2020届高三专题复习雅礼试卷讲评
2020年长沙市雅礼实验中学高三语文期末试卷及答案解析

2020年长沙市雅礼实验中学高三语文期末试卷及答案解析一、现代文阅读(36分)(一)现代文阅读I(9分)阅读下面的文字,完成下面小题。
路,低着头陈宝全我家存着的一张黑白照片,大约是我两岁的时候拍的。
灰白的背景里,我和姐姐像那个年月营养不良的麦捆,松松垮垮地挨在-起,眼睛里充满了胆怯和对眼下生活的懵懂。
身后的一棵小树,瘦弱得分辨不出季节。
一条狭窄的村道从眼前逶迤而过,看不见来处,也不知道伸向何方。
后来,我沿着这条路去了更远的地方求学、生活。
而我姐姐留了下来,嫁给路旁一户李姓人家。
而今我似乎明白,那些路,是村庄的符号,更是生活变迁的标识。
年轻的时候,对这些村庄的路不屑一顾,确定它们对我没有多少意义。
双脚带风,走到哪里,那里就是宽阔的大道。
人到上有老下有小的中年,再看村子里的路,心头有点酸楚:它们,大路拉着小路,就像大人牵着孩子。
有了这些路,村庄与外面的世界才慢慢融合在一起。
而仍然是这些路,又成为村庄的精神胎记。
我家后囤子有条不足一米宽的小路,是父亲开的。
我家在村庄的最北端,属于村庄的偏僻之地,这条路走的人自然也少。
只有父亲和母亲经常到菜园子里春播秋收。
我最喜欢这条路的春天,草木疯长,路边开满了野花,迎风微笑,蜜蜂嗡嗡叫,过着蜜汁的生活。
我站着,路就躺下,我躺下,路就直直地立了起来,这么反复间,我才不觉得累。
我三岁时,父亲刚刚从老宅分了出来,新院子就在后囤子旁边,但院墙还没有打起来,建起的两间房子尚没有门窗。
擅长木匠的父亲便一个人在房子里做着门窗。
好几天里,我经常提着瓦罐,沿着这条路给父亲送饭。
有一天,下着小雨,路滑滑的,我一脚没踩稳就顺着陡坡往下滚,父亲一眼看见了,惊叫着从路的另一头跑了过来。
可惜的是,瓦罐未碎饭却倒了个精光,父亲没有责怪我,只顾看我受伤了没有。
时至今日,我仍然记着一条路给我的人间温暖。
有一条年代久远的路横贯村庄,把村庄一分为二。
远看,它更像一条曲转的河流。
原先,这条路上没有贼匪,没有黑社会,没有飞驰的大车,人们走得跟羊一样自在。
2020年长沙市雅礼雨花中学高三英语第一次联考试题及答案解析

2020年长沙市雅礼雨花中学高三英语第一次联考试题及答案解析第一部分阅读(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A、B、C、D四个选项中选出最佳选项AThere have been many great painters in the rich history of Chinese art. Here are four of the greatest painters from China.Li Cheng (919—967, Five Dynasties and early Song Dynasty)Li Cheng contributed greatly to one of the golden ages of landscape paintings in world history. During his time, he was considered the best landscape painter ever. He is remembered especially for the winter landscapes he created and for simple compositions of tall, old evergreens set against a dry landscape. Several of his paintings are in thin ink which gives them a foggy appearance.Fan Kuan (990—1020 , Song Dynasty)Fan Kuan began his career by modeling Li Cheng's work but later created his own style, claiming that the only true teacher was nature. His finest workTravelers among Mountains and Streamsis a masterpiece of landscape painting and many future artists turned to it for inspiration.Qi Baishi (1864-1957)One of the greatest contemporary Chinese painters, Qi Baishi is known for not being influenced by Western styles like most painters of his time. He can be considered as the last great traditional painter of China. He painted almost everything from insects to landscapes. He is regarded highly in Chinese art for the freshness that he brought to the familiar types of birds and flowers, insects and grass.Wu Guanzhong (1919—2010)Widely considered as the founder of modern Chinese painting , Wu Guanzhong has painted various aspects of China, like its architecture, plants, animals, people and landscapes. Wu went on to combine Western and Chinese styles to create a unique form of modem art. In 1992, he became the first living Chinese artist whose work was exhibited at the British Museum.1.What do we know about Li Cheng?A.He loved landscape paintings.B.He copied many artists' work.C.His work gained worldwide recognition.D.He was considered as Fan Kuan's teacher.2.What is the main feature of Qi Baishi's paintings?A.They have foggy appearances.B.They lack diversity in the theme.C.They come under Western influence.D.They show advanced traditional painting skills.3.What did the four Chinese painters have in common?A.They were all modern painters.B.They all created landscape paintings.C.They were all impacted by Western art.D.They were all pioneers intraditional art history.BIn recent years, with the development of technology, it is common to see robots into our homes in the form of toys and vacuums (吸尘器) without question. Children’s toys that rely on robotics for both entertainment and education are becoming more popular and more easily accessible. Robot vacuums, too, are so popular that the Roomba has even earned itself a name in popular television.A selection of other household wares can be purchased and owned for a reasonable cost, but they all look and perform like vacuums. Our domestic helpers are currently designed to vacuuming, mopping, sweeping and mowing.Of course the best known of these is the Roomba, but there is actually some competition amongst the autonomous vacuum manufacturers. Roomba, made by iRobot, now comes in several designs and has been through many versions of improvements. The basic model is just about $400, but there’s a more attractive version, complete with a wireless command center to control the robot from anywhere in the home. Samsung also has a vacuum, called the Smart Tango, which makes increased improvements on the Roomba by including brushes for cleaning corners. It’s possible that some of the less well-known vacuums might be even more exciting, like the Neato Robotics XV that takes on a square shape to better clean corners. Incremental (增值的) improvements are what drive the home robot industry.Now that domestic robots are becoming more popular, the near future should be exciting. If we are going to welcome robots into our lives and our homes, we should expect them to work for it, and work hard. It’s the dawn of the robot revolution, whether we like it or not, and it would be great to see that technology put to use in moreinnovative(创新的) ways.4. What does Roomba in the passage mean?A. A man’s name who has bought a domestic robot.B. A name of a big company who makes robots.C. A type of vacuum whose name is Roomba.D. A place where people can buy proper robots.5. The Smart Tango is different from other types in that ________.A. it has a wireless command centerB. it has brushes for cleaning cornersC. it is more famous and excitingD. it is driven without electricity6. What can we infer according to the last paragraph?A. Domestic robots will free us from housework.B. Domestic robots will become cheaper and cheaper.C. Domestic robots have brought environmental revolution.D. Domestic robots takes the lead in science and technology.7. What does the author intend to tell in the passage?A. Domestic robots are closely related to our daily life.B. It is interesting to see domestic robots at home.C. Domestic robots are too expensive for most people.D. It is convenient to buy a domestic robot.CFrom Mozart to Metallica, tons of people enjoy listening to various types of music while they paint, write, or draw. Most believe that music helps increase creativity, but an international study conducted by English and Swedish researchers is challenging that view. The study results wereechoedby scientists fromLancasterUniversity, and theUniversityofGavle, saying their findings show music actually weakens creativity.To reach their conclusions, researchers had volunteers complete verbal problems designed to inspire creativity while sitting in a quiet room, and then again while music played in the background. They found that background music significantly weakened the volunteers’ ability to complete tasks connected with verbal creativity. The team also tested background noises like those commonly heard in a library, but found that such noises had no influence on subjects’ creativity.The tasks were simple word games. For example, volunteers were given three words, such as dress, rise, and flower. Then, they were asked to find a single word connected with all three that could be combined to form acommon phrase or word. The single word, in this case, would be “sun” (sundress, sunrise, sunflower). Volunteers completed the tasks in either a quiet room, or while listening to two different types of music: rock music or light music“We found strong evidence of weakened performance when playing background music in comparison to quiet background conditions,” says co-author D. Neil McLatchie ofLancasterUniversity. He and his colleagues find that music negatively influences the verbal working memory processes of the brain, preventing creativity. Also, as far as the library background noises having seemingly no effect, the study’s authors believe that was the case because library noises create a “regular state” environment that doesn’t affect concentration.