2006年育才中学高三三月份月考
育才中学推选优秀学生参加高校选拔的暂定办法
育才中学推选优秀学生参加高校选拔的暂定办法一、指导思想：努力体现以德育为核心，以创新精神和实践能力为重点的素质教育要求，体现我校的办学方针，鼓励学生一马当先、积极主动地追求发展。
二、具体办法：实行本人申请、全面考核、累积积点、师生互评、学校批准、择优推荐、公示的方法。
从以下几方面进行测评，累计积点：1、资格认定：①品学兼优，遵守学校各项规章制度，品德评定为良以上。
“品行鉴定意见”应由班主任出具，经年级组认定、学工部审核后生效。
有重大违纪的不予推荐，有关具体内容见“附件1”。
②重视体锻，通过体锻标准。
达标成绩由高三体育备课组出具。
2、积点方法：①所计高一高二科目均以学程考试成绩进行核算，高三科目将考试成绩转换成标准分进行核算。
②计算科目：高一、高二8科：语、数、外、物、化、生、政、史、地、计高三3+1：历次月考成绩（不包括高三第一学期摸底考）③计算方法：高一、高二由“育才中学课程管理系统”给出的积点/25高三：（3+1第一次月考+……+八校联考+……+区统考）标准分总分*23、如高中阶段成绩测评累计积点相同的，则以高三原始成绩总分之和高低排序，择优推荐。
4、参加国家、市、区级竞赛：高中阶段参加省（市）发明创造奖获得者或者学科单项竞赛优胜者：全国一等奖 3市一等奖全国二等奖 2全国三等奖市二等奖 1市三等奖区一等奖0.5实施办法：①以上加分仅限于教育部、上海市教委认可的高考加分竞赛项目（具体见“附件2”），具体认定以教育管理部的意见为准。
②同类竞赛获奖，积分时取高不取低。
5、各类综合表彰：高中阶段获市三好学生、市优秀学生（团）干部 2区三好学生、区优秀学生（团）干部 1每学年学校三好学生、优秀团员0.5每周一星0.5每学程每门学科免修0.5实施办法：以上荣誉须经学生工作部认定之后，方予加分。
6、凡作出特殊成绩与贡献者，由本人申请，经校务会议讨论可适当加1——3分；在学校、年级或班级里担任干部，工作负责，踏实肯干，确实做出一定贡献的，经年级讨论，并报学生工作部认定，可适当加0.5——1分。
重庆市育才中学校2022-2023学年高一下学期3月月考数学试题(解析版)
重庆市育才中学校高2025届2022-2023学年（下）3月月考数学试题本试卷为第I 卷（选择题）和第II 卷（非选择题）两部分，共150分，考试时间120分钟.注意事项：1.答卷前，考生务必将自己的姓名､准考证号填写在答题卡上.2.回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑.如需改动，用橡皮擦干净后，再选涂其他答案标号.回答非选择题时，将答案写在答题卡上，写在本试卷上无效.3.考试结束后，将本试卷和答题卡一并交回.第I 卷一､选择题：本题共8小题，每小题5分，共40分.在每个小题给出的四个选项中，只有一项是符合题目要求的.1.已知平面向量()()()1,0,1,,2,1a b k c ==-=，若()2a b c+ ∥，则k =（）A.1B.1- C.14-D.14【答案】C 【解析】【分析】求出2a b + 的坐标，根据()2a b c + ∥，列出方程，计算可得.【详解】因为()()1,0,1,a b k ==-，所以()()()1,021,12,2a k k b =+-=-+，因为()2//a b c +，()2,1c = ，所以()11220k -⨯-⨯=，解得14k =-故选：C.2.已知α是第二象限角，则点tan ,sin22P αα⎛⎫⎪⎝⎭位于（）A.第一象限B.第二象限C.第三象限D.第四象限【答案】D 【解析】【分析】已知α是第二象限角，求2α和2α终边所在位置，判断tan 2α和sin 2α的符号，确定点tan,sin22P αα⎛⎫⎪⎝⎭所在象限.【详解】α是第二象限角，则()π2ππ2πZ 2k k k α+<<+∈，()ππππZ 422k k k α+<<+∈，2α的终边在一三象限，tan 02α>，()π4π22π4πZ k k k α+<<+∈，2α的终边在三四象限和y 轴非负半轴，sin 20α<，则点tan ,sin22P αα⎛⎫⎪⎝⎭位于第四象限.故选：D3.如图，60C 是一种碳原子簇，它是由60个碳原子构成的，其结构是以正五边形和正六边形面组成的凸32面体，这60个C 原子在空间进行排列时，形成一个化学键最稳定的空间排列位置，恰好与足球表面格的排列一致，因此也叫足球烯.根据杂化轨道的正交归一条件，两个等性杂化轨道的最大值之间的夹角()0180θθ<≤满足：233153cos cos cos cos 02222αβθγθδθθ⎛⎫⎛⎫++-+-= ⎪ ⎪⎝⎭⎝⎭，式中,,,αβγδ分别为杂化轨道中,,,s p d f 轨道所占的百分数.60C 中的杂化轨道为等性杂化轨道，且无,d f 轨道参与杂化，碳原子杂化轨道理论计算值为 2.28sp ，它表示参与杂化的,s p 轨道数之比为1:2.28，由此可计算得一个60C 中的凸32面体结构中的五边形个数和两个等性杂化轨道的最大值之间的夹角的余弦值分别为（）A.2520,57-B.2520,57C.2512,57-D.2512,57【答案】C 【解析】【分析】设60C 中的凸32面体结构中共有x 个五边形，y 个六边形，列方程即可求解,x y ，再根据所给公式求出cos θ.【详解】设一个60C 中的凸32面体结构中共有x 个五边形，y 个六边形，因为每个顶点都是三个面的公共顶点，所以56603x y+=，又因为32x y +=，解得12,20x y ==，所以共有12个正五边形；又因为1 2.28,,03.28 3.28αβγδ====，所以1 2.28cos 03.28 3.28θ+=，解得25cos 57θ=-，故选：C.4.已知175sin cos ,π,π134ααα⎛⎫+=-∈ ⎪⎝⎭，则sin cos αα-=（）A.213 B.213-C.713D.713-【答案】C 【解析】【分析】根据5π,π4α⎛⎫∈ ⎪⎝⎭，sin cos αα＞，运用同角关系计算.【详解】()2222222171717sin cos ,sin ,sin cos 2sin cos 131313αααααααα+=-∴+=++=，21202sin cos 13αα=，()222224949sin cos 2sin cos ,sin cos 1313αααααα+-=-=，5π7π,,sin cos ,sin cos 0,sin cos 413ααααααα⎛⎫∈--= ⎪⎝⎭＞＞；故选：C.5.已知非零向量,a b满足()()()()7,2211a b a b a b a b -⊥-+⊥- ，则sin ,a b =（）A.35B.45C.513D.1213【答案】A 【解析】【分析】由已知向量的垂直，根据数量积为0，列方程组求解.【详解】()()7a b a b -⊥- ，则()()227870a b a b a a b b -⋅-=-⋅+=，①()()2211a b a b +⊥- ，则有()()22221127220a b a b aa b b +⋅-=-⋅-=，②78⨯⨯①-②，得2292250a b -= ，则有5a b = ，代入①式，2222540cos ,70b b a b b -+=，解得4cos ,5a b = ，由[],0,π∈ a b ，得3sin ,5a b =.故选：A6.已知 1.5241,log 3,sin 12a b c ⎛⎫=== ⎪⎝⎭，则,,a b c 的大小关系为（）A.a b c <<B.b c a <<C.c a b <<D.a c b<<【答案】D 【解析】【分析】通过和中间数13,24比大小即可.【详解】 1.51212a ⎛⎫⎪⎝<=⎭；443log 3log 4b =>=；2221ππ3=sin sin 1sin 2434c <=<=；所以a c b <<故选：D7.如图，在梯形ABCD 中，112AD DC AB ===且,AB AD P ⊥为以A 为圆心AD 为半径的14圆弧上的一动点，则()PD PB PC ⋅+ 的最小值为（）A.3-B.3-C.3-D.3-【答案】B 【解析】【分析】建立平面直角坐标系，利用向量的坐标运算及三角函数的性质求解.【详解】以A 为原点，AB 为x 轴，AD 为y 轴，建立如图所示的平面直角坐标系，则有()0,0A ，()2,0B ，()1,1C ，()0,1D ，设()πcos ,sin 02P ⎛⎫≤≤⎪⎝⎭ααα，得()cos ,1sin PD =-- αα，()2cos ,sin PB =-- αα，()1cos ,1sin PC =--αα，则()()()cos ,1sin 32cos ,12sin PD PB PC ⋅+=--⋅--αααα222cos 3cos 2sin 3sin 1=-+-+ααααπ34⎛⎫=-+ ⎪⎝⎭α由π02α≤≤，当π4α=时，()PD PB PC ⋅+ 有最小值3-.故选：B8.设函数()()2sin 1(0)f x x ωϕω=+->，若对任意实数(),f x ϕ在区间[]0,π上至少有3个零点，至多有4个零点，则ω的取值范围是（）A.810,33⎡⎫⎪⎢⎣⎭B.10,43⎡⎫⎪⎢⎣⎭C.144,3⎡⎫⎪⎢⎣⎭D.1416,33⎡⎫⎪⎢⎣⎭【答案】B 【解析】【分析】由题可转化为研究函数2sin 1y x ω=-在任意一个长度为π的区间上的零点问题，求出函数2sin 1y x ω=-在y 轴右侧靠近坐标原点处的零点，得到相邻四个零点之间的最大距离，相邻五个零点之间的距离，结合条件列式即得.【详解】因为ϕ为任意实数，故函数()f x 的图象可以任意平移，从而研究函数()f x 在区间[]0,π上的零点问题，即研究函数2sin 1y x ω=-在任意一个长度为π0π-=的区间上的零点问题，令2sin 1y x ω=-0=，得1sin 2x ω=，则它在y 轴右侧靠近坐标原点处的零点分别为π6ω，5π6ω，13π6ω，17π6ω，25π6ω，L ，则它们相邻两个零点之间的距离分别为2π3ω，4π3ω，2π3ω，4π3ω，L，故相邻四个零点之间的最大距离为10π3ω，相邻五个零点之间的距离为4πω，所以要使函数()f x 在区间[]0,π上至少有3个零点，至多有4个零点，则需相邻四个零点之间的最大距离不大于π，相邻五个零点之间的距离大于π，即10ππ34ππωω⎧≤⎪⎪⎨⎪>⎪⎩，解得1043ω≤<.故选：B.二､多选题：本题共4小题，每小题5分，共20分.在每小题给出的选项中，有多项符合题目要求，全部选对的得5分，部分选对的得2分，有选错的得0分.9.已知在同一平面内的向量,,a b均为非零向量，则下列说法中正确的有（）A.若,a b b c∥∥，则a c∥B.若a c a b ⋅=⋅ ，则b c= C.()()a b c a b c⋅⋅=⋅⋅ D .若a b 且a c ⊥，则()c a b ⋅+= 【答案】AD 【解析】【分析】平面向量共线的传递性判断A ，由向量数量积的定义可判断B ，根据数量积及共线向量的概念可判断C ，根据向量垂直及向量数量积的概念可判断D.【详解】对A ，在同一平面内的向量,,a b c 均为非零向量，若//a b 且//b c ，则//a c ，即A 正确；对B ，若a c a b ⋅=⋅ ，则cos ,cos ,a c a c a b a b ⋅=⋅ ，又0a ≠ ，所以cos ,cos ,b a b c c =，因为,b c 与a 的夹角不一定相等，所以b c =不一定成立，即B 错误；对C ，因为()a b c ⋅⋅ 与c 共线，()a b c ⋅⋅与a 共线，所以()()a b c a b c ⋅⋅=⋅⋅ 不一定成立，即C 错误；对D ，若//a b 且a c ⊥ ，则c b ⊥ ，()0c a b c a c b ⋅+=⋅+⋅= ，即D 正确.故选：AD ．10.函数()()sin 0,0,2πf x A x A ωϕωϕ⎛⎫=+>>< ⎪⎝⎭的部分图象如图所示，则下列说法中正确的有（）A.2ω=B.7π,012⎛⎫-⎪⎝⎭为函数()f x 的一个对称中心点C.117π,π63⎡⎤⎢⎥⎣⎦为函数()f x 的一个递增区间D.可将函数cos2x 向右平移1π6个单位得到()f x 【答案】ABD 【解析】【分析】根据函数图像可求出A 、ω、ϕ的值，可得()f x 的解析式，利用三角函数的性质对各选项进行判断可得答案.【详解】由题可得得，1A =，2ππ2π36T ⎛⎫=⨯-=⎪⎝⎭，则2π2πω==，故A 正确；又π16f ⎛⎫= ⎪⎝⎭，所以ππ22π(Z)62k k ϕ⨯+=+∈，又π2ϕ<，所以π6ϕ=，所以()πsin 26f x x ⎛⎫=+ ⎪⎝⎭，对于B ，当7π12=-x 时，7π7ππsin 2012126f ⎛⎫⎛⎫-=-⨯+= ⎪ ⎪⎝⎭⎝⎭，所以函数图象关于点7π,012⎛⎫- ⎪⎝⎭对称，故B 正确；对于C ，由πππ2π22π,Z 262k x k k -+≤+≤+∈，可得ππππ,Z 36k x k k -+≤≤+∈，令2k =，可得5π13π36x ≤≤，所以117π,π63⎡⎤⎢⎥⎣⎦不是函数()f x 一个递增区间，故C 错误；对于D ，将函数cos2x 向右平移1π6个单位得到()πππππcos2cos 2sin 2sin 263326y x x f x x x ⎛⎫⎛⎫⎛⎫⎛⎫=-=-=-+=+= ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭，故D 正确.故选：ABD.11.已知()(),f x g x 分别是定义在R 上的奇函数和偶函数，且()()e x f x g x +=，则下列说法中正确的有（）A.()01g = B.22()()1f xg x -=C.()()()22f x f x g x =⋅ D.若()()20f m f m ++>，则1m >-【答案】ACD 【解析】【分析】()(),f x g x 分别是定义在R 上的奇函数和偶函数，由()()e x f x g x +=可得()()e xf xg x --+=，可解出e e ()2x x g x -+=，e e ()2x xf x --=，再逐个验证选项即可.【详解】函数()(),f x g x 分别是定义在R 上的奇函数和偶函数，且满足()()e xf xg x +=可得()()e x f x g x --+-=，即()()e x f x g x --+=，与()()e x f x g x +=联立，可得e e ()2x xg x -+=，e e ()2x x f x --=，()00e e 20122g +===，A 选项正确；2222e e e e 22()()1224x x x xf xg x --⎛⎫⎛⎫-+---=-==- ⎪ ⎪⎝⎭⎝⎭，故B 选项错误；()22e e 22x xf x --=，()()22e e e e e e 22222x x x x x xf xg x ----+-⋅=⨯⋅=，()()()22f x f x g x =⋅，C 选项正确；函数e e ()2x xf x --=是定义在R 上的奇函数，且在R 上单调递增，若()()20f m f m ++>，则()()()2f m f m f m +>-=-，有2m m +>-，所以1m >-，D 选项正确.故选∶ACD ．12.已知两个不相等的非零向量,a b，两组向量12345,,,,x x x x x 和12345,,,,y y y y y 均由3个a和2个b 排列而成，记1122334455min ,S x y x y x y x y x y S =⋅+⋅+⋅+⋅+⋅ 表示S 所有可能取值中的最小值，则下列命题正确的是（）A.S 有3个不同的值B.22min 22S a a b b=+⋅+ C.若//a b ，则min S 与b 无关D.若2min ||2||,4||a b S b == ，则a b⊥ 【答案】AD 【解析】【分析】求出S 的三种结果，得出min S ，对选项进行分析得出答案.【详解】2(,134.5i i x y i =，，，）均由3个a和2个b排列而成，所以1122334455S x y x y x y x y x y =⋅+⋅+⋅+⋅+⋅ 可能情况有三种︰22132S a b =+ ；2222S a a b b =+⋅+ ；234S a b a =⋅+ ，故A 选项正确；()222221223220S S S S a b a b a b a b a b-=-=+-⋅≥+-=-≥.则S 中最小为234S a b a =⋅+ ，即2min 4S a b a =⋅+ ，B 选项错误；若//a b 则2min 4S a b a =⋅+ 与b 有关，故C 选项错误；若2a b = ，222min 4444S a b a a b b b =⋅+=⋅+= ，有0a b ⋅= ，则a b ⊥ ，D 选项正确.故选：AD ．第II 卷三､填空题：本大题共4小题，每小题5分，共20分.13.已知点(1,2)A ，点(4,5)B ，若2AP PB =，则点P 的坐标是________．【答案】P （3,4）【解析】【详解】试题分析：设(),P x y ，代入2AP PB= 得()()1,224,53,3x y x y x y --=--∴==()3,3P ∴考点：向量的坐标运算14.已知()2023πsin 2023π2sin 2αα⎛⎫-=+⎪⎝⎭，则2sin2cos αα+=__________.【答案】35-##-0.6【解析】【分析】利用诱导公式化简可得tan 2α=，然后根据二倍角公式及同角关系式转化为齐次式即得.【详解】由()2023πsin 2023π2sin 2αα⎛⎫-=+ ⎪⎝⎭，得sin 2cos αα=-，则cos 0α≠，所以tan 2α=-，所以22222cos 2sin cos 12tan 143sin2cos sin cos tan 1415ααααααααα++-+====-+++.故答案为：35-.15.写出一个同时满足下列三个条件的函数()f x =__________.①()f x 不是常数函数②()1f x +为奇函数③()()22f x f x +=-【答案】cos 2x π（答案不唯一）．【解析】【分析】写出符合要求的三角函数即可【详解】分析函数的性质，可考虑三角函数，函数的对称轴为2x =，对称中心()1,0，周期可以为4，()10f =，函数解析式可以为()πcos2f x x =（答案不唯一）．故答案为：πcos2x （答案不唯一）．16.已知函数()11ππcos2cos ,,2222f x x x x ⎡⎤=--∈-⎢⎥⎣⎦（1）()f x 的值域为__________.（2）设()()3sin 4cos g x a x x =+，若对任意的1ππ,22x ⎡⎤∈-⎢⎥⎣⎦，总存在[]20,πx ∈，使得()()12f x g x =，则实数a 的取值范围为__________.