2021年人教版4模块综合检测卷

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人教新课标 必修四 模块综合测验卷

人教新课标 必修四 模块综合测验卷

必修四模块综合第I卷第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分30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. When can the man see the headmaster?A. At 9:30.B. At 11:45.C. At 12:40.2. Why does the man want to keep the window shut?A. Because he is ill.B. Because he wants to open it himself.C. Because the air inside is fresh enough.3. What is Mike?A. A teacher.B. A student.C. A writer.4. What has made working at home possible?A. Personal computers.B. Communication industry.C. Living far from companies.5. Where is the woman?A. In a soap factory.B. In her house.C. At an information desk.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。

听每段对话或独白,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

每段对话或独白听两遍。

听第6段材料,回答第6至7小题。

6. Where does the conversation most probably take place?A. At home.B. On a bus.C. In the bank.7. Why do the two speakers want to buy a car?A. They have a lot of money.B. The man lives too far away from his office.C. The woman’s office is too far away from her home.听第7段材料,回答第8至9小题。

2020学年高中英语模块综合检测(含解析)新人教版必修4(2021-2022学年)

2020学年高中英语模块综合检测(含解析)新人教版必修4(2021-2022学年)

模块综合检测(时间:100分钟满分:120分)选择题部分Ⅰ。

阅读理解(共10小题;每小题2.5分,满分25分)阅读下列短文,从每题所给的A、B、C和D四个选项中,选出最佳选项。

AWhat do you think aboutthevery firstthing in the morning? Your thoughts duringthe firsthalf。

hour of the morning will greatly influencethe entireday. You may not realize this, but it is a fact. If you set out with worry and depression, you are giving the key note to a day of discord (不和谐) and misfortunes。

If you think about hopeand happiness, you are sounding a note of harmony and success.Control your morning thoughts。

The firstmoment on waking, no matterwhat your moodis, say to yourself, “I will get all the comfort andpleasure possible out of thisday, and I will do something to add to the pleasure of the world's happinessor well­being (安乐). I will controlmyself when caused to be unhappy. I willlook for the bright side of everyevent。

2021版七年级语文下册 第四单元综合检测试卷(含解析) 新人教版

2021版七年级语文下册 第四单元综合检测试卷(含解析) 新人教版

单元综合检测(90分钟100分)一、积累与运用(20分)1.根据拼音写汉字或给加点的字注拼音。

(2分)戏曲yuán( )远流长,是我国的经典,它会让我们赏心悦目。

戏曲是我们的国宝,是我们的国cuì( ),是我们的国剧。

当我们细心地欣赏戏曲时,就会发现它们犹如一个个精彩的历史故事,将历史的画卷一一展现在我们的面前,将我们带入了历史的隧.( )道,让我们身临其境,让我们感慨.( )不已。

2.下列句子中加点成语运用有误的一项是(2分) ( )A.他穿上鞋子,大口喝了一口浓茶,于是大彻大悟....地自言自语起来。

B.三年来,民间投资政策落实虽然取得积极进展,但许多方面仍存在不少问题。

主要问题是政策落实参差不齐....,重点领域政策执行困难。

C.2013年10月21日,2013—2014赛季CBA体测在四川金强篮球训练基地开始,尽管球员使出浑身解数....,但众多名将还是倒在两分钟强度投篮的测试中。

D.十一假期,天气晴朗,同学们纷纷相约来到长江畔,或戏水玩沙,或放风筝,在大自然中尽情享受天伦之乐....。

3.(2013·黄冈中考)下列各句中,没有语病的一项是(2分) ( )A.营造健康文明的网络文化环境,清除不健康信息已成为新时期精神文明建设的迫切需要。

B.在阅读文学名著的过程中,常常能够使我们明白许多做人的道理,悟出人生的真谛。

C.会不会用心观察,能不能重视积累,是提高写作水平的基础。

D.他上课认真听讲,下课一有工夫不是看语文、数学等书,就是看报纸,全班同学没有一个不说他学习不积极。

4.下列说法不正确的一项是(2分) ( )A.《社戏》选自《鲁迅全集》,是鲁迅写的一篇回忆性散文。

B.《安塞腰鼓》是当代作家刘成章写的一篇著名散文。

C.《竹影》的作者是我国现代著名画家、散文家丰子恺。

D.《观舞记》的作者冰心,原名谢婉莹,《繁星》《春水》是她的代表作。

5.(2013·德州中考)下列各句中,标点符号使用正确的一项是(2分) ( )A.“干什么呀!”他变了脸色,“你又不是老师,凭什么批评我?”B.我不知道这条路是否能走通?但我仍然要坚定不移地走下去。

2021年秋高中英语必修四人教版单元测评:模块综合测评

2021年秋高中英语必修四人教版单元测评:模块综合测评

模块综合测评(时间:120分钟总分值:150分)第一局部听力(共两节,总分值30分)第一节(共5小题;每题1.5分,总分值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三个选项中选出最正确选项。

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 bank?A.Next to the bookstore.B.Behind the bookstore.C.Far from the bookstore.答案:A2.What does the man mean?A.There are too many shopping centres already.B.They aren’t going to build one.C.He hasn’t been to the other shopping centres.答案:A3.What does the man mean?A.Bob said nothing at the meeting.B.Something is wrong with Bob’s ears.C.Bob doesn’t listen to him.答案:C4.What is the man going to do?A.See the woman smile.B.Take the woman’s photo.C.Take out the film.答案:B5.What’s the relationship between the man and the woman?A.They’re friends.B.They’re mother and son.C.They’re husband and wife.答案:A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。

高中数学 模块综合检测(C)(含解析)新人教A版必修4(2021年整理)

高中数学 模块综合检测(C)(含解析)新人教A版必修4(2021年整理)

高中数学模块综合检测(C)(含解析)新人教A版必修4编辑整理:尊敬的读者朋友们:这里是精品文档编辑中心,本文档内容是由我和我的同事精心编辑整理后发布的,发布之前我们对文中内容进行仔细校对,但是难免会有疏漏的地方,但是任然希望(高中数学模块综合检测(C)(含解析)新人教A版必修4)的内容能够给您的工作和学习带来便利。

同时也真诚的希望收到您的建议和反馈,这将是我们进步的源泉,前进的动力。

本文可编辑可修改,如果觉得对您有帮助请收藏以便随时查阅,最后祝您生活愉快业绩进步,以下为高中数学模块综合检测(C)(含解析)新人教A版必修4的全部内容。

模块综合检测(C)(时间:120分钟满分:150分)一、选择题(本大题共12小题,每小题5分,共60分)1.若角600°的终边上有一点(-4,a),则a的值是()A.4错误! B.-4错误!C。

错误! D.-错误!2.若向量a=(3,m),b=(2,-1),a·b=0,则实数m的值为()A.-32B。

错误! C.2 D.63.设向量a=(cos α,错误!),若a的模长为错误!,则cos 2α等于( )A.-错误! B.-错误! C.错误! D。

错误!4.平面向量a与b的夹角为60°,a=(2,0),|b|=1,则|a+2b|等于( )A.错误! B.2错误! C.4 D.125.tan 17°+tan 28°+tan 17°tan 28°等于()A.-错误! B.错误! C.-1 D.16.若向量a=(1,1),b=(2,5),c=(3,x),满足条件(8a-b)·c=30,则x等于()A.6 B.5 C.4 D.37.要得到函数y=sin x的图象,只需将函数y=cos(x-错误!)的图象()A.向右平移错误!个单位B.向右平移错误!个单位C.向左平移π3个单位D.向左平移错误!个单位8.设函数f(x)=sin(2x+π3),则下列结论正确的是( )A.f(x)的图象关于直线x=错误!对称B.f(x)的图象关于点(错误!,0)对称C.把f(x)的图象向左平移错误!个单位,得到一个偶函数的图象D.f(x)的最小正周期为π,且在[0,错误!]上为增函数9.已知A,B,C是锐角△ABC的三个内角,向量p=(sin A,1),q=(1,-cos B),则p与q 的夹角是()A.锐角 B.钝角C.直角 D.不确定10.已知函数f(x)=(1+cos 2x)sin2x,x∈R,则f(x)是()A.最小正周期为π的奇函数B.最小正周期为错误!的奇函数C.最小正周期为π的偶函数D.最小正周期为错误!的偶函数11.设0≤θ≤2π,向量错误!=(cos θ,sin θ),错误!=(2+sin θ,2-cos θ),则向量错误!的模长的最大值为( )A。

