17天一大联考卷子
浙江省英语高考试卷2017年及英语联考考试试卷
浙江省英语高考试卷2017年及英语联考考试试卷浙江省英语高考试卷2017年及英语联考考试试卷拼一载春秋，搏一生无悔。
高考加油!下面是店铺为大家推荐的浙江省英语高考试卷2017年，仅供大家参考!英语联考考试试卷第I卷(70分) 听力部分略注意事项：1.答第I卷前，考生务必将自己的姓名、准考证号填写在答题卡上。
2.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。
不能答在本试卷上，否则无效。
第—部分阅读理解(共两节，满分40分)第一节(共15小题;每小题2分，满分30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。
ABill Gates was born in 1955. He grew up in Seattle, Washington. When he was young, he was good at science and maths. And he wanted to be a scientist.Bill started to play with computers when he was 13. At that time, computers were very large machines. Once he was interested in a very old computer. He and some of his friends spent lots of time doing unusual (不寻常的) things with it. In the end, they worked out a software (软件) program with the old machine. Bill sold it for 4200 dollars when he was only 17.In 1973, Bill went to Harvard University (哈佛大学). He developed the Basic language for the first microcomputer (微型电子计算机). In his third year, he left Harvard. Bill began his company in 1975 with his friend Paul Allen. They thought that thecomputer would be a very important tool in every office and in every home, so they began developing software for personal computers. They developed the software to make it easier for people to use computers.In 1999, Bill wrote a book. In the book, he told people how computer technology could solve business problems in new ways. It was one of the best-sellers on The New York Times' list. Bill Gates has many hobbies. He enjoyed reading very much. He also enjoys playing golf and bridge.21. When did Bill begin to play with computer?A. In 1965.B. In 1966.C. In 1968.D. In 1972.22. How old was Bill when he went to Harvard University?A.17 years oldB. 18 years oldC. 19 years oldD. 20 years old23. Bill wrote a book about _____.A. how to do unusual thingsB. how to play computer games.C. how to work out a software programD.how to solve business problems with the help of computers in new ways24. We can learn from the text that _____.A. Bill worked for Microsoft before he went to Harvard UniversityB. Bill sold his first software program for 420 dollarsC. Bill began his company in 1976 with his friendD. reading is one of Bill's hobbiesBShakespeare’s Birthplace and Exhibition of Shakespeare’s WordOPENING TIMES20 Mar to 19 OctMon to Sat: 9:00am to 5:00pmSun: 9:30am to 5:00pm20 Oct to 19 MarMon to Sat: 9:00am to 4:00pmSun: 10:00 am to 4:00pmWelcome to the world-famous house where William Shakespeare was born in 1564 and where he grew up.The property (房产) remained in the ownership of Shakespeare’s family until 1806.The House has welcomed visitors traveling from all over the world, for over 250 years.◆Enter through the Visitors’ Centre and see the highly-praised exhibition Shakespeare’s Word, a lively and full introduction to the life and work of Shakespeare.◆Stand in the rooms where Shakespeare grew up.◆Discover examples of furniture and needle work from Shakespe are’s period.◆Enjoy the traditional (传统的) English garden, planted with trees and flowers mentioned in poet’s works.■The Birthplace is within easy walking distance of all the car parks shown on the map; nearest is Windsor Street(3 minutes’ walk).■The House may be difficulties to wheelchair users. But the Visitors’ Centre, its exhibition, and the garden are accessible (可用的) to wheelchair users.The Shakespeare Coffee House (opposite the Birthplace) .25. How much is the price for a family of two grown-ups and four children?A.￡9.80B. ￡12.00C. ￡14.20D. ￡16.4026.Where is the nearest parking place to Shakespeare’s Birthplace?A. Behind the exhibition hall.B. Opposite the Visitors’ Centre.C. At Windsor Street.D. Near the Coffee House.27. A wheelchair user may need help to enter_____________.A. the houseB. the gardenC. the Visitors’ CentreD. the exhibition hallCIf you do not use your arms or your legs for some time, they become weak; when you start using them again, they slowly become strong again. Everybody knows that. Yet many people do not seem to know that memory works in the same way.When someone says that he has a good memory, he really means that he keeps his memory in practice by using it. When someone else says that his memory is poor, he really means that he does not give it enough chance to become strong. If a friend says that his arms and legs are weak, we know that it is his own fault(过错). But if he tells us that he has a poor memory, many of us think that his parents are to blame(受责备), and few of us know that it is just his own fault.Have you ever found that some people can’t read or write but usually they have better memories? This is because they cannot read or write and they have to remember things; they cannot write down in a little notebook. They have to remember days, names, songs and stories; so their memory is the whole time being exercised. So if you want to have a good memory, learn from the people: Practice remembering.28. The main reason for one’s poor memory is that _______.A. his father or mother may have a poor memoryB. He does not use his name or legs for some timeC. his memory is not often usedD. he can’t read or write29. Which of the following is NOT true?A. Your memory works in the same way as your arms or legs.B. Your memory, like your arms or legs, becomes weak if you don’t give it enoughchance for practice.C. Don’t learn how to read and write if you want to have a better memory.D. A good memory comes from more practice.30. Some people can’t read or write, but they usually have better memories, because ___.A. they have saved much troubleB. they have saved much time to remember thingsC. they have to use their memories all the timeD. they can’t write everything in a little notebook31. Which is the best title for this passage?A. Don’t Stop Using Your Arms Or LegsB. How To Have a Good MemoryC. Strong Arms And Good MemoriesD. Learn From the PeopleDIt’s school time again! You’re probably feeling excited and maybe a little sad for the vacation is over. Some kids feel nervous or a little scared on the first day of school because of all the new things: new teachers, new friends, and maybe even a new school. Luckily, these “new” worries only stick around for a little while. Let’s find out more about go ing back to school.Most teachers kick off the school year by introducing themselves and talking about what you’ll be doing that year. Some teachers give students a chance to tell something aboutthemselves to the rest of the class.When teachers do the talking on the first day, they often go over classroom rules so you’ll know what’s allowed and what’s not. Pay close attention so you’ll know if you need to raise your hand to ask a question and what the rules are about visiting the restroom.You might already know a lot of people in your class on the first day. But it’s a great day to make new friends, so try to say “hello” to the kids you know and new ones that you don’t. Make the first move and you’ll be glad about what you did and so will your new friends!Most teachers let you pick your own seat on the first day, but by the second or third morning, they’ll have mapped out a seating plan. It’s a good idea to write down where your seat is in your notebook so you don’t forget.Here are a few final tips for a fantastic(奇异的) first day: Get enough sleep. Have a healthy breakfast. Try your best. Develop good work habits, like writing down your assignments(作业) and handing in your homework on time. Take your time with schoolwork. If you don’t understand something, ask the teacher.32. In the second paragraph, the underlined phrase ”kick off ” most probably means”_____”.A. jump overB. startC. talk aboutD. finish33. Which of the following is TRUE according to the passage?A. All students feel excited about a new term.B. All teachers let students pick their seats.C. New school worries do not last very long.D. Being hard-working helps make new friends.34. If students have any questions, it’s suggested that they should ________.A. turn to their teachersB. write them downC. ask their parentsD. help with each other35. What’s the author’s purpose in writing this passage?A. To offer students some advice on starting a new school term.B. To encourage students to work harder in a new school term.C. To remind children of the importance of starting school.D. To introduce a way to improve students’ learning ability.第二节(共5小题;每小题2分，满分10分)根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。
贵州省贵阳市第一中学2024-2025学年高三上学期第一次联考(9月月考) 数学试卷[含答案]
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数学试卷注意事项：1.答题前，考生务必用黑色碳素笔将自己的姓名､准考证号､考场号､座位号在答题卡上填写清楚.2.每小题选出答案后，用2B 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､在试题卷上作答无效.3.考试结束后，请将本试卷和答题卡一并交回.满分150分，考试用时120分钟.一､单项选择题（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.已知集合，则（）{}{}2230,1,2,3,4A x x x B =-->=∣A B ⋂=A.B.C.D.{}1,2{}1,2,3{}3,4{}42.下列函数在其定义域内单调递增的是（）A.B.1y x =-2ln y x=C. D.32y x =e xy x =3.已知等差数列满足，则（）{}n a 376432,6a a a a +=-=1a =A.2B.4C.6D.84.已知点是抛物线上一点，若到抛物线焦点的距离为5，且到轴的距离为A ()2:20C y px p =>A A x 4，则（ ）p =A.1或2 B.2或4 C.2或8 D.4或85.已知函数的定义域为.记的定义域为集合的定义域为集合.则“()23f x -[]2,3()f x (),21x A f -B ”是“”的（ ）x A ∈x B ∈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6.已知函数的定义域为.设函数，函数.若是偶函数，()f x R ()()e xg x f x -=+()()5e xh x f x =-()g x 是奇函数，则的最小值为（）()h x ()f x A. B.C.D.e2e7.从的二项展开式中随机取出不同的两项，则这两项的乘积为有理项的概率为（）51x ⎫+⎪⎭A. B. C. D.253513238.已知圆，设其与轴､轴正半轴分别交于，两点.已知另一圆的半径221:220C x y x y +--=x y M N 2C为，且与圆相外切，则的最大值为（）1C22C M C N ⋅A.20B.C.10D.二､多项选择题（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，有多个选项是符合题目要求的，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.离散型随机变量的分布列如下表所示，是非零实数，则下列说法正确的是（ ）X ,m n X 20242025Pm nA. B.服从两点分布1m n +=X C.D.()20242025E X <<()D X mn=10.已知函数，下列说法正确的是（ ）()()214log 21f x ax ax =-+A.的定义域为，当且仅当()f x R 01a <<B.的值域为，当且仅当()f x R 1a C.的最大值为2，当且仅当()f x 1516a =D.有极值，当且仅当()f x 1a <11.设定义在上的可导函数和的导函数分别为和，满足R ()f x ()g x ()f x '()g x '，且为奇函数，则下列说法正确的是（）()()()()11,3g x f x f x g x --=''=+()1g x +A.B.的图象关于直线对称()00f =()g x 2x =C.的一个周期是4 D.()f x 20251()0k g k ==∑三､填空题（本大题共3小题，每小题5分，共15分）12.过点作曲线且的切线，则切点的纵坐标为__________.