第二学期13周周考试卷及答案

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七年级语文下册月考试卷及答案 (2)

七年级语文下册月考试卷及答案 (2)

第二学期阶段练习七年级语文(满分:150分;考试时间:150分钟;将答案写在答题纸上)一、积累运用(42分)1.下列加点字注音全部正确..的一项是()(3分)A.迸.溅(bèng) 菌.子(jūn)妥帖.(tiē)忍俊不禁.(jīn) B.孱.头(zàn) 竹篾.(miè)祈祷(qí)叱咤风云(zhà)C.挑.逗(tiāo) 猥.琐(wěi)校.(jiào)对血(xuě)气方刚D.震悚.(sǒng) 拖沓(tà)修葺(qì)荒草萋萋..( qī)2.下列句中加点的成语使用恰当..的一项是()(3分)A.无论是高深莫测的星空;还是不值一提的灰尘;都是大自然栩栩如生....的艺术品。

B.这部影片刚刚上映;便成了茶余饭后大庭广众....们的热门话题。

C.一切伟大的行动和思想;都有一个微不足道....的开始。

“天宫二号”的成功发射;科学家们以无所不为....的勇气;去完成科研任务。

3.下列解说有误..的一项是()(3分)A.买三碗雄伟壮丽诗人谈诗统筹方法解说:这几个短语的类型各不相同。

B.为迎接“烟花三月”国际经贸旅游节;让旅客们在航班上感受扬州本土文化;在飞往厦门的航线上打造了“烟花三月下扬州”主题航班。

解说:这句话没有语病。

C.“吃了.用了.人家的东西;不说清楚还行?”和“他说;住那儿多年了.。

”解说:两句话中的“了”分别是动态助词和语气助词。

D.“父母在;不远游;游必有方。

”这句出自《论语》的话;如今已成为许多家有高龄父母的子女的生活信条。

解说:这句话的标点符号使用没有错误。

4.下列关于文学作品内容及常识的表述;完全正确....的一项是()(3分)A. 小说《驿路梨花》以围绕“小茅屋的主人到底是谁”;设置了两次误会、三个悬念;分两个层次;刻画了一组人物;展示了他们助人为乐的美好品格。

B.《爱莲说》作者是唐代的周敦颐;“说”是一种文体;可以直接表明作者的见解;也可以寄寓一定的道理。

2020年浙江省宁波市鄞州中学第二学期测试试题含答案

2020年浙江省宁波市鄞州中学第二学期测试试题含答案

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(完整版)中职《三角函数》试卷精选全文

(完整版)中职《三角函数》试卷精选全文

可编辑修改精选全文完整版东莞市电子科技学校2013~2014学年第二学期13级期末考试试卷《数学》 13级计算机部(广告班除外)班级: 姓名: 学号 : 成绩: 一、选择题:(本大题共15小题,每小题4分,共60分) 1.60-︒角的终边在 ().A 、第一象限B 、第二象限C 、第三象限D 、第四象限 2.与角30︒终边相同的角是 ( ).A 、60-︒B 、390︒C 、-300︒D 、390-︒ 3.150︒= ( ).A 、34πB 、23πC 、56πD 、32π 4.3π-=( ).A 、30︒B 、60-︒C 、60︒D 、90︒ 5.下列各角中不是界限角的是()。

A 、0180-B 、0280C 、090D 、0360 6.正弦函数sin y α=的最小正周期是 ( )A 、4πB 、3πC 、2πD 、π7.如果∂角是第四象限的角,则角α-是第几象限的角 ( )A 、第一象限B 、第二象限C 、第三象限D 、第四象限 8.求值5cos1803sin902tan06sin 270︒-︒+︒-︒=( )A 、-2B 、2C 、3D 、-39.已知角α的终边上的点P 的坐标为(-3,4),则sin α=( )。

A 、35- B 、45C 、34-D 、43-10.与75︒角终边相同的角的集合是( ).A 、75,}k z ββ=︒⨯︒∈{|+k 360 B 、75,}k z ββ=︒⨯︒∈{|+k 180 C 、75,}k z ββ=︒⨯︒∈{|+k 90 D 、75,}k z ββ=︒⨯︒∈{|+k 270 11.已知sin 0,θ<且tan 0,θ>则角θ为( )A 、 第一象限B 、第二象限C 、第三象限D 、第四象限 12.下列各选项中正确的是( )A 、终边相同的角一定相等B 、第一象限的角都是锐角C 、锐角都是第一象限的角D 、小于090的角都是锐角 13.下列等式中正确的是( )A 、sin(720)sin αα+︒=-B 、cos(2)cos απα+=C 、sin(360)sin αα-︒=-D 、tan(4)tan απα+=-14.已知α为第一象限的角,化简tan = ( )A 、 tan αB 、tan α-C 、sin αD 、cos α 15.下列各三角函数值中为负值的是( )A 、sin115︒B 、cos330︒C 、tan(120)-︒D 、sin80︒ 二、填空题:(本大题共4小题,每小题4分,共16分) 16.60︒= 150︒= (角度化弧度)23π= 12π= (弧度化角度) 17.若tan 0θ>,则θ是第 象限的角。

海淀区2024届高三二模数学试题及答案

海淀区2024届高三二模数学试题及答案

海淀区2023-2024学年第二学期期末练习高三数学 2024.5本试卷共6页,150分。

考试时长120分钟。

考生务必将答案答在答题纸上,在试卷上作答无效。

考试结束后,将本试卷和答题纸一并交回。

第一部分(选择题 共40分)一、选择题共10小题,每小题4分,共40分。

在每小题列出的四个选项中,选出符合题目要求的一项。

(1)已知集合,{|3}B x a x =≤<. 若A B ⊆,则a 的最大值为(A )2 (B )0 (C )1-(D )2-(2)在52()x x-的展开式中,x 的系数为(A )10- (B )40- (C )10 (D )40 (3)函数3, 0,()1(),03x x x f x x ⎧≤⎪=⎨>⎪⎩是(A )偶函数,且没有极值点 (B )偶函数,且有一个极值点(C )奇函数,且没有极值点 (D )奇函数,且有一个极值点(4)已知抛物线24x y =的焦点为F ,点A 在抛物线上,||6AF =. 则线段AF 的中点的纵坐标为(A )52 (B )72(C )3 (D )4(5)在ABC △中,4AB =,5AC =,3cos 4C =,则BC 的长为 (A )6或32(B )6 (C) (D )3(6)设,a b ∈R ,0ab ≠,且a b >,则{1,0,1,2}A =-(A )b a a b< (B )2b aa b+> (C )sin()a b a b -<- (D )32a b >(7)在ABC △中,π2C ∠=,CA CB ==点P 满足(1)CP CA CB λλ=+-,且4C P A B ⋅=,则λ=(A )14- (B )14(C )34- (D )34(8)设{}n a 是公比为q (1q ≠-)的无穷等比数列,n S 为其前n 项和,10a >.则“0q >”是“n S 存在最小值”的 (A )充分而不必要条件 (B )必要而不充分条件(C )充分必要条件(D )既不充分也不必要条件(9)设函数()f x 的定义域为D ,对于函数()f x 图象上一点00(,)x y ,若集合00{|()(),}k k x x y f x x D ∈-+≤∀∈R 只有1个元素,则称函数()f x 具有性质0x P .下列函数中具有性质1P 的是(A )()|1|f x x =- (B )()lg f x x =(C )3()f x x =(D )π()sin2f x x =- (10)设数列{}n a 的各项均为非零的整数,其前n 项和为n S . 若j i -(i ,*j ∈N )为正偶数,均有2j i a a ≥,且20S =,则10S 的最小值为 (A )0 (B )22 (C )26 (D )31第二部分(非选择题 共110分)二、填空题共5小题,每小题5分,共25分。

