四川成都实验高级中学2018届高三上学期12月月考

2023届四川省成都市实验外国语学校高三模拟考试(二)英语试题一、听力选择题1. What language does the woman know a little?A．Italian.B．French.C．German.2.A．He is not lazy at all.B．He prefers online shopping.C．He’s on the way to a physical store.D．He considers shopping offline a waste of time.3. When will the game begin?A．At 8: 00.B．At 8: 30.C．At 9: 00.4. What’s John doing now ?A．Playing football.B．Watching TV.C．Studying.5. Why didn’t the woman have breakfast?A．She is busy with her paper.B．She gets used to skipping it.C．She doesn’t think it’s important.二、听力选择题6. 听下面一段较长对话，回答以下小题。

1. What day is it today?A．Wednesday.B．Thursday.C．Friday.2. What has stopped the woman taking the table by the window?A．It’s already been reserved.B．It’s in the smoking area.C．It costs more.3. What does the man suggest the woman do?A．Try another restaurant.B．Choose an outdoor table.C．Make a reservation another day.7. 听下面一段较长对话，回答以下小题。

成都经开区实验中学2018届高三（上）12月月考语文试题注意事项：1．本试卷分为第Ⅰ卷（阅读题）和第Ⅱ卷（表达题）两部分，考生务必将自己的姓名、准考证号、填写在答题卡上。

2．作答时将答案填写在答题卡上，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卷阅读题（70分）一、现代文阅读(35分)(一)论述类文本阅读(本题共3小题,9分)阅读下面的文字,完成1～3题。

关于中国古文字的产生时代这一问题始终缺乏具体确切的答案。

就是世界各国的文字学专家，对于文字产生的具体时代也有着不同的见解和学派。

一种说法，认定文字的起源并不比语言产生晚，认为人类早在旧石器时代或新时器时代初期就产生了文字。

持这种观点的人，主要把人类最早的图画、图示和各种刻记纳入了文字的范围。

他们把原始人的许多岩画谓之‚图画文字‛；也有把纹文时期的纹记视作正式的文字。

这主要是对文字的真正定义和内涵缺乏一个正确的认定。

当然我们很容易想象人类没有形成语言的时候，他们要表达某种意念或某种信息，只好用手势比划或在地上画图表示。

但是文字终归是文字，它和那些原始的辅助表意手段有着根本的区别，这个根本的区别是不管世界上哪个民族的文字，没有一个不是他们民族语言的符号，尽管世界文字之林中有的是表意文字，有的是表词文字，有的是标声文字，但都是和他们自己民族的语言一致的，凡是语言能表达的，文字就能表达。

所以可以说文字尽管在某种功能上超出了语言，但仍然是语言的辅助工具。

这样的文字，自然是产生在语言产生以后。

还有一种观点也是不能使人认同的。

那就是拘泥于实证主义的研究方法，即没有见到完整的实证材料，就认定事物的不存在，就对没有被认识或没有被发现的甚至对理应存在和实际有过但已消失的事物持一完全否定的态度。

这种实证主义的方法并非是科学的态度。

比如说，在中国古文字起源的研究中，很多人认定殷商时期的甲骨文就是最古老的汉字，可在甲骨文被发现以前，人们长期认为商周时期的金文是中国最古老的文字。

四川省成都实验中学2021届高三化学上学期12月月考试题（含解析)（化学部分）(解析）本试卷分选择题和非选择题两部分,共40题，满分300分，考试时间150分钟。

可能用到的相对原子质量：H 1 C 12 O 16 Na 23 S 32第Ⅰ卷（选择题共126分)一、选择题（每小题6分，本大题共13小题。

每小题给出的四个选项中,只有一项是符合题目要求的.)7．中国传统文化对人类文明贡献巨大，古代文献中充分记载了古代化学研究成果。

下列关于KNO3的古代文献,对其说明不合理的是(）草经集注》C 提纯“（KNO3)所在山泽，冬月地上有霜，扫取以水淋汁后，乃煎炼而成”——《开宝本草》溶解，蒸发结晶D 性质“（火药）乃焰消(KNO3）、硫黄、杉木炭所合，以为烽燧铳机诸药者”-—《本草纲目》利用KNO3的氧化性8．设N A为阿伏加德罗常数的值。

下列说法正确的是（) A．14 g聚丙烯中含有的碳原子数为3N AB．23 g Na缓慢氧化和在空气中燃烧转移电子数均为N AC．11。

2 L(标准状况）H2含有的中子数为N AD．常温下，pH＝13的NaOH溶液中含有的OH－数目为0。

1N A9．W、X、Y、Z四种短周期元素，它们在周期表中位置如表所示,下列说法不正确的是（）A．W、Y、X三种元素的原子半径依次减小B．Z元素的气态氢化物的热稳定性比Y的强C．W元素与氢元素可能会形成W2H6D．Y元素的单质能从NaZ溶液中置换出Z元素的单质10．实验室中某些气体的制取、收集及尾气处理装置如图所示（省略夹持和净化装置），下列选项中正确的是（）选项a中的物质b中的物质c中收集的气体d中的物质A浓氨水CaO NH3H2OB浓硫酸Na2SO3SO2NaOH溶液C浓硝酸Cu NO2H2OD浓盐酸MnO2Cl2NaOH溶液11．某防晒产品中含水杨酸乙基己酯(结构简式如图）、丙烯酸（CH2===CH-COOH）、甘油（丙三醇)、水等物质。

成都市实验外国语学校⾼2018届零诊模拟考试数学及答案成都市实验外国语学校⾼2018届零诊模拟考试数学试题及答案命题⼈：赵光明第Ⅰ卷（选择题共60分）⼀、选择题(本⼤题共12⼩题，每⼩题5分，共60分, 在每⼩题给出的四个选项中，只有⼀项是符合题⽬要求的.) 1、已知集合{||2}A x x =<，2{430}B x x x =-+<，则A B 等于( B ).A {21}x x -<< .B {12}x x << .C {23}x x <<.D {23}x x -<<2、设复数2zi =+，则z z -=( C ).A 4.B 0.C 2.D3、在等差数列{}n a 中，39a a =且公差0d <，则使前n 项和n S 取得最⼤值时的n 的值为( B ).A 4或5.B 5或6 .C 6或7 .D 不存在 4、某公司的班车在7:00,8:00,8:30发车,⼩明在7:50⾄8:30之间到达发车站乘坐班车,且到达发车站的时刻是随机的,则他等车时间不超过10分钟的概率是（ B ）（A ）13 （B ）12 （C ）23 （D ）345、P 是双曲线22219x y a -=上⼀点，双曲线的⼀条渐近线为320x y -=，12F F 、分别是双曲线的左、右焦点，若16PF =，则2PF =( A ).A 2或10 .B 2.C 10.D 9 6、某⼏何体的三视图如右图所⽰，其中俯视图为扇形，则该⼏何体的体积为( D ) .A 23π.B 3π.C 29π.D 169π7、已知实数x ，y 满⾜21y x x y a x ≥+??+≤??≥?，其中320(1)a x dx =-?，则实数1y x +的最⼩值为（ B ）A ．32B ．43C ．23D ．52（⽂科）已知实数,x y 满⾜3,2,2.x y x y y +≥??-≤??≤? 那么2z x y =+的最⼩值为（B ）（A ）5（B ）4（C ）3（D ）28、阅读程序框图，为使输出的数据为31，则①处应填的数字为( B ).A 4.B 5.C 6.D 7俯视图侧视图9、函数()f x 在定义域R内可导，若()(2)f x f x =-，且(1)()0x f x '-<，若(0),a f =1()2b f =,(3)c f =，则,,a b c 的⼤⼩关系是( B ).A a b c >>.B b a c >> .C c b a >> .D a c b >>10、如图，抛物线2:4W y x =与圆22:(1)25C x y -+=交于,A B 两点，点P 为劣弧AB 上不同于,A B 的⼀个动点，与x 轴平⾏的直线PQ 交抛物线W 于点Q ，则PQC 的周长的取值范围是( B )A ( 9,11) B(10,12) C(12,14) D (10,14)11、在平⾏四边形ABCD 中，0AB BD ?= ，22240AB BD +-=，若将其沿BD 折成直⼆⾯⾓ A BD C --，则三棱锥A BDC -的外接球的表⾯积为( A ) .A 4π.B 8π .C 16π .D 2π 12、设函数32()f x ax bx cx d =+++有两个极值点12,x x ，若点11(,())P x f x 为坐标原点，点22(,())Q x f x 在圆22:(2)(3)1C x y -+-=上运动时，则函数()f x 图象的切线斜率的最⼤值为( D )A.3+2+23第Ⅱ卷（⾮选择题共90分）⼆、填空题（本⼤题共4⼩题,每⼩题5分,共20分,把答案填在答题卷中相应的横线上.）13、平⾯向量a 与b的夹⾓为23π，且()1,0a =，1b = 则2a b + 14、若抛物线px y 22=的焦点与椭圆1522=+y x 的右焦点重合，则p =4_____. 15、已知数列错误！未找到引⽤源。