【答案】①.5,14⎡⎤--⎢⎥⎣⎦②.15,,416∞∞⎛⎤⎡⎫--⋃+ ⎪⎥⎢⎝⎦⎣⎭【解析】【分析】利用倍角公式化简函数解析式，由定义域求函数值域；由题意，()g x 的值域包含()f x 的值域，分类讨论解不等式即可.【详解】()221115cos2cos cos cos 1cos 2224f x x x x x x ⎛⎫=--=--=-- ⎪⎝⎭，由,22ππx ⎡⎤∈-⎢⎥⎣⎦，有[]cos 0,1x ∈，则当1cos 2x =时，()f x 有最小值54-，当cos 0x =或cos 1x =时，()f x 有最大值1-，所以()f x 的值域为5,14⎡⎤--⎢⎥⎣⎦.()15,14f x ⎡⎤∈--⎢⎥⎣⎦，()()()3sin 4cos 5sin g x a x x a x ϕ=+=+，其中3cos 5ϕ=，4sin 5ϕ=，π0,2ϕ⎛⎫∈ ⎪⎝⎭，[]20,πx ∈，[]2,π+x ϕϕϕ+∈，因为对任意的1ππ,22x ⎡⎤∈-⎢⎥⎣⎦，总存在[]20,πx ∈，使得()()12f x g x =，所以()1f x 的值域是()2g x 的值域的子集，0a =时()0g x =不合题意，0a >时，当π+x ϕϕ+=，()g x 有最小值，则有()455sin 545+4πa a a ϕ⎛⎫=⨯-=-≤- ⎪⎝⎭，解得516a ≥，此时π2x ϕ+=时，()g x 有最大值50a >，0a <时，当π2x ϕ+=，()g x 有最小值，则有π55sin 524a a =≤-，解得14a -≤，此时π+x ϕϕ+=时，()g x 有最大值40a ->，则实数a 的取值范围为15,,416∞∞⎛⎤⎡⎫--⋃+ ⎪⎥⎢⎝⎦⎣⎭.故答案为：5,14⎡⎤--⎢⎥⎣⎦；15,,416∞∞⎛⎤⎡⎫--⋃+ ⎪⎥⎢⎝⎦⎣⎭.四､解答题：本大题6个小题，共70分，解答时应写出必要的文字说明､演算步骤或推理过程，并答在答题卡相应的位置上.17.已知平面向量,,a b c满足()()π2,0,1,,R ,,3a b c a tb t a b ===-∈= .（1）求b 在a上的投影向量的坐标；（2）当c最小时，求b 与c 的夹角.【答案】（1）1,02⎛⎫⎪⎝⎭（2）π2【解析】【分析】（1）利用投影向量的公式计算即可；（2）c a tb =- ，两边同时平方，c 最小时，求得1t =，b与c的夹角即b 与a b -的夹角，利用向量数量积计算即可.【小问1详解】由题意，||2,||1a b == ，设a e a =，b 在a 上的投影向量为11cos ,122b e a b e e ⋅⋅=⨯= ，所以b 在a 上的投影向量的坐标为1,02⎛⎫⎪⎝⎭.【小问2详解】c ====≥（1t =时等号成立），则c 最小时，c a b =- ，所以()22cos ,cos ,0b a b b a b b a a b b b c b a b b a b b a b⋅-⋅-⋅-====⋅-⋅-⋅-，因为0,π,b c ≤≤ 所以当c 最小时，b 与c 的夹角的大小为π2.法二：()ππ13332,0,cos ,sin ,,,332222a b c a b ⎛⎫⎛⎛⎫==±=±=-= ⎪ ⎪ ⎪ ⎝⎭⎝⎭⎝⎭，13332222cos ,0b c b c b c⎛⎛⨯+±⨯ ⋅==⋅ ，得所求夹角为π2.18.如图，在平面直角坐标系xOy 中，角α的终边与单位圆的交点为()11,A x y ，角π6α+终边与单位圆的交点为()22,B x y .（1）若π0,2α⎛⎫∈ ⎪⎝⎭，求12x y +的取值范围；（2）若点B 的坐标为1,33⎛⎫- ⎪ ⎪⎝⎭，求点A 的坐标.【答案】（1）32⎛⎝（2）1,66A ⎛⎫+ ⎪ ⎪⎝⎭.【解析】【分析】（1）由三角函数定义求点,A B 的坐标，根据三角恒等变换用α表示12x y +，结合正弦函数性质求其取值范围；（2）由三角函数定义可得π1π22cos ,sin 6363αα⎛⎫⎛⎫+=-+=⎪ ⎪⎝⎭⎝⎭，根据两角差正弦和余弦公式求cos ,sin αα可得点A 的坐标.【小问1详解】由题意()ππcos ,sin ,cos ,sin 66A B αααα⎛⎫⎛⎫⎛⎫++ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭，所以12π1cos sin cos 622x y ααααα⎛⎫+=++=++ ⎪⎝⎭1213πsin cos 223x y ααα⎫⎛⎫+=+=+⎪ ⎪⎪⎝⎭⎭，由π0,2α⎛⎫∈ ⎪⎝⎭，可得ππ5336π,α⎛⎫+∈ ⎪⎝⎭，所以π1sin ,132α⎛⎫⎛⎤+∈ ⎪ ⎥⎝⎭⎝⎦，所以12x y +的取值范围是2⎛ ⎝.【小问2详解】由1,33B ⎛⎫- ⎪ ⎪⎝⎭，得π1π22cos ,sin 6363αα⎛⎫⎛⎫+=-+= ⎪ ⎪⎝⎭⎝⎭，ππππππcos cos cos cos sin sin 666666αααα⎡⎤⎛⎫⎛⎫⎛⎫=+-=++ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎝⎭⎣⎦，11cos 32α⎛⎫=-=⎪⎝⎭ππππππsin sin sin cos cos sin 666666αααα⎡⎤⎛⎫⎛⎫⎛⎫=+-=+-+ ⎪ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎝⎭⎣⎦，11sin 332α⎛⎫=-⨯= ⎪⎝⎭所以点A 的坐标为1,66⎛⎫⎪ ⎪⎝⎭.19.已知平面向量,OM ON 不共线，由平面向量基本定理知，对于该平面内的任意向量OP，都存在唯一的有序实数对(),x y ，使得OP xOM yON =+.（1）证明：,,P M N 三点共线的充要条件是1x y +=；（2）如图，ABC 的重心G 是三条中线,,AD BE CF 的交点，证明：重心为中线的三等分点.【答案】（1）证明见解析；（2）证明见解析.【解析】【分析】（1）根据共线向量基本定理结合充要条件的概念即得；（2）根据向量共线定理及推论可得()1AG y AB y AE =-+ ，AG AD λ=，进而23AG AD =，即证；或利用平面几何知识即得.【小问1详解】证明：必要性，,,P M N 三点共线，不妨设MP yMN =，可得()OP OM y ON OM -=- ，()1OP y OM yON =-+，又OP xOM yON =+ ，所以1x y =-，得1x y +=，得证；充分性：,1OP xOM yON x y =++=，()1OP y OM yON ∴=-+，即()OP OM y ON OM -=- ，MP yMN ∴= ，又MP 与MN有公共点M ，所以,,P M N 三点共线；所以,,P M N 三点共线的充要条件是1x y +=；【小问2详解】法一（向量法）ABC 的重心G 是三条中线,,AD BE CF 的交点，可设()1AG y AB y AE =-+ ，111222AD AB AC AB AE =+=+，因为,,A G D 三点共线，可设AG AD λ=，则()1y AB y AE -+ 2AB AE λλ=+，所以12y y λλ⎧-=⎪⎨⎪=⎩，解得23y λ==，所以23AG AD =，G ∴为AD 的三等分点，同理可证G 为,BE CF 的三等分点，∴重心为中线的三等分点.法二（几何法）：连接EF ，,E F 为,AC AB的中点，1//,2EF BC EF BC ∴=，12EF FG EG BC GC GB ∴===，所以13FG EG FC EB ==，同理可得13EG DG EB DA ==，所以重心为中线的三等分点.20.已知向量cos ,sin ,cos ,sin 22222x x x x x a b ⎛⎫⎛⎫=+=- ⎪ ⎪⎝⎭⎝⎭ ，函数()f x a b =⋅ .（1）求函数()f x 的单调增区间和对称轴；（2）若关于x 的方程()0f x m -=在0,2π⎡⎤⎢⎥⎣⎦上有两个不同的解，记为,αβ.①求实数m 的取值范围；②证明：()2cos 12m αβ-=-.【答案】（1）ππ22,2,Z 33ππk k k ⎡⎤-+∈⎢⎥⎣⎦，对称轴为3ππ，Zx k k =+∈（2）①)2；②证明见解析【解析】【分析】（1）根据向量点乘和三角函数恒等变换公式化简()f x ，利用整体代入法计算出单调增区间和对称轴；（2）根据()f x 范围求实数m 的取值范围；根据,αβ是()0f x m -=两个不同解可知()()f f αβ=，根据图象可得2π23αβα-=-，利用倍角公式计算即可.【小问1详解】()πcos cos sin sin cos 2sin222226x x x x x f x x x x ⎛⎫⎛⎫⎛⎫=++-=+=+ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭，令πππ2ππ2π2π,,2π2π,Z 26233k x k k Z k x k k -≤+≤+∈-≤≤+∈此时函数()f x 单调递增，∴函数()f x 单调递增区间为ππ22,2,Z 33ππk k k ⎡⎤-+∈⎢⎥⎣⎦.令πππ62x k +=+得()ππ,Z 3x k k =+∈，所以函数()f x 的对称轴为()ππ,Z 3x k k =+∈；【小问2详解】①π0,2x ⎡⎤∈⎢⎥⎣⎦ ，ππ2π,663x ⎡⎤∴+∈⎢⎥⎣⎦，由图象分析得()f x m =，有两个不同的解，则3ππsin 1,2sin 2266x x ⎛⎫⎛⎫≤+<≤+< ⎪ ⎪⎝⎭⎝⎭，)2m ∴∈.②因为,αβ是方程π2sin 6x m ⎛⎫+= ⎪⎝⎭的两个根，所以ππ2sin ,2sin 66m m αβ⎛⎫⎛⎫+=+= ⎪ ⎪⎝⎭⎝⎭，由图象分析得，2π2π2π,,2333αββααβα+==--=-，()2222πππcos cos 2cos 22sin 121133622m m αβααα⎛⎫⎛⎫⎛⎫⎛⎫-=-=-+=+-=⨯-=- ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭.21.已知R a ∈，函数()()22log 3f x x x a =-+.（1）若函数()f x 的图象经过点()3,1，求不等式()1f x <的解集；（2）设2a >，若对任意[]3,4t ∈，函数()f x 在区间[],1t t +上的最大值与最小值的差不超过1，求a 的取值范围.【答案】（1）{01xx <<∣或23}x <<；（2）[)4,+∞.【解析】【分析】（1）将点()3,1代入()()22log 3f x x x a =-+可求出a ，然后根据函数的单调性即得；（2）由复合函数的单调性知()()22log 3f x x x a =-+在区间[],1t t +上单调递增，进而得到最大值与最小值，再由题可得252a t t ≥-+-对任意[]3,4t ∈恒成立，构造新函数，求最值可得出答案.【小问1详解】由题可得()()223log 3331f a =-⨯+=，解得2a =，即()()22log 32f x x x =-+由()()222log 321log 2f x x x =-+<=，可得22320322x x x x ⎧-+>⎨-+<⎩，解得01x <<或23x <<，所以不等式()1f x <的解集为{01x x <<∣或23}x <<；【小问2详解】因为()()22log 3f x x x a =-+是复合函数，设()23p x x x a =-+，()2log ()f x p x =，因为[]3,4t ∈，()23p x x x a =-+在区间[],1t t +单调递增，()2log ()f x p x =单调递增，故函数()f x 在区间[],1t t +上单调递增，又2a >，所以()223390p x x x a a a =-+>-+=>，所以()()max min ()1,()f x f t f x f t =+=，由题意，()()11f t f t +-≤，即()()2222log (1)31log 23t t a t t a ⎡⎤+-++≤-+⎣⎦，对任意[]3,4t ∈恒成立，故()()22(1)3123t t a t t a +-++≤-+，对任意[]3,4t ∈恒成立，整理得：252a t t ≥-+-，令()252g t t t =-+-，[]3,4t ∈，只需max ()g t a ≤即可，因为()252g t t t =-+-的对称轴为52t =，图象是开口向下的抛物线，故()252g t t t =-+-在[]3,4t ∈上单调递减，故()max ()34g t g ==，所以4a ≥，即a 的取值范围是[)4,+∞.22.设n 次多项式()()1211210,0nn n n n n T x a x a xa x a x a a --=+++++≠ ，若其满足()cos cos n T n θθ=，则称这些多项式()n T x 为切比雪夫多项式.例如：由2cos22cos 1θθ=-可得切比雪夫多项式()2221T x x =-.（1）求切比雪夫多项式()3T x ；（2）求sin18 的值；（3）已知方程38610x x --=在()1,1-上有三个不同的根，记为123,,x x x ，求证：1230x x x ++=.【答案】（1）()3343T x x x=-（2）51sin184-=（3）证明见解析【解析】【分析】（1）根据两角和余弦公式和二倍角余弦公式利用cos θ表示cos3θ，由此可得()3T x ；（2）由诱导公式可得cos54sin36= ，根据（1）和二倍角正弦公式和平方关系可求sin18 ；（3）方法一：由已知314302x x --=，设cos x θ=，由（1）可求θ，再根据两角和差余弦公式证明1230x x x ++=；方法二：由已知()()()3123143402x x x x x x x x --=---=，根据整式性质可得1230x x x ++=.【小问1详解】因为()cos3cos 2cos2cos sin2sin θθθθθθθ=+=-所以()()2232cos32cos 1cos 2sin cos 2cos cos 21cos cos θθθθθθθθθ=--=---所以3cos34cos 3cos θθθ=-，所以()3343T x x x =-；【小问2详解】因为cos54sin36= ，所以34cos 183cos182sin18cos18-= ，又cos180> ，所以24cos 1832sin18-= ，所以()241sin 1832sin18--=即24sin 182sin1810+-= ，因为sin180> ，解得1sin18,4-=（14-舍去）；【小问3详解】由题意，314302x x --=，法一：设cos x θ=，代入方程得到3114cos 3cos 0cos322θθθ--=⇒=，解三角方程得ππ32π,32π,Z 33k k k θθ=+=-+∈，不妨取123π5π7π,,999θθθ===，123π5π7ππ4π2πcoscos cos cos cos cos 999999x x x ⎛⎫++=++=-+ ⎪⎝⎭，而4π2π3ππ3πππcoscos cos cos cos 9999999⎛⎫⎛⎫+=++-= ⎪ ⎪⎝⎭⎝⎭，综上1230x x x ++=.法二：令()()()3123143402x x x x x x x x --=---=即()()323123122313123144302x x x x x x x x x x x x x x x x x ⎡⎤-+++++-=--=⎣⎦依据多项式系数对应相等得到1230x x x ++=.综上1230x x x ++=.【点睛】“新定义”主要是指即时定义新概念、新公式、新定理、新法则、新运算五种，然后根据此新定义去解决问题，有时还需要用类比的方法去理解新的定义，这样有助于对新定义的透彻理解.但是，透过现象看本质，它们考查的还是基础数学知识，所以说“新题”不一定是“难题”，掌握好三基，以不变应万变才是制胜法宝.第21页/共21页。
上海市育才中学2024-2025学年高一上学期第一次月考英语试题(含答案)
英语阶段性练习（Unit1Book1）（65分钟）I. Listening ComprehensionSection A1. A. At a bus station. B. At a gas station.C. At a military museum.D. At a mobile phone shop.2. A.3. B.5. C.8. D.11.3. A. He had difficulty going to sleep at night.B. He stayed up late watching TV series.C. He produced an attractive TV series.D. He had to buy a new sleeping pillow.4. A. The woman should clean up the closet.B. The woman has bought too many new clothes.C. The woman should put on some weight.D. The woman should buy more clothes.5. A. The sunlight will burn up the jacket. B. The beach is too cold.C. She is proud of her dark skin color.D. The sunlight is helpful to her.6. A. Many activities were arranged in the bookshop.B. Many people were listening to a writer’s lecture.C. The bookshop was popular because of the book resources.D. People went to the bookshop to get free books.7. A. Customers can pay cash. B. There are more varieties of goods.C. Customers can get a bargain price.D. Customers can receive the goods faster.8. A. Peter isn’t in his hometown himself.B. Peter isn’t their good friend.C. Peter is unlikely to appear to pick them up.D. Peter owns an empty house in his hometown.9. A. Do a part-time job on campus.B. Borrow money from the woman.C. Obtain financial assistance from school.D. Study a