华师大附属东昌中学
2021年华东师范大学附属东昌中学高三英语模拟试卷及答案解析
2021年华东师范大学附属东昌中学高三英语模拟试卷及答案解析第一部分阅读（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ADive with Big SharksOur shark dive adventures make use of hookah systems and shark cages. A hookah system is a system of providing air from the surface to divers down below. Cage divers breathe by using a regulator connected to an air hose.Is SharkDiving Dangerous?Yes. You could get sunburnt. You could hit your head on the top bunk getting out of bed. You could fall overboard. As for a shark attack,according to the International Shark Attack File,you are far more likely to be killed by a dog or a deer.Pricing & DetailsOne day Cage Diver Adventure S 875Our expert shark diver team will accompany you to the best viewing areas within the Marine Sanctuary.There,we'll drop our cage and prepare to provide you with a view you'll never forget.No dive experience is necessary.Our cages sit just below the surface.You'll be able to breathe comfortably from your snorkel or air hose while you move about the cage,taking photos and having fun.Top Shark Adventure S 375If you want to see great white sharks but prefer them a little further away,we offer great top-side shark viewing from our observation deck. Help scan the horizon for fins and watch for sharks attacking their prey（猎物）.Important NoteThere's No Shark GuaranteeAlthough we go to the best places at the best time of year, we cannot guarantee you'll see sharks. We've been very successful in past shark seasons and expect another incredible year. However, if we see nosharks, there is no refund.1.Which of the following isTRUEabout the two adventures?A.Top Shark Adventure makes use of hookah systems.B.Cage Diver Adventure offers you a view of the bottom of the sea.C.Cage Diver Adventure is less interesting than the other.D.Top Shark Adventure is suitable for those worried about danger.2.What is most likely to happen according to the advertisement?A.you fail to achieve your purpose of the trip.B.you are out of breath deep down in the sea.C.you are hurt by a shark while diving there,D.you suffer from lack of skill in shark diving.3.It can be inferred from the advertisement that shark diving is ________A.difficult but excitingB.challenging and tiringC.amazing and enjoyableD.expensive but popularBDisease-carrying mosquitoes can spread diseases without affecting themselves. Nearly 700 million people get a mosquito-borne illness each year, which results in over one million deaths. Humans experience continuous pressures from disease-carrying mosquitoes in many parts of the world, so we have to find ways to fight against those insects because they keep getting scarier.Even though DEET remains the most commonly used, and most powerful, mosquito repellent ever developed, scientists are actively pursuing effective products based entirely on plant oils. While DEET is an effective contact repellent, many people dislike the oily feel and smell on their skin, and sometimes some people are sensitive to it. Consumers are always interested in alternatives to DEET and other synthetic repellents, so there are numerous natural repellents on the market.In his lab atIowaStateUniversity, Dr. Joel Coats and his team have successfully tested these repellents against three species of dangerous mosquitoes. The first group of the new repellents act through the air. These chemicals have a vapor action that provides protection, and they are called “spatial” repellents, since they act through space. These are potentially most useful in backyards, parks, and houses. The other group are the classic ones that stop insects from standing on a treated surface, such as human skin, clothing or tents; collaborators at the USDA-ARS and BioGents have conducted testing with humans to confirm the effectiveness and identify the very best ones.The new repellents were designed and made from the natural materials in plant essential oils. They maintain many of the advantages of the natural repellents: They are fully biodegradable, with no ecological concerns or environmental wastes, and generally considered safe like the thousands of types of plant essential oils used in the flavor and perfume industries. However, thorough testing will be conducted to determine if they are trulynon-poisonous because there is still no enough evidence.4. What does the author mainly want to show in paragraph 1?A. The way mosquitoes spread disease.B. The high death rate of mosquito-borne illness.C. The difficulty of fighting disease-carrying mosquitoes.D. The urgency of finding tools to fight against mosquitoes.5. What is a disadvantage of DEET?A. It won’t be effective for long.B. It can’t be applied universally.C. It causes discomfort to the users.D. It greatly harms people’s health.6. What can we learn about the second group of the new repellents?A. They can kill mosquitoes indirectly.B. They are mainly used in the open air.C. They are more effective on human skin.D. They can prevent mosquitoes from contacting users.7. What’s the author’s attitude to the new repellents?A. Subjective.B. Objective.C. Doubtful.D. Disapproving.COne-year-old Tallulah turned purple and stopped moving after the sweet became stuck in her throat. Her mum Leigh-Anne said the drama began during a visit to her grandma’s house when her grandparents gave her older kids some sweets.“Then at about 4:45 pm, Tallulah started to choke—we all went into a panic.”“It seemed like it went on for ages. Not one of us knew what to do.”“I rang an ambulance while my grandma and granddad tried to get the sweet to come up.”“Tallulah was panicking at first but then she started to go purple—she almost had no oxygen left in her.”With her daughter limp (无力的) and time running out, Leigh—Anne knew she couldn’t afford to wait for the ambulance to arrive.“The only thing I could think was to go out into the street.” She said.