(新学期课堂同步精炼)高二下学期开学考测试(附答案)

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高二下学期开学考试物理试题(含答案和解析)

高二下学期开学考试物理试题(含答案和解析)

四川省成都外国语学校【精品】高二下学期开学考试物理试题学校:___________姓名:___________班级:___________考号:___________一、单选题1.下列说法正确的是( )A .卡文迪许通过扭秤实验测出了静电力常量B .电流的磁效应是法拉第发现的C .摩擦起电是在摩擦的过程中创造了电荷D .FE q=是通过比值法来定义物理量 2.真空中两个点电荷相距r 时,静电力为F ,如果保持它们的电量不变,而将距离增大为2r 时,则静电力将变为A .2F B .F C .4FD .2F3.如图所示,一质量为M 的木质框架放在水平桌面上,框架上悬挂一劲度系数为k 的轻质弹簧,弹簧下端拴接一质量为m 的铁球。

用手向下拉一小段距离后释放铁球。

铁球便上下做谐运动,则( )A .弹簧处于原长时的位置是铁球做简谐运动的平衡位置B .在铁球向平衡位置运动的过程中,铁球的位移、回复力、加速度都逐渐减小,速度和小球重力势能增大C .若弹簧振动过程的振幅可调,则当框架对桌面的压为零时,弹簧的压缩量为mgkD .若弹簧振动过程的振幅可调,且保证木质框架不会离开桌面,则铁球的振幅最大是()m M gk+ 4.如图所示,虚线a 、b 、c 代表电场中的三个等势面,相邻等势面之间的电势差相等,即U ab=U bc,实线为一带负电的点电荷仅在电场力作用下通过该区域时的运动轨迹,P、R、Q是这条轨迹上的三点,R在等势面b上,据此可知()A.三个等势面中,c的电势最低B.该点电荷在R点的加速度方向垂直于等势面bC.Q点的场强大于P点场强D.该点电荷在P点的电势能比在Q点的小5.图中a、b、c、d为四根与纸面垂直的长直导线,其横截面位于正方形的四个顶点上,导线中通有大小相同的电流,方向如图所示.可以判断出a、b、c、d四根长直导线在正方形中心O处产生的磁感应强度方向是( )A.向左B.向右C.向上D.向下6.如图所示,a、b是两个匀强磁场边界上的两点,左边匀强磁场的磁感线垂直纸面向里,右边匀强磁场的磁感线垂直纸面向外,右边的磁感应强度大小是左边的磁感应强度大小的2倍。

高二语文下学期开学考试题(有答案)

高二语文下学期开学考试题(有答案)

高二语文下学期开学考试题(有答案)注意:1、本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,满分150 分。

考试时间:150分钟。

答卷前,考生务必将自己的姓名和考号填写或填涂在答题卷指定的位置。

2、选择题答案用2B铅笔把答题卡上对应题目的答案标号涂黑;如需改动,用橡皮擦干净后,再选涂其他答案;不能答在试题卷上。

3、主观题必须用黑色字迹的钢笔或签字笔在答题卷上作答,答案必须写在答题卷各题目指定区域内的相应位置上,超出指定区域的答案无效;如需改动,先划掉原来的答案,然后再写上新的答案。

第I卷阅读题一、现代文阅读(一)(9分,每小题3分)阅读下面的文字,完成1~3题。

人们常说“小说是讲故事的艺术”,但故事不等于小说,故事讲述人与小说家也不能混为一谈。

就传统而言,讲故事的讲述亲身经历或道听途说的故事,口耳相传,把它们转化为听众的经验;小说家则通常记录见闻传说,虚构故事,经过艺术处理,把它们变成小说交给读者。

除流传形式上的简单差异外,早起小说和故事的本质区别并不明显,经历和见闻是它们的共同要素,在传统较为落后的过去,作为远行者的商人和水手最适合充当故事讲述人的角色,故事的丰富程度与远行者的游历成正比。

受此影响,国外古典小说也常以人物的经历为主线组织故事,《荷马史诗》《一千零一夜》都是描述某种特殊的经历和遭遇,《唐吉可德》中的故事是唐吉可德的行侠奇遇和所见所闻,17世纪欧洲的流浪汉小说也体现游历见闻的连缀。

在中国民间传说和历史故事为志怪类的小说提供了用之不竭的素材,话本等古典小说形式也显示出小说和传统故事的亲密关系。

虚构的加强使小说和传统之间的区别清晰起来。

小说中的故事可以来自想象。

不一定是作者的亲历亲闻。

小说家常闭门构思,作品大多诞生于他们离群索居的时候,小说家可以闲坐在布宜诺斯艾利斯的图书馆中,或者在巴黎一间终年不见阳光的阁楼里,杜撰他们想象中的历险故事,但是,一名水手也许历经千辛万苦才能把在东印度群岛听到的故事带回伦敦;一个匠人漂泊一生,积攒下无数的见闻、典故或趣事,当他晚年坐在火炉旁给孩子们讲述这一切的时候,他本人就是故事的一部分,传统故事是否值得转述,往往只取决于故事本身的趣味性和可流传性,与传统的故事方式不同,小说家一般并不单纯转述故事,他是在从事故事的制作和生产,有深思熟虑的讲述目的。

学年下学期高二开学考试化学试题(附答案)

学年下学期高二开学考试化学试题(附答案)

广西桂林市第十八中学学年下学期高二开学考试化学试题注意:1、本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,满分100 分。

考试时间:90分钟。

答卷前,考生务必将自己的姓名和考号填写或填涂在答题卷指定的位置。

2、选择题答案用2B铅笔把答题卡上对应题目的答案标号涂黑;如需改动,用橡皮擦干净后,再选涂其他答案;不能答在试题卷上。

3、主观题必须用黑色字迹的钢笔或签字笔在答题卷上作答,答案必须写在答题卷各题目指定区域内的相应位置上,超出指定区域的答案无效;如需改动,先划掉原来的答案,然后再写上新的答案。

可能用到的相对原子质量:H—1 C—12 N—14 O—16第I卷(选择题共54 分)一、选择题(本题包括20小题,每小题3分,共60分,每小题只有一个正确答案)1.下列叙述正确的是( )A.电能是二次能源 B. 水力是二次能源C.天然气是二次能源 D. 水煤气是一次能源2.下列说法中有明显错误的是()A.压缩容器体积,增大压强,活化分子百分数不变,化学反应速率增大B.升高温度,活化分子百分数增大,化学反应速率增大C.加入反应物,活化分子百分数增大,化学反应速率增大D.使用催化剂,降低了反应所需的活化能,活化分子百分数增大,化学反应速率增大3.反应4NH3(g)+5O2(g) 4NO(g)+6H2O(g)在5L密闭容器中进行,半分钟后NO的物质的量增加了0.3 mol,则此反应的平均速率v为()A.v(O2)=0.01 mol/(L·s)B.v(NO)=0.08 mol/(L·s)C.v(H2O)=0.003 mol/(L·s)D.v(NH3)=0.001 mol/(L·s)4.下列事实,不能用勒夏特列原理解释的是()A.氨水中加酸,NH+4的浓度增大B.合成氨工业中不断从反应混合物中液化分离出氨气C.实验室中常用排饱和食盐水的方法收集Cl2D.合成氨控制在500 ℃左右的温度5.下列方法对2SO2(g)+O2(g) 2SO3(g)的反应速率没有影响的是()A.加入SO3B.容积不变,充入N2 C.压强不变,充入N2D.降低温度6.1mol碳完全燃烧后可放出393.5KJ的热量,下列热化学方程式正确的是()A.C(s)+O2(g)=CO2(g)△H = +393.5 kJ/molB.C(s)+1/2O2(g)=CO(g)△H = -393.5 kJ/molC.C + O2 = CO2 △H = -393.5 kJ/molD.C(s)+O2(g)=CO2(g)△H = -393.5 kJ/mol7.根据以下3个热化学方程式:2H2S(g)+3O2(g)=2SO2(g)+2H2O(l) △H =-Q1 kJ/mol2H2S(g)+O2(g)=2S (s)+2H2O(l) △H =-Q2 kJ/mol 2H2S(g)+O2(g)=2S (s)+2H2O(g) △H =-Q3 kJ/mol判断Q1、Q2、Q3三者关系正确的是()A. Q1>Q2>Q3B. Q1>Q3>Q2C. Q3>Q2>Q1D. Q2>Q1>Q38. 下列各图中,表示正反应是吸热反应的图是( )9.已知反应CO(g)=C(s)+O2(g)的△H为正值,△S为负值。