practical case on campus.10. A. The man had a bad time eating with his boss.B. The man had a stomachache after lunch.C. The man enjoys eating with co-workers.D. The man is particular about the food he eats.Section BQuestions 11 through 13 are based on the following passage.11. A. Users can choose height range.B. Users can improve their eyesight.C. Users can select the SPF of the cream.D. Users can get a full exposure to the cream.12. A. The laziness of the inventor. B. The needs of the hotels.C. The fund the inventor obtained.D. The online usage video.13. A. Inserting coins into the equipment.. Paying cash to the inventor.C. Buying Snappy Screen card.D. Paying for the hotel room.Questions 14 through 16 are based on the following passage.14. A. He got a medal for courage. B. He would become a French citizen.C. He got a position in the fire service.D. He received a large sum of money.15. A. He was playing a game. B. He was sleeping in the balcony.C. He was staying with his family.D. He was injured.16. A. The balcony was not strong enough.B. The boy's father left him alone at home.C. The house was on fire.D. Firemen broke into the house.Questions 17 through 20 are based on the following conversation.17. A. The origin of AI technology.B. The bad influences AI might bring.C. The changes AI brings to humans.D. The waterproof feature of AI technology.18. A. Too few people he has interviewed.B. The inferior AI products on the market.C. The lack of resources on AI theory.D. The lack of fund for the study.19. A. It is located in the downtown area.B. It has professional staff.C. It offers great online resource.D. It opens to general public free of charge.20. A. To prove that the applicant has legal income.B. To guarantee applicant’s ability to pay overdue fine.C. To use the bank card as the membership card.D. To pay the membership fee.II. VocabularyA. aliveB. ashamedC. clipD. livingE. keenF. mattersG. objectH. recordingI. reflectionJ. routine K. typicalIn today's digital era, the act of clipping and recording moments from our lives has become a ___31___ . It's not just about preserving memories; it's a way to keep those moments __32___. What truly _____ 33 _____ is theability to reflect on these recordings later, revisiting those _____ 34_____ days filled with joy or sorrow.Many are _____ 35_____ on capturing every _____ 36_____ and view that catches their eye, whether it is a sunset's colors or a child's innocent laughter. Such practices not only enrich our personal archives but also serve as reminders of the beauty and complexity of life.Occasionally, when we run across an old _____ 37_____ in the phone or computer, it can stimulate emotions we thought were long forgotten. It's in these moments that we realize how __38___ we might have been to let certain experiences go unnoticed. ____39___ , therefore, becomes a powerful tool for self- ____40___ and appreciation of the present, ensuring that our stories remain vivid and our memories forever fresh.III. ClozeThe United Nations is an international organization founded in 1945 and committed to maintaining international peace and security, developing friendly relations among nations; promoting social progress, better living standards and human rights.Due to its unique international ____41______ , the Organization takes action on a wide range of issues, and provides a forum for its 193 Member States to express their views, through the General Assembly, the Security Council, the Economic and Social Council and other bodies and committees.The work of the United Nations ___42___ every comer of the globe and focuses on a broad range of fundamental issues, like sustainable development, protection of the environment and refugees and disaster relief.In September 2015, world leaders gathered at the United Nations Headquarters in New York and adopted the 2030 Agenda for Sustainable Development, a new universal standard for development that will ____43____ no one is left behind. 17 Sustainable Development Goals（SDGs）___ 44___ the cornerstone of the historic Agenda. Building on the success of the Millennium Development Goals （MDGs）, the SDGs aim to go further to end all forms of poverty. Universal and indivisible, the SDGs call for action by all countries -- developed, developing and middle-income - to promote ___45___ while protecting the planet over the next fifteen years.The SDGs officially came into ___46__ on 1 January 2016. Gove mm ents, businesses and civil society have started to make efforts to achieve the goals, by adopting plans to achieve them. If you are ___47___ to take concrete action for global peace and development, take a look at the United Nations Volunteers （UNV）programme.Every year, up to 8,000 qualified and experienced women and men of some 160 different nationalities volunteer at least six months of their lives to help others. These UN Volunteers work in about 130 countries promoting peace, __48__ to disasters, empowering communities and helping to build sustainable livelihoods and lasting development.UN Volunteers come from ___49__ professional backgrounds but all of them are pushing for the positive changes. They are ____50___ to be creative and entrepreneurial, and foster volunteer is m for peace and development both within and beyond their assignments. They work at the heart of communities in __51____ with governments, United Nations entities ad civil society.Being a UN Volunteer is not a career, but it is rich with opportunities and experience and offers huge personal ____52____ .As a UN Volunteer, you receive a Volunteer Living Allowance（津贴）（VLA）, which ___ 53 ___ basic needs, housing and utilities. ___ 54 __, UNV will provide a settling-in-grant, life, health, and permanentdisability __55___, return airfares and a nominal resettlement allowance.41. A. character B. quantity C. cooperation D. celebrity42. A. moves B. reveals C. touches D. passes43. A. assure B. set C. measure D. ensure44. A. undergo B. sign C. represent D. reduce45. A. poverty B. prosperity C. property D. popularity46. A. force B. time C. usage D. attention47. A. delivered B. organized C. challenged D. inspired48. A. progressing B. responding C. applying D. serving49. A. similar B. unique C. typical D. various50. A. encouraged B. developed C. entitled D.made51. A. partnership B. citizenship C. relationship D. hardship52. A. awards B. presents C. rewards D. supplies53. A. takes B. covers C. engages D. respects54. A. However B. Therefore C. Nonetheless D. Furthermore55. A. insurance B. guarantee C. maintenance D. assuranceⅣ. Reading（A）‘A cannon（大炮）sounded as Jonny Fry took off on his horse through the streets of St. Joseph, Missouri. Crowds lined the streets, cheering and waving. Fry rode west for 90 miles at full speed. He then passed a leather bag full of letters to another rider. Eventually a chain of riders carried the letters across nearly 2,000 miles of wilderness, and the Pony Express was born.Between California and the rest of the U.S. lay a vast wilderness. Mail service was slow and often delayed. As war among the states seemed likely, people wanted news more quickly.In December 1859, William Russell, owner of a transport company, decided to meet that need. He and partners William Waddell and Alexander Majors worked constantly for months. They selected a route from St. Joseph, Missouri, to Sacramento, California. Then they purchased 400 horses, hired workers and set up relay stations （导站）every 10 to 15 miles. Riders would change horses at these stations, and every 90 to 120 miles a new rider would take over.Shortly before Jonny Fry set out, another Pony Express rider set off from Sacramento for St. Joseph. These first deliveries reached their destinations in close to the 10 days Russell had promised.But the route was far from safe. On April 18, 1860, a rider’s horse stumbled （绊倒）and fell on the rider, killing him. Riders faced rivers, mountains, extreme temperatures-and very unfriendly natives.In May 1860,tensions between the settlers and Native Americans turned into fighting. Native Americans attacked several Pony Express stations, forcing some of them to