湖南省雅礼中学2014届高三第六次月考(物理)试题及答案

雅礼中学2025届高三月考试卷（一）物理本试题卷分选择题和非选择题两部分，共8页。

时量75分钟，满分100分。

一、单选题（本题共6小题，每小题4分，共24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. 每次看到五星红旗冉冉升起，我们都会感到无比自豪和骄傲，在两次升旗仪式的训练中，第一次国旗运动的图像如图中实线所示，第二次国旗在开始阶段加速度较小，但跟第一次一样，仍能在歌声结束时到达旗杆顶端，其运动的图像如图中虚线所示，下列图像可能正确的是（ ）A. B. C. D.2. 无缝钢管制作原理如图所示，竖直平面内，管状模型置于两个支承轮上，支承轮转动时通过摩擦力带动管状模型转动，铁水注入管状模型后，由于离心作用，紧紧地覆盖在模型的内壁上，冷却后就得到无缝钢管。

已知管状模型内壁半径R ，则下列说法正确的是（ ）A. 铁水是由于受到离心力作用才覆盖在模型内壁上B. 模型各个方向上受到的铁水的作用力相同C. 管状模型转动的角速度D. 若最上部的铁水恰好不离开模型内壁，此时仅重力提供向心力3. 一物块静止在粗糙程度均匀的水平地面上，在0~4s 内所受水平拉力F 随时间t 的变化关系图像如图甲所示，在0~2s 内的速度图像如图乙所示，最大静摩擦力大于滑动摩擦力，下列说法正确的是（ ）的的的v t -v t -ωA. 物块的质量为2kgB. 在4s 内物块的位移为8mC. 在4s 内拉力F 做功为18JD. 在4s 末物块的速度大小为4m/s4. 磷是构成DNA 的重要元素，2023年科学家在土卫二的海洋中检测到磷。

此发现意味着土卫二有可能存在生命。

目前探测器已经测出了土卫二的密度为，现发射一颗贴近土卫二表面的人造卫星对土卫二进一步观测，已知万有引力常量为G ，则根据题中所给数据可以计算出（ ）A. 人造卫星的周期B. 土卫二的质量C. 人造卫星的向心加速度大小D. 人造卫星与土卫二之间的万有引力大小5. 如图所示，水平轻细线bc 两端拴接质量均为m 的小球甲、乙，a 、d 为两侧竖直墙壁上等高的两点，小球甲，乙用轻细线ab 和轻弹簧cd 分别系在a 、d 两点，轻细线ab 、轻弹簧cd 与竖直方向的夹角均为，现将水平拉直的轻细线bc 剪断，在剪断瞬间，轻细线ab 上的拉力与轻弹簧cd 上的弹力之比为，小球甲、乙的加速度大小之比为，，，则下列说法正确的是（ ）A. ，B. ，C. ，D. ，6. 如图所示，在某次跳台滑雪比赛中，运动员以初速度从跳台顶端A 水平飞出，经过一段时间后落在倾斜赛道上的B 点，运动员运动到P 点时离倾斜赛道最远，P 点到赛道的垂直距离为PC ，P点离赛道的竖直ρ37θ= :F F T 12:a a sin 370.6= cos370.8= :4:5F F =T 12:16:5a a =:4:5F F =T 12:4:25a a =T :16:25F F =12:16:5a a =T :16:25F F =12:4:5a a =0v高度为PD ，赛道的倾角为，重力加速度为g ，空气阻力不计，运动员（包括滑雪板）视为质点。

