【英语】福建省厦门市2016-2017学年高二上学期期末质量检测试题及答案
厦门市2016-2017学年度第一学期高二年级质量检测英语试题第一部分听力（共两节，满分30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1.5分，满分7.5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。
例:How much is the shirt?A.￡19.15.B.￡9.18.C.￡9.15.答案是C。
1.Why did the woman wake up at night?A.Because of the bed.B.Because of the light.C.Because of the mouse.2.Where is the second house they saw?A.On Orange Street.B.On Bridge StreetC.On Broadway Street3.When will Tom arrive home?A.At10:00.B.At10:55.C.At11:00.4.What does the man suggest the woman do?A.Ask Amy to wait.B.Visit the art gallery.C.Write the paper.5.How does the man feel in the end?A.Angry.B.Excited.C.Worried.第二节（共15小题；每小题1.5分，满分22.5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三个选项中选出最佳选项，并标在试卷的相应位置。
听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟；听完后，各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听第6段材料，回答第6至7题。
6.When did the two speakers last meet?A.Eight months ago.B.Two years ago.C.Three years ago.7.What did the woman do when she was in China?A.She was a journalist.B.She was an editor.C.She was a programmer.听第7段材料，回答第8至10题。
8.What is the man going to do this afternoon?A.Go to movies.B.Watch a match.C.Play tennis.9.What does Kate think about tennis?A.She has no interest in it.B.She likes it very much.C.She finds it cool.10.Why is watching movies on TV so difficult for the man?A.Too many noises.B.Too many channels.C.Too many advertisements.听第8段材料，回答第11至13题。
11.What are the two speakers talking about?A.Ordering a meal.B.Asking for advice.C.Making coffee.12.How would the man like his steak?A.Rare.B.Medium rare.C.Medium.13.Which kind of salad dressing does the man choose at last?A.Oil and Vinegar.B.Blue Cheese.C.French.听第9段材料,回答第14至16题。
14.Who went to the bookstore with Ted?A.Mary.B.Paul.C.Jack.15.What happened to Paul?A.He missed the movie.B.He ran a fever.C.He felt sick.16.What can be learned from the conversation?A.Mary feels sorry.B.Paul is grounded.C.Jack dislikes museum.听第10段材料，回答第17至20题。
17.How did Barley get to the college?A.By bus.B.By bike.C.On foot.18.Where did Barley live before the police found him?A.In a tent.B.At a motel.C.In a dormitory.19.Who started GoFundMe page for Barley?A.Richard.B.The other officer.C.Casey.20.What can best describe Barley?A.He is helpful.B.He is strong-willed.C.He is weak.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。
ADear strangers,I remember you.Eighteen months ago,when my cell phone rang,you were walking into Whole Foods prepared to do your grocery shopping,just as I had been only minutes before you. But I had already abandoned my cart full of groceries in the entryway.My brother was on the other end of the line telling me my father had taken his own life early that morning.I started to cry and scream as my whole body trembled.I fell to the floor,my knees bending under the weight of what I had just learned.You could have simply stopped and had a brief look, but you didn’t.Instead,you surrounded me as I yelled through my sobs,“My father killed himself. He’s dead.”I remember one of you asked for my phone and whom you should call.What was my password?You needed my husband’s name as you searched through my contacts.I remember that I could hear your words as you tried to reach my husband for me,leaving an urgent message or him to call me.I recall hearing you discuss among yourselves who would drive me home in mycar and who would follow that person back to the store.You didn’t even know one another,but it didn’t matter.You met me,a stranger,in the worst moment of my life,and you were around me with common purpose—to help.In my fog,I told you that I had a friend who worked at Whole Foods,and one of you brought her to me.And I even recall as I sat with her,one of you sent over a gift card to Whole Foods; though you didn’t know me,you wanted to let me know that you would be thinking of me.That gift card helped to feed my family when the idea of cooking was so far beyond my emotional reach.I never saw you after that.But I know this to be true:Because you reached out to help,you offered a ray of light in the darkest moment I’ve ever experienced.You may not remember me. But I will never,ever forget you.21.How did the author react to the news?A.She broke down.B.She lost her memory.C.She argued with her brother.D.She dropped her phone.22.What did one of the strangers do for the author?A.He helped get her friend.B.He lent her a gift card.C.He brought her husband to her.D.He called her brother.23.What can we learn from the underlined part in Paragraph4?A.I had no idea how to cook.B.I was not in the mood of cooking.C.I couldn’t reach for any groceries.D.I had no way to support the family.24.What’s the author’s purpose in writing the text?To express thanks. B.To call for help.C.To tell a true story.D.To show her sorrow.BFor centuries China has been associated with rich culture and there’s no shortage of books. But that raises a problem:How do you find the book you want?In ancient time the books were made of bamboo and wooden chips.After somebodyfinished reading a book,they would roll the book from the tail end and the top end would be left ually,two empty chips would be put in the top end and the chapter name written on the back of the empty chips.So when somebody took the book,they would know the content without opening it.This became the embryonic form(雏形)of the earliest bookmark.During the Warring States Period(475-221BC),silk books gradually replaced bamboo chips. To make it easier for readers to find the book,people would tie a small plate at one end of the roller marked with the book name or other information.The bookmark during that time was always made with bamboo,with flower patterns on the bamboo as decoration.Wealthy families used ivory as bookmark or hung the ivory bookmarks on the bookshelves.In Tang Dynasty,printing made paper books become popular.The silk strip with book name was used as bookmark and stuck on the book cover.To distinguish(区分)with the bookmark of scroll(卷轴),the bookmark was called“floating mark”During the Qing Dynasty(1644-1911), hardcover and paperback books became thicker and the name