()0,0(0x y a a =>1)a ≠13.今年暑期旅游旺季，贵州以凉爽的气候条件和丰富的旅游资源为依托，吸引了各地游客前来游玩.由安顺黄果树瀑布､荔波小七孔､西江千户苗寨､赤水丹霞､兴义万峰林､铜仁梵净山6个景点谐音组成了贵州文旅的拳头产品“黄小西吃晚饭”.小明和家人计划游览以上6个景点，若铜仁梵净山不安排在首末位置，且荔波小七孔和西江千户苗寨安排在相邻位置，则一共有__________种不同的游览顺序方案.（用数字作答）14.已知函数若存在实数且，使得，()223,0,ln ,0,x x x f x x x ⎧++=⎨>⎩ 123,,x x x 123x x x <<()()()123f x f x f x ==则的最大值为__________.()()()112233x f x x f x x f x ++四､解答题（共77分.解答应写出文字说明，证明过程或演算步骤）15.（本小题满分13分）下图中的一系列三角形图案称为谢尔宾斯基三角形.图（1）是一个面积为1的实心正三角形，分别连接这个正三角形三边的中点，将原三角形分成4个小正三角形，并去掉中间的小正三角形得到图（2），再对图（2）中的每个实心小正三角形重复以上操作得到图（3），再对图（3）中的每个实心小正三角形重复以上操作得到图（4），…，依此类推得到个图形.记第个图形中实心三角形的个数为，第n 个图形n n n a 中实心区域的面积为.nb （1）写出数列和的通项公式；{}n a {}n b （2）设，证明.121121n n n n n c a b a b a b a b --=++++ 43n n n a c a <16.（本小题满分15分）如图，在三棱台中，和都为等腰直角三角形，111A B C ABC -111A B C ABC 为线段的中点，为线段上的点.111112,4,90,CC C A CA ACC BCC CBA G ∠∠∠====== AC HBC （1）若点为线段的中点，求证：平面；H BC 1A B ∥1C GH （2）若平面分三棱台所成两部分几何体的体积比为，求二面角1C GH 111A B C ABC -2:5的正弦值.11C GH B --17.（本小题满分15分）已知双曲线与双曲线的离心率相同，且经过点()2222:10,0x y M a b a b -=>>2222:12x y N m m -=M 的焦距为.()2,2,N （1）分别求和的方程；M N （2）已知直线与的左､右两支相交于点，与的左､右两支相交于点，D ，，判断l M ,A B N C ABCD=直线与圆的位置关系.l 222:O x y a +=18.（本小题满分17分）为了检测某种抗病毒疫苗的免疫效果，需要进行动物与人体试验.研究人员将疫苗注射到200只小白鼠体内，一段时间后测量小白鼠的某项指标值，按分组，绘制频率分[)[)[)[)[]0,20,20,40,40,60,60,80,80,100布直方图如图所示.试验发现小白鼠体内产生抗体的共有160只，其中该项指标值不小于60的有110只.假设小白鼠注射疫苗后是否产生抗体相互独立.（1）填写下面的列联表，并根据列联表及的独立性检验，判断能否认为注射疫苗后小白鼠22⨯0.01α=产生抗体与指标值不小于60有关；单位：只指标值抗体小于60不小于60合计有抗体没有抗体合计（2）为检验疫苗二次接种的免疫抗体性，对第一次注射疫苗后没有产生抗体的40只小白鼠进行第二次注射疫苗，结果又有20只小白鼠产生抗体.（i ）用频率估计概率，求一只小白鼠注射2次疫苗后产生抗体的概率；P （ii ）以（i ）中确定的概率作为人体注射2次疫苗后产生抗体的概率，进行人体接种试验，记100个人P 注射2次疫苗后产生抗体的数量为随机变量.求及取最大值时的值.X ()E X ()P X k =k参考公式：（其中为样本容量）()()()()22()n ad bc a b c d a c b d χ-=++++n a b c d =+++参考数据：α0.1000.0500.0100.005x α2.7063.8416.6357.87919.（本小题满分17分）三角函数是解决数学问题的重要工具.三倍角公式是三角学中的重要公式之一，某数学学习小组研究得到了以下的三倍角公式：①；②.3sin33sin 4sinθθθ=-3cos34cos 3cos θθθ=-根据以上研究结论，回答：（1）在①和②中任选一个进行证明；（2）已知函数有三个零点且.()323f x x ax a =-+123,,x x x 123x x x <<（i ）求的取值范围；a （ii ）若，证明：.1231x x x =-222113x x x x -=-贵阳第一中学2025届高考适应性月考卷（一）数学参考答案一､单项选择题（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）题号12345678答案DCBCBCAA【解析】1.由题，或，则，故选D.{1A xx =<-∣{}3},1,2,3,4x B >={}4A B ⋂=2.对于A 选项，的定义域为，该函数在和上单调递增，在定义1y x =-()(),00,∞∞-⋃+(),0∞-()0,∞+域内不单调；对于B 选项，的定义域为，该函数在上单调递减，在2ln y x =()(),00,∞∞-⋃+(),0∞-上单调递增，在定义域内不单调；对于C 选项，的定义域为，该函数在定()0,∞+32y x==[)0,∞+义域上单调递增；对于D 选项，的定义域为，当时，；当e x y x =().1e xy x =+'R (),1x ∞∈--0y '<时，，在上单调递减，在上单调递增，因此该函数在定()1,x ∞∈-+0y '>xe y x ∴=(),1∞--()1,∞-+义域内不单调，故选C.3.，故选B.53756415232,16,26,3,44a a a a d a a d a a d =+===-===-= 4.设点，则整理得，解得或，故选C.()00,A x y 200002,5,24,y px p x y ⎧=⎪⎪+=⎨⎪=⎪⎩582p p ⎛⎫-= ⎪⎝⎭2p =8p =5.的定义域为.当时，的定义域为，()23f x - []2,323x ()1233,x f x -∴ []1,3即.令，解得的定义域为，即.[]1,3A =1213x- ()12,21x x f ∴- []1,2[]1,2B =“”是“”的必要不充分条件，故选B.,B A ⊆∴ x A ∈x B ∈6.由题，解得，所以()()()()()()()(),e e ,5e 5e ,x xx xg x g x f x f x h x h x f x f x --⎧⎧=-+=-+⎪⎪⇒⎨⎨=---=--+⎪⎪⎩⎩()3e 2e x x f x -=+，当且仅当，即时，等号成立，()3e2e xxf x -=+3e 2e x x -=12ln 23x =C.min ()f x ∴=7.设的二项展开式的通项公式为，51x ⎫+⎪⎭53521551C C ,0,1,2kkk k kk T x k x --+⎛⎫=== ⎪⎝⎭，所以二项展开式共6项.当时的项为无理项；当时的项为有理项.两项乘积为有3,4,50,2,4k =1,3,5k =理数当且仅当此两项同时为无理项或同时为有理项，故其概率为，故选A.223326C C 2C 5+=8.由题，，即圆心为，且，为的221:(1)(1)2C x y -+-=()11,1C()()2,0,0,2M N MN 1C 直径.与相外切，由中线关系，有1C 2C 12C C ∴==，当且()()2222222222121222218240,202C M C NC M C N C C C MC M C N ++=+=⨯+=∴⋅=仅当时，等号成立，所以的最大值为20，故选A.22C M C N=22C M C N⋅二､多项选择题（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，选对但不全的得部分分，有选错的得0分）题号91011答案ACDBCBCD【解析】9.对于A 选项，由分布列性质可知正确；对于B 选项，由两点分布定义可知错误；对于C 选项，，正确；()()()202420252024120252024.01,20242025E X m n n n n n E X =+=-+=+<<∴<< 对于D 选项，令，则服从两点分布，，2024Y X =-Y ()()1D Y n n mn=-=，正确，故选ACD.()()()2024D X D Y D Y mn∴=+==10.令，对于A 选项，的定义域为或()2221,Δ44g x ax ax a a =-+=-()f x 0a ⇔=R ，故A 错误；对于B 选项，的值域为在定义域内的值域为0,01Δ0a a >⎧⇔<⎨<⎩ ()f x ()g x ⇔R ，故B 正确；对于C 选项，的最大值为在定义域内的最小值()0,0,1Δ0a a ∞>⎧+⇔⇔⎨⎩ ()f x ()2g x ⇔为，故C 正确；对于D 选项，有极值在定义域内有极值()0,11511616116a a g >⎧⎪⇔⇔=⎨=⎪⎩()f x ()g x ⇔且，故D 选项错误，故选BC.()0,110a a g ≠⎧⇔⇔<⎨>⎩0a ≠11.对于A 选项，因为为奇函数，所以，又由，可得()1g x +()10g =()()11g x f x --=，故A 错误；对于B 选项，由可得()()()101,01g f f -==-()()3f x g x '=+'为常数，又由，可得，则()()3,f x g x C C=++()()11g x f x --=()()11g x f x --=，令，得，所以，所以()()131g x g x C --+-=1x =-()()221g g C --=1C =-的图象关于直线对称，故B 正确；对于C 选项，因为为奇函数，()()()13,g x g x g x -=+2x =()1g x +所以，所以，所以()()()311g x g x g x +=-=-+()()()()()2,42g x g x g x g x g x +=-+=-+=是一个周期为4的周期函数，，()g x ()()()()()()31,47131f x g x f x g x g x f x =+-+=+-=+-=所以也是一个周期为4的周期函数，故C 正确；对于D 选项，因为为奇函数，所以()f x ()1g x +，又，又是周期为4的周期函数，所以()()()()10,204g g g g ==-=-()()310g g ==()g x ，故D 正确，故选BCD.20251()(1)0k g k g ===∑三､填空题（本大题共3小题，每小题5分，共15分）题号121314答案e14433e 6-【解析】12.设切点坐标为切线方程为.将代入得，可得(),,ln ,txt a y a a ='∴ ln xy a a x =⋅(),tt a ln t ta a t a ⋅=切点纵坐标为.1log e,ln a t a ==∴elog e t a a a==13.先对小七孔和千户苗寨两个相邻元素捆绑共有种方法，再安排梵净山的位置共有种方法，再排其22A 13C 余元素共有种排法，故共有种不同的方案.44A 214234A C A 144⋅⋅=14.设，由的函数图象知，，又，()()()123f x f x f x t===()f x 23t < 1232,ln x x x t +=-=.令()()()3112233e ,2e t tx x f x x f x x f x t t =∴++=-+在上单调递增，则()()()()2e ,23,1e 20,t t t t t t t t t ϕϕϕ'=-+<=+->∴ (]2,3，的最大值为.()3max ()33e 6t ϕϕ==-()()()112233x f x x f x x f x ∴++33e 6-四､解答题（共77分.解答应写出文字说明，证明过程或演算步骤）15.（本小题满分13分）（1）解：数列是首项为1，公比为3的等比数列，因此；{}n a 11133n n n a --=⨯=数列是首项为1，公比为的等比数列，因此，.{}n b 341133144n n n b --⎛⎫⎛⎫=⨯= ⎪⎪⎝⎭⎝⎭（2）证明：由（1）可得1210121121121333333334444n n n n n n n n n c a b a b a b a b ------⎛⎫⎛⎫⎛⎫⎛⎫=++++=⋅+⋅++⋅+⋅ ⎪⎪⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭12101111134444n n n ---⎡⎤⎛⎫⎛⎫⎛⎫⎛⎫=++++⎢⎥⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎢⎥⎣⎦121114134311414n nn n --⎡⎤⎛⎫⋅-⎢⎥ ⎪⎡⎤⎝⎭⎢⎥⎛⎫⎣⎦=⋅=⋅⋅-⎢⎥⎪⎝⎭⎢⎥⎣⎦-因为，2114314411334n n n nn nc a --⎡⎤⎛⎫⋅⋅-⎢⎥ ⎪⎡⎤⎝⎭⎢⎥⎛⎫⎣⎦==-⎢⎥⎪⎝⎭⎢⎥⎣⎦所以，所以.413n n c a <43n n na c a < 16.（本小题满分15分）（1）证明：如图1，连接，设，连接，1A C 11A C C G O⋂=1,HO A G三棱台，则，又，111A B C ABC -11A C ∥AC 122CG AC ==四边形为平行四边形，∴11A C CG 则.1CO OA =点是的中点，H BC .1BA ∴∥OH 又平面平面，OH ⊂11,C HG A B ⊄1C HG 平面.1A B ∴∥1C HG （2）解：因为平面分三棱台所成两部分几何体的体积比为，1C GH 111A B C ABC -2:5所以，11127C GHC AB V V B C ABC-=-即，()1111121373GHC ABC AB C S CC S S CC ⋅⋅=⋅⋅++⋅ 化简得，12GHC ABC S S =此时点与点重合.H B ，1190C CA BCC ∠∠== 且都在平面，则平面，11,,C C BC CC AC BC AC C ∴⊥⊥⋂=ABC 1CC ⊥ABC 又为等腰直角三角形，则.ABC BG AC ⊥又由（1）知，则平面，1A G ∥1CC 1A G ⊥ABC 建立如图2所示的坐标系,G xyz -则，()()()()2,0,0,0,2,0,0,0,0,0,2,0H A G C -()()110,2,2,1,1,2C B --设平面的法向量，1C HG ()()()1,,,0,2,2,2,0,0n x y z GC GH ==-= 则令，解得，220,20,y z x -+=⎧⎨=⎩1y =()0,1,1n = 设平面的法向量，1B GH ()()1,,,1,1,2m a b c GB ==- 则令，解得.20,20,a b c a -+=⎧⎨=⎩2b =()0,2,1m = 设二面角的平面角为，11C GH B --θ，cos cos ,m n m n m n θ⋅=<>=== 所以，sin θ==所以二面角.11C GH B --17.（本小题满分15分）解：（1）由题意可知双曲线的焦距为N =解得，即双曲线.21m =22:12y N x -=因为双曲线与双曲线的离心率相同，M N 不妨设双曲线的方程为，M 222y x λ-=因为双曲线经过点，所以，解得，M ()2,242λ-=2λ=则双曲线的方程为.M 22124x y -=（2）易知直线的斜率存在，不妨设直线的方程为l l ，()()()()11223344,,,,,,,,y kx t A x y B x y C x y D x y =+联立消去并整理得22,,2y kx t y x λ=+⎧⎪⎨-=⎪⎩y ()2222220,k x ktx t λ----=此时可得，()()222222Δ44220,20,2k t k t t k λλ⎧=+-+>⎪⎨--<⎪-⎩22k <当时，由韦达定理得；2λ=212122224,22kt t x x x x k k --+==--当时，由韦达定理得，1λ=234342222,22kt t x x x x k k --+==--则，ABCD====化简可得，222t k +=由（1）可知圆，22:2O x y +=则圆心到直线的距离，Ol d ====所以直线与圆相切或相交.l O 18.（本小题满分17分）解：（1）由频率分布直方图知，200只小白鼠按指标值分布为：在内有（只）；[)0,200.00252020010⨯⨯=在）内有（只）；[20,400.006252020025⨯⨯=在）内有（只）；[40,600.008752020035⨯⨯=在）内有（只）；[60,800.025********⨯⨯=在内有（只）[]80,1000.00752020030⨯⨯=由题意，有抗体且指标值小于60的有50只；而指标值小于60的小白鼠共有（只），10253570++=所以指标值小于60且没有抗体的小白鼠有20只，同理，指标值不小于60且没有抗体的小白鼠有20只，故列联表如下：单位：只指标值抗体小于60不小于60合计有抗体50110160没有抗体202040合计70130200零假设为：注射疫苗后小白鼠产生抗体与指标值不小于60无关联.0H 根据列联表中数据，得.220.01200(502020110) 4.945 6.6351604070130x χ⨯⨯-⨯=≈<=⨯⨯⨯根据的独立性检验，没有充分证据认为注射疫苗后小白鼠产生抗体与指标值不小于60有关.0.01α=（2）（i ）令事件“小白鼠第一次注射疫苗产生抗体”，事件“小白鼠第二次注射疫苗产生抗体”A =B =，事件“小白鼠注射2次疫苗后产生抗体”.C =记事件发生的概率分别为，则，,,A B C ()()(),,P A P B P C ()()160200.8,0.520040P A P B ====.()1P C =-()()10.20.50.9P A P B =-⨯=所以一只小白鼠注射2次疫苗后产生抗体的概率.0.9P =（ii ）由题意，知随机变量，()100,0.9X B ~所以.()1000.990E X np ==⨯=又，设时，最大，()()C 0.90.10,1,2,,k k n k n P X k k n -==⨯⨯= 0k k =()P X k =所以00000000000010011910010010011101100100C 0.90.1C 0.90.1,C 0.90.1C 0.90.1,k k k k k k k k k k k k -++-----⎧⨯⨯≥⨯⨯⎪⎨⨯⨯≥⨯⨯⎪⎩解得，因为是整数，所以.089.990.9k 0k 090k =19.（本小题满分17分）（1）若选①，证明如下：()()22sin3sin 2sin2cos cos2sin 2sin cos 12sin sin θθθθθθθθθθθ=+=+=+-()()2232sin 1sin 12sin sin 3sin 4sin θθθθθθ=-+-=-若选②，证明如下：()()22cos3cos 2cos2cos sin2sin 2cos 1cos 2sin cos θθθθθθθθθθθ=+=-=--.()3232cos cos 21cos cos 4cos 3cos θθθθθθ=---=-（2）（i ）解：，()233f x x a =-'当时，恒成立，所以在上单调递增，至多有一个零点；0a ()0f x ' ()f x (),∞∞-+当时，令，得；令，得0a >()0f x '=x =()0f x '<x <<令，得()0f x '>x <x>所以在上单调递减，在上单调递增.()f x ((),,∞∞-+有三个零点，则即解得，()fx (0,0,f f ⎧>⎪⎨<⎪⎩2220,20,a a ⎧+>⎪⎨-<⎪⎩04a <<当时，，04a <<4a +>且，()()()()32224(4)3445160f a a a a a a a a a+=+-++=++++>所以在上有唯一一个零点，()fx )4a +同理()2220,g a -<-=-=-<所以在上有唯一一个零点.()f x (-又在上有唯一一个零点，所以有三个零点，()f x (()f x 综上可知的取值范围为.a ()0,4（ii ）证明：设，()()()()321233f x x ax a x x x x x x =-+=---则.()212301f a x x x ==-=又，所以.04a <<1a =此时，()()()()210,130,110,230f f f f -=-<-=>=-<=>方程的三个根均在内，3310x x -+=()2,2-方程变形为，3310x x -+=3134222x x ⎛⎫=⋅-⋅ ⎪⎝⎭令，则由三倍角公式.ππsin 222x θθ⎛⎫=-<< ⎪⎝⎭31sin33sin 4sin 2θθθ=-=因为，所以.3π3π3,22θ⎛⎫∈- ⎪⎝⎭7ππ5π7ππ5π3,,,,,666181818θθ=-=-因为，所以，123x x x <<1237ππ5π2sin ,2sin ,2sin 181818x x x =-==所以222221π7ππ7π4sin 4sin 21cos 21cos 181899x x ⎛⎫⎛⎫-=-=--- ⎪ ⎪⎝⎭⎝⎭137ππ5π7π2cos 2cos 2sin 2sin 991818x x =-=--=-。
山东省潍坊市大联考2024-2025学年高三上学期10月月考英语试题
