{高中试卷}高三月考2
高三年级物理第一学期第2次月考
高三年级物理第一学期第2次月考试卷B命题：费宏 考试时刻100分钟 本卷满分120分 一、 单选题(每题3分共18分，每题只有一个选项是正确)1．如图所示，三角形ABC 三边中点分别为D 、E 、F ， 在三角形中任取一点O ，假如OE 、OF 、DO 三个矢量代表 三个力，那么这三个力的合力为 【 】A .OAB .OBC .OCD .DO2．春天有许多游客放风筝，会放风筝的人，可使风筝静止在空中，以下四幅图中AB 代表风筝截面，OL 代表风筝线，风向水平，风筝可能静止的是 【 】3．如图所示为一质量分布平均的光滑圆柱体的横截面，O 为圆心，半径为R ．圆柱体横卧在两个光滑台阶A 和B 上，A 和B 分别为圆柱体和两个台阶的接触点，∠AOB ＝105°．若圆柱体对台阶A 的压力的大小正好等于圆柱体所受重力的大小，则A 、O 两点间的水平距离为 【 】A ．RB ．R 23C ．R 22D ．R 21 4．如图所示，AB 和BC 是两段相切于B 点顺连在一起的完全相同的粗糙的圆弧形路面，且A 、B 、C 在同一水平面上．质量为m 的小物块以初速度v 0从A 端沿路面滑到C 端时的速度大小为v 1；而以同样大小的初速度v 0从C 端沿路面滑到A 端时的速度大小为v 2，则 【 】A ．v 1＞v 2B ．v 1＜v 2C ．v 1＝v 2D ．无法比较v 1、v 2的大小风向A B L AO 风向 AB L O 风向AB O BL 风向 A B L D O ABOR ABCv 05．两木块自左向右运动，现用高速摄影机在同一底片上多次曝光，记录下木块每次曝光时的位置，如图所示，连续两次曝光的时刻间隔是相等的，由图可知 【 】A ．在时刻t 2以及时刻t 5两木块速度相同B ．在时刻t 1两木块速度相同C ．在时刻t 3和时刻t 4之间某瞬时两木块速度相同D ．在时刻t 4和时刻t 5之间某瞬时两木块速度相同6．如图所示，四个小球在离地面不同高度处，同时由静止开释，不计空气阻力，从某一时刻起，每隔相等的时刻间隔小球依次碰到地面。
高三上学期第二次月考英语试题
高三第二次月考试卷英语时量:120分钟PART ONE: LISTENING COMPREHENSIONSECTION A (22.5 points)Directions: In this section, you will hear 6 conversations between 2 speakers. For each conversation, there are several questions and each question is followed by 3 choices. Listen to the conversations carefully and then answer the questions by marking the corresponding letter(A、B or C) on the QUESTION BOOKLET. You will hear each conversation TWICE.Conversation 11.How late was the delivery?A. 3 hoursB. 4 hoursC. 5 hours2. What does the woman promise to do?A. Refund the man his money.B. Deliver on time next time.C. Offer the man a free delivery. Conversation 23.What does the woman do on weekends?A. She does some reading.B. She helps her father.C. She goes to the beach.4.When does the man sing and tell stories in the camp?A. In the morning.B. In the afternoon.C. In the evening. Conversation 35.What is the woman’s temperature?A. 37.5℃B. 38℃C. 38.5℃6.What is the doctor’s suggestion?A. To stay in bed and h a ve a good rest.B. To take some medicine and drink more water.C. To stay in hospital for a few days.Conversation 47.What did dinosaurs live on?A. Plants or meat.B. Other animals.C. Birds.8.How many kinds of early animals are mentioned in the conversation?A. 2.B. 3.C. 4.9.What do we know about early lizards?A. They were very ugly.B. They lived in the sea.C. They laid eggs. Conversation 510.Where is the woman going?A. To the classroom.B. To the library.C. To the bookstore.11.What does the man advise the woman to do?A. Learn reading skills from the native speakers.B. Talk with the native students out of class.C. Avoid looking up words in the dictionary too often.12.What is the woman trying to do?A. Adjust the situation there.B. Avoid attending class discussions.C. Learn more about reading skills.Conversation 613.What does the woman ask the man about?A. His school life.B. His development plan.C. The size of his class.14.How many hobby groups are mentioned in the conversation?A. 5B. 6C. 715.When does the man have sports training?A. Every Tuesday and Wednesday afternoon.B. Every Tuesday and Friday afternoon.C. Every Wednesday and Friday afternoon.Section BDirections:In this section, you’ll hear a mini-talk. Listen carefully and then fill in the numbered blanks with the information you’ve got. Fill in each blank with NO MORE THAN 3 WORDS.You will hear the mini-talk TWICE.PART TWO: LANGUAGE KNOWLEDGE (45 points)SECTION A (15 points)Directions:Beneath each of the following sentences are 4 choices marked A, B, C and D. Choose the answer that best completes the sentence.21. Iran’s first nuclear power plant started supplying electricity to the national grid on Saturday, ____ on stream after years of delays, Iran’s ISNA news agency reported.A. to comeB. comeC. having comeD. coming22. Across North China _____ in the world, namely the Great Wall.A. the longest wall runsB. runs the longest wallC. does run the longest wallD. does the longest wall run23. Look, ____ fashionable clothes is she wearing that all the eyes are fixed on her!A. soB. suchC. howD. what24. The split between humans and living apes is thought by some scholars ____ 15 to 20 millionyears ago.A. to occurB. occurringC. to have occurredD. having occurred25. ____ a certain doubt among the people as to the practical value of the project.A. There remainsB. It hasC. They haveD. It remains26. The traffic problem we are looking forward to seeing _____ should have attracted the localgovernment’s attention.A. solvingB. solvedC. to solveD. solve27. With April 18’s railway speedup, highway and air transport will have to compete with _____ service for passengers.A. goodB. bestC. betterD. the best28. ---How was your trip abroad?---Great. The only pity is that I _____ to buy some gifts for my friends.A. forgotB. forgetC. would forgetD. was forgetting29. ____ the flight to Beijing will be delayed is ____ I’m especially worried about.A. If; whatB. Whether; thatC. When; thatD. Whether; what30. Many questions have been answered by John. He must have previewed the lessons last night, _____ he?A. needn’tB. hasn’tC. mustn’tD. didn’t31. — I’ll be able to come to see your performance at 8:30 tomorrow evening.— I’m sorry, by then my performance ___ and I ____ reporters in the meeting room.A. will be ended; will meetB. is to end; will meetC. will have ended; will be meetingD. will end; am going to meet32. Being physically examined for free twice a year is what every clerk _____ be ensured in any company of the state.A. mustB. ought toC. shallD. need33. I had just got up and was about to cook my breakfast _____ the telephone rang loudly, but it hung up _____ I could answer it.A. when; beforeB. while; beforeC. when; untilD. as; after34. If a computer crashes, you will lose the file you ______ on if you don’t save it early enough.A. workB. are workingC. will workD. worked35. — We shall take the conditions into careful account ____ you have attached to this contract.— Thanks. Hope for further cooperation.A. asB. whatC. whereD. whichSECTION B (18 points)Directions: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with a word or phrase that best fits the context.I teach clay classes to children and adults. One cold afternoon a family came in to 36 a clay class together. It was a one-hour class that was meant to allow the family some fun at making their own artistic creations.Three children aged nine, seven and three sat in their seats 37 to begin their little artistic creations. The warm clay and various tools made their eyes full of 38. They were so excited to begin. Every word I said made them feel great in the clay class and so they began to create. The young artists enjoyed their 39 to touch, feel and express themselves until their parents 40 and controlled their every moment. They were told to do this, do that, use this, not that ... I 41 watched as the little blooms (花朵) of creations were taken from their small hands.The three-year-old rebelled (反抗) and was told she was 42. I wanted to ask how the parents would feel if they were enjoying themselves creating a piece of art and I took it from them and totally controlled the 43 of its creation. I asked the controlling father if he would like to have a cupof coffee and let the three-year-old just 44, explaining she would have fun and stop 45 him. He did and she continued focusing on her work with the most beautiful smile.In this hurried world, I 46 that some parents and teachers are not letting children 47 as they want. It is better to let children create works of art in their own way rather than tell them to do in ways that please us as adults.36. A. give B. teach C. enjoy D. conduct37. A. afraid B. ready C. unwilling D. careful38. A. confusion B. terror C. expectation D. anxiety39. A. habit B. determination C. responsibility D. freedom40. A. took over B. gave up C. went out D. calmed down41. A. hopefully B. curiously C. sorrowfully D. excitedly42. A. lazy B. brave C. proud D. naughty43. A. aim B. process C. time D. reason44. A. play B. cry C. sit D. watch45. A. asking B. annoying C. pleasing D. fighting46. A. find B. suggest C. agree D. doubt47. A. behave B. exercise C. study D. createSECTION C (12 points)Directions: Complete the following passage by using ONE word that best fits the context.In many countries today, laws protect wildlife. In India, the need for such protection was realized centuries ago. About 300 B.C,48 Indian writer described forests that were somewhat like national parks today. The killing of game beasts was carefully supervised. Some animals were fully protected. 