您的位置：360文档中心› 2025年在全球新能源汽车市场中的比例将接近50%,2030年将达到39%

2025年在全球新能源汽车市场中的比例将接近50%,2030年将达到39%

合集下载

新高考全国卷2025届高三英语下学期5月冲刺考试试题一

新高考全国卷2025届高三英语下学期5月冲刺考试试题一

（新高考，全国卷）2025届高三英语下学期5月冲刺考试试题（一）留意事项：1.答卷前，考生务必将自己的姓名、考生号等填写在答题卡上。

2.回答选择题时，选岀每小题答案后，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.考试结束后，将本试卷和答题卡一并交回。

考试时间100分钟，满分120分第一部分阅读（共两节，满分50分）第一节（共15小题；每小题2.5分，满分37.5分）阅读下列短文，从每题所给的四个选项（A、B、C和D）中，选出最佳选项。

并在答题卡上将该选项涂黑。

AHave you ever imagined having a museum all to yourself, just like in the movie starring Ben Stiller? Well, maybe not quite that exact experience. The American Museum of Natural History has been bringing arts, culture, and education to the public since 1869. Though its doors are currently closed, you can still visit the amazing exhibits from the comfort of your own home.The museum offers virtual tours, live learning, and countless resources for kids and families of all ages. Mark your calendars! This Thursday at 2P ET on the American Museum of Natural History Facebook page, there will be a Facebook Live event that features a pre-record tour of the Hall of Human Origins guided by anthropological Ashley Hammond. Get a special tour of this popular exhibition hall and hear about early human relatives, including Lucy, the most famous Australopithecus parenthesis.Every Tuesday and Thursday at 2 PMEST, you can hop on a Live Virtual Museum Tour with a museum tour guide or expert to explore a featured exhibit via Facebook Live. Missed Tuesday's Hall of Planet Earth Tour on Tuesday? You can go back and check out the previous tours posted to Facebook earlier.Looking to get out of this world? On Friday, April 3rd, head over to the museum's YouTube channel to catch a live guided tour of the universe with Director of Industrialization, Carter Em mart.Want to wander around the halls on your own? Google Arts and Culture allows you to do just that. From the 94-foot-long model of a Blue Whale in the Hall of Ocean Life to the Tyrannosaurus Rex fossil in the Hall of Saurian Dinosaurs, you have the place to yourself to explore!Go ahead and start your adventure.1.If you want to know something about human beings, we can .A.wander around the halls on your ownB.watch a Facebook Live event at 2P ET this ThursdayC.hop on a Live Virtual Museum Tour at 2 PMEST every Tues day and ThursdayD.go to the museum's YouTube channel to catch a live guided tour on Friday, April 3rd2.If you miss Tuesday's Hall of Plant Earth Tour on Tuesday, you can .A.go to Hall of Ocean LifeB.enjoy a Facebook live on ThursdayC.hop on replays posted to Facebook earlierD.ask Carter Em mart for help3.Where can you probably find this passage?A.In a magazine.B. In a history book.C. In a novel.D. In a scientific report.BWhen I left my country Trinidad and Tobago for China to study Chinese language and culture, never in my wildest imagination did I think that twenty-one years later I would still be here. Two decades is a long time, yet I often tell my friends that it feels like only two years. That is because China is an exciting place to be with lots of opportunities for begetters or those who seek adventure. It so happens that I fall into both categories.At first, I was not sure whether I would like my experience in China, so I initially planned to study for only one year when I left Trinidad for Beijing. But as fate would have it, I fell in love with China, its people, and its culture. I extended my course for another three years to learn all I could about this fascinating country, Being in China has allowed me the rare chance to meet many distinguished people. It has also allowed me to do things which I think I would not otherwise have had the chance to do.Years later, I published my first book, “The Emperor, His Bride and the Dragon Robe. " However, to my unexpected delight, this book was officially presented to China as one of state gifts. My mother, full of exhilaration, told me that it was televised on the local news. To this day, I feel highly honored. Among my to-do list is a musical play for children based on the book. Through this book and the play, I hope to share my love for China and engender in children around the world a similar love for Chinese culture.I believe people should have dreams no matter how big or small. However, most importantly, I believe that people must strive to make more opportunities for themselves to fulfill their dreams rather than wait around for a few opportunities to come along by chance. In this regard, I like to use one of Francis Bacon's quotes: "A wise man will create more opportunities than he finds. ”4.Why does two decades feel like only two years to the author?A.Because he has been seeking for opportunities and strive for his dream.B.Because China is an exciting place with lot of opportunities.C.Because he has many friends who are always with him.D.Because he has a fancy for China its people, and its culture.5.What can we infer from Paragraph 2?A.The author was certain he would fall for the experience in China.B.After studying for four years in China, the author fell in love with it.C.It was rare for the author to meet distinguished people in China.D.Being in China, the author has done things which he couldn't do in his hometown.6.What does the underlined word "engender" in paragraph 3 mean?A.Spread.B. Arouse.C. Conclude.D. Emit.7.What is the best title for the passage?A.My Love for ChinaB.My Special Experience in ChinaC.Ambition is the Key to the SuccessD.China, a Dream Man's ResortCMelanie Clap ham has spent the last three years catching images of grizzlies （灰熊） at Knight Inlet, on the B.C. coast, using small camera traps housed in metal and attached securely to the forest branches.Three years and thousands of images later, the behavioral researcher and postdoctoral student at the University of Victoria has partnered with two software developers living in Silicon Valley and a grizzly research centre in Alaska to develop facial recognition technology used to identify the bears."They don't have distinctive markings on their bodies, n said Clap ham, whose interest in this technology sprang from the need to "identify and recognize individual bears over time" as part of her behavioral research over the last 11 years.Bears grow and shrink a lot depending on the season, and their appearance changes frequently during their 20-to 25-year-long lifespans. Clap ham began to wonder if AJ. might be able to solve her problem in the same way the technology recognize people's faces.Now, she says, the open-source Bear ID software can be used and adapted by anyone and could have huge hints for understanding the animals* behaviour and avoiding bear-human encounters.In relation to the technology based on human facial recognition, Ed Miller and his partner Mary Nguyen are the software developers from California who connected with Clap ham in an online forum for conservation technology in late 2024.The pair were looking for photos of bears ”for fun" as a way to learn more about recognition software, and so they connected with Clap ham to offer their expertise （专业学问） in adapting artificial intelligence."The technology Pete using is based on the same software used to recognize humans, n said Miller, who added that human identification is far easier, as there are millions of images the software can learn from.We need lots of images of individual animals to tell the system which bear is which, ” said Clap ham, who explained "deep learning" as the process where the software trains itself to recognize certain bears more accurately with more pictures it gets. Clamshell says Beard currently has an 84 percent accuracy rate. This is especially important, given that a bear will look dramatically different throughout the year as its appearance, especially fur and weight, changes a lot.8.What is Clamp's purpose of catching images of grizzlies?A.To monitor the changes of some specific bears.B.To collect data for facial recognition technology.C.To finish his postdoctoral research.D.To meet his interest in the technology.9.What do we know about Clap ham?A.Clap ham uses Al to track bears.B.Clap ham has been in her behavioral research for about 25 years.C.Clap ham cooperates with others to identify the bears.D.Clap ham majors in software developing at the University of Victoria.10.Why did Ed Miller and his partner contact Clap ham?A.They provided her with their specific knowledge in Al.B.They wanted to know more about Al.C.They were eager to attend the online forum.D.They were using the same Bear ID software.11.What is the purpose of this text?A.To call on more people to protect grizzlies.B.To describe the living habits of grizzlies.C.To introduce the technology used to identify grizzlies.D.To record a friendship between Melanie and her partners.DCalifornia likes to think of itself as the state where the future happens, and in 2008, its voters decided the future was high-speed rail. So they approved a $9 billion bond issue to begin an incredible government infrastructure project： a bullet train connecting San Francisco and Los- Angeles, at a cost of $33 billion.For years, the optimists have imagined Californians will travel quickly, comfortably and environmentally between the state*s two major population centers. The pessimists, meanwhile, have watched the project costs a lot. At last count, the estimates had traveled up to $75 billion, even were still climbing.On Tuesday, Gov. Gavin News om （D ） in his state speech called for the state to transform the project to a less costly rail that would run through the Central Valley, which attracted voters' elsewhere attention,because what happened in California illustrates the fact that any U.S. rail project may take a risk.Distance. In other places of the world, major population centers are much closer to each other. And big cities that are reasonably close together is pretty much an essential condition for high-speed rail, which is why they have it and we don't. Imagine what it would take to build a line from New York City to Los Angeles 一 or to Chicago, Houston or Phoenix.Wealth. Of course, the United States does have a few cities that look ripe for rail. And instead of high-speed rail between these cities, we have the express, which takes eight hours to travel from Washington to Boston. Why haven't we built something better? Because truly high-speed rail needs to travel in a fairly straight line. Building newer, better, straighter rail lines would require the government to buy all the land between Point A and Point B and tear down anything that happened to be in the way. However, what's between Point A and Point B is a great deal of highly valuable real estate that will be very expensive to purchase.California displays all these problems totally. The part of the rail line that was reasonably cheap to build didn't go anywhere near where the people were ； it ran through the Central Valley where land was reasonably cheap and the lobbies wererelatively few.12.In paragraph 1, the author intends toA.prove the point of the passageB.explain a government projectC.introduce a topic for discussionD.present the background information13.What do we know about the bullet train project?A.All people are not in favor of the bullet train project.B.The cost of the project is approximately 75 billion.C.Other states have to risk building their own bullet program.D.The project will make Americans travel rapidly, cozily and environmentally.14.What mainly leads to bullet train program's failure?ernors in California tend to leave the project behind.B.The distance is the main reason to limit the bullet project in America.C.None of cities in America can afford to build a bullet train.D.Bullet train is not as popular as express in America.15.What attitude does the author hold towards the high-speed rail in California?A. Indifferent.B. Neutral.C. Negative.D. Optimistic.其次节（共5小题；每小题2.5分，满分12.5分）依据短文内容，从短文中的选项中选出能填入空白处的最佳选项。

