六盘山三模试卷
2020年宁夏六盘山高级中学高三英语三模试卷及答案解析
2020年宁夏六盘山高级中学高三英语三模试卷及答案解析第一部分阅读(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项AInformation on school visits to Kew GardensEnjoy yourselves in a wonderland of science with over 50,000 living plants and a variety of educational events or amusing activities. Here is essential information about planning a school visit to Kew.Educational course pricesYou can plan a self-led visit or book one of our educational courses. Students will take part in the educational courses in groups of 15. Prices vary according to different situations.EYFS (Early Years Foundation Stage) to Key Stage 4:45-minute course: 35/group 90-minute course: 70/groupKey Stage 5:Half day (one course): 80/group Full day (two courses): 160/groupTeachers and adults:Up to required key stage proportions (比例): FreeAdults needed for 1:1 special educational needs support: FreeAdults above the required proportions: 11/personThe payment will due within 28 calendar days of making the booking.Health and safetyRequired supervising (监护) adult-student proportions:Key Stage 1: 1:5 Key Stage 2: 1:8Key stage 3: 1:10 Key Stage 4: 1:12Key Stage 5: 1:12The group sizes should be controlled if you are visiting potentially busy areas such as the glasshouse and other attractions. The maximum number of students visiting the glasshouses is 15 per group and each group to Kew shops should include no more than 10 students.If there is an emergency, please contact the nearest Kew staff member or call Constabulary on 0208 32 3333 for direct and quick support. Please do not call 999.Planning your visitYour tickets and two planning passes will be sent to you upon receipt of your payment. You can complete your risk assessment with the passes, ensure you bring your tickets and the receipt document and show them to the staff members at the gate on the day of your visit.Recommended timingsThe Kew Gardens opens at 10 am. You are recommended to spend at least three to five hours on your visit. The closing time varies throughout the year. But the earliest is 3:30 pm. We have a fixed schedule for educational courses, which is from 10:30 am to 2:20 pm.1.How much should a group of 15 Key Stage I students and 4 teachers pay for a 45-minute course?A.35B.46C.57D.812.What should one do in an emergency?A.Check the risk assessment.B.Call 999 immediately.C.Ask adults or teachers for help.D.Seek help from the staff member nearby.3.What is the purpose of the text?A.To introduce Kew Gardens.B.To give tips on visiting Kew Gardens.C.To attract potential visitors to Kew Gardens.D.To inform coming activities in Kew Gardens.BImagine the feeling of swinging at a baseball going 100 miles per hour—without leaving your living room, or being in race car as it roars down the track, while you are sitting on the couch.These are just some of the ways that sports business leaders say virtual reality (VR) will revolutionize how people train for and experience sports. Virtual and augmented(增强的)realities are together known as mixed reality (MR). “American footballers are already using VR to better train their minds andread the field,” Ludden said. “This can allow players to perfect their skills without risking injury.”Canadian company D-BOX Technologies designs and produces moving seats found in cinema and theme parks. It is now moving into sports, and shows its Formula One (F1) racing simulator(模拟器). The seats stimulate the force of gravity, speed and every shaking as Fl champion Lewis Hamilton zips around city streets.A simulation seat uses pre-programed data now. Someday, though, it could use real-time information sent by the car. “They couldbroadcast live content through a network in pop-up theaters around the world,” Ludden said. Say you want to experience the true stress of a batter being up against major-league baseball pitcher. “You can have a heartbeat added to the sensation on the seat and then you can feel it, boom, boom,” Maheu explained. “When he swings and hits the ball, you can have an impact.”One day, fans around the world could physically experience every game from their favorite player in real time. Ludden said that current and near-future technology could create “augmented stadiums” for live audiences. Panasonic launched its “Smart Venue” plans which included the overlaying of graphics, advertisements, player statistics and replays on the field of play at a pro football game. “If you are seated in the cheap seats, you can see this really useful.” “Fans may someday join in stadium wide games, using the field as a virtual gaming platform,” Ludden added.4. What does the underlined phrase “read the field” in paragraph 2 mean?A. Get off the playing field.B. Build up a football court.C. Judge the situation on the field.D. Ask players to play on the spot.5. What does Maheu think audiences can do in the future baseball game?A. Enjoy live content in any theater.B. Program the simulation seats in advance.C. Control the force and speed of the baseball.D. Experience the real time game with the player.6. What does Ludden mainly describe in the last paragraph?A. The origin of VR.B. A future stadium.C. An advertisement platform.D. The expectations of audiences.7. What is the main idea of the text?A. VR can improve players' skills for sports.B. VR increases fans' joy in the baseball game.C. VR can improve sports experience for players and fans.D. VR promises a new future for football players and games.CMany teens may feel anxious sometimes. It’s the kind of nervousness that makes you bite your nails before a big test. We spend more time online than we should. We feel good about ourselves or bad based on how manyLikes and Followers we get on social media. Young people are developing a false view of life.On the screen, we see what people want to show us. People usually only post photos where they are looking their best. They are surrounded by friends and seem that they are having a great time. No one seems sad or lonely. In short, life isfabulous. But sooner or later, our young people compare their real life to it. They find that theirs doesn’t seem as fun or exciting and grow worried that they may be missing out.No wonder teachers are reporting more anxious students. It’s reported that a lot more college students feel ―overwhelming anxiety. The percentage jumped from 50% in 2011 to 62% in 2016. Anxiety is now the most common mental-health problem in my country. It affects nearly one-third of teens and adults.Certainly, we can’t blame it on social media alone. We expect toomuch from our children and a lot of these expectations aren’t reasonable. Their schedules are packed with sports, clubs and homework. They don’t have enough free time. We want our children to succeed, and we don’t care how much it costs.As parents, we must have more balance. On one hand, we push too hard, and on the other hand, we make life too easy for children. We shouldn’t and can’t promise our children that they will always be happy. We shouldn’t try to protect them from the problems of everyday life. Let them solve the problems in person.8. What is the text mainly about?A. What causes teens’ nervousness.B. How to deal with teens’ anxiety.C. What a view of life means to people.D. How to treat social media appropriately.9. What does the underlined word “fabulous” in paragraph 2 mean?A. Wonderful.B. Encouraging.C. Anxious.D. Doubtful.10. Why does the author mention the numbers in paragraph 3?A. To draw teachers’ attention.B. To show teachers’ mental problems.C. To present the seriousness of teens’ anxiety.D. To show adults have more problems than teens.11. What should parents do to help their children out?A. Try to meet their expectations.B. Help them with their homework.C. Give them more free time to play sports.D. Allow them to solve their own problems in life.DWhat about your emotions? How do they help you to understand what you are reading?In Jane Yolan'sOwl Moon,a girl explores with her father on a snowy night. She longs for this special night.And she's amazed when she sees an owl.Have you felt longing before? Amazement? Well, if you have, it helps you have a sense of agreement. When we can put ourselves inside a