上海市2023-2024学年高二下学期期中考试 数学(A卷)含答案

上海市2023-2024学年高二下学期期中考试 数学(A卷)含答案

2023学年第二学期高二年级数学期中考试试卷(A )(答案在最后)时间:120分钟满分:150分注:请将试题的解答全部写在答题纸的相应位置,写在试卷上无效.一、填空题(本大题共有12小题,第1-6题每题4分,第7-12题每题5分,满分54分)考生应在答题纸的相应位置直接填写结果.1.设随机变量X 服从二项分布19,3B ⎛⎫ ⎪⎝⎭,则[]D X =_________.2.8位选手参加射击比赛,最终的成绩(环数)分别为42,38,45,43,41,47,44,46,这组数据的第75百分位数是_________.参考表格:3.在一个22⨯列联表中,通过数据计算28.325χ=,则这两个变量间有关的可能性为________.参考表格:()20P x χ≥0.050.0250.0100.0010x 3.841 5.024 6.63510.8284.曲线()ln f x x x =+在1x =处的切线方程是________.5.某同学在一次考试中,8道单选题中有6道有思路,2道没思路,有思路的有90%的可能性能做对,没思路的有25%的可能性做对,则他在8道题中随意选择一道题,做对的概率是__________.6.“守得住经典,当得了网红”,这是时下人们对国货最高的评价,网络平台的发展让越来越多的消费者熟悉了国货品牌的优势,使得各大国货品牌都受到高度关注,销售额迅速增长,已知某国货品牌2023年8-12月在D 网络平台的月销售额y (单位:百万元)与月份x 具有线性相关关系,并根据这5个月的月销售额,求得回归方程为 4.23ˆy x =+,则该国货品牌2023年8-12月在D 网络平台的总销售额为______百万元.7.今天星期三,再过1天是星期四,那么再过20242天是星期_________.8.已知52345012345(23)x a a x a x a x a x a x +=+++++,则123452345a a a a a -+-+=________.(用数字作答)9.双曲线具有如下光学性质:从一个焦点发出的光线经双曲线反射后,反射光线的反向延长线一定经过另一个焦点.已知双曲线()2222:10x yC a ba b-=>,,如图从C的一个焦点F射出的光线,经过P Q,两点反射后,分别经过点M和N.若12cos13PM PQ PM PQ PQN∠+=-=-,,则C的离心率为_________. 10.函数()11,03ln,0x xf xx x⎧+≤⎪=⎨⎪>⎩,若方程()0f x ax-=恰有3个根,则实数a的取值范围为______.11.一只蜜蜂从蜂房A出发向右爬,每次只能爬向右侧相邻的两个蜂房(如图),例如:从蜂房A只能爬到1号或2号蜂房,从1号蜂房只能爬到2号或3号蜂房........此类推,用na表示蜜蜂爬到n号蜂房的方法数.设集合{}232025S a a a=,,,,集合B是集合S的非空子集,则B中所有元素之和为奇数的概率为________.12.现有6根绳子,共有12个绳头,每个绳头只打一次结,且每个结仅含两个绳头,所有绳头打结完毕视为结束.则这6根绳子恰好能围成一个圈的概率为______.二、选择题(本大题共有4小题,第13-14题每远4分,第15-16题每题5分,满分18分)毎题有且只有一个正确选项.考生应在答題纸的相应位置,将正确选项用2B铅笔涂黑.13.要调查下列问题,适合采用全面调查(普查)的是()A.某城市居民3月份人均网上购物的次数B.某品牌新能源汽车最大续航里程C.检测一批灯泡的使用寿命D.调查一个班级学生每周的体育锻炼时间14.对两个变量的三组数据进行统计,得到以下散点图,关于两个变量相关系数的比较,正确的是()A.123r r r >>B.231r r r >>C.132r r r >>D.321r r r >>15.江先生每天9点上班,上班通常开私家车加步行或乘坐地铁加步行,私家车路程近一些,但路上经常拥堵,所需时间(单位:分钟)服从正态分布2(38,7)N ,从停车场步行到单位要6分钟;江先生从家到地铁站需要步行5分钟,乘坐地铁畅通,但路线长且乘客多,所需间(单位:分钟)服从正态分布2(44,2)N ,下地铁后从地铁站步行到单位要5分钟,从统计的角度出发,下列说法中合理的有()参考数据:若2()~(,)P Z N μσ,则()0.6826P Z μσμσ-<<+=,(22)0.9544P Z μσμσ-<<+=,(33)0.9974P Z μσμσ-<<+=A .若8:00出门,则开私家车不会迟到B.若8:02出门,则乘坐地铁上班不迟到的可能性更大C.若8:06出门,则乘坐地铁上班不迟到的可能性更大D.若8:12出门,则乘坐地铁几乎不可能上班不迟到16.n S 是数列{}n a 前n 项和,11243,41n n a a a n +==--,给出以下两个命题:命题211212:2n p a a a a a a n n +++=+ ;命题q :对任意正整数n ,不等式()ln 21n S n n >++恒成立.下列说法正确的是()A.命题p q 、都是真命题B.命题p 为真命题,命题q 为假命题C.命题p 为假命题,命题q 为真命题D.命题p q 、都是假命题三、解答题(本大题共5题,满分78分)解答下列各题须在答题纸的相应位置写出必要的步骤.17.如图所示,在棱长为2的正方体1111ABCD A B C D -中,,E F 分别为线段1,DD BD 的中点.(1)求异面直线EF 与BC 所成的角;(2)求三棱锥11C B D F -的体积.18.已知函数2()6ln(1),f x ax x a =-+为常数.(1)若()y f x =在1x =处有极值,求a 的值并判断1x =是极大值点还是极小值点;(2)若()y f x =在[]23,上是增函数,求实数a 的取值范围.19.本市某区对全区高中生的身高(单位:厘米)进行统计,得到如下的频率分布直方图.(1)若数据分布均匀,记随机变量X 为各区间中点所代表的身高,写出X 的分布列及期望.(2)现从身高在区间[)170,190的高中生中分层抽样抽取一个160人的样本.若身高在区间[)170,180中样本的均值为176厘米,方差为10;身高在区间[)180,190中样本的均值为184厘米,方差为16,试求这160人身高的方差.20.已知椭圆()22:11x C y t t+=>的左、右焦点分别为12F F 、,直线():0l y kx m m =+≠与椭圆C 交于M N 、两点(M 点在N 点的上方),与y 轴交于点E .(1)当3t =时,点A 为椭圆C 上除顶点外任一点,求12AF F △的周长;(2)当4t =且直线l 过点()10D -,时,设EM DM EN DN λμ== ,,求证:λμ+为定值,并求出该值;(3)若椭圆C 的离心率为223,当k 为何值时,22OM ON +恒为定值;并求此时MON △面积的最大值.21.对于有穷数列()12,,,3m a a a m ≥ ,若存在等差数列{}n b ,使得11221m m m b a b a b a b +≤<≤<<≤< ,则称数列{}n a 是一个长为m 的“弱等差数列”.(1)证明:数列124,,是“弱等差数列”;(2)设函数()sin f x x x =,()f x 在()0,2024内的全部极值点按从小到大的顺序排列为12,,,m a a a ,证明:12,,,m a a a 是“弱等差数列”;(3)证明:存在长为2024的“弱等差数列”{}n a ,且{}n a 是等比数列.2023学年第二学期高二年级数学期中考试试卷(A )时间:120分钟满分:150分注:请将试题的解答全部写在答题纸的相应位置,写在试卷上无效.一、填空题(本大题共有12小题,第1-6题每题4分,第7-12题每题5分,满分54分)考生应在答题纸的相应位置直接填写结果.1.设随机变量X 服从二项分布19,3B ⎛⎫ ⎪⎝⎭,则[]D X =_________.【答案】2【解析】【分析】根据给定条件,利用二项分布的方差公式计算得解.【详解】依题意,11[]9(1)233D X =⨯⨯-=.故答案为:22.8位选手参加射击比赛,最终的成绩(环数)分别为42,38,45,43,41,47,44,46,这组数据的第75百分位数是_________.参考表格:【答案】45.5【解析】【分析】先排序,再由875%6⨯=,可取第6和第7个数之和的一半即可得解.【详解】先排序可得38,41,42,43,44,45,46,47,由875%6⨯=,所以第75百分位数是454645.52+=.故答案为:45.53.在一个22⨯列联表中,通过数据计算28.325χ=,则这两个变量间有关的可能性为________.参考表格:()20P x χ≥0.050.0250.0100.0010x 3.841 5.024 6.63510.828【答案】99%##0.99【解析】【分析】根据独立性检验的知识确定正确答案.【详解】由于28.325 6.635χ=>,所以两个变量之间有关系的可能性为99%.故答案为:99%4.曲线()ln f x x x =+在1x =处的切线方程是________.【答案】21y x =-【解析】【分析】求出函数的导函数,把1x =代入即可得到切线的斜率,然后根据(1,1)和斜率写出切线的方程即可.【详解】解:由函数ln y x x =+知1'1y x=+,把1x =代入'y 得到切线的斜率112k =+=则切线方程为:12(1)y x -=-,即21y x =-.故答案为:21y x =-【点睛】本题考查导数的几何意义,属于基础题.5.某同学在一次考试中,8道单选题中有6道有思路,2道没思路,有思路的有90%的可能性能做对,没思路的有25%的可能性做对,则他在8道题中随意选择一道题,做对的概率是__________.【答案】5980【解析】【分析】根据全概率公式求解即可.【详解】设事件A 表示“考生答对”,设事件B 表示“考生选到有思路的题”则小明从这8道题目中随机抽取1道做对的概率为:3159()()()()(0.90.254480P A P B P A B P B P A B =+=⨯+⨯=∣∣.故答案为:5980.6.“守得住经典,当得了网红”,这是时下人们对国货最高的评价,网络平台的发展让越来越多的消费者熟悉了国货品牌的优势,使得各大国货品牌都受到高度关注,销售额迅速增长,已知某国货品牌2023年8-12月在D 网络平台的月销售额y (单位:百万元)与月份x 具有线性相关关系,并根据这5个月的月销售额,求得回归方程为 4.23ˆyx =+,则该国货品牌2023年8-12月在D 网络平台的总销售额为______百万元.【答案】225【解析】【分析】根据样本中心点()x y 在回归直线上的性质,先计算出x ,代入回归方程求得y ,再用y 代表月平均销售额,即可算得总销售额.【详解】依题意,89101112105x ++++==,因样本中心点()x y 在回归直线上,代入得:4.210345y =⨯+=,所以该国货品牌2023年8-12月在D 网络平台的总销售额为545225⨯=百万元.故答案为:225.7.今天星期三,再过1天是星期四,那么再过20242天是星期_________.【答案】天(或日)【解析】【分析】首先由67432642026474724284(71)⨯+==⨯=+,再利用二项展开式即可得解.【详解】由()()6742024674326740674167367467467467422484714C 7C 7C ⨯+==⨯=+=⋅+⋅++ ()067416736736746746744C 7C 7C 74=⋅+⋅+⋅+ ,所以20242除7余4,所以再过20242天是星期天.故答案为:天(或日).8.已知52345012345(23)x a a x a x a x a x a x +=+++++,则123452345a a a a a -+-+=________.(用数字作答)【答案】15【解析】【分析】根据条件,两边求导得到12342345415(23)2345x a a x a x a x a x +=++++,再取=1x -,即可求出结果.【详解】因为52345012345(23)x a a x a x a x a x a x +=+++++,两边求导可得12342345415(23)2345x a a x a x a x a x +=++++,令=1x -,得到23454115(23)2345a a a a a -=-+-+,即12345234515a a a a a -+-+=,故答案为:15.9.双曲线具有如下光学性质:从一个焦点发出的光线经双曲线反射后,反射光线的反向延长线一定经过另一个焦点.已知双曲线()2222:10x y C a b a b-=>,,如图从C 的一个焦点F 射出的光线,经过P Q ,两点反射后,分别经过点M 和N .