天一大联考2024-2025学年高三上学期第二次联考(10月)数学试题

天一大联考2024-2025学年高三上学期第二次联考(10月)数学试题

“天一大联考·齐鲁名校联盟”2024—2025学年高三年级第二次联考数学一､单项选择题：本题共8小题，每小题5分，共40分.1.已知集合{}1,2,3,4,5,6U =，{}13,5A =,，{}1,2,3B =，则()U A B =ð（）A.{}2,4,5,6 B.{}4,6 C.{}2,4,6 D.{}2,5,62.已知0,0m n >>，且3m n +=的最大值为（）A.8B. C. D.2+3.函数)()(e e x x f x x -=-的图象大致为（）A. B. C. D.4.一块扇形薄铁板的半径是30，圆心角是120 ，把这块铁板截去一个半径为15的小扇形后，剩余铁板恰好可作为一个圆台的侧面，则该圆台的体积为（）A.π9B.1750π9C.π3D.5.设等比数列{}n a 的前n 项和为n S ，则“数列{}n S 为递增数列”是“321a a a >>”的（）A.充分不必要条件B.必要不充分条件C.充分必要条件D.既不充分也不必要条件6.函数221,2()2,2x x f x x x ⎧-<-=⎨-≥-⎩的最小值为（）A .4- B.2- C.3D.57.已知数列{}n a 满足：11a =，点()1,n n n a a ++在函数1y kx =+的图象上，其中k 为常数()0k ≠，且124,,a a a 成等比数列，则k 的值为（）A.2B.3C.4D.58.已知定义在R 上的函数()f x 满足()1(1)f x f x =--，若函数442x x y =+与函数()y f x =的图象的交点为112220252025(),),(,),,(,x y x y x y ，则20251)(i i i x y =+=∑（）A.0B.20252C.2025D.60752二､多项选择题：本题共3小题，每小题6分，共18分.9.下列说法正确的是（）A.若,a b c >∈R ，则22ac bc >B.若22,a b c c c>∈R ，则a b >C.若a b >，则22a b >D.函数2sin sin y x x=+的最小值为10.如图，有一列曲线012,,, P P P ，已知0P 所围成的图形是面积为1的等边三角形，1(0,1,2,3,)k P k += 是对k P 进行如下操作得到的：将k P 的每条边三等分，以每边中间部分的线段为边，向外作等边三角形，再将中间部分的线段去掉，记k S 为曲线k P 所围成图形的面积，则（）A.3P 的边数为128B.24027S =C.n P 的边数为34n⨯ D.834()559nn S =-⋅11.已知函数()32,f x x ax a =-+∈R ，则（）A.()f x 的图象关于点()0,2对称B.(),a f x ∃∈R 仅有一个极值点C.当1a =时，()f x 图象的一条切线方程为240x y -+=D.当3a <时，()f x 有唯一的零点三､填空题：本题共3小题，每小题5分，共15分.12.已知集合*2{13,{|(2)20}|}A x x B x ax a x =∈≤<=-++=N ，若“x B ∈”是“x A ∈”的充分不必要条件，则实数a 的所有取值组成的集合是______.13.蜜蜂被举为“天才的建筑师”，蜂巢结构是一种在一定条件下建筑用材最少的结构.如图是一个蜂房的立体模型，底面ABCDEF 是正六边形，棱,,,,,AG BH CI DJ EK FL 均垂直于底面ABCDEF ，上顶由三个全等的菱形,,PGHI PIJK PKLG 构成，10928GPI IPK KPG θ'∠=∠=∠=≈ ，设1BC =，则上顶的面积为______.（参考数据：1cos ,tan32θθ=-=）14.已知函数()ln f x x x =，则()f x 的最小值为______；设函数()()2g x x af x =-，若()g x 在()0,∞+上单调递增，则实数a 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.已知数列{}n a 满足()2*112,1n n n a a a a n +==-+∈N.（1）比较20242026,a a 的大小，并写出过程；（2）设数列1n a ⎧⎫⎨⎬⎩⎭的前n 项和为n S ，证明：1n S <.16.已知函数()f x 与其导函数()f x '的定义域均为R ，且()f x 为奇函数，当0x >时，()()()2,10f x f x f ->='.（1）判断()y f x '=的奇偶性；（2）解不等式()0f x >.17.如图，在四棱锥P ABCD -中，侧棱PA ⊥底面,ABCD AB BC ⊥，且2,PA AB BC AD CD =====（1）证明：BD ⊥平面PAC ；（2）求平面PBC 与平面PAD 夹角的正弦值.18.设函数()ln(1)(0)f x x k x k =+-≠.（1）讨论()f x 的单调区间.（2）已知直线l 是曲线()y f x =在点(,())(2)t f t t >处的切线.（i ）求直线l 的方程；（ii ）判断直线l 是否经过点(2,2).19.设数阵111202122x x X x x ⎛⎫=⎪⎝⎭，其中{}11122122,,,1,2,3,4,5,6x x x x ∈.设{}{}12,,,1,2,3,4,5,6k B n n n =⊆ ，其中*12,k n n n k <<<∈N 且6k ≤.定义变换t M 为“对于数阵的每一列，若其中有t 或t -，则将这一列中所有数均保持不变；若其中没有t 且没有t -，则这一列中每个数都乘以()121,,,k t n n n -= ”，()0B M X 表示“将0X 经过1n M 变换得到1X ，再将1X 经过2n M 变换得到2,X ，以此类推，最后将1k X -经过k n M 变换得到k X ”.记数阵k X 中四个数的和为()0B T X .（1）若{}021,2,534X B ⎛⎫==⎪⎝⎭，写出0X 经过2M 变换后得到的数阵1X ，并求()0B T X 的值；（2）若{}012321,,,34X B n n n ⎛⎫== ⎪⎝⎭，求所有()0B T X 取值的和；（3）对任意确定的一个数阵0X ，证明：所有()0B T X 取值的和不大于8-；（4）如果01336X ⎛⎫=⎪⎝⎭，其他条件不变，你研究（1）后得出什么结论？“天一大联考·齐鲁名校联盟”2024—2025学年高三年级第二次联考数学一､单项选择题：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.已知集合{}1,2,3,4,5,6U =，{}13,5A =,，{}1,2,3B =，则()U A B =ð（）A.{}2,4,5,6 B.{}4,6 C.{}2,4,6 D.{}2,5,6【答案】A 【解析】【分析】由集合的交集运算、补集运算即可求解.【详解】由题意集合{}1,2,3,4,5,6U=，{}13,5A =,，{}1,2,3B =，则{}1,3A B = ，(){}2,4,5,6U A B = ð.故选：A.2.已知0,0mn >>，且3m n +=，则的最大值为（）A.8B.C.D.2【答案】B 【解析】【分析】根据给定条件，利用配凑法及基本不等式求出最大值.【详解】由0,0mn >>，3m n +=，得6(2)(1)m n =+++≥，当且仅当213m n +=+=，即1,2m n ==时取等号，==≤的最大值为故选：B3.函数)()(e e x x f x x -=-的图象大致为（）A. B. C. D.【答案】B 【解析】【分析】利用函数()f x 奇偶性排除两个选项，再利用0x >时，函数值的正负判断即可.【详解】函数)()(e e x x f x x -=-的定义域为R ，()()(e )e x x f x x f x -=-=--，因此函数()f x 是偶函数，其图象关于y 轴对称，排除AC ；当0x >时，0e e 1x x -<<<，则()0f x <，排除D ，选项B 符合题意.故选：B4.一块扇形薄铁板的半径是30，圆心角是120 ，把这块铁板截去一个半径为15的小扇形后，剩余铁板恰好可作为一个圆台的侧面，则该圆台的体积为（）A.π9B.1750π9C.π3D.【答案】C 【解析】【分析】根据给定条件，求出原扇形及截去的小扇形围成的圆锥体积，再利用圆台的定义求出圆台体积.【详解】半径为30，圆心角为120 的扇形围成圆锥的底面圆半径r ，则2π2π303r =⋅，解得10r =，该圆锥的高h=2211ππ10π333V r h ==⋅⋅=，截去半径为15的小扇形围成圆锥的底面圆半径0r ，则02π2π153r =⋅，解得05r =，该圆锥的高0h==2200011ππ5π333V r h ==⋅⋅=，所以该圆台的体积为0π27π31π33VV -=-=.故选：C5.设等比数列{}n a 的前n 项和为n S ，则“数列{}n S 为递增数列”是“321a a a >>”的（）A.充分不必要条件B.必要不充分条件C.充分必要条件D.既不充分也不必要条件【答案】D 【解析】【分析】由321a a a >>可得10,01a q <<<或10,1a q >>，由{}n S 递增得出0n a >恒成立，再由充分条件、必要条件的定义判断即可.【详解】令等比数列{}n a 的公比为q ，由321a a a >>，得1112a a a q q >>，则10,01a q <<<或10,1a q >>，由数列{}n S 为递增数列，得110n n n a S S ++=->，即N n *∀∈，10n a q >，因此10,0a q >>，所以“数列{}n S 为递增数列”是“321a a a >>”的既不充分也不必要条件.故选：D6.函数221,2()2,2x x f x x x ⎧-<-=⎨-≥-⎩的最小值为（）A.4- B.2- C.3 D.5【答案】B 【解析】【分析】根据给定条件，分段探讨函数()f x 的单调性，进而求出最小值.【详解】当2x <-时，函数()21x f x =-在(,2)-∞-上单调递增，31()4f x -<<-；当2x ≤-时，函数2()2f x x =-在[2,0]-上单调递减，在[0,)+∞上单调递增，()(0)2f x f ≥=-，所以当0x =时，min ()2f x =-.故选：B7.已知数列{}n a 满足：11a =，点()1,n n n a a ++在函数1y kx =+的图象上，其中k 为常数()0k ≠，且124,,a a a 成等比数列，则k 的值为（）A.2B.3C.4D.5【答案】A 【解析】【分析】根据递推公式求出2a ，4a ，再根据124,,a a a 成等比数列，可求k 的值.【详解】因为点()1,n n n a a ++在函数1y kx =+的图象上，所以11n n a a kn ++=+⇒11n n kn a a +=+-，所以11a =，211k ka a =+-=，32211a k k a =+-=+，43312k k a a =+-=，因为124,,a a a 成等比数列，所以212k k =⨯⇒2k =或0k =（舍去）.故选：A8.已知定义在R 上的函数()f x 满足()1(1)f x f x =--，若函数442x x y =+与函数()y f x =的图象的交点为112220252025(),),(,),,(,x y x y x y ，则20251)(i i i x y =+=∑（）A.0B.20252C.2025D.60752【答案】C 【解析】【分析】根据给定条件，求出函数()f x 及442x xy =+的图象的对称中心，再结合中心对称图形的性质计算即得.【详解】依题意，由()1(1)f x f x =--，得()(1)1f x f x +-=，则函数()y f x =的图象关于点11(,)22对称，令4()42xxg x =+，则114444()(1)1424242424x x x x x x x g x g x --+-=+=+=++++⋅，因此函数()y g x =的图象关于点11(,)22对称，显然函数()y f x =与()y g x =的图象对称中心相同，则函数()y f x =与()y g x =的图象的交点关于点11(,22对称，不妨令点(,)i i x y 与20262026(,)(1,2,3,,2025)i i x y i --= 关于点11(,)22对称，则202620261,1i i i i x x y y --+=+=，20262026()()2i i i i x y x y --+++=，所以202512(202520252)i i i x y =+=⨯=∑.故选：C 【点睛】结论点睛：函数()y f x =的定义域为D ，x D ∀∈，①存在常数a ，b 使得()(2)2()()2f x f a x b f a x f a x b +-=⇔++-=，则函数()y f x =图象关于点(,)a b 对称.②存在常数a 使得()(2)()()f x f a x f a x f a x =-⇔+=-，则函数()y f x =图象关于直线x a =对称.二､多项选择题：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.下列说法正确的是（）A.若,ab c >∈R ，则22ac bc > B.若22,a b c c c>∈R ，则a b >C.若ab >，则22a b > D.函数2sin sin y x x=+的最小值为【答案】BC 【解析】【分析】对A 举反例即可；对B 根据不等式性质即可判断；对C ，利用指数函数单调性即可判断；对D 举反例即可.【详解】对A ，当0c=时，22ac bc =，故A 错误；对B ，当22a b c c >，则20c >，则a b >，故B 正确；对C ，根据指数函数2x y =在R 上单调递增，且a b >，则22a b >，故C 正确；对D ，当sin 1x =-时，2sin 3sin y x x=+=-<D 错误.故选：BC.10.如图，有一列曲线012,,,P P P ，已知0P 所围成的图形是面积为1的等边三角形，1(0,1,2,3,)k P k += 是对k P 进行如下操作得到的：将k P 的每条边三等分，以每边中间部分的线段为边，向外作等边三角形，再将中间部分的线段去掉，记k S 为曲线kP 所围成图形的面积，则（）A.3P 的边数为128 B.24027S =C.n P 的边数为34n⨯ D.834()559n n S =-⋅【答案】BCD 【解析】【分析】根据给定信息，归纳可得n P 的边数判断AC ；依次计算归纳得n P 所围图形的面积判断BD.【详解】依题意，令0P 图形的边长为a，214a =，边数是3；根据图形规律，1P 图形边长为3a，边数为0P 边数的4倍，即34⨯；2P 图形边长为23a ，边数为234⨯；依此类推，n P 图形边长为3n a，边数为34n ⨯，C 正确；3P 的边数为334192⨯=，A 错误；由图形规律知曲线n P 所围图形的面积n S 等于曲线1n P -所围面积加上每一条边增加的小等边三角形的面积，而每一个边增加的小等边三角形面积为2()43n ⨯，则121(34)()43n nn n aSS --=+⨯⨯，整理得1114()39n n n S S ---=⨯，数列1{}nn S S --是等比数列，1P图形的面积21413()433a S =+⨯⨯=，121321144[1(]4183499()433559()9()()1n n n n n S S S S S S S S ---=+⨯-=+-+--⨯++=- ，D 正确；2831640558127S =-⨯=，B 正确.故选：BCD 11.已知函数()32,f x x ax a =-+∈R ，则（）A.()f x 的图象关于点()0,2对称B.(),a f x ∃∈R 仅有一个极值点C.当1a=时，()f x 图象的一条切线方程为240x y -+= D.当3a <时，()f x 有唯一的零点【答案】ACD 【解析】【分析】根据函数的奇偶性判断A ，根据三次函数的性质判断B ，根据导数的意义求切线判断C ，利用极值点的符号判断D.【详解】对A ：设()3g x x ax =-，则函数()g x 为奇函数，图象关于原点()0,0对称，将()3g x x ax =-的图象向上平移2个单位，得函数()32f x x ax =-+的图象，故函数()f x 的图象关于点()0,2对称，A 正确；对B ：由三次函数的性质可知，函数()f x 要么有2个极值点，要么没有极值点，所以B 错误；对C ：当1a=时，()32f x x x =-+，()231f x x '=-.由()2f x '=⇒2312x -=⇒1x =或1x =-.若1x =，则2y =，所以()f x 在1x =处的切线方程为：即2y x =；若1x =-，则2y =，所以()f x 在1x =-处的切线方程为：()221y x -=+即240x y -+=.故C 正确；对D ：因为()23f x x a '=-，若0a ≤，则()0f x '≥在(),-∞+∞上恒成立，则()f x 在(),-∞+∞上单调递增，由三次函数的性质可知，此时函数()f x 只有一个零点；若0a >，由()0f x '<⇒33x -<<，由()0f x '>⇒3x <-或3x >.所以函数()f x 在,3⎛-∞-⎝⎭和,3⎛⎫+∞ ⎪ ⎪⎝⎭上单调递增，在,33⎛⎫- ⎪ ⎪⎝⎭上单调递减，要使函数()f x 只有1个零点，须有03f ⎛⎫> ⎪ ⎪⎝⎭（因为()02f =，所以03f ⎛⎫-< ⎪ ⎪⎝⎭不成立），即32033a ⎛⎫-⋅+> ⎪ ⎪⎝⎭⇒3a <，得0<<3a .综上可知：当3a <时，函数()f x 有唯一的零点，故D 正确.故选：ACD 【点睛】方法点睛：本题可以结合三次函数的图象和性质进行分析.三､填空题：本题共3小题，每小题5分，共15分.12.已知集合*2{13,{|(2)20}|}A x x B x ax a x =∈≤<=-++=N ，若“x B ∈”是“x A ∈”的充分不必要条件，则实数a 的所有取值组成的集合是______.【答案】{0,2}【解析】【分析】用列举法表示集合A ，利用充分不必要条件的定义，借助集合的包含关系分类求解即得.【详解】依题意，{1,2}A =，{|(2)(1)0}B x ax x =--=，显然B ≠∅，由“x B ∈”是“x A ∈”的充分不必要条件，得BA ，当0a=时，{1}B =，符合题意，当0a ≠时，方程2(2)20ax a x -++=的根为1和2a，显然22a ≠，否则B A =，不符合题意，因此21a=，解得2a =，此时{1}B =，符合题意，所以实数a 的所有取值组成的集合是{0,2}.故答案为：{0,2}13.蜜蜂被举为“天才的建筑师”，蜂巢结构是一种在一定条件下建筑用材最少的结构.如图是一个蜂房的立体模型，底面ABCDEF 是正六边形，棱,,,,,AG BH CI DJ EK FL 均垂直于底面ABCDEF ，上顶由三个全等的菱形,,PGHI PIJK PKLG 构成，10928GPI IPK KPG θ'∠=∠=∠=≈ ，设1BC =，则上顶的面积为______.（参考数据：1cos ,tan 232θθ=-=）【答案】924【解析】【分析】根据蜂房的结构特征，即可根据锐角三角函数以及三角形面积公式求解.【详解】依题意，由10928GPIIPK KPG θ'∠=∠=∠=≈ ，得10928GHI θ'∠=≈ ，在菱形PGHI 中，连接G I 并取其中点O，连接OH ，则2224tan2GOOH GO GI θ===，由正六边形ABCDEF 的边长1BC =，得2sin 603AC AB == ，由蜂巢结构特征知，AG CI =，又,AG CI都垂直于平面ABCDEF ，则//AG CI ，于是四边形ACIG 是平行四边形，有=3GI AC =，则26=44OH GI =，因此一个菱形的面积为1632223244GHISGI OH =⋅⋅=⨯ =，所以上顶的面积为3292344⨯=.故答案为：92414.已知函数()ln f x x x =，则()f x 的最小值为______；设函数()()2g x x af x =-，若()g x 在()0,∞+上单调递增，则实数a 的取值范围是______.【答案】①.1e-②.[]0,2【解析】【分析】空1，直接求导利用()f x 的单调性去求其最小值即可；空2，利用导数与单调性的关系建立不等式，利用不等式的恒成立解决参数范围即可.【详解】由题可知()ln f x x x =定义域为()0,∞+()ln 1f x x ='-显然，当10,e x ⎛⎫∈ ⎪⎝⎭时，′<0，()f x 单调递减；当1,+e x ∞⎛⎫∈ ⎪⎝⎭时，′>0，()f x 单调递增；所以()f x 的最小值为11e e f ⎛⎫=- ⎪⎝⎭；由题可知，()()22ln g x x af x x ax x=-=-所以()2ln g x x a x a =--'由题可知()2ln 0g x x a x a '=--≥恒成立，当0a <，显然当0x →时，()g x ∞'→-，故不成立；当0a=时，()2g x x '=，因为∈0,+∞，所以()20g x x '=>，故成立；当0a >时，由2ln 0x a x a --≥恒成立，得21ln xa x+≥恒成立，即max 21ln x a x +⎛⎫≥ ⎪⎝⎭不妨令()1ln x h x x +=，所以()2ln xh x x -='所以显然当∈0,1时，ℎ′>0，ℎ单调递增；当()1,+x ∞∈时，ℎ′<0，ℎ单调递减；所以()()max 11h x h ==，即2102a a ≥⇒<≤综上所述：[]0,2a ∈故答案为：1e-；0,2【点睛】关键点点睛，当不等式化简时，不要在不等式两边去随意乘或者除以一个未知数，要保证知道其正或负，再去作乘除计算.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.已知数列{}n a 满足()2*112,1n n n a a a a n +==-+∈N .（1）比较20242026,a a 的大小，并写出过程；（2）设数列1n a ⎧⎫⎨⎬⎩⎭的前n 项和为n S ，证明：1n S <.【答案】（1）20242026a a <（2）证明见解析.【解析】【分析】（1）证明数列的单调性，可比较给出的两项的大小.（2）先根据统计得到111111n n n a a a +=---，再求n S 进行判断即可.【小问1详解】因为211n n n a a a +=-+⇒()2212110n n n n n a a a a a +-=-+=-≥，所以1n n a a +≥.若1n n a a +=，则211n n n n a a a a +=-+=⇒1n a =，这与12a =矛盾.所以1n n a a +>.故20242026a a <.【小问2详解】由211n n n a a a +=-+⇒()2111n nn n n a a a a a +-=-=-，所以()11111111n n n n n a a a a a +==----⇒111111n n n a a a +=---.所以11111111nnn i i i i i S a a a ==+⎛⎫==- ⎪--⎝⎭∑∑1111111111n n a a a ++=-=----.由（1）可知：12n a +>，所以1n S <.16.已知函数()f x 与其导函数()f x '的定义域均为R ，且()f x 为奇函数，当0x >时，()()()2,10f x f x f ->='.（1）判断()y f x '=的奇偶性；（2）解不等式()0f x >.【答案】（1）偶函数，理由见解析（2）(1,0)(1,)-+∞ 【解析】【分析】（1）对()()f x f x -=-两边同时求导即可证明；（2）构造函数2()()ex f x h x =，求导得到其单调性即可得到()f x 在(1,)+∞上大于零，在(0,1)上小于零，再根据其为奇函数即可得到答案.【小问1详解】因为()f x 为奇函数，定义域为R ，所以()()f x f x -=-，两边同时求导可得()()f x f x ''--=-，即()()f x f x ''-=，所以()y f x '=为偶函数.【小问2详解】因为当0x >时，()2()f x f x '->，所以()2()f x f x '>.构造函数2()()e x f x h x =，则2()2()()e xf x f x h x '-'=，所以当0x >时，()0,()h x h x >'在(0,)+∞上单调递增，又因为(1)0f =，所以(1)0,()h h x =在(1,)+∞上大于零，在(0,1)上小于零，又因为2e 0x>，所以()f x 在(1,)+∞上大于零，在(0,1)上小于零，因为()f x 是定义域为R 的奇函数，所以(0)0,()f f x =在(,1)∞--上小于零，在(1,0)-上大于零，综上所述，()0f x >的解集为(1,0)(1,)-+∞ .17.如图，在四棱锥P ABCD -中，侧棱PA ⊥底面,ABCD AB BC ⊥，且2,PA AB BC AD CD =====（1）证明：BD ⊥平面PAC ；（2）求平面PBC与平面PAD 夹角的正弦值.【答案】（1）证明见解析（2）5【解析】【分析】（1）首先证明AC BD ⊥，再利用线面垂直的性质得PA BD ⊥，最后线面垂直的判定即可证明；（2）建立合适的空间直角坐标系，求出相关平面的法向量，最后根据面面角的空间向量求法即可得到答案.【小问1详解】记AC BD O = ，如图.因为,AB BC AD CD ==，BD BD =，所以ABD CBD ≅ ，所以ADOCDO ∠=∠，由等腰三角形三线合一知90AOD COD ︒∠=∠=，即AC BD ⊥，又PA ⊥底面,ABCD BD ⊂平面ABCD ，所以PA BD ⊥，因为AC PA A ⋂=，且AC ⊂平面,PAC PA ⊂平面PAC ，所以BD ⊥平面PAC .【小问2详解】取PC 的中点M，连接OM ，则//OM PA ，所以OM ⊥平面ABCD ，所以,,OC OD OM 三条直线两两互相垂直，以,,OC OD OM 所在的直线分别为x ，y ，z 轴建立如图所示的空间直角坐标系Oxyz ，由题意及（1）知1,2OAOD ==，则(1,0,0),(0,1,0),(1,0,0),(0,2,0),(1,0,2)A B C D P ---，所以(1,2,2),(1,2,0),(1,1,2),(1,1,0)PD AD PB BC =-==--=，设平面PAD 的法向量为()111,,m x y z =，同理设平面PBC的法向量为()222,,n x y z =，则2222220n PB x y z n BC x y ⎧⋅=--=⎪⎨⋅=+=⎪⎩ ，可取(1,1,1)n =- .所以cos ,5m n m n m n ⋅===-⋅，所以平面PBC 与平面PAD 夹角的余弦值为5，所以平面PBC 与平面PAD 夹角的正弦值为5.【点睛】18.设函数()ln(1)(0)f x x k x k =+-≠.（1）讨论()f x 的单调区间.（2）已知直线l 是曲线()y f x =在点(,())(2)t f t t >处的切线.（i ）求直线l 的方程；（ii ）判断直线l 是否经过点(2,2).【答案】（1）答案见解析；（2）（i ）(1)ln(1)11k kty x k t t t =++----；（ii ）不经过.【解析】【分析】（1）求出函数()f x 的导数，再按0k <和0k >分类求出()f x 的单调区间.（2）（i ）由（1）结合导数的几何意义求出切线l 的方程；（ii ）令2x =，求出y 的值并判断与2的大小.【小问1详解】函数()ln(1)f x x k x =+-的定义域为(1,)+∞，求导得(1)()111k x k f x x x --'=+=--，当0k<时，11k ->，由()0f x '<，得11x k <<-；由()0f x '>，得1x k >-，函数()f x 在(1,1)k -上单调递减，在(1,)k -+∞上单调递增，当0k>时，11k -<，则恒有()0f x '>，函数()f x 在(1,)+∞上单调递增，所以当0k <时，函数()f x 的单调递减区间是(1,1)k -，单调递增区间是(1,)k -+∞；当0k>时，函数()f x 的单调递增区间是(1,)+∞，无递减区间.【小问2详解】（i ）由（1）知，()11kf t t '=+-，而()ln(1)f t t k t =+-，则直线l 的方程为ln(1)](1))1[(y k t k t x t t +--=+--，即(1ln(1)11k kty x k t t t =++----.（ii ）由（i ）知，直线l 的方程为(1)ln(1)11k kty x k t t t =++----，当2x =时，22(1)ln(1)2[ln(1)]111k kt ty k t k t t t t -=++--=++----，令21()ln(1)1ln(1)11t g t t t t t -=+-=-+---，而2t >，求导得22112()0(1)1(1)t g t t t t -'=-+=>---，函数()g t 在(2,)+∞上单调递增，因此()(2)0g t g >=，即2t ∀>，()0g t ≠，而0k ≠，于是22[ln(1)]21tk t t -++-≠-，所以直线l 不经过点(2,2).19.设数阵111202122x x X x x ⎛⎫= ⎪⎝⎭，其中{}11122122,,,1,2,3,4,5,6x x x x ∈.设{}{}12,,,1,2,3,4,5,6k B n n n =⊆ ，其中*12,k n n n k <<<∈N 且6k ≤.定义变换t M 为“对于数阵的每一列，若其中有t 或t -，则将这一列中所有数均保持不变；若其中没有t 且没有t -，则这一列中每个数都乘以()121,,,k t n n n -= ”，()0B M X 表示“将0X 经过1n M 变换得到1X ，再将1X 经过2n M 变换得到2,X ，以此类推，最后将1k X -经过k n M 变换得到k X ”.记数阵k X 中四个数的和为()0B T X .（1）若{}021,2,534X B ⎛⎫== ⎪⎝⎭，写出0X 经过2M 变换后得到的数阵1X ，并求()0B T X 的值；（2）若{}012321,,,34X B n n n ⎛⎫== ⎪⎝⎭，求所有()0B T X 取值的和；（3）对任意确定的一个数阵0X ，证明：所有()0B T X 取值的和不大于8-；（4）如果01336X ⎛⎫= ⎪⎝⎭，其他条件不变，你研究（1）后得出什么结论？【答案】（1）（2）40（3）证明见解析（4）()013BTX =【解析】【分析】（1）先写出12134X -⎛⎫= ⎪-⎝⎭，再计算得22134X -⎛⎫= ⎪-⎝⎭，最后相加即可；（2）分{1,2,3,4}B ⊆和{}32,3,B n =或{}331,4,,{5,6}B n n =∈以及{}11,5,6,{1,2,3,4}B n n =∈讨论即可；（3）分若1121x x ≠和1121x x =两大类讨论即可；（4）直接代入计算得11336X --⎛⎫= ⎪--⎝⎭，21336X ⎛⎫= ⎪⎝⎭即可得到答案.【小问1详解】因为021,{2,5}34X B ⎛⎫== ⎪⎝⎭，0X 经过2M 变换后得到数阵12134X -⎛⎫= ⎪-⎝⎭，1X 经过5M变换后得到数阵22134X -⎛⎫= ⎪-⎝⎭，所以()021340BT X =-+-+=.【小问2详解】若{1,2,3,4}B ⊆，则32134X -⎛⎫= ⎪-⎝⎭或32134X -⎛⎫= ⎪-⎝⎭，可得()00,4BT X =种情况；若{}32,3,B n =或{}331,4,,{5,6}B n n =∈，则32134X --⎛⎫= ⎪--⎝⎭，可得()010,4B T X =-种情况；若{}123,,B n n n =，从{1,4}和{2,3}中各取出一个元素a ，b ，12min{,},max{,},{5,6}n a b n a b n ==∈，则32134X ⎛⎫= ⎪⎝⎭，可得()010,8BT X =种情况；若{}11,5,6,{1,2,3,4}B n n =∈，则32134X -⎛⎫= ⎪-⎝⎭或32134X -⎛⎫= ⎪-⎝⎭，可得()00,4B T X =种情况.综上，所有()0BT X 取值的和为404(10)8104040⨯+⨯-+⨯+⨯=.【小问3详解】若1121x x ≠，在{1,2,3,4,5,6}的所有非空子集中，①含有11x且不含21x 的子集共42个，其中含有奇数个元素的集合有8个，经过变换后第一列均仍为1121,x x ，其中含有偶数个元素的集合有8个，经过变换后第一列均变为1121,x x --；②含有21x 且不含11x 的子集共42个，其中含有奇数个元素的集合有8个，经过变换后第一列均仍为1121,x x ，其中含有偶数个元素的集合有8个，经过变换后第一列均变为1121,x x --；③同时含有11x和21x 的子集共42个，其中含有奇数个元素的集合有8个，经过变换后第一列均变为1121,x x --，其中含有偶数个元素的集合有8个，经过变换后第一列均仍为1121,x x ；④不含11x也不含21x 的子集共421-个，其中含有奇数个元素的集合有8个，经过变换后第一列均变为1121,x x --，其中含有偶数个元素的集合有7个，经过变换后第一列均仍为1121,x x .若1121x x =，在{1,2,3,4,5,6}的所有非空子集中，①含有11x的子集共52个，其中含有奇数个元素的集合有16个，经过变换后第一列均仍为1121,x x ，其中含有偶数个元素的集合有16个，经过变换后第一列均变为1121,x x --；②不含11x的子集共521-个，其中含有奇数个元素的集合有16个，经过变换后第一列均变为1121,x x --，其中含有偶数个元素的集合有15个，经过变换后第一列均仍为1121,x x ；综上，经过变换后，所有k X 的第一列数的和为()()()112111211121(88881616)(88871615)2x x x x x x +++++--+++++++=--同理，经过变换后所有k X 的第二列数的和为()12222x x --.所以所有()0BT X 取值的和为()112112222x x x x ----，又因为11122122,,,{1,2,3,4,5,6}x x x x ∈，所以所有()0B T X 取值的和不超过8-.【小问4详解】如果01336X ⎛⎫= ⎪⎝⎭，其他条件不变，0X 经过2M 变换后得到数阵11336X --⎛⎫= ⎪--⎝⎭，1X 经过5M 变换后得到数阵21336X ⎛⎫=⎪⎝⎭，则（1）中()013B T X =.【点睛】关键点点睛：本题第三问的关键是利用分类讨论的思想，分1121x x ≠和1121x x =讨论即可.。

