2016普通高等学校招生全国统一考试(新课标I)

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新课标卷Ⅰ2016高考真题福建答案

新课标卷Ⅰ2016高考真题福建答案

2016年普通高等学校招生全国统一考试新课标I卷参考答案第一部分听力1.C 2.B 3.A 4.C 5.A 6.C 7.B 8.A 9.A 10.C 11.B 12.B 13.C 14.A 15.B 16.B 17.A 18.C 19.A 20.B第二部分阅读理解(共两节,满分40 分)第一节(共15 小题:每小题 2 分,满分30 分)A21.A 22.C 23.D 24.CB25.A 26.D 27.C 28.AC29.B 30.D 31.BD32.C 33.A 34.D 35.B第二节(共 5 小题:每小题 2 分,满分10 分)36.D 37.E 38.G 39.F 40.A第三部分英语知识运用(共两节,满分45 分)第一节完形填空(共20 小题:每小题 1.5 分,满分30 分)41.C 42.D 43.C 44.B 45.A 46.B 47.A 48.D 49.C 50.A 51.D 52.B 53.D 54.A 55.C 56.B 57.B 58.D 59.A 60.C第二节英语知识运用(共10 小题:每小题 1.5 分,满分15 分)61.Attraction 62.was allowed 63.Officially 64.To 65.when 66.Permitted 67.Introducing 68.Its 69.Days 70.the 第四部分写作(共两节,满分35 分)第一节短文改错(共10 小题:每小题 1 分,满分10 分)71.that → where 72.but去掉73.had → have 74.honest → honesty 75.or→ and 76.using → used 77.becoming 前加of 78.the → a79.our → his 80.stead →steadily第二节书面表达(满25 分)略Dear Ms Jenkins,I’m Li Hua from your English writing class last term.I’m writing to ask for your help.I’m applying for a part-time job at a foreign company in my city during the summer vacation,and I have just completed my application letter and resume.However,I am not quit e sure of the language and the format I’ve used.I know you have a very busy schedule,but I’d be very grateful if you could take some time to go through them and make necessary changes.Please find my application letter and resume in the attachment.Thank you for your kindness!Yours,Li Hua 2016高考江苏卷英语听力原文及答案解析Text 1W:What are we going to get for Lydia’s birthday?M:How about a pair of running shoes?W:You know she hates doing exercise.M:Then I guess we can buy her a birthday cake.Text 2M:Excuse me.Do you have time to help take a picture of us?W:Oh,I’m sorry.I’ve got to catch a bus.M:That’s OK.Text 3M:It really annoys me when Kate calls her friends during office hours.W:If I were you,I would tell her to stop.M:Maybe you’re right.I will talk to her sometime.Text 4M:Here’s the menu,Madame.Would you like something to drink?W:Yes,please.May I see the wine list?M:Certainly.Here you are.Text 5M:We need to have some fresh air.Do you mind my opening the window?W:As a matter of fact,I’m feeling a bit cold.Text 6M:Guess what I’ll be doing this summer?W:What?M:I’m going to work at the Riverside Hotel.W:What exactly will you be doing?M:Let’s see.I’ll be doing some small repairs inside and outside the hotel.I’ll be cutting grass an d taking care of the flowers.W:Sounds interesting.What’s the pay?M:Well,uh…about fifteen dollars an hour,five hours a day,and Sunday free.W:That’s good money.What are you going to do with it?M:I’ll pay for the textbooks for next term.Text 7M:Hi,Sue.How’s it going?W:Oh,hi,Frank,just fine.How are your classes?M:Pretty good.I’m glad this is my last term here,though.W:Why is that? I thought you were enjoying school.M:I was.But now I’m getting tired of it.I’m ready for the real world.W:What are you planning to do when you graduate?M:First,I want to get a job as a computer programmer,and then after five years or so,I’d like to start my own business.W:Sounds good.I still have three terms to go until I’m done.M:You’ll make it for sure.Well,see you later.W:Bye!Text 8M:Hello,Milton Hotel Reservations.How may I assist you?W:Hi,I’m calling to make some changes to an existing reservation.M:Certainly.Do you have the reservation number?W:Sure,it’s 219.M:That’s a reservation for Sally Menk el.Is that right?W:Yes,that’s right.I’d like to change the check-in date from September 15 to September 16.M:Certainly.I can make that change for you.Is that the only change?W:No,the check-out date will also change from the 23rd to the 24th.M:No problem.We have you arriving on the 16th of September and leaving the 24th of September — altogether,eight nights.Will there be anything else?W:Yes.Instead of a courtyard room,I’d like a room with a view,preferably on an upper floor.M:I can certainly change that for you,but there will be a change in the room rate.The new rate is $199 per night,instead of the original $179.W:That’s OK.These are all the changes.Thank you very much.M:You’re welcome.Have a nice day!Text 9W:Gordon,I’m go ing to Keswick in the Lake District this weekend.M:Really?W:Yeah,five of us.Why don’t you join us? We’re getting to Keswick on Friday.Then we’re going boating on Saturday.And on Sunday,we’re going to do some shopping.Then I will take some time to visit my aunt Lucy.M:You’re not going to camp,are you? Isn’t it a bit cold?W:No,we’re not.It is a bit cold.We’re staying in a country inn.It’s not like five-star hotels or anything.But it’s really comfortable.M:Hmm,sounds interesting.You know,it’s the boating that I don’t like.W:Well,what are you up to?M:Sit on my sofa all weekend and watch the DVDs I’ve just bought.And that will be a busy weekend.I may finish watching Friends.W:I don’t k now how you can do that all weekend,Gordon.I’d get bored.M:I know,I know.But that’s really what I want to do.Text 10M:Hello.Welcome to the program.In America,May and June are the traditional months for graduations.A listener in China,Wang Ming,who is about to get an engineering degree,wants to know how American college graduates find jobs.Right now,the answer is:not very easily.A latest study on the college job market showed that employers wanted to hire 22% fewer graduates this year than last.The study also showed that just 20% of those who looked for jobs before graduation have found one by now.This is compared to half of students who had looked for a job by this time two years ago.But one difference:fewer of this year’s graduates have started to search for jobs.Engineering graduates were more likely to have started their job search already,and to have accepted a job.This is among the best-paid professions for people with just a college degree.On average,engineering majors expect to start at about $62,000 a year.。

2016年高考文科数学全国卷1-答案

2016年高考文科数学全国卷1-答案

作出二元一次不等式组①表示的平面区域,即可行域,如图中阴影部分所示.
7z77z
【提示】(Ⅰ)设E 是AB 的中点,证明60AOE ∠=︒;
(Ⅱ)设'O 是A B C D ,,,四点所在圆的圆心,作直线'OO ,证明'OO AB ⊥,'OO CD ⊥,由此可证明//AB CD .
【考点】四点共圆、直线与圆的位置关系及证明
23.【答案】(Ⅰ)圆,222sin 10a ρρθ-+-=
(Ⅱ)1
【解析】(Ⅰ)消去参数t 得到1C 的普通方程222(1)x y a +-=.
1C 是以(0,1)为圆心,a 为半径的圆.
将cos ,sin x y ρθρθ==代入1C 的普通方程中,得到1C 的极坐标方程为
222sin 10a ρρθ-+-=.
(Ⅱ)曲线12,C C 的公共点的极坐标满足方程组222sin 10,
4cos ,a ρρθρθ⎧-+-=⎨
=⎩
若0ρ≠,由方程组得2216cos 8sin cos 10a θθθ-+-=,由已知tan 2θ=,可得216cos 8sin cos 0θθθ-=,从而210a -=,解得1a =-(舍去),1a =.
1a =时,极点也为12,C C 的公共点,在3C 上.所以1a =.
【提示】(Ⅰ)把cos 1sin x a t
y a t =⎧⎨=+⎩
化为普通方程,再化为极坐标方程;
(Ⅱ)通过解方程组可以求得.
【考点】参数方程,极坐标方程与直角坐标方程的互化
11/ 11。