49 the forest, nobody was allowed to cut timber, burn wood for charcoal, 50 catch animals for their furs. Animals 51 became dangerous to human visitors were caught or killed outside the park, so that other animals would not become uneasy.The need for wildlife protection is 52 now than ever before. About a thousand species of animals are 53 danger of dying out, and the rate at which 54 are being destroyed has increased. 55 we took no measures to protect wildlife, some day our children would see no living creatures except man himself.PART THREE READING COMPREHENSIONDirections: 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.AA water bearer in India had two large pots．One of the pots had a crack in it, and while the other pot was perfect and always delivered a full load of water at the end of the long walk from the stream to the master's house, the cracked pot arrived only half full．For two years this went on daily．The perfect pot was proud of its accomplishment．Of course，the poor cracked pot was ashamed of its own imperfection．After two years of this sense of bitter failure, it spoke one day to the water bearer by the stream．“I am ashamed of myself, and I want to apologize to you．”“Why?” asked the bearer．“What are you ashamed of?”“I have been able for these last two years to deliver only half my load, because this crack inmy side causes water to leak out all the way back．And you do not get full value for your efforts” the pot explained．The water bearer, hearing this, said, “As we return to the master's house, I want you to notice the beautiful flowers along the path．”As they went up the hill, the cracked pot took notice of the sun warming the beautiful wild flowers on the side of the path, and this cheered it a little．The bearer said, “Did you notice that there were flowers only on your side of the path, but not on the other pot's side ?” That is because I have known about you, and I took advantage of it．I planted flower seeds on your side of the path, and every day while we walk back from the stream, you have watered them．For two years I have been able to pick those beautiful flowers to decorate my master's table．Without you being just the way you are, he would not have this beauty in his house．”56．What's the moral of the passage?A．No perfect man exists in this world．B．We should learn to be kind to ourselves．C．We sometimes don't have to mind too much the way we are．D．Once we have shortcomings, we should try to overcome them．57．Why did the cracked pot feel ashamed?A．Because it didn't hold water．B．Because the water bearer didn't like it．C．Because it didn't have a perfect appearance．D．Because it could only accomplish half of its load．58．What can we know from the passage?A．The water bearer felt sorry for the cracked pot．B．The cracked pot was more useful than the perfect one．C．The cracked pot watered the flowers along the path．D．The water bearer preferred the perfect pot to the cracked one．59．How would the cracked pot feel at the end of the story?A．Much happier with itself．B．Disappointed with itself．C．Prouder than the perfect pot．D．Still ashamed of itself．60．Which of the following proverbs can best go with this passage?A．Every dog has his day．B．Every man has his price．C．Every picture tells a story．D．Every cloud has a silver lining．BWhat I’ve noticed about some very successful business owners is that they are always learning and growing.For your business to be better you need to be better and, you don’t get better by doing nothing!You are the average of the five PEOPLE you hang around with most. Surround yourself with people whom you want to be like.To do this you may have to get right out of your comfort zone and join in a coaching program, go along to seminars(专题讨论会) where you can meet other like-minded people or those playing at a higher level of business than you.There’s no excuse not to get off your butt and learn your way to success. Regardless of whereyou are located，events are taking place all around the country regularly as well as online. Saying you have “no money” or “no time” is not a good enough excuse.For many business owners they spend their whole lives using these excuses and going nowhere. Don’t focus on where you are now, but where you want to be. You’ll find the money and the time if you seriously want to change.You ongoing learning can take you from struggling along to soaring high in all areas of your life.It’s been nearly six months since you made those New Year decisions to make changes in your business and your life.Are you getting the results you wanted?Is your business more profitable?Do you have more free time?Is the quality of your life improving?Is your answer “yes”, then take a look at why that is. What have you been doing to get these results? And if you’re not moving ahead, seriously look at why that is. When’s the last time you read a book, listened to a CD, got coached or attended a seminar to learn how to do things better? The people I know who are getting great results in their business and personal lives are always learning and growing.You are the Secret to Your Success, so do everything within your power to invest in yourself.61. All the following ways can contribute to your business EXCEPT ___________.A. getting in touch with successful peopleB. not dreaming of living a comfortable lifeC. avoiding finding unnecessary excusesD. deciding to change your present life62. The underlined phrase “get off your butt” in Paragraph 5 probably means _________.A. stop being lazy and start doing something usefulB. leave for a place where you want to goC. do something as you likeD. shoot an object63.According to the passage, the most important thing to success is __________.A. making a plan ahead of timeB. attending important meetings oftenC. making changes when necessaryD. keeping learning all the time64.Why does the writer write the word “people” in capital letters in Paragraph Three?A. People are supposed to set higher goals.B. It’s sometimes difficult to change a person.C. Average people are easy to get along with.D. You don’t get better by doing something.65.What would be the best title for this passage?A. Failure is the Mother of SuccessB. Successful Business OwnersC. Learn Your Way to SuccessD. No More any ExcusesCStudies have shown that the fewer medicines a person has to take the more likely he or she will take them. Last week, a study was released about a new treatment that combines 5 medicines for heart disease in one pill. Salim Yusuf of McMaster University in Hamilton, Ontario, Canada was the lead investigator. He presented the findings at the American College of Cardiology Conference in Orlando, Florida.The experimental drug is known as Polycap. It contains aspirin, a drug to lower cholesterol (胆固醇)and three medicines to lower blood pressure. The study was carried out at 50 health centers across India. More than two thousand people between the ages of 45 and 80 took part in the study. All had at least one risk factor for heart disease. These include high blood pressure, high cholesterol, diabetes or being severely overweight.The people were divided into 9 groups of about 200 people each. One