“I rushed out and screamed for someone to help while my grandma rushed out crying with Tallulah.”At exactly the moment, Caitlin, who is studying public services atRedcarCollege, was passing byQueen Street.She said, “I was waiting to go to work when I heard someone screaming for help, so I ran straight over.”The 17-year-old girl added, “Something just clicked and I went into auto mode. The little girl was completely limp, so I checked her airways and tilted (使倾斜) her over and started hitting her back. I turned her round and tapped on her chest, then after what felt like forever she coughed up the sweet and spat it out.As soon as she started crying I felt a huge relief. I was just so pleased I was able to help.”Caitlin was taught her lifesaving skills when she joined the Army Cadets four years ago.8. When did Tallulah get choked?A. While eating sweets.B. While enjoying a drama.C. While having a meal.D. While taking some medicine.9. Why did the family go out into the street?A. To buy some needed tools.B. To search for timely help.C. To get a breath of fresh air.D. To wait for the ambulance to arrive.10. Which of the following can best describe Caitlin?A. Brave and selfless.B. Kind and energetic.C. Determined and generous.D. Quick-thinking and helpful.11. What may be the best title for the text?A. First aid skill sounds important.B. Screaming for help makes sense.C. Eating sweets endangers baby girl.D. Heroic teenager saves baby girl’s life.DThe mass death of flying foxes in extreme heat in North Queensland last month underlines the importance ofUniversityofQueenslandwildliferesearch released today.The UQ research sheds light on how various species have responded to major climate events.A study led by UQ School of Earth and Environmental Science researcher Dr Sean Maxwell has spent more than 70 years quantifying the responses of various species.“The growing frequency and intensity of extreme weather events such as cyclones, droughts and floods is causing unpredictable and immediate changes to ecosystems and blocking existing management efforts,” Dr Maxwell said.“Some of the negative responses we found were quite concerning, including more than 100 cases of dramatic population declines and 31 cases of local population extinction following an extreme event.”"Populations of critically endangered bird species inHawaii, such as the palia, have been annihilated due to drought, leaving none of its kind, and populations of lizard species have been wiped out due to cyclones intheBahamas."Cyclones were the most common extreme event for birds, fish, plants and reptiles, while mammals and amphibians were most responsive to drought events, with drought leading to 12 cases of major population decline in mammals.Drought also led to 13 cases of breeding declines in bird populations and 12 cases of changes in the composition of invertebrate communities.UQ Centre for Biodiversity and Conservation Science director Professor James Watson said the detailed information would help inform ecosystem management.“The research clearly shows species will respond, often negatively, to extreme events,” Professor Watson said.“As climate change continues to ensure extreme climate and weather events are more and more common,we now need to act to ensure species have the best chance to survive.Wherever possible, high quality and intact habitat areas should be retained, as these are the places where species are most resilient(易恢复的) to increasing exposure to extreme events.”12. How was the UQ researchconducted?A. By observing extreme weather events.B. By protecting the endangered species.C. By recording reactions of animals to extreme climate.D. By analyzing the reason whymass animal death happened.13. What does the underlined word “annihilated” in paragraph 4 probably mean?A. destroyedB. defeatedC. decreasedD. disappeared14. Which of the following sentences is true about extreme weather events?A. Drought caused 13 cases of distinctionin bird populations.B. Drought caused 12 cases of population decline in mammals.C. Birds and mammals are most responsive to cyclones.D. Cyclones wiped out populations of lizard.15. What can we infer from Professor Watson’s words?A. Animals often show negative responses to extreme events.B. The existing management ways for wildlife protection are limited.C. Different methods should be adopted to ensure the survival of different species.D. Complete and undamaged habitats are of great importance to species’ survival.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。
上海市华东师范大学附属东昌中学2023-2024学年高二下学期期中物理试卷(学生版)
华东师范大学附属东昌中学2023-2学期高二年级期中考试物理试卷（等级）考生注意1．试卷满分100分，考试时间60分钟。
2．本考试分设试卷和答题纸，答题前，务必在答题纸上填写姓名、报名号、考场号和座位号，并将核对后的条形码贴在指定位置上。
作答必须写在答题纸上，在试卷上作答一律不得分。
3．本试卷标注“多选”的试题，每小题应选两个及以上的选项，但不可全选：未特别标注的选择类试题，每小题只能选一个选项。
4．本试卷标注“计算”、“简答”、“论证”的试题，在列式计算、逻辑推理以及回答问题的过程中，须给出必要的图示、文字说明、公式、演算等。
简谐运动振动是自然界中普遍存在的运动形式，简谐运动是最简单、最基本的振动。
1. 如图甲是单摆振动的情形，O 是运动中的最低位置，M 、N 是摆球所能到达的最远位置。
设摆球向右方向为正方向，如图乙是该单摆的振动图像。
已知当地重力加速度大小，。
（1）关于单摆及其振动，以下认识正确的是（）A ．将某个物体用绳子悬挂起来就构成一个单摆B ．将金属小球用长度不变的细长轻绳悬挂起来，就构成一个单摆C ．单摆在任意摆角下，在竖直面内的自由摆动是简谐运动D ．单摆在人的不断敲击下做摆幅越来越大的运动是简谐运动（2）关于单摆的运动，下列说法正确的是（ ）A ．经过平衡位置时所受的合力为零B ．经过平衡位置时所受的回复力不为零C ．回复力是重力和摆线拉力的合力D ．回复力是重力沿圆弧切线方向的分量（3）根据图乙所示的振动图像，下列说法正确的是（）A ．单摆的振幅是0.07m ，振动的周期是2s 210m/s =g 2π10=B ．若仅将摆球质量变大，单摆周期将变小C ．单摆的摆长为1mD ．当时摆球在M 点2. 图甲为共振筛的示意图，电动偏心轮每转一周，给筛子一个驱动力，该共振筛振幅A 随频率f 的变化关系如图乙所示。
已知增加筛子的质量，筛子的固有周期会变大。
则：（1）筛子固有频率为______Hz ，共振状态下偏心轮的转速是______r/min ；（2）当偏心轮转速为54r/min 时，为使筛子再次发生共振，需将筛子质量______（选填“增大”或“减小”）。
东昌中学构建金融素养培育特色课程体系的实践研究
•中小学政治课教学研究•东昌中学构建金融素养培育特色课程体系的实践研究杨文强【摘要】本论文是东昌中学创建金融素养培育特色课程体系的实践和行动研究，回顾金融素养培育课程体系的开发和实施过程，阐述对金融素养和金融素养课程体系的认知，总结和阐述金融素养课程体系的构成与分类；探索实施金融素养培育课程体系后对学校发展、教师专业发展特别是对学生学习生活和未来规划的影响。
【关键词】金融素养;金融素养课程；特色高中一、创建金融素养培育特色课程体系的背景1.区位优势华东师范大学附属东昌中学地处浦东陆家嘴国际金融贸易区的核心区，学校周边不仅有多家国有控股银行、外资银行、国内地域性和专业性银行,更有上海证券交易所、期货交易所等上千家现代金融机构。
这种区域优势,不仅为开展学生金融素养培育提供了丰富的人才资源，也为培育学生金融素养提供了广阔的实践平台，更为增强学生金融意识、金融素养提供了良好的文化氛围。
2.政策优势国家和上海市中长期教育规划中提到：“促进高中办学体制多样化，扩大优质资源；推进培养模式多样化，满足不同潜质学生的发展需要；鼓励普通高中办出特色。
”2011年上海市教委推出了“特色普通高中建设推进项目”，旨在加强普通高中的特色建设，树立上海普通高中优质特色发展的典型。
推进高中的特色化办学，是当前基础教育发展的一大趋势，也是深化教育改革的_个热点。
这对于全面推进素质教育、深化教育教学改革有着重要意义，也是加强学校管理、提升学校品位、打造学校品牌的重要举措。
3.学校目标近年来,东昌中学结合浦东“金融中心”建设，逐步思考将校本课程研发、学生社会实践转向以学生“金融素养”培育和金融企业（机构）实践为主体，使东昌中学的课程建设逐步聚焦于特色高中建设需要，努力争创上海市有品牌有影响的金融素养培育特色高中。
二、构建实施金融素养培育课程体系的过程东昌中学的金融素养培育课程体系建设是一个循序渐进、逐步完善到逐步稳定的过程。
开发金融课程源于笔者的一个梦想:怎么让孩子们未来能够健康和幸福地参与经济生活？第一阶段，利用东昌中学的区位优势，开设一门具有东昌特色的校本经济课程。
2021届华东师范大学附属东昌中学高三语文月考试题及答案
2021届华东师范大学附属东昌中学高三语文月考试题及答案一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的文字，完成下面小题。
我的第一个上级(节选)马烽①去年夏天，我在省水利学校毕业以后，被暂时安排到防汛指挥部去协助工作，农建局田局长是我们的领导。
这个人面色苍白，走路总是低着头，背着手，慢吞吞地迈着八字步，讲话也是少气无力，好像什么事都不能使他激动。
②一天夜里，山洪暴发了。
我接到电话，一口气跑到农建局，撞开田副局长的房门：“老田，永安河发洪水！安乐庄决口了！”他一只手支起半个身子问道：“安乐庄什么地方决口了？”我说：“汽车路东。
”他听完，竟又躺下了，不紧不慢地说：“没甚要紧。
”我又急又气：“你知道有多么大流量？一百多个！”“那更没办法！反正堵也堵不住，任由它流吧。
”③正在这时，办公室小秦慌慌急急跑进门，大声说道：“三岔河也发洪了！”他的话音刚落，老田就像中了电似地“呼”一下坐了起来回道：“多大流量？”小秦说：“水漫到龙王庙背后了。
”老田说：“那至少有九十个。
”他一面急忙穿衣服，一面说：“赶快通知海口村、田家庄，全体上堤。
快！”④我刚跑回办公室，老田就大踏步跑进来，一手拿着件雨衣，精神抖擞，满面红光。
他把雨衣一扔，抓起耳机就开始给各村打电话。
“马上接杜村，上舍……听着，把三支渠的闸拔开一孔……什么？已经全拔开了？我就怕你们来这么一手，马上闸住两孔……”他放下这个耳机，马上又抓起另一个，详细地指示：要防守哪段河堤，开哪个支渠闸，闭哪个支渠闸……我忙把河流渠道图铺在他面前的桌子上，他根本没看一眼，继续讲他的。
⑤老田打完电话，回头对我说：“咱俩到海门去，恐怕那里南堤要出问题。
”在车上，我问他，为什么安乐庄决了口，他一点都不着急；而三岔河只有九十多个流量，他就急成那个样子？他说：“永安河坡度大，洪水来源少。
别看来势猛，顶多四个钟头水就退了；再说，汽车路东种的都是高秆作物，过一下水也淹不死。
三岔河不一样，洪水来源多，同时愈往下游坡度愈小。
华东师范大学附属东昌中学_企业报告(业主版)
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上海国际招标有限公司关于华东 师范大学附属东昌中学改扩建工 程自来水泵房及室外管线工程 （ 项 目 编 号 ： 1904024034 ） 的 成...