2021年人教教七年级上第四章章末综合检测试卷含解析

2021年人教教七年级上第四章章末综合检测试卷含解析

章末综合检测(时间:90分钟满分:12021一、选择题(每小题3分,共30分)1. 下列第一行的四个图形,每个图形均是由四种简单的图形a,b,c,d(圆、直线、三角形、长方形)中的两种组成.例如,由a,b组成的图形记作a⊙b,那么由此可知,下列选项的图形,可以记作a⊙d 的是( )2. 如图4-1,该几何体从正面看得到的平面图形是( )图4-13. 对于直线AB、线段CD、射线EF,其中能相交的图是( )4. 下列现象:(1)用两个钉子就可以把木条固定在墙上;(2)从A地到B地架设电线,总是尽可能沿着线段AB架设;(3)植树时,只要确定两棵树的位置,就能确定同一行树所在的直线;(4)把弯曲的公路改直,就能缩短路程.其中能用“两点确定一条直线”来解释的现象是( )A.(1)(2)B.(1)(3)C.(2)(4)D.(3)(4)5. 如图4-2,AB=12,C为AB的中点,点D在线段AC上,且AD∶CB=1∶3,则线段DB的长度为( )图4-2A.4B.6C.8D.106. 已知线段AB和点P,如果P A+PB=AB,那么( )A.P为AB的中点B.点P在线段AB上C.点P在线段AB外D.点P在线段AB的延长线上7. 学校、书店、邮局在平面图上的标点分别是A,B,C,书店在学校的正东方向,邮局在学校的南偏西25°,那么平面图上的∠CAB 等于( )A.25°B.65°C.115°D.155°8. 若∠1=40.4°,∠2=40°4′,则∠1与∠2的关系是( )A.∠1=∠2B.∠1>∠2C.∠1<∠2D.以上都不对图4-39. 如图4-3,∠AOB=130°,射线OC是∠AOB内部任意一条射线,OD,OE分别是∠AOC,∠BOC的平分线,下列叙述正确的是( )A.∠DOE的度数不能确定B.∠AOD+∠BOE=∠EOC+∠COD=∠DOE=65°C.∠BOE=2∠COD∠EOCD.∠AOD=1210. 如图4-4,OD⊥AB于点O,OC⊥OE,图中与∠AOC互补的角有( )图4-4A.1个B.2个C.3个D.4个二、填空题(每小题4分,共32分)11.夏天,快速转动的电扇叶片,给我们一个完整的平面的感觉,说明_____.12.如图4-5,C,D是线段AB上的两点,若AC=4,CD=5,DB=3则图中所有线段长度的和是_____.图4-513.已知∠A=100°,那么∠A的补角是_____.14.时钟上3点40分时分针与时针夹角的度数为____.15.如图4-6,O在直线AB上,∠AOD=90°,∠COE=90°,则图中相等的锐角有_____对.图4-616.已知∠AOC和∠BOD都是直角,如果∠AOB=150°,那么∠COD 的度数为_____.17.如图4-7,一个正方体的表面上分别写着连续的6个整数,且每两个相对面上的两个数的和都相等,则这6个整数的和为_____.图4-718.平面内有四个点A,B,C,D,过其中每两个点画直线可以画出的直线有_____.三、解答题(共58分)19.(8分)计算:(1)22°18′×5;(2)90°-57°23′27″.20218分)把图4-8的展开图和它们的立体图形连起来.图4-821.(10分)如图4-9,已知线段a,b,c,用圆规和直尺画图.(不用写作法,保留画图痕迹)(1)画线段AB,使得AB=a+b-c;(2)在直线AB外任取一点K,画射线AK和直线BK;(3)反向延长AK至点P,使AP=KA,画线段PB,比较所画图形中线段P A与BK长度的和与线段AB长度的大小.图4-922.(10分)如图4-10,已知线段AB和CD的公共部分BD=13AB=14CD,线段AB,CD的中点E,F之间的距离是10 cm,求线段AB,CD的长度.图4-1023.(10分)如图4-11(1),已知直角三角形两直角边的长分别为3和4,斜边的长为5.(1)试计算该直角三角形斜边上的高;(2)按如图4-11(2),4-11(3),4-11(4)三种情形计算该直角三角形绕某一边旋转得到的立体图形的体积.(结果保留π)图4-1124.(12分)如图4-12,O为直线AB上一点,∠AOC=50°,OD平分∠AOC,∠DOE=90°.(1)请你数一数,图中有多少个小于平角的角;(2)求出∠BOD的度数;(3)请通过计算说明OE是否平分∠BOC.图4-12答案一、1.A 解析:根据题意,知a代表长方形,d代表直线,所以记作a⊙d的图形是长方形和直线的组合.故选A.2. A3. B 解析:A.直线AB与线段CD不能相交,故此选项不符合题意;B.直线AB与射线EF能相交,故此选项符合题意;C.射线EF与线段CD不能相交,故此选项不符合题意;D.直线AB与射线EF不能相交,故此选项不符合题意.故选B.4. B 解析:(1)用两个钉子就可以把木条固定在墙上,根据是两点确定一条直线;(2)从A地到B地架设电线,总是尽可能沿着线段AB架设,根据是两点之间,线段最短;(3)植树时,只要确定两棵树的位置,就能确定同一行树所在的直线,根据是两点确定一条直线;(4)把弯曲的公路改直,就能缩短路程,根据是两点之间,线段最短.故选B.5. D 解析:因为C为AB的中点,AB=12,所以AC=BC=12AB=12×12=6.因为AD∶CB=1∶3,所以AD=2,所以DB=AB-AD=12-2=10.故选D.6. B 解析:如图D4-1.因为P A+PB=AB,所以点P在线段AB上.故选B.图D4-17. C 解析:如图D4-2.由图可知,∠CAB=∠1+∠2=25°+90°=115°.故选C.图D4-28. B 解析:因为∠1=40.4°=40°24′,∠2=40°4′,所以∠1>∠2.故选B.9. B 解析:因为OD,OE分别是∠AOC,∠BOC的平分线,所以∠AOD=∠COD,∠EOC=∠BOE.又因为∠AOD+∠BOE+∠EOC+∠COD=∠AOB=130°,所以∠AOD+∠BOE=∠EOC+∠COD=∠DOE=65°.故选B.10. B 解析:根据题意,得(1)因为∠AOC+∠BOC=180°,所以∠BOC 与∠AOC互补.(2)因为OD⊥AB,OC⊥OE,所以∠EOD+∠DOC=∠BOC+∠DOC=90°,所以∠EOD=∠BOC,所以∠AOC+∠EOD=180°,所以∠EOD与∠AOC互补,所以图中与∠AOC互补的角有2个.故选B.二、11.线动成面12. 41 解析:AD=AC+CD=9,AB=AC+CD+DB=12,CB=CD+DB=8,故题图中所有线段长度的和为AC+AD+AB+CD+CB+DB=41.13. 80°14. 130°解析:3点40分时分针与时针夹角的度数为30°×=130°.4+1315. 2 解析:因为∠AOD=90°,所以∠AOC+∠COD=90°.因为∠COE=90°,所以∠COD+∠DOE=90°,所以∠AOC=∠DOE.因为∠BOD=180°-∠AOD=90°,所以∠DOE+∠BOE=90°,所以∠BOE=∠COD.故图中相等的锐角有2对.16. 30°或150°解析:如图D4-3(1),因为∠BOD=90°,∠AOB=150°,所以∠AOD=60°.又因为∠AOC=90°,所以∠COD=30°.如图D4-3(2),因为∠BOD=90°,∠AOC=90°,∠AOB=150°,所以∠AOD=60°,所以∠COD=150°.综上所述,∠COD的度数为30°或150°.图D4-317. 51 解析:因为正方体的表面展开图,相对的面一定相隔一个正方形,所以6若不是最小的数,则6与9是相对面.因为6与9相邻,所以6是最小的数,所以这6个整数的和为6+7+8+9+10+11=51. 18. 1条、4条或6条解析:如果A,B,C,D四点在同一直线上,那么只能确定一条直线,如图D4-4(1);如果4个点中有3个点(不妨设点A,B,C)在同一直线上,而第4个点,点D不在此直线上,那么可以确定4条直线,如图D4-4(2);如果4个点中,任何3个点都不在同一直线上,那么点A分别和点B,C,D确定3条直线,点B 分别与点C,D确定2条直线,最后点C,D确定一条直线,这样共确定6条直线,如图D4-4(3).综上所述,过其中每2个点可以画1条、4条或6条直线.(1)(2)(3)图D4-4三、19.解:(1)22°18′×5=110°90′=111°30′.(2)90°-57°23′27″=32°36′33″.2021解:如图D4-5.图D4-521. 分析:(1)首先作射线CE在射线CE上截取CD=a,BD=b,再在CB上截取AC=c,则可得出AB=a+b-c;(2)根据射线和直线的概念过点K即可作出;(3)根据AP=AK,利用两点之间线段最短即可得出答案.解:(1)如图D4-6(1).(2)如图D4-6(2).(1)(2)(3)图D4-6(3)如图D4-6(3).因为AP=KA,所以线段P A与BK长度的和大于线段AB的长度.22.解:设BD=x cm,则AB=3x cm,CD=4x cm,AC=6x cm.因为E,F分别为线段AB,CD的中点,所以AE=12AB=1.5x(cm),CF=12CD=2x(cm).所以EF=AC-AE-CF=6x-1.5x-2x=2.5x(cm). 因为EF=10 cm,所以2.5x=10,解得x=4. 所以AB=12 cm,CD=16 cm.23. 解:(1)三角形的面积为12×5h=12×3×4,解得h= 12/5.(2)在图4-11(2)中,所得立体图形的体积为13π×32×4=12π;在图4-11(3)中,所得立体图形的体积为13π×42×3=16π;在图4-11(4)中,所得立体图形的体积为13π×(125)2×5=485π.24. 解:(1)图中小于平角的角有∠AOD,∠AOC,∠AOE,∠DOC,∠DOE,∠DOB,∠COE,∠COB,∠EOB,共9个.(2)因为∠AOC=50°,OD平分∠AOC,所以∠DOC=1/2∠AOC=25°,∠BOC=180°-∠AOC=130°.所以∠BOD=∠DOC+∠BOC=155°.(3)因为∠DOE=90°,∠DOC=25°,所以∠COE=∠DOE-∠DOC=90°-25°=65°.又因为∠BOE=∠BOD-∠DOE=155°-90°=65°,所以∠COE=∠BOE,即OE平分∠BOC.。

