2020-2021学年江苏省连云港市、徐州市、宿迁市高考年级第三次模拟考试英语试题及答案

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江苏省徐州市、宿迁市、连云港市2022届高三第三次模拟考试(三模) 英语 图片版含答案

江苏省徐州市、宿迁市、连云港市2022届高三第三次模拟考试(三模) 英语 图片版含答案

高三班级第三次模拟考试英语试题参考答案第一部分听力(共两节,满分20分)1-5 CCBAB 6-10 ABAAC 11-15 BBCBB 16-20 ABCAB其次部分英语学问运用(共两节,满分35分)第一节单项填空(共15题;每小题1分,满分15分)21 – 25 DCBDA 26 – 30 ABDAB 31 – 35BCBCB其次节完型填空(共20小题;每小题1分,满分20分)36-40 ABDAB 41-45 ABCDA 46-50 CACDB 51-55 CABDB第三部分阅读理解(共15小题;每小题2分,满分30分)56 – 57 DB 58 – 60 ADC 61-64BADD 65 – 70 BDCCDC第四部分任务型阅读(共10小题;每小题l分,满分l0分)71. Finding 72. possibility/probability 73. participating 74. Factors 75. emotionally76. confidence/willingness 77. stronger 78. access 79. leaving/quitting80. instead第五部分:书面表达(满分 25分)这篇小短文有204个字。

本篇小文章是选自《美国语文经典读本》其次册第八课。

[Sample]To have her kittens stay in the proper place—attic, Puss tried several times but was always stopped by the servant. Thanks for a strange and kind cat’s help, she made it. (32words)I’m deeply moved by what the strange cat did. He helped Puss selflessly asking for nothing in return, which made me think of what happened to me recently. It was on Monday morning, when several teachers attended our English lesson. Among them, a middle-aged teacher sat beside me, leaving her wallet on my desk. But she forgot to take it after class. So I rushed out, gave it back to her and ran away.As far as I am concerned, the spirit of being ready to help others should be advocated. Helping others is a traditional virtue in China. If you help others today, maybe others will help you in the future. In this way, we can construct a harmonious society. (121words)听力原文Text1M: Welcome home, Mary! Have you had any problems at university since last September?W: Merry Christmas, Uncle! Well, nothing to speak of.M: No more worries? No complaints? Great!Text2M: Excuse me. Can you tell me where the Shanghai Grand Theater is? I’m completely lost.W: Oh, just walk along this street and turn right at the corner. You will see the Shanghai Museum on People’s Avenue. The Shanghai Grand Theater is opposite that.Text3W: I heard you got full marks in the math exam. Congratulations!M: Thanks! I’m sure you also did a good job.Text4M: I don’t know why people are just mad about David Beckham. Is there really anything so fascinating with him? W: He appeals to the fans thanks to two things, his bending kicks and his good looks.Text5 W: Rod, I hear you’ll be leaving at the end of this month. Is it true?M: Yeah. I’ve been offered a much better position with another firm. I’d be a fool to turn it down.Text6W: What do you plan to do during the holiday?M: I want to visit China. Can you recommend any interesting places I should visit there?W: I think Shanghai will interest you. It is a big city.M: Good idea. Where are you going to spend your holiday?W: I will stay at home. I want to spend my time studying French. It is said that French is a very popular language because it is the most beautiful language in the world. I think it’s a very useful language, even though it isn’t very easy.M: You really work hard all the time. You should really do some sports. For example, you can play basketball or football.W: That’s a good idea, but I think I’d rather play table tennis better. That’s a fun sport.Text7W: Hi, Tom! What are you doing?M: Oh, I’m studying for an English test.W: Have you made any progress?M: No, as a matter of fact, v ery little. I really don’t know how to study all this stuff! How is your English? How do you study?W: Well, you might try using the Internet to help you, you can listen to English songs Online, watch films in English, and find lots of other things that can increase your interest and help you learn English. That really worked. Many of my classmates also made great progress. The Internet helps you not only learn English but also expand your knowledge of Western cultures.M: That’s really wonderful. I never t hought English language learning could be so much fun.W: Yes, in fact, learning English is not as difficult as you think.M: Next time, can you show me how to practice English online?W: Sure. Well, I have to return some books. Good luck in your examination!M: Thank you! Bye.Text8W: Doctor, I’ve got a little pain and numbness in my ears.M: Let me see. Well, do you often listen to loud music with earphones?W: Yes. I usually listen to music or practice English listening two or three hours per day.M: That’s the very problem. Listening to loud music more than 90 minutes a day can damage your hearing according to a new study.W: How can I improve my hearing, doctor?M: Well, stop listening to loud music with earphones immediately.W: Can’t I listen to music with earphones in the future?M: Oh, yes, but later. Try to reduce the time you use earphones and never use your ears too much.W: By the way, are there any user-friendly earphones to protect my ears?M: No, in the future we may get some. From now on you should concern yourself about your hearing loss, or it may cause a terrible result.W: I see. Thank you, doctor.Text9W: Good afternoon. I’d like some information about the train, please.M: Yes, Madam. Where are you going?W: To London. You see, I have a sister there and she studies in…M: So your question is“When is the next train to London?”. Is that right?W: Yes, that’s right. When’s the next train to London?M: At half past four. That’s in about six minutes.W: Thank you very much. Oh! Can I get something to eat on the train? I always have something to eat when I travel. I find that a cup of tea and a cake always calm me down.M: Yes, Madam. You can get what you want on the train.W: Oh, good! Err, how much does a cup of tea cost?M: I’m not s ure, Madam. Fifty pence, I think.W: Oh, dear! Things are getting so expensive.M: Yes, Madam. Your train is going to leave in five minutes from Platform 13.W: Platform 13! Oh, dear! I never travel on trains that leave from Platform 13! 13 is an unlucky n umber. When’s the next train after the 4:30 one?Text10M: Attention please! Thank you. Now, welcome to the Spring Hotel. We hope you have a wonderful holiday. Let me tell you about some of the services we have here. On the ground floor, you will find reception. Here we can answer your questions and help you with any general problems. Please leave your room key here whenever you go out. On the sixth floor is a restaurant where you can have breakfast from half past seven to half past nine each morning. The bedrooms are on the fourth and fifth floors. If you want to use a phone, there are telephones available on the second floor. Any questions? No? Fine. You can go shopping on the first floor. For people who like sports, if you want to swim, there’s a nice n ew pool on the seventh floor, where you can swim and enjoy the warm sunshine. We also have some tennis courts behind the hotel. Oh, and if you want to watch your favorite program, the televisions are on the third floor. Any questions? No? Well, good. Have a nice holiday.My office is on the sixth floor, just next to the pub. lf you’re in need of any help, please don’t hesitate to come and ask.。

2020年江苏省徐州市、淮安市、南通市、泰州市、扬州市、连云港市、宿迁市高考数学三模试卷 (解析版)

2020年江苏省徐州市、淮安市、南通市、泰州市、扬州市、连云港市、宿迁市高考数学三模试卷 (解析版)