永胜县第二中学校2019-2020学年上学期高二数学12月月考试题含解析班级__________ 姓名__________ 分数__________一、选择题1． 设()f x 是偶函数，且在(0,)+∞上是增函数，又(5)0f =，则使()0f x >的的取值范围是（ ）A ．50x -<<或5x >B ．5x <-或5x >C ．55x -<<D ．5x <-或05x <<2． 抛物线y=﹣x 2上的点到直线4x+3y ﹣8=0距离的最小值是（ ）A ．B ．C ．D ．33． 已知函数f （2x+1）=3x+2，且f （a ）=2，则a 的值等于（ ） A ．8B ．1C ．5D ．﹣14． 若命题“p 或q ”为真，“非p ”为真，则（ ）A ．p 真q 真B ．p 假q 真C ．p 真q 假D ．p 假q 假5． 若函数f （x ）=3﹣|x ﹣1|+m 的图象与x 轴没有交点，则实数m 的取值范围是（ ） A ．m ≥0或m ＜﹣1B ．m ＞0或m ＜﹣1C ．m ＞1或m ≤0D ．m ＞1或m ＜06． 在△ABC 中，sinB+sin （A ﹣B ）=sinC 是sinA=的（ ）A ．充分非必要条件B ．必要非充分条件C ．充要条件D ．既不充分也非必要条件7． 已知正方体ABCD ﹣A 1B 1C 1D 1中，点E 为上底面A 1C 1的中心，若+，则x 、y 的值分别为（ ）A ．x=1，y=1B ．x=1，y=C ．x=，y=D ．x=，y=18． 线段AB 在平面α内，则直线AB 与平面α的位置关系是（ ）A ．AB ⊂αB ．AB ⊄αC ．由线段AB 的长短而定D ．以上都不对9． 已知命题p ：∃x ∈R ，cosx ≥a ，下列a 的取值能使“￢p ”是真命题的是（ ）A ．﹣1B ．0C ．1D ．210．已知函数，函数，其中b ∈R ，若函数y=f （x ）﹣g （x ）恰有4个零点，则b 的取值范围是（ ）A ．B ．C ．D ．11．设复数z 满足z （1+i ）=2（i 为虚数单位），则z=（ ） A ．1﹣i B ．1+i C ．﹣1﹣iD ．﹣1+i12．已知α，β为锐角△ABC 的两个内角，x ∈R ，f （x ）=（）|x ﹣2|+（）|x ﹣2|，则关于x 的不等式f （2x ﹣1）﹣f （x+1）＞0的解集为（ ）A ．（﹣∞，）∪（2，+∞）B ．（，2）C ．（﹣∞，﹣）∪（2，+∞）D ．（﹣，2）二、填空题13．下列关于圆锥曲线的命题：其中真命题的序号 ．（写出所有真命题的序号）．①设A ，B 为两个定点，若|PA|﹣|PB|=2，则动点P 的轨迹为双曲线；②设A ，B 为两个定点，若动点P 满足|PA|=10﹣|PB|，且|AB|=6，则|PA|的最大值为8； ③方程2x 2﹣5x+2=0的两根可分别作椭圆和双曲线的离心率； ④双曲线﹣=1与椭圆有相同的焦点．14．对于集合M ，定义函数对于两个集合A ，B ，定义集合A △B={x|f A （x ）fB （x ）=﹣1}．已知A={2，4，6，8，10}，B={1，2，4，8，12}，则用列举法写出集合A △B 的结果为 ．15．在极坐标系中，直线l 的方程为ρcos θ=5，则点（4，）到直线l 的距离为 ．16．若直线：012=--ay x 与直线2l ：02=+y x 垂直，则=a .17．【2017-2018学年度第一学期如皋市高三年级第一次联考】已知函数()ln 4f x x x =+-的零点在区间()1k k +，内，则正整数k 的值为________． 18．在ABC ∆中，已知角C B A ,,的对边分别为c b a ,,，且B c C b a sin cos +=，则角B为 .三、解答题19．电视传媒公司为了解某地区观众对某类体育节目的收视情况，随机抽取了100名观众进行调查，其中女性有55名．下面是根据调查结果绘制的观众日均收看该体育节目时间的频率分布直方图：将日均收看该体育节目时间不低于40分钟的观众称为“体育迷”，已知“体育迷”中有10名女性．（1）根据已知条件完成下面的2×2列联表，并据此资料你是否认为“体育迷”与性别有关？非体育迷体育迷合计男女总计（2）将日均收看该体育节目不低于50分钟的观众称为“超级体育迷”，已知“超级体育迷”中有2名女性，若从“超级体育迷”中任意选取2名，求至少有1名女性观众的概率．附：K2=P（K2≥k0）0.50 0.40 0.25 0.15 0.10 0.05 0.0250.010 0.005 0.001k00.455 0.708 1.323 2.072 2.706 3.84 5.024 6.63520．已知A（﹣3，0），B（3，0），C（x0，y0）是圆M上的三个不同的点．（1）若x0=﹣4，y0=1，求圆M的方程；（2）若点C是以AB为直径的圆M上的任意一点，直线x=3交直线AC于点R，线段BR的中点为D．判断直线CD与圆M的位置关系，并证明你的结论．21．【2017-2018学年度第一学期如皋市高三年级第一次联考】已知函数()()3231312f x x k x kx =-+++，其中.k R ∈ （1）当3k =时，求函数()f x 在[]0,5上的值域； （2）若函数()f x 在[]1,2上的最小值为3，求实数k 的取值范围.22．如图，M 、N 是焦点为F 的抛物线y 2=2px （p ＞0）上两个不同的点，且线段MN 中点A 的横坐标为，（1）求|MF|+|NF|的值；（2）若p=2，直线MN 与x 轴交于点B 点，求点B 横坐标的取值范围．23．（本小题满分12分）已知平面向量(1,)a x =，(23,)b x x =+-，()x R ∈. （1）若//a b ，求||a b -；（2）若与夹角为锐角，求的取值范围.24．设椭圆C ：+=1（a ＞b ＞0）过点（0，4），离心率为．（1）求椭圆C 的方程；（2）求过点（3，0）且斜率为的直线被椭圆所截得线段的中点坐标．永胜县第二中学校2019-2020学年上学期高二数学12月月考试题含解析（参考答案）一、选择题1． 【答案】B考点：函数的奇偶性与单调性．【思路点晴】本题主要考查函数的单调性、函数的奇偶性，数形结合的数学思想方法.由于函数是偶函数，所以定义域关于原点对称，图象关于y 轴对称，单调性在y 轴两侧相反，即在0x >时单调递增，当0x <时，函数单调递减.结合(5)0f =和对称性，可知(5)0f ±=，再结合函数的单调性，结合图象就可以求得最后的解集.12． 【答案】A【解析】解：由，得3x 2﹣4x+8=0．△=（﹣4）2﹣4×3×8=﹣80＜0．所以直线4x+3y ﹣8=0与抛物线y=﹣x 2无交点．设与直线4x+3y ﹣8=0平行的直线为4x+3y+m=0联立，得3x 2﹣4x ﹣m=0．由△=（﹣4）2﹣4×3（﹣m ）=16+12m=0，得m=﹣．所以与直线4x+3y ﹣8=0平行且与抛物线y=﹣x 2相切的直线方程为4x+3y ﹣=0．所以抛物线y=﹣x 2上的一点到直线4x+3y ﹣8=0的距离的最小值是=．故选：A ．【点评】本题考查了直线与圆锥曲线的关系，考查了数学转化思想方法，训练了两条平行线间的距离公式，是中档题．3． 【答案】B【解析】解：∵函数f （2x+1）=3x+2，且f （a ）=2，令3x+2=2，解得x=0，∴a=2×0+1=1．故选：B．4．【答案】B【解析】解：若命题“p或q”为真，则p真或q真，若“非p”为真，则p为假，∴p假q真，故选：B．【点评】本题考查了复合命题的真假的判断，是一道基础题．5．【答案】A【解析】解：∵函数f（x）=3﹣|x﹣1|+m的图象与x轴没有交点，∴﹣m=3﹣|x﹣1|无解，∵﹣|x﹣1|≤0，∴0＜3﹣|x﹣1|≤1，∴﹣m≤0或﹣m＞1，解得m≥0或m＞﹣1故选：A．6．【答案】A【解析】解：∵sinB+sin（A﹣B）=sinC=sin（A+B），∴sinB+sinAcosB﹣cosAsinB=sinAcosB+cosAsinB，∴sinB=2cosAsinB，∵sinB≠0，∴cosA=，∴A=，∴sinA=，当sinA=，∴A=或A=，故在△ABC中，sinB+sin（A﹣B）=sinC是sinA=的充分非必要条件，故选：A7．【答案】C【解析】解：如图，++（）．故选C．8．【答案】A【解析】解：∵线段AB在平面α内，∴直线AB上所有的点都在平面α内，∴直线AB与平面α的位置关系：直线在平面α内，用符号表示为：AB⊂α故选A．【点评】本题考查了空间中直线与直线的位置关系及公理一，主要根据定义进行判断，考查了空间想象能力．公理一：如果一条线上的两个点在平面上则该线在平面上．9．【答案】D【解析】解：命题p：∃x∈R，cosx≥a，则a≤1．下列a的取值能使“￢p”是真命题的是a=2．故选；D．10．【答案】D【解析】解：∵g（x）=﹣f（2﹣x），∴y=f（x）﹣g（x）=f（x）﹣+f（2﹣x），由f（x）﹣+f（2﹣x）=0，得f（x）+f（2﹣x）=，设h（x）=f（x）+f（2﹣x），若x≤0，则﹣x≥0，2﹣x≥2，则h（x）=f（x）+f（2﹣x）=2+x+x2，若0≤x≤2，则﹣2≤﹣x≤0，0≤2﹣x≤2，则h（x）=f（x）+f（2﹣x）=2﹣x+2﹣|2﹣x|=2﹣x+2﹣2+x=2，若x＞2，﹣x＜﹣2，2﹣x＜0，则h（x）=f（x）+f（2﹣x）=（x﹣2）2+2﹣|2﹣x|=x2﹣5x+8．作出函数h （x ）的图象如图：当x ≤0时，h （x ）=2+x+x 2=（x+）2+≥，当x ＞2时，h （x ）=x 2﹣5x+8=（x ﹣）2+≥，故当=时，h （x ）=，有两个交点，当=2时，h （x ）=，有无数个交点，由图象知要使函数y=f （x ）﹣g （x ）恰有4个零点，即h （x ）=恰有4个根，则满足＜＜2，解得：b ∈（，4），故选：D ．【点评】本题主要考查函数零点个数的判断，根据条件求出函数的解析式，利用数形结合是解决本题的关键．11．【答案】A【解析】解：∵z （1+i ）=2，∴z===1﹣i ．故选：A ．【点评】本题考查了复数的运算法则、共轭复数的定义，属于基础题．12．【答案】B【解析】解：∵α，β为锐角△ABC 的两个内角，可得α+β＞90°，cos β=sin （90°﹣β）＜sin α，同理cos α＜sin β，∴f （x ）=（）|x ﹣2|+（）|x ﹣2|，在（2，+∞）上单调递减，在（﹣∞，2）单调递增，由关于x的不等式f（2x﹣1）﹣f（x+1）＞0得到关于x的不等式f（2x﹣1）＞f（x+1），∴|2x﹣1﹣2|＜|x+1﹣2|即|2x﹣3|＜|x﹣1|，化简为3x2﹣1x+8＜0，解得x∈（，2）；故选：B．二、填空题13．【答案】②③．【解析】解：①根据双曲线的定义可知，满足|PA|﹣|PB|=2的动点P不一定是双曲线，这与AB的距离有关系，所以①错误．②由|PA|=10﹣|PB|，得|PA|+|PB|=10＞|AB|，所以动点P的轨迹为以A，B为焦点的图象，且2a=10，2c=6，所以a=5，c=3，根据椭圆的性质可知，|PA|的最大值为a+c=5+3=8，所以②正确．③方程2x2﹣5x+2=0的两个根为x=2或x=，所以方程2x2﹣5x+2=0的两根可分别作椭圆和双曲线的离心率，所以③正确．④由双曲线的方程可知，双曲线的焦点在x轴上，而椭圆的焦点在y轴上，所以它们的焦点不可能相同，所以④错误．故正确的命题为②③．故答案为：②③．【点评】本题主要考查圆锥曲线的定义和性质，要求熟练掌握圆锥曲线的定义，方程和性质．14．【答案】{1，6，10，12}．【解析】解：要使f A（x）f B（x）=﹣1，必有x∈{x|x∈A且x∉B}∪{x|x∈B且x∉A}={6，10}∪{1，12}={1，6，10，12，}，所以A△B={1，6，10，12}．故答案为{1，6，10，12}．【点评】本题是新定义题，考查了交、并、补集的混合运算，解答的关键是对新定义的理解，是基础题．15．【答案】3．【解析】解：直线l的方程为ρcosθ=5，化为x=5．点（4，）化为． ∴点到直线l 的距离d=5﹣2=3．故答案为：3．【点评】本题考查了极坐标化为直角坐标、点到直线的距离，属于基础题．16．【答案】1 【解析】试题分析：两直线垂直满足()02-12=⨯+⨯a ，解得1=a ，故填：1. 考点：直线垂直【方法点睛】本题考查了根据直线方程研究垂直关系，属于基础题型，当直线是一般式直线方程时，0:1111=++c y b x a l ，0:2222=++c y b x a l ，当两直线垂直时，需满足02121=+b b a a ，当两直线平行时，需满足01221=-b a b a 且1221c b c b ≠，或是212121c c b b a a ≠=，当直线是斜截式直线方程时，两直线垂直121-=k k ，两直线平行时，21k k =，21b b ≠.117．【答案】2 【解析】18．【答案】4π 【解析】考点：正弦定理．【方法点晴】本题考查正余弦定理,根据正弦定理，将所给的含有边和角的等式化为只含有角的等式，再利用三角形的三角和是︒180，消去多余的变量，从而解出B 角.三角函数题目在高考中的难度逐渐增加，以考查三角函数的图象和性质，以及三角形中的正余弦定理为主，在2016年全国卷（ ）中以选择题的压轴题出现.三、解答题19．【答案】【解析】解：（1）由频率分布直方图中可知：抽取的100名观众中，“体育迷”共有（0.020+0.005）×10×100=25名．可得2×2列联表：非体育迷体育迷合计男30 15 45女45 10 55总计75 25 100将2×2列联表中的数据代入公式计算可得K2的观测值为：k==≈3.030．∵3.030＜3.841，∴我们没有理由认为“体育迷”与性别有关．（2）由频率分布直方图中可知：“超级体育迷”有5名，从而一切可能结果所组成的基本事件空间Ω={（a1，a2），（a1，a3），（a2，a3），（a1，b1），（a1，b2），（a2，b1），（a2，b2），（a3，b1），（a3，b2），（b1，b2）}，其中a i（i=1，2，3）表示男性，b j （j=1，2）表示女性．设A表示事件“从“超级体育迷”中任意选取2名，至少有1名女性观众”，则事件A包括7个基本事件：（a1，b1），（a1，b2），（a2，b1），（a2，b2），（a3，b1），（a3，b2），（b1，b2）．∴P（A）=．【点评】本题考查了“独立性检验基本原理”、古典概率计算公式、频率分布直方图及其性质，考查了推理能力与计算能力，属于中档题．20．【答案】【解析】解：（1）设圆的方程为x2+y2+Dx+Ey+F=0圆的方程为x2+y2﹣8y﹣9=0…（2）直线CD与圆M相切O、D分别是AB、BR的中点则OD∥AR，∴∠CAB=∠DOB，∠ACO=∠COD，又∠CAO=∠ACO，∴∠DOB=∠COD又OC=OB，所以△BOD≌△COD∴∠OCD=∠OBD=90°即OC⊥CD，则直线CD与圆M相切．…（其他方法亦可）21．【答案】（1）[]1,21;（2）2k ≥.【解析】试题分析：（1）求导，再利用导数工具即可求得正解；（2）求导得()'f x =()()31x x k --，再分1k ≤和1k >两种情况进行讨论；试题解析：（1）解：3k = 时，()32691f x x x x =-++则()()()23129313f x x x x x =-+=--' 令0f x '=得121,3x x ==列表由上表知函数()f x 的值域为[]1,21（2）方法一：()()()()2331331f x x k x k x x k =-++=--'①当1k ≤时，[]()1,2,'0x f x ∀∈≥，函数()f x 在区间[]1,2单调递增 所以()()()min 31113132f x f k k ==-+++= 即53k =（舍） ②当2k ≥时，[]()1,2,'0x f x ∀∈≤，函数()f x 在区间[]1,2单调递减所以()()()min 28613213f x f k k ==-++⋅+= 符合题意③当12k <<时,当[)1,x k ∈时，()'0f x <()f x 区间在[)1,k 单调递减 当(],2x k ∈时，()'0f x >()f x 区间在(],2k 单调递增所以()()()322min 313132f x f k k k k k ==-+++= 化简得：32340k k -+= 即()()2120k k +-=所以1k =-或2k =（舍）注：也可令()3234g k k k =-+则()()23632g k k k k k =='-- 对()()1,2,0k g k ∀∈'≤()3234g k k k =-+在()1,2k ∈单调递减所以()02g k <<不符合题意综上所述：实数k 取值范围为2k ≥方法二：()()()()2331331f x x k x k x x k =-++=--'①当2k ≥时，[]()1,2,'0x f x ∀∈≤，函数()f x 在区间[]1,2单调递减 所以()()()min 28613213f x f k k ==-++⋅+= 符合题意 …………8分 ②当1k ≤时，[]()1,2,'0x f x ∀∈≥，函数()f x 在区间[]1,2单调递增所以()()min 23f x f <=不符合题意③当12k <<时,当[)1,x k ∈时，()'0f x <()f x 区间在[)1,k 单调递减 当(],2x k ∈时，()'0f x >()f x 区间在(],2k 单调递增 所以()()()min 23f x f k f =<=不符合题意综上所述：实数k 取值范围为2k ≥ 22．【答案】【解析】解：（1）设M （x 1，y 1），N （x 2，y 2），则x 1+x 2=8﹣p ，|MF|=x 1+，|NF|=x 2+， ∴|MF|+|NF|=x 1+x 2+p=8；（2）p=2时，y 2=4x ，若直线MN 斜率不存在，则B （3，0）；若直线MN 斜率存在，设A （3，t ）（t ≠0），M （x 1，y 1），N （x 2，y 2），则代入利用点差法，可得y 12﹣y 22=4（x 1﹣x 2）∴k MN =，∴直线MN 的方程为y ﹣t=（x ﹣3），∴B 的横坐标为x=3﹣，直线MN 代入y 2=4x ，可得y 2﹣2ty+2t 2﹣12=0△＞0可得0＜t 2＜12，∴x=3﹣∈（﹣3，3），∴点B 横坐标的取值范围是（﹣3，3）． 【点评】本题考查抛物线的定义，考查点差法，考查学生分析解决问题的能力，属于中档题．23．【答案】（1）2或2）(1,0)(0,3)-．【解析】试题分析：（1）本题可由两向量平行求得参数，由坐标运算可得两向量的模，由于有两解，因此模有两个值；（2）两向量,a b 的夹角为锐角的充要条件是0a b ⋅>且,a b 不共线，由此可得范围．试题解析：（1）由//a b ，得0x =或2x =-， 当0x =时，(2,0)a b -=-，||2a b -=， 当2x =-时，(2,4)a b -=-，||25a b -=.（2）与夹角为锐角，0a b ∙>，2230x x -++>，13x -<<，又因为0x =时，//a b ， 所以的取值范围是(1,0)(0,3)-.考点：向量平行的坐标运算，向量的模与数量积．【名师点睛】由向量的数量积cos a b a b θ⋅=可得向量的夹角公式，当为锐角时，cos 0θ>，但当cos 0θ>时，可能为锐角，也可能为0（此时两向量同向），因此两向量夹角为锐角的充要条件是0a b a b⋅>且,a b 不同向，同样两向量夹角为钝角的充要条件是0a b a b⋅<且,a b 不反向．24．【答案】【解析】解：（1）将点（0，4）代入椭圆C 的方程得=1，∴b=4，…由e==，得1﹣=，∴a=5，…∴椭圆C的方程为+=1．…（2）过点（3，0）且斜率为的直线为y=（x﹣3），…设直线与椭圆C的交点为A（x1，y1），B（x2，y2），将直线方程y=（x﹣3）代入椭圆C方程，整理得x2﹣3x﹣8=0，…由韦达定理得x1+x2=3，y1+y2=（x1﹣3）+（x2﹣3）=（x1+x2）﹣=﹣．…由中点坐标公式AB中点横坐标为，纵坐标为﹣，∴所截线段的中点坐标为（，﹣）．…【点评】本题考查椭圆的方程与几何性质，考查直线与椭圆的位置关系，考查韦达定理的运用，确定椭圆的方程是关键．。

英语试题第一部分：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。

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 is the woman doing?A. Repairing a computer.B. Making a payment.C. Requesting a refund.2. Why does the man come to the woman?A. To invite her to dinner.B. To give her a present.C. To seek some advice.3. What is the man going to do first?A. Make reservations.B. Check with his wife.C. Work out a plan.4. What is the woman's opinion on the new building?A. Unattractive.B. Pretty.C. Unique.5. What is the probable relationship between the speakers?A. Salesman and customer.B. Householder and renter.C. Colleagues.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。

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 are the speakers discussing?A. The accommodations.B. The destination.C. The activities.7. What is the man's attitude towards the woman's words?A. Supportive.B. Disapproving.C. Unclear.听第7段材料，回答第8、9题。

