高二下学期期中复习测试题

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北京市2023-2024学年高二下学期期中考试化学试题含答案

北京市2023-2024学年高二下学期期中考试化学试题含答案

北京市2023-2024学年第二学期期中测试高二化学(答案在最后)试卷说明:试卷分值100,考试时间90分钟,I卷为选择题,共22个小题,II卷为主观题,包括第23至第27题可能用到的相对原子质量:H1B11C12N14O16Cu64I卷一.选择题(共22个小题,每题2分,共44分。

每小题只有一个正确选项,请选择正确答......案填在机读卡相应的题号处............)1.下列变化过程只需要破坏共价键的是A.碘升华B.金刚石熔化C.金属钠熔融D.氯化钠溶于水【答案】B【解析】【详解】A.碘升华破坏的是分子间作用力,A错误;B.金刚石中碳碳之间是共价键,融化的时候,需要破坏共价键,B正确;C.金属钠属于金属晶体,融化的时候破坏的是金属键,C错误;D.氯化钠中存在着钠离子和氯离子之间的离子键,溶于水时破坏的是离子键,D错误;故选B。

2.某粗苯甲酸样品中含有少量氯化钠和泥沙。

用重结晶法提纯苯甲酸的实验步骤中,下列操作未涉及的是A. B. C. D.【答案】C【解析】【详解】A.图中加热溶解,便于分离泥沙,故A正确;B.冷却结晶可析出苯甲酸晶体,故B正确;C.重结晶实验中不涉及萃取、分液,故C错误;D.苯甲酸在水中溶解度随温度降低而减小,需要趁热过滤,防止损失,故D正确;故选:C。

3.下列物质的类别与所含官能团都正确的是A.醛类—CHOB.羧酸—COOHC.酚类—OH D.CH 3OCH 3酮类—O—【答案】B【解析】【详解】A .属于酯类,官能团为-COO-,A 错误;B .属于羧酸,官能团为-COOH ,B 正确;C .属于醇类,官能团为-OH ,C 错误;D .CH 3OCH 3属于醚类,官能团为醚键:-O-(与氧原子直接相连的原子为碳原子),D 错误;故选B 。

4.下列物质的一氯代物只有一种的是A.乙烷B.丙烷C.邻二甲苯D.对二甲苯【答案】A【解析】【详解】A .乙烷只有一种位置的H 原子,因此其一氯取代产物只有一种,A 符合题意;B .丙烷有2种不同位置的H 原子,因此其一氯取代产物有2种,B 不符合题意;C .邻二甲苯有3种不同位置的H 原子,因此其一氯代物有3种不同结构,C 不符合题意;D .对二甲苯有2种不同位置的H 原子,因此其一氯代物有2种不同结构,D 不符合题意;故合理选项是A 。

高二下学期期中考试复习一

高二下学期期中考试复习一

知识方法·要点1、掌握磁通量概念及其意义,能够正确判断磁通量的变化情况。

2、了解电磁感应现象,掌握发生电磁感应现象,产生感应电动势、产生感应电流的条件。

3、掌握右手定则和楞次定律,并能灵活运用于感应电流方向的判断。

4、掌握法拉第电磁感应定律,明确tE ∆φ∆=和E=LvB 两种表述形式的适用条件和适用范围,并能运用法拉第电磁感应定律熟练地计算电磁感应现象中所产生的感应电动势。

5、对导体棒旋转切割磁感线时所产生的感应电动势能够灵活地运用法拉第电磁感应定律做出正确的计算。

6、了解自感现象,掌握自感现象中的基本特征。

要点整合要点一 楞次定律的理解和应用1.楞次定律解决的问题是感应电流的方向问题,它涉及到两个磁场,______________ (新产生的磁场)和____________________ (原来就有的磁场),前者和后者的关系不是“同向”和“反向”的简单关系,而是前者“阻碍”后者“变化”的关系.2.对“阻碍意义的理解” (1)阻碍原磁场的变化.“阻碍”不是阻止,而是“延缓”,感应电流的磁场不会阻止原磁场的变化,只能使原磁场的变化被________或者说被_______了,原磁场的变化趋势不会改变,不会发生逆转.(2)阻碍的是原磁场的变化,而不是原磁场本身,如果原磁场不变化,即使它再强,也不会产生___________.(3)阻碍不是相反,当原磁通量减小时,感应电流的磁场与原磁场_____,以阻碍其_____;当磁体远离导体运动时,导体运动将和磁体运动_______,以阻碍其___________.(4)由于“阻碍”,为了维持原磁场的变化,必须有外力克服这一“阻碍”而做功,从而导致___________转化为_____,因而楞次定律是_____________和_____________在电磁感应中的体现.3.运用楞次定律处理问题的思路 (1)判定感应电流方向问题的思路运用楞次定律判定感应电流方向的基本思路可以总结为“一原、二感、三电流.........”. ①明确..原磁场:弄清原磁场的方向以及磁通量的变化情况. ②确定..感应磁场:即根据楞次定律中的“阻碍”原则,结合原磁场磁通量变化情况,确定出感应电流产生的感应磁场的方向.③判定..电流方向:即根据感应磁场的方向,运用安培定则判断出感应电流方向. (2)判断..闭合电路(或电路中可动部分导体)相对运动类问题的分析策略 在电磁感应问题中,有一类综合性较强的分析判断类问题,主要是磁场中的闭合电路在一定条件下产生了_______,而此电流又处于磁场中,受到_____力作用,从而使闭合电路或电路中可动部分的导体发生了运动.要点二 电磁感应中的力学问题通过导体的感应电流在磁场中将受到安培力作用,从而引起导体速度、加速度的变化. (1)基本方法①用法拉第__________定律和_______定律求出感应电流的____和感应电动势的_____. ②求出回路中________的大小.③分析研究导体______情况(包括安培力,用______定则确定其方向) ④列出________方程或者_______方程求解. 感应电流→感应电流→通电导体受到安培力→合外力变化→加速度变化→速度变化,周而复始的循环,循环结束时,加速度为____,导体达稳定状态,速度达到______.要点三 电磁感应中的电路问题在电磁感应中,切割磁感线的导体或磁通量变化的回路将产生感应电动势,该导体或回路相当于电源.解决电路问题的基本方法:①用法拉第电磁感应定律或楞次定律确定感应电动势的____和_____ ②画出_____电路图.③运用闭合电路欧姆定律、串并联电路性质,电功率等公式进行求解. 要点四 电磁感应中的图象问题电磁感应中常常涉及磁感应强度、磁通量、感应电动势、感应电流、安培力或外力随时间变化的图象.这些图象问题大体上可以分为两类:①由给定的电磁感应过程选出或画出正确图象.②由给定的有关图象分析电磁感应过程,求解相应的物理量.不管是何种类型,电磁感应中的图象问题往往需要综合_______定则、_______定则、_____定律和____________定律等规律分析解决.要点五 电磁感应中的能量问题电磁感应过程中产生的感应电流在磁场中必定受到______作用,要维持感应电流的存在,必须有“外力”克服安培力做功,此过程中,其它形式的能量转化为电能,“外力”克服安培力做了多少功,就有多少其它形式的能转化为______.当感应电流通过用电器时,电能又转化为其它形式的能量,安培力做功的过程,是_______转化为_______________________的过程,安培力做了多少功,就有多少电能转化为其它形式的能量.考 点考点1 感应电流的产生和方向1、如图,金属棒ab 置于水平放置的U 形光滑导轨上,在ef 右侧存在有界匀强磁场B ,磁场方向垂直导轨平面向下,在ef 左侧的无磁场区域cdef 内有一半径很小的金属圆环L ,圆环与导轨在同一平面内。

山东省青岛第二中学2023-2024学年高二下学期期中考试数学试卷(含简单答案)

山东省青岛第二中学2023-2024学年高二下学期期中考试数学试卷(含简单答案)

青岛第二中学2023-2024学年高二下学期期中考试数学试卷一、选择题:本题共8小题,每小题5分,共40分.在每小题给出的四个选项中,只有一项是符合题目要求的.1. 设集合,则( )A. B. C. D.2. 若关于的不等式成立的充分条件是,则实数的取值范围是( )A B. C. D.3. 下列有关一元线性回归分析的命题正确的是( )A. 若两个变量的线性相关程度越强,则样本相关系数就越接近于1B. 经验回归直线是经过散点图中样本数据点最多的那条直线C. 在经验回归方程中,若解释变量增加1个单位,则预测值平均减少0.5个单位D. 若甲、乙两个模型的决定系数分别为0.87和0.78,则模型乙的拟合效果更好4. 已知,则下列命题为真命题的是( )A. 若,则 B. 若,则C. 若,则 D. 若,则5. 7名同学到甲、乙、丙三个场馆做志愿者,每名同学只去1个场馆,甲场馆安排3名,乙场馆安排2名,丙场馆安排2名,则不同的安排方法共有( )A. 210种B. 420种C. 1260种D. 630种6. 已知一组样本数据的方差为9,且,则样本数据的方差为( )A. 9.2B. 8.2C. 9.8D. 97. 若不等式的解集为,则不等式解集为( )A B. ..{1,2,3,4,5},{1,3,5},{1,2,5}U T S ===()U S T = ð{2}{1,2}{2,4}{1,2,4}x |1|x a +<04x <<a 1a ≤-5a >1a <-5a ≥r ˆ20.5yx =-x ˆy 2R ,,R a b c ∈a b >ac bc>0a b >>0.40.4a b -->a b >1122a cb c++⎛⎫⎛⎫< ⎪⎪⎝⎭⎝⎭0,0a b c >>>b b c a a c+>+125,,,x x x 1324x x x x +=+123451,1,1,1,x x x x x -+-+20ax bx c ++≥[]1,30ax ccx b+≥+(]4,3,3∞∞⎡⎫--⋃+⎪⎢⎣⎭(]4,3,3∞∞⎛⎫--⋃+⎪⎝⎭C. D. 8. 某人在次射击中击中目标的次数为,其中,击中偶数次为事件A ,则( )A. 若,则取最大值时B. 当时,取得最小值C. 当时,随着的增大而减小 D. 当的,随着的增大而减小二、选择题:本题共3小题,每小题6分,共18分.在每小题给出的选项中,有多项符合题目要求.全部选对的得6分,部分选对的得部分分,有选错的得0分.9. 在的展开式中,下列说法正确的是( )A. 各二项式系数的和为64 B. 常数项是第3项C. 有理项有3项D. 各项系数的绝对值的和为72910. 已知位于第一象限的点在曲线上,则( )A. B. C. D.11. 二次函数是常数,且的自变量与函数值的部分对应值如下表:…-1012……22…且当时,对应的函数值.下列说法正确的有( )A. B. C. 关于的方程一定有一正、一负两个实数根,且负实数根在和0之间D. 和在该二次函数的图象上,则当实数时,三、填空题:本题共3小题,每小题5分,共15分.12. 函数定义域是______.13. 已知集合,,若中恰有一个整数,的43,3⎡⎤-⎢⎥⎣⎦43,3⎡⎫-⎪⎢⎣⎭n ,~(,)X X B n p N*,01n p ∈<<10,0.8n p ==()P X k =9k =12p =()D X 112p <<()P A n 102p <<()P A n 61x ⎛- ⎝(,)a b 111x y+=(1)(1)1a b --=-228a b +≥23a b +≥+221223a b +≥2,(,y ax bx c a b c =++0)a ≠x y x ym n32x =0y <0abc >1009mn >x 20ax bx c ++=12-()112,P t y +()222,P t y -12t <12y y >()ln(21)f x x =+-{}2|60M x x x =+->{}2|230,0N x x ax a =-+≤>M N ⋂则的最小值为_________.14. 已知函数,若对于恒成立,则实数的取值范围是______.四、解答题:本题共5小题,共77分.解答应写出必要的文字说明、证明过程或演算步骤.15. 2024年4月25日,神舟十八号载人飞船发射升空,并于北京时间2024年4月26日3时32分,成功对接于空间站天和核心舱径向端口,整个自主交会对接过程历时约6.5小时!奔赴星辰大海,中国人探索浪漫宇宙脚步驰而不息,逐梦太空的科学探索也不断向前。