高二下学期开学考试试题1 7

高二下学期开学考试试题1 7

卜人入州八九几市潮王学校铁人二零二零—二零二壹高二语文下学期开学考试试题试题说明:1、本试题总分值是150分,答题时间是150分钟。

2、请将答案填写上在答题卡上,在在考试完毕之后以后只交答题卡。

一、现代文阅读〔36分〕〔一〕阐述类文本阅读〔此题一共3小题,9分〕阅读下面的文字,完成1-3题。

①当今的艺术仿佛在兴致勃勃地享受一场技术的盛宴。

戏曲舞台上眼花缭乱的灯光照射,3D电影院里上下左右晃动的座椅,魔术师利用各种光学仪器制造观众的视觉误差,摄影师借助计算机将一张平庸的面容修饰得貌假设天仙……总之,从声光电的全面介入到各种闻所未闻的机械设备,技术的开展速度令人吃惊。

然而,有多少人考虑过这个问题:技术到底赋予了艺术什么?关于世界,关于历史,关于神秘莫测的人心——技术增添了哪些发现?在许多贪大求奢的文化工程、文艺演出中,我们不难看到技术崇拜正在形成。

②技术是艺术消费的组成局部,艺术的创作与传播从来没有分开技术的支持。

但即便如此,技术也从未扮演过艺术的主人。

史记、窦娥冤、红楼梦……这些之所以成为经典,是因为它们的思想光与艺术魅力,而不是因为书写于竹简,上演于舞台,或者者印刷在书本里。

然而,在现代社会,技术的日新月异造就了人们对技术的盲目崇拜,以致于许多人没有觉察艺术消费正在出现一个颠倒:许多时候,技术植入艺术的真正原因其实是工业社会的技术消费,而不是艺术演变的内在冲动。

换言之,这时的技术无形中晋升为领跑者,艺术更像是技术创造力图开拓的场。

③中国艺术的“简约〞传统隐含了对于“炫技〞的不屑。

古代思想家认为,繁杂的技术具有炫目的迷惑性,目迷五色可能干扰人们对于“道〞的持续注视。

他们众口一辞地告诫“文胜质〞可能导致的危险,这是古代思想家的人文情怀。

当然,这并非号召艺术回绝技术,而是敦促文化消费审慎地考虑技术的意义:假设不存在震撼人心的主题,繁杂的技术只能沦为虚有其表的形式。

④这种虚有其表的形式在当下并不少见,光怪陆离的外观往往掩盖了内容的苍白。

《新学期课堂同步精炼》高二下学期开学考测试(含答案)

《新学期课堂同步精炼》高二下学期开学考测试(含答案)

参考答案听力1-5 ABCAC6-10 BACAC 11-15 BBACC 16-20 ABCBA阅读理解21-25CBDCB 26-30ACBBB 31-35DBDCB36-40 FEDBG完形填空41-45ADBCB 46-50CBCDC 51-55ABDCC 56-60BCADC语篇填空61. to tell; 62. about; 63. It;64. costs;65. The66. for;67. that;68. called;69. to do;70. hottest短文改错I live I a village in the north of Henan Province. It’s the government’s policy for building roads to every village that has brought great changes to my hometown.Before the policy was carried out, there was no road or bus line near the village, so traffic was very inconvenient. Local specialities couldn’t be transported out to other places for sale. Large areas of wasteland and many pools were deserted. Children had to walk on narrow, muddy roads to go to school. But now everything is quite different. There have been wide roads and bus service, and traffic is very convenient. The local specialities can be transported to different places, which has greatly increased the income of the villages. Nowadays children can go to school either by bus or by bike. Meanwhile, the villagers have developed eco-tourism by using wasteland and pools, which attracts many tourists from cities. People’s life is becoming better and better.书面表达Class is where we study and struggle for a bright future. It's the stage on which we show our talents and also a community to develop our social potential. In this sense, a sound class is the assurance of a reliable and enjoyable environment for us. To build a favorite class, I advocate the actions as follows.First, all hands make a big fire. thus, I will contribute every bit of my effort to public benefts. For example,I will volunteer to clean the blackboard, empty the dustbin, or anything else if no one is on duty. Second, each individual mirros a community. if I were a broken window, there would be no safety and harmony in class. Therefore, I will behave myself well for the betterment of my class.Last but not least, I wish everyone could take the class as our family, love each other and cherish every moment in school.听力材料(Text 1)W: Oh, is this a photo of your graduation? Who is the girl standing next to you? M: That’s Jane. She became a doctor after graduation.(Text 2)M: How about going camping with us tomorrow?W: I’d like to, but I’ve promised Tim I’d go downtown with him tomorrow. He needs some new books.(Text 3)M: This is the last time I’m telling you. Turn the radio off or leave.W: Why don’t you read somewhere else?M: Because I need to use the dictionaries in this classroom for my writing. (Text 4)M: How I wish I were Adam!W: Why?M: He’s the funniest guy I’ve ever met. He makes everybody laugh.W: Well, people are different. You’re very considerate and friendly.(Text 5)W: Did you feed Jim today?M: Yes, and I also gave him a bath.W: Thanks, honey. You’re such a good father.M: Well, believe it or not, I’ve been reading a lot online about how to be a good dad.(Text 6)W: Hey, Mark, I’ve been driving for hours. Would you mind driving for a while? I want to take a rest.M: Sorry. I would like to, but I don’t know how to drive.W: What? You haven’t learned how to drive?M: No. I usually go to work by bus or subway. It’s very convenient because I don’t live far from my office.W: Have you considered learning how to drive?M: No. There are already too many cars on the streets. Besides, I don’t really need to use a car all that much.(Text 7)W: Is this The Mona Lisa by Leonardo da Vinci? It looks a bit strange.M: Well, Mona Lisa looks much younger in this painting. This is called The Isleworth Mona Lisa.W: Ah, the smile on her face is bigger than the original Mona Lisa. Who painted this one?M: It is believed to have also been painted by Leonardo da Vinci in the early 16thcentury.W: Really? I thought it must have been painted by a modern-day artist.M: Hmm… Th e description says it was first discovered shortly before World War1. It belongs to a private family and they haven’t sold it to any museum yet. (Text 8)M: What’s your plan for the summer, Lisa?W: I’m going to spend a few days in Beijing. My cousin is teaching English in a language school there. I’ll travel around the city with her. What about you? M: I’m going to take a trip to Cairo. I want to visit the Egyptian Museum there. W: Is it the museum’s famous paintings that you want to see?M: Actually I’m more interested in seeing the museum’s coin display. I read that the first floor has a collection of coins used by the ancient Egyptians.W: Ah, that’s right. You like collecting coins. What else are you interested in seeing there?M: Well, I know that the second floor has some amazing things from the final two dynasties of ancient Egypt, many things which were used by the Egyptian kings.W: So how many floors does it have?M: Just two floors. It’s not very big, but I think it’s really worth a visit.(Text 9)W: What are you reading, Sam?M: The Kite Runner. It’s a great book.W: What is it about?M: It’s about a boy who grows up in Afghanistan during the 1980s. It’s a very touching story and also a little sad. Do you like such books?W: I used to like exciting adventure stories, but now I love reading books like that. M: Jenny, would you be interested in joining a book club?W: Sure. Do you know of one?M: Yes, I’m in one now. If you’d like to join, we meet every other week on Saturday evening and discuss the books that we’ve read. We begin at 8:00 pm and chat for about one and a half hours. We usually learn a lot and it’s a lot of fun. W: I’m really interested, but I can’t go for the next several weeks. I’m concentrating on working out in the gym for the next six weeks, but I’ll have time after that. M: Great. You can join us then.W: Yeah. I really look forward to meeting other people who also love reading. (Text 10) WThere were several reasons behind our decision to move to Flemington. The process started about 18 months ago. We were still living in a one-bedroom apartment then, but at that time we decided to move into a larger house before we had a baby. We began to look at houses at reasonable prices. Then, much sooner than expected, I was having a baby. Soon after that, Mark’s company announced that they were relocating to Flemington, about 90 miles south of us. Mark’s company had be en good to him, and it was one of the few around withexcellent benefits, family-friendly policies, and a child-care center on site. With a baby on the way, these things were essential for us. Since employees had to find their own housing, we began looking at houses in Flemington. We also did some research on our school’s website and asked my colleagues for advice. We were very excited about what we found — reasonable housing costs, great schools, and a lively town. We’ll be moving about a month before the baby is due. Let’s hope he doesn’t decide to come early.。