若12cos 13PM PQ PM PQ PQN ∠+=-=- ,,则C 的离心率为_________.【答案】3【解析】【分析】作出MP ,QN 的反向延长线交于双曲线的左焦点1F ,由已知可得PM PQ ⊥ ,112cos 13PQF ∠=,设1||13||12,F Q t PQ t ==,可得||2,23,PF t a t ==由勾股定理可求得1||,F F =进而可求C 的离心率.【详解】由双曲线的光学性质可知MP ,QN 的反向延长线交于双曲线的左焦点1F ,如图所示:由||||PM PQ PM PQ +=- ,两边平方可得222222PM PM PQ PQ PM PM PQ PQ ++=-+ ,所以0PM PQ = ,所以PM PQ ⊥ ,所以190∠=︒F PF ,又12cos 13PQN ∠=-,所以112cos 13PQF ∠=,设1||13||12,F Q t PQ t ==,则1||5PF t =,设||FQ m =,则||12FP t m =-,根据双曲线定义,可得11||||||||2PF PF QF QF a -=-=,所以5(12)132t t m t m a --=-=,解得10m t =,所以||2,23,PF t a t ==在1Rt F PF 中,222211||||||29,F F PF PF t =+=所以1||,F F =所以C的离心率为3c e a ==.故答案为:3.10.函数()11,03ln ,0x x f x x x ⎧+≤⎪=⎨⎪>⎩,若方程()0f x ax -=恰有3个根,则实数a 的取值范围为______.【答案】11,3e ⎡⎫⎪⎢⎣⎭【解析】【分析】画出()11,03ln ,0x x f x x x ⎧+≤⎪=⎨⎪>⎩的图象,再分析()y f x =与直线y ax =的交点个数即可.【详解】画出函数()f x 的图象,如图所示:由题意可知0a >,先求y ax =与ln y x =相切时的情况,由图可得此时ln y x =,1y x'=设切点为()00,ln x x ,则0001ln a x x ax ⎧=⎪⎨⎪=⎩,解得0e x =,1e a =,此时直线e x y =,此时直线e x y =与()y f x =只有两个公共点,所以1e a <,又斜率11e 3>,又当13a =时13y x =与11,(0)3y x x =+≤平行,13y x =与()y f x =有三个公共点,而当13a <,直线y ax =与()y f x =有四个交点,故11,3e a ⎡⎫∈⎪⎢⎣⎭.故答案为:11,3e ⎡⎫⎪⎢⎣⎭11.一只蜜蜂从蜂房A 出发向右爬,每次只能爬向右侧相邻的两个蜂房(如图),例如:从蜂房A 只能爬到1号或2号蜂房,从1号蜂房只能爬到2号或3号蜂房........此类推,用n a 表示蜜蜂爬到n 号蜂房的方法数.设集合{}232025S a a a = ,,,,集合B 是集合S 的非空子集,则B 中所有元素之和为奇数的概率为________.【答案】20232024221-【解析】【分析】根据题意,得到数列{}n a 满足12n n n a a a --=+,求得在{}232025S a a a = ,,,偶数项共有675项,奇数项为1349项,得到S 中有202421-的非空子集,以及B 中所有元素之和为奇数的个数,结合古典概型的概率计算公式,即可求解.【详解】由题意知,该蜜蜂爬到1号蜂房的路线数为1,第2号蜂房的路线数为2,第3号蜂房的路线数为3,第4号蜂房的路线数为5,第5号蜂房的路线数为8, ,则第n 号蜂房的路线数为12(3,N )n n n a a a n n *--=+≥∈,所以54575686713,21,34,a a a a a a a a a =+==+==+= ,即数列{}n a 为1,2,3,5,8,13,21,34, ,其中25811,,,,a a a a 为偶数,所以在{}232025S a a a = ,,,偶数项共有675项,奇数项为1349项,又由{}232025S a a a = ,,,,可得S 中有202421-的非空子集,若B 中元素之和为奇数,则B 中的奇数共有奇数个,偶数可以随意,所以满足条件的B 的个数为:01267513134867513482023675675675675134913491349(C C C C )(C C C )222+++++++=⋅= ,所以B 中所有元素之和为奇数的概率为20232024221P =-.故答案为:20232024221-.12.现有6根绳子,共有12个绳头,每个绳头只打一次结,且每个结仅含两个绳头,所有绳头打结完毕视为结束.则这6根绳子恰好能围成一个圈的概率为______.【答案】256693【解析】【分析】直接根据圆排列及古典概型计算.【详解】依题意,环排列有:11111108642C C C C C 3840⋅⋅⋅⋅=种,总的连接方式有:22222121086466C C C C C 66452815610395A 720⋅⋅⋅⋅⨯⨯⨯⨯==种,所以恰好能围成一个圈的概率为384025610395693P ==.故答案为:256693.二、选择题(本大题共有4小题,第13-14题每远4分,第15-16题每题5分,满分18分)毎题有且只有一个正确选项.考生应在答題纸的相应位置,将正确选项用2B 铅笔涂黑.13.要调查下列问题,适合采用全面调查(普查)的是()A.某城市居民3月份人均网上购物的次数B.某品牌新能源汽车最大续航里程C.检测一批灯泡的使用寿命D.调查一个班级学生每周的体育锻炼时间【答案】D 【解析】【分析】结合普查和抽查的适用条件即可求解.【详解】A ,B 选项中要调查的总体数量和工作量都较大,适合采用抽查;C 选项的检测具有毁损性,适合抽查;D 选项要调查的总体数量较小,工作量较小,适合采用普查,故选:D.14.对两个变量的三组数据进行统计,得到以下散点图,关于两个变量相关系数的比较,正确的是()A.123r r r >>B.231r r r >>C.132r r r >>D.321r r r >>【答案】C 【解析】【分析】根据散点图中点的分布的特征,确定3个图对应的相关系数的正负以及大小关系,可得答案.【详解】由散点图可知第1个图表示的正相关,故10r >;第2,3图表示的负相关,且第2个图中的点比第3个图中的点分布更为集中,故23,0r r <,且23r r >,故230r r <<,综合可得231r r r <<,即132r r r >>,故选:C15.江先生每天9点上班,上班通常开私家车加步行或乘坐地铁加步行,私家车路程近一些,但路上经常拥堵,所需时间(单位:分钟)服从正态分布2(38,7)N ,从停车场步行到单位要6分钟;江先生从家到地铁站需要步行5分钟,乘坐地铁畅通,但路线长且乘客多,所需间(单位:分钟)服从正态分布2(44,2)N ,下地铁后从地铁站步行到单位要5分钟,从统计的角度出发,下列说法中合理的有()参考数据:若2()~(,)P Z N μσ,则()0.6826P Z μσμσ-<<+=,(22)0.9544P Z μσμσ-<<+=,(33)0.9974P Z μσμσ-<<+=A.若8:00出门,则开私家车不会迟到B.若8:02出门,则乘坐地铁上班不迟到的可能性更大C.若8:06出门,则乘坐地铁上班不迟到的可能性更大D.若8:12出门,则乘坐地铁几乎不可能上班不迟到【答案】D 【解析】【分析】对于A ,由(59)0.0013P Z ≥=即可判断;对于BC ,分别计算开私家车及乘坐地铁不迟到的概率即可判断;对于D ,计算(38)0.0013P Z ≤=即可判断【详解】对于A ,当满足1(1759)10.9974(59)0.001322P Z P Z -<≤-≥===时,江先生仍旧有可能迟到,只不过发生的概率较小,故A 错误;对于B ,若8:02出门,①江先生开私家车,当满足1(2452)(52)(2452)0.97722P Z P Z P Z -<<≤=+<<=时,此时江先生开私家车不会迟到;②江先生乘坐地铁,当满足()()().1P 40Z 48P Z 48P 40Z 48097722-<<≤=+<<=时,此时江先生乘坐地铁不会迟到;此时两种上班方式,江先生不迟到的概率相当,故B 错误;对于C ,若8:06出门,①江先生开私家车,当满足1(3145)(48)(45)(3145)0.84132P Z P Z P Z P Z -<<≤>≤=+<<=时,此时江先生开私家车不会迟到;②江先生乘坐地铁,当满足().1P Z 44052≤==时,此时江先生乘坐地铁不会迟到;此时两种上班方式,显然江先生开私家车不迟到的可能性更大,故C 错误;对于D ,若8:12出门,江先生乘坐地铁上班,当满足()().1P 38Z 50P Z 38000132-<<≤==时,江先生乘坐地铁不会迟到,此时不迟到的可能性极小,故江先生乘坐地铁几乎不可能上班不迟到,故D 正确.故选:D.【点睛】关键点点睛:本题解决的关键是分别分析得江先生使用不同交通工具在路上所花时间,结合正态分布的对称性求得其对应的概率,从而得解.16.n S 是数列{}n a 前n 项和,11243,41n n a a a n +==--,给出以下两个命题:命题211212:2n p a a a a a a n n +++=+ ;命题q :对任意正整数n ,不等式()ln 21n S n n >++恒成立.下列说法正确的是()A.命题p q 、都是真命题B.命题p 为真命题,命题q 为假命题C.命题p 为假命题,命题q 为真命题D.命题p q 、都是假命题【答案】A 【解析】【分析】由题意可求出12n a a a 的表达式,利用等差数列的求和公式可判断命题p ;证明出当01x <≤时,ln 1≤-x x ,可得出212ln2121n n n +≤--,再结合放缩法可判断命题q .【详解】因为()()11244223,4121212121n n n n a a a a a n n n n n +⎛⎫==-=-=-- ⎪--+-+⎝⎭,所以()12221121n n a a n n +-=-+--,所以,数列221n a n ⎧⎫-⎨⎬-⎩⎭为常数列,则122121n a a n -=-=-,所以22112121n n a n n +=+=--;所以123521211321n n a a a n n +=⨯⨯⨯=+- ,令21n b n =+,则12n n b b +-=,所以数列{}n b 为首项为3,公差为2的等差数列,因此()()2112123213572122n n n a a a a a a n n n +++++=+++++=+ ,即命题p 正确;设()1ln x x x ϕ=--,其中01x <≤,则()111xx x xϕ'-=-=,当01x <<时,()0x ϕ'<,()x ϕ单调递减,则()()10x ϕϕ≥=,即ln 1≤-x x ,当且仅当1x =时,等号成立,所以21212ln1212121n n n n n ++<-=---,即2235212ln ln ln 3211321n n S n n n n +⎛⎫=++++>++++ ⎪--⎝⎭ ,则()3521ln ln 211321n n S n n n n +⎛⎫=+⨯⨯⨯=++ ⎪-⎝⎭,所以命题q 正确.故选:A.三、解答题(本大题共5题,满分78分)解答下列各题须在答题纸的相应位置写出必要的步骤.17.如图所示,在棱长为2的正方体1111ABCD A B C D -中,,E F 分别为线段1,DD BD 的中点.(1)求异面直线EF 与BC 所成的角;(2)求三棱锥11C B D F -的体积.【答案】(1)arccos 3.(2)43.【解析】【分析】(1)分别以1,,DA DC DD 为x 轴,y 轴,z 轴,建立空间直角坐标系,利用向量法能求出异面直线EF 与BC 所成的角.(2)先求出11C B D S ,再由向量法求出点F 到平面11D B C 的距离,由此根据1111C B D F F B D C V V --=即可求出三棱锥11C B D F -的体积.【小问1详解】以D 为坐标原点,分别以1,,DA DC DD 为x 轴,y 轴,z 轴,建立空间直角坐标系,∵在棱长为2的正方体1111ABCD A B C D -中,,E F 分别为线段1,DD BD 的中点,∴(0,0,1),(1,1,0),(2,2,0),(0,2,0)E F B C ,∴(1,1,1),(2,0,0)EF BC =-=-,设异面直线EF 与BC 所成的角为π,(0]2θθ∈,,则|||2|3cos cos ,|3||||32|EF BC EF BC EF BC θ⋅=〈〉==⋅⨯,∴异面直线EF 与BC 所成的角为3arccos 3.