辽宁省凌源市第三中学2019-2020学年高二下学期第三次线上月考数学word版

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凌源市第三中学2018-2019学年高三上学期第三次月考试卷数学含答案

凌源市第三中学2018-2019学年高三上学期第三次月考试卷数学含答案班级__________ 座号_____ 姓名__________ 分数__________一、选择题（本大题共12小题，每小题5分，共60分.每小题给出的四个选项中，只有一项是符合题目要求的.）1．已知正三棱柱111ABC A B C -的底面边长为4cm ，高为10cm ，则一质点自点A 出发，沿着三棱柱的侧面，绕行两周到达点1A 的最短路线的长为（）A ．16cmB ．123cmC ．243cmD ．26cm2．一个几何体的三个视图如下，每个小格表示一个单位, 则该几何体的侧面积为（）A.4πB.25πC. 5πD. 25π+π【命题意图】本题考查空间几何体的三视图，几何体的侧面积等基础知识，意在考查学生空间想象能力和计算能力．3．如图，AB 是半圆O 的直径，AB ＝2，点P 从A 点沿半圆弧运动至B 点，设∠AOP ＝x ，将动点P 到A ，B 两点的距离之和表示为x 的函数f （x ），则y ＝f （x ）的图象大致为（）4．底面为矩形的四棱锥P -ABCD 的顶点都在球O 的表面上，且O 在底面ABCD 内，PO ⊥平面ABCD ，当四棱锥P -ABCD 的体积的最大值为18时，球O 的表面积为（）A ．36πB ．48πC ．60πD ．72π5．在ABC ∆中，若60A ∠=，45B ∠=，BC =AC =（）A ．B ． C.D ．26．如图所示，在三棱锥P ABC -的六条棱所在的直线中，异面直线共有（）111]A ．2对B ．3对C ．4对D ．6对7．已知函数x x x f 2sin )(-=，且)2(),31(log ),23(ln 3.02f c f b f a ===，则（） A ．c a b >> B ．a c b >> C ．a b c >> D ．b a c >>【命题意图】本题考查导数在单调性上的应用、指数值和对数值比较大小等基础知识，意在考查基本运算能力． 8．圆心在直线2x ＋y ＝0上，且经过点（－1，－1）与（2，2）的圆，与x 轴交于M ，N 两点，则|MN |＝（） A ．4 2 B ．4 5 C ．2 2D ．2 59．一个空间几何体的三视图如图所示，其中正视图为等腰直角三角形，侧视图与俯视图为正方形，则该几何体的体积为（）A ．64B ．32C ．643 D ．32310．下列命题正确的是（）A ．已知实数,a b ，则“a b >”是“22a b >”的必要不充分条件B ．“存在0x R ∈，使得2010x -<”的否定是“对任意x R ∈，均有210x ->”C ．函数131()()2xf x x =-的零点在区间11(,)32内D ．设,m n 是两条直线，,αβ是空间中两个平面，若,m n αβ⊂⊂，m n ⊥则αβ⊥11．函数()2cos()f x x ωϕ=+（0ω>，0ϕ-π<<）的部分图象如图所示，则 f （0）的值为（）A.32-B.1-C.D.【命题意图】本题考查诱导公式，三角函数的图象和性质，数形结合思想的灵活应用.12．已知双曲线2222:1(0,0)x y C a b a b-=>>，12,F F 分别在其左、右焦点，点P 为双曲线的右支上的一点，圆M 为三角形12PF F 的内切圆，PM 所在直线与轴的交点坐标为(1,0)，与双曲线的一条渐近线平行且距离为2，则双曲线C 的离心率是（）A B ．2 C D ．2二、填空题（本大题共4小题，每小题5分，共20分.把答案填写在横线上）13．设某双曲线与椭圆1362722=+y x 有共同的焦点，且与椭圆相交，其中一个交点的坐标为 )4,15(，则此双曲线的标准方程是 .14．设函数32()(1)f x x a x ax =+++有两个不同的极值点1x ，2x ，且对不等式12()()0f x f x +≤恒成立，则实数的取值范围是．15．设x ，y 满足约束条件，则目标函数z=2x ﹣3y 的最小值是．16．在△ABC 中，角A ，B ，C 所对的边分别为a ，b ，c ，若△ABC 不是直角三角形，则下列命题正确的是（写出所有正确命题的编号）①tanA •tanB •tanC=tanA+tanB+tanC②tanA+tanB+tanC 的最小值为3③tanA ，tanB ，tanC 中存在两个数互为倒数 ④若tanA ：tanB ：tanC=1：2：3，则A=45°⑤当tanB ﹣1=时，则sin 2C ≥sinA •sinB ．三、解答题（本大共6小题，共70分。

辽宁省凌源市第三中学2019-2020学年高二下学期第三次线上月考英语试卷

英语试题第一部分阅读（共两节, 满分35分）第一节（共10小题；每小题2.5分, 满分25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。