2020-2021版高中数学模块综合测评新人教A版选修4

模块综合测评(时间:120分钟满分:150分)一、选择题(本大题共12小题,每小题5分,共60分)1.若a>b>c,则的值()A.大于0B.小于0C.小于或等于0D.大于或等于0解析因为a>b>c,所以a-c>b-c>0.所以,所以>0,故选A.答案A2.不等式|x+3|+|x-2|<5的解集是()A.{x|-3≤x<2}B.RC.⌀D.{x|x<-3或x>2}解析令f(x)=|x+3|+|x-2|=则f(x)的图象如图,由图可知,f(x)<5的解集为⌀.故原不等式的解集是⌀.答案C3.若P=(x>0,y>0,z>0),则P与3的大小关系是()A.P≤3B.P<3C.P≥3D.P>3解析因为1+x>0,1+y>0,1+z>0,所以=3,即P<3.答案B4.不等式>a的解集为M,且2∉M,则a的取值范围为()A. B.C. D.解析由已知2∉M,可得2∈∁R M,于是有≤a,即-a≤≤a,解得a≥,故应选B.答案B5.某人要买房,随着楼层的升高,上、下楼耗费的体力增多,因此不满意度升高,设住第n层楼,上、下楼造成的不满意度为n;但高处空气清新,嘈杂音较小,环境较为安静,因此随楼层升高,环境不满意度降低,设住第n层楼时,环境不满意程度为,则此人应选()A.1楼B.2楼C.3楼D.4楼解析设第n层总的不满意程度为f(n),则f(n)=n+≥2=2×3=6,当且仅当n=,即n=3时等号成立.答案C6.若关于x的不等式|x-1|+|x-3|≤a2-2a-1在R上的解集为⌀,则实数a的取值范围是()A.a<-1或a>3B.a<0或a>3C.-1<a<3D.-1≤a≤3解析|x-1|+|x-3|的几何意义是数轴上与x对应的点到1,3对应的两点距离之和,则它的最小值为2.∵原不等式的解集为⌀,∴a2-2a-1<2,即a2-2a-3<0,解得-1<a<3.故选C.答案C7.已知x+3y+5z=6,则x2+y2+z2的最小值为()A. B.C. D.6解析由柯西不等式,得x2+y2+z2=(12+32+52)(x2+y2+z2)×≥(1×x+3×y+5×z)2×=62×.答案C8.设函数f(n)=(2n+9)·3n+1+9,当n∈N+时,f(n)能被m(m∈N+)整除,猜想m的最大值为()A.9B.18C.27D.36解析当n=1时,f(1)=(2×1+9)·31+1+9=108.当n=2时,f(2)=(2×2+9)·32+1+9=360.故猜想m的最大值为36.(1)当n=1时,猜想成立.(2)当n=k(k≥1)时猜想成立,即f(k)=(2k+9)·3k+1+9能被36整除.当n=k+1时,f(k+1)=[2(k+1)+9]·3k+2+9=(2k+9+2)·3·3k+1+9=3[(2k+9)·3k+1+9]+6·3k+1-18=3[(2k+9)·3k+1+9]+18(3k-1).∵(2k+9)·3k+1+9,18(3k-1)均能被36整除,∴猜想成立.综上,m的最大值为36.答案D9.(2021 山东淄博一模)设向量=(1,-2),=(a,-1),=(-b,0),其中O为坐标原点,a>0,b>0,若A,B,C三点共线,则的最小值为()A.4B.6C.8D.9解析=(a-1,1),=(-b-1,2),∵A,B,C三点共线,∴2(a-1)-(-b-1)=0,整理,得2a+b=1.又a>0,b>0,则=(2a+b)=4+≥4+2=8,当且仅当b=2a=时,等号成立.故选C.答案C10.用反证法证明“△ABC的三边长a,b,c的倒数成等差数列,求证B<”,假设正确的是()A.B是锐角B.B不是锐角C.B是直角D.B是钝角答案B11.实数a i(i=1,2,3,4,5,6)满足(a2-a1)2+(a3-a2)2+(a4-a3)2+(a5-a4)2+(a6-a5)2=1,则(a5+a6)-(a1+a4)的最大值为()A.3B.2C. D.1解析因为[(a2-a1)2+(a3-a2)2+(a4-a3)2+(a5-a4)2+(a6-a5)2](1+1+1+4+1)≥[(a2-a1)×1+(a3-a2)×1+(a4-a3)×1+(a5-a4)×2+(a6-a5)×1]2=[(a6+a5)-(a1+a4)]2,所以[(a6+a5)-(a1+a4)]2≤8,即(a6+a5)-(a1+a4)≤2.答案B12.已知x,y,z,a,b,c,k均为正数,且x2+y2+z2=10,a2+b2+c2=90,ax+by+cz=30,a+b+c=k(x+y+z),则k=()A. B.C.3D.9解析因为x2+y2+z2=10,a2+b2+c2=90,ax+by+cz=30,所以(a2+b2+c2)(x2+y2+z2)=(ax+by+cz)2,又(a2+b2+c2)(x2+y2+z2)≥(ax+by+cz)2,当且仅当=k时,等号成立,则a=kx,b=ky,c=kz,代入a2+b2+c2=90,得k2(x2+y2+z2)=90,于是k=3.答案C二、填空题(本大题共4小题,每小题5分,共20分)13.已知关于x的不等式2x+≥7在x∈(a,+∞)上恒成立,则实数a的最小值为.解析2x+=2(x-a)++2a≥2+2a=2a+4≥7(当且仅当(x-a)2=1时,等号成立),则a≥,即实数a的最小值为.答案14.不等式|x-4|+|x-3|≤a有实数解的充要条件是.解析不等式a≥|x-4|+|x-3|有解⇔a≥(|x-4|+|x-3|)min=1.答案a≥115.设x,y,z∈R,2x+2y+z+8=0,则(x-1)2+(y+2)2+(z-3)2的最小值为.解析由柯西不等式可得(x-1)2+(y+2)2+(z-3)2(22+22+12)≥[2(x-1)+2(y+2)+(z-3)]2=(2x+2y+z-1)2=81,所以(x-1)2+(y+2)2+(z-3)2≥9当且仅当,即x=-1,y=-4,z=2时,等号成立.答案916.导学号26394074对于任意实数a(a≠0)和b,不等式|a+b|+|a-b|≥|a||x-1|恒成立,则实数x的取值范围是.解析依题意只需不等式的左边的最小值≥|a||x-1|,由绝对值三角不等式得|a+b|+|a-b|≥|(a+b)+(a-b)|=|2a|=2|a|,故只需求解2|a|≥|a||x-1|即可,解得-1≤x≤3.答案[-1,3]三、解答题(本大题共6小题,共70分)17.(本小题满分10分)已知x,y均为正数,且x>y,求证2x+≥2y+3.证明因为x>0,y>0,x-y>0,所以2x+-2y=2(x-y)+=(x-y)+(x-y)+≥3=3,所以2x+≥2y+3.18.(本小题满分12分)已知m>1,且关于x的不等式m-|x-2|≥1的解集为[0,4].(1)求m的值;(2)若a,b均为正实数,且满足a+b=m,求a2+b2的最小值.解(1)∵m>1,不等式m-|x-2|≥1可化为|x-2|≤m-1,∴1-m≤x-2≤m-1,即3-m≤x≤m+1.∵其解集为[0,4],∴解得m=3.(2)由(1)知a+b=3.(方法一:利用基本不等式)∵(a+b)2=a2+b2+2ab≤(a2+b2)+(a2+b2)=2(a2+b2),∴a2+b2≥,∴a2+b2的最小值为.(方法二:利用柯西不等式)∵(a2+b2)·(12+12)≥(a×1+b×1)2=(a+b)2=9,∴a2+b2≥,∴a2+b2的最小值为.(方法三:消元法求二次函数的最值)∵a+b=3,∴b=3-a.∴a2+b2=a2+(3-a)2=2a2-6a+9=2,∴a2+b2的最小值为.19.(本小题满分12分)用数学归纳法证明:>n!(n>1,n∈N+).(n!=n×(n-1)×…×2×1)证明(1)当n=2时,>2!=2,不等式成立.(2)假设当n=k(k≥2)时不等式成立,即>k!.当n=k+1时,=+…+(k+1)·=(k+1)·>(k+1)·k!=(k+1)!,所以当n=k+1时不等式成立.由(1)(2)可知,对n>1的一切自然数,不等式成立.20.(本小题满分12分)已知x+y>0,且xy≠0.(1)求证:x3+y3≥x2y+y2x;(2)如果恒成立,试求实数m的取值范围.(1)证明因为x3+y3-(x2y+y2x)=x2(x-y)-y2(x-y)=(x+y)(x-y)2,且x+y>0,(x-y)2≥0,所以x3+y3-(x2y+y2x)≥0,故x3+y3≥x2y+y2x.(2)解①若xy<0,则等价于.又因为=-3,即<-3,因此m>-6.②若xy>0,则等价于.因为=1,即≥1(当且仅当x=y时,等号成立),故m≤2.综上所述,实数m的取值范围是(-6,2].21.导学号26394075(本小题满分12分)设函数f(x)=|x+2|-|x-2|.(1)解不等式f(x)≥2;(2)当x∈R,0<y<1时,求证:|x+2|-|x-2|≤.(1)解由已知可得,f(x)=故f(x)≥2的解集为{x|x≥1}.(2)证明由(1)知,|x+2|-|x-2|≤|(x+2)-(x-2)|=4.∵0<y<1,∴0<1-y<1.∴[y+(1-y)]=2+≥4,当且仅当,即y=时,等号成立.∴|x+2|-|x-2|≤.22.(本小题满分12分)已知a,b,c为非零实数,且a2+b2+c2+1-m=0,+1-2m=0.(1)求证:;(2)求实数m的取值范围.(1)证明由柯西不等式得(a2+b2+c2)≥,即(a2+b2+c2)≥36.∴.(2)解由已知得a2+b2+c2=m-1,=2m-1,∴(m-1)(2m-1)≥36,即2m2-3m-35≥0,解得m≤-或m≥5.又a2+b2+c2=m-1>0,=2m-1>0,∴m≥5,即实数m的取值范围是[5,+∞).【感谢您的阅览,下载后可自由编辑和修改,关注我每天更新】。

