高二上学期期中测试题 2
新人教版高二英语上学期期中测试题(非常棒) 2
大店中学高二年级上学期第二次月考英语试题第二部分：单项选择1. The film ______ him ______ what he had seen in China.A. reminded; toB. remembered; ofC. recalled; withD. reminded; of2. I _____ sight of an empty seat at the back of the bus and went directly there.A. lost B caught C. looked D. took3. He asked us to him _____ carrying through their plan.A. assist; withB. help; toC. assist; inD. help; with4. His report was so exciting that it was interrupted by applause（掌声）.A. constantlyB. constantC. seldomD. never5. ______ in a friendly way, their quarrel came to an end.A. Being settledB. SettledC. SettlingD. Having settled6. A driver should______ the road when ______ .A. concentrate on; driveB. concentrate in; drivingC. concentrate to; droveD. concentrate on; driving7. We must work hard to a good knowledge of English.A. takeB. acquireC. catchD. hold8. Tom kept quiet about the accident ______ lose his job.A. so not as toB. so as not toC. so as to notD. not so as to9. Not until the early years of the 19th century ______ what heat was.A. man did knowB. did man knewC. didn't man knowD. did man know10. The stone bridge ______last year is very beautiful.A . built B. was built C . being built D .to be built11.The poor man , ______，ran out of the dark cave.A. tiring and frightenedB. tired and frightenedC. tired and frighteningD. tiring and frightening12.Don‟t be discouraged.__ things as they are and you will enjoy every day of your life.A. TakingB. To takeC. TakeD. Taken13.These articles are written in simple language, _____ makes it easy to read.A. thatB. thisC. whichD. it14.After the war, a new school building was put up ______there used to be a theatre.A. thatB. whereC. whichD. when15. Tom‟s mother kept telling him that he should work harder, but _____ didn‟t help.A. heB. whichC. sheD. it16.Along with the letter was his promise____ he would visit me this coming Christmas.A. whichB. thatC. whatD. whether17. Victor apologized for_____ to inform me of the change in the plan.A. his being not ableB. him not to be ableC. his not being ableD. him to be not able18. The thief admitted over 10 motorbikes and now he has been arrestedA. to stealB. to have stolenC. stealingD. having stolen19.---The novel gone with wind is said _____ into several language.A. to translateB. being translatedC. having been translatedD. to have been translated20. _____ is no need .A. It ; complainingB. That; complainingC. There ; to complainD. This; to complain21.Did he find hard to learn a foreign language ?A. thatB. himC. itD. himself22.The population of China is larger than ______ of the United States.A. thisB. thatC. theseD. those23．You may send me an e-mail or just give me a call. will do．A．Neither B．Each C．Any D. Either24. He‟s got himself into a dangerous situation he is likely to be accusedof meeting someone.A. whereB. whichC. whileD. why25. The water ______ cool when I joined into the pool for morning exercise.A. was feltB. is feltC. feelsD. Felt第三节：完形填空Michel is a young girl who works for the police 26 a handwriting expert. She has helped 27 many criminals by using her special talents.When she was fourteen, Michel was already 28 interested in the differences in her friends' 29 that she would spend hours 30 them. After 31 college she went to France for a 32 two-year class in handwriting at the School of Police Science.Michel says that it is 33 for people to hide their handwriting. She can discover 34 of what she needs to know simply 35 looking at the writing with her own eyes, 36 she also has machines 37 help her make 38 different kinds of paper and ink. This knowledge is often 39 great help to the police.Michel believes that handwriting is a good 40 of what kind of person the 41 is. "I wouldn't go out with a fellow 42 I didn't like his handwriting. " She says. But she 43 she fell in love with her future husband, a young policeman 44 she studied his handwriting. It is later proved to be 45 , however.26 A. with B. by C. like D. as27 A. search B. follow C. catch D. judge28 A. so B. too C. quite D. extra29 A. books B. letter C. tongues D. handwriting30 A. writing B. studying C. settling D. uncovering31 A. attending B. finishing C. starting D. stepping into32 A. powerful B. natural C. special D. common33 A. main B. safe C. easy D. impossible34 A. most B. nothing C. little D. sight35 A. with B. by C. of D. about36 A. so B. for C. thus D. but37 A. they B. in which C. that D. those38 A. up B. out C. for D. into39 A. of B. to C. with D. for40 A. test B. sign C. means D. habit41 A. thief B. criminal C. writer D. policeman42 A. whether B. unless C. if D. after43 A. adds B. tells C. repeats D. cries44 A. before B. after C. so D. and45 A. necessary B. all right C. important D. quite easy第四部分：阅读理解AA traveller was staying in an Egyptian village. One day, she held up her camera to take pictures of the children. Suddenly the young ones began to shout at her. The traveller's face turned red and she apologized to the head for what she was doing, and told him she had forgotten that people in some places believed a person would lose his soul if his picture was taken. She explained to him the operation of a camera for a long time. Several times the head tried to say something, but he couldn't. When she believed that the head didn't fear any longer, the traveller then let him speak. With a smile, he said, "The children were trying to tell you that you forgot to take off the lens(镜头) cap!" 46. The children shouted when the traveller was taking pictures ofthem because _____.A. they didn't want to stop playingB.the traveller forgot to take off the cap on her headC. they didn't want to have their pictures takenD. the traveller was not doing well with her camera47. The traveller made an apology to the head because _____.A.she thought it was not right to take people's pictures without telling them beforehand(事先)B. the children would lose their soulsC. she had stayed in the village too longD. she didn't take a picture of the head first48. The traveller explained how to use a camera to the head because _____.A. the head was very interested in her cameraB. the head wanted to learn to take picturesC. she was afraid of the headD. she wanted the head not to worry about what she was doing49. When the head smiled, it's clear that _____.A. the children wanted to play with herB. the traveller didn't know what the children meantC. he wanted the traveller to tell him something elseD. the traveller didn't let him speak50. Which of the following is NOT right?A. The traveller knew something about people in some countries.B. The children wouldn't mind if the traveller took pictures of them.C. The head was afraid that the traveller's camera would hurt the children.D. The traveller didn't understand why the children shouted.BPopeye the Sailor first became a popular cartoon in the 1930s.The sailor in that cartoon ate lots of spinach to make him strong. People watched him, and they began to buy and eat a lot more spinach. Popeye helped sell 33 percent more spinach than before! Spinach became a necessary part of many people‟s diets. Even some children who hated the taste began to eat the vegetable.Many people thought that the iron in spinach made Popeye strong, but this is not true. Spinach does not have any more iron than any other green vegetable.People only thought spinach had a lot of iron because the people who studied the food made a mistake. In the 1890s, a group of people studied what was inside vegetables. This group said that spinach had ten times more iron than it did. The group wrote the number wrong, and everyone accepted it.Today, we know that the little iron there is in spinach cannot make a difference in how strong a person is. However, spinach does have something else which the body needs—folic acid.It is interesting to point out that folic acid can help make a person strong. Maybe it was really the folic acid that made Popeye strong all along.51.A good title for this reading passage is______.A. Popeye the SailorB. The Truth About SpinachC.A Mistake with NumbersD. Folic Acid Makes You Strong52.Why did many people eat spinach after they saw Popeye the Sailor?A. They thought spinach made them strong.B. They thought Popeye was funny.C. Spinach had a lot of iron.D. People liked folic acid.53.A research group told people that spinach______.A. made Popeye strongB. was a green vegetableC.had less iron than other green vegetablesD.had more iron than other green vegetables54.The reading passage says that perhaps Popeye got his strength from______.A. ironB. folic acidC. spinachD. exercise55.Folic acid is ______.A. something in foodB. a vegetableC. dangerousD. a certain kind of spinachCBody and FoodYour body, which has close relations with the food you eat, is the most important thing you own, so it needs proper treatment and proper nourishment (营养).The old saying “An apple a day keeps the doctor away ”is not as silly as some people think.The body needs fruit and vegetables because they contain vitamin C.Many people take extra vitamins in pill form, believing that these will make them healthy.But a good diet is made up of nourishing food and this gives all the vitamins you need. The body doesn‟t need or use extra vitamins, so why waste money on them?In the modern western world, many people are too busy to bother about eating properly.They throw anything into their stomachs, eating hurriedly and carelessly. The list of illnesses caused or made worse by bad eating habits is frightening,56.