ANotice for Tourists to Beijing Expo 2019The 2019 Beijing Horticultural Expo (园博会) from April 29 to October 7 is an opportunity for international discovery, where cultural elements of different places in China are on display, as well as gardens and buildings about each country’s wonderful culture and history. Here is a notice for whoever plans to come.Quick Entry by Showing ID CardsTo make sure your quick and orderly entry into the Expo Site, please wait in lines and get your ID card ready for pre-check, ticket check and security inspection before entering. During the ticket check, those who hold discounted tickets must provide the materials about discount.Safe Tour to the ExpoConsidering your personal safety, please don’t take anything that may disturb other visitors or the order of the Expo Site, such as pets, kites, and speakers. Except for wheelchairs for the elderly and the disabled and strollers (婴儿推车) for children, no vehicles(机动车辆) are allowed to enter the Expo Site.Taking Action to Protect the EnvironmentTo create a green, beautiful environment for yourself and others, please follow the travel rules, protect public buildings, keep environment clean and care plants growing in the Expo Site. Behaviors such as climbing or destroying structures and exhibits are not permitted.Orderly Travel by Limiting the NumberFor the safety of visitors, sometimes staff members may limit the number of tourists into some gardens and buildings in the event of overcrowding. Thank you for your understanding.Enjoying Service and Having FunTourist service centers in the Expo Site provide services including information inquiry (查询),storage of personal belongings, wheelchair and stroller renting, lost found service, and search for missing persons. We hope you have a good time at the Expo.You can call the service hotline at 86-10-86484017 for inquiry, suggestion or complaint. Thanks for your attention.1. Which of the following is allowed to enter the Expo Site?A. Wheelchairs.B. Dogs.C. Trucks.D. Cars.2. Which service is NOT provided by the tourist service centers in the Expo Site?A. Stroller renting.B. Information inquiry.C. Selling the souvenirs.D. Lost found service.3. What can we learn from the passage?A. Visitors can buy tickets onlineB. The number of the tourists is not limited.C. Whoever climbs the structure will get fined.D. Visitors should show ID card before entering.BFive years ago, when I taught art at a school in Seattle, I used Tinkertoys as a test at the beginning of a term to find out something about my students. I put a small set of Tinkertoys in front of each student, and said: “Make s omething out of the Tinkertoys. You have 45 minutes today—and 45 minutes each day for the rest of the week.”A few students hesitated to start. They waited to see what the rest of the class would do. Several others checked the instructions and made something according to one of the model plans provided. Another group built something out of their own imaginations.Once I had a boy who worked experimentally with Tinkertoys in his free time. His constructions filled a shelf in the art classroom and a good part of his bedroom at home. I was delighted at the presence of such a student. Here was an exceptionally creative mind at work. His presence meant that I had an unexpected teaching assistant in class whose creativity would infect(感染)other students.Encouraging this kind of thinking has a downside. I ran the risk of losing those students who had a different style of thinking. Without fail one would declare, “But I’m just not creative.”“Do you dream at night when you’re asleep?”“Oh, sure.”“So tell me one of your most interesting dreams.”The student would tell something wildly imaginative. Flying in the sky or in a time machine or growing three heads. “That’s pretty creative. Who does that for you?”“Nobody. I do it.”“Really—at night, when you’re asleep?”“Sure.”“Try doing it in the daytime, in class, okay?”4. The teacher used Tinkertoys in class in order to.A. know more about the studentsB. make the lessons more excitingC. raise the students’ interest in artD. teach the students about toy design5. What do we know about the boy mentioned in Paragraph 3?A. He liked to help his teacher.B. He preferred to study alone.C. He was active in class.D. He was imaginative.6. What does the underlined word “downside” in Paragraph 4 probably mean?A. Mistake.B. Drawback.C. Difficulty.D. Burden.CWe have heard some interesting ways that 5G wireless technology might change our lives in the future.5G, short for the 5th generation mobile communication technology, promises Internet speeds between 50 to 100 times faster than current 4G systems. While 5G is set to be used in some limited areas of America this year, much of the world is not expected to receive widely available service until 2023.One project in Britain, however, is already testing this superfast technology on an unlikely group of Internet users-cows. The project was developed by American technology company Cisco Systems. It also receives money from the British government. Cisco says the program seeks to explore the future of 5G connectivity in rural areas around the world.Testing areas were set up at farms in three rural areas of England. The cows are equipped with 5G-connected devices (装置) that link up to a robotic milking system, which uses sensors and machine learning to fully automate the process. System designers say technology takes over after acow feels ready to be milked and walks toward an automatic gate. The device is designed to recognize each individual cow. It then positions equipment to the right body position for milking. During the process, machines release food for the cow as a reward.Other 5G technology tools include automated brushes that turn on when the cow rubs up against them. Sensors also control the amount of light to the cows’ living areas depending on the weather. And, an automatic feeding system makes sure the animals always get enough to eat.Duncan Forbes, head of the project, told Reuters that the project shows the farm’s cow operations can be greatly improved with 5G technology and that the experiment provides strong evidence that 5G technology can be widely used in the future, not just on farms in Britain, but in rural communities across the world.7. What is the purpose of Cisco Systems’ program?A. To win financial support from British government.B. To test the effects of 5G technology on animalsC. To promote its technological development in BritainD. To expand the future use of 5G in rural communities.8. What does the underlined word “It” in Paragraph 3 refer to?A. The project.B. The company.C. The technology .D. The group.9. What can we learn about 5G according to the text?A. It is no worse than 4G in terms of speed.B. It is already widely available in the world.C. It enables cows to control their own milking.D. It is based on sensors and machine learning.10. In which section of a newspaper may this text appear?A. Entertainment.B. Lifestyle.C. Education.D. Technology.第二节七选五（每小题2分，共10分）Children art education is something that is much encouraged for children’s creative growt h. This is necessary because a child can have a sense of appreciation of the arts and along with all the other things that they learn in school. 11Children art education should be designed from a very early age so that they are allowed to express themselves freely in whatever way that they wish to. ___12___However, the task of children art education is not a very easy one, because you have to find the right way in which you can help them get interested in what you are trying to teach them. ___13 The physical space where the art education is to be carried out should also be carefully decorated to make it attractive to the children. Use of colors should be made in abundance(丰富) as that is what attracts a child first to it.Displaying examples of artwork that you think would be important enough to influence the children is also a good idea. Of course, a photograph of a painting of the Madonna(圣母马利亚) is something that children might not appreciate. 14 This will be a source of inspiration for the children to try them out as well.The creative process is something that you must not interfere with(干涉) during a children art education class. 15 But after that, you must wait for the children themselves to come up with something meaningful to their ability. As long as the result is connected with the topic, every child’s art work is worth praising.A. You can just give them a topic for drawing.B. Instead, put up paintings that children will be fond of.C. The reason for this could be the lack of interest in the teacher.D. A trip to an art museum can be an inspiration to many students.E. They also have a place where they can express all their feelings.F. This is the only way that art can be appreciated as children grow up.G. So creating the right environment for children art education is very important.第二部分语言运用（共两节, 满分45分）第一节完形填空（每小题1.5分，共30分）Back in the 15th century, in a tiny village near Nuremberg lived a family with several children. In order to 16 these children, the father worked hard. Despite their seemingly 17 condition, two of the children had a 18 . They both wanted to pursue their talent for 19 , but their father would never be 20 able to send either of them to Nuremberg to study at the Academy.After long 21 , the two boys finally came to an agreement. They would toss(投掷) a coin. The loser would work in the nearby _22 to support his brother while he attended the Academy. Then, when the winner completed his studies, he would support the other either with _23_ of his artwork or by laboring in the mines.They tossed a coin as they had 24 . Albrecht Durer won the toss and 25 to NuRemberg. Albert went to work in the dangerous mines and financed his brother. Albrecht’s26 were far better than those of most of his professors, and by the time he 27 , he was beginning to earn lots of money for his works.When the young artist returned to his village, he thanked his beloved 28 for the years of sacrifice that had made it 29 to achieve his ambition, and said, “Now, it is your 30 to pu rsue your dream, and I will support you.”31 , Albert refused Albrecht and said softly,“It is too 32 for me. The bones in every finger have been broken. I cannot even hold a glass, 33 hold a pen or a brush.”One day, to show gratitude(感激) to Albert, Albrecht _34 his brother’s abused hands as a gift and called his powerful drawing simply “Hands”, but the entire world renamed his 35 of love “The Praying Hands”.16.A. educate B. raise C. praise D. blame17.A. excellent B. reasonable C. hopeless D. attractive18.A. quarrel B. fight C. secret D. dream19.A. art B. music C. acting D. writing20.A. physically B. financially C. mentally D. psychologically21.A. expressions B. preparation C. discussions D. unemployment22.A. cities B. academies C. factories D. mines23.A. sales B. copies C. models D. displays24.A. pursued B. agreed C. repeated D. predicted25.A. took off B. drove away C. went off D. ran away26.A. reports B. lessons C. articles D. works27.A. graduated B. married C. retired D. died28.A. father B. brother C. professors D. villagers29.A. convenient B. important C. possible D. flexible30.A. turn B. luck C. dignity D. honor31.A. However B. Moreover C. Therefore D. Otherwise32.A. early B. late C. exciting D. challenging33.A. rather than B. more than C. or rather D. much less34.A. drew B. cured C. washed D. tested35.A. game B. story C. gift D. idea第二节语法填空（共10小题，每小题1.5分，满分15分）阅读下面材料，在空白处填入适当的内容(1个单词)或括号内单词的正确形式。

辽宁省凌源市第三中学2019-2020学年高二下学期第三次线上月考英语试卷(含解析)

英语试题第一部分阅读（共两节, 满分35分）第一节（共10小题；每小题 2.5分, 满分25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。