上海腾景建设工程 有限公司
关于其他制冷空调设备的协议供 货馆
上海潜航空调有限 公司
关于其他制冷空调设备的协议供 货馆
关于打印/复印纸的协议供货馆
上海神霖贸易有限 公司
1900.7 1303.7 1060.0
2023-07-14 2023-08-03 2023-07-17
（6）办公消耗用品及类似物品（3）
重点项目
项目名称
中标单位
中标金额（万元） 公告时间
本报告于 2023 年 08 月 17 日 生成
5/19
重点项目
项目名称
中标单位
中标金额（万元） 公告时间
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2023-03-10
（4）印刷服务（4）
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TOP1 关于印刷服务的定点采购馆
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上海倚丰彩印包装 有限公司
19.7
2023-06-25
TOP2
华东师范大学附属东昌中学定点 采购馆项目
上海临浦印刷有限 公司
13.8
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上海倚丰彩印包装 有限公司
1.1 总体指标 ...........................................................................................................................1 1.2 需求趋势 ...........................................................................................................................1 1.3 项目规模 ...........................................................................................................................2 1.4 行业分布 ...........................................................................................................................3 二、采购效率 .................................................................................................................................7 2.1 节支率分析 .......................................................................................................................7 2.2 项目节支率列表 ................................................................................................................7 三、采购供应商 .............................................................................................................................8 3.1 主要供应商分析 ................................................................................................................8 3.2 主要供应商项目 ................................................................................................................9 四、采购代理机构........................................................................................................................10 4.1 主要代理机构分析 ..........................................................................................................10 4.2 主要代理机构项目 ..........................................................................................................11 五、信用风险 ...............................................................................................................................12 附录 .............................................................................................................................................13
上海华东师范大学附属东昌中学南校数学高一上期中经典题(含答案解析)
一、选择题1．(0分)[ID ：11822]函数()2312x f x x -⎛⎫=- ⎪⎝⎭的零点所在的区间为（ ） A ．()0,1B ．()1,2C ．()2,3D ．()3,42．(0分)[ID ：11818]已知函数f (x )＝23,0{log ,0x x x x ≤>那么f 1(())8f 的值为( )A ．27B ．127C ．－27D ．－1273．(0分)[ID ：11815]若偶函数()f x 在区间(]1-∞-，上是增函数，则( ) A ．3(1)(2)2f f f ⎛⎫-<-< ⎪⎝⎭B ．3(1)(2)2f f f ⎛⎫-<-< ⎪⎝⎭C ．3(2)(1)2f f f ⎛⎫<-<- ⎪⎝⎭D ．3(2)(1)2f f f ⎛⎫<-<- ⎪⎝⎭4．(0分)[ID ：11809]不等式()2log 231a x x -+≤-在x ∈R 上恒成立，则实数a 的取值范围是（ ） A ．[)2,+∞B ．(]1,2C ．1,12⎡⎫⎪⎢⎣⎭D ．10,2⎛⎤ ⎥⎝⎦5．(0分)[ID ：11807]如图，点O 为坐标原点，点(1,1)A ，若函数xy a =及log b y x =的图象与线段OA 分别交于点M ，N ，且M ，N 恰好是线段OA 的两个三等分点，则a ，b 满足．A ．1a b <<B ．1b a <<C ．1b a >>D ．1a b >>6．(0分)[ID ：11801]设集合{|32}M m m =∈-<<Z ，{|13}N n n M N =∈-≤≤⋂=Z ，则A ．{}01，B ．{}101-，，C ．{}012，，D ．{}1012-，，， 7．(0分)[ID ：11799]已知(31)4,1()log ,1a a x a x f x x x -+<⎧=⎨≥⎩是(,)-∞+∞上的减函数，那么a 的取值范围是（ ）A ．(0,1)B ．1(0,)3C ．11[,)73D ．1[,1)78．(0分)[ID ：11784]1()xf x e x=-的零点所在的区间是（ ） A ．1(0,)2B ．1(,1)2C ．3(1,)2D ．3(,2)29．(0分)[ID ：11779]已知()f x 是定义域为(,)-∞+∞的奇函数,满足(1)(1)f x f x -=+.若(1)2f =，则(1)(2)(3)(50)f f f f ++++=（ ）A ．50-B ．0C ．2D ．5010．(0分)[ID ：11756]函数()111f x x =--的图象是（ ） A ． B ．C ．D ．11．(0分)[ID ：11787]已知函数21(1)()2(1)ax x f x x x x x ⎧++>⎪=⎨⎪-+≤⎩在R 上单调递增，则实数a 的取值范围是 A ．[]0,1B ．(]0,1C ．[]1,1-D ．(]1,1-12．(0分)[ID ：11748]已知定义在R 上的函数()21()x mf x m -=-为实数为偶函数,记0.5(log 3),af 2b (log 5),c (2)f f m ,则,,a b c ,的大小关系为（ ）A ．a b c <<B ．c a b <<C ．a c b <<D ．c b a <<13．(0分)[ID ：11742]已知0.80.820.7,log 0.8, 1.1a b c ===，则,,a b c 的大小关系是（ ） A ．a b c << B ．b a c << C ．a c b << D ．b c a <<14．(0分)[ID ：11733]设0.60.3a =，0.30.6b =，0.30.3c =，则a ，b ，c 的大小关系为（ ） A ．b a c <<B ．a c b <<C ．b c a <<D ．c b a <<15．(0分)[ID ：11781]函数2xy x =⋅的图象是（ ）A ．B ．C ．D ．二、填空题16．(0分)[ID ：11919]已知函数241,0()3,0x x x x f x x ⎧--+≤=⎨>⎩，则函数(())3f f x =的零点的个数是________.17．(0分)[ID ：11908]设函数21()ln(1||)1f x x x=+-+，则使得()(21)f x f x >-成立的x 的取值范围是_____. 