2020年扬州市、徐州市、南通市、泰州市、淮安市、连云港市、宿迁市高考数学三模试卷一、填空题(共14小题).1.已知集合A={﹣1,0,1},B={0,2},则A∪B=.2.设复数z满足(3﹣i)z=,其中i为虚数单位,则z的模是.3.如图是一个算法流程图,则输出的k的值是.4.某校高一、高二、高三年级的学生人数之比为4:4:3,为了解学生对防震减灾知识的掌握情况,现采用分层抽样的方法抽取n名学生进行问卷检测.若高一年级抽取了20名学生,则n的值是.5.今年我国中医药选出的“三药三方”对治疗新冠肺炎均有显著效果,功不可没.“三药”分别为金花清感颗粒、连花清瘟胶囊、血必净注射液;“三方”分别为清肺排毒汤、化湿败毒方、宜肺败毒方,若某医生从“三药三方”中随机选出2种,则恰好选出1药1方的概率是.6.在平面直角坐标系xOy中,已知抛物线y2=4x的准线是双曲线=1(a>0)的左准线,则实数a的值是.7.已知cos(α+β)=,sinβ=,α,β均为锐角,则sinα的值是.8.公园里设置了一些石凳供游客休息,这些石凳是经过正方体各棱的中点截去8个一样的四面体得到的(如图所示).设石凳的体积为V1,正方体的体积为V2,则的值是.9.已知x>1,y>1,xy=10,则的最小值是.10.已知等比数列{a n}的前n项和为S n,若4S2,S4,﹣2S3成等差数列,且a2+a3=2,则a6的值是.11.海伦(Heron,约公元1世纪)是古希腊亚历山大时期的数学家,以他的名字命名的“海伦公式”是几何学中的著名公式,它给出了利用三角形的三边长a,b,c计算其面积的公式S△ABC=,其中p=,若a=5,b=6,c=7,则借助“海伦公式”可求得△ABC的内切圆的半径r的值是.12.如图,△ABC为等边三角形,分别延长BA,CB,AC到点D,E,F,使得AD=BE =CF.若,且DE=,则的值是.13.已知函数f(x)=,若函数g(x)=f(﹣x)+f(x)有且仅有四个不同的零点,则实数k的取值范围是.14.在平面直角坐标系xOy中,过点P(2,﹣6)作直线交圆O:x2+y2=16于A,B两点,C(x0,y0)为弦AB的中点,则的取值范围是.二、解答题(本大题共6小题,共计90分.请在答题纸指定区域内作答,解答应写出文字说明,证明过程或演算步骤.)15.△ABC中,角A,B,C所对的边分别为a,b,c.若.(1)求cos C的值;(2)若A=C,求sin B的值.16.如图,在直三棱柱ABC﹣A1B1C1中,AC⏊BC,D,E分别是A1B1,BC的中点.求证:(1)平面ACD⊥平面BCC1B1;(2)B1E∥平面ACD.17.某单位科技活动纪念章的结构如图所示,O是半径分别为1cm,2cm的两个同心圆的圆心,等腰△ABC的顶点A在外圆上,底边BC的两个端点都在内圆上,点O,A在直线BC的同侧.若线段BC与劣弧所围成的弓形面积为S1,△OAB与△OAC的面积之和为S2,设∠BOC=2θ.(1)当θ=时,求S2﹣S1的值;(2)经研究发现当S2﹣S1的值最大时,纪念章最美观,求当纪念章最美观时,cosθ的值.(求导参考公式:(sin2x)'=2cos2x,(cos2x)'=﹣2sin2x)18.(16分)如图,在平面直角坐标系xOy中,已知椭圆=1(a>b>0)的左、右焦点分别为F1,F2,过点F2的直线交椭圆于M,N两点.已知椭圆的短轴长为2,离心率为.(1)求椭圆的标准方程;(2)当直线MN的斜率为时,求F1M+F1N的值;(3)若以MN为直径的圆与x轴相交的右交点为P(t,0),求实数t的取值范围.19.(16分)已知{a n}是各项均为正数的无穷数列,数列{b n}满足b n=a n•a n+k(n∈N*),其中常数k为正整数.(1)设数列{a n}前n项的积,当k=2时,求数列{b n}的通项公式;(2)若{a n}是首项为1,公差d为整数的等差数列,且b2﹣b1=4,求数列的前2020项的和;(3)若{b n}是等比数列,且对任意的n∈N*,a n•a n+2k=a n+k2,其中k≥2,试问:{a n}是等比数列吗?请证明你的结论.20.(16分)已知函数f(x)=,g(x)=,其中e是自然对数的底数.(1)若函数f(x)的极大值为,求实数a的值;(2)当a=e时,若曲线y=f(x)与y=g(x)在x=x0处的切线互相垂直,求x0的值;(3)设函数h(x)=g(x)﹣f(x),若h(x)>0对任意的x∈(0,1)恒成立,求实数a的取值范围.【选做题】本题包括21,22,23三小题,请选定其中两题作答,每小题10分,共计20分,解答时应写出文字说明,证明过程或演算步骤.[选修4-2:矩阵与变换]21.已知m∈R,=是矩阵M=的一个特征向量,求M的逆矩阵M﹣1.[选修4-4:坐标系与参数方程]22.在极坐标系中,圆C的方程为ρ=2r sinθ(r>0).以极点为坐标原点,极轴为x轴正半轴建立平面直角坐标系,直线l的参数方程为(t为参数).若直线l与圆C恒有公共点,求r的取值范围.[选修4-5:不等式选讲]23.已知x>1,y>1,且x+y=4,求证:≥8.【必做题】第24题、第25题,每题10分,共计20分,解答时应写出文字说明,证明过程或演算步骤.24.某“芝麻开门”娱乐活动中,共有5扇门,游戏者根据规则开门,并根据打开门的数量获取相应奖励.已知开每扇门相互独立,且规则相同,开每扇门的规则是:从给定的6把钥匙(其中有且只有1把钥匙能打开门)中,随机地逐把抽取钥匙进行试开,钥匙使用后不放回.若门被打开,则转为开下一扇门;若连续4次未能打开,则放弃这扇门,转为开下一扇门;直至5扇门都进行了试开,活动结束.(1)设随机变量X为试开第一扇门所用的钥匙数,求X的分布列及数学期望E(X);(2)求恰好成功打开4扇门的概率.25.如图,在平面直角坐标系xOy中,已知抛物线y2=2px(p>0)的焦点为F,准线与x 轴的交点为E.过点F的直线与抛物线相交于A,B两点,EA,EB分别与y轴相交于M,N两点,当AB⊥x轴时,EA=2.(1)求抛物线的方程;(2)设△EAB的面积为S1,△EMN面积为S2,求的取值范围.参考答案一、填空题(本大题共14小题,每小题5分,共计70分.不需要写出解答过程,请将答案填写在答题卡相应的位置上.)1.已知集合A={﹣1,0,1},B={0,2},则A∪B={﹣1,0,1,2}.【分析】进行并集的运算即可.解:∵A={﹣1,0,1},B={0,2},∴A∪B={﹣1,0,1,2}.故答案为:{﹣1,0,1,2}.2.设复数z满足(3﹣i)z=,其中i为虚数单位,则z的模是1.【分析】把已知等式变形,再由商的模等于模的商求解.解:由(3﹣i)z=,得z=,∴|z|=||=.故答案为:1.3.如图是一个算法流程图,则输出的k的值是5.【分析】由已知中的程序语句可知:该程序的功能是利用循环结构计算并输出变量k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:模拟程序的运行,可得k=1不满足条件k2﹣4k>0,执行循环体,k=2不满足条件k2﹣4k>0,执行循环体,k=3不满足条件k2﹣4k>0,执行循环体,k=4不满足条件k2﹣4k>0,执行循环体,k=5此时,满足条件k2﹣4k>0,退出循环,输出k的值为5.故答案为:5.4.某校高一、高二、高三年级的学生人数之比为4:4:3,为了解学生对防震减灾知识的掌握情况,现采用分层抽样的方法抽取n名学生进行问卷检测.若高一年级抽取了20名学生,则n的值是55.【分析】先求出高一年级学生占的比例,再根据比例即可求解结论.解:高一年级学生占的比例为=,故应满足:=⇒n=55人,故答案为:55.5.今年我国中医药选出的“三药三方”对治疗新冠肺炎均有显著效果,功不可没.“三药”分别为金花清感颗粒、连花清瘟胶囊、血必净注射液;“三方”分别为清肺排毒汤、化湿败毒方、宜肺败毒方,若某医生从“三药三方”中随机选出2种,则恰好选出1药1方的概率是.【分析】某医生从“三药三方”中随机选出2种,基本事件总数n==15,恰好选出1药1方包含的基本事件个数m==9.由此能求出恰好选出1药1方的概率.解:“三药三方”对治疗新冠肺炎均有显著效果,功不可没.“三药”分别为金花清感颗粒、连花清瘟胶囊、血必净注射液,“三方”分别为清肺排毒汤、化湿败毒方、宜肺败毒方,若某医生从“三药三方”中随机选出2种,基本事件总数n==15,恰好选出1药1方包含的基本事件个数m==9.∴恰好选出1药1方的概率是p===.故答案为:.6.在平面直角坐标系xOy中,已知抛物线y2=4x的准线是双曲线=1(a>0)的左准线,则实数a的值是.【分析】求出抛物线的准线方程,求出双曲线的左准线方程,得到关系式,求解即可.解:抛物线y2=4x的准线是双曲线=1(a>0)的左准线,可得:﹣1=﹣=﹣,解得a=.故答案为:.7.已知cos(α+β)=,sinβ=,α,β均为锐角,则sinα的值是.【分析】由α,β的范围得出α+β的范围,然后利用同角三角函数间的基本关系,由cos (α+β)和sinβ的值,求出sin(α+β)和cosβ的值,然后由α=(α+β)﹣β,把所求的式子利用两角差的正弦函数公式化简后,将各自的值代入即可求出值.解:解:由cos(α+β)=,sinβ=,根据α,β∈(0,),得到α+β∈(0,π),所以sin(α+β)==,cosβ==,则sinα=sin[(α+β)﹣β]=sin(α+β)cosβ﹣cos(α+β)sinβ=×﹣×=.故答案为:.8.公园里设置了一些石凳供游客休息,这些石凳是经过正方体各棱的中点截去8个一样的四面体得到的(如图所示).设石凳的体积为V1,正方体的体积为V2,则的值是.【分析】设正方体的棱长为2a,求出正方体的体积,再由正方体的体积减去8个三棱锥的体积得石凳的体积,则答案可求.解:设正方体的棱长为2a,则正方体的体积.由题意可得,石凳的体积为V1=8a3﹣=.∴=.故答案为:.9.已知x>1,y>1,xy=10,则的最小值是9.【分析】利用“乘1法”与基本不等式的性质即可得出.解:因为x>1,y>1,xy=10,所以lgx+lgy=1,则=()(lgx+lgy)=5+=9,当且仅当时即lgy=2lgx且xy=10即x=,y=时取等号,故答案为:9.10.已知等比数列{a n}的前n项和为S n,若4S2,S4,﹣2S3成等差数列,且a2+a3=2,则a6的值是﹣32.【分析】等比数列{a n}的公比设为q,运用等差数列的中项性质和等比数列的通项公式,解方程可得首项和公比,进而得到所求值.解:等比数列{a n}的公比设为q,前n项和为S n,若4S2,S4,﹣2S3成等差数列,则2S4=4S2﹣2S3,可得2(a1+a1q+a1q2+a1q3)=4(a1+a1q)﹣2(a1+a1q+a1q2),化为2+q=0,可得q=﹣2,由a2+a3=2,可得﹣2a1+4a1=2,解得a1=1,则a6=1•(﹣2)5=﹣32,故答案为:﹣32.11.海伦(Heron,约公元1世纪)是古希腊亚历山大时期的数学家,以他的名字命名的“海伦公式”是几何学中的著名公式,它给出了利用三角形的三边长a,b,c计算其面积的公式S△ABC=,其中p=,若a=5,b=6,c=7,则借助“海伦公式”可求得△ABC的内切圆的半径r的值是.【分析】利用S△ABC==pr,代入即可得出.解:∵a=5,b=6,c=7,∴p===9.则S△ABC==r×(5+6+7),可得:r=.故答案为:.12.如图,△ABC为等边三角形,分别延长BA,CB,AC到点D,E,F,使得AD=BE =CF.若,且DE=,则的值是.【分析】设AD=BE=CF=x,由于,所以BA=2AD=2x=AC=BC,BD=3x.在△BDE中,由余弦定理知,,代入数据可解得x=1,从而有AF=3,CE=3,然后结合平面向量数量积的运算即可得解.解:设AD=BE=CF=x,∵,∴BA=2AD=2x=AC=BC,∴BD=BA+AD=3x,在△BDE中,由余弦定理知,,即,解得x=1.∴AF=3,CE=3,∴=.故答案为:.13.已知函数f(x)=,若函数g(x)=f(﹣x)+f(x)有且仅有四个不同的零点,则实数k的取值范围是(27,+∞).【分析】表示出函数g(x),分k=0,k<0及k=0讨论,易知当k=0及k<0时均不合题意,而观察解析式可知,问题可化为有且仅有两个不同的零点,故利用导数研究函数g(x)在(0,+∞)上的最小值小于0即可.解:依题意,,当k=0时,原函数有且只有一个零点,不合题意,故k≠0;观察解析式,易知函数g(x)为偶函数,则函数g(x)有且仅有四个不同的零点,可转化为有且仅有两个不同的零点,当k<0时,函数g(x)在(0,+∞)上递增,最多一个零点,不合题意;当k>0时,,令g′(x)>0,解得,令g′(x)<0,解得,故函数g(x)在上递减,在上递增,要使g(x)在(0,+∞)上有且仅有两个不同的零点,则,解得k>27.故答案为:(27,+∞).14.在平面直角坐标系xOy中,过点P(2,﹣6)作直线交圆O:x2+y2=16于A,B两点,C(x0,y0)为弦AB的中点,则的取值范围是[,).【分析】作出图象,根据条件可求得点C的运动轨迹为x2+y2﹣2x+6y=0,的取值范围可转化为求点C与点Q(﹣1,3)的距离范围,数形结合即可解:如图所示,由圆的性质知:PC⊥OC,∴•=0,又∵=(x0﹣2,y0+6),=(x0,y0),则•=x0(x0﹣2)+y0(y0+6)=x02+y02﹣2x0+6y0=0∴点C的轨迹方程为圆:x2+y2﹣2x+6y=0即(x﹣1)2+(y+3)2=10,圆心(1,﹣3),半径r=则的取值范围可转化为求点C与点Q(﹣1,3)的距离范围如图所示,因为点C在圆O内,故只需求出OQ和QM或QN的长度即可,易得OQ==,联立,整理得2x﹣6y﹣16=0即直线MN方程为x﹣3y﹣8=0,再联立,解得,,即M(,),N(,),故QM==QN==,因为C取不到M或N点,故的取值范围是[,).故答案为:[,).二、解答题(本大题共6小题,共计90分.请在答题纸指定区域内作答,解答应写出文字说明,证明过程或演算步骤.)15.△ABC中,角A,B,C所对的边分别为a,b,c.若.(1)求cos C的值;(2)若A=C,求sin B的值.【分析】(1)利用正弦定理转化条件,利用余弦定理求得cos C的值;(2)利用三角函数的内角和定理与三角恒等变换,即可求出sin B的值.解:(1)△ABC中,,由正弦定理得=,整理得5(a2+b2﹣c2)=8ab,由余弦定理得cos C===;(2)由(1)知cos C=,C是△ABC的内角,所以sin C==;又A=C,所以sin B=sin(π﹣A﹣C)=sin(A+C)=sin2C=2sin C cos C=2××=.16.如图,在直三棱柱ABC﹣A1B1C1中,AC⏊BC,D,E分别是A1B1,BC的中点.求证:(1)平面ACD⊥平面BCC1B1;(2)B1E∥平面ACD.【分析】(1)推导出AC⊥CC1,AC⊥平面BCC1B1,由此能证明平面ACD⊥平面BCC1B1.(2)取AC中点F,连结EF,DF,推导出四边形B1DFE为平行四边形,从而B1E∥DF,由此能证明B1E∥平面ACD.【解答】证明:(1)直三棱柱ABC﹣A1B1C1中,CC1⊥底面ABC,又AC⊂底面ABC,∴AC⊥CC1,∵AC⊥BC1,CC1∩BC=C,∴AC⊥平面BCC1B1,∵AC⊂平面ACD,∴平面ACD⊥平面BCC1B1.(2)取AC中点F,连结EF,DF,∵E,F分别为BC,AC中点,∴EF∥AB,EF=,直三棱柱ABC﹣A1B1C1中,AB A1B1,∵D为A1B1中点,∴B1D∥AB,B1D=,∴EF B1D,∴四边形B1DFE为平行四边形,∴B1E∥DF,∵DF⊂平面ACD,B1E⊄平面ACD,∴B1E∥平面ACD.17.某单位科技活动纪念章的结构如图所示,O是半径分别为1cm,2cm的两个同心圆的圆心,等腰△ABC的顶点A在外圆上,底边BC的两个端点都在内圆上,点O,A在直线BC的同侧.若线段BC与劣弧所围成的弓形面积为S1,△OAB与△OAC的面积之和为S2,设∠BOC=2θ.(1)当θ=时,求S2﹣S1的值;(2)经研究发现当S2﹣S1的值最大时,纪念章最美观,求当纪念章最美观时,cosθ的值.(求导参考公式:(sin2x)'=2cos2x,(cos2x)'=﹣2sin2x)【分析】(1)结合弓形面积公式及三角形的面积公式分别求出S2,S1,然后结合三角函数的性质即可求解;(2)结合(1)的面积表示,结合导数与单调性的关系可求.解:(1)由题意可知,∠BOC=2θ∈(0,π),故,S1==θ﹣sinθcosθ=,S2=﹣sin2θ=﹣sin2θ=2sinθ,当时,S1=,S2=,故S2﹣S1=(cm2),(2)S2﹣S1=2sinθ+sin2θ﹣θ,,令f(θ)=2sinθ+sin2θ﹣θ,,则f′(θ)=2cosθ+cos2θ﹣1=2cos2θ+2cosθ﹣2,令f′(θ)=0可得,cosθ=(舍负),记cosθ0=,,当θ∈(0,θ0)时,f′(θ)>0,函数单调递增,当时,f′(θ)<0,函数单调递减,故当θ=θ0时,即cosθ=时,f(θ)取得最大值,即S2﹣S1取得最大值.18.(16分)如图,在平面直角坐标系xOy中,已知椭圆=1(a>b>0)的左、右焦点分别为F1,F2,过点F2的直线交椭圆于M,N两点.已知椭圆的短轴长为2,离心率为.(1)求椭圆的标准方程;(2)当直线MN的斜率为时,求F1M+F1N的值;(3)若以MN为直径的圆与x轴相交的右交点为P(t,0),求实数t的取值范围.【分析】(1)设焦距为2c,运用离心率公式,可得a,b,c的方程,解方程可得a,b,进而得到椭圆方程;(2)由(1)可得c=2,即F1(﹣2,0),F2(2,0),联立直线方程和椭圆方程,求得M,N,即可得到所求和;(3)方法一、讨论直线MN的斜率不存在,求得|MN|,可得t的值;MN的斜率存在时,设MN:y=k(x﹣2),M(x1,y1),N(x2,y2),联立椭圆方程,运用韦达定理和中点坐标公式,弦长公式,结合圆的方程和换元,运用函数的单调性可得所求范围;方法二、运用直径所对的圆周角为直角,结合向量的数量积的性质和坐标表示,化简整理,可得t的不等式组,解得t的范围.解:(1)设焦距为2c,则2b=2,b2=a2﹣c2,e==,解得a=,b=,则椭圆的方程为+=1;(2)由(1)可得c=2,即F1(﹣2,0),F2(2,0),由可得或,即M(,),N(,﹣)或N(,),M(,﹣),因此|F1M|+|F1N|=+=;(3)方法一、①MN的斜率不存在时,MN:x=2,|MN|=,以MN为直径的圆的方程为(x﹣2)2+y2=,其与x轴相交的右交点为P(2+,0),即t=2+;②MN的斜率存在时,设MN:y=k(x﹣2),M(x1,y1),N(x2,y2),由,可得(1+3k2)x2﹣12k2x+12k2﹣6=0,△=(12k2)2﹣4(1+3k2)(12k2﹣6)=24(k2+1)>0恒成立,x1+x2=,x1x2=,|x1﹣x2|===,y1+y2=k(x1+x2)﹣4k=k•﹣4k=﹣,则MN的中点为(,﹣),|MN|=•|x1﹣x2|=•=,故以MN为直径的圆的方程为(x﹣)2+(y+)2=,令y=0,可得x=,由题意可得t=,可令1+3k2=m(m≥1),则k2=,t=2﹣+,可令x=,x∈(0,2],可得t=2﹣x+,可令f(x)=2﹣x+,x∈(0,2),由于(x+)2<x2+x+,则f′(x)=<0,故f(x)在(0,2)递减,f(0)=2+,f(2)=,因此f(x)∈[,2+),综上可得t∈[,2+].方法二、x1x2=,则y1y2=k2(x1﹣2)(x2﹣2)=k2[x1x2﹣2(x1+x2)+4]=k2[﹣2•+4]=﹣,P在以MN为直径的圆上,则•=0,(x1﹣t)(x2﹣t)+y1y2=0,x1x2﹣t(x1+x2)+t2+y1y2=0,即﹣+t2﹣=0,化为(3t2﹣12t+10)k2=6﹣t2,由于P为右交点,故t>2,因此,解得t∈[,2+].19.(16分)已知{a n}是各项均为正数的无穷数列,数列{b n}满足b n=a n•a n+k(n∈N*),其中常数k为正整数.(1)设数列{a n}前n项的积,当k=2时,求数列{b n}的通项公式;(2)若{a n}是首项为1,公差d为整数的等差数列,且b2﹣b1=4,求数列的前2020项的和;(3)若{b n}是等比数列,且对任意的n∈N*,a n•a n+2k=a n+k2,其中k≥2,试问:{a n}是等比数列吗?请证明你的结论.【分析】(1)直接利用关系式的变换的应用求出数列通项公式.(2)首先求出数列的通项公式,进一步利用裂项相消法在数列求和中的应用求出结果.(3)利用等比数列的的定义的应用求出结果.解:(1)因为,所以(n≥2),两式相除得:=2n﹣1(n≥2),当n=1时,,符号上式,∴(n∈N*),当k=2时,b n=a n•a n+2=2n﹣1•2n+1=4n;(2)由于b n=a n a n+1,且a1=1,所以b1=a1a k+1=a k+1,b2=a2a k+2=(d+1)(a k+1+d).所以=4,由于d和k都为正整数,所以d≥1,所以a k+1≥a2=1+d≥2,所以d2+d(a k+1+1)=4≥d2+3d.解得d≤1,所以d=1,即a n=n.所以d2+d(a k+1+1)=4=a k+1+2,即a k+1=2,解得k=1.所以b n=a n+1a n=n(n+1),所以.则:,所以.(3){b n}是等比数列,公比为,且对任意的n∈N*,所以=q2k.a n•a n+2k=a n+k2,所以,所以,所以=,则,所以.故数列{a n}是等比数列.20.(16分)已知函数f(x)=,g(x)=,其中e是自然对数的底数.(1)若函数f(x)的极大值为,求实数a的值;(2)当a=e时,若曲线y=f(x)与y=g(x)在x=x0处的切线互相垂直,求x0的值;(3)设函数h(x)=g(x)﹣f(x),若h(x)>0对任意的x∈(0,1)恒成立,求实数a的取值范围.【分析】(1)由题意可知a>0,先对f(x)求导,分析单调性,得到极大值,让其等于,即可解得a的值.(2)分别求出f(x),g(x)在x=x0处切线的斜率,让它们乘积等于1,即可解得x0的值.(3)问题可以转化为,对任意x∈(0,1)恒成立,设H(x)=,由(1)可知,H(x)在(0,1)上单调递增,且当x∈(1,+∞)时,H(x)>0,当x∈(0,1)时,H(x)<0,可得ae x>x,也就是ae x>x对任意x∈(0,1)恒成立,即,设G(x)=(x∈(0,1)),只要G(x)max≤a,即可得出答案.解:(1)因为f(x)=,则f′(x)==,因为g(x)=,所以a>0,则当x∈(0,e)时,f′(x)>0,f(x)单调递增,当x∈(e,+∞)时,f′(x)<0,f(x)单调递减,所以当x=e时,f(x)的极大值f(e)==,解得a=1.(2)当a=e时,f(x)=,g(x)=,则f′(x)=,g′(x)=,由题意可知,f′(x0)g′(x0)=•=﹣1,整理得x0e+elnx0=e,设φ(x)=xe x+elnx,则φ′(x)=(x+1)e x+>0,所以φ(x)单调递增,因为φ(1)=e,所以x0=1.(3)由题意可知,>0,对任意x∈(0,1)恒成立,整理得,对任意x∈(0,1)恒成立,设H(x)=,由(1)可知,H(x)在(0,1)上单调递增,且当x∈(1,+∞)时,H(x)>0,当x∈(0,1)时,H(x)<0,若ae x≥1>x,则H(ae x)≥0>H(x),若0<ae x<1,则H(ae x)>H(x),且H(x)在(0,1)上单调递增,所以ae x>x,综上可知,ae x>x对任意x∈(0,1)恒成立,即,设G(x)=(x∈(0,1)),则G′(x)=>0,所以G(x)单调递增,所以G(x)<G(1)=≤a,即a的取值范围为[,+∞).【选做题】本题包括21,22,23三小题,请选定其中两题作答,每小题10分,共计20分,解答时应写出文字说明,证明过程或演算步骤.[选修4-2:矩阵与变换]21.已知m∈R,=是矩阵M=的一个特征向量,求M的逆矩阵M﹣1.【分析】由=是属于特征值n的一个特征向量,得M=n,然后求出m,得到矩阵M,再设矩阵的逆矩阵M﹣1=,由MM﹣1=,求出M的逆矩阵M﹣1.解:由=是属于特征值n的一个特征向量,得M=n,∵M==,=n=,∴1+m=3=n,解得m=2,∴矩阵M=,设矩阵的逆矩阵M﹣1=,则MM﹣1===,∴,解得a=﹣,b=,c=,d=﹣,解得M﹣1=.[选修4-4:坐标系与参数方程]22.在极坐标系中,圆C的方程为ρ=2r sinθ(r>0).以极点为坐标原点,极轴为x轴正半轴建立平面直角坐标系,直线l的参数方程为(t为参数).若直线l与圆C恒有公共点,求r的取值范围.【分析】求出圆的直角坐标方程,把直线的参数方程化为普通方程,利用点到直线的距离与半径列出不等式求解即可.解:由ρ=2r sinθ得ρ2=2rρsinθ,∴圆C的方程为x2+y2﹣2ry=0,把参数方程为(t为参数),消去参数t,可得:普通方程:x﹣y﹣2=0,直线与圆有公共点,可得:d=≤r,解得r≥2.∴实数r的取值范围为[2,+∞).[选修4-5:不等式选讲]23.已知x>1,y>1,且x+y=4,求证:≥8.【分析】设x﹣1=m,y﹣1=n,则m>0,n>0,且m+n=2,再利用基本不等式即可得证.【解答】证明:设x﹣1=m,y﹣1=n,又x>1,y>1,则m>0,n>0,且m+n=x+y ﹣2=2,∴=,当且仅当m=n=1,即x=y=2时,等号成立,故原命题得证.【必做题】第24题、第25题,每题10分,共计20分,解答时应写出文字说明,证明过程或演算步骤.24.某“芝麻开门”娱乐活动中,共有5扇门,游戏者根据规则开门,并根据打开门的数量获取相应奖励.已知开每扇门相互独立,且规则相同,开每扇门的规则是:从给定的6把钥匙(其中有且只有1把钥匙能打开门)中,随机地逐把抽取钥匙进行试开,钥匙使用后不放回.若门被打开,则转为开下一扇门;若连续4次未能打开,则放弃这扇门,转为开下一扇门;直至5扇门都进行了试开,活动结束.(1)设随机变量X为试开第一扇门所用的钥匙数,求X的分布列及数学期望E(X);(2)求恰好成功打开4扇门的概率.【分析】(1)根据互斥事件概率公式计算X的可能取值对应的概率,得出分布列和数学期望;(2)根据二项分布的概率公式计算概率.解:(1)X的可能取值为1,2,3,4,P(X=1)=,P(X=2)==,P(X=3)==,P(X=4)==,∴X的分布列是:X1234PE(X)=1×+2×+3×+4×=3.(2)每扇门被打开的概率为=,设被打开的门的数量为ξ,则ξ~B(5,),∴恰好成功打开4扇门的概率为:P(ξ=4)=•()4•=.25.如图,在平面直角坐标系xOy中,已知抛物线y2=2px(p>0)的焦点为F,准线与x 轴的交点为E.过点F的直线与抛物线相交于A,B两点,EA,EB分别与y轴相交于M,N两点,当AB⊥x轴时,EA=2.(1)求抛物线的方程;(2)设△EAB的面积为S1,△EMN面积为S2,求的取值范围.【分析】(1)求得抛物线的焦点坐标,以及E的坐标,运用两点间的距离公式,解得p,进而得到抛物线的方程;(2)设AB:x=my+,A(x1,y1),B(x2,y2),联立抛物线的方程,运用韦达定理,以及直线方程,求得M,N的坐标,化简整理,运用三角形的面积公式,化简整理,结合韦达定理,即可得到所求范围..解:(1)抛物线y2=2px(p>0)的焦点为F(,0),准线与x轴的交点为E(﹣,0),当AB⊥x轴时,A的横坐标为,所以y A2=2px A=P2,所以|EA|===2,解得p=,所以抛物线的方程为y2=2x;(2)设AB:x=my+,A(x1,y1),B(x2,y2),联立抛物线的方程y2=2x,消去x,可得y2﹣2my﹣2=0,则y1+y2=2m,y1y2=﹣2,直线AE的斜率为k AE=,则AE的方程为y=(x+),令x=0,可得y=•,即M(0,•),同理可得N(0,•),===2(x1+)(x2+)=2[x1x2+(x1+x2)+]=2x1x2+(x1+x2)+1=+(+)+1=1+[(y1+y2)2﹣2y1y2]+1=(y1+y2)2+4=4m2+4≥4.(当m=0时,取得等号).即的取值范围为[4,+∞).。

