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      2014年北京市高考英語(yǔ)答案

      時(shí)間:2019-05-14 18:35:45下載本文作者:會(huì)員上傳
      簡(jiǎn)介:寫寫幫文庫(kù)小編為你整理了多篇相關(guān)的《2014年北京市高考英語(yǔ)答案》,但愿對(duì)你工作學(xué)習(xí)有幫助,當(dāng)然你在寫寫幫文庫(kù)還可以找到更多《2014年北京市高考英語(yǔ)答案》。

      第一篇:2014年北京市高考英語(yǔ)答案

      2014 年普通高等學(xué)校招生全國(guó)統(tǒng)一考試

      英語(yǔ)(北京卷)答案

      第一部分:聽(tīng)力理解

      第一節(jié)

      1-5: CABBC

      第二節(jié)

      6-10: CACBA

      11-15: ACABB

      第三節(jié)

      16.Dance17.middle18.Hinde19.10420.post

      第二部分

      第一節(jié) 單項(xiàng)填空

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

      第二節(jié) 完形填空

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

      第三節(jié)

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

      第二節(jié)

      71-75: GDFBA

      第四部分

      第一節(jié)

      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

      第二節(jié)

      內(nèi)容要點(diǎn)

      1.點(diǎn)明問(wèn)題;

      2.討論問(wèn)題;

      3.提交建議;

      4.看到變化。

      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年北京市英語(yǔ)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山東高考英語(yǔ)答案

      2014年普通高等學(xué)校招生全國(guó)統(tǒng)一考試(山東)

      英語(yǔ)試題參考答案

      (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)

      第一節(jié)

      參考答案

      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.第二節(jié)

      (略)

      第四篇:北京市中考英語(yǔ)試卷及答案2013年

      2013年北京市高考英語(yǔ)作文:描述照顧生病母親的過(guò)

      第一節(jié) 情景作文(20 分)

      假設(shè)你是紅星中學(xué)高三(1)班的學(xué)生李華,請(qǐng)按照以下四幅圖的先后順序,用英文寫一篇周記。記述爸爸出 差期間,媽媽生病,你照顧她的過(guò)程。

      注意: 1.周記的開(kāi)頭已經(jīng)為你寫好。

      2.詞數(shù)不少于 60.Last Monday,書面表達(dá)(共兩節(jié),15 分)

      【參考范文】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.【參考范文】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 假設(shè)你是紅星中學(xué)高三(1)班的學(xué)生李華,校報(bào)英文版正在開(kāi)展“續(xù)寫雷鋒日記”活動(dòng)。請(qǐng)根據(jù)以下四幅圖的先后順序,將你所做的一件好事以日記形式記述下來(lái),向校報(bào)投稿。

      注意:1.日記的開(kāi)頭已為你寫好。

      2.詞數(shù)不少于60。

      Saturday,June2

      Fine

      This morning_______________________________________________ _____________________________________________________________ _____________________________________________________________

      [參考范文及解析] 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幅圖的要點(diǎn)分別是“偶遇游客”、“得知目的地”、“帶領(lǐng)乘坐公交車”、“送行、致謝”

      寫作時(shí)先確定時(shí)態(tài)為過(guò)去時(shí),然后分段:第一段是“偶遇游客”,第二段經(jīng)過(guò)和結(jié)果,第三段抒情,方才完成記敘文“記敘+抒情”的寫作思路。

      范文中運(yùn)用了非謂語(yǔ)動(dòng)詞作狀語(yǔ),非限定性定語(yǔ)從句等句式,用appropriated、satisfaction和puzzled等對(duì)人物情緒做了合理聯(lián)想和表達(dá)。

      給考生的啟示:有限的寫作時(shí)間內(nèi),用有限的單詞表現(xiàn)對(duì)于高級(jí)句式和詞匯的掌握程度,并以第一人稱的角度對(duì)人物的感情做合理聯(lián)想,使文章更為深刻。

      第二節(jié) 開(kāi)放作文(15分)

      請(qǐng)根據(jù)下面提示,寫一篇短文。詞數(shù)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.[參考范文及解析]

