黑龙江省大庆市大庆实验中学2020-2021学年高三上学期实验三部第一次线上教学质量检测理综生物试卷

2020-2021学年黑龙江省大庆实验中学高三英语上学期期中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ACharlie Thorne and the Last Equationby Stuart GibbsThe CIA is on a task to find an equation (方程式) called Pandora, which could destroy the world if the wrong people get it. For help, they turn to Charlie, a 12-year-old girl who's as smart as Albert Einstein. People who like action-packed mysteries will enjoy reading this exciting book.AstroNutsby Jon Scieszka and Steven WeinbergIn AstroNuts, the Earth has been destroyed by humans for thousands of years. Four animals set out from Mount Rushmore, the headquarters (总部) of NNASA. Their task is to find a new planet fit for human life. Finally, they discover one: Plant Planet. The story's theme (主题) is simple: Don't harm the planet. Readers who love fantasy will enjoy AstroNuts.Stargazingby Jen WangChristine hears that Moon, who's new in town, is the kind of kid who beats people up for fun. But Moon and her mum come to live with Christine's family, and the two kids become best friends. Moon even shares a big secret with Christine. Stargazing is based on author Jen Wang's experiences as a child. The story is about the power of friendship and how people are able to change.Roll with Itby Jamie SumnerRoll with It is a story about a 12-year-old girl named Ellie. She has difficulty walking on her own and uses a wheelchair. When Ellie and her mum move to another state to take care of Ellie's grandpa, she must learn to navigate (处理) a new school and new friendships. This page-turner is a must-read for everyone. It's a heartwarming story that really shows the value of familyand how being different is special.1. Which book tells readers to protect the place we live in?A.AstroNutsB.Stargazing.C.Roll with It.D.Charlie Thorne and the Last Equation.2. What makes Stargazing different from the other three books?A. It talks about friendship.B. It tells stories about animals.C. It contains lots of scientific knowledge.D. It was written according to the author's experiences.3. What happened to Ellie?A. She had difficulty in making friends.B. She had an accident which left her in a wheelchair.C. She went to a new school and had to start all over again.D. She lost her mum and was taken care of by her grandpa.BMany of us were delighted to learn that a high school senior Kwasi Enin was accepted to all eight Ivy League universities. To our surprise, he wasn't excited as expected, but appeared extra calm. He announced that he would revisit the universities to find the best suitable in music or medicine. He also wanted to compare their financial aid packages.Kwasi's success story is rare, but his reaction is not. After the admission letters arrive at home, students have 30 days to really think about what kind of school would help them grow as a person, which school would best prepare them for the future, and at which school they would be happiest. And they also have to think about whether they can afford the school they choose.But how to answer the questions about which school is the best suitable university? Some young people are attracted to large universities with great school spirit and a list of offerings. But besides those advantages, many of these universities focus on graduate work and research, with undergraduates taught mostly by part-time instructors. Others are attracted to smaller boarding schools with discussion-based classes. But some of these schools will have much limitation for students who want a high-energy city life experience.Many students today seem to think they should pick the university where they will get the diploma that will help them get the most highly paid job. This is a sad misunderstanding of what a college education should provide.A good college education should prepare them to overcome any difficulty andthrivein society. It helps them to form the habit of creative mind and spirit that will continue to develop far beyond their university years. So when you choose college, you should consider if it is filled with useful learning to help create new spaces for different possibilities of growth.4. What can we know about Kwasi Enin from paragraph 1?A. He was from a very poor family.B. He would choose the top university.C. He was too excited to calm himself at the good news.D. He considered his interests when choosing his university.5. What can you infer from paragraph 2?A. Few students can be admitted to university.B. Many students face the choices like Kwasi.C. Top universities are the first choice for most students.D. American students can afford their university by themselves.6. Which of the following can best explain the underlined word “thrive” in paragraph 4?A. FailB. SucceedC. ResearchD. Work7. What should the best university be like according to the text?A. Very large and have good instructors.B. Small boarding schools with discussion-based classes.C. It will offerthe diploma to get the most highly paid job.D. It will help continue to develop far beyond university years.CWhile space travel still gets lot of attention, not enough attention has been paid to the exploration of oceans, about which we know much less than the dark side of the moon.Ninety percent of the ocean floor has not even been recorded and while we have been to the moon, the technology to explore the ocean's floors is still being developed. For example, a permanent partially-underwater sea exploration station, called the Sea Orbiter, is currently in development.The oceans play a major role in controlling our climate. But we have not learned yet how to use them to cool us off rather than contribute to our overheating. Ocean organisms are said to hold the promise of cures for a wide of the unique eyes of skate (ray fish) led to advances in conquering blindness, the horseshoe crab was important indeveloping a test for bacterial pollution, and sea urchins helped in the development of test-tube fertilization(人工授精). The toadfish's' ability to regenerate its central nervous system is of much interest to neuroscientists. A recent Japanese study concluded that the drug Eribulin, which was taken from sea sponges, is effective in fighting with breast, colon, and Urinary cancer.Given the approaching crisis of water insufficiency, we badly need to improve current methods, of desalinating(淡化) ocean water and make them more efficient and less costly. By 2025, 1.8 billion people are expected to suffer from severe water shortage, with that number jumping to 3. 9 billion by 2050-well over a third of the entire global population.If the oceansdo not make your heart go beating faster, how about engineering a bacterium that eats carbon dioxide — and thus helps protect the world from overheating — and produces fuel which will allow us to drive our cars and machines, without oil? I cannot find any evidence that people young or old, Americans or citizens of other nations would be less impressed or less inspired with such a breakthrough than with one more set of photos of a faraway galaxy or a whole Milky Way full of stars.8. What does the author think about the ocean exploration?A. It is equal to the space exploration.B. It is well developed.C. It deserves more attention and devotion.D. It is beyond our knowledge.9. What technology has been developed to make use of the oceans?A. Curing human diseases with ocean organisms.B. Preventing the world getting warmer.C. Mapping the global ocean floor.D. Removing salt from sea water.10. What does the author imply in the last paragraph?A. The temperature rise will be overcome by a bacterium.B. Solving the existing problems is more significant.C. The space exploration is worth the efforts.D. The ocean exploration is not inspiring.11. What is the best title of the passage?A. Oceans, the Last Hope.B. Oceans, the Hidden Treasure.C. Space, the Final Frontier.D. Space, the Faraway Dream.DAncient Dunhuang manuscripts housed abroad have been edited and published by the Institute for Overseas National Literature of Northwest Minzu University since 2006. Up till now, 9 manuscripts kept in the British Library and22 inthe National Library of France have been finished, the institute said on April 24, 2018.Tens of thousands of valuable ancient documents and cultural relics, discovered in the Mogao Grottos in Dunhuang, Gansu province, were scattered overseas in the early 20th century. Dunhuang manuscripts currently in the British Library and the National Library of France are the most important ancient national documents housed abroad.Co-edited by Institute for Overseas National Literature of Northwest Minzu University, Shanghai Classics Publishing House, the British Library and the National Library of France, these Dunhuang manuscripts return home in publication form for the first time. The institute is also preparing an online database of the manuscripts.According to Cai Rang, associate director of the institute, Dunhuang manuscripts scattered overseas in Russia, Britain, France and Japan have rich contents, including Buddhism law, social contract, history,linguisticsand art. The institute has edited and published 31 manuscripts over the past 13 years, but the work has not been finished. It plans to publish 15 from the British Library and over 30 from the National Library of France all together. In addition, it will also publish manuscripts collected by other countries.“Some manuscripts are hard to read because of the indecipherable words. So we read carefully and understand them by comparing with Buddhism documents handed down from ancient times,” Cai said. “Next, document classification and compilation will be our key work for further research.”The work done by the institute is helpful to study the history and culture of Tubo(present-day Tibet) during the period of 8th to 11th century and the history of national cultural exchanges at that time.12. When did so many valuable ancient documents, discovered in the Mogao Grottos, were scattered overseas?A. In the late 19th century.B. In the middle of the 19th century.C. At the beginning of the 19th century.D. At the beginning of the 20th century.13. How do the members of the institute understand some manuscripts that are hard to read?A. By using modern technology.B. By asking other famous experts.C. By comparing them with Buddhism documents.D. By studying the history and culture of Tubo.14. The possible meaning of the underlined word “linguistics” in paragraph 4 is “______”.A. the scientific study of languageB. the opinion that people have about someone or somethingC. something that people may have as part of their characterD. a system or method for carrying passengers or goods from one place to another15. What is the theme of the news report?A. Dunhuang manuscripts scattered overseas have rich contents.B. China publishes Dunhuang manuscripts housed overseas.C. High value of ancient documents and cultural relics in Mogao Grottos.D. Prepare an online database of Dunhuang manuscripts housed overseas.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。

黑龙江省大庆实验中学2020-2021学年高一物理上学期第一次线上教学质量检测（1月底）试题考试时间：90分钟满分：110分一、选择题（本题共14小题，每小题4分，共56分。

在每小题给出的四个选项中，第1题～第8题只有一个选项是符合题目要求的；第9题～第14题有多个选项是符合题目要求的，全部选对的得4分，选对但不全的得2分，有选错的得0分）1.下面说法中正确的是（）A．加速度变化的运动一定是曲线运动B．做曲线运动的物体速度方向一定变化C．速度方向变化的运动一定是曲线运动D．加速度恒定的运动一定不会是曲线运动2.物体从静止开始做匀加速直线运动，第3s内通过的位移是3m，则（）A．第3s内的平均速度是1m/s B．物体的加速度是1.2m/s2C．前3s内的位移是6m D．3s末的速度是4m/s3.如图所示，质量为m的正方体滑块，在水平力作用下，紧靠在竖直墙上，处于静止状态。

若要滑块静止，水平力的最小值为2.5mg（已知最大静摩擦力等于滑动摩擦力），则滑块与竖直墙间的动摩擦因数为（）A．0.4 B．0.5C．0.6 D．0.74.如图所示，在竖直平面内建立直角坐标系xOy，该平面内有AM、BM、CM三条光滑固定轨道，其中A、C两点处于同一个圆上，C是圆上任意一点，A、M分别为此圆与y、x 轴的切点．B点在y轴上且∠BMO＝60°，O′为圆心．现将a、b、c三个小球分别从A、B、C点同时由静止释放，它们将沿轨道运动到M点，如所用时间分别为t A、t B、t C，则t A、t B、t C大小关系是()A．t A＜t C＜t BB．t A＝t C＜t BC．t A＝t B＝t CD．由于C点的位置不确定，无法比较时间大小关系5. A 、B 两个球用轻弹簧连接，A 球质量为2m ，B 球质量为m ，小球A 由轻绳悬挂在天花板上O 点，两球处于平衡状态，如图所示。