三月的原野阅读答案

三月的原野阅读答案

三月的原野阅读答案篇一:宿迁市、徐州市、连云港市高三语文三模试卷和答案宿迁市20届高三第三次模拟考试语文试题一、语言文字运用(15分)1.在下面一段话空缺处依次填入词语,最恰当的一组是(3分)网络上大规模▲ 的“挖掘机哪家强?”造句热,使蓝翔技校▲ 。

事实上,现在“蓝翔”已经和“土豪”等词语一样,成了某种文化的象征。

“蓝翔”两个字早已不再是蓝翔技校的代名词,成了类似“高深莫测,咋咋呼呼”的含义。

A.爆发名声大噪演绎B.爆发名声大噪演化C.暴发声名鹊起演化D.暴发声名鹊起演绎2.下列各句中,没有语病的一项是(3分)..A.电影《失孤》无疑是三月份最受关注的华语片,但部分观众认为细节的缺乏、情节的破碎,是导致了《失孤》在飘渺中流于形式化概念的主要原因。

B.有人观看了《穹顶之下》后,撰文反击柴静,同时指出每个老百姓都是雾霾的制造者,但是治理、消除、管控雾霾,责任在政府。

C.保持文化的蓬勃生机,要求文艺工作者不仅要具有广阔的视野和博大的胸怀,而且和自己学术观点不一样的同行也要相互学习,切磋技艺,取长补短。

D.生态文明建设是建设美丽中国的必然要求,对于满足人民群众对良好生态环境新期待、形成人与自然和谐发展的现代化建设新格局,具有十分重要的意义。

3.下列交际用语使用得体的一项是(3分)A.通知:兹定于6月5日下午3时在报告厅召开高考考务会,请全体工作人员按时参加。

B.书信:毕业之后,学生垂念师恩。

值此春节到来之际,谨祝恩师节日快乐,万事如意!C.询问:家严大人今年高寿?多年不见,甚为牵挂,过两天我一定登门看望。

D.请柬:新居落成,我明天搬迁,为答谢您的祝贺,特于府上备下薄酒,恭请光临。

4.下列诗句,与“忽如一夜春风来,千树万树梨花开”所写景物季节相同的一项是(3分)A.晴日暖风生麦气,绿阴幽草胜花时。

B.六出飞花入户时,坐看青竹变琼枝。

C.菡萏香销翠叶残,西风愁起绿波间。

D.寂寞空庭春欲晚,梨花满地不开门。

2020-2021学年江苏省名校联盟高三下学期地理开学摸底考试卷含详解

2020-2021学年江苏省名校联盟高三下学期地理开学摸底考试卷含详解

江苏省名校联盟2020-2021学年高三下学期开学考试地理试题(新高考)一、单项选择题:共22题,每题2分,共44分。

每题只有一个选项最符合题意。

彝族土掌房为彝族先民的传统民居,距今已有500多年的历史,层层叠落,相互连通,远远看去甚是壮观,后期彝汉混居,融合了部分汉族民居的特点,逐步形成具有鲜明地方特色的民居建筑,堪称民居建筑文化与建造技术发展史上的“活化石”。

以石为墙基,用土坯砌墙或用土筑墙,墙上架梁,梁上铺木板、木条或竹子,上面再铺一层土,经洒水抿捶,形成平台房顶,不漏雨水。

房顶又是晒场。

有的大梁架在木柱上,担上垫木,铺茅草或稻草,草上覆盖稀泥,再放细土捶实而成。

多为平房,部分为二屋或三层。

据此完成下面小题。

1.土掌房的效果类似于A.内蒙蒙古包B.陕西窑洞C.傣族竹楼D.客家族土楼2.土掌房反应的气候特点A.湿冷B.湿热C.干旱少雨D.干热3.7月,小明去该地旅游,他发现当地A.正午太阳高度逐渐变大B.昼长夜短C.正午时影长变短D.昼渐变长某月,在江汉平原某城市(32°N,113°E)上学的小明早上6:10左右在经过上学途中的同一路段时,发现出现在正前方的太阳的高度逐渐降低到地平线附近。

下图示意小明上学的线路。

据此完成下列小题。

4.小明发现这一现象是在()A.2月B.4月C.6月D.8月5.小明6:10左右经过的路段最可能是()A.甲B.乙C.丙D.丁6.此时小明所在城市()A.多晴天B.多大风C.气温低D.多大雾河漫滩是指河谷底部在洪水期才被淹没的部分,由河流的横向迁移和洪水的沉积作用形成,下图为某河漫滩东西方向剖面示意图,该剖面位于自南向北河流的平直河段上。

据此完成下面小题。

7.该河流位于()A.东半球B.南半球C.西半球D.北半球8.据材料分析,河漫滩发育较好的河流是()A.山区水位季节变化小的河流B.山区水位季节变化大的河流C.平原水位季节变化小的河流D.平原水位季节变化大的河流9.若图中河床各处岩性相同,则河道中水流侵蚀作用最强的位置是()A.甲B.乙C.丙D.丁某城市公园在河道或沟渠边坡建造植草沟来处理雨水。