2024-2025学年度高三上期数学10月阶段性测试（答案在最后）（考试时间：120分钟；满分150分）第Ⅰ卷（选择题，共58分）一、单项选择题：本题共8小题，每小题5分，共40分．1．已知集合{{},21x A x y B y y ====+，则A B = （）A ．(]0,1B ．(]1,2C ．[]1,2D ．[]0,22．已知复数z 满足23i z z +=+，则3iz+=（）A ．12i+B ．12i-C ．2i+D ．2i-3．已知向量,a b 满足222a b a b -=-= ，且1b = ，则a b ⋅=（）A ．14B ．14-C ．12D ．12-4．如图为函数()y f x =在[]6,6-上的图象，则()f x 的解析式只可能是（）A ．())lncos f x x x =+B ．())lnsin f x x x =+C ．())ln cos f x x x =-D ．())ln sin f x x x=-5．已知()()cos f x x a x =+为奇函数，则曲线()y f x =在点()()π,πf 处的切线方程为（）A ．ππ0x y +-=B ．ππ0x y -+=C ．π0x y -+=D ．0x y +=6．在体积为12的三棱锥A BCD -中，,AC AD BC BD ⊥⊥，平面ACD ⊥平面ππ,,34BCD ACD BCD ∠=∠=，若点,,,A B C D 都在球O 的表面上，则球O 的表面积为（）A ．12πB ．16πC ．32πD ．48π7．若()()sin cos2sin αβααβ+=-，则()tan αβ+的最大值为（）A ．62B ．64C ．22D ．248．设202420230.2024log 2023,log 2022,log 0.2023a b c ===，则（）A ．c a b<<B ．b c a<<C ．b a c<<D ．a b c<<二、多项选择题：本题共3小题，每小题6分，共18分．9．设等比数列{}n a 的公比为q ，其前n 项和为n S ，前n 项积为n T ，并满足条件：2024120242025202511,1,01a a a a a ->><-，下列结论正确的是（）A ．20242025S S <B ．202420261a a <C ．2024T 是数列{}n T 中的最大值D ．数列{}n T 无最大值10．透明的盒子中装有大小和质地都相同的编号分别为1,2,3,4的4个小球，从中任意摸出两个球．设事件1A =“摸出的两个球的编号之和小于5”，事件2A =“摸出的两个球的编号都大于2”，事件3A =“摸出的两个球中有编号为3的球”，则（）A ．事件1A 与事件2A 是互斥事件B ．事件1A 与事件3A 是对立事件C ．事件1A 与事件3A 是相互独立事件D ．事件23A A 与事件13A A 是互斥事件11．已知6ln ,6e n m m a n a =+=+，其中e nm ≠，则e nm +的取值可以是（）A ．eB ．2eC ．23eD ．24e第Ⅱ卷（非选择题，共92分）三、填空题：本题共3小题，每小题5分，共15分，第14题第一个空3分，第二个空2分．12．若1sin 3α=-，则()cos π2α-=______．13．设n S 是数列{}n a 的前n 项和，点()()*,n n a n ∈N在直线2y x =上，则数列1n S ⎧⎫⎨⎬⎩⎭的前n 项和为______．14．已知点()()2,0,1,4,A B M N 、是y 轴上的动点，且满足4,MN AMN =△的外心P 在y 轴上的射影为Q ，则点P 的轨迹方程为______，PQ PB +的最小值为______．四、解答题：本题共5小题，共77分．15．（13分）设ABC△的内角,,A B C的对边分别为,,a b c，且()()()sin sin sin sin b a ABC BAC c ABC C +∠-∠=∠-，,BC AC 边上的两条中线,AD BE 相交于点P．（1）求BAC ∠；（2）若2,cos 14AD BE DPE ==∠=，求ABC △的面积．16．（15分）如图，在三棱锥D ABC -中，ABC △是以AB 为斜边的等腰直角三角形，ABD △是边长为2的正三角形，E 为AD 的中点，F 为DC 上一点，且平面BEF ⊥平面ABD ．（1）求证：AD ⊥平面BEF ；（2）若平面ABC ⊥平面ABD ，求平面BEF 与平面BCD 夹角的余弦值．17．（15分）为研究“眼睛近视是否与长时间看电子产品有关”的问题，对某班同学的近视情况和看电子产品的时间进行了统计，得到如下的列联表：近视情况每天看电子产品的时间合计超过一小时一小时内近视10人5人15人不近视10人25人35人合计20人30人50人附表：α0.10.050.010.0050.001x α2.7063.8416.6357.87910.828()()()()22()n ad bc a b c d a c b d χ-=++++．（1）根据小概率值0.05α=的2χ独立性检验，判断眼睛近视是否与长时间看电子产品有关；（2）在该班近视的同学中随机抽取3人，则至少有两人每天看电子产品超过一小时的概率是多少？（3）以频率估计概率，在该班所在学校随机抽取2人，记其中近视的人数为X ，每天看电子产品超过一小时的人数为Y ，求()P X Y =的值．18．（17分）已知函数()()ln 1f x x =+．（1）求曲线()y f x =在3x =处的切线方程；（2）讨论函数()()()F x ax f x a =-∈R 的单调性；（3）设函数()()1111g x x f f x x ⎛⎫⎛⎫=+-+ ⎪ ⎪⎝⎭⎝⎭．证明：存在实数m ，使得曲线()y g x =关于直线x m =对称．19．（17分）已知椭圆C 的对称中心在坐标原点，以坐标轴为对称轴，且经过点)和2,3⎛- ⎝⎭．（1）求椭圆C 的标准方程；（2）过点()2,0M 作不与坐标轴平行的直线l 交曲线C 于,A B 两点，过点,A B 分别向x 轴作垂线，垂足分别为点D ，E ，直线AE 与直线BD 相交于P 点．①求证：点P 在定直线上；②求PAB △面积的最大值．2024-2025学年度高三上期数学10月阶段性测试（参考答案）一、单项选择题：BAACDDDC8．【解】由对数函数的性质知0.20240.2024log 0.2023log 0.20241c =>=，2024202420242023202320230log 1log 2023log 20241,0log 1log 2022log 20231=<<==<<=，所以1,01,01c a b ><<<<；当2n >时，()()ln 1ln ln 10n n n +>>->，所以()()()()222ln 1ln 1ln 1ln 1(ln )(ln )2n n n n n n ++-⎡⎤+⋅--<-⎢⎥⎣⎦()()()2222222222ln 1ln 11ln (ln )(ln )(ln )(ln )(ln )0222n n n n n n n n n ⎡⎤-+-⎡⎤⎛⎫=-=-<-=-=⎢⎥ ⎪⎢⎢⎥⎝⎭⎣⎦⎣⎦，取2023n =，则2lg2022lg2024(lg2023)0⋅-<，所以220232024lg2022lg2023lg2022lg2024(lg2023)log 2022log 20230lg2023lg2024lg2023lg2024b a ⋅--=-=-=<⋅，即b a <，综上，b a c <<．二、多项选择题：ABC ACD CD ．11．【解】令()6ln f x x x =-，则()661xf x x x-=-='，故当()0,6x ∈时，()()0,f x f x '>单调递增，当()6,x ∈+∞时，()()0,f x f x '<单调递减，()()6ln ,66lne e ,e n n n m m a n a f m f =+==+∴= ，又e n m ≠，不妨设06e n m <<<，解法一：记12,e nx m x ==，设()()()()12,0,6g x f x f x x =--∈，则()()()()2662(6)1201212x x x g x f x f x x x x x ---=---=-=<--'''在()0,6上恒成立，所以()g x 在()0,6上单调递减，所以()()()()()1260,0,6g x f x f x g x =-->=∈，则()()()11212f x f x f x ->=，又因为()1212,6,x x -∈+∞，且()f x 在()6,+∞上单调递减，所以1212x x -<，则1212x x +>，所以e 12n m +>．解法二：由6ln ,66lne e nnm m a n a =+==+，两式相减，可得e 6ln e n nm m =-，令e (1)n t t m=>，则()()61ln 6ln 6ln 6ln 1,,e ,e 111n n t t t t tt m t m mt m t t t +=-===∴+=---；令()()()1ln 21,1g t t t t t =+-->，则()11ln 2ln 1t g t t t t t+=+-=+-'，令1ln 1(1)y t t t =+->，则221110t y t t t-=-=>'在()1,+∞上恒成立，所以()g t '在()1,+∞上单调递增，因为()()10g t g ''>=在()1,+∞上恒成立，所以()g t 在()1,+∞上单调递增，则()()10g t g >=，即()1ln 21t tt +>-，所以()61ln e 121n t tm t ++=>-．解法三：6ln ,66lne e nnm m a n a =+==+ ，两式相减得e 6lne ln n nmm-=-，212121ln ln 2x x x xx x -+<<-，可得e 12n m +>，三、填空题：79-1n n +24y x =；314．【解】设点()0,M t ，则()0,4)N t -根据点P 是AMN 的外心，(),2P x t -，而22||PM PA =，则2224(2)(2)x x t +=-+-，所以2(2),24t x y t -==-从而得到点P 的轨迹为24y x =，焦点为()1,0F 由抛物线的定义可知1PF PQ =+，因为4,14PF PB BF PF PB PQ PB +≥=+=++≥，即3PQ PB +≥，当点P 在线段BF 上时等号成立．四、解答题：15．【解】（1）因为()()()sin sin sin sin b a ABC BAC c ABC C +∠-∠=∠-，所以由正弦定理得222b c a bc +-=，由余弦定理得2221cos 22b c a BAC bc +-∠==，又0πBAC <∠<，所以π3BAC ∠=．（2）因为P 是,BC AC 边上的两条中线AD 与BE 的交点，所以点P 是ABC △的重心．又7,2,AD BE APB DPE ==∠=∠，所以在ABP △中，由余弦定理22222cos c AB PA PB PA PB APB==+-⋅∠2227474724333314⎛⎛⎫=+-⨯⨯⨯= ⎪⎝⎭⎝⎭，所以2c =，又π2,3BE BAC =∠=，所以2AE BE ==，所以24b AE ==，所以ABC △的面积为1π42sin 2323⨯⨯⨯=．16．【解】（1）ABD △是边长为2的正三角形，E 为AD 的中点，则BE AD ⊥．且平面BEF ⊥平面ABD ，平面BEF 平面,ABD BE AD =⊂平面ABD ，则AD ⊥平面BEF ．（2）由于底面ABC △为等腰直角三角形，ABD △是边长为2正三角形，可取AB 中点O ，连接OD ，则,OD AB OC AB ⊥⊥．且平面ABC ⊥平面ABD ，且平面ABC 平面ABD AB =，则OD ⊥平面ABC ．因此,,OC OA OD 两两垂直，可以建立空间直角坐标系O xyz -．ABD △是边长为2的正三角形，则可求得高3OD =．底面ABC △为等腰直角三角形，求得1OC OA OB ===．可以得到关键点的坐标()()()(0,1,0,0,1,0,1,0,0,0,0,3A B C D -由第（1）问知道平面BEF 的法向量可取(0,3AD =-．设平面BCD 的法向量为(),,m x y z =，且()(1,1,0,1,0,3BC CD ==- ，则m BC m CD ⎧⋅=⎪⎨⋅=⎪⎩，则030x y x z +=⎧⎪⎨-+=⎪⎩，解得()3,3,1m = ．则2321cos ,727m AD m AD m AD⋅〈〉==⨯⋅ ．则平面BEF 与平面BCD 夹角的余弦值为217．17．【解】（1）零假设0H 为：学生患近视与长时间使用电子产品无关．计算可得，220.0550(1025105)4006.349 3.8411535203063x χ⨯⨯-⨯==≈>=⨯⨯⨯，根据小概率值0.05α=的2χ独立性检验，推断0H 不成立，即患近视与长时间使用电子产品的习惯有关．（2）每天看电子产品超过一小时的人数为ξ，则()()()21310510331515C C C 45512069223C C 45591P P P ξξξ⨯+≥==+==+==，所以在该班近视的同学中随机抽取3人，则至少有两人每天看电子产品超过一小时的概率是6991．（3）依题意，()()1111110,22245525P X Y P X Y ===⨯====⨯=，事件1X Y ==包含两种情况：①其中一人每天看电子产品超过一小时且近视，另一人既不近视，每天看电子产品也没超过一小时；②其中一人每天看电子产品超过一小时且不近视，另一人近视且每天看电子产品没超过一小时，于是()1122111161C C 2551025P X Y ===⨯⨯+⨯⨯=，所以()()()()1165301242525100P X Y P X Y P X Y P X Y ====+==+===++=．18．【解】（1）切点为()3,ln4．因为()11f x x '=+，所以切线的斜率为()134k f ='=，所以曲线()y f x =在3x =处的切线方程为()1ln434y x -=-，化简得48ln230x y -+-=；（2）由题意可知()()ln 1F x ax x =-+，则()F x 的定义域为()1,-+∞，()()11,1,,11ax a F x a x x x +-=-=∈-'+∞++当0a ≤时，()101F x a x '=-<+，则()F x 在()1,-+∞上单调递减；当0a >时，令()0F x '=，即10ax a +-=，解得11x a=-，若()11111,01a ax a x F x a a x '-+--<≤=-=≤+；若()111,01ax a x F x a x +--'>=>+，则()F x 在11,1a ⎛⎤-- ⎥⎝⎦上单调递减，在11,a ⎛⎫-+∞ ⎪⎝⎭上单调递增．综上所述，当0a ≤时，()F x 在()1,-+∞上单调递减；当0a >时，()F x 在11,1a ⎛⎤-- ⎥⎝⎦上单调递减，在11,a ⎛⎫-+∞ ⎪⎝⎭上单调递增；（3）证明：函数()()111ln 1ln 2g x x x x ⎛⎫⎛⎫=++-+ ⎪ ⎪⎝⎭⎝⎭，函数()g x 的定义域为()(),10,-∞-+∞ ．若存在m ，使得曲线()y g x =关于直线x m =对称，则()(),10,-∞-+∞ 关于直线x m =对称，所以12m =-由()()111ln 1ln 211g x x x x ⎛⎫⎛⎫--=-+-+ ⎪ ⎪----⎝⎭⎝⎭21121lnln ln ln 111x x x x x x x x x x +++=--=-+++()()()11211211ln ln ln 1ln ln 1x x x x x x x g x x x x x x+++++=+--=+-=+．可知曲线()y g x =关于直线12x =-对称．19．【解】（1）设椭圆C 的方程为221(0,0,)mx ny m n m n +=>>≠，代入已知点的坐标，得：312413m n m n +=⎧⎪⎨+=⎪⎩，解得1612m n ⎧=⎪⎪⎨⎪=⎪⎩，所以椭圆C 的标准方程为22162x y +=．（2）如图：①设直线l 的方程为()20x my m =+≠，并记点()()()112200,,,,,A x y B x y P x y，由222,162x my x y =+⎧⎪⎨+=⎪⎩消去x ，得()223420m y my ++-=，易知()()222Δ16832410m m m =++=+>，则12122242,33m y y y y m m --+==++．由条件，()()12,0,,0D x E x ，直线AE 的方程为()1212y y x x x x =--，直线BD 的方程为()2121y y x x x x =--，联立解得()()2112211212012121222223my y my y x y x y my y x y y y y y y ++++====++++，所以点P 在定直线3x =上．②0212121121111312222PAB S AD x x y x y my y my y =⋅-=⋅-=⋅-=-△，而121212my y y y =+，所以()121212my y y y =+，则1211211224PABy y S y y y +=-=-=△令t =，则1t >，所以21222224PAB t S t t t=⋅=⋅≤++△，当且仅当t =时，等号成立，所以PAB △面积的最大值为4．。