现突然剪断轻绳OA ，让小球下落，在剪断轻绳的瞬间，设小球A 、B 的加速度分别用a 1和a 2表示，则（ ） A ．a 1=g ，a 2=g B ．a 1=0，a 2=2gC ．a 1=g ，a 2=0D ．a 1=1.5g ，a 2=06. 如图所示，在倾角为α的斜面上，放一质量为m 的小球，小球和斜坡及挡板间均无摩擦，当挡板绕O 点逆时针缓慢地转向水平位置的过程中，则有（ ） A ．斜面对球的支持力逐渐增大 B ．斜面对球的支持力不变 C ．挡板对小球的弹力先减小后增大 D ．挡板对小球的弹力先增大后减小 7. 如图所示，小船以大小为v(船在静水中的速度)、方向与上游河岸成θ的速度从O 处过河，经过一段时间，正好到达正对岸的O ＇处。

大庆实验中学2021年实验三部第一次线上教学质量检测高三英语学科试题说明：1. 本套试题答题时间100分钟，总分为150分。

2. 本套试题共分三部分：阅读理解、语言知识运用、写作。

第一部分阅读理解（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的四个选项(A、B、C、D) 中，选出最佳选项。

AKings CampsAbout Our CampsKings Camps is part of the Kings Active Foundation and is devoted to helping young people reach their potential. We provide sports camps and summer camps at over 40 places across the UK for children aged 4-17. We bring together the very best aspects of sports and holiday clubs to provide friendship, fun and exciting adventures for young people from the UK and around the world.Why Choose Kings Camps* Creative, inspiring camps: We have a strong belief that sport has an important role to play in a happy childhood and our not-for-profit status enables us to provide some places and invest in new and creative ways to inspire.* Learn important life skills: Kings Camps equip children with important life skills and understanding that being active is vital to health and well-being.* International students are welcome: We welcome children from outside of the UK who will make friends here by communicating with kids from different backgrounds, but we do request that they at least understand English to guarantee their safety and that of others, and of course their enjoyment.CommentsRyan has enjoyed every aspect. When we’ve asked about his day, he’s said it was “amazing and fantastic”. All the staff are extremely friendly, enthusiastic and have a true belief in what they are doing.-Tracy Lee Fantastic & friendly staff! It’s an action-packed week of multiple sports. My sweetheart made new friends, overcame swimming fears & came away more confident too! I cannot recommend it enough!-Kay Court 21．What does Kings Camps aim to do?A. Combine sports and holiday clubs.B. Get kids to diversity.C. Provide adventurous and challenging camps.D. Encourage kids to develop their potential. 22．What’s necessary for international students?A. Awareness of safety.B. Love for outdoor sports.C. Basic English ability.D. Good communicating skills.23．Who probably made the comments?A. Parents.B. Teachers.C. Campers.D. Staff.答案：DCABMy mom is about to have a spinal (脊柱的) operation. The operation is relatively minor, but does carry a risk of paralysis. Friends and family have reacted to this news by taking in such pessimistic terms that Mom has come to label this kind of talk as “psychological theft”. It occurs when other people increase your anxiety rather than provide comfort.Last week Mom went to the post office and ran into Geoff who works for the local school. “How are you?” he asked. “Not great,” she replied. “I’ve been having some trouble with my back and I’m going to need an operation.” “Oh, the back is the most dangerous place to operate on!” he responded. “My mom had that and she was in terrible pain. Make sure you get all your affairs in order before you go under the knife—it takes months to get over it!” Geoff’s intention had been benign(美好的). He’d given his own mother’s story to show sympathy. However, Mom only heard pain, danger, knife, and months.In the past few weeks, Mom’s mates and colleagues have told her stories about how their Auntie Trisha, who had just received an operation, was left speechless after hearing that she required a rapid follow-up operation; and how their neighbor, who had only a minor operation, never walked again.It’s really not that difficult to think of alternative things they could have said that would be equally true, but more beneficial to patients to h ear. “The specialists in our hospitals are among the best in the world. It’s amazing what they can do these days!”—that’s a good one. “You’re going to feel much better afterward.”—that’s another.I’m not suggesting patients should be sheltered from the reality of the risks they’re taking. But if the decision to have an operation can’t really be avoided, what’s the purpose of underlining the drawbacks? It’s just common sense to say: “Get well soon, and how can I help?24. What does “psychological theft” in Paragraph 1 refer to?A. The negative comments.B. Psychological disorder.C. The comforting words.D. Physical disability.25. How might the author’s mom have felt after hearing what Geoff said?A. SympatheticB. RelievedC. MovedD. Worried26. How did the author explain his idea in the text?A. By using examplesB. By analyzing causesC. By following time order.D. By discussing research findings27. What lesson does the author want to teach us?A. Treat patients with adoration.B. Communicate more with other patientsC. Be well-prepared for unavoidable operations.D. Find an appropriate way to comfort patients.答案：ADADCThe fashion industry urgently needs to transition to a “slow” fashion model to reduce its heavy environmental damage, according to researchers at Aalto University, Finland.The environmental impact of fashion’s global supply chain continues to rise, they say in a paper published in the journal Nature. The industry currently produces over 92 million tons of waste and consumes 79 trillion liters of water per year. It causes around 10 percent of the world’s greenhouse gas emissions (排放) and is also a high source chemical pollution. Developing countries tend to suffer most from the industry, since production often occurs there.The fashion industry is also projected to grow significantly in the coming decades. Fashion brands are already producing almost twice the amount of clothing they did 20 years ago, while global consumption of textiles (纺织品) is projected to increase by over 60 percent to 102 million tons by 2030, the paper said.“It’s critical that we, as consumers, accept that these cheap clothes are not possible if the environmental impacts are really all taken care of,” said Dr. Kirsi Niinimaiki. “In the future we should produce less. If we buy less, there will be less waste.”People in the UK buy more clothes per person than any other country in Europe. Last year, a cross-party parliamentary committee in the UK called on the government to add a 1 penny “producer responsibility charge” on each item of clothing to pay for better collection and recycling of unwanted clothes.Dr. Mark Sumner, a lecturer in sustainability in retail and fashion at the University of Leeds, said that the paper gave a good overview of the material aspects of the industry, but missed the opportunity to highlight efforts by more responsible brands to address these environmental problems.“There are definitely success stories out there in terms of what industry has done,” said Sumner, adding that not enough brands are using that best practice. He also said it’s important not to see “slow” fashion as the only answer.28．What is one of the major concerns about the fast fashion industry?A．Emission of greenhouse gases. B．Waste of clothing materials.C．A sharp increase in diseases. D．A fierce competition in fashion.29．What is the purpose of paragraph 5?A．To support the idea of less clothing production.B．To illustrate how fast fashion industry develops.C．To prove that British people buy more clothes.D．To show a measure to reduce clothes purchase. 30．What do Dr. Mark Sumner’s words imply?A．“Slow” fashion is a must. B．Consumers should buy less.C．A new approach is needed. D．Industry has done well enough.31．What does the text mainly talk about?A．Future of the fashion industry. B．Needs to deal with fast fashion.C．Reduction of clothing production. D．Changes to sustainability in fashion.答案：ADCBDLikely to suffer from loss of memory from time to time? Smart cameras can now remind you.Khai Truong at the University of Toronto in Canada and his colleagues have created a smartphone app that records interactions with household objects. The system involves barcode-like(像条形码的) markers that the user sticks to objects whose use they would like to track.With the smartphone worn around your neck, the app automatically records a short video clip(片段) when a marked object comes into view. “The user is able to look through the application and see the last time they interacted with it,” says Truong. The app can help people track the state of objects—such as whether they locked a door or switched a light off—as well as routine actions. At present, it successfully records about 75 percent of interactions, but only works for fixed objects.A similar but separate system can solve the problem. E. Akin Sisbot and Jonathan Connell at IBM Research in New York have invented a ceiling-mounted(安装在天花板上的) camera that monitors objects and people. It continuously watches an area, such as a tabletop in your home, tracking the placement of objects in relation to one another. It also remembers who first brought an object into the field of view as well as if anyone moved it afterwards. When asked, “Where is my wallet?” the system might respond, “It is next to the vase, un der the magazines.”The camera could also be used in factories or operating theaters to track vital tools, says Sisbot. For now, the camera uses a depth sensor to spot things. It is limited to detecting objects thicker than 3 centimeters, meaning that it has trouble with thin objects such as a closed laptop placed flat on a table.The accuracy of such smart camera systems may need to improve before they are widely adopted. “You’ve got to trust the technology for it to be of any comfort or reassurance” says Geoffrey Ward at the University of Essex in the UK.32.How does the smartphone app mentioned in paragraph 2 work?A. By switching off electricity automatically.B. By scanning barcodes of household objects.C. By recording the movement of marked objects.D. By informing owners of potential dangers.33. What is the limitation of the ceiling-mounted camera?A. It is unlikely to make a sound.B. It is unable to recognize movable objects.C. It fails to find objects thinner than 3 centimetres.D. It hardly senses objects without barcode-like makers.34. What’s Geoffrey Ward’s attitude towards the smart camera systems?A. ConcernedB. SupportiveC. SkepticalD. Ambiguous35. What’s the best title for the text?A. New Smartphones Make Life EasierB. Camera Designers Face New ChallengesC. Smart Cameras Help Increase Home SecurityD. New Systems Help People with Memory Problems答案：CCBD第二节(共5小题；每小题2分，满分10分)根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。

2020-2021学年黑龙江大庆实验中学高三上期第一次月考物理卷（解析版）姓名:_____________ 年级:____________ 学号:______________题型选择题填空题解答题判断题计算题附加题总分得分1. （知识点：滑动摩擦力f=μN）(8分）如图所示，质量为m的物体，放在一固定斜面上，物体与斜面间的动摩擦因数μ=当斜面倾角为θ时物体恰能沿斜面匀速下滑，此时再对物体施加一个大小为F的水平向右的恒力，物体可沿斜面匀速向上滑行。