高二下学期开学考试试题_3 (2)(共24页)

高二下学期开学考试试题_3 (2)(共24页)

2021-2021学年(xuénián)高二语文下学期开学考试试题考试时间是是:150分钟注意:本套试卷包含Ⅰ、Ⅱ两卷。

第一卷为选择题,所有答案必须需要用2B铅笔涂在答题卡中相应的位置。

第二卷为非选择题,所有答案必须填在答题卷的相应位置。

答案写在试卷上均无效,不予记分。

一、现代文阅读〔一〕阐述类文本阅读〔此题一共 3 小题,9 分〕阅读下面的文字,完成 1~3 题。

一个没有英雄的历史是寂寞的无声的历史,没有英雄的民族是孱弱的民族。

中国人的民族性格是在特定的经济消费方式和制度下的文化的凝结,而文化精华又与广阔人民哺育了中国历史的和现实的出色人物。

他们堪称民族的脊梁,国家的栋梁。

中华民族历史和现实中的人物,就是中华文化根本精神的人格化,也是中国人民的出色的儿子。

他们既是文化和人民的产儿,又是具有文化传承和民族鼓励力量的样板。

中华文化的根本精神是中华民族文化的精粹,是中华民族精神的主轴和珍贵的精神财富。

我们不能否认传统文化中存在的糟粕需要批判,人民中受其影响而产生的落后的东西需要不断改良。

我们不赞美三寸金莲,不能赞美纳妾等一切与近代文明相悖的东西。

但也应该相信没有永久不变的中国人,没有永久不变的民族性格。

在旧的经济制度和政治制度下形成的中国人的某些缺点会发生变化。

人的本质是社会关系的总和。

阿Q是旧式农民形象,而不是中国农民永久的形象。

没有天性丑陋的中国人。

任何对国民性和所谓民族劣根性的鞭挞,最终假设不指向旧的经济制度和政治制度,只停留在文化层面,那么是难中腠里的。

中华文化的根本精神具有世代延续的价值。

可是假如没有高度兴旺的先进消费力、先进的消费方式和先进的政治制度,传统文化是不能单独发生作用的。

中国鸦片战争以后一百多年的民族屈辱史已证明了这一点。

当时的孔子只是孔庙中的圣人,当时的经典只能是藏书楼里的典籍。

当年黑格尔非常轻视孔子的思想,说?论语?“里面所讲的是一种常识道德,这种常识道德我们在哪里都找得到,在哪一个民族都找得到,可能还要好些,这是些毫无出色之点的东西〞。

高二下学期开学考试试题含解析 3

卜人入州八九几市潮王学校高级二零二零—二零二壹高二语文下学期开学考试试题〔含解析〕答题时间是:150分钟总分值是:150分一、语言运用〔总分值是9分〕1.以下各句中加点成语使用,全都不正确的一项为哪一项哪一项〔〕①寒假自习期间,少数同学自律较差,出现了上课讲话、看玄幻小说、睡觉等现象,虽经教师教育,但仍不以为训....。