Battery Tester BT3554-50说明书

Battery Tester BT3554-50说明书

Recording and judgmentAudio guidance BATTERY TESTER BT3554-50When Z3210 is installedMeasurement parameters2ment data for management purposes.Simply follow the audio guidance to measure, record, and organize data.Measurement navigator that keeps you from going back.Product bundlesRegister site informations in advance.Register profile information for each measurement siteusing GENNECT Cross or GENNECT One and load it on the instrument.Register profilesLoad profilesThird-floor informationProfile 31Choose from 1 to 100.Building name, floor,etc.UPS or battery name,etc.1500Managing battery information+Carrying Case C1014Protector Z5041Neck strap0 Adj BoardUSB cableGENNECT One Software CDFuse Set Second-floor informationProfile 2First-floor informationProfile 1Profile 1Profile 2Wireless Adapter Z3210For Bluetooth® wireless communicationsBT3554-92BT3554-91BT3554-52BT3554-51BT3554-50Pin Type Lead L2020Pin Type Lead 9465-10Pin Type Lead L2020Pin Type Wireless Adapter Z321033Record data automatically while probing.Judgment results (PASS, WARNING, or FAIL) relative to comparator threshold values are recorded by theinstrument along with measured values and transferred to your mobile device.Noise resistance that lets you measure even when the UPS is in operationMeasured values fail to stabilize Normal instrumentApp and software functionalityPin Type Lead L2020A: 70 mm (2.76") (red),150 mm (5.91") (black, max. 630 mm [24.80"])B: 164 mm (6.46")L: 1941 mm (76.42") (red)Pin Type Lead 9465-10A: 45 mm (1.77") (red), 400 mm (15.75") (black max.) B: 177 mm (6.97") L: 1925 mm (75.79") (red)Temperature Probe 9451SL: 100 mm (3.94")9451-01Temperature Probe 9451L: 1500 mm (59.06")Carrying Case C1014Hard caseProtector Z5041For BT3554 and BT3554-50Set of 5Fuse Set Z5050For BT3554, BT3554-50Separate surface fastener required if affixing to carrying case0 Adj Board Z5038For L2020, 9465-10, and 9772Pin Type Lead 9772A: 45 mm (1.77") (red), 400 mm (15.75") (black max.) B: 173 mm (6.81")L: 1921 mm (75.63") (red)Remote Control Switch 9466Hold and save measured val-ues by pressing the button. Cable length: approx. 2 m (78.74")Clip Type Leadwith Temperature Sensor 9460A: 300 mm (11.81")B: 106 mm (4.17")L: 2268 mm (89.29")Large Clip Type Lead 9467A: 300 mm (11.81") B: 131 mm (5.16") L: 1350 mm (53.15") tip φ 28 mm (1.10")OptionsRegarding probe length4Manage data easily.Measurement data is linked to profile information and saved. This approach lets you reduce the number of man-hours spent managing measured batteries.Transfer measurement data to a smartphone Transfer internal memory data to a computerAccess QR code for details.USBTip Pin 9465-90L2020/9465-10 tip pin replacementTip Pin 9772-909772 tip pin replacementThe effects of inverter noise Noise reduction technologyOrder codeHEADQUARTERS 81 Koizumi,Ueda, Nagano 386-1192 Japan https:///All information correct as of Nov. 24, 2022. Contents are subject to change without notice.BT3554-50E8-2YE Printed in JapanDISTRIBUTED BYNote: Company names and product names appearing in this brochure are trademarks or registered trademarks of various companies.*The Bluetooth ® word mark and logos are registered trademarks owned by Bluetooth SIG, Inc. and any use of such marks by HIOKI E.E. CORPORATION is under license.Operating precautionsPass/fail judgment threshold values vary with factors including the battery’s manufacturer, type, and capacity. The internal resistance and terminal voltage of a new or known-good battery must be measured first. It may be difficult to determine the deterioration state of traditional open type (liquid) lead-acid or alkaline batteries which demonstrate smaller changes in internal resistance than sealed lead acid batteries.Specifications。

山东省青岛2024-2025学年高三上学期期中考试数学试卷(含解析)

山东省青岛2024-2025学年高三上学期期中考试数学试卷(含解析)

2024-2025学年度第一学期期中考试解析-高三上数学一、选择题：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，i 为虚数单位，为z 的共轭复数，则（ ）A.B. 4C.D.2. 已知集合，，则（ ）A. B.C. D. 3. （ ）A.B.D.24.已知向量，，其中，若，则（ ）A. 40B. 48C.D. 625. 已知的内角A ，B ，C 的对边a ，b ，c 成等比数列，则的最大值为（ ）A.B.C.D.6. 若定义在上的偶函数在上单调递增，则，，的大小关系为（）A. B.C. D.7. 已知a ，且，，，则（ ）A.B. C.D. 8. 已知当时，恒成立，则实数a 的取值范围为（ ）A. B. C. D. 二、选择题：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.44i z =+z z i -=5(){}2024log 20250M x y x ==-<{}2026x N y y ==M N = (2024,2025)(,2025)-∞(0,)+∞(2025,)+∞4log 50.5=1215-()1,54a λ=+ ()2,8b λ=+ 0λ≥a b ∥ ()b a b ⋅-=34-ABC △B 6π3π2π23πR ()f x [)0,+∞1πf ⎛⎫- ⎪⎝⎭31f ⎛⎫- ⎪⎝⎭127f -⎛⎫⎪⎝⎭12117π3f f f -⎛⎫⎛⎫⎛⎫->>- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭1211π73f f f -⎛⎫⎛⎫⎛⎫>->- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭1211π73f f f -⎛⎫⎛⎫⎛⎫->-> ⎪⎪ ⎪⎝⎭⎝⎭⎝⎭121173πf f f -⎛⎫⎛⎫⎛⎫->>- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭b ∈R 0b ≠1a b ≠-1sin 21a b a bα-=+ab =1cos 1cos αα-+πtan 4α⎛⎫+⎪⎝⎭1sin 1sin αα-+2πtan 4α⎛⎫+ ⎪⎝⎭0x >ln e ln x x ex x a -≥(],0-∞(20,e ⎤⎦(],1-∞[)e,+∞9. 已知且，则（ ）A. B. C.2D.10. 已知幂函数的图象经过点，下列结论正确的有（ ）A. B.是偶函数C. D.若，则11. 已知函数，则下列说法正确的有（ ）A. 的定义域为B. 有解C. 不存在极值点D. 三、填空题：本题共3小题，每小题5分，共15分.12. 曲线在点处的切线方程为______.13. 数列共有5项，前三项成等比数列，公比为q . 后三项成等差数列，公差为d ，且若第5项为1，第2项与第4项的和为18，第1项与第3项的和为35，则____________.14. 在中，若，，三点分别在边，，上（均不在端点上），则，，的外接圆交于一点O ，称为密克点.在梯形ABCD 中，，，M 为CD 的中点，动点P 在BC 边上（不包含端点），与的外接圆交于点Q （异于点P ），则BQ 的最小值为______.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.（13分）已知单位向量，满足．（1）求；（2）求在上的投影向量（用表示）．16.（15分）0xy >21x y +=0y <102x <<42xy+≥22log log 0x y +<()f x 14,16⎛⎫⎪⎝⎭()00f =()f x ()12f '-=()()321f x f x ->+233,,4322x ⎛⎫⎛⎫∈⎪ ⎪⎝⎭⎝⎭()(1)log x f x x +=()f x ()0,+∞()2f x =()f x ()()()11f x f x x <+>21e x y x x -=-()1,0{}n a dq +=111A B C △1M 1N 1P 11A B 11B C 11C A 111A M P △111B M N △111C N P △60B C ∠=∠=︒22AB AD ==ABP △CMP △1e 2e121()23e e e ⋅+= 1232e e -125e e - 1e1e定义三阶行列式运算：，其中（i ，）.关于x 的不等式的解集为M .（1）求M ；（2）已知函数在实数集单调递增，求a 的取值范围.17．（15分）函数（，，）的部分图象如图，和均在函数的图象上，且Q 是图象上的最低点．（1）求函数的单调递增区间；（2）若，，求的值．18. （15分）已知数列是首项为2，公比为4的等比数列，数列满足.（1）求数列和的通项公式；（2）求数列的前n 项和.19.（17分）已知函数（1）求的值；111213212223112233122331132132132231122133112332313233a a a a a a a a a a a a a a a a a a a a a a a a a a a =++---ij a ∈R {}1,2,3j ∈10100001x x x->()()241,,e 22,x x a x x Mf x a x M⎧-+∈=⎨--∈⎩R ðR ()()sin f x A x ωϕ=+0A >0ω>π2ϕ<()1,0P ()4,2Q -()f x ()f x ()056f x =058,33x ⎡⎤∈⎢⎥⎣⎦0cos x π{}2na {}nb 321212222n n b b b b n -++++= {}n a {}n b n n a b ⎧⎫⎨⎬⎩⎭n S ()3f x x x =-()0f（2）设，当时，记在区间上的最大值为M ，最小值为m ，求的取值范围.高三期中考试题 数学参考答案1. D 【解析】由，可得.故选D.2. C 【解析】由可得，则；，故，则.故选C.3. C 【解析】由题意得.故选C 项.4. D 【解析】因为，，且，故，解得或（舍去），经检验当时，，故.故选D.5. B 【解析】由题意可得，由余弦定理可得，，，.故选B.6. .B 【解析】因为是定义在上的偶函数，所以，，又()()()ln 0g x a x x f f x =-+-(]1,3a ∈-()g x []1,e Mm -44i z =+45i z i -==-=()2024log 20250x -<020241x <-<()2024,2025M =20260x y =>()0,N =+∞()2024,2025M N = 444222log 5111log loglog log 5log 552510.522222-⎛⎫======⎪⎝⎭()1,54a λ=+ ()2,8b λ=+ a b ∥ ()()54218λλ++=⨯0λ=145-0λ=a b ∥ ()()()2,81,4124834b a b ⋅-=⋅--=-⨯-⨯=-2b ac =2222221cos 2222a cb ac b ac ac B ac ac ac +---=≥==0B π<< 03B π∴<≤()f x R 3113f f ⎛⎫⎛⎫-= ⎪ ⎪⎝⎭⎝⎭11ππf f ⎛⎫⎛⎫-= ⎪ ⎪⎝⎭⎝⎭127-=，在上单调递增，所以.故选B 项.7. D 【解析】由题意可得，解得.故选D.8. A 【解析】由对恒成立，令，则，令，得，当时，，当时，，所以在上单调递减，在上单调递增，所以，即.令，，，当时，；当时，，所以在上单调递减，在上单调递增，所以，所以.故选A.9. BCD 【解析】由且，得，解得，同理得，故A 项错误，B 项正确；对于C 项，，当且仅当时，取等，故C项正确；对于D项，，故D 项正确.故选BCD 113π>>()f x [)0,+∞1211π73f f f -⎛⎫⎛⎫⎛⎫>->- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭1sin 21a b a bα-=+22221sin 2sin cos 2sin cos 1sin 2sin cos 2sin cos a b αααααααααα+++==-+-()()()()22222sin cos 1tan πtan 4sin cos 1tan ααααααα++⎛⎫==+ ⎝--=⎪⎭ln e ln x x x x a -≥0x >()ln f x x x =()ln 1f x x ='+()0f x '=1ex =10e x <<()0f x '<1e x >()0f x '>()f x 10,e ⎛⎫ ⎪⎝⎭1,e ⎛⎫+∞ ⎪⎝⎭()11e ef x f ⎛⎫≥=- ⎪⎝⎭1ln ex x ≥-ln tx x =()1e et g t et t ⎛⎫=-≥- ⎪⎝⎭()e t g t e '=-11e t -≤<()0g t '<1t >()0g t '>()g t 1,1e ⎡⎫-⎪⎢⎣⎭()1,+∞()()min 01g t g ==0a ≤0xy >210x y +=>(12)0x x ->102x <<01y <<422x y +==>…14x =12y =()22222222121log log log log log log 302822x y x y x y xy ⎡⎤⋅+⎛⎫+====-<⎢⎥ ⎪⎝⎭⎢⎥⎣⎦…项.10. BCD 【解析】设幂函数，由，得，所以，所以无意义，故A 项错误；，所以是偶函数，故B 项正确；由，得，故C 项正确；因为是偶函数，且在上单调递减，所以由，得，即且解得且，故D 项正确.故选BCD 项.11. ACD 【解析】对于A 选项，由函数的定义知的定义域为，故A 正确.对于B 选项，令，则，即，判别式，无实数解，故B 错误.对于C 选项，，可知，设函数，可知，令，解得，则在上单调递减，在上单调递增，且在上，则的图象为的图象向左平移一个单位长度，易得两者无交点，则无零点，即不存在极值点，故C 正确.对于D 选项，方法一：由的单调性可知，D 正确.方法二：作差有，且，故，D 正确.故选ACD.12. 【解析】，故时，，故曲线在点处()a f x x =()14416af ==2α=-()2f x x -==()0f ()()f x f x -=()f x ()32f x x -=-'()12f '-=()f x ()0,+∞()()321f x f x ->+321x x -<+22(32)(1)x x -<+320,10,x x -≠⎧⎨+≠⎩243x <<32x ≠()f x ()0,+∞(1)log 2x x +=2(1)x x +=2403x x +=+70∆=-<()(1)ln log ln(1)x x xf x x +==+()2211ln(1)ln (1)ln(1)ln 1ln 1)(1)ln )((1x xx x x x x x f x x x x x +-++-+==+++'()ln g x x x =()ln 1g x x ='+()0g x '=1e x =()g x 10,e ⎛⎫⎪⎝⎭1,e⎛⎫+∞ ⎪⎝⎭()0,1()0g x <()()1ln 1y x x =++()g x ()f x '()f x ()f x ()()()(1)21lo (1)g log x x x f x x f x ++-+=+-()()()2ln ln 2ln 1ln(2)ln 1x x x x x ⋅+-++⋅+=()()()()222ln ln 22ln 1ln ln 2ln 122x x x x x x ⎡⎤⎡⎤+++⋅+<<=+⎢⎥⎢⎥⎣⎦⎣⎦()()()11f x f x x <+>22y x =-()212e 1x y x x -'+-=1x =2y '=21e x y x x -=-()1,0的切线方程为.13. 5【解析】由题意得该数列的项可设为，，，，1，又即从而，即，即，解得所以.14.【解析】如图，延长BA ，CD 交于点E ，则为正三角形.由题设结论，，，的外接圆有唯一公共点，该公共点即为题中的点Q ，故点Q 在的外接圆上.由题意得，，则是直角三角形，故其外接圆半径.在中，由余弦定理可知，，当Q 在线段BD 上，且时，BQ 取得最小值.15. 解：（1）．……6分（2）在上的投影向量为．……13分16．（15分）解：（1），（3分）所以，所以原不等式的解集.（6分）（2）由（1）知，所以（7分）22y x =-()212d q +()12d q +12d +1d +()()211218,121235,d d q d d q ⎧+++=⎪⎨+++=⎪⎩()()221217,2234,q d q q d q ⎧+=-⎪⎨+=-⎪⎩()()()()2212341722q q q q +-=-+232334682343422q q q q q q -+-=+--235700q q -=2,3,q d =⎧⎨=⎩5q d +=1-EBC △ABP △CMP △AME △AME △120BAD ∠=︒90BAM ∠=︒AME △1R AD ==ABD △BD ==1QD =11232e e -==125e e -1e ()121111352e e e e e e -⋅⋅=-()()1010110001x x x x x xx x x=-=->-1x >{|1}M x x =>{|1}M x x =>()()241,1e 22,1xx a x x f x a x ⎧-+>=⎨--⎩…在实数集上单调递增，，又因为当时，是单调增函数，所以当时，，解得（10分）综上，a 的取值范围是.17. 解：（1）由题得，，故，．由，得，，故，，，故，故．，即单调递增区间为，．……9分（2）由，即，又，则，故，．……15分18.解：（1）由题意得，（2分）所以.（3分）由，得当时，，（5分）所以，即.（6分）又当时，也符合，()f x R 4112a +∴≤14a ∴≤1x ≤()f x 1x =224e a a --≤-12ea ≤-,1(]2e -∞-2A =334T =4T =π2ω=2113f ⎛⎫=- ⎪⎝⎭π113π2π232k ϕ⨯+=+k Z ∈π2π3k ϕ=-+k Z ∈π2ϕ<π3ϕ=-()ππ2sin 23f x x ⎛⎫=- ⎪⎝⎭ππππ152π2π44223233k x k k x k -+≤-≤+⇒-+≤≤+()f x 154,433k k ⎡⎤-++⎢⎥⎣⎦k Z ∈()056f x =0ππsin 2335x ⎛⎫-= ⎪⎝⎭058,33x ⎡⎤∈⎢⎥⎣⎦0πππ,π232x ⎛⎫⎛⎫-∈ ⎪ ⎪⎝⎭⎝⎭04ππcos 235x ⎛⎫-=- ⎪⎝⎭0000ππππππ1ππcos cos cos sin 223323223x x x x ⎡⎤⎛⎫⎛⎫⎛⎫⎛⎫=-+=-⋅--= ⎪ ⎪ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎝⎭⎝⎭⎣⎦()22000ππcos cos(2c )2)1os (2(122x x x π=-∴=⨯=⨯-=1212422na n n --=⨯=21n a n =-32122222n b b b b n ++++= 2n …()31212222122n n b b b b n --++++=- 122nn b -=2n n b =1n =12b =所以.（7分）（2）设，则，（8分）（9分）两式作差得，（10分）即，（12分）所以.19.（17分）已知函数（1）求的值；（2）设，当时，记在区间上的最大值为M ，最小值为m ，求的取值范围.解：（1）由，（2分）所以，所以，（4分）所以.（5分）（2）由（1）可得，（6分）2nn b =()1212nn n n a c n b ⎛⎫==- ⎪⎝⎭()21221111()32221nn n S c c c n ⎛⎫⎛⎫=+++=⨯+⨯++- ⎪ ⎪⎝⎭⎝⎭ ()231111132112222n n S n +⎛⎫⎛⎫⎛⎫=⨯+⨯++-⋅ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭()231111112211222222221nn n S n +⎛⎫⎛⎫⎛⎫⎛⎫=⨯+⨯+⨯++⨯-- ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭()1121722222212111272111n n n n n S n ++⎡⎤⎛⎫-⎢⎥ ⎪⎝⎭+⎢⎥⎛⎫⎣⎦=-+--=- ⎪⎝⎭-2277n nn S +=-()3f x x x =-+()0f ()()()ln 0g x a x x f f x =-++-(]1,3a ∈-()g x []1,e Mm -()3f x x x =-+()()332223031()2e f f x x e x -'-'=-+()()30012f f =-+''()02f '=()3f x x x =-()206f e =()32ln 6g x x a x e =-++，（7分）①当时，，， 在区间上单调递减， （8分）所以的最小值.（9分）的最大值，（10分），（11分）这时的取值范围为.（12分）②当时，，，在区间上，， 在区间上单调递减，（13分）所以的最小值.（14分）的最大值，（15分），这时的取值范围为.（16分）综上所述，当时，取值范围为；当时，取值范围为.（17分）变式：已知函数.（1）求曲线在处的切线方程；（2）设，当时，记在区间上的最大值为M ，最小值为m ，求的取值范围.()32333x a x a g x x x '⎛⎫- ⎪=-+=- ⎪ ⎪⎝⎭10a -<≤03a ->()0g x '≤()g x [1,]e ()g x ()326m g e a e e ==-++()g x ()2161M g e ==-()232333(1)6161(1,]M m g g e e a e e a e e e -=-=-+--=--∈-Mm -33(1,]e e -03a <≤013a<≤01<≤[1,]e ()0g x '≤()g x [1,]e ()g x ()326m g e a e e ==-++()g x ()2161M g e ==-()232333(1)6161[4,1)M m g g e e a e e a e e e -=-=-+--=--∈--Mm -33[4,1)e e --10a -<≤M m -33(1,]e e -03a <≤Mm -33[4,1)e e --()3f x x x =-+()y f x =0x =()()()20g x ax x f f x =+--10a -<<()g x []1,0-M m -解：（1）由，（2分）所以，所以，（4分）所以，所以.（5分）所以在处的切线方程为（6分）化为.（7分）（2）由（1）可得，（8分）所以，，两零点为 （9分）-+单调递减单调递增（11分）因为，（12分）所以时，，（13分）()3f x x x =-+()()3223031(1)2f f x x x -'-'=-+()()30012f f =-+''()02f '=()3f x x x =-+()06f =()y f x =0x =()620y x -=-260x y -+=()()()()22332006g x ax x f f x ax x f x x x ax =-++-⎛=-+--+ ⎝=-++()22323()3a g x x ax x x =-+=--'10a -<<1222,0,033a x x ⎛⎫=∈-= ⎪⎝⎭x 21,3a ⎡⎤-⎢⎥⎣⎦2,03a ⎡⎤⎢⎥⎣⎦()g x '()g x ()60g =()()7106g a g =+>=-[]1,0x ∈-()()max 17M g x g a==-=+（14分）所以设，（15分）（16分）所以在上单调递增，因为，所以的取值范围为.（17分）()n 33mi 3238462742769a m g x g a a a ⎛⎫== ⎪⎝=-++=+⎭()33472741276h a M m a a a a -=-==+--++10a -<<()22449433'1()()()0994922h a a a a a =-+=--=-+->()h a 10a -<<()4127h -=()01h =M m -4,127⎛⎫⎪⎝⎭。