试求：（1）斜面倾角θ；（2）水平向右的恒力F的大小。

【答案】（1）θ=30°（2）【解析】试题分析：（1）分析物体受力如图所示，物体匀速下滑时，mgsinθ=f （1分）FN=mgcosθ（1分）又f=μFN（1分）解得θ=30°（1分）(直接列mgsinθ=μmgcosθ也给分)评卷人得分（2）分析物体受力如图所示，物体沿斜面匀速上升，沿斜面方向：Fcos30°=mgsin30°+f1（1分）垂直于斜面方向：FN1=mgcos30°+Fsin30°（1分）又f1=μFN1 （1分）联立得：=（1分）考点：物体的平衡正交分解法（11分）如图甲所示，有一块木板静止在足够长的粗糙水平面上，木板质量为M=4kg，长为L=1.4m,木板右端放着一小滑块，小滑块质量为m=1kg，现用水平恒力F作用在木板M右端，恒力F取不同数值时，小滑块和木板的加速度分别对应不同数值，两者的a-F图象如图乙所示，取g=10m/s2，求:(1)小滑块与木板之间的滑动摩擦因数μ1，以及木板与地面的滑动摩擦因数μ2 ；(2)若水平恒力F=27.8N，且始终作用在木板M上，当小滑块 m从木板上滑落时，经历的时间为多长。

【答案】（1）（2）t=2s【解析】试题分析：（1）由乙图可知，当恒力时，小滑块与木板将出现相对滑动，由图可知小滑块的加速度以小滑块为研究对象，根据牛顿第二定律有（1分）解得（1分）以木板为研究对象，根据牛顿第二定律有（1分）得（1分）结合图象可得（1分）解得（1分）（2）设m在M上滑动的时间为t,当水平恒力F=27.8N时由（1）可知滑块的加速度为而滑块在时间t内的位移为（1分）由（1）可知滑块的加速度为（1分）而木板在时间t内的位移为（1分）由题可知（1分）联立解得t=2s(1分) 考点：牛顿第二定律匀加速直线运动【名师点睛】（1）当恒力时，小滑块与木板相对静止，而当恒力时，小滑块与木板将出现相对滑动，小滑块在滑动摩擦力f作用下向右匀加速运动的加速度，木板在拉力F和滑动摩擦力f作用下向右匀加速运动的加速度(2)当木板的加速度大于小物块的加速度时，小物块在木板上发生滑动；（3）使m恰能从M上面滑落下来的临界条件是：（17分）如图所示，倾角为45°的粗糙斜面AB底端与半径R=0.4m的光滑半圆轨道BC平滑相接，O为轨道圆心，BC为圆轨道直径且处于竖直平面内，A、C两点等高。

名篇名句默写专题黑龙江省实验中学第一次阶段测试（三）名篇名句默写（本题共1小题，6分）20．补写出下列句子中的空缺部分。

（1）白居易在《琵琶行》中，自述在江州的居住环境潮湿荒凉的句子是：“_______，_______。

”（2）李白在《蜀道难》一诗中，用“_______”，一句表示蜀地与秦地之间少有往来的情形，用“_______”进一步表明蜀地与秦地之间没有人可以通行的道路。

（3）屈原的《离骚》中，“_______”一句与“不忘初心”表达的情怀相同，“_______”一句点明作者坚信自己纯洁的品质并没有亏损。

【分析】此题考查了默写常见的名句名篇，能力层级为A，高考时，以《考试说明》规定的篇目为主，文体侧重于诗歌和散文，完成此类型题目，主要是靠同学们平时的积累，同时也要注意突破关键字（生僻字，通假字，同义异形字，语气助词等），避免错别字的出现，做题时，书写要工整清晰，留意语句的出处和具体的语境（情景默写的方式增加了对学生理解能力的考查，学生要在理解文意的基础上进行识记）。

【解答】故答案为：（1）住近湓江地低湿黄芦苦竹绕宅生（重点字：湓）（2）不与秦塞通人烟西当太白有鸟道（重点字：塞）（3）退将复修吾初服唯昭质其犹未亏（重点字：昭）【点评】《琵琶行》中的名句辑录：1．“嘈嘈切切错杂弹，大珠小珠落玉盘”由琴声想到珠玉声，是声音的类比联想。

2．描写琵琶女犹豫不决而出场的诗句是：千呼万唤始出来，犹抱琵琶半遮面。

3．“同是天涯沦落人，相逢何必曾相识”是全诗的主旨，更是诗人与琵琶女感情的共鸣。

4．既交代秋天的背景又蕴含离别之意的句子是：浔阳江头夜送客，枫叶荻花秋瑟瑟。

鹤岗一中第一次月考(三)名篇名句默写(9分)16. 补写出下面名句中的空缺部分。

(每空1分，共9分)(1) .李白在《蜀道难》一诗中引用神话传说为其增添了浪漫气息，如引用“五丁开山”一神话的句子是：“_______________，___________________。

大庆实验中学2020—2021学年度上学期第一次月考高三英语试题第二部分阅读理解（共两节，满分40 分）第一节（共15 小题; 每小题2 分，满分30 分）阅读下列短文，从每题所给的四个选项（A、B、C和D）中，选出最佳选项，并在答题卡上将该项涂黑。

ABelow you will find a profile of each gap year program. If there is a specific program you are interested in or a question you have regarding these programs, please feel free to contact them directly.Youth InternationalSince 1997, Youth International has been providing many people with the most exciting and educational experience of their life. It is a learning program that combines international travel, inter-cultural exchange, home stays, volunteer community service work and outdoor adventures.Phone: 720-270-3323Carpe Diem EducationThrough volunteer service, travel and cultural exchange, students of Carpe Diem Education receive a personal insight into themselves and their cultures. We specialize in guiding summer and semester programs for high school students. Each course is carefully crafted to develop leadership and self-awareness within our students, who return home better prepared to be leaders in thought and action.Phone: 503-285-1800Center for Interim ProgramsFounded in 1980, Center for Interim Programs is the first independent gap-year organization in America. Based on students’ interests and budget, our e xperienced teachers help make individual course choices including: low cost volunteer options, language programs, research trips, and so on. We are committed to helping students find more experiences of formal education and work.Phone: 413-585-0980EF Gap YearEF Gap Year offers students the ability to explore several countries where you will learn a language, volunteer for a good cause, gain international work experience. Choose between a12-week semester or 26-week academic year. You can then personalize your program to meet your learning goals and interests by deciding where, when, and-how you wish to study on your own. Our well-structured curriculum even allows you to place out of college-level classes and earn advanced credits toward graduation.1. Which phone number can you dial if you want to improve your leadership?A. 720-270-3323.B. 503-285-1800.C. 413-585-0980.D. 800-726-9746.2. What is EF Gap Year’s unique feature?A. Its reasonable fee.B. Its experienced teachers.C. Its personalized program.D. Its diverse curriculum choices.3. What do the four programs have in common?A. They pay attention to volunteer service work.B. They combine education with home stays.C. They mainly offer cross-cultural education.D. They raise students’ interest i n adventure.【答案】1. B 2. C 3. A【解析】这是一篇应用文。

大庆实验中学2020—2021学年度上学期期中考试高三物理试卷考试时间：90分钟满分：110分第I卷（选择题共56分）选择题共56分一、单选题（每题4分，共28分）1. 如图所示，在粗糙的水平地面上放着一左侧截面是半圆的柱状物体B，在B与竖直墙之间放置一光滑小球A，整个装置处于静止状态。

现用水平力F推动B缓慢向左移动一小段距离后，它们仍处于静止状态，在此过程中，下列判断正确的是（）A. 小球A对物体B的压力逐渐减小B. 墙面对小球A的支持力先增大后减小C. 地面对物体B的摩擦力逐渐增大D. 水平力F逐渐增大【答案】A【解析】【详解】AB．对A球受力分析，当推动B缓慢向左移动一小段距离后，F N1与竖直方向的夹角变小，整个过程动态变化如下图所示由图可知，F N1减小，根据牛顿第三定律，则小球A对物体B的压力逐渐减小；F N2也减小，则墙面对小球A的支持力一直减小，综上所述，A正确，B错误；C．以整体为研究对象，竖直方向受力平衡，则地面对B的支持力等于总重力，根据牛顿第三定律可得，B对地面的压力F N 不变，则地面对物体B 的摩擦力F f 也不变，C 错误； D ．整体水平方向受力平衡，根据平衡条件有2N f F F F =+由于静摩擦力不变，F N 2逐渐减小，则F 逐渐减小，D 错误。

故选A 。

2. 如图所示，从倾角为θ的斜面顶端分别以v 0和2v 0的速度水平抛出a 、b 两个小球，若两个小球都落在斜面上且不发生反弹，不计空气阻力，则a 、b 两球A. 水平位移之比为1∶2B. 下落的高度之比为1∶2C. 在空中飞行的时间之比为1∶2D. 落到斜面时速度方向与斜面夹角之比为1∶2 【答案】C 【解析】【详解】因为两个小球均落到斜面上，所以二者的位移偏转角相同，又由于初速度之比为1∶2，所以根据位移偏转角的正切值0tan 2gtv θ=，所以运动时间之比为1∶2，C 正确；再结合0x v t =，可得水平位移之比为1：4，A 错误；再根据212h gt =，下落的高度之比为1：4，B 错误；再根据速度偏转角的正切值是位移偏转角正切值的两倍可知，速度偏转角相同，速度方向与斜面夹角之比为1∶1，D 错误3. 如图所示，水平转台上有一个质量为m 的物块，用长为L 的细绳将物块连接在转轴上，细线与竖直转轴的夹角为θ角，此时绳中张力为零，物块与转台间动摩擦因数为μ（μ＜tanθ），最大静摩擦力等于滑动摩擦力，物块随转台由静止开始缓慢加速转动，则下列说法正确的是（ ）A. 转台一开始转动，细绳立即绷直对物块施加拉力B. 当绳中出现拉力时，转台对物块做的功为μmgL sinθC.D. 【答案】D 【解析】【详解】A ．转台一开始转动时，物体与转台之间的静摩擦力提供向心力，随转速的增加，静摩擦力逐渐变大，当达到最大静摩擦力时物块开始滑动，此时细绳绷直对物块施加拉力，选项A 错误； B ．对物体受力分析知物块离开圆盘前合力cos N T mg θ+= ②根据动能定理知当弹力T =0，r =L sinθ④由①②③④解得CD ．当N =0，f =0，由①②③知C错误，D 正确。