ANotice for Tourists to Beijing Expo 2019The 2019 Beijing Horticultural Expo (园博会) from April 29 to October 7 is an opportunity forinternational discovery, where cultural elements of different places in China are on display, as wellas gardens and buildings about each country’s wonderful culture and history. Here is a notice for whoever plans to come.Quick Entry by Showing ID CardsTo make sure your quick and orderly entry into the Expo Site, please wait in lines and get yourID card ready for pre-check, ticket check and security inspection before entering. During the ticketcheck, those who hold discounted tickets must provide the materials about discount.Safe Tour to the ExpoConsidering your personal safety, please don’t take anything that may disturb other visitors or the order of the Expo Site, such as pets, kites, and speakers. Except for wheelchairs for the elderlyand the disabled and strollers (婴儿推车) for children, no vehicles(机动车辆) are allowed to enterthe Expo Site.Taking Action to Protect the EnvironmentTo create a green, beautiful environment for yourself and others, please follow the travel rules,protect public buildings, keep environment clean and care plants growing in the Expo Site.Behaviors such as climbing or destroying structures and exhibits are not permitted.Orderly Travel by Limiting the NumberFor the safety of visitors, sometimes staff members may limit the number of tourists into somegardens and buildings in the event of overcrowding. Thank you for your understanding.Enjoying Service and Having FunTourist service centers in the Expo Site provide services including information inquiry (查询),storage of personal belongings, wheelchair and stroller renting, lost found service, and search formissing persons. We hope you have a good time at the Expo.You can call the service hotline at 86-10-86484017 for inquiry, suggestion or complaint. Thanksfor your attention.1. Which of the following is allowed to enter the Expo Site?A. Wheelchairs.B. Dogs.C. Trucks.D. Cars.2. Which service is NOT provided by the tourist service centers in the Expo Site?A. Stroller renting.B. Information inquiry.C. Selling the souvenirs.D. Lost found service.3. What can we learn from the passage?A. Visitors can buy tickets onlineB. The number of the tourists is not limited.C. Whoever climbs the structure will get fined.D. Visitors should show ID card before entering.BFive years ago, when I taught art at a school in Seattle, I used Tinkertoys as a test at thebeginning of a term to find out something about my students. I put a small set of Tinkertoys infront of each student, and said: “Make something out of the Tinkertoys. You have 45 minutestoday—and 45 minutes each day for the rest of the week.”A few students hesitated to start. They waited to see what the rest of the class would do.Several others checked the instructions and made something according to one of the model plansprovided. Another group built something out of their own imaginations.Once I had a boy who worked experimentally with Tinkertoys in his free time. Hisconstructions filled a shelf in the art classroom and a good part of his bedroom at home. I wasdelighted at the presence of such a student. Here was an exceptionally creative mind at work. Hispresence meant that I had an unexpected teaching assistant in class whose creativity wouldinfect(感染)other students.Encouraging this kind of thinking has a downside. I ran the risk of losing those students whohad a different style of thinking. Without fail one would declare, “But I’m just not creative “Do you dream at night when you’re asleep?”“Oh, sure.”“So tell me one of your most interesting dreams.”The student would tell something wildly imaginative. Flying in the sky or in a time machine or growing three heads. “That’s pretty Who does that for you?”“Nobody. I do it.”“Really—at night, when you’re asleep?”“Sure.”“Try doing it in the daytime, in class, okay?”4. The teacher used Tinkertoys in class in order to.A. know more about the studentsB. make the lessons more excitingC. raise the students’ interest in artD. teach the students about toy design5. What do we know about the boy mentioned in Paragraph 3?A. He liked to help his teacher.B. He preferred to study alone.C. He was active in class.D. He was imaginative.6. What does the underlined word “downside” in Paragraph 4 probably mean?A. Mistake.B. Drawback.C. Difficulty.D. Burden.CWe have heard some interesting ways that 5G wireless technology might change our lives inthe future.5G, short for the 5th generation mobile communication technology, promises Internet speedsbetween 50 to 100 times faster than current 4G systems. While 5G is set to be used in some limitedareas of America this year, much of the world is not expected to receive widely available serviceuntil 2023.One project in Britain, however, is already testing this superfast technology on an unlikelygroup of Internet users-cows. The project was developed by American technology company CiscoSystems. It also receives money from the British government. Cisco says the program seeks toexplore the future of 5G connectivity in rural areas around the world.Testing areas were set up at farms in three rural areas of England. The cows are equipped with5G-connected devices (装置) that link up to a robotic milking system, which uses sensors andmachine learning to fully automate the process. System designers say technology takes over after acow feels ready to be milked and walks toward an automatic gate. The device is designed torecognize each individual cow. It then positions equipment to the right body position for milking.During the process, machines release food for the cow as a reward.Other 5G technology tools include automated brushes that turn on when the cow rubs upagainst them. Sensors also control the amount of light to the cows’ living areas depending on the weather. And, an automatic feeding system makes sure the animals always get enough to eat.Duncan Forbes, head of the project, told Reuters that the project shows the farm’s cow operations can be greatly improved with 5G technology and that the experiment provides strongevidence that 5G technology can be widely used in the future, not just on farms in Britain, but inrural communities across the world.7. What is the purpose of Cisco Systems’ program?A. To win financial support from British government.B. To test the effects of 5G technology on animalsC. To promote its technological development in BritainD. To expand the future use of 5G in rural communities.8. What does the underlined word “It” in Paragraph 3 refer to?A. The project.B. The company.C. The technology .D. The group.9. What can we learn about 5G according to the text?A. It is no worse than 4G in terms of speed.B. It is already widely available in the world.C. It enables cows to control their own milking.D. It is based on sensors and machine learning.10. In which section of a newspaper may this text appear?A. Entertainment.B. Lifestyle.C. Education.D. Technology.第二节七选五（每小题2分，共10分）h.Children art education is something that is much encouraged for children’s creative growt This is necessary because a child can have a sense of appreciation of the arts and along with all theother things that they learn in school. 11Children art education should be designed from a very early age so that they are allowed toexpress themselves freely in whatever way that they wish to. ___12___However, the task of children art education is not a very easy one, because you have to findthe right way in which you can help them get interested in what you are trying to teach them. ___13 The physical space where the art education is to be carried out should also be carefully decorated to make it attractive to the children. Use of colors should be made in abundance(丰富) as that is what attracts a child first to it.Displaying examples of artwork that you think would be important enough to influence the children is also a good idea. Of course, a photograph of a painting of the Madonna(圣母马利亚) is something that children might not appreciate. 14 This will be a source of inspiration for the children to try them out as well.The creative process is something that you must not interfere with(干涉) during a children art education class. 15 But after that, you must wait for the children themselves to come up with something meaningful to their ability. As long as the result is connected with the topic, every child’s art work is worth praising.A. You can just give them a topic for drawing.B. Instead, put up paintings that children will be fond of.C. The reason for this could be the lack of interest in the teacher.D. A trip to an art museum can be an inspiration to many students.E. They also have a place where they can express all their feelings.F. This is the only way that art can be appreciated as children grow up.G. So creating the right environment for children art education is very important.第二部分语言运用（共两节, 满分45分）第一节完形填空（每小题 1.5分，共30分）Back in the 15th century, in a tiny village near Nuremberg lived a family with several children. In order to 16 these children, the father worked hard. Despite their seemingly 17 condition, two of the children had a 18 . They both wanted to pursue their talent for 19 , but their father would never be 20 able to send either of them to Nuremberg to study at the Academy.After long 21 , the two boys finally came to an agreement. They would toss(投掷) a coin. The loser would work in the nearby _22 to support his brother while he attended the Academy. Then, when the winner completed his studies, he would support the other either with _23_ of his artwork or by laboring in the mines.They tossed a coin as they had 24 . Albrecht Durer won the toss and 25 to NuRemberg.26 were farAlbert went to work in the dangerous mines and financed his brother. Albrecht’ｓbetter than those of most of his professors, and by the time he 27 , he was beginning to earn lots ofmoney for his works.When the young artist returned to his village, he thanked his beloved 28 for the years ofsacrifice that had made it 29 to achieve his ambition, and said, “Now, i t is your 30 topursue your dream, and I will support you.”31 , Albert refused Albrecht and said softly,“Itis too 32 for me. The bones in everyfinger have been broken. I cannot even hold a glass, 33 hold a pen or a brush.”One day, to show gratitude(感激) to Albert, Albrecht _34 his brother’s abused hands as a gift and called his powerful drawing simply “Hands”, but the entire world renamed his35 oflove “The Praying Hands”．16.A. educate B. raise C. praise D. blame17.A. excellent B. reasonable C. hopeless D. attractive18.A. quarrel B. fight C. secret D. dream19.A. art B. music C. acting D. writing20.A. physically B. financially C. mentally D. psychologically21.A. expressions B. preparation C. discussions D. unemployment22.A. cities B. academies C. factories D. mines23.A. sales B. copies C. models D. displays24.A. pursued B. agreed C. repeated D. predicted25.A. took off B. drove away C. went off D. ran away26.A. reports B. lessons C. articles D. works27.A. graduated B. married C. retired D. died28.A. father B. brother C. professors D. villagers29.A. convenient B. important C. possible D. flexible30.A. turn B. luck C. dignity D. honor31.A. However B. Moreover C. Therefore D. Otherwise32.A. early B. late C. exciting D. challenging33.A. rather than B. more than C. or rather D. much less34.A. drew B. cured C. washed D. tested35.A. game B. story C. gift D. idea第二节语法填空（共10小题，每小题 1.5分，满分15分）阅读下面材料，在空白处填入适当的内容(1个单词)或括号内单词的正确形式。