story we can understand it better. Our brain tells us,"Oh, this girl's experience is a bit like mine."And boom! We can relate to her.But this skill is not born in us.So young kids have to learn it.Pictures and images help young readers to understand and recognize feelings.Readers feel joy when seeing the smiling faces of friends.They feel fear when turning the page to find a scary monster. They are just pictures,but the feelings are real. This skill, to understand the thinking and feeling of others, is what researchers call "theory of mind".For example, think about the faces of people and animals in stories. The Big Bad Wolf's scary teeth. The 'o' shape of a surprised character's mouth. Or big, wide eyes like the girl inOwl Moon.By noticing the faces,readers can start to figure out what it feels like to be that character.And that helps to figure out how people feel and think in real life.But most young readers don't go into deep,scary woods.And some may not go to the beach or play basketball.In the bookYo!Yes?two kids meet and play ball. The kids start the story on opposite pages.But as the story goes on,they get closer until they are together.Some young readers might not play basketball, but they can read the clues on the page to figure out how the kids are feeling.And some readers might not like the game,but they can feel excited for the characters because of how the characters look and move.12. What can we learn from the second paragraph?A.Owl Moontalks about the exploration of the Moon.B. The feeling of"amazement"is a sense of agreement.C. Readers are blessed with the ability to interpret others.D. Readers with similar experience understand the story better.13. Which of the following might help readers develop"theory of mind"?A. True feelings.B. Smiling faces.C. Picture books.D. Reading skills.14. Young readers can go deeper into a story by________.A. tracking the plotB. developing new skillsC. sharing similar hobbiesD. analyzing the background15. What is the text mainly about?A. How thinking influences reading.B. How readers improve reading skills.C. How emotions help enhance reading.D. How kids figure out the clues of stories.第二节(共5小题;每小题2分,满分10分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。
宁夏六盘山高级中学高三第三次模拟考试语文试题 扫描含答案
宁夏六盘山高级中学2016-2017学年第二学期高三期中试卷参考答案1、B.(“因其有损于鲁迅的形象”有误,原文受到激烈批评的应该是鲁迅作品,或者作品中的人物形象)2、D.“严重误导青少年”在文中指的是商家对鲁迅作品深刻意义的瓦解和对鲁迅形象的歪曲。
3、A.“所有文学作品中的人物形象”有误,如《西游记》《三国演义》等改变为游戏并未构成与原著完全相反的观感。
4、C D(A.“通过相互善意的问候”有误,女孩并没有问候男人。
B.“隐藏在美丽外表下的丑恶灵魂”有误。
文中只是对男人关心女孩的一种本能反应,说法有点过头。
E.小说没有运用象征手法。
)5、①敏感谨慎。
②孤独无助(伤心)。
③心地善良(善解人意)④渴望父爱(答出两点,并作简要分析即可得分)6、①在情节方面,小说以“暖”为线索,温暖贯穿全篇,女孩渴望温暖,男人的体温给寒冷中的女孩传递了一份温暖,读者从其中感受到了人间温暖。
②在人物塑造方面,男人传递温暖,女孩重新唤起她对亲人的思念和对未来美好生活的向往,让我们看到了男人美好的心灵和女孩面对生活的勇气。
③在小说主题方面,善解人意的旅人,清澈如水的美好缘份,暖如春阳的娓娓交心,无不泛起一份人间温情,小说以“暖”为题,让读者的心灵得到一次融化与洗礼,深化了小说主题。
(大意对,有简要分析即可得分)7、D(A“生活节俭”无此意。
B “对载人航天器的研究”有误,原文是对火箭的研究。
C “因为火箭发射失败”有误,原文是“本来发射正常的火箭被风吹倒了”)8、①王希季在没技术,没资料,没外援,只有人的情况下,靠土办法,摸着石头过河,既无前人经验,也无现实把握,在十分艰苦的环境中开始了火箭研究。
②随着中国航天事业完成了从无到有再到强的奇迹蜕变,王希季也从火箭的“门外汉”成为航天界泰斗。
③中国18种探空火箭中,有12种是由王希季担任型号负责人研制出来的。
他的创业团队在60年代的生物火箭上做过狗的试验,为中国的载人航天研究节省了很多时间。
宁夏六盘山高级中学2020届高三第三次模拟考试文科综合试题与答案
地理部分答案一.选择题:1C 2B 3B 4D 5C 6D 7B 8A 9C 10A 11C二.综合题36.(1)国土面积小,发展电信产业的建设成本低;(2分)资金雄厚,经济实力强;(2分)国家政策扶持,积极推动该产业发展;(2分)居民收入水平高,购买力强(2分)(2)国家狭小,自然资源匮乏;(2分)第一第二产业基础薄弱,工业体系不完整(2分);地理位置优越,交通便利,海外市场广阔(2分);第三产业的投资小、效益好、就业容量大(2分)(3)有利于拓展欧洲市场;(2分)打造民族品牌,提升国际市场竞争;(2分)加强国际间的交流与合作,提高国际地位;(2分)利于提高生产技术,进行产业升级(2分)37.(1)位于城郊,对主城区人口密集处影响小,且有绿化带、湖泊等缓冲隔离;临近湖泊,风景优美,利于病人康复;院区占地面积大,郊区后备土地资源充足;临近主干道,交通便利,利于建材运输。
(8分)(2)我国及武汉市拥有雄厚的工业基础;武汉市交通、通信等基础设施完善,有利于人员、物资、信息流通;拥有先进的设计及建筑技术;国家政策的大力支持;全民参与抗疫的热情和决心。
(任答3点得6分,其他答案合理也可酌情给分)(3)远离湖北主疫区,交通不便;藏历新年比春节晚近一个月,此时返藏人员较少;冬季为旅游淡季,出行西藏的外地游客较少;人口密集程度小,不利于病毒传染扩散。
(8分)选做题43.选修3:旅游地理困难:高原反应;气温降低;滑坡、泥石流等自然灾害;野生动物威胁(每点1分,4分)意义:有利于增加就业机会,带动了沿线经济的发展;带动酒店、餐饮等相关产业的发展;促进了不同区域的文化交流;绿色低碳的出行方式有利于保护环境;有利于增强国民素质,提升生活质量。
(3点即可,6分)44.选修6:环境保护影响:食物来源减少;栖息地缩小并割裂,种群基因交流减少,种群退化,数量下降(4分)措施:建立亚洲象自然保护区;在各孤岛之间的建设生态廊道;禁止毁林开放,合理规划各种工程建设;加强生态教育,建立保护基金。
宁夏银川市六盘山高级中学2023届高三三模数学(文)试题
一、单选题二、多选题1.曲线表示( )A .椭圆B .双曲线C .抛物线D .圆2. 已知集合,,则A.B.C.D.3. 已知,则的最小值是( )A .2B.C.D .34.若为数列的前项和,且,则( )A.B.C.D .305. 三人制足球(也称为笼式足球)以其独特的魅力,吸引着中国众多的业余足球爱好者.在某次三人制足球传球训练中,队有甲、乙、丙三名队员参加,甲、乙、丙三人都等可能地将球传给另外两位队友中的一个人.若由甲开始发球(记为第一次传球),则第四次仍由甲传球的概率是( )A.B.C.D.6.若某单位员工每月网购消费金额(单位:元)近似地服从正态分布,现从该单位任选10名员工,记其中每月网购消费金额恰在500元至2000元之间的人数为,则的数学期望为( )参考数据:若随机变量服从正态分布则,则,,.A .2.718B .6.827C .8.186D .9.5457. 柏拉图多面体是因柏拉图及其追陮者对正多面体的研究而得名.如图是棱长均为的柏拉图多面体,点,,,分别为,,,的中点,则异面直线与所成角的余弦值为()A.B.C.D.8.已知数列满足,,若表示不超过的最大整数,则( )A .1B .2C .3D .59.已知函数,则在有两个不同零点的充分不必要条件可以是( )A.B.C.D.10.已知曲线,且,则下列结论正确的是( )A.若曲线为椭圆或双曲线,则其焦点坐标为(,0)宁夏银川市六盘山高级中学2023届高三三模数学(文)试题宁夏银川市六盘山高级中学2023届高三三模数学(文)试题三、填空题四、解答题B.若曲线是椭圆,则C .若且,则曲线是双曲线D .直线与曲线恒有两个交点11.如图,在正四棱柱中,,E ,F ,N分别是棱,,的中点,则()A.B .直线BE 与平面相交C .平面D .直线NC与平面的交点是的重心12.已知等比数列的公比为,且,则下列选项正确的是( )A.B.C.D.13.函数,当时,的零点个数为_____________;若恰有4个零点,则的取值范围是______________.14.的展开式中,的系数是的系数与的系数的等差中项.若实数,那么___________.15.已知正项等比数列的前项和为,公比为,若,则的值为___________.16.已知正方形的相对顶点和,求顶点和的坐标.17.在中,,,且,再从条件①、条件②中选择一个作为已知.条件①:;条件②:.(1)求b 的值;(2)求的面积.18. 如图,在四棱锥中,为等边三角形,边长为2,为等腰直角三角形,,,,平面平面ABCD.(1)证明:平面PAD ;(2)求平面PAD 与平面PBC 所成锐二面角的余弦值;(3)棱PD 上是否存在一点E,使得平面PBC ?若存在,求出的值;若不存在,请说明理由.19. 已知函数.(1)若存在极值,求的取值范围;(2)若,已知方程有两个不同的实根,,证明:.(其中是自然对数的底数)20. 如图,已知三棱柱中,侧面底面为等腰直角三角形,.(1)若O为的中点,求证:;(2)求直线与平面所成角的正弦值.21. 近年来,各平台短视频、网络直播等以其视听化自我表达、群圈化分享推送、随时随地传播、碎片化时间观看等特点深受人们喜爱,吸引了眼球赚足了流量,与此同时,也存在功能失范、网红乱象、打赏过度、违规营利、恶意营销等问题.为促使短视频、网络直播等文明、健康,有序发展,依据《网络短视频平台管理规范》《网络短视频内容审核标准细则》等法律法规,某市网信办、税务局、市场监督管理局联合对属地内短视频制作、网络直播进行审查与监管.(1)对短视频、网络直播的整体审查包括总体规范、账户管理、内容管理等三个环节,三个环节均通过审查才能通过整体审查.设某短视频制作团队在这三个环节是否通过审查互不影响,且各环节不能通过审查的概率分别为.①求该团能通过整体审查的概率:②设该团队通过整体审查后,还要进入技术技能检测环节,若已知该团队最终通过整体审查和技术技能检测的概率为35%,求该团队在已经通过整体审查的条件下通过技术技能检测的概率;(2)某团队为提高观众点击其视频的流量,通过观众对其视频的评论分析来优化自己的创作质量,现有100条评论数据如下表:对视频作品否满意时间合计改拍前视频改拍后视频满意285785不满意12315合计4060100试问是否有99.9%的把握可以认为观众对该视频的满意度与该视频改拍相关程度有关联?参考公式:,0.10.050.010.0050.0012.7063.8416.6357.87910.828不。
2023届宁夏银川市六盘山高级中学高三三模语文试题
2023届宁夏银川市六盘山高级中学高三三模语文试题学校:___________姓名:___________班级:___________考号:___________一、论述类文本阅读阅读下面的文字,完成下面小题。
在传统语境中,“才子”一般用于称赞富有文学才华之人,又为文学批评家所常用,已经成为中国文学批评的一个重要范畴。
在多数情况下,“才子”之“才”均作“文才”理解,鲁迅将明代人情小说分出“才子佳人”一类,并指出其中“所谓才者,惟在能诗”就是典型的例子。
但是“才子”一词最早出现时,“才”却并非专指“文才”。
《左传·文公十八年》载高阳氏和高辛氏各有“才子八人”,合称“八元八恺”,是传世文献中“才子”概念的最早用例。
《左传》中“八元八恺”的“才”并非在文学方面,而是表现为“齐圣广渊,明允笃诚”“忠肃共懿,宣慈惠和”的德行与政才。
此外,按照杜预注,这里的“子”也不是男子的通称,而是“苗裔”之意。
结合上下文,《左传》中“才子”之义可以理解为“某家族中具有政治德行之才的苗裔”,与后世“才子”的义涵有较大差别。
在汉代,人物品鉴中多见“才子”的称呼,其含义基本与《左传》相同。
那么,从何时开始,“才子”与《左传》中的原义脱离了关系呢?这最早可追溯到西晋潘岳的《西征赋》。
潘岳作《西征赋》述行怀古,历数西汉名臣,其中特别举出“终童山东之英妙,贾生洛阳之才子。
”贾谊出身甚微,家世无闻,不符合“世家子弟”的条件,此处将贾谊称为“才子”,显然与《左传》中的意思有所区别。
不过,《西征赋》中“才子”的用法并未完全脱离政才的范畴,更非偏指文才,真正使“才子”义涵转向“文才”的关键,是《西征赋》将“才子”与“贾谊”绑定的做法。
到了南朝以后,贾谊则越来越多地被当作知名文人来看待,他的“才”也越来越多地被指向文才。
在这样的语境下,用以形容贾谊的“才子”之“才”,也开始指向“文才”。
南朝以“才子”表示“文才之士”的例子中,最令人瞩目的是梁武帝对“才子”评语的频繁使用。
宁夏六盘山高级中学文综三摸答案
三模地理答案一、选择题(44分)(2)乙(2分)甲处海拔高,距离上水库近,水流势能小;丙处和丁处为抽水管线,不能发电;乙处与上水库高差大,水流势能大,发电作用强。
(12分)(3)其他事业 (或. 生活用电)(2分)节约电能,提高利用率;降低成本;稳定电压,提高电网的稳定性和可靠性(6分)。
37.(20分)(1)大于200毫米。
(2分)城市“热岛效应”使近地面气流上升,降水较多;城市大气中悬浮颗粒物较多,凝结核多,降水较多。
(4分)(2)银川平原,地势平坦,土壤肥沃;灌溉水源充足;阴雨天气少,光热较充足,气温日较差大,有利于农作物养分的积累。
(8分)(3)增加空气湿度,改善大气环境。
(2分) 60年代主要是随人口增长要解决吃饭问题,围湖造田发展农业;90年代城市建设加快,围湖造陆。
(4分)42.(10分)(1)坡度15°左右滑坡发生的可能性最大;(2分)砂质土壤发生滑坡的可能性最大。
(2分)(2)封山育林;植树造林;岩土体改造工程、疏排水工程,加固稳定变形土体;加强监测与预警预报;加强宣传教育,提高防灾减灾意识等。
(6分)43.(10分)(1)工业废水、生活污水。
(4分)(2)水污染综合指数降低,水质渐好;(2分)原因是工业用地比例降低,工业废水排放减少;城市绿地比例增加,城市地表径流减小;污水处理量增大;人们的环境保护意识加强等。
(4分)历史第三次模拟答案选择题24 C 25 B 26D 27B 28 B 29 D 30 B 31D 32 D 33 C 34D 35A40.(37分)(1)(7分)特点:精耕细作,成就辉煌(1分);小农经济,分散经营(1分)经济资源协调能力差(1分)。
影响:积极方面:推动了农业技术的提高,促进了农业和社会经济发展;有利于社会稳定,巩固了封建统治。
(2分)消极方面:规模小、经营分散的小农经济不利于资源的合理使用;缺乏抵抗自然灾害的能力;限制了生产技术的进一步提高。
高三第三次模拟考试试题 试题 (2)
宁夏六盘山高级中学 2021届高三第三次模拟考试试题制卷人:打自企;成别使;而都那。
审核人:众闪壹;春壹阑;各厅……日期:2022年二月八日。
语文本套试卷分第I卷〔阅读题〕和第II卷〔表达题〕两局部,其中第I卷第三、四题为选考题,其他题为必考题。
考生答题时,将答案答在答题卡上,在套本套试卷上答题无效。
在在考试完毕之后以后,将本套试卷和答题卡一起交回。
考前须知:1、在答题之前,所有考生必须先将本人的姓名、准考证号填写上在答题卡上,认真核对条形码上的姓名、准考证号,并将条形码粘贴在答题卡的规定的正确位置上。
2、答题时使用毫米的黑色中性〔签字〕笔或者碳素笔书写,字体工整、笔迹清楚。
3、请按照题号在各题的答题区域〔黑色线框〕内答题,超出答题区域书写之答案无效。
4、保持卡面清洁,不折叠,不破损。
5、做选考题时,考生按照题目要求答题,并需要用2B铅笔在答题卡上把所选的题目对应的题号涂黑。
第一卷阅读题甲必考题一、现代文阅读〔9分,每一小题3分〕阅读下面的文字,完成1~3题。
崇高作为一种庄严的美和雄浑的美,与优美的区别是非常明显的。
崇高与优美一样都是人的本质力量在对象世界的感性显现,其区别在于,优美表达了人的本质力量与客体在对象世界中的和谐统一,崇高那么表达了这种本质力量与客体在对象世界中的矛盾冲突的统一。
