2018年上海闵行区高三一模试题附答案
闵行区上海高三年级质量调研考试英语试卷-讲解稿

闵行区2018学年第一学期高三年级质量调研考试英语试卷I. Listening ComprehensionSection A1. A. By car. B. On foot. C. By bus. D. By bike.2. A. A policeman. B. A postman. C. A doctor. D. A teacher.3. A. He didn’t sleep well last night. B. He did too much work last night.C. He went to bed late last night.D. He worked late with his work.4. A. The man has just missed his flight. B. The plane is delayed due to bad weather.C. The plane will leave at 9:14.D. The departure time is unknown.5. A. Offering a suggestion. B. Starting an argument.C. Stopping a fight.D. Correcting a mistake.6. A. Apply for a discount. B. Read the agreement.C. Fill in the application form.D. Buy a certain product.7. A. The woman was too tired to see the TV programme.B. The man missed part of the TV programme.C. Both of the speakers found the TV programme boring.D. The man regretted wasting time on the TV programme.8. A. Select the data. B. Revise the report. C. Collect more data. D. Present the report.9. A. Go to bed earlier. B. Learn how to be attentive.C. Spend more time outdoors.D. Take her job more seriously.10. A. He feels sorry for the students. B. He is strongly against the punishment.C. He approves of the professor’s action.D. He offers an option to stop cheating. Section BQuestions 11 through 13 are based on the following passage.11. A. Volunteer work and study. B. Work and relaxation.C. Adventurous trip and project.D. Cultural study and local visiting.12. A. It offers ways to learn a new language. B. It helps broaden your horizons.C. It guarantees you to earn money.D. It might take you to unusual places13. A. Volunteering in foreign countries. B. Ways to spend a working holiday.C. Great places for a working holiday.D. The concept of a working holiday. Questions 14 through 16 are based on the following passage.14. A. Committed. B. Flexible. C. Independent. D. Agreeable.15. A. They easily get wounded when doing exercise.B. They feel uncomfortable when they are observed.C. They are serious about their exercise plan.D. They are suitable to take up co-operative sports.16. A. To help people understand what personality type they are.B. To explain how personal characteristics affect exercise habits.C. To identify the dangers of doing the wrong type of exercise.D. To describe different types of exercises available at present.Questions 17 through 20 are based on the following conversation.17. A. Student opinions on a biology program. B. The present situation of biology education.C. The treatment and status of biology professors.D. The quality of biology laboratory equipment.18. A. Incapable teaching staff. B. Inaccessible laboratory.C. Insufficient budget.D. Unmotivated students.19. A. It’s unsuitable for complex experiments. B. It’s too complicated to use.C. It’s more than satisfactory.D. It’s much better than expected.20. A. The professors should be more devoted to the program.B. Some professors may go elsewhere to teach.C. Some professors can’t get budge to conduct experiments.D. The professors aren’t academ ically recognized.II. Grammar and vocabularySection AWe want our children to succeed in school and, perhaps even more importantly, in life. But the paradox(悖论) is that our children can only truly succeed (21) when/if______ they first learn how to fail. Consider the finding that world-class figure skaters fall over more often in practice than low-level figure skaters. Why are the really good skaters falling over the most?The reason is actually quite simple. Top skaters are constantly challenging themselves in practice. (22) To stretch______ (stretch) their limitations, they keep trying their best. They fall over so often, but it is precisely (exactly)why they learn so fast. Lower-level skaters have a quite different approach. They are always attempting jumps they can already do very easily, (23) remaining ______ (remain) within their comfort zone. This is why they don’t fall over. In a superficial sense,从表面上看they look successful, because they are always on their feet胸有成竹. Never (24) failing______ (fail) in practice prevents them from making progress.(25) _What_____ is true of skating is also true of life. James Dyson worked through 5,126 prototypes (原型) for his newest vacuum before coming up with the design(26) which/that______ made his fortune.致富These failures were essential to the pathway of learning. As Dyson put(27) it正如他所说______: “You can’t develop new technology unless you test new ideas and learn when things go wrong. Failure is essential to invention.”In healthcare, however, things are very different. Clinicians don’t like to admit (to)failure, partly because they have strong egos (自我) —particularly the senior doctors—and partly because they fear litigation (诉讼). The consequence is that (28) _instead of_____ ______ learning from failure, healthcare often covers up failure. The direct consequence is that the same mistakes (29) are repeated ______ (repeat). According to the Journal of Patient Safety, 400,000 people die every year in American hospitals alone due to preventable error.(30) ______Until/Unless healthcare learns to respond positively to failure, things will not improve.21. if / when 22. To stretch 23. remaining 24. failing25. What 26.which / that 27. it 28.instead of29. are repeated 30. Until / Unlessagree on is that what the kids use and expect from their world has changed rapidly. And it’s all because of technology.devices set this new , even from their Millennial (千禧年的) elders, who are quite familiar with technology.The younger group was so evident to psychologist Larry a new generation in a new book, Rewired: Understanding the ingeneration and the Way They Learn, out next month. Rosen says the technically G. dominated 35 life experience技术主导的生活体验of those born since the early 1990s is so different from the Millennial elders he wrote about in his 2007 book, Me, MySpace and I: Parenting the Net Generation, that they distinguish themselves as a new generation, which he hasRosen says portability is the key. 可移植性是关键。
精选上海市闵行区2018学年度第一学期高三数学(一模)期末质量监控试题(精品解析)

上海市闵行区2018学年度第一学期高三数学(一模)期末质量监控试卷一、选择题(本大题共4小题,共12.0分)1.若a,b为实数,则“”是“”的A. 充要条件B. 充分非必要条件C. 必要非充分条件D. 既非充分必要条件【答案】B【解析】【分析】根据充分条件和必要条件的概念,即可判断出结果.【详解】解不等式得或;所以由“”能推出“或”,反之不成立,所以“”是“”的充分不必要条件.故选B【点睛】本题主要考查充分条件与必要条件的概念,熟记概念即可,属于基础题型.2.已知a,b为两条不同的直线,,为两个不同的平面,,,则下面结论不可能成立的是A. ,且B.C. ,且D. b与,都相交【答案】D【解析】【分析】由点线面的位置关系,结合题中条件,即可分析出结果.【详解】因为a,b为两条不同的直线,,为两个不同的平面,,,所以有以下三种情况:(1)若,则;(2)若,则;(3)若且,则且;因此不可能b与,都相交.故选D【点睛】本题主要考查空间中线面位置关系,由线线平行,分类讨论线面关系即可,属于基础题型.3.已知函数,与其反函数有交点,则下列结论正确的是A. B.C. D.a与b的大小关系不确定【答案】B【解析】【分析】由函数与其反函数有交点,可得函数与直线有交点,进而可得出结果.【详解】因为函数,与其反函数有交点,所以函数与直线有交点,即方程有实根,整理得,所以,又,所以.故选B【点睛】本题主要考查反函数的概念,原函数与反函数有交点,必然与直线有交点,由此即可求解,属于基础题型.4.在平面直角坐标系中,已知向量,O是坐标原点,M是曲线上的动点,则的取值范围A. B. C. D.【答案】A【解析】【分析】先设,由M是曲线上的动点,得到,再由向量数量积运算的坐标表示,即可求出结果.【详解】设,则,因为M是曲线上的动点,所以,又,所以;因为,所以的取值范围是.故选A【点睛】本题主要考查向量数量积的坐标运算,熟记公式即可,属于常考题型.二、填空题(本大题共12小题,共36.0分)5.已知全集,集合,则______.【答案】【解析】【分析】解不等式得到集合,进而可求出结果.【详解】解不等式得或,所以集合或,因为,所以.故答案为【点睛】本题主要考查补集的运算,熟记概念即可,属于基础题型.6.______.【答案】【解析】【分析】在原式的基础上,分子分母同除以,进而可求出结果.【详解】因为.故答案为【点睛】本题主要考查型极限,只需分子分母同除以即可得出结果,属于基础题型.7.若复数z满足是虚数单位,则______.【答案】【解析】【分析】由先得到,再由复数的除法运算即可得出结果.【详解】因为,所以.故答案为【点睛】本题主要考查复数的运算,熟记除法运算法则即可,属于基础题型.8.方程的解为______.【答案】【解析】【分析】方程可化为,求解即可.【详解】由得即,解得.故答案为【点睛】本题主要考查矩阵,由矩阵的运算转化为含指数的方程,即可求解,属于基础题型. 9.等比数列中,,,则______.【答案】256【解析】【分析】先设等比数列的公比为,根据题中条件求出,进而可求出结果.【详解】设等比数列的公比为,因为,,所以,因此,所以.故答案为256【点睛】本题主要考查等比数列的性质,熟记等比数列性质即可,属于基础题型.10.的展开式中项的系数为___.(用数字表示)【答案】【解析】试题分析:由得:项的系数为.考点:二项展开式定理求特定项11.已知两条直线:,:,则与的距离为______.【答案】【解析】【分析】将:化为,再由平行线间的距离公式即可求出结果.