group took Polycap. The other groups took either a single drug or different combinations of the medicines in the Polycap pill. The study showed that Polycap lowered blood pressure and cholesterol without many side effects. Doctor Y usuf said the single pill, taken once a day, could reduce the average person's risk of heart disease and stroke by about half. The maker of Polycap, Cadila Pharmaceuticals of India, paid for the study.Cardiovascular diseases（心血管疾病）of the heart and blood vessels are the number one cause of death around the world. These diseases kill more than seventeen million people every year.80 percent of them are in low and middle income countries. Doctor Yusuf said the single pill treatment could revolutionize heart disease prevention. People would be more likely to take one pill a day than many pills. And one pill would cost less than several pills.Other heart doctors say heart disease prevention is important but not necessarily with pills. They say patients might be able to get the same results with changes in diet and exercise.Doctors say that more research on Polycap is needed. They say the drug should be tested on thousands more people, including those in different risk, age and ethnic grou ps.66. The underlined sentence in Paragragh 1 implies that people _______.A. don’t like to take medicinesB. tend to take fewer medicinesC. should take medicines if it is necessaryD. tend to fall sick if they take fewer medicines67. Which of the following about Doctor Y usuf’s experiment on Polycap in India is True?A. Polycap was tested for its five different combinations.B. Polycap was proved basically safe and effective.C. All the participatiors had at least one symptom of heart disease.D. Participators were required to come from different ages and ethnic groups.68. From the passage, we can infer that Polycap ________.A. is generally accepted by heart doctorsB. mainly consists of medicines to lower blood pressureC. is certain to play important role in heart disease preventionD. will be cheaper than other common medicines to treat heart disease69. According to the passage, the following people are more likely to get Cardiovascular disease except ______.A. old peopleB. low-income peopleC. overweight peopleD. diabetic people70. What would be the best title for the passage?A. Different opinions about heart disease prevention.B. A combination pill may cut heart disease risk in half.C. A combination pill cures blood pressure and bad cholesterol.D. Cardiovascular diseases are the number one killer in the world.PART FOUR WRITING (45%)SECTION A (10%)Directions: Read the following passage. Complete the diagram by using the information from the passage.Write NO MORE THAN THREE WORDS for each answer.An altimeter（测高计）is an instrument used in an airplane to tell the pilot how high he is flying.The altimeter used in most airplanes is a kind of barometer（气压计）,which, like barometers found in ordinary homes, is a measuring instrument for air pressure. The weight of the atmosphere presses downwards everywhere. At sea level this pressure is more than 14 pounds on every square inch of the surface. The higher you go into the air, the lower the air pressure is. An altimeter measures this air pressure to show the altitude of the airplane, or, how high it is abo ve sea level.But the altimeter does not show how high the plane is above the ground. A plane might be flying at an altitude of 15,000 feet, but it would be only a thousand feet or so above the ground if the land in that area happened to be 14,000 feet above sea level.The pilot adjusts his altimeter to the actual sea-level pressure before setting out from the airport and then corrects it in flight according to new information given to him by radio. He flies high enough to be above any mountain he may pass on his course. The barometer altimeter is correct within about 300 feet.Another kind of altimeter, the radio altimeter, makes use of radio reflections. It calculates the height of the flying pl ane by sending out electrical signals to the surface below and measuring the time required f or them to bounce back（反弹）. It is correct within 15 feet over water, but is not reliable over land. Big planes usually have both kinds of altimeter.SECTION B (10%)Directions: Read the following passage. Answer the questions according to the information given in the passage and the required words limit. Write your answers on your answer sheet.Image that you are sitting in a park and suddenly remember that you need to send an urgent email. You have your laptop computer with you and you connect it to a wireless access point (WAP) and access the Internet. Then you send your email. Wireless Internet has tricked(慢慢进入) into our existence in so many ways that we don’t even realize what a necessary part it has become of our day-to-day lives.How do we connect to a wireless Internet service for laptop computers? There are two ways to do this. They are Wi-Fi connection and bluetooth connection. Wireless networking is very easy. You just switch on the Wi-Fi button that has been provided in the laptop for Wi-Fi browsing. Once that is done, the computer will make you connect to the Internet.An increasing fact about Wi-Fi networking is that you can create a network between similar wireless devices. This is especially easy in Wi-Fi compatible (兼容的) laptops. Consider a situation where you and your friend both have Wi-Fi laptops, and there is some data you want to exchange. But the beauty of Wi-Fi networking technology is that you can set up a network between you and your friend’s laptop, and share all the data you want.Broadband Internet connection is typically preferred by any user as it provides better speed for Internet browsing. Broadband wireless Internet for laptops is similar to Wi-Fi technology. Wherever you find a Wi-Fi hot spot, its backbone is always broadband Internet. Up until the invention of 3G mobile phones, Internet speed on mobile phone was very slow. 3G has the promise of changing the entire face of broadband technology. Just imagine browsing at speeds of more than 2MB per second. To give you a measure of the speed, 3G enables you to watch high quality live videos on the Internet.It is amazing to see how technology changes and the rate at which it changes. There may come a time when you may be able to browse using mobile wireless Internet for laptops anywhere in the world.81. What’s the author’s purpose by giving an example at the beginning of the passage? (No more than 5 words.)____________________________________________________________________.82. If you want to share data with your friends through Wi-Fi networking, what should you do? ( No more than 6 words.)____________________________________________________________________.83. What does the passage mainly talk about? ( No more than 5 words.)____________________________________________________________________.84. What is the greatest advantage of 3G mobile phones? ( No more than 5 words.)____________________________________________________________________.SECTION C (25%)Directions: Write an English letter according to the instructions given below in Chinese.下图描绘的是当今学生中存在的一种现象。
天津市南开中学2023-2024学年高三上学期第二次月考数学试卷
答案第31 页，共22 页
【详解】设{an
-
n}
的公比为
q
，则
q
=
a2 a1
-2 -1
=
11- 2 4 -1
=
3
，
所以 an - n = (a1 -1) × qn-1 = (4 -1) ×3n-1 = 3n ，则 an = n + 3n ，
所以 a4 = 4 + 34 = 85 ，
所以落在区间[4,85] 内的偶数共有 41 个，故t (a4 ) = 41 .
11．在
æ çè
3x2
-
2 x
ö5 ÷ø
的展开式中，
x
的系数是
．
三、双空题
12．已知直线 l : y = kx - 2(k > 0) 与圆 x2 + y2 = 1 相切，且被圆 x2 + ( y + a)2 = 4(a > 0) 截
得的弦长为 2 3 ，则 k = ； a = ．
四、填空题
13．锐角a
(2)求数列{anbn} 的前 n 项和 Sn ；
å (3)若数列{dn} 满足 d1 = 1 ， dn + dn+1 = bn ，记Tn =
n
dk
m ．是否存在整数 ，使得对
b k =1 2k
任意的 n Î N * 都有1 £
mTn
-
dn b2n
<
m 2 成立？若存在，求出
的值；若不存在，说明理由．
故选：C. 9．B
【分析】根据三角函数的变换规则求出 g ( x) 的解析式，再根据正弦函数的性质判断
A、C、D，利用诱导公式判断 B.