辽宁省凌源市第三中学2019-2020学年高二物理上学期第三次月考试题(含解析)

辽宁省凌源市第三中学2019-2020学年高二物理上学期第三次月考试题（含解析）一．选择题（1--8单选，9--12多选。

每题4分，共48分）1.电动机正常工作时A. 电能全部转化为内能B. 电能全部转化为机械能C. 电能主要转化为机械能，只有一小部分转化为内能D. 电功率大于机械功率和热功率之和【答案】C【解析】【详解】ABC.电动机正常工作时，消耗的电能主要转化为机械能和小部分内能，故选项A、B不符合题意，C符合题意；D.根据能量转化和守恒定律得知电功率等于机械功率和热功率之和，故选项D不符合题意。

2.如图所示，小磁针正上方的直导线与小磁针平行，当导线中通向右的电流时，小磁针会发生偏转。

首先观察到这个实验现象的物理学家和观察到的现象是( )A. 物理学家安培，小磁针的S极垂直转向纸内B. 物理学家安培，小磁针的N极垂直转向纸内C. 物理学家奥斯特，小磁针的S极垂直转向纸内D. 物理学家奥斯特，小磁针的N极垂直转向纸内【答案】D【解析】【详解】发现电流周围存在磁场的科学家是奥斯特，根据安培定则可知该导线下方磁场方向垂直纸面向里，因此小磁针的N极垂直转向纸内，故ABC错误，D正确。

故选D。

3.如图所示，P点在螺线管的正上方，当螺线管通以恒定电流，不计其他磁场的影响，螺线管左边的小磁针静止时，N极的指向水平向左．则下面判断中正确的是( )A. 电源的甲端是正极，P点的磁感应强度方向向右B. 电源的乙端是正极，P点的磁感应强度方向向右C. 电源的甲端是负极，P点的磁感应强度方向向左D. 电源的乙端是负极，P点的磁感应强度方向向左【答案】B【解析】【分析】根据小磁针的指向判断螺线管的N、S极；根据安培定则判断通电螺线管的电流方向。

【详解】根据小磁针的指向可知螺线管的左端为N极；根据安培定则可知通电螺线管的乙端接电源的正极；在螺线管外部的磁感线从N极到S极，可知P点的磁感应强度方向向右；故选B。

4.在图中，已标出磁场B的方向，通电直导线中电流I的方向，以及通电直导线所受磁场力F的方向，其中正确的是（）A. B. C. D.【答案】B【解析】【分析】熟练应用左手定则是解决本题的关键，在应用时可以先确定一个方向，然后逐步进行，如可先让磁感线穿过手心，然后通过旋转手，让四指和电流方向一致或让大拇指和力方向一致，从而判断出另一个物理量的方向，用这种程序法，防止弄错方向.【详解】根据左手定值可知：A图中的安培力应该垂直导线竖直向下，故A错误；B图中安培力方向垂直于导线斜向下，故B正确；C图中的导线不受安培力，故C错误；D图中的安培力应该垂直于电流方向斜向右下方，故D错误。

2019-2020学年高二数学下学期第三次月考试题文

2019-2020学年高二数学下学期第三次月考试题文一、选择题（每题5分，共60分）1.已知集合,则( )A. B. C. D.2.的值为( )A．B．C．D．3．已知命题则为（）A． B．C． D．4.最小值为( )A.-1B.C.D.15.函数的定义域是( )A. B. C. D.6.在等比数列中，，则＝( )A．4 B．2 C．±4D．±27.要得到函数的图象，只需将函数的图象( ) A．向右平移个单位B．向左平移个单位C．向左平移个单位D．向右平移个单位8.已知为等差数列，若，则( )A. 24B. 27C. 36D. 549.已知数列满足,且, 则 ( )A. 8B. 9C. 10D. 1110.已知内角A，B，C的对边分别为a，b, c, 且 ,则—定为（）A.等腰三角形B.钝角三角形C.锐角三角形D.等腰直角三角形11.设P是圆上的一点，则点P到直线的距离的最小值是( )A.2B.3C.4D.612.“”是“直线与直线平行”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二、填空题（每题5分，共20分）13．设函数,则________.14.若数列满足则=________.15.已知中, 则的面积为________16.如图所示的是函数的图象，由图中条件写出该函数的解析式为__________________.三、解答题（每题13分，共70分）17、(本小题满分13分)已知, .(1)求的值;(2)求的值;18. (本小题满分13分)已知圆C的圆心为，直线与圆C相切.(1)求圆C的标准方程；(2)若直线过点，且被圆C所截得的弦长为2，求直线的方程.19. (本小题满分13分)设的角所对边的长分别为，且．（1）求角的大小；（2）若，求的面积20.（本小题满分13分）已知等差数列的前项和，，.(1)求等差数列的通项公式；(2)求21. （本小题满分13分）在平面直角坐标系中,直线的参数方程为(为参数),以直角坐标系的原点为极点,以轴的非负半轴为极轴建立极坐标系,曲线的极坐标方程为.(1)求直线的极坐标方程和曲线的直角坐标方程;(2)已知直线与曲线交于两点,试求两点间的距离.22、延展题（本小题满分5分）已知函数，给出下列四个结论：①函数的最小正周期是②函数在区间上是减函数③函数的图象关于点对称④函数的图象可由函数的图象向左平移个单位得到其中正确结论是______．数学试卷解析一、选择题（每小题5分，本题共60分）1. 答案：C2.答案：A3. 答案：A4.答案：B5.答案：B6.答案：B7. 答案：A8.答案：C9.答案：C10.答案：A11.答案：A12.答案：D1.已知集合,则( )A. B. C. D.1.答案：C解析：依题意，，故.2.的值为( )A．B．C．D．2.答案：A解析：3．已知命题则为（）A． B．C． D．3.答案：A4.最小值为( )A.-1B.C.D.14.答案：B5.函数的定义域是( )A. B. C. D.5.答案：B解析：∵函数，∴；解得，∴函数的定义域是.故选：B6.在等比数列中，，则＝( )A．4 B．2 C．±4D．±26.答案：B7.要得到函数的图象，只需将函数的图象( ) A．向右平移个单位B．向左平移个单位C．向左平移个单位D．向右平移个单位7.答案：A8.已知为等差数列，若，则( )A. 24B. 27C. 36D. 548.答案：C9.已知数列满足,且, 则 ( )A. 8B. 9C. 10D. 119.答案：C10.已知的内角A，B，C的对边分别为a，b, c, ,则—定为（）A.等腰三角形B.钝角三角形C.锐角三角形D.等腰直角三角形10.答案：A解析：由结合正弦定理得，,从而11.设P是圆上的一点，则点P到直线的距离的最小值是( )A.2B.3C.4D.611.答案：A解析：由圆的标准方程可得圆心，所以圆心C到直线的距离为.又圆C的半径长为2，所以圆C上任一点P到直线的最小距离是.故选A.12.“”是“直线与直线平行”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12.答案：D解析：当，；两直线方程分别为：与直线此时两直线重合，充分性不成立．若直线与直线平行，则当时，两直线方程分别为或，此时两直线不平行，当，若两直线平行，则，即且，解得即必要性不成立，故选D二、填空题（每小题5分本题共20分）13．设函数,则________.答案：1514.若数列满足则=________.14.答案：解析：由,∴是以的等比数列,故.15.已知中, 则的面积为________ 15.答案：解析：由正弦定理得解得所以.16.如图所示的是函数的图象，由图中条件写出该函数的解析式为__________________.16.答案：解析：将函数的图象沿x轴向左平移个单位长度，就得到本题的图象，故所求函数为.三、解答题（共70分，17题—21题，每题的第一问满分6分，第二问满分7分）17、(本小题满分13分)已知, .(1)求的值;(2)求的值;17.答案：（1）4/3（2）（-根号2）/1018. (本小题满分13分)已知圆C的圆心为，直线与圆C相切.(1)求圆C的标准方程；(2)若直线过点，且被圆C所截得的弦长为2，求直线的方程.18.答案：（1）（2）或19. (本小题满分13分)设的角所对边的长分别为，且．（1）求角的大小；（2）若，求的面积19. 答案：(1)∵中，∴由正弦定理可得，∴，又,∴，由可得；(2)由余弦定理可得，将代入上式可得，∴的面积20.（本小题满分13分）已知等差数列的前项和，，.(1)求等差数列的通项公式；(2)求20.答案：(1) 由题可知从而有. （6分）(2) 由(1)知，从而. （12分）21. (本小题满分13分)在平面直角坐标系中,直线的参数方程为(为参数),以直角坐标系的原点为极点,以轴的非负半轴为极轴建立极坐标系,曲线的极坐标方程为.(1)求直线的极坐标方程和曲线的直角坐标方程;(2)已知直线与曲线交于两点,试求两点间的距离. 21.答案：(1)直线,即;曲线,即,曲线的普通方程为.(2)将直线的参数方程代入得即或,两点间的距离22、(延展题) (本小题满分5分)已知函数，给出下列四个结论：①函数的最小正周期是②函数在区间上是减函数③函数的图象关于点对称④函数的图象可由函数的图象向左平移个单位得到其中正确结论是______．22.答案：①③解析：函数，①因为，则的最小周期，结论正确；②当时，在上不是单调函数，结论错误；③因为，函数图象的一个堆成中心为，结论正确；④函数的图象可由函数的图象向左平移个单位得到，结论错误。