【详解】因为:可化为,所以与的距离为.故答案为【点睛】本题主要考查两条平行线间的距离公式,熟记公式即可,属于基础题型.12.已知函数,的值域为,则的取值范围是______.【答案】【解析】【分析】由作出其图像,由值域为,即可求出结果.【详解】因为,作出其图像如下:因为函数,的值域为,所以由图像可得,;所以.故答案为【点睛】本题主要考查函数的性质,根据函数的值域求参数范围,通常需要作出函数图像,由数形结合的思想来处理,属于常考题型.13.如图,在过正方体的任意两个顶点的所有直线中,与直线异面的直线的条数为______.【答案】12【解析】【分析】由异面直线的概念,一一列举出与异面的直线即可.【详解】由题中正方体可得与异面的直线有:,,,,,;,,,,,,共12条.故答案为12【点睛】本题主要考查异面直线,熟记概念即可,属于基础题型.14.在中,角A,B,C的对边分别为a,b,c,面积为S,且,则______.【答案】0【解析】【分析】由三角形面积公式和余弦定理可将化为,进而可求出结果.【详解】因为,余弦定理,又,所以有,即,所以,因此或,所以或,因为C三角形内角,所以,故.故答案为0【点睛】本题主要考查解三角形,熟记余弦定理和三角形面积公式即可求出结果,属于常考题型.15.已知向量,,且,若向量满足,则的最大值为______.【答案】【解析】【分析】先由题中条件求出,再由即可求出结果.【详解】因为,,且所以,所以,因此.故的最大值为【点睛】本题主要考查向量的模的最值问题,根据向量模的几何意义,即可求解,属于常考题型.16.若无穷数列满足:,当,时.其中表示,,,中的最大项,有以下结论:若数列是常数列,则若数列是公差的等差数列,则;若数列是公比为q的等比数列,则则其中正确的结论是______写出所有正确结论的序号【答案】【解析】【分析】根据题中条件,逐项判断即可.【详解】若数列是常数列,则有,所以,又,所以,故,又,所以,即.故正确;若数列是公差的等差数列,若,则数列是递增数列,则,则,,不能满足数列为公差的等差数列;若,则数列是递减数列,则,所以满足题意;故正确;若数列是公比为q的等比数列,若q>1,由可知数列是递增数列,所以,所以,即q=2满足题意;若0<q<1,由可知数列是递减数列,所以,所以,故,因为0<q<1,所以显然不成立,故0<q<1不满足题意;若q<0,则数列是摆动数列,不能满足题意;综上q>1,故正确.故答案为【点睛】本题主要考查数列的应用,灵活运用数列的性质是解题的关键,难度较大.三、解答题(本大题共5小题,共60.0分)17.如图,正三棱柱的各棱长均为2,D为棱BC的中点.求该三棱柱的表面积;求异面直线AB与所成角的大小.【答案】(1);(2).【解析】【分析】根据棱柱的表面积公式直接求解即可;先取AC中点E,连结DE,,根据题意可得是异面直线AB与所成角,解三角形即可. 【详解】解:正三棱柱的各棱长均为2,该三棱柱的表面积:.取AC中点E,连结DE,,为棱BC的中点,,,是异面直线AB与所成角或所成角的补角,,,,异面直线AB与所成角的大小为.【点睛】本题主要考查几何体的表面积公式以及异面直线所成的角,在几何体中作出异面直线所成的角即可,属于基础题型.18.已知抛物线C:.若C上一点到其焦点的距离为3,求C的方程;若,斜率为2的直线l交C于两点,交x轴的正半轴于点M,O为坐标原点,求点M的坐标.【答案】(1);(2).【解析】【分析】根据抛物线的定义,由C上一点到其焦点的距离为3,可求出,进而可求出抛物线方程;由先求出抛物线方程,再设直线l:,代入抛物线方程,设,,结合韦达定理和判别式,根据求出的值即可.【详解】解:由抛物线的定义得:,解得:,所以抛物线C的方程为:;时,抛物线C:,设直线l:,并代入抛物线C:得:,,解得设,,则,,,解得或当时,不在x轴正半轴上,舍去;当时,故点M的坐标为【点睛】本题主要考查抛物线的方程与简单性质,通常需要联立直线与抛物线方程,结合韦达定理和题中条件求解,属于常考题型.19.在股票市场上,投资者常根据股价每股的价格走势图来操作,股民老张在研究某只股票时,发现其在平面直角坐标系内的走势图有如下特点:每日股价元与时间天的关系在ABC段可近似地用函数的图象从最高点A到最低点C的一段来描述如图,并且从C 点到今天的D点在底部横盘整理,今天也出现了明显的底部结束信号.老张预测这只股票未来一段时间的走势图会如图中虚线DEF段所示,且DEF段与ABC段关于直线l:对称,点B,D的坐标分别是.请你帮老张确定a,,的值,并写出ABC段的函数解析式;如果老张预测准确,且今天买入该只股票,那么买入多少天后股价至少是买入价的两倍?【答案】(1),,,,;(2)16.【解析】【分析】由B,D的坐标确定的值,和C的坐标,进而确定周期,求出,再由C的坐标,求出,即可得出函数解析式;(2)由(1)线求出DEF的解析式,令,求出即可.【详解】解:因为B,D的坐标分别是,且DEF段与ABC段关于直线l:对称,所以,所以,,,,由可得,,.由题意得DEF的解析式为:,由,得,故买入天后股价至少是买入价的两倍.【点睛】本题主要考查三角函数的应用,熟记三角函数的图像和性质即可,属于常考题型.20.对于函数,若函数是增函数,则称函数具有性质A.若,求的解析式,并判断是否具有性质A;判断命题“减函数不具有性质A”是否真命题,并说明理由;若函数具有性质A,求实数k的取值范围,并讨论此时函数在区间上零点的个数.【答案】(1),具有性质A;(2)假命题;(3)详见解析.【解析】【分析】由,结合即可得出解析式,和单调性,进而可得出结果;判断命题“减函数不具有性质A”,为假命题,举出反例即可,如;若函数具有性质A,可知在为增函数,进而可求出实数k的取值范围;再令,则在区间上零点的个数,即是的根的个数,结合k 的取值范围,即可求出结果.【详解】解:,,在R上递增,可知具有性质A;命题“减函数不具有性质A”,为假命题,比如:,在R上递增,具有性质A;若函数具有性质A,可得在递增,可得,解得;由,可得,即,可得,时显然成立;时,,由在递减,且值域为,时,或1,有三解,3个零点;当时,,即,可得,1个零点;当时,,t有一解,x两解,即两个零点;当,且时,无解,即x无解,无零点.【点睛】本题主要考查函数的解析式与函数的单调性,以及函数零点问题,按照题中条件结合函数的性质分析即可,属于常考题型.21.对于数列,若存在正数p,使得对任意都成立,则称数列为“拟等比数列”.已知,且,若数列和满足:,且,.若,求的取值范围;求证:数列是“拟等比数列”;已知等差数列的首项为,公差为d,前n项和为,若,,,且是“拟等比数列”,求p的取值范围请用,d表示.【答案】(1)详见解析;(2).【解析】【分析】由即可求出结果;根据题中“拟等比数列”的定义,由,结合条件推出存在正数,使得有成立即可;由题中条件,,,先求出的范围;再根据是“拟等比数列”,分类讨论和,即可得出结果.【详解】解:,,且,,,.由题意得,当且时,,对任意,都有,即存在,使得有,数列数列是“拟等比数列”;,,,,,,由得,从而解得,又是“拟等比数列”,故存在,使得成立,当时,,,由得,由图象可知在时递减,故,当时,,,由得,由图象可知在时递减,故,由得p的取值范围是.【点睛】本题主要考查数列的应用,根据题中的新定义,结合条件,分类讨论即可求出结果,过程较繁琐,难度较大.。
上海市闵行区2018学年第一学期高三年级期末(一模)质量调研考试语文试卷(含答案)

闵行区2018学年第一学期高三年级质量调研考试语文试卷1、本考试设试着和答题纸两部分,试卷包括试题与答题要求,所有答题必须写在答题纸上,做在试着上一律不得分2、答题纸与试卷在试题编号上是一一对应的,答题时应注意,不能错位。
3、考试时150分钟。
试卷满分150分。
一、积累与运用(10分)1、按要求填空(5分)(1)分野中峰变,___________________。
(王维《终南山》)(2)郴江幸自绕郴山,___________________。
(秦观《_________・郴州旅舍》)(3)李白《登金陵凤凰台》一诗中,暗示皇帝被奸邪包围,自己报国无门,心情沉痛的句子是“___________________,___________________。
”2、按要求选择(5分)(1)下列选项中,名句运用不恰当的一项是()。
(2分)A.生活中,老李是个很直率的人,和人交往向来是“知无不言,言无不尽”。
B.时代的洪流滚滚向前,改变了千年来“日出而作,日落而息”的生活方式。
C.对待传统文化,我们应如鲁迅先生所说的,做到“取其精华,去其糟粕”。
D.不少同学作文时往往是“下笔千言,离题万里”,搜肠刮肚也找不到材料。
(2)将下列编号的语句依次填入语段的空白处,语意连贯的一项是()。
(3分)规则意识的淡漠或者缺失,不仅仅是关乎个人素质的问题,__________,没有成熟的规则意识,同样会影响人与人之间的交流成本,让合作变得艰难。
①增加了个体在社会运行中不可预料的风险②因为没有规则意识契约精神就不可能成为社会主流③往大处说是阻碍社会文明进步的硬伤④往小处说将直接影响每个人的生活⑤没有契约精神何谈文明A.④①⑤③②B.③②⑤④①C.④①③②⑤ D.③①④②⑤二、阅读(70分)(一)阅读下文,完成第3-7题(16分)海洋,孕育上海海派文化时平(1)上海文化以引领时尚著称,在国际上也是一门引人关注的显学,而对上海海洋文化的研究却属于新生事物。
年上海市闵行区高考数学一模试卷

2018年上海市闵行区高考数学一模试卷一.填空题(本大题共12题,1-6每题4分,7-12每题5分,共54分)1.(4分)集合P={x|0≤x<3,x∈Z},M={x|x2≤9},则P∩M=. 2.(4分)计算=.3.(4分)方程的根是.4.(4分)已知是纯虚数(i是虚数单位),则= .(4分)已知直线l的一个法向量是,则l的倾斜角的大小是.5.6.(4分)从4名男同学和6名女同学中选取3人参加某社团活动,选出的3人中男女同学都有的不同选法种数是(用数字作答)7.(5分)在(1+2x)5的展开式中,x2项系数为(用数字作答)8.(5分)如图,在直三棱柱ABC﹣A1B1C1中,∠ACB=90°,AC=4,BC=3,AB=BB1,则异面直线A1B与B1C1所成角的大小是(结果用反三角函数表示)9.(5分)已知数列{an}、{b n}满足b n=lna n,n∈N*,其中{b n}是等差数列,且,则b1+b2+…+b1009= .10.(5分)如图,向量与的夹角为120°,,,P是以O为圆心,为半径的弧上的动点,若,则λμ的最大值是.11.(5分)已知F1、F2分别是双曲线(a>0,b>0)的左右焦点,过F1且倾斜角为30°的直线交双曲线的右支于P,若PF2⊥F1F2,则该双曲线的渐近线方程是.12.(5分)如图,在折线ABCD中,AB=BC=CD=4,∠ABC=∠BCD=120°,E、F分别是AB、CD的中点,若折线上满足条件的点P至少有4个,则实数k 的取值范围是.二.选择题(本大题共4题,每题5分,共20分)13.(5分)若空间中三条不同的直线l1、l2、l3,满足l1⊥l2,l2∥l3,则下列结论一定正确的是( )A.l1⊥l3ﻩB.l1∥l3C.l1、l3既不平行也不垂直 D.l1、l3相交且垂直14.(5分)若a>b>0,c<d<0,则一定有()A.ad>bc B.ad<bcﻩC.ac>bdD.ac<bd15.(5分)无穷等差数列{an}的首项为a1,公差为d,前n项和为S n(n∈N*),则“a1+d>0”是“{Sn}为递增数列”的( )条件.A.充分非必要B.必要非充分C.充要D.既非充分也非必要16.(5分)已知函数(n<m)的值域是[﹣1,1],有下列结论:①当n=0时,m∈(0,2];②当时,;③当时,m∈[1,2];④当时,m∈(n,2];其中结论正确的所有的序号是()A.①②ﻩB.③④C.②③ D.②④三.解答题(本大题共5题,共14+14+14+16+18=76分)17.(14分)已知函数(其中ω>0).(1)若函数f(x)的最小正周期为3π,求ω的值,并求函数f(x)的单调递增区间;(2)若ω=2,0<α<π,且,求α的值.18.(14分)如图,已知AB是圆锥SO的底面直径,O是底面圆心,,AB=4,P是母线SA的中点,C是底面圆周上一点,∠AOC=60°.(1)求圆锥的侧面积;(2)求直线PC与底面所成的角的大小.19.(14分)某公司举办捐步公益活动,参与者通过捐赠每天的运动步数获得公司提供的牛奶,再将牛奶捐赠给留守儿童,此活动不但为公益事业作出了较大的贡献,公司还获得了相应的广告效益,据测算,首日参与活动人数为10000人,以后每天人数比前一天都增加15%,30天后捐步人数稳定在第30天的水平,假设此项活动的启动资金为30万元,每位捐步者每天可以使公司收益0.05元(以下人数精确到1人,收益精确到1元).(1)求活动开始后第5天的捐步人数,及前5天公司的捐步总收益;(2)活动开始第几天以后公司的捐步总收益可以收回启动资金并有盈余?20.(16分)已知椭圆的右焦点是抛物线Γ:y2=2px的焦点,直线l与Γ相交于不同的两点A(x1,y1)、B(x2,y2).(1)求Γ的方程;(2)若直线l经过点P(2,0),求△OAB的面积的最小值(O为坐标原点);(3)已知点C(1,2),直线l经过点Q(5,﹣2),D为线段AB的中点,求证:|AB|=2|CD|.21.(18分)对于函数y=f(x)(x∈D),如果存在实数a、b(a≠0,且a=1,b=0不同时成立),使得f(x)=f(ax+b)对x∈D恒成立,则称函数f(x)为“(a,b)映像函数”.(1)判断函数f(x)=x2﹣2是否是“(a,b)映像函数”,如果是,请求出相应的a、b 的值,若不是,请说明理由;(2)已知函数y=f(x)是定义在[0,+∞)上的“(2,1)映像函数”,且当x∈[0,1)时,f(x)=2x,求函数y=f(x)(x∈[3,7))的反函数;},使得当x∈[a n,an+1)(n∈N*)时,2(3)在(2)的条件下,试构造一个数列{anx+1∈[an,an+2),并求x∈[a n,a n+1)(n∈N*)时,函数y=f(x)的解析式,及y+1=f(x)(x∈[0,+∞))的值域.ﻬ2018年上海市闵行区高考数学一模试卷参考答案与试题解析一.填空题(本大题共12题,1-6每题4分,7-12每题5分,共54分)1.(4分)集合P={x|0≤x<3,x∈Z},M={x|x2≤9},则P∩M={0,1,2} .【解答】解:∵集合P={x|0≤x<3,x∈Z}={0,1,2},M={x|x2≤9}={x|﹣3≤x≤3},∴P∩M={0,1,2}.故答案为:{0,1,2}.2.(4分)计算=.【解答】解:===,故答案为:.3.(4分)方程的根是10 .【解答】解:∵,即1+lgx﹣3+lgx=0,∴lgx=1,∴x=10.故答案为:10.4.(4分)已知是纯虚数(i是虚数单位),则=.【解答】解:∵是纯虚数,∴,得sin且cos,∴α为第二象限角,则cos.∴=sinαcos+cosαsin=.故答案为:﹣.5.(4分)已知直线l的一个法向量是,则l的倾斜角的大小是.【解答】解:设直线l的倾斜角为θ,θ∈[0,π).设直线的方向向量为=(x,y),则=x﹣y=0,∴tanθ==,解得θ=.故答案为:.6.(4分)从4名男同学和6名女同学中选取3人参加某社团活动,选出的3人中男女同学都有的不同选法种数是96 (用数字作答)【解答】解:根据题意,在4名男同学和6名女同学共10名学生中任取3人,有C103=120种,其中只有男生的选法有C43=4种,只有女生的选法有C63=20种则选出的3人中男女同学都有的不同选法有120﹣4﹣20=96种;故答案为:96.7.(5分)在(1+2x)5的展开式中,x2项系数为40(用数字作答)【解答】解:设求的项为T r+1=C5r(2x)r,今r=2,∴T3=22C52x2=40x2.∴x2的系数是408.(5分)如图,在直三棱柱ABC﹣A1B1C1中,∠ACB=90°,AC=4,BC=3,AB=BB1,则异面直线A1B与B1C1所成角的大小是arccos(结果用反三角函数表示)【解答】解:∵在直三棱柱ABC﹣A1B1C1中,∠ACB=90°,AC=4,BC=3,AB=BB1, BC∥B1C1,∴∠ABC是异面直线A1B与B1C1所成角,1∵A1B===5,A1C===,∴cos∠A1BC===.∴∠A1BC=arccos.∴异面直线A1B与B1C1所成角的大小是arccos.故答案为:arccos.9.(5分)已知数列{an}、{bn}满足b n=lna n,n∈N*,其中{b n}是等差数列,且,则b1+b2+…+b1009=2018.【解答】解:数列{a n}、{bn}满足bn=lnan,n∈N*,其中{b n}是等差数列,∴b n+1﹣bn=lnan+1﹣lna n=ln=常数t.∴=常数e t=q>0,}为等比数列.因此数列{an且,∴a1a1009=a2a1008==….则b1+b2+…+b1009=ln(a1a2…a1009)==lne2018=2018.故答案为:2018.10.(5分)如图,向量与的夹角为120°,,,P是以O为圆心,为半径的弧上的动点,若,则λμ的最大值是.【解答】解:如图建立平面直角坐标系,设P(cosθ,sinθ),,,.∵,∴,sinθ=.∴,∴λμ=﹣+=+,故答案为:11.(5分)已知F1、F2分别是双曲线(a>0,b>0)的左右焦点,过F1且倾斜角为30°的直线交双曲线的右支于P,若PF2⊥F1F2,则该双曲线的渐近线方程是y=±x .|=m,|PF2|=n,|F1F2|=2c,【解答】解:设|PF1在直角△PF1F2中,∠PF1F2=30°,可得m=2n,则m﹣n=2a=n,即a=n,2c=n,即c=n,b==n,可得双曲线的渐近线方程为y=±x,即为y=±x,故答案为:y=±x.12.