②好的文章都是直抵人心的,它一定会呈现自己的简洁,表达自己的真情实感,而绝不会拾.人牙慧...。

③8月19日,全城瞩目的万达终于开业了,人倾城而动,万人空巷....,齐聚在这一梦幻之地。

④乐视公司在贾跃亭的主导之下,大力推行多元开展之路,不想资金链断裂,难免阪上走...丸.,公司易主。

⑤如今越来越多的人重视给孩子取名字,但很多人找先生批八字,看五行,过于拘泥,不免有胶柱鼓瑟....之嫌。

⑥6月,印度HY 队侵入中印边境我方洞朗地区,阻挠我方正常修路活动,印方这一火中取栗....行为,必将付出应有的代价。

A.①②⑥B.①③⑤C.②③④D.①④⑥【答案】D【解析】试题分析:不以为训:不值得作为效法的准那么或者典范。

坂上走丸:比喻形势开展迅速成或者工作进展顺利。

火中取栗:比喻冒危险给别人出力,自己却上大当,一无所得。

2.以下各句中,没有语病的一项为哪一项哪一项〔〕A.经过大家几天来耐心细致的教育,终于使他充分地认识到了自己的缺点及其危害,并下定决心彻底改正。

B.作为家长,我们要努力进步孩子的兴趣爱好,而不是盲目跟风,让孩子参加各种特长班,剥夺孩子自主选择的权利。

C.上班顶峰时段,机动车道上,各种轿车、摩托车、巴士、机动车辆往来穿梭;人行道上,人们行色匆匆,络绎不绝。

D.腾讯公司称,除夕当天全国微信红包收发超过10亿个,不过它没有给出整个春节长假的总体数字。

【答案】D【解析】【详解】此题考察辨析并修改病句的才能。

解答此类题目,应先阅读选项,排除有明显错误标志的句子,然后按照主谓宾的成分压缩句子,先观察主HY分,是否存在搭配不当、残缺等问题,再分析修饰成分。

高二下期语文开学考试测试题

高2021届高二(下)开学考试语文测试卷(答案附后)一、现代文阅读(36分)(一)论述类文本阅读(本题共3小题,9分)阅读下面的文字,完成后面下列小题。

灵感既然是突如其来,突然而去,不由自主,那不就无法可以用人力来解释么?从前人大半以为灵感非人力,以为它是神灵的感动和启示。

在灵感之中,仿佛有神灵凭附作者的躯体,暗中驱遣他的手腕,他只是坐享其成。

但是从近代心理学发现潜意识活动之后,这种神秘的解释就不能成立了。

什么叫做“潜意识”呢?我们的心理活动不尽是自己所能觉到的。

自己的意识所不能察觉到的心理活动就属于潜意识。

意识既不能察觉到,我们何以知道它存在呢?变态心理中有许多事实可以为凭。

比如说催眠,受催眠者可以谈话、做事、写文章、做数学题,但是醒过来后对于催眠状态中所说的话和所做的事往往完全不知道。

此外还有许多精神病人现出“两重人格”。

例如一个人乘火车在半途跌下,把原来的经验完全忘记,换过姓名在附近镇市上做了几个月的买卖。

有一天他忽然醒过来,发现身边事物都是不认识的,才自疑何以走到这么一个地方。

旁人告诉他说他在那里开过几个月的店,他绝对不肯相信。

心理学家根据许多类似事实,断定人于意识之外又有潜意识,在潜意识中也可以运用意志、思想,受催眠者和精神病人便是如此。

在通常健全心理中,意识压倒潜意识,只让它在暗中活动。

在变态心理中,意识和潜意识交替来去。

它们完全分裂开来,意识活动时潜意识便沉下去,潜意识涌现时,便把意识淹没。

灵感就是在潜意识中酝酿成的情思猛然涌现于意识。

它好比伏兵,在未开火之前,只是鸦雀无声地准备,号令一发,它乘其不备地发动总攻击,一鼓而下敌。

在没有侦探清楚的敌人(意识)看,它好比周亚夫将兵从天而至一样。

据心理学家的实验,在进步停顿时,你如果索性不练习,把它丢开去做旁的事,过些时候再起手来写,字仍然比停顿以前较进步。

这是什么道理呢?就因为在意识中思索的东西应该让它在潜意识中酝酿一些时候才会成熟。

功夫没有错用的,你自己以为劳而不获,但是你在潜意识中实在仍然于无形中收效果。

2023-2024学年北京九中高二下学期2月开学考数学试题及答案

2024北京九中高二(下)开学考数 学(考试时间120分钟 满分150分)一、单选题(共80分)45120135150.已知双曲线轴上,且其中一条渐近线的方程为,则双曲线中,ABC 和ABP 均为正三角形,PABC 的大小为60,则异面直线PB 与AC 所成角的余弦值是(A .18− B .18.1− D .145,再从条件①12AB AC k =−.参考答案ME BC MF AD,且ME,EMF为异面直线AD与中,1,==ME MF EF,由于ABC和ABP均为等边三角形,所以P AB C的平面角,即60POC=,POC∴为等边三角形,故,进而3 ON PC=设正方体的棱长为1,则),1F a ,(E b ,故(0,AF a =,(CE b =−与直线CE 所成角为θ, 1,AF CE =CERt Rt≅,所以OFG OFHML=,1==++AB GM OG OL LM故答案为:3.π11BD DD D =⊂平面BDD 1AB A =,且在PCD中,在菱形ABCDAB DC,所以所以ADB为正三角形.因为AD则()(3,0,,3,2EF t EC =−=−根据条件,可得平面FCD 的法向量为(11,0,0n =设平面EFC 的法向量为(2,n x y =, 2200n EF n EC ⎧⋅=⎪⎨⋅=⎪⎩,所以30x tz y +==2t =,则y ,所以(22,n t =的大小为45,121212,3n n n n n n ⋅==><PD 的中点,所以PD ,即ADB 为正三角形.因为ADB 为正三角形且()0,0,(t t >则()(3,0,,3,2EF t EC =−=−根据条件,可得平面FCD 的法向量为(11,0,0n =的法向量为(2,n x y =,2200n EF n EC ⎧⋅=⎪⎨⋅=⎪⎩,所以0y =2t =,则y ,所以(22,n t =的大小为45,121212,3n n n n n n ⋅==><PD 的中点,所以PD 2)原问题等价于0OM ON ⋅=,联立方程,利用韦达定理即可得到结果)设()00,B x y 则C 2所以12OM ON x x ⋅=()2221212k k −+=+所以,OM ON ⊥,即∠所以,以线段MN为直径的圆经过原点.【点睛】圆锥曲线中定点问题的常见解法(1)假设定点坐标,根据题意选择参数,建立一个直线系或曲线系方程,而该方程与参数无关,故得到一个关于定点坐标的方程组,以这个方程组的解为坐标的点即所求定点;(2)从特殊位置入手,找出定点,再证明该点符合题意.。

山东省高二下学期开学考试数学试题(解析版)