began to be printed on the cover.So people began to place the bookmark inside the book to mark the reading process.The bookmark we currently use come in various shapes and materials and are creatively designed.25.What was a bookmark originally designed for?A.To decorate a book.B.To distinguish social class.C.To show the content of a book.D.To mark the reading process.26.What appeared in the Warring States Period?A.Floating mark.B.The bookmark of scroll.C.Bookmarks with flower patterns.D.Empty chips with chapter names.27.What is the passage mainly about?A.The function of the bookmark in China.B.Creatively-designed bookmark in China.C.Typical Chinese culture—floating markD.The evolution of bookmark in ancient ChinaCIs virtual reality sexist?Women experience more motion sickness than men whileusing VR,and researchers have suggested a novel theory,which is contrary to the commonly held view that motion sickness is caused when your senses provide conflicting information.“Women are more likely to be influenced than men by motion sickness in general,”says Thomas Stoffregen at the University of Minnesota.“We wanted to know whether that was also the case with VR headsets.”Stoffregen and his team ran experiments in which36people-half of them men,half of them women-played two VR games using the Facebook-owned Oculus Rift headset.The first game in which players had to push a small glass ball around a maze(迷宫)only made a few people feel sick.But the second game that involved taking a virtual walk around a ghost house set off feelings of sickness in14out of18women and only six out of18men.Why were women more likely to be affected?Stoffregen’s answer is grounded in his broader ideas on motion sickness.He thinks that it is linked to“postural sway”-the slight movements our bodies make when we stand or sit still.People who sway more,he says,will be more likely to feel sick.“Women tend to be smaller than men,they have a different body shape,and they have smaller feet than men of comparable height,”he says.“In a purely physical sense,there’s reduced stability(稳定)in the female body,so there’s an increased likelihood that any sort of disturbing motion stimulus will contribute to instability.”But not everyone agrees.28.What can we learn from Stoffregen’s experiments?A.Women are more afraid of ghosts.B.VR headsets amaze all the people.C.The virtual walk makes more women feel sick.D.Oculus Rift is specially designed for the experiment.29.What is Stoffregen’s theory?A.Women tend to be smaller than men.B.Reduced stability leads to motion sickness.C.Women sway less when standing or sitting still.D.Motion sickness occurs when one’s senses conflict.30.What would probably be discussed in the following paragraph?A.Ways to measure postural sway.B.Different opinions on the new theory.C.The influence of motion sickness.D.More examples to support Stoffregen’s theory.DA multipotentialite is someone with many interests and creative pursuits(追求).It is easy to see your multipotentiality as a shortcoming that you need to overcome.And you might feel like something is wrong with you.But in fact there are some great strengths to being this way.Here are two super powers.One:rapid learning.When multipotentialites become interested in something,we go hard. We observe everything we can get our hands on.We are also used to being beginners,because we’ve been beginners so many times in the past,and this means that we’re less afraid of trying new things and stepping out of our comfort zones.What’s more,we bring everything we’ve learned to every new area,so we’re hardly starting from scratch.Second:adaptability.It is the ability to change into whatever you need to be in a given situation.Abe Cajudo is sometimes a video director,sometimes a web designer,and sometimes James Bond.He’s valuable because he can take on various roles.Fast Company magazine identified adaptability as the single most important skill to develop in the21st century.The economic world is changing so quickly and unpredictably that only multipotentialites are really able to meet the needs of the market.We have a lot of complex problems in the world right now, and we need creative,out-of-the box thinkers to solve them.In fact,some of the best teams are made up of a specialist and multipotentialite together.The specialist can dive in deep ideas,while the multipotentialite brings much knowledge of the project.It’s a beautiful partnership.So to the multipotentialites,I say:Follow your curiosity down those rabbit holes.Embracing your many passions leads to a happier life.And perhaps more importantly-multipotentialites,the world needs you.31.What may be the best title for the text?A.Think out of the boxB.Take advantage of your strengthC.What you are is a multipotentialiteD.Be proud of being a multipotentialite32.Why are multipotentialites needed now?A.To satisfy people’s curiosity.B.To strengthen social support.C.To make up a great partnership.D.To deal with the changing world.33.What does the underlined phrase in Paragraph2probably mean?A.Full of confidence.B.Out of curiosity.C.Without previous knowledge.D.With much experience.34.Why is Abe Cajudo mentioned in the passage?A.To prove his popularity.B.To share his personal affairs.C.To display his various talents.D.To show his great adaptability.Which of the following can best describe multipotentialites?A.Adventurous and flexible.B.Creative and considerate.C.Talented and stubborn.D.Enthusiastic and smart.第二节（共5小题；每小题2分，满分l0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。
福建省厦门市2024_2025学年九年级语文上学期期末质量检测试题新人教版
福建省厦门市2024-2025学年上学期九年级期末质量检测语文试题(试卷满分：150分考试时间：120分钟)考生留意：1．全卷分三个部分，共23题；2．答案一律写在答题卡上，否则不能得分。
一积累与运用(20分)（一）语言积累（12分）1．请依据提示填写相应的古诗文。
（12分）（1）关关雎鸠，。
（《诗经·关雎》）（1分）（2），天际若比邻。
（王勃《送杜少府之任蜀州》）（1分）（3）闲来垂钓碧溪上，。
（李白《行路难（其一）》）（1分）（4）安得广厦千万间，。
（杜甫《茅屋为秋风所破歌》）（1分）（5）满面尘灰烟火色，。
（白居易《卖炭翁》）（1分）（6），蜡炬成灰泪始干。
（李商隐《无题》）（1分）（7），月有阴晴圆缺，此事古难全。
（苏轼《水调歌头》）（1分）（8）北冥有鱼，。
（庄子《北冥有鱼》）（1分）（9），教然后知困。
（《虽有嘉肴》）（1分）（10）山水之乐，。
（欧阳修《醉翁亭记》）（1分）（11）《酬乐天扬州初逢席上见赠》中运用典故写出人事变迁之怅惘的句子是：“，”。
（2分）（二）语言运用（8分）2. 下列文学常识说法正确．．的一项是（）（2分）A．范仲淹是北宋政治家、文学家，其作品《岳阳楼记》朗朗上口，千古传颂。
B．雨果是德国大文豪，代表作有《巴黎圣母院》《凄惨世界》《九三年》《醇厚人》。
