山东省邹城市第一中学2016届高三英语4月模拟考试试题
016届高三迎三模模拟卷(4月)英语(附答案)

江苏省溧水高级中学三模模拟试卷英语(2016年4月18日)注意事项:1. 本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,总分120分。
答题前,考生务必将自己的姓名、考生号填写在答题卡上。
2. 回答第Ⅰ卷时,每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑,如需改动,用橡皮擦干净后,再选涂其他答案标号。
写在试卷上无效。
3. 回答第Ⅱ卷时,将答案填写在答题卡上,写在试卷上无效。
第Ⅰ卷(共85分)第一部分听力(共两节,满分20分)做题时,先将答案标在试卷上。
录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。
第一节(共5小题;每小题1分,满分5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有15秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. What season is it now?A. Summer.B. Autumn.C. Winter.2. How might the woman feel?A. Satisfied.B. Annoyed.C. Excited.3. What does the woman want to know?A. What the time is now.B. Where the bar is.C. Who the lady is.4. What will the woman probably do next?A. Clean the car.B. Brush her teeth.C. Leave the house.5. What does the man offer to do?A. Fix the car for the woman.B. Give the woman a ride.C. Call a taxi for the woman.第二节(共15小题;每小题1分,满分15分)听下面5段对话或独白,每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
2016届山东省邹城一中高三下学期4月模拟语文(解析版)资料

2016届山东省邹城一中高三下学期4月模拟语文阅读下面一段文字,完成小题。
我看到了不屈不挠、生命飞扬的自然景象!,,。
,。
刹.那间,我感动不已。
我在(琢磨/捉摸)中再次抬起头,仰望古塔上的一团绿荫以及翱翔在它周围的鸟儿,我分明感觉飞鸟才是给这树、这塔以生存的施与者。
因为塔内堆积的鸟粪给盘根错结的根蔓提供了(必要/必须)的养分,使宛若蛇虬的根茎不断(延伸/延续)。
而且推而远之,这存活塔顶的绿色华盖,也不知道是仰仗何年何月何只鸟雀,衔落于塔顶的树种造就了这棵树的最初生命。
日复一日,由于树的根须不断渗入塔的缝隙吸取大地的精华,于是又营造了塔内湿润的空间。
鸟、树和塔是如此的相互厮守、依存,难割难舍,最终在天地之间顶托了一道秾丽的风景与和谐统一的煌煌气象,这是天意?这是缘分?我按捺不住激动的心情,几乎要双手合十向这座古塔顶礼模拜了。
1.文中加横线字的注音和字形都不正确的一项是A.刹(shà)那间厮守B.施与(yú)盘根错结C.蛇虬(qiú)翱翔D.秾(nóng)丽顶礼模拜2.依次选用文中括号里的词语,最恰当的一项是A.琢磨必须延续B.捉摸必须延伸C.捉摸必要延续D.琢磨必要延伸3.依次填入原文横线处的语句,衔接最恰当的一组是①这本无生命的石头恐怕早已坍塌于荒野下②你中有我,我中有你,古塔完全被树根抬举起来③终于觉察古塔是被树根簇拥着、裹挟着④感叹之余,围绕着古塔,我转了一圈⑤假如没有众多的树根以顽强之力护卫古塔A.⑤①④③② B.③①②⑤④C.④③②⑤① D.②⑤③④①4.下列各句中,加横线的成语使用正确的一项是A.2016年,北京22家市属医院将全部实施非急诊全面预约制度,不过,为打击号贩子而全部取消现场放号,又有因噎废食之嫌。
B.山水画家施贞泉先生,其作画时轻松若定、恣意挥洒,而且他作画神速,倚马可待,令人叹为观止。
C.从红门开始,我与朋友踏着一级又一级台阶,历尽艰辛,登上泰山极顶,放眼望去,“天无涯兮地无边”,自己显得那么渺小,登高自卑之感油然而生。
山东省邹城市第一中学2015届高三4月高考模拟语文试题

2015届山东省邹城市第一中学高三4月高考模拟语文试题本试卷分五大题。
满分150分。
考试用时150分钟。
(参评试题)一、古代诗文阅读(27分)(一)默写常见的名句名篇(6分)1.补写出下列名句名篇中空缺的部分。
(6分)(1)锲而不舍,。
(2),洪波涌起。
(3)一去紫台连朔漠,。
(4)风急天高猿啸哀,。
,不尽长江滚滚来。
((5)?烟波江上使人愁。
(6),两山排闼送青来。
(二)文言文文阅读(15分)国之所以治乱者三,杀戮刑罚,不足用也。
国之所以安危者四,城郭险阻,不足守也。
国之所以富贫者五,轻税租,薄赋敛,不足恃也。
治国有三本,而安国有四固,而富国有五事。
五事,五经也。
君之所审者三:一曰:德不当其位;二曰:功不当其禄;三曰:能不当其官。
此三本者,治乱之原也。
故国有德义未明于朝者,则不可加于尊位;功力未见于国者,则不可授与重禄;临事不信于民者,则不可使任大官。
故德厚而位卑者,谓之过;德薄而位尊者,谓之失。
宁过于君子,而毋失于小人。
过于君子,其为怨浅;失于小人,其为祸深。
是故国有德义未明于朝而处尊位者则良臣不进有功力未见于国而有重禄者则劳臣不劝有临事不信于民而任大官者则材臣不用。
三本者审,则下不敢求;三本者不审,则邪臣上通,而便辟制威。
如此,则明塞于上,而治壅于下,正道捐弃,而邪事日长。
三本者审,则便辟无威于国,道涂无行禽,疏远无蔽狱,孤寡无隐治。
故曰:刑省治寡,朝不合众。
君之所慎者四:一曰:大德不至仁,不可以授国柄。
二曰:见贤不能让,不可与尊位。
三曰:罚避亲贵,不可使主兵。
四曰:不好本事,不务地利,而轻赋敛,不可与都邑。
此四务者,安危之本也。
故曰:卿相不得众,国之危也;大臣不和同,国之危也;兵主不足畏,国之危也;民不怀其产,国之危也。
故大德至仁,则操国得众;见贤能让,则大臣和同;罚不避亲贵,则威行于邻敌;好本事,务地利,重赋敛,则民怀其产。
君之所务者五:一曰:山泽不救于火,草木不植成,国之贫也。
二曰:沟渎不遂于隘,鄣水不安其藏,国之贫也。
2016届山东省邹城市第一中学高三10月月考英语试题 及答案

高三英语阶段检测2015.10命题:冯再杰校对:汤金霞本试卷分第I卷(选择题)和第II卷(非选择题)两部分。
考试结束后,将本试卷和答案卡一并交回。
注意事项:1.答第I卷前考生务必将自己的姓名、准考证号填写在答题卡上。
2.选出每小题答案前,用2B铅笔把答题卡上对应题目的答案标号涂黑。
如需改动, 用橡皮擦干净后, 再选涂其他答案标号框, 不能答在本试卷上,否则无效。
第一部分听力(共两节,满分30分)做题时,先将答案标在试卷上。
录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。
第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话,每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. Why did the woman have to go home?A. To get her check.B. To get some cashC. To fetch her ID card2. What does the woman mean?A. She likes to share books with the man.B. She agrees to talk with the man frequently.C. She wants to borrow some books from the man.3. What does the man offer to serve as for the woman?A. Her driverB. Her body guard.C. Her tour guide.4. When is the pop show?A. At 8:30B. At 8:20C. At 7:305. What are the speakers mainly talking about?A. A paperB. A courseC. A teacher 第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白,每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
高级中学2016届高三上学期第四次月考英语试题(附答案) (1)

平坝第一高级中学2015—2016学年度第一学期1月高三英语月考试卷本试卷分为第I卷(选择题)和第II卷(非选择题)两个部分,满分120分,考试时间100分钟。
第I卷第一部分阅读理解(共两节,满分40分)第一节(共15小题,每小题2分,满分30分)阅读下列短文,从每题所给的四个选项(A、B、C和D)中,选出最佳选项。
并在答题卡上将该选项涂黑.AInstead of hitting the beach, fourteen high school students traded swimming suits for lab coats last summer and turned their attention to scientific experiments.The High School Research Program offers high school students guidance with researchers in Texas A&M’s College of Agriculture and Life Sciences.Jennifer Funkhou ser, academic adviser for the Department of Rangeland Ecology and Management, dirests this four-week summer program designed to increase understanding of research and its career potential(潜能)。
Several considerations go into selecting students, including grades, school involvement and interest in science and agriculture. And many students come from poorer school districts, Funkhouser says. “This is their chance to learn techniques and do experiments they never would have a chance to do in high school.Warner Ervin of Houston is interested in animal science and learned how to tell male from female mosquitoes(蚊子).His adviser, Craig Coates, studies the genes of mosquitoes that allow them to fight against malaria and yellow fever. Coates thought this experience would be fun and helpful to the high school students.The agricultural research at A&M differs from stereotypes. It’s “molecular(分子)science on the cutting edge,” Funkhouser says. The program broadened students’ knowledge. Victor Garcia of Rio Grande City hopes to become a biology teacher and says he learned a lot about chemistry from the program.At the end of the program, the students presented papers on their research.They’re also paid$600 for their work-another way this program differs from others, which often charge a fee.Fourteen students got paid to learn that science is fun, that agriculture is a lot more than milking and plowing and that research can open many doors.1.The research program is chiefly designed for________. .A. high school advisers from HoustonB. college