高一下学期期中考试
2023~2024学年度第二学期高一年级期中考试数学试卷

唐山市第三十六中学2023-2024学年高一下学期期中考试数学试卷一、选择题1.判断下列各命题的真假:①向量与平行,则与的方向相同或相反;②两个有共同起点而且相等的向量,其终点必相同;③零向量是没有方向的;④向量就是有向线段.其中假命题的个数为( )A .2B .3C .4D .52.如图,分别是长方体的棱的中点,则等于( )A .B .C .D .3.已知,,为非零平面向量,则下列说法正确的是( )A .B .若,则C .若,则,D .4.已知向量,,且,则实数的值为( )A .B .3C .8D .125.已知单位向量,的夹角为,则( )A .1BCD .36.在中,角A ,B ,C 所对边分别为,,,,则值等于( )a b a b E F ,ABCD A B C D '-'''AB CD ,AB CF + AD 'AC ' DE AE a b c()()a b c a b c ⋅⋅=⋅⋅ a c b c ⋅=⋅ a b =//a bλR ∃∈λb a = ||||||a b a b ⋅=⋅ (2,4)a = (,6)b m =- //a bm 3-a b 2π3a b -= ABC V ,,a b c π3A =2b =8c =22a b c sinA sinB sinC -+-+AB .CD7.已知复数在复平面内对应的点在第四象限,则实数的取值范围是( )A .B .C .D .8.在三棱锥P -ABC 中,PA ⊥底面ABC ,PA =2,底面ABC 是边长为的正三角形,M 为AC 的中点,球O 是三棱锥P -ABM 的外接球.若D 是球0上一点,则三棱锥D -PAC 的体积的最大值是( )A.2B .CD二、多项选择题9.在△ABC 中,下列说法正确的是( )A .若,则B .若,则C .若,则D .若,则10.若关于 方程 ( 是实数)有两个不等复数根 ,其中 ( 是虚数单位),下面四个选项正确的有( )A .B.C .D .11.如图,在直三棱柱中,,,E 为的中点,过AE 的截面与棱BB 、分别交于点F 、G ,则下列说法中正确的是( )(2)(1)i z m m =+++m (2,1)--(,2)(1,)⋃-∞--+∞(1,)-+∞(,2)-∞-A B C >>sinA sinB sinC>>A B C >>222sin A sin B sin C>>A B C >>cosA cosB cosC<<A B C >>222cos A cos B cos C<<x 的20x px q ++=p q ,αβ和12α=-+i 1αβ⨯=21αβ=2αβ=332αβ+=111ABC A B C -90ACB ∠=︒12AC BC CC ===11B C 11A CA .当点F 为棱中点时,截面B .线段长度的取值范围是C .当点F 与点B 重合时,三棱锥的体积为D .存在点F ,使得三、填空题12.已知平面和直线,给出条件:①;②;③;④;⑤.(1)当满足条件 时,有;(2)当满足条件 时,有.(填所选条件的序号)13.下列说法正确的序号为 .①若复数,则;②若全集为复数集,则实数集的补集为虚数集;③已知复数,,若,则,均为实数;④复数的虚部是1.14.如图,在四边形 中,对角线 与 相交于点 .已知 ,, ,且 是 的中点,若 ,则 的值为 .四、解答题15.如图,在平面四边形ABCD 中,已知,,△ABC 为等边三角形,记.1BB AFEG 3++1C G []01,C AEF -431A F AE ⊥αβ,m αm P αm ⊥αm ⊂αβ⊥αβP βm P βm ⊥3i z =+13i 1010z =-1z 2z 12z z >1z 2z 3i 1z =-+ABCD AC BD O AC BC =AC BC ⊥AD BD ⊥O AC 2AD AB CD CB ⋅-⋅= AC BD ⋅ 1AD =2CD =αADC ∠=(1)若,求△ABD 的面积;(2)若,求△ABD 的面积的取值范围.16.已知向量.(1)当时,求的值;(2)设函数,且,求 的最大值以及对应的的值.17.已知是关于x 的实系数一元二次方程.(1)若a是方程的一个根,且,求实数k 的值;(2)若,是该方程的两个实根,且,求使的值为整数的所有k 的值.18.如图,多面体 中,底面 是菱形, ,四边形 是正方形且 平面 .(1)求证:平面 ;(2)若 ,求多面体 的体积 .19.如图,两个相同的正四棱锥底面重合组成一个八面体,可放入一个底面为正方形的长方体内,且长方体的正方形底面边长为2,高为4,已知重合的底面与长方体的正方形底面平行,八面体的各顶点均在长方体的表面上.πα3=πα,π2⎛⎫∈⎪⎝⎭)1cos 12a x x b ⎛⎫==- ⎪ ⎪⎝⎭,a b ⊥ tan x ()()f x a b b =+⋅ π02x ⎡⎤∈⎢⎥⎣⎦,()f x x 24410kx kx k -++=1a =1x 2x Z k ∈1221x x x x +ABCDEF ABCD 60BCD ∠=︒BDEF DE ⊥ABCD //CF ADE AE =ABCDEF V(2)求该八面体表面积S的取值范围.。
高一下学期期中考试语文试题及答案

高一下学期期中考试语文试题及答案高一语文(考试用时:150分钟总分:150分)注意事项:所有试题的答案均填写在答题卡上,答案写在试卷上的无效。
一、现代文阅读(29分)(一)现代文阅读I(本题共5小题,19分)阅读下面的文字,完成下面小题。
材料一:汤显祖年轻的时候怀有满腔经天纬地的济世热情。
汤显祖的政治理想,说来也很简单,就是经世致用,造福百姓,使人人冻馁无虞,安居而乐业;使人人乐于向善,民德归于淳厚。
在他看来,士大夫要先正其身,然后才能正人,而政治是否清明,老百姓日子是否好过,最终决定于士大夫和官员的德性和修养。
他曾经代拟过一篇《为士大夫喻东粤守令文》:“清吏之法法身,浊吏之法法人也。
”他所提出的“清吏”与“浊吏”两个概念,并不新鲜,但是,他的“法身”和“法人”的说法,却极为深刻,精辟地揭示了中国自古以来封建统治者的一种普遍德性:他们置身于法律之上,拿自己当法外的“特选之民”,高人一等,飞扬跋扈;他们的道德绳墨,只是用来裁制百姓,他们法律的刀斧,也只是用来宰割人民---他们教别人不要“利己”,自己却很少“利人”,教别人要“利人”,自己却专门“利己”。
在汤显祖看来,吃饭是老百姓的头等大事,而农业生产则是一个官员必须关心的头等大事。
他做官期间,曾多次下到乡里,劝农励耕。
他曾经在诗里记录过自己这方面的活动,在《丙申平昌迎春,晓云如金,有喜》里,他这样写道:“仙县春来仕女前,插花堂上领春鞭。
青郊一出同人笑,黄气三书有大年。
”可见,为了督促和鼓励农民耕地种田,他确实用了一番心思。
汤显祖的政绩,不仅当时就赢得了人民的敬意,为他建了“生祠”,直到清代顺治年间,遂昌知县缪之弼还为他建了“遗爱祠”。
如果说,汤显祖早期的两部剧作的主题,在探讨“至情”,那么,他晚年的写作,则在强化了反讽力度的同时,致力于寻求精神出路-解决自己的精神困境,回答那些与“生活哲学”有关的重大问题。
也就是说,汤显祖后期的“二梦”,是一种缘于精神焦虑的写作。
湖北省宜昌市部分省级示范高中2023-2024学年高一下学期期中联考英语试题含答案

宜昌市部分省级示范高中2024年春季学期期中考试高一年级英语试卷(答案在最后)命题学校:考试时间:120分钟满分:150分第一部分:听力(共两节,满分30分)第一节(共5小题;每小题 1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1.What would the man like to order?A.Some coffeeB.Some apple juice.C.A cup of tea.2.Which program does the man like best?A.SportsB.History.C.News.3.What is the probable relationship between the speakers?A.Salesgirl and customer.B.Wife and husband.C.Passenger and driver.4.Where does the conversation probably take place?A.In a ticket officeB.In a bankC.In a department store5.When did the fire probably break out?A.At about7.B.At about8.C.At about9.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。
听每段对话或独白前,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。
每段对话或独白读两遍。
听第6段材料,回答第6、7题。
6.What does the man’s friend want to drink?A.Orange juice.k with sugar.C.Coffee with milk.7.How much should the man pay?A.10yuan.B.90yuan.C.100yuan.听第7段材料,回答第8、9题。
高一下学期期中考试语文试卷含答案(共3套)

高一下学期期中考试语文试卷含答案(共3套)高一第二学期期中考试语文试题(满分:150分;考试时间:150分钟)(一)论述类文本阅读(本题共3小题,9分)文化软实力,是指一个国家或地区基于文化而具有的凝聚力、生命力、创新力、传播力和影响力。
“文化软实力”的说法源自XXX的软实力理论。
一般来说,软实力是一种隐形的力量,蕴含在文化、政治价值观、外交政策和国际形象四个载体中。
在这四个载体中,文化是核心,其他三个组成部分也都深深地烙上了文化的影子。
甚至有人直接把软实力解释成文化力。
基于此,文化软实力就有了广义和狭义之分,广义的文化软实力就是指“软实力”;狭义的文化软实力,则是构成软实力的文化要素。
文化软实力的形成必须依赖先进的文化,而这种文化只有与时俱进才能更好地服务于相应的时代和社会,才能更好地促进个人全面自由的发展,才能体现出强大的吸引力和感染力。
文化软实力的作用,主要体现在国内和国际两个方面。
在国内,它通过文化建设不断增强本国文化的认同感,抵御国外一些敌对文化理念的侵袭,增强国内民众的凝聚力。
通过吸收国外先进文化元素和不断改造本国文化中落后的成分,使本国文化更加适应当前形势,更好地指导经济建设,更好地彰显本国文化的强劲生命力。
在国际政治舞台上,兼容并蓄、富有活力的本国文化必将为国外受众所认可,使本国所奉行的理念得到传播,从而提升国家形象和影响力。
文化软气力产生于一定的文化资本。
这些资本包括国家价值寻求、社会理念、宗教崇奉、品德规范,还包括风俗惯、民族精神、国民素质、文学艺术等,还与教育、科技、文化财产的开展水平密切相干。
文化软气力产生的根本是人们对本国中心价值体系的认同和接受。
与传统手段相比,非强制手段是文化软气力完成的手段,而国家的综合国力是文化软气力的力量施展阐发形式。
在现实社会中,往往存在重器不重道的现象。
它表现在国家综合实力的建设上,就是重视提升硬实力而不重视提升文化软实力。
重视提升硬实力是对的,文化软实力也一定要以硬实力为基础。
江西省南昌市第二中学2023-2024学年高一下学期5月期中英语试题(解析版)

南昌二中2023-2024学年度下学期高一期中考试英语试卷第一部分:听力理解(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话,每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题,每段对话仅读一遍。
1.What are the speakers talking about?A.A floodB. A fireC. A car accident2.How many tickets does the woman need to buy?A.7B.6C.53.What do we know about the man?A. He just bought a new house.B. He doesn’t get along with his roommates.C. He is a man of bad temper.4.What time is it now?A.9:00B. 9:10C.9:405.What does the man say about the woman?A. She should have been more active.B. She was very talkative.C. She danced very well.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。
听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,每小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听第6段材料,回答第6、7题。
6.Where does the woman come from?A. Australia.B. EnglandC. Brazil.7.What does the woman do ?A. She’s a trader.B. She’s a student.C. She’s a receptionist.听第7段材料,回答第8至9题。
人教版高一下学期期中考试数学试卷及答案解析(共五套)