大庆实验中学2020-2021学年度上学期期中考试高三理科数学答案1．C 2．C 3.D 4.A 5.C 6.B 7.A 8.B 9. A 10.D 11. B 12.A 13.)3,1(- 14.3 15.52π 16．2317.（Ⅰ）3A π=（Ⅱ）33解：（Ⅰ）由2tan tan tan B bA B c=+及正弦定理可知， sin 2sin cos sin sin sin cos cos BB B A BC A B∴=+()2sin cos cos sin cos sin sin B A B B B A B C⋅∴⋅=+，所以2cos 1A =，又()0,A π∈，所以3A π=（Ⅱ）由余弦定理2222cos a b c bc A =+-， 得21393c c =+-，所以2340c c --=，即()()410c c -+=，所以4c =，从而113sin 343322ABCSbc A ==⨯⨯⨯= 18.(1)证明见解析；(2)60°.解析:（1）连结PD （P A=PB （PD AB （//DE BC （BCAB （DE AB （又PD DE D ⋂=（AB 平面PDE （PE ⊂平面PDE （∴AB PE （（2）法一（平面P AB 平面ABC，平面P AB 平面ABC=AB，PD AB，PD 平面ABC （ 则DE PD,又ED AB，PD 平面AB=D （DE 平面P AB,过D 做DF 垂直PB 与F ，连接EF ，则EF PB （∠DFE 为所求二面角的平面角（DE=32（DF =32，则3DE tan DFE DF∠==，故二面角的A PB E --大小为60︒法二：平面P AB 平面ABC，平面P AB 平面ABC=AB，PD AB，PD平面ABC （如图，以D 为原点建立空间直角坐标系（ B (1（0（0)（P (0（0（)（E (0（32（0)（ PB =(1（0（3-)（PE =(0（32（3-（（ 设平面PBE 的法向量()1,,n x y z =（30,330,2x z y z ⎧-=⎪⎨-=⎪⎩令3z =，得()13,2,3n =（DE ⊥平面P AB （∴平面P AB 的法向量为()20,1,0n =（ 设二面角的A PB E --大小为，由图知，1212121,2n n cos cos n n n n θ⋅===⋅（ 所以60,θ=︒即二面角的A PB E --大小为60︒. 19.（1）70.5分；（2）634人；（3）0.499 （1）由题意知： 中间值 45 55 65 75 85 95 概率0.10.150.20.30.150.1∴450.1550.15650.2750.3x =⨯+⨯+⨯+⨯ 850.15950.170.5+⨯+⨯=， ∴4000名考生的竞赛平均成绩x 为70.5分. （2）依题意z 服从正态分布()2,N μσ，其中70.5x μ==，2204.75D σξ==，14.31σ=，∴z 服从正态分布()()22,70.5,14.31N N μσ=，而()(56.1984.81)0.6826P z P z μσμσ-<<+=<<=，∴()10.682684.810.15872P z -≥==.∴竞赛成绩超过84.81分的人数估计为0.158********.8⨯=人634≈人.（3）全市竞赛考生成绩不超过84.81分的概率10.15870.8413-=.而()4,0.8413B ξ~，∴()()44431410.8413P P C ξξ≤=-==-⋅ 10.5010.499=-=. 20.（1）证明见解析，21nn a =-；（2）11202.（1）证明：因为n ，n a ，n S 成等差数列，所以2n n S n a +=，① 所以()1112n n S n a --+-=()2n ≥.②①－②，得1122n n n a a a -+=-，所以()1121n n a a -+=+()2n ≥. 又当1n =时，1112S a +=，所以11a =，所以112a +=， 故数列{}1n a +是首项为2，公比为2的等比数列，所以11222n n n a -+=⋅=，即21nn a =-.（2）根据（1）求解知，()22log 121121nn b n =+--=-，11b =，所以12nnb b ，所以数列{}n b 是以1为首项，2为公差的等差数列.又因为11a =，23a =，37a =，415a =，531a =，663a =，7127a =，8255a =，64127b =，106211b =，107213b =，所以()()1210012107127c c c b b b a a a +++=+++-+++()()7127212107(1213)107214222772212-⨯+⨯⎡⎤=-+++-=-+⎣⎦-281072911202=-+=.21．（Ⅰ）见解析；（Ⅱ）.1k解析（（I（()()21ln '1xx f x x e x -=++（ 易知()'f x 在()0e ，上为正，因此()f x 在区间()01，上为增函数，又1210ee ef e e-⎛⎫=< ⎪⎝⎭（0f I e =>（）因此10f f I e ⎛⎫< ⎪⎝⎭（），即()f x 在区间()01，上恰有一个零点（由题可知()0f x >在()1+∞，上恒成立，即在()1+∞，上无零点， 则()f x 在()0+∞，上存在唯一零点（ （II（设()f x 的零点为0x ，即0000ln 0x x x e x +=（原不等式可化为ln 1x xe x k x--≥（令()ln 1x xe x g x x--=（则()ln 'x x xe x g x x+=（由（I（可知()g x 在()00x ，上单调递减， 在()0x ，+∞上单调递增（故只求()0g x （，设00x x e t =（下面分析0000ln 0x x x e x +=（设00x x e t =，则ln x t x =-（ 可得0000lnx tx lnx x lnt =-⎧⎨+=⎩，即()01ln x t t -=若1t >，等式左负右正不相等，若1t <（等式左正右负不相等（只能1t =（因此()0000000ln 1ln 1x x e x xg x x x --==-=，即1k 求所求（ 22. （1）S 的普通方程为：)040(0422≥≤<=-+y x x y x ，或)0,0(≥>y x 或)0,0(≥≠y x方程写标准式也可S 的极坐标方程为：)20(cos 4πθθρ<≤=（不写范围扣2分）（2）]3,0[πα∈23.（1）见证明；（2）35[,]22-. 【详解】解：（1）由柯西不等式得22222)11x x ⎡⎤⎛⎡⎤+≥⋅ ⎢⎥⎣⎦⎣⎦⎝+． ∴()22243()3x yx y +⨯≥+，当且仅当3x y =时取等号． ∴22334x y +≥；（2）1111()224y x x y x y x y x y ⎛⎫+=++=++≥+= ⎪⎝⎭， 要使得不等式11|2||1|a a x y+≥-++恒成立，即可转化为|2||1|4a a -++≤， 当2a ≥时，421a -≤，可得522a ≤≤， 当1a 2-<<时，34≤，可得1a 2-<<，当1a ≤-时，214a -+≤，可得312a -≤≤-，∴a 的取值范围为：35[,]22-．。

