最新-高二年级第一学期期中考试试题精品

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高二年级第一学期期中试卷题目

⾼⼆年级第⼀学期期中试卷题⽬我们学习好了英语对我们出社会和考⼤学都是⾮常的有⽤的，今天⼩编就给⼤家整理⼀下英语⾼⼆英语，希望的来阅读哦描述⾼⼆英语上学期期中联考试题第⼀部分听⼒ (共两节，满分20分) 做题时，先将答案标在试卷上。

录⾳内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第⼀节 (共5⼩题;每⼩题1分，满分5分) 听下⾯5段对话。

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

听完每段对话后，你都有10秒钟的时间来回答有关⼩题和阅读下⼀⼩题。

每段对话仅读⼀遍。

1. What is the woman doing now? A. Watching TV. B. Taking part in an activity. C. Preparing for an exam. 2. Where does the conversation probably take place? A. In an office. B. In a store. C. In a hotel. 3. When will the speakers meet? A. At 6:20. B. At 6:10. C. At 5:40. 4. In which country does Jane want to spend her holiday? A. America. B. Korea. C. Japan. 5. What do we know about the woman? A. She is fired. B. She didn’t work hard. C. She can take a day off tomorrow. 第⼆节(共15⼩题;每⼩题1分，满分15分) 听下⾯5段对话或独⽩。

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

天津市2024-2025学年高二上学期期中考试英语试题(含答案,无听力原文及音频)

天津市2024-2025学年度第一学期期中学情调研高二年级英语学科本试卷分共100分，考试时间为100分钟。

答卷前，请务必将自己的姓名、考号、座位号填写在答题卡上相应位置。

答卷时，务必将答案涂写在答题卡上，答在试卷上的无效。

考试结束后，将答题卡和答题纸一并收回。

祝各位同学考试顺利！第Ⅰ卷 (共65分)第一部分：听力理解 (共15 小题；每小题0.5分，满分7.5分)第一节听下面5段对话。

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

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. What does the man want to know?A. Where the woman works out.B. How the woman stays fit.C. How to stay healthy.2. What is the man interested in?A. Whether people in China bargain everywhere.B. How to get a better price when doing the shopping in China.C. Where Chinese people usually go shopping.3. What's the most probable relationship between the two speakers?A. Old friends.B. Boss and secretary.C. Colleagues.4. What do we know about the woman?A. She is severely stressed.B. She is the man's doctor.C. She falls asleep easily.5. When was the woman scheduled to go to China at first?A. This Friday.B. This Saturday.C. This Sunday第二节听下面几段材料。

【高二上物理】浙江省杭州北斗联盟2023-2024学年高二上学期期中联考物理试题(解析版)

2023学年第一学期杭州北斗联盟期中联考高二年级物理学科试题考生须知：1.本卷共8页满分100分，考试时间90分钟。

2.答题前，在答题卷指定区域填写班级、姓名、考场号、座位号及准考证号并填涂相应数字。

3.所有答案必须写在答题纸上，写在试卷上无效。

4.考试结束后，只需上交答题纸。

选择题部分一、选择题I（本题共13小题，每小题3分，共39分。

每小题列出的四个备选项中只有一个是符合题目要求的，不选、多选、错选均不得分）1. 下列物理量中为标量的是（）A. 加速度B. 库仑力C. 电势D. 电场强度【答案】C【解析】【详解】A．加速度是既有大小又有方向的矢量，故A错误；B．库仑力是既有大小又有方向的矢量，故B错误；C．电势是只有大小没有方向的标量，故C正确；D．电场强度是既有大小又有方向的矢量，故D错误。

故选C。

2. 用国际单位制的基本单位表示电场强度的单位，下列正确的是（）A. N/CB. V/mC. kg•m/（C•s2）D. kg•m/（A•s3）【答案】D【解析】【详解】电场强度FEq，电场力的单位为N，电量的单位为C，所以电场强度的单位是N/C，而23/1/11/kg m s N C kg m A s A s⋅==⋅⋅⋅，D 正确．3. 杭州第19届亚运会将于2023年9月23日-10月8日在杭州举行，下列有关运动项目的描述正确的是（）A. 自行车比赛中，研究自行车车轮的转动时自行车可视为质点B. 篮球比赛中，篮球在空中飞行和进篮时惯性不变C. 跳水运动员下落时，运动员看到水面迎面扑来，是选择水面为参考系的缘故D. 蹦床比赛中，运动员刚接触蹦床时的速度最大【答案】B 【解析】【详解】A ．自行车比赛中，研究自行车车轮的转动时，自行车大小形状不可忽略，不可视为质点，故A 错误；B ．篮球比赛中，篮球在空中飞行和进篮时质量不变，惯性不变，故B 正确；C ．跳水运动员下落时，运动员看到水面迎面扑来，是自己为参考系的缘故，故C 错误；D ．蹦床比赛中，运动员刚接触蹦床时重力大于蹦床的弹力，继续下降，当其重力等于蹦床的弹力时，速度最大，故D 错误。

高二上学期期中考试语文试卷及参考答案

高二上学期期中考试语文试卷及参考答案第一学期期中考试高二年级语文试卷分值：150分时间：150分钟一、名篇名句默写（每空1分，共10分）1.补写出下列句子中的空缺部分。