students majoring in agricultureC. high school students from different placesD. researchers at the College of Agriculture and Life Sciences2 .It can be inferred from the text that the students in poorer areas________. .A. had little chance to go to collegeB. could often take part in the programC. found the program useful to their futureD. showed much interest in their high school experiments3 When the program was over, the students_____________.A. entered that collegeB. wrote research reportsC. paid for their researchD. found way to make money4 .What would be the best title for the text?A.A Program for Agricultural Science StudentsB. A Program for Animal Science StudentsC. A Program for Medical Science LoversD. A Program for Future Science LoversBMa Weidu is the founder and curator (馆长) of Guanfu Museum, which is the first private museum in China. With mainly Chinese artifacts off exhibition, the museum was founded on Jan 18,l 997. It was not well. known by the public until 2008.On Jan l,2008,Ma Weidu was invited to be the guest in the flagship CCTV program "Lecture Room".He talked about his museum and more importantly, he introduced Chinese artifacts and their underlying historical, cultural significance.In 2009,Ma Weidu initiated the Beijing Guanfu Cultural Foundation. Creating a cuulturefoundation had been Ma's dream for many years. The Beijing Guanfu Cultural Foundation is committed to spreading Chinese traditional culture, funding the development of Guanfu Museum, heritage research andconservation projects, building a platform for public culture, promoting and guiding the public in the spirit of"charity&culture sponsorship (赞助)".Guanfu Museum is currently located at N0.18 Jinnan Road, Zhangwanfen, Dashanzi, Chaoyang district, Beijing. But due to the city planning, it has to move. Guanfu Museum,which has experienced relocationfor three times,is now looking for a new place. According to Mr. Ma,it should be around 20,000-30,000square meters.5 According to this passage, which of the followings is true?A. Guanfu Museum was built in 2008.B. Ma Weidu was invited to lecture by Guanfu Museum.C. The public were attracted by Chinese artifacts on exhibition.D. Guanfu Museum, the first private museum in China was founded last century6 Guanfu Museum, according to the city government,_______ .A. will be paintedB. has to hold a new exhibitionC. has to move to a new placeD. has to be enlarged7. The proper title of this passage should be_____.A. Ma's HopesB. MuseumsC. Culture RelicsD. Chinese CultureCThere was a wonder-filled little girl who was hurt so badly that when she entered adult life, she thought she was so bad that others just couldn’t be nice to her.But she had a lot of curiosity, which kept her going. She sought many wise people to help her understand why she was so bad that her mother hurt her and why she was unable to be betterso that men wouldn’t hurt her.She was on a journey that she thought was to h elp her be “better”. She carried a big bag with her everywhere she went. Inside it were all the hurts she had experienced. Because she was so eager to please those who offered their wisdom, she willingly agreed with what they said about forgiving. But she held that bag of hurts tightly.After 50 years of carrying that bag around and showing it to all she met as if it were proof that she was a good person, she decided to open it up and just see what happened.When she looked inside it was filled with bits of paper. At one time they had words on them, detailing the hurts. But time had faded (使褪色) the words and all that was left was some useless paper. She had been struggling to carry this bag that held nothing but the image of something that once was.She saw the absurdity of carrying that bag around. It made her laugh. By letting go of the paper and having pity for those that hurt her, this little girl was able to become a woman with beautiful hair. The woman looked in the mirror and said, “I am lovable.”Although it took many years, she was finally open to bringing love and respect into her life.8 .When she grew up, the little girl wondered .A.where she should goB.why she couldn’t be lovedC.how she could be a sweet girlD.why she was hurt so deeply9 What advice did girl get from wise people?A.To please those who hurt her.B.To keep carrying the bag of hurts.C.To record the hurts she suffered.D.To forgive those who hurt her.10 The woman laughed in the end because .A.she had beautiful hairB.she became lovableC.she realized her sillinessD.she became a wise woman11 What lesson can we learn from the story?A.You will walk fast without carrying a load.B.You will get peace after forgiving others.C.You will become young if you forget your hurts.D.You will be loved after changing your image.DA new generation addiction is quickly spreading all over the world. Weboholism, a twentieth century disease, affects people from different ages. They surf the Net, use email and speak in chat rooms. They spend many hours on the computer, and it becomes a compulsive habit. They cannot stop, and it affects their lives.Ten years ago, no one thought that using computers could become compulsive behavior that could affect the social and physical life of computer users. This obsessional behavior has affected teenagers and college students. They are likely to log on computers and spend long hours at different websites.They become hooked on computers and gradually their social and school life is affected by this situation. They spend all free time surfing and don't concentrated on homework, so this addiction influences their grades and success at school. Because they can find everything on the websites, theyhang out there. Moreover, this addiction to websites influences their social life.They spend more time in front of computers than with their friends. The relation with their friendschanges. The virtual life becomes more important than their