海淀二模试卷讲评

2022-2023学年北京市海淀区高三（第2次）模拟考试物理试卷1. 如图所示，固定在水平地面上的汽缸内密封一定质量的理想气体，活塞处于静止状态。

若活塞与汽缸的摩擦忽略不计，且气体与外界环境没有热交换。

将活塞从A处缓慢地推至B 处的过程中，下列说法正确的是( )A. 气体的内能保持不变B. 气体的内能逐渐增大C. 气体的压强保持不变D. 气体的压强逐渐减小2. 正弦交变电源与电阻R、理想交流电压表按照图1方式连接，已知电阻，交流电压表示数为。

图2是通过电阻R的电流i随时间t变化的图像。

下列说法正确的是( )A. 电阻R两端的电压u随时间t变化的规律是B. 电阻R两端的电压u随时间t变化的规律是C. 通过电阻R的电流i随时间t变化的规律是D. 通过电阻R的电流i随时间t变化的规律是3. 关于电磁波的应用，下列说法正确的是( )A. 雷达可以利用反射电磁波定位，是因为其工作波段的电磁波衍射效应较为明显B. 移动电话选择微波作为工作波段，是因为微波比其它波段的电磁波的波速更快C. X射线衍射能探测晶体内原子的排列结构，是因为X射线的波长与原子间距相近D. 工程上用射线探测金属内部缺陷，是因为射线具有频率高、波动性强的特点4. 如图所示为光电效应实验中某金属的遏止电压与入射光的频率的关系图像。

已知元电荷e。

根据该图像不能得出的是( )A. 饱和光电流B. 该金属的逸出功C. 普朗克常量D. 该金属的截止频率5. 物理课上，老师演示了一个实验：如图所示，水平粗糙木板上放置两个物块，其间有一个处于拉伸状态的弹簧。

将木板抬至空中保持水平，两物块相对木板保持静止，然后将整个装置无初速释放，下落过程中可能观察到的现象是( )A. 两物块依旧相对木板保持静止B. 两物块相对木板运动且彼此靠近C. 质量大的物块与木板保持相对静止，质量小的物块靠近质量大的物块D. 质量小的物块与木板保持相对静止，质量大的物块靠近质量小的物块6. 关于做自由落体运动的物体，下列说法正确的是( )A. 动能随时间t变化的快慢随时间均匀增大B. 动量p随时间t变化的快慢随时间均匀增大C. 重力势能随位移x变化的快慢随时间均匀减小D. 机械能E随位移x变化的快慢随时间均匀减小7. 如图所示，圆盘可在水平面内绕通过O点的竖直轴转动俯视，圆盘上距轴r处有一质量为m的物块可视为质点。