山东省潍坊市大联考2024-2025学年高三上学期10月月考英语试题一、阅读理解Mathematics for Computer ScienceThis subject offers an introduction to discrete mathematics(离散数学)oriented toward computer science and engineering.Course Meeting TimesLectures:3 sessions/week, 1. 5 hour/session Problem Sets (psets)Problem sets account for 20% of the final grade. Making a reasonable effort on the problem sets is, for most students, crucial for mastering the course material. Problem sets are designed to be completed in at most 3 hours; the time is monitored through student reports. Online Feedback ProblemsOnline problems to be completed before most class meetings are posted on the class website. These consist of straightforward questions that provide useful feedback about the assigned material. Some students prefer to try the online problems before reading the text or watching videos as an advance guide to going over the material; that’s fine. Watching designated videos, or at least looking at the lecture-slide handouts, is generally helpful but optional.Like team problem-solving in class, online problems are graded only on participation: Students receive full credit as long as they try the problem, even if their answer is wrong. Online feedback problems account for 10% of the final grade. Midterm ExamsThree 80-minute midterm exams will be given. The midterm exams each account for 15%of the final grade.Midterm questions will typically be variations of prior problems from class and psets, and the best way to prepare is to review on the published solutions to these problems. The first exam covers all previous weeks’ material;subsequent exams focus on the material after the previous exam. Final ExamThere will be a three-hour final exam. This exam is worth 25% of the final class grade. The final exam will cover the entire subject with somewhat greater emphasis on material from after Midterm 3. Most exam questions will be variants of problems assigned during the term(psets,class, midterm, and online). It may include a few questions which combine topics that were originally covered separately.1．What determines students’ grade in Online Feedback Problems?A．Active involvement.B．Submission time.C．Accuracy of answers.D．Completion of assignments2．What is the main focus of the final exam?A．Variants of problem sets.B．Combined topics in class.C．The content after midterm exams.D．The whole-term course materials. 3．Which of the following weighs the most in the final grade?A．Final Exam.B．Midterm Exams.C．Problem Sets.D．Online Feedback ProblemsIt all started with a simple question;“Can I paint your portrait (肖像)?”In the summer of 2015, Brian Peterson was reading the book Love Does, about the power of love in action, when his quiet was disturbed by a homeless man. Inspired by the book, Peterson made an unexpected decision: He was going to introduce himself. In that first conversation, Peterson learned that the man’s name was Matt Faris who failed to pursue a career in music and ended up being homeless.“I saw beauty on the face of a man who hadn’t shaved in probably a year, had overgrown fingernails, and probably hadn’t had a shower in close to a year. ”Even though Peterson hadn’t pioked up a paintbrush in about eight years, he asked if he could paint Faris’s portrait. Faris said yes.Peterson’s connection with Faris led him to form Faces of Santa Ana, a nonprofit organization focused on befriending and painting portraits of members of the community who are unhoused. Working from a black-and-white photo of the subject, Peterson chooses colors inspired by the subject’s personality and life story, creating an impressive portrait.Peterson sells the striking artwork, signed by both subject and artist, dividing the proceeds and putting half into what he calls a“love account”for his model. He then helps people use the money to get back on their feet. Many of Peterson’s new friends use the donations to secure immediate necessities. But Peterson has learned not to make assumptions about what a personneeds most. “I’ve made a mistake thinking I knew what people wanted, ” he says, “but why don’t we just ask them?”Peterson has discovered that there’s more to the finished products than the money they bring to someone who’s down and out. The buyers tend to connect to the story of the person in the painting, finding similarities and often friendship with someone they might have otherwise overlooked. “People often tell me, ‘I was the one that would cross the street. But I see homeless people differently now, ’ ”Peterson says.4．What brought Peterson and Faris together?A．Beauty on Faris’ face.B．Inspiration from a book.C．Peterson’s passion for art.D．Faris’ suffering in his life.5．What does the underlined word “mistake” in paragraph 5 refer to?A．Selling the homeless’ portraits for profits.B．Giving instant necessities to the homeless.C．Asking the homeless for their needs directly.D．Taking what the homeless want for granted.6．What does Peterson imply in the last paragraph?A．The homeless are gaining more concern.B．The life of the homeless is different now.C．Buyers value friendship with the homeless.D．Figures in Peterson’s paintings are popular.7．What can we learn from this text?A．A good model is key to a fine artwork.B．An expected decision makes a great artist.C．A picture really is worth a thousand words.D．Each unfortunate person has his own misfortune.The Malagasy baobab tree, whose thick trunks and tiny branches dot Madagascar’s landscape, should not, by rights, have survived to the present day. Scientists believe that its large seeds were once spread by the giant tortoises and lemur monkeys that wandered the island. When these species went extinct over one thousand years ago owing to human activity, the baobab treeshould have disappeared too. It did not. Seheno Andriantsaralaza at the University of Antananarivo and Onja Razafindratsima at the University of California, now think they may know the reason why.Together with their colleagues, the scientists monitored 15 tree canopies (树冠) in a western region of Madagascar, to identify any animals that might have claimed the role of baobab-seed spread. The researchers also set up camera traps around seed-containing fruits lying on the ground, and searched any faeces (粪便) that they encountered along the way for the presence of seeds.They report in the journal Biotropica that a native rodent (啮齿动物) known as the western bunch-tailed rat was caught on camera handling whole fruits on four occasions. Although there was no footage of the rat breaking the fruits open, the team did chance upon 13 fruits that had been chewed into and had their seeds removed. Though the bite marks were not clear enough to identify an initiator, this was clear evidence that a seed- distributing animal was out there. They then found the ecological equivalent of a smoking gun:baobab seeds in seven different piles of bush-pig faces.While the finding is important in its own right, it also provides valuable evidence that introduced species may not be entirely harmful. Madagascar’s pigs, for example, though not native, have made themselves essential to the survival of truly native species. Similar relationships are suspected to hold in South America between rabbits native to Europe and plants with no seed distributors. For Dr Andriantsaralaza, that suggests the full ecological role of introduced species should be considered before talk of extinction begins.8．What made scientists think the baobab tree should have disappeared?A．The extinction of its seed spreader.B．The destruction of human activitiesC．The inadaptation of the baobab tree.D．The increase of its natural enemies. 9．What are camera traps used for?A．Seeking seed-containing fruits.B．Tracking the footprints of seeds.C．Recording the animals’ activities.D．Monitoring the baobab tree canopies. 10．Which is most probably the seed-eater according to the report?A．The giant tortoise B．The bunch-tailed rat.C．The European rabbit.D．The bush-pig.11．What is the author’s purpose in writing the text?A．To introduce a new way to protect the baobab tree.B．To show non-native species are not always harmful.C．To compare different kinds of seed-distributing methods.D．To explain the importance of protecting endangered species.The concepts of delayed satisfaction, self-control, and self-regulation are often used interchangeably and inconsistently. The ability to delay an impulse (冲动) for an immediate reward to receive a more favorable reward at a later time is the standard definition of delayed satisfaction.Studies show that delayed satisfaction is one of the most effective personal characteristics of successful people. People who learn how to manage their need to be satisfied in the moment develop more in their careers, relationships, health, and finances than people who give in to it.Being able to delay satisfaction isn’t the easiest skill to acquire. It involves feeling dissatisfied, which is why it seems impossible for people who haven’t learned to control their impulses. Choosing to have something now might feel good, but making the effort to have discipline and manage your impulses can result in bigger or better rewards in the future. Over time, delayed satisfaction will improve your self-control and ultimately help you achieve your long-term goals faster.The Seinfeld Strategy is one of several helpful self-satisfaction techniques you can use to put off satisfaction for longer periods of time. Every day that you delay satisfaction and avoid temptation (诱惑) , you cross it off your calendar. After a few days, this creates a chain. This strategy works well for people who enjoy gamification (游戏化). If you find it satisfying to keep the chain going, you’re less likely to give in to temptation.Do you find yourself going back to your temptations without thinking about it?If this is the case, you can practice mindfulness to become more aware of what you do. When you notice yourself doing something out of habit, stop for a moment. Ask yourself why you’re doing what you’re doing. Take some time to analyze how you’re feeling. Pay attention to the details. Take a moment of mindulness to interrupt your autopilot every time this happens. The more you practice this, the more you’ll break the habit of going for instant satisfaction.12．What is delayed satisfaction?A．Making quick decisions.B．Giving in to desires instantly.C．Immediate rewards for impulses.D．Postponing rewards for better outcomes. 13．Why is delayed satisfaction hard to attain?A．It can lead to missed opportunities B．It’s bad for achieving long-term goals.C．It’s impossible to resist the inner needs D．It’s a comfort to possess something at once. 14．What does paragraph 4 mainly talk about?A．The complexity of the Seinfeld Strategy B．An example of self-discipline techniques.C．A method of practicing delayed satisfaction.D．The effectiveness of keeping the chain going.15．What does the author suggest people do for returning temptations?A．Ignore useless details.B．Get rid of old bad habits.C．