雅礼中学2014年上学期期高三英语第六次月考试卷本试卷分为四个部分，包括听力理解、语言知识运用、阅读和书面表达。

考试结束后，将答题卷和答题卡一并交回。

时量120分钟。

满分150分。

PARTⅠLISTENING COMPREHENSION (30 marks)Section A（22.5 marks）Directions：In this section, you’ll hear six conversations between two speake rs. For each conversation, there are several questions and each question is followed by three choices marked A, B and C. Listen carefully and then choose the best answer for each question.You will hear the short passage TWICE.EXAMPLE:When will the magazine probably arrive?A. WednesdayB. ThursdayC. FridayThe answer is B.略PARTⅡLAGUAGE KNOWLEDGE (45 marks)Section A (15 marks)Directions: For each of the following unfinished sentences there are four choices marked A, B, C and D. Choose the one that best completes the sentence.21. Peter has gone to Beijing, but I’m wondering why he ________ in such a hurry.A. leftB. has leftC. leavesD. was leaving一、________ Mr. Smith was the last man I wanted to see, I did all in my power to help him.A. BecauseB. Now thatC. WhileD. As23. In this victory speech ________ before a huge crowd of his supporters, Obama declared that“Change has come to America”.A. deliveredB. to be deliveredC. having deliveredD. delivering24. ________ matters much ________ our headmaster supports our plan.A. As; thatB. It; whetherC. That; ifD. What; whether25. ---- Jack, you ________ the window quickly, will you?---- Ok! Oh! The window ________ broken.A. are shutting; wasB. shut; isC. have shut; hasD. will shut; has been26.----Will you take up teaching as a career after graduation?---- Hard to say. I ________ go abroad for higher education instead, but it depends.A. mustB. shouldC. mightD. shall27.________ the kids had when they experienced the first snowfall--in the South.A. How funB. What funC. How funnyD. What a fun28. Now there is just one point ________ I wish you to make quite clear.A. whereB. whichC. whatD. when29. No one really knows for sure ________ makes a person become right-handed rather thanleft-handed.A. whateverB. whoC. whatD. whoever30. It’s reported that many a house ________ by the heavy snow so far in New York.A. have been destroyedB. has been destroyedC. was destroyedD. were destroyed31.----Why does Jane always ask you for help?---- There is no one else ________, is there?A. who to turn toB. she can turn toC. for whom to turnD. for her to turn32.People laugh at jokes, but seldom ________ about how they work.A. they thinkB. think theyC. they do thinkD. do they think33.It is the only time in history that two Nobel Prizes ________ to the same person.A. have been givenB. had been givenC. have givenD. will give34.---- How is the man injured in the accident?---- The doctor said if ________ in a proper way, he was likely to be saved.A. treatedB. treatingC. is treatedD. to be treated35.---- What happened to you?---- I ________ in the street when someone patted me on the shoulder, which frightened me.A. was about to walkB. was walkingC. had walkedD. walkSection B (18 marks)Directions: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in the blank with the word or phrase that best fits the context.I have a cousin who’s a farmer. If you saw him on market day with his old boots and worn hat, you might think he was a tramp (流浪汉). But he’s either buying or selling 36 for thousands of pounds. He doesn’t have a luxurious lifestyle, but the farmhouse, dozens of acres of land, machinery, cattle, chicken and crops must be worth a fortune. You might describe him as a 37 man, but he doesn’t look like that.My father owned a shop selling newspapers, cigarettes, birthday cards, toys, books and magazines. The shop 38 a comfortable life for us, but we weren’t rich. The only luxury was that my father owned a car when most people in our town couldn’t afford one, 39professionals such as doctors and lawyers.My point is that 40 can be misleading. You often can’t 41 the rich from the way they look.I once met Richard Branson, the Virgin Airlines tycoon (巨头).He’d 42 a hot balloon across the Atlantic and landed in the Irish sea. When he was rescued, he was wearing jeans, a T-shirt and a sneaker. You’d 43 have guessed he was a multimillionaire.Andrew Bruce, the 11th Earl (伯爵) of Elgin, 44 a group of press people to his luxurious home in Fife, Scotland. His 45 , the 7th Earl of Elgin was famous for removing and transporting to Britain priceless sculptures from the Parthenon, a very famous temple in Greece. The so-called Elgin Marbles can be seen in the British Museum. The Earl’s castle was 46 cold, despite a fire burning in the hearth (壁炉). He explained to us,“I’m sorry it’s so cold, but you should know that a gentleman’s house is always cold.”There you have it: 47 doesn’t always bring comfort.36. A. eggs B. foods C. cattle D. fruits37. A. wealthy B. poor C. good D. successful38. A. offered B. provided C. supplied D. gave39. A. besides B. including C. as well as D. apart from40. A. things B. facts C. appearances D. faces41. A. say B. tell C. recognize D. judge42. A. made B. driven C. bought D. flown43. A. perhaps B. neither C. never D. rather44. A. allowed B. permitted C. demanded D. invited45. A. forefather B. father C. grandfather D. father-in-law46. A. frozen B. freezing C. surprising D. surprised47. A health B. wealth C. luck D. povertySection C (12 marks)Directions: Complete the following passage by filling in each blank with one word that best fits the context.Lazy people are often seen as useless. However, are they really being lazy 48 are they acting lazy? There is a huge difference between saying you act lazy sometimes and you are a lazy person.So 49 causes a person to be inactive? The answer is 50 lack of goals and purpose. If you give someone a good enough reason to do something, they will do it. People 51 don’t seem to do anyt hing just haven’t found a good enough reason to do anything. Lazy students don’t study because they don’t see the point in studying. If you give them a reason, a strong enough reason that is, they will take action. For example, if you are 52 lazy to go to the gym, would you go if someone offered you a million dollars to go? If you are too lazy to clean out the garage,would someone pointing a gun to your head help you take action? The reason can be positive or negative as long as 53 is strong enough to induce(引起) action.Action will lead to success while inaction will lead to failure. 54 you take action though, you need a compelling reason to do so. 55 motivate someone who is lazy, what you need to do is help them find purpose and enough reasons to work towards a certain goal.PART ⅢREADI NG COMPREHENTION (30 marks) Directions: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage.AI remember the first time that I was extremely happy. I was about 8 years old when for the first time, there was a computer in the classroom. I remember that my teacher allowed each student to take turns to play various educational games on the computer. One day, I found the source code(编码)for one of these games. Without knowing or being taught any programming language, I was able to figure out some of the BASIC code. I just gave myself an infinite number of lives in the game, so I could continue playing it forever. This was also my first introduction to algebra, and I didn't even know it at the time. This was a decisive moment in my life. I was quite excited because of what I was learning and what I was able to do. As a result, I was enthusiastic for the rest of my life about self learning and computers, and I was quite happy doing them too.I've noticed that people who are truly content with life are enthusiastic about what they do. This enthusiasm, along with good health, is the key to being happy. It also leads to self-confidence and content in life too. It may also lead to success, wealth, and achievements.Success, wealth, or achievements can also bring some people happiness, yet I know plenty of rich people who are unhappy. I know many people with successful businesses that are not happy with what they are doing. I know people who continuously buy themselves new toys, such as cars, computers, and televisions, yet never seem content for too long. Please remember, happiness is the journey of life, not the destination.56. What can we know from Paragraph 1?A. The author has a great talent for algebra.B. Creative thinking is necessary for every child.C. The BASIC code of the computer is not difficult.D. The author's experience in his childhood changed his life.57. The underlined word "infinite" meansA. bigB. limitlessC. normalD. small58. The author wants to tell us through his experience in the school that ________.A. interest is the best teacherB. children are the hope of the futureC. young people are fearlessD. where there’s a will, there’s a way59. What is the secret of happiness in the author's opinion?A. Success and wealth.B. Gifts and self-confidence.C. Enthusiasm and good health.D. Knowledge and achievements.60. We can infer from the last paragraph that ________.A. people who are rich and successful in career generally feel unhappyB. wealth can’t bring people any happiness and comfortC. one will feel unhappy once he has gained all the things that he wantsD. being enthusiastic about what you do is more important than wealthBParents are fuelling bad behavior among their children by attempting to "buy" their love with expensive gifts nowadays.Over recent decades we seem to have created a "must have" culture among our young people. Many mothers and fathers believe they are "failing as parents" if they are unable to ensure that their children have the latest toy, electronic devices (the lap-top, cellphones, Game Boy, etc.) along with their friends. In many cases, families also feel pressured to enroll (使加入) children in "'every interest club or after-school activity that is available" to fill up their time like most of the other children.But experts warned that the move might affect "precious family time" negatively. Graham Gorton, chairman of the Independent Schools Association, said that parents spent too much time filling their children's lives, which had a series of negative effects on "the very precious family time that exists"."It seems that those times when a whole weekend without planned work was seen as a luxury (奢侈) and a perfect opportunity to spend time together and share those valuable moments of childhood are long gone," Mr. Gorton said. "As a child I only once said that phrase that parents feared 'I'm bored'. Immediately my mother took action and produced a list of jobs and then insisted that I complete every one of them. Though l didn't think cleaning all the floors could really get rid of my boredom, I enjoyed the feeling of staying at home with my mother and brothers."Earlier this year, some researchers suggested that relatively wealthy parents were sometimes guilty of failing to teach basic social skills to children. "Often, it’s the rich middle classes that buy off their children through the computer and the TV. That then sets them apart from their family, and then the parents are surprised when their child isn’ t coming to school."61. The second paragraph mainly tells us that ________.A. today’s children have little time to playB. today’s children depend on electronic tools too muchC. today’s parents feel pressured by the "'must have" cultureD. being qualified parents becomes harder for today's young people62. What does the "must have" phenomenon refer to based on the passage?A. That children must have what other children have.B. That children must have proper pressure to work hard.C. That children must have special skills to ensure a better future.D. That parents must have patience to know their children better.63. Which of the following is the best advice you could give to parents based on Gorton’s statement?A. Plan fewer activities for their children to improve family time.B. Buy their children fewer things that they are fond of.C. Give their children more housework to do.D. Leave their children alone when the kids feel bored.64. Mr. Gorton tended to think that in the past ________.A. children enjoyed doing houseworkB. children never thought life was boringC. children liked to spend time with their familyD. children often had some planned work after school65. From the passage we can infer that ________.A. wealthy parents don’t like teaching basic social skills to childrenB. parents should be much more strict with their children at homeC. children should not be brought up in a wealthy and pleasant environmentD. only satisfying children’s material needs is not a good way of parentingCThere is some unwelcome news for students preparing for exams and officers putting in long hours-----you don't need the break as much as you may think that makes you feel less tired.Scientists have long assumed that willpower (意志力) is a limited resource, which is why you feel the need to have a rest, have a snack and come back to a task when you're feeling better. They argued that the only way to restore willpower was by rest, food or entertainment.But psychologists have challenged this theory, saying weak willpower is all in your head. They found that people's beliefs in willpower determine how long and how well they'll be able to work on a tough mental exercise. "If you think of willpower as something that's limited, you're more likely to be tired when you perform a difficult task," said Prof.Veronika Job. "'But if you think of willpower as something that is not easily used up, you can go on and on."The researchers designed four experiments to test students'-beliefs in willpower. After a tiring task, those, who believed or were led to believe that willpower is a limited resource, performed worse on standard concentration tests than those who thought of willpower as something they had more control over. They also found that leading up to final exam week, students who believed the limited resource theory ate junk food 24 percent more often than those who believed they hadmore control in resisting temptation (诱惑).Mr. Job said. "The theory that willpower is a limited resource is interesting, but it has had unintended consequences. Students who may already have trouble studying are being told that their power of concentration is limited, and they need to take frequent breaks. But a belief in willpower as a non-limited resource makes people stronger in their ability to work through challenges.'"The findings could help people who are" battling temptation. Willpower isn't driven by a biologically based process as much as we used to think. The belief in it is what influences your behavior.66. The theory that willpower is limited supports that _________.A. people must eat snacks when they feel tiredB. people do need a break to restore their willpowerC. t here’s no way to strengthen people’s willpowerD. weak willpower doesn’t affect people’s life much67. What have the scientists long believed regarding willpower?A. It is in the charge of people.B. It is a limited resource.C. There is no way to restore willpower.D. It doesn’t easily run out.68. Which of the following best helps the students to prepare better for their exams?A. Push themselves even if they want to take a break.B. Don’t eat fast food while studying.C. Remind themselves willpower is not limited.D. Stay in a comfortable and quiet place.69. The following groups can benefit from the findings exceptA. patients following strict dietsB. children liking to watch TVC. smokers trying to give up smokingD. employees facing a new but well-paid task70. What’s the best title for the passage?A. A new theory about willpowerB. How to build strong willpowerC. The great influence of willpowerD. Willpower doesn’t last longPART ⅣWRITING (45 marks)Section A (10 marks)Directions: Read the following passage. Fill in the numbered blankets by using the information from the passage. Write NO MORE THAN THREE WORDS for each answer.Before the 1850’s, the United States had a numbe r of small colleges, most of them dating from colonial（殖民的）days. They were small, church connected institutions whose primary concern was to shape the moral character of their students.Throughout Europe, institutions of higher learning had developed, bearing the ancient name of university. In German university was concerned primarily with creating and spreadingknowledge, not morals. Between mid-century and the end of the 1800’s, more than nine thousand young Americans, dissatisfied with their training at home, went to Germany for advanced study. Some of them return to become presidents of admired colleges－Harvard, Yale, Columbia－and transform them into modern universities. The new presidents broke all ties with the churches and brought in a new kind of faculty. Professors were hired for their knowledge of a subject, not because they were of the proper faith and had a strong arm for disciplining students. The new principle was that a university was to create knowledge as well as pass it on, and this called for a faculty made up of teacher-scholars. Drilling and learning by memorization were replaced by the G erman method of lecturing, in which the professor’s own research was presented in class. Graduate training leading to the Ph.D., an ancient German degree representing the highest level of advanced scholarly achievement, was introduced. With the establishment of the seminar system, graduate student learned to question, analyze, and conduct their own research.At the same time, the new university greatly expanded in size and course offerings, breaking completely out of the old, restricted curriculum of mathematics, classics, rhetoric, and music. The president of Harvard pioneered the elective system, by which students were able to choose their own course of study. The new goal was to make the university relevant(相关的) to the real pursuits（追求）of the world.Title The development of the 71 in the USASection B (10 marks)Directions: Read the following passage. Answer the questions according to the information given in the passage and required words limit. Write your answers on your answer sheet.A 10-year-old boy decided to study judo despite the fact that he had lost his left arm in a terrible car accident.The boy began lessons with an old Japanese judo master. The boy was doing well, so he couldn't understand why, after three months of training, the master had taught him only one move. "Sensei," the boy finally said, "shouldn't I be learning more moves?" "This is the only move you know, but this is the only move you'll ever need to know," the sensei replied. Not quite understanding, but believing in his teacher, the boy kept training.Several months later, the sensei took the boy to his first tournament.Surprising himself, the boy easily won his first two matches. The third match proved to be more difficult, but after some time, his opponent became impatient and charged; the boy deftly used his one move to win the match. Still amazed by his success, the boy was now in the finals.This time, his opponent was bigger, stronger, and more experienced. For a while, the boy appeared to be overmatched. Concerned that the boy might get hurt, the referee called a timeout. He was about to stop the match when the sensei intervened."No," the sensei insisted, "let him continue." Soon after the match resumed, his opponent made a big mistake: he dropped his guard. Instantly, the boy had won the match and the tournament. He was the champion.On the way home, the boy and sensei reviewed every move in each and every match.Then the boy gathered the courage to ask what was really on his mind. "Sensei, how did I win the tournament with only one move?" You won for two reasons," the sensei answered. "First, you've almost mastered one of the most difficult throws in all of judo. Second, the only known defense for that move is for your opponent to grab your left arm."81. How did the judo master teach the boy? (no more than 8 words)__________________________________________________________________________ 82. Why did the referee call a timeout? (no more than 9 words)__________________________________________________________________________ 83. What were the reasons for the boy’s winning the champion? (no more than 20 words)__________________________________________________________________________ 84. What can we learn from the short story? (no more than 8 words)__________________________________________________________________________Section C (25 marks)Directions: Write an English composition according to the instructions given below inChinese.请仔细观察图片，描述图片内容，并以生活中发生的一件事情说明图片所说明的内涵。

