)含答案解析
2019年全国普通高等学校招生统一考试英语(全国卷3参考版)【含答案及解析】
2019年全国普通高等学校招生统一考试英语（全国卷3参考版）【含答案及解析】姓名___________ 班级____________ 分数__________题号一二三四五六总分得分一、阅读理解1. 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。
AMusicOpera at Music Hall: 1243 Elm Street. The season runsJune through August, with additional performances in March andSeptember. The Opera honors Enjoy the Arts membershipdiscounts. Phone: 241-2742. .Chamber Orchestra: The Orchestra plays at Memorial Hall at1406 Elm Street, which offers several concerts from Marchthrough June. Call 723-1182 for more information. .Symphony Orchestra: At Music Hall and Riverbend. For ticketsales, call 381-3300. Regular season runs September throughMay at Music Hall in summer at Riverbend./home.asp .College Conservatory of Music (CCM): Performances are on themain campus( 校园 ) of the university, usually at PatriciaCobbett Theater. CCM organizes a variety of events, includingperformances by the well-known LaSalle Quartet, CCM’s Philharmonic Orchestra, and various groups of musicianspresenting Baroque through modern music. Students with I.D.cards can attend the events for free. A free schedule ofevents for each term is available by calling the boxoffice at 556-4183. /events/calendar .Riverbend Music Theater: 6295 Kellogg Ave. Large outdoor theater with theclosest seats under cover (price difference).Big name shows all summer long!Phone:232-6220. .1. Which number should you call if you want to see anopera?A. 241-2742.B. 723-1182.C. 381-3300.________D. 232-6220.2. When can you go to a concert by Chamber Orchestra?A. February. ____________________B. May.C. August. ______________________________D. November.3. Where can students go for free performances with theirI.D. cards?A. Music Hall.B. Memorial Hall.C. Patricia Cobbett Theater.D. Riverbend Music Theater.4. How is Riverbend Music Theater different from the otherplaces?A. It has seats in the open air.B. It gives shows all year round.C. It offers membership discounts.D. It presents famous musical works.2. 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。
人教版小学三年级英语真题含答案及解析
人教版小学三年级英语真题含答案及解析(共50道题)下面有答案和解题分析一、综合题1.The sun __________ (shine) brightly today. It __________ (be) a perfect day for a picnic. We __________ (bring) sandwiches, juice, and fruit to the park.2.I _______ (not) like to swim in winter.3.Which animal can live both on land and in water?A. TigerB. FishC. FrogD. Lion4.I _______ (not/like) to eat spicy food.5.Every day, I __________ (1) to school by bus. It __________ (2) about 20 minutes to get there. After I __________ (3) to school, I __________ (4) my friends and we__________ (5) to our classrooms. I __________ (6) math and English in the morning, and in the afternoon, we __________ (7) science and art.6.It __________ (be) my birthday yesterday. I __________ (invite) my friends to my house. We __________ (have) a party and __________ (eat) cake. My best friend__________ (give) me a nice present. I __________ (be) so happy!7.Which of these is a part of the plant?A. LeafB. PlateC. KnifeD. Fork8.I _______ (be) sick yesterday.9.He _______ (not/like) playing video games.10.I _______ (do) my homework right now.11.What color is a ripe banana?A. RedB. YellowC. GreenD. Blue12.I _______ (like) playing basketball with my friends after school. Sometimes we _______ (play) at the park, and other times we _______ (play) at the school gym. 13.Which animal has wings?A. DogB. BirdC. ElephantD. Cat14.What is the opposite of "hard"?A. SoftB. HeavyC. TallD. Slow15.He _______ (doesn’t / don’t / didn’t) like to swim.16.I _______ (go) to the zoo last weekend. We _______ (see) many animals like lions, tigers, and bears. It _______ (be) a sunny day, so we _______ (take) lots of photos. We _______ (have) so much fun!17.I _______ (help) my parents clean the house.18.Which one is a fruit?A. CucumberB. LettuceC. StrawberryD. Onion19.There are ______ apples in the basket.A. muchB. manyC. someD. few20.I __________ (go) to the library every week. I __________ (love) reading books about animals and space. Last week, I __________ (borrow) a book about planets. It__________ (be) very interesting.21.Which one is a shape?A. TriangleB. SpoonC. DeskD. Pencil22.This summer, my family is going to ______ for our holiday. We plan to visit famous places like the ______ and the ______. I am most excited about seeing the______, because I love learning about ______. At night, we will eat ______ at a local restaurant.23.My best friend ______ (be) a very good singer. She ______ (practice) singing every day, and last week she ______ (perform) at the school talent show. Everyone ______ (be) amazed by her voice, and she ______ (win) first place!24.Which of these animals is known for its trunk?A. LionB. ElephantC. DogD. Cat25.Which one is a fruit?A. PotatoB. CarrotC. AppleD. Cucumber26.She _______ (eat) lunch with her friends every day.28.She _______ (go) to the library every Saturday.29.Which of the following objects is best for writing on paper?A. PenB. EraserC. BrushD. Knife30.Which one is the opposite of "cold"?A. FastB. SmallC. WarmD. Tall31.He _______ (is / are / am) always happy.32.Yesterday, I ______ (go) to the supermarket to buy some fruit. I ______ (buy) apples, bananas, and oranges.33.Which of these animals can jump?A. DogB. KangarooC. FishD. Elephant34.I _______ (not) know his name.35.He _______ (study / studies / studied) in the library every afternoon.36.We _______ (play) tennis next Saturday.37.Which one is a season of the year?A. SummerB. MondayC. JanuaryD. Breakfast38.I _______ (like) reading books.39.We _______ (study) English in the afternoon.40.We __________ (have) a test in math today. I __________ (study) very hard for it. Yesterday, I __________ (review) all the chapters, and I __________ (practice) some problems. I __________ (feel) ready for the test.41.I _______ (see) a movie with my friends last weekend.42.Which of these is the color of the sky on a sunny day?A. BlueB. GreenC. YellowD. Red43.Which sentence uses "an" correctly?A. I saw an apple.B. I saw a apple.C. I saw an dog.D. I saw a elephant.45.Yesterday, my family and I ______ (go) to the zoo. We ______ (see) many animals, like lions, tigers, and monkeys. I ______ (take) a lot of photos. My little brother ______ (like) the monkeys the most because they ______ (be) very funny. After visiting the animals, we ______ (eat) lunch at the zoo café. It ______ (be) a fun day.46.You are at the playground with your friends. You see a swing, a slide, and a seesaw. You decide to play on the swing. What are you playing on?A. The seesawB. The slideC. The swingD. The monkey bars47.He _______ (have) a sister.48.We _______ (eat) lunch together every day.49.They _______ a party last night.50.Which animal is known for its trunk?A. LionB. ElephantC. DogD. Cat(答案及解释)。
库存管理练习试卷1(题后含答案及解析)
库存管理练习试卷1(题后含答案及解析)题型有：1. 判断题 2. 单选题判断题1．企业维持库存的原因主要表现在企业对客户、生产、运力等资源的平衡利用上。
A．正确B．错误正确答案：A 涉及知识点：库存管理2．如果订货的没有客户所需要的物料可供发货，就可能失去销售的机会或可能增加额外的费用，通常称这种费用为机会成本。
A．正确B．错误正确答案：B 涉及知识点：库存管理3．平均库存值指某一时刻内全部库存所占有的资金总和。
A．正确B．错误正确答案：B 涉及知识点：库存管理4．供应链上存在着需求与供给的不确定性，即向供应商订货量的方差会大于向其顾客销售量的方差，并且这种被动会沿着供应链向上不断扩大。
A．正确B．错误正确答案：A 涉及知识点：库存管理5．订货批量越大，每一计划期需要的订货次数就越少，订货总成本也就越低。
A．正确B．错误正确答案：A 涉及知识点：库存管理6．在确定订货批量时，必须考虑运输费率对总成本的影响。
A．正确B．错误正确答案：A 涉及知识点：库存管理7．供应链的基本要素包括供应商、销售商和流通代理企业。
A．正确B．错误正确答案：B 涉及知识点：库存管理8．标准成本的计算是以每种产品一年计划的准备次数与一年中合计生产的单位数之间的商为基础的。
A．正确B．错误正确答案：B 涉及知识点：库存管理9．链一般包括物资流通、商业流通、信息流通、资金流通四个流程。
四个流程的流通方向都是单向的。
A．正确B．错误正确答案：B 涉及知识点：库存管理10．分析法是一种决定哪些物品应该考虑采用集中仓储的方法。
A．正确B．错误正确答案：A 涉及知识点：库存管理11．供应链管理的策略就是通过致力整个供应链上信息的快速、准确的流动，来减少不可预料情况的发生，从而避免不合理的采购和不需要的库存。
A．正确B．错误正确答案：A 涉及知识点：库存管理12．ABC分析法中C类物品可以每一周进行库存检查比较合适。
A．正确B．错误正确答案：B 涉及知识点：库存管理13．模型和固定批量模型，在需求速度和提前期都是稳定不变的条件下良种模型所起作用是完全相同的。
小学4年级下册英语试题含答案及解析
小学4年级下册英语试题含答案及解析(共50道题)下面有答案和解题分析一、综合题1.I _______ (eat) breakfast at 7:00 am.2.Which is the correct question for "你好吗?"A. What is your name?B. How are you?C. Where are you from?D. What time is it?3.I _______ (not/not) like tomatoes.4.He _______ (eat) breakfast at 7:00 every morning.5.I _______ (watch) TV when you called.6.Which one is used for eating soup?A. SpoonB. KnifeC. ForkD. Plate7.We __________ (visit) a museum last weekend. The museum __________ (have) many interesting exhibits. My favorite exhibit __________ (be) the dinosaur skeleton. I __________ (learn) a lot about ancient animals.8.Which color is the sun?A. GreenB. YellowC. BlueD. Red9.I’m going to visit my grandparents this __________. We are going to __________ together and play __________. My grandma is going to cook some __________ for us. I’m very excited because we will also have a __________ in the garden. I can’t wait for the weekend!10.Which of these is a body part?A. BookB. TableC. HandD. Chair11.What is the name of the orange fruit?A. StrawberryB. AppleC. BananaD. Orange12.My dad ______ (work) in an office. He ______ (leave) for work at 8:00 every morning. After work, he sometimes ______ (go) to the gym to exercise. I ______ (like) to spend time with him, so on weekends we ______ (play) basketball together.13.Which of these is a fruit?A. TomatoB. CarrotC. PotatoD. Onion14.What do you call a place where we buy books?A. SupermarketB. LibraryC. BookstoreD. Playground15.They _______ (not) watch TV in the evening.16.He _______ (not/like) to get up early.17.They ________ (play) football now.18.What is the opposite of "happy"?A. SadB. AngryC. SleepyD. Excited19.I __________ (eat) breakfast at 7:00 AM every day.20.She _______ (have) a dog.21.They _______ (play) basketball now.22.My family ______ (go) on a vacation to the mountains last summer. We ______ (arrive) there in the afternoon and ______ (set) up our tents. It ______ (rain) a little, but we ______ (still) have fun. We ______ (go) hiking every day and ______ (see) many beautiful animals and plants. It ______ (be) a wonderful trip.23.Which of these is a part of a tree?A. LeafB. TableC. CarD. Chair24.Which of these is a drink?A. WaterB. BreadC. CarrotD. Spoon25.Which one is a color?A. RedB. DogC. CatD. Chair26.She _______ (not/be) here right now.27.We _______ (run) in the park every morning.28.Which of these is the opposite of "hot"?A. ColdB. FastC. BrightD. Slow29.We _______ (go) to the library every Saturday. I _______ (like) reading books about animals. My little brother _______ (prefer) picture books.30.We _______ (be) friends.31.It ______ (be) very cold this morning, so I ______ (wear) my warm jacket. I______ (leave) the house at 8:00 AM and ______ (catch) the bus to school. During class, we ______ (learn) about animals and plants. After school, I ______ (meet) my friends at the park and we ______ (play) soccer together.32.Which of these is a color?A. SpoonB. RedC. ChairD. Paper33.He _______ (talk/talks) to his friends during break.34.What is the opposite of "happy"?A. SadB. TallC. LightD. Hot35.We _______ (was / were / is) at school yesterday.36.We usually ______ (go) to the library on Saturdays. Last Saturday, we ______ (borrow) some new books, and I ______ (read) one of them in the afternoon. This week, I ______ (choose) a book about space because I ______ (find) it fascinating.37.They _______ to the park every Saturday.38.Which one is used for eating?A. ForkB. KnifeC. SpoonD. Plate39.We _______ (study / studies / studied) English every day.40.What do we use to write?A. BookB. PenC. EraserD. Chair41.She _______ (do / does / did) not like oranges.42.I _______ (not) like doing homework.43.We _______ (read) books every evening.44.Which animal says "woof"?A. DogB. CowC. CatD. Elephant45.We _______ (not) go to bed early.46.Which of these is a piece of furniture?A. BedB. SpoonC. PlateD. Chair47.He _______ (eat) breakfast at 7 a.m.48.Jake and his friends are going on a field trip to the museum. They are excited to see the __________ exhibit. There are many __________ and paintings on the walls. Jake learns about different __________ and their history. He takes a lot of notes in his__________.49.I _______ (take) my dog for a walk every morning.50.They _______ (go / goes) to school by bike.(答案及解释)。
