湖南省长郡中学2013-2014学年高二下学期学业水平摸底(一)英语试题(扫描版)
湖南省长郡中学2023-2024学年高二上学期期中考试英语试题
湖南省长郡中学2023-2024学年高二上学期期中考试英语试题学校:___________姓名：___________班级：___________考号：___________一、短对话1．What is the weather like probably?A．Sunny.B．Cloudy.C．Rainy.2．How will the speakers go to the cinema?A．By taxi.B．By bus.C．By underground. 3．How does the woman feel about the musical?A．Great.B．So-so.C．Terrible.4．What makes the man feel good about the new job?A．The salary.B．The traveling chances.C．The working time. 5．What do we know about the man?A．He won’t go to the party.B．He will attend the party alone.C．He will take friends to the party.二、长对话听下面一段较长对话，回答以下小题。
6．How soon will the woman go back home?A．In about one week.B．In about half a month C．In about a month. 7．What’s the probable relationship between the speakers?A．Teacher and student B．Former colleagues.C．Primary classmates.听下面一段较长对话，回答以下小题。
8．How might the relationship between the woman and her parents be?A．Good.B．Bad.C．Distant.9．How old might the woman be?A．16.B．15.C．18.10．Which of the following is right?A．The woman has no pocket money.B．The woman has much freedom.C．The woman wants to go on holidays with her parents.听下面一段较长对话，回答以下小题。
2024届湖南省长沙市长郡中学英语高三第一学期期末学业水平测试试题含解析
2024届湖南省长沙市长郡中学英语高三第一学期期末学业水平测试试题注意事项：1．答题前，考生先将自己的姓名、准考证号填写清楚，将条形码准确粘贴在考生信息条形码粘贴区。
2．选择题必须使用2B铅笔填涂；非选择题必须使用0．5毫米黑色字迹的签字笔书写，字体工整、笔迹清楚。
3．请按照题号顺序在各题目的答题区域内作答，超出答题区域书写的答案无效；在草稿纸、试题卷上答题无效。
4．保持卡面清洁，不要折叠，不要弄破、弄皱，不准使用涂改液、修正带、刮纸刀。
第一部分（共20小题，每小题1.5分，满分30分）1．To work from home, which one could hardly imagine, has been made with the development of computer technology.A．possible B．it possible C．possibly D．to be possible2．Sorry I’m so late, but you cannot imagine ________ great trouble I took to find your house.A．which B．howC．what D．that3．Being happy is a skill that can be learned, and one way to ________ ourselves to be happy is to write down the little things that cheer us up each day.A．convey B．appeal C．train D．attempt4．We need a spiritual faith, or a philosophy, it should include this truth: if you choose to find the positive in every situation, you will be blessed, and if you choose to find the awful, you will be cursed. As with happiness itself, this is ________your decision to make.A．absolutely B．totallyC．exactly D．largely5．I can ________ Diana’s thoughts from the changes in her facial expressions.A．read B．notice C．count D．watch6．That student admitted in the math exam, that he would never do that again in the future.A．to have cheated, promising B．cheating, promisedC．to cheating, promised D．having cheated, promising7．With a travelling speed of up to 350 kilometres per hour, the railway to be built between Beijing and Shanghai _______ the journey time from 12 hours to 5 hours．A．cuts B．will cut C．is cutting D．has cut8．People believe that the China Dream is not very difficult ________ so long as the whole nation works hard for it.A．realizing B．to be realizedC．realized D．to realize9．People who drink and drive are ________ danger both to themselves and to others. They are in ________ danger of losing their lives.A．the; the B．a; a C．a; / D．/; /10．China’s Chang’e 4 robotic probe entered lunar orbit on Wednesday, ________ a major step in its mission to make a soft la nding on the moon’s far side.A．marking B．to markC．having marked D．marked11．_____ the lawyers, volunteers from the Libyan Red Cross Society also joined the efforts in helping the Chinese go home safely.A．Except for B．In spite of C．Apart from D．Instead of12．It was John who broke the window. Why are you talking to me as if I it? A．had done B．have done C．did D．am doing13．The witness an important detail when describing the accident.A．brought out B．kept offC．left out D．ran into14．Our class held a fierce ________ as to whether to reduce the amount of homework or not.A．bargain B．competition C．debate D．campaign15．The coat I bought yesterday is not expensive at all. As a matter of fact, I would gladly have paid ______ for it.A．as much twice B．much as twiceC．as twice much D．twice as much16．Wild swans’ ________ in the area is a good indication of a better environment. A．exhibition B．escapeC．absence D．appearance17．Our country has launched a campaign to ban smoking in public places, which with some heavy smokers.A．concerns B．was concernedC．concerned D．is concerned18．Oh！I can feel something _____ up my leg！It must be an insect．A．to climb B．climbingC．climb D．climbed19．James, I don’t mind lending you the money ____ you pay it back within a month. A．although B．now thatC．unless D．as long as20．Jess was sad and her friend helped her ___ the first awful weeks after her husband Bill died.A．break through B．break downC．get through D．get rid of第二部分阅读理解（满分40分）阅读下列短文，从每题所给的A、B、C、D四个选项中，选出最佳选项。
2024-2025学年湖南省长郡中学高三上学期一调英语试题及答案
长郡中学 2025 届高三第一次调研考试英语本试题卷共10页。
时量 120分钟，满分150分。
注意事项：1. 答卷前，考生务必将自己的姓名、准考证号填写在答题卡上。
2. 回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。
如需改动，用橡皮擦干净后，再选涂其他答案标号。
回答非选择题时，将答案写在答题卡上。
写在本试卷上无效。
3. 考试结束后，将本试卷和答题卡一并交回。
第一部分听力 (共两节，满分30分)做题时，先将答案标在试卷上。
录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。
第一节 (共5小题; 每小题1.5分, 满分7.5分)听下面5段对话。
每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项。
听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. What kind of sport does David like best?A. Football.B. Tennis.C. Basketball.2. Where are the speakers?A. In a hotel.B. In a shop.C. In a restaurant.3. When will Amy and Andrew meet?A. At 6:40.B. At 7:20.C. At 8:00.4. What is the probable relationship between the speakers?A. Mother and son.B. Doctor and patient.C. Teacher and student.5. What are the speakers mainly talking about?A. Class arrangements.B. Grading policies.C. Exam results.第二节 (共15小题; 每小题1.5分, 满分22.5分)听下面5段对话或独白。
湖南省长沙市长郡中学2024届高三模拟考试(一)英语试题
长郡中学2024届高三模拟考试(一)英语第一部分听力(共两节，满分30分)做题时，先将答案标在试卷上。
录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。
第一节(共5小题；每小题1.5分，满分7.5分)听下面5段对话。
每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项。
听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1.Who is the woman probably talking to?A.Her co-worker.B.Her brother.C.Her husband.2.When did the man see the film?A.On TuesdayB.On Thursday.C.On Saturday3.Which part of the movie disappoints the woman?A.The movie's plotB.The actors'clothes.C.The acting skills4.Why does the woman talk to John?A.To ask for advice.B.To buy a car from him.C.To borrow money5.What is the probable relationship between the speakers?A.Driver and passengerB.Guide and tourist.C.Shopkeeper and salesgirl.第二节(共15小题；每小题1.5分，满分22.5分)听下面5段对话或独白。