辽宁省朝阳市凌源第三高级中学2019-2020学年高二数学文模拟试卷含解析

辽宁省朝阳市凌源第三高级中学2019-2020学年高二数学文模拟试卷含解析一、选择题：本大题共10小题，每小题5分，共50分。

在每小题给出的四个选项中，只有是一个符合题目要求的1. 的展开式中的系数为()A. 80B. 40C. 20D. 10参考答案：B【分析】利用二项式展开式的通项公式得到系数.【详解】展开式的通项公式为：当时，的系数为故答案选B【点睛】本题考查了二项式定理，意在考查学生的计算能力.2. 在体积为15的斜三棱柱ABC－A1B1C1中，S是C1C上的一点，S－ABC的体积为3，则三棱锥S－A1B1C1的体积为（）A．1 B． C．2D．3参考答案：C3. ．设集合，，则=A． B． C． D．参考答案：B4. 若偶函数在上是增函数，则下列关系式中成立的是（）A、 B、C、 D、参考答案：D略5. 随机变量X～B（6，），则P（X=3）=（）A．B．C．D．参考答案：C【考点】CN：二项分布与n次独立重复试验的模型．【分析】X～B（6，）表示6次独立重复试验，每次实验成功概率为，P（X=3）表示6次试验中成功三次的概率．【解答】解：P（X=3）==故选C6. 下列关于棱柱的一些叙述正确的有（）①侧棱都相等，侧面是平行四边形；②两个底面与平行于底面的截面是全等的多边形；③过不相邻的两条侧棱的截面是平行四边形。

Ａ①②；Ｂ①③；Ｃ②③；Ｄ①②③参考答案：D略7. 已知，则函数的最小值是A．B．C． D．参考答案：C8. 设函数的最小正周期为，且，则（）A．在单调递减 B．在单调递增C．在单调递增 D．在单调递减参考答案：A9. 圆的圆心和半径分别为A. 圆心(1,3)，半径为2B. 圆心(1,－3)，半径为2C. 圆心(－1,3)，半径为4D. 圆心(1,－3)，半径为4参考答案：B10. 已知点P在曲线y=上，a为曲线在点P处的切线的倾斜角，则a的取值范围是（）A.[0,)B.C. D.参考答案：D二、填空题:本大题共7小题,每小题4分,共28分11. 方程表示的直线可能是__________．(填序号)参考答案：略12. 经过两直线2x﹣3y﹣12=0和x+y﹣1=0的交点，并且在两坐标轴上的截距相等的直线方程为__________．参考答案：2x+3y=0；或x+y+1=0考点：直线的截距式方程；两条直线的交点坐标．专题：计算题；方程思想；分类法；直线与圆．分析：联解两条直线的方程，得到它们的交点坐标（﹣3，﹣1）．再根据直线是否经过原点，分两种情况加以讨论，即可算出符合题意的两条直线方程．解答：解：由解得∴直线2x﹣3y﹣12=0和x+y﹣1=0的交点坐标为（3，﹣2）①所求直线经过原点时，满足条件方程设为y=kx，可得3k=﹣2，解得k=﹣，此时直线方程为y=﹣x，即2x+3y=0；②当所求直线在坐标轴上的截距不为0时，方程设为x+y=a，可得3﹣2=a，解之得a=1，此时直线方程为x+y﹣1=0综上所述，所求的直线方程为2x+3y=0；或x+y+1=0．点评：本题给出经过两条直线，求经过两条直线的交点且在轴上截距相等的直线方程．着重考查了直线的基本量与基本形式、直线的位置关系等知识，属于基础题13. 是的___________________条件；参考答案：充分不必要14. 某校有老师200人，男学生1200人，女学生1000人，现用分层抽样的方法从所有师生中抽取一个容量为n的样本，已知从女学生中抽取的人数为80人，则n= .参考答案：19215. 已知点M（0，﹣1），N（2，3）．如果直线MN垂直于直线ax+2y﹣3=0，那么a等于．参考答案：1【考点】直线的一般式方程与直线的垂直关系；直线的斜率．【分析】利用相互垂直的直线的斜率之间关系即可得出．【解答】解：∵点M（0，﹣1），N（2，3），∴k MN==2，∵直线MN垂直于直线ax+2y﹣3=0，∴2×=﹣1，解得a=1．故答案为1．【点评】本题考查了相互垂直的直线的斜率之间关系，属于基础题．16. 直线(a+1)x－(2a+5)y－6=0必过一定点，定点的坐标为参考答案：（－4，－2）17. 已知二面角为120，且则CD的长为 -------------参考答案：2a略三、解答题：本大题共5小题，共72分。

2020年辽宁省朝阳市凌源第三高级中学高二数学文模拟试卷含解析

2020年辽宁省朝阳市凌源第三高级中学高二数学文模拟试卷含解析一、选择题：本大题共10小题，每小题5分，共50分。

在每小题给出的四个选项中，只有是一个符合题目要求的1. 执行下面的程序框图，若，则输出的等于【】.A. B. C. D.参考答案：B2. 若椭圆+=1（a＞b＞0）和圆x2+y2=（+c）2，（c为椭圆的半焦距），有四个不同的交点，则椭圆的离心率e的取值范围是（）A．B．C．D．参考答案：A【考点】圆与圆锥曲线的综合．【分析】由题设知，由，得2c＞b，再平方，4c2＞b2，；由，得b+2c＜2a，．综上所述，．【解答】解：∵椭圆和圆为椭圆的半焦距）的中心都在原点，且它们有四个交点，∴圆的半径，由，得2c＞b，再平方，4c2＞b2，在椭圆中，a2=b2+c2＜5c2，∴；由，得b+2c＜2a，再平方，b2+4c2+4bc＜4a2，∴3c2+4bc＜3a2，∴4bc＜3b2，∴4c＜3b，∴16c2＜9b2，∴16c2＜9a2﹣9c2，∴9a2＞25c2，∴，∴．综上所述，．故选A．3. 下列命题中，正确命题的个数是（）①②③④⑤⑥A 2B 3C 4D 5参考答案：D略4. 执行图中程序框图，若输入x1=2，x2=3，x3=7，则输出的T值为（）A．3 B．4 C．D．5参考答案：B【考点】程序框图．【分析】先弄清该算法功能，S=0，i=1，满足条件i≤3，执行循环体，依此类推，当i=4，不满足条件i≤3，退出循环体，输出所求即可．【解答】解：S=0，i=1，满足条件i≤3，执行循环体，S=2，T=，i=2满足条件i≤3，执行循环体S=2+3=5，T=，i=3，满足条件i≤3，执行循环体，S=5+7=12，T=4，i=4，不满足条件i≤3，退出循环体，则T=4．故选：B．5. 过圆上一动点作圆的两条切线，切点分别为，设向量的夹角为，则的取值范围为( )(A) ； (B) ； (C) ； (D)．参考答案：A6. 函数的单调递增区间是A. B.(0,3) C.(1,4) D. w.w.w.参考答案：D略7. 已知,其中为虚数单位,则 ( )A.-1B.1C.2D.3参考答案：B略8. 函数在处取到极值，则的值为( )A. B. C.0 D.参考答案：A9. 如图1，M、N、P为正方体AC1的棱AA1、A1B1、A1D1的中点，现沿截面MNP切去锥体A1-MNP，则剩余几何体的侧视图（左视图）为（）参考答案：B略10. 已知函数f(x)=，若，则k的取值范围是A、0≤k<B、0<k<C、k<0或k>D、0<k≤参考答案：A二、填空题:本大题共7小题,每小题4分,共28分11. 设、是平面直角坐标系（坐标原点为）内分别与轴、轴正方向相同的两个单位向量，且，，则的面积等于 .参考答案：12. 命题“，”的否定是.参考答案：，略13. 如果f(a＋b)＝f(a)·f(b)，f(1)＝2，则________.参考答案：402614. 已知数列的各项都是正整数，且若存在，当且为奇数时，恒为常数，则．参考答案：1或5略15. 不等式≤0的解集为．参考答案：{x|x＜0，或x≥1 }【考点】其他不等式的解法．【分析】不等式即即，由此求得x的范围．【解答】解：不等式≤0，即≥0，即，求得 x＜0，或x≥1，故答案为：{x|x＜0，或x≥1 }．16. 用“秦九韶算法”计算多项式，当x=2时的值的过程中，要经过次乘法运算和次加法运算。