人教版高一下学期期中考试数学试卷(一)注意事项:本试卷满分150分,考试时间120分钟,试题共22题.答卷前,考生务必用0.5毫米黑色签字笔将自己的姓名、班级等信息填写在试卷规定的位置.一、选择题(本大题共8小题,每小题5分,共40分)在每小题所给出的四个选项中,只有一项是符合题目要求的.1.点C是线段AB靠近点B的三等分点,下列正确的是()A.B.C.D.2.已知复数z满足z(3+i)=3+i2020,其中i为虚数单位,则z的共轭复数的虚部为()A.B.C.D.3.如图,▱ABCD中,∠DAB=60°,AD=2AB=2,延长AB至点E,且AB=BE,则•的值为()A.﹣1 B.﹣3 C.1 D.4.设i是虚数单位,则2i+3i2+4i3+……+2020i2019的值为()A.﹣1010﹣1010i B.﹣1011﹣1010iC.﹣1011﹣1012i D.1011﹣1010i5.如图,在正方体ABCD﹣A1B1C1D1中,异面直线A1B与CD所成的角为()A.30°B.45°C.60°D.135°6.在△ABC中,角A,B,C所对的边分别为a,b,c,若(a﹣2b)cos C=c(2cos B﹣cos A),△ABC的面积为a2sin,则C=()A.B.C.D.7.在正方体ABCD﹣A1B1C1D1中,下列四个结论中错误的是()A.直线B1C与直线AC所成的角为60°B.直线B1C与平面AD1C所成的角为60°C.直线B1C与直线AD1所成的角为90°D.直线B1C与直线AB所成的角为90°8.如图,四边形ABCD为正方形,四边形EFBD为矩形,且平面ABCD与平面EFBD互相垂直.若多面体ABCDEF的体积为,则该多面体外接球表面积的最小值为()A.6πB.8πC.12πD.16π二、多选题(本大题共4小题,每小题5分,选对得分,选错、少选不得分)9.在△ABC中,角A,B,C的对边分别为a,b,c,若a2=b2+bc,则角A可为()A.B.C.D.10.如图,四边形ABCD为直角梯形,∠D=90°,AB∥CD,AB=2CD,M,N分别为AB,CD的中点,则下列结论正确的是()A.B.C.D.11.下列说法正确的有()A.任意两个复数都不能比大小B.若z=a+bi(a∈R,b∈R),则当且仅当a=b=0时,z=0C.若z1,z2∈C,且z12+z22=0,则z1=z2=0D.若复数z满足|z|=1,则|z+2i|的最大值为312.如图,已知ABCD﹣A1B1C1D1为正方体,E,F分别是BC,A1C的中点,则()A.B.C.向量与向量的夹角是60°D.异面直线EF与DD1所成的角为45°三、填空题(本大题共4小题,每小题5分,共20分.不需写出解答过程,请把答案直接填写在横线上)13.已知正方形ABCD的边长为2,点P满足=(+),则||=;•=.14.若虛数z1、z2是实系数一元二次方程x2+px+q=0的两个根,且,则pq=.15.已知平面四边形ABCD中,AB=AD=2,BC=CD=BD=2,将△ABD沿对角线BD折起,使点A到达点A'的位置,当A'C=时,三棱锥A﹣BCD的外接球的体积为.16.已知一圆锥底面圆的直径为3,圆锥的高为,在该圆锥内放置一个棱长为a 的正四面体,并且正四面体在该几何体内可以任意转动,则a的最大值为.四、解答题(本大题共6小题,共70分.请在答题卡指定区域内作答,解答时应写出文字说明、证明过程或演算步骤)17.在四边形ABCD中,AB∥CD,AD=BD=CD=1.(1)若AB=,求BC;(2)若AB=2BC,求cos∠BDC.18.(1)已知z1=1﹣2i,z2=3+4i,求满足=+的复数z.(2)已知z,ω为复数,(1+3i)﹣z为纯虚数,ω=,且|ω|=5.求复数ω.19.如图,墙上有一壁画,最高点A离地面4米,最低点B离地面2米.观察者从距离墙x(x>1)米,离地面高a(1≤a≤2)米的C处观赏该壁画,设观赏视角∠ACB=θ.(1)若a=1.5,问:观察者离墙多远时,视角θ最大?(2)若tanθ=,当a变化时,求x的取值范围.20.如图,已知复平面内平行四边形ABCD中,点A对应的复数为﹣1,对应的复数为2+2i,对应的复数为4﹣4i.(Ⅰ)求D点对应的复数;(Ⅱ)求平行四边形ABCD的面积.21.如图所示,等腰梯形ABFE是由正方形ABCD和两个全等的Rt△FCB和Rt△EDA组成,AB=1,CF=2.现将Rt△FCB沿BC所在的直线折起,点F移至点G,使二面角E﹣BC﹣G的大小为60°.(1)求四棱锥G﹣ABCE的体积;(2)求异面直线AE与BG所成角的大小.22.如图,四边形MABC中,△ABC是等腰直角三角形,AC⊥BC,△MAC是边长为2的正三角形,以AC为折痕,将△MAC向上折叠到△DAC的位置,使点D在平面ABC内的射影在AB上,再将△MAC向下折叠到△EAC的位置,使平面EAC⊥平面ABC,形成几何体DABCE.(1)点F在BC上,若DF∥平面EAC,求点F的位置;(2)求直线AB与平面EBC所成角的余弦值.参考答案一、选择题(本大题共8小题,每小题5分,共40分)在每小题所给出的四个选项中,只有一项是符合题目要求的.1.点C是线段AB靠近点B的三等分点,下列正确的是()A.B.C.D.【答案】D【分析】根据共线向量的定义即可得结论.【解答】解:由题,点C是线段AB靠近点B的三等分点,=3=﹣3,所以选项A错误;=2=﹣2,所以选项B和选项C错误,选项D正确.故选:D.【知识点】平行向量(共线)、向量数乘和线性运算2.已知复数z满足z(3+i)=3+i2020,其中i为虚数单位,则z的共轭复数的虚部为()A.B.C.D.【答案】D【分析】直接利用复数代数形式的乘除运算化简,然后利用共轭复数的概念得答案.【解答】解:∵z(3+i)=3+i2020,i2020=(i2)1010=(﹣1)1010=1,∴z(3+i)=4,∴z=,∴=,∴共轭复数的虚部为,故选:D.【知识点】复数的运算3.如图,▱ABCD中,∠DAB=60°,AD=2AB=2,延长AB至点E,且AB=BE,则•的值为()A.﹣1 B.﹣3 C.1 D.【答案】C【分析】利用图形,求出数量积的向量,然后转化求解即可.【解答】解:由题意,▱ABCD中,∠DAB=60°,AD=2AB=2,延长AB至点E,且AB=BE,可知=+=,=﹣=﹣2,所以•=()•(﹣2)=﹣2﹣2=1.故选:C.【知识点】平面向量数量积的性质及其运算4.设i是虚数单位,则2i+3i2+4i3+……+2020i2019的值为()A.﹣1010﹣1010i B.﹣1011﹣1010iC.﹣1011﹣1012i D.1011﹣1010i【答案】B【分析】利用错位相减法、等比数列的求和公式及其复数的周期性即可得出.【解答】解:设S=2i+3i2+4i3+ (2020i2019)∴iS=2i2+3i3+ (2020i2020)则(1﹣i)S=i+i+i2+i3+……+i2019﹣2020i2020.==i+==﹣2021+i,∴S==.故选:B.【知识点】复数的运算5.如图,在正方体ABCD﹣A1B1C1D1中,异面直线A1B与CD所成的角为()A.30°B.45°C.60°D.135°【答案】B【分析】易知∠ABA1即为所求,再由△ABA1为等腰直角三角形,得解.【解答】解:因为AB∥CD,所以∠ABA1即为异面直线A1B与CD所成的角,因为△ABA1为等腰直角三角形,所以∠ABA1=45°.故选:B.【知识点】异面直线及其所成的角6.在△ABC中,角A,B,C所对的边分别为a,b,c,若(a﹣2b)cos C=c(2cos B﹣cos A),△ABC的面积为a2sin,则C=()A.B.C.D.【答案】C【分析】先利用正弦定理将已知等式中的边化角,再结合两角和公式与三角形的内角和定理,可推出sin B=2sin A;然后利用三角形的面积公式、正弦定理,即可得解.【解答】解:由正弦定理知,==,∵(a﹣2b)cos C=c(2cos B﹣cos A),∴(sin A﹣2sin B)cos C=sin C(2cos B﹣cos A),即sin A cos C+sin C cos A=2(sin B cos C+cos B sin C),∴sin(A+C)=2sin(B+C),即sin B=2sin A.∵△ABC的面积为a2sin,∴S=bc sin A=a2sin,根据正弦定理得,sin B•sin C•sin A=sin2A•sin,化简得,sin B•sin cos=sin A•cos,∵∈(0,),∴cos>0,∴sin==,∴=,即C=.故选:C.【知识点】正弦定理、余弦定理7.在正方体ABCD﹣A1B1C1D1中,下列四个结论中错误的是()A.直线B1C与直线AC所成的角为60°B.直线B1C与平面AD1C所成的角为60°C.直线B1C与直线AD1所成的角为90°D.直线B1C与直线AB所成的角为90°【答案】B【分析】连接AB1,求出∠ACB1可判断选项A;连接B1D1,找出点B1在平面AD1C上的投影O,设直线B1C与平面AD1C所成的角为θ,由cosθ=可判断选项B;利用平移法找出选项C和D涉及的异面直线夹角,再进行相关运算,即可得解.【解答】解:连接AB1,∵△AB1C为等边三角形,∴∠ACB1=60°,即直线B1C与AC所成的角为60°,故选项A正确;连接B1D1,∵AB1=B1C=CD1=AD1,∴四面体AB1CD1是正四面体,∴点B1在平面AD1C上的投影为△AD1C的中心,设为点O,连接B1O,OC,则OC=BC,设直线B1C与平面AD1C所成的角为θ,则cosθ===≠,故选项B错误;连接BC1,∵AD1∥BC1,且B1C⊥BC1,∴直线B1C与AD1所成的角为90°,故选项C正确;∵AB⊥平面BCC1B1,∴AB⊥B1C,即直线B1C与AB所成的角为90°,故选项D正确.故选:B.【知识点】直线与平面所成的角、异面直线及其所成的角8.如图,四边形ABCD为正方形,四边形EFBD为矩形,且平面ABCD与平面EFBD互相垂直.若多面体ABCDEF的体积为,则该多面体外接球表面积的最小值为()A.6πB.8πC.12πD.16π【答案】A【分析】由题意可得AC⊥面EFBD,可得V ABCDEF=V C﹣EFBD+V A﹣EFBD=2V A﹣EFBD,再由多面体ABCDEF 的体积为,可得矩形EFBD的高与正方形ABCD的边长之间的关系,再由题意可得矩形EFBD的对角线的交点为外接球的球心,进而求出外接球的半径,再由均值不等式可得外接球的半径的最小值,进而求出外接球的表面积的最小值.【解答】解:设正方形ABCD的边长为a,矩形BDEF的高为b,因为正方形ABCD,所以AC⊥BD,设AC∩BD=O',由因为平面ABCD与平面EFBD互相垂直,AC⊂面ABCD,平面ABCD∩平面EFBD=BD,所以AC⊥面EFBD,所以V ABCDEF=V C﹣EFBD+V A﹣EFBD=2V A﹣EFBD=2•S EFBD•CO'=•a•b•a =a2b,由题意可得V ABCDEF=,所以a2b=2;所以a2=,矩形EFBD的对角线的交点O,连接OO',可得OO'⊥BD,而OO'⊂面EFBD,而平面ABCD⊥平面EFBD,平面ABCD∩平面EFBD=BD,所以OO'⊥面EFBD,可得OA=OB=OE=OF都为外接球的半径R,所以R2=()2+(a)2=+=+=++≥3=3×,当且仅当=即b=时等号成立.所以外接球的表面积为S=4πR2≥4π•3×=6π.所以外接球的表面积最小值为6π.故选:A.【知识点】球的体积和表面积二、多选题(本大题共4小题,每小题5分,选对得分,选错、少选不得分)9.在△ABC中,角A,B,C的对边分别为a,b,c,若a2=b2+bc,则角A可为()A.B.C.D.【答案】BC【分析】由已知利用余弦定理整理可得cos A=,对于A,若A=,可得b=<0,错误;对于B,若A=,可得b=>0,对于C,若A=,可得b=>0,对于D,若A=,可得c=0,错误,即可得解.【解答】解:因为在△ABC中,a2=b2+bc,又由余弦定理可得:a2=b2+c2﹣2bc cos A,所以b2+bc=b2+c2﹣2bc cos A,整理可得:c=b(1+2cos A),可得:cos A=,对于A,若A=,可得:﹣=,整理可得:b=<0,错误;对于B,若A=,可得:=,整理可得:b=>0,对于C,若A=,可得:cos==,整理可得:b=>0,对于D,若A=,可得:cos=﹣=,整理可得:c=0,错误.故选:BC.【知识点】余弦定理10.如图,四边形ABCD为直角梯形,∠D=90°,AB∥CD,AB=2CD,M,N分别为AB,CD的中点,则下列结论正确的是()A.B.C.D.【答案】ABC【分析】由向量的加减法法则、平面向量基本定理解决【解答】解:由,知A正确;由知B正确;由知C正确;由N为线段DC的中点知知D错误;故选:ABC.【知识点】向量数乘和线性运算、平面向量的基本定理11.下列说法正确的有()A.任意两个复数都不能比大小B.若z=a+bi(a∈R,b∈R),则当且仅当a=b=0时,z=0C.若z1,z2∈C,且z12+z22=0,则z1=z2=0D.若复数z满足|z|=1,则|z+2i|的最大值为3【答案】BD【分析】通过复数的基本性质,结合反例,以及复数的模,判断命题的真假即可.【解答】解:当两个复数都是实数时,可以比较大小,所以A不正确;复数的实部与虚部都是0时,复数是0,所以B正确;反例z1=1,z2=i,满足z12+z22=0,所以C不正确;复数z满足|z|=1,则|z+2i|的几何意义,是复数的对应点到(0,﹣2)的距离,它的最大值为3,所以D正确;故选:BD.【知识点】复数的模、复数的运算、虚数单位i、复数、命题的真假判断与应用12.如图,已知ABCD﹣A1B1C1D1为正方体,E,F分别是BC,A1C的中点,则()A.B.C.向量与向量的夹角是60°D.异面直线EF与DD1所成的角为45°【答案】ABD【分析】在正方体ABCD﹣A1B1C1D1中,建立合适的空间直角坐标系,设正方体的棱长为2,根据空间向量的坐标运算,以及异面直线所成角的向量求法,逐项判断即可.【解答】解:在正方体ABCD﹣A1B1C1D1中,以点A为坐标原点,分别以AB,AD,AA1为x 轴、y轴、z轴建立空间直角坐标系,设正方体的棱长为2,则A(0,0,0),A1(0,0,2),B(2,0,0),B1(2,0,2),C (2,2,0),D(0,2,0),D1(0,2,2),所以,故,故选项A正确;又,又,所以,,则,故选项B正确;,所以,因此与的夹角为120°,故选项C错误;因为E,F分别是BC,A1C的中点,所以E(2,1,0),F(1,1,1),则,所以,又异面直线的夹角大于0°小于等于90°,所以异面直线EF与DD1所成的角为45°,故选项D正确;故选:ABD.