每段对话或独白后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项。
听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟，听完后，各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
湖南省长沙市长郡中学2024-2025学年高二上学期10月月考数学试题(含答案)
2024—2025第一次阶段性检测数学时量：120分钟 满分：150分得分______一、选择题：本大题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的.1.已知复数，则（ ）C.3D.52.无论为何值，直线过定点（ ）A. B. C. D.3.在平行四边形中，，，，则点的坐标为（ ）A. B. C. D.4.已知，则（ ）A. B.C. D.5.直线关于对称的直线方程为（）A. B. C. D.6.已知椭圆：，则（ ）A. B.C.8或2D.87.已知实数满足，则的范围是（ ）A. B. C. D.8.已知平面上一点，若直线上存在点使，则称该直线为点的“相关直线”，下列直线中不是点的“相关直线”的是（ ）A. B. C. D.3i1iz +=+z =λ()()()234210x y λλλ++++-=()2,2-()2,2--()1,1--()1,1-ABCD ()1,2,3A -()4,5,6B -()0,1,2C D ()5,6,1--()5,8,5-()5,6,1-()5,8,5--π1sin 33α⎛⎫+= ⎪⎝⎭πcos 23α⎛⎫- ⎪⎝⎭79-7929-292410x y --=0x y +=4210x y ++=4210x y +-=4210x y --=4210x y -+=C ()22104x y m m +=>m =,x y ()22203y x x x =-+ (4)1y x ++[]2,6(][),26,-∞+∞ 92,4⎡⎤⎢⎥⎣⎦(]9,2,4⎡⎫-∞+∞⎪⎢⎣⎭()5,0M l P 4PM =()5,0M ()5,0M 3y x =-2y =430x y -=210x y -+=二、选择题：本大题共3小题，每小题6分，共18分.在每小题给出的四个选项中，有多项符合题目要求，全部选对的得6分，部分选对的得部分分，有选错的得0分.9.已知直线：，圆：，为坐标原点，下列说法正确的是（ ）A.若圆关于直线对称，则B.点到直线C.存在两个不同的实数，使得直线与圆相切D.存在两个不同的实数，使得圆上恰有三个点到直线的距离为10.已知圆：与圆：的一个交点为，动点的轨迹是曲线，则下列说法正确的是（ ）A.曲线的方程为B.曲线的方程为C.过点且垂直于轴的直线与曲线相交所得弦长为D.曲线上的点到直线11.在边长为2的正方体中，为边的中点，下列结论正确的有（ ）A.与B.过，，三点的正方体的截面面积为3C.当在线段上运动时，的最小值为3D.若为正方体表面上的一个动点，，分别为的三等分点，则的最小值为三、填空题：本题共3小题，每小题5分，共15分.12.通过科学研究发现：地震释放的能量E （单位：焦耳）与地震里氏震级M 之间的关系为.已知2011年甲地发生里氏9级地震，2019年乙地发生里氏7级地震，若甲、乙两地地震释放的能量分别为，，则______.13.直线的倾斜角的取值范围是______l 20x y λλ+--=C 221x y +=O C l 2λ=-O l λl C λC l 121F ()()222328x y m m ++=……2F ()()222310x y m -+=-M M C C 22110064x y +=C 2212516x y +=1F x C 325C 4510x ++=ABCD A B C D '-'''M BC AM D B ''A M D 'ABCD A B C D '-'''P A C 'PB PM '+Q B C C B ''EF A C 'QE QF +lg 4.8 1.5E M =+1E 2E 12E E =()243410ax ay +-+=14.如图，设，分别是椭圆的左、右焦点，点P 是以为直径的圆与椭圆在第一象限内的一个交点，延长与椭圆交于点，若，则直线的斜率为______.四、解答题：本题共5小题，共77分.解答应写出文字说明、证明过程或演算步骤.15.（13分）已知两圆和.求：（1）m 取何值时两圆外切？（2）当时，两圆的公共弦所在直线的方程和公共弦长.16.（15分）在中，内角A ，B ，C 的对边分别为a ，b ，c ，已知.（1）求的值；（2）若，，求的面积.17.（15分）如图，在四棱锥中，平面，，四边形满足，，，点为的中点，点为棱上的动点.（1）求证：平面；（2）是否存在点，使得平面与平面所成角的余弦值为？若存在，求出线段的长度；若不存在，说明理由.18.（17分）某校高一年级设有羽毛球训练课，期末对学生进行羽毛球五项指标（正手发高远球、定点高远球、吊球、杀球以及半场计时往返跑）考核，满分100分.参加考核的学生有40人，考核得分的频率分布直方图如图所示.1F 2F ()222210x y a b a b+=>>12F F 2PF Q 222PF F Q =1PF 222610x y x y +---=2210120x y x y m +--+=45m =A B C △()()cos 2cos 2cos A C b c a B -=-sin sin CA1cos 4B =2b =A BC △P ABCD -PA ⊥ABCD 2PA AB AD ===ABCDAB AD ⊥B C A D ∥4BC =M PC E BC DM ∥PAB E PDE ADE 23BE（1）由频率分布直方图，求出图中t 的值，并估计考核得分的第60百分位数；（2）为了提升同学们的羽毛球技能，校方准备招聘高水平的教练.现采用分层抽样的方法（样本量按比例分配），从得分在内的学生中抽取5人，再从中挑出两人进行试课，求两人得分分别来自和的概率；（3）若一个总体划分为两层，通过按样本量比例分配分层随机抽样，各层抽取的样本量、样本平均数和样本方差分别为：，，；，，.记总的样本平均数为，样本方差为，证明：19.（17分）已知动直线与椭圆：交于，两点，且的面积为坐标原点.（1）证明：和均为定值；（2）设线段的中点为，求的最大值；（3）椭圆上是否存在三点D ，E ，G，，使得？若存在，判断的形状；若不存在，请说明理由.[)70,90[)70,80[)80,90m x 21s n y 22s w 2s ()(){}22222121s m s x w n s y w m n ⎡⎤⎡⎤=+-++-⎢⎥⎢⎥⎣⎦⎣⎦+l C 22132x y +=()11,P x y ()22,Q x y OPQ △OPQ S △O 2212x x +2212y y +P Q M OM PQ ⋅C ODE ODG OEG S S S ===△△△D E G △长沙市第一中学2024—2025学年度高二第一学期第一次阶段性检测数学参考答案一、二、选择题题号1234567891011答案BAAACCADABDBCDAC1.B 【解析】∵，∴. .故选B.2.A 【解析】由得：，由得∴直线恒过定点.故选A.3.A【解析】设，则，，得.故选A.4.A 【解析】，又，所以.故选A.5.C 【解析】取直线关于对称的直线上任意一点，易知点关于直线对称的点的坐标为，由点在直线上可知，即.故选C.6.C 【解析】椭圆：的离心率为，，解得或.故选C.7.A 【解析】表示函数图象上的点与的连线的斜率，结合图象可知，斜率分别在与（相切时）处取最大值和最小值，()()()()23i 1i 3i 33i i i 2i 1i 1i 1i 2z +-+-+-====-++-z ==()()()234210x y λλλ++++-=()()223420x y x y λ++++-=220,3420x y x y ++=⎧⎨+-=⎩2,2,x y =-⎧⎨=⎩()()()234210x y λλλ++++-=()2,2-(),,D x y z ()5,7,3AB =- (),1,2DC x y z =---()5,6,1D --22πππ17cos 2cos 212sin 1233399ααα⎡⎤⎛⎫⎛⎫⎛⎫+=+=-+=-⨯= ⎪ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎝⎭⎣⎦π2π22π33αα⎛⎫-=+- ⎪⎝⎭π2π2π7cos 2cos 2πcos 23339ααα⎡⎤⎛⎫⎛⎫⎛⎫-=+-=-+=- ⎪ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎝⎭⎣⎦2410x y --=0x y +=()00,P x y P 0x y +=()00,Q y x --Q 2410x y --=002410y x -+-=004210x y --=C ()22104x y m m +=>==8m =2m =41y x ++()22203y x x x =-+……()1,4--()0,2()2,2所以的范围是.故选A.8.D 【解析】根据题意，当点到直线的距离时，该直线上存在点使得，此时直线为点的“相关直线”，对于A ，，即，点到直线的距离，该直线是点的“相关直线”；对于B ，，点到直线的距离，该直线是点的“相关直线”；对于C ，，点到直线的距离，该直线是点的“相关直线”；对于D，，点到直线的距离，该直线不是点的“相关直线”.故选D.9.ABD 【解析】直线：过定点，圆：，圆心，半径，对选项A ：直线过圆心，则，解得，故选项A 正确；对选项B ：点O 到直线l的距离的最大值为B 正确；对选项C ：直线与圆相切，则圆心到直线的距离，解得，故选项C 错误；对选项D ：当圆上恰有三个点到直线的距离为时，圆心到直线的距离，解得，故选项D 正确.故选ABD.10.BCD 【解析】对A 选项与B 选项，由题意知圆与圆交于点，则，，所以，所以点的轨迹是焦点在轴上的椭圆，且，，即，，所以，所以曲线的方程为，故A 选项错误，B 选项正确；41y x ++[]2,6M l 4d …P 4PM =l()5,0M 30y x =-=30x y --=M l 4d <()5,0M 2y =M l 0224d =-=<()5,0M 430x y -=M l 4d ==()5,0M 210x y -+=M l 4d ()5,0M l 20x y λλ+--=()2,1P C 221x y +=()0,0C 1r =20λ--=2λ=-PC =l C 1d 34λ=-C l 12C l 12d λ=1F 2F M 1MF m =210MF m =-1212106MF MF F F +=>=M x 210a =26c =5a =3c =4b =C 2212516x y +=对C 选项，通径的长度为，故C 选项正确；对D 选项，设与直线平行的直线为，，将与联立得，令，解得，此时直线与椭圆相切，当时，切点到直线的距离最大，直线的方程为，故曲线上的点到直线D 选项正确.故选BCD.11.AC 【解析】以为坐标原点，，，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，则，，，，，∴，，∴，∴与A 正确；取的中点，连接，，，则，故梯形为过点，，的该正方体的截面，∵，，∴梯形，1632255⨯=4510x ++=l 40x t ++=51t ≠40x t ++=2212516x y +=221004000y t ++-=()22Δ3004004000tt =--=40t =±l 40t =-4510x ++=l 4400x +-=C 4510x ++=A 'A D ''A B ''A A '()0,0,2A ()1,2,2M ()2,0,0D '()0,2,0B '()2,2,0C '()1,2,0AM = ()2,2,0DB''=-cos ,AM D B AM D B AM D B '⋅'''''⋅==AM D B ''C C 'N M N D N 'AD 'M N BC AD ''∥∥M N D A 'A M D 'MN AD '=AM D N ='=M N D A '=∴梯形的面积为，故B 错误；由对称性可知，，故，又由于，，，四点共面，故，当为与的交点时等号成立，故C 正确，设点关于平面的对称点为，连接，当与平面的交点为时，最小，过点作的平行线，过点作的平行线，两者交于点，此时，D 错误.故选AC.三、填空题12.1000 【解析】由题知，.13. 【解析】设直线的倾斜角为，当时，直线为，；当时，，当且仅当时取等号， ∴；当时，，当且仅当时取等号， ∴，综上可得.14.【解析】连接，，由点在以为直径的圆上，故.M N D A '1922⨯+=PB PD '='PB PM PD PM '++'=A 'B C D '3PB PM PD PM D M +=+'''=…P A C 'D M 'F B C C B ''F 'EF 'EF 'B C C B ''Q QE QF QE QF +=+'E AD 'F AB G 13EG AD =='2G F '=EF =='11112222lg 4.8 1.59,lg lg 3lg 31000lg 4.8 1.57E E EE E E E E =+⨯⎧⇒-=⇒=⇒=⎨=+⨯⎩π2π,33⎡⎤⎢⎥⎣⎦()243410a x ay +-+=α0α=310x +=π2α=0α>2433tan 44a k a a a α+===+= (3)4a a =ππ,32α⎡⎫∈⎪⎢⎣⎭0α<24333tan 444a k a a a a a α+⎛⎫===+=--+-= ⎪-⎝⎭ (3)4a a -=-π2π,23α⎛⎤∈ ⎥⎝⎦π2π,33α⎡⎤∈⎢⎥⎣⎦121PF 1QF P 12F F 12PF PF ⊥又，在椭圆上，故有，.设，则，，，.在中，由勾股定理得，解得，于是，，故.四、解答题15.【解析】（1）由已知化简两圆的方程为标准方程分别为：，，则圆心分别为，，，解得.（2）当，则，所以两圆相交，则两圆的公共弦所在直线的方程为：，即，圆心到直线的距离，所以公共弦长.16.【解析】（1）由正弦定理得，所以，所以，化简得，又，所以，因此.（2）由，得，由余弦定理及，又，得，解得，从而.