      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.立意可以是人生的價(jià)值在于過(guò)去所做的事情,而不在于生命所處的階段,或者是新手應(yīng)該尊重前輩,并向前輩學(xué)習(xí)。

      范文中運(yùn)用插入語(yǔ),非謂語(yǔ)動(dòng)詞對(duì)圖片進(jìn)行描述后,又對(duì)短鉛筆的想法做了合理聯(lián)想,最后自然過(guò)渡到議論,比較自然。

      給考生的啟示:語(yǔ)言短小精悍,立意貼合圖片,思路清晰連貫。

      第四部分:書面表達(dá)(共兩節(jié),35分)第一節(jié) 情景作文(20分)假設(shè)你是紅星中學(xué)高二(1)班的學(xué)生李華,下面四幅圖描述了近期發(fā)生在你們班的一個(gè)真實(shí)故事,請(qǐng)根據(jù)圖片的先后順序,為??坝⒄Z(yǔ)園地”寫一篇短文,詞數(shù)不少于60。

      (請(qǐng)務(wù)必將情景作文寫在答題卡指定區(qū)域內(nèi))

      第二節(jié) 開(kāi)放作文(15分)請(qǐng)根據(jù)下面提示,寫一篇短文。詞數(shù)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.(請(qǐng)務(wù)必將情景作文寫在答題卡指定區(qū)域內(nèi))

      2011年普通高等學(xué)校招生全國(guó)統(tǒng)一考試

      第四部分:書面表達(dá)(共兩節(jié),35分)第一節(jié) 情景作文(20分)

      一、內(nèi)容要點(diǎn):

      1.老師滑倒

      2.學(xué)生送老師看病 3.學(xué)生看望老師 4.老師回班上課

      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.第二節(jié) 開(kāi)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:書面表達(dá)(共兩節(jié),共35分)

      第一節(jié):情景作文(20分)

      假設(shè)你是紅星中學(xué)高二一班學(xué)生李華,利用上周哦的時(shí)間幫助祖父母安排了去北戴河的旅行。請(qǐng)根據(jù)以下四幅圖的先后順序,寫一篇英文周記,敘述你從準(zhǔn)備到送行的全過(guò)程。

      注意: 詞數(shù)不少于60。提示詞:郊區(qū) suburbs

      賓館北戴河地圖

      Last weekend, I helped my grandparents prepare their trip to Beidaihe.第二節(jié) 開(kāi)放作文(15分)

      請(qǐng)根據(jù)下面提示,寫一篇短文。詞數(shù)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:書面表達(dá)(共兩節(jié),35分)第一節(jié) 情景作文(20分)

      一、內(nèi)容要點(diǎn): 1.查詢信息 2.買票 3.準(zhǔn)備行裝 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.第二節(jié) 開(kāi)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:北京市高考數(shù)學(xué)試卷(理科)「附答案解析」

      2018年北京市高考數(shù)學(xué)試卷(理科)一、選擇題共8小題,每小題5分,共40分。在每小題列出的四個(gè)選項(xiàng)中,選出符合題目要求的一項(xiàng)。

      1.(5分)已知集合A={x||x|<2},B={﹣2,0,1,2},則A∩B=()A.{0,1} B.{﹣1,0,1} C.{﹣2,0,1,2} D.{﹣1,0,1,2} 2.(5分)在復(fù)平面內(nèi),復(fù)數(shù)的共軛復(fù)數(shù)對(duì)應(yīng)的點(diǎn)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)執(zhí)行如圖所示的程序框圖,輸出的s值為()A. B. C. D. 4.(5分)“十二平均律”是通用的音律體系,明代朱載堉最早用數(shù)學(xué)方法計(jì)算出半音比例,為這個(gè)理論的發(fā)展做出了重要貢獻(xiàn),十二平均律將一個(gè)純八度音程分成十二份,依次得到十三個(gè)單音,從第二個(gè)單音起,每一個(gè)單音的頻率與它的前一個(gè)單音的頻率的比都等于.若第一個(gè)單音的頻率為f,則第八個(gè)單音的頻率為()A.f B.f C.f D.f 5.(5分)某四棱錐的三視圖如圖所示,在此四棱錐的側(cè)面中,直角三角形的個(gè)數(shù)為()A.1 B.2 C.3 D.4 6.(5分)設(shè),均為單位向量,則“|﹣3|=|3+|”是“⊥”的()A.充分而不必要條件 B.必要而不充分條件 C.充分必要條件 D.既不充分也不必要條件 7.(5分)在平面直角坐標(biāo)系中,記d為點(diǎn)P(cosθ,sinθ)到直線x﹣my﹣2=0的距離.當(dāng)θ、m變化時(shí),d的最大值為()A.1 B.2 C.3 D.4 8.(5分)設(shè)集合A={(x,y)|x﹣y≥1,ax+y>4,x﹣ay≤2},則()A.對(duì)任意實(shí)數(shù)a,(2,1)∈A B.對(duì)任意實(shí)數(shù)a,(2,1)?A C.當(dāng)且僅當(dāng)a<0時(shí),(2,1)?A D.當(dāng)且僅當(dāng)a≤時(shí),(2,1)?A   二、填空題共6小題,每小題5分,共30分。