全国卷年全国高考英语试题及答案

全国卷年全国高考英语试题及答案

绝密★启封前2016年普通高等学校招生全国统一考试(新课标I)英语注意事项:1.答题前,先将自己的姓名、准考证号填写在试题卷和答题卡上,并将准考证号条形码粘贴在答题卡上的指定位置。

用2B铅笔将答题卡上试卷类型A后的方框涂黑。

2.选择题的作答:每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑,写在试题卷、草稿纸和答题卡上的非答题区域均无效。

3.非选择题的作答:用签字笔直接答在答题卡上对应的答题区域内。

写在试题卷、草稿纸和答题卡上的非答题区域均无效。

4.考试结束后,请将本试题卷和答题卡一并上交。

第Ⅰ卷第一部分听力(共两节,满分30 分)做题时,现将答案标在试卷上,录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节(共 5 小题;每小题分,满分分)听下面 5 段对话,每段对话后有一个小题。

从题中所给的A,B,C 三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10 秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

例:How much is the shirtA.£B.£C.£答案是C。

are the speakers talking aboutA. Having a birthday party.B. Doing some exercise.C. Getting Lydia a gift2. What is the woman going to doA. Help the man.B. Take a bus.C. Get a camera3. What does the woman suggest the man doA. Tell Kate to stop.B. Call Kate, s friends.C. Stay away from Kate.4. Where does the conversation probably take placeA. In a wine shop.B. In a supermarket.C. In a restaurant.5. What does the woman meanA. Keep the window closed.B. Go out for fresh air.C. Turn on the fan.听第6段材料,回答第6、7题。

全国卷高考英语试题及答案-全国卷1

全国卷高考英语试题及答案-全国卷1

绝密★启封前2016年普通高等学校招生全国统一考试(新课标I)英语注意事项:1.答题前,先将自己的姓名、准考证号填写在试题卷和答题卡上,并将准考证号条形码粘贴在答题卡上的指定位置。

用2B铅笔将答题卡上试卷类型A后的方框涂黑。

2.选择题的作答:每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑,写在试题卷、草稿纸和答题卡上的非答题区域均无效。

3.非选择题的作答:用签字笔直接答在答题卡上对应的答题区域内。

写在试题卷、草稿纸和答题卡上的非答题区域均无效。

4.考试结束后,请将本试题卷和答题卡一并上交。

第Ⅰ卷第一部分听力(共两节,满分 30 分)做题时,现将答案标在试卷上,录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节(共 5 小题;每小题分,满分分)听下面 5 段对话,每段对话后有一个小题。

从题中所给的 A,B,C 三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有 10 秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

are the speakers talking aboutA. Having a birthday party.B. Doing some exercise.C. Getting Lydia a gift2. What is the woman going to doA. Help the man.B. Take a bus.C. Get a camera3. What does the woman suggest the man doA. Tell Kate to stop.B. Call Kate, s friends.C. Stay away from Kate.4. Where does the conversation probably take placeA. In a wine shop.B. In a supermarket.C. In a restaurant.5. What does the woman meanA. Keep the window closed.B. Go out for fresh air.C. Turn on the fan.听第6段材料,回答第6、7题。

2016高考全国卷1理综化学含答案及解析

2016高考全国卷1理综化学含答案及解析

2016年普通高等学校招生全国统一考试(新课标I卷)理科综合(化学部分)7.化学与生活密切相关。

下列有关说法错误的是()A、用灼烧的方法可以区分蚕丝和人造纤维B、食用油反复加热会产生稠环芳烃等有害物质C、加热能杀死流感病毒是因为蛋白质受热变性D、医用消毒酒精中乙醇的浓度为95%【答案】D【解析】A、蚕丝的主要成分为蛋白质,灼烧时会有烧焦羽毛的气味,而人造纤维由纤维素改性得到,灼烧时有刺激性气味,可由此区分二者,故A正确。

B、食用油反复加热,碳链会变成环状,产生稠环芳烃等有害物质,故B正确。

C、加热、强酸碱、重金属盐均可以使蛋白质变性,因此加热可杀死流感病毒,故C正确。

D、医用酒精中乙醇的浓度为75%,工业酒精中乙醇的浓度为95%,故D错误。

因此,本题选D。

8.设N A为阿伏加德罗常数值。

下列有关叙述正确的是()A、14 g乙烯和丙烯混合气体中的氢原子数为2N AB、1 mol N2与4 mol H2反应生成的NH3分子数为2N AC、1 mol Fe溶于过量硝酸,电子转移数为2N AD、标准状况下,2.24L CCl4含有的共价键数为0.4N A【答案】A【解析】9.下列关于有机化合物的说法正确的是()A、2-甲基丁烷也称为异丁烷B、由乙烯生成乙醇属于加成反应C、C4H9Cl有3种同分异构体D、油脂和蛋白质都属于高分子化合物【答案】B【解析】10.作能达到实验目的的是()A、用长颈漏斗分离出乙酸与乙醇反应的产物B、用向上排空气法收集铜粉与稀硝酸反应产生的NOC、配制氯化铁溶液时,将氯化铁溶解在较浓的盐酸中再加水稀释D 、将Cl 2与HCl 混合气体通过饱和食盐水可得到纯净的Cl 2【答案】C 【解析】11.三室式电渗析法处理含Na 2SO 4废水的原理如图所示,采用惰性电极,ab 、cd 均为离子交换膜,在直流电场的作用下,两膜中间的Na +和SO 42–可通过离子交换膜,而两端隔室中离子被阻挡不能进入中间隔室。

全国卷高考英语试题及答案-全国卷

全国卷高考英语试题及答案-全国卷

绝密★启封前2016年普通高等学校招生全国统一考试(新课标I)英语注意事项:1.答题前,先将自己的姓名、准考证号填写在试题卷和答题卡上,并将准考证号条形码粘贴在答题卡上的指定位置。

用2B铅笔将答题卡上试卷类型A后的方框涂黑。

2.选择题的作答:每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑,写在试题卷、草稿纸和答题卡上的非答题区域均无效。

3.非选择题的作答:用签字笔直接答在答题卡上对应的答题区域内。

写在试题卷、草稿纸和答题卡上的非答题区域均无效。

4.考试结束后,请将本试题卷和答题卡一并上交。

第Ⅰ卷第一部分听力(共两节,满分 30 分)做题时,现将答案标在试卷上,录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节(共 5 小题;每小题分,满分分)听下面 5 段对话,每段对话后有一个小题。

从题中所给的 A,B,C 三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有 10 秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

are the speakers talking aboutA. Having a birthday party.B. Doing some exercise.C. Getting Lydia a gift2. What is the woman going to doA. Help the man.B. Take a bus.C. Get a camera3. What does the woman suggest the man doA. Tell Kate to stop.B. Call Kate, s friends.C. Stay away from Kate.4. Where does the conversation probably take placeA. In a wine shop.B. In a supermarket.C. In a restaurant.5. What does the woman meanA. Keep the window closed.B. Go out for fresh air.C. Turn on the fan.听第6段材料,回答第6、7题。

2016全国卷1化学试题答案+解析

2016年普通高等学校招生全国统一考试化学(新课标I )一、选择题:本大题共13小题,每小题6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。

7.化学与生活密切相关,下列有关说法错误的是A.用灼烧的方法可以区分蚕丝和人造纤维B.食用油反复加热会产生稠环芳香烃等有害物质C.加热能杀死流感病毒是因为蛋白质受热变性D.医用消毒酒精中乙醇的浓度为95%【答案】D考点:考查化学在生活的应用的知识。