2020-2021学年江苏省三市联考高考数学三模试卷及答案解析

2020-2021学年江苏省三市联考高考数学三模试卷及答案解析

江苏省徐州市、连云港市、宿迁市高考数学三模试卷一、填空题:本大题共14小题,每小题5分,计70分.不需写出解答过程,请把答案写在答题卡的指定位置上.1.已知集合A={x|x=2k+1,k∈Z},B={x|0<x<5},则A∩B= .2.已知复数z满足(3+i)z=10i(其中i为虚数单位),则复数z的共轭复数是.3.如图是一次摄影大赛上7位评委给某参赛作品打出的分数的茎叶图.记分员在去掉一个最高分和一个最低分后,算得平均分为91分,复核员在复核时,发现有一个数字(茎叶图中的x)无法看清,若记分员计算无误,则数字x应该是.4.甲、乙、丙三人一起玩“黑白配”游戏:甲、乙、丙三人每次都随机出“手心(白)”、“手背(黑)”中的某一个手势,当其中一个人出示的手势与另外两人都不一样时,这个人胜出;其他情况,不分胜负.则一次游戏中甲胜出的概率是.5.执行如图所示的算法流程图,则输出k的值为.6.已知点F为抛物线y2=4x的焦点,该抛物线上位于第一象限的点A到其准线的距离为5,则直线AF的斜率为.7.已知公差为d的等差数列{a n}的前n项和为S n,若=3,则= .8.已知圆锥的母线长为10cm,侧面积为60πcm2,则此圆锥的体积为cm3.9.若实数x,y满足约束条件,则|3x﹣4y﹣10|的最大值为.10.已知函数f(x)=sinx(x∈[0,π])和函数g(x)=tanx的图象交于A,B,C三点,则△ABC 的面积为.11.若点P,Q分别是曲线y=与直线4x+y=0上的动点,则线段PQ长的最小值为.12.已知,,是同一平面内的三个向量,其中,是相互垂直的单位向量,且()•(﹣)=1,||的最大值为.13.已知对满足x+y+4=2xy的任意正实数x,y,都有x2+2xy+y2﹣ax﹣ay+1≥0,则实数a的取值范围为.14.已知经过点P(1,)的两个圆C1,C2都与直线l1:y=x,l2:y=2x相切,则这两圆的圆心距C1C2等于.二、解答题:本大题共6小题,计90分.解答应写出必要的文字说明,证明过程或演算步骤,请把答案写在答题卡的指定区域内.15.如图,在梯形ABCD中,已知AD∥BC,AD=1,BD=2,∠CAD=,tan∠ADC=﹣2,求:(1)CD的长;(2)△BCD的面积.16.如图,在直三棱柱ABC﹣A1B1C1中,已知AB=AC,M,N,P分别为BC,CC1,BB1的中点.求证:(1)平面AMP⊥平面BB1C1C;(2)A1N∥平面AMP.17.在平面直角坐标系xOy中,已知点P(1,)在椭圆C:=1(a>b>0)上,P到椭圆C的两个焦点的距离之和为4.(1)求椭圆C的方程;(2)若点M,N是椭圆C上的两点,且四边形POMN是平行四边形,求点M,N的坐标.18.经市场调查,某商品每吨的价格为x(1<x<14)百元时,该商品的月供给量为y1万吨,y1=ax+a2﹣a(a>0);月需求量为y2万吨,y2=﹣x2﹣x+1.当该商品的需求量大于供给量时,销售量等于供给量;当该商品的需求量不大于供给量时,销售量等于需求量.该商品的月销售额等于月销售量与价格的乘积.(1)若a=,问商品的价格为多少时,该商品的月销售额最大?(2)记需求量与供给量相等时的价格为均衡价格,若该商品的均衡价格不低于每吨6百元,求实数a的取值范围.19.已知函数f(x)=,g(x)=ax﹣2lnx﹣a (a∈R,e为自然对数的底数).(1)求f(x)的极值;(2)在区间(0,e]上,对于任意的x0,总存在两个不同的x1,x2,使得g(x1)=g(x2)=f(x0),求a的取值范围.20.在数列{a n}中,已知a1=1,a2=2,a n+2=(k∈N*).(1)求数列{a n}的通项公式;(2)求满足2a n+1=a n+a n+2的正整数n的值;(3)设数列{a n}的前n项和为S n,问是否存在正整数m,n,使得S2n=mS2n﹣1?若存在,求出所有的正整数对(m,n);若不存在,请说明理由.三.[选做题]本题包括A、B、C、D四小题,请选定其中两小题,并在相应的答题区域内作答.解答时应写出文字说明、证明过程或演算步骤.A.[选修4-1:几何证明选讲](本小题满分10分)21.如图,AB是圆O的直径,弦BD,CA的延长线相交于点E,过E作BA的延长线的垂线,垂足为F.求证:AB2=BE•BD﹣AE•AC.B.[选修4-2:矩阵与变换](本小题满分0分)22.已知矩阵A=,向量=,计算A5.C.[选修4-4:坐标系与参数方程](本小题满分0分)23.在极坐标系中,直线l的极坐标方程为,以极点为原点,极轴为x轴的正半轴建立平面直角坐标系,曲线C的参数方程为(α为参数),求直线l与曲线C 的交点P的直角坐标.D.[选修4-5:不等式选讲](本小题满分0分)24.已知a、b∈R,a>b>e(其中e是自然对数的底数),求证:b a>a b.(提示:可考虑用分析法找思路)四.[必做题]第22、23题,每小题0分,计20分.请把答案写在答题卡的指定区域内.25.已知甲箱中装有3个红球、3个黑球,乙箱中装有2个红球、2个黑球,这些球除颜色外完全相同.某商场举行有奖促销活动,设奖规则如下:每次分别从以上两个箱中各随机摸出2个球,共4个球.若摸出4个球都是红球,则获得一等奖;摸出的球中有3个红球,则获得二等奖;摸出的球中有2个红球,则获得三等奖;其他情况不获奖.每次摸球结束后将球放回原箱中.(1)求在1次摸奖中,获得二等奖的概率;(2)若连续摸奖2次,求获奖次数X的分布列及数学期望E(X).26.在集合A={1,2,3,4,…,2n}中,任取m(m≤n,m,n∈N*)个元素构成集合A m.若A m 的所有元素之和为偶数,则称A m为A的偶子集,其个数记为f(m);若A m的所有元素之和为奇数,则称A m为A的奇子集,其个数记为g(m).令F(m)=f(m)﹣g(m).(1)当n=2时,求F(1),F(2),F(3)的值;(2)求F(m).参考答案与试题解析一、填空题:本大题共14小题,每小题5分,计70分.不需写出解答过程,请把答案写在答题卡的指定位置上.1.已知集合A={x|x=2k+1,k∈Z},B={x|0<x<5},则A∩B= {1,3} .【考点】交集及其运算.【分析】由A与B,求出两集合的交集即可.【解答】解:∵A={x|x=2k+1,k∈Z},B={x|0<x<5},∴A∩B={1,3},故答案为:{1,3}.2.已知复数z满足(3+i)z=10i(其中i为虚数单位),则复数z的共轭复数是1﹣3i .【考点】复数代数形式的乘除运算.【分析】利用复数的运算法则、共轭复数的定义即可得出.【解答】解:∵(3+i)z=10i,∴(3﹣i)(3+i)z=10i(3﹣i),∴10z=10(3i+1),化为:z=1+3i,则复数z的共轭复数是1﹣3i.故答案为:1﹣3i.3.如图是一次摄影大赛上7位评委给某参赛作品打出的分数的茎叶图.记分员在去掉一个最高分和一个最低分后,算得平均分为91分,复核员在复核时,发现有一个数字(茎叶图中的x)无法看清,若记分员计算无误,则数字x应该是 1 .【考点】茎叶图.【分析】根据讨论x>4时,求出平均分不是91分,显然x≤4,表示出平均分,得到关于x的方程,解出即可.【解答】解:若x>4,去掉一个最高分(90+x)和一个最低分86后,平均分为(89+91+92+92+94)=91.6分,不合题意,故x≤4,最高分是94,去掉一个最高分94和一个最低分86后,故平均分是(89+92+90+x+91+92)=91,解得x=1,故答案为:1.4.甲、乙、丙三人一起玩“黑白配”游戏:甲、乙、丙三人每次都随机出“手心(白)”、“手背(黑)”中的某一个手势,当其中一个人出示的手势与另外两人都不一样时,这个人胜出;其他情况,不分胜负.则一次游戏中甲胜出的概率是.【考点】列举法计算基本事件数及事件发生的概率.【分析】根据题意,分析可得甲、乙、丙出的方法种数都有2种,由分步计数原理可得三人进行游戏的全部情况数目,进而可得甲胜出的情况数目,由等可能事件的概率,计算可得答案.【解答】解:一次游戏中,甲、乙、丙出的方法种数都有2种,所以总共有23=8种方案,而甲胜出的情况有:“甲黑乙白丙白”,“甲白乙黑丙黑”,共2种,所以甲胜出的概率为=,故答案为:.5.执行如图所示的算法流程图,则输出k的值为 3 .【考点】程序框图.【分析】根据框图的流程模拟运行程序,直到满足条件n=1,跳出循环,确定输出k的值.【解答】解:n=13是奇数,n==6>1,不符,此时k=1,n=6是偶数,n=3>1,不符,此时k=2,n=3是奇数,n=1=1,符合,此时k=3,故答案为:3.6.已知点F为抛物线y2=4x的焦点,该抛物线上位于第一象限的点A到其准线的距离为5,则直线AF的斜率为.【考点】抛物线的简单性质.【分析】求出抛物线的焦点坐标,设出A,利用抛物线上位于第一象限的点A到其准线的距离为5,求出A的横坐标,然后求解斜率.【解答】解:由题可知焦点F(1,0),准线为x=﹣1设点A(x A,y A),∵抛物线上位于第一象限的点A到其准线的距离为5,∴x A+=5,∴x A=4,∴y A=4,∴点A(4,4),∴直线AF的斜率为=,故答案为:.7.已知公差为d的等差数列{a n}的前n项和为S n,若=3,则= .【考点】等差数列的前n项和.【分析】设出等差数列的首项,由=3得到首项和公差的关系,代入等差数列的通项公式可得.【解答】解:设等差数列{a n}的首项为a1,则,由=3,得,即d=4a1,∴=.故答案为:.8.已知圆锥的母线长为10cm,侧面积为60πcm2,则此圆锥的体积为96πcm3.【考点】旋转体(圆柱、圆锥、圆台).【分析】根据侧面积计算圆锥的底面半径,根据勾股定理得出圆锥的高,代入圆锥的体积公式计算体积.【解答】解:设圆锥的底面半径为r,则S侧=π×r×10=60π,解得r=6.∴圆缀的高h==8,∴圆锥的体积V===96π.故答案为:96π.9.若实数x,y满足约束条件,则|3x﹣4y﹣10|的最大值为.【考点】简单线性规划.【分析】由题意作平面区域,而根据点到直线的距离公式可知转化为求阴影内的点到直线l的距离最大,从而解得.【解答】解:由题意作平面区域如下,,直线l的方程为3x﹣4y﹣10=0,点A到直线l的距离最大,由解得,A(,),故点A到直线l的距离d==,故|3x﹣4y﹣10|的最大值为×5=;故答案为:.10.已知函数f(x)=sinx(x∈[0,π])和函数g(x)=tanx的图象交于A,B,C三点,则△ABC 的面积为π.【考点】正切函数的图象;正弦函数的图象.【分析】根据题意,令sinx=tanx,结合x∈[0,π]求出x的值,得出三个点A、B、C的坐标,即可计算△ABC的面积.【解答】解:根据题意,令sinx=tanx,即sinx(1﹣)=0,解得sinx=0或1﹣=0,即sinx=0或cosx=;又x∈[0,π],所以x=0或x=π或x=;所以点A(0,0),B(π,0),C(,);所以△ABC的面积为S=|AB|h=×π×=π.故答案为:π.11.若点P,Q分别是曲线y=与直线4x+y=0上的动点,则线段PQ长的最小值为.【考点】两点间距离公式的应用.【分析】求出原函数的导函数,得到与直线4x+y=0平行的曲线的切线方程,由平行线间的距离公式求得线段PQ长的最小值.【解答】解:由y==1+,得y′=,由,得x2=1,∴x=±1.当x=1时,y=5,则与4x+y=0且与曲线y=相切的直线方程为y﹣5=﹣4(x﹣1),即4x+y﹣9=0.此时两平行线间的距离为;当x=﹣1时,y=﹣3,则与4x+y=0且与曲线y=相切的直线方程为y+3=﹣4(x+1),即4x+y+7=0.此时两平行线间的距离为.∴曲线y=与直线4x+y=0上两动点PQ距离的最小值为.故答案为:.12.已知,,是同一平面内的三个向量,其中,是相互垂直的单位向量,且()•(﹣)=1,||的最大值为1+.【考点】平面向量数量积的运算.【分析】不妨设=(1,0),=(0,1),设=(x,y),根据向量的坐标运算和数量积运算得到(x﹣)2+(y﹣)2=2,结合图形即可求出最大值.【解答】解:∵,是相互垂直的单位向量,不妨设=(1,0),=(0,1),设=(x,y),∴=(1﹣x,﹣y),﹣=(﹣x,﹣y),∵()•(﹣)=1,∴﹣(1﹣x)x﹣y(﹣y)=1,∴x2﹣x+y2﹣y=1,∴(x﹣)2+(y﹣)2=2,∴向量的轨迹为以(,)为圆心,以为半径的圆,∴圆心到原点的距离为1,∴||的最大值为1+故答案为:1+13.已知对满足x+y+4=2xy的任意正实数x,y,都有x2+2xy+y2﹣ax﹣ay+1≥0,则实数a的取值范围为(﹣∞,] .【考点】基本不等式.【分析】依题意,由正实数x,y满足x+y+4=2xy,可求得x+y≥4,由x2+2xy+y2﹣ax﹣ay+1≥0恒成立可求得a≤x+y+恒成立,利用双钩函数的性质即可求得实数a的取值范围.【解答】解:因为正实数x,y满足x+y+4=2xy,而4xy≤(x+y)2,代入原式得(x+y)2﹣2(x+y)﹣8≥0,解得(x+y)≥4或(x+y)≤﹣2(舍去)由x2+2xy+y2﹣ax﹣ay+1≥0可得a(x+y)≤(x+y)2+1,即a≤x+y+令t=x+y∈[4,+∞),则问题转化为a≤t+,因为函数y=t+在[4,+∞)递增,所以y min=4+=,所以a≤故答案为:(﹣∞,].14.已知经过点P(1,)的两个圆C1,C2都与直线l1:y=x,l2:y=2x相切,则这两圆的圆心距C1C2等于.【考点】直线与圆的位置关系.【分析】设圆心坐标为(x,y),由于圆与直线l1:y=x,l2:y=2x都相切,根据点到直线的距离公式得圆心只能在直线y=x上,设C1(a,a),C2(b,b),推导出a,b是方程(1﹣x)2+()2=的两根,由此能求出.这两圆的圆心距CC2.1【解答】解:设圆心坐标为(x,y),由于圆与直线l1:y=x,l2:y=2x都相切,根据点到直线的距离公式得:,解得y=x,∴圆心只能在直线y=x上,设C1(a,a),C2(b,b),则圆C1的方程为(x﹣a)2+(y﹣a)2=,圆C2的方程为(x﹣b)2+(y﹣b)2=,将(1,)代入,得:,∴a,b是方程(1﹣x)2+()2=,即=0的两根,∴,ab=,∴|C1C2|==•=•=.故答案为:.二、解答题:本大题共6小题,计90分.解答应写出必要的文字说明,证明过程或演算步骤,请把答案写在答题卡的指定区域内.15.如图,在梯形ABCD中,已知AD∥BC,AD=1,BD=2,∠CAD=,tan∠ADC=﹣2,求:(1)CD的长;(2)△BCD的面积.【考点】解三角形的实际应用.【分析】(1)根据tan∠ADC=﹣2计算sin∠ADC,得出sin∠ACD,在△ACD中使用正弦定理求出CD;(2)根据∠ADC+∠BCD=180°求出sin∠BCD,cos∠BCD,在△BCD中使用余弦定理解出BC,则=.S△BCD【解答】解:(1)∵tan∠ADC=﹣2,∴sin∠ADC=,cos∠ADC=﹣.∴sin∠ACD=sin(∠CAD+∠ADC)=sin∠CADcos∠ADC+cos∠CADsin∠ADC==.在△ACD中,由正弦定理得,即,解得CD=.(2)∵AD∥BC,∴∠ADC+∠BCD=180°,∴sin∠BCD=sin∠ADC=,cos∠BCD=﹣cos∠ADC=.在△BCD中,由余弦定理得BD2=CD2+BC2﹣2BC•CDcos∠BCD,即40=5+BC2﹣2BC,解得BC=7或BC=﹣5(舍).=BC•CDsin∠BCD==7.∴S△BCD16.如图,在直三棱柱ABC﹣A1B1C1中,已知AB=AC,M,N,P分别为BC,CC1,BB1的中点.求证:(1)平面AMP⊥平面BB1C1C;(2)A1N∥平面AMP.【考点】直线与平面平行的判定;平面与平面垂直的判定.【分析】(1)由已知条件推导出AM⊥BC,AM⊥BB1,从而AM⊥平面BB1C1C,由此能证明平面AMP⊥平面BB1C1C.