      9.(5分)設(shè){an}是等差數(shù)列,且a1=3,a2+a5=36,則{an}的通項(xiàng)公式為  ?。? 10.(5分)在極坐標(biāo)系中,直線ρcosθ+ρsinθ=a(a>0)與圓ρ=2cosθ相切,則a=  ?。? 11.(5分)設(shè)函數(shù)f(x)=cos(ωx﹣)(ω>0),若f(x)≤f()對(duì)任意的實(shí)數(shù)x都成立,則ω的最小值為  ?。? 12.(5分)若x,y滿足x+1≤y≤2x,則2y﹣x的最小值是  ?。? 13.(5分)能說(shuō)明“若f(x)>f(0)對(duì)任意的x∈(0,2]都成立,則f(x)在[0,2]上是增函數(shù)”為假命題的一個(gè)函數(shù)是  ?。? 14.(5分)已知橢圓M:+=1(a>b>0),雙曲線N:﹣=1.若雙曲線N的兩條漸近線與橢圓M的四個(gè)交點(diǎn)及橢圓M的兩個(gè)焦點(diǎn)恰為一個(gè)正六邊形的頂點(diǎn),則橢圓M的離心率為   ;

      雙曲線N的離心率為   .   三、解答題共6小題,共80分。解答應(yīng)寫出文字說(shuō)明,演算步驟或證明過(guò)程。

      15.(13分)在△ABC中,a=7,b=8,cosB=﹣.(Ⅰ)求∠A;

      (Ⅱ)求AC邊上的高. 16.(14分)如圖,在三棱柱ABC﹣A1B1C1中,CC1⊥平面ABC,D,E,F(xiàn),G分別為AA1,AC,A1C1,BB1的中點(diǎn),AB=BC=,AC=AA1=2.(Ⅰ)求證:AC⊥平面BEF;

      (Ⅱ)求二面角B﹣CD﹣C1的余弦值;

      (Ⅲ)證明:直線FG與平面BCD相交. 17.(12分)電影公司隨機(jī)收集了電影的有關(guān)數(shù)據(jù),經(jīng)分類整理得到下表:

      電影類型 第一類 第二類 第三類 第四類 第五類 第六類 電影部數(shù) 140 50 300 200 800 510 好評(píng)率 0.4 0.2 0.15 0.25 0.2 0.1 好評(píng)率是指:一類電影中獲得好評(píng)的部數(shù)與該類電影的部數(shù)的比值. 假設(shè)所有電影是否獲得好評(píng)相互獨(dú)立.(Ⅰ)從電影公司收集的電影中隨機(jī)選取1部,求這部電影是獲得好評(píng)的第四類電影的概率;

      (Ⅱ)從第四類電影和第五類電影中各隨機(jī)選取1部,估計(jì)恰有1部獲得好評(píng)的概率;

      (Ⅲ)假設(shè)每類電影得到人們喜歡的概率與表格中該類電影的好評(píng)率相等.用“ξk=1”表示第k類電影得到人們喜歡.“ξk=0”表示第k類電影沒(méi)有得到人們喜歡(k=1,2,3,4,5,6).寫出方差Dξ1,Dξ2,Dξ3,Dξ4,Dξ5,Dξ6的大小關(guān)系. 18.(13分)設(shè)函數(shù)f(x)=[ax2﹣(4a+1)x+4a+3]ex.(Ⅰ)若曲線y=f(x)在點(diǎn)(1,f(1))處的切線與x軸平行,求a;