【知识点】异面直线及其所成的角三、填空题(本大题共4小题,每小题5分,共20分.不需写出解答过程,请把答案直接填写在横线上)13.已知正方形ABCD的边长为2,点P满足=(+),则||=;•=.【分析】根据向量的几何意义可得P为BC的中点,再根据向量的数量积的运算和正方形的性质即可求出.【解答】解:由=(+),可得P为BC的中点,则|CP|=1,∴|PD|==,∴•=•(+)=﹣•(+)=﹣2﹣•=﹣1,故答案为:,﹣1.【知识点】平面向量数量积的性质及其运算14.若虛数z1、z2是实系数一元二次方程x2+px+q=0的两个根,且,则pq=.【答案】1【分析】设z1=a+bi,则z2=a﹣bi,(a,b∈R),根据两个复数相等的充要条件求出z1,z2,再由根与系数的关系求得p,q的值.【解答】解:由题意可知z1与z2为共轭复数,设z1=a+bi,则z2=a﹣bi,(a,b∈R 且b≠0),又,则a2﹣b2+2abi=a﹣bi,∴(2a+b)+(a+2b)i=1﹣i,∴,解得.∴z1=+i,z2=i,(或z2=+i,z1=i).由根与系数的关系,得p=﹣(z1+z2)=1,q=z1•z2=1,∴pq=1.故答案为:1.【知识点】复数的运算15.已知平面四边形ABCD中,AB=AD=2,BC=CD=BD=2,将△ABD沿对角线BD折起,使点A到达点A'的位置,当A'C=时,三棱锥A﹣BCD的外接球的体积为.【分析】由题意画出图形,找出三棱锥外接球的位置,求解三角形可得外接球的半径,再由棱锥体积公式求解.【解答】解:记BD的中点为M,连接A′M,CM,可得A′M2+CM2=A′C2,则∠A′MC=90°,则外接球的球心O在△A′MC的边A′C的中垂线上,且过正三角形BCD的中点F,且在与平面BCD垂直的直线m上,过点A′作A′E⊥m于点E,如图所示,设外接球的半径为R,则A′O=OC=R,,A′E=1,在Rt△A′EO中,A′O2=A′E2+OE2,解得R=.故三棱锥A﹣BCD的外接球的体积为.故答案为:.【知识点】球的体积和表面积16.已知一圆锥底面圆的直径为3,圆锥的高为,在该圆锥内放置一个棱长为a的正四面体,并且正四面体在该几何体内可以任意转动,则a的最大值为.【分析】根据题意,该四面体内接于圆锥的内切球,通过内切球即可得到a的最大值.【解答】解:依题意,四面体可以在圆锥内任意转动,故该四面体内接于圆锥的内切球,设球心为P,球的半径为r,下底面半径为R,轴截面上球与圆锥母线的切点为Q,圆锥的轴截面如图:则OA=OB=,因为SO=,故可得:SA=SB==3,所以:三角形SAB为等边三角形,故P是△SAB的中心,连接BP,则BP平分∠SBA,所以∠PBO=30°;所以tan30°=,即r=R=×=,即四面体的外接球的半径为r=.另正四面体可以从正方体中截得,如图:从图中可以得到,当正四面体的棱长为a时,截得它的正方体的棱长为a,而正四面体的四个顶点都在正方体上,故正四面体的外接球即为截得它的正方体的外接球,所以2r=AA1=a=a,所以a=.即a的最大值为.故答案为:.【知识点】旋转体(圆柱、圆锥、圆台)四、解答题(本大题共6小题,共70分.请在答题卡指定区域内作答,解答时应写出文字说明、证明过程或演算步骤)17.在四边形ABCD中,AB∥CD,AD=BD=CD=1.(1)若AB=,求BC;(2)若AB=2BC,求cos∠BDC.【分析】(1)直接利用余弦定理的应用求出结果;(2)利用余弦定理的应用建立等量关系式,进一步求出结果.【解答】解:(1)在四边形ABCD中,AD=BD=CD=1.若AB=,所以:cos∠ADB==,由于AB∥CD,所以∠BDC=∠ABD,即cos∠BDC=cos∠ABD=,所以BC2=BD2+CD2﹣2•BD•CD•cos∠BDC==,所以BC=.(2)设BC=x,则AB=2BC=2x,由余弦定理得:cos∠ADB==,cos∠BDC===,故,解得或﹣(负值舍去).所以.【知识点】余弦定理18.(1)已知z1=1﹣2i,z2=3+4i,求满足=+的复数z.(2)已知z,ω为复数,(1+3i)﹣z为纯虚数,ω=,且|ω|=5.求复数ω.【分析】(1)把z1,z2代入=+,利用复数代数形式的乘除运算化简求出,进一步求出z;(2)设z=a+bi(a,b∈R),利用复数的运算及(1+3i)•z=(1+3i)(a+bi)=a﹣3b+(3a+b)i为纯虚数,可得,又ω==i,|ω|=5,可得,即可得出a,b,再代入可得ω.【解答】解:(1)由z1=1﹣2i,z2=3+4i,得=+==,则z=;(2)设z=a+bi(a,b∈R),∵(1+3i)•z=(1+3i)(a+bi)=a﹣3b+(3a+b)i为纯虚数,∴.又ω===i,|ω|=5,∴.把a=3b代入化为b2=25,解得b=±5,∴a=±15.∴ω=±(i)=±(7﹣i).【知识点】复数的运算19.如图,墙上有一壁画,最高点A离地面4米,最低点B离地面2米.观察者从距离墙x(x>1)米,离地面高a(1≤a≤2)米的C处观赏该壁画,设观赏视角∠ACB=θ.(1)若a=1.5,问:观察者离墙多远时,视角θ最大?(2)若tanθ=,当a变化时,求x的取值范围.【分析】(1)首项利用两角和的正切公式建立函数关系,进一步利用判别式确定函数的最大值;(2)利用两角和的正切公式建立函数关系,利用a的取值范围即可确定x的范围.【解答】解:(1)如图,作CD⊥AF于D,则CD=EF,设∠ACD=α,∠BCD=β,CD=x,则θ=α﹣β,在Rt△ACD和Rt△BCD中,tanα=,tanβ=,则tanθ=tan(α﹣β)==(x>0),令u=,则ux2﹣2x+1.25u=0,∵上述方程有大于0的实数根,∴△≥0,即4﹣4×1.25u2≥0,∴u≤,即(tanθ)max=,∵正切函数y=tan x在(0,)上是增函数,∴视角θ同时取得最大值,此时,x==,∴观察者离墙米远时,视角θ最大;(2)由(1)可知,tanθ===,即x2﹣4x+4=﹣a2+6a﹣4,∴(x﹣2)2=﹣(a﹣3)2+5,∵1≤a≤2,∴1≤(x﹣2)2≤4,化简得:0≤x≤1或3≤x≤4,又∵x>1,∴3≤x≤4.【知识点】解三角形20.如图,已知复平面内平行四边形ABCD中,点A对应的复数为﹣1,对应的复数为2+2i,对应的复数为4﹣4i.(Ⅰ)求D点对应的复数;(Ⅱ)求平行四边形ABCD的面积.【分析】(I)利用复数的几何意义、向量的坐标运算性质、平行四边形的性质即可得出.(II)利用向量垂直与数量积的关系、模的计算公式、矩形的面积计算公式即可得出.【解答】解:(Ⅰ)依题点A对应的复数为﹣1,对应的复数为2+2i,得A(﹣1,0),=(2,2),可得B(1,2).又对应的复数为4﹣4i,得=(4,﹣4),可得C(5,﹣2).设D点对应的复数为x+yi,x,y∈R.得=(x﹣5,y+2),=(﹣2,﹣2).∵ABCD为平行四边形,∴=,解得x=3,y=﹣4,故D点对应的复数为3﹣4i.(Ⅱ)=(2,2),=(4,﹣4),可得:=0,∴.又||=2,=4.故平行四边形ABCD的面积==16.【知识点】复数的代数表示法及其几何意义21.如图所示,等腰梯形ABFE是由正方形ABCD和两个全等的Rt△FCB和Rt△EDA组成,AB=1,CF=2.现将Rt△FCB沿BC所在的直线折起,点F移至点G,使二面角E﹣BC﹣G的大小为60°.(1)求四棱锥G﹣ABCE的体积;(2)求异面直线AE与BG所成角的大小.【分析】(1)推导出GC⊥BC,EC⊥BC,从而∠ECG=60°.连接DG,推导出DG⊥EF,由BC⊥EF,BC⊥CG,得BC⊥平面DEG,从而DG⊥BC,进而DG⊥平面ABCE,DG是四棱锥G ﹣ABCE的高,由此能求出四棱锥G﹣ABCE的体积.(2)取DE的中点H,连接BH、GH,则BH∥AE,∠GBH既是AE与BG所成角或其补角.由此能求出异面直线AE与BG所成角的大小.【解答】解:(1)由已知,有GC⊥BC,EC⊥BC,所以∠ECG=60°.连接DG,由CD=AB=1,CG=CF=2,∠ECG=60°,有DG⊥EF①,由BC⊥EF,BC⊥CG,有BC⊥平面DEG,所以,DG⊥BC②,由①②知,DG⊥平面ABCE,所以DG就是四棱锥G﹣ABCE的高,在Rt△CDG中,.故四棱锥G﹣ABCE的体积为:.(2)取DE的中点H,连接BH、GH,则BH∥AE,故∠GBH既是AE与BG所成角或其补角.在△BGH中,,,则.故异面直线AE与BG所成角的大小为.【知识点】异面直线及其所成的角、棱柱、棱锥、棱台的体积22.如图,四边形MABC中,△ABC是等腰直角三角形,AC⊥BC,△MAC是边长为2的正三角形,以AC为折痕,将△MAC向上折叠到△DAC的位置,使点D在平面ABC内的射影在AB上,再将△MAC向下折叠到△EAC的位置,使平面EAC⊥平面ABC,形成几何体DABCE.(1)点F在BC上,若DF∥平面EAC,求点F的位置;(2)求直线AB与平面EBC所成角的余弦值.【分析】(1)点F为BC的中点,设点D在平面ABC内的射影为O,连接OD,OC,取AC 的中点H,连接EH,由题意知EH⊥AC,EH⊥平面ABC,由题意知DO⊥平面ABC,得DO∥平面EAC,取BC的中点F,连接OF,则OF∥AC,从而OF∥平面EAC,平面DOF∥平面EAC,由此能证明DF∥平面EAC.(2)连接OH,由OF,OH,OD两两垂直,以O为坐标原点,OF,OH,OD所在直线分别为x,y,z轴,建立空间直角坐标系,利用向量法能求出直线AB与平面EBC所成角的余弦值.【解答】解:(1)点F为BC的中点,理由如下:设点D在平面ABC内的射影为O,连接OD,OC,∵AD=CD,∴OA=OC,∴在Rt△ABC中,O为AB的中点,取AC的中点H,连接EH,由题意知EH⊥AC,又平面EAC⊥平面ABC,平面EAC∩平面ABC=AC,∴EH⊥平面ABC,由题意知DO⊥平面ABC,∴DO∥EH,∴DO∥平面EAC,取BC的中点F,连接OF,则OF∥AC,又OF⊄平面EAC,AC⊂平面EAC,∴OF∥平面EAC,∵DO∩OF=O,∴平面DOF∥平面EAC,∵DF⊂平面DOF,∴DF∥平面EAC.(2)连接OH,由(1)可知OF,OH,OD两两垂直,以O为坐标原点,OF,OH,OD所在直线分别为x,y,z轴,建立如图所示空间直角坐标系,则B(1,﹣1,0),A(﹣1,1,0),E(0,1,﹣),C(1,1,0),∴=(2,﹣2,0),=(0,2,0),=(﹣1,2,﹣),设平面EBC的法向量=(a,b,c),则,取a=,则=(,0,﹣1),设直线与平面EBC所成的角为θ,则sinθ===.∴直线AB与平面EBC所成角的余弦值为cosθ==.【知识点】直线与平面平行、直线与平面所成的角人教版高一下学期期中考试数学试卷(二)注意事项:本试卷满分150分,考试时间120分钟,试题共22题.答卷前,考生务必用0.5毫米黑色签字笔将自己的姓名、班级等信息填写在试卷规定的位置.一、选择题(本大题共8小题,每小题5分,共40分)在每小题所给出的四个选项中,只有一项是符合题目要求的.(2﹣i)z对应的点位于虚轴的正半轴上,则复数z对应的点位于()1.已知复平面内,A.第一象限B.第二象限C.第三象限D.第四象限2.平行四边形ABCD中,点E是DC的中点,点F是BC的一个三等分点(靠近B),则=()A.B.C.D.3.已知向量=(6t+3,9),=(4t+2,8),若(+)∥(﹣),则t=()A.﹣1 B.﹣C.D.14.已知矩形ABCD的一边AB的长为4,点M,N分别在边BC,DC上,当M,N分别是边BC,DC的中点时,有(+)•=0.若+=x+y,x+y=3,则线段MN的最短长度为()A.B.2 C.2D.25.若z∈C且|z+3+4i|≤2,则|z﹣1﹣i|的最大和最小值分别为M,m,则M﹣m的值等于()A.3 B.4 C.5 D.96.已知球的半径为R,一等边圆锥(圆锥母线长与圆锥底面直径相等)位于球内,圆锥顶点在球上,底面与球相接,则该圆锥的表面积为()A.R2B.R2C.R2D.R27.农历五月初五是端午节,民间有吃粽子的习惯,粽子又称粽籺,俗称“粽子”,古称“角黍”,是端午节大家都会品尝的食品,传说这是为了纪念战国时期楚国大臣、爱国主义诗人屈原.小明在和家人一起包粽子时,想将一丸子(近似为球)包入其中,如图,将粽叶展开后得到由六个边长为4的等边三角形所构成的平行四边形,将粽叶沿虚线折起来,可以得到如图所示的粽子形状的六面体,则放入丸子的体积最大值为()A.πB.πC.πD.π8.已知半球O与圆台OO'有公共的底面,圆台上底面圆周在半球面上,半球的半径为1,则圆台侧面积取最大值时,圆台母线与底面所成角的余弦值为()A.B.C.D.二、多选题(本大题共4小题,每小题5分,选对得分,选错、少选不得分)9.下列有关向量命题,不正确的是()A.若||=||,则=B.已知≠,且•=•,则=C.若=,=,则=D.若=,则||=||且∥10.若复数z满足,则()A.z=﹣1+i B.z的实部为1 C.=1+i D.z2=2i11.如图,在平行四边形ABCD中,E,F分别为线段AD,CD的中点,AF∩CE=G,则()A.B.C.D.12.已知正方体ABCD﹣A1B1C1D1,棱长为2,E为线段B1C上的动点,O为AC的中点,P 为棱CC1上的动点,Q为棱AA1的中点,则以下选项中正确的有()A.AE⊥B1CB.直线B1D⊥平面A1BC1C.异面直线AD1与OC1所成角为D.若直线m为平面BDP与平面B1D1P的交线,则m∥平面B1D1Q三、填空题(本大题共4小题,每小题5分,共20分.不需写出解答过程,请把答案直接填写在横线上)13.已知向量=(m,1),=(m﹣6,m﹣4),若∥,则m的值为.14.将表面积为36π的圆锥沿母线将其侧面展开,得到一个圆心角为的扇形,则该圆锥的轴截面的面积S=.15.