高中化学高三模拟12月天津市汉沽一中0809学年高三第四次月考（化学）高中化学化学2018年12月本试卷共14 页, 29小题, 总分值150分。

考试用时120分钟。

相对原子质量；H 1 C 12 N 14 O 16 S 32 F 19 Cl 35.5 Al 27 Ca 40 Cu 64Fe 56 K 39 Mg 24 Na 23 La 139 Zn 65 M n 55 N i 59一、选择题(此题包括10小题, 每题3分, 共30分。

每题只有一个选项符合题意。

)1. 材料: ①今年元月, 韩国某冷库在聚氨四旨发泡作业时, 因可燃性油气挥发而引发爆炸；②同月, 我市龙岗一旅社食堂因液化石油气泄露引发爆炸。

以下讲法中正确的选项是A．聚氨酶属于有机高分子化合物, 是纯洁物B. 可燃性气体与空气混合遇火一定会发生爆炸C. 液化石油气的要紧成分是COD. 可燃性气体在空气中燃烧属于氧化还原反应2．以下物质中, 既含有离子键, 又含有共价键的是A. MgCl2B. Na2O2C. Na2OD. Al2O33. 分类是化学研究中常用的方法。

以下分类方法中, 不正确的选项是A. 依据分子组成中含有氢原子的数目, 将酸分为一元酸、二元酸等B. 依据有否电子转移, 将化学反应分为氧化还原反应和非氧化还原反应C. 依据能量的变化, 将化学反应分为放热反应和吸热反应D. 依据组成元素的种类, 将纯洁物分为单质和化合物4．甲、乙、丙为二、三周期的元素, 原子序数依次增大, 甲和乙同周期, 甲和丙同族, 甲、乙原子序数之和与丙的原子序数相等, 甲、丙原子的最外层电子数之和与乙原子的电子总数相等。

以下讲法中, 不正确的选项是A. 乙是地壳中含量最多的元素B. 丙的氢化物比甲的氢化物热稳固性强C. 乙与丙形成的化合物可制作光导纤维D. 甲、乙两种元素形成的化合物中一定含有共价键5．最近媒体报道了一些化学物质, 如:爆炸力极强的N5.结构类似白磷的N4.比黄金还贵的18O2.太空中的甲醇气团等。

崇礼区实验中学2018-2019学年上学期高二数学12月月考试题含解析班级__________ 姓名__________ 分数__________一、选择题1．如果（m∈R，i表示虚数单位），那么m=（）A．1B．﹣1C．2D．02．有一学校高中部有学生2000人，其中高一学生800人，高二学生600人，高三学生600人，现采用分层抽样的方法抽取容量为50的样本，那么高一、高二、高三年级抽取的人数分别为（）A．15，10，25B．20，15，15C．10，10，30D．10，20，203．设函数f′（x）是奇函数f（x）（x∈R）的导函数，f（﹣2）=0，当x＞0时，xf′（x）﹣f（x）＜0，则使得f （x）＞0成立的x的取值范围是（）A．（﹣∞，﹣2）∪（0，2）B．（﹣∞，﹣2）∪（2，+∞）C．（﹣2，0）∪（2，+∞）D．（﹣2，0）∪（0，2）4．抛物线y=﹣8x2的准线方程是（）A．y=B．y=2C．x=D．y=﹣25．如图，正方体ABCD﹣A1B1C1D1的棱线长为1，线段B1D1上有两个动点E，F，且EF=，则下列结论中错误的是（）A．AC⊥BEB．EF∥平面ABCDC．三棱锥A﹣BEF的体积为定值D．异面直线AE，BF所成的角为定值6．某几何体的三视图如图所示，则该几何体的表面积为（）A．8+2B．8+8C．12+4D．16+47．复数z=（m∈R，i为虚数单位）在复平面上对应的点不可能位于（）A．第一象限B．第二象限C．第三象限D．第四象限8．已知直线l1经过A（﹣3，4），B（﹣8，﹣1）两点，直线l2的倾斜角为135°，那么l1与l2（）A．垂直B．平行C．重合D．相交但不垂直9．已知向量=（﹣1，3），=（x，2），且，则x=（）A．B．C．D．10．设数集M={x|m≤x≤m+}，N={x|n﹣≤x≤n}，P={x|0≤x≤1}，且M，N都是集合P的子集，如果把b﹣a叫做集合{x|a≤x≤b}的“长度”，那么集合M∩N的“长度”的最小值是（）A．B．C．D．11．如右图，在长方体中，=11，=7，=12，一质点从顶点A射向点，遇长方体的面反射（反射服从光的反射原理），将次到第次反射点之间的线段记为，，将线段竖直放置在同一水平线上，则大致的图形是（）AB CD12．在△ABC 中，角A ，B ，C 所对的边分别是a ，b ，c ，若﹣+1=0，则角B 的度数是（）A ．60°B ．120°C ．150°D ．60°或120°二、填空题13．【2017-2018第一学期东台安丰中学高三第一次月考】在平面直角坐标系中，直线与函数xOy l 和均相切（其中为常数），切点分别为和()()2220f x x a x =+>()()3220g x x a x =+>a ()11,A x y ，则的值为__________．()22,B x y 12x x +14．已知直线l 过点P （﹣2，﹣2），且与以A （﹣1，1），B （3，0）为端点的线段AB 相交，则直线l 的斜率的取值范围是 ．15．已知函数f （x ）=x 3﹣ax 2+3x 在x ∈[1，+∞）上是增函数，求实数a 的取值范围 ．16．命题“对任意的x ∈R ，x 3﹣x 2+1≤0”的否定是 ．　17．已知f （x ）=，x ≥0，若f 1（x ）=f （x ），f n+1（x ）=f （f n （x ）），n ∈N +，则f 2015（x ）的表达式为 ．18．下列说法中，正确的是 ．（填序号）①若集合A={x|kx 2+4x+4=0}中只有一个元素，则k=1；②在同一平面直角坐标系中，y=2x 与y=2﹣x 的图象关于y 轴对称；③y=（）﹣x 是增函数；④定义在R 上的奇函数f （x ）有f （x ）•f （﹣x ）≤0．　三、解答题19．啊啊已知极坐标系的极点在直角坐标系的原点，极轴与x轴的正半轴重合，直线l的参数方程为（t为参数），圆C的极坐标方程为p2+2psin（θ+）+1=r2（r＞0）．（Ⅰ）求直线l的普通方程和圆C的直角坐标方程；（Ⅱ）若圆C上的点到直线l的最大距离为3，求r值．20．（本小题满分10分）选修4-1：几何证明选讲．如图，AB是⊙O的直径，AC是⊙O的切线，BC交⊙O于E，过E的切线与AC交于D.（1）求证：CD＝DA；（2）若CE＝1，AB＝，求DE的长．221．已知函数f（x）=a x（a＞0且a≠1）的图象经过点（2，）．（1）求a的值；（2）比较f（2）与f（b2+2）的大小；（3）求函数f（x）=a（x≥0）的值域．22．已知在平面直角坐标系中的一个椭圆，它的中心在原点，左焦点为，且过点D（2，0）．（1）求该椭圆的标准方程；（2）设点，若P是椭圆上的动点，求线段PA的中点M的轨迹方程．23．某运动员射击一次所得环数X的分布如下：X0～678910P00.20.30.30.2现进行两次射击，以该运动员两次射击中最高环数作为他的成绩，记为ξ．（I）求该运动员两次都命中7环的概率；（Ⅱ）求ξ的数学期望Eξ．24．数列{a n}满足a1=，a n∈（﹣，），且tana n+1•cosa n=1（n∈N*）．（Ⅰ）证明数列{tan2a n}是等差数列，并求数列{tan2a n}的前n项和；（Ⅱ）求正整数m，使得11sina1•sina2•…•sina m=1．　崇礼区实验中学2018-2019学年上学期高二数学12月月考试题含解析（参考答案）一、选择题1．【答案】A【解析】解：因为，而（m∈R，i表示虚数单位），所以，m=1．故选A．【点评】本题考查了复数代数形式的乘除运算，考查了复数相等的概念，两个复数相等，当且仅当实部等于实部，虚部等于虚部，此题是基础题．2．【答案】B【解析】解：每个个体被抽到的概率等于=，则高一、高二、高三年级抽取的人数分别为800×=20，600×=15，600×=15，故选B．【点评】本题主要考查分层抽样的定义和方法，用每层的个体数乘以每个个体被抽到的概率等于该层应抽取的个体数，属于基础题．3．【答案】A【解析】解：设g（x）=，则g（x）的导数为：g′（x）=，∵当x＞0时总有xf′（x）﹣f（x）＜0成立，即当x＞0时，g′（x）＜0，∴当x＞0时，函数g（x）为减函数，又∵g（﹣x）====g（x），∴函数g（x）为定义域上的偶函数，∴x＜0时，函数g（x）是增函数，又∵g（﹣2）==0=g（2），∴x＞0时，由f（x）＞0，得：g（x）＜g（2），解得：0＜x＜2，x＜0时，由f（x）＞0，得：g（x）＞g（﹣2），解得：x＜﹣2，∴f（x）＞0成立的x的取值范围是：（﹣∞，﹣2）∪（0，2）．故选：A．4．【答案】A【解析】解：整理抛物线方程得x2=﹣y，∴p=∵抛物线方程开口向下，∴准线方程是y=，故选：A．【点评】本题主要考查抛物线的基本性质．解决抛物线的题目时，一定要先判断焦点所在位置．5．【答案】D【解析】解：∵在正方体中，AC⊥BD，∴AC⊥平面B1D1DB，BE⊂平面B1D1DB，∴AC⊥BE，故A正确；∵平面ABCD∥平面A1B1C1D1，EF⊂平面A1B1C1D1，∴EF∥平面ABCD，故B正确；∵EF=，∴△BEF的面积为定值×EF×1=，又AC⊥平面BDD1B1，∴AO为棱锥A﹣BEF的高，∴三棱锥A﹣BEF的体积为定值，故C正确；∵利用图形设异面直线所成的角为α，当E与D1重合时sinα=，α=30°；当F与B1重合时tanα=，∴异面直线AE、BF所成的角不是定值，故D错误；故选D．6．【答案】D【解析】解：根据三视图得出该几何体是一个斜四棱柱，AA1=2，AB=2，高为，根据三视图得出侧棱长度为=2，∴该几何体的表面积为2×（2×+2×2+2×2）=16，故选：D【点评】本题考查了空间几何体的三视图，运用求解表面积，关键是恢复几何体的直观图，属于中档题．　7．【答案】C【解析】解：z====+i，当1+m＞0且1﹣m＞0时，有解：﹣1＜m＜1；当1+m＞0且1﹣m＜0时，有解：m＞1；当1+m＜0且1﹣m＞0时，有解：m＜﹣1；当1+m＜0且1﹣m＜0时，无解；故选：C．【点评】本题考查复数的几何意义，注意解题方法的积累，属于中档题．8．【答案】A【解析】解：由题意可得直线l1的斜率k1==1，又∵直线l2的倾斜角为135°，∴其斜率k2=tan135°=﹣1，显然满足k1•k2=﹣1，∴l1与l2垂直故选A9．【答案】C【解析】解：∵，∴3x+2=0，解得x=﹣．故选：C．【点评】本题考查了向量共线定理、方程的解法，考查了推理能力与计算能力，属于中档题．10．【答案】C【解析】解：∵集M={x|m≤x≤m+}，N={x|n﹣≤x≤n}，P={x|0≤x≤1}，且M，N都是集合P的子集，∴根据题意，M的长度为，N的长度为，当集合M∩N的长度的最小值时，M与N应分别在区间[0，1]的左右两端，故M∩N的长度的最小值是=．故选：C．11．【答案】C【解析】根据题意有：A的坐标为：（0，0，0），B的坐标为（11，0，0），C的坐标为（11，7，0），D的坐标为（0，7，0）；A1的坐标为：（0，0，12），B1的坐标为（11，0，12），C1的坐标为（11，7，12），D1的坐标为（0，7，12）；E的坐标为（4，3，12）（1）l1长度计算所以：l1=|AE|==13。