1）___在《陈情表》中，陈述自己家没有什么亲属、仆人的实情的句子是“臣本无父。

”2）___在《陈情表》中以乌鸦反哺比喻要守孝道的句子是“以孝治国。

”3）___在《游褒禅山记》里认为，古人看各种风景，都有心得，其原因是“心之所往。

”4）___在《报任安书》中，为表明只有卓越超群不同一般的人才能名传后世，举了很多历史人物的事例，其中举___和___的例子的两句是“世之奇伟、瑰怪、非常之名。

帝王将相。

”5）我们一般用___《报任安书》“天下英雄所见略同”的三句话来概括___写《史记》的目的。

二、语言文字运用（每小题3分，共12分）2.下面加点词语注音全都正确的一项是（）（3分）A。

可汗 kè hán 瞋目chēn mù 床蓐 rù / 覆校无疑 jiào wú yí/ 老校退卒 xiàoB。

拔擢 zhuó砧板 z hēn 彘肩 zhì / 剌谬 miào / 戮力 lù lìC。

揾泪 wèn 刀俎jǔ 玉玦quē 目眦尽裂 zì / 恣卒为暴 zì cùwéi bàoD。

籼稻shān dào 车骑 jì / 盘桓 huán / 责臣逋慢bū / 吾未晡食bǔ3.下面加点词语运用不正确的一项是（）（3分）A。

跨进新年，深圳正以勃勃英姿，在改革开放的道路上阔步前进。

B。

他根据自己的实践，以科学家的胆识和眼光断定杂交水稻研究具有光辉的前景，决心义无反顾地坚持研究。

C。

除了你的善良的精神以外，身无长物，我恭维了你又有什么好处呢？D。

上外版英语高二上学期期中试卷与参考答案

上外版英语高二上学期期中模拟试卷与参考答案一、听力第一节（本大题有5小题，每小题1.5分，共7.5分）1、Listen to the conversation and choose the best answer to the question you hear. Question: What is the main topic of the conversation?A. The man’s favorite sports team.B. The upcoming school sports day.C. The woman’s participation in a sports club.Answer: BExplanation: The conversation revolves around the woman’s plan to participate in the school sports day, indicating that the main topic is indeed the upcoming event.2、Listen to the dialogue and answer the question you hear.Question: How does the man feel about the weather?A. He is excited about the sunny weather.B. He is not happy with the rainy weather.C. He doesn’t mind the weather as long as it’s not too hot.Answer: BExplanation: The man expresses his disappointment in the rainy weather, suggesting that he is not happy with it.3、What does the woman imply when she says “You know, it’s not as easyas it looks.”A)The man should not try to do it.B)The task is more complicated than it seems.C)It will take longer than they expect.D)It’s not worth the effort.Answer: BExplanation: The phrase “not as easy as it looks” suggests that something is more difficult than it appears to be. Therefore, the woman implies that the task is more complicated than it seems.4、Why does the man say he will have to pass the exam even if he is not a student?A)He needs the credit for his college degree.B)He is taking the exam for his job.C)He is supporting his family through his studies.D)He is attending the exam as a hobby.Answer: AExplanation: The man’s statement about passing the exam implies that it is part of a requirement for his college degree. Therefore, he needs the credit for his academic progress.5、You will hear a conversation between two students about their weekend plans. Listen carefully and answer the question.Question: How many hours will the girl spend on her part-time job on Saturday?Answer: 7 hoursExplanation: In the conversation, the girl mentions that she has to work for 7 hours on Saturday, which is the duration of her part-time job.二、听力第二节（本大题有15小题，每小题1.5分，共22.5分）1、Listen to the conversation and answer the question.A. The man is a student in his first year at Shanghai International Studies University.B. The woman is the student’s tutor at the university.C. They are discussing the student’s performance in the first semester.D. They are planning a trip to Shanghai Disneyland.Question: Who is the woman in the conversation?Answer: B. The woman is the student’s tutor at the university.Explanation: The key to this question is in the context of the conversation. The woman addresses the man as “you” and refers to herself as “your tutor,” which indicates that she is in a mentoring role, likely a tutor.2、Listen to the dialogue and complete the sentence with the missing word.W: I can’t believe it’s already November. It feels like just yesterday we were saying goodbye to the summer.M: Yeah, time flies. Speaking of which, how are you doing with your midterm exams?W: (pause) I’m feeling a bit nervous about them. I’m not sure if I’ve prepared well enough.Question: How does the woman feel about the midterm exams?Answer: Nervous.Explanation: The woma n explicitly states, “I’m feeling a bit nervous about them,” which directly answers the question about her feelings regarding the exams.3.You will hear a conversation between two students discussing their weekend plans. Listen to the conversation and answer the following question.Question: What does the male student plan to do on Saturday?A. Go to the movies.B. Visit a museum.C. Stay at home and relax.Answer: C. Stay at home and relax.Explanation: In the conversation, the male student mentions that he doesn’t have any plans for Saturday and would rather stay at home and relax.4.You will hear a short interview with a famous author. Listen to the interview and answer the following question.Question: Why does the author prefer to write in the morning?A. Because she is more creative in the morning.B. Because she has less distractions in the morning.C. Because she is less tired in the morning.Answer: B. Because she has less distractions in the morning.Explanation: During the interview, the author explains that she prefers towrite in the morning because it’s when she has fewer distractions, allowing her to focus better on her writing.5.You hear a conversation between two students, Alice and Bob, discussing their midterm exam preparation. Listen and answer the following question: Question: How does Alice feel about the midterm exam?A. ExcitedB. NervousC. IndifferentD. ConfidentAnswer: B. NervousExplanation: In the conversation, Alice mentions, “I’m really nervous about the midterm,” indicating her anxiety regarding the exam.6.You hear a teacher, Mr. Green, explaining a concept to his class before the midterm. Listen and answer the following question:Question: What is Mr. Green’s main advice to the students for the upcoming exam?A. To focus on the most difficult topics.B. To review the material regularly.C. To study late into the night.D. To avoid studying altogether.Answer: B. To review the material regularly.Explanation: Mr. Green says, “The most important thing is to review thematerial regularly so you’re not overwhelmed on the day of the exam,” which suggests that consistent review is his recommended approach.7.You will hear a conversation between two students discussing a school project. Listen to the conversation and answer the question.What is the main topic of the school project?A. A science fair experimentB. A literature review on a famous novelC. A research paper on climate changeD. A historical reenactmentAnswer: BExplanation: The conversation mentions that they are working on a “comprehensive literature review of the various interpretations of the novel,” which indicates that the main topic is a literature review on a famous novel.8.You will hear a short lecture about the importance of teamwork in the workplace. Listen to the lecture and answer the question.Which of the following is NOT mentioned as a benefit of teamwork according to the lecture?A. Increased creativityB. Improved problem-solving abilitiesC. Enhanced communication skillsD. Reduced individual workloadAnswer: DExplanation: The lecture discusses how teamwork leads to increased creativity, improved problem-solving, and enhanced communication skills. However, it does not mention that teamwork reduces the individual workload as a benefit.9.You will hear a conversation between two students discussing a school project. Listen and answer the question.Question: What is the project about?A) A science fairB) A cultural exchangeC) A history presentationD) A music festivalAnswer: B) A cultural exchangeExplanation: Th e conversation mentions “we are planning a cultural exchange project,” which indicates the topic of the project is a cultural exchange.10.You will hear a short announcement about a school event. Listen and answer the question.Question: When will the event take place?A) This Friday at 2 PMB) Next Tuesday at 3 PMC) This Saturday at 4 PMD) Next Wednesday at 2 PMAnswer: A) This Friday at 2 PMExplanation: The announcement clearly states, “The event will be held this Friday at 2 PM in the school auditoriu m.”11.You will hear a conversation between two students discussing their weekend plans. Listen and answer the following question.What does the student say they will do on Saturday?A)Go shopping with a friend.B)Attend a sports event.C)Stay at home and watch movies.Answer: C) Stay at home and watch movies.Explanation: In the conversation, the student mentions that they have no plans for Saturday and would rather stay at home and watch movies.12.You will hear a short lecture about the effects of climate change on wildlife. Listen and answer the following question.What is one significant effect of climate change mentioned in the lecture?A)Increased food availability for wildlife.B)The migration patterns of animals shifting.C) A decrease in the number of polar bears.Answer: B) The migration patterns of animals shifting.Explanation: The lecture discusses how climate change is causing the migration patterns of animals to change, which can disrupt their natural habitats and food sources.13.You will hear a conversation between two students discussing their weekend plans. Listen carefully and answer the question.What activity does the second student plan to do on Saturday afternoon?A)Go to the movies.B)Visit a friend.C)Go for a hike.Answer: B) Visit a friend.Explanation: In the conversation, the second student mentions that they have plans to visit a friend on Saturday afternoon.14.You will hear a short lecture about the importance of exercise. Listen carefully and answer the question.According to the lecture, what is one of the benefits of regular exercise?A)Improved memory.B)Better sleep.C)Increased appetite.Answer: B) Better sleep.Explanation: The lecturer discusses how regular exercise can lead to better sleep quality, which is one of the benefits mentioned.15.How many students are studying at the university library on a sunny weekend?A. 200B. 300C. 400D. 500Answer: BExplanation: The conversation between two students discussing their weekend plans mentions that there are about 300 students at the library, which is a common spot for studying on sunny weekends. Therefore, the correct answer is B. 300.