高中高三物理上学期第二次月考试卷(含解析)-人教版高三全册物理试题
河南省洛阳市伊川实验高中2015届高三上学期第二次月考物理试卷一、选择题〔14×3分=42分，本大题中2、5、6、7、8小题多项选择，其他题单项选择〕1．关于功和能的关系，如下说法正确的答案是〔〕A．物体受拉力作用向上运动，拉力做的功是1J，如此物体重力势能的增加量也是1J B．一个重10N的物体，在15N的水平拉力的作用下，分别在光滑水平面和粗糙水平面上发生一样的位移，拉力做的功相等C．一辆汽车的速度从10km/h加速到20km/h，或从50km/h加速到60km/h，两种情况下牵引力做的功一样多D．“神舟十号〞载人飞船的返回舱在大气层以外向着地球做无动力飞行的过程中，机械能增大2．如下列图，实线为一匀强电场的电场线，一个带电粒子射入电场后，留下一条从a到b 虚线所示的径迹，重力不计，如下判断正确的答案是〔〕A． b点电势高于a点电势B．粒子在a点的动能大于在b点的动能C．粒子在a点的电势能小于在b点的电势能D．场强方向向左3．质量为m的汽车，其发动机额定功率为P．当它开上一个倾角为θ的斜坡时，受到的阻力为车重力的k倍，如此车的最大速度为〔〕A． B．C． D．4．一物体静止在升降机的地板上，在升降机匀加速上升的过程中，地板对物体的支持力所做的功等于〔〕A．物体抑制重力所做的功B．物体动能的增加量C．物体动能增加量与重力势能增加量之差D．物体动能增加量与重力势能增加量之和5．质量为m的物体，从静止开始以2g的加速度竖直向下运动的位移为h，空气阻力忽略不计，如下说法正确的答案是〔〕A．物体的重力势能减少mgh B．物体的重力势能减少2mghC．物体的动能增加2mgh D．物体的机械能保持不变6．水平传送带匀速运动，速度大小为v，现将一小工件放到传送带上，设工件初速度为零，当它在传送带上滑动一段距离后速度达到v而与传送带保持相对静止．设工件质量为m，它与传送带间的动摩擦因数为μ，如此在工件相对传送带滑动的过程中正确的说法是〔〕 A．滑动摩擦力对工件做的功为B．工件的动能增量为C．工件相对于传送带滑动的路程大小为D．传送带对工件做的功为零7．在平直的公路上，汽车由静止开始做匀加速运动，当速度达到V m，立即关闭发动机而滑行直到停止，v﹣t 图线如图，汽车的牵引力大小为F1，摩擦力大小为F2，全过程中，牵引力做功为W1，抑制摩擦力做功为W2，如此〔〕A． F1：F2=1：3 B． F1：F2=4：1 C． W1：W2=1：1 D． W1：W2=1：38．如下列图，一轻弹簧固定于O点，另一端系一重物，将重物从与悬点O在同一水平面且弹簧保持原长的A点无初速地释放，让它自由摆下，不计空气阻力，在重物由A点摆向最低点的过程中〔〕A．重物的重力势能减少 B．重物的重力势能增大C．重物的机械能不变 D．重物的机械能减少9．一物块沿倾角为θ的斜坡向上滑动．当物块的初速度为v时，上升的最大高度为H，如下列图；当物块的初速度为时，上升的最大高度记为h．重力加速度大小为g．物块与斜坡间的动摩擦因数和h分别为〔〕A．tanθ和 B．〔﹣1〕tanθ和C．tanθ和 D．〔﹣1〕tanθ和10．如图是安装在列车车厢之间的摩擦缓冲器结构图，图中①和②为楔块，③和④为垫块，楔块与弹簧盒、垫块间均有摩擦，在车厢相互撞击时弹簧压缩过程中〔〕A．缓冲器的机械能守恒B．摩擦力做功消耗机械能C．垫块的动能全部转化成内能D．弹簧的弹性势能全部转化为动能11．如图，两根一样的轻质弹簧，沿足够长的光滑斜面放置，下端固定在斜面底部挡板上，斜面固定不动．质量不同、形状一样的两物块分别置于两弹簧上端．现用外力作用在物体上，使两弹簧具有一样的压缩量，假设撤去外力后，两物块由静止沿斜面向上弹出并离开弹簧，如此从撤去外力到物块速度第一次减为零的过程，两物块〔〕A．最大速度一样 B．最大加速度一样C．上升的最大高度不同 D．重力势能的变化量不同12．如下列图，有三个斜面a、b、c，底边的长分别为L、L、2L，高度分别为2h、h、h．某物体与三个斜面间的动摩擦因数都一样，这个物体分别沿三个斜面从顶端由静止下滑到底端．三种情况相比拟，如下说法正确的答案是〔〕A．物体损失的机械能△E c=2△E b=4△E aB．因摩擦产生的热量2Q a=2Q b=Q cC．物体到达底端的动能E ka=2E kb=2E kcD．因摩擦产生的热量4Q a=2Q b=Q c13．取水平地面为重力势能零点，一物块从某一高度水平抛出，在抛出点其动能与重力势能恰好相等．不计空气阻力，该物块落地时的速度方向与水平方向的夹角为〔〕A． B． C． D．14．如下列图，D、A、B、C四点水平间距相等，DA、AB、BC竖直方向高度差之比为1：3：5．在A、B、C三点分别放置一样的小球，释放三个压缩的弹簧，小球沿水平方向弹出，小球均落在D点，不计空气阻力，如此如下关于A、B、C三点处的小球说法正确的答案是〔〕A．三个小球在空中运动的时间之比为1：2：3B．三个小球弹出时的动能之比为1：4：9C．三个小球在空中运动过程中重力做功之比为1：3：5D．三个小球落地时的动能之比为2：5：10二、填空题与实验题〔16分，15题4分16题12分〕15．某同学把附有滑轮的长木板平放在实验桌上，将细绳一端拴在小车上，另一端绕过定滑轮，挂上适当的钩码，使小车在钩码的牵引下运动，以此定量探究绳拉力做功与小车动能变化的关系，此外还准备了打点计时器与配套的电源、导线、复写纸、纸带、小木块等，组装的实验装置如下列图．〔1〕假设要完成该实验，必需的实验器材还有哪些．实验开始时，他先调节木板上定滑轮的高度，使牵引小车的细绳与木板平行，他这样做的目的是如下的哪个〔填字母代号〕A．防止小车在运动过程中发生抖动B．可使打点计时器在纸带上打出的点迹清晰C．可以保证小车最终能够实现匀速直线运动D．可在平衡摩擦力后使细绳拉力等于小车受的合力〔3〕平衡摩擦力后，当他用多个钩码牵引小车时，发现小车运动过快，致使打出的纸带上点数较少，难以选到适宜的点计算小车速度，在保证所挂钩码数目不变的条件下，请你利用本实验的器材提出一个解决方法：．〔4〕他将钩码重力做的功当做细绳拉力做的功，经屡次实验发现拉力做功总是要比小车动能增量大一些，这一情况可能是如下哪些原因造成的〔填字母代号〕．A．在接通电源的同时释放了小车B．小车释放时离打点计时器太近C．阻力未完全被小车重力沿木板方向的分力平衡掉D．钩码做匀加速运动，钩码重力大于细绳拉力．16．验证机械能守恒定律的实验采用重物自由下落的方法：〔1〕用公式mv2=mgh时对纸带上起点的要求是为此目的，所选择的纸带第1，2两点间距应接近．假设试验中所用重锤质量m=1kg，打点纸带的记录如图1所示，打点时间间隔为0.02s，如此记录B点时，重锤的速度v B=，重锤的动能E KB=．从开始下落至B点，重锤的重力势能减少量是，因此可能得出的结论是．〔3〕即使在实验操作规范，数据测量与数据处理很准确的前提下，该实验求得的△E P也一定略△E k〔填大于或小于〕，这是实验存在系统误差的必然结果，该系统误差产生的主要原因是．