real life. They have a new language that they speak in the chat rooms and it causes cultural changes in society.Because of the change in their behavior, they begin to isolate themselves from the society and live with their virtual friends. They share their emotions and feelings with friends who they have never met in their life. Although they feel confident on the computer, they are not confident with real live friends they have known all their life. It is a problem for the future. This addictive behavior is beginning to affectthe entire world.12 The author's attitude towards weboholism is that of being ______.A. optimisticB.disapprovingC. positiveD. acceptable13 The main idea of the passage is about ______.A. the cause of weboholismB. the advantage of weboholismC. the popularity of weboholismD. the influence of weboholism14 The underlined world "obsessional" in the second paragraph probably means ______.A. attractiveB. addictiveC. professionalD. potential15 We can infer from the passage that ______.A. weboholism has the greatest effect on teenagersB. students can hardly balance real and virtual lifeC. people are addicted to games on the InternetD. virtual life is more vivid and attractive anyway第二节(共5小题;每小题2分,满分10分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。
山东省滨州市邹平双语学校2016届高三上学期第一次月考

2015-2016第一学期三区高三英语月考试题( 时间:120分钟满分:150分)第Ⅰ卷第一部分听力(共两节,满分30分)第一节(共5小题;每小题 1.5分,满分7.5分)1. What did the woman do yesterday evening?A. She ate out.B. She watched TV.C. She watched a match.2. What does the woman suggest?A. The man should try harder.B. The man should give up.C. She could help the man.3. Where is the talk probably taking place?A. In a clothing shop.B. In a restaurant.C. In a library.4. What kind of person is Richard?A. A trouble-maker.B. A selfish person.C. A helpful person.5. What might be the man's problem?A. He has caught a bad cold.B. He has heart trouble.C. His stomach aches. 第二节(共15小题;每小题1.5分,满分22.5分)听第 6 段材料,回答第 6-8 题。
6. What are the two speakers talking about?A. A book.B. A story.C. A character.7. What does the woman think about the persons in the story?A. They are poor.B. They are active.C. They have no life.8. What does the man think about the books he read written by that writer?A. Good.B. Bad.C. Some good, some bad.听第 7 段材料,回答第 9-10 题。
2016年山东省济宁市邹城一中高考数学模拟试卷(理科)(4月份)(解析版)
2016年山东省济宁市邹城一中高考数学模拟试卷(理科)(4月份)一、选择题(本大题共10个小题,每小题5分,共50分,在每小题给出的四个选项中,只有一项是符合题目要求的)1.若集合A={1,2,3,4,5},集合B={x|x(4﹣x)<0},则图中阴影部分表示()A.{1,2,3,4}B.{1,2,3}C.{4,5}D.{1,4}2.设i为虚数单位且z的共轭复数是,若z+=4,z=8,则z的虚部为()A.±2B.±2iC.2D.﹣23.已知ξ服从正态分布N(1,σ2),a∈R,则“P(ξ>a)=0.5”是“关于x的二项式的展开式的常数项为3”的()A.充分不必要条件B.必要不充分条件C.既不充分又不必要条件D.充要条件4.已知函数y=f(x)+x是偶函数,且f(2)=1,则f(﹣2)=()A.﹣1B.1C.﹣5D.55.若正数a,b满足+=1,则+的最小值为()A.3B.4C.5D.66.执行如图所示的程序框图,如果输入的x∈[﹣1,3],则输出的y属于()A.[0,2]B.[1,2]C.[0,1]D.[﹣1,5]7.若实数x,y满足,则z=的最小值为()A.﹣2B.﹣3C.﹣4D.﹣58.将4本完全相同的小说,1本诗集全部分给4名同学,每名同学至少1本书,则不同分法有()A.24种B.28种C.32种D.16种9.如图,将绘有函数f(x)=2sin(ωx+φ)(ω>0,<φ<π)部分图象的纸片沿x轴折成直二面角,若AB之间的空间距离为,则f(﹣1)=()A.﹣2B.2C.D.10.如图,F1、F2是双曲线=1(a>0,b>0)的左、右焦点,过F1的直线l与双曲线的左右两支分别交于点A、B.若△ABF2为等边三角形,则双曲线的离心率为()A.4B.C.D.二、填空题:本大题共5小题,每小题5分,共25分,把答案填在答题卷的横线上..11.(x2+)6的展开式中x3的系数是.(用数字作答)12.已知某几何体的三视图如图,正(主)视图中的弧线是半圆,根据图中标出的尺寸,可得这个几何体的表面积是(单位:cm2).13.在Rt△ABC中,∠A=90°,AB=AC=2,点D为AC中点,点E满足,则=.14.已知P、A、B、C是球O球面上的四点,△ABC是正三角形,三棱锥P﹣ABC的体积为,且∠APO=∠BPO=∠CPO=30°,则球O的表面积为.15.已知函数f(x)=,存在x1<x2<x3,f(x1)=f(x2)=f(x3),则的最大值为.三、解答题:本大题共6小题,满分75分,解答应写出文字说明、证明过程或演算步骤16.已知函数f(x)=2cosx(sinx﹣cosx)+m(m∈R),将y=f(x)的图象向左平移个单位后得到y=g(x)的图象,且y=g(x)在区间[0,]内的最大值为.(Ⅱ)在△ABC中,内角A、B、C的对边分别是a、b、c,若g(B)=l,且a+c=2,求△ABC的周长l的取值范围.17.如图,AC是圆O的直径,点B在圆O上,∠BAC=30°,BM⊥AC交AC于点M,EA⊥平面ABC,FC∥EA,AC=4,EA=3,FC=1.(1)证明:EM⊥BF;(2)求平面BEF与平面ABC所成的锐二面角的余弦值.18.某学校为了解高三年级学生寒假期间的学习情况,抽取甲、乙两班,调查这两个班的学生在寒假期间每天平均学习的时间(单位:小时),统计结果绘成频率分布直方图(如右图).已知甲、乙两班学生人数相同,甲班学生每天平均学习时间在区间[2,4]的有8人.(1)图中a的值为;(2)用各组时间的组中值代替各组平均值,估算乙班学生每天学习的平均时长;(3)从甲、乙两个班每天平均学习时间大于10个小时的学生中任取4人参加测试,设4人中甲班学生的人数为ξ,求ξ的分布列和数学期望.19.已知正项数列{a n},若前n项和S n满足8S n=a n2+4a n+3,且a2是a1和a7的等比中项(1)求数列{a n}的通项公式;(2)符号[x]表示不超过实数x的最大整数,记b n=[log2()],求b1+b2+b3+….20.已知椭圆C1:的离心率为,焦距为,抛物线C2:x2=2py(p>0)的焦点F是椭圆C1的顶点.(Ⅰ)求C1与C2的标准方程;(Ⅱ)C1上不同于F的两点P,Q满足,且直线PQ与C2相切,求△FPQ的面积.21.已知函数(a∈R)在其定义域内有两个不同的极值点.(Ⅱ)记两个极值点分别为x1,x2,且x1<x2.已知λ>0,若不等式恒成立,求λ的范围.2016年山东省济宁市邹城一中高考数学模拟试卷(理科)(4月份)参考答案与试题解析一、选择题(本大题共10个小题,每小题5分,共50分,在每小题给出的四个选项中,只有一项是符合题目要求的)1.若集合A={1,2,3,4,5},集合B={x|x(4﹣x)<0},则图中阴影部分表示()A.{1,2,3,4}B.{1,2,3}C.{4,5}D.{1,4}【考点】V enn图表达集合的关系及运算.【分析】化简B={x|x(4﹣x)<0}={x<0或x>4},而图中阴影部分表示的集合是A∩∁R B,从而解得.【解答】解:由图中阴影部分表示的集合是A∩∁R B∵B={x|x(4﹣x)<0}={x<0或x>4},∴∁R B={x|0≤x≤4},∵集合A={1,2,3,4,5},∴A∩∁R B={1,2,3,4}故选:A2.设i为虚数单位且z的共轭复数是,若z+=4,z=8,则z的虚部为()A.±2B.±2iC.2D.﹣2【考点】复数代数形式的混合运算.【分析】设z=a+bi,a、b∈R;利用z的共轭复数是=a﹣bi,列出方程组求出a、b的值即可.【解答】解:设z=a+bi,a、b∈R;∴z的共轭复数是=a﹣bi,又z+=2a=4,∴a=2;z=a2+b2=4+b2=8,∴b=±2;∴z的虚部为±2.故选:A.3.已知ξ服从正态分布N(1,σ2),a∈R,则“P(ξ>a)=0.5”是“关于x的二项式的展开式的常数项为3”的()A.充分不必要条件B.必要不充分条件C.既不充分又不必要条件D.充要条件【考点】必要条件、充分条件与充要条件的判断.【分析】根据充分条件和必要条件的定义结合正态分布已经二项式定理的内容进行判断即可.【解答】解:若P(ξ>a)=0.5,则a=1,若关于x的二项式的展开式的常数项为3,则通项公式T k+1==•a3﹣k•x3﹣3k,由3﹣3k=0,得k=1,即常数项为=3a2=3,解得a=1或a=﹣1,即“P (ξ>a )=0.5”是“关于x 的二项式的展开式的常数项为3”的充分不必要条件,故选:A4.已知函数y=f (x )+x 是偶函数,且f (2)=1,则f (﹣2)=( ) A .﹣1B .1C .﹣5D .5【考点】函数奇偶性的性质;抽象函数及其应用.【分析】根据函数y=f (x )+x 是偶函数,可知f (﹣2)+(﹣2)=f (2)+2,而f (2)=1,从而可求出f (﹣2)的值.【解答】解:令y=g (x )=f (x )+x , ∵f (2)=1,∴g (2)=f (2)+2=1+2=3,∵函数g (x )=f (x )+x 是偶函数, ∴g (﹣2)=3=f (﹣2)+(﹣2),解得f (﹣2)=5. 故选D .5.若正数a ,b 满足+=1,则+的最小值为( )A .3B .4C .5D .6【考点】基本不等式在最值问题中的应用. 【分析】首先判断>0,>0;再由基本不等式确定最小值即可.【解答】解:∵a >0,b >0, +=1; ∴a >1,b >1,a+b=ab ;∴>0,>0,∴+≥2=2=4;(当且仅当=,即a=,b=3时,等号成立). 故选:B .6.执行如图所示的程序框图,如果输入的x ∈[﹣1,3],则输出的y 属于( )A.[0,2]B.[1,2]C.[0,1]D.[﹣1,5]【考点】程序框图.【分析】根据程序框图,分析程序的功能,结合输出自变量的范围条件,利用函数的性质即可得到结论.【解答】解:模拟执行程序框图,可得程序框图的功能是计算并输出y=的值.若﹣1≤x<0,则不满足条件输出y=2﹣x﹣1∈(0,1],若0≤x≤3,则满足条件,此时y=log2(x+1)∈[0,2],输出y∈[0,2],故选:A.7.若实数x,y满足,则z=的最小值为()A.﹣2B.﹣3C.﹣4D.﹣5【考点】简单线性规划.【分析】作出不等式组对应的平面区域.,利用分式函数的意义以及直线的斜率进行求解即可【解答】解:作出不等式组对应的平面区域如图:z===1+,设k=,则k的几何意义为区域内的点到定点D(2,﹣2)的斜率,由图象知AD的斜率最小,由得,即A(1,2),此时AD的斜率k=,则z=1+k=1﹣4=﹣3,即z=的最小值为﹣3,故选:B8.