黑龙江省哈尔滨德强高级中学2023-2024学年高二下学期期中测试英语试题

黑龙江省哈尔滨德强高级中学2023-2024学年高二下学期期中测试英语试题

黑龙江省哈尔滨德强高级中学2023-2024学年高二下学期期中测试英语试题一、阅读理解At Beaver Creek, The Extraordinary Awaits You Are no two snowflakes alike? The snowflakes we see in the winter are most likely completely unique from one other.Beaver Creek is a great place to experience the beauty of the snow, with programs for everyone—from children, teens, and women-only lessons to small groups and private-guided experiences.First Track, from Beaver Creek Reserve, lets you be the first on the mountain, with an adventure that begins at 7:30 a.m. when you are met by ski professionals and taken on a private, guided tour—before the mountain is open to the public. Once you have skied, you are treated to a delicious breakfast at Allie’s Cabin.If you are looking for a higher level of comfort there is the White Carpet Club, from Beaver Creek Reserve. Located in the heart of Beaver Creek Village, it maximizes your time on the mountain by streamlining your access to it. At the club, there are private lockers and boot dryers, along with preferred self-parking and a slope-side ski waiter. A receptionist can assist with lift tickets, pass purchases, dinner reservations, and activity recommendations.Of course, there is more to explore during the winter in Beaver Creek as well. There is ice skating, snowshoeing, shopping, and spas—you name it, Beaver Creek has it. It is the perfect place to take advantage of the snow and be in the moment, in the mountains, together.The extraordinary is a rare combination of one-of-a-kind experiences designed to be shared with service that exceeds expectation. The extraordinary brings you closer to one another and offers a special place to belong together. Belong in The Extraordinary.1.First Track can offer visitors ________.A.an early visit B.an ice skating showC.a tasty lunch D.a free skiing lesson2.What is the White Carpet Club special for?A.Skillful trainers.B.Quiet living experience.C.Thoughtful service.D.Good views over the mountain.3.The passage is written to ________.A.attract visitors B.compare different programsC.appeal for sports D.introduce training coursesPeople are often surprised by my fear of blood and needles. Working in a public health unit, I probably have more daily interaction with syringes (注射器) than the average person. But the truth is, having my blood drawn scares me.When I was 9, I had an extremely high temperature. A nurse rudely stuck the needle into my arm, causing a sharp pain. I avoided having blood taken from my body for the next years. A few months before I turned 30, my doctor ordered a blood test as part of my physical exam. I panicked. Sensing my fear, she gently suggested I see a psychiatrist (精神科医生). “Dr. M helped my other patients with the same fear,” she said.When Dr. M called to set up an appointment, I hesitated. Talking about personal problems to a medically licensed stranger was uncommon in my family. “A psychiatrist?” Mother asked. “He’ll just put you on drugs. Can’t you get over this yourself?” I almost listened. But I went to Dr. M’s clinic finally. “Phobias (恐惧症) are pretty easy to treat,” he said. I wasn’t convinced of his words. But despite my doubts, I was running out of options.We started with basic exposure treatment. I watched videos of blood being drawn every day. As I started to feel more at ease with needles, Dr. M suggested drawing blood. As he gently pressed the needle into my flesh, I felt dizzy and breathed deeply. After a few minutes, I looked at the needle. It wasn’t so bad. I didn’t pass out. “You’re doing well,” Dr. M said.I still find it unpleasant getting blood drawn, but thanks to several months of treatment, I’m less scared. I hate to admit it, but Dr. M is right — phobias can be treated. Sometimes, we have to leave our comfort zone and face our fears. And this can lead to freedom in the end.4.What caused the author’s phobia?A.Her sensitivity to pains.B.Her awful performance in study.C.Her bad medical experience.D.Her regular contact with syringes.5.What was the author’s mother’s attitude toward her appointment with Dr. M?A.Conservative.B.Disapproving.C.Supportive.D.Optimistic.6.What can we know about the author’s treatment?A.It cost the author a great deal of money.B.It eased the author’s pain totally.C.It centered on watching videos.D.It was smoother than expected.7.Which can be a suitable title for the text?A.Bravery brings us a happier lifeB.Walk toward what scares usC.Ups and downs make one strongD.Nurse our physical health carefullySports can help you keep fit and get in touch with nature. However, whether you are on the mountains, in the waves, or on the grassland, you should be aware that your sport of choice might have great influence on the environment.Some sports are resource-hungry. Golf, as you may know, eats up not only large areas of countryside, but also tons of water. Besides, all sorts of chemicals and huge amounts of energy are used to keep its courses (球场) in good condition. This causes major environmental effects. For example, in the dry regions of Portugal and Spain, golf is often held responsible for serious water shortage in some local areas.There are many environment-friendly sports. Power walking is one of them that you could take up today. You don’t need any special equipment except a good pair of shoes; and you don’t have to worry about resources and your purse. Simple and free, power walking can also keep you fit. If you walk regularly, it will be good for your heart and bones. Experts say that 20 minutes of power walking daily can make you feel less anxious, sleep well and have better weight control.Whatever sport you take up, you can make it greener by using environment-friendly equipment and buying products made from recycled materials. But the final goal should be “green gyms”. They are better replacements (代替物) for traditional health clubs and modern sports centers. Members of green gyms play sports outdoors, in the countryside or other open spaces. There is no special requirement for you to start your membership. And best of all, it’s free. 8.What is the meaning of the underlined words in paragraph 2?A.resource-consuming B.resource-developingC.resource-protecting D.resource-controlling9.Which of the following is NOT the advantage of power walking according to the passage?A.It is environmentally-friendly.B.It improves our health.C.It is simple and free.D.It can make you put on weight. 10.Which sport is the writer most likely to advise us to do?A.Playing basketball in a gym.B.Motor racing in the desert.C.Cycling around a lake.D.Swimming in a sports center.11.What is the main purpose of writing the passage?A.To show us some major outdoor sports.B.To encourage us to go in for green sports.C.To discuss the influence of some popular sports.D.To introduce some environmentally-friendly sports.In February, news broke that Salesforce CEO Marc Benioff had taken a “digital detox(脱瘾)”: 10 tech-free days at a French Polynesian resort. For a small group of people, taking a step back from devices is an achievable dream, but for most, it’s an impossibility, especially.A digital detox requires dismissing technology almost entirely: taking a break from screens, social media and video conferences for multiple days. The goals—reducing stress or anxiety, andreconnecting with the physical world—are well-intentioned. However, experts say a digital detox isn’t practical anymore for most people.“Technology is very much a part of us now. We bank with an app, read restaurant menus on phones and even sweat with exercise instructors through a screen,” says Seattle-based consultant Emily Cherkin, who specialises in screen-time management. “It’s so embedded(嵌入式的) in our lives, we’re setting ourselves up for failure if we’re going to go phone-free for a week.”As people become increasingly interdependent on technology, doing a digital detox no longer seems like a reasonable goal. But there may be a more realistic solution that will lessen our tech obsession(着迷), without forcing us to totally disconnect.Rather than cutting out technology altogether, practice digital mindfulness. Make sure the use of technology is purposeful. Instead of a full detox, digital mindfulness may be more practical for some people: less worry about cutting tech out entirely, and more focus on being intentional with its use.The goal shouldn’t be to cut off technology or to put a full stop. People still need to send an email, but can do so without getting distracted by the various online contents.This approach is called “grey detoxing”—you’re not totally immersed(沉浸的) or totally cut off from technology. Instead of causing ourselves more anxiety by attempting to live without our phones for a week, we can approach unavoidable screen time in a way that feels right for our individual lives.12.Why does the author mention Marc Benioff in paragraph 1?A.To criticize the idea of digital detox.B.To urge readers to take a digital detox.C.To bring digital detox up for discussion.D.To recommend a French Polynesian resort. 13.Why is a digital detox impossible for most people nowadays?A.They are more stressed and anxious.B.They are deeply influenced by tech.C.They are unwilling to go phone-free.D.They tend to be more pessimistic. 14.What might be a solution to tech obsession?A.Going on a holiday to Polynesia.B.Avoiding tech altogether.C.Disconnecting occasionally.D.Using tech purposefully.15.What does the author think of less screen time in modern society?A.It is achievable.B.It is unreasonable.C.It is ridiculous.D.It is unrealistic.How to Overcome Your FearsHere are five practical strategies that will help you overcome your fears and face a brighter future.Identify your fearsTake a moment to explore inward and identify the specific sources of your fears. Recognize what it is that is holding you back. Ask yourself, “What are you genuinely afraid of?” Spend a few quiet moments observing your thoughts and emotions. 16 .Understand the root causeSpend some time exploring the nature of your fears. Begin reflective thinking by asking yourself, “Why am I experiencing fear?” and “What is causing me to feel anxious?” 17 , you will gain an understanding of the factors behind your fears. When you deeply understand the sources of your fears, you are in a better position to overcome them.Practice acceptanceThe journey of overcoming fear requires acceptance. 18 . Instead, it is a catalyst(催化剂)that guides you on the path of personal development. By recognizing that fear is a universal human emotion that everyone faces at certain times, you can employ its energy for positive purposes.19To overcome your fears effectively, you need to set clear and achievable goals. That’s because you can use a sense of purpose to drive you forward on your journey. In addition, breaking down your broader goals into smaller ones is important.Take action20 . Start small by dealing with the least frightening aspect of your fear and gradually work your way up to more challenging steps. As you go through this journey of action, unexpected opportunities for personal growth will arise. Ultimately, you’ll strengthen your self-confidence as you reconsider your beliefs about life and yourself.A.Establish clear goalsB.Remember all your goalsC.If you take action at onceD.When you answer these questionsE.Then write down in detail what comes up for youF.Accepting fear is not a barrier to self-improvementG.At this point on your journey to overcome fear, it’s time to take action二、完形填空A few years ago, I walked into Panera and placed my order. After I paid and filled my plastic cup with water, I walked further into the cafe. As I sat down, I 21 an older man eating his soup alone. The chair in his front was 22 : he wasn’t saving the seat. He was genuinely eating all alone. As this fact settled in, a feeling of sadness began to 23 me. Why was he eating alone? Was he lonely?This wasn’t the first time I’d felt 24 when spotting someone eating alone. I 25 assume they’re lonely and need someone to be there for them. For some reason, eating with others is the 26 . Modern society has evolved to the point 27 most people do almost every activity together. If we’re getting lunch after class, we’d rather 28 to see if someone will come with us. But is it possible we just don’t want to appear lonely?Many people are always 29 someone else, and that may make people think we always need someone with us to feel 30 about ourselves. But that’s not true. We can enjoy being alone— not everyone needs to be 31 surrounded by friends to be happy. And we shouldn’t be 32 to eat alone if that’s what we want to do.I don’t think I’ll ever not get sad if I see someone eating alone, but I’ll 33 in mind that maybe they just want a 34 from the world, or maybe they prefer it that way. It’s important to 35 seeing someone doing something alone doesn’t always mean they’re lonely.21.A.recognized B.noticed C.examined D.disturbed 22.A.delicate B.unique C.valuable D.empty 23.A.respond to B.belong to C.reflect on D.wash over 24.A.sad B.desperate C.dizzy D.exited25.A.precisely B.automatically C.cruelly D.selfishly 26.A.liberation B.misery C.norm D.solution 27.A.which B.where C.what D.when 28.A.break down B.leave behind C.get away D.ask around 29.A.against B.beyond C.around D.within 30.A.better B.prouder C.nobler D.gentler 31.A.constantly B.dynamically C.decently D.fortunately 32.A.blank B.determined C.delighted D.afraid 33.A.refresh B.keep C.absorb D.decide 34.A.guarantee B.reputation C.break D.reward 35.A.realize B.promote C.recall D.convey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。

