2014年北京市高考英语答案

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第一篇:2014年北京市高考英语答案

2014 年普通高等学校招生全国统一考试

英语(北京卷)答案

第一部分:听力理解

第一节

1-5: CABBC

第二节

6-10: CACBA

11-15: ACABB

第三节

16.Dance17.middle18.Hinde19.10420.post

第二部分

第一节 单项填空

21-25:ACBAD26-30: BADCB31-35: CDADB

第二节 完形填空

36-40: ABCCD41-45: CBDDC46-50: BDAAB51-55:DABCA

第三节

56-60: ADBCB61-65: ACBDZ66-70: CBACD

第二节

71-75: GDFBA

第四部分

第一节

One possible version

Dear Chris,How is everything going?

I remember we planned to visit Yunnan in July but now bucause my left foot was injured, I cannot go with you as planned.I’m sorry about it.Can I suggest that we put it off until early august? I wish you could understand.I’m looking forward to hearing from you soon.Yours,Joe

第二节

内容要点

1.点明问题;

2.讨论问题;

3.提交建议;

4.看到变化。

One possible version

There was a problem with the parking place for bikes in our school, I noticed the entrance was small and almost blocked.So my classmates and I had a discussion and wrote a report.Then we went to meet the schoolmaster in his office and gave the report to him.He accepted our suggfestions.Soon afterwards, a second entrance was opened to the parking place.Now it is easy to park our bikes there.

第二篇:2011年北京市英语高考完形填空

I used to hate being called upon in class mainly because.I didn’t like attention drawn to myself.Andotherwise assigned(指定)a seat by the teacher.I always to sit at the back of the classroom.All this after I joined a sports team.It began when a teacher suggested.I try out for the basketball team.At first I thought it was a crazybecause I didn’t have a good sense of balance nor did I have theto keep pace with the others on the team and they would tease me.But for the teacher who kept insisting on my“ for it”, I wouldn’t have decided to give it a try.Getting up the courage to go to the tryouts was only theof it.When I first startedthe practice sessions.I didn’t even know the rules of the game much what I was doing.Sometimes I’d getand take a shot at the wrong direction—which made me ’t the only one “new” at the game, so I decided toon learning the game, do my best at each practice session, and not be too hard on myself for the things I didn’t“just yet”.I practiced and practiced.Soon I knew the and the “moves” Being part of a team was fun and motivating.Very soon the competitivein me was winning over my lack of confidence.With time, I learned how to play and made friends in the — friends who respected my efforts to work hard and be a team player.I never had so much fun!

With my self-confidence comes more praise from teachers and classmates.I have gone from “” in the back of the classroom and not wanting to call attention to myself.raising my hand — even when I sometimes wasn’t 100 percent I had the right answer.Now I have more self-confidence in myself.36.A.as B.until C.unless 37.A.hoped B.agreed C.meant

38.A.continued 39.A.idea 40.A.right 41.A.going 42.A.point 43.A.enjoying 44.A.less 45.A.committed 46.A.Interestingly 47.A.focus 48.A.want 49.A.steps 50.A.role 51.A.process 52.A.expressed 53.A.dreaming 54.A.by 55.A.lucky

B changed B.plan B.chance B.looking B.half B.preparingB.later B.motivated B.Fortunately B.act B.do B.orders B.part

B.operation B.improved B.playing B.for B.happy

C.settled C.belief C.ability C.cheering C.rest C.attending C.worse C.embarrassedC.Obviously C.rely C.support C.rules C.mind

C.movement C.preserved C.relaxing C.with C.sure

D.though D.chose D.started D.saying D.patience D.applying D.basis D.watching D.further D.confused D.Hopefully D.try D.know D.games D.value D.situation D.recognized D.hiding D.to

D.satisfied

36.C37.D38.B39.A40.C

41.A42.B43.C44.A45.D

46.B47.A48.D49.C50.B

51.A52.B53.D54.D55.C

第三篇:2014山东高考英语答案

2014年普通高等学校招生全国统一考试(山东)

英语试题参考答案

(第一卷)

第一、二部分— 5:DAACB6 —10:ABDCB11—15:CADCB16—20:CDABD 21—25:CBDAB26—30:DCDAC31—35:ABCDB36—40:DACAB 41—45:ABDCB46—50:CDBBD51—55:ACBAC56—60:ACBDA 61—65:DBCAC

(第二卷)

第一节

参考答案

66、Thereduced/lowered cost of a new car and better road conditions.67、They should park outside the town, and take the bus to the centre.68、Their incomes dropped because fewer people drove past.69、encouraging people to use public transport more

encouraging people to use their cars less

relieving the problem(of overowning)

70、Because public transport is not reliable/satisfactory.第二节

(略)