2020-2021学年黑龙江省大庆实验中学高三（上）期中数学试卷（文科）一、单选题（本大题共12小题，共60.0分）1. 设集合A ={x ∈N||x|≤2}，B ={y|y =1−x 2}，则A ∩B =( )A. {x|−2≤x ≤1}B. {0,1}C. {1,2}D. {x|0≤x ≤1}2. 已知复数z 满足z(1−i)=2i ，则复数z 在复平面内对应的点所在象限为( )A. 第一象限B. 第二象限C. 第三象限D. 第四象限3. 若0<b <1，则“a >b 3”是“a >b ”的( )A. 充分不必要条件B. 必要不充分条件C. 充要条件D. 既不充分也不必要条件4. 函数f(x)=x 2+ln|x|x的图象大致为( )A.B.C.D.5. 若m ，n 是两条不同的直线，α，β是两个不同的平面，则下列命题正确的是( )A. 若α⊥β，m ⊥β，则m//αB. 若m//α，n ⊥m ，则n ⊥αC. 若m//α，n//α，m ⊂β，n ⊂β，则α//βD. 若m//β，m ⊂α，α∩β=n ，则m//n6. 设x ，y 满足约束条件{x +y ≤43x +y ≥2x −y ≤2，则z =3x −y 的最大值为( )A. 4B. 6C. 8D. 107. 在边长为2的正方形ABCD 中，E 为CD 的中点，AE 交BD 于F.若AF ⃗⃗⃗⃗⃗ =x AB ⃗⃗⃗⃗⃗ +3y AD⃗⃗⃗⃗⃗⃗ ，则x +y =( )A. 1B. 59 C. −13 D. −598. 已知函数f(x)=ln(x +√x 2+1)，若正实数a ，b 满足f(4a)+f(b −1)=0，则1a +1b的最小值为( )A. 4B. 8C. 9D. 139. 已知tanθ+1tanθ=4，则cos 2(θ+π4)=( )A. 12B. 13C. 14D. 1510. 如图，四边形ABCD 是边长为1的正方形，MD ⊥ABCD ，NB ⊥ABCD.且MD =NB =1.则下列结论中：①MC ⊥AN②DB//平面AMN ③平面CMN ⊥平面AMN④平面DCM//平面ABN 所有假命题的个数是( )A. 0B. 1C. 2D. 311. 已知函数f(x)=√3cos(2x −π2)−cos2x ，若要得到一个奇函数的图象，则可以将函数f(x)的图象( )A. 向左平移π6个单位长度 B. 向右平移π6个单位长度 C. 向左平移π12个单位长度D. 向右平移π12个单位长度12. 已知函数f(x)=xlnx +ae x 有两个极值点，则实数a 的取值范围是( )A. (−∞,1e )B. (0,1e )C. (−1e ,+∞)D. (−1e ,0)二、单空题（本大题共4小题，共20.0分）13. 已知A(−1,0)，B(1,2)，C(1,t)，若AB ⃗⃗⃗⃗⃗ //BC ⃗⃗⃗⃗⃗ ，则t 的值是______ ．)的14.函数f(x)=Asin(ωx+φ)(A>0,ω>0,|φ|<π2部分图象如图所示，则f(x)=______ ．15.等差数列{a n}中，其前n项和为S n，满足a3+a4=12，a5=9，则S7的值为______ ．16.中国古代数学经典《九章算术》中，将四个面都为直角三角形的三棱锥称之为鳖臑(biēnào).若三棱锥P−ABC为鳖臑，且PA⊥平面ABC，PA=2，AB=3，AB⊥BC，该鳖臑的外接球的表面积为29π，则该鳖臑的体积为______．三、解答题（本大题共6小题，共70.0分）17.等差数列{a n}的前n项和为S n，若a4+a5=16，S6=36．(1)求{a n}的通项公式；(2)设b n=1，求{b n}的前n项和T n．a n⋅a n+118.△ABC的内角A，B，C的对边分别为a，b，c，已知△ABC的面积为b2．3sinB(1)求sin A sin C；(2)若cosAcosC=1，b=3，求a+c的值．619.如图，在正三棱柱ABC−A1B1C1中，AB=AA1=2，E，F分别为AB，B1C1的中点．(Ⅰ)求证：B1E//平面ACF；(Ⅱ)求三棱锥B1−ACF的体积．20.在数列{a n}中，a1=1，a2=3.a n+2=3a n+1−2a n−n+1．(1)证明{a n+1−a n−n}为等比数列；(2)求a n．21.为了缓解日益拥堵的交通状况，不少城市实施车牌竞价策略，以控制车辆数量.某地车牌竞价的基本规则是：①“盲拍”，即所有参与竞拍的人都要网络报价一次，每个人不知晓其他人的报价，也不知道参与当期竞拍的总人数；②竞价时间截止后，系统根据当期车牌配额，按照竞拍人的出价从高到低分配名额.某人拟参加2020年11月份的车牌竞拍，他为了预测最低成交价，根据竞拍网站的数据，统计了最近5个月参与竞拍的人数(见表)：月份 2020.6 2020.07 2020.08 2020.09 2020.10 月份编号t 1 2 3 4 5 竞拍人数y(万人)0.50.611.41.7(1)由收集数据的散点图发现，可用线性回归模型拟合竞拍人数y(万人)与月份编号t 之间的相关关系.请用最小二乘法求y 关于t 的线性回归方程y ̂=b ̂t +a ，并预测2020年11月份参与竞拍的人数．(2)某市场调研机构从拟参加2020年11月份车牌竞拍人员中，随机抽取了200人，对他们的拟报价价格进行了调查，得到如下频数分布表和频率分布直方图： 报价区间(万元) [1,2) [2,3) [3,4)[4,5) [5,6] [6,7) [7,8] 频数1030a 60302010(ⅰ)求a 、b 的值及这200位竟拍人员中报价大于5万元的人数；(ⅰ)若2020年11月份车牌配额数量为3000，假设竞拍报价在各区间分布是均匀的，请你根据以上抽样的数据信息，预测(需说明理由)竞拍的最低成交价． 参考公式及数据：①y ̂=b ̂x +a ̂，其中b ̂=∑x i n i=1y i −nx −y−∑x i 2n i=1−nx−2，a ̂=y −−b ̂x −；②∑t i 25i=1=55，∑t i 5i=1y i =18.8．22.已知函数f(x)=lnx+ax (a∈R)的图象在x=1e处的切线斜率为−e．(1)求实数a的值，并讨论f(x)的单调性；(2)若g(x)=e x⋅f(x)，证明：g(x)>1．答案和解析1.【答案】B【解析】解：∵集合A={x∈N||x|≤2}={x∈N|−2≤x≤2}={0,1，2}，B={y|y=1−x2}={y|y≤1}，∴A∩B={0,1}．故选：B．分别求出集合A和B，由此能求出A∩B．本题考查交集的求法，考查交集、不等式性质等基础知识，考查运算求解能力和思维能力，考查函数与方程思想，是基础题．2.【答案】B【解析】解：由z(1−i)=2i，得z=2i1−i =2i(1+i)(1−i)(1+i)=−1+i，∴复数z在复平面内对应的点的坐标为(−1,1)，所在象限为第二象限．故选：B．把已知等式变形，利用复数代数形式的乘除运算化简，求出z的坐标得答案．本题考查复数代数形式的乘除运算，考查复数的代数表示法及其几何意义，是基础题．3.【答案】B【解析】解：根据0<b<1推知b>b3，由此可推“a>b3”是“a>b”的必要不充分条件．故选：B．根据0<b<1推知b与b3的大小关系，由此可推“a>b3”是“a>b”的关系．本题考查充分必要条件的判断，考查基本的推理能力，属于基础题．4.【答案】C【解析】解：函数的定义域为(−∞,0)∪(0,+∞)，当x>0时，f(x)=x2+lnxx ，f(1e)=1e2+ln1e1e=1e2−e>0，故AD不符合，当x <0时，f(x)=x 2+ln(−x)x，f(−1)=1>0，当x →−∞时，f(x)→+∞，故A 不符合， 故选：C ．根据函数值的正负即可判断．本题考查了函数图象的识别，关键掌握函数值的正负，属于基础题．5.【答案】D【解析】解：若α⊥β，m ⊥β，则m 与α可能平行也可能相交，故A 错误； 若m//α，n ⊥m ，则n ⊂α或n//α或n 与α相交，故B 错误； 若m//α，n//α，m ⊂β，n ⊂β，则α//β或α与β相交，故C 错误； 若m//β，m ⊂α，α∩β=n ，则m//n ，故D 正确． 故选：D ．根据空间线面位置关系的判定或定义进行判断．本题考查了空间线面位置关系的判断与性质，属于中档题．6.【答案】C【解析】解：由x ，y 满足约束条件{x +y ≤43x +y ≥2x −y ≤2作出可行域如图，联立{x +y =4x −y =2，解得A(3,1)，化目标函数z =3x −y 为y =3x −z ， 由图可知，当直线y =3x −z 过A 时，直线在y 轴上的截距最小，z 有最大值为8． 故选：C ．由约束条件作出可行域，化目标函数为直线方程的斜截式，数形结合得到最优解，联立方程组求得最优解的坐标，代入目标函数得答案．本题考查简单的线性规划，考查数形结合的解题思想方法，是中档题．7.【答案】B【解析】解：解法一，以A 为原点，AB 为x 轴，建立平面直角坐标系，则AB ⃗⃗⃗⃗⃗ =(2,0)，AD ⃗⃗⃗⃗⃗⃗ =(0,2)， 所以x AB⃗⃗⃗⃗⃗ +3y AD ⃗⃗⃗⃗⃗⃗ =(2x,6y)； 根据题意知，FEAF =DEAB =12，所以AE ⃗⃗⃗⃗⃗ =(1,2)，AF ⃗⃗⃗⃗⃗ =23AE ⃗⃗⃗⃗⃗ =(23,43)， 所以2x =23，6y =43， 解得x =13，y =29， 所以x +y =13+29=59．解法二：根据题意知，FEAF =DEAB =12， 所以DF =12FB ，DF =13DB ，所以AF ⃗⃗⃗⃗⃗ =AD ⃗⃗⃗⃗⃗⃗ +DF ⃗⃗⃗⃗⃗ =AD ⃗⃗⃗⃗⃗⃗ +13DB ⃗⃗⃗⃗⃗⃗ =AD ⃗⃗⃗⃗⃗⃗ +13(AB ⃗⃗⃗⃗⃗ −AD ⃗⃗⃗⃗⃗⃗ )=13AB ⃗⃗⃗⃗⃗ +23AD ⃗⃗⃗⃗⃗⃗ ， 所以AF ⃗⃗⃗⃗⃗ =x AB ⃗⃗⃗⃗⃗ +3y AD ⃗⃗⃗⃗⃗⃗ ， 即{x =133y =23， 解得{x =13y =29，所以x +y =13+29=59． 故选：B ．解法一，建立适当的平面直角坐标系，利用坐标表示向量，根据平面向量的坐标表示与线性运算，列方程求出x 、y 的值，再求和．解法二：根据平面向量的线性表示与运算，利用平面向量的基本定理列出方程求得x 、y的值，再求和．本题考查了平面向量的线性表示与运算问题，也考查了运算求解能力，是中档题．8.【答案】C【解析】【分析】本题考查函数的最值及其几何意义，考查函数单调性、奇偶性的性质及应用，训练了利用基本不等式求最值，是难题．证明函数f(x)为奇函数且是定义域内的增函数，由已知可得4a+b=1，然后利用“1的代换”结合基本不等式求最值．【解答】解：函数f(x)=ln(x+√x2+1)的定义域为R，且f(−x)=ln(−x+√x2+1)=x+√x2+1=−ln(x+√x2+1)=−f(x)，∴f(x)为R上的奇函数，又x+√x2+1是[0,+∞)上的增函数，由复合函数的单调性可知f(x)在[0,+∞)上是增函数，则f(x)是定义域上的增函数，又f(4a)+f(b−1)=0，得f(4a)=−f(b−1)=f(1−b)，∴4a=1−b，即4a+b=1．又a>0，b>0，∴1a +1b=(1a+1b)(4a+b)=5+ba+4ab≥5+2√ba⋅4ab=9，当且仅当b=2a，即a=16,b=13时取等号．故选：C．9.【答案】C【解析】解：由tanθ+1tanθ=4，得sinθcosθ+cosθsinθ=4，即sin2θ+cos2θsinθcosθ=4，∴sinθcosθ=14，∴cos2(θ+π4)=1+cos(2θ+π2)2=1−sin2θ2=1−2sinθcosθ2=1−2×142=14．故选：C．由已知求得sinθcosθ的值，再由二倍角的余弦及诱导公式求解cos2(θ+π4)的值．本题考查三角函数的化简求值，考查了同角三角函数基本关系式及诱导公式的应用，是基础题．10.【答案】B【解析】解：∵四边形ABCD是边长为1的正方形，MD⊥平面ABCD，NB⊥平面ABCD，且MD=BN=1，∴将题中的几何体放在正方体ABCD−A′NC′M中，如图所示．对于①，MC与AN是棱长为1的正方体中，位于相对面内的异面的面对角线，因此可得MC、AN所成角为90°，可得MC⊥AN，故①正确；对于②，∵DB//MN，MN⊂平面AMN，DB⊄平面AMN，∴DB//平面AMN，故②正确；对于③，正方体ABCD−A′NB′M中，由二面角A−MN−C的大小不是直角，得面CMN⊥面AMN不成立，故③不正确；对于④，∵面DCM与面ABN分别是正方体ABCD−A′NC′M的内外侧面所在的平面，∴面DCM//面ABN成立，故④正确．