(2)取B1C1中点E,连结A1E、NE、B1C,推导出平面A1NE∥平面APM,由此能证明A1N∥平面AMP.【解答】证明:(1)∵在直三棱柱ABC﹣A1B1C1中,AB=AC,M是BB1的中点,∴AM⊥BC,AM⊥BB1,∵BC∩BB1=B,∴AM⊥平面BB1C1C,∵AM⊂平面AMP,∴平面AMP⊥平面BB1C1C.(2)取B1C1中点E,连结A1E、NE、B1C,∵M,N,P分别为BC,CC1,BB1的中点,∴NE∥BC1∥PM,A1E∥AM,∵PM∩AM=M,A1E∩NE=E,PM、AM⊂平面APM,A1E、NE⊂平面A1EN,∴平面A1NE∥平面APM,∵A1N⊂平面A1NE,∴A1N∥平面AMP.17.在平面直角坐标系xOy中,已知点P(1,)在椭圆C:=1(a>b>0)上,P到椭圆C的两个焦点的距离之和为4.(1)求椭圆C的方程;(2)若点M,N是椭圆C上的两点,且四边形POMN是平行四边形,求点M,N的坐标.【考点】椭圆的简单性质.【分析】(1)由点P(1,)在椭圆上,P到椭圆C的两个焦点的距离之和为4,列出方程组求出a,b,由此能求出椭圆C的方程.(2)由题意设直线AB:y=,A(x1,y1),B(x2,y2),联立,消去y,得:3x2+3mx+m2﹣3=0,由此利用韦达定理、弦长公式、平行四边形性质,结合已知条件能求出M、N的坐标.【解答】解:(1)∵点P(1,)在椭圆C:=1(a>b>0)上,P到椭圆C的两个焦点的距离之和为4,∴,解得a=2,b=,∴椭圆C的方程为.(2)由题意设直线MN:y=,M(x1,y1),N(x2,y2),联立,消去y,得:3x2+3mx+m2﹣3=0,△>0,,∵四边形POMN是平行四边形,∴|MN|==,解得m=±3,当m=3时,解方程:3x2+9x+6=0,得M(﹣1,),N(﹣2,0);当m=﹣3时,解方程:3x2﹣9x+6=0,得M(1,),N(2,6).18.经市场调查,某商品每吨的价格为x(1<x<14)百元时,该商品的月供给量为y1万吨,y1=ax+ a2﹣a(a>0);月需求量为y2万吨,y2=﹣x2﹣x+1.当该商品的需求量大于供给量时,销售量等于供给量;当该商品的需求量不大于供给量时,销售量等于需求量.该商品的月销售额等于月销售量与价格的乘积.(1)若a=,问商品的价格为多少时,该商品的月销售额最大?(2)记需求量与供给量相等时的价格为均衡价格,若该商品的均衡价格不低于每吨6百元,求实数a的取值范围.【考点】函数模型的选择与应用.【分析】(1)利用商品的月销售额等于月销售量与价格的乘积,分类讨论,即可求解商品的价格为多少时,该商品的月销售额最大?(2)设f(x)=y1﹣y2=ax+a2﹣a﹣(﹣x2﹣x+1)=x2+(+a)x+a2﹣a﹣1,因为a>0,所以f(x)在区间(1,14)上是增函数,若该商品的均衡价格不低于6百元,即函数f(x)在区间[6,14)上有零点,即可得出结论.【解答】解:(1)若a=,y1=x﹣,y2>y1,即﹣x2﹣x+1>x﹣,∵1<x<14,∴1<x<6,月销售量为y1=x﹣,商品的月销售额等于(x﹣)x,在(1,6)上单调递增,(x﹣)x<;y2≤y1,即﹣x2﹣x+1≤x﹣,∵1<x<14,∴6≤x<14,月销售量为y2=﹣x2﹣x+1,商品的月销售额等于y=(﹣x2﹣x+1)x,y′=﹣(x﹣8)(3x+28),∴函数在(6,8)上单调递增,(8,14)上单调递减,x=8时,取得最大值>,∴商品的价格为8元时,该商品的月销售额最大;(2)设f(x)=y1﹣y2=ax+a2﹣a﹣(﹣x2﹣x+1)=x2+(+a)x+a2﹣a﹣1因为a>0,所以f(x)在区间(1,14)上是增函数,若该商品的均衡价格不低于6百元,即函数f(x)在区间[6,14)上有零点,所以f(6)≤0,f(14)>0,所以0<a≤.19.已知函数f(x)=,g(x)=ax﹣2lnx﹣a (a∈R,e为自然对数的底数).(1)求f(x)的极值;(2)在区间(0,e]上,对于任意的x0,总存在两个不同的x1,x2,使得g(x1)=g(x2)=f(x0),求a的取值范围.【考点】利用导数研究函数的极值;利用导数研究函数的单调性.【分析】(1)求出f(x)的导数,得到函数的单调区间,从而求出函数的极值即可;(2)求出当x∈(0,e]时,函数f(x)的值域,通过讨论a的范围结合g(x)的单调性,求出a的具体范围即可.【解答】解:(1)因为f(x)=,所以f′(x)=,…令f′(x)=0,得x=1.…当x∈(﹣∞,1)时,f′(x)>0,f(x)是增函数;当x∈(1,+∞)时,f′(x)<0,f(x)是减函数.所以f(x)在x=1时取得极大值f(1)=1,无极小值.…(2)由(1)知,当x∈(0,1)时,f(x)单调递增;当x∈(1,e]时,f(x)单调递减.又因为f(0)=0,f(1)=1,f(e)=e•e1﹣e>0,所以当x∈(0,e]时,函数f(x)的值域为(0,1].…当a=0时,g(x)=﹣2lnx在(0,e]上单调,不合题意;…当a≠0时,g′(x)=,x∈(0,e],故必须满足0<<e,所以a>.…此时,当x 变化时,g′(x),g(x)的变化情况如下:x (0,)(,e]g′(x)﹣0 +g(x)单调减最小值单调增所以x→0,g(x)→+∞,g()=2﹣a﹣2ln,g(e)=a(e﹣1)﹣2,所以对任意给定的x0∈(0,e],在区间(0,e]上总存在两个不同的x1,x2使得g(x1)=g(x2)=f(x0),当且仅当a满足下列条件,即,…令m(a)=2﹣a﹣2ln,a∈(,+∞),m′(a)=﹣,由m′(a)=0,得a=2.当a∈(2,+∞)时,m′(a)<0,函数m(a)单调递减;当a∈(,2)时,m′(a)>0,函数m(a)单调递增.所以,对任意a∈(,+∞)有m(a)≤m(2)=0,即2﹣a﹣2ln≤0对任意a∈(,+∞)恒成立.由a(e﹣1)﹣2≥1,解得a≥,综上所述,当a∈[,+∞)时,对于任意给定的x0(0,e],在区间(0,e]上总存在两个不同的x1,x2,使得g(x1)=g(x2)=f(x0).…20.在数列{a n}中,已知a1=1,a2=2,a n+2=(k∈N*).(1)求数列{a n}的通项公式;(2)求满足2a n+1=a n+a n+2的正整数n的值;(3)设数列{a n}的前n项和为S n,问是否存在正整数m,n,使得S2n=mS2n﹣1?若存在,求出所有的正整数对(m,n);若不存在,请说明理由.【考点】数列的求和;数列递推式.【分析】(1)由题意可得数列{a n}的奇数项是以1为首项,公差为2的等差数列;偶数项是以2为首项,公比为3的等比数列.分别利用等差数列与等比数列的通项公式即可得出.(2)①当n为奇数时,由2a n+1=a n+a n+2可得:=n+n+2,化为:=n+1,令f(x)=2×﹣x﹣1(x≥1),利用导数研究函数的单调性即可得出.②当n为偶数时,由2a n+1=a n+a n+2可得:2(n+1)=2+2×,化为:n+1=+,即可判断出不成立.(3)S2n=(a1+a3+…+a2n﹣1)+(a2+a4+…+a2n)=3n+n2﹣1,n∈N*.S2n﹣1=S2n﹣a2n=3n﹣1+n2﹣1.假设存在正整数m,n,使得S2n=mS2n﹣1,化为3n﹣1(3﹣m)=(m﹣1)(n2﹣1),可得1,2,3.分类讨论即可得出.【解答】解:(1)由a1=1,a2=2,a n+2=(k∈N*).可得数列{a n}的奇数项是以1为首项,公差为2的等差数列;偶数项是以2为首项,公比为3的等比数列.∴对任意正整数k,a2k﹣1=1+2(k﹣1)=2k﹣1;a2k=2×3k﹣1.∴数列{a n}的通项公式a n=,k∈N*.(2)①当n为奇数时,由2a n+1=a n+a n+2可得:=n+n+2,化为:=n+1,令f(x)=2×﹣x﹣1(x≥1),由f′(x)=××ln﹣1≥﹣1=ln3﹣1>0,可知f(x)在[1,+∞)上是增函数,∴f(x)≥f(1)=0,∴当且仅当n=1时,满足=n+1,即2a2=a1+a3.=a n+a n+2可得:2(n+1)=2+2×,②当n为偶数时,由2an+1化为:n+1=+,上式左边为奇数,右边为偶数,因此不成立.综上,满足2a n+1=a n+a n+2的正整数n的值只有1.(3)S2n=(a1+a3+…+a2n﹣1)+(a2+a4+…+a2n)=+=3n+n2﹣1,n∈N*.S2n﹣1=S2n﹣a2n=3n﹣1+n2﹣1.假设存在正整数m,n,使得S2n=mS2n﹣1,则3n+n2﹣1=m(3n﹣1+n2﹣1),∴3n﹣1(3﹣m)=(m﹣1)(n2﹣1),(*)从而3﹣m≥0,∴m≤3,又m∈N*,∴m=1,2,3.①当m=1时,(*)式左边大于0,右边等于0,不成立.②当m=3时,(*)式左边等于0,∴2(n2﹣1)=0,解得n=1,∴S2=3S1.③当m=2时,(*)式可化为3n﹣1=(n+1)(n﹣1),则存在k1,k2∈N*,k1<k2,使得n﹣1=,n+1=,且k1+k2=n﹣1,从而==2,∴﹣=2,=1,∴k1=0,k2﹣k1=1,于是n=2,S4=2S3.综上可知,符合条件的正整数对(m,n)只有两对:(2,2),(3,1).三.[选做题]本题包括A、B、C、D四小题,请选定其中两小题,并在相应的答题区域内作答.解答时应写出文字说明、证明过程或演算步骤.A.[选修4-1:几何证明选讲](本小题满分10分)21.如图,AB是圆O的直径,弦BD,CA的延长线相交于点E,过E作BA的延长线的垂线,垂足为F.求证:AB2=BE•BD﹣AE•AC.【考点】与圆有关的比例线段.【分析】连接AD,利用AB为圆的直径结合EF与AB的垂直关系,通过证明A,D,E,F四点共圆知,BD•BE=BA•BF,再利用△ABC∽△AEF得到比例式,最后利用线段间的关系即求得AB2=BE•BD ﹣AE•AC.【解答】证明:连接AD,因为AB为圆的直径,所以∠ADB=90°,又EF⊥AB,∠AFE=90°,则A,D,E,F四点共圆,∴BD•BE=BA•BF,又△ABC∽△AEF,∴,即AB•AF=AE•AC∴BE•BD﹣AE•AC=BA•BF﹣AB•AF=AB•(BF﹣AF)=AB2.B.[选修4-2:矩阵与变换](本小题满分0分)22.已知矩阵A=,向量=,计算A5.【考点】特征向量的意义.【分析】令f(λ)==λ2﹣5λ+6=0,解得λ=2或3.分别对应的一个特征向量为;.设=m++n.解得m,n,即可得出.【解答】解:∵f(λ)==λ2﹣5λ+6,由f(λ)=0,解得λ=2或3.当λ=2时,对应的一个特征向量为α1=;当λ=3时,对应的一个特征向量为α2=.设=m++n.解得.∴A5=2×25+1×35=.C.[选修4-4:坐标系与参数方程](本小题满分0分)23.在极坐标系中,直线l的极坐标方程为,以极点为原点,极轴为x轴的正半轴建立平面直角坐标系,曲线C的参数方程为(α为参数),求直线l与曲线C 的交点P的直角坐标.【考点】简单曲线的极坐标方程.【分析】先利用直角坐标与极坐标间的关系,即利用ρcosθ=x,ρsinθ=y,ρ2=x2+y2,进行代换将极坐标方程化成直角坐标方程.再利用消去参数的方法化参数方程为直角坐标方程,通过直角坐标方程求出交点即可.【解答】解:因为直线l的极坐标方程为所以直线l的普通方程为,又因为曲线C的参数方程为(α为参数)所以曲线C的直角坐标方程为,联立解方程组得或,根据x的范围应舍去,故P点的直角坐标为(0,0).D.[选修4-5:不等式选讲](本小题满分0分)24.已知a、b∈R,a>b>e(其中e是自然对数的底数),求证:b a>a b.(提示:可考虑用分析法找思路)【考点】分析法和综合法.【分析】直接利用分析法的证明步骤,结合函数的单调性证明即可.【解答】证明:∵b a>0,a b>0,∴要证:b a>a b只要证:alnb>blna只要证.(∵a>b>e)取函数,∵∴当x>e时,,∴函数在上是单调递减.∴当a>b>e时,有,即.得证.四.[必做题]第22、23题,每小题0分,计20分.请把答案写在答题卡的指定区域内.25.已知甲箱中装有3个红球、3个黑球,乙箱中装有2个红球、2个黑球,这些球除颜色外完全相同.某商场举行有奖促销活动,设奖规则如下:每次分别从以上两个箱中各随机摸出2个球,共4个球.若摸出4个球都是红球,则获得一等奖;摸出的球中有3个红球,则获得二等奖;摸出的球中有2个红球,则获得三等奖;其他情况不获奖.每次摸球结束后将球放回原箱中.(1)求在1次摸奖中,获得二等奖的概率;(2)若连续摸奖2次,求获奖次数X的分布列及数学期望E(X).【考点】离散型随机变量的期望与方差;离散型随机变量及其分布列.【分析】(1)设“在1次摸奖中,获得二等奖”为事件A,利用互斥事件概率计算公式能求出在1次摸奖中,获得二等奖的概率.(2)设“在1次摸奖中,获奖”为事件B,先求出P(B),由题意可知X的所有可能取值为0,1,2.分别求出相应的概率,由此能求出X的分布列和E(X).【解答】解:(1)设“在1次摸奖中,获得二等奖”为事件A,则P(A)==.…(2)设“在1次摸奖中,获奖”为事件B,则获得一等奖的概率为=,获得三等奖的概率为P3==,所以P(B)==.…由题意可知X的所有可能取值为0,1,2.P(X=0)=(1﹣)2=,P(X=1)==,P(X=2)=()2=.所以X的分布列是X 0 1 2P所以E(X)=0×+2×=.…26.在集合A={1,2,3,4,…,2n}中,任取m(m≤n,m,n∈N*)个元素构成集合A m.若A m 的所有元素之和为偶数,则称A m为A的偶子集,其个数记为f(m);若A m的所有元素之和为奇数,则称A m为A的奇子集,其个数记为g(m).令F(m)=f(m)﹣g(m).(1)当n=2时,求F(1),F(2),F(3)的值;(2)求F(m).【考点】子集与真子集;元素与集合关系的判断.【分析】(1)根据已知条件利用列举法能F(1),F(2),F(3);(2)分m为奇数和m为偶数两种情况,再根据二项式定理和排列组合的知识即可求出答案.【解答】解:(1)当n=2时,集合为{1,2,3,4},当m=1时,偶子集有{2},{4},奇子集有{1},{3},f(1)=2,g(1)=2,F(1)=0;当m=2时,偶子集有{2,4},{1,3},奇子集有{1,2},{1,4},{2,4},{3,4},f(2)=2,g(2)=4,F(2)=﹣2;当m=3时,偶子集有{1,2,3},{1,3,4},奇子集有{1,2,4},{2,3,4},f(3)=2,g(3)=2,F(3)=0;(2)当m为奇数时,偶子集的个数f(m)=C n0C n m+C n2C n m﹣2+C n4C n m﹣4+…+C n m﹣1C n1,奇子集的个数g(m)=C n1C n m﹣1+C n3C n m﹣3+…+C n m C n0,所以f(m)=g(m),F(m)=f(m)﹣g(m)=0.当m为偶数时,偶子集的个数f(m)=C n0C n m+C n2C n m﹣2+C n4C n m﹣4+…+C n m C n0,奇子集的个数g(m)=C n1C n m﹣1+C n3C n m﹣3+…+C n m﹣1C n1,所以F(m)=f(m)﹣g(m)=C n0C n m﹣C n1C n m﹣1+C n2C n m﹣2﹣C n3C n m﹣3+…﹣C n m﹣1C n1+C n m C n0,一方面,(1+x)m(1﹣x)m=(C m0+C m1x+C m2x2+…+C m m x m)[C m0﹣C m1x+C m2x2+…+(﹣1)m C m m x m]所以,(1+x)m(1﹣x)m中x m的系数为C m0C m m﹣C m1C m m﹣1+C m2C m m﹣2﹣C m3C m m﹣3+…﹣C m m﹣1C m1+C m m C m0,另一方面,(1+x)m(1﹣x)m=(1﹣x2)m,(1﹣x2)m中x m的系数为(﹣1),故f(m)=(﹣1),综上,F(m)=。