2019-2020年高二下学期第三次月考数学（理）含答案

2019-2020年⾼⼆下学期第三次⽉考数学（理）含答案2019-2020年⾼⼆下学期第三次⽉考数学（理）含答案★友情提⽰：要把所有答案都写在答题卷上，写在试卷上的答案⽆效。

⼀、选择题（本题共10个⼩题，每⼩题5分，共50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．计算ii +3的值为（） A ．i 31+ B ．i 31-- C ．i 31- D ．i 31+－2．下表是降耗技术改造后⽣产甲产品过程中记录的产量x(吨)与相应的⽣产能耗y （吨标准煤）的⼏组对应数据，根据表中提供的数据，可求出y 关于x 的线性回归⽅程?y=0.7x+0.35，那么表中m 的值为（）A.4B.3.15C.4.5D.33. 5位同学报名参加两个课外活动⼩组，每位同学限报其中的⼀个⼩组，则不同的报名⽅法共有（）A ．10种B ．20种C ．25种D ．32种4．⼆项式12)2(x x +展开式中的常数项是( )A ．第7项B ．第8项C ．第9项D ．第10项5.如图，由函数()x f x e e =-的图象，直线2x =及x 轴所围成的阴影部分⾯积等于（）A ．221e e --B ．22e e -C ．22e e - D ．221e e -+6.已知===，。