又因为，且，所以.P Q 122PF PF a +=122QF QF a +=2QF m =22PF m =122PF a m =-12QF a m =-3PQ m =1Rt PQF △()()()2223222m a m a m +-=-3a m =223a PF =143a PF =1121tan 2PF k PF F ∠==()()221311x y -+-=()()()22566161x y m m -+-=-<()1,3M ()5,6N =+25m =+45m =4=44<<+()22222611012450x y x y x y x y +----+--+=43230x y +-=()1,3M 43230x y +-=2d l ==()()cos 2cos sin 2sin sin cos A C B C A B -=-cos sin 2cos sin 2sin cos sin cos A B C B C B A B -=-cos sin sin cos 2cos sin 2sin cos A B A B C B C B +=+()()sin 2sin A B B C +=+πA B C ++=sin 2sin C A =sin 2sin CA=sin 2sin C A =2c a =2222cos b a c ac B =+-1cos 4B =2b =22214444a a a =+-⨯1a =2c =1cos 4B =0πB <<sin B =因此.17.【解析】（1）因为平面，，平面，所以，，又，所以，，两两垂直.以为坐标原点，所在直线为轴，所在直线为轴，所在直线为轴，建立空间直角坐标系，如下图所示，则，，，，因为点为中点，所以，，又，，所以，所以，，为共面向量，则在平面内存在直线与平面外的直线平行，所以平面.（2）设，，，，依题意可知，平面的法向量为，设平面的法向量为，则令，则.因为平面与平面所成角的余弦值为，所以，解得或，所以存在点使得平面与平面所成角的余弦值为，或.18.【解析】（1）由题意得：，解得，11sin 1222ABC S ac B ==⨯⨯=△PA ⊥ABCD A D AB ⊂ABCD PA AD ⊥PA AB ⊥AB AD ⊥PA AB A D A AB x A D y AP z ()0,0,2P ()2,0,0B ()0,2,0D ()2,4,0C M PC ()1,2,1M ()1,0,1DM =()0,0,2AP = ()2,0,0AB =1122DM AP AB =+ DM ,AP A BPAB l PAB DM DM ∥PAB ()2,,0E a 04a ……()0,2,2DP =- ()2,2,0DE a =-ADE ()0,0,2AP =PDE (),,n x y z =()220,220,DP n y z DE n x a y ⎧⋅=-+=⎪⎨⋅=+-=⎪⎩1z =2,1,12a n -⎛⎫= ⎪⎝⎭ PDE ADE 232cos ,3AP n AP n AP n ⋅==⋅23=1a =3a =E PDE ADE 231BE =3BE =()100.010.0150.020.0251t ⨯++++=0.03t =设第60百分位数为，则，解得，即第60百分位数为85.（2）由题意知，抽出的5位同学中，得分在的有人，设为，，在的有人，设为a ，b ，c .则样本空间为，.设事件“两人分别来自和”，则，，因此，所以两人得分分别来自和的概率为. （3）由题得：①；②略19.【解析】（1）（ⅰ）当直线的斜率不存在时，，两点关于轴对称，所以，，因为在椭圆上，所以，①又因为，所以由①②得，，此时，.（ⅱ）当直线的斜率存在时，设直线的方程为，由题意知，将其代入得，其中，即，（*）又，，所以，x ()0.01100.015100.02100.03800.6x ⨯+⨯+⨯+⨯-=85x =[)70,8085220⨯=A B [)80,90125320⨯=()()()()()()()()()(){}Ω,,,,,,,,,,,,,,,,,,,A B A a A b A c B a B b B c a b a c b c =()Ω10n =M =[)70,80[)80,90()()()()()(){},,,,,,,,,,,M A a A b A c B a B b B c =()6n M=()()()63Ω105n M P M n ===[)70,80[)80,9035mx ny m n w x y m n m n m n+==++++l P Q x 21x x =21y y =-()11,P x y 2211132x y +=OPQ S =△11x y ⋅=1x =11y =22123x x +=22122y y +=l l y kx m =+0m ≠22132x y +=()()222236320k x kmx m +++-=()()2222Δ36122320k m k m =-+->2232k m +>122623km x x k +=-+()21223223m x x k -=+PQ ==因为点到直线的距离为，所以又，整理得，且符合（*）式，此时，，综上所述，，，结论成立。
13级英语试题
2014-2015下学期高二期末考试英语试题(13级)I. 选择题（本大题共30小题，每小题2分，共60分）从每小题所给的四个选项中，选出一个可以填入题中空白处的最佳选项。
未选、错选或多选均不得分。
1．I want to know ______ is wrong with you.A. thatB. whatC. whichD. when2. -- Tell us __________ you broke your leg.-- On the playground.A. whereB. whenC. howD. why3. This box is really pretty. But there is no name on it. I’ve forgotten ________ gave this gift to me.A. thatB. whomC. whoD. whose4.Yesterday afternoon Sam didn’t feel well. He said ______ he had a stomachache.A. whatB. whyC. howD. that5. He asked ________ for the pills in the clinic..A. did I pay how muchB. I paid how muchC. how much did I payD. how much I paid6. He asked, “How are you getting along?”→He asked _______.A. how am I getting alongB. how are you getting alongC. how I was getting alongD. how was I getting along7. She asked me if I knew ______.A. whose pen is itB. whose pen it wasC. whose pen it isD. whose pen was it8. -- I’d like to make an appointment ______ the doctor.-- The doctor is free _________ 10:00 this morning.A. for; inB. to; onC. with; atD. and; to9. -- __________ have you felt this way?-- For about one day.A. How longB. How soonC. How oftenD. How much10. Sophie looked pale and kept __________ all the time.A. coughB. to coughC. coughedD. coughing11. Excuse me, but can you ____________ my bag?A. keep eye onB. keep an eye onC. keep an eye inD. keep an eye12. Paul has changed his job three times______ he graduated from the university.A. beforeB. whenC. sinceD. after13. Turn off the light and lock the door________ you leave the room.A. whenB. while D. until D. after14. ______ the teacher shouted “go”, all the students started to run.A. WhileB. OnceC. IfD. Before15. Hello, Myra, we’ll be on holiday next week. What about_________ together?A. to travelB. travelC. travelledD. travelling 16. You can join our computer club. We can introduce more useful websites ____ you.A. onB. forC. toD. with17. ________ technology develops, our future will be more convenient.A. As long asB. UnlessC. BeforeD. As soon as18. We will go to visit the museum if it _________ tomorrow.A. won’t rainB. isn’t rainC. doesn’t rainD. didn’t rain19. The pyramid ________ about five thousand years ago.A. was buildedB. was builtC. buildsD. has been built20. We can't enter the room because its door_________ .A. lockedB. locksC. is lockedD. is locking21. --When _____ this kind of computers______?--Last year.A. did; useB. was; usedC. is; usedD. are; used22. ________cotton（棉花）______ in the southeast of China?A. Is; grownB. Are; grownC. Does; growsD. Do; grow23. --Kate, this is Susan Jones, my old school friend.--Nice to meet you.--_________.A. Nice to see youB. Nice to meet you, tooC. It’s my pleasure to meet youD. Thank you24. -- Thanks for the lovely and delicious food.-- _______.A. No thanksB. Never mindC. All rightD. My pleasure25. -- Good morning! ________-- No thanks, I’m just lookin g around.A. How do you do?B. What can I do for you?C. How are you?D. Would you like to help me?26. English _____________ in Canada.A. speaksB. is speakedC. is spokenD. is speaking27. Most of the bottles were made _________ China. They look very beautiful.A. inB. fromC. ofD. by28. Johnson usually __________ much time playing football.A. paysB. takesC. costsD. spends29. Do you have any flights________ Sydney next Tuesday morning?A. onB. toC. inD. for30. We’ll plant trees tomorrow, but I don’t know _______ Tom will come and join us .A. ifB. whichC. whatD. whereII.阅读理解（本大题共15小题，每小题2分，共30分）阅读下列短文，每篇文章后面有五个小题，请从每小题所给的四个选项中，选出最佳答案。
2023-2024学年湖南省长沙市长郡中学高二(上)期中数学试卷【答案版】