将4本完全相同的小说,1本诗集全部分给4名同学,每名同学至少1本书,则不同分法有()A.24种B.28种C.32种D.16种【考点】计数原理的应用.【分析】分二类,有一个人分到一本小说和一本诗集,有一个人分到两本小说,根据分类计数原理可得【解答】解:第一类,每位同学各分1本小说,再把1本诗集全部分给4名同学任意一个,共有4种方法,第二类,这本诗集单独分给其中一位同学,4相同的小说,分给另外3个同学,共有C41C31=12种,根据分类计数原理,共有4+12=16种,故选:D.9.如图,将绘有函数f(x)=2sin(ωx+φ)(ω>0,<φ<π)部分图象的纸片沿x轴折成直二面角,若AB之间的空间距离为,则f(﹣1)=()A.﹣2B.2C.D.【考点】点、线、面间的距离计算.【分析】根据图象过点(0,1),结合φ的范围求得φ的值,再根据A、B两点之间的距离为=,求得T的值,可得ω的值,从而求得函数的解析式,从而求得f(﹣1)的值.【解答】解:由函数的图象可得2sinφ=1,可得sinφ=,再根据<φ<π,可得φ=.再根据A、B两点之间的距离为=,求得T=6,再根据T==6,求得ω=.∴f(x)=2sin(x+),f(﹣1)=2sin(﹣+)=2,故选:B.10.如图,F1、F2是双曲线=1(a>0,b>0)的左、右焦点,过F1的直线l与双曲线的左右两支分别交于点A、B.若△ABF2为等边三角形,则双曲线的离心率为()A.4B.C.D.【考点】双曲线的简单性质.【分析】由双曲线的定义,可得F1A﹣F2A=F1A﹣AB=F1B=2a,BF2﹣BF1=2a,BF2=4a,F1F2=2c,再在△F1BF2中应用余弦定理得,a,c的关系,由离心率公式,计算即可得到所求.【解答】解:因为△ABF2为等边三角形,不妨设AB=BF2=AF2=m,A为双曲线上一点,F1A﹣F2A=F1A﹣AB=F1B=2a,B为双曲线上一点,则BF2﹣BF1=2a,BF2=4a,F1F2=2c,由,则,在△F1BF2中应用余弦定理得:4c2=4a2+16a2﹣2•2a•4a•cos120°,得c2=7a2,则.故选:B.二、填空题:本大题共5小题,每小题5分,共25分,把答案填在答题卷的横线上..11.(x2+)6的展开式中x3的系数是20.(用数字作答)【考点】二项式系数的性质.【分析】先求出二项式展开式的通项公式,再令x的系数等于3,求得r的值,即可求得展开式中x3的系数.【解答】解:由于(x2+)6的展开式的通项公式为T r+1=•x12﹣3r,令12﹣3r=3,解得r=3,故展开式中x3的系数是=20,故答案为:20.12.已知某几何体的三视图如图,正(主)视图中的弧线是半圆,根据图中标出的尺寸,可得这个几何体的表面积是3π+4(单位:cm2).【考点】由三视图求面积、体积.【分析】由三视图知几何体是半个圆柱,由三视图求出几何元素的长度,由圆柱的表面积公式求出几何体的表面积.【解答】解:根据三视图可知几何体是半个圆柱,且正视图是底面,∴底面圆的半径是1cm,母线长是2cm,∴几何体的表面积S=π×12+π×1×2+2×2=3π+4(cm2),故答案为:3π+4.13.在Rt△ABC中,∠A=90°,AB=AC=2,点D为AC中点,点E满足,则=﹣2.【考点】平面向量数量积的运算.【分析】由已知画出图形,结合向量的加法与减法法则把用表示,展开后代值得答案.【解答】解:如图,∵,∴=,又D为AC中点,∴,则===.故答案为:﹣2.14.已知P、A、B、C是球O球面上的四点,△ABC是正三角形,三棱锥P﹣ABC的体积为,且∠APO=∠BPO=∠CPO=30°,则球O的表面积为16π.【考点】球的体积和表面积.【分析】设△ABC的中心为S,球O的半径为R,△ABC的边长为2a,由已知条件推导出a=R,再由三棱锥P﹣ABC的体积为,求出R=2,由此能求出球O的表面积.【解答】解:如图,P,A,B,C是球O球面上四点,△ABC是正三角形,设△ABC的中心为S,球O的半径为R,△ABC的边长为2a,∵∠APO=∠BPO=∠CPO=30°,OB=OP=R,∴OS=,BS=,∴=,解得a=R,2a=R,∵三棱锥P﹣ABC的体积为,∴=,解得R=2,∴球O的表面积S=4πR2=16π.故答案为:16π.15.已知函数f(x)=,存在x1<x2<x3,f(x1)=f(x2)=f(x3),则的最大值为.【考点】分段函数的应用.【分析】先确定1<x2<e3,再令y=,求出函数的最大值,即可得出结论.【解答】解:由题意,0<lnx2<3,∴1<x2<e3,又=,故令y=,则y′=,∴x∈(1,e),y′>0,x∈(e,e3),y′<0,∴函数在(1,e)上单调递增,在(e,e3)上单调递减,∴x=e时,函数取得最大值,∴的最大值为.故答案为:.三、解答题:本大题共6小题,满分75分,解答应写出文字说明、证明过程或演算步骤16.已知函数f(x)=2cosx(sinx﹣cosx)+m(m∈R),将y=f(x)的图象向左平移个单位后得到y=g(x)的图象,且y=g(x)在区间[0,]内的最大值为.(Ⅰ)求实数m的值;(Ⅱ)在△ABC中,内角A、B、C的对边分别是a、b、c,若g(B)=l,且a+c=2,求△ABC的周长l的取值范围.【考点】函数y=Asin(ωx+φ)的图象变换;三角函数中的恒等变换应用;余弦定理.【分析】(Ⅰ)利用三角函数的倍角公式进行化简,结合三角函数的单调性进行求解即可求实数m的值;(Ⅱ)根据余弦定理结合基本不等式的关系进行求解.【解答】解:(Ⅰ)f(x)=2cosx(sinx﹣cosx)+m=sin2x﹣cos2x﹣1+m=sin(2x﹣)﹣1+m,∴g(x)=sin[2(x+)﹣]﹣1+m=sin(2x+)﹣1+m,∵x∈[0,],∴2x+∈[,],∴当2x+=时,即x=时,函数g(x)取得最大值+m﹣1=,则m=1.(Ⅱ)∵g(x)=sin(2x+),且g(B)=sin(B+)=l,即sin(B+)=,∵0<B<,∴<B+<,∴当B+=,即B=,∵a+c=2,∴由余弦定理得b2=a2+c2﹣2accosB=a2+c2﹣ac=(a+c)2﹣3ac≥(a+c)2﹣,当且仅当a=c=1时等号成立,又b<a+c=2,∴1≤b<2,∴△ABC的周长l=a+b+c∈[3,4),故△ABC的周长l的取值范围是[3,4).17.如图,AC是圆O的直径,点B在圆O上,∠BAC=30°,BM⊥AC交AC于点M,EA⊥平面ABC,FC∥EA,AC=4,EA=3,FC=1.(1)证明:EM⊥BF;(2)求平面BEF与平面ABC所成的锐二面角的余弦值.【考点】用空间向量求平面间的夹角;直线与平面垂直的性质;二面角的平面角及求法.【分析】(1)根据线面垂直得到线与线垂直,根据直径所对的圆周角是直角,得到两个三角形是等腰直角三角形,有线面垂直得到结果.(2)做出辅助线,延长EF交AC于G,连BG,过C作CH⊥BG,连接FH.,做出∠FHC为平面BEF与平面ABC所成的二面角的平面角,求出平面角.【解答】解:(1)证明:∵EA⊥平面ABC,BM⊂平面ABC,∴EA⊥BM.又∵BM⊥AC,EA∩AC=A,∴BM⊥平面ACFE,而EM⊂平面ACFE,∴BM⊥EM.∵AC是圆O的直径,∴∠ABC=90°.又∵∠BAC=30°,AC=4,∴,AM=3,CM=1.∵EA⊥平面ABC,FC∥EA,∴FC⊥平面ABC.∴△EAM与△FCM都是等腰直角三角形.∴∠EMA=∠FMC=45°.∴∠EMF=90°,即EM⊥MF(也可由勾股定理证得).∵MF∩BM=M,∴EM⊥平面MBF.而BF⊂平面MBF,∴EM⊥BF.(2)延长EF交AC于G,连BG,过C作CH⊥BG,连接FH.由(1)知FC⊥平面ABC,BG⊂平面ABC,∴FC⊥BG.而FC∩CH=C,∴BG⊥平面FCH.∵FH⊂平面FCH,∴FH⊥BG,∴∠FHC为平面BEF与平面ABC所成的二面角的平面角.在Rt△ABC中,∵∠BAC=30°,AC=4,∴,由,得GC=2.∵,又∵△GCH∽△GBM,∴,则.∴△FCH是等腰直角三角形,∠FHC=45°,∴平面BEF与平面ABC所成的锐二面角的余弦值为.18.某学校为了解高三年级学生寒假期间的学习情况,抽取甲、乙两班,调查这两个班的学生在寒假期间每天平均学习的时间(单位:小时),统计结果绘成频率分布直方图(如右图).已知甲、乙两班学生人数相同,甲班学生每天平均学习时间在区间[2,4]的有8人.(1)图中a的值为0.0375;(2)用各组时间的组中值代替各组平均值,估算乙班学生每天学习的平均时长;(3)从甲、乙两个班每天平均学习时间大于10个小时的学生中任取4人参加测试,设4人中甲班学生的人数为ξ,求ξ的分布列和数学期望.【考点】离散型随机变量的期望与方差;频率分布直方图;离散型随机变量及其分布列.【分析】(1)由频率分布直方图的性质得能求出a.(2)由频率分布直方图能估算乙班学生每天学习的平均时长.(3)由甲班学习时间在区间[2,4]的有8人,甲、乙两班学生人数相同,求出甲、乙两班学生人数都为40人,从而得在两班中学习埋单大于10小时的同学共有7人,ξ的所有可能取值为0,1,2,3,分别求出相应的概率,由此能求出ξ的分布列和Eξ.【解答】解:(1)由频率分布直方图的性质得:(a+0.0875+0.1+0.125+0.15)×2=1,解得a=0.0375.(2)由频率分布直方图估算乙班学生每天学习的平均时长为:=3×0.05+5×0.15+7×0.35+9×0.35+11×0.1=7.6.(3)∵甲班学习时间在区间[2,4]的有8人,∴甲班的学生人数为=40,∵甲、乙两班学生人数相同,∴甲、乙两班学生人数都为40人,∴甲班学习时间在区间(10,12]的有40×0.0375×2=3人,乙班学习时间在区间(10,12]的有40×0.05×2=4人,∴在两班中学习埋单大于10小时的同学共有7人,∴ξ的所有可能取值为0,1,2,3,P (ξ=0)==,P (ξ=1)==,P (ξ=2)==,P (ξ=3)==,∴ξE ξ==.19.已知正项数列{a n },若前n 项和S n 满足8S n =a n 2+4a n +3,且a 2是a 1和a 7的等比中项 (1)求数列{a n }的通项公式;(2)符号[x ]表示不超过实数x 的最大整数,记b n =[log 2()],求b 1+b 2+b 3+….【考点】数列的求和;数列递推式. 【分析】(1)由8S n =a n +4a n +3,得,从而得到a n ﹣a n ﹣1=4,(n ≥2,n ∈N ),由此利用a 2是a 1和a 3的等比中项,能求出数列{a n }的通项公式.(2)由b n =[log 2()]=[log 2n ],令S=b 1+b 2+b 3+…,得到S=1×2+2×22+3×23+…+(n ﹣1)×2n ﹣1+n ,由此利用错位相减法能求出b 1+b 2+b 3+….【解答】解:(1)∵正项数列{a n },前n 项和S n 满足8S n =a n +4a n +3,①∴,(n ≥2,n ∈N ),②由①﹣②,得8a n =(a n ﹣a n ﹣1)(a n +a n ﹣1)+4a n ﹣4a n ﹣1, 整理,得(a n ﹣a n ﹣1﹣4)•2a n =0,(n ≥2,n ∈N ), ∵{a n }是正数数列,∴a n +a n ﹣1>0,∴a n ﹣a n ﹣1=4,(n ≥2,n ∈N ), ∴{a n }是公差为4的等差数列, 由8a 1=,得a 1=3或a 1=1,当a 1=3时,a 2=7,a 7=27,不满足a 2是a 1和a 3的等比中项, 当a 1=1时,a 2=5,a 7=25,满足a 2是a 1和a 3的等比中项, ∴a n =1+(n ﹣1)×4=4n ﹣3. (2)∵a n =4n ﹣3,∴b n =[log 2()]=[log 2n ],由符号[]表示不超过实数x 的最大整数,知当2m ≤n ≤2m+1时,[loh 2n ]=m ,∴令S=b 1+b 2+b 3+…=[log 21]+[log 22]+[log 23]+…+[]=0+1+1+2+…+3+…+4+…+n ﹣1+…+n∴S=1×2+2×22+3×23+…+(n ﹣1)×2n ﹣1+n ,③2S=1×22+2×22+3×23+…+(n ﹣1)×2n +2n ,④③﹣④,得﹣S=2+22+23+24+…+2n ﹣1﹣(n ﹣1)•2n ﹣1 =﹣(n ﹣1)•2n ﹣n=(2﹣n )•2n ﹣n ﹣2, ∴S=(n ﹣2)•2n +n+2.20.已知椭圆C 1:的离心率为,焦距为,抛物线C 2:x 2=2py (p >0)的焦点F 是椭圆C 1的顶点.(Ⅰ)求C 1与C 2的标准方程;(Ⅱ)C 1上不同于F 的两点P ,Q 满足,且直线PQ 与C 2相切,求△FPQ 的面积.【考点】椭圆的简单性质.【分析】(I )设椭圆C 1的焦距为2c ,依题意有,,由此能求出椭圆C 1的标准方程;又抛物线C 2:x 2=2py (p >0)开口向上,故F 是椭圆C 1的上顶点,由此能求出抛物线C 2的标准方程. (II )设直线PQ 的方程为y=kx+m ,设P (x 1,y 1),Q (x 2,y 2),则,,联立,得(3k 2+1)x 2+6kmx+3m 2﹣12=0,由此利用根的判别式、韦达定理、弦长公式,结合已知条件能求出△FPQ 的面积.【解答】解:(I )设椭圆C 1的焦距为2c ,依题意有,,解得,b=2,故椭圆C 1的标准方程为.…又抛物线C 2:x 2=2py (p >0)开口向上,故F 是椭圆C 1的上顶点, ∴F (0,2),∴p=4,故抛物线C 2的标准方程为x 2=8y .…(II )由题意得直线PQ 的斜率存在.设直线PQ 的方程为y=kx+m ,设P (x 1,y 1),Q (x 2,y 2),则,,∴,…即(*)联立,消去y整理得,(3k2+1)x2+6kmx+3m2﹣12=0(**).