〔4〕根据纸带算出相关各点的速度υ，量出下落的距离h，如此以为纵轴，以h为横轴画出的图线应是图2所示图中的四、解答题〔10+8+12+12=42分，写出必要的演算过程、解题步骤与重要关系式，并得出结果〕17．如图，真空中xOy平面直角坐标系上的ABC三点构成等边三角形，边长L=2.0m，假设将电荷量均为q=+2.0×10﹣6C的两点电荷分别固定在A、B点，静电力常量k=9.0×109N•m2/C2，求：〔1〕两点电荷间的库仑力大小；C点的电场强度的大小和方向．18．如下列图，m A=4kg，m B=1kg，A与桌面间的动摩擦因数μ=0.2，B与地面间的距离s=0.8m，A、B间绳子足够长，A、B原来静止，求：〔1〕B落到地面时的速度为多大；〔用根号表示〕B落地后，A在桌面上能继续滑行多远才能静止下来．〔g取10m/s2〕19．如下列图为“嫦娥三号〞探测器在月球上着陆最后阶段的示意图，首先在发动机作用下，探测器受到推力在距月面高度为h1处悬停〔速度为0，h1远小于月球半径〕，接着推力改变，探测器开始竖直下降，到达距月面高度为h2处的速度为v，此后发动机关闭，探测器仅受重力下落至月面．探测器总质量为m〔不包括燃料〕，地球和月球的半径比为k1，质量比为k2，地球外表附近的重力加速度为g，求：〔1〕月球外表附近的重力加速度大小与探测器刚接触月球时的速度大小；从开始竖直下降到接触月面时，探测器机械能的变化．20．图为某游乐场内水上滑梯轨道示意图，整个轨道在同一竖直平面内，外表粗糙的AB段轨道与四分之一光滑圆弧轨道BC在B点水平相切．点A距水面的高度为H，圆弧轨道BC的半径为R，圆心O恰在水面．一质量为m的游客〔视为质点〕可从轨道AB的任意位置滑下，不计空气阻力．〔1〕假设游客从A点由静止开始滑下，到B点时沿切线方向滑离轨道落在水面上的D点，OD=2R，求游客滑到B点时的速度v B大小与运动过程轨道摩擦力对其所做的功W f；假设游客从AB段某处滑下，恰好停在B点，又因受到微小扰动，继续沿圆弧轨道滑到P点后滑离轨道，求P点离水面的高度h．〔提示：在圆周运动过程中任一点，质点所受的向心力与其速率的关系为F向=m〕河南省洛阳市伊川实验高中2015届高三上学期第二次月考物理试卷参考答案与试题解析一、选择题〔14×3分=42分，本大题中2、5、6、7、8小题多项选择，其他题单项选择〕1．关于功和能的关系，如下说法正确的答案是〔〕A．物体受拉力作用向上运动，拉力做的功是1J，如此物体重力势能的增加量也是1J B．一个重10N的物体，在15N的水平拉力的作用下，分别在光滑水平面和粗糙水平面上发生一样的位移，拉力做的功相等C．一辆汽车的速度从10km/h加速到20km/h，或从50km/h加速到60km/h，两种情况下牵引力做的功一样多D．“神舟十号〞载人飞船的返回舱在大气层以外向着地球做无动力飞行的过程中，机械能增大考点：功能关系；功的计算．分析： A、重力势能的增量等于抑制重力所做的功，与外力做功无关；B、根据做功公式W=Fs，可以判断拉力F对物体做功的情况；C、根据动能定理即可分析两种情况下牵引力做功的大小；D、载人飞船的返回舱在大气层以外向着地球做无动力飞行的过程中，只有地球对飞船的引力做功，机械能守恒．解答：解：A、重力势能的增量等于抑制重力所做的功，与外力做功无关，故A错误；B、由W=Fs知，拉力的大小一样，木块的位移也一样，所以拉力对两木块做的功一样多，故B正确；C、根据动能定理可知：W﹣W f=，所以速度从10km/h加速到20km/h，或从50km/h加速到60km/h的两种情况下牵引力做的功不同，故C错误；D、载人飞船的返回舱在大气层以外向着地球做无动力飞行的过程中，只有地球对飞船的引力做功，机械能守恒，故D错误．应当选：B点评：此题主要考查了重力势能的影响因素，恒力做功公式、动能定理等的直接应用，难度不大，属于根底题．2．如下列图，实线为一匀强电场的电场线，一个带电粒子射入电场后，留下一条从a到b 虚线所示的径迹，重力不计，如下判断正确的答案是〔〕A． b点电势高于a点电势B．粒子在a点的动能大于在b点的动能C．粒子在a点的电势能小于在b点的电势能D．场强方向向左考点：电势能；电势．专题：电场力与电势的性质专题．分析：由运动的轨迹与电场线确定出受力向，根据力的方向与速度方向的夹角确定电场力做功的正负，从而判断出能量的大小关系．解答：解：A、D、由曲线运动的知识可知：带电粒子所受的电场力向左，因为带电粒子带电性质不确定，所以场强的方向也不能确定，从而不能判断ab两点电势的上下，故AD错误；B、C、带电粒子从a到b点过程中，电场力做负功，电荷的电势能增大，由动能定理，粒子的动能减小，即粒子在a点的动能大于在b点的动能，粒子在a点的电势能小于在b点的电势能，故BC正确；应当选：BC点评：在电场中跟据带电粒子运动轨迹和电场线关系判断电场强度、电势、电势能、动能等变化是对学生根本要求，也是重点知识，要重点掌握．3．质量为m的汽车，其发动机额定功率为P．当它开上一个倾角为θ的斜坡时，受到的阻力为车重力的k倍，如此车的最大速度为〔〕A． B．C． D．考点：功率、平均功率和瞬时功率．专题：功率的计算专题．分析：当牵引力等于阻力时，汽车达到最大速度，汽车匀速向上运动时，对汽车受力分析，汽车处于受力平衡状态，由此可以求得汽车在上坡情况下的牵引力的大小，由P=Fv分析可得出结论．解答：解：当牵引力等于阻力时，汽车达到最大速度，汽车匀速运动时，受力平衡，由于汽车是沿倾斜的路面向上行驶的，对汽车受力分析可知，汽车的牵引力F=f+mgsinθ=kmg+mgsinθ=mg〔k+sinθ〕，由功率P=Fv，所以上坡时的速度：，故D正确应当选：D点评：汽车的功率不变，但是在向上运动和向下运动的时候，汽车的受力不一样，牵引力减小了，P=Fv可知，汽车的速度就会变大，分析清楚汽车的受力的变化情况是解决此题的关键．4．一物体静止在升降机的地板上，在升降机匀加速上升的过程中，地板对物体的支持力所做的功等于〔〕A．物体抑制重力所做的功B．物体动能的增加量C．物体动能增加量与重力势能增加量之差D．物体动能增加量与重力势能增加量之和考点：功能关系；功的计算；动能和势能的相互转化．分析：对物体进展受力分析，运用动能定理研究在升降机加速上升的过程，表示出地板对物体的支持力所做的功．知道重力做功量度重力势能的变化．解答：解：物体受重力和支持力，设重力做功为W G，支持力做功为W N，运用动能定理研究在升降机加速上升的过程得：W G+W N=△E kW N=△E k﹣W G由于物体加速上升，所以重力做负功，设物体抑制重力所做的功为：W G′，W G′=﹣W G所以：W N=△E k﹣W G=W N=△E k+W G′．根据重力做功与重力势能变化的关系得：w G=﹣△E p，所以有：W N=△E k﹣W G=W N=△E k+△E p．应当选：D．