福建省泉州市2025届高中毕业班模拟检测(一)数学试题(含答案)

福建省泉州市2025届高中毕业班模拟检测(一)数学试题(含答案)

2025届高中毕业班模拟检测（一）2024.09高三数学本试卷共19题 满分150分 考试时间：120分钟注意事项：1．答题前，考生先将自己的姓名、准考证号填写在答题卡上．2．考生作答时，将答案答在答题卡上．请按照题号在各题的答题区域（黑色线框）内作答，超出答题区域书写的答案无效．在草稿纸、试题卷上答题无效．3．选择题答案使用2B 铅笔填涂，如需改动，用橡皮擦干净后，再选涂其它答案标号；非选择题答案使用0.5毫米的黑色中性（签字）笔或碳素笔书写，字体工整、笔迹清楚．4．保持答题卡卡面清洁，不折叠、不破损．考试结束后，将本试卷和答题卡一并交回．一、选择题：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．已知集合，则（ ）A ．B ．C ．D ．2．若复数满足（其中是虚数单位，），则“”是“”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3．等差数列的首项为2，公差不为0．若成等比数列，则公差为（ ）A ．B ．C ．1D ．4．若，则（ ）A ．B ．C ．D ．5．已知圆柱的底面直径为2的球面上，该圆柱的侧面积为（ ）A ．B ．C ．D ．6．已知，若与的夹角为，则在上的投影向量为（ ）A ．B ．C ．D ．{}{}*240,12A x x x B x x =∈-≤=∈-≤N Z A B = {}0,1,2{}0,1,2,3{}1,2,3{}1,2,3,4z ()1i i z a +=-i R a ∈1z =1a ={}n a 245,,a a a 2525-1-π4sin 125α⎛⎫+=⎪⎝⎭5πcos 26α⎛⎫-= ⎪⎝⎭1225-725-72512258π6π5π4π2b a =a b 60︒2a b - b 12b12b- 32b- 32b7．已知函数的定义域为，且，记，则（ ）A ．B ．C ．D ．8．已知函数，若不等式的解集中恰有两个不同的正整数解，则实数的取值范围是（ ）A ．B ．C ．D ．二、选择题：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．全部选对得6分，部分选对的得部分分，有选错的得0分．9．某校从某年级中随机抽取了100名学生的成绩，整理得到如图所示的频率分布直方图．为进一步分析高分学生的成绩分布情况，计算得到这100名学生中，成绩位于）内的学生成绩方差为12，成绩位于内的同学成绩方差为10．则（）A ．B ．估计该年级学生成绩的中位数约为77.14C ．估计该年级成绩在80分及以上的学生成绩的平均数为87.50D ．估计该年级成绩在80分及以上的学生成绩的方差为3210．已知展开式中共有8项．则该展开式结论正确的是（ ）A ．所有项的二项式系数和为128B ．所有项的系数和为C ．系数最大项为第2项D ．有理项共有4项11．设函数，则（）A ．当时，有三个零点()f x ()0,+∞()()()()(),1e x y f x y xyf x f y f ++==()()1,2,32a f b f c f ⎛⎫=== ⎪⎝⎭a b c<<b a c<<a c b <<c b a<<()2ln f x x mx x =-+()0f x >m 2ln23ln3,89++⎡⎫⎪⎢⎣⎭3ln32ln2,94++⎛⎫⎪⎝⎭3ln32ln2,94++⎡⎫⎪⎢⎣⎭2ln23ln3,89++⎛⎫⎪⎝⎭[80,90[)90,1000.004a=()*nx n ⎛+∈ ⎝N 832⎛⎫⎪⎝⎭()32231f x x ax =-+1a >()f xB ．当时，是的极大值点C ．存在，使得为曲线的对称轴D ．存在，使得点为曲线的对称中心三、填空题：本题共3小题，每小题5分，共15分．12．已知随机变量，若，则实数a 的值为______________．13．圆的圆心与抛物线的焦点重合，A 为两曲线的交点，则原点到直线的距离为______________．14．数列满足，且，则数列的前2024项和为______________．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．（本小题13分）记的内角、、的对边分别为．已知．（1）证明：；（2）若，求的面积．16．（本小题15分）如图，在四棱锥中，，平面平面为中点．（1）求证：平面；（2）点在棱上，与平面所成角的正弦值为，求平面与平面夹角的余弦值．17．（本小题15分）已知点为圆上任意一点，，线段的垂直平分线交直线于点，设点的轨迹为曲线．0a <0x =()f x ,a b x b =()y f x =a ()()1,1f ()y f x =()22,3N ξ~(3)(21)P a P a ξξ<-=>+22(1)25x y -+=22(0)y px p =>F AF {}n a 11a =()*11n n a a n n +=++∈N 1n a ⎧⎫⎨⎬⎩⎭ABC △A B C a b c 、、223cos cos 222C A a c b +=sin sin 2sin A C B +=2,3b AB AC =⋅=ABC △P ABCD -12,,902PD PC CB BA AD AD CB CPD ABC =====∠=∠=︒∥PCD ⊥,ABCD E PD PD ⊥PCA Q PA CQPDC PCD CDQ P 22:(2)4C x y -+=()2,0A -PA PC M M H（1）求曲线的方程；（2）若过点的直线1与曲线的两条渐近线交于，两点，且为线段的中点．（ⅰ）证明：直线1与曲线有且仅有一个交点；（ⅱ）求的取值范围．18．（本小题17分）已知函数．（1）当时，恒成立，求实数a 的取值范围；（2）证明：当时，曲线与曲线总存在两条公切线；（3）若直线是曲线与的两条公切线，且的斜率之积为1，求的关系式．19．（本小题17分）已知无穷数列，给出以下定义：对于任意的，都有，则称数列为“数列”；特别地，对于任意的，都有，则称数列为“严格数列”．（1）已知数列的前项和分别为，且，试判断数列，数列是否为“数列”，并说明理由；（2）证明：数列为“数列”的充要条件是“对于任意的，当时，有；”（3）已知数列为“严格数列”，且对任意的．求数列的最小项的最大值．参考答案1．【分析】分别求出两个集合后根据交集定义求解．【详解】；；．故选：C ．2．【分析】由复数的运算结合模长公式求出，再由充分必要条件定义判断．【详解】由得，H M H S T M ST H 21OS OT+()()()e ,ln ,xf x ag x x b a b ==+∈R 1b =()()f x g x ≥1e ,1a b -=<()yf x =()yg x =12,l l ()y f x =()y g x =12,l l ,a b {}n a *n ∈N 212n n n a a a +++≥{}n a T *n ∈N 212n n n a a a +++>{}n a T {}{},n n a b n ,n n A B 121,2n n n a n b -=-=-{}n A {}n B T {}n a T *,,k m n ∈N k m n <<()()()k n m n m a m k a n k a -+-≥-{}n b T *1128,,8,8n n b b b ∈∈=-=-N Z {}n b {}{}{}*2*40041,2,3,4A x x x x x =∈-≤=∈≤≤=N N {}{}{}{}12212131,0,1,2,3B x x x x x x =∈-≤=∈-≤-≤=∈-≤≤=-Z Z Z {}1,2,3A B = a ()1i i z a +=-()()()()i 1i i 11i,11i 1i 1i 22a a a a z z ----+===-=++-，解得或．故“”是“”的必要不充分条件．故选：B3．【分析】根据等比中项可得，结合等差数列的通项公式运算求解．【详解】设等差数列的公差为，若成等比数列，则，即，整理可得，解得或（舍去），所以公差为．故选：D ．4．【分析】根据诱导公式以及二倍角公式即可代入求解．【详解】故选：C ．5．【分析】利用球的体积公式求出球的半径，结合圆柱半径可得圆柱的高，然后可解．【详解】球的体积为，可得其半径2，半径为，在轴截面中，可知圆柱的高为，所以圆柱的侧面积为．故选：A ．6．【分析】应用向量的数量积及运算律，结合投影向量公式计算即可得解．【详解】因为与的夹角为，所以，则2211122a a -+⎛⎫⎛⎫∴+-= ⎪ ⎪⎝⎭⎝⎭1a =1a =-1z =1a =2425a a a =⋅{}n a 0d ≠245,,a a a 2425a a a =⋅()()2(23)224d d d +=++2520d d +=25d =-0d =25-34π3R =R =1r =4h ==2π8πrh =2,b a a =b 60︒21cos6022a b a b a a a ⋅=︒=⨯⨯=所以在上的投影向量为．故选：B ．7．【分析】根据函数满足的表达式以及，利用赋值法即可计算出的大小．【详解】由可得，令，代入可得，即，令，代入可得，即，令，代入可得，即；由可得，显然可得．故选：A8．【分析】不等式可化为，利用导数分析函数的单调性，作函数的图象，由条件结合图象列不等式求的取值范围．【详解】函数的定义域为，不等式化为：．令故函数在上单调递增，在上单调递减．当时，，当时，，当时，，当时，，当，且时，，()222222242a b b a b b a a a-⋅=⋅-=-=- 2a b - b ()222||1222a b b b a b b a a b b-⋅-⨯=⨯=-()f x ()1e f =,,a b c ()()()()(),1e x y f x y xyf x f y f ++==12x y ==()21111e 222f f ⎛⎫=⨯= ⎪⎝⎭12a f ⎛⎫==± ⎪⎝⎭1x y ==()()22221e f f ==()2e 22b f ==1,2x y ==()()()23e 332122e e 2f f f ==⨯=()3e 33c f ==e 2.71828≈ 23e e 23±<<a b c <<()0f x >ln 1x mx x -<()ln xg x x=()()ln 1,xh x mx g x x=-=m ()2ln f x x mx x =-+()0,+∞()0f x >ln 1x mx x-<()()()2ln 1ln 1,,x x h x mx g x g x x x -='=-=，()g x ()0,e ()e,+∞1x >()0g x >1x =()0g x =01x <<()0g x <x →+∞()0g x →0x >0x →()g x →-∞画出及的大致图象如下，因为不等式的解集中恰有两个不同的正整数解，故正整数解为1，2．故，即．故．故选：C ．9．【分析】利用小长方形面积和为1得A 项错误；面积等于0．5的值即为中位数，可知B 正确；利用直方图中平均数和方差公式可得C 正确，D 错误．【详解】A 项，，A 项错误；B 项，内频率为：内频率为：，则中位数在内，设中位数为，则，则，B 正确；成绩在80分及以上的同学的成绩的平均数为分，方差为，C 正确，D 错误．故选：BC ．10．【分析】先根据展开式的项数确定的值，根据二项式系数的性质判断A 的真假，令可得所有项的系数和，判断B 的真假，利用二项展开式的通项公式可判断CD 的真假．【详解】因为的展开式共有8项，所以．所以所有项的二项式系数和为，故A 正确；对B ：令，可得所有项的系数和为，故B 错误；()g x ()h x ()0f x >()()()()2233h g h g ⎧<⎪⎨≥⎪⎩ln2212ln3313m m ⎧-<⎪⎪⎨⎪-≥⎪⎩3ln32ln294m ++≤<()23762101,0.005a a a a a a ++++⨯=∴=[]50,70[]50.005100.250.5,50,80⨯⨯=<120.005100.60.5⨯⨯=>[]70,80x ()0.257070.0050.5x +-⨯⨯=77.14x =31859587.544⨯+⨯=223112(87.585)10(87.595)30.2544⎡⎤⎡⎤⨯+-+⨯+-=⎣⎦⎣⎦n 1x =nx ⎛+ ⎝7n =72128=1x =7813122⎛⎫⎛⎫+≠ ⎪ ⎪⎝⎭⎝⎭因为二项展开式的通项公式为：．对C ：设，由，所以第3项的系数最大，故C 错误；对D ：由为整数，且可得，的值可以为：0，2，4，6，所以二项展开式中，有理项共有4项，故D 正确．故选：AD11．【分析】A 选项，先分析出函数的极值点为，根据零点存在定理和极值的符号判断出在上各有一个零点；B 选项，根据极值和导函数符号的关系进行分析；C 选项，假设存在这样的，使得为的对称轴，则为恒等式，据此计算判断；D 选项，若存在这样的，使得为的对称中心，则，据此进行计算判断，亦可利用拐点结论直接求解．【详解】A 选项，，由于，故时，故在上单调递增，时，单调递减，则在处取到极大值，在处取到极小值，由，则，根据零点存在定理在上有一个零点，又，则，则在上各有一个零点，于是时，有三个零点，A 选项正确；B 选项，时，单调递减，37721771C C 2r rr r r r r T x x--+⎛⎫=⋅⋅=⋅⋅ ⎪⎝⎭71C 2nnn a ⎛⎫=⋅ ⎪⎝⎭()()()()()()117711117717!7!118C C 2!7!1!8!22327!17!511C C !7!21!6!322n n n n n n n n n n n n n n n n n a a n a a n n n n n ---+++⎧⎧⎧⎫⎫⎛⎛⎧⋅≥⋅≥⋅⎪⎪≤ ⎪ ⎪⎪⋅---⎪≥⎪⎪⎝⎝⎪⎪⎭⎭⇒⇒⇒⇒=⎨⎨⎨⎨≥⎫⎫⎛⎛⎪⎪⎪⎪≥≥⋅≥⋅ ⎪ ⎪⎪⎪⎪⎪⋅-+-⎩⎩⎝⎝⎭⎭⎩⎩372r-0,1,2,,7r = r 0,x x a ==()f x ()()()1,0,0,,,2a a a -,a b x b =()f x ()()2f x f b x =-a ()1,33a -()f x ()()266f x f x a +-=-()()2666f x x ax x x a '=-=-1a >()(),0,x a ∈-∞+∞ ()0f x '>()f x ()(),0,,a -∞+∞()0,x a ∈()()0,f x f x '<()f x 0x =x a =()()3010,10f f a a =>=-<()()00f f a <()f x ()0,a ()()31130,2410f a f a a -=--<=+>()()()()100,20f f f a f a -<<()f x ()()1,0,,2a a -1a >()f x ()()6,0f x x x a a -'=<()()(),0,0,x a f x f x <'∈时单调递增，此时在处取到极小值，B 选项错误；C 选项，假设存在这样的，使得为的对称轴，即存在这样的使得，即，根据二项式定理，等式右边展开式含有的项为，于是等式左右两边的系数都不相等，原等式不可能恒成立，于是不存在这样的，使得为的对称轴，C 选项错误；D 选项，方法一：利用对称中心的表达式化简，若存在这样的，使得为的对称中心，则，事实上，，于是即，解得，即存在使得是的对称中心，D 选项正确．方法二：直接利用拐点结论任何三次函数都有对称中心，对称中心的横坐标是二阶导数的零点，，由，于是该三次函数的对称中心为，由题意也是对称中心，故，即存在使得是的对称中心，D 选项正确．故选：AD【点睛】结论点睛：（1）的对称轴为关于对称()0,x ∈+∞()()0,f x f x '>()f x 0x =,a b x b =()f x ,a b ()()2f x f b x =-32322312(2)3(2)1x ax b x a b x -+=---+3(2)b x -3x 303332C (2)()2b x x -=-3x ,a b x b =()f x ()133f a =-a ()1,33a -()f x ()()266f x f x a +-=-()()()()3232222312(2)3(2)112612241812f x f x x ax x a x a x a x a +-=-++---+=-+-+-()()26612612241812a a x a x a-=-+-+-126012240181266a a a a -=⎧⎪-=⎨⎪-=-⎩2a =2a =()()1,1f ()f x ()()()322231,66,126f x x ax f x x ax f x x a =-+='='--'()02a f x x =⇔='',22a a f ⎛⎫⎛⎫ ⎪ ⎪⎝⎭⎝⎭()()1,1f 122aa =⇔=2a =()()1,1f ()f x ()f x ()()()22x b f x f b x f x =⇔=-；（）(),a b；（3）任何三次函数都有对称中心，对称中心是三次函数的拐点，对称中心的横坐标是的解，即是三次函数的对称中心12．【分析】根据正态分布的对称性求解．【详解】由题意得，，解得．故答案为：213．【分析】先求出圆心坐标，从而可求焦准距，再联立圆和抛物线方程，求A 及的方程，从而可求原点到直线的距离．【详解】圆的圆心为，故即，由可得，故或（舍），故，故直线即或，故原点到直线的距离为，故答案为：14．【分析】由运用迭代法求出，则，利用裂项相消法即可求得的前2024项和．【详解】由可得，则，则，故数列的前2024项和为．故答案为：．()()22f x f a x b ⇔+-=()32f x ax bx cx d =+++()0f x ''=,33bb f a a ⎛⎫⎛⎫-- ⎪ ⎪⎝⎭⎝⎭32122a a -++=⨯2a =AF AF 22(1)25x y -+=()1,0F 12p=2p =222(1)254x y y x⎧-+=⎨=⎩22240x x +-=4x =6x =-()4,4A ±()4:13AF y x =±-4340x y --=4340x y +-=AF 4455d ==4511n n a a n +=++()12n n n a +=()1211211n a n n n n ⎛⎫==- ⎪++⎝⎭1n a ⎧⎫⎨⎬⎩⎭11n n a a n +=++11n n a a n +-=+()()()()()11221111212n n n n n n n a a a a a a a a n n ---+=-+-++-+=+++-+=()1211211n a n n n n ⎛⎫==- ⎪++⎝⎭1n a ⎧⎫⎨⎬⎩⎭11111140482122212232024202520252025⎛⎫⎛⎫⎛⎫⎛⎫-+-++-=-=⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭ 4048202515．【分析】（1）利用三角恒等变换结合正弦定理化简可证得结论成立；（2）利用平面向量数量积的定义可得出，结合余弦定理以及可求得的值，由此可求得的面积．【详解】（1）因为，则，即，由正弦定理可得因此，．（2）因为，由正弦定理可得，由平面向量数量积的定义可得，所以，，可得，即，所以，，则所以，，则为锐角，且，因此，．16．【分析】（1）应用面面垂直性质定理证明线面垂直；（2）先应用空间向量法计算线面角得出参数，再计算二面角即可．【详解】（1）由题意：，同理又．而，即又平面平面，平面平面平面，平面平面，又，且面面平面．（2）以为原点，建立如图所示的空间直角坐标系，cos 3bcA =24ac b +==a c 、ABC △223coscos 222C A a c b +=()()1cos 1cos 322a C c Ab +++=cos cos 3ac a C c A b +++=()()3sin sin sin sin cos cos sin sin sin sin B A C A C A C A C A C =+++=+++()sin sin sin sin sin sin A C B A C B π=++-=++，sin sin 2sin A C B +=sin sin 2sin A C B +=24a c b +==cos 3AB AC cb A ⋅==2222242322b c a c a c bc +-+-⋅==222c a -=()()()42c a c a c a -+=-=12c a -=97,44c a ==，332cos 9324A bc ===⨯A sin A ===1119sin 22224ABC S bc A ===⨯⨯=△2,90,BC AB ABC AC ==∠=︒∴==CD =2224,,AD CD AC AD CD AC =∴+=∴⊥CD ==PC PD⊥PCD ⊥ABCD PCD ,ABCD CD AC =⊂ABCD AC ∴⊥,PCD PD ⊂,PCD PD AC ∴⊥PC PD ⊥PC ⊂,PCA AC ⊂,,PCA PC AC C PD =∴⊥ PCA C则，，设，有，取面的一个法向量，则，故．令是平面的一个法向量，则，即令，有，则故平面与平面．17．【分析】（1）由双曲线的定义进行求解；（2）（ⅰ）设，求出，由直线1与曲线方程进行求解；（ⅱ）由，则利用基本不等式求解．【详解】（1）为的垂直平分线上一点，则，则点的轨迹为以为焦点的双曲线，且，()()()0,0,0,,C A D P ()(,,CD CP PA ∴===(01)PQ PA λλ=<<)))11CQ CP PA λλλ=+=-- PCD ()0,1,0m =1cos ,2CQ m λ=== CQ = (),,n x y z = CDQ 00n CD n CQ ⎧⋅=⎪⎨⋅=⎪⎩00x z ⎧=++=1y =()0,1,2n =-cos ,n m n m n m ⋅==PCD CDQ ()()()001122,,,,,M x y S x y T x y 03ST x k y =H 12220034443OS OT x x x y ⋅===⨯=-2124OS OS OT OS +=+M PA MP MA =24MA MC MP MC AC -=-=<=∴M ,A C 22,2a c ==故点的轨迹方程为．（2）（ⅰ）设，双曲线的渐近线方程为：，如图所示：则①，②，①+②得，，①-②得，，，得由题可知，则，得，即，直线的方程为，即，又点在曲线上，则，得，将方程联立，得，得，由，可知方程有且仅有一个解，得直线1与曲线有且仅有一个交点．（ⅱ）由（ⅰ）联立，可得，同理可得，，M 22:13y H x -=()()()001122,,,,,M x y S x y T x y y =11y =22y =)1212y y x x +=-)1212y y x x -=+=()121212123x x y yx x y y -+=+-MS MT =1201202,2x x x y y y +=+=()1200123x x y x y y -=-003ST xk y =∴ST ()00003x y y x x y -=-22000033x x y y x y -=- M H 220033x y -=0033x x y y -=22001333y x x x y y ⎧-=⎪⎨⎪-=⎩()222200003630y x x x x y -+--=22003630x x x x -+-=()()()2200Δ64330x x =-⨯-⨯-=H 0033y x x y y ⎧=⎪⎨-=⎪⎩1x =2x =则，故当且仅当，即时取等号．故的取值范围为．【点睛】关键点点睛：第二问中的第2小问中，先要计算，再由基本不等式求解范围．18．【分析】（1）参变量分离可得，设，利用导数求出的最大值，从而可得的取值范围；（2）设两个函数的切点，由点斜式求解切线方程，利用公切线联立可得，再构造函数，利用导数即可证明，即可求证；（3）根据公切线得，化简整理可得，题目转化为有两个不等实根，且互为倒数，不妨设两根为，由可得的关系，代入中，可得有两个不等实根，代入化简即可求解．【详解】（1）由得，则，设，由于均为上的单调递减函数，12220034443OS OT k x x y ⋅===⨯=-2124OS OS OT OS +=+≥24OS OS =OS =21OS OT+)+∞4OS OT ⋅=ln 1e x x a +≥()ln 1e xx F x +=()F x a 111ln ln 1x b x x =-+()ln ln 1xh x x x=-+1b <()()()()f sg t f s g t s t'-='=-()ln ln ln 1ln 1a t s t t t b t =--=--++-()()ln 1ln 1ln p t t t t b t a =--++-=1,m m ()1h m h m ⎛⎫= ⎪⎝⎭,a b ()h t 11ln 1tb t t --=⋅+()()f x g x ≥e ln 1x a x ≥+ln 1e xx a +≥()()1ln 1ln 1,e e x xx x x F x F x --='+=1,ln y y x x==-()0,+∞故为上的单调递减函数，结合在为正，在为负，故在上单调递增，在单调递减，，则，即的取值范围是．（2）设直线是的公切线，设的切点为的切点为，所以切线方程为，因此且结合，故，故，进而可得，令，故，由于为单调递减函数，且，故当在单调递增；当在单调递减；故又当，且，故总有两个不相等的实数根，因此直线有两条，（3）由题意得：存在实数，使在处的切线和在处的切线重合，1ln 1y x x=--()0,+∞()10F '=()F x ∴'()0,1()1,+∞()F x ()0,1()1,+∞()max ()1F x F ∴=()11ea F ≥=a 1,e ⎡⎫+∞⎪⎢⎣⎭l ()(),f x g x ()g x ()()11,ln ,x x b f x +()22,e xx a ()()1e ,x f x a g x x''==()()2211211ln ,e e x x y x x x b y a x x a x =-++=-+211ex a x =2212ln 1e e x xx b a ax +-=-1e a -=212111e1ln x x x x -=⇒-=-()21121ln ln 1e 1x x x b a x x +-=-=111ln ln 1x b x x =-+()ln ln 1x h x x x =-+()21ln x x h x x-'-=1ln y x x =--()10h '=()()()0,1,0,x h x h x >'∈()0,1()()()1,,0,x h x h x '∈+∞<()1,+∞()()11h x h ≤=，(),x h x →+∞→-∞()0,x h x →→-∞111ln ln 1x b x x =-+l ,s t ()f x x s =()g x x t =，即，则，又，题目转化为有两个不等实根，且互为倒数，不妨设两根为，则由得，化简得，，【点睛】关键点点睛：由公切线得，进而得，利用斜率互倒数，利用代入化简．19．【分析】（1）根据等差等比的求和公式可得，即可利用定义以及作差法求解，（2）利用累加法，结合放缩法可得，即可求证必要性，取即可求证充分性，（3）根据定义可得为单调递增数列，且，进而得，即可根据单调性得最小值为，结合放缩法和等差求和公式可得，即可求解．()()()()f s g t f s g t s t -∴=='-'1ln 1e ln e sst b a t b t a t s t s t----===--()1ln ,1ln 1s t t t bt s t t b t -=--=---1e ln ln sa a s t t=⇒+=- ，()ln ln ln 1ln 1a t s t t t b t ∴=--=--++-()()ln 1ln 1ln p t t t t b t a =--++-=1m m，()1p m p m ⎛⎫=⎪⎝⎭()()1111ln 1ln 1ln 1m m m b m b m m m m --++-=-++-()()()()2211111ln 111212b m b m m m m b m m m m m⎛⎫--⎪--+⎝⎭===-+--+-()()()()()ln 1ln 111111a m m b m b m b m b ∴=--+-=----+-=-ln b a∴=-1ln 1e ln e sst ba tb t a t s t s t----===--()ln ln ln 1ln 1a t s t t t b t =--=--++-()1p m p m ⎛⎫= ⎪⎝⎭()212112,12212nn n nn n A n B+--===-=--()()()()11,n m m m m k m m a a n m a a a a m k a a +--≥---≤--1,2m k n k =+=+{}n c Z n c ∈128112712610b b c c c -=+++= {}n b m b ()()()1271281min 8,822m m m m m b ⎧⎫---≤----⎨⎬⎩⎭【详解】（1）由于为等差数列，所以，为等比数列，，任意的，都有，故，所以数列是为“数列”，任意的，都有，故，所以数列不是为“数列”，（2）先证明必要性：因为为“数列”，所以对任意的，都有，即，所以对任意的，当时，有所以，又，所以，又故，即故，再证明充分性：对于任意的，当时，有，即，对于任意的，则有，即可，所以为“数列”，21n a n =-()21212n n n A n +-==12n n b -=-121212nn n B -=-=--*n ∈N 222212(2)2(1)20n n n A A A n n n +++-=++-+=>212n n n A A A +++>{}n A T *n ∈N 21212222220n n n n n n n B B B +++++-=--+⨯=-<212n n n B B B +++<{}n B T {}n a T *n ∈N 212n n n a a a +++≥211n n n n a a a a +++-≥-*,,k m n ∈N k m n <<()()()()()11211n m n n n n m m m m a a a a a a a a n m a a ---++-=-+-++-≥-- 1n mm m a a a a n m+-≥--()()()()()11211m k m m m m k k m m a a a a a a a a m k a a ---+--=-+-++-≤-- 1m km m a a a a m k--≤--111,m km m m m m ma a a a a a a a m k-++--≤-≤--1m k n m m m a a a a a a m k n m +--≤-≤--m k n ma a a a m k n m--≤--，()()()k n m n m a m k a n k a -+-≥-*,,k m n ∈N k m n <<()()()k n m n m a m k a n k a -+-≥-m k n ma a a a m k n m--≤--*N ,1,2k m k n k ∈=+=+12111k k k k a a a a +++--≤212k k k a a a +++≥{}n a T（3）数列为“严格数列”，且对任意的，有，即，设，则为单调递增数列，且，所以因为．所以，所以存在时，，所以，当，数列为单调递减数列，当，因此存在最小值，且最小值为，由于，所以，且，所以，即，即所以，当时，，当时，，当时，所以当时，的最大值为，此时，因为，所以数列的最小项的最大值为{}n b T *n ∈N 212n n n b b b +++>211n n n n b b b b +++->-1n n n c b b +=-{}n c Z n c ∈()()()111221121n n n n n n n b b b b b b b b c c c ------=-+-+-=+++ 11288,8b b =-=-128112712610b b c c c -=+++= *N ,2127m m ∈≤≤10,0m m c c -<≥*1,N ,0n n n m n b b -≤∈-<{}n b *1,N ,0n n n m n b b +≥∈-≥{}n b m b Z n c ∈11270,1,,127m m c c c m +≥≥≥- 1211,2,,1m m c c c m --≤-≤-≤-+ ()112112m m m m m b b c c c ----=+++≤-()182m m m b -≤--()()1281271261271282m m m m b b c c c ---=+++≥()()12712882mm m b--≤--()()()1271281min 8,822m m m m m b ⎧⎫---≤----⎨⎬⎩⎭()()()()11271281276422m m m m m ----+=-64m =()()()12712818822m m m m -----=--64m >()()()12712818822m m m m ----->--064m <<()()()12712818822m m m m -----<-64m =m b ()1820242m m ---=-64,1,2,3,,127n c n n =-= 6465640c b b =-={}n b 65642024b b ==-【点睛】关键点点睛：由得，利用累加法和放缩法得是证明第（2）问的关键．由，设，则为单调递增数列，且，由，得存在时，，所以，当，数列为单调递减数列，当，是第（3）问的求解关键．212n n n a a a +++≥211n n n n a a a a +++-≥-()()()()()11211n m n n n n m m m m a a a a a a a a n m a a ---++-=-+-++-≥-- 211n n n n b b b b +++->-1n n n c b b +=-{}n c Z n c ∈128112712610b b c c c -=+++= *N ,2127m m ∈≤≤10,0m m c c -<≥*1,N ,0n n n m n b b -≤∈-<{}n b *1,N ,0n n n m n b b +≥∈-≥。