8.设N A为阿伏加德罗常数值。

下列有关叙述正确的是A.14 g乙烯和丙烯混合气体中的氢原子数为2N AB.1 molN2与4 mol H2反应生成的NH3分子数为2N AC.1 molFe溶于过量硝酸,电子转移数为2N AD.标准状况下,2.24 LCCl4含有的共价键数为0.4N A【答案】A【解析】试题分析:A、乙烯和丙烯的最简式相同,均是CH2,14 g乙烯和丙烯混合气体中的氢原子数为14214AN⨯⨯=2N A,正确;B、N2与H2反应生成的NH3的反应是可逆反应,反应物不能完全转化为生成物。

1 molN2与4 mol H2反应生成的NH3分子数小于2N A,B错误;C.1 molFe溶于过量硝酸生成硝酸铁,电子转移数为3N A,错误;D、标准状况下四氯化碳是液态,不能利用气体摩尔体积计算物质的量,错误。

考点:考查阿伏加德罗常数计算的知识。

9.下列关于有机化合物的说法正确的是A.2-甲基丁烷也称异丁烷B.由乙烯生成乙醇属于加成反应C.C4H9Cl有3种同分异构体D.油脂和蛋白质都属于高分子化合物【答案】B【解析】试题分析:A.2-甲基丁烷也称异戊烷,错误;B.乙烯与水发生加成反应生成乙醇,正确;C.C4H9Cl有4种同分异构体,错误;D.油脂不是高分子化合物,错误。

考点:考查有机物结构和性质判断的知识。

10.下列实验操作能达到实验目的的是A.用长颈漏斗分离出乙酸与乙醇反应的产物B.用向上排空气法收集铜粉与稀硝酸反应产生的NOC.配制氯化铁溶液时,将氯化铁溶解在较浓的盐酸中再加水稀释D.将Cl2与HCl混合气体通过饱和食盐水可得到纯净的Cl2【答案】C考点:考查化学实验基本操作的知识。

2016年安徽高考英语试题及答案

2016普通高等学校招生全国统一考试(新课标I)英语第Ⅰ卷第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话,每段对话后有一个小题。

从题中所给的A,B,C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1.Whatarethespeakerstalkingabout?A.Havingabirthdayparty.B.Doingsomeexercise.C.GettingLydiaagift2.Whatisthewomangoingtodo?A.Helptheman.B.Takeabus.C.Getacamera3.Whatdoesthewomansuggestthemando?A.TellKatetostop.B.CallKate,sfriends.C.StayawayfromKate.4.Wheredoestheconversationprobablytakeplace?A.Inawineshop.B.Inasupermarket.C.Inarestaurant.5.Whatdoesthewomanmean?A.Keepthewindowclosed.B.Gooutforfreshair.C.Turnonthefan.听第6段材料,回答第6、7题。

6.Whatisthemangoingtodothissummer?A.Teachacourse.B.Repairhishouse.C.Workatahotel.7.Howwillthemanusethemoney?A.Tohireagardener.B.Tobuybooks.C.Topayforaboattrip.听第7段材料,回答第8、9题。

8.Whatistheprobablerelationshipbetweenthespeakers?A.Schoolmates.B.Colleagues.C.Roommates.9.WhatdoesFrankplantodorightaftergraduation?A.Workasaprogrammer.B.Travelaroundtheworld.C.Starthisownbusiness.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。

2016年高考全国2卷英语试题及答案

绝密★启封前2016普通高等学校招生全国统一考试(新课标I)英语试卷类型A注意事项:1.答题前,先将自己的姓名、准考证号填写在试题卷和答题卡上,并将准考证号条形码粘贴在答题卡上的指定位置。

用2B铅笔将答题卡上试卷类型A后的方框涂黑。

2.选择题的作答:每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑,写在试题卷、草稿纸和答题卡上的非答题区域均无效。

3.非选择题的作答:用签字笔直接答在答题卡上对应的答题区域内。

写在试题卷、草稿纸和答题卡上的非答题区域均无效。

4.考试结束后,请将本试题卷和答题卡一并上交。

第Ⅰ卷第一部分听力(共两节,满分30 分)做题时,现将答案标在试卷上,录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节(共 5 小题;每小题 1.5 分,满分7.5 分)听下面 5 段对话,每段对话后有一个小题。

从题中所给的A,B,C 三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10 秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

例:How much is the shirt?A.£ 19.15B.£ 9.18C.£ 9.15答案是C。

1.What are the speakers talking about?A. Having a birthday party.B. Doing some exercise.C. Getting Lydia a gift2.What is the woman going to do?A. Help the man.B. Take a bus.C. Get a camera3.What does the woman suggest the man do?A. Tell Kate to stop.B. Call Kate, s friends.C. Stay away from Kate.4.Where does the conversation probably take place?A. In a wine shop.B. In a supermarket.C. In a restaurant.5.What does the woman mean?A. Keep the window closed.B. Go out for fresh air.C. Turn on the fan.听第6段材料,回答第6、7题。