崇高是以力量、气势见胜的,属于阳刚美、动态美;优美那么以静态的气韵见长,属于阴柔美、静态美。
如自然界中的暴风骤雨、崇山峻岭等,都给人以崇高感;而春光明媚、山清水秀那么给人以优美感。
前者以力量、气势见胜,后者以神采、气韵见胜。
而且优美的事物都具有较强的形式美,它常表现为光滑、娟秀、匀称。
而崇高的事物往往不拘泥于形式美,它常常表现为凹凸不平、不匀称,甚至在形式上有时还表现出几分怪、几分丑。
除了在形式上优美与崇高有着明显的区别以外,在美感方面崇高感与优美感也是有区别的。
优美感侧重于平静和谐的愉悦,得到的是赏心悦目的快乐,即在感受优美的时候,我们的精神是通体愉快的,我们的心境是单纯一致的,我们的感情是松弛舒畅的。
宁夏银川市六盘山高级中学2023届高三三模数学(文)试题(含答案与解析)
宁夏六盘山高级中学2023届高三年级第三次模拟考试数学(文科)注意事项:1.答题前,考生务必将自己的姓名、准考证号填写在答题卡相应的位置,并将核对后的条形码贴在答题卡条形码区域内.2.选择题答案使用2B 铅笔填涂,非选择题答案使用0.5毫米的黑色中性(签字)笔或碳素笔书写,字体工整,笔迹清楚.3.做答时,务必将答案写在答题卡相应位置上,写在本试题上、超出答题区域或非题号对应区域的答案一律无效.4.考试结束后,将本试题和答题卡一并交回.一、选择题:本大题共12小题,每小题5分,满分60分.1. 已知集合{}4,3,2,1,0,1,2,3,4A =----,{}2B x x =>,则A B = ()A. {}4,3,3,4--B. ()(),22,∞∞--⋃+C. {}2,1,0,1,2--D. [2,2]-2. 已知12z i =-,且0z az b ++=,其中a ,b 为实数,则( ) A. 1,2a b ==-B. 1,2a b =-=C. 1,2a b ==D. 1,2a b =-=-3. 已知命题p :对任意x ∈R ,总有210x x -+≥;q :若22a b <,则a b <.则下列命题为真命题的是( ) A. p q ⌝∧B. p q ∧⌝C. p q ⌝∧⌝D. p q ∧4. 中国古代数学著作《九章算术》是人类科学史上应用数学的最早巅峰.书里记载了这样一个问题“今有女子善织,日自倍,五日织五尺.问日织几何?”译文是“今有一女子很会织布,每日加倍增长,5天共织5尺,问每日各织布多少尺?”,则该女子第二天织布( ) A.531尺 B.1031尺 C.1516尺 D.516尺 5. 若点M 是圆22:40C x y x +-=上的任一点,直线:20l x y ++=与x 轴、y 轴分别相交于A 、B 两点,则MAB ∠的最小值为( ) A.π12B.π4C.π3D.π66. 已知2log 5a =,5log 3b =,0.72c =,则( ) A. b a c <<B. a c b <<C. c a b <<D. b c a <<7. 明朝著名易学家来知德以其太极图解释一年、一日之象的图式,一年气象图将二十四节气配以太极图,说明一年之气象,来氏认为“万古之人事,一年之气象也,春作夏长秋收冬藏,一年不过如此”.如图是来氏太极图,其大圆半径为4,大圆内部的同心小圆半径为1,两圆之间的图案是对称的,若在大圆内随机取一点,则该点落在黑色区域的概率为( )A.12B.25C.716D.15328. 已知奇函数()f x 在R 上单调递增,且(1)2f =,则()2xf x <的解集为( ) A. ()0,1B. [)0,1C. ()1,1-D. ()1,0-9. 已知函数()π2sin 216f x x ⎛⎫=-- ⎪⎝⎭,以下说法中,正确的是( ) ①函数()f x 关于点π,012⎛⎫⎪⎝⎭对称; ②函数()f x 在ππ,66⎡⎤-⎢⎥⎣⎦上单调递增; ③当π2π,63x ⎛⎫∈⎪⎝⎭时,()f x 的取值范围为()2,0-; ④将函数()f x 的图像向右平移π12个单位长度,所得图像对应的解折式为()2sin21g x x =-. A. ①②B. ②③④C. ①③D. ②10. 如图,1F ,2F 分别为椭圆()222210x y a b a b+=>>的左、右焦点,点P 在椭圆上,2 POF是面积为e 的值是( )A.1-B.1-C.D. 4-11. 已知函数()()sin f x A x ωϕ=+的部分图象如图所示,其中π0,0,02A ωϕ->><<.在已知21x x 的条件下,则下列选项中可以确定其值的量为( )A.ω B. ϕC.φωD. sin A ϕ12. 已知集合{|()0}M f αα==,{|()0}N g ββ==.若存在M α∈,N β∈,使||n αβ-<,则称函数()f x 与()g x 互为“n 度零点函数”.若函数2()e 1x f x -=-与函数2()e x g x x a =-互为“1度零点函数”.则实数a 的取值范围为( ) A. 214,e e ⎛⎤⎥⎝⎦B. 2214,e e ⎛⎤⎥⎝⎦C. 242,e e ⎡⎫⎪⎢⎣⎭D. 3212,e e ⎡⎫⎪⎢⎣⎭二、填空题:本大题共4小题,每小题5分,共20分.13. 已知双曲线以两坐标轴为对称轴,且它的一个顶点为(2,0)A ,它的一条渐近线方程为12y x =,则双曲线的标准方程为________.14. 设向量()2,1a =r ,()1,b x =-,若()a b a ⊥- ,则b = ___________.15. 设n S 是等差数列{}n a 的前n 项和,27a =-,512S a =,当n S 取得最小值时,n =______. 16. 如图所示为某几何体的三视图,则该几何体外接球的表面积为____________.二、解答题:共70分,解答应写出文字说明、证明过程或演算步骤.第17~21题为必考题,每个试题考生都必须作答.第22、23题为选考题,考生根据要求作答. (一)必考题:共60分.17. 某电视台在一次对收看文艺节目和新闻节目观众抽样调查中,随机抽取了100名电视观众,相关的数据如下表所示:文艺节目 新闻节目 总计 20至40岁40 18 58 大于40岁 15 27 42 总计 5545100(1)由表中数据直观分析,收看新闻节目的观众是否与年龄有关?(2)用分层抽样方法在收看新闻节目观众中随机抽取5名,大于40岁的观众应抽取几名? (3)在上述抽取的5名观众中任取2名,求恰有1名观众的年龄为20至40岁的概率. 18. ()sin a C C =+;②sin sin2B C a C c +=;③1cos 2a C c b +=,这三个条件中任选一个,补充在下面问题中,然后解答补充完整的题目.在ABC 中,内角A ,B ,C 的对边分别为a ,b ,c .已知______.(1)求角A ;(2)若1b =,3c =,求BC 边上的中线AD 的长.注:若选择多个条件分别进行解答,则按第一个解答进行计分.19. 如图,在直角梯形ABCD 中,AD BC ∥,AD CD ⊥,四边形CDEF 为平行四边形,平面CDEF ⊥平面ABCD ,2BC AD =.(1)证明:DF 平面ABE ;(2)若1AD =,2CD ED ==,π3FCD ∠=,求三棱锥B ADE -的体积. 的的20. 已知0ab ≠,曲线23()xf x a x =-在1x =处的切线方程为630x by +-=.(1)求a ,b 的值;(2)证明:当(0,1]x ∈时,()tan f x x <.21. 已知中心为坐标原点,焦点在坐标轴上的椭圆C经过点M,)1N-.(1)求C 方程;(2)已知点()3,0D ,直线():3,0l x ty n n t =+≠≠与C 交于,A B 两点,且直线,DA DB 斜率之和为1t,证明:点(),t n 在一条定抛物线上. (二)选考题:共10分.请考生在第22、23题中选定一题作答,并用2B 铅笔在答题卡上将所选题目对应的题号方框涂黑.按所涂题号进行评分,不涂、多涂均按所答第一题评分;多答按所答第一题评分. [选修4-4:坐标系与参数方程]22. 在直角坐标系xOy 中,曲线C 的参数方程为1cos 1sin x y αα=+⎧⎨=+⎩(α为参数),直线:40l x y +--,以坐标原点O 为极点,x 轴的正半轴为极轴建立极坐标系. (1)求直线l 和曲线C 的极坐标方程;(2)若直线()0:R l θβρ=∈与直线l 相交于点A ,与曲线C 相交于不同的两点M ,N .求OM ON OA ++的最小值. [选修4-5:不等式选讲]23. 已知关于x 函数()223(R)f x x x x =-++∈. (1)求关于x 的不等式()7f x ≥的解集.(2)若函数()f x 的最小值为m 、且实数a ,b 满足222a b m +=,求2a b +的最大值.参考答案一、选择题:本大题共12小题,每小题5分,满分60分.1. 已知集合{}4,3,2,1,0,1,2,3,4A =----,{}2B x x =>,则A B = ()A. {}4,3,3,4--B. ()(),22,∞∞--⋃+C. {}2,1,0,1,2--D. [2,2]-【答案】A的的的【解析】【分析】求出集合B ,利用交集的定义可求得集合A B ⋂.【详解】因为{}{22B x x x x =>=<-或}2x >,{}4,3,2,1,0,1,2,3,4A =----, 因此,{}4,3,3,4A B =-- . 故选:A.2. 已知12z i =-,且0z az b ++=,其中a ,b 为实数,则( ) A. 1,2a b ==- B. 1,2a b =-=C. 1,2a b ==D. 1,2a b =-=-【答案】A 【解析】【分析】先算出z ,再代入计算,实部与虚部都为零解方程组即可 【详解】12z i =-12i (12i)(1)(22)i z az b a b a b a ++=-+++=+++-由0z az b ++=,结合复数相等的充要条件为实部、虚部对应相等,得10220a b a ++=⎧⎨-=⎩,即12a b =⎧⎨=-⎩故选:A3. 已知命题p :对任意x ∈R ,总有210x x -+≥;q :若22a b <,则a b <.则下列命题为真命题的是( ) A. p q ⌝∧ B. p q ∧⌝C. p q ⌝∧⌝D. p q ∧【答案】B 【解析】【分析】先判断命题p ,命题q 的真假,在判断选项的真假详解】由22131(024x x x -+=-+> 所以命题p 真命题令0,1a b ==-,则22a b <,但是a b > 所以命题q 为假命题 故p q ∧⌝为真 故选:B.4. 中国古代数学著作《九章算术》是人类科学史上应用数学的最早巅峰.书里记载了这样一个问题“今有女子善织,日自倍,五日织五尺.问日织几何?”译文是“今有一女子很会织布,每日加倍增长,5天共织5尺,问每日各织布多少尺?”,则该女子第二天织布( )【为A.531尺 B.1031尺 C.1516尺 D.516尺 【答案】B 【解析】【分析】由题得每日织布尺数成公比为2的等比数列,根据前5项和得第二天织布数. 【详解】由题,设每日织布数的数列为{}n a ,则{}n a 为以2为公比的等比数列,由题知51(12)512a -=-,得1531a =,所以第二天织布尺数为251023131a =⨯=. 故选:B.5. 若点M 是圆22:40C x y x +-=上的任一点,直线:20l x y ++=与x 轴、y 轴分别相交于A 、B 两点,则MAB ∠的最小值为( ) A.π12B.π4C.π3D.π6【答案】A 【解析】【分析】作出图形,分析可知当直线AM 与圆C 相切,且切点位于x 轴下方时,MAB ∠取最小值,求出OAB ∠、CAM ∠的大小,可求得MAB ∠的最小值.【详解】如下图所示:直线l 的斜率为1-,倾斜角为3π4,故π4OAB Ð=,圆C 的标准方程为()2224x y -+=,圆心为()2,0C ,半径为2r =, 易知直线l 交x 轴于点()2,0A -,所以,4AC =,由图可知,当直线AM 与圆C 相切,且切点位于x 轴下方时,MAB ∠取最小值, 由圆的几何性质可知CM AM ⊥,且122CM AC ==,则π6CAM ∠=, 故ππππ64612MAB OAB ∠≥∠-=-=. 故选:A.6. 已知2log 5a =,5log 3b =,0.72c =,则( ) A. b a c << B. a c b <<C. c a b <<D. b c a <<【答案】D 【解析】【分析】容易得出0.725log 52log 31122><<<,,,从而得出a ,b ,c 的大小关系.【详解】225log log 42a =>= ,55log 3log 51b =<=,00.7122212<<==,即12c <<,所以b<c<a . 故选:D .7. 明朝著名易学家来知德以其太极图解释一年、一日之象的图式,一年气象图将二十四节气配以太极图,说明一年之气象,来氏认为“万古之人事,一年之气象也,春作夏长秋收冬藏,一年不过如此”.如图是来氏太极图,其大圆半径为4,大圆内部的同心小圆半径为1,两圆之间的图案是对称的,若在大圆内随机取一点,则该点落在黑色区域的概率为( )A.12B.25C.716D.1532【答案】D 【解析】【分析】求出大圆,小圆面积,进而求出阴影部分面积,利用几何概型求概率公式得到答案.【详解】设大圆面积为1S ,小圆面积2S ,则21π416πS =⨯=,22π1πS =⨯=.则黑色区域的面积为()12115π22S S ⨯-=,所以落在黑色区域的概率为()121115232S S P S -==, 故选:D8. 已知奇函数()f x 在R 上单调递增,且(1)2f =,则()2xf x <的解集为( ) A. ()0,1 B. [)0,1C. ()1,1-D. ()1,0-【答案】C【解析】【分析】判断()()F x xf x =奇偶性和单调性,由此求得不等式的解集. 【详解】令()()F x xf x =, 依题意()f x 是R 上递增的奇函数,所以()()()()F x xf x xf x F x -=--==,即()F x 为偶函数, 任取120x x >>,则()()()1200f x f x f >>=, 则()()1122x f x x f x >,所以()()()()1211220F x F x x f x x f x -=->, 故()F x 在()0,∞+上递增,在(),0∞-上递减,由于(1)2f =,所以()()()()2()111xf x xf x f F x F <⇔<⋅⇔<, 所以11x -<<.所以()2xf x <的解集为()1,1-. 故选:C【点睛】判断函数的奇偶性,可根据奇偶性的定义,判断()()f x f x -=-或()()f x f x -=来确定. 9. 已知函数()π2sin 216f x x ⎛⎫=-- ⎪⎝⎭,以下说法中,正确的是( ) ①函数()f x 关于点π,012⎛⎫⎪⎝⎭对称; ②函数()f x 在ππ,66⎡⎤-⎢⎥⎣⎦上单调递增; ③当π2π,63x ⎛⎫∈⎪⎝⎭时,()f x 的取值范围为()2,0-; ④将函数()f x 的图像向右平移π12个单位长度,所得图像对应的解折式为()2sin21g x x =-. A. ①② B. ②③④C. ①③D. ②【答案】D 【解析】【分析】利用正弦函数的性质,解决函数图像的对称中心、单调区间、值域和平移问题. 【详解】由题意可得,()f x π2sin 216x ⎛⎫=-- ⎪⎝⎭,令()π2πZ 6x k k -=∈,则()ππZ 212k x k =+∈, 所以()f x 图像的对称中心为()ππ,1Z 212k k ⎛⎫+-∈⎪⎝⎭,说法①错误; 的当ππ,66x ⎡⎤∈-⎢⎥⎣⎦,则πππ2,626x ⎡⎤-∈-⎢⎥⎣⎦,ππ,26⎡⎤-⎢⎥⎣⎦是函数2sin 1y x =-单调递增区间,说法②正确;当π2π,63x ⎛⎫∈⎪⎝⎭时,ππ7π2,666x ⎛⎫-∈ ⎪⎝⎭,所以π1sin 2,162x ⎛⎫⎛⎤-∈- ⎪ ⎥⎝⎭⎝⎦,则()f x 的取值范围为(]2,1-,说法③错误;将函数()f x 的图像向右平移π12个单位长度,所得图像对应的解折式为()πππ2sin 212sin 212316g x x x ⎡⎤⎛⎫⎛⎫=---=-- ⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦,说法④错误.故选:D10. 如图,1F ,2F 分别为椭圆()222210x y a b a b+=>>的左、右焦点,点P 在椭圆上,2 POF 是面积为e 的值是( )A.1-B.1-C.D. 4-【答案】B 【解析】【分析】根据正三角形可得c 及点P 坐标,将点P 代入椭圆方程,可得a ,b ,进而可得离心率.【详解】由于2 POF 是面积为 过点P 作PH x ⊥轴于H ,则H 为2OF 的中点,所以12P x c =,=P y ,所以22POF S c ==△4c =,所以(2,P ,将点P 的坐标代入椭圆方程,得224121a b+=, 即22412116b b+=+,解得2b =216a =+,2224c e a ∴===-,1e ∴=-,故选:B .11. 已知函数()()sin f x A x ωϕ=+的部分图象如图所示,其中π0,0,02A ωϕ->><<.在已知21x x 的条件下,则下列选项中可以确定其值的量为( )A.ω B. ϕC.φωD. sin A ϕ【答案】B 【解析】【分析】根据函数图象可知,12,x x 是函数()f x 的两个零点,即可得21πx x ϕϕ-=-,利用已知条件即可确定ϕ的值.【详解】根据图象可知,函数()f x 的图象是由sin y A x ω=向右平移ϕω-个单位得到的; 由图可知12()()0f x f x ==,利用整体代换可得120,πx x ωϕωϕ+=+=,所以21πx x ϕϕ-=-,若21x x 为已知,则可求得21π1x x ϕ=-. 故选:B12. 已知集合{|()0}M f αα==,{|()0}N g ββ==.若存在M α∈,N β∈,使||n αβ-<,则称函数()f x 与()g x 互为“n 度零点函数”.若函数2()e 1x f x -=-与函数2()e x g x x a =-互为“1度零点函数”.则实数a 的取值范围为( ) A. 214,e e ⎛⎤⎥⎝⎦B. 2214,e e ⎛⎤⎥⎝⎦ C. 242,e e ⎡⎫⎪⎢⎣⎭D. 3212,e e ⎡⎫⎪⎢⎣⎭【答案】A 【解析】【分析】由2()e 10x f x -=-=,得2x =,设20()e x g x x a ==-的解为0x ,根据函数2()e 1x f x -=-与函数2()e x g x x a =-互为“1度零点函数”,由021x -<,得到013x <<,再由2e x x a =,转化为2ex x a =,在()1,3x ∈时有解求解.