，若= , （,a b ∈R) , 则（）A.a =5,b =24B.a =6,b =24C.a =6,b =35D.a =5,b =357．袋中装有6个不同的红球和4个不同的⽩球，不放回地依次摸出2个球，在第1次摸出红球的条件下，第2次摸出的也是红球的概率为（）A ．59B ．49C ．29D ．238．函数)(x f y =的图像如图所⽰，下列数值排序正确的是（）A. )2()3()3()2(0f f f f -<'<'<B. )2()2()3()3(0f f f f '<-<'<C. )2()3()2()3(0f f f f -<'<'<D. )3()2()2()3(0f f f f '<'<-<9. 箱⼦⾥有５个⿊球，４个⽩球，每次随机取出⼀个球，若取出⿊球，则放回箱中，重新取球；若取出⽩球，则停⽌取球，那么在第４次取球之后停⽌的概率为（）A ．315445C C C B ．354()99? C ．5194?D ．13454()99C ?? 10. 如图所⽰，()f x 是定义在区间[, ]c c -（0c >）上的奇函数，令()()g x a f x b =+，并有关于函数()g x 的四个论断：①若0a >，对于[1, 1]-内的任意实数, m n （m n <），()()0g n g m n m->-恒成⽴；②函数()g x 是奇函数的充要条件是0b =；③若1a ≥，0b <，则⽅程()0g x =必有3个实数根；④a R ?∈，()g x 的导函数)(x g '有两个零点；其中所有正确结论的序号是( )．A. ①②B. ①②③C. ①④D. ②③④⼆、填空题(本题共5⼩题，每⼩题4分，共20分)11．设随机变量ξ服从正态分布(,9)N u ，若 (3)(1)p p ξξ>=<，则u =12. 若X ～B (20,p)，当p=21且P(X=k)取得最⼤值时，k=________. 13. 现有⼀个关于平⾯图形的命题,如图所⽰,同⼀个平⾯内有两个边长都是a 的正⽅形,其中⼀个的某顶点在另⼀个的中⼼,则这两个正⽅形重叠部分的⾯积恒为42a .类⽐到空间,有两个棱长均为 a 的正⽅体,其中⼀个的某顶点在另⼀个的中⼼,则这两个正⽅体重叠部分的体积恒为.14. 设复数z=cos θ+i sin θ，0θπ≤≤，则1+z 的最⼤值为 .15. 已知数组：1()1，12(,)21，123(,,)321，1234(,,,)4321，，1231(,,,,,),1221n n n n n --- 记该数组为：123456(),(,),(,,),,a a a a a a 则2012a = ．三、解答题:本⼤题共6⼩题,共80分.解答应写出⽂字说明,证明过程或演算步骤.16. (本题满分13分)已知甲袋装有1个红球，4个⽩球；⼄袋装有2个红球，3个⽩球，所有球⼤⼩都相同，现从甲袋中任取2个球，⼄袋中任取2个球.(1)求取到的4个球全是⽩球的概率；(2)求取到的4个球中红球个数不少于⽩球个数的概率.17．(本题满分13分)某同学参加科普知识竞赛，需回答3个问题.竞赛规则规定：答对第⼀、⼆、三个问题分别得100分、100分、200分，答错得零分.假设这名同学答对第⼀、⼆、三个问题的概率分别为0.8,0.7,0.6，且各题答对与否相互之间没有影响.(1)求这名同学得300分的概率;(2)求这名同学⾄少得300分的概率.18．(本题满分13分)设函数32()2338f x x ax bx c =+++在1x =及2x =处取得极值．（Ⅰ）求a 、b 的值；（Ⅱ）当2c =-时，求函数()f x 在区间[03]，上的最⼤值．19．（本⼩题满分13分）当*n N ∈时，111111234212n S n n=-+-++-- ， 11111232n T n n n n =+++++++．（Ⅰ）求1S ，2S ，1T ，2T ；（Ⅱ）猜想n S 与n T 的⼤⼩关系，并⽤数学归纳法证明．20．（本题满分14分）张先⽣家住H ⼩区，他在C 科技园区⼯作，从家开车到公司上班有L 1，L 2两条路线（如图），L 1路线上有A 1，A 2，A 3三个路⼝，各路⼝遇到红灯的概率均为21；L 2路线上有B 1，B 2两个路⼝，各路⼝遇到红灯的概率依次为43，53．（1）若⾛L 1路线，求最多遇到1次红灯的概率；（2）若⾛L 2路线，求遇到红灯次数X 的数学期望；（3）按照“平均遇到红灯次数最少”的要求，请你帮助张先⽣从上述两条路线中选择⼀条最好的上班路线，并说明理由．21. (本⼩题满分14分)已知函数1()ln f x a x x=-，a ∈R ．（1）若曲线()y f x =在点(1,(1))f 处的切线与直线20x y +=垂直，求a 的值；（2）求函数()f x 的单调区间；（3）当1a =，且2x ≥时，证明：(1)25f x x --≤．“华安、连城、永安、漳平⼀中、龙海⼆中、泉港⼀中”六校联考 2011-2012学年下学期第三次⽉考⾼⼆数学（理科）试题参考答案题号 12 3 4 5 6 7 8 9 10 答案C D D C B C A B BA⼆、填空题(本题共5⼩题，每⼩题4分，共20分)11. 2 12. 10 13. 38a 1４. 2 15．559三、解答题:本⼤题共6⼩题,共80分.解答应写出⽂字说明,证明过程或演算步骤.16.(本题满分13分)解：基本事件为“从甲袋中任取2个球，⼄袋中任取2个球”，故基本事件的总数N =2255C C ?;………………2分(1)设“取到的4个球全是⽩球”为事件A ，则事件A 中包含的基本事件数为n 1=2243C C ?；………………4分∴P(A)=1n 9N 50=. ………………6分(2)设“取到的4个球中红球个数不少于⽩球个数”为事件B ，则事件B 中包含的基本事件数为：1111221122142342142n C C C C C C C C C 34,=++=………………10分∴P(B)=2n 17N 50= ………………13分.17（本⼩题满分13分）解：17.设事件A 为“答对第⼀题”，事件B 为“答对第⼆题”，事件C 为“答对第三题”，则P(A)=0.8，P(B)=0.7,P(C)=0.6. ………………2分(1)这名同学得300分可表⽰为(A ∩B ∩C)∪(A ∩B ∩C).所以P((A ∩B ∩C)∪(A ∩B ∩C))， =P(A ∩B ∩C)+P(A ∩B ∩C)=P(A )·P(B)·P(C)+P(A)·P(B )·P(C)=(1-0.8)×0.7×0.6+0.8×(1-0.7)×0.6=0.228. ………………7分(2)这名同学⾄少得300分包括得300分或400分.该事件表⽰为(A ∩B ∩C)∪(A ∩B ∩C)∪(A ∩B ∩C)，所以P((A ∩B ∩C)∪(A ∩B ∩C)∪(A ∩B ∩C)) =P((A ∩B ∩C)∪(A ∩B ∩C))+P(A ∩B ∩C)=0.228+0.8×0.7×0.6=0.564.………13分18. （本⼩题满分13分）解：①解： 2()663f x x ax b '=++，………………1分因为函数()f x 在1x =及2x =取得极值，则有(1)0f '=，(2)0f '=．………………3分即6630241230a b a b ++=??++=?，．………………4分解得3a =-，4b =．………………6分②由（Ⅰ）可知，32()29128f x x x x c =-++，2()618126(1)(2)f x x x x x '=-+=--．………………7分当(01)x ∈，时，()0f x '>；当(12)x ∈，时，()0f x '<；当(23)x ∈，时，()0f x '>．………………9分所以，当1x =时，()f x 取得极⼤值(1)58f c =+，⼜(0)8f c =，(3)98f c =+．………………11分则当[]03x ∈，时，()f x 的最⼤值为(3)987f c =+=-．…13分19. （本⼩题满分13分）解：(1) 1112S T ==，22712S T ==；………………2分（2）猜想：n n S T =（*n N ∈）………………4分证明：(1)当1n =时，11S T =；（2）假设当n k =时，k k S T =，即11111111112342121232k k k k k k -+-++-=++++-+++，………………6分当1n k =+时 111111112342122122k k k k -+-++-+--++111111()12322122k k k k k k =+++++-+++++………………8分 111111()()12223221k k k k k k =-++++++++++………………10分 111112322122k k k k k =+++++++++，即11k k S T ++=，………………12分结合（1）（2），可知*n N ∈，n n S T =成⽴.………………………13分20. （本⼩题满分14分）解: （Ⅰ）设⾛L 1路线最多遇到1次红灯为A 事件，则0312331111()=()()2222P A C C ?+??=． ………………3分所以⾛L 1路线，最多遇到1次红灯的概率为21．………4分（II ）依题意，X 的可能取值为0，1，2………………5分)0(=X p =(1-43))531(-?=101 )531(43)1(-?==X P +20953)431(=?- 2095343)2(=?==X P ………………7分随机变量X 的分布列为： 1. X2.3. 4. 5.P6. 1017. 2098. 209 2027220912090101=?+?+?=EX …………9分（Ⅲ）设选择L 1路线遇到红灯次数为Y ，随机变量Y 服从⼆项分布，1(3,)2Y B ，…11分所以13322EY =?=． ………………13分因为EX EY <，所以选择L 2路线上班最好． ………………14分21. （本⼤题满分14分）解：（1）函数()f x 的定义域为{}|0x x >，21()a f x x x '=+．⼜曲线()y f x =在点(1,(1))f 处的切线与直线20x y +=垂直，所以(1)12f a '=+=，即1a =．--------- 4分（2）由于21()ax f x x +'=．--------- 5分当0a ≥时，对于(0,)x ∈+∞，有()0f x '>在定义域上恒成⽴，即()f x 在(0,)+∞上是增函数．--------- 7分当0a <时，由()0f x '=，得1(0,)x a=-∈+∞．当1(0,)x a∈-时，()0f x '>，()f x 单调递增；当1(,)x a ∈-+∞时，()0f x '<，()f x 单调递减．------- 10分（3）当1a =时，1(1)ln(1)1f x x x -=---，[)2,x ∈+∞．令1()ln(1)251g x x x x =---+-．2211(21)(2)()21(1)(1)x x g x x x x --'=+-=----．当2x >时，()0g x '<，()g x 在(2,)+∞单调递减．⼜(2)0g =，所以()g x 在(2,)+∞恒为负． ------- 12分所以当[2,)x ∈+∞时，()0g x ≤．即1ln(1)2501x x x ---+-≤．故当1a =，且2x ≥时，(1)25f x x --≤成⽴．----- 14分。

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数学试卷一．选择题（5＇×12＝60分）1．已知全集U ＝R ，集合A ＝{x |x ＜-2或x ＞2}，则∁U A ＝ ( ) A ．(-2，2) B ．(-∞，-2)∪(2，+∞) C ．[-2，2] D ．(-∞，-2]∪[2，+∞)2．设x ∈R ，则“2-x ≥0”是“0≤x ≤2”的 ( ) A ．充分而不必要条件 B ．必要而不充分条件 C ．充要条件 D ．既不充分也不必要条件3．命题“∀x ∈M ，f(-x)＝-f(x)”的否定是 ( ) A ．∃x 0∈M ，f(-x 0)＝-f(x 0) B ．∀x ∈M ，f(-x)≠-f(x) C ．∀x ∈M ，f(-x)＝f(x) D ．∃x 0∈M ，f(-x 0)≠-f(x 0)4．为评估一种农作物的种植效果，选了n 块地作试验田．这n 块地的亩产量(单位：kg)分别为x 1，x 2，…，x n ，下面给出的指标中可以用来评估这种农作物亩产量稳定程度的是 ( ) A ．x 1，x 2，…，x n 的平均数 B ．x 1，x 2，…，x n 的标准差 C ．x 1，x 2，…，x n 的最大值 D ．x 1，x 2，…，x n 的中位数 5．已知a >b >0，c <0，下列不等关系中正确的是 ( ) A. ac >bc B. a c >b c C. a-c <b-c D. a+c >b+c6．已知53sin -=α，且)23,2(ππα∈，则=αtan ( ) A ．34 B ．43 C ．43- D ．43±7．圆x 2＋y 2 -2x ＋4y -4＝0的圆心，半径分别为 ( )A. (-1,2) ; 9B. (1,2) ; 3C. (-1,2) ; 3D. (1,-2) ; 38．直线x-y +3=0被圆(x +2)2＋(y-2)2＝2截得的弦长等于 ( ) A ．6 B ．3 C． D ．269．函数y ＝sin 2x ＋cos 2x 的最小正周期为 ( ) A ．4π B ．2π C ．π D.10．在等差数列{a n }中，若a 2，a 10是方程x 2+12x-8=0的两个根，那么a 6的值为 ( )A ．-6B ．-12C ．12D ．611. 椭圆131222=+y x 的一个焦点为F 1，点P 在椭圆上，如果线段PF 1的中点M 在y 轴上，那么点M 的纵坐标是 ( )A ．43± B ．23± C ．22± D ．43±12. 在等比数列{a n }中，a n >0，且a 3a 5+a 2a 10+2a 4a 6=100，则a 4+a 6=( ) A ．25 B ．20 C ．10 D ．30 二、填空题（5＇×4＝20分）请将答案填在答题纸上． 13．已知向量)6,2(=a，),1(λ-=b 。