新人教版2021届第四单元测试卷

新人教版【最新】第四单元测试卷学校:___________姓名:___________班级:___________考号:___________一、单选题1.You can’t talk in classroom.A.the B.aC.an D./2.Can you watch TV school nights?A.at B.onC.in D./3.Don’t talk eat in class.A.and B.orC.so D.but4.Do you have lots of to follow in your school?A.friends B.questionsC.rules D.dreams5.—is the new rule?—Well, we can’t fight with our classmates.A.Where B.HowC.What D.Which6.away this dirty shirt and me a clean one.A.Bring; bring B.Take; takeC.Bring; take D.Take; bring7.you go home now?A.Do; have to B.Does; have toC.Have; to D.Has; to8.I’m unhappy. He has questions to ask me.A.too much B.too manyC.much too D.many too9.On Sunday I have to go to the farm my parents.A.to help B.helpC.helping D.helped10.Her bike is broken. She walk to school.A.have to B.has toC.has D.is11.He can’t,but he often to the beach to play.A.swimming; go B.swim;goesC.swims; goes D.swim; go12.—Do I have to early in the morning,Dad?—Sure.A.make dinner B.make lunchC.go to bed D.get up13.It’s raining all day, so my brother stay at home.A.must B.have toC.must to D.has to14.A lot of good teachers their students.A.are strict with B.is strict inC.are strict in D.is strict with15.—Don’t play basketball in class.—,I won’t do it again.A.Excuse me B.SorryC.Thank you D.You’re welcome二、完型填空Every school has its own rules. In some schools in America, students 16 chew gum(口香糖).Some schools in Britain don’t let their17 have strange hairstyles. However, one 18 is very popular around the world: students 19 to wear the uniforms on school days, but many students don’t like to wear the school uniforms. They think the uniforms are the ugliest 20 in the world, but schools don’t let their students wear21 own clothes at school. Some students draw cartoons or some famous singers on their 22 .They think it is very 23 that everyone wears the same clothes. What do you think 24 it? Do you 25 your school uniform?16.A.can’t B.have to C.can D.don’t have to 17.A.classmates B.students C.boys D.girls18.A.school B.class C.rule D.uniform 19.A.has B.must C.should D.have 20.A.clothes B.shirts C.skirts D.cloth 21.A.their B.they C.them D.themselves 22.A.schoolbags B.books C.uniforms D.desks 23.A.bored B.boring C.interesting D.interested 24.A.about B.over C.by D.for 25.A.like B.make C.wash D.listen三、补全对话7选5A: Hi, my name is Jane.B: Hi, Jane. I'm Connie. Nice to meet you. Are you a new student here?A: Yeah.26.B: Sure. Don't shout or run in the hallways.A: Aha, I don't think I'll do that. Can we eat food in class?B: No, we can't.A: OK.27.What else?B: 28.If you don't, the teachers on duty at the gate won't let you in.A: I see.29.B: When you go to the library you should take it.A: 30.B: Try to work hard at all your subjects(科目).If you fail(不及格) any of them, the teachers will call your parents.A: Thank you, Connie. I think if I work really hard, I won't fail any subjects.A.What can I do for you?B.I'll remember that.C.The teacher will punish(惩罚) you if you do that.D.Could you tell me some of the school rules?E.When do we have to use our student ID cards?F.Is there anything you think the most important at this school?G.Remember to wear your school uniform every day.四、阅读单选Every school has its rules. There are some rules in Singapore’s schools, too. Read the school rules and try to be a good student.National song Students and teachers in Singapore must sing the national song in the morning every weekday.School uniform Students should wear school uniforms at school. They can’t wear their own clothes at school.Be on time Students should get to school on time every weekday. In the morning, students should be in school by 7:15 a.m. In the afternoon, they must be in school by 12:35 p.m. But on Wednesday they do sports, so they can get to school by 1:15 p.m.Leave school Students can’t go out of school alone(单独).If someone has to leave school, he must be accompanied(陪伴) by a parent or an authorized(经授权的) person.Visitors to the school Visitors to the school should report to the General Office first. They can’t get into the school without permission(批准).31.How many school rules are mentioned(提及) in this article?A.Three. B.Four.C.Five. D.Six.32.How often do students and teachers in Singapore sing the national song in a week? A.Twice. B.Three times.C.Four times. D.Five times.33.Students should wear at school.A.their own clothes B.white T-shirtsC.blue trousers D.school uniforms34.What time should students usually get to school in the morning?A.By 1:15 p.m. B.By 7:15 a.m.C.By 12:35 p.m. D.By 8:15 a.m.Hi, boys and girls! Welcome to our museum. It’s free. You don’t have to pay any money. But we have some rules for you. Please remember them and do as I say. Firstly, don’t have food or drink here. You may make our museum dirty. Secondly, you can take photos here, but don’t touch(触摸) the things here. Thirdly ,keep quiet in the museum. Don’t talk loudly. Finally, t he museum is not open after five o’clock in the afternoon. Please leave before five. Have a goodtime here! Thank you!35.There are some rules for a .A.store B.parkC.school D.museum36.What does the underlined word “free” mean in Chinese?A.免费的B.迅速的C.高兴的D.自由的37.Students can in the museum.A.have food and drink B.take photosC.touch the things D.talk loudly38.We can learn from the article that .A.students don’t want to pay any moneyB.the museum has five rules for the studentsC.the speaker who tells the students the rules is a teacherD.the speaker hopes the students have a good timeI’m Bob and I’m a student of No.2 Middle School. We have many rules in our school. I think some of them are good for us, but others are not so good.We have to wear our school uniforms. I think it’s good. If we don’t have this rule, some students will think more about clothes, but not study. We can’t be late for school and we have to listen to the teacher carefully in class. All of these rules are good for our study and I like them.We can’t take mobile phones(手机) to school. I don’t think it’s a good idea. Sometimes our parents are busy and can’t get home early. They need to tell us about that. If we don’t take phones, how can they call us? Also, we can’t go to the movies on weekends. I know we should study hard, but we need to relax a little, too.39.What does Bob think of his school rules?A.He can’t stand most of the rules.B.He thinks the rules are strict.C.He thinks some of the rules are good.D.He doesn’t mind them.40.Which rule does Bob like?A.Listen to the teacher carefully in class.B.Can’t eat outside.C.Can’t go to the movies on weekends.D.Wear sports shoes to school.41.Why doesn’t Bob like the rule of no mobile phones at scho ol?A.Because he can’t play games and listen to music on it.B.Because his parents can’t call him when they are busy.C.Because he can’t call his friends when he wants to see them.D.Because he can’t call his parents when he can’t go home.42.Which of the following is TRUE?A.Bob studies in No.12 Middle School.B.Bob has to wear sports clothes at school.C.The parents don’t ask their children to take mobile phones.D.Bob thinks students should not spend all their time on their study.五、回答问题My name is Dave. I’m from the USA.I don’t have much to say about clothes. Buying clothes at a clothes store is really boring for me. I’m only a middle school student, so I never think about what to wear. I often wear the school uniform on weekdays, because it’s the school rule. We must follow it. I have two jackets and I often wear them on weekends.I have two T-shirts. One is a birthday present(礼物) from my father. It’s white. The other(另一个) is my favorite. It is from my best friend Jack. It is red. There is a blue picture on it. Jack says he can find me easily when I wear the T-shirt at school.Amy is my sister. She is seventeen years old. She likes all kinds of colorful and beautiful clothes. Her favorite is a purple skirt. She says beautiful clothes can make her happy and relaxed.43.What does Dave think of buying clothes at a clothes store?(不超过5个词)__________________________________________________________________________ 44.What does Dave’s favorite T-shirt look like?(不超过10个词)__________________________________________________________________________ 45.Why does Amy like beautiful clothes?(不超过15个词)__________________________________________________________________________六、根据首字母填空46.Please _______(到达) at school on time every day.47.They always play football _______(在外面).48.My mother cleans the _______(厨房) every day.49.I can’t ________(记得) his name and phone number.50.You can’t go home ________(在……以前) you finish your homework.七、材料作文51.请你用英语向你们班的新同学Mary介绍一下你们班的班规,并告诉她请遵守这些班规,词数60左右。