统计基础习题（含答案解析）
统计基础习题（含答案解析）第⼀章总论⼀、判断题：１、社会经济统计的研究对象是社会经济现象总体的各个⽅⾯。
２、在统计调查过程中所采⽤的⼤量观察法，是必须对研究对象的所有单位进⾏调查。
３、在全国⼯业普查中，全国⼯业企业数是统计总体，每个⼯业企业是总体单位。
４、总体单位是标志的承担者，标志是依附于总体单位的。
５、当对品质标志的标志表现所对应的单位进⾏总计时就形成统计指标。
６、因为统计指标都是⽤数值表⽰的，所以数量标志就是统计指标。
７、统计指标及其数值可以作为总体。
８、所有的统计指标和可变的数量标志都是变量。
⼆、填空题：1、“统计”⼀词的含义是、和。
它们之间的关系是统计⼯作的成果，和统计⼯作是理论和实践的关系。
2、统计学的研究对象是。
3、标志是说明特征的，分标志和品质标志两种，前者的具体表现是，后者的具体表现是。
4、当我们要研究⼯业企业⽣产经营状况时，全部⼯业企业就构成，⽽每⼀个⼯业企业则是。
5、⼯⼈的年龄、⼯资、⼯龄属于标志，⽽⼯⼈的性别、民族、⼯种属于标志。
6、设备台数、⼯⼈⼈数属于变量，⽽⾝⾼、年龄、体重属于变量。
７、统计研究的基本⽅法是、、综合指标法。
８、要了解某⼀个企业的产品⽣产情况，总体是，总体单位是。
９、性别是标志，标志表现则具体为。
10、在⼈⼝总体中，总体单位是，“⽂化程度”是总体单位的标志。
三、单项选择题：1、统计总体的基本特征表现为：（）A、同质性、⼴泛性、社会性B、同质性、⼤量性、变异性C、同质性、综合性、⼤量性2、研究某市⼯业企业⽣产设备使⽤状况，那么，统计总体为( )A、该市全部⼯业企业B、该市每⼀个⼯业企业C、该市全部⼯业企业每⼀台设备D、该市⼯业企业的全部⽣产设备3、某组五名学⽣的考试得分分别为：60、70、80、85、90，这五个数字是：( )A、指标B、标志C、变量值D、变量4、要了解某班40名学⽣的学习情况，则总体单位是：( )A、40个学⽣B、每⼀个学⽣的成绩C、每⼀个学⽣D、40个学⽣的学习成绩5、数量指标的表现形式为：( )A、相对数B、绝对数C、平均数6、某学⽣某门课考试成绩为80分，则成绩是：( )A、品质标志B、变量C、变量值D、标志值7、某店有50名职⼯，把他们的⼯资加起来除以50，这是：( )A、对50个变量求平均B、对50个变量值求平均C、对50个标志求平均D、对50个指标求平均8、某市全部商店作为总体，每⼀个商店为总体单位，则该市全部商品零售额是：( )A、品质标志B、质量指标C、数量指标D、变量值９、了解某地区⼯业企业职⼯情况，下⾯哪个是统计指标（）Ａ、该地区所有职⼯的⼯资⽔平Ｂ、该地区⼯业企业职⼯的⽂化程度Ｃ、该地区⼯业企业职⼯的⼯资总额Ｄ、该地区职⼯所从事的⼯种１０、统计⼯作的过程不包括（）Ａ、统计调查Ｂ、统计分布Ｃ、统计整理Ｄ、统计分析四、多项选择题：1、要研究某局所属30个企业职⼯的⼯资⽔平，则：( )A、总体是某局B、总体是某局所属30个企业C、总体是30个企业全部职⼯D、总体是30个企业全部职⼯的⼯资E、总体单位是每⼀个企业F、总体单位是每⼀名职⼯2、下列指标中哪些是质量指标:( )A、⼯⼈劳动⽣产率B、设备利⽤率C、新产品数量D、单位产品⼯时消耗量E、废品量F、利润额3、下列变量中哪些是连续变量：( )A、⾝⾼B、⼈数C、体重D、年龄E、设备台数F、企业数4、以某市⼯业企业为总体则下列各项中哪些是指标:( )A、某市某⼚职⼯⼈数B、全市⼯业企业总产值C、全市⼯业企业职⼯平均⼯资D、全市⼯业企业数５、要了解某地区全部成年⼈⼝的就业情况，那么（）Ａ、全部成年⼈是研究的总体Ｂ、成年⼈⼝总数是统计指标Ｃ、成年⼈⼝变业率是统计标志Ｄ、“职业”是每个⼈的特征，是数量标志Ｅ、某⼈职业是“教师”，这⾥的“教师”是标志表现６、国家统计系统的功能或统计的职能是（）Ａ、信息职能Ｂ、咨询职能Ｃ、监督职能Ｄ、决策职能Ｅ、协调职能７、在⼯业普查中（）Ａ、⼯业企业总数是统计总体Ｂ、每⼀个⼯业企业是总体单位Ｃ、固定资产总额是统计指标Ｄ、机器台数是连续变量Ｅ、职⼯⼈数是离散变量８、下列各项中属于统计指标的有（）Ａ、2009年全国⼈均总产值Ｂ、某台机床使⽤年限Ｃ、某市年供⽔量Ｄ、某地区原煤⽣产量Ｅ、某学员平均成绩第⼆章统计设计和统计调查⼀、判断题：1、全⾯调查和⾮全⾯调查是根据调查结果所得的资料是否全⾯来划分的。
二年级二年级下册阅读理解解题技巧和训练方法及练习题（含答案）含解析
二年级下册阅读理解解题技巧和训练方法及练习题(含答案)含解析 一、二年级语文下册阅读理解练习1．阅读下文，回答下面的问题。
羊妈妈收菜羊妈妈带着小羊到菜园去收菜。
他们走到萝卜地里。
羊妈妈拔了一个萝卜。
小羊要吃萝卜叶子。
羊妈妈说：“萝卜的根最好吃。
”他们走到白菜地里。
羊妈妈拔了一棵小白菜。
小羊要吃白菜的根。
羊妈妈说：“白菜的叶子才好吃呢！”他们走到西红柿地里。
小羊要吃西红柿的叶子。
羊妈妈说：“要吃西红柿的果实呀！” （1）短文主要写________带着________到________去收菜的事。
（2）读一读，选一选。
①根 ②②叶子 ③③果实西红柿的________好吃，萝卜的________好吃，白菜的________好吃。
【答案】 （1）羊妈妈；小羊；菜园（2）③；①；②【解析】2．读儿歌，完成练习。
春天到了春天到，春天到，花儿露出了笑脸（liǎn niǎn ），小草伸长（cháng cháng zhǎng zhǎng ）了腰。
春天到，春天到，燕子搭新窝，喜鹊喳喳叫。
（1）为划线字选择正确的读音。
笑脸________（liǎnniǎn ） 伸长________（cháng cháng zhǎng zhǎng ） （2）“花儿露出了笑脸”指________。
（①花开了；②花笑了）“小草伸长了腰”指________。
（①小草刚睡醒；②小草长高了）（3）儿歌中的燕子，喜鹊在春天里忙着做什么呢？选一选。
（填序号）①搭新窝 ②②喳喳叫在________，在________。
【答案】 （1）liǎn ；cháng（2）①；②（3）①；②【解析】 3．阅读短文，完成练习。
两只羊一天，一只白羊从南面上了独（dú）木桥，一只黑羊从北面上了独木桥。
它们同时来到桥中间，白羊说：“你退（tuì）回去，让我先过桥！”黑羊说：“你退回去，让我先过桥！” 它们谁也不肯（kěn ）让谁，就打了起来。
高一英语阅读理解(5篇)(含答案及解析)
高一英语阅读理解（5篇）（含答案及解析）一、阅读理解阅读下列短文，从各题所给的四个选项(A、B、C和D)中，选出最佳选项。
1.A woman who had been diagnosed with cancer had been given3 months to live. So she told her pastor which songs she wanted sung at the service，what scriptures she would like read，and what she wanted to be wearing. The woman also told her pastor that she wanted to be buried with her favorite bible.“There’s one more thing.” she said excitedly.“I want to be buried with a fork in my right hand.”The pastor stood looking at the woman not knowing quite what to say.The woman explained，“In all my years of attending church socials and functions where food was involved；my favorite part was when whoever was clearing away the dishes of the main course would lean over and say ‘you can keep your fork’. It was my favorite part because I knew that something better was coming. So I just want people to see me there in that casket with a fork in my hand and I want them to wonder ‘What’s with the fork？’ Then I want you to tell them，‘Something better is coming，so keep your fork too.’”1/ 15The pastor’s eyes were welled up with tears of joy as he hugged the woman goodbye. He knew this would be one of the last times he would see her before her death. But he also knew that the woman had a better grasp of heaven than he did. She knew that something better was coming.At the funeral the pastor told the people how he could not stop thinking about the fork and told them that they probably would not be able to stop thinking about it either.So the next time you reach down for your fork，let it remind you so gently that there is something better coming.Keep your fork. The best is yet to come.1. Why did the woman have a talk with the pastor？Because ________.A. she wouldn’t like to dieB. she was anxious about deathC. she wanted to discuss certain aspects of her final deathD. she begged him to help her to get rid of her illness2. When hearing the woman’s last wish，the pastor felt ________.A. shockedB. excitedC. entertainedD. amazed3. Someone says to you “Keep your fork”，which suggests ________.2/ 15A. the main course will be servedB. your favorite is comingC. the service will begin soonD. the service will be finished at once4. Why was the pastor so pleased when he heard her expression？Because________.A. he understood the woman betterB. he realized she has understood death completelyC. the woman didn’t ask for too muchD. he could carry out her last wishes2.The last few days before Christmas passed quickly and it was soon Christmas Eve. That night when everyone went to bed, Bunny couldn't sleep. He still couldn't think of what he wanted his Christmas gift to be. He wondered how Father Christmas would know what to bring him if he didn't know himself.As he was sitting up in bed, Bunny heard a big noise on the roof (屋顶) and a sound downstairs. It was Father Christmas, he realized. Bunny jumped out of bed and raced down the hall to the stairs hoping to have a look at the old man with his own eyes.3/ 15By the time Bunny reached the bottom of the stairs, though, everything was again silent. Beautiful gifts were piled (堆积) under the Christmas tree, but Father Christmas was gone. He looked for him for a few minutes, but it was already too late. Bunny turned to climb back upstairs when he heard a cry.“Hello,” said Bunny. “Is somebody there?”He was answered by another cry. Bunny looked around the big pile of gifts to see what was making the noise. Right under the tree was a funny looking brown animal with big feet and sad eyes. Bunny might have mistaken it for a dog, if it hadn't been for the antlers (鹿角) on its head.“Are you a reindeer (驯鹿)?” asked Bunny.“Yes,” replied the animal, “my name is Ralph.”“And you were pulling Father Christmas' sled (雪橇)?”“I was until I got airsick,” replied Ralph, “I'm afraid I wasn't cut out for the job. Now I'm stuck here and I don't know how to get back to the North Pole.”“Well, if you like, you can stay with us as a friend,” said Bunny. As he made the offer, Bunny suddenly realized the gift he wanted from Father Christmas was a new friend!1. Why couldn't Bunny fall asleep on Christmas Eve?4/ 15A. He had a lot of things to do.B. He was disturbed by a big noise.C. He was thinking of what gift he would get.D. He wanted to have a look at the reindeer.2. As soon as Bunny came downstairs, he _____.A. was probably very sadB. found what he wantedC. ran into a reindeerD. heard a loud cry3. The underlined part “cut out for” in the text probably means “_____”.A. fit forB. afraid ofC. proud ofD. interested in4. What would be the best title for the text?A. A strange Christmas treeB. A special Christmas giftC. A quiet but smart boyD. A lovely reindeer3.Online shopping has become more and more popular these years. Women have jumped ahead of men for the first time in using the Internet to do their holiday shopping, according to a study published last week in5/ 15the US.For years men have been more likely to shop on the Internet than women, but during the 2013 holiday season 58 percent of those shopping online were women.“It shows how popular the Internet is becoming,” said Lee Rainie, director of the Pew Internet and American Life Project group, which carried out the study. Rainie said it was only a matter of time before women shoppers caught up with men. This is because women traditionally make decisions about spending.Users were more likely to shop online to save time. Internet users between the ages 18 and 29 were responsible for some of the surprising increase in the online gift- buying population this time around.However, three-quarters of the US Internet users did not buy holiday gifts online in 2013. They worried about credit card security(安全），or just compared online prices with off-line prices, then dashed off to the shops to get the best deals.“But even if shoppers don’t buy online, websites are becoming promotion(促销）tools for stores，’，said Dan Hess’vice president of Com Score Network Inc. Hess said that actually most stores，websites can make customers fully believe the security of their credit card numbers. And most are able to ensure that gifts arrive on time.6/ 15It’s all about making the shopping experience more efficient, more reliable and more comfortable,” Hess said.(1) Which of the following statements is true?A. There were fewer women online shoppers than men in 2013.B. More women shopped online than men in 2013.C. Most of the Internet users between the ages 18 and 29 were women.D. people in the US were more likely to buy gifts online.(2) What does the underlined part “dashed off” probably mean? _A. 关闭B. 推迟C. 匆忙D. 起飞(3) According to Dan Hess, shopping online is ________.A. unsafeB. convenientC. a waste of moneyD. cheaper4.Most people know the feeling when you walk into a lift(电梯）with other people. A study has found that where people stand is based on their social position on entering the lift.7/ 15Rebekah Rousi，a Ph. D. student, did a study of lift behavior in two of the tallest office buildings in Adelaide, Australia. As part of her research, she took a total of 30 lift rides in the two buildings, and discovered there was a fixed order about where people chose to stand.In her research paper, she wrote that more senior men seemed to walk straight towards the back of the lift. She said, “In front of them were younger men, and in front of them were women of all ages.” She also noticed there was a difference in the direction where people looked during the ride. “Men watched the monitors, looked in the side mirrors (in one building) to see themselves, and in the door mirrors (in the other building) to watch others. Women would watch the monitors and avoid looking into others，eyes (unless in conversations) and the mirrors. ”Rebekah Rousi concluded that shyer people stand toward the front，where they can’t see other passengers，while fearless people stand in the back, where they have a good view of everyone else.(1) According to the study, where people stand in a lift is decided by ________.A. their social positionB. the monitorsC. other passengersD. others’ position8/ 15(2) Who are most likely to go to the back of the lift?A. Shyer people.B. Senior men.C. Younger men.D. Women.(3) Which is true according to the passage?A. The order in which people stand in a lift is fixed.B. Few people feel embarrassed with strangers in a lift.C. Women like watching themselves in the side mirrors.D. Fearless people stand in the back to avoid seeing others.(4) The passage is probably taken from ________.A. a lift instructionB. a storybookC. a travel guideD. a newspaper5. 七选五根据短文内容，从短文后的选项中选出能填人空白处的最佳选项。