(5分)如图,在折线ABCD中,AB=BC=CD=4,∠ABC=∠BCD=120°,E、F 分别是AB、CD的中点,若折线上满足条件的点P至少有4个,则实数k 的取值范围是(﹣,﹣2) .【解答】解:以BC的垂直平分线为y轴,以BC为x轴,建立如图所示的平面直角坐标系,∵AB=BC=CD=4,∠ABC=∠BCD=120°,∴B(﹣2.0),C(2,0),A(﹣4,2),D(4,2),∵E、F分别是AB、CD的中点,∴E(﹣3,),F(3,),设P(x,y),﹣4≤x≤4,0≤y≤2,∵,∴(﹣3﹣x,﹣y)(3﹣x,﹣y)=x2+(y﹣)+9=k,即x2+(y﹣)﹣9=k+9,当k+9>0时,点P的轨迹为以(0,)为圆心,以为半径的圆,当圆与直线DC相切时,此时圆的半径r=,此时点有2个,当圆经过点C时,此时圆的半径为r==,此时点P有4个,∵满足条件的点P至少有4个,结合图象可得,∴<k+9<7,解得﹣<k<﹣2,故实数k的取值范围为(﹣,﹣2),故答案为:(﹣,﹣2)二.选择题(本大题共4题,每题5分,共20分)13.(5分)若空间中三条不同的直线l1、l2、l3,满足l1⊥l2,l2∥l3,则下列结论一定正确的是()A.l1⊥l3ﻩB.l1∥l3C.l1、l3既不平行也不垂直ﻩD.l1、l3相交且垂直【解答】解:∵空间中三条不同的直线l、l2、l3,满足l1⊥l2,l2∥l3,1∴l1⊥l3,故选:A.14.(5分)若a>b>0,c<d<0,则一定有()A.ad>bcﻩB.ad<bcﻩC.ac>bdD.ac<bd【解答】解:∵c<d<0,∴﹣c>﹣d>0.又a>b>0,则一定有﹣ac>﹣bd,可得ac<bd.故选:D.15.(5分)无穷等差数列{an}的首项为a1,公差为d,前n项和为Sn(n∈N*),则“a1+d>0”是“{Sn}为递增数列”的()条件.A.充分非必要ﻩB.必要非充分C.充要 D.既非充分也非必要【解答】解:等差数列{an}的首项为a1,公差为d,前n项和为Sn=na1+d,则Sn+1=(n+1)a1+,则S n+1﹣S n=(n+1)a1+﹣na1﹣d=a1+nd,若{Sn}为递增数列,a1+nd>0,∵S2﹣S1=a1+d>0,∴a1+nd>0不能推出a1+d>0但a1+d能推出a1+nd,故a1+d>0”是“{Sn}为递增数列必要非充分,故选:B16.(5分)已知函数(n<m)的值域是[﹣1,1],有下列结论:①当n=0时,m∈(0,2];②当时,;③当时,m∈[1,2];④当时,m∈(n,2];其中结论正确的所有的序号是( )A.①②B.③④C.②③D.②④【解答】解:当x>1时,x﹣1>0,f(x)=22﹣x+1﹣3=23﹣x﹣3,单调递减,当﹣1<x<1时,f(x)=22+x﹣1﹣3=21+x﹣3,单调递增,∴f(x)=22﹣|x﹣1|﹣3在(﹣1,1)单调递增,在(1,+∞)单调递减,∴当x=1时,取最大值为1,∴绘出f(x)的图象,如图:①当n=0时,f(x)=,由函数图象可知:要使f(x)的值域是[﹣1,1],则m∈(1,2];故①错误;②当时,f(x)=,f(x)在[﹣1,]单调递增,f(x)的最大值为1,最小值为﹣1,∴;故②正确;③当时,m∈[1,2];故③正确,④错误,故选C.三.解答题(本大题共5题,共14+14+14+16+18=76分)17.(14分)已知函数(其中ω>0).(1)若函数f(x)的最小正周期为3π,求ω的值,并求函数f(x)的单调递增区间;(2)若ω=2,0<α<π,且,求α的值.【解答】解:(1)函数=sin(ωx),∵函数f(x)的最小正周期为3π,即T=3π=∴ω=那么:,由,k∈Z,得:∴函数f(x)的单调递增区间为,k∈Z;(2)函数=sin(ωx),∵ω=2∴f(x)=sin(2x),,可得sin(2α)=∵0<α<π,∴≤(2α)≤2α=或解得:α=或α=.18.(14分)如图,已知AB是圆锥SO的底面直径,O是底面圆心,,AB=4,P是母线SA的中点,C是底面圆周上一点,∠AOC=60°.(1)求圆锥的侧面积;(2)求直线PC与底面所成的角的大小.【解答】解:(1)∵AB是圆锥SO的底面直径,O是底面圆心,,AB=4, P是母线SA的中点,C是底面圆周上一点,∠AOC=60°.∴r==2,l===4,∴圆锥的侧面积S=πrl=π×2×4=8π.(2)过点P作PE⊥圆O,交AO于E,连结CE,则E是AO中点,∴PE=PO=,CE==,∴∠PCE是直线PC与底面所成角,∵PE=CE,PE⊥CE,∴,∴直线PC与底面所成的角为.19.(14分)某公司举办捐步公益活动,参与者通过捐赠每天的运动步数获得公司提供的牛奶,再将牛奶捐赠给留守儿童,此活动不但为公益事业作出了较大的贡献,公司还获得了相应的广告效益,据测算,首日参与活动人数为10000人,以后每天人数比前一天都增加15%,30天后捐步人数稳定在第30天的水平,假设此项活动的启动资金为30万元,每位捐步者每天可以使公司收益0.05元(以下人数精确到1人,收益精确到1元).(1)求活动开始后第5天的捐步人数,及前5天公司的捐步总收益;(2)活动开始第几天以后公司的捐步总收益可以收回启动资金并有盈余?【解答】解:(1)设第x天的捐步人数为x,则f(x)=.∴第5天的捐步人数为f(5)=10000•(1+15%)4=17490.由题意可知前5天的捐步人数成等比数列,其中首项为10000,公比为1.15,∴前5天的捐步总收益为×0.05=3371;(2)设活动第x天后公司捐步总收益可以回收并有盈余,①若1≤x≤30,则×0.05>300000,91≈32.3(舍).解得x>log1.15②若x>30,则[+10000•1.1529•(x﹣30)]•0.05>300000,解得x>32.87.∴活动开始后第33天公司的捐步总收益可以收回启动资金并有盈余. 20.(16分)已知椭圆的右焦点是抛物线Γ:y2=2px的焦点,直线l与Γ相交于不同的两点A(x1,y1)、B(x2,y2).(1)求Γ的方程;(2)若直线l经过点P(2,0),求△OAB的面积的最小值(O为坐标原点);(3)已知点C(1,2),直线l经过点Q(5,﹣2),D为线段AB的中点,求证:|AB|=2|C D|.【解答】(1)解:由椭圆,得a2=10,b2=9,则c=1.∴椭圆的右焦点,即抛物线Γ:y2=2px的焦点为(1,0),则,p=2,∴Γ的方程为y2=4x;(2)解:设直线l:x=my+2,联立,得y2﹣4my﹣8=0.则y1+y2=4m,y1y2=﹣8.∴==,即△OAB的面积的最小值为;(3)证明:当AB所在直线斜率存在时,设直线方程为y+2=k(x﹣5),即y=kx﹣5k ﹣2.联立,可得ky2﹣4y﹣20k﹣8=0.,.=.===.∵C(1,2),∴,,则=(x1﹣1)(x2﹣1)+(y1﹣2)(y2﹣2)=x1x2﹣(x1+x2)+1+y1y2﹣2(y1+y2)+4=,当AB所在直线斜率不存在时,直线方程为x=5,联立,可得A(5,﹣),B(5,2),,,有,∴CA⊥CB,又D为线段AB的中点,∴|AB|=2|CD|.21.(18分)对于函数y=f(x)(x∈D),如果存在实数a、b(a≠0,且a=1,b=0不同时成立),使得f(x)=f(ax+b)对x∈D恒成立,则称函数f(x)为“(a,b)映像函数”.(1)判断函数f(x)=x2﹣2是否是“(a,b)映像函数”,如果是,请求出相应的a、b的值,若不是,请说明理由;(2)已知函数y=f(x)是定义在[0,+∞)上的“(2,1)映像函数”,且当x∈[0,1)时,f(x)=2x,求函数y=f(x)(x∈[3,7))的反函数;(3)在(2)的条件下,试构造一个数列{an},使得当x∈[a n,a n+1)(n∈N*)时,2x+1∈[a n+1,a n+2),并求x∈[an,a n+1)(n∈N*)时,函数y=f(x)的解析式,及y=f(x)(x∈[0,+∞))的值域.【解答】解:(1)由f(x)=x2﹣2,可得f(ax+b)=(ax+b)2﹣2=a2x2+2abx+b2﹣2,由f(x)=f(ax+b),得x2﹣2=a2x2+2abx+b2﹣2,则,∵a≠0,且a=1,b=0不同时成立,∴a=﹣1,b=0.∴函数f(x)=x2﹣2是“(﹣1,0)映像函数”;(2)∵函数y=f(x)是定义在[0,+∞)上的“(2,1)映像函数”,∴f(x)=f(2x+1),则f(0)=f(1)=f(3),f(1)=f(3)=f(7),∴f(0)=f(3),f(1)=f(7),而当x∈[0,1)时,f(x)=2x,∴x∈[3,7)时,设f(x)=2sx+t,由,解得s=,t=﹣.∴x∈[3,7)时,f(x)=.令y=(3≤x<7),得,∴x=(1≤y<2),∴函数y=f(x)(x∈[3,7))的反函数为y=(1≤x<2);(3)由(2)可知,构造数列{an},满足a1=0,a n+1=2an+1,+1=2(an+1),则an+1∴数列{an+1}是以1为首项,以2为公比的等比数列,则,即.当x∈[a n,an+1)=[2n﹣1﹣1,2n﹣1).令,解得s=21﹣n,t=21﹣n﹣1.∴x∈[a n,a n+1)(n∈N*)时,函数y=f(x)的解析式为f(x)=.当x∈[0,+∞)时,函数f(x)的值域为[1,2).。
2018-2019学年上海闵行区高三一模数学试卷及答案

高三年级质量调研考试数学试卷 第1页共11页闵行区2018学年第一学期高三年级质量调研考试数 学 试 卷(满分150分,时间120分钟)考生注意:1.答卷前,考生务必在答题纸上将学校、班级、考生号、姓名等填写清楚.2.请按照题号在答题纸各题答题区域内作答,超出答题区域书写的答案无效;在草稿纸、试题卷上答题无效. 3.本试卷共有21道试题.一、填空题(本大题共有12题,满分54分)考生应在答题纸上相应编号的空格内直接填写结果,第1~6题每个空格填对得4分,第7~12题每个空格填对得5分,否则一律得零分. 1. 已知全集U =R ,集合2{30}A x x x =-≥,则U A ð= .2. 2221lim 331n n n n →∞-=++ . 3. 若复数z 满足(12)43i z i +=+(i 是虚数单位),则z = . 4. 方程110322x=-的解为 .5. 等比数列}{n a 中,121=+a a ,5616a a +=,则910a a += .6. ()512x -的展开式中3x 项的系数为 .(用数字作答)7. 已知两条直线12:4230:2+10l x y l x y +-=+=和,则12l l 与的距离为 . 8. 已知函数[]()|1|(1),,f x x x x a b =-+∈的值域为[]0 8,,则a b +的取值范围是 . 9. 如图,在过正方体1111ABCD A B C D -的任意两个顶点的所有直线中,与直线1AC 异面的直线的条数为 . 10. 在ABC △中,角 A B C 、、的对边分别为 a b c 、、,面积为S ,且224()S a b c =+-,则cos C = .高三年级质量调研考试数学试卷 第2页共11页11. 已知向量()()cos ,sin ,cos ,sin a b ααββ==,且3παβ-=,若向量c 满足1c a b --=,则c 的最大值为 .12. 若无穷数列{}n a 满足:10a ≥,当*,2n n ∈≥N 时,{}1121max ,,,n n n a a a a a ---=(其中{}121max ,,,n a a a -表示121,,,n a a a -中的最大项),有以下结论:①若数列{}n a 是常数列,则()*0n a n =∈N ; ②若数列{}n a 是公差0d ≠的等差数列,则0d <; ③若数列{}n a 是公比为q 的等比数列,则1q >;④若存在正整数T ,对任意*n ∈N ,都有n T n a a +=,则1a 是数列{}n a 的最大项.则其中的正确结论是 .(写出所有正确结论的序号)二、选择题(本大题共有4题,满分20分)每题有且只有一个正确答案,考生应在答题纸的相应编号上,将代表答案的小方格涂黑,选对得5分,否则一律得零分. 13. 若,a b 为实数,则“1a <-”是“11a>-”的 ( )(A) 充要条件 (B) 充分非必要条件(C) 必要非充分条件 (D) 既非充分也非必要条件 14. 已知a 、b 为两条不同的直线,α、β为两个不同的平面, a αβ=, //a b ,则下列结论不可能...成立的是 ( ) (A) //b b βα,且Ü (B) //b b αβ,且Ü (C) // //b b αβ,且 (D) b 与 αβ、都相交15.已知函数(),0,0y x a a b =≥>>与其反函数有交点,则下列结论正确的是 ( )(A) a b = (B) a b < (C) a b > (D)a b 与的大小关系不确定 16. 在平面直角坐标系中,已知向量(1,2)a =,O 是坐标原点,M 是曲线22x y +=上的动点,则a OM ⋅的取值范围为 ( ) (A) []2,2-(B) ⎡⎣(C) ⎡⎢⎣⎦(D)⎡⎢⎣高三年级质量调研考试数学试卷 第3页共11页三、解答题(本大题满分76分)本大题共有5题,解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤.17. (本题满分14分)本题共有2个小题,第1小题满分6分,第2小题满分8分. 如图,正三棱柱111ABC A B C -的各棱长均为2,D 为棱BC 的中点. (1)求该三棱柱的表面积;(2)求异面直线AB 与1C D 所成角的大小.18. (本题满分14分)本题共有2个小题,第1小题满分6分,第2小题满分8分.已知抛物线Γ:22(0)y px p =≠.(1)若Γ上一点(1,)M t 到其焦点的距离为3,求Γ的方程;(2)若2p =,斜率为2的直线l 交Γ于两点A B 、,交x 轴的正半轴于点M ,O 为坐标原点,0OA OB ⋅=,求点M 的坐标.高三年级质量调研考试数学试卷 第4页共11页19. (本题满分14分)本题共有2个小题,第1小题满分8分,第2小题满分6分. 在股票市场上,投资者常根据股价(每股的价格)走势图来操作.股民老张在研究某只股票时,发现其在平面直角坐标系内的走势图有如下特点:每日股价()y 元与时间()x 天的关系在ABC 段可近似地用函数()()sin 200,0,0y a x a ωϕωϕπ=++>><<的图像从最高点A 到最低点C 的一段来描述(如右图),并且从C 点到今天的D 点在底部横盘整理,今天也出现了明显的底部结束信号.老张预测这只股票未来一段时间的走势图会如图中虚线DEF 段所示,且DEF 段与ABC 段关于直线:34l x =对称,点B D 、的坐标分别是()()12,2044,12、. (1)请你帮老张确定,,a ωϕ的值,并写出ABC 段的函数解析式;(2)如果老张预测准确,且今天买入该只股票,那么买入多少天后股价至少是买入价的两倍?高三年级质量调研考试数学试卷 第5页共11页20. (本题满分16分)本题共有3个小题,第1小题满分4分,第2小题满分6分,第3小题满分6分.对于函数()y f x =,若函数()(1)()F x f x f x =+-是增函数,则称函数()y f x =具有性质A .(1)若2()2xf x x =+,求()F x 的解析式,并判断()f x 是否具有性质A ; (2)判断命题“减函数不具有性质A ”是否真命题,并说明理由;(3)若函数23()(0)f x kx x x =+≥具有性质A ,求实数k 的取值范围,并讨论此时函数()(sin )sin g x f x x =-在区间[]0,π上零点的个数.高三年级质量调研考试数学试卷 第6页共11页21. (本题满分18分)本题共有3个小题,第1①小题满分4分,第1②小题满分6分,第2小题满分8分.对于数列{}n a ,若存在正数p ,使得1n n a pa +≤对任意*n ∈N 都成立,则称数列{}n a 为“拟等比数列”.(1)已知0 0a b >>,,且a b >,若数列{}n a 和{}n b满足:11 2a ba b +==,且*11 )2n nn n a b a b n +++==∈N ,. ①若11a =,求1b 的取值范围;②求证:数列*{}()n n a b n -∈N 是“拟等比数列”.(2)已知等差数列{}n c 的首项为1c ,公差为d ,前n 项和为n S ,若10c >,403540360,0S S ><,且{}n c 是“拟等比数列”,求p 的取值范围(请用1,c d 表示).高三年级质量调研考试数学试卷 第7页共11页闵行区2018学年第一学期高三年级质量调研考试数学试卷参考答案与评分标准一. 填空题 1.()0,3; 2.23; 3.2i -; 4.2log 5x =; 5.256; 6.80-; 7.2;8.[]2 4,;9.12; 10.0; 11.1+12.①②③④. 二. 选择题 13.B ; 14.D ; 15.B ; 16.A . 三. 解答题17.[解] (1)22=32+22S ⨯ ………6分 (2)取AC 的中点E ,联结1DE C E 、,则//ED AB ,所以1C DE ∠(或其补角)是异面直线AB 与1C D 所成的角.…8分 在1C DE △中,11C D C E =1DE =,所以2221111cos 210C D DE C E C DE C D DE +-∠===⋅⋅.………12分 所以AB 与1C D所成的角的大小为. …………14分 18.