答题：解答应写出文字说明、证明过程或演算步骤. 17．（本小题满分10分）等差数列{}n a 中，71994,2,a a a ==（I ）求{}n a 的通项公式； （II ）设{}1,.n n n nb b n S na =求数列的前项和 【测量目标】等差数列的通项公式、裂项相消法求数列的前n 项和.【考查方式】（1）根据等差数列的通项公式求出首项和公差，进而求出等差数列的通项公式.（2）已知通项公式，利用裂项相消法求和.【试题解析】（1）设等差数列{}n a 的公差为d ，则()11.n a a n d =+-因为71994,2,a a a =⎧⎨=⎩所以()11164,1828.a d a d a d +=⎧⎨+=+⎩(步骤1)解得11,1.2a d =⎧⎪⎨=⎪⎩所以{}n a 的通项公式为1.2n n a +=(步骤2) （2）因为()222,11n b n n n n ==-++所以2222222.122311n n S n n n ⎛⎫⎛⎫⎛⎫=-+-+⋅⋅⋅+-=⎪ ⎪ ⎪++⎝⎭⎝⎭⎝⎭(步骤3) 18．（本小题满分12分）设ABC △的内角,,A B C 的对边分别为,,a b c ，()()a b c a b c ac ++-+=.（I ）求B（II）若1sin sin 4A C =，求C . 【测量目标】余弦定理解三角形，三角恒等变换公式及其应用.【考查方式】已知三角形的三边及三边关系.（1）由已知关系式展开，利用余弦定理求角. （2）三角形内角和得出A C +,由给出的sin sin A C 的形式，联想构造与已知条件相匹配的余弦公式,求出角C .【试题解析】（1）因为()(),a b c a b c ac ++-+=所以222.a c b ac +-=-(步骤1)由余弦定理得2221cos ,22a cb B ac +-==-因此120.B =(步骤2)（2）由（1）知60,A C +=所以()cos cos cos sin sin A C A C A C -=+cos cos sin sin 2sin sin A C A C A C =-+()11cos 2sin sin 2242A C A C =++=+⨯=(步骤1) 故30A C -=或30,A C -=- 因此15C =或45.C =(步骤2) 19．（本小题满分12分）如图，四棱锥P-ABCD 中，==90ABC BAD ∠∠，BC =2AD ,△P AB 与△PAD 都是边长为2的等边三角形. 图（1）（I ）证明：;PB CD ⊥（II ）求点.A PCD 到平面的距离【测量目标】空间垂直关系的证明和点到平面距离的求解.第19题图【考查方式】已知四棱锥，底面为特殊的直角梯形，侧面为特殊三角形（1）借助线线、线面垂直求解.（2）通过做辅助线将点面距离转化为图形中的线段，再求解.【试题解析】（1）证明：取BC 的中点E ，连接DE ,则四边形ABCD 为正方形. 过点P 作PO ABCD ⊥平面，垂足为O .连接OA ,OB,OD ,OE . 图（2） 由PAB △和PAD △都是等边三角形知,PA PB PD ==(步骤1)所以,O A O B O D ==即O 为正方形ABED 对角线的交点，故 ,OE BD ⊥从而.P B O E ⊥(步骤2)因为O 是BD 的中点，E 是BC 的中点，所以OE //CD .因此.PB CD ⊥(步骤3)(2)解：取PD 的中点F ，连接OF ，则//.OF PB 由（1）知，,PB CD ⊥故.OF CD ⊥(步骤4)又12OD BD ==OP ==故POD △为等腰三角形，(步骤5) 因此.OF PD ⊥又,PD CD D = 所以.OF PCD ⊥平面(步骤6)因为//,AE CD CD PCD ⊂平面，,AE PCD ⊄平面所以//.AE PCD 平面(步骤7) 因此点O 到平面PCD 的距离OF 就是点A 到平面PCD 的距离，(步骤8) 而112OF PB ==，所以点A 到平面PCD 的距离为1. (步骤9) 20．