给2025年人们的一封信英语作文

给2025年人们的一封信英语作文

给2025年人们的一封信英语作文Dear Future Citizens of 2025,。

As you read this letter, I am writing from a world that is rapidly changing, evolving, and facing an array of challenges and opportunities. The year is 2023, and while we cannot predict everything that the future holds, we can make informed guesses and hope to shape a better world for you.First and foremost, I hope that by the time you read this, our planet is in a state of harmonious coexistence with nature. The climate crisis is real, and it is upon us. The decisions we make today will determine the future of our planet for generations to come. I pray that we have集体行动起来, reduced our carbon emissions, and embraced renewable energy sources. I hope that our cities are designed with sustainability at their core, and that we have moved towards a circular economy, minimizing waste and maximizing resource efficiency.Technology, undoubtedly, will have marched forward at an unprecedented pace. I envision a world where artificial intelligence and machine learning have been harnessed not just for profit, but for the betterment of society.或许, we will have made strides in eradicating diseases and extending the human lifespan. The internet and digital connectivity will be even more pervasive, connecting every corner of the globe and breaking down barriers of communication and access to information.Education, too, will have transformed. The traditional classroom setting may give way to more personalized and immersive learning experiences, where students can explore and engage with content in ways that are both exciting and effective. Lifelong learning will be the norm, and access to education will be equitable, regardless of geography or socio-economic status.Economies will have diversified and become more resilient. The focus will shift towards sustainable practices and ethical business models, with companies andorganizations prioritizing the welfare of their employees, customers, and the environment. Global cooperation and collaboration will be essential in addressing the challenges we face, such as climate change, poverty, and inequality.Politics and governance will also undergo significant changes. Democratic institutions will be strengthened, and the voice of the people will be heard and respected.或许, we will have moved towards a more decentralized and inclusive system of governance, where decision-making is more transparent and participatory.Culture and society will continue to evolve as well. The world will become more interconnected culturally, and we will celebrate our differences while also finding common ground. Art, music, and film will reflect the diverse experiences and perspectives of humanity, and they will serve as a powerful force for unity and understanding.Finally, I hope that by 2025, we have made progress in building a more just and equitable world. Racism, genderdiscrimination, and other forms of oppression will still exist, but I pray that we have taken significant steps towards dismantling these systems of inequality. We will need to continue to learn, to listen, and to act with empathy and compassion towards each other if we are to create a truly inclusive and sustainable future.In conclusion, as you read this letter from a worldthat is both familiar and alien to you, I hope that it inspires you to dream big, to act boldly, and to believe in the power of human potential. Remember that everyindividual action has the potential to shape the future, and that your choices and efforts are invaluable.With hope and optimism, I leave you with these words: Build the future you want to see, and remember that the power to change the world lies within each and every one of us.Sincerely,。