[解] (1)由条件知1342pp +=⇒=, ………………4分 所以Γ的方程为28y x =. ………………………6分 (2)设点M A B 、、的坐标分别为(),0(0)m m >、()11,A x y ,()22,B x y , 则直线l 的方程为2()y x m =-; ………………………8分{222()2404y x m y y m y x=-⇒--==, ………………………10分 2221212124,16y y y y m x x m ⇒=-== ………………………12分21212404OA OB x x y y m m m ⋅=+=-+=⇒=,高三年级质量调研考试数学试卷 第8页共11页所以M 点的坐标为()4,0. ………………………14分19.[解] (1)因为C D 、两点关于直线l 对称,所以点C 的坐标为()24,12,……2分又点B 恰在平衡位置,C 为最低点,得2448248124ππω==⇒=⇒=T T …4分 将()12,20B 代入解析式可得:sin 120cos 024πϕϕ⎛⎫⋅+=⇒= ⎪⎝⎭,∴2πϕ=, …………………………6分再结合C 点是最低点,可得8a =. ∴ABC 段的解析式为[]8sin +20,0,24242y x x ππ⎛⎫=+∈⎪⎝⎭……………………8分 (2)由对称性得,DEF 段的解析式为:()()[]8sin 68+208cos 68+20,44,6824224y x x x πππ⎡⎤⎡⎤=-+=-∈⎢⎥⎢⎥⎣⎦⎣⎦,…10分若股价至少是买入价的两倍,则()8cos 68+202424x π⎡⎤-≥⎢⎥⎣⎦ ………………………………12分()1cos 68242x π⎡⎤⇒-≥⎢⎥⎣⎦解得6068x ≤≤所以买入16天后,股价至少是买入价的两倍. …………………………14分 20.[解] (1)212()(1)()(1)22221x x x F x f x f x x x x +=+-=++--=++ ……2分而()221xF x x =++在(),-∞+∞上是增函数,所以()f x 是否具有性质A . ………………………………4分 (2)假命题. ………………………………6分如函数()0)f x x =≥是减函数, ………………………………8分高三年级质量调研考试数学试卷 第9页共11页()F x ==函数()F x 在[0 )+∞,上单调递增,∴()0)f x x =≥具有性质P .∴命题是假命题. ………………………………10分 (3) 23232()(1)(1)3(23)1(0)F x k x x kx x x k x k x =+++--=++++≥, 因为函数23()(0)f x kx x x =+≥具有性质A , 所以233062k k +-≤⇒≥- . ………………………………12分 23()(sin )sin sin sin sin g x f x x k x x x =-=+-,由()0g x =得23sin sin sin 0sin 0k x x x x +-=⇒=或2sin sin 1k x x =-+0x ⇒=或x π=或1sin sin k x x ⎛⎫=-- ⎪⎝⎭()0,x π∈. …………………14分设sin x t =,则(]1,0,1k t t t ⎛⎫=--∈ ⎪⎝⎭由函数(]1,0,1k t t t ⎛⎫=--∈ ⎪⎝⎭的图像可知当3,02k ⎡⎫∈-⎪⎢⎣⎭时,11t t =>,1sin x t =无解;当0k =时,11t =,sin 1x =⇒2x π=;当()0,k ∈+∞时,()10,1t t =∈,1sin x t =在()0,π上有两个解;综上所述:当3,02k ⎡⎫∈-⎪⎢⎣⎭时,()g x 在区间[]0,π上零点的个数为2; 当0k =时,()g x 在区间[]0,π上零点的个数为3;当()0,k ∈+∞时,()g x 在区间[]0,π上零点的个数为4.………………16分 21.[解] (1)①∵0 0a b >>,,且a b >,112a ba +==,∴11b <,高三年级质量调研考试数学试卷 第10页共11页∴1(0 1)b ∈,, ……………………………………4分②依题意得:112a ba b +=>= 所以,当* 2n n ∈≥N ,时,1102n n n n a b a b --+-=>,……………6分 所以对任意*n ∈N ,都有111()222n n n n n n n n a b a b a b a b ++++-=<=-, ………………8分 即存在12p =,使得11()n n n n a b p a b ++-<-,∴数列*{}(N )n n a b n -∈是“拟等比数列”.……………………………………10分(2)()201840351403640364035004036002c S c c S ⋅>⎧>⎧⎪⇒⎨⎨+⋅<<⎩⎪⎩…………………12分 201820181201820192019100201700020180c c c d c c c c d >>+>⎧⎧⎧⇒⇒⇒⎨⎨⎨+<<+<⎩⎩⎩ 由10c >可知0d <,从而解得120182017c d-<<-, …………………14分 又{}n c 是“拟等比数列”,故存在0p >,使得1n n c pc +≤1︒当2018n ≤时,0n c >,()()+11111111111n n c c n d dp c c c n d c n dn d +⋅≥==+=++-⋅+-⋅⎛⎫-- ⎪⎝⎭由1120182017201812019c cd d-<<-⇒<-<, 由图像可知1111c n d +⎛⎫-- ⎪⎝⎭在2018n ≤时递减,故211201620171,20172018c d p c c ⎛⎫≥=+∈ ⎪⎝⎭; ………………………16分高三年级质量调研考试数学试卷 第11页共11页2︒当2019n ≥时,0n c <,()()+11111111111n n c c n d d p c c c n d c n d n d +⋅≤==+=++-⋅+-⋅⎛⎫-- ⎪⎝⎭ 由1120182017201812019c c d d-<<-⇒<-<, 由图像可知1111c n d +⎛⎫-- ⎪⎝⎭在2019n ≥时递减,故1p ≤; 由12︒︒可得,此时p 的取值范围是111d c ⎡⎤+⎢⎥⎣⎦, ………………………18分。
2018年上海市闵行区高考数学一模试卷及答案

2018年上海市闵行区高考数学一模试卷一.填空题(本大题共12题,1-6每题4分,7-12每题5分,共54分)1.(4分)集合P={x|0≤x<3,x∈Z},M={x|x2≤9},那么P∩M= .2.(4分)计算= .3.(4分)方程的根是.4.(4分)已知是纯虚数(i是虚数单位),那么= .5.(4分)已知直线l的一个法向量是,那么l的倾斜角的大小是.6.(4分)从4名男同窗和6名女同窗当选取3人参加某社团活动,选出的3人中男女同窗都有的不同选法种数是(用数字作答)7.(5分)在(1+2x)5的展开式中,x2项系数为(用数字作答)8.(5分)如图,在直三棱柱ABC﹣A1B1C1中,∠ACB=90°,AC=4,BC=3,AB=BB1,那么异面直线A1B与B1C1所成角的大小是(结果用反三角函数表示)9.(5分)已知数列{a n}、{b n}知足b n=lna n,n∈N*,其中{b n}是等差数列,且,那么b1+b2+…+b1009= .10.(5分)如图,向量与的夹角为120°,,,P是以O 为圆心,为半径的弧上的动点,假设,那么λμ的最大值是.11.(5分)已知F1、F2别离是双曲线(a>0,b>0)的左右核心,过F1且倾斜角为30°的直线交双曲线的右支于P,假设PF2⊥F1F2,那么该双曲线的渐近线方程是.12.(5分)如图,在折线ABCD中,AB=BC=CD=4,∠ABC=∠BCD=120°,E、F别离是AB、CD的中点,假设折线上知足条件的点P至少有4个,那么实数k的取值范围是.二.选择题(本大题共4题,每题5分,共20分)13.(5分)假设空间中三条不同的直线l1、l2、l3,知足l1⊥l2,l2∥l3,那么以下结论必然正确的选项是()A.l1⊥l3B.l1∥l3C.l1、l3既不平行也不垂直D.l1、l3相交且垂直14.(5分)假设a>b>0,c<d<0,那么必然有()A.ad>bc B.ad<bc C.ac>bd D.ac<bd15.(5分)无穷等差数列{an }的首项为a1,公差为d,前n项和为Sn(n∈N*),那么“a1+d>0”是“{Sn}为递增数列”的()条件.A.充分非必要 B.必要非充分C.充要D.既非充分也非必要16.(5分)已知函数(n<m)的值域是[﹣1,1],有以下结论:①当n=0时,m∈(0,2];②当时,;③当时,m∈[1,2];④当时,m∈(n,2];其中结论正确的所有的序号是()A.①②B.③④C.②③D.②④三.解答题(本大题共5题,共14+14+14+16+18=76分)17.(14分)已知函数(其中ω>0).(1)假设函数f(x)的最小正周期为3π,求ω的值,并求函数f(x)的单调递增区间;(2)假设ω=2,0<α<π,且,求α的值.18.(14分)如图,已知AB是圆锥SO的底面直径,O是底面圆心,,AB=4,P是母线SA的中点,C是底面圆周上一点,∠AOC=60°.(1)求圆锥的侧面积;(2)求直线PC与底面所成的角的大小.19.(14分)某公司举行捐步公益活动,参与者通过捐赠天天的运动步数取得公司提供的牛奶,再将牛奶捐赠给留守儿童,此活动不但为公益事业作出了较大的奉献,公司还取得了相应的广告效益,据测算,首日参与活动人数为10000人,以后天天人数比前一天都增加15%,30天后捐步人数稳固在第30天的水平,假设此项活动的启动资金为30万元,每位捐步者天天能够使公司收益0.05元(以下人数精准到1人,收益精准到1元).(1)求活动开始后第5天的捐步人数,及前5天公司的捐步总收益;(2)活动开始第几天以后公司的捐步总收益能够收回启动资金并有盈余?20.(16分)已知椭圆的右核心是抛物线Γ:y2=2px的核心,直线l与Γ相交于不同的两点A(x1,y1)、B(x2,y2).(1)求Γ的方程;(2)假设直线l通过点P(2,0),求△OAB的面积的最小值(O为坐标原点);(3)已知点C(1,2),直线l通过点Q(5,﹣2),D为线段AB的中点,求证:|AB|=2|CD|.21.(18分)关于函数y=f(x)(x∈D),若是存在实数a、b(a≠0,且a=1,b=0不同时成立),使得f(x)=f(ax+b)对x∈D恒成立,那么称函数f(x)为“(a,b)映像函数”.(1)判定函数f(x)=x2﹣2是不是是“(a,b)映像函数”,若是是,请求出相应的a、b的值,假设不是,请说明理由;(2)已知函数y=f(x)是概念在[0,+∞)上的“(2,1)映像函数”,且当x∈[0,1)时,f(x)=2x,求函数y=f(x)(x∈[3,7))的反函数;(3)在(2)的条件下,试构造一个数列{an },使适当x∈[an,an+1)(n∈N*)时,2x+1∈[an+1,an+2),并求x∈[an,an+1)(n∈N*)时,函数y=f(x)的解析式,及y=f(x)(x∈[0,+∞))的值域.2018年上海市闵行区高考数学一模试卷参考答案与试题解析一.填空题(本大题共12题,1-6每题4分,7-12每题5分,共54分)1.(4分)集合P={x|0≤x<3,x∈Z},M={x|x2≤9},那么P∩M= {0,1,2} .【解答】解:∵集合P={x|0≤x<3,x∈Z}={0,1,2},M={x|x2≤9}={x|﹣3≤x≤3},∴P∩M={0,1,2}.故答案为:{0,1,2}.2.(4分)计算= .【解答】解:===,故答案为:.3.(4分)方程的根是10 .【解答】解:∵,即1+lgx﹣3+lgx=0,∴lgx=1,∴x=10.故答案为:10.4.(4分)已知是纯虚数(i是虚数单位),那么= .【解答】解:∵是纯虚数,∴,得sin且cos,∴α为第二象限角,那么cos.∴=sinαcos+cosαsin=.故答案为:﹣.5.(4分)已知直线l的一个法向量是,那么l的倾斜角的大小是.【解答】解:设直线l的倾斜角为θ,θ∈[0,π).设直线的方向向量为=(x,y),那么=x﹣y=0,∴tanθ==,解得θ=.故答案为:.6.(4分)从4名男同窗和6名女同窗当选取3人参加某社团活动,选出的3人中男女同窗都有的不同选法种数是96 (用数字作答)【解答】解:依照题意,在4名男同窗和6名女同窗共10名学生中任取3人,有C103=120种,其中只有男生的选法有C43=4种,只有女生的选法有C63=20种那么选出的3人中男女同窗都有的不同选法有120﹣4﹣20=96种;故答案为:96.7.(5分)在(1+2x)5的展开式中,x2项系数为40 (用数字作答)【解答】解:设求的项为Tr+1=C5r(2x)r,今r=2,∴T3=22C52x2=40x2.∴x2的系数是408.(5分)如图,在直三棱柱ABC﹣A1B1C1中,∠ACB=90°,AC=4,BC=3,AB=BB1,那么异面直线A1B与B1C1所成角的大小是arccos(结果用反三角函数表示)【解答】解:∵在直三棱柱ABC﹣A1B1C1中,∠ACB=90°,AC=4,BC=3,AB=BB1,BC∥B1C1,∴∠A1BC是异面直线A1B与B1C1所成角,∵A1B===5,A1C===,∴cos∠A1BC===.∴∠A1BC=arccos.∴异面直线A1B与B1C1所成角的大小是arccos.故答案为:arccos.9.(5分)已知数列{an }、{bn}知足bn=lnan,n∈N*,其中{bn}是等差数列,且,那么b1+b2+…+b1009= 2018 .【解答】解:数列{an }、{bn}知足bn=lnan,n∈N*,其中{bn}是等差数列,∴bn+1﹣bn=lnan+1﹣lnan=ln=常数t.∴=常数e t=q>0,因此数列{an}为等比数列.且,∴a1a1009=a2a1008==….那么b1+b2+…+b1009=ln(a1a2…a1009)==lne2018=2018.故答案为:2018.10.(5分)如图,向量与的夹角为120°,,,P是以O 为圆心,为半径的弧上的动点,假设,那么λμ的最大值是.【解答】解:如图成立平面直角坐标系,设P(cosθ,sinθ),,,.∵,∴,sinθ=.∴,∴λμ=﹣+=+,故答案为:11.(5分)已知F1、F2别离是双曲线(a>0,b>0)的左右核心,过F 1且倾斜角为30°的直线交双曲线的右支于P,假设PF2⊥F1F2,那么该双曲线的渐近线方程是y=±x .【解答】解:设|PF1|=m,|PF2|=n,|F1F2|=2c,在直角△PF1F2中,∠PF1F2=30°,可得m=2n,那么m﹣n=2a=n,即a=n,2c=n,即c=n,b==n,可得双曲线的渐近线方程为y=±x,即为y=±x,故答案为:y=±x.12.(5分)如图,在折线ABCD中,AB=BC=CD=4,∠ABC=∠BCD=120°,E、F别离是AB、CD的中点,假设折线上知足条件的点P至少有4个,那么实【解答】解:以BC的垂直平分线为y轴,以BC为x轴,成立如下图的平面直角坐标系,∵AB=BC=CD=4,∠ABC=∠BCD=120°,∴B(﹣2.0),C(2,0),A(﹣4,2),D(4,2),∵E、F别离是AB、CD的中点,∴E(﹣3,),F(3,),设P(x,y),﹣4≤x≤4,0≤y≤2,∵,∴(﹣3﹣x,﹣y)(3﹣x,﹣y)=x2+(y﹣)+9=k,即x2+(y﹣)﹣9=k+9,当k+9>0时,点P的轨迹为以(0,)为圆心,以为半径的圆,当圆与直线DC相切时,现在圆的半径r=,现在点有2个,当圆通过点C时,现在圆的半径为r==,现在点P有4个,∵知足条件的点P至少有4个,结合图象可得,∴<k+9<7,解得﹣<k<﹣2,故答案为:(﹣,﹣2)二.选择题(本大题共4题,每题5分,共20分)13.(5分)假设空间中三条不同的直线l1、l2、l3,知足l1⊥l2,l2∥l3,那么以下结论必然正确的选项是()A.l1⊥l3B.l1∥l3C.l1、l3既不平行也不垂直D.l1、l3相交且垂直【解答】解:∵空间中三条不同的直线l1、l2、l3,知足l1⊥l2,l2∥l3,∴l1⊥l3,应选:A.14.(5分)假设a>b>0,c<d<0,那么必然有()A.ad>bc B.ad<bc C.ac>bd D.ac<bd【解答】解:∵c<d<0,∴﹣c>﹣d>0.又a>b>0,那么必然有﹣ac>﹣bd,可得ac<bd.应选:D.15.(5分)无穷等差数列{an }的首项为a1,公差为d,前n项和为Sn(n∈N*),那么“a1+d>0”是“{Sn}为递增数列”的()条件.A.充分非必要 B.必要非充分C.充要D.既非充分也非必要【解答】解:等差数列{an }的首项为a1,公差为d,前n项和为Sn=na1+d,那么Sn+1=(n+1)a1+,那么Sn+1﹣Sn=(n+1)a1+﹣na1﹣d=a1+nd,若{Sn }为递增数列,a1+nd>0,∵S2﹣S1=a1+d>0,∴a1+nd>0不能推出a1+d>0但a1+d能推出a1+nd,故a1+d>0”是“{Sn}为递增数列必要非充分,应选:B16.(5分)已知函数(n<m)的值域是[﹣1,1],有以下结论:①当n=0时,m∈(0,2];②当时,;③当时,m∈[1,2];④当时,m∈(n,2];其中结论正确的所有的序号是()A.①②B.③④C.②③D.②④【解答】解:当x>1时,x﹣1>0,f(x)=22﹣x+1﹣3=23﹣x﹣3,单调递减,当﹣1<x<1时,f(x)=22+x﹣1﹣3=21+x﹣3,单调递增,∴f(x)=22﹣|x﹣1|﹣3在(﹣1,1)单调递增,在(1,+∞)单调递减,∴当x=1时,取最大值为1,∴绘出f(x)的图象,如图:①当n=0时,f(x)=,由函数图象可知:要使f(x)的值域是[﹣1,1],那么m∈(1,2];故①错误;②当时,f(x)=,f(x)在[﹣1,]单调递增,f(x)的最大值为1,最小值为﹣1,∴;故②正确;③当时,m∈[1,2];故③正确,④错误,应选C.