三、阅读第一节（第1题7.5分，其余每题10分，总37.5分）第一题Reading PassageIn the small coastal town of Porthkerry, the local community has always valued its rich history and natural beauty. The town is known for its stunning cliffs, picturesque beaches, and a vibrant local culture. One of the most iconic landmarks in the town is the Porthkerry Castle, which has stood for over 800 years.The castle was built in the 13th century by the de Braose family, a prominent noble family at the time. Over the centuries, it has been home to many different owners, each leaving their mark on the structure and the surrounding area. The castle has witnessed battles, royal visits, and the daily lives of ordinary people.Today, the Porthkerry Castle is a museum and a popular tourist destination. It houses a collection of historical artifacts, including weaponry, furniture, and paintings. The castle grounds are also home to a variety of plants and animals, making it a haven for nature lovers.1.What is the main purpose of the Porthkerry Castle today?A. It is a residential home for a noble family.B. It is a museum and a tourist destination.C. It is a center for educational workshops.D. It is a military base.2.Who built the Porthkerry Castle in the 13th century?A. The de Braose familyB. The King of EnglandC. The local villagersD. A famous architect3.What has the Porthkerry Castle witnessed over the centuries?A. Only battles and royal visitsB. The daily lives of ordinary peopleC. The construction of new buildingsD. The expansion of the town4.What is one of the attractions for nature lovers at the Porthkerry Castle?A. The historical artifactsB. The paintings in the museumC. The variety of plants and animalsD. The stunning cliffsAnswers:1.B2.A3.B4.C第二题Passage:The following is a story about the importance of teamwork.In a small town, there was a race scheduled between two teams: the Lightning Team and the Thunder Team. Both teams were known for their competitiveness and were equally matched. The race was set to take place on a sunny Saturday morning, and the entire town was buzzing with excitement.The Lightning Team had a star runner, Alex, who was known for his speed and determination. The Thunder Team had a strong team spirit, with each member contributing to the overall effort. The team captain, Jamie, was a master strategist and knew how to maximize the team’s potential.On the day of the race, the weather was perfect. The crowd gathered at the starting line, eager to see who would win. The race began, and the Lightning Team took an early lead. However, the Thunder Team was not to be outdone. Jamie had planned the strategy well, and each member of the team played their part perfectly.As the race progressed, the teams exchanged leads several times. The crowd cheered loudly for both teams, but it was clear that the race was coming down to the wire. With only a few meters to go, the Lightning Team seemed to have the upper hand. But then, disaster struck. Alex tripped over his own shoelaceand fell to the ground.The crowd gasped in shock. The Lightning Team was in disarray, and it looked like the Thunder Team would win. But instead of celebrating, the Thunder Team members rushed to help Alex. They formed a human chain and lifted him to his feet. With renewed determination, Alex and the Thunder Team member who had helped him continued the race.In the end, the Thunder Team won the race by a mere second. The crowd erupted in cheers, and the entire town was filled with a sense of camaraderie and joy. The Lightning Team was gracious in defeat, and the Thunder Team was hailed as heroes.This race taught the town an important lesson: teamwork can overcome individual setbacks and lead to unexpected victories.Questions:1、What was the main theme of the story?A) The importance of speed in a raceB) The power of teamworkC) The importance of individual talentD) The unpredictability of a race2、Who was the star runner on the Lightning Team?A) JamieB) AlexC) The team captainD) A member of the Thunder Team3、What strategy did Jamie use to help the Thunder Team win the race?A) He focused on improving the speed of his team membersB) He encouraged individual efforts from each team memberC) He planned for potential setbacks and had backup strategiesD) He tried to distract the Lightning Team4、How did the Thunder Team’s victory affect the town?A) It caused a lot of conflict and argumentB) It brought the town closer togetherC) It led to increased competition between the teamsD) It resulted in a decrease in town spiritAnswers:1、B2、B3、C4、B第三题Read the following passage and answer the questions that follow.In the small coastal town of Amble, the local library has been a hub of community activity for decades. Known for its vast collection of books and friendly staff, the library has played a significant role in the lives of its residents. However, with the advent of digital technology and e-books, thelibrary’s future seemed uncertain.Last year, the library’s director, Sarah Miller, noticed a decline in the number of visitors. She decided to take action and launched a new initiative aimed at attracting young people to the library. The program, called “Read & Win,” offered prizes for students who r ead a certain number of books during the school year. The prizes ranged from small gadgets to gift cards for local bookstores.Sarah’s initiative paid off. The number of young people visiting the library increased significantly. Many students expressed their gratitude for the program, saying it not only encouraged them to read more but also helped them connect with other book lovers in the community.1、What was the main concern of the library director, Sarah Miller?A. The lack of digital technology in the libraryB. The declining number of visitorsC. The outdated collection of booksD. The high cost of maintaining the library2、What was the purpose of the “Read & Win” program?A. To reduce the library’s budgetB. To encourage students to read moreC. To promote the sale of e-booksD. To attract tourists to the town3、How did the “Read & Win” program affect the library?A. It decreased the number of visitorsB. It increased the number of visitorsC. It improved the library’s financial situationD. It led to the closure of the library4、What did many students say about the “Read & Win” program?A. It was too expensive for themB. It did not encourage them to read moreC. It helped them connect with other book loversD. It was not as effective as expectedAnswers:1、B2、B3、B4、CFourth QuestionRead the passage carefully and choose the best answer to each of the questions that follow. You will find the answers immediately after each question.Passage:In an age where technology seems to be taking over every aspect of our lives, there remains one activity that has not been overshadowed by digital advancements: reading traditional books. Despite the popularity of e-books andaudiobooks, many people still prefer the tactile experience of holding a physical book and turning its pages. This preference for traditional books is not only due to nostalgia but also because of the unique benefits they offer.A recent study conducted by the National Reading Agency found that individuals who read physical books tend to have a better memory of the content compared to those who read the same material on digital devices. Furthermore, there is a growing concern about the impact of screen time on sleep patterns, with research suggesting that reading from screens before bedtime can negatively affect the quality of sleep. Physical books, on the other hand, do not emit blue light which can disrupt melatonin production, thus helping readers get a good night’s rest.Moreover, libraries and bookstores continue to thrive as communal spaces where people gather to share ideas and engage in discussions about literature. The atmosphere in these places fosters a sense of community and a love for reading that cannot be replicated online. While digital media offers convenience and access to a vast amount of information, it does not replace the unique experience of visiting a library or bookstore.In conclusion, while technology offers many conveniences, the tradition of reading physical books remains strong. It provides a way to disconnect from the digital world, improve memory retention, and foster social connections within communities.End of PassageQuestions:1、According to the passage, what is one benefit of reading physical books over digital devices?Answer: Better memory retention of the content.2、What potential negative effect on health is associated with reading from screens before bedtime?Answer: Disruption of sleep quality due to blue light affecting melatonin production.3、How do libraries and bookstores contribute to community life according to the text?Answer: They serve as communal spaces where people gather to share ideas and engage in discussions about literature.4、What does the passage suggest about the future of physical books despite technological advances?Answer: Physical books will remain significant as they offer ways to disconnect, improve memory, and foster social connections.