一、单选题1.已知是关于x 的方程的根,则实数( ) 2i +250x ax ++==a A . B . C .2 D .42i -4-【答案】B【解析】依题意知方程的根互为共轭复数,结合韦达定理可求得结果. 【详解】因为是关于x 的方程的根,则另一根为 2i +250x ax ++=2i -由韦达定理得,所以 ()()22i i a ++-=-4a =-故选:B2.直线的斜率为( ) 10x +=A B .C D .【答案】C【解析】可化为. 10x +=y x =+【详解】可化为10x +=y x =+k =故选:C【点睛】本题主要考查了已知直线方程求斜率,属于基础题.3.已知动点在直线上运动,动点在直线上运动,且,则P 1:3410l x y -+=Q 2:640l x my ++=12l l //的最小值为( )PQ A .B .C .D .3531015110【答案】C【解析】根据两平线上任意两点距离的最小值即为平行线间的距离求解. 【详解】因为, 12l l //所以,解得, 64341m =≠-8m =-化简得 2:3420l x y -+=设间的距离为,则,12,l l d 15d ==由平行线的性质知的最小值为,PQ 15故选:C4.已知在一个二面角的棱上有两个点、,线段、分别在这个二面角的两个面内,并且A B AC BD都垂直于棱,,,, ) AB 5AB =3AC =4BD =CD =A . B .C .D .30︒45︒90︒150︒【答案】C【解析】设这个二面角的度数为,由题意得,从而得到,由此能求出结αCD CA AB BD =++cos α果.【详解】解:设这个二面角的度数为,α由题意得,CD CA AB BD =++, 22222||||cos()CD CA AB BD CA BD πα∴=+++⋅-,292516234cos α∴=++-⨯⨯⨯解得, cos 0α=∴,90α=︒∴这个二面角的度数为, 90︒故选:C.【点睛】本题考查利用向量的几何运算以及数量积研究面面角,属于中档题. 5.如图,在正四面体中,是的中点,则与所成角的余弦值是OABC D OA BD OCA .BCD 12【答案】B【分析】取的中点,连接,,可得就是与所成的角, 设,可得AC E DE BE BDE ∠BD OC OA a =,,利用余弦定理可得的值,可得答案.BD BE ==12DE a =cos BDE ∠【详解】解:如图: ,取的中点,连接,,可得就是与所成的角, AC E DE BE BDE ∠BD OC设,则,,OA a =BD BE ==12DE a =222cos 2BD DE BE BDE BD DE +-∠==⋅故选: B.【点睛】本题主要考查异面直线所成得角的余弦值的求法,注意余弦定理的灵活运用,属于基础题.6.若关于的方程的取值范围是( ) x 3kx k =+-k A .B .C .D .4,3⎛⎫+∞ ⎪⎝⎭43,32⎛⎤⎥⎝⎦40,3⎛⎫ ⎪⎝⎭43,32⎛⎫ ⎪⎝⎭【答案】B【解析】转化为函数与函数(1)3y k x =-+y =的斜率可求得结果.,MA MB【详解】因为关于的方程恰有两个实数根,x 3kx k =+-所以函数与函数与半圆(1)3y k x =-+y =(1)3y k x =-+y =如图:直线经过定点,(1)3y k x =-+(1,3)M 当直线与半圆时,(1)3y k x =-+y =A ,解得, 1=43k =当直线经过点时,,(1)3y k x =-+(1,0)B -32k=所以满足函数与函数的范围为.(1)3y k x =-+y =k 43,32⎛⎤⎥⎝⎦故选:B【点睛】方法点睛:已知函数有零点(方程有根)求参数值(取值范围)常用的方法: (1)直接法:直接求解方程得到方程的根,再通过解不等式确定参数范围; (2)分离参数法:先将参数分离,转化成求函数的值域问题加以解决;(3)数形结合法:先对解析式变形,进而构造两个函数,然后在同一平面直角坐标系中画出函数的图象,利用数形结合的方法求.7.若等差数列的前项和为,首项,,,则满足成{}n a n n S 10a >202020210a a +>202020210a a ⋅<0n S >立的最大正整数是( ) n A . B .C .D .4039404040414042【答案】B【解析】由等差数列的,及得数列是递减的数列,因此可确定,10a >202020210a a ⋅<202020210,0a a ><然后利用等差数列的性质求前项和,确定和的正负. n n S 【详解】∵,∴和异号,202020210a a ⋅<2020a 2021a 又数列是等差数列,首项,∴是递减的数列,, {}n a 10a >{}n a 202020210,0a a ><由,所以,202020210a a +>140404040202020214040()2020()02a a S a a +==+>,14041404120214041()404102a a S a +==<∴满足的最大自然数为4040. 0n S >n 故选:B .【点睛】关键点睛:本题求满足的最大正整数的值,关键就是求出,时成0n S >n 100n n S S +><,立的的值,解题时应充分利用等差数列下标和的性质求解,属于中档题. n 8.已知数列的各项均为正数,,,若数列的前项和为{}n a 12a =114n n n n a a a a ++-=+11n n a a +⎧⎫⎨⎬+⎩⎭n5,则( ) n =A .119 B .121 C .120 D .122【答案】C【解析】根据题设条件化简得到,结合等差数列的通项公式,求得2214n n a a +-=n a=,结合裂项法,求得数列的前项和,列出方程,即可求解.1112n n a a +=+n 【详解】由题意,数列的各项均为正数,,,{}n a 12a =114n n n na a a a ++-=+可得,所以数列是以4首项,公差为4的等差数列,2214n n a a +-={}2n a 所以,可得24n a n =n a=又由,111122n n a a +==+前项和,n )111122n S ==令,解得.)1152=120n =故选:C.【点睛】裂项求和的方法与注意点:1、裂项相消法求和:把数列的通项公式拆成两项的差,在求和时中间的一些项可以相互抵消,从而求得数列的前项和;n 2、使用裂项相消法求和时,要注意正负项相消时,消去了哪些项,保留了哪些项,且不可漏写未被消去的项,未被消去的项有前后对称的特点.二、多选题9.已知数列,下列结论正确的有( ) {}n a A .若,,则. 12a =11n n a a n +++=20211a =B .若则11132n n a a a ++=,=,71457a =C .若,则数列是等比数列12n n S =3+{}n a D .若,则 11212n n n a a a a ++=,=()*n N ∈15215a =【答案】AB【解析】直接利用叠加法可判断选项A ,从而判断,利用构造新数列可求出B,D 中数列的通项公式,可判断,选项C 求出数列的前3项从而可判断. 【详解】选项A. 由,即 11n n a a n +=++11n n a a n +-=+则()()()()19191818120207121a a a a a a a a a a =-+-+-++-+ 20191822211=+++++= 故A 正确.选项B. 由得 132n n a a +=+,()1311n n a a +=++,所以数列是以为首项,3为公比的等比数列. {}1n a +112a +=则,即,所以,故B 正确.1123n n a -+=⨯1231n n a -=⨯-672311457a =⨯-=选项C. 由,可得当时,12n n S =3+1n =11722a =+=3当时,得,2n =2211193622a S S ⎛⎫⎛⎫=-=+-+= ⎪ ⎪⎝⎭⎝⎭当时,得,3n =332112791822a S S ⎛⎫⎛⎫=-=+-+= ⎪ ⎪⎝⎭⎝⎭显然,所以数列不是等比数列,故C 错误.2213a a a ≠{}n a 选项D. 由,可得122n n n a a a +=+11112n n a a +-=所以数列是以1为首项,为公差的等差数列.1n a ⎧⎫⎨⎬⎩⎭12所以,则,即,故D 错误. ()1111122n n n a +=+-=1511826a ==1518a =故选:AB【点睛】关键点睛:本题考查利用递推关系求数列的通项公式,解答的关键是掌握求数列通项公式的常见方法,由叠加法可得,利用构造新()()()()19191818120207121a a a a a a a a a a =-+-+-++-+ 数列解决问题,属于中档题.()1311n n a a +=++,11112n n a a +-=10.关于双曲线与双曲线下列说法正确的是( )221:132x y C -=222:123y x C -=A .它们的实轴长相等 B .它们的渐近线相同 C .它们的离心率相等 D .它们的焦距相等【答案】BD【解析】根据两个双曲线分别求解四个选项中的性质,再比较,判断选项.【详解】双曲线,,,实轴长,渐近线方程221:132x y C -=223,2a b ==2225c a b =+=2a =,离心率y x ==c e a ===2c =双曲线,,,实轴长222:123y x C -=222,3a b ==2225c a b =+=2a =y x ==,离心率 c e a ===2c =综上比较,可知两个双曲线的渐近线,焦距相等. 故选:BD11.已知圆和圆的公共点为,,则( )221:1C x y +=222:40C x y x +-=A B A . B .