Gigaset_2025C

Gigaset_2025C

Keys:1 Repertory keys2 Recall key3 Redial key4 Mute key5 Loudspeaker key (handsfree)6 Quiet key7 Loud key8 Set key9 Up key 10 Down key 11 OK key (dial number displayed)Display symbols:In setting modeNo memory content Ready for useMicrophoneoffBaby Call (direct call) number setRecall key pressedWhen trying to connect: Entry mark when settingtelephone locked (e.g.volume) PauseTelephone locked "*" key pressedPhone number "#" key pressed(Flashing) A new message New call in calls listhas arrived1567891011oDAn2AWlcDLTrRDDL Recommended installation of telephoneDo not expose the telephone to direct sunlight or other sources of heat.Operate at temperatures between +5°C and +40°C.Maintain a distance of at least one metre between the telephone and radioequipment, e.g. wireless telephones, wireless pagers or TV sets. Otherwise,telephone communication could be impaired.Do not install the telephone in dusty areas as this can shorten the servicelife of the telephone.To clean, wipe with a moist or antistatic cloth only, never with a dry cloth (riskof electrostatic charging and discharge) or harsh cleaning agent.Furniture lacquer and polish can be damaged by contact with parts of theunit (e.g. device feet).ConnectingInsert the short end of the coiled handset cord in the socket provided in the handsetand the long end in the socket marked with an on the underside of the device.Plug the phone cord into the outlet on the wall and the other end into the socketmarked with a on the underside of the device. Then your telephone is ready touse.Setting the internal clockInitiatefunction.The current set time is displayed in 24-hourmode(defaultsetting:00:00).Enter the correct time in hours <hh> andminutes<mm>(24-hourmode).PresstheSetkey.2+...Oror...Dialling a numberLift handset, dial number.Dial number (you can delete wrongly entered digits with and then enter the correct digit) and press the OK key. Lift the handset if you want to talk via the handset.Any digits after the 20th digit are moved from right to left in the display. The Lastnumber redial key can be used to insert pauses when dialling, starting from the 2nd digit.Activating/deactivating handsfree talkingWith this telephone you can also make calls (handsfree mode) or settings without lifting the handset. The best distance from the microhone for speaking is about 50 cm.Activating during a call via the handset With the Loudspeaker key pressed replace the handset.Activating before dialling the numberBefore dialling: hold down the Loudspeaker key until you can hear the dialling tone.Deactivating handsfree talking Lift the handset during the call.Last number redialThe last 5 numbers dialled (up to 32 digits) are saved automatically. You can retrieve these numbers from the last number redial list and dial them again:Redialling the last number dialledLift the handset and press the last number redial key. The last number dialled is redialled immediately.Dialling from the last number redial listWithout lifting the handset: Press the last number redial key.The last number dialled is displayed. (You can leave the last number redial list at any time by pressing .)Using the Down/Up key, scroll through the list to the desired number.Press OK orkey.Lift the handset if you want to talk via the handset.Dialling using the repertory keysLift handset.Press repertory key.Press repertory key.Press OK orKey.Lift the handset if you want to talk via the handset.3,kt . t o......ATcTTPTSTleSTTTSTleDifferent telephone settings must be completed with the Set key so that they are available permanently.In order to cancel without saving, e. g. after an incorrect entry, replace the handsetinstead of pressing at the end. The original setting is then retained.Saving repertory numbersYou can save up to 10 phone numbers on the repertory keys.Press the Set key.Press the repertory key under which you want to save therepertory number.A number that has already been saved isshown in the display. If the memory for this repertory key isempty, appears in the display.Enter the number to be saved (up to 32 digits).or ...Saving from the calls list (if service is available)Press the Down/Up key until the desired number appearsin the display.or ...Saving from the redial listPress the last number redial key.If necessary, press the Down/Up key until the desirednumber appears in the display.To terminate the process:PresstheSetkey.Saving during a callYou can also save phone numbers in the same way during a call (notebookfunction).Note:After the 20th digit, any following digits are moved from right to left in thedisplay.Delete all repertory numbersInitiatefunction.PresstheSetkey.4..................Adjusting the ringer volumeThe volume of the ringer can be set to one of 4 levels and can also be deactivated completely (default setting: level 4).To adjust when the telephone is idle:Initiatefunction.Adjust the volume with the Loud/Quiet key.PresstheSetkey.To adjust while the telephone is ringing:Press before lifting the handset.The last value set is saved.Setting the ringer frequencyThe frequency (speed) of the ringer can be set to one of threelevels (default setting: level3).Initiate function.Press one of the keys 1...3:1-highest frequency; 3-lowest frequency.Press the Set key.Setting the ringtoneTen different ringtones can be set for the ringer (default setting: ringtone9).To set with the telephone is idle:Initiatefunction.Press one of the keys 0 (9)PresstheSetkey.To set while the telephone is ringing:Press one of the keys 0 to 9.Setting the handset volumeThe volume on the handset can be set to one of 3 levels and saved (default setting: level1).Initiatefunction.Adjust the volume with the Loud/Quiet key.PresstheSetkey.5ess.../ATn p WATm dSIpCC I t dTMuteYou can deactivate the telephone's handset and microphone. Then the other party can no longer hear you. During a call: Press the Mute key Activate: Press the key again.Activating/deactivating the mute melodyYou can activate and deactivate the mute melody for the mute (default setting: on). Initiate function. 0: deactivate; 1: activate. Press the Set key.Adjusting the loudspeakerWhile in handsfree mode, you can set the volume to one of 7 levels and save it (default setting: 1).Adjust the volume with the Loud/Quiet key.Baby call (direct call) number settingIf a phone number is set for the Baby Call function, it is dialled automatically whenthe telephone is locked (s."Activating/deactivating the telephone lock") after you lift the handset (or press the Loudspeaker key) and press anykey (except,,, ). The telephone does not allow other phone numbers to be dialled, but incoming calls can be received.Initiate function. Enter direct call number (max. 32 digits). Press the Set key.When the baby call function is activated, you will see and in the display.DeactivatingInitiate function. Press the Set key.Deleting a baby call phone number Initiate function. Press the Set key.6///...Activating/deactivating the telephone lockThe telephone can be locked completely (with the exception of a baby call phone number if set) so that it cannot dial any phone numbers, including the baby call phone number. You can continue to receive incoming calls. Initiate function. 0: unlock; 1: lock. Press the Set key.When the telephone is locked, you will see in the display.Activate/deactivate call duration displayThe telephone can display the approximate call duration in the display (up to 99 min. 59 sec., default setting: on). If the call duration display is activated, the time display begins 8seconds after the last digit is dialled. Initiate function. 0: deactivate; 1: activate. Press the Set key.Setting the pause timeIf necessary, the pause time of the Pause key (pause function from second input position) can be changed to 1, 3 or 6 seconds (default setting: 3 seconds). Initiate function. 0: 1 sec.; 1: 3 sec.; 2: 6 sec.. Press the Set key.Changing the dialling modeInitiate function. 1: Tone dialling. 2: Pulse dialling without fl ash function. Press the Set key.Changing dialling modes during a callIf your telephone is set to pulse dialling and you want to use functions that require tone dialling (e.g. remote access to an answering machine), you can change the dialling mode during a call. With an open connection: press the star key. Enter digits for remote control/data transferThe telephone is reset to the original dialling mode by hanging up the handset.7ytnu t...UUt I b n r dCT a i "u I a tVI t i8Enter/delete an external line pre fi xeIf you are using your telephone on a PABX, you may have to save one or more external line prefixes. A dialling pause is set automatically with the external line pre fi xe. You can save up to three external line pre fi xes. Initiate function. If no external line pre fi xe has been set in a location,appears in the display. If several external line pre fi xes are saved, these are displayed next to one another; the current external line pre fi xe to be edited fl ashes. If necessary, use the key to toggle between the external line pre fi xes displayed, to edit them as follows: To delete an external line pre fi x: Press the Set key. To enter an external line pre fi xe: Enter a one- to three-digit code. Enter additional external line pre fi xes: Press the Redial key, enter next external line pre fi xe. Press the Set key.Recall key (on a PABX)During an external call, you can make a recall or divert the call. To do this, press the Recall key . The subsequent procedure depends on your PABX. By default, the Recall key is preset to a fl ash time (interruption time) of 100 ms for use of new features on public telephone systems. With a connection to a PABX , it may be necessary to change the fl ash time (e.g. 600 ms). To do this, please refer to the operating instructions for your PABX.Setting the fl ash time for the Recall key Initiate function. Enter desired fl ash code: 0: 100 ms (default setting) ; 2: 270 ms; 1: 120 ms; 3: 375 ms; 4: 600ms. Press the Set key....9Using Calling Line Identi fi cationUnder the following conditions, the phone numbers of incoming calls are shown on the display (up to the fi rst 20 digits) and automatically saved in a calls list:Both your network provider and the caller's network provider must offerCalling Line Identi fi cation.Both you and the caller must have requested this service (if necessary)from your respective network providers.If these conditions are met, but a caller has suppressed Calling Line Identi fi cation before making the call, you will see ---P--- on the display instead of the phone number. If it has not been possible to identify the number for other (technical) reasons, you will see ---0--- or ---E--- on the display. Longer phone (over 20 digits) numbers are shown with the end of the number shortened.Calls ListThe calls list includes up to 50 entries. If the same subscriber calls more than once, a new entry is not created. Up to 99 repeated calls are displayed in the additional information. The most recent call is displayed in the fi rst position (position number "01"). Older entries are moved downwards with each new incoming phone number until they are deleted from the list.In the calls list you can (a)scroll through the phone numbers displayed, (b)query additional information, (c)delete entries, (d)save phone numbers from the calls list to a repertory key(see page 4), (e)dial a displayed phone number directly.Viewing caller histories and dialling phone numbersIf there are entries in the calls list that have you have not yet seen, fl ashes in the display. You can retrieve the calls list, scroll through its entries, view additional information about a call and call back directly: Press the Down key.The phone number of the most recent call appears in the display. Its position number "01" is displayed to the left of the number. If the phone number contains more than 17 digits, the fi rst 17 digits will be displayed fi rst, then the rest after one second. (You can leave the calls list at any timeby pressing .)To scroll through the calls list: Press the Down/Up key. To view additional information:Press key. The date and time of the displayed entry is displayed with the number of calls (on the right).Press key again to back to the calls list.Dial the phone number: Press OK or key. After 2s the number is dialled.Lift the handset if you want to talk via the handset.eels , w e e/...AINLtDdPmPTaRTIC Activating/deactivating display of the area codeYou can set the telephone so that the area code is not displayed for incoming calls.This can be useful, for example, if you cannot see the last digits of a long phone number on the display.You can deactivate the display for 2 area codes (up to 5 digits each).Initiatefunction.If no number has been set in a storage location appearsin the display.If both storage locations are in use, these are displayedside by side; the current storage location to be editedfl ashes. If necessary, use the key to toggle between thestorage locations displayed to edit them as follows:Delete a number:PresstheSetkey.Enter a number:Enteraonetofi ve digit number.Enter another prefi x:Press the last number redial key, enter next prefi x.PresstheSetkey."You have a message" displayDifferent network providers offer services for saving messages (e.g. voice mail services).If one or more newly received messages have been stored with the network provider, it may inform you via your telephone. In the display, this notificationfrom the network provider is displayed with the flashing symbol and the phone number.You can now dial the corresponding phone number in order to check the stored messages. If there are no more new messages for you, the symbol disappears. Deleting entries in the calls listDelete all entries at the same timePresstheDown/Upkey.Press the Set key, star key and Set key one after another.The calls list is deleted and closed.DeleteindividualentriesScroll to the desired entry.Press the Set key twice.Theentryisdeleted.10AppendixIf your telephone is not working perfectlyNo signal tone: The tone volume might be set to 0.Lift handset, no dialling tone: Is the connecting lead correctly plugged into the telephone and the telephone socket?Dialling tone audible but telephone will not dial: The connection is OK. Is the dialling mode set correctly (see page 7)?PABX only: No connection or incorrect connection when dialling from memory (e.g. redial, speed dial):Program external line pre fi xe.The other party cannot hear you: Have you pressed the Mute key? Press the key again. Is the plug from the handset cord inserted correctly?Recall key does not work: Set appropriate fl ash time.The caller's phone number and the calls list are not displayed:Is the adapter connected (see page 2)?Can Calling Line Identi fi cation be used (see page 9)?. esddelkned。

MIC2075-2YM;MIC2025-1YMM;MIC2075-1YM;MIC2025-2YM;MIC2025-1YM;中文规格书,Datasheet资料

MIC2075-2YM;MIC2025-1YMM;MIC2075-1YM;MIC2025-2YM;MIC2025-1YM;中文规格书,Datasheet资料

Single-Channel Power Distribution Switch MM8®General Description The MIC2025 and MIC2075 are high-side MOSFET switches optimized for general-purpose power distribution requiring circuit protection.The MIC2025/75 are internally current limited and have thermal shutdown that protects the device and load. The MIC2075 offers “smart” thermal shutdown that reduces cur-rent consumption in fault modes. When a thermal shutdown fault occurs, the output is latched off until the faulty load is removed. Removing the load or toggling the enable input will reset the device output.Both devices employ soft-start circuitry that minimizes inrush current in applications where highly capacitive loads are em-ployed. A fault status output flag is provided that is asserted during overcurrent and thermal shutdown conditions.The MIC2025/75 is available in the MM8® 8-lead MSOP and 8-lead SOP .Typical ApplicationV CCFeatures• 140mΩ maximum on-resistance • 2.7V to 5.5V operating range• 500mA minimum continuous output current • Short-circuit protection with thermal shutdown• Fault status flag with 3ms filter eliminates false asser -tions• Undervoltage lockout• Reverse current flow blocking (no “body diode”)• Circuit breaker mode (MIC2075) reduces power consumption• Logic-compatible input • Soft-start circuit• Low quiescent current• Pin-compatible with MIC2525•UL File # E179633Applications• USB peripherals• General purpose power switching • ACPI power distribution • Notebook PCs • PDAs•PC card hot swapMicrel, Inc. • 2180 Fortune Drive • San Jose, CA 95131 • USA • tel + 1 (408) 944-0800 • fax + 1 (408) 474-1000 • MM8 is a registered trademark of Micrel, Inc.UL Recognized ComponentPin DescriptionPin NumberPin NamePin Function1 EN Switch Enable (Input): Active-high (-1) or active-low (-2).2FLGFault Flag (Output): Active-low, open-drain output. Indicates overcurrent or thermal shutdown conditions. Overcurrent condition must exceed t D in order to assert FLG. 3 GND Ground4 NC not internally connected5 NC not internally connected6, 8 OUT Supply (Output): Pins must be connected together.7INSupply Voltage (Input).Pin ConfigurationOUT IN OUT NCEN FLG GND NC 8-Lead SOIC (BM) 8-Lead MSOP (BMM)Ordering InformationPart NumberEnableTemperature RangePackageStandard Pb-Free MIC2025-1BM MIC2025-1YM Active High -40°C to +85°C 8-Lead SOIC MIC2025-2BM MIC2025-2YM Active Low -40°C to +85°C 8-Lead SOIC MIC2025-1BMM MIC2025-1YMM Active High -40°C to +85°C 8-Pin MSOP MIC2025-2BMM MIC2025-2YMM Active Low -40°C to +85°C 8-Pin MSOP MIC2075-1BM MIC2075-1YM Active High -40°C to +85°C 8-Lead SOIC MIC2075-2BM MIC2075-2YM Active Low -40°C to +85°C 8-Lead SOIC MIC2075-1BMM MIC2075-1YMM Active High -40°C to +85°C 8-Pin MSOP MIC2075-2BMMMIC2075-2YMMActive Low-40°C to +85°C8-Pin MSOPElectrical CharacteristicsV IN = +5V; T A = 25°C, bold values indicate –40°C ≤ T A ≤ +85°C; unless notedSymbol Parameter Condition Min Typ Max Units I DD Supply Current MIC20x5-1, V EN ≤ 0.8V,(switch off), 0.75 5 µAOUT = openMIC20x5-2, V EN ≥ 2.4V,(switch off), 0.75 5 µAOUT = openMIC20x5-1, V EN ≥ 2.4V,(switch on), 160 µAOUT = openMIC20x5-2, V EN ≤ 0.8V,(switch on), 160 µAOUT = openV EN Enable Input Voltage low-to-high transition 2.1 2.4 Vhigh-to-low transition 0.8 1.9 V Enable Input Hysteresis 200 mV I EN Enable Input Current V EN = 0V to 5.5V –10.01 1 µAControl Input Capacitance 1 pF R DS(on)Switch Resistance V IN = 5V, I OUT = 500mA 90 140 mΩV IN = 3.3V, I OUT = 500mA 100 160 mΩOutput Leakage Current MIC2025/2075 (output off) 10 µAOFF Current in Latched MIC2075 50 µAThermal Shutdown (during thermal shutdown state)t ON Output Turn-On Delay R L = 10Ω, C L = 1µF, see “Timing Diagrams” 1 2.5 6ms t R Output Turn-On Rise Time R L = 10Ω, C L = 1µF, see “Timing Diagrams” 0.5 2.3 5.9ms t OFF Output Turnoff Delay R L = 10Ω, C L = 1µF, see “Timing Diagrams” 50 100 µs t F Output Turnoff Fall Time R L = 10Ω, C L = 1µF, see “Timing Diagrams” 50 100 µs I LIMIT Short-Circuit Output Current V OUT = 0V, enabled into short-circuit. 0.5 0.7 1.25 ACurrent-Limit Threshold ramped load applied to output, Note 40.60 0.85 1.25 AShort-Circuit Response Time V OUT = 0V to I OUT = I LIMIT 24 µs(Short applied to output)t D Overcurrent Flag Response V IN = 5V, apply V OUT = 0V until FLG low 1.5 3 7 ms Delay VIN= 3.3V, apply V OUT = 0V until FLG low 1.5 3 8 ms Undervoltage Lockout V IN rising 2.2 2.5 2.7 VThreshold VIN falling 2.0 2.3 2.5VAbsolute Maximum Ratings (Note 1)Supply Voltage (V IN)..........................................–0.3V to 6V Fault Flag Voltage (V FLG) ..............................................+6V Fault Flag Current (I FLG).............................................25mA Output Voltage (V OUT)...................................................+6V Output Current (I OUT) ...............................Internally Limited Enable Input (I EN).....................................–0.3V to V IN +3V Storage Temperature (T S) ........................–65°C to +150°C ESD Rating, Note 3Operating Ratings (Note 2)Supply Voltage (V IN) ...................................+2.7V to +5.5V Ambient Temperature (T A)..........................–40°C to +85°C Junction Temperature (T J)........................Internally Limited Thermal ResistanceSOP (θJA) ..........................................................160°C/W MSOP(θJA) ........................................................206°C/WTest CircuitV OUTTiming DiagramsVOutput Rise and Fall TimesV VActive-Low Switch Delay Times (MIC20x5-2)V V Active-High Switch Delay Times (MIC20x5-1)Symbol Parameter ConditionMin Typ Max Units Error Flag Output I L = 10mA, V IN = 5V 8 25 Ω ResistanceI L = 10mA, V IN = 3.3V 11 40 Ω Error Flag Off Current V FLAG = 5V 10 µA Overtemperature Threshold T J increasing 140 °CT J decreasing120°CNote 1. Exceeding the absolute maximum rating may damage the device.Note 2. The device is not guaranteed to function outside its operating rating.Note 3. Devices are ESD sensitive. Handling precautions recommended.Note 4. See “Functional Characteristics: Current-Limit Response” graph.C U R R E N T (µA )TEMPERATURE (°C)S upply On-C urrent vs.T emperatureO N -R E S I S T A N C E (m Ω)TEMPERATURE (°C)On-R es is tance vs.T emperatureR E S I S T A N C E (m Ω)INPUT VOLTAGE (V)On-R es is tance vs.Input V oltageR I S E T I M E (m s )TEMPERATURE (°C)T urn-On R is e T ime vs.T emperatureC U R R E N T (µA )INPUT VOLTAGE (V)S upply On-C urrent vs.Input V oltage20040060080010001200C U R R E N T L I M I T T H R E S H O L D (m A )TEMPERATURE (°C)C urrent-L imit T hres holdC U R R E N T L I M I T (m A )TEMPERATURE (°C)S hort-C ircuit C urrent-L imitvs.T emperature1.02.03.04.05.0R I S E T I M E (m s )INPUT VOLTAGE (V)T urn-On R is e T ime vs.Input V oltage100200300400500600700800C U R R E N T L I M I T (m A )INPUT VOLTAGE (V)S hort-C ircuit C urrent-L imit100200300400500600700800900100011001200C U R R E N T L I M I T T H R E S H O L D (m A )INPUT VOLTAGE (V)C urrent-L imit T hres hold0.51.01.52.02.5E N A B L E T H R E S H O L D (V )INPUT VOLTAGE (V)E nable T hres holdE N A B L E T H R E S H O L D (V )TEMPERATURE (°C)E nable T hres hold vs.T emperatureD E L A Y T I M E (m s )TEMPERATURE (°C)F lag Delay vs.T emperatureD E L A Y T I M E (m s )INPUT VOLTAGE (V)F lag Delay vs.Input V oltageU V L O T H R E S H O L D (V )TEMPERATURE (°C)UV L O T hres hold vs.T emperatureFunctional CharacteristicsBlock DiagramENFunctional DescriptionInput and OutputIN is the power supply connection to the logic circuitry and the drain of the output MOSFET. OUT is the source of the output MOSFET. In a typical circuit, current flows from IN to OUT toward the load. If V OUT is greater than V IN, current will flow from OUT to IN since the switch is bidirectional when enabled. The output MOSFET and driver circuitry are also designed to allow the MOSFET source to be externally forced to a higher voltage than the drain (V OUT > V IN) when the switch is disabled. In this situation, the MIC2025/75 avoids undesirable current flow from OUT to IN.Thermal ShutdownThermal shutdown is employed to protect the device from damage should the die temperature exceed safe margins due mainly to short circuit faults. Each channel employs its own thermal sensor. Thermal shutdown shuts off the output MOSFET and asserts the FLG output if the die temperature reaches 140°C. The MIC2025 will automatically reset its output should the die temperature cool down to 120°C. The MIC2025 output and FLG signal will continue to cycle on and off until the device is disabled or the fault is removed. Figure 2 depicts typical timing. If the MIC2075 goes into thermal shutdown, its output will latch off and a pull-up current source is activated. This allows the output latch to automatically reset when the load (such as a USB device) is removed. The output can also be reset by toggling EN. Refer to Figure 1 for details. Depending on PCB layout, package, ambient temperature, etc., it may take several hundred milliseconds from the in-cidence of the fault to the output MOSFET being shut off. The worst-case scenario of thermal shutdown is that of a short-circuit fault and is shown in the in the “Function Char-acteristics: Thermal Shutdown Response” graph.Power DissipationThe device’s junction temperature depends on several fac-tors such as the load, PCB layout, ambient temperature and package type. Equations that can be used to calculate power dissipation of each channel and junction temperature are found below.P D = R DS(on)× I OUT2Total power dissipation of the device will be the summation of P D for both channels. To relate this to junction temperature, the following equation can be used:T J = P D×θJA + T Awhere:T J = junction temperatureT A = ambient temperatureθJA = is the thermal resistance of the package Current Sensing and LimitingThe current-limit threshold is preset internally. The preset level prevents damage to the device and external load but still allows a minimum current of 500mA to be delivered to the load.The current-limit circuit senses a portion of the output MOSFET switch current. The current-sense resistor shown in the block diagram is virtual and has no voltage drop. The reaction to an overcurrent condition varies with three scenarios: Switch Enabled into Short-CircuitIf a switch is enabled into a heavy load or short-circuit, the switch immediately enters into a constant-current mode, reducing the output voltage. The FLG signal is asserted indicating an overcurrent condition. See the Short-Circuit Response graph under Functional Characteristics.Short-Circuit Applied to Enabled OutputWhen a heavy load or short-circuit is applied, a large transient current may flow until the current-limit circuitry responds. Once this occurs the device limits current to less than the short-cir-cuit current limit specification. See the Short-Circuit Transient Response graph under Functional Characteristics.Current-Limit Response—Ramped LoadThe MIC2025/75 current-limit profile exhibits a small foldback effect of about 200mA. Once this current-limit threshold is exceeded the device switches into a constant current mode. It is important to note that the device will supply current until the current-limit threshold is exceeded. See the Current-Limit Response graph under Functional Characteristics.Fault FlagThe FLG signal is an N-channel open-drain MOSFET output. FLG is asserted (active-low) when either an overcurrent or thermal shutdown condition occurs. In the case where an overcurrent condition occurs, FLG will be asserted only after the flag response delay time, t D , has elapsed. This ensures that FLG is asserted only upon valid overcurrent conditions and that erroneous error reporting is eliminated. For example, false overcurrent conditions can occur during hot-plug events when a highly capacitive load is connected and causes a high transient inrush current that exceeds the current-limit threshold. The FLG response delay time t D is typically 3ms.Undervoltage LockoutUndervoltage lockout (UVLO) prevents the output MOS -FET from turning on until V IN exceeds approximately 2.5V. Undervoltage detection functions only when the switch is enabled.V EN VI Load Removed V I I DCFigure 1. MIC2075-2 Timing: Output Reset by Removing LoadV V I V I I Figure 2. MIC2025-2 Timing分销商库存信息:MICRELMIC2075-2YM MIC2025-1YMM MIC2075-1YMMIC2025-2YM MIC2025-1YM MIC2075-2YMM MIC2025-2YMM MIC2075-1YMM MIC2025-1YM TR MIC2025-2YM TR MIC2025-1YMM TR MIC2025-2YMM TR MIC2075-1YM TR MIC2075-2YM TR MIC2075-1YMM TR MIC2075-2YMM TR MIC2025-1BM MIC2025-2BMMIC2025-1BMM MIC2025-1BMM TR MIC2025-1BM TR MIC2025-2BMM MIC2025-2BMM TR MIC2025-2BM TR MIC2075-1BM MIC2075-1BMM MIC2075-1BMM TR MIC2075-1BM TR MIC2075-2BM MIC2075-2BMM MIC2075-2BMM TR MIC2075-2BM TR。