18．(0分)[ID ：11888]若42x ππ<<，则函数3tan 2tan y x x =的最大值为 ．19．(0分)[ID ：11869]如果函数221xx y a a =+-（0a >，且1a ≠）在[]1,1-上的最大值是14，那么a 的值为__________. 20．(0分)[ID ：11858]10343383log 27()()161255---+=__________．21．(0分)[ID ：11856]定义在[3,3]-上的奇函数()f x ，已知当[0,3]x ∈时，()34()x x f x a a R =+⋅∈，则()f x 在[3,0]-上的解析式为______．22．(0分)[ID ：11843]关于函数()2411x x f x x -=--的性质描述，正确的是__________.①()f x 的定义域为[)(]1,00,1-；②()f x 的值域为()1,1-；③()f x 的图象关于原点对称；④()f x 在定义域上是增函数.23．(0分)[ID ：11842]非空有限数集S 满足：若,a b S ∈，则必有ab S ∈．请写出一个．．满足条件的二元数集S ＝________．24．(0分)[ID ：11835]甲、乙、丙、丁四个物体同时从某一点出发向同一个方向运动，其路程()(1,2,3,4)i f x i =关于时间(0)x x ≥的函数关系式分别为1()21x f x =-，22()f x x =，3()f x x =，42()log (1)f x x =+，有以下结论：①当1x >时，甲走在最前面； ②当1x >时，乙走在最前面；③当01x <<时,丁走在最前面，当1x >时，丁走在最后面； ④丙不可能走在最前面，也不可能走在最后面； ⑤如果它们一直运动下去，最终走在最前面的是甲．其中，正确结论的序号为 （把正确结论的序号都填上,多填或少填均不得分）．25．(0分)[ID ：11864]已知函数()266,34,x x f x x ⎧-+=⎨+⎩ 00x x ≥<，若互不相等的实数1x ，2x ，3x 满足()()()123f x f x f x ==，则123x x x ++的取值范围是__________．三、解答题26．(0分)[ID ：12026]某家庭进行理财投资，根据长期收益率市场预测，投资债券等稳健型产品的收益()f x 与投资额x 成正比，且投资1万元时的收益为18万元，投资股票等风险型产品的收益()g x 与投资额x 的算术平方根成正比，且投资1万元时的收益为0.5万元，（1）分别写出两种产品的收益与投资额的函数关系；（2）该家庭现有20万元资金，全部用于理财投资，问：怎样分配资金能使投资获得最大收益，其最大收益为多少万元？27．(0分)[ID ：11996]小张经营某一消费品专卖店，已知该消费品的进价为每件40元，该店每月销售量（百件）与销售单价x （元/件）之间的关系用下图的一折线表示，职工每人每月工资为1000元，该店还应交付的其它费用为每月10000元.（1）把y 表示为x 的函数；（2）当销售价为每件50元时，该店正好收支平衡（即利润为零），求该店的职工人数； （3）若该店只有20名职工，问销售单价定为多少元时，该专卖店可获得最大月利润？（注：利润=收入-支出）28．(0分)[ID ：11950]函数f(x)=2x −a2x 是奇函数． (1)求f(x)的解析式；(2)当x ∈(0,+∞)时，f(x)>m ⋅2−x +4恒成立，求m 的取值范围． 29．(0分)[ID ：11938]设ABC ∆的内角A ，B ，C 的对边分别为a ，b ，c ，tan a b A =，且B 为钝角. （1）证明：2B A π-=； （2）求sin sin A C +的取值范围.30．(0分)[ID ：11931]已知函数()f x A ，函数()0(11)2xg x x ⎫-⎛=⎪⎭≤ ≤⎝的值域为集合B .（1）求AB ；（2）若集合{}21C x a x a =≤≤-，且CB B =，求实数a 的取值范围.【参考答案】2016-2017年度第*次考试试卷 参考答案**科目模拟测试一、选择题 1．B 2．B 3．D 4．C 5．A 6．B 7．C 8．B 9．C10．B11．C12．B13．B14．B15．A二、填空题16．4【解析】【分析】根据分段函数的解析式当时令则解得当时做出函数的图像即可求解【详解】当时令则解得当时令得作出函数的图像由图像可知与有两个交点与有一个交点则的零点的个数为4故答案为：4【点睛】本题考查17．【解析】试题分析：由题意得函数的定义域为因为所以函数为偶函数当时为单调递增函数所以根据偶函数的性质可知：使得成立则解得考点：函数的图象与性质【方法点晴】本题主要考查了函数的图象与性质解答中涉及到函数18．-8【解析】试题分析：设当且仅当时成立考点：函数单调性与最值19．3或【解析】【分析】令换元后函数转化为二次函数由二次函数的性质求得最大值后可得但是要先分类讨论分和求出的取值范围【详解】设则对称轴方程为若则∴当时解得或（舍去）若则∴当时解得或（舍去）答案：3或【点20．【解析】21．f（x）＝4﹣x﹣3﹣x【解析】【分析】先根据计算再设代入函数利用函数的奇偶性得到答案【详解】定义在﹣33上的奇函数f（x）已知当x∈03时f（x）＝3x+a4x（a∈R）当x＝0时f（0）＝0解得22．①②③【解析】【分析】由被开方式非负和分母不为0解不等式可得f（x）的定义域可判断①；化简f（x）讨论0＜x≤1﹣1≤x＜0分别求得f（x）的范围求并集可得f（x）的值域可判断②；由f（﹣1）＝f（23．{01}或{－11}【解析】【分析】因中有两个元素故可利用中的元素对乘法封闭求出这两个元素【详解】设根据题意有所以必有两个相等元素若则故又或所以（舎）或或此时若则此时故此时若则此时故此时综上或填或【24．③④⑤【解析】试题分析：分别取特值验证命题①②；对数型函数的变化是先快后慢当x=1时甲乙丙丁四个物体又重合从而判断命题③正确；指数函数变化是先慢后快当运动的时间足够长最前面的动物一定是按照指数型函数25．【解析】【分析】画出分段函数的图像由图像结合对称性即可得出【详解】函数的图像如下图所示不妨设则关于直线对称所以且满足则故的取值范围是【点睛】解决本题的关键是要会画分段函数的图像由图像结合对称性经过计三、解答题26．27．28．29．30．2016-2017年度第*次考试试卷参考解析【参考解析】**科目模拟测试一、选择题1．B解析：B【解析】【分析】判断函数()2 312xf x x-⎛⎫=- ⎪⎝⎭单调递增，求出f（0）=-4，f（1）=-1，f（2）=3＞0，即可判断．【详解】∵函数()2 312xf x x-⎛⎫=- ⎪⎝⎭单调递增，∴f（0）=-4，f（1）=-1，f（2）=7＞0，根据零点的存在性定理可得出零点所在的区间是()1,2，故选B ． 【点睛】本题考查了函数的单调性，零点的存在性定理的运用，属于容易题．2．B解析：B 【解析】 【分析】利用分段函数先求f （1)8）的值，然后在求出f 1(())8f 的值． 【详解】 f＝log 2＝log 22－3＝－3，f＝f (－3)＝3－3＝.【点睛】本题主要考查分段函数求值以及指数函数、对数函数的基本运算，属基础题．3．D解析：D 【解析】 【分析】函数()f x 为偶函数，则()()f x f x =-则()()22f f =-，再结合()f x 在(]1-∞-，上是增函数，即可进行判断. 【详解】函数()f x 为偶函数,则()()22f f =-.又函数()f x 在区间(]1-∞-，上是增函数. 则()()3122f f f ⎛⎫<-<- ⎪⎝⎭-，即()()3212f f f ⎛⎫<-<- ⎪⎝⎭故选：D. 【点睛】本题考查函数奇偶性和单调性的应用，考查化归与转化的思想，属于基础题.4．C解析：C 【解析】 【分析】由()2223122-+=-+≥x x x 以及题中的条件，根据对数函数的单调性性，对a 讨论求解即可. 【详解】由()2log 231a x x -+≤-可得()21log 23log -+≤a ax x a，当1a >时，由()2223122-+=-+≥x x x 可知2123-+≤x x a无实数解，故舍去； 当01a <<时，()2212312-+=-+≥x x x a在x ∈R 上恒成立，所以12a ≤，解得112a ≤<. 故选：C 【点睛】本题主要考查对数函数的单调性，涉及到复合函数问题，属于中档题.5．A解析：A 【解析】 【分析】由,M N 恰好是线段OA 的两个三等分点，求得,M N 的坐标，分别代入指数函数和对数函数的解析式，求得,a b 的值，即可求解. 【详解】由题意知(1,1)A ，且,M N 恰好是线段OA 的两个三等分点，所以11,33M ⎛⎫ ⎪⎝⎭，22,33N ⎛⎫ ⎪⎝⎭， 把11,33M ⎛⎫ ⎪⎝⎭代入函数xy a =，即1313a =，解得127a =，把22,33N ⎛⎫ ⎪⎝⎭代入函数log b y x =，即22log 33b =，即得3223b ⎛⎫== ⎪⎝⎭，所以1a b <<. 故选A. 【点睛】本题主要考查了指数函数与对数函数的图象与性质的应用，其中解答熟练应用指数函数和对数函数的解析式求得,a b 的值是解答的关键，着重考查了推理与运算能力，属于基础题.6．B解析：B 【解析】试题分析：依题意{}{}2,1,0,1,1,0,1,2,3,M N =--=-∴{}1,0,1M N ⋂=-. 考点：集合的运算7．C解析：C 【解析】 【分析】要使函数()f x 在(,)-∞+∞上为减函数，则要求①当1x <，()(31)4f x a x a =-+在区间(,1)-∞为减函数，②当1x ≥时，()log a f x x =在区间[1,)+∞为减函数，③当1x =时，(31)14log 1a a a -⨯+≥，综上①②③解方程即可.【详解】令()(31)4g x a x =-+，()log a h x x =.要使函数()f x 在(,)-∞+∞上为减函数，则有()(31)4g x a x =-+在区间(,1)-∞上为减函数，()log a h x x =在区间[1,)+∞上为减函数且(1)(1)g h ≥,∴31001(1)(31)14log 1(1)a a a g a a h -<⎧⎪<<⎨⎪=-⨯+≥=⎩,解得1173a ≤<. 