直线的方程是 12||2C C =AB 14x =C .D .12AC AC ⊥||AB =【答案】ABD【解析】两圆相减就是直线的方程,再利用圆心距,判断C ,利用弦长公式求. AB AB 【详解】圆的圆心是,半径,圆,圆心,,1C ()0,011r =()222:24C x y -+=()2,022r =,故A 正确;122C C ∴=两圆相减就是直线的方程,两圆相减得,故B 正确; AB 1414x x =⇒=,,,,所以不正确,故C 不正确;11AC =22AC =122C C =2221212AC AC C C +≠12AC AC ⊥圆心到直线的距离,D 正确. ()0,014x =14d =AB ===故选:ABD【点睛】关键点点睛:本题关键选项是B 选项,当两圆相交,两圆相减后的二元一次方程就是相交弦所在直线方程.12.已知正方体的棱长为,点,在平面内,若1111ABCD A B C D -2E F 1111D C B A ||AE =,则( )AC DF ⊥A .点的轨迹是一个圆 EB .点的轨迹是一个圆 FC .的最小值为EF 1D .与平面AE 1A BD 【答案】ACD【分析】对于A 、B 、C 、D 四个选项,需要对各个选项一一验证.选项A :由,分析得E 的轨迹为圆; ||AE ==1||1A E =选项B :由,而点F 在上,即F 的轨迹为线段,; AC DBF ⊥11B D 11B D 选项C :由E 的轨迹为圆,F 的轨迹为线段,可分析得; 11B D min ||EF d r =-选项D :建立空间直角坐标系,用向量法求最值.【详解】对于A:,即点E 为在面||AE ===1||1A E =内,以为圆心、半径为1 的圆上;故A 正确;1111D C B A 1A 对于B: 正方体中,AC ⊥BD ,又,且BD ∩DF=D ,所以,所1111ABCD A B C D -AC DF ⊥AC DBF ⊥以点F 在上,即F 的轨迹为线段,故B 错误; 11B D 11B D 对于C:在平面内,1111D C B A到直线的距离为当点,落在上时,;故C 正确;1A 11BD d =EF 11AC min ||1EF =对于D:建立如图示的坐标系,则()()()()10,0,0,2,0,0,0,0,2,0,2,0A B A D 因为点E 为在面内,以为圆心、半径为1 的圆上,可设1111D C B A 1A ()cos ,sin ,2E θθ所以 ()()()1cos ,sin ,2,2,0,2,2,2,0,AE A B BD θθ==-=-设平面的法向量,则有 1A BD (),,n x y z = 1·220·220n BD x y n A B x z ⎧=-+=⎪⎨=-=⎪⎩不妨令x =1,则,()1,1,1n =设与平面所成角为α,则:AE 1A BD||sin |cos ,|||||n AE n AE n AE α====⨯ A 当且仅当时,4πθ=sin α=故D 正确 故选:CD【点睛】多项选择题是2020年高考新题型,需要要对选项一一验证.三、填空题13.若直线与直线互相垂直,则实数的值为__________. 10x y -+=310mx y +-=m 【答案】3【解析】直接利用两直线垂直,求出m .【详解】因为直线与直线互相垂直, 10x y -+=310mx y +-=所以,解得: 30m -=3m =故答案为:3【点睛】若用一般式表示的直线,不用讨论斜率是否存在,只要A 1A 2+B 1 B 2=0,两直线垂直.14.已知,,且与的夹角为钝角,则实数的取值范围为____.()1,1,0a =r ()1,0,2b =-r ka b + 2a b -k 【答案】()7,22,5⎛⎫-∞-⋃- ⎪⎝⎭【分析】结合向量的坐标运算,两向量夹角为钝角需满足数量积为负,且夹角不为平角.【详解】,与的夹角为钝角,则()()1,,2,23,2,2ka b k k a b +=--=-ka b + 2a b -,即.()()2570ka b a b k +⋅-=-<75k <又当与的夹角为平角时,有,得. ka b + 2a b - 12322k k -==-2k =-故实数的取值范围为且. k 75k <2k =-故答案为:()7,22,5⎛⎫-∞-⋃- ⎪⎝⎭15.若数列{a n }的前n 项和为S n =a n +,则数列{a n }的通项公式是a n =______.2313【答案】;1(2)n n a -=-【详解】试题分析:解:当n=1时,a 1=S 1=a 1+,解得a 1=1,当n≥2时,a n =S n -S n-1=()-23132133n a +()=-整理可得a n =−a n−1,即=-2,故数列{a n }是以1为首项,-2为公12133n a -+23n a 123n a -13231n n a a -比的等比数列,故a n =1×(-2)n-1=(-2)n-1故答案为(-2)n-1. 考点:等比数列的通项公式.16.若为直线上一个动点,从点引圆的两条切线,(切P 40x y -+=P 2240y x C x +-=:PM PN点为,),则的最小值是________.M N MN【解析】根据题意得当的长度最小时,取最小值,进而根据几何关系求解即可.||MN ||PC 【详解】如图,由题可知圆C 的圆心为,半径.(2,0)C 2r =要使的长度最小,即要最小,则最小.||MN MCN ∠MCP ∠因为, ||||tan 2PM PM MCP r ∠==所以当最小时,最小因为,||PM ||MN PM =∣所以当最小时,最小.||PC ||MN因为 min ||PC ==所以 cos MCP ∠==所以 sin MCP ∠=由于 1in 2s 2MCP MN ∠=所以. min ||MN =【点睛】本题解题的关键是根据已知当的长度最小,即要最小,进而得当最小||MN MCN ∠||PC时,最小.由于的最小值为点到直线,故考查化归转化思||MN ||PC C 40x y -+=min ||PC =想和运算能力,是中档题.四、解答题17.在①圆与轴相切,且与轴正半轴相交所得弦长为C y x ②圆经过点和;C ()4,1A ()2,3B ③圆与直线相切,且与圆相外切这三个条件中任选一个,补充在C 210x y --=22:(2)1Q x y +-=下面的问题中,若问题中的圆存在,求出圆的方程;若问题中的圆不存在,说明理由. C C C 问题:是否存在圆,______,且圆心在直线上. C C 12y x =注:如果选择多个条件分别解答,按第一个解答计分.【答案】答案见解析.【解析】选择①、②、③,分别用待定系数法求圆的方程;E 【详解】选择条件①:设圆心的坐标为,圆的半径为C (),a b C r 因为圆心在直线上,所以 C 12y x =12b a =因为圆与轴相切,且与轴正半轴相交所得弦长为C y x 所以,,且0a >0b >2r a b ==由垂径定理得解得,223r b =+1b =所以,2a =2r =所以圆的方程为C 22(2)(1)4x y -+-=选择条件②:设圆心的坐标为,圆的半径为C (),a b C r 因为圆心在直线上,所以 C 12y x =12b a =因为圆经过点和,的中点C ()4,1A ()2,3B AB ()3,2M 所以的中垂线方程为AB 1y x =-联立直线 12y x =解得 21x y =⎧⎨=⎩即,,2a =1b =2r =所以圆的方程为C 22(2)(1)4x y -+-=选择条件③:设圆心的坐标为,圆的半径为C (),a b C r 因为圆心在直线上,所以 C 12y x =2a b =, r =所以与圆相外切, r =C Q所以||1CQ r =+1r =+可得:,因为该方程,所以方程无解 2540b b -=∆<0故不存在满足题意的圆.C 【点睛】“结构不良问题”是2020年新高考出现的新题型:题目所给的三个可选择的条件是平行的,即无论选择哪个条件,都可解答题目,而且,在选择的三个条件中,并没有哪个条件让解答过程比较繁杂,只要推理严谨、过程规范,都会得满分.18.已知数列满足,.{}n a 11a =13(1)n n na n a +=+(1)设,求证:数列是等比数列; n n a b n={}n b (2)求数列的前项和.{}n a n n S 【答案】(1)证明见解析;(2). (21)3144n n n S -=+【解析】(1)将变形为,得到为等比数列, 13(1)n n na n a +=+131n n a a n n+=+{}n b (2)由(1)得到的通项公式,用错位相减法求得{}n a n S 【详解】(1)由,,可得, 11a =13(1)n n na n a +=+131n n a a n n +=+因为则,,可得是首项为,公比为的等比数列, n n a b n=13n n b b +=11b ={}n b 13(2)由(1),由,可得, 13n n b -=13n n a n-=13n n a n -=⋅,01211323333n n S n -=⋅+⋅+⋅++⋅ ,12331323333n n S n =⋅+⋅+⋅++⋅ 上面两式相减可得:0121233333n n n S n --=++++-⋅ , 13313nn n -=-⋅-则. (21)3144n n n S -=+【点睛】数列求和的方法技巧:(1)倒序相加:用于等差数列、与二项式系数、对称性相关联的数列的求和.(2)错位相减:用于等差数列与等比数列的积数列的求和.(3)分组求和:用于若干个等差或等比数列的和或差数列的求和.(4) 裂项相消法:用于通项能变成两个式子相减,求和时能前后相消的数列求和.19.