∴所有假命题的个数是1个．故选：B．由题意将题中的几何体放在正方体ABCD−A′NC′M中，再根据正方体的性质和空间垂直、平行的有关定理，对四个命题逐一判断得答案．本题考查命题的真假判断与应用，考查空间想象能力与思维能力，是中档题．11.【答案】C【解析】解：f(x)=√3cos(2x −π2)−cos2x =√3sin2x −cos2x =2sin(2x −π6)， 将函数f(x)2=sin(2x −π6)的图象向左平移π12个单位， 可得y =2sin[2(x +π12)−π6]=2sin2x 的图象， 显然，y =sin2x 为奇函数， 故选：C ．利用辅助角公式化积，结合y =Asin(ωx +φ)的图象变换规律及正弦函数、余弦函数的奇偶性得出结论．本题主要考查y =Asin(ωx +φ)的图象变换规律，正弦函数、余弦函数的奇偶性，是中档题．12.【答案】D【解析】解：∵f′(x)=1+lnx +ae x ，由题意，f′(x)=1+lnx +ae x =0有两个不同的实根， 即y =−a 和y =1+lnx e x在(0,+∞)上有两个交点，令g(x)=1+lnx e x ，∴g′(x)=1x−lnx−1e x．记ℎ(x)=1x −lnx −1，ℎ(x)在(0,+∞)上单调递减，且ℎ(1)=0，所以当x ∈(0,1]时，ℎ(x)≥0，g′(x)≥0，所以g(x)在(0,1]上单调递增； 当x ∈(1,+∞)时，ℎ(x)<0，g′(x)<0，所以g(x)在(1,+∞)上单调递减， 故g(x)max =g(1)=1e .当x →0时，g(x)→−∞；当x →+∞时，g(x)→0， 当0<−a <1e ，即−1e <a <0时，y =−a 和y =1+lnx e x在(0,+∞)上有两个交点，故选：D ．求出f′(x)=1+lnx +ae x ，由题意可得y =−a 和y =1+lnx e x在(0,+∞)上有两个交点，令g(x)=1+lnx e x，g′(x)=1x−lnx−1e x.记ℎ(x)=1x −lnx −1，ℎ(x)在(0,+∞)上单调递减，g(x)在(0,1]上单调递增；求解函数的最值，然后推出结果．本题考查了函数的单调性、最值问题，考查导数的应用以及转化思想，是一道中档题．13.【答案】2【解析】解：A(−1,0)，B(1,2)，C(1,t)， 所以AB ⃗⃗⃗⃗⃗ =(2,2)，BC ⃗⃗⃗⃗⃗ =(0,t −2)， 又AB ⃗⃗⃗⃗⃗ //BC ⃗⃗⃗⃗⃗ ，所以2(t −2)−2×0=0， 解得t =2． 故答案为：2．由平面向量的坐标表示与共线定理，列方程求出t 的值．本题考查了平面向量的坐标表示与共线定理应用问题，是基础题．14.【答案】2sin(2x −π3)【解析】解：由图象可得：34T =5π12+π3，且A =2，则T =π，所以T =2πω=π，即ω=2，所以f(x)=2sin(2x +φ)，又f(x)max =f(5π12)=2sin(2×5π12+φ)=2， 所以φ+5π6=π2+2kπ，k ∈Z ，则φ=2kπ−π3，k ∈Z ，又|φ|<π2， 所以φ=−π3，所以函数f(x)是解析式为：f(x)=2sin(2x −π3)， 故答案为：f(x)=2sin(2x −π3).利用图象求出A 和周期，进而可以求出ω，再利用最值求出φ的值即可． 本题考查了三角函数的图象性质，涉及到周期和最值问题，属于基础题．15.【答案】49【解析】解：∵a 3+a 4=12，a 5=9， ∴2a 1+5d =12，a 1+4d =9， 解得a 1=1，d =2， ∴S 7=7a 1+7×62×d =7+42=49，故答案为：49．利用等差数列的通项公式和前n 项和公式求解．本题考查等差数列的通项公式和前n项和公式的合理运用，是基础题．解题时要认真审题，仔细解答．16.【答案】4【解析】解：由题意，三棱锥P−ABC为鳖臑，且PA⊥平面ABC，PA=2，AB=3，如图所示，∠PAB=∠PAC=∠ABC=∠PBC=90°，又该鳖臑的外接球的表面积为29π，可知PC为外接球的直径，则R2=29π4π=294，∴BC=√4R2−PA2−AB2=√4×294−22−32=4，则该鳖臑的体积为：V=13×12×3×2×4=4．故答案为：4．由题意知PC是三棱锥P−ABC的外接球直径，由外接球直径求得BC的长，再计算三棱锥的体积．本题考查了三棱锥外接球的体积计算问题，是中档题．17.【答案】解：(1)由题意得{a4+a5=2a1+7d=16, S6=6a1+6×52d=36, 解得{a1=1, d=2, 所以a n=1+2(n−1)=2n−1；(2)b n=1a n⋅a n+1=1(2n−1)(2n+1)=12(12n−1−12n+1)，所以T n=12(1−13+13−15+⋯+12n−1−12n+1)=12(1−12n+1)=n2n+1．故{b n }的前n 项和T n 为：n2n+1．【解析】本题考查等差数列通项公式的应用，数列求和的方法，考查计算能力，属于中档题．(1)利用已知条件求出数列的首项与公差，然后求解通项公式； (2)化简通项公式利用裂项相消法求解数列的和即可．18.【答案】解：(1)△ABC 的内角A ，B ，C 的对边分别为a ，b ，c ，∵△ABC 的面积为b 23sinB，∴12ac ⋅sinB =b 23sinB ，即2b 2=3ac ⋅sinBsinB ． 再利用正弦定理可得2sin 2B =3sinAsinC ⋅sin 2B ， 因为sinB >0， ∴sinAsinC =23．(2)cosAcosC =16，b =3，sinAsinC =23，∴cosAcosC −sinAsinC =−12=cos(A +C)=−cosB ，∴cosB =12，∴B =π3． 由正弦定理，asinA =bsinB =c sinC =2R =2√3， ∴sinAsinC =a 2R ⋅c 2R=ac 4R2=ac12=23，ac =8，再根据余弦定理，b 2=9=a 2+c 2−2ac ⋅cosB =(a +c)2−3ac ， ∴(a +c)2=9+3ac =33，∴a +c =√33．【解析】(1)由题意利用正弦定理求得sin A sin C 的值．(2)由题意利用两角差的余弦公式求得cos B 的值，可得B 的值，再利用正弦定理求得ac 的值，利用余弦定理求得a +c 的值．本题主要考查正弦定理、余弦定理的应用，属于中档题．19.【答案】(Ⅰ)证明：取AC 的中点M ，连结EM ，FM ，在△ABC 中，∵E 、M 分别为AB ，AC 的中点，∴EM//BC 且EM =12BC ，又F为B1C1的中点，B1C1//BC，∴B1F//BC且B1F=12BC，即EM//B1F且EM=B1F，故四边形EMFB1为平行四边形，∴B1E//FM，又MF⊂平面ACF，B1E⊄平面ACF，∴B1E//平面ACF；(Ⅱ)解：设O为BC的中点，∵棱柱底面是正三角形，AB=2，∴有AO=√3，且AO⊥平面BCC1B1，于是V B1−ACF =V A−B1CF=13×S△B1CF×AO=13×12×2×√3=√33．【解析】(Ⅰ)取AC的中点M，连结EM，FM，由三角形中位线定理可得EM//BC且EM=12BC，结合已知得到EM//B1F且EM=B1F，则四边形EMFB1为平行四边形，可得B1E//FM，再由线面平行的判定可得B1E//平面ACF；(Ⅱ)设O为BC的中点，由已知得到AO⊥平面BCC1B1，然后利用等积法求三棱锥B1−ACF的体积．本题考查直线与平面平行的判定，考查空间想象能力与思维能力，训练了利用等积法求多面体的体积，是中档题．20.【答案】解：(1)由a n+2=3a n+1−2a n−n+1得a n+2−a n+1−(n+1)=2(a n+1−a n−n)，又a2−a1−1=1，所以{{a n+1−a n−n}是以1为首项，2为公比的等比数列；(2)由(1)可得a n+1−a n−n=2n−1，∴a n−a n−1−(n−1)=2n−2，即a n−a n−1=2n−2+n−1，∴a n−1−a n−2=2n−3+(n−2)，…a2−a1=1+1，用累加法即可的a n−1=2n−2+2n−3+⋯+1+(n−1)+(n−2)+⋯+1，即a n−1=2n−1−1+n(n−1)2，∴a n=2n−1+n2−n2，又a1=1也满足上式，∴a n=2n−1+n2−n2．【解析】(1)由a n+2=3a n+1−2a n −n +1得a n+2−a n+1−(n +1)=2(a n+1−a n −n)， 又a 2−a 1−1=1，所以可得{{a n+1−a n −n}是以1为首项，2为公比的等比数列； (2)由(1)可得a n+1−a n −n =2n−1，∴a n −a n−1−(n −1)=2n−2，即a n −a n−1=2n−2+n −1， 最后用累加法即可求出a n ，同时验证a 1=1是否满足即可．本题第一问考查通过变形将已知条件变为等比数列的形式，再求出首项，第二问利用累加法分组求和从而计算出通项公式，设计巧妙，值得借鉴．本题属于难题．21.【答案】解：(1)由题意求出t −=13(1+2+3+4+5)=3，y −=15(0.5+0.6+1.4+1.7)=1.04，由∑t i 25i=1=55，∑t i 5i=1y i =18.8， b ̂=∑t i 5i=1y i −5t −⋅y−∑t i 25i=1−5t2=18.8−5×3×1.0455−5×32=3.210=0.32，a ̂=y −−b ̂t −=1.04−0.32×3=0.08， 从而得到回归直线方程为y ̂=0.32t +0.08， 当t =6时，可得y =0.32×6+0.08=2(万)， (2)(i)由a200=0.5，可得a =40，由频率和为1，得(0.05×2+0.10+2c +0.20+0.30)×1=1，解得b =0.15， 200位竟拍人员报价大于5万元的人数为(0.05+0.10+0.15)×200=60人， (ii)2020年11月份车牌配额数量为3000，根据竞价规则，报价在最低成交价以上人数占总人数的比例为300020000×100%=15%，又由于频率分布直方图和竞拍报价大于6万元的频率为0.05+0.10=0.15， 所以根据统计思想(样本估计总体)可预测竞拍的最低成交价为6万元．【解析】(1)由题意求出t −=3，y −=1.04，代入公式求值，从而得到回归直线方程； (2)(i)根据频数等于总数与频率的乘积可得a ，根据频率面积之和为1，可得b ，再根频率等于频数除以总数可得结果，(ii)先求出报价在最低成交价以上的人数占总人数的比例，再对应频率分布直方图的频率可得结果．本题考查线性回归方程的求法，考查样本数据的平均值与方程的求法，考查频率分布直方图，考查计算能力，是中档题．22.【答案】解：(1)f(x)的定义域是(0,+∞)，f′(x)=1x −a x 2，故f′(1e )=e −ae 2=−e ，解得：a =2e ，故f′(x)=1x −2ex 2，令f′(x)>0，解得：x >2e ，令f′(x)<0，解得：0<x <2e ， 故f(x)在(0,2e )递减，在(2e ,+∞)递增； (2)证明：函数g(x)的定义域是(0,+∞)， 要证g(x)>1，即证e x (lnx +2ex )>1， 即证xlnx >xe x −2e ，设ℎ(x)=xlnx(x >0)，则ℎ′(x)=lnx +1， ∵ℎ′(1e )=ln 1e +1=0，故x ∈(0,1e )时，ℎ′(x)<0，x ∈(1e ,+∞)时，ℎ′(x)>0， 故ℎ(x)在(0,1e )递减，在(1e ,+∞)递增， 故ℎ(x)min =ℎ(1e )=−1e ， 设t(x)=xe x −2e (x >0)，则t′(x)=1−x e x，故x ∈(0,1)时，t′(x)>0，t(x)递增， x ∈(1,+∞)时，t′(x)<0，t(x)递减， 故t(x)max =t(1)=−1e ，综上，在(0,+∞)上恒有ℎ(x)>t(x)成立， 故g(x)>1．【解析】(1)求出函数的导数，根据f′(1e )=−e ，求出a 的值，求出函数的单调区间即可； (2)问题转化为证明xlnx >xe x −2e ，设ℎ(x)=xlnx(x >0)，t(x)=xe x −2e (x >0)，根据函数的单调性证明结论成立即可．本题考查了切线斜率问题，考查函数的单调性，最值问题，考查导数的应用以及不等式的证明，考查转化思想，是一道综合题．。