“Your body has close relations with the food you eat.”It really means that ______.A.all kinds of food you eat can be made into your bodyB.your body is made up of the food you eatC.what you eat has great effect on your healthD.the more you eat, the fitter you will feel57.The old saying referred to in the passage tells us that ______.A.eating apples regularly does lots of good to our healthB.the apple is the best among all kinds of fruitsC.apples can take the place of doctorsD. an apple is a sure cure for illness58.In the second paragraph, the writer tries to let us know ______.A.our bo dies need food or we can‟ t liveB.often eating apples is a good habitC.taking extra vitamin pills is completely uselessD.a good diet is of great importance for our health59.In the modern western countries ______.A.people don …t want to pay more attention to their eatingB.lots of people‟ s illnesses are caused or made worse by bad eating habitsC.people throw everything into their stomachs without chewingD.people are only too busy to cook meals for themselves60.From the passage we can draw a conclusion that if we want to keep healthy,we should ____.A. only eat an apple a dayB. eat properlyC. take as many vitamin pills as possibleD.throw something into our stomachs slowly and carefullyDAt midnight on New Year‟s Eve, people in Rome, Ita ly, throw out all the things they no longer want. The streets are filled with old chairs, beds, clothes and dishes. In Madrid, Spain, the new year comes in more quietly. People flock to the main square. Each holds a bag of grapes. As the clock strikes twelve, the people eat the grapes— one for each stroke.In Tokyo people eat noodles on New Year‟s Eve. This food is said to bring long life. Early the next morning, some Japanese families climb Mount Fuji. There they watch the first sunrise of the new year.61. This story is about New Year‟s Eve in ____.A. Italy and SpainB. China and JapanC. JapanD. Italy ,Spain and Japan62. In Rome, Italy, the streets are filled with old things on New Year‟s Eve because __ .A.people throw out all the things they no longer wantB.the city has never been cleanC.people want to change back what they wantD.they want to see the sun come up63. In Tokyo, people eat noodles on New Year‟s Eve _____.A.because they like eating noodles very muchB.so that the daytime could be longerC.in order that they could live longerD.though they don‟t eat themEA lot of people think Scotland is a part of England , but, as any Scotsman will tell you , it certainly is not. In fact, until the eighteenth century Scotland was an independent country, with a parliament of its own. The English had tried many times over many centuries to bring Scotland under their rule. They succeeded at last in 1707, and some Scots have never forgiven them.Scotland is now governed from London, but in some ways it is still a separate nation. It has its own capital city(Edinburgh),its own law, and its own stamps, it even has a language of its own, spoken now by only a few people in the islands.In some ways Southern Scotland is like England, with his good farmland and low green hills. Central and Northern (the Highlands) have high mountains and deep valleys, fast rivers and cold lakes. These days, of course, there are good roads and railways all through Scotland. Aberdeen, the northeast city where the oil from the North Sea comes to land, is especially easy to reach. But it can still be quite different to travel in the winter when the hills are covered with snow. It always takes a long time to visit the beautiful but far-off islands on the west coast. One reason why Scotland has stayed so different from England is the wildness of the land. It has always been difficultto get around there.63. From the history of the Great Britain we can know Scotland________.A.had been an independent country by the 19th centuryB.had been a dependent country by the 18th centuryC.was a separate country before the 18th centuryD.was a dependent country before 18th century64. The first paragraph tell us _______.A.the Scots used to fight against the rules from England many timesB.the Scots defeated the English at all the battlesC.the Scots never defeated the English at all the battlesD.it was quite easy for the English to occupy Scotland65. Scotland has a language of its own,________.A.which is spoken by all the ScotsB.which not many people speak nowC.which is almost the same as English languageD.which the English prevent from being spoken66. From the third paragraph we can infer ______.A.there are no good roads in Scotland because of high mountainsB.there are no railways in Scotland because of deep valleysC.England has not any rivers and lakes in the center and the northD.England has good farmland and low green hills英语试题第一节：单项选择（共25小题，1分/题，共25分）1—5 DBCAB 6—10 DBBDA 11—15 BCCBD 16—20 BCDDC 21—25 CBDAD第二节完形填空（共20小题；每小题1分，满分20分）26、D 27、C 28、A 29、D 30、B31、B 32、C 33、D 34、A 35、B36、D 37、C 38、B 39、A 40、B41、C 42、C 43、A 44、A 45、B第三部分：阅读理解（共20小题，每题2分，满分40分）46-50 DADCC51--55 BADBA 56--60 CADBB61---63 DAC。
江西省高二上学期期中考试数学试题(解析版)
一、单选题1．已知直线的图像如图所示，则角是（ ）sin cos :y x l θθ=+θA ．第一象限角B ．第二象限角C ．第三象限角D ．第四象限角【答案】D【分析】本题可根据直线的斜率和截距得出、，即可得出结果. sin 0θ<cos 0θ>【详解】结合图像易知，，， sin 0θ<cos 0θ>则角是第四象限角， θ故选：D.2．的展开式中的系数为（ ） ()()8x y x y -+36x y A ． B ．C ．D ．2828-5656-【答案】B【分析】由二项式定理将展开，然后得出，即可求出的系数. 8()x y +8()()x y x y -+36x y 【详解】由二项式定理：8()()x y x y -+080171808888()(C C C )x y x y x y x y =-+++080171808080171808888888(C C C )(C C C )x x y x y x y y x y x y x y =+++-+++090181818081172809888888(C C C )(C C C )x y x y x y x y x y x y =+++-+++ 观察可知的系数为. 36x y 6523888887876C C C C 2821321⨯⨯⨯-=-=-=-⨯⨯⨯故选：B.3．已知条件：，条件：表示一个椭圆，则是的（ ） p 0mn >q 221x y m n+=p q 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【答案】B【分析】根据曲线方程，结合充分、必要性的定义判断题设条件间的关系.【详解】由，若，则表示一个圆，充分性不成立；0mn >0m n =>221x y m n +=而表示一个椭圆，则成立，必要性成立. 221x y m n+=0mn >所以是的必要不充分条件. p q 故选：B4．两平行平面分别经过坐标原点O 和点，且两平面的一个法向量，则两,αβ()1,2,3A ()1,0,1n =-平面间的距离是（ ）A B C D ．【答案】A【分析】由空间向量求解【详解】∵两平行平面分别经过坐标原点O 和点，,αβ(1,2,3),(1,2,3)A OA =且两平面的一个法向量，(1,0,1)n =-∴两平面间的距离 ||||n OA d n ⋅=== 故选：A5．2022年遂宁主城区突发“920疫情”，23日凌晨2时，射洪组织五支“最美逆行医疗队”去支援遂宁主城区，将分派到遂宁船山区、遂宁经开区、遂宁高新区进行核酸采样服务，每支医疗队只能去一个区，每区至少有一支医疗队，若恰有两支医疗队者被分派到高新区，则不同的安排方法共有（ ） A ．30种 B ．40种 C ．50种 D ．60种【答案】D【分析】先从5支医疗队中选取2支医疗队去高新区，再将剩下的3支医疗队分配到船山区与经开区，最后根据分步乘法原理求解即可.【详解】解：先从5支医疗队中选取2支医疗队去高新区，有种不同的选派方案，25C 10=再将剩下的3对医疗队分配到船山区与经开区，有种不同的选派方案，2232C A 6=所以，根据分步乘法原理，不同的安排方案有种.222532C C A 60=故选：D6．已知圆：，直线：，为上的动点，过点作圆的两条切线C 2220x y x +-=l 10x y ++=P l P C 、，切点分别、，当最小时，直线PC 的方程为（ ）PA PB A B ·PC ABA ．B ．C ．D ．+=0x y 10x y --=2210x y -+=2210x y ++=【答案】B【分析】根据圆的切线的有关知识，判断出最小时，直线与直线垂直，进而可得直·PC AB l PC 线的方程.PC 【详解】圆的标准方程为，圆心为，半径为.C ()2211x y -+=()1,0C =1r 依圆的知识可知，四点P ，A ，B ，C 四点共圆，且AB ⊥PC ， 所以，而14422PAC PC AB S PA AC PA ⋅==⨯⨯⋅=△当直线时，最小，此时最小， PC l ⊥PA PC AB ⋅所以此时，即. :=1PC y x -10x y --=故选：B.7．某奥运村有，，三个运动员生活区，其中区住有人，区住有人，区住有人A B C A 30B 15C 10已知三个区在一条直线上，位置如图所示奥运村公交车拟在此间设一个停靠点，为使所有运动员..步行到停靠点路程总和最小，那么停靠点位置应在（ ）A ．区B ．区C ．区D ．，两区之间A B C A B 【答案】A【分析】分类讨论，分别研究停靠点为区、区、区和，两区之间时的总路程，即可得出A B C A B 答案．【详解】若停靠点为区时，所有运动员步行到停靠点的路程和为：米； A 15100103004500⨯+⨯=若停靠点为区时，所有运动员步行到停靠点的路程和为：米； B 30100102005000⨯+⨯=若停靠点为区时，所有运动员步行到停靠点的路程和为：米； C 303001520012000⨯+⨯=若停靠点为区和区之间时，设距离区为米，所有运动员步行到停靠点的路程和为：A B A x ， 30151001010020054500x x x x +⨯-+⨯+-=+（）（）当取最小值，故停靠点为区． 0x =A 故选：A8．已知是双曲线上的三个点，经过原点，经过右焦点，若,,A B C 22221(0,0)x y a b a b -=>>AB O AC F 且，则该双曲线的离心率是（ ）BF AC ⊥2AF CF =A ．B C D ．5394【答案】B【分析】根据题意，连接，构造矩形；根据双曲线定义表示出各个边长，由直角','AF CF 'FAF B 三角形勾股定理求得 的关系，进而求出离心率．a c 、【详解】设左焦点为， ，连接F'AF m =','AF CF 则 ， ， ， 2FC m ='2AF a m =+'22CF a m =+'2FF c =因为，且经过原点 BF AC ⊥AB O 所以四边形 为矩形'FAF B 在Rt △中， ，代入'AF C 222'+'AF AC F C =()()()2222+3=22a m m a m ++化简得 23a m =所以在Rt △中，，代入 'AF F 222'+'AF AF F F =()222222233a a a c ⎛⎫⎛⎫++= ⎪ ⎪⎝⎭⎝⎭化简得 ，即 22179c a =e =所以选B【点睛】本题考查了双曲线的综合应用，根据条件理清各边的相互关系，属于中档题．二、多选题9．下列结论正确的是（ ）A ．“”是“直线与直线互相垂直”的充要条件1a =-210a x y -+=20x ay --=B ．已知，O 为坐标原点，点是圆外一点，直线的方程是，0ab ≠(,)P a b 222x y r +=m 2ax by r +=则与圆相交m C ．已知直线和以，为端点的线段相交，则实数的取值范围为10kx y k ---=(3,1)M -(3,2)N k 1322k -≤≤D ．直线的倾斜角的取值范围是sin 20x y α++=θπ3π0,,π44⎡⎤⎡⎫⎪⎢⎥⎢⎣⎦⎣⎭ 【答案】BD【分析】由题意利用直线的倾斜角和斜率、直线的方程，直线与圆的位置关系，逐一判断各个选项是否正确，从而得出结论．【详解】解：对于A ，由直线与直线互相垂直，210a x y -+=20x ay --=，化为，解得或，21(1)()0a a ∴⨯+-⨯-=20a a +==0a 1- “”是“直线与直线互相垂直”的充分但不必要条件，故A 错误；∴1a =-210a x y -+=20x ay --=对于B ，因为点是圆外一点，所以，所以圆心到直线的距离(,)P a b 222x y r +=222a b r +>m，可得与圆相交，故B 正确；||d r =m 对于C ，已知直线和以，为端点的线段相交，则、两个点在直10kx y k ---=(3,1)M -(3,2)N M N 线的两侧或直线上，10kx y k ---=则有，解可得或，故C 错误； (311)(321)0k k k k -------≤12k ≤-32k ≥对于D ，设直线的倾斜角，则,， sin 20x y α++=θtan sin [1θα=-∈-1]故的取值范围是，故D 正确. θ3[0,[,)44πππ 故选：BD .10．已知的展开式中第3项与第5项的系数之比为，则下列结论成立的是（ ） 2(n x 314A ．B ．展开式中的常数项为45 10n =C ．含的项的系数为210D ．展开式中的有理项有5项5x【答案】ABC【分析】根据二项式的展开式的通项公式，结合第3项与第5项的系数之比为()52211C r n rr r n T x-+=-，可得.再根据公式逐个选项判断即可. 31410n =【详解】二项式的展开式的通项为，由于第3项与第5项的()()5222221C 11C rr n r rrn r r r n nT xx x---+=-=-系数之比为，则，故，得. 31424C 3C 14n n=()()()()1312123141234n n n n n n -⨯=---⨯⨯⨯25500n n --=∴(n ＋5)(n －10)＝0，解得n ＝10，故A 正确；则，令，解得， ()52021101C rr r r T x-+=-52002r-=8r =则展开式中的常数项为，故B 正确； 810C 45=令，解得，则含的项的系数为，故C 正确； 52052r -=6r =5x ()66101C 210-=令，则r 为偶数，此时，故6项有理项． 520Z 2r-∈0,2,4,6,8,10r =故选：ABC11．2022年2月5日晩，在北京冬奥会短道速滑混合团体接力决赛中，中国队率先冲过终点，为中国体育代表团拿到本届奥运会首枚金牌.赛后，武大靖，任子威，曲春雨，范可欣，张雨婷5名运动员从左往右排成一排合影留念，下列结论正确的是（ ） A ．武大靖与张雨婷相邻，共有48种排法 B ．范可欣与曲春雨不相邻，共有72种排法 C ．任子威在范可欣的右边，共有120种排法D ．任子威不在最左边，武大靖不在最右边，共有78种排法 【答案】ABD【分析】利用分步乘法计数原理结合排列与排列数，逐项分析判断即可.【详解】解：A 项中，武大靖与张雨婷相邻，将武大靖与张雨婷排在一起有种排法， 22A 再将二人看成一个整体与其余三人全排列，有种排法，44A 由分步乘法计数原理得，共有（种）排法，故选项A 正确；2424A A 48=B 项中，范可欣与曲春雨不相邻，先将其余三人全排列，有种排法， 33A 再将范可欣与曲春雨插入其余三人形成的4个空位中，有种排法，24A由分步乘法计数原理得，共有（种）排法，故选项B 正确；3234A A =72C 项中，任子威在范可欣的右边，先从五个位置中选出三个位置排其余三人，有种排法， 35A 剩下两个位置排任子威、范可欣，只有1种排法，所以任子威在范可欣的右边，共有（种）排法，故选项C 错误；35A =60D 项中，武大靖，任子威，曲春雨，范可欣，张雨婷5人全排列，有种排法， 55A 任子威在最左边，有种排法，武大靖在最右边，有种排法， 44A 44A 任子威在最左边，且武大靖在最右边，有种排法，33A 所以任子威不在最左边，武大靖不在最右边，共有（种）排法，故选项D 正确. 543543A -2A +A =78故选：ABD.12．为庆祝党的二十大胜利召开，由南京市委党史办主办，各区委党史办等协办组织的以“喜迎二十大 永远跟党走 奋进新征程”为主题的庆祝中共南京地方组织成立周年知识问答活动正在进100行，某党支部为本次活动设置了一个冠军奖杯，奖杯由一个铜球和一个托盘组成，如图①，已知球的体积为，托盘由边长为的正三角形铜片沿各边中点的连线垂直向上折叠而成，如图②.则32π38下列结论正确的是（ ）A ．经过三个顶点的球的截面圆的面积为 ,,ABC 43πB ．异面直线与所成的角的余弦值为AD BE 916C ．连接，构成一个八面体，则该八面体的体积为 ,,AB BC CA ABCDEF ABCDEF 18D ．点 D 2【答案】ACD【分析】对A ：经过三个顶点的球的截面圆即为的外接圆，运算求解；对B ：建系，,,A B C MNG △利用空间向量处理异面直线夹角问题；对C ：八面体由三个全等的四棱锥ABCDEF和直棱柱组合而成，结合相关体积公式运算求解；,,D ACGM E ABNM F BCGN ---ABC MNG -对D ：点到球面上的点的最小距离为，结合球的性质运算求解.D OD R -【详解】如图1，取的中点分别为，连接 ,,DE EF DF ,,M NG ,,,,,AM BN CG MN NG GM 根据题意可得：均垂直于平面，可知 ,,AM BN CG DEF ABC MNG ≅△△∵的边长为2，设的外接圆半径为r ，则MNG △MNG △sin MN 2r MGN ==∠∴的外接圆面积为r =MNG △4ππ32r =∴经过三个顶点的球的截面圆的面积为，A 正确； ,,A B C 43π八面体由三个全等的四棱锥和直棱柱组合ABCDEF ,,D ACGM E ABNM F BCGN ---ABC MNG -而成直棱柱的底面边长为2，高ABC MNG -AM =12262ABC MNG V -=⨯⨯=设，则为的中点 EN MN H = H MN ∵平面，平面 AM ⊥DEF EH ⊂DEF ∴AM EH ⊥又∵为等边三角形且为的中点，则EMN A H MN MN EH ⊥，平面 AM MN M = ,AM MN ⊂ABNM ∴平面EH ⊥ABNM即四棱锥的高为E ABNM -EH =1243E ABNM V -=⨯=∴八面体的体积为，C 正确；ABCDEF 318E ABNM ABC MNG V V V --=+=设的中心分别为，球的球心为，由题意可得其半径 ,ABC MNG △△12,O O O =2R 则可知三点共线，连接 12,,O O O 1,O B OD则可得：212112O D O O O O O O O O OD ===+==点，D 正确；D 2-如图2，以G 为坐标原点建立空间直角坐标系则有：((()(),,2,0,0,0,A B D E -∴((,DA BE =-=- 又∵ 5cos ,8DA BE DA BE DA BE⋅==-∴异面直线与所成的角的余弦值为，B 错误；AD BE 58故选：ACD.【点睛】1.对于多面体体积问题，要理解几何体的结构特征，并灵活运用割补方法； 2.对于球相关问题，主要根据两个基本性质：①球的任何截面都是圆面；②球心和截面圆心的连线与截面垂直.三、填空题13．若，则______．2213C P x xx -+=x =【答案】5【分析】将排列数、组合数按照公式展开，即可解出x 的值.【详解】因为，， ()22313C 3C 2x x x x x --==21P (1)x x x +=+所以，由可得，3(x -1)=2(x +1)2213C P x x x -+=解得，x =5.故答案为：5.14．各数位数字之和等于8(数字可以重复) 的四位数个数为_____． 【答案】120【分析】四个数位数字分别为，则，应用插空法求四位数个数. 1234,,,a a a a 12348a a a a +++=【详解】设对应个位到千位上的数字，则，且， 1234,,,a a a a *4N a ∈N(1,2,3)i a i ∈=1234a a a a +++8=相当于将3个表示0的球与8个表示1的球排成一排，即10个空用3个隔板将其分开，故共种.310C 120=故答案为：12015．