2020-2021学年最新高考总复习数学(文)第三次高考模拟训练试题及答案解析一

2020-2021学年最新高考总复习数学(文)第三次高考模拟训练试题及答案解析一

最新高考数学三模试卷(文科)一、选择题1.设集合A={x|x (x ﹣3)<0},B={x|x ﹣2≤0},则A ∩B=( )A .(0,2]B .(0,2)C .(0,3)D .[2,3)2.设z 满足i (1+z )=2+i ,则|z|=( )A .B .C .2D .13.设命题p :∀x >0,xe x >0,则¬p 为( )A .∀x ≤0,xe x ≤0B .∃x 0≤0,x 0e x0≤0C .∀x >0,xe x ≤0D .∃x 0>0,x 0e x0≤04.从3名男生和2名女生中任意推选2名选手参加辩论赛,则推选出的2名选手恰好是1男1女的概率是( )A .B .C .D .5.如图所示的程序框图的算法思路源于我国古代数字著作《数书九章》,称为“秦九韶算法”.执行该程序框图,若输入x=2,n=5,则输出的v=( )A .26B .48C .57D .646.一个圆柱挖去一部分后,剩余部分的三视图如图所示,则剩余部分的表面积等于( )A .39πB .48πC .57πD .63π7.已知x ,y 满足约束条件,则的最大值是( )A .﹣2B .﹣1C .D .28.已知函数f (x )=Asin (ωx+φ)(A >0,ω>0)的图象与直线y=b (0<b <A )相交,其中一个交点P 的横坐标为4,若与P 相邻的两个交点的横坐标为2,8,则函数f (x )( )A .在[0,3]上是减函数B .在[﹣3,0]上是减函数C .在[0,π]上是减函数D .在[﹣π,0]上是减函数9.设函数f (x )=e x +ax 在(0,+∞)上单调递增,则实数a 的取值范围为( )A .[﹣1,+∞)B .(﹣1,+∞)C .[0,+∞)D .(0,+∞)10.正三棱柱的底面边长为,侧棱长为2,且三棱柱的顶点都在同一球面上,则该球的表面积为( )A .4πB .8πC .12πD .16π11.已知定义在R 上的函数f (x )是奇函数,且f (x )在(﹣∞,0)上是减函数,f (2)=0,g (x )=f (x+2),则不等式xg (x )≤0的解集是( )A .(﹣∞,﹣2]∪[2,+∞)B .[﹣4,﹣2]∪[0,+∞)C .(﹣∞,﹣4]∪[﹣2,+∞)D .(﹣∞,﹣4]∪[0,+∞)12.已知抛物线C :y 2=2px (p >0)的焦点为F ,点A ,B 在C 上,且点F 是△AOB 的重心,则cos ∠AFB 为( )A .﹣B .﹣C .﹣D .﹣二、填空题13.若和是两个互相垂直的单位向量,则|+2|=_______.14.已知α为锐角,cos α=,则sin (﹣α)=_______.15.在△ABC 中,∠A ,∠B ,∠C 所对的边长分别是x+1,x ,x ﹣1,且∠A=2∠C ,则△ABC 的周长为_______.16.已知圆C :(x ﹣a )2+y 2=1(a >0),过直线l :2x+2y+3=0上任意一点P 作圆C 的两条切线PA ,PB ,切点分别为A ,B ,若∠APB 为锐角,则a 的取值范围为_______.三、解答题17.设S n 是数列{a n }的前n 项和,且S n =2a n ﹣1.(1)证明:数列{a n }是等比数列;(2)求数列{na n }的前n 项和T n .18.在四棱锥P ﹣ABCD 中,底面ABCD 是菱形,AB=2,∠BAD=60°,PC ⊥BD .(1)证明:PB=PD ;(2)若平面PBD ⊥平面ABCD ,且∠DPB=90°,求点B 到平面PDC 的距离.19.PM2.5是指空气中直径小于或等于2.5微米的细颗粒物,它对人体健康和大气环境质量的影响很大.2012年2月,中国发布了《环境空气质量标准》,开始大力治理空气污染.用x=1,2,3,4,5依次表示2013年到2017年这五年的年份代号,用y 表示每年3月份的PM2.5指数的平均值(单位:μg/m 3).已知某市2013年到2016年每年3月份PM2.5指数的平均值的折线图如图:(1)根据折线图中的数据,完成表格:年份2013 2014 2015 2016年份代号(x) 1 2 3 4PM2.5指数(y)(2)建立y关于x的线性回归方程;(3)在当前治理空气污染的力度下,预测该市2017年3月份的PM2.5指数的平均值.附:回归直线方程=x+中参数的最小二乘估计公式;=, =﹣.20.已知椭圆C: +=1(a>b>0)的离心率为,以该椭圆上的点和椭圆的两个焦点为顶点的三角形的周长为6.(1)求椭圆C的方程;(2)设过点C的左焦点F的直线l交C于A,B两点,是否存在常数λ,使||=λ•恒成立,若存在,求出λ的值;若不存在,请说明理由.21.已知函数f(x)=+b在x=1处的切线方程为x+y﹣3=0.(1)求a,b.(2)证明:当x>0,且x≠1时,f(x)>.[选修4-1:几何证明选讲]22.如图,E为⊙O上一点,点A在直径BD的延长线上,过点B作⊙O的切线交AE的延长线于点C,CE=CB.(1)证明:AE2=AD•AB.(2)若AE=4,CB=6,求⊙O的半径.[选修4-4:坐标系与参数方程选讲]23.已知曲线C的极坐标方程是ρsin2θ﹣8cosθ=0,以极点为平面直角坐标系的原点,极轴为x轴的正半轴,建立平面直角坐标系xOy.在直角坐标系中,倾斜角为α的直线l过点P(2,0).(1)写出曲线C的直角坐标方程和直线l的参数方程;(2)设点Q和点G的极坐标分别为(2,),(2,π),若直线l经过点Q,且与曲线C 相交于A,B两点,求△GAB的面积.[选修4-5:不等式选讲]24.已知函数f(x)=.(1)求函数f(x)的值域;(2)若函数f(x)的值域是[m,n],且a2+b2=m,c2+d2=n,求ac+bd的取值范围.参考答案与试题解析一、选择题1.设集合A={x|x (x ﹣3)<0},B={x|x ﹣2≤0},则A ∩B=( )A .(0,2]B .(0,2)C .(0,3)D .[2,3)【考点】交集及其运算.【分析】求出A 与B 中不等式的解集分别确定出A 与B ,找出两集合的交集即可.【解答】解:由A 中不等式解得:0<x <3,即A=(0,3),由B 中不等式解得:x ≤2,即B=(﹣∞,2],则A ∩B=(0,2],故选:A .2.设z 满足i (1+z )=2+i ,则|z|=( )A .B .C .2D .1【考点】复数求模.【分析】根据复数的四则运算求出z ,然后利用复数的模长公式进行求解即可.【解答】解:由i (1+z )=2+i ,得1+z==1﹣2i ,则z=﹣2i ,则|z|=2,故选:C3.设命题p :∀x >0,xe x >0,则¬p 为( )A .∀x ≤0,xe x ≤0B .∃x 0≤0,x 0e x0≤0C .∀x >0,xe x ≤0D .∃x 0>0,x 0e x0≤0【考点】命题的否定.【分析】根据全称命题的否定是特称命题进行判断.【解答】解:命题是全称命题,则命题的否定是特称命题,则¬p :∃x 0>0,x 0e x0≤0,故选:D4.从3名男生和2名女生中任意推选2名选手参加辩论赛,则推选出的2名选手恰好是1男1女的概率是( )A .B .C .D .【考点】古典概型及其概率计算公式.【分析】本题是一个等可能事件的概率,试验发生所包含的事件数是C 52种结果,满足条件的事件是抽到的2名学生恰好是1男1女,有C 31C 21,进而得到概率.【解答】解:从3名男生和2名女生中任意推选2名选手参加辩论赛,共有C 52=10种选法, 选出的2名选手恰好是1男1女有C 31C 21=6种,故推选出的2名选手恰好是1男1女的概率是=,故选:C .5.如图所示的程序框图的算法思路源于我国古代数字著作《数书九章》,称为“秦九韶算法”.执行该程序框图,若输入x=2,n=5,则输出的v=( )A.26 B.48 C.57 D.64【考点】程序框图.【分析】根据已知的程序框图可得,该程序的功能是利用循环结构计算并输出变量v的值,模拟程序的运行过程,可得答案.【解答】解:模拟程序的运行,可得x=2,n=5,v=1,k=2执行循环体,v=4,k=3满足条件k<5,执行循环体,v=11,k=4满足条件k<5,执行循环体,v=26,k=5不满足条件k<5,退出循环,输出v的值为26.故选:A.6.一个圆柱挖去一部分后,剩余部分的三视图如图所示,则剩余部分的表面积等于()A.39π B.48π C.57π D.63π【考点】由三视图求面积、体积.【分析】根据三视图可知该几何体是:一个圆柱在上底面挖去了一个同底等高的圆锥,由三视图求出几何元素的长度,由圆柱、圆锥的侧面积公式求出剩余部分的表面积.【解答】解:根据三视图可知该几何体是:一个圆柱在上底面挖去了一个同底等高的圆锥,且圆柱底面圆的半径为3,母线长是4,则圆锥的母线长是=5,∴剩余部分的表面积S=π×32+2π×3×4+π×3×5=48π,故选:B.7.已知x,y满足约束条件,则的最大值是()A.﹣2 B.﹣1 C.D.2【考点】简单线性规划.【分析】作出不等式组对应的平面区域,利用直线的斜率公式,结合数形结合进行求解即可.【解答】解:作出不等式组对应的平面区域如图,则的几何意义是区域内的点到原点的斜率,由图象知OA的斜率最大,由得,即A(2,4),此时的最大值是,故选:D8.已知函数f(x)=Asin(ωx+φ)(A>0,ω>0)的图象与直线y=b(0<b<A)相交,其中一个交点P的横坐标为4,若与P相邻的两个交点的横坐标为2,8,则函数f(x)()A.在[0,3]上是减函数B.在[﹣3,0]上是减函数C.在[0,π]上是减函数D.在[﹣π,0]上是减函数【考点】正弦函数的图象.【分析】先根据正弦函数的图象的对称性可得函数f(x)的图象的相邻的两条对称轴分别为x=3和x=6,且函数f(x)在[3,6]上单调递减,故f(x)在[0,3]上是增函数,在[﹣3,0]上是减函数,从而得出结论.【解答】解:∵函数f(x)=Asin(ωx+φ)(A>0,ω>0)的图象与直线y=b(0<b<A)相交,其中一个交点P的横坐标为4,若与P相邻的两个交点的横坐标为2,8,则函数f(x)的图象的相邻的两条对称轴分别为x=3和x=6,且函数f(x)在[3,6]上单调递减,故f(x)在[0,3]上是增函数,在[﹣3,0]上是减函数,故选:B.9.设函数f(x)=e x+ax在(0,+∞)上单调递增,则实数a的取值范围为()A.[﹣1,+∞)B.(﹣1,+∞)C.[0,+∞)D.(0,+∞)【考点】利用导数研究函数的单调性.【分析】函数f(x)=e x+ax在区间(0,+∞)上单调递增⇔函数f′(x)=e x+a≥0在区间在区间(0,+∞)上成立.(0,+∞)上恒成立⇔a≥[﹣e x]min【解答】解:f′(x)=e x+a,∵函数f(x)=e x+ax在区间(0,+∞)上单调递增,∴函数f′(x)=e x+a≥0在区间(0,+∞)上恒成立,∴a≥[﹣e x]在区间(0,+∞)上成立,min∵在区间(0,+∞)上﹣e x<﹣1,∴a≥﹣1,故选:A.10.正三棱柱的底面边长为,侧棱长为2,且三棱柱的顶点都在同一球面上,则该球的表面积为()A.4πB.8πC.12π D.16π【考点】球的体积和表面积.【分析】根据正三棱柱的对称性,它的外接球的球心在上下底面中心连线段的中点.再由正三角形的性质和勾股定理,结合题中数据算出外接球半径,用球表面积公式即可算出该球的表面积.【解答】解:设三棱柱ABC﹣A′B′C′的上、下底面的中心分别为O、O′,,根据图形的对称性,可得外接球的球心在线段OO′中点O1∵OA=AB=1,OO=AA′=11A=∴O1因此,正三棱柱的外接球半径R=,可得该球的表面积为S=4πR2=8π故选:B.11.已知定义在R上的函数f(x)是奇函数,且f(x)在(﹣∞,0)上是减函数,f(2)=0,g(x)=f(x+2),则不等式xg(x)≤0的解集是()A.(﹣∞,﹣2]∪[2,+∞)B.[﹣4,﹣2]∪[0,+∞)C.(﹣∞,﹣4]∪[﹣2,+∞)D.(﹣∞,﹣4]∪[0,+∞)【考点】奇偶性与单调性的综合.【分析】由题意可得g(x)关于点(﹣2,0)对称,g(0)=f(2)=0,g(﹣4)=f(﹣2)=0,画出g(x)的单调性示意图,数形结合求得不等式xg(x)≤0的解集.【解答】解:由题意可得g(x)的图象是把f(x)的图象向左平移2个单位得到的,故g(x)关于点(﹣2,0)对称,g(0)=f(2)=0,g(﹣4)=f(﹣2)=0,它的单调性示意图,如图所示:根据不等式xg(x)≤0可得,x的符号和g(x)的符号相反,∴xg(x)≤0的解集为(﹣∞,﹣4]∪[﹣2,+∞),故选:C.12.已知抛物线C:y2=2px(p>0)的焦点为F,点A,B在C上,且点F是△AOB的重心,则cos∠AFB为()A.﹣ B.﹣ C.﹣D.﹣【考点】抛物线的简单性质.【分析】设A(m,)、B(m,﹣),则=,p=,可得A的坐标,求出AF,利用二倍角公式可求.【解答】解:由抛物线的对称性知,A、B关于x轴对称.设A(m,)、B(m,﹣),则=,∴p=.∴A(m, m),∴AF=m,∴cos∠AFB==,∴cos∠AFB=2cos2∠AFB﹣1=﹣.故选:D.二、填空题13.若和是两个互相垂直的单位向量,则|+2|= .【考点】平面向量数量积的运算.【分析】计算()2,然后开方即可.【解答】解:∵和是两个互相垂直的单位向量,∴,.∴()2==5,∴||=.故答案为:.14.已知α为锐角,cosα=,则sin(﹣α)= .【考点】两角和与差的正弦函数.【分析】由已知利用同角三角函数基本关系式可求sinα,利用特殊角的三角函数值及两角差的正弦函数公式化简所求即可计算得解.【解答】解:∵α为锐角,cosα=,∴sin==,∴sin(﹣α)=sin cosα﹣cos sinα=﹣×=.故答案为:.15.在△ABC中,∠A,∠B,∠C所对的边长分别是x+1,x,x﹣1,且∠A=2∠C,则△ABC 的周长为15 .【考点】余弦定理.【分析】由已知及正弦定理,二倍角的正弦函数公式可得:cosC=,又由余弦定理可得:cosC=,从而可得=,解得x,即可得解三角形的周长.【解答】解:∵∠A,∠B,∠C所对的边长分别是x+1,x,x﹣1,且∠A=2∠C,∴由正弦定理可得:,∴,可得:cosC=,又∵由余弦定理可得:cosC=,∴=,整理即可解得x=5,∴△ABC的周长为:(x+1)+x+(x﹣1)=3x=15.故答案为:15.16.已知圆C:(x﹣a)2+y2=1(a>0),过直线l:2x+2y+3=0上任意一点P作圆C的两条切线PA,PB,切点分别为A,B,若∠APB为锐角,则a的取值范围为(,+∞).【考点】圆的切线方程.【分析】作出直线l和圆C,PA,PB为圆的两条切线,连接AC,BC,PC,由∠APB为锐角,可得0<∠APC<,运用解直角三角形可得可得1<PA恒成立,由勾股定理可得PA2=PC2﹣1,求得PC的最小值,可得PA的最小值,解不等式即可得到所求a的范围.【解答】解:作出直线l和圆C,PA,PB为圆的两条切线,连接AC,BC,PC,由圆心C(a,0)到直线l的距离为d=>>1,可得直线和圆相离.由∠APB为锐角,可得0<∠APC<,即0<tan∠APC<1,在Rt△APC中,tan∠APC==,可得1<PA恒成立,由勾股定理可得PA2=PC2﹣1,当PC⊥l时,PC取得最小值,且为,即有1<,解得a>.故答案为:(,+∞).三、解答题17.设S n 是数列{a n }的前n 项和,且S n =2a n ﹣1.(1)证明:数列{a n }是等比数列;(2)求数列{na n }的前n 项和T n .【考点】数列的求和;等比数列的通项公式.【分析】(1)由S n =2a n ﹣1.可得当n=1时,a 1=2a 1﹣1,解得a 1.当n ≥2时,a n =S n ﹣S n ﹣1,化为:a n =2a n ﹣1.利用等比数列的通项公式即可得出.(2)由(1)可得:a n =2n ﹣1.na n =n •2n ﹣1.利用“错位相减法”与等比数列的前n 项和公式即可得出.【解答】(1)证明:∵S n =2a n ﹣1.∴当n=1时,a 1=2a 1﹣1,解得a 1=1.当n ≥2时,a n =S n ﹣S n ﹣1=2a n ﹣1﹣(2a n ﹣1﹣1),化为:a n =2a n ﹣1.∴数列{a n }是等比数列,首项为1,公比为2.(2)解:由(1)可得:a n =2n ﹣1.na n =n •2n ﹣1.∴数列{na n }的前n 项和T n =1+2×2+3×22+…+n •2n ﹣1,2T n =2+2×22+…+(n ﹣1)•2n ﹣1+n •2n ,∴﹣T n =1+2+22+…+2n ﹣1﹣n •2n =﹣n •2n =(1﹣n )•2n ﹣1,∴T n =(n ﹣1)•2n +1.18.在四棱锥P ﹣ABCD 中,底面ABCD 是菱形,AB=2,∠BAD=60°,PC ⊥BD .(1)证明:PB=PD ;(2)若平面PBD ⊥平面ABCD ,且∠DPB=90°,求点B 到平面PDC 的距离.【考点】点、线、面间的距离计算.【分析】(1)如图所示,连接AC 交BD 于点O ,连接OP .利用菱形的性质可得AC ⊥BD ,利用线面垂直的判定与性质定理可证明BD ⊥PO .又O 是BD 的中点,可得PB=PD .(2)底面ABCD 是菱形,AB=2,∠BAD=60°,可得△PBD 与△BCD 都是等边三角形.由平面PBD ⊥平面ABCD ,平面PBD ∩平面ABCD=BD ,PO ⊥BD .可得PO ⊥平面ABCD ,因此PO ⊥AC ,又AC⊥BD,可建立如图所示的空间直角坐标系.设平面PCD的法向量=(x,y,z),则,利用点B到平面PDC的距离d=即可得出.【解答】(1)证明:如图所示,连接AC交BD于点O,连接OP.∵四边形ABCD是菱形,∴AC⊥BD,又PC⊥BD,且PC∩AC=C,∴BD⊥平面PAC.则BD⊥PO.又O是BD的中点,∴PB=PD.(2)解:底面ABCD是菱形,AB=2,∠BAD=60°,∴△PBD与△BCD都是等边三角形.∵平面PBD⊥平面ABCD,平面PBD∩平面ABCD=BD,PO⊥BD.∴PO⊥平面ABCD,∴PO⊥AC,又AC⊥BD,可建立如图所示的空间直角坐标系.∵∠DPB=90°,PB=PD,BD=2,∴PO=1,∴P(0,0,1),B(1,0,0),D(﹣1,0,0),C(0,,0),=(﹣1,0,﹣1),=(0,,﹣1),=(1,﹣,0),设平面PCD的法向量=(x,y,z),则,∴,取=,则点B到平面PDC的距离d===.19.PM2.5是指空气中直径小于或等于2.5微米的细颗粒物,它对人体健康和大气环境质量的影响很大.2012年2月,中国发布了《环境空气质量标准》,开始大力治理空气污染.用x=1,2,3,4,5依次表示2013年到2017年这五年的年份代号,用y表示每年3月份的PM2.5指数的平均值(单位:μg/m3).已知某市2013年到2016年每年3月份PM2.5指数的平均值的折线图如图:(1)根据折线图中的数据,完成表格:年份2013 2014 2015 2016年份代号(x) 1 2 3 4PM2.5指数(y)(2)建立y关于x的线性回归方程;(3)在当前治理空气污染的力度下,预测该市2017年3月份的PM2.5指数的平均值.附:回归直线方程=x+中参数的最小二乘估计公式;=, =﹣.【考点】线性回归方程.【分析】(1)根据折线图中的数据,完成表格即可;(2)计算线性回归方程中的系数,可得线性回归方程;(3)x=5代入线性回归方程,可得结论.【解答】解:(1)年份2013 2014 2015 2016年份代号(x) 1 2 3 4PM2.5指数(y)90 88 70 64(2)=2.5, =78,(xi ﹣)(yi﹣)=﹣48,=5,==﹣9.6, =﹣=102,∴y关于x的线性回归方程是: =﹣9.6x+102;(3)2017年的年份代号是5,当x=5时, =﹣9.6×5+102=54,∴该市2017年3月份的PM2.5指数的平均值的预测值是54μg/m3.20.已知椭圆C: +=1(a>b>0)的离心率为,以该椭圆上的点和椭圆的两个焦点为顶点的三角形的周长为6.(1)求椭圆C的方程;(2)设过点C的左焦点F的直线l交C于A,B两点,是否存在常数λ,使||=λ•恒成立,若存在,求出λ的值;若不存在,请说明理由.【考点】椭圆的简单性质.【分析】(1)由=,2a+2c=6,a2=b2+c2,联立解出即可得出椭圆C的方程.(2)F(﹣1,0),设A(x1,y1),B(x2,y2).当直线l的斜率不存在时,x1=﹣1,不妨取y1=,可得λ==﹣.当直线l的斜率存在时,设直线l的方程为y=k(x+1),代入椭圆方程整理为:(4k2+3)x2+8k2x+4k2﹣12=0,△>0,利用根与系数的关系可得=,•=(x1+1)(x2+1)+y1y2,计算即可得出.【解答】解:(1)∵=,2a+2c=6,a2=b2+c2,解得a=2,c=1,b2=3.∴椭圆C的方程为=1.(2)F(﹣1,0),设A(x1,y1),B(x2,y2).当直线l的斜率不存在时,x1=﹣1,不妨取y1=,||=3, =, =.•=,则λ===﹣.当直线l的斜率存在时,设直线l的方程为y=k(x+1),则,整理为:(4k2+3)x2+8k2x+4k2﹣12=0,△=64k4﹣4(4k2+3)(4k2﹣12)=122(1+k2)>0,x 1+x2=,x1x2=.==,=(x1+1,y1),=(x2+1,y2)..• =(x1+1)(x2+1)+y1y2=(k2+1)[x1x2+(x1+x2)+1]=,则==﹣.综上所述:可得存在常数λ=﹣,使||=λ•恒成立.21.已知函数f(x)=+b在x=1处的切线方程为x+y﹣3=0.(1)求a,b.(2)证明:当x>0,且x≠1时,f(x)>.【考点】利用导数求闭区间上函数的最值;利用导数研究曲线上某点切线方程.【分析】(1)求出函数的导数,根据f(1)=2,f′(1)=﹣1,求出a,b的值即可;(2)问题转化为(x﹣﹣2lnx)>0,令g(x)=x﹣﹣2lnx,(x>0),求出g(x)的单调区间,从而证出结论即可.【解答】解:(1)f(x)的定义域是(0,+∞),f(x)=+b,切点是(1,2),∴f(1)=b=2,f′(x)=,∴f′(1)=a=﹣1,故a=﹣1,b=2;(2)证明:由(1)得:f(x)=+2,f(x)>,∴(x﹣﹣2lnx)>0,令g(x)=x﹣﹣2lnx,(x>0),则g′(x)=(x﹣1)2>0,∴g(x)在(0,1)递增,在(1,+∞)递增,∵g(1)=0,∴g(x)>0⇔x>1,g(x)<0⇔0<x<1,∴x>1时, g(x)>0,0<x<1时, g(x)>0,x>0且x≠1时,(x﹣﹣2lnx)>0,∴当x>0,且x≠1时,f(x)>.[选修4-1:几何证明选讲]22.如图,E为⊙O上一点,点A在直径BD的延长线上,过点B作⊙O的切线交AE的延长线于点C,CE=CB.(1)证明:AE2=AD•AB.(2)若AE=4,CB=6,求⊙O的半径.【考点】与圆有关的比例线段.【分析】(1)证明AC是⊙O的切线,根据切割线定理可得:AE2=AD•AB.(2)根据切割线定理求出AD,即可求⊙O的半径.【解答】(1)证明:∵过点B作⊙O的切线交AE的延长线于点C,∴∠CBO=∠CBE+∠OBE=90°.∵CE=CB,OE=OB,∴∠CEB=∠CBE,∠OEB=∠OBE,∴∠CEO=∠CEB+∠OEB=∠CBE+∠OBE=90°,∴CE⊥OE,∵OE是⊙O的半径,∴AC是⊙O的切线,根据切割线定理可得AE2=AD•AB.(2)解:∵CE=CB=6,AE=4,∴AC=10,∴AB=8∵AE2=AD•AB,AE=4,∴42=AD•8,∴AD=2,∴BD=8﹣2=6,∴⊙O的半径为3.[选修4-4:坐标系与参数方程选讲]23.已知曲线C的极坐标方程是ρsin2θ﹣8cosθ=0,以极点为平面直角坐标系的原点,极轴为x轴的正半轴,建立平面直角坐标系xOy.在直角坐标系中,倾斜角为α的直线l过点P(2,0).(1)写出曲线C的直角坐标方程和直线l的参数方程;(2)设点Q和点G的极坐标分别为(2,),(2,π),若直线l经过点Q,且与曲线C相交于A,B两点,求△GAB的面积.【考点】简单曲线的极坐标方程;参数方程化成普通方程.【分析】(1)ρsin2θ﹣8cosθ=0,化为ρ2sin2θ﹣8ρcosθ=0,令,即可得出直角坐标方程.直线l的参数方程为:(t为参数).(2)点Q和点G的极坐标分别为(2,),(2,π),分别化为:Q(0,﹣2),G(﹣2,0).kl=1,倾斜角为,可得直线l的参数方程:(t为参数).将参数方程代入曲线C的方程可得:t2﹣8t﹣32=0,设t1与t2为此方程的两个实数根,可得|AB|=|t1﹣t2|=.点G到直线l的距离d.即可得出S△GAB=|BA|•d.【解答】解:(1)ρsin2θ﹣8cosθ=0,化为ρ2sin2θ﹣8ρcosθ=0,∴直角坐标方程为:y2=8x.直线l的参数方程为:(t为参数).(2)点Q和点G的极坐标分别为(2,),(2,π),分别化为:Q(0,﹣2),G(﹣2,0),kl==1,倾斜角为,直角坐标方程为:y=x﹣2.可得直线l的参数方程:(t为参数).将参数方程代入曲线C的方程可得:t2﹣8t﹣32=0,△=128+4×32>0,设t1与t2为此方程的两个实数根,可得:t1+t2=,t1t2=﹣32.∴|AB|=|t1﹣t2|===16.点G到直线l的距离d==2.∴S △GAB=|BA|•d==16.[选修4-5:不等式选讲]24.已知函数f(x)=.(1)求函数f(x)的值域;(2)若函数f(x)的值域是[m,n],且a2+b2=m,c2+d2=n,求ac+bd的取值范围.【考点】函数的最值及其几何意义.【分析】(1)记g(x)=|x+3|﹣|x﹣1|+5,分类讨论求得g(x)=,从而求值域;(2)由柯西不等式知(a2+b2)(c2+d2)≥(ac+bd)2,从而求取值范围.【解答】解:(1)记g(x)=|x+3|﹣|x﹣1|+5,则g(x)=,故g(x)∈[1,9],故f(x)∈[1,3].(2)由(1)知,a2+b2=1,c2+d2=3,由柯西不等式知,(a2+b2)(c2+d2)≥(ac+bd)2,(当且仅当ad=bc时,取等号;)即(ac+bd)2≤3,故﹣≤ac+bd≤,故ac+bd的取值范围为[﹣,].2016年9月12日。

2020-2021学年江苏省连云港市、徐州市、宿迁市高考年级第三次模拟考试英语试题及答案

2020-2021学年江苏省连云港市、徐州市、宿迁市高考年级第三次模拟考试英语试题及答案

高考英语模拟试题高三年级第三次模拟考试(三)英语本试卷共12页,满分120分,考试时间120分钟。

第一部分听力(共两节,满分20分)第一节(共5小题;每小题1分,满分5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有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. What does the man mean?A. She should take more exercise.B. She'd better have a few days' rest.C. She is badly ill.( ) 2. When will Mr. White be free?A. This Saturday.B. Next Friday.C. Next Sunday.( ) 3. Where does the conversation most probably take place?A. At a bookstore.B. At a post office.C. At a supermarket.( ) 4. How many people are mentioned in the dialogue?A. At least four.B. Only three.C. More than five.( ) 5. How much may the man spend on the chair in the end?A. $15.B. $ 25.C. $ 20.第二节(共15小题;每小题1分,满分15分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

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。

2021年高考英语一轮温习Unit1Friendship(练)(含解析)新人教版必修1

2021年高考英语一轮温习Unit1Friendship(练)(含解析)新人教版必修1

Unit 1 FriendshipⅠ.词义辨析1.She refused to offer any help,which quite ________ all the people present.A.ignored B.settled C.upset D.suffered2.“If you ________ your diet,trouble will follow,” my doctor warned me.A.pack B.ignore C.recover D.concern3.He devoted himself ________ to his research work,having no time for play.A.exactly B.gratefully C.loosely D.entirelyCBDD4.I’m afraid it’s not within my ________ to get the work done in such a short time.A.energy B.strength C.power D.force5.She drew back the ________ at nine in the morning to let some sunlight in.A.curtains B.suitcases C.partners D.teenagers6.I can’t express how ________ I am for all you’ve done for me.A.grateful B.loose C.calm D.entire7.In order to build the dam,they moved the local people and________ them in another place. A.recovered B.ignored C.concerned D.settled8.The doctors are delighted to find that the patient is beginning to________ from heart trouble.A.uncover B.discover C.recover D.coverCAADC9.The train arrived at ________ 8 o’clock,neither earlier nor later.A.clearly B.exactly C.widely D.nearly10.Steve gave me some useful ________ on how to take good pictures.A.powers B.tips C.items D.thundersBBⅡ.名词性从句专练1. 【2021届北京市东城区高三5月综合练习(二模)】Agatha didn’t tell me ______ she wouldpick up her son from school.A. whichB. whenC. whereD. what【答案】B2. 【2021届北京市东城区高三5月综合练习(二模)】 It is well known _____Confucius was bornin the city of Qufu in Shandong province, China.A. whetherB. whereC. thatD. what【答案】C【解析】考查连接词。

