广东省四校2023-2024学年高三上学期11月联考英语试卷
2024届广东省联考联盟高三英语第一学期期末学业水平测试模拟试题含解析
2024届广东省联考联盟高三英语第一学期期末学业水平测试模拟试题注意事项：1．答卷前，考生务必将自己的姓名、准考证号填写在答题卡上。
2．回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其它答案标号。
回答非选择题时，将答案写在答题卡上，写在本试卷上无效。
3．考试结束后，将本试卷和答题卡一并交回。
第一部分（共20小题，每小题1.5分，满分30分）1．Don’ t worry. A number of efforts are being ma de ______ the whole system operating normally.A．being kept B．keptC．keeping D．to keep2．We completed one third of the project, and the loan _______ in place, we had to delay the rest till the next month.A．not arranged B．was not arrangedC．not arranging D．had not been arranged3．All of us have the desire to visit the three main temples in Athens, especially ______ that contains a gold and ivory statue of Athena.A．the one B．one C．the ones D．those4．Countries which continue importing huge quantities of waste will have to____ the issue of pollution.A．maintain B．simplify C．overlook D．address5．语音知识（共5小题；每小题1分，满分5分）从A、B、C、D四个选项中，找出其划线部分与所给单词的划线部分读音相同的选项。
超实用备战高考英语考试易错题——语法填空：有提示词之非谓语动词(6大陷阱) (原卷版)
易错点21 语法填空之非谓语动词目录01 易错陷阱（6大陷阱）02 举一反三【易错点提醒一】非谓语动词与谓语动词辨析易混易错点【易错点提醒二】非谓语动词作定语易混易错点【易错点提醒三】非谓语动词作状语易混易错点【易错点提醒四】非谓语动词作宾语易混易错点【易错点提醒五】非谓语动词作补语易混易错点【易错点提醒六】固定句式易混易错点03 易错题通关养成良好的答题习惯，是决定高考英语成败的决定性因素之一。
做题前，要认真阅读题目要求、题干和选项，并对答案内容作出合理预测;答题时，切忌跟着感觉走，最好按照题目序号来做，不会的或存在疑问的，要做好标记，要善于发现，找到题目的题眼所在，规范答题，书写工整;答题完毕时，要认真检查，查漏补缺，纠正错误。
易错陷阱1：非谓语动词与谓语动词辨析易混易错点。
【分析】首先应找到谓语动词，这若句中已有谓语动词，还需观察是否有连词表示平行的逻辑关系。
主语后所跟的动词不一定是谓语，常常在设空处出现非谓语充当的后置定语，或是定语从句中的谓语，需要整体对句子结构进行分析，找到真正的谓语动词或主句中的谓语动词或并列的谓语动词，而剩下就很可能是非谓语动词。
易错陷阱2：非谓语动词作定语易混易错点。
【分析】非谓语作后置定语时，容易被误判为谓语动词，故应当审查全句。
后置定语重在判断非谓语动词与所修饰名词之间的主、被动关系以及不规则动词的词形变化。
易错陷阱3：非谓语动词作状语易混易错点。
【分析】首先应当判断非谓语动词是否表示目的，目的在于只能用动词不定式充当。
若充当条件、方式、伴随等状语，则主要判断其与主语之间的主、被动关系。
目的状语用于句中时，不能用逗号，句首则可以。
作结果状语时，不定式表示出乎意料的结果，分词表示自然、可想而知的结果。
易错陷阱4：非谓语动词作宾语易混易错点。
【分析】牢记在以下动词后，只能跟动词的-ing形式作宾语。
1.consider, suggest, advise, admit, delay, practise, deny, finish, enjoy, appreciate, forbid, imagine, risk, mind, allow, permit, escape等。
广东省四校2023-2024学年高三上学期11月联考语文试题【含答案】
广东省四校2023-2024学年高三上学期11月联考语文试题总分:150分考试时间:150分钟一、现代文阅读（35分）(一)现代文阅读I（本题共5小题，19分）材料一:家庭在西洋是一种界限分明的团体。
如果有一位朋友写信给你说他将要“带了他的家庭”一起来看你，他很知道要和他一同来的是哪几个人。
在中国，这句话含糊得很。
在英美，家庭包括他和他的妻以及未成年的孩子。
如果他只和他太太一起来，就不会用“家庭”。
在我们中国“阖第光临”虽则常见，但是很少人能说得出这个“第”字究竟应当包括些什么人。
提到了我们的用字，这个“家”字可以说最能伸缩自如了。
“家里的”可以指自己的太太一个人，“家门”可以指伯叔侄子一大批;“自家人”可以包罗任何要拉入自己的圈子，表示亲热的人物。
自家人的范围是因时因地可伸缩的，大到数不清，真是天下可成一家。
为什么我们这个最基本的社会单位的名词会这样不清不楚呢?在我看来却表示了我们的社会结构本身和西洋的格局是不相同的，我们的格局不是一捆一捆扎清楚的柴，而是好像把一块石头丢在水面上所发生的一圈圈推出去的波纹，愈推愈远，也愈推愈薄。
每个人都是他社会影响所推出去的圈子的中心。
被圈子的波纹所推及的就发生联系。
每个人在某一时间某一地点所动用的圈子是不一定相同的。
我们社会中最重要的亲属关系就是这种丢石头形成同心圆波纹的性质。
亲属关系是根据生育和婚姻事实所发生的社会关系。
从生育和婚姻所结成的网络，可以一直推出去包括无穷的人，过去的、现在的和未来的人物。
我们俗语里有“一表三千里”，就是这个意思，其实三千里者也不过指其广袤的意思而已。
这个网络像个蜘蛛的网，有一个中心，就是自己。
我们每个人都有这么一个以亲属关系布出去的网，但是没有一个网所罩住的人是相同的。
在一个社会里的人可以用同一个体系来记认他们的亲属，所同的只是这体系罢了。
体系是抽象的格局，或是范畴性的有关概念。
当我们用这体系来认取具体的亲亲戚戚时，各人所认的就不同了。
高中英语试题-广东省四校2023-2024学年高三上学期联考(二)英语答案
2023－2024学年第一学期高三四校联考（二）英语试题参考答案第二部分阅读第一节21-23CAC24-27DABD28-31BCDC32-35CADC第二节36-40FDGEB第三部分语言知识运用第一节41-45DBCDC46-50ABDAC51-55ADACB第二节56.uncovered57.Dating58.as59.where60.to preventid62.original63.technologically64.the65.tools第四部分写作第一节应用文写作Dear Jim,How is everything going?Regarding your request for book recommendations in Chinese culture,I’d be more than happy to assist you.My first recommendation is the type of books centering on Chinese Civilization.These books provide a comprehensive overview of China’s rich history,including its ancient civilization,dynasties,and cultural developments.Moreover,books on“Chinese Traditions and Customs”are highly recommended,which delves into the unique Chinese customs,traditions,and rituals that shape Chinese society.By reading them,readers will definitely gain valuable insights into China’s cultural heritage,and its societal norms.Hopefully,my recommendation will greatly contribute to your school’s collection and provide students with a broader perspective on Chinese culture.Yours,Li Hua第二节读后续写Everett did as he was told.At first,Everett stood on the side of the road in a daze.As a five-year-old boy,he did not know how to stop the passing car.However,thinking of his father and brother struggling in the well,he gathered his courage again and shouted bravely for help.Although a couple didn‘t quite understand what Everett was saying,the repeated words like“danger”“well”“help”quickly attracted their attention.They realized the seriousness of the problem.They called119immediately and drove Everett to the farm.The couple hurried to the well with Everett.When they reached the top of the well,they were shocked by what they saw.In the well,Brandon's hand was shaking constantly,and Louie was crying with fear.The couple kept encouraging them and went to the trunk to get a rope to help them out.Hardly had they used the rope to pull Louie out when the rescuers arrived.After getting out,Brandon expressed his gratitude to every rescuer.They said that if it hadn’t been for Everett’s efforts,they would not have been able to come to their assistance in time.Of course, God helps those who help themselves,what really saved them was their own spirit of not giving up in the face of difficulties.2023－2024学年第一学期高三四校联考（二）英语试题答案详解阅读A篇【导语】这是一篇应用文。
广东省部分学校2024-2025学年高三上学期11月联考英语试题(含解析)
2025届高三年级11月份联考英语试题本试题卷共8页。
全卷满分120分。
考试用时120分钟。