【小问2详解】∵在棱长为2的正方体1111ABCD A B C D -中,11112B D B C D C ===,∴111322222322B DC S ⨯== ∵112,2,2),0,0,2),(0,2,0),(1,1,0)((B D C F ,∴1111(2,20),(0,22),(1,1,2)D B D D C F ==-=-,,,设平面11D B C 的法向量(,,)n x y z =,则1110n D B n D C ⎧⋅=⎪⎨⋅=⎪⎩,∴220220x y y z +=⎧⎨-=⎩,令1x =,则可取(1,1,1)n =--r ,∴点F 到平面11D B C 的距离1||33||3n D F d n ⋅== ,∴三棱锥11C B D F -的体积11111111234233333C BD F F B D C B D C V V S d --==⨯=⨯⨯= .18.已知函数2()6ln(1),f x ax x a =-+为常数.(1)若()y f x =在1x =处有极值,求a 的值并判断1x =是极大值点还是极小值点;(2)若()y f x =在[]23,上是增函数,求实数a 的取值范围.【答案】(1)32a =;1x =是()y f x =的极小值点(2)实数a 的取值范围为)1,2∞⎡+⎢⎣【解析】【分析】(1)先根据函数在1x =处有极值求出a 的值,将a 值代入原函数求导进行判断函数在1x =左右的导函数正负号即可得到结果;(2)()y f x =在[]23,上是增函数,转化成()0f x '≥在[]23x ∈,恒成立,进而分离参数转化成23a x x≥+在[]23x ∈,恒成立进行求解即可得到结果.【小问1详解】()f x 的定义域为[)1,∞-+,则6()21f x ax x-'=+;由题意,()y f x =在1x =处有极值,即()01f '=,即230a -=;∴32a =;∴63(2)(1)()311x x f x x x x+-=-'=++,∴当1x >时,()0f x '>,()f x 为增函数;当11x -<<时,()0f x '<,()f x 为减函数;∴1x =是()y f x =的极小值点.【小问2详解】∵()y f x =在[]23,上是增函数,∴()0f x '≥在[]23x ∈,恒成立,即有6201ax x-≥+,23a x x ∴≥+在[]23x ∈,恒成立,只需求2max3a x x ⎛⎫≥ ⎪+⎝⎭;[]23x ∈ ,,[]22116,1224x x x ⎛⎫∴+=+-∈ ⎪⎝⎭,2311,42x x ⎡⎤∴∈⎢⎥+⎣⎦;12a ∴≥,∴a 的取值范围为)1,2∞⎡+⎢⎣.19.本市某区对全区高中生的身高(单位:厘米)进行统计,得到如下的频率分布直方图.(1)若数据分布均匀,记随机变量X 为各区间中点所代表的身高,写出X 的分布列及期望.(2)现从身高在区间[)170,190的高中生中分层抽样抽取一个160人的样本.若身高在区间[)170,180中样本的均值为176厘米,方差为10;身高在区间[)180,190中样本的均值为184厘米,方差为16,试求这160人身高的方差.【答案】(1)分布列见详解,期望为171.7(2)27.25【解析】【分析】(1)依据分布列和期望的定义即可求得X 的分布列及期望;(2)依据方差的定义去求这160人的方差.【小问1详解】由(0.0270.0250.0220.010.001)101x +++++⨯=,解得0.015x =,所以X 的分布列为:X155165175185195205P0.220.270.250.150.10.01()0.221550.271650.251750.151850.11950.01205171.7E X =⨯+⨯+⨯+⨯+⨯+⨯=.【小问2详解】由于身高在区间[)170,180,[)180,190的人数之比为5:3,所以分层抽样抽取160人,区间[)170,180,[)180,190内抽取的人数分别为100人与60人.在区间[)170,180中抽取的100个样本的均值为176,方差为10,即176x =,2110s =,在区间[)180,190中抽取的60个样本的均值为184,方差为16,即184y =,2216s =,所以这160人身高的均值为10017660184179160z ⨯+⨯==,从而这160人身高的方差为2s 22221210060()()160100s x z s y z ⎡⎤⎡⎤=+-++-⎣⎦⎣⎦221006010(176179)16(184179)27.25160160⎡⎤⎡⎤=⨯+-+⨯+-=⎣⎦⎣⎦,因此这160人身高的方差为27.25.20.已知椭圆()22:11x C y t t+=>的左、右焦点分别为12F F 、,直线():0l y kx m m =+≠与椭圆C 交于M N 、两点(M 点在N 点的上方),与y 轴交于点E .(1)当3t =时,点A 为椭圆C 上除顶点外任一点,求12AF F △的周长;(2)当4t =且直线l 过点()10D -,时,设EM DM EN DN λμ==,,求证:λμ+为定值,并求出该值;(3)若椭圆C 的离心率为223,当k 为何值时,22OM ON +恒为定值;并求此时MON △面积的最大值.【答案】(1)(2)83(3)32【解析】【分析】(1)根据椭圆定义求解三角形周长;(2)联立:(0)l y kx m m =+≠与22:19x C y +=,得到两根之和两根之积,由,EM DM EN DN λμ== 得到121211x x x x λμ+=+++,结合两根之和,两根之积求出答案;(3)先由离心率得到椭圆方程,联立直线方程,得到两根之和,两根之积,表达出()()()2222222919121691k m k OM ON k -+++=+⨯+,结合22||||OM ON +为定值得到13k =±,并求出此时MN ,和点O 到直线l 的距离d ,利用基本不等式得到32MON S ≤.【小问1详解】当3t =时,椭圆方程为22:13x C y +=,故a =c =,由椭圆定义可得,12AF F △的周长为22a c +=;【小问2详解】4t =时,椭圆方程为22:14x C y +=,故联立:(0)l y kx m m =+≠与22:14x C y +=可得,()222418440k x kmx m +++-=设()()1122,,,M x y N x y ,则2121222844,4141km m x x x x k k --+==++,因为直线l 过点()10D -,,所以0k m =-+,即k m =,所以22121222844,4141k k x x x x k k --+==++因为()10D -,,设()0,E E y ,所以()11,E x EM y y =- ,()111,D x M y =+ ,()22,E x EN y y =- ,()221,x DN y =+ ,又因为,EM DM EN DN λμ== ,所以()()11221,1x x x x λμ=+=+,所以111x x λ=+,221x x μ=+,所以121212*********x x x x x x x x x x λμ+++=+=-+++++222222841844413224218231k k k k k k +=--=+--+++=++,所以λμ+为定值83.【小问3详解】由题意得3=,解得9t =,椭圆方程2219x y +=,联立2299y kx m x y =+⎧⎨+=⎩,消元得()2229118990k x kmx m +++-=,当()()2222Δ324369110k m k m =-+->,即22910k m -+>时,设()()1122,,,M x y N x y ,则1221891km x x k -+=+,21229991m x x k -⋅=+,又因为M 、N 在椭圆上,则121219y x =-,222219y x =-,则22222212121199x x OM ON x x +=+-++-()()2221112222228899x x x x x x ⎡⎤=++=+-⋅⎣+⎦()()()()()2222222222299191912162169191k m m k k m k k k ⎡⎤-++-++⎢⎥=+=+⨯⎢⎥++⎣⎦当22OM ON +为定值时,即与2m 无关,故2910k -=,得13k =±,此时219MN k ===+又点O 到直线l的距离d ==所以12MON S d MN =⨯⨯=△()222333322222m m +-==≤⋅=,当且仅当m =1m =±时,等号成立,因为2291k m ∆=-+,经检验,此时Δ0>成立,所以MON △面积的最大值为32.【点睛】方法点睛:圆锥曲线中最值或范围问题的常见解法(1)几何法,若题目的条件和结论能明显体现几何特征和意义,则考虑利用几何法来解决;(2)代数法,若题目的条件和结论能体现某种明确的函数关系,则可首先建立目标函数,再求这个函数的最值或范围.21.对于有穷数列()12,,,3m a a a m ≥ ,若存在等差数列{}n b ,使得11221m m m b a b a b a b +≤<≤<<≤< ,则称数列{}n a 是一个长为m 的“弱等差数列”.(1)证明:数列124,,是“弱等差数列”;(2)设函数()sin f x x x =,()f x 在()0,2024内的全部极值点按从小到大的顺序排列为12,,,m a a a ,证明:12,,,m a a a 是“弱等差数列”;(3)证明:存在长为2024的“弱等差数列”{}n a ,且{}n a 是等比数列.【答案】(1)证明见解析(2)证明见解析(3)证明见解析【解析】【分析】(1)找到一个符合条件的数列{}n b 即可证明;(2)令()0f x '=得到极值点符合的等式关系,即为y x =-和tan y x =图象交点的横坐标,再结合二者图象的特点找到交点的位置,确定数列{}n b 即可证明;(3)先构造一个等比数列{}n a ,其通项公式为()()1202411,2,,2024n n n a k k n --=+= ,证明存在一个正整数k ,使其为长为2024的“弱等差数列”即可.【小问1详解】存在数列21125,,3,366是等差数列,且211251234366<<<<<<,所以数列124,,是“弱等差数列”.【小问2详解】()sin cos f x x x x +'=,令()0f x '=得tan x x -=,所以极值点即为y x =-和tan y x =图象交点的横坐标,由y x =-和tan y x =在()0,∞+内的图象可知,在每个周期都有一个交点,所以令12n n b -=π,则1n n n b a b +<<,所以12,,,m a a a 是“弱等差数列”.【小问3详解】构造正整数等比数列{}n a ,()()1202411,2,,2024n n n a k k n --=+= ,其中k 是待定正整数,下面证明:存在正整数k ,使得等比数列{}n a 是长为2024的“弱等差数列”.取20242024202320231,1,b a b a =-=-若存在这样的正整数k 使得()()()()2202220232023202220211232023202420251111b k b k k b k k b k k b k b ≤<≤+<≤+<<≤+<≤+< 成立,所以()()()2023202220222024202320242023111d b b a a k k k k =-=-=+-+=+,由()()1202411,2,,2024n n n a k k n --=+= ,得()()()()112022202412024202311111n n n n n n n n a a k k k k k k k d ------+-=+-+=+<+=,于是()()()202420232024120242024n n n n a a a a a a b n d b -=+-++->--= ()12023n ≤≤,又因为2024202420241b a a =-<,所以当1,2,,2024n = 时,<n n b a ,而()()()1211211n n n n a a a a a a b n d b -+=+-++-<+-= ,所以1122202420242025b a b a b a b ≤<≤<<≤< ,最后说明存在正整数k 使得12a b <,由()()()()()202320222022122024202211211320241m b b d k m k k k k -=-=+---+=++-->,上式对于充分大的k 成立,即总存在满足条件的正整数k .所以,存在长为2024的“弱等差数列”{}n a ,且{}n a 是等比数列.【点睛】思路点睛:新定义题目解题策略:(1)依据新定义取特殊值证明其成立;(2)如果有多个条件,先假设符合其中一个条件,再证明其余的条件也符合.。