三.解答题(本大题共5题,共14+14+14+16+18=76分)17.(14分)已知函数(其中ω>0).(1)假设函数f(x)的最小正周期为3π,求ω的值,并求函数f(x)的单调递增区间;(2)假设ω=2,0<α<π,且,求α的值.【解答】解:(1)函数=sin(ωx),∵函数f(x)的最小正周期为3π,即T=3π=∴ω=那么:,由,k∈Z,得:∴函数f(x)的单调递增区间为,k∈Z;(2)函数=sin(ωx),∵ω=2∴f(x)=sin(2x),,可得sin(2α)=∵0<α<π,∴≤(2α)≤2α=或解得:α=或α=.18.(14分)如图,已知AB是圆锥SO的底面直径,O是底面圆心,,AB=4,P是母线SA的中点,C是底面圆周上一点,∠AOC=60°.(1)求圆锥的侧面积;(2)求直线PC与底面所成的角的大小.【解答】解:(1)∵AB是圆锥SO的底面直径,O是底面圆心,,AB=4,P是母线SA的中点,C是底面圆周上一点,∠AOC=60°.∴r==2,l===4,∴圆锥的侧面积S=πrl=π×2×4=8π.(2)过点P作PE⊥圆O,交AO于E,连结CE,那么E是AO中点,∴PE=PO=,CE==,∴∠PCE是直线PC与底面所成角,∵PE=CE,PE⊥CE,∴,∴直线PC与底面所成的角为.19.(14分)某公司举行捐步公益活动,参与者通过捐赠天天的运动步数取得公司提供的牛奶,再将牛奶捐赠给留守儿童,此活动不但为公益事业作出了较大的奉献,公司还取得了相应的广告效益,据测算,首日参与活动人数为10000人,以后天天人数比前一天都增加15%,30天后捐步人数稳固在第30天的水平,假设此项活动的启动资金为30万元,每位捐步者天天能够使公司收益0.05元(以下人数精准到1人,收益精准到1元).(1)求活动开始后第5天的捐步人数,及前5天公司的捐步总收益;(2)活动开始第几天以后公司的捐步总收益能够收回启动资金并有盈余?【解答】解:(1)设第x天的捐步人数为x,那么f(x)=.∴第5天的捐步人数为f(5)=10000•(1+15%)4=17490.由题意可知前5天的捐步人数成等比数列,其中首项为10000,公比为1.15,∴前5天的捐步总收益为×0.05=3371;(2)设活动第x天后公司捐步总收益能够回收并有盈余,①假设1≤x≤30,那么×0.05>300000,91≈32.3(舍).解得x>log1.15②假设x>30,那么[+10000•1.1529•(x﹣30)]•0.05>300000,解得x>32.87.∴活动开始后第33天公司的捐步总收益能够收回启动资金并有盈余.20.(16分)已知椭圆的右核心是抛物线Γ:y2=2px的核心,直线l与Γ相交于不同的两点A(x1,y1)、B(x2,y2).(1)求Γ的方程;(2)假设直线l通过点P(2,0),求△OAB的面积的最小值(O为坐标原点);(3)已知点C(1,2),直线l通过点Q(5,﹣2),D为线段AB的中点,求证:|AB|=2|CD|.【解答】(1)解:由椭圆,得a2=10,b2=9,那么c=1.∴椭圆的右核心,即抛物线Γ:y2=2px的核心为(1,0),则,p=2,∴Γ的方程为y2=4x;(2)解:设直线l:x=my+2,联立,得y2﹣4my﹣8=0.那么y1+y2=4m,y1y2=﹣8.∴==,即△OAB的面积的最小值为;(3)证明:当AB所在直线斜率存在时,设直线方程为y+2=k(x﹣5),即y=kx ﹣5k﹣2.联立,可得ky2﹣4y﹣20k﹣8=0.,.=.===.∵C(1,2),∴,,则=(x1﹣1)(x2﹣1)+(y1﹣2)(y2﹣2)=x1x2﹣(x1+x2)+1+y1y2﹣2(y1+y2)+4=,当AB所在直线斜率不存在时,直线方程为x=5,联立,可得A(5,﹣),B(5,2),,,有,∴CA⊥CB,又D为线段AB的中点,∴|AB|=2|CD|.21.(18分)关于函数y=f(x)(x∈D),若是存在实数a、b(a≠0,且a=1,b=0不同时成立),使得f(x)=f(ax+b)对x∈D恒成立,那么称函数f(x)为“(a,b)映像函数”.(1)判定函数f(x)=x2﹣2是不是是“(a,b)映像函数”,若是是,请求出相应的a、b的值,假设不是,请说明理由;(2)已知函数y=f(x)是概念在[0,+∞)上的“(2,1)映像函数”,且当x∈[0,1)时,f(x)=2x,求函数y=f(x)(x∈[3,7))的反函数;(3)在(2)的条件下,试构造一个数列{an },使适当x∈[an,an+1)(n∈N*)时,2x+1∈[an+1,an+2),并求x∈[an,an+1)(n∈N*)时,函数y=f(x)的解析式,及y=f(x)(x∈[0,+∞))的值域.【解答】解:(1)由f(x)=x2﹣2,可得f(ax+b)=(ax+b)2﹣2=a2x2+2abx+b2﹣2,由f(x)=f(ax+b),得x2﹣2=a2x2+2abx+b2﹣2,则,∵a≠0,且a=1,b=0不同时成立,∴a=﹣1,b=0.∴函数f(x)=x2﹣2是“(﹣1,0)映像函数”;(2)∵函数y=f(x)是概念在[0,+∞)上的“(2,1)映像函数”,∴f(x)=f(2x+1),那么f(0)=f(1)=f(3),f(1)=f(3)=f(7),∴f(0)=f(3),f(1)=f(7),而当x∈[0,1)时,f(x)=2x,∴x∈[3,7)时,设f(x)=2sx+t,由,解得s=,t=﹣.∴x∈[3,7)时,f(x)=.令y=(3≤x<7),得,∴x=(1≤y<2),∴函数y=f(x)(x∈[3,7))的反函数为y=(1≤x<2);(3)由(2)可知,构造数列{an },知足a1=0,an+1=2an+1,那么an+1+1=2(an+1),∴数列{an+1}是以1为首项,以2为公比的等比数列,则,即.当x∈[an ,an+1)=[2n﹣1﹣1,2n﹣1).令,解得s=21﹣n,t=21﹣n﹣1.∴x∈[an ,an+1)(n∈N*)时,函数y=f(x)的解析式为f(x)=.当x∈[0,+∞)时,函数f(x)的值域为[1,2).。
2018届上海市闵行区高考一模英语试题及答案

上海市闵行区2018届高考一模英语试题第I卷 (共 107 分)I. Listening ComprehensionSection ADirections: In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A. At a grocery. B. At a laundry. C. At a restaurant. D. At a post office.2. A. Generous. B. Considerate. C. Diligent.D. Impersonal.3. A. Travelling plan. B. Personal interest.C. Overseas study.D. Job opportunity.4. A. Having a break. B. Continuing themeeting.C. Moving on to the next item.D. Waiting a little longer.5. A. Take exercises. B. See a doctor. C. Havea test. D. Give a speech.6. A. It is a routine offer. B. It is quite healthy.C. It is new on the menu.D. It is a good bargain.7. A. She is driving fast to the airport. B. She may be late for the football game.C. She is worried about missing her flight.D. She is currently caught in a traffic jam.8. A. The man can stay in her brother’s apartment.B. Her brother can help the man find a cheaper hotel.C. Her brother can find an apartment for the man.D. The man should have booked a less expensive hotel.9. A. He was looking forward to seeing the giraffes.B. He enjoyed watching the animal performance.C. He got home too late to see the TV special.D. He fell asleep in the middle of the TV program.10. A. The man should consider his privacy first.B. The man will choose a low-rent apartment.C. The man is not certain if he can find a quieter place.D. The man is unlikely to move out of the dormitory. Section BDirections: In Section B, you will hear two short passages, and you will be asked three questions on each of the passages. The passages will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. The storm. B. The flood. C. The wind. D. The rain.12. A. Cold and dry. B. Cloudy and rainy.C. Warm and sunny.D. Windy and cool.13. A. On Friday morning. B. On Friday night.C. On Saturday morning.D. On Saturday night.Questions 14 through 16 are based on the following passage.14. A.It’ll allow them to receive free medical treatment.B.It’ll protect them from possible financial crise s.C.It’ll enable them to enjoy the best medical care.D.It’ll prevent the doctors from overcharging them .15.A.They may not be able to receive timely medical trea tment.B.They can only visit doctors who speak their native languages.C.They have to go through very complicated applicati on procedures.D.They can’t immediately get back the money paid fo r their medical cost.16. A. How to obtain student health insurance before going abroad.B. How to save money when you are buying overseas insurance.C. The importance of obtaining insurance when studying abroad.D. The advantages of student insurance over international travel insurance.Section CDirections: In Section C, you will hear two longer conversations. The conversations will be read twice. After you hear each conversation, you are required to fill in the numbered blanks with the information you have heard. Write your answers on your answer sheet. Blanks 17 through 20 are based on the following conversation.Complete the form. Write ONE WORD for each answer.Blanks 21 through 24 are based on the following conversation.Complete the form. Write no more than THREE WORDS for each answer.II. Grammar and VocabularySection ADirections: After reading the passage below, fill in the blanks to make the passage coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks,use one word that best fits each blank.(A)One of the most important skills in reading is the knowledge of words. Since Vocabulary Building and Word Study, another book in this series, (25) ______ (devote) to vocabulary development, we are not going to deal with this skill directly, but you should be aware of the basic fact (26) ______ the number of words you know determines the difficulty and complexity of the material you can read and understand. (27) ______ you have an extremely limited reading vocabulary, you will be able to read only very simple material. For example, children just beginning to read in first grade must use books that only have a few dozen words in them. As they progress through school their reading vocabulary keepsincreasing (28) ______ they are adults. Adults typically have reading vocabularies of over 50,000 words.Failure (29) ______(develop) an extensive reading vocabulary will definitely hold you back in your efforts to improve comprehension. Most people learn new words by encountering them in reading or conversation, (30) ______ the meaning is usually clear from the context. A few words are learned by looking them up in the dictionary.If you feel that vocabulary is a problem for you, you may wish to do something systematic about it. There are many good books (31) ______ to help develop vocabulary, including the one in this series. (32) ______(begin) with any of them will help your vocabulary.