雅礼中学2023届高三月考试卷（三）物理命题人：张睿智审题人：李仪辉得分：___________本试题卷分选择题和非选择题两部分，共8页。

时量75分钟，满分100分。

一、单选题（本题共6小题，每小题4分，共24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.单板大跳台是一项紧张刺激的项目。

2022年北京冬奥会期间，一观众用手机连拍功能拍摄运动员从起跳到落地的全过程，合成图如图所示。

忽略空气阻力，且将运动员视为质点。

则运动员A．在空中飞行过程是变加速曲线运动B．在斜向上飞行到最高点的过程中，其动能全部转化为重力势能C．运动员从起跳后到落地前，重力的瞬时功率先减小后增大D．运动员在空中飞行过程中，动量的变化率在不断变化2.如图甲所示，某电场中的一条电场线恰好与直线AB重合，以A点为坐标原点，向右为正方向建立直线坐标系，B点的坐标为0.06Bx m，若一α粒子仅在电场力的作用下由A点运动至B点，其电势能增加60 eV，该电场线上各点的电场强度大小E随位移x的变化规律如图乙所示，若A点电势为15V，下列说法正确的是A．x轴上各点的电场强度方向都沿x轴正方向B ．该a 粒子沿x 轴正方向做匀减速直线运动C ． B 点电势是A 点电势的2倍D ．图乙中E 0应为212.510V m -⨯⋅3.九重之际向天问，天宫掠影惊苍穹。

“天宫”空间站中三名宇航员正环绕地球运行，与此同时，“天问”探测器在环绕火星运行。

假设它们的运行轨道都是圆轨道，地球与火星的质量之比为p ，“天宫”与“天问”的轨道半径之比为q 。

关于“天宫”空间站与“天问”探测器，下列说法不正确的是A 3q p B p q C ．加速度之比为2p q D ．动能之比为p q4.江南多雨，屋顶常常修成坡度固定的“人”字形，“人”字形的尖顶屋可以看做由两个斜面构成。

如图所示，斜面与水平方向的夹角均为α，房屋长度2x 为一定值，将雨滴从“人”字形坡顶开始的下滑过程简化为雨滴从光滑斜面顶端由静止下滑。

湖南省长沙市雅礼中学2023届高三下学期月考(六)数学试题含答案湖南省长沙市雅礼中学2023届高三下学期月考(六)数学试题含答案1. 选择题1.1. 单选题1. 下列四个不等式中，正确的是：A. |x + 2| > 3B. |2x - 4| ≥ 0C. |3x - 1| ≤ -1D. |4x + 2| < 22. 解答题2.1. 简答题（1）解方程x² + 5x + 6 = 0。

解：首先，我们可以尝试因式分解，将方程进行因式分解，得到：x² + 3x + 2x + 6 = 0(x + 3)(x + 2) = 0由此可知，方程的解为x = -3或x = -2。

（2）已知集合A = {2, 3, 4}，集合B = {3, 4, 5}，求A和B的交集与并集。

解：集合A和B的交集为A∩B = {3, 4}，即A和B共有的元素是3和4。

集合A和B的并集为A∪B = {2, 3, 4, 5}，即A和B的所有元素的集合。

2.2. 计算题铁路物流公司欲购买一辆装货能力为200t的货车，现有两种型号的货车可供选择：型号A的装货能力为50t，每辆售价200万元；型号B的装货能力为40t，每辆售价160万元。

为了降低成本，铁路物流公司希望选择装载能力和价格成比例的货车。

那么，应选择购买多少辆型号A和多少辆型号B的货车？解：设购买型号A的货车辆数为x，购买型号B的货车辆数为y，则有以下两个约束条件：50x + 40y = 200200x + 160y = C由第一个等式可得：x = (200 - 40y)/50将x的表达式代入第二个等式中，得到：200(200 - 40y)/50 + 160y = C根据题意，装载能力和价格成比例，可以得到比例关系：(200 - 40y)/50 = 0.4y求解得到：200 - 40y = 0.4y * 50200 - 40y = 20y200 = 60yy = 200/60 = 10/3 ≈ 3.33将y的值代入x的表达式中，得到：x = (200 - 40 * 3.33)/50 ≈ 1.34所以，应选择购买1辆型号A和3辆型号B的货车，以满足装载能力与价格成比例的要求。