20212021学年高中数学模块综合评价新人教A版选修44

模块综合评价(时间:120分钟 满分:150分)一、选择题(本大题共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的)1.点M 的直角坐标是(-1,3),则点M 的极坐标为( )A.⎝ ⎛⎭⎪⎫2,π3 B.⎝⎛⎭⎪⎫2,-π3C.⎝⎛⎭⎪⎫2,2π3D.⎝⎛⎭⎪⎫2,2k π+π3(k ∈Z)解析:点M 的极径是2,点M 在第二象限,故点M 的极坐标是⎝⎛⎭⎪⎫2,2π3.答案:C2.极坐标方程cos θ=32(ρ∈R)表示的曲线是( )A .两条相交直线B .两条射线C .一条直线D .一条射线解析:由cos θ=32,解得θ=π6或θ=116π,又ρ∈R,故为两条过极点的直线.答案:A3.曲线ρcos θ+1=0关于直线θ=π4对称的曲线的方程是( )A .ρsin θ+1=0B .ρcos θ+1=0C .ρsin θ=2D .ρcos θ=2 解析:因为M (ρ,θ)关于直线θ=π4的对称点是N ⎝ ⎛⎭⎪⎫ρ,π2-θ,从而所求曲线方程为ρcos ⎝ ⎛⎭⎪⎫π2-θ+1=0,即ρsin θ+1=0.答案:A4.直线⎩⎪⎨⎪⎧x =1+12t ,y =-33+32t (t 为参数)和圆x 2+y 2=16交于A ,B 两点,则AB 的中点坐标为( )A .(3,-3)B .(-3,3)C .(3,-3)D .(3,-3)解析:将x =1+t2,y =-33+32t 代入圆方程,得⎝ ⎛⎭⎪⎫1+t 22+⎝ ⎛⎭⎪⎪⎫-33+32t 2=16, 所以t 2-8t +12=0,则t 1=2,t 2=6, 因此AB 的中点M 对应参数t =t 1+t 22=4,所以x =1+12×4=3,y =-33+32×4=-3,故AB 中点M 的坐标为(3,-3).答案:D5.化极坐标方程ρ2cos θ-ρ=0为直角坐标方程为( ) A .x 2+y 2=0或y =1 B .x =1 C .x 2+y 2=0或x =1 D .y =1解析:ρ(ρcos θ-1)=0,ρ=x 2+y 2=0或ρcos θ=x =1.答案:C6.直线⎩⎪⎨⎪⎧x =1+2t ,y =2+t(t 为参数)被圆x 2+y 2=9截得的弦长为( )A.125B.1255C.955D.9510解析:把⎩⎪⎨⎪⎧x =1+2t ,y =2+t化为标准形式为⎩⎪⎨⎪⎧x =1+25t ′,y =2+15t ′将其代入x 2+y 2=9,整理得t ′2+85t ′-4=0,由根与系数的关系得t ′1+t ′2=-85,t ′1t ′2=-4.故|t ′1-t ′2|=(t ′1+t ′2)-4t ′1t ′2=⎝ ⎛⎭⎪⎪⎫-852+16=125·5,所以弦长为125 5. 答案:B 7.已知圆M :x 2+y 2-2x -4y =10,则圆心M 到直线⎩⎪⎨⎪⎧x =4t +3,y =3t +1(t 为参数)的距离为( )A .1B .2C .3D .4解析:由题意易知圆的圆心M (1,2),由直线的参数方程化为一般方程为3x -4y -5=0,所以圆心到直线的距离为d =|3×1-4×2-5|32+42=2.答案:B8.点M ⎝ ⎛⎭⎪⎫1,7π6关于直线θ=π4(ρ∈R)的对称点的极坐标为( )A.⎝ ⎛⎭⎪⎫1,4π3B.⎝⎛⎭⎪⎫1,2π3C.⎝ ⎛⎭⎪⎫1,π3 D.⎝⎛⎭⎪⎫1,-7π6 解析:点M ⎝ ⎛⎭⎪⎫1,7π6的直角坐标为⎝ ⎛⎭⎪⎫cos 7π6,sin 7π6=⎝⎛⎭⎪⎪⎫-32,-12,直线θ=π4(ρ∈R),即直线y =x ,点⎝ ⎛⎭⎪⎪⎫-32,-12关于直线y =x 的对称点为⎝ ⎛⎭⎪⎪⎫-12,-32,再化为极坐标为⎝⎛⎭⎪⎫1,4π3.答案:A9.极坐标方程(ρ-1)(θ-π)=0(ρ≥0)和参数方程⎩⎪⎨⎪⎧x =tan θ,y =2cos θ(θ为参数)所表示的图形别离是( )A .直线、射线和圆B .圆、射线和双曲线C .两直线和椭圆D .圆和抛物线解析:因为(ρ-1)(θ-π)=0,所以ρ=1或θ=π(ρ≥0),ρ=1表示圆,θ=π(ρ≥0)表示一条射线,参数方程⎩⎪⎨⎪⎧x =tan θ,y =2cos θ(θ为参数)化为普通方程为y24-x 2=1,表示双曲线. 答案:B10.已知直线l 的参数方程为⎩⎪⎨⎪⎧x =at ,y =a 2t -1(t 为参数),椭圆C 的参数方程为⎩⎪⎨⎪⎧x =1+cos θ,y =2sin θ(θ为参数),且它们总有公共点.则a 的取值范围是( )A.⎣⎢⎡⎭⎪⎫-32,0∪(0,+∞) B .(1,+∞)C.⎣⎢⎡⎭⎪⎫-32,+∞D.⎣⎢⎡⎭⎪⎫-32,4 解析:由已知得⎩⎪⎨⎪⎧at =1+cos θ,a 2t -1=2sin θ,则4(at -1)2+(a 2t -1)2=4, 即a 2(a 2+4)t 2-2a (a +4)t +1=0, Δ=4a 2(a +4)2-4a 2(a 2+4)=16a 2(2a +3). 直线l 与椭圆总有公共点的充要条件是Δ≥0, 即a ≥-32.答案:C11.已知直线l 过点P (-2,0),且倾斜角为150,以直角坐标系的原点为极点,x 轴的正半轴为极轴成立极坐标系,曲线C 的极坐标方程为ρ2-2ρcos θ=15.若直线l 交曲线C 于A ,B 两点,则|PA |·|PB |的值为( )A .5B .7C .15D .20解析:易知直线l 的参数方程为⎩⎪⎨⎪⎧x =-2-32t ,y =12t(t 为参数),把曲线C 的极坐标方程ρ2-2ρcos θ=15化为直角坐标方程是x 2+y 2-2x =15.将直线l 的参数方程代入曲线C 的直角坐标方程,得t 2+33t -7=0.设A ,B 两点对应的参数别离为t 1,t 2,则t 1t 2=-7, 故|PA |·|PB |=|t 1|·|t 2|=|t 1t 2|=7. 答案:B12.过椭圆C :⎩⎪⎨⎪⎧x =2cos θ,y =3sin θ(θ为参数)的右核心F 作直线l 交C 于M ,N 两点,|MF |=m ,|NF |=n ,则1m +1n的值为( )A.23 B.43C.83D .不能肯定解析:曲线C 为椭圆x 24+y 23=1,右核心为F (1,0),设l :⎩⎪⎨⎪⎧x =1+t cos θ,y =t sin θ(t 为参数),代入椭圆方程得(3+sin 2θ)t 2+6t cos θ-9=0,设M 、N 两点对应的参数别离为t 1,t 2,则t 1t 2=-93+sin 2θ,t 1+t 2=-6cos θ3+sin 2θ,所以1m +1n =1|t 1|+1|t 2|=|t 1-t 2||t 1t 2|=(t 1+t 2)2-4t 1t 2|t 1t 2|=43.答案:B二、填空题(本大题共4小题,每小题5分,共20分.把答案填在题中横线上) 13.已知直线l :⎩⎪⎨⎪⎧x =-1+32t ,y =12t(t 为参数)过定点P ,曲线C 的极坐标方程为ρ=2sin θ,直线l 与曲线C 交于A ,B 两点,则|PA |·|PB |的值为________.解析:将直线l :⎩⎪⎨⎪⎧x =-1+32t ,y =12t(t 为参数)代入曲线C :ρ=2sin θ的直角坐标方程x 2+y 2-2y =0,整理,得t 2-(3+1)t +1=0,设直线l与曲线C 的交点A ,B 的对应的参数别离为t 1,t 2,则t 1t 2=1,即|PA |·|PB |=|t 1t 2|=1.答案:114.已知圆的渐开线的参数方程⎩⎪⎨⎪⎧x =3cos φ+3φsin φ,y =3sin φ-3φcos φ(φ为参数),当φ=π4时,对应的曲线上的点的坐标为________.