2016高考全国1数学试卷及解析

2016年普通高等学校招生全国统一考试(新课标Ⅰ卷)理科数学第Ⅰ卷一.选择题(共12小题)1.设集合A={x|x2﹣4x+3<0},B={x|2x﹣3>0},则A∩B=()A.(﹣3,﹣)B.(﹣3,)C.(1,)D.(,3)2.设(1+i)x=1+yi,其中x,y是实数,则|x+yi|=()A.1 B.C.D.23.已知等差数列{a n}前9项的和为27,a10=8,则a100=()A.100 B.99 C.98 D.974.某公司的班车在7:00,8:00,8:30发车,小明在7:50至8:30之间到达发车站乘坐班车,且到达发车站的时刻是随机的,则他等车时间不超过10分钟的概率是()A.B.C.D.5.已知方程﹣=1表示双曲线,且该双曲线两焦点间的距离为4,则n的取值范围是()A.(﹣1,3)B.(﹣1,) C.(0,3) D.(0,)6.如图,某几何体的三视图是三个半径相等的圆及每个圆中两条相互垂直的半径.若该几何体的体积是,则它的表面积是()A.17πB.18πC.20πD.28π7.函数y=2x2﹣e|x|在[﹣2,2]的图象大致为()A.B.C.D.8.若a>b>1,0<c<1,则()A.a c<b c B.ab c<ba cC.alog b c<blog a c D.log a c<log b c9.执行如图的程序框图,如果输入的x=0,y=1,n=1,则输出x,y的值满足()A.y=2x B.y=3x C.y=4x D.y=5x10.以抛物线C的顶点为圆心的圆交C于A、B两点,交C的准线于D、E两点.已知|AB|=4,|DE|=2,则C的焦点到准线的距离为()A.2 B.4 C.6 D.811.平面α过正方体ABCD﹣A1B1C1D1的顶点A,α∥平面CB1D1,α∩平面ABCD=m,α∩平面ABB1A1=n,则m、n所成角的正弦值为()A.B.C.D.12.已知函数f(x)=sin(ωx+φ)(ω>0,|φ|≤),x=﹣为f(x)的零点,x=为y=f(x)图象的对称轴,且f(x)在(,)上单调,则ω的最大值为()A.11 B.9 C.7 D.5二.填空题(共4小题)13.设向量=(m,1),=(1,2),且|+|2=||2+||2,则m=.14.(2x+)5的展开式中,x3的系数是.(用数字填写答案)15.设等比数列{a n}满足a1+a3=10,a2+a4=5,则a1a2…a n的最大值为.16.某高科技企业生产产品A和产品B需要甲、乙两种新型材料.生产一件产品A需要甲材料1.5kg,乙材料1kg,用5个工时;生产一件产品B需要甲材料0.5kg,乙材料0.3kg,用3个工时,生产一件产品A的利润为2100元,生产一件产品B的利润为900元.该企业现有甲材料150kg,乙材料90kg,则在不超过600个工时的条件下,生产产品A、产品B的利润之和的最大值为元.三.解答题(共7小题)17.△ABC的内角A,B,C的对边分别为a,b,c,已知2cosC(acosB+bcosA)=c.(Ⅰ)求C;(Ⅱ)若c=,△ABC的面积为,求△ABC的周长.18.如图,在以A,B,C,D,E,F为顶点的五面体中,面ABEF为正方形,AF=2FD,∠AFD=90°,且二面角D﹣AF﹣E与二面角C﹣BE﹣F都是60°.(Ⅰ)证明平面ABEF⊥平面EFDC;(Ⅱ)求二面角E﹣BC﹣A的余弦值.19.某公司计划购买2台机器,该种机器使用三年后即被淘汰.机器有一易损零件,在购进机器时,可以额外购买这种零件作为备件,每个200元.在机器使用期间,如果备件不足再购买,则每个500元.现需决策在购买机器时应同时购买几个易损零件,为此搜集并整理了100台这种机器在三年使用期内更换的易损零件数,得如图柱状图:以这100台机器更换的易损零件数的频率代替1台机器更换的易损零件数发生的概率,记X表示2台机器三年内共需更换的易损零件数,n表示购买2台机器的同时购买的易损零件数.(Ⅰ)求X的分布列;(Ⅱ)若要求P(X≤n)≥0.5,确定n的最小值;(Ⅲ)以购买易损零件所需费用的期望值为决策依据,在n=19与n=20之中选其一,应选用哪个?20.设圆x2+y2+2x﹣15=0的圆心为A,直线l过点B(1,0)且与x轴不重合,l 交圆A于C,D两点,过B作AC的平行线交AD于点E.(Ⅰ)证明|EA|+|EB|为定值,并写出点E的轨迹方程;(Ⅱ)设点E的轨迹为曲线C1,直线l交C1于M,N两点,过B且与l垂直的直线与圆A交于P,Q两点,求四边形MPNQ面积的取值范围.21.已知函数f(x)=(x﹣2)e x+a(x﹣1)2有两个零点.(Ⅰ)求a的取值范围;(Ⅱ)设x1,x2是f(x)的两个零点,证明:x1+x2<2.22.在直角坐标系xOy中,曲线C1的参数方程为(t为参数,a>0).在以坐标原点为极点,x轴正半轴为极轴的极坐标系中,曲线C2:ρ=4cosθ.(Ⅰ)说明C1是哪种曲线,并将C1的方程化为极坐标方程;(Ⅱ)直线C3的极坐标方程为θ=α0,其中α0满足tanα0=2,若曲线C1与C2的公共点都在C3上,求a.23.已知函数f(x)=|x+1|﹣|2x﹣3|.(Ⅰ)在图中画出y=f(x)的图象;(Ⅱ)求不等式|f(x)|>1的解集.2018年04月22日fago的高中数学组卷参考答案与试题解析一.选择题(共12小题)1.设集合A={x|x2﹣4x+3<0},B={x|2x﹣3>0},则A∩B=()A.(﹣3,﹣)B.(﹣3,)C.(1,)D.(,3)【分析】解不等式求出集合A,B,结合交集的定义,可得答案.【解答】解:∵集合A={x|x2﹣4x+3<0}=(1,3),B={x|2x﹣3>0}=(,+∞),∴A∩B=(,3),故选:D.【点评】本题考查的知识点是集合的交集及其运算,难度不大,属于基础题.2.设(1+i)x=1+yi,其中x,y是实数,则|x+yi|=()A.1 B.C.D.2【分析】根据复数相等求出x,y的值,结合复数的模长公式进行计算即可.【解答】解:∵(1+i)x=1+yi,∴x+xi=1+yi,即,解得,即|x+yi|=|1+i|=,故选:B.【点评】本题主要考查复数模长的计算,根据复数相等求出x,y的值是解决本题的关键.3.已知等差数列{a n}前9项的和为27,a10=8,则a100=()A.100 B.99 C.98 D.97【分析】根据已知可得a5=3,进而求出公差,可得答案.【解答】解:∵等差数列{a n}前9项的和为27,S9===9a5.∴9a5=27,a5=3,又∵a10=8,∴d=1,∴a100=a5+95d=98,故选:C.【点评】本题考查的知识点是数列的性质,熟练掌握等差数列的性质,是解答的关键.4.某公司的班车在7:00,8:00,8:30发车,小明在7:50至8:30之间到达发车站乘坐班车,且到达发车站的时刻是随机的,则他等车时间不超过10分钟的概率是()A.B.C.D.【分析】求出小明等车时间不超过10分钟的时间长度,代入几何概型概率计算公式,可得答案.【解答】解:设小明到达时间为y,当y在7:50至8:00,或8:20至8:30时,小明等车时间不超过10分钟,故P==,故选:B.【点评】本题考查的知识点是几何概型,难度不大,属于基础题.5.已知方程﹣=1表示双曲线,且该双曲线两焦点间的距离为4,则n的取值范围是()A.(﹣1,3)B.(﹣1,) C.(0,3) D.(0,)【分析】由已知可得c=2,利用4=(m2+n)+(3m2﹣n),解得m2=1,又(m2+n)(3m2﹣n)>0,从而可求n的取值范围.【解答】解:∵双曲线两焦点间的距离为4,∴c=2,当焦点在x轴上时,可得:4=(m2+n)+(3m2﹣n),解得:m2=1,∵方程﹣=1表示双曲线,∴(m2+n)(3m2﹣n)>0,可得:(n+1)(3﹣n)>0,解得:﹣1<n<3,即n的取值范围是:(﹣1,3).