故选：C. 【点睛】考查分段函数求参数的问题.其中一次函数y ax b =+，当0a <时，函数y ax b =+在R 上为减函数，对数函数log ,(0)a y x x =>，当01a <<时，对数函数log ay x =在区间(0,)+∞上为减函数.8．B解析：B 【解析】 函数f （x ）=e x ﹣1x 是（0，+∞）上的增函数，再根据f （12）2＜0，f （1）=e ﹣1＞0，可得f （12）f （1）＜0，∴函数f （x ）=e x ﹣1x 的零点所在的区间是（12，1），故选B ．点睛：判定函数的零点所在区间，只需计算区间端点处的函数值，并判断是否异号，只要异号，则区间内至少有一个零点存在.9．C解析：C 【解析】分析：先根据奇函数性质以及对称性确定函数周期，再根据周期以及对应函数值求结果. 详解：因为()f x 是定义域为(,)-∞+∞的奇函数，且(1)(1)f x f x -=+， 所以(1)(1)(3)(1)(1)4f x f x f x f x f x T +=--∴+=-+=-∴=, 因此(1)(2)(3)(50)12[(1)(2)(3)(4)](1)(2)f f f f f f f f f f ++++=+++++，因为(3)(1)(4)(2)f f f f =-=-，，所以(1)(2)(3)(4)0f f f f +++=，(2)(2)(2)(2)0f f f f =-=-∴=，从而(1)(2)(3)(50)(1)2f f f f f ++++==，选C.点睛：函数的奇偶性与周期性相结合的问题多考查求值问题，常利用奇偶性及周期性进行变换，将所求函数值的自变量转化到已知解析式的函数定义域内求解．10．B解析：B【解析】【分析】 把函数1y x =先向右平移一个单位，再关于x 轴对称，再向上平移一个单位即可. 【详解】 把1y x =的图象向右平移一个单位得到11y x =-的图象， 把11y x =-的图象关于x 轴对称得到11y x =--的图象， 把11y x =--的图象向上平移一个单位得到()111f x x =--的图象， 故选：B .【点睛】 本题主要考查函数图象的平移，对称，以及学生的作图能力，属于中档题.11．C解析：C【解析】x ⩽1时,f (x )=−(x −1)2+1⩽1，x >1时,()()21,10a a f x x f x x x=++'=-在(1,+∞)恒成立， 故a ⩽x 2在(1,+∞)恒成立，故a ⩽1，而1+a +1⩾1，即a ⩾−1，综上,a ∈[−1,1]，本题选择C 选项.点睛：利用单调性求参数的一般方法：一是求出函数的单调区间，然后使所给区间是这个单调区间的子区间，建立关于参数的不等式组即可求得参数范围；二是直接利用函数单调性的定义：作差、变形，由f (x 1)－f (x 2)的符号确定参数的范围，另外也可分离参数转化为不等式恒成立问题．12．B解析：B【解析】由()f x 为偶函数得0m =,所以0,52log 3log 32121312,a =-=-=-=2log 521514b =-=-=,0210c =-=,所以c a b <<,故选B.考点：本题主要考查函数奇偶性及对数运算.13．B解析：B【解析】【分析】根据指数函数的单调性以及对数函数的单调性分别判断出a b c 、、的取值范围，从而可得结果.【详解】0.8000.70.71a <=<=，22log 0.8log 10b =<=，0.801.1 1.11c =>=，b ac ∴<<，故选B.【点睛】本题主要考查对数函数的性质、指数函数的单调性及比较大小问题，属于难题.解答比较大小问题，常见思路有两个：一是判断出各个数值所在区间（一般是看三个区间 ）；二是利用函数的单调性直接解答；数值比较多的比大小问题也可以两种方法综合应用.14．B解析：B【解析】【分析】根据指数函数的单调性得出0.60.30.30.3<，而根据幂函数的单调性得出0.30.30.30.6<，从而得出a ，b ，c 的大小关系．【详解】解：0.3x y =在定义域上单调递减，且0.360.<，0.60.30.30.3∴<， 又0.3y x ∴=在定义域上单调递增，且0.360.<，0.30.30.30.6∴<，0.60.30.30.30.30.6∴<<，a cb ∴<<故选：B ．【点睛】考查指数函数和幂函数的单调性，以及增函数和减函数的定义．15．A解析：A【解析】【分析】先根据奇偶性舍去C,D,再根据函数值确定选A.【详解】 因为2xy x =⋅为奇函数，所以舍去C,D;因为0x >时0y >,所以舍去B ，选A.【点睛】有关函数图象识别问题的常见题型及解题思路(1)由解析式确定函数图象的判断技巧：（1）由函数的定义域，判断图象左右的位置，由函数的值域，判断图象的上下位置；②由函数的单调性，判断图象的变化趋势；③由函数的奇偶性，判断图象的对称性；④由函数的周期性，判断图象的循环往复．(2)由实际情景探究函数图象．关键是将问题转化为熟悉的数学问题求解，要注意实际问题中的定义域问题．二、填空题16．4【解析】【分析】根据分段函数的解析式当时令则解得当时做出函数的图像即可求解【详解】当时令则解得当时令得作出函数的图像由图像可知与有两个交点与有一个交点则的零点的个数为4故答案为：4【点睛】本题考查解析：4【解析】【分析】根据分段函数的解析式当0x ≤时，令()3f x =，则2413x x --+=，解得2x =-±0x >时，()31x f x =>，1x =，做出函数()f x ，1,22y y y ==-=--.【详解】241,0()3,0x x x x f x x ⎧--+≤=⎨>⎩， ∴当0x ≤时，()()2241255f x x x x =--+=-++≤，令()3f x =，则2413x x --+=，解得2x =-±120,423,-<-+<-<--当0x >时，()31xf x =>， 令()3f x =得1x =，作出函数()f x ，1,22,22y y y ==-=--由图像可知，()f x 与1y =有两个交点，与22y =-+则(())3f f x =的零点的个数为4.故答案为：4【点睛】本题考查了分段函数的零点个数，考查了数形结合的思想，属于基础题.17．【解析】试题分析：由题意得函数的定义域为因为所以函数为偶函数当时为单调递增函数所以根据偶函数的性质可知：使得成立则解得考点：函数的图象与性质【方法点晴】本题主要考查了函数的图象与性质解答中涉及到函数 解析：1(1)3， 【解析】 试题分析：由题意得，函数21()ln(1)1f x x x =+-+的定义域为R ，因为()()f x f x -=，所以函数()f x 为偶函数，当0x >时，21()ln(1)1f x x x =+-+为单调递增函数，所以根据偶函数的性质可知：使得()(21)f x f x >-成立，则21x x >-，解得113x <<.考点：函数的图象与性质.【方法点晴】本题主要考查了函数的图象与性质，解答中涉及到函数的单调性和函数的奇偶性及其简单的应用，解答中根据函数的单调性与奇偶性，结合函数的图象，把不等式()(21)f x f x >-成立，转化为21x x >-，即可求解，其中得出函数的单调性是解答问题的关键，着重考查了学生转化与化归思想和推理与运算能力，属于中档试题.18．-8【解析】试题分析：设当且仅当时成立考点：函数单调性与最值 解析：-8【解析】 试题分析：2tan 1tan 1,42x x x ππ∴∴设2tan t x =()()()2221412222142248111t t t y t t t t -+-+∴==-=----≤-⨯-=----当且仅当2t =时成立考点：函数单调性与最值19．3或【解析】【分析】令换元后函数转化为二次函数由二次函数的性质求得最大值后可得但是要先分类讨论分和求出的取值范围【详解】设则对称轴方程为若则∴当时解得或（舍去）若则∴当时解得或（舍去）答案：3或【点 解析：3或13 【解析】【分析】令x t a =，换元后函数转化为二次函数，由二次函数的性质求得最大值后可得a ．但是要先分类讨论，分1a >和01a <<求出t 的取值范围．【详解】设0x t a =>，则221y t t =+-，对称轴方程为1t =-.若1,[1,1]a x >∈-，则1,x t a a a ⎡⎤=∈⎢⎥⎣⎦， ∴当t a =时，2max 2114y a a =+-=，解得3a =或5a =-（舍去）. 若01a <<，[1,1]x ∈-，则1,x t a a a ⎡⎤=∈⎢⎥⎣⎦ ∴当1t a =时，2max 112114y a a ⎛⎫=+⨯-= ⎪⎝⎭解得13a =或15a =-（舍去） 答案：3或13【点睛】本题考查指数型复合函数的最值，本题函数类型的解题方法是用换元法把函数转化为二次函数求解．注意分类讨论．20．【解析】解析：11【解析】1383log1255-⎛⎫⎛⎫--+=⎪ ⎪⎝⎭⎝⎭35181122+-+=21．f（x）＝4﹣x﹣3﹣x 【解析】【分析】先根据计算再设代入函数利用函数的奇偶性得到答案【详解】定义在﹣33上的奇函数f（x）已知当x∈03时f（x）＝3x+a4x（a∈R）当x＝0时f（0）＝0解得解析：f（x）＝4﹣x﹣3﹣x【解析】【分析】先根据()00f=计算1a=-，再设30x≤≤﹣，代入函数利用函数的奇偶性得到答案.【详解】定义在[﹣3，3]上的奇函数f（x），已知当x∈[0，3]时，f（x）＝3x+a4x（a∈R），当x＝0时，f（0）＝0，解得1+a＝0，所以a＝﹣1．故当x∈[0，3]时，f（x）＝3x﹣4x．当﹣3≤x≤0时，0≤﹣x≤3，所以f（﹣x）＝3﹣x﹣4﹣x，由于函数为奇函数，故f（﹣x）＝﹣f（x），所以f（x）＝4﹣x﹣3﹣x.故答案为：f（x）＝4﹣x﹣3﹣x【点睛】本题考查了利用函数的奇偶性求函数解析式，属于常考题型.22．①②③【解析】【分析】由被开方式非负和分母不为0解不等式可得f （x）的定义域可判断①；化简f（x）讨论0＜x≤1﹣1≤x＜0分别求得f（x）的范围求并集可得f（x）的值域可判断②；由f（﹣1）＝f（解析：①②③【解析】【分析】由被开方式非负和分母不为0，解不等式可得f（x）的定义域，可判断①；化简f（x），讨论0＜x≤1，﹣1≤x＜0，分别求得f（x）的范围，求并集可得f（x）的值域，可判断②；由f（﹣1）＝f（1）＝0，f(x)不是增函数，可判断④；由奇偶性的定义得f（x）为奇函数，可判断③．