第四篇:北京市中考英语试卷及答案2013年

2013年北京市高考英语作文:描述照顾生病母亲的过

第一节 情景作文(20 分)

假设你是红星中学高三(1)班的学生李华,请按照以下四幅图的先后顺序,用英文写一篇周记。记述爸爸出 差期间,妈妈生病,你照顾她的过程。

注意: 1.周记的开头已经为你写好。

2.词数不少于 60.Last Monday,书面表达(共两节,15 分)

【参考范文】1

Last Monday, my father said goodbye to my mother and me and went on a business trip.He would be away for three days.Just the next morning I found my mother wasn’t feeling well.She had a cold.I immediately went to get her some medicine and then prepared some noodles for her.With my special care, my

mother recovered quickly.When my father came back home, my mother told him what had happened.He praised me for what I had done.I feel very happy that I have done something for my mother.【参考范文】2

Last Monday, my father would be on a business trip for five days.Having gotten my promise of being great at home and taking care of my mother, he rest assured and put his luggage into the trunk of his car.Seeing my father’s driving away, my mother and I waved our hands and said goodbye to him.For a moment, I began to miss my father, wishing that he would be safe and well the next days.I thought everything would go well, as this was not the first time that my father had been away for several days.Yet, unfortunately, my mother caught a cold the next morning.Looking at her pale face, I experienced high levels of anxiety.However, I told myself that I had to calm down and look after my mum, as I promised to my dad.The moment I got my mum to sleep, I put cold towel on her forehead, found pills in the medicine box, and made some noodles for her.Luckily, she waked up and felt better, after taking the pills and the noodles.That her fever was gone relieved and satisfied me a lot.In the next four days, I was taking her body temperature twice a day, ensuring that she was completely well.To our delight, my dad went back home safely and healthily on Saturday.On hearing what I had done to my mum, he, as well as my mum, beamed at me and gave me a big thumb.Hearing their praises and seeing

bright smiles on their faces, I really felt overjoyed and thrilled.Conceivably, taking care of my parents was, indeed and definitely, my mission and obligation.Hello, Jim!Look at this interesting picture.There is a man,whose head is almost bald,sitting on a large rock fishing in the river.He is thinking of catching a large fish,but when he pulls up the fishing line, he has caught a small fish instead.From the way the man turns down the corner of his mouth, it’s not difficult to see that he is not pleased by what he had caught.Of course, no one is pleased when the results turn out to be the opposite of what we have been expecting.But we should be grateful for what we have achieved.The man in the picture, indeed tried his best to catch a large fish, but he has failed.Instead of complaining what had luck he is having.He should be grateful and smile for the small fish he has caught, for there are many fishermen who often come back empty handed.Therefore,in our daily life, it we are all grateful for the smallest thing we achieve, no matter how much effort we have put into it, we will have a happy feeling.So do keep in mind that if we smile at life, life will smile back to us.In the way, we will be in a good mood everyday and it will make our life much more better.(2)情景作文(20分)2012 假设你是红星中学高三(1)班的学生李华,校报英文版正在开展“续写雷锋日记”活动。请根据以下四幅图的先后顺序,将你所做的一件好事以日记形式记述下来,向校报投稿。

注意:1.日记的开头已为你写好。

2.词数不少于60。

Saturday,June2

Fine

This morning_______________________________________________ _____________________________________________________________ _____________________________________________________________

[参考范文及解析] This morning, when I was walking on the street, I saw that two travelers were reading a map, looking puzzled.It seemed they were lost.I went up to them and asked how I could help.They told me they were looking for the Temple of Heaven.I led them to the nearby bus stop and advised them to take Bus No.20, the bus came.We waved good-bye to each other.Seeing them on the bus, I felt a kind of satisfaction.4幅图的要点分别是“偶遇游客”、“得知目的地”、“带领乘坐公交车”、“送行、致谢”

写作时先确定时态为过去时,然后分段:第一段是“偶遇游客”,第二段经过和结果,第三段抒情,方才完成记叙文“记叙+抒情”的写作思路。

范文中运用了非谓语动词作状语,非限定性定语从句等句式,用appropriated、satisfaction和puzzled等对人物情绪做了合理联想和表达。

给考生的启示:有限的写作时间内,用有限的单词表现对于高级句式和词汇的掌握程度,并以第一人称的角度对人物的感情做合理联想,使文章更为深刻。

第二节 开放作文(15分)

请根据下面提示,写一篇短文。词数不少于50。

You are discussing the following picture with your English friend Jim.Now you are telling him how you understand the picture and what makes you think so.[参考范文及解析]

You are discussing the following picture with your English friend Jim.Now you are telling him how you understand the picture and what makes you think so.One possible version:

I think the white pencil, looking proud and delighted, is laughing at the black pencil by saying “You're nearing the end!” The black pencil, on the other hand, remains calm.The white pencil's words let him fall into deep thoughts.He clearly remembers he has been used in writing and drawing.It's true that he's approaching the end but he has been living a memorable life and there has rarely been a dull moment.In reality, we should learn from the black pencil: not to mind what others say so long as we are confident in what we have done.立意可以是人生的价值在于过去所做的事情,而不在于生命所处的阶段,或者是新手应该尊重前辈,并向前辈学习。

范文中运用插入语,非谓语动词对图片进行描述后,又对短铅笔的想法做了合理联想,最后自然过渡到议论,比较自然。

给考生的启示:语言短小精悍,立意贴合图片,思路清晰连贯。

第四部分:书面表达(共两节,35分)第一节 情景作文(20分)假设你是红星中学高二(1)班的学生李华,下面四幅图描述了近期发生在你们班的一个真实故事,请根据图片的先后顺序,为校刊“英语园地”写一篇短文,词数不少于60。

(请务必将情景作文写在答题卡指定区域内)

第二节 开放作文(15分)请根据下面提示,写一篇短文。词数不少于50。

In your spoken English class, your teacher shows you the following picture.You are asked to describe the picture and explain how you understand it.(请务必将情景作文写在答题卡指定区域内)

2011年普通高等学校招生全国统一考试

第四部分:书面表达(共两节,35分)第一节 情景作文(20分)

一、内容要点:

1.老师滑倒

2.学生送老师看病 3.学生看望老师 4.老师回班上课

二、One possible version: Last Monday, we were having our Chinese class when the teacher suddenly slipped and fell.We were all worried about her.One of the boys carried her on the back, with the help of some others, to the clinic immediately.It turned out that her right leg was broken and she had to stay in the hospital.The following day, we went to visit her with flower and fruit.Seeing her lying in bed with leg wrapped in bandages, we felt sorry for her and hoped that she would recover soon.This Thursday she returned to the class on a wheelchair to give us lessons.We were all deeply moved and proud of having such a responsible teacher.第二节 开放作文(15分)One possible version: In the picture, there stands a tree full of fruit on one side of the stream.Across the stream, a

man is trying to reach out on the edge of the band for the fruit with a net attached to a pole.Not far away there is a bridge that can lead him to the tree for more fruit.The message conveyed in the picture is clear.In pursuing a dream, we might focus on only one say of making it come true, forgetting that there may be alternatives.As indicated in the picture, if the man is willing to look for other possibilities, he can find a better and more rewarding way to achieve his goal.All he has to do is to turn around, cross the bridge and walk to the tree.2010 第四部分:书面表达(共两节,共35分)

第一节:情景作文(20分)

假设你是红星中学高二一班学生李华,利用上周哦的时间帮助祖父母安排了去北戴河的旅行。请根据以下四幅图的先后顺序,写一篇英文周记,叙述你从准备到送行的全过程。

注意: 词数不少于60。提示词:郊区 suburbs

宾馆北戴河地图

Last weekend, I helped my grandparents prepare their trip to Beidaihe.第二节 开放作文(15分)

请根据下面提示,写一篇短文。词数不少于50。

In your spoken English class, your teacher shows you the following picture.You are asked to describe the picture and explain how you understand it.第四部分:书面表达(共两节,35分)第一节 情景作文(20分)

一、内容要点: 1.查询信息 2.买票 3.准备行装 4.送行

二、One possible version: Last weekend, I helped my grandparent prepare their trip to Beidaihe.On Saturday morning, together with my grandparents, I searched the Internet for the train schedule, the weather in Beidaihe, and some hotel information.In the afternoon, I went to the train station and managed to buy two tickets for my grandparents although there was a long queue.After dinner, I packed into the suitcase the tings my grandparents need, such as clothes, glasses, an umbrella, and a map.The next morning, I went to the station to see them off.Waving goodbye to them on the platform, I felt happy for them and wished them a safe journey.第二节 开放作文(15分)One possible version: In the picture, between two closely-located buildings grows a big tree.Unlike most trees, this one bends in the middle, struggling all the way up to get more sunshine.The picture reminds me of those how succeed in unfavorable conditions.Faced with difficulties, they never give up but try their best to find a way out.Life can be hard.But if we have the courage and determination, we will finally get the sunshine we want as the tree in the picture does.