（本小题满分12分）甲、乙、丙三人进行羽毛球练习赛，其中两人比赛，另一人当裁判，每局比赛结束时，负的一方在下一局当裁判，设各局中双方获胜的概率均为1,2各局比赛的结果都相互独立，第1局甲当裁判.（I ）求第4局甲当裁判的概率；（II ）求前4局中乙恰好当1次裁判概率. 【测量目标】相互独立事件同时发生的概率，互斥事件概率加法公式的应用.【考查方式】（1）直接利用独立事件的概率公式求解.（2）由已知，直接利用互斥事件的加法公式求解.【试题解析】（1）记1A 表示事件“第2局结果为甲胜”，2A 表示“第3局甲参加比赛时，结果为甲负”，A 表示事件“第4局甲当裁判”.则12.A A A = ()()()()12121.4P A P A A P A P A === (步骤1)(2)记1B 表示事件“第1局比赛结果为乙胜”，2B 表示事件“第2局乙参加比赛，结果为乙胜”，3B 表示事件“第3局中乙参加比赛时，结果为乙胜”，B 表示事件“前4局中乙恰好当1次裁判”， 则1312312.B B B B B B B B =++ （步骤2）()()1312312P B P B B B B B B B =++=()()()1312312P B B P B B B P B B ++=()()()()()()()1312312P B P B P B P B P B P B P B ++=111+484+ =5.8(步骤3) 21．（本小题满分12分）已知函数()32=33 1.f x x ax x +++（I ）求();a f x =的单调性； （II ）若[)()2,0,x f x ∈+∞时，…求a 的取值范围. 【测量目标】导数在研究函数中的应用.【考查方式】已知含未知数a 的函数()f x （1）对()f x 求导，得出()f x =0时的根，根据导数性质讨论函数单调性.(2)利用特殊值法和放缩法求a 的范围.【试题解析】（1）当a =()3231,f x x x =-++()23 3.f x x '=-+(步骤1)令()0,f x '=得121, 1.x x ==(步骤2)当()1x ∈-∞时，()0,f x '>()f x 在()1-∞上是增函数；当)1x ∈时，()0,f x '<()f x 在)1上是减函数；当)1,x ∈+∞时，()0,f x '>()f x 在)1,+∞上是增函数. (步骤3) （2）由()20f …得4.5a -…当45a -…，()2,x ∈+∞时， ()()225321312f x x ax x ⎛⎫'=++-+ ⎪⎝⎭… =()1320,2x x ⎛⎫--> ⎪⎝⎭所以()f x 在()2,+∞上是增函数，(步骤4)于是当[)2+x ∈∞，时，()()20f x f 厖.综上，a 的取值范围是4,.5⎡⎫-+∞⎪⎢⎣⎭(步骤5) 22．（本小题满分12分） 已知双曲线()221222:10,0x y C a b F F a b-=>>的左、右焦点分别为，，离心率为3，直线2y C =与（I ）求,;a b（II ）2F l C A B 设过的直线与的左、右两支分别相交于、两点，且11,AF BF = 证明：22AF AB BF 、、成等比数列.【测量目标】双曲线的方程、性质，直线与双曲线的位置关系，等比中项等性质.【考查方式】(1)由双曲线与直线的位置关系、双曲线的几何性质求出a,b 值.(2)由直线方程和双曲线方程，利用双曲线与直线的位置关系及两点间距离公式证明线段的等比关系.【试题解析】（1）解：由题设知3,c a =即2229,a b a+=故228.b a = 所以C 的方程为22288.x y a -=(步骤1)将y=2代入上式，求得x =(步骤2)由题设知，=解得2 1.a =所以1,a b ==(步骤3)（2）证明：由（1）知，()()123,0,3,0,F F -C 的方程为2288.x y -=○1(步骤4)由题设可设l 的方程为()3,y k x k =-<将其代入○1并化简，得 ()222286980.k x k x k --++=(步骤5)设()1122,,(,),A x y B x y 则22121212226981,1,,.88k k x x x x x x k k +-+==--剠(步骤6)于是()1131,AF