依题意,x1,x2是方程(**)的两根,△=144k2﹣12m2+48>0,∴,,…将x1+x2和x1•x2代入(*)得m2﹣m﹣2=0,解得m=﹣1,(m=2不合题意,应舍去).…联立,消去y整理得,x2﹣8kx+8=0,令△'=64k2﹣32=0,解得.…经检验,,m=﹣1符合要求.此时,,∴.…21.已知函数(a∈R)在其定义域内有两个不同的极值点.(Ⅰ)求a的取值范围;(Ⅱ)记两个极值点分别为x1,x2,且x1<x2.已知λ>0,若不等式恒成立,求λ的范围.【考点】利用导数研究函数的极值;利用导数研究函数的单调性.【分析】(Ⅰ)由导数与极值的关系知可转化为方程f′(x)=lnx﹣ax=0在(0,+∞)有两个不同根;再转化为函数y=lnx与函数y=ax的图象在(0,+∞)上有两个不同交点,或转化为函数与函数y=a 的图象在(0,+∞)上有两个不同交点;或转化为g(x)=lnx﹣ax有两个不同零点,从而讨论求解;(Ⅱ)可化为1+λ<lnx1+λlnx2,结合方程的根知1+λ<ax1+λax2=a(x1+λx2),从而可得;而,从而化简可得,从而可得恒成立;再令,t∈(0,1),从而可得不等式在t∈(0,1)上恒成立,再令,从而利用导数化恒成立问题为最值问题即可.【解答】解:(Ⅰ)由题意知,函数f(x)的定义域为(0,+∞),方程f′(x)=0在(0,+∞)有两个不同根;即方程lnx﹣ax=0在(0,+∞)有两个不同根;(解法一)转化为函数y=lnx与函数y=ax的图象在(0,+∞)上有两个不同交点,如右图.可见,若令过原点且切于函数y=lnx图象的直线斜率为k,只须0<a<k.令切点A(x0,lnx0),故,又,故,解得,x0=e,故,故.(解法二)转化为函数与函数y=a的图象在(0,+∞)上有两个不同交点.又,即0<x<e时,g′(x)>0,x>e时,g′(x)<0,故g(x)在(0,e)上单调增,在(e,+∞)上单调减.=g(e)=;故g(x)极大又g(x)有且只有一个零点是1,且在x→0时,g(x)→﹣∞,在在x→+∞时,g(x)→0,故g(x)的草图如右图,可见,要想函数与函数y=a的图象在(0,+∞)上有两个不同交点,只须.(解法三)令g(x)=lnx﹣ax,从而转化为函数g(x)有两个不同零点,而(x>0),若a≤0,可见g′(x)>0在(0,+∞)上恒成立,所以g(x)在(0,+∞)单调增,此时g(x)不可能有两个不同零点.若a>0,在时,g′(x)>0,在时,g′(x)<0,所以g(x)在上单调增,在上单调减,从而=,又因为在x→0时,g(x)→﹣∞,在在x→+∞时,g(x)→﹣∞,>0,即,所以.于是只须:g(x)极大综上所述,.(Ⅱ)因为等价于1+λ<lnx1+λlnx2.由(Ⅰ)可知x1,x2分别是方程lnx﹣ax=0的两个根,即lnx1=ax1,lnx2=ax2所以原式等价于1+λ<ax1+λax2=a(x1+λx2),因为λ>0,0<x1<x2,所以原式等价于.又由lnx1=ax1,lnx2=ax2作差得,,即.所以原式等价于,因为0<x1<x2,原式恒成立,即恒成立.令,t∈(0,1),则不等式在t∈(0,1)上恒成立.令,又=,当λ2≥1时,可见t∈(0,1)时,h′(t)>0,所以h(t)在t∈(0,1)上单调增,又h(1)=0,h(t)<0在t∈(0,1)恒成立,符合题意.当λ2<1时,可见t∈(0,λ2)时,h′(t)>0,t∈(λ2,1)时h′(t)<0,所以h(t)在t∈(0,λ2)时单调增,在t∈(λ2,1)时单调减,又h(1)=0,所以h(t)在t∈(0,1)上不能恒小于0,不符合题意,舍去.综上所述,若不等式恒成立,只须λ2≥1,又λ>0,所以λ≥1.2016年7月19日21页。
山东省淄博市第七中学2016届高三4月月考英语试题
高三阶段测试英语试题2016.4 说明:本试卷分第I卷(选择题)和第II卷(非选择题),共12页。
满分150分。
考试用时120分钟。
第I卷答案涂写在答题卡上,第II卷在答题纸上做答,考试结束后,考生只交答题卡和答题纸。
第I卷第一部分听力(共两节,满分30分)做题时,先将答案标在试卷上。
录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。
第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. Why hasn’t the woman finished her homework?A. Her hands are injured.B. She has a low fever.C. It is too cold.2. What do you know about the man?A. He is a taxi driver.B. He is an airport clerk.C. He is a computer operator.3. Where has the woman been to recently?A. Germany.B. West Africa.C. France.4. What does the woman want the man to do?A. To get a haircut.B. To do some shopping.C. To attend a party.5. How does the man know the advertisement?A. By calling the manager.B. By visiting the hotel.C. By reading today’s newspaper.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。
精校解析word版---山东省邹城市第一中学高三上学期期中考试英语试题
山东省邹城市第一中学高三年级上学期期中考试英语试题第一部分听力(共两节,满分30分)做题时,先将答案标在试卷上。
录音内容结束后,你将有两分钟的时间将试卷上的答案转写到答题卡上。
第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
第二部分阅读理解(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的四个选项(A、B、C和D)中,选出最佳选项,并在答题卡上将该项涂黑。
AWhether you have a piece of land or a few pots on a balcony, plant a family garden: You’ll all live healthier! Gardening is an easy activity to share and you’ll harvest benefits along with your fresh ve getables, colorful flowers, and aromatic herbs. Even better, you don’t have to wait for your plants to bloom to see those benefits.When parents and kids work together to plant and care for a garden, they can all enjoy these benefits:Physical activity: Gardening and yard work are mild exercise, which we all need every day (for at least 30 minutes). While tending your family garden doesn’t require the intense activity of running or playing singles tennis, it’s still beneficial to your health. For one thing, research shows that once you start gardening, you usually continue for more than 30 minutes. And gardening combines delicate skill strengthening and stretching.Lower stress, better mood: Gardening is excellent stress relief for good reasons: enjoying fresh air and sunlight, performance of relaxing tasks, and even contact with harmless bacteria in the soil that helps release serotonin (血清素) in the brain.Outdoor time: Children prefer spending a lot of time indoors, which can negatively affect their behavior and health. A family garden gets them outside enjoying and experiencing the natural world.1. What does the author think of gardening?A. Too strong workB. Proper exerciseC. Time wasteD. Special skills required2. Which of the following isn’t th e reason for gardening to relieve stress?A. Keeping one’s brain fit.B. Enjoying fresh air.C. Getting farm experienceD. Performing simple and easy act3. What can be learned from the passage?A. Those who have plants on a balcony don’t need a gard en.B. Some bacteria help to build up one’s brain.C. Gardening helps to drive cars.D. Bad behavior results from staying indoors too long.【答案】1. B 2. C 3. B【解析】本文介绍园艺给人带来的好处。
山东省邹城市第一中学高三数学4月模拟考试试题 文
山东省邹城市第一中学2016届高三数学4月模拟考试试题 文本试卷分第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. 注意事项:1.答卷前,考生务必用2B 铅笔和0.5毫米黑色签字笔(中性笔)将姓名、准考证号、考试科目、试卷类型填涂在答题卡规定的位置上.2.第I 卷每小题选出答案后,用2B 铅笔把答题卡上对应的答案标号涂黑;如需改动,用橡皮擦干净后,再选涂其他答案标号.答案不能答在试题卷上.3.第Ⅱ卷必须用0.5毫米黑色签字笔(中性笔)作答,答案必须写在答题卡各题目指定区域内相应的位置,不能写在试题卷上;如需改动,先划掉原来的答案,然后再写上新的答案;不准使用涂改液、胶带纸、修正带.不按以上要求作答的答案无效.第I 卷(选择题 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,只有一项是符合题目要求的. 1.在复平面内,复数31ii--对应的点的坐标为 A . (2,1) B . (1,2)- C . (1,2) D . (2,1)-2. 设平面向量()()1,2,2,m n b =-=u r r,若//m n u r r ,则m n -u r r 等于( )A . 5B . 10C . 13D . 35 3. 设集合⎭⎬⎫⎩⎨⎧<<=16241x xA ,{})3ln(2x x y x B -==,从集合A 中任取一个元素,则这个元素也是集合B 中元素的概率是 A .61 B .31 C .21 D .324.甲几何体(上)与乙几何体(下)的组合体的三视图如下图所示,甲、乙的体积分别为1V 、2V ,则12:V V 等于( )A . 1:4B . 1:3C . 2:3D . 1:π5.函数3log x xy x⋅=的图象可能是A. B. C. D.6. 设()sinf x x x=+()x R∈,则下列说法错误..的是A.()f x是奇函数 B.()f x在R上单调递增C.()f x的值域为R D. ()f x是周期函数7. 执行右图所示的程序框图,输出的x值为A.5B.6C.7D.88. 函数2sin12xyπ=+的部分图象如下图所示,则()2OA OB AB+⋅=u u u r u u u r u u u rA.10- B.5-C.5D.109. 设x,y满足条件1,3,,x yy my x+-⎧⎪⎨⎪⎩≤≤≥若3z x y=+的最大值与最小值的差为7,则实数m=A.32 B.32-C.14 D.14-10.已知函数()1g x x=-,函数()f x满足()()121f x f x+=--,当(]0,1x∈时,()2f x x x=-,对于(]11,2x∀∈,2x R∀∈,则()()()()221212x x f x g x-+-的最小值为A.12B.49128C.81128D.125128第II卷(非选择题共100分)二、填空题:本大题共5小题,每小题5分,共25分.11.