点评：解这类问题的关键要熟悉功能关系，也就是什么力做功量度什么能的变化，并能建立定量关系．动能定理的应用范围很广，可以求速度、力、功等物理量．5．质量为m的物体，从静止开始以2g的加速度竖直向下运动的位移为h，空气阻力忽略不计，如下说法正确的答案是〔〕A．物体的重力势能减少mgh B．物体的重力势能减少2mghC．物体的动能增加2mgh D．物体的机械能保持不变考点：功能关系．分析：重力势能的变化量等于重力对物体做的功．只有重力对物体做功，物体的机械能才守恒．根据动能定理研究动能的变化量．根据动能的变化量与重力的变化量之和求解机械能的变化量．解答：解：A、由质量为m的物体向下运动h高度时，重力做功为mgh，如此物体的重力势能减小mgh．故A正确B错误．C、合力对物体做功W=ma•h=2mgh，根据动能定理得知，物体的动能增加2mgh．故C正确．D、由上物体的重力势能减小mgh，动能增加2mgh，如此物体的机械能增加mgh．故D错误．应当选：AC．点评：此题考查分析功能关系的能力．几对功能关系要理解记牢：重力做功与重力势能变化有关，合力做功与动能变化有关，除重力和弹力以外的力做功与机械能变化有关．6．水平传送带匀速运动，速度大小为v，现将一小工件放到传送带上，设工件初速度为零，当它在传送带上滑动一段距离后速度达到v而与传送带保持相对静止．设工件质量为m，它与传送带间的动摩擦因数为μ，如此在工件相对传送带滑动的过程中正确的说法是〔〕 A．滑动摩擦力对工件做的功为B．工件的动能增量为C．工件相对于传送带滑动的路程大小为D．传送带对工件做的功为零考点：动能定理的应用．专题：动能定理的应用专题．分析：工件在传送带上运动时先做匀加速运动，后做匀速运动，物体和传送带要发生相对滑动，滑动摩擦力对传送带要做功．根据功能关系求解工件机械能的增量．由运动学公式求解相对位移．解答：解：A、工件从静止开始在摩擦力作用下加速达到v，摩擦力对工件做正功，使工件的动能增加了mv2，根据动能定理知，摩擦力对工件做的功W=mv2，A、B正确D错误．C、工件从开始运动到与传送带速度一样的过程中，工件相对传送带向后运动，设这段时间为t，t==，相对位移l=vt﹣t=t=，C正确．应当选：ABC．点评：当物体之间发生相对滑动时，一定要注意物体的动能增加的同时，内能也要增加，这是解此题的关键地方．7．在平直的公路上，汽车由静止开始做匀加速运动，当速度达到V m，立即关闭发动机而滑行直到停止，v﹣t 图线如图，汽车的牵引力大小为F1，摩擦力大小为F2，全过程中，牵引力做功为W1，抑制摩擦力做功为W2，如此〔〕A． F1：F2=1：3 B． F1：F2=4：1 C． W1：W2=1：1 D． W1：W2=1：3考点：动能定理的应用；匀变速直线运动的图像．专题：动能定理的应用专题．分析：由动能定理可得出汽车牵引力的功与抑制摩擦力做功的关系，由功的公式可求得牵引力和摩擦力的大小关系；解答：解：对全过程由动能定理可知W1﹣W2=0，故W1：W2=1：1，故C正确，D错误；W1=FsW2=fs′由图可知：s：s′=1：4所以F1：F2=4：1，故A错误，B正确应当选BC点评：此题要注意在机车起动中灵活利用功率公式与动能定理公式，同时要注意图象在题目中的应用．8．如下列图，一轻弹簧固定于O点，另一端系一重物，将重物从与悬点O在同一水平面且弹簧保持原长的A点无初速地释放，让它自由摆下，不计空气阻力，在重物由A点摆向最低点的过程中〔〕A．重物的重力势能减少 B．重物的重力势能增大C．重物的机械能不变 D．重物的机械能减少考点：机械能守恒定律；功率、平均功率和瞬时功率．专题：机械能守恒定律应用专题．分析：根据重力做功，判断重力势能的变化，在整个运动的过程中，有重力和弹簧的弹力做功，系统机械能守恒，通过系统机械能守恒判断重物机械能的变化．解答：解：A、重物由A点摆向最低点的过程中，重力做正功，重力势能减小．故A正确，B错误．C、在整个运动的过程中，只有重力和弹簧的弹力做功，系统机械能守恒，而弹簧的弹性势能增加，如此重物的机械能减小．故C错误，D正确．应当选AD．点评：解决此题的关键掌握重力做功和重力势能的关系，知道系统机械能包括重力势能、弹性势能和动能的总和保持不变．9．一物块沿倾角为θ的斜坡向上滑动．当物块的初速度为v时，上升的最大高度为H，如下列图；当物块的初速度为时，上升的最大高度记为h．重力加速度大小为g．物块与斜坡间的动摩擦因数和h分别为〔〕A．tanθ和 B．〔﹣1〕tanθ和C．tanθ和 D．〔﹣1〕tanθ和考点：牛顿第二定律．专题：牛顿运动定律综合专题．分析：两次上滑过程中，利用动能定理列式求的即可；解答：解：以速度v上升过程中，由动能定理可知以速度上升过程中，由动能定理可知联立解得，h=故D正确．应当选：D．点评：此题主要考查了动能定理，注意过程的选取是关键；10．如图是安装在列车车厢之间的摩擦缓冲器结构图，图中①和②为楔块，③和④为垫块，楔块与弹簧盒、垫块间均有摩擦，在车厢相互撞击时弹簧压缩过程中〔〕A．缓冲器的机械能守恒B．摩擦力做功消耗机械能C．垫块的动能全部转化成内能D．弹簧的弹性势能全部转化为动能考点：功能关系；弹性势能；机械能守恒定律．分析：通过抑制摩擦力做功，系统的机械能向内能转化，结合能量守恒定律分析即可．解答：解：A、通过抑制摩擦力做功，系统的机械能向内能转化，故机械能减小，故A错误；B、通过抑制摩擦力做功，系统的机械能向内能转化，故B正确；C、垫块的动能转化为弹性势能和内能，故C错误；D、弹簧的弹性势能转化为动能和内能，故D错误．应当选：B．点评：此题关键是明确缓冲器通过摩擦将局部动能转化为内能，还会储存局部弹性势能，再次向内能和动能转化，根底问题．11．如图，两根一样的轻质弹簧，沿足够长的光滑斜面放置，下端固定在斜面底部挡板上，斜面固定不动．质量不同、形状一样的两物块分别置于两弹簧上端．现用外力作用在物体上，使两弹簧具有一样的压缩量，假设撤去外力后，两物块由静止沿斜面向上弹出并离开弹簧，如此从撤去外力到物块速度第一次减为零的过程，两物块〔〕A．最大速度一样 B．最大加速度一样C．上升的最大高度不同 D．重力势能的变化量不同考点：功能关系；弹性势能．分析：使两弹簧具有一样的压缩量，如此储存的弹性势能相等，根据能量守恒判断最后的重力势能．解答：解：A、物块受力平衡时具有最大速度，即：mgsinθ=k△x如此质量大的物块具有最大速度时弹簧的压缩量比拟大，上升的高度比拟低，即位移小，而运动过程中质量大的物块平均加速度较小，v2﹣02=2ax加速度小的位移小，如此最大速度v较小，故A错误；B、开始时物块具有最大加速度，开始弹簧形变量一样，如此弹力一样，根据牛顿第二定律：a=可见质量大的最大加速度较小，故B错误；CD、由题意使两弹簧具有一样的压缩量，如此储存的弹性势能相等，物块上升到最大高度时，弹性势能完全转化为重力势能，如此物块最终的重力势能mgh相等，重力势能的变化量相等，而两物块质量不同，如此上升的最大高度不同，故C正确D错误．