四、阅读第二节（12.5分）Title: The Art of MeditationReading Passage:Meditation has been a part of human culture for thousands of years, serving as a means for relaxation, stress reduction, and spiritual growth. Its originscan be traced back to ancient India, where it was practiced by sages and monks to achieve enlightenment. Over time, meditation has spread across the world, gaining popularity in various forms, such as mindfulness, transcendental meditation, and yoga.One of the most common types of meditation is mindfulness meditation, which involves paying attention to the present moment and accepting it without judgment. This form of meditation can be practiced anywhere, at any time, and does not require any special equipment. It is particularly useful for individuals looking to improve their focus, reduce anxiety, and increase their overall sense of well-being.Mindfulness meditation i s based on the concept of being aware of one’s thoughts, feelings, and bodily sensations. It encourages individuals to observe their experiences without getting caught up in them. By practicing mindfulness, one can learn to let go of negative thoughts and emotions, leading to a more peaceful and balanced state of mind.Here is a brief guide on how to practice mindfulness meditation:1.Find a quiet and comfortable place to sit or lie down.2.Close your eyes and take a few deep breaths to relax.3.Focus on your breath, noticing the sensation of air entering and leaving your nostrils.4.When your mind starts to wander, gently bring your attention back to your breath.5.Continue this practice for 5 to 10 minutes, or as long as you feel comfortable.Question:What is the primary purpose of mindfulness meditation?A)To achieve enlightenmentB)To improve focus and reduce anxietyC)To practice yogaD)To increase spiritual growthAnswer: B) To improve focus and reduce anxiety五、语言运用第一节 _ 完形填空（15分）Cloze Test (完形填空)Directions: For this part, you are going to read a passage with 15 blanks. Choose the best word or phrase from the four choices marked A, B, C, and D for each blank. There is only one correct answer for each blank.Passage:The world of the internet has transformed our lives in countless ways. It has made information more accessible than ever before, connecting people across the globe in real-time. However, it also poses new challenges and risks. One of these is the increasing amount of misinformation that can be found online. While the internet is undoubtedly a valuable tool for learning and communication, it requires users to develop critical thinking skills to discern fact fromfiction. This ability is particularly important for young people who have grown up with easy access to technology but may lack the experience necessary to judge the 1 of the information they encounter.In addition to being able to evaluate the credibility of online content, digital literacy involves understanding the importance of privacy and security measures when using the web. Sharing personal details on social media platforms can make individuals vulnerable to identity theft and other forms of cybercrime. Therefore, it is essential to educate children about the risks associated with 2 too much personal information online.Another challenge of the digital age is the potential for addiction to electronic devices and the internet itself. The constant connectivity can lead to a feeling of being overwhelmed by the constant influx of 3. This is why many experts recommend setting boundaries around screen time and encouraging alternative activities that promote mental well-being.Despite these challenges, the internet remains a powerful force for good, providing opportunities for collaboration, innovation, and learning. By teaching students how to use it responsibly, we can help ensure they benefit from its advantages while avoiding its pitfalls.Now choose the best option to complete the passage:1.A. quantity B. quality C. quantity D. availability2.A. sharing B. hiding C. collecting D. deleting3.A. emails B. letters C. information D. stimuliKey:1.B. quality2.A. sharing3.D. stimuliThis exercise tests students’ comprehension of the passage as well as their vocabulary and grammar usage. The context here relates to the impact of the internet on modern life, which is a relevant topic for today’s youth.六、语言运用第二节 _ 语法填空（15分）Language Usage Section II: Grammar CompletionReading Passage:When it comes to learning a new language, many people struggle to find the right balance between memorization and practice. One effective method is to immerse yourself in the language by watching movies, listening to music, and reading books. This not only helps you get accustomed to the pronunciation and intonation but also exposes you to a wide range of vocabulary and expressions.However, it is important to note that simply listening and watching is not enough. You need to actively engage with the language by speaking and writing. Try to find a language exchange partner or join a conversation group to practice speaking. This will not only improve your speaking skills but also help you gain confidence in using the language in real-life situations.When it comes to writing, start by keeping a daily journal in the targetlanguage. This will help you practice your writing skills and keep track of your progress. Don’t worry if you make mistakes; they are an essential part of the learning process. The key is to be consistent and not be afraid to make mistakes.In conclusion, the key to mastering a new language is to find the right balance between memorization and practice. By immersing yourself in the language, actively engaging with it, and being patient and persistent, you will eventually achieve your goals.Grammar Completion:Choose the most appropriate word or phrase to complete the following sentences. Write the letter corresponding to your answer in the blanks.1.One effective method is to________yourself in the language by watching movies, listening to music, and reading books.a)immerseb)immersec)immersedd)immersing2.This not only helps you get________to the pronunciation and intonation but also exposes you to a wide range of vocabulary and expressions.a)accustomedb)accustomedc)accustomedd)accustomed3.Try to find a language exchange partner or join a________to practice speaking.a)conversation groupb)conversation groupc)conversation groupd)conversation group4.This will not only improve your speaking skills but also help you gain________in using the language in real-life situations.a)confidenceb)confidencec)confidenced)confidence5.When it comes to writing, start by keeping a daily journal in the target language. This will help you practice your________skills and keep track of your progress.a)readingb)readingc)readingd)reading6.Don’t worry if you make mistakes; they are an essential part of the________process.a)learningb)learningc)learningd)learning7.The key to mastering a new language is to find the right balancebetween________and practice.a)memorizationb)memorizationc)memorizationd)memorization8.By immersing yourself in the language, actively engaging with it, and being patient and ________, you will eventually achieve your goals.a)persistentb)persistentc)persistentd)persistent9.This method is especially useful for those who are________in learning a new language.a)motivatedb)motivatedc)motivatedd)motivated10.It is important to note that simply listening and watching is not enough; you need to actively engage with the language by________and writing.a)speakingb)speakingc)speakingd)speakingAnswers:1.a) immerse2.b) accustomed3.a) conversation group4.b) confidence5.a) reading6.a) learning7.a) memorization8.d) persistent9.b) motivated10.a) speaking七、写作第一节 _ 应用文写作（15分）Writing Part I: Practical WritingDirections: Write an appropriate response to the following situation in no less than 80 words. Your writing should be clear, well-organized, and appropriate to the context.Scenario:Your school has recently organized a cultural exchange program with a sister school in the UK. You have been chosen to host a student from the UK for one week. Write an email to your guest introducing yourself and offering some suggestions for activities during their stay.To:SarahThompson(*****************************.uk)From:。