注意事项：1. 答题前，先将自己的姓名、准考证号填写在答题卡上，并将准考证号条形码粘贴在答题卡上的指定位置。
2. 选择题的作答：每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑，写在试题卷、草稿纸和答题卡上的非答题区域均无效。
3. 非选择题的作答：用签字笔直接写在答题卡上对应的答题区域内。
写在试题卷、草稿纸和答题卡上的非答题区域均无效。
4. 考试结束后，请将本试题卷和答题卡一并上交。
第一部分阅读(共两节，满分50分)第一节(共15小题；每小题2.5分，满分37.5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。
ASeattle, a lively city in the United States, awaits your exploration! Here’re its four famous attractions.The Space NeedleThis is an absolute must-visit. Standing tall against the skyline, it offers breathtaking 360-degree views of the city. You can take a lift ride to the sightseeing stand. Tickets start from$30 for adults. There’re also dining options available at the top, allowing you to taste a delicious meal while taking in beautiful scenery. Before visiting it, check the weather forecast, as clear days provide the best views.Discovery ParkIt’s Seattle’s largest park, offering miles of hiking tracks through forests, grassland, and along the coastline. You can spot wildlife such as eagles, deer, and various seabirds. The park also has beautiful beaches where you can relax and take in the ocean views. It’s free to access. Stay on designated tracks to avoid disturbing the natural habitat. In case of encountering wildlife, keep a safe distance and do not feed them.Chihuly Garden and GlassThe exhibition showcases the splendid works of Dale Chihuly, with his complex and colorful glass art installations (装置). The garden combines nature and art well, with glass sculptures placed among the greenery. Tickets are about $35. Photography is allowed, but be careful not to touch the glass art, as it’s very delicate.Seattle Art MuseumHousing an extensive collection of art from different cultures and time periods, this museum is a cultural treasure. From ancient sculptures to modern paintings, there’s something to inspire every visitor. Special exhibitions are often held. Admission prices vary depending on the exhibits. Keep in mind that some exhibits may have restrictions on photography.1. What is special about the Space Needle?A. It provides a bird’s eye view of Seattle.B. It sends weather messages to tourists.C. It serves tourists with free meals at the top.D. It offers free lift rides to the sightseeing stand.2. What are tourists expected to do while visiting Discovery Park?A. Buy tickets on the spot.B. Keep to the marked paths.C. Avoid meeting wildlife.D. Stay away from the coastline.3. What do Chihuly Garden and Glass and Seattle Art Museum have in common?A. Their admission prices are the same.B. They ban tourists from taking pictures.C. They especially appeal to art lovers.D. Their works mainly focus on nature.BDemis Hassabis, one of the recipients of the 2024 Nobel Prize in Chemistry, was born on July 27th, 1976, in London.He started learning chess at 4 and won the London Under-8 Championship two years later. Another 7 years later, he achieved the second place globally in the under-14 age group chess competition. This was the first time that he’d shocked the world. In 1992, he was admitted to the Computer Science program at the University of Cambridge. He entered University College London in 2005 to pursue a Ph.D. in cognitive neuroscience. In 2011, he founded DeepMind, which is a world-leading artificial intelligence (AI) research group. After 5 years, the AI program AlphaGo he created defeated the world’s top Go player, Lee Sedol. In 2020, DeepMind’s AI system AlphaFold participated in a competition organized by the Critical Assessment of Structure Prediction (CASP) on calculating the 3D structure of protein molecules (分子) and achieved an unexampled level of prediction accuracy.Hassabis was considered one of the “smartest humans on Earth” by the great British physicist Stephen Hawking, who served as a professor at the University of Cambridge before he passed away in 2018. In 2014, DeepMind was acquired by Google for 600 million dollars even before it had publicly released any products and had only 20 technicians. Since then, Hassabis and his team have influenced Google’s development direction for the next decade, guiding the tech giant from a mobile-first approach to an AI-first one.Demis Hassabis, along with David Baker and John M. Jumper, was awarded the 2024 Nobel Prize in Chemistry for their contributions to protein structure prediction. Their AlphaFold has solved a 50-year-old problem by being able to predict the complex structures of approximately 200 million known proteins. The success of AlphaFold not only lies in its accuracy of prediction but also in its broad application prospects. Through this model, scientists can gain a deeper understanding of the structure and function of proteins, providing more accurate information for new drug research and development and disease treatment.4. At what age did Hassabis surprise the world for the first time?A.4.B.6.C.13.D.16.5. What is paragraph 2 mainly about regarding Hassabis?A. His hobbies as a child.B. His growing-up experiences.C. His studies on protein molecules.D. His purposes of inventing AlphaGo.6. Why does the author mention Stephen Hawking?A. To compare him with Hassabis.B. To recall the birth of DeepMind.C. To stress Hassabis’ bond with him.D. To show Hassabis’ being recognized.7. What does Hassabis’ success mainly