      (Ⅱ)若f(x)在x=2處取得極小值,求a的取值范圍. 19.(14分)已知拋物線C:y2=2px經(jīng)過(guò)點(diǎn)P(1,2),過(guò)點(diǎn)Q(0,1)的直線l與拋物線C有兩個(gè)不同的交點(diǎn)A,B,且直線PA交y軸于M,直線PB交y軸于N.(Ⅰ)求直線l的斜率的取值范圍;

      (Ⅱ)設(shè)O為原點(diǎn),=λ,=μ,求證:+為定值. 20.(14分)設(shè)n為正整數(shù),集合A={α|α=(t1,t2,…tn),tk∈{0,1},k=1,2,…,n},對(duì)于集合A中的任意元素α=(x1,x2,…,xn)和β=(y1,y2,…yn),記 M(α,β)=[(x1+y1﹣|x1﹣y1|)+(x2+y2﹣|x2﹣y2|)+…(xn+yn﹣|xn﹣yn|)](Ⅰ)當(dāng)n=3時(shí),若α=(1,1,0),β=(0,1,1),求M(α,α)和M(α,β)的值;

      (Ⅱ)當(dāng)n=4時(shí),設(shè)B是A的子集,且滿足:對(duì)于B中的任意元素α,β,當(dāng)α,β相同時(shí),M(α,β)是奇數(shù);

      當(dāng)α,β不同時(shí),M(α,β)是偶數(shù).求集合B中元素個(gè)數(shù)的最大值;

      (Ⅲ)給定不小于2的n,設(shè)B是A的子集,且滿足:對(duì)于B中的任意兩個(gè)不同的元素α,β,M(α,β)=0,寫出一個(gè)集合B,使其元素個(gè)數(shù)最多,并說(shuō)明理由.   2018年北京市高考數(shù)學(xué)試卷(理科)參考答案與試題解析   一、選擇題共8小題,每小題5分,共40分。在每小題列出的四個(gè)選項(xiàng)中,選出符合題目要求的一項(xiàng)。

      1.(5分)已知集合A={x||x|<2},B={﹣2,0,1,2},則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},則A∩B={0,1},故選:A.   2.(5分)在復(fù)平面內(nèi),復(fù)數(shù)的共軛復(fù)數(shù)對(duì)應(yīng)的點(diǎn)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:復(fù)數(shù)==,共軛復(fù)數(shù)對(duì)應(yīng)點(diǎn)的坐標(biāo)(,﹣)在第四象限. 故選:D.   3.(5分)執(zhí)行如圖所示的程序框圖,輸出的s值為()A. B. C. D. 【解答】解:在執(zhí)行第一次循環(huán)時(shí),k=1,S=1. 在執(zhí)行第一次循環(huán)時(shí),S=1﹣=. 由于k=2≤3,所以執(zhí)行下一次循環(huán).S=,k=3,直接輸出S=,故選:B.   4.(5分)“十二平均律”是通用的音律體系,明代朱載堉最早用數(shù)學(xué)方法計(jì)算出半音比例,為這個(gè)理論的發(fā)展做出了重要貢獻(xiàn),十二平均律將一個(gè)純八度音程分成十二份,依次得到十三個(gè)單音,從第二個(gè)單音起,每一個(gè)單音的頻率與它的前一個(gè)單音的頻率的比都等于.若第一個(gè)單音的頻率為f,則第八個(gè)單音的頻率為()A.f B.f C.f D.f 【解答】解:從第二個(gè)單音起,每一個(gè)單音的頻率與它的前一個(gè)單音的頻率的比都等于. 若第一個(gè)單音的頻率為f,則第八個(gè)單音的頻率為:=. 故選:D.   5.(5分)某四棱錐的三視圖如圖所示,在此四棱錐的側(cè)面中,直角三角形的個(gè)數(shù)為()A.1 B.2 C.3 D.4 【解答】解:四棱錐的三視圖對(duì)應(yīng)的直觀圖為:PA⊥底面ABCD,AC=,CD=,PC=3,PD=2,可得三角形PCD不是直角三角形. 所以側(cè)面中有3個(gè)直角三角形,分別為:△PAB,△PBC,△PAD. 故選:C.   6.(5分)設(shè),均為單位向量,則“|﹣3|=|3+|”是“⊥”的()A.充分而不必要條件 B.必要而不充分條件 C.充分必要條件 D.既不充分也不必要條件 【解答】解:∵“|﹣3|=|3+|” ∴平方得||2+9||2﹣6?=||2+9||2+6? 則?=0,即⊥,則“|﹣3|=|3+|”是“⊥”的充要條件,故選:C.   7.(5分)在平面直角坐標(biāo)系中,記d為點(diǎn)P(cosθ,sinθ)到直線x﹣my﹣2=0的距離.當(dāng)θ、m變化時(shí),d的最大值為()A.1 B.2 C.3 D.4 【解答】解:由題意d==,tanα=﹣,∴當(dāng)sin(θ+α)=﹣1時(shí),dmax=1+≤3. ∴d的最大值為3. 故選:C.   8.(5分)設(shè)集合A={(x,y)|x﹣y≥1,ax+y>4,x﹣ay≤2},則()A.對(duì)任意實(shí)數(shù)a,(2,1)∈A B.對(duì)任意實(shí)數(shù)a,(2,1)?A C.當(dāng)且僅當(dāng)a<0時(shí),(2,1)?A D.當(dāng)且僅當(dāng)a≤時(shí),(2,1)?A 【解答】解:當(dāng)a=﹣1時(shí),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},顯然(2,1)不滿足,﹣x+y>4,x+y≤2,所以A,C不正確;