C. 苏轼词风婉约，他使词摆脱了对音乐的依附，拓宽了词的题材、意境、手法。
D．《儒林外史》《三国演义》均为我国明清时期闻名的章回体长篇历史演义小说。
3.阅读下面的文字，按要求作答。
（6分）阅读让你的世界草长莺飞，繁花似锦；让你身居斗室，漫游广①mào（）宇宙。
通过阅读，使你的心灵犹如秋雨中的池塘，渐渐丰盈。
阅读更是写作灵感之泉不涸的秘谛。
一个人（甲）（A．纵然 B．虽然）有支生花妙笔，（乙）（A．尽管 B．倘如）没有经过阅读的浸润②zī（）养，也唯恐会有江郎才尽之时。
（1）文中①②处依据拼音写汉字。
福建省厦门市2023-2024学年高二下学期7月期末英语试题(含答案)
厦门市2023—2024学年第二学期高二年级质量检测英语试题本试卷分五部分，共12页。
满分150分。
考试用时120分钟。
本试题附有答题卡。
注意事项：1. 答卷前，考生务必将自己的姓名、准考证号等信息填写在答题卡和试卷指定位置上。
2. 回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。
如需改动，用橡皮擦干净后，再选涂其他答案标号。
回答非选择题时，将答案写在答题卡上，写在本试卷上无效。
3. 考试结束，考生将本试卷和答题卡一并交回。
第一部分听力（共两节，满分30分）第一节（共5小题；每小题1.5分，满分7.5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。
例：How much is the shirtA. 19.15.B. 9.18.C. 9.15.答案是C。
1. Who probably has the informationA. Kate.B. John.C. Ginger.2. What should the man doA. Keep on practising.B. Control his anger.C. Adjust the schedule.3. What is the man going to do nextA. Perform magic.B. Do hairdressing.C. Teach painting.4. How is the man feeling nowA. Confused.B. Ashamed.C. Anxious.5. Why does Steven still live in the old flatA. He can't afford a new one.B. He has friendly neighbors.C. He wants to save money.第二节（共15小题；每小题1.5分，满分22.5分）听下面5段对话或独白。
大连二十中2016-2017学年高二英语期末试卷及答案
2016～2017 学年度上学期期末考试高二英语试卷考试时间120 分钟试题分数 150第一部分：听力第一节1.Whose wallet probably is it?A.the woman’sB. The man’sC. Gina’s2.What is the date today?A.May 1stB. May 2ndC. May 3rd3.Who made the woman’s personal web page?A.She herselfB. Her friend.C. The man4.What are the speakers mainly talking about?A. A townB. An accidentC. A murder5.What is the probable relationship between the speakers?A.Relatives.B. Colleagues.C. Neighbors.第二节（共15 小题；每小题1.5 分，满分22.5 分）6.What is the man doing?A.Watching TVB. Listening to the radio.C. Reading the newspaper7.Why are the rich and famous targets for being talked about according to the man?A.The public are interested in them.B. The public want to laugh at them.C. The public want to make them famous.听第7 段材料，回答第8-10 题。
8.Why does the man look sad? A. He had to stop his study. B. He failed his test again.C. He couldn’t understand the foreigners.9.How did the woman improve her French?A.By reading French newspapersB. By watching French films.C. By speaking to herself.10 . What does the woman suggest the man do?A.Be confident.B. Talk with foreignersC. Ask the teacher for advice.听第8 段材料，回答第11 至13 题。
2016-2017学年八年级上学期第一次学情调研考试英语试题(含答案)
2016-2017学年上学期第一次学情调研初二英语试卷（考试时间：100分钟考试形式：闭卷考试总分：100分）第Ⅰ卷（选择题，共45 分）一、单项选择（共15 小题，每小题 1 分，计15 分）从A 、B 、C、D 四个选项中选出一个可以填入句中空白处的最佳答案。
1. Tom is _______ honest boy, and he has ______ good sense of humour.A. a, aB. an, /C. a, /D. an, a2. My best friend Jack looks happy, and he always _____ a smile ________ his face.A. wears, onB. put, inC. wear, onD. wears, in3.I didn’t have breakfast this morning. Do you have _____________?A. anything eatB. something eatingC. anything to eatD. something eat4. They young man wears glasses because he spends too much time _____ the computer games.A. to playB. playingC. to workD. working5. — Health is money.— But I think it is money.A. as important asB. more important thanC. so important thanD. the same as6. There is a big cake. ____________ share it.A. May be we canB. We may beC. Maybe we canD. Maybe can we7.We all think he is one of the _____________ in our school.A. good teacherB. best teachersC. better teachersD. best teacher8. —Peter is _______ than you, right?—Yes, but he is ________ runner in our class.A. heavier, the bestB. heavier, bestC. heavyer, the bestD. heavy, better9. —What is your brother like? —_______________.A. He likes watching TV.B. He is a worker in a factory.C. He is very helpful and friendly.D. He works on a farm.10. “_________” is another way of saying “cookie”.A. BiscuitB. EraserC. HolidayD. Maths11.Times seems __________ when we are playing computer games.A. go fasterB. to go fasterC. to going fastD. go fast12.Why don’t you _________ an English Club to practice _________English?A. to join, speakingB. to join, to speakC. join, to speakD. join, speaking13. I have __________ money than you, but I have __________ friends than you.A. more, moreB. less, moreC. fewer, moreD. more, less14. The Changjiang River is ________ river in China.A. the longestB. the longerC. the second longerD. the second longest15. —Would you like some bread？—______________. I’m hungry now.A. Yes, I would.B. No, I don’tC. Yes, please.D.No, thanks.二、完形填空（共10 小题，每小题 1 分，计10 分）阅读下面短文，掌握其大意，然后从各题所给的四个选项中选出一个最佳答案。
2019-2020学年福建省厦门市高二下学期期末考试英语试题及答案
2019-2020学年福建省厦门市高二下学期期末考试试题及答案第一部分听力理解(共两节，满分30分)第一节(共5小题，每小题1.5分，满分7.5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.Where is Helen going?A.To the library.B.To a history class.C.To a friend’s house.2.What is the man doing?A.Asking for help.B.Giving directions.C.Opening an account.3.Who will the woman introduce to the man?A.Barbara.B.Steve.C.Mary.4.When does the wedding start?A.At7:15.B.At7:30.C.At8:00.5.What is the probable relationship between the speakers?A.Teacher and student.B.Husband and wife.C.Mother and son.第二节(共15小题，每小题1.5分，满分22.5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三个选项中选出最佳选项，并标在试卷的相应位置。
听每段对话前，你将有时间阅读各个小题，每小题5秒钟。
听完后，各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听第6段材料，回答第6、7题。
6.Where can the man buy cheaper tickets?A.In the hotel.B.At the gate.C.On the Internet.7.What is the weather like on the day of the Derby?A.Rainy.B.Cool.C.Uncertain.听第7段材料，回答第8至10题。
福建省厦门市、泉州市五校2024-2025学年高二上学期11月期中联考试题 数学含答案