当焦点在y轴上时,可得:﹣4=(m2+n)+(3m2﹣n),解得:m2=﹣1,无解.故选:A.【点评】本题主要考查了双曲线方程的应用,考查了不等式的解法,属于基础题.6.如图,某几何体的三视图是三个半径相等的圆及每个圆中两条相互垂直的半径.若该几何体的体积是,则它的表面积是()A.17πB.18πC.20πD.28π【分析】判断三视图复原的几何体的形状,利用体积求出几何体的半径,然后求解几何体的表面积.【解答】解:由题意可知三视图复原的几何体是一个球去掉后的几何体,如图:可得:=,R=2.它的表面积是:×4π•22+=17π.故选:A.【点评】本题考查三视图求解几何体的体积与表面积,考查计算能力以及空间想象能力.7.函数y=2x2﹣e|x|在[﹣2,2]的图象大致为()A.B.C.D.【分析】根据已知中函数的解析式,分析函数的奇偶性,最大值及单调性,利用排除法,可得答案.【解答】解:∵f(x)=y=2x2﹣e|x|,∴f(﹣x)=2(﹣x)2﹣e|﹣x|=2x2﹣e|x|,故函数为偶函数,当x=±2时,y=8﹣e2∈(0,1),故排除A,B;当x∈[0,2]时,f(x)=y=2x2﹣e x,∴f′(x)=4x﹣e x=0有解,故函数y=2x2﹣e|x|在[0,2]不是单调的,故排除C,故选:D.【点评】本题考查的知识点是函数的图象,对于超越函数的图象,一般采用排除法解答.8.若a>b>1,0<c<1,则()A.a c<b c B.ab c<ba cC.alog b c<blog a c D.log a c<log b c【分析】根据已知中a>b>1,0<c<1,结合对数函数和幂函数的单调性,分析各个结论的真假,可得答案.【解答】解:∵a>b>1,0<c<1,∴函数f(x)=x c在(0,+∞)上为增函数,故a c>b c,故A错误;函数f(x)=x c﹣1在(0,+∞)上为减函数,故a c﹣1<b c﹣1,故ba c<ab c,即ab c >ba c;故B错误;log a c<0,且log b c<0,log a b<1,即=<1,即log a c>log b c.故D错误;0<﹣log a c<﹣log b c,故﹣blog a c<﹣alog b c,即blog a c>alog b c,即alog b c<blog a c,故C正确;故选:C.【点评】本题考查的知识点是不等式的比较大小,熟练掌握对数函数和幂函数的单调性,是解答的关键.9.执行如图的程序框图,如果输入的x=0,y=1,n=1,则输出x,y的值满足()A.y=2x B.y=3x C.y=4x D.y=5x【分析】由已知中的程序框图可知:该程序的功能是利用循环结构计算并输出变量x,y的值,模拟程序的运行过程,分析循环中各变量值的变化情况,可得答案.【解答】解:输入x=0,y=1,n=1,则x=0,y=1,不满足x2+y2≥36,故n=2,则x=,y=2,不满足x2+y2≥36,故n=3,则x=,y=6,满足x2+y2≥36,故y=4x,故选:C.【点评】本题考查的知识点是程序框图,当循环的次数不多,或有规律时,常采用模拟循环的方法解答.10.以抛物线C的顶点为圆心的圆交C于A、B两点,交C的准线于D、E两点.已知|AB|=4,|DE|=2,则C的焦点到准线的距离为()A.2 B.4 C.6 D.8【分析】画出图形,设出抛物线方程,利用勾股定理以及圆的半径列出方程求解即可.【解答】解:设抛物线为y2=2px,如图:|AB|=4,|AM|=2,|DE|=2,|DN|=,|ON|=,x A==,|OD|=|OA|,=+5,解得:p=4.C的焦点到准线的距离为:4.故选:B.【点评】本题考查抛物线的简单性质的应用,抛物线与圆的方程的应用,考查计算能力.转化思想的应用.11.平面α过正方体ABCD﹣A1B1C1D1的顶点A,α∥平面CB1D1,α∩平面ABCD=m,α∩平面ABB1A1=n,则m、n所成角的正弦值为()A.B.C.D.【分析】画出图形,判断出m、n所成角,求解即可.【解答】解:如图:α∥平面CB1D1,α∩平面ABCD=m,α∩平面ABA1B1=n,可知:n∥CD1,m∥B1D1,∵△CB1D1是正三角形.m、n所成角就是∠CD1B1=60°.则m、n所成角的正弦值为:.故选:A.【点评】本题考查异面直线所成角的求法,考查空间想象能力以及计算能力.12.已知函数f(x)=sin(ωx+φ)(ω>0,|φ|≤),x=﹣为f(x)的零点,x=为y=f(x)图象的对称轴,且f(x)在(,)上单调,则ω的最大值为()A.11 B.9 C.7 D.5【分析】根据已知可得ω为正奇数,且ω≤12,结合x=﹣为f(x)的零点,x=为y=f(x)图象的对称轴,求出满足条件的解析式,并结合f(x)在(,)上单调,可得ω的最大值.【解答】解:∵x=﹣为f(x)的零点,x=为y=f(x)图象的对称轴,∴,即,(n∈N)即ω=2n+1,(n∈N)即ω为正奇数,∵f(x)在(,)上单调,则﹣=≤,即T=≥,解得:ω≤12,当ω=11时,﹣+φ=kπ,k∈Z,∵|φ|≤,∴φ=﹣,此时f(x)在(,)不单调,不满足题意;当ω=9时,﹣+φ=kπ,k∈Z,∵|φ|≤,∴φ=,此时f(x)在(,)单调,满足题意;故ω的最大值为9,故选:B.【点评】本题考查的知识点是正弦型函数的图象和性质,本题转化困难,难度较大.二.填空题(共4小题)13.设向量=(m,1),=(1,2),且|+|2=||2+||2,则m=﹣2.【分析】利用已知条件,通过数量积判断两个向量垂直,然后列出方程求解即可.【解答】解:|+|2=||2+||2,可得•=0.向量=(m,1),=(1,2),可得m+2=0,解得m=﹣2.故答案为:﹣2.【点评】本题考查向量的数量积的应用,向量的垂直条件的应用,考查计算能力.14.(2x+)5的展开式中,x3的系数是10.(用数字填写答案)【分析】利用二项展开式的通项公式求出第r+1项,令x的指数为3,求出r,即可求出展开式中x3的系数.【解答】解:(2x+)5的展开式中,通项公式为:T r==25﹣+1r,令5﹣=3,解得r=4∴x3的系数2=10.故答案为:10.【点评】本题考查了二项式定理的应用,考查了推理能力与计算能力,属于基础题.15.设等比数列{a n}满足a1+a3=10,a2+a4=5,则a1a2…a n的最大值为64.【分析】求出数列的等比与首项,化简a1a2…a n,然后求解最值.【解答】解:等比数列{a n}满足a1+a3=10,a2+a4=5,可得q(a1+a3)=5,解得q=.a1+q2a1=10,解得a1=8.则a1a2…a n=a1n•q1+2+3+…+(n﹣1)=8n•==,当n=3或4时,表达式取得最大值:=26=64.故答案为:64.【点评】本题考查数列的性质数列与函数相结合的应用,转化思想的应用,考查计算能力.16.某高科技企业生产产品A和产品B需要甲、乙两种新型材料.生产一件产品A需要甲材料1.5kg,乙材料1kg,用5个工时;生产一件产品B需要甲材料0.5kg,乙材料0.3kg,用3个工时,生产一件产品A的利润为2100元,生产一件产品B的利润为900元.该企业现有甲材料150kg,乙材料90kg,则在不超过600个工时的条件下,生产产品A、产品B的利润之和的最大值为216000元.【分析】设A、B两种产品分别是x件和y件,根据题干的等量关系建立不等式组以及目标函数,利用线性规划作出可行域,通过目标函数的几何意义,求出其最大值即可;【解答】解:(1)设A、B两种产品分别是x件和y件,获利为z元.由题意,得,z=2100x+900y.不等式组表示的可行域如图:由题意可得,解得:,A(60,100),目标函数z=2100x+900y.经过A时,直线的截距最大,目标函数取得最大值:2100×60+900×100=216000元.