正项数列的前n 项和Sn 满足:{}n a 222(1)()0n n S n n S n n -+--+= (1)求数列的通项公式;{}n a n a (2)令,数列{bn}的前n 项和为Tn ,证明:对于任意的n ∈N*,都有Tn < . 221(2)n n n b n a +=+564【答案】(1)(2)见解析2;n a n =【详解】(1)因为数列的前项和满足:,所以当时,, 即解得或, 因为数列都是正项, 所以, 因为, 所以, 解得或, 因为数列都是正项, 所以, 当时,有, 所以, 解得,当时,,符合所以数列的通项公式,; (2)因为,所以, 所以数列的前项和为:, 当时, 有, 所以, 所以对于任意,数列的前项和.20.如图,在四棱锥中,为矩形,,平面P ABCD -ABCD AD PA PB ===PA PB ⊥平面.PAB ⊥ABCD(1)证明:平面平面;PAD ⊥PBC (2)若为中点,求平面与平面的夹角的余弦值.M PC AMD BMD【答案】(1)证明见解析;(2【解析】(1)利用平面,得到,又有,,得到平面DA ⊥PAB DA PB ⊥PA PB ⊥DA PA A = PB ⊥,从而平面平面;PAD PAD ⊥PBC (2)建立空间直角坐标系,利用向量法求平面与平面的夹角的余弦值.AMD BMD 【详解】(1)证明:因为平面平面,平面平面,PAB ⊥ABCD PAB ⋂ABCD AB =矩形中,,所以平面ABCD DA AB ⊥DA ⊥PAB 因为平面,所以PB ⊂PAB DA PB ⊥又因为,,平面,PA PB ⊥DA PA A = DA ⊂PAD 平面PA ⊂PAD 所以平面.PB ⊥PAD 因为平面,所以,平面平面.PB ⊂PBC PAD ⊥PBC (2)解:由(1)知平面,取中点,连结,则,DA ⊥PAB AB O PO PO AB ⊥以为原点,建立如图所示的空间直角坐标系,O O xyz-则,,,,(2,0,0,)P (0,2,0)A -(0,2,0)B (0,2,D-M 则,,,(0,0,DA =-(1,3,DM =(0,4,DB =- 设平面的一个法向量为,AMD (,,)n x y z = 则即 00n DA n DM ⎧⋅=⎨⋅=⎩030z x y =⎧⎪⎨+=⎪⎩令,则,,所以1y =-3x =0z =(3,1,0)n =- 同理易得,平面的一个法向量为BMD (m =- 所以.cos ,||||m n m n m n ⋅<>===⋅ 由图示,平面与平面所成夹角为锐角,AMD BMD所以平面与平面 AMD BMD 【点睛】立体几何解答题的基本结构: (1)第一问一般是几何关系的证明,用判定定理;(2)第二问是计算,求角或求距离(求体积通常需要先求距离),通常可以建立空间直角坐标系,利用向量法计算.21.如图,在三棱锥中,,为的中点.P ABC -AB BC ==4PA PB PC AC ====O AC(1)证明:平面;PO ⊥ABC (2)若点在棱上,且二面角为,求与平面所成角的正弦值.M BC M PA C --30°PC PAM【答案】(1)证明见解析;(2. 【分析】(1)根据等腰三角形性质得PO 垂直AC ,再通过计算,根据勾股定理得PO 垂直OB ,最后根据线面垂直判定定理得结论;(2)方法一:根据条件建立空间直角坐标系,设各点坐标,根据方程组解出平面PAM 一个法向量,利用向量数量积求出两个法向量夹角,根据二面角与法向量夹角相等或互补关系列方程,解得M 坐标,再利用向量数量积求得向量PC 与平面PAM 法向量夹角,最后根据线面角与向量夹角互余得结果.【详解】(1)因为,为的中点,所以,且 4AP CP AC ===O AC OP AC ⊥OP =连结.OB因为,所以为等腰直角三角形, AB BC ==ABC A 且 ,由知. 1,22OB AC OB AC ⊥==222OP OB PB +=PO OB ⊥由知,平面.,OP OB OP AC ⊥⊥PO ⊥ABC (2)[方法一]:【通性通法】向量法如图,以为坐标原点,的方向为轴正方向,建立空间直角坐标系 .O OB x O xyz -由已知得(0,0,0),(2,0,0),(0,2,0),(0,2,0),(0,0,(0,2,O B A C P AP -= 取平面的法向量.PAC (2,0,0)OB = 设,则.(,2,0)(02)M a a a -<≤(,4,0)AM a a =- 设平面的法向量为.PAM (,,)n x y z =由得 , 0,0AP n AM n ⋅=⋅= 2=0+(4)=0y ax a y -⎧⎪⎨⎪⎩可取2,)n a a =--所以.由已知得 . cos OB n 〈⋅〉= cos OB n 〈⋅〉=.解得(舍去), . =4a =-43a =所以 . 43n ⎛⎫=- ⎪ ⎪⎝⎭又 ,所以 . (0,2,PC =- cos ,PC n 〈〉=所以与平面. PC PAM [方法二]:三垂线+等积法由(1)知平面,可得平面平面.如图5,在平面内作,垂PO ⊥ABC PAC ⊥ABC ABC MN AC ⊥足为N ,则平面.在平面内作,垂足为F ,联结,则,故MN ⊥PAC PAC NF AP ⊥MF MF AP ⊥为二面角的平面角,即. MFN ∠M PA C --30MFN ∠=︒设,则,在中,.在中,由MN a =,4NC a AN a ==-Rt AFN△)FN a =-Rt MFN △,得,则.设点C 到平面的距离为h,由(4)a a =-43a =823FM a ==PAM ,得,解得与平面所成角的正弦值M APC C APM V V --=2141184433323h ⋅=⋅⋅⋅⋅h =PC PAM [方法三]:三垂线+线面角定义法由(1)知平面,可得平面平面.如图6,在平面内作,垂PO ⊥ABC PAC ⊥ABC ABCMN AC ⊥足为N ,则平面.在平面内作,垂足为F ,联结,则,故MN ⊥PAC PAC NF AP ⊥MF MF AP ⊥为二面角的平面角,即.同解法1可得. MFN ∠M PA C--30MFN ∠=︒43MN a ==在中,过N 作,在中,过N 作,垂足为G ,联结.在APC △NE PC ∥FNM △NG FM ⊥EG 中,.因为,所以. Rt NGM △43NG NM ===NE PC ∥843NE NA a ==-=由平面,可得平面平面,交线为.在平面内,由,PA ⊥FMN PAM ⊥FMN FM FMN NG FM ⊥可得平面,则为直线与平面所成的角.NG ⊥PAM NEG ∠NE PAM 设,则,所以直线与平面所成角的正弦NEG α∠=sin NG NE α===NE PC ∥PC PAM [方法四]:【最优解】定义法如图7,取的中点H ,联结,则C 作平面的垂线,垂足记为T (垂足T PA CH CH =PAM 在平面内).联结,则即为二面角的平面角,即,得PAM HT CHT ∠M PA C --30CHT ∠=︒CT =联结,则为直线与平面所成的角.在中,PT CPT ∠PC PAM Rt PCT △4,PC CT ==sin CPT ∠=【整体点评】(2)方法一:根据题目条件建系,由二面角的向量公式以及线面角的向量公式硬算即可求出,是该类型题的通性通法;方法二:根据三垂线法找到二面角的平面角,再根据等积法求出点到面的距离,由定义求出线面角,是几何法解决空间角的基本手段;方法三:根据三垂线法找到二面角的平面角,再利用线面角的等价转化,然后利用定义法找到线面角解出,是几何法解决线面角的基本思想,对于该题,略显麻烦;方法四:直接根据二面角的定义和线面角的定义解决,原理简单,计算简单,是该题的最优解.22.已知椭圆的左右顶点分别为,2222:1(0)x y E a b a b +=>>A B D .(1)求椭圆的标准方程;E (2)过点作与轴不重合的直线与椭圆相交于,两点(在,之间).证明:直()4,0P x l E M N N P M 线与直线的交点的横坐标是定值.MB NA 【答案】(1);(2)证明见解析. 2214x y +=【解析】(1)待定系数法求椭圆标准方程; (2)用“设而不求法”表示出M 、N ,,从而表示出直线MB ,NA , 证明直线与直线的交点的MB NA 横坐标是定值.【详解】(1)因为 c e a ==12b a =椭圆过点,D 所以, 2221142b b +=所以,,2a =1b =所以椭圆的方程为 E 2214x y +=(2)设直线,设,:4l x my =+()11,M x y ()22,N x y 联立得: 22414x my x y =+⎧⎪⎨+=⎪⎩()2248120m y my +++=,或2161920m ∆=->m >m <-由韦达定理得:, 12284m y y m -+=+122124y y m =+由题意得:直线,直线 11:(2)2y MB y x x =--22:(2)2y NA y x x =++所以()()12212(2)2(2)y x x y x x +-=-+即 ()()12112212121262262x my y y my y y my y y y my y +--=+++整理得,()()121221622226x y y my y y y -=++即()()121221622326x y y y y y y -=-+++⎡⎤⎣⎦即()()12126262x y y y y -=-若,则,此时,213y y =1m =±2161920m ∆=-<所以12620y y -≠所以1x =【点睛】(1)待定系数法是求二次曲线的标准方程的常用方法;(2)“设而不求”是一种在解析几何中常见的解题方法,可以解决直线与二次曲线相交的问题.。