四川省双流县中学2018级一诊模拟考试数学（文科）试题本试卷分选择题和非选择题两部分。

第I 卷（选择题）第Ⅱ卷（非选择题）。

本试卷满分150分，考试时间120分钟。

第I 卷（选择题，满分60分）一、选择题：（本大题共12小题，每小题5分，共60分.在每小题列出的四个选项中，只有一项是符合题目要求的.）1.已知集合}02{2≤--=x x x A ，}01{2>-=x x B ，则A B =（ ）A.[2,1)-B. (1,1)-C. (1,2]D. (2,1)(1,2]-- 2.已知复数z 满足(1)5i z i -=+，则z =（ ）A. 23i +B. 23i -C. 32i +D. 32i -3.已知等差数列}{n a 的公差为2，若842,,a a a 成等比数列，则}{n a 的前n 项和=n S （ ）A ．)1(+n nB ．)1(-n nC ．2)1(+n n D ．2)1(-n n 4.双曲线)0,0(1:2222>>=-b a b y a x C 的离心率为25，则C 的渐近线方程为（ ）A ．14y x =±B ．13y x =±C ．12y x =± D ．y x =± 5.若1||,3||==b a 且(3)2a b b +⋅=-，则cos ,a b <>=（ ） A.63-B.31- C ．33- D ．636.如图为某几何体的三视图，则其体积为（ ）A.243π+ B.243π+ C.43π+ D.43π+7.已知,x y 满足2303301x y x y y +-≤⎧⎪+-≥⎨⎪≤⎩，y x z +=2的最大值为m ，若正数,a b 满足a b m +=，则ba 41+的最小值为（ ）A. 9B. 32C.34 D.528.已知m ，n 是两条不同的直线，α，β是两个不同的平面，有以下几个命题，其中正确的个数是（ ）①若//αβ，n α⊂，m β⊂，则//m n ；②若m n ⊥，m β⊥，n α⊥，则n β⊥ ③若m n ⊥，m α⊥，n β⊥，则αβ⊥；④若//m α，//n β，αβ⊥，则m n ⊥ ⑤若αβ⊥，m α⊂，n β⊂，则m n ⊥ A ．1 B ．2 C ．3 D ．4 9.函数)11sin(ln)(+-=x x x f 的图象大致为（ ）10.已知点D C B A ,,,在同一个球的球面上，2,2===AC BC AB 若四面体ABCD 中球心O 恰好在侧棱DA 上，14=DB ，则这个球的表面积为（ ） A.254πB.4πC. 16πD. 8π 11．已知n S 是数列{}n a 的前n 项和，3,2,1321===a a a ，．若数列}{21++++n n n a a a 是以2为公比的等比数列，则26S 的值为（ ）A ．7)12(327-B ．7)22(327-C ．7)12(326- D ．7)22(326-12.O 为坐标原点，F 为抛物线x y C 4:2=的焦点，过F 的直线交C 于B A ,且BF FA 2=，则OAB ∆的面积为（ ）A ．4B ．2C ．322D ．22第I 卷（非选择题，满分90分）二．填空题（每小题5分，共20分）13.已知函数()121x af x =++（a R ∈）为奇函数，则=a . 14.如图，若4n =时，则输出的结果为 .15．设α为锐角，若4cos 65πα⎛⎫+= ⎪⎝⎭，则sin 212a π⎛⎫+ ⎪⎝⎭的值为__________．16.已知()()21e xf x x x =-+-（e是自然对数的底数）的图象与()321132g x x x m =++的图象有3个不同的交点，则m 的取值范围是__________ 三．解答题（本大题共6小题，共70分．解答应写出文字说明，证明过程或演算步骤．）17．（本小题满分12分）设函数n m x f ⋅=)(，其中向量)1,cos 2(x m =，)2sin 3,(cos x x n =． （1）求函数)(x f 的最小正周期和单调递减区间；（2）在A B C ∆中，c b a ,,分别是角C B A ,,的对边，已知ABC b A f ∆==,1,2)(的面积为433，求ABC ∆外接圆的半径R ； ______________________________________▲___________________________________18.（本小题满分12分）为了普及法律知识，达到“法在心中”的目的，某市法制办组织了一次普法知识竞赛．统计局调查队从甲、乙两单位中各随机抽取了5名职工的成绩，如下： 甲单位职工的成绩（87 88 91 91 93分）乙单位职工的成绩（分）85 89 91 92 93（1）根据表中的数据，分别求出样本中甲、乙两单位职工成绩的平均数和方差，并判断哪个单位职工对法律知识的掌握更为稳定；（2）用简单随机抽样的方法从乙单位的5名职工中抽取2名，他们的成绩组成一个样本，求抽取的2名职工的成绩之差的绝对值至少是4分的概率；______________________________________▲______________________________19（本小题满分12分）如图，三棱锥ABC O -的三条侧棱OC OB OA ,,两两垂直，且2===OC OB OA ，ABC ∆为正三角形，M 为ABC ∆内部一点，点P 在OM 的延长线上，且MP OM 31=，PB PA =．（1）证明：POC AB 平面⊥；（2）求三棱锥PBC A -的体积； ______________________________________▲______________20.（本小题满分12分）已知椭圆)0(1:2222>>=+b a by a x C 的焦点)0,1(),0,1(21F F -，且经过点)231(，P ． （1）求椭圆C 的方程；（2）设过1F 的直线l 与椭圆C 交于B A ,两点，问在椭圆C 上是否存在一点M ，使MOCBAP四边形2AMBF 为平行四边形，若存在，求出直线l 的方程；若不存在，请说明理由；______________________________________▲___________________________________21.（本小题满分12分） 已知函数xe kx x f +=ln )(（k 为常数， 2.71828e =⋅⋅⋅是自然对数的底数），曲线)(x f y =在点())1(,1f 处的切线与x 轴平行．（1）求k 的值；（2）求)(x f 的单调区间；（3）设()()g x xf x'=，其中()f x '为()f x 的导函数．证明：对任意0x >，()21g x e -<+；______________________________________▲___________________________________ 请考生在22、23二题中任选一题作答，如果多做，则按所做的第一题记分.22.（本小题满分10分）选修4-4：坐标系与参数方程已知曲线12cos :1sin x t C y t =-+⎧⎨=+⎩ (t 为参数)，24cos :3sin x C y θθ=⎧⎨=⎩ (θ为参数)．（Ⅰ）化1C ，2C 的方程为普通方程，并说明它们分别表示什么曲线； （Ⅱ）过曲线2C 的左顶点且倾斜角为4π的直线l 交曲线1C 于,A B 两点，求AB ； ______________________________________▲___________________________________23.（本小题满分10分）已知函数()32f x a x x =--+． (Ⅰ)若2a =，解不等式()3f x ≤；（Ⅱ）若存在实数x ，使得不等式()12|2|f x a x ≥-++成立，求实数a 的取值范围．四川省双流县中学2018级一诊模拟考试答案一．选择题:CBACC, DBBBC AC 二．填空题13，2-；14,94；15，50217；16，-()1,611--e三．解答题 17.解：（I ）由题意得：1)62sin(212sin 32cos 2sin 3cos 2)(2++=++=+=πx x x x x x f函数的最小正周期为π=T ...........3分 由Z k k x k ∈+≤+≤+,2236222πππππ得函数)(x f 的单调减区间是Z k k k ∈++],32,6[ππππ........6分 （II ）21)62sin(2,2)(=++∴=πA A f又π<<A 0，解得3π=A ，由ABC ∆的面积为433sin 211433==A bc b 得， ,3=∴c 再由余弦定理7,cos 2222=-+=a A bc c b a 解得321,sin 2=∴=R A a R 18.解：（I ）90939191888751=++++=）（甲x ，90939291898551=++++=）（乙x ....2分524]909390919091)9088()9087[(51222222=-+-+-+-+-=）（）（）（甲s8]909390929091)9089()9085[(51222222=-+-+-+-+-=）（）（）（乙s ........4分∴<8524 甲单位职工对法律知识的掌握更为稳定........5分 （II ）设抽取的2名职工的成绩只差的绝对值至少是4分为事件A ，所有基本事件有：(85,89),(85,91),(85,92)(85,93),(89,85),(89,91),(89,92),(89,93),(91,85),(91,89),(91,92),(91,93),(92,85),(92,89),(92,91)(92,93),(93,85),(93,89),(93,91),(93,92),共20个..........8分事件A 包含的基本事件有：(85,89),(85,91),(85,92),(85,93),(89,85),(89,93),(91,85),(92,85),(93,85), (93,89),共10个.......10分212010)(==∴A P19. （1）因为OA ，OB ，OC 两两垂直，所以,OC OA OC OB OC OA OB O OA OB OAB ⊥⎫⎪⊥⎪⇒⊥⎬=⎪⎪⊂⎭平面平面OAB ，而AB ⊂平面OAB ，所以AB OC ⊥，取AB 中点D ，连接OD ，PD ，因为OA OB =，PA PB =，所以,AB OD AB PD AB OD PD D OD PD POD ⊥⎫⎪⊥⎪⇒⊥⎬=⎪⎪⊂⎭平面平面POD ，而PO ⊂平面POD ，所以AB PO ⊥，所以,AB OC AB PO AB OC OP O OC OP POC ⊥⎫⎪⊥⎪⊥⎬=⎪⎪⊂⎭平面平面POC ………………6分（2）由已知可得，C OAB OAB V S OC -∆=⋅=⨯⨯⨯⨯=11122223323，又A B A CB C ===2，所以sin ABC S ∆=⨯⨯⨯=1226032，设点O ，P 到平面ABC 的距离分别为h 1，h 2，由O A B C C O A BV V --=，得ABC S h ∆⋅=11233，则h =163，因为h OM h MP ==1213，所以h =26，所以A PBC P ABC ABC V V S h --∆==⋅=⨯⨯=21136233． …………………………………………………………………………………………………12分 20.（1）因为c =1，a b+=221914，a b c =+222，所以a =2，b =3，所以椭圆C 的标准方程为x y +=22143………………5分（2）假设存在符合条件的点(),M x y 00，设直线l 的方程为x my =-1，联立x my x y =-⎧⎨+=⎩2213412，消去x 得，()m y my +--=2234690，有条件知∆>0，设(),A x y 11，(),B x y 22，则m y y m +=+122634，所以AB 的中点为,m m m ⎛⎫- ⎪++⎝⎭22433434，因为四边形AMBF 2为平行四边形，所以AB 的中点与MF 2重合，即x m y m m +⎧=-⎪⎪+⎨⎪=⎪+⎩0202142343234，所以,m m M m m ⎛⎫+- ⎪++⎝⎭22231263434，把点M 的坐标代入椭圆的方程得m m --=422724800，解得m =2209，所以存在符合条件的直线l ，其方程为()y x =±+35110……………12分 21（1）()ln xx kx f x e --'=1，由已知()k f e-'==110，所以k =1…………………3分 （2）函数()f x 的定义域为(),+∞0，由（1）知，()ln xx x f x e --'=11，设()ln k x x x=--11，则()k x x x'=--<2110，即()k x 在(),+∞0上是减函数，由()k =10知，当x <<01时，()k x >0，从而()f x '>0，当x >1时，()k x <0，从而()f x '<0，综上可知，()f x 的单调递增区间是(),01，单调递减区间是(),+∞1（3）由（2）知，当x ≥1时，()()g x xf x e -'=≤<+201，故只需证明()g x e <+21在x <<01时成立·当x <<01时，x e >1，且()g x >0，所以()ln ln xx x xg x x x x e--=<--11·设()ln F x x x x =--1，(),x ∈01，则()()ln F x x '=-+2，当(),x e -∈20时，()F x '>0，()F x 为增函数，当(),x e -∈21时，()F x '<0，()F x 是减函数，所以当x e -=2时，()F x 取得最大值()F ee --=+221，所以()()g x F x e -<≤+21．综上，对任意x >0，()g x e -<+21．……………………12分22.解：⑴222212:(2)(1)1,: 1.169x y C x y C ++-=+=曲线1C 为圆心是(2,1)-，半径是1的圆．曲线2C 为中心是坐标原点，焦点在x 轴上，长轴长是8，短轴长是6的椭圆．……4分⑵曲线2C 的左顶点为(4,0)-，则直线l 的参数方程为)(22424为参数t t y t x ⎪⎪⎩⎪⎪⎨⎧=+-= 将其代入曲线1C 整理可得：04232=+-t t ，设,A B 对应参数分别为21,t t ，则4,232121==+t t t t所以2t 4t -)t -(t |t -t |||2122121===AB 2121212||||()42AB s s s s s s =-=+-=. ……………10分方法二，直线方程为4y +=x ，圆心到直线4y +=x 的距离为21=d 22112||=-=AB。