高二下学期期中考试物理测试卷

高二下学期期中考试物理测试卷

高二下学期期中考试物理试卷学校:___________姓名:___________班级:___________考号:___________一、单选题1.如图甲所示为医护人员利用“彩超”测定血管中血液的流速,图乙为波源S连续振动,形成的水波,靠近桥墩P时浮出水面的叶片A静止不动。

下列说法不正确的是()A.图甲中,测定血管中血液的流速,利用了多普勒效应的原理B.若波源与仪器相互靠近,仪器接收到超声波的频率变大C.乙图中,为使水波能带动叶片振动,可降低波源振动的频率D.乙图中,为使水波能带动叶片振动,可减小波源距桥墩的距离2.a、b两束相互平行的单色光,以一定的入射角照射到平行玻璃砖上表面,经平行玻璃砖折射后汇聚成一束复色光c,从平行玻璃砖下表面射出,如图所示。

下列说法正确的是()A.b光波长比a光波长短B.a光在玻璃砖中的传播速度比b光在玻璃砖中的传播速度大C.用同一装置做双缝干涉实验时,a光照射得到的条纹间距小D.增大入射角,a光在玻璃砖下表面可发生全反射3.如图甲所示,弹簧振子以O点为平衡位置,在A、B两点之间做简谐运动,取向右为正方向,振子的位移x随时间t的变化如图乙所示。

下列说法正确的是()A .0.1s t =时,振子的速度方向向左B .0.2s t =时,振子的加速度为正向最大C .振子做简谐运动的表达式为()12sin 5cm 2x t ππ⎛⎫=- ⎪⎝⎭ D .在0.1~0.2s 时间内,振子的动能逐渐增大4.主动降噪耳机能根据环境中的噪声(纵波)产生相应的降噪声波,如图所示,降噪声波与环境噪声同时传入人耳,两波相互叠加,达到降噪的目的。

下列说法正确的是( )A .主动降噪技术应用了波的衍射原理B .降噪声波与环境噪声的波长必须相等C .降噪声波和环境噪声发生干涉,耳膜振动加强D .环境噪声频率越高,从耳机传播到耳膜的速度越大5.图甲是某一交流发电机的示意图,两磁极N 、S 间存在可视为水平向右的匀强磁场,V 为理想电压表,电阻9R =Ω,线圈内阻1r =Ω。

2020-2021学年江西省南昌市洪都中学高二下学期期中数学复习卷(含答案解析)

2020-2021学年江西省南昌市洪都中学高二下学期期中数学复习卷(含答案解析)