已知分别为双曲线的左、右顶点，点为双曲线上任意一点，12,A A 2222:1(0)x y C a b a b -=>>P C 记直线，直线的斜率分别为，若，则双曲线的离心率为__________. 1PA 2PA 12,k k 122k k ⋅=C【分析】设，应用斜率两点式得到，根据为双曲线上一点即可得双曲线参()00,P x y 22202y x a=-P C 数关系，进而求其离心率【详解】依题意，设，则，，又()()12,0,,0A a A a -()00,P x y 0012002y y k k x a x a ⋅=⋅=+-22202y x a∴=-，，故，即()2222220220000222211b x a x y x y b a b a a -⎛⎫-=⇒=-= ⎪⎝⎭222b a ∴=22213b e a =+=e =16．在棱长为1的正方体中，M 是棱的中点，点P 在侧面内，若1111ABCD A B C D -1AA 11ABB A ，则的面积的最小值是________．1D P CM ⊥PBC △【分析】建立空间直角坐标系，利用空间向量、三角形的面积公式、二次函数进行求解.【详解】如图，以点D 为空间直角坐标系的原点，分别以DA ，DC ，所在直线为x ，y ，z 轴， 1DD 建立空间直角坐标系，则点，所以， ()1,,,[01]P y z y z ∈、，()10,0,1D ()11,,1D P y z =-因为，所以，()10,1,0,1,0,2C M ⎛⎫ ⎪⎝⎭11,1,2CM =-⎛⎫ ⎪⎝⎭ 因为，所以，所以，1D P CM ⊥ ()11102y z -+-=21z y =-因为，所以， ()1,1,0B ()0,1,21BP y y =--，=因为，所以当时， 01y ≤≤35y =min BP =因为正方体中，平面平面，故， BC ⊥11,ABB A BP ⊂11ABB A BC BP ⊥所以()min 1=12PBC S ⨯A四、解答题17．已知的顶点. ABC A ()()()2,64,2,2,0A B C -，(1)求边的中垂线所在直线的方程； BC (2)求的面积. ABC A 【答案】(1)； 340x y +-=(2)14.【分析】（1）求出直线的斜率，再由垂直关系得出直线边的中垂线的斜率，最后由点斜式BC BC 写出所求方程；（2）求出直线的方程，再求出点到直线的距离以及，最后由三角形面积公式计算即AB C AB AB 可.【详解】（1）直线的斜率为，直线边的中垂线的斜率为，BC 2014(2)3-=--BC 3-又的中点为，BC ()1,1边的中垂线所在直线的方程为：，即； BC ()131y x -=--340x y +-=（2）直线的方程为：，即， AB 626(2)24y x --=--2100x y +-=点到直线的距离 C AB d=故的面积为. ABC A 1142S AB d =⋅=18．已知展开式的二项式系数和为512，且()(2)n f x x =-.2012(2)(1)(1)(1)n n n x a a x a x a x -=+-+-+⋅⋅⋅+-(1)求的值； 123n a a a a +++⋅⋅⋅⋅⋅⋅+(2)求被除的余数. ()20f 17【答案】(1) 1(2) 1【分析】（1）根据题意，得到，求得，结合展开式，分别令和，求得2512n =9n =1x =2x =和，即可求解；01a =-012390a a a a a ++++⋅⋅⋅⋅⋅+=⋅（2）由，结合二项式的展开式，即可求解.999(20)(2021817)(1)f ==+=-【详解】（1）解：由展开式的二项式系数和为，可得，解得，(2)n x -5122512n =9n =则，9290129(2)(1)(1)(1)x a a x a x a x -=+-+-+⋅⋅⋅+-令，可得，1x =90(12)1a =-=-令，可得，2x =012399(22)0a a a a a ++++⋅⋅⋅⋅=-⋅+=⋅所以， 12390(1)1a a a a +++⋅⋅⋅⋅⋅=--+=⋅即.1231n a a a a +++⋅⋅⋅⋅⋅+=⋅（2）解：由题意，可得，999(20)(2021817)(1)f ==+=-又由，90918890081789999999(171)1717171717(1717)1C C C C C C C +=⋅+⋅++⋅+⋅=⋅⋅+⋅+++ 所以被除的余数为.()20f 17119．如图，在四棱锥中，已知四边形是梯形，P ABCD -ABCD ，是正三角形．,,22⊥===∥AB CD AD AB AB BC CD PBC △(1)求证：；BC PA ⊥(2)当四棱锥体积最大时，二面角的大小为，求的值. P ABCD -B PA C --θcos θ【答案】(1)证明见解析； (2). 15【分析】（1）取BC 的中点O ，连接AO ，可证明，由线面垂直的判定定理可证AO BC ⊥PO BC ⊥明平面PAO ，即得证；BC ⊥（2）分析可知当平面平面ABCD 时，四棱锥体积最大，建立空间直角坐标系，PBC ⊥P ABCD -由二面角的向量公式，计算即可.【详解】（1）证明：如图，取AB 的中点E ，连接CE ，A C ．∵，， 2AB CD =AB CD ∥∴CD 与AE 平行且相等， ∴四边形AECD 是平行四边形，又，∴四边形AECD 是矩形，∴． AD AB ⊥CE AB ⊥∴，∴是等边三角形． =AC BC AB =ABC A 取BC 的中点O ，连接AO ，则． AO BC ⊥连接PO ，∵，∴， PB PC =PO BC ⊥∵，平面PAO ，=PO AO O ⋂PO AO ⊂，∴平面PAO ，∵PA 平面PAO ，∴； BC ⊥⊂BC PA ⊥（2）由（1）知，是等边三角形，∴， ABC A CE =∴梯形ABCD 的面积为定值， S =故当平面平面ABCD 时，四棱锥体积最大． PBC ⊥P ABCD -∵，平面平面ABCD ，平面 PO BC ⊥PBC ⋂BC =PO ⊂PBC ∴平面ABCD ，平面ABCD ，∴.PO ⊥,OA OB ⊂,PO OA PO OB ⊥⊥∵OP ，OA ，OB 两两互相垂直，∴以O 为坐标原点，OA ，OB ，OP 分别为x 轴、y 轴和z 轴的正方向，建立如图所示的空间直角坐标系，则． (0,1,0),(0,1,0),A B C P -∴，，=(0,1,PA PB -- =(0,1,CP --设平面PAB 的法向量为，则，取，则． ()111,,n x y z =1111=0==0PA n PB n y ⋅-⋅-⎧⎪⎨⎪⎩ 111x z ==n = 同理设平面PAC 的法向量为，则，取，则． (,,)m x y z ===0=0CP m y PA m ⋅--⋅-⎧⎪⎨⎪⎩ 1x z ===(1,m - 设平面PAB 与平面PAD 的夹角为，则，θ1cos =|cos<,>|=||=||||5m n m n m n ⋅θ即为所求二面角的余弦值.B PAC --20．如图，某海面上有､､三个小岛（面积大小忽略不计），岛在岛的北偏东方向O A B A O 45︒处，岛在岛的正东方向处.B O 20km(1)以为坐标原点，的正东方向为轴正方向，为单位长度，建立平面直角坐标系，写出O O x 1km A 、的坐标，并求、两岛之间的距离；B A B (2)已知在经过、、三个点的圆形区域内有未知暗礁，现有一船在岛的南偏西方向距O A B O 30°O 岛处，正沿着北偏东行驶，若不改变方向，试问该船有没有触礁的危险？ 20km 60︒【答案】(1)，， ()40,40A ()20,0B (2)该船有触礁的危险【分析】（1）结合图像，易得的坐标，再利用两点距离公式即可得解；,A B （2）先由待定系数法求得过、、三点的圆的方程，再求得该船航线所在直线的方程，利用O A B 点线距离公式可知该船航线与圆的位置关系，据此可解.【详解】（1）∵在的东北方向处，在的正东方向处， AO B O 20km ∴，， ()40,40A ()20,0B 由两点间的距离公式得；=（2）设过、、三点的圆的方程为，O A B 220x y Dx Ey F ++++=将､､代入上式得，解得，()0,0O ()40,40A ()20,0B 222=040+40+40+40+=020+20+=0F D E F D F ⎧⎪⎨⎪⎩=20=60=0D E F --⎧⎪⎨⎪⎩所以圆的方程为，即，故圆心为，半径2220600x y x y +--=()()2210301000x y -+-=()10,30r =设船起初所在的位置为点，则，且该船航线所在直线的斜率为C (10,C --， ()tan 6030tan 30︒-︒=︒=由点斜式得该船航线所在直线的方程：，l 200x -=所以圆心到：的距离为l 200x -=d+由于， 2(5700+=+21000700=>+即， 5d =+<所以该船有触礁的危险.21．已知椭圆的右焦点，离心率为，且点在椭圆上．2222:1(0)x y C a b a b +=>>F 1231,2M ⎛⎫ ⎪⎝⎭C (1)求椭圆的标准方程；C (2)过的直线不与轴重合与椭圆相交于、两点，不在直线上且F (x )C A B P AB ，是坐标原点，求面积的最大值．()2OP OA OB λλ=+-O PAB △【答案】(1)22143x y +=(2) 32【分析】（1）依题意得到方程组，解得，，即可求出椭圆方程；2a 2b （2）设直线的方程为，，，，联立直线与椭圆方程，消AB 1x my =+()11,A x y ()22,B x y ()00,P x y 元、列出韦达定理，即可表示出，再表示出点到直线的距离，根据面积公式及基本不等AB P AB 式计算可得.【详解】（1）解：由题意，又，解得，， 221=2914+=1c a a b⎧⎪⎪⎨⎪⎪⎩222c a b =-24a =23b =的方程为；C ∴22143x y +=（2）解：设直线的方程为，，，，AB 1x my =+()11,A x y ()22,B x y ()00,P x y 则，消元整理得， 22=+1+=143x my x y ⎧⎪⎨⎪⎩()2234690m y my ++-=所以，，122634my y m +=-+122934y y m =-+，()2212+13+4m m -由， ()2OP OA OB λλ=+-得，()()()()001212,2,2x y x x y y λλλλ=+-+-()()()()()0121212212122x x x my my my my λλλλλλ∴=+-=++-+=+-+， ()0122yy y λλ=+-到直线的距离P ∴ABh22112(+1)=×23+4PAB m S m ∴A 设，而在时递增，t =13y t t=+1t ≥当，即时，的最大值为．∴=1t 1=0m =PAB S A 3222．如图，已知抛物线的焦点F ，且经过点，．()2:20C y px p =>()()2,0A p m m >5AF =(1)求p 和m 的值；(2)点M ，N 在C 上，且．过点A 作，D 为垂足，证明：存在定点Q ，使得AM AN ⊥AD MN ⊥DQ 为定值．【答案】(1)，； 2p =4m =(2)证明见解析.【分析】（1）由抛物线定义有求，由在抛物线上求m 即可. ||252pAF p =+=p A （2）令，，，联立抛物线得到一元二次方程，应用韦达定理，根据:MN x ky n =+11(,)M x y 22(,)N x y 及向量垂直的坐标表示列方程，求k 、n 数量关系，确定所过定点，再由AM AN ⊥MN B 易知在以为直径的圆上，即可证结论. AD MN ⊥D AB 【详解】（1）由抛物线定义知：，则， ||252pAF p =+=2p =又在抛物线上，则，可得. ()()4,0A m m >244m =⨯4m =（2）设，，由（1）知：，11(,)M x y 22(,)N x y (4,4)A 所以，，又，11(4,4)AM x y =-- 22(4,4)AN x y =--AM AN ⊥所以，121212121212(4)(4)(4)(4)4()4()320x x y y x x x x y y y y --+--=-++-++=令直线，联立，整理得，且，:MN x ky n =+2:4C y x =2440y ky n --=216160k n ∆=+>所以，，则，124y y k +=124y y n =-21212()242x x k y y n k n +=++=+，222121212()x x k y y kn y y n n =+++=综上，， 2216121632(48)(44)0n k n k n k n k ---+=--+-=当时，过定点；84n k =+:(4)8MN x k y =++()8,4B -当时，过定点，即共线，不合题意； 44n k =-:(4)4MN x k y =-+(4,4),,A M N 所以直线过定点，又，故在以为直径的圆上， MN ()8,4B -AD MN ⊥D AB而中点为，即为定值，得证.AB ()6,0Q 2AB DQ ==。
2024-2025学年高二上学期期中模拟考试数学试题02(直线与圆 圆锥曲线)含解析
2024-2025学年高二数学上学期期中模拟卷注意事项：1．答卷前，考生务必将自己的姓名、准考证号等填写在答题卡和试卷指定位置上。
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4．测试范围：人教A版（2019）选择性必修第一册第一章~第三章（空间向量与立体几何+直线与圆+圆锥曲线）。
5．难度系数：0.65。
第一部分（选择题共58分）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．5．如图，在平行六面体ABCD 则AC'的长为（）A．98562+B．【答案】A-'【解析】平行六面体ABCD A故选：A7．已知椭圆的方程为2 9 x+的周长的最小值为（）A．8B 【答案】C则由椭圆的中心对称性可知可知12AF BF 为平行四边形，则可得2ABF △的周长为2AF A ．0B ．【答案】D二、选择题：本题共3小题，每小题6分，共18分．在每小题给出的选项中，有多项符合题目要求．全部选对的得6分，部分选对的得部分分，有选错的得0分．则21242||222y y m HC ++===12||4||22yy p AB HM ++===所以||2sin ||2(HC m HMN HM m ∠==因为20m ≥，所以212(1)m ∈三、填空题：本题共3小题，每小题5分，共15分．则11,22BN BA BD DM =+ 所以1122BN DM BA ⎛⋅=+ ⎝ 1144BA BC BD BC =⋅+⋅-uu r uu u r uu u r uu u r四、解答题：本题共5小题，共77分．解答应写出文字说明、证明过程或演算步骤．15．（13分）已知两直线1:20l x y ++=和2:3210l x y -+=的交点为P ．(1)直线l 过点P 且与直线310x y ++=平行，求直线l 的一般式方程；(2)圆C 过点()1,0且与1l 相切于点P ，求圆C 的一般方程．【解析】（1）直线l 与直线310x y ++=平行，故设直线l 为130x y C ++=，（1分）联立方程组203210x y x y ++=⎧⎨-+=⎩，解得11x y =-⎧⎨=-⎩．（3分）∴直线1:20l x y ++=和2:3210l x y -+=的交点()11P --，．16．（15分）在正四棱柱1111ABCD A B C D -中，124AA AB ==，点E 在线段1CC 上，且14CC CE = ，点F 为BD 中点.(1)求点1D 到直线EF 的距离；(2)求证：1A C ⊥面BDE .【解析】（1）如图,以D 为原点，以,DA DC 正四棱柱111ABCD A B C -()()(10,0,4,0,2,1,1,1,0D E F ∴则点1D 到直线EF 的距离为：17．（15分）18．（17分）如图，在四棱锥P ABCD -中，M 为棱PC 的中点.(1)证明：BM ∥平面PAD ；(2)若5PC =，1AB =，（2）1AB = ，2DC ∴=，又PD 222PC PD DC ∴=+，则PD DC ⊥又平面PDC ⊥平面ABCD ，平面PD ∴⊥平面ABCD ，（7分）19．（17分）416（2）（i ）由题意知直线l 的方程为联立221416x y ⎧-=⎪⎨，化简得(4m 2（ii ）1212232,41m y y y y m -+=-直线AD 的方程为11y y x =+。
2024-2025学年广西高二数学上学期期中调研测试卷附答案解析
2024-2025学年广西高二数学上学期期中调研测试卷一、单选题（本大题共8小题）1．直线40x +=的倾斜角为（）A ．30︒B ．60︒C ．120︒D ．150︒2．椭圆22154x y +=的焦距为（）A ．1B ．2C ．3D ．43．已知12(1,(,2,n x n x ==-分别是平面,αβ的法向量，若αβ⊥，则x =（）A ．1B ．7C ．2-D ．24．已知直线1:20l x my ++=和直线2:(23)20l mx m y ++-=平行，则m 的值为（）A ．3B ．3或1-C ．1-D ．3-5．如图，在四面体ABCD 中，E 为DC 的中点，F 为BE 的中点，设,,AB a AC b AD c === ，则AF =（）A ．111422a b c-++ B ．111244a b c ++C ．111242a b c+- D ．111442a b c++ 6．已知,A B 是抛物线22y x =上的两点，且线段AB 的中点为(1,1)，则直线AB 的方程为（）A ．210x y --=B ．10x y +-=C ．0x y -=D ．210x y -+=7．图1为一种卫星接收天线，其曲面与轴截面的交线为拋物线的一部分，已知该卫星接收天线的口径AB =1MO =，信号处理中心F 位于焦点处，以顶点O 为坐标原点，建立如图2所示的平面直角坐标系xOy ，若P 是该抛物线上一点，点Q 是圆22(3)(2)1x y -+-=上一点，则||||PF PQ +的最小值为（）A ．4B ．3C ．5D ．58．已知双曲线2222:1(0,0)x y C a b a b-=>>的左、右焦点分别12,.F F A 是C 上的一点（在第一象限），直线2AF 与y 轴交于点B ，若11AF BF ⊥，且2232AF F B =，则C 的离心率为（）A ．305B ．32C ．6D ．355二、多选题（本大题共3小题）9．已知圆22:(6)16C x y ++=，设点(,)P x y 为圆上的动点，则下列选项正确的是（）A ．点P 到原点O 的距离的最小值为2B ．过点(3,0)A -的直线与圆C 截得的最短弦长为6C ．yx的最大值为1D ．过点(1,0)B -作圆的切线有2条10．如图，在正方体1111ABCD A B C D -中，点P 在线段1BC 上运动，则下列四个结论正确的有（）A ．1BC 与AC 所成角为60oB ．三棱锥1A D PC -的体积不变C ．//DP 平面11ABD D ．1DP BC ^11．已知12,F F 分别为椭圆22:143x yC +=的左、右焦点，若点12,A A 分别为椭圆C 的左、右顶点，P是椭圆C 上一动点，下列结论中正确的有（）A ．12PF PF ⋅的范围为[2,3]B ．若12F F P 为直角三角形，则12F F P 的面积为3C ．若点(1,1)B，则2PB PF +的最大值为4D ．直线12,PA PA 的斜率之积为34-三、填空题（本大题共3小题）12．若直线220mx y +-=经过两直线53170x y --=和50x y --=的交点，则m =.13．已知直线:0l x y --=，点P 为椭圆22:14y C x +=上的一个动点，则点P 到直线l 的距离的最小值为.14．一动圆与圆221:10240C x y y +++=和222:10240C x y y +--=都外切，则动圆的圆心的轨迹方程为.四、解答题（本大题共5小题）15．已知抛物线2:2(0)C y px p =>的焦点与双曲线222:1(0)4x y E a a -=>的右焦点重合，双曲线E 的渐近线方程为20x =.(1)求抛物线C 的标准方程和双曲线E 的标准方程;(2)若斜率为2且纵截距为1的直线l 与抛物线C 交于M ，N 两点，F 为抛物线C 的焦点，求FMN 的面积.16．为增强市民的节能环保意识，某市面向全市征召义务宣传志愿者.从符合条件的500名志愿者中随机抽取100名志愿者，其年龄频率分布直方图如图所示，其中年龄的分组区间是：第1组[20,25)、第2组[25,30)、第3组[30,35)、第4组[35,40)、第5组[40，45].(1)求图中x 的值并根据频率分布直方图估计这500名志愿者中年龄在[30,35)的人数；(2)估计抽出的100名志愿者年龄的第61百分位数；(3)若在抽出的第1组、第2组和第4组志愿者中，采用按比例分配分层抽样的方法抽取6名志愿者参加中心广场的宣传活动，再从这6名中采用简单随机抽样方法选取2名志愿者担任主要负责人.求抽取的2名志愿者中恰好来自不同一组的概率.17．在ABC V 中，内角A B C ，，的对边分别为a b c ，，，且ABC V 的外接圆半径R 满足sin cos (cos cos )R A A c B b C =+.(1)求角A ；(2)若a =ABC V 面积的最大值.18．如图，在四棱锥P ABCD -中，平面PAD ⊥平面ABCD ，PA PD ⊥，AB AD ⊥，PA PD =，1AB =，2AD =，AC CD ==(1)求证：PD ⊥平面PAB ；(2)求直线PA 与平面PCD 所成角的余弦值；(3)在棱PB 上是否存在点M ，使得//AM 平面PCD ？若存在，求出BMBP的值；若不存在，请说明理由.19．在平面直角坐标系xOy 中，若在曲线1E 的方程0(),F x y =中，以(,)x y λλ（λ为非零的正实数）代替(,)x y 得到曲线2E 的方程(,)0F x y λλ=，则称曲线12E E 、关于原点“伸缩”，变换(,)(,)x y x y λλ→称为“伸缩变换”，λ称为伸缩比.如果曲线221:243E x y +=经“伸缩变换”后得到曲线2E ，射线1(0)2y x x =>与12E E 、分别交于两点A ，B 且||2AB =.(1)求2E 的方程;(2)若M ，N 在2E 上，,,BM BN BD MN D ⊥⊥为垂足，求证：存在定点Q ，使得|DQ |为定值.2024-2025学年广西高二数学上学期期中调研测试卷1．【答案】A【详解】40x +=的斜率为3，故倾斜角为30︒，故选：A 2．【答案】B【详解】由题意得1c ==，则其焦距为2.故选：B.3．【答案】D【详解】由于αβ⊥，所以12n n ⊥ ，故12260n n x x ⋅=+-=，解得2x =，故选：D 4．【答案】A 【详解】由题意可得12232m m m =≠+-，则223m m +=，2230m m --=，即()()310m m -+=，解得3m =或1-，当3m =时，132392=≠-，显然成立，符合题意；当1m =-时，112112-==--，不符合题意.故选：A.5．【答案】B【详解】由F 是BE 的中点，则12BF BE = ，由E 为CD 的中点，则12DE DC = ，在ABD △中，BD AD AB =-，在ACD 中，DC AC AD =- ，()11112222AF AB BF AB BE AB BD DE AB AD AB DC ⎛⎫=+=+=++=+-+ ⎪⎝⎭()111111111224244244AB AD AC AD AB AD AC a b c =++-=++=++.故选：B.6．【答案】C【详解】设1,1,2,2,则2211222,2y x y x ==，故221212121212222y y y x x x x y y y --=-⇒=-+，由于AB 的中点为(1,1)，故122y y +=，因此12121221AB y y k x x y y -===-+,故直线方程为11y x =-+，即0x y -=，经检验，直线0x y -=与抛物线相交，满足条件.故选：C 7．【答案】A【详解】由题意设抛物线的方程为22(0)y px p =>，因为AB =，1MO =，所以点(1,B -在抛物线上，将B 的坐标代入到抛物线的方程中，可得82p =，故4p =，所以抛物线的方程为28y x =，所以抛物线的焦点F 的坐标为(2,0)，准线方程为2x =-，圆22(3)(2)1x y -+-=的圆心位()3,2H ,半径位1R =，可知圆在抛物线内部，如图：如图，过点P 作PP '与准线垂直，P '为垂足，点H 作HN 与准线垂直，N 为垂足，则||||PF PP '=，所以3214PF PQ PP PQ P Q NH R +=+≥≥-='+-='，当且仅当P ，H ，P '三点共线时，所以||||PF PQ +的最小值为4．