      當(dāng)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},顯然(2,1)在可行域內(nèi),滿足不等式,所以B不正確;

      故選:D.   二、填空題共6小題,每小題5分,共30分。

      9.(5分)設(shè){an}是等差數(shù)列,且a1=3,a2+a5=36,則{an}的通項(xiàng)公式為 an=6n﹣3?。? 【解答】解:∵{an}是等差數(shù)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}的通項(xiàng)公式為an=6n﹣3. 故答案為:an=6n﹣3.   10.(5分)在極坐標(biāo)系中,直線ρcosθ+ρsinθ=a(a>0)與圓ρ=2cosθ相切,則a= 1+?。? 【解答】解:圓ρ=2cosθ,轉(zhuǎn)化成:ρ2=2ρcosθ,進(jìn)一步轉(zhuǎn)化成直角坐標(biāo)方程為:(x﹣1)2+y2=1,把直線ρ(cosθ+sinθ)=a的方程轉(zhuǎn)化成直角坐標(biāo)方程為:x+y﹣a=0. 由于直線和圓相切,所以:利用圓心到直線的距離等于半徑. 則:=1,解得:a=1±.a(chǎn)>0 則負(fù)值舍去. 故:a=1+. 故答案為:1+.   11.(5分)設(shè)函數(shù)f(x)=cos(ωx﹣)(ω>0),若f(x)≤f()對(duì)任意的實(shí)數(shù)x都成立,則ω的最小值為. 【解答】解:函數(shù)f(x)=cos(ωx﹣)(ω>0),若f(x)≤f()對(duì)任意的實(shí)數(shù)x都成立,可得:,k∈Z,解得ω=,k∈Z,ω>0 則ω的最小值為:. 故答案為:.   12.(5分)若x,y滿足x+1≤y≤2x,則2y﹣x的最小值是 3?。? 【解答】解:作出不等式組對(duì)應(yīng)的平面區(qū)域如圖:

      設(shè)z=2y﹣x,則y=x+z,平移y=x+z,由圖象知當(dāng)直線y=x+z經(jīng)過(guò)點(diǎn)A時(shí),直線的截距最小,此時(shí)z最小,由得,即A(1,2),此時(shí)z=2×2﹣1=3,故答案為:3   13.(5分)能說(shuō)明“若f(x)>f(0)對(duì)任意的x∈(0,2]都成立,則f(x)在[0,2]上是增函數(shù)”為假命題的一個(gè)函數(shù)是 f(x)=sinx . 【解答】解:例如f(x)=sinx,盡管f(x)>f(0)對(duì)任意的x∈(0,2]都成立,當(dāng)x∈[0,)上為增函數(shù),在(,2]為減函數(shù),故答案為:f(x)=sinx.   14.(5分)已知橢圓M:+=1(a>b>0),雙曲線N:﹣=1.若雙曲線N的兩條漸近線與橢圓M的四個(gè)交點(diǎn)及橢圓M的兩個(gè)焦點(diǎn)恰為一個(gè)正六邊形的頂點(diǎn),則橢圓M的離心率為;

      雙曲線N的離心率為 2?。? 【解答】解:橢圓M:+=1(a>b>0),雙曲線N:﹣=1.若雙曲線N的兩條漸近線與橢圓M的四個(gè)交點(diǎn)及橢圓M的兩個(gè)焦點(diǎn)恰為一個(gè)正六邊形的頂點(diǎn),可得橢圓的焦點(diǎn)坐標(biāo)(c,0),正六邊形的一個(gè)頂點(diǎn)(,),可得:,可得,可得e4﹣8e2+4=0,e∈(0,1),解得e=. 同時(shí),雙曲線的漸近線的斜率為,即,可得:,即,可得雙曲線的離心率為e==2. 故答案為:;

      2.   三、解答題共6小題,共80分。解答應(yīng)寫出文字說(shuō)明,演算步驟或證明過(guò)程。

      15.(13分)在△ABC中,a=7,b=8,cosB=﹣.(Ⅰ)求∠A;

      (Ⅱ)求AC邊上的高. 【解答】解:(Ⅰ)∵a<b,∴A<B,即A是銳角,∵cosB=﹣,∴sinB===,由正弦定理得=得sinA===,則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),則AC邊上的高h(yuǎn)=csinA=3×=.   16.(14分)如圖,在三棱柱ABC﹣A1B1C1中,CC1⊥平面ABC,D,E,F(xiàn),G分別為AA1,AC,A1C1,BB1的中點(diǎn),AB=BC=,AC=AA1=2.(Ⅰ)求證:AC⊥平面BEF;

      (Ⅱ)求二面角B﹣CD﹣C1的余弦值;

      (Ⅲ)證明:直線FG與平面BCD相交. 【解答】(I)證明:∵E,F(xiàn)分別是AC,A1C1的中點(diǎn),∴EF∥CC1,∵CC1⊥平面ABC,∴EF⊥平面ABC,又AC?平面ABC,∴EF⊥AC,∵AB=BC,E是AC的中點(diǎn),∴BE⊥AC,又BE∩EF=E,BE?平面BEF,EF?平面BEF,∴AC⊥平面BEF.(II)解:以E為原點(diǎn),以EB,EC,EF為坐標(biāo)軸建立空間直角坐標(biāo)系如圖所示:

      則B(2,0,0),C(0,1,0),D(0,﹣1,1),∴=(﹣2,1,0),=(0,﹣2,1),設(shè)平面BCD的法向量為=(x,y,z),則,即,令y=2可得=(1,2,4),又EB⊥平面ACC1A1,∴=(2,0,0)為平面CD﹣C1的一個(gè)法向量,∴cos<,>===. 由圖形可知二面角B﹣CD﹣C1為鈍二面角,∴二面角B﹣CD﹣C1的余弦值為﹣.(III)證明:F(0,0,2),(2,0,1),∴=(2,0,﹣1),∴?=2+0﹣4=﹣2≠0,∴與不垂直,∴FG與平面BCD不平行,又FG?平面BCD,∴FG與平面BCD相交.   17.(12分)電影公司隨機(jī)收集了電影的有關(guān)數(shù)據(jù),經(jīng)分類整理得到下表:

      電影類型 第一類 第二類 第三類 第四類 第五類 第六類 電影部數(shù) 140 50 300 200 800 510 好評(píng)率 0.4 0.2 0.15 0.25 0.2 0.1 好評(píng)率是指:一類電影中獲得好評(píng)的部數(shù)與該類電影的部數(shù)的比值. 假設(shè)所有電影是否獲得好評(píng)相互獨(dú)立.(Ⅰ)從電影公司收集的電影中隨機(jī)選取1部,求這部電影是獲得好評(píng)的第四類電影的概率;