某单位为了了解用电量y度与气温x C o之间的关系,随机统计了某4天的用电量与当天气温,并制作了对照表.由表中数据得回归直线方程ˆˆˆybx a =+中ˆ2b =-,预测当气温为4-C o 时,用电量的度数是__12.已知函数()22f x x bx =+的图象在点()()0,0A f 处的切线l 与直线30x y -+=平行,若数列()1f n ⎧⎫⎪⎪⎨⎬⎪⎪⎩⎭的前n 项和为n S ,则2016S = 13.若42log (34)log a b ab +=,则a b +的最小值是 .14.直线0ax by c ++=与圆22:1O x y +=相交于,A B 两点,且3AB =,则OA OB ⋅u u u r u u u r =________15.已知双曲线()222210,0x y a b a b-=>>的半焦距为c ,过右焦点且斜率为1的直线与双曲线的右支交于两点,若抛物线24y cx =的准线被双曲线截得的弦长是2223be (e 为双曲线的离心率),则e 的值为 .三、解答题:本大题共6小题,共75分,解答时应写出必要的文字说明、证明过程或演算步骤. 16.(本小题满分12分)已知(3sin ,2)m x =u r ,2(2cos ,cos )n x x =r ,函数.(1)求函数()f x 的值域;(2)在△ABC 中,角,,A B C 和边,,a b c 满足()2,2,sin 2sin a f A B C ===,求边c . 17.(本小题满分12分)某高中为了选拔学生参加“全国中学生英语能力竞赛(NEPCS )”,先在本校进行初赛(满分150分),若该校有100名学生参加初赛,并根据初赛成绩得到如图所示的频率分布直方图.(1)根据频率分布直方图,计算这100名学生参加初赛成绩的中位数;(2)该校推荐初赛成绩在110分以上的学生代表学校参加竞赛,为了了解情况,在该校推荐参加竞赛的学生中随机抽取2人,求选取的两人的初赛成绩在频率分布直方图中处于不同组的概率.气温(C o ) 1813 101-用电量(度)2434 38 6418.(本小题满分12分)在四棱锥E ABCD -中,底面ABCD 是正方形,AC 与BD 交于点O ,EC ⊥底面ABCD ,G 、F 分别为EO 、EB 中点,且2AB CE =. (Ⅰ)求证://DE 平面ACF ; (Ⅱ)求证:CG ⊥平面BDE ;(Ⅲ)若1AB =,求三棱锥F ACE -的体积.19. (本小题满分12分)已知数列{}n a 是公差不为零的等差数列,12482,,a a a a =,且成等比数列. (Ⅰ)求数列{}n a 的通项;(Ⅱ)设(){}1nn n b a --是等比数列,且257,71b b ==,求数列{}n b 的前n 项和n T .20.(本小题满分13分) 已知函数()ln b f x x ax x =-+,对任意的(0,)x ∈+∞,满足1()()0f x f x+=,其中,a b 为常数.A B D C O E F G(Ⅰ)若()f x 的图像在1x =处的切线经过点(0,5)-,求a 的值;(Ⅱ)已知01a <<,求证2()02a f >;(Ⅲ)当()f x 存在三个不同的零点时,求a 的取值范围. 21.(本小题满分14分)已知椭圆C :)0(12222>>=+b a by a x 的离心率为23,点A 在椭圆C 上.(Ⅰ)求椭圆C 的方程;(Ⅱ)设动直线l 与椭圆C 有且仅有一个公共点,判断是否存在以原点O 为圆心的圆,满足此圆与l 相交两点1P ,2P (两点均不在坐标轴上),且使得直线1OP ,2OP 的斜率之积为定值?若存在,求此圆的方程;若不存在,说明理由.邹城一中高三数学(文)试题参考答案一.选择题 ADCBB,DBDCB 二.填空题11.68 12.20162017 13. 14. 12- 15. 2三.解答题16.解:(I )()2cos 2cos f x m n x x x =⋅=+u r r 2cos 21x x =++2sin(2)16x π=++.........................3分1sin(2)16x π-≤+≤Q ,则函数()f x 的值域为[]1,3-;. ........................5分(II )()2sin(2)126f A A π=++=Q ,1sin(2)62A π∴+=,......6分又132666A πππ<+<,5266A ππ∴+=,则3A π=,.........................8分由sin 2sin B C =得2b c =,已知2a =,.........................10分 由余弦定理2222cos a b c bc A =+-得233c =..........................12分 17.(I )设初赛成绩的中位数为x ,则:()()0.0010.0040.009200.02700.5x ++⨯+⨯-=.........................3分解得81x =,所以初赛成绩的中位数为81;..... ....................5分(II )该校学生的初赛分数在[)110,130有4人,分别记为A ,B ,C ,D ,分数在[)130,150有2人,分别记为a ,b ,在则6人中随机选取2人,总的事件有(A ,B ),(A ,C ),(A ,D ), (A ,a ),(A ,b ),(B ,C ),(B ,D ),(B ,a ),(B ,b ),(C ,D ),(C ,a ),(C ,b ),(D ,a ),(D ,b ),(a ,b )共15个基本事件,. ..................9分 其中符合题设条件的基本事件有8个..............11分故选取的这两人的初赛成绩在频率分布直方图中处于不同组的概率为815P =..........12分18.(本小题满分12分) (Ⅰ)证明: 连结OF ,在正方形ABCD 中,AC 与BD 交于点O ,则O 为BD 的中点,又∵F 是EB 中点,∴OF 是BDE ∆的中位线,∴//OF DE , ………………2分 ∵DE ⊄平面ACF ,OF ⊂平面ACF ,∴//DE 平面ACF ; ………………4分 (Ⅱ)证明∵EC ⊥底面ABCD , BD ⊂平面ABCD , ∴EC BD ⊥,∵BD AC ⊥,且AC CE C =I ,∴BD ⊥平面ACE ,∵CG ⊂平面ACE ,∴CG BD ⊥, ………………6分在正方形ABCD 中,AC 与BD 交于点O ,且2AB CE =,∴12CO AC CE ==, 在OCE ∆中,G 是EO 中点,∴CG EO ⊥,∵EO BD E =I ,∴CG ⊥平面BDE ; ………………9分(Ⅲ)解:∵1AB =,∴22EC =, ∵F 是EB 中点,且EC ⊥底面ABCD ,AB DC O E F G∴1111112211222362224F ACE B ACE E ABC ABC V V V S CE ---∆===⨯⋅=⨯⨯⨯⨯=…12分 19.试题解析:(Ⅰ)设数列{}n a 的公差为(0)d d ≠12a =Q ,且248,,a a a 成等比数列2(32)(2)(72)d d d ∴+=++ ………………2分解得2d =,故1(1)2n a a n d n =+-= ………………4分(Ⅱ)令(1)nn n n c b a =--,设{}n c 的公比为q257,71,2n b b a n ===Q 22253,81c b a c ∴=-==35227,3cq q c ∴===2123n n n c c q --∴== ………………8分从而13(1)2n nn b n -=+- ………………9分12n n T b b b =+++L011(333)(246(1)2n n n -=++++-+-++-L L当n 为偶数时,3212n n n T +-=当n 为奇数时,3232n n n T --= ………………12分204分8分11分13分21.(Ⅰ)解:由题意,得3c a =,222a b c =+, 又因为点3(1,)A 在椭圆C 上,所以221314ab+=,解得2a =,1b =,3c =,所以椭圆C 的方程为1422=+y x . ………………5分 (Ⅱ)结论:存在符合条件的圆,且此圆的方程为225x y +=. ………………6分 证明如下:假设存在符合条件的圆,并设此圆的方程为222(0)x y r r +=>.当直线l 的斜率存在时,设l 的方程为m kx y +=. ………………7分由方程组22,1,4y kx m x y =+⎧⎪⎨+=⎪⎩ 得0448)14(222=-+++m kmx x k , ………………8分 因为直线l 与椭圆C 有且仅有一个公共点,所以2221(8)4(41)(44)0km k m ∆=-+-=,即2241m k =+. ………………9分由方程组222,,y kx m x y r =+⎧⎨+=⎩ 得2222(1)20k x kmx m r +++-=, ………………10分则22222(2)4(1)()0km k m r ∆=-+->.设111(,)P x y ,222(,)P x y ,则12221km x x k -+=+,221221m r x x k -⋅=+, 设直线1OP ,2OP的斜率分别为1k ,2k , 所以221212121212121212()()()y y kx m kx m k x x km x x m k k x x x x x x +++++=== 222222222222222111m r km k km m m r k k k m r m r k --⋅+⋅+-++==--+, ………………12分将2241m k =+代入上式,得221222(4)14(1)r k k k k r -+⋅=+-.要使得12k k 为定值,则224141r r -=-,即25r =,验证符合题意. 所以当圆的方程为225x y +=时,圆与l 的交点12,P P 满足12k k 为定值14-. ………………13分当直线l 的斜率不存在时,由题意知l 的方程为2x =±, 此时,圆225x y +=与l 的交点12,P P 也满足1214k k =-. 综上,当圆的方程为225x y +=时,圆与l 的交点12,P P 满足斜率之积12k k 为定值14-. 21 ………………14分。
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2015-2016学年度高三下学期模拟考试英语2016.4本试卷分第I卷(选择题)和第II卷(非选择题)两部分。
考试结束后,将本试卷和答案卡一并交回。
注意事项:1.答第I卷前考生务必将自己的姓名、准考证号填写在答题卡上。
2.选出每小题答案前,用2B铅笔把答题卡上对应题目的答案标号涂黑。
如需改动, 用橡皮擦干净后, 再选涂其他答案标号框, 不能答在本试卷上,否则无效。
第一部分听力(共两节,满分30分)做题时,先将答案标在试卷上。
录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。
第一节(共5个小题:每小题1. 5分)听下面5段对话。
每段对话后有一道小题,从每题所给的A B C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. How much did they pay for the repair of the bike?A. 25 yuan.B. 50 yuan.C. 100 yuan.2. When did Henry probably leave home?A. At 9:00.B. At 9:30.C. At 10:00.3. What is the man trying to do?A. To make an apology.B. To make a decision.C. To make a suggestion.4. When did the alarm go off?A. At 8:30.B. At 8:00.C. At 7:30.5. Who was the best among the three?A. The man speaker.B. The woman speaker.C. Linda.第二节(共15小题;每小题1.5分)听下面5段对话或独白。
每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听第6段材料,回答第6和第8题。
6. Why will the woman go to Paris?A. To visit a friend.B. To do business.C. To gosightseeing.7. When will the speakers meet?A. On Tuesday.B. On Wednesday.C. On Thursday.8. How will the woman inform Steve?A. By phone.B. By e-mail.C. By fax.听第7 段材料,回答第9和第11题。
9. When does the man usually get to the bus stop from his house?A. At 7:20.B. At 6:40.C. At 7:00.10. What does the man often do after dinner firstly?A. Watching TV or talking with his family.B. Playing and reading with his family.C. Working on his website.11. Which of the following is NOT true according to the conversation?A. The man often gets off work at about 5 o’clock.B. The man often goes to bed after one or two at night.C. The man created his website in his office.听第8段材料,回答第12至14题。