【高二上数学】浙江省9+1高中联盟2023-2024学年高二上学期期中考试数学试题(解析版)

2023学年第一学期浙江省9+1高中联盟高二年级期中考试数学考生须知：1．本卷满分150分，考试时间120分钟；2．答题前，在答题卷指定区域填写班级、姓名、考场、座位号及准考证号并核对条形码信息；3．所有答案必须写在答题卷上，写在试卷上无效，考试结束后，只需上交答题卷； 4．参加联批学校的学生可关注“启望教育”公众号查询个人成绩分析.一．单项选择题（本大题共8个小题，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. 直线310x +−=的倾斜角是（） A.π3B.2π3C.π6D.5π6【答案】D 【解析】【分析】将直线方程化为斜截式，从而得到直线的斜率与倾斜角. 【详解】直线310x −=，即3333y x =−+，则直线的斜率33k =−，所以倾斜角为5π6. 故选：D2. 若复数z 满足：()12i 8i z +=+，则复数z 的虚部为（） A. 3− B. 2C. 3D. 3i −【答案】A 【解析】【分析】先根据复数的除法运算求解出z ，然后判断出z 的虚部即可. 【详解】因为()12i 8i z +=+，所以()()()()8i 12i 8i 816i i 223i 12i 12i 12i 5z +−+−++====−++−，所以z 的虚部为3−，故选：A.3. “1x <”是“ln 0x <”成立的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【答案】B 【解析】【详解】由ln 0x < ，解得01x << ，所以“1x <”是“ln 0x <”成立的必要不充分条件.故选B. 4. 若函数()()cos 2f x x φ=+的图象关于直线56πx =−对称，则ϕ的最小值是（） A.4π3B.2π3C. π3 D. π6【答案】C 【解析】【分析】利用余弦函数的对称轴列式，计算即可得解.【详解】由题意555cos π1ππ,Z ππ,Z 333k k k k ϕϕϕ⎛⎫−+=±⇒−+=∈⇒=+∈ ⎪⎝⎭ϕ⇒=⋅⋅⋅，4π3−，π3−，2π3，5π3，…，则ϕ的最小值是π3，故选：C.5. 在直三棱柱111ABCA B C 中，1,,,AB BC AB BC AA D E ⊥==分别为,AC BC 的中点，则异面直线1C D 与1B E 所成角的余弦值为（）A.33B.5 C.1010D.3010【答案】D 【解析】【分析】设2AB =，取11A B 的中点F ，连接1,,C F DF DE ，则可得1C DF ∠为异面直线1C D 与1B E 所成的角或补角，然后在1C DF 中求解即可.【详解】设2AB =，取11A B 的中点F ，连接1,,C F DF DE ，则11112B F A B = 因为,D E 分别为,AC BC 的中点，所以DE ∥AB ，12DE AB =，因为11A B ∥AB ，11A B AB =，所以DE ∥1B F ，1B F DE =，所以四边形1DEB F 为平行四边形，所以DF ∥1B E ，所以1C DF ∠为异面直线1C D 与1B E 所成的角或补角.因为1,,2,AB BC AB BC AA D E =⊥==分别为,AC BC 的中点，所以()222222111125,125,226DF B E C F C D ==+==+==+=，所以11163022cos 5C DC DF DF ∠===. 故选：D6. 若关于x 的不等式()2190x m x −++≤在[]1,4上有解，则实数m 的最小值为（）A. 9B. 5C. 6D.214【答案】B 【解析】【分析】先通过分离参数得到91m x x +≥+，然后利用基本不等式求解出9x x+的最小值，则m 的最小值可求.【详解】因为()2190x m x −++≤在[]1,4上有解，所以91m x x+≥+在[]1,4上有解，所以[]()min 911,4m x x x ⎛⎫+≥+∈⎪⎝⎭，又因为9926x x x x+≥⋅=，当且仅当9x x =即3x =时取等号，所以16m +≥，所以5m ≥，即m 的最小值为5，故选：B.7. 设椭圆1C ：()222210x y a b a b +=>>与双曲线2C ：22221x ya b−=的离心率分别为1e ，2e ，且双曲线2C 的渐近线的斜率小于155，则21e e 的取值范围是（）A. ()1,4B. ()4,+∞C. ()1,2D. ()2,+∞【答案】C 【解析】【分析】由双曲线的渐近线的斜率小于155，即可得出22305b a <<，由此即可求出21e e 的取值范围，从而求解【详解】由题意得，221c a b =−222c a b =+所以22221112221c c a b b e a a a a −====−22222222221c c a b b e a a a a+====+又因为双曲线的渐近线的斜率小于155，得222305b k a <=<，所以222212101b e a e b a+=>−，即()2222211211,411e k e k k ⎛⎫+==−+∈ ⎪−−⎝⎭，得()211,2e e ∈，故C 正确. 故选：C.8. 如图，四棱锥P ABCD −中，//AB CD ，22AB CD ==，ACD 是正三角形，PA AC ⊥，平面PAC ⊥平面PBC ，若点F 是PAD 所在平面内的动点，且满足2FA FD +=，点E 是棱PC （包含端点）上的动点，则当直线AE 与CD 所成角取最小值时，线段EF 的长度不可能为（）A.5 B.62C.264D.72【答案】A 【解析】【分析】由三余弦定理确定直线AE 与CD 所成角取最小值时点E 的位置，根据椭圆定义确定F 点的轨迹，在平面PAD 内，以O 为坐标原点，DA 为x 轴建立平面直角坐标系，求椭圆方程，求OF 范围；因为EO ⊥平面PAD ，所以EO OF ⊥，根据勾股定理求67,22EF. 【详解】三余弦定理：如图直线AB 与平面BOC 相交于点B ，过A 作AO ⊥平面BOC ,垂足为O ，BC 为平面BOC 内一直线, 过O 向BC 引垂线且垂足为C ，连结BO ，因为AO ⊥平面BOC ，AO BO ⊥,AO BC ⊥ 又因为BC OC ⊥,且AO OC O =,所以BC⊥平面AOC ，所以BC AC ⊥所以AOB 90∠=,90OCB ∠=,90ACB ∠=, 设ABO α∠=,ABC β∠=,CBO,cosBCAB ,cos BOAB ,cos BCBO, 所以cos cos cos βαγ=⋅；因为ACD 是正三角形，所以1DC AC ==,60ACD ∠=, 又因为//AB CD ，所以60CAB ∠=,在ABC 中，1AC =,2AB =,60CAB ∠=，由余弦定理有：2222cos 60BC AC AB AC AB ，解得3BC =,满足222AB AC BC =+,所以BC AC ⊥, 过A 作AH PC ⊥于点H ,因为平面PAC ⊥平面PBC ，平面PAC 平面PBC PC =,由面面垂直的性质可知AH BC ⊥, 又AHAC A =,所以BC ⊥平面PAC ；因为AE 与CD 所成的角等于AE 与AB 所成的角设为θ，即EAB θ=∠，由三余弦定理得：11cos cos cos cos 22EAC CAB EAC θ=∠⋅∠=∠≤，此时E 与C 重合，设AD 的中点为O ，因为ACD 是正三角形，⊥EO AD , 则222213122EOEAAO，根据已知条件，点F 的轨迹满足椭圆定义，设椭圆方程()2222100x y a b a b +=>>，，因22FAFDa ，所以1a =，因为12AD c ，所以12c =，因为a c >，所以点F 的轨迹是椭圆，222a b c =+，所以32b =, 在平面PAD 内，以O 为坐标原点，DA 为x 轴建立平面直角坐标系，为为椭圆方程为22413y x +=，设()00,F x y ，则2200413y x ，又因为PA AE ⊥,PA BE ⊥,AE BE E =,所以PA ⊥平面ABCD ,PA EO ⊥,PA AD A ⋂=, 所以EO ⊥平面PAD ，因为EO ⊥平面PAD ，所以EO OF ⊥, 所以222222000371443EFOE OF x y y ，又因为20304y ，所以267174434y ，所以67,22EF， 2426626727284424244故选：A【点睛】三余弦定理的应用，利用椭圆方程求OF 的范围，利用垂直关系转化边长求EF 范围.二．多项选择题（本大题共4个小题，每小题5分，共20分．在每个小题给出的四个选项中，有多项符合题目要求，全部选对的得5分，选对但不全的得2分，有选错或不选的得0分）9. 下列命题正确的是（） A. 集合{},,A a b c =的子集共有8个B. 若直线1l ：10x ay +−=与2l ：210a x y −+=垂直，则1a =C. 若221x y +=（x ，R y ∈），则34x y −的最大值为5D. 长、宽、高分别为1、2、3的长方体的外接球的表面积是14π【答案】ACD 【解析】【分析】根据子集的概念求出子集判断A ，利用两直线垂直的公式列式计算判断B ，换元法利用余弦函数的最值判断C ，根据长方体的外接球的直径为体对角线求解半径，代入球的表面积公式计算判断D . 【详解】集合{},,A a b c =的子集有∅，{}a ，{}b ，{}c ，{},a b ，{},a c ，{},b c ，{},,a b c 共8个，故A 正确；因为直线1l ：10x ay +−=与2l ：210a x y −+=垂直，则20a a −=，即()2110a a ⨯+⨯−=，解得0a =或1，故B 错误；由221x y +=设cos x θ=，sin y θ=，则()343cos 4sin 5cos 5x y θθθϕ−=−=+≤，故C 正确；由长方体的体对角线为其外接球的直径知：222212314R =++=，所以142R =，所以长方体的外接球的表面积是24π14πS R ==，故D 正确；故选：ACD10. 已知向量()2,cos a θ=−，()sin ,1b θ=，则下列命题正确的是（） A. 不存在R θ∈，使得//a b B. 当2tan 2θ=时，a b ⊥ C. 对任意R θ∈，都有a b ≠D. 当3a b ⋅=时，a 在b 方向上的投影向量的模为355【答案】ABD 【解析】【分析】根据向量间运算与三角恒等变换逐项判断即可. 