2023_2024学年四川省成都市高二上学期12月月考英语模拟试题第I卷（共80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,从试题所给的A、B、C三个选项中选出最佳选项,并标在试题的相应位置.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 happened to the man?A. He was locked out.B. He missed the train.C. He lost his keys.2. Where are the speakers?A. At a gym.B. At a restaurant.C. At a cinema.3. What does the woman dislike about her trip?A. The weather.B. The traffic.C. The scenery.4. When will the conference begin?A. At 7:30.B. At 8:30.C. At 9:00.5. What are the speakers talking about?A. A job position.B. A fellow worker.C. A new office.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,并标在试题的相应位置.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 seems to be the man’s problem?A. He doesn’t sleep well.B. He has no more pills.C. He can’t focus properly.7. What is the man advised to do?A. Stay home from work.B. Have a check-up.C. Stop feeling anxious.听第7段材料,回答第8至10题.8. How did the man get to work today?A. By bike.B. By bus.C. On foot9. What is the man’s major concern about driving a car?A. Expenses.B. Health.C. Environment.10. What does the woman think of using a car?A. It’s costly.B. It’s dangerous.C. It’s convenient.听第8段材料,回答第11至13题.11. Who reached the top of Mount Qomolangma?A. Ellen.B. Jonathan.C. Heather.12. How long did the man spend on his adventure travel?A. 96 days.B. Four months.C. One year.13. What does the woman think about doing in the future?A. Walking through a national park.B. Cycling along a country road.C. Camping out in the mountains.听第9段材料,回答第14至17题.14. What is Julian doing?A. Planning her budget.B. Conducting an interview.C. Giving advice on studies.15. Which costs Matt most each month?A. Food.B. Clothes.C. Books.16. How much does Matt spend on sports a month?A. About $15.B. About $30.C. About $50.17. What is the probable relationship between the speakers?A. Strangers.B. Schoolmates.C. Co-workers.听第10段材料,回答第18至20题.18. Which colour is unsuitable for office walls?A. Green.B. Red.C. Yellow.19. What effect does natural light have on office workers?A. It is good for their eyes.B. It helps them concentrate.C. It makes them feel cheerful.20. Who are most likely to be interested in the talk?A. Managers.B. Painters.C. Teachers.第二部分　阅读理解(共两节,满分50分)第一节阅读下列短文,从每题所给的四个选项（A、B、C、D）中,选出最佳选项,并在答题卡上将该项涂黑.（共15小题；每小题2.5分,共37.5分）AWhen you stand in the starlight under the Milky Way, the night sky is one of the most incredible wonders many of us have not seen, which is under threat due to the constant march of human progress and consequent light pollution. Below are some of my favorite dark sky destinations around the world.The Aoraki Mackenzie International Dark Sky Reserve, New ZealandThe Aoraki Mackenzie International Dark Sky Reserve is one of the best places in the country to view the night sky. Comprised of Aoraki/Mt Cook National Park and the Mackenzie Basin of Aotearoa New Zealand's South Island, the Dark Sky Reserve was certified in 2012 to continue protecting the dark skies in the area.Stargazing(天体观察)enthusiasts should plan a visit to Mt John Observatory, tours of which must be booked in advance.The Wild Atlantic Way, IrelandMost visitors go to Ireland's western Wild Atlantic Way to explore beyond hot spots like Dublin and Cork, but the route is also fast becoming a great region for stargazing, thanks to the efforts of local communities along the Way.If planning a trip, keep in mind that Ireland does have wet months where overcast skies are more likely to affect your stargazing prospects.Wadi Rum, JordanWadi Rum is also sometimes called "The Valley of the Moon", though you're more likely to compare it to Mars if you visit it. Famous for its otherworldly landscapes in movies like Prometheus, Rogue One: A Star Wars Story and The Martian, Wadi Rum's towering red rock formations are extremely interesting to explore by day, and offer protection from any distant light pollution once the sun sets.Stargazing in Wadi Rum is a breathtaking experience and reminds us how our ancestors have experienced the night sky for thousands of years.21. What might be a disadvantage of stargazing inIreland?A. Conservation efforts.B. Heavy tourism.C. Customer service.D. Local climate.22. What sets Wadi Rum apart from the other dark sky destinations?A. Unusual landscapes.B. Ancestors' experiences.C. Its geographical location.D. Its local film studios.23. What is the purpose of the text?A. To compare things.B. To inform readers.C. To warn readers.D. To announce something.BLee hyun-seok grew up in South Korea addicted to Japanese manga (漫画) series such as “Dragon Ball” and “Slam Dunk”. As soon as he could, he migrated to Tokyo to build a successful career as a manga artist and editor. Then in the early 2000s came “webtoons”（网络漫画）, a South Korean cartoon innovation optimized for smartphones. Mr. Lee was at first unimpressed. Compared with manga’s inventive graphic styles and profound plots, he found webtoons just the opposite.Yet Japanese manga is being eclipsed（黯然失色）by Korean webtoons. Last year the manga print market shrank by 2.3% to ¥265bn ($1.9bn). The size of the global webtoons market was meanwhile valued at $3.7bn. Manga is going digital slowly, in part because it is still designed for print, so awkward to read on smartphones. Seeing which way the wind was blowing, Mr. Lee abandoned manga for the webtoon industry in 2014.Though webtoons such as “Itaewon Class” and “Solo Levelling” have become popular among Japanese consumers, most Japanese publishers have stuck stubbornly to manga. “The Japanese industry is very conservative,” sighs Mr. Lee. The manga industry’s business model, in which stories are first published in weekly magazines and then in books, has hardly changed since the 1960s. Webtoons have grown so fast, in part because they can be read more easily. Other recent South Korean exports, such as the Netflix sensation “Squid Game” and BTS, a boy band, have taken the world by storm thanks to the same combination of innovation and smart marketing behind webtoons.Some are concerned about the future. Japan’s manga fans are, like all its population, aging. The average reader of the Weekly Shonen Magazine, a manga for children launched in 1989, is now over 30. “Manga could end up as old people’s culture,” warns Mr Lee. “Children these days are viewing through webtoons on their smartphones. Why not make something that suits their taste?”24. What did Mr Lee think of “webtoons” in the early 2000s?A. He considered it as inventive.B. He considered it as attractive.C. He considered it as original.D. He considered it as shallow.25. Why does the author mention numbers in paragraphs 2 ?A. To confirm the advantages of smartphones.B. To share the popularity of Japanese manga.C. To introduce the influence of webtoons industry on Japanese manga.D. To compare Mr. Lee’s career as an artist and an editor.26. What can we learn from Japanese manga?A. Japanese manga can be read more easily.B. Since the 1960s, the manga has grown so fast.C. The manga industry is unwilling to transform.D. “Squid Game” was adopted from Japanese manga.27. What is Mr. Lee’s attitude towards manga’s future?A. Worried.B. Aggressive.C. Confident.D. Annoyed.CAs athletes get stronger and faster, the pace of play continues to increase. The burden of making sure games are played according to the rules and that the officiating (裁判) is accurate is now being taken out of human hands and falling more and more into the lap of technology. It's called the video replay.The National Football League is expanding its replay system this upcoming season to include pass interference (传球干扰). Major League Baseball now relies on it for safe-or-out and home run calls. If you've been watching the FIFA World Cup, you may have noticed that the Video Assistant Referee (VAR) played a key role in almost every game. And in the Kentucky Derby, a horse was disqualified for knocking another horse. No one knew why until a video replay confirmed the call and controversy was avoided.However, many purists—those who want people to follow rules carefully and do things in the traditional way—especially in soccer, argue it's not the way the game was invented, and that the video replay is tainting the sport. But don't you want to see the proper application of the rules throughout the games? I know I do. Yes, it can slow the game down, but I feel it is worth it. If technological advancements allow fans watching from home to spot mistakes instantly, those sameviews need to be available to the officiating crews. Another example occurred in the most recent National Football Conference (NFC) Championship Game between the Los Angeles Rams and the New Orleans Saints. When obvious pass interference was committed by the Los Angeles Rams player Nickell Robey-Coleman, with just 109 seconds to play, no flag was raised on the field. It weakened the New Orleans Saints spirits. The Los Angeles Rams won a 26-23 overtime victory. The no-call deeply angered the public. The video replay showed the referees had just missed one of the most apparent pass interference calls.There are no easy answers regarding replay technology and whether it is a curse (魔咒). But for me, keeping the officiating honest and on task is the right step in limiting controversy.28. What trend in sports can be observed in paragraph 2?A. The video replay has been widely used.B. League games have become competitive.C. Rules of professional games are becoming stricter.D. People are showing more interest in sports than before.29. What does the underlined word “tainting” in paragraph 3 mean?A. Tricking.B. Promoting.C. Damaging.D. Restoring.30. What might the New Orleans Saints think of the referees in the NFC Championship Game?A. They relied a lot on the video replay.B. They cared too much about details.C. They were definitely stressed out.D. They were terribly disqualified.31. What would be the best title for the text?A. Video replays: high-end technology in sportsB. Is technology like VAR a blessing in sports?C. Officiating: a duty that requires honestyD. What do qualified referees really mean?DOne key element of human language is semantics(语义). Scientists had long thought that unlike our words, animal vocalizations (发声) were involuntary, reflecting the emotional state of the animal without conveying any other information. But over the last four decades, numerous studieshave shown that various animals have distinct calls with specific meanings.Many bird species use different alarm calls. Japanese tits, which nest in tree holes, have one call that causes their baby birds to get down to avoid being pulled out of the nest by crows, and another call for tree snakes that sends them jumping out of the nest entirely. Siberian jays vary their calls depending on whether an enemy is seen looking for food or actively attacking—and each call gets a different response from other nearby birds.Two recent studies suggested that the order of some birds' vocalizations may impact their meaning. Though the idea is still controversial, this could represent a basic form of the rules governing the order and combination of words and elements in human language known as syntax (句法), as illustrated by the classic "dog bites man" vs. "man bites dog" example.Even if some birds share basic aspects of human language, we still know very little about what’s actually going on in their minds. Most animal communication research has focused on describing signals and behavior, which on the surface can look a lot like human behavior. Determining if the underlying cognitive(认知的) processes driving the behavior are also similar is much more challenging, as at the heart of this question is intentionality: Are animals merely reacting to their environment, or do they intend to convey information to one another?32. What was scientists’ long-held belief about animal vocalizations?A. They conveyed no emotion.B. They were semantically related.C. They varied greatly with species.D. They expressed no intended meaning.33. How does the author develop paragraph 2?A. By listing data.B. By giving examples.C. By providing definition.D. By making comparison.34. What does the underlined word “this” in paragraph 3 refer to?A. What birds’ vocalizations mean.B. How rules govern human language.C. What the two recent studies indicate.D. How bird’s vocalizations are combined.35. What does the last paragraph mainly tell us?A. Shared aspects of human and birds’ languages.B. Focus of most animal communication research.C. Underlying cognitive processes of birds’ vocalizations.D. Insufficient knowledge about birds’ communication intentionally.第二节根据短文内容,从短文后的选项中选出能填入空白处的最佳选项,选项中有两项为多余选项.（共5小题,每小题2.5分,共12.5分）Revenge bedtime procrastination (报复性睡眠拖延症) refers to the decision to delay sleep in response to stress or a lack of free time earlier in the day. The addition of the word “revenge”, by the Chinese, to the concept of bedtime procrastination became popular on social media. 36Both language versions reflect frustration tied to long, stressful work hours that left little time for personal enjoyment.People who engage in bedtime procrastination know and generally want to receive enough sleep.37 This is known as an intention-behavior gap.Bedtime procrastination can cause sleep deprivation (缺乏). 38 Not getting enough sleep affects thinking, memory, and decision-making. Sleep deprivation also raises the risk of daytime sleepiness, which can harm productivity and academic achievement while increasing the risks of sleeping driving.39 It causes difficulties in regulating emotions. It’s also been connected to mental health disorders, such as depression and anxiety. It is also discovered that sleep deprivation worsens physical health, making people more easily to suffer from illnesses. Ongoing sleep loss consequences can build up over time. Therefore, it contributes to significant long-term health problems.The best solution for bedtime procrastination is creating good sleep patterns and a good sleeping environment. 40A. But they fail to actually do so.B. The Chinese are noted for working long hours.C. A lack of sleep is tied to mental and physical health.D. But remember that it will take more than one night to truly get into good sleep habits.E. We struggle to stay productive during the day to make up for the loss.F. Without enough hours of sleep, the mind and body can’t properly recharge.G. The English term “revenge bedtime procrastination” appeared from a translation of this expression.第三部分语言知识运用(共三节,满分30分)第一节完形填空(共15小题,每小题1分,满分15分)阅读下面短文,从短文各题所给的A、B、C 和D四个选项中,选出可填入空白处的最佳选项.Dr Max is a thoughtful and devoted expert. I couldn’t agree more with Dr Max’s 41 that the empathy (移情) shown by a doctor is key in a patient’s evaluation of the 42 received.While having a colonoscopy (结肠镜检查) without sedation(镇静剂) I was working hard to keep myself 43 by using techniques learned to 44 panic attacks. As the camera 45 the twists and turns of my bowel, the specialist 46 I was quietly humming a tune to myself. He asked what the 47 was and I replied, “Puff the Magic Dragon”, however, I 48 that I only knew two 49 of the song.The specialist then started 50 along with my humming and agreed that he also only knew two lines. He asked the student observing to google the lyrics. When the student 51 he explained that anything they can do to help a patient through a difficult 52 was worth doing.53 with the words we then sang along for the rest of the procedure!Not only do I remember this years later, but when 54 by the student afterwards my advice was to learn from the specialist’s bedside manner as it was so 55 .41. A. devotion B. warning C. belief D. mission42. A. treatment B. operation C. condition D. recovery43. A. quiet B. alarmed C. sensible D. calm44. A. generate B. maintain C. rid D. control45. A. witnessed B. measured C. explored D. outlined46. A. concluded B. noticed C. justified D. evaluated47. A. matter B. tune C. singer D. poet48. A. explained B. whispered C. implied D. revealed49. A. editions B. composers C. lines D. titles50. A. surfing B. dancing C. working D. whistling51. A. hesitated B. resisted C. complained D. replied52. A. infection B. procedure C. operation D. event53. A. Linked B. Confronted C. Armed D. Flooded54. A. interviewed B. followed C. accompanied D. remarked55. A. demanding B. lacking C. meaningful D. important第2节语法填空(共10小题,每小题1.5分,满分15分)阅读下面短文,在空白处填入1个适当的单词或括号内单词的正确形式.Animated Film “Chang An” Love Letter to Golden Age of PoetryA Chinese animated (动画）film, Chang An, featuring some of China’s most well-known poets has been a surprise hit at the box office. Since its release on July 8, the film 56. ___________(fetch) over 700 million yuan, leading the domestic box office over the weekend. It is also a success by 57. _______________(arouse) enthusiasm for poetic romance in the hearts of moviegoers.618A.D.toA.D.et by in the Tang dynasty, 58. ____________is often seen as a golden age in Chinese history from 618A.D.toA.D. A.D. due to its 59. ____________(economy) and cultural prosperity, describing the stories of Gao Shi and Li Bai, their struggles to better themselves and society in the Tang dynasty.Sitting in cinemas 60. ______________(crowd) with moviegoers of all age groups, the audience eagerly recited poetry alongside the actors on screen as they headed down memory lane. The movie is undoubtedly a masterpiece showing 61. ____________characteristics of Chinese culture to the world. The62.____________realartThe62.____________realart of the film is realizing that these familiar names were not just great poets 62.____________real people with flesh and blood,” Liao Kun, a Guangdong resident who watched the movie Saturday, told Sixth Tone.63.___________ chief director, 63.___________(stress) after the film’s debut (首秀）, “We created Chang’an to explore into the limitless treasures of Chinese culture through animation.” Xie Junwei, who co-directed the movie with Zou Jing, said the film was aimed at increasing 64. ____________(aware) among young Chinese of traditional Chinese culture and history. “We want to tell the stories of the brilliant figures to young people through an animated film so that they can have a deeper understanding and more enthusiasm 65. _____________our history and classic works,” Xie told domestic media Beijing Daily.第4部分写作(共两节,满分40分)第1节应用文写作（20分）假定你是校英文报的记者李华,上周五你校学生会在体育场刚举办了一场“慈善义卖”(Charity Sale)活动,请你写一篇英文报道,要点如下：1. 活动目的和过程；2. 活动反响及意义.注意：1. 词数100词左右；2. 可适当增加细节,以使行文连贯.________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ ___________________________________________________________________________第二节：读后续写(满分20分)阅读下面短文,根据所给情节进行第二自然段续写,使之构成一个完整的故事.Andy, a 15-year-old boy, and his mother Emily didn’t have much in life, living daily with just enough to eat three meals and pay the bills. Without his mother knowing, Andy had been saving up his pocket money for months to buy her a birthday present. He had collected a total of $25.As a cleaner at a nearby park, Emily never took a bus, choosing to walk to and from work to save money. One day, Andy noticed that her mother came home wet to the skin. She told him her umbrella broke because of the strong wind and she wanted to get home in time to have dinner with him.The following day, his mother woke up with a fever. “Mom, I’m sure you have to stay in bed today” Andy told her. “I’ll look after you. Here is your medicine,” he said, handing her a glass of water. Then, Andy left the house and walked towards the nearby supermarket to buy vegetables with which to cook some so up for his mother.It occurred to him that his mother’s birthday was in a couple of days, so he decided to get his mother an umbrella first and spend the rest of his money on food.“What is your strongest umbrella here, sir?” he asked the shop assistant. “I am looking for one that won’t break easily.” The man showed Andy a variety of umbrellas. Andy believed the purple one would be the best choice, his mother’s favorite color. It was $22. Andy almost jumped with joy! The boy was happy that he managed to lend his mom a hand by finding her something she really needed. Also, he even had some money to re!Stepping out of the shop with the gift and vegetables in his shopping bag, Andy realized it was starting to rain. He was running to a nearby bus stop to seek shelter with other people when he noticed an old lady walking slowly and struggling for a balance. It was pouring down. Without hesitation, he took the umbrella out of his bag and ran over.Paragraph1：The old lady looked up and realized an umbrella was sheltering her from the rain. She cast him an appreciative look with a beam of delight." Thank you. And by the way, it’s wise to get an umbrella before the rain started pouring," she said. Andy shook his head, telling her that his mom’s umbrella was broken, and he wanted to get her a new one she could use for years." I didn’t expect that. But it must be very nice to have a caring boy like you," she replied, helping Andy to wipe off the raindrops on the umbrella. They exchanged a knowing look and waved goodbye after the rain stopped.注意：1. 续写词数应为80词左右;2.请按如下格式在答题卡的相应位置作答.3.续写第一自然段已给出.Paragraph 2:Anxious about his sick mom, Andy rushed home, all over wet.____________________________________________________________________________ ____________________________________________________________________________ ____________________________________________________________________________ ____________________________________________________________________________ ____________________________________________________________________________ __________________________________________________________________高二英语答案听力理解（1-20小题,每小题1.5分,共30分）1-5 ACBCB 6-10 CAABA 11-15 BBBAC 16-20 ACBAC阅读理解（21-35小题,每小题2.5分,共37.5分）21-23 DAB24-27 DCCA28-31 ACDB32-35 DBCD补全对话（七选五）（36-40小题,每小题2.5分,共12.5分）36-40 GAFCD完形填空（41-55小题,每小题1分,共15分）文本解读：本文是记叙文,主要讲述了作者在进行结肠镜检查时,医生所表现出的同理心,让作者很是温暖.多年以后,作者对这一幕都记忆犹新,认为医生对病人的同理心很重要.41.C.考查名词.根据文本大意作答.句意：我非常认同Max医生的信念：医生所显示出的同理心对于病人评估自己所接受的治疗很关键.mission使命.42.A.考查名词.解答此题的关键是后置定语received.病人在医院里接受的应该是治疗.43.D.考查形容词.根据本句作答.句意：因为没有使用镇静剂,在进行结肠镜检查时,我使用学到的控制疼痛的法子尽力使自己不紧张.sensible明智的.44.D.考查动词.根据本句作答.generate产生；maintain维持；rid 去除.45.C.考查动词.根据本句作答.句意：在仪器检查我的肠部时,医生注意到我在哼着小曲.outline 概述；勾勒轮廓.46.B.......据本句作答.justify证明......合理.47.B.考查名词.根据上句作答.句意：医生问我哼的什么曲子,我说是Puff the Magic Dragon,并解释说只会其中的两句歌词.48.A.考查动词.根据本句作答.whisper低声说,耳语.49.C.考查名词.根据下句作答.eition版本；title标题.50.D.考查动词.根据本句作答.句意：伴随着我哼的小曲,医生吹起了口哨,说自己也只记得两句.51.A.考查动词.根据本句作答.句意：他让实习生去搜索歌词.实习生表现出犹豫时,他解释说任何能帮助病人走过艰难过程的事情都是值得一做的.52.B.考查名词.根据本句作答.infection感染53.C.考查动词.根据上句及本句作答.句意：（装备）现在有了歌词,在余下的检查时间里,我们一起唱这首歌.confront面对.54.A.考查动词.根据本句作答.句意：多年以后我不仅记得这一幕,当被采访时,我的建议是学习这位医生对待病人的态度,因为同理心很重要.55.D.考查形容词.根据文本大意和第一节第二句作答.demanding要求高的.语法填空（56-65小题,每小题1.5分,共15分）56. has fetched 57. arousing 58. which 59. economic 60. crowded61. the 62. but 63. stressed 64. awareness 65. about书面表达（40分）第一节：应用问写作（20分）【范文】Aiming to help the students from poverty-stricken areas, the Students’ Union launched a grand Charity Sale last Friday at our School’s stadium, which witnessed the kindness and generosity of all the students and teachers present.p.m.cheduled, the activity began at 3:00 p.m., lasting three hours. What came first was a moving speech delivered by a volunteer, mirroring/ reflecting the authentic/ real life of the needy students and kindling/ arousing the kindness of us students. What subsequently （随后）followedwas the exciting sales section where many second-hand items, ranging from electronic devices to old fancy clothes, were sold at a pretty low price. So enthusiastic were our schoolmates and teachers that all goods were snapped up (抢购）within hours. All the money raised would be sent to the Red Cross and distributed to those needy students.This Charity Sale proved/ turned out to be successful and meaningful. Not only did we promote our social responsibility, but we also understood every bit of our kindness could pile up enormous strength/ make a great difference.第二节：读后续写（20分）【范文】Anxious about his sick mom, Andy rushed home, all over wet. Andy arrived home and his mother asked why he had been out for so long. Andy briefed her on his experience of his way home without speaking of the umbrella." That’s my boy," beamed Mom with a proud smile broadening from side to side. Andy giggled shyly, putting a cold towel on Mom’s forehead. When his mother’s birthday arrived, Andy excitedly woke up to give his mom the present box." Happy birthday! Mom!" exclaimed Andy, his eyes rkling with thrill. Touched by her son’s efforts, Emily opened the gift and was surprised to see the umbrella inside. Tears began to gather in her eyes and she went to hug Andy tightly.。