imply?A. AI aids scientific progress.B. Teamwork makes a difference.C. Opportunities are multiple.D. Prediction is the key point.CScurvy has long been associated with early explorers who lacked access to fresh fruits and vegetables while they traveled around the globe for years at a time. But scurvy, which is caused by Vitamin C deficiency (缺乏), isn’t an illness that has gone away.Doctors recently diagnosed (诊断) scurvy in two patients living in distant parts of the planet, one in Canada and one in Australia. The Australian case centers on a 51-year-old man. Doctors ran their first series of tests to check for internal bleeding, as well as blood disorders. But none of their diagnostic tools offered any clues as to what was causing the man’s illness. Doctors learned that the man was unemployed and living alone. He’d been eating mostly processed foods, and he had begun skipping meals more frequently in the weeks leading to his hospital visit. He received a weight loss surgery eight years earlier but to save money, he stopped taking the nutrients that the doctor told him to take.Armed with this information, doctors ordered a new round of tests, which showed that the man had no detectable levels of Vitamin C in his system. Eventually, doctors diagnosed him with scurvy.Doctors in Canada described a similar experience this month in the Canadian Medical Association Journal. A 65-year-old woman came to a Toronto hospital with leg weakness, and poor mobility.In both cases, doctors didn’t routinely test for scurvy because they thought humans had got rid of scurvy. “This is based on the condition that there is plentiful Vitamin C in our modern food supply, so deficiency should not occur,” says Lauren Ball, a community health researcher. Fortunately, Vitamin C deficiency is easy to treat.While doctors diagnosed 8.2 cases per 100,000 children in 2016, that number had increased to 26.7 per 100,000 by 2020. The average age of patients with scurvy was 2 years old. And in an analysis of nearly 13,000 Vitamin C tests, 29.9 percent of patients had a modest deficiency and 24.5 percent had a significant deficiency.8. What did the doctors find after conducting their first series of tests on the man?A. The man once suffered from scurvy.B. The man used to eat irregularly.C. Their diagnostic tools went wrong.D. Their test methods didn’t work at all.9. What’s the doctors’ initial attitude to the possibility of developing scurvy in both cases?A. Concerned.B. Dismissive.C. Doubtful.D. Unclear.10. What are the statistics in the last paragraph about concerning scurvy?A. Its features.B. Its variety.C. Its trend.D. Its results.11. What’s the best title for the text?A. Scurvy: A New Problem?B. Scurvy: A Forgotten Illness?C. Scurvy: A Permanent Memory?D. Scurvy: An Easily-treated Disease?DMislabeling is a worldwide issue, and it occurs when the species of fish you’re buying is not the one you actually receive.Fish products often pass through multiple countries before they finally reach our table. Along the way,products can be misidentified as another species or intentionally renamed to make more profit. For instance, a cheap fish like tilapia may be given the name of a more expensive fish, like red snapper, or an endangered species might be passed off as an alternative that is doing better in numbers. These make it extremely hard to monitor their logistics (物流) processes.To investigate this issue in Canada, we examined mislabeling market names in finfish (鳍鱼) products in Calgary between 2014 and 2020. We sampled 347 finfish products from Calgary restaurants. These samples were then genetically tested by using a species-specific marker called a DNA barcode.In Canada, the Canadian Food Inspection Agency (CFIA) maintains a Fish List that provides the acceptable common names for the labeling of fish. A seafood product was considered mislabeled if it was sold by using a name not found on the Fish List for the DNA-identified species. For instance, there is only one species that can be sold under the name salmon (鲑鱼) : Atlantic salmon. If sockeye salmon was sold as salmon without any other qualifier, it was considered mislabeled.We’ve discovered that mislabeling runs rampant in Calgary and that certain product names are more likely to hide species of conservation concern. The result: up to one in five finfish was not as advertised. Several examples of mislabeling involved substituting an expensive product fora cheaper species: tilapia for snapper, rainbow trout for Atlantic salmon. While companies in places like Miami and Mississippi have faced fines for such practices, the global nature of fisheries makes legal action difficult. If you eat seafood, there is a chance that you could be misled as a consumer and end up eating threatened species. You can actually do something to reduce these possibilities.12. What can we say about tracking fish products?A. It’s unnecessary.B. It’s low-cost.C. It’s difficult.D. It’s time-bound.13. What can be learned about the Fish List of the CFIA?A. It provides fish options for restaurants.B. It lists all Canadian endangered species.C. It encourages the mislabeling of fish.D. It has strict Canadian fish naming rules.14. What does the underlined phrase “runs rampant” in the last paragraph mean?A. Remains serious.B. Stops suddenly.C. Becomes expensive.D. Disappears soon.15. What might the author continue talking about?A. Ways to avoid buying mislabeled fish.B. More mislabeled endangered fish species.C. History of mislabeling finfish products.D. Companies’ practices of mislabeling fish.第二节(共5小题；每小题2.5分，满分12.5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。