【详解】解:由2()e 10x f x -=-=,得2x =, 由20()e x g x x a ==-,得2e x x a =,设其解为0x ,因为函数2()e 1x f x -=-与函数2()e x g x x a =-互为“1度零点函数”, 所以021x -<,解得013x <<,由2e xx a =,得2ex x a =,()1,3x ∈时有解,令()2ex x h x =,则()22exx x h x -'=,13x <<, 当12x <<时,()0h x '>,()h x 单调递增,当23x <<时,()0h x '<,()h x 单调递减, 所以()()()()23max 2,419e e 1,3e h x h h h ====, 所以实数a 的取值范围为214,e e ⎛⎤⎥⎝⎦, 故选:A二、填空题:本大题共4小题,每小题5分,共20分.13. 已知双曲线以两坐标轴为对称轴,且它的一个顶点为(2,0)A ,它的一条渐近线方程为12y x =,则双曲线的标准方程为________.【答案】2214x y -=【解析】【分析】由双曲线的一个顶点为(2,0)A ,可得焦点在x 轴上,且2a =,根据渐近线方程求出b 的值即可得双曲线的标准方程 【详解】由双曲线一个顶点为(2,0)A , 所以可得双曲线的焦点在x 轴上,且2a =, 由双曲线的一条渐近线方程为12y x =所以1,2b a =所以1b = 所以双曲线的标准方程为2214x y -=.故答案为:2214x y -=.14. 设向量()2,1a =r ,()1,b x =-,若()a b a ⊥- ,则b = ___________.【答案】【解析】【分析】由平面向量数量积的坐标运算求解【详解】()3,1b a x -=--,由题意得()0a b a ⋅-= ,即610x -+-=,得7x =b ==.故答案为:15. 设n S 是等差数列{}n a 的前n 项和,27a =-,512S a =,当n S 取得最小值时,n =______. 【答案】8 【解析】【分析】先利用题给条件求得等差数列{}n a 的首项与公差,进而求得n S 的表达式,再利用单调性去求n S 的最小值即可解决.【详解】等差数列{}n a 中,27a =-,512S a =, 则111+75+102a d a d a =-⎧⎨=⎩,解之得1310d a =⎧⎨=-⎩,则313n a n =-,则(10313)323()223n n n S n n -+-==-当231N 6n n ≤≤∈,即13N n n ≤≤∈,时,n S 单调递增; 的当2323N63n n≤≤∈,即47Nn n≤≤∈,时,n S单调递减;当23N3n n≥∈,即8Nn n≥∈,时,n S单调递增,又1110S a==,73237(7)7 23S=⨯⨯-=,83238(8)4 23S=⨯⨯-=,则当n S取得最小值时,8n=.故答案为:8.16. 如图所示为某几何体的三视图,则该几何体外接球的表面积为____________.【答案】20π【解析】【分析】作出原几何体的直观图,找出该几何体的外接球球心,计算出外接球的半径,结合球体体积公式可得结果.【详解】由三视图还原原几何体如下图所示,由图可知,原几何体为三棱锥-P ABC,且平面PAC⊥平面ABC,取AC 的中点D ,连接PD 、BD ,则AD CD ==,3BD PD ==,由三视图可知BD AC ⊥,PD AC ⊥,BD PD D ⋂=Q ,则AC ⊥平面PBD ,由勾股定理可得AB BC PA PC AC ======,则ABC 、PAC △均为正三角形,因为平面PAC ⊥平面ABC ,平面PAC 平面ABC AC =,PD AC ⊥,PD ⊂平面PAC ,PD ∴⊥平面ABC ,BD ⊂Q 平面ABC ,PD BD ∴⊥,过ABC 的外心E 在平面PBD 内作EO BD ⊥,过PAC △的外心F 在平面PBD 内作FO PD ⊥,设EO FO O =I , 因为AC ⊥平面PBD ,EO ⊂平面PBD ,则EO AC ⊥,因为EO BD ⊥,AC BD D = ,EO ∴⊥平面ABC ,同理,FO ⊥平面PAC , 所以,O 为三棱锥-P ABC 的外接球球心, 因为E 为等边ABC 的外心,则113DE BD ==,同理1DF =, 在平面PBD 内,因为OF DF ⊥,DE DF ⊥,OE DE ⊥,DE DF =, 所以,四边形OEDF 为正方形,所以,1OF DE ==,因为2PF PD DF =-=,所以,OP ==因此,该几何体外接球的表面积为2420OP ππ⋅=. 故答案为:20π.二、解答题:共70分,解答应写出文字说明、证明过程或演算步骤.第17~21题为必考题,每个试题考生都必须作答.第22、23题为选考题,考生根据要求作答. (一)必考题:共60分.17. 某电视台在一次对收看文艺节目和新闻节目观众的抽样调查中,随机抽取了100名电视观众,相关的数据如下表所示:文艺节目 新闻节目 总计 20至40岁40 18 58 大于40岁152742总计 5545100(1)由表中数据直观分析,收看新闻节目的观众是否与年龄有关?(2)用分层抽样方法在收看新闻节目的观众中随机抽取5名,大于40岁的观众应抽取几名? (3)在上述抽取的5名观众中任取2名,求恰有1名观众的年龄为20至40岁的概率. 【答案】(1)有关 (2)3(3)35【解析】【分析】(1)通过数据的直观分析即可得到结论; (2)利用分层抽样的方式进行抽取即可求解;(3)通过列举法举出所有的取法,和恰有1名观众年龄为20至40岁的取法,即可求解 【小问1详解】因为在20至40岁的58名观众中有18名观众收看新闻节目,而大于40岁的42名观众中有27名观众收看新闻节目.所以,经直观分析,收看新闻节目的观众与年龄是有关的. 【小问2详解】应抽取大于40岁的观众人数为2745×5=35×5=3(名) 【小问3详解】用分层抽样方法抽取的5名观众中,20至40岁有2名(记为12,Y Y ),大于40岁有3名(记为123,,A A A ),5名观众中任取2名,共有10种不同取法:12111213212223121323,,,,,,,,,YY Y A Y A Y A Y A Y A Y A A A A A A A . 设A 表示随机事件“5名观众中任取2名,恰有1名观众年龄为20至40岁”. 则A 中的基本事件有6种:111213212223,,,,,Y A Y A Y A Y A Y A Y A , 故所求概率为()610P A ==35.18. ()sin a C C =+;②sin sin2B C a C c +=;③1cos 2a C c b +=,这三个条件中任选一个,补充在下面问题中,然后解答补充完整的题目.在ABC 中,内角A ,B ,C 的对边分别为a ,b ,c .已知______.(1)求角A ;(2)若1b =,3c =,求BC 边上的中线AD 的长.注:若选择多个条件分别进行解答,则按第一个解答进行计分.【答案】(1)任选一个,答案均为3π(2. 【解析】【分析】(1)选①,由正弦定理化边为角,然后由诱导公式,两角和的正弦公式,商数关系求得A ; 选②,由正弦定理化边为角,由诱导公式、二倍角公式变形可求得A ; 选③,由余弦定理化角为边,再由余弦定理求得A ;(2)在ABD △和ACD 中分别应用余弦定理后相加可得AD . 【小问1详解】()sin a C C =+,sin (sin )B A C C =+,)sin sin cos A C A c A C +=+,cos cos sin )sin sin cos A C A C A C A C +=+,sin sin sin A C A C =,三角形中sin 0C ≠,所以tan A =,又(0,)A π∈, 所以3A π=;选②sin sin2B Ca C c += 由正弦定理得sin sin sin sin sin cos 22B C AA C C C +==,三角形中sin 0C ≠, 所以2sincos cos 222A A A =,又三角形中cos 02A ≠,所以1sin 22A =,(0,)A π∈,所以26A π=,即3A π=;选③1cos 2a C cb +=, 由余弦定理得222122a b c c b b +-+=,整理得222b c a bc +-=,所以2221cos 22b c a A bc +-==,而(0,)A π∈,3A π=;小问2详解】由(1)2222cos 19213cos73a b c bc A π=+-=+-⨯⨯=,a =【由余弦定理得:2222cos b AD CD AD CD CDA =+-⋅∠2222cos c AD BD AD BD BDA =+-⋅∠,又BD CD =,cos cos CDA BDA ∠=-∠,所以22222221222b c AD BD CD AD a +=++=+,所以21113(197)224AD =+-⨯=,AD =. 19. 如图,在直角梯形ABCD 中,AD BC ∥,AD CD ⊥,四边形CDEF 为平行四边形,平面CDEF ⊥平面ABCD ,2BC AD =.(1)证明:DF 平面ABE ;(2)若1AD =,2CD ED ==,π3FCD ∠=,求三棱锥B ADE -的体积. 【答案】(1)证明见解析(2【解析】【分析】(1)连接CE 交DF 于点H ,取BE 的中点G ,连接,AG GH ,根据条件证明四边形ADHG 为平行四边形,然后得到//DH AG 即可;(2)取CD 的中点为O ,连接OF ,依次证明OF ⊥平面ABCD 、//EF 平面ABCD ,然后可求出点E 到平面ABCD 的距离,然后根据B ADE E ABD V V --=算出答案即可. 【小问1详解】证明:连接CE 交DF 于点H ,取BE 的中点G ,连接,AG GH , 因为四边形CDEF 为平行四边形,所以H 为CE 的中点, 所以1//,=2GH BC GH BC , 因为AD BC ∥,2BC AD =,所以//,=GH AD GH AD , 所以四边形ADHG 为平行四边形,所以//DH AG ,即//DF AG , 因为AG ⊂平面ABE ,DF ⊄平面ABE ,所以DF 平面ABE , 【小问2详解】取CD 的中点为O ,连接OF , 因为2CD ED ==,π3FCD ∠=,所以CDF 为等边三角形,所以OF =,OF CD ⊥,因为平面CDEF ⊥平面ABCD ,平面CDEF 平面ABCD CD =,OF ⊂平面CDEF ,所以OF ⊥平面ABCD ,所以点F 到平面ABCD 的距离为OF =,因为//EF CD ,EF ⊄平面ABCD ,CD ⊂平面ABCD , 所以//EF 平面ABCD ,所以点E 到平面ABCD 的距离为OF =,因为ABCD 是直角梯形,AD BC ∥,AD CD ⊥,1AD =,2CD =, 所以112ABD S AD CD =⋅⋅= ,所以113B ADE E ABD V V --==⨯=. 20. 已知0ab ≠,曲线23()xf x a x=-在1x =处的切线方程为630x by +-=. (1)求a ,b 的值;(2)证明:当(0,1]x ∈时,()tan f x x <.【答案】(1)3,2a b ==-(2)证明见解析 【解析】【分析】(1)根据切点和斜率求得,a b .(2)化简()tan f x x <,利用构造函数法,结合导数证得不等式成立. 【小问1详解】 由题可知33(1)1f a b-==-,即1b a =-. 又()()()22222223633()a x x a x f x a x a x -++==--',所以2336(1)(1)a f a b+==-'-,解得32a b =⎧⎨=-⎩,即3,2a b ==-.【小问2详解】23()3xf x x=-,(0,1]x ∈, 要证()tan f x x <,23sin 3cos x xx x<-, 只需证23sin 3cos sin 0x x x x x -->, 令2()3sin 3cos sin g x x x x x x =--,则2()3cos 3cos 3sin 2sin cos (sin cos )g x x x x x x x x x x x x x =-+--=-', 令()sin cos ,(0,1]h x x x x x =-∈,则()cos cos sin sin 0h x x x x x x x '=-+=>, 所以()h x 在(0,1]上单调递增,所以()(0)0h x h >=,即()0g x '>,所以()g x 在(0,1]上单调递增,则()(0)0g x g >=,即当(0,1]x ∈时,()tan f x x <. 21. 已知中心为坐标原点,焦点在坐标轴上的椭圆C经过点M,)1N-.(1)求C 的方程;(2)已知点()3,0D ,直线():3,0l x ty n n t =+≠≠与C 交于,A B 两点,且直线,DA DB 的斜率之和为1t,证明:点(),t n 在一条定抛物线上. 【答案】(1)22193x y +=(2)证明见解析 【解析】【分析】(1)根据椭圆标准方程求法,列方程组解决即可;(2)设直线,DA DB 的斜率分别为1k ,2k ,()11,A x y ,()22,B x y .将x ty n =+代入22193x y+=,得()2223290ty tny n +++-=, ()22121230k k t y y n +=--=,根据韦达定理化简得223n t =+即可解决.【小问1详解】依题意设C 的方程为221px qy +=, 因为C经过点M,)1N-,所以32161p q p q +=⎧⎨+=⎩,解得1913p q ⎧=⎪⎪⎨⎪=⎪⎩,故C 的方程为22193x y +=.【小问2详解】证明:设直线,DA DB 的斜率分别为1k ,2k ,()11,A x y ,()22,B x y .将x ty n =+代入22193x y +=,得()2223290t y tny n +++-=.由题设可知()2212390t n ∆=-+>,12223tn y y t +=-+,212293n y y t -=+,所以()()()()()()()()1221122112121212213333333333y x y x y ty n y ty n y yk k x x x x ty n ty n -+-+-++-+=+==----+-+- ()()()()1212221212231(3)3ty y n y y tt y y t n y y n +-+==+-++-, 所以()221230t y y n --=,所以()()()22222293333033n n t n n t n t t -+⎡⎤--=---=⎢⎥++⎣⎦. 因为3n ≠, 所以()223303n tn t +--=+, 所以223n t =+,故点(),t n 在抛物线223y x =+上,即点(),t n 在一条定抛物线上.(二)选考题:共10分.请考生在第22、23题中选定一题作答,并用2B 铅笔在答题卡上将所选题目对应的题号方框涂黑.按所涂题号进行评分,不涂、多涂均按所答第一题评分;多答按所答第一题评分. [选修4-4:坐标系与参数方程]22. 在直角坐标系xOy 中,曲线C 的参数方程为1cos 1sin x y αα=+⎧⎨=+⎩(α为参数),直线:40l x y +--,以坐标原点O 为极点,x 轴的正半轴为极轴建立极坐标系. (1)求直线l 和曲线C 的极坐标方程;(2)若直线()0:R l θβρ=∈与直线l 相交于点A ,与曲线C 相交于不同的两点M ,N .求OM ON OA ++的最小值.【答案】(1)cos sin 40ρθρθ+-=,()22cos sin 10ρθθρ-++=;(2).【解析】【分析】(1)直接利用转换关系,把参数方程极坐标方程和直角坐标方程之间进行转换;(2)利用极径的应用和三角函数关系式的变换及正弦型函数的性质的应用和基本不等式的应用求出结果. 【详解】(1)由直线:40l x y +-=得其极坐标方程为cos sin 40ρθρθ+-=. 由1cos :1sin x C y αα=+⎧⎨=+⎩,(α为参数).得222210x y x y +--+=,又222x y ρ=+,cos x ρθ=,sin y ρθ=,则其极坐标方程为()22cos sin 10ρθθρ-++=. (2)由题意,设()1,M ρβ,()2,N ρβ,()3,A ρβ, 把θβ=代入()22cos sin 10ρθθρ-++=,得()22cos sin 10ρββρ-++=,∴()12π2cos sin 4OM ON ρρβββ⎛⎫+=+=+=+ ⎪⎝⎭,由θβ=与曲线C 相交于不同的两点M ,N ,可知π02β<<. 把θβ=代入cos sin 40ρθρθ+-=得34cos sin OP ρββ===+.∴π1sin π4sin 4OM ON OA ββ⎡⎤⎥⎛⎫⎥++=++≥ ⎪⎛⎫⎝⎭⎥+ ⎪⎢⎥⎝⎭⎣⎦当且仅当π1sin π4sin 4ββ⎛⎫+= ⎪⎛⎫⎝⎭+ ⎪⎝⎭,π02β<<, 即π4β=时,等号成立,OM ON OA ++的最小值为. 【点睛】本题主要考查了参数方程极坐标方程和直角坐标方程之间的转换,三角函数关系式的恒等变换,一元二次方程根和系数关系式的应用,主要考查学生的运算能力和转换能力及思维能力,属于中档题.[选修4-5:不等式选讲]23. 已知关于x 的函数()223(R)f x x x x =-++∈. (1)求关于x 的不等式()7f x ≥的解集.(2)若函数()f x 的最小值为m 、且实数a ,b 满足222a b m +=,求2a b +的最大值. 【答案】(1){2x x ≤-或}2x ≥(2)【解析】【分析】(1)写出()f x 的分段形式,分类讨论,求出不等式的解集; (2)利用(1)中分段函数的单调性求出4m =,设2cos ,a b θθ==,[]0,2πθ∈,利用辅助角公式求出2a b +的最值. 【小问1详解】31,3()2235,3131,1x x f x x x x x x x --<-⎧⎪=-++=-+-≤≤⎨⎪+>⎩,当3x <-时,317x --≥,解得:83x ≤-,与3x <-取交集得:3x <-, 当31x -<<时,57x -+≥,解得:2x ≤-,与31x -<<取交集得:32x -<≤-, 当1x >时,317x +≥,解得:2x ≥,与1x >取交集得:2x ≥, 综上:不等式()7f x ≥的解集为{2x x ≤-或}2x ≥; 【小问2详解】()f x 为连续函数,且当1x ≤时,()f x 单调递减,当1x >时,()f x 单调递增,故当1x =时,()f x 取得最小值,()()min 14f x f ==, 则2224a b +=,设2cos ,a b θθ==,[]0,2πθ∈,则()24cos a b θθθϕ+==+,其中1cos ,sin 3ϕϕ==当()sin 1θϕ+=时,2a b +取得最大值为。
[VIP专享]六盘山三模试卷
宁夏六盘山高中2008-2009学年高三第三次模拟试卷学科;英语测试时间:120分钟满分:150分命题人:毛亚莉注意事项:1.答第一卷之前,考生务必将自己的姓名、准考证号、考试科目、试卷类型(A),用2B铅笔规范、准确地涂写在答题卡上。
2.每小题选出答案后,用铅笔将答题卡上对应题目的答案标号涂黑。