福建省泉州市五校2024-2025学年高三上学期11月期中联考试题 历史(含答案)

福建省泉州市五校2024-2025学年高三上学期11月期中联考试题 历史(含答案)

福建省泉州市2025届高三五校联考历史试卷本试卷满分100分，考试用时75分钟。

2024.11注意事项：1.答题前，考生务必将自己的姓名、考生号、考场号、座位号填写在答题卡上。

2.回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。

如需改动，用橡皮擦干净后，再选涂其他答案标号。

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.考试结束后，将本试卷和答题卡一并交回。

一、单选题1．1886年初，李鸿章在《遵议维持商局疏》中提出：“轮船招商局之设以与外洋轮船相争衡，故此呈请免北洋三口出口税工成……所有本届商局轮船运漕不再区分扣减，亦不扣海运局公费。

”这一主张有益于（）A．保护中国航运业的利权B．遏制外国资本在华输出C．增强清政府的军事实力D．改变政府财政收入结构2．下表所示为19世纪三四十年代欧洲三大工人运动的简况。

该表可用来研究（）时间运动政府的应对结果1831年法国里昂工人武装起义军事镇压失败1836—1848年英国宪章运动拒绝接受请愿书，解散全国宪章派协会失败1844年德意志西里西亚织工起义军事镇压失败A．资产阶级的妥协性B．民主革命方式的多样性C．工人运动的联动性D．科学理论诞生的必要性3．法老宣称自己是太阳神之子，拨巨款修筑神庙，赋予神职人员公职。

法老之下地位最高的官员往往由大祭司担任。

这表明古代埃及（）A．神权与王权关系密切B．人民对神灵的崇拜C．法老重视来世的生活D．实行君主专制统治4．如表为东汉后期和唐朝前期黄河流域、长江流域县城数量表（单位：座），此表反映的变化最能说明（）时间区域东汉后期唐朝前期黄河流域765669长江流域342611A．长江流域已占据经济优势B．唐朝政治制度比东汉完善C．南北方经济差距逐渐缩小D．唐朝统治疆域在不断扩大5．万里茶道是17世纪末至20世纪初，继丝绸之路之后在欧亚大陆兴起的又一条重要国际商道。

它南起中国福建武夷山，延伸至俄罗斯圣彼得堡，全长1.3万公里，是历史上跨越陆地距离最长的商贸通道。

2025年人教版高考英语试题与参考答案

2025年人教版高考英语试题与参考答案

2025年人教版英语高考复习试题与参考答案一、听力第一节（本大题有5小题，每小题1.5分，共7.5分）1、What is the relationship between the two speakers?A) Teacher and studentB) Doctor and patientC) Shop assistant and customerD) Manager and employeeAnswer: C) Shop assistant and customerExplanation: In the conversation, one speaker asks about the size and color options available for a particular item of clothing, indicating that they are shopping. The other speaker responds with details about the stock and offers assistance, which is char acteristic of a shop assistant’s role.2、Where does the conversation most likely take place?A) At a hospitalB) In a classroomC) At a clothing storeD) In an officeAnswer: C) At a clothing storeExplanation: Given the context of the conversation, with one party inquiring about product details and the other providing information typical of salesinteractions, it is reasonable to conclude that the conversation takes place at a clothing store.3.You will hear a conversation between two friends discussing their weekend plans. Listen and choose the best answer to the question.Question: What does the woman plan to do this weekend?A. Go to the beach.B. Visit her parents.C. Have a picnic with her friends.Answer: B. Visit her parents.Explanation: In the conversation, the woman mentions that she is going to visit her parents this weekend. The other options are not mentioned in the conversation.4.You will hear a short lecture about the importance of exercise. Listen and answer the following question.Question: According to the lecture, what is one of the main benefits of regular exercise?A. Improved concentration.B. Faster metabolism.C. Better sleep.Answer: A. Improved concentration.Explanation: The lecture emphasizes that regular exercise can help improve concentration and mental alertness. The other options are mentioned as benefitsof exercise but not as the main one in this context.5、What time does the train leave?A. 7:00 a.m.B. 8:00 a.m.C. 9:00 a.m.Answer: B. 8:00 a.m.Explanation: In the conversation, the man asks about the departure time of the train, and the woman replies that it leaves at 8 o’clock in the morning. The key information is often found in the response to the inquiry, and in this case, the listener must pay attention to the specific time mentioned.二、听力第二节（本大题有15小题，每小题1.5分，共22.5分）1、A. What are the speakers mainly discussing?B. Why did the woman feel sad last night?C. What does the man suggest doing next?D. How do the speakers know each other?Answer: CExplanation: The man suggests going out for dinner to celebrate, which indicates that they know each other well enough to engage in casual conversation abouta celebration.2、A. What is the main purpose of the call?B. Why did the man miss the appointment?C. What does the woman propose to do next?D. How does the woman feel about the situation?Answer: AExplanation: The woman asks the man if he called to cancel the appointment, which suggests that the main purpose of the call is to discuss the appointment. The man’s response to the question confirms this.3、What does the woman imply about the apartment?A) It’s overpriced.B) It’s very spacious.C) It’s in high demand.D) It’s poorly maintained.Answer: C) It’s in high demand.Explanation: The woman mentions that several people are interested in the apartment and that it might go quickly, indicating that the apartment is popular among potential renters.Question 44、Why does the man want to see the apartment as soon as possible?A) He needs a place to live urgently.B) He is worried about the rent being too high.C) He is concerned about the condition of the apartment.D) He wants to negotiate the terms of the lease.Answer: A) He needs a place to live urgently.Explanation: The man says he needs the apartme nt “as soon as possible,” which suggests that he is in urgent need of a place to live.Please remember that real exam questions would be accompanied by audio recordings, and the conversations would be more complex. The above examples are for illustrative purposes only.5.You will hear a conversation between two students discussing their weekend plans. Listen and answer the question.Question: What activity are the students planning to do together on Sunday afternoon?A) Go to the cinema.B) Have a picnic in the park.C) Visit an art gallery.Answer: B) Have a picnic in the park.Explanation: In the conversation, the students mention that they want to go for a picnic in the park, which suggests that option B is the correct answer.6.You will hear a monologue about the importance of exercise for students. Listen and answer the question.Question: According to the speaker, what is the main benefit of regular exercise for students?A) Improved academic performance.B) Better concentration.C) More energy.Answer: B) Better concentration.Explanation: The speaker emphasizes that regular exercise helps students to concentrate better in their studies, which makes option B the correct answer.7、What does the woman suggest the man do regarding the post office visit?A. Go there immediately.B. Avoid going there after 6 PM.C. Visit after 4 PM to avoid the crowd.D. Take a bus to get there faster.Answer: C. Visit after 4 PM to avoid the crowd.Explanation: The woman suggests visiting the post office after 4 PM because it is less crowded then. She mentions that it’s usually very crowded before 4 PM, implying that it would be better to go after that time.8、When and where are Tom and Lisa planning to meet before their shopping trip?A. At 3 o’clock at the usual café.B. At 3 o’clock at the store.C. At 4 o’clock at the usual café.D. At 4 o’clock at the store.Answer: A. At 3 o’clock at the usual café.Explanation: Tom and Lisa agree to meet at 3 o’clock, and Lisa confirms that she will meet him at the usual café before they start shopping. The meeting place is clearly stated as the café, not the store, and the time is set for 3o’clock.This sample provides a typical format for the listening comprehension part of the English Gaokao, including practical dialogues and relatedmultiple-choice questions.9.You will hear a conversation between two students, Tom and Lisa, discussing their weekend plans. Listen and choose the best answer to the question you will hear.Question: What does Tom decide to do on Sunday?A. He will go to the gym.B. He will visit his grandparents.C. He will go hiking.Answer: BExplanation: In the conversation, Lisa mentions that her grandparents are visiting and Tom responds, “Oh, cool! I’ll join you guys on Sunday.”10.You will hear a short talk about the benefits of exercise. Listen and answer the following question.Question: According to the talk, which of the following is NOT a benefit of regular exercise?A. Improved moodB. Better sleepC. Increased risk of injuryAnswer: CExplanation: The speaker discusses how regular exercise can improve mood, enhance sleep, and boost overall health. However, they do not mention an increased risk of injury as a benefit.11、What is the main topic of the conversation?A. The benefits of morning exercise.B. How to improve one’s diet.C. A discussion about popular sports.D. The importance of a good night’s sleep.Correct Answer: A.Explanation: The speakers mainly discuss the advantages of exercising in the morning, such as improved metabolism and better mood throughout the day. They do not focus on diet improvement, sports popularity, or sleeping habits.12、According to the man, what is an additional benefit of morning workouts besides physical health?A. Enhanced social skills.B. Increased productivity at work.C. Better academic performance.D. Improved cooking abilities.Correct Answer: B.Explanation: During the conversation, the man mentions that he finds himself more productive during his workday when he has exercised in the morning. There is no mention of social skills enhancement, academic performance, or cookingabilities in the context of morning workouts.End of Passage 1Please prepare to listen to Passage 2.Please remember this is a simulated example designed for illustrative purposes only. Real exam questions would come directly from the curriculum and reflect actual topics covered in class.13.You will hear a conversation between two friends discussing their weekend plans. Listen to the conversation and answer the question.Question: What activity do the friends decide to do together this weekend?A) Go hikingB) Visit a museumC) Go swimmingD) Have a picnicAnswer: B) Visit a museumExplanation: In the conversation, the friends talk about their plans for the weekend. They mention that they want to go somewhere educational and fun, which leads them to decide on visiting a museum.14.You will hear a short lecture about the importance of exercise. Listen to the lecture and answer the question.Question: According to the lecture, what is the main benefit of regular exercise?A) Improved memoryB) Better sleepC) Increased productivityD) Lower stress levelsAnswer: D) Lower stress levelsExplanation: In the lecture, the speaker discusses the various benefits of exercise. Although all the options are mentioned as positive outcomes of regular exercise, the main focus is on how it can help lower stress levels, making it the correct answer.15.How much money does the man plan to spend on the new laptop?A.$800.B.$1200.C.$1600.D.$2000.Answer: C.$1600.Explanation: In the conversation, the man mentions that he has saved$800 for the laptop and he needs to borrow another$800 from his friend. Therefore, the total amount he plans to spend is$1600.三、阅读第一节（第1题7.5分，其余每题10分，总37.5分）第一题Read the following passage and answer the questions that follow.In the heart of bustling New York City, there’s a small, quaint bookstorethat has been a staple in the neighborhood for over a century. Known as “The Corner Bookstore,” it is a sanctuary for book lovers, a place where the scent of aged paper and ink intertwines with the sounds of rustling pages and gentle laughter.The store, which occupies a corner of a narrow street, is lined with wooden bookshelves filled with an eclectic mix of books ranging from classic literature to contemporary fiction, poetry, and non-fiction. The owner, Mrs. Thompson, who is in her late sixties, has been running the store for the past forty years. Her passion for literature is evident in every nook and cranny of the shop.One sunny afternoon, a young woman named Emily strolled into the bookstore. She had heard about The Corner Bookstore from a friend and was eager to explore its contents. As she walked through the aisles, she couldn’t help but marvel at the store’s charm and the sense of community that seemed to emanate from every corner.1、What is the primary purpose of the passage?A. To introduce the owner of The Corner BookstoreB. To describe the unique atmosphere of The Corner BookstoreC. To discuss the popularity of bookstores in New York CityD. To explain the history of The Corner Bookstore2、What is the age of Mrs. Thompson?A. 50 years oldB. 60 years oldC. 65 years oldD. 70 years old3、What type of books can be found in The Corner Bookstore?A. Only contemporary fictionB. Only classic literatureC. A mix of classic literature, contemporary fiction, poetry, andnon-fictionD. Only poetry4、Why did Emily visit The Corner Bookstore?A. To buy a new bookB. To meet Mrs. ThompsonC. To experience the unique atmosphereD. To find a specific bookAnswers:1、B2、C3、C4、CQuestion 2:Read the following passage and answer the questions that follow.Passage:In recent years, the Internet has become an integral part of our daily lives.It has revolutionized the way we communicate, access information, and conduct business. However, along with its numerous benefits, the Internet also brings about several challenges.One of the most significant challenges is the issue of online security. With the increasing number of cyber attacks and data breaches, it has become crucial for individuals and organizations to take necessary precautions to protect their personal and sensitive information. This includes using strong passwords, keeping software up to date, and being cautious of phishing scams.Another challenge is the issue of online addiction. Many people spend excessive amounts of time online, which can lead to a range of health problems, including sleep disorders, depression, and anxiety. It is important for individuals to maintain a healthy balance between their online and offline lives.Furthermore, the Internet has given rise to a phenomenon known as “information overload.” With the vast amount of information available online, it can be difficult for individuals to discern what is accurate and reliable. This can lead to misinformation and confusion.Despite these challenges, the Internet continues to be a powerful tool that can greatly enhance our lives. It is essential for us to navigate it responsibly and make the most of its benefits while mitigating its risks.Questions:1、What is the main purpose of the passage?A. To discuss the benefits of the Internet.B. To highlight the challenges of the Internet.C. To argue against the use of the Internet.D. To present a balanced view of the Internet.2、Which of the following is NOT mentioned as a challenge of the Internet?A. Online security issues.B. Health problems due to excessive internet use.C. The availability of free information.D. Misinformation and confusion.3、According to the passage, what is the solution to the problem of online addiction?A. To use the Internet less frequently.B. To maintain a healthy balance between online and offline lives.C. To rely on technology to control internet use.D. To avoid using the Internet altogether.4、The passage suggests that how should individuals navigate the Internet?A. By avoiding it altogether.B. By using it without any precautions.C. By using it responsibly and making the most of its benefits while mitigating its risks.D. By focusing only on the positive aspects of the Internet.Answers:1、B2、C3、B4、C第三题Reading Section ARead the following passage carefully and then answer the questions below.The rise of online shopping has transformed the retail industry. No longer are consumers confined to purchasing goods from physical stores. The convenience and wide variety of products available online have made it a popular choice for many. However, this shift has also brought about several challenges for retailers and consumers alike.One of the major challenges is the issue of counterfeit goods. With the ease of online shopping, it has become increasingly difficult for consumers to distinguish between genuine and fake products. This has led to a significant loss in revenue for both retailers and manufacturers. In response, many companies have started implementing stricter quality control measures and using advanced technology to detect counterfeit items.Another challenge is the environmental impact of online shopping. The massive increase in packaging waste and carbon emissions from deliveries have raised concerns about sustainability. Retailers are now looking for ways to reduce their carbon footprint by offering more sustainable packaging optionsand optimizing delivery routes.Despite these challenges, online shopping continues to grow. One of the reasons for its popularity is the convenience it offers. Consumers can shop from the comfort of their own homes at any time of the day or night. Additionally, online retailers often provide better prices and a wider selection of products compared to brick-and-mortar stores.1、What is the main topic of the passage?A. The benefits of online shoppingB. The challenges of online shoppingC. The history of online shoppingD. The impact of online shopping on the retail industry2、What is a significant challenge mentioned in the passage regarding online shopping?A. The high cost of shippingB. The difficulty in distinguishing between genuine and fake productsC. The lack of customer serviceD. The limited variety of products3、How are companies addressing the issue of counterfeit goods?A. By increasing the price of genuine productsB. By implementing stricter quality control measuresC. By reducing the number of online retailersD. By banning online shopping altogether4、What is one of the reasons for the popularity of online shopping mentioned in the passage?A. The high cost of goodsB. The difficulty in finding parking in physical storesC. The convenience it offersD. The limited availability of productsAnswers:1、B2、B3、B4、C第四题Reading PassageIn the small coastal town of Gull’s Bay, there is a legend that has been passed down through generations. The legend speaks of a hidden treasure buried deep in the sands of the beach. Many adventurers have come to the town in search of this treasure, but none have succeeded. The legend is said to be true, and the treasure is believed to be worth millions of dollars.The story goes like this: A long time ago, a wealthy merchant named Richard sailed from his hometown to Gull’s Bay with a treasure chest filled with gold, jewels, and precious stones. He planned to bury the chest on the beach, but was chased by pirates. In a desperate attempt to escape, Richard buried the chestunder a large rock and ran away, leaving a clue for future generations to find it.The clue was a riddle that reads: “I lie hidden beneath the sand, where the waves come and go. Look to the east and the north, where the sun does not stand. Underneath the rock, you will find, the treasure that I’ve given.”The legend says that the treasure is buried in the exact spot where the sun is not standing, which means it is located at the point where the sun rises and sets at the same time during the summer solstice. This is a very specific spot, and it is said to be near the lighthouse.Questions:1、What is the legend of Gull’s Bay about?A) A treasure buried on the beach.B) A pirate shipwreck.C) A mysterious lighthouse.D) An ancient temple.2、Why did Richard hide the treasure?A) To protect it from the government.B) To hide it from his enemies.C) To leave it for future generations.D) To use it to start a new business.3、What is the clue to finding the treasure?A) A map showing the exact location.B) A riddle that reveals the spot.C) A message left by a previous adventurer.D) A hidden symbol on the beach.4、Where is the treasure believed to be located?A) Near the lighthouse.B) Under the beach sand.C) In the ocean waves.D) On top of the lighthouse.Answers:1、A) A treasure buried on the beach.2、B) To hide it from his enemies.3、B) A riddle that reveals the spot.4、A) Near the lighthouse.四、阅读第二节（12.5分）Reading SectionPassage:The following is an excerpt from a travel article about a famous tourist attraction in China.The Forbidden City, located in the heart of Beijing, was the imperial palace of the Ming and Qing dynasties. It is one of the most well-preserved ancient palaces in the world and is now a museum that showcases the rich history andculture of China. The palace covers an area of 720,000 square meters and has over 870 buildings, including halls, pavilions, and gardens. The walls of the Forbidden City are 860 meters long, and the moat surrounding it is 52 meters wide. The complex was built in 1406 and took 14 years to complete. The Forbidden City is a symbol of the power and wealth of the Chinese emperors and is a must-visit destination for any traveler to China.Questions:1.What is the Forbidden City?A. A modern museum in BeijingB. The imperial palace of the Ming and Qing dynastiesC. A famous restaurant in ChinaD. A popular shopping center in Beijing2.What is the main purpose of the Forbidden City now?A. To serve as the official residence of the Chinese presidentB. To be used as a military baseC. To house the emperors of the Ming and Qing dynastiesD. To be a museum showcasing the rich history and culture of China3.How large is the Forbidden City?A. 720,000 square metersB. 860 meters longC. 52 meters wideD. 14 years to complete4.Who built the Forbidden City?A. The Ming and Qing dynastiesB. The Chinese peopleC. The tourists who visit itD. The United Nations5.What is the Forbidden City a symbol of?A. The beauty of Chinese architectureB. The power and wealth of the Chinese emperorsC. The love of the Chinese people for their historyD. The importance of education in ChinaAnswers:1.B2.D3.A4.A5.B五、语言运用第一节 _ 完形填空（15分）Title: A Day at the ZooThe zoo was a place full of excitement and adventure. It was a sunny Saturday morning, and the park was bustling with visitors. The animals seemed particularly lively that day, and the children were running around in awe oftheir new surroundings.As we entered the zoo, the first exhibit we came across was the big cats. The majestic lions lay in the shade, their golden fur glistening in the sunlight. The cubs, however, were full of energy and played with each other, leaping and chasing.1.The zoo was a place full of excitement and ___________.A. surpriseB. adventureC. sadnessD. boredomWe then moved on to the birdhouse. The colorful birds were perched on the branches, chirping and singing their songs. The peacocks spread their beautiful tails, displaying their vibrant colors for the onlookers.2.The birds in the birdhouse were especially___________that day.A. lazyB. activeC. quietD. shyAs we continued our journey, we reached the reptile house. The snakes slithered gracefully along the ground, and the crocodiles lay still, watching us with their cold, calculating eyes. The children, however, were more interested in the tortoises, who moved at a much slower pace.3.The reptiles in the reptile house were known fortheir___________movements.A. quickB. slowC. energeticD. loudNext, we visited the primates. The apes were playful and intelligent, and the monkeys were acrobats, swinging from branch to branch with ease. The children were amazed by the agility of the primates and enjoyed watching them interact with each other.4.The primates at the zoo were known for their___________behavior.A. calmB. aggressiveC. playfulD. silentAfter a while, we decided to take a break and have some lunch. The zoo provided a picnic area where families could enjoy a meal while watching the animals. We sat under a large tree, eating our sandwiches and watching the zebras graze peacefully nearby.5.The picnic area in the zoo was perfect for visitors to ___________.A. play gamesB. watch animalsC. take photosD. have a mealAs the day came to an end, we made our way back to the entrance. The zoo had been a memorable experience, and we couldn’t wait to return. The animals had taught us about their behaviors and habits, and the children had gained a deeper appreciation for wildlife.6.Our visit to the zoo was a___________experience for us all.A. boringB. thrillingC. frustratingD. forgettableAnswer Key:1.B. adventure2.B. active3.B. slow4.C. playful5.D. have a meal6.B. thrilling六、语言运用第二节 _ 语法填空（15分）Read the following passage. For each blank, choose the best answer from the four choices given.One of the most fascinating aspects of the English language is its ability to evolve and adapt. Over the centuries, it has absorbed words from many different languages, including Latin, Greek, French, and German. This process is known as 1.. It’s not just words that change, 2.. Grammar and pronunciation also adapt to the needs of the speakers. For example, the word “neighbor” comes from Old English, but its pronunciation has changed over time. Similarly, the verb “to do” has several different past forms, such as “did,” “done,” and “done.” This 3. reflects the language’s flexibility and its responsiveness to the way people actually speak.One of the most interesting examples of language change is the 4. of the word “OK.” It originated in the United States in the 1830s and was created as an abbreviation for “oll korrect.” Over time, it has become a universally recognized term of approval. The 5. of this word demonstrates how a simple abbreviation can gain widespread acceptance and become an integral part of the language.Language change is also evident in the way we use 6.. For instance, the word “email” used to be written as “e-mail,” but now it is commonly written without the hyphen. This change reflects the evolving conventions of written English.The 7. of language change can be attributed to several factors. One of the main reasons is the 8. of people and ideas. As societies become more interconnected, languages naturally borrow words and phrases from one another.Another factor is the 9. of technology, which allows for the rapid spread of new words and phrases. Lastly, the 10. of language itself is another reason for its continuous evolution.Choose the best answer for each blank:1.A) borrowingB) adaptationC) evolutionD) transformation2.A) its structureB) its meaningC) its originD) its pronunciation3.A) diversityB) consistencyC) simplicityD) complexity4.A) originB) pronunciationC) usageD) spelling5.A) acceptanceB) rejectionC) evolutionD) pronunciation6.A) nounsB) verbsC) adjectivesD) pronouns7.A) processB) impactC) reasonD) challenge8.A) movementB) exchangeC) influenceD) development9.A) innovationB) improvementC) expansionD) advancement10.A) adaptabilityB) reliabilityC) stabilityD) flexibilityAnswer key:1.C) evolution2.A) its structure3.A) diversity4.A) origin5.A) acceptance6.C) adjectives7.A) process8.B) exchange9.D) advancement10.A) adaptability七、写作第一节 _ 应用文写作（15分）写作题目：You have recently returned from a short trip to a famous tourist destination. Write a letter to a friend who is planning to visit the same place soon. In your letter, include the following points:1.Briefly describe your visit.2.Share your favorite experience or place.3.Offer suggestions for activities or places to visit.4.Mention any tips or advice you have for them.Writing Example:Dear [Friend’s Name],I hope this letter finds you well! I just wanted to share with you my recent trip to [Tourist Destination], which was absolutely incredible. I’m sure you’ll be excited to hear about it, especially since you’re planning your own trip there soon.I arrived in [Tourist Destination] on a sunny Saturday morning. The city was bustling with activity, and the scenery was breathtaking. My favorite experience was visiting the [Famous Landmark or Attraction], which is a must-see for anyone visiting the area. The architecture was stunning, and the views from the top were just phenomenal.I spent the afternoon exploring the nearby [Local Market or Park], where I tried some delicious local cuisine and bought some unique souvenirs. One of the highlights was trying the [Specific Food or Drink], which was both tasty and a cultural experience in itself.For activities, I would highly recommend taking a hike up [Mountain or Hill], as the trail offers some of the most spectacular views of the city and surrounding landscape. If you’re interested in history, the [Historical Museum or Site] is a great place to learn about the area’s past.Here are a few tips I’ve gathered:1.Don’t miss out on the [Specific Event or Festival] if it’s happening during your visit; it’sa fantastic way to experience the local culture.2.Always carry a map or have a GPS on hand, as some areas can be quite hilly and the streets can be confusing.3.Stay hydrated and wear comfortable shoes, as you’ll be doing a lot of。