若b a //，则λ＝_________14．已知a ，b 为正实数，且141=+ba ，则a+b 的最小值为________15．在∆ABC 中，A=60。

，AC=4，BC=32，则∆ABC 面积为__________16．在等比数列{a n }中，其前n 项和为S n 。

已知S 3＝47，S 6＝463，则a 8＝____________三、解答题:（计70分）请将答案填在答题纸上．17．已知函数f(x)=2sinx (sinx+cosx )，求函数f(x)的单调区间。

18．在△ABC 中，a =3，b-c =2，cos B =21-（1）求b，c的值；（2）求sin(B+C)的值19．已知等差数列{a n}的前n项和为S n，等比数列{b n}的前n项和为T n，a1＝-1，b1＝1，a2＋b2＝2。

（1）若a3＋b3＝5，求{b n}的通项公式（2）若T3＝21，求S320．如图所示，在四棱锥P-ABCD中底面ABCD是矩形，△PCD为正三角形，平面PCD⊥平面ABCD，PB⊥AC，E为PD中点。

（1）求证：PB∥平面AEC（2）求二面角E-AC-D的大小21．某大学艺术专业400名学生参加某次测评，根据男女学生人数比例，使用分层抽样的方法从中随机抽取了100名学生，记录他们的分数，将数据分成7组：[20，30)，[30，40)，…，[80，90]，并整理得到如下频率分布直方图：（1）从总体的400名学生中随机抽取一人，估计其分数小于70的概率；（2）已知样本中分数小于40的学生有5人，试估计总体中分数在区间[40，50)内的人数；（3）已知样本中有一半男生的分数不小于70，且样本中分数不小于70的男女生人数相等，试估计总体中男生和女生人数的比例．22．已知椭圆2222:1x yCa b+=的右焦点为(1，0)，且经过点A(0，1)。

（1）求椭圆C的方程；（2）设O 为原点，直线:(1)l y kx t t =+≠±与椭圆C 交于两个不同点P ，Q ，直线AP 与x 轴交于点M ，直线AQ 与x 轴交于点N ，若|OM |·|ON |=2，求证：直线l 经过定点。

答案一.选择题1.C2.B3.D4.B5.D6.B7.D8.A9.C 10.A 11.A 12.C 二.填空题13.-3 14.9 15.32 16.32 三.解答题17.解：f(x)=2sin 2x+2sinxcosx=x x2sin 22cos 12+-⋅=sin2x-cos2x +1 =1)42sin(2+-πx令Z k k x k ∈+≤-≤+-,224222πππππ，解得Z k k x k ∈+≤≤+-,838ππππ令Z k k x k ∈+≤-≤+,2234222πππππ，解得Z k k x k ∈+≤≤+,8783ππππ∴函数f(x)的增区间为Z k k k ∈++-],83,8[ππππ减区间为Z k k k ∈++],87,83[ππππ18.解：（1）由余弦定理2222cos b a c ac B =+-，得2221323()2b c c =+-⨯⨯⨯-．因为2b c =+，所以2221(2)323()2c c c +=+-⨯⨯⨯-．解得5c =．所以7b =．（2）由1cos 2B =-得sin B =．由正弦定理得sin sin a A B b ==．在ABC △中，B C A +=π-．所以sin()sin B C A +==19.解：设{a n }的公差为d ，{b n }的公比为q ，则a n ＝－1＋(n －1)d ，b n ＝q n －1. 由a 2＋b 2＝2得d ＋q ＝3.① (1)由a 3＋b 3＝5得2d ＋q 2＝6.② 联立①和②解得(舍去)或因此{b n }的通项公式为b n ＝2n －1. (2)由b 1＝1，T 3＝21得q 2＋q －20＝0，解得q ＝－5或q ＝4.当q ＝－5时，由①得d ＝8，则S 3＝21；当q ＝4时，由①得d ＝－1，则S 3＝－6. 20.解（1）连OE O AC BD 连结于点交,PB OE BD O PD E //,,∴中点为中点为ΘAEC OE 平面⊂Θ，AEC ，PB 平面⊄AEC PB 平面//∴（2）解法一：设b AD a CD ==,，过.,,BH H CD PH P 连结垂足为作⊥ABCD PCD 平面平面⊥Θ ⊥∴PH 平面ABCD ， AC BH AC PB ⊥∴⊥,Θ取HD 中点G ,连结EG 、OG ，,21//PH EG 则AC OG BH OG ⊥∴,21// EO ，PB //ΘAC EO AC PB ⊥∴⊥,的平面角为二面角D AC E EOG --∠∴ACB BHC AC BH ∠=∠∴⊥,Θb a baa b BC AB CH BC 2,2,==∴=∴b EO b PH EG 23,4621===4,22sin π=∠∴=∠∴EOG EOG 4π的大小为二面角D AC E --∴解法二：设b AD a CD ==,，过,,H CD PH P 垂足为作⊥ABCD PCD 平面平面⊥Θ ⊥∴PH 平面ABCD ，又是矩形底面ABCD Θ 故可以分别以OH 、HC 、HP 所在直线为x 轴、y 轴、z 轴建立空间直角坐标系H-xyz 。

由已知得H （0，0，0）,A （a,-b,0）,B(a,b,0),C(0,b,0),D(0,-b,0),P(0,0, b 3),E()2320b b ，，（- )0,2,(),3,,(b a AC b b a PB -=-=,0222=+-=⋅∴⊥b a AC PB AC PB Θ解得b a 2=，)0,2,2(b b AC -=∴，)23,23,0(b b EC -= ⎪⎩⎪⎨⎧=-=⋅=+-=⋅∴=02323022),,(bz by EC n by bx AC n EAC z y x n 的一个法向量，是平面设取y=1,得的一个法向量为平面ACD b HP )3,0,0(),3,1,2(==⋅Θ22363,cos =⨯<∴bb HP n 4π的大小为二面角D AC E --∴21．解：(1)根据频率分布直方图可知，样本中分数不小于70的频率为(0.02＋0.04)×10＝0.6，所以样本中分数小于70的频率为1-0.6＝0.4.所以从总体的400名学生中随机抽取一人，其分数小于70的概率估计为0.4.(2)根据题意，样本中分数不小于50的频率为(0.01＋0.02＋0.04＋0.02)×10＝0.9，分数在区间[40，50)内的人数为100-100×0.9-5＝5. 所以总体中分数在区间[40，50)内的人数估计为400×＝20.(3)由题意可知，样本中分数不小于70的学生人数为(0.02＋0.04)×10×100＝60，所以样本中分数不小于70的男生人数为60×＝30.所以样本中的男生人数为30×2＝60，女生人数为100－60＝40，男生和女生人数的比例为60∶40＝3∶2.所以根据分层抽样原理，总体中男生和女生人数的比例估计为3∶2. 22．解：（1）由题意得，b 2=1，c =1．所以a 2=b 2+c 2=2．所以椭圆C 的方程为2212x y +=（2）设P （x 1，y 1），Q （x 2，y 2），则直线AP 的方程为1111y y x x -=+．令y =0，得点M 的横坐标111M x x y =--．又11y kx t =+，从而11||||1M x OM x kx t ==+-．同理，22||||1x ON kx t =+-．由22,12y kx t x y =+⎧⎪⎨+=⎪⎩得222(12)4220k x ktx t +++-=．则122412kt x x k +=-+，21222212t x x k -=+．所以1212||||||||11x x OM ON kx t kx t ⋅=⋅+-+-()12221212||(1)(1)x x k x x k t x x t =+-++-22222222212||224(1)()(1)1212t k t kt k k t t k k -+=-⋅+-⋅-+-++12||1t t+=-．又||||2OM ON ⋅=，所以12||21tt+=-．解得t =0，所以直线l 经过定点(0，0)。