雅礼中学2025届高三月考试卷（三）化学得分：______本试题卷分选择题和非选择题两部分，共8页。

时量75分钟，满分100分。

可能用到的相对原子质量：H~1 C~12 N~14 O~16 Na~23 S~32 Cl~35.5 Zr~91第Ⅰ卷（选择题 共42分）一、选择题（本题共14小题，每小题3分，共42分，每小题只有一个选项符合题意。

）1．化学和我们的生活有十分密切的联系，下列表述不正确的是（）A.在碳素钢中加入Cr 和Ni 制得不锈钢可以增强钢的强度以及抗腐蚀能力B.改变铝制品表面氧化膜的厚度可以影响染料着色从而产生美丽的颜色C.焰色试验选择Fe 作为载体是因为铁元素受热不发生电子跃迁、不产生发射光谱D.半导体材料氮化镓是一种新型无机非金属材料2．下列化学用语表示不正确的是（）A.2，2-二甲基丁烷的结构简式：B.三氟化硼分子的空间填充模型：C.次氯酸分子的电子式：D.基态溴原子的简化电子排布式：3．2024年诺贝尔化学奖表彰了三位科学家在蛋白质设计和结构预测领域作出的贡献，中国科学家颜宁在这方面也做了大量的工作，以下相关说法不正确的是（ ）A.蛋白质分散在水中形成的分散系可以产生丁达尔效应B.要使蛋白质晶化得到较大的蛋白质晶体需要快速结晶C.通过X 射线衍射可以得到高分辨率的蛋白质结构D.蛋白质复杂结构的形成与极性键、非极性键、氢键、范德华力等有关4．以下实验方案正确且能达到实验目的的是（ ）选项实验目的实验方案A 制备少量硝酸边加强热边向饱和硝酸钠溶液中滴加浓硫酸B验证晶体的自范性将形状不规则的蔗糖块放入饱和蔗糖溶液中静置一段时间后取出C 验证C 和Si 的非金属性强弱将焦炭和石英砂混合加强热（1800~2000℃），检验气体产物以证明反应发生D测定中和热将稍过量的NaOH 固体投入装有一定量稀盐酸的烧杯中并测33253CH |CH C C H |CH --H :O:Cl :[]25Ar 4s 4p量其温度变化5．某化学小组在实验室尝试用氨气制备硝酸，过程如下：。