故答案为:216000.【点评】本题考查了列二元一次方程组解实际问题的运用,二元一次方程组的解法的运用,不等式组解实际问题的运用,不定方程解实际问题的运用,解答时求出最优解是解题的关键.三.解答题(共7小题)17.△ABC的内角A,B,C的对边分别为a,b,c,已知2cosC(acosB+bcosA)=c.(Ⅰ)求C;(Ⅱ)若c=,△ABC的面积为,求△ABC的周长.【分析】(Ⅰ)已知等式利用正弦定理化简,整理后利用两角和与差的正弦函数公式及诱导公式化简,根据sinC不为0求出cosC的值,即可确定出出C的度数;(2)利用余弦定理列出关系式,利用三角形面积公式列出关系式,求出a+b的值,即可求△ABC的周长.【解答】解:(Ⅰ)∵在△ABC中,0<C<π,∴sinC≠0已知等式利用正弦定理化简得:2cosC(sinAcosB+sinBcosA)=sinC,整理得:2cosCsin(A+B)=sinC,即2cosCsin(π﹣(A+B))=sinC2cosCsinC=sinC∴cosC=,∴C=;(Ⅱ)由余弦定理得7=a2+b2﹣2ab•,∴(a+b)2﹣3ab=7,∵S=absinC=ab=,∴ab=6,∴(a+b)2﹣18=7,∴a+b=5,∴△ABC的周长为5+.【点评】此题考查了正弦、余弦定理,三角形的面积公式,以及三角函数的恒等变形,熟练掌握定理及公式是解本题的关键.18.如图,在以A,B,C,D,E,F为顶点的五面体中,面ABEF为正方形,AF=2FD,∠AFD=90°,且二面角D﹣AF﹣E与二面角C﹣BE﹣F都是60°.(Ⅰ)证明平面ABEF⊥平面EFDC;(Ⅱ)求二面角E﹣BC﹣A的余弦值.【分析】(Ⅰ)证明AF⊥平面EFDC,利用平面与平面垂直的判定定理证明平面ABEF⊥平面EFDC;(Ⅱ)证明四边形EFDC为等腰梯形,以E为原点,建立如图所示的坐标系,求出平面BEC、平面ABC的法向量,代入向量夹角公式可得二面角E﹣BC﹣A的余弦值.【解答】(Ⅰ)证明:∵ABEF为正方形,∴AF⊥EF.∵∠AFD=90°,∴AF⊥DF,∵DF∩EF=F,∴AF⊥平面EFDC,∵AF⊂平面ABEF,∴平面ABEF⊥平面EFDC;(Ⅱ)解:由AF⊥DF,AF⊥EF,可得∠DFE为二面角D﹣AF﹣E的平面角;由ABEF为正方形,AF⊥平面EFDC,∵BE⊥EF,∴BE⊥平面EFDC即有CE⊥BE,可得∠CEF为二面角C﹣BE﹣F的平面角.可得∠DFE=∠CEF=60°.∵AB∥EF,AB⊄平面EFDC,EF⊂平面EFDC,∴AB∥平面EFDC,∵平面EFDC∩平面ABCD=CD,AB⊂平面ABCD,∴AB∥CD,∴CD∥EF,∴四边形EFDC为等腰梯形.以E为原点,建立如图所示的坐标系,设FD=a,则E(0,0,0),B(0,2a,0),C(,0,a),A(2a,2a,0),∴=(0,2a,0),=(,﹣2a,a),=(﹣2a,0,0)设平面BEC的法向量为=(x1,y1,z1),则,则,取=(,0,﹣1).设平面ABC的法向量为=(x2,y2,z2),则,则,取=(0,,4).设二面角E﹣BC﹣A的大小为θ,则cosθ===﹣,则二面角E﹣BC﹣A的余弦值为﹣.【点评】本题考查平面与平面垂直的证明,考查用空间向量求平面间的夹角,建立空间坐标系将二面角问题转化为向量夹角问题是解答的关键.19.某公司计划购买2台机器,该种机器使用三年后即被淘汰.机器有一易损零件,在购进机器时,可以额外购买这种零件作为备件,每个200元.在机器使用期间,如果备件不足再购买,则每个500元.现需决策在购买机器时应同时购买几个易损零件,为此搜集并整理了100台这种机器在三年使用期内更换的易损零件数,得如图柱状图:以这100台机器更换的易损零件数的频率代替1台机器更换的易损零件数发生的概率,记X表示2台机器三年内共需更换的易损零件数,n表示购买2台机器的同时购买的易损零件数.(Ⅰ)求X的分布列;(Ⅱ)若要求P(X≤n)≥0.5,确定n的最小值;(Ⅲ)以购买易损零件所需费用的期望值为决策依据,在n=19与n=20之中选其一,应选用哪个?【分析】(Ⅰ)由已知得X的可能取值为16,17,18,19,20,21,22,分别求出相应的概率,由此能求出X的分布列.(Ⅱ)由X的分布列求出P(X≤18)=,P(X≤19)=.由此能确定满足P (X≤n)≥0.5中n的最小值.(Ⅲ)法一:由X的分布列得P(X≤19)=.求出买19个所需费用期望EX1和买20个所需费用期望EX2,由此能求出买19个更合适.法二:解法二:购买零件所用费用含两部分,一部分为购买零件的费用,另一部分为备件不足时额外购买的费用,分别求出n=19时,费用的期望和当n=20时,费用的期望,从而得到买19个更合适.【解答】解:(Ⅰ)由已知得X的可能取值为16,17,18,19,20,21,22,P(X=16)=()2=,P(X=17)=,P(X=18)=()2+2()2=,P(X=19)==,P(X=20)===,P(X=21)==,P(X=22)=,∴X的分布列为:X16171819202122 P(Ⅱ)由(Ⅰ)知:P(X≤18)=P(X=16)+P(X=17)+P(X=18)==.P(X≤19)=P(X=16)+P(X=17)+P(X=18)+P(X=19)=+=.∴P(X≤n)≥0.5中,n的最小值为19.(Ⅲ)解法一:由(Ⅰ)得P(X≤19)=P(X=16)+P(X=17)+P(X=18)+P(X=19)=+=.买19个所需费用期望:EX1=200×+(200×19+500)×+(200×19+500×2)×+(200×19+500×3)×=4040,买20个所需费用期望:EX2=+(200×20+500)×+(200×20+2×500)×=4080,∵EX1<EX2,∴买19个更合适.解法二:购买零件所用费用含两部分,一部分为购买零件的费用,另一部分为备件不足时额外购买的费用,当n=19时,费用的期望为:19×200+500×0.2+1000×0.08+1500×0.04=4040,当n=20时,费用的期望为:20×200+500×0.08+1000×0.4=4080,∴买19个更合适.【点评】本题考查离散型随机变量的分布列和数学期望的求法及应用,是中档题,解题时要认真审题,注意相互独立事件概率乘法公式的合理运用.20.设圆x2+y2+2x﹣15=0的圆心为A,直线l过点B(1,0)且与x轴不重合,l 交圆A于C,D两点,过B作AC的平行线交AD于点E.(Ⅰ)证明|EA|+|EB|为定值,并写出点E的轨迹方程;(Ⅱ)设点E的轨迹为曲线C1,直线l交C1于M,N两点,过B且与l垂直的直线与圆A交于P,Q两点,求四边形MPNQ面积的取值范围.【分析】(Ⅰ)求得圆A的圆心和半径,运用直线平行的性质和等腰三角形的性质,可得EB=ED,再由圆的定义和椭圆的定义,可得E的轨迹为以A,B为焦点的椭圆,求得a,b,c,即可得到所求轨迹方程;(Ⅱ)设直线l:x=my+1,代入椭圆方程,运用韦达定理和弦长公式,可得|MN|,由PQ⊥l,设PQ:y=﹣m(x﹣1),求得A到PQ的距离,再由圆的弦长公式可得|PQ|,再由四边形的面积公式,化简整理,运用不等式的性质,即可得到所求范围.【解答】解:(Ⅰ)证明:圆x2+y2+2x﹣15=0即为(x+1)2+y2=16,可得圆心A(﹣1,0),半径r=4,由BE∥AC,可得∠C=∠EBD,由AC=AD,可得∠D=∠C,即为∠D=∠EBD,即有EB=ED,则|EA|+|EB|=|EA|+|ED|=|AD|=4,故E的轨迹为以A,B为焦点的椭圆,且有2a=4,即a=2,c=1,b==,则点E的轨迹方程为+=1(y≠0);(Ⅱ)椭圆C1:+=1,设直线l:x=my+1,由PQ⊥l,设PQ:y=﹣m(x﹣1),由可得(3m2+4)y2+6my﹣9=0,设M(x1,y1),N(x2,y2),可得y1+y2=﹣,y1y2=﹣,则|MN|=•|y1﹣y2|=•=•=12•,A到PQ的距离为d==,|PQ|=2=2=,则四边形MPNQ面积为S=|PQ|•|MN|=••12•=24•=24,当m=0时,S取得最小值12,又>0,可得S<24•=8,即有四边形MPNQ面积的取值范围是[12,8).