劳动法题库(含答案及解析版)
劳动法题库（含答案及解析版）2020年3月8日翟天野题库说明：劳动法题库共计146题，其中单选题115题，判断题31题。
题目难度分为三个等级，简单难度包括法律基本知识概念、法条原文、法律知识普及等116题；中等难度包括实践常见法律风险及防范等25题；困难难度包括法律条文理解与实践运用等5题。
题库适用法律范围包括《中华人民共和国劳动法》、《中华人民共和国劳动合同法》、《女职工劳动保护特别规定》、《工资支付条例》、《劳动争议调解仲裁法》。
一、单项选择题1.针对因生产特点、工作特殊需要或职责范围的关系，无法按标准工作时间衡量或需要机动作业的职工所采用的工时形式是( )。
A.标准工时B.计件工时制C.不定时工作制D.综合计算工时制【答案】C【解析】企业因生产特点不能实行本法第三十六条、第三十八条规定的，经劳动行政部门批准，可以实行其他工作和休息办法。
2.我国劳动法规定，国家对女职工实行特护劳动保护。
下面的做法不符合这一规定的是( )。
A.某砖厂女职工董某怀孕期间，厂里安排她简单打扫清洁卫生，不再做搬运工。
B.某企业为完成全年生产任务，便要求每个职工每天加班1个小时，怀孕达6个月的女职工刘某也不例外。
C.某公司通知其女职工周某，鉴于她的孩子已满13个月，公司决定恢复她的“三班倒”的工作制。
D.某矿山女职工肖某被安排到井下工作。
【答案】D【解析】禁止安排女职工从事矿山井下、国家规定的第四级体力劳动强度的劳动和其他禁忌从事的劳动。
3.下列选项有关劳动争议仲裁的表述，哪一项是错误的,( )A.劳动争议发生后，当事人就争议的解决有仲裁协议的，可以进行仲裁B.仲裁是劳动争议解决的必经程序，未经仲裁不得诉讼C.劳动争议仲裁委员会解决劳动争议时可以依法进行调解，仲裁调解书具有法律效力D.因签订集体合同发生的争议不能采用仲裁的方式解决，但履行集体合同发生的争议则可以仲裁【答案】A【解析】当事人就争议的解决有仲裁协议的，应当进行仲裁4.下列哪项权利不属于劳动监察机构及劳动监察员享有的权利?( )A.调查权B.决策权C.处分权D.检查权【答案】B【解析】参照《劳动法》85、86、87、88条。
人教版初中九年级英语试题(答案及解析)
人教版初中九年级英语试题(答案及解析)(共50道题)下面有答案和解题分析一、综合题1.Which word is the correct plural form of "woman"?A. WomensB. WomanesC. WomenD. Womans2.Which of the following is the correct use of “too”?A. She is too tired to go to the party.B. She is too tired for go to the party.C. She is too tired for going to the party.D. She is tired too much to go to the party.3.Which of the following sentences is in the correct past simple tense?A. He was goes to the market.B. He went to the market.C. He is going to the market.D. He goes to the market.4.By the time I __________ (finish) my homework, my parents __________ (arrive) homeI __________ (feel) tired, but I __________ (not/finish) my work yetI __________ (decide) to take a short break before continuing.5.Which of the following sentences uses the correct article?A. I saw a elephant in the zoo.B. I saw an elephant in the zoo.C. I saw the elephant in the zoo.D. I saw elephant in the zoo.st weekend, I __________ (1) __________ (visit) my cousin in the countryside. She __________ (2) __________ (live) in a small village surrounded by mountains. We __________ (3) __________ (go) for a walk in the forest near her house. It __________ (4) __________ (be) a beautiful day, and the weather __________ (5) __________ (be) perfect for hiking. We __________ (6) __________ (see) many animals and __________ (7) __________ (take) lots of photos. I __________ (8) __________ (have) a great time.7.Which of the following sentences is in the past perfect tense?A. I had eaten lunch before you called.B. I ate lunch before you called.C. I have eaten lunch before you called.D. I eat lunch before you called.8.Choose the correct preposition: "The dog is hiding _______ the table."A. inB. underC. onD. at9.Which of the following sentences is correct?A. She can sings well.B. She can sing well.C. She sing well.D. She sings well.10.Which sentence correctly uses the adverb "never"?A. She never goes to school on Sundays.B. She never go to school on Sundays.C. She never went to school on Sundays.D. She never to go school on Sundays.11.I __________ (never, see) such a beautiful painting beforeIt __________ (be) a work of art that __________ (make) by an artist who __________ (live) in the 18th centuryMany people __________ (come) to admire it every year, and it __________ (become) more and more famous over time.12.Which of the following sentences uses the correct form of the verb "to be"?A. I am a teacher.B. I is a teacher.C. I be a teacher.D. I are a teacher.13.Which sentence uses “many” correctly?A. He has many books on the shelf.B. He has many money in his wallet.C. He has many milk in his fridge.D. He has many furniture in his house.14.This time last year, I __________ (study) in New YorkI __________ (have) a great time, and I __________ (make) many new friendsI __________ (learn) a lot about American culture, and I __________ (travel) to several interesting placesI __________ (feel) very happy during my stay there.15.She __________ (always, help) others when they __________ (need) itLast week, she __________ (give) some money to a poor familyEveryone __________ (be) very impressed by her kindness.16.Which of the following sentences is correct?A. I have less books than you.B. I have fewer books than you.C. I have less book than you.D. I have few books than you.17.Choose the correct question: "________ she like to swim?"A. DoB. DoesC. IsD. Are18.Which sentence is in the correct form of the future continuous tense?A. She will be reading when you arrive.B. She is reading when you arrive.C. She will read when you arrive.D. She read when you arrive.19.Which sentence contains an error?A. She is playing the piano now.B. He does not like apples.C. They can speaks English.D. I like to read books.20.Choose the correct word to complete the sentence: "I __________ to the library every Saturday."A. goB. wentC. goingD. goes21.Choose the correct answer: He _______ to the gym every morning.A. goesB. is goingC. goD. went22.We ______ (not/see) each other for a long time, so when I ______ (meet) him yesterday, I ______ (feel) very happyHe ______ (tell) me that he ______ (study) in another city nowHe ______ (invite) me to visit him next holiday, and I ______ (agree) to go.23.I ________ (1) (go) to the cinema with my friends tomorrow. We ________ (2) (watch) a new movie, and I ________ (3) (look) forward to it because I ________ (4) (hear) it is very exciting.24.By the time I __________ (finish) my dinner, my friends __________ (arrive) at the party. I __________ (already, eat) most of the food, and I __________ (feel) quite full.25.Which of the following sentences is correct?A. I have already seen that movie.B. I already have seen that movie.C. I seen that movie already.D. I have see that movie already.26.What is the correct form of the verb in the sentence: "They _____ swimming every Saturday"?A. GoB. GoesC. GoingD. Gone27.If they ______ (not hurry), they ______ (miss) the train.28.The teacher __________ (give) us a quiz last weekI __________ (study) hard, so I __________ (do) well on itMy friend, however, __________ (not, study), and she__________ (not, do) as well as me.29.Choose the correct answer: I _______ the homework already.A. doB. am doingC. didD. have done30.They __________ to the park every Sunday.A. goB. goesC. is goingD. went31.Which of the following sentences contains an example of indirect speech?A. He said, "I am going to the park."B. He said he is going to the park.C. He said that he was going to the park.D. He says he goes to the park.32.Choose the correct sentence:A. He is more taller than me.B. He is the more taller than me.C. He is taller than me.D. He is tallest than me.33.Which sentence correctly uses the word "during"?A. I met him during the lunch break.B. I met him during lunch break.C. I met him during of lunch break.D. I met him during at lunch break.34.Which sentence is in the correct form of the future tense?A. I will going to the concert tomorrow.B. I will go to the concert tomorrow.C. I will goes to the concert tomorrow.D. I will gone to the concert tomorrow.35.Which sentence uses the correct plural form of "child"?A. There are two childs in the park.B. There are two children in the park.C. There are two childs in park.D. There are two childrens in the park.36.Choose the correct form of the verb to complete the sentence: “She __________ her homework already.”A. doesB. is doingC. didD. has done37.Which of the following is the correct plural form of "child"?A. ChildrenB. ChildsC. ChildernD. Childes38.Choose the correct sentence using "there is" or "there are."A. There is many people in the room.B. There are many people in the room.C. There are a lot of people in the room.D. There is a lot of people in the room.39.Choose the correct question:A. What does she doing?B. What is she do?C. What is she doing?D. What she is doing?40.Choose the correct word to complete the sentence: I usually _______ to school by bus.A. goB. goesC. goingD. gone41.Which sentence is correct when talking about a repeated action in the past?A. I used to go swimming every summer.B. I use to go swimming every summer.C. I used going swimming every summer.D. I used going to swim every summer.42.Which sentence correctly uses "much"?A. How much people are in the room?B. How much books do you have?C. How much rice do you need?D. How much chairs do you need?43.Which of the following sentences is correct?A. She is the tall girl in the class.B. She is the taller girl in the class.C. She is the tallest girl in the class.D. She is the most tall girl in the class.44.Which of the following is an example of an article?A. TheB. HeC. RunD. Quickly45.Which sentence correctly uses "few"?A. There are few apples left in the basket.B. There are a few apples left in the basket.C. There are few water in the bottle.D. There are a few water in the bottle.46.Choose the correct sentence:A. She has already finish her homework.B. She has already finished her homework.C. She already has finished her homework.D. She already finished her homework.47.Which sentence is grammatically correct?A. There is five students in the classroom.B. There are five students in the classroom.C. There is five student in the classroom.D. There are five student in the classroom.48.Which of the following sentences correctly uses "a" or "an"?A. I saw a elephant in the zoo.B. I saw an elephant in the zoo.C. I saw a apple in the zoo.D. I saw an apple in the zoo.49.Choose the correct answer to complete the sentence: The train _______ at 10:00 tomorrow.A. arrivesB. will arriveC. arrivedD. is arriving50.It __________ (1) (rain) a lot last week. Every day, we __________ (2) (stay) at home because it __________ (3) (be) too wet to go outside. On Monday, we __________ (4) (watch) a movie, and on Tuesday, we __________ (5) (play) board games. Although we __________ (6) (not go) outside, we __________ (7) (have) a lot of fun together. Theweather __________ (8) (be) much better this week, so we __________ (9) (go) out more often.(答案及解释)。
【部编版】七年级上册语文：课内现代文阅读理解及答案（含答案和解析）
【部编版】七年级上册语⽂：课内现代⽂阅读理解及答案（含答案和解析）七年级上册现代⽂（课内）阅读汇编《春》（⼀）⼩草偷偷地⼟⾥钻出来，嫩嫩的，绿绿的。
园⼦⾥，⽥野⾥，瞧去，⼀⼤⽚⼀⼤⽚满是的。
坐着，躺着，打两个滚，踢⼏脚球，赛⼏趟跑，捉⼏回迷藏，风轻悄悄的，草软绵绵。
桃树、杏树、梨树，你不让我，我不让你，都开满了花赶趟⼉。
红的像⽕，粉的像霞光，⽩的像雪。
花⾥带着甜味⼉；闭了眼，树上仿佛已经满是桃⼉、杏⼉、梨⼉。
花下成千成百的蜜蜂嗡嗡地闹着，⼤⼩的蝴蝶飞来飞去。
野花遍地是：杂样的，有名字的，没名字的，散在草丛⾥像眼睛，像星星，还眨呀眨的。
1、想像⼀下，下⾯的诗句中哪⼀句描写的画⾯与“⼩草偷偷地从⼟⾥钻出来”最接近？（）A、春风⼜绿江南岸（王安⽯《泊船⽠州》）B、浅草才能没马蹄（⽩居易《钱塘湖春⾏》）C、草⾊遥看近却⽆（韩愈《早春呈⽔部张⼗⼋员外》）D、风吹草低见⽜⽺（《敕勒歌》）2、第⼆段写春花的顺序是从到，写出了春花繁密茂盛，、的特征。
3、写“花下成千成百的蜜蜂嗡嗡地闹着，⼤⼩的蝴蝶飞来飞去”。
这⼀句对写春花有什么作⽤？4、作者由花朵想到了果实，“闭了眼，树上仿佛已经满是桃⼉、杏⼉、梨⼉”。
那么，你由“成千成百的蜜蜂嗡嗡地闹着，⼤⼩的蝴蝶飞来飞去”想到了什么？⽤⼏句话把它描写出来。
5、景物本⾝没有思想感情的，但我们可以把它当作有思想感情的⼈来写。
⽂中有这样的例⼦，你认为写得好不好？请说说你的看法。
6、第①段描述春天景象，先总写______________，后分写____________________。
7、“红的像⽕，粉的像霞，⽩的像雪。
”是写花的__________，其特点是_____________。
8、“闭了眼，树上仿佛已经满是桃⼉、杏⼉、梨⼉。
”作者通过（），使描述更加（）。
9、⽂中写蜜蜂“闹”、蝴蝶“飞来飞去”是⽤来衬托______________________。
10、本⽂段共⽤了下列哪些修辞⽅法？（）A、⽐喻、拟⼈、借代、引⽤B、拟⼈、⽐喻、排⽐、夸张C、⽐喻、夸张、排⽐、引⽤D、拟⼈、借代、⽐喻、夸张（⼆）⾬是最寻常的，⼀下就是三两天。
2022年高考真题：数学(新高考II卷)【含答案及解析】