Concentrate on true inner wants.D．Cancel temptations from the calendar.Facing hard things is, well, hard. Sometimes we are forced to simply shift, such as with a loss or failure; however, much of the time, we may recognize the difficulties underneath that need attention but feel too overwhelmed to address them. 16 The difficulties might be an outdated habit, a troubling memory or a long- ignored conflict.Some individuals fool themselves into thinking there really is not a problem. It’s like someone trying to hold active mice under a blanket by holding down the edges of the moving blanket. Pulling back the blanket to let the mice out is needed, despite being terrifying. Once the mice are released, there may be further challenges to get them out of the house. 17 People often encounter similar problems. Avoiding or leaving the situation appears to fix the problem but fails to tackle the underlying cause. Rather than repeatedly cutting weeds, getting down on the ground and pulling the roots is more effective. Facing issues is like uncovering roots for new beginnings to blossom. 1819 It feels like a balled-up mess of wires. Pulling hard at only one wire actually tightens the knot. We have to take a calmer look and pull apart each wire one at a time. We can’t expect the wires to loosen or unwind themselves. 20 Nevertheless, once they have been freed, they can sometimes be separated, put aside, and used as needed. A．They are just there and part of the chaos.B．And humans are masters of avoidance and denial.C．The overall benefits from tough work enable growthD．Making changes in one’s life is hard and complicated.E．Sadly, it often worsens when we approach it with anger.F．However, problems cannot be faced if they are not first recognized.G．Quick fixes might provide relief in the short term but often not in the long term.二、完形填空Madeline sat on her bed and tried to write. Tears dropped onto the page, making the ink 21 . Her best friend was moving, and her heart 22 as she penned how much she would miss him.She heard her mother speaking downstairs, but the words were 23 . This was a 24 , as her mother said, “Kids her age don’t know what love is!”Uncle Joe responded in a low tone. 25 , Madeline couldn’t catch what was said. She continued to write until she reached the end and 26 , she laid her head on the pillow and sobbed. Then, she felt a 27 hand on her shoulder. “It’s going to be okay. You can 28 each other, right?” Madeline 29 her head. “Mom says I’m not old enough for a phone.”Uncle Joe answered 30 , “But I’m friends with Leo’s dad. Your mom said you can talk to Leo on my phone. Plus, I gave her his number.” He didn’t tell her that her mom had thought it all silly and unnecessary.Madeline 31 , throwing her arms around Uncle Joe’s neck. “Thank you!”“No problem.” His eyes fell on the 32 . “Want me to ensure this gets to Leo?” Madeline nodded “Feel like playing a game? I promise I won’t mess around.”Madeline smiled — small but 33 .Uncle Joe wouldn’t read the letter; she 34 him. A sure thing was that his sister was mistaken. Though only eight, Madeline and Leo shared a pure, innocent bond — untouched by 35 .21．A．emerge B．bleed C．leak D．escape 22．A．raced B．hesitated C．ached D．melted23．A．indistinct B．sharp C．pale D．impolite 24．A．burden B．warning C．bonus D．blessing 25．A．Instead B．Anyway C．Again D．Moreover 26．A．worried B．confused C．bored D．exhausted 27．A．comforting B．firm C．smooth D．trembling 28．A．greet B．call C．miss D．visit 29．A．dropped B．shook C．covered D．touched 30．A．proudly B．slowly C．quietly D．casually 31．A．sat up B．turned over C．looked up D．bent down 32．A．toy B．phone C．note D．pen 33．A．friendly B．genuine C．unique D．tight 34．A．refused B．begged C．respected D．trusted 35．A．romance B．doubt C．mood D．status三、语法填空阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。
炎德英才大联考长沙市一中2025届高三月考试卷(三)数学答案
时量：120分钟满分：150分一、选择题（本大题共8个小题，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炎德英才大联考长沙市一中2025届高三月考试卷（三）数学）1.若复数z 满足1i34i z +=-，则z =（）A.5B.25C.5D.2【答案】C 【解析】【分析】根据复数的除法运算求出复数z ，计算其模，即得答案.【详解】由1i34i z+=-可得()()()()1i 34i 1i 17i 34i 34i 34i 25z +++-+===--+，则25z =,故选：C2.已知数列{}n a 的前n 项和2n S n n =-，则345a a a ++等于（）A.12B.15C.18D.21【答案】B 【解析】【分析】利用52S S -即可求得345a a a ++的值.【详解】因为数列{}n a 的前n 项和22n S n n =-，所以34552=a a a S S ++-()2252522215=-⨯--⨯=.故选：B.3.抛物线24y x =的焦点坐标为（）A.(1,0)B.(1,0)-C.1(0,16-D.1(0,)16【答案】D 【解析】【分析】先将抛物线方程化为标准方程，从而可求出其焦点坐标【详解】解：由24y x =，得214x y =，所以抛物线的焦点在y 轴的正半轴上，且124p =，所以18p =，1216p =，所以焦点坐标为1(0,16，故选：D4.如图是函数()sin y x ωϕ=+的部分图象，则函数的解析式可为（）A.πsin 23y x ⎛⎫=-⎪⎝⎭B.πsin 3y x ⎛⎫=+⎪⎝⎭C.πsin 26y x ⎛⎫=+ ⎪⎝⎭D.5πcos 26y x ⎛⎫=-⎪⎝⎭【答案】A 【解析】【分析】观察图象，确定函数()sin y x ωϕ=+的周期，排除B ，由图象可得当5π12x =时，函数取最小值，求ϕ由此判断AC ，结合诱导公式判断D.【详解】观察图象可得函数()sin y x ωϕ=+的最小正周期为2ππ2π36T ⎛⎫=-=⎪⎝⎭，所以2ππω=，故2ω=或2ω=-，排除B ；观察图象可得当π2π5π63212x +==时，函数取最小值，当2ω=时，可得5π3π22π+122k ϕ⨯+=，Z k ∈，所以2π2π+3k ϕ=，Z k ∈，排除C ；当2ω=-时，可得5ππ22π122k ϕ-⨯+=-，Z k ∈，所以π2π+3k ϕ=，Z k ∈，取0k =可得，π3ϕ=，故函数的解析式可能为πsin 23y x ⎛⎫=-⎪⎝⎭，A 正确；5ππππcos 2cos 2sin 26233y x x x ⎛⎫⎛⎫⎛⎫=-=+-=-- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭，D 错误故选：A.5.1903年，火箭专家、航天之父康斯坦丁・齐奥尔科夫斯基就提出单级火箭在不考虑空气阻力和地球引力的理想情况下的最大速度v 满足公式：1201lnm m v v m +=，其中12,m m 分别为火箭结构质量和推进剂的质量，0v 是发动机的喷气速度.已知某单级火箭结构质量是推进剂质量的2倍，火箭的最大速度为8km /s ，则火箭发动机的喷气速度为（）（参考数据：ln20.7≈，ln3 1.1,ln4 1.4≈≈）A.10km /sB.20km /sC.80km /s 3D.40km /s【答案】B 【解析】【分析】根据实际问题，运用对数运算可得.【详解】由题意122m m =，122200122lnln 82m m m m v v v m m ++===，得03ln 82v =，故0888203ln3ln 2 1.10.7ln 2v ==≈=--，故选：B6.若83cos 5αβ=，63sin 5αβ=，则()cos αβ+的值为（）A.4-B.4C.4-D.4【答案】C 【解析】【分析】已知两式平方相加，再由两角和的余弦公式变形可得．【详解】因为83cos 5αβ+=，63sin 5αβ-=，所以25(3cos 4)62αβ=，2(3sin )2536αβ=，即所以2259cos co 6s 1042cos ααββ++=，229sin sin +10sin 2536ααββ-=，两式相加得9)104αβ+++=，所以10cos()4αβ+=-，故选：C ．7.如图，一个质点从原点O 出发，每隔一秒随机向左或向右移动一个单位长度，向左的概率为23，向右的概率为13，共移动4次，则该质点共两次到达1的位置的概率为（）A.427B.827C.29D.49【答案】A 【解析】【分析】根据该质点共两次到达1的位置的方式有0101→→→和0121→→→，且两种方式第4次移动向左向右均可以求解.【详解】共移动4次，该质点共两次到达1的位置的方式有0101→→→和0121→→→，且两种方式第4次移动向左向右均可以，所以该质点共两次到达1的位置的概率为211124333332713⨯⨯+⨯⨯=.故选：A.8.设n S 为数列的前n 项和，若121++=+n n a a n ，且存在*N k ∈，1210k k S S +==，则1a 的取值集合为（）A.{}20,21-B.{}20,20-C.{}29,11- D.{}20,19-【答案】A 【解析】【分析】利用121++=+n n a a n 可证明得数列{}21n a -和{}2n a 都是公差为2的等差数列，再可求得()2=21n S n n +，有了这些信息，就可以从k 的取值分析并求解出结果.【详解】因为121++=+n n a a n ，所以()()()()()()212342123+41=++++++37+41=212n n n n nS a a a a a a n n n --⋅⋅⋅=++⋅⋅⋅-=+，假设()2=21=210n S n n +，解得=10n 或21=2n -（舍去），由存在*N k ∈，1210k k S S +==，所以有19k =或20k =，由121++=+n n a a n 可得，+1223n n a a n ++=+，两式相减得：22n n a a +-=，当20k =时，有2021210S S ==，即210a =，根据22n n a a +-=可知：数列奇数项是等差数列，公差为2，所以()211+11120a a =-⨯=，解得120a =-，当19k =时，有1920210S S ==，即200a =，根据22n n a a +-=可知：数列偶数项也是等差数列，公差为2，所以()202+10120a a =-⨯=，解得218a =-，由已知得123a a +=，所以121a =.故选：A.二、选择题（本大题共3小题，每小题6分，共18分.在每小题给出的选项中，至少有两项是符合题目要求，若全部选对得6分，部分选对得部分分，选错或不选得0分）9.如图，在正方体1111ABCD A B C D -中，点E ，F 分别为1AD ，DB 的中点，则下列说法正确的是（）A.直线EF 与11D B 为异面直线B.直线1D E 与1DC 所成的角为60oC.1D F AD ⊥D.//EF 平面11CDD C 【答案】ABD 【解析】【分析】直接根据异面直线及其所成角的概念可判断AB ，利用反证法可判断C ，利用线面平行判定定理可判断D.【详解】如图所示，连接AC ，1CD ，EF ，由于E ，F 分别为1AD ，DB 的中点，即F 为AC 的中点，所以1//EF CD ，EF ⊄面11CDD C ，1CD ⊆面11CDD C ，所以//EF 平面11CDD C ，即D 正确；所以EF 与1CD 共面，而1B ∉1CD ，所以直线EF 与11D B 为异面直线，即A 正确；连接1BC ，易得11//D E BC ，所以1DC B ∠即为直线1D E 与1DC 所成的角或其补角，由于1BDC 为等边三角形，即160DC B ∠=，所以B 正确；假设1D F AD ⊥，由于1AD DD ⊥，1DF DD D = ，所以AD ⊥面1D DF ，而AD ⊥面1D DF 显然不成立，故C 错误；故选：ABD.10.已知P 是圆22:4O x y +=上的动点，直线1:cos sin 4l x y θθ+=与2:sin cos 1l x y θθ-=交于点Q ，则（）A.12l l ⊥ B.直线1l 与圆O 相切C.直线2l 与圆O 截得弦长为23 D.OQ 17【答案】ACD 【解析】【分析】选项A 根据12l l ⊥，12120A A B B +=可判断正确；选项B 由圆心O 到1l 的距离不等半径可判断错误；选项C 根据垂直定理可得；选项D 先求出()4sin cos ,4cos sin Q θθθθ-+，根据两点间的距离公式可得.【详解】选项A ：因()cos sin sin cos 0θθθθ+-=，故12l l ⊥，A 正确；选项B ：圆O 的圆心O 的坐标为()0,0，半径为2r =，圆心O 到1l 的距离为12244cos sin d r θθ-==>+，故直线1l 与圆O 相离，故B 错误；选项C ：圆心O 到1l 的距离为()22211sin cos d θθ-==+-，故弦长为222223l r d =-=，故C 正确；选项D ：由cos sin 4sin cos 1x y x y θθθθ+=⎧⎨-=⎩得4cos sin 4sin cos x y θθθθ=+⎧⎨=-⎩，故()4cos sin ,4sin cos Q θθθθ+-，故OQ ==，故D 正确故选：ACD11.已知三次函数()32f x ax bx cx d =+++有三个不同的零点1x ，2x ，()3123x x x x <<，函数()()1g x f x =-也有三个零点1t ，2t ，()3123t t t t <<，则（）A.23b ac>B.若1x ，2x ，3x 成等差数列，则23bx a=-C.1313x x t t +<+D.222222123123x x x t t t ++=++【答案】ABD 【解析】【分析】对于A ，由题意可得()0f x '=有两个不同实根，则由0∆>即可判断；对于B ，若123,,x x x 成等差数列，则()()22,x f x 为()f x 的对称中心，即可判断；对于C ，结合图象，当0a >和0a <时，分类讨论即可判断；对于D ，由三次函数有三个不同的零点，结合韦达定理，即可判断.【详解】因为()32f x ax bx cx d =+++，则()232f x ax bx c '=++，0a ≠，对称中心为,33b b f a a ⎛⎫⎛⎫-- ⎪ ⎪⎝⎭⎝⎭，对于A ，因为()f x 有三个不同零点，所以()f x 必有两个极值点，即()2320f x ax bx c '=++=有两个不同的实根，所以2Δ4120b ac =->，即23b ac >，故A 正确；对于B ，由123,,x x x 成等差数列，及三次函数的中心对称性，可知()()22,x f x 为()f x 的对称中心，所以23bx a=-，故B 正确；对于C ，函数()()1g x f x =-，当()0g x =时，()1f x =，则1y =与()y f x =的交点的横坐标即为1t ，2t ，3t ，当0a >时，画出()f x 与1y =的图象，由图可知，11x t <，33x t <，则1313x x t t +<+，当0a <时，则1313x x t t +>+，故C 错误；对D ，由题意，得()()()()()()32123321231a x x x x x x ax bx cx da x t x t x t ax bx cx d ⎧---=+++⎪⎨---=+++-⎪⎩，整理，得123123122331122331b x x x t t t ac x x x x x x t t t t t t a ⎧++=++=-⎪⎪⎨⎪++=++=⎪⎩，得()()()()2212312233112312233122x x x x x x x x x t t t t t t t t t ++-++=++-++，即222222123123x x x t t t ++=++，故D 正确.故选：ABD.【点睛】关键点点睛：本题D 选项的关键是利用交点式得到三次方程的韦达定理式再计算即可.三、填空题（本大题共3个小题，每小题5分，共15分）12.已知随机变量X 服从二项分布(),B n p ，若()3E X =，()2D X =，则n =_____.【答案】9【解析】【分析】根据二项分布的期望、方差公式，即可求得答案.【详解】由题意知随机变量X 服从二项分布(),B n p ，()3E X =，()2D X =，则()3,12np np p =-=，即得1,93p n ==，故答案为：913.已知平面向量a ，b 满足2a = ，1= b ，且b 在a上的投影向量为14a - ，则ab + 为______.【答案】【解析】【分析】由条件结合投影向量公式可求a b ⋅ ，根据向量模的性质及数量积运算律求a b +.【详解】因为b 在a上的投影向量为14a - ，所以14b a a a aa ⋅⋅=- ，又2a = ，所以1a b ⋅=-，又1= b ，所以a b +==14.如图，已知四面体ABCD 的体积为32，E ，F 分别为AB ，BC 的中点，G ，H 分别在CD ，AD 上，且G ，H 是靠近D 点的四等分点，则多面体EFGHBD 的体积为_____.【答案】11【解析】【分析】连接,EG ED ，将多面体EFGHBD 被分成三棱锥G EDH -和四棱锥E BFGD -，利用题设条件找到小棱锥底面面积与四面体底面面积的数量关系，以及小棱锥的高与四面体的高的数量关系，结合四面体的体积即可求得多面体EFGHBD 的体积.【详解】如图，连接,EG ED ，则多面体EFGHBD 被分成三棱锥G EDH -和四棱锥E BFGD -.因H 是AD 上靠近D 点的四等分点，则14DHE AED S S = ，又E 是AB 的中点，故11114428DHE AED ABD ABD S S S S ==⨯= ，因G 是CD 上靠近D 点的四等分点，则点G 到平面ABD 的距离是点C 到平面ABD 的距离的14，故三棱锥G EDH -的体积1113218432G EDH C ABD V --=⨯=⨯=；又因点F 是BC 的中点，则133248CFG BCD BCD S S S =⨯= ，故58BFGD BCD S S = ，又由E 是AB 的中点知，点E 到平面BCD 的距离是点A 到平面BCD 的距离的12，故四棱锥E BFGD -的体积51532108216E BFGD A BCD V V --=⨯=⨯=，故多面体EFGHBD 的体积为11011.G EDH E BFGD V V --+=+=故答案为：11.