2023-2024学年湖南省长沙市长郡中学高二（上）期中数学试卷一、选择题（本大题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的）1．若两个不同平面α，β的法向量分别为u →=（1，2，﹣1），v →=（﹣3，﹣6，3），则（ ） A ．α∥βB ．α⊥βC ．α，β相交但不垂直D ．以上均不正确2．《莱因德纸草书》（RhindPapyrus ）是世界上最古老的数学著作之一，书中有一道这样的题目，请给出答案：把100个面包分给5个人，使每人所得面包个数成等差数列，且使较大的三份之和的17是较小的两份之和，则最小的一份为（ ） A ．53B ．103C ．56D ．1163．若直线y ＝kx ﹣2与直线y ＝3x 垂直，则k ＝（ ） A ．3B ．13C ．﹣3D ．−134．直三棱柱ABC ﹣A 1B 1C 1中，∠BCA ＝90°，M ，N 分别是A 1B 1，A 1C 1的中点，BC ＝CA ＝CC 1，则BM 与NA 所成的角的余弦值为（ ） A ．−√3010 B ．√306C ．√3010D ．√225．双曲线C 与椭圆x 29+y 24=1有相同的焦点，一条渐近线的方程为x ﹣2y ＝0，则双曲线C 的标准方程为（ ） A ．x 24−y 2=1 B ．y 29−x 236=1C ．x 29−y 236=1 D ．y 24−x 2=16．已知抛物线E ：x 2＝4y 和圆F ：x 2+（y ﹣1）2＝1，过点F 作直线l 与上述两曲线自左而右依次交于点A ，C ，D ，B ，则|AC |与|BD |的乘积为（ ） A ．1B ．2C ．3D ．√27．已知数列{a n }满足2a n+1a n +a n+1−3a n =0(n ∈N ∗)且a 1＞0．若{a n }是递增数列，则a 1的取值范围是（ ） A ．(0，12) B ．(12，1)C ．（0，1）D ．(0，√2−1)8．已知椭圆x 2a 2+y 2b 2=1（a ＞b ＞0）上一点A 关于原点的对称点为B 点，F 为其右焦点，若AF ⊥BF ，设∠ABF ＝α，且α∈（π4，π3），则该椭圆的离心率的取值范围是（ ）A ．(√22，√3−1) B ．(√22，1)C ．(√22，√32)D ．(√33，√63)二、选择题（本大题共4小题，每小题5分，共20分，在每小题给出的四个选项中，有多项符合题目要求，全部选对的得5分，部分选对的得2分，有选错的得0分）9．已知m →=(1，a +b ，a −b)（a ，b ∈R ）是直线l 的方向向量，n →=(1，2，3)是平面α的法向量，则下列结论正确的是（ ） A ．若l ∥α，则5a ﹣b +1＝0 B ．若l ∥α，则a +b ﹣1＝0C ．若l ⊥α，则a +b ﹣2＝0D ．若l ⊥α，则a ﹣b ﹣3＝010．已知数列{a n }是等比数列，那么下列数列一定是等比数列的是（ ） A ．{1a n}B ．{a n a n +1}C ．{lg （a n 2）}D ．{a n +a n +1}11．已知p ∈R ，直线l 1：x ﹣py +p ﹣2＝0过定点A ，l 2：px +y +2p ﹣4＝0过定点B ，l 1与l 2交于点M ，则下列结论正确的是（ ） A ．l 1⊥l 2B ．MA •MB 的最大值是25C ．点M 的轨迹方程是x 2+y 2﹣5x ＝0D ．MA +2MB 的最大值为5√512．已知抛物线C ：y 2＝2px （p ＞0）的焦点为F ，直线l 与C 交于A （x 1，y 1），B （x 2，y 2）两点，其中点A 在第一象限，点M 是AB 的中点，作MN 垂直于准线，垂足为N ，则下列结论正确的是（ ） A ．若以AB 为直径作圆M ，则圆M 与准线相切B ．若直线l 经过焦点F ，且OA →⋅OB →=−12，则p ＝4C ．若AF →=3FB →，则直线l 的倾斜角为π3D ．若以AB 为直径的圆M 经过焦点F ，则|AB||MN|的最小值为√2三、填空题（本大题共4小题，每小题5分，共20分）13．已知⊙M 的圆心为M （3，﹣5），且与直线x ﹣7y +2＝0相切，则圆C 的面积为 ． 14．如图，在三棱锥O ﹣ABC 中，OA ，OB ，OC 两两垂直，OA ＝OC ＝3，OB ＝2，则直线OB 与平面ABC 所成角的正弦值为 ．15．已知双曲线C ：x 2a 2−y 2b 2=1(a ＞0，b ＞0)的左、右焦点分别为F 1，F 2，P 为C 上一点，且∠F 1PF 2＝60°，则当C 的离心率e ＝ 时，满足sin ∠PF 2F 1＝3sin ∠PF 1F 2．16．已知数列1，1，2，1，2，4，1，2，4，8，1，2，4，8，16，…，其中第一项是20，接下来的两项是20，21，再接下来的三项是20，21，22，依此类推． （1）这个数列的第100项为 ；（2）整数N 满足条件：N ＞1000且该数列的前N 项和为2的整数幂，则最小整数N ＝ ． 四、解答题（本大题共6小题，共70分.解答时应写出必要的文字说明、证明过程或演算步骤）17．（10分）已知数列{a n }各项均为正数，且a 1＝2，a n+12−2a n+1=a n 2+2a n ．（1）证明：{a n }为等差数列，并求出通项公式； （2）设b n =(−1)n a n ，求b 1+b 2+b 3+⋯+b 20．18．（12分）四棱锥P ﹣ABCD 中，BC ∥AD ，BC ⊥平面P AB ，P A ＝AB ＝BC ＝2AD ＝2，E 为AB 的中点，且PE ⊥EC ．（1）求证：BD ⊥平面PEC ； （2）求二面角E ﹣PC ﹣D 的正弦值．19．（12分）已知圆M ：x 2+（y ﹣2）2＝1，Q 是x 轴上的动点，QA ，QB 分别切圆M 于A ，B 两点． （1）求四边形QAMB 面积的最小值； （2）若|AB |=4√23，求Q 点的坐标．20．（12分）设抛物线y 2＝2px （p ＞0）的准线为l ，A 、B 为抛物线上两动点，AA '⊥l 于A '，定点K （0，1）使|KA |+|AA '|有最小值√2． （1）求抛物线的方程；（2）当KA →=λKB →（λ∈R 且λ≠1）时，是否存在一定点T 满足TA →⋅TB →为定值？若存在，求出T 的坐标和该定值；若不存在，请说明理由．21．（12分）已知数列{a n}，a1＝2，a n+1=2−1a n ，数列{b n}满足b1＝1，b2nb2n−1=b2n+1b2n=a n．（1）求证：数列{1a n−1}为等差数列，并求出数列{a n}的通项公式；（2）求b2n+1的表达式；（3）求证：1b2+1b4+⋯+1b2n＜1．22．（12分）已知椭圆C：x2a2+y2b2=1(a＞b＞0)的左、右焦点分别为F1、F2，焦距为2，上、下顶点分别为B1、B2，A为椭圆上的点，且满足k AB1⋅k AB2=−34．（1）求椭圆C的标准方程；（2）过F1、F2作两条相互平行的直线l1，l2交C于M，N和P，Q，顺次连接构成四边形PQNM，求四边形PQNM面积的取值范围．2023-2024学年湖南省长沙市长郡中学高二（上）期中数学试卷参考答案与试题解析一、选择题（本大题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的）1．若两个不同平面α，β的法向量分别为u →=（1，2，﹣1），v →=（﹣3，﹣6，3），则（ ） A ．α∥βB ．α⊥βC ．α，β相交但不垂直D ．以上均不正确解：∵v →=−3u →，∴v →∥u →．故α∥β． 故选：A ．2．《莱因德纸草书》（RhindPapyrus ）是世界上最古老的数学著作之一，书中有一道这样的题目，请给出答案：把100个面包分给5个人，使每人所得面包个数成等差数列，且使较大的三份之和的17是较小的两份之和，则最小的一份为（ ） A ．53B ．103C ．56D ．116解：把100个面包分给5个人，使每人所得面包个数成等差数列， 使较大的三份之和的17是较小的两份之和，设分的面包，从小到大依次为a 1，a 2，a 3，a 4，a 5， 依题意得17(a 3+a 4+a 5)=a 1+a 2，故3a 1+9d ＝7（2a 1+d ），2d ＝11a 1， 由S 5＝5a 3＝5（a 1+2d ）＝100， 得a 1+2d ＝12a 1＝20， 解得a 1=53． 故选：A ．3．若直线y ＝kx ﹣2与直线y ＝3x 垂直，则k ＝（ ） A ．3B ．13C ．﹣3D ．−13解：∵直线y ＝kx ﹣2与直线y ＝3x 垂直， ∴3k ＝﹣1，解得k =−13．故选：D ．4．直三棱柱ABC ﹣A 1B 1C 1中，∠BCA ＝90°，M ，N 分别是A 1B 1，A 1C 1的中点，BC ＝CA ＝CC 1，则BM 与NA 所成的角的余弦值为（ ） A ．−√3010B ．√306C ．√3010D ．√22解：直三棱柱ABC ﹣A 1B 1C 1中，∠BCA ＝90°， M ，N 分别是A 1B 1，A 1C 1的中点， 如图，BC 的中点为O ，连结ON ，MN ∥B 1C 1且MN =12B 1C 1＝OB ，则MNOB 是平行四边形， BM 与AN 所成角就是∠ANO ， ∵BC ＝CA ＝CC 1，设BC ＝CA ＝CC 1＝2，∴CO ＝1，AO =√5，AN =√5，MB =√B 1M 2+BB 12=√(√2)2+22=√6， 在△ANO 中，由余弦定理可得：cos ∠ANO =AN 2+NO 2−AO 22AN⋅NO =62×√5×√6=√3010．故选：C ． 5．双曲线C 与椭圆x 29+y 24=1有相同的焦点，一条渐近线的方程为x ﹣2y ＝0，则双曲线C 的标准方程为（ ） A ．x 24−y 2=1 B ．y 29−x 236=1C ．x 29−y 236=1D ．y 24−x 2=1解：由题意双曲线C 与椭圆x 29+y 24=1有相同的焦点，知c =√5，设双曲线的方程为x 2﹣4y 2＝λ（λ＞0），∴x 2λ−y 2λ4=1，∴λ+λ4=5，∴λ＝4．则双曲线C 的标准方程为x 24−y 2=1．故选：A ．6．已知抛物线E ：x 2＝4y 和圆F ：x 2+（y ﹣1）2＝1，过点F 作直线l 与上述两曲线自左而右依次交于点A ，C ，D ，B ，则|AC |与|BD |的乘积为（ ） A ．1B ．2C ．3D ．√2解：由抛物线E ：x 2＝4y 和圆F ：x 2+（y ﹣1）2＝1，可知抛物线焦点为F （0，1）， 设A （x 1，y 1），B （x 2，y 2），设直线的方程为x ＝m （y ﹣1），由{x =m(y −1)x 2=4y ，得m 2y 2﹣（2m 2+4）y +m 2＝0， 则y 1y 2＝1，由抛物线的定义可知|AF |＝y 1+1，|BF |＝y 2+1， ∴|AC |＝y 1，|BD |＝y 2， ∴|AC |×|BD |＝y 1y 2＝1，当且仅当y 1＝2y 2，即y 1=√2，y 2=√22时取等号．故选：A ．7．已知数列{a n }满足2a n+1a n +a n+1−3a n =0(n ∈N ∗)且a 1＞0．若{a n }是递增数列，则a 1的取值范围是（ ） A ．(0，12)B ．(12，1)C ．（0，1）D ．(0，√2−1)解：根据2a n+1a n +a n+1−3a n =0(n ∈N ∗)， 可得a n+1=3a n2a n +1， 所以1a n+1=13⋅1a n+23，所以1a n+1−1=13⋅(1a n−1)，从而可得数列{1a n+1−1}是以1a 1−1为首项，13为公比的等比数列，所以1a n−1=(1a 1−1)⋅(13)n−1，整理有a n =11+(1a 1−1)(13)n−1，因为a n +1＞a n ＞0， 所以11+(1a 1−1)(13)n＞11+(1a 1−1)(13)n−1＞0，整理得：(1a 1−1)⋅13＜1a 1−1，即0＜a 1＜1， 故选：C ． 8．已知椭圆x 2a 2+y 2b 2=1（a ＞b ＞0）上一点A 关于原点的对称点为B 点，F 为其右焦点，若AF ⊥BF ，设∠ABF ＝α，且α∈（π4，π3），则该椭圆的离心率的取值范围是（ ） A ．(√22，√3−1) B ．(√22，1)C ．(√22，√32) D ．(√33，√63)解：椭圆x 2a 2+y 2b 2=1（a ＞b ＞0）上一点A 关于原点的对称点为B 点，F 为其右焦点，设左焦点为F ′．所以|AF ′|+|AF |＝2a ，根据对称关系：四边形AF ′BF 为矩形． 所以|AB |＝|FF ′|＝2c ， 由于AF ⊥BF ，设∠ABF ＝α， 所以|AF |＝2c sin α，|AF ′|＝2c cos α， 所以2c sin α+2c cos α＝2a ， 所以ca =1sinα+cosα=√2sin(α+π4)，由于α∈（π4，π3），故α+π4∈(π2，7π12)， 所以√2+√64＜sin(α+π4)＜1， 所以√2sin(α+π4)∈(√22，√3−1)，即离心率的范围． 故选：A ．二、选择题（本大题共4小题，每小题5分，共20分，在每小题给出的四个选项中，有多项符合题目要求，全部选对的得5分，部分选对的得2分，有选错的得0分）9．已知m →=(1，a +b ，a −b)（a ，b ∈R ）是直线l 的方向向量，n →=(1，2，3)是平面α的法向量，则下列结论正确的是（ ） A ．若l ∥α，则5a ﹣b +1＝0 B ．若l ∥α，则a +b ﹣1＝0C ．若l ⊥α，则a +b ﹣2＝0D ．若l ⊥α，则a ﹣b ﹣3＝0解：根据题意，m →=(1，a +b ，a −b)（a ，b ∈R ）是直线l 的方向向量，n →=(1，2，3)是平面α的法向量，若l ∥α，则m →⊥n →，则有m →⋅n →=0，即1+2（a +b ）+3（a ﹣b ）＝0，即5a ﹣b +1＝0，A 正确，B 错误； 若l ⊥α，则m →∥n →，则有11=a+b 2=a−b 3，变形可得a +b ﹣2＝0且a ﹣b ﹣3＝0，C 、D 正确． 故选：ACD ．10．已知数列{a n }是等比数列，那么下列数列一定是等比数列的是（ ） A ．{1a n}B ．{a n a n +1}C ．{lg （a n 2）}D ．{a n +a n +1}解：根据题意，{a n }为等比数列，设其公比为q （q ≠0）； 对于A ，1a n =1a 1q n−1=1a 1⋅(1q)n−1，∴数列{1a n}是以1a 1为首项，1q为公比的等比数列，故A 正确；对于B ，a n+1a n+2a n a n+1=a n+2a n=q 2，∴数列{a n a n +1}是以a 1a 2为首项，q 2为公比的等比数列，故B 正确；对于C ，当a n ＝1时，lg(a n 2)=0，数列{lg(a n 2)}不是等比数列，故C 错误；对于D ，当q ＝﹣1时，a n +a n +1＝0，数列{a n +a n +1}不是等比数列，故D 错误． 故选：AB ．11．已知p ∈R ，直线l 1：x ﹣py +p ﹣2＝0过定点A ，l 2：px +y +2p ﹣4＝0过定点B ，l 1与l 2交于点M ，则下列结论正确的是（ ） A ．l 1⊥l 2B ．MA •MB 的最大值是25C ．点M 的轨迹方程是x 2+y 2﹣5x ＝0D ．MA +2MB 的最大值为5√5解：对于A ，1•p +（﹣p ）•1＝0，∴l 1⊥l 2，A 正确； 对于B ，l 1恒过定点A （2，1），l 2恒过定点B （﹣2，4），由选项A 正确可推得，MA 2+MB 2＝AB 2＝25≥2MA •MB ，MA ＝MB 时等号成立，∴MA •MB 的最大值是252，B 错误；对于C ，设M （x ，y ），则MA ⊥MB ，MA 2+MB 2＝AB 2＝52，（x ﹣2）2+（y ﹣1）2+（x +2）2+（y ﹣4）2＝25，化简有x 2+y 2﹣5y ＝0，C 错误；对于D ，设∠MAB ＝θ，θ∈(0，π2)，则MA ＝5cos θ，MB ＝5sin θ，∴MA +2MB =5(cosθ+2sinθ)=5√5sin(θ+φ)≤5√5，即MA +2MB 的最大值为5√5，D 正确． 故选：AD ．12．