2023年北京市海淀区中考英语二模试卷1. This is Mary's English book.______is at home.（）A. IB. MyC. MeD. Mine2. Our school library opens______8 a.m.on weekdays.（）A. atB. forC. inD. of3. - ______was your trip to Yunnan，Tom？-It was wonderful.（）A. HowB. WhatC. WhenD. Where4. -John，______you please take out the trash？-OK，Mum.I'll do it right away.（）A. mustB. shouldC. couldD. need5. Read these books，______you'll get to know more about Chinese tea culture.（）A. thoughB. unlessC. andD. but6. Linda is______in art than her sister Lily.（）A. creativeB. more creativeC. most creativeD. the most creative7. We______a survey about students' reading habits last weekend.（）A. doB. didC. have doneD. will do8. Millions of tourists______the Great Wall every year.（）A. visitB. will visitC. visitedD. have visited9. I______a video call with Grandma when Dad arrived home.（）A. haveB. will haveC. am havingD. was having10. Time flies.Mr.Smith______ English in our school for three years.We all enjoy his classes.（）A. teachesB. will teachC. has taughtD. was teaching11. The graduation party______ in the school hall next month.（）A. was heldB. will be heldC. heldD. will hold12. -Jack，do you know______？-Sure.Once a week would be OK.（）A. how long should I water the plantsB. how long I should water the plantsC. how often should I water the plantsD. how often I should water the plantsThe alarm clock went off for the second time.Tired from staying up late to watch a movie，Maddie unwillingly dragged herself out of bed.She loved movies，and she often stayed up late to watch them on TV.She（1）______ got dressed，took a bite of her breakfast，and raced to school.During the first lesson，Maddie had a（2）______ time focusing.She was staring out of the window when her teacher，Ms.Lorenz called her."Maddie，I think you've forgotten your homework again，" Ms.Lorenz said. "How about finishing it today at school"Maddie agreed.After school，she thought about the（3）______ for a while.Then she decided to write about a girl who loved movies.To make the story more fun，she decided to give the girl the ability to jump into the action of any movie she wanted.Maddie became so（4）______ in writing her story that at first she didn't hear Ms.Lorenz tell her she could go home.When she had finished it，she gave it to Ms.Lorenz. When Ms.Lorenz returned the story，there was green ink everywhere.Maddie（5）______ her way of feedback--Ms.Lorenz used green ink to make corrections because she thought students connected red ink with bad feelings.That night，Maddie examined her story.She（6）______ when she reached the last page and saw the words，"Nice job，Maddie!Your plot is exciting.I wish I could jump into movies!Write another piece to correct your mistakes and give it to me next week." The words made Maddie want to do better.Right away，she rewrote the story，taking care to（7）______ her mistakes.The next morning，she put her story into the homework box.And she got more encouraging words and instructions from Ms.Lorenz.Little by little，Maddie got better about completing homework and turning it in on time.She also worked on new movie ideas and discussed them with Ms.Lorenz.Her writing and grades were gradually improving. Her love for movies，along with her teacher's（8）______ ，opened up a new world of learning.13. A. angrily B. carefully C. hurriedly D. properly14. A. hard B. busy C. fair D. cool15. A. goal B. task C. plan D. list16. A. experienced B. disappointed C. surprised D. involved17. A. liked B. missed C. improved D. questioned18. A. froze B. smiled C. dreamt D. regretted19. A. mark B. prove C. correct D. accept20. A. comfort B. promise C. decision D. guidance21.A.Pet and Feed AnimalsB.Pick Your OwnC.Grow VegetablesD.Buy Fresh Productsneed.If you are looking for a place for you and your kids to spend some quality time，a local farm is a good choice.Here are the activities you can do there.（1）______Whether it is tomatoes or beans，your kids can grow them with the help of a farmer.They can dig their hands into the soil，plant seeds （种子），water the plants and finally get full-grown vegetables.They can have much fun while learning the basics of farm work.（2）______Have your kids ever tried picking fruit or vegetables directly from the plants It's a hands--onactivity that can help kids experience the hard work of farmers.They can also learn about different kinds of produce.（3）______A local farm usually has animals that your children can feed or pet，such as pigs，ducks，andchickens.Kids can touch the animals，play with them，and feed them.Even adults can enjoy the pleasant experience.OIt was a school trip.My friends and I were taking photos in the street of Vienna，trying to record the beauty of a foreign country.The change in climate had made my contacts （隐形眼镜）dry out quite often，so I was wearing my glasses.But I didn't think the glasses looked good on me，so for any photo I was in，I would take them off.When I took them off in front of Vienna State Opera，I heard a tiny"ping!"-- a screw （螺丝）fell off my glasses and they separated in half.Hurriedly，my friends and I began to search for the tiny screw.We looked everywhere，but it was nowhereto be found.I was close to tears.How could I enjoy the rest of my trip if I could see nothing clearly! Just at that moment，my friend saw a store with a sign in German whose windows were full of various pairs of glasses.To try our luck，we raced into the store.Not speaking any German，we did our best to explain the situation.A friendly old man smiled gently and took my glasses into the back of the store.After a while，he came back into the room.Smiling softly，he handed me my glasses，fully fixed.I tried to ask the man how much it would cost.However，the man just smiled at me and said，"For you？I will do it for free."That is the kindest thing a stranger has ever done for me.While I was in a foreign country and unable to even speak his language，this man，with a business to run，still gifted me that tiny，necessary screw--thus the ability to see for the rest of my trip and also a memory to be treasured.22. What happened in front of Vienna State Opera？______A. The writer fell to the ground.B. The writer's contacts dried out.C. The writer made new friends.D. The writer's glasses broke apart.23. How did the writer feel when she found the store？______A. Regretful.B. Hopeful.C. Interested.D. Disappointed.24. The writer will treasure the memory because ______ .A. she was moved by a strangerB. she was gifted a set of screwsC. she liked the view in ViennaD. she picked up a new languagePMost kids know it's wrong to shout at or hit someone.But what if that someone is Alexa-- an AI （Artificial Intelligence人工智能）speaker，or Roomba--a robot cleaner？A new study by Teresa Flanagan，a developmental psychologist from Duke University，found that kids aged four to eleven felt neither Roomba nor Alexa should be shouted at or attacked.However，that feeling weakened as kids grew into their teens.Flanagan was inspired to do the research after watching some movies about robots. "In those movies，we see adults interacting with robots in terrible ways，" said Flanagan. "But how would kids interact with them？"Flanagan invited 127 children aged four to eleven to take part in the study.The kids were asked to watch a video of the two devices （设备）and then answer a few questions，like whether it was wrong to attack AI devices and how smart and sensitive they thought Alexa was compared to Roomba.Flanagan studied the survey data and found something encouraging.Overall，kids decided that both Alexa and Roomba，unlike humans，probably weren't ticklish （怕痒）and wouldn't feel pain when they got hit.However，they gave Alexa，but not Roomba，high marks formental and emotional abilities，such as being able to think or getting upset when someone was mean to it. "Young children think Alexa，even without a body，has emotions and a mind，" Flanagan said. "But they don't think Roomba does - maybe that has something to do with Alexa's ability to have verbal （语音的）communication."Although they believed the two intelligent devices had different abilities，children across all ages agreed it was wrong to hit or shout at the machines.However，the older kids got，the more they reported it would be more acceptable to attack devices.The findings offer new ideas about the changing relationship between children and intelligent devices and raise important questions about how we are supposed to treat them in general and as parents.Should adults，for example，model good behavior for their kids by thanking AI devices for their help？For now，Flanagan is trying to understand why children think it is wrong to attack intelligent devices.25. Why does Flanagan mention some movies in Paragraph 3______A. To present how robots influence children.B. To stress the importance of AI technology.C. To explain where the research idea came from.D. To show the popularity of children's TV shows.26. What can we learn from the passage？______A. Adults set a good example on treating AI.B. Flanagan studied the data with the help of AI.C. Alexa has more physical feelings than Roomba.D. Younger kids hold a friendlier attitude to AI devices.27. What will Flanagan probably do next？______A. Examine what good manners kids show toward AI.B. Test whether Alexa is much smarter than Roomba.C. Survey how adults interact with intelligent devices.D. Study why kids think AI is worthy of nice treatment.QAwe is the feeling of amazement and respect mixed with surprise.Research shows that awe experiences decrease stress and anxiety and increase positive （积极的）emotions and overall satisfaction in life.It can also improve our relationships，making us more likely to help others and more supported.Most of us connect awe with something unusual and beautiful：nature，music or a spiritual experience.But ordinary people can bring about awe too.Research shows we can be awed by our nearest and dearest- the people sitting next to us，or talking on the other end of thephone.Psychologists call this interpersonal awe.Often，this interpersonal awe is how people respond （反应）to life's big changes，like seeing a baby's first steps or a friend fighting against cancer.Yet interpersonal awe happens in our everyday life，too.John Bargh，a psychologist，said he was "truly awestruck" by his then-5-year-old daughter while dining in a McDonald's years ago.When she heard another child crying across the restaurant，she took the toy from her Happy Meal，walked over and handed it to him.We can't make someone else behave in an awesome way，but we can get prepared to notice it when they do and take steps to strengthen the emotion's positive effects.To increase your chances of feeling awed by people around you，Marianna Graziosi，an expert in positive psychology，suggests you direct your attention to people's positive sides and catch them at their best.You may think your brother or friend is selfish；there may be a little truth to that，but it's never the whole story. "Try to find examples of him helping others or doing something great.In otherwords，become a field scientist，like Jane Goodall，" Graziosi said.To help you recognize and remember a special experience，say out loud "Wow，that was awesome!" when awe strikes you.Enjoy it in the moment and tell others about it.This will strengthen your positive emotions.And recall it or write about it later.Studies show awe can be drawn again simply by remembering an awe experience.When you feel awed by people around you，thank them and let them know just how much their actions affect you.This makes the other person feel good and can give your relationship an improvement.And you'll also have high levels of happiness and psychological well-being.Awe could be anywhere.Perhaps awe，while an ordinary response to the extraordinary，is also an extraordinary response to the ordinary.28. You are likely to feel interpersonal awe when ______ .A. you listen to a CD of a popular singerB. you read a book about a famous personC. you get a surprise help from your familyD. you watch a beautiful sunset with friends29. By saying the underlined words in Paragraph 5，Graziosi suggests that we should ______ .A. read about how to get on with othersB. try to get a full picture of other peopleC. watch carefully how others do their jobsD. learn from an expert in positive psychology30. The writer would probably agree that ______ .A. awe often comes from senses of satisfactionB. awe moments are usually very difficult to createC. nature provides the most awe in our everyday lifeD. recalling awe experiences strengthens relationships31. Which of the following would be the best title for the passage？______A. Respond to Awe in Everyday LifeB. Spread Awe to People AroundC. Express Thanks to Awesome PeopleD. Take down Awesome Moments32. It's easy to see the fun in buying a huge ice cream or going shopping with friends，but saving money can be even more exciting.That's because saving creates opportunities and opens the door to all kinds of adventures.The key to successful saving is setting a goal.Get a pen and paper，and write down how much it costs，then work out how much you can afford to save each week and how long it will take you to reach your goal.It's important to be realistic because you'll soon get bored if you plan to save every single penny.Some weeks you might decide to save a little more than planned so that you can reach your goals even faster.To avoid accidentally spending your savings，it's a good idea to keep them separate from the rest of your money.You could do this by using a money box or a savings account （账户）.When you save your money in a savings account，you have the opportunity to earn interest.This is money that a bank pays you for leaving the money there -- and the longer you leave money in a savings account，the more interest you will earn.Some savings accounts allow you to take out money whenever you like，but some will keep your money locked away for a certain time.Some parents set up a savings account because they want to put some money away for their kids' future.Once you get the hang of saving，you can set some more bigger goals.Perhaps you want to start saving for university，or you'd like to travel the world when you finish school.Remember，you can save for more than one thing at a time，as some of your goals might take months--or even years- to achieve.If you're new to saving，start off by thinking of some fun goals，like taking a vacation or going to a concert.Every penny you save brings you one step closer to reaching your goal--and what could be more exciting than that(1) What is the key to successful saving______(2) How could you avoid accidentally spending your savings______(3) Why do some parents set up a savings account for kids______(4) If you get pocket money from your parents every week，will you consider setting up a savings account or spending it on things you like？Why？（Give at least two reasons）______ ______33. 根据中文和英文提示，完成一篇不少于50词的文段写作。