(B)Conversation is an important part of life to everyone. However some people are afraid of starting a casual conversation with a stranger (33) ______ ______ a fear of not having anything interesting to say.Fear of rejection is also a reason for keeping silent. Small talk in a conversation can serve as a way of warming up and getting to know each other. Most people, (34) ______ ______ ______ successful they are at work, will find the process of making small talk uncomfortable, even pointless. However, you (35) ______ (have) a pretty small world if you refuse to let others in. Starting off with (36) ______ usual comment or question will make others comfortable, and even just greeting them with a simple “hello” will often be enough.Once you have made the first move, there is no turning back. You need to keep things going before you can gracefully say goodbye. Think about what you would like to share with a new friend and that is (37) ______ you can ask the other person about. Work, family, hobbies or interests are some general points of discussion. (38) ______ (keep) the ball rolling, you can show that you are really interested in what they are saying. Once you find common points of interest, things will definitely take off from there.Small talk (39) ______ turn out to be a pleasantexperience. The conversation should be brief and casual without turning into a long and boring discussion. Do not let yourself control the conversation. Let (40) ______ talk too. Stop worrying about how big of a fool you may appear to be. You may even find out later on that you have the “talent” of getting people to open up to you by engaging them in small talk.Section BDirections: Complete the following passage by using the words in the box. Each word can only be used once. Note that there is one word more than you need.Job seekers should know the rules of labor market before they try to find a proper job. Most career organizations 41 three stages for graduates to follow the process of securing a suitable career:recognizing abilities, matching these to available jobs and presenting them well to possible employers.Job seekers have to make a careful assessment of their abilities. One area of assessment should be of their 42 qualifications, which include special skills within their subject area. Graduates should also consider their own values and attitudes. An honest assessment of personal interests and abilities such as creative skills, or skills acquired from work experience, should also be given careful 43 .The second stage is to study the opportunities which are available for employment. To do this, graduates can study job and 44 information in newspapers, or they can pay a visit to a careers office, write to friends or relatives who may already be 45 in a particular profession. After studying all the various 46 , they should be in a position to make informed comparisons between various careers.Good personal 47 is essential in the search for a good career. Job application forms should be filled in carefully and correctly, without grammar of spelling errors. They should also prepare properly byfinding out all they can about the possible employer. When 48information is asked for, job seekers should describe their abilities and work experience in more depth, as well as 49 their own abilities with the employer’s needs, explain why the y are interested in a career with the 50 company and try to show that they already know something about the company and its activities. Interviewees should try to give positive answers and not be afraid of asking questions about anything they are unsure about.III. Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.People read for different purposes. Sometimes people read for information while on other occasions, for understanding. Therefore, we can know that the word “reading” has two51 senses.The first sense is the one in which we read newspapers, magazines, and so on. We can get access to the content of those materials 52 . Such materials may increase our 53of information, but they cannot improve our understanding. And clearly we don’t have any difficulty in gaining the new information, for our understanding is 54 to them before we start. Otherwise, we would have felt the shock of puzzlement.The second sense is the one in which we read something that at first we do not completely 55 . Here the thing to be read is at the first sight better or higher than the reader. The writer is communicating something that can increase the reader’s understanding. Such 56 between unequals –people who know and people who don’t know – must be made possible. Otherwise one person could never learn from another. Here “learning” means understanding more, not 57 more information.What are the 58 in this kind of reading? First, there is inequality in understanding. The writer mustbe “ 59 ”to the reader in understanding. Besides, his book must convey something he possesses while his potential readers 60 . Second, the reader must be able to 61 this inequality in some degree. And he should always try to reach the same level of understanding with the writer. If the 62 is approached, success of communication is achieved.Besides gaining information and understanding, there’s another goal of reading:63 . It is the least 64 and requires the least amount of effort. Everyone who knows how to read can read for entertainment if he wants to. 65 , any book that can be read for understanding or information can probably be read for entertainment as well.51. A. meaningful B. broad C. separateD. informative52. A. easily B. wholly C. brieflyD. highly53. A. analysis B. comprehension C. appreciation D. store54. A. equal B. contrary C. inferiorD. devoted55. A. agree B. understand C. approve D. enjoy56. A. contact B. relationship C. methodD. communication57. A. selecting B. ignoring C. rememberingD. creating58. A. conditions B. procedures C. approachesD. purposes59. A. humble B. superior C. kind D. generous60. A. know B. acquire C. lack D. deny61. A. recognize B. follow C. neglect D. overcome62. A. principle B. equality C. informationD. content63. A. knowledge B. ability C. entertainment D. culture64. A. enjoyable B. reasonable C. flexibleD. demanding65. A. In fact B. On the contrary C. In addition D. On the other handSection BDirections: Read the following passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)Dennis Sinar, 51, a doctor from New York, is quick to explain why he took a year-long break from his job. “I was pretty burned out after practicing medicine for 26 years. I needed a recharge.” So he took a “gap year”, from July 2011 to June 2018, to explore things like ancient buildings, antique restoration, archaeology and traditional Eastern medicine, in locations including Alaska, Nepal and Romania.“Taking a break from work is an excellent way for adults to go into a new career or refresh an old one,”said Holly Bull, president of Princeton, N, J. “In recent years, mid-career breaks have been gaining more interest,” she said. A report on adult gap years published this year by a market research company also described the potential American market for gap yearsas a “sleeping giant.”