详解】对于A ，若//a b ，则有sin cos 2sin 2221θθθθ=−⇒=−<−⇒不存在，故A 正确；对于B ，若ab ⊥，则【202cos 0tan 2a b θθθ⋅=⇒−+=⇒=，故B 正确；若22222cos sin 1cos 21a b θθθ=⇒+=+⇒=−，存在θ，故C 不正确；()22sin cos 333,33a b θθθθθϕ⎫⋅=−+=+=+=⎪⎪⎭其中3cos ,sin ,363ϕϕ== 所以()()cos 12π,k Z k θϕθϕ+=⇒+=∈222sin sin 3θϕ⇒==， 2333cos 35551sin a b a bθθ⋅====+，故D 正确；故选：ABD11. 已知直线l ：()()1120x y λλλ++−+=，C ：2240x y y +−=，则下列结论正确的是（）A. 直线l 恒过定点()2,4−B. 直线l 与C 必定相交C.C 与1C ：2240x y x +−=公共弦所在直线方程y x =D. 当0λ=时，直线l 与C 的相交弦长是2【答案】BC 【解析】【分析】求出直线l 过的定点判断A ；由点与圆的位置关系判断B ；求出公共弦所在直线方程判断C ；利用圆的弦长公式计算判断D.【详解】依题意，直线l ：()()20x y x y λ−+++=，由200x y x y −+=⎧⎨+=⎩，解得11x y =−⎧⎨=⎩，直线l 恒过定点()1,1−，A 错误；显然点()1,1−在C 内，则直线l 与C 必定相交，B 正确；C 的圆心(0,2)C ，半径2r =，1C 的圆心1(2,0)C ，半径12r =，111||22(,)CC r r r r =−+，即C 与1C 相交，把两个圆的方程相减得公共弦所在直线方程440y x −+=，即y x =，C 正确；为当0λ=时，直线l ：0x y +=，点()0,2C 到直线l 的距离，0222d +==，因此直线l 与C 的相交弦长为22222r d −=，D 错误.故选：BC12. 设椭圆C ：2214x y +=的左、右焦点分别为1F 、2F ，椭圆C 的右顶点为A ，点P 、Q 都在椭圆C 上且P 、Q 关于原点对称，直线x m =与椭圆C 相交于点M 、N ，则下列说法正确的是（） A. 四边形12PFQF 不可能是矩形 B.2PQF 周长的最小值为6C. 直线P A ，QA 的斜率之积为定值14−D. 当2F MN 的周长最大时，2F MN 3 【答案】BCD 【解析】【分析】A ：先判断出四边形12PFQF 是平行四边形，然后根据对角线长度的关系判断即可； B ：利用椭圆的定义以及PQ 的范围求解出2PQF 周长的最小值；C ：利用坐标表示出斜率关系，然后根据点在椭圆上化简运算，从而求得结果；D ：将点M 设为(),2πcos ,in 2s πθθθ⎛⎫⎛⎫∈⎪ ⎪⎝⎭⎝⎭，然后表示出2F MN 的周长，结合三角形函数确定出周长最小时θ的值，从而可求面积.【详解】对于A ：因为点O 平分12,PQ F F ，所以四边形12PFQF 是平行四边形，又因为2a =，1b =且[]2,2PQ b a ∈，所以[]221,2,4c a b PQ =−=∈，所以123F F =12PQ F F =有可能成立，故A 不正确；对于B ：因为四边形12PFQF 是平行四边形，所以21QF PF=，所以2PQF 周长为2221246PF QF PQ PF PF PQ a PQ PQ ++=+=+=+≥+，故B 正确；对于C ：因为()2,0A ，设()11,P x y ，所以()11,Q x y −−，所以21211122111141422444AP AQx y y y k k x x x x −−−⋅=⋅===−−−−−−，故C 正确；对于D ：由题意可知()2,0m ∈−，设()π2cos ,πsin ,2M θθθ⎛⎫⎛⎫∈⎪ ⎪⎝⎭⎝⎭，)23,0F ，所以()()()222222cos 3sin 03cos 43cos 43cos 223MF θθθθθθ=−+−=−+=−=，所以2F MN 的周长为π4232sin 44sin 83θθθ⎛⎫−+=+−≤ ⎪⎝⎭，当且仅当πsin 13θ⎛⎫−= ⎪⎝⎭，即ππ5π326θθ−=⇒=时取等号，所以2112sin 2cos 3123322F MN S θθ=⨯⨯=⨯⨯=△，故D 正确；故选：BCD.【点睛】关键点点睛：本题考查椭圆性质的综合运用，其中涉及到焦点三角形、定值等问题，着重考查学生的转化与计算能力，难度较大.C 项的解答关键在于表示完斜率乘积后利用点所满足的椭圆方程进行化简计算，D 项的解答关键在于将点的坐标设为三角函数形式，利用三角形函数的取值范围进行分析求解.三．填空题（本大题共4个小题，每小题5分，共20分．把答案填在答题卡相应的横线上）13. 若双曲线221691440x y −−=上一点M 与它的一个焦点的距离为9，则点M 与另一个焦点的距离为________．【答案】15或3 【解析】【分析】化双曲线方程为标准方程，利用双曲线定义求解.【详解】因为221916x y −=，所以3a =，4b =，5c =，设点M 与另一个焦点的距离为x ，则由双曲线的定义得，926x a −==，解得15x =或3x =. 故答案为：15或314. 已知一个圆锥的侧面积为6π，它的侧面展开图是一个半圆，则此圆锥的体积为___________. 【答案】3π 【解析】【分析】设圆锥的底面半径为r ，母线长为l ，分析得出2l r =，由圆锥的侧面积计算出l 、r 的值，可求得圆锥的高，再利用圆锥的体积公式可求得结果.【详解】设圆锥的底面半径为r ，母线长为l ，则圆锥的底面圆周长为r l 2π=π，可得2l r =，圆锥的侧面积为226rl r πππ==，解得3r =，23l =，所以，圆锥的高为223h l r =−=，因此，该圆锥的体积为21133333V r h πππ==⨯⨯=. 故答案为：3π.15. 若直线l ：0x y m ++=与曲线C ：29y x =−只有一个公共点，则实数m 的取值范围是________．【答案】(]{}3,332−−【解析】【分析】先对曲线C 进行变形，可知其表示圆的上半部分，画出曲线C 及直线l ，采用数形结合即可求得结果．【详解】因为曲线2:9C y x =−，可化为()2290x y y +=≥，所以曲线C 是以(0,0)为圆心，3为半径的圆的上半部分，直线:l y x m =−−的斜率为1−，在y 轴上的截距为m −，画图如下：由于直线与曲线只有一个公共点，由图得：[)(]3,33,3m m −∈−⇒∈−，当直线l 与圆相切时，则3322m d m ==⇒=±，由图可知32m =−综上：(]3,3m ∈−或32m =−．故答案为：(]{}3,332−−．16. 已知扇形OPQ 中，半径2r =，圆心角为π02θθ⎛⎫<<⎪⎝⎭，若要在扇形上截取一个面积为1的矩形ABCD ，且一条边在扇形的一条半径上，如图所示，则tan θ的最小值为________．【答案】43【解析】【分析】连接CO ，设COP α∠=，分别用含α的三角函数表示,AB BC ，表示出矩形ABCD 的面积，由矩形面积为1求得tan θ的最小值.【详解】连接CO ，设COP α∠=，则2sin AD BC α==，2cos OB α=，2sin tan tan AD OA αθθ==，2sin 2cos tan AB OB OA ααθ=−=−，则2sin 2cos 2sin 1tan ABCD S AB BC αααθ⎛⎫=⋅=−⋅= ⎪⎝⎭，则24sin 4sin cos 1tan αααθ−=，即24sin 4sin cos 1tan αααθ=−，即24sin tan 4sin cos 1αθαα=−24cos cos 41sin sin αααα=⎛⎫−− ⎪⎝⎭， ∴当cos 12tan sin 2ααα=⇒=时，()min 4tan 3θ=，故答案为：43四．解答题（本大题共6个小题，共70分．解答应写出文字说明、证明过程或演算步骤）17. 已知ABC 的内角A ，B ，C 的对边分别是a ，b ，c ，且sin 3cos 0b A a B +=．（1）求角B 的大小；（2）若2a =，AC 边上的中线3BD =，求ABC 的面积S ．【答案】（1）2π3（2）23【解析】【分析】（1）由正弦定理统一为角的三角函数，化简即可得解；（2）利用中线的向量性质()12BD BA BC =+，结合余弦定理求出4c =，用面积公式求ABC 的面积【小问1详解】sin sin 3cos 0sin 3tan 3B A A B B B B =⇒=−⇒=−，因为()0,πB ∈，所以2π3B = 【小问2详解】()2211134222804242BD BA BC c c c c c ⎡⎤⎛⎫=+⇒=++⋅−⇒−−=⇒= ⎪⎢⎥⎝⎭⎣⎦113sin 2423222S ac B ⇒==⨯⨯⨯= 18. 亚洲运动会简称亚运会，是亚洲规模最大的综合性运动会，由亚洲奥林匹克理事会的成员国轮流主办，每四年举办一届．1951年第1届亚运会在印度首都新德里举行，七十多年来亚洲运动员已成为世界体坛上一支不可忽视的力量，而中国更是世界的体育大国和亚洲的体育霸主．第19届亚运会于2023年9月23日至10月8日在杭州举办，为普及体育知识，增强群众体育锻炼意识，衢州举办了亚运知识竞赛活动．活动分为男子组和女子组进行，最终决赛男女各有40名选手参加，下图是其中男子组成绩的频率分布直方图（成绩介于85到145之间），（1）求图中缺失部分的直方图的高度，并估算男子组成绩排名第8的选手分数：（2）若计划从男子组中105分以下的选手中随机抽样调查2个同学的答题状况，则抽到的选手中至多有1位是95分以下选手的概率是多少？（3）若女子组40位选手的平均分为117，标准差为11，试求所有选手的平均分和方差．【答案】（1）0.025；131 （2）1415（3）118；146 【解析】【分析】（1）先求出所有矩形的面积和为1，从而可求缺失部分的面积，根据矩形面积可求得第8名的成绩位于区间125分至135分之间，从而求解；（2）求得105以下合计6个人，对这6人编号后，利用列举法求解；（3）利用平均数和方差的定义求解即可. 【小问1详解】根据题意得：0.050.20.20.3101h ++++=，得：0.025h =，所以：图中缺失部分的直方图的高度0.025h =；因为分数位于135分至145分人数为：0.1404⨯=人，分数位于125分至135分人数：0.254010⨯=，设第8名选手的分数为x ，则：13541010x −=，得：131x =，所以可估算排名第8名选手的分数为131. 