北京市海淀区2022届高三二模语文试题（原卷解析）一、非连续性文本阅读阅读下面材料，完成下面小题。

材料一国泰民安是人民最基本、最普遍的诉求和愿望，对生命财产安全、生活稳定以及免遭痛苦、威胁或疾病的安全需求也是人类的本能欲望。

这种安全需求在人的生命历程中处于基础性地位，是人类在漫长的生存奋斗中积累和凝聚的普遍性观念。

安全是人生存发展的首要条件，在一定意义上，人类发展史也是人类安全需求发展的历史。

现代安全的意义体现着人类的进步精神，特别是在国家与社会层面，人类越来越把安全看作文明进步的基础。

安全感和安全观念促使人类社会作为一个整体不断向前发展，不断追求安宁、和谐与幸福，努力消除冲突、混乱与痛苦。

正如黑格尔所讲，安全作为一种普遍精神，使人类在和平的时候尽量谋求彼此福利的增进，即使在战争的时候也要尽量减少破坏。

安全犹如空气和阳光，受益而不觉，失之则难回，它源于人类的基本需求，成为人类的终极需要和根本利益。

随着人类文明的演进，人类在自身的进化中不断促进着安全内涵和外延的扩大。

人的安全感首先来自于国家保障的安全。

中国古代思想家主张，“国以民为本，社稷亦为民而立”。

人民民主专政是国家安全的内涵所在，社会主义国家把实行人民民主专政作为保证人民当家作主的基本职能。

国家安全主要是国家政权、主权、领土统一和完整，人民福祉、经济社会可持续发展、国家其他重大利益相对处于没有危险和不受内外威胁的状态，以及保障其持续处于安全状态的能力。

这种国家安全不仅是全体人民追求的目标，也是国家发展的重要基石和人民福祉的根本保障。

“利莫大于治，害莫大于乱”的中国古训，当今世界一些国家和地区深陷战火和苦难的悲惨景象，近年来恐怖主义带来的巨大危害和当下新冠肺炎疫情蔓延给各国人民造成的深重灾难，都深刻地启示我们：没有国家安全的基础和社会稳定的环境，任何美好蓝图都是空中楼阁，任何人都将一事无成。

这也同样印证了古希腊哲学家德漠克利特所言，“一个治理得很好的国家是最可靠的庇护所，其中有着一切。

海淀区九年级第二学期期末练习语文参考答案及评分标准2022.05 一、基础·运用（共14分）1．（1）示例：指记述典章制度、军事、哲学、文学、艺术等方面成就的珍贵文献（2）示例：①柳宗元②小石潭记（共4分。

共2小题，每小题2分）2．（1）C（2）B（共4分。

共2小题，每小题2分）3．（1）D（2）C（共4分。

共2小题，每小题2分）4．B（2分）二、古诗文阅读（共16分）（一）（共4分）5．柳暗花明又一村（1分。

有错该空不得分）6．予独爱莲之出淤泥而不染（1分。

有错该空不得分）7．示例1：长风破浪会有时直挂云帆济沧海示例2：知困然后能自强也（共2分。

共2空，每空1分，每句中允许有一个不会写的字用拼音替代，有错该空不得分）（二）（共5分）8．①竦峙②示例：宏伟壮阔（共2分。

共2空，每空1分）9．示例：曹操的诗句借助想象表现大海吞吐日月星辰的气势，表达了诗人统一天下的雄心壮志和博大胸襟。

王安石的诗句写不畏视线被浮云遮挡，表明自己在政治上的高瞻远瞩，以及拨云见日的豪迈气概。

王冕的诗句盛赞墨梅的高风亮节，不求人夸只留清香的美德，表达诗人不媚世俗的高尚情操。

（共3分。

曹操诗句，1分；王安石诗句，1分；王冕诗句，1分）10．C（2分）11．乙示例：文中孟子在阐释大丈夫得志与否时，说的是“得志，与民由之；不得志，独行其道”，即得志的时候和百姓一同遵循正道而行，不得志的时候要独自走自己的道路。

乙句的表述与文意不一致。

（共2分。

选择正确，1分；理由，1分，言之成理即可）12．示例：夏完淳兵败被囚，面对洪承畴释放、归顺做官的允诺，不为所动；嘲讽怒骂叛贼，誓死不降；厉声劝导岳丈钱栴慷慨赴死，舍生取义。