2022年普通⾼等学校招⽣全国统⼀考试（新⾼考全国Ⅱ卷）数学一、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.已知集合{}{}1,1,2,4,11A B x x =-=-£，则A B =I （ ）A.{1,2}- B.{1,2}C.{1,4}D.{1,4}-2.(22i)(12i)+-=（ ）A.24i-+ B.24i-- C.62i+ D.62i-3.中国的古建筑不仅是挡风遮雨的住处，更是美学和哲学的体现．如图是某古建筑物的剖面图，1111,,,DD CC BB AA 是举，1111,,,OD DC CB BA 是相等的步，相邻桁的举步之比分别为11111231111,0.5,,DD CC BB AA k k k OD DC CB BA ====，若123,,k k k 是公差为0.1的等差数列，且直线OA 的斜率为0.725，则3k =（ ）A. 0.75B. 0.8C. 0.85D. 0.94.已知(3,4),(1,0),t ===+r r r r r a b c a b ，若,,<>=<>r r r ra cbc ，则t =（ ）A.6- B.5- C. 5D. 65.有甲乙丙丁戊5名同学站成一排参加文艺汇演，若甲不站在两端，丙和丁相邻的不同排列方式有多少种（ ）A. 12种B. 24种C. 36种D. 48种6.角,a b 满足sin()cos()sin 4p a b a b a b æö+++=+ç÷èø，则（ ）A .tan()1a b += B.tan()1a b +=-C.tan()1a b -= D.tan()1a b -=-7.正三棱台高为1，上下底边长分别为，所有顶点在同一球面上，则球的表面积是（ ）A.100πB.128πC.144πD.192π8.若函数()f x 的定义域为R ，且()()()(),(1)1f x y f x y f x f y f ++-==，则221()k f k ==å（ ）A.3- B.2- C. 0D. 1二、选择题：本题共4小题，每小题5分，共20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5分，部分选对的得2分，有选错的得0分.9.函数()sin(2)(0π)f x x j j =+<<的图象以2π,03æöç÷èø中心对称，则（）A.y =()f x 在5π0,12æöç÷èø单调递减B.y =()f x 在π11π,1212æö-ç÷èø有2个极值点C.直线7π6x =是一条对称轴D.直线2y x =-是一条切线10.已知O 为坐标原点，过抛物线2:2(0)C y px p =>的焦点F 的直线与C 交于A ，B 两点，点A 在第一象限，点(,0)M p ，若||||AF AM =，则（）A. 直线AB 的斜率为B.||||OB OF =C.||4||AB OF > D.180OAM OBM Ð+Ð<°11.如图，四边形ABCD 为正方形，ED ^平面ABCD ，,2FB ED AB ED FB ==∥，记三棱锥E ACD -，F ABC -，F ACE -的体积分别为123,,V V V ，则（）A.322V V =B.312V V =C.312V V V =+ D.3123V V =12.对任意x ，y ，221+-=x y xy ，则（）A.1x y +£ B.2x y +³-C.222x y +£D.221x y +³三、填空题：本题共4小题，每小题5分，共20分.13.已知随机变量X 服从正态分布()22,N s，且(2 2.5)0.36P X <£=，则( 2.5)P X >=____________．14.写出曲线ln ||y x =过坐标原点的切线方程：____________，____________．15.已知点(2,3),(0,)A B a -，若直线AB 关于y a =的对称直线与圆22(3)(2)1x y +++=存在公共点，则实数a 的取值范围为________．16.已知椭圆22163x y +=，直线l 与椭圆在第一象限交于A ，B 两点，与x 轴，y 轴分别交于M ，N 两点，且||||,||MA NB MN ==l 的方程为___________．四、解答题：本题共6小题，共70分．解答应写出文字说明，证明过程或演算步骤．17.已知{}n a 为等差数列，{}n b 是公比为2的等比数列，且223344a b a b b a -=-=-．（1）证明：11a b =；（2）求集合{}1,1500k m k b a a m =+££中元素个数．18.记ABC V 的三个内角分别为A ，B ，C ，其对边分别为a ，b ，c ，分别以a ，b ，c 为边长的三个正三角形的面积依次为123,,S S S ，已知1231,sin 23S S S B -+==．（1）求ABC V 的面积；（2）若sin sin 3A C =，求b ．19.在某地区进行流行病调查，随机调查了100名某种疾病患者的年龄，得到如下的样本数据频率分布直方图．（1）估计该地区这种疾病患者的平均年龄（同一组中的数据用该组区间的中点值作代表）；（2）估计该地区一人患这种疾病年龄在区间[20,70)的概率；（3）已知该地区这种疾病的患病率为0.1%，该地区年龄位于区间[40,50)的人口占该地区总人口的16%，从该地区任选一人，若此人年龄位于区间[40,50)，求此人患该种疾病的概率．（样本数据中的患者年龄位于各区间的频率作为患者年龄位于该区间的概率，精确到0.0001）20.如图，PO 是三棱锥P ABC -的高，PA PB =，AB AC ^，E 是PB 的中点．（1）求证：//OE 平面PAC ；（2）若30ABO CBO Ð=Ð=°，3PO =，5PA =，求二面角C AE B --的正弦值．21.设双曲线2222:1(0,0)x y C a b a b -=>>的右焦点为(2,0)F ，渐近线方程为y =．（1）求C 的方程；（2）过F 的直线与C 的两条渐近线分别交于A ，B 两点，点()()1122,,,P x y Q x y 在C 上，且1210,0x x y >>>．过P 且斜率为Q M ，请从下面①②③中选取两个作为条件，证明另外一个条件成立：①M 在AB 上；②PQ AB ∥；③||||MA MB =．注：若选择不同的组合分别解答，则按第一个解答计分.22.已知函数()e e ax x f x x =-．（1）当1a =时，讨论()f x 的单调性；（2）当0x >时，()1f x <-，求a 的取值范围；（3）设n *ÎNln(1)n ++>+L ．答案及解析一、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.已知集合{}{}1,1,2,4,11A B x x =-=-£，则A B =I （）A.{1,2}- B.{1,2}C.{1,4}D.{1,4}-【答案】B 【解析】【分析】求出集合B 后可求A B I .【详解】{}|02B x x =££，故{}1,2A B =I ，故选：B.2.(22i)(12i)+-=（）A.24i -+ B.24i-- C.62i+ D.62i-【答案】D 【解析】【分析】利用复数的乘法可求()()22i 12i +-.【详解】()()22i 12i 244i 2i 62i +-=+-+=-，故选：D.3.中国的古建筑不仅是挡风遮雨的住处，更是美学和哲学的体现．如图是某古建筑物的剖面图，1111,,,DD CC BB AA 是举，1111,,,OD DC CB BA 是相等的步，相邻桁的举步之比分别为11111231111,0.5,,DD CC BB AA k k k OD DC CB BA ====，若123,,k k k 是公差为0.1的等差数列，且直线OA 的斜率为0.725，则3k =（）A. 0.75B. 0.8C. 0.85D. 0.9【答案】D 【解析】【分析】设11111OD DC CB BA ====，则可得关于3k 的方程，求出其解后可得正确的选项.【详解】设11111OD DC CB BA ====，则111213,,CC k BB k AA k ===，依题意，有31320.2,0.1k k k k -=-=，且111111110.725DD CC BB AA OD DC CB BA +++=+++，所以30.530.30.7254k +-=，故30.9k =，故选：D4.已知(3,4),(1,0),t ===+rrrrra b c a b ，若,,<>=<>r rr ra cbc ，则t =（）A.6- B.5- C. 5D. 6【答案】C 【解析】【分析】利用向量的运算和向量的夹角的余弦公式的坐标形式化简即可求得【详解】解：()3,4c t =+r ,cos ,cos ,a c b c =r r r,即931635t t c c+++=r r ,解得5t =,故选：C5.有甲乙丙丁戊5名同学站成一排参加文艺汇演，若甲不站在两端，丙和丁相邻的不同排列方式有多少种（）A. 12种 B. 24种C. 36种D. 48种【答案】B 【解析】【分析】利用捆绑法处理丙丁，用插空法安排甲，利用排列组合与计数原理即可得解【详解】因为丙丁要在一起，先把丙丁捆绑，看做一个元素，连同乙，戊看成三个元素排列,有3!种排列方式；为使甲不在两端，必须且只需甲在此三个元素的中间两个位置任选一个位置插入，有2种插空方式；注意到丙丁两人的顺序可交换，有2种排列方式，故安排这5名同学共有：3!2224´´=种不同的排列方式，故选：B6.角,a b 满足sin()cos()sin 4p a b a b a b æö+++=+ç÷èø，则（）A.tan()1a b += B.tan()1a b +=-C.tan()1a b -= D.tan()1a b -=-【答案】D 【解析】【分析】由两角和差的正余弦公式化简，结合同角三角函数的商数关系即可得解.【详解】由已知得：()sin cos cos sin cos cos sin sin 2cos sin sin a b a b a b a b a a b ++-=-,即：sin cos cos sin cos cos sin sin 0a b a b a b a b -++=，即：()()sin cos 0a b a b -+-=,所以()tan 1a b -=-,故选：D7.正三棱台高为1，上下底边长分别为，所有顶点在同一球面上，则球的表面积是（）A.100πB.128πC.144πD.192π【答案】A 【解析】【分析】根据题意可求出正三棱台上下底面所在圆面的半径12,r r ，再根据球心距，圆面半径，以及球的半径之间的关系，即可解出球的半径，从而得出球的表面积．【详解】设正三棱台上下底面所在圆面的半径12,r r ，所以122,2sin 60sin 60r r ==o o，即123,4r r ==，设球心到上下底面的距离分别为12,d d ，球的半径为R ，所以1d =2d =121d d -=或121d d +=1=或1+=，解得225R =符合题意，所以球的表面积为24π100πS R ==．故选：A ．8.若函数()f x 的定义域为R ，且()()()(),(1)1f x y f x y f x f y f ++-==，则221()k f k ==å（）A.3-B.2-C.0D.1【答案】A 【解析】【分析】根据题意赋值即可知函数()f x 的一个周期为6，求出函数一个周期中的()()()1,2,,6f f f L 的值，即可解出．【详解】因为()()()()f x y f x y f x f y ++-=，令1,0x y ==可得，()()()2110f f f =，所以()02f =，令0x =可得，()()()2f y f y f y +-=，即()()f y f y =-，所以函数()f x 为偶函数，令1y =得，()()()()()111f x f x f x f f x ++-==，即有()()()21f x f x f x ++=+，从而可知()()21f x f x +=--，()()14f x f x -=--，故()()24f x f x +=-，即()()6f x f x =+，所以函数()f x 的一个周期为6．因为()()()210121f f f =-=-=-，()()()321112f f f =-=--=-，()()()4221f f f =-==-，()()()5111f f f =-==，()()602f f ==，所以一个周期内的()()()1260f f f +++=L ．由于22除以6余4，所以()()()()()221123411213k f k f f f f ==+++=---=-å．故选：A ．二、选择题：本题共4小题，每小题5分，共20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5分，部分选对的得2分，有选错的得0分.9.函数()sin(2)(0π)f x x j j =+<<的图象以2π,03æöç÷èø中心对称，则（）A.y =()f x 在5π0,12æöç÷èø单调递减B.y =()f x 在π11π,1212æö-ç÷èø有2个极值点C. 直线7π6x =是一条对称轴D. 直线2y x =-是一条切线【答案】AD 【解析】【分析】根据三角函数的性质逐个判断各选项，即可解出．【详解】由题意得：2π4πsin 033f j æöæö=+=ç÷ç÷èøèø，所以4ππ3k j +=，k ÎZ ，即4ππ,3k k j =-+ÎZ ，又0πj <<，所以2k =时，2π3j =，故2π()sin 23f x x æö=+ç÷èø．对A ，当5π0,12x æöÎç÷èø时，2π2π3π2,332x æö+Îç÷èø，由正弦函数sin y u =图象知()y f x =在5π0,12æöç÷èø上是单调递减；对B ，当π11π,1212x æöÎ-ç÷èø时，2ππ5π2,322x æö+Îç÷èø，由正弦函数sin y u =图象知()y f x =只有1个极值点，由2π3π232x +=，解得5π12x =，即5π12x =为函数的唯一极值点；对C ，当7π6x =时，2π23π3x +=，7π()06f =，直线7π6x =不是对称轴；对D ，由2π2cos 213y x æö¢=+=-ç÷èø得：2π1cos 232x æö+=-ç÷èø，解得2π2π22π33x k +=+或2π4π22π,33x k k +=+ÎZ ,从而得：πx k =或ππ,3x k k =+ÎZ ,所以函数()y f x =在点0,2æöç÷ç÷èø处的切线斜率为02π2cos 13x k y ==¢==-，切线方程为：(0)2y x -=--即2y x =-．故选：AD ．10.已知O 为坐标原点，过抛物线2:2(0)C y px p =>的焦点F 的直线与C 交于A ，B 两点，点A 在第一象限，点(,0)M p ，若||||AF AM =，则（）A. 直线AB的斜率为 B.||||OB OF =C.||4||AB OF > D.180OAM OBM Ð+Ð<°【答案】ACD 【解析】【分析】由AF AM =及抛物线方程求得3(,)42p A ，再由斜率公式即可判断A 选项；表示出直线AB的方程，联立抛物线求得(,)33p B -，即可求出OB 判断B 选项；由抛物线的定义求出2512pAB =即可判断C 选项；由0OA OB ×<u u u r u u u r ，0MA MB ×<u u u r u u u r 求得AOB Ð，AMB Ð为钝角即可判断D 选项.【详解】对于A ，易得(,0)2pF ，由AF AM =可得点A 在FM 的垂直平分线上，则A 点横坐标为3224p pp +=，代入抛物线可得2233242p y p p =×=，则3(,)42p A ，则直线AB的斜率为2342p p =-，A 正确；对于B，由斜率为可得直线AB的方程为x =，联立抛物线方程得220y py p -=，设11(,)B x y，则126p y p +=，则13y =-，代入抛物线得2123p x æö-=×ç÷ç÷èø，解得13p x =，则(,)33p B -，则=，B 错误；对于C ，由抛物线定义知：325244312p p p AB p p OF =++=>=，C 正确；对于D，2333(,)(,)0423343234p p p p p OA OB æö×=×-=×+×-=-<ç÷ç÷èøuu u r u u u r ，则AOB Ð为钝角，又2225(,)(,)0423343236p p p p p MA MB æöæö×=-×--=-×-+×-=-<ç÷ç÷ç÷èøèøuu u r uu u r ，则AMB Ð为钝角，又360AOB AMB OAM OBM Ð+Ð+Ð+Ð=o ，则180OAM OBM Ð+Ð<o ，D 正确.故选：ACD.11.如图，四边形ABCD 为正方形，ED ^平面ABCD ，,2FB ED AB ED FB ==∥，记三棱锥E ACD -，F ABC -，F ACE -的体积分别为123,,V V V，则（）A.322V V = B.312V V =C.312V V V =+D.3123V V =【答案】CD 【解析】【分析】直接由体积公式计算12,V V ，连接BD 交AC 于点M ，连接,EM FM ，由3A EFM C EFM V V V --=+计算出3V ，依次判断选项即可.【详解】设22AB ED FB a ===，因为ED ^平面ABCD ，FB ED P ，则()2311114223323ACD V ED S a a a =××=×××=V ，()232111223323ABC V FB S a a a =××=×××=V ，连接BD 交AC 于点M ，连接,EM FM ，易得BD AC ^，又ED ^平面ABCD ，AC Ì平面ABCD ，则ED AC ^，又ED BD D =I ，,ED BD Ì平面BDEF ，则AC ^平面BDEF ，又12BM DM BD ===，过F 作FG DE ^于G ，易得四边形BDGF 为矩形，则,FG BD EG a ===，则,EM FM ====，3EF a ==，222EM FM EF +=，则EM FM ^，2122EFM S EM FM a =×=V ，AC =，则33123A EFM C EFM EFM V V V AC S a --=+=×=V ，则3123V V =，323V V =，312V V V =+，故A 、B 错误；C 、D 正确.故选：CD.12.对任意x ，y ，221+-=x y xy ，则（）A.1x y +£B.2x y +³-C.222x y +£D.221x y +³【答案】BC 【解析】【分析】根据基本不等式或者取特值即可判断各选项的真假．【详解】因为22222a b a b ab ++æö££ç÷èø（,a b ÎR ），由221+-=x y xy 可变形为，()221332x y x y xy +æö+-=£ç÷èø，解得22x y -£+£，当且仅当1x y ==-时，2x y +=-，当且仅当1x y ==时，2x y +=，所以A 错误，B 正确；由221+-=x y xy 可变形为()222212x y x y xy ++-=£，解得222x y +£，当且仅当1x y ==±时取等号，所以C 正确；因为221+-=x y xy 变形可得223124y x y æö-+=ç÷èø，设cos ,sin 22y x y q q -==，所以cos ,x y q q q ==，因此22225cos sin cos 13x y q q q q =+=+++42π2sin 2,23363q æöéù=+-Îç÷êúèøëû，所以当,33x y ==-时满足等式，但是221x y +³不成立，所以D 错误．故选：BC ．三、填空题：本题共4小题，每小题5分，共20分.13.已知随机变量X 服从正态分布()22,N s，且(2 2.5)0.36P X <£=，则( 2.5)P X >=____________．【答案】0.14##750．【解析】【分析】根据正态分布曲线的性质即可解出．【详解】因为()22,X N s:，所以()()220.5P X P X <=>=，因此()()()2.522 2.50.50.360.14P X P X P X >=>-<£=-=．故答案为：0.14．14.写出曲线ln ||y x =过坐标原点的切线方程：____________，____________．【答案】①.1ey x =②.1ey x =-【解析】【分析】分0x >和0x <两种情况，当0x >时设切点为()00,ln x x ，求出函数的导函数，即可求出切线的斜率，从而表示出切线方程，再根据切线过坐标原点求出0x ，即可求出切线方程，当0x <时同理可得；【详解】解：因为ln y x =，当0x >时ln y x =，设切点为()00,ln x x ，由1y x¢=，所以001|x x y x =¢=，所以切线方程为()0001ln y x x x x -=-，又切线过坐标原点，所以()0001ln x x x -=-，解得0e x =，所以切线方程为()11e e y x -=-，即1ey x =；当0x <时()ln y x =-，设切点为()()11,ln x x -，由1y x¢=，所以111|x x y x =¢=，所以切线方程为()()1111ln y x x x x --=-，又切线过坐标原点，所以()()1111ln x x x --=-，解得1e x =-，所以切线方程为()11e e y x -=+-，即1ey x =-；故答案为：1e y x =；1ey x =-15.已知点(2,3),(0,)A B a -，若直线AB 关于y a =的对称直线与圆22(3)(2)1x y +++=存在公共点，则实数a 的取值范围为________．【答案】13,32éùêúëû【解析】【分析】首先求出点A 关于y a =对称点A ¢的坐标，即可得到直线l 的方程，根据圆心到直线的距离小于等于半径得到不等式，解得即可；【详解】解：()2,3A -关于y a =对称的点的坐标为()2,23A a ¢--，()0,B a 在直线y a =上，所以A B ¢所在直线即为直线l ，所以直线l 为32a y x a -=+-，即()3220a x y a -+-=；圆()()22:321C x y +++=，圆心()3,2C --，半径1r =，依题意圆心到直线l 的距离1d =£，即()()2225532a a -£-+，解得1332a ££，即13,32a éùÎêúëû；故答案为：13,32éùêúëû16.已知椭圆22163x y +=，直线l 与椭圆在第一象限交于A ，B 两点，与x 轴，y 轴分别交于M ，N 两点，且||||,||MA NB MN ==l 的方程为___________．【答案】0x +-=【解析】【分析】令AB 的中点为E ，设()11,A x y ，()22,B x y ，利用点差法得到12OE AB k k ×=-，设直线:AB y kx m =+，0k <，0m >，求出M 、N 的坐标，再根据MN 求出k 、m ，即可得解；【详解】解：令AB 的中点为E ，因为MA NB =，所以ME NE =，设()11,A x y ，()22,B x y ，则2211163x y +=，2222631x y +=，所以2222121206633x x y y -+-=，即()()()()12121212063x x x x y y y y -++-+=所以()()()()1212121212y y y y x x x x +-=--+，即12OE AB k k ×=-，设直线:AB y kx m =+，0k <，0m >，令0x =得y m =，令0y =得m x k =-，即,0m M k æö-ç÷èø，()0,N m ，所以,22m m E k æö-ç÷èø，即1222mk m k´=--，解得2k =-或2k =（舍去），又MN =，即MN ==，解得2m =或2m =-（舍去），所以直线:22AB y x =-+，即0x +-=；故答案为：0x -=四、解答题：本题共6小题，共70分．解答应写出文字说明，证明过程或演算步骤．17.已知{}n a 为等差数列，{}n b 是公比为2的等比数列，且223344a b a b b a -=-=-．（1）证明：11a b =；（2）求集合{}1,1500k m k b a a m =+££中元素个数．【答案】（1）证明见解析；（2）9．【解析】【分析】（1）设数列{}n a 的公差为d ，根据题意列出方程组即可证出；（2）根据题意化简可得22k m -=，即可解出．【小问1详解】设数列{}n a 的公差为d ，所以，()11111111224283a d b a d b a d b b a d +-=+-ìí+-=-+î，即可解得，112db a ==，所以原命题得证．【小问2详解】由（1）知，112d b a ==，所以()1111121k k m b a a b a m d a -=+Û´=+-+，即122k m -=，亦即[]221,500k m -=Î，解得210k ££，所以满足等式的解2,3,4,,10k =L ，故集合{}1|,1500k m k b a a m =+££中的元素个数为10219-+=．18.记ABC V 的三个内角分别为A ，B ，C ，其对边分别为a ，b ，c ，分别以a ，b ，c 为边长的三个正三角形的面积依次为123,,S S S，已知1231,sin 23S S S B -+==．（1）求ABC V 的面积；（2）若sin sin 3A C =，求b ．【答案】（1）8（2）12【解析】【分析】（1）先表示出123,,S S S，再由1232S S S -+=求得2222a c b +-=，结合余弦定理及平方关系求得ac ，再由面积公式求解即可；（2）由正弦定理得22sin sin sin b acB A C=，即可求解.【小问1详解】由题意得22221231,,22444S a a S b S c =××===，则2221234442S S S a c -+=-+=，即2222a c b +-=，由余弦定理得222cos 2a c b B ac+-=，整理得cos 1ac B =，则cos 0B >，又1sin 3B =，则cos 3B ==，1cos 4ac B ==，则1sin 28ABC S ac B ==V ；【小问2详解】由正弦定理得：sin sin sin b a cB A C==，则22sin sin sin sin sin 3b ac ac B A C A C =×==，则3sin 2b B =，31sin 22b B ==.19.在某地区进行流行病调查，随机调查了100名某种疾病患者的年龄，得到如下的样本数据频率分布直方图．（1）估计该地区这种疾病患者的平均年龄（同一组中的数据用该组区间的中点值作代表）；（2）估计该地区一人患这种疾病年龄在区间[20,70)的概率；（3）已知该地区这种疾病的患病率为0.1%，该地区年龄位于区间[40,50)的人口占该地区总人口的16%，从该地区任选一人，若此人年龄位于区间[40,50)，求此人患该种疾病的概率．（样本数据中的患者年龄位于各区间的频率作为患者年龄位于该区间的概率，精确到0.0001）【答案】（1）44.65岁；（2）0.89；（3）0.0014．【解析】【分析】（1）根据平均值等于各矩形的面积乘以对应区间的中点值的和即可求出；（2）设A ={一人患这种疾病的年龄在区间[20,70)}，根据对立事件的概率公式()1()P A P A =-即可解出；（3）根据条件概率公式即可求出．【小问1详解】平均年龄(50.001150.002250.012350.017450.023x =´+´+´+´+´550.020650.012750.006850.002)1044.65+´+´+´+´´=（岁）．【小问2详解】设A ={一人患这种疾病的年龄在区间[20,70)}，所以()1()1(0.0010.0020.0060.002)1010.110.89P A P A =-=-+++´=-=．