厦泉五校2024-2025学年高二年级第一学期期中联考数学试题（答案在最后）（考试时间：120分钟满分：150分命题人：）试卷分第I 卷（选择题）和第II 卷（非选择题）两部分第I 卷（选择题，共58分）一、单选题:（本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的.）1.已知(1,1,0),(1,1,2)a b ==- ，则a b ⋅= （）A.1- B.0C.1D.22.椭圆221259x y +=上一点P 到左焦点的距离为6，则P 到右焦点的距离为（）A.5B.6C.4D.123.“3a =”是“直线20x y +-=与圆C ：()()228x a y a -+-=相切”的（）A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分条件也不必要条件4.下列命题中，不正确的命题是（）A.空间中任意两个向量一定共面B.若a b ∥，则存在唯一的实数λ，使得a bλ=C.对空间中任一点O 和不共线的三点A ，B ，C ，若243OP OA OB OC =-+，则P ，A ，B ，C 四点共面D.若{},,a b c 是空间的一个基底，m a c =+，则{},,a b m 也是空间的一个基底5.平行六面体1111ABCD A B C D -的底面ABCD 是边长为2的正方形，且1160A AD A AB ∠=∠=︒，13AA =，M 为11A C ，11B D 的交点，则线段BM 的长为（）A.B.C.3D.6.在平面直角坐标系xOy 中，直线l ：0mx y m +-=被圆M ：224210x y x y +--+=截得的最短弦的长度为（）A.B.2C. D.47.已知12,F F 分别为椭圆()2222:10x y E a b a b+=>>的两个焦点，P 是椭圆E 上的点，12PF PF ⊥，且122PF PF =，则椭圆E 的离心率为（）A.102B.104 C.53D.568.如图是一个棱数为24是由一个正方体截去八个一样的四面体所得.若点E 为线段BC 上的动点，则直线DE 与直线AF 所成角的余弦值的取值范围是（）A.132⎡⎢⎣⎦，B.132⎡⎢⎣⎦，C.122⎡⎢⎣⎦， D.1，22⎡⎢⎢⎥⎣⎦二、多选题：（本题共3小题，每小题6分，共18分．在每小题给出的选项中，有多项符合题目要求．全部选对的得6分，部分选对的得部分分，有选错的得0分．）9.空间直角坐标系O xyz -中，已知()1,2,2A -，()0,1,1B ，下列结论正确的有（）A.()1,1,3AB =--B.点A 关于xOy 平面对称的点的坐标为()1,2,2-C.若()2,1,1m =，则⊥m ABD.若(),2,6n a =- ，n BA∥，则2a =-10.如图，在棱长为2的正方体1111ABCD A B C D -中，E ，F ，G 分别为棱11A D ，1AA ，CD 的中点，则（）A.直线BE 与CD 所成角的余弦值为53B.点F 到直线BE 的距离为1C.1B G ⊥平面BEFD.点1A 到平面BEF 的距离为4311.已知椭圆22:1259x y C +=，12,F F 分别为它的左右焦点，A ，B 分别为它的左右顶点，点P 是椭圆上的一个动点，下列结论中正确的有（）A.存在P 使得12π2F PF ∠=B.椭圆C 的弦MN 被点()1,1平分，则925MN k =-C.12PF PF ⊥，则12F PF 的面积为9D.直线PA 与直线PB 斜率乘积为定值925第Ⅱ卷（非选择题，共92分）三、填空题:(本题共3小题，每小题5分，共15分.)12.已知直线l 的一个方向向量为()1,5-，则直线l 的斜率为_______．13.已知F 为椭圆22:14x C y +=的一个焦点，点M 在C 上，O 为坐标原点，若||||OM OF =，则OMF的面积为________．14.古希腊数学家阿波罗尼奥斯（约公元首262～公元前190年）的著作《圆锥曲线论》是古代世界光辉的科学成果，著作中这样一个命题：平面内与两定点距离的比为常数k （0k >且1k ≠）的点的轨迹是圆，后人将这个圆称为阿波罗尼斯圆，已知点()1,0A -，()2,0B ，圆()()221:24C x y m -+-=()0m >，在圆上存在点P 满足2PA PB =，则实数m 的取值范围是______．四、解答题:(本题共5小题，共77分．解答应写出文字说明、证明过程或演算步骤.)15.已知ABC V 的顶点()4,3A ，AB 边上的高所在直线为30x y --=，D 为AC 中点，且BD 所在直线方程为370x y +-=.（1）求AB 边所在的直线方程；（2）求顶点B 的坐标.16.已知空间三点()2,0,1A ，()2,4,3B -，()1,1,1C ．（1）求向量AB 与AC夹角的余弦值；（2）求ABC V 的面积.17.已知圆C 的圆心M 在直线2y x =-上，并且经过点(0,1)P -，与直线10x y --=相切．（1）求圆C 的方程；（2）经过点(2,1)的直线l 与圆C 相交于A ，B 两点，若||2AB =，求直线l 的方程．18.已知椭圆C 的中心在坐标原点，左焦点为F10），点12M ⎫⎪⎭在椭圆上.（1）求椭圆C 的标准方程；（2）过点P （1，0）的直线l 交椭圆C 于两个不同的点A 、B ，若△AOB （O 是坐标原点）的面积S =45，求直线AB 的方程.19.已知O 为坐标原点，圆O ：221x y +=，直线l ：y x m =+（01m ≤<），如图，直线l 与圆O 相交于A （A 在x 轴的上方），B 两点，圆O 与x 轴交于,M N 两点（M 在N 的左侧），将平面xOy 沿x 轴折叠，使y 轴正半轴和x 轴所确定的半平面（平面AMN ）与y 轴负半轴和x 轴所确定的半平面（平面BMN ）互相垂直，再以O 为坐标原点，折叠后原y 轴负半轴，原x 轴正半轴，原y 轴正半轴所在直线分别为x ，y ，z 轴建立如图所示的空间直角坐标系．（1）若0m =．（ⅰ）求三棱锥A BMN -的体积；（ⅱ）求二面角A BN M --的余弦值．（2）是否存在m ，使得AB 折叠后的长度与折叠前的长度之比为306？若存在，求m 的值；若不存在，请说明理由．厦泉五校2024-2025学年高二年级第一学期期中联考数学试题（考试时间：120分钟满分：150分命题人：）试卷分第I 卷（选择题）和第II 卷（非选择题）两部分第I 卷（选择题，共58分）一、单选题:（本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的.）1.已知(1,1,0),(1,1,2)a b ==- ，则a b ⋅= （）A.1- B.0C.1D.2【答案】B 【解析】【分析】根据空间向量数量积的坐标运算求值即可.【详解】因为()()()1,1,01,1,21111020a b ⋅=⋅-=⨯-+⨯+⨯=.故选：B2.椭圆221259x y +=上一点P 到左焦点的距离为6，则P 到右焦点的距离为（）A.5 B.6 C.4D.12【答案】C 【解析】【分析】根据椭圆的定义求解即得.【详解】由225a =，则5a =，所以210a =，根据椭圆的定义，点P 到右焦点的距离为1064-=.故选：C.3.“3a =”是“直线20x y +-=与圆C ：()()228x a y a -+-=相切”的（）A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分条件也不必要条件【答案】A 【解析】【分析】根据相切关系可得1a =-或3，结合充分、必要条件分析判断.【详解】圆C ：()()228x a y a -+-=的圆心为(,)C a a ，半径为，若直线20x y +-=与圆C =，解得1a =-或3，且{}3是{}1,3-的真子集，所以“3a =”是“直线20x y +-=与圆C ：()()228x a y a -+-=相切”的充分不必要条件．故选：A.4.下列命题中，不正确的命题是（）A.空间中任意两个向量一定共面B.若a b ∥，则存在唯一的实数λ，使得a bλ= C.对空间中任一点O 和不共线的三点A ，B ，C ，若243OP OA OB OC =-+，则P ，A ，B ，C 四点共面D.若{},,a b c 是空间的一个基底，m a c =+，则{},,a b m 也是空间的一个基底【答案】B 【解析】【分析】根据共面向量、向量平行、四点共面、基底等知识对选项进行分析，从而确定正确答案.【详解】A 选项，空间中任意两个向量可以通过平移的方法平移到同一个平面，所以空间中任意两个向量一定共面，A 选项正确.B 选项，若a b ∥，可能a 是非零向量，b是零向量，此时不存在λ，使a b λ=，所以B 选项错误.C 选项，对于243OP OA OB OC =-+，有()2431+-+=，所以,,,P A B C 四点共面，所以C 选项正确.D 选项，若{},,a b c 是空间的一个基底，m a c =+，假设m xa yb =+ ，(),1a c xa yb c x a yb +=+=-+，则,,a b c 共面，与已知矛盾，所以,,a b m不共面，所以{},,a b m是基底，所以D 选项正确.故选：B5.平行六面体1111ABCD A B C D -的底面ABCD 是边长为2的正方形，且1160A AD A AB ∠=∠=︒，13AA =，M 为11A C ，11B D 的交点，则线段BM 的长为（）A.B.C.3D.【答案】A 【解析】【分析】由11122BM AA AD AB =+-平方即可求解.【详解】由题意可知：()11111111111112222BM BB B D BB A D A B AA AD AB =+=+-=+-，则2222211111111122442BM AA AD AB AA AD AB AA AD AA AB AB AD⎛⎫=+-=+++⋅-⋅-⋅ ⎪⎝⎭11911323201122=+++⨯⨯-⨯⨯-=，所以BM =.故选：A.6.在平面直角坐标系xOy 中，直线l ：0mx y m +-=被圆M ：224210x y x y +--+=截得的最短弦的长度为（）A.B.2C. D.4【答案】C 【解析】【分析】先求出直线l 过定点()1,0A ，由圆的几何性质可知，当AM ⊥直线l 时，弦长最短，求解即可.【详解】直线l ：0mx y m +-=过定点()1,0A ，圆M ：()()22214x y -+-=，圆心()2,1M ，半径2R =因为点()1,0A 在圆M 内，由圆的几何性质可知，当AM ⊥直线l 时，弦长最短为==故选：C7.已知12,F F 分别为椭圆()2222:10x y E a b a b+=>>的两个焦点，P 是椭圆E 上的点，12PF PF ⊥，且122PF PF =，则椭圆E 的离心率为（）A.2B.4 C.3D.6【答案】C 【解析】【分析】利用椭圆的定义结合勾股定理，易得等式求出离心率.【详解】由椭圆定义得：122PF PF a +=，又因为122PF PF =，所以解得：1242,33PF a PF a ==，再由于12PF PF ⊥，122F F c =，结合勾股定理可得：()22242233a a c ⎛⎫⎛⎫+= ⎪ ⎪⎝⎭⎝⎭，解得2259c a =，所以椭圆E 的离心率为3，故选：C.8.如图是一个棱数为24是由一个正方体截去八个一样的四面体所得.若点E 为线段BC 上的动点，则直线DE 与直线AF 所成角的余弦值的取值范围是（）A.1232⎡⎢⎣⎦， B.1332⎡⎢⎣⎦， C.1222⎡⎢⎣⎦， D.13，22⎡⎢⎢⎥⎣⎦【答案】C 【解析】【分析】在原正方体中建立空间直角坐标系，由空间向量求解【详解】由题意得该几何体有6282的等边三角形，在原正方体中建立如图所示的空间直角坐标系，原正方体边长为2，则(2,1,0)A ，(2,2,1)F ，(1,2,2)D ，设(,1,2),01E t t t -≤≤，(0,1,1)AF = ，(1,1,0)DE t t =---，则直线DE 与直线AF 所成角的余弦值2222212112cos 121212(1)(1)t t t t t t t θ++===+++⋅-+--，而01t ≤≤，故[]220,11t t ∈+，12cos [,]22θ∈，故选： C.二、多选题：（本题共3小题，每小题6分，共18分．在每小题给出的选项中，有多项符合题目要求．全部选对的得6分，部分选对的得部分分，有选错的得0分．）9.空间直角坐标系O xyz -中，已知()1,2,2A -，()0,1,1B ，下列结论正确的有（）A.()1,1,3AB =--B.点A 关于xOy 平面对称的点的坐标为()1,2,2-C.若()2,1,1m =，则⊥m AB D.若(),2,6n a =- ，n BA∥，则2a =-【答案】ACD 【解析】【分析】根据空间向量的坐标运算计算判断A ．由对称点的性质判断B ，由向量的数量积是否为0判断C ，由向量平行的坐标表示求参判断D ，【详解】由题意(0,1,1)(1,2,2)(1,1,3)AB =--=--，A 正确；关于xOy 平面对称的点的坐标,x y 坐标相同，z 坐标相反，因此点A 关于xOy 平面对称的点的坐标为(1,2,2)，B 错，若()2,1,1m = ，则2130m AB ⋅=--+= ，所以⊥ m AB ，C 正确；若(),2,6n a =- 且n BA∥，则23116a -==--，解得2a =-，D 正确，故选：ACD ．10.如图，在棱长为2的正方体1111ABCD A B C D -中，E ，F ，G 分别为棱11A D ，1AA ，CD 的中点，则（）A.直线BE 与CD 所成角的余弦值为3B.点F 到直线BE 的距离为1C.1B G ⊥平面BEFD.点1A 到平面BEF 的距离为43【答案】BC 【解析】【分析】建系，利用空间向量求异面直线夹角、点到线的距离、判断线面垂直以及点到面的距离.【详解】如图，以D 为坐标原点建立空间直角坐标系，则()()()()()()()1112,0,0,2,2,0,0,2,0,0,0,0,2,0,22,2,2,0,0,2A B C D A B D ，且E ，F ，G 分别为棱11A D ，1AA ，CD 的中点，可知()()()1,0,2,2,0,1,0,1,0E F G ，可得()()()()()111,2,2,0,2,1,0,2,0,2,1,2,0,0,1=--=-===uur uu u r uuu r uuu r uuu rBE BF DC GB FA ，对于选项A ：因为42cos ,323⋅-===-⨯⋅uur uuu ruur uuu r uur uuu r BE DC BE DC BE DC，所以直线BE 与CD 所成角的余弦值为23，故A 错误；对于选项B ：因为BF 在BE方向上的投影向量的模长为2⋅=uu u r uuruur BF BE BE，且5BF = 点F 到直线BE ()22521-=，故B 正确；对于选项C ：因为110GB BE GB BF ⎧⋅=⎪⎨⋅=⎪⎩，可得11GB BE GB BF ⊥⎧⎨⊥⎩，且BE BF B = ，,BE BF ⊂平面BEF ，所以1B G ⊥平面BEF ，故C 正确；对于选项D ：因为平面BEF 的法向量可以为()12,1,2=uuu rGB ，点1A 到平面BEF 的距离为11123⋅=uuu r uuu r uuur FA GB GB ，故D 错误；故选：BC.11.已知椭圆22:1259x y C +=，12,F F 分别为它的左右焦点，A ，B 分别为它的左右顶点，点P 是椭圆上的一个动点，下列结论中正确的有（）A.存在P 使得12π2F PF ∠=B.椭圆C 的弦MN 被点()1,1平分，则925MN k =-C.12PF PF ⊥，则12F PF 的面积为9D.直线PA 与直线PB 斜率乘积为定值925【答案】ABC 【解析】【分析】根据余弦定理结合余弦定理求出12F PF ∠的范围判断A ；根据点差法求中点弦的斜率判定B ；根据勾股定理和面积公式求解判断C ；根据斜率公式及点P 在椭圆上求解斜率之积判断D.