夏完淳的表现正体现了大丈夫的“富贵不能淫，贫贱不能移，威武不能屈”。

（共3分。

共3件事，每件事，1分）三、名著阅读（5分）13．示例：在《骆驼祥子》中，作者通过描写恶劣的环境来表现人物。

当天气炎热到令人难以忍受的程度，祥子为了生计不得不拉车挣钱；当大雨滂沱令人难以行走的时候，祥子仍在雨中挣命。

海淀区九年级第二学期期末练习2024.05学校姓名准考证号考生须知1.本试卷共10页，共五道大题，22道小题，满分100分。

考试时间150分钟。

2.在试卷和答题卡上准确填写学校名称、姓名和准考证号。

3.试题答案一律填涂或书写在答题卡上，在试卷上作答无效。

4.在答题卡上，选择题用2B铅笔作答，其他试题用黑色字迹签字笔作答。

5.考试结束，请将本试卷、答题卡和草稿纸一并交回。

一、基础·运用(共14分)初三年级组织“与你相约，续梦前行”毕业季系列展示活动。

语文老师组织同学们开展了戏剧表演和诗歌朗诵两个专题活动，邀请你参与到活动筹备与总结的工作中，帮助完成以下任。

【活动筹备】“成长有‘戏’”戏剧表演1.同学们为自己表演的剧目《范进中举》设计了海报，请你协助完成下面的任务。

(1)在海报的方框里用正楷字书写“与你相约，续梦前行”八个字，补上毕业季系列展示活动主题。

(1分)(2)将海报上关于作家的相关信息补充完整。

(1分)(3)同学们要在海报上加入原著中范进的一句话，与画面相配。

请选择最恰当的一项并说明理由。

(2分)【甲】是了。

我也记得是中的第七名。

【乙】噫!好!我中了!2.下面是同学们根据《孙权劝学》改编的课本剧剧本，请帮助作出选择。

孙权劝学时间建安十五年的某天傍晚地点孙权府邸人物孙权、吕蒙(孙权不停地踱步，时而摸摸胡子，时而背手低头沉思，神色严肃)吕蒙 (轻步上前，向孙权行礼，神态恭谨) 主公，有何吩咐?）孙权 (面对吕蒙，言辞恳切) 尊君如今已是吴国掌事之人，不可以不学习呀!吕蒙 ()可是，臣实在是军务繁忙，学习……这，再等等吧。

孙权 ()孤难道是想让你精研儒家经典，成为学官吗? 只是希望你粗通文略，了解历史罢了!你说军务繁多，比孤还多?孤常常读书，自感获益良多啊!吕蒙 (躬身行礼致谢)诺!(1)剧本中有两处舞台说明尚待完成，选出设计最贴合角色特点的一项(2分)A.吕蒙面露难色孙权轻哼一声，语含失望B.吕蒙面露难色孙权轻叹一声，语重心长C.吕蒙面带疑惑孙权轻哼一声，语含失望D.吕蒙面带疑惑孙权轻叹一声，语重心长(2)孙权与吕蒙的对话中使用了多种称谓，其中不恰当的一项是(1分)A.主公B.尊君C.臣D.孤“青春如诗”诗歌朗诵会3.下面是某同学为诗歌朗诵会创作的一首散文诗，请你帮助完成以下工作。

2023-2024学年北京市海淀区首都师范大学附属中学九年级下学期中考二模数学试题一、选择题：本题共8小题，每小题3分，共24分。

在每小题给出的选项中，只有一项是符合题目要求的。

1.如图是某几何体的三视图，则该几何体是【】A.圆锥B.圆柱C.三棱柱D.三棱锥2.据报道，2024年春节假期北京接待游客约1750万人次，旅游收入同比增长近四成．将17500000用科学记数法表示应为()A. B.C.D.3.如图，，，若，则的大小为()A. B. C. D.4.实数a 在数轴上的对应点的位置如图所示，下列结论中正确的是()A.B.C. D.5.每一个外角都是的正多边形是() A.正四边形B.正六边形C.正七边形D.正九边形6.关于x 的一元二次方程有两个相等的实数根，则m 的值为()A.B.1C.D.47.2024年央视春晚的主题为“龙行龘龘，欣欣家国”．“龙行龘龘”寓意中华儿女奋发有为、昂扬向上的精神风貌．将分别印有“龙”“行”“龘”“龘”四张质地均匀、大小相同的卡片放入盒中，从中随机抽取一张不放回，再从中随机抽取一张，则抽取的两张卡片上恰有一张印有汉字“龘”的概率为()A.B.C.D.8.如图，在等边三角形ABC中，有一点P，连接PA、PB、PC，将BP绕点B逆时针旋转得到BD，连接PD、AD，有如下结论：①≌；②是等边三角形；③如果，那么以上结论正确的是()A.①②B.①③C.②③D.①②③二、填空题：本题共8小题，每小题3分，共24分。

9.若代数式在实数范围内有意义，则x的取值范围是__________.10.分解因式__________.11.方程的解为__________.12.在平面直角坐标系xOy中，若函数的图象经过点和则的值为__________.13.如图，在矩形ABCD中，M，N分别为BC，CD的中点，则的值为__________.14.某实验基地为全面掌握“无絮杨”树苗的生长规律，定期对2000棵该品种树苗进行抽测．近期从中随机抽测了100棵树苗，获得了它们的高度单位：数据经过整理后绘制的频数分布直方图如右图所示．若高度不低于300cm的树苗为长势良好，则估计此时该基地培育的2000棵“无絮杨”树苗中长势良好的有__________棵．15.如图，AB是的直径，点C在上，过点B作的切线与直线AC交于点若，则__________16.小明是某蛋糕店的会员，他有一张会员卡，在该店购买的商品均按定价打八五折．周末他去蛋糕店，发现店内正在举办特惠活动：任选两件商品，第二件打七折，如果两件商品不同价，则按照低价商品的价格打折，并且特惠活动不能使用会员卡．小明打算在该店购买两个面包，他计算后发现，使用会员卡与参加特惠活动两者的花费相差元，则__________花费较少直接填写序号：①使用会员卡；②参加特惠活动；两个面包的定价相差__________元．三、计算题：本大题共2小题，共12分。