【小问3详解】设{B =任选一人年龄位于区间}[40,50)，{C =任选一人患这种疾病}，则由条件概率公式可得()0.1%0.023100.0010.23(|)0.00143750.0014()16%0.16P BC P C B P B ´´´====»．20.如图，PO 是三棱锥P ABC -的高，PA PB =，AB AC ^，E 是PB 的中点．（1）求证：//OE 平面PAC ；（2）若30ABO CBO Ð=Ð=°，3PO =，5PA =，求二面角C AE B --的正弦值．【答案】（1）证明见解析（2）1113【解析】【分析】（1）连接BO 并延长交AC 于点D ，连接OA 、PD ，根据三角形全等得到OA OB =，再根据直角三角形的性质得到AO DO =，即可得到O 为BD 的中点从而得到//OE PD ，即可得证；（2）过点A 作//Az OP ，如图建立平面直角坐标系，利用空间向量法求出二面角的余弦值，再根据同角三角函数的基本关系计算可得；【小问1详解】证明：连接BO 并延长交AC 于点D ，连接OA 、PD ，因为PO 是三棱锥P ABC -的高，所以PO ^平面ABC ，,AO BO Ì平面ABC ，所以PO AO ^、PO BO ^，又PA PB =，所以POA POB @△△，即OA OB =，所以OAB OBA Ð=Ð，又AB AC ^，即90BAC Ð=°，所以90OAB OAD Ð+Ð=°，90OBA ODA Ð+Ð=°，所以ODA OADÐ=Ð所以AO DO =，即AO DO OB ==，所以O 为BD 的中点，又E 为PB 的中点，所以//OE PD ，又OE Ë平面PAC ，PD Ì平面PAC ，所以//OE 平面PAC【小问2详解】解：过点A 作//Az OP ，如图建立平面直角坐标系，因为3PO =，5AP =，所以4OA ==，又30OBA OBC Ð=Ð=°，所以28BD OA ==，则4=AD，AB =，所以12AC =，所以()2,0O，()B，()2,3P ，()0,12,0C，所以32E æöç÷èø，则32AE æö=ç÷èøuu u r，()AB =u u ur ，()0,12,0AC =uu u r ，设平面AEB 的法向量为(),,n x y z =r，则3020n AE y z nAB ì×=++=ïíï×==îu u uv v u u u v v ，令2z =，则3y =-，0x =，所以()0,3,2n =-r；设平面AEC 的法向量为(),,m a b c =u r，则302120m AE b c m AC b ì×=++=ïíï×==îuu u v v uu u v v，令a =6c =-，0b =，所以)6m =-u r；所以cos ,13n m n m n m×===-r u rr u r r u r 设二面角C AE B --为q ，由图可知二面角C AE B --为钝二面角，所以cos 13q =-，所以11sin 13q ==故二面角C AE B --的正弦值为1113；21.设双曲线2222:1(0,0)x y C a b a b -=>>的右焦点为(2,0)F，渐近线方程为y =．（1）求C 的方程；（2）过F 的直线与C 的两条渐近线分别交于A ，B 两点，点()()1122,,,P x y Q x y 在C 上，且1210,0x x y >>>．过P且斜率为QM ，请从下面①②③中选取两个作为条件，证明另外一个条件成立：①M 在AB 上；②PQ AB ∥；③||||MA MB =．注：若选择不同的组合分别解答，则按第一个解答计分.【答案】（1）2213y x -=（2）见解析【解析】【分析】（1）利用焦点坐标求得c 的值，利用渐近线方程求得,a b 的关系，进而利用,,a b c 的平方关系求得,a b 的值，得到双曲线的方程；（2）先分析得到直线AB 的斜率存在且不为零，设直线AB 的斜率为k , M (x 0,y 0),由③|AM |=|BM |等价分析得到200283k x ky k +=-；由直线PM 和QM 的斜率得到直线方程，结合双曲线的方程，两点间距离公式得到直线PQ 的斜率03x m y =，由②//PQ AB 等价转化为003ky x =，由①M 在直线AB 上等价于()2002ky k x =-，然后选择两个作为已知条件一个作为结论，进行证明即可.【小问1详解】右焦点为(2,0)F ，∴2c =,∵渐近线方程为y =，∴ba=b =，∴222244c a b a =+==，∴1a =，∴b =∴C 的方程为：2213y x -=；【小问2详解】由已知得直线PQ 的斜率存在且不为零，直线AB 的斜率不为零，若选由①②推③或选由②③推①：由②成立可知直线AB 的斜率存在且不为零；若选①③推②，则M 为线段AB 的中点，假若直线AB 的斜率不存在，则由双曲线的对称性可知M 在x 轴上，即为焦点F ,此时由对称性可知P 、Q 关于x 轴对称，与从而12x x =，已知不符；总之，直线AB 的斜率存在且不为零.设直线AB 的斜率为k ,直线AB 方程为()2y k x =-,则条件①M 在AB 上，等价于()()2000022y k x ky k x =-Û=-；两渐近线的方程合并为2230x y -=,联立消去y 并化简整理得：()22223440k x k x k --+=设()()3334,,,A x y B x y ,线段中点为(),N N N x y ,则()2342226,2233N N N x x k kx y k x k k +===-=--,设()00,M x y ,则条件③AM BM =等价于()()()()222203030404x x y y x x y y -+-=-+-,移项并利用平方差公式整理得：()()()()3403434034220x x x x x y y y y y éùéù--++--+=ëûëû，()()3403403434220y y x x x y y y x x -éùéù-++-+=ëûëû-,即()000N N x x k y y -+-=,即200283k x ky k +=-；由题意知直线PM的斜率为, 直线QM∴由))10102020,y y x x y y x x -=--=-,∴)121202y y x x x -=+-,所以直线PQ的斜率)1201212122x x x y y m x x x x +--==---,直线)00:PM y x x y =-+,即00y y =,代入双曲线的方程22330x y --=,即)3yy +-=中，得：()()00003y y éù+-=ëû,解得P的横坐标：100x y æö=++÷÷ø,同理：200x y æö=+÷÷ø，∴120,x x x æö-=∴03x m y =,∴条件②//PQ AB 等价于003m k ky x =Û=，综上所述：条件①M 在AB 上，等价于()2002ky kx =-；条件②//PQ AB 等价于003ky x =；条件③AM BM =等价于200283k x ky k +=-；选①②推③:由①②解得：2200002228,433k k x x ky x k k =\+==--,∴③成立；选①③推②：由①③解得：20223k x k =-，20263k ky k =-，∴003ky x =，∴②成立；选②③推①：由②③解得：20223k x k =-，20263k ky k =-，∴02623x k -=-，∴()2002ky kx =-，∴①成立.22.已知函数()e e ax x f x x =-．（1）当1a =时，讨论()f x 的单调性；（2）当0x >时，()1f x <-，求a 的取值范围；（3）设n *ÎNln(1)n ++>+L ．【答案】（1）()f x 的减区间为(),0-¥，增区间为()0,+¥.（2）12a £（3）见解析【解析】【分析】（1）求出()f x ¢，讨论其符号后可得()f x 的单调性.（2）设()e e 1axxh x x =-+，求出()h x ¢¢，先讨论12a >时题设中的不等式不成立，再就102a <£结合放缩法讨论()h x ¢符号，最后就0a £结合放缩法讨论()h x 的范围后可得参数的取值范围.（3）由（2）可得12ln t tt<-对任意的1t >恒成立，从而可得()ln 1ln n n +-<任意的*n N Î恒成立，结合裂项相消法可证题设中的不等式.【小问1详解】当1a =时，()()1e x f x x =-，则()e xf x x ¢=，当0x <时，()0f x ¢<，当0x >时，()0f x ¢>，故()f x 的减区间为(),0-¥，增区间为()0,+¥.【小问2详解】设()e e 1axxh x x =-+，则()00h =，又()()1e e axxh x ax ¢=+-，设()()1e e axxg x ax =+-，则()()22e e axxg x a a x ¢=+-，若12a >，则()0210g a ¢=->，因为()g x ¢为连续不间断函数，故存在()00,x Î+¥，使得()00,x x "Î，总有()0g x ¢>，故()g x 在()00,x 为增函数，故()()00g x g >=，故()h x 在()00,x 为增函数，故()()01h x h >=-，与题设矛盾.若102a <£，则()()()ln 11e e ee ax ax ax xx h x ax ++¢=+-=-，下证：对任意0x >，总有()ln 1x x +<成立，证明：设()()ln 1S x x x =+-，故()11011x S x x x-¢=-=<++，故()S x 在()0,+¥上为减函数，故()()00S x S <=即()ln 1x x +<成立.由上述不等式有()ln 12e e e e e e 0ax ax x ax ax x ax x +++-<-=-£，故()0h x ¢£总成立，即()h x 在()0,+¥上为减函数，所以()()01h x h <=-.当0a £时，有()e e e1100axxaxh x ax ¢=-+<-+=，所以()h x 在()0,+¥上为减函数，所以()()01h x h <=-.综上，12a £.【小问3详解】取12a =，则0x ">，总有12e e 10xx x -+<成立，令12ex t =，则21,e ,2ln x t t x t >==，故22ln 1t t t <-即12ln t t t<-对任意的1t >恒成立.所以对任意的*n N Î，有2ln<整理得到：()ln 1ln n n +-<，()ln 2ln1ln 3ln 2ln 1ln n n+>-+-+++-L L ()ln 1n =+，故不等式成立.【点睛】思路点睛：函数参数的不等式的恒成立问题，应该利用导数讨论函数的单调性，注意结合端点处导数的符号合理分类讨论，导数背景下数列不等式的证明，应根据已有的函数不等式合理构建数列不等式.。
【英语】中考英语名词试题(有答案和解析)含解析
【英语】中考英语名词试题(有答案和解析)含解析一、初中英语名词1．—Mr.Li,I feel a little nervous before the coming exam.—You'd better take a break from studies and relax yourself.A.restB.breathC.walk【答案】A【解析】【分析】句意： ---李老师，考试前我感觉有点紧张。
---你最好从学习中休息一下，放松自己。
break中断；休息；rest休息；breath呼吸；walk步行。
故答案为A。
【点评】考查词义辨析，理解句意，弄清划线的单词和备选项的意思，即可得出答案。
2．---_______fathers didn't come to the meeting.Why?---Because they have gone to Beijing.A.Jeff's and Amy'sB.Jeff and AmyC.Jeff's and AmyD.Jeff and Amy's 【答案】A【解析】【分析】句意:一一杰夫和艾米的父亲没有来开会。
为什么?一一因为他们去北京了。
当两个人共同拥有同一个人或物时，只在最一个人后加所有格符号;当两个人拥有同类事物中的不同个体时，都要加所有格符号。
本题中fathers是复数，说明两人父亲不同，两人都得加所有格符号，故选A。
3．—What would you like to eat?—Some_________,please.A.breadB.cakeC.coffeeD.tea【答案】A【解析】【分析】句意：——你想要吃什么？——请来一些面包。
bread面包；cake蛋糕；coffee咖啡；tea茶；根据What would you like to eat?可知此处介绍吃的食物，有some修饰，故此处用bread或cakes，故选A。
2021年高考真题：数学(理科)(全国甲卷)【含答案及解析】
2021年普通⾼等学校招⽣全国统⼀考试（甲卷）理科数学一、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.设集合{}104,53M x x N xx ìü=<<=££íýîþ，则M N =I （）A.103x x ìü<£íýîþ B.143x x ìü£<íýîþC.{}45x x £< D.{}05x x <£2.为了解某地农村经济情况，对该地农户家庭年收入进行抽样调查，将农户家庭年收入的调查数据整理得到如下频率分布直方图：根据此频率分布直方图，下面结论中不正确的是（）A.该地农户家庭年收入低于4.5万元的农户比率估计为6%B.该地农户家庭年收入不低于10.5万元的农户比率估计为10%C.估计该地农户家庭年收入的平均值不超过6.5万元D.估计该地有一半以上的农户，其家庭年收入介于4.5万元至8.5万元之间3.已知2(1)32i z i -=+，则z =（）A .312i --B.312i -+ C.32i -+ D.32i --4.青少年视力是社会普遍关注的问题，视力情况可借助视力表测量．通常用五分记录法和小数记录法记录视力数据，五分记录法的数据L 和小数记录表的数据V 的满足5lg L V =+．已知某同学视力的五分记录法的数据为4.9，则其视力的小数记录法的数据为（）（ 1.259»）A.1.5B.1.2C.0.8D.0.65.已知12,F F 是双曲线C 的两个焦点，P 为C 上一点，且121260,3F PF PF PF Ð=°=，则C 的离心率为（）A.2B.2C.D.6.在一个正方体中，过顶点A 的三条棱的中点分别为E ，F ，G ．该正方体截去三棱锥A EFG -后，所得多面体的三视图中，正视图如图所示，则相应的侧视图是（）A. B. C. D.7.等比数列{}n a 的公比为q ，前n 项和为n S ，设甲：0q >，乙：{}n S 是递增数列，则（）A.甲是乙的充分条件但不是必要条件B.甲是乙的必要条件但不是充分条件C.甲是乙的充要条件D.甲既不是乙的充分条件也不是乙的必要条件8.2020年12月8日，中国和尼泊尔联合公布珠穆朗玛峰最新高程为8848.86（单位：m ），三角高程测量法是珠峰高程测量方法之一．如图是三角高程测量法的一个示意图，现有A ，B ，C 三点，且A ，B ，C 在同一水平面上的投影,,A B C ¢¢¢满足45A C B Т¢¢=°，60A B C ¢¢Ð¢=°．由C 点测得B 点的仰角为15°，BB ¢与CC ¢的差为100；由B 点测得A 点的仰角为45°，则A ，C 两点到水平面A B C ¢¢¢的高度差AA CC ¢¢-约为1.732»）（）A.346B.373C.446D.4739.若cos 0,,tan 222sin p a a a a æöÎ=ç÷-èø，则tan a =（）A.15B.C.3D.310.将4个1和2个0随机排成一行，则2个0不相邻的概率为（）A.13B.25C.23D.4511.已如A ，B ，C 是半径为1的球O 的球面上的三个点，且,1AC BC AC BC ^==，则三棱锥O ABC -的体积为（）A.12 B.12C.4D.412.设函数()f x 的定义域为R ，()1f x +为奇函数，()2f x +为偶函数，当[]1,2x Î时，2()f x ax b =+．若()()036f f +=，则92f æö=ç÷èø（）A.94-B.32-C.74D.52二、填空题：本题共4小题，每小题5分，共20分．13.曲线212x y x -=+在点()1,3--处的切线方程为__________．14.已知向量()()3,1,1,0,a b c a kb ===+r r r r r ．若a c ^r r ，则k =________．15.已知12,F F 为椭圆C ：221164x y+=的两个焦点，P ，Q 为C 上关于坐标原点对称的两点，且12PQ F F =，则四边形12PFQF 的面积为________．16.已知函数()2cos()f x x w j =+的部分图像如图所示，则满足条件74()()043f x f f x f p p æöæöæöæö--->ç÷ç÷ç÷ç÷èøèøèøèø的最小正整数x 为________．三、解答题：共70分．解答应写出交字说明、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：共60分．17.甲、乙两台机床生产同种产品，产品按质量分为一级品和二级品，为了比较两台机床产品的质量，分别用两台机床各生产了200件产品，产品的质量情况统计如下表：一级品二级品 合计甲机床15050200乙机床12080200合计270130400（1）甲机床、乙机床生产的产品中一级品的频率分别是多少?（2）能否有99%的把握认为甲机床的产品质量与乙机床的产品质量有差异?附：22()()()()()n ad bc K a b c d a c b d -=++++()2P K k ³0.0500.0100.001k3.8416.63510.82818.已知数列{}n a 的各项均为正数，记n S 为{}n a 的前n 项和，从下面①②③中选取两个作为条件，证明另外一个成立．①数列{}n a 是等差数列：②数列是等差数列；③213aa =．注：若选择不同的组合分别解答，则按第一个解答计分．19.已知直三棱柱111ABC A B C -中，侧面11AA B B 为正方形，2AB BC ==，E ，F 分别为AC 和1CC 的中点，D 为棱11A B 上的点．11BF A B ^（1）证明：BF DE ^；（2）当1B D 为何值时，面11BB C C 与面DFE 所成的二面角的正弦值最小?20.抛物线C 的顶点为坐标原点O ．焦点在x 轴上，直线l ：1x =交C 于P ，Q 两点，且OP OQ ^．已知点()2,0M ，且M e 与l 相切．（1）求C ，M e 的方程；（2）设123,,A A A 是C 上的三个点，直线12A A ，13A A 均与M e 相切．判断直线23A A 与M e 的位置关系，并说明理由．21.已知0a >且1a ¹，函数()(0)ax x f x x a=>．（1）当2a =时，求()f x 的单调区间；（2）若曲线()y f x =与直线1y =有且仅有两个交点，求a 的取值范围．（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．[选修4-4：坐标系与参数方程]（10分）22.在直角坐标系xOy 中，以坐标原点为极点，x 轴正半轴为极轴建立极坐标系，曲线C 的极坐标方程为r q =．（1）将C 的极坐标方程化为直角坐标方程；（2）设点A 的直角坐标为()1,0，M 为C 上的动点，点P 满足AP =u u u ru u u r，写出Р的轨迹1C 的参数方程，并判断C 与1C 是否有公共点．[选修4-5：不等式选讲]（10分）23.已知函数()2,()2321f x x g x x x =-=+--．（1）画出()y f x =和()y g x =的图像；（2）若()()f x a g x +³，求a 的取值范围．答案及解析一、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.设集合{}104,53M x x N xx ìü=<<=££íýîþ，则M N =I （）A.103x x ìü<£íýîþB.143x x ìü£<íýîþC.{}45x x £< D.{}05x x <£【答案】B 【解析】【分析】根据交集定义运算即可【详解】因为1{|04},{|5}3M x x N x x =<<=££，所以1|43M N x x ìüÇ=£<íýîþ,故选：B.【点睛】本题考查集合的运算，属基础题，在高考中要求不高，掌握集合的交并补的基本概念即可求解.2.为了解某地农村经济情况，对该地农户家庭年收入进行抽样调查，将农户家庭年收入的调查数据整理得到如下频率分布直方图：根据此频率分布直方图，下面结论中不正确的是（）A.该地农户家庭年收入低于4.5万元的农户比率估计为6%B.该地农户家庭年收入不低于10.5万元的农户比率估计为10%C.估计该地农户家庭年收入的平均值不超过6.5万元D.估计该地有一半以上的农户，其家庭年收入介于4.5万元至8.5万元之间【答案】C 【解析】【分析】根据直方图的意义直接计算相应范围内的频率，即可判定ABD,以各组的中间值作为代表乘以相应的频率，然后求和即得到样本的平均数的估计值，也就是总体平均值的估计值，计算后即可判定C.【详解】因为频率直方图中的组距为1，所以各组的直方图的高度等于频率.样本频率直方图中的频率即可作为总体的相应比率的估计值.该地农户家庭年收入低于4.5万元的农户的比率估计值为0.020.040.066%+==,故A 正确；该地农户家庭年收入不低于10.5万元的农户比率估计值为0.040.0230.1010%+´==,故B 正确；该地农户家庭年收入介于4.5万元至8.5万元之间的比例估计值为0.100.140.2020.6464%50%++´==>,故D 正确；该地农户家庭年收入的平均值的估计值为30.0240.0450.1060.1470.2080.2090.10100.10110.04120.02130.02140.027.68´+´+´+´+´+´+´+´+´+´+´+´=(万元)，超过6.5万元，故C 错误.综上，给出结论中不正确的是C.故选：C.【点睛】本题考查利用样本频率直方图估计总体频率和平均值，属基础题，样本的频率可作为总体的频率的估计值，样本的平均值的估计值是各组的中间值乘以其相应频率然后求和所得值，可以作为总体的平均值的估计值.注意各组的频率等于´频率组距组距.3.已知2(1)32i z i -=+，则z =（）A.312i --B.312i -+C.32i -+ D.32i --【答案】B 【解析】【分析】由已知得322iz i+=-，根据复数除法运算法则，即可求解.【详解】2(1)232i z iz i -=-=+，32(32)23312222i i i i z i i i i ++×-+====-+--×.故选：B.4.青少年视力是社会普遍关注的问题，视力情况可借助视力表测量．通常用五分记录法和小数记录法记录视力数据，五分记录法的数据L 和小数记录表的数据V 的满足5lg L V =+．已知某同学视力的五分记录法的数据为4.9，则其视力的小数记录法的数据为（）（ 1.259»）A. 1.5 B. 1.2C. 0.8D. 0.6【答案】C 【解析】【分析】根据,L V 关系，当 4.9L =时，求出lg V ，再用指数表示V ，即可求解.【详解】由5lg L V =+，当 4.9L =时，lg 0.1V =-，则10.11010100.8V --===».故选：C .5.已知12,F F 是双曲线C 的两个焦点，P 为C 上一点，且121260,3F PF PF PF Ð=°=，则C 的离心率为（）A.2 B.2C. D.【答案】A 【解析】【分析】根据双曲线的定义及条件，表示出12,PF PF ，结合余弦定理可得答案.【详解】因为213PF PF =，由双曲线的定义可得12222PF PF PF a -==，所以2PF a =，13PF a =；因为1260F PF Ð=°,由余弦定理可得2224923cos60c a a a a =+-´××°，整理可得2247c a =，所以22274a c e ==，即2e =.故选：A【点睛】关键点睛：双曲线的定义是入手点，利用余弦定理建立,a c 间的等量关系是求解的关键.6.在一个正方体中，过顶点A 的三条棱的中点分别为E ，F ，G ．该正方体截去三棱锥A EFG -后，所得多面体的三视图中，正视图如图所示，则相应的侧视图是（）A. B. C. D.【答案】D 【解析】【分析】根据题意及题目所给的正视图还原出几何体的直观图，结合直观图进行判断.【详解】由题意及正视图可得几何体的直观图，如图所示，所以其侧视图为故选：D7.等比数列{}n a 的公比为q ，前n 项和为n S ，设甲：0q >，乙：{}n S 是递增数列，则（）A. 甲是乙的充分条件但不是必要条件B. 甲是乙的必要条件但不是充分条件C. 甲是乙的充要条件D. 甲既不是乙的充分条件也不是乙的必要条件【答案】B 【解析】【分析】当0q >时，通过举反例说明甲不是乙的充分条件；当{}n S 是递增数列时，必有0n a >成立即可说明0q >成立，则甲是乙的必要条件，即可选出答案．【详解】由题，当数列为2,4,8,---L 时，满足0q >，但是{}n S 不是递增数列，所以甲不是乙的充分条件．若{}n S 是递增数列，则必有0n a >成立，若0q >不成立，则会出现一正一负的情况，是矛盾的，则0q >成立，所以甲是乙的必要条件．故选：B ．【点睛】在不成立的情况下，我们可以通过举反例说明，但是在成立的情况下，我们必须要给予其证明过程．8.2020年12月8日，中国和尼泊尔联合公布珠穆朗玛峰最新高程为8848.86（单位：m ），三角高程测量法是珠峰高程测量方法之一．如图是三角高程测量法的一个示意图，现有A ，B ，C 三点，且A ，B ，C 在同一水平面上的投影,,A B C ¢¢¢满足45A C B Т¢¢=°，60A B C ¢¢Ð¢=°．由C 点测得B 点的仰角为15°，BB ¢与CC ¢的差为100；由B 点测得A 点的仰角为45°，则A ，C 两点到水平面A B C ¢¢¢的高度差AA CC ¢¢-约为1.732»）（）A.346B.373C.446D.473【答案】B 【解析】【分析】通过做辅助线，将已知所求量转化到一个三角形中，借助正弦定理，求得''A B ，进而得到答案．【详解】过C 作'CH BB ^，过B 作'BD AA ^，故()''''''100100AA CC AA BB BH AA BB AD -=--=-+=+，由题，易知ADB △为等腰直角三角形，所以AD DB =．所以''100''100AA CC DB A B -=+=+．