【点睛】方法点睛：本题主要考查多面体的体积求法，属于较难题.一般的求法有两种：（1）分割法：即将多面体通过连线，作面的垂线等途径，将其分成若干可以用公式求解；（2）补形法：即将多面体通过辅助线段构造柱体，锥体或台体，利用整体体积减去个体体积等间接方法求解.四、解答题（本大题共5个小题，共77分.解答应写出文字说明、证明过程或演算步骤）15.设ABC V 的内角A ，B ，C 的对边分别为a ，b ，c ，已知sin cos 0a B A -=.（1）求A ；（2）若sin sin 2sin B C A +=，且ABC V 的面积为a 的值.【答案】（1）π3A =（2）2a =【解析】【分析】（1）利用正弦定理的边角变换得到tan A =，从而得解；（2）利用正弦定理的边角变换，余弦定理与三角形面积公式得到关于a 的方程，解之即可得解.【小问1详解】因为sin cos 0a B A -=,即sin cos a B A =，由正弦定理得sin sin cos A B B A ⋅=⋅，因为sin 0B ≠，所以sin A A =,则tan A =，又()0,πA ∈，所以π3A =.【小问2详解】因为sin sin 2sin B C A +=，由正弦定理得2b c a +=，因为π3A =，所以113sin 222ABC S bc A bc ==⨯= 4bc =，由余弦定理2222cos a b c bc A =+-⋅，得224b c bc +-=，所以()234b c bc +-=，则()22344a -⨯=，解得2a =.16.设()()221ln 2f x x ax x x =++，a ∈R .（1）若0a =，求()f x 在1x =处的切线方程；（2）若a ∈R ，试讨论()f x 的单调性.【答案】（1）4230--=x y （2）答案见解析【解析】【分析】（1）由函数式和导函数式求出(1)f 和(1)f '，利用导数的几何意义即可写出切线方程；（2）对函数()f x 求导并分解因式，根据参数a 的取值进行分类讨论，由导函数的正负推得原函数的增减，即得()f x 的单调性.【小问1详解】当0a =时，()221ln 2f x x x x =+，()2(ln 1)f x x x =+'，因1(1),(1)22f f '==，故()f x 在1x =处的切线方程为12(1)2y x -=-，即4230--=x y ；【小问2详解】因函数()()221ln 2f x x ax x x =++的定义域为(0,)+∞，()(2)ln 2(2)(ln 1)f x x a x x a x a x =+++=++'，①当2a e ≤-时，若10e x <<，则ln 10,20x x a +<+<，故()0f x '>，即函数()f x 在1(0,)e上单调递增；若1e x >，由20x a +=可得2a x =-.则当1e 2a x <<-时，20x a +<，ln 10x +>，故()0f x '<，即函数()f x 在1(,)e 2a -上单调递减；当2a x >-时，ln 10,20x x a +>+>，故()0f x '>，即函数()f x 在(,)2a -+∞上单调递增；②当2e a >-时，若1e x >，则ln 10,20x x a +>+>，故()0f x '>，即函数()f x 在1(,)e +∞上单调递增；若12e a x -<<，则ln 10,20x x a +<+>，故()0f x '<，即函数()f x 在1(,)2e a -上单调递减；若02a x <<-，则ln 10,20x x a +<+<，故()0f x '>，即函数()f x 在(0,)2a -上单调递增，当2e a =-时，()0f x '≥恒成立，函数()f x 在()0,+∞上单调递增，综上，当2e a <-时，函数()f x 在1(0,)e 上单调递增，在1(,)e 2a -上单调递减，在(,)2a -+∞上单调递增；当2e a =-时，函数()f x 在()0,+∞上单调递增；当2e a >-时，函数()f x 在(0,2a -上单调递增，在1(,2e a -上单调递减，在1(,)e+∞上单调递增.17.已知四棱锥P ABCD -，底面ABCD 为菱形，,PD PB H =为PC 上的点，过AH 的平面分别交,PB PD 于点,M N ，且BD ∥平面AMHN ．（1）证明：MN PC ⊥；（2）当H 为PC 的中点，,PA PC PA ==与平面ABCD 所成的角为60︒，求平面PAM 与平面AMN 所成的锐二面角的余弦值．【答案】（1）证明见详解（2）3913【解析】【分析】（1）根据线面垂直可证BD ⊥平面PAC ，则BD PC ⊥，再根据线面平行的性质定理可证BD ∥MN ，进而可得结果；（2）根据题意可证⊥PO 平面ABCD ，根据线面夹角可知PAC 为等边三角形，建立空间直角坐标系，利用空间向量求面面夹角.【小问1详解】设AC BD O = ，则O 为,AC BD 的中点，连接PO ，因为ABCD 为菱形，则AC BD ⊥，又因为PD PB =，且O 为BD 的中点，则PO BD ⊥，AC PO O = ，,AC PO ⊂平面PAC ，所以BD ⊥平面PAC ，且PC ⊂平面PAC ，则BD PC ⊥，又因为BD ∥平面AMHN ，BD ⊂平面PBD ，平面AMHN 平面PBD MN =，可得BD ∥MN ，所以MN PC ⊥.【小问2详解】因为PA PC =，且O 为AC 的中点，则PO AC ⊥，且PO BD ⊥，AC BD O = ，,AC BD ⊂平面ABCD ，所以⊥PO 平面ABCD ，可知PA 与平面ABCD 所成的角为60PAC ∠=︒，即PAC 为等边三角形，设AH PO G =I ，则,G AH G PO ∈∈，且AH⊂平面AMHN ，PO ⊂平面PBD ，可得∈G 平面AMHN ，∈G 平面PBD ，且平面AMHN 平面PBD MN =，所以G MN ∈，即,,AH PO MN 交于一点G ，因为H 为PC 的中点，则G 为PAC 的重心，且BD ∥MN ，则23PM PN PG PB PD PO ===，设2AB =，则11,32PA PC OA OC AC OB OD OP ========，如图，以,,OA OB OP 分别为,,x y z 轴，建立空间直角坐标系，则)()22,0,0,3,0,,1,0,,133A P M N ⎛⎫⎛⎫- ⎪ ⎪⎝⎭⎝⎭，可得()24,1,0,,0,33AM NM AP ⎛⎫⎛⎫=== ⎪ ⎪⎝⎭⎝⎭uuu r uuur uuu r ，设平面AMN 的法向量()111,,x n y z =，则1111203403n AM y z n NM y ⎧⋅=++=⎪⎪⎨⎪⋅==⎪⎩，令11x =，则110,y z ==，可得(n = ，设平面PAM 的法向量()222,,m x y z =，则2222220330m AM y z m AP z ⎧⋅=++=⎪⎨⎪⋅=+=⎩，令2x =，则123,1y z ==，可得)m =u r，可得39cos ,13n m n m n m ⋅===⋅r u r r u r r u r ，所以平面PAM 与平面AMN 所成的锐二面角的余弦值3913.18.已知双曲线22:13y x Γ-=的左、右焦点为1F ，2F ，过2F 的直线l 与双曲线Γ交于A ，B 两点.（1）若AB x ⊥轴，求线段AB （2）若直线l 与双曲线的左、右两支相交，且直线1AF 交y 轴于点M ，直线1BF 交y 轴于点N .（i ）若11F AB F MN S S = ，求直线l 的方程；（ii ）若1F ，2F 恒在以MN 为直径的圆内部，求直线l 的斜率的取值范围.【答案】（1）线段AB 的长为6；（2）（i ）直线l的方程为221x y =±+；（ii ）直线l的斜率的取值范围为33(,)(,7447-- .【解析】【分析】（1）直接代入横坐标求解纵坐标，从而求出的值；（2）（i ）（ii ）先设直线和得到韦达定理，在分别得到两个三角形的面积公式，要求相等，代入韦达定理求出参数的值即可.【小问1详解】由双曲线22:13y x Γ-=的方程，可得221,3a b ==，所以1,2a b c ===，所以1(2,0)F -，2(2,0)F ，若AB x ⊥轴，则直线AB 的方程为2x =，代入双曲线方程可得(2,3),(2,3)A B -，所以线段AB 的长为6；【小问2详解】（i）如图所示，若直线l 的斜率为0，此时l 为x 轴，,A B 为左右顶点，此时1,,F A B 不构成三角形，矛盾，所以直线l 的斜率不为0，设:2l x ty =+，1122()A x y B x y ,,(,)，联立22132y x x ty ⎧-=⎪⎨⎪=+⎩，消去x 得22(31)1290t y ty ++=，t 应满足222310Δ14436(31)0t t t ⎧-≠⎨=-->⎩，由根与系数关系可得121222129,3131t y y y y t t +=-=--，直线1AF 的方程为110(2)2y y x x -=++，令0x =，得1122y y x =+，点112(0,2y M x +，直线1BF 的方程为220(2)2y y x x -=++，令0x =，得2222y y x =+，点222(0,2y N x +，121122221111|||||2||2|F F F B A A F B F S y F S S F y y y -=⨯-==- ，111212221||||||222F M N M F MNN S y y x y y y y x x =-=-=-++ 12122112212121212222(4)2(4)8()||||||44(4)(4)4()16y y y ty y ty y y ty ty ty ty t y y t y y +-+-=-==+++++++，由11F AB F MN S S = ，可得1212212128()||2||4()16y y y y t y y t y y -=-+++，所以21212|4()16|4t y y t y y +++=，所以222912|4(16|43131t t t t t ⨯+-+=--，解得22229484816||431t t t t -+-=-，22916||431t t -=-，解得22021t =，经检验，满足222310Δ14436(31)0t t t ⎧-≠⎨=-->⎩，所以t =所以直线l 的方程为221x y =±+；（ii ）由1F ，2F 恒在以MN 为直径的圆内部，可得2190F MF >︒∠，所以110F F N M < ，又112211,22(2,)(2,22F y y N x x M F =+=+ ，所以1212224022y y x x +⨯<++，所以121210(2)(2)y y x x +<++，所以1221212104()16y y t y y t y y +<+++，所以2222931109124()163131t t t t t t -+<⨯+-+--，所以22970916t t -<-，解得271699t <<，解得433t <<或433t -<<-，经检验，满足222310Δ14436(31)0t t t ⎧-≠⎨=-->⎩，所以直线l的斜率的取值范围为33(,(,7447-- .【点睛】方法点睛：圆锥曲线中求解三角形面积的常用方法：（1）利用弦长以及点到直线的距离公式，结合12⨯底⨯高，表示出三角形的面积；（2）根据直线与圆锥曲线的交点，利用公共底或者公共高的情况，将三角形的面积表示为12211||||2F F y y ⨯-或121||||2AB x x ⨯-.19.已知{}n a 是各项均为正整数的无穷递增数列，对于*k ∈N ，设集合{}*k i B i a k =∈<N ∣，设k b 为集合k B 中的元素个数，当k B =∅时，规定0k b =.（1）若2n a n =，求1b ，2b ，17b 的值；（2）若2n n a =，设n b 的前n 项和为n S ，求12n S +；（3）若数列{}n b 是等差数列，求数列{}n a 的通项公式.【答案】（1）12170,1,4b b b ===（2）1(1)22n n +-⨯+（3）n a n=【解析】【分析】（1）根据集合新定义，利用列举法依次求得对应值即可得解；（2）根据集合新定义，求得12,b b ，121222i i i b b b i +++==== ，从而利用分组求和法与裂项相消法即可得解.（3）通过集合新定义结合等差数列性质求出11a =，然后利用反证法结合数列{}n a 的单调性求得11n n a a +-=，利用等差数列定义求解通项公式即可；【小问1详解】因为2n a n =，则123451,4,9,16,25a a a a a =====，所以{}*11i B i a =∈<=∅N ∣，{}*22{1}i B i a =∈<=N ∣，{}*1717{1,2,3,i B i a =∈<=N ∣，故12170,1,4b b b ===.【小问2详解】因为2n n a =，所以123452,4,8,16,32a a a a a =====，则**12{|1},{|2}i i B i a B i a =∈<=∅=∈<=∅N N ，所以10b =，20b =，当122i i k +<≤时，则满足i a k <的元素个数为i ，故121222i i i b b b i +++==== ，所以()()()1112345672122822n n n n S b b b b b b b b b b b ++++=++++++++++++ 1212222n n =⨯+⨯++⨯ ，注意到12(1)2(2)2n n n n n n +⨯=-⨯--⨯，所以121321202(1)21202(1)2(2)2n n nS n n ++=⨯--⨯+⨯-⨯++-⨯--⨯ 1(1)22n n +=-⨯+.【小问3详解】由题可知11a ≥，所以1B =∅，所以10b =，若12a m =≥，则2B =∅，1{1}m B +=，所以20b =，11m b +=，与{}n b 是等差数列矛盾，所以11a =，设()*1n n n d a a n +=-∈N ，因为{}n a 是各项均为正整数的递增数列，所以*n d ∈N ，假设存在*k ∈N 使得2k d ≥，设k a t =，由12k k a a +-≥得12k a t ++≥，由112k k a t t t a +=<+<+≤得t b k <，21t t b b k ++==，与{}n b 是等差数列矛盾，所以对任意*n ∈N 都有1n d =，所以数列{}n a 是等差数列，1(1)n a n n =+-=.【点睛】方法点睛：求解新定义运算有关的题目，关键是理解和运用新定义的概念以及元算，利用化归和。
2024-2025学年高三一轮复习联考(三)_全国卷文数(含答案)
2024届高三一轮复习联考（三）全国卷文科数学试题注意事项：1.答卷前，考生务必将自己的姓名､考场号､座位号､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2B 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，考试时间为120分钟，满分150分一､选择题：本题共12小题，每小题5分，共60分.在每小题给出的四个选项中，只有一项是符合题目要求的.1.已知集合{}{}212,1A xx B x x =<<=∣∣，则A B ⋃=（）A.[)1,2-B.(),2∞-C.[)1,3- D.[]1,2-2.命题2:,220p x R x x ∀∈+-<的否定p ⌝为（）A.2000,220x R x x ∃∈+->B.2,220x R x x ∀∈+-C.2,220x R x x ∀∈+->D.2000,220x R x x ∃∈+-3.3.已知复数2(1i)z =+（i 为虚数单位），则复数z 的虚部为（）A.2B.2- C.2iD.2i-4.若函数()222,0,log ,0,x x x f x x x ⎧-=⎨>⎩则()2f f ⎡⎤-=⎣⎦（）A.2- B.2 C.3- D.35.已知1sin 62πα⎛⎫-= ⎪⎝⎭，则2cos 23πα⎛⎫+= ⎪⎝⎭（）A.14-B.14C.12-D.126.函数()21x xe ef x x --=+在[]3,3-上的大致图象为（）A.B.C. D.7.函数2sin cos21y x x=-+的最小值是（）A.3-B.1-C.32- D.12-8.已知数列{}n a的前n项和22nS n n m=-++，且对任意*1,0n nn N a a+∈-<，则实数m 的取值范为是（）A.()2,∞-+ B.(),2∞--C.()2,∞+ D.(),2∞-9.已知等比数列()*a满足4221,m nq a a a≠=，（其中,*m n N∈），则91m n+的最小值为（）A.6 B.16 C.32 D.210.已知函数()cos3f x xπ⎛⎫=+⎪⎝⎭，若()f x在[]0,a上的值域为11,2⎡⎤-⎢⎥⎣⎦，则实数a的取值范为（）A.40,3π⎛⎤⎥⎝⎦B.24,33ππ⎡⎤⎢⎥⎣⎦C.2,3π∞⎡⎫+⎪⎢⎣⎭ D.25,33ππ⎡⎤⎢⎥⎣⎦11.设4sin1,3sin2,2sin3a b c===，则（）A.a b c<< B.c b a<<C.c a b<< D.a c b<<12.已矨,,A B C均在球O的球面上运动，且满足3AOBπ∠=，若三棱锥O ABC-体积的最大值为6，则球O的体积为（）A.12πB.48πC.D.二､填空题：本题共4小题，每小题5分，共20分.13.已知()(1,,a k b==，若a b⊥，则k=__________.14.已知{}n a是各项不全为零的等差数列，前n项和是n S，且2024S S=，若()2626nS S m=≠，则正整数m=__________.15.设,m n为不重合的直线，,,αβγ为不重合的平面，下列是αβ∥成立的充分条件的有（）（只填序号）.①,m a m β⊂∥②,,m n n m αβ⊂⊥⊥③,αγβγ⊥⊥④,m m αβ⊥⊥16.已知函数()14sin ,01,2,1,x x x f x x x π-<⎧=⎨+>⎩若关于x 的方程()()()2[]210f x m f x m --+-=恰有5个不同的实数解，则实数m 的取值集合为__________.三､解答题：共70分.解答应写出文字说明､证明过程或演算步骤.第17-21题为必考题，每个试题考生都必须作答.第22､23题为选考题，考生根据要求作答.（一）必考题：60分.17.（12分）已知数列{}n a 满足12122,log log 1n n a a a +==+，（1）求数列{}n a 的通项公式；（2）求(){}32nn a -的前n 项和nS.18.（12分）已知ABC 中，三个内角,,A B C 的对边分别为,,,,cos cos 2cos 4a b c C a A c C b B π=+=.（1）求tan A ；（2）若c =，求ABC 的面积.19.（12分）如图，在四棱锥P ABCD -中，底面ABCD 是矩形，O 是BC 的中点，PB PC ==，22PD BC AB ===.（1）求证：平而PBC ⊥平面ABCD ；（2）求点A 到平面PCD 的距离.20.（12分）已知数列()n a 满足()21112122222326n n n n n a a a a n -+-++++=-⋅+ .（1）求{}n a 的通项公式；（2）若2n an n b a =+，求数列n b 的前n 项和T .21.（12分）已知函数()ln x af x ex x -=-+.（1）当1a =时，求曲线()f x 在点()()1,1f 处的切线方程，（2）当0a 时，证明，()2f x x >+.（二）选考题：共10分.请考生在第22，23题中任选一题作答.如果多做，则按所做的第一题计分.22.[选修4-4：坐标系与参数方程]（10分）在平面直角坐标系，xOy 中，直线l的参数方程为2,21,2x a y t ⎧=+⎪⎪⎨⎪=⎪⎩（t 为参数），以O 为极点，x 轴的非负半轴为极轴建立极坐标系，曲线C 的极坐标方程为22413sin ρθ=+.（1）求直线l 和曲线C 的直角坐标方程；（2）若曲线C 经过伸缩变换,2,x x y y ⎧=⎪⎨⎪='⎩'得到曲线C '，若直线l 与曲线C '有公共点，试求a 的取值范围.23.[选修4-5：不等式选讲]（10分）已知函数()22(0)f x x x t t =++->，若函数()f x 的最小值为5.（1）求t 的值；（2）若,,a b c 均为正实数，且2a b c t ++=，求1412a b c++的最小值.2024届高三一轮复习联考（三）全国卷文科数学参考答案及评分意见1.A【解析】由21x ，即()()110x x -+，解得11x -，所以{}11B xx =-∣，所以{12}A B xx ⋃=-<∣.故选A .2.D 【解析】2,220x x x ∀∈+-<R 的否定为：2000,220x x x ∃∈+-R ，故选D.3.A 【解析】2(1i)2i z =+=，即复数z 的虚部为2，故选A .4.D【解析】()()()222(2)228,8log 83f f -=--⨯-===，故选D.5.C 【解析】因为1sin 62πα⎛⎫-= ⎪⎝⎭，所以2211cos 2cos 2cos 22sin 11366622ππππααπαα⎡⎤⎡⎤⎛⎫⎛⎫⎛⎫⎛⎫+=-+=--=--=-=- ⎪ ⎪ ⎪ ⎪⎢⎥⎢⎥⎝⎭⎝⎭⎝⎭⎝⎭⎣⎦⎣⎦.故选C.6.A 【解析】()()2e e 1x xf x f x x ---==-+，所以函数()y f x =是奇函数，排除B 选项，又()22e e 215f --=>，排除C ，D 选项，故选A.7.D 【解析】由题意，函数22sin cos212sin 2sin y x x x x =-+=+，令[]sin 1,1t x =∈-，可得221122222y t t t ⎛⎫=+=+- ⎪⎝⎭，当12t =-，即1sin 2x =-时，函数取得最小值，最小值为12-.故选D.8.A【解析】因为10n n a a +-<，所以数列{}n a 为递减数列，当2n 时，()2212(1)2123n n n a S S n n m n n m n -⎡⎤=-=-++---+-+=-+⎣⎦，故可知当2n 时，{}n a 单调递减，故{}n a 为递减数列，只需满足21a a <，即112m m-+⇒-.故选A .9.D【解析】由等比数列的性质，可得()911911918,10102888m n m n m n m n m n n m ⎛⎛⎫⎛⎫+=+=++=+++= ⎪ ⎪ ⎝⎭⎝⎭⎝，当且仅当6,2m n ==时，等号成立，因此，91m n +的最小值为2.故选D.10.B 【解析】()cos 3f x x π⎛⎫=+⎪⎝⎭，结合图象，()f x 的值域是11,,0,2333x a x a πππ⎡⎤-++⎢⎣⎦，于是533a πππ+，解得2433aππ，所以实数a 的取值范围为24,33ππ⎡⎤⎢⎥⎣⎦.故选B.11.B 【解析】设()()2sin cos sin ,x x x xf x f x x x -=='，令()()cos sin ,sing x x x x g x x x =-'=-，当()0,x π∈时，()0g x '<，故()g x 在()0,π上递减，()()()00,0g x g f x <=∴<'，故()sin xf x x=在()0,π上递减，023π<<< .