close temporarily.Despite the risks, the Pony Express lost only one delivery during its one and a half years in operation. And when Abraham Lincoln was elected president, the Pony Express carried the news to California in only five days. It also spread news of the beginning of America's Civil War.The Pony Express service ended in October 1861 after a telegraph line across the U.S. was completed.Yet the riders' courage and accomplishments still hold a place in legend（传奇）.56. The passage is mainly about the _____ of the Pony Express.A. functionB.influenceC. historyD. operation57. To _____, William Russell decided to set up the Pony Express.A. shorten the time of sending mailB. deliver mail farther and more safelyC. meet people's different needsD. help people keep in contact in war58. What might Russell have promised about the first deliveries?A. To make the destinations closer.B. To make the deliveries within 10 days.C. To ensure the deliveries with two riders.D. To complete the deliveries before the war.59. What can be learned from the passage?A. The Pony Express had an insignificant position in the development of mail service.B. The fighting between the settlers and Native Americans ended the Pony Express.C. The Pony Express became successful as a result of the dangers on the route.D. The Pony Express not only delivered mail but spread news as well.（B）Please Choose Cloud ServicesFinding the Cloud service can be difficult as there are so many available.Consumers need to consider and compare each service’s features, storage possibilities, free space and best price. You may be unsure, or you do not know where to begin. Trying every Cloud service would take a lot of time and work. But, the website might help.CloudDrop-box Google-Drive One-DriveBoxBest For LightweightUsersTeams AndCollaborationDevotedWindowsUsersEnterpriseSolutionsFree Storage Space2GB15GB15GB10GBCheapest PremiumOption1TB100GB100GB100GB One valuable feature is the protection of digital files, including photos, videos, documents, music and more. If the worst happens and your computer crashes, or gets lost or damaged, your files can still be found in the Cloud.Cloud services let you access your files from almost anywhere. No need to worry about a file stored on your computer when you are away from your desk. If you have an Internet connection, you can open your files from any computer, or from a phone, when they are stored using a Cloud service.Using a Cloud service makes sharing files easier. If you send documents to a group using emails, you may then wonder which version everyone is working with. When sharing files using a Cloud service, everyone will see the same document and the latest version.The first thing to consider is the amount of storage space you need. Check how much space you are currently using on your computer and mobile devices. If your computer or phone is filled with photos and you have little free space, you may want to move your photos to the Cloud. Some Cloud services are free. But, if you need a lot of space, you will probably need one that costs money.Even if everyone you have ever met is a Windows user, you still probably want a Cloud service that works with many platforms. You might become friends with anAndroid user or start a job with a company that computes on Apple!60. Which of the following is true according to the passage?A. Google-Drive users can obtain 100GB for free.B. Most Cloud services can send files to both the Cloud and the local computer.C. It's hard to transfer your file when you turn on the computer if it is stored in the Cloud.D. Never worry about finding your files in the Clouds if the computer system crashes.61. What will you have to do if you need more space to store more files in the Cloud?A. To use your own Cloud services.B. To share Cloud space with others.C. To spend money buying space.D. To buy the latest version of the service.62. How can consumers find the Cloud service that suits them best?A. By learning more information about the space occupied by files.B. By comparing Cloud’s function, storage, available space and prices.C. By sending files to a group of people via e-mail inquiring about the Cloud service.D.By making friends with Android users who use Apple computers.（C）Every day, Lucy Dong and her best friend Amy Zhu wake at 7 a.m. -- 7.10 a.m.If they are lucky---rush through their breakfast of steamed buns and noodles, and head off to what may be the best schooling system in the world.The 10-year-olds, who are natives of Shanghai, study in 35-minute bursts from around 8 a.m. to 4 p.m., with a small break for lunch -- and a class meeting --- sandwiched in the middle.Outside school hours, the girls' lives are lots of extra-curricular activities: English class, flute class, drumming class, handwriting class, calligraphy class, Taekwondo training, modelling lessons and choir practice.Shanghai was crowned --- for the second time …the champion of the Programme for International Student Assessment （PISA., which compares the maths, reading and science skills of some 510,000 secondary school students around the world）.Some experts question the value of comparing cities and countries. Others point out that Shanghais relatively well-funded schools and well-paid teachers are not representative of the Chinese education system as a whole.Even so, the latest results are likely to see more and more educators flock east in search of the mega-city's magic formula.Professor Kong Lingshuai of the College of Education at Shanghai Normal University has studied the city'sPISA successes. He says that the secret is a mix of "traditional elements and modern elements". The former is related to the high expectations of "tiger" parents, and a belief in Chinese children from a young age that effort is crucial to gaining a good education. The "modern elements" include Shanghai's willingness to constantly adapt its curriculum and teaching practices; its focus on improving under-achieving schools by pairing them with those that excel; its openness to foreign ideas; and the introduction of performance-related pay.An obsession with training has also been the key, says Prof. Kong. As of last year, new teachers have to undergo a standardized, one-year training course before starting in the classroom. Once qualified, they are required to complete at least 240 hours' training in their first five years, including online learning, paper reading, essay writing and so on. Teachers are also encouraged to attend each other's classes to promote a culture of "idea sharing, exchanging and positive competition".Outsiders often dismiss China's education system as a pressure-cooker-style craze of exams that places too much emphasis on rote-learning and does little to stimulate creativity. But in Shanghai at least, that may be starting to change. Authorities are attempting to move away from testing that relies too heavily on memorizing facts and figures, and some schools are also giving students more time to play, rather than just study.63. The author mentions Lucy Dong and Amy Zhu in the first 3 paragraphs to ________.A. praise the industriousness of these two girlsB. give readers the whole picture of Chinese educationC. illustrate what education in Shanghai looks likeD. criticize the burden the education brings on them64. Why do some experts challenge the result that Shanghai ranked 1st in PISA test?A. Because the value of PISA, which only tests 510,000 students globally, is cast doubt on.B.Because Shanghai has drawn more attention and resources from the central government.C. Because only maths, reading and science skills are compared in PISA test, which is not enough.D. Because Shanghai has a better system of financial support for schools as well as for teachers.65. Which of the following is closest in meaning to "dismiss"in the last paragraph?A. To consider something inferior.B. To stop thinking about something.C. To give the credit for something.D. To claim the features of something.66. Which of the following statements is true according to the passage?A. Strict parents contribute little to the academic success of their children.B. Students have more free time to spend on their interest across China.C. Many education researchers are getting to Shanghai to study the phenomenon.D. Under-achieving schools are usually substituted by those academically superior schools.