数学参考答案一、选择题：本题共12小题，每小题5分，共60分．1.【答案】D2.【答案】B3.【答案】B4.【答案】C5.【答案】C6.【答案】B7.【答案】C8.【答案】A9.【答案】D10.【答案】A11.【答案】C12.【答案】C二、填空题：本题共4小题，每小题5分，共20分。

13.【答案】214.【答案】115.【答案】①④16.【答案】5+17.（本小题满分10分【解】（1）由22sin (cos )()sin cos sin cos ααf ααααα×-==-×+，所以(sin cos 666f πππ=-=；（2）由12()25f α=得，12sin cos 025αα×=-<，又(0,)απ∈，所以(,)2παπ∈，所以sin cos 0αα->，又21249(sin cos )=12sin cos =1+22525αααα--´=，所以7sin cos 5αα-=所以sin 42πα-（）=72(sin cos )10αα-=.18．（本小题满分12分【解】（1）()1cos 112πsin 22224x f x x x +⎛⎫=+-=+ ⎪⎝⎭，所以()f x 的最小正周期2π2π1T ==，由ππ=π,42x k k Z ++∈，得π=π,4x k k Z +∈，所以对称轴方程为π=π,4x k k Z +∈．（2）由π610f α⎛⎫+= ⎪⎝⎭，得5πsin 21210α⎛⎫+= ⎪⎝⎭，即5π3sin 125α⎛⎫+= ⎪⎝⎭，令5π12t α=+，则5π12t α=-，且3sin 5t =，所以2π5π2π3πsin 2sin 2sin 2cos23632t t t α⎛⎫⎛⎫⎛⎫-=--=-= ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭2712sin 25t =-=．19.（本小题满分12分【解】（1）由图象可知2A =，422433T πππ⎡⎤⎛⎫=--= ⎪⎢⎥⎝⎭⎣⎦，所以212T πω==，所以1()2sin 2f x x ϕ⎛⎫=+ ⎪⎝⎭，由图可求出最低点的坐标为,23π⎛⎫-⎪⎝⎭，所以2sin 236f ππϕ⎛⎫⎛⎫=+=- ⎪ ⎪⎝⎭⎝⎭，所以262k ππϕπ+=-+，所以22,3k k Z πϕπ=-+∈，因为||ϕπ<，所以23πϕ=-，所以12()2sin 23f x x π⎛⎫=- ⎪⎝⎭，π411==T f .（2）由题意知，函数22()2sin 2()2sin 2233g x x m x m ππ⎡⎤⎛⎫=+-=-+ ⎪⎢⎥⎣⎦⎝⎭，因为()g x 的图象关于直线512x π=对称，所以5222,1232m k k Z ππππ⨯-+=+∈，即,62k m k Z ππ=+∈，因为02m π<<，所以6m π=，所以()2sin 23g x x π⎛⎫=- ⎪⎝⎭.当5,126x ππ⎡⎤∈⎢⎥⎣⎦时，42,363x πππ⎡⎤-∈-⎢⎥⎣⎦，可得sin 2,132x π⎡⎤⎛⎫-∈-⎢⎥ ⎪⎝⎭⎣⎦，所以2sin 2[2]3x π⎛⎫-∈ ⎪⎝⎭，即函数()g x 的值域为⎡⎤⎣⎦.20.（本小题满分12分【解】（1）由已知可得()sin (0)6f x x πωω⎛⎫=+> ⎪⎝⎭的最小正周期为π，即2T ππω==∴2,()sin(26f x x πω==+，令222,262k x k k Z πππππ-≤+≤+∈解得,36k x k k Z ππππ-≤≤+∈∴()f x 的单调递增区间是,,36k k k Z ππππ⎡⎤-+∈⎢⎥⎣⎦（2）在ABC 中，若12A f ⎛⎫= ⎪⎝⎭，由（1）得，()sin 26f x x π⎛⎫=+⎪⎝⎭，所以sin 16A π⎛⎫+= ⎪⎝⎭因为0,A π<<所以62A ππ+=，即3A π=23sin sin sin sin sin cos3226B C B B B B B ππ⎛⎫⎛⎫+=+-=+=+ ⎪ ⎪⎝⎭⎝⎭因为203B π<<，所以5666B πππ<+<；所以1sin 1,2626B B ππ⎛⎫⎛⎫<+≤<+≤ ⎪ ⎪⎝⎭⎝⎭所以sin sin B C +的取值范围2⎛ ⎝21.（本小题满分12分【解】（1）0a > ，对任意的x ∈R ，120x a ++>，所以，函数()f x 的定义域为R ，由于函数()f x 为奇函数，则()00f =，即()112002a f a-==+，解得2a =，所以，()()1212122221x x x x f x +--==++，下面验证函数()()21221x x f x -=+为奇函数，函数()()21221x x f x -=+的定义域为R ，()()()()()()22121122212221212x x x x x x x x f x f x --------====-+⋅++，综上所述，函数()()21221x x f x -=+为奇函数，因此，2a =；（2）()()()()2122111221221221x x x x x f x +--===-+++为R 上的增函数，证明如下：任取1x 、2x R ∈且12x x >，则12220x x >>，()()()()121221121211111122022122121212121x x x x x x x x f x f x -⎛⎫⎛⎫-=---=-=> ⎪ ⎪++++++⎝⎭⎝⎭，所以，()()12f x f x >，所以，函数()f x 为R 上的增函数；（3）()()22sin cos 0f m m x f x -+≥ ，所以，()()()2cos 2sin sin 2f x f m m x f m x m ≥--=-，可得2cos sin 2x m x m ≥-，即21sin sin 2x m x m -≥-，即2sin sin 210x m x m +--≤，令[]sin 1,1t x =∈-，()221g t t mt m =+--，则不等式()0g t ≤对任意的[]1,1t ∈-恒成立，所以，()()13010g m g m ⎧-=-≤⎪⎨=-≤⎪⎩，解得0m ≥.因此，实数m 的取值范围是[)0,+∞.22.（本小题满分12分【解】（1）当2t e =-时，不等式()0f x ≥，即为()()210x x e e e +-≥，也就是2x e e ≥，解得2x ≥，所以，不等式()0f x ≥的解集为[)2,+∞；（2）不等式()1()141x x x f x ee e <++-+即为()21(1)141x x x x x t e t e e e e +++<++-+，化简，即()()21411x x t e e <-++对任意()0+x ∈∞，恒成立，记()()214()(0)11x x h x x e e =->++.由于当()0+x ∈∞，时，110,12x e ⎛⎫∈ ⎪+⎝⎭，则213()24(,0)14x h e x ⎡⎤=--∈-⎢⎥+⎣⎦.所以，max 34t =-.（3）由于函数()2()1()1111x x x x f x e t t g x e e e +-===++++是“可构造三角形函数”，首先，必有0t ≥才能保证()0>g x ；其次，必需max min ()2()g x g x <，而当01t ≤<时，1()111x x x e t t g x e e +-==+++是R 上的增函数，则()g x 的值域为(),1t ，由11212t t ≤⇒≤<；当1t =时，()1g x =，符合题意；而当1t >时，1()111x x x e t t g x e e +-==+++是R 上的减函数，则()g x 的值域为()1,t ，由212t t ≤⇒<≤；综上，1,22t ⎡⎤∈⎢⎥⎣⎦.。