()()sin3sin232,,2sin33sin232f f ∴<<<，故c b <，()()()sin 2012,sin1,sin22sin1,3sin232sin14sin12ππππππ-<<-<<<-<-<-，故b a <，故c b a <<，故选B.12.C 【解析】如图所示，当点C 位于垂直于面AOB 的直径端点时，三棱锥O ABC -的体积最大，设球O 的半径为R ，此时231133632212O ABC C AOB V V R R --==⨯⨯⨯==，故3R =O 的体积为343R V π==，故选C.13.3-【解析】0a b a b ⊥⇔⋅=，所以()(1,10,3k k ⋅=+==-.14.18【解析】设等差数列{}n a 的首项和公差分别为1,a d ，则2122n d d S n a n ⎛⎫=+- ⎪⎝⎭，所以n S 可看成关于n 的二次函数，由二次函数的对称性及202426,m S S S S ==，可得20242622m++=，解得18m =.15.④【解析】根据线面的位置关系易知，①②③中面α和面β可能相交也可能平行，④：若m α⊥且m β⊥，根据面面平行的判定可知垂直于同一直线的两平面互相平行，故④正确.16.()3,1--【解析】作出函数()f x 的大致图象，如图所示，令()t f x =，则()()()2[]210f x m f x m --+-=可化为()()()221110t m t m t m t --+-=-+-=，则11t =或21t m =-，则关于x 的方程()()()2[]210f x m f x m --+-=恰有5个不同的实数解等价于()t f x =的图象与直线12,t t t t ==的交点个数之和为5个，由图可得函数()t f x =的图象与直线1t t =的交点个数为2，所以()t f x =的图象与直线2t t =的交点个数为3个，即此时214m <-<，解得31m -<<-.17.【解析】（1）在数列{}n a 中，已知12122log log log 1n n n na a a a ++-==，所以12n na a +=，.即{}n a 是首项为12a =，公比为2的等比数列，所以()1*222n n n a n -=⨯=∈N .（2）由()()32322nn n a n -=-⨯，故()()231124272352322n n n S n n -=⨯+⨯+⨯++-⨯+-⨯ ，所以()()23412124272352322nn n S n n +=⨯+⨯+⨯++-⨯+-⨯ ，则()23123222322n n n S n +⎡⎤-=+⨯+++--⨯⎣⎦，()()()11212433221053212n n n n n ++-=-+⨯--⨯=-+-⋅-，故()110352n n S n +=+-⋅.18.【解析】（1）解法一：由题，cos cos 2cos a A c C b B +=，由正弦定理得，sin2sin cos sin cos B A A C C =+，.3,,sin2sin 2sin 2cos2422C A B C B A A A ππππ⎛⎫⎛⎫=++==-=-=- ⎪ ⎪⎝⎭⎝⎭，所以1cos2sin cos 2A A A -=+，221sin cos sin cos 2A A A A --=22tan 1tan 1tan 12A A A --=+，化简得2tan 2tan 30A A --=，解得tan 3A =或tan 1A =-（舍去）.解法二：由题，cos cos 2cos a A c C b B +=，由正弦定理得，2sin2sin2sin2B A C =+，即()()()()2sin2sin sin B A C A C A C A C ⎡⎤⎡⎤=++-++--⎣⎦⎣⎦，即()()sin2sin cos B A C A C =+-，又A B C π++=，故()sin sin A C B +=，所以()2sin cos sin cos B B B A C =-，又0B π<<，故sin 0B ≠，所以()2cos cos B A C =-，又A B C π++=，故()cos cos B A C =-+，化简得sin sin 3cos cos A C A C =，因此tan tan 3A C =且tan 1C =，所以tan 3A =.（2）由（1）知tan 3A =，因此()tan tan tan tan 21tan tan A CB AC A C+=-+=-=-，.所以sin 10A =，sin 5B =2sin 2C =，因为,6sin sin a c a A C==，.所以1125sin 612225ABC S ac B ==⨯⨯= .19.【解析】（1）因为,PB PC O =是BC 的中点，所以PO BC ⊥，在直角POC 中，1PC OC ==，所以PO =，在矩形ABCD 中，1,2AB BC ==，所以DO =，又因为2PD =，所以在POD 中，222PD PO OD =+，即PO OD ⊥.而,,BC OD O BC OD ⋂=⊂平面ABCD ，所以PO ⊥平面ABCD ，而PO ⊂平面PBC ，所以平面PBC ⊥平面ABCD .'（2）由（1）平面PBC ⊥平面ABCD ，且DC BC ⊥，所以DC ⊥平面PBC ，所以DC PC ⊥，即PCD 是直角三角形，因为1PC CD ==，所以13122PDC S =⨯=，又知11212ACD S =⨯⨯= ，PO ⊥平面ABCD ，设点A 到平面PCD 的距离为d ，则A PCD P ACD V V --=，即1133PCD ACD S d S PO ⨯⨯=⨯⨯ ，即1311323d ⨯⨯=⨯⨯所以263d =，所以点A 到平面PCD 的距离为3..20.【解析】（1）由题当1n =时，()111223262a +=-⋅+=，即11a =.()21112122222326n n n n n a a a a n -+-++++=-⋅+ ①当2n 时，()211212222526n n n a a a n --+++=-⋅+ ②.①-②得()()()1223262526212nn n n n a n n n +=-⋅+--⋅-=-⋅，所以21n a n =-..（2）由（1）知，212221n an n n b a n -=+=+-，则()()()()3521212325221n n T n -=++++++++- ()()3521222213521n n -=+++++++++-⋅()()212214121232..1423nn n n n +⨯-+-+-=+=-21.【解析】（1）当1a =时，()()111e ln ,e 1x xf x x x f x x--=-+=-+'，所以()()12,11f f '==，.则切线方程为()211y x -=⨯-，.即10x y -+=曲线()f x 在点()()1,1f 处的切线方程为10x y -+=.（2）证明：要证()2f x x >+，即证e ln 2x a x -->，设()eln ,0x aF x x x -=->，即证()2F x >，当0a 时，()()1e 1e ln ,ex a x ax ax F x x F x x x----=-=-='在()0,∞+上为增函数，且()e1x ah x x -=-中，()()0100e 110,1e 1e 10a a h h --=⨯-=-=-->.故()0F x '=在()0,∞+上有唯一实数根0x ，且()00,1x ∈..当()00,x x ∈时，()0F x '<，当()0,x x ∞∈+时，()0F x '>，从而当0x x =时，()F x 取得最小值.由()00F x '=，得001ex ax -=，故()()000001eln 2x aF x F x x x a a x -=-=+->.综上，当0a 时，()2F x >即()2f x x >+.22.【解析】（1）由题2,21,2x a t y t ⎧=+⎪⎪⎨⎪=⎪⎩（t 为参数），消去参数t得直线:20l x a -=，.22413sin ρθ=+，即2224cos 4sin ρθθ=+，即曲线C 的直角坐标方程为2214x y +=.（2）由,2,x x y y ⎧=⎪⎨⎪='⎩'得2,,x x y y =⎧⎨=''⎩又2214x y +=，所以()()22214x y +'='，即'2'21x y +=，所以曲线C '的方程是221x y +=，.由1d =得11a -.所以a 的取值范围是[]1,1-.23.【解析】（1）()222f x x x t x x t x t =++-=++-+-，()2222y x x tx x t t t =++-+--=+=+，当2x t -时等号成立，.⋅又知当x t =时，x t -取得最小值，所以当x t =时，()f x 有最小值，此时()min ()25f x f t t ==+=，所以3t =..（2）由（1）知，23a b c ++=，()22141114111162(121)232333a b c a b c a b c ⎛⎫++=++++=++= ⎪⎝⎭，当且仅当333,,824a b c ===时取等号，所以1412a b c ++的最小值为163.。
2020-2021学年北京市第十七中学高三英语第一次联考试题及参考答案
2020-2021学年北京市第十七中学高三英语第一次联考试题及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30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AOur Teen Summer Spanish Program is two weeks of fun, educational excitement that helps students learn Spanish fast. Our Spanish summer program allows our students to learn from highly trained, certified teachers and be absorbed in the language and theculture of Costa Rica.Features include:* Intensive(强化的) daily Spanish classes* Extracurricular classes in dance, cooking, music, and handiwork* Outdoor activities including hiking, camping, rafting, and ziplining(高空滑索)* Homestay with a local Costa Rican family* Volunteer work in needy neighborhoodsOur Teaching Methods:We are proud to use TPRS---Total Physical Response Storytelling---in our curriculum. This innovative method uses strange and amusing stories to teach new vocabulary, increase fluency, and get students involved by giving them the opportunity to alter the details themselves. Because of the silliness, creativity, and repetition involved, TPRS allows students to learn easily and remember information effortlesslyMemorizing vocabulary and listening to lectures on grammar are slow, inefficient ways to learn a new language. The best way to truly learn and commit new material to memory is through conversation. In our Spanish classes, students can expect to speak up to 80% of each class. By speaking in the new language freely and consistently, students can see progress faster because they are using the new grammar and vocabulary that they have learned at the same time. This helps the brain remember the new words and grammar structures for future use, making it much easier to progress.1.What does the program do?A.It offers weekly Spanish classesB.It focuses more on outdoor activitiesC.It gives teachers a chance to receive trainingD.It provides activities about the Spanish culture2.What is the best way to learn a language according to the text?A.Memorizing a larger vocabularyB.Speaking more in the new language.C.Mastering more grammar structuresD.Writing stories to share with others3.What is the purpose of the text?A.To employexperienced Spanish teacherB.To hire foreign volunteers for a programC.To attract teen foreigners to a programD.To introduce language learning methodsBSince I was born and brought up in a rural town, I have a great interest in nature. Using the chance of studying abroad in my second year at college, I decided to go toCanadajust because I wanted to see the beautiful phenomena there So after I finished the study program, I went toYellowknifein theNorthwest Territories.I clearly remember the sixth night inYellowknife. Suddenly my host mother came to my room around 8 p.m. and told me to change clothes and go outside quickly carrying her camera.The northern lights were flickering (闪烁) in the sky! I was shocked and just stood there with my mouth open.I forgot to take pictures of the mysterious lights.Since that night, whenever it was sunny, I went outside at night and looked at the sky. It was so cold that I lost all feeling in myhands and feet.As I took pictures of the northern lights, I came to find a characteristic movement of the lights. They first appear in the north part of the sky and then they gradually come down to the south part of the sky. After that, suddenly, they come in the middle of the north and south only for a while, which is the time when the best northern lights can be seen. Since it is only a few seconds for the northern lights to come down to the middle of the sky, it is very hard to get good pictures.The stronger the sun acts, the better and stronger the northern lights flicker in the sky. That’s because they come about from the collisions (碰撞) between atmospheric gases and the solar wind. Much more solar wind comes to the earth when the sun is active, whichleads to the best northern lights. And the color1 s of the northern lights depend on the height of the collisions and the kinds of gases.4. Why did the host mother ask the author to go out?A. She wanted to take a picture of him.B. She wanted to take a walk with him.C. She wanted to tell him something important.D. She wanted him to see the northern lights.5. The author forgot to take pictures after going out because ______.A. the host mother didn’t remind him to take the cameraB. he was shocked by the wonderful sightC. the lights flickering in the sky disappeared too soonD. he lost all feeling in his hands and feet6. When is the best time to see the northern lights?A. When they appear in the north part of the sky.B. When they come down to the south part of the sky.C. When they are between the north and south.D. When they rise in the east part of the sky.7. What does the last paragraph mainly tell us?A. Waysto take good pictures.B. The relationship between the sun and the northern lights.C.The color1 s of the northern lights.D. The time of the best northern lights.CThere are 195 countries in the world today but almost none of them have purple on their national flag. So what’s wrong with purple? It’s such a popular color1 today. Why would no country use it in their flag? The answer is really quite simple. Purple was just for too expensive.The color1 purple has been associated with royalty power and wealth for centuries. Queen Elizabeth I forbade anyone except close members of the royal family to wear it. Purple’s high status comes from the rarity and cost of the dye (染料)originally used to produce it. Fabric traders got the dye from a small sea snail (海螺)that was only found in the Tyre region of the Mediterranean. More than 10,000 snails were needed to create just one gram of purple; not to mention a lot of work went into producing the dye, which made purple dye so expensive.Since only wealthy rulers could afford to buy and wear the color1 , it became associated with the royal family.Sometimes, however, the dye was too expensive even for royalty. Third century Roman Emperor Aurelio famously wouldn’t allow his wife to buy a scarf made from purple silk because it cost three times its weight