【详解】对于A .由余弦定理知()2222212122112211212122cos 22PF PF PF PF F F PF PF F F F PF PF PF PF PF +-⋅-+-∠==()()22222122112212212122711122522PF PF F F PF PF F F b PF PF a PF PF +-+-=-≥-=-=-⎛⎫+ ⎪⎝⎭，当且仅当11PF PF =时，等号成立，因为cos y θ=在(0,π)上递减，所以此时21F PF ∠为钝角最大，所以存在P 使得12π2F PF ∠=，所以A 正确；对于B .当直线MN 的斜率不存在，即直线1x =时，1,,1,55M N ⎛⎫⎛- ⎪ ⎪ ⎝⎭⎝⎭，()1,1不是线段MN 的中点，所以直线MN 的斜率存在.设()()1122,,,M x y N x y ，则222211221,1259259x y x y +=+=，两式相减并化简得12121212925y y y y x x x x +--=⋅+-，所以1212925MN y y k x x -==--，所以B 正确；对于C .12210PF PF a +==，1228F F c ==，因为12PF PF ⊥，所以222121264PF PF F F +==，因为()2221212122PF PF PF PF PF PF +=++，解得1218PF PF =.因为12PF PF ⊥，所以1212192F PF S PF PF ==△，所以C 正确；对于D .()()5,0,5,0A B -，设()()000,5P x y x ≠±，则22001259x y +=，整理得()202092525x y -=-，可得直线PA ，PB 的斜率分别为0000,55PA PB y yk k x x ==+-，所以()20200022000092592555252525PA PB x y y yk k x x x x --⋅=⨯===-+---，所以D 错误.故选：ABC.第Ⅱ卷（非选择题，共92分）三、填空题:(本题共3小题，每小题5分，共15分.)12.已知直线l 的一个方向向量为()1,5-，则直线l 的斜率为_______．【答案】5-【解析】【分析】根据直线的方向向量与直线斜率之间的关系分析求解即可.【详解】由题意可知，直线l 的斜率为551k ==--.故答案为：5-.13.已知F 为椭圆22:14x C y +=的一个焦点，点M 在C 上，O 为坐标原点，若||||OM OF =，则OMF的面积为________．【答案】12##0.5【解析】【分析】法一：直接设出M 坐标为00(,)x y ，利用M 在椭圆上以及||||OM OF =得到关于00,x y 的方程组，从而解出0||y ，进而求出OMF 的面积.法二：利用||||OM OF =，推导出M 与两焦点构成的三角形为直角三角形，再求出该直角三角形面积，从而利用几何关系得出OMF 的面积.【详解】法一：设椭圆上()00,M x y ，则2203x y +=，又220014x y +=，联立解得2013y =，03y =，则01112232OMF S OF y === ．法二：设椭圆的另一焦点F '，OF OF OM '==，则焦点F MF '△为直角三角形，设||FM m =，||F M n '=，则24m n a +==，2222||12m n FF '+===，解得2mn =，所以112F MF S mn '==△.则1122OMFF MF S S '==△△．故答案为：1214.古希腊数学家阿波罗尼奥斯（约公元首262～公元前190年）的著作《圆锥曲线论》是古代世界光辉的科学成果，著作中这样一个命题：平面内与两定点距离的比为常数k （0k >且1k ≠）的点的轨迹是圆，后人将这个圆称为阿波罗尼斯圆，已知点()1,0A -，()2,0B ，圆()()221:24C x y m -+-=()0m >，在圆上存在点P 满足2PA PB =，则实数m 的取值范围是______．【答案】52122⎤⎢⎥⎣⎦【解析】【分析】设s ，根据2PA PB =求出点P 的轨迹方程，根据题意可得两个圆有公共点，根据圆心距大于或等于半径之差的绝对值小于或等于半径之和，解不等式即可求解.【详解】设s ，因为点()1,0A -，()2,0B ，2PA PB =，=22650x y x +-+=，所以()2234x y -+=，可得圆心()3,0，半径2R =，由圆()()221:24C x y m -+-=可得圆心()2,C m ，半径12r =，因为在圆C 上存在点P 满足2PA PB =，所以圆()2234x y -+=与圆()()221:24C x y m -+-=有公共点，所以112222-≤≤+，整理可得：2925144m ≤+≤，解得22m ≤≤，所以实数m的取值范围是,22⎣⎦，故答案为：,22⎣⎦.四、解答题:(本题共5小题，共77分．解答应写出文字说明、证明过程或演算步骤.)15.已知ABC V 的顶点()4,3A ，AB 边上的高所在直线为30x y --=，D 为AC 中点，且BD 所在直线方程为370x y +-=.（1）求AB 边所在的直线方程；（2）求顶点B 的坐标.【答案】（1）70x y +-=；（2）(0,7)B .【解析】【分析】（1）利用垂直关系求出直线AB 的斜率，进而求出其方程.（2）求出直线的交点坐标即可.【小问1详解】由AB 边上的高所在直线30x y --=的斜率为1，得直线AB 的斜率为1-，又直线AB 过()4,3A ，所以直线AB 的方程为()34y x -=--，即70x y +-=.【小问2详解】由直线BD 的方程为370x y +-=，而顶点B 为直线AB 与直线BD 的交点，由37070x y x y +-⎧⎨+-⎩＝＝，解得07x y ⎧⎨=⎩＝，所以点()0,7B .16.已知空间三点()2,0,1A ，()2,4,3B -，()1,1,1C ．（1）求向量AB 与AC夹角的余弦值；（2）求ABC V 的面积.【答案】（1）5-（2【解析】【分析】（1）根据空间向量数量积与模长的坐标表示可得向量夹角余弦值；（2）根据夹角余弦值可得正弦值，进而可得三角形面积.【小问1详解】由()2,0,1A ，()2,4,3B -，()1,1,1C ，则()0,4,2AB =- ，()1,1,0AC =- ，AB = ，AC = ，所以cos ,AB AC == ；【小问2详解】由（1）得cos cos ,5A AB AC ==- ，则sin 5A ==，所以11sin 225ABC S AB AC A =⋅⋅=⨯ 17.已知圆C 的圆心M 在直线2y x =-上，并且经过点(0,1)P -，与直线10x y --=相切．（1）求圆C 的方程；（2）经过点(2,1)的直线l 与圆C 相交于A ，B 两点，若||2AB =，求直线l 的方程．【答案】（1）222430x y x y +-++=（2）4350x y --=或2x =【解析】【分析】（1）设圆C 的方程为222()()(0)x a y b r r -+-=>，由题意，列出方程组，求解得,,a b r 的值，即可写出圆C 的方程；（2）分直线的斜率是否存在进行讨论，斜率不存在时，联立方程求出点,A B 的坐标，计算弦长验证，斜率存在时，设l 的方程为1(2)y k x -=-，由圆心到直线的距离等于半径求出k 的值即得.【小问1详解】设圆C 的方程为222()()(0)x a y b r r -+-=>，由已知得2222(1),b a a b r r ⎧⎪=-⎪⎪+--=⎨⎪⎪=⎪⎩，解得1a =，2b =-，r =所以圆C 的方程为22(1)(2)2x y -++=，即222430x y x y +-++=；【小问2详解】①若直线l 有斜率，可设l 的方程为1(2)y k x -=-，即(12)0kx y k -+-=，由已知，则圆心(1,2)M -到直线l=解得43k =，此时，直线l 的方程为41(2)3y x -=-，即4350x y --=；②若直线l 没有斜率，则l 的方程为2x =，将其代入22(1)(2)2x y -++=，可得1y =-或=3y -，即得(2,1)A -，(2,3)B -，满足条件||2AB =，综上所述，直线l 的方程为4350x y --=或2x =．18.已知椭圆C 的中心在坐标原点，左焦点为F 10），点12M ⎫⎪⎭在椭圆上.（1）求椭圆C 的标准方程；（2）过点P （1，0）的直线l 交椭圆C 于两个不同的点A 、B ，若△AOB （O 是坐标原点）的面积S =45，求直线AB 的方程.【答案】（1）24x +y 2=1；（2）x +y ﹣1=0或x ﹣y ﹣1=0.【解析】【分析】（1）由已知可得椭圆的左、右焦点坐标，而点12M ⎫⎪⎭在椭圆上，所以|MF 1|+|MF 2|=2a ，从而可求出a 的值，再由222b a c =-可求出b ，从而可求得椭圆C 的标准方程；（2）设1122()A x y B x y ,,(,)，由题可设直线AB 的方程为x =my +1，然后将直线方程与椭圆方程联立方程组，消去x ，利用根与系数的关系，从而可表示出△AOB 的面积，列方程可求出m 的值，进而可得直线AB 的方程.【详解】解：（1）根据题意，设椭圆C 的方程为2222x y a b+=1（a ＞b ＞0），因为椭圆的左焦点为F 10），设椭圆的右焦点为F 20），由椭圆的定义知|MF 1|+|MF 2|=2a ，所以2a =4，所以a =2，所以1b ===，所以椭圆C 的方程为24x +y 2=1，（2）设1122()A x y B x y ,,(,)，由题可设直线AB 的方程为x =my +1.联立直线与椭圆的方程，22141x y x my ⎧+=⎪⎨⎪=+⎩，消去x 得（4+m 2）y 2+2my ﹣3=0，则有12122223,44m y y y y m m -+==-++，所以1212S OP y y =-==24m =+又由S =45，即2445m =+解得m 2=1，即m =±1.故直线AB 的方程为x =±y +1，即x +y ﹣1=0或x ﹣y ﹣1=019.已知O 为坐标原点，圆O ：221x y +=，直线l ：y x m =+（01m ≤<），如图，直线l 与圆O 相交于A （A 在x 轴的上方），B 两点，圆O 与x 轴交于,M N 两点（M 在N 的左侧），将平面xOy 沿x 轴折叠，使y 轴正半轴和x 轴所确定的半平面（平面AMN ）与y 轴负半轴和x 轴所确定的半平面（平面BMN ）互相垂直，再以O 为坐标原点，折叠后原y 轴负半轴，原x 轴正半轴，原y 轴正半轴所在直线分别为x ，y ，z 轴建立如图所示的空间直角坐标系．（1）若0m =．（ⅰ）求三棱锥A BMN -的体积；（ⅱ）求二面角A BN M --的余弦值．（2）是否存在m ，使得AB 折叠后的长度与折叠前的长度之比为6？若存在，求m 的值；若不存在，请说明理由．【答案】（1）（ⅰ）16；（ⅱ）7（2）存在，2m =【解析】【分析】（1）（ⅰ）由已知，可得22,22A ⎛⎫ ⎪ ⎪⎝⎭，22,22B ⎛⎫-- ⎪ ⎪⎝⎭，即可求得求三棱锥A BMN -的体积；（ⅱ）求出平面ANB 的一个法向量()11n =+和平面BMN 的一个法向量()0,0,1m =，利用向量的坐标运算即可求得二面角A BN M --的余弦值．（2）分别求出AB 折叠前的长度与折叠后的长度，比为6时，求得22m =，可得答案.【小问1详解】（ⅰ）若0m =，折叠前直线l 的方程为y x =，联立221y x x y =⎧⎨+=⎩，解得22x y ⎧=⎪⎪⎨⎪=⎪⎩或22x y ⎧=-⎪⎪⎨⎪=-⎪⎩，可得,22A ⎛⎫ ⎪ ⎪⎝⎭，,22B ⎛⎫-- ⎪ ⎪⎝⎭，圆O ：221x y +=，与x 轴交于,M N 两点，则2MN =，折叠后三棱锥A BMN -的体积为111232226⨯⨯⨯⨯=．（ⅱ）由（ⅰ）及已知，则0,,22A ⎛⎫ ⎪ ⎪⎝⎭，,,022B ⎛⎫- ⎪ ⎪⎝⎭，()0,1,0M -，()0,1,0N，0,1,22AN ⎛⎫=-- ⎪ ⎪⎝⎭，,1,022BN ⎛⎫=-+ ⎪ ⎪⎝⎭．设平面ANB 的一个法向量为(),,n x y z =，则00AN n BN n ⎧⋅=⎪⎨⋅=⎪⎩，即10221022y z x y ⎧⎛⎫--=⎪ ⎪ ⎪⎪⎝⎭⎨⎛⎫⎪-++= ⎪⎪ ⎪⎝⎭⎩，令1y =，则1x =+1z =，所以()11n =+．易知()0,0,1m =为平面BMN 的一个法向量，设二面角A BN M --的大小为θ，由题可知θ为锐角，所以cos cos ,m n m n m n θ⋅==7===故二面角ABN M --的余弦值为7-．【小问2详解】设折叠前1,1，2,2，圆心()0,0到直线l 的距离d =，则AB==，直线l与圆O方程联立22,1,y x mx y=+⎧⎨+=⎩得222210x mx m++-=，即12x x m+=-，21212mx x-=．设A，B在新图形中的对应点分别为,A B''，()110,,A x y'，()22,,0B y x-'，A B=''=．若AB折叠后的长度与折叠前的长度之比为6，306=，解得2m=，故当2m=时，AB折叠后的长度与折叠前的长度之比为6．【点睛】关键点点睛：（2）由几何法求出AB折叠前的长度，直曲联立，消元后得韦达定理，利用弦长公式求得折叠后的长度，令它们之比为6时，求得2m=，可得答案.。
2016-2017学年福建省福州高级中学高二年级上学期期末考试英语学科试卷(Word版)