故选：A8．【答案】D【详解】设1BF m =，如下图所示：由题意可得2BF m =，2122,233AF m AF m a ==+；又22AF F AB B =+，由11AF BF ⊥可得22211AF BF AB +=，即22222233m a m m m ⎛⎫⎛⎫++=+ ⎪ ⎪⎝⎭⎝⎭，解得3m a =；所以2112,4,3AF a AF a BF a ===；因为111190,90AF O BF O BF O F BO ∠+∠=∠+∠=，所以11AF O F BO ∠=∠；即11cos cos AFO F BO ∠=∠，可得222112211212AF F F AF OB AF F F BF +-=，即22216442423a c a a c a+-=⨯⨯，解得c a =故选：D9．【答案】AD【详解】由题意可知：圆22:(6)16C x y ++=的圆心为()6,0-，半径4r =，对于选项A ：点P 到原点O 的距离的最小值为2PO r -=，故A 正确；对于选项B ：因为3CA r =<，可知点(3,0)A -在圆C 内，所以最短弦长为=B 错误；对于选项C ：因为yx表示直线OP 的斜率，当OP 与圆C 相切时，此时OP =，yx取到最大值255r OP ==，故C 错误；对于选项D ：因为5CB r =>，可知点B 在圆C 外，所以过点(1,0)B -作圆的切线有2条，故D 正确；故选：AD.10．【答案】ABC【详解】对于A 选项，连接AC 、11AC 、1A B ，则1111A B A C BC ==，所以，11A BC V 是等边三角形，所以1160A C B ∠=，因为11//AA CC ，11AA CC =，所以，四边形11AAC C 为平形四边形，所以，11//AC AC ，所以，异面直线1BC 与AC 所成的角等于1160A C B ∠=，A 对；对于B 选项，因为11//AB C D ，11AB C D =，所以，四边形11ABC D 为平行四边形，所以，11//BC AD ，因为1BC ⊄平面1ACD ，1AD ⊂平面1ACD ，所以，1//BC 平面1ACD ，因为1P BC ∈，所以，点P 到平面1ACD 的距离等于点B 到平面1ACD 的距离，为定值，又因为1ACD △的面积为定值，故三棱锥1P ACD -的体积为定值，B 对；对于C 选项，由B 选项可知，11//BC AD ，因为1BC ⊄平面11AB D ，1AD ⊂平面11AB D ，所以，1//BC 平面11AB D ，同理可证//BD 平面11AB D ，因为1BC BD B = ，1BC 、BD ⊂平面1BC D ，所以，平面1//BC D 平面11AB D ，因为DP ⊂平面1BC D ，所以，//DP 平面11AB D ，C 对；对于D 选项，若1DP B C ⊥，且四边形11BB C C 为正方形，则11B C BC ⊥，因为1DP BC P = ，DP 、1BC ⊂平面1BC D ，则1B C ⊥平面1BC D ，又因为AB ⊥平面11BB C C ，1B C ⊂平面11BB C C ，则1B C AB ⊥，因为11B C BC ⊥，1AB BC B =I ，AB 、1BC ⊂平面11ABC D ，所以，1B C ⊥平面11ABC D ，又因为过点P 有且只有一个平面与直线1B C 垂直，矛盾，假设不成立，D 错.故选：ABC.11．【答案】ACD【详解】对于A ，()()121,0,1,0F F -，设()00,P x y ，2200143x y +=，则22003(4)4y x =-，故()()2222212000000000311,1,1(4)1244PF PF x y x y y x x x x ⋅=---⋅--=+-=-+-=+ ，由于2004x ≤≤，故[]2120122,34PF PF x ⋅=+∈ ，A 正确，对于B ，当212PF F F ⊥时，此时31,2P ⎛⎫± ⎪⎝⎭，故12F F P 的面积为12113322222P F F y =⨯⨯=，故B 错误，对于C ，由于1(1,0)F -，又(1,1)B ，所以1||BF =所以21114444PB PF PB PF PB PF BF +=+-=+-≤+=+当且仅当1,,P B F 三点共线时，且1F 在,P B 之间时取等号，故C 正确．对于D ，由椭圆22:143x y C +=，得()12(2,0),2,0A A -，设()00,P x y ，则1220002000224PA PA y y y k k x x x =⨯=+--，又2200143x y +=，则22003(4)4y x =-，所以12220022003(4)34444PA PAx y k k x x -===---，故D 正确；故选：ACD12．【答案】10【详解】联立5317050x y x y --=⎧⎨--=⎩，解得14x y =⎧⎨=-⎩，将点()1,4-代入到直线220mx y +-=，得820m --=，故10m =.故答案为：10.13．【答案】【详解】由点P 在椭圆22:14y C x +=上，设(cos ,2sin ),R P θθθ∈，则点P 到直线l的距离d =，其中锐角ϕ由1tan 2ϕ=确定，而1sin()1θϕ--≤≤，则当sin()1θϕ-=-时，min d =所以点P 到直线l的距离的最小值为故答案为：14．【答案】221(0)916y x y -=<【详解】圆221:(5)1C x y ++=的圆心1(0,5)C -，半径11r =，圆222:(5)49C x y +-=的圆心2(0,5)C ，半径27r =，设动圆的圆心(,)P x y ，半径为r ，依题意，1217PC rPC r ⎧=+⎪⎨=+⎪⎩，则2112||||610||PC PC C C -=<=，因此动圆的圆心P 的轨迹是以12,C C 为焦点，实轴长为6的双曲线下支，实半轴长3a =，半焦距5c =，虚半轴长4b ==，方程为221(0)916y x y -=<.故答案为：221(0)916y x y -=<15．【答案】(1)212y x =，22154x y -=；.【详解】（1）双曲线222:1(0)4x y E a a -=>的渐近线方程为20x ay ±=，而双曲线E的渐近线方程为20x =，则a =，双曲线E 的方程为22154x y -=，双曲线E 的右焦点坐标为(3,0)，而抛物线2:2(0)C y px p =>的焦点为(,0)2p，于是32p=，解得6p =，所以抛物线C 的标准方程为212y x =.（2）直线l 的方程为21y x =+，由22112y x y x=+⎧⎨=⎩消去x 得2660y y -+=，2646120∆=-⨯=>，设1122(,),(,)M x y N x y ，则12126,6y y y y +==，12||y y -==令直线l 与x 轴的交点为A ，1(,0)2A -，由（1）知(3,0)F ，所以FMN的面积12117||||2222FMN S AF y y =-=⨯⨯=.16．【答案】(1)0.07x =，175(2)33(3)1115【详解】（1）()0.020.060.040.0151x ++++⨯=，解得0.07x =，5000.075175⨯⨯=，估计这500名志愿者中年龄在[30,35)的人数为175.（2）设第61百分位数为y ，由()0.020.060.0750.750.61++⨯=>，()0.020.0650.40.61+⨯=<，则[)30,35y ∈，可得()()0.020.0650.07300.61y +⨯+⨯-=，解得33y =.（3）第1组、第2组和第4组的人数之比为0.02:0.06:0.041:3:2=，抽取的6人中第1组、第2组和第4组的人数分别为1,3,2，从这6名中抽取的2名志愿者中恰好来自不同一组的概率111111133212222666C C C C C C 11++C C C 15P ==.17．【答案】(1)π3A =；(2)334.【详解】（1）在ABC V 中，由正弦定理2sin sin sin a b c R A B C ===及sin cos (cos cos )R A A c B b C =+，得()()1sin cos cos cos 2cos sin cos sin cos 2a R A A c Bb C R C C B B C ==+=+()2cos sin 2cos sin cos R A B C R A A a A =+==，解得1cos 2A =，又0πA <<，所以π3A =.（2）由（1）知，π3A =，a =由余弦定理得2222232cos a b c bc A b c bc bc ==+-=+-≥，当且仅当b c ==因此1333sin 244ABC S bc A ==≤ ，所以ABC V 面积的最大值为334.18．【答案】(1)证明见解析(2)13(3)存在，且13BM BP =【详解】（1）证明：因为平面PAD ⊥平面ABCD ，且平面PAD ⋂平面ABCD AD =，且AB AD ⊥，AB ⊂平面ABCD ，所以，AB ⊥平面PAD ，因为PD ⊂平面PAD ，所以，AB PD ⊥，因为PD PA ⊥，PA AB A = ，PA 、AB ⊂平面PAB ，所以，PD ⊥平面PAB .（2）解：取AD 中点为O ，连接OC 、OP ，又因为PA PD =，则PO AD ⊥，则1AO PO ==，因为AC CD ==，则OC AD ⊥，则2CO =，在平面ABCD 内，因为OC AD ⊥，AB AD ⊥，则//OC AB ，因为AB ⊥平面PAD ，则OC ⊥平面PAD ，以点O 为坐标原点，OC 、OA 、OP 所在直线分别为x 、y 、z 轴建立如图所示的空间直角坐标系O xyz -，则0,0,1、()1,1,0B 、()0,1,0D -、()0,1,0A 、()2,0,0C ，则()0,1,1PA =- ，()0,1,1DP = ，()2,1,0DC = ，设平面PCD 的法向量为(),,n x y z = ，则020n DP y z n DC x y ⎧⋅=+=⎪⎨⋅=+=⎪⎩ ，取1x =，可得()1,2,2n =- ，设PA 与平面PCD 的夹角为θ，则sin cos ,3n PA n PA n PA θ⋅====⋅ ，则1cos 3θ==，所以，直线PA 与平面PCD 所成角的余弦值为13.（3）解：设()()1,1,1,,BM BP λλλλλ==-=- ，其中01λ≤≤，则()()()1,0,0,,1,,AM AB BM λλλλλλ=+=+-=+- ，因为//AM 平面PCD ，则122130AM n λλλλ⋅=+--=-= ，解得13λ=，因此，在棱PB 上存在点M ，使得//AM 平面PCD ，且13BM BP =.19．【答案】(1)22163x y +=(2)详见解析.【详解】（1）解：设伸缩比为λ，则曲线2E 的方程为2222243x y +=λλ.由221,(0)2243y x x x y ⎧=>⎪⎨⎪+=⎩解得112x y =⎧⎪⎨=⎪⎩，即1(1,)2A ,由22221,(0)2243y x x x y λλ⎧=>⎪⎨⎪+=⎩解得112x y λλ⎧=⎪⎪⎨⎪=⎪⎩，即11(,)2B λλ,因为||AB =224111(1)()225-+-=λλ，解之得12λ=，所以曲线2E 的方程为22163x y +=（2）证明：当直线MN 的斜率存在且不为0时，设直线MN 方程为y kx m =+（k 为斜率），联立方程得22163y kx mx y =+⎧⎪⎨+=⎪⎩，消去y 得，222(12)4260k x kmx m +++-=，直线MN 与椭圆交于两点1122(,),(,)M x y N x y ，所以0∆>，即228(63)0k m -+>，由韦达定理可得，122412km x x k -+=+，21222612m x x k -=+，因为(2,1)B 且BM BN ⊥，所以0BM BN ⋅= ，则1212(2)(2)(1)(1)0x x y y --+--=，即121212122()()50x x x x y y y y -++-++=，其中22121212()y y k x x km x x m =+++，1212()2y y k x x m +=++，所以221212(1)(2)()250k x x km k x x m m ++--++-+=，于是可得22222264(1)()(2)()2501212m km k km k m m k k --++--+-+=++化简整理可得22483210k km m m ++--=，即(231)(21)0k m k m +++-=.所以2310k m ++=或210k m +-=，经检验两式均能使0∆>.当2310k m ++=时，直线MN 方程为3312y kx k =--，则直线BD 方程为1(2)1y x k =--+，设点D 的坐标为(,)x y ，则由33121(2)1y kx ky x k =--⎧⎪⎨=--+⎪⎩消去参数k ，可得22338230x y x y +--+=，即22418()()339x y -+-=，此时存在定点41(,)33Q 使得|DQ |为定值3；当210k m +-=时，直线MN 方程为12y kx k =+-，则直线BD 方程为1(2)1y x k =--+，设点D 的坐标为(,)x y ，则由121(2)1y kx ky x k =+-⎧⎪⎨=--+⎪⎩消去参数k ，可得224250x y x y +--+=，即22(2)(1)0x y -+-=，所以点(2,1)D 与点(2,1)B 重合，不符合题意，故舍去.当0k =时，可由2310k m ++=求得，13m =-，所以1(2,)3D -，可验证点1(2,)3D -在圆22418()()339x y -+-=上，此时存在定点41(,)33Q 使|DQ |为定值当直线MN 的斜率不存在时，不妨设直线MN 方程为x n =，由22260x nx y =⎧⎨+-=⎩可解得点(M n，(,N n ，由0BM BN ⋅= 可得：(2)(2)1)(1)0n n --+=，解之得23n =（2n =舍去），所以点2(,1)3D ，可验证点2(,1)3D 在圆22418()()339x y -+-=上，此时存在定点41(,)33Q 使|DQ |为定值3.综上所述，存在定点41(,)33Q 使|DQ |为定值.。
浙江省宁波市镇海中学2024-2025学年高二上学期期中测试数学试卷
浙江省宁波市镇海中学2024-2025学年高二上学期期中测试数学试卷一、单选题1．在等差数列{}n a 中，已知12a =，315S =，则4a 等于（ ）A ．11B ．13C ．15D ．162．若椭圆2212x y m +=的右焦点与抛物线24y x =的焦点重合，则m 的值为（ ）A ．1B ．3C ．4D ．53．若点P 到直线1x =-和它到点()1,0的距离相等，则点P 的轨迹方程为（ ）A ．2x y=B ．2y x=C ．24x y=D ．24y x=4．任取一个正整数，若是奇数，就将该数乘3再加上1；若是偶数，就将该数除以2.反复进行上述两种运算，经过有限次步骤后，必进入循环圈1421→→→.这就是数学史上著名的“冰雹猜想”（又称“角谷猜想”等）.已知数列{}n a 满足：11a =，1,231,nn n n n a a a a a +⎧⎪=⎨⎪+⎩当为偶数当为奇数，则2024S =（ ）A ．4720B ．4722C ．4723D ．47255．已知函数()f x 是奇函数，函数()g x 是偶函数，且当0x >时，()0f x '>，()0g x '>，则0x <时，以下说法正确的是（ ）A ．()()0f x g x ''+>B ．()()0f xg x ''->C ．()()0f x g x ''>D ．()()0f x g x ''>6．若函数()211kx f x x +=+在[)2,+∞上单调递增，则k 的取值范围为（ ）A ．43k ≥-B ．1k ≤-C ．1k ≤D ．43k ≤-7．已知2023log 2024a =，2024log 2025b =，2025log 2026c =，则（ ）A ．a b c>>B ．a c b>>C ．c b a>>D ．c a b>>8．已知椭圆22:13627x y C +=，左焦点为F ，在椭圆C 上取三个不同点P 、Q 、R ，且2π3PFQ QFR RFP ∠=∠=∠=，则123FP FQ FR ++的最小值为（ ）A．43B．43C．43D．43二、多选题9．下列选项正确的是（ ）A ．1y x=，21y x '=-B ．2x y =，2ln2x y '=C ．ln y x =，1y x'=D ．cos2y x =，sin2y x=-'10．已知抛物线2:4C y x =，F 为其焦点，直线l 与抛物线交C 于()11,M x y ，()22,N x y 两点，则下列说法正确的是（ ）A ．若点A 为抛物线上的一点，点B 坐标为()3,1，则AF AB +的最小值为3B ．若直线l 过焦点F ，则以MN 为直径的圆与1x =-相切C ．若直线l 过焦点F ，当MN OF ⊥时，则5OM ON ⋅=D ．设直线MN 的中点坐标为()()000,0x y y ≠，则该直线的斜率与0x 无关，与0y 有关11．数列{}n a 满足11a =，22a =，21n n n a a a ++>+，则下列结论中一定正确的是（ ）A ．1050a >B ．20500a <C ．10100a <D ．20500a >三、填空题12．已知1n a +=11a =，则100a =.13．已知双曲线22221x y a b -=与直线1y x =-相交于A ，B 两点，其中AB 中点的横坐标为23-，则该双曲线的离心率为 .14．已知函数()()()5e ln 155xf x a x a x =++-+-，若()0f x ≥在()0,∞+上恒成立，则实数a的取值范围为 .四、解答题15．已知函数()e xf x x =.(1)求()f x 的最小值；(2)求()f x 在点()1,e 处的切线方程.16．设等比数列{}n a 的前n 项和为n S ，且11a =-，122n n n S S S ++=+.(1)求数列{}n a 的通项公式.(2)求数列()1nn n a ⎧⎫-⋅⎪⎪⎨⎬⎪⎪⎩⎭的前n 项和n T .17．已知双曲线22:13y C x -=(1)求双曲线C 的渐近线方程；(2)已知点()0,4P 、()2,0Q ，直线PQ 与双曲线C 交于A 、B 两点，1PQ QA λ=，2PQ QB λ=，求12λλ+的值.18．已知函数()()21ln f x mx x m x=+-∈R ，()21e 1x g x x x x =---，其中()f x 在1x =处取得极值(1)求m 的值；(2)求函数()f x 的单调区间；(3)若()()nx g x f x ≤-恒成立，求实数n 的取值范围.19．在必修一中，我们曾经学习过用二分法来求方程的近似解，而牛顿（Issac Newton ，1643-1727）在《流数法》一书中给出了“牛顿切线法”求方程的近似解.具体步骤如下：设r 是函数y =f (x )的一个零点，任意选取0x 作为r 的初始近似值，曲线y =f (x )在点(x 0,f (x 0))处的切线为1l ，设1l 与x 轴交点的横坐标为1x ，并称1x 为r 的1次近似值；曲线y =f (x )在点(x 1,f (x 1))处的切线为2l ，设2l 与x 轴交点的横坐标为2x ，称2x 为r 的2次近似值.一般地，曲线y =f (x )在点()()(),N n n x f x n ∈处的切线为1n l +，记1n l +与x 轴交点的横坐标为1n x +，并称1n x +为r 的1n +次近似值.不断重复以上操作，在一定精确度下，就可取n x 为方程()0f x =的近似解.现在用这种方法求函数()22f x x =-的大于零的零点r 的近似值，取02x =.(1)求1x 和2x ；(2)求n x 和1n x -的关系并证明()*N n ∈；(3)()1*1N i i nx n ∑=<<+∈.。
上学期高二期中考试试卷试题 2
广西四中2021-2021学年度上学期高二语文期中考试试卷时间是：150分钟，卷面总分：150分，交卷时只交答题卡和作文纸，请保管好试卷以便讲评。
温馨提示：亲爱的同学，通过半个学期的学习，今天你展示才能的时机到了，只要你认真审题，放松自己，将答案写在答题纸上，你一定会有出色的表现，相信自己，你一定是最棒的！一、选择题〔每一小题2分，一共18分〕1. 以下加点词中读音有误的一项是哪一项：A. 愆．期(qiān) 遗．施(yí) 机杼．(zhù) 阡陌．(mò)B. 苗裔．(yì) 悯．然(mǐn) 豆蔻．(kòu) 荠．麦(jì)C. 謇謇．(jiǎn) 契．阔(qì) 初霁．(jì) 黍．离(shǔ)D. 纨．素(wán) 吐哺．(bǔ) 浣．女(huàn) 怆．然(chuàng)2. 以下词语中有错别字的一项是哪一项：A. 羽扇纶巾深思慎取披荆斩棘如弃草芥B. 犹劳兴国文过是非金戈铁马惊涛拍岸C. 夙兴夜寐智力孤危晓风残月雕栏玉砌D. 逸豫亡身理固宜然舞榭歌台漂沦憔悴3．以下各句中，加点的成语使用恰当的一句是：创作；朱本晓A．许多农民巧妙地将服装厂剪裁后丢弃的“下脚料〞做成帘子，当做蔬菜大棚的“棉被〞，这真是一念之差．．．．，变废为宝。
B．王大伯非常喜欢小动物,只要见到漂泊的小猫小狗，他都想方法把它们喂饱，有的人对此感到不解，他却乐此不疲．．．．C．文艺演出现场,身着盛装的表演者光着脚,微笑着，一边跳着傣族舞，一边向人们泼水致意，在场群众纷纷拍手称快．．．．。
D．厂长动情地说：“为了改变目前不利的场面，我们采用一种新的对策，希望大家一共同努力，功败垂成．．．．，在此一举。
〞4．以下各句中，没有语病的一句是：A．HY环境的好坏，效劳质量的优劣，政府公务人员素质的上下，都是地区经济安康开展的重要保证。
高二语文上学期期中考试试题及答案
高二上学期期中考试试题说明：本试卷分第I卷（选择题）和第II卷（非选择题）两部分，全卷共8页，满分150分，考试时间150分钟。
第I卷选择题（36分）一、选择题（每小题3分）1、下列加点字的注音全部正确的一项是（）A、氓．（māng）之蚩蚩公姥．（mǔ）周公吐哺．（pǔ）渚．（zhǔ）清沙白B、纤．（qiān）细洞天石扉．（fèi）商贾．（gǔ）迁谪．（zhāi ）C、虾．（há）蟆红绡．（qiào）玉簟．（diàn）祠．（cí）堂D、纶．（guān）巾怆．（chuàng）然雕栏玉砌．（qì）浣．（huàn）女2、下列词语字形书写无误的一项（）A、春秋代叙主薄葳蕤赍钱B、慷慨气概耳濡目染绣腰襦C、苍海惘然寂漠萧萧D、仓皇良晨雀桥隐秘3、下列词语中加点解释正确的一项（）A、匪我愆．（拖延）期女也不爽．（痛快）日月忽．（迅速）其不淹兮B、好自相扶．将（服侍）便言多令．（美好）才何日可掇．（摘取，捨取）C、势拔．五岳（超出）事．权贵（做事）开济．．（开创、匡济）D、繁．（多）霜鬓豆蔻词工．（精巧）随意春芳歇．（歇息）4、下列句子中，加点成语使用正确的一句是( )A、这里有良好的水土条件，又有一个团结向上的领导班子，因此人民的生活安居乐业．．．．。
B、一些不法商贩将早已过期的食品在包装上改头换面．．．．，又拿出来销售，这种不法行为必须严厉打击。
C、近二十年来，我们村子的面貌有了很大变化，但与先进地区相比，就黯然失色．．．．了。
D、李峡被敌人抓住，投入监狱，虽然全身被打得遍体鳞伤．．．．，但仍坚持斗争。
5、下列各句没有语病的一句是( )A、我国发射的第一艘试验飞船在完成了空间飞行试验，于今天3时41分在内蒙古中部地区着陆。
B、战胜了实力强劲的大连实德队，再一次证明鲁能泰山队不可动摇的冠军地位。
C、很多家用电器，国产的并不比洋货差。
天津市部分区2023-2024学年高二上学期期中考试英语试题2
My new coat is old now. It is in too bad of shape to even give away to___28___.I wonder if I should buy a new one soon but I think I will___29___a while. I don’t really need a new one and maybe I can find a few more things to give to the charity___30___.