如需改动,用橡皮擦干净后,再选涂其它答案标号,不能答在试卷上。
3.71-75小题选E涂AB,选F涂AC,选G涂AD。
I卷(客观题,共115分)第一部分:听力(满分30 分)。
做题时,先将答案划在试卷上。
录音结束后,你有两分钟的时间将答案转涂到答题卡上。
第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. What will the woman do tomorrow?A. Take an examination.B. Return home.C. Buy a new dress.2. What are the two speakers talking about?A. How to improve their English.B. An English newspaper.C. Their progress in learning English.3.What does the man think of Yao Ming?A. He is the tallest basketball player in the world.B. He is famous all over the world.C. He is only well-known in China.4. What will the woman do for the man?A. Wash his clothes.B. Take him to the supermarket.C. Get him some fruit.5. What does the woman want to do?A. She wants to buy some stamps.B. She wants to sell a stamp to the man.C. She wants the man to give her a stamp.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。
宁夏六盘山高级中学2021届高三第三次模拟考试语文试题
宁夏六盘山高级中学2021年高三第三次模拟考试语文试题学校:___________姓名:___________班级:___________考号:___________一、现代文阅读阅读下面的文字,完成小题。
不能拿鲁迅当商业噱头周俊生刚刚过去的2021年,是鲁迅先生逝世80周年。
文艺界、媒体界、出版界、戏剧界、展览界等,以各自的方式参与了纪念鲁迅的活动,向这位思想文化巨擘致敬。
但是,有的人却将鲁迅视作噱头、看成商机,一款名为“鲁迅群侠传”的手机游戏产品,将鲁迅作品中的一些人物,如孔乙己、阿Q装扮成游侠,闹得不亦乐乎。
鲁迅作品,包括作品中的这些人物形象早已深入人心,这款打着“纪念鲁迅”名头的游戏产品一经面世就受到激烈批评,现已下线。
用电子游戏来演绎鲁迅可不可以?答案是肯定的。
电子游戏出现以后,一直很有市场,现已成为吸金量可观的产业。
一些游戏公司挑选经典文学作品中的一些文学形象改编成游戏作品,不仅丰富了游戏题材,也有助于这些作品的传播。
但是,将鲁迅作品中的人物形象植入游戏产品,需要保持慎重的态度。
鲁迅的小说以深刻揭露旧社会的黑暗和国民性中的弱点见长,其中的一些人物如阿Q、祥林嫂、孔乙己,不仅毫无侠性,而且是悲剧人物。
让他们以侠客形象出现在游戏中,如果尊重原著题旨,这款游戏就玩不下去,如果按侠客的套路来重塑这些人物,就构成了与鲁迅原著完全相反的观感。
这样的游戏产品实际上是对鲁迅作品的颠覆,是对鲁迅的不尊重。
作为具有娱乐性质的活动,电子游戏进行经典文学作品二度开发,将合适的文学元素嫁接到游戏中本无不可,但需要注意的是,这种改编必须是在尊重原著形象的基础上完成的,不能对原著形象进行颠覆。
以中国的几部古典名著而论,《三国演义》《西游记》比较适合改编成游戏产品,它们的历史战争和神仙魔怪题材与电子游戏所要达到的竞技、魔怪效果比较切合,而《红楼梦》和《儒林外史》就不大适合改编成游戏产品,因为这两部名著所表现的儿女情长和社会风俗与游戏所要达到的效果有很大距离。
高三数学第三次模拟考试试题理 2
宁夏六盘山高级中学2021届高三数学第三次模拟考试试题 理制卷人:歐陽文化、歐陽理複;制卷時間:二O 二二年二月七日考前须知:1.在答题之前,考生先将本人的姓名、准考证号码填写上在本试题相应的位置、涂清楚。
2.选择题必须使需要用2B 铅笔填涂;非选择题必须使用0.5毫米黑色字迹的签字笔书写,字体工整、笔迹清楚。
3.请按照题号顺序在答题卡各题目的答题区域内答题,超出答题区域书写之答案无效;在草稿纸、试卷上答题无效。
4.作图可先使用铅笔画出,确定后必须用黑色字迹的签字笔描黑。
5.保持卡面清洁,不要折叠,不要弄破、弄皱,不准使用涂改液、修正带、刮纸刀。
一、选择题:此题一共12小题,每一小题5分,一共60分.在每一小题给出的四个选项里面,只有一项是哪一项符合题目要求的. 1.集合}13|{3<=+x x A ,}0124|{2>--=x x x B ,那么=B A C R )(( )A 、]3,(--∞B 、)2,3[--C 、),6()2,3[+∞--D 、),6()2,3(+∞--2.在复平面内,复数1z 和2z 对应的点分别是)1,2(A 和)1,0(B ,那么=21z z ( ) A 、i 21-- B 、i 21+- C 、i 21- D 、i 21+3.如图是某高三年级的三个班在一学期内的六次数学测试的平均成绩y 关于测试序 号x 的函数图像,为了容易看出一个班级的成绩变化,将离散的点用虚线连接,根据 图像,给出以下结论:①一班成绩始终高于年级平均程度,整体成绩比拟好;②二班成绩不够稳定,波动程度较大;③三班成绩虽然屡次低于年级平均程度,但在稳步提升。
其中正确结论的个数为( )A 、0B 、1C 、2D 、34.?孙子算经?是中国古代重要的数学著作,书中有一道题为:今有出门望见九堤,堤有 九木,木有九枝,枝有九巢,巢有九禽,禽有九雏,雏有九毛,毛有九色,问各几何?假设记堤与枝的个数分别为,m n ,现有一个等差数列{}n a ,其前n 项和为n S ,且2a m =, 6S n =,那么4a =〔 〕 A 、84B 、159C 、234D 、2435. 抛物线2:(0)C y ax a =>的焦点F 是双曲线223312y x -=的一个焦点,那么a =〔 〕A 、2B 、4C 、12D 、146. 设ΔABC 的内角A ,B ,C 所对的边分别为a ,b ,c ,2cos cos cos c B b A a B +=-, 那么∠B =〔 〕A 、6π B 、3π C 、56πD 、23π7.等边ABC ∆的边长为2,假设BE BC 3=,DC AD =,那么=⋅AE BD ( )A 、2-B 、310-C 、2D 、3108.直三棱柱111C B A ABC -中,120=∠ABC ,2=AB ,11==CC BC ,那么异面直线1AB 与1BC 所成角的余弦值为( )A 、510 B 、515 C 、23 D 、33 9. 假设函数()sin()f x A x ωϕ=+〔其中0A >,||)2πϕ<图象的一个对称中心为(3π,0),其相邻一条对称轴方程为712x π=,该对称轴处所对应的函数值为1-,为了得到 ()cos2g x x =的图象,那么只要将()f x 的图象( )A 、向右平移6π个单位长度 B 、向左平移12π个单位长度 C 、向左平移6π个单位长度 D 、向右平移12π个单位长度10. α满足sin()46πα+=,那么2tan 12tan αα+=〔 〕A 、98B 、98-C 、3D 、3-11.函数)(x f 是定义在R 上的偶函数,设函数)(x f 的导函数为)(x f ',假设对任意 的0>x 都有0)()(2>'⋅+x f x x f 恒成立,那么( )A 、)3(9)2(4f f <-B 、)3(9)2(4f f >-C 、)2(3)3(2->f fD 、)2(2)3(3->-f f12.)0,1(F 为抛物线P :px y 22=(0>p )的焦点,过点F 且斜率为k 的直线l 与 曲线P 交于B 、C 两点,过O 与BC 中点M 的直线与曲线P 交于N 点,那么OBNOMCS S ∆∆的取值范围是( )A 、)41,0(B 、)21,0(C 、)22,0( D 、)23,0( 二、填空题:此题一共4小题,每一小题5分,一共20分.13. 函数2log ()(0)()31(0)xx x f x x -<⎧=⎨-≥⎩,且()()10f a f +=,那么实数a 的值等于______. 14. 很多网站利用验证码来防止恶意登录,以提升网络平安. 某马拉松赛事报名网站的登录 验证码由0,1,2,⋯,9中的四个数字随机组成,将从左往右数字依次增大的验证码 称为“递增型验证码〞(如0123),某人收到了一个“递增型验证码〞,那么该验证码的首位数字是1的概率为___________.15. 如下图,半径为4的球O 中有一内接圆柱,当圆柱的侧面积 最大时,球的外表积与该圆柱的侧面积之差是___________. 16. 函数a x e x x f x--⋅+=2)1()(,假设0)(<x f有且仅有一个整数根,那么实数a 的取值范围是___________.三、解答题:一共70分。
宁夏六盘山高级中学届高三英语第三次模拟考试试题
宁夏六盘山高级中学2021 年第二学期高三英语第三次模拟试卷英语测试时间:120 分钟总分值: 150 分第一卷30 分〕第一局部:听力〔共两节,总分值7.5 分〕第一节〔共 5 小题,每题 1.5 分,总分值听下面 5 段对话。
每段对话后有一个小题,从题中所给的A、B、C 三个选项中选出最正确选项,并标在试卷的相应位置。
听完每段对话后,你都有 10 秒钟的时间来答复有关小题和阅读下一小题。
每段对话仅读一遍。
1. Where is the woman going now?A. To the library.B. To a coffee shop.C. To thesupermarket.2.What does the man care about most?A. Earning some extra money.B. Keeping the environment clean.C. Helping out the corner shop.3.When is the woman ’s birthday?4.What is the man doing?A.Playing with his daughter.B. Playing a joke on the woman.C.Playing a game on his smartphone.5.Why is the woman ’s French so good?A.She has been studying for ten years.B. She was born in France.C.She works hard at it.第二节〔共 15 小题,每题 1.5 分,总分值 22.5 分〕听下面 5 段对话或独白。
每段对话或独白后有几个小题,从题中所给的A、B、C 三个选项中选出最正确选项,并标在试卷的相应位置。
听每段对话或独白前,你将有时间阅读各个小题,每题 5 秒钟;听完后,各小题将给出 5 秒钟的作答时间。
宁夏银川市六盘山高级中学2023届高三三模数学(文)试题(1)
一、单选题1. 龙洗,是我国著名的文物之一,因盆内有龙纹故称龙洗,为古代皇宫盥洗用具,其盆体可以近似看作一个圆台.现有一龙洗盆高15cm ,盆口直径40cm ,盆底直径20cm .现往盆内倒入水,当水深6cm 时,盆内水的体积近似为()A.B.C.D.2. 以下判断正确的是A .命题“若,则”为真命题B .命题“”的否定是“”C .“”是“函数是偶函数”的充要条件D .命题“在中,若,则”为假命题3. 已知,,则( )A .0B .1C .3D .54. 现有四个函数:①;②;③;④的图象(部分)如下,但顺序被打乱,则按照从左到右将图象对应的函数序号安排正确的一组是A .④①②③B .①④③②C .①④②③D .③④②①5. 2021年10月16日0时23分,搭载神舟十三号载人飞船的长征二号F 遥十三运载火箭,在酒泉卫星发射中心按照预定时间精准点火发射,顺利将翟志刚、王亚平,叶光富3名航天员送入太空,飞行乘组状态良好,发射取得圆满成功,火箭在发射时会产生巨大的噪音,已知声音的声强级(单位:)与声强x (单位:)满足.若人交谈时的声强级约为,且火箭发射时的声强与人交谈时的声强的比值约为,则火箭发射时的声强级约为( )A.B.C.D.6.定义在上的奇函数满足,当,,则( )A.B.C.D.7. 挪威画家爱德华·蒙克于1893年创作的《呐喊》是表现主义绘画的代表作品,刻画了一个极其痛苦的表情.画作局部如下图所示,人像的脸近似为一个椭圆,下巴近似为一个圆,圆心在椭圆的下顶点上,椭圆与圆有两个交点,,椭圆的两焦点与圆的圆心在同一直线上,记椭圆的中心为.连接直线,,,经测量发现与圆相切,圆的半径为,.记该椭圆的离心率为,为不超过的最大整数,则的值为()A .2B .4C .6D .88.设为抛物线的焦点,为该抛物线上三点.若,则( )A .9B .6C .4D .3宁夏银川市六盘山高级中学2023届高三三模数学(文)试题(1)宁夏银川市六盘山高级中学2023届高三三模数学(文)试题(1)二、多选题三、填空题四、解答题9. 将函数向左平移个单位,得到函数,下列关于的说法正确的是( )A .关于对称B.当时,关于对称C .当时,在上单调递增D .若在上有三个零点,则的取值范围为10. 已知二项式的展开式中所有项的系数的和为64,则( )A.B .展开式中的系数为C .展开式中奇数项的二项式系数的和为32D.展开式中二项式系数最大的项为11. 已知函数f (x )的定义域为A ,若对任意,都存在正数M使得总成立,则称函数是定义在A 上的“有界函数”.则下列函数是“有界函数”的是( )A.B.C.D.12. 如图,在四面体中,,,,为的中点,点是棱的中点,则()A .平面B.C .四面体的体积为D .异面直线与所成角的余弦值为13. 已知向量,,,则______.14. 若满足,满足,则_____.15.设等差数列的前项和为,若,,则________.16.如图,四棱锥中,,,平面.点M 是的中点,且平面平面.(1)证明:平面;(2)求直线与平面所成角的正弦值.17. 已知数列满足,,为的前n项和.(1)求的通项公式;(2)设,数列的前n项和满足对一切正奇数n恒成立,求实数m的取值范围.18. 已知椭圆的右焦点为,点,在椭圆上运动,且的最小值为;当点不在轴上时点与椭圆的左、右顶点连线的斜率之积为.(1)求椭圆的方程;(2)已知直线与椭圆在第一象限交于点,若的内角平分线的斜率不存在.探究:直线的斜率是否为定值,若是,求出该定值;若不是.请说明理由.19. 中日围棋擂台赛是由中国围棋队与日本围棋队各派若干名棋手,以擂台制形式举行的围棋团体赛.这是中国和国外开设的最早的围棋对抗赛,由中国围棋协会、日本棋院和中国《新体育》杂志社联合举办,日本电器公司(NEC)赞助,因此也称NEC杯中日围棋擂台赛.该赛事从1984年开始至1996年停办,共进行了11届,结果中国队以7比4的总比分获胜.该赛事对中国围棋甚至世界围棋发展产生了很大影响,被认为是现代围棋最成功的比赛之一.中日围棋擂台赛由中日双方各派同样数量的若干名棋手组成队伍,两队各设一名主帅,采用打擂台的形式,决出最后的胜负.比赛事先排定棋手的上场顺序(主帅最后上场),按顺序对局,胜者坐擂,负方依次派遣棋手打擂,直至一方“主帅”被击败为止.设中、日两国围棋队各有名队员,按事先排好的顺序进行擂台赛,中国队的名队员按出场的先后顺序记为;日本队的名队员按出场的先后顺序记为.假设胜的概率为(为常数).(1)当时,若每个队员实力相当,求中国队有四名队员被淘汰且最后战胜日本队的概率;(2)记中国队被淘汰人且中国队获得擂台赛胜利的概率为,求的表达式;(3)写出中国队获得擂台赛胜利的概率的表达式(不用说明理由).20. 鲜虾是在日常生活中常能吃到的一种水产品,鲜虾肉肥嫩鲜美,在生活中很多人都喜欢吃鲜虾,而且鲜虾有很高的营养价值.某超市为了解本店鲜虾的日销售情况,对过去20天鲜虾的日销售量(单位:千克)进行了统计,得如图所示的条形图.(1)求这20天鲜虾的日销售量的平均值.(2)该超市每天提供的鲜虾有罗氏虾和基围虾两种,假设接下来的几个月,每天提供的鲜虾总量为这20天日销售量的平均值,这两种虾的日销售率(某种虾当天的销量与该种虾当天供货量的比值)、进价、售价如下表:日销售率进价/(元/千克)售价/(元/千克)罗氏虾0.93245基围虾0.952432已知当日没有售完的罗氏虾和基围虾统一按照售价的一半全部处理给内部员工.若该超市每天销售鲜虾的利润不低于1400元,罗氏虾每天的进货量与当日鲜虾总进货量的比值为t,求实数t的最小值.(结果精确到小数点后两位数)21. 某工厂生产一种精密仪器,由第一、第二和第三工序加工而成,三道工序的加工结果相互独立,每道工序的加工结果只有两个等级.三道工序的加工结果直接决定该仪器的产品等级:三道工序的加工结果均为级时,产品为一等品;第三工序的加工结果为级,且第一、第二工序至少有一道工序加工结果为级时,产品为二等品;其余均为三等品.每一道工序加工结果为级的概率如表一所示,一件产品的利润(单位:万元)如表二所示:表一工序第一工序第二工序第三工序概率表二等级一等品二等品三等品利润2385(1)用表示一件产品的利润,求的分布列和数学期望;(2)因第一工序加工结果为级的概率较低,工厂计划通过增加检测成本对第一工序进行改良,假如改良过程中,每件产品检测成本增加万元(即每件产品利润相应减少万元)时,第一工序加工结果为级的概率增加.问该改良方案对一件产品利润的期望是否会产生影响?并说明理由.。
高中物理 2022年宁夏六盘山高级中学高考物理三模试卷
2022年宁夏六盘山高级中学高考物理三模试卷一、选择题(每小题4分,共48分.在每小题给出的四个选项中,至少有一个选项是正确的,全部选对得4分,对而不全得2分.)A .垂直于斜面,做功为零B .垂直于斜面,做功不为零C .不垂直于斜面,做功为零D .不垂直于斜面,做功不为零1.(4分)如图所示,表面光滑的斜面体A 静止在光滑水平地面上。
物体B 从静止开始沿斜面下滑。
设斜面对物体的作用力为F ,则相对地面,作用力F ( )A .合外力对质点做的功为零,则质点的动能、动量都一定不改变B .合外力对质点施的冲量不为零,则质点的动量、动能都一定改变C .某质点受到的合外力不为零,其动量、动能都一定改变D .某质点的动量、动能都改变了,它所受到的合外力一定不为零2.(4分)下列说法正确的是( )A .B .C .D .3.(4分)一辆小车在水平面上做匀速直线运动,从某时刻起,小车所受牵引力和阻力随时间变化的规律如图所示,则作用在小车上的牵引力F 的功率随时间变化的规律是图中的( )A .平抛过程较大B .竖直上抛过程较大C .竖直下抛过程较大D .三者一样大4.(4分)在距地面某一高度,同时以相等速率分别平抛,竖直上抛,竖直下抛质量相等的物体m ,当它们从抛出到落地前瞬时,不计空气阻力,比较它们的动量的增量△p ,有( )A .Mv 1−Mv 2M −m 向东B .Mv 1M −m 向东5.(4分)如图,质量为m 的人立于平板车上,人与车的总质量为M ,人与车以速度v 1在光滑水平面上向东运动。