in gold. A single pound of dye cost three pounds of gold, which equals 56,000 dollars today. Therefore, even the richest countries couldn’t spend that much having purple on their flags.The dye became more accessible to lower-class about a century and a half ago. In 1856, 18-year-old English chemist William Henry Perkin accidentally created a man-made purple compound (化合物)while attemptingto produce an anti-malaria drug. He noticed that the compound could be used to dye fabrics, so he patented the dye, manufactured it and got rich. Purple dye was then mass-produced so everybody could afford it.Till now, a handful of new national flags have been designed and a few of them have chosen to use purple in their flag. So don’t be making any bets just yet.8. Why was color1 purple expensive in the past?A. Because only royal families were allowed to wear purple.B. Because it took a long time to get purple dye from gold.C. Because purple was worth as much as its weight in gold.D. Because purple dye used to be rare and hard to produce.9. Why did Roman Emperor Aurelio forbid his wife to buy a purple scarf?A. Because of poor quality.B. Because of long tradition.C. Because of bad taste.D. Because of high price.10. What is purple's situation now?A. Purple has been widely used on national flags.B. Purple dye is now affordable to ordinary people.C. Royal family stop using purple because it’s toocommon.D. Fewer snails are used to produce purple dye than before.11. Which of the following would be the best title for the passage?A. No Purple Flags?B. Purple vs GoldC. How to Produce Purple Dye?D. The Birth of Purple ColorDThe measurable threat to the environment has been worsened by the spread of COVID-19 that increases the need for plastic protective equipment. Most plastic is made from fossil fuels. Millions of tons of greenhouse gas are released from the development of these resources and plastic production and burning.The end life of plastic is just worrisome. Less than 10% of the plastic produced has been recycled. Even more of it has been burned. But the vast majority of plastic has been buried inland, and it is increasingly polluting the environment. We hear mostly about ocean plastic and the harm done to sea creatures that mistake plastic bags and bits for food. But microplastic is even more worrisome. Plastic doesn't break down biologically but instead breaks down into tiny particles（a very small piece of something）, which have been found in every corner of the planet, on land and in the air, in drinking water and food sources.Yet the public has not given this global environmental disaster the attention it requires. Instead, they have viewed single-use plastic—which makes up about 40% of plastic used each year—as a litter issue that can be solved through better recycling and waste management. That attitude must change because the recent global breakdown of the market for recycling has made it clear that it has never been, nor ever will be, able to keep up with plastic trash use.California has been the forerunner of plastic waste reduction—it was the first state to ban single-use plastic bags and may be the first state to transform the way goods are packaged. The state also came close to passing an act which would have required that products sold in plastic packaging in the state have a proven recycling rate of 75% by 2032. California, though influential, can't solve this crisis alone. The US has long been producing a great amount of plastic trash and it should engage in reducing the use of plastic as well.12. Why does the author mention the release of greenhouse gas in paragraph 1?A. To show the harm of plasticB. To warn of the climate change.C. To call for the development of fossil fuels.D. To highlight the importance of plastic equipment.13. What's the author's attitude towards the public opinion on single-use plastic?A. Favorable.B. Tolerant.C. Curious.D. Opposed.14. What's California's role in reducing plastic waste?A. A pioneer.B. A failure.C. An objector.D. A predictor.15. What can be the best title for the text?A. Microplastic Products Are HarmfulB. Waste Recycling Is an Urgent MatterC. Plastic Waste Pollution Is a Wake-up CallD. Global Environmental Disasters Are Increasing第二节（共5小题；每小题2分，满分10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。
炎德英才大联考2025届高三上学期月考数学试卷(三)含答案
大联考长郡中学2025届高三月考试卷（三）数学本试卷共8页。
时量120分钟。
满分150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．设集合{1,2}A =，{2,3}B =，{1,2,3,4}C =，则（ ） A ．A B =∅B ．A BC =C ．A C C =D ．A C B =2．在复平面内，复数1z 对应的点和复数212i z =+对应的点关于实轴对称，则12z z =（ ）A ．5B C ．34i −−D ．34i −+3．已知向量a ，b满足3a = ，b = ，且()a ab ⊥+ ，则b 在a 方向上的投影向量为（ ）A ．3B ．3−C ． 3a −D ．a −4．已知函数()f x 的定义域为()(),54,3f f x =+R 是偶函数，12,x x ∀∈[3,)+∞，有()()12120f x f x x x −>−，则（ ） A ．()04f <B ．()14f =C ．()24f >D ．()30f <5．若正四棱锥的高为8，且所有顶点都在半径为5的球面上，则该正四棱锥的侧面积为（ ） A ．24B ．32C ．96D ．1286．已知曲线e x y =在1x =处的切线l 恰好与曲线ln y a x =+相切，则实数a 的值为（ ） A ．1B ．2C ．3D ．47．在直角坐标系中，绕原点将x 轴的正半轴逆时针旋转角π02αα<<交单位圆于A 点、顺时针旋转角ππ42ββ <<交单位圆于B 点，若A 点的纵坐标为1213，且OAB △，则B 点的纵坐标为（ ）A ．B ．C ．D ．8．已知双曲线()2222:10,0x y C a b a b−=>>的左顶点为(),,0A F c 是双曲线C 的右焦点，点P 在直线2x c =上，且tan APF ∠，则双曲线C 的离心率是（ ）． B ．C ．4+ D ．2二、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．函数()()π3sin 0,2f x x ωϕωϕ=+><的部分图象如图所示，则下列匀选项中正确的有（ ）A ．()f x 的最小正周期为2πB ．2π3f是()f x 的最小值 C ．()f x 在区间π0,2上的值域为33,22−D ．把函数()y f x =的图象上所有点向右平移π12个单位长度，可得到函数3sin 2y x =的图象 10．在长方体1111ABCD A B C D −中，1222AB AA AD ===，点P 满足AP AB AD λµ=+，其中[][]0,1,0,1λµ∈∈，则（ ）A ．若1B P 与平面ABCD 所成的角为π4，则点P 的轨迹长度为π4B ．当λµ=时，1B P ∥平面11ACD C ．当12λ=时，有且仅有一个点，使得1A P BP ⊥D ．当2µλ=时，1A P DP +11．在2024年巴黎奥运会艺术体操项目集体全能决赛中，中国队以69.800分的成绩夺得金牌，这是中国艺术体操队在奥运会上获得的第一枚金牌．艺术体操的绳操和带操可以舞出类似四角花瓣的图案，它可看作由抛物线()2:20C y px p =>绕其顶点分别逆时针旋转90,180,270°°°后所得三条曲线与C 围成的（如图阴影区域），,A B 为C 与其中两条曲线的交点，若1p =，则（ ）．开口向上的抛物线的方程为212y x = B ．4AB =C ．直线x y t +=截第一象限花瓣的弦长的最大值为34D ．阴影区域的面积大于4三、填空题（本大题共3小题，每小题5分，共15分．）12．若()523450123451x a a x a x a x a x a x −=+++++，则2a =_____．13．已知函数()24,1,ln 1,1,x x a x f x x x ++<=+≥ 若函数()2y f x =−有3个零点，则实数a 的取值范围是_____． 14．设n T 为数列{}n a 的前n 项积，若n n T a m +=，其中常数0m >，数列1n T为等差数列，则m =_____． 四、解答题（本大题共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．（本小题满分13分）记ABC △的内角,,A B C 所对的边分别为,,a b c ，已知()()b c a b c a bc +−++=． （1）求A ；（2）若D 为BC 边上一点，3,4,BAD CAD AC AD ∠=∠=sin B ． 16．（本小题满分15分）如图，三棱柱111ABC A B C −中，11160,,,1,2A AC AC BC A C AB AC AA ∠=°⊥⊥==．（1）求证：1A C ⊥平面ABC ；（2）若直线1BA 与平面11BCC B ，求平面11A BB 与平面11BCC B 夹角的余弦值． 17．（本小题满分15分）人工智能（AI ）是一门极富挑战性的科学，自诞生以来，理论和技术日益成熟．某公司研究了一款答题机器人，参与一场答题挑战．若开始基础分值为()*m m ∈N 分，每轮答2题，都答对得1分，仅答对1题得0分，都答错得－1分．若该答题机器人答对每道题的概率均为12，每轮答题相互独立，每轮结束后机器人累计得分为X ，当2X m =时，答题结束，机器人挑战成功，当0X =时，答题也结束，机器人挑战失败．（1）当3m =时，求机器人第一轮答题后累计得分X 的分布列与数学期望； （2）当4m =时，求机器人在第6轮答题结束且挑战成功的概率． 18．（本小题满分17分）．已知椭圆()2222:10x y C a b a b +=>>3．,A B 是椭圆的左、右顶点，过,A B 分别做椭圆的切线，取椭圆上x 轴上方任意两点,P Q （P 在Q 的左侧），并过P ，Q 两点分别作椭圆的切线交于R 点，直线RP 交点A 的切线于I ，直线RQ 交点B 的切线于J ，过R 作AB 的垂线交IJ 于K ．（1）求椭圆C 的标准方程；（2）若()1,2R ，直线RP 与RQ 的斜率分别为1k 与2k ，求12k k 的值； （3）求证：IK IA JKJB=．19．（本小题满分17分）对于函数()f x ，若实数0x 满足()00f x x =，则称0x 为()f x 的不动点．已知0a ≥，且()21ln 12f x x ax a =++−的不动点的集合为A ．以min M 和max M 分别表示集合M 中的最小元素和最大元素．（1）若0a =，求A 的元素个数及max A ； （2）当A 恰有一个元素时，a 的取值集合记为B ． （i ）求B ；（ii ）若min a B =，数列{}n a 满足()112,n n nf a a a a +==，集合n C =*141,,3nk k a n = −∈∑N ．求证：*4,max 3n n C ∀∈=N ．长郡中学2025届高三月考试卷（三）数学参考答案题号 1 2 3 4 5 6 7 8 91011答案CADBCBBDBD BCD ABD一、选择题（本大题共8小题，每小题5分，共40分．）1．C 【解析】由题意，{2},{1,2,3},{1,2,3,4},{1,2}A B A B A C C A C ===== ，对比选项可知只有C 选项符合题意．2．A 【解析】因为复数1z 对应的点和复数212i z =+对应的点关于实轴对称，所以112i z =−，所以()()1212i 12i 5z z =−+=． 3．D 【解析】因为()a ab ⊥+，则()290a a b a a b a b ⋅+=+⋅=+⋅= ，故9a b ⋅=− ，所以b 在a 方向上的投影向量为299a b a a a a⋅−⋅=⋅=−．4．B 【解析】因为12,[3,)x x ∀∈+∞，有()()12120f x f x x x −>−，所以()f x 在[3,)+∞上单调递增，又()3f x +是偶函数，则()3f x+的图象关于0x =对称，所以()f x 的图象关于3x =对称，则()()()0654f f f =>=，故选项A 错误；()()154f f ==，故选项B 正确；()()()2454f f f =<=，故选项C 错误；()3f的正负不能确定，故选项D 错误．5．C 【解析】如图，设P 在底面的投影为G ，易知正四棱锥P ABCD −的外接球球心在PG 上， 由题意，球O 的半径5,853PO AO OG ====−=，所以4,8AG PA AB === 故PAB △中，边AB所以该正四棱锥的侧面积为14962××=．．B 【解析】由e x y =得e xy ′=，又切点为（1，e ），故e k =，切线l 为e y x =， 设l 与曲线ln y a x =+的切点为()001,e ,x x y x ′=，所以01e x =，解得切点为1,1e， 所以1ln11ea a +=−=，解得2a =． 7．B 【解析】由A 点的纵坐标为1213，得125sin ,cos 1313αα==，显然ππ42α<<， 而()111sin 2AOB S αβ=×××+=△()sin αβ+，又ππ42β<<， 因此ππ2αβ<+<，3π4αβ+=，有3π4βα=−，)3π512sin sin cos sin 41313βααα=−=+=+=显然B 点在第四象限，所以B 点的纵坐标为 8．D 【解析】如图，设直线2x c =与x 轴交于点,H PH m =， 则tan ,tan 2m mPFH PAH c a c∠=∠=+． 因为APF PFH PAH ∠=∠−∠，所以()tan tan tan tan 1tan tan PFH PAHAPF PFH PAH PFH PAH∠−∠∠=∠−∠=+∠⋅∠()22222212m m m a c a c c a c m m ac c ac c m m c a c m−+++==++++⋅++．因为22ac c m m++≥m =时，等号成立，所以tan APF ∠≤，整理得22430c ac a −−=，则2430e e −−=，解得2e =+．二、选择题（本大题共3小题，每小题6分，共18分．）9．BD 【解析】∵()()3sin f x x ωϕ=+，由题图知33π44T =，∴πT =，2ω=，故A 错误； ∵π2π623T +=，∴可得2π3f是()f x 的最小值，故B 正确； ∵ππ3sin 2366f ϕ=×+=，∴πsin 13ϕ+=，∴π2π6k ϕ=+，k ∈Z ， 又π2ϕ<，∴π6ϕ=，∴()π3sin 26f x x =+，∵π0,2x∈ ，∴ππ7π2,666x +∈ ， ∴()π33sin 2,362f x x=+∈−，故C 错误； 将()f x 的图象向右平移π12个单位长度得到的图象为πππ3sin 23sin 212126f x x x−=−+=，故D 正确．10．BCD 【解析】对于A 中，连接BP ，在长方体1111ABCD A B C D −中，可得1BB ⊥平面ABCD ，所以1B PB ∠即为1B P 与平面ABCD 所成的角，即1π4B PB ∠=，在直角1BB P △中，可得11BP BB ==，所以点P 的轨迹为以B 为圆心，半径为1的14圆，其周长为1π2π142××=，所以A 错误；对于B 中，当λµ=时，因为1222AB AA AD ===，且点P 满足AP AB AD λµ=+，所以点P 在线段AC 上，连接11,,AC AB B C ，在长方体1111ABCD A B C D −中，可得1111,AC A C B C A D ∥∥，因为AC ⊄平面11AC D ，且11A C ⊂平面11A C D ，所以AC ∥平面11AC D ，同理可证1B C ∥平面11A C D ，又因为AC B C C = ，且1,AC B C ⊂平面1AB C ，所以平面1AB C ∥平面11A C D ，因为1B P ⊂平面1AB C ，所以1B P ∥平面11A C D ，所以B 正确；对于C 中．当12λ=时，因为1222AB AA AD ===，且点P 满足AP AB AD λµ=+ ，取,AB CD 的中点,E F ，过接,,EF AF BF ，可得点P 在线段EF 上运动，若1A P BP ⊥，因为1AA ⊥平面ABCD 且BP ⊂平面ABCD ，所以111111,,,AA BP A P A A A A P A A ⊥=⊂ 平面1A AP 、故BP ⊥平面1A AP ，又AP ⊂平面1A AP ，故BP AP ⊥，所以点P 在以AB 为直径的圆上，又因为22AB AD ==，可得线段EF 与以AB 为直径的圆只有一个交点F ，所以当点P 与F 重合时，即当且仅当P 为CD 的中点时，能使得1A P BP ⊥，所以C 正确；对于D 中，当2µλ=时，因为1222AB AA AD ===，且点P 满足AP AB AD λµ=+ ，取,AB CD 的中点,E F ，连接,AF EF ，可得点P 在线段AF 上运动，沿着AF 将直角1AA F △和平面ADF △展开在一个平面上，如图所示，在1AA D △中，113π1,1,4AA AD A AD ==∠=，由余弦定理得2221113π2cos24A D AA AD AA AD =+−⋅=+，所以1A D =1A P DP +的最小值为，所以D 正确．．