北京市海淀区2023-2024学年高三下学期期末练习（二模）数学试卷学校：___________姓名：___________班级：___________考号：___________一、选择题1．已知集合，.若，则a 的最大值为( )A.2 B.0 C. D.-22．在的展开式中，x 的系数为( )A.40 B.10 C. D.3．函数是( )A.偶函数，且没有极值点B.偶函数，且有一个极值点C.奇函数，且没有极值点D.奇函数，且有一个极值点4．已知抛物线，则线段的中点的纵坐标为( )C.3D.45．在中，，，的长为( )6．设a ，，，且，则( )C. D.7．在中，，且，则( )A.C.8．设是公比为的无穷等比数列，为其前n 项和，.则“”是“存在最小值”的( ){}1,0,1,2A =-{3}B xa x =≤<∣A B ⊆1-52(x x-40-10-()3,01,03x x x f x x ⎧≤⎪=⎨⎛⎫>⎪ ⎪⎝⎭⎩24x =6AF ABC △4AB =5AC =cos C =+b ∈R 0ab ≠a b ><2a b >()sin a b a b -<-32a b>ABC △C ∠=CB ==()1CA CB λλ=+- 4CP AB ⋅= λ={}n a ()1q q ≠-n S 10a >0q >n SA.充分而不必要条件B.必要而不充分条件C.充分必要条件D.既不充分也不必要条件9．设函数的定义域为D ，对于函数图象上一点，若集合只有1个元素，则称函数具有性质.下列函数中具有性质的是( )A.C. D.10．设数列的各项均为非零的整数，其前n 项和为.若为正偶数，均有，且，则的最小值为( )A.0B.22C.26D.31二、填空题11．若，则12．已知函数.①若，则函数②若函数在区间上的最小值为，则实数13．二维码是一种利用黑､白方块记录数据符号信息的平面图形.某公司计划使用一款由个黑白方块构成的二维码门禁，现用一款破译器对其进行安全性测试，已知该破译器每秒能随机生成个不重复的二维码，为确保一个二维码在114．如图，在正方体中，P 为棱上的动点，平面，Q 为垂足.给出下列四个结论：①；()f x ()f x ()00,x y ()(){}00,k k x x y f x x D ∈-+∀∈≤R ∣()f x 0x P 1P ()f x x =-()lg f x x =()3f x x =()πsin 2f x x =-{}n a n S ()*,j i i j -∈N 2j i a a ≥20S =10S ()2(i)2i x x +=∈R x ()2cos sin f x x a x =+0a =(f x ()f x ()0,π2-a ()2*n n ∈N n n ⨯162n n ⨯1111ABCD A B C D -AB DQ ⊥1D PC 1D Q CQ =②线段的长随线段的长增大而增大；③存在点P ，使得；④存在点P ，使得平面.三、双空题15．已知双曲线四、解答题16．已知函数，从条件①､条件②､条件③这三个条件中选择一个作为已知，使函数存在且唯一确定.（1）求的值；（2）若不等式在区间内有解，求m 的取值范围.条件①：；条件②：的图象可由的图象平移得到；条件③：在区间内无极值点，且.17．在三棱锥中，，M 为的中点.（1）如图1，若N 为棱上一点，且，求证：平面平面；（2）如图2，若O 为延长线上一点，且平面，直线与平面与平面所成角的正弦值.DQ AP AQ BQ ⊥//PQ 1D DA 22:4x C y -=2()2cos (0)2xf x x ωωω=>()f x ω()2f x <()0,m (2π)3f =()y f x =2cos2y x =()f x ππ(,36-ππ(2(263f f -=-+P ABC -2AB PB ==AP PC MN AP ⊥BMN ⊥PAC CA PO ⊥ABC 2AC ==PB ABC PBC18．图象识别是人工智能领域的一个重要研究方向.某中学人工智能兴趣小组研发了一套根据人脸照片识别性别的程序.在对该程序的一轮测试中，小组同学输入了200张不同的人脸照片作为测试样本，获得数据如下表（单位：张）：（1）从这200张照片中随机抽取一张，已知这张照片的识别结果为女性，求识别正确的概率；（2）在新一轮测试中，小组同学对3张不同的男性人脸照片依次测试，每张照片至多测一次，当首次出现识别正确或3张照片全部测试完毕，则停止测试.设X 表示测试的次数，估计X 的分布列和数学期望；（3）为处理无法识别的照片，该小组同学提出上述程序修改的三个方案：方案一：将无法识别的照片全部判定为女性；方案二：将无法识别的照片全部判定为男性；方案三：将无法识别的照片随机判定为男性或女性（即判定为男性的概率为50%，判定为女性的概率为）.现从若干张不同的人脸照片（其中男性､女性照片的数量之比为）中随机抽取一张，分别用方案一､方案二､方案三进行识别，其识别正确的概率估计值分别记为，，.试比较，，的大小.（结论不要求证明）19．已知椭圆E 的焦点在x 轴上，中心在坐标原点.以E的一个顶点和两个焦点为顶点的三角形是等边三角形，且其周长为（1）求栯圆E 的方程；（2）设过点的直线l （不与坐标轴垂直）与椭圆E 交于不同的两点A ，C ，与直EX 50%1:11p 2p 3p 1p 2p 3p ()2,0M线交于点P .点B 在y 轴上，D 为坐标平面内的一点，四边形是菱形.求证：直线过定点.20．已知函数.（1）若，①求曲线在点处的切线方程；②求证：函数恰有一个零点；（2）若对恒成立，求a 的取值范围.21．设正整数，，，，这里，2，…，n .若，且，则称，，…，具有性质P .（1）当时，若，，具有性质P ，且，，，令，写出m 的所有可能值；（2）若，，…，具有性质P ：①求证：；②求16x =ABCD PD ()()ln 0)f x x a a =-+>1a =()y f x =()()2,2f ()f x ()ln 2f x a a ≤+(),3x a a ∈2n ≥i a *i d ∈N (){}1,1,2,i i i A x x a k d k ==+-= 1i =*12n A A A =N ()1i j A A i j n =∅≤<≤ 1A 2A n A 3n =1A 2A 3A 11a =22a =33a =123m d d d =1A 2A n A ()1,2,,i i a d i n ≤= 1n i =参考答案1．答案：C 解析：由于，所以，故a 的最大值为，故选：C.2．答案：A 解析：设的通项，则，化简得，令，则x 的系数为，即A 正确.故选：A.3．答案：B 解析：当时，，则，当时，，则，所以函数是偶函数，由图可知函数有一个极大值点.故选：B.4．答案：C 解析：抛物线的焦点，解得，故线段.故选：C.5．答案：A解析：由余弦定理可得，A B ⊆1a ≤-1-52(x x-1k T +()5115C 2k k k k T x x --+=-()5215C 2k k k k T x -+=⋅-⋅2k =()225C 240-=0x ≤0x ->1()()3()3x x f x f x --===0x >0x -<1()3()()3x x f x f x --===()f x ()f x 24x y =(0,1F 6A y +=5A y =AF 3=222222543cos 2104AC CB ABCB C AC BC BC +-+-===⋅故选：A.6．答案：C解析：对于A ，取，，故A 错误，对于B ，，，故B 错误，对于C ，由于，，故在单调递减，故，因此，，由于，所以，故，C 正确，对于D ，，，则故选：C.7．答案：B 解析：由题可知，，故，故，解得故选：B.8．答案：A 解析：若且公比，则，所以单调递增，存在最小值，故充分条件成立.若且，当n 为奇数时，，单调递减，故最大值为时，，而15180BC BC +=⇒=2a =b =-122a b =->=-1a =b =-2a b=()sin 0y x x x =->cos 10y x '-≤=sin y x x =-()0,+∞sin 0x x -<sin x x <()0,x ∈+∞a b >0a b ->()sin a b a b -<-3a =-4b =-13227a b =<0CA CB ⋅= CP AB⋅ ()()()()2211881168CA CB CB CA CA CB λλλλλλλ⎡⎤=+-⋅-=-+-=-+-=-+⎣⎦ 1684λ-+=λ=10a >0q >110n n a a q -=>n S n S 1S 10a >q =11112211013212n n n a a ⎡⎤⎛⎫--⎢⎥ ⎪⎡⎤⎝⎭⎢⎥⎛⎫⎣⎦==-->⎢⎥ ⎪⎛⎫⎝⎭⎢⎥⎣⎦-- ⎪⎝⎭121132n n S a ⎡⎤⎛⎫=+⎢⎥ ⎪⎝⎭⎢⎥⎣⎦n S 1n =11S a =，当n 为偶数时，，单调递增，故最小值为，所以，即由，存在最小值得不到公比，故必要性不成立.