第五篇:北京市高考数学试卷(理科)「附答案解析」

2018年北京市高考数学试卷(理科)一、选择题共8小题,每小题5分,共40分。在每小题列出的四个选项中,选出符合题目要求的一项。

1.(5分)已知集合A={x||x|<2},B={﹣2,0,1,2},则A∩B=()A.{0,1} B.{﹣1,0,1} C.{﹣2,0,1,2} D.{﹣1,0,1,2} 2.(5分)在复平面内,复数的共轭复数对应的点位于()A.第一象限 B.第二象限 C.第三象限 D.第四象限 3.(5分)执行如图所示的程序框图,输出的s值为()A. B. C. D. 4.(5分)“十二平均律”是通用的音律体系,明代朱载堉最早用数学方法计算出半音比例,为这个理论的发展做出了重要贡献,十二平均律将一个纯八度音程分成十二份,依次得到十三个单音,从第二个单音起,每一个单音的频率与它的前一个单音的频率的比都等于.若第一个单音的频率为f,则第八个单音的频率为()A.f B.f C.f D.f 5.(5分)某四棱锥的三视图如图所示,在此四棱锥的侧面中,直角三角形的个数为()A.1 B.2 C.3 D.4 6.(5分)设,均为单位向量,则“|﹣3|=|3+|”是“⊥”的()A.充分而不必要条件 B.必要而不充分条件 C.充分必要条件 D.既不充分也不必要条件 7.(5分)在平面直角坐标系中,记d为点P(cosθ,sinθ)到直线x﹣my﹣2=0的距离.当θ、m变化时,d的最大值为()A.1 B.2 C.3 D.4 8.(5分)设集合A={(x,y)|x﹣y≥1,ax+y>4,x﹣ay≤2},则()A.对任意实数a,(2,1)∈A B.对任意实数a,(2,1)∉A C.当且仅当a<0时,(2,1)∉A D.当且仅当a≤时,(2,1)∉A   二、填空题共6小题,每小题5分,共30分。

9.(5分)设{an}是等差数列,且a1=3,a2+a5=36,则{an}的通项公式为   . 10.(5分)在极坐标系中,直线ρcosθ+ρsinθ=a(a>0)与圆ρ=2cosθ相切,则a=   . 11.(5分)设函数f(x)=cos(ωx﹣)(ω>0),若f(x)≤f()对任意的实数x都成立,则ω的最小值为   . 12.(5分)若x,y满足x+1≤y≤2x,则2y﹣x的最小值是   . 13.(5分)能说明“若f(x)>f(0)对任意的x∈(0,2]都成立,则f(x)在[0,2]上是增函数”为假命题的一个函数是   . 14.(5分)已知椭圆M:+=1(a>b>0),双曲线N:﹣=1.若双曲线N的两条渐近线与椭圆M的四个交点及椭圆M的两个焦点恰为一个正六边形的顶点,则椭圆M的离心率为   ;

双曲线N的离心率为   .   三、解答题共6小题,共80分。解答应写出文字说明,演算步骤或证明过程。

15.(13分)在△ABC中,a=7,b=8,cosB=﹣.(Ⅰ)求∠A;

(Ⅱ)求AC边上的高. 16.(14分)如图,在三棱柱ABC﹣A1B1C1中,CC1⊥平面ABC,D,E,F,G分别为AA1,AC,A1C1,BB1的中点,AB=BC=,AC=AA1=2.(Ⅰ)求证:AC⊥平面BEF;

(Ⅱ)求二面角B﹣CD﹣C1的余弦值;

(Ⅲ)证明:直线FG与平面BCD相交. 17.(12分)电影公司随机收集了电影的有关数据,经分类整理得到下表:

电影类型 第一类 第二类 第三类 第四类 第五类 第六类 电影部数 140 50 300 200 800 510 好评率 0.4 0.2 0.15 0.25 0.2 0.1 好评率是指:一类电影中获得好评的部数与该类电影的部数的比值. 假设所有电影是否获得好评相互独立.(Ⅰ)从电影公司收集的电影中随机选取1部,求这部电影是获得好评的第四类电影的概率;

(Ⅱ)从第四类电影和第五类电影中各随机选取1部,估计恰有1部获得好评的概率;

(Ⅲ)假设每类电影得到人们喜欢的概率与表格中该类电影的好评率相等.用“ξk=1”表示第k类电影得到人们喜欢.“ξk=0”表示第k类电影没有得到人们喜欢(k=1,2,3,4,5,6).写出方差Dξ1,Dξ2,Dξ3,Dξ4,Dξ5,Dξ6的大小关系. 18.(13分)设函数f(x)=[ax2﹣(4a+1)x+4a+3]ex.(Ⅰ)若曲线y=f(x)在点(1,f(1))处的切线与x轴平行,求a;

(Ⅱ)若f(x)在x=2处取得极小值,求a的取值范围. 19.(14分)已知抛物线C:y2=2px经过点P(1,2),过点Q(0,1)的直线l与抛物线C有两个不同的交点A,B,且直线PA交y轴于M,直线PB交y轴于N.(Ⅰ)求直线l的斜率的取值范围;