2020-2021学年江西省南昌市洪都中学高二下学期期中数学复习卷一、单选题(本大题共12小题,共60.0分)(a∈R)是纯虚数,则a=()1.设i是虚数单位,若复数a+1+i1−iA. −2B. −1C. 0D. 12.某产品的广告费用x与销售额y的不完整统计数据如表:广告费用x(万元)345销售额y(万元)2228m若已知回归直线方程为ŷ=9x−6,则表中m的值为()A. 40B. 39C. 38D. 373.给出以下四个问题:①输入一个正数x,求它的常用对数值;②求面积为6的正方形的周长;③求三个数a,b,c中的最大数;④求函数的函数值.其中不需要用条件语句来描述其算法的有()A. 1个B. 2个C. 3个D. 4个4.设a是直线,α是平面,那么下列选项中,可以推出a//α的是()A. 存在一条直线b,a//b,b⊂αB. 存在一条直线b,a⊥b,b⊥αC. 存在一个平面β,a⊂β,α//βD. 存在一个平面β,a⊥β,α⊥β5.一个几何体的三视图及其尺寸如图,则该几何体的表面积为()A. 24πB. 15πC. 15D. 246.在三棱柱ABC−A1B1C1中,侧棱垂直于底面,∠ACB=90°,∠BAC=30°,BC=1,且三棱柱ABC−A1B1C1的体积为3,则三棱柱ABC−A1B1C1的外接球的表面积为()A. 16πB. 12πC. 8πD. 4π7.若某班从4名男生、2名女生中选出3人参加志愿者服务,则至少选出2名男生的概率为()A. 15B. 25C. 35D. 458.一个长方体,过同一个顶点的三个面的面积分别是√6,√3,√2,则长方体的对角线长为()A. 2√3B. 3√2C. 6D. √69.棱长为2√3的正四面体内切一球,然后在正四面体和该球形成的空隙处各放入一个小球,则这些球的最大半径为()A. √2B. √22C. √24D. √2610.通过随机询问11名性别不同的大学生是否爱好某项运动,得到如下的列联表:参照附表,得到的正确结论是()A. 有99%以上的把握认为“爱好该项运动与性别有关”B. 有99%以上的把握认为“爱好该项运动与性别无关”C. 在犯错误的概率不超过0.1%的前提下,认为“爱好该项运动与性别有关”D. 在犯错误的概率不超过0.1%的前提下,认为“爱好该项运动与性别无关”11.平面内有n个圆,其中每两个圆都相交于两点,且每三个圆都不共点,用f(n)表示这n个圆把平面分割的区域数,那么f(n+1)与f(n)之间的关系为()A. f(n+1)=f(n)+nB. f(n+1)=f(n)+2nC. f(n+1)=f(n)+n+1D. f(n+1)=f(n)+n−112.如图,在单位正方体ABCD−A1B1C1D1中,点P在线段AD1上运动,给出以下四个命题:①三棱锥D−BPC1的体积为定值;②异面直线C1P与直线CB1所成的角为定值;③二面角P−BC1−D的大小为定值;④AP⊥平面A1B1CD.其中真命题有()A. 1个B. 2个C. 3个D. 4个二、单空题(本大题共4小题,共20.0分)13.若m,n∈{1,2,3,4,5},且m≠n,则函数f(x)=mx2−nx+2在(−∞,1]上是减函数的概率是______ .14.16.已知椭圆具有性质:若是椭圆上关于原点对称的两个点,点是椭圆上任意一点,当直线的斜率都存在时,记为,则.试对双曲线写出类似的结论15.观察下列等式据此规律,第n个等式可为______ .16.如图,正方体的棱长为1,P为BC的中点,Q为线段上的动点,过点A,P,Q的平面截该正方体所得的截面记为S.则下列命题正确的是_____(写出所有正确命题的编号).①当时,S为四边形;②当时,S为等腰梯形;③当时,S与的交点R满足;④当时,S为六边形;⑤当时,S的面积为.三、解答题(本大题共6小题,共70.0分)17.如图1,在四边形ABCD中,AD//BC,∠D=90°,BC=3,AD=DC=1.把△ACD沿着AC翻折至△ACD1的位置,D1∉平面ABC,连结BD1,如图2.(1)当BD1=2√2时,证明:平面ACD1⊥平面ABD1;(2)当三棱锥D1−ABC的体积最大时,求点B到平面ACD1的距离,18.在复平面内,复数1−i对应的点在第几象限?i19.如图所示,在四棱锥P−ABCD中,△PAD是等边三角形,平面PAD⊥平面ABCD,AD//BC,BC=6,AD=4,∠ABC=∠DCB=60°,O为AD的中点.(1)求证:AB⊥平面POC;(2)过点O作平面α//平面PCD,平面α与棱PB交于点Q,求三棱锥P−COQ的体积.20.为推行“高中新课程改革”,某数学老师分别用“传统教学”和“新课程”两种不同的教学方式,在甲、乙两个平行班级进行教学实验,为了比较教学效果.期中考试后,分别从两个班级中各随机抽取20名学生的成绩进行统计,结果如下表:记成绩不低于120分者为“成绩优良”.分数[100,109)[110,119)[120,129)[130,139)[140,150]甲班频数75431乙班频数12557(1)从以上统计数据填写下面2×2列联表,并判断能否犯错误的频率不超过0.01的前提下认为“成绩优良与教学方式有关”?甲班乙班总计成绩优良成绩不优良总计P(K2≥k0)0.100.050.0250.010 k0 2.706 3.841 5.024 6.635,其中n=a+b+c+d.临界值表如上表:附:K2=n(ad−bc)2(a+b)(c+d)(a+c)(b+d)(2)现从上述40人中,学校按成绩是否优良采用分层抽样的方法抽取8人进行考核,在这8人中,记成绩不优良的乙班人数为X,求X的分布列及数学期望.21.如图,PA垂直矩形ABCD所在的平面,AD=PA=2,CD=2,点E,F分别是AB,PD的中点.求证:AF//平面PCE.求证:平面PCE⊥平面PCD.求四面体PECF的体积.22.某种产品的广告费支出与销售额(单位:万元)之间有如下对应数据:245683040605070(1)求回归直线方程;(2)试预测广告费支出为10万元时,销售额多大?(3)在已有的五组数据中任意抽取两组,求至少有一组数据其预测值与实际值之差的绝对值不超过5的概率.(参考数据:参考公式:线性回归方程系数:,)【答案与解析】1.答案:C解析:解:∵a+1+i1−i =a+(1+i)(1+i)(1−i)(1+i)=a+i,若复数a+1+i1−i(a∈R)是纯虚数,则a=0,故选:C.利用复数的运算法则与纯虚数的定义即可得出.本题考查了复数的运算法则与纯虚数的定义,属于基础题.2.答案:A解析:本题考查了线性回归方程,属于基础题.根据回归直线过样本中心点,易得x=4,代入回归方程得y,从而可得m的值.解:回归方程ŷ=9x−6过样本点的中心(x,y),可求出x=4,代入得:y=36−6=30,则30=22+28+m3,∴m=40.故选A.3.答案:B解析:本题考查算法适宜用条件结构的问题,是在解决时需要讨论的问题,对于选项①,②值,代入相应的公式求即可,对于选项③,④值域代入相应的公式时需要分类讨论,故要用到条件语句来描述其算法,属于基础题.解:对于①输入一个正数x,求它的常用对数值,代入lg x求即可;对于②,求面积为6的正方形的周长,代入a2求即可;对于③,求三个数a ,b ,c 中的最大数,必须先进行大小比较,要用条件语句;对于④,求函数f(x)={x −1,x ≥0x +2,x <0的函数值,必须对所给的x 进行条件判断,也要用条件语句.其中不需要用条件语句来描述其算法的有2个. 故选B .4.答案:C解析:解:由线面平行的判定定理,必须指明直线a 在平面α外,故排除A ,a ⊥b ,b ⊥α,则a 可能在平面α内,故排除B ,由面面平行的定义可知若两个平面平行,则其中一个平面内的任意一条直线都平行于另一个平面,故C 正确;垂直于同一平面的一条直线与一个平面可能在一个面内,故排除D , 故选:C .若平面外一条直线与平面内一条直线平行,则这条直线平行于这个平面,特别注意此直线为“平面外直线”,由此即可排除ABD 三个选项,若两个平面平行,则其中一个平面内的任意一条直线都平行于另一个平面,这也是证明线面平行的常用结论本题考察了线面平行的判定定理的准确应用,以及由面面平行证明线面平行的方法,解题时要对定理准确理解,避免出错5.答案:A解析:解:根据几何体的三视图,得;该几何体是底面圆的直径为6,母线长为5的圆锥体,该圆锥的表面积为S 表面积=π×32+π×3×5=24π.故选:A .根据几何体的三视图,得出该几何体是圆锥,求出它的表面积即可.本题考查了利用空间几何体的三视图求几何体的表面积的应用问题,是基础题目.6.答案:A解析:本题考查了求三棱柱的外接球的表面积,利用三棱柱的结构特征求得外接球的半径是关键,属于中档题.根据棱柱的体积公式求得棱柱的侧棱长,再利用三棱柱的底面是直角三角形可得外接球的球心为上、下底面直角三角形斜边中点连线的中点O,从而求得外接球的半径R,代入球的表面积公式计算.解:∵∠ACB=90°,∠BAC=30°,BC=1,∴AC=√3.∵AA1⊥底面ABC,∴三棱柱ABC−A1B1C1的体积V=12×1×√3⋅CC1=3,得CC1=2√3,∴三棱柱ABC−A1B1C1的外接球半径R=12√1+(√3)2+(2√3)2=2,∴S表=4π×22=16π.故选:A.7.答案:D解析:解:从4名男生、2名女生中选出3人,有C63=20种方法;其中至少选出2名男生的选法有C43+C42×C21=16种方法,∴1620=45.故选D.求出从4名男生、2名女生中选出3人的选法数,再求出至少选出2名男生的选法数,代入古典概型概率公式计算.本题考查了古典概型的概率计算及排列组合的应用,关键是利用分类计算原理求至少选出2名男生的选法种数.8.答案:D解析:本题考查棱柱的结构特征,考查计算能力,空间想象能力,是基础题.设出长方体的三边,利用面积公式求出三边,然后求出对角线的长.解:设长方体三度为x,y,z,则xy=√6,yz=√3,xz=√2,∴x=√2,y=√3,z=1.∴长方体的对角线长为√2+3+1=√6故选:D.9.答案:C解析:解:由题意,此时的球与正四面体相切,由于棱长为2√3的正四面体,故四个面的面积都是12×2√3×2√3sin∠60°=3√3又顶点A到底面BCD的投影在底面的中心G,此G点到底面三个顶点的距离都是高的23倍,又高为2√3sin∠60°=3,故底面中心G到底面顶点的距离都是2由此知顶点A到底面BCD的距离是√(2√3)2−22=2√2此正四面体的体积是13×2√2×3√3=2√6,又此正四面体的体积是13×r×3√3×4,故有r=√64√3=√22.上面的三棱锥的高为√2,原正四面体的高为2√2,所以空隙处放入一个小球,则这球的最大半径为a,a √2=√222√2,∴a=√24.故选C.棱长为2√3的正四面体内切一球,那么球O与此正四面体的四个面相切,即球心到四个面的距离都是半径,由等体积法求出球的半径,求出上面三棱锥的高,利用相似比求出上部空隙处放入一个小球,求出这球的最大半径.本题考查球的体积和表面积,用等体积法求出球的半径,熟练掌握正四面体的体积公式及球的表面积公式是正确解题的知识保证.相似比求解球的半径是解题的关键.10.答案:A解析:解:由题意知本题所给的观测值,k2=110×(40×30−20×20)260×50×60×50≈7.8∵7.8>6.635,∴这个结论有0.01=1%的机会说错,即有99%以上的把握认为“爱好该项运动与性别有关”故选:A.根据条件中所给的观测值,同题目中节选的观测值表进行检验,得到观测值对应的结果,得到结论有99%以上的把握认为“爱好该项运动与性别有关”.本题考查独立性检验的应用,考查对于观测值表的认识,这种题目一般运算量比较大,主要要考查运算能力.11.答案:B解析:解:∵一个圆将平面分为2份两个圆相交将平面分为4=2+2份,三个圆相交将平面分为8=2+2+4份,四个圆相交将平面分为14=2+2+4+6份,…平面内n个圆,其中每两个圆都相交于两点,且任意三个圆不相交于同一点,则该n个圆分平面区域数f(n)=2+(n−1)n=n2−n+2∴f(n+1)=f(n)+2n,故选:B我们由两个圆相交将平面分为4分,三个圆相交将平面分为8分,四个圆相交将平面分为14部分,我们进行归纳推理,易得到结论.本题主要考查了进行简单的合情推理.归纳推理的一般步骤是:(1)通过观察个别情况发现某些相同性质;(2)从已知的相同性质中推出一个明确表达的一般性命题(猜想).12.答案:D解析:本题考查了异面直线所成角以及直线与平面和二面角的应用问题,也考查了三棱锥的体积计算问题,是综合题.根据线、面的位置关系及二面角的计算逐个进行判断即可.解:对于①,由V D−BPC1=V P−DBC1知,S△DBC1面积一定,且P∈AD1,AD1//平面BDC1,∴点A到平面DBC1的距离即为点P到该平面的距离,三棱锥的体积为定值,①正确;对于②,棱长为1的正方体ABCD−A1B1C1D1中,点P在线段AD1上运动,∴B1C⊥平面ABC1D1,而C1P⊂平面ABC1D1,∴B1C⊥C1P,异面直线C1P与CB1所成的角为定值90°,②正确;对于③,二面角P−BC1−D的大小,是平面ABC1D1与平面BDC1所成的二面角,这两个平面为固定的平面,它们的夹角为定值,③正确;对于④,点P在线段AD1上运动,AD1⊥A1D,AD1⊥CD,且A1D∩CD=D,A1D,CD⊂平面A1B1CD,∴AD1⊥平面A1B1CD,∴AP⊥平面A1B1CD,④正确;综上,正确的命题序号是①②③④.故选:D.13.答案:310解析:由二次函数的性质得n≥2m,先求出基本事件总数,再求出满足条件的基本事件个数,由此能求出函数f(x)=mx2−nx+2在(−∞,1]上是减函数的概率.本题考查概率的求法,是基础题,解题时要认真审题,注意等可能事件概率计算公式和二次函数的性质的合理运用.解:∵函数f(x)=mx2−nx+2在(−∞,1]上是减函数,∴n2m≥1,即n≥2m,∵m,n∈{1,2,3,4,5},且m≠n,∴基本事件总数为5×4=20,满足条件的基本事件有:(1,2),(1,3),(1,4),(1,5),(2,4),(2,5),共6个,∴函数f(x)=mx2−nx+2在(−∞,1]上是减函数的概率p=620=310.故答案为:310.14.答案:解析:由椭圆到双曲线进行类比,不难写出关于双曲线的结论:若M,N是双曲线上关于原点对称的两个点,证明如下:15.答案:1−12+13−14+⋯+12n−1−12n=1n+1+1n+2+⋯+12n解析:解:根据等式,左边有2n项,右边第一项分母为n+1,最后一项分母为n+n=2n.据此规律,第n个等式可为1−12+13−14+⋯+12n−1−12n=1n+1+1n+2+⋯+12n.故答案为:1−12+13−14+⋯+12n−1−12n=1n+1+1n+2+⋯+12n.根据等式,左边有2n项,右边第一项分母为n+1,最后一项分母为n+n=2n,即可得出结论.本题是规律探究题,考查观察与归纳推理的能力,比较基础.16.答案:①②③⑤解析:试题分析:当时,截面为等腰梯形,②正确;当向移动时,,在上取,使,可知①正确;当时,延长使,连结交于,连接交于,连接,,③正确,由③可知时,点上移,截面为五边形,④错误;⑤中当时与重合,取中点,连接,截面为菱形,所以正确考点:1.命题真假;2.正方体截面问题17.答案:(1)证明:在图1中,∠ADC=90°,则图2中,CD1⊥AD1,在△BD1C中,∵BD1=2√2,CD1=1BC=3,∴BD12+CD12=BC2,可得CD1⊥BD1,又AD1∩BD1=D1,∴CD1⊥平面ABD1,∵CD1⊂平面ACD1,∴平面ACD1⊥平面ABD1;(2)解:当三棱锥D1−ABC的体积最大时,平面AD1C⊥平面ABC,过D1作D1E⊥AC,则D1E⊥平面ABC,并求得D1E=√22.S△ABC=12×3×1=32,S△AD1C=12×1×1=12.设点B到平面ACD1的距离为h,由V D1−ABC =V B−AD1C,得13×32×√22=13×12ℎ,即ℎ=3√22.故点B到平面ACD1的距离为3√22.解析:(1)由已知可得CD1⊥AD1,求解三角形证明CD1⊥BD1,由线面垂直的判定可得CD1⊥平面ABD1,进一步得到平面ACD1⊥平面ABD1;(2)当三棱锥D1−ABC的体积最大时,平面AD1C⊥平面ABC,过D1作D1E⊥AC,则D1E⊥平面ABC,并求得D1E=√22,然后利用等积法求B到平面ACD1的距离.本题考查平面与平面垂直的判定,考查空间想象能力与思维能力,训练了利用等积法求多面体的体积,是中档题.18.答案:解:1−ii =i−i2i2=−1−i,其对应的点(−1,−1)在第三象限.解析:直接由已知的复数得到其在复平面内对应点的坐标得答案.本题考查了复数的代数表示法及其几何意义,是基础题.19.答案:(1)证明:过O作OE//AB交BC于E,则四边形ABEO为是平行四边形,∴BE=AO=OD=2,OE=AB=CD=6−42cos60∘=2,CE=6−2=4,OC=√4+4−2⋅2⋅2⋅cos120°=2√3,满足OC2+OE2=CE2,∴OE⊥OC,则AB⊥OC,∵△PAD是等边三角形,O为AD的中点,∴PO⊥AD,又平面PAD⊥平面ABCD,平面PAD∩平面ABCD=AD,∴PO⊥平面ABCD,则PO⊥AB,又PO∩OC=O,∴AB⊥平面POC;(2)解:过O作OF//CD交BC于F,过F作FQ//PC交PB于点Q,则平面OFQ即为平面α,得CF=DO=2,∴PQPB =CFCB=26=13,由(1)可得点Q到平面POC的距离为点B到平面POC的距离的13,∴V P−COQ=V Q−POC=13V B−POC=13V P−BOC=13⋅13S△BOC⋅OP=19×12×6×√3×2√3=2.解析:(1)过O作OE//AB交BC于E,则四边形ABEO为是平行四边形,求解三角形可得OE⊥OC,则AB⊥OC,再由△PAD是等边三角形,O为AD的中点,得PO⊥AD,结合平面PAD⊥平面ABCD,可得PO⊥平面ABCD,则PO⊥AB,由线面垂直的判定得AB⊥平面POC;(2)过O作OF//CD交BC于F,过F作FQ//PC交PB于点Q,可得平面OFQ即为平面α,得到点Q到平面POC的距离为点B到平面POC的距离的13,再由等积法求三棱锥P−COQ的体积.本题考查直线与平面垂直的判定,考查空间想象能力与思维能力,训练了利用等积法求多面体的体积,是中档题.20.答案:解:(1)由统计数据得2×2列联表:根据2×2列联表中的数据,得K2的观测值为K2=40(8×3−12×17)220×20×25×15=8.64>6.635,所以能在犯错概率不超过0.01的前提下认为“成绩优良与教学方式有关”.(2)由表可知,8人中成绩不优良的人数为7+5+1+240×8=3,则X的可能取值为0,1,2,3.40人中,成绩不优良的人数共有15人,其中甲班12人,乙班3人,P(X=0)=C123C153=4491;P(X=1)=C122C31C153=198455;P(X=2)=C121C32C153=36455;P(X=3)=C33C153=1455.所以X的分布列为:所以E(X)=0×4491+1×198455+2×36455+3×1455=273455.解析:(1)先根据频数分布表填写2×2列联表,然后利用K2的公式即可得解;(2)40人中,成绩不优良的人数共有15人,其中甲班12人,乙班3人,由此求得8人中成绩不优良的人数为3人,于是X的可能取值为0,1,2,3,再分别求出每个X取值对应的概率即可得解.本题考查统计和概率,涉及统计案例中的独立性检验,超几何分布的分布列与期望,考查学生分析问题的能力和运算能力,属于基础题.21.答案:【小题1】见解析【小题2】见解析【小题3】V=S △PCF·EG=解析:【小题1】试题分析:设G为PC的中点,连接FG,EG.因为F为PD的中点,E为AB的中点,所以FG CD,AE CD,所以FG AE,所以四边形AEGF为平行四边形,所以AF//GE.因为GE平面PEC,AF⊄平面PEC,所以AF//平面PCE.【小题2】试题分析:因为PA=AD=2,所以AF⊥PD.又因为PA⊥平面ABCD,CD平面ABCD,所以PA⊥CD.因为AD⊥CD,PA∩AD=A,所以CD⊥平面PAD,因为AF平面PAD,所以AF⊥CD,因为PD∩CD=D,所以AF⊥平面PCD,所以GE⊥平面PCD.因为GE平面PEC,所以平面PCE⊥平面PCD.【小题3】试题分析:由(2)知GE⊥平面PCD,所以EG为四面体PECF的高,又EG=AF=,CD=2,S△PCF=PF·CD=2,所以四面体PECF的体积V=S△PCF·EG=.22.答案:(1)(2)销售收入大约为82.5万元(3)解析:试题分析:(1)首先求出x,y的平均数,利用最小二乘法做出线性回归方程的系数,根据样本中心点满足线性回归方程,代入已知数据求出a的值,写出线性回归方程.(2)当自变量取10时,把10代入线性回归方程,求出销售额的预报值,这是一个估计数字,它与真实值之间有误差.(3)利用列举法计算基本事件数及事件发生的概率.本题考查回归分析的初步应用,考查求线性回归方程,考查预报y的值,是一个综合题目,解此类题,关键是理解线性回归分析意义,这种题目是新课标的大纲要求掌握的题型,是一个典型的题目,在近年的高考中频率有增高的趋势,此类题运算量大,解题时要严谨防止运算出错.试题解析:(1)解:,[2分]又已知,于是可得:,[4分]因此,所求回归直线方程为:[6分] (2)解:根据上面求得的回归直线方程,当广告费支出为10万元时,(万元)即这种产品的销售收入大约为82.5万元.[9分](3)解:24568304060507030.543.55056.569.5基本事件:(30,40),(30,60),(30,50),(30,70),(40,60),(40,50),(40,70),(60,50),(60,70),(50,70)共10个两组数据其预测值与实际值之差的绝对值都超过5:(60,50)[12分]所以至少有一组数据其预测值与实际值之差的绝对值不超过5的概率为[14分]考点:1.回归分析的初步应用;2.列举法计算基本事件数及事件发生的概率。