已知抛物线C ：y 2＝2px （p ＞0）的焦点为F ，直线l 与C 交于A （x 1，y 1），B （x 2，y 2）两点，其中点A 在第一象限，点M 是AB 的中点，作MN 垂直于准线，垂足为N ，则下列结论正确的是（ ） A ．若以AB 为直径作圆M ，则圆M 与准线相切B ．若直线l 经过焦点F ，且OA →⋅OB →=−12，则p ＝4C ．若AF →=3FB →，则直线l 的倾斜角为π3D ．若以AB 为直径的圆M 经过焦点F ，则|AB||MN|的最小值为√2解：由抛物线C ：y 2＝2px （p ＞0）的焦点为F ，直线l 与C 交于A （x 1，y 1），B （x 2，y 2）两点，其中点A 在第一象限，点M 是AB 的中点，作MN 垂直于准线，垂足为N ，对于A ，当以AB 为直径作圆M ，且AB 经过焦点F 时，|MN|=12(|AF|+|BF|)=12|AB|，此时圆M 与准线相切，当直线l 不经过焦点F 时，圆M 不一定与准线相切，故A 错误；对于B ，过F 的直线l 的方程设为x =my +p2，把直线方程与抛物线方程联立，可得y 2﹣2pmy ﹣p 2＝0，y 1y 2=−p 2，x 1x 2=(y 1y 2)24p2=14p 2，OA →⋅OB →=x 1x 2+y 1y 2=14p 2−p 2=−12(p ＞0)，解得p ＝4，B 正确；对于C ，AF →=(p2−x 1，−y 1)，FB →=(x 2−p2，y 2)，AF →=3FB →，可得p 2−x 1=3(x 2−p2)，﹣y 1＝3y 2，又y 1y 2=−p 2，x 1x 2=(y 1y 2)24p 2=14p 2，可求出A(3p 2，√3p)，B(p6，−√3p 3)，k 1=y 1−y 2x 1−x 2=√3p+√3p33p 2−p 6=√3，∴直线l 的倾斜角为π3，C 正确；对于D ，设AF ＝a ，BF ＝b ，由抛物线的定义可得|MN|=12(|AF|+|BF|)=12(a +b)，以AB 为直径的圆M 经过焦点F ，∴AF ⊥BF ，|AB|=√a 2+b 2，|AB||MN|=√a 2+b 212(a+b)=√(a+b)2−2ab12(a+b)≥√(a+b)2−(a+b)2212(a+b)=√2，当且仅当a ＝b 时，即AF ＝BF 时等号成立，D 正确． 故选：BCD ．三、填空题（本大题共4小题，每小题5分，共20分）13．已知⊙M 的圆心为M （3，﹣5），且与直线x ﹣7y +2＝0相切，则圆C 的面积为 32π ． 解：因为圆M 与直线．x ﹣7y +2＝0相切，所以点M （3，﹣5）到直线：x ﹣7y +2＝0的距离即为圆M 的半径， 所以r =|3−7×(−5)+2|√1+(−7)=405√2=4√2，圆C 的面积为π×(4√2)2=32π． 故答案为：32π．14．如图，在三棱锥O ﹣ABC 中，OA ，OB ，OC 两两垂直，OA ＝OC ＝3，OB ＝2，则直线OB 与平面ABC 所成角的正弦值为3√1717．解：如图所示，以点O 为坐标原点，建立空间直角坐标系O ﹣xyz ，则A （0，0，3），B （2，0，0），C （0，3，0）， 直线OB 的方向向量OB →=(2，0，0)， 由于AB →=(2，0，−3)，AC →=(0，3，−3)， 若m →=(x ，y ，z)是平面ABC 的一个法向量，则{AB →⋅m →=2x −3z =0AC →⋅m →=3y −3z =0， 据此可得m →=(32，1，1)， ∴|cos ＜OB →，m →＞|=|OB →⋅m →|OB →||m →||=32×172=3√1717， 故直线OB 与平面ABC 所成角的正弦值为3√1717． 故答案为：3√1717． 15．已知双曲线C ：x 2a 2−y 2b 2=1(a ＞0，b ＞0)的左、右焦点分别为F 1，F 2，P 为C 上一点，且∠F 1PF 2＝60°，则当C 的离心率e ＝ √72时，满足sin ∠PF 2F 1＝3sin ∠PF 1F 2． 解：双曲线C ：x 2a 2−y 2b 2=1(a ＞0，b ＞0)的左、右焦点分别为F 1，F 2，P 为C 上一点，由sin ∠PF 2F 1＝3sin ∠PF 1F 2得|PF 1|＝3|PF 2|， 由双曲线的定义可得|PF 1|﹣|PF 2|＝2|PF 2|＝2a ， 所以|PF 2|＝a ，|PF 1|＝3a ；因为∠F 1PF 2＝60°，由余弦定理可得4c 2＝9a 2+a 2﹣2×3a •a •cos60°， 整理可得4c 2＝7a 2，所以e 2=c 2a 2=74，即e =√72．故答案为：√72． 16．已知数列1，1，2，1，2，4，1，2，4，8，1，2，4，8，16，…，其中第一项是20，接下来的两项是20，21，再接下来的三项是20，21，22，依此类推． （1）这个数列的第100项为 256 ；（2）整数N 满足条件：N ＞1000且该数列的前N 项和为2的整数幂，则最小整数N ＝ 1897 ． 解：对数列进行分组如下： 第一组：20，1个数， 第二组：20，21，2个数， 第三组：20，21，22，3个数， ……，第k +1组：20，21，22，…，2k ，k +1个数； （1）由1+2+3+⋯+k =k(k+1)2≤100可得k ≤13，且1+2+3+⋯+13＝91，所以该数列的第100项在第14组的第9个数，即28＝256． （2）该数列前k 组的项数和为1+2+3+⋯+k =k(k+1)2， 由题意可知N ＞1000，即k(k+1)2＞1000，解得k ≥45，n ∈N *，即N 出现在第44组之后． 又第k 组的和为20+21+⋯+2k−1=1×(1−2k)1−2=2k −1，所以前k 组的和为1+（1+2）+⋯+（1+2+⋯+2k ﹣1）＝（21﹣1）+（22﹣1）+⋯ +（2k ﹣1）＝（21+22+⋯+2k ）﹣k ＝2k +1﹣k ﹣2， 设满足条件的N 在第k +1（k ∈N *）组（k ≥44）， 且第N 项为第k +1组的第m （m ∈N *）个数， 第k +1组的前m 项和为1+2+22+⋯+2m ﹣1＝2m ﹣1，要使该数列的前N 项和为2的整数幂， 即2m ﹣1与﹣k ﹣2互为相反数， 即2m ﹣1＝2+k ， 所以k ＝2m ﹣3，由k ≥44，所以2m ﹣3≥44，解之得m ≥6， 取最小值m ＝6，此时k ＝26﹣3＝61， 对应满足的最小条件为N =61(61+1)2+6=1897． 故答案为：256；1897．四、解答题（本大题共6小题，共70分.解答时应写出必要的文字说明、证明过程或演算步骤）17．（10分）已知数列{a n }各项均为正数，且a 1＝2，a n+12−2a n+1=a n 2+2a n ．（1）证明：{a n }为等差数列，并求出通项公式； （2）设b n =(−1)n a n ，求b 1+b 2+b 3+⋯+b 20．解：（1）证明：因为a 1＝2，a n+12−2a n+1=a n 2+2a n ， 所以a n+12−a n 2=(a n+1−a n )(a n+1+a n )=2(a n+1+a n )，因为数列{a n }各项均为正数，即a n +1+a n ＞0， 所以a n +1﹣a n ＝2，即数列{a n }为等差数列，公差为d ＝2，首项为a 1＝2． 所以a n ＝2+（n ﹣1）×2＝2n ；（2）由（1）知a n＝2n，其公差为d＝2，所以b n=(−1)n a n=(−1)n⋅2n，所以b1+b2+b3+⋯+b20＝（﹣a1+a2）+（﹣a3+a4）+⋯+（﹣a19+a20）＝10d＝20．18．（12分）四棱锥P﹣ABCD中，BC∥AD，BC⊥平面P AB，P A＝AB＝BC＝2AD＝2，E为AB的中点，且PE⊥EC．（1）求证：BD⊥平面PEC；（2）求二面角E﹣PC﹣D的正弦值．（1）证明：因为BC⊥平面P AB，PE⊂平面P AB，所以BC⊥PE，因为PE⊥EC，EC∩BC＝C，EC，BC⊂平面BCD，所以PE⊥平面BCD，又BD⊂平面BCD，所以PE⊥BD，因为tan∠ABD=ADAB=12，tan∠BCE=BEBC=12，所以∠ABD＝∠BCE，因为∠BCE+∠CEB＝90°，所以∠ABD+∠CEB＝90°，即BD⊥CE，又PE∩CE＝E，PE，CE⊂平面PEC，所以BD⊥平面PEC．（2）解：由（1）得PE⊥AB，因为E为AB的中点，且P A＝AB＝2，所以PB＝2，以E为坐标原点，EB，EP所在直线分别为x轴，z轴，过点E作BC的平行线为y轴，建立空间直角坐标系E﹣xyz，则E （0，0，0），P(0，0，√3)，C （1，2，0），D （﹣1，1，0），B （1，0，0）， 所以PC →=(1，2，−√3)，PD →=(−1，1，−√3)，PE →=(0，0，−√3)， 设平面PCD 的法向量为m →=(x ，y ，z)， 由PC →⋅m →=0，PD →⋅m →=0得，{x +2y −√3z =0−x +y −√3z =0，令x ＝1，则y ＝﹣2，z =−√3，所以m →=(1，−2，−√3)， 由（1）知，平面PCE 的一个法向量为BD →=(−2，1，0)， 所以cos ＜m →，BD →＞=m →⋅BD →|m →||BD →|=8×5=−√105，所以二面角E ﹣PC ﹣D 的正弦值为√155． 19．（12分）已知圆M ：x 2+（y ﹣2）2＝1，Q 是x 轴上的动点，QA ，QB 分别切圆M 于A ，B 两点． （1）求四边形QAMB 面积的最小值； （2）若|AB |=4√23，求Q 点的坐标．解：（1）∵圆M ：x 2+（y ﹣2）2＝1，Q 是x 轴上的动点，QA ，QB 分别切圆M 于A ，B 两点，∴MA ⊥AQ ，∴S 四边形MAQB =|MA|⋅|QA|=|QA|=√|MQ|2−|MA|2=√|MQ|2−1≥√|MO|2−1=√3． ∴四边形QAMB 面积的最小值为√3．（2）设AB 与MQ 交于P ，则MP ⊥AB ，MB ⊥BQ ， ∴|MP|=√1−(223)2=13．在Rt △MBQ 中，|MB |2＝|MP ||MQ |，即1=13|MQ|， ∴|MQ |＝3，设Q （x ，0），则x 2+22＝9， ∴x =±√5， ∴Q(±√5，0)．20．（12分）设抛物线y 2＝2px （p ＞0）的准线为l ，A 、B 为抛物线上两动点，AA '⊥l 于A '，定点K （0，1）使|KA |+|AA '|有最小值√2． （1）求抛物线的方程；（2）当KA →=λKB →（λ∈R 且λ≠1）时，是否存在一定点T 满足TA →⋅TB →为定值？若存在，求出T 的坐标和该定值；若不存在，请说明理由．解：（1）不妨设抛物线焦点为F ， 此时F(p2，0)，因为A 、B 为抛物线上两动点，AA '⊥l 于A '， 所以|AA '|＝|AF |，又定点K （0，1）使|KA |+|AA '|有最小值√2， 此时|KA|+|AA′|=|KA|+|AF|≥|KF|=√2， 即|KF|=√(p 2−0)2+(0−1)2=√2， 解得p ＝2或p ＝﹣2（舍去）， 则抛物线的方程为y 2＝4x ； （2）因为KA →=λKB →， 所以K ，A ，B 三点共线，不妨设直线AB 方程为x ＝t （y ﹣1），A （x 1，y 1），B （x 1，y 1），T （m ，n ）， 联立{y 2=4x x =t(y −1)，消去x 并整理得y 2﹣4y +4t ＝0，此时Δ＝（4t ）2﹣4×4t ＞0， 解得t ＜0或t ＞1，由韦达定理得y 1+y 2＝4t ，y 1y 2＝4t ， 所以x 1＝t （y 1﹣1），x 2＝t （y 2﹣1），此时TA →⋅TB →=(x 1−m)(x 2−m)+(y 1−n)(y 2−n)，因为TA →⋅TB →=[ty 1﹣（m +t ）][ty 2﹣（m +t ）]+（y 1﹣n ）（y 2﹣n ） ＝（t 2+1）y 1y 2﹣[t （m +t ）+n ]（y 1+y 2）+（m +t ）2+n 2 ＝4t （1﹣4m ）2﹣4t [t （m +t ）+n ]+（m +t ）2+n 2 ＝（1﹣4m ）t 2+2（2﹣2n +m ）t +m 2+n 2， 若存在一定点T 满足TA →⋅TB →为定值， 此时1﹣4m ＝0且2﹣2n +m ＝0， 解得m =14，n =98，此时T(14，98)，此时TA →⋅TB →=8564． 21．（12分）已知数列{a n }，a 1＝2，a n+1=2−1a n ，数列{b n }满足b 1＝1，b 2n b 2n−1=b 2n+1b 2n=a n ．（1）求证：数列{1a n −1}为等差数列，并求出数列{a n }的通项公式； （2）求b 2n +1的表达式； （3）求证：1b 2+1b 4+⋯+1b 2n＜1．（1）证明：由a 1＝2，a n+1=2−1a n 可知a n+1−1=a n −1a n ， ∴1a n+1−1=a n a n −1=1+1a n −1故1a n+1−1−1a n −1=1，又1a 1−1=1，∴数列{1a n −1}是以1为公差，1为首项的等差数列，∴1a n −1=n ，即a n =n+1n ． （2）解：由b 2n b 2n−1=b 2n+1b 2n=a n ，有b 2n+1b 2n−1=a n2=(n+1n)2，∴b2n+1=b2n+1b2n−1×b2n−1b2n−3×...×b3b1×b1=(n+1n)2×(nn−1)2×⋯×(21)2×1＝（n+1）2，∴b2n+1=(n+1)2．（3）证明：由（2）可得：b2n=b2n+1a n=(n+1)2n+1n=n(n+1)，∴1b2+1b4+⋯+1b2n=11×2+12×3+⋯+1n(n+1)=1−12+12−13+⋯+1n−1n+1=1−1n+1＜1．22．（12分）已知椭圆C：x2a2+y2b2=1(a＞b＞0)的左、右焦点分别为F1、F2，焦距为2，上、下顶点分别为B1、B2，A为椭圆上的点，且满足k AB1⋅k AB2=−34．（1）求椭圆C的标准方程；（2）过F1、F2作两条相互平行的直线l1，l2交C于M，N和P，Q，顺次连接构成四边形PQNM，求四边形PQNM面积的取值范围．解：（1）由于焦距为2，则c＝1，设A（x0，y0），则x02a2+y02b2=1，又B1（0，b），B2（0，﹣b），k AB1⋅k AB2=−34，则k AB1⋅k AB2=y0−bx0⋅y0+bx0=y02−b2x02=−b2a2=−34，∴a2=43b2=43(a2−1)，∴a＝2，b=√3．即椭圆C 的标准方程为x 24+y 23=1．（2）由对称性可知，四边形PQNM 为平行四边形， 设MN ：x ＝my +1，M （x 1，y 1），N （x 2，y 2），将直线MN 的方程与椭圆方程联立得：（3m 2+4）y 2+6my ﹣9＝0． 由根与系数的关系可得，y 1+y 2=−6m 3m 2+4，y 1y 2=−93m 2+4， 则|MN|=√1+m 2|y 1−y 2|=√1+m 2√36m 2(3m 2+4)2+363m 2+4=12(1+m 2)3m 2+4， 设点F 2（1，0）到直线l 1的距离为d ，则d =2√1+m 2，所以四边形PQNM 面积为：S =|MN|d =24√1+m 23m 2+4．设√m 2+1=t ≥1，则S =24t 3t 2+1=243t+1t在t ∈[1，+∞）单调递减，所以S 的取值范围为（0，6]．。