因为15BCH Ð=°，所以100''tan15CH C B ==°在'''A B C V 中，由正弦定理得：''''100100sin 45sin 75tan15cos15sin15A B C B ===°°°°°，而sin15sin(4530)sin 45cos30cos 45sin 304°=°-°=°°-°°=，所以1004''1)273A B ´´==+»，所以''''100373AA CC A B -=+»．故选：B ．【点睛】本题关键点在于如何正确将''AA CC -的长度通过作辅助线的方式转化为''100A B +．9.若cos 0,,tan 222sin p a a a a æöÎ=ç÷-èø，则tan a =（）A.15B.5C.3 D.3【答案】A 【解析】【分析】由二倍角公式可得2sin 22sin cos tan 2cos 212sin a a a a a a ==-，再结合已知可求得1sin 4a =，利用同角三角函数的基本关系即可求解.【详解】cos tan 22sin aa a=-Q 2sin 22sin cos cos tan 2cos 212sin 2sin a a a aa a a a\===--，0,2p a æöÎç÷èøQ ，cos 0a \¹，22sin 112sin 2sin a a a \=--，解得1sin 4a =，cos 4a \==，sin tan cos 15a a a \==.故选：A.【点睛】关键点睛：本题考查三角函数的化简问题，解题的关键是利用二倍角公式化简求出sin a .10.将4个1和2个0随机排成一行，则2个0不相邻的概率为（）A.13B.25C.23 D.45【答案】C 【解析】【分析】采用插空法，4个1产生5个空，分2个0相邻和2个0不相邻进行求解.【详解】将4个1和2个0随机排成一行，可利用插空法，4个1产生5个空，若2个0相邻，则有155C =种排法，若2个0不相邻，则有2510C =种排法，所以2个0不相邻的概率为1025103=+.故选：C.11.已如A ，B ，C 是半径为1的球O 的球面上的三个点，且,1AC BC AC BC ^==，则三棱锥O ABC -的体积为（）A.12 B.12C.4D.4【答案】A 【解析】【分析】由题可得ABC V 为等腰直角三角形，得出ABC V 外接圆的半径，则可求得O 到平面ABC 的距离，进而求得体积.【详解】,1AC BC AC BC ^==Q ，ABC \V 为等腰直角三角形，AB \=，则ABC V 外接圆的半径为2，又球的半径为1，设O 到平面ABC 的距离为d ，则2d ==，所以11111332212O ABC ABC V S d -=×=´´´´=V .故选：A.【点睛】关键点睛：本题考查球内几何体问题，解题的关键是正确利用截面圆半径、球半径、球心到截面距离的勾股关系求解.12.设函数()f x 的定义域为R ，()1f x +为奇函数，()2f x +为偶函数，当[]1,2x Î时，2()f x ax b =+．若()()036f f +=，则92f æö=ç÷èø（）A.94-B.32-C.74D.52【答案】D 【解析】【分析】通过()1f x +是奇函数和()2f x +是偶函数条件，可以确定出函数解析式()222f x x =-+，进而利用定义或周期性结论，即可得到答案．【详解】因为()1f x +是奇函数，所以()()11f x f x -+=-+①；因为()2f x +是偶函数，所以()()22f x f x +=-+②．令1x =，由①得：()()()024f f a b =-=-+，由②得：()()31f f a b ==+，因为()()036f f +=，所以()462a b a b a -+++=Þ=-，令0x =，由①得：()()()11102f f f b =-Þ=Þ=，所以()222f x x =-+．思路一：从定义入手．9551222222f f f f æöæöæöæö=+=-+=-ç÷ç÷ç÷ç÷èøèøèøèø1335112222f f f f æöæöæöæö-=-+=-+=-ç÷ç÷ç÷ç÷èøèøèøèø511322=2222f f f f æöæöæöæö-=-+=--+-ç÷ç÷ç÷ç÷èøèøèøèø所以935222f f æöæö=-=ç÷ç÷èøèø．思路二：从周期性入手由两个对称性可知，函数()f x 的周期4T =．所以91352222f f f æöæöæö==-=ç÷ç÷ç÷èøèøèø．故选：D ．【点睛】在解决函数性质类问题的时候，我们通常可以借助一些二级结论，求出其周期性进而达到简便计算的效果．二、填空题：本题共4小题，每小题5分，共20分．13.曲线212x y x -=+在点()1,3--处的切线方程为__________．【答案】520x y -+=【解析】【分析】先验证点在曲线上，再求导，代入切线方程公式即可．【详解】由题，当1x =-时，3y =-，故点在曲线上．求导得：()()()()222221522x x y x x +--==++¢，所以1|5x y =-=¢．故切线方程为520x y -+=．故答案为：520x y -+=．14.已知向量()()3,1,1,0,a b c a kb ===+r r r r r ．若a c ^r r ，则k =________．【答案】103-.【解析】【分析】利用向量的坐标运算法则求得向量c r的坐标，利用向量的数量积为零求得k 的值【详解】()()()3,1,1,0,3,1a b c a kb k ==\=+=+r r r r Q r,(),33110a c a c k ^\=++´=Q n r r r r ，解得103k =-,故答案为：103-.【点睛】本题考查平面向量的坐标运算，平面向量垂直的条件，属基础题，利用平面向量()()1122,,,p x y q x y ==r r垂直的充分必要条件是其数量积12120x x y y +=.15.已知12,F F 为椭圆C ：221164x y+=的两个焦点，P ，Q 为C 上关于坐标原点对称的两点，且12PQ F F =，则四边形12PFQF 的面积为________．【答案】8【解析】【分析】根据已知可得12PF PF ^，设12||,||PF m PF n ==，利用勾股定理结合8m n +=，求出mn ，四边形12PFQF 面积等于mn ，即可求解.【详解】因为,P Q 为C 上关于坐标原点对称的两点，且12||||PQ F F =，所以四边形12PFQF 为矩形，设12||,||PF m PF n ==，则228,48m n m n +=+=，所以22264()2482m n m mn n mn =+=++=+，8mn =，即四边形12PFQF 面积等于8.故答案为：8.16.已知函数()2cos()f x x w j =+的部分图像如图所示，则满足条件74()()043f x f f x f p p æöæöæöæö--->ç÷ç÷ç÷ç÷èøèøèøèø的最小正整数x 为________．【答案】2【解析】【分析】先根据图象求出函数()f x 的解析式，再求出7(),(43f f p 4p-的值，然后求解三角不等式可得最小正整数或验证数值可得.【详解】由图可知313341234T p p p =-=，即2T pp w==，所以2w =；由五点法可得232p p j ´+=，即6p j =-；所以()2cos 26f x x p æö=-ç÷èø.因为7()2cos 143f p 11p æö-=-=ç÷èø，()2cos 032f 4p 5p æö==ç÷èø；所以由74(()())(()(043f x f f x f p p--->可得()1f x >或()0f x <；因为()12cos 22cos 1626f p p p æöæö=-<-=ç÷ç÷èøèø，所以，方法一：结合图形可知，最小正整数应该满足()0f x <，即cos 206x p æö-<ç÷èø，解得,36k x k k p 5p p +<<p +ÎZ ，令0k =，可得536x <<pp ，可得x 的最小正整数为2.方法二：结合图形可知，最小正整数应该满足()0f x <，又(2)2cos 406f p æö=-<ç÷èø，符合题意，可得x 的最小正整数为2.故答案为：2.【点睛】关键点睛：根据图象求解函数的解析式是本题求解的关键，根据周期求解w ，根据特殊点求解j .三、解答题：共70分．解答应写出交字说明、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：共60分．17.甲、乙两台机床生产同种产品，产品按质量分为一级品和二级品，为了比较两台机床产品的质量，分别用两台机床各生产了200件产品，产品的质量情况统计如下表：一级品二级品 合计甲机床15050200乙机床12080200合计270130400（1）甲机床、乙机床生产的产品中一级品的频率分别是多少?（2）能否有99%的把握认为甲机床的产品质量与乙机床的产品质量有差异?附：22()()()()()n ad bc K a b c d a c b d -=++++()2P K k ³0.0500.0100.001k3.8416.63510.828【答案】（1）75%；60%；（2）能.【解析】【分析】本题考查频率统计和独立性检验，属基础题，根据给出公式计算即可【详解】（1）甲机床生产的产品中的一级品的频率为15075%200=,乙机床生产的产品中的一级品的频率为12060%200=.（2）()22400150801205040010 6.63527013020020039K ´-´==>>´´´,故能有99%的把握认为甲机床的产品与乙机床的产品质量有差异.18.已知数列{}n a 的各项均为正数，记n S 为{}n a 的前n 项和，从下面①②③中选取两个作为条件，证明另外一个成立．①数列{}n a 是等差数列：②数列是等差数列；③213aa =．注：若选择不同的组合分别解答，则按第一个解答计分．【答案】答案见解析【解析】，结合,n n a S 的关系求出n a ，利用{}n a 是等差数列可证213a a =；选②③作条件证明①时，an b =+，结合,n n a S 的关系求出n a ，根据213a a =可求b ，然后可证{}n a 是等差数列.【详解】选①②作条件证明③：(0)an b a =+>，则()2n S an b =+，当1n =时，()211a S a b ==+；当2n ³时，()()221n n n a S S an b an a b -=-=+--+()22a an a b =-+；因为{}n a 也是等差数列，所以()()222a b a a a b +=-+，解得0b =；所以()221n aa n =-，所以213a a =.选①③作条件证明②：因为213a a =，{}n a 是等差数列，所以公差2112d a a a =-=，所以()21112n n n S na d n a -=+==，)1n -=+=所以是等差数列.选②③作条件证明①：(0)an b a =+>，则()2n S an b =+，当1n =时，()211a S a b ==+；当2n ³时，()()221n n n a S S an b an a b -=-=+--+()22a an a b =-+；因为213a a =，所以()()2323a a b a b +=+，解得0b =或43a b =-；当0b =时，()221,21n a a a a n ==-，当2n ³时，2-1-2n n a a a =满足等差数列的定义，此时{}n a 为等差数列；当43a b =-4=3an b an a =+-03a=-<不合题意，舍去.综上可知{}n a 为等差数列.【点睛】这类题型在解答题中较为罕见，求解的关键是牢牢抓住已知条件，结合相关公式，逐步推演，等差数列的证明通常采用定义法或者等差中项法.19.已知直三棱柱111ABC A B C -中，侧面11AA B B 为正方形，2AB BC ==，E ，F 分别为AC 和1CC 的中点，D 为棱11A B 上的点．11BF A B ^（1）证明：BF DE ^；（2）当1B D 为何值时，面11BB C C 与面DFE 所成的二面角的正弦值最小?【答案】（1）见解析；（2）112B D =【解析】【分析】通过已知条件，确定三条互相垂直的直线，建立合适的空间直角坐标系，借助空间向量证明线线垂直和求出二面角的平面角的余弦值最大，进而可以确定出答案．【详解】因为三棱柱111ABC A B C -是直三棱柱，所以1BB ^底面ABC ，所以1BB AB ^因为11//A B AB ，11BF A B ^，所以BF AB ^，又1BB BF B Ç=，所以AB ^平面11BCC B ．所以1,,BA BC BB 两两垂直．以B 为坐标原点，分别以1,,BA BC BB 所在直线为,,x y z 轴建立空间直角坐标系，如图．所以()()()()()()1110,0,0,2,0,0,0,2,0,0,0,2,2,0,2,0,2,2B A C B A C ，()()1,1,0,0,2,1E F ．由题设(),0,2D a （02a ££）．（1）因为()()0,2,1,1,1,2BF DE a ==--uu u v u uu v，所以()()0121120BF DE a ×=´-+´+´-=uu u v uu u v，所以BF DE ^．（2）设平面DFE 的法向量为(),,m x y z =u r，因为()()1,1,1,1,1,2EF DE a =-=--uu u v u u u v，所以00m EF m DE ì×=í×=îu u u v v u u u v v ，即()0120x y z a x y z -++=ìí-+-=î．令2z a =-，则()3,1,2m a a =+-v因为平面11BCC B 的法向量为()2,0,0BA =u u u r，设平面11BCC B 与平面DEF 的二面角的平面角为q ，则cos m BA m BA q ×===×u u u v v u u u v v ．当12a =时，2224a a -+取最小值为272，此时cos q3=．所以()minsin 3q ==，此时112B D =．【点睛】本题考查空间向量的相关计算，能够根据题意设出(),0,2D a （02a ££），在第二问中通过余弦值最大，找到正弦值最小是关键一步．20.抛物线C 的顶点为坐标原点O ．焦点在x 轴上，直线l ：1x =交C 于P ，Q 两点，且OP OQ ^．已知点()2,0M ，且M e 与l 相切．（1）求C ，M e 的方程；（2）设123,,A A A 是C 上的三个点，直线12A A ，13A A 均与M e 相切．判断直线23A A 与M e 的位置关系，并说明理由．【答案】（1）抛物线2:C y x =，M e 方程为22(2)1x y -+=；（2）相切，理由见解析【解析】【分析】（1）根据已知抛物线与1x =相交，可得出抛物线开口向右，设出标准方程，再利用对称性设出,P Q 坐标，由OP OQ ^，即可求出p ；由圆M 与直线1x =相切，求出半径，即可得出结论；（2）先考虑12A A 斜率不存在，根据对称性，即可得出结论；若121323,,A A A A A A 斜率存在，由123,,A A A 三点在抛物线上，将直线121223,,A A A A A A 斜率分别用纵坐标表示，再由1212,A A A A 与圆M 相切，得出2323,y y y y +×与1y 的关系，最后求出M 点到直线23A A 的距离，即可得出结论.【详解】（1）依题意设抛物线200:2(0),(1,),(1,)C y px p P y Q y =>-，20,1120,21OP OQ OP OQ y p p ^\×=-=-=\=uu u r uu u r Q ，所以抛物线C 的方程为2y x =，(0,2),M M e 与1x =相切，所以半径为1，所以M e 的方程为22(2)1x y -+=；（2）设111222333(),(,),(,)A x y A x y A x y 若12A A 斜率不存在，则12A A 方程为1x =或3x =，若12A A 方程为1x =，根据对称性不妨设1(1,1)A ，则过1A 与圆M 相切的另一条直线方程为1y =，此时该直线与抛物线只有一个交点，即不存在3A ，不合题意；若12A A 方程为3x =，根据对称性不妨设12(3,A A 则过1A 与圆M 相切的直线13A A为(3)3y x -=-，又1313313131,03A A y y k y x x y y -====\=-+，330,(0,0)x A =，此时直线1323,A A A A 关于x 轴对称，所以直线23A A 与圆M 相切；若直线121323,,A A A A A A 斜率均存在，则121323121323111,,A A A A A A k k k y y y y y y ===+++，所以直线12A A 方程为()11121y y x x y y -=-+，整理得1212()0x y y y y y -++=，同理直线13A A 的方程为1313()0x y y y y y -++=，直线23A A 的方程为2323()0x y y y y y -++=，12A A Q 与圆M相切，1=整理得22212121(1)230y y y y y -++-=，13A A 与圆M 相切，同理22213131(1)230y y y y y -++-=所以23,y y 为方程222111(1)230y y y y y -++-=的两根，2112323221123,11y y y y y y y y -+=-×=--，M 到直线23A A的距离为：2123|2|y -+=221==，所以直线23A A 与圆M 相切；综上若直线1213,A A A A 与圆M 相切，则直线23A A 与圆M 相切.【点睛】关键点点睛：（1）过抛物线上的两点直线斜率只需用其纵坐标（或横坐标）表示，将问题转化为只与纵坐标（或横坐标）有关；（2）要充分利用1213,A A A A 的对称性，抽象出2323,y y y y +×与1y 关系，把23,y y 的关系转化为用1y 表示.21.已知0a >且1a ¹，函数()(0)ax x f x x a=>．（1）当2a =时，求()f x 的单调区间；（2）若曲线()y f x =与直线1y =有且仅有两个交点，求a 的取值范围．【答案】（1）20,ln2æùçúèû上单调递增；2,ln2éö+¥÷êëø上单调递减；（2）()()1,,e e È+¥.【解析】【分析】（1）求得函数的导函数，利用导函数的正负与函数的单调性的关系即可得到函数的单调性；（2）利用指数对数的运算法则，可以将曲线()y f x =与直线1y =有且仅有两个交点等价转化为方程ln ln x a x a =有两个不同的实数根，即曲线()y g x =与直线ln ay a=有两个交点，利用导函数研究()g x 的单调性，并结合()g x 的正负，零点和极限值分析()g x 的图象，进而得到ln 10a a e<<，发现这正好是()()0g a g e <<，然后根据()g x 的图象和单调性得到a 的取值范围.【详解】（1）当2a =时，()()()()22222ln 2222ln 2,242xx x x x x x x x x x f x f x ¢--===n n n ,令()'0f x =得2ln 2x =,当20ln 2x <<时，()0f x ¢>,当2ln 2x >时，()0f x ¢<,∴函数()f x 在20,ln2æùçúèû上单调递增；2,ln2éö+¥÷êëø上单调递减；（2）()ln ln 1ln ln a x a x x x af x a x x a a x a x a==Û=Û=Û=,设函数()ln x g x x =,则()21ln xg x x-¢=,令()0g x ¢=，得x e =,在()0,e 内()0g x ¢>,()g x 单调递增；在(),e +¥上()0g x ¢<,()g x 单调递减；()()1max g x g e e\==,又()10g =，当x 趋近于+¥时，()g x 趋近于0，所以曲线()y f x =与直线1y =有且仅有两个交点，即曲线()y g x =与直线ln ay a=有两个交点的充分必要条件是ln 10a a e<<,这即是()()0g a g e <<,所以a 的取值范围是()()1,,e e È+¥.【点睛】本题考查利用导数研究函数的单调性，根据曲线和直线的交点个数求参数的取值范围问题，属较难试题，关键是将问题进行等价转化，分离参数，构造函数，利用导数研究函数的单调性和最值，图象，利用数形结合思想求解.（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．[选修4-4：坐标系与参数方程]（10分）22.在直角坐标系xOy 中，以坐标原点为极点，x 轴正半轴为极轴建立极坐标系，曲线C 的极坐标方程为r q =．（1）将C 的极坐标方程化为直角坐标方程；（2）设点A 的直角坐标为()1,0，M 为C 上的动点，点P 满足AP =u u u r u u u r，写出Р的轨迹1C 的参数方程，并判断C 与1C 是否有公共点．【答案】（1）(222x y -+=；（2）P 的轨迹1C 的参数方程为32cos 2sin x y qqì=-+ïí=ïî（q 为参数），C 与1C 没有公共点.【解析】【分析】（1）将曲线C 的极坐标方程化为2cos r q =，将cos ,sin x y r q r q ==代入可得；（2）设(),P x y ，设)Mq q +，根据向量关系即可求得P 的轨迹1C 的参数方程，求出两圆圆心距，和半径之差比较可得.【详解】（1）由曲线C 的极坐标方程r q =可得2cos r q =，将cos ,sin x y r q r q ==代入可得22x y +=，即(222x y -+=，即曲线C 的直角坐标方程为(222x y +=；（2）设(),P x y ，设)Mq qQAP =u u u r u u u r ，())()1,22cos x y q q q q \-=+-=+，则122cos 2sin x y q q ì-=+ïí=ïî，即32cos 2sin x y qq ì=+ïí=ïî，故P 的轨迹1C 的参数方程为32cos 2sin x y qq ì=+ïí=ïî（q 为参数）Q曲线C 的圆心为)，曲线1C 的圆心为()3-，半径为2，则圆心距为3-，32-<-Q ，\两圆内含，故曲线C 与1C 没有公共点.【点睛】关键点睛：本题考查参数方程的求解，解题的关键是设出M 的参数坐标，利用向量关系求解.[选修4-5：不等式选讲]（10分）23.已知函数()2,()2321f x x g x x x =-=+--．（1）画出()y f x =和()y g x =的图像；（2）若()()f x a g x +³，求a 的取值范围．【答案】（1）图像见解析；（2）112a ³【解析】【分析】（1）分段去绝对值即可画出图像；（2）根据函数图像数形结和可得需将()y f x =向左平移可满足同角，求得()y f x a =+过1,42A æöç÷èø时a的值可求.【详解】（1）可得2,2 ()22,2x xf x xx x-<ì=-=í-³î，画出图像如下：34,231()232142,2214,2xg x x x x xxì-<-ïïï=+--=+-£<íïï³ïî，画出函数图像如下：（2）()|2|f x a x a+=+-，如图，在同一个坐标系里画出()(),f xg x图像，()y f x a=+是()y f x=平移了a个单位得到，则要使()()f x ag x+³，需将()y f x=向左平移，即0a>，当()y f x a =+过1,42A æöç÷èø时，1|2|42a +-=，解得112a =或52-（舍去），则数形结合可得需至少将()y f x =向左平移112个单位，112a \³.【点睛】关键点睛：本题考查绝对值不等式的恒成立问题，解题的关键是根据函数图像数形结合求解.。
任务型阅读(含(答案)解析)
任务型阅读知识清单一、考点分析任务型阅读是“阅读理解”的另一种形式,综合考查学生归纳概括能力和语篇结构理解能力。
其内容涉及广泛,要求学生在阅读理解的基础上,完成一项任务或解决一个问题。
所以“任务型阅读”是介于阅读理解和写作之间,其任务已不同于阅读理解中的选择题或书面表达,而是在理解文字的基础上,完成相应的图表或文字练习,从而有效地测试学生用英语“做事”的能力。
根据任务类型,常见题型有以下四种:1. 完成表格型此类任务型阅读要求我们在理解文本信息的基础上,根据材料提供的直接信息或由我们推理、提炼后的间接信息完成题目要求的任务。
其阅读内容更贴近学生的生活实际,任务的设置变化多样,不光有简单信息的捕捉,而且有阅读短文,通过对短文信息的归纳,加工处理,运用语言逻辑推理和思维能力来完成表格。
2. 回答问题型此类任务型阅读要求我们根据短文、表格、图片或图文结合的材料回答命题者设定的问题,所设置的任务通过事实或细节的查找就能完成,与普通阅读理解的解题方法相似,只是题目设计采用了主观题形式,没有给出选项,需要我们从材料中寻求信息,以一个完整的句子,或者是其适当的缩略形式作答。
从问题所涉及的内容看,考查文本表层理解多于深层理解,其设计的问题多为五W或一般疑问句的细节性问题,而涉及推理判断、文章主旨、写作意图及作者态度、感受等的题目则少之又少。
此类题型是学生失分较多的题型之一,要求学生有较扎实的语言基础和较强的综合运用英语的能力。
3. 句子还原型还原短文型“阅读理解”题有两种形式:即选择句子还原短文和排列段落还原短文。
第一种形式要求考生根据短文内容,从文后所给的句子中选出适当的句子填入短文空白处。
第二种形式是给出一篇200~300个词的短文,要求考生根据短文内容和结构,将顺序打乱的段落重新排序,有时首段或尾段的位置已给出。
这种题型旨在考查考生对短文整体结构的理解能力,要求考生从短文的篇章结构的层面上把握短文,了解其大概意思和结构,分清句子或段落之间的逻辑关系,然后还原成短文的原貌。
中考数学题库(含答案和解析)
解得：
在数轴上表示其解集如下：
故选B
【点睛】本题考查的是一元一次不等式的解法.在数轴上表示不等式的解集.掌握“小于向左拐”是解本题的关键．
6.“方胜”是中国古代妇女的一种发饰.其图案由两个全等正方形相叠组成.寓意是同心吉祥．如图.将边长为2cm的正方形ABCD沿对角线BD方向平移1cm得到正方形 .形成一个“方胜”图案.则点D. 之间的距离为（）
13.小曹同学复习时将几种三角形的关系整理如图.请帮他在横线上____填上一个适当的条件．
中考数学题库(含答案和解析)
一、选择题（本题有10小题）
1.若收入3元记为+3.则支出2元记为（）
A.1B.-1C.2D.-2
【答案】D
【解析】
【分析】根据正负数的意义可得收入为正.收入多少就记多少即可．
【详解】解：∵收入3元记 +3.