如图,已知有两个以O为圆心的同心圆,小圆的半径为1,大圆的半径为2,点A 为小圆上的动点,点P,Q是大圆上的两个动点,且•=1,则||的最大值是.16.如图,在三棱锥A﹣BCD的平面展开图中,已知四边形BCED为菱形,BC=1,BF=,若二面角A﹣CD﹣B的余弦值为﹣,M为BD的中点,则CD=,直线AD与直线CM所成角的余弦值为.四、解答题(本大题共6小题,共70分.请在答题卡指定区域内作答,解答时应写出文字说明、证明过程或演算步骤)17.已知,.(1)若与同向,求;(2)若与的夹角为120°,求.18.已知a、b、c是△ABC中∠A、∠B、∠C的对边,a=4,b=6,cos A=﹣.(1)求c;(2)求cos2B的值.19.已知:复数z1与z2在复平面上所对应的点关于y轴对称,且z1(1﹣i)=z2(1+i)(i为虚数单位),|z1|=.(Ⅰ)求z1的值;(Ⅱ)若z1的虚部大于零,且(m,n∈R),求m,n的值.20.(Ⅰ)在复数范围内解方程|z|2+(z+)i=(i为虚数单位)(Ⅱ)设z是虚数,ω=z+是实数,且﹣1<ω<2.(1)求|z|的值及z的实部的取值范围;(2)设,求证:μ为纯虚数;(3)在(2)的条件下求ω﹣μ2的最小值.21.如图,直三棱柱A1B1C1﹣ABC中,AB=AC=1,,A1A=4,点M为线段A1A 的中点.(1)求直三棱柱A1B1C1﹣ABC的体积;(2)求异面直线BM与B1C1所成的角的大小.(结果用反三角表示)22.如图所示,在正方体ABCD﹣A1B1C1D1中,点G在棱D1C1上,且D1G=D1C1,点E、F、M分别是棱AA1、AB、BC的中点,P为线段B1D上一点,AB=4.(Ⅰ)若平面EFP交平面DCC1D1于直线l,求证:l∥A1B;(Ⅱ)若直线B1D⊥平面EFP.(i)求三棱锥B1﹣EFP的表面积;(ii)试作出平面EGM与正方体ABCD﹣A1B1C1D1各个面的交线,并写出作图步骤,保留作图痕迹.设平面EGM与棱A1D1交于点Q,求三棱锥Q﹣EFP的体积.答案解析一、选择题(本大题共8小题,每小题5分,共40分)在每小题所给出的四个选项中,只有一项是符合题目要求的.(2﹣i)z对应的点位于虚轴的正半轴上,则复数z对应的点位于()1.已知复平面内,A.第一象限B.第二象限C.第三象限D.第四象限【答案】B【分析】直接利用复数的运算和几何意义的应用求出该点所表示的位置.【解答】解:设z=a+bi(a,b∈R),所以(2﹣i)(a+bi)=2a+b+(2b﹣a)i,由于对应的点在虚轴的正半轴上,所以,即,所以a<0,b>0.故该点在第二象限.故选:B.【知识点】复数的代数表示法及其几何意义2.平行四边形ABCD中,点E是DC的中点,点F是BC的一个三等分点(靠近B),则=()A.B.C.D.【答案】D【分析】利用平行四边形的性质以及向量相等的概念,再利用平面向量基本定理进行转化即可.【解答】解:因为ABCD为平行四边形,所以,故.故选:D.【知识点】平面向量的基本定理3.已知向量=(6t+3,9),=(4t+2,8),若(+)∥(﹣),则t=()A.﹣1 B.﹣C.D.1【答案】B【分析】根据平面向量的坐标表示和共线定理,列方程求出t的值.【解答】解:向量=(6t+3,9),=(4t+2,8),所以+=(6t+3,11),﹣=(4t+2,5).又(+)∥(﹣),所以5(6t+3)﹣11(4t+2)=0,解得t=﹣.故选:B.【知识点】平面向量共线(平行)的坐标表示4.已知矩形ABCD的一边AB的长为4,点M,N分别在边BC,DC上,当M,N分别是边BC,DC的中点时,有(+)•=0.若+=x+y,x+y=3,则线段MN的最短长度为()A.B.2 C.2D.2【答案】D【分析】先根据M,N满足的条件,将(+)•=0化成的表达式,从而判断出矩形ABCD为正方形;再将+=x+y,左边用表示出来,结合x+y =3,即可得NC+MC=4,最后借助于基本不等式求出MN的最小值.【解答】解:当M,N分别是边BC,DC的中点时,有(+)•===,所以AD=AB,则矩形ABCD为正方形,设,,则=.则x=2﹣λ,y=2﹣μ.又x+y=3,所以λ+μ=1.故NC+MC=4,则MN==(当且仅当MC=NC=2时取等号).故线段MN的最短长度为2.故选:D.【知识点】平面向量数量积的性质及其运算5.若z∈C且|z+3+4i|≤2,则|z﹣1﹣i|的最大和最小值分别为M,m,则M﹣m的值等于()A.3 B.4 C.5 D.9【答案】B【分析】由题意画出图形,再由复数模的几何意义,数形结合得答案.【解答】解:由|z+3+4i|≤2,得z在复平面内对应的点在以Q(﹣3,﹣4)为圆心,以2为半径的圆及其内部.如图:|z﹣1﹣i|的几何意义为区域内的动点与定点P得距离,则M=|PQ|+2,m=|PQ|﹣2,则M﹣m=4.故选:B.【知识点】复数的运算6.已知球的半径为R,一等边圆锥(圆锥母线长与圆锥底面直径相等)位于球内,圆锥顶点在球上,底面与球相接,则该圆锥的表面积为()A.R2B.R2C.R2D.R2【答案】B【分析】设圆锥的底面半径为r,求得圆锥的高,由球的截面性质,运用勾股定理可得r,由圆锥的表面积公式可得所求.【解答】解:如图,设圆锥的底面半径为r,则圆锥的高为r,则R2=r2+(r﹣R)2,解得r=R,则圆锥的表面积为S=πr2+πr•2r=3πr2=3π(R)2=πR2,故选:B.【知识点】球内接多面体、旋转体(圆柱、圆锥、圆台)7.农历五月初五是端午节,民间有吃粽子的习惯,粽子又称粽籺,俗称“粽子”,古称“角黍”,是端午节大家都会品尝的食品,传说这是为了纪念战国时期楚国大臣、爱国主义诗人屈原.小明在和家人一起包粽子时,想将一丸子(近似为球)包入其中,如图,将粽叶展开后得到由六个边长为4的等边三角形所构成的平行四边形,将粽叶沿虚线折起来,可以得到如图所示的粽子形状的六面体,则放入丸子的体积最大值为()A.πB.πC.πD.π【答案】A【分析】先根据题意求得正四面体的体积,进而得到六面体的体积,再由图形的对称性得,内部的丸子要是体积最大,就是丸子要和六个面相切,设丸子的半径为R,则,由此求得R,进而得到答案.【解答】解:由题意可得每个三角形面积为,由对称性可知该六面体是由两个正四面体合成的,可得该四面体的高为,故四面体的体积为,∵该六面体的体积是正四面体的2倍,。
四川省内江市重点中学2022-2023学年高一下学期期中数学试题及参考答案

内江市重点中学2022—2023学年高一(下)期中考试数学试题考试时间:120分钟 满分:150分第I 卷选择题(满分60分)一、单选题(每题5分,共40分)1.已知向量a ,b 满足5a =,6b =,6a b ⋅=-,则cos ,a b =( ) A. 45-B.45C.15D. 15-2.已知1sin 123πθ⎛⎫-= ⎪⎝⎭,则5cos 12πθ⎛⎫+= ⎪⎝⎭( ) A.13B. 13-C.223D. 223-3.在ABC △中,()310AE AB AC =+,D 为BC 边的中点,则( ) A. 37AE ED = B. 73AE ED = C. 23AE ED =D. 32AE ED =4.数学与音乐有着紧密的关联,我们平时听到的乐音一般来说并不是纯音,而是由多种波组合而成的复合音.如图为某段乐音的图象,则该段乐音对应的函数解析式可以为( )A. 11sin sin 2sin 323y x x x =++ B. 11sin sin 2sin 323y x x x =-- C. 11sin cos 2cos323y x x x =++D. 11cos cos 2cos323y x x x =++5.已知函数()()cos 22sin cos R 344f x x x x x πππ⎛⎫⎛⎫⎛⎫=--++∈ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭,现给出下列四个结论,其中正确的是( ) A.函数()f x 的最小正周期为2πB.函数()f x 的最大值为2C.函数()f x 在,66ππ⎡⎤-⎢⎥⎣⎦上单调递增 D.将函数()f x 的图象向右平移12π个单位长度;所得图象对应的解析式为()sin 2g x x = 6.若函数()2sin 6f x x πω⎛⎫=- ⎪⎝⎭在区间,43ππ⎡⎤-⎢⎥⎣⎦上存在最小值-2,则非零实数ω的取值范围是( ) A. (],1-∞-B. 4,3⎡⎫+∞⎪⎢⎣⎭C. (]4,1,3⎡⎫-∞-⋃+∞⎪⎢⎣⎭D. (][),12,-∞-⋃+∞7.已知向量a ,b 满足2a =,2b =,且2a b ⋅=-,c 为任意向量,则()()a cb c-⋅-的最小值为( ) A.-2B. 52-C.-3D. 72-8.人脸识别技术应用在各行各业,改变着人类的生活,而所谓人脸识别,就是利用计算机分析人脸视频或者图像,并从中提取出有效的识别信息,最终判别人脸对象的身份.在人脸识别中为了检测样本之间的相似度主要应用距离的测试,常用的测量距离的方式有曼哈顿距离和余弦距离.假设二维空间中有两个点()11,A x y ,()22,B x y ,O 为坐标原点,余弦相似度similarity 为向量OA ,OB 夹角的余弦值,记作()cos ,A B ,余弦距离为()1cos ,A B -.已知()sin ,cos P αα,()sin ,cos Q ββ,()sin ,cos R αα-,若P ,Q 的余弦距离为13,Q ,R 的余弦距离为12,则tan tan αβ⋅=( ) A.7 B. 17 C.4 D. 14二、多选题(全选对得5分,少选得2分,选错不得分,每题5分,共20分)9.下列说法正确的是( ) A.若AB BC ⊥,则0AB BC ⋅= B.零向量与任意向量平行C. 2AB BC CD AD ++=D.在正六边形ABCDEF 中,AB DE =10.若函数()()()sin 0,0,0f x A x A ωϕωϕπ=+>><<在一个周期内的图象如图所示,则( )A. ()f x 的最小正周期为3πB. ()f x 的增区间是()53,344k k k Z ππππ⎡⎤-+∈⎢⎥⎣⎦C. ()()50f x f x π-+-=D.将2sin 3y x π⎛⎫=+ ⎪⎝⎭的图象上所有点的横坐标变为原来的32倍(纵坐标不变)得到()f x 的图象11. ABC △的内角A ,B ,C 的对边分别为a ,b ,c ,则下列命题为真命题的是( ). A.若A B >,则sin sin A B >B.若222sin sin sin A B C +<,则ABC △是钝角三角形 C.若cos cos a A b B =,则ABC △为等腰三角形D.若22AB =,45B =︒,3AC =,则满足条件的三角形有且只有一个12.筒车是我国古代发明的一种水利灌溉工具,因其经济环保,至今还在农业生产中使用.如图,一个半径为6米的筒车逆时针匀速转动,其圆心O 距离水面3米,已知筒车每分钟转动1圈,如果当筒车上一盛水桶M (视为质点)从水中浮现时(图中点0P )开始计时,经过t 秒后,盛水桶M 运动到P 点,则下列说法正确的是( )A.当10t =秒时,063PP =米;B.在转动一周内,盛水桶M 到水面的距离不低于6米的持续时间为20秒;C.当[]55,75t ∈时,盛水桶M距水面的最大距离为3+D.盛水桶M 运动15秒后筒车上另一盛水桶恰好露出水面,则转动中两盛水桶高度差的最大值为.第Ⅱ卷 非选择题(满分90分)三、填空题(每题5分,共20分)13.已知在ABC △中,7813sin sin sin A B C==,则角C 的大小为________. 14.已知()1,1A ,()2,1B ,()1,2C --,()3,4D ,则向量AB 在CD 方向上的投影向量的坐标为_____.15.已知ABC △的内角A 、B 、C 的对边分别为a 、b 、c ,若1cos 3A =,23b c =,且ABC △,则a =___________. 16.下列命题:①若()3,4OA =-,()6,3OB =-,()5,3OC m m =---,ABC ∠为锐角,则实数m 的取值范围是34m >-②若非零向量0AB AC BC AB AC ⎛⎫⎪+⋅= ⎪⎝⎭,且12AB AC AB AC ⋅=,则ABC △为等边三角形 ③若单位向量1e ,2e 的夹角为60°,则当()122e te t R +∈取最小值时,1t =④已知O 是平面上一定点,A ,B ,C 是平面上不共线的三个点,动点P 满足()OP OA AB AC λ=++,()0,λ∈+∞,则动点P 一定通过ABC △的重心⑤如果ABC △内接于半径为R 的圆,且())222sin sin sinR A C b B -=-,则ABC △2其中正确的序号为_______________________.四、解答题(17题10分,其余每题12分,共70分)17.已知a ,b ,c 是同一平面内的三个向量,其中()1,3a =-,()2,4b =-,()2,c m =. (1)若()a b c ⊥+,求c ;(2)若ka b +与2a b -共线,求k 的值.18.已知函数()()23cos cos 2cos 2f x x x x ππ⎛⎫=---⎪⎝⎭. (1)求函数()f x 的对称轴和对称中心; (2)若0,2πθ⎡⎤∈⎢⎥⎣⎦,32310f θπ⎛⎫+= ⎪⎝⎭,求tan 4πθ⎛⎫+ ⎪⎝⎭的值. 19.已知函数()sin 2sin 23f x x x π⎛⎫=+-⎪⎝⎭. (1)求()f x 的最大值及相应的x 值; (2)设函数()4g x f x π⎛⎫=⎪⎝⎭,如图,点P ,M ,N 分别是函数()y g x =图像的零值点、最高点和最低点,求cos MPN ∠的值.20.已知平面向量2sin 2,26m x π⎛⎫⎛⎫=-+- ⎪ ⎪⎝⎭⎝⎭,()21,sin n x =,()f x m n =⋅, (1)求函数()f x 在0,2π⎡⎤⎢⎥⎣⎦上的单调递增区间;(2)将函数()f x 的图象所有的点向右平移12π个单位,再将所得图象上各点横坐标缩短为原来的12(纵坐标不变),再向下平移1个单位得到()g x 的图象,若()g x m =在5,824x ππ⎡⎤∈-⎢⎥⎣⎦上恰有2个解,求m 的取值范围.21.