江苏省盐城市2023-2024学年八年级下学期期末数学试卷(含答案详解)

江苏省盐城市2023-2024学年八年级下学期期末数学试卷(含答案详解)

2023-2024学年度第二学期期终考试八年级数学试题注意事项:1、本试卷考试时间为100分钟,试卷满分120分,考试形式闭卷。

2、本试卷中所有试题必须作答在答题纸上规定的位置,否则不给分。

3、答題前,务必将自己的学校、班组、姓名、准考证号填写在答题纸上相应位置。

一、选择题(本大题共有8小题,每小题3分,共24分.在每小题所给出的四个选项有一项是符合题目要求的,请将正确选项的字母代号填写在答题纸上相应位置)1.以下调查中,适宜普查的是( )A .了解全班同学每周体育锻炼的时间B .了解夏季冷饮市场上冰淇淋的质量C .了解串场河中鱼的种类D .了解一批洗衣机的使用寿命2.反比例函数的图像一定经过的点( )A .(-3,2)B.(2,3)C .(-2,3)D .(2,-3)3.下列二次根式中,属于最简二次根式的是( )A BC D 4.菱形具有矩形不一定具有的性质是( )A .对边相等B .对边平行C .对角线互相平分D .对角线互相垂直5.若分式中x 、y 的值都变为原来的3倍,则分式的值( )A .不变B .是原来的3倍C .是原来的D .是原来的6.估计 )A .2和3B .3和4C .4和5D .5和67.顺次连接四边形四边中点所得的四边形一定是( )A .平行四边形B .矩形C .菱形D .正方形8.照相机成像时,照相机镜头的焦距f ,物体到镜头的距离u ,胶片(像)到镜头的距离满足.已知f 、v .则( )A .B .C .D .6y x =33x x y -1319()111v f f u v=+≠u =fvf v -f vfv -fvv f -v ffv-二、填空题(本大题共有8小题,每小题3分,共24分,不需写出解答过程,请将答案直接写在答题纸上相应位置)9.若有意义,则x 的取值范围是___________.10___________.11___________.12.抛掷一枚质地均匀的正方体骰子一次,下列3个事件:①向上一面的点数是奇数;②向上一面的点数是3的倍数:③向上一面的点数不小于3.其中发生的可能性最小的事件是___________.(填序号)13.在平面直角坐标系中,若点,在反比例函数的图像上,则___________.(填 “”“”或“”).14.如图,菱形的面积为24,若,则___________.15.已知,且,则的值为___________.16.如图,在矩形纸片中,,,E 是边上一点,先将沿折叠,点B 落在点处,与交于点F ;再折叠矩形纸片,使得点C 与点重合,点D 落在点处,折痕为.则___________.三、解答题(本大题共有9小题,共72分,请在答题纸指定区域内作答,解答时应写出文字说明、推理过程或演算步骤)1722x -=()11,A y ()22,B y ()0k y k x=<1y 2y >=<ABCD 8AC =BD =111x y -=2x y ≠2xy x x y--ABCD 4AB =16BC =BC ABE AE B 'EB 'AD ABCD B 'D ¢EG FG =18.解分式程:.19.先化简,再求值,其中.20.密闭容器内有一定质量的二氧化碳,当容器的体积V (单位:)变化时,气体的密度(单位:)随之变化.已知密度与体积V 是反比例函数关系,它的图象如图所示,当时,.(1)求密度ρ关于体积V 的函数表达式;(2)当时,求二氧化碳密度ρ的值.21.为了解某初中校学生最喜爱的球类运动项目,给学校提出更合理的配置体育运动器材和场地的建议.兴趣小组随机抽取部分学生进行问卷调查,被调查学生须从“篮球、乒乓球、足球、排球、羽毛球”中选择自己最喜爱的一个球类运动项目,根据调查结果绘制了如下所示的不完整的统计图.根据统计图信息,解答下列问题:(1)在扇形统计图中,“乒乓球”所在扇形的圆心角为________.(2)将条形统计图补充完整;(3)估计该校800名初中生中最喜爱篮球项目的人数;23122x x x--=--2121121a a a a a +⎛⎫+÷ ⎪--+⎝⎭1a +3m ρ3kg/m ρ32.5m V =34kg /m ρ=35m V =(4)根据调查结果,请你向学校提一条合理建议.22.观察下列等式:,…解答下列问题:(1)根据上面3个等式的规律,写出第⑤个等式:_______;(2)用含n (n 为正整数)的等式表示上面各个等式的规律,并加以证明.23.四边形是平行四边形,E 、F 分别是、上的点,连接.(1)如图1,对角线、相交于点O ,若经过点O ,求证:.(2)在如图2中,仅用无刻度的直尺作线段,使它满足:①点M 、N 分别在、上;②.(不写画法,保留画图痕迹)24.定义图形如图1,在四边形中,M 、N 分别是边、的中点,连接.若两侧的图形面积相等,则称为四边形的“对中平分线”===ABCD AD BC EF AC BD EF OE OF =MN AD BC MN EF =ABCD AD BC MN MN MN ABCD提出问题有对中平分线的四边形具有怎样的性质呢?分析问题(1)如图2,为四边形的“对中平分线”,连接,,由M 为的中点,知与的面积相等,则,有怎样的位置关系呢?请说明理由.(2)在(1)的基础上,小明提出了下列三个命题,其中假命题的是_____(请把你认为假命题的序号都填上)①若,则四边形是平行四边形;②若,则四边形是菱形;③若,则四边形是矩形.深入探究如图3,四边形有两条对中平分线,分别是,,且相交于点O ,若.请探索四边形的形状并说明理由.25.如图,直线轴于点H ,且与反比例函数及反比例函数与的图像分别交于点A 、B .(1)若,,连接、.①的面积为_______;②当时,求点B 的坐标.(2)若点,过点A 作x 轴的平行线,与一次函数的图像交于点D ,点D 在直线l 的左侧,若和变化时,的值始终不变,求对应k 的值.MN ABCD AN DN AD AMN DMN AD BC MN AB ABCD MN AB =ABCD MN BC ⊥ABCD ABCD MN EF MN EF =ABCD l x ⊥()110,0k y k x x =>>2k y x=()200k x ,18k =22k =-OA OB ABO OA OB ⊥()20H ,()2102y kx k k =+≠1k 2k +AB AD参考答案1.解:A 、了解全班同学每周体育锻炼的时间,适合普查,故本选项符合题意;B 、了解夏季冷饮市场上冰淇淋的质量,适合抽样调查,故本选项不符合题意;C 、了解串场河中鱼的种类,适合抽样调查,故本选项不符合题意;D 、了解一批洗衣机的使用寿命,适合抽样调查,故本选项不符合题意;故选:A .2解:反比例函数中,A 、∵,∴此点不在函数图象上,故本选项不符合题意;B 、∵,此点在函数图象上,故本选项符合题意;C 、∵,∴此点不在函数图象上,故本选项不合题意;D 、∵,∴此点不在函数图象上,故本选项不符合题意.故选:B .3,选项A 、B、C 都不是最简二次根式,故选:D .4.解:菱形的性质有:对边平行且相等;对角相等,邻角互补;对角线互相垂直平分;矩形的性质有:对边平行且相等;四个角都是直角;对角线互相平分;根据菱形和矩形的性质得出:菱形具有而矩形不一定具有的性质是对角线互相垂直;故选:D .5.解:∵分式中的、的值都变为原来的倍.∴,∴此分式的值不变.故选:A .6又∵,,∴,∴4和5两个整数之间,6y x=6k =()3266-⨯=-≠236⨯=2366-⨯=-≠()2366⨯-=-≠===33x x y-x y 3()()333333333x y x x x x y y x x x y --=--===162025<<<<45<<故选:C .7.解:如图,∵为中点,为中点,∴,,同理,∴,∴四边形是平行四边形.故选:A .8.解:∵,∴,∴,故选:C .9.解:由题意得:,解得:,故答案为:.10.1112.解:①“向上一面的点数是奇数”的概率为,②“向上一面的点数是3的倍数”的概率为,③“向上一面的点数不小于”的概率为,,故其中发生的可能性最小的事件是②,故答案为:②.E ADF AB 12EF BD =EF BD ∥GH BD GH BD =,∥EF GH EF GH =∥,EFGH ()111v f f u v =+≠111v f u v fvf -=-=fv u v f =-20x -≠2x ≠2x ≠==1213323231123>>13.