12. Where are the two speakers?A. In a tailor’s shop.B. In a clothing shop.C. In a supermarket.13. Which of the following statements is easy to change according to the woman?A. Style.B. Value.C. Quality.14. What color has the man finally chosen?A. Dark grey.B. Light grey.C. Navy blue.听第9段材料,回答第15至17题。
15. What is the man’s office number?A. 646-49966.B. 646-42266.C. 633-25646.16. What does the man book?A. Three bottles of red wine.B. A table in the general hall.C. A private room without a KTV.17. When will the man get to the restaurant?A. About 7:30 p.m. on Friday.B. About 7 p.m. on Saturday.C. About 7:30 p.m. on Saturday.听第10段独白,回答第18至20题。
18. How long did it take Tom to find his parents by the Internet?A. Only one day.B. About one year.C. 26 years.19. When did Tom meet his mother Silvia?A. On New Year’s Day.B. On Teachers’ Day.C. On Mother’s Day.20. What can we learn from the text?A. Tom’s birth information.B. Where Tom’s father lives.C. Why Tom’s parents left him.第二部分阅读理解(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的四个选项(A、B、C和D)中,选出最佳选项,并在答题卡上将该项涂黑。
A.us National Nature Reserve has had an eventful life! From its days near theNote:* All under 18's must be accompanied by an adult* Pets allowed: No* Cost: Free* Phone: 01674830736* Location: SNH Visitor Centre, St Cyrus National Nature Reserve21.Which of the following events may help you learn about the local people's way of life?A. Ice & Fire.B. Scary Stories.C. The Sands of Time.D. Between the Tides22.The underlined word “spooky” in the text probably means ________.A. interestingB. funnyC. realD. frightening23.What do the four events have in common?A. They are held in August.B. People can't take their pets with them.C. They are held in the evening.D. People need to be guided during the journey.24.According to the text, ________.A. you need to pay to join in the activitiesB. St Cyrus National Nature Reserve is famous for salmonC. a tenyearold child can attend these events only in adults' companyD. St Cyrus National Nature Reserve was set up not long agoB.Spring is coming and it is time for those about to graduate to look for jobs. Competition is tough, so job seekers must carefully consider their perso nal choices. Whatever we are wearing,our family and friends may accept us, but the workplace may not.A high school newspaper editor said it is unfair for companies to discourage visible tattoos (纹身),nose rings, or certain dress styles. It is true you can’t judge a book by its cover, yet peo ple do “cover” themselves in order to convey (传递)certain messages. What we wear, including tattoos and nose rings, is an expression of who we are. Just as people convey messages about themselves with their appearances, so do companies. Dress standards exist in the business world for a number of reasons, but the main concern is often about what custome rs accept.Others may say how to dress is a matter of personal freedom, but for businesses it is more about whether to make or lose money. Most employers do care about the personal appearances of their employees (雇员), because those people represent the companies to their customers.As a hiring manager I am paid to choose the people who would make the best impression on our customers. There are plenty of well-qualified candidates, so it is not wrong to reject someone who might disappoint my customers. Even though I am open-minded, I can’t expect all our customers are.There is nobody to blame but yourself if your set of choices does not match that of your preferred employer. No company should have to change to satisfy a candidate simply because he or she is unwilling to respect its standards, as long as its standards are legal.25.Which of the following is the newspaper editor’s opin ion according to Paragraph 2?A. People’s appearances carry messages about themselves.B. Customers’ choices influence dress standards in companies.C. Candidates with tattoos or nose rings should be fairly treated.D. Strange dress styles should no t be encouraged in the workplace.26.What can be inferred from the text?A. What to wear is not a matter of personal choice for companies.B. Hiring managers make the best impression on their candidates.C. Candidates must wear what companies prefer for an interview.D. Companies sometimes have to change to respect their candidates.27.Which of the following would be the best title for the text?A. Hiring Managers MatterB. Personal Choices MatterC. Appearances MatterD. Employees Matter28.The author’s attitude towards strange dress styles in the workplace may best be described as _________.A. enthusiasticB. positiveC. sympatheticD. negative C.“Mama, when I grow up, I’m going to be one of those!” I said this after seeing the Capital Dancing Company perform when I was three. It was the first time that my dream had taken on a vivid form and acted as something important to start my training. As I grew older and was exposed to more, my interests in the world of dance certainly varied but that little girl’s dream of someday becoming a dancer in the company never left me. In the summer of 2005 when I was 18, I received the phone call which made that dream a reality: I became a member of the company dating back to 1925.As I look back on that day now, it surely lacks any sense of reality. I believe I stayed in a state of pleasant disbelief until I was halfway through rehearsals (排练) on my first day. I never actually expect to get the job. After being offered the position, I was completely astonished. I remember shaking with excitement.Though I was absolutely thrilled with the chance, it did not come without its fair share of challenge. Through the strict rehearsal period of dancing six days a week, I found