在①()22cos cos 2c ab c a B b A b +=-+,②()()()sin sin sin sin b c B C a A B +⋅-=--,③)sin 3cos b C a c B =-⋅这三个条件中任选一个,补充在下面的问题中,并解答.记ABC △的内角A ,B ,C 的对边分别为a ,b ,c ,已知______. (1)求角C 的大小;(2)若23c =,求4sin B a -的取值范围.注:如果选择多个条件分别解答,那么按第一个解答计分.22.如图,有一景区的平面图是一个半圆形,其中O 为圆心,直径AB 的长为2km ,C ,D 两点在半圆弧上,且BC CD =,设COB θ∠=; (1)当12πθ=时,求四边形ABCD 的面积;(2)若要在景区内铺设一条由线段AB ,BC ,CD 和DA 组成的观光道路,则当θ为何值时,观光道路的总长l 最长,并求出l 的最大值.数学参考答案1.D2.B3.C4.A5.C6.C7.B8.A9.AB 10.ABD 11.ABD 12.BCD12.【详解】解:以水轮所在平面为坐标平面,以水轮轴心O 为坐标原点,以平行于水面的直线为x 轴建立平面直角坐标系,点P 距离水面的高度h 关于时间t 的函数为()()sin h f t A t B ωϕ==++.则93A B A B +=⎧⎨-+=-⎩,解6A =,3B =,又水轮每分钟转动一周,则2ππ6030ω==,∴()π6sin 330f t t ϕ⎛⎫=++⎪⎝⎭, 由()06sin 30f ϕ=+=,得1sin 2ϕ=-,∴π6ϕ=-,则()ππ6sin 3306f t t ⎛⎫=-+⎪⎝⎭. 对于A :当10t =时,0π3POP ∠=,又6OP =,∴0ππ6sin 6sin 666PP =+=,故A 错误;对于B :令()ππ6sin 36306f t t ⎛⎫=-+≥⎪⎝⎭,则ππ1sin 3062t ⎛⎫-≥ ⎪⎝⎭,所以πππ5π63066t ≤-≤, 解得1030t ≤≤,则在转动一圈内,盛水桶M 到水面的距离不低于6米以上的持续时间为301020-=秒,故B 正确;对于C ,当[]55,75t ∈,则ππ5π7π30633,t ⎡⎤∈⎢⎣-⎥⎦,则ππsin 306t ⎡∈⎢⎭⎣⎛⎫- ⎪⎝⎦, 所以()3f t ⎡⎤∈-⎣⎦,故C 正确;设盛水桶M 运动时间为1t ,则另一桶为115t -, 所以()()()1111ππππ6sin 36s 3151in 3063056h f t f t t t ⎛⎫⎡⎤-+---⎪⎢⎥⎝⎭=⎣--⎦=-△1111πππππππππ6sin 6sin 6sin 6cos 3063062306306t t t t ⎡⎤⎛⎫⎛⎫⎛⎫⎛⎫-----+- ⎪ ⎪ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎝⎭⎝⎭⎣=⎦=11πππππ30643012t t ⎛⎫⎛⎫-++ ⎪ ⎪=⎝⎭⎝⎭=,当1ππsin 13012t ⎛⎫+=⎪⎝⎭时()max h =△,故D 正确;故选:BCD. 13.120°/23π14. 46,1313⎛⎫⎪⎝⎭15.2 16.②④⑤ 16【详解】对于①,由()()()3,4,6,3,5,3OA OB OC m m =-=-=---,得()3,1BA OA OB =-=--,()1,BC OC OB m m =-=---,因为ABC ∠为锐角,故0BA BC ⋅>且,BA BC 不共线,所以()()310310m m m m ---+>⎧⎪⎨+--≠⎪⎩,解得34m >-且12m ≠,故①错误;对于②,因为非零向量0AB AC BC AB AC ⎛⎫⎪+⋅= ⎪⎝⎭,所以BAC ∠的平分线与BC 垂直,ABC △为等腰三角形,又12AB AC ABAC⋅=,所以3BAC π∠=,所以ABC △为等边三角形,故②正确; 对于③,()222222212112212444424132e te e te e t e t t t t t +=+⋅+=+⨯+=++=++,当1t =-时,()122e te t R +∈取得最小值,故③错误;对于④,已知O 是平面上一定点,A ,B ,C 是平面上不共线的三个点,动点P 满足()OP OA AB AC λ=++,()0,λ∈+∞,记BC 中点为E ,则2AB AC AE +=,则2AP AE λ=,AE 直线过重心,故P 一定过重心;故④正确对于⑤∵())2sin 2sin 2s in R A C b B -=-,∴根据正弦定理,得)222a c b b b -=-=-,可得222a b c +-=∴cos 2C =,∵角C 为三角形的内角,∴角C 的大小为4π∵n 2si 4c R π==∴由余弦定理2222cos c a b a b C =+⋅-,可得(222222R a b b ab ab =+⋅≥=-,当且仅当a b =时等号成立∴(222ab R ≤=+∴(22111sin 22222ABC S ab C R R =≤⋅=△,即ABC △面积的最大值为212R ;故⑤正确, 故答案为②④⑤17.(1)(2)-2.【详解】(1)因为()2,4b =-,()2,c m =,∴()0,4b c m +=+, ∵()a b c ⊥+,()1,3a =-,∴()()340a b c m ⋅+=+=, ∴4m =-,∴()2,4c =-,∴25c =;(2)由已知:()2,34ka b k k +=--+,()20,2a b -=, ∴()()220340k k --⨯-⨯+=,∴2k =-. 18.(1)23k x ππ=+,k Z ∈,1,2122k ππ⎛⎫+-⎪⎝⎭(2)7 【详解】解:(1)由已知1cos 2111()cos 2cos 2sin 2222262x f x x x x x x π+⎛⎫=-=--=-- ⎪⎝⎭, 令262x k πππ-=+,则23k x ππ=+,k Z ∈. 所以()f x 的对称轴为23k x ππ=+,k Z ∈. 令26x k ππ-=,则212k x ππ=+,k Z ∈ 所以函数()f x 的对称中心为1,2122k ππ⎛⎫+-⎪⎝⎭,k Z ∈ (2)21113sin sin cos 2336222210f θπθπθθππ⎛⎫⎛⎫⎛⎫+=+--=+-=-=⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭,∴4cos 5θ= 又0,2πθ⎡⎤∈⎢⎥⎣⎦,∴4sin ,tan 335θθ==, 1tan 1tan 741tan 34413θπθθ++⎛⎫+=== ⎪-⎝⎭-.19.(1)1;,Z 12x k k ππ=+∈(2)5【详解】(1)解:由题意,函数()11sin2cos2sin2sin2sin 222223f x x x x x x x π⎛⎫=+-=+=+ ⎪⎝⎭ 所以函数()f x 的最大值为()max 1f x =,此时2232x k πππ+=+,即,Z 12x k k ππ=+∈(2)由题意,函数()sin 2sin 4323g x x x ππππ⎡⎤⎛⎫⎛⎫=+=+ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦,过D 作MD x ⊥轴于D ,因为1PD DM ==所以90PMN ∠=︒,可得PM MN PN ===在直角MPN △中,可得cos PM MPN PN ∠===20.(1),32ππ⎡⎤⎢⎥⎣⎦(2)1,12⎡⎫⎪⎢⎣⎭【详解】(1)解:因为2sin 2,26m x π⎛⎫⎛⎫=-+- ⎪ ⎪⎝⎭⎝⎭,()21,sin n x =且()f x m n =⋅, 所以()22sin 22sin 6f x m n x x π⎛⎫=⋅=-+- ⎪⎝⎭()122cos 21cos 22x x x ⎫=-+--⎪⎪⎝⎭1cos 221cos 2123x x x π⎛⎫=+=++ ⎪⎝⎭, 即()cos 213f x x π⎛⎫=++ ⎪⎝⎭, 令2223k x k ππππ-≤+≤,Z k ∈,解得236k x k ππππ-≤≤-,Z k ∈, 又因为0,2x π⎡⎤∈⎢⎥⎣⎦,所以函数()f x 的单调增区间为:,32ππ⎡⎤⎢⎥⎣⎦. (2)解:因为()cos 213f x x π⎛⎫=++ ⎪⎝⎭, 所以将函数()f x 的图象所有的点向右平移12π个单位得到cos 21cos 21121236f x x x ππππ⎡⎤⎛⎫⎛⎫⎛⎫-=-++=++ ⎪ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎝⎭⎣⎦,将所得图象上各点横坐标缩短为原来的12(纵坐标不变)再向下平移1个单位得到()cos 46g x x π⎛⎫=+ ⎪⎝⎭,又因为5,824x ππ⎡⎤∈-⎢⎥⎣⎦,所以4,63t x πππ⎡⎤=+∈-⎢⎥⎣⎦,令4036x ππ-≤+≤,解得824x ππ-≤≤-,令046x ππ≤+≤,解得52424x ππ-≤≤,即函数()g x 在,824ππ⎡⎤--⎢⎥⎣⎦上单调递增,在5,2424ππ⎡⎤-⎢⎥⎣⎦上单调递减,且1cos 832g ππ⎛⎫⎛⎫-=-= ⎪ ⎪⎝⎭⎝⎭,作出cos 3y t t ππ⎛⎫=-≤≤ ⎪⎝⎭图像可得:所以m 的取值范围1,12⎡⎫⎪⎢⎣⎭.21.(1)π3C =(2)(23,23-【详解】(1)选择条件①.由余弦定理得()()22222222211cos cos 2222c ab ac B bc A b a c b b c a b +=-+=+--+-+.整理得222a b c ab +-=,所以由余弦定理得2221cos 22a b c C ab +-==.又因为()0,πC ∈,所以π3C =.选择条件②.由正弦定理得()()()b c b c a a b +-=--,整理得222a b c ab +-=,由余弦定理得2221cos 22a b c C ab +-==.又因为()0,πC ∈,所以π3C =选择条件③.由正弦定理得)sin sin 3sin sin cos B C A C B -.整理得()sin sin cos B C C B A B C ==+,所以sin sin cos B C B C =.因为sin 0B >,所以sin C C =.显然cos 0C ≠,所以tan C =又因为()0,πC ∈,所以π3C =. (2)因为c =,π3C =,所以由正弦定理得4sin sin a c A C==,即4sin a A =. 因为πA B C ++=,所以2π3B A =-,所以2ππ4sin 4sin 4sin 2sin 4cos 36B a A A A A A ⎛⎫⎛⎫ ⎪ ⎪=⎝⎭⎝-=---+⎭=. 因为2π03A <<,所以ππ5π666A <+<,所以π4cos 6A -⎛⎫ ⎪⎝⎭<+<, 故4sin B a -的取值范围是(-. 22.【答案】(1)144+;(2)5 【详解】(1)连结OD ,则5,126COD AOD ππ∠=∠= ∴四边形ABCD的面积为1151211sin 11sin 2122644ππ⨯⨯⨯⨯+⨯⨯⨯=+ (2)由题意,在BOC △中,2OBC πθ-∠=,由正弦定理1sin sin()cos 22BC OB πθθθ==-,∴sin 2sin 2cos 2BC CD θθθ=== 同理在AOD △中,,2OAD DOA θπθ∠=∠=-,由正弦定理sin(2)sin DA OD πθθ=-,∴sin 22cos sin DA θθθ== ∴224sin 2cos 24sin 212sin ,02222l θθθπθθ⎛⎫=++=++-<< ⎪⎝⎭令sin 022t t θ⎛⎫=<< ⎪ ⎪⎝⎭,∴()222124212444452l t t t t t ⎛⎫=++-=+-=--+ ⎪⎝⎭∴12t=时,即3πθ=,l的最大值为5。
下学期高一语文期中考试

下学期高一语文期中考试一、选择题(11×2分=22分)1、下列词语中加点的字,读音有错误的一组是()A、琐屑.(xiè)磋.(cuō)商挈.(qiè)妇将雏B、狂吠.(fèi)涅.(liè)惴.(zhuì)惴不安C、丰腴.(yú)台.(tāi)州延口残喘.(chuǎn)D、炫.(xuàn)目偈.(jì)子竦.(sǒng)身一摇2、下列词语中,没有错别字的一组是()A、旷野清幽瞬息万变B、回溯阅历脑羞成怒C、札记审慎精美绝伧D、禁锢永决矫揉造作3、下列文段中横线处应填入的最恰当的关联词是()但他和我一同走路的时候,可就走得近了,简直是扶住我,_______怕我被汽车或电车撞死;我这面也为他近视而又要照顾别人担心,大家都仓皇失措的愁一路,______ _____不是万不得已,我是不大和他一同出去的,我实在看得他吃力,_______自己也吃力。
A、为了因此除非所以B、因为因此除非因而C、因为所以倘因而D、为了所以倘所以4、填入下列空缺处最恰当的一组词语是()如果我们回想一下,我党在______时期,我们对于马克思列宁主义的认识和对于中国革命的认识是何等______,何等______,则现在我们对于这些的认识是_______得多,_______得多了。
A、幼稚肤浅贫乏丰富深刻B、幼年肤浅贫乏丰富深刻C、幼年肤浅贫乏深刻丰富D、幼稚贫乏肤浅深刻丰富5、下面这段文字空缺句排列正确的一项是()在这种态度下,就是不要割断历史。
_______;_______;_______。
①不但要懂得外国革命史,还要懂得中国革命史②不但要懂得中国的今天,还要懂得中国的昨天和前天③不单是懂得希腊就行了,还要懂得中国。
A、③②①B、②③①C、③①②D、①②③6、下列各句中,加点的成语使用不恰当的一句是()A、本来还不错的一篇文章,让你们这样改来改去,反而改得不三不四....了。