广东省四校(佛山一中、广州六中、金山中学、中山一中)2024届高三上学期11月联考数学试题(解析版)
2024届高三级11月四校联考数学试题 佛山市第一中学、广州市第六中学 汕头市金山中学、中山市第一中学试卷总分：150分，考试时间：120分钟.注意事项：1.答卷前，考生务必将自己的姓名、准考证号填写在答题卡上.本次考试采用特殊编排考号，请考生正确填涂.2.回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑.如需改动用橡皮擦干净后，再选涂其他答案标号.回答非选择题时，将答案写在答题卡上.写在本试卷上无效.第一部分 选择题（共60分）一、单选题（本大题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的）1. 已知集合{}lg 0A x x =≤，{}11B x x =−≤，则A B = （ ）A. AB. BC. R AD. B R【答案】A 【解析】【分析】根据对数函数的性质、绝对值的性质确定集合,A B ，再由交集定义计算．【详解】由已知{|01}A x x =<≤，02{}|B x x ≤≤=， 所以{|01}A B x x =< ≤=A ， 故选：A2. 已知向量()3,a m =−，()1,2b =− ，若()//b a b −，则m 的值为（ ）A. 6−B. 4−C. 0D. 6【答案】D 【解析】【分析】根据向量的坐标运算结合向量平行的坐标表示运算求解.【详解】由题意可得：()4,2−=−+a b m，若()//b a b −，则28m +=，解得6m =. 故选：D.3. 若函数 ()3,4,4,4x a x f x ax x − ≥= −+< （0,1a a >≠）是R 上的单调函数, 则a 的取值范围为（ ）A. ()50,11,4 ∪B. 51,4C. 4,15D. 40,5【答案】D 【解析】【分析】根据指数函数和一次函数的单调性，结合分割点处函数值的大小关系，列出不等式，求解即可.【详解】因为 4y ax =−+是减函数，且()f x 是R 上的单调函数， 根据题意，()f x 为R 上的单调减函数；故可得 01,,44a a a <<≤−+ 解得405a <≤，即a 的取值范围为40,5 ． 故选：D .4. 若复数z 满足()1i 1i z +=+，则z 的虚部为 （ ）A. B. C.D.【答案】D 【解析】【分析】先根据复数的模及除法运算求出复数z ，进而得到z ，从而求解.【详解】由()1i 1i z +=+=得z =，所以z=，即z 故选：D ．5. 数列{}n a 满足12019a =，且对*n ∀∈N ，恒有32n n n a a +=+，则7a =（ ） A. 2021 B. 2023C. 2035D. 2037【答案】D【解析】【分析】由已知可依次求出47,a a 的值，即可得出答案.【详解】由已知可得，14112202a a =+=，47472203a a =+=. 故选：D.6. 如图，已知圆锥的顶点为S ，AB 为底面圆的直径，点M ，C 为底面圆周上的点，并将弧AB 三等分，过AC 作平面α，使SB α∥，设α与SM 交于点N ，则SMSN的值为（ ）A.43B.32C.23D.34【答案】B 【解析】【分析】连接MB 交AC 于点D ，连接,,ND NA NC ，根据线面平行得性质证明SB DN ∥，再根据MC AB ∥可得DM MCDB AB=，进而可得出答案. 【详解】连接MB 交AC 于点D ，连接,,ND NA NC ，则平面NAC 即为平面α，因为SB α∥，平面SMB DN α∩=，SB ⊂平面SMB ，所以SB DN ∥， 因为AB 为底面圆的直径，点M ，C 将弧AB 三等分，所以30ABM BMC MBC BAC ∠=∠=∠=∠=°，12MCBC AB ==，所以MC AB ∥且12MC AB =，所以12DM MC DB AB ==， 又SB DN ∥，所以12MNDM SNDB ==，所以32SM SN =. 故选：B .7. 已知函数()f x 及其导函数()f x ′的定义域均为ππ,22 − ，且()f x 为偶函数，π26f =−，()()3cos sin 0f x x f x x ′+>，则不等式3π1cos 024f x x+−>的解集为（ ）A. π,03−B. ππ,32C. 2ππ,33−D. 2π,03−【答案】D 【解析】【分析】构建()()3ππsin ,,22=∈− g x f x x x ，求导，利用导数判断原函数单调性，结合单调性解不等式.【详解】令()()3ππsin ,,22=∈−g x f x x x ，则()()()()()2323sin co 3cos s sin si sin n ′′=+=′+ g x f x x x f x x f x x f x x x ，因为ππ,22x∈−，则sin 0x >，且()()3cos sin 0f x x f x x ′+>， 可知()0g x ′>，则()g x 在ππ,22−上单调递增， 又因为()f x 为偶函数，ππ266f f −==−， 可得3πππ1sin 6664−=−−= g f 令()1π46>=−g x g ，可得ππ62x −<<， 注意到33ππππsin cos 2222g x f x x f x x+=++=+，不等式3π1cos 024f x x +−>，等价于ππ26+>−g x g ， 可得πππ622−<+<x ，解得2π03−<<x ， 所以不等式3π1cos 024f x x+−>的解集为2π,03 −. 故选：D.【点睛】关键点睛：构建函数()()3ππsin ,,22 =∈−g x f x x x ，利用单调性解不等式()14g x >，利用诱导公式可得3π1cos 024f x x +−>，等价于ππ26+>− g x g ，即可得结果. 8.已知函数21()sin 0)22xf x x ωωω=+>，若()f x 在3,22ππ上无零点，则ω的取值范围是（ ）A. 280,,99+∞B. 228(0,][,]939C. 28(0,][,1]99D. [)28,991,∞+ 【答案】B 【解析】【分析】先结合二倍角公式和辅助角公式将函数进行化简,得到 ()sin 3f x x πω=−,由题可得323232T ωππωπππω −−−≤=和233(1)23k k ωπππωπππ ≤− +≥−,结合0ω>即可得解.【详解】因为211()sin 0)cos )sin 222xf x x x x ωωωωω+>−+−1sin sin 23x x x πωωω==−若322x ππ<<，则323323x ωπππωππω−<−<−，∴323232T ωππωπππω −−−≤=， 则21ω≤，又0ω>，解得01ω<≤.又233(1)23k k ωπππωπππ ≤−+≥− ，解得2282()339k k k Z ω+≤≤+∈. 228233928039k k k +≤+ +> ，解得4132k −<≤，k Z ∈ ，0k ∴=或1−.当0k =时，2839ω≤≤；当1k =−时，01ω<≤，可得209ω<≤.∴2280,,939ω∈. 故选B.【点睛】本题主要考查三角函数的图象与性质，还涉及二倍角公式和辅助角公式，考查学生数形结合的思想、逻辑推理能力和运算能力，属于中档题．二、多选题（本大题共4小题，每小题5分，共20分．在每小题给出的选项中，有至少两项符合题目要求.全部选对的得2分，有选错的得0分）9. 若{}n a 是公比为q 的等比数列，记n S 为{}n a 的前n 项和，则下列说法正确的是（ ） A. 若{}n a 是递增数列，则1q > B. 若10a >，01q <<，则{}n a 是递减数列 C. 若0q >，则4652S S S +> D. 若1n nb a =，则{}n b 是等比数列 【答案】BD 【解析】【分析】对于AC ：举反例分析判断；对于B ：根据数列单调性的定义结合等比数列通项公式分析判断；对于D ：根据等比数列定义分析判断.【详解】对于选项A ：例如111,2a q =−=，则112n n a − =−，可知数列{}n a 是递增数列，但1q <，故A 错误；对于选项B ：因为()1111111n n n n n a a a q a qa q q −−+−=−=−，若10a >，01q <<，则110,0,10−>>−<n a q q ，可得10n n a a +−<，即1n n a a +<， 所以数列{}n a 是递减数列，故B 正确；对于选项C ：例如1q =，则11461541026=++==a a S S a S ， 即4652S S S +=，故C 错误； 对于选项D ：因为{}n a 是公比为q 的等比数列，则0n a ≠，则111111n n n n n nb a a b a q a +++===，所以数列{}n b 是以公比为1q 的等比数列，故D 正确； 故选：BD.10.已知(a = ，若1b = ，且π6,a b = ，则（ ）A. a b b −=B. b 在a方向上投影向量的坐标为 C. ()2a a b ⊥−D. ()23b a b ⊥−【答案】ACD 【解析】【分析】根据模长公式判断A 选项,根据投影向量公式判断B 选项,根据数量积公式结合向量垂直计算判断C,D 选项.【详解】(,a a =∴=，1a b −=, A 选项正确；b 在a方向上投影向量的坐标为π1cos 162a b a ⋅=×=， B 选项错误；()22π2=22cos 32106a a b a a b a a b ⋅−−⋅=−⋅=−×= ，()2a a b ∴⊥− ，C 选项正确；()22π23=232cos 321306b a b a b b a b b ⋅−⋅−=⋅−=×−= ，D 选项正确； 故选：ACD.11. 定义{}max ,a b 为a ，b 中较大的数，已知函数(){}max sin ,cos f x x x =，则下列结论中正确的有（ ）A. ()f x 的值域为[]1,1−B. ()f x 是周期函数C. ()f x 图像既有对称轴又有对称中心D. 不等式()0f x >的解集为π2π2ππ,2x k x k k−+<<+∈Z 【答案】BD 【解析】【分析】做出函数()f x 的图像，利用图像确定出值域，周期，单调区间，即可求解.【详解】做出函数()f x 的图像，如图所示：令sin cos x x =π04x−=，则ππ4x k −=，k ∈Z ，解得ππ4x k =+，k ∈Z ，当5π2π4xk =+，k ∈Z 时，()f x =由图可知，()f x 的值域为，故A 错误； 且()f x 是以2π为最小正周期的周期函数，故B 正确；由图可知函数()f x 有对称轴，但是没有对称中心，故C 错误； 由图可知，()π2π2ππ2k x k k −+<<+∈Z 时，()0f x >，故D 正确. 故选：BD.12. 定义在()1,1−上的函数()f x 满足()()1x y f x f y f xy−−=−，且当()1,0x ∈−时，()0f x <，则下列结论中正确的有（ ） A. ()f x 奇函数 B. ()f x 是增函数 C. 112243f f f+=D. 111342f f f+<【答案】ABC 【解析】【分析】对于A ：根据题意结合奇函数的定义分析判断；对于B ：根据题意结合函数单调性分析判断；对于C ：根据题意令21,34==xy 代入运算即可；对于D ：令11,24x y ==，结合函数单调性分析判断. 