x ==-+123 1.BF x ==+(步骤7)由11,AF BF =得()123131,x x -+=+(步骤8) 即2122262,,383k x x k +=-=--故 解得212419,.59k x x ==-从而(步骤9)由于2113,AF x ===-2231,BF x ===- 故()2212234,AB AF BF x x =-=-+=(步骤10)()221212=39116,AF BF x x x x +--= 因而222,AF BF AB = 所以22AF AB BF 、、成等比数列(步骤11).。
湖南省衡阳市八中2014届高三物理第六次月考试题 新人教版
湖南省衡阳市八中2014届高三物理第六次月考试题新人教版命题人：刘龙华审题人：唐忠涛一、选择题（本题共12小题，每题4分，共48分。
其中1-8题为单选题，9-12题为多选题，全对得4分，选对但不全的得2分，选错不给分）1. 密绕在轴上的一卷地膜用轻绳一端拴在轴上，另一端悬挂在墙壁上A点，如图所示，当逆时针缓慢向下用力F抽出地膜时，整卷地膜受的各个力要发生变化，不计地膜离开整卷时对地膜卷的粘扯拉力和地膜卷绕轴转动时的摩擦力，但在D点地膜与墙壁间有摩擦力，随着地膜的不断抽出，下述分析正确的是( )A．悬挂地膜的轻绳上的拉力在增大B．地膜对墙的压力在增大C．地膜与墙壁间的摩擦力在减小D．地膜卷受墙的支持力与轻绳的拉力的合力不变。
2. 如图甲所示为一个质量为m、带电荷量为+q的圆环，可在水平放置的足够长的粗糙细杆上滑动，细杆处于磁感应强度为B的匀强磁场中。
现给圆环向右初速度v0，其中，0mgvqB>在以后的运动过程中，圆环运动的速度图象可能是图乙中的（）3.如图所示，假设地球是个半径为R的标准的球体，其表面的重力加速度为g，有一辆汽车沿过两极的圆周轨道沿地面匀速率行驶，不计空气阻力，则下列说法中正确的是（）A．重力和地面的支持力是一对平衡力B．汽车的机械能保持不变C．汽车在北极处于超重状态，在南极处于失重状态D．若汽车速率为v=4. 如图甲所示，轻质弹簧的下端固定在水平面上，上端放置一小物体(物体与弹簧不连接)，初始时物体处于静止状态。
现用竖直向上的拉力F作用在物体上，使物体开始向上做匀加速直线运动，拉力F与物体位移x的关系如图乙所示(g=10m/s2)，则下列结论正确的是( )A.物体与弹簧分离时，弹簧处于压缩状态B.物体的质量为3 kgC.物体的加速度大小为5 m/s2D.弹簧的劲度系数为7. 5 N/cm5．如图所示，AB为半圆环ACB的水平直径，C为环上的最低点，环半径为R。
一个小球从A点以速度v水平抛出，不计空气阻力，则下列判断正确的是（）A. 0v 越大，小球落在圆环时的时间越长B.即使0v 取值不同，小球掉到环上时的速度方向和水平 方向之间的夹角也相同C.若0v 取值适当，可以使小球垂直撞击半圆环D.无论0v 取何值，小球都不可能垂直撞击半圆环6. 如图，在水平面(纸面)内有三根相同的均匀金属棒ab 、Ac 和MN 其中ab 、ac 在a 点接触，构成“v ”字型导轨。
湖南省衡阳八中2014-2015学年高二10月月考(六科联赛)数学(文)试题 Word版含答案
高二10月月考（六科联赛）数学（文）试题时量 120分钟 满分 100分一、选择题 (本大题共10小题，每小题3分，共30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．命题“若p 则q ”的逆命题是 A ．若q 则p B ．若¬p 则¬q C ．若¬q 则¬p D ．若p 则¬q 2．若p 是真命题，q 是假命题，则A ．p ∧q 是真命题B ．p ∨q 是假命题C ．¬p 是真命题D ．¬q 是真命题 3.“1a =”是“(1)(2)0a a --=”成立的A . 充分非必要条件 .B 必要非充分条件.C 充要条件 .D 既不充分也不必要条件4．一个命题与它的逆命题、否命题、逆否命题这四个命题中A ．真命题与假命题的个数相同B ．真命题的个数一定是奇数C ．真命题的个数一定是偶数D ．真命题的个数可能是奇数，也可能是偶数 5．