福州高级中学2016-2017学年第一学期期末考试高二英语试卷（完卷时间：120分钟满分：150分）编辑人：胡红梅审核人：福州教管部本试卷分第一卷（100分）和第二卷（50分），全卷满分150分，考试时间120分钟第一部分：听力（略）第二部分：阅读理解（共两节，满分40分）第一节阅读理解（共15小题；每小题2分，.满分30分）AMotherhood may make women smarter and may help prevent dementia（痴呆） in old age by bathing the brain in protective hormones（荷尔蒙）, U．S．researchers reported on Thursday．Tests on rats show that those who raise two or more litters of pups do considerably better in tests of memory and skills than rats who have no babies, and their brains show changes that suggest they may be protected against diseases such as Alzheimer’s（早老痴呆症）．University of Richmond psychology professor Craig Kinsley believes his findings will translate into humans．“Our research shows that the hormones of pregnancy（怀孕） are protecting the brain, including estrogen（雌激素）, which we know has many neuroprotective （保护神经的） effects,” Kinsley said．“It’s rat data but humans are mammals just like these animals are mammals,” he added in a telephone interview．“They go through pregnancy and hormonal changes．”Kinsley said he hoped public health officials and researchers will look to see if having had children protects a woman from Alzheimer’s and other forms of age-related brain decline．“When people think about pregnancy, they think about what happens to babies and the mother from the neck down,” said Kinsley, who presented his findings to the annual meeting of the Society of Neuroscience in Orlando, Florida．“They do not realize that hormones are washing on the brain．If you look at female animals that have never gone through pregnancy, they act differently towardyoung．But if she goes through pregnancy, she will sacrifice her life for her infant —that is a great change in her behavior that showed in genetic alterations（改变） to the brain．”21:How do scientists know “Motherhood may make women smarter”?A．Some researchers have told them．B．Many women say so．C．They know it by experimenting on rats．D．They know it through their own experience．22:What does the phrase “litters of pups” mean in the second paragraph? A．Baby rats．B．Animals．C．Old rats．D．Grown-up rats．23: Which title is the best for this passage?A．Do You Want to Be Smarter?B．Motherhood Makes Women SmarterC．Mysterious HormonesD．An Important StudyBAdult ￡10.00Child(5-15) ￡5.70Family tickets2 adults with up to 2 children ￡25.701 adults with up to2 children ￡15.70Concessions (ID required)Senior (60 or over) ￡8.00Student ￡8.00Unemployed ￡8.00Disabled ￡9.00Groups of 15 or moreAdult ￡7.20Child ￡5.10ConcessionsAre you a Brighton ＆Hove resident living within postcodes BN1, BN2, BN3 or BN41 or a student studying at Brighton or Sussex University? If so, you can enjoy reduced admission to the Royal Pavilion all year round:From 1 April 2012 to 31 March 2013: ￡5.00 per adult with up to four accompanying children free; proof of Brighton ＆ Hove residency is required.Student registered at either Brighton or Sussex University are also entitled to residents’ rate, even if they live outside of the city. A student card confirming current undergraduate or postgraduate study is required.From 1 April 2013 to 31 March 2014 : ￡ 5.25 per adult with up to four accompanying children free; proof of Brighton ＆Hove residency is required.You can enjoy free admission to the Royal Pavilion all year round with membership of the Royal Pavilion ＆Museums Foundation. This is the charity that supports the work of the Royal Pavilion and Museums.Opening timePlease note: Closed for essential maintenance 21-25 January 2013Open dailyOctober to March: 10 a.m. to 5:15 p. m (last tickets at 4:30 p.m.)April to September: 9 a.m. to 5:45 p.m.(last tickets at 5 p.m.)Closed 24 December (from 2:30 p.m.)and all day on 25 and 26 December24. To enter the Royal Pavilion, a group of 20 junior middle school boys will pay_______.A. ￡180.00B. ￡125.70C. ￡102.00D. ￡135.7025. To enjoy free admission to the Royal Pavilion all year round,_______A. you must study at Brighton or Sussex UniversityB. you must have a student cardC. you must be a member of the Royal Pavilion ＆Museums FoundationD. you must be a local adult with at most four children26. If you visit the Royal Pavilion in June,______.A. you can’t come in after 4:30 p.m.B. you can’t come in after 5p.m.C. you will come in for freeD. you can’t enjoy reduced admissionCWe sometimes think global warming or climate change is a problem very far away from our lives, and that only the government needs to worry about it. But it's hardly possible to completely stay out of it scientists are now 95 percent sure that humans have been the “dominant cause” of global warming trends since the 1950s.One of the conclusions of a report released on Sept 27 by the United Nations says that human activities have caused global temperatures to rise by 0.89 0C between 1901 and 2012. That might not seem like a lot, but the truth is that a major part of that heat has been absorbed into the oceans, which is not surprising given that they cover two thirds of Earth's surface. Also, water has a much greater capacity (容量) to absorb heat than the air does, according to The Economist.While many greenhouse gases occur naturally and are needed to keep the Earth warm enough to support life, humans' use of fossil fuels is the main source of excess (多余的) greenhouse gases. According to CNN, by driving cars, using electricity produced by burning coal and oil or heating our homes with coal or natural gas, we release a large amount of greenhouse gases into the atmosphere.Your body can barely feel a difference of 0.89 0C, but the Earth's ecosystems are so sensitive that even small changes can disturb them. It often starts with the smallest creatures at the bottom of the food chain, eventually affecting bigger animals, many of which could end up becoming extinct.Global warming is also linked to an increase in extreme weather. A larger amount of carbon dioxide traps more energy inside the atmosphere. This changes the patterns of storms and rainfall in many regions and can lead to droughts and floods. Worse still, melting sea ice in warmer oceans is causing sea levels to rise at a speed of more than three millimeters per year, according to The Guardian, which also increases the risk of flooding.“The report should serve as another wakeup call that our activities today have a profound impact on society, not only for us, but for many generations to come，”French Michel Jarraud, secretary-general of the World Meteorological Organization, said at a news conference.27: The underlined word “dominant” in the first paragraph means “ ________”．