(Ⅱ)设O为原点,=λ,=μ,求证:+为定值. 20.(14分)设n为正整数,集合A={α|α=(t1,t2,…tn),tk∈{0,1},k=1,2,…,n},对于集合A中的任意元素α=(x1,x2,…,xn)和β=(y1,y2,…yn),记 M(α,β)=[(x1+y1﹣|x1﹣y1|)+(x2+y2﹣|x2﹣y2|)+…(xn+yn﹣|xn﹣yn|)](Ⅰ)当n=3时,若α=(1,1,0),β=(0,1,1),求M(α,α)和M(α,β)的值;

(Ⅱ)当n=4时,设B是A的子集,且满足:对于B中的任意元素α,β,当α,β相同时,M(α,β)是奇数;

当α,β不同时,M(α,β)是偶数.求集合B中元素个数的最大值;

(Ⅲ)给定不小于2的n,设B是A的子集,且满足:对于B中的任意两个不同的元素α,β,M(α,β)=0,写出一个集合B,使其元素个数最多,并说明理由.   2018年北京市高考数学试卷(理科)参考答案与试题解析   一、选择题共8小题,每小题5分,共40分。在每小题列出的四个选项中,选出符合题目要求的一项。

1.(5分)已知集合A={x||x|<2},B={﹣2,0,1,2},则A∩B=()A.{0,1} B.{﹣1,0,1} C.{﹣2,0,1,2} D.{﹣1,0,1,2} 【解答】解:A={x||x|<2}={x|﹣2<x<2},B={﹣2,0,1,2},则A∩B={0,1},故选:A.   2.(5分)在复平面内,复数的共轭复数对应的点位于()A.第一象限 B.第二象限 C.第三象限 D.第四象限 【解答】解:复数==,共轭复数对应点的坐标(,﹣)在第四象限. 故选:D.   3.(5分)执行如图所示的程序框图,输出的s值为()A. B. C. D. 【解答】解:在执行第一次循环时,k=1,S=1. 在执行第一次循环时,S=1﹣=. 由于k=2≤3,所以执行下一次循环.S=,k=3,直接输出S=,故选:B.   4.(5分)“十二平均律”是通用的音律体系,明代朱载堉最早用数学方法计算出半音比例,为这个理论的发展做出了重要贡献,十二平均律将一个纯八度音程分成十二份,依次得到十三个单音,从第二个单音起,每一个单音的频率与它的前一个单音的频率的比都等于.若第一个单音的频率为f,则第八个单音的频率为()A.f B.f C.f D.f 【解答】解:从第二个单音起,每一个单音的频率与它的前一个单音的频率的比都等于. 若第一个单音的频率为f,则第八个单音的频率为:=. 故选:D.   5.(5分)某四棱锥的三视图如图所示,在此四棱锥的侧面中,直角三角形的个数为()A.1 B.2 C.3 D.4 【解答】解:四棱锥的三视图对应的直观图为:PA⊥底面ABCD,AC=,CD=,PC=3,PD=2,可得三角形PCD不是直角三角形. 所以侧面中有3个直角三角形,分别为:△PAB,△PBC,△PAD. 故选:C.   6.(5分)设,均为单位向量,则“|﹣3|=|3+|”是“⊥”的()A.充分而不必要条件 B.必要而不充分条件 C.充分必要条件 D.既不充分也不必要条件 【解答】解:∵“|﹣3|=|3+|” ∴平方得||2+9||2﹣6•=||2+9||2+6• 则•=0,即⊥,则“|﹣3|=|3+|”是“⊥”的充要条件,故选:C.   7.(5分)在平面直角坐标系中,记d为点P(cosθ,sinθ)到直线x﹣my﹣2=0的距离.当θ、m变化时,d的最大值为()A.1 B.2 C.3 D.4 【解答】解:由题意d==,tanα=﹣,∴当sin(θ+α)=﹣1时,dmax=1+≤3. ∴d的最大值为3. 故选:C.   8.(5分)设集合A={(x,y)|x﹣y≥1,ax+y>4,x﹣ay≤2},则()A.对任意实数a,(2,1)∈A B.对任意实数a,(2,1)∉A C.当且仅当a<0时,(2,1)∉A D.当且仅当a≤时,(2,1)∉A 【解答】解:当a=﹣1时,集合A={(x,y)|x﹣y≥1,ax+y>4,x﹣ay≤2}={(x,y)|x﹣y≥1,﹣x+y>4,x+y≤2},显然(2,1)不满足,﹣x+y>4,x+y≤2,所以A,C不正确;

当a=4,集合A={(x,y)|x﹣y≥1,ax+y>4,x﹣ay≤2}={(x,y)|x﹣y≥1,4x+y>4,x﹣4y≤2},显然(2,1)在可行域内,满足不等式,所以B不正确;