“A gap year is a challenge for the older individual to step out of a comfort zone and take a risk. I enjoyed that side most.” said Dr. Sinar, who kept a daily blog about his experience. His time studying Eastern medicine “assured the reasons I went into health care,”said Dr. Sinar, who returned to practice medicine at his old job, although he works fewer days. “I use those experiences to provide my patients with more care,” he added. “And I listen better than I did before.”George Garritan, chairman of the Department of Leadership and Human Capital Management at New York University, certainly agrees with Dr. Sinar. He said a gap-year experience could be worthwhile for employees and companies. For employees, investing in themselves and improving skill sets is a move that will benefit throughout their career. He added that returning employees feel refreshed and have given more thought to their career. For companies, offering unpaid leaves makes good sense for attracting and keeping talented employees.66. Dr. Sinar took a gap year because he ________.A. had lost his old jobB. wanted to refresh after 26 years’ workC. had a desire for travellingD. became interested in historical research67. The phrase “sleeping giant” (in 2nd paragraph) indicates that ________.A. it’s too early for people to accept the conception of gap yearB. the effect of gap year policy remains to be seenC. it’s difficult to foresee the gap year marketD. more American people will accept the gap year policy68. What’s George Garritan’s at titude toward the “gap year”?A. Positive.B. DoubtfulC. Uninterested.D. Uncertain.69. What’s the passage mainly about?A. How an adult plans a mid-career gap year.B. Why a gap year is worthwhile for adults.C. Whether a gap year is popular with adults.D. Why a gap year is challenging for individuals.(B)Submitting Assignment OnlineIn order to upload an assignment to the system properly, you must save the assignment using one of the following applications: Microsoft Word, Corel WordPerfect, or Microsoft Excel.After you complete an assignment, it is important to save your work. This ensures that assignment being uploaded to the system is the most updated version. Your word processing program may attempt to save the assignment to a folder on your computer. We recommend creating a new folder, named after your course, in a location that is easy to remember, such as “My Documents.”File Name RequirementsSelect a file name for the assignment that is easy to remember. The file name must NOT contain spaces. Any spaces used in the file name will prevent the file from uploading to the system. File names using the extensions .rtf, .html, .zip, .jpg, or .exe are NOT allowed.Assignment Upload ProcedureWhen the file is ready to upload, follow these steps:1. Sign in the system.2. Enter your address information and click the CONTINUE button. You will be routed to the “Directions” screen.3. Review the directions and click the CONTINUE button. You will be routed to the “Special Assignment Upload” screen.4. Click the SEARCH button. The “Choose File” window pops up.5. Find the location on your computer where you saved your assignment, and select the file.After you have selected the file, click the UPLOAD button to upload your assignment to the system. Assignment Submission DatesAssignments must be submitted by midnight on the specified due date. Submit your assignment on time so that it reaches the system on or before the due date. No late assignments will be accepted or marked. Assignments delivered after the due date will not be assessed and will be failed. Please manage your timecarefully as family and work demands will not be accepted as excuses for late or non-submissions. Tutors and administrators do not have the authority to grant extensions.70. Which of the following file names is acceptable for submitting assignment?A. research paper.docB. research paper.pdfC. researchpaper.docD. researchpaper.rtf71. Students are advised to save an assignment to a specific folder in order to ________.A. sign in the systemB. remember where the document isC. choose the appropriate applicationD. remember the document name quickly72. The phrase “pops up” in the passage probably means “________”.A. appearsB. eliminatesC. scansD. browses73. If you summit your assignment after the due time, you ________.A. can have a second chanceB. should ask a tutor for helpC. may apply for an extensionD. will get a fail for the work(C)Even in a weak job market, the old college try is n’t the answer for everyone. A briefing paper from the Brookings Institution warns that “we may have overdone the message” on college, senior fellow Isabel Sawhill said.“We’ve been telling students and their families for years that college is the only way to succeed in the economy and of course there’s a lot of truth to that,” Ms. Sawhill said. “On average it does pay off… But if you load up on a whole lot of student debt and then you don’t graduate, that is a very bad situation.”One comment that people often repeat among the years of slow job growth has been the value of education for landing a job and advancing in a career. April’s national unemployment rate stood at 7.5%, according to the Labor Department. The unemployment rate forhigh-school gradua tes over 25 years old who hadn’t attended college was 7.4%, compared with 3.9% for those with a bachelor’s degree or more education. The difference is even bigger among those aged 16-24. The jobless rate for those with only a high school diploma in that age group is about 20%. At the same time, recent research by Canadian economists cautions that a college degree is no guarantee of promising employment.Ms. Sawhill pointed out that among the aspects that affect the value of a college education is the field of one’s major: Students in engineering or other sciences end up earning more than ones who major in the arts or education. The cost of tuition and the availability of financial aid are other considerations, with public institutions generally a better financial bargain than private ones.She suggested two ways for improving the situation: increasing vocational-technical training programs and taking a page from Europe’s focus on early education rather than post-secondary learning. “The European countries put a little more attention to getting people prepared in the primary grades,” she said. “Then theyhave a higher bar for whoever goes to college—but once you get into college, you’re more likely to be highly subsidized (资助).”She also is a supporter of technical training—to teach students how to be plumbers, welders and computer programmers—because “employers are desperate” for workers with these skills.74. According to the passage, people usually think that ________.A. the cost of technical schooling is a problemB. one will not succeed without a college degreeC. technical skills are most important for landing a jobD. there is an increased competition in getting intoa college75. The underlined part “taking a page from”probably means “ ________ ”A. revisingB. promotingC. definingD. adopting76. What can we infer from the passage?A. Public institutions charge more for education.B. European universities are stricter with students.C. Students with certain skills are in great demand.D. Canadian students prefer to major in engineering.77. Ms. Sawhill may probably agree that ________.A. too much stress has been put on the value of college degreesB. technical training is more important than college educationC. a college degree will ensure promising employmentD. it’s easier for art students to find favorite jobsSection CDirections: Read the passage carefully. Then answer the questions or complete the statements in the fewest possible words.Skeptics are a strange lot. Some of them refuse to admit the serious threat of human activities to the environment, and they are tired of people who disagree with them. Those people, say skeptics, spread nothing but bad news about the environment. The “eco-guilt”brought on by the