福建省福州市2023-2024学年高二下学期期中联考试题 数学含答案

福建省福州市2023-2024学年高二下学期期中联考试题 数学含答案

2023-2024学年第二学期期中质量检测高二数学试卷(答案在最后)(满分:150分;考试时间:120分钟)注意事项:1.本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分.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.回答第Ⅰ卷时,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.写在本试卷上无效.3.回答第Ⅱ卷时,将答案写在答题卡上.写在本试卷上无效.4.测试范围:选择性必修第二册第五章、选择性必修第三册第六章、第七章第Ⅰ卷一、单项选择题:本题共8小题,每小题5分,共40分.在每小题给出的四个选项中,只有一个选项是符合题目要求的.1.计算52752+C A 的值是()A.62B.102C.152D.5402.下列导数运算正确的是()A.cos sin x x x '⎛⎫=- ⎪⎝⎭B.()21log ln 2x x '=C.()22xx'= D.()32e 3exxx x '=3.若9290129(2)x a a x a x a x -=++++L ,则129a a a +++ 的值为()A.1- B.1 C.511- D.5124.若2()f x x bx c =++的图象的顶点在第二象限,则函数()f x '的图象是()A. B.C. D.5.曲线()(22e 21xf x x x =--+-在0x =处的切线的倾斜角是()A.2π3B.5π6C.3π4 D.π46.现有完全相同的甲,乙两个箱子(如图),其中甲箱装有2个黑球和4个白球,乙箱装有2个黑球和3个白球,这些球除颜色外完全相同.某人先从两个箱子中任取一个箱子,再从中随机摸出一球,则摸出的球是黑球的概率是()A.1115B.1130C.115D.2157.有7种不同的颜色给下图中的4个格子涂色,每个格子涂一种颜色,且相邻的两个格子颜色不能相同,若最多使用3种颜色,则不同的涂色方法种数为()A.462B.630C.672D.8828.已知函数()e 2xx k f x =-,若0x ∃∈R ,()00f x ≤,则实数k 的最大值是().A.1eB.2eC.12eD.e e二、多项选择题:本题共3小题,每小题6分,共18分.在每小题给出的选项中,有多项符合题目要求,全部选对的得6分,部分选对的得部分分,有选错的得0分.9.已知1)nx+*(N )n ∈展开式中常数项是2C n ,则n 的值为().A.3B.4C.5D.610.高中学生要从必选科目(物理和历史)中选一门,再在化学、生物、政治、地理这4个科目中,依照个人兴趣、未来职业规划等要素,任选2个科目构成“1+2选考科目组合”参加高考.已知某班48名学生关于选考科目的结果统计如下:选考科目名称物理化学生物历史地理政治选考该科人数36392412a b下面给出关于该班学生选考科目的四个结论中,正确的是()A.33a b +=B.选考科目组合为“历史+地理+政治”的学生可能超过9人C.在选考化学的所有学生中,最多出现6种不同的选考科目组合D.选考科目组合为“历史+生物+地理”的学生人数一定是所有选考科目组合中人数最少的11.若不等式e ln 0x ax a -<在[)2,x ∞∈+时恒成立,则实数a 的值可以为()A.3eB.2eC.eD.2第Ⅱ卷三、填空题:本题共3小题,每小题5分,共15分.12.某气象台统计,该地区下雨的概率为415,刮四级以上风的概率为215,既刮四级以上的风又下雨的概率为110,设A 为下雨,B 为刮四级以上的风,则()P B A =___________.13.某校一次高三数学统计,经过抽样分析,成绩X 近似服从正态分布()2110,N σ,且P (90110)X ≤≤0.3=,该校有1000人参加此次统考,估计该校数学成绩不低于130分的人数为________.14.将4名志愿者分配到3个不同的北京冬奥场馆参加接待工作,每个场馆至少分配一名志愿者的方案种数为________.(用数字作答)四、解答题(本大题共5题,共77分,解答时应写出文字说明,证明过程或演算步骤)15.已知函数3()ln (R)f x x ax a =+∈,且(1)4f '=.(1)求a 的值;(2)设()()ln g x f x x x =--,求()y gx =过点(1,0)的切线方程.16.已知n⎛⎝在的展开式中,第6项为常数项.(1)求n ;(2)求含2x 的项的系数;(3)求展开式中所有的有理项.17.如图,有三个外形相同的箱子,分别编号为1,2,3,其中1号箱装有1个黑球和3个白球,2号箱装有2个黑球和2个白球,3号箱装有3个黑球,这些球除颜色外完全相同.小明先从三个箱子中任取一箱,再从取出的箱中任意摸出一球,记事件i A (123i =,,)表示“球取自第i 号箱”,事件B 表示“取得黑球”.(1)求()P B 的值:(2)若小明取出的球是黑球,判断该黑球来自几号箱的概率最大?请说明理由.18.为普及空间站相关知识,某部门组织了空间站模拟编程闯关活动,它是由太空发射、自定义漫游、全尺寸太阳能、空间运输等10个相互独立的程序题目组成.规则是:编写程序能够正常运行即为程序正确.每位参赛者从10个不同的题目中随机选择3个进行编程,全部结束后提交评委测试,若其中2个及以上程序正确即为闯关成功.现已知10个程序中,甲只能正确完成其中6个,乙正确完成每个程序的概率为0.6,每位选手每次编程都互不影响.(1)求乙闯关成功的概率;(2)求甲编写程序正确的个数X 的分布列和期望,并判断甲和乙谁闯关成功的可能性更大.19.已知曲线()31:3C y f x x ax ==-.(1)求函数()313f x x ax =-()0a ≠的单调递增区间;(2)若曲线C 在点()()3,3f 处的切线与两坐标轴围成的三角形的面积大于18,求实数a 的取值范围.2023-2024学年第二学期期中质量检测高二数学试卷(满分:150分;考试时间:120分钟)注意事项:1.本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分.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.回答第Ⅰ卷时,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.写在本试卷上无效.3.回答第Ⅱ卷时,将答案写在答题卡上.写在本试卷上无效.4.测试范围:选择性必修第二册第五章、选择性必修第三册第六章、第七章第Ⅰ卷一、单项选择题:本题共8小题,每小题5分,共40分.在每小题给出的四个选项中,只有一个选项是符合题目要求的.1.计算52752+C A 的值是()A.62 B.102C.152D.540【答案】A 【解析】【分析】利用组合和排列数公式计算【详解】5275762254622C A =+´+创=故选:A2.下列导数运算正确的是()A.cos sin x x x '⎛⎫=- ⎪⎝⎭B.()21log ln 2x x '=C.()22xx'= D.()32e 3exxx x '=【答案】B 【解析】【分析】利用常见函数的导数可以判断B 、C 的真假,利用积的导数的运算法则判断D 的真假,利用商的导数的运算法则判断A 的真假.【详解】∵()22cos cos cos sin cos x x x x x x x x x x x ''⋅-⋅--⎛⎫== ⎪⎝'⎭,故A 错误;∵()21log ln 2x x '=,故B 正确;∵()22ln 2x x '=,故C 错误;∵()()()33323e e e 3e e x x x x x x x x x x ⋅'''=⋅+=+,故D 错误.故选:B.3.若9290129(2)x a a x a x a x -=++++L ,则129a a a +++ 的值为()A.1- B.1 C.511- D.512【答案】C 【解析】【分析】根据题意,分别令1x =与0x =代入计算,即可得到结果.【详解】当1x =时,20911a a a a ++++=L ;当0x =时,0512a =所以,1211511a a a +++=-L 故选:C4.若2()f x x bx c =++的图象的顶点在第二象限,则函数()f x '的图象是()A.B.C.D.【答案】C 【解析】【分析】求导后得到斜率为2,再由极值点是导数为零的点小于零,综合直线的特征可得正确答案.【详解】因为()2f x x b '=+,所以函数()f x '的图象是直线,斜率20k =>;又因为函数()f x 的顶点在第二象限,所以极值点小于零,所以()f x '的零点小于零,结合直线的特征可得C 符合.故选:C5.曲线()(22e 21xf x x x =--+-在0x =处的切线的倾斜角是()A.2π3B.5π6C.3π4 D.π4【答案】A 【解析】【分析】利用导数的几何意义求得切线斜率,即可求得切线的倾斜角.【详解】()()2e 22,0xf x x f =--∴'-'= ,设切线的倾斜角为[),0,πθθ∈,则tan θ=,即2π3θ=,故选:A .6.现有完全相同的甲,乙两个箱子(如图),其中甲箱装有2个黑球和4个白球,乙箱装有2个黑球和3个白球,这些球除颜色外完全相同.某人先从两个箱子中任取一个箱子,再从中随机摸出一球,则摸出的球是黑球的概率是()A.1115B.1130C.115D.215【答案】B 【解析】【分析】根据条件概率的定义,结合全概率公式,可得答案.【详解】记事件A 表示“球取自甲箱”,事件A 表示“球取自乙箱”,事件B 表示“取得黑球”,则()()()()1212,,2635P A P A P B A P B A =====,由全概率公式得()()()()111211232530P A P B A P A P B A +=⨯+⨯=.故选:B .7.有7种不同的颜色给下图中的4个格子涂色,每个格子涂一种颜色,且相邻的两个格子颜色不能相同,若最多使用3种颜色,则不同的涂色方法种数为()A.462B.630C.672D.882【答案】C 【解析】【分析】根据题意,按使用颜色的数目分两种情况讨论,由加法原理计算可得答案.【详解】根据题意,分两种情况讨论:若用两种颜色涂色,有27C 242⨯=种涂色方法;若用三种颜色涂色,有()37C 3221630⨯⨯⨯+=种涂色方法;所以有42630672+=种不同的涂色方法.故选:C.8.已知函数()e 2xx k f x =-,若0x ∃∈R ,()00f x ≤,则实数k 的最大值是().A.1eB.2eC.12eD.e e【答案】B 【解析】【分析】将问题转化为002e x x k ≤在0x ∈R 上能成立,利用导数求2()exxg x =的最大值,求k 的范围,即知参数的最大值.【详解】由题设,0x ∃∈R 使02e x x k ≤成立,令2()exxg x =,则()21e x g x x ⋅-'=,∴当1x <时()0g x '>,则()g x 递增;当1x >时()0g x '<,则()g x 递减;∴2()(1)e g x g ≤=,故2e k ≤即可,所以k 的最大值为2e.故选:B.二、多项选择题:本题共3小题,每小题6分,共18分.在每小题给出的选项中,有多项符合题目要求,全部选对的得6分,部分选对的得部分分,有选错的得0分.9.已知1)nx+*(N )n ∈展开式中常数项是2C n ,则n 的值为().A.3B.4C.5D.6【答案】AD 【解析】【分析】根据二项式展开式得到321C n r r r nT x-+=,再令302n r-=,则得到123C C n n n =,解出即可.【详解】展开式的通项为131221C ()()C n r r n rr rr nnT x x x---+==,若要其表示常数项,须有302n r-=,即13r n =,又由题设知123C C n n =,123n \=或123n n -=,6n ∴=或3n =.故选:A D .10.高中学生要从必选科目(物理和历史)中选一门,再在化学、生物、政治、地理这4个科目中,依照个人兴趣、未来职业规划等要素,任选2个科目构成“1+2选考科目组合”参加高考.已知某班48名学生关于选考科目的结果统计如下:选考科目名称物理化学生物历史地理政治选考该科人数36392412ab下面给出关于该班学生选考科目的四个结论中,正确的是()A.33a b +=B.选考科目组合为“历史+地理+政治”的学生可能超过9人C.在选考化学的所有学生中,最多出现6种不同的选考科目组合D.选考科目组合为“历史+生物+地理”的学生人数一定是所有选考科目组合中人数最少的【答案】AC 【解析】【分析】结合统计结果对选项逐一分析即可得.