故选:D.   二、填空题共6小题,每小题5分,共30分。

9.(5分)设{an}是等差数列,且a1=3,a2+a5=36,则{an}的通项公式为 an=6n﹣3 . 【解答】解:∵{an}是等差数列,且a1=3,a2+a5=36,∴,解得a1=3,d=6,∴an=a1+(n﹣1)d=3+(n﹣1)×6=6n﹣3. ∴{an}的通项公式为an=6n﹣3. 故答案为:an=6n﹣3.   10.(5分)在极坐标系中,直线ρcosθ+ρsinθ=a(a>0)与圆ρ=2cosθ相切,则a= 1+ . 【解答】解:圆ρ=2cosθ,转化成:ρ2=2ρcosθ,进一步转化成直角坐标方程为:(x﹣1)2+y2=1,把直线ρ(cosθ+sinθ)=a的方程转化成直角坐标方程为:x+y﹣a=0. 由于直线和圆相切,所以:利用圆心到直线的距离等于半径. 则:=1,解得:a=1±.a>0 则负值舍去. 故:a=1+. 故答案为:1+.   11.(5分)设函数f(x)=cos(ωx﹣)(ω>0),若f(x)≤f()对任意的实数x都成立,则ω的最小值为. 【解答】解:函数f(x)=cos(ωx﹣)(ω>0),若f(x)≤f()对任意的实数x都成立,可得:,k∈Z,解得ω=,k∈Z,ω>0 则ω的最小值为:. 故答案为:.   12.(5分)若x,y满足x+1≤y≤2x,则2y﹣x的最小值是 3 . 【解答】解:作出不等式组对应的平面区域如图:

设z=2y﹣x,则y=x+z,平移y=x+z,由图象知当直线y=x+z经过点A时,直线的截距最小,此时z最小,由得,即A(1,2),此时z=2×2﹣1=3,故答案为:3   13.(5分)能说明“若f(x)>f(0)对任意的x∈(0,2]都成立,则f(x)在[0,2]上是增函数”为假命题的一个函数是 f(x)=sinx . 【解答】解:例如f(x)=sinx,尽管f(x)>f(0)对任意的x∈(0,2]都成立,当x∈[0,)上为增函数,在(,2]为减函数,故答案为:f(x)=sinx.   14.(5分)已知椭圆M:+=1(a>b>0),双曲线N:﹣=1.若双曲线N的两条渐近线与椭圆M的四个交点及椭圆M的两个焦点恰为一个正六边形的顶点,则椭圆M的离心率为;

双曲线N的离心率为 2 . 【解答】解:椭圆M:+=1(a>b>0),双曲线N:﹣=1.若双曲线N的两条渐近线与椭圆M的四个交点及椭圆M的两个焦点恰为一个正六边形的顶点,可得椭圆的焦点坐标(c,0),正六边形的一个顶点(,),可得:,可得,可得e4﹣8e2+4=0,e∈(0,1),解得e=. 同时,双曲线的渐近线的斜率为,即,可得:,即,可得双曲线的离心率为e==2. 故答案为:;

2.   三、解答题共6小题,共80分。解答应写出文字说明,演算步骤或证明过程。

15.(13分)在△ABC中,a=7,b=8,cosB=﹣.(Ⅰ)求∠A;

(Ⅱ)求AC边上的高. 【解答】解:(Ⅰ)∵a<b,∴A<B,即A是锐角,∵cosB=﹣,∴sinB===,由正弦定理得=得sinA===,则A=.(Ⅱ)由余弦定理得b2=a2+c2﹣2accosB,即64=49+c2+2×7×c×,即c2+2c﹣15=0,得(c﹣3)(c+5)=0,得c=3或c=﹣5(舍),则AC边上的高h=csinA=3×=.   16.(14分)如图,在三棱柱ABC﹣A1B1C1中,CC1⊥平面ABC,D,E,F,G分别为AA1,AC,A1C1,BB1的中点,AB=BC=,AC=AA1=2.(Ⅰ)求证:AC⊥平面BEF;

(Ⅱ)求二面角B﹣CD﹣C1的余弦值;

(Ⅲ)证明:直线FG与平面BCD相交. 【解答】(I)证明:∵E,F分别是AC,A1C1的中点,∴EF∥CC1,∵CC1⊥平面ABC,∴EF⊥平面ABC,又AC⊂平面ABC,∴EF⊥AC,∵AB=BC,E是AC的中点,∴BE⊥AC,又BE∩EF=E,BE⊂平面BEF,EF⊂平面BEF,∴AC⊥平面BEF.(II)解:以E为原点,以EB,EC,EF为坐标轴建立空间直角坐标系如图所示:

则B(2,0,0),C(0,1,0),D(0,﹣1,1),∴=(﹣2,1,0),=(0,﹣2,1),设平面BCD的法向量为=(x,y,z),则,即,令y=2可得=(1,2,4),又EB⊥平面ACC1A1,∴=(2,0,0)为平面CD﹣C1的一个法向量,∴cos<,>===. 由图形可知二面角B﹣CD﹣C1为钝二面角,∴二面角B﹣CD﹣C1的余弦值为﹣.(III)证明:F(0,0,2),(2,0,1),∴=(2,0,﹣1),∴•=2+0﹣4=﹣2≠0,∴与不垂直,∴FG与平面BCD不平行,又FG⊄平面BCD,∴FG与平面BCD相交.   17.(12分)电影公司随机收集了电影的有关数据,经分类整理得到下表:

电影类型 第一类 第二类 第三类 第四类 第五类 第六类 电影部数 140 50 300 200 800 510 好评率 0.4 0.2 0.15 0.25 0.2 0.1 好评率是指:一类电影中获得好评的部数与该类电影的部数的比值. 假设所有电影是否获得好评相互独立.(Ⅰ)从电影公司收集的电影中随机选取1部,求这部电影是获得好评的第四类电影的概率;

(Ⅱ)从第四类电影和第五类电影中各随机选取1部,估计恰有1部获得好评的概率;

(Ⅲ)假设每类电影得到人们喜欢的概率与表格中该类电影的好评率相等.用“ξk=1”表示第k类电影得到人们喜欢.“ξk=0”表示第k类电影没有得到人们喜欢(k=1,2,3,4,5,6).写出方差Dξ1,Dξ2,Dξ3,Dξ4,Dξ5,Dξ6的大小关系. 【解答】解:(Ⅰ)设事件A表示“从电影公司收集的电影中随机选取1部,求这部电影是获得好评的第四类电影”,总的电影部数为140+50+300+200+800+510=2000部,第四类电影中获得好评的电影有:200×0.25=50部,∴从电影公司收集的电影中随机选取1部,求这部电影是获得好评的第四类电影的频率为:

P(A)==0.025.(Ⅱ)设事件B表示“从第四类电影和第五类电影中各随机选取1部,恰有1部获得好评”,第四类获得好评的有:200×0.25=50部,第五类获得好评的有:800×0.2=160部,则从第四类电影和第五类电影中各随机选取1部,估计恰有1部获得好评的概率:

P(B)==0.35.(Ⅲ)由题意知,定义随机变量如下:

ξk=,则ξk服从两点分布,则六类电影的分布列及方差计算如下:

第一类电影:

ξ1 1 0 P 0.4 0.6 E(ξ1)=1×0.4+0×0.6=0.4,D(ξ1)=(1﹣0.4)2×0.4+(0﹣0.4)2×0.6=0.24. 第二类电影:

ξ2 1 0 P 0.2 0.8 E(ξ2)=1×0.2+0×0.8=0.2,D(ξ2)=(1﹣0.2)2×0.2+(0﹣0.2)2×0.8=0.16. 第三类电影:

ξ3 1 0 P 0.15 0.85 E(ξ3)=1×0.15+0×0.85=0.15,D(ξ3)=(1﹣0.15)2×0.15+(0﹣0.85)2×0.85=0.1275. 第四类电影:

ξ4 1 0 P 0.25 0.75 E(ξ4)=1×0.25+0×0.75=0.15,D(ξ4)=(1﹣0.25)2×0.25+(0﹣0.75)2×0.75=0.1875. 第五类电影:

ξ5 1 0 P 0.2 0.8 E(ξ5)=1×0.2+0×0.8=0.2,D(ξ5)=(1﹣0.2)2×0.2+(0﹣0.2)2×0.8=0.16. 第六类电影:

ξ6 1 0 P 0.1 0.9 E(ξ6)=1×0.1+0×0.9=0.1,D(ξ5)=(1﹣0.1)2×0.1+(0﹣0.1)2×0.9=0.09. ∴方差Dξ1,Dξ2,Dξ3,Dξ4,Dξ5,Dξ6的大小关系为:

Dξ6<Dξ3<Dξ2=Dξ5<Dξ4<Dξ1.   18.(13分)设函数f(x)=[ax2﹣(4a+1)x+4a+3]ex.(Ⅰ)若曲线y=f(x)在点(1,f(1))处的切线与x轴平行,求a;

(Ⅱ)若f(x)在x=2处取得极小值,求a的取值范围. 【解答】解:(Ⅰ)函数f(x)=[ax2﹣(4a+1)x+4a+3]ex的导数为 f′(x)=[ax2﹣(2a+1)x+2]ex. 由题意可得曲线y=f(x)在点(1,f(1))处的切线斜率为0,可得(a﹣2a﹣1+2)e=0,解得a=1;