当此人相对于车以速度v 2竖直跳起时,车的速度变为( )C .Mv 1+Mv 2M −m 向东D .v 1 向东A .振子所受的弹力大小为5N ,方向指向x 轴的负方向B .振子的速度方向指向x 轴的正方向C .在0~4s 内振子作了1.75次全振动D .在0~4s 内振子通过的路程为0.35cm ,位移为06.(4分)劲度系数为20N /cm 的弹簧振子,它的振动图象如图所示,在图中A 点对应的时刻( )A .摆球从A 运动到B 的过程中重力做的功为12mv 2B .摆球从A 运动到B 的过程中重力的平均功率为mv2TC .摆球运动到B 时重力的瞬时功率是mgvD .摆球运动到B 时重力的瞬时功率是零7.(4分)如图所示,单摆摆球的质量为m ,做简谐运动的周期为T ,摆球从最大位移A 处由静止开始释放,摆球运动到最低点B 时的速度为v ,则( )A .若△t =T 4,则在△t 时间内振子经过的路程为一个振幅B .若△t =T 2,则在△t 时间内振子经过的路程为两个振幅C .若△t =T 2,则在t 时刻和(t +△t )时刻振子的位移一定相同D .若△t =T ,则在t 时刻和(t +△t )时刻振子的速度一定相同8.(4分)一弹簧振子做简谐运动,周期为T ,则下列说法正确的是( )A .弹力做的功一定为零B .弹力做的功可能是零到12mv 2之间的某一值C .弹力的冲量大小可能是零到2mv 之间的某一值D .弹力的冲量大小一定不为零9.(4分)做简谐运动的弹簧振子,质量为m ,最大速率为v .从某时刻算起,在半个周期内( )二、填空题(共19分.把正确答案填写在题中的横线上,或按题目要求作答.)A .两弹簧振子完全相同B .两弹簧振子所受回复力最大值之比F 甲:F 乙=2:1C .振子甲速度为零时,振子乙速度最大D .振子的振动频率之比f 甲:f 乙=1:210.(4分)甲、乙两弹簧振子,振动图象如图所示,则可知( )A .此水波的传播速度是2m /sB .此水波的周期是2sC .此水波的波长是4mD .当O 点处在波峰位置时,B 点正通过平衡位置向下运动11.(4分)一兴趣小组的同学们观察到湖面上一点O 上下振动,振幅为0.2m ,以O 点为圆心形成圆形水波,如图所示,A 、B 、O 三点在一条直线上,OA 间距离为4.0m ,OB 间距离为2.4m .某时刻O 点处在波峰位置,观察发现2s 后此波峰传到A 点,此时O 点正通过平衡位置向下运动,OA 间还有一个波峰.将水波近似为简谐波.则以下说法正确的是( )A .当F =mgtanθ时,质点的机械能守恒B .当F =mgsinθ时,质点的机械能守恒C .当F =mgtanθ时,质点的机械能一定增大D .当F =mgsinθ时,质点的机械能可能增大也可能减小12.(4分)如图所示为竖直平面内的直角坐标系.一个质量为m 的质点,在恒力F 和重力的作用下,从坐标原点O 由静止开始沿直线OA 斜向下运动,直线OA 与y 轴负方向成θ角(θ<45°).不计空气阻力,则以下说法正确的是( )13.(8分)如图所示,气垫导轨是常用的一种实验仪器。
宁夏银川市六盘山高级中学2023届高三三模数学(理)试题
一、单选题二、多选题1.已知函数,则函数f [f (x )]的定义域为( )A .{x |x ≠1}B .{x |x ≠2}C .{x |x ≠1或x ≠2}D .{x |x ≠1且x ≠2}2. 已知集合,,若,则( )A .-1B .-2C .0D .13. 已知为虚数单位,复数,则A.B.C.D.4. 已知,,则( )A .2B.C.D.5. 已知圆锥SO 的母线长为2,AB 是圆O 的直径,点M 是SA的中点.若侧面展开图中,为直角三角形,则该圆锥的侧面积为( )A.B.C.D.6. 已知,分别是首项为1的等差数列{}和首项为1的等比数列{}的前n 项和,且满足4=,9=8,则的最小值为A .1B.C.D.7. 已知函数,则( )A .1B.C.D.8. 已知幂函数在上是减函数,则的值为( )A .3B.C .1D.9. 已知函数,则下列结论中正确的是( )A.函数的一个周期为B.函数在上单调递增C .直线是函数图象的一条对称轴D .函数的值域为10.函数及其导函数的定义域均为R ,且是奇函数,设,,则以下结论正确的有( )A.函数的图象关于直线对称B.若的导函数为,定义域为R,则C .的图象存在对称中心D.设数列为等差数列,若,则11. 下列说法正确的是( )A .相关系数可衡量两个变量之间线性关系的强弱,的值越接近于1,线性相关程度越强B .在对两个分类变量进行独立性检验时,计算出的观测值为,已知,则可以在犯错误的概率不超过的前提下认为两个分类变量无关宁夏银川市六盘山高级中学2023届高三三模数学(理)试题宁夏银川市六盘山高级中学2023届高三三模数学(理)试题三、填空题四、解答题C .一组容量为100的样本数据,按从小到大的顺序排列后第50,51个数据分别为13,14,则这组数据的中位数为D.相关指数可用来刻画一元回归模型的拟合效果,回归模型的越大,拟合效果越好12.如图,在长方体中,,,是棱上的一点,点在棱上,则下列结论正确的是()A .若,,,四点共面,则B .存在点,使得平面C .若,,,四点共面,则四棱锥的体积为定值D.若为的中点,则三棱锥的外接球的表面积是13.如图所示,四边形中,,,,则______;______.14.若,则__________.15. 已知,函数,,若存在一条直线与曲线和均相切,则使不等式恒成立的最小整数的值是__________.16. 为了弘扬体育精神,某校组织秋季运动会,在一项比赛中,学生甲和乙各自进行了8组投篮,得分情况如下:甲10887968乙791057688(1)求出乙的平均得分;(2)如果学生甲的平均得分为8分,那么这组数据的第75百分位数是多少.17. 如图,在四面体中,是正三角形,是直角三角形,.(1)求证:;(2)已知点E 在棱上,且,设,若二面角的余弦值为,求.18. 直角三角形中,是的中点,是线段上一个动点,且,如图所示,沿将翻折至,使得平面平面.(1)当时,证明:平面;(2)是否存在,使得与平面所成的角的正弦值是?若存在,求出的值;若不存在,请说明理由.19. 设函数,为的导函数.(1)当时,①若函数的最大值为0,求实数的值;②若存在实数,使得不等式成立,求实数的取值范围.(2)当时,设,若,其中,证明:.20. 已知函数,.(1)若在上的最大值为,求实数的值;(2)若对任意,都有恒成立,求实数的取值范围;(3)在(1)的条件下,设,对任意给定的正实数,曲线上是否存在两点、,使得是以(为坐标原点)为直角顶点的直角三角形,且此三角形斜边中点在轴上?请说明理由.21. 如图1,平面四边形中,和均为边长为的等边三角形,现沿将折起,使,如图2.(1)求证:平面平面;(2)求点到平面的距离.。
宁夏六盘山高级中学2022届高三英语第三次模拟考试试题(含解析)
C. She is going to the mountains with her friends.
15. What does the man do in the morning at camp?
第二节(共15小题;每小题1.5分,满分22.5分)
请听下面5段对话或独白,每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。在听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,每个小题将给出5秒钟的作答时间。每段对话或独白读两遍。
听第6段材料,回答第6、7题。
(1) be 18 years of age or older and in good health;
(2) have the professional skills and job experience required for the intended employment;
(3) have no criminal record;
A. A teacher. B. A scholar. C. A doctor.
听第8段材料,回答第10至12题。
10. Who probably won’t meet the target this month?
A. The man. B. The womahink of Jenny?
(1) Foreigners who want to work in China should first get in touch with a valid Chinese employer who has an employment license for foreigners issued by a labor administrative bureau.
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宁夏六盘山高中2008-2009学年高三第三次模拟试卷学科;英语测试时间:120分钟满分:150分命题人:毛亚莉注意事项:1.答第一卷之前,考生务必将自己的姓名、准考证号、考试科目、试卷类型(A),用2B铅笔规范、准确地涂写在答题卡上。
2.每小题选出答案后,用铅笔将答题卡上对应题目的答案标号涂黑。
如需改动,用橡皮擦干净后,再选涂其它答案标号,不能答在试卷上。
3.71-75小题选E涂AB,选F涂AC,选G涂AD。
I卷(客观题,共115分)第一部分:听力(满分30 分)。
做题时,先将答案划在试卷上。
录音结束后,你有两分钟的时间将答案转涂到答题卡上。
第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. What will the woman do tomorrow?A. Take an examination.B. Return home.C. Buy a new dress.2. What are the two speakers talking about?A. How to improve their English.B. An English newspaper.C. Their progress in learning English.3.What does the man think of Y ao Ming?A. He is the tallest basketball player in the world.B. He is famous all over the world.C. He is only well-known in China.4. What will the woman do for the man?A. Wash his clothes.B. Take him to the supermarket.C. Get him some fruit.5. What does the woman want to do?A. She wants to buy some stamps.B. She wants to sell a stamp to the man.C. She wants the man to give her a stamp.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。
每段对话或独白后有几个小题,从每题所给的A、B、C.三个选项中选出最佳选项,并标在试卷的相应位置。
听每段对话或独白前,你将有时间阅读各小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。
每段对话或独白读两遍听第6段材料,回答第6、7题。
6. Why are there many people in the waiting room?A. Their plane has been delayed.B. They have caught a cold.C. They are waiting for a train.7. What does the man do?A. He is a doctor.B. He is a patient.C. He is a passenger.听第7段材料,回答第8至10题。
8. What’s the relationship between the two speakers?A. Wife and husband.B. Clerk and customer.C. Saleswoman and customer.9. On which floor will the man stay?A. On the second floor.B. On the tenth floor.C. On the fifteenth floor.10. How much should the man pay in total?A. $700.B. $900.C. $1,050.听第8段材料,回答第11至14题。
11. What is the man doing at the room?A. He is having a meeting.B. He is on holiday.C. He is meeting his friend.12. What does the man’s friend advise him to do?A. To hire a car.B. To see the beach.C. To stay a few more days.13. How much should the man pay in all?A. $1,000.B. $1,500.C. $2,000.14. When will the two speakers meet?A. Tomorrow afternoon.B. This evening.C. Tomorrow morning.听第9段材料,回答第15至17题。
15. What’s the relationship between the two speakers?A. Neighbors.B. Classmates.C. Strangers.16. What’s the woman doing?A. She is working in an international trade company.B. She is working in a university.C. She is a tour guide.17. What did the man do last summer?A. He attended a class reunion.B. He met the woman at the gathering.C. He began to work in the university.听第10段材料,回答第18至20题。
18. What can we conclude about the professor?A. He does n’t like his students.B. He is very strict.C. He’s new in the college.19. Why did so many students fail this course last term?A. They didn’t like the teacher.B. They didn’t follow the teacher’s rules.C. They didn’t know how to spell.20. When is the professor in his office?A. On Tuesdays.B. On Sundays.C. On Fridays.第二部分英语知识运用(共两节,满分45分)第一节单项填空(共15小题;每小题1分,满分15分)21. – Is Mr. Smith rich?–Y es, very rich. When he worked in town, he earned a lot. Now he has a big farm in country.A. a; theB. the; aC. 不填; theD. the; 不填22. Be patient with your students it takes time for them to adapt themselves to the new school life.A. orB. soC. asD. and23. – Have you finished your research paper?– Not yet, I have to go online to search for more information to it.A. relevantB. evidentC. availableD. current24. In those days the boy a lot with his friends, which worried his parents.A. hung onB. hung outC. hung offD. hung up25. – What do you think made the woman so upset?– _______ weight .A. As she put onB. Put onC. Putting onD. Because of putting on26. The conference was attended by more than 11,000 people, it the largest UN climate change gathering ever held.A. makingB. to makeC. madeD. to be making27. We'll be free tomorrow, so I suggest ______ to the history museum.A. to visitB. visitingC. we should visitD. a visit28. Mr. Smith is ______ a good teacher ______ we all respect.A. such;thatB. such;asC. so;thatD. so;as29. – Y ou haven't been to Beijing, have you?–______. And how I wish to go there again!A. Y es, I haveB. Y es, I haven'tC. No, I haveD. No, I haven't30.— Are you satisfied with what she has done?