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安徽省六安一中高二英语阶段性考题2013-4-18出题人郑德俊Ⅰ.单项填空21.It is necessary for us to________these interesting old customs.A.prevent B.present C.prepare D.preserve22.I can hardly________very loud noise while I am studying.A.keep away from B.put up with C.have efforts on D.get rid of23.She________some old photographs in a drawer.A.came across B.came about C.came after D.came at24.To their great relief, the missing child returned home,________,after an absence of two weeks.A.felt tired and sound B.tiring and soundlyC.feeling tired but soundly D.tired but sound25.________you are grown up, you should not rely on your parents.A.Now that B.Before C.Although D.As well as26.Entry in this competition is restricted________high school students.A.for B.with C.to D.over27.The girl witnessed________the man burst into the bank on a rainy night.A.to see B.seeing C.to seeing D.having seen28.The cruel man________his wife and child.A.abandoned B.involved C.appointed D.tolerated29.--- I am terribly sorry,sir.I thought it was parking place.---________.Here is your ticket.A.It doesn’t matter.B.You are right.C.I can’t agree with youD.Sorry,that’s no excuse.30. —Do you regret bringing up your opinion at the meeting?—No.________,I think it is very necessary.A.On the contrary B.First of all C.On the other hand D.In a word31.Holidays is _____time for leisure.But making good use of_____time is not an easy task for most people.A.the;theB.a;theC.the;aD.不填;a32.Actually,_______works hard can be _________they want to be,whether it is a pilot;an engineer or manager.A.anyone;whoeverB.no matter who;whoeverC.whoevver;no matter whatD..whoever;whatever33.---Which of those electronic dictionaries do you like most?---________.They are both expensive and of little use.A.BothB.NeitherC.NoneD.No one34.We are all born_______a sense of direction,but it is not properly understood how it works.A.ofB.withC.byD.into35.You can’t hardly imagine the great trouble they took______those injured in Japan earthquake.A.rescuingB.to rescueC.rescuedD.have rescuedⅡ.完形填空One day I took a bus with my girlfriend. It was so36__that we stood for several stops until a vacant seat was__37__for her.Then a pretty girl rushed towards me, saying, “Hi, where are you going?”I was so struck by the stranger that I had a hard time trying to __38__ her. Clearly she had taken me __39__ somebody else. I returned her greeting with politeness, __40__ to give explanation to my girlfriend later.Noticing my dialogue with somebody else, my girlfriend__41__ her eyes and found she was pretty. She asked jealously, “Who's she?”The pretty girl, quite__42__of the situation, spokeout first, “Hi, let me__43__myself. I' m Nancy, used to be __44__. Very gl ad to meet you.” She behaved very __45__.But I was __46__ to search in my memory for someone called Nancy among my neighbors. I was worrying how to explain this to my girlfriend __47__ the pretty girl again turned to me, “Will you give me your cell phone number so that we __48__ keep contacts(联系) later?”I had to submit to her __49__.Then the girl got off the bus at the next stop.A minute later I got a short __50__ on my phone-from a stranger. My girlfriend __51__ my phone and read the note. It was from the girl, who said, “Just now, two thieves tried to __52__ your pocket. I had to act as an acquaintance to draw your attention. I should have left at the __53__ stop but gave up as I noticed these two thieves also __54__ to leave at the same stop. Because of the __55__ I delayed my departure. Now you' ll understand all I have done to you. Wish you luck.”36.A.tiring B.slow C.crowded D.messy37.A.ready B.available C.special D.suitable38.A.identify B.realize C.remind D.judge39.A.on B.by C.with D.for40.A.preferring B.planning C.managing D.promising41.A.fixed B.opened C.raised D.focused42.A.proud B.careful C.afraid D.aware43.A.explain B.enjoy C.introduce D.help44.A.partners B.neighbors C.friends D.colleagues45.A.naturally B.generously C.cautiously D.normally46.A.in a hurry B.at a loss C.at ease D.in time47.A.while B.before C.whe D.until48.A.would B.can C.should D.must49.A.request B.question C.command D.opinion50.A.suggestion B.warning C.notice D.message51.A.removed B.seized C.found D.caught52.A.steal B.reach C.get D.pick53.A.terminal B.former C.previous D.latter54.A.pretending B.intending C.demanding D.announcing55.A.incident B.matter C.accident D.affairⅢ.阅读理解AAmelia Earhart's (1897-1937) childhood was not the typical girl's.She liked nothing better than to explore the banks of the Missouri River, where she climbed trees and hunted rats with her rifle (步枪).She also built her own roller coaster(过山车).The exciting life of the American aviation (航空)pioneer is dramatized in the movie Amelia, which is due to come out on October 23.After Earhart paid a pilot $1 to take her up for a short fly in his plane, aviation became her love and career.As she later explained, “Never do things others can do and will do, if there are things others cannot do or will not do.”Earhart found herself a flying teacher and started to learn to fly.She took all sorts of jobs to pay for the lessons and to buy a second-hand plane on her 24th birthday.In 1932, Earhart flew solo (单人的) across the Atlantic.She became the first woman to make the solo crossing.She also made a flying suit for women and went on to design other clothes for women who led active lives.“Now and then women should do for themselves what men havealready done—occasionally what men have not done —thereby establishing themselves as persons, and perhaps encouraging other women toward greater independence of thought and action, ”she said.When she was nearly 40, Earhart was ready for a final challenge—to be the first woman to fly around the world.However, in midflight, she and her navigator (导航员) disappeared in bad weather.Earhart will be forever remembered as a brave pioneer for both aviation and women. 