故公比“”是“存在最小值”的充分不必要条件.故选： A.9．答案：D解析：根据题意，要满足性质，则的图象不能在过点的直线的上方，且这样的直线只有一条；对A ：的直线，如下所示：数形结合可知，过点的直线有无数条都满足题意，故A 错误；对B ：的图象，以及过点的直线，如下所示：数形结合可知，不存在过点的直线，使得的图象都在该直线的上方，故B 错误；对C ：的图象，以及过点的直线，如下所示：123n S a <121132n n S a ⎡⎤⎛⎫=-⎢⎥ ⎪⎝⎭⎢⎥⎣⎦n S 2n =2S =S 110a >n S 0q >10a >0q >n S 1P ()f x ()()1,1f ()f x x =-)1,0()1,0()lg f x x =()1,0()1,0()f x ()3f x x =()1,1数形结合可知，不存在过点的直线，使得的图象都在该直线的上方，故C 错误；对D ：的图象，以及过点的直线，如下所示：数形结合可知，存在唯一的一条过点的直线，即，满足题意，故D 正确.故选：D.10．答案：B 解析：因为，所以，互为相反数，不妨设，，为了取最小值，取奇数项为正值，取偶数项为负值，且各项尽可能小，由题意知：满足，取的最小值；满足，因为，，故取的最小值；满足，取的最小值；同理，取的最小值；所以，满足，取的最小值；()1,1()f x ()πsin 2f x x =-()1,1-()1,1-1y =-0k =2120S a a =+=1a 2a 10a >20a <10S 3a 312a a ≥3a 12a 5a 51531224a a a a a ≥⎧⎨≥≥⎩10a >1142a a >5a 14a 7a 717317531224248a a a a a a a a a ≥⎧⎪≥≥⎨⎪≥≥≥⎩7a 18a 9a 116a 135791111112481631a a a a a a a a a a a ++++=++++=4a 422a a ≥4a 22a满足，因为，所以，取的最小值；满足，因为，所以，取的最小值；同理，取的最小值；所以，所以，因为数列的各项均为非零的整数，所以当时，有最小值22.故选：B.11．答案：1解析：因为，所以，即，所以，解得.故答案为：1.12．答案：①π；②解析：当时，，当时，，且二次函数开口向下，要使得在区间上的最小值为，则需要，且当时取最小值，故，解得，故答案为：π，.13．答案：7解析：由题意可知的二维码共有个，，故，6a 62642224a a a a a ≥⎧⎨≥≥⎩20a <2224a a >6a 12a 8a 828418641224248a a a a a a a a a ≥⎧⎪≥≥⎨⎪≥≥≥⎩20a <222482a a a >>8a 12a 10a 12a 24681022222222229a a a a a a a a a a a ++++=++++=101211131931922S a a a a a =+=-={}n a 11a =10S 2(i)2i x +=222i i 2i x x ++=212i 2i x x -+=21022x x ⎧-=⎨=⎩1x =2-0a =()2cos f x x ==2ππ2==()222cos sin sin sin 1sin 12a f x x a x x a x x ⎛⎫=+=-++=--++ ⎪⎝⎭()0,πx ∈(]sin 0,1x ∈()f x ()0,π2-1022a a -≥-sin 1x =112a -++=-2a =-2-n n ⨯22n ≤221615316022602n n -⨯⨯≤⇒≤2231637n n -≥⇒≥由于，所以，故答案为：7.14．答案：①②④解析：在正方体中，令，以点D 为原点，建立如图所示的空间直角坐标系，设，则，，，，，，令平面的法向量，则，取，得，由平面于Q ，得，即，，显然，解得于是，对于①，，①正确；对于②，上单调递增，②正确；对于③，而，，，，若，显然，即不存在，使得，③错误；对于④，平面的一个法向量，而，*n ∈N 7n ≥1111ABCD A B C D -1AB =(01)AP t t =≤≤(0,0,0)D (0,1,0)C 1(0,0,1)D (1,,0)P t 1(0,1,1)CD =-(1,1,0)CP t =-1D PC (,,)n x y z = 10(1)0n CD y z n CP x t y ⎧⋅=-+=⎪⎨⋅=+-=⎪⎩1y =(1,1,1)n t =-DQ ⊥1D PC ((1),,)DQ n t λλλλ==-((1),,)Q t λλλ-((1),1,)CQ t λλλ=-- 2(1)10CQ n t λλλ⋅=-+-+=λ=222111(,,)(1)2(1)2(1)2t Q t t t --+-+-+1||||D Q CQ ===||DQ ==(1,0,0)A (1,1,0)B ((1)1,,)AQ t λλλ=-- ((1)1,1,)BQ t λλλ=---2222[(1)1](1)(23)(32)10AQ BQ t t t t λλλλλλ⋅=--+-+=-+--+=22(32)4(23)430t t t t ∆=---+=--<[0,1]t ∈0AQ BQ ⋅= 1D DA (0,1,0)DC = ((1)1,,)PQ t t λλλ=---由，得，即，令，，显然函数在上的图象连续不断，而，，因此存在，使得，此时平面，因此存在点P ，使得平面，④正确.所以所有正确结论的序号是①②④.故答案为：①②④.；或（）解析：，即，故焦点与到，则以C 的一个焦点为圆心，且与双曲线C 的渐近线相切的圆的方程为或，或（）.16．答案：（1）条件选择见解析，（2）解析：（1）依题意，，选条件①，由，得，即，，，显然的值不唯一，因此函数不唯一，不符合题意.选条件②，的图象可由的图象平移得到，因此的最小正周期为函数的最小正周期π，而，0PQ DC t λ⋅=-=t λ=t =322310t t -+-=32()231f t t t t =-+-[0,1]t ∈()f t [0,1](0)10f =-<(1)10f =>(0,1)t ∈()0f t =PQ ⊄1D DA //PQ 1D DA 22(1x y ++=22(1x y +=22:4x C y -==12y x =20x y -=)()2x y -=122(1x y ++=22(1x y -+=221x y +=22(1x y -+=2ω=π(,)3+∞π()cos 12cos(13f x x x x ωωω=+=-+(2π)3f =ππ2cos()1233ω-+=ππcos(33ω-=ππ2π33k -=+k ∈N ππ2π33k -=-+k ∈*N ω()f x ()y f x =2cos 2y x =()y f x =2cos 2y x =ω>π=所以.选条件③，在区间内无极值点，且，则，即函数分别在时取得最大值､最小值，于是的最小正周期，由在区间内无极值点，得的最小正周期，因此，而，所以.（2）由（1）知，由，得，由不等式在区间内有解，即内有解，则有所以m 的取值范围是.17．答案：（1）证明见解析解析：（1）连接，，.因为，M 为的中点，所以.又，，，平面，所以平面.因为平面，所以平面平面.（2）因为平面，平面，平面，所以，，为直线与平面所成的角.2ω=()f x ππ(,36-ππ(2(263f f -=-+ππ(()463f f --=()f x x =π3x =-()f x ππ2[()]π63T ≤⨯--=()f x ππ(,)36-()f x ππ2[(π63T ≥⨯--=πT =0ω>2π2Tω==π()2cos(213f x x =-+(0,)x m ∈πππ2(,2333x m -∈--()2f x <(0,)m πcos(23x -<)m π23m ->>π(,)3+∞BM MN BN AB PB =AP BM AP ⊥MN AP ⊥MN BM M = MN BM ⊂BMN AP ⊥BMN AP ⊂PAC BMN ⊥PAC PO ⊥ABC OB ⊂ABC OC ⊂ABC PO OB ⊥PO OC ⊥PBO ∠PB ABC因为直线与平面所以因为，所以，，所以.又，故.所以.如图建立空间直角坐标系.则，，，，.所以，，.设平面的法向量为，则即令，则.设与平面所成角为，则所以直线与平面.