命题“∈∃x R,0123=+-x x ”的否定是A ．∈∃x R,0123≠+-x xB ．不存在∈x R, 0123≠+-x xC ．∈∀x R,0123=+-x xD ．∈∀x R, 0123≠+-x x6. 椭圆的中心在坐标原点，对称轴是坐标轴，且经过点Ａ（－３，０），Ｂ（０，22）， 则椭圆的标准方程是A.18922=+y xB.19822=+y xC.122322=+y xD. 122322=+x y7.设双曲线)0,0(12222>>=-b a b y a x 的虚轴长为6，焦距为10，则双曲线的实轴长为 A. 8 B. 6 C. 4 D. 2 8.下列曲线中离心率为62的是A.22124x y -=B.22142x y -=C.22146x y -=D.221410x y -=9.已知点P 是椭圆5922y x +=1上的动点，过P 作椭圆长轴的垂线，垂足为M ，则PM 中点的轨迹方程为 A 、159422=+y x B 、154922=+y x C 、120922=+y x D 、53622y x +=110.已知椭圆2222:1(0)x y E a b a b+=>>的右焦点为(3,0)F ,过点F 的直线交椭圆于,A B 两点.若AB 的中点坐标为(1,1)-,则E 的方程为 A ．2214536x y += B ．2213627x y +=C ．2212718x y +=D ．221189x y +=二.填空题(每小题3分,共15分)11.命题“若a ≥b ，则a 3≥b 3”的否命题是 ．12.椭圆22192x y +=的焦点为12,F F ，点P 在椭圆上，若1||4PF =，则2||PF = . 13.若双曲线2x 4-22y b=1(b>0)的渐近线方程式为y=1x 2±，则ｂ等于 ;14. 命题“∈∀x R,x 2+2x +m >0”是真命题,则实数m 的取值范围 ．15.已知椭圆22221(0)x y a b a b+=>>的左、右焦点分别为12(,0),(,0)F c F c -，若椭圆上存在一点P 使1221sin sin a cPF F PF F =，则该椭圆的离心率的取值范围为 .三、解答题(本大题共6小题，共55分.解答应写出文字说明、证明过程或演算步骤。
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正视图侧视图俯视图2013年衡阳市八中第六次阶段检测文科数学试题命题人 廖洪波 审题人 蒋金元 参考公式：样本数据n x x x ,,,21 的标准差；x x x x x x x ns n 其中],)()()[(122221-+-+-=为样本平均数； 柱体体积公式：为底面面积其中S Sh V ,=、h 为高；锥体体积公式：h S Sh V ,,31为底面面积其中=为高； 球的表面积、体积公式：,34,432R V R S ππ==其中R 为球的半径。
一、选择题：本大题共9小题，每小题5分，共45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.设集合M ＝{－1,0,1}，N ＝{x |x 2≤x }，则M ∩N ＝( )A ．{0}B ．{0,1}C ．{－1,1}D ．{－1,0,1}2.已知,x y R ∈，i 为虚数单位，且x i +y=1+i 则(1)x yi ++的值为（ ）A. 2B.2i - C. 4- D. 2i3．“函数x x f a log )(=在区间（0，＋∞）上为增函数”是“a =3A ．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4．若某多面体的三视图如图所示，则此多面体的体积是（ ）A ．2B ．4C ．6D ．125根据上表，可得回归方程y=bx+a 中的b 为9.4，据此模型，预报使用肥料量为6吨时产量是( ) A ．72.0吨 B ．67.7吨 C ．65.5吨 D ．63.6吨6．已知双曲线与椭圆2211664x y +=有共同的焦点，且它的一条渐近线方程为0x y +=，则这双曲线的方程为（ ）A ．