大庆实验中学2020-2021学年度上学期高三期末英语试题第一部分阅读理解（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）AArt,Culture&Performances in ArizonaArt on the Wild SideThree of Arizona’s favorite painters will exhibit their latest works inspired by the natural world in their unique styles.Through this exhibit,guests can engage in the Zoo’s mission to care for the natural world.Date:February1-April31,2020Location:455N Galvin Pkwy,Phoenix,AZ85008Price:$90Always...Patsy ClineAlways...Patsy Cline is a musical play,complete with country humor,true emotion,and even some audience participation,includes many of Patsy’s unforgettable hits such as“Crazy”,“Fall to Pieces”,“Sweet Dreams,and Walking After Midnight”…27songs in all!Date:June14-21,2020Location:Hale Theater,50West Page Avenue,Gilbert,AZ85233Price:$65The Two Gendemen of VeronaBest friends Valentine and Proteus choose different paths in life only to run into each other again when they both fall in love with the same woman.The Two Gentlemen of Verona is one of Shakespeare's earliest plays,his first comedy and also one of the most rarely performed plays in the canon.Date:July18-25,2020Location:Scottsdale Desert Stages Theatre,7014E Camelback Rd,Scottsdale,AZ85251 Price:$28Mandy MooreMandy Moore is giving a concert in our city!She is touring to support her upcoming album.Mandy's seventh studio work Silver Landings set to be released on September6.Date:September27-October3,2020Location:111N3rd St Phoenix,AZ85004Price:$421.When can you go to an exhibition of paintings?A.In March.B.In June.C.In July.D.In September.2.Where can you go if you are interested in Shakespeare’s plays?A.Art on the Wild Side.B.Always...Patsy Cline.C.The Two Gentlemen of Verona.D.Mandy Moore.3.How much will you pay if you want to enjoy a concert?A.$28.B.$42.C.$65.D.$90.BMy Grandpa Forgets Who I AmA few days ago I visited my grandfather in hospital.He has Alzhemier’s—a degenerative disease that usually starts slowly and gets worse over time.I thought I was prepared to see him.I knew chances were slim that he’d actually recognize me.He didn’t.As a matter of fact,he had no idea that he even had grandchildren.But he was excited that somebody came to visit him.I tried to explain to him who I was.But after he told me multiple times that he didn’t have grandchildren,I gave up.And my heart broke into a million little pieces.I was tired of explaining things to him.So I just smiled.He smiled back.It’s a genuine smile.Like a long time ago,when he’d take me by the hand and made this big world a little bit less scary for me.Now I have to take his hand.We sat in silence for a little while,before he told me to call my grandma.This was the first time I had tried so hard to hold back tears.My grandma died four years ago and he didn’t remember.He thought she was stuck on her way to pick him up.My grandpa used to be a strong,hard-working man.He was the person you turned to when you needed your car fixed,your tires changed or something heavy to be carried.Sadly, that man left this world a long time ago,and left behind a man that is lost and scared.I want to help him.I want to make him feel better.I want to tell him about his old life, and how great it was.So I sat with him and I held his hand,and every once in a while I told him how good he looked and how much I liked the color of his shirt and how it brought out the blue in his eyes.I told him that my grandma was on her way whenever he asked about her,and I made sure the glass in his hand was always filled with water.I can’t take away his pain.I can’t help him remember.I can’t make the disease go away. All I can do is hold on to the memories—hold on for both of us.4.When the author first saw her grandpa in hospital______.A.she gave up on himB.they were both excitedC.he didn’t recognize herD.they talked about the past5.The author was close to tears because______.A.grandma died about four years agoB.grandpa needed to be taken care ofC.grandma didn’t make it to the hospitalD.grandpa believed grandma was still alive6.Which of the following best describe the author?A.Tolerant and merciful.B.Considerate and patient.C.Warm-hearted and grateful.D.Strong-minded and generous.7.The author wrote this passage to______.A.show pity towards her grandpaB.record memories of her grandpaC.express deep love for her grandpaD.call on further study on Alzheimer’sCNow Hear ThisWhat do former American president Bill Clinton and rock musician Pete Townshend have in common?Both men have hearing damage from exposure to loud music,and both now wear hearing aids as a consequence.As a teenager,Clinton played saxophone in a band.Townshend, who has the more severe hearing loss,was a guitarist for a band called the Who.He is one of the first rock musicians to call the public’s attention to the problem of hearing loss from exposure to loud music.Temporary hearing loss can happen after only15minutes of listening to loud music.One early warning sign is when your ears begin to feel warm while you listen to music at a rock concert or through headphones.One later is that an unusual sound or a ringing is sometimes produced in your head after the concert.“What happens is that the hair cells in the inner ear are damaged,but they’re not dead,”says physician and ear specialist Dr.Sam Levine.According to Dr.Levine,if you avoid further exposure to loud noise,it’s possible to recondition the cells somewhat.However,he adds,“Eventually,over a long period of time,hair cells are permanently damaged.”And this is no small problem.What sound level is dangerous?According to Dr.Levine,regular exposure to noise above 85decibels(分贝)is considered dangerous.The chart below offers a comparison of decibel levels to certain sounds.Here’s another measurement you can use.If you’re at a rock concert and the music is so loud that you have to shout to make yourself heard,you’re at risk for hearing loss.That’s when wearing protective devices such as earplugs becomes critical.The facts are pretty frightening.But are rock bands turning down the volume?Most aren’t.“Rock music is supposed to be loud,”says drummer Andrew Sather.“I wouldn’t have it any other way.And neither would the real fans of rock.”Continued exposure to loud music and the failure to wear earplugs can lead to deafness, according to Dr.Levine.He states,“There’s no cure for hearing loss.Your ears are trying to tell you something.That ringing is the scream of your hair cells dying.Each time that happens, more and more damage is done.”Levels of Common NoisesNormal conversation................................................50-65dBFood blender.................................................................88dBJet plane flying above a person standing outside.........103dBRock band during a concert..........................................110-140dB8.From Paragraph1,we can learn that_______.A.loud music is a major cause of hearing lossB.famous people tend to have hearing problemsC.teenagers should stay away from school bandsD.the problem of hearing damage is widely known9.In Paragraph3,the underlined word“recondition”means______.A.not to be seenB.to make good againC.to fill with soundD.to become larger in size10.The purpose of the chart at the end of the article is to show_______.A.relationship between daily activities and hearing lossB.the noise levels of familiar soundsC.the effect of rock concertsD.a list of harmful sounds11.Which of the following statements will Dr.Sam Levine probably agree?A.Doctors know how to cure hearing loss.B.Many are taking the risk of losing hearing.C.Drummer Andrew Sather gives good advice.D.When your ears feel warm,your hair cells are dead.DIn the picture Landscape with Diogenes by the17th century French artist Poussin,the ancient philosopher Diogenes is described casting away his last possession,a drinking bowl.He realizes he doesn’t need it after seeing a youth cupping a hand to drink from a river.The significance for us is that Diogenes’spiritual descendant(后代)known as“new minimalists”are now everywhere,if not as completely possession-free as he was.There are hundreds of websites praising the virtues of tidy living.Everyone is trying to cut down on things these days.People are trying to reduce their carbon footprints,their waistlines, and their monthly outgoings.What’s more,there’s a general fear that people are becoming choked by their possessions,and this is fueled by the knowledge that the leading hobby these days seems to be shopping.It’s true,sales of e-readers and e-books go beyond those of paperbacks.As a result,the need for bookshelves is cut out.However,today’s new minimalists don’t urge us to burn our books and destroy our CDs, but just make sure we have them as digital files.So,for example,I have digitised versions of some of my old vinyl LP(黑胶)records and haven’t,as yet,stimulated myself to take the LPs to the nearest charity shop–and I admit I shall probably go on keeping them.Technology has gone beyond our dreams and there is always the doubt that our hard drives will crash and all will be lost.Far more important,however,is the fact that our memories are so inseparably tied to our possessions that we can’t get rid of stuff.We are not exactly suffering withdrawal symptoms(症状)as we try to break our addiction to objects.We are just acquiring new stuff, which means we can bin or recycle our old stuff.I’m happy to have found another website which seems to solve a whole lot of problems at once–a thriving online advice service offering storage solutions.The interior(室内的) designer responsible for this does not suggest getting rid of stuff,but rather recommends buying more stuff such as elegant flexible baskets or colourful lidded containers to hide the first lot of stuff from view.I love this philosophy–convince yourself you’ve got your desire forpossessions under control,without having to lose a thing.After all,wearen’t mercilessenough to follow Diogenes and cast away all our possessions.12.Why does the author mention a picture by the artist Poussin?A.It illustrates a modern trend.B.It describes a wise philosopher.C.Its meaning is only now becoming clear.D.Its message is not as simple as it appears.13.The author believes minimalism may not succeed mainly because of people’s______.A.resistance to media pressureziness in the face of changeck of faith in digital hardwareD.strong bond with physical objects14.According to the author,people invest in smart new storage in order to_____.A.satisfy their desire to make purchasesB.make attractive additions to their homesC.provide a temporary solution to a problemD.ease their conscience over having too many things.15.Which of the following would be the best title for the passage?A.Less is MoreB.Low Carbon Is an AttitudeC.Treasure What You HaveD.Psychology of Overconsumption.第二节(共5小题；每小题2分，满分10分)根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。

黑龙江省大庆实验中学2020-2021学年高一上学期第一次线上教学质量检测（1月底）物理考试时间：90分钟满分：110分一、选择题（本题共14小题，每小题4分，共56分。

在每小题给出的四个选项中，第1题～第8题只有一个选项是符合题目要求的；第9题～第14题有多个选项是符合题目要求的，全部选对的得4分，选对但不全的得2分，有选错的得0分）1.下面说法中正确的是（）A．加速度变化的运动一定是曲线运动B．做曲线运动的物体速度方向一定变化C．速度方向变化的运动一定是曲线运动D．加速度恒定的运动一定不会是曲线运动2.物体从静止开始做匀加速直线运动，第3s内通过的位移是3m，则（）A．第3s内的平均速度是1m/s B．物体的加速度是1.2m/s2C．前3s内的位移是6m D．3s末的速度是4m/s3.如图所示，质量为m的正方体滑块，在水平力作用下，紧靠在竖直墙上，处于静止状态。

若要滑块静止，水平力的最小值为 2.5mg（已知最大静摩擦力等于滑动摩擦力），则滑块与竖直墙间的动摩擦因数为（）A．0.4B．0.5C．0.6D．0.74.如图所示，在竖直平面内建立直角坐标系xOy，该平面内有AM、BM、CM三条光滑固定轨道，其中A、C两点处于同一个圆上，C是圆上任意一点，A、M分别为此圆与y、x轴的切点．B点在y轴上且∠BMO＝60°，O′为圆心．现将a、b、c三个小球分别从A、B、C点同时由静止释放，它们将沿轨道运动到M点，如所用时间分别为t A、t B、t C，则t A、t B、t C大小关系是()A．t A＜t C＜t BB．t A＝t C＜t BC．t A＝t B＝t CD．由于C点的位置不确定，无法比较时间大小关系5.A、B两个球用轻弹簧连接，A球质量为2m，B球质量为m，小球A由轻绳悬挂在天花板上O点，两球处于平衡状态，如图所示。

现突然剪断轻绳OA，让小球下落，在剪断轻绳的瞬间，设小球A、B的加速度分别用a1和a2表示，则（）A．a1=g，a2=g B．a1=0，a2=2gC．a1=g，a2=0D．a1=1.5g，a2=06.如图所示，在倾角为α的斜面上，放一质量为m的小球，小球和斜坡及挡板间均无摩擦，当挡板绕O点逆时针缓慢地转向水平位置的过程中，则有（）A．斜面对球的支持力逐渐增大B．斜面对球的支持力不变C．挡板对小球的弹力先减小后增大D．挡板对小球的弹力先增大后减小7. 如图所示，小船以大小为v(船在静水中的速度)、方向与上游河岸成θ的速度从O 处过河，经过一段时间，正好到达正对岸的O ＇处。