（2）分布列见解析；（3）PB ABC PBO ∠=2PB =1PO =OB =2=1OA =2AB =222AB OB OA =+OB OA ⊥O xyz -()0,1,0A )B()0,3,0C ()0,0,1P 110,,22M ⎛⎫⎪⎝⎭()0,3,1PC =- ()BC = 510,,22MC ⎛⎫=- ⎪⎝⎭PBC (),,n x y z =00n PC n BC ⎧⋅=⎪⎨⋅=⎪⎩3030y z y -=⎧⎪⎨+=⎪⎩1y =)n = CM PBC θsin cos ,MC n MC n MC nθ⋅====⋅ CM PBC ()2116E X =231p p p >>解析：（1）根据题中数据，共有张照片被识别为女性，其中确为女性的（2）设事件输入男性照片且识别正确.根据题中数据，由题意知X 的所有可能取值为1，2，3.所以X 的分布列为.（3）.（2）证明见解析，.因为以E 的一个顶点和两个焦点为顶点的三角形是等边三角形，且其周长为所以所以..（2）设直线l 的方程为，令，得.由得.206080+==:A ()P A =()1P X ==()13244X ==⨯=()11344P X ==⨯=312123161616⨯+⨯=231p p p >>216y +=221(0)y a b b+=>>222c a b =-22a c +==a ==26=216y +=()20x ty t =+≠16x =y =1416,t ⎛⎫ ⎪⎝⎭2234242x y x ty ⎧+=⎨=+⎩()223412120t y ty ++-=设，，则设的中点为，则所以因为四边形为菱形，所以N 为的中点，.所以直线的斜率为.所以直线的方程为.令得.设点D 的坐标为，则，即.所以直线的方程为，即.所以直线过定点.20．答案：（1）①；②证明见解析()11,A x y ()22,C x y 12y y +=12y y =AC ()33,N x y 1232y y y +==332x ty =+=ABCD BD AC BD ⊥BD t -BD 22683434t y t x t t ⎛⎫+=-- ⎪++⎝⎭0x =22863434t t y t t =-=++220,34t t ⎛⎫ ⎪+⎝⎭()44,x y 43216234x x t ==+4322234t y y t =-=+221614,3434t D t t ⎛⎫- ⎪++⎝⎭PD ()221414143416161634t t t y x t t ++-=--+()746y x t=-PD ()4,02y =（2）解析：（1）当时，.①所以，.所以曲线在点处的切线方程为.②由①知，.当，所以；当.所以在区间上单调递增，在区间上单调递减.因为，，，所以函数恰有一个零点.（2）由设，，则.所以是上的减函数.因为，，所以存在唯一，.所以与的情况如下：[)1,+∞1a =()()ln 1f x x =-+()11f x x =-'()22f =()20f '=()y f x =()()2,2f 2y =()()ln 1f x x =-+(]1,3x ∈()11f x x =-'()20f '=()1,2x ∈1>>()0f x '>(2,3x ∈1<<()0f x '<()f x ()1,2()2,3()22f =()3ln20f =>()31e 330f -+=-+<-+<()f x ()()ln f x x a =-+()f x '=()()g x x a =--(),3x a a ∈()10g x -'=<()g x (),3a a ()0g a =>()320g a a =-<()0,3x a a ∈()()000g x x a =-=()f x '()f x.当时，因为，所以.所以.所以，符合题意.当时，因为，所以.所以，不合题意.综上所述，a 的取值范围是.21．答案：（1）27或32引理1：若，，…，具有性质P ，则.引理1的证明：假设结论不成立.不妨设，则正整数，但，故一定属于某个，不妨设为.则由知存在正整数k ，使得.这意味着对正整数，有，，但，矛盾.所以假设不成立，从而一定有，从而引理1获证.引理2：若，，…，具有性质P ，则，且证明：取集合.注意到关于正整数k 的不等式等价于而由引理1有，即.()()()()0000ln ln 2f x x a x a x a =-+=-+-1a ≥()20g a a =-≤02x a ≤()()()0ln 222ln 2f x a a a a a a ≤-+-=+()()0ln 2f x f x a a ≤≤+01a <<()20g a a =>02x a >()()()0ln 222ln 2f x a a a a a a >-+-=+[)1,+∞1A 2A n A ()1,2,,i i a d i n ≤= ()1,2,,i i a d i n ≤= 11a d >111a d A -∉*12n A A A =N 11a d -()2i A i n ≤≤2A 112a d A -∈()11221a d a k d -=+-1112c a d d d =-+()111212111c a d d d a d d A =-+=+-∈()()11122212212211c a d d d a k d d d a k d d A =-+=+-+=++-∈12A A =∅ ()1,2,,i i a d i n ≤= 1A 2A n A 111ni i d ==∑1ni i ia d ==∑{}121,2,...,...n T d d d =()1201...i i n a k d d d d <+-≤11i i i i a a k d d -<≤-i i a d ≤011iia d ≤-<这意味着数列而，，…，两两之间没有公共元素，且并集为全体正整数，故T 中的元素属于且仅属于某一个，这就证明了引理2的第一个结论；再考虑集合T 中全体元素的和.一方面，直接由另一方面，，公差为的等差数列.所以的所有元素之和为.最后，再将这n 个集合的全部元素之和相加，得到T 中全体元素的和为.，所以有综上，引理2获证.1i i i i a a k d d -<≤-+12...n i d d d k d ≤()(11,2,...k i i x a k d k =+-=1A 2A n A (1i A i n ≤≤12...A T A T +++ 12......nT T A T A T d +=+++ 211...1nd d +++={121,2,...,...n T d d d =T A i i d i T A 11122i i i i i i i i TT TT T a a d T d d d d d ⎛⎫⎛⎫⋅+-=+- ⎪ ⎪⎝⎭⎝⎭()1,2,...,i T A i n = 112ni i i i T T a T d d =⎛⎫⎛⎫+- ⎪ ⎪ ⎪⎝⎭⎝⎭∑112ni i i i T T a T d d =⎛⎫⎛⎫=+- ⎪ ⎪ ⎪⎝⎭⎝⎭∑()22111111122222nnn ni i i i i i i i i iiiT T T T Tn TTa a a T TT d d d d d ====⎛⎫+⎛⎫=+-=+-=+- ⎪ ⎪ ⎪⎝⎭⎝⎭∑∑∑∑1n i i ia d ==∑1ni i i a d ==回到原题.将，，从小到大排列为，则，.若，则，所以每个不等号都取等，从而，故；情况1：若，矛盾；情况2：若.此时如果，矛盾；如果，故；如果，由于，设，，则，.故对于正整数对，有，从而，这与矛盾.综上，m 的取值只可能是27或32.当时，；当时，.所以的所有可能取值是27和32.（2）①由引理1的结论，即知；②由引理2的第二个结论，即知1d 2d 3d 123r r r ≤≤123123m d d d r r r ==23123111111r r d d d +=++=13r ≥1231111111111311r r r r r r r =++≤++=≤1233r r r ===12327m r r r ==1r =311110r r +=-=1r =31111r r =-=23221111r r r r =+≤+=24≤2r =21102r =-=2r =2112r =-=34=12332m r r r ==23r =12r =()()123123,,,,i i i r r r d d d ={}{}123,,1,2,3i i i =12i d =23i d =()()2121212112331212211i i i i i i i i k a a a a k a a a a ⎧=+--+--⎪⎨=+--+--⎪⎩2112231i i k k a a -=--12121223i i i i a k a k A A +=+∈ 12i i A A =∅ ()()123,,3,3,3d d d =27m =()()123,,4,2,4d d d =32m =123m d d d =()1,2,,i i a d i n ≤= 1ni i i a d ==∑。