雅礼中学2025届高三月考试卷（三）数学命题人：审题人：得分：________本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，共8页.时量120分钟，满分150分.第Ⅰ卷一、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.命题“存在，”的否定是A.存在，B.不存在，C.任意，D.任意，2.若集合（i 是虚数单位），，则等于A. B. C. D.3.已知奇函数，则A.-1B.0C.1D.4.已知，是两条不同的直线，，是两个不同的平面，则下列可以推出的是A.，， B.，，C.，， D.，，5.已知函数图象的一个最高点与相邻的对称中心之间的距离为5，则A.0B. C.4D.x ∈Z 220x x m ++…x ∈Z 220x x m ++>x ∈Z 220x x m ++>x ∈Z 220x x m ++…x ∈Z 220x x m ++>{}2341,i ,i ,i A ={}1,1B =-A B ⋂{}1-{}1{}1,1-∅()()22cos x x f x m x -=+⋅m =12m l αβαβ⊥m l ⊥m β⊂l α⊥m l ⊥l αβ⋂=m α⊂m l m α⊥l β⊥l α⊥m l m β()()4cos (0)f x x ωϕω=+>6f ϕπ⎛⎫-=⎪⎝⎭2ϕ2ϕ6.已知是圆上一个动点，且直线与直线（，，）相交于点，则的取值范围为A. B.C. D.7.是椭圆上一点，，是的两个焦点，，点在的角平分线上，为原点，，且.则的离心率为A.8.设集合，那么集合中满足条件“”的元素个数为A.60B.90C.120D.130二、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.9.如图为某地2014年至2023年的粮食年产量折线图，则下列说法正确的是A.这10年粮食年产量的极差为16B.这10年粮食年产量的第70百分位数为35C.这10年粮食年产量的平均数为33.7D.前5年的粮食年产量的方差小于后5年粮食年产量的方差10.已知函数满足，，并且当时，，则下列关于函数说法正确的是M 22:1C x y +=1:30l mx ny m n --+=2:30l nx my m n +--=m n ∈R 220m n +≠P PM 1,1⎤-+⎦1⎤-⎦1,1⎤-+⎦1⎤+⎦P 2222:1(0)x y C a b a b+=>>1F 2F C 120PF PF ⋅= Q 12F PF ∠O 1OQPF OQ b =C 12(){}{}{}12345,,,,|1,0,1,1,2,3,4,5iAx x x x x x i ∈-=A 1234513x x x x x ++++……()f x ()()22f x f x ππ+=-()()0fx f x ππ++-=()0,x π∈()cos f x x =()f xA. B.最小正周期C.的图象关于直线对称D.的图象关于对称11.若双曲线，，分别为左、右焦点，设点是在双曲线上且在第一象限的动点，点为的内心，，则下列说法不正确的是A.双曲线的渐近线方程为B.点的运动轨迹为双曲线的一部分C.若，，则D.不存在点，使得取得最小值答题卡题号1234567891011得分答案第Ⅱ卷三、填空题：本题共3小题，每小题5分，共15分.12.的展开式中的系数为________.13.各角的对应边分别为，，，满足，则角的取值范围为________.14.对任意的，不等式（其中e 是自然对数的底）恒成立，则的最大值为________.四、解答题：本题共5小题，共77分.请在答题卡指定区域内作答.解答时应写出文字说明、证明过程或演算步骤.15.（本小题满分13分）设为正项等比数列的前项和，，.（1）求数列的通项公式；（2）数列满足，，求数列的前项和.302f π⎛⎫=⎪⎝⎭2T π=()f x x π=()f x (),0π-22:145x y C -=1F 2F P I12PF F △()0,4A C 045x y±=I 122PF PF =12PI xPF yPF =+ 29y x -=P 1PA PF +523x x ⎛⎫+ ⎪⎝⎭4x ABC △a b c 1b ca c a b+++…A *n ∈N 11e 1nan n n ⎛⎫⎛⎫+⋅ ⎪ ⎪+⎝⎭⎝⎭…a n S {}n a n 21332S a a =+416a ={}n a {}n b 11b =1222log log n nn n b a b a ++={}n b n n T16.（本小题满分15分）如图，在四棱锥，，，，点在上，且，.（1）若为线段的中点，求证：平面；（2）若平面，求平面与平面所成夹角的余弦值.17.（本小题满分15分）已知函数有两个极值点为，，.（1）当时，求的值；（2）若（e 为自然对数的底数），求的最大值.18.（本小题满分17分）已知抛物线的焦点为，为上任意一点，且的最小值为1.（1）求抛物线的方程；（2）已知为平面上一动点，且过能向作两条切线，切点为，，记直线，，的斜率分别为，，，且满足.①求点的轨迹方程；②试探究：是否存在一个圆心为，半径为1的圆，使得过可以作圆的两条切线，，切线，分别交抛物线于不同的两点，和点，，且为定值？若存在，求圆的方程，不存在，说明理由.19.（本小题满分17分）对于一组向量，，，…，（且），令，如果存在，使得，那么称是该向量组的“长向量”.（1）设，且，若是向量组，，的“长向量”，求实数的取值范P ABCD -BCAD 1AB BC ==3AD =E AD PE AD ⊥2DE PE ==F PE BFPCD AB ⊥PAD PAB PCD ()21ln 2f x x x ax =+-1x ()212x x x <a ∈R 52a =()()21f x f x -21e x x …()()21f x f x -2:2(0)E x py p =>F H E HF E P P E M N PM PN PF 1k 2k 3k 123112k k k +=P ()0,(0)Q λλ>P Q 1l 2l 1l 2l E ()11,A s t ()22,B s t ()33,C s t ()44,D s t 1234s s s s Q 1a 2a 3a n a N n ∈3n …123n n S a a a a =++++{}()1,2,3,,p a p n ∈ p n p a S a - …p a(),2n a n x n =+n ∈N 0n >3a 1a 2a 3ax围；（2）若，且，向量组，，，…，是否存在“长向量”？给出你的结论并说明理由；（3）已知，，均是向量组，，的“长向量”，其中，.设在平面直角坐标系中有一点列，，，…，，满足为坐标原点，为的位置向量的终点，且与关于点对称，与（且）关于点对称，求的最小值.sin,cos 22n n n a ππ⎛⎫= ⎪⎝⎭n ∈N 0n >1a 2a 3a 7a 1a 2a 3a 1a2a3a()1sin ,cos a x x =()22cos ,2sin a x x = 1P 2P 3P n P 1P 2P 3a 21k P +2k P 1P 22k P +21k P +k ∈N 0k >2P10151016P P参考答案一、二、选择题题号1234567891011答案DCADCBCDACDADABD1.D2.C 【解析】集合，，.故选C.3.A【解析】是奇函数，，，，，.故选A.4.D 【解析】有可能出现，平行这种情况，故A 错误；会出现平面，相交但不垂直的情况，故B 错误；，，，故C 错误；，，又由，故D 正确.故选D.5.C 【解析】设的最小正周期为，函数图象的一个最高点与相邻的对称中心之间的距离为5，则有，得，则有，解得，所以，所以.故选C.6.B 【解析】依题意，直线恒过定点，直线恒过定点，显然直线，因此，直线与交点的轨迹是以线段为直径的圆，其方程为：，圆心，半径，而圆的圆心，半径，如图：，两圆外离，由圆的几何性质得：，{}i,1,1,i A =--{}1,1B =-{}1,1A B ⋂=-()f x ()()22cos x x f x m x -=+⋅()()()2222x x x xf x f x m --⎡⎤∴+-=+++⎣⎦cos 0x =()()122cos 0x x m x -∴++=10m ∴+=1m =-αβαβm l m α⊥l βαβ⊥⇒ l α⊥m l m α⇒⊥ m βαβ⇒⊥ ()f x T 224254T ⎛⎫+= ⎪⎝⎭12T =212πω=6πω=()4cos 6f x x πϕ⎛⎫=+ ⎪⎝⎭664cos 4cos046f ϕϕπϕππ⎛⎫⎛⎫-=-⨯+== ⎪ ⎪⎝⎭⎝⎭()()1:310l m x n y ---=()3,1A ()()2:130l n x m y -+-=()1,3B 12l l ⊥1l 2l P AB 22(2)(2)2x y -+-=()2,2N 2r =C ()0,0C 11r =12NC r r =>+12min1PMNC r r =--=-，所以的取值范围为.故选B.7.C 【解析】如图，设，，延长交于点，由题意知，为的中点，故为中点，又，即，则，又由点在的角平分线上得，则是等腰直角三角形，故有化简得即代入得，即，又，所以，所以，.故选C.8.D 【解析】因为或，所以若，则在中至少有一个，且不多于3个.所以可根据中含0的个数进行分类讨论.①五个数中有2个0，则另外3个从1，-1中取，共有方法数为，②五个数中有3个0，则另外2个从1，-1中取，共有方法数为，③五个数中有4个0，则另外1个从1，-1中取，共有方法数为，所以共有种.故选D.9.ACD 【解析】将样本数据从小到大排列为26，28，30，32，32，35，35，38，39，42，这10年的粮食年产量极差为，故A 正确；，结合A 选项可知第70百分位数为第7个数和第812max1PMNC r r =++=+PM 1⎤-+⎦1PF m =2PF n =OQ 2PF A 1OQ PF O 12F F A 2PF 120PF PF ⋅= 12PF PF ⊥2QAP π∠=Q 12F PF ∠4QPA π∠=AQP △2222,4,11,22m n a m n c b n m ⎧⎪+=⎪+=⎨⎪⎪+=⎩2,2,m n b m n a -=⎧⎨+=⎩,,m a b n a b =+⎧⎨=-⎩2224m n c +=222()()4a b a b c ++-=2222a b c +=222b a c =-2223a c =223e =e =0i x =1i x =1234513x x x x x ++++……()1,2,3,4,5i x i =1i x =i x 2315C 2N =⋅3225C 2N =⋅435C 2N =⋅23324555C 2C 2C 2130N =⋅+⋅+⋅=422616-=1070%7⨯=个数的平均数，即，故B 不正确；这10年粮食年产量的平均数为，故C 正确；结合图形可知，前5年的粮食年产量的波动小于后5年的粮食产量波动，所以前5年的粮食年产量的方差小于后5年的粮食年产量的方差，故D 正确.故选ACD.10.AD 【解析】由于时，，并且满足，则函数的图象关于直线对称.由于，所以，故，故，故函数的最小正周期为，根据，知函数的图象关于对称.由于时，，，故A 正确，由于函数的最小正周期为，故B 错误；由函数的图象关于对称，易知的图象不关于直线对称，故C 错误；根据函数图象关于点对称，且函数图象关于直线对称，知函数图象关于点对称，又函数的最小正周期为，则函数图象一定关于点对称，故D 正确.故选AD.11.ABD 【解析】双曲线，可知其渐近线方程为，A 错误；设，，的内切圆与，，分别切于点，，，可得，，，由双曲线的定义可得：，即，又，解得，则点的横坐标为，由点与点的横坐标相同，即点的横坐标为，故在定直线上运动，B 错误；由，且，解得，，，，则，同理可得：，设直线，直线，联立方程得，设的内切圆的半径为，则，解得，即，353836.52+=()13232302835384239263533.710⨯+++++++++=()0,x π∈()cos f x x =()()22f x f x ππ+=-()f x 2x π=()()0fx f x ππ++-=()()fx f x ππ+=--()()()()()22f xf x f x f x ππππ--+=+=--=-()()()24f x f x f x ππ=-+=+4π()()0fx f x ππ++-=()f x (),0π()0,x π∈()cos f x x =3cos 022222f f ff πππππππ⎛⎫⎛⎫⎛⎫⎛⎫=+=--=-=-=⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭4π()f x (),0π()f x x π=(),0π2x π=()3,0π4π(),0π-22:145x y C -=02x =1PF m =2PF n =12PF F △1PF 2PF 12F F S K T PS PK =11F S FT =22F T F K =2m n a -=12122F S F K FT F T a -=-=122FT F T c +=2F T c a =-T a I T I 2a =I 2x =122PF PF =1224PF PF a -==18PF =24PF =1226F F c ==126436167cos 2868PF F ∠+-∴==⨯⨯12sin PF F ∠==12tan PF F ∠∴=21tan PF F ∠=)1:3PF y x =+)2:3PF y x =-(P 12PF F △r ()12118684622PF F S r =⨯⨯=⨯++⋅△r =I ⎛ ⎝，，，由，可得解得，，故，C 正确；，，当且仅当，，三点共线取等号，易知，故存在使得取最小值，D 错误.故选ABD.三、填空题：本题共3小题，每小题5分，共15分.12.90 【解析】展开式的通项公式为，令，解得，所以展开式中的系数为.13. 【解析】从所给条件入手，进行不等式化简，观察到余弦定理公式特征，进而利用余弦定理表示，由可得，可得.14. 【解析】对任意的，不等式（其中e 是自然对数的底）恒成立，只需恒成立，只需恒成立，只需恒成立，2,PI ⎛∴=- ⎝ (17,PF =- (21,PF =- 12PI xPF yPF =+ 27,,x y -=--⎧⎪⎨=⎪⎩29x =49y =29y x -=1224PF PF a -== 12244PA PF PA PF AF ∴+=+++…A P 2F ()1min549PA PF +=+=P 1PA PF +523x x ⎛⎫+ ⎪⎝⎭()()521031553C C 3rr rrr r r T x x x --+⎛⎫=⋅⋅=⋅⋅ ⎪⎝⎭1034r -=2r =4x 225C 310990⋅=⨯=0,3π⎛⎤⎥⎝⎦()()1b c b a b c a c a c a b+⇒+++++……()()222a c a b b c a bc ++⇒++…cos A 222b c a ac +-…2221cos 22b c a A bc +-=…0,3A π⎛⎤∈ ⎥⎝⎦11ln2-*n ∈N 11e 1n an n n ⎛⎫⎛⎫+⋅ ⎪ ⎪+⎝⎭⎝⎭…11e n an +⎛⎫+ ⎪⎝⎭…()1ln 11n a n ⎛⎫++ ⎪⎝⎭…11ln 1a n n -⎛⎫+ ⎪⎝⎭…构造，，，.