【详解】①，由240110x xx⎧-≥⎪⎨--≠⎪⎩，解得﹣1≤x≤1且x≠0，可得函数()11f x x =--的定义域为[﹣1，0）∪（0，1]，故①正确；②，由①可得f （x ，即f （x ，当0＜x ≤1可得f （x 1，0]；当﹣1≤x ＜0可得f （x [0，1）．可得f （x ）的值域为（﹣1，1），故②正确；③，由f （x ）＝﹣||x x的定义域为[﹣1，0）∪（0，1]，关于原点对称，f （﹣x ）＝|x x＝﹣f （x ），则f （x ）为奇函数，即有f （x ）的图象关于原点对称，故③正确．④，由f （﹣1）＝f （1）＝0，则f （x ）在定义域上不是增函数，故④错误；故答案为：①②③【点睛】本题考查函数的性质和应用，主要是定义域和值域的求法、单调性的判断和图象的特征，考查定义法和分类讨论思想，以及化简运算能力和推理能力，属于中档题．23．{01}或{－11}【解析】【分析】因中有两个元素故可利用中的元素对乘法封闭求出这两个元素【详解】设根据题意有所以必有两个相等元素若则故又或所以（舎）或或此时若则此时故此时若则此时故此时综上或填或【解析：{0,1}或{－1,1}，【解析】【分析】因S 中有两个元素，故可利用S 中的元素对乘法封闭求出这两个元素．【详解】设{}(),S a b a b =<，根据题意有22,,a ab b S ∈，所以22,,a b ab 必有两个相等元素．若22a b =，则=-a b ，故2ab a =-，又2a a =或2a b a ==-，所以0a =（舎）或1a =或1a =-，此时{}1,1S =-．若 2a ab =，则0a =，此时2b b =，故1b = ，此时{}0,1S =．若2b ab =，则0b =，此时2a a =，故1a =，此时{}0,1S =．综上，{}0,1S =或{}1,1S =-，填{}0,1或{}1,1-．【点睛】集合中元素除了确定性、互异性、无序性外，还有若干运算的封闭性，比如整数集，对加法、减法和乘法运算封闭，但对除法运算不封闭（两个整数的商不一定是整数），又如有理数集，对加法、减法、乘法和除法运算封闭，但对开方运算不封闭．一般地，若知道集合对某种运算封闭，我们可利用该运算探究集合中的若干元素．24．③④⑤【解析】试题分析：分别取特值验证命题①②；对数型函数的变化是先快后慢当x=1时甲乙丙丁四个物体又重合从而判断命题③正确；指数函数变化是先慢后快当运动的时间足够长最前面的动物一定是按照指数型函数 解析：③④⑤【解析】试题分析：分别取特值验证命题①②；对数型函数的变化是先快后慢，当x=1时甲、乙、丙、丁四个物体又重合，从而判断命题③正确；指数函数变化是先慢后快，当运动的时间足够长，最前面的动物一定是按照指数型函数运动的物体，即一定是甲物体；结合对数型和指数型函数的图象变化情况，可知命题④正确．解：路程f i （x ）（i=1，2，3，4）关于时间x （x≥0）的函数关系是： ，，f 3（x ）=x ，f 4（x ）=log 2（x+1），它们相应的函数模型分别是指数型函数，二次函数，一次函数，和对数型函数模型． 当x=2时，f 1（2）=3，f 2（2）=4，∴命题①不正确；当x=4时，f 1（5）=31，f 2（5）=25，∴命题②不正确；根据四种函数的变化特点，对数型函数的变化是先快后慢，当x=1时甲、乙、丙、丁四个物体又重合，从而可知当0＜x ＜1时，丁走在最前面，当x ＞1时，丁走在最后面， 命题③正确；指数函数变化是先慢后快，当运动的时间足够长，最前面的动物一定是按照指数型函数运动的物体，即一定是甲物体，∴命题⑤正确．结合对数型和指数型函数的图象变化情况，可知丙不可能走在最前面，也不可能走在最后面，命题④正确．故答案为③④⑤．考点：对数函数、指数函数与幂函数的增长差异．25．【解析】【分析】画出分段函数的图像由图像结合对称性即可得出【详解】函数的图像如下图所示不妨设则关于直线对称所以且满足则故的取值范围是【点睛】解决本题的关键是要会画分段函数的图像由图像结合对称性经过计 解析：11(,6)3【解析】【分析】画出分段函数的图像，由图像结合对称性即可得出。
2020届华东师范大学附属东昌中学高三英语期中考试试题及答案
2020届华东师范大学附属东昌中学高三英语期中考试试题及答案第一部分阅读（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ADo you want to get home from work knowing you have made a real difference in someone’s life? If yes, don’t care about sex or age! Come and join us, then you’ll make it!Position:Volunteer Social Care Assistant (No Pay with Free Meals)Place:ManchesterHours:Part TimeWe are now looking for volunteers to support people with learning disabilities to live active lives! Only 4 days left. Don’t miss the chance of lending your warm hands to help others!Role:You will provide people with learning disabilities with all aspects of their daily lives. You will help them to develop new skills. You will help them to protect their rights and their safety. But your primary concern is to let them know they are valued.Skills and Experience Required:You will have the right values and great listening skills. You will be honest and patient. You will have the ability to drive a car and to communicate in fluent written and spoken English since you’ll have to help those people with different learning disabilities. Previous care-related experience will be a great advantage for you.1.The text is meant to_________.A.carry an adB.send an invitationC.present a documentD.leave a note2.The volunteers’ primary responsibility is to help people with learning disabilities__________.A.to learn new living skillsB.to get some financial supportC.to properly protect themselvesD.to realize their own importance3.Which of the following can first be chosen as a volunteer?A.The one who can drive a car.B.The one who can speak English fluently.C.The one who has relevant work experience.D.The one who has the patience to listen to others.BRemember when your mom told you not to eat too many candy bars or sweets because they can cause tooth decay (蛀牙)? However, it turns out that chocolate can be moresalutaryto your teeth than you might expect. Recent studies show that chocolate can effectively fight against tooth decay, as if we need another excuse to eat chocolate.Chocolate offers protection like fluoride, a main ingredient in most household toothpastes. Not only does chocolate protect our teeth, but it can do so very effectively. Studies show that chocolate has compounds that provide strong protection for teeth. One of the compounds in chocolate, CBH, is shown to protect even more effectively than fluoride.Tooth decay happens when bacteria work to turn sugar into acids in our mouth. This is why eating foods with high sugar