【详解】对于选项A ：因为()()1x y f x f y f xy −−=−，令0xy ==，则()()()000f f f −=，可得()00f =， 令y x =−得：22()()1x f x f x f x −−= +，再以x −代x ，得：22()()1x f x f x f x −−−=+，两式相加得：2222011x x f f x x −+=++，即222211x x f f x x −=− ++ ， 令()()22,1,11=∈−+x g x x x ，则()()()2222101−′=>+x g x x 对任意()1,1x ∈−恒成立， 可知()g x 在()1,1−上单调递增，且()()11,11g g −=−=， 所以()g x 在()1,1−内的值域为()1,1−， 由222211x x f f x x −=−++，()1,1x ∈−，即()()f x f x −=−，()1,1x ∈−， 是所以定义在(1,1)−上的函数()f x 为奇函数，故A 正确；对于选项B ：因为函数()f x 为定义在(1,1)−上的奇函数，且当(1,0)x ∈−时，()0f x <，不妨设1211x x −<<<，则121212()()1x x f x f x f x x−−=−，因为1211x x −<<<，则121201x x x x −<−且12121212(1)(1)1011x x x x x x x x −+−+=>−− 可知1212101x x x x −−<<−，所以121201x x f x x−< −， 则12())0(f x f x −<，即12()()f x f x <， 故函数()f x 在(1,1)−上为增函数，B 正确；对于选项C ，令21,34==x y ，且()()1x y f x f y f xy −−=−， 则211342−=f f f ，即112243f f f+=，故C 正确； 对于选项D ：令11,24x y ==，且()()1x y f x f y f xy −−= −， 则112247−=f f f ， 因为2173<，且函数()f x 在(1,1)−上为增函数，可得2173<f f ， 即111243−<f f f ，所以111342+>f f f ，故D 错误. 故选：ABC.【点睛】方法点睛：函数的性质主要是函数的奇偶性、单调性和周期性以及函数图象的对称性，在解题中根据问题的条件通过变换函数的解析式或者已知的函数关系，推证函数的性质，根据函数的性质解决问题．第二部分 非选择题（共90分）三、填空题（本大题共4小题，每小题5分，共20分）13. 已知()2y f x x =−为奇函数，且()13f =，则()1f −=________.【答案】1− 【解析】【分析】由题意()()2y g x f x x ==−为奇函数，所以由奇函数的性质有()()()()111120g g f f +−=+−−=，结合()13f =即可求解. 【详解】由题意()()2y g x f x x ==−为奇函数，所以由奇函数的性质可得()()()()()()()221111111120g g f f f f +−=−+−−−=+−−=，又因为()13f =，所以解得()11f −=−. 故答案为：1−.14. 设n S 是数列{}n a 的前n 项和，且2cos π3=−nnS n ，则6a =________. 【答案】212##10.5 【解析】【分析】根据n a 与n S 之间的关系，结合诱导公式运算求解.【详解】因为2cos π3=−n n S n ，则255ππ15cos π25cos 2π25cos 253332 =−=−−=−=−S ， 266cos 2π36135−−S ，所以665121352522=−=−−=a S S 故答案为：212. 15. 在ABC 中，角A ，B ，C 所对的边分别为a ，b ，c .120ABC ∠=°，ABC ∠的平分线交AC 于点D ，且1BD =，则43a c +的最小值为________.【答案】7+【解析】【分析】利用等面积法可得ac a c =+，从而111a c+=，再利用乘“1”法及基本不等式可求解． 【详解】因为ABCABD BCD S S S =+△△△， 所以111sin1201sin 601sin 60222ac c a ⋅°=××°+××°，所以ac a c =+，可得111a c+=． 所以()41134773437a c a c c a a c a c=+=+++≥+=++ .（当且仅当34c a a c=，即1a =+，1c =+．故答案为：7+16. 设()()ln ,024,24x x f x f x x <≤= −<<，若方程() f x m =恰有三个不相等的实根，则这三个根之和为________；若方程() f x m =有四个不相等的实根()1,2,3,4i x i =，且1234x x x x <<<，则()2221234x x x x +++的取值范围为______. 【答案】 ①. 6 ②. 45(22,)2【解析】【分析】由函数解析式知函数图象关于直线2x =对称，作出图象，可知212x <<，234x x +=，144x x +=，即可求得12348x x x x +++=，同时把()2221234x x x x +++用2x 表示，利用换元法，函数的单调性求得其范围．【详解】()(4)f x f x =−，因此()f x 的图象关于直线2x =对称，作出函数()f x 的图象，如图，作直线y m =，若是三个根，则1m =，12317,2,22x x x ===，1236x x x ++=， 若是四个根，由图可知212x <<，234x x +=，144x x +=，所以12348x x x x +++=， 12ln ln x x -=，因此121=x x ，()222222222123422222221121()(4)(4)28()34x x x x x x x x x x x x =++−+−=+−+++++22222112()8()30x x x x =+−++，令221t x x =+，则()222123422(2)22t x x x x +=+−++， 对函数1(12)y m m m=+<<，设1212m m <<<，1212121212111()(1)y y m m m m m m m m −=+−−=−−， 因为1212m m <<<，所以120m m −<，12110m m −>，所以120y y −<，即12y y <， 即1(12)y m m m=+<<是增函数，所以522y <<，因素2215(2,)2t x x =+∈，22(2)22y t =−+在5(2,)2t ∈时递增， 所以2452(2)22(22,)2y t =−+∈． 故答案为：6；45(22,)2．四、解答题（本大题共6小题，共70分.解答应写出文字说明、证明过程或演算步骤）17. 若()()πsin 0,0,2y f x A x A ωϕωϕ+>><的部分图象如图所示．（1）求函数()y f x =的解析式；（2）将()y f x =图象上所有点向左平行移动(0)θθ>个单位长度，得到()y g x =的图象；若()y g x =图象的一个对称中心为5π06，，求θ的最小值． 【答案】（1）()π2sin 26f x x=+（2）π12【解析】【分析】（1）由函数的图象的顶点坐标求出A ，由周期求出ω，由代入点法求出ϕ的值，从而可得函数的解析式． （2）根据函数sin()yA x ωϕ+的图象变换规律求得()g x 的解析式，再利用整体代换法与正弦函数的对称性得到θ关于k 的表达式，从而求得θ的最小值． 【小问1详解】根据()f x 的部分图象易知其最大值为2，又0A >，故2A =，周期11πππ1212T=−−=，则2ππω=，又0ω>，所以2ω=， 所以()()2sin 2f x x ϕ=+， 又π,012−在图象上，所以π2sin 06ϕ −+=，故11π2π,6k k ϕ−+=∈Z ，则11π2π,6k k ϕ=+∈Z ， 又π2ϕ<，所以π6ϕ=， 所以()π2sin 26f x x=+. 【小问2详解】 将()y f x =图象上所有点向左平行移动(0)θθ>个单位长度，得到()()ππ2sin 22sin 2266y g x x x θθ==++=++的图象， 因为()y g x =图象的一个对称中心为5π06，，所以5ππ22π,66k k θ×++=∈Z ，即π11π,212k k θ=−∈Z ， 因为0θ>，所以π11π0212k −>，则116k >，又k ∈Z ，所以当2k =时，θ取得最小值为π12． 18. 已知数列{}n a 是公差不为零的等差数列，12a =，且139,,a a a 成等比数列. （1）求数列{}n a 的通项公式； （2）数列{}n b 满足11111,2n n n b a b b −−，求数列{}n b 的前n 项和n S . 【答案】（1）2n a n =；（2）1n n S n =+. 【解析】【分析】（1）设{}n a 的公差为d ，由等比中项的性质有()2(22)228d d +=+可求d ，进而写出{}n a 的通项公式；（2）应用累加法求{}n b 的通项公式，再由裂项相消法求{}n b 的前n 项和n S .【详解】（1）设数列{}n a 的公差为d ，由12a =，2319a a a =有：()2(22)228d d +=+，解得2d =或0d =(舍去)∴2n a n =. （2）1112n n n b b −−=， ∴()112211111112,21,,22n n n n n n b b b b b b −−−−=−=−−=× ，将它们累加得：2111 2.n n n b b −=+− ∴21n b n n=+，则()111111223111n n S n n n n =+++=−=××+++ . 19. 如图，在四棱锥P ABCD −中，底面ABCD 是边长为2的菱形，60BAD ∠=°，侧面PAD 为等边三角形．（1）求证：AD PB ⊥；（2）若P AD B −−的大小为120°，求A PB C −−的正弦值．【答案】（1）证明见解析；（2. 【解析】【分析】(1)取AD 的中点O ，连接OB ，OP ，BD ，证明AD ⊥平面POB 即得；(2)在平面POB 内过O 作Oz OB ⊥，以射线OA ，OB ，Oz 分别为x ，y ，z 轴非负半轴建立空间直角坐标系，借助空间向量推理计算即可得解.【详解】（1）取AD 的中点O ，连接OB ，OP ，BD ，如图，因PAD 为正三角形，则OP AD ⊥，又底面ABCD 是菱形，且60BAD ∠=°，则ABD △是正三角形，于是得OB AD ⊥，而OP OB O = ，,OP OB ⊂平面POB ，则AD ⊥平面POB ，又PB ⊂平面POB ， 所以AD PB ⊥；（2）由(1)知P AD B −−的平面角为POB ∠，即120POB ∠=°，==OP OB ，显然平面POB ⊥平面ABD POB 内过O 作Oz OB ⊥，平面POB 平面ABD OB =，则Oz ⊥平面ABD ，如图，以O 为原点建立空间直角坐标系，则(0,0,0)O ，(1,0,0)A ，B ，(C −，3(0,)2P ，(AB − ，3)2PB =− ，(2,0,0)CB = ，设平面PAB 的法向量为1111(,,)n x y z =，则1111113020n PB y z n AB x ⋅=−= ⋅=−+= ，令11y =，得1n =，设平面PBC 的法向量为2222(,,)n x y z =，则2222220302n CB x n PBz ⋅==⋅=−=，令21y =，得2n =，121212cos ||||n n n n n n ⋅〈⋅〉==⋅，设A PB C −−的大小为θ，从而得sin θ=， 所以A PB C −−. 20. 已知()()1ln 0f x x ax a x=−≥，e 为自然对数的底数. （1）若函数()f x 在e x =处的切线平行于x 轴，求函数()f x 的单调区间； （2）若函数()f x 在1,e e上有且仅有两个零点，求实数a 的取值范围. 