“To conclude, the findings here challenge the popular view that music increases creativity, and instead show that music, whatever type it is, is always a disadvantage for creative performance in problem solving,” the study reads.8. The underlined word in Paragraph 1 can be replaced by ________.A. challengedB. acceptedC. doubtedD. heard9. What were the volunteers asked to do in the study?A. To play music.B. To combine given words.C. To create new words.D. To connect words with music.10. What can we infer from the study?A. Quiet background inspires creativity best.B. Library noise does no harm to creativity.C. Music has a bad effect on language ability.D. Music types matter in creative performance.11. Which is the main idea of the passage?A. Quiet Environments Prevent Concentration.B. Background NoiseAffects Concentration.C. Composing Music Weakens Creativity.D. Listening to Music Reduces Creativity.DI was at my parent’s dinner table. Before me was a worn journal of thin and discolor1 ed pages. It was my grandfather’s journal and now belonged to my father. My grandfather had passed away in the months leading up to my birth. I never got to visit the places he had frequented and the people who had been a part of his life’s journey.I was now about to enter his world, through the words he had left behind. Within minutes, I wascaptivatedby the power of the written words. In the magical script (笔迹) before me, I was transported to another age when food was an everyday art, planned, prepared and enjoyed in the company of others, and a time when people hadthe heart to pause their own lives to embrace (拥抱) each other’s struggles. All this was conveyed to me in the beauty of the words that flowed together to connect with the writer’s mind and understand the world they lived in.That kind of writing seems to be lost on us today. We have gotten used to writing in bite-sized pieces for a public looking for entertainment, and hungry for information. No wonder, there are nearly 200 million bloggers on the Internet and a new blog is created somewhere in the world every half a second. Instead of adding to our collective wisdom, most of these writings reflectthe superficiality (肤浅) and impatience of our day and age.This not only robs us of the skill of writing impressive essays, it also prevents us from exploring what is indeed important. Writing humbles (使谦卑) us in a way that is vital for our character growth, by reminding us about the limits of the self and our appropriate place in the vast flow of life. Writing frees us by helping us explore the unknown so that we really open up to magic of the world around us. I saw all of this in the writing of my grandfather. And I’ve seen it again and again in the writings of the greatest thinkers of humanity. Their writing reflect deep thought on issues of human importance.12. The underlined word “captivated” in the second paragraph can be replaced by “________”.A. puzzledB. frightenedC. attractedD. defeated13. In the author’s grandfather’s age, people ________.A. lived a hard lifeB. cared about each otherC. were fond of writingD. treated food as an art14. The author begins the text with her grandfather’s journal in order to ________.A. show her respect to her grandfatherB. present the importance of good writingC. express her interest in reading as well as writingD. raise the problems with today’s writing15. In the last paragraph, the author is trying to _________.A. discuss what good writing is likeB. express her strong desire to learn writing skillsC. stress the effects of her grandfather’s journal on herD. show her admiration for her grandfather’s writing第二节(共5小题;每小题2分,满分10分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。
2020年长沙市雅礼实验中学高三语文期末试卷及答案解析

2020年长沙市雅礼实验中学高三语文期末试卷及答案解析一、现代文阅读(36分)(一)现代文阅读I(9分)阅读下面的文字,完成下面小题。
动画是想象力艺术,通过离奇的设定,变幻莫测的人物、场景和道具,把观众带入一个想象的世界。
不论这个世界多么神奇,动画最终传递的是照进现实的力量,作用于真实的社会生活。
近年来,国产动画在制作上取得了长足进步,但如何让想象力有效地照进现实,给予观众正向的价值,仍有较大提升空间。
优秀的动画片总会唤起人们对生活的热爱。
阿达的《三个和尚》,在充满中国韵味的画面中讲述团结协作的故事。
宫崎骏的《千与千寻》,观众随着千寻的脚步在神秘世界冒险,收获的不仅是放飞想象力的畅快,更是对勤劳、真诚、善良的认可。
随着网络动画的兴起,近年来涌现了一批适合成年人观看的国产动画。
这对拓宽动画受众群、拓展动画市场是好事,但一些成人动画为了抓人眼球,把焦点放在“爽”的观感上,而非“美”的价值。
在一些修仙题材动画中,主人公不用怎么努力就可以成仙,虚构的是实力至上、强者为尊的世界,崇尚的是弱肉强食的价值观,不仅与社会主义核心价值观背道而驰,与以和为贵的中华文化传统理念也大相径庭。
精心塑造外形漂亮的主人公,着力描绘酷炫的打斗场面,想象力没问题,但这样的动画很难让观众带着正能量回到现实中。
国外的一项调查表明,低幼儿童会显性地受到动漫人物行为的影响,并且效仿这些行为。
4岁左右的儿童可能并不明白为什么这么做,但会单纯地把影片中主人公的行为拿到现实中尝试。
这就是有的动画情节具有危险性的原因。
我们的创作要格外留意何种行为会被儿童带回现实。
与童话和幻想题材动画相比,现实题材取材自现实生活,人物场景距离观众很近,小观众受到的影响是直观直接的。
外国动画中有不少运动、校园、职场等现实题材作品,在日本电视动画中,现实题材占比大约为1/5。
中国动画中现实题材的内容相对较少,电视动画中的比例在5%上下,网络动画则更低,但也能看到不少亮色。
湖南雅礼中学2020届高三月考试卷(六)文科数学试题答案解析与点睛(18页)

湖南雅礼中学2020届高三月考试卷(六)文科数学试题数学(文科)试题注意事项:1. 本试卷分第Ⅰ卷(选择题)和第Ⅰ卷(非选择题)两部分.答卷前,考生务必将自己的姓名、准考证号填写在答题卡上.2. 回答第Ⅰ卷时,选出每小题答案后,用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.写在本试卷上无效.3. 回答第Ⅰ卷时,将答案写在答题卡上,写在本试卷上无效.4. 考试结束后,将本试卷和答题卡一并交回.第Ⅰ卷一、选择题:本大题共12个小题,每小题5分,共60分.在每小题给出的四个选项中,只有一个选项是符合题目要求的.1.集合{}13A x x =<<,集合{}2,B y y x x A ==-∈,则集合A B =I ( ) A. {}13x x << B. {}13x x -<<C. {}11x x -<<D. ∅【答案】D 【解析】 【分析】求出集合B ,利用交集的定义可求得集合A B I .【详解】因为{}13A x x =<<,{}{}2,11B y y x x A y y ==-∈=-<<,所以A B =∅I , 故选:D.【点睛】本题考查交集的计算,考查计算能力,属于基础题. 2.复数12z i =-的虚部为( ) A. 2i B. 2i -C. 2D. -2【答案】D 【解析】 【分析】根据复数的概念可知复数12z i =-的虚部.【详解】形如(,)a bi a R b R +∈∈的数叫做复数,a 和b 分别叫它的实部和虚部,所以复数12z i =-的虚部为-2. 故选:D.【点睛】考查复数的概念,知识点较为基础.3.已知()f x 是定义在R 上的偶函数,且在(],0-∞上是减函数,设()20.3a f =,()2log 5b f =,()0.32c f =,则,,a b c 的大小关系是() A. b c a << B. a b c <<C. c b a <<D. a c b <<【答案】D 【解析】 【分析】根据偶函数的对称性可知()f x 在[)0,+∞上为增函数;通过临界值比较出自变量的大小关系,根据单调性可得结果.【详解】()f x Q 是R 上的偶函数,且在(],0-∞上为减函数 ()f x ∴在[)0,+∞上为增函数0.30222log 5log 422210.30>=>>=>>Q()()()0.322log 520.3f f f ∴>>,即a c b <<本题正确选项:D【点睛】本题考查根据函数单调性比较函数值大小的问题,关键是能够利用奇偶性的性质得到函数在自变量所在区间内的单调性,通过自变量大小关系的比较得到函数值的大小关系. 4.若实数x ,y 满足x +y >0,则“x >0”是“x 2>y 2”的( ) A. 充分不必要条件 B. 必要不充分条件 C. 充分必要条件 D. 既不充分也不必要条件【答案】B 【解析】 【分析】根据充分条件、必要条件的判定方法,结合不等式的性质,即可求解,得到答案. 【详解】由题意,实数x ,y 满足x +y >0,若x >0,则未必有x 2>y 2, 例如x =1,y =2时,有x 2<y 2;反之,若x 2>y 2,则x 2﹣y 2>0,即(x +y )(x ﹣y )>0; 由于x +y >0,故x ﹣y >0,∴x >y 且x >﹣y ,∴x >0成立;所以当x +y >0时,“x >0”推不出“x 2>y 2”,“x 2>y 2”⇒“x >0”; ∴“x >0”是“x 2>y 2”的必要不充分条件. 答案:B .【点睛】本题主要考查了不等式的性质,以及充分条件、必要条件的判定,其中解答中熟记充分条件、必要条件的判定方法,结合不等式的性质求解是解答的关键,着重考查了推理与论证能力,属于基础题. 5.在长方形ABCD 中,2AB =,1AD =,点E 为BC 的中点,点F 为CD 的中点,则AE BF ⋅=u u u v u u u v( ) A. 1-B. 32-C. 2-D. 52-【答案】B 【解析】 【分析】根据题意,得到12=+=+u u u r u u u r u u u r u u u r u u u r AE AB BE AB AD ,12BF BC CF AD AB =+=-u u u r u u u r u u u r u u u r u u u r,再由向量数量积的运算法则,直接计算,即可得出结果.【详解】因为在长方形ABCD 中,2AB =,1AD =,点E 为BC 的中点,点F 为CD 的中点,所以12=+=+u u u r u u u r u u u r u u u r u u u r AE AB BE AB AD ,12BF BC CF AD AB =+=-u u u r u u u r u u u r u u u r u u u r1122⎛⎫∴⋅=+⋅⎛⎫ ⎪⎝-+ ⎪⎝⎭⎭u u u u u u r u u u r u u u r u r u u u r u u r AE BF A D A B A AB D 2211313222422AB AD AB AD =-++⋅=-+=-u u u r u u u r u u u r u u u r .故选:B【点睛】本题主要考查向量的数量积运算,熟记运算法则即可,属于常考题型.6.一只小虫在边长为2的正方形内部爬行,到各顶点的距离不小于1时为安全区域,则小虫在安全区域内爬行的概率是( ) A. 14π-B.4πC. 16π-D.6π 【答案】A 【解析】 【分析】作出正方形,并作出安全区域,将安全区域的面积与正方形的面积相除可得出所求事件的概率. 【详解】如下图所示,由于小虫到每个顶点的距离不小于1为安全区域,则安全区域为以正方形每个顶点为圆心半径为1的扇形弧以及扇形以外的部分,为图中阴影部分,其面积22214S ππ=⨯-⨯=-,故概率4144P ππ-==-. 故选:A.【点睛】本题为平面区域型几何概率问题,确定事件所围成的区域是解题的关键,考查数形结合思想与计算能力,属于中等题.7.