— Not in the least.It couldn’t be __________.A.any worse B.any better C.so bad D.so good 31.Any applicant form ______ properly will not be accepted by the company.A.not to be filled B.not filledC.not being filled D.not having been filled32.— Where is Lucy?—I can’t say for sure w here she is, but she ______ be out shopping.A.can B.should C.may D.must33.Unluckily, when I dropped in , Dr. Smith ____, so we only had time for a few words.A.was just leaving B.has just leftC.had just left D.just left34.Our government took urgent measures ______ the terrible weather disaster at the beginning of 2008.A.in terms of B.in case ofC.in response to D.in honor of35. – What do you think of this one?–, too modern and unusual.A. It’s really not my cup of teaB. Don’t mention itC. I couldn’t find a better oneD. It’s just so-so第二节完形填空(共20小题;每小题1.5分,满分30分)When I come across a good article in reading newspapers, I often want to cut and keep it. But just as I am about to do so, I find the article on the 36 side is as much interesting. It may be a discussion of the way to 37 in good health, or 38 about how to behave and conduct yourself in society. If I cut the front article, the opposite one is likely to suffer 39 , leaving one half of it or keeping the text 40 the title. Therefore, the scissors(剪刀)would stay before they start, 41 the cutting would be halfway done when I find out the 42 result.Sometimes two things are to be done at the same time, both worth your 43 . Y ou can only take up one of them; the other has to wait or be 44 up. But you know the future is unpredictable – the changed situation may not 45 you to do what is left behind. Thus you are 46 in a difficult position and feel sad. How come nice 47 and clever ideas should gather around all at once? It may happen that your life 48 greatly on your preference of your one choice to the other.In fact that is what 49 is like; we are often 50 with the two opposite sides of a thing which are both desirable(引人的) 51 a newspaper cutting. It often occurs that our attention is drawn to the thing only 52 we get into another. The 53 may be more important than the latter and give rise to a divided mind. I 54 remember a philosopher’s remarks: ―When one door shuts, another opens in life.‖ So a casual(不经意的) 55 may not be a bad one.36. A. same B. opposite C. either D. front37. A. get B. bring C. lead D. keep38. A. advice B. news C. theory D. report39. A. damage B. destroy C. hurt D. injury40. A. on B. for C. without D. off41. A. or B. but C. so D. for42. A. satisfying B. regrettable C. surprising D. impossible43. A. courage B. patience C. strength D. attention44. A. given B. picked C. held D. made45. A. persuade B. agree C. allow D. tell46. A. filled B. struck C. caught D. attracted47. A. chances B. conditions C. wishes D. ways48. A. progresses B. goes C. changes D. improves49. A. study B. life C. society D. nature50. A. supplied B. connected C. fixed D. faced51. A. to B. like C. as D. by52. A. as B. until C. before D. after53. A. following B. former C. above D. next54. A. still B. also C. almost D. once55. A. treatment B. action C. choice D. remark第三部分阅读理解(共20小题;每小题2分,满分40分)ACasual Fridays are days when the usual dress code of an officeis relaxed. Since Fridays mark the start of the weekend in manycountries, behavior in general tends to be more relaxed on Fridays,with some employers not being as particular about standards, andsometimes this is formalized in a casual Friday policy. Truly casualclothing like sweatpants and t-shirts is usually not satisfying.The casual Friday concept was one of the first signs that strict dress codes for offices were starting to relax. The principle of Dress-Down Fridays seems to have appeared around the 1950s, probably in response to changing attitudes about careers and the workplace. Over time, some offices have relaxed their dress codes even more, adopting a business casual dress code every day.On casual Fridays, employees may wear clothing which is classified as ―business casual,‖ meaning that it still obeys certain standards. Clothing must be clean and in good repair, with no spots or discolorations(变色). Men are generally allowed to wear casual trousers and bright-colored clothes, and in some offices they may be allowed to wear button down shirts or polo shirts, with or without a tie. Women tend to wear neat skirts or dresses; casual trousers are common and generally perfectly acceptable, except in very conservative workplaces.As a general rule, extremely casual clothing like jeans, shorts, t-shirts, tank tops(吊带衫), and so forth is discouraged on casual Fridays. Employees are still expected to make an effort to look professional, out of respect to their offices and their customers.Many people feel that casual Fridays are good for employees’ mood, encouraging employees to express their individuality and feel more comfortable in the office. Others feel that business casual dress and the concept of casual Fridays devalues the workplace, by taking a sense of formality and respect away. Approaches to casual Fridays vary, depending on the industry and the nation; the technology industry, for example, is well known for casual dress in general, while major financial companies tend to dislike casual dress.56. From the second paragraph, we can infer _______________.A. the casual Friday concept is out of fashionB. it’s forbidden to wear casual clothing in offices beforeC. the idea of casual Fridays has originated from the early 1990sD. employees can wear all kinds of casual clothing every day now.57. The underlined word ―devalue‖ in the last paragraph means ____________.A. think little ofB. think highly ofC. estimate highlyD. take the place of58. Generally speaking, on casual Fridays as a worker in a company you can wear _____.A. jeansB. tank topsC. shortsD. polo shirts59. The article most probably appears in the newspaper Column ___________.A. SportsB. EntertainmentC. CultureD. CustomsB"Good morning, Discovery Center. We're happy to be here with you. I'm Barb Morgan. And we are ready for your first question. "That was teacher-turned-astronaut Barbara Morgan, speaking from more than 320 kilometers above Earth. She was greeting the students in the northwestern state of Idaho. They gathered at the Discovery Center in Boise on Tuesday to ask the astronauts questions by video link.Barbara Morgan taught elementary school in McCall, Idaho, before she trained for space. She and six other astronauts arrived Friday on the shuttle(航天飞机)Endeavour to bring supplies and new equipment to the international station. Barbara Morgan is fifty-five years old. She taught for many years before she became an astronaut.One student asked, "Hi, I'm Sarah Blum. How does being a teacher relate with being an astronaut on this mission?" Barbara Morgan said "Well, actually, astronauts and teachers actually do the same things. We discover and we share. The great thing about being a teacher is that you get to do that with students. And the great thing about being an astronaut is you get to do it in space. And those are absolutely wonderful jobs."Barbara Morgan first prepared for a shuttle flight more than twenty years ago. She trained in case NASA needed a substitute for Christa McAuliffe, its choice to become the first teacher in space.Then, in 1986, Christa McAuliffe died with the Challenger crew when the shuttle exploded shortly after launch. After the disaster, NASA officials barred other civilians from shuttle flights. But in 1998, they created a new position for teachers to become fully trained astronauts. Barbara Morgan is NASA's first "educator astronaut" launched into orbit.One of her first tasks was to operate Endeavour's robotic arm to inspect the shuttle for any launch-related damage. Cameras showed a small area hit by a piece of protective foam (泡沫)that fell off the fuel tank. NASA officials say the damage is not a safety threat but they are deciding what to do about it.60. Barbara Morgan is different from other astronauts on the shuttle in that _____.A. she used to be a woman pilotB. she used to be a teacherC. she was chosen to become the first teacher in spaceD. she was the first woman to travel in space61. The chief aim of the space flight on the shuttle