下证，再构造函数，，，，设，，，令，，，，在时，，单调递减，，即，所以递减，，即，所以递减，并且，所以有，，所以，所以在上递减，所以的最小值为.，即的最大值为.四、解答题：本题共5小题，共77分.请在答题卡指定区域内作答.解答时应写出文字说明、证明过程或演算步骤.15.【解析】（1）因为是正项等比数列，所以，公比，因为，所以，即，则，解得（舍去）或，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（3分）又因为，所以，所以数列的通项公式为.∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（6分）（2）依题意得，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（7分）当时，，所以，因为，所以，当时，符合上式，所以数列的通项公式为.∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（10分）()()11ln 1m x x x =-+(]0,1x ∈()()()()()22221ln 11ln 1x x x m x x x x ++-=++'(]0,1x ∈()(]22ln 1,0,11x x x x+<∈+()()22ln 11x h x x x =+-+(]0,1x ∈()()()2221ln 12(1)x x x xh x x ++-'-=+(]0,1x ∈()()()221ln 12F x x x x x =++--()()2ln 12F x x x =+-'(]0,1x ∈()()2ln 12G x x x =+-(]0,1x ∈()21xG x x=-+'(]0,1x ∈(]0,1x ∈()0G x '<()G x ()()00G x G <=()0F x '<()F x ()()00F x F <=()0h x '<()h x ()00h =()22ln 11x x x+<+(]0,1x ∈()0m x '<()m x (]0,1x ∈()m x ()111ln2m =-11ln2a ∴-…a 11ln2-{}n a 10a >0q >21332S a a =+()121332a a a a +=+21112320a q a q a --=22320q q --=12q =-2q =3411816a a q a ===12a ={}n a 2n n a =1222222log log 2log log 22n n n n n n b a nb a n +++===+2n …()324123112311234511n n b b b b n b b b b n n n --⨯⋅⋅⋅=⨯⨯⨯⨯=++ ()121n b b n n =+11b =()21n b n n =+1n =1n b ={}n b ()21n b n n =+因为，所以.∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（13分）16.【解析】（1）设为的中点，连接，，因为是中点，所以，且，因为，，，，所以四边形为平行四边形，，且，所以，且，即四边形为平行四边形，所以，因为平面平面，所以平面.∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（6分）（2）因为平面，所以平面，又，所以，，相互垂直，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（7分）以为坐标原点，建立如图所示的空间直角坐标系，则，，，，，所以，，，，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（9分）设平面的一个法向量为，则取，则，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（11分）设平面的一个法向量为，()211211n b n n n n ⎛⎫==- ⎪++⎝⎭1111112212221223111n n T n n n n ⎛⎫⎛⎫⎛⎫⎛⎫=-+-++-=-=⎪ ⎪ ⎪ ⎪+++⎝⎭⎝⎭⎝⎭⎝⎭M PD FM CM F PE FMED 12FM ED =AD BC 1AB BC ==3AD =2DE PE ==ABCE BC ED 12BC ED =FM BC FM BC =BCMF BFCM BF ⊄,PCD CM ⊂PCD BF PCD AB ⊥PAD CE ⊥PAD PE AD ⊥EP ED EC E ()0,0,2P ()0,1,0A -()1,1,0B -()1,0,0C ()0,2,0D ()1,0,0AB = ()0,1,2AP = ()1,0,2PC =- ()1,2,0CD =-PAB ()111,,m x y z =1110,20,m AB x m AP y z ⎧⋅==⎪⎨⋅=+=⎪⎩ 11z =-()0,2,1m =- PCD ()222,,n x y z =则取，则，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（13分）设平面与平面所成夹角为，则∙∙∙∙∙∙∙∙∙∙∙（15分）17.【解析】（1）函数的定义域为，则，当时，可得，，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（2分）当或时，；当时，；所以在区间，上单调递增，在区间上单调递减；∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（4分）所以和是函数的两个极值点，又，所以，；所以，即当时，.∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（6分）（2）易知，又，所以，是方程的两个实数根，则且，，所以，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（9分）所以，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（11分）设，由，可得，令，，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（13分）则，所以在区间上单调递减，222220,20,n PC x z n CD x y ⎧⋅=-=⎪⎨⋅=-+=⎪⎩ 21z =()2,1,1n = PAB PCD θcos θ=()21ln 2f x x x ax =+-()0,+∞()211x ax f x x a x x -+=+-='52a =()()2152122x x x x f x x x'⎛⎫---+ ⎪⎝⎭==10,2x ⎛⎫∈ ⎪⎝⎭()2,x ∈+∞()0f x '>1,22x ⎛⎫∈ ⎪⎝⎭()0f x '<()f x 10,2⎛⎫ ⎪⎝⎭()2,+∞1,22⎛⎫ ⎪⎝⎭12x =2x =()f x 12x x <112x =22x =()()()211115152ln225ln 2ln222848f x f x f f ⎛⎫⎛⎫-=-=+--+-=- ⎪ ⎪⎝⎭⎝⎭52a =()()21152ln28f x f x -=-()()()()22221212111ln2x f x f x x x a x x x -=+---()21x ax f x x-+='1x 2x 210x ax -+=2Δ40a =->120x x a +=>121x x =2a >()()()()()()()2222222121212112211111lnln 22x x f x f x x x a x x x x x x x x x x -=+---=+--+-()()222222221212111121121111lnln ln 222x x x x x x x x x x x x x x x x ⎛⎫=--=-⋅-=-- ⎪⎝⎭21x t x =21e x x (21)e x t x =…()11ln 2g t t t t ⎛⎫=-- ⎪⎝⎭e t …()222111(1)1022t g t t t t-⎛⎫=-+=-< ⎪⎝⎭'()g t [)e,+∞得，故的最大值为.∙∙∙∙∙∙∙∙∙∙∙∙∙∙（15分）18.【解析】（1）设抛物线的准线为，过点作直线于点，由抛物线的定义得，所以当点与原点重合时，，所以，所以抛物线的方程为.∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（4分）（2）①设，过点且斜率存在的直线，联立消去，整理得：，由题可知，即，所以，是该方程的两个不等实根，由韦达定理可得∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（6分）又因为，所以，，由，有，所以，因为，，，所以点的轨迹方程为.②由①知，设，，且，∙∙∙∙∙∙∙∙∙（9分）联立消去，整理得，又，，，，由韦达定理可得，同理可得，所以，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（11分）又因为和以圆心为，半径为1的圆相切，，即.同理，所以，是方程的两个不等实根，()()11e 1e 1e 12e 22eg t g ⎛⎫=--=-+ ⎪⎝⎭…()()21f x f x -e 1122e -+E l 2py =-H 1HH ⊥l 1H 1HF HH =H O 1min 12pHH ==2p =E 24x y =(),P m n P ():l y k x m n =-+()24,,x y y k x m n ⎧=⎪⎨=-+⎪⎩y 24440x kx km n -+-=()2Δ164440k km n =--=20k mk n -+=1k 2k 1212,,k k m k k n +=⎧⎨=⎩()0,1F 31n k m -=0m ≠123112k k k +=121232k k k k k +=21m m n n =-0m ≠12n n -=1n ∴=-P ()10y x =-≠(),1P m -()14:1l y k x m =--()25:1l y k x m =--1m ≠±0m ≠()244,1,x y y k x m ⎧=⎪⎨=--⎪⎩y 2444440x k x k m -++=()11,A s t ()22,B s t ()33,C s t ()44,D s t 12444s s k m =+34544s s k m =+()()()212344515454444161616s s s s k m k m k k m m k k =++=+++1l ()0,(0)Q λλ>1()()2224412120m k m k λλλ-++++=()()2225512120m k m k λλλ-++++=4k 5k ()()22212120m k m k λλλ-++++=所以由韦达定理可得∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（14分）所以，若为定值，则，又因为，所以，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（16分）所以圆的方程为.∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（17分）19.【解析】（1）由题意可得：，解得.∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（3分）（2）存在“长向量”，且“长向量”为，，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（5分）理由如下：由题意可得，若存在“长向量”，只需使，又，故只需使，即，即，当或6时，符合要求，故存在“长向量”，且“长向量”为，.∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（8分）（3）由题意，得，，即，即，同理，，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（10分）三式相加并化简，得，即，，所以，设，由得∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（12分）设，则依题意得：∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（13分）()452245221,12,1m k k m k k m λλλ⎧++=-⎪⎪-⎨+⎪=⎪-⎩()()()22222123445452216161616162221621611m m s s s s k k m m k k m m λλλλ=+++=+--+=-+--1234s s s s 220λ-=0λ>λ=Q 22(1x y +=312a a a +…40x -……2a 6a1n a ==p a1n p S a - …()()712371010101,01010100,1S a a a a =++++=+-+++--+++-+=-71p S a -=== 022cos12p π+ (1)1cos 22p π--……2p =2a 6a123a a a + (2)2123a a a + …()22123a a a +...222123232a a a a a ++⋅ (2)22213132a a a a a ++⋅ …222312122a a a a a ++⋅…2221231213230222a a a a a a a a a +++⋅+⋅+⋅…()21230a a a ++…1230a a a ++ …1230a a a ++=()3,a u v = 1220a a a ++= sin 2cos ,cos 2sin ,u x x v x x =--⎧⎨=--⎩(),n n n P x y ()()()()()()212111222222222121,2,,,,2,,,k k k k k k k k x y x y x y x y x y x y ++++++⎧=-⎪⎨=-⎪⎩得，故，，所以，，当且仅当时等号成立，∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（16分）故.∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙（17分）()()()()2222221122,2,,,k k k k x y x y x y x y ++⎡⎤=-+⎣⎦()()()()2222221122,2,,,k k x y k x y x y x y ++⎡⎤=-+⎣⎦()()()()2121221122,2,,,k k x y k x y x y x y ++⎡⎤=--+⎣⎦()()()212222212221221112,4,,4k k k k k k P P x x y y k x y x y k PP ++++++⎡⎤=--=-=⎣⎦22212(sin 2cos )(cos 2sin )58sin cos 54sin21PP x x x x x x x =--+--=+=+ …()4x t t ππ=-∈Z 10151016min1014420282P P =⨯=。