已知函数()2sin()(0,0)f x x ωϕωϕπ=+><<的最小正周期为π,若将()f x 的图象向左平移3π个单位后得到函数()g x 的图象关于y 轴对称,则函数()f x 的图象( ) A. 关于直线2x π=对称B. 关于直线3x π=对称C. 关于点(,0)2π对称 D. 关于点(,0)3π对称【答案】B 【解析】 【详解】由条件知22,w wππ=⇒= 2()2sin(2)()2sin(2())2sin(2)33f x xg x x x ππϕϕϕ=+⇒=++=++ 关于y 轴对称,可得(0)2g =±,可得2,6k k z πϕπ=-+∈ ,0ϕπ<<,所以56πϕ=,故得5()2sin(2)6f x x π=+,当,() 2.3x f x π==-对称中心为:5,0212k k z ππ⎛⎫-∈ ⎪⎝⎭C,D ,均不正确. 故选B.点睛:此题考查的是函数图像的平移和对称,周期性,先根据周期的公式得到2w =, 再根据平移公式得到()g x ,根据轴对称性得到56πϕ=,故得5()2sin(2)6f x x π=+,可以根据选项代入表达式,比如B 选项,可以带入函数判断函数值是否为最值;8.已知实数x ,y 满足521802030x y x y x y +-≤⎧⎪-≥⎨⎪+-≥⎩,若直线10kx y -+=经过该可行域,则实数k 的最大值是( ) A. 1 B.32C. 2D. 3【答案】B 【解析】【分析】先根据约束条件画出可行域,再利用直线20kx y -+=过定点()0,1,再利用k 的几何意义,只需求出直线10kx y -+=过点()2,4B 时,k 值即可. 详解】直线20kx y -+=过定点()0,1, 作可行域如图所示,,由5218020x y x y +-=⎧⎨-=⎩,得()2,4B . 当定点()0,1和B 点连接时,斜率最大,此时413202k -==-, 则k 的最大值为:32故选:B .【点睛】本题主要考查了简单的线性规划,以及利用几何意义求最值,属于基础题. 9.两个等差数列{}n a 和{}n b ,其前n 项和分别为n S ,n T ,且723n n S n T n +=+,则220715a ab b +=+( ) A.49B.378C.7914 D.14924【答案】D 【解析】 【分析】根据等差数列的性质前n 项和的性质进行求解即可. 【详解】因为等差数列{}n a 和{}n b ,所以2201111715111122a a a a b b b b +==+,又211121S a =,211121T b =,【故令21n =有2121721214921324S T ⨯+==+,即1111211492124a b =,所以111114924a b = 故选D.【点睛】本题主要考查等差数列的等和性质:若{}n a 是等差数列,且(,,,*)m n p q m n p q N +=+∈,则m n p q a a a a +=+ 与等差数列{}n a 前n 项和n S 的性质*21(21),()n n S n a n N -=-∈10.已知三个实数2、b 、8成等比数列,则双曲线22219y x b-=的渐近线方程为( )A. 340±=x yB. 430x y ±=20y ±=D. 9160x y ±=【答案】A 【解析】 【分析】根据等比中项的定义求得2b 的值,可得出双曲线的标准方程,进而可求得双曲线的渐近线方程. 【详解】由题意,三个实数2、b 、8成等比数列,可得216b =,即双曲线221916y x -=的渐近线方程为340±=x y ,故选:A.【点睛】本题考查双曲线渐近线方程的求解,解答的关键就是求出双曲线的标准方程,考查计算能力,属于基础题.11.定义在R 上的偶函数()f x 在[)0,∞+单调递增,且()f 21-=,则()f x 21-≤的x 的取值范围是( ) A. []0,4 B. (][),22,∞∞--⋃+ C. (][),04,∞∞-⋃+ D. []2,2-【答案】A 【解析】 【分析】先得()21f =,再根据偶函数化简()21f x -≤,即为()()22f x f -≤,由单调性可得22x -≤,运用绝对值不等式的解法可得x 的取值范围.【详解】定义在R 上偶函数()f x 在[)0,+∞单调递增,且()21f -=,可得()()221f f =-=,()21f x -≤,即为()()22f x f -≤,可得22x -≤, 即222x -≤-≤, 解得04x ≤≤,即x 的取值范围是[]0,4,故选A.【点睛】首先根据函数的性质把不等式转化为(())(())f g x f h x >的形式,然后根据函数的单调性去掉“f ”,转化为具体的不等式(组),此时要注意()g x 与()h x 的取值应在外层函数的定义域内.12.已知函数(),()ln 1xf x e eg x x =-=+,若对于1x ∀∈R ,()20x ∃∈+,∞,使得()()12f x g x =,则12x x -的最大值为( ) A. e B. 1-eC. 1D. 11e-【答案】D 【解析】 【分析】不妨设f(1x )=g(2x )=a ,从而可得12x x -的表达式,求导确定函数的单调性,再求最小值即可. 【详解】不妨设f(1x )=g(2x )=a , ∴1x e e -=21lnx +=a , ∴1x =ln(a+e),2x =1a e -, 故12x x -=ln(a+e)-1a e -,(a >-e ) 令h (a )=ln(a+e)-1a e -,h ′(a )11a e a e-=-+, 易知h ′(a )在(-e ,+∞)上是减函数, 且h ′(0)=0,故h (a )在a 0=处有最大值, 即12x x -的最大值为11e-;故选D .【点睛】本题考查了函数的性质应用及导数的综合应用,考查了指对互化的运算,属于中档题.第Ⅰ卷二、填空题:本大题共4小题,每小题5分,共20分.13.若1sin()43πα-=,则cos()4πα+=__________.【答案】13;【解析】由题意得,1()cos()sin()424443πππππαααα+=--⇒+=-=.14.已知向量a r ,b r的夹角为34π,()3,4,10a a b =-⋅=-v v v ,则b r 的模长是______.【答案】 【解析】 【分析】由平面向量模的运算及数量积的运算得:由向量,的夹角为,=(-3,4),=-10,得=||||cos =-10,即||==2,得解.【详解】由向量,的夹角为,=(-3,4),=-10,得=||||cos=-10,即||==2,故答案为2.【点睛】本题考查了平面向量模的运算及数量积的运算,属中档题.15.直角ABC V 的三个顶点都在球O 的球面上,2AB AC ==,若球O 的表面积为12π,则球心O 到平面ABC 的距离等于__________.【答案】1 【解析】直角ABC V 的斜边CB 为ABC V 所在截面小圆的直径,则该截面小圆的半径为r =12π可得球的半径R =,球心O 到平面ABC 的距离1d ==.16.设(),()f x g x 是定义在R 上两个周期函数,()f x 的周期为4,()g x 的周期为2,且()f x 是奇函数.当2(]0,x ∈时,()f x =(2),01()1,122k x x g x x +<≤⎧⎪=⎨-<≤⎪⎩,其中0k >.若在区间(0]9,上,关于x 的方程()()f x g x =有8个不同的实数根,则k 的取值范围是_____.【答案】1,34⎡⎫⎪⎢⎪⎣⎭. 【解析】 【分析】分别考查函数()f x 和函数()g x 图像的性质,考查临界条件确定k 的取值范围即可. 【详解】当(]0,2x ∈时,()f x =即()2211,0.x y y -+=≥又()f x 为奇函数,其图象关于原点对称,其周期为4,如图,函数()f x 与()g x 的图象,要使()()f x g x =在(]0,9上有8个实根,只需二者图象有8个交点即可.当1g()2x =-时,函数()f x 与()g x 的图象有2个交点; 当g()(2)x k x =+时,()g x 图象为恒过点()2,0-的直线,只需函数()f x 与()g x 的图象有6个交点.当()f x 与()g x 图象相切时,圆心()1,0到直线20kx y k -+=的距离为11=,得4k =,函数()f x 与()g x 的图象有3个交点;当g()(2)x k x =+过点1,1()时,函数()f x 与()g x 的图象有6个交点,此时13k =,得13k =. 综上可知,满足()()f x g x =在(]0,9上有8个实根的k 的取值范围为134⎡⎢⎣⎭,.【点睛】本题考点为参数的取值范围,侧重函数方程的多个实根,难度较大.不能正确画出函数图象的交点的而致误,根据函数的周期性平移图象,找出两个函数图象相切或相交的临界交点个数,从而确定参数的取值范围.三、解答题:本大题共70分.解答应写出文字说明、证明过程或演算步骤.第17~21题为必考题,每个试题考生都必须作答,第22、23题为选考题,考生根据要求作答. (一)必考题:共60分.17.在ABC ∆中,,,a b c 分别为角,,A B C 的对边,且满足274cos cos 2()22A B C -+= (Ⅰ)求角A 的大小;(Ⅱ)若3b c +=,求a 的最小值. 【答案】(Ⅰ)60o A ∴= (Ⅱ)32【解析】(Ⅰ)A B C π++=Q ,2274cos cos 2()2(1cos )cos 22cos 2cos 322A B C A A A A ∴-+=+-=-++=, 212cos 2cos 02A A ∴-+=.1cos 2A ∴=,0A π<<Q ,60o A ∴=.(Ⅱ)由余弦定理222cos 2b c a A bc+-=,得222bc b c a =+-.2229()39393()24b c a b c bc bc +∴=+-=-≥-=, 32a ∴≥.所以a 的最小值为32, 当且仅当32b c ==时取等号.18.2019年2月13日《烟台市全民阅读促进条例》全文发布,旨在保障全民阅读权利,培养全民阅读习惯,提高全民阅读能力,推动文明城市和文化强市建设.某高校为了解条例发布以来全校学生的阅读情况,随机调查了200名学生每周阅读时间X (单位:小时)并绘制如图所示的频率分布直方图.(1)求这200名学生每周阅读时间的样本平均数x和中位数a(a的值精确到0.01);(2)为查找影响学生阅读时间的因素,学校团委决定从每周阅读时间为[6.5,7.5),[7.5,8.5)的学生中抽取9名参加座谈会.(i)你认9个名额应该怎么分配?并说明理由;(ii)座谈中发现9名学生中理工类专业的较多.请根据200名学生的调研数据,填写下面的列联表,并判断是否有95%的把握认为学生阅读时间不足(每周阅读时间不足8.5小时)与“是否理工类专业”有关?附:22()()()()()n ad bcKa b c d a c b d-=++++(n a b c d=+++).临界值表:【答案】(1)平均数9,中位数8.99;(2)(i)按照1:2进行名额分配;理由见详解;(ii )有. 【解析】 【分析】(1)根据平均数,中位数的定义进行求解即可(2)完成列联表,计算2K 的观测值,结合独立性检验的性质进行判断即可. 【详解】(1)该组数据的平均数60.0370.180.290.35100.19x =⨯+⨯+⨯+⨯+⨯110.09120.049+⨯+⨯=,因为0.030.10.20.350.680.5+++=>,所以中位数[8.5,9.5)a ∈, 由0.030.10.2(8.5)0.350.5a +++-⨯=,解得0.50.338.58.990.35a -=+≈;(2)(i )每周阅读时间为[6.5,7.5)的学生中抽取3名,每周阅读时间为[7.5,8.5)的学生中抽取6名. 理由:每周阅读时间为[6.5,7.5)与每周阅读时间为[7.5,8.5)是差异明显的两层,为保持样本结构与总体结构的一致性,提高样本的代表性,宜采用分层抽样的方法抽取样本;因为两者频率分别为0.1,0.2,所以按照1:2进行名额分配.(ii )由频率分布直方图可知,阅读时间不足8.5小时的学生共有200(0.030.10.2)66⨯++=人,超过8.5小时的共有20066134-=人. 于是列联表为:2K 的观测值2200(40742660)4.432 3.84166134100100k ⨯⨯-⨯=≈>⨯⨯⨯, 所以有95%的把握认为学生阅读时间不足与“是否理工类专业”有关.【点睛】本题主要考查独立性检验的应用,根据数据计算出K 2的观测值是解决本题的关键.考查学生的计算能力.19.如图1,等腰梯形ABCD 中,AD BC ∥,2AB AE BE CD ====,4BC ED ==,O 为BE 中点,F 为BC 中点.将ABE △沿BE 折起到A BE 'V 的位置,如图2.(1)证明:CD ⊥平面AOF ';(2)若平面A BE '⊥平面BCDE ,求点F 到平面A EC '的距离.【答案】(1)证明见解析;(2)2. 【解析】 【分析】(1)先证CD EC ⊥,接着证CD OF ⊥,根据已知条件得AO CD '⊥,即可得结论;(2)点F 到平面A EC '的距离转化为点B 到平面A EC '的距离的一半,取A E '的中点记为H ,证明BH ⊥平面A EC ',求出BH ,即可得结论.【详解】(1)EC =222BE EC BC +=,即BE EC ⊥, ∵CD BE P ,∴CD EC ⊥O 为BE 中点,F 为BC 中点.∴OF EC ∥,∴CD OF ⊥∵A B A E ''=,O 为BE 中点,∴AO BE '⊥,∴AO CD '⊥ 而AO OF O '⋂=,∴CD ⊥平面AOF'.(2)OF EC ∥∴点F 到平面AEC 的距离即为点O 到平面A EC '的距离, 即点B 到平面A EC '的距离的一半.取A E '的中点记为H ,连结BH ,则BH A E '⊥∵平面A BE '⊥平面BCDE ,且交线为BE , 由(1)知EC BE ⊥,∴EC ⊥平面A BE ',∴EC BH ⊥, 又EC A E E '⋂=∴BH ⊥平面A EC ',BH = ∴B 到平面A EC '∴点F 到平面A EC '【点睛】本题考查了平面立体转化的问题,运用好折叠之前,之后的图像,考查线面垂直以及点的面的距离,解题的关键是对空间直线与平面的位置关系定理要熟练,属于中档题.20.已知椭圆()2222:10x y C a b a b +=>>过点()2,1,且离心率2e =. (1)求椭圆C 的方程; (2)已知斜率为12的直线l 与椭圆C 交于两个不同点,A B ,点P 的坐标为()2,1,设直线PA 与PB 的傾斜角分别为,αβ,证明:αβπ+=.【答案】(1)22:182x y C +=(2)证明见解析【解析】 【分析】(1)根据题意,由椭圆的几何性质可得,解可得a 、b 的值,将a 、b的值代入椭圆的方程即可得答案;22411a b e ⎧+=⎪⎪⎨⎪==⎪⎩(2)证明αβπ+=即证明直线PA 与PB 的斜率120k k +=,根据题意,设直线1:2l y x m =+,联立直线与椭圆的方程,将韦达定理代入1211221122y x k k y x +--+=--变形即可证明.【详解】()1由题意得224112a b e ⎧+=⎪⎪⎨⎪==⎪⎩解得228,2a b ==,所以椭圆的方程为:22:182x y C += ()2设直线1:2l y x m =+,由2212182y x m x y ⎧=+⎪⎪⎨⎪+=⎪⎩,消去y 得,222240x mx m ++-=,2248160m m =-+>V解得22m -<<,当0m =时,12y x =(舍) 设()()1122,,,A x y B x y ,则212122,24x x m x x m +=-=-g 由题意,易知PA 与PB 的斜率存在,所以,2παβ≠,设直线PA 与PB 的斜率分别为12,k k则1tan k α=,2tan k β=,要证αβπ+=,即证()tan tan tan απββ=-=-,只需证120k k +=12121211,,22y y k k x x --==--Q 故()()()()()()1221121122121212112222y x y x y y x x x x k k --+----+=-=---+又111,2y x m =+2212y x m =+所以()()()()12211212y x y x --+--=()()122111121222x m x x m x ⎛⎫⎛⎫+--++-- ⎪ ⎪⎝⎭⎝⎭, ()()()1212241x x m x x m =+-+--g ()()()2122422410x x m m m m =-+----=g120,k k ∴+=αβπ+=【点睛】本题考查椭圆的方程,考查直线与椭圆的位置关系。
2020届湖南省长沙市雅礼中学高三下学期第七次月考数学(文)试卷及解析