Endeavour was to_____.A. bring supplies and new equipment to the international stationB. inspect the shuttle for any launch-related damageC. give a music lesson by Barbara Morgan from spaceD. launch a communication satellite62. We can infer that _____.A. Barbara Morgan prefers a teacher to an astronautB. Barbara Morgan prefers an astronaut to a teacherC. Barbara Morgan like both a teacher and an astronautD. Barbara Morgan like neither a teacher nor an astronaut63. If Christa McAuliffe had not been on the shuttle the Challenger in 1986, what would have happened?A. Christa McAuliffe would have become a famous actress.B. Barbara Morgan would have died.C. nothing would have happened to the Challenger.D. Barbara Morgan would have become a NASA official.64. Which of the following is the right order about Barbara Morgan?a. answered the questions in spaceb. prepared for a shuttle flight more than twenty years agoc. taught elementary school in McCall, Idahod. became a substitute for Christa McAuliffeA. a, b, c, dB. a, c, b, dC. c, b, d, aD. d, b, c, aCThe campaign is over. The celebrations have ended. And the workfor US president-elect Barack Obama has begun.The 47-year-old politician rose to the highest post because of hisstand against the war in Iraq and his plans to fix a weak economy. Butwhat will the first 47-year-old African-American president do for race relations?Obama’s victory appears to have given blacks and other minorities a true national role model. For years, many looked to athletes and musicians for inspiration. As Darius Turner, an African-American high school student in Los Angeles, told the Los Angeles Times, ―Kobe doesn’t have to be everybody’s role model anymore.‖Recent polls(民意测验)also suggest that Obama’s victory has given Americans new optimism about race relations. For example, a USA Today poll found that two-thirds of Americans believe relations between blacks and whites ―will finally be worked out‖. This is the most hopeful response since the question was first asked during the civil rights revolution in 1963.However, it’s still too early to tell whether Obama’s presidency will begin to solve many of the social problems facing low-income black communities.Although blacks make up only 13 percent of the US population, 55 percent of all prisoners are African-American. Such numbers can be blamed on any number of factors on America’s racist past, a failure of government policy and the collapse(瓦解)of the family unit in black communities.It is unlikely that Obama will be able to reverse (扭转) such trends overnight. However, Bill Bank, an expert of African-American Studies, says that eventually young blacks need to find role models in their own communities. ―That’s not Martin Luther King, and not Barack Obama,‖ he told the Los Angeles Times. ―It’s actually the people closest to them. Barack only has so much influence.‖In the opinion of black British politician Trevor Phillips, Obama’s rise will contribute more to multiculturalism than to race relations in the US.―When the G8 meets, the four most important people in the room will be the president of China, the prime minister of India, the prime minister of Japan and Barak Obama,‖ he told London’s The Times newspaper. ―It will be the first time we’ve seen that on our television screens. That will be a huge psychological shift (心理转变) for both the white people and the colored ones in the world.‖65. According to Bill Bank, ____________.A. it’s better for young blacks to f ind role models in those who are close to themB. young blacks should not be so much influenced by ObamaC. blacks should find other role models because Obama is far from their realityD. Obama is not the proper role model for African-Americans66. What would be the best title for this passage?A. The First African-American PresidentB. America’s New Role ModelC. Obama-- A Successful Black .D. Choosing a Right Role Model.67. What will be the huge psychological shift Trevor mentioned at the end of the passage?A. The other three leaders all support Obama.B. Obama is an African-American president.C. None of the four leaders is white.D. The other three leaders except Obama are from Asian countries.DIn the push to improve education, test scores get most of the attention. Many educators say, "Test scores shadow other ways in which children can be inspired to learn." Here is a fresh method to ensure students to make the grade.●The SchoolFairview Elementary School is located in the south side of Modesto, California, where the crime rate is high. Fairview, with about 1,000 students from kindergarten through to sixth grade, has long suffered from discipline problems, poor test scores, and a nearly total lack of communication with their parents. These problems are not surprising because many of the parents —immigrants who work on farms or in factories —speak little or no English.●The MethodSince 2002, Fairview Elementary School has been on a five-year program: to keep America strong, children must be trained to respect many points of view, and understand the five freedoms, such as freedom of speech. As students learn the rights and responsibilities of a good citizen, they will develop the skills and attitudes needed to excel in study.The program has no set classes. Fairview students enjoy "freedoms" other kids might envy (they voted not to wear school uniforms, for example). They accept such responsibilities as speaking up during class discussions, and keeping the school clean and safe (Fairview is considered the cleanest of 33 schools in its district). Different from tradition, there is no hand-raising in class. "Instead," says teacher Deborah Supnet, "we teach them to listen until the other child stops talking." It is called an exercise in respect.●_________________Last year, the number of students who performed well in math doubled, from 15% to 30%. Suspension(停学) rates dropped by 50%. What is particularly encouraging to headmaster Rob Williams is that the school now has an active parents’group: Parent s with a V oice. One of those parents, Laura Malagon, believes that the program allows her to play a more active role in her children’s school life.68. The passage mainly talks about ______.A. the disadvantages immigrants faceB. a change in an immigrant schoolC. many ways to improve students’ qualitiesD. a fresh idea that can improve education69. According to the passage, which was NOT a problem at the Fairview Elementary School?A. The bad behavior of students.B. The poor performance of their students.C. The lack of freedom to speak.D. The lack of contact with parents.70. We can add a subheading to the last paragraph in the blank with ________.A. Signs of SuccessB. Future of the SchoolC. Effort of the ParentsD. Parents with a V oiceEDuring your lifetime, the person you’ll spend most time with is you! 71.When you are going upwards in life, you tend to overestimate yourself. It seems that everything you seek for is within your reach; luck and opportunities will come your way and are overjoyed that you think they constitute part of your worth.72.To get a thorough understanding of yourself is to gain a correct view of yourself—be aware of both your strengths and weaknesses. Y ou may look forward hopefully to the future but be sure not to expect too much, for not all ideals can be fully realized.73 . In time of sadness, do yourself a favor by sharing it with your friends so as to change a gloomy mood into a cheerful one; in time of tiredness, do yourself a favor by getting a good sleep or taking a tonic(恢复精力的东西). 74.To get a thorough understanding of yourself is to get full control of yourself. 75.A.When you are going downhill, you tend to underestimate yourself, mistaking difficulties for your own incompetence.B.In time of trouble, turn to your teacher for help.C.If you fall ill, it’s up to you to take good care of yourself.D.Having done so, you will find your life full of color and flavor.E.To know yourself better, you should learn more.F.To get a thorough understanding of yourself is to know when to do yourself a favor.G.But do you really understand yourself?宁夏六盘山高中2008-2009学年高三第三次模拟试卷答题卡II卷(主观题)35分第四部分写作(共二节,满分35分)第一节短文改错。