听下面一段长对话，回答小题。
6 What did the man do in Mr. Turner’s shop last week?
A. He bought a suit.B. He ordered a suit.C. He returned a suit.
长郡中学2023-2024学年高二上学期期中考试物理试题答案 (2)
长郡中学 2023年下学期高二期中考试物理时量: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。
每小题给出的四个选项中，只有一个选项正确)1.下列有关光学现象说法正确的是A.甲中荷叶上的露珠显得特别“明亮”是由于光的全反射B.乙中看到的全息照片利用的是光的全反射原理而制成的C.丙中用涂有增透膜的相机拍照，可以拍摄清楚汽车内部的情景D.丁中肥皂在阳光下呈现彩色条纹是光的衍射现象2.为了交通安全，常在公路上设置如图所示的减速带，减速带使路面稍微拱起以达到使车辆减速的目的。
一排等间距设置的减速带，可有效降低车速。
如果某路面上有一排减速带，每相邻两减速带之间的距离为1.5m，一辆轴距(前后轮转轴中心间的距离)也为 1.5m 、固有频率为 2 Hz的汽车匀速驶过这排减速带，下列说法正确的是A.当汽车以5m/s的速度行驶时，其振动频率为2 HzB. 当汽车以3m/s的速度行驶时最不颠簸C.当汽车以3m/s的速度行驶时颠簸最厉害D.汽车速度越大，颠簸就越厉害3.如图所示，O为弹簧振子的平衡位置，t=0时刻把小球向左拉到某位置静止释放。
以水平向右为正方向，下列描述小球相对O点的位移x、小球的速度 v、小球的加速度a 和小球所受回复力 F变化的图像中，正确的是4.消除噪声污染是当前环境保护的一个重要课题，如图所示的消声器可以用来削弱高速气流产生的噪声。
波长为λ的声波沿水平管道自左向右传播，在声波到达a处时，分成上下两束波，这两束声波在b处相遇时可削弱噪声。
已知上下两束波从a处到b处的路程分别为s₁和s₂，下列说法正确的是A.该消声器是根据波的衍射原理设计的B.该消声器是根据波的多普勒效应原理设计的C. s₁ 和s₂ 关系满足ss1−ss2=nnnn(nn=0,1,2,3))时，消声效果最好D. s₁ 和s₂ 关系满足ss1−ss2=(2nn+1)nn2(nn=0,1,2.3⋯)时，消声效果最好5.本届杭州亚运会跳水项目上，中国跳水队实现“十全十美”，所有项目的金牌全部收入囊中。
2020-2021学年松江二中高二上期中考试试卷
2020-2021学年松江二中高二上期中考试试卷松江二中2020学年第一学期期中考试（高二）英语试题Ⅱ.Grammar and VocabularySection BDirections: After reading the passage below, fill in the blanks to. make the passage coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.Looking for tips on how to be more interesting? With a little effort, your popularity will soar.Don't ramble(漫谈）and stay positive.This simple advice comes from Scott Adams,creator of Dilbert,"Brevity(简洁）will slow the inevitable decline in your popularity(21)_____(cause) by talking. And saying something positive as often as possible will be a mood booster to (22) is in the room with you. If you set a positive tone, it will pass on to others."Act like your heroes. Write down the names of the ten most interesting people you know Then list three speaker Scott Ginsberg on his blog. Then look for patterns and figure out a way(24)_____(include) those characteristics into your own public persona.Nail that "tell me about yourself"question. You know the details well, but distilling(渗透the most interesting parts of your life into a chat can be a challenge. Stanford Graduate School of Business professor Chip Heath suggests (25)________(apply) one of these three plots to you life story: either the challenge plot (you overcame an obstacle to get to where you are); the creativityplot (you decided not to follow a traditional path);(26)_____the connection plot(you did something similar to the person asking).Be engaged with what you do. Though it may not seem like it, being interesting is the same exact thing as eventually become bored with you, and you will become bored with your life. Think of yourself asa sponge, and each and every day, (28)______(absorb) as much as you can of what is happening around you. Ask, and listen. Start with the other person's hobbies, family or upcoming travel plans, and go from there. If you know about the subject, conversation (29)______(flow). If you don't continue asking questions to understand more. Remember to pay special attention(30)___the answers.Section B(15%)Directions: Complete the following passage by using the words in the box. Each word can only be used only once. Note that there is one word more than you need.TRA VELYOUNG,TRA VELFARDear Young Dreamer,You wrote to me about your problem. You dream of travelling, and you__ 31__ how,as a teenager, you can make it happen.I understand that you are not very happy about taking the __32__ that don't interest you. You don't think you will ever need much of what you are studying right now. While some of your lessons may not seem directly __33 __to your needs at the moment, you are learning valuable study skills. You do need these skills to __34__ the world around you, to proces information, and above all, to think for yourself. These exact skills will help youovercomeobstacles that stand between you and your travel dreams.I say dream big but stay practical. However, it's the practical part that most people__35____. Even the most pleasant life has its obstacles, but a setback won't prevent you from reaching your dreams if you stay _36___in the practical, in the action you can take to bring yourself closer to your dreams. Work hard, chart a course ahead and then _37___stick to it. Travel was a goal which I believed would finally _38___ that I had made it through to the other side of my troubled background. I made it there. You can, too.As a teen, you have some limitations when making your travel plans. Your parents have a__39__ in where and how you travel. The amount of money they can spend on your travel is also a factor. Cost will determine your experience-whether you visit Europe, participate in a student exchange programme, or stay and work to help with family finances.Given the limitations that have shaped your life until now, I hope you will maintain your travel dreams. A plan for travel acts like a silent ship running alongside your life as you take yo first solo steps into the world. It is there waiting for you, always inviting you to step on__40____.Sincerely yours,ShannonⅢ. Reading ComprehensionSection ALet's face it: while we go to the office to work, no one is expected to keep their head down the entire time they're there. Yet companies expect 41 out of their employees: that's why they're paying you to be there. No talking or socializing at all isone ____42_, while the other is being able to talk to whomever you want, whenever you want, about whatever you want, even if you never quite manage to make it to your desk until 11:00 a.m. Both examples are ____43____unrealistic. So first we have the entire gray area in between those two, and that's just during office time.The phrase "people-oriented"is generally used to ___44____ something that vaguely represents a company where everyone is pleasant and happy. It's fun to work there, everyone likes his or her job and each employee is ___45______ well and fairly. Know any companies like that?People-oriented is a traditional and unclear phrase that needs ___46____ to make sure your definition is the same as the company's and that you can spot a ___47_____ if there is one.So what do you mean by people-oriented? Do you want a company that promotes from within and doesn't __48____gathering at the water cooler? A company where management makes a practice of being __49___? A place where customers are of great __50_____ both in philosophy and actuality, or one that is involved in its community and requires each __51____to join or participate in a specific event once or twice each year? When you look closer at what this phrase means to you, you'll discover that some aspects are more important to you than others.Knowing what you mean by this phrase gives you the power to discover if the company's___52___ is the same as yours.Why bother leaving a message that may not be ____53____ for days when you can reach someone instantly with a text or instant message? That seems to be increasingly dominant viewpoint , anyway . When texting and instant messaging for ____54____ needs , keep in mind some tips from workplace andcareer experts.Marla Harr, a business etiquette consultant and trainer, says that when relying on texting and messaging for work, you should __55____ that the person you want to send a text message wants to receive and communicate in that way.C.qualificationD.contribution41.A.efficiency B.enthusiasm C. qualification D. contribution42.A.phenomenon B. issue C. extreme D. disadvantage43.A.illegally B.subjectively C.regularly D.equally44.A.exchange B.convey C.provide D.promote45.A.trained B.honored plimented D.treated46.A.defining B.restricting C.analyzing D.summarizing47.A.feature B.difference C.distance D.principle48.A.concern about B.look into C.disapprove of D.charge with49.A.attractive B.accessible" C.aggressive D.accurate50.A.importance B.benefit C.emphasis D.conscience51.A.employer B.employee C.customer D.participant52.A.regulation B.explanation C.definition D.opinionunched B.released C.revised D.checked54.A.business B.privacy C.public D.secret55.A.recall B.ensure rm D.remindSection B. Passages(A)Massive changes in all of the world's deeply cherished sporting habits are underway.Whether it's one of London's parks full of people playing softball, and Russians taking up rugby,or the Superbowl rivaling the British Football Cup Final as a televised spectator event in Britain,the patterns of players and spectators are changingbeyond recognition. We are witnessing a globalization of our sporting culture.That annual bicycle race, the Tour de France, much loved by the French is a good case in point. Just a few years back it was a strictly continental affair with France, Belgium and Holland,Spain and Italy taking part. But in recent years it has been dominated by Colombian mountain climbers, and American and Irish riders. The people who really matter welcome the shift toward globalization. Peugeot, Michelin and Panasonic are multi-national corporations that want worldwide returns for the millions they invest in teams. So it does them literally a world of good to see this unofficial world championship become just that.This is undoubtedly an economy-based revolution we are witnessing here, one made possible by communications technology, but made to happen because of marketing considerations.Sell the game and you can sell Coca Cola or Budweiser as well.a American football has been sold to Europe is a good example ofhow all sports will develop. The aim of course is not really to spread the sport for its own sake, but to increase the number of people interested in the major money-making events. The economics of the Superbowl are already astronomical. With seats at US$125, gate receipts alone were a staggering $10, 000, 000. The most important statistic of the day, however, was the $100, 000, 000in TV advertising fees. Imagine how much that becomes when the eyes of the world are watching.So it came as a terrible shock, but not really as a surprise, to learn that some people are nov suggesting that soccer change from being a game of two 45-minute halves, to one of fou25-minute quarters. The idea isunashamedly to capture more advertising revenue, without giving any thought for the integrity of a sport which relies for its essence on the flowing nature of the action.Moreover, as sports expand into world markets, and as our choice of sports as consumers also grows, so we will demand to see them played at a higher and higher level. In boxing we have already seen numerous, dubious world title categories because people will not pay to see anything less than a "World Title" fight, and this means that the title fights have to be held in differe countries around the world!56.Globalization of sporting culture means that___________A.more people are taking up sportsB traditional sports are getting popularC.many local sports are becoming internationalD. foreigners are more interested in local sports57. Which of the following is NOT related to the massive changes in the world's deeply cherished sporting habits?A.Good economic returns.B.Revival of traditional games.C. Communications technology.D. Marketing strategies.58. The author's attitude towards the suggestion to change soccer into one of four 25-minut quarters is______.A. favorableB. neutralC. reservedD. critical59. This passage mainly discusses_.A.the commercialization of sporting cultureB.the worldwide popularization of sportsC.the speculation in sporting eventsD.the restoration of sportsmanship(B)We asked you to________in 50 words or fewer. Once again, you didn't disappoint us.Here are our favourites, Which one do you like the best? Write and let us know.[1] My uncle has a 1974 Ferrarl Dino 308-'a thing of beauty,a joy forever', a poet once said. It's the perfect balance of engineering and design. The noise it makes as it pulls away, sets my heart racing. A flash of Ferrari red and it's gone.