【答案】（1）()f x 的单调递增区间为()0,e ，单调递减区间为()e,+∞（2）211e 2ea << 【解析】【分析】（1）求出()f x ′，利用导数的几何意义得到0a =，再利用导数与函数性质的关系即可得解； （2）构造函数()2ln xF x x=，将问题转化为()F x 与y a =的图象有两个交点，利用导数分析()F x 的性质，结合图象即可得解. 【小问1详解】 因为()()1ln 0f x x ax a x=−≥，所以()21ln x f x a x −′=−， 的又函数()f x 在e x =处的切线平行于x 轴，则()e 0f ′=，即21ln e0ea −−=，解得0a =， 此时()21ln xf x x−′=，令()0f x ′=，解得e x =， 当0e x <<时，()0f x '>，()f x 单调递增， 当e x >时，()0f x ′<，()f x 单调递减，所以()f x 的单调递增区间为()0,e ，单调递减区间为()e,+∞. 【小问2详解】因()f x 在1,e e上有且仅有两个零点，令()0f x =，则1ln 0x ax x −=，即2ln x a x =在1,e e上有且仅有两个零点，令()2ln x F x x =，1,e e x∈，则问题转化为()F x 与y a =的图象有两个交点， 又()312ln xF x x−′=，当1ex ∈ 时，()0F x ′>，)F x 单调递增，当)x ∈时，()0F x ′<，()F x 单调递减，所以()F x在x =12eF=， 又201e e F =− <，()2e e 1F =， 作出()F x 与y a =的大致图象，如图，为结合图象可得211e 2ea <<， 所以实数a 的取值范围为211e 2ea <<. 21. 某单位为端正工作人员仪容，在单位设置一面平面镜.如图，平面镜宽BC 为2m ，某人在A 点处观察到自己在平面镜中所成的像为A ′.当且仅当线段AA ′与线段BC 有异于B ，C 的交点D 时，此人能在镜中看到自己的像.已知π3BAC ∠=.（1）若在A 点处能在镜中看到自己的像，求ACAB的取值范围； （2）求某人在A 处与其在平面镜中的像的距离AA ′的最大值. 【答案】（1）1,22（2） 【解析】【分析】（1）设ACB θ∠=，则ππ62θ<<，利用正弦定理结合三角恒等变换可得)sin AC θθ=+，AB θ=，进而整理可得12AC AB =，结合正切函数运算求解；（2）根据（1）中结果结合三角恒等变换整理得π26AA θ=−+′，结合正弦函数分析求解. 【小问1详解】设ACB θ∠=，由题意可知ABC 为锐角三角形，则π022ππ032θθ<<<−<，可得ππ62θ<<，由正弦定理sin sin sinAC AB BCABC ACB BAC===∠∠∠，可得)πsin3AC ABCθθθ=∠=+=+，AB ACBθ=∠=，则12ACAB=+，因为ππ62θ<<，则tanθ>，可得1tanθ<<，即32<<，所以1,22ACAB∈.【小问2详解】由（1）可知：)sinACθθ=+，ABθ=，由题意可知：A A BC′⊥，AD AA=′，利用等面积法可得)1112sin222AAθθθ××=+′整理得2π4sin cos2sin2226 AAθθθθθθ==−−′，因为ππ62θ<<，则ππ5π2,666θ−∈，当ππ262θ−=，即π3θ=时，AA′取到最大值.22. 设()2cos1f x ax x=+−，a∈R.（1）当1πa=时，求函数()f x的最小值；（2）当12a≥时，证明：()0f x≥；（3）证明：()*1114coscos cos ,1233+++>−∈>n n n nN . 【答案】22. π14− 23. 证明见解析 24. 证明见解析【解析】【分析】（1）由题意可知：()f x 为偶函数，所以仅需研究0x ≥的部分，求导，分π2x >和π02x ≤<两种情况，利用导数判断原函数的单调性和最值；（2）由题意可知：()f x 为偶函数，所以仅需研究0x ≥的部分，求导，利用导数判断原函数的单调性和最值，分析证明；（3）由（2）可得：()211cos12>−≥n n n ，分2n =和3n ≥两种情况，利用裂项相消法分析证明； 【小问1详解】因为()f x 的定义域为R ，且()()()()22cos 1cos 1−−+−−+−f x a x x ax x f x ，所以()f x 为偶函数，下取0x ≥， 当1πa =时，()21cos 1π=+−f x x x ，则()2sin π′=−f x x x ， 当π2x >时，则()2sin 1sin 0π′=−>−≥f x x x x ，可知()f x 在π,2∞ +内单调递增， 当π02x ≤≤时，令()()g x f x ′=，则()2cos π′=−g x x ， 可知()g x ′在π0,2内单调递增， 因为201π<<，则0π0,2x ∃∈ ，使得02cos πx =， 当[)00,x x ∈时，()0g x ′<；当0π,2x x ∈ 时，()0g x ′>； 所以()g x 在[)00,x 上单调递减，在0π,2x上单调递增，且()π002g g == ，则()()0f x g x ′=≤在π0,2 内恒成立，可知()f x 在π0,2内单调递减； 综上所述：()f x 在π0,2 内单调递减，在π,2∞ + 内单调递增， 所以()f x 在[)0,∞+内的最小值为ππ124f =−， 又因为()f x 为偶函数，所以()f x 在R 内的最小值为π14−. 【小问2详解】由（1）可知()f x 为定义在R 上的偶函数，下取0x ≥，可知()2sin f x ax x ′=−，令()()2sin ϕ′==−x f x ax x ， 因12a ≥，则()2cos 1cos 0x a x x ϕ≥−′=−≥， 则()x ϕ在[)0,∞+内单调递增，可得()()00x ϕϕ≥=， 即()0f x ′≥在[)0,∞+内恒成立，可知()f x 在[)0,∞+内单调递增，所以()f x 在[)0,∞+内的最小值为()00f =，结合偶函数性质可知：()0f x ≥.【小问3详解】由（2）可得：当1a =时，()2cos 10=+−≥f x x x ，当且仅当0x =时，等号成立， 即2cos 1≥−x x ，令*1,2,=≥∈x n n nN ，则211cos 1>−n n ， 当2n =时，211324cos 1222433>−=>=−，不等式成立； 当3n ≥时，222114411cos 111124412121 >−=−>−=−− −−+n n n n n n ， 即111cos 122121 >−− −+n n n ，则有： 111cos 12235 >−− ，111cos 12357 >−− ，⋅⋅⋅，111cos 122121 >−− −+n n n ， 相加可得：()()11111425cos cos cos 12233213321− +++>−−−=−− ++n n n n n n ， 为因为3n ≥，则()250321−>+n n ，所以1114cos cos cos 233+++>− n n ； 综上所述：()*1114cos cos cos ,1233+++>−∈>n n n nN . 【点睛】方法点睛：利用导数证明不等式的基本步骤（1）作差或变形；（2）构造新的函数()f x ；（3）利用导数研究()f x 的单调性或最值；（4）根据单调性及最值，得到所证不等式；特别地：当作差或变形构造的新函数不能利用导数求解时，一般转化为分别求左、右两端两个函数的最值问题．。
易错点25 语法填空：无提示词之连词(4大陷阱)-备战2024年高考英语考试易错题(解析版)
易错点25 无提示词之连词目录01 易错陷阱（4大陷阱）02 举一反三【易错点提醒一】并列连词易混易错点【易错点提醒二】关系词易混易错点【易错点提醒三】名词性从句引导词易混易错点【易错点提醒四】状语从句引导词易混易错点03 易错题通关易错陷阱1：并列连词易混易错点。
【分析】并列连词解答出错时，主要原因是长难句引起句子结构分析出错。
其次由于句子逻辑意思理解出错而混淆and, but, or, 或not...but, not only...but also...，neither...nor等的用法。
同时并列结构存在不同层次，不仅仅是句子的并列，也可以是词与词、词组与词组、分句与分句的并列。
所连接的部分构成并列平行关系，认识这一点对解题尤为关键。
易错陷阱2：关系词易混易错点。
【分析】关系代词和关系副词也属于连词的范畴。
出错原因主要是关系代词和关系副词的基本用法不清。
需掌握以下考查要点的基础知识。
1.定语从句中缺少主语、宾语、表语和定语时用关系代词。
2.定语从句中缺少状语时才用关系副词。
This is the factory_which/that__ he visited yesterday.This is the factory _where/in which he worked last year.3.whose作定语，表示“先行词的...”，后面加名词。
4.介词+which/whom的区别。
5.that与which的区别。
6.as与which的区别。
易错陷阱3：名词性从句引导词易混易错点。
【分析】what引导名词性从句时，在从句中作主语、宾语、表语和定语。
而that引导名词性从句时，在从句中不充当句子成分，但引导主语从句、表语从句和同位语从句时通常不能省略。
易错陷阱4：状语从句引导词易混易错点。
【分析】考生对于常用从属连词一般掌握较好。
但一些特别的词本不属于从属连词，也可以引导时间、条件、原因等状语从句的用法需牢记。
超实用备战高考英语考试易错题——语法填空：有提示词之形容词和副词(3大陷阱) (解析版)
易错点22 有提示词之形容词和副词目录01 易错陷阱（3大陷阱）02 举一反三【易错点提醒一】词性、词形转换类易混易错点【易错点提醒二】级别类易混易错点【易错点提醒三】用法类易混易错点03 易错题通关养成良好的答题习惯，是决定高考英语成败的决定性因素之一。
做题前，要认真阅读题目要求、题干和选项，并对答案内容作出合理预测;答题时，切忌跟着感觉走，最好按照题目序号来做，不会的或存在疑问的，要做好标记，要善于发现，找到题目的题眼所在，规范答题，书写工整;答题完毕时，要认真检查，查漏补缺，纠正错误。
易错陷阱1：词性、词形转换类易混易错点。
【分析】形容词一般在词尾加-ly变为副词，但也有不规则变化形式需牢记。
易错陷阱2：级别类易混易错点。
【分析】易错陷阱3：用法类易混易错点。
【分析】形容词作定语用于修饰名词，常谓语名词之前；分词形容词作表语时，-ing类常修饰事物，如：exciting, surprising, moving, puzzling等；-ed类副词形容词常修饰人或人的表情，如：excited, surprised, moved, puzzled等。
形容词作状语，修饰主语，与主语构成逻辑上的主系表关系。
副词作状语，修饰动词、形容词、副词和句子。
【易错点提醒一】词性转换类易混易错点【例1】（广东省深圳市红岭中学2023-2024学年高三统考试题）It is an ancient _________ (architecture) complex with a history of more than 600 years. In the Ming and Qing Dynasties, twenty-four emperors lived here ruling China for nearly 500 years.【答案】architectural【解析】考查形容词。
句意：它是一个有600多年历史的古代建筑群。
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2023－2024学年第一学期高三四校联考（三）英语试卷命题学校东莞市第六高级中学命题人劳玉梅张衍文审题人劳玉梅张衍文说明：本试卷共8页，满分120分，考试用时120分钟。
注意事项: 1. 答卷前，考生务必将自己的姓名、班别、考生号、考场号和座位号填写在答题卡上。
因笔试不考听力，选择题从第二部分的“阅读”开始，试题序号从“21”开始。
2. 作答选择题时，选出每小题答案后，用2B铅笔在答题卡上对应题目选项的答案信息点涂黑；如需改动，用橡皮擦干净后，再选涂其他答案。