每日一题逆温高考频度：★★★☆☆难易程度：★★★☆☆典例在线安吉白茶茶园早春气温垂直分布特征明显，茶园昼夜温差大，逆温天气频繁出现，凌晨5：00左右逆温最强，霜冻天气来临时，防霜风扇（左图）对茶园增温效果明显，能有效预防消除茶园霜冻害，下面右图为茶园某时气温垂直分布示意图。

据此回答1—2题。

1．下列关于风扇的作用原理说法正确的是①风扇吹出的热风可以使茶树免受霜冻影响②风扇可以把高处的暖空气往下吹到茶树上③风扇的转动可以增强近地面对流运动程度④风扇转动可带动空气流动，削弱上层冷空气下沉A．①③B．②④C．②③D．①④2．“防霜冻风扇”最适宜的安装高度为A．9—10米B．7—8米C．2—3米D．4米以下【参考答案】1．B 2．B2．右图显示，在4—8米之间随着高度的升高，气温升高，有逆温层存在。

“防霜冻风扇”安装在7—8米之间，有利于将高处的暖空气往下吹到茶树上，从而减轻茶树受霜冻影响或使茶树免受霜冻影响，B对。

其他高度不能有效防霜冻危害，A、C、D错。

解题必备逆温现象及其影响1．逆温现象的表现分析一般情况下，气温随海拔升高而降低，大约每升高1 000米，气温下降6℃，但有时会出现气温随海拔升高而上升的现象，即出现了逆温现象(如下图所示B、C之间)，根据其成因分为辐射逆温、锋面逆温、平流逆温和地形逆温等。

2．辐射逆温的产生、消失过程分析(1)图a为正常气温垂直分布情况。

在晴朗无云或少云的夜晚，地面很快冷却，贴近地面的大气层也随之降温。

离地面越近降温越快，离地面越远降温越慢，因而形成了自地面开始的逆温(图b)。

(2)随着地面冷却的加剧，逆温逐渐向上扩展，黎明时达到最强(图c)。

(3)日出后，太阳辐射逐渐增强，地面很快增温，逆温层便逐渐自下而上消失(图d、e)。

辐射逆温厚度从数十米到数百米，在大陆上常年都可出现。

夏季夜短，逆温层较薄，消失较快；冬季夜长，逆温层较厚，消失较慢。

3．逆温的类型和成因类型成因特点在晴朗无云或少云的夜晚，地面辐射强，冷却辐射逆温大陆上常年均可出现，尤以冬季最强快，离地面越近，降温越快暖空气水平移动到冷的地面或水面上而发生的愈近地表，降温愈快平流逆温冷接触作用锋面逆温冷暖气团温度差异显著，暖气团位于锋面上部出现于锋面附近地形逆温冷空气沿斜坡向低谷和盆地流动出现于山谷或盆地4成雾早晨易出现多雾天气，降低大气能见度，影响人们的出行，易出现交通事故大气污染逆温使空气垂直对流受阻，造成近地面污染物不能及时扩散(如雾霾)，从而危害人体健康，如果位于盆地内，将会更加严重沙尘暴逆温时不利于沙尘扬起航空低空逆温造成的多雾天气给飞机起降带来麻烦，而高空逆温对飞机飞行极为有利。

四川省成都市成华区某校2023-2024学年高二上学期12月月考英语试题学校:___________姓名：___________班级：___________考号：___________一、阅读选择When you stand in the starlight under the Milky Way, the night sky is one of the most incredible wonders many of us have not seen, which is under threat due to the constant march of human progress and consequent light pollution. Below are some of my favorite dark sky destinations around the world.The Aoraki Mackenzie International Dark Sky Reserve, New ZealandThe Aoraki Mackenzie International Dark Sky Reserve is one of the best places in the country to view the night sky. Comprised of Aoraki/Mt Cook National Park and the Mackenzie Basin of Aotearoa New Zealand’s South Island, the Dark Sky Reserve was certified in 2012 to continue protecting the dark skies in the area.Stargazing (天体观察) enthusiasts should plan a visit to Mt John Observatory, tours of which must be booked in advance.The Wild Atlantic Way, IrelandMost visitors go to Ireland’s western Wild Atlantic Way to explore beyond hot spots like Dublin and Cork, but the route is also fast becoming a great region for stargazing, thanks to the efforts of local communities along the Way.If planning a trip, keep in mind that Ireland does have wet months where overcast skies are more likely to affect your stargazing prospects.Wadi Rum, JordanWadi Rum is also sometimes called “The Valley of the Moon”, though you’re more likely to compare it to Mars if you visit it. Famous for its otherworldly landscapes in movies like Prometheus, Rogue One: A Star Wars Story and The Martian, Wadi Rum’s towering red rock formations are extremely interesting to explore by day, and offer protection from any distant light pollution once the sun sets.Stargazing in Wadi Rum is a breathtaking experience and reminds us how our ancestors have experienced the night sky for thousands of years.1．What might be a disadvantage of stargazing in Ireland?A．Conservation efforts.B．Heavy tourism.C．Customer service.D．Local climate.2．What sets Wadi Rum apart from the other dark sky destinations?A．Unusual landscapes.B．Ancestors’ experiences.C．Its geographical location.D．Its local film studios.3．What is the purpose of the text?A．To compare things.B．To inform readers.C．To warn readers.D．To announce something.Lee hyun-seok grew up in South Korea addicted to Japanese manga (漫画) series such as “Dragon Ball” and “Slam Dunk”. As soon as he could, he migrated to Tokyo to build a successful career as a manga artist and editor. Then in the early 2000s came “webtoons” (网络漫画), a South Korean cartoon innovation optimized for smartphones. Mr. Lee was at first unimpressed. Compared with manga’s inventive graphic styles and profound plots, he found webtoons just the opposite.Yet Japanese manga is being eclipsed (黯然失色) by Korean webtoons. Last year the manga print market shrank by 2.3% to ¥265bn ($1.9bn). The size of the global webtoons market was meanwhile valued at $3.7bn. Manga is going digital slowly, in part because it is still designed for print, so awkward to read on smartphones. Seeing which way the wind was blowing, Mr. Lee abandoned manga for the webtoon industry in 2014.Though webtoons such as “Itaewon Class” and “Solo Levelling” have become popular among Japanese consumers, most Japanese publishers have stuck stubbornly to manga. “The Japanese industry is very conservative,” sighs Mr. Lee. The manga industry’s business model, in which stories are first published in weekly magazines and then in books, has hardly changed since the 1960s. Webtoons have grown so fast, in part because they can be read more easily. Other recent South Korean exports, such as the Netflix sensation “Squid Game” and BTS, a boy band, have taken the world by storm thanks to the same combination of innovation and smart marketing behind webtoons.Some are concerned about the future. Japan’s manga fans are, like all its population, aging. The average reader of the Weekly Shonen Magazine, a manga for children launched in 1989, is now over 30. “Manga could end up as old people’s culture,” warns Mr Lee. “Children these days are viewing through webtoons on their smartphones. Why not make something that suits their taste?”4．What did Mr Lee think of “webtoons” in the early 2000s?A．He considered it as inventive.B．He considered it as attractive.C．He considered it as original.D．He considered it as shallow.5．Why does the author mention numbers in paragraphs 2 ?A．To confirm the advantages of smartphones.B．To share the popularity of Japanese manga.C．To introduce the influence of webtoons industry on Japanese manga.D．To compare Mr. Lee’s career as an artist and an editor.6．What can we learn from Japanese manga?A．Japanese manga can be read more easily.B．Since the 1960s, the manga has grown so fast.C．The manga industry is unwilling to transform.D．“Squid Game” was adopted from Japanese manga.7．What is Mr. Lee’s attitude towards manga’s future?A．Worried.B．Aggressive.C．Confident.D．Annoyed.As athletes get stronger and faster, the pace of play continues to increase. The burden of making sure games are played according to the rules and that the officiating (裁判) is accurate is now being taken out of human hands and falling more and more into the lap of technology. It’s called the video replay.The National Football League is expanding its replay system this upcoming season to include pass interference (传球干扰). Major League Baseball now relies on it for safe-or-out and home run calls. If you’ve been watching the FIFA World Cup, you may have noticed that the Video Assistant Referee (V AR) played a key role in almost every game. And in the Kentucky Derby, a horse was disqualified for knocking another horse. No one knew why until a video replay confirmed the call and controversy was avoided.However, many purists—those who want people to follow rules carefully and do things in the traditional way—especially in soccer, argue it’s not the way the game was invented, and that the video replay is tainting the sport. But don’t you want to see the proper application of the rules throughout the games? I know I do. Yes, it can slow the game down, but I feel it is worth it. If technological advancements allow fans watching from home to spot mistakes instantly, those same views need to be available to the officiating crews. Another exampleoccurred in the most recent National Football Conference (NFC) Championship Game between the Los Angeles Rams and the New Orleans Saints. When obvious pass interference was committed by the Los Angeles Rams player Nickell Robey-Coleman, with just 109 seconds to play, no flag was raised on the field. It weakened the New Orleans Saints spirits. The Los Angeles Rams won a 26-23 overtime victory. The no-call deeply angered the public. The video replay showed the referees had just missed one of the most apparent pass interference calls.There are no easy answers regarding replay technology and whether it is a curse (魔咒). But for me, keeping the officiating honest and on task is the right step in limiting controversy. 8．What trend in sports can be observed in paragraph 2?A．The video replay has been widely used.B．League games have become competitive.C．Rules of professional games are becoming stricter.D．People are showing more interest in sports than before.9．What does the underlined word “tainting” in paragraph 3 mean?A．Tricking.B．Promoting.C．Damaging.D．Restoring. 10．What might the New Orleans Saints think of the referees in the NFC Championship Game?A．They relied a lot on the video replay.B．They cared too much about details.C．They were definitely stressed out.D．They were terribly disqualified.11．What would be the best title for the text?A．Video replays: high-end technology in sportsB．Is technology like V AR a blessing in sports?C．Officiating: a duty that requires honestyD．What do qualified referees really mean?One key element of human language is semantics (语义). Scientists had long thought that unlike our words, animal vocalizations (发声) were involuntary, renecting the emotional state of the animal without conveying any other information. But over the last four decades, numerous studies have shown that various animals have distinct calls with specific meanings.Many bird species use different alarm calls. Japanese tits, which nest in tree holes, have one call that causes their baby birds to get down to avoid being pulled out of the nest by crows, and another call for tree snakes that sends them jumping out of the nest entirely. Siberian jays vary their calls depending on whether an enemy is seen looking for food or actively attacking — and each call gets a different response from other nearby birds.Two recent studies suggest that the order of some birds’ vocalizations may impact their meaning. Though the idea is still controversial, this could represent a basic form of the rules governing the order and combination of words and elements in human language known as syntax (句法), as illustrated by the classic “dog bites man” vs. “man bites dog” example.Even if some birds share basic aspects of human language, we still know very little about what’s actually going on in their minds. Most animal communication research has focused on describing signals and behavior, which on the surface can look a lot like human behavior. Determining if the underlying cognitive (认知的) processes driving the behavior are also similar is much more challenging, as at the heart of this question is intentionality: Are animals merely reacting to their environment, or do they intend to convey information to one another?12．What was scientists’ long-held belief about animal vocalizations?A．They conveyed no emotion.B．They were semantically related.C．They varied greatly with species.D．They expressed no intended meaning. 13．How does the author develop paragraph 2?A．By listing data.B．By giving examples.C．By providing definition.D．By making comparisons.14．What does the underlined word ”this” in paragraph 3 refer to?A．What birds’ vocalizations mean.B．How rules govern human language.C．What the two recent studies indicate.D．How bird’s vocalizations are combined. 15．What does the last paragraph mainly tell us?A．Shared aspects of human and birds’ languages.B．Focus of most animal communication research.C．Underlying, cognitive processes of birds’ vocalizations.D．Insufficient knowledge about birds’ communication intentionality.Revenge bedtime procrastination (报复性睡眠拖延症) refers to the decision to delaysleep in response to stress or a lack of free time earlier in the day. The addition of the word “revenge”, by the Chinese, to the concept of bedtime procrastination became popular on social media. 16Both language versions reflect frustration tied to long, stressful work hours that left little time for personal enjoyment.People who engage in bedtime procrastination know and generally want to receive enough sleep. 17 This is known as an intention-behavior gap.Bedtime procrastination can cause sleep deprivation (缺乏). 18 Not getting enough sleep affects thinking, memory, and decision-making. Sleep deprivation also raises the risk of daytime sleepiness, which can harm productivity and academic achievement while increasing the risks of sleeping driving.19 It causes difficulties in regulating emotions. It’s also been connected to mental health disorders, such as depression and anxiety. It is also discovered that sleep deprivation worsens physical health, making people more easily to suffer from illnesses. Ongoing sleep loss consequences can build up over time. Therefore, it contributes to significant long-term health problems.The best solution for bedtime procrastination is creating good sleep patterns and a good sleeping environment. 20A．But they fail to actually do so.B．The Chinese are noted for working long hours.C．A lack of sleep is tied to mental and physical health.D．But remember that it will take more than one night to truly get into good sleep habits.E．We struggle to stay productive during the day to make up for the loss.F．Without enough hours of sleep, the mind and body can’t properly recharge.G．The English term “revenge bedtime procrastination” appeared from a translation of this expression.二、完形填空received.While having a colonoscopy (结肠镜检查) without sedation (镇静剂) I was working hard to keep myself 23 by using techniques learned to 24 panic attacks. As the camera 25 the twists and turns of my bowel, the specialist 26 I was quietly humming a tune to myself. He asked what the 27 was and I replied, “Puff the Magic Dragon”, however, I 28 that I only knew two 29 of the song.The specialist then started 30 along with my humming and agreed that he also only knew two lines. He asked the student observing to google the lyrics. When the student 31 he explained that anything they can do to help a patient through a difficult 32 was worth doing. 33 with the words we then sang along for the rest of the procedure!Not only do I remember this years later, but when 34 by the student afterwards my advice was to learn from the specialist’s bedside manner as it was so 35 . 21．A．devotion B．warning C．belief D．mission 22．A．treatment B．operation C．condition D．recovery 23．A．quiet B．alarmed C．sensible D．calm 24．A．generate B．maintain C．rid D．control 25．A．witnessed B．measured C．explored D．outlined 26．A．concluded B．noticed C．justified D．evaluated 27．A．matter B．tune C．singer D．poet 28．A．explained B．whispered C．implied D．revealed 29．A．editions B．composers C．lines D．titles 30．A．surfing B．dancing C．working D．whistling 31．A．hesitated B．resisted C．complained D．replied 32．A．infection B．procedure C．operation D．event 33．A．Linked B．Confronted C．Armed D．Flooded 34．A．interviewed B．followed C．accompanied D．remarked 35．A．demanding B．lacking C．meaningful D．important三、语法填空阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。