discouraging news about our planetgives rise to the popularity of skeptics as people search for more comforting worldviews.Perhaps that explains why a new book by Bjorn Lomborg received so much publicity. That book, The Skeptical Environmentalist, declares that it measures the “real state of the world” as fine. Of course , another explanation is the deep pockets of some big businesses with special interests. Indeed, Mr. Lomborg's views are similar to those of some industry- funded organizations, which start huge activities through the media to confuse the public about issues like global warming.So it was strange to see Mr. Lomborg's book go largely unchallenged in the media though his beliefs were contrary to most scientific opinions. One national newspaper in Canada ran a number of articles and reviews full of words of praise, even with the conclusion that “After Lomborg, the environmental movement will begin to die down.”Such one-sided views should have immediately been challenged. But only a different review appeared in Nature, a respected science magazine with specificreadership. The review remarked that Mr. Lomborg's “preference for unexamined materials is incredible(不可信的)”.A critical (批判的)eye is valuable, and the media should present information in such a way that could allow people to make informed decisions. Unfortunately, that is often inaccessible as blocked by the desire to be shocking or to defend some special interests. People might become half-blind before a world partially exhibited by the media. That's a shame, because matters concerning the health of the planet are far too important to be treated lightly.78. Skeptics are searching for more comfortingworldviews to argue that the environment isn’t threatened by _________________.79. What’s the result of Mr. Lomborg’s views?80. The author mentioned “articles and reviews fullof words of praise” to illustrate that Mr.Lomborg’s views ____________________.81. According to the last paragraph, only if________________ can people make right decisions.第II 卷 (共 43 分)I. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.1. 请你方便的时候打电话给我。
上海市闵行区2018年高三第一学期期末(一模)学科质量检测历史试题及答案(word版)

闵行区2017-2018学年第一学期期末考质量监测高三年级历史学科试卷2018.1考生注意:1.考试时间60分钟,试卷满分100分。
2.本考试设试卷和答题纸两部分,试卷包括试题与答题要求;所有答题必须写在答题纸上;做在试卷上一律不得分。
3.答题前,务必在答题纸上填写准考证号和姓名。
4.答题纸与试卷在试题编号上是一一对应的,答题时应特别注意,不能错位。
一、选择题(共40分,每小题2分。
每小题只有一个正确答案)1.回顾人类从早期游徙不定到定居生活的转变过程,最具决定性意义的一步是A.谷物种植B.房屋建造C.陶器制作D.牲畜饲养2.“构成中国的内聚性的又一重要因素是,存在着一种可追溯到数千年前、最古老的商朝的书面语。
”句中“商朝的书面语”具体指A.甲骨文B.大篆C.小篆D.隶书3.“周幽王烽火戏诸侯”、“孔子周游列国”,以下与“诸侯”、“列国”相关的制度是A.禅让制B.世袭制C.分封制D.郡县制4.孔子思想中主张以爱人之心调解和谐社会人际关系的是①“仁者爱人”②贵贱有“序”③“己所不欲,勿施于人”④“有教无类”A.①②B.①③C.①④D.②④5.秦始皇建立的封建专制主义中央集权制度的基本特征是A.确立皇权至高无上B.最高统治者称皇帝C.地方实行郡县制D.中央三公九卿等官职6.通经致用,普及儒家教化并获得巨大成功的第一人是A.孔子B.孟子C.秦始皇D.汉武帝7.图示法是历史学习的一种好方法。
下面那幅图直观反应了三国鼎立的形势A B C D8.“它排除了丞相个人专断,相权过大威胁皇权而出现的政治危机,而且增强了决策施政的程序性和合理性,提高了行政效率。
”材料中的“它”是指A.西汉郡国并行制度B.唐朝三省六部制度C.元朝的行省制度D.明朝的内阁制度9.下图为《元朝疆域图》,其中宣政院管辖的地区是A.①B.②C.③D.④10.“利玛窦是天主教在中国传教的开拓者之一,同时也是18世纪以来第一位阅读中国文学并对中国典籍进行钻研的西方学者,被尊称为‘泰西儒士’”。
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2018年闵行区高三一模语文试题一积累运用(10分)1.按要求填空。
(5分)(1)________________________,不尽长江滚滚来。
(杜甫《登高》)(2)昔我往矣,杨柳依依。
今我来思,_______________。
(《诗经·________·采薇》)(3)王驾《雨晴》中有“蜂蝶纷纷过墙去,却疑春色在邻家”两句。
将蜂蝶拟人化,富有情趣;周邦彦《苏幕遮》中类似的句子是“___________________,___________________。
”2.按要求选择。
(5分)(1)小强因期中考试成绩不理想而有些低沉,你作为好朋友想送他一句话。
下面最合适的一句是()A.只有在天空最暗的时候,才可以看到天上的星星。
B.旁观者的姓名永远爬不上比赛的计分板上。
C.前面是堵墙,用微笑面时,就变成一座桥。
D.山涧的泉水经过一路曲折,才唱出一支美妙的歌。
(2)填入下面语段空白处的语句,最恰当的一项是( )(3分)人类命运共同体是一个关于思维形式和规则的科学理念。
它__________________,因此它具有无限的科学理论和魅力。
①揭示事物世界的本质②形成概念、判断、推理③通过比较、分析、综合、抽象、概括、演绎、归纳等方法④反映实践现实⑤依据人类在认识过程中的客观实践的事实A.①⑤③④②B.④①⑤②③C.⑤③②④①D.③⑤②①④二阅读(70分)(一)阅读下文,完成第3-7题。
(16分)网络文艺处在“雅化”关键期董阳①截至2017年6月,我国网络文学用户规模达到3.53亿,根据网络小说改编的电影、电视剧、网络游戏铺天盖地。
据清华大学课题组发布的《2016中国IP产业报告》,中国IP 影响力排名前100位,网络小说就占了61部。
这意味着,无论你读不读网络小说,将来你看的电影、电视剧,听的歌曲,玩的游戏,很可能都跟网络小说有关。
可以说,我们正处在“网络文学+”——一个由网络文艺“接管..”大众文化的时代。
②这不是危言耸听,从文化发展史角度看也并不奇怪。
一时代有一时代之文学,网络文学其实就是互联网时代的通俗文学。
今天被我们奉为经典的元杂剧、明清小说,都是通俗文学,都是在底层文人大量的民间创作基础上,于勾栏瓦舍的频繁演出中涌现出来的大众文艺精品。
远的不说,金庸武侠本来就是在报纸上连载的通俗小说,它继承了晚清民国以来通俗文学传统,最终融入主流文化,而我们看到的许多武侠小说衍生的电影、电视剧、流行音乐、网络游戏、漫画,其核心的创意正是通俗小说本身。
今天我们将《西厢记》、四大名著奉为经典,把金庸小说放在很高的位置,将来某部网络小说被奉为新名著,某部由网络小说改编的电影被奉为新经典,完全是有可能的。
③有人说,网络小说怪力乱神、子虚乌有,怎么能够担当这样的重任?在我看来,网络小说在发展初期,有过放任自流的阶段,确实存在泥沙俱下的问题,有的还很严重。
但“风物长宜放眼量”,今天正是包括网络小说在内的网络文艺从亚文化向主流文化转换的关键阶段。
对此,我们既要有信心、有心胸,也要对问题和难度有足够的清醒意识。
④目前,“网络文学”在数量上已经“+”得够多了,“IP”主要是指网络小说的授权改编和衍生。
中国网络文学体量实在是太庞大了,中国年轻人的想象力和创造力大量投入网络文学中,其规模在全世界首屈一指。
而且随着网络视听和传统影视市场不断扩容,各路资本纷纷介入,大量收购IP,网络小说身价水涨船高,据说有的网络小说IP估值几个亿。
之后就是我们今天看到的现象,海量网络小说改编项目上马,大有“狂轰滥炸”之势。
⑥我们还要认识到,相比于网络小说,影视和游戏作品影响更为广泛,网络小说改编过程中,要在价值观表达上具有清醒意识。
一部网络小说动辄上千万字,更新速度极快,文字水平参差不齐,价值表达未经深思,此类问题普遍存在,即便所谓“大神级”作品也不能免俗。
影视改编不能停留在照搬的层次,而应当在文化品质和价值内涵上做出有效提升。
⑦网络小说读者主要是青少年群体,他们正处在价值观形成阶段,并不具有成熟判断力,而且由于这个群体相对封闭,小说中存在的价值观问题往往不容易察觉和公开。
一旦推送到大银幕和小荧屏上,其价值观冲突就格外激烈,比如某些“宫斗”作品所宣扬的“丛林法则”,某部“穿越”作品出现的“乱伦”问题,等等,都曾引起社会舆论激烈争议。
要强调的是,这种争议并不意味着社会不宽容,我们要清醒地意识到,网络小说在“+”的过程中,不但需要艺术表达形式的优化与提升,同时也需要改编者在思想表达上应当具有底线意识,给人以积极健康的思想影响。
⑧大众文艺的兴盛是文艺高峰形成的广大基础,对大众文艺进行吸纳和提炼,正是文艺高峰形成的必由之路。
我们不仅要从产业的角度去看它的体量之“庞大”,更要从文化的角度看它的影响之深刻。
事物发展往往“起于青萍之末”,这个事实越早看到,我们就越有文化自觉,就越能顺势而为,引导创作,从而繁荣社会主义文艺,筑就新时代文艺高峰。
(选自2017年11月22日《人民日报》,有删改) 3.解释第①段中“接管”的含义。
(2分)4.概括第②段的主要内容。
(2分)5.第④段中作者提出“‘网络文学’在数量上已经“+”得够多了”的观点,并用“IP”的例子加以证明,请具体分析这一例子是如何证明观点的。
(4分)6.下列与文意相符的一项是()(3分)A.本文作者否定了“网络小说怪力乱神、子虚乌有”的创作状况。
B. 文中的“风物长宜放眼量”,要我们看到网络小说的发展未来。
C. 作者从文化产业的角度否定了那种“得IP者得天下”的观点。
D. 结尾的“起于青萍之末”就是要我们看到网络文艺发展的未来。
7.本文标题是“网络文艺处在‘雅化’关键期”,作者在文中提出了哪些“雅化”的建议。
(5分)(二)阅读下文,完成第8-11题。
(16分)轻盈的浮动的朱以撒①站在自家的院子里就可以看到徐缓的山坡上一大片的芦苇正在迎着来风。
一个人对于居住的选择,有时是游移在具体的房屋之外的,面积、质量、形制这些可以测量的部分被忽略了,而感觉、视觉站出来说话。
我看上这套山居,正是因为在一个蜻蜓乱飞的黄昏,我与满山坡的芦苇相遇..。
它们在夕阳下闪动着银子般的光泽,使人下定决心不再犹豫。
②芦花是柔软之物,一阵风起,芦花悠悠,有的就飘进了院子,落在案上或者我的肩上。
宋人黄庭坚曾说兰花之香是国香,生于深山里,不为人知却照样芳香。
看来天下万千植物都是如此,顺天适性,和人是毫无关系的。
它们与人不同的就在于自然而然,何所来何所去,尽随风来雨往。
一朵芦花落在我的袖子上,不是因为它的重量,而是它的柔软被我感受到——柔软往往是使人感到温暖的一种形态,毛茸茸的、蓬松的,使人放心。
那些敛约的神情,优雅的姿势,朴素的色调,都在人们乐意接受之列,只是后来坚硬越发突兀,放纵恣肆,攘袖瞠目,也就离柔软渐渐远去。
一个人老迈时,坚硬的牙齿全脱落了,又以坚硬的假牙来替代,而一条柔软的舌头却完好如新——道理是可以讲得通的,可是在现实中,大多数人还是争当牙齿,以坚硬面对世界。
③自然之力也难以摧折一丛芦苇——这是台风过后的景象,纤细之秆的韧性显露出来,在随势俯仰中成为一道委婉的弧线。
芦苇就是野草一般的植物,与荆榛莽葛一样,由于过于野而有贱气,人们会在院子里种一片竹子以示高洁,却不会种一丛芦苇。
由于不为人栽种,它的野性形成的内部力量越发有劲道,只是向来重外表的人浑然无知。
④让芦苇入画的人当然有。
这也预示着这个人要有与之相契合的心性,有野的一面,也有细如牛毛的一面。
以工笔来再现一丛芦苇,需要几个月的时间,或者更长——人们往往用时间的长短来衡量劳作的难易,不仅是手上功夫,更是内心的契合,能否把这种植物,从秆的坚韧圆劲到花的迷蒙、缥缈、清虚都表现出来。
如果不行,说明与这种植物内在缺乏一条相通的路径。
那么就不要坚持了,可以改为画竹,竹的硬朗实在比芦苇的虚无柔和更易于把握,而细微正是这个时代的人最难触及的。
细而不弱,功力见矣。
情不知何起,一往而深,那么,就慢慢来吧,毫厘不爽,纤缕必见,最后连画家也成了一秆芦苇。
唐子西说:“山静似太古,日长如小年。
”为何会有度日如年之感?只缘于置身于植物之间,与植物居。
植物的生长是看不到的,只能积多了时日才见出不同。
那么,一个人终日可见芦苇,他也会多一些徐徐的闲雅,慢慢地做一件事,把它做好。
慢的可靠性可以从慢生长态的植物中体现出来,由于内在储存了大量的时日,质地厚实强大,它们成了植物中的精品。
⑤夕阳敛约光线时,芦苇丛中都是声响,归巢的鸟掠过,又停留其间,聒噪聒噪,反而显得芦苇的静谧安详。
杨树皮白光洁,一阵风来就哗哗作响悲怆不已,这种与宗教有所关联的植物使人听其声而不安。
芭蕉偌大的叶片发出的啪啪巨响,似乎要掀动屋瓦,在空旷里生出寒意来。
没有人会听到芦苇的声响,这是一种不出声响的植物。
是这个世界太嘈杂了,淹没了它的私语。
这也使芦苇的气息素来都往下走,温和、素淡、清凉,还有些许薄薄的寂寥。
永远是那般的细腻修长,像极了旧日里清瘦的文人,轻轻地来,轻轻地往,静静地翻书,静静地行文,少与人交接而乐于自处。
⑥像水边的人终日可以看到流水,像山里的人终日可以开门见山,时日久了也就成了山水的一个部分。
⑦坐对青山,日子悄然而缓慢,如同满山草木天生天养。
如果一个人住在三环边上,可以看到千百汽车蟑螂般地穿梭,声浪向上翻卷,进入房内,心紧了起来,动作紧了起来。
谁有当年陶渊明的淡定,心远地自偏呢,把繁华的喧嚣视同安宁的桃源。
每一种感受都是很个人的,我相信人都有从坚硬过渡到柔软的一个过程,它是一个朴素的回归,它可以是一家人围在一起的一次晚餐,可以是一次安然无忧的入睡,可以是一次负暄时的陶陶然的心情。
一个人在注视芦花的轻柔时,他对于寻常日子的寻常要求,也倾向于如此。
(选自2017年6月9日《光明日报》,有删改)8.分析第①段加点词“相遇”的表达效果。
(3分)9.解释第④段画线句的含义。
(4分)10.第⑤段中描摹了哪些声响?其用意是什么?(4分)11.阅读全文,谈谈你对本文标题“轻盈的浮动的”的理解。
(5分)(三)阅读下面作品,完成第12-14题。
(8分)隔浦莲近拍·中山县圃姑射亭避暑作周邦彦新篁摇动翠葆,曲径通深窈。
夏果收新脆,金丸落,惊飞鸟。
浓翠迷岸草。
蛙声闹,骤雨鸣池沼。
水亭小,浮萍破处,帘花檐影颠倒。
纶巾羽扇,困卧北窗清晓。
屏里吴山梦自到。
惊觉,依然身在江表。
【注】①翠葆:原指饰有翠鸟羽毛的车盖。
江表:这里指江宁(今南京)、溧水一带。
12.周邦彦,北宋著名词人,字美成,号_____________。
(1分)13.下列对本词的理解恰当的一项是()。
(2分)A.“新篁”句写翠绿的新竹在车盖上招摇。
B. “夏果”句写果子落地发出清脆的响声。
C.下片中的“困卧”对应了题中的“避暑”。