应当选：C．点评：此题考查了弹簧问题，注意平衡位置不是弹簧的原长处，而是受力平衡的位置．12．如下列图，有三个斜面a、b、c，底边的长分别为L、L、2L，高度分别为2h、h、h．某物体与三个斜面间的动摩擦因数都一样，这个物体分别沿三个斜面从顶端由静止下滑到底端．三种情况相比拟，如下说法正确的答案是〔〕A．物体损失的机械能△E c=2△E b=4△E aB．因摩擦产生的热量2Q a=2Q b=Q cC．物体到达底端的动能E ka=2E kb=2E kcD．因摩擦产生的热量4Q a=2Q b=Q c考点：功能关系；动能和势能的相互转化．分析：损失的机械能转化成摩擦产生的内能．物体从斜面下滑过程中，重力做正功，摩擦力做负功，根据动能定理可以比拟三者动能大小，注意物体在运动过程中抑制摩擦力所做功等于因摩擦产生热量，据此可以比拟摩擦生热大小．解答：解：设斜面和水平方向夹角为θ，斜面长度为X，如此物体下滑过程中抑制摩擦力做功为：W=mgμXcosθ，Xcosθ即为底边长度．A、物体下滑，除重力外有摩擦力做功，根据能量守恒，损失的机械能转化成摩擦产生的内能．。
高三月考二试题A卷
试卷类型：A 卷高三月考二试题文科数学说明：全卷共计三个大题23个小题，分I 、II 卷，其中第I 卷答案用2B 铅笔按规则涂在答题卡上的对应位置上，第II 卷用黑色钢笔或者0.5mm 签字笔工整地填写在答题卡的对应位置处。
第Ⅰ卷一、 选择题：（每小题只有一个正确答案，每小题5分，共计12个小题，共60分） 1、 设集合= ，全集为. 则.2、 已知复数,则其共轭复数=. ABCD3、 已知命题:p 实数,,a b c 成等比数列；2:q b ac =.则p 是q 的条件. A 充分不必要B 必要不充分C 既不充分也不必要4、 已知ABC ∆的三边,,a b c 分别对应于角,,A B C ，，则()C =.ABC5、 不等式()2lg 21x x --≤的解集为().A ()3,4-B []3,4-C ()[]3,12,4--D [)(]3,12,4--6、 已知α为锐角，且，则()sin α=.ABCD 7、 在ABC ∆中，90,2,4OB BC AB ∠===,点,DE 分别为边,BC AC 的中点，则向量AD 与BE 的数量积BE AD = .8、则当ω取最小值时函数的图像可以由函数的图像而得到.9、，则()()()2223 (22018)f f f+++A 2017B 22017 C 4034 D10、定义域为实数集的偶函数()f x是周期为2的周期函数，且在区间[]0,1上()xf x e=，则A B C 19e D 1911、已知实数,x y满足102x yx yy-+≥+≤≥⎧⎪⎨⎪⎩，设点(),P x y和()1,1Q-，则(PQ=A B C D 112、定义在实数集上的偶函数()f x的导函数为()/f x，若对任意实数x都有则是关于x的不等式()()2211x f x f x-<-成立的实数x的取值范围为().B ()1,1- C ()()1,00,1- D ()(),11,-∞-+∞第Ⅱ卷二、 填空题：（每小题5分，共计4个小题，共20分）13、设()2xf x x e =+,则曲线()y f x =在点()0,1处的切线方程为 .14、已知“在边长为a 的正三角形内任意一点到三边的距离之和为定值类比于此可得：“在棱长为a 的正四面体中任意一点到四个面的距离之和为定值.”则这个定值为 .15、设函数()(){314,1log ,x 1a a x a x x f x -+<≥=满足对任意12,x x R ∈当12x x <时总有()()120f x f x ->成立，那么实数a 的取值集合为 .16、下列命题中真命题的序号为（少填或错填均不得分） .○1若一个球的半径缩小为原来的一半，则其体积缩小为原来的八分之一； ○2若两组数据的平均值相等，则它们的标准差也相等； ○3直线10x y ++=与圆 ○4若两个平面都垂直于同一个平面，则这两个平面平行. 三、 解答题：（共7个小题，共计70分，其中第二十二和二十三题为选做题，从中选择一个并涂黑相应位置符号再作答，其余均为必做题）17、 （本题12分）已知△OAB 中，点D 在线段OB 上，且OD=2DB ，延长BA 到C ，使BA=AC ．设,OA a OB b ==． （1）用,a b →→表示向量,OC DC ；（2）若向量OC 与()OA k DC k R +∈共线，求k 的值．18、 （本题12函数()f x a b →→=⋅（1）求函数()f x 的最小正周期及单调递减区间； （2）求函数()f x 在区间19、 （本题12分）已知命题p ：“曲线C 1x 轴上的椭圆”，命题q ：“曲线C 2”．（1）若命题p 是真命题，求m 的取值范围； （2）若p 是q 的必要不充分条件，求t 的取值范围．20、 （本题12分）已知ABC ∆中，（1）求角A 的大小； （2）求周长l 的取值范围.21、 （本题12分）已知函数()x x f ln x =.（1）判断()x f 的单调性并求其极值；（2）若不等式()()1x +≥x k f 对任意恒成立，求实数k 的取值范围.选做题22、 （本题10分）在平面直角坐标系中直线l 的参数方程为 (t 为参数).以坐标原点O 为极点，x 轴的非负半轴为极轴的极坐标系中，曲线C:.（1）写出直线l 的普通方程及曲线C 的直角坐标方程；（2）若直线l 与曲线C 相交于点A 、B ，求三角形AOB 的面积S. 23、 （本题10分）设函数.（1）求不等式()5f x ≤的解集; （2）若存在实数,使得()204f x a a +< 成立，求实数a 的取值集合.。
高三第二次月考试卷
一．选择题（每题5分，共60分。
请把正确的答案填写在上面表中）1、集合{}R x x x A x ∈=--=,012|2的所有子集个数为（ ） A 、4 B 、3 C 、2 D 、12、若一个命题的否命题是真命题，则其逆命题（ ）A 、 不一定是真命题B 、一定是真命题C 、一定是假命题D 、不一定是假命题3、一个容量为n 的样本，分成若干组，已知某组的频数和频率分别是40，0.125，则n 为（ ）A 、640B 、320C 、240D 、1604、函数13)(3+-=x x f x 在闭区间[—3，0]上的最大值、最小值分别是A 、1，1B 、1，-17C 、3，-17D 、9，-195、“b a >”是“ba 11<”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6、已知函数==--++=)2(,10)2(8)(35f f qx p x f x x 则满足 A 、10 B 、—10 C 、—26 D 、—187、函数)1(22-<+=x x y x 的反函数是A 、11-+=x y )1(-<xB 、11-+=x y )1(->xC 、11-+-=x y )1(-<xD 、11-+-=x y )1(->x8、函数 x x y 22)31(-= 的值域是 A 、]0,3[- B 、]3,0(C 、]3,(-∞D 、），∞+3[ 9、函数)32(log 22-+=x x y 的单调减区间是A 、)3,(--∞B 、)1,(--∞C 、),1(+∞-D 、），∞+1(10、将函数)1(log 2+=x y 的图象A 、先向左平移1个单位B 、先向右平移1个单位C 、先向上平移1个单位D 、先向下平移1个单位再作关于直线的图象对称的图象，可得2xy x y ==。