2020届湖南省长沙市雅礼中学高三下学期第七次月考数学(文)试卷★祝考试顺利★(解析版)一、选择题:本大题共12小题,每小题5分,共60分.每小题所给的四个选项中只有一个是正确的.1.设集合{}|lg A y y x ==,集合{|B x y ==,则A B =( )A. []0,1B. (]0,1C. [)0,+∞D. (],1-∞ 【答案】D【解析】∵{}|lg =A y y x R ==,{(]|=1B x y ==-∞,,∴(],1A B ⋂=-∞,故选D. 2.已知(,)a bi a b R +∈是11i i -+的共轭复数,则a b +=( ) A. 1- B. 12- C. 12 D. 1 【答案】D【解析】 首先计算11i i-+,然后利用共轭复数的特征计算,a b 的值. 【详解】21(1)21(1)(1)2i i i i i i i ---===-++-, ()a bi i i ∴+=--=,0,1,1a b a b ∴==∴+=.故选:D.3.空气质量指数AQI 是反映空气质量状况的指数,AQI 指数值越小,表明空气质量越好,其对应关系如表:空气质量优良轻度污染中度污染重度污染严重污染如图是某市10月1日-20日AQI指数变化趋势:下列叙述错误的是()A. 这20天中AQI指数值的中位数略高于100B. 这20天中的中度污染及以上的天数占14C. 该市10月的前半个月的空气质量越来越好D. 总体来说,该市10月上旬的空气质量比中旬的空气质量好【答案】C【解析】根据所给图象,结合中位数的定义、AQI指数与污染程度的关系以及古典概型概率公式,对四个选项逐一判断即可.【详解】对A,因为第10天与第11天AQI指数值都略高100,所以中位数略高于100,正确;对B,中度污染及以上的有第11,13,14,15,17天,共5天占14,正确;对C,由图知,前半个月中,前4天的空气质量越来越好,后11天该市的空气质量越来越差,错误;对D,由图知,10月上旬大部分AQI指数在100以下,10月中旬大部分AQI指数在100以上,所以正确,故选C.4.阅读如图所示的程序框图,运行相应的程序,输出的S的值等于()。
2020届长沙市雅礼中学高三英语第一次联考试题及答案解析

2020届长沙市雅礼中学高三英语第一次联考试题及答案解析第一部分阅读(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A、B、C、D四个选项中选出最佳选项AFilms to watch in MarchGretaA lonely young waitress finds a handbag on aNew Yorksubway train. Luckily, the address is inside, so she returns it to the piano teacher who left it there. She then discovers that the piano teacher makes a habit of dropping bags around the city on purpose to make new friends. Directed by Neil Jordan, the actors are attractive. Grace Moretz is the waitress and Isabelle Huppert is the one who admires her. Thanks to them, Greta winds up being far more enjoyable than it has any right to be.The AftermathThe Aftermath is one of the few World WarⅡ-related films. Based on Rhidian Brook’s novel,this touching romantic movie is starred by Keira Knightley, who is the wife of a British officer. She hates the Germans because her son was killed in an air raid. But is there a chance that a tall, dark andhandsome man might persuade her to overcome her hate?Captain MarvelIt took Marvel Studios a decade to finally make a female superhero. It’s also the first Marvel film to be directed by a woman, who has directed Half Nelson and Missisippi Grind. Captain Marvel is set in the 1990s. The Oscar-winning Brie Larson stars as Carol Danvers, a US Air Force fighter pilot, while Samuel L Jackson, as usual, plays Nick Fury.DumboDumbo is a classic Disney cartoon, who has big ears and there have been plenty of those kind of movies in recent years.Burtonalways has inspirations to direct such kind of movies from Edward Scissorhands to Miss Peregrine’s Home for Peculiar Children. Magical figures have always been his favorite characters.1. What makes Captain Marvel special?A. Winning Oscar.B. Its female director.C. Time setting.D. A woman pilot.2. Which film is better for a 7-year-old kid to watch?A. The AftermathB. Captain MarvelC. GreteD. Dumbo3. Where can you most probably read the text?A. Reader’s DigestB. NatureC. Scientific AmericanD. National GeographicBWilliam had worked hard to pass the exams to enter high school. When the term began, however, his father told him that there was no money to pay for his school fees because of the summer drought. Still, William wanted to learn.He made the village library his school. One day, he found a book calledUsing Energy. On the book’s cover was a picture of windmills, tall steel towers with blades (叶片) spinning like giant fans. He learned that wind—something of which his hometown had plenty—could produce electricity.He couldn’t help picturing his own windmill in mind, but collecting the parts and tools he needed would take months. In a junkyard, he dug through piles of used metal, old cars, and worn-out tractors, searching for anything that might help him construct his machine. He made four-foot-long blades from plastic pipe, which he melted over a fire, flattened out, and hardened with bamboo poles.Earning some money, he paid a blacksmith to attach the piston (活塞) to the pedal sprocket (踏板链轮) of an old bicycle frame. This would be the axle (轮轴) of the windmill. When the wind blew, the spinning blades would turn the bicycle wheel and spin a small dynamo (发电机), donated by his friend.When he had collected all the parts, William began putting them together. He fixed the blades to the tractor fan he found, using washers (垫圈) he’d made from bottle caps. Next he pushed the fan onto the piston joint to the bicycle frame. With the help of his two best friends, Williambuilt a 16-foot-tall tower from trunks of trees and lifted the ninety-pound windmill to the top.The big moments eventually came. He climbed up the tower and connected two wires that held a small bulb. As the wind whipped around him, the blades began to turn, slowly at first, then faster and faster. The light bulb flickered (闪烁), then flashed to life. The crowd cheered from below. “Wachitabwina (well done)!”William’s machine now powered his house. And the story of the boy who’d built a power-generating windmill to rescue his family from the drought spread across the Internet.4. After readingUsing Energy, what did William decide to do?A. Make use of wind.B. Enter a high school.C. Build a village library.D. Learn to survive a drought.5. According to Para.3, which of the following word can best describe William?A. Humorous.B. Determined.C. Cooperative.D. Friendly.6. What can you learn about William’s machine?A. It was built by villagers.B. It worked at the first attempt.C. It took him years to complete it.D. It was made from metal materials.7. What is the besttitle for the passage?A. Winds Of HopeB. Ideas Worth SpreadingC. Learning from ExperienceD. Windmills for VillagesCJon Pedley is making a big change. He is giving up his life as a businessman for a life of helping others. He is trading his beautiful farmhouse in England for life in a mud hut in Uganda, East Africa.Pedley admits that he has notalways led a very positive life. At times he drank too much and got in trouble with the law. “I’ve always put the pursuit of money in front of everything else. As long as I was all right, I didn’t care who I was hurting, ” says Pedley.But a visit to Uganda in 2007 gave Pedley a new outlook on life. He was amazed at what he saw and how much the people there appreciated the work he was doing. “I worked there for a few days and these people who have nothing were thanking me by giving me bags of potatoes, which are a fortune for them,” he said.Now Pedley is selling his business, his $1.5 million farmhouse, and his expensive car — and moving into a hut made of mud and boards in a small Ugandan village. There he will help run an organization that hopes to improve the quality of life for people in the village of Kigazi. He will help to build schoolrooms for children and tanks to hold clean water for villagers. Today, people in Kigazi must walk two miles to a hospital, so Pedley will help to build doctors’ offices, too.Pedley’s organization will also work with English teenagers who are in trouble. The teens will be sent to a “camp” in Uganda that Pedley will run. The teens will live in mud huts and help to build water, health, and education facilities for kids in Kigazi, many of whom have lost their parents to poverty or disease. Pedley hopes the teens will see a side of life that might help them turn around their own lives and set them on a new and more positive path.8. Which of the following best describes Pedley’s life in the past?A. Negative.B. Colorful.C. Independent.D. Selfish.9. What will Pedley do in the small Ugandan village?A. Do business with the local people.B. Help farmers increase potato output.C. Assist villagers with construction work.D. Introduce tools to improve English teaching.10. Why will Pedley work with English teenagers in trouble?A. To encourage them to make friends with locals.B. To inspire them to live a more positive life.C. To train them to become doctors in the future.D. To