      (Ⅱ)從第四類電影和第五類電影中各隨機(jī)選取1部,估計(jì)恰有1部獲得好評(píng)的概率;

      (Ⅲ)假設(shè)每類電影得到人們喜歡的概率與表格中該類電影的好評(píng)率相等.用“ξk=1”表示第k類電影得到人們喜歡.“ξk=0”表示第k類電影沒(méi)有得到人們喜歡(k=1,2,3,4,5,6).寫出方差Dξ1,Dξ2,Dξ3,Dξ4,Dξ5,Dξ6的大小關(guān)系. 【解答】解:(Ⅰ)設(shè)事件A表示“從電影公司收集的電影中隨機(jī)選取1部,求這部電影是獲得好評(píng)的第四類電影”,總的電影部數(shù)為140+50+300+200+800+510=2000部,第四類電影中獲得好評(píng)的電影有:200×0.25=50部,∴從電影公司收集的電影中隨機(jī)選取1部,求這部電影是獲得好評(píng)的第四類電影的頻率為:

      P(A)==0.025.(Ⅱ)設(shè)事件B表示“從第四類電影和第五類電影中各隨機(jī)選取1部,恰有1部獲得好評(píng)”,第四類獲得好評(píng)的有:200×0.25=50部,第五類獲得好評(píng)的有:800×0.2=160部,則從第四類電影和第五類電影中各隨機(jī)選取1部,估計(jì)恰有1部獲得好評(píng)的概率:

      P(B)==0.35.(Ⅲ)由題意知,定義隨機(jī)變量如下:

      ξk=,則ξk服從兩點(diǎn)分布,則六類電影的分布列及方差計(jì)算如下:

      第一類電影:

      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. 第二類電影:

      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. 第三類電影:

      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. 第四類電影:

      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. 第五類電影:

      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. 第六類電影:

      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的大小關(guān)系為:

      Dξ6<Dξ3<Dξ2=Dξ5<Dξ4<Dξ1.   18.(13分)設(shè)函數(shù)f(x)=[ax2﹣(4a+1)x+4a+3]ex.(Ⅰ)若曲線y=f(x)在點(diǎn)(1,f(1))處的切線與x軸平行,求a;

      (Ⅱ)若f(x)在x=2處取得極小值,求a的取值范圍. 【解答】解:(Ⅰ)函數(shù)f(x)=[ax2﹣(4a+1)x+4a+3]ex的導(dǎo)數(shù)為 f′(x)=[ax2﹣(2a+1)x+2]ex. 由題意可得曲線y=f(x)在點(diǎn)(1,f(1))處的切線斜率為0,可得(a﹣2a﹣1+2)e=0,解得a=1;

      (Ⅱ)f(x)的導(dǎo)數(shù)為f′(x)=[ax2﹣(2a+1)x+2]ex=(x﹣2)(ax﹣1)ex,若a=0則x<2時(shí),f′(x)>0,f(x)遞增;

      x>2,f′(x)<0,f(x)遞減. x=2處f(x)取得極大值,不符題意;

      若a>0,且a=,則f′(x)=(x﹣2)2ex≥0,f(x)遞增,無(wú)極值;

      若a>,則<2,f(x)在(,2)遞減;

      在(2,+∞),(﹣∞,)遞增,可得f(x)在x=2處取得極小值;

      若0<a<,則>2,f(x)在(2,)遞減;

      在(,+∞),(﹣∞,2)遞增,可得f(x)在x=2處取得極大值,不符題意;

      若a<0,則<2,f(x)在(,2)遞增;

      在(2,+∞),(﹣∞,)遞減,可得f(x)在x=2處取得極大值,不符題意. 綜上可得,a的范圍是(,+∞).   19.(14分)已知拋物線C:y2=2px經(jīng)過(guò)點(diǎn)P(1,2),過(guò)點(diǎn)Q(0,1)的直線l與拋物線C有兩個(gè)不同的交點(diǎn)A,B,且直線PA交y軸于M,直線PB交y軸于N.(Ⅰ)求直線l的斜率的取值范圍;