江苏省徐州市2020-2021学年度第二学期九年级一模英语试卷(含答案)

江苏省徐州市2020-2021学年度第二学期九年级一模英语试卷(含答案)

2020- 2021 学年度初中毕业、升学第一次模拟检测九年级英语试题(全卷共110分,考试时间100分钟)一、选择填空(每小题1分,共15分)从A、B、C. D四个选项中,选出可以填入空白处的最佳选项。

1. Look at the picture on the right What might this fireman say to himself?A. Why not?B. What's up? C Anything else? D. Feelbetter2. When are you arriving? I'll pick you up________ the station.A.atB. toC. onD. off3. We have to say goodbye now,________ Our friendship will last forever.A. andB. butC. orD. so4. Zhaozhou Bridge is one of _______stone bridges in the world.A. oldB. olderC. oldestD. the oldest5. Maria is always full of_______ because she takes exercise every day.A. energyB. talentC. humourD. wealth6. We are supposed to ________ smart phones and take more exercise instead.A. put upB. put awayC. put onD. put out7. It ________dark. Shall I turn on the light?A. getsB. gotC. is gettingD. was getting8. Ken was ___________ late for school. The bell rang right after he entered the classroom.A. stillB. alwaysC. alreadyD. almost9. _______ hopes for a sweet home as it provides us with warmth and trust.A. NoneB. EveryoneC. NobodyD. Somebody10. Bob, dinner is ready. Please wash your hands___________ you eat.A. until .B. afterC. whileD. before11. -- Mum, why do I have to wash hands so many times a day?--For your health, you _______ be too careful.A: can't B. shouldn't C; mustn't ' D. needn't12. China has formed a deep friendship with countries around the world. We are working______ together in many fields.A. closelyB. easilyC. clearlyD. quickly13. --Dear friends, do you still remember why you came here three years ago?--To ________ our dreams!A. copyB. saveC. imagineD. achieve14.--I'm going to sell all your old books, They,re taking up too much space.--___________ ! They're treasures. You sell them, you take my life!A.What a pityB. Sounds greatC. You can't be seriousD. Couldn't agree more15. Susie Sunbeam was not her real name; that was Susan Brown. But everyone called her Susie Sunbeam, because she always brought brightness with her wherever she came. Her grandfather gave her this name, and it seemed to fit the little girl so nicely that soon it took the place of her own The story mainly tells us_________.A.'what Susie Sunbeam stood forB. how beautiful Susie Sunbeam wasC. how she got the name Susie SunbeamD. how much her grandfather loved her二、完形填空(每小患1分,共15分)根据短文内容,从各题所给的A、B、C、D四个选项中选出最佳选项。

江苏省连云港市2020-2021学年下学期八年级英语期中模拟试卷三(有听力、有答案)

江苏省连云港市2020-2021学年下学期八年级英语期中模拟试卷三(有听力、有答案)

连云港市2020-2021学年下学期八年级英语期中模拟试卷三满分:150分时间:90分姓名:__________ 得分:__________一、听力(共20小题,1-10小题每题1分,11-20小题每题2分,满分30分)I.听对话,选择正确答案。

每组对话读两遍。

1. What place of interest may he visit?A.B.C.2. How many girls are there in the class?A.B.C.3. Where are the two speakers talking now?A. B. C.4.How is Linda going to the cinema tonight?A. B. C.5. Which city did Sandy visit last?A. Paris.B. London.C. New York.6. When will Jim’s birthday party begin?A. At 5:30.B. At 5:00.C. At 4:30.7. What is Jack’s sister?A. A nurse.B. A teacher.C. A student.8. Who does the red pen belong to?A. Betty.B. Mike.C. Lily.9. What can we learn about this young lady?A. She hasn’t got married.B. She is very shy.C. She is homorous.10. What's the date today?A. Sep 9th.B. Sep 10th .C. Sep 11th.II.听对话或短文,选择正确答案。

每段对话或短文读两遍。

听一段对话,完成11-12小题。

11. When is hi s father’s birthday?A. This Sunday.B. This Thursday.C. ThisSaturday.12. What are they going to do for his father’s birthday?A. They’re going to have a party.B. They’re going to have a picnic.C. They’re going to have a big dinner.Plan for summer vacationTom ●doesn’t want to stay at home●plans to go 13David ●is going to visit Harbin●will stay there for 14 daysAlice ●will learn how to play the 15●wants to be a musician13. A. fishing B. camping C. swimming14. A. 5 B. 6 C. 715. A. guitar B. violin C. piano听第二篇短文,完成16-20 题。

【附20套高考模拟试题】江苏省徐州、连云港、宿迁三市2020届高三第三次模拟英语试题含答案

【附20套高考模拟试题】江苏省徐州、连云港、宿迁三市2020届高三第三次模拟英语试题含答案

江苏省徐州、连云港、宿迁三市2020届高三第三次模拟英语试题第一部分(共20小题每,小题1.5分,满分30分)1.That children ______meet with setbacks is a matter of necessity as they_____, so parents don’t worry about that.A.shall; grew up B.must;grew up C.can; grow up D.will; grow up2.—It’s so humid t hese days!—Don’t worry! The rain ________ to stop from tomorrow.A.will expect B.expectsC.will be expected D.is expected3.We the sunshine in Sanya now if it were not for the delay of our flight.A.were enjoying B.would have enjoyedC.would be enjoying D.will enjoy4.—Can I have a day off tomorrow? I need to visit my grandma in the hospital.—__________. I can manage without you.A.Forget it B.Of courseC.It depends D.I’m afraid not5.A Chinese proverb has it that a tower is built when soil on earth _________, and a river is formed when streams come together.A.accumulates B.accelerates C.collapses D.loosens6.Mary felt from the outside world, since she lacked an Internet connection and couldn’t receive any e-mail.A.cut down B.cut in C.cut off D.cut out7.We __________back in the hotel now if you didn’t lose the map.A.are B.wereC.would be D.will be8.Which do you prefer, tea or coffee?—_______. I really don’t mind.A.Both B.None C.Neither D.Either9.Never before ________ the famous museum was just a stone's thro away from their school, so out________.A.had they known; went all they B.they had known; went all theyC.had they known; they all went D.they had known; they all went10.Thanks to the “sugar tax”,food factories have reduced sugar in their products, ________ about 45 million kilograms of sugar.A.to save B.savedC.saving D.having saved11.He is confident, ________________ in my opinion, is most important in society.A.how B.whichC.that D.what12.---Tom has failed again in the math exam. I am really let down.---______ He has already done his best.A.Don’t push him. B.No kidding.C.He should be to blame. D.No wonder.13.--You know Mr. Green has been ill for days?--Yes, I wonder if he is ______ better now.A.any B.some C.any D.no14.Office furniture like chairs and sofas should be attractive as well as comfortable.A.manually B.visuallyC.physically D.securely15.I have often thought it would be a blessing if each human being _______ blind and deaf for a few days at some time in his life.A.has been stricken B.were strickenC.had been D.would be16.—Where can we park car?—Don’t worry. There’s sure to be parking lot nearby.A.the; the B.the; aC.不填; a D.不填; the17.The teacher came into the classroom _______ by his students.A.following B.to be following C.followed D.having followed.18.The two pens are the same, but the red one cost _____________ that one.A.as much twice as B.twice as much asC.much as twice as D.as twice much as19.Not until I went up further ________ that under the tree ________, obviously sound asleep.A.that I saw; did a boy lie B.I saw; lay a boyC.did I see; did a boy lie D.did I see; lay a boy20.________ who are able to work through the struggle are the ________ who are going to be successful. A.Someone; one B.Anyone; oneC.He; ones D.Those; ones第二部分阅读理解(满分40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。

2020-2021江苏省徐州市中考英语模拟试题(含答案)

2020-2021江苏省徐州市中考英语模拟试题(含答案)

江苏省徐州市2020年中考英语模拟试题满分100分,考试时间100分钟一、选择填空(共15小题,每小题1分,满分15分)1. 一Which place do you prefer to live in, a city or a countryside? 一I don't care. ____ is fine.A. AllB. EitherC. NeitherD. None2. Cindy shut the door heavily and rushed out. No one in the office knew _____she was so angry.A. whyB. whetherC. thatD. where3. 一_______ is the nearest theme park from here?一It will take us half an hour to get there by bike.A. How muchB. How longC. How soonD. How far4. Sally took a photo of her friends while they _____computer games.A. playB. are playingC. were playingD. have played5. Which of the following underlined parts is different in pronunciation with others?A. Don't put your feet on the seat.B. We reached Los Angeles late at night.C. Could I have fish instead of ham?D. What's the meaning of this?6.The 2022 Winter Olympics _____ in Beijing and towns in Hebei Province.A. holdB. will holdC. are heldD. will be held7.Harry,the tallest boy in our class, ____ easily reach the books on the top shelf of our library.A. mustB. shouldC. canD. need8.―What do you think of the action movie Mulan?―It comes from an old Chinese story. Mulan dresses up as a boy and takes her father's place in the army.A. to fightB. fightingC. fightsD. fought9. Not only my classmates but also our English teacher ____ Chinese poems,so we often share their favourite ones.A. likeB. likedC. likesD. liking10.一Alex is planning to climb the mountain without even using a rope.一What? I've never heard of ____ idea before.A. a crazierB. the crazierC. a craziestD. the craziest11. Sallie rolls up her new trousers carefully _____ them from dust.A. protectB. protectingC. to protectD. to protecting12. This new kind of energy costs very little and will never _______________.A. turn outB. run out.C. take outD. put out13. He began to wonder ________.A. that there was a road to the top of the hillB. if was there a road to the top of the hillC. whether there was a road to the top of the hillD. how could he get to the top of the hill14. 一All the citizens should work together to make our city more beautiful. 一That's right.A. Practice makes perfectB. Put all your eggs in your basketC. When in Rome, do as the Romans doD. Many hands make light work15. 一I'm sure the traffic is heavy now. 一____ . I have to reach the airport before 9 a. m.A. I hope notB. I don't think soC. I'm afraid notD. I don't care二、完形填空(共15 小题,每小题1分,满分15分)I had to stay at Dallas Airport that night, waiting for a transferring flight(转机).I 16 to sleep in a chair at the boarding gate. As I took my left shoe off to relax my wounded leg wrapped(缠绕)in a long piece of cloth, a lady 17 came up to me. "Are you OK, Miss?" she asked. "Yeah, it's all right, I 18 my leg," I answered coldly, as I was on alert(警惕的)then.Anyway, the lady went on, "Oh, you need to rest. We can help you. We can 19 a room for you to sleep in."Tons of 20 came into my mind. "How can I trust her? Does she 21 in the airport?" I began looking for an ID card on her while holding my phone in my pocket. "Don't ___22 .I work here." She smiled at me and showed me her ID card. I still didn't 23 her until she asked a man in uniform to get a wheelchair for me. Then she wheeled me to the Lost and Found where the airport takes care of 24 kids. It's a cute room with a tiny comfortable bed. "Good night, " she 25 at me again. Her blue eyes helped me to relax. "Good night, truly, thank you! " I said to her.I had a good sleep indeed. The next morning someone else there sent me to the boarding gate.The 26 I got at Dallas Airport made me feel 27 for a lot of things-that kind lady, the help of the airport and, 28 , the care from strangers. No matter where we go, someone in the world is 29 to help us. I have come to realize that the best 30 of thankfulness is to pass on this kindness to the people who need my help.16. A. waited B. asked C. planned D. offered17. A. finally B. carefully C. nervously D. suddenly18. A. lost B. hurt C. confirmed D. beat19. A. provide B. afford C. choose D. buy20. A. words B. thoughts C. notes D. memories21. A. live B. charge C. sleep D. work22. A. hurry B. refuse C. worry D. cheat23. A. need B. trust C. understand D. admire24. A. naughty B. young C. missing D. lonely25. A. smiled B. pointed C. shook D. laughed26. A. care B. guide C. advice D. encouragement27. A. satisfied B. patient C. guilty D. thankful28. A. at first B. at least C. after all D. above all29. A. willing B. afraid C. free D. eager30. A. communication B. conversation C. expression D. impression三、阅读理解(共12小题,每小题2分,满分24分)AKing's College Summer School is an annual(每年的)training program for all the middle school students who want to improve their English. Courses are given by the teachers of King's College and other colleges in New York. Trips to museums and culture centers are also organized. This year's summer school will be from June 1 to July 15. More information is as follows:Application(申请)date?Students in New York should send their applications before April 18,2019?Students of other cities should send their applications before April 16,2019?Foreign students should send their applications before April 10,2019 Courses?English LanguageSpoken English:22 hours Reading and Writing:10 hours ?American History:16 hours ?American Culture:16 hoursSteps?A letter of self﹣introduction?A letter of recommendation(推荐)from a teacher★The letters should be written in English with all the necessary information. Cost?Daily lesson:$200?Sports and activities:$100?Travels:$200?Hotel service:$400★You may choose to live with your friends or relatives in the same city.Please write to:Thompson Sanders1026 King's StreetNew York,NY10016,USAE﹣mail:KC﹣Summer﹣School@hotmail. com31. According to the article,who will give the courses?A. The teachers of King's College and the middle schools in America.B. The teachers of Summer School and other colleges in the United States.C. The teachers of the middle schools and Summer School in New York.D. The teachers of King's College and other colleges in New York.32. When should foreign students send their applications?A. Before April 10,2019B. Before June 1,2019C. Before April 16,2019D. Before July 15,201933. If you want to live with your friends in New York,you will have to pay the school____.A. ﹩300B. ﹩400C. ﹩500D. ﹩90034. Which of the following is TRUE about King's College Summer School?A. Only top students can take part in the program.B. Visits to museums and culture centers are part of the program.C. King's College Summer School is not organized every year.D. You can get in touch with the school by telephone or by e﹣mail.BWho is the greatest teacher in Chinese history? You can' t think of anyone else when you hear people read Confucius' 2,500-year-old words:“All study but no thinking makes people puzzled. All thinking but no study makes people lazy.September 28th was Confucius' birthday. He was born in the Kingdom of Lu,in today's Shandong Province. He had a hard child. His father died when he was only 3. His mother brought him up, As a child, he had to work to help his mother, but young Confucius didn’t give up studying. He visited many famous teachers and learned music, history, poetry and sports.Later, he became a teacher and started the first public school in Chinese history. At that time,only children from rich families could go to school if they wanted to learn. He had about 3,000 students in his lifetime.Chinese look upon Confucius as the greatest thinker and also the greatest teacher. For more than 2 , 500 years,Confucius' ideas have been around in people's everyday lives. Now they have gone far into Western countries and South Asia. People can still hear them today.Why are his ideas so popular? Confucius' most important ideas are kindness and good manners. Confucius said young people should take care of the old. A kind person should care for others. Be strict wish yourself, but be kind to others. As a great teacher, Confucius said that all should go to school if they wanted to learn. People use his ideas to help themselves and society. Now, morethan 100 Internet websites are teaching people about Confucius and his ideas.35 .Why do we still remember Confucius today?A. Because he had about 3,000 students in his lifetime.B. Because he lived a poor life in his childhood.C. Because he had wise thoughts.D. Because he started the first public school in Chinese history.36. The underlined word "them" in the fourth paragraph refers to .A. the greatest thinker and teacherB. Confucius' studentsC. people in Western countries and South AsiaD. Confucius' ideas37. Which of the following is NOT Confucius' words?A. People should get education.B. The young should take care of the old.C. Be strict with yourself, but be kind to others.D. All study but no thinking makes people lazy.CFew words are spoken more often every day on the streets of Britain than "I'm sorry". This phrase has become such a common response(反应)that it has taken on a lot of meanings.Saying "Sorry" means to apologize(道歉). This is simple and easy to understand. We learn it both as a native speaker and as a student of foreign languages.But in Britain,it takes on another meaning. It is a cultural expression. Imagine this:a manwalks down the street,looking down at his phone. A woman is walking in the opposite direction,towards the man. She sees him,but she can't get out of the way in time. The man bumps (碰撞)into the woman. Who should say sorry? Naturally,the man should say sorry,because it was he who wasn't looking where he was going. Yet in Britain,it is common for bothto apologize.It is known that British people,like most people,do not enjoy conflict(冲突). So to quickly calm the situation,British people will apologize to each other.Other times it may sound funny to hear "sorry". Some of my friends say it at restaurants,as they ask the waiter:"Sorry,but can I order another drink? " It is not to apologize,but just to express that we need the waiter. In Britain,sorry doesn't always mean exactly what you think.38. According to the passage,"saying sorry" is a cultural expression in _____.A. the USAB. the UKC. the WWFD. the UN39. What does the underlined phrase "take on" mean in Paragraph 1 ?A. 从事B. 雇佣C. 呈现D. 拿起40. The example in the third paragraph is used to_____.A. describe a situation that people should avoidB. describe how "sorry" has another meaning in BritainC. explain why people should say sorry to each otherD. show how polite British people are41. The restaurant example shows that "sorry" can be used to_____.A. apologizeB. calm a situation downC. explain what you're thinking aboutD. ask a waiter to bring something42. What might be a good title for this passage?A. "I'm Sorry" Is More Than just an ApologyB. Traditional British MannersC. How to Best Catch Others' AttentionD. Finding a Way out of a Difficult Situation四、词语运用(共16小题,每小题1分,满分16分)A)选用方框内的单词或词组填空,其中有一个单词或词组是多余的。