56.According to the passage, which of the following statements is true?A. A movie about Earhart's exciting life has come out.B. Earhart never did things others could and would do.C. Earhart's love for aviation came after she flew a plane for a short time.D. In Earhart's opinion, women should think and act more independently.57.The underlined word in paragraph 2 means________.A.presented B.adopted C.published D.created58.From the passage we can know that ________.A.Earhart showed most interest in adventure trips to the Missouri River during her childhood B.Earhart's ambition was to fly across the AtlanticC.Earhart challenged herself constantlyD.Besides flying, Earhart also designed planes and clothes59.You can probably find this article in________.A.a booklet B.a newspaper C.a guide book D.an advertisementBSpace is a dangerous place, not only because of meteors(流星), but also because of rays from the sun and other stars. The atmosphere again acts as our protective blanket on earth. Light gets through, and this is necessary for plants to make the food which we eat. Heat, too, makes our environment endurable(可忍的). Various kinds of rays come through the air from outer space, but enormous(巨大的) quantities of radiation from the sun are screened off. As soon as men leave the atmosphere they are exposed to this radiation but their spacesuits or the walls of their spacecraft, if they are inside, do prevent a lot of radiation damage.Radiation is the greatest known danger to explorers in space. The unit of radiation is called “rem”. Scientists have reason to think that a man can put up with far more radiation than 0.1 rem without being damaged; the figure of 60 rems has been agreed on. The trouble is that it is extremely difficult to be sure about radiation damage a person may feel perfectly well, but the cells of this or her sex organs may be damaged, and this will not be discovered until the birth of deformed(畸形的) children or even grandchildren. Missions of the Apollo flights have had to cross belts of high radiation and, during the outward and return journeys, the Apollo crew accumulated(积累) a large amount of rems. So far, no dangerous amounts of radiation have been reported, but the Apollo missions have been quite short. We simply do not know yet how men are going to get on when they spend weeks and months outside the protection of the atmosphere, working in a space laboratory. Drugs might help to decrease the damage done by radiation, but no really effective ones have been found so far.60. According to the first paragraph, the atmosphere is necessary to man because it _______.A. protects him against the harmful rays from spaceB. provides sufficient light for plant growthC. supplies the heat necessary for human survivalD. screens off the falling meteors61. We know from the passage that _________.A. exposure to even tiny amounts of radiation is fatal(致命的)B. the effect of exposure to radiation is slow in comingC. radiation is avoidable in space explorationD. astronauts in spacesuits needn’t worry about radiation damage62. The harm radiation that has been done to the Apollo crew members ________.A. is insignificantB. is enormousC. seems overestimated(过高估计的)D. remains unknown63 It can be inferred from the passage that _________.A. the Apollo mission was very successfulB. protection from space radiation is no easy jobC. astronauts will have deformed children or grandchildrenD. radiation is not a threat to well-protected space explorersCNo one knows exactly how many disabled people there are in the world,but estimates suggest the figure is over 450 million. The number of disabled people in India alone is probably more than double the total population of Canada.In the United Kingdom,about one in ten people have some disability. Disability is not just something that happens to other people:as we get older,many of us will become less mobile,hard of hearing or have falling eyesight.Disablement can take many forms and occur at any time of life. Some people are born with disabilities.Many others become disabled as they get older. There are many progressive disabling diseases.The longer time goes on,the worse they become. Some people are disabled in accidents.Many others may have a period of disability in the form of a mental illness. All are affected by people’s attitude towards them.Disabled people face many physical barriers.Next time you go shopping or to work or to visit friends,imagine how you would manage if you could not get up steps, or onto buses and trains. How would you cope if you could not see where you were going or could not hear the traffic?But there are other barriers:prejudice(偏见)can be even harder to break down and ignorance(无知)inevitably represents by far the greatest barrier of all. It is almost impossible for the able-bodied to fully appreciate what the severely disabled go through,so it is important to draw attention to these barriers and show that it is the individual person and their ability,not their disability,which counts.64.The key word in Paragraph 4 is__________.A. disabilityB. ignoranceC. prejudiceD. barriers65.The last word of the passage“ counts” most probably means__________.A. being most importantB. being consideredC. being includedD. being numbered 66.Which of the following statements is NOT true according to the passage?A. Even the able-bodied may lose some of their body functions when they get older.B. There are about 10 percent disabled persons in the UK.C. The whole society should pay due attention to the barriers faced by the disabled people.D. There no longer exists prejudice against the disabled.67 .It can be concluded from the passage that______ .A. we should try our best to prevent disablementB. both physical and mental barriers are hard to break downC. we must take a proper attitude towards the disabledD. the able-bodied people will never fully understand the disabledD EVENTSA. China National Art MuseumB. Beizhan TheatreC. Beijing Concert HallD. 