make them learn about different cultures.11. What is the best title for the text?A. From millionaire to mud hutB. A life-changing adventureC. A rich man becoming homelessD. More money, more worriesDKids often admire well-known celebrities, putting posters of their favorite musicians, movie stars and athletes on their bedroom walls. But rarely does a young person get to meet or talk to their idol. Yet for one young tennis player - Coco Gauff - her chance to do just that happened in an amazing way!Coco was born on March 13, 2004. At the age of 4, she developed an interest in tennis after watching Venus Williams win the Australian Open on TV. Coco began playing at 7 and showed a real talent for the sport. When young Coco turned 10, she began training at a tennis centre run by Venus's coach, Patrick Mouratoglou. He still remembers the first time he met Coco. He says, “She impressed me with her determination and fighting spirit. ” Convinced of her talent, Mouratoglou sponsored(资助)Coco to attend his academy in France. While she was preparing to go to the academy in 2019, she received a call that would change her plans and her life!Wimbledon(温布尔顿网球比赛)organizers called and offered Gauff entry to the tournament(联赛)as a wildcard(外卡选手).This madeher one of the youngest players to ever qualify. Before she knew it she was on her way to London. After arriving, she received another surprise. For her Wimbledon debut(首次登场),she would be playing her lifelong hero, Venus Williams! The tennis legend is 24years older than Gauff.The world watched with amazement as young Gauff beat Venus in two straight sets! Afterwards, Gauff shook Venus's hand, thanked her and said, “I wouldn't be here if it wasn't for you. " After the match, Venus said of Gauff, “I think the sky is the limit;it really is. " Venus said, “I feel honored that I was on her wall [as a poster] at some point in her life. Soon she will be on other girls' walls. It's nice because it will keep going from the next generation to the next generation. "12. What can we learn about Coco from the text?A. She took up tennis as career at the age of 4.B. She went to academy at the age of 7.C. She had played against William before 10.D. She beat William at the age of 1513. Which of the following can best describe Coco?A. Talented and modest.B. Lucky and responsible.C. Proud and hard-working.D. Respectful and cheerful.14. What can we infer from Venus's words?A. Coco had reached her limits.B. Coco would rise to fame after the match.C. Coco's poster would be passed on.D. Coco had once visited her home.15. What can be a suitable title for the text ?A. Coco Gauff:Tennis's Next Superstar.B. Coco Gauff:A Poster on the Wall.C. What Posters Mean to a Young Girl.D. The Significance of Admiring an Idol.第二节(共5小题;每小题2分,满分10分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。
2020年长沙市雅礼中学高三英语第二次联考试卷及答案解析
2020年长沙市雅礼中学高三英语第二次联考试卷及答案解析第一部分阅读(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A、B、C、D四个选项中选出最佳选项AObesity (肥胖症) is becoming a problem in our busy society, and almost one in three American adults is now considered to be obese. Children obesity is alsoat an all-time high.Obesity means being very overweight. If you are obese, you have too much bodyfat. If you eat more food than your body can use, this will make you put on weight. Food that your body does not need will be stored as fat by your body.The following are the major factors that increase the risk of obesity.What you eat plays a major role in weight gain. Eating a lot of fast food such as hamburgers, sweet drinks, ice creams and other sweet food can increase the risk of becoming obese.If you do not do enough exercise, you will put on weight as the food you eat is not being used to make energy for physical activities.The chances of you being obese are greater if your parents are obese.There are many psychological factors that cause people to eat too much. People who are worried, unhappy or bored will often eat to make themselves feel better. This is known as comfort eating.Age is another factor, as you tend (趋于) to be less active when you get older. When you get older, you need to eat less, and if you do not eat less, you will put on weight. Obesity can cause many health problems such as heart problems, high blood pressure and many other serious medical conditions.1. The underlined sentence in paragraph 1 means that ____.A. obesity does not do harm to health.B. there are more obese children than before.C. all the American children are obese.D. there are less obese children in the USA.2. According to the passage, there are ____ major factors that increase the risk of obesity.A. threeB. fourC. fiveD. six3. What will the writer most probably talk about after the last paragraph?A. How to avoid obesity.B. How to live in the busy USA.C. What illnesses are caused by obesity.D. How doctors treat heart problems.BA company called Neuralink has shared a video where a monkey is playing a video game. That' s fairly unusual, but what makes the video even stranger is that the monkey is playing the video game with just his mind.The monkey in the video is called Pager who has two of Neuralink's special "Link" devices(装置)inside his brain. The devices planted in Pager's brain are connected to 2,048 wires which lead to the parts of Pager's brain that control movements of the arms and hands.Scientists taught Pager to play a video game. At first, Pager controlled the video game using a joystick it, which is a normal gaming controller. But as Pager played, his Link devices wirelessly sent out information about the signals his brain was using to control his arms and hands. Neuralink's scientists recorded all of these signals.Then they used computers to match the signals from Pager's brain to the movements that his hands were actually doing. This was the most difficult work and the scientists counted on artificial intelligence ( AI) to help them decode(解码)Pager's brain signals.The final step was to have a computer make moves inthe video game as if Pager had actually moved the joystick. If Pager thought about moving the joystick up, the computer would send an “up” signal to the video game.At first, the researchers let Pager keep moving the joystick with his hand, even though it was no longer connected to the computer. But soon Pager was able to play the video game using just his brain.Even though Neuralink's work right now focuses on animals and video games, there's a very serious purpose behind it. Neuralink wants to make it possible for humans who have lost the ability to make physical movements to interact with the world around them.4. What are “Link” devices used to do?A. To pick up the arms' and hands' signals.B. To link the computer to the monkey's brain.C. To send out information about the brain's signals.D. To control movements of the arms and hands.5. What challenged scientists most in the study?A. Recording and sending out body signals.B. Training Pager to use the joystick correctly.C. Planting "Link" devices into Pager's brain.D. Matching brain signals to body movements.6. What is Neuralink's real purpose of the study?A. To test artificial intelligence.B. To help those without arms or legs.C. To study how animals play video games.D. To develop more complex video games.7. What can be the best title for the text?A. Video Games for Animals Are Developed.B. Science Proves the Intelligence of Monkeys.C. Monkey Plays Video Games Using His Mind.D. Neuralink Is Leading the World in Technology.CConcrete is the world's most consumed material