【点评】本题考查轨迹方程的求法,注意运用椭圆和圆的定义,考查直线和椭圆方程联立,运用韦达定理和弦长公式,以及直线和圆相交的弦长公式,考查不等式的性质,属于中档题.21.已知函数f(x)=(x﹣2)e x+a(x﹣1)2有两个零点.(Ⅰ)求a的取值范围;(Ⅱ)设x1,x2是f(x)的两个零点,证明:x1+x2<2.【分析】(Ⅰ)由函数f(x)=(x﹣2)e x+a(x﹣1)2可得:f′(x)=(x﹣1)e x+2a (x﹣1)=(x﹣1)(e x+2a),对a进行分类讨论,综合讨论结果,可得答案.(Ⅱ)设x1,x2是f(x)的两个零点,则﹣a==,令g(x)=,则g(x1)=g(x2)=﹣a,分析g(x)的单调性,令m>0,则g (1+m)﹣g(1﹣m)=,设h(m)=,m>0,利用导数法可得h(m)>h(0)=0恒成立,即g(1+m)>g(1﹣m)恒成立,令m=1﹣x1>0,可得结论.【解答】解:(Ⅰ)∵函数f(x)=(x﹣2)e x+a(x﹣1)2,∴f′(x)=(x﹣1)e x+2a(x﹣1)=(x﹣1)(e x+2a),①若a=0,那么f(x)=0⇔(x﹣2)e x=0⇔x=2,函数f(x)只有唯一的零点2,不合题意;②若a>0,那么e x+2a>0恒成立,当x<1时,f′(x)<0,此时函数为减函数;当x>1时,f′(x)>0,此时函数为增函数;此时当x=1时,函数f(x)取极小值﹣e,由f(2)=a>0,可得:函数f(x)在x>1存在一个零点;当x<1时,e x<e,x﹣2<﹣1<0,∴f(x)=(x﹣2)e x+a(x﹣1)2>(x﹣2)e+a(x﹣1)2=a(x﹣1)2+e(x﹣1)﹣e,令a(x﹣1)2+e(x﹣1)﹣e=0的两根为t1,t2,且t1<t2,则当x<t1,或x>t2时,f(x)>a(x﹣1)2+e(x﹣1)﹣e>0,故函数f(x)在x<1存在一个零点;即函数f(x)在R是存在两个零点,满足题意;③若﹣<a<0,则ln(﹣2a)<lne=1,当x<ln(﹣2a)时,x﹣1<ln(﹣2a)﹣1<lne﹣1=0,e x+2a<e ln(﹣2a)+2a=0,即f′(x)=(x﹣1)(e x+2a)>0恒成立,故f(x)单调递增,当ln(﹣2a)<x<1时,x﹣1<0,e x+2a>e ln(﹣2a)+2a=0,即f′(x)=(x﹣1)(e x+2a)<0恒成立,故f(x)单调递减,当x>1时,x﹣1>0,e x+2a>e ln(﹣2a)+2a=0,即f′(x)=(x﹣1)(e x+2a)>0恒成立,故f(x)单调递增,故当x=ln(﹣2a)时,函数取极大值,由f(ln(﹣2a))=[ln(﹣2a)﹣2](﹣2a)+a[ln(﹣2a)﹣1]2=a{[ln(﹣2a)﹣2]2+1}<0得:函数f(x)在R上至多存在一个零点,不合题意;④若a=﹣,则ln(﹣2a)=1,当x<1=ln(﹣2a)时,x﹣1<0,e x+2a<e ln(﹣2a)+2a=0,即f′(x)=(x﹣1)(e x+2a)>0恒成立,故f(x)单调递增,当x>1时,x﹣1>0,e x+2a>e ln(﹣2a)+2a=0,即f′(x)=(x﹣1)(e x+2a)>0恒成立,故f(x)单调递增,故函数f(x)在R上单调递增,函数f(x)在R上至多存在一个零点,不合题意;⑤若a<﹣,则ln(﹣2a)>lne=1,当x<1时,x﹣1<0,e x+2a<e ln(﹣2a)+2a=0,即f′(x)=(x﹣1)(e x+2a)>0恒成立,故f(x)单调递增,当1<x<ln(﹣2a)时,x﹣1>0,e x+2a<e ln(﹣2a)+2a=0,即f′(x)=(x﹣1)(e x+2a)<0恒成立,故f(x)单调递减,当x>ln(﹣2a)时,x﹣1>0,e x+2a>e ln(﹣2a)+2a=0,即f′(x)=(x﹣1)(e x+2a)>0恒成立,故f(x)单调递增,故当x=1时,函数取极大值,由f(1)=﹣e<0得:函数f(x)在R上至多存在一个零点,不合题意;综上所述,a的取值范围为(0,+∞)证明:(Ⅱ)∵x1,x2是f(x)的两个零点,∴f(x1)=f(x2)=0,且x1≠1,且x2≠1,∴﹣a==,令g(x)=,则g(x1)=g(x2)=﹣a,∵g′(x)=,∴当x<1时,g′(x)<0,g(x)单调递减;当x>1时,g′(x)>0,g(x)单调递增;设m>0,则g(1+m)﹣g(1﹣m)=﹣=,设h(m)=,m>0,则h′(m)=>0恒成立,即h(m)在(0,+∞)上为增函数,h(m)>h(0)=0恒成立,即g(1+m)>g(1﹣m)恒成立,令m=1﹣x1>0,则g(1+1﹣x1)>g(1﹣1+x1)⇔g(2﹣x1)>g(x1)=g(x2)⇔2﹣x1>x2,即x1+x2<2.【点评】本题考查的知识点是利用导数研究函数的极值,函数的零点,分类讨论思想,难度较大.22.在直角坐标系xOy中,曲线C1的参数方程为(t为参数,a>0).在以坐标原点为极点,x轴正半轴为极轴的极坐标系中,曲线C2:ρ=4cosθ.(Ⅰ)说明C1是哪种曲线,并将C1的方程化为极坐标方程;(Ⅱ)直线C3的极坐标方程为θ=α0,其中α0满足tanα0=2,若曲线C1与C2的公共点都在C3上,求a.【分析】(Ⅰ)把曲线C1的参数方程变形,然后两边平方作和即可得到普通方程,可知曲线C1是圆,化为一般式,结合x2+y2=ρ2,y=ρsinθ化为极坐标方程;(Ⅱ)化曲线C2、C3的极坐标方程为直角坐标方程,由条件可知y=x为圆C1与C2的公共弦所在直线方程,把C1与C2的方程作差,结合公共弦所在直线方程为y=2x可得1﹣a2=0,则a值可求.【解答】解:(Ⅰ)由,得,两式平方相加得,x2+(y﹣1)2=a2.∴C1为以(0,1)为圆心,以a为半径的圆.化为一般式:x2+y2﹣2y+1﹣a2=0.①由x2+y2=ρ2,y=ρsinθ,得ρ2﹣2ρsinθ+1﹣a2=0;(Ⅱ)C2:ρ=4cosθ,两边同时乘ρ得ρ2=4ρcosθ,∴x2+y2=4x,②即(x﹣2)2+y2=4.由C3:θ=α0,其中α0满足tanα0=2,得y=2x,∵曲线C1与C2的公共点都在C3上,∴y=2x为圆C1与C2的公共弦所在直线方程,①﹣②得:4x﹣2y+1﹣a2=0,即为C3 ,∴1﹣a2=0,∴a=1(a>0).【点评】本题考查参数方程即简单曲线的极坐标方程,考查了极坐标与直角坐标的互化,训练了两圆公共弦所在直线方程的求法,是基础题.23.已知函数f(x)=|x+1|﹣|2x﹣3|.(Ⅰ)在图中画出y=f(x)的图象;(Ⅱ)求不等式|f(x)|>1的解集.【分析】(Ⅰ)运用分段函数的形式写出f(x)的解析式,由分段函数的画法,即可得到所求图象;(Ⅱ)分别讨论当x≤﹣1时,当﹣1<x<时,当x≥时,解绝对值不等式,取交集,最后求并集即可得到所求解集.【解答】解:(Ⅰ)f(x)=,由分段函数的图象画法,可得f(x)的图象,如右:(Ⅱ)由|f(x)|>1,可得当x≤﹣1时,|x﹣4|>1,解得x>5或x<3,即有x≤﹣1;当﹣1<x<时,|3x﹣2|>1,解得x>1或x<,即有﹣1<x<或1<x<;当x≥时,|4﹣x|>1,解得x>5或x<3,即有x>5或≤x<3.综上可得,x<或1<x<3或x>5.则|f(x)|>1的解集为(﹣∞,)∪(1,3)∪(5,+∞).【点评】本题考查绝对值函数的图象和不等式的解法,注意运用分段函数的图象的画法和分类讨论思想方法,考查运算能力,属于基础题.。