解析:当φ=π4时,代入渐开线的参数方程,得⎩⎪⎨⎪⎧x =3cos π4+3·π4·sin π4,y =3sin π4-3·π4·cos π4,x =322+32π8,y =322-32π8,所以当φ=π4时,对应的曲线上的点的坐标为⎝ ⎛⎭⎪⎪⎫322+32π8,322-32π8.答案:⎝ ⎛⎭⎪⎪⎫322+32π8,322-32π8 15.若直线l 的极坐标方程为ρcos ⎝ ⎛⎭⎪⎫θ-π4=32,曲线C :ρ=1上的点到直线l 的距离为d ,则d 的最大值为________.解析:直线的直角坐标方程为x +y -6=0,曲线C 的方程为x 2+y 2=1,为圆;d 的最大值为圆心到直线的距离加半径,即为d max =|0+0-6|2+1=32+1.答案:32+116.在直角坐标系Oxy 中,椭圆C 的参数方程为⎩⎪⎨⎪⎧x =a cos θ,y =b sin θ(θ为参数,a >b >0).在极坐标系中,直线l 的极坐标方程为ρcos ⎝ ⎛⎭⎪⎫θ+π3=32,若直线l 与x 轴、y 轴的交点别离是椭圆C 的右核心、短轴端点,则a =________.解析:椭圆C 的普通方程为x 2a 2+y 2b 2=1(a >b >0),直线l 的直角坐标方程为x -3y -3=0,令x =0,则y =-1,令y =0,则x =3,所以c =3,b =1,所以a 2=3+1=4,所以a =2. 答案:2三、解答题(本大题共6小题,共70分.解承诺写出文字说明、证明进程或演算步骤)17.(本小题满分10分)在平面直角坐标系xOy 中,直线l 的参数方程为⎩⎪⎨⎪⎧x =t +1,y =2t (t为参数),曲线C 的参数方程为⎩⎪⎨⎪⎧x =2tan 2 θ,y =2tan θ(θ为参数).试求直线l 和曲线C 的普通方程,并求出它们的公共点的坐标.解:因为直线l 的参数方程为⎩⎪⎨⎪⎧x =t +1,y =2t(t 为参数),由x =t +1,得t =x -1,代入y =2t ,取得直线l 的普通方程为2x -y -2=0.同理取得曲线C 的普通方程为y 2=2x .联立方程组⎩⎪⎨⎪⎧y =2(x -1),y 2=2x ,解得公共点的坐标为(2,2),⎝ ⎛⎭⎪⎫12,-1.18.(本小题满分12分)已知某圆的极坐标方程为ρ2-42ρcos ⎝ ⎛⎭⎪⎫θ-π4+6=0,求:(1)圆的普通方程和参数方程;(2)圆上所有点(x ,y )中,xy 的最大值和最小值. 解:(1)原方程可化为ρ2-42ρ⎝⎛⎭⎪⎫cos θcos π4+sin θsin π4+6=0,即ρ2-4ρcos θ-4ρsin θ+6=0.① 因为ρ2=x 2+y 2,x =ρcos θ,y =ρsin θ, 所以①可化为x 2+y 2-4x -4y +6=0,即(x -2)2+(y -2)2=2,即为所求圆的普通方程.设⎩⎪⎨⎪⎧cos θ=2(x -2)2,sin θ=2(y -2)2,所以参数方程为⎩⎪⎨⎪⎧x =2+2cos θ,y =2+2sin θ(θ为参数).(2)由(1)可知xy =(2+2cos θ)(2+2sin θ)=4+22(cos θ+sin θ)+2cos θsin θ= 3+22(cos θ+sin θ)+(cos θ+sin θ)2.设t =cos θ+sin θ,则t =2sin ⎝ ⎛⎭⎪⎫θ+π4,t ∈[-2,2].所以xy =3+22t +t 2=(t +2)2+1.当t =-2时,xy 有最小值1;当t =2时,xy 有最大值9.19.(本小题满分12分)已知曲线C 的极坐标方程是ρ=2cos θ,以极点为平面直角坐标系的原点,极轴为x 轴的正半轴,成立平面直角坐标系,直线l 的参数方程是⎩⎪⎨⎪⎧x =32t +m ,y =12t (t 为参数). (1)求曲线C 的直角坐标方程和直线l 的普通方程;(2)当m =2时,直线l 与曲线C 交于A 、B 两点,求|AB |的值. 解:(1)由ρ=2cos θ,得:ρ2=2ρcos θ,所以x 2+y 2=2x ,即(x -1)2+y 2=1, 所以曲线C 的直角坐标方程为(x -1)2+y 2=1.由⎩⎪⎨⎪⎧x =32t +m ,y =12t 得x =3y +m ,即x -3y -m =0,所以直线l 的普通方程为x -3y -m =0.(2)设圆心到直线l 的距离为d ,由(1)可知直线l :x -3y -2=0,曲线C :(x -1)2+y 2=1,圆C 的圆心坐标为(1,0),半径1, 则圆心到直线l 的距离为d =|1-3×0-2|1+(3)2=12. 所以|AB |=2 1-⎝ ⎛⎭⎪⎫122= 3.因此|AB |的值为3.20.(本小题满分12分)已知圆C 1的参数方程为⎩⎪⎨⎪⎧x =2cos φ,y =2sin φ(φ为参数),以坐标原点O 为极点,x 轴的正半轴为极轴成立极坐标系,圆C 2的极坐标方程为ρ=4sin ⎝ ⎛⎭⎪⎫θ+π3.(1)将圆C 1的参数方程化为普通方程,将圆C 2的极坐标方程化为直角坐标方程; (2)圆C 1,C 2是不是相交?若相交,请求出公共弦长;若不相交,请说明理由.解:(1)由⎩⎪⎨⎪⎧x =2cos φ,y =2sin φ(φ为参数),得圆C 1的普通方程为x 2+y 2=4.由ρ=4sin ⎝ ⎛⎭⎪⎫θ+π3,得ρ2=4ρ⎝⎛⎭⎪⎫sin θcos π3+cos θsin π3,即x 2+y 2=2y +23x ,整理得圆C 2的直角坐标方程为(x -3)2+(y -1)2=4.(2)由于圆C 1表示圆心为原点,半径为2的圆,圆C 2表示圆心为(3,1),半径为2的圆,又圆C 2的圆心(3,1)在圆C 1上可知,圆C 1,C 2相交,由几何性质易知,两圆的公共弦长为23.21.(本小题满分12分)在直角坐标系xOy 中,以O 为极点,x 轴正半轴为极轴成立极坐标系,圆C 的极坐标方程为ρ=22cos ⎝ ⎛⎭⎪⎫θ+π4,直线l 的参数方程为⎩⎪⎨⎪⎧x =t ,y =-1+22t(t为参数),直线l 与圆C 交于A ,B 两点,P 是圆C 上不同于A ,B 的任意一点.(1)求圆心的极坐标;(2)求△PAB 面积的最大值.解:(1)圆C 的直角坐标方程为x 2+y 2-2x +2y =0,即(x -1)2+(y +1)2=2.所以圆心坐标为(1,-1),圆心极坐标为⎝⎛⎭⎪⎫2,7π4. (2)直线l 的普通方程为22x -y -1=0, 圆心到直线l 的距离d =|22+1-1|3=223, 所以|AB |=22-89=2103, 点P 到直线AB 距离的最大值为2+223=523,故最大面积S max =12×2103×523=1059.22.(本小题满分12分)在直角坐标系xOy 中,曲线C 1的参数方程为⎩⎪⎨⎪⎧x =a cos t ,y =1+a sin t (t 为参数,a >0).在以坐标原点为极点、x 轴正半轴为极轴的极坐标系中,曲线C 2:ρ=4cos θ.(1)说明C 1是哪一种曲线,并将C 1的方程化为极坐标方程;(2)直线C 3的极坐标方程为θ=α0,其中α0知足tan α0=2,若曲线C 1与C 2的公共点都在C 3上,求a .解:(1)消去参数t 取得C 1的普通方程为x 2+(y -1)2=a 2,则C 1是以(0,1)为圆心,a 为半径的圆.将x =ρcos θ,y =ρsin θ代入C 1的普通方程中,取得C 1的极坐标方程为ρ2-2ρsin θ+1-a 2=0.(2)曲线C 1,C 2的公共点的极坐标知足方程组 ⎩⎪⎨⎪⎧ρ2-2ρsin θ+1-a 2=0,ρ=4cos θ.若ρ≠0,由方程组得16cos 2θ-8sin θcos θ+1-a 2=0, 由已知tan θ=2,得16cos 2θ-8sin θcos θ=0, 从而1-a 2=0,解得a =-1(舍去)或a =1. 当a =1时,极点也为C 1,C 2的公共点,且在C 3上. 所以a =1.。