2250x y -=B ．2224x y -=C ．2250x y -=-D ．2224x y -=-7．已知非零向量a 、b ，满足a b ⊥，则函数2()()f x ax b =+ (R)x ∈是( )A. 既是奇函数又是偶函数B. 非奇非偶函数C. 偶函数D. 奇函数 8．在ABC ∆中,若bBa A cos sin =,则B 的值为 ( ) A. 300B. 900C. 600D. 4509．已知()f x 是R 上最小正周期为2的周期函数，且当02x ≤<时,3()f x x x =-,则函数()y f x =的图象在区间[0,6]上与x 轴的交点的个数为（ ）（A ）6 （B ）7 （C ）8 （D ）9二、填空题，本大题共7小题，考生作答6小题.每小题5分共30分，把答案填在答题卡中对应题号后的横线上. （一）选做题，（请考生在第10，11两题中任选一题作答，如果全做 ，则按前一题记分） 10.在极坐标系中，过点)4π作圆4sin ρθ=的切线，则切线的极坐标方程是 。
11.用0.618法确定试点，则经过 次试验后，存优范围缩小为原来的4618.0倍. (二)必做题（12~16题）12．若0,0,x y >>且x y +=2，则11x y+的最小值是 13．某大学对1000名学生的自主招生水平测试成绩进行统计，得到样本频率分布直方图如下图所示，现规定不低于70分为合格，则合格人数是14．某程序的框图如图所示，执行该程序，若输入10，则输出的S 为 ． 第14题第13题 1115.曲线y =x (3ln x +1)在点)1,1(处的切线方程为 .16．定义：数列{}n x ：n n n n x x x x x ++==+231123,1，数列{}n y ：23211nn n x x y ++=；数列{}n z ：232132nn nn x x x z +++=；则=+11z y ；若{}n y 的前n 项的积为P ，{}n z 的 前n 项的和为Q ，那么P+Q= ．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. 17.(本题满分12分)某市地铁全线共有四个车站，甲、乙两人同时在地铁第1号车站（首发站）乘车，假设每人自第2号站开始，在每个车站下车是等可能的，约定用有序实数对(,)x y 表示“甲在x 号车站下车，乙在y 号车站下车”（1）用有序实数对把甲、乙两人下车的所有可能的结果列举出来； （2）求甲、乙两人同在第3号车站下车的概率； （3）求甲、乙两人在不同的车站下车的概率．18.（本小题满分12分）已知函数21()cos cos 2f x x x x =--，.x R ∈（Ⅰ）求函数()f x 的最大值和最小正周期；（Ⅱ）设ABC ∆的内角,,A B C 的对边分别,,,a b c 且3c =,()0f C =，若sin()2sin ,A C A +=求,a b 的值．19.（本小题满分12分） 如图，四棱锥S －ABCD 的底面是矩形，SA ⊥底面ABCD ，P 为BC 边的中点，AD ＝2，AB=1．SP与平面ABCD 所成角为4π． （1）求证：平面SPD ⊥平面SAP; （2）求三棱锥S －APD 的体积，20、（本小题满分13分）在等差数列{}n a 中，2723a a +=-，3829a a +=-． （Ⅰ）求数列{}n a 的通项公式；（Ⅱ）设数列{}n n a b +是首项为1，公比为c 的等比数列，求{}n b 的前n 项和n S ．21.（本小题满分13分）一动圆P 过定点(),F 10且与直线x =-1相切，圆心P 的轨迹为曲线C ，过F 作曲线C 两条互相垂直的弦,AB CD ，设,AB CD 的中点分别为M 、N . （1）求曲线C 的方程；（2）求证：直线MN 必过定点.22.（本小题满分13分）己知函数()ln 1f x x ax =-+在2x =处的切线斜率为1.2-(I)求实数a 的值及函数()f x 的单调区间；(II) 设xkkx x x g ++=2)(2，对)0,(),,0(21-∞∈∃+∞∈∀x x 使得)()(21x g x f ≤成立，求正实数k 的取值范围；正视图侧视图俯视图2013年衡阳市八中第六次阶段检测文科数学试题命题人 廖洪波 审题人 蒋金元 参考公式：样本数据n x x x ,,,21 的标准差；x x x x x x x ns n 其中],)()()[(122221-+-+-=为样本平均数； 柱体体积公式：为底面面积其中S Sh V ,=、h 为高；锥体体积公式：h S Sh V ,,31为底面面积其中=为高； 球的表面积、体积公式：,34,432R V R S ππ==其中R 为球的半径。