2020年海淀二模语文试卷解析基础运用1、2题解析一、基础·运用1(1)B对比分析：难度相当难易度：低试题解析：髓[suǐ]只有这一个读音绚烂：指浓烈繁华，绚丽多彩。

炫：指光明照耀。

“文明之花“为主语，所以”绚烂“正确。

寻：上下结构，笔顺：横折、横、横、横、竖钩、点学易感悟：考察字音字形笔顺，非常基础。

1(2)3对比分析：难度相当难易度：低试题解析：历久弥新：指经历长久的时间而更加鲜活，更加有活力，更显价值。

此类题型如果遇到不熟悉的词语，要结合前后文语境，带入所选义项验证一下。

学易感悟：常见词语，很简单。

1(3)乙对比分析：难度相当难易度：中试题解析：考察了句子特点和朗读技巧。

乙句是对偶句，所以朗读节奏应该前后一致并且对应。

原句说“读出明显对比变化“不恰当。

学易感悟：掌握朗读相关技巧。

2(1)A对比分析：难度相当难易度：低试题解析：第一处联系上下文语境，一直在围绕中华文化的“多元开放“讨论。

并且”君子和而不同“与”追求共性“不符。

第二处关注后文”社会理想“也就是全体社会成员共同的目标，因此”天下为公“更合适。

学易感悟：仔细读文段，结合前后文。

2(2)“源远流长“改为”多元开放“对比分析：难度相当难易度：中试题解析：注意题干要求“作为总结句”，也就是画线句要起到总结上文的作用。

这一段最主要谈论的是中华文化的包容性，即“多元开放”。

显然画线句没有出现“多元开放”。

把“源远流长”改成“多元开放即可。

学易感悟：认真审题，抓住文段主要内容和中心句。

基础·运用3题解析第3题难易度：中等试题解析：（1）消息、新闻类题型。

该题型一般包括：新闻标题拟写类、新闻信息提取概括类、新闻材料看法类以及综合学习类等题型。

同时可结合书法、对联、病句、成语修辞赏析等其他基础题型进行综合考查。

本题属于根据新闻稿内容补充导语总起句，考查新闻相关内容，属于八年级常规题型，在近三年的海淀二模中首次出现。

导语是消息的核心，导语需要突出文稿中的关键信息，集中呈现最有价值的新闻事实，需要提取文稿的主要信息对比三个选项得出答案为B。