content can lead to more tooth decay. The compounds in chocolate, however, are anti-bacteria and can fight against bacteria in your mouth. The CBH compound in particular also works to strengthen tooth enamel (牙釉质), andprotects against tooth decay.Does this mean you can cat as much chocolate as you want without worrying about your teeth? It depends on the types of chocolate that you like. The protective effect of chocolate is most effective when you chew on cocoa beans. Of course, this option is not very appealing to; most people. A more tasty option is to choose dark chocolate with little sugar content, ideally no more than 6 to 8 grams per serving. For other types of chocolate with higher sugar content, the effect will be lessened. However, because of the protective compounds, it is still better for your teeth than other sweets and desserts containing the same amount of sugar.4. The word “salutary” in paragraph 1 means?A. Beneficial.B. Harmful.C. Familiar.D. Useless.5. What can we know about the compound CBH in chocolate?A. It can help chocolate cure tooth decay.B. It can effectively stop teeth from decaying.C. It may protect teeth better than toothpastes do.D. It may soon replace most household toothpastes.6. How does chocolate fight tooth decay?A. By breaking down acids.B. By building up compounds.C. By fixing up tooth enamel.D. By fighting against bacteria.7. What's the main idea of the text?A. Chocolate plays the role of toothpaste.B. Chocolate protects against tooth decay.C. Chocolate is the best choice for teeth protection.D. Chocolate is healthier to teeth than other sweets.CThereare two days that set you on your path in life: the day you’re born, and the day you realize why you were born.Growing up south of Chicago in Harvey, Illinois, most people just had their heads down trying to make it from point A to point B. I was the same way, just going with the flow. I played basketball in high school because I was good at it and because other people thought I should until I discovered my talent.I give up basketball and started doing speeches. It wasn’t a popular decision but my grandfather told me to do what made me happy. I fell in love with comedy and performing. And when I discovered the passion, I realized why I was born.I knew I had something to offer —I knew that not only am I powerful, but I can make a difference.I realized a long time ago that my dream is not to be famous or rich. My talent is to entertain. But it’s more than that. I have the chance to reach people, to brighten days, to bring laughter and positive energy into lives and inspire. And I am grateful forit.Acting putting myself out there and having doors closed on me time and time again has taught me a lot about myself. I have learned to trust what I have to offer the world over momentary doubt. I’ve learned to put my faith over my feelings. And I've grown a tough skin. More importantly, I have learned there is a long way towards our goals and that when we put our talents and passion to work, we determine our value.Like a lot of places across the country, there’s poverty, crime, violence and unemployment in Harvey. And growing up there, a lot of people have tragically low expectations for life. But I know that with the right opportunity and with help along the way, everyone can find their passion and go after it. My life is proof.8. What was the author born to do according to the text?A. Be a basketball player.B. Act and perform.C. Make speeches.D. Teach people.9. What does the underlined word “it” in Paragraph 5 refer to?A. Chance.B. Energy.C. Days.D. Laughter.10. What is the author’s purpose of writing this text?A. To help others find their talents.B. To prove his decision was right.C. To inspire people to follow their dreams.D. To encourage people to set a goal.11. What can be the best tile for the text?A. Success Lies in Hard Work.B. How to Achieve the Dream Is Important.C. The Two Important Days in Life.D. The Day I Realized What I Was Born to Do.DIt is universally acknowledged that renewable energy sources such as solar, wind and hydropower are all much better for the climate than fossil fuels.It is true for wind and solar. However, the picture for hydropower is more complicated than we think.A new study by the Environmental Defense Fund analyzed the climate impacts of 1,500 hydropower facilities across the globe. That accounts for about half of hydropower generation worldwide. The researchers looked at whether the facilities behave as a greenhouse gas sink or as a source. To figure this out, they looked into all the different components that help determine a hydropower facility's greenhouse gasemissions (排放).