[2] Fading light. The last gasps of the day. On Earth, colours lose their hold as the sky above drains them. Crimson, rose, blood red. On Earth, I watch in amazement. The sun takes its farewell and dips behind the horizon. I leave, wondering what new beauties the day to come will bring[3] He turns, he stops, he turns and runs, he stops and turns againNo two moves are the same.He looks up, he looks down.He looks forward, he looks round.Will he pass? Will he shoot? Will he turn again?[4] As I watch my cat feed her hungry kittens, all eight of them, each fighting for a space, I think about the miracle of life---how well it all works, how well it all fits together. Does anything get perfect?[5] I'm sitting here, head in hand. There is no room left in my head. It's ready to explode.And then the bell sounds, loud and welcome. The sound echoes... echoes through the corridors.Happy children put downtheir pens and wait for the words, simple but beautiful:'Off you go.'[6] I live by the sea. Where else is there to live? The sea fills my heart with joy. A big blu expanse of happiness, The crashing waves, obeying no one, come and then go. I always return to the sea. It knows the answers before I even ask the questions[7]'Hello,I'm home.'It brings a smile to my face when I hear my wife's voice.I look up and she's there. Has she walked out of a painting? She is beauty,' is all I think.60. Which of the following phrases fits best in the blank of the first paragraph?A.define beautyB.write a poemC.recommend a songD.describe creativity61.Which two posts above both talk about nature?A.Post 1 and Post 7.B.Post 2 and Post 6.C.Post 2 and Post 3.D.Post 4 and Post 7.62. In Post 5, the writer is actually talking about_.A.his/her noisy kidsB.the power of musicC.a terrible explosionD.the time when school is over(C)For the most part, it seems, workers in richer countries have little to fear from globalization. And a lot to gain. But is the same thing true for workers in poor countries? The answer is that they are even more likely than their rich-country counterparts tobenefit，because they have less to lose and more to gain Traditional economics takes an optimistic line on integration(整合）and the developing countries. Openness to foreign trade and investment should encourage capital to flow to poor economies. In the developing world, capital is scarce, so the returns on investment there should be higher than in the industrialized countries, where the best opportunities to make money by adding capital to labour have already been used up. If poor countries lower their barriers to trade and investment, as the theory goes, rich foreigners will want to send over some of their capital.If this inflow of resources arrives in the form of loans or portfolio investment(组合投资）,it will top up domestic savings and loosen the financial restriction on additional investment by local companies. If it arrives in the form of new foreign-controlled operations, FDI, so much the better this kind of capital brings technology and skills from abroad packaged along with it, with less financial risk as well. In either case, the addition to investment ought to push incomes up, partly by raising the demand for labour and partly by making labour more productive.This is why workers in FDI-receiving countries should be in an even better position to profit from integration than workers in FDI-sending countries. Also, with or without inflows of foreign capital, the same gains from trade should apply in developing countries as in rich ones. The gains from trade logic often arouse suspicion, because the benefits seem to come from nowhere. Surely one side or the other must lose. Not so. The benefits that a rich country gets through trade do not come at the expense of its poor country trading partners, or vice versa. Recall that according to the theory, trade is a positive sum game. In all thesetrades, both sides-exporters and importers,borrowers and lenders, shareholders and workers can gain.63. Why are workers in poor countries more likely to benefit from the process of globalization?A. They can get more financial aidB . They have nothing to loseC. They have less to lose and more to gain.D. They can get more chances to gain a good job.64. What can be the final result of the inflow of the resource?A. It will top up domestic savings.B. It will loosen the financial restriction.C.It will push people's incomes up.D. It will bring technology and skills from abroad.65. What can we know from the last paragraphA.Poor countries get the most profit during the process of trade.B. Rich countries get profit from trade at poor countries' expense.C.Poor countries get more profit from trade than rich ones.D.All aspects involved in the trade can get benefit.66. Which can be the most appropriate title for this passage?A.Benefited or HurtB.Who Benefits the MostC.Helping the PoorD.The Inflow of ResourcesSection C(8%)Directions: Complete the following passage by using the sentences given below. Each sentence can be used only once. Note that there are two more sentences than you need.A. What we wear can reflect our skin colour.B. Other participants wore their normal clothes.C. The clothes we wear can affect our behavior.D. Some experts believe that the participants with white coats are more likely to be more creative.E. For example, we should choose a look that makes us feel more creative and happy.F. Galinsky and Adams think that the white coats made the participants feel more confident and careful .The science of styleDeciding what to wear in the moming is a challenge for some people. We often worry about what others will think of us because of our clothes. But researchers are beginning to think that our wearing has an equally powerful effect on how we see ourselves.Scientists Adam Galinsky and Hajo Adams report that there is science behind our style. In their research, Galinsky and Adams had some participants wear white lab coats similar to the ones scientists or doctors wear. ____67______The participants took a test that measured theirability to pay attention. The people wearing the white coats performed better than the people in regular clothes.____68____The researchers also believe that other kinds of "symbolic" clothescan influence the behavior of the people wearing them. A police officer's uniform or a judge's robe, for example, increases the wearer's feeling of power or confidence. And in workplaces that have a dress code,"symbolic"clothes may also affect how well employees do their jobs.Fashion blogger Jessica Quirk believes that our clothes greatly affect how we feel about ourselves. On her blog called"What I Wore," Quirk posts a photo of what she's wearing every day.She says that people should pay more attention to what they wear_____69______ One way to do this at work is to "dress up for the exciting job we want," she says, "not the boring job we have."Another way is to wear special clothes that mean something to us. For example, Quirk wears one pair of shoes to all her important meetings. They are her "lucky" shoes.As Quirk points out-and as researchers have discovered-our clothes are "like our second skin."___70____They can also tell the world something about us, without us having to say a word.IV.Summary Writing(10%)Directions: Read the following passage. Summarize in no more than 60 words the main idea of the passage and how it is illustrated. Use your own words as far as possible.It's not piano lessons or dance classes. Nowadays, the biggest extra-curricular activity is going to a tutor. "I spend about 800 Canadian dollars a month on tutors. It's costly," says Pat, a mother in Canada. However, she adds, "after finding out half my daughter's class had tutors, I felt like my child was going to fall behind because everyone else seemed to be ahead."Shelley, a mother of three, also has tutors constantly coming in and out of her home. "When I used to sit down with my children, it was hard to get them focused. I was always yelling. When I got a tutor once a week, they became focused for one entire hour and could get most of their homework done."Tutoring isn't simply a private school phenomenon. Nor is it geared(适应，满足）only toward lower-achieving students. In Canada alone, seven percent of high school students reported using a tutor in 2010. That increased to 15 percent last year.Overall, parents hire tutors because they are worried schoolsare not meeting their expectations, but there is also a cultural shift. A special value is placed on education in Asia, where tutoring is viewed as an extension of the school day. As a large number of Asians emigrated to the West over the recent years, their attitudes towards education have had an impact.Another reason for the growth in business is parental frustration and their packed schedules. "A lot of parents just don't have time to help their children with homework," says Julie Diamond,president of an American tutoring company."Others couldn't help their children after Grade 3."There has been a shift in the attitudes, too. "Children used to get bullied for having a tutor,"Diamond says. "Now it's becoming the norm to have one."Children don't seem to mind that they have a tutor. One parent feels surprised that so many of her child's classmates have tutors. "For the amount we pay in tuition, they should have as much extra help as they need," she says. Still, she's now thinking of getting a tutor. Why? Her daughter has actually asked for one.V.Translation(15%)Directions: Translate the following sentences into English, using the words given in the brackets.72、你越努力，收获就会越大。
江苏南京2024年高二上学期期中调研测试物理试卷+答案
南京市2024-2025学年度第一学期期中学调研测试高二物理一、单项选择题：本题共11小题，每小题4分，共计44分.每小题只有一个....选项符合题意.1.“中国天眼”是世界最大的单口径射电望远镜，能接受来自宇宙深处的电磁波，帮助人类探索宇宙。
下列关于电磁波的说法正确的是（）A.变化的磁场产生电场，变化的电场产生磁场B.不同电磁波在真空中传播速度不同C.电磁波不能在水中传播D.微波炉加热食物是利用了红外线的热效应2.某同学在探究感应电流产生的条件时，进行了如图所示的实验。
下面几种情况中，线圈B中没有感应电流产生的是（）A.开关闭合或开关断开的瞬间B.开关闭合稳定后，快速移动滑动变阻器滑片C.开关闭合稳定后，将线圈A从线圈B中拔出D.开关断开状态下，将线圈A从线圈B中拔出3.下列关于静电现象的说法不正确．．．的是（）A.图甲中，当摇动起电机，烟雾缭绕的塑料瓶顿时变得清澈透明，利用了静电吸附的原理B.图乙中，优质话筒线的外层包裹着金属编织层，利用了静电屏蔽的原理C.图丙中，在加油站给汽车加油前触摸一下手形静电释放器，利用了感应起电的原理D.图丁中，燃气灶电子点火器的放电电极是针尖形，利用了尖端放电的原理4.某同学将两个相同的灵敏电流表G分别改装成一个电流表和一个电压表，如图甲、乙所示，则下列说法正确的是（）A.甲表是电压表，R 减小时量程减小B.甲表是电流表，R 减小时量程增大C.乙表是电流表，R 增大时量程增大D.乙表是电压表，R 增大时量程减小5.如图所示，带正电的金属球C 放在绝缘支架上，置于铁架台旁，把系在绝缘细线上带正电的铝箔小球a 挂在P 点下方，小球a 静止时细线与竖直方向的夹角为θ。
若用一个与球C 完全相同的不带电的金属球D 触碰一下球C ，移走球D 后小球a 再次静止时（ ）A.球C 所带的电荷量变为原来的一半B.细线上的拉力大小变为原来的一半C.细线与竖直方向的夹角θ变为原来的一半D.球C 与小球a 之间的静电力大小变为原来的一半6.如图甲所示为科研人员使用透射电子显微镜观察生物样品的情景，电子显微镜利用电场控制电子运动，镜中电场的电场线分布如图乙所示，虚线为一个电子在此电场中运动的轨迹，若电子从A 运动到B ，则此运动过程中（ ）A. A 点的电场强度比B 点的电场强度大B.电子在B 点的加速度比在A 点的加速度大C.电子的电势能不断增大D.电子受到的电场力的功率越来越小7.如图所示，某实验小组用如图甲所示的装置做“验证动量守恒定律”的实验。
2022-2023学年山东省泰安市高二上学期期中考试数学试题(解析版)