(Ⅱ)f(x)的导数为f′(x)=[ax2﹣(2a+1)x+2]ex=(x﹣2)(ax﹣1)ex,若a=0则x<2时,f′(x)>0,f(x)递增;

x>2,f′(x)<0,f(x)递减. x=2处f(x)取得极大值,不符题意;

若a>0,且a=,则f′(x)=(x﹣2)2ex≥0,f(x)递增,无极值;

若a>,则<2,f(x)在(,2)递减;

在(2,+∞),(﹣∞,)递增,可得f(x)在x=2处取得极小值;

若0<a<,则>2,f(x)在(2,)递减;

在(,+∞),(﹣∞,2)递增,可得f(x)在x=2处取得极大值,不符题意;

若a<0,则<2,f(x)在(,2)递增;

在(2,+∞),(﹣∞,)递减,可得f(x)在x=2处取得极大值,不符题意. 综上可得,a的范围是(,+∞).   19.(14分)已知抛物线C:y2=2px经过点P(1,2),过点Q(0,1)的直线l与抛物线C有两个不同的交点A,B,且直线PA交y轴于M,直线PB交y轴于N.(Ⅰ)求直线l的斜率的取值范围;

(Ⅱ)设O为原点,=λ,=μ,求证:+为定值. 【解答】解:(Ⅰ)∵抛物线C:y2=2px经过点 P(1,2),∴4=2p,解得p=2,设过点(0,1)的直线方程为y=kx+1,设A(x1,y1),B(x2,y2)联立方程组可得,消y可得k2x2+(2k﹣4)x+1=0,∴△=(2k﹣4)2﹣4k2>0,且k≠0解得k<1,且k≠0,x1+x2=﹣,x1x2=,故直线l的斜率的取值范围(﹣∞,0)∪(0,1);

(Ⅱ)证明:设点M(0,yM),N(0,yN),则=(0,yM﹣1),=(0,﹣1)因为=λ,所以yM﹣1=﹣yM﹣1,故λ=1﹣yM,同理μ=1﹣yN,直线PA的方程为y﹣2=(x﹣1)=(x﹣1)=(x﹣1),令x=0,得yM=,同理可得yN=,因为+=+=+===== =2,∴+=2,∴+为定值.   20.(14分)设n为正整数,集合A={α|α=(t1,t2,…tn),tk∈{0,1},k=1,2,…,n},对于集合A中的任意元素α=(x1,x2,…,xn)和β=(y1,y2,…yn),记 M(α,β)=[(x1+y1﹣|x1﹣y1|)+(x2+y2﹣|x2﹣y2|)+…(xn+yn﹣|xn﹣yn|)](Ⅰ)当n=3时,若α=(1,1,0),β=(0,1,1),求M(α,α)和M(α,β)的值;

(Ⅱ)当n=4时,设B是A的子集,且满足:对于B中的任意元素α,β,当α,β相同时,M(α,β)是奇数;

当α,β不同时,M(α,β)是偶数.求集合B中元素个数的最大值;

(Ⅲ)给定不小于2的n,设B是A的子集,且满足:对于B中的任意两个不同的元素α,β,M(α,β)=0,写出一个集合B,使其元素个数最多,并说明理由. 【解答】解:(I)M(a,a)=2,M(a,β)=1.(II)考虑数对(xk,yk)只有四种情况:(0,0)、(0,1)、(1,0)、(1,1),相应的分别为0、0、0、1,所以B中的每个元素应有奇数个1,所以B中的元素只可能为(上下对应的两个元素称之为互补元素):

(1,0,0,0)、(0,1,0,0)、(0,0,1,0)、(0,0,0,1),(0,1,1,1)、(1,0,1,1)、(1,1,0,1)、(1,1,1,0),对于任意两个只有1个1的元素α,β都满足M(α,β)是偶数,所以四元集合B={(1,0,0,0)、(0,1,0,0)、(0,0,1,0)、(0,0,0,1)}满足 题意,假设B中元素个数大于等于4,就至少有一对互补元素,除了这对互补元素之外还有至少1个含有3个1的元素α,则互补元素中含有1个1的元素β与之满足M(α,β)=1不合题意,故B中元素个数的最大值为4.(Il)B={(0,0,0,…0),(1,0,0…,0),(0,1,0,…0),(0,0,1…0)…,(0,0,0,…,1)},此时B中有n+1个元素,下证其为最大. 对于任意两个不同的元素α,β,满足M(α,β)=0,则α,β中相同位置上的数字不能同时为1,假设存在B有多于n+1个元素,由于α=(0,0,0,…,0)与任意元素β都有M(α,β)=0,所以除(0,0,0,…,0)外至少有n+1个元素含有1,根据元素的互异性,至少存在一对α,β满足xi=yi=l,此时M(α,β)≥1不满足题意,故B中最多有n+1个元素.   — END —

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