after water. Because it already surrounds us in the built environment, researchers have been exploring the idea of using concrete to store electricity—turning buildings into giant batteries. The idea has been gaining ground as we have come to increasingly rely on renewable energy from the wind and sun: rechargeable batteries are necessary when the breeze dies down or darkness falls.Experimental concrete batteries have only managed to hold a small part of what a traditional battery does. But one team now reports in Buildings that it has developed a rechargeable original model that could represent a more than 900 percent increase in stored charge, compared with earlier attempts.A live-in concrete battery might sound unlikely. Still, "you can make a battery out of a potato," notes Aimee Byrne. In a future where sustainability is key, she likes the idea of buildings that avoid waste by providing shelter and powering electronics.Although the new design stores more than 10 times as much power as earlier attempts, it still has a long way to go: 200 square meters of it "can provide about 8 percent of the daily electricity consumption" of a typical U.S. home, Zhang says.This is not enough to compete with today's rechargeable devices. "We're getting milliamps (毫安) out of concrete batteries—we're not getting amps (安培), "Byrne says." We're getting hours as opposed to days of charge." But she adds that" concrete batteries are completely in their childhood, compared to other batterydesigns." The earliest batteries were simple andbulky. Researchers experimented with new materials and designs for more than a century to develop today's small devices. Byrne suggests concrete-based energy storage could undergo a similar evolution. "The whole idea is that we're looking far into the future," she says. "We're playing the long game with it."8. What can we learn about the concrete batteries?A. They become increasingly renewable.B. They are the most consumed batteries.C. They are being developed by researchers.D. They will replace energy from the wind and sun.9. Why does Byrne mention a battery out of a potato?A. To show it is easy to build concrete batteries.B. To argue it is possible to develop concrete batteries.C. To make her statement more interesting.D. To call on people to protect the environment.10. What does the underlined word "bulky" in Paragraph 5 mean?A. HeavyB. CheapC. EfficientD. Small.11. What doesByrne think of concrete batteries?A. They beat today's rechargeable devices.B. They are simple and bulky.C. They have a doubtful future.D. They have a long way to go.DHenry Cavill: Bring Superman to LifeHenry Cavill knew that he wanted to be a star at 16 years of age, after a chance meeting with movie star Russell Crowe who inspired hispassion for acting. But for the British-born actor, the bright lights and attraction ofHollywoodwere a long way away. Supported by his secretary mother and stockbroker father, he decided to study drama during high school. His journey to super star began.Before gaining the international recognition he has now, Cavill tried out for roles in the Harry Potter and Twilight series but failed to get either. He would have to keep waiting for his big chance.Determined as ever, Cavill took any acting jobs he could get his hands on and appeared in several low-budget horror movies and TV shows in hopes of getting noticed. It almost worked. In the early 2000s, at just 22 years old, he narrowly missed out on becoming the new James Bond. Finally, in 2007, his hard work paid off. He won a leading role as the first Duke of Suffolk in the period showThe Tudors. The TV show was very popular and helped to raise Cavill's popularity inAmerica.In 2011, Cavil landed his breakout role, playing Superman in the DC Extended Universe. He hasn't looked backand has since starred in many hit films, such asMission: Impossible- Fallout.More recently, he stepped back on to the small screen. Since 2019, he has starred in the popular seriesThe Witcher, adapted from the book series and video games of the same name. In the TV show, Cavill played a brave monster hunter named Geralt of Rivia, which was the perfect role for Cavill because he was a fan of the video games. Cavill also got a chance to play a classic English character — master detective Sherlock Holmes — in 2020'sEnola Holmes.However, Cavill isn't just a good guy on screen. His charity work also makes him a real-life hero. In 2014, he took part in the Ice Bucket Challenge while wearing his full Superman suit to support the ALS Association. Currently, he is an ambassador for the UK's Royal Marines Charity, which supports war veterans (退伍军人). Why does he do it? He love to make people feel good and bring smiles to people' faces. Indeed, Henry Cavill in living proof that you don't always need to wear a cape (斗篷) to act like a hero.12. Why did Cavil act in low-budget film and TV works early in his career?A. He was too polite to refuse.B. He was hoping to get noticed.C. He was encouraged to do so by his parents.D. He was friends with the directors of the projects.13. The role of the monster hunter was the perfect for Cavill because ________ .A. he had experienced hunting monstersB. he had played the same role in a movieC. he knew the writer of the books personallyD. he enjoyed the video games that the show was rooted in14. Which of the following words can best describe Cavill?A. Modest and friendly.B. Determined and kind.C. Talented and faithful.D. Honest and considerate.15. What made Cavill a real-life hero?A. Being a successful actor.B. Playing Superman on screen.C. Devoting to charities.D. Wearing a cape to take part in activities.第二节(共5小题;每小题2分,满分10分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。
湖南省长沙市雅礼中学2020届高三下学期月考(六)物理试卷含解析《含高考15套》
湖南省长沙市雅礼中学2020届高三下学期月考(六)物理试卷一、单项选择题:本题共6小题,每小题4分,共24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、下列说法正确的是()A.中子与质子结合成氘核时吸收能量B.卢瑟福的 粒子散射实验证明了原子核是由质子和中子组成的C.入射光照射到某金属表面发生光电效应,若仅减弱该光的强度,则不可能发生光电效应D.根据玻尔理论,氢原子的电子由外层轨道跃迁到内层轨道,原子的能量减少,电子的动能增加2、两根相互平行的水平放置长直导线分别通有方向相同的电流I1和I2,且I1>I2;有一电流元IL与两导线均平行,且处于两导线的对称面上,导线某一横截面所在平面如图所示,则下列说法正确的是()A.电流元所处位置的磁场方向一定竖直向下B.电流元所处位置的磁场方向一定水平向右C.要使电流元平衡,可在电流元正上方某位置加一同向通电导线D.如果电流元在所处位置受的安培力为F,则两导线在该处的磁感应强度大小为B=F IL3、质量为m、初速度为零的物体,在不同变化的合外力作用下都通过位移0x.下列各种情况中合外力做功最多的是()A.B.C.D.4、在“用单摆测定重力加速度”的实验中,实验时用拉力传感器测得摆线的拉力大小F随时间t变化的图象如图所示,则该单摆的周期为()A.t B.2t C.3t D.4t5、下列关于物质结构的叙述中不正确...的是A.天然放射性现象的发现表明了原子核内部是有复杂结构的B.质子的发现表明了原子核是由质子和中子组成的C.电子的发现表明了原子内部是有复杂结构的D. 粒子散射实验是原子核式结构模型的实验基础6、如图所示为一简易起重装置,(不计一切阻力)AC是上端带有滑轮的固定支架,BC为质量不计的轻杆,杆的一端C用铰链固定在支架上,另一端B悬挂一个质量为m的重物,并用钢丝绳跨过滑轮A连接在卷扬机上。
开始时,杆BC与AC的夹角∠BCA>90°,现使∠BCA缓缓变小,直到∠BCA=30°。
2020届湖南省长沙市雅礼中学高三高考数学理模拟一试题A卷解析版
2020届湖南省长沙市雅礼中学高三高考数学(理)模拟(一)试题(A 卷)一、单选题1.若复数z 的共轭复数z 满足:31i z =+,则复数z 对应的点在( ) A .第一象限 B .第二象限C .第三象限D .第四象限【答案】A【解析】根据虚数单位的幂的运算化简后,根据共轭复数的概念写出z 的结果,进而判定对应点所在的象限. 【详解】1i 1i z z =-⇒=+,故z 对应的点在第一象限.故选:A . 【点睛】本题考查虚数单位的幂的运算,共轭复数的概念,复数的几何意义,属基础题. 2.已知集合2{|log (1)1}P x x =-<,2|1Q x x ⎧<=⎫⎨⎬⎩⎭,则()R P Q ⋂等于( ) A .(1,2] B .[0,2]C .(1,2)D .(0,3]【答案】A【解析】化简集合,P Q ,求出Q 的补集,再结合交集的定义求解结论即可. 【详解】2{|log (1)1}{|012}(1P x x x x =-<=<-<=,3), 2|1(Q x x ⎧⎫=<=-∞⎨⎬⎩⎭,0)(2⋃,)+∞,故[0RQ =,2];故()(1R P Q ⋂=,2]. 故选:A . 【点睛】本题主要考查集合的基本运算,考查了对数函数的定义域以及分式不等式的求解,比较基础.3.某商家统计了去年P,Q两种产品的月销售额(单位:万元),绘制了月销售额的雷达图,图中A点表示P产品2月份销售额约为20万元,B点表示Q产品9月份销售额约为25万元.根据图中信息,下面统计结论错误..的是()A.P产品的销售额极差较大B.P产品销售额的中位数较大C.Q产品的销售额平均值较大D.Q产品的销售额波动较小【答案】B【解析】由图示中P产品的销售额的波动较大,Q产品的销售额的波动较小,再根据极差、中位数、平均值的概念,可得选项.【详解】据图求可以看出,P产品的销售额的波动较大,Q产品的销售额的波动较小,并且Q产品的销售额只有两个月的销售额比25万元稍小,其余都在25万元至30万元之间,所以P产品的销售额的极差较大,中位数较小,Q产品的销售的平均值较大,销售的波动较小,故选:B.【点睛】本题考查识别统计图的能力,会根据图示得出其数字特征的大小关系,属于基础题. 4.《九章算术 衰分》中有如下问题:“今有甲持钱五百六十,乙持钱三百五十,丙持钱一百八十,凡三人俱出关,关税百钱.欲以钱数多少衰出之,问各几何?”翻译为“今有甲持钱560,乙持钱350,丙持钱180,甲、乙、丙三个人一起出关,关税共计100钱,要按个人带钱多少的比例交税,问三人各应付多少税?”则下列说法中错误的是()A.甲付的税钱最多B.乙、丙两人付的税钱超过甲C.乙应出的税钱约为32D.丙付的税钱最少【答案】B【解析】通过阅读可以知道,A D说法的正确性,通过计算可以知道,B C说法的正确性.【详解】甲付的税钱最多、丙付的税钱最少,可知,A D 正确:乙、丙两人付的税钱占总税钱的3511002<不超过甲。