（D）A. Holidays are when my family gathers and has a good time together,B. School education has civilized me with a mind for peace and calm.C. My parents raised me with a lot of beliefs that have had an effect on who I am today.D. Summer is the time when we always organize some special events as a big family.E. However, I know that it is important to follow the golden rule and to treat others with dignity and respect.F. We take care of each other and try to treat others with respect.What Makes Me "Me"?My culture comes from where I am from and where I was raised. I am an American.I was born and raised in Alaska. My ethnic background is German, English, and Scottish. Culture is all about family. It is a family's belief systems, the cultural traditions that are celebrated, and the special holidays and events that occur in the year._____67_____These are the beliefs that don't just belong to my culture, but they come straight from the values of my family. I believe that women are more intelligent than men because my mother is a strong woman. I was raised to believe that through hard work I could become successful and have a good life. I strive to do this every day and to be good at my life and my jobs. I'm not really religious. _____ 68_____ I believe that if you are a good person to others you will be rewarded for it in many ways.There are a lot of cultural traditions that my family has. Some are from my cultural background and some are from the place where I was raised. I raise my son to have manners and to be well behaved. I raise him to know that hard work is important and that he is expected to put his best effort into things._____69_____We gather every summer in Haines so that all the cousins can grow up together. Birthdays were always important events while growing up. My mom was excited to celebrate every holiday. Trick or treating on Halloween, hiding baskets on Easter, passing out cards to classmates on Valentine's Day, wearing green on St. Patrick's Day, and of course loads of presents under the tree on Christmas morning. My parents always made holidays a big deal so now my son also gets to be excited for them every year.My culture makes me a person that celebrates life. My family has a lot of good friends and we are usually a good part of the community. ______70____ We like to go outdoors and have adventures. My belief systems, cultural traditions, and special events have made me a happy person who is fun to be around and who can have a positive impact on others.Keys31-35 JAFKE36-40 GCBHICloze41-45 ACDCB46-50 ADBDA 51-55 ACBDAReadingA:56-59 CABD B:60-62 DCB C:63-66 CDAC D:67-70 CEAF。
2024届重庆育才中学高三下学期3月联考(文理)数学试题
2024届重庆育才中学高三下学期3月联考（文理）数学试题注意事项：1．答卷前，考生务必将自己的姓名、准考证号填写在答题卡上。
2．回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其它答案标号。
回答非选择题时，将答案写在答题卡上，写在本试卷上无效。
3．考试结束后，将本试卷和答题卡一并交回。
一、选择题：本题共12小题，每小题5分，共60分。
在每小题给出的四个选项中，只有一项是符合题目要求的。
1．△ABC 中，AB ＝3，BC 13=，AC ＝4，则△ABC 的面积是（ ）A ．33B ．332C ．3D ．322．函数的图象可能是下面的图象( )A ．B ．C ．D ．3．在101()2x x-的展开式中，4x 的系数为( ) A ．-120B ．120C ．-15D ．154．设复数z 满足12z zz +=+，z 在复平面内对应的点的坐标为(),x y 则（ ） A ．221x y =+ B ．221y x =+ C ．221x y =-D ．221y x =-5．已知向量,a b 满足||1,||3a b ==,且a 与b 的夹角为6π,则()(2)a b a b +⋅-=( ) A ．12B ．32-C ．12-D ．326．已知正三角形ABC 的边长为2，D 为边BC 的中点，E 、F 分别为边AB 、AC 上的动点，并满足2AE CF =，则DE DF ⋅的取值范围是（ ） A ．11[,]216- B ．1(,]16-∞ C ．1[,0]2-D ．(,0]-∞7．已知抛物线2:2(0)C y px p =>的焦点为F ，过点F 的直线l 与抛物线C 交于A ，B 两点（设点A 位于第一象限），过点A ，B 分别作抛物线C 的准线的垂线，垂足分别为点1A ，1B ，抛物线C 的准线交x 轴于点K ，若11||2||A KB K =，则直线l 的斜率为 A ．1B ．2C ．22D ．38．若实数,x y 满足不等式组2,36,0,x y x y x y +≥⎧⎪-≤⎨⎪-≥⎩则3x y +的最小值等于（ ）A ．4B ．5C ．6D ．79．定义在R 上的奇函数()f x 满足()()330f x f x --+-=，若()11f =，()22f =-，则()()()()1232020f f f f ++++=（ ）A ．1-B ．0C ．1D ．210．已知函数()ln f x x =，()()23g x m x n =++，若()0,x ∀∈+∞总有()()f x g x ≤恒成立.记()23m n +的最小值为(),F m n ，则(),F m n 的最大值为（ ）A ．1B ．1eC ．21eD ．31e11．已知棱锥的三视图如图所示，其中俯视图是等腰直角三角形，则该三棱锥的四个面中，最大面积为（ ）A ．22B ．23C ．4D ．2612．已知m ，n 为异面直线，m ⊥平面α，n ⊥平面β，直线l 满足l ⊥m ，l ⊥n ，,l α⊄,l β⊄则 （ ）A ．α∥β且l ∥αB ．α⊥β且l ⊥βC ．α与β相交，且交线垂直于lD ．α与β相交，且交线平行于l二、填空题：本题共4小题，每小题5分，共20分。
安徽省滁州市定远县育才学校2021-2022学年高二(上)第三次月考物理试题(普通班)(解析版)
C．复印机复印文件资料，就是利用静电墨粉成在纸上，属于静电利用，C错误；
D．静电除尘时除尘器中的空气被电离，烟雾颗粒吸附电子而带负电，颗粒向电源正极移动，属于静电应用，D错误。
C. 滑动变阻器消耗的功率P最大时，电源输出功率也最大
D. 无论如何调整滑动变阻器Rr的阻值，都无法使电源的输出功率达到4.5W
【10题答案】
【答案】ABD
【解析】
【详解】A.将 看成电源的内阻，据外电路电阻等于电源内阻时电源输出功率最大知，当等效电源的内外电阻相等时滑动变阻器消耗的功率最大，由图乙知，当 时，滑动变阻器消耗的功率 最大，最大功率为
B. 由x1到x3电场强度减小
C. 由x1运动到x4的过程中电势能一直增大
D. 由x1运动到x4的过程中电场力先减小后增大
【3题答案】
【答案】C
【解析】
【详解】A．从x1向右场强为 轴负方向，则从 到 处逆着电场线方向移动，电势升高，故A错误；
B．图像中E的正负只表示方向，不代表大小，根据图像由 到 电场强度增加，故B错误；
B．由于油滴先向下减速，后向上加速，故电场力大于重力，故从P到Q，合力对其做正功，动能增大，故油滴在Q点的动能比它在P点的大，B正确；
C．同理从P到Q，电场力对其做正功，其电势能减小，故油滴在Q点的电势能比它在P点的小，C错误；
D．由于油滴所在电场为匀强电场，故所受的电场力为恒力，所受合力为恒力，故加速度处处相等，D错误。
B．在B处放置一个电荷量为 的点电荷，负电荷在O点产生的电场强度的方向也是从O点指向B点，所以它们的合场强也不等于零，B错误；
2022-2023学年宁夏育才中学高一下学期3月月考地理试题
2022-2023学年宁夏育才中学高一下学期3月月考地理试题1. 读世界人口密度分布示意图，完成下面小题。
1．世界人口最稠密的地区分布于（）A．中高纬度近海的高原地区B．中低纬度近海的平原地区C．中高纬度内陆的河谷地区D．中低纬度河谷的雨林地区2．图中甲地区人口密度每平方千米不足10人，其影响的主要原因是（）A．地势高峻崎岖B．气候炎热干旱C．战争频繁发生D．生育率较低2. 下图为影响人口迁移的主要因素示意图，读图完成下面小题。
1．我国东南沿海地区吸引大量民工迁入的主要原因是（）A．①B．②C．③D．⑥2．近期，乌克兰居民大量外迁的主要原因是（）A．①B．②C．④D．⑤3. S国位于亚洲，石油储量和产量均居世界前列，是世界上最富裕的国家之一。
2020年该国人口总数约为3481万，男女比例严重失衡。
下图为该国2020年人口结构统计图。
据此完成下面小题。
1．该国男女比例严重失衡，主要原因是（）A．老龄化严重B．科技发达C．生育政策D．资源开发2．该国最可能位于（）A．东亚B．南亚C．西亚D．中亚4. “胡焕庸线”是我国人口分布重要分界线。
读我国人口密度及胡焕庸线示意图，完成下面小题。
1．我国人口分布说法正确的是（）A．平原多山地少B．东北多西南少C．内陆多沿海少D．农村多城市少2．乙省环境人口容量小的主要影响因素是（）A．科技水平B．水源C．对外开放程度D．受教育程度5. 人口净流入是指常住人口与户籍人口的差值，常住人口大于户籍人口，说明该城市处于净流入，反之则为净流出。
图为我国部分重点城市近年人口年均净流入情况。
完成下面小题。
1．图中数据表明（）①中西部城市人口回流明显②东部城市人口以净流出为主③东部城市出现人口再集聚④中西部城市人口以净流出为主A．①②B．①③C．③④D．②④2．对比图中人口流动数据，下列说法正确的是（）A．西安市场商品流通变慢B．长沙人口压力得到缓解C．南昌经济发展后劲优势明显D．杭州人口流入可缓解老龄化6. 人户分离是指中华人民共和国境内公民的经常居住地和常住户口登记地二者不一致，“人户分离”现象包括“有户（籍）无人”和“有人无户（籍）”两种形式。
2022-2023学年吉林省白城市通榆县育才学校、四井子学校三年级(下)月考数学试卷(3月份)
2022-2023学年吉林省白城市通榆县育才学校、四井子学校三年级（下）月考数学试卷（3月份）一、仔细填一填。
（每空1分，共29分）1．（4分）根据每组第一个算式，直接写出其他各式的得数。
（1）6÷3＝260÷3＝600÷3＝（2）20÷2＝10200÷2＝2000÷2＝2．（6分）填表。
算式商的位数估算的结果准确值901÷9282÷73．（6分）在横线里填上“＞”“＜”或“＝”。
69÷3 20160÷4 45350÷5 210÷380÷4 30960÷3 320640÷8 64÷8 4．（2分）360是6的倍，4的倍是420。
5．（2分）黄山有“天下第一奇山”“天开图画”“松海云川”之称。
君君和爸爸一起登上黄山山顶观看日出，当他们面向太阳升起的方向，他们的左边是方，右边是方。
6．（5分）看图填一填。
（1）小猴家在小马家的面，小鹿家在小马家的面。
（2）小猴家在水池的方向，小兔家在水池的方向，小鹿家在水池的方向。
7．（1分）赵叔叔骑车旅行，5天一共骑行448千米。
赵叔叔平均每天大约骑行千米。
8．（2分）一个水杯9元，王阿姨带了132元，最多可以买个这样的水杯，还剩元。
9．（1分）在互联网技术高速发展的背景下，地方政府“直播带货”体现了“互联网+农业”的新发展理念：某村村长直播带货，为当地的带销农产品打开了销路，9分钟一共可以卖出千克菠萝。
二、公正判一判。
（对的打“√”，错的打“x”）10．（1分）三位数除以一位数，商最多是两位数。
11．（1分）东与西相对，北与南相对。
12．（1分）0除以任何不是0的数，都得0。
13．（1分）笔算除法时，除的过程中每一步的余数必须小于除数。
14．（1分）在□÷8＝13……△中，当△最大时，□是112。
宁夏回族自治区银川市宁夏育才中学2025届高三月考3物理试题含解析
宁夏回族自治区银川市宁夏育才中学2025届高三月考3物理试题注意事项：1．答题前，考生先将自己的姓名、准考证号填写清楚，将条形码准确粘贴在考生信息条形码粘贴区。
2．选择题必须使用2B铅笔填涂；非选择题必须使用0．5毫米黑色字迹的签字笔书写，字体工整、笔迹清楚。
3．请按照题号顺序在各题目的答题区域内作答，超出答题区域书写的答案无效；在草稿纸、试题卷上答题无效。
4．保持卡面清洁，不要折叠，不要弄破、弄皱，不准使用涂改液、修正带、刮纸刀。
一、单项选择题：本题共6小题，每小题4分，共24分。
在每小题给出的四个选项中，只有一项是符合题目要求的。
1、我国计划在2020年发射首个火星探测器，实现火星环绕和着陆巡视探测。
假设“火星探测器”贴近火星表面做匀速圆周运动，测得其周期为T。
已知引力常量为G，由以上数据可以求得（）A．火星的质量B．火星探测器的质量C．火星的第一宇宙速度D．火星的密度2、甲、乙两车某时刻由同一地点、沿同一方向开始做直线运动。
甲、乙两车的位置x随时间t的变化如图所示，则（）A．0时刻，甲车速度比乙车小B．t2时刻，甲乙两车速度相等C．0~t1时间内，甲车的平均速度比乙车大D．0~t2时间内，甲车通过的距离大3、光滑水平面上，一质量为m的滑块以速度v与质量为M的静止滑块相碰，碰后两者粘在一起共同运动。
设碰撞过∆。
下列说法正确的是（）程中系统损失的机械能为E∆变大A．若保持M、m不变，v变大，则E∆变小B．若保持M、m不变，v变大，则E∆变小C．若保持m、v不变，M变大，则E∆变小D．若保持M、v不变，m变大，则E4、为了人民的健康和社会的长远发展，我国环保部门每天派出大量的洒水车上街进行空气净化除尘，已知某种型号的洒水车的操作系统是由发动机带动变速箱，变速箱带动洒水泵产生动力将罐体内的水通过管网喷洒出去，假设行驶过程中车受到的摩擦阻力跟其质量成正比，受到的空气阻力跟车速成正比，当洒水车在平直路面上匀速行驶并且匀速洒水时，以下判断正确的是（）A．洒水车的动能保持不变B．发动机的功率保持不变C．牵引力的功率要随时间均匀减小D．牵引力大小跟洒水时间成反比5、某同学用单摆测当地的重力加速度．他测出了摆线长度L和摆动周期T，如图(a)所示．通过改变悬线长度L，测出对应的摆动周期T，获得多组T与L，再以T2为纵轴、L为横轴画出函数关系图像如图(b)所示．由此种方法得到的重力加速度值与测实际摆长得到的重力加速度值相比会（）A．偏大B．偏小C．一样D．都有可能6、如图所示，斜面体ABC固定在水平地面上，斜面的高AB为2m，倾角为θ=37°，且D是斜面的中点，在A点和D 点分别以相同的初速度水平抛出一个小球，结果两个小球恰能落在地面上的同一点，则落地点到C点的水平距离为（）A．42m3B．22m3C．32m4D．4m3二、多项选择题：本题共4小题，每小题5分，共20分。
湖北武汉市育才高级中学2022-2023学年高一下学期3月月考英语试题(原卷版)
武汉市育才高中2022-2023学年3月月考高一英语试卷第二部分阅读(共两节，满分50分)第一节(共15小题;每小题2. 5分，满分37. 5分)阅读下列短文，从每题所给的A、B、C和D四个选项中，选出最佳选项。