A．common B．indirectC．secondary D．leading28: Which of the following statements is TRUE according to the passage? A．Greenhouse gases have been proven to harm merely the Earth .B．There is little we can do to prevent global warming.C．Global warming can increase the chances of droughts and floods.D．The hugeness of oceans makes them better absorbers of heat than the air. 29: The rise in global temperatures between 1901 and 2012 ________.A．is more harmful to smaller creatures than bigger onesB．is not serious because most of the extra heat has been absorbed by the oceans C．has made the Earth's ecosystems more sensitive to changesD．is likely to cause many species to become extinct30: What is Michel Jarraud's attitude toward the report released by the UN? A．Negative. B．Supportive.C．Doubtful. D．Uninterested.DAutumn means different things to different people. “It all depends on your personality.” said British naturalist Richard Mabey. “Personality shapes your view of the season,” he said. “You may see it as a fading-away, a packing-up (结束), or as a time of packing in another sense --- the excited gathering of resources before a long journey.”If this is true, perhaps it tells us a little about, for instance, Thomas Hood, the 19th Century English poet. About November, he wrote:No warmth, no cheerfulness, no healthful easeNo shade, no shine, no butterflies, no beesNovember!On the other hand, another English poet John Keats, already sensing he was seriously ill, was inspired by a late September day to pen one of the most famous poems in the English language, To Autumn. He wrote to a friend afterwards that there was something comforting and healing about it.According to Richard Mabey, Keats has the biological evidence on his side. Autumn is not a time of slowing down, but a time of new beginnings and great movements of creatures. For example, just at the moment that Keats's “gathering swallows” (in To Autumn) are departing for Africa, millions of creatures are fleeing from the frozen north like Iceland, Greenland and Russia to winter along the east and south coasts of Britain. According to scientists, before falling, the leaves transfer their chlorophyll (叶绿素) and carbohydrates (碳水化合物) into the woody parts of the tree for safe-keeping over winter. What remains are the natural antioxidants (防老剂) in the leaves: the yellow and orange carotenoids (类胡萝卜素), and another protective chemical specially produced for autumn, the bright-red anthocyanin (花青素). High color is not a signal of deterioration (退化) and decline, but of detox (排毒的) ability and good health.A century after Keats, the American poet Loren Eiseley wrote in his journal: “Suppose we saw ourselves burning like maples in a golden autumn. And that we could break up like autumn leaves…dropping their substance like chlorophyll. Would not our attitude towards death be different?”31. From Thomas Hood’s poem, we may infer that _______.A. he suffered a lot from cold NovemberB. he missed the shining summer days very muchC. he had a negative attitude towards autumnD. he enjoyed butterflies and bees very much32. In autumn, leaves turn yellow before falling because ______.A. they can’t bear the freezingB. they can’t get enough water from the wood partC. chlorophyll and carbohydrates have been lost through leavesD. chlorophyll and carbohydrates have come back to the wood part33. From the passage we can learn that________.A. autumn has different faces in different people’s eyesB. John Keats was a good biologist as well as a poetC. all creatures move from the cold north to Britain for winterD. The three poets were only known for their poems about autumn34. Which word can best describe Loren Eiseley’s attitude towards autumn?A. FearfulB. OptimisticC. DoubtfulD. Realistic.35. What does the underlined sentence mean?A. Man can never live long, just as leaves must leave the tree annually.B. Man is different from autumn leaves, which will come again the next spring.C. Man should have a positive attitude towards death, quite different from autumn leaves.D. Man should treat death calmly, just like autumn leaves fall to the ground.第二节第二节（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。
高二上学期期末英语汇编：应用文写作(含解析)
福建省漳州市2023-2024学年高二上学期期末英语试题（含解析）
47.假定你是李华，你将参加校英语报举办的主题为“I want to be a person of value.”的演讲比赛。请用英语写一份演讲稿，内容包括：
1.你的榜样；
2.你的行动。
注意：
1.词数80词左右；
One of the key reasons is Chinese cuisine diversity. Each region in China has its own unique cooking styles and flavors, resulting in a wide variety of dishes. In addition, traditional Chinese meals often include a combination of colors, flavors and tastes, creating a great dining experience.
高二上学期期末英语汇编：应用文写作（含解析）
福建省厦门市2023-2024学年高二上学期期末英语质检卷（含解析）
第一节(பைடு நூலகம்分15分)
46.英语课上，同学们就“Do smart homes make life easier?”进行了分组讨论。请你作为小组代表发言，内容包括：
1．阐述观点;
2．说明理由。
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厦门市2016—2017学年度第二学期高二下文科质检答案及评分标准(1)
厦门市2016—2017学年度第二学期高二年级质量检测数学（文科）参考答案一、选择题：本大题共12小题，每小题5分，共60分．1-6 BAADCC 7-12 DABCCB二、填空题：本大题共4小题，每小题5分，共20分．13．01,2>+∈∀x R x 14. 5 15. 2≤b 16. ),338()52,4(+∞Y 三、解答题：本大题共6小题，共70分．17. 本题考查函数的极值及闭区间上函数的最值等基础知识，考查学生运算与分析问题能力，考查化归与转化思想．【解析】：由已知'2()3f x x m =+............................................1分因为()f x 在点2x =处取得极值，所以'(2)0f =, 即 120m +=解得12m =-.....3分经检验12m =-符合题意...................................................4分所以，3()124f x x x =-+ '2()3123(2)(2)f x x x x =-=+-,(3,3)x ∈-.................................5分当'()0f x >，得32x -<<-或23x <<；当'()0f x <，得22x -<<.................................................6分可知，()f x 在[)2,3--上单调递增，在(2,2)-上单调递减，在(]3,2上单调递增.....8分所以()f x 极小值为(2)12f =-，又(3)13f -=，..............................9分所以，12)(min -=x f .....................................................10分18.本小题主要考查线性回归方程的应用知识;考查数学抽象、数学建模思想，考查运算求解能力,实际应用能力．