解:∵,∴反比例函数的图象在二、四象限,∵,∴点,在第四象限,y 随x 的增大而增大,∴.故答案为:.14.解:∵四边形是菱形,面积为24,且,∴.故答案为:6.15.解:∵,∴,∴,故答案为:.16.解:∵四边形为矩形,∴,,,,根据折叠可知:,,,,,,∴,∵,∴,∵,∴,∴,∴,设,则,在中,根据勾股定理得:,即,解得:,∴,0k <()0k y k x=<210>>()11,A y ()22,B y 21y y ><ABCD 8AC =2426BD AC ⨯==111x y-=xy y x =-21222xy x y x x y x x y x y x y----===----1-ABCD 4AB DC ==16AD BC ==90B C D ∠=∠=∠=︒AD BC ∥BE B E '=CE B E '=4AB AB '==AEB AEB '∠=∠90AB F B '∠=∠=︒CEG B EG '∠=∠BE CE =16BE CE BC +==8BE CE B E '===AD BC ∥AEB EAF ∠=∠AEB EAF '∠=∠AF EF =EF AF x ==8B F x '=-Rt AB F '△222AF B F AB ''=+()22248x x =+-5x =5EF =∵,∴,∴,∴.故答案为:5.17.18.解:,去分母得:,整理得:,此方程无解,∴原方程无解.19.解:,把代入得:原式.20.(1)解:∵密度与体积V 是反比例函数关系,∴设,∵当时,.∴,∴,∴密度关于体积V 的函数解析式为:;(2)解:把代入得:,AD BC ∥AGE CEG ∠=∠AGE GEF ∠=∠5FG EF ==5=-5=23122x x x--=--232x x +-=-12x x -=-2121121a a a a a +⎛⎫+÷ ⎪--+⎝⎭()2112111a a a a a a +-⎛⎫+÷ =⎪--⎝⎭-()21212a a a a -=⋅-1a =-1a =11=-=ρ()0,0k V k Vρ=>≠32.5m V =34kg /m ρ=4 2.5k =2.5410k =⨯=ρ()100V Vρ=>5V =()100V V ρ=>1025ρ==当时,求二氧化碳密度ρ的值为.21.(1)解:在扇形统计图中,“乒乓球”所在扇形的圆心角为:.(2)解:被抽查的总人数为:(名),∴被抽查的100人中最喜爱羽毛球的人数为:(名),被抽查的100人中最喜爱篮球的人数为:(名),补全图形如图所示:(3)解:(名),答:估计该校800名初中生中最喜爱篮球项目的人数为320名.(4)解:因为喜欢篮球的学生较多,建议学校多配置篮球器材、增加篮球场地等.(答案不唯一)22.(1(2)解:第1个等式中分母为,第2个等式中分母为,第3个等式中分母为,第4个等式中分母为,35m V =32kg /m 36030%108︒÷=︒3030%100÷=1005%5⨯=∴100301015540----=40800320100⨯==1=======2211=+2521=+21031=+21741=+得第个等式中分母为应为:∴第∵左边右边∴左边右边.23.(1)证明:∵四边形为平行四边形,∴,,∴,,∴,∴.(2)解:如图,即为所求作的线段;∵四边形为平行四边形,∴,,∴,,∴,∴,同理可得:,∴,∴,即,∵,∴四边形为平行四边形,∴.24.解:分析问题:(1);理由如下:过点A 作于点E ,过点D 作于点F ,如图所示:n 21n +n ======ABCD OA OC =AD BC ∥AEO CFO ∠=∠EAO FCO ∠=∠AOE COF △≌△OE OF =MN ABCD OA OC =AD BC ∥AMO CFO ∠=∠MAO FCO ∠=∠AOM COF ≌AM CF =AOE CON ≌△△AE CN =AM AE CF CN -=-ME FN =ME FN ∥MNFE MN EF =AD BC ∥AE BC ⊥DF BC ⊥∵,,∴,∵为四边形的“对中平分线”,∴,∵M 是的中点,∴,∴,∴,∴,∵N 是的中点,∴,∴,∴四边形为平行四边形,∴,即;(2)①∵,∴,∵,∴四边形为平行四边形,∴,∵M 、N 分别是边、的中点,∴,,∴,∵,AE BC ⊥DF BC ⊥AE DF ∥MN ABCD ABNM CDMN S S =四边形四边形AD AMN DMN S S = AMN DMN ABNM CDMN S S S S -=- 四边形四边形ABN DCN S S =V V 1122BN AE CN DF ⨯=⨯BC BN CN =AE DF =AEFD AD EF ∥AD BC ∥AD BC ∥AM BN ∥MN AB ABNM AM BN =AD BC 12AM AD =12BN BC =AD BC =AD BC ∥∴四边形为平行四边形,故①是真命题;②当四边形为平行四边形时,,,∵M 、N 分别是边、的中点,∴,,∴,∵,∴四边形为平行四边形,∴,∴当四边形为平行四边形,而不是菱形时,,故②是假命题;③当四边形为等腰梯形时,延长、交于点E ,如图所示:∵四边形为等腰梯形,∴,∴,∵点N 为的中点,∴,∴,∵,∴,∴,∵,,∴,ABCD ABCD AD BC ∥AD BC =AD BC 12AM AD =12BN BC =AM BN =AM BN ∥ABNM AB MN =ABCD AB MN =ABCD BA CD ABCD B C ∠=∠EB EC =BC EN BC ⊥90BNE ∠=︒AD BC ∥90AME BNE ∠=∠=︒EM AD ⊥EB EC =EA ED =EB AB EC CD -=-即,∴,∴四边形为等腰梯形,,∴时,四边形不一定是矩形,故③是假命题;综上分析可知:真命题为①.(3)四边形为菱形;理由如下:∵四边形有两条对中平分线,分别是,,∴根据解析(1)可得:,,∴四边形为平行四边形,∴,∵M 、N 分别是边、的中点,∴,,∴,∵,∴四边形为平行四边形,∴,同理可得:四边形为平行四边形,∴,∵,∴,∴四边形为菱形.25.(1)解:①∵,,直线轴于点H ,∴,,∴;EA ED =AM DM =ABCD MN BC ⊥MN BC ⊥ABCD ABCD ABCD MN EF AD BC ∥AB CD ∥ABCD AD BC =AD BC 12AM AD =12BN BC =AM BN =AM BN ∥ABNM AB MN =EBCF EF BC =MN EF =AB BC =ABCD 18k =22k =-l x ⊥1118422AOH S k ==⨯= 2112122OBH S k ==⨯-= 415AOB AOH OBH S S S =+=+=②设,则,,,,∵,∴为直角三角形,∴,∴,解得:,负值舍去,∴点B 的坐标为;(2)解:∵点,∴,,∴,∵过点A 作x 轴的平行线,与一次函数的图像交于点D ,∴把代入得:,解得:,∴,∴,∴,∵和变化时,的值始终不变,∴为定值,∴为定值,∴,∴.()2,0B m m m -⎛⎫> ⎪⎝⎭8A m m ⎛⎫ ⎪⎝⎭,2224OB m m =+22264OA m m =+22282100AB m m m ⎛⎫=+= ⎪⎝⎭OA OB ⊥AOB 222AB OA OB =+22222100644m m m m m =+++2m =()2,1-()20H ,12,2k A ⎛⎫ ⎪⎝⎭22,2k B ⎛⎫ ⎪⎝⎭122k k AB -=()2102y kx k k =+≠12k y =()2102y kx k k =+≠()121022k kx k k =+≠122k k x k-=121,22k k k D k -⎛⎫ ⎪⎝⎭1222k k AD k -=-1212222k k k AB AD k k ---+=+1k 2k +AB AD 1212222k k k k k ---+()()()()121212121212222222k k k k k k k k k k k k k k k -------+-=+=+10k -=1k =。