成都2023-2024学年度高一(上)12月阶段考试英语（答案在最后）注意事项：1.答题前，务必将自己的姓名、考号填写在答题卡规定的位置上。

2.答选择题时，必须使用2B铅笔将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。

3.答非选择题时，必须使用0.5毫米黑色笔迹的签字笔，将答案书写在答题卡规定的位置上。

4.所有题目必须在答题卡上作答，在试题卷上答题无效。

5.考试结束后，只将答题卡交回。

第一部分听力(共两节，满分30分)做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

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 does the man want to drink?A.Tea.B.Juice.C.Beer.2.What will the man do today?A.Go hiking.B.Do his work.C.Phone a club.3.Where does the conversation take place?A.In a library.B.At a restaurant.C.In a school.4.What are the speakers mainly talking about?A.European’s breakfast.B.European’s table manners.C.European’s favorite vegetables.5.Why does the woman refuse to eat the cake?A.She has eaten a lot.B.She is on a diet.C.She dislikes it.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。

成都市实验外国语学校高三10月月考数学试题总分：150考试时间：120分钟一、单选题：本题共8小题，每小题5分，共40分。

在每小题给出的选项中，只有一项是符合题目要求的。

1．命题“，使”的否定是（ ）A ．，使B ．不存在，使C ．，D ．，2．已知等差数列的前项和为，若，且，则（ ）A ．60B ．72C ．120D ．1443．若，则（ ）A ．3B ．4C ．9D ．164，侧面展开图的扇形圆心角为的圆锥侧面积为（ ）A ．B ．C ．D ．5．小王每次通过英语听力测试的概率是，且每次通过英语听力测试相互独立，他连续测试3次，那么其中恰有1次通过的概率是（ ）A ．B ．C ．D ．6．已知，是方程的两个根，则（ ）A ．B ．C ．D ．7．当阳光射入海水后，海水中的光照强度随着深度增加而减弱，可用表示其总衰减规律，其中是消光系数，（单位：米）是海水深度，（单位：坎德拉）和（单位：坎德拉）分别表示在深度处和海面的光强．已知某海域5米深处的光强是海面光强的，则该海域消光系数的值约为（参考数据：，）（）A ．0.2B ．0.18C ．0.1D ．0.148．已知函数，方程有四个不同根，，，，且满足，则的取值范围是（ ）x ∃∈R 210x x +-=x ∃∈R 210x x +-≠x ∈R 210x x +-=x ∀∉R 210x x +-≠x ∀∈R 210x x +-≠{}n a n n S 21024a a +=36a =8S =24log log 2m n +=2m n =2π39π6π23292273949tan 23︒tan 37︒2230x mx +-=m =--0eKDD I I -=K D D I 0I D 40%K ln 20.7≈ln 5 1.6≈()22log ,012,04x x f x x x x ⎧>⎪=⎨++≤⎪⎩()f x a =1x 2x 3x 4x 1234x x x x <<<221323432x x x x x x +-A ．B ．C ．D ．二、多选题：本题共3小题，共18分。

四川省成都实验中学2024学年全国高三统一第一次网上联考数学试题测试题注意事项：1．答卷前，考生务必将自己的姓名、准考证号、考场号和座位号填写在试题卷和答题卡上。

用2B 铅笔将试卷类型（B ）填涂在答题卡相应位置上。

将条形码粘贴在答题卡右上角"条形码粘贴处"。

2．作答选择题时，选出每小题答案后，用2B 铅笔把答题卡上对应题目选项的答案信息点涂黑；如需改动，用橡皮擦干净后，再选涂其他答案。

答案不能答在试题卷上。

3．非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内相应位置上；如需改动，先划掉原来的答案，然后再写上新答案；不准使用铅笔和涂改液。

不按以上要求作答无效。

4．考生必须保证答题卡的整洁。

考试结束后，请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．已知直线1:240l ax y ++=，2:(1)20l x a y +-+=，则“1a =-”是“12l l ”的A ．充分不必要条件B ．必要不充分条件C ．充分必要条件D ．既不充分也不必要条件2．已知角α的终边经过点P(0sin 47,cos 47)，则sin(013α-)= A ．12B ．32C ．12-D ．32-3．总体由编号为01，02，...，39，40的40个个体组成.利用下面的随机数表选取5个个体，选取方法是从随机数表（如表）第1行的第4列和第5列数字开始由左到右依次选取两个数字，则选出来的第5个个体的编号为（ ）A ．23B ．21C ．35D ．324．如图，已知三棱锥D ABC -中，平面DAB ⊥平面ABC ，记二面角D AC B --的平面角为α，直线DA 与平面ABC 所成角为β，直线AB 与平面ADC 所成角为γ，则（ ）A ．αβγ≥≥B ．βαγ≥≥C ．αγβ≥≥D ．γαβ≥≥5．已知函数有三个不同的零点（其中），则 的值为( )A ．B ．C ．D ．6．已知函数()1ln 11xf x x x+=++-且()()12f a f a ++>，则实数a 的取值范围是( ) A ．11,2⎛⎫--⎪⎝⎭B ．1,02⎛⎫-⎪⎝⎭C ．10,2⎛⎫ ⎪⎝⎭D ．1,12⎛⎫ ⎪⎝⎭7．若,x y 满足约束条件02636x y x y ≤+≤⎧⎨≤-≤⎩，则2z x y =+的最大值为（ ）A ．10B ．8C ．5D ．38．某几何体的三视图如图所示，则该几何体的最长棱的长为（ ）A ．5B ．4C ．2D ．229．已知函数()ln ln(3)f x x x =+-，则（ ） A ．函数()f x 在()0,3上单调递增 B ．函数()f x 在()0,3上单调递减 C ．函数()f x 图像关于32x =对称 D ．函数()f x 图像关于3,02⎛⎫⎪⎝⎭对称 10．已知1F 、2F 是双曲线22221(0,0)x y a b a b-=>>的左右焦点，过点2F 与双曲线的一条渐近线平行的直线交双曲线另一条渐近线于点M ，若点M 在以线段12F F 为直径的圆外，则双曲线离心率的取值范围是（ ） A ．(2,)+∞B ．3,2)C ．2,3)D ．2)11．已知某超市2018年12个月的收入与支出数据的折线图如图所示：根据该折线图可知，下列说法错误的是（ ） A ．该超市2018年的12个月中的7月份的收益最高 B ．该超市2018年的12个月中的4月份的收益最低C ．该超市2018年1-6月份的总收益低于2018年7-12月份的总收益D ．该超市2018年7-12月份的总收益比2018年1-6月份的总收益增长了90万元12．以下四个命题：①两个随机变量的线性相关性越强，相关系数的绝对值越接近1；②在回归分析中，可用相关指数2R 的值判断拟合效果，2R 越小，模型的拟合效果越好； ③若数据123,,,,n x x x x 的方差为1，则1232+1,2+1,2+1,,2+1n x x x x 的方差为4；④已知一组具有线性相关关系的数据()()()11221010,,,,,,x y x y x y ，其线性回归方程ˆˆˆy bx a =+，则“()00,x y 满足线性回归方程ˆˆˆybx a =+”是“1210010x x x x +++= ，1210010y y y y ++=”的充要条件；其中真命题的个数为( ) A ．4B ．3C ．2D ．1二、填空题：本题共4小题，每小题5分，共20分。

2018高考化学易错点3：NA（知识讲解）含解析易错点三NA瞄准高考1．（2017课标Ⅱ）阿伏加德罗常数的值为N A。

下列说法正确的是A．1L0.1mol·NH4Cl溶液中，的数量为0.1N AB．2.4gMg与H2SO4完全反应，转移的电子数为0.1 N AC．标准状况下，2.24LN2和O2的混合气体中分子数为0.2 N AD．0.1mol H2和0.1mol I2于密闭容器中充分反应后，其分子总数为0.2 N A【答案】D2．（2017课标Ⅲ）为阿伏加德罗常数的值。

下列说法正确的是A．0.1 mol 的中，含有个中子B．pH=1的H3PO4溶液中，含有个C．2.24L（标准状况）苯在O2中完全燃烧，得到个CO2分子D．密闭容器中1 mol PCl3与1 mol Cl2反应制备PCl5（g），增加个P-Cl键【答案】A【解析】A．11B中含有6个中子，0.1mol 11B含有6N A个中子，A正确；B．溶液体积未定，不能计算氢离子个数，B错误；C．标准状况下苯是液体，不能利用气体摩尔体积计算22.4L 苯的物质的量，则无法判断其完全燃烧产生的CO2分子数目，C错误；D．PCl3与Cl2反应生成PCl5的反应是可逆反应，反应物不可能完全转化为生成物，则所1molPCl3与1molCl2反应生成的PCl5小于1mol，增加的P－Cl键的数目小于2N A个，D错误；答案选A。

3．（2016课标Ⅰ）设N A为阿伏加德罗常数值。

下列有关叙述正确的是A．14 g乙烯和丙烯混合气体中的氢原子数为2N AB．1 mol N2与4 mol H2反应生成的NH3分子数为2N AC．1 mol Fe溶于过量硝酸，电子转移数为2N AD．标准状况下，2.24 L CCl4含有的共价键数为0.4N A 【答案】A锁定考点小题快练1．（2018届山东省济宁市高三上学期期末考试）设N A为阿伏伽德罗常数的值。

下列说法正确的是A. 标准状况下，5.6L.Cl2 与足量NaOH 反应转移的电子数为0.25N AB. 室温下，1LpH=13 的NaOH溶液中，由水电离的OH-离子数目为0.1N AC. 氢氧燃料电池正极消耗22.4L( 标准状况)气体时，电路中通过的电子数目为2N AD. 5NH4NO32HNO3 +4N2↑+9H2O 反应中,生成56gN2 时，转移的电子数目为3.75N A【答案】A【解析】Cl2 与足量NaOH 反应生成氯化钠和次氯酸钠，1mol氯气转移1mol电子，所以标准状况下，5.6LCl2 与足量NaOH反应转移的电子数为0.25N A，故A正确；室温下，pH=13的NaOH溶液中，由水电离的OH-离子的浓度是，1LpH=13 的NaOH溶液中，由水电离的OH-离子数目为N A，故B错误；氢氧燃料电池正极是氧气得电子，1mol氧气得4mol电子，所以消耗22.4L( 标准状况)气体时，电路中通过的电子数目为4N A，故C错误；根据方程式5NH4NO32HNO3 +4N2↑+9H2O，生成4molN2 转移15mol电子，所以生成56gN2 时，转移的电子数目为7.5N A，故D错误。

浙江省杭州市西湖高级中学2017-2018学年高一语文上学期12月月考试题（含解析）考试时间：120分钟卷面总分：150卷I 总分100分一、语言运用。

1. 下列各组词语中，加点字注音全都正确的一组是（）A. 坍圮．（pǐ）窸窸窣．窣(sū) 熨．帖(yùn) 隽．永(juàn)B. 田凫．(fú) 慰藉．(jiè) 老饕．(tāo) 椋．鸟(liáng)C. 跌宕．(dàng) 锡镴．(là) 玉墀．(xī) 狙．击(zǔ)D. 湔．雪(qián) 蓊蓊．郁郁(wěng) 倩．影(qiàn) 敛裾．(jū)【答案】B【解析】试题分析：本题考查字音。

A.熨yù；C.墀chí，狙jū；D.湔jiān。

2. 下列各组词语中没有错别字的一组是（）A. 朱拓顿响寒喧百无聊赖B. 桌帷蹙缩炮烙泅水C. 拮据糍粑剌造纨绔子弟D. 踌躇谬种俨然沸反盈天【答案】D【解析】试题分析：本题考查字形。

A.钝响，寒暄；B.桌帏；C.敕造。

3. 填入下列各句横线处的词语，最恰当的一组是（）①游客不仅可以游走在清新可人的桂花树下，还可以到本土明星献上的文艺表演。

②精彩绝伦的文博会既让我们领略了文化的博大，也把一大批民间艺人推到了前台。

③对不少新生而言，大学生教官的出现让军训的和可怕一扫而光。

A. 欣赏精妙苍白B. 鉴赏精深苍白C. 欣赏精深枯燥D. 鉴赏精妙枯燥【答案】C【解析】试题分析：本题考查词语的运用。

鉴赏是对文物、艺术品等的鉴定和欣赏：开放古老的房屋供旅游者鉴赏；欣赏则是对一个人的人品或者才能很是满意和赞同，也可以是对一部电影和戏曲的满意。

二者的主要区别就是鉴赏一般是物而不针对人，而欣赏不仅包括人，也包括物。

所以第一空选“欣赏”，排除BD。

精妙：主要指思想或者想法等精神层面的东西，寓意是恰到好处。

精深，是一个形容词，释义有精熟深通、精微深奥。

成都高2026届12月阶段性测试物理试卷（答案在最后）考试时间：90分钟满分：100分注意事项：答题前，用蓝色或黑色签字笔将自己的姓名、班级、准考证号填写在答题卡上的指定位置，并用2B铅笔把准考证号对应的标号涂黑；选择题的作答：用2B铅笔把答题卡上对应题目的答案标号涂黑；非选择题的作答：用蓝色或黑色签字笔直接答在答题卡上对应的区域内。

第I卷选择题（共44分）一、单项选择题（本题8个小题，每题只有一个选项符合题意，每题3分，共24分）1.在国际单位制中，下列说法中正确的是（）A.力的单位是根据牛顿第三定律定义的B.Hz、N、2m/s都是国际单位制中的导出单位C.千克是国际单位制中力学的三个基本物理量之一D.质量和时间都是国际单位制中力学的基本单位【答案】B【解析】【详解】A．力的单位是根据牛顿第二定律定义的，故A错误；B．国际单位制中的基本单位有m、kg、s、A、K、cd、mol等七个。

Hz、N、2m/s都是国际单位制中的导出单位，故B正确；C．千克是国际单位制中力学的三个基本单位之一，故C错误；D．质量和时间的单位都是国际单位制中力学的基本单位，故D错误。

故选B。

2.图像法能简洁描述物体的运动情况，下列所给的图像中不能反映做直线运动物体回到初始位置的是A. B.C. D.【答案】B【解析】-图像，x t-图像都只能表示直线运动【详解】v tA．由图可知，物体开始和结束时的纵坐标均为0，说明物体又回到了初始位置，故A不符合题意；B．由图可知，物体一直沿正方向运动，位移不断增大，故不可能回到初始位置，故B符合题意；C．物体在前1s沿正向做匀速直线运动，位移大小为==⨯=x vt12m2m物体在后1s沿负向做匀速直线运动，位移大小为==⨯=x vt12m2m2s末时物体的总位移为零，故2s末物体回到初始位置，故C不符合题意；D．物体第1s内沿正方向，位移大小为112m1mx=⨯⨯=2第2s内位移为1m，沿负方向，2s末时物体的总位移为零，故2s末物体回到初始位置，故D不符合题意。