湖南省长沙市长郡中学2023-2024学年高二下学期期中考试英语试题(含答案)
长郡中学2023-2024学年高二下学期期中考试英语时量：120分钟满分：150分第一部分听力（共两节，满分30分）略第二部分阅读（共两节，满分50分）第一节（共15小题；每小题2.5分，满分37.5分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项。
AFrom December 1st, 2023 to November 30th, 2024, visitors can stay in China for up to 15 days without a visa. Below are several easy steps for planning a first China tour.Decide How Many Days to StayWe suggest you take at least a week for your first trip to see the highlights in the top three cities: Beijing (3–4 days), Xi’an (2 days), and Shanghai (1–2 days).To discover more of China, like charming Guilin and lovely Chengdu pandas, you would need a few more days.Consider When to Travel to ChinaThings to do in Beijing, Xi’an, and Shanghai are seldom affected by the seasons.Spring (April–May) and autumn (September–October) are generally the most comfortable and recommended times for a China tour. They are neither too hot nor too cold, but fall is generally drier and warmer than spring.A more ideal travel time for you could be March and early April or September when there are smaller crowds, favorable prices, and still good weather.China is a good summer holiday destination too.Consider Your BudgetThe biggest cost could be international airfares. The cost of airfares from the US or Europe to China varies a lot depending on when you fly and which airline you use, from around US$1,200 to US$3,000 for an economy round trip.The biggest price difference between the off and peak seasons is in the price of hotels and airfares. Prices in peak seasons can go up by 50 to 100%.For a private tour, the average cost per day is about US$220–350 per person, including flights/trains within China, 4- or 5-star hotels, lunches, attractions, guides, and private transport.We Believe Private and Tailor-Made Tours Are BestWith our private tours, you would have much more personal choice in how your tour goes. You could have more hand-picked and interactive experiences, like visiting a local family with your own local guide.With private guiding and transport, we would make full use of your time. You could focus on the sightseeing you want to do, skipping what’s not of interest and the long queues in the most crowded attractions.21. If you have a tour in China in winter, which places are suitable for you?A. Beijing and Guilin.B. Xi’an and Chengdu.C. Shanghai and Chengdu.D. Beijing and Xi’an.22. What can we know from the text?A. A visa is needed for a 12-day tour.B. The highest expense is the accommodation fee.C. Travelling in March can save tourists money.D. Prices in peak seasons usually go up by 150%.23. Which of the following is NOT the reason why a tailor-made tour is recommended?A. You have more choices about your route.B. Your time will be maximized.C. You are likely to interact with locals in person.D. You will spend less on the tour.BThree years into my postdoc(博士后), I started to wonder whether I needed a new career plan. After applying for more than two dozen teaching jobs, I hadn’t landed a single interview.I had once considered going to art school but had put that idea to the side when I decided to pursue chemistry as an undergraduate. In the years that followed, I kept up my interest in art by taking painting classes at night. My family was bursting with mathematicians, computer programmers, and engineers, so it felt natural to have my daily life center around science.But in the spring after my failed job search, that started to change after a friend excitedly showed me proofs of a review article. She was astonished by what the journal’s scientific illustrator had done with her fundamental sketches(速写). “That would be such a fun job.” I thought.I decided to test out a new career direction by volunteering to create similar illustrations for my institute’s newsletters. I spent my nights and weekends reading scientific papers and thinking about how to illustrate the results. It was a fun task. I felt I was perhaps on the right path. But could I make a full-time career?Searching online, I tracked down people who had that kind of job. I found many had training through scientific illustration master’s degree programs. After living on graduate student and postdoctoral salaries for years, I didn’t have enough money saved up for the programs, so I decided to get a certificate in digital design.I now work as a visual designer at a biomedical research institute where I spend my days working with research to communicate their work visually. I love the fact that I get to combine my scientific and artistic sides.24. Why did the author attend classes at night?A. To please her family.B. To pass her undergraduate tests.C. To pursue her hobby.D. To complete optional courses in art.25. What effect did the proofs have on the author?A. They shook her belief in science.B. They tested out what she learned in class.C. They gave her inspiration for her scientific paper.D. They motivated her to find a new career direction.26. What prevented the author seeking a scientific illustration master’s degree?A. Her busy schedule.B. Her financial difficulties.C. Her new interest in digital design.D. Her lack of confidence.27. How does the author feel about her current job?A. Pressured.B. Desperate.C. Curious.D. Satisfied.CAdministrators of the Mogao Caves in Dunhuang, Gansu province, are striving to harmonize tourists’ exploration of the site with the need to safeguard murals and artifacts, through innovative measures such as digital presentations.Sandstorms, rainfall and tourist visits constitute the most severe threats to the UNESCO World Heritage Site, said Wang Xiaowei, director of the Dunhuang Grottoes Monitoring Center at the Dunhuang Academy.Since the Mogao Caves opened to the public in 1979, the number of visitors has been growing at an average annual rate of around 20 percent, reaching 2.15 million in 2019 before the outbreak of the COVID-19 pandemic. Thisyear, the site is expected to receive a record 3 million visitors.