答案不能答在试卷上。
3. 非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内相应位置上；如需改动，先划掉原来的答案，然后再写上新答案；不准使用铅笔和涂改液。
不按以上要求作答无效。
4. 考生必须保持答题卡的整洁。
考试结束后，将答题卡交回。
第二部分阅读（共两节，满分50分）第一节（共15小题；每小题2.5分，满分37.5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。
AYou've probably tried easy centercity Beijing rides if you enjoy riding. Now it's time to level up your rides, so they will take you out into the countryside and attractive mountains.Death ValleyApproximate distance:100km loop (环线)It may sound terrifying, but Beijing's cyclists named this northern route “Death Valley” on account of the huge tomb it passes, not the number of onroad accidents it has witnessed. In reality, it's the closest, mostly carfree loop to the city center.Tuesday/Thursday TTTApproximate distance: 54km loopThis Team Time Trial is very popular among the Beijing cycling munity as an early morning ride before work. While it's not particularly scenic, there are few traffic lights, and the roads are not too busy and have adequate bike lanes (车道) on each side, making it a relatively troublefree ride. If you'd like to ride with a group, seek out early morning Beijing rider groups and join them on Tuesday or Thursday mornings at 5 am or 6 am depending on the season.The VerseApproximate distance: 120km loopIt passes a café with good (although expensive) food and even highend acmodation, but the best part is their super weling attitude towards cyclists. It's a bit of hike to get there but well worth the effort, as you are rewarded with amazing views of the Great Wall upon arrival, and a ride back that is mostly downhill.Tongzhou Grand Canal Forest ParkApproximate distance: 70km loopThis park is the closest place for riders to enjoy the Grand Canal in Beijing. The Tongzhou Grand Canal is a true wonder in China, and one of the greatest civil engineering projects in history. The oldest parts of it date back hundreds of years BC. It will be a fun ride here.21. Why was the northern route named“Death Valley”?A. It has deadly loops for cars.B. It is located in a distant place.C. There is a large tomb on the route.D. Numerous accidents had happened along the road.22. Which of the following covers the shortest distance?A. Death ValleyB. Tuesday / Thursday TTTC. The VerseD. Tongzhou Grand Canal Forest Park23. What do the last two routes have in mon?A. They both offer a view of historical sites.B. They both have an adequate bike lanes.C. They both pass a secondtonone café.D. They both offer a downhill ride.BI live in Xizhou in Yunnan Province, on the historic Tea Horse Road. I have to admit that when I first heard that Paul Salopek was going to walk the entire globe on his own two feet, I was blown away. I couldn't imagine that there could be such an unusual person in the world.Last May, I met Paul. He told me that it was his first time in China. He talked to me with great excitement about the history, migrations, and discoveries in my region of China. He spoke of the Shu Yandu Dao (the Southern Silk Road), the travels of the 17thcentury Chinese explorer Xu Xiake, the Tea Horse Road and the early 20thcentury American botanist Joseph Rock. He also talked of Xuanzang. Paul considered many of them heroes and in a sense Chinese pioneers of slow journalism.I decided to acpany Paul on his walk toward Yunnan. On September 28, 2021, we set out. Our days were simple: walk, eat, sleep, and repeat. We woke up at sunrise, set off in high spirits, and rested at sunset, dragging ourselves into exhausted sleep.We met many people on the road. Some were curious, surrounding us and watching us; some gave us directions; some invited us into their home to take a rest; some spoke of the charm of their hometown. We met many beautiful souls, simple souls and warm souls. We were walking with our minds.Together, we were impressed by the biodiversity of the Gaoligong Mountains. As I walked on ancient paths through mountains, I seemed to hear the antique voices of past travelers urging me to be careful on the road.Looking back on the more than 200 miles I walked with Paul, I came to a realization. Walking for its own sake, while healthy and admirable, is only a small part of the benefit of moving with our feet. A deeper reward is rediscovering the world around us, shortening the distance between each other, and sharing each other's cultures.24. How did the writer first respond to Paul's travel plan?A. Puzzled.B. Scared.C. Surprised.D. Disappointed.25. What can we learn about Paul Salopek from paragraph 2?A. He had a knowledge of China.B. He was a western journalist.C. He came to China several times.D. He was Joseph Rock's acquaintance.26. What does paragraph 4 tell us about the writer and Paul?A. They built bonds with people.B. They satisfied the locals' curiosity.C. They set off in high spirits.D. They honored the ancestors.27. What is the main purpose of the writer's writing the text?A. To suggest a new way of travel.B. To share and reflect on a journey.C. To advocate protection of biodiversity.D. To introduce and promote Chinese culture.COne key element of human language is semantics(语义). Scientists had long thought that unlike our words, animal vocalizations (发声) were involuntary, reflecting the emotional state of the animal without