广东省揭阳市2024高三冲刺(高考物理)人教版考试(冲刺卷)完整试卷

广东省揭阳市2024高三冲刺(高考物理)人教版考试(冲刺卷)完整试卷

广东省揭阳市2024高三冲刺（高考物理）人教版考试(冲刺卷)完整试卷一、单项选择题（本题包含8小题，每小题4分，共32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）(共8题)第(1)题将一小球以的初速度水平抛出，抛出后3s小球并未落地，空气阻力忽略不计，则第1s内、第2s内、第3s内小球的重力的功率的比值为（ ）A．1：2：3B．1：3：5C．D．第(2)题如图，光滑水平面上三个光滑金属立柱a、b、c固定，a、b、c两两间距均为L，a、c和b、c分别用一拉直金属线连接，a、b间连接长度为的轻质柔软细金属丝，初始细金属丝处于松弛状态，现在空间中加一竖直向上的磁场，磁感应强度B随时间t变化的规律为，随着磁场的变化，细金属丝逐渐绷紧，已知当时，a、b间柔软金属丝组成的几何图形形状保持不变，设整个回路的总电阻为R，其中。

则时，柔软金属丝所受安培力大小为（）A．B．C．D．第(3)题州一中运动会中，某同学立定跳远脚蹬地起跳瞬间的受力示意图正确的是（）A．B．C．D．第(4)题如图所示，分别用频率为v、2v的光照射某光电管，对应的遏止电压之比为1：3，普朗克常量用h表示，则（）A ．用频率为的光照射该光电管时有光电子逸出B．该光电管的逸出功为C．用2v的光照射时逸出所有光电子的初动能比用v的光照射时逸出所有光电子的初动能大D．加正向电压时，用2v的光照射时饱和光电流一定大第(5)题过去几千年中，人类对行星的认识与研究仅限于太阳系内，行星“51Pegb”的发现拉开了研究太阳系外行星的序幕。

“51Pegb”绕其中心恒星做匀速圆周运动，周期约为4天，轨道半径约为地球绕太阳运动半径的，已知太阳的质量约为，则该中心恒星的质量约为（）A．B．C．D．第(6)题如图所示，在水平线某竖直平面内，距地面高度为，一条长为的轻绳两端分别系小球A和B，小球在水平线上，竖直向上的外力作用在A上，A和B都处于静止状态。

现从上另一点静止释放小球1，当小球1下落至与小球B等高位置时，从上静止释放小球A和小球2，小球2在小球1的正上方。

山东省泰安市泰安实验中学2025届高考数学五模试卷含解析

山东省泰安市泰安实验中学2025届高考数学五模试卷含解析

山东省泰安市泰安实验中学2025届高考数学五模试卷注意事项：1．答卷前，考生务必将自己的姓名、准考证号填写在答题卡上。

2．回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其它答案标号。

回答非选择题时，将答案写在答题卡上，写在本试卷上无效。

3．考试结束后，将本试卷和答题卡一并交回。

一、选择题：本题共12小题，每小题5分，共60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．若不等式22ln x x x ax -+对[1,)x ∈+∞恒成立，则实数a 的取值范围是（ ）A ．(,0)-∞B ．(,1]-∞C ．(0,)+∞D ．[1,)+∞2．已知集合2{|log (1)2},,A x x B N =-<=则A B =（ ）A ．{}2345，，，B ．{}234，，C ．{}1234，，， D ．{}01234，，，， 3．已知等差数列{}n a 的前13项和为52，则68(2)a a +-=（ ） A ．256 B ．-256 C ．32D ．-32 4．某几何体的三视图如图所示，其中正视图是边长为4的正三角形，俯视图是由边长为4的正三角形和一个半圆构成，则该几何体的体积为（ ）A ．438π+B ．238π+C ．434π+D ．834π+ 5．若i 为虚数单位，则复数22sincos 33z i ππ=-+的共轭复数z 在复平面内对应的点位于（ ） A ．第一象限 B ．第二象限 C ．第三象限 D ．第四象限6．若双曲线22214x y a -=3，则双曲线的焦距为（ ） A ．26B ．25C ．6 D ．87．已知复数z 满足()14i z i -=，则z =（ ）A ．22B ．2C ．4D ．38．已知集合M ＝{y ｜y ＝，x ＞0}，N ＝{x ｜y ＝lg （2x －）}，则M∩N 为（ ）A ．（1，＋∞）B ．（1，2）C ．[2，＋∞）D ．[1，＋∞）9．已知函数()2x f x x x ln a ⎛⎫=- ⎪⎝⎭，关于x 的方程f （x ）＝a 存在四个不同实数根，则实数a 的取值范围是（ ） A ．（0，1）∪（1，e ） B ．10e ⎛⎫ ⎪⎝⎭， C ．11e ⎛⎫ ⎪⎝⎭， D ．（0，1）10．已知0x =是函数()(tan )f x x ax x =-的极大值点，则a 的取值范围是A ．(,1)-∞-B ．(,1]-∞C ．[0,)+∞D ．[1,)+∞11．已知实数x ，y 满足约束条件202201x y x y x +-≤⎧⎪--≤⎨⎪≥⎩，则目标函数21y z x -=+的最小值为 A ．23-B ．54-C ．43-D ．12- 12．若x yi +(,)x y ∈R 与31i i +-互为共轭复数，则x y +=（ ） A ．0 B ．3C ．－1D ．4 二、填空题：本题共4小题，每小题5分，共20分。

2025届吉林省长春市第一五〇中学高三冲刺模拟英语试卷含解析

2025届吉林省长春市第一五〇中学高三冲刺模拟英语试卷含解析

2025届吉林省长春市第一五〇中学高三冲刺模拟英语试卷考生请注意：1．答题前请将考场、试室号、座位号、考生号、姓名写在试卷密封线内，不得在试卷上作任何标记。

2．第一部分选择题每小题选出答案后，需将答案写在试卷指定的括号内，第二部分非选择题答案写在试卷题目指定的位置上。

3．考生必须保证答题卡的整洁。

考试结束后，请将本试卷和答题卡一并交回。

第一部分（共20小题，每小题1.5分，满分30分）1．—Let’s have a game of tennis; the loser has to trea t the other to an ice-cream.— ________.A．I’m afraid so B．It’s a deal C．You’ve got a point D．I suppose not2．Jane’s pale face suggested that she ______ ill and her parents suggested that she ______ a medical exam.A．be; should have B．was; haveC．should be; had D．was; had3．—What does the sign over there read?—"No person_______ smoke or carry a lighted cigarette, cigar or pipe in this area．"A．shall B．may C．must D．Will4．You’d better write down the phone number of that restaurant for future ________.A．purpose B．reference C．memory D．assessment5．----Can I park my car here?----Sure not, we don’t allow here.A．to parking B．park C．to park D．parking6．I was caught in a traffic jam for over an hour, otherwise, I ________ you waiting for such a long time.A．had not kept B．will not keepC．would not have kept D．have not kept7．I certainly expect to be elected best student of the year. It’s really ________!A．a good Samaritan B．a wet blanketC．a feather in my cap D．a piece of cake8．. The study of natural history is not something to be left to biologists. In fact, their capacity _____ the time they can spend away from their offices is very limited.A．in terms of B．in search ofC．in view of D．in spite of9．She did not feel a bit nervous though it was the first time she ___________ in public.A．spoke B．have spokenC．had spoken D．were speaking10．Most of the money for the reconstruction of the quake-stricken town has been allocated by the government, the rest______ from the coming charity concerts.A．to be collected B．having been collectedC．being collected D．to have been collected11．—What’s that noise?—Oh，I forget to tell you.The new machine________.A．is testing B．was being testedC．is being tested D．has been tested12．It's said that the power plant is now large as what it was.A．twice as B．as twiceC．twice much D．much twice13．________ online payment is safe, people will be more likely to link their bank cards to WeChat.A．Even though B．As thoughC．Ever since D．As long as14．It is usually thought a little child says is truth.A．What that B．that what C．which D．that15．—The terrible accident is under investigation.—Actually, quicker action _________ those workers trapped in the mine.A．might have saved B．must have saved C．should have saved D．could have saved16．If it _____ earlier, the printing machine would not have broken down．A．has been repaired B．is repairedC．had been repaired D．was repaired17．--- Where is your new home now?--- In the new developed zone. But I ______ downtown for five years.A．lived B．had livedC．have lived D．was living18．Please remind your grandpa to take medicine on time, for a man of his age ____be very forgetful．A．need B．must C．shall D．can19．The suggestion came from the chairman ______ the new rule ______.A．what; was developed B．that; was developedC．what; be developed D．that; be developed20．__________him not to do so, he wouldn’t have made such a serious mistake.A．Did I persuade B．If I persuadeC．If I should persuade D．Had I persuaded第二部分阅读理解（满分40分）阅读下列短文，从每题所给的A、B、C、D四个选项中，选出最佳选项。

重载铁路再生制动能量利用方案研究

重载铁路再生制动能量利用方案研究

第16卷第1期2021年3月电气工程学报Vol.16 No.1Mar. 2021DOI：10.11985/2021.01.020重载铁路再生制动能量利用方案研究刘华伟1耿安琪2何正友2胡海涛2张宏伟2(1. 神华包神铁路集团有限责任公司包头014010；2. 西南交通大学电气工程学院成都611756)摘要：重载铁路运输作为铁路的重要发展方向之一，具有效率高、成本低且运能大的特点。

近年来，我国重载铁路运能不断提高的同时，也使得能源消耗问题日益凸显。

针对如何实现重载铁路的节能降耗，提出了一种基于混合储能的再生制动能量利用方案，通过利用蓄电池和超级电容器在性能上的互补性，实现混合储能系统对重载铁路再生制动能量的高效利用。

结合神朔铁路的实测数据，对该条线路的负荷情况进行了分析，并针对混合储能系统设计了有效的能量管理策略，最后在实测数据的基础上对系统的经济性进行了评估。

分析结果验证了提出能量管理策略的有效性，以及再生制动能量利用方案具有很好的经济性。

关键词：重载铁路；再生制动能量；混合储能；经济性中图分类号：TM711Research on Energy Utilization Scheme of Regenerative Braking forHeavy Haul RailwayLIU Huawei1GENG Anqi2HE Zhengyou2HU Haitao2ZHANG Hongwei2(1. Shenhua Baoshen Railway Group Co. Ltd., Baotou 014010;2. School of Electrical Engineering, Southwest Jiaotong University, Chengdu 611756)Abstract：As one of the important development directions of railroad, heavy-duty railroad transportation has the characteristics of high efficiency, low cost and large capacity. In recent years, while China heavy-duty railroad capacity has been increasing, it also makes the problem of energy consumption increasingly prominent. A hybrid energy storage based regenerative braking energy utilization scheme is proposed to realize the efficient utilization of regenerative braking energy for heavy-duty railroads by using the complementary performance of storage battery and supercapacitor. The load conditions of the line are analyzed with the measured data of the Shenshuo railroad, and an effective energy management strategy is designed for the hybrid energy storage system, and finally the economics of the system is evaluated based on the measured data. The analysis results verify the effectiveness of the proposed energy management strategy and the good economics of the regenerative braking energy utilization scheme.Key words：Heavy-haul railway；regenerative braking energy；hybrid energy storage system；economy1 引言2019年，全国铁路货运总发送量完成43.98亿吨，增长7.2%[1]。

CONTROL DEVICE FOR VEHICLE

CONTROL DEVICE FOR VEHICLE

专利名称：CONTROL DEVICE FOR VEHICLE发明人：IGARASHI MASA,五十嵐 政申请号：JP2009144288申请日：20090617公开号：JP2011000931A公开日：20110106专利内容由知识产权出版社提供专利附图：摘要：PROBLEM TO BE SOLVED: To provide a control device for a vehicle, which has a simple structure and does not require complicate control content, and can improve a braking force or driving force while applicable tires are not limited to keep theversatilely.SOLUTION: A control device 1 for the vehicle includes a classification detecting means 2a for detecting the classification of a tire which constitute a wheel, a camberangle setting means 2b for setting a predetermined angle range of a wheel camber angle where the road-contacting shape of a tread included in the tire is maintained in a trapezoid shape, so as to be associated with the classification, and an instructiondetecting means 2c for detecting a driving instruction or a braking instruction of the vehicle. The control device further includes a camber angle controlling means 2d for controlling the camber angle within the predetermined angle range when the instruction detecting means 2c detects the driving instruction or the braking instruction.申请人：TOYOTA MOTOR CORP,トヨタ自動車株式会社地址：愛知県豊田市トヨタ町１番地国籍：JP代理人：伊東 忠彦更多信息请下载全文后查看。

  1. 1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
  2. 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
  3. 3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。

2025 年在全球新能源汽车市场中的比例将接近50
％，2030 年将达到39％
导读：据预测，随着新能源汽车对内燃机汽车在成本方面逐渐建立起领先优势，到2025 年全球新能源汽车的销量将从2017 年的110 万辆增至1100 万辆，而2030 年将继续攀升至3000 万辆。

5 月21 日，彭博新能源财经（BNEF）在2018 年全球新能源汽车市场展望报告中指出，受益于电池大规模量产和价格快速下降，公路交通的电动化
进程将在2025 至2030 年间进入高速发展模式。

到2030 年，全球新能源汽车在乘用车和公交车两个市场的新车销售占比将分别飙升至28％和84％。

BNEF 预测，随着新能源汽车对内燃机汽车在成本方面逐渐建立起领先优势，到2025 年全球新能源汽车的销量将从2017 年的110 万辆增至1100 万辆，而2030 年将继续攀升至3000 万辆。

值得一提的是，中国将引领这场变革，2025 年在全球新能源汽车市场中
的比例将接近50％，2030 年将达到39％。

相关文档
最新文档