∴支出2元记为-2．
故选：D
【点睛】本题考查正、负数的意义；在用正负数表示向指定方向变化的量时.通常把向指定方向变化的量规定为正数.而把向指定方向的相反方向变化的量规定为负数．
【答案】D
【解析】
【分析】根据同底数幂的乘法法则进行运算即可．
【详解】解：
故选D
【点睛】本题考查的是同底数幂的乘法.掌握“同底数幂的乘法.底数不变.指数相加”是解本题的关键．
4.如图.在⊙O中.∠BOC＝130°.点A在 上.则∠BAC的度数为（ ）
A.55°B.65°C.75°D.130°
【答案】B
12.不透明的袋子中装有5个球.其中有3个红球和2个黑球.它们除颜色外都相同．从袋子中随机取出1个球.它是黑球的概率是_____．
【答案】
【解析】
初三英语阅读理解解题技巧及经典题型及练习题(含答案)含解析
初三英语阅读理解解题技巧及经典题型及练习题(含答案)含解析一、阅读理解1．阅读理解I'm writing this article in China, far away from my home in the United States. You might wonder what I do to remember my mom on Mother's Day. Well, l certainly call her to say "Happy Mother's Day" and promise that I will visit home soon.I still remember the first time I forget Mother's Day. When I finally figured it out, I asked, "Why does mom get a special holiday? Why isn't there a Children's Day for us?My mom explained, in that way that only moms seem to be able in explain, "Because every day is Children's Day!"I knew that I had messed up. I thought about all the time and love my mom had given me. I thought about the food she had made, the toys she had brought and the long hours she had spent with me. There might not be a perfect mom, but there is mom's love, which can fix anything. After that day, for 364 days, I was looking forward to the next Mother's Day.That was the first year l forgot Mother's Day. It was also the last year I forgot.Are you struggling to consider what to do for your mom on Mother's Day?Common things are to write a message that thanks for the love she has given you, send her a card or buy her a gift. But the most important thing is -DON'T FORGET!（1）How does the writer celebrate Mother's Day this year?A. He calls his mom.B. He send his mom a card.C. He visits home to see his mom.（2）What might the underlined phrase "figure out" means?A. 弄清楚B. 预料到C. 归纳出（3）Why did the writer begin to expect Mother's Day?A. Because he missed his mom far in the United States.B. Because he wanted to spend more time with his mom.C. Because he realized he had got much care from his mom.（4）What is the purpose of the passage?A. To explain why there isn't a Children's DayB. To advise us not to forget to keep our own promises.C. To remind us to remember the love mom has given us.（5）Which is the best title for the passage?A. An Interesting TalkB. An Important DayC. An Expensive Gift【答案】（1）A（2）A（3）C（4）C（5）B【解析】【分析】本文讲述了作者理解了为什么过母亲节，提醒大家要给妈妈过母亲节。
心理学自考题-11_真题(含答案与解析)-交互
心理学自考题-11(总分100, 做题时间90分钟)一、单选题1.教师想了解学生对作文的态度，让学生填写了问卷，与部分学生座谈，与个别学生谈话，找到了问题及原因。
他用的是______• A.观察法• B.调查法• C.测验法• D.实验法SSS_SIMPLE_SINA B C D分值: 2答案:B[解析] 本题考查心理学研究方法中的调查法。
调查法分为问卷法和谈话法两种方式。
本题题干中的教师，正是运用了问卷与谈话两种方式来了解学生对作文的态度。
考生在掌握调查法内容的同时，应注意它与观察法、测验法和实验法的区别。
答案为B。
2.在四种典型的生物节律中，对人的心理状态影响最大的是______• A.年• B.月•**分钟D.日SSS_SIMPLE_SINA B C D分值: 2答案:D[解析] 在生物节律中，年、月、90分钟三种循环对人的心理状态影响不算太大，而日周期则有重大影响。
答案为D。
3.在催眠状态下，有些被催眠者可能会根据催眠者的指示，去做那些在一般情况下依据社会准则不能做的事情，该催眠状态下的心理特征是______• A.感觉麻痹• B.感觉扭曲和幻觉• C.解除抑制• D.对催眠经验的记忆消失SSS_SIMPLE_SINA B C D分值: 2答案:C[解析] 一般情况下，那些依据社会准则不能做的事情是受到抑制的，人们不可能让被试去做。
但是在催眠状态下，抑制被解除，他就可能根据催眠者的指示去做。
该催眠状态下的心理特征是解除抑制。
答案为C。
4.知觉恒常性通常包括三类，下面不属于知觉恒常性的是______• A.大小恒常性• B.重量恒常性• C.形状恒常性• D.颜色恒常性SSS_SIMPLE_SINA B C D分值: 2答案:B[解析] 本题用排除法。
知觉恒常性包括大小恒常性、形状恒常性与颜色恒常性。
答案为B。
5.下列关于长时记忆的说法不正确的是______• A.容量大• B.保持时间长• C.信息编码是相对深度水平加工的结果• D.扮演着意识的角色，使我们知道在接收什么、做什么SSS_SIMPLE_SINA B C D分值: 2答案:D[解析] 长时记忆的容量大，保存时间长，一般被认为是无限的。
2023年高考新课标全国Ⅱ卷英语真题+答案解析(不含听力)
2023年高考新课标全国Ⅱ卷英语真题+答案解析（不含听力）一、阅读理解Yellowstone National Park offers a variety of ranger programs throughout the park, and throughout the year. The following are descriptions of the ranger programs this summer.Experiencing Wildlife in Yellowstone (May 26 to September 2)Whether you’re hiking a backcountry trail (小径), camping, or just enjoying the park’s amazing wildlife from the road, this quick workshop is for you and your family. Learn where to look for animals and how to safely enjoy your wildlife watching experience. Meet at the Canyon Village Store.Junior Ranger Wildlife Olympics (June 5 to August 21)Kids can test their skills and compare their abilities to the animals of Yellowstone. Stay for as little or as long as your plans allow. Meet in front of the Visitor Education Center.Canyon Talks at Artist Point (June 9 to September 2)From a classic viewpoint, enjoy Lower Falls, the Yellowstone River, and the breathtaking colors of the canyon (峡谷) while learning about the area’s natural and human history. Discover why artists and photographers continue to be drawn to this special place. Meet on the lower platform at Artist Point on the South Rim Drive for this short talk.Photography Workshops (June 19 & July 10)Enhance your photography skills — join Yellowstone’s park photographer for a hands-on program to inspire new and creative ways of enjoying the beauty and wonder of Yellowstone.6/19 — Waterfalls &Wide Angles: meet at Artist Point.7/10 — Wildflowers &White Balance: meet at Washburn Trailhead in Chittenden parking area. 1．Which of the four programs begins the earliest?A．Photography Workshops.B．Junior Ranger Wildlife Olympics.C．Canyon Talks at Artist Point.D．Experiencing Wildlife in Yellowstone.2．What is the short talk at Artist Point about?A．Works of famous artists.B．Protection of wild animals.C．Basic photography skills.D．History of the canyon area.3．Where will the participants meet for the July 10 photography workshop?A．Artist Point.B．Washburn Trailhead.C．Canyon Village Store.D．Visitor Education Center.Turning soil, pulling weeds, and harvesting cabbage sound like tough work for middle and high schoolkids. And at first it is, says Abby Jaramillo, who with another teacher started Urban Sprouts, a school garden program at four low-income schools. The program aims to help students develop science skills, environmental awareness, and healthy lifestyles.Jaramillo’s students live in neighborhoods where fresh food and green space are not easy to find and fast food restaurants outnumber grocery stores. “The kids literally come to school with bags of snacks and large bottles of soft drinks,” she says. “They come to us thinking vegetables are awful, dirt is awful, insects are awful.” Though some are initially scared of the insects and turned off by the dirt, most are eager to try something new.Urban Sprouts’ classes, at two middle schools and two high schools, include hands-on experiments such as soil testing, flower-and-seed dissection, tastings of fresh or dried produce, and work in the garden. Several times a year, students cook the vegetables they grow, and they occasionally make salads for their entire schools.Program evaluations show that kids eat more vegetables as a result of the classes. “We have students who say they went home and talked to their parents and now they’re eating differently,” Jaramillo says.She adds that the program’s benefits go beyond nutrition. Some students get so interested in gardening that they bring home seeds to start their own vegetable gardens. Besides, working in the garden seems to have a calming effect on Jaramillo’s special education students, many of whom have emotional control issues. “They get outside,” she says, “and they feel successful.”4．What do we know about Abby Jaramillo?A．She used to be a health worker.B．She grew up in a low-income family.C．She owns a fast food restaurant.D．She is an initiator of Urban Sprouts.5．What was a problem facing Jaramillo at the start of the program?A．The kids’ parents distrusted her.B．Students had little time for her classes.C．Some kids disliked garden work.D．There was no space for school gardens.6．Which of the following best describes the impact of the program?A．Far-reaching.B．Predictable.C．Short-lived.D．Unidentifiable.7．What can be a suitable title for the text?A．Rescuing School Gardens B．Experiencing Country LifeC．Growing Vegetable Lovers D．Changing Local LandscapeReading Art: Art for Book Lovers is a celebration of an everyday object — the book, represented here inalmost three hundred artworks from museums around the world. The image of the reader appears throughout history, in art made long before books as we now know them came into being. In artists’ representations of books and reading, we see moments of shared humanity that go beyond culture and time.In this “book of books,” artworks are selected and arranged in a way that emphasizes these connections between different eras and cultures. We see scenes of children learning to read at home or at school, with the book as a focus for relations between the generations. Adults are portrayed (描绘) alone in many settings and poses —absorbed in a volume, deep in thought or lost in a moment of leisure. These scenes may have been painted hundreds of years ago, but they record moments we can all relate to.Books themselves may be used symbolically in paintings to demonstrate the intellect (才智), wealth or faith of the subject. Before the wide use of the printing press, books were treasured objects and could be works of art in their own right. More recently, as books have become inexpensive or even throwaway, artists have used them as the raw material for artworks — transforming covers, pages or even complete volumes into paintings and sculptures.Continued developments in communication technologies were once believed to make the printed page outdated. From a 21st-century point of view, the printed book is certainly ancient, but it remains as interactive as any battery-powered e-reader. To serve its function, a book must be activated by a user: the cover opened, the pages parted, the contents reviewed, perhaps notes written down or words underlined. And in contrast to our increasingly networked lives where the information we consume is monitored and tracked, a printed book still offers the chance of a wholly private, “off-line” activity.8．Where is the text most probably taken from?A．An introduction to a book.B．An essay on the art of writing.C．A guidebook to a museum.D．A review of modern paintings.9．What are the selected artworks about?A．Wealth and intellect.B．Home and school.C．Books and reading.D．Work and leisure.10．What do the underlined words “relate to” in paragraph 2 mean?A．Understand.B．Paint.C．Seize.D．Transform.11．What does the author want to say by mentioning the e-reader?A．The printed book is not totally out of date.B．Technology has changed the way we read.C．Our lives in the 21st century are networked.D．People now rarely have the patience to read.As cities balloon with growth, access to nature for people living in urban areas is becoming harder to find. If you’re lucky, there might be a pocket park near where you live, but it’s unusual to find places in a city that are relatively wild.Past research has found health and wellness benefits of nature for humans, but a new study shows that wildness in urban areas is extremely important for human well-being.The research team focused on a large urban park. They surveyed several hundred park-goers, asking them to submit a written summary online of a meaningful interaction they had with nature in the park. The researchers then examined these submissions, coding (编码) experiences into different categories. For example, one participant’s experience of “We sat and listened to the waves at the beach for a while” was assigned the categories “sitting at beach” and “listening to waves.”Across the 320 submissions, a pattern of categories the researchers call a “nature language” began to emerge. After the coding of all submissions, half a dozen categories were noted most often as important to visitors. These include encountering wildlife, walking along the edge of water, and following an established trail.Naming each nature experience creates a usable language, which helps people recognize and take part in the activities that are most satisfying and meaningful to them. For example, the experience of walking along the edge of water might be satisfying for a young professional on a weekend hike in the park. Back downtown during a workday, they can enjoy a more domestic form of this interaction by walking along a fountain on their lunch break.“We’re trying to generate a language that helps bring the human-nature interactions back into our daily lives. And for that to happen, we also need to protect nature so that we can interact with it,” said Peter Kahn, a senior author of the study.12．What phenomenon does the author describe at the beginning of the text?A．Pocket parks are now popular.B．Wild nature is hard to find in cities.C．Many cities are overpopulated.D．People enjoy living close to nature.13．Why did the researchers code participant submissions into categories?A．To compare different types of park-goers.B．To explain why the park attracts tourists.C．To analyze the main features of the park.D．To find patterns in the visitors’ summaries.14．What can we learn from the example given in paragraph 5?A．Walking is the best way to gain access to nature.B．Young people are too busy to interact with nature.C．The same nature experience takes different forms.D．The nature language enhances work performance.15．