conveying any other information. But over the last four decades, numerous studies have shown that various animals have distinct calls with specific meanings.Many bird species use different alarm calls. Japanese tits, which nest in tree holes, have one call that causes their baby birds to get down to avoid being pulled out of the nest by crows, and another call for tree snakes that sends them jumping out of the nest entirely. Siberian jays vary their calls depending on whether an enemy is seen looking for food or actively attacking—and each call gets a different response from other nearby birds.Two recent studies suggested that the order of some birds' vocalizations may impact their meaning. Though the idea is still controversial,this could represent a basic form of the rules governing the order and bination of words and elements in human language known as syntax (句法), as illustrated by the classic "dog bites man" vs. "man bites dog" example.Even if some birds share basic aspects of human language, we still know very little about what’s actually going on in their minds. Most animal munication research has focused on describing signals and behavior, which on the surface can look a lot like human behavior. Determining if the underlying cognitive(认知的) processes driving the behavior are also similar is much more challenging, as at the heart of this question is intentionality: Are animals merely reacting to their environment, or do they intend to convey information to one another?28.What was scientists’ longheld belief about animal vocalizations?A.They conveyed no emotion.B. They were semantically related.C. They varied greatly with species.D. They expressed no intended meaning.29. How does the author develop paragraph 2?A. By listing data.B. By giving examples.C. By providing definition.D. By making parison.30. What does the underlined word “this” in paragraph 3 refer to?A. What birds’ vocalizations mean.B. How rules govern human language.C. What the two recent studies indicate.D. How bird’s vocalizations are bined.31. What does the last paragraph mainly tell us?A. Shared aspects of human and birds’ languages.B. Focus of most animal munication research.C. Underlying cognitive processes of birds’ vocalizations.D. Insufficient knowledge about birds’ munication intentionally.DAs athletes get stronger and faster, the pace of play continues to increase. The burden of making sure games are played according to the rules and that the officiating (裁判) is accurate is now being taken out of human hands and falling more and more into the lap of technology. It's called the video replay.The National Football League is expanding its replay system this uping season to include pass interference (传球干扰). Major League Baseball now relies on it for safeorout and home run calls. If you've been watching the FIFA World Cup, you may have noticed that the Video Assistant Referee (V AR) played a key role in almost every game. And in the Kentucky Derby, a horse was disqualified for knocking another horse. No one knew why until a video replay confirmed the call and controversy was avoided.However, many purists—those who want people to follow rules carefully and do things in the traditional way—especially in soccer, argue it's not the way the game was invented, and that the video replay is tainting the sport. But don't you want to see the proper application of the rules throughout the games? I know I do. Yes, it can slow the game down, but I feel it is worth it. If technological advancements allow fans watching from home to spot mistakes instantly, those same views need to be available to the officiating crews. Another example occurred in the most recent National Football Conference (NFC) Championship Game between the Los Angeles Rams and the New Orleans Saints. When obvious pass interference was mitted by the Los Angeles Rams player Nickell RobeyColeman, with just 109 seconds to play, no flag was raised on the field. It weakened the New Orleans Saints spirits. The Los Angeles Rams won a 2623 overtime victory. The nocall deeply angered the public. The video replay showed the referees had just missed one of the most apparent pass interference calls.There are no easy answers regarding replay technology and whether it is a curse (魔咒). But for me, keeping the officiating honest and on task is the right step in limiting controversy.32. What trend in sports can be observed in paragraph 2?A. The video replay has been widely used.B. League games have bee petitive.C. Rules of professional games are being stricter.D. People are showing more interest in sports than before.33. What does the underlined word “tainting” in paragraph 3 mean?A. Tricking.B. Promoting.C. Damaging.D. Restoring.34. What might the New Orleans Saints think of the referees in the NFC Championship Game?A. They relied a lot on the video replay.B. They cared too much about details.C. They were definitely stressed out.D. They were terribly disqualified.35. What would be the best title for the text?A. Video replays: highend technology in sportsB. Is technology like V AR a blessing in sports?C. Officiating: a duty that requires honestyD. What do qualified referees really mean?第二节（共 5 小题；每小题 2.5 分，满分12.5 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。