AHere’s good news for those English beginners. King’s College Summer School is an annual training program for high school students at all levels who want to improve their English. Courses are given by the teachers of King’s College and other colleges in New York. Trips to museums and culture centers are also organized. This year’s summer school will be from July 25 to August 15.More information is as follows:1. You can most probably read the text in __________.A. a telephone bookB. a travel guideC. a textbookD. a newspaper2. If you are to live with your friends in New York, you will have to pay the school __________.A. $900B. $600C. $500D. $3003. What information can you get from the text?A. The program will last two months.B. You can write to Thompson in your mother tongue.C. As a Chinese student, you can send your application on July 8, 2007.D. You can get in touch with the school by e-mail or by telephone.BLike many children, young Lotte Reiniger was crazy about movies, which at the time were a completely new invention. She taught herself how to cut paper silhouettes (剪影), of people, animals, and objects. I could cut silhouettes almost as soon as I could manage to hold a pair of scissors,” Reiniger said. “I could paint, too, and read...But everybody was surprised at the scissor cuts”At first, Reiniger wanted to be an actress, but her skill at making silhouettes drew the attention of the German film industry. Before 1927, films were silent. To help the audience understand the film, title cards with printed text appeared during the film between scenes. Reiniger helped create title cards for films, using her silhouettes. In 1918, she was asked to provide stop-motion animation (定格动面), in which objects are photographed (拍摄)in a series of slightly different positions and then replayed at high speed so that the objects appear to move on their own, for wooden rats (老鼠)in the movie The Pied Piper of Hamelin. It was a breakthrough that led to her own films, first short films and then, in 1926, The Adventures of Prince Achmed, the first full-length animated film.Although Reiniger once described herself as “a primitive (原始的)caveman artist”, her work is not simple. She carefully cut bits of card, paper, and wire, creating wonderful shapes, and then made them move and dance by hand. The black shapes were then placed on colorful backgrounds. She made more than 60 films, around 40 of which survive, all cut by her own hands. Most were based on timeless fairy tales, like Cinderella and Sleeping Beauty.Reiniger was truly a pioneer both in animation and for women in film-making. Though her last film came out in 1980, her style is still influential and can often be seen in today’s films.4. What do we know about Reiniger’s cuts?A. People silhouettes were her favorite.B. Painting and reading helped her cuts.C. She had a gift for cutting silhouettes.D. Movies excited her interest in silhouettes.5. What did Reiniger do to help the audience understand the silent films?A. She replayed the film between scenes.B. She cut silhouettes for title cards in films.C. She broke up long films into shorter ones.D. She made photos of the objects in the film.6. What does the underlined word “It” in Paragraph 2 refer to?A. Creating title cards for films.B. Making wooden rats for a film.C. Producing The Adventures of Prince Achmed.D. Bringing stop-motion animation into a film.7. Which of the following words can best describe Reiniger?A. Gifted and generous.B. Creative and productive.C. Honest and self-confident.D. Traditional and hard-working.CTechnology use seems to be the new wave of addiction hitting people of all ages. Its extreme use can be compared to the use of drugs, which is called Plugin Heroin.The next time you’re in a crowded public place, look at the people around you. It’s hard to find someone who isn’t glued to the tiny screen, fingers moving at lightning speeds, texting their friends, emailing co-workers or listening to music. It may seem ridiculous that someone is that addicted to such a small object. I’ve seen people who seem to be at their wits’ end (不知所措) if their phone or iPad has been taken away, lost or left at home.Some people may ask, “What’s wrong with technology use? It’s a way for people to communicate.” While this is true, the overuse of technology isn’t always appropriate in certain settings. Schools are becoming stricter about the use of cell phones, iPads and other electronics in classrooms. Various workplaces have signs hanging on their walls warning employees that “Cell phone use is not permitted” or “Cell phones are forbidden.”Electronics may be a way for people to communicate and stay in touch with each other, but the disadvantages may outweigh the benefits. People are losing the ability to hold face-to-face conversations with others. However, it’s hard to avoid electronics in this day and age because almost everything is turning into an electronic format. Books, originally meant for paper design, are now being transferred (转存) to electronic forms. Photo albums, and even yearbooks, can now be viewed via the internet. With this growing trend, future generations are bound to become even more addicted to technology.Is there a cure for electronic addiction? Simply turning electronics off for an hour or two a day may help to an extent, but it will not completely rid electronic addiction. There only seems to be one cure left, and it may be the hardest: self-control.8. What is Plug-in Heroin?A. The growing popularity of electronics.B. The serious consequences of technology useC. The unreasonable dependence on electronics.D. The future possibility of technology development.9. The second paragraph is intended to __________.A. tell people it is ridiculous to use electronicsB. persuade people to quit electronicsC. prove electronics are harmful to peopleD. show people’s addiction to electronics10. Why is it difficult to avoid electronics nowadays?A. Too many things are available in electronic formsB. The design of paper books is less interesting.C. It is a must to use electronics to keep in touch.D. Fewer choices are left for communication.11. In the author’s opinion, what is the best way to cure electronic addiction?A. Turn off the electronics for an hour or two.B. Be stricter about the use of cell phones, iPads and other electronics in classroomsC. Learn to control yourself.D. Cell phones are forbidden in the workplaces.DThe well-known U. S. coffee shop Starbucks just opened its largest store in the land of tea: China! The Shanghai Starbucks is 2,700 square meters and employs more than 400 people.But is coffee good for us?There have been many studies on the health benefits of coffee. Recently, researchers at the University of Southampton’s Faculty of Medicine in Great Britain looked at results from 201 observational coffee studies and 17 clinical trials of coffee. They discovered that coffee drinkers had a lower risk of heart and liver disease and some cancers. However, their findings are uncertain. The researchers could not prove coffee was the cause of these lowered risks. One of the researchers Robin Poole notes that their research included mainly observational data. Therefore, they could not prove any cause and effect relationship. In observational uncontrolled study, researchers simply watch what happens to a series of people in one group.Still, researchers found that the benefits of moderate coffee drinking seem to outweigh the risks. Their report says that drinking coffee “was more often associated with benefit than harm.” But he warns that the coffee story is much more complicated. He advises people not to drink more than four cups a day. And not everyone should drink so much coffee.Researchers found that:-Too much coffee during pregnancy can be dangerous-People, especially women, whose bones break easily, should limit how much coffee they drink-People with abnormal heart beat patterns are advised to drink coffee without caffeine.In addition, the scientists point out that the research was only about coffee. Yet many coffee drinkers don’t just drink coffee. They put sugar into it. They add milk or cream. They may have a baked treat on the side.Poole warns that to get the full health benefits of coffee, keep it simple and don’t eat the pastries that often come with a cup of coffee.12. What is the writer’s purpose in writing this passage?A. To advertise the coffee shop Starbucks.B. To persuade people to drink coffee.C. To present the findings of a research.D. To compare coffee and tea.13. Why were the findings of the research uncertain?A. The research was an uncontrolled study.B. The research was based only on clinical trials.C. The samples were not enoughD. More than one group took part in the research.14. What does the underlined word “outweigh” mean in paragraph 4?A. to have a weight ofB. to be equal toC. to be less thanD. to be more than15. What do we know about coffee drinking __________.A. Coffee is of benefit to all kinds of peopleB. Coffee is healthier if drunk together with cream.C. Proper amounts of coffee reduce many disease risks.D. Coffee helps bone-breakers recover.第二节(共5小题;每小题2. 5分，满分12. 5分)阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