2021_2020学年高中英语模块综合检测(A)新人教版必修4

模块综合检测(A)(时间:120分钟总分值:150分)第一局部听力(共两节,总分值30分)第一节(共5小题;每题1.5分,总分值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。

1What may be one of the woman’s choices?A.To work for a small company.B.To start her own business.C.To work for a large company.答案:B2What does the man like?A.Calm water.B.Being sad.C.A stormy ocean.答案:C3What can we learn about Mr White?A.He’s a good teacher.B.He’s not a good teacher.C.He’s one of the students.答案:A4What can we know about the conversation?A.The man refuses the woman’s offer.B.The woman will help the man.C.They are talking about the result of the exam.答案:A5What was the man doing just now?A.Making a telephone call.B.Trying to talk to the operator.C.Operating a computer.答案:B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。

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2021年人教版必修4模块综合检测卷学校:___________姓名:___________班级:___________考号:___________一、阅读选择Queen Victoria was monarch of Great Britain from 1837 until her death in 1901. This period is often called the Victorian Age.Queen Victoria was a stern and serious woman. One reason she was so serious was that she had suffered a great loss. When she was twenty years old, she married a German prince named Albert. Victoria and Albert were deeply in love, and their marriage was extremely happy. In 1861, after they had been married for twenty one years, Albert died, leaving Queen Victoria heart broken. For the rest of her life, the lonely Victoria mourned his loss. It was customary in those days for a widow to dress in black for a short time after the death of her husband. But Queen Victoria dressed in black for forty years. And for forty years, as another sign of her grief, she wrote her letters on white paper edged in black.Even before Prince Albert died, Queen Victoria was known as a very serious woman. She had a stron g sense of duty and worked very hard at all her tasks. In her diary she wrote, “I love to be employed; I hate to be idle.” She never forgot that she was Britain’s queen and always acted with great dignity(尊严). Victoria had high ideals and moral standards (道德标准) that sometimes made her seem stuffy(古板的).She was also very sure of herself. She always thought that she was right, and expected everyone to agree with her.1.Which of the following statements about Queen Victoria is NOT true according to the passage?A.She had great confidence in herself.B.She ruled Great Britain for sixty-four years.C.She enjoyed her marriage to a German prince.D.She became a serious woman after her beloved husband died.2.Queen Victoria wrote her letters on white paper edged in black because . A.she was a stern womanB.black was her favourite colourC.that was one way to show her feeling of sadnessD.it was a custom among monarchs of Great Britain3.All of the following characteristics EXCEPT can be used to properly describe Queen Victoria.A.confident B.lonesomeC.workaholic D.indecisive (优柔寡断的)Often people use laptops (笔记本电脑) on trains and airplanes, in airports and hotels. These laptops connect people to their workplace. In the United States today, laptops also connect students to their classrooms.Westlake College in Virginia will start a laptop computer program that allows students to do schoolwork anywhere they want. Within five years, each of the 1,500 students at the college will receive a laptop. The laptops are part of a $10 million computer program at Westlake, a 110-year-old college. The students with laptops will also have connection with the Internet. Besides, they will be able to use email to “speak” with their teachers, their classmates, and their families. However, the most important part of the laptop program is that students will be able to use computers without going to computer labs. They can work with it at home, in a fast-food restaurant or under the trees_anywhere at all!Because of the many changes in computer technology, laptop use in higher education, such as colleges and universities, is workable (可行的). As laptops become more powerful, they become more similar to desktop computers. Also, the portable computers can connect students to not only the Internet, but also libraries and other resources. State higher education officials are studying how laptops can help students. State officials are also testing laptop programs at other universities, too.4.What does this passage mainly talk about?A.Laptops are very popular in universities of America.B.More laptops are being used in universities of America.C.People like using laptops everywhere, including in universities.D.Laptops will be used in Westlake College in Virginia.5.The main purpose of the laptop program is to give each student a laptop to ________. A.do their schoolworkB.have access to the InternetC.send emailsD.connect them to libraries6.The underlined word “speak” in the second paragraph most probably means ________. A.talk with speakersB.talk with one’s mouthC.communicateD.use the computer language7.Which of the following is TRUE about Westlake College?A.It is an old college in America.B.1,500 students have laptops.C.All students use computers.D.Students there can do everything.Dr Dian Fossey, one of the world’s leading women scientists, had a remarkable career. The work she devoted her life to protecting and studying the mountain gorillas (大猩猩) of Africa—has proved highly effective and has resulted in the steady (平稳的) increase of this most endangered great apes.Fossey made her first trip to Africa in 1963. Three years later, she returned to Africa to begin a long term study of the mountain gorillas. She set up camp in the Democratic Republic of Congo but moved to Rwanda because of political reasons in 1967. She established her “Karisoke” Research Centre camp on September 24,1967.Fossey’s aims were to study gorilla ecology(生态学) and social organization. She found that in order to achieve this, she needed to recognize individual gorillas, which required that the gorillas get used to her presence(出现). By copying gorillas’ behaviour and sounds, Fossey began to gain their trust, and in 1970 an adult male gorilla she had named “Peanuts” reached out to touch her hand.Close observations over thousands of hours enabled Fossey to gain the gorillas’ trust and bring forth new knowledge about their behaviour. Stories and photographs of her work were published in National Geographic Magazine and elsewhere.In 1977, one of Fossey’s favorite gorillas, Digit, was killed by poachers and she established the Digit Fund to help raise money for gorilla protection efforts in the same year.On December 26, 1985, Fossey was murdered while going back to her house in Karisoke. Her body was discovered near the research centre. Most probably? Dian Fossey had been killed by the poachers she’d fought against. On her tombstone (墓碑): “No one loved gorillas more...”In 1988, the life and the work of Fossey were made into a movie based on her book. 8.Why does the author say Dr Dian Fossey “had a remarkable career”?A.Because she travelled all over the world.B.Because she liked to play with gorillas.C.Because she studied gorilla ecology.D.Because she made great apes increase steadily.9.The underlined word “poachers” in Paragraph 5 probably refers to people who . A.sell drugs against the lawB.hunt