一、选择题：本大题共9小题，每小题5分，共45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.设集合M ＝{－1,0,1}，N ＝{x |x 2≤x }，则M ∩N ＝( B )A ．{0}B ．{0,1}C ．{－1,1}D ．{－1,0,1} 2.已知,x y R ∈，i 为虚数单位，且x i +y=1+i 则(1)x yi ++的值为（ D ）A. 2B.2i - C. 4- D. 2i3．“函数x x f a log )(=在区间（0，＋∞）上为增函数”是“a =3”的（ B ）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 4．若某多面体的三视图如图所示，则此多面体的体积是（ A ） A ．2 B ．4C ．6D ．125．某农场农作物使用肥料量x 与产量y 的统计数据如下表：根据上表，可得回归方程y=bx+a 中的b 为9.4，据此模型，预报使用肥料量为6吨时产量是( C ) A ．72.0吨 B ．67.7吨 C ．65.5吨 D ．63.6吨6．已知双曲线与椭圆2211664x y +=有共同的焦点，且它的一条渐近线方程为0x y +=，则这双曲线的方程为（ D ）A ．2250x y -=B ．2224x y -=C ．2250x y -=-D ．2224x y -=-7．已知非零向量a 、b ，满足a b ⊥，则函数2()()f x ax b =+ (R)x ∈是( C )A. 既是奇函数又是偶函数B. 非奇非偶函数C. 偶函数D. 奇函数 8．在ABC ∆中,若bBa A cos sin =,则B 的值为 ( D )A. 300B. 900C. 600D. 4509．已知()f x 是R 上最小正周期为2的周期函数，且当02x ≤<时,3()f x x x =-,则函数()y f x =的图象在区间[0,6]上与x 轴的交点的个数为（ B ）（A ）6 （B ）7 （C ）8 （D ）9二、填空题，本大题共7小题，考生作答6小题.每小题5分共30分，把答案填在答题卡中对应题号后的横线上. （一）选做题，（请考生在第10，11两题中任选一题作答，如果全做 ，则按前一题记分） 10.在极坐标系中，过点)4π作圆4sin ρθ=的切线，则切线的极坐标方程是ρcos θ=2 。
11.用0.618法确定试点，则经过 5 次试验后，存优范围缩小为原来的4618.0倍. (二)必做题（12~16题）12．若0,0,x y >>且x y +=2，则11x y+的最小值是 2 13．某大学对1000名学生的自主招生水平测试成绩进行统计，得到样本频率分布直方图如下图所示，现规定不低于7014．某程序的框图如图所示，执行该程序，若输入10，则输出的S 为 1033 ． 15.曲线y =x (3ln x +1)在点)1,1(处的切线方程为34-=x y .16．定义：数列{}n x ：n n n n x x x x x ++==+231123,1，数列{}n y ：23211nn n x x y ++=；数列{}n z ：232132nn nn x x x z +++=；则=+11z y 1 ；若{}n y 的前n 项的积为P ，{}n z 的 前n 项的和为Q ，那么P+Q= 1 ．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. 17.