雅礼中学2014届高三第六次月考试卷历史第Ⅰ卷选择题（48分）一、选择题（本大题共24小题，每小题2分，共计48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．陈苏镇在《汉代政治与〈春秋〉学》说：“一方面必须“承秦”，包括承秦之制，另一方面又必须尊重东方社会之习俗，特别是楚、齐、赵人之俗。

这是历史对刘邦的苛刻要求，也是汉初实行郡国并行制的深层背景。

”材料说明西汉初年实行郡国并行是基于A．为了打败刘邦的需要B．吸取秦朝灭亡的原因C．东西方文化的差异D．刘邦的中庸思想【解析】C 这是新的学术观点，根据材料可以得出。

2．《晋书·王询传》记载:“魏氏给公卿以下租牛客户数各有差(不同)。

白后小人惮(害怕)役(服劳役)，多乐为之，贵势之门，动有百数。

又太原诸部亦以匈奴胡人为田客，多者数千。

”这主要表明A.两晋民族对峙严重，民族关系紧张B.晋代农民与地主的租佃关系得到发展C.魏国实行赋税制度减轻了农民的负担D.两晋察举选拔逐渐以士族门阀为依据【解析】B A项不正确，材料中只有“以匈奴胡人为田客”的表述，没有反映出民族关系对峙严重的信息。

B项正确，从材料第一句话“租牛客户数”、“以匈奴胡人为田客”，都反映了租佃关系的存在和发展。

C项不正确，材料中反映的是民众害怕“服劳役”，所以愿意到地主贵族的租田佃户。

D项在材料中没有体现。

3．唐太宗时，全国分为十个道，每个道包括若干个州，以便于皇帝派官员视察各州的行政工作。

与“道”职能相当的是A．秦朝的郡B．西汉的州C．宋朝的谏院D．元朝的中书省【解析】B 职掌规谏朝政缺失，西汉的州开始和唐朝的道一样，开始是监察地方的机构，后来才演变为行政区。

4.一位正担任中央大臣者，回忆其早年入仕过程，虽顺利通过礼部考试，因未能通过吏部测试，不得派任官职，只好暂时接受地方政府首长聘任，担任僚佐。

根据你的历史知识，这个官僚可能身处那个时期？A汉代B唐代C元代D清代【解析】B 唐代科举采用资格考试与任官考试分途的方法，考生州、县试合格后，再到礼部按科别参加考试，及格后仅能取得任官资格，考生还要再通过吏部考试，才能分配任官.5. 法国学者加奈隆说：“民众支配雅典，演说支配民众。

炎德·英才大联考雅礼中学2014届高三月考试卷（四）物 理高三物理备课组组稿【考试范围：必修1、必修2、选修3-1静电场，恒定电流，磁场】本试卷分选择题和非选择题两部分，共8页。

时量90分钟，满分100分。

第Ⅰ卷 选择题（共48分）一、选择题（本题包含12小题，每小题4分，共48分，其中1～8小题只有一个选项正确，9～12小题有多个选项正确，全部选对的得4分，选对但不全的得2分，错选或不选得0分，将选项填在答题卷）1、如图1所示，放置在水平地面上的直角劈M 上有一个质量为m 的物体，若m 在直角劈上面匀加速下滑，Ｍ仍保持静止，那么正确的说法是（ D ）A.Ｍ对地面的压力等于（M+m ）gB.Ｍ对地面的压力大于（M+m ）gC.地面对Ｍ没有摩擦力D.地面对Ｍ有向左的摩擦力2、如图2所示，A 、B 是带等量同种电荷的小球，电荷量是105 C ，A 固定在竖直放置的10 cm 长的绝缘支杆上，B 平衡于倾角为30°的绝缘粗糙斜面上时，且与A等高，若B 的质量为310Kg ，则B 受到斜面的静摩擦力为（ D ）(g 取l02/m s ) A ．15 N B ．310 N C ．315 N D ．335 N3、如图3所示，小球从高处下落到竖直放置的轻弹簧上，那么小球从接触弹簧开始到将弹簧压缩到最短的过程中（弹簧一直保持竖直），下列说法中正确的是（ Ｃ ）Ａ．小球在最低点受到的弹力等于小球重力的2倍Ｂ．小球的速度一直在增大Ｃ．小球的动能先增大后减小Ｄ．小球的加速度最大值等于重力加速度4、 如图4所示，发射地球同步卫星时，先将卫星发射至近地圆轨道1上，然后经点火，卫星沿椭圆轨道2运行，最后再次点火，将卫星送入轨道3，轨道1、2相切于Q 点，轨道2、3相切于P 点，则当卫星分别在1、2、3轨道上正常运动时。

下列说法正确的是（D ）A. 将卫星直接发射到轨道3上的发射速度可能等于发射到轨道1上的发射速度的2倍B. 卫星在轨道3上的周期小于在轨道1上的周期C. 卫星在轨道1上经过Q 点时的加速度大于它在轨道2上经过Q 点时的加速度D. 卫星从轨道1到轨道3动能的变化量小于势能的变化量5 、如图所示，虚线a 、b 和c 是某点电荷形成的电场中的三个间距相等的 图2等势面，它们的电势为a b c U U U 、、，一带正电的粒子射入电场中，其运动轨迹如实线KLMN 所示，由图可知（ D ）A ．当粒子从M 到N 的过程中，动能减少，电势能增加B ．当粒子从K 到L 的过程中，电场力做正功，机械能增加C ．形成电场的点电荷一定带负电，且相邻的电势差ab bc U U =D ．形成电场的点电荷一定带正电，且相邻的电势差ab bc U U >6．如图6所示，相距为d 的水平金属板M 、N 的左侧有一对竖直金属板P 、Q ，板P 上的小孔S 正对板Q 上的小孔O ，M 、N 间有垂直于纸面向里的匀强磁场，在小孔S 处有一带负电粒子，其重力 和初速度均不计，当滑动变阻器的滑片在AB 的中点时，带负电粒子恰能在M 、N 间做直线运动，当滑动变阻器的滑片滑到A 点后 ( D )A ．粒子在PQ 间运动过程中，带负电粒子电势能能一定增大B ．粒子在MN 间运动过程中，带负电粒子动能一定增大C ．粒子在MN 间运动过程中，带负电粒子轨迹向上极板偏转弯曲,机械能一定增大D ．粒子在MN 间运动过程中，带负电粒子轨迹向下极板偏转弯曲,电势能不变7、如图7所示，在竖直平面内有一半径为R 的圆弧轨道，半径OA 水平、OB 竖直，一个质量为m 的小球自A 的正上方P 点由静止开始自由下落，小球沿轨道到达最高点B 时对轨道压力为2mg 。