湖南省长沙市雅礼中学2020届高三上学期月考试卷(一)数学(理)试题 Word版含解析
雅礼中学2020届高三月考试卷(一)数学(理科)本试卷分第I 卷(选择题)和第I 卷(非选择题)两部分,共8页.时量120分钟,满分150分.第Ⅰ卷(共60分)一、选择题:本大题共12个小题,每小题5分,共60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.若复数z 的共轭复数z 满足:()12i z i -=,则复数z 等于( ) A. 1i + B. 1i -+ C. 1i - D. 1i --【答案】D 【解析】 【分析】由()12i z i -=得出21iz i=-,利用复数的除法法则求出z ,利用共轭复数的概念可求出复数z .【详解】()12i z i -=Q ,()()()()2121211112i i i i z i i i i +-∴====-+--+,因此,1i z =--, 故选:D.【点睛】本题考查复数的除法运算,同时也考查了共轭复数计算,考查计算能力,属于基础题.2.已知集合{}{}/10,/A x x B x x a =-<<=≤,若A B ⊆,则a 的取值范围为( ) A. (,0]-∞ B. [0,)+∞C. (),0-∞D. ()0,∞+【答案】B 【解析】 【分析】画出集合,A B 的数轴表示,利用数轴解题.【详解】画出集合A,B 的数轴表示,因为A B ⊆,所以0a ≥,故选B. 考点:集合包含关系判断及其应用3.在ABC △中,(BC uuu r +BA u u u r )·AC u u u r =|AC u u u r|2,则ABC △的形状一定是 A. 等边三角形 B. 等腰三角形 C. 直角三角形 D. 等腰直角三角形【答案】C 【解析】由(BC uuu r +BA u u u r)·AC u u u r =|AC u u u r |2,得AC u u u r ·(BC uuu r +BA AC -u u u r u u u r )=0,即AC u u u r ·(BC uuu r +BA u u u r +CA u u u r )=0,∴2AC u u u r ·BA u u u r =0,∴AC u u u r ⊥BA u u u r,∴A =90°.即ABC V 的形状一定是直角三角形. 本题选择C 选项.4.我国古代数学名著《九章算术》中割圆术有:“割之弥细,所失弥少,割之又割,以至于不可割,则与圆周合体而无所失矣.”其体现的是一种无限与有限的转化过程,比如在222+++⋅⋅⋅x ,这可以通过方程2x x +=确定出来2x =,类似的不难得到11111+=++⋅⋅⋅( )51-- 51- 51+ 51-+ 【答案】C 【解析】分析:通过类比推理的方法,得到求值的方法:列方程,求解(舍去负根)即可.详解:由已知代数式求值方法,列方程,求解,舍负根. 可得 11(0)x x x+=> 解得x x ==(舍) 故选C.点睛:类比推理方法的前提是两种对象部分有共同属性,由特殊点向特殊点推理.通过类比推理考核研究问题的深度、思维散发情况和观察的仔细程度.5.()6211x x x ⎛⎫-- ⎪⎝⎭展开式中的常数项为( ) A. 35- B. 5-C. 5D. 35【答案】A 【解析】 【分析】将二项式()6211x x x ⎛⎫-- ⎪⎝⎭表示为()666221111x x x x x x x x ⎛⎫⎛⎫⎛⎫--=--- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭,得出其通项,令x 的指数为零,求出参数的值,再将参数的值代入通项可得出展开式中的常数项.【详解】()666221111x x x x x x x x ⎛⎫⎛⎫⎛⎫--=--- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭Q , 展开式通项为()()626628266661111krk r k k r r k k rr C x x C x C x C x x x ----⎛⎫⎛⎫⋅⋅--⋅⋅-=⋅-⋅-⋅-⋅ ⎪ ⎪⎝⎭⎝⎭,令620820k r -=⎧⎨-=⎩,得34k r =⎧⎨=⎩,因此,二项式()6211x x x ⎛⎫-- ⎪⎝⎭展开式中的常数项为346635C C --=-,故选:A.【点睛】本题考查二项式展开式中指定项系数的计算,解题的关键就是写出二项展开式的通项,根据指数求出参数的值,进而求解,考查计算能力,属于中等题.6.给出三个命题:①直线上有两点到平面的距离相等,则直线平行平面;②夹在两平行平面间的异面直线段的中点的连线平行于这个平面;③过空间一点必有唯一的平面与两异面直线平行.正确的是( ) A. ②③ B. ①②C. ①②③D. ②【答案】D 【解析】 【分析】通过举反例可判断出命题①的正误;利用平面与平面平行的性质定理以及直线与平面平行的性质定理可判断出命题②的正误;通过实例判断出命题③的正误.【详解】对于命题①,如果这两点在该平面的异侧,则直线与该平面相交,命题①错误; 对于命题②,如下图所示,平面//α平面β,A α∈,C α∈,B β∈,D β∈,且E 、F 分别为AB 、CD 的中点,过点C 作//CG AB 交平面β于点G ,连接BG 、DG .设H 是CG 的中点,则//EH BG ,BG ⊂Q 平面β,EH ⊄平面β,//EH ∴平面β. 同理可得//HF 平面β,EH HF H =Q I ,∴平面//EFH 平面β. 又Q 平面//α平面β,∴平面//EFH 平面α,EF ⊂Q 平面EFH ,//EF ∴平面α,//EF 平面β,命题②正确;对于命题③,如下图所示,设AB 是异面直线a 、b 的公垂线段,E 为AB 上一点,过点E 作//a a ',//b b ',当点E 不与点A 或点B 重合时,a '、b '确定的平面α即为与a 、b 都平行的平面;若点E 与点A 或点B 重合时,则a α⊂或b α⊂,命题③错误.故选:D.【点睛】本题考查线线、线面、面面平行关系的判定与性质,解题时要注意这三种平行关系的相互转化,考查推理能力与空间想象能力,属于中等题.7.执行如图所示的程序框图,若输出k的值为8,则判断框内可填入的条件是()A. s≤34?B. s≤56?C. s≤11 12?D. s≤25 24?【答案】C 【解析】试题分析:模拟执行程序框图,k的值依次为0,2,4,8,,因此1111124612s=++=(此时6k=),因此可填1112s≤,故选C.考点:程序框图及循环结构.8.若01,22x ⎡⎤∃∈⎢⎥⎣⎦,使得200210x x λ-+<成立是假命题,则实数λ的取值范围是( )A. (-∞B. (⎤⎦C. 92⎡⎤⎢⎥⎣⎦D. {}3【答案】A 【解析】 【分析】由题意得知,全称命题“1,22x ⎡⎤∀∈⎢⎥⎣⎦,2210x x λ-+≥”是真命题,利用参变量分离法得出12x x λ≤+,然后利用基本不等式求出12x x+的最小值,可得出实数λ的取值范围.【详解】因为01,22x ⎡⎤∃∈⎢⎥⎣⎦,使得200210x x λ-+<成立是假命题,所以1,22x ⎡⎤∀∈⎢⎥⎣⎦,2210x x λ-+≥恒成立是真命题, 即1,22x ⎡⎤∀∈⎢⎥⎣⎦,12x xλ≤+恒成立是真命题, 当1,22x ⎡∈⎤⎢⎥⎣⎦时,由基本不等式得12x x +≥=1,222x ⎡⎤=∈⎢⎥⎣⎦时,等号成立,λ∴≤λ的取值范围是(-∞,故选:A.【点睛】本题考查利用特称命题的真假求参数的取值范围,在求参数的取值范围时,可灵活利用参变量分离法,转化为函数的最值求解,考查运算求解能力,属于中等题.9.圆锥的母线长为2,其侧面展开图的中心角为θ弧度,过圆锥顶点的截面中,面积的最大值为2,则θ的取值范围是( )A. ),2πB. π⎡⎤⎣⎦C.}D. ,2π⎫⎪⎪⎣⎭【答案】A 【解析】 【分析】设轴截面的中心角为α,过圆锥顶点的截面的顶角为β,且βα≤,由过圆锥顶点的截面中,面积的最大值为2,明确β能取到2π,从而明确轴截面的中心角为α的范围,进而得到结果. 【详解】设轴截面的中心角为α,过圆锥顶点的截面的顶角为β,且βα≤ 过圆锥顶点的截面的面积为:122sin β2sin β2⨯⨯⨯=, 又过圆锥顶点的截面中,面积的最大值为2, 故此时β2π=,故απ2π≤<圆锥底面半径r )2sin22α=∈ ∴侧面展开图的中心角为θ弧度2sin222πsin22απα⨯⨯==∈),2π 故选:A.【点睛】本题考查圆锥侧面展开图扇形圆心角的计算,解题时要弄清楚圆锥底面圆的周长与侧面展开图扇形的互相相等来建立等量关系,考查空间想象能力,属于中等题.10.已知()f x 是定义在实数集R 上的奇函数,a 为非正的常数,且当0x >时,()2f x ax x =-.若存在实数m n <,使得()f x 的定义域与值域都为[],m n ,则实数a 的取值范围是() A. ∞(-,1) B. (]1,0-C. (],0∞-D. ∅【答案】B 【解析】 【分析】由题意得出函数()y f x =在R 上单调递减,结合题意得出0m n <<,由题意得出22am m nan n m⎧+=⎨-=⎩,两式相加得出0m n +=,可得出1a m =--,从而可得出实数a 的取值范围. 【详解】Q 函数()y f x =为R 上的奇函数,则()00f =,适合()2f x ax x =-.当0a ≤且0x ≥时,函数()2f x ax x =-为减函数.设0x <,则0x ->,()()()22f x a x x x ax -=⋅---=--, 此时,()()2f x f x x ax =--=+,且该函数在(),0-∞上单调递增,所以,函数()y f x =在实数集R 上单调递减,由题意可得()()f m nf n m ⎧=⎪⎨=⎪⎩,则点(),m n 和点(),n m 在函数()y f x =的图象上,且这两点关于直线y x =对称.若0m n <<,则这两点均为第二象限,都在直线y x =的上方,不可能关于直线y x =对称; 若0n m >>,则这两点均为第四象限,都在直线y x =的下方,不可能关于直线y x =对称. 因此,0m n <<.由()()f m n f n m ⎧=⎪⎨=⎪⎩,得22am m n an n m⎧+=⎨-=⎩,两式相加得()()()220a m n m n m n ++--+=, 即()()10m n a m n ++--=,10a n m ∴=-+>(舍去)或0m n +=,则n m =-. 代入2am m n +=,得2am m m +=-,11a m ∴=-->-,又0a ≤Q ,10a ∴-<≤. 因此,实数a 的取值范围是(]1,0-,故选:B.【点睛】本题考查函数单调性的应用,考查函数的定义域与值域问题,解题时要分析出函数的单调性及其他基本性质进行求解,考查分析问题和解决问题的能力,属于中等题.11.椭圆与双曲线共焦点1F 、2F ,它们的交点P 对两公共焦点1F 、2F 的张角为122F PF θ∠=,椭圆与双曲线的离心率分别为1e 、2e ,则( )A.222212cos sin 1e e θθ+= B. 222212sin cos 1e e θθ+=C. 2212221cos sin e e θθ+=D. 2212221sin cos e e θθ+=【答案】B 【解析】 【分析】设椭圆的长轴长为12a ,双曲线的实轴长为22a ,并设1PF m =,2PF n =,利用椭圆和双曲线的定义以及余弦定理可得出1a 、2a 关于c 的等式,从而可得出1e 、2e 的关系式. 【详解】设椭圆的长轴长为12a ,双曲线的实轴长为22a ,并设1PF m =,2PF n =,焦距为2c ,在12PF F ∆中,由余弦定理得()2222cos22m n mn c θ+-=, 由椭圆和双曲线的定义得1222m n a m n a +=⎧⎨-=⎩,解得1212m a a n a a =+⎧⎨=-⎩.代入()2222cos22m n mn c θ+-=,得()()()()222121212122cos 24a a a a a a a a c θ++--+-=,即()222221221cos 22a a a a c θ++-=,()()222121cos21cos22a a c θθ∴-++=,即22222122sin 2cos 2a a c θθ+=,22221222sin cos 1a a c c θθ∴+=,因此,222212sin cos 1e e θθ+=. 故选:B.【点睛】本题考查共焦点和共交点的椭圆和双曲线的综合问题,要充分结合椭圆、双曲线的定义以及余弦定理列等式求解,考查分析问题和解决问题的能力,属于中等题.12.在ABC ∆sin sin A B C +的最大值为( )12B. 2【答案】B 【解析】 【分析】 解法1:利用()sin sin A C B =+,得出sin sin A B C +=)sin sin cos C C B C B +,然后利用辅助角公式以sin sin A B C +的最大值;解法2:sin sin A B C +=()()cos cos 2B C B C A --++,然后利用()cos 1B C -≤sin sin A B C +的最大值. 【详解】法1:()sin sin sin sin A B C C B B C +=++cos sin sin sin C B C B B C =+)sin sin cos C C B C B =++≤=2=≤=,当且仅当sin sin B C ==sin A 时,等号成立,sin sin A B C +的最大值为2,故选:B ; 法2:()()cos cossin sin 2B C B C A B C A --++=+1cos 111cos 22222A A A A ++=++≤=≤,当且仅当sin sin 3B C ==,sin 3A =时,等号成立,sin sin A B C +的最大值为2,故选:B.【点睛】本题考查三角形中的最值的求解,涉及到三角恒等变换中的一些变形技巧,解题时要注意化异角为同角,充分利用辅助角公式来求解,考查运算求解能力,属于难题.第Ⅱ卷(共90分)本卷包括必考题和选考题两部分.第13~21题为必考题,每个试题考生都必须作答.第22~23题为选考题,考生根据要求作答.二、填空题:本大题共4小题,每小题5分,共20分.13.函数sin xy e x =的图象在原点处的切线方程是__________. 【答案】0x y -= 【解析】 【分析】易知原点在函数sin xy e x =的图象上,利用导数求出切线的斜率,然后写出直线的点斜式方程,可得出所求切线方程.【详解】易知原点在函数sin xy e x =的图象上,()sin cos xy e x x '=+,当0x =时,1y '=.因此,所求切线方程为y x =,即0x y -=,故答案为:0x y -=.【点睛】本题考查导数的几何意义,考查利用导数求函数图象的切线方程,解题时要熟悉导数求切线方程的基本步骤,考查计算能力,属于中等题.14.已知抛物线24y x =的焦点为F ,准线与x 轴的交点为,M N 为抛物线上的一点,且满足32NF MN =,则NMF ∠为 . 【答案】【解析】【详解】过N 作NH 垂直准线,垂足为H , 则|NF|=|NH|,因为32NF =, 所以32NH =, 3cos cos NH NMF MNH MN∴∠=∠==6NMF π∴∠=,故答案为6π.15.已知函数()()2sin 16f x x x R πω⎛⎫=-+∈ ⎪⎝⎭的图象的一条对称轴为x π=,其中ω为常数,且()1,2ω∈,则函数()f x 的最小正周期为__________.【答案】65π 【解析】 【分析】 由题意得出()62k k Z ππωππ-=+∈,可得出ω的表达式,结合()1,2ω∈可求出ω的值,然后利用正弦型函数的周期公式2T πω=可得出函数()y f x =的最小正周期.【详解】由函数()()2sin 16f x x x R πω⎛⎫=-+∈ ⎪⎝⎭图象的一条对称轴为x π=. 可得62k ππωππ-=+,k Z ∈,23k ω∴=+,又()1,2ω∈,53ω∴=.因此,函数()y f x =的最小正周期为26553T ππ==,故答案为:65π. 【点睛】本题考查利用正弦型函数的对称轴求参数,同时也考查了正弦型函数周期的计算,要结合题意得出参数的表达式,结合参数的取值范围求出参数的值,考查分析问题和解决问题的能力,属于中等题.16.已知实数a 、b 、c 满足a b c <<,6.9a b c ab bc ca ++=⎧⎨++=⎩下列命题中:①01a <<;②13b <<;③34c <<;④()()55b c --的最小值是154,所有真命题为__________. 