“If you enter the caves during the peak tourism months of July, August and September, you’ll find it hard to breathe,” Wang said. The carbon dioxide and moisture exhaled by visitors accumulate inside the caves and cause damage to the murals, Wang said.To preserve the caves, the duration of visits is limited and sometimes stopped during rain or dust storms. To try and ensure visitors aren’t disappointed when restrictions are in place, the center provides a digital exhibition, he said.Currently, the center is being expanded to cater for an additional 3,000 visitors on top of the existing capacity of 6,000.The Dunhuang Academy began digitally recording and storing images of murals and painted sculptures over 30 years ago. The digitization project has successfully covered over 200 caves, with a dedicated team of 110 experts currently undertaking the work.The Mogao Caves are immovable, and transporting them is impossible, according to Su Bomin, head of the Dunhuang Academy. And he added, “However, with digitization, we can perfectly replicate Dunhuang art and showcase it worldwide, introducing Eastern culture to the world.”In 2016, the Digital Dunhuang repository went live, sharing high-definition images and panoramic tours of the most exquisite 30 caves globally. Currently, visitors from 78 countries have accessed the repository, totaling over 16.8 million visits.Su said Dunhuang can provide diverse cultural exchanges through its cultural relics. “By digitizing these relics, we enable people worldwide to understand Dunhuang’s culture, thereby gaining a deeper appreciation for China’s historical commitment to diverse cultural exchanges — that is, an ethos of inclusivity, mutual learning and a shared future,” he said.28. Which of the following is NOT the reason for providing a digital exhibition?A. The factors related to COVID-19 put the caves in grave danger.B. The increasing number of tourists visiting Dunhuang might harm the caves.C. The authority is aimed to balance tourism and relics conservation.D. The duration of visits is limited and sometimes stopped during rain or dust storms.29. What does the underlined word “replicate” probably mean?A. Copy.B. Safeguard.C. Access.D. Transport.30. What does Su Bomin think of digitization?A. It records and stores images of murals and painted sculptures.B. It shares high definition and panoramic tours of the most exquisite caves.C. It allows for an international exchange of cultures through the relics.D. It enables people to appreciate the lasting beauty of the murals.31. What can be the best title for the text?A. The Significance of the Mural PaintingsB. The Restoration in Mogao CavesC. The Innovation on Mogao Caves’ PreservationD. The Dunhuang Spirit in Chinese CultureDThe road to Mars is long and fraught with peril. One challenge is getting humans to the red planet; another is ensuring that once they’ve arrived, they’ll be able to manage life there.To prepare astronauts for an extended stay on Mars, NASA’s latest simulated mission, CHAPEA — Crew Health and Performance Exploration Analog — will isolate four people inside a mock-Mars base in Texas for 378 days — roughly the time a manned mission to Mars would spend on the surface.Once inside they will adopt a pre-planned schedule taking part in simulated activities and science work, eating like astronauts, and dealing with maintenance and equipment failures, while undergoing strenuous psychological and physiological testing.The first simulation will begin in June, and will be followed by two more, each with a different crew in identical conditions, with the last simulation starting in 2026.“We’ve built a high-accuracy Mars surface mission scenario,” says Scott M. Smith, co-investigator for CHAPEA. The participants will experience a 22-minute delay in external communications, as astronauts would on Mars. Ambient noise will be played through speakers around the base, ensuring no outside sounds can be heard by participants.Aiming for accuracy has resulted in a habitat that could be feasibly built on Mars, Smith adds. The base, called “Mars Dune Alpha”, is a custom design by Bjarke Ingels Group and 3D-printing company ICON, and resides inside a hangar at the Johnson Space Center in Houston, Texas. Printed in a month from ICON’s concrete formula dubbed “Lavacrete”, on Mars, the idea is to build using Martian soil.“NASA has evaluated a tremendous number of options for off-world habitat construction — repurposed rockets and landers, inflatables, assembled buildings, etc.,” explains ICON CEO Jason Ballard. “They’ve come to believe what we believe: that when you evaluate it from a financial, safety and flexibility standpoint, robotic construction using local materials is far and away the best option.”32. What’s the purpose of NASA’s latest simulated mission?A. To get astronauts to Mars.B. To isolate four people inside a base in Mars.C. To help astronauts to do experiments in Mars.D. To prepare astronauts for managing life in Mars.33. Which of the following is TRUE according to the passage?A. The last simulation will end in 2026.B. Each stimulation has a different crew in the same conditions.C. The participants can hear outside sounds.D. The participants will do things different from those that astronauts do.34. What’s Smith’s attitude to the simulated mission?A. Indifferent.B. Pessimistic.C. Optimistic.D. Skeptical.35. What is the most commonly used technique in the text?A. Making comparison.B. Giving examples.C. Analyzing causes and effects.D. Listing figures.第二节（共5小题；每小题2.5分，满分12.5分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
湖南省长郡中学2022-2023学年高二下学期普通高中学业水平合格性考试模拟试题(含解析)