高三上学期第三次月考数学试卷(附答案解析)
高三上学期第三次月考数学试卷(附答案解析)考试时间：120分钟；总分：150分学校:___________姓名：___________班级：___________第I卷（选择题）一、单选题（本大题共8小题，共40分。
在每小题列出的选项中，选出符合题目的一项）1. 已知集合A={−1,0,1,2,},B={x∈Z|x−2x≤0}，则A∩B=( )A. {0,1}B. {1,2}C. {−1,1,2}D. {0,1,2}2. 若复数z=a+2i2−i(a∈R)为纯虚数，则a=( )A. −4B. −2C. −1D. 13. 已知向量a=(1,−1),b=(1,t)，若〈a,b〉=π3，则t=( )A. 2−3B. 2+3C. 2+3或2−3D. −14. 若函数f(x)=1−cosxsinx(x∈[π3,π2])，则f(x)的值域为( )A. [3,+∞)B. [33,+∞)C. [1,3]D. [33,1]5. 正四面体S−ABC内接于一个半径为R的球，则该正四面体的棱长与这个球的半径的比值为( )A. 64B. 33C. 263D. 36. 在给某小区的花园绿化时，绿化工人需要将6棵高矮不同的小树在花园中栽成前后两排，每排3棵，则后排的每棵小树都对应比它前排每棵小树高的概率是( )A. 13B. 16C. 18D. 1127. 如图，圆内接四边形ABCD中，DA⊥AB，∠D=45°，AB=2，BC=22，AD=6.现将该四边形沿AD旋转一周，则旋转形成的几何体的体积为( )A. 84π3B. 30πC. 92π3D. 40π8. 函数f(x)的定义域为R，且f(x)−f(x+4)=0，当−2≤x<0时，f(x)=(x+1)2，当0≤x<2时，f(x)=1−x，则n=12022f(n)=( )A. 1010B. 1011C. 1012D. 1013二、多选题（本大题共4小题，共20分。
重庆市育才中学校2023-2024学年高一下学期3月月考数学试题
重庆市育才中学校2023-2024学年高一下学期3月月考数学试题学校:___________姓名：___________班级：___________考号：___________一、单选题1．已知集合{}ln A x y x ==，集合12B x x ⎧⎫=<⎨⎬⎩⎭，则A B =I （ ）A ．1,2⎛⎫-∞ ⎪⎝⎭B ．(),ln2-∞-C ．RD ．10,2⎛⎫ ⎪⎝⎭2．“π3x =”是sin x =的（ ） A ．充分必要条件 B ．必要不充分条件 C ．充分不必要条件D ．既不充分也不必要条件3．已知角α的顶点与坐标原点重合，始边与x 轴的非负半轴重合，终边与单位圆交于点34,55P ⎛⎫- ⎪⎝⎭，那么cos(π)α+等于（ ）A ．45- B ．35C ．45D ．35-4．函数()()22cos e e 1x x x f x +=+的部分图像大致为（ ） A ． B ．C ．D ．5．已知cos sin 2cos sin θθθθ-=+，则πtan 4θ⎛⎫-= ⎪⎝⎭（ ）A ．2-B ．2C ．12- D ．126．等边ABC V 的边长为3，若2AD DC =u u u r u u u r ，BF FD =u u u r u u u r，则AF =u u u r （ ）A ．2B C D 7．已知函数()sin (0)f x x ωω=>在区间2π5π[,]36-上是增函数，且在区间[0,π]上恰好取得一次最大值1，则ω的取值范围是（ ） A ．3(0,]5B ．13[,]25C ．13[,]24D ．15[,)228．已知平面向量a r ，b r满足||2a =r ，π,6b a b 〈+〉=r r r ，则a b -r r 的最大值为（ ）A ．4B．2 C．2 D ．6二、多选题9．已知向量(2,1)a =r ，(3,1)b =-r，则下列说法正确的是（ ）A ．()//a b a +r r rB ．向量a r 在向量b r 上的投影向量为12b -rC ．a r 与a b -r rD．若c =⎝⎭r ，则a c ⊥r r 10．已知函数()()()sin 0,0,0πf x A x A ωϕωϕ=+>><<的部分图象如图所示，则下列结论正确的是（ ）A ．函数()f x 的最小正周期为πB ．函数π()6y f x =+是偶函数C ．点5π(,0)6是()f x 图象的一个对称中心 D ．函数()f x 在区间π[,0]2-上单调递增11．已知对任意角,αβ均有公式()()sin 2sin 22sin cos αβαβαβ+=+-.设ABC V 的内角,,A B C 满足1sin 2sin()sin()2A ABC C A B +-+=--+，面积S 满足12S ≤≤.记,,a b c 分别为,,A B C 所对的边，则下列等式或不等式一定成立的是（ ）A ．1sin sin sin 8A B C =B．2sin aA≤≤C．8abc ≤≤D ．()8bc b c +>三、填空题12．已知()1,2a =r ，()2,3b =-r ，()1,c x =r ，()a b c +⊥r r r，则c r =.13．如图所示，设Ox ，Oy 是平面内相交成60︒角的两条数轴，1e r，2e u u r分别是与x 轴，y轴正方向同向的单位向量，若向量()12,R OP xe ye x y =+∈u u u r u r u u r，则把有序数对(),x y 叫做向量OP u u u r 在坐标系xOy 中的坐标.设()0,3OM =u u u u r ，()4,0ON =u u u r ，则OM ON ⋅=u u u u r u u u r.14．重天市育才中学为美化校园将一个半圆形空地改造为一个穿梭花园.如图所示，O 为圆心，半径为1千米，点A 、B 都在半圆弧上，设2AON AOB θ∠=∠=，其中π04θ<<.若在花园内铺设一条参观的线路，由线段NA 、AB 、BM 三部分组成，要使参观的线路最长，则θ=.（答案请用使用弧度制表示）四、解答题15．设向量()2,3OA =-uu r，()2,1OB =u u u r ，(),5OC x =u u u r . (1)若A 、B 、C 三点共线，求实数x 的取值；(2)若OB u u u r ，OC u u u r的夹角为锐角，求实数x 的取值范围.16．在ABC V 中，角A ，B ，C 的对边分别为a ，b ，c ，tan 2cos cos tan tan BA C A C=+.(1)求角B ；(2)若2b =，求a c +的最大值.17．函数()f x 对任意的实数a ，b ，都有()()()3f a b f a f b +=+-，且当0x >时，()3f x >.(1)求()0f 的值；(2)求证：()f x 是R 上的增函数；(3)解关于实数x 的不等式()()1591236x x f f --+⋅+-⋅>. 18．已知向量33(cos ,sin ),(cos ,sin )2222x x x x a b ==-r r ，函数()||1f x a b m a b =⋅-++r rr r ，ππ[,]34x ∈-，R m ∈.(1)当0m =时，求π()6f 的值；(2)若m ()f x 的最小值； (3)是否存在实数m ，使函数224()()49g x f x m =+，ππ[,]34x ∈-有四个不同的零点?若存在，求出m 的取值范围；若不存在，说明理由. 19．已知非空实数集S ，T 满足：任意x S ∈，均有1x S x-∈；任意y T ∈，均有11y T y -∈+． (1)直接写出S 中所有元素之积的所有可能值； (2)若T 由四个元素组成，且所有元素之和为3，求T ；(3)若S T I 非空，且由5个元素组成，求S T U 的元素个数的最小值．。
育才中学教学质量检测制度
育才中学教学质量检测制度为了保证学校期中、期末考试的公正性、严肃性以及学生成绩的真实性，使全部任课教师及所有学生都有一个公平竞争的环境，依据富乐一中“绩效工资方案”特制定以下教学质量检测制度：教师部分一、试题（卷）：1、月考考题原则上由各备课组命制，教务处审核，期中、期末考题一般由县教研室提供。
2、从试题到试卷，采取密封保存的方法。
试卷由非教学人员到县教育局领取，并由试卷管理者送保密室入柜保密。
3、各接触人员必须严格保密，不得泄露只言片语。
若发现试题严重泄露将一查到底，追究直接责任人责任。
4、分发试卷必须严格保密，要在多人监督下拆封，随考随取，不得过早拆封。
5、承担打印试卷的同志要准确无误，反复检查，打错一处扣考评分1分。
6、试卷有问题现场处理，教务处核实，请教县教研室后统一更正。
二、监考：1、监考人员必须按时到岗，不得迟到、旷会，违者按一般会议扣考评分0.5分。
2、监考表中各考场以人名的第一人为主监考，第二人为副监考，主监考在前，副监考在后，要一静一动，来回巡察，主监考负责领取试卷，写考场纪实，并交送考场记实，强调考场纪律；副监考负责清理考场。
3、监考时，不能睡觉，若被场外监考发现或有学生反映，一经查实，扣考评分1分。
4、监考时，不能读书看报，若被场外监考发现或有学生反映，一经查实，扣考评分1分。
5、监考时，不能两监考在一起说闲话，若被场外监考发现扣考评分1分。
6、监考时，不能随便串考场（更正试题错误除外）一经发现扣考评分1分，也不能站在门口张望，一经发现，扣考评分0.5分。
7、监考时，不得接待亲友，贻误监考，一经发现，按离岗对待。
扣考评分2分。
8、监考请假按事、病假扣考评分。
9、监考时，不得借故离开考场，凡检查不在岗位的一律视为离岗，扣考评分2分。
10、监考时，凡检查有一名学生睡觉，扣监考教师0.1分。
11、收卷时，丢失一份试卷，扣考评分1分。
12、收卷时，将试卷装订颠倒的，每份试卷扣考评分0.1分，收错试卷每份扣考评分1分。
育才中学考试试卷
育才中学考试试卷一、选择题（每题2分，共20分）1. 请选出下列单词中拼写错误的一个：A. beautifulB. importantC. difficultD. belif2. 下列哪个选项是正确的英语表达？A. 他每天学习英语。
B. He studies English everyday.C. He study English every day.D. He studies English every day.3. 根据题目所给的语境，选择最合适的单词填空：A. The _______ of the meeting has been postponed to next week.A. dateB. timeC. placeD. subject4. 下列哪个句子使用了正确的时态？A. I am reading a book now.B. I read a book yesterday.C. I have read a book last week.D. I was read a book yesterday.5. 根据题目所给的语境，选择最合适的短语填空：A. I can't believe you _______ the exam without studying.A. passedB. failedC. missedD. skipped6. 根据题目所给的语境，选择最合适的单词填空：A. The _______ of the company is to provide high-quality products.A. purposeB. resultC. reasonD. effect7. 下列哪个选项是正确的英语表达？A. 他喜欢在公园里散步。
B. He likes walk in the park.C. He likes to walking in the park.D. He likes to walk in the park.8. 根据题目所给的语境，选择最合适的短语填空：A. The _______ of the building is very impressive.A. designB. planC. ideaD. thought9. 下列哪个句子使用了正确的时态？A. She is going to the store.B. She goes to the store.C. She went to the store.D. She will go to the store.10. 根据题目所给的语境，选择最合适的单词填空：A. The _______ of the project was a success.A. completionB. beginningC. processD. failure二、填空题（每空1分，共10分）11. The _______ of the book is very interesting.（答案：story）12. He _______ to the meeting yesterday.（答案：attended）13. The _______ of the project is to improve the environment.（答案：goal）14. She is _______ for the job interview.（答案：preparing）15. The _______ of the company is to provide excellent service.（答案：mission）16. I _______ to the concert last night.（答案：went）17. The _______ of the building is very modern.（答案：architecture）18. He _______ to the proposal.（答案：agreed）19. The _______ of the new product is next month.（答案：launch）20. She _______ the book and started reading it.（答案：opened）三、阅读理解（每题2分，共20分）阅读下面的文章，回答后面的问题。
2022-2023学年湖北省咸宁市私立育才中学高三化学月考试题含解析
2022-2023学年湖北省咸宁市私立育才中学高三化学月考试题含解析一、单选题（本大题共15个小题，每小题4分。
在每小题给出的四个选项中，只有一项符合题目要求，共60分。
）1. 下列无色溶液中的离子能大量共存的是( )A．K+、Na+、MnO4-、SO42- B．H+、K+、NO3-、SO32-C．Na+、Ba2+、Cl-、Br- D．Fe3+、NH4+、SCN-、HCO3-参考答案：C略2. 下列实验方案中，不能测定Na2CO3和NaHCO3混合物中Na2CO3质量分数的是( ) A．取a克混合物充分加热，减重b克B．取a克混合物与足量稀盐酸充分反应，加热、蒸干、灼烧，得b克固体C．取a克混合物与足量稀硫酸充分反应，逸出气体用碱石灰吸收，增重b克D．取a克混合物与足量Ba(OH)2溶液充分反应，过滤、洗涤、烘干，得b克固体参考答案：C解析：A．NaHCO3受热易分解生成碳酸钠、水和二氧化碳，所以通过加热分解利用差量法即可计算出Na2CO3质量分数，故A不选； B．Na2CO3和NaHCO3均可与盐酸反应生成水、二氧化碳和氯化钠，所以bg固体是氯化钠，利用守恒法可计算出Na2CO3质量分数，故B不选； C．混合物与足量稀硫酸充分反应，也会生成水和二氧化碳，所以逸出的气体是二氧化碳，但会混有水蒸气，即碱石灰增加的质量不是二氧化碳的质量，不能测定含量，故C 选； D．Na2CO3和NaHCO3都与Ba（OH）2反应，反应的方程式为CO32-+Ba2+=BaCO3↓、HCO3-+OH-+Ba2+=H2O+BaCO3↓，因此最后得到的固体是BaCO3，所以可以计算出Na2CO3质量分数，故D不选。