【详解】对A :由3924482a b +++=⨯,则33a b +=,故A 正确;对B :由选择化学的有39人,选择物理的有36人,故至少有三人选择化学并选择了历史,故选考科目组合为“历史+地理+政治”的学生最多有9人,故B 错误;对C :确定选择化学后,还需在物理、历史中二选一,在生物、地理、政治中三选一,故共有236⨯=种不同的选考科目组合,故C 正确;对D :由于地理与政治选考该科人数不确定,故该说法不正确,故D 错误.故选:AC.11.若不等式e ln 0x ax a -<在[)2,x ∞∈+时恒成立,则实数a 的值可以为()A.3eB.2eC.eD.2【答案】BCD 【解析】【分析】构造函数()ex xf x =,将e ln 0x ax a -<恒成立问题转化为()()ln f x f a <恒成立问题,求导,研究()e xxf x =单调性,画出其图象,根据图象逐一验证选项即可.【详解】由e ln 0x ax a -<得ln ln ln e ex a x a aa <=,设()e x x f x =,则()1ex xf x ='-,当1x <时,()0f x '>,()f x 单调递增,当1x >时,()0f x '<,()f x 单调递减,又()00f =,()11e f =,当0x >时,()0ex xf x =>恒成立,所以()ex xf x =的图象如下:,ln ln e ex a x a<,即()()ln f x f a <,2x ≥,对于A :当3e a =时,ln ln 31>2a =+,根据图象可得()()ln f x f a <不恒成立,A 错误;对于B :当2e a =时,()ln ln 211,2a =+∈,根据图象可得()()ln f x f a <恒成立,B 正确;对于C :当e a =时,ln 1a =,根据图象可得()()ln f x f a <恒成立,C 正确;对于D :当2a =时,ln ln 2a =,又()()ln 22ln 212ln 2ln 2,2e 2ef f ===,因为221263ln 23ln 2e e ⨯-⨯=,且2e,e 6>>,即26ln 1,1e ><,所以221263ln 23ln 02e e⨯-⨯=->,即()()ln 22f f >,根据图象可得()()ln f x f a <恒成立,D 正确;故选:BCD.【点睛】关键点点睛:本题的关键将条件变形为ln ln e e x ax a <,通过整体结构相同从而构造函数()e x x f x =来解决问题.第Ⅱ卷三、填空题:本题共3小题,每小题5分,共15分.12.某气象台统计,该地区下雨的概率为415,刮四级以上风的概率为215,既刮四级以上的风又下雨的概率为110,设A 为下雨,B 为刮四级以上的风,则()P B A =___________.【答案】38【解析】【分析】利用条件概率的概率公式()()()P AB P B A P A =即可求解.【详解】由题意可得:()415P A =,()215P B =,()110P AB =,由条件概率公式可得()()()13104815P AB P B A P A ===,故答案为:38.13.某校一次高三数学统计,经过抽样分析,成绩X 近似服从正态分布()2110,N σ,且P (90110)X ≤≤0.3=,该校有1000人参加此次统考,估计该校数学成绩不低于130分的人数为________.【答案】200【解析】【分析】根据X 近似服从正态分布()2110,N σ,且P (90110)X ≤≤0.3=,求得(130)p X ≥即可.【详解】因为X 近似服从正态分布()2110,N σ,且P (90110)X ≤≤0.3=,所以()()113012901300.22P X P X ⎡⎤≥=-≤≤=⎣⎦,又该校有1000人参加此次统考,估计该校数学成绩不低于130分的人数为10000.2200⨯=人.故答案为:200.14.将4名志愿者分配到3个不同的北京冬奥场馆参加接待工作,每个场馆至少分配一名志愿者的方案种数为________.(用数字作答)【答案】36【解析】【分析】先将4人分成2、1、1三组,再安排给3个不同的场馆,由分步乘法计数原理可得.【详解】将4人分到3个不同的体育场馆,要求每个场馆至少分配1人,则必须且只能有1个场馆分得2人,其余的2个场馆各1人,可先将4人分为2、1、1的三组,有211421226C C C A =种分组方法,再将分好的3组对应3个场馆,有336A =种方法,则共有6636⨯=种分配方案.故答案为:36四、解答题(本大题共5题,共77分,解答时应写出文字说明,证明过程或演算步骤)15.已知函数3()ln (R)f x x ax a =+∈,且(1)4f '=.(1)求a 的值;(2)设()()ln g x f x x x =--,求()y g x =过点(1,0)的切线方程.【答案】(1)1(2)22y x =-【解析】【分析】(1)利用导数求解参数即可.(2)先设切点,利用导数表示斜率,建立方程求出参数,再写切线方程即可.【小问1详解】定义域为,()0x ∈+∞,21()3f x ax x'=+,而(1)13f a '=+,而已知(1)4f '=,可得134a +=,解得1a =,故a 的值为1,【小问2详解】3()()ln g x f x x x x x =--=-,设切点为0003(,)x x x -,设切线斜率为k ,而2()31g x x '=-,故切线方程为300200()(31)()y x x x x x --=--,将(1,0)代入方程中,可得3200000()(31)(1)x x x x --=--,解得01x =(负根舍去),故切线方程为22y x =-,16.已知n ⎛ ⎝在的展开式中,第6项为常数项.(1)求n ;(2)求含2x 的项的系数;(3)求展开式中所有的有理项.【答案】(1)10n =;(2)454;(3)2454x ,638-,245256x.【解析】【分析】(1)求出n⎛ ⎝的展开式的通项为1r T +,当=5r 时,指数为零,可得n ;(2)将10n =代入通项公式,令指数为2,可得含2x 的项的系数;(3)根据通项公式与题意得1023010r Zr r Z -⎧∈⎪⎪≤≤⎨⎪∈⎪⎩,求出r 的值,代入通项公式并化简,可得展开式中所有的有理项.【详解】(1)n ⎛ ⎝的展开式的通项为233311122r rn r r n r r r r n n T C x x C x ----+⎛⎫⎛⎫=-=- ⎪ ⎪⎝⎭⎝⎭,因为第6项为常数项,所以=5r 时,有203n r -=,解得10n =.(2)令223n r -=,得()()116106222r n =-=⨯-=,所以含2x 的项的系数为221014524C ⎛⎫-= ⎪⎝⎭.(3)根据通项公式与题意得1023010r Zr r Z -⎧∈⎪⎪≤≤⎨⎪∈⎪⎩,令()1023r k k Z -=∈,则1023r k -=,即352r k =-.r Z ∈,∴k 应为偶数.又010r ≤≤,∴k 可取2,0,-2,即r 可取2,5,8.所以第3项,第6项与第9项为有理项,它们分别为2221012C x ⎛⎫- ⎪⎝⎭,551012C ⎛⎫- ⎪⎝⎭,8821012C x -⎛⎫- ⎪⎝⎭,即2454x ,638-,245256x .【点睛】关键点点睛:本题考查二项式展开式的应用,考查二项式展开式的通项公式以及某些特定的项,解决本题的关键点是求解展开式的有理项时,令()1023r k k Z -=∈,由r Z ∈以及010r ≤≤,求出k 的值,进而得出r 的值,代入通项公式化简可得有理项,考查了学生计算能力,属于中档题.17.如图,有三个外形相同的箱子,分别编号为1,2,3,其中1号箱装有1个黑球和3个白球,2号箱装有2个黑球和2个白球,3号箱装有3个黑球,这些球除颜色外完全相同.小明先从三个箱子中任取一箱,再从取出的箱中任意摸出一球,记事件i A (123i =,,)表示“球取自第i 号箱”,事件B 表示“取得黑球”.(1)求()P B 的值:(2)若小明取出的球是黑球,判断该黑球来自几号箱的概率最大?请说明理由.【答案】(1)712(2)可判断该黑球来自3号箱的概率最大.【解析】【分析】(1)因先从三个箱子中任取一箱,再从取出的箱中任意摸出一球为黑球,其中有三种可能,即黑球取自于1号,2号或者3号箱,故事件B 属于全概率事件,分别计算出()i P A 和(|),1,2,3i P B A i =,代入全概率公式即得;(2)由“小明取出的球是黑球,判断该黑球来自几号箱”是求条件概率(|),1,2,3i P A B i =,根据条件概率公式分别计算再比较即得.【小问1详解】由已知得:1231()()()3P A P A P A ===,12311(|),(|),(|)1,42P B A P B A P B A ===而111111()(|)(),4312P BA P B A P A =⋅=⨯=222111()(|)(),236P BA P B A P A =⋅=⨯=33311()(|)()1.33P BA P B A P A =⋅=⨯=由全概率公式可得:1231117()()()().126312P B P BA P BA P BA =++=++=【小问2详解】因“小明取出的球是黑球,该黑球来自1号箱”可表示为:1A B ,其概率为111()112(|)7()712P A B P A B P B ===,“小明取出的球是黑球,该黑球来自2号箱”可表示为:2A B ,其概率为221()26(|)7()712P A B P A B P B ===,“小明取出的球是黑球,该黑球来自3号箱”可表示为:3A B ,其概率为331()43(|)7()712P A B P A B P B ===.综上,3(|)P A B 最大,即若小明取出的球是黑球,可判断该黑球来自3号箱的概率最大.18.为普及空间站相关知识,某部门组织了空间站模拟编程闯关活动,它是由太空发射、自定义漫游、全尺寸太阳能、空间运输等10个相互独立的程序题目组成.规则是:编写程序能够正常运行即为程序正确.每位参赛者从10个不同的题目中随机选择3个进行编程,全部结束后提交评委测试,若其中2个及以上程序正确即为闯关成功.现已知10个程序中,甲只能正确完成其中6个,乙正确完成每个程序的概率为0.6,每位选手每次编程都互不影响.(1)求乙闯关成功的概率;(2)求甲编写程序正确的个数X 的分布列和期望,并判断甲和乙谁闯关成功的可能性更大.【答案】(1)0.648(2)分布列见解析,期望为95,甲比乙闯关成功的概率要大.【解析】【分析】(1)根据题意,直接列出式子,代入计算即可得到结果;(2)根据题意,由条件可得X 的可能取值为0,1,2,3,然后分别计算其对应概率,即可得到分布列,然后计算甲闯关成功的概率比较大小即可.【小问1详解】记事件A 为“乙闯关成功”,乙正确完成每个程序的概率为0.6,则()()2233C 0.610.6(0.6)0.648;P A =⨯⨯-+=【小问2详解】甲编写程序正确的个数X 的可能取值为0,1,2,3,()()()()211233464664333310101010C C C C C C 13110,1,2,3C 30C 10C 2C 6P X P X P X P X ============,故X 的分布列为:X0123P 1303101216故()1311901233010265E X =⨯+⨯+⨯+⨯=,甲闯关成功的概率1120.648263P =+=>,故甲比乙闯关成功的概率要大.19.已知曲线()31:3C y f x x ax ==-.(1)求函数()313f x x ax =-()0a ≠的单调递增区间;(2)若曲线C 在点()()3,3f 处的切线与两坐标轴围成的三角形的面积大于18,求实数a 的取值范围.【答案】(1)答案见解析(2)()()0,99,18U 【解析】【分析】(1)求出函数的导函数,分0a >、a<0两种情况讨论,分别求出函数的单调递增区间;(2)利用导数的几何意义求出切线方程,再令0x =、0y =求出在坐标轴上的截距,再由面积公式得到不等式,解得即可.【小问1详解】∵()313f x x ax =-定义域为R ,且()2f x x a '=-,①当a<0时,()20f x x a '=->恒成立,∴()f x 在R 上单调递增;②当0a >时,令()20f x x a '=->,解得x <x >,∴()f x 在(,∞-,)∞+上单调递增,综上:当a<0时,()f x 的单调递增区间为(),-∞+∞;当0a >时,()f x 的单调递增区间为(,∞-,)∞+.【小问2详解】由(1)得()2339f a a =-=-',又∵()393f a =-,∴切线方程为()()()9393y a a x --=--,依题意90a -≠,令0x =,得18y =-;令0y =,得189x a=-,切线与坐标轴所围成的三角形的面积11816218299S a a =⨯⨯=--,依题意162189a >-,即919a>-,解得09a <<或918<<a ,即实数a 的取值范围为()()0,99,18⋃.。