2022-2023学年山东省泰安市高二上学期期中考试数学试题一、单选题1．经过()1A ，()3,1B -两点的直线的倾斜角为（ ）A ．π6B ．π3C ．2π3D ．5π6【答案】D【分析】利用倾斜角与斜率关系即可求解.【详解】因为直线经过()1A ，()3,1B -，则直线斜率为k ==α，则()tan 0,ααπ=∈，此时5π6α=. 故选：D2．若()2,4,1a =-与()2,,1b m =-共线，则m =（ ） A ．－4 B ．－2C ．2D ．4【答案】A【分析】依题意可得b a λ=，即可得到方程组，解得即可. 【详解】解：因为()2,4,1a =-与()2,,1b m =-共线，所以b a λ=，即()()2,,12,4,1m λ-=-，即2241m λλλ-=⎧⎪=⎨⎪=-⎩，解得14m λ=-⎧⎨=-⎩. 故选：A3．已知圆M 的方程为222410x y x y ++-+=，则圆心M 的坐标为（ ） A ．1,2 B ．1,2C ．()2,4-D ．()2,4-【答案】B【分析】先化成标准式，即得圆心坐标.【详解】()()22222410124++-+=∴++-=x y x y x y ， 因此圆心坐标为()1,2-M . 故选：B.4．两条平行直线l ：3460x y -+=与l ：3490x y --=间的距离为（ ）A ．13B ．35C ．3D ．5【答案】C【分析】直接利用两条平行直线间的距离公式求解即可． 【详解】两条平行直线1l ：3460x y -+=与2l ：3490x y --=1535==． 故选：C ．5．已知平面α的一个法向量为()1,2,2n =--，点()0,1,0A 为α内一点，则点1,0,1P 到平面α的距离为（ ） A ．4 B ．3 C ．2 D ．1【答案】 D【分析】利用空间向量的数量积以及点到面的距离向量求法即可求解. 【详解】因为()1,1,1AP =-，()1,2,2n =--， 所以1223AP n ⋅=-++=，143n =++=， 则点P 到平面α的距离1nAP n d ⋅==.故选：D6．已知圆M ：()2224x y -+=内有点()3,1P ，则以点P 为中点的圆M 的弦所在直线方程为（ ） A ．20x y +-= B ．20x y --= C ．40x y +-= D ．20x y -+=【答案】C【分析】由圆M 的标准方程得出圆心和半径，连接PM ，作PM 的垂线，交圆M 于A ，B 两点，以点P 为中点的圆M 的弦即为AB ，求出直线MP 的斜率，利用两直线垂直关系，则可求出直线AB 的斜率，用点斜式方程即可求出直线AB .【详解】由圆M 的标准方程()2224x y -+=，可知圆心()2,0M ，半径2r =，如图，连接MP ，作MP 的垂线，交圆M 于A ，B 两点，以点P 为中点的圆M 的弦即为AB ， 10132MP k -==-，MP AB ⊥ 11ABMPk k ∴=-=-所以直线AB 的方程为：()113y x -=--，整理得40x y +-=， 故选：C.7．已知a ，b 为两条异面直线，在直线a 上取点1A ，E ，在直线b 上取点A ，F ，使1AA a ⊥，且1AA b ⊥（称1AA 为异面直线a ，b 的公垂线）.已知12A E =，3AF =，5EF =，132AA =，则异面直线a ，b 所成的角为（ ）A ．6πB ．3π C ．23π D ．56π 【答案】B【分析】由题可设异面直线a ，b 所成的角为θ，利用向量可得cos θ的值，即求. 【详解】设异面直线a ，b 所成的角为θ，(0,]2πθ∈∵1AA a ⊥，且1AA b ⊥，12A E =，3AF =，5EF =，132AA = ∴11EF EA A A AF =++∴2222111111222EF EA A A AF EA A A A A AF EA AF =+++⋅+⋅+⋅∴1cos 2θ=±，又(0,]2πθ∈∴3πθ=.故选：B.8．若直线0kx y k ++=与曲线212y x x =+-仅有一个公共点，则实数k 的取值范围是（ ） A ．{}11,03⎡⎫--⎪⎢⎣⎭B ．{}11,03⎛⎫--⋃ ⎪⎝⎭C ．141,33⎡⎤⎧⎫--⋃-⎨⎬⎢⎥⎣⎦⎩⎭D ．141,33⎛⎤⎧⎫--⋃-⎨⎬ ⎥⎝⎦⎩⎭【答案】D【分析】首先确定曲线的形状，然后结合直线恒过定点考查临界情况结合图像即可确定实数k 的取值范围．【详解】曲线212y x x =+-即22(1)20(1)x y x y +--=，即22(1)(1)1(1)x y y -+-=，表示(1,1)M 为圆心，1r =为半径的圆的上半部分， 直线0kx y k ++=即(1)y k x =-+恒过定点(1,0)-， 作出直线与半圆的图象，如图，考查临界情况：当直线过点(0,1)时，直线的斜率1k -=，此时直线与半圆有两个交点， 当直线过点(2,1)时，直线的斜率13k -=，此时直线与半圆有1个交点， 当直线与半圆相切时，圆心(1,1)M 到直线0kx y k ++=的距离为1，且0k ->， 211k =+，解得：43k =-，(0k =舍去）． 据此可得，实数k 的取值范围是14(1,]33⎧⎫---⎨⎬⎩⎭．故选：D ．二、多选题9．已知()1,2A ，()3,4B -，()2,0C -，则（ ） A ．直线0x y -=与线段AB 有公共点 B ．直线AB 的倾斜角大于135︒C ．ABC 的边BC 上的高所在直线的方程为470x y -+=D ．ABC 的边BC 上的中垂线所在直线的方程为480x y ++= 【答案】BC【分析】A 选项，画出图像即可看出有无交点；B 选项用先用直线斜率公式求出斜率，再比较倾斜角与135︒的大小；C 选项ABC 的边BC 上的高所在直线过点A ，且斜率和直线BC 的斜率乘积为1-，用点斜式写出边BC 上的高所在直线；D 选项ABC 的边BC 上的中垂线经过BC 的中点，且斜率和直线BC 的斜率乘积为1-，从而利用点斜式写出中垂线所在直线的方程； 【详解】如图所示：所以直线0x y -=与线段AB 无公共点，A 错误；因为421312AB k -==---1>-，所以直线AB 的倾斜角大于135︒，B 正确． 因为4432BC k ==--+，且边BC 上的高所在直线过点A ， 所以ABC 的边BC 上的高所在直线的方程为12(1)4y x -=-，即470x y -+=，C 正确，因为线段BC 的中点为5,22⎛⎫- ⎪⎝⎭，且直线BC 的斜率为40432-=--+， 所以BC 上的中垂线所在直线的方程为15242y x ⎛⎫-=+ ⎪⎝⎭，即28210x y -+=，故D 错误. 故选：BC.10．已知直线l ：1ax by +=，圆C ：221x y +=，点(),M a b ，则（ ） A ．若M 在圆上，直线l 与圆C 相切 B ．若M 在圆内，直线l 与圆C 相离 C ．若M 在圆外，直线l 与圆C 相离 D ．若M 在直线l 上，直线l 与圆C 相切【答案】ABD【分析】根据点与圆的位置关系，得,a b 的关系，即可确定直线l 与圆C 的关系来判断A ，B ，C 选项；根据点与直线的位置关系，得得,a b 的关系，即可确定直线l 与圆C 的关系来判断D 选项. 【详解】解：圆C ：221x y +=，圆心()0,0C ，半径1r =对于A ，若M 在圆上，则221MC a b r =+==，圆心到直线l 的距离为：221111d r a b -====+，则直线l 与圆C 相切，故A 正确；对于B ，若M 在圆内，则221MC a b =+<，圆心到直线l 的距离为：2211d r a b-=>=+，则直线l 与圆C 相离，故B 正确；对于C ，若M 在圆外，则221MC a b =+>，圆心到直线l 的距离为：2211d r a b-=<=+，直线l 与圆C 相交，故C 错误；对于D ，若M 在直线l 上，则221a b +=，圆心到直线l 的距离为：221111d r a b -====+，则直线l与圆C 相切，故D 正确. 故选：ABD.11．如图，四棱柱1111ABCD A B C D -的底面ABCD 是正方形，O 为底面中心，1A O ⊥平面ABCD ，12AB AA ==．以O 为坐标原点，建立如图所示的空间直角坐标系，则（ ）A ．(12,0,2OB =B ．1AC ⊥平面1OBBC ．平面1OBB 的一个法向量为()0,1,1n =-D ．点B 到直线1A C 3【答案】BCD作答.【详解】依题意, ABCD 是正方形, AC BD ⊥,AC 与BD 的交点O 为原点,12AB AA ==,在给定的空间直角坐标系中,)()()(1,,0,,B C A A ,而()112,AB AB ==,则点1B,(12,OB =,故A 错误;()2,0,0OB=,(12,OB =,设平面1OBB 的法向量(),,n x y z =,则12020n OB x n OB x ⎧⋅==⎪⎨⋅==⎪⎩, 令1y =,得()0,1,1n =-,故C 正确;()10,2,22AC n =-=,即1A C ⊥平面1OBB ,故B 正确; (10,AC =,(12,0,A B =,1111A B AC d A C⋅==,B 到1AC 的距离221h A B d =-==故D 正确故选:BCD12．古希腊著名数学家阿波罗尼奥斯（约公元前262-前190）发现：平面内到两个定点A ，B 的距离之比为定值()1λλ≠的点的轨迹是圆．后来，人们将这个圆以他的名字命名，称为阿波罗尼奥斯圆，简称阿氏圆．在平面直角坐标系xOy 中，已知()1,0A -，()2,0B ，动点C 满足12CA CB=，直线l ：10mx y m -++=，则（ ）A ．直线l 过定点()1,1-B ．动点C 的轨迹方程为()2224xy ++= C ．动点C 到直线l 1D ．若直线l 与动点C 的轨迹交于P ，Q 两点，且PQ =，则1m =- 【答案】ABD【分析】设(,)C x y , 由题意求出点C 的轨迹以及轨迹方程, 利用直线与圆的位置关系, 依次判断四个选项即可.【详解】对于A, 直线l ：10mx y m -++=，(1)10m x y +-+=，101101x x y y +==-⎧⎧⇒⎨⎨-+==⎩⎩，直线l 过定点()1,1-，故选项A 正确;对于B,设(,)C x y ,因为动点C 满足||1||2CA CB =,所以 12=, 整理可得2240x y x ++=, 即22(2)4x y ++=,所以动点C 的轨迹是以(2,0)N -为圆心，2r =为半径的圆, 动点C 的轨迹方程为22(2)4x y ++=,故选项B 正确;对于 C, 当直线l 与MN 垂直时, 动点C 到直线l 的距离最大, 且最大值为2故选项C 错误; 对于D, 记圆心N 到直线l 的距离为d ,则d =因为 ()222||4PQ r d =-,则 ()2248r d -=,因为 2r =,所以 d =即=解得 1m =-, 故选项D 正确.故选: ABD.三、填空题13．已知直线1l ：210x y +-=，2l ：210x ay +-=，若1l ∥2l ，则a 的值是________． 【答案】4【分析】由两直线平行可得1221A B A B =，代入相关数据计算即可. 【详解】解：因为1l ∥2l ， 所以224a =⨯=. 故答案为：4.14．写出过()4,0M ，()0,4N 两点，且半径为4的圆的一个标准方程：________． 【答案】2216x y +=（或()()224416x y -+-=）【分析】设所求圆的标准方程为：()()2216x a y b -+-=，代入M ，N 两点的坐标求解即可. 【详解】解：设所求圆的标准方程为：()()2216x a y b -+-=，则有()()2222416416a b a b ⎧-+=⎪⎨+-=⎪⎩， 解得00a b =⎧⎨=⎩或44a b =⎧⎨=⎩，所以所求圆的标准方程为：2216x y +=或()()224416x y -+-=. 故答案为：2216x y +=或()()224416x y -+-=.15．在中国古代数学著作《就长算术》中，鳖臑（biēnào ）是指四个面都是直角三角形的四面体.如图，在直角ABC ∆中，AD 为斜边BC 上的高，3AB =，4AC =，现将ABD ∆沿AD 翻折AB D '∆，使得四面体AB CD '为一个鳖臑，则直线B D '与平面ADC 所成角的余弦值是______.【答案】916【分析】作'B M CD ⊥于交CD 于M ,可证明'B M ⊥平面ACD ,则'B DM ∠即为B D '与平面ADC 的夹角.根据线段关系即可求解.【详解】作'B M CD ⊥于交CD 于M因为,'AD CD AD DD ⊥⊥ 且'CD DD D ⋂= 所以AD ⊥平面'DB C 而AD ⊂平面ACD所以平面ACD ⊥平面'DB C又因为平面ACD 平面'DB C DC =,且'B M CD ⊥所以'B M ⊥平面ACD则'B DM ∠即为B D '与平面ADC 的夹角 因为直角ABC ∆中,3AB =,4AC =所以5BC ===341255AB AC AD BC ⨯⨯===则165DC ===所以169'555DB BC DC =-=-= 在直角三角形'B DC 中,9'95cos 'cos '16165DB B DM B DC DC ∠=∠=== 故答案为:916【点睛】本题考查了空间几何体中直线与平面的夹角求法,直线与平面垂直关系的判定,对空间想象能力和计算能力要求较高,属于中档题.16．已知()111,,a x y z =，()222,,b x y z =，且2a =，3b =，6a b ⋅=-，则111222x y z x y z ++=++________．【答案】23-【分析】由2a =，3b =，6a b ⋅=-，可得向量a 与b 平行，且23=-a b ，从而可得结果. 【详解】∵||2a =，||3b =，6a b ⋅=-，所 以23cos ,6,,[0,π],,πa b a b a b ⨯⨯<>=-<>∈∴<>=. ∴ 向量a 与b 平行，且23=-a b ， 所以1112222(,,)(,,)3x y z x y z =-，所以1223x x =-. ∴1112221223x y x x z z x y +==-+++．故答案为：23-.四、解答题17．已知直线1l ：112y x =-，2l ：y kx b =+，且12l l ⊥． (1)求k 的值； (2)若直线1l 与2l 的交点的直线y x =上，求直线2l 的方程．【答案】(1)2k =-(2)26y x =--【分析】(1)根据两直线垂直的条件即可求解；(2)联立两直线方程求出交点坐标，代入直线2l 的方程即可求解.【详解】（1）直线1l 的斜率为12，直线2l 的斜率为k ．因为12l l ⊥，所以112k ⨯=-， 故2k =-．（2）由题意可知：联立两直线方程可得：112y x y x⎧=-⎪⎨⎪=⎩，解得22x y =-⎧⎨=-⎩． 将点()2,2--代入2l 的方程得()()222b -=-⨯-+，解得6b =-，所以直线2l 的方程为26y x =--．18．已知()1,3,4A ，()1,5,4B -，()1,2,1C -．(1)求,AB BC ；(2)求AC 在BC 上的投影向量．【答案】(1)2π3(2)()0,2,2--【分析】（1）由向量夹角余弦公式,分别计算向量数量积和向量的模,再根据夹角范围,确定夹角的值. （2）根据投影向量定义分别计算两个向量的数量积和模,再求出向量BC 的同方向单位向量,计算即可得到投影向量.【详解】（1）解：因为()2,2,0AB =-，()0,3,3BC =--，所以6AB BC ⋅=-，22AB =，32BC =，所以61cos ,23222AB BC AB BC AB BC ⋅-===-⨯⋅． 因为0,πAB BC ≤≤，所以2π,3AB BC =． （2）因为()2,1,3AC =---，()0,3,3BC =--，所以1227cos ,71432AC BC ==⨯． 因为220,,22BCBC ⎛⎫=-- ⎪ ⎪⎝⎭， 所以AC 在BC 上的投影向量为()2722cos ,140--722=0,2,2BC AC AC BC BC ⎛⎫=⨯⋅ ⎪ ⎪⎝⎭⋅--，，．19．如图，在平行六面体1111ABCD A B C D -中，4AB AD ==，15AA =，1160DAB BAA DAA ∠=∠=∠=︒，M ，N 分别为11D C ，11C B 中点．(1)求1AC 的长；(2)证明：1MN AC ⊥．【答案】(1)1113AC(2)证明见解析.【分析】（1）设AB a =，AD b =，1AA c =，将1AC 用,,a b c 表示出来，根据向量的模长公式即可得到结果.（2）将1,MN AC ，分别用,,a b c 表示出来，根据10MN AC ⋅=，即可证明1MN AC ⊥.【详解】（1）设AB a =，AD b =，1AA c =，则4a b ==，5c =，8a b ⋅=，10a c b c ⋅=⋅=，111122MN MC C N a b =+=- 11AC AB BC CC a b c =++=++．因为()22AC a b c =++()2222a b c a b b c c a =+++⋅+⋅+⋅ ()222445281010=+++++113=，所以1AC （2）证明：因为()11122MN AC a b a b c ⎛⎫⋅=-⋅++ ⎪⎝⎭ 22211112222a c ab bc =+⋅--⋅ 2211114104102222=⨯+⨯-⨯-⨯ 0=，所以1MN AC ⊥．20．已知圆M ：()()222125x y -+-=，圆N ：2214520x y x my +--+=，过圆M 的圆心M 作圆N 的切线，切线长为5．(1)求m 的值，并判断圆M 与圆N 的位置关系；(2)过圆N 的圆心N 作圆M 的切线l ，求l 的方程．【答案】(1)4m =，圆M 与圆N 相交(2)7x =或125940x y +-=，【分析】（1）先用配方法确定圆N 的圆心和半径，然后根据切线长公式计算出m 的值，再根据圆心距和半径之间的大小关系判断位置关系；（2）过圆外一点可作圆的两条切线，在我们求解的过程中需要对直线的斜率是否存在进行讨论.【详解】（1）由题意知，()2,1M ，7,2m N ⎛⎫ ⎪⎝⎭，圆N 的半径N r ==， 由勾股定理得2225N MN r =+，即()2222212721522m m ⎛⎫-⎛⎫-+-=+ ⎪ ⎪ ⎪⎝⎭⎝⎭， 解得4m =．所以()()22722126MN =-+-=，1N r =，6M N r r +=，4M N r r -=．因为M N M N r r MN r r -<<+，所以圆M 与圆N 相交；（2）当l 的斜率不存在时，l 的方程为7x =．检验知满足相切.当l 的斜率存在时，设l 的方程为()27y k x -=-，即720kx y k --+=，因为l 与圆M 相切，所以2217251k k k --+=+，解得125k =-， 所以l 的方程为()12275y x -=--，即125940x y +-=． 综上所述，l 的方程为7x =或125940x y +-=，21．如图，圆柱上，下底面圆的圆心分别为O ，1O ，该圆柱的轴截面为正方形，三棱柱111ABC A B C 的三条侧棱均为圆柱的母线，且1306AB AC OO ==，点P 在轴1OO 上运动．(1)证明：不论P 在何处，总有1BC PA ⊥；(2)当P 为1OO 的中点时，求平面1A PB 与平面1B PB 夹角的余弦值．【答案】(1)证明见解析11【分析】（1）证明线面垂直，进而证明线线垂直；（2）利用空间向量的坐标运算方法求面面角的余弦值.【详解】（1）证明：连接AO 并延长，交BC 于M ，交圆柱侧面于N ．因为AB AC =，OB OC =，所以AOB AOC △≌△，所以BAM CAM ∠=∠，所以ABM ACM ≌，所以M 为BC 中点，所以OA BC ⊥．又在圆柱1OO 中，1AA ⊥平面ABC ，BC ⊂平面ABC ，1AA BC ⊥，1AO AA A =，1,AO AA ⊂平面11AOO A ,所以BC ⊥平面11AOO A ．因为不论P 在何处，总有1PA ⊂平面11AOO A ，所以1BC PA ⊥．（2）设11(0)OO AA AN a a ===>，则AB AC ==． 在ABC 中，5cos 6AC AM AC CAM AC a AN =∠=⨯=， 则13OM a =．所以CM BM =． 如图，建立空间直角坐标系1O xyz -，其中11//B C x 轴，y 轴是11B C 的垂直平分线， 则110,,02A a ⎛⎫- ⎪⎝⎭，11,,0)3B a，1,,3B a a ⎫⎪⎪⎝⎭，10,0,2P a ⎛⎫ ⎪⎝⎭， 所以155(,,)66A B a a =，111(0,,)22A P a a =，1(0,0,)B B a =，111(,,)632B P a a =--． 设平面1APB 的一个法向量为(),,m x y z =，则50611022ay az ay az ++=⎨⎪+=⎪⎩，取1x =，得(1,5,m =． 设平面1BPB 的一个法向量为(),,n b c d =，则011032ad ac ad =⎧⎪⎨-+=⎪⎩，取2b =，得()2,5,0n =-． 设平面1A PB 与平面1B PB 的夹角为θ，则 11cos |cos ,|||||11m nm n m n θ⋅=<>==，所以平面1A PB 与面1B PB 夹角（锐角）的余弦值为1111．22．已知线段AB 的端点B 的坐标是()6,4，端点A 的运动轨迹是曲线C ，线段AB 的中点M 的轨迹方程是()()22421x y -+-=．(1)求曲线C 的方程；(2)已知斜率为k 的直线l 与曲线C 相交于异于原点O 的两点,,E F 直线,OE OF 的斜率分别为1k ，2k ，且122k k =．若BD EF ⊥，D 为垂足，证明：存在定点Q ，使得DQ 为定值．【答案】(1)()2224x y -+=(2)证明见解析【分析】（1）利用中点坐标公式以及求轨迹方程的方法求解；（2）利用韦达定理结合题意求解.【详解】（1）设(),A x y ，00(,)M x y ，由中点坐标公式得006,242x x y y +⎧=⎪⎪⎨+⎪=⎪⎩． 因为点M 的轨迹方程是()()22421x y -+-=，所以2264(4)(2)122x y ++-+-=， 整理得曲线C 的方程为()2224x y -+=．（2）设直线l 的方程为y kx m =+，()11,E x y ，()22,F x y ，120x x ≠，由22(2)4y kx m x y =+⎧⎨-+=⎩，得222(1)2(2)0k x km x m ++-+=， 所以1222(2)1km x x k -+=-+，21221m x x k=+，所以()()()221212121212121212kx m kx m k x x km x x m y y k k x x x x x x +++++=== 222222(2)41121km km m k k k m mk -++=+=+=+， 所以4m k =，且0∆>即2224(2)4(1)0km k m --+>，即2440m km +-<，所以直线l 的方程为()4y k x =+，即直线l 过定点()4,0P -． 因为BP 为定值，且BDP △为直角三角形，BP 为斜边， 所以当点Q 是BP 的中点时，1||2QD BP =为定值． 因为()6,4B ，()4,0P -，所以由中点坐标公式得()1,2Q ． 所以存在定点()1,2Q 使得DQ 为定值．。
宁波镇海中学2024-2025学年高二上学期期中测试数学试卷
镇海中学2024学年第一学期期中考试高二数学试题卷一、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，只有一个选项是正确的.1.在等差数列{}n a 中，已知12a =，315S =，则4a 等于（ ） A.11 B.13 C.15 D.162.若椭圆2212x y m +=的右焦点与抛物线24y x =的焦点重合，则m 的值为（ ） A.1 B.3 C.4 D.53.若点P 到直线1x =−和它到点(1,0)的距离相等，则点P 的轨迹方程为（ ）A.2x y =B.2y x =C.24x y =D.24y x =4.任取一个正整数，若是奇数，就将该数乘3再加上1；若是偶数，就将该数除以2.反复进行上述两种运算，经过有限次步骤后，必进入循环圈1421→→→.这就是数学史上著名的“冰雹猜想”（又称“角谷猜想”等）.已知数列{}n a 满足：11a =，1,231,n n n n n a a a a a + = + 当为偶数当为奇数，则2024S =（ ） A.4720B.4722C.4723D.4725 5.已知函数()f x 是奇函数，函数)g x 是偶函数，且当0x >时，()0f x ′>，()0g x ′>，则0x <时，以下说法正确的是（ ） A.()()0f x g x ′+>′ B.()()0f x g x ′−>′C.()()0f x g x ′′>D.()()0f x g x ′′> 6.若函数()211kx f x x +=+在[)2,+∞上单调递增，则k 的取值范围为（ ） A.43k ≥− B.1k ≤− C.1k ≤ D.43k ≤− 7.已知2023log 2024a =，2024log 2025b =，2025log 2026c =，则（ ）A.a b c >>B.a c b >>C.c b a >>D.c a b >>8.已知椭圆22:13627x y C +=，左焦点为F ，在椭圆C 上取三个不同点P ，Q ，R ，且23FFQ QFR RFP π∠∠∠===，则123FP FQ FR ++的最小值为（ ）A.43B.43C.43D.43 二、选择题：本题共3小题，每小题6分，共18分.在每小题给出的选项中，有多项符合题目要求，全部选对得6分，部分选对的得部分分，有选错的得0分.9.下列选项正确的是（ ） A.1y x=，21y x ′=− B.2x y =，2ln2x y ′= C.ln y x =，1y x ′= D.cos2y x =，sin2y x =−′10.已知抛物线2:4C y x =，F 为其焦点，直线l 与抛物线交C 于()11,M x y ，()22,N x y 两点，则下列说法正确的是（ ） A.若点A 为抛物线上的一点，点B 坐标为(3,1)，则AF AB+的最小值为3 B.若直线l 过焦点F ，则以MN 为直径的圆与1x =−相切C.若直线l 过焦点F ，当MN OF ⊥时，则5OM ON ⋅= D.设直线MN 的中点坐标为()()000,0x y y ≠，则该直线的斜率与0x 无关，与0y 有关 11.数列{}n a 满足11a =，22a =，21n n n a a a ++>+，则下列结论中一定正确的是（ ） A.1050a >B.20500a <C.10100a <D.20500a > 三、填空题：本题共3小题，每小题5分，共15分. 12.已知1n a +=11a =，则100a =__________.13.已知双曲线22221x y a b −=与直线1y x =−相交于A ，B 两点，其中AB 中点的横坐标为23−，则该双曲线的离心率为_____.14.已知函数()()()5ln 155x f x e a x a x =++−+−，若()0f x ≥在()0,+∞上恒成立，则实数a 的取值范围为__________.四、解答题：本题共5小题，共77分.解答应写出文字说明、证明过程或演算步骤.15.已知函数()x f x xe =. （1）求()f x 的最小值；（2）求()f x 在点()1,e 处的切线方程.16.设等比数列{}n a 的前n 项和为n S ，且11a =−，122n n n S S S ++=+. （1）求数列{}n a 的通项公式.（2）求数列()1n n n a −⋅的前n 项和n T . 17.已知双曲线22:13y C x −= （1）求双曲线C 的渐近线方程；（2）已知点()0,4P ，()2,0Q ，直线PQ 与双曲线C 交于A ，B 两点，1PQ QA λ= ，2PQ QB λ=，求12λλ+的值.18.已知函数()()21ln f x mx x m R x =+−∈，()211x g x xe x x =−−−，其中()f x 在1x = （1）求m 的值；（2）求函数()f x 的单调区间； （3）若()()nx g x f x ≤−恒成立，求实数n 的取值范围.19.在必修一中，我们曾经学习过用二分法来求方程的近似解，而牛顿（Issac Newton ，1643-1727）在《流数法》一书中给出了“牛顿切线法”求方程的近似解.具体步骤如下：设r 是函数()yf x =的一个零点，任意选取0x 作为r 的初始近似值，曲线()y f x =在点()()00,x f x 处的切线为1l ，设1l 与x 轴交点的横坐标为1x ，并称1x 为r 的1次近似值；曲线()y f x =在点()()11,x f x 处的切线为2l ，设2l 与x 轴交点的横坐标为2x ，称2x 为r 的2次近似值.一般地，曲线()yf x =在点()()(),n n x f x n ∈N 处的切线为1n l +，记1n l +与x 轴交点的横坐标为1n x +，并称1n x +为r 的1n +次近似值.不断重复以上操作，在一定精确度下，就可取n x 为方程()0f x =的近似解.现在用这种方法求函数()22f x x =−的大于零的零点r 的近似值，取02x =.（1）求1x 和2x ；（2）求n x 和1n x −的关系并证明()*n N ∈；（3()*11n i i x n N =<<+∈∑.。
江苏省扬州市江都区2024-2025学年高二上学期11月期中测试数学试题