“There are so many contributors to greenhouse gas emissions from hydropower — but essentially, the majority of greenhouse gas emissions arise from the reservoir (水库) itself, as vegetation and soils are submerged underwater in the dam thatis used for the hydropower generation.” said lissa Ocko, a senior climate scientist at the Environmental Defense Fund and co-author of the study. As the submerged vegetation breaks down, it releases greenhouse gases such as CO2.“The larger the surface area of the reservoir, the more greenhouse gases are going to be given out from that reservoir. Also, the temperature plays a role as well how warm the reservoir is will affect how much greenhouse gases are produced and given out from the reservoir.” added Ocko.Through their analysis, Ocko and her co-author Steven Hamburg, also with the Environmental Defense Fund, discovered that the climate impacts of hydropower cover a complete range. The good news is that some facilities perform just as well as wind and solar. But shockingly, more than 100 facilities are actually worse for the climate than fossil fuels. The study is in the journalEnvironmentalScience and Technology.This finding doesn't mean we should forget about hydropower. “But we just need to be careful to make sure that we have climate benefits. There are a lot of situations where hydropower can be equal to wind and solar. So it really depends on the specific facility." said Ocko.12. How do scientists prove hydropower facilities' effect on the climate?A. By making comparisons.B. By conducting experiments.C. By referring to previous studies.D. By analyzing causes and effects.13. What are the main sources of greenhouse gases from hydropower?A. Vegetation and soils.B. Heat and sunlight.C. Pollutants in the reservoir.D. Components of the generators.14. What have Ocko and her co-author Steven Hamburg found?A The surface area of a reservoir decides the climate.B. Hydropower often influences the climate in all aspects.C. Some facilities perform much better than wind and solar.D. Fossil fuels are worse for the climate than over 100 facilities.15. What is the text mainly about?A. Scientists urge an end to hydropower.B. Hydropower is not necessarily green.C. Hydropower is worse than fossil fuels.D. Renewable energy is a complicated issue.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。
上海华东师范大学附属东昌中学南校学生留学程序
上海华东师范大学附属东昌中学南校介绍：上海市卢湾区教师进修学院附属中山学校位于卢湾区南端，与卢浦大桥和2010年世博会会址毗邻。
2002年卢湾区政府斥资一亿元人民币，对学校进行了全面改建和扩建。
新校舍于2003年正式竣工，一流的硬件设施使学校面貌发生了根本的改变。
目前学校占地面积15655平方米，建筑面积14815平方米。
除36个普通教室外，还配备了理化生、电脑、史地、劳技、语音、音美、唱游、阅览和电子阅览等28个专用教室，另有5个支撑课改、培养学生实践能力的创新实验室已基本建成。
此外，还有大型形体训练室和一个可容纳400余人的阶梯教室，是卢湾区规模较大、设施一流的九年一贯制学校，中学部设立部分寄宿班级。
良好的硬件设施为提高我校教育教学质量提供了巨大的保障和广阔的发展空间。
经过近几年的发展，学校办学规模进一步扩大，目前，学校一至九年级共有36个班级，学生900余名。
2003年，作为区组织人事制度改革试点单位，学校进行了教师队伍的全面优化。
经过改革，教师队伍的学历层次有了较大提升，职称和年龄结构趋向合理。
学校始终以“自主实践”为办学主线，不断完善管理体制、积极投身课改研究、致力促进教师成长、努力提升学校内涵。
学校先后成为上海市课程改革一期、二期实验基地，上海市素质教育实验学校和上海市小班化试点学校。
通过前三年的发展，学校以二期课改为契机，在实践中求创新、在课改中谋发展，各方面取得了令人瞩目的成绩：连续三届获评为卢湾区文明单位。
连续三年被评为A类学校，被评为上海市安全文明校，上海市学生行为规范示范校，上海市红十字达标校，上海市航模特色学校，被评为卢湾区德育先进集体、卢湾区科技特色学校、区健康促进校，06年被评为卢湾区教学成绩表彰单位。
作为上海市二期课改实验基地，卢教院附属中山学校全面贯彻新课程理念，以学生发展为本，以学校传统品牌和龙头课题——“学生实践能力的培养”为抓手，构建充满生命力的课程结构新体系，并建设了陶瓷艺术、科技创新、模型制作、数字化等实验环境，将实践教育引入学科教学，使学科领域的知识在实践活动中获得延伸、综合、重组与提升，倡导和谐互动的师生关系，培养学生“自主实践、全面发展”的能力，在此基础上，我校还实施符合我校发展实际的小班化教育实验，在教育中强调精致化，推进学生个性化发展。
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华东师范大学附属东昌中学
华东师范大学附属东昌中学是区属重点中学，原名上海市东昌中学，是上海市浦东新区的一所区级实验性示范性高中。
创建于1954年5月，1993 年和华东师范大学联合办学改为现名。
学校位于世纪大道边、地铁东昌路站北侧，交通便捷。
校内环境优雅，教育教学设施齐全。
现有38个教学班，1881名学生。
学校有一个锐意、改革的领导班子和一支敬业爱生的师资队伍。
学校概况
学校坚持“行以至诚”的校训，在教育教学过程中贯彻“诚而自律、合而共进”的办学理念，以“诚信教育”为抓手，弘扬学校“诚信、合作”学校文化，使“立人以诚”成为东昌人的共同追求。
校内建有千兆校园网络，多功能厅、图书馆、电视台、广播台、DIS理化生实验室、音乐室、美术室、舞蹈房、科技活动室、学生机房及 100 座的电子阅览室等各类专用教室，高一、二年级每个教室都安装了互动电子白板，同时学校建有教学资源库、师生发展性诚信档案网络评估系统等管理软件，校内教育教学和管理均已基本实现数字化。
学校活动及社团
东昌中学不仅硬件设施卓越，同时也很重视软件设备。
学校每年除了举办体育节、艺术节、科技节等系列活动另外也开展了五项团体赛、足球比赛、篮球比赛、十佳歌手比赛、以及每年年底举办的东昌Ever派对等等。
彩藤心理社、论坛水务局、SF社、羽毛球社、cosplay社、月海美学社、理学社、魔术社、音乐剧社、彗星乐队等众多社团也是人才济济，无一不体现着学生的特长与特色。
各个社团也会在Ever派对上展示。
学校努力培养学生成为一个不是只会“死读书”而是有自己特色与特长的综合性人才。
校园一角。