2020-2021学年湖北省部分重点中学高二下学期期中数学复习卷(含答案解析)

2020-2021学年湖北省部分重点中学高二下学期期中数学复习卷(含答案解析)

2020-2021学年湖北省部分重点中学高二下学期期中数学复习卷一、单选题(本大题共12小题,共60.0分)1.下列说明正确的是()A. “若a>1,则a2>1”的否命题是“若a>1,则a2≤1”B. {a n}为等比数列,则“a1<a2<a3”是“a4<a5”的既不充分也不必要条件C. ∃x0∈(−∞,0),使3x0<4x0成立D. “tanα≠√3”必要不充分条件是“a≠π3”2.设复数z1=1−i,z2=2+i,其中i为虚数单位,则z1⋅z2的虚部为()A. −1B. 1C. −iD. i3.“∃x0∈R,x02+1<0”的否定是()A. ∀x∈R,x2+1≥0B. ∀x∈R,x2+1<0C. ∃x0∈R,x02+1≥0D. ∃x0∈R,x2+1<04.若,则等于()A. −2B. −4C. 2D. 05.已知椭圆x2a2+y2b2=1(a>b>0)的左顶点A的斜率为k的直线交椭圆C于另一点B,且点B的在x轴上的射影恰好为右焦点F,若椭圆的离心率为23,则k的值为()A. −13B. 13C. ±13D. ±126.方程|x|+|y|=1所表示的图形在直角坐标系中所围成的面积是()A. 2B. 1C. 4D. √27.直线l1:y=mx+1,直线l2的方向向量为a⃗=(1,2),且l1⊥l2,则m=A. 12B. −12C. 2D. −28.已知F1,F2是双曲线E:x2a2−y2b2=1的左,右焦点,点M在E上,MF1与x轴垂直,sin∠MF2F1=13,则双曲线E的渐近线方程为()A. y=±12x B. y=±x C. y=±√3 D. y=±2x9. 已知A 、B 两点均值焦点为F 的抛物线y 2=2px(p >0)上,若|AF ⃗⃗⃗⃗⃗ |+|BF ⃗⃗⃗⃗⃗ |=4,线段AB 的中点到直线x =p2的距离为1,则p 的值为( )A. 1B. 1或3C. 2D. 2或610. 已知命题p :f(x)=12+12x −1为奇函数;命题q :∀x ∈(0,π2),sinx <x <tanx ,则下面结论正确的是( )A. p ∧(¬q)是真命题B. (¬p)∨q 是真命题C. p ∧q 是假命题D. p ∨q 是假命题11. 直线y =kx 与函数f(x)=|x 2−1|x−1图象有两个交点,则k 的范围是( )A. (0,√3)B. (0,1)∪(1,√3)C. (1,√3)D. (0,1)∪(1,2)12. 抛物线y =ax 2的准线方程为y =−1,则实数a =( )A. 4B. 14C. 2D. 12二、单空题(本大题共4小题,共20.0分)13. 已知M(2,0),N(3,0),P 是抛物线C :y 2=3x 上一点,则PM⃗⃗⃗⃗⃗⃗ ⋅PN ⃗⃗⃗⃗⃗⃗ 的最小值是______ . 14. 已知函数f(x)=x 3+ax 2+bx +c ,下列结论中正确的序号是______ .①∃x 0∈R ,使f(x 0)=0;②若x 0是f(x)的极值点,则f′(x 0)=0;③若x 0是f(x)的极小值点,则f(x)在区间(−∞,x 0)上单调递减; ④函数y =f(x)的图象是中心对称图形. 15. 设F 1、F 2为曲线C 1:的焦点,P 是曲线:与C 1的一个交点,则△PF 1F 2的面积为_______________________. 16. 设函数,若在区间内的图象上存在两点,在这两点处的切线相互垂直,则实数的取值范围是 . 三、解答题(本大题共6小题,共70.0分) 17. (12分)(I)求函数图象上的点处的切线方程;(Ⅱ)已知函数,其中是自然对数的底数,对于任意的,恒成立,求实数的取值范围。