地理:宁波市2023-2024学年高二第二学期期末试卷及答案

地理:宁波市2023-2024学年高二第二学期期末试卷及答案

宁波市2023学年第二学期期末考试高二地理试卷考生须知:1.本试题卷分选择题和非选择题两部分,共8页,满分100分,考试时间90分钟。

2.答题前,在答题卷指定区域填写班级、姓名、考场号、座位号及准考证号。

3.所有答案必须写在答题卷上,写在试卷上无效。

选择题部分一、选择题I(本大题共25小题,每小题2分,共50分。

每小题列出的四个备选项中只有一个是符合题目要求的,不选、多选、错选均不得分)“未来工厂”是以数字设计、智能生产、绿色制造为基础,以核心竞争力提升为目标的现代化工厂。

宁波某石化企业作为第一批未来工厂试点企业力图通过未来工厂建设,构建新智造模式。

完成1、2题。

1.该石化企业布局在宁波的主导区位因素是()A.科技B.市场C.原料D.劳动力2.该石化企业经改造提升后()A.原料来源广泛化B.环境污染分散化C.生产流程机械化D.产品研发高效化增殖放流是以人工方式向海洋、湖泊等水域投放鱼、虾等幼苗的活动。

浙江省自2016年至今已累计增殖放流各类幼苗390亿尾。

下表为浙江省三个年份海产品产量统计。

完成第3题。

第3题表年份天然生产(万吨)人工养殖(万吨)2020年256.86137.232018年287.39120.92016年331.4597.193.浙江省海产品产量及其变化是()A.2020年总产量最高B.2018年天然生产产量占比最高C.总产量持续下降D.人工养殖增幅变大秦岭主峰太白山位于陕西省中南部,海拔3771米,垂直自然带谱复杂。

完成4、5题。

4.秦岭北坡从山麓至山顶的自然带为()A.常绿阔叶林带、落叶阔叶林带、山地针叶林带、高山草甸带B.常绿阔叶林带、常绿硬叶林带、针阔混交林带、高山草甸带C.落叶阔叶林带、山地针叶林带、高山灌丛带、高山草甸带D.落叶阔叶林带、针阔混交林带、高山苔原带、高山草甸带5.太白山山顶的植被特征通常()A.多见茎花B.生命周期短C.多革质叶片D.茎秆粗壮目前的垃圾处理方式为填埋和焚烧发电,下图为某垃圾焚烧发电厂生产工艺流程示意图,但我国推广垃圾焚烧发电存在诸多困难。

2024年上海青浦区初三二模语文试卷和答案

2024年上海青浦区初三二模语文试卷和答案

2024年上海青浦区九年级第二学期学业质量调研语文试卷(时间100分钟,满分150分)2024.04考生注意:本卷共有22题,请将所有答案写在答题纸上。

写在试卷上一律不计分。

一、文言文(34分)(一)默写与运用(13分)1.纤纤擢素手,__________________。

(《古诗十九首》)2.____________________,洪波涌起。

(曹操《观沧海》)3.马作的卢飞快,________________。

(辛弃疾《破阵子·为陈同甫赋壮词以寄之》)4.青西郊野公园的水杉茂密成荫,鸟儿嘤嘤鸣唱,不由让人联想起欧阳修《醉翁亭记》中的句子:“______________,______________”。

(二)阅读下面选文,完成5—10题(21分)【甲】山不在高,有仙则名。

水不在深,有龙则灵。

斯是陋室,惟吾德馨。

苔痕上阶绿,草色入帘青。

谈笑有鸿儒..,往来无白丁。

可以调素琴,阅金经。

无丝竹之乱耳,无案牍之劳形。

南阳诸葛庐,西蜀子云亭。

孔子云:何陋之有?【乙】唐荆川①性俭素。

冬不炉,夏不扇,岁衣一布,月食一肉。

结庐陈渡②,不蔽风雨。

时往来乡郭,乘小舟,低头盘膝,见者不知为贵人。

即遭凌侮,不较也。

冬则加草以为温,有老友见之泪下,为市.一床,而终身无厚茵褥③。

门生子弟,从公游处,不堪其苦,而公独安之,曰:“不如是,何以袚除欲根?”【注释】①唐荆川:唐顺之,号荆川。

明朝大臣。

②陈渡:地名。

③茵褥:床垫。

5.【甲】文的作者是_______(人名)。

(2分)6.解释下列句中的加点词。

(4分)(1)谈笑有鸿儒..()(2)为市.一床()7.用现代汉语翻译下面的句子。

(3分)可以调素琴,阅金经。

8.【甲】【乙】两文都描写了住所的环境。

【甲】文“_____________,_____________”写出其住所___________的特点;【乙】文“_____________”则凸显出唐荆川住所简陋的特点。

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2013—2014年度高三第二学期十三周周考试题英语I 语言知识及应用(共两节,满分45分)第一节完形填空(共15小题;每小题2分,满分30分)阅读下面短文,掌握其大意,然后从1~15各题所给的A、B、C和D项中,选出最佳选项,并在答题卡上将该项涂黑。

Advertising is about creating images, and this is especially true when advertising food and drinks. What the 1 looks like is more important than what it tastes like. If companies hope to sell food successfully, the food must 2 appetizing.Television advertising of food often uses 3 . Apparently, food looks especially appetizing if it moves. Chocolate sauce looks much more 4 when you see it being poured over ice cream than it does just sitting in a jar.5 effects also help to sell food: sausages sizzling(咝咝响) in a frying pan are mouth-watering.A TV 6 for a brand of coffee had the sound of coffee percolating(过滤) in the background. The commercial was so 7 that it lasted five years.The 8 of food and its packaging is also very important. If the color looks 9 , people won’t eat it. Nobody would normally eat blue bread or drink blue beer. Other 10 food colors are purple, gray, and in some cases, white.How people expect something to taste often influences how it 11 does taste. Researchers gave some mineral water to two groups of people. They told one group that the water was mineral water, and almost all people said, “It tastes 12 .”Then the researchers told the other group that the water was tap water. This group said the water tasted a little funny. The word tap created a(n)13 image of chlorine(氯).It is the same with 14 . A food manufacturer gave a group of people the same 15 in a glass jar and in a can and asked them to taste it. They all claimed that the product in the glass jar tasted better.So it seems to be true. Image is everything.1. A. company B. food C. image D. milk2. A. look B. taste C. sound D. smell3. A. sound B. sense C. movement D. imagination4. A. beautiful B. natural C. clean D. delicious5. A. Action B. Television C. Music D. Sound6. A. advertisement B. program C. show D. speech7. A. noisy B. expensive C. successful D. long8. A. price B. name C. brand D. color9. A. right B. great C. interesting D. wrong10. A. unknown B. unpopular C. practical D. famous11. A. actually B. especially C. elegantly D. seriously12. A. strange B. funny C. good D. salty13. A. perfect B. unusual C. unpleasant D. unbelievable 14. A. advertising B. packaging C. producing D. tasting15. A. product B. water C. meat D. coffee第二节语法填空(共10小题; 每小题1.5分,满分15分)阅读下面短文,按照句子结构的语法性和上下文连贯的要求,在空格处填入一个适当的词或使用括号中词语的正确形式填空,并将答案填写在答题卷标号为16~25的相应位置上。

How can we know that the birds we see in the South in the winter are the same ones that come north in the spring? Once John J. Audubon, a bird 16 (love), wondered about this. Every year he 17 (watch) a pair of little phoebes nesting in the same place. He decided to put tiny silver bands(箍) on 18 legs. The next spring, the birds 19 the bands came back in the very same place. The phoebe, it was learned, spent winter 20 it was warm enough to find food. Today there are hundreds of birdbanders all over America.The government of the U.S. has a special birdbanding department 21 makes all the birdbands. The bands do not hurt the birds, as they are made 22 aluminium and are very light. Each band has23 special number. On each band are these words: “Inform Fish and Wildlife Service, Washington,D.C”.Anyone who finds a dead bird with a band on its legs 24 (ask) to send the band to Washington with a note 25 (tell) where the bird was found. In this way naturalists add to their knowledge of the habits and needs of birds.II 阅读(共两节,满分50分)第一节阅读理解(共20小题;每小题2分,满分40分)阅读下列短文,从每题所给的A、B、C和D项中,选出最佳选项,并在答题卡上将该项涂黑。