11、如果不等式123->a ax 对任意]1,1[-∈x 总成立，则a 的取值范围是 A 、511<<-a B 、51>a C 、151-<>a a 或 D 、1-<a 12、已知)(x f 是定义在R 上的增函数，其图象经过点A （0，-1）和点B （3，1）则不等式1)(<x f 的解集是A 、)0,(-∞B 、)3(∞+，C 、)3,0(D 、），∞+⋃-∞3[)0,( 二、填空题。
2021年高三上学期第二次月考数学(理)试题(学生版)
2021年高三上学期第二次月考数学（理）试题（学生版）学生版本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，其中第Ⅱ卷第22－24题为选考题，其它题为必考题．考生作答时，将答案答在答题卡上．在本试卷上答题无效．考试结束后，将本试卷和答题卡一并交回．注意事项：1．答题前，考生务必先将自己的姓名、准考证号填写在答题卡上;2．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;3．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。
一、选择题（本大题共12小题，每小题5分，共60分．在每小题给出的四个选项中，只有一个是符合题目要求）1．已知集合,,则（）A．B．C．D．2．命题，函数，则（）A．是假命题；，B．是假命题；，C．是真命题；，D．是真命题；，3．“非空集合M不是P的子集”的充要条件是（）A．B．C．又D．4．若35sin,,0,cos524aπααπ⎛⎫⎛⎫=-∈-+⎪ ⎪⎝⎭⎝⎭则= （）A．B．C．D．5．已知01,log ,log ,c c ca b a b c m n r a <<<<===，则的大小关系是（ ）A ．B ．C ．D ．6．若的内角所对的边满足，且，则的最小值为（ ） A ． B ． C ． D ．7．如图，设是图中边长为的正方形区域，是内函数图象下方的点构成的区域．向中随机投一点，则该点落入中的概率为 （ ） A ． B ． C ． D ．8．下列区间中，函数，在其上为增函数的是 （ ） A ． B ． C ． D ．9．将函数的图形按向量平移后得到函数g （x ）的图形，满足g （－x ）=g （+x ）和g （－x ）+g （x ）=0，则向量的一个可能值是 （ ） A ． B ． C ． D ．10．已知是上的可导函数，对于任意的正实数，都有函数在其定义域内为减函数，则函数的图象可能为下图中 （ ）11．定义一种运算，若函数，是方程的解，且，则的值（ ） A ．恒为正值 B ．等于 C ．恒为负值 D ．不大于12．关于的方程，给出下列四个命题： ( ) ①存在实数，使得方程恰有2个不同的实根； ②存在实数，使得方程恰有4个不同的实根； ③存在实数，使得方程恰有5个不同的实根； ④存在实数，使得方程恰有8个不同的实根； 其中假．命题的个数是 A ．0 B ．1 C ．2 D ．3第Ⅱ卷本卷包括必考题和选考题两部分．第13题～第21题为必考题，每个题考生都必须作答。
2021年高三上学期月考2理科数学试题 含答案
2021年高三上学期月考2理科数学试题 含答案一：选择题（本大题12小题，每题5分，共60分） 1. 集合2|40,|5,M N |3Mx x x Nx m x x x n m n，则（ ） A ．6 B ．7 C ．8 D ．92. 已知2,1bibR i b i为虚数单位，若为实数，则（ ） A ． B ． C ．1 D ．23.已知081:,242,q :(0,),222x xxP x R x ，则下列判断正确的是（ ）A ．是假命题B ．是真命题C ．是真命题D ．是真命题 4.已知，则 （ ）A ．0B ．1C ．D 。
5.等比数列中，452,5,lg 8n a a a 则数列的前项和为（ ）A ．6B ．5C ．4D 36. 一几何体的三视图如右所示，则该几何体的体积为（ ）A.200+9πB. 200+18πC. 140+9πD. 140+18π7.已知实数10,,(0,0)230x y x y z ax by a b x y 满足当在该约束条件下取到最小值4时，则的最大值为（ ） A ．2 B ．4 C ． 1 D ． 88.将函数的图像沿轴向左平移个单位，得到一个偶函数的图像，则的一个可能取值为（ ） A ． B ．C ．D ．9.已知点P 是所在平面内一点，，现将一粒黄豆随机撒在内，则黄豆落在内的概率为（ ） A. B. C. D.10.已知函数的部分图像如下图所示，则函数的解析式为（ ） A ． B ． C ． D二．填空题：共4小题，每题5分，共20分13.已知定义在R 上的可导函数的图像在点处的切线方程为，则 14.已知的最大值是1，则实数=15.若2013220130122013(12)x a a x a x a x ，则16.已知，点O 是坐标原点，点 ，则32015121232015cos cos coscos sinsinsinsin=三．解答题，共6题，共70分17（12分）,,,ABC A B C a b c 内角，，所对的边分别为且（1）求角B 的大小。
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20XX年高中测试
高
中
试
题
试
卷
科目：
年级：
考点：
监考老师：
日期：
九月月考试题
文科数学试卷20XX .10.8
命题人：审题人：
试题时间：120分钟总分：
150分
一、选择题（本大题共12个小题,每小题5分,共60分.在每小题
给出的四个选项中,只有一项是符合要求的，找到正确答案并涂在在答题卡上）。
二、填空题（本大题共4个小题,每小题5分,共20分.答案写在答题卡上）
九月月考试题
文科数学试卷20211 .10.8
命题人：审题人：
试题时间：120分钟总分：
150分
二、填空题（本大题共4个小题,每小题5分,共20分.答案写在答题卡上）
三、解答题（本大题共6个小题,共70分.）。
19（本小题满分12分）
20、（本小题满分12分）
(三选一：在22，23，24三个大题中选一题，其他两题不做，做也不得分，如做两题，以前一题为准)
高考试卷仅供参考
11。