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高二下学期期中复习测试题(时间:120分钟,满分:135分)Ⅰ. 语言知识及应用(共两节,满分45分)第一节完形填空(共15小题;每小题2分,满分30分)Fiona Famous was a very popular girl at school. She felt 1 as no other girl had so many friends at school and in the neighborhood.But everything 2 on National Friendship Day. On that day in class everyone had to make three presents to give to their three best friends. Fiona 3 the task of choosing three from all the dozens of her friends. However, she was the only one in her class who did not 4 a present! She felt terrible, and spent hours 5 . Everyone came and comforted her for a while.But each one only stayed for a short time before leaving.This was 6 what Fiona had done so many times to others.She 7 that she was a good companion and acquaintance (泛泛之交), but she had not been a true friend to anyone. She had tried to be close to everyone, but now she found that was not enough to create8 friendships.When she got home that night, Fiona told her mother about the whole thing. Her mother said, "You can't be a close friend to everybody.It's only 9 to have a few true friends.The others will just be playmates or acquaintances.”Hearing this, Fiona decided to change her way so that she could finally have some true friends. That night, in 10 , she thought about what she could do to get them. She thought about her mother. Her mother was always willing to 11 her, she put up with all of Fiona's dislikes and 12 , she always forgave her, she loved her a great deal ... That was what makes a 13 !And Fiona 14 , realizing that she already had the best friend that anyone could 15 .She fell asleep and had a good dream.1.A.tired B.lucky C.strange D.surprised2.A.mattered B.developed C.changed D.appeared3.A.enjoyed B.improved C.failed D.ruined4.A.receive B.show plete D.make5.A.remaking B.thinking C.crying D.checking6.A.especially B.gradually C.briefly D.exactly7.A.forgot B.realized C.doubted plained8.A.new B.true C.long D.smart9.A.possible B.fair C.important D.natural10.A.chair B.bed C.school D.table11.A.correct B.follow C.help D.find12.A.dreams B.ideas C.stories D.problems13.A.playmate B.parent C.friend D.mother14.A.smiled B.suffered C.screamed D.hesitated(犹豫)pare B.protect C.remember D.want第二节语法填空(共10小题;每小题1.5分,满分15分)I enjoy doing physical training in my free time. Yesterday, when I was working out at a gym on my lunch hour, I saw 16 elderly lady there riding on a bike. She couldn't walk very well, 17 she had a cane (手杖) near her, but she was still there18 (work) hard.After I finished my exercising, I told her that I was going to wait for her 19 she got done and that I would walk her out to her car. It was a little far to her car and it was a 20 (fog) day yesterday in Missouri. I walked her to the car and opened the car door for her, 21 won me her big smile and gratitude.I know 22 is a small thing, but I felt so good inside for helping her. 23 my help, she could have fallen so 24 (easy) out in the wet parking lot, and no one would have known she had fallen. I 25 (know) she needed my help, and her smile was enough to make my day.Ⅱ. 阅读(共两节, 满分50分)第一节阅读理解(共20小题;每小题2分,满分40分)AArthur Miller was born on October 17th, 1915 and died on February 10th, 2005.Over the course of seven decades(十年)of literature career(文学生涯), Arthur Miller created some of the most memorable stage plays(舞台剧)in American Literature. He is the author of Death of a Salesman and The Crucible. Born and raised in Manhattan, Miller went through the best and the worst of American society.Arthur Miller's childhood:His father was a productive shop-keeper and clothin g manufacturer(制造商) until the Great Depression(大萧条时期) dried up nearly all business opportunities.Yet, despite being faced with poverty, Miller made the best of his childhood. He was a very active young man, in love with such sports as football and baseball. When he wasn't playing outside, he enjoyed reading adventure stories. He was also kept busy by his many boyhood jobs. He often worked alongside his father. During other times in his life, he delivered bakery goods and worked as a clerk in a car parts warehouse.College life: In 1934, Miller left the east coast to attend the University of Michigan. He was accepted into their school of journalism. His experiences during the Depression made him skeptical (怀疑的) about religion. Politically, he began leaning towards the "Left”.And since the theater was the cutting edge way for socio-economic liberals (自由主义者) to express their views, he decided to enter the Hopwood Drama competition. His first play, No Villain, received an award from the university.It was an impressive beginning for the young playwright; he had never studied plays or playwriting, and he had written his play in just five days!Miller's later years: In 1987, his autobiography was published. Many of his later plays dealt with personal experience. In particular, his final play, Finishing the Picture mirrors the last days of his marriage to Marilyn Monroe.In 2005,Arthur Miller passed away at the age of 89.26.What can we know about Arthur Miller from Paragraph 2?A.He was born into a poor family.B.His father wasn't good at business.C.He couldn't adjust himself to poverty.D.He had to do lots of jobs to make a living.27.Arthur Miller began to write plays ________.A.when he was in collegeB.when he was a small boyC.because his father encouraged him to do soD.because it was one of his school assignments(作业)28.We can infer from Paragraph 3 that ________.A.the University of Michigan is on the east coast of the USAler's university education made him doubt religionler learned playwriting all by himselfD.socio-economic liberals were probably not politically "left”29.Which of the following plays tells the story of Arthur Miller and Marilyn Monroe?A.Death of a Salesman.B.The Crucible.C.No Villain.D.Finishing the Picture.30.The passage is intended to________.A.introduce Arthur Miller's playsB.tell us about Arthur Miller's childhoodC.explain how Arthur Miller started to write playsD.give us a brief introduction to Arthur Miller's lifeBTabb doesn't look like a typical music teacher. But every weekday evening in the French Quarter New Orleans, he beats out the rhythm on his music stand as students play their chosen instruments.“I'm doing my best to take young people away from harmful things,” said Tabb. His program, "The Roots of Music”, offers free music education to more than 100 students. He struggles to keep young people on the straight and narrow in the city with the nation's highest murder rate(凶杀率).Tabb chose to target 9-to14-year-olds with his program. “That's a very important time in your life,” he said. “If I catch them then, I can hold onto them for at least four or five years and guide them the way that will lead them to success.”Students meet from 4 pm to 7 pm every weekday, all year round. They work with tutors(助教) on schoolwork, practice their music and eat a hot meal before heading home. With the money provided by some people, Tabb's group is able to provide bus transportation, instruments and food for free. He calls it his “no excuse” policy. “You have no excuse why you're not here,” Tabb said. Tabb owes the success in part to the nature of music. “You're always learning something new,”he said. “That's what keeps the kids coming back every day.”But the program isn't only about fun. “Music is about discipline (纪律),”said Tabb. He insists on good behavior and keeps kids in order with threatsof sit-ups(仰卧起坐), pushups(俯卧撑) or tasks like picking up grains of rice —but these measures aren't just punishment. Tabb wants young people to realizethat music can help th em build a better future. “I don't say that I'm saving lives,” he said. “I say I'm giving life —a whole different life of music.”31.The underlined phrase "keep young people on the straight and narrow” may mean ________.A.keep young people standing straightB.keep young people on the correct life track(道路)C.keep young people busy performing musicD.keep young people away from the dangerous parts of the city32.Which of the following kids may NOT be included in Tabb's program “ The Roots of Music”?A.Jack, 8 years old.B.Tom, 9 years old.C.John, 11 years old.D.Linda, 13 years old.33.What attracts children to join in the program to learn music?A.The free food and transportation.B.The famous music teacher.C.The chance to learn new things.D.The strict discipline rules.34.By saying “music is about discipline”, Tabb means ________.A.keeping discipline is more important than learning musicB.obeying(遵守) rules is important in playing music wellC.music is also connected with kids' gradesD.kids can learn how to behave through music35.What is the main idea of this passage?A.Tabb's program offers young people help.B.Kids improve grades through music learning.C.Tabb offers kids free food to learn music.D.Tabb performs music for street children.CThe United States of America is the most culturally diverse(多样化) country in the world in terms of culture religion, ethnicity(种族) and sexual orientation. As a combination of various races and cultures, America is home to all. The culture here is so unique that citizens can be just as proud of their original cultural heritage(遗产) as they are to be American.What is now the US was initially inhabited(居住) by native people until the land was settled by various European groups and African slaves. Since the 20th century, the country has become a heaven for people from all over the globe(全球).The arrival of immigrants(移民) has shifted populations from rural(农村的) areas into cities because immigrants tend to settle in urban areas. At present, 81 percent of the inhabitants in the US live in cities.Cultural and ethnic diversity adds a unique flavor to cities that is expressed through distinct(独特的)neighborhoods, restaurants, places of worship, museums, nightlife and multicultural learning environments.Unique musical forms, such as jazz, rock and roll, Chicano music, and the blues, grow in the US by mixing a variety of culturally distinct musical traditionsto create a new form.At the executive (行政的) level, the country is headed by a mixed-race president; two posts(职位) on the Supreme Court are held by members of the country's two largest minorities.There is also diversity in state and local governments.Without its rich mixture of races and cultures, America would not be the nation that it is today. Founded upon the basis of equality and freedom, America acts as a stage where different cultures not only co-exist peacefully, but develop well.36.According to the passage, America is a country________.A.that welcomes people from all over the worldB.where citizens take more pride in their original cultureC.that is mostly settled by Europeans and AfricansD.where 81% of the population are immigrants37.Where did most people live before many immigrants came to America?A.In the urban areas.B.In the rural areas.C.In the southern part of America.D.In the northern part of America.38.We can learn from the sixth paragraph that________.A.diversity(多样化) has existed in America in almost every aspect(方面)B.most posts in American state governments are held by mixed-race peopleC.equality and freedom make America a fair countryD.diversity has greatly affected the American political field39.What's the best title for this passage?A.The effects of culture and diversity on America.B.The influence of globalization on America.C.The political development of America.D.The bright future of America.40.We would most probably read this passage in a book about________.A.scienceB.historyC.cultureD.amusementDDo you know what your child is going to do when the school bell rings at the end of the day? More than 14 million students leave school every afternoon and have nowhere to go, since they do not have access to(使用) affordable after-school opportunities. According to the National Youth Violence Prevention Resource Center (NYVPRC), 90% of the Americans think all youths should haveaccess to after-school programs(课外活动项目), but two-thirds of parents say they have trouble finding programs locally. The bad news is that the situation may be getting worse.After-school hours are the peak time for juvenile crimes and risky behaviors, including alcohol(酒精) and drug(药物) use. NYVPRC states that children who do not spend any time in after-school activities are 49 percent more likely to use drugs and 37 percent more likely to become a teen parent. Kids are also at the highest risk of becoming a victim of violence after school, particularly between the hours of 2 p.m. and 6 p.m. The highest amount of juvenile crime occurs between 3 p.m. and 4 p.m., when most children are dismissed from school(放学).The NYVPRC defines after-school programs as safe and structured activities that offer children opportunities to learn new skills. The skills students learn can range from technology and math to reading and art. Some programs also offer opportunities for internship (实习), community service, or mentoring. These programs have been shown to improve academic achievement, as well as relieve the stresses(减压) on working families. A report by the U.S. Department of Education and the U.S Department of Justice shows that students in after-school programs have fewer behavioral problems and more self-confidence, and can handle conflicts better than students who are not involved with these programs. In addition, according to the Harvard Family Research Project, after-school programs help students from low income families overcome the inequities (不公平) they face in the school system.41.Which of the following is TRUE according to the first paragraph?A.Most parents don't believe in after-school programs.B.Students are not willing to attend after-school programs.C.It's difficult for parents to find after-school programs for their kids.D.Parents don't care about where their children go after school.42.The underlined word “juvenile” in Paragraph 2 has a similar meaning t o “_____”.A.studentB.teenagerC.adultD.campus43.We learn from the second paragraph that________.A.the teachers should watch over kids after schoolB.children are dismissed from school too lateC.after-school hours are a risky time for childrenD.children should go home immediately school is over44..The author of the passage probably________.A.fully supports after-school programsB.doubts the effects of after-school programsC.believes structured activities are useless for childrenD.thinks students today are too stressed45.What is the theme of the passage?A.Prevention of juvenile crimes.B.Risks kids face after school.C.A research report on the stresses of the students.D.The benefits of after-school programs.第二节信息匹配(共5小题;每小题2分,满分10分)首先,请阅读以下专家的解答:A.First, let' s talk about your boyfriend.If he' s laughing at you about it and won' t stop (even after you've talked to him about the fact that it hurts and annoys you) then you might need to stop being friends with him. If he really doesn't care about how he is hurting you, then he's not good enough to be your boyfriend. Second, if that girl is bothering you and threatening to beat you up you really need to turn to an adult for help. If you can't feel safe at school, youcan't concentrate and will get poor grades, which will affect your future.B.It' s true that your first year in high school is important. It's a fresh startwith your grades and it affects the rest of your high school years, but if you just stay calm, optimistic, and focused, you'll do fine. There's really no need to worry about grades as long as you work hard. Try to stay happy and confident. I think a good way to get an extra confidence boost is to wear your favorite clothes.C.You' re already on the right track by knowing what you want to change about yourself and why you feel the way you do. Understanding yourself is a big step! One good thing to do is to hang out with people a lot like you. If you have more in common with someone, then it is much easier to talk to them. Usually when you feel comfortable around a group of people, your confidence increases a lot.D.What I've discovered from my years of school changing is that confidence is the key to a good first day. Make sure you're not too threatening to your new classmates in any way, and it usually helps to be twice as nice as usual. Just smile and try to have fun. Also, about the clothes, wear something cute with bright colors. Then you'll look good, but you won' t have too many eyes on you all at once. Good luck and have fun!ually it helps your grades if your family life is less of a challenge. If you get better grades and quit running up high phone bills, you would get along better with your parents. Getting along with your parents is pretty important.They've been through a lot and are usually good people to go to for help. So try your best to get along with your family.F.If you don't feel pretty, remember that everyone looks different. There are different kinds of "pretty”, believe it or not. Some people have great smiles, some have great hair, and some have nice eyes. Pay more attention to the parts of you that you like and pay less attention to what you don't like.阅读下列人物的困惑,然后匹配相应的专家解答:46.I am in eighth grade and am about to start high school.I am super nervous about going to high school and afraid that I won' t fit in and get good grades. It is all I think about nowadays and it is ruining my life. What should I do? Please help me!47.I had a fight with a girl at school and ended up almost getting forced to leave school. Everybody makes fun of me and now my boyfriend is joining in with them. I have told him that it hurts my feelings, but he says that it is just a joke and that he may do it again. I feel scared to go to school, because I knocked her down at school and she'll get all her friends to beat me up the next time she sees me. I run as fast as I can to attend classes. Please help me!48.My dad and mom always get me mad. I have bad grades.My cell phone's bill is quite high. I think it' s because of my anger at my dad and mom. What should I do to get along well with my family?49. My family will move to a new city and I will have to start a new school. I'm really nervous about being a new kid. I don' t know what kids wear in a completely new school, how they act, or anything else. I don't want to make a bad impression on my first day. Oh, I will be in 10th grade and I've never changed schools before.50.When I was eight my parents divorced(离婚).I used to be willing to do something and talk to someone I didn't know without being nervous. After thedivorce I switched(转变) to many different schools and got really shy.My problem is that I'm shy around boys. I have to get to know them a lot to open up more and I think I might have ruined it a couple of times. I want my old self back. How do I open up even more without feeling scared?第II卷(共两节)II.读写任务(25分)阅读下面的短文,然后按照要求写一篇150词左右的英语作文。

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