【答案】①②③④ 【解析】 【分析】构造函数()()()()3269x f x x a x b x c x x abc =-+=----,利用导数分析函数()y f x =的单调性,可得出()()()3 0f x f f ==极小值,()()()14f x f f ==极大值,再由a 、b 、c 为函数()y f x =的三个零点可判断出命题①、②、③的正误,由题中条件得出6b c a +=-,()()2963bc a a a =--=-,代入()()55b c --可判断出命题④的正误.【详解】令()()()()f x x a x b x c =---,则()3269f x x x x abc =-+-.()()() '313f x x x =--,()()()3 0f x f abc f ==-=极小值,()()()14 4f x f abc f ==-=极大值,如下图所示:易知函数() y f x =的三个零点分别为a 、b 、c ,由于a b c <<,由图象可知,01a <<,13b <<,34c <<,则命题①、②、③正确;由题中条件可知6b c a +=-,()()2963bc a a a =--=-. 因此()()()()()2255253562545b bc b c a a a a c -=-++=---+=-+-=211515244a ⎛⎫ ⎪+≥⎝⎭-,命题④也为真命题,故答案为:①②③④.【点睛】本题考查不等式真假的判断,解题的关键就是根据等式结构构造新函数求判断,并将参数转化为函数的零点,在考查函数的综合问题时,要充分利用导数研究函数的单调性,考查函数方程思想,属于难题.三、解答题:共70分.解答应写出文字说明、证明过程或演算步骤.第17~21题为必考题,每个试题考生都必须作答,第22.23题为选考题,考生根据要求作答. (一)必考题:60分.17.已知数列{}n a 是首项为1,公比为12的等比数列,12n n S a a a =+++L . (1)若n S 、98、1n a -成等差数列,求n 的值; (2)证明*n N ∀∈,有312112231222112n n n n a a a S S S S S S ++++++<-L . 【答案】(1)3n =;(2)详见解析. 【解析】 【分析】(1)先利用等比数列的通项公式和前n 项和公式分别求出n a 、n S ,由题意条件得出194n n S a -+=,即为111292224n n ---+=,从而解出n 的值; (2)将112k k k a S S ++裂项为()111112222k k k k k k k k k S S a S S S S S S +++++-==-,利用裂项法求出31212231222n n n a a a S S S S S S +++++L ,再利用放缩法可得出所证不等式. 【详解】(1)由等比数列的通项公式得1111122n n n a --⎛⎫=⨯=⎪⎝⎭, 由等比数列的前n 项和公式得11111221212n n n S -⎛⎫⨯- ⎪⎝⎭==--, n S Q 、98、1n a -成等差数列,所以,194n n S a -+=,即121192224n n ---+=,化简得11124n -=,解得3n =;(2)()()1111122221,2,3,kkkk k k k k kS SakS S S S S S+++++-==-=⋅⋅⋅Q,且11212nn nS++-=,因此,31212231122311122222222222nn n n n na aaS S S S S S S S S S S S S S++++++⋅⋅⋅+=-+-++-=-L111121121121212nn n n++++=-=-<---.【点睛】本题考查等比数列通项公式与求和公式,同时也考查了裂项求和法,解题时要熟悉裂项求和法对数列通项结构的要求,考查运算求解能力,属于中等题.18.已知在正方体1111ABCD A B C D-中E,F分别是1,DD BD的中点,G在棱CD上,且14CG CD=.(1)求证:1EF B C⊥;(2)求二面角1F EG C--的余弦值.【答案】(1)见解析;(2)二面角1F EG C--的余弦值为1414-.【解析】【详解】试题分析:(1)如图建立空间直角坐标系O xyz-,设正方体棱长为4,则求出相应点和相应向量的坐标可证1EF B C⊥;(2)平面11D DCC的一个法向量为()4,0,0BC=-u u u v,设并求出平面EFG的一个法向量(),,n x y z=v,应用向量的夹角公式,最后由图可知,二面角为钝角,可得到二面角1F EG C--的余弦值.试题解析:(1)如图建立空间直角坐标系O xyz -,设正方体棱长为4,则()()()()()()()110,0,2,2,2,0,0,4,0,4,4,0,0,4,4,4,4,4,0,3,0E F C B C B G ()()12,2,2,4,0,4EF B C =-=--u u u v u u u v,∴()()()12420240EF B C ⋅=⨯-+⨯+-⨯-=u u u v u u u v∴1EF B C ⊥u u u v u u u v,∴1EF B C ⊥(2)平面11D DCC 的一个法向量为()4,0,0BC =-u u u v设平面EFG 的一个法向量为(),,n x y z =v∴00n EF n FG ⎧⋅=⎨⋅=⎩u u u v v u u u v v 即020x y z x y +-=⎧⎨-+=⎩∴23y x z x =⎧⎨=⎩令1x =,则2,3y z ==,∴可取()1,2,3n =v∴14cos ,144n BC n BC n BC⋅===-⨯⋅u u u v v u u u v vu u u v v 如图可知,二面角为钝角,∴二面角1F EG C --的余弦值为1414-19.某健身机构统计了去年该机构所有消费者的消费金额(单位:元),如下图所示:(1)将去年的消费金额超过 3200 元的消费者称为“健身达人”,现从所有“健身达人”中随机抽取 2 人,求至少有 1 位消费者,其去年的消费金额超过 4000 元的概率; (2)针对这些消费者,该健身机构今年欲实施入会制,详情如下表:预计去年消费金额在(] 0,1600内的消费者今年都将会申请办理普通会员,消费金额在(]1600,3200内的消费者都将会申请办理银卡会员,消费金额在(] 3200,4800内的消费者都将会申请办理金卡会员. 消费者在申请办理会员时,需-次性缴清相应等级的消费金额.该健身机构在今年底将针对这些消费者举办消费返利活动,现有如下两种预设方案:方案 1:按分层抽样从普通会员, 银卡会员, 金卡会员中总共抽取 25 位“幸运之星”给予奖励: 普通会员中的“幸运之星”每人奖励 500 元; 银卡会员中的“幸运之星”每人奖励 600 元; 金卡会员中的“幸运之星”每人奖励 800 元.方案 2:每位会员均可参加摸奖游戏,游戏规则如下:从-个装有 3 个白球、 2 个红球(球只有颜色不同)的箱子中, 有放回地摸三次球,每次只能摸-个球.若摸到红球的总数消费金额/元为 2,则可获得 200 元奖励金; 若摸到红球的总数为 3,则可获得 300 元奖励金;其他情况不给予奖励. 规定每位普通会员均可参加 1 次摸奖游戏;每位银卡会员均可参加 2 次摸奖游戏;每位金卡会员均可参加 3 次摸奖游戏(每次摸奖的结果相互独立) . 以方案 2 的奖励金的数学期望为依据,请你预测哪-种方案投资较少?并说明理由. 【答案】(1)1933(2)预计方案2投资较少.详见解析 【解析】 【分析】(1)由题意,随机变量X 的可能值为“0,1,2”,得(1)(1)(2)P X P X P X ≥==+=,即可求解。
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• 间者发兵,但用玺书,或以诏令;如有奸人诈伪,无由 知觉。愚以为军旅尚兴,贼虏未殄,征兵郡国,宜有慎 重,可立虎符,以绝奸端。昔魏之公子,威倾邻国,犹 假兵符以解赵围,若无如姬之仇,则其功不显。
• 先前发兵,只是用玺书,如果有奸邪的人欺骗作假, 也无法知道。我认为军事行动还在进行,贼人尚未消 灭,从各郡国征调军队,应该慎重,可以设置虎符帝 王授予臣属兵权和调发军队的信物,用铜铸成虎形, 分两半,两半相符才能生效。,以断绝奸人作恶之源。 先前魏国的公子无忌,威风令邻国倾倒,仍要借助兵 符,才能解救对赵国的包围,如果如姬没有杀父之仇, 那末他的功绩也就不会显露。
之事,寄寓自己壮志不能伸的惆怅。
答案:
A项,两个“沙鸥”意象特征不同,杜甫诗中“沙 鸥”是“凄苦孤独”的意象。
15.这首诗表现了诗人矛盾的内心世界,试结合全诗作 简要分析。(6分)
答案 ①本诗写秋日所感,表现了诗人本已闲居可 以享受闲适而又不甘于闲居闲适的矛盾心情;诗 人虽以沙鸥闲适自由自比,但又说闲居度日如年。 ②虽说事大如天,但醉后亦休;虽说醉可休事, 但一闻砧杵之声而生感慨,一见梧桐叶落而思故 园。
• 光武帝爱惜他的才能,所以没有同意这个请求。杜诗喜欢推举贤 才,多次举荐名人士清河人刘统以及鲁阳董崇等人。当初,法令 制度还非常简单,朝廷还只是以皇帝的玺书玺,印章最早的名称。 秦以前,无论官、私印都称“玺”;统一六国后规定为皇帝印章 调兵,没有虎符作凭信,杜诗上疏说:“臣下听说军队是国家的凶 器,圣人对此极慎重。依照旧制,发兵都以虎符作凭信,其余的 征调,仅用竹制的令牌。符信符合无误,作为重要信用,这是用 来证明国家的命令,保持威严与重要地位的。
• ②制造悬念,吸引读者的注意力;
• ③丰富了小说的内容,使故事情节有序展开,人物、事 件接连出场;
• ④增强结果的戏剧效应,丰富作品主题的内涵。 《牲 畜林》 , 多次使用“延迟法”, 其中较大的有六次。这 样的“延迟”使得原本紧张激烈的情节节奏舒缓下来, 牲畜林里的各种动物可以自由地登台表演, 战争的阴影 被更有生命力的和谐自然挤到一边。
• 位置是不可替代的稀缺资源,知道这一点的不只是房地产开发商, ① 。调整菜单上菜品的顺序会对顾客的选择产生较大的影响。 ② ,会让后面的菜品价格看起来比较合理。研究人员还发现 顾客看菜单的方式很固定,顾客通常会看右上角这个最佳位置。 黄金位置总是有限的,在有限的空间里放置过多的菜品会适得其 反,因为 ③ 。
七年,迁南阳太守。性节俭而政治清平,以诛暴立威,善于计略, 省爱民役。造作水排,铸为农器,用力少,见功多,百姓便之。又 修治陂池,广拓土田,郡内比室殷足。时人方于召信臣,故南阳为 之语曰:“前有召父,后有杜母”。诗自以无劳,不安久居大郡, 求欲降避功臣。
建武七年,升任汝南太守。他生活俭朴,施政清正平和,
• 译文 利欲驱使人东奔西走,如同万头火牛奔突一样, 倒不如做个江湖之人,像沙鸥鸟那样自由自在。
无所事事的时候感觉一日长似一年,即使是天大 的事,喝醉了也就无事了。
在捣衣棒的敲击声中,深巷里的明月渐渐西沉, 井边的梧桐树忽然摇动叶落,方知故乡也是秋天 了。
想极目远眺,苦于没有登高的地方,哪能像陈登 站在百尺楼上,高论天下大事呢。
• 世祖召见,赐以棨戟,重使之河东,诛降逆贼杨 异等。诗到大阳,闻贼欲北度,乃与长史急焚其 船,部勒郡兵,将突骑趁击,斩异等,贼遂剪灭。 拜成皋令,视事三岁,举政尤异。再迁为沛郡都 尉,转汝南都尉,所在称治。
• 光武帝召见了他,赐给他棨戟,有缯衣或油漆的木戟, 官吏出行作前导的仪仗。后常用以称贵宾或好友远道 光临。仪仗,又派他到河东巡视,去招降和诛杀反叛 的贼人杨异等。杜诗抵达大阳,听说贼人企图北渡, 于是与长史迅速烧毁船只,调集部署郡中士兵,带领 骑兵突击贼兵,杀死了杨异等人,贼兵因此被消灭。 拜授为成臯令,做了三年官,政绩十分突出。又升迁
长期担任大郡长官而感到不安),于是请求降职,把职位
让给功臣。
• 帝惜其能,遂不许之。诗雅好推贤,数进知名士清河刘统及鲁阳 董崇等。初,禁网尚简,但以玺书发兵,未有虎符之信,诗上疏 曰:“臣闻兵者国之凶器,圣人所慎。旧制发兵,皆以虎符,其 余征调,竹使而已。符第合会,取为大信,所以明著国命,敛持 威重 11 12 14 17 18 19
C CB ADCDDCADCA
16.补写出下列句子中的空缺部分。(6分)
(1)有一首诗说“时人莫小池中水,浅处无妨有卧龙”,
《陋室铭》中与这句诗意思相近的句子
是:
,
。
(2)荀子《劝学》中以“雕刻打磨”为喻,从正面来论
证坚持学习的重要性的两句是“
• 材料一 近年来,长沙、西安、合肥、深圳等地纷纷出 台人才引进新政。落户门槛、资格条件一降再降,从海 外人才、研究生到大学生、中专生,各地都希望获得大 量新鲜的人才“血液”,给城市发展注入新动能。
• 材料二 2月19日,国家发改委发布《关于培育发展现 代化都市圈的指导意见》:要放开放宽除个别超大城市 (如北京、上海)外的城市落户限制,加快消除城乡区 域间户籍壁垒,统筹推进本地人口和外来人口市民化, 促进人口有序流动。
,
。
(3)世界上最美好的东西往往是无法用语言表达的,只
能用心去感受它。这一哲理可以用《饮酒》中
“
,
”来表达。
• 答案 (1)水不在深 有龙则灵 • (2)锲而不舍 金石可镂 • (3)此中有真意 欲辨已忘言
• 20.在下面一段文字横线处各补写一句话,使整段文字语意完整 连贯,逻辑严密。每处不超过15个字。(6分)
③诗人渴望扶世救民与年老体弱的矛盾。种种矛 盾在心中纠缠,使其郁闷难释。(每点3分,答出 任意两点给6分 )
• 杜诗,字君公,河内汲人也,少有才能,仕郡功 曹,有公平称。更始时,辟大司马府。建武元年 岁中三迁为侍御史安集洛阳时将军萧广放纵兵士 暴横民间百姓惶扰诗敕晓不改遂格杀广还以状闻。
• 杜诗,字君公,是河内汲县人。年轻时很有才能,在 郡里为官担任公曹,以公平著称。更始年间,被大司 马府征召。建武元年,一年之中连升三次官职到侍御 史,负责安定洛阳。当时将军萧广放纵士兵干不法之 事,在城中横行霸道,百姓惶恐惊扰。杜诗晓逾肖广, 他仍不改悔,于是杜诗就杀掉了萧广,回京后把情况 奏报给光武帝。
• 事有烦而不可省,费而不得已,盖此谓也。”书奏,从之。诗身 虽在外,尽心朝廷,谠言善策,随事献纳。视事七年,政化大行。 十四年,坐遣客为弟报仇被征,会病卒。司隶校尉鲍永上书言诗 贫困无田宅,丧无所归。诏使治丧郡邸,赙绢千匹。
• 事情有复杂不可以减省的,繁费出于不得已,说的就是 这个道理。”杜诗上书奏报,建议被朝廷采纳。杜诗虽然 在外地做官,仍然尽心于朝廷大事,有忠正的言论,好 的建议,都随时进献上奏。任官七年,政令教化大大推 行。建武十四年,因为 坐,古人可能因某事获罪,也可 能因受亲戚或朋友的牵连而获罪,即“连坐”。派宾客 为弟弟报仇,被征召问罪,正巧因病去世。司隶校尉鲍 永上奏说,杜诗家中贫困,没有田宅,死后没有地方下 葬。皇上下诏在郡中官邸治丧,并赐一千匹绢办理丧事。
• (2)符第合会,取为大信,所以明著国令, 敛持威重也。
符信符合无误,作为重要信用,这是用来证明 国家的命令,保持威严与重要地位的。 【解析】计分点:“所以”,用以,用来……的。 “明著”,表明,彰显。“敛持”,保持。 “威重”,威严与重要地位。共4分。句意1分。
• 22.阅读下面的材料,根据要求写作。(60分)
• 秋 思①
• 陆游
• 利欲驱人万火牛②,江湖浪迹一沙鸥。
• 日长似岁闲方觉,事大如天醉亦休。
• 砧杵敲残深巷月,井梧摇落故园秋。
• 欲舒老眼无高处,安得元龙③百尺楼。
• 【注】:①陆游作本诗时年已七十九岁,闲居故 乡山阴。②火牛:古代火攻战具,牛的双角绑上 利刃,尾巴绑上易燃物点燃令牛冲向敌军。③元 龙:陈元龙,即陈登,三国时人,素有扶世救民 的志向。④欲:欲望。 ⑤驱:赶逐。⑥浪迹: 到处漫游,行踪不定。⑦休:此处作“忘了”解。 ⑧井梧:水井边的梧桐树。
②从请假的理由来看,哈利、董事长和杰封都是请病假, 但身体都很健康,说明日复一日的繁重的工作已经使人 的心灵不堪重负,轻松愉悦的理想生活难以实现,已经 造成一种心灵的疾病。
③从小说的主题来看,三人都意外地碰到了一起,而且 心照不宣,都知道了彼此的秘密。说明生活的压力使人 们难以忍受,却又无法摆脱,只好用撒谎的办法来追求 自己想要的生活,表达了对现实生活的抗拒和无奈。
• 13.把文中画横线的句子翻译成现代汉语。(10 分)
(1)诗自以无劳,不安久居大郡,求欲降避功臣。
杜诗认为自己没有功劳,不能安于长期担任大郡 长官(为长期担任大郡长官而感到不安),于是 请求降职,把职位让给功臣。 【解析】计分点:“以”,认为;“不安”,不能 安于(为……感到不安);“降”,降职; “避”,避开,回避,联系前文来看,此指让出 职位。共4分。句意1分。
8.小说在以线性结构层层推进时,也常常“延迟”小说的进展,故 意给故事情节或人物心理的进展设置障碍。试举出《晴朗的夏日》 中一个“延迟”的例子,并说说其好处。(6分)
答案 “延迟”1:哈利出门钓鱼,并没直接遇到董 事长,而是12点转换地方时才恰好碰到。(2分) 这样安排的作用:①使得哈利有充足的时间独处, 充分感受逃离工作、融入自然的美好;(人物) ②故事节奏更缓和,氛围更轻松。(情节) ③使 “抗拒繁冗事务,追求快乐生活”的主题更鲜明。 (主旨)④董事长的“延迟”出场起到“意料之外, 情理之中”的效果,(效果)
14.下列对这首诗的分析与鉴赏,不恰当的一项是( ) A.首联第二句和杜甫《旅夜书怀》中“天地一沙鸥”句, 都借“沙鸥”意象表现诗人的闲适淡泊。 B.颔联写自己闲暇无所事事的时候才感觉度日如年,即 使是天大的事,喝醉了也就放下了。 C.颈联借助寒砧、深巷、残月、梧桐等诸多意象,描绘 出一幅凄清冷寂的画面,融情于景。 D.尾联用典,借志在扶世救民的陈登登上高楼极目远眺