湖南省长郡中学2022-2023学年高二下学期普通高中学业水平合格性考试模拟试题学校:___________姓名：___________班级：___________考号：___________一、阅读理解Do you know what to do when there is an emergency? By calling the police, you can protect yourself and those around you.Call the police in all of the following emergencies:◆ A crime, such as a theft, especially if it is still in progress.◆ A car accident, especially if someone is injured.◆ Domestic violence, such as a child being mistreated.◆ Anything else that seems like an emergency.You may also call the police when you see something suspicious(可疑的)in your neighborhood:◆ Someone you don’t know is frequently walking around in your neighborhood. This could be a sign that the person is trying to break into a house.◆ Someone is trying to open the doors of a car. This could be a sign that the person is trying to steal the car.When you see something suspicious, do not suppose someone else has already called the police. People often hesitate to call the police for fear of danger. However, the police want to help prevent crime.What should you do when you call the police?◆Dial 911 (the U.S. emergency number; the number varies from one country to another—in China, you dial 110 to call the police). Stay calm when calling and give your name, address and phone number. Then, tell the person why you are calling (What happened? Where did it happen? When did it happen? Is it still in progress?). Follow any instructions you are given. For example, the dispatcher (调度员) might say, “Stay on the line,” or “Leave the building.”◆ If you dial the emergency number by mistake.do not hang up. Doing so could make the dispatcher think an emergency really exists. Instead，just tell the person that you called by mistake.Most police departments have a communication center. The communication center staffreach police officers by radio. Police officers carry headsets. like earphones, to stay in touch with the communication center.1．When calling the police, you DON’T need to give ______ to the dispatcher.A．your name B．your phone numberC．your ID card number D．some details of the emergency 2．What should you do if you dial the emergency number by mistake?A．Power off your smartphone.B．Hang up your phone at once.C．Tell the dispatcher you called by mistake.D．Go to a police station to explain your mistake.3．What do the communication center staff in police departments do?A．Monitor police officers.B．Answer emergency calls.C．Tell people what to do in an emergency.D．Reach police officers when there is an emergency.4．What is the purpose of the passage?A．To tell people when and how to call the police.B．To introduce a police officer’s general duties.C．To share the author’s experience of calling the police.D．To thank the police for trying to prevent crimes.Anderson Carey is 12 years old. One day, he saw a magazine article that interested him.It was about prosthetics(假肢), which can be used to replace a hand, arm or leg.The article said people are using 3-D printers to build these devices. Anderson thought this was very cool. He wanted to learn more about it. So Anderson talked to his science teacher, Dr. Holly Martin. He asked if they could build a prosthetic together. The timing(时机的把握) was perfect. Martin had just heard about a group called Enabling the Future. This group asks volunteers to help to build robotic arms and legs. The volunteers build them for people who share their stories on the website.Anderson and Martin looked through the website together. They decided to help a man from the country of Romania. His name is Cornel Crismaru, who lost his leg, hand and part of his arm.In February, Anderson and Martin got to work. Building the robotic arm was not easy. Anderson ran into some problems along the way. He had hoped to use a 3-D Printer at hisschool. One of the pieces for the arm was bigger than the size of the printer, though.Soon Anderson had an idea to solve this problem. He reached out to a 3-D printing company in Woodstock, Georgia. The company agreed to help. Anderson and Martin could use their big 3-D printers. After that, Anderson worked on the arm for about three months.Anderson and Martin sent the arm to Crismaru in May. In August, they received a notice. It is from Crismaru’s son. He thanked Anderson and Martin for their help.Martin said she hopes children and grown-ups who hear about Andersons projects will realize that it may be hard to change the world, but they can start with small acts. Some of these can help a person in a huge way.5．Anderson talked to his science teacher about_______.A．starting a website together B．buying a 3-D printerC．building a prosthetic together D．studying robots6．Anderson and Dr. Martin learned from the website that Crismaru______.A．lost some body parts B．wanted to be a volunteerC．was homeless D．was interested in robots7．How did Anderson solve his problem?A．He made a new 3-D printer.B．Hе took Dr.Martin’s advice.C．He worked together with his school.D．He got help from a 3-D printing company.8．What can we learn from the text?A．All roads lead to Rome.B．Failure is the mother of success. C．Those who help others help themselves.D．Small acts make a big difference.Have you ever wondered what wild animals do when no one is watching? Scientists have been able to record the “private” moments of wildlife with leading-edge technology. Low-cost, dependable and small modern cameras are of big help.Cameras placed in hard-to-reach places have taken videos of everything from small desert cats to later snow-loving felines (猫科动物) in the northern Rocky Mountains. These cameras are important tools to learn new information on wildlife.Some videos help scientists see the effects of climate change. For example, the desert animal javelina (矛牙野猪) and the tree-loving coatimundi (南美浣熊) have been caught on cameras north of their normal home. This could mean global warming is enlarging their二、其他下面文章中有3处需要添加小标题。
湖南省长沙市长郡中学2023-2024学年高二下学期期末考试物理试题(含解析)
长郡中学2023-2024学年高二下学期期末考试物理时量：75分钟满分：100分得分________一、单选题（本大题共6小题，每小题4分，共24分，每小题有且仅有一个选项符合题意）1．2021年3月23日，考古工作者对三星堆遗址新发现的6个“祭祀坑”的73份碳屑样本使用年代检测方法进行分析，初步判定其中4号坑最有可能属于商代晚期。
发生的是衰变，半衰期年。
则下列说法正确的是A ．发生衰变产生的新核是B ．释放的射线，就是核外电子C ．10个原子核经过5730年一定有5个发生了衰变D ．随着全球变暖，生物体中的衰变会加快2．下列关于分子动理论的说法正确的是A ．物体温度升高，物体内每一个分子的运动速率都变大B ．布朗运动就是分子的无规则运动C ．水和酒精混合后总体积减小，说明分子间有空隙D ．只有气体和液体有扩散现象，固体没有3．下列关于固体和液体性质的说法正确的是A ．彩色液晶显示器利用了液晶的光学性质具有各向同性的特点B ．液体表面张力产生的原因是液体表面层分子间距离比较大，分子力表现为斥力C ．非晶体的物理性质具有各向同性而晶体的物理性质都是各向异性D ．农民使用“松土保墒”进行耕作，通过把地面的土壤锄松，破坏土壤里的毛细管，使得土壤下面的水分不容易被输送到地表，从而保存地下的水分4．2024年3月20日，长征八号火箭成功发射，将鹊桥二号直接送入预定地月转移轨道后离开火箭。
如图所示，鹊桥二号进入近月点P 、远月点A 的月球捕获椭圆轨道，开始绕月球飞行。
经过多次轨道控制，鹊桥二号最终进入近月点P 和远月点B 、周期为24小时的环月椭圆轨道。
关于鹊桥二号的说法正确的是A．离开火箭时速度大于地球的第二宇宙速度14C 14C β5730T =14C β147N14C β14C 14CB ．在捕获轨道运行的周期大于24小时C ．在捕获轨道上经过P 点时，需要点火加速，才可能进入环月轨道D ．经过A 点的加速度比经过B 点时大★5．如图所示，用三根轻质细线a 、b 、c 将两个小球1和2连接并悬挂，两小球处于静止状态时，细线a 与竖直方向的夹角为37°，细线c 水平，已知两个小球的质量分别为和，重力加速度为g ，取，，下列说法正确的是A ．细线a 的拉力大小为5mgB ．细线c 的拉力大小为4mgC ．细线b 的拉力大小为4mgD ．细线b 与竖直方向的夹角为53°6．如图所示，电动公交车做匀减速直线运动进站，进站前连续经过R 、S 、T 三点，已知ST 间的距离等于RS 间距离，RS 段的平均速度是12m/s ，ST 段的平均速度是6m/s ，则公交车经过S 点时的瞬时速度为A ．11m/sB ．10m/sC ．9m/sD ．8m/s二、多选题（本题共4个小题，每小题5分，共20分，每小题有多个选项符合题意，全部选对得5分，选不全得3分，选错或不答的得0分）7．自热锅因其便于加热和方便携带，越来越受到户外驴友的欢迎。