animals against the lawC.hate successful peopleD.like to do harm to people10.From the words on Fosseys tombstone, we can infer that ________.A.Fossey was the person who loved gorillas mostB.after Fossey died, no one loved gorillasC.Fossey was the first one to study gorillasD.everybody loved gorillas as Fossey did11.According to the passage, Fossey’s life can be described as ________.A.sad and unluckyB.lonely and boringC.simple and happyD.challenging and successfulJenny had a hearing problem when she was four and a half years old. When she was seven, she had juvenile rheumatoid arthritis (幼年型类风湿性关节炎). She could not put pressure on the heels of her feet, so she walked on tiptoes. All through her school, Jenny suffered, yet never complained. She wore a smile on her face, a song on her lips, and love and acceptance of others.Jenny is a beautiful blonde (金发女郎), with warm brown eyes. She never competed in a sport. She could not even take part in a gym class.She was totally acceptable, popular, and funny. She tried her best to attend every football game, and cheer the team on. Wherever she went, she was carrying a pillow which could ease the pain when she sat down. Then came her senior year and she would be considered forscholarships. However, school activities, especially sports, could often mean the difference between receiving an award or losing out.So Jenny came to a decision; she began to apply for the high school football coach. She begged, pleaded and promised. Finally the coach gave in. Jenny even became the manager of the Garrett High School Football Team at last.She carried big buckets of water to her teammates. She bandaged knees and ankles before every game. She gave encouraging talks. It turned out to be one of the best years for Garrett High School Football Team, in its twenty-five-year history.When asked why the team was winning all the games, one linebacker (中后卫) explained, “Well, when you’ve been knocked down, and you can’t seem to move, you look up and see Jenny Lewis, walking hard across the field, carrying her buckets and pillow. It makes anything the rest of us may suffer seem pretty insignifican t.”At the Senior Awards ceremony, Jenny received a number of scholarships to College of Charleston. She was praised to be the first successful girl written in Garrett High School history.12.It is implied in the first paragraph that ________.A.Jenny often complained about her misfortunesB.Jenny lost her hearing when she was seven years oldC.Jenny was an optimistic girl though she had disabilitiesD.Jenny couldn’t stand as a result of juvenile rheumatoid arthritis13.According to the linebacker, the secret of their winning all the games was that ________. A.Jenny had a significant influence on all the team membersB.Jenny played excellently in every football matchC.the Garrett High School Football Team had the best playersD.the linebacker performed significantly in all the games14.According to the passage, which of the following statements is TRUE?A.Jenny finally persuaded the president to be the manager of the football team.B.Jenny received a number of scholarships to Princeton University.C.The Garrett High School Football Team has existed for more than 30 years.D.Jenny was highly praised for her contribution to the Garrett High School.15.What might be the best title for the passage?A.How to Be a Football PlayerB.A Special AthleteC.A Different ScholarshipD.How to Be a Football Coach二、七选五Effective time management is the primary means to a less stressful life. High school, especially during your senior year, can be frustrating. This is the time of your life when you are preparing yourself for college and real world. 16.Plan each day.Planning your day can help you accomplish more and feel more in control of your life. Write a to—do list, putting the most important task at the top. 17.Prioritize (按重要性排列) your weekly schedule as a student.Prioritizing tasks will ensure that you spend your time and energy on those that are truly important to you. 18.Friends will want to hang out with you on the weekends, but they will understand if you explain to them that you need to study or catch up on college-related work.19.Keep a diary of everything you do for free days to determine how you are spending your time. Look for time that can be used more wisely. For example, if you take a bus to school, you can use the time to catch up on reading. Thus, you can free up some time to exercise or spend with your family or friends.Get plenty of sleep, eat a healthy diet and exercise regularly.20.It will help improve efficiency so that you can complete your work in less time. A.A healthy lifestyle can improve your focus and concentration.B.How do you manage your time doing all your activities without being overly stressed? C.Take a break before you need one.D.Any academic studies must come first, then extra curriculum activities, and then social life.E.Evaluate how you’re spending your time.F.You need to try every possible means to save time.G.Every daily activity should be considered seriously.三、完形填空When she walked in, his eyes lit up. They had both 21 someone they loved.So this day was a 22 one. She was 23 , lonely and afraid. He knew how she felt. There was an obvious 24 in his heart, too. 25 what strengthened their love for each other was heartache.She stood there. He motioned to her to sit next to him. She 26 for a moment, but gave in. In fact, she 27 did. Then she held his hand. He looked at her and placed his hand on hers. Suddenly he reached 28 the pillow, pulled out a large 29 and nervously handed it to her. She was a bit in 30 . Then he handed her a small box, wrapped 31 paper of hearts and flowers. She couldn’t 32 what was happening.“Now,” he said, “read the card.”She opened the envelope. It 33 as follows:I know that this has been a hard year for 34 of us. I know that Valentine’s Day is a35 day for people in love. I want you to know that…I love you. I know that Valentines are supposed to get 36 . But I went to the store, only to get the 37 piece.She paused and tears formed in each corner. 38 she unwrapped the box, finding a chocolate heart that was broken into pieces along with a 39 :The lady in the store said all she had left was a 40 heart. I told her so did we. I am so sorry that Dad left us, Mom. But I just wanted you to know we still have each other.Happy Valentine’s Day,Love, Your son,Adam 21.A.missed B.forgot C.lost D.respected 22.A.wonderful B.difficult C.happy D.usual 23.A.hurt B.wounded C.injured D.harmed 24.A.disappointment B.surprise C.excitementD.pain25.A.But B.If C.So D.While 26.A.searched B.hesitated C.walked D.cried 27.A.always B.never C.seldom D.sometimes 28.A.beside B.into C.under D.on 29.A.box B.envelope C.card D.handkerchief 30.A.surprise B.peace C.silence D.shock 31.A.on B.in C.by D.with32.A.remember B.consider C.wonder D.believe 33.A.read B.looked C.wrote D.seemed 34.A.all B.neither C.both D.none 35.A.special B.terrible C.common D.formal 36.A.turkey B.eggs C.chocolate D.trees 37.A.first B.last C.smallest D.best 38.A.Hurriedly B.Quickly C.Happily D.Slowly 39.A.word B.note C.rose D.notice 40.A.sweet B.mended C.cheap D.broken四、用单词的适当形式完成短文阅读下面材料,在空白处填入适当的内容(1个单词)或括号内单询的正确形式。

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