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参考答案听力1-5 ABCAC6-10 BACAC 11-15 BBACC 16-20 ABCBA阅读理解21-25CBDCB 26-30ACBBB 31-35DBDCB36-40 FEDBG完形填空41-45ADBCB 46-50CBCDC 51-55ABDCC 56-60BCADC语篇填空61. to tell; 62. about; 63. It;64. costs;65. The66. for;67. that;68. called;69. to do;70. hottest短文改错I live I a village in the north of Henan Province. It’s the government’s policy for building roads to every village that has brought great changes to my hometown.Before the policy was carried out, there was no road or bus line near the village, so traffic was very inconvenient. Local specialities couldn’t be transported out to other places for sale. Large areas of wasteland and many pools were deserted. Children had to walk on narrow, muddy roads to go to school. But now everything is quite different. There have been wide roads and bus service, and traffic is very convenient. The local specialities can be transported to different places, which has greatly increased the income of the villages. Nowadays children can go to school either by bus or by bike. Meanwhile, the villagers have developed eco-tourism by using wasteland and pools, which attracts many tourists from cities. People’s life is becoming better and better.书面表达Class is where we study and struggle for a bright future. It's the stage on which we show our talents and also a community to develop our social potential. In this sense, a sound class is the assurance of a reliable and enjoyable environment for us. To build a favorite class, I advocate the actions as follows.First, all hands make a big fire. thus, I will contribute every bit of my effort to public benefts. For example,I will volunteer to clean the blackboard, empty the dustbin, or anything else if no one is on duty. Second, each individual mirros a community. if I were a broken window, there would be no safety and harmony in class. Therefore, I will behave myself well for the betterment of my class.Last but not least, I wish everyone could take the class as our family, love each other and cherish every moment in school.听力材料(Text 1)W: Oh, is this a photo of your graduation? Who is the girl standing next to you? M: That’s Jane. She became a doctor after graduation.(Text 2)M: How about going camping with us tomorrow?W: I’d like to, but I’ve promised Tim I’d go downtown with him tomorrow. He needs some new books.(Text 3)M: This is the last time I’m telling you. Turn the radio off or leave.W: Why don’t you read somewhere else?M: Because I need to use the dictionaries in this classroom for my writing. (Text 4)M: How I wish I were Adam!W: Why?M: He’s the funniest guy I’ve ever met. He makes everybody laugh.W: Well, people are different. You’re very considerate and friendly.(Text 5)W: Did you feed Jim today?M: Yes, and I also gave him a bath.W: Thanks, honey. You’re such a good father.M: Well, believe it or not, I’ve been reading a lot online about how to be a good dad.(Text 6)W: Hey, Mark, I’ve been driving for hours. Would you mind driving for a while? I want to take a rest.M: Sorry. I would like to, but I don’t know how to drive.W: What? You haven’t learned ho w to drive?M: No. I usually go to work by bus or subway. It’s very convenient because I don’t live far from my office.W: Have you considered learning how to drive?M: No. There are already too many cars on the streets. Besides, I don’t really need to use a car all that much.(Text 7)W: Is this The Mona Lisa by Leonardo da Vinci? It looks a bit strange.M: Well, Mona Lisa looks much younger in this painting. This is called The Isleworth Mona Lisa.W: Ah, the smile on her face is bigger than the original Mona Lisa. Who painted this one?M: It is believed to have also been painted by Leonardo da Vinci in the early 16thcentury.W: Really? I thought it must have been painted by a modern-day artist.M: Hmm… Th e description says it was first discovered shortly before World War1. It belongs to a private family and they haven’t sold it to any museum yet. (Text 8)M: What’s your plan for the summer, Lisa?W: I’m going to spend a few days in Beijing. My cousin is teaching Engl ish in a language school there. I’ll travel around the city with her. What about you? M: I’m going to take a trip to Cairo. I want to visit the Egyptian Museum there. W: Is it the museum’s famous paintings that you want to see?M: Actually I’m more interested in seeing the museum’s coin display. I read that the first floor has a collection of coins used by the ancient Egyptians.W: Ah, that’s right. You like collecting coins. What else are you interested in seeing there?M: Well, I know that the second floor has some amazing things from the final two dynasties of ancient Egypt, many things which were used by the Egyptian kings.W: So how many floors does it have?M: Just two floors. It’s not very big, but I think it’s really worth a visit.(Text 9)W: What are you reading, Sam?M: The Kite Runner. It’s a great book.W: What is it about?M: It’s about a boy who grows up in Afghanistan during the 1980s. It’s a very touching story and also a little sad. Do you like such books?W: I used to like exciting adventure stories, but now I love reading books like that. M: Jenny, would you be interested in joining a book club?W: Sure. Do you know of one?M: Yes, I’m in one now. If you’d like to join, we meet every other week on Saturday evening and discuss the boo ks that we’ve read. We begin at 8:00 pm and chat for about one and a half hours. We usually learn a lot and it’s a lot of fun. W: I’m really interested, but I can’t go for the next several weeks. I’m concentrating on working out in the gym for the next si x weeks, but I’ll have time after that. M: Great. You can join us then.W: Yeah. I really look forward to meeting other people who also love reading. (Text 10) WThere were several reasons behind our decision to move to Flemington. The process started about 18 months ago. We were still living in a one-bedroom apartment then, but at that time we decided to move into a larger house before we had a baby. We began to look at houses at reasonable prices. Then, much sooner than expected, I was having a baby. Soon after that, Mark’s company announced that they were relocating to Flemington, about 90 miles south of us. Mark’s company had been good to him, and it was one of the few around withexcellent benefits, family-friendly policies, and a child-care center on site. With a baby on the way, these things were essential for us. Since employees had to find their own housing, we began looking at houses in Flemington. We also did some research on our school’s website and asked my colleagues for advice. We were very excited about what we found — reasonable housing costs, great schools, and a lively town. We’ll be moving about a month before the baby is due. Let’s hope he doe sn’t decide to come early.。

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