【小问2详解】分数105以下人数有：85分至95分人数：0.05402⨯=人，95分至105分人数：0.1404⨯=人，总共：6人，将6人依次编号为1，2，3，4，5，6（95分以下人编号为1，2），任选2个人的方法如下：列举出所有样本点：12，13，14，15，16，23，24，25，26，34，35，36，45，46，56共计15种，至多有1位是95分以下的选手有14种，所以概率为：1415P =. 【小问3详解】男子组40位选手的平均分:0.05900.11000.21100.31200.251300.1140119y =⨯+⨯+⨯+⨯+⨯+⨯=，所有选手的平均分：1171191182z +==，女子组的方差：2121xS =，男子组的方差：()2222222901190.05190.190.210.3110.25210.1169y S =−⨯+⨯+⨯+⨯+⨯+⨯=()()222222214014011171214012111740x S x x x x =+⋅⋅⋅+−=⇒+⋅⋅⋅+=+， ()()222222214014011191694016911940y S y y y y =+⋅⋅⋅+−=⇒+⋅⋅⋅+=+，所有选手的方差：()222222222140140112111716911921182901191181171181181468022zS x x y y +++−⨯++−−=+⋅⋅⋅+++⋅⋅⋅+−===综述：所有选手的平均分118z =，所有选手的方差2146z S =.19. 已知双曲线C 的渐近线方程是3y x =±，点()2,3M 在双曲线C 上. （1）求双曲线C 的离心率e 的值；（2）若动直线l ：1y kx =+与双曲线C 交于A ，B 两点，问直线MA ，MB 的斜率之和是否为定值？若是，求出该定值；若不是，请说明理由. 【答案】（1）2 （2）是，3 【解析】【分析】（1）根据双曲线的渐近线设出方程，将点的坐标代入求解方程，利用离心率公式直接求解即可；（2）联立方程，韦达定理，代入两斜率之和表达式化简即可求解. 【小问1详解】的由双曲线C 的渐近线方程是3y x =±，故设C ：223x y λ−=，因为()2,3M 在双曲线C 上，所以1293λ=−=，所以C ：2213y x −=，所以1a =，3b =222c a b =+=，所以2ce a==；【小问2详解】设()11,A x y ，()22,B x y ，联立22331x y y kx ⎧−=⎨=+⎩得()223240k x kx −−−=，则248120k ∆=−>得24k <且23k ≠，12223kx x k +=−，12243x x k −=−，又111113132222222MA y kx k k k k k x x x −+−−+−===+−−−， 222223132222222MB y kx k k k k k x x x −+−−+−===+−−−，所以()121122222MA MBk k k k x x ⎛⎫+=+−+ ⎪−−⎝⎭()()()212121222244322122142424233kx x k k k k k k x x x x k k −+−−=+−=+−−+−++−−−()()()()()()()22222232124262212121341244221k k k k k k k k k k k k k k k k k k +−−++−=+−=−−=−−=−+−−+−+−.即直线MA ，MB 的斜率之和是3.20. 如图，四棱锥P ABCD −中，底面ABCD 为矩形，4BC =，2PC PD CD ===，M 为AD 的中点．（1）若BM PC ⊥，求证：BM PM ⊥；（2）若二面角P CD A −−的余弦值为33，求直线PB 与平面PAD 所成角θ的正弦值．【答案】（1）证明见解析（2）23【解析】【分析】（1）证明出BM ⊥平面PCM ，利用线面垂直的性质可证得结论成立；（2）设CD 的中点为N ，AB 的中点为E ，连接PN 、PE 、NE ，过点P 在平面PNE 内作PO NE ⊥，垂足为点O ，分析可知，二面角P CD A −−的平面角为PNE ∠，根据已知条件求出ON 、PN 的长，推导出PO ⊥平面ABCD ，以点O 为坐标原点，DC 、ON 、OP 的方向分别为x 、y 、z 轴的正方向建立空间直角坐标系，利用空间向量法可求得sin θ的值. 【小问1详解】证明：因为四边形ABCD 为矩形，则4AD BC ==，因为M 为AD 的中点，则122AM AD ==，又因为2AB =，AB AM ⊥，则ABM 为等腰直角三角形，所以，45AMB ∠=，同理可证45CMD ∠=，所以，18090BMC AMB CMD ∠=−∠−∠=，即BM CM ⊥，因为BM PC ⊥，PC CM C ⋂=，PC 、CM ⊂平面PCM ，所以，BM ⊥平面PCM ，因为PM ⊂平面PCM ，所以，BM PM ⊥. 【小问2详解】证明：设CD 的中点为N ，AB 的中点为E ，连接PN 、PE 、NE ，过点P 在平面PNE 内作PO NE ⊥，垂足为点O ，因为2PC PD CD ===，且N 为CD 的中点，则PCD 为等边三角形，且PN CD ⊥，2222213PN PD DN =−=−=因为四边形ABCD 为矩形，则//AB CD 且AB CD =，因为N 、E 分别为CD 、AB 的中点，所以，//AE DN 且AE DN =，且AD DN ⊥，所以，四边形ADNE 为矩形，所以，CD NE ⊥，所以，二面角P CD A −−的平面角为PNE ∠，则3cos 3PNE ∠=，因为PO NE ⊥，则3cos 313ON PN PNE =∠==，则22312PO PN ON =−=−=因为CD NE ⊥，PN CD ⊥，PN NE N =，PN 、NE ⊂平面PNE ，所以，CD ⊥平面PNE ，因为PO ⊂平面PNE ，则PO CD ⊥，因为PO NE ⊥，CDNE N =，CD 、NE ⊂平面ABCD ，所以，PO ⊥平面ABCD ，以点O 为坐标原点，DC 、ON 、OP 的方向分别为x 、y 、z 轴的正方向建立如下图所示的空间直角坐标系，则()1,3,0A −−、()1,1,0D −、()1,3,0B −、(2P ，则()0,4,0AD =，(2AP =，(2BP =−，设平面PAD 的法向量为(),,n x y z =，则40320n AD y n AP x y z ⎧⋅==⎪⎨⋅=+=⎪⎩，取2x =，则()2,0,1n =−，所以，222sin cos ,3323n BP n BP n BPθ⋅====⨯⋅，因此，直线PB 与平面PAD 所成角θ的正弦值为23. 21. 已知函数()()232f x x x a x a =−−−．（1）当0a =时，求函数()f x 的值域；（2）若不等式()33f x ≥对x ∈R 恒成立，求实数a 的最小值．【答案】（1）[)0,∞+ （2）215a ≥ 【解析】【分析】（1）根据分段函数分别求各段()f x 的取值范围，然后取其并集即得. （2）首先去绝对值，分别求出0a ≤和0a >时，()f x 的最小值，结合恒成立条件解不等式即得. 【小问1详解】（1）()222,00325,0x x a f x x x x x x ⎧≥=⇒=−=⎨<⎩，①()[)200,x f x x ≥⇒=∈+∞；②()()2050,x f x x <⇒=∈+∞；综上：函数()f x 的值域是[)0,∞+；【小问2详解】（2）去绝对值得()22223,53,x ax a x af x x ax a x a⎧+−≥=⎨−+<⎩，当x a ≥时，方程2230x ax a +−=的21130a ∆=≥，()2222313324f x x ax a x a a ⎛⎫=+−=+− ⎪⎝⎭，当x a <时，方程22530x ax a −+=的22110a ∆=−≤，()222235553510100f x x ax a x a a ⎛⎫=−+=−+ ⎪⎝⎭，①2313430022a a a f a a ⎪−⎛⎫≤⇒≤−⇒−=< ⎝⎭，不符题意，∴0a ≤舍去； ②302a a a >⇒>−，()2min 3355331010100a a a f x f a ⎛⎫>⇒==≥ ⎪⎝⎭， 260215a a ⇒≥⇒≥；综上：215a ≥22. 已知椭圆C 的中心在原点，一个焦点为()1,0F 2倍．（1）求椭圆C 的标准方程；（2）设过焦点F 的直线l 与椭圆C 交于A 、B 两点，1F 是椭圆的另一个焦点，若1ABF 内切圆的半径23r =，求直线l 的方程．【答案】（1）2212x y += （2）1x y =±+【解析】【分析】（1）由题意可求得1c =，2a b =，并且222a b c =+，求得a ，b ，c ，代入椭圆标准方程可得解；（2）设出直线l 方程与椭圆方程联立，根据韦达定理可得12y y +，12y y ，可求得112212112ABF S F F y y y y =⋅⋅−=−△，再根据内切圆半径可表示出1142ABF S a r =⋅⋅△，由此求得答案. 【小问1详解】由题可得1c =，焦点在x 轴上，222a b=2a b =， )2221b b ∴=+，解得21b =，22a =，所以椭圆C ：2212x y +=. 【小问2详解】设()11,,A x y ，()22,B x y ，设直线l 的方程为1x ty =+，()22222222101x y t y ty x ty ⎧+=⇒++−=⎨=+⎩的根为1y ，2y ， 12222t y y t +=−+，12212y y t −=+，且2880t ∆=+>，又∵()12221211212212212422ABF t S c y y y y y y y y t +=⋅⋅−=−=+−=+△，111244422233ABF S a r =⋅⋅=⨯=△， 2221413t t ⋅+=⇒=±，所以直线l 的方程为：1x y =±+.【点睛】思路点睛：本题第二小问属于直线与圆锥曲线综合性问题，设出过点F 的直线l 与椭圆联立，由韦达定理可得12y y +，12y y ，可求出1122112ABF S F F y y =⋅⋅−△，另根据三角形内切圆半径和面积的关系可求得1142ABF S a r =⋅⋅△，由此可求得直线l 的方程.。