ABD 【解析】由题意，开口向右的抛物线方程为2:2C y x =，顶点在原点，焦点为11,02F，将其逆时针旋转90°后得到的抛物线开口向上，焦点为210,2F，则其方程为22x y =，即212y x =，故A 正确； 对于B ，根据A 项分析，由222,2y x x y = =可解得0x =或2x =，即2A x =，代入可得2A y =， 由图象对称性，可得()()2,2,2,2A B −，故4AB =，即B 正确； 对于C ，如图，设直线x y t +=与第一象限花瓣分别交于点,M N ，由2,2,y x t y x =−+ =解得11,M M x t y =+− =− 由2,2,y x t x y =−+ =解得1,1N N x y t = =+− ，即得()11,1,1M t N t +−−+， 则弦长为2MN =−由图知，直线x y t +=经过点A 时t 取最大值4，经过点O 时t 取最小值0，即在第一象限部分满足04t <<， 不妨设u=13u <<，且212u t −=，代入得，)()2222113MNuu =+−=−−<<， 由此函数的图象知，当2u =时，MN 取得最大值为，即C 错误；对于D ，根据对称性，每个象限的花瓣形状大小相同，故可以先求18部分面积的近似值．如图：在抛物线()2102yx x ≥上取一点P ，使过点P 的切线与直线OA 平行，由1y x ′==可得切点坐标为11,2P，因为:0OA l x y −=，则点P 到直线OA的距离为d =于是1122OPA S ==△，由图知，半个花瓣的面积必大于12，故原图中的阴影部分面积必大于1842×=，故D 正确． 三、填空题（本大题共3小题，每小题5分，共15分．）12．10−【解析】()51x −的展开式通项是：()55C 1kk k x −−，依题意，得52k −=，即3k =，所以()3325C 110a =−=−． 13．（－3，6）【解析】函数()24,1,ln 1,1,x x a x f x x x ++<=+≥ 当1x ≥时，方程ln 12x +=，解得e x =，函数()2y f x =−有一个零点，则当1x <时，函数()2y f x =−须有两个零点，即242x x a ++=在1x <时有两个解．设()242g x x x a =++−，对称轴为()2,x g x =−在(),2−∞−上单调递减，在()2,−+∞上单调递增，∴()10g >，且()20g −<，即1420,4820,a a ++−> −+−< 解得36a −<<，所以a 的取值范围是（－3，6）．14．1或2【解析】当2n ≥时，111,11n n n n n n n n m mT a T a a m a T m a −−−+=+===++−， 所以()1211111111111121n n n n n n n n a n m T T m a m a m a m ma m m a −−−−−−−−=−=−=≥−−−−−+−．由数列1n T为等差数列，则1211n n a m ma −−−−为常数d ，①若0d =，则()112n a n −=≥恒成立，即()11n a n =≥恒成立，∴2m =；②若0d ≠，则2111n n a dm dma −−−=−，∴21,1,dm dm = = 解得1,1,m d = =综上所述，1m =或2m =．四、解答题（本大题共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．【解析】（1）()()()222222b c a b c a b c a b bc c a bc +−++=+−=++−=，则222b c a bc +−=−，所以2221cos 22b c a A bc +−==−，因为0πA <<，所以2π3A =． （2）由（1）得，2π3A =，因为3BAD CAD ∠=∠，所以π6CAD ∠=，如图，在ACD △中，由余弦定理，得2222cos 31647CD AD AC AD AC DAC =+−⋅∠=+−=，即CD =，在ACD △中，由正弦定理sin sin CD AD DAC C =∠=，所以sin C =，因为π03C <<，故cos C ，在ABC △中，()1sin sin sin cos cos sin 2B A C A C A C =+=+=−．16．【解析】（1）在1A AC △中，由余弦定理可得2221111cos 2AC A A A C A AC AC A A +−∠=⋅⋅，则222112cos 60212A C +−°=××，解得213A C =， 由22211A C AC A A +=，则在1A AC △中，1A C AC ⊥，因为1,,A C AB AC AB ⊥⊂平面,ABC AC AB A = ，所以1A C ⊥平面ABC ．（2）易知1,,A C AC BC 两两相互垂直，分别以1,,CA CB CA 为,,x y z 轴建立空间直角坐标系，如图，设BC k =，则(()()(11,0,,0,0,0,0,,A B k C C −(()(110,,0,,0,,BA k CB k CC −=−设平面11BCC B 的法向量(),,n x y z = ，则10,0,n CB n CC ⋅= ⋅=可得0,0,ky x = −+=令x =0,1y z ==，所以平面11BCC B的一个法向量)n =， 设直线1BA 与平面11BCC B 所成的角为θ，则11sin BA n BA n θ⋅=⋅，可得=1k =，易知(11BB CC ==−，设平面11A BB 的法向量()000,,m x y z = ，则110,0m BA m BB ⋅=⋅=可得00000,0,y x −+= −+=令01z =，则00x y =， 所以平面11A BB的一个法向量)m =，设平面11A BB 与平面11BCC B 的夹角为α，则cos n m n mα⋅==⋅17．【解析】（1）当3m =时，第一轮答题后累计得分X 所有取值为4，3，2，()()()1111111114,32,2,224222224P X P X P X ==×===××===×=所以第一轮答题后累计得分X 的分布列为：所以()1114323424E X =×+×+×=． （2）当4m =时，设“第六轮答题后，答题结束且挑战成功”为事件A ，此时情况有2种，分别为： 情况①：前5轮答题中，得1分的有3轮，得0分的有2轮，第6轮得1分； 情况②：前4轮答题中，得1分的有3轮，得—1分的有1轮，第5、6轮都得1分，所以()3232335411111111C C 4244441024P A =××+××= ． 18．【解析】（1）由题意：22222,2,3, 1.a b a a c bc a b c ==+=⇒= = =+ 所以椭圆C 的标准方程为22143x y +=．4分（2）设过点R 的切线方程为()21y k x −=−，即()2y kx k =+−， 由()222,1,43y kx k x y =+− += 消去y ，整理得()()()222438242120k x k k x k ++−+−−=， 由()()()222206424434212kk k k ∆=⇒−=+−− ，整理得23410k k +−=，所以1213k k =−．（3）设()()000,0,R x y y RK >的延长线交x 轴于K ′点，如图：因为AI K K JB ′∥∥，则022IKAK x JKBK x ′+==′−． 设P ，Q 两点处切线斜率分别为12,k k ，过R 点的椭圆的切线方程为()00y y k x x −=−，即()00y kx y kx =+−，由()0022,143y kx y kx x y =+−+= 消去y ，化简整理，得()()()22200004384120kx k kx y x kx y +−−+−−=，由0∆=，得()()()2222000064443412kkx y k kx y −=+−−，化简整理，得()22200004230x k x y k y −−+−=， 由韦达定理，得20001212220023,44x y y k k k k x x −+==−−，所以()()1002002,2l J y k x y y k x y =−−+=−+， 所以要证明IK IA JKJB=，只需证明()()100002002222k x y x x k x y −−++=−−+，即()()()()()()()()22222000100012001201200042424242,k x y x k x y x k k x y k k x k k x x y −++=−+−⇔++=+⇔+−=因为00122024x y k k x +=−，所以上式成立，即IK IA JK JB =成立． 19．【解析】（1）当0a =时，()1ln 12f x x =+，其定义域为()0,+∞． 由()f x x =得1ln 102x x −+=． 设()1ln 12g x x x =−+，则()122xg x x −′=， 当10,2x∈ 时，()0g x ′>；当1,2x ∈+∞ 时，()0g x ′<， 所以()g x 在10,2上单调递增；在1,2 +∞上单调递减， 注意到()10g =，所以()g x 在1,2+∞上恰有一个零点1x =，且()1102g g>=， 又()22e e 0g −−=−<，所以()21e 02g g −<，所以()g x 在10,2 上恰有一个零点0x ， 即()f x 在1,2 +∞上恰有一个不动点1,x =在10,2上恰有一个不动点0x x =， 所以{}0,1A x =，所以A 的元素个数为2，又因为01x <，所以max 1A =． （2）（i ）当0a =时，由（1）知，A 有两个元素，不符合题意； 当0a >时，()21ln 12f x x ax a =++−，其定义域为()0,+∞， 由()f x x =得21ln 102x ax x a +−+−=． 设()()21ln 1,0,2h x x ax x a x =+−+−∈+∞，则()214212122ax x h x ax x x −+′=+−=， 设()2421F x ax x =−+，则416a ∆=−，①当14a ≥时，()()0,0,0F x h x ′∆≤≥≥，所以()h x 在()0,+∞上单调递增， 又()10h =，所以()h x 在()0,+∞上恰有一个零点1x =， 即()f x 在()0,+∞上恰有一个不动点1x =，符合题意； ②当104a <<时，0∆>，故()F x 恰有两个零点()1212,x x x x <． 又因为()()010,1410F F a =>=−<，所以1201x x <<<， 当()10,x x ∈时，()()0,0F x h x ′>>； 当()12,x x x ∈时，()()0,0F x h x ′<<； 当()2,x x ∈+∞时，()()0,0F x h x ′>>，所以()h x 在()10,x 上单调递增，在()12,x x 上单调递减，在()2,x +∞上单调递增，注意到()10h =，所以()h x 在()12,x x 上恰有一个零点1x =，且()()()()1210,10h x h h x h >=<=， 又0x →时，()h x →−∞，所以()h x 在()10,x 上恰有一个零点0x ′，从而()f x 至少有两个不动点，不符合题意；所以a 的取值范围为1,4 +∞ ，即集合1,4B=+∞ ．（ii ）由（i ）知，1,4B=+∞ ，所以1min 4aB =， 此时，()()22113113ln ,ln 244244f x x x h x x x x +++−+， 由（i ）知，()h x 在()0,+∞上单调递增，所以，当1x >时，()()10h x h >=，所以()f x x >，即()1f x x>，故若1n a >，则11n a +>，因为，若存在正整数N 使得1N a ≤，则11N a −≤，从而21N a −≤，重复这一过程有限次后可得11a ≤，与12a =矛盾，从而，*,1n n a ∀∈>N ， 下面我们先证明当1x >时，()3ln 12x x <−， 设()()33ln ,1,22G x x x x =−+∈+∞，所以()1323022x G x x x ′−=−=<， 所以()G x 在()1,+∞上单调递减，所以()()10G x G <=，即当1x >时，()3ln 12x x <−，从而当1x >时，2211311ln 24444x x x x x ++−<−， 从而()2113ln 1244114xx x x x ++−<−，即()()1114f x x x −<−，故()()1114n nn f a a a −<−， 即()11114n n a a +−<−，由于11,1n n a a +>>， 所以110,10n n a a +−>−>，故11114n n a a +−<−，故2n ≥时，121211111111114444n n n n n a a a a −−−−−<−<−<<−= ，所以*1111114144,111434314n n nk k n k k n a −==− ∀∈−≤==−< −∑∑N ，故4max 3n C =．。
2024天一大联考语文作文
2024天一大联考语文作文
《我的天一大联考语文作文体验》
嘿呀，说起 2024 天一大联考语文作文，那可真是一次让我印象深刻的经历呀！记得当时一拿到卷子，看到作文题目，我的脑袋就“嗡”地一下。
题目是关于梦想的，咱就说这梦想谁还没有啊，可真要写起来，还真有点犯难。
我就开始寻思，我自己的梦想是啥呢？想来想去，想到我小时候特别痴迷看星星。
那时候啊，晚上只要没啥事，我就会跑到院子里，找个舒服的小板凳一坐，仰着头看那满天的星星。
我就想啊，那些星星上是不是有外星人呢？它们是不是也像我们一样生活着？哎呀，我那叫一个好奇。
为了能更清楚地看星星，我还专门让我爸给我买了个小望远镜。
你还别说，通过那个望远镜看星星，感觉就是不一样。
我能看到星星更清晰的样子，有时候还能看到一些移动的小点，我就坚信那就是外星人的飞船。
我甚至还想象着自己以后能坐着飞船去探索那些遥远的星球，去发现更多的未知。
然后我就决定把这个事情写进作文里。
我就开始详细描述我看星星的过程，怎么找小板凳，怎么找个好角度，望远镜拿在手里的感觉，看到星星那
一刻的兴奋，还有我那些奇奇怪怪的幻想。
写着写着，我感觉自己好像又回到了那个充满好奇和幻想的小时候。
等我写完作文，我自己都笑了，没想到看星星这么一件小事，也能写成一篇作文。
现在再想想那次天一大联考语文作文，还真是挺有趣的一次经历呢！感觉就像是把自己生活中的一个小片段给放大了一样。
嘿嘿，这就是我的 2024 天一大联考语文作文体验啦！
以上作文仅供参考，你可以根据实际情况进行调整。
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24．自汉高祖至景帝的60多年间，赋体文学在创作上大体继续着楚辞的余绪，以骚体赋为
主流。有文学史家称，汉初“只可算是辞的时代，而不是赋的时代”。汉初赋体文学形成此特
点的原因主要是
A．西汉初年经济凋敝 B．道家无为思想的制约
C．战国楚辞体的影响 D．政府文化政策的控制
25．司马睿（晋元帝）从东渡到登基，主要依赖了王导、王敦兄弟的大力支持。在此期间，
王导联合南北士族，运筹帷幄，政令己出，王敦则总掌兵权，专任征伐。这说明当时
A．外戚独揽国家大权 B．门阀政治制约皇权
C．中央集权受到削弱 D．国家处于四分五裂
26．宋神宗熙宁时，长安商税每年八万余贯，北方其他州商税超过八万贯的只有高密、历城、
东平、沧州等处。南方的杭州、楚州商税在十万贯以上，扬、潭州超过九万贯，苏、湖、婆
州在七万贯以上，不包括四川，南方商税已占全国之半。这反映了当时
A．商业发展比较迅速 B．商税是财政的主要来源
C．政府鼓励商业发展 D．南北方商业发展不平衡
27.图8为明代仍在使用的唐代出现的“木牛”，即人力代耕架。对此解读正确的是，明代
A．人力获得了解放 B．农业生产出现了分工合作
C．传统农业精细化 D．农业技术没有革命性突破
28.有人指出，“自强求富”的洋务运动，师法俄日的戊戌变法，晚清最后十年的“新政”，在辛
亥革命前的近半个世纪，中国在磕磕绊绊之中已艰难地启动了这一新的历史进程。贯穿“这
一新的历史进程”的主题是
A．实现国家的富强民主 B．抗击西方列强侵略
C．使社会走向开放文明 D．维护清朝统治秩序
29．图9是中国抗击日本作战情况统计。
图9反映出
A．中国的抗战艰难困苦 B．国共两党通力合作抗战
C．两个政府和战场抗战 D．中国实行全面抗战路线
30. 20世纪70年代后期发生的两件不大不小的事件，使中国开启了新的纪元，这就是广东
农民的逃港潮和安徽农民的包产到户。这两件事表明
A．时代需要对外开放 B．改革开放的序幕被揭开
C．对内改革势在必行 D．国家以经济建设为重心
31. 2005年，中国政府提出了“大国是重点，周边是关键，发展中国家是基础”的外交方针。
此外交方针的提出说明中国
A．采取灵活外交策略 B．推行全方位对外开放
C．实行多边外交战略 D．实施独立自主不结盟
32.在古罗马人的意识中，有法与法律之分。法来自正义，指向公正、自然和神意；法律来
自权威机关制定出来的规则，包括“限制和禁令”。西塞罗说：“如果某个规则不应当被确定
为法．确认该规则的法律则是无效的。”这反映出罗马法
A．包括法与法律两部分 B．需维护公平社会秩序
C．是法律与道德的综合 D．不是由权威机关制定
33.1871年以后，在俾斯麦的参与下，德国出台商业法、营业自由法、关税法、货币法等，
还将境内7个不同的货币区和33个具有钞票发行权的银行统一，以马克作为法定货币单位。
这些措施
A．是德意志帝国宪法的本质反映 B．表明帝国首相具有制定法律权利
C．暴露了帝国政府在经济上的反动 D．客观上推动了德国工业革命进程
34.1923年，苏联把对农民征收的多种税收改为统一的农业税，并把实物税改为货币税。上
述做法
A．有利于苏联农业产品商品化 B．标志着苏联农业税制度的成熟
C．是苏联长期推行的税收制度 D．有利于工农联盟和苏联的建立
35．早在2010年年底，为改善全球经济治理，国际货币基金组织(IMF)董事会通过重大改革
方案。此后五年，在绝大多数国家相继批准的情况下，美国却成了唯一阻挡IMF改革的国
家。美国能阻挡IMF改革的主要原因是，美国
A．是国际货币基金组织的创始国 B．综合国力一直位居世界第一
C．持国际货币基金组织股份最多 D．是当今世界上第一大经济体
24．答案A
命题透析本题以汉初文学特点为背景材料，旨在考查考生获取和解读信息、调动和运用知识
的能力。
思路点拨由材料可知，汉初，汉赋没有形成汉中期以后气势磅礴、词藻华丽、大力铺陈的特
点，主要在于社会经济处丁恢复期，制约了汉赋的发展，A项符合题意。B、C两项者都不
是主要原因；汉初文化政策相对比较宽松，汉武帝“罢黜百家、独尊儒术”后，文化政策趋于
严厉，排除D项。
25．答案B
命题透析本题以东晋王导兄弟对政权的把控为背景材料，旨在考查考生获取和解读信息、调
动和运用知识的能力。
思路点拨由材料可知，作为士族的王氏兄弟既掌握“政令”，又“掌兵权”，权倾内外，表明当
时皇权受到门阀士族的限制，B项正确。王导兄弟不是外戚，排除A项；王导兄弟是朝臣，
不能体现中央对地方控制的削弱和国家处于分裂状态，排除C、D两项。
26．答案D
命题透析本题以宋神宗时的商税收入状况为背景材料，旨在考查考生获取和解读信息、调动
和运用知识的能力。
思路点拨材料表明，南方商税多的城市和商税总额均超过北方，说明南方商业发展比北方快，
这反映出南北方商业发展不平衡，D项正确。材料反映的是宋代商税收入情况，无法得出商
业发展速度和各税种所占比重，排除A、B两项；材料没有体现政府的商业政策，排除C项。
27．答案D
命题透析本题以唐代“木牛”在明代的使用为背景材料，旨在考查考生获取和解读信息、调动
和运用知识的能力。
思路点拨材料中“木牛”在唐代已出现，明代仍使用，说明明代耕作农具没有质的发展，农业
技术没有革命性突破，D项正确。“木牛”使用的是人力，排除A项；仅从图片无法得出农业
生产分工合作，排除B项；精细化是由一系列生产环节组成，仅凭材料无法体现，排除C
项。
28．答案C
命题透析本题以中国近代前期的近代化运动为背景材料，旨在考查考生获取和解读信息、调
动和运用知识的能力。
思路点拨洋务运动没有民主色彩，排除A项；清末“新政”不具有抗击西方列强侵略的性质，
排除B项；戊戌变法要求改变清朝统治秩序，排除D项。洋务运动、戊戌变法、清末“新政”
都有利于加强中国与西方国家的联系，推动中国近代社会文明的发展，C项正确。
29．答案A
命题透析本题以中国抗击日本作战情况统计为背景材料，旨在考查考生获取和解读信息、调
动和运用知识的能力。
思路点拨由材料中的统计数据可知，中国在抗击日本侵略过程中充满了艰难困苦，A项正确。
材料是对敌后战场与正面战场作战情况统计，不能反映出国共合作抗战情况和实行全面抗战
路线，排除B、D两项；当时只有一个政府即巾华民国，排除C项。
30．答案C
命题透析本题以20世纪70年代后期中国发生的事件为背景材料，旨在考查考生获取和解读
信息、调动和运用知识的能力。
思路点拨联系所学知识可知，广东农民的逃港潮是因为生活困难，安徽农民包产到户是为了
促进牛产，解决生活困难，这表明当时小国必须进行改革，C项正确。安徽农民的包产到户
不能体现需要对外开放，排除A项；十一届三中全会的召开揭开了改革开放的序幕，排除B
项；虽然安徽农民的包产到户是经济事件，但不能由此得出国家以经济建设为重心，排除D
项。
31．答案C
命题透析本题以新时期中国外交方针为背景材料，旨在考查考生获取和解读信息、调动和运
用知识的能力。
思路点拨材料表明，中国政府既重点发展与大国的关系，又注重发展与周边国家和第三世界
国家的关系，这是新时期多边外交方针的体现，C项正确。A项中“灵活”材料无法体现；材
料体现的是外交方针，不是对外开放，排除B项；材料体现了独立自主，但没有体现不结盟，
排除D项。
32．答案B
命题透析本题以罗马法为背景材料，旨在考查考生获取和解读信息、调动和运用知识的能力。
思路点拨分析材料可知，法律必须包含着法即正义，否则无效，B项正确。材料叙述的是人
们的法律意识，并不是岁马法律与道德，排除A、C两项；D项与材料中“法律来自权威机关
制定出来的规则”相悖。
33．答案D
命题透析本题以德国统一后采取的促进经济发展的措施为背景材料，旨在考查考生获取和解
读信息、调动和运用知识的能力。
思路点拨材料体现的是德国统一后采取的经济措施，无法体现德意志帝国宪法的本质，A项
错误；材料只能说明俾斯麦参与了立法和法律的颁布，不能说明帝国首相具有立法权，B项
错误；材料中俾斯麦政府的做法，是为了适应统一之后德国社会的发展要求，C项错误。联
系所学知识可知，经济法规的制定及银行和货币的统一，有利于德国国内统一市场的形成，
推动德国资本主义经济的发展和工业革命的开展，D项正确。
34．答案A
命题透析本题以1923年苏联农业税改革为背景材料，旨在考查考生获取和解读信息、调动
和运用知识的能力。
思路点拨由材料可知，苏联把实物税改为货币税，农民要交税，需要把农产品出卖转化为货
币，这推动了苏联的农产品商品化，A项正确。这些做法在1927年后逐步取消，B、C两项
错误；苏联建立于1922年12月，D项错误。
35．答案C
命题透析本题以美国对IMF改革的阻挠为背景材料，旨在考查考生获取和解读信息、调动
和运用知识的能力。
思路点拨联系所学知识可知，国际货币基金组织( IMF)投票权是由成员国的配额多少所决定
的，所占的股份越多，投票权也就越大。美国在国际货币基金组织成立时拥有最多股权，因
而美国也就拥有国际货币基金组织决策的否决权，C项正确。