What should be done before we can interact with nature according to Kahn?A．Language study.B．Environmental conservation.C．Public education.D．Intercultural communication.二、七选五As an artist who shares her journey on social media, I’m often asked by curious followers how to begin an art journey. Unfortunately, there is no magic list I can offer. I do remember, though, what it was like to be a complete beginner. So I’ve put together some good tips for starting an art journey.·Start small. I suggest using a sketchbook (素描本) for small studies. These small studies provide inspiration and may be a springboard for more complex works in the future. 16 You’ll want to look back on your journey to see how far you’ve come.·Paint often and paint from life. There’s no better way to improve than to put in those brush miles. Whether you paint still lifes, portraits, or landscapes, paint from life as much as possible. 17 ·Continually challenge yourself to try something new. 18 Artistic growth can be a bit painful. Welcome to the club；we’ve all been there. I love taking on challenges. I once took up a challenge to create a painting every day for a month and post the works online.· 19 Seeking and accepting constructive feedback (反馈) is crucial to growth. I post my work on social media and, in turn, have met some of the kindest people. They make me feel valued and respected, no matter my level of artistic ability.The journey you’re on won’t follow a straight path. 20 Push through, give it time and put in the effort. You will harvest the rewards of an artistic life.A．Get out of your comfort zone.B．Make career plans and set goals.C．Don’t throw away your beginner art.D．Share your work if you feel comfortable doing so.E．You’ll hit roadblocks, and you’ll feel discouraged at times.F．Evaluate your performance and, if needed, redefine your role.G．You’ll develop that painting muscle memory that only comes with repetition.三、完形填空In April last year, I saw a post on the PNP (Pilots N Paws) website from a family in Topeka. They had to move to Virginia but they were on a very tight 21 . They could not afford to pay for 22 for their dog, Tiffy, and 23 wanted to take her with them.It just 24 that I was planning another PNP flight with another pilot, Karen, who 25 to take Tiffy from Kansas City to Virginia. What I was to do was fly to Topeka to 26 Tiffy.When I met Tiffy’s owners, they seemed very 27 . George, the husband, was trying to be calm, but I could tell this was 28 for him, having to leave his dog to a 29 and trust that everything would 30 .After some goodbyes, I asked George and his wife to help me 31 Tiffy into the plane. I promised to take care of Tiffy and 32 them as soon as we got to Kansas City.The flight was 33 , and Tiffy was a great passenger. The next day, she 34 with Karen and made it back to George in Virginia within a few days. He was so 35 and sent me a nice e-mail with pictures. It felt great to know that I had helped bring this family together again. 21．A．turn B．budget C．schedule D．connection22．A．food B．shelter C．medicine D．transportation 23．A．desperately B．temporarily C．secretly D．originally24．A．appeared B．proved C．happened D．showed25．A．waited B．offered C．hurried D．failed26．A．see off B．look for C．hand over D．pick up27．A．confused B．nervous C．annoyed D．curious28．A．hard B．fine C．common D．lucky29．A．coworker B．passenger C．stranger D．neighbor30．A．speed up B．work out C．come back D．take off31．A．feed B．follow C．change D．load32．A．call B．join C．leave D．serve33．A．unnecessary B．unexpected C．unavoidable D．uneventful34．A．returned B．fought C．flew D．agreed35．A．thankful B．generous C．proud D．sympathetic四、用单词的适当形式完成短文阅读下面短文, 在空白处填入1个适当的单词或括号内单词的正确形式。
2022年高考真题：物理(全国甲卷)【含答案及解析】
2022年全国⾼考甲卷物理试题二、选择题1.北京2022年冬奥会首钢滑雪大跳台局部示意图如图所示。
运动员从a 处由静止自由滑下，到b 处起跳，c 点为a 、b 之间的最低点，a 、c 两处的高度差为h 。
要求运动员经过一点时对滑雪板的压力不大于自身所受重力的k 倍，运动过程中将运动员视为质点并忽略所有阻力，则c 点处这一段圆弧雪道的半径不应小于（）A.1hk + B.h kC.2h kD.21h k -2.长为l 的高速列车在平直轨道上正常行驶，速率为v 0，要通过前方一长为L 的隧道，当列车的任一部分处于隧道内时，列车速率都不允许超过v （v <v 0）。
已知列车加速和减速时加速度的大小分别为a 和2a ，则列车从减速开始至回到正常行驶速率v 0所用时间至少为（）A.02v v L l a v -++ B.02v v L l a v -++ C.()032v v L la v-++ D.()032v v L la v-++3.三个用同样的细导线做成的刚性闭合线框，正方形线框的边长与圆线框的直径相等，圆线框的半径与正六边形线框的边长相等，如图所示。
把它们放入磁感应强度随时间线性变化的同一匀强磁场中，线框所在平面均与磁场方向垂直，正方形、圆形和正六边形线框中感应电流的大小分别为12I I 、和3I 。
则（ ）A .132I I I << B.132I I I >> C.123I I I => D.123I I I ==4.两种放射性元素的半衰期分别为0t 和02t ，在0=t 时刻这两种元素的原子核总数为N ，在02t t =时刻，尚未衰变的原子核总数为3N，则在04t t =时刻，尚未衰变的原子核总数为（ ）A.12N B.9N C.8N D.6N 5.空间存在着匀强磁场和匀强电场，磁场的方向垂直于纸面（xOy 平面）向里，电场的方向沿y 轴正方向。
一带正电的粒子在电场和磁场的作用下，从坐标原点O 由静止开始运动。
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【最新】江西省高三第一次联考物理试题（新余市渝水区新余市第四中学、宜春市上高县上高二中）学校:___________姓名：___________班级：___________考号：___________一、单选题1．火星探测器绕火星近地做圆周轨道飞行，其线速度和相应的轨道半径为v 0和R 0，火星的一颗卫星在圆轨道上的线速度和相应的轨道半径为v 和R ，则下列关系正确的是（ ）A ．lg （0v v ）=12lg （0R R ） B ．lg （0v v ）=2lg （0R R ） C ．lg （0v v ）=12lg （0R R） D ．lg （0v v ）=2lg （0R R） 2．如图所示，在匀强磁场区域中有一光滑斜面体，在斜面体上放了一根长为L 、质量为m 的导线，当通以如图方向的电流后，导线恰好能保持静止，则磁感应强度B 满足（ ）A ．mgB IL =，方向竖直向上 B ．sin mg B ILθ=，方向垂直纸面向外 C ．cos mg B ILθ=，方向沿斜面向上 D ．tan mg B ILθ=，方向竖直向下 3．如图电路中，电源的内电阻为r ，R 1、R 3、R 4均为定值电阻，电表均为理想电表． 闭合电键S ，当滑动变阻器R 2的滑动触头向右滑动时，下列说法中正确的是（ ）A．电压表的示数变小B．电流表的示数变大C．电流表的示数变小D．R1中电流的变化量一定大于R4中电流的变化量4．【最新】11月16日，第26届国际度量衡大会上，经过60个成员国代表投票表决，重新定义了千克．新定义以普朗克常数为基准，从而使1千克脱离了实际物体．你可能不能推导出千克的新定义，但可以通过一定的物理分析，分析下列新千克的定义（公式中h为普朗克常数，ν为铯原子共振频率，C为光速）表示式为（）A．1kg= 1.36×1018hν/C2B．1kg= 1.36×1018h2ν/CC．1kg= 1.36×1018hν2/CD．1kg= 1.36×1018h2ν2/C25．水平放置的平行金属板A、B连接一恒定电压，两个质量相等的电荷M和N同时分别从极板A的边缘和两极板的正中间沿水平方向进入板间电场，两电荷恰好在板间某点相遇，如图所示．若不考虑电荷的重力和它们之间的相互作用，则下列说法正确的是( )A．电荷M的比荷大于电荷N的比荷B．两电荷在电场中运动的加速度相等C．从两电荷进入电场到两电荷相遇，电场力对电荷M做的功等于电场力对电荷N做的功D．电荷M进入电场的初速度大小与电荷N进入电场的初速度大小一定相同二、多选题6．如图所示，质量为m的物体，放在一固定斜面上，当斜面倾角为30°时恰能沿斜面匀速下滑．对物体施加一大小为F的水平向右恒力，物体可沿斜面匀速向上滑行．设最大静摩擦力等于滑动摩擦力，当斜面倾角增大并超过某一临界角θ0时，不论水平恒力F 多大，都不能使物体沿斜面向上滑行，则下列说法正确的是（）ABC．这一临界角θ0的大小30°D．这一临界角θ0的大小60°7．如图所示，实线是一质子仅在电场力作用下由a点运动到b点的运动轨迹，虚线可能是电场线，也可能是等差等势线，则（）A．若虚线是电场线，则质子在a点的电势能大，动能小B．若虚线是等差等势线，则质子在a点的电势能大，动能小C．质子在a点的加速度一定大于在b点的加速度D．a点的电势一定高于b点的电势8．如图所示，两根光滑平行的金属导轨，放在倾角为θ的斜面上，导轨的左端接有电阻R，导轨自身电阻不计，斜面处在一匀强磁场中，方向垂直斜面向上，一质量为m、电阻不计的金属棒，在沿斜面并与棒垂直的恒力F作用下沿导轨匀速上滑，并上升了h高度，则在上滑h的过程中（）A．金属棒所受合外力所做的功等于mgh与电阻R上产生的热量之和B．恒力F与重力的合力所做的功等于电阻R上产生的热量C．金属棒受到的合外力所做的功为零D．恒力F与安培力的合力所做的功为mgh9．下列说法中正确的是（）A．布朗运动是悬浮在液体中固体分子所做的无规则运动B．叶面上的小露珠呈球形是由于液体表面张力的作用C．液晶显示器利用了液晶对光具有各向异性的特点D．当两分子间距离大于平衡位置的间距r0时，分子间的距离越大，分子势能越小E.温度升高时，分子热运动的平均动能一定增大，但并非所有的分子的速率都增大三、实验题10．利用气垫导轨验证机械能守恒定律，实验装置如图甲所示，水平桌面上固定一倾斜的气垫导轨，导轨上A点处有一带长方形遮光片的滑块，其总质量为M，左端由跨过轻质光滑定滑轮的细绳和一质量为m的小球相连；遮光片两条长边与导轨垂直，导轨上B 点有一光电门，可以测量遮光片经过光电门时的挡光时间t，用d表示A点到光电门B 处的距离，b表示遮光片的宽度，将遮光片通过光电门的平均速度看做滑块通过B点时的瞬时速度，实验时滑块在A处由静止开始运动．(1)某次实验测得倾角θ＝30°，重力加速度用g表示，滑块从A处到达B处时m和M组成的系统动能增加量可表示为ΔE k＝_________，系统的重力势能减少量可表示为ΔE p ＝________，在误差允许的范围内，若ΔE k＝ΔE p，则可认为系统的机械能守恒．(用题中字母表示)(2)在上述实验中，某同学改变A、B间的距离，作出的v2－d图象如图乙所示，并测得M＝m，则重力加速度g＝________m/s2.11．太阳能电池板在有光照时，可以将光能转化为电能，在没有光照时，可以视为一个电学器件．某实验小组根据测绘小灯泡伏安特性曲线的实验方法，探究一个太阳能电池板在没有光照时(没有储存电能)的I－U特性．所用的器材包括：太阳能电池板，电源E，电流表A，电压表V，滑动变阻器R，开关S及导线若干．(1)为了达到上述目的，实验电路应选用图甲中的________(填“a”或“b”)．(2)该实验小组根据实验得到的数据，描点绘出了如图乙所示的I－U图象．由图可知，当电压小于2.00 V时，太阳能电池板的电阻________(填“很大”或“很小”)；当电压为2.80 V时，太阳能电池板的电阻为________ Ω.(3)当有光照射时，太阳能电池板作为电源，其路端电压与总电流的关系如图丙所示，分析该曲线可知，该电池板作为电源时的电动势为________ V．若把它与阻值为1 kΩ的电阻连接构成一个闭合电路，在有光照射情况下，该电池板的效率是________%.(结果保留三位有效数字)四、解答题12．如图所示，相距x＝4m、质量均为M的两个完全相同的木板A、B置于水平地面上，一质量也为M、可视为质点的物块C置于木板A的左端．已知物块C与木板A、B之间的动摩擦因数均为μ1＝0.40，木板A、B与水平地面之间的动摩擦因数均为μ2＝0.10，最大静摩擦力可以认为等于滑动摩擦力，开始时，三个物体均处于静止状态．现给物块C施加一个水平向右的恒力F，且F＝0.3Mg，已知木板A、B碰撞后立即粘连在一起（g 取10 m/s2）．(1)通过计算说明A与B碰前A与C是一起向右做匀加速直线运动．(2)求从物块C开始运动到木板A与B相碰所经历的时间．(3)已知木板A、B的长度均为L＝0.2m，请通过分析计算后判断：物块C最终会不会从木板上掉下来？13．图所示，灯丝K可以连续逸出不计初速度的电子，在KA间经大小为U的加速电压加速后，从A板中心小孔射出，再从M、N两极板的正中间以平行极板的方向进入偏转电场．M 、N 两极板长为L ，间距为3L ．如果在两板间加上如图b 所示的电压U MN ，电子恰能全部射入如图所示的匀强磁场中．不考虑极板边缘的影响，电子穿过平行板的时间极短，穿越过程可认为板间电压不变，磁场垂直纸面向里且范围足够大，不考虑电场变化对磁场的影响．已知电子的质量为m ，电荷量为e ，不计电子的重力及它们之间的相互作用力．求:（1）偏转电场电压U MN 的峰值；（2）已知4T t =在时刻射入偏转电场的电子恰好能返回板间，求匀强磁场的磁感应强度B 的大小；（3）从电子进入偏转电场开始到离开磁场的最短时间是多少．14．如图所示的装置中,A 、B 和C 为三支内径相等的玻璃管,它们都处于竖直位置,A 、B 两管的上端等高,管内装有水,A 管上端封闭,内有气体,B 管上端开口与大气相通,C 管中水的下方有活塞顶住,A 、B 、C 三管由内径很小的细管连接在一起,开始时,A 、B 两管中气柱的长度均为L ＝3.9m,C 管中水柱的长度L 0＝5.2m,整个装置处于平衡状态．现将活塞缓慢向上顶,直到C 管中的水全部被顶到上面的管中, 环境温度不变．已知大气压强p 0＝1.0×105Pa ．g 取10m/s 2，33=110/kg m 水ρ⨯求：（1）判断水是否会从B 管溢出（2）此时A 管中的气柱的长度L 1′15．图示为用玻璃做成的一块棱镜的截面图，其中ABOD 是矩形，OCD 是半径为R 的四分之一圆弧，圆心为O ．一条光线从AB 面上的某点入射，入射角θ1＝45°，它进入棱镜后恰好以临界角射在BC面上的O点，光路图如图所示，求：①求该棱镜的折射率n；②求光线在该棱镜中传播的速度大小v（已知光在空气中的传播速度c＝3.0×108m/s）．五、填空题16．为一列简谐横波在t=0 时的波形图，P 是平衡位置在x=0.5m 处的质点，Q 是平衡位置在x=2.0m 处的质点；图（b）为质点Q 的振动图象．下列说法正确的是（____）A．这列简谐波沿x 轴正方向传播B．这列简谐波沿x 轴负方向传播C．波的传播速度为20m/sD．从t=0 到t=0.25s，波传播的距离为50cmE．在t=0.10s 时，质点Q的加速度方向与y 轴正方向相同参考答案1．C【解析】【详解】 火星探测器和火星的卫星绕火星做圆周运动所需的向心力是由火星的万有引力提供的，则02020Mm v G m R R =探探 ① ，22Mm v G m R R卫卫= ② 由①②得：2020R v v R = ③，取对数得：002lg lg R v v R ⎛⎫⎛⎫= ⎪ ⎪⎝⎭⎝⎭，即001lg lg 2R v v R ⎛⎫⎛⎫= ⎪ ⎪⎝⎭⎝⎭，故选C ．【点睛】本题主要考查万有引力的应用．2．D【解析】 试题分析：磁场垂直向外和向里时，由于电流与磁场方向平行，故安培力为零，故不可能平衡，故B 错误；磁场方向竖直向上时，根据左手定则，安培力水平向右，导体棒还受到重力和支持力，三个力不可能平衡，故A 错误；磁场方向沿斜面向上时，安培力垂直斜面向上，则重力、支持力和安培力也不可能平衡，选项C 错误；磁场竖直向下时，安培力水平向左，与重力、弹力平衡，有tan BIL mg θ=，解得tan mg B ILθ=故D 正确；故选D ．考点：左手定则；安培力；物体的平衡3．C【详解】设R 1、R 2、R 3、R 4的电流分别为I 1、I 2、I 3、I 4，电压分别为U 1、U 2、U 3、U 4．干路电流为I 总，路端电压为U ，电流表电流为I ．A ．当滑动变阻器R 2的滑动触头P 向右滑动时，R 2变大，外电路总电阻变大，I 总变小，由U =E -Ir 可知，U 变大，则电压表示数变大．U 变大，I 3变大，故A 错误；BC ．因I 4=I 总-I 3，则I 4变小，U 4变小，而U 1=U -U 4，U 变大，U 4变小，则U 1变大，I 1变大.又I 总=I +I 1，I 总变小，I 1变大，则I 变小．所以R 1两端的电压变大，电流表的示数变小．故B 错误，C 正确．D ．由I 4=I 1+I 2，I 4变小，I 1变大，则I 2变小，则|△I 1|＜|△I 2|，|△I 2|＞|△I 4|，则不能确定R 1中电流的变化量与R 4中电流的变化量的大小．故D 错误．【点睛】本题是电路的动态分析问题；解题时按“局部→整体→局部”的顺序进行分析，采用总量的方法分析电流表示数的变化．4．A【解析】【详解】光子的能量为h ν，根据爱因斯坦质能方程可知2h mc ν=，则质量2h m c ν=，故A 正确． 5．A【解析】 若两板间电压为U ，间距为d ，长为L ．则由题意：N M v t v t L +=①2211222N M N MUq Uq d t t dm dm +=②，由①式分析得v M 、v N 不一定相同，D 错误；由②式分析得：>N M M N q q m m ，A 正确，由>N M M Nq q m m 进一步分析可得两个电荷的加速度M N a a >，电场力对电荷所做的功M N W W >，BC 错误．6．AD【解析】试题分析：物体恰能匀速下滑，满足mgsin30°=μmgcos30°，解得θ，物体沿斜面匀速上滑，满足：Fcosθ=mgsinθ+μ（mgcosθ+Fsinθ），解得mgsin mgcos F cos sin θμθθμθ+-＝，当cosθ-μsinθ=0，F →∞， 解得01tan μθ＝，θ0＝600，故AD 正确；BC 错误；故选AD ．考点：物体的平衡【名师点睛】本题是力平衡问题，关键是分析物体的受力情况，作出受力的示意图，要培养良好的作图习惯，注意结合正交分解法列方程求解．7．BC【解析】试题分析：若虚线是电场线，粒子所受的电场力沿电场线向左，质子由a点运动到b点的过程中，电场力做负功，质子的电势能增大，动能减小，故质子在a点的电势能小，动能大，故A错误．若虚线是等差等势线，根据电场线与等势线垂直，可知电场力大致向下，质子由a点运动到b点的过程中，电场力对质子做正功，质子的电势能减小，动能增大，则质子在a点的电势能大，动能小，故B正确．电场线的疏密表示场强的大小，等差等势线越密，场强越大，则知a点的场强一定大于b点的场强，由牛顿第二定律得qE=ma，则质子在a 点的加速度一定大于在b点的加速度，故C正确．若虚线是电场线，电场线方向向左，b点的电势高于a点的电势．若虚线是等差等势线，电场线向下，a点的电势高于b点的电势，故D错误．故选BC．考点：带电粒子在电场中的运动8．BCD【解析】试题分析：以金属棒为研究对象分析受力可知，其受到恒力F、重力、安培力，由合外力做的功就为三力做功之和，有外力做功、克服重力做功mgh、克服安培力做的功（即电路产生的焦耳热），由能量守恒合功可知，所以选项BCD正确；考点：能量守恒、功、动能定理9．BCE【解析】布朗运动是悬浮在液体中固体颗粒所做的无规则运动，不是固体分子的运动，选项A错误；叶面上的小露珠呈球形是由于液体表面张力的作用，选项B正确；液晶显示器利用了液晶对光具有各向异性的特点，选项C正确；当两分子间距离大于平衡位置的间距r0时，分子间的距离越大，分子势能越大，选项D错误；温度升高时，分子热运动的平均动能一定增大，但并非所有分子的速率都增大，选项E正确，故选BCE．10．22()2M m b t + ()2M m gd - 9.6【详解】（1）滑块从A 处到达B 处的速度:bv t＝则系统动能的增加量：()()222122k M m b E M m v t++=＝ 系统重力势能的减小量：△E p =mgd-Mgdsin30°=（m−2M）gd （2）根据系统机械能守恒的：12（M+m ）v 2＝（m−2M）gd则：()22m M gdv M m-+＝图线的斜率：2 2.40.5m M k g M m -=+＝解得：g=9.6m/s 211．a 很大 1.0×103 2.8 64 【详解】（1）测绘伏安特性曲线，电压和电流需从零开始测起，滑动变阻器采用分压式接法，应选图a 所示实验电路． （2）由U I R=，得电阻阻值U R I =，由图乙所示图象可知，在电压小于2.00V 时，电流I很小，所以太阳能电池的电阻很大．由图丙所示图象可知，当电压为2.80V 时，电流I=2.80×10-3A ，电阻332.80V 1.010Ω 1.00k Ω2.8010AU R I -===⨯=⨯． （3）由图可知，图象与纵坐标的交点为电源的电动势，故电动势为2.8V ；若与1kΩ的电阻连接构成一个闭合电路；在U I -图中作出对应的电阻的伏安特性曲线，如图所示；图象的交点为电源的工作点，则由图可知电源的工作电压为1.8V ，则电源的效率1.8100%64.3%2.8η=⨯= ； 12．（1）见解析（2）4s （3）见解析 【解析】 【分析】判断物体是否能够一起运动，要通过判断速度和加速度来判断．初速度相同的情况下，加速度相同才能够一起运动，由此能够得出本题的力的关系． 【详解】（1）设木板 A 与物块 C 之间的滑动摩擦力大小为1f F木板 A 与水平地面之间的滑动摩擦力大小为2f F 有 110.4f F Mg Mg μ==()220.2f F Mg Mg Mg μ=+=设AC 能够保持相对静止的最大拉力为1F 则对A 有：12f f A F F Ma -= ，解得0.2A a g = 则0.2C A a a g == 对C 有：11f C F F Ma -= 解得10.6F Mg =可见210.3f F F Mg F <=<故可知在木板 A 、B 相碰前，在 F 的作用下，木板 A 与物块 C 一起水平向右做匀加速直线运动．（2）设 A 与 C 一起向右匀加速过程中它们的加速度为 a ，运动时间为 t ，与木块 B 相碰时的速度为 v ，有()2f F F M M a -=+ ，212x at = ，v at = ， 解得4t s = ，2/v m s =（3）碰撞后瞬间，物块 C 的速度不变，设 A 、 B 碰后速度为 v '， 由动量守恒定律有：2Mv Mv ='解得木板 A 、 B 共同运动的初速度 2v v '=此后，物块 C 在木板上滑动时的加速度为121/f C F F a m s M-==-物块 C 在木板上滑动时，木板 A 、 B 共同运动的加速度为122f f ABF F a M'-=其中()2220.3f F Mg Mg Mg μ=+='解得20.5/AB a m s =若木板 A 、B 很长，则物块 C 不会掉下来，设物块 C 再运动时间 1t 后，三者的速度相同，有112C AB vv a t a t +=+，解得123t s = 在此过程中，物块 C 的位移为 211111029C C x v t a t m =+=木板 A 、 B 的位移为21117229AB AB v x t a t m =+=由于120.43AB C x x m L m -=<=可见，物块 C 与木板 A 、B 达到共同速度时还在木板上． 进一步分析，由于20.3f f F F Mg F '==<可知达到共同速度后物块 C 将与木板 A 、B 一起做匀速直线运动 可见物块 C 将不会从木板上掉下来 【点睛】本题主要考查牛顿运动定律与运动学、动量的综合应用题． 13．（1）（2）（3）min (13)29m t LeU π=+ 【解析】试题分析：（1）电子在经过加速电场过程中，根据动能定理可得① （2分）由题意可知在偏转电压出现峰值时进入的电子恰好沿极板边缘飞出电场② （1分）③（1分）④（1分）联立上式可得⑤（1分）（2）设在时刻进入偏转电场的电子离开电场时速度大小为v，v与v0之间夹角为θ，⑥所以θ=300 （2分）v0=vcosθ ⑦（1分）电子垂直进入磁场洛伦兹力充当向心力⑧（2分）根据几何关系⑨（1分）解得⑩（3）电子在偏转电场中运动历时相等，设电子在磁场中圆周运动周期为T，经N板边缘飞出的电子在磁场中运动时间最短，在磁场中飞行时间为联立①④⑦⑨可得min03L Ttv=+（2分）（其它表示方法同样得分）min t =+ （结果正确的其它表示方法同样得分）（2分） 考点：考查了带电粒子在电磁场中的运动 14．（1）会溢出 （2）3m 【解析】试题分析：设A 管内的水柱上升x ，此时水能到达B 的顶端，则对A 内气体，根据玻意耳定律：01()P LS P L x S =-，其中10()P P g L x ρ=+- 代入数据解得：x=0．9m因为0．9m+3．9m<5．2m ，故此时水会从B 中溢出，此时A 管中的气柱的长度3．9m-0．9m=3．0m 考点：玻意耳定律：【名师点睛】此题是对气体状态方程的考查；解题时采用假设的思想，通过分析A 气体的状态参量判断水柱能否溢出． 15．①2②×108m/s 。