(本题满分12分)某市地铁全线共有四个车站，甲、乙两人同时在地铁第1号车站（首发站）乘车，假设每人自第2号站开始，在每个车站下车是等可能的，约定用有序实数对(,)x y 表示第14题 第13题11“甲在x 号车站下车，乙在y 号车站下车”（1）用有序实数对把甲、乙两人下车的所有可能的结果列举出来； （2）求甲、乙两人同在第3号车站下车的概率； （3）求甲、乙两人在不同的车站下车的概率． 解：（1）甲、乙两人下车的所有可能的结果有：（2，2）、（2，3）、（2，4）、（3，2）、 （3，3）、（3，4）、（4，2）、（4，3）、（4，4）共9种 ………………… 4分（2）设甲、乙两人同时在第3节车站下车为事件A 8分 （3）甲、乙两人在同一地铁站下车的基本事件有（2，2）、（3，3）、（4，4）共3 12分18.（本小题满分12分）已知函数21()cos cos 2f x x x x =--，.x R ∈（Ⅰ）求函数()f x 的最大值和最小正周期；（Ⅱ）设ABC ∆的内角,,A B C 的对边分别,,,a b c 且3c =,()0f C =，若sin()2sin ,A C A +=求,a b 的值．解：（1）1cos 21()2sin(2)12226x f x x x π+=--=--……．3分 则()f x 的最大值为0，最小正周期是22T ππ==…………………6分 （2）()sin(2)106f C C π=--=则sin(2)16C π-= 1100222666C C C πππππ<<∴<<∴-<-<2623C C πππ∴-=∴= sin()2sin A C A += 由正弦定理得12a b =①……………………9分 由余弦定理得2222cos3c a b ab π=+- 即229a b ab +-=②由①②解得a =b =…………………………12分 19.（本小题满分12分） 如图，四棱锥S －ABCD 的底面是矩形，SA ⊥底面ABCD ，P 为BC 边的中点，AD ＝2，AB=1．SP与平面ABCD 所成角为4π． （1）求证：平面SPD ⊥平面SAP;（2）求三棱锥S －APD 的体积，20、（本小题满分13分）在等差数列{}n a 中，2723a a +=-，3829a a +=-． （Ⅰ）求数列{}n a 的通项公式；（Ⅱ）设数列{}n n a b +是首项为1，公比为c 的等比数列，求{}n b 的前n 项和n S ． 解：（Ⅰ）设等差数列{}n a 的公差是d ．依题意 3827()26a a a a d +-+==-，从而3d =-． ………2分 所以 2712723a a a d +=+=-，解得 11a =-． ………4分所以数列{}n a 的通项公式为 23+-=n a n ． …6分 （Ⅱ）由数列{}n n a b +是首项为1，公比为c 的等比数列，得 1-=+n n n cb a ，即123-=++-n n cb n ，所以 123-+-=n n cn b ． ……8分所以 21[147(32)](1)n n S n c c c -=++++-+++++21(31)(1)2n n n c c c --=+++++ ． ……10分 从而当1=c 时，2(31)322n n n n nS n -+=+=； ……11分 当1≠c 时，(31)121n n n n c S c--=+-． ………13分 21.（本小题满分13分）一动圆P 过定点(),F 10且与直线x =-1相切，圆心P 的轨迹为曲线C ，过F 作曲线C 两条互相垂直的弦,AB CD ，设,AB CD 的中点分别为M 、N . （1）求曲线C 的方程；（2）求证：直线MN 必过定点.解：（1）设(),P x y x =+1，化简得y x =24（2）设():AB y k x =-1 ，代入y x =24得()k x k x k -++=2222220，A B M x x k x k ++==2222，()My k x k =-=21, 故,k M k k ⎛⎫ ⎪+⎝⎭2222因为AB CD ⊥，所以将点M 坐标中的k 换成k-1，即得(),N k k +-2212。