3. 用N A表示阿伏加德罗常数的值，下列说法正确的是（）A．常温常压下，11.2 L己烷中含分子数为0.5N AB．标准状况下，22.4L Cl2与水完全反应，转移的电子数为2N A C．常温常压下，氧气和臭氧的混合物16g中含有N A个氧原子D．1L2mol·L-1的Al（NO3）3溶液中含Al3+个数为2N A参考答案：C略4. 已知K sp(AgCl)＝1.8×10－10，K sp(AgI)＝1.0×10－16.下列关于不溶物之间转化的说法中错误的是A．AgCl不溶于水，不能转化为AgIB．两种不溶物的K sp相差越大，不溶物就越容易转化为更难溶的不溶物C．AgI比AgCl更难溶于水，所以AgCl可以转化为AgID．常温下，AgCl若要在NaI溶液中开始转化为AgI，则NaI的浓度必须不低于参考答案：A略5. 下列有关离子方程式书写正确的A.在强碱溶液中次氯酸钠与Fe(0H)3反应生成Na2FeO4B．从酸化的海带灰浸出液中提取碘：C．以金属银为阳极电解饱和硫酸铜溶液：D．过氧化氢能使酸性KMnO4溶液褪色：参考答案：D6. 25℃、101 kPa下，碳、氢气、甲烷和葡萄糖的燃烧热依次是393.5 kJ?mol-1、285.8 kJ?mol-1、890.3 kJ?mol-1、2 800 kJ?mol-1，则下列热化学方程式正确的是（）A. C(s)+O2(g)===CO(g) △H＝-393.5 kJ?mol-1B. 2H2(g)+O2(g)===2H2O(g) △H＝+571.6 kJ?mol-1C. CH4(g)+2O2(g)===CO2(g)+2H2O(g) △H＝-890.3 kJ?mol-1D. C6H12O6（s）+3O2(g)===3CO2(g) +3H2O(l) △H＝-1 400 kJ?mol-1参考答案：D略7. 设N A为何伏加德罗常数的数值，下列说法不正确的是（）A．标况下22.4 LCl2通入足量H2O中充分反应，转移电子数为N AB．常温常压下，2gD2O中含有的电子总数为N AC．常温常压下，46gNO2和N2O4混合气体中含有的原子数为3N AD．1mol K与足量O2反应，生成K2O、K2O2和KO2的混合物时转移的电子数为N A参考答案：A考点：阿伏加德罗常数．.分析：A、依据n=计算物质的量，结合氯气和水反应是可逆反应不能进行完全；B、依据n=计算物质的量，结合分子式计算电子数；C、46gNO2和N2O4混合气体最简式相同为NO2，依据n=计算NO2中物质的量得到原子数；D、钾全部反应计算电子转移；解答：解：A、依据n=计算物质的量==1mol，结合氯气和水反应是可逆反应不能进行完全，转移电子数小于N A，故A错误；B、依据n=计算物质的量==0.1mol，结合分子式计算电子数为N A，故B正确；C、46gNO2和N2O4混合气体最简式相同为NO2，依据n=计算NO2中物质的量得到原子数==1mol，含有的原子数为3N A，故C正确；D、钾全部反应计算电子转移，1mol K与足量O2反应，生成K2O、K2O2和KO2的混合物时转移的电子数为N A，故D正确；故选A．点评：本题考查了阿伏伽德罗常数的分析应用，主要是气体摩尔体积条件分析判断，物质的量和微粒数的计算，掌握概念实质是关键，题目较简单．8. 下列叙述正确的是（）A．将CO2通入BaCl2溶液中至饱和，无沉淀产生；再通入SO2，产生沉淀B．在稀硫酸中加入铜粉，铜粉不溶解；再加入Cu（NO3）2固体，铜粉仍不溶解C．向AlCl3溶液中滴加氨水，产生白色沉淀；再加入过量NaHSO4溶液，沉淀消失D．纯锌与稀硫酸反应产生氢气的速率较慢；再加入少量CuSO4固体，速率不改变参考答案：C试题分析：A、CO2和SO2都不与BaCl2反应，没有沉淀产生，A错误；B、铜和稀硫酸不反应，加入Cu（NO3）2固体，NO3-具有强氧化性，在酸性条件下与Cu发生氧化还原反应，铜可溶解，B错误；C、向AlCl3溶液中滴加氨水，生成Al（OH）3，Al（OH）3不溶于弱碱，但可溶于强碱和强酸，NaHSO4可电离出氢离子，则再加入过量NaHSO4溶液，沉淀消失，C正确；D、加入少量CuSO4固体，锌置换出铜，形成原电池，反应速率变大，D错误。
2006年肇庆实验中学高三第一次月考
肇庆实验中学2005~2006学年高三第一次月考物理试卷本试卷分选择题和非选择题两部分,共8页。
考试用时120分钟第一部分选择题（共40分）一、本题共10小题，每小题4分，共40分。
在每小题给出的四个选项中，有的小题只有一个选项正确，有的小题有多个选项正确。
全部选对的得4分，选不全的得2分，有选错或不答的得0分。
1．如图1所示，质量为ｍ的物体，在沿斜面向上的拉力Ｆ作用下，沿质量为Ｍ的斜面匀速下滑，此过程中斜面仍静止，则水平面对斜面 ［ ］图1Ａ．有水平向左的摩擦力 Ｂ．无摩擦力 Ｃ．支持力为（Ｍ＋ｍ）ｇＤ．支持力小于（Ｍ＋ｍ）ｇ2．用三根轻绳将质量为m 的物块悬挂在空中,如图所示.已知ac 和bc 与竖直方向的夹角分别为030和60，则ac 绳和bc 绳中的拉力分别为:A.1,22m g m gB.1,22m g m gC.1,42m g m gD.1,24m g m g3．一杂技演员,用一只手抛球.他每隔0.40s 抛出一球,接到球便立即把球抛出,已知除抛、接球的时刻外，空中总有四个球，将球的运动看作是竖直方向的运动，球到达的最大高度是（高度从抛球点算起，取210/g m s ）mA. 1.6mB. 2.4mC. 3.2mD.4.0m4．短跑运动员在100m 竞赛中，测得7s 末的速度是9m/s ，10s 末到达终点时的速度是10.2m/s,则运动员在全程的平均速度为A ．9m/s B.9.6m/s C.10m/s D.10.2m/s 5.如图，竖直放置的圆环O 为圆心，A 为最高点，将物体从A 点释放经t 1落到B 点，沿光滑斜面物体从C 点由静止释放经t 2落到B 点，沿光滑斜面将物体从D 点由静止释放经t 3落到B 点，关于t 1、t 2、t 3的大小，以下说法中正确的是：（ ） A.t 1＞t 2＞t 3 B 、t 1＝t 2＝t 3C.t 1＞t 2＝t 3 D 、以上答案均不正确6．由同种材料制成的物体A 和B 放在长木板上，随长木板一起以速度v 向右做匀速直线运动，已知m A ＞m B ，某时刻木板停止运动，设木板足够长，则下列说法正确的是:（ ）A.若木板光滑，由于A 的惯性较大，A 、B 间的距离将增大B.若木板粗糙，由于B 的惯性较小，A 、B 间的距离将减小C.若木板光滑，A 、B 间距离保持不变D.若木板粗糙，A 、B 间距离保持不变7. 如图13所示，一向右运动的车厢项上悬挂两个单摆M 和N ，它们只能在图示的平面内摆动。
大学4年逆袭的励志小故事
大学4年逆袭的励志小故事大学4年逆袭的励志故事(一)24岁的林宇第一次当伴郎。
他悉心置办了一身正装，颜色灰暗、款式规矩的衬衫加西裤。
当初找工作时，也不曾有这样拾掇自己的兴致。
在外地的大学室友要结婚了，特意叮嘱他：“不要穿你的万年球服过来。
”林宇对室友的老成一番嘲弄，但还是郑重地答应了。
套上装成熟的衣服，林宇感到一阵别扭，却是欣喜的烦恼。
他人的婚礼，自我的规范，对这个大学结业生来说，生活似乎要进入正常化的轨道。
林宇属于社会所解读的“自我中心”的一代。
消费主义的风潮裹挟了他们从童年至青年的全部过程。
同时，他们又处在一个剧烈变化、理想主义远去的时代，必须在最短的时间内完成从单纯到世故、从无知到精明的火箭般成长。
他们的青春期，注定要比任何时候都短暂。
被现实被自己挫败2003年以后，大学扩招导致就业潮持续高涨，读书的性价比劣势改变了高等教育的传统功能。
昔日的象牙塔被混乱的价值观打破，在去理想化和保持出世精神之间徘徊。
培养精英的意义不再，整个社会在慨叹中国的高校已沦为职业培训场而又不够实用的时候，普遍的落差感便成了这代人的心理共鸣。
“毕业即失业”并不是夸大其词的恐慌制造，是大部分人必须面对的残酷现实。
而与此形成反差的是另一种参照：“官二代”“富二代”的飞扬跋扈和不劳而获。
普通的年轻人甚至无须从网络和报端上获取信息，因为同龄的“官二代”“富二代”或许就在他们身边。
“只有身份才能改变身份”的社会价值判断，深深烙在他们尚未成熟的价值体系中。
“即使按部就班完成学业，也不一定会有一个好的出路。
对普通人来讲，这是一个在起跑线上就已输掉的时代。
”林宇说。
也许正是这种想法，让他选择了“混”.大一的第二个学期，林宇挂科了。
学校把成绩单寄到家里的时候，他和父母都不以为意。
这个家庭还没有从儿子考上北京重点大学的喜悦中冷却下来。
2006年，大学生的贬值和就业的困难已经被社会热议，却并不影响一个西部县城对高等教育的过度期望，进入大学在那里仍然是一个神话式的前程。
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2006年育才中学高三三月份月考理科综合能力试卷 2006.3本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，满分300分，考试时间150分钟。
本卷可能用到的相对原子质量：H-1 C-12 N-14 O-16 Mn-55 K-39第Ⅰ卷（选择题，共21题，共126分）一、选择题（本题包括13小题。
每小题只有一个选项最符合题意）1．如图表示酶、激素、蛋白质三者之间的关系。
下列叙述中正确的是 ( )A．物质1是激素，不一定是由专门器官产生的B．物质2是蛋白质，其合成部位是细胞中的核糖体C．物质3是酶，其基本组成单位是氨基酸D．能产生物质2的细胞肯定能产生32．细胞中复制、转录和翻译的主要场所依次是（）A．细胞核、细胞核、细胞质 B．细胞核、细胞质、核糖体C．细胞核、细胞核、高尔基体 D．细胞核、细胞质、线粒体3．将一株成熟植物体的全部叶片均剪去一半后，下列哪个过程不会立即受到影响（）A．蒸腾作用 B．水分的吸收和运输 C．光合作用 D．矿质元素的吸收4．右图所示的食物网存在于某相对封闭的生态系统中，若甲（生产者）固定的太阳能为N，能量传递效率为10%，则丁获得的能量是（）A．等于0.1N B．等于0.01NC．少于0.1N D．多于0.1N5．下列关于“非典”的病原体SARS病毒及其疫苗的叙述，正确是（）A．SARS病毒的遗传物质组成含有5种碱基、8种核苷酸B．接种SARS疫苗能增强人体免疫力是因为接种了SARS抗体C．可用含碳源、氮源、生长因子、水、无机盐的培养基培养SARS病毒D．表现抗原特异性的是SARS病毒的衣壳6．“可燃冰”是天然气在海底的低温和高压条件下，与水分子形成的结晶(CH4·nH2O)，在我国的东海和南海海域发现了储量丰富的“可燃冰”。
但是开采“可燃冰”必须科学而慎重，否则会产生严重的环境问题，其可能是（）A．臭氧层空洞B．光化学烟雾C．酸雨D．温室效应7．下列指定粒子的个数比为2 : 1的是（）A．BaO2（过氧化钡）固体中的阴离子和阳离子B ．21H 原子中的中子和质子 C ．NaHCO 3晶体中的阳离子和阴离子D ．Be 2+中的质子和电子8．在体积都为1 L ，pH 都等于2的盐酸和醋酸溶液中，投入0.65 g 锌粒，则下图所示比较符合客观事实的是（ ）9．下列离子方程式正确的是 （ ）A 、过量的NaHSO 4与Ba(OH)2溶液反应 H ＋＋SO 42－＋Ba 2＋＋OH －==BaSO 4↓＋H 2OB 、硫酸亚铁溶液中加入稀硝酸 Fe 2+ + 2H + + NO 3— == Fe 3+ + H 2O + NO 2↑C ．向KAl(SO 4)2溶液中加入Ba(OH)2溶液使SO 42-完全沉淀 Al 3+ + 2SO 42- + 2Ba 2+ + 4OH - = AlO 2-+ 2BaSO 4↓+ 2H 2OD ．向溴化亚铁溶液中通入过量氯气 2 Fe 2+ + Cl 2 == 2 Fe 3+ + 2Cl -10．用N A 表示阿伏加德罗常数，下列说法正确的是 （ ）A ．1.0 mol Ar 4018和1.1 molCa 4020中含有的中子数相同B ．1 mol 铁被氧化，失电子数为2N AC ．22.4 L NH 3中含有的分子数为N AD ．11.7 g 食盐晶体中含有0.2 N A NaCl 分子11．在给定的四种溶液中，加入以下各种离子，各离子有可能在原溶液中较大量共存的 是 （ ） A ．pH 为1的溶液：Ca 2+、Na +、Mg 2+、NO -3B ．滴加石蕊试液显红色的溶液：Fe 3+、NH +4、Cl – 、SCN –C ．与铝反应放出氢气的溶液：K +、HCO -3、Br – 、Ba 2+D ．所含溶质为Na 2S 2O 3的溶液：K +、SO -24、NO -3、H +12．下列关于实验的描述不正确的是 （ ）A ．在用Fe(OH)3胶体溶液做电泳实验时，阴极区红色加深，可知Fe(OH)3胶体粒子带正电荷B ．在镀件上电镀铜时，可用金属铜作阳极C ．Cu —Zn 原电池(稀硫酸作电解液)中，铜片上出现气泡D ．用铂电极电解CuS04溶液，当Cu 2＋浓度降低至原来一半时，停止通电，若加入Cu(OH)2能使溶液恢复成原来浓度．13．将2 mol A 和2 mol B 充入某密闭容器中发生反应：2A(g) + B(g)x C(g)，达到化学平衡后，C 的体积分数为a 。
假设该反应的条件分别和下列各选项的条件相同，下列判断正确的是（）A．若在恒温恒容下，当x = 1时，按1 mol B、2 mol C作为起始物质，达到平衡后，C的体积分数仍为aB．若在恒温恒容下，当x = 2时，将2 mol C作起始物质，达到平衡后，C的体积分数大于aC．若在恒温恒容下，当x = 3时，2 mol B、6 mol C作起始物质，达到平衡后，C的体积分数不等于aD．若在恒温恒压下，按3 mol A、3 mol B作起始物质，达到平衡后，C的体积分数仍为a，则x可为任意正整数二、选择题（本题包括8小题，每小题6分。
每小题给出的四个选项中，有的只有一个选项正确，有的有多个选项正确，全部选对得6分，选对但不全的得3分，有选错的得0分）14. 物体A放在斜面体的斜面上，和斜面体一起向右做加速运动，如图所示，若物体与斜面体保持相对静止，物体A受到斜面对它的支持力和摩擦力的合力的方向可能是：A．向右斜上方；B．水平向右；C．向右斜下方；D．上述三种情况都不可能。
15.下列说法正确的是A.不考虑漏气和摩擦，内燃机也不可能把内能全部转化为机械能而不引起其他变化B.布朗运动反映了悬浮在液体中的固体小颗粒内部分子的无规则运动C.两个分子间距离增大时，分子力一定减小D.气体的内能等于气体内所有分子热运动的动能之和16. 一列简谐横波在某一时刻的波形图象如图所示，A、B、C是介质中的三个质点。
已知波向x轴正方向传播，波速为v=20m/s，下列说法正确的是A．这列波的波长是10m；B．质点B此时向y轴正方向运动；C．质点A的振幅为零；D．质点C再经过0.15s通过平衡位置。
17. 如图所示，闭合金属导线框水平放置在竖直向上的匀强磁场中，匀强磁场的磁感应强度增加时，则A．线框中的感应电流一定增大B．线框中的感应电流可能减小C．线框中的感应电流方向从上向下看一定沿顺时针方向D．线框中的感应电流方向从上向下看可能沿逆时针方向18.在研究物理问题时，为了能抓住问题的实质，经常需要忽略次要因素，抽象出理想化的物理模型来进行研究和推理，从而得出重要的结论。
我们在推导第一宇宙速度时，就需要作出一些近似的、简单化的假设，下列的假设中不符合．．．该推导的是A.忽略大气阻力，卫星作匀速圆周运动B.卫星的运转周期等于地球自转的周期C.卫星的轨道半径等于地球半径D.卫星需要的向心力完全由地球对它的万有引力提供19. 如图所示，在足够大的粗糙水平绝缘面上固定着一个带负电的点电荷Q，将一个质量为m、带电量为q的小金属块（金属块可视为质点）在水平面上由静止释放，金属块将在水平面上沿远离Q的方向开始运动．则在金属块从开始运动到停下的整个过程中A．金属块的加速度一直减小B．金属块的电势能先减小后增大C．电场力对金属块做的功的值等于金属块增加的机械能D．电场力对金属块所做功的数值一定等于摩擦产生的热AC21.有两个带电量相等的平行板电容器A和B，它们的正对面积分别为S A和，，两个电子以相同的初速沿板长方向(与场强垂直穿两板间距分别为d A和dB过电场，下列结论正确的是S．两电子在电场中加速度之比等于A22. （17分）（1）①用螺旋测微器测量一小粒珠子的直径如图所示，则其直径为_________mm．②图示为多用表欧姆挡刻度．当多用表用×100挡时，欧姆表读数为_______Ω． （2）在测定金属的电阻率的实验中： ①写出用测量值表示电阻率的表达式= ____________．②在各测量量中，对测量结果影响最大的是___________，因此用精密度较高的仪器________进行测量；其他测量仪器还有围23.m =与水平面(sin α，放在水平方向的匀强磁场和匀强电场中，电场强度，方向垂直于纸面向里，电量q ，由静止开始（26. ：请回答下列问题：（1）写出G分子的电子式：；（（（（27分）某化学课外兴趣小组为探究铜跟浓硫酸的反应情况，用下图所示装置先进行了有关实验：是用来收集实验中产生气体的装置，但未将导管画全，请把导管补充完整。
）实验中取6.4g铜片和12mL 18mol · L – 1浓硫酸在圆底烧瓶中共热，直到反应完毕，最后发现烧瓶中还有铜片剩余，该小组学生根据所学的化学知识认为还有一定量的①请写出铜跟浓硫酸反应的化学方程式：_________________________。
②为什么有一定量的余酸但未能使铜片完全溶解，你认为原因是：____________________________③下列药品中能够用来证明反应结束后的烧瓶中确有余酸的是_______（填写字母编号）。
A．锌片B．BaCl2溶液C．银粉 D．Na2CO3溶液28．(18分)有机物参加的反应往往比较复杂，常有副反应发生。
如下图所示，乙烯跟溴水反应既可生成A，同时又生成C2H5Br和B。
已知G由碳、氢、氧三种元素组成，且质量比为m(C)∶m(H)∶m(O) = 6∶1∶12 ，H为高分子化合物，I为六元环状化合物。
请根据图示信息回答：(1)试管牛和克隆牛的培育过程中均用到的细胞工程技术有。
（写出一种即可）(2)用来促进B牛多排卵的激素可能是分泌的促性腺激素。
E牛的性别是。
(3)产生F牛的理论依据是，其生殖方式属于。
(4)要培育高产奶率的转基因牛，则目的基因的受体细胞一般是，转基因牛乳腺细胞中含有较多的参与分泌物加工的等细胞器。
31．Duchenne肌营养不良是由法国神经学家Duchenne在19世纪60年代发现的遗传病，多见于男孩，通（A和a（（_________ __________________（4）Duchenne肌营养不良的根本治疗途径即基因治疗，是将编码抗肌萎缩蛋白的正常基因，通过质粒或_________，这是治疗成功与否的关键。
24（18分）25（20分）26（（14分）（1）_____________________________________________________(2) _________________________________________________________________(3) _________________________________________________________________(4)___________________________________________________________________(5) ______________________________27（16分）(1）B是用来收集实验中产生气体的装置，但未将导管画全，请把导管补充完整。