it vital to pick up the material fast with every last bit of concentration. It is that extreme attention to detail (细节) and stress on practice that set us apart. To then follow those high-energy rehearsals with a busy show schedule of up to five performances a day, I discovered a new meaning of the words “hard work.” What I thought were my physical boundaries were pushed much further than I thought possible. I learned to make each performance better than the last.Today, when I look at the unbelievable company that I have the great honor of being a part of, not only as a member, but as a dance captain, I see a tradition that has inspired not only generations of little girls but a splendid company that continues to develop and grow-and inspires people every day to follow their dreams.29. How many years has the Capital Dancing Company existed when the author receivedthe phone call from it?A. 180B. 1925C. 80D. 200530. How did the author feel when she look back on that day now?A. strangeB. indifferentC. unrealisticD. lucky31. Which of the following statement can best interpret the underlined sentence inparagraph 3?A. Though I was excited, I should share the chance with others.B. Though I was excited, I know clearly where there is chance there is challenge.C. Though I was excited, it’s a challenge for me to share the chance with others.D. Though I was excited, it’s fair to share the chance when there is challenge.D.There is some unwelcome news for students preparing for exams and officers putting in long hours-----you don't need the break as much as you may think that makes you feelless tired.Scientists have long assumed that willpower (意志力) is a limited resource, which is why you feel the need to have a rest, have a snack and come back to a task when you're feeling better. They argued that the only way to restore willpower was by rest, foo d or entertainment.But psychologists have challenged this theory, saying weak willpower is all in your head. They found that people's beliefs in willpower determine how long and how well they'll be able to work on a tough mental exercise. "If you think of willpower as something that's limited, you're more likely to be tired when you perform a difficult task," said Prof. Veronika Job. "'But if you think of willpower as something that is not easily used up, you can go on and on."The researchers designed four experiments to test students' beliefs in willpower. After a tiring task, those, who believed or were led to believe that willpower is a limited resource, performed worse on standard concentration tests than those who thought of willpower as something they had more control over. They also found that leading up to final exam week, students who believed the limited resource theory ate junk food 24 percent more often than those who believed they had more control in resisting temptation (诱惑).Mr. Job said. "The theory that willpower is a limited resource is interesting, but it has had unintended consequences. Students who may already have trouble studying are being told that their power of concentration is limited, and they need to take frequent breaks. But a belief in willpower as a non-limited resource makes people stronger in their ability to work through challenges.'"The findings could help people who are battling tem ptation. Willpower isn't driven by a biologically based process as much as we used to think. The belief in it is what influences your behavior.32. The theory that willpower is limited supports that _________.A. people must eat snacks when they feel tiredB. t here’s no way to strengthen people’s willpowerC. people do need a break to restore their willpowerD. weak willpower doesn’t affect people’s life much33. What have the scientists long believed regarding willpower?A. It is in the charge of people.B. There is no way to restorewillpower.C. It is a limited resource.D. It doesn’t easily run out.34. Which of the following best helps the students to prepare better for their exams?A. Push themselves even if they want to take a break.B. Don’t eat fast food while studying.C. Stay in a comfortable and quiet place.D. Remind themselves willpower is not limited.35. What does the passage want to tell us?A. How to build strong willpowerB. A new theory about willpowerC. The great influence of willpowerD. Willpower doesn’t last long 第二节:(共5小题;每小题2分,满分10分)根据短文的内容,从短文后的选项中选出能填入空白处的最佳选项,选项中有两项为多余选项High school can be quite the stressful time for any student.There are numerous stressesto deal with and the pressure can be more intense as you enter your senior year.36.The answer is as follows.37. Adjust your approach to your specific situation.For instance, if you’re active in your community and your school and have a large family, you may feel depressed by having all these people involved in your life on a daily basis. 38. You just may need a moment to be alone and collect your thoughts before moving on to the challenges that face you.39. If your stress pers ists and you can’t figure out a way to handle it, you may want to try speaking to your school counselor(顾问).If you don’t feel comfortable speaking to your counselor, try getting some other types of counseling.Realize your limits, and plan around them.Don’t take on more than you can handle.If you take on too many things, you will be spread too thin and won’t be able to perform at your best in anything. 40. You will definitely feel more at ease!A.Seek professional counseling.B.Why does the stress come into being?C.Try thinking of alternative ways to deal with stress.D.When you feel relieved, you can have a happier life.E.Taking some time out to be alone may be the best way to handle such stress.F.But how on earth can you reduce some of the stress?G.Evaluate what tasks and activities are most important and leave others behind.第三部分英语知识运用(共三节,满分45分)第一节完形填空(共20小题;每题1. 5分,满分30分)阅读下面短文,从短文后各题所给的四个选项(A, B, C和D)中,选出可以填入空白处的最佳选项,并在答题卡上将该项涂黑。