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高考真题及答案 高考真题解析 2016普通高等学校招生全国统一考试(新课标I) 英 语 试卷类型A

第Ⅰ卷 第一部分 听力(共两节,满分 30 分) 做题时,现将答案标在试卷上,录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。 第一节(共5小题;每小题1.5分,满分7.5分) 听下面5段对话,每段对话后有一个小题。从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。每段对话仅读一遍。 例:How much is the shirt? A. £ 19. 15 B. £ 9. 18 C. £ 9. 15 答案是 C。 1. What are the speakers talking about? A. Having a birthday party. B. Doing some exercise. C. Getting Lydia a gift. 2. What is the woman going to do? A. Help the man. B. Take a bus. C. Get a camera. 3. What does the woman suggest the man do? A. Tell Kate to stop. B. Call Kate, s friends. C. Stay away from Kate. 4. Where does the conversation probably take place? A. In a wine shop. B. In a supermarket. C. In a restaurant. 5. What does the woman mean? A. Keep the window closed. B. Go out for fresh air. C. Turn on the fan. 听第6段材料,回答第6、7题。 6. What is the man going to do this summer? A. Teach a course. B. Repair his house. C. Work at a hotel. 7. How will the man use the money? A. To hire a gardener. B. To buy books. C. To pay for a boat trip. 听第7段材料,回答第8、9题。 8. What is the probable relationship between the speakers? A. Schoolmates. B. Colleagues. C. Roommates. 9. What does Frank plan to do right after graduation? A. Work as a programmer. B. Travel around the world. C. Start his own business. 第二节 (共15小题;每小题1. 5分,满分22. 5分) 听下面5段对话或独白。每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。每段对话或独白读两遍。 听第8段材料,回答第10至12题 10. Why does the woman make the call? A. To book a hotel room. B. To ask about the room service C. To make changes ti a reservation 高考真题及答案 高考真题解析 11. When will the women arrive at the hotel? A. On September 15 B. On September 16 C. On September 23 12. How much will the woman pay her room per night? A. $179 B. $199 C. $219 听第9段材料,回答第13至16题。 13. What is the woman’s plan for Saturday? A. Going shopping B. Going camping C. Going boating 14. Where will the woman stay in Keswick? A. In a country inn B. In a five-star hotel C. In her aunt’s home 15. What will Gordon do over the weekend? A. Visit his friends B. Watch DVDs C. Join the woman 16. What does the woman think of Gordon’s coming weekend? A. Relaxed B. Boring C. Busy. 听第10段材料,回答第17至20题 17. Who is Wang Ming? A. A student B. An employer C. An engineer 18. What does the speaker say about the college job market this year? A. It’s unpredictable B. It’s quite stable C. It’s not optimistic 19. What percentage of student job seekers have found a job by now? A. 20% B. 22% C. 50% 20. Why are engineering graduates more likely to accept a job? A. They need more work experience B. The salary is usually good C. Their choice is limited. 第二部分 阅读理解(共两节,满分40分) 第一节(共15小题:每小题2分,满分30分) 阅读下列短文,从每题所给的四个选项(A、B、C和D)中选出最佳选项,并在答题卡上将该项涂黑。 A You probably know who Marie Curie was, but you may not have heard of Rachel Carson. Of the outstanding ladies listed below, who do you think was the most important woman of the past 100 years? Jane Addams (1860-1935) Anyone who has ever been helped by a social worker has Jane Addams to thank. Addams helped the poor and worked for peace. She encouraged a sense of community(社区) by creating shelters and promoting education and services for people in need In 1931,Addams became the first American woman to win the Nobel Peace Prize. Rachel Carson (1907-1964) If it weren’t for Rachel Carson, the environmental movement might not exist today. Her popular 1962 book Silent Spring raised awareness of the dangers of pollution and the harmful effects of chemicals on humans and on the world’s lakes and oceans. Sandra Day O’Connor (1930-present) When Sandra Day O’Connor finished third in her class at Stanford Law School, in 1952, she could not find work at a law firm because she was a woman. She became an Arizona 高考真题及答案 高考真题解析 state senator(参议员) and ,in 1981, the first woman to join the U. S. Supreme Court. O’Connor gave the deciding vote in many important cases during her 24 years on the top court. Rosa Parks (1913-2005) On December 1,1955, in Montgomery, Alabama, Rasa Parks would not give up her seat on a bus to a passenger. Her simple act landed Parks in prison. But it also set off the Montgomery bus boycott. It lasted for more than a year, and kicked off the civil-rights movement. “The only tired I was, was tired of giving in,” said Parks. 21. What is Jane Addams noted for in history? A. Her social work. B. Her lack of proper training in law. C. Her efforts to win a prize. D. Her community background. 22. What is the reason for O’Connor’s being rejected by the law firm? A. Her lack of proper training in law. B. Her little work experience in court. C. The discrimination against women. D. The poor financial conditions. 23. Who made a great contribution to the civil-rights movement in the US? A. Jane Addams. B. Rachel Carson. C. Sandra Day O’Connor. D. Ross Parks. 24. What can we infer about the women mentioned in the text? A. They are highly educated. B. They are truly creative. C. They are pioneers. D. They are peace-lovers. B Grandparents Answer a Call As a third generation native of Brownsville, Texas, Mildred Garza never pleased move away. Even when her daughter and son asked her to move to San Antonio to help their children, she politely refused. Only after a year of friendly discussion did Ms. Gaf finally say yes. That was four years ago. Today all three generations regard the move to a success, giving them a closer relationship than they would have had in separate cities. No statistics show the number of grandparents like Garza who are moving closer to the children and grandchildren. Yet there is evidence suggesting that the trend is growing. Even President Obama’s mother-in-law, Marian Robinson, has agreed to leave Chicago and into the White House to help care for her granddaughters. According to a study grandparents com. 83 percent of the people said Mrs. Robinson ‘s decision will influence the grandparents in the American family. Two-thirds believe more families will follow the example of Obama’s family. “In the 1960s we were all a little wild and couldn’t get away from home far enough fast enough to prove we could do it on our own,” says Christine Crosby, publisher of Grate magazine for grandparents. We now realize how important family is and how important to be near them, especially when you’re raining children.” Moving is not for everyone. Almost every grandparent wants to be with his or her grandchildren and is willing to make sacrifices, but sometimes it is wiser to say no and

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