ANot long ago the movie 2012 came into screen. The people were threatened by those scenes of destroying flood, severe earthquake, terrifying hurricane and constructions representing human civilization being destroyed and even swallowed by disaster.Luckily, they are just the director' s imagination, but the present situation is not heading a positive direction, either.Take my own experience in Alaska as an example. Once I took a trip to the glacier. Along the way there stood signposts marking the snow lines of different years. They started from the foot of the mountain, but it was at the top when I finally saw melting glaciers(融化的冰川). My heart ached seeing the beautiful blue ice melting at every second.Sad but true, they are the effects of global warming and the result of our human impact. Furthermore, each year the rising sea level will kill 56 million people, and that' s about the population of the entire Italy. According to studies, if the temperature keeps on rising like this, by the year 2050, some islands and coastal cities including New York, Shanghai, Tokyo and Sydney will be drowned in water.Our fortune is in our own hands. It depends on us to shape our future, to reduce future human impact and find ways to form a peaceful relationship with our environment. Therefore, it's time for actions to be taken right now. Contribute a little to energy saving by using more efficient light bulbs and less hot water. Let recycle become our habit by thinking twice before throwing something away. Let us take public transportation as our first choice when going to a certain place. It might cost more time for now, but it' sto the benefit of a permanent future. Take care of every tree and grass around us by watering them or simply just avoid destroying them.In a word, small drops of water make a big ocean. The earth does not belong to us. On the contrary, we belong to the earth. Please bear in mind that the earth is our home. It is our responsibility to build a brighter and better future of our planet and prevent what happened in the movie 2012 from becoming reality.26.Why does the author talk about the movie 2012 in the passage ?A.To give example B.To lead into the topicC.To make prediction D.To provide the evidence27.How did the author feel when he took a trip to the glacier?A.Worried B.Puzzled C.Scared D.Bored28.It can be inferred from the third paragraph that .A.56 million people in Italy have been killed owing to the rising sea levelB.Some islands and coastal cities will be drowned in water by the year of 2050C.Human being will be in danger if we don't take actions to prevent the global warmingD.It is certain that what happened in the movie 2012 will come into reality29.According to the passage, you are advised to .A.drive our own private cars instead of taking buses to some placeB.recycle everything that is used C.go to see the movie 2012 at onceD.work together to take good care of our planet30.What does the author mean by saying "small drops of water make a big ocean"?A.Think twice before taking action, B.It's our duty to protect the ocean.C.Everyone together can make a difference.D.It's important to save every drop of water.BGroup buying is one of the fastest growing trends in South Africa today. Industry leaders are confident the growth potential remains strong since group buying is location-specific(区域性的). Start-up costs are low and profit room remains high, so many sites continue to receive invested money despite widespread cr iticism and Facebook’s decision to phase out of the deals business due to privacy concerns.In the early stages of all industries, some companies fail because they cannot compete with stronger companies in difficult economic conditions. To deal with difficult conditions, an alarming number of businesses are developing group buying websites in places like China and India, so the increase of group buying in South Africa is nothing more than a natural progression into the international mainstream.The group buying concept is fairly new and consumers have accepted this concept because they can now make full use of the rich information available on the Internet. Group buying is convenient and easy so it works. Anyone can view a site, join a mailing list, subscribe to RSS or print out a coupon(优惠券). The current group buying structure offered by the industry leaders works although there are still challenges to overcome.Perhaps, the future of group buying is tied to the joining together of social media and mobile devices. Mobile devices are with us wherever we go and almost everyone is using some type of social media site like Facebook or Twitter to stay informed. Using GPS and social media technology to provide real time location-specific promotions would be beneficial to every consumer looking for the best deals in town.Pause for a moment and think about it! What is better than signing on to your phone while having fun in town and you receive a real time information that your favorite shop across the street is offering a killer deal?The future of the group buying in South Africa is bright and we can expect to see more advanced approaches to this concept in the future. In addition to the technological advances consumers will see the range of promotions expand to include new products and services.31. What does the underlined phrase “phase out of” mean in the passage?A. Gradually stop.B. Gradually increase.C. Begin to develop.D. Continue to enlarge.32. The author sets China and India as examples to show that _________.A. China and India are powerful countriesB. China and India are in difficult economic conditionsC. group buying is successful worldwideD. group buying is an international trend33. Which of the following is true according to the passage?A. People have accepted group buying because it’s a new concept.B. Social media and mobile devices have been joined together for group buying.C. GPS and social media technology will be helpful in group buying.D. Shops usually offer a killer deal when their customers are having fun.34. The author’s attitude towards the future of group buying is __________.A. pessimisticB. optimisticC. objectiveD. subjective35. What does the passage mainly tell us?A. The history of group buying.B. Group buying in South Africa.C. Growing trends in South Africa.D. The group buying concept.CIt was Thanksgiving morning and in the crowded kitchen of my small home I was busy preparing the traditional Thanksgiving turkey when the doorbell rang. I opened the front door and saw two small children in rags huddling together inside the storm door on the top step.“Any old papers, lady?” asked one of them.I was busy. I wanted to say “no”until I looked down at their feet. They were wearing thin little sandals, wet with heavy snow.“Come in and I’ll make you a cup of hot cocoa.”They walked over and sat down at the table. Their wet sandals left marks upon the floor. I served them cocoa and bread with jam to fight against the cold outside. Then I went back to the kitchen and started again on my household budget.The silence in the front room struck me. I looked in. The girl held the empty cup in her hands, looking at it. The boy asked in a flat voice, “Lady, are you rich?”I looked at my shabby slipcovers. The girl put her cup back in its saucer carefully and said, “Your cups match your saucers.” Her voice was hungry with a need that no amount of food could supply. They left after that, holding their bundles of papers against the wind. They hadn’t said “Thank you.”They didn’t need to. They had reminded me that I had so much for which to be grateful. Plain blue china cups and saucers were only worth five pence. But they matched.I tasted the potatoes and stirred the meat soup. Potatoes and brown meat soup, a roof over our heads, my man with a good steady job—these matched, too.I moved the chairs back from the fire and cleaned the living room. The muddy prints of small sandals were still wet upon my floor. Let them be for a while, I thought, just in case I should begin to forget how rich I am.36. Two children came to the writer’s front door because _________________.A. it was Thanksgiving DayB. they were beggarsC. they wanted old papersD. they wanted a cup of cocoa37. Why did the writer let the children in?A. She showed great pity on themB. She had old papers to sellC. She wanted to invite them to her Thanksgiving feastD. She wanted them to see how rich she was38. The girl thought the writer was rich perhaps because ________________.A. she saw that the lady’s room was comfortableB. she saw the cups matched the saucersC. the writer’s slipcovers were very newD. the writer was preparing a big meal while she was too hungry.39. From the passage, we can infer that whether you are rich depends on ____________.A. how much money you have hadB. how you feel about your lifeC. how you have helped othersD. what job your husband is doing40. The writer left the muddy prints of small sandals on the floor for a while to ___________.A. show her husband that someone had comeB. remind her that she had helped two childrenC. remind her that she was very rich in the neighborhoodD. remind her how life should beDStage fright or performance anxiety is the anxiety, or fear which may occur in an individual by the requirement to perform in front of an audience. It is most commonly seen in school situations, like stand-up projects and class speeches. It has numerous forms: heart beating fast, trembling hands and legs, sweaty hands, dry mouth etc.In fact, most of the fear occurs before you step on stage. Once you’re up there, it usually goes away. Thus, it is a phenomenon that you must learn to control. Try to think of stage fright in a positive way. It heightens your energy, adds color to your cheeks. With these good side effects you will actually look healthier and more physically attractive.Many of the top performers in the world get stage fright so you are in good company. Stage fright may come and go or decrease, but it usually does not disappear permanently. You must concentrate on getting the feeling out and present what you have prepared calmly.Remember “Nobody” ever died from stage fright. But, according to surveys, many people would rather die than give a speech. If that applies to you, and you are an unlucky guy who is with stage fright the whole time, try out some of the strategies(策略) as follows to help get yourself under control. Realize that you may never overcome stage fright, but you can learn to control it, and use it to your advantage.Strategies are as follows when the program begins:1) If legs are trembling, lean on table or shift legs or move.2) Don’t hold notes. The audience can see them shake. Use three-by-five cards instead.3) Use eye contact. Look at the friendliest faces in the audience.Remember nervousness doesn’t show one-tenth as much as it feels. Before each presentation, make a short list of the items you think will make you feel better. Don’t be afraid to experiment with different combinations. You never know which ones will work best until you try. Use these steps to control stage fright so it doesn’t control you. Once you are used to stage fright, you will find you on the road to a great speech-maker.41. Someone may be most likely to suffer from stage fright when he/she is ______.A. attending an English classB. standing in a classroomC. watching a performanceD. talking in front of people42. By thinking of stage fright in a positive way, one could ______.A. learn to control stage frightB. get rid of stage frightC. calm down before stepping on stageD. become more physically attractive43. Which of the following is true?A. Top performers usually suffer from stage fright.B. Stage fright may stay with a person for a life time.C. Nobody would rather die than give a speech.D. No one can overcome or control stage fright.44. The author advises people with stage fright to ______.A. show one-tenth of their nervousnessB. experiment with different kinds of stage frightC. refer to the strategies whenever they feel the needD. use one of the strategies each time45. The passage mainly talks about ______.A. how to deal with stage frightB. what stage fright is likeC. when stage fright occursD. why people have stage fright第二节信息匹配(共5小题;每小题2分,满分10分)阅读下列应用文及相关信息,并按照要求匹配信息。

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