陕西省咸阳市礼泉县2024-2025学年高二上学期期中考试语文试题(含答案)

礼泉县2024～2025学年度第一学期期中学科素养评价质量调研高二年级语文注意事项：1．本试题共10页，满分150分，时间150分钟。

2．答卷前，考生务必将自己的姓名和准考证号填写在答题卡上。

3．回答选择题时，选出每小题答案后，用2B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。

4．考试结束后，监考员将答题卡按顺序收回，装袋整理；试题不回收。

第Ⅰ卷（阅读题共70分）一、现代文阅读（35分）（一）现代文阅读Ⅰ（本题共5小题，19分）阅读下面的文字，完成1～5题。

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事实证明，伟大的事业激励人才、造就人才、成就人才，广大人才只有投身自主创新的生动实践，才能充分释放自身的才华和能量，担负起时代赋予的使命职责。

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当前比历史上任何时期更接近实现中华民族伟大复兴的宏伟目标，也比历史上任何时期都更加渴求人才。

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廊坊八中2018—2018学年第一学期期中考试
高二物理
一、选择题：(每小题4分，10小题，共40分.下列各题中有一个或几个答案是正确的，请选出正确答案，选全的得4分，对而不全的得2分，有选错的或不选的得0分)
1．某物质的质量为m、密度为ρ、阿伏伽德罗常数为N A、摩尔质量为M，用下面那组数
据，可以计算出该物质的一个分子的质量（）
A．m、N A B．M、N A C．ρ、N A D．m、ρ、N A
2．有两个分子，用r来表示它们的距离。

当r=r0时，两分子间的引力和斥力大小相等，使两分子从相距很远逐渐靠近，直至不能靠近为止。

在整个过程中两分子间相互作用的势能（）
A．一直增加 B.一直减小
C．先增加后减小D．先减小后增大
3．关于物体内能的变化，以下说法中正确的是（）
A．物体吸收热量，内能一定增加
B．物体吸收热量，同时对外做功，内能可能不变
C．物体对外做功，内能一定减小
D．物体放出热量，同时对外做功，内能可能不变
4．关于温度，以下说法正确的是（）
A．温度是表示物体冷热程度的物理量
B．温度是物体内大量分子平均速率大小的标志
C．温度是物体内大量分子平均动能大小的标志
D．温度是表示物体所含热量多少的标志
5．一根一端封闭的玻璃管开口向下插入水银槽中，内封一定质量的气体，管内水银面低于
管外。

在温度不变时，将玻璃管插入一些，如图所示，下列说法Array正确的是（）
A．玻璃管内气体体积减小
B．玻璃管内气体体积增大
C．管内外水银面高度差减小
D．管内外水银面高度差增大
6．热力学第二定律指出（）
A ．不可能使热量由低温物体传递到高温物体
B ．不可能从单一热源吸收热量并把它全部用来做功
C ．热机效率η≤1
D ．大量分子参与的宏观过程具有方向性
7．关于电场线的叙述，以下说法正确的是（）
A ．电场线上任意一点的切线方向就是电荷在该点的运动方向
B ．电场线为曲线的是非匀强电场，电场线为直线的是匀强电场
C ．顺着电场线的方向移动电荷，电荷的电势能一定降低
D ．只受电场力作用的正电荷的加速度方向一定和它所在处的电场强度方向一致
8．在如图所示的实验装置中，平行板电容器的极板B 与一灵敏静电计相接，极板A 接地。

若极板A 稍向上移动一点，由观察到的静电计指针变化作出平行板电容器电容变小的结论的依据是（）
A ．两极板间的电压不变，极板上的电量变小
B ．两极板间的电压不变，极板上的电量变大
C ．极板上的电量几乎不变，两极板间的电压变大
D ．极板上的电量几乎不变，两极板间的电压变小
9．如图所示，实线是一个电场中的电场线，虚线是一个负
检验电荷在这个电场中的轨迹，若电荷是从a 处运动到b 处，
以下判断正确的是（）
A ．电荷从a 到b 加速度减小
B ．b 处电势能大
C ．b 处电势高
D ．电荷在b 处速度最大
10．两个大小相同、带等量异种电荷的导体小球A 和B ，彼此间的引力为F ．另一个不带电的与A 、B 大小相同的导体小球C ，先与A 接触，再与B 接触，然后移开，这时A 和B 之间的作用力为F'，则F 与F'之比为
A ．8:3
B ．8:1
C ．1:8
D ．4:1 二、填空题：(6小题，每空2分，共30分)
11．在点电荷Q 形成的电场中有一点A ，其电势为1.2×118V （无限远处的电势为零）。

现
在A 点固定一个电量为3×10－4C 的正电荷，而将Q 由静止释放，在Q 移到无穷远的过程
中，电场力做功为 J ，电荷Q 的动能改变了 J 。

12．电场中有abc三点，2×10-8C的正电荷从a到b电场力做功4×10-6J，3×10-8C的负电荷从b到c电场力作功9×10-6J，则U ab＝V，U bc＝V，U ab＝V。

13．已知将电量q＝2.0×10-7C的电荷从电场中的M 点移到N 时，电场力做功5×10-5J，将该电荷从N点移到无穷远点时，电场力做功1.5×10-5 J，则M点的电势为，N点的电势为。

14．在电场中放一电量为q=-3.2×10-8C的点电荷，受到电场力大小F=8.0×10-6N，该力的方向跟P点电场强度的方向，P点的场强大小为；如果在P点放另一带电量为q=6.4×10-8库仑的点电荷，则点电荷所受电场力大小为；方向跟P 点场强方向；如果P点无电荷，则P点的场强应为。

15．现有3kg的水，则含水分子数个，每个水分子质量为kg，若水分子是一个挨一个排列，则每个水分子的体积为m3．
三、实验题：（6分）
16．在做《用油膜法估测分子大小》的实验中，用油酸酒精溶液的为每118mL溶液中有纯油酸6mL注射器测得1mL述溶液为75滴．把一滴该溶液滴人盛水的浅盘里，待水面稳定
玻璃板放在坐标纸上，其形状和尺寸如图所示．坐标中正方形方格的
边长为1cm，则：
(1)油酸膜的面积是cm2．
(2)每滴油酸酒精中含有纯油酸的体积；
(3)按以上实验数据估测出油酸分子的直径．
四、计算题：(3小题，17题8分，18题6分，19题10分，共24分，要求有必要的文字说明、方程式和重要的演算步骤，只有结果没有过程的不能得分，有数值计算的必须写出数值和单位)
17．（8分）取一个横截面积为S=3×10-2m2的不高的圆筒，筒内装有质量为m=0.6kg的水，用它来测量射到地面上的太阳能，某天中午在太阳光下照射2min后，水的温度升高了△t=1℃．已知射到大气顶层的太阳能只有43％到达地面，另外57％被大气吸收和反射，而未到达地面，地球到太阳的距离r=1.5×1011m，水的比热容c=4.2×103J/kg，试估算太阳辐
射的功率P．
18．（6分）空中有竖直向下的电场,电场强度的大小处处相等.一个质量为m=2.0×10 kg 的带电微粒，其带电量是6.0×10-8C，它在空中下落的加速度为0.1g．设微粒的带电量不变，空气阻力不计，取g = 10m /s2，求空中电场强度的大小。

19．（10分）如图所示，两平行板A、B相距为d，电势差为U，A板电势比B板高。

在A 板中央有一小块放射性物质，它能不断地向各个方向放射出质量为m、电量为q的带正电的粒子，粒子的最大速度为v0。

由于B板内侧涂有荧光粉，当粒子到达B板后能被B板吸收并发出荧光。

（不计粒子的重力）求B板上的发光面积。