江苏省扬州市、宿迁市、连云港市2021年4月新高考适应性考试日语试题(含答案)

江苏省扬州市、宿迁市、连云港市2021年4月新高考适应性考试日语试题(含答案)

参考答案
一、听力(15*2=30)
12345678910 C A C A C B A C A A 1112131415
B B A B A
二、知识运用(40*1=40)
16171819202122232425 A C D B D A B C D C 26272829303132333435 C A D B D B C C A D 36373839404142434445 C A C D B A C B B D 46474849505152535455
C A
D B A B
B D A C
三、阅读(20*2.5=50)
56575859606162636465 B D C A D A B C C D 66676869707172737475
C B A
D B A D B C B
四、作文(30)
◎档次标准
第六档(26~30分)写出“写作要点”的全部内容,语言准确流畅,句型及表达形式丰富。

第五档(20~25分)写出“写作要点”的全部内容,语言表达恰当。

第四档(15~19分)写出“写作要点”的大部分内容,语言表达通顺。

第三档(10~14分)写出“写作要点”的一部分内容,语言表达基本通顺。

第二档(5~9分)写出“写作要点”的少部分内容,语言表达欠通顺。

第一档(0~4分)写出“写作要点”的很少内容,语言表达不通顺或字数少于100字。

◎评分说明
①少于300字者,每少写一行扣1.5分。

②每个用词或书写错误扣0.5分。

③每个语法错误(活用、时态、助词、句型等)扣1分。

④标点符号及格式错误扣分总值不超过2分。

1。

2020-2021学年连云港市平明中学高三英语第三次联考试卷及参考答案

2020-2021学年连云港市平明中学高三英语第三次联考试卷及参考答案

2020-2021学年连云港市平明中学高三英语第三次联考试卷及参考答案第一部分阅读(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AThe OrchardI had a very nice meal at the Orchard restaurant.The food was lovely and the service was quite good.We chose to eat in the garden which was full of beautiful flowers and very relaxing.The only disappointment was the dessert — the apple piewas far too sweet and it was cold too.Apart from that it was an enjoyable evening.As for the price — what a bargain,excellent value for money!Park InnThe best thing about Park Inn was the service — it was excellent.There was a warm welcome when we arrived and the waiters were very helpful all evening.However,the food wasn’t as good.The menu looked interesting but the meals were rather tasteless.It’s not a cheap restaurant and I wasn’t happy to pay so much for boring food.Richard’s PlaceWhen we enteredthe restaurant,we were surprised by the lovely interior (内部的) design of it.So stylish,so modern!This is one of the most popular restaurants in town and it’s very easy to see why.The food was great and excellent value for money but it was ruined by very,very poor service!TheRiversideIt was lovely sitting outside and looking over the river.There were lots of little lanterns (灯笼) and color1 ed lights everywhere and it all looked very pretty.The food was good,nothing very special but quite tasty.The service was OK; we didn’t have to wait too long for our food but the waiter never looked very happy!It’s quite an expensive place but with the view I think it’s quite good value for money.1. In which restaurant can customers eat in the garden?A. Park Inn.B. TheOrchard.C. The Riverside.D. Richard’s Place.2. What makes the customers of Park Inn most satisfied?A. The food.B. The price.C. The service.D. The environment.3. In Richard’s Place,customers can ________.A. receive good serviceB. enjoy its moderndesignC. listen to wonderful musicD. enjoy the beauty of a riverBHumans have found an easy way to tell if others are lying. Recent research shows that the best way so far is being clever at how you ask questions and listen to the answers.Much of this research is based on the idea that telling a lie is simply harder mental work than telling the truth. Making up a story takes more effort than simply recording something that happened. And like a writer, a liar has to keep all the unreal details in his memory and sound believable when he explains them.One method that seems to work is asking them to tell their story in reverse order. This is harder when the story isn't true and makes it easier for you to tell they are lying. An even more basic way that helps is to just ask more questions, especially unexpected ones. Truth-tellers can easily find more to say, but it's a challenge for a liar to come up with something that's not in his prepared story.Researchers suggest that you shouldn't lay all your cards on the table at the start, but only gradually present what proof you have. The liars' stories may not agree with that proof, making it clear that they're lying.So it looks like there are ways to increase the chances of catching a liar; we've just been basing our methods on the wrong stories. Low-tech ways of causing people to make mistakes in conversation seem to work better than any science about eye movement or machines used to recognize a liar. To find a liar, watch less and listen more.4. Why does the author mention the writer?A. To show it's hard to make up lies.B. To show it's hard to recognise a liar.C. To show writers know liars best.D. To show writers are very clever.5. What do we know about liars?A. They often have much to say.B. They often ask many questions.C. They usually prepare a made-up story.D. They usually feel good about themselves.6. What advice is given to help people catch liars?A. Asking them to set their stories down.B. Presenting your proof one by one.C. Telling different stories to them.D. Letting them ask questions.7. Which can be the best title for the text?A. Why People LieB. How to Stop People LyingC. Low-tech Ways to Find a LiarD. LiarsAre Smarter than ThoughtCWith graduation days being celebrated all over the country, a student who has to use a wheelchair honored his mother on his graduation day in a special way. Easley High School graduate, Alex Mays surprised people present when he got up and walked across the stage at Clemson's Littlejohn Coliseum.“I was really happy—it made me feel good,” Alex said.Alex was not given a chance to live right from his birth. He was born at 25 weeks and weighed just 1 pound, 10 ounces at birth. When he was very young, he had a disease and lost the ability to walk. After his mother's death in 2013, Alex had several other difficult life changes until he came to live with his grandparents, Dousay and her husband, Dewayne. Dousay said that when Alex came to live with them, they decided to bring him up in the best possible way they could.Last fall, Alex said that he would walk across the stage to get his diploma to honor his late mother. He practiced hard and worked with a physical therapist for 9 months to complete his plan.The only help Alex got was from his mom's best friend, Tonya Johnson, who pushed his wheelchair to the stage wearing one of his mother's favorite shirts. “I had support from my family. I couldn't have done it without them,” Alex said.“Alex made everyone in the building feel encouraged that day” Pickens County School District public information specialist John Eby said. “The school teachers knew he was going to get up to get his diploma, but the distance he walked was a surprise, even to them,” Eby said.“Some of life's most important tests aren’t given in a classroom; Alex tested himself and passed with flying color1 s,” Eby added.8. In what way did Alex honor his late mother on his graduation day?A. By dressing like her.B. By saying sorry to her.C. By inviting her best friend.D. By walking to get his diploma.9. What can we learn from Paragraph 3?A. Alex was born healthy.B. Alex went through a lot.C. Alex had a purpose in life as a child.D. Alex has lived with his grandparents all the time.10. What did Alex also express on his graduation day?A. His big regret in life.B. His feelings for hisschool.C. His thanks for his family.D. His will to complete his study.11. Which of the following words can best describe Alex?A. Strong-minded.B. Warm-hearted.C. Cool-headed.D. Easy-going.DThe founder of Earth Day was Gaylord Nelson, a U.S. Senator fromWisconsin. During the late 1960s, Americans witnessed the uninvited side effects of high productivity. Factories and power plants were sending out smoke and industrial waste while Americans were using petrol for their massive(大量的) cars, making air pollution almostsynonymous withthe nation’s development.What moved Senator Nelson to action was the 1969 massive oil spill inCalifornia, the largest in theUnited Statesat that time. The spill proved to be an environmental nightmare as it had a significant effect on marine life, killing about 3,500 sea birds, as well as marine animals such as dolphins, elephant seals and sea lions, fueling public anger. Inspired by the student antiwar movement at that period of time, Nelson found it an appropriate time to direct the energy of the students towards a fight for environmental protection. He decided that it was time to educate the Americans on the need to protect the environment. Thus Earth Day was born in 1970, and public environmentalawareness took centre stage.On 22ndApril 1970, millions of Americans took to the street and thousands of students marched to appeal for a healthy, sustainable environment. There was now a new synergy(协同作用) among different groups which had previously been fighting their causes related to the environment. Their fight for environmental conservation became so overwhelming that affected businesses were forced to follow environmental standards if they wanted to continue their operations.As it became more apparent that environmental issues were not just localized ones but a global concern, the year 1990 saw Earth Day reach out to many more around the world. Earth Day 1990 helped pave the way for the 1992 United Nations Earth Summit inRio de Janeiro, bringing together many nations for a united effort towards protecting the environment.12. Which of the following can replace the underlined phrase “synonymous with” in paragraph 1?A. familiar withB. opposite toC. different fromD. equal to13. Why did Nelson found Earth Day?A. To support students’ antiwar movement.B. To draw people’s attention to the seriousness of the oil spill.C. To arouse American’s awareness of environmental conservation.D. To educate Americans to protect marine life threatened by oil spill events.14. What can be inferred from the passage?A. Businesses would like to follow environmental standards.B. Earth Day united people to fight for environmental protection.C. It was the side effects of high productivity that led Nelson to take action.D. The 1992 United Nations Earth Summit made Earth Day known to more countries.15. The passage mainly talks about_______.A. how Earth Day came into beingB. why Earth Day was so significantC. who the founder of Earth Day wasD. what Earth Day meantto the world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。

江苏省徐州市2020-2021学年学业水平合格性考试模拟卷

江苏省徐州市2020-2021学年学业水平合格性考试模拟卷

徐州市高二年级合格考第一次模拟考试地理注意事项∶1. 本试卷满分100 分,考试用时90 分钟。

2.答题前,请考生务必将自己的学校、班级、姓名写在密封线内。

一、选择题∶本大题共40小题,每小题2分,共计80分。

在每小题的四个选项中,只有一个选项符合题目要求。

科幻电影《流浪地球》讲述了地球因太阳"氨闪"而被迫逃离太阳系寻找新家园的故事。

"流浪地球"计划分为三步∶第一步,终止地球自转;第二步,将地球推入木星轨道,借助木星引力,弹射出太阳系;第三步,地球经历2500年的星际流浪,抵达新家园距离太阳最近的恒星——比邻星。

图1为"地球流浪过程示意图"。

读图,回答1~3 题。

1.地球在图中a、b、c三个位置对应的天体类型依次是A.行星--卫星--行星B. 行星--行星--行星C.行星--卫星--卫星D. 行星--卫星--恒星2.地球进入比邻星宜居轨道后,下列推测最可信的是A. 比邻星位于河外星系B. 比邻星位于银河系C.比邻星不会发光发热D.比邻星绕太阳运动3.比邻星能成为地球新家园是因为比邻星可以为地球提供A. 充足的食物B. 适宜的大气C.稳定的光照条件D.大量的液态水1859年9月1日,英国天文爱好者卡林顿观测到日面上出现两道极其明亮的白光,其亮度迅速增加,远远超过光球背景,明亮的白光仅维持几分钟就很快消失了,这是人类第一次观测到该现象。

据此回答4~5 题。

4.卡林顿观测到的现象是A.耀斑B. 黑子C. 磁暴D.极光5.该现象剧烈爆发时,对地球的影响是A.为地球提供光热资源B. 全球许多国家发生强烈地震C.干扰无线电短波通信D.引起高层大气出现云雨天气图2为"某经线附近莫霍界面的深度分布示意图"。

读图,回答6~7题。

6.下列叙述正确的是( )A.从图中可以看出地壳厚度不均B. 莫霍界面以上为岩石圈C. 大洋地壳一定比大陆地壳厚D.莫霍界面是岩浆发源地7.2019年9月24日,巴基斯坦发生5.8级地震,震源深度10km。

(word)2021年江苏七市(南通 泰州 扬州 徐州 淮安 连云港 宿迁 )第三次模拟英语及答案

(word)2021年江苏七市(南通 泰州 扬州 徐州 淮安 连云港 宿迁 )第三次模拟英语及答案

南通市2021届高三第三次调研测试英语(含答案)注意事项考生在答题前请认真阅读本注意事项及各题答题要求1. 答题前,考生务必将自己的姓名、考生号、考场号、座位号填写在答题卡上。

2. 回答选择题时,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,请用橡皮擦干净后,再选涂其他答案标号。

回答非选择题时,将答案写在答题卡上,写在本试卷上无效。

3. 考试结束后,将本试卷和答题卡一并交回。

第一部分听力 (共两节,满分30分)做题时, 先将答案标在试卷上。

录音内容结束后, 你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节 (共5小题;每小题1. 5分,满分7. 5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项。

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. What is the weather like today?A. Dry.B. Rainy.C. Sunny.2. When did the woman first call?A. In the morning.B. In the afternoon.C. In the evening.3. What color is the car?A. Black.B. Grey.C. Blue.4. When did the boy start playing football?A. At 7 p. m.B. At 8 p. m.C. At 9 p. m.5. What are the speakers probably doing?A. Learning a line.B. Shooting a film.C. Buying a camera.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三选出最准选项。

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高考英语模拟试题高三年级第三次模拟考试(三)英语本试卷共12页,满分120分,考试时间120分钟。

第一部分听力(共两节,满分20分)第一节(共5小题;每小题1分,满分5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有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. What does the man mean?A. She should take more exercise.B. She'd better have a few days' rest.C. She is badly ill.( ) 2. When will Mr. White be free?A. This Saturday.B. Next Friday.C. Next Sunday.( ) 3. Where does the conversation most probably take place?A. At a bookstore.B. At a post office.C. At a supermarket.( ) 4. How many people are mentioned in the dialogue?A. At least four.B. Only three.C. More than five.( ) 5. How much may the man spend on the chair in the end?A. $15.B. $ 25.C. $ 20.第二节(共15小题;每小题1分,满分15分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

听第6段材料,回答第6至7题。

( ) 6. What does the woman think of the London taxi?A. Rather slow.B. Comfortable.C. Too expensive.( ) 7. In the woman's opinion, what is the best way to travel in London?A. By bus.B. By underground.C. On foot.听第7段材料,回答第8至10题。

( ) 8. Who are the two speakers?A. A man and his servant.B. A man and his wife.C. A man and his sister.( ) 9. What does the man always do?A. He often helps with the housework.B. He often makes the room dirty.C. He often enters the room without taking off his shoes.( )10. Why does the woman go away?A. The man doesn't give her money for her work.B. The man drops cigarette ash everywhere in the house.C. The man doesn't think housework is important.听第8段材料,回答第11至13题。

( )11. What's the man looking for?A. Some records of guitar music.B. Some tapes of modern music.C. Some records of piano music.( )12. Why does the man want to buy some records of Julian Bream?A. He wanted to buy the records for his father as a gift.B. He damaged the records his father gave him in his suitcase.C. He saw them in the shop window the week before.( )13. How many records will the man buy?A. Three.B. Four.C. Five.听第9段材料,回答第14至16题。

( )14. Who is busy?A. Mrs. Brown.B. The man's wife.C. The man himself.( )15. Who's going to look after the man's daughter?A. The nurses.B. The man.C. The man's wife.( )16. What's the possible relationship between them?A. Colleagues.B. Employer and employee.C. Neighbors.听第10段材料,回答第17至20题。

( )17. How do Nancy and Harry feel without their children?A. Frightened.B. Nice.C. Worried.( )18. What do Nancy and Harry do after they do the dishes?A. Watch TV or see a movie.B. Drive to town and do some shopping.C. Watch TV or play cards.( )19. Why do they drive to town sometimes?A. To see a film.B. To take a walk.C. To visit friends.( )20. How far is the state park from their house?A. One mile.B. Half a mile.C. A few miles.第二部分:英语知识运用(共两节,满分35分)第一节:单项填空(共15小题;每小题1分,满分15分)请认真阅读下面各题,从题中所给的A、B、C、D四个选项中,选出最佳选项。

( )21. “ ________ I went through ups and downs in life,”Gordon said, “I never found the importance of being self­disciplined as well as the significance of life.”A. IfB. SinceC. UntilD. Unless( )22. Instead of hiding behind walls, a defender sometimes must ________ engage and disable enemy forces before they strike.A. flexiblyB. activelyC. primitivelyD. conservatively( )23. The draft regulation on the online protection of minors, ________ for public opinions by the cyberspace authorities, has drawn wide public concern.A. releasingB. having releasedC. to releaseD. released( )24. Any reform of the government should be ________ the popular mood and times.A. in tune withB. in preference toC. in light ofD. in terms of( )25. We sell a lot of products offshore and the opportunity to open up markets in regions ________ we don't currently sell a lot to is a great one.A. whereB. thatC. whatD. when( )26. Ladies and gentlemen, on behalf of our government, I'd like to ________ a sincerewelcome and heartfelt gratitude.A. exploitB. exposeC. expandD. extend( )27. New energy vehicle­sharing projects ________ in dozens of cities across the country to fuel China's sharing economy in the next few years.A. are to carry outB. are being carried outC. were carried outD. will have been carried out( )28. The British government published an official policy document ________ its plans to bring the UK out of the European Union.A. letting outB. putting outC. setting outD. working out( )29. —I know it is really a lot to ask, but can I use your apartment during the summer?—________. I happen to be out of town. It is all for your taking.A. Behave yourselfB. Be my guestC. Have funD. Take care( )30. —Did you pass your driving test?—Yes, otherwise I ________ to the picnic next month.A. couldn't driveB. couldn't have drivenC. won't driveD. wouldn't have driven( )31. In the Name of People, a popular TV drama, shows a deep ________ on the fight against corruption.A. argumentB. themeC. reflectionD. impression( )32. —What a consequence!—Yes. I ________ him about it, but without success.A. will remindB. would remindC. was remindingD. had reminded( )33. It is estimated that China has over 770 million 4G users, which doubles ________ it was in 2015.A. thatB. whichC. whatD. how( )34. —What's wrong with him? He seemed upset.—He had to give up his drawing, not because he wanted ________ that way but because he had to be.A. thisB. oneC. itD. such( )35. I am at a loss why Mike is always ________ every time I meet him. Believe it or not, we used to be best friends.A. giving me the cold shoulderB. making my dayC. following suitD. beating around the bush第二节:完形填空(共20小题;每小题l分,满分20分)请认真阅读下面短文,从短文后各题所给的A、B、C、D四个选项中,选出最佳选项。

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