1 Wusi Dajie69. Menashe Kadishrnan is famous for painting __________.A. sheepB. birdsC. deerD. flowers70. Which of the following is TRUE?A. The folk concert will last three days.B. The ballet “The Nutcracker” will be put on once.C. The Israeli paintings will be displayed for a month.D. China National Art Museum lies in Xicheng District.71. If you dial(拨) the telephone number 6500-3388 on Dec. 8, you can _________.A. go to the folk concertB. visit the Art MuseumC. watch the balletD. none of the aboveEResearch on the human brain has been attracting more and more scientists in recent years, just like the booming hi-tech industry. One of the latest research topics is how to change the humanbrain or combine the computer and the human brain, i.e. to transplant a chip into o human brain. This innovation(改革) may make everyone’s dream come true. If we compare a human brain to a hard disc, what the scientists are doing is to enlarge the applicable capacity of the hard disc. For the time being, there are some difficulties in such transplant experiments, but scientists never give up.Experiments have started on animals. In 1996, a transplant experiment performed at the Defense and Military Physiology Research Institute in the U.S. turned a bear into a dolphin.The dolphin was named Ted, and the bear was named Tallin. Using the most advanced technology available, deep and detailed images were made of the memory area in Ted’s brain containing information about swimming by the scientists. They obtained a series of useful information, the signals transferred by the nervous system. Such information was saved into a button-sized chip, which was then transplanted into the action memory area in Tallin’s brain. The information saved on the chip was released by means of electric power.Recently, another comprehensive memory transplant was performed at the Motor Never Research Institute of the University of California. The comprehensive memory transplanted in the experiment included actions, moods, logic, words, images, etc. The experiment involved an entire transplant of the memory area. This was the largest such experiment done so far.The transplant was performed making a transfer from a dog named “Genius” to a dog named “Idiot”. “Genius” could understand and follow up to 100 gestures and orders made by its master. It was a real genius in memorizing. “Idiot” was the younger brother of “Genius”. It had no contact with people at all since its birth. It became an animal with nothing in its brain, without any memory.The operation was a complete success. When the two dogs woke up. “Idiot” had grasped all the abilities “Genius” possessed; it was good at memorizing and sensible. It could follow every gesture and any command given by its master. But “Genius” gave no response to its master, and in fact did not recognize him at all.72. The purpose of the experiment is _________.A. to combine the computer and the human brainB. to make bear swimC. to make some stupid dog turn cleverD. to enlarge the applicable capacity of human brain73. Which of the following statements is TRUE?A. The scientists transferred a button-sized chip wit useful information in Tallin’s brain.B. The comprehensive memory transplant was performed at the Motor Nerve Research University.C. The second experiment wasn’t an entire transpla nt of the memory area.D. The Idiot was an animal with memory before brain experiment74. What does the underlined word “success” refer to?A. The two dogs woke up.B. They were both good at memorizing.C. The Idiot grasped all the abilities of Genius.D. The Genius grasped all the abilities of Idiot.75. According to the text, we can infer ________.A. a person can know more after the experimentB. a bear can swim after being transferred a chip with relevant useful informationC. a dog can become clever after entire transplant of the memory areaD. it is really good for animals to have been involved in the transplant experimentⅣ.任务型读写You can’t avoid a conflict at work. If you can avoid a conflict, it means you will win what you want regardless of what the other person wants. Since the potential issue has not been removed, it will simply reappear later. Here are the necessary steps to effectively get rid of conflicts at work.★Realize that some conflicts are unavoidable at work. On numerous occasions, conflict and disagreement are likely to happen. But when a conflict happens it’s not the end of the world. On the contrary, it can be the beginning of an interesting learning process. Conflicts mean that people care enough to disagree strongly. The trick is not to allow the conflict to go on forever.★Handle conflicts sooner rather than later. Solve a conflict when it starts, as it only gets worse with time going by. Conflicts at work arise not from something that was said, but from something that wasn’t said! Everyone’s waiting for the other to admit he’s wrong and gets more unpleasant after the conflict has lasted a while. It’s essential to interrupt the “waiting game” before it gets to that point.★Ask nicely. If somebody has done something that made you angry, or if you don’t understand their viewpoint or actions, simply asking nicely about it can make a world of difference. Never assume that people do what they do to annoy or hurt you. Sometimes there’s a good reason why that person does what he or she does, and a potential conflict disappears right there. Do remember to make an inquiry, not an accusation of any sort.★Appreciate. Praise the other part in the conflict. Tell them why it’s worth it to you to solve the conflict. This can be difficult as few people find it easy to praise and appreciate a person they disagree strongly with, but it’s a great way to move forward.假设在英语课上老师给你们欣赏了电影《太空大灌篮》的主题曲I Believe I Can Fly.这是一首催人奋进的励志歌曲.请你根据以下要点以“I Believe I Can Fly”为题,用英语写一篇短文,参加某英语杂志社的写作比赛.内容要点;1.这首歌告诉我们自信的重要性;2.自信给人勇气尝试,使人成功;3.梦想和自信使人勇于面对困难;4.没有梦想和自信,不会获得成功;5.你的感悟(考生自拟,至少一点).注意:1.词数90左右.短文的开头已为你写好,不计入总词数;2.参考词汇:自信—confidence(n.);3.短文须包括所有要点,不要逐词翻译,可适当发挥,使短文连贯、通顺,文中不得出现真实的人名、校名等相关信息.I Believe I Can FlyI Believe I Can Fly is a nice song by R.Kelly._______________Key:I.单选:21—35 DBADA CCADA BDBBAⅡ.完型:本文是一篇记叙文,讲述了在一辆公共汽车上,有两个小偷想偷我钱包里的钱,却被一个漂亮女孩发现以及这个女孩如何帮“我”摆脱这两个小偷的故事。