      (Ⅱ)設(shè)O為原點(diǎn),=λ,=μ,求證:+為定值. 【解答】解:(Ⅰ)∵拋物線C:y2=2px經(jīng)過(guò)點(diǎn) P(1,2),∴4=2p,解得p=2,設(shè)過(guò)點(diǎn)(0,1)的直線方程為y=kx+1,設(shè)A(x1,y1),B(x2,y2)聯(lián)立方程組可得,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=,故直線l的斜率的取值范圍(﹣∞,0)∪(0,1);

      (Ⅱ)證明:設(shè)點(diǎn)M(0,yM),N(0,yN),則=(0,yM﹣1),=(0,﹣1)因?yàn)?λ,所以yM﹣1=﹣yM﹣1,故λ=1﹣yM,同理μ=1﹣yN,直線PA的方程為y﹣2=(x﹣1)=(x﹣1)=(x﹣1),令x=0,得yM=,同理可得yN=,因?yàn)?=+=+===== =2,∴+=2,∴+為定值.   20.(14分)設(shè)n為正整數(shù),集合A={α|α=(t1,t2,…tn),tk∈{0,1},k=1,2,…,n},對(duì)于集合A中的任意元素α=(x1,x2,…,xn)和β=(y1,y2,…yn),記 M(α,β)=[(x1+y1﹣|x1﹣y1|)+(x2+y2﹣|x2﹣y2|)+…(xn+yn﹣|xn﹣yn|)](Ⅰ)當(dāng)n=3時(shí),若α=(1,1,0),β=(0,1,1),求M(α,α)和M(α,β)的值;

      (Ⅱ)當(dāng)n=4時(shí),設(shè)B是A的子集,且滿足:對(duì)于B中的任意元素α,β,當(dāng)α,β相同時(shí),M(α,β)是奇數(shù);

      當(dāng)α,β不同時(shí),M(α,β)是偶數(shù).求集合B中元素個(gè)數(shù)的最大值;

      (Ⅲ)給定不小于2的n,設(shè)B是A的子集,且滿足:對(duì)于B中的任意兩個(gè)不同的元素α,β,M(α,β)=0,寫出一個(gè)集合B,使其元素個(gè)數(shù)最多,并說(shuō)明理由. 【解答】解:(I)M(a,a)=2,M(a,β)=1.(II)考慮數(shù)對(duì)(xk,yk)只有四種情況:(0,0)、(0,1)、(1,0)、(1,1),相應(yīng)的分別為0、0、0、1,所以B中的每個(gè)元素應(yīng)有奇數(shù)個(gè)1,所以B中的元素只可能為(上下對(duì)應(yīng)的兩個(gè)元素稱之為互補(bǔ)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),對(duì)于任意兩個(gè)只有1個(gè)1的元素α,β都滿足M(α,β)是偶數(shù),所以四元集合B={(1,0,0,0)、(0,1,0,0)、(0,0,1,0)、(0,0,0,1)}滿足 題意,假設(shè)B中元素個(gè)數(shù)大于等于4,就至少有一對(duì)互補(bǔ)元素,除了這對(duì)互補(bǔ)元素之外還有至少1個(gè)含有3個(gè)1的元素α,則互補(bǔ)元素中含有1個(gè)1的元素β與之滿足M(α,β)=1不合題意,故B中元素個(gè)數(shù)的最大值為4.(Il)B={(0,0,0,…0),(1,0,0…,0),(0,1,0,…0),(0,0,1…0)…,(0,0,0,…,1)},此時(shí)B中有n+1個(gè)元素,下證其為最大. 對(duì)于任意兩個(gè)不同的元素α,β,滿足M(α,β)=0,則α,β中相同位置上的數(shù)字不能同時(shí)為1,假設(shè)存在B有多于n+1個(gè)元素,由于α=(0,0,0,…,0)與任意元素β都有M(α,β)=0,所以除(0,0,0,…,0)外至少有n+1個(gè)元素含有1,根據(jù)元素的互異性,至少存在一對(duì)α,β滿足xi=yi=l,此時(shí)M(α,β)≥1不滿足題意,故B中最多有n+1個(gè)元素.   — END —

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