雅礼中学2010届高三月考试卷(六)化学试卷考生须知：1．本试卷考查内容：选修5占30%，其他内容占70%。

2．本试卷共5页（含答卷），共21小题，满分为100分，考试时量90分钟。

3．本卷可能用到的相对原子质量：H－1 N－14 O－16 Cl—35.5 C—12 Si—28 Na－23 Fe—56 Mg—24 Al—27 Cu—64第Ⅰ卷（本卷包含16个小题，共48分）一．选择题（每小题只有一个符合题意的选项，每小题3分，共48分）：1．下列说法中不正确的是（）①三聚氰胺（C3N6H6）俗称“蛋白精”，加入到奶粉中的目的是提高含氮量，以提高蛋白质的含量，增加奶粉的营养，对身体有益；②医疗上可用碳酸钡作X射线透视肠胃的内服药；③用新制备的Cu(OH)2悬浊液与病人尿液共热，可检验病人尿液中是否含有葡萄糖；④铝和铜具有良好的导电性，所以电工操作时，可以把铜线和铝线绞接在一起；⑤明矾可以用于净水，主要是由于铝离子可以水解得到氢氧化铝；⑥“水滴石穿”主要是溶解了CO2的雨水与CaCO3长期作用生成了可溶性的Ca(HCO3)2的缘故。

A. ②③⑤B. ①②④C.③④⑥D. ③④⑤【答案】B 2．某炔烃经催化加氢后可得到2-甲基丁烷，则该炔烃的名称是（）A.2-甲基-1-丁炔B.2-甲基-3-丁炔C.3-甲基-1-丁炔D.3-甲基-2-丁炔【答案】C 3．设N A为阿伏加德罗常数的数值，则下列说法正确的是（）A．1mol甲基（CH3-）所含有的电子总数为10N AB. 铁、铜和稀硫酸构成原电池，当转移N A个电子，产生的气体体积一定为11.2 LC. 足量的单质铁与1mol氯气充分反应后转移的电子总数为3N AD．对于反应：C2H2 (g)+5/2O2(g) == 2CO2(g)+H2O(l) △H=－1300 kJ·mol-1，若有4N A个碳氧双键形成时，则可放出1300 kJ的能量【答案】D 4．某酯的结构可表示为：C m H2m＋1COOC n H2n＋1，其中m＋n＝5，若该酯在酸性条件下水解得到的一种水解产物经催化氧化可最终转化成它的另一种水解产物，则原来的酯是（）A.乙酸丙酯B.乙酸乙酯C. 丙酸丙酯D. 丙酸异丙酯【答案】C 5．下列各组离子在指定溶液中一定能够大量共存的是（）A．pH=12的溶液中：K+，Na+，AlO2—，S2—，SO32—B．无色溶液中：K+，Cl—，MnO4—，PO43—，SO42—C．水电离的H+浓度C(H+)=10—12mol·L—1的溶液中：ClO-，SO42-，NO3-，NH4+，Na+D．某强酸性溶液中：Fe2+，Al3+，NO3—，I—，Cl—【答案】A6．把固体Ca(OH)2放入适量的蒸馏水中，一段时间后达到溶解平衡：Ca(OH)2（s）Ca2+ (aq)+ 2OH-(aq)，则下列说法中正确的是（）A．给溶液加热，溶液中的Ca2+浓度一定增大B．恒温条件下向溶液中加入少量CaO，溶液的pH升高C．向溶液中加入少量CH3COONa晶体，则Ca(OH)2固体的质量增多D．向溶液中加入少量冰醋酸，溶液中的OH-浓度增大【答案】C 7．已知某有机物结构简式为：若将Na、NaOH、NaHCO3分别与等物质的量的该有机物恰好完全反应时，则消耗Na、NaOH、NaHCO3的物质的量之比为（）A．3∶3∶2 B．3∶2∶1 C．3∶2∶2 D．3∶1∶1 【答案】B 8．一定量的甲烷燃烧后产物为CO、CO2和H2O（g），共重144 g，此混合气体缓慢通过浓H2SO4后，浓H2SO4增重72 g，则燃烧产物中CO2的质量为（）A. 72gB. 66 gC. 48 gD. 44 g 【答案】D 9．现有在固定容积的密闭容器中进行的可逆反应：4L（g）2M（g）+N（g）△H＞0，下图表示温度，则图中Y轴可以表示的涵义是（）和压强对该平衡状态的影响，同时已知有：p p12A. 体系中物质L的百分含量B. 体系中总的气体分子数目C. 气体混合物的平均相对分子质量D. 气体混合物的平均密度【答案】C10．右图中为直流电源，为浸透饱和氯化钠溶液和酚酞试液的滤纸，为电镀槽，通电后发现上的c点附近显红色，则下列叙述中正确的是（）A．e极上发生还原反应B．a为直流电源的负极C．若要实现铁上镀锌，则f电极为锌板D．c极上发生的反应：2H+ ＋2e－＝H2↑【答案】D11．将CO2持续通入下列八种饱和溶液：①Na2CO3②K2SiO3 ③NaAlO2 ④C6H5ONa ⑤Ca(ClO)2⑥BaCl2⑦NH3和NaCl ⑧Ca(OH)2，最终能够得到沉淀或析出晶体的是（）A．②⑤⑥⑧B．③④⑤⑧C．①④⑥⑦D．①②③⑦【答案】D 12．现有1.0 mol/L的NaOH溶液0.2L，若通入4.48L（标准状况）SO2气体使其充分反应后，则所得溶液中各粒子浓度大小关系正确的是（）A．c(SO32―) + c( OH―) = c(H＋) + c(H2SO3)B．c(Na+) = c(H2SO3) + c(HSO3―) + c ( H+)C．c(Na+)+c(H＋) = c(HSO3―) +c(SO32―) + c( OH―)D．c(Na+)＞c(HSO3―)＞c( OH―)＞c(H2SO3)＞c(SO32―)＞c(H＋) 【答案】A 13．霉酚酸酯(MMF)是器官移植中抑制细胞增殖最常用的药物，下列关于MMF的说法正确的是( )A．MMF所有碳原子一定处在同一平面B．1molMMF能与含3molNaOH溶液完全反应C．MMF的分子式为C24H31O7N D．MMF能发生取代、氧化、加成和消去反应【答案】B 14．在浓度为C mol/L的AgNO3溶液VmL中，加入一定量的pH=1的盐酸时恰好沉淀完全，此时得到pH=2的溶液100mL，则C约为（设反应前后溶液体积变化忽略不计）（）A、0.011B、0.022C、0.01D、0.1 【答案】A 15．分子式为C4H7Cl的属于链状化合物的可能结构共有（包含顺反异构和对映异构）（）A．14种B．12种C．11种D．8种【答案】B 16.现有一定量的铁粉和铜粉的混合物，将其平均分成四等份，分别加入同浓度不同体积的稀硝酸，充分反一种)：根据上表中的数据计算分析，下列推断正确的是( )A．①中溶解了5.6 g Fe B．②中溶解了9.6 g CuC．硝酸的浓度为4 mol/L D．④中V = 6720 【答案】C第Ⅱ卷（本卷包括5小题，共52分）二、填空题（每空2分，共52分）：17．（8分）已知：氨水的电离程度与醋酸的电离程度在同条件下几乎相等。