【解析】：（1）由所给数据计算得： 1(123456) 3.56x =+++++= .........................................1分 1(10.211.112.113.315.515.8)136y =+++++= ............................2分 616294.227321.2i ii x y x y =-=-=∑g622169173.517.5i i xx =-=-=∑.............................................3分616221621.2ˆ 1.2117.56i ii i i x y x y b xx ==-==≈-∑∑g ..........................................4分 $13 1.21 3.58.77ay bx =-=-⨯≈$ ..........................................6分 所求的回归方程为$1.218.77y x =+.........................................7分(2)由（1）知回归方程为$1.218.77y x =+的相关指数6221621ˆ()0.761110.030.9726.4()i i i ii y y R y y ==-=-=-≈-=-∑∑.............................10分 因为0.970.86>，所以线性回归模型拟合效果更好...........................12分19.本题考查抛物线的定义，直线与抛物线的位置关系等知识；考查分类,化归与转化数学思想；考查推理论证能力、运算求解能力.【解析】：（1）由题意可得422=+p ............................................2分 解得4=p ............................................................3分所以抛物线方程为x y 82=....................................................4分（2）方法一： 设),(11y x A ，),(22y x B ，),(00y x C AB 的中点联立方程⎩⎨⎧=+=xy kx y 812 整理得01)82(22=+-+x k x k ............................5分 (ⅰ) 当0=k 时，直线l 与抛物线只有一个交点，不符合题意....................6分(ⅱ) 当0≠k 时，04)82(22>--=∆k k 解得2<k 02≠<∴k k 且..........7分 221042kk x x x -=+=.......................................................8分 kk k k kx y 4141200=+-⋅=+=...............................................9分 又由),(00y x C AB 的中点在02=+y x 上得04282=+-kk k .......................................................11分 4k =- 经检验，满足条件...................................................12分方法二： 设),(11y x A ，),(22y x B ，),(00y x C AB 的中点联立方程⎩⎨⎧=+=xy kx y 812 整理得01)82(22=+-+x k x k ............................5分 (ⅰ) 当0=k 时，直线l 与抛物线只有一个交点，不符合题意..................6分(ⅱ)当0≠k 时，04)82(22>--=∆k k 解得2<k 02≠<∴k k 且.........7分因为1218x y = ① ，2228x y =② 由①-② 得：)(8))((212121x x y y y y -=-+ 又因为12120,0x x y y -≠+≠所以k y y y 42210=+=........................................9分 20041kk k y x -=-= ........................................10分 又由),(00y x C AB 的中点在02=+y x 上得04282=+-kk k .......................................................11分 4k =- 经检验，满足条件...................................................12分20.本题考察函数与导数的知识；考察化归与转化、数学建模的思想；考察运算求解、应用导数知识解决实际问题的能力.【解析】：（I ）点(144,120)H ，所以120=，得10a =.....................2分又AM t =，所以(P t ，所以S 关于t 的函数关系式为()(150(144)S t t t =⋅-+-⋅15020(0144)t t t =-<< ........................5分（II ）方法一： 3122'()(15020)'(150201440)'S t t t t t t =-+=-+112215030720150t t-=-+=-............................7分== ..........................8分 '()064S t t =⇒=；'()0064S t t >⇒<<；'()064144S t t <⇒<<所以()S t 在区间(0,64)上单调递增，在区间(64,144)上单调递减................10分 所以当64t =时，()S t 取到最大值，为10880平方米. ’答：S 的最大值为10880平方米. ............................................12分 方法二：令m 23150201440(012)S m m m m =-+<<，则 2'30060144060(3)(8)S m m m m =-+=-+-...........................8分 '08S m =⇒=；'008S m >⇒<<；'0812S m <⇒<<所以S 在区间(0,8)上单调递增，在区间(8,12)上单调递减.....................10分 所以当8m =时，S 取到最大值，为10880平方米. ’答：S 的最大值为10880平方米.............................................12分21. 本题考查直线与椭圆的位置关系，直线过定点，椭圆标准方程等基础知识；考查推理论证能力、运算求解能力和化归与转化等数学思想．【解析】：(Ⅰ) 方法一：依题意得直线l 过定点(1,0)P ，..........................2分P 在椭圆内部，..............................................................3分 所以直线l 与椭圆的位置关系是相交 ...........................................4分(Ⅱ) 设11(,)A x y ，22(,)B x y ，因为11(,)A x y 关于x 轴对称点为111(,)Ax y - 联立⎪⎩⎪⎨⎧=++=134122y x my x 得22(3m 4)y 6my 90++-= ........................5分122634m y y m -+=+，122934y y m -=+......................................6分因为直线1A B 的斜率2121y y k x x +=-12()x x ≠ 所以直线1A B 的方程为211121()y y y y x x x x ++=--..................................8分 令0y =得121212121212(1)(1)y x x y y my my y x y y y y ++++==++121221my y y y =++..............10分2292.3414634m m mm -+=+=-+ 所以直线1A B 与x 轴的交点坐标为定点(4,0)..................................12分解法二：联立⎪⎩⎪⎨⎧=++=134122y x my x 得22(3m 4)y 6my 90++-= .........................2分 222364(3m 4)(9)1441440m m ∆=-+-=+>..................................3分 所以直线l 与椭圆的位置关系是相交 ...........................................4分 (Ⅱ) 设11(,)A x y ，22(,)B x y ，因为11(,)A x y 关于x 轴对称点为111(,)Ax y - 122634m y y m -+=+，122934y y m -=+.....................................6分 因为直线1A B 的斜率2121y y k x x +=-12()x x ≠ 所以直线1A B 的方程为211121()y y y y x x x x ++=--..................................8分 令0y =得121212121212(1)(1)y x x y y my my y x y y y y ++++==++121221my y y y =++..............10分 2292.3414634m m mm -+=+=-+ 所以直线1A B 与x 轴的交点坐标为定点(4,0)..................................12分22．本小题考查学生运用函数与导数的知识,研究函数的单调性、极值,和函数的图像,考查函数与方程思想、转化与化归思想、分类与整合思想；考查运算求解能力、推理论证能力．【解析】：（Ⅰ）()()'()2ln 32f x x a x x a x a =-+--+()()2ln 1x a x =--........2分 因为02),,0(,0>-+∞∈≤a x x a 所以 , 令'()0f x =得x e =, 且()0,x e ∈时, '()0f x <,()f x 单调递减,(),x e ∈+∞时, '()0f x >,()f x 单调递增,所以()f x 的极小值为()212f e ae e =-,无极大值...... ...........4分（Ⅱ）方程()0f x =在()1,+∞上有两个不等实根,即函数()y f x =在()1,+∞上有两个零点, ①当0a ≤时,由（Ⅰ）可知, ()f x 在()1,e 单调递减, 在(),e +∞单调递增,又因为()3102f a =-<,不合题意, 舍去 ............................5分 ②当()0,2a e ∈时, ()0,,2a x e ⎛⎫∈⋃+∞ ⎪⎝⎭时, '()0f x >, ,2a x e ⎛⎫∈ ⎪⎝⎭时,'()0f x <, ()f x 单调递增区间为0,2a ⎛⎫ ⎪⎝⎭和(),e +∞，()f x 单调递减区间为,2a e ⎛⎫ ⎪⎝⎭, 要使函数()y f x =在()1,+∞上有两个零点，必须()10,2f e e a e ⎛⎫=-< ⎪⎝⎭ 得12a e <，所以()f x 在()1,e 单调递减, 在(),e +∞单调递增, 所以()3120,2f a =->得34a >，又因为02)(42>=e e f 所以342e a << ...................................................8分 ③ 2a e =时，()f x 在()1,+∞单调递增,不合题意；.....................9分 ④当()2,a e ∈+∞时, ()1,,2a x e ⎛⎫∈⋃+∞ ⎪⎝⎭时, '()0f x >, ,2a x e ⎛⎫∈ ⎪⎝⎭时 ,'()0f x <, ()f x 单调递增区间为()1,e 和,2a ⎛⎫+∞⎪⎝⎭，()f x 单调递减区间为,2a e ⎛⎫ ⎪⎝⎭, 因为()3120,2f a =-> 要使函数()y f x =在()1,+∞上有两个零点， 则2152ln 0,282a a f a ⎛⎫⎛⎫=-< ⎪ ⎪⎝⎭⎝⎭ 得522a e >，又()20,2a f a => 综上所述，523,2,42e a e ⎛⎫⎛⎫∈⋃+∞ ⎪ ⎪⎝⎭⎝⎭................................12分。
高二英语上学期期末考试试题(附答案)
海南中学2016-2017学年第一学期期末考试高二英语试题第一部分听力（共两节，满分30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1.5分，满分7.5分)听下面5段对话。
每段对话后有一个小题，从试题所给的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. How does the man feel？A. Worried。
B. Satisfied。
C。
Regretful.2。
What was the regular price of one bag?A. 300 dollars.B. 600 dollars。
C. 450 dollars.3。
Why hasn’t the woman seen the man recently？A. He has changed his job。
B。
He has been too busy。
C. He has been out of town。
4. Who might be promoted to the president of the company？A. Alden。
B. Bruce. C。
Kevin.5. What does the woman want the man to do？A. Carry the box for her. B。
Hold the door open for her。
C. Open the box for her。
第二节（共15小题；每小题1。
5分，满分22.5分)听下面5段对话。
每段对话后有几个小题，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。
听每段对话前，你将有时间阅读各个小题，每小题5秒钟；听完后，各小题将给出5秒钟的作答时间。