江苏省扬州市江都区2024-2025学年高二上学期11月期中测试数学试题一、单选题1．经过两点()()2,,,4A m B m -的直线l 的倾斜角为135 ，则m 的值为（）A ．-2B ．1C ．3D ．42．对于任意的实数m ，直线()130m x y m +++=恒过定点（）A ．()3,3B ．()3,3-C ．()3,3--D ．()3,3-3．双曲线22:1y C x m-=的焦点坐标为(0,，则m =（）A ．2B ．4CD ．14．已知圆1C ：224x y +=和圆2C ：224470x y x y +--+=，则两圆的位置关系为（）A ．外离B ．外切C ．相交D ．内切5．点()2,3P 关于直线20x y ++=的对称点的坐标为（）A ．()3,2--B ．()2,3--C ．()5,4--D ．()4,5--6．若双曲线经过点()，且它的两条渐近线方程是3y x =±，则双曲线的方程是（）．A ．2219y x -=B ．2219x y -=C ．221273y x -=D ．221273x y -=7．直线10x y -+=分别与x 轴，y 轴交于,A B 两点，点P 在圆()2222x y -+=上，则ABP 面积的取值范围是（）A ．[]1,3B ．[]2,6C ．15,22⎡⎤⎢⎥⎣⎦D ．17,22⎡⎤⎢⎥⎣⎦8．设椭圆2222:1x y C a b+=（0a b >>）的左焦点为F ，O 为坐标原点，过点F 且斜率为3的直线与C 的一个交点为Q （点Q 在x 轴上方），且OF OQ =，则C 的离心率为（）A ．2B ．13C ．4D ．4二、多选题9．已知直线1l ：340ax y ++=，2l ：()2250x a y a +-+-=，则下列结论正确的是（）A ．1l 在y 轴上的截距为43-B ．若12//l l ，则1a =-或3a =C ．若12l l ⊥，则32a =-D ．若1l 不经过第二象限，则0a ≤10．已知圆C ：()()22111x y -+-=，点()2,3P ，则下列结论正确的是（）A ．点P 在圆C 外B ．圆C 上动点Q到点P 2C ．过点P 作圆C 的切线，则切线方程为2x =或3460x y -+=D ．过点P 作圆C 的切线，切点为A ，B ，则直线AB 的方程为240x y --=11．如图，P 是椭圆1C ：22194x y +=与双曲线2C ：22221x ym n-=（0m >，0n >）在第一象限的交点，且1C ，2C 共焦点1F ，2F ，12F PF θ∠=，2C 的离心率为e ，则下列结论正确的是（）A ．13PF m =+，23PF m =-B ．若双曲线2C 的方程是22114x y-=，则90θ=︒C ．若60θ=︒，则e =D ．12PF F 的面积为2n三、填空题12．若方程222410x y mx y m +++++=表示圆，则实数m 的取值范围为.13．已知直线1:3470l x y -+=与直线()2:6110l x m y m -++-=平行，则1l 与2l 之间的距离为.14．已知椭圆22:143x y C +=的左、右焦点分别为1F ，2F ，M 为C 上任意一点，N 为圆()()22:961E x y -+-=上任意一点，则1MN MF -的最小值为.四、解答题15．在ABC V 中，()5,2A -，()7,4B ，()5,4C --.(1)求ABC V 中，BC 边上的中线所在直线的方程；(2)求ABC V 中，BC 边上的高所在直线的方程.16．已知圆C 的圆心在直线10x y --=上，且过()2,2A -，()3,3B 两点.(1)求圆C 的标准方程；(2)若过点()3,4Q 的直线l 被圆C 截得的弦长为6，求直线l 的方程.17．已知椭圆C ：22221x y a b+=（0a b >>）经过点()3,1P ，焦距为()2,4Q 且斜率为1的直线l 与椭圆C 相交于M ，N 两点.(1)求椭圆C 的方程；(2)求PMN 的面积.18．在平面直角坐标系中，已知()13,0F -，()23,0F ，动点P 满足12PF PF -=P 的轨迹为曲线C .(1)求C 的方程；(2)过点1F 的直线l （斜率存在且不为0）与曲线C 相交于M ，N 两点.①若MN 的中点为Q ，设直线MN 和OQ 的斜率分别为1k ，2k ，求12k k ⋅的值；②满足112MF F N =，求直线l 方程.19．如图，已知椭圆C ：22221x y a b +=（0a b >>）的上顶点为0,3点A 作圆M ：()2223x y r ++=（03r <<）的两条切线分别与椭圆C 相交于点B ，D （不同于点A ）.(1)求椭圆C的方程；k k 为定值；(2)设直线AB和AD的斜率分别为1k，2k，求证:12(3)求证：直线BD过定点.。
吉林市普通高中2024-2025学年高二上学期期中调研测试英语试题
吉林市普通高中2024-2025学年高二上学期期中调研测试英语试题一、听力选择题1．What nationality is Sasha?A．Russian.B．American.C．Chinese.2．Why is the ceremony put off?A．The host got stuck in traffic.B．The important guest hasn’t arrived.C．The band needs time to set up their instruments.3．What are the speakers doing?A．Cooking a meal at the man’s house.B．Working in a restaurant’s kitchen.C．Ordering food at a café.4．What does the woman ask Tony to do?A．Say sorry to his brother.B．Make up with his sister.C．Control his feelings.5．What are the speakers talking about?A．A science TV series.B．A fact about the girl’s friend.C．Information from a school lesson.听下面一段较长对话，回答以下小题。
6．What does the man plan to leave at home?A．His camera.B．His phone.C．His computer.7．How many sweaters is the man going to pack?A．Four.B．Three.C．Two.听下面一段较长对话，回答以下小题。
8．Why is the woman so excited?A．It’s her first time to Hawaii.B．She is enjoying a wonderful view.C．The man has been driving far to meet her.9．When does the conversation probably take place?A．In the early morning.B．In the late afternoon.C．At midnight.听下面一段较长对话，回答以下小题。
高二上学期期中考试数学试卷含答案(共5套)
高二上学期期中考试数学试题本卷分Ⅰ（选择题）、Ⅱ卷（非选择题）两部分，其中Ⅰ卷1至2页，第二卷2至4页，共150分，考试时间120分钟。
第Ⅰ卷（选择题，共60分）一、单选题：本题共12个小题，每小题5分1．“”是“”的（）A．充分不必要条件B．必要不充分条件C．充分必要条件D．既不充分也不必要条件2．有下列四个命题：（1）“若，则，互为倒数”的逆命题；（2）“面积相等的三角形全等”的否命题；（3）“若，则有实数解”的逆否命题；（4）“若，则”的逆否命题.其中真命题为（）A．（1）（2）B．（2）（3）C．（4）D．（1）（2）（3）3．若则为（）A．等边三角形 B．等腰直角三角形C．有一个内角为30°的直角三角形 D．有一个内角为30°的等腰三角形4．已知.若“”是真命题，则实数a的取值范围是A．(1，+∞)B．(－∞，3)C．(1，3)D．5．的内角，，的对边分别为，，，若，，，则的面积为A．B．C．D．6．已知中，，则等于（）A．B．或C．D．或7．等差数列的前项和为，若，则等于（）A．58B．54C．56D．528．已知等比数列中，，，则（）A．2B．C．D．49．已知，则z=22x+y的最小值是A．1 B．16 C．8 D．410．若关于的不等式的解集为，则的取值范围是（）A．B．C．D．11．当ａ＞０，关于代数式，下列说法正确的是（）A．有最小值无最大值B．有最大值无最小值C．有最小值也有最大值D．无最小值也无最大值12．在△ABC中，AB＝2，C＝，则AC＋BC的最大值为A．B．3C．4D．2第Ⅱ卷（非选择题，共90分）二、填空题：共4个小题，每小题5分，共20分13．命题的否定是______________．14．已知的三边长构成公差为2的等差数列，且最大角的正弦值为，则这个三角形的周长为________.15．已知数列{a n}的前n项和为S n，a1＝1，当n≥2时，a n＋2S n－1＝n，则S2 017的值____ ___ 16．已知变量满足约束条件若目标函数的最小值为2，则的最小值为__________．三、解答题：共6题，共70分，解答应写出必要的文字说明、证明过程或演算步骤。
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高二上学期期中测试题第一卷（选择题共85分）第一部分:英语知识运用(共两节，满分45分)第一节:单项填空(共15小题；每小题1分，满分15分)1. It is difficult for us to learn a lesson in life ____we’ve actually had that lesson.A. untilB. afterC. sinceD. when2. Peter received a letter just now ________ his grandma would come to see him soon. A．said B．says C．saying D．to say3. ---Could you turn the TV down a little bit?---______. Is it disturbing you?A. Take it easy.B. I’m sorry.C. Not a bitD. It depends4. The students in the class were________four groups before they started the game.A divided intoB separated intoC divided fromD separated into5. Most of the people __________ to the party were famous scientists.A. invitedB. to invitedC. being invitedD. inviting6. As a result of the serious flood, two-thirds of the buildings in the area _____.A. need repairingB. needs to repairC.needs repairingD.need to repair7. ---―Did you have any difficulty in today’s homework?‖---―No, in fact I found __________.‖A. it very easy to doB. it very easy doneC. very easy for doingD. very easy to do it8. There was a traffic jam. ___________，he was late.A As a resultB As a result ofC As the resultD As result9. The _____of the clock _________me that I was late.A view;remindedB sight; recalledC sight; remindedD view; recalled10 I will __________ Jane today, as she can’t come to the meeting herself.A take placeB take place ofC take the placeD take the place of11. ----They haven’t finish ed all the work up to now.----They ____________.A shouldB ought toC are supposed toD are supposed to have12．Come and see me whenever ___________.A you are convenientB you will be convenientC it is convenient to youD it will be convenient to you13. Tom is good at his work, but he seems to _____ confidence.A lackB lack ofC lacking inD short of14. The class __________ 80 students is hard to manage.A consisting ofB consists ofC making up ofD made up15. Only then _________the importance of my health.A did I realizeB had I realizedC I realizedD I had realized第二节：完形填空（共20小题；每小题1.5分，满分30分）阅读下面短文，掌握其大意，然后从36-55各题所给的四个选项（A、B、C和D）中，选出最佳选项，并在答题卡上将该项涂黑。
―Where would we be right now if I did not have the restaurant? How else would I be able to provide for our family? Who give me, a man without 16，a job?‖ My father 17 say these things on our car rides to the stores that supply our 18.He would tell me stories of long, hard hours spent 19 in the hot kitchen and helping all the 20. Now I 21 late on weekends to welcome him when he comes home，22 that he will tell me something 23 .He always knows how to inspire me. I could never understand 24 he grew to be so wise without even having the opportunity to complete high school.―Make sure you get a job 25 you have to wear a tie!‖ he’d always say with a smile. They were like a paintbrush(画笔) that began 26 this beautiful picture in my head that I have entitled(有资格的)my 27 in life.My father has worked day in and day out，with few days off, for 20 years, never 28 .He would explain how this is 29 he had to do to make my 30 better. My father worked so I 31 go to the best schools. He worked to supply me with 32 that would provide a job where I had to wear a tie.My father is a 33 in which I can only hope to catch a reflection(反省)of myself. He is the one I strive to model myself after. 34 , when I make a great name for myself as a doctor,someone will ask, ―Where did you get such a great work ethic (道德) ?‖―My father!‖ I’ ll say, adjusting(调整) my 35 and I’ ll smile.16. A work B education C money D an education17. A used to B would C always D might18. A family B kitchen C restaurant D hotel19. A to work B work C works D working20. A friends B cooks C customers D waiters21. A hold up B keep up C stand up D stay up22. A wished B wishing C hoped D hoping23. A wonderful B interesting C big D wise24. A why B how C where D when25. A when B where C which D that26. A drawing B describing C imagining D painting27. A job B success C goal D future28. A complained B complains C complaining D complaint29. A what B that C which D why30. A job B life C days D future31. A would B could C should D should32. A opportunities B hopes C occasions D possibilities33. A model B mirror C hero D example34. A The other day B Another day C Some days D One day35. A tie B suit C dress D shirt第三部分：阅读理解（共20小题；每小题2分，满分40分）阅读下列短文，从每题所给的四个选项（A、B、C和D）中，选出最佳选项，并在答题卡上将该项涂黑。
AI first went to Harrow in the summer term. The school had the biggest swimming pool I had ever seen. It was a good joke to come up behind a naked boy, and push him into the pool.I made quite a habit of this with boys of my own size or less.One day I saw a boy wrapped in a towel on the side of the pool. He was no bigger than Iwas, so I thought him a fair game. Coming secretly behind, I pushed him in, holding on to his towel so that it would not get wet, I was surprised to see an angry face come out from the water, and a being of great strength masking its way by face strokes (猛力地划)to the shore. I fled, but in vain. He overtook me, seized me violently, and threw me into the deepest part of the pool. I soon climbed out on the other side, and found myself surrounded by a crowd of younger boys.‖ Do you know what you have done?‖ the y said, ―It’s Amery; he is in Grade Six. He is champion at gym; he has got his football honor.‖I was frightened and felt ashamed. How could I tell his position when be was wrapped in a bath towel and so small.‖ He didn’t seem pleased at all, so I added in a most brilliant word,‖My father, who is a great man, is also small.‖ At this be laughed, and after some general words about my rude behavior and how I had better be careful in the future, signified the incident was closed.36. The writer thought Amery‖ a fair game‖ because the boy .A. looked like an animalB. was fond of gamesC.was of similar sizeD.was good at sports37. The writer felt‖ ashamed‖ because.A. he was laughed at by other boysB.Amery turned out to be in the same gradeC.he pushed Amery hard and hurt himD.he played a joke on an outstanding athlete38. By saying ―My father, who is a great man, is also small‖, the write r .A. tried to please AmeryB.challenged AmeryC.threatened AmeryD.admired his father39. Which of the following is TRUE?A. The writer could run faster than Amery.B.The writer liked playing on boys of all sizes.C.Amery was a student in Grade Four.D.Amery forgave the writer for his rude behavior.BHuntingThe days of the hunter are almost over in India. This is partly because there is practically nothing left to kill, and partly because some steps have been taken, mainly by banningtiger-shooting, to protect those animals which still survive.Some people say that Man is naturally a hunter. I disagree with this view. Surely out earliest forefathers, who at first possessed no weapons, spent their time digging for roots, and were no doubt themselves often hunted by meat-eating animals.I believe the main reason why the modern hunter kills is that he thinks people will admire his courage in overpowering dangerous animals. Of course, there are some who truly believe that the killing is not really the important thing, and that the chief pleasure lies in the joy of the hunt and the beauties of the wild countryside. There are also those for whom hunting in fact offers a chance to prove themselves and risk death by design; these men go out after dangerous animals like tigers, even if they say they only do it to rid the countryside of